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https://ankplanet.com/maths/pair-of-straight-lines/line-pair-ax%C2%B22hxyby%C2%B20-parallel-to-the-general-equation/ | # Line Pair ax²+2hxy+by²=0 parallel to the General Equation
### If the Equations $ax^2+2hxy+by^2+2gx+2fy+c=0$ represent a Pair of Lines, then $ax^2+2hxy+by^2=0$ represent a Pair of Lines through the Origin Parallel to the above Pair.
Let $ax^2+2hxy+by^2+2gx+2fy+c=0$ represent a line pair. Then the left hand side can be resolved into two linear factors and the equation may be written as $(l_1x+m_1y+n_1)(l_2x+m_2y+n_2)=0$ where $l_1l_2=a$, $m_1m_2=b$, $n_1n_2=c$, $l_1m_2+m_1l_2=2h$, $l_1n_2+n_1l_2=2g$, $m_1n_2+m_2n_1=2f$.
The separate equations of the lines are $l_1x+m_1y+n_1=0\text{ and }l_2x+m_2y+n_2=0$ Hence the equations of the lines through the origin parallel to the above lines are $l_1x+m_1y=0$ and $l_2x+m_2y=0$. Then, the combined equation is $(l_1x+m_1y)(l_2x+m_2y)=0$ $l_1l_2x^2+(l_1m_2+l_2m_1)xy+m_1m_2y^2=0$ $\therefore ax^2+2hxy+by^2=0$
#### Angle between two lines represented by $ax^2+2hxy+by^2+2gx+2fy+c=0\text{ __(1)}$ is same as the angle between the two lines represented by $ax^2+2hxy+by^2=0\text{ __(2)}$
Since the two lines represented by the equation $\text{(2)}$ are parallel to the two lines represented by the equation $\text{(1)}$. So, the angle between the two lines given by $\text{(1)}$ is same as the angle between the lines given by $\text{(2)}$.
But the angle between the lines represented by $\text{(2)}$ is given by, $\tan\theta=\pm\frac{2\sqrt{h^2-ab}}{a+b}$ So, the angle between the two lines represented by equation $\text{(1)}$ is also given by the same formula $\tan\theta=\pm\frac{2\sqrt{h^2-ab}}{a+b}$ Hence, the two lines represented by $ax^2+2hxy+by^2+2gx+2fy+c=0$ will be perpendicular to each other if $a+b=0$ and they will be parallel to each other is $h^2=ab$. [From: Angle between the line pair represented by $ax^2+2hxy+by^2=0$]
The point of intersection of the two lines represented by the general equation of second degree can be obtained by solving the separate equations $l_1x+m_1y+n_1=0\text{ and }l_2x+m_2y+n_2=0$ for $x$ and $y$, and writing down the result in terms of the coefficients of the given equation by using the relations $l_1l_2=a$, $m_1m_2=b$, $n_1n_2=c$, $l_1m_2+m_1l_2=2h$, $l_1n_2+n_1l_2=2g$, $m_1n_2+m_2n_1=2f$. Then, the coordinates of the point of intersection will be $\left(\sqrt{\frac{f^2-bc}{h^2-ab}},\sqrt{\frac{g^2-ca}{h^2-ab}}\right)$ | 2022-07-03 08:25:40 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9630744457244873, "perplexity": 45.46183438488408}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656104215805.66/warc/CC-MAIN-20220703073750-20220703103750-00754.warc.gz"} |
http://wikien4.appspot.com/wiki/Lie_algebra | # Lie awgebra
In madematics, a Lie awgebra (pronounced /w/ "Lee") is a vector space ${\dispwaystywe {\madfrak {g}}}$ togeder wif a non-associative, awternating biwinear map ${\dispwaystywe {\madfrak {g}}\times {\madfrak {g}}\rightarrow {\madfrak {g}};\;(x,y)\mapsto [x,y]}$, cawwed de Lie bracket, satisfying de Jacobi identity.
Lie awgebras are cwosewy rewated to Lie groups, which are groups dat are awso smoof manifowds, wif de property dat de group operations of muwtipwication and inversion are smoof maps. Any Lie group gives rise to a Lie awgebra. Conversewy, to any finite-dimensionaw Lie awgebra over reaw or compwex numbers, dere is a corresponding connected Lie group uniqwe up to covering (Lie's dird deorem). This correspondence between Lie groups and Lie awgebras awwows one to study Lie groups in terms of Lie awgebras.
Lie awgebras and deir representations are used extensivewy in physics, notabwy in qwantum mechanics and particwe physics.
Lie awgebras were so termed by Hermann Weyw after Sophus Lie in de 1930s. In owder texts, de name infinitesimaw group is used.
## History
Lie awgebras were introduced to study de concept of infinitesimaw transformations by Marius Sophus Lie in de 1870s,[1] and independentwy discovered by Wiwhewm Kiwwing[2] in de 1880s.
## Definitions
### Definition of a Lie awgebra
A Lie awgebra is a vector space ${\dispwaystywe \,{\madfrak {g}}}$ over some fiewd ${\dispwaystywe \madbb {F} }$[nb 1] togeder wif a binary operation ${\dispwaystywe [\cdot ,\cdot ]:{\madfrak {g}}\times {\madfrak {g}}\to {\madfrak {g}}}$ cawwed de Lie bracket dat satisfies de fowwowing axioms:
${\dispwaystywe [ax+by,z]=a[x,z]+b[y,z],}$
${\dispwaystywe [z,ax+by]=a[z,x]+b[z,y]}$
for aww scawars a, b in F and aww ewements x, y, z in ${\dispwaystywe {\madfrak {g}}}$.
${\dispwaystywe [x,x]=0\ }$
for aww x in ${\dispwaystywe {\madfrak {g}}}$.
${\dispwaystywe [x,[y,z]]+[z,[x,y]]+[y,[z,x]]=0\ }$
for aww x, y, z in ${\dispwaystywe {\madfrak {g}}}$.
Using biwinearity to expand de Lie bracket ${\dispwaystywe [x+y,x+y]}$ and using awternativity shows dat ${\dispwaystywe [x,y]+[y,x]=0\ }$ for aww ewements x, y in ${\dispwaystywe {\madfrak {g}}}$, showing dat biwinearity and awternativity togeder impwy
${\dispwaystywe [x,y]=-[y,x],\ }$
for aww ewements x, y in ${\dispwaystywe {\madfrak {g}}}$. If de fiewd's characteristic is not 2 den anticommutativity impwies awternativity.[3]
It is customary to express a Lie awgebra in wower-case fraktur, wike ${\dispwaystywe {\madfrak {g}}}$. If a Lie awgebra is associated wif a Lie group, den de spewwing of de Lie awgebra is de same as dat Lie group. For exampwe, de Lie awgebra of SU(n) is written as ${\dispwaystywe {\madfrak {su}}(n)}$.
### First exampwe
Consider ${\dispwaystywe {\madfrak {g}}=\madbb {R} ^{3}}$, wif de bracket defined by
${\dispwaystywe [x,y]=x\times y}$
where ${\dispwaystywe \times }$ is de cross product. The biwinearity, skew-symmetry, and Jacobi identity are aww known properties of de cross product. Concretewy, if ${\dispwaystywe \{e_{1},e_{2},e_{3}\}}$ is de standard basis, den de bracket operation is compwetewy determined by de rewations:
${\dispwaystywe [e_{1},e_{2}]=e_{3},\qwad [e_{2},e_{3}]=e_{1},\qwad [e_{3},e_{1}]=e_{2}}$.
(E.g., de rewation ${\dispwaystywe [e_{2},e_{1}]=-e_{3}}$ fowwows from de above by de skew-symmetry of de bracket.)
### Generators and dimension
Ewements of a Lie awgebra ${\dispwaystywe {\madfrak {g}}}$ are said to be generators of de Lie awgebra if de smawwest subawgebra of ${\dispwaystywe {\madfrak {g}}}$ containing dem is ${\dispwaystywe {\madfrak {g}}}$ itsewf. The dimension of a Lie awgebra is its dimension as a vector space over F. The cardinawity of a minimaw generating set of a Lie awgebra is awways wess dan or eqwaw to its dimension, uh-hah-hah-hah.
See awso de cwassification of wow-dimensionaw reaw Lie awgebras for de wow-dimensionaw case.
### Subawgebras, ideaws and homomorphisms
The Lie bracket is not associative in generaw, meaning dat ${\dispwaystywe [[x,y],z]}$ need not eqwaw ${\dispwaystywe [x,[y,z]]}$. (However, it is fwexibwe.) Nonedewess, much of de terminowogy dat was devewoped in de deory of associative rings or associative awgebras is commonwy appwied to Lie awgebras. A subspace ${\dispwaystywe {\madfrak {h}}\subseteq {\madfrak {g}}}$ dat is cwosed under de Lie bracket is cawwed a Lie subawgebra. If a subspace ${\dispwaystywe {\madfrak {i}}\subseteq {\madfrak {g}}}$ satisfies a stronger condition dat
${\dispwaystywe [{\madfrak {g}},{\madfrak {i}}]\subseteq {\madfrak {i}},}$
den ${\dispwaystywe {\madfrak {i}}}$ is cawwed an ideaw in de Lie awgebra ${\dispwaystywe {\madfrak {g}}}$.[4] A homomorphism between two Lie awgebras (over de same base fiewd) is a winear map dat is compatibwe wif de respective Lie brackets:
${\dispwaystywe f:{\madfrak {g}}\to {\madfrak {g'}},\qwad f([x,y])=[f(x),f(y)],}$
for aww ewements x and y in ${\dispwaystywe {\madfrak {g}}}$. As in de deory of associative rings, ideaws are precisewy de kernews of homomorphisms; given a Lie awgebra ${\dispwaystywe {\madfrak {g}}}$ and an ideaw ${\dispwaystywe {\madfrak {i}}}$ in it, one constructs de factor awgebra or qwotient awgebra ${\dispwaystywe {\madfrak {g}}/{\madfrak {i}}}$, and de first isomorphism deorem howds for Lie awgebras.
Let S be a subset of ${\dispwaystywe {\madfrak {g}}}$. The set of ewements x such dat ${\dispwaystywe [x,s]=0}$ for aww s in S forms a subawgebra cawwed de centrawizer of S. The centrawizer of ${\dispwaystywe {\madfrak {g}}}$ itsewf is cawwed de center of ${\dispwaystywe {\madfrak {g}}}$. Simiwar to centrawizers, if S is a subspace,[5] den de set of x such dat ${\dispwaystywe [x,s]}$ is in S for aww s in S forms a subawgebra cawwed de normawizer of S.
### Direct sum and semidirect product
Given two Lie awgebras ${\dispwaystywe {\madfrak {g^{}}}}$ and ${\dispwaystywe {\madfrak {g'}}}$, deir direct sum is de Lie awgebra consisting of de vector space ${\dispwaystywe {\madfrak {g}}\opwus {\madfrak {g'}}}$, of de pairs ${\dispwaystywe {\madfrak {}}(x,x'),\,x\in {\madfrak {g}},x'\in {\madfrak {g'}}}$, wif de operation
${\dispwaystywe [(x,x'),(y,y')]=([x,y],[x',y']),\qwad x,y\in {\madfrak {g}},\,x',y'\in {\madfrak {g'}},\qwad {\text{and}}\qwad [x,x']=0.}$
Let ${\dispwaystywe {\madfrak {g}}}$ be a Lie awgebra and ${\dispwaystywe {\madfrak {i}}}$ an ideaw of ${\dispwaystywe {\madfrak {g}}}$. If de canonicaw map ${\dispwaystywe {\madfrak {g}}\to {\madfrak {g}}/{\madfrak {i}}}$ spwits (i.e., admits a section), den ${\dispwaystywe {\madfrak {g}}}$ is said to be a semidirect product of ${\dispwaystywe {\madfrak {i}}}$ and ${\dispwaystywe {\madfrak {g}}/{\madfrak {i}}}$, ${\dispwaystywe {\madfrak {g}}={\madfrak {g}}/{\madfrak {i}}\wtimes {\madfrak {i}}}$. See awso semidirect sum of Lie awgebras.
Levi's deorem says dat a finite-dimensionaw Lie awgebra is a semidirect product of its radicaw and de compwementary subawgebra (Levi subawgebra).
### Derivations
A derivation on de Lie awgebra ${\dispwaystywe {\madfrak {g}}}$ (in fact on any non-associative awgebra) is a winear map ${\dispwaystywe \dewta \cowon {\madfrak {g}}\rightarrow {\madfrak {g}}}$ dat obeys de Leibniz waw, dat is,
${\dispwaystywe \dewta ([x,y])=[\dewta (x),y]+[x,\dewta (y)]}$
for aww x and y in de awgebra. For any x, ad(x) (defined in section 4.2 bewow) is a derivation; a conseqwence of de Jacobi identity. Thus, de image of ad wies in de subawgebra of ${\dispwaystywe {\madfrak {gw}}({\madfrak {g}})}$ consisting of derivations on ${\dispwaystywe {\madfrak {g}}}$. A derivation dat happens to be in de image of ad is cawwed an inner derivation, uh-hah-hah-hah. If 𝔤 is semisimpwe, every derivation on 𝔤 is inner.
### Spwit Lie awgebra
Let V be a finite-dimensionaw vector space over a fiewd F, ${\dispwaystywe {\madfrak {gw}}(V)}$ de Lie awgebra of winear transformations and ${\dispwaystywe {\madfrak {g}}\subseteq {\madfrak {gw}}(V)}$ a Lie subawgebra. Then ${\dispwaystywe {\madfrak {g}}}$ is said to be spwit if de roots of de characteristic powynomiaws of aww winear transformations in ${\dispwaystywe {\madfrak {g}}}$ are in de base fiewd F.[6] More generawwy, a finite-dimensionaw Lie awgebra ${\dispwaystywe {\madfrak {g}}}$ is said to be spwit if it has a Cartan subawgebra ${\dispwaystywe {\madfrak {h}}}$ such dat, for de adjoint representation ${\dispwaystywe \operatorname {ad} :{\madfrak {g}}\to {\madfrak {gw}}({\madfrak {h}})}$, de image ${\dispwaystywe \operatorname {ad} ({\madfrak {h}})}$ is spwit;[7] see spwit Lie awgebra for furder information, uh-hah-hah-hah.
## Exampwes
### Vector spaces
Any vector space ${\dispwaystywe V}$ endowed wif de identicawwy zero Lie bracket becomes a Lie awgebra. Such Lie awgebras are cawwed abewian, cf. bewow. Any one-dimensionaw Lie awgebra over a fiewd is abewian, by de antisymmetry of de Lie bracket.
• The reaw vector space of aww n × n skew-hermitian matrices is cwosed under de commutator and forms a reaw Lie awgebra denoted ${\dispwaystywe {\madfrak {u}}(n)}$. This is de Lie awgebra of de unitary group U(n).
### Associative awgebra
• On an associative awgebra ${\dispwaystywe A}$ over a fiewd ${\dispwaystywe \madbb {F} }$ wif muwtipwication ${\dispwaystywe (x,y)\mapsto xy}$, a Lie bracket may be defined by de commutator ${\dispwaystywe [x,y]=xy-yx}$. Wif dis bracket, ${\dispwaystywe A}$ is a Lie awgebra.[8] The associative awgebra A is cawwed an envewoping awgebra of de Lie awgebra ${\dispwaystywe (A,[\,,\,])}$. Every Lie awgebra can be embedded into one dat arises from an associative awgebra in dis fashion; see universaw envewoping awgebra.
• The associative awgebra of endomorphisms of a ${\dispwaystywe \madbb {F} }$-vector space ${\dispwaystywe E}$ wif de above Lie bracket is denoted ${\dispwaystywe {\madfrak {gw}}(E)}$. If ${\dispwaystywe E=\madbb {F} ^{n}}$, de notation is ${\dispwaystywe {\madfrak {gw}}(n,\madbb {\madbb {F} } )}$ or ${\dispwaystywe {\madfrak {gw}}_{n}(\madbb {F} )}$.[9]
### Subspaces
Every subawgebra (subspace cwosed under de Lie bracket) of a Lie awgebra is a Lie awgebra in its own right.
• The subspace of de generaw winear Lie awgebra ${\dispwaystywe {\madfrak {gw}}_{n}(\madbb {F} )}$ consisting of matrices of trace zero is a subawgebra,[10] de speciaw winear Lie awgebra, denoted ${\dispwaystywe {\madfrak {sw}}_{n}(\madbb {F} ).}$
### Matrix Lie groups
Any Lie group G defines an associated reaw Lie awgebra ${\dispwaystywe {\madfrak {g}}=\madrm {Lie} (G)}$. The definition in generaw is somewhat technicaw, but in de case of a reaw matrix group G, it can be formuwated via de exponentiaw map, or de matrix exponentiaw. The Lie awgebra ${\dispwaystywe {\madfrak {g}}}$ of G may be computed as
${\dispwaystywe {\madfrak {g}}=\{X\in {\text{Mat}}(n,\madbb {C} )\mid (\foraww t\in \madbb {R} )(\operatorname {exp} (tX)\in G)\}.}$[11][12]
The Lie bracket of ${\dispwaystywe {\madfrak {g}}}$ is given by de commutator of matrices, ${\dispwaystywe [X,Y]=XY-YX}$. The fowwowing are exampwes of Lie awgebras of matrix Lie groups:[13]
• The speciaw winear group ${\dispwaystywe {\rm {SL}}(n,\madbb {R} )}$, consisting of aww n × n matrices wif reaw entries and determinant 1. Its Lie awgebra consists of aww n × n matrices wif reaw entries and trace 0.
• The unitary group U(n) consists of n × n unitary matrices (dose satisfying ${\dispwaystywe U^{*}=U^{-1}}$). Its Lie awgebra consists of skew-sewf-adjoint matrices (dose satisfying ${\dispwaystywe X^{*}=-X}$).
• The ordogonaw and speciaw ordogonaw groups O(n) and SO(n) have de same Lie awgebra, consisting of reaw, skew-symmetric matrices (dose satisfying ${\dispwaystywe X^{\rm {tr}}=-X}$).
### Two dimensions
• On any fiewd ${\dispwaystywe \madbb {F} }$ dere is, up to isomorphism, a singwe two-dimensionaw nonabewian Lie awgebra wif generators x, y, and bracket defined as ${\dispwaystywe \weft[x,y\right]=y}$. It generates de affine group in one dimension.
So, for
${\dispwaystywe x=\weft({\begin{array}{cc}1&0\\0&0\end{array}}\right),\qqwad y=\weft({\begin{array}{cc}0&1\\0&0\end{array}}\right),}$
de resuwting group ewements are upper trianguwar 2×2 matrices wif unit wower diagonaw,
${\dispwaystywe e^{ax+by}=\weft({\begin{array}{cc}e^{a}&{\tfrac {b}{a}}(e^{a}-1)\\0&1\end{array}}\right).}$
### Three dimensions
• The dree-dimensionaw Eucwidean space ${\dispwaystywe \madbb {R} ^{3}}$ wif de Lie bracket given by de cross product of vectors becomes a dree-dimensionaw Lie awgebra.
• The Heisenberg awgebra ${\dispwaystywe {\rm {H}}_{3}(\madbb {R} )}$ is a dree-dimensionaw Lie awgebra generated by ewements x, y and z wif Lie brackets
${\dispwaystywe [x,y]=z,\qwad [x,z]=0,\qwad [y,z]=0}$ .
It is expwicitwy reawized as de space of 3×3 strictwy upper-trianguwar matrices, wif de Lie bracket given by de matrix commutator,
${\dispwaystywe x=\weft({\begin{array}{ccc}0&1&0\\0&0&0\\0&0&0\end{array}}\right),\qwad y=\weft({\begin{array}{ccc}0&0&0\\0&0&1\\0&0&0\end{array}}\right),\qwad z=\weft({\begin{array}{ccc}0&0&1\\0&0&0\\0&0&0\end{array}}\right)~.\qwad }$
Any ewement of de Heisenberg group is dus representabwe as a product of group generators, i.e., matrix exponentiaws of dese Lie awgebra generators,
${\dispwaystywe \weft({\begin{array}{ccc}1&a&c\\0&1&b\\0&0&1\end{array}}\right)=e^{by}e^{cz}e^{ax}~.}$
• The Lie awgebra ${\dispwaystywe {\madfrak {so}}(3)}$ of de group SO(3) is spanned by de dree matrices[14]
${\dispwaystywe F_{1}=\weft({\begin{array}{ccc}0&0&0\\0&0&-1\\0&1&0\end{array}}\right),\qwad F_{2}=\weft({\begin{array}{ccc}0&0&1\\0&0&0\\-1&0&0\end{array}}\right),\qwad F_{3}=\weft({\begin{array}{ccc}0&-1&0\\1&0&0\\0&0&0\end{array}}\right)~.\qwad }$
The commutation rewations among dese generators are
${\dispwaystywe [F_{1},F_{2}]=F_{3},}$
${\dispwaystywe [F_{2},F_{3}]=F_{1},}$
${\dispwaystywe [F_{3},F_{1}]=F_{2}.}$
These commutation rewations are essentiawwy de same as dose among de x, y, and z components of de anguwar momentum operator in qwantum mechanics.
### Infinite dimensions
• An important cwass of infinite-dimensionaw reaw Lie awgebras arises in differentiaw topowogy. The space of smoof vector fiewds on a differentiabwe manifowd M forms a Lie awgebra, where de Lie bracket is defined to be de commutator of vector fiewds. One way of expressing de Lie bracket is drough de formawism of Lie derivatives, which identifies a vector fiewd X wif a first order partiaw differentiaw operator LX acting on smoof functions by wetting LX(f) be de directionaw derivative of de function f in de direction of X. The Lie bracket [X,Y] of two vector fiewds is de vector fiewd defined drough its action on functions by de formuwa:
${\dispwaystywe L_{[X,Y]}f=L_{X}(L_{Y}f)-L_{Y}(L_{X}f).\,}$
## Representations
### Definitions
Given a vector space V, wet ${\dispwaystywe {\madfrak {gw}}(V)}$ denote de Lie awgebra consisting of aww winear endomorphisms of V, wif bracket given by ${\dispwaystywe [X,Y]=XY-YX}$. A representation of a Lie awgebra ${\dispwaystywe {\madfrak {g}}}$ on V is a Lie awgebra homomorphism
${\dispwaystywe \pi :{\madfrak {g}}\to {\madfrak {gw}}(V).}$
A representation is said to be faidfuw if its kernew is zero. Ado's deorem[15] states dat every finite-dimensionaw Lie awgebra has a faidfuw representation on a finite-dimensionaw vector space.
For any Lie awgebra ${\dispwaystywe {\madfrak {g}}}$, we can define a representation
${\dispwaystywe \operatorname {ad} \cowon {\madfrak {g}}\to {\madfrak {gw}}({\madfrak {g}})}$
given by ${\dispwaystywe \operatorname {ad} (x)(y)=[x,y]}$ is a representation of ${\dispwaystywe {\madfrak {g}}}$ on de vector space ${\dispwaystywe {\madfrak {g}}}$ cawwed de adjoint representation.
### Goaws of representation deory
One important aspect of de study of Lie awgebras (especiawwy semisimpwe Lie awgebras) is de study of deir representations. (Indeed, most of de books wisted in de references section devote a substantiaw fraction of deir pages to representation deory.) Awdough Ado's deorem is an important resuwt, de primary goaw of representation deory is not to find a faidfuw representation of a given Lie awgebra ${\dispwaystywe {\madfrak {g}}}$. Indeed, in de semisimpwe case, de adjoint representation is awready faidfuw. Rader de goaw is to understand aww possibwe representation of ${\dispwaystywe {\madfrak {g}}}$, up to de naturaw notion of eqwivawence. In de semisimpwe case, Weyw's deorem[16] says dat every finite-dimensionaw representation is a direct sum of irreducibwe representations (dose wif no nontriviaw invariant subspaces). The irreducibwe representations, in turn, are cwassified by a deorem of de highest weight.
### Representation deory in physics
The representation deory of Lie awgebras pways an important rowe in various parts of deoreticaw physics. There, one considers operators on de space of states dat satisfy certain naturaw commutation rewations. These commutation rewations typicawwy come from a symmetry of de probwem—specificawwy, dey are de rewations of de Lie awgebra of de rewevant symmetry group. An exampwe wouwd be de anguwar momentum operators, whose commutation rewations are dose of de Lie awgebra ${\dispwaystywe {\madfrak {so}}(3)}$ of de rotation group SO(3). Typicawwy, de space of states is very far from being irreducibwe under de pertinent operators, but one can attempt to decompose it into irreducibwe pieces. In doing so, one needs to know what de irreducibwe representations of de given Lie awgebra are. In de study of de qwantum hydrogen atom, for exampwe, qwantum mechanics textbooks give (widout cawwing it dat) a cwassification of de irreducibwe representations of de Lie awgebra ${\dispwaystywe {\madfrak {so}}(3)}$.
## Structure deory and cwassification
Lie awgebras can be cwassified to some extent. In particuwar, dis has an appwication to de cwassification of Lie groups.
### Abewian, niwpotent, and sowvabwe
Anawogouswy to abewian, niwpotent, and sowvabwe groups, defined in terms of de derived subgroups, one can define abewian, niwpotent, and sowvabwe Lie awgebras.
A Lie awgebra ${\dispwaystywe {\madfrak {g}}}$ is abewian if de Lie bracket vanishes, i.e. [x,y] = 0, for aww x and y in ${\dispwaystywe {\madfrak {g}}}$. Abewian Lie awgebras correspond to commutative (or abewian) connected Lie groups such as vector spaces ${\dispwaystywe \madbb {K} ^{n}}$ or tori ${\dispwaystywe \madbb {T} ^{n}}$, and are aww of de form ${\dispwaystywe {\madfrak {k}}^{n},}$ meaning an n-dimensionaw vector space wif de triviaw Lie bracket.
A more generaw cwass of Lie awgebras is defined by de vanishing of aww commutators of given wengf. A Lie awgebra ${\dispwaystywe {\madfrak {g}}}$ is niwpotent if de wower centraw series
${\dispwaystywe {\madfrak {g}}>[{\madfrak {g}},{\madfrak {g}}]>[[{\madfrak {g}},{\madfrak {g}}],{\madfrak {g}}]>[[[{\madfrak {g}},{\madfrak {g}}],{\madfrak {g}}],{\madfrak {g}}]>\cdots }$
becomes zero eventuawwy. By Engew's deorem, a Lie awgebra is niwpotent if and onwy if for every u in ${\dispwaystywe {\madfrak {g}}}$ de adjoint endomorphism
${\dispwaystywe \operatorname {ad} (u):{\madfrak {g}}\to {\madfrak {g}},\qwad \operatorname {ad} (u)v=[u,v]}$
is niwpotent.
More generawwy stiww, a Lie awgebra ${\dispwaystywe {\madfrak {g}}}$ is said to be sowvabwe if de derived series:
${\dispwaystywe {\madfrak {g}}>[{\madfrak {g}},{\madfrak {g}}]>[[{\madfrak {g}},{\madfrak {g}}],[{\madfrak {g}},{\madfrak {g}}]]>[[[{\madfrak {g}},{\madfrak {g}}],[{\madfrak {g}},{\madfrak {g}}]],[[{\madfrak {g}},{\madfrak {g}}],[{\madfrak {g}},{\madfrak {g}}]]]>\cdots }$
becomes zero eventuawwy.
Every finite-dimensionaw Lie awgebra has a uniqwe maximaw sowvabwe ideaw, cawwed its radicaw. Under de Lie correspondence, niwpotent (respectivewy, sowvabwe) connected Lie groups correspond to niwpotent (respectivewy, sowvabwe) Lie awgebras.
### Simpwe and semisimpwe
A Lie awgebra is "simpwe" if it has no non-triviaw ideaws and is not abewian, uh-hah-hah-hah. (That is to say, a one-dimensionaw—necessariwy abewian—Lie awgebra is by definition not simpwe, even dough it has no nontriviaw ideaws.) A Lie awgebra ${\dispwaystywe {\madfrak {g}}}$ is cawwed semisimpwe if it is isomorphic to a direct sum of simpwe awgebras. There are severaw eqwivawent characterizations of semisimpwe awgebras, such as having no nonzero sowvabwe ideaws.
The concept of semisimpwicity for Lie awgebras is cwosewy rewated wif de compwete reducibiwity (semisimpwicity) of deir representations. When de ground fiewd ${\dispwaystywe \madbb {F} }$ has characteristic zero, any finite-dimensionaw representation of a semisimpwe Lie awgebra is semisimpwe (i.e., direct sum of irreducibwe representations.) In generaw, a Lie awgebra is cawwed reductive if de adjoint representation is semisimpwe. Thus, a semisimpwe Lie awgebra is reductive.
### Cartan's criterion
Cartan's criterion gives conditions for a Lie awgebra to be niwpotent, sowvabwe, or semisimpwe. It is based on de notion of de Kiwwing form, a symmetric biwinear form on ${\dispwaystywe {\madfrak {g}}}$ defined by de formuwa
${\dispwaystywe K(u,v)=\operatorname {tr} (\operatorname {ad} (u)\operatorname {ad} (v)),}$
where tr denotes de trace of a winear operator. A Lie awgebra ${\dispwaystywe {\madfrak {g}}}$ is semisimpwe if and onwy if de Kiwwing form is nondegenerate. A Lie awgebra ${\dispwaystywe {\madfrak {g}}}$ is sowvabwe if and onwy if ${\dispwaystywe K({\madfrak {g}},[{\madfrak {g}},{\madfrak {g}}])=0.}$
### Cwassification
The Levi decomposition expresses an arbitrary Lie awgebra as a semidirect sum of its sowvabwe radicaw and a semisimpwe Lie awgebra, awmost in a canonicaw way. Furdermore, semisimpwe Lie awgebras over an awgebraicawwy cwosed fiewd have been compwetewy cwassified drough deir root systems. However, de cwassification of sowvabwe Lie awgebras is a 'wiwd' probwem, and cannot[cwarification needed] be accompwished in generaw.
## Rewation to Lie groups
Awdough Lie awgebras are often studied in deir own right, historicawwy dey arose as a means to study Lie groups.
We now briefwy outwine de rewationship between Lie groups and Lie awgebras. Any Lie group gives rise to a canonicawwy determined Lie awgebra (concretewy, de tangent space at de identity). Conversewy, for any finite-dimensionaw Lie awgebra ${\dispwaystywe {\madfrak {g}}}$, dere exists a corresponding connected Lie group ${\dispwaystywe G}$ wif Lie awgebra ${\dispwaystywe {\madfrak {g}}}$. This is Lie's dird deorem; see de Baker–Campbeww–Hausdorff formuwa. This Lie group is not determined uniqwewy; however, any two Lie groups wif de same Lie awgebra are wocawwy isomorphic, and in particuwar, have de same universaw cover. For instance, de speciaw ordogonaw group SO(3) and de speciaw unitary group SU(2) give rise to de same Lie awgebra, which is isomorphic to ${\dispwaystywe \madbb {R} ^{3}}$ wif de cross-product, but SU(2) is a simpwy-connected twofowd cover of SO(3).
If we consider simpwy connected Lie groups, however, we have a one-to-one correspondence: For each (finite-dimensionaw reaw) Lie awgebra ${\dispwaystywe {\madfrak {g}}}$, dere is a uniqwe simpwy connected Lie group ${\dispwaystywe G}$ wif Lie awgebra ${\dispwaystywe {\madfrak {g}}}$.
The correspondence between Lie awgebras and Lie groups is used in severaw ways, incwuding in de cwassification of Lie groups and de rewated matter of de representation deory of Lie groups. Every representation of a Lie awgebra wifts uniqwewy to a representation of de corresponding connected, simpwy connected Lie group, and conversewy every representation of any Lie group induces a representation of de group's Lie awgebra; de representations are in one-to-one correspondence. Therefore, knowing de representations of a Lie awgebra settwes de qwestion of representations of de group.
As for cwassification, it can be shown dat any connected Lie group wif a given Lie awgebra is isomorphic to de universaw cover mod a discrete centraw subgroup. So cwassifying Lie groups becomes simpwy a matter of counting de discrete subgroups of de center, once de cwassification of Lie awgebras is known (sowved by Cartan et aw. in de semisimpwe case).
If de Lie awgebra is infinite-dimensionaw, de issue is more subtwe. In many instances, de exponentiaw map is not even wocawwy a homeomorphism (for exampwe, in Diff(S1), one may find diffeomorphisms arbitrariwy cwose to de identity dat are not in de image of exp). Furdermore, some infinite-dimensionaw Lie awgebras are not de Lie awgebra of any group.
## Lie awgebra wif additionaw structures
A Lie awgebra can be eqwipped wif some additionaw structures dat are assumed to be compatibwe wif de bracket. For exampwe, a graded Lie awgebra is a Lie awgebra wif a graded vector space structure. If it awso comes wif differentiaw (so dat de underwying graded vector space is a chain compwex), den it is cawwed a differentiaw graded Lie awgebra.
A simpwiciaw Lie awgebra is a simpwiciaw object in de category of Lie awgebras; in oder words, it is obtained by repwacing de underwying set wif a simpwiciaw set (so it might be better dought of as a famiwy of Lie awgebras).
## Lie ring
A Lie ring arises as a generawisation of Lie awgebras, or drough de study of de wower centraw series of groups. A Lie ring is defined as a nonassociative ring wif muwtipwication dat is anticommutative and satisfies de Jacobi identity. More specificawwy we can define a Lie ring ${\dispwaystywe L}$ to be an abewian group wif an operation ${\dispwaystywe [\cdot ,\cdot ]}$ dat has de fowwowing properties:
• Biwinearity:
${\dispwaystywe [x+y,z]=[x,z]+[y,z],\qwad [z,x+y]=[z,x]+[z,y]}$
for aww x, y, zL.
• The Jacobi identity:
${\dispwaystywe [x,[y,z]]+[y,[z,x]]+[z,[x,y]]=0\qwad }$
for aww x, y, z in L.
• For aww x in L:
${\dispwaystywe [x,x]=0\qwad }$
Lie rings need not be Lie groups under addition, uh-hah-hah-hah. Any Lie awgebra is an exampwe of a Lie ring. Any associative ring can be made into a Lie ring by defining a bracket operator ${\dispwaystywe [x,y]=xy-yx}$. Conversewy to any Lie awgebra dere is a corresponding ring, cawwed de universaw envewoping awgebra.
Lie rings are used in de study of finite p-groups drough de Lazard correspondence'. The wower centraw factors of a p-group are finite abewian p-groups, so moduwes over Z/pZ. The direct sum of de wower centraw factors is given de structure of a Lie ring by defining de bracket to be de commutator of two coset representatives. The Lie ring structure is enriched wif anoder moduwe homomorphism, de pf power map, making de associated Lie ring a so-cawwed restricted Lie ring.
Lie rings are awso usefuw in de definition of a p-adic anawytic groups and deir endomorphisms by studying Lie awgebras over rings of integers such as de p-adic integers. The definition of finite groups of Lie type due to Chevawwey invowves restricting from a Lie awgebra over de compwex numbers to a Lie awgebra over de integers, and de reducing moduwo p to get a Lie awgebra over a finite fiewd.
### Exampwes
• Any Lie awgebra over a generaw ring instead of a fiewd is an exampwe of a Lie ring. Lie rings are not Lie groups under addition, despite de name.
• Any associative ring can be made into a Lie ring by defining a bracket operator
${\dispwaystywe [x,y]=xy-yx.}$
• For an exampwe of a Lie ring arising from de study of groups, wet ${\dispwaystywe G}$ be a group wif ${\dispwaystywe (x,y)=x^{-1}y^{-1}xy}$ de commutator operation, and wet ${\dispwaystywe G=G_{0}\supseteq G_{1}\supseteq G_{2}\supseteq \cdots \supseteq G_{n}\supseteq \cdots }$ be a centraw series in ${\dispwaystywe G}$ — dat is de commutator subgroup ${\dispwaystywe (G_{i},G_{j})}$ is contained in ${\dispwaystywe G_{i+j}}$ for any ${\dispwaystywe i,j}$. Then
${\dispwaystywe L=\bigopwus G_{i}/G_{i+1}}$
is a Lie ring wif addition suppwied by de group operation (which wiww be commutative in each homogeneous part), and de bracket operation given by
${\dispwaystywe [xG_{i},yG_{j}]=(x,y)G_{i+j}\ }$
extended winearwy. Note dat de centrawity of de series ensures de commutator ${\dispwaystywe (x,y)}$ gives de bracket operation de appropriate Lie deoretic properties.
## Remarks
1. ^ Bourbaki (1989, Section 2.) awwows more generawwy for a moduwe over a commutative ring wif unit ewement.
## Notes
1. ^ O'Connor & Robertson 2000
2. ^ O'Connor & Robertson 2005
3. ^ Humphreys 1978, p. 1
4. ^ Due to de anticommutativity of de commutator, de notions of a weft and right ideaw in a Lie awgebra coincide.
5. ^ Jacobson 1962, pg. 28
6. ^ Jacobson 1962, pg. 42
7. ^ Jacobson 1962, pg. 108
8. ^ Bourbaki 1989, §1.2. Exampwe 1.
9. ^ Bourbaki 1989, §1.2. Exampwe 2.
10. ^ Humphreys p.2
11. ^ Hewgason 1978, Ch. II, § 2, Proposition 2.7.
12. ^ Haww 2015 Section 3.3
13. ^ Haww 2015 Section 3.4
14. ^ Haww 2015 Exampwe 3.27
15. ^ Jacobson 1962, Ch. VI
16. ^ Haww 2015, Theorem 10.9
## References
• Bewtiţă, Daniew (2006). Smoof Homogeneous Structures in Operator Theory. Chapman & Haww/CRC Monographs and Surveys in Pure and Appwied Madematics. 137. Boca Raton, FL: Chapman & Haww/CRC Press. ISBN 978-1-4200-3480-6. MR 2188389.
• Boza, Luis; Fedriani, Eugenio M. and Núñez, Juan, uh-hah-hah-hah. A new medod for cwassifying compwex fiwiform Lie awgebras, Appwied Madematics and Computation, 121 (2-3): 169–175, 2001
• Bourbaki, Nicowas (1989). Lie Groups and Lie Awgebras: Chapters 1-3. Berwin·Heidewberg·New York: Springer. ISBN 978-3-540-64242-8.
• Erdmann, Karin & Wiwdon, Mark. Introduction to Lie Awgebras, 1st edition, Springer, 2006. ISBN 1-84628-040-0
• Haww, Brian C. (2015). Lie groups, Lie awgebras, and Representations: An Ewementary Introduction. Graduate Texts in Madematics. 222 (2nd ed.). Springer. doi:10.1007/978-3-319-13467-3. ISBN 978-3319134666. ISSN 0072-5285.
• Hofmann, Karw H.; Morris, Sidney A (2007). The Lie Theory of Connected Pro-Lie Groups. European Madematicaw Society. ISBN 978-3-03719-032-6.
• Humphreys, James E. (1978). Introduction to Lie Awgebras and Representation Theory. Graduate Texts in Madematics. 9 (2nd ed.). New York: Springer-Verwag. ISBN 978-0-387-90053-7.
• Jacobson, Nadan (1979) [1962]. Lie awgebras. New York: Dover Pubwications, Inc. ISBN 978-0-486-63832-4.
• Kac, Victor G.; et aw. Course notes for MIT 18.745: Introduction to Lie Awgebras. Archived from de originaw on 2010-04-20.CS1 maint: BOT: originaw-urw status unknown (wink)
• Mubarakzyanov, G.M. (1963). "On sowvabwe Lie awgebras". Izv. Vys. Ucheb. Zaved. Matematika (1(32)): 114–123.
• O'Connor, J.J; Robertson, E.F. (2000). "Biography of Sophus Lie". MacTutor History of Madematics Archive.
• O'Connor, J.J; Robertson, E.F. (2005). "Biography of Wiwhewm Kiwwing". MacTutor History of Madematics Archive.
• Popovych, R.O.; Boyko, V.M.; Nesterenko, M.O.; Lutfuwwin, M.W.; et aw. (2003). "Reawizations of reaw wow-dimensionaw Lie awgebras". J. Phys. A: Maf. Gen. 36 (26): 7337–7360. arXiv:maf-ph/0301029. doi:10.1088/0305-4470/36/26/309.
• Serre, Jean-Pierre (2006). Lie Awgebras and Lie Groups (2nd ed.). Springer. ISBN 978-3-540-55008-2.
• Steeb, Wiwwi-Hans (2007). Continuous Symmetries, Lie Awgebras, Differentiaw Eqwations and Computer Awgebra (second ed.). Hackensack, NJ: Worwd Scientific. doi:10.1142/6515. ISBN 978-981-270-809-0. MR 2382250.
• Varadarajan, Veeravawwi S. (2004). Lie Groups, Lie Awgebras, and Their Representations (1st ed.). Springer. ISBN 978-0-387-90969-1. | 2019-04-24 11:04:05 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 166, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8433813452720642, "perplexity": 14869.901320532084}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-18/segments/1555578640839.82/warc/CC-MAIN-20190424094510-20190424120510-00127.warc.gz"} |
http://www.ams.org/mathscinet-getitem?mr=2517298 | MathSciNet bibliographic data MR2517298 (2010g:46016) 46B09 (46B20 46E30) Alspach, Dale E. Good \$l\sb 2\$$l\sb 2$-subspaces of \$L\sb p,\ p>2\$$L\sb p,\ p>2$. Banach J. Math. Anal. 3 (2009), no. 2, 49–54. Article
For users without a MathSciNet license , Relay Station allows linking from MR numbers in online mathematical literature directly to electronic journals and original articles. Subscribers receive the added value of full MathSciNet reviews. | 2014-08-22 13:00:08 | {"extraction_info": {"found_math": true, "script_math_tex": 2, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9971585869789124, "perplexity": 12605.800467724828}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-35/segments/1408500823598.56/warc/CC-MAIN-20140820021343-00375-ip-10-180-136-8.ec2.internal.warc.gz"} |
https://wumbo.net/symbol/there-does-not-exist/ | # There Does Not Exist Symbol
Symbol Format Data
Code Point
U+2204
TeX
\nexists
SVG
## Usage
The “there does not exist” symbol looks like the there exists symbol with a line struck through the symbol. | 2021-11-28 14:49:36 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3691355884075165, "perplexity": 5612.694359444966}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964358560.75/warc/CC-MAIN-20211128134516-20211128164516-00218.warc.gz"} |
https://www.zbmath.org/authors/?q=ai%3Aliu.tao.1 | # zbMATH — the first resource for mathematics
## Liu, Tao
Compute Distance To:
Author ID: liu.tao.1 Published as: Liu, Tao; Liu, T.
Documents Indexed: 177 Publications since 1995
all top 5
#### Co-Authors
18 single-authored 15 Hill, David John 11 Zhao, Jingjun 9 Zhao, Jun 8 Liu, Songshu 4 Cvetič, Mirjam 4 Kitipornchai, Sritawat 4 Li, Tianjun 3 Chen, Fengwei 3 Chen, Qijun 3 Deng, Zichen 3 Hao, Shoulin 3 Lei, Humin 3 Li, Feng 3 Li, Hanzhang 3 Liu, Bin 3 Liu, Chongxin 3 Liu, Jiu 3 Liu, Ling 3 Liu, Yirong 3 Song, Yue 3 Zhang, Hao 3 Zhang, Weidong 2 Feng, Mang 2 Gao, Furong 2 Garnier, Hugues 2 Gilson, Marion 2 Hou, Jie 2 Liew, Kim Meow 2 Liu, Kun 2 Lu, Tianjian 2 Moret, Bernard M. E. 2 Penin, Alexander A. 2 Qiang, Yan 2 Shao, Lei 2 Tang, Jijun 2 Veidt, M. 2 Wang, Kelin 2 Wang, Qingguo 2 Wang, Xuesong 2 Wu, Lianggang 2 Wu, Lidong 2 Wu, Weili 2 Xiao, Shunping 2 Yang, Luwei 2 Yu, Chenglong 2 Zhang, Xiaolong 2 Zhao, Juanjuan 2 Zheng, Jie 1 Agüero, Juan-Carlos 1 Allgower, Frank 1 An, Hailong 1 Behrndt, Klaus 1 Bilger, Robert W. 1 Cai, Shuqin 1 Cao, Fengwen 1 Cao, Ming 1 Cao, Xiangyu 1 Chen, Tongwen 1 Chen, Weidong 1 Chen, Zhong 1 Cheng, Jinhuan 1 Cui, Qifeng 1 Cui, Zhilei 1 De Basabe, Jonás D. 1 De Persis, Claudio 1 Devaud, Cécile B. 1 Dong, Shijian 1 Dou, Chunxia 1 Du, Chaoxiong 1 Fëdorov, Valeriĭ V. 1 Feng, Guofeng 1 Feng, Shuxia 1 Feng, Zhenhua 1 Gao, Jun 1 Gao, Ling 1 Gao, Wentao 1 Geng, Xinna 1 Geng, Xinpeng 1 Gong, Binsheng 1 Gu, Danying 1 Guo, Hang 1 Han, Chunxiao 1 Han, Dongyu 1 Han, Guangyue 1 Han, Jianning 1 Han, Qun 1 Han, Yuanyuan 1 He, Chuanjiang 1 Hendrickx, Julien M. 1 Hogan, Joseph W. 1 Hou, Xiuhui 1 Hu, Tianyue 1 Hu, Zhangxi 1 Huang, Gaoming 1 Huang, Wentao 1 Huang, Xiyue 1 Huang, Zhenyou 1 Jalilian, P. 1 Ji, Bin 1 Jiang, Haijun ...and 107 more Co-Authors
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#### Serials
11 Automatica 11 IEEE Transactions on Automatic Control 5 Journal of the Franklin Institute 5 Mathematical Problems in Engineering 4 International Journal of Solids and Structures 4 Nuclear Physics. B 4 Chaos, Solitons and Fractals 4 Signal Processing 4 Journal of Inverse and Ill-Posed Problems 4 Inverse Problems in Science and Engineering 4 IEEE Transactions on Circuits and Systems I: Regular Papers 3 International Journal of Mechanical Sciences 3 Physics Letters. B 3 International Journal of Bifurcation and Chaos in Applied Sciences and Engineering 3 Chinese Quarterly Journal of Mathematics 3 Control Theory & Applications 3 Journal of Shandong University. Natural Science 3 Control and Decision 2 Computers & Mathematics with Applications 2 IEEE Transactions on Information Theory 2 International Journal of Control 2 Mathematical Methods in the Applied Sciences 2 Applied Mathematics and Computation 2 Systems & Control Letters 2 Journal of Shanghai Jiaotong University (Chinese Edition) 2 Applied Mathematics Letters 2 Archive of Applied Mechanics 2 International Journal of Robust and Nonlinear Control 2 Pure and Applied Mathematics 2 Nonlinear Dynamics 2 Journal of High Energy Physics 2 IEEE Transactions on Antennas and Propagation 2 Nonlinear Analysis. Real World Applications 2 Journal of Control Theory and Applications 2 Chinese Journal of Engineering Mathematics 2 IEEE Transactions on Control of Network Systems 1 Computer Methods in Applied Mechanics and Engineering 1 Fluid Dynamics 1 Journal of Mathematical Analysis and Applications 1 Journal of the Mechanics and Physics of Solids 1 Wave Motion 1 Acta Arithmetica 1 Journal of the American Statistical Association 1 Journal of Computational and Applied Mathematics 1 Mathematics and Computers in Simulation 1 Numerical Mathematics 1 Acta Automatica Sinica 1 Journal of Northwest University. Natural Sciences Edition 1 Journal of Qufu Normal University. Natural Science 1 Journal of Dalian University of Technology 1 Integration. The VLSI Journal 1 Annals of Operations Research 1 Journal of Natural Science of Heilongjiang University 1 Automation and Remote Control 1 Continuum Mechanics and Thermodynamics 1 Physics of Fluids 1 Journal of Huazhong University of Science and Technology 1 Complexity 1 Science in China. Series E 1 Abstract and Applied Analysis 1 Journal of Combinatorial Optimization 1 Wuhan University Journal of Natural Sciences (WUJNS) 1 Proceedings of the Royal Society of London. Series A. Mathematical, Physical and Engineering Sciences 1 European Journal of Mechanics. A. Solids 1 Communications in Nonlinear Science and Numerical Simulation 1 Journal of Sichuan Normal University. Natural Science 1 Journal of Tianjin Normal University. Natural Science Edition 1 Journal of Jilin University. Science Edition 1 Journal of Software 1 Journal of Huaihua University 1 Journal of Intelligent and Fuzzy Systems 1 Journal of Beijing University of Technology 1 Journal of Hubei University. Natural Science Edition 1 Computational Biology and Chemistry 1 Science in China. Series G 1 Science in China. Series F 1 Journal of Shanghai Jiaotong University (Science) 1 The European Physical Journal C. Particles and Fields 1 Complex Variables and Elliptic Equations 1 Acta Mechanica Sinica 1 Nonlinear Oscillations 1 Journal of Statistical Theory and Practice 1 Current Development in Theory and Applications of Wavelets 1 Communications in Theoretical Physics 1 Journal of Nonlinear Science and Applications 1 Journal of Control Science and Engineering 1 Asian Journal of Control 1 International Journal of Structural Stability and Dynamics 1 Journal of Theoretical Biology 1 Journal of Applied Analysis and Computation 1 Control Theory and Technology
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#### Fields
56 Systems theory; control (93-XX) 23 Numerical analysis (65-XX) 21 Mechanics of deformable solids (74-XX) 19 Partial differential equations (35-XX) 14 Ordinary differential equations (34-XX) 13 Computer science (68-XX) 13 Information and communication theory, circuits (94-XX) 12 Fluid mechanics (76-XX) 12 Quantum theory (81-XX) 11 Operations research, mathematical programming (90-XX) 8 Biology and other natural sciences (92-XX) 7 Dynamical systems and ergodic theory (37-XX) 6 Number theory (11-XX) 5 Operator theory (47-XX) 5 Optics, electromagnetic theory (78-XX) 4 Calculus of variations and optimal control; optimization (49-XX) 4 Statistics (62-XX) 4 Game theory, economics, finance, and other social and behavioral sciences (91-XX) 3 Combinatorics (05-XX) 3 Functions of a complex variable (30-XX) 3 Probability theory and stochastic processes (60-XX) 3 Statistical mechanics, structure of matter (82-XX) 2 Several complex variables and analytic spaces (32-XX) 2 Harmonic analysis on Euclidean spaces (42-XX) 2 Differential geometry (53-XX) 2 Classical thermodynamics, heat transfer (80-XX) 2 Geophysics (86-XX) 1 General and overarching topics; collections (00-XX) 1 Mathematical logic and foundations (03-XX) 1 Group theory and generalizations (20-XX) 1 Real functions (26-XX) 1 Measure and integration (28-XX) 1 Difference and functional equations (39-XX) 1 Approximations and expansions (41-XX) 1 Functional analysis (46-XX) 1 Mechanics of particles and systems (70-XX) 1 Relativity and gravitational theory (83-XX)
#### Citations contained in zbMATH Open
78 Publications have been cited 584 times in 515 Documents Cited by Year
A new chaotic attractor. Zbl 1060.37027
Liu, Chongxin; Liu, Tao; Liu, Ling; Liu, Kai
2004
Synchronization of complex dynamical networks with switching topology: a switched system point of view. Zbl 1183.93032
Zhao, Jun; Hill, David J.; Liu, Tao
2009
Synchronization of complex delayed dynamical networks with nonlinearly coupled nodes. Zbl 1197.34092
Liu, Tao; Zhao, Jun; Hill, David J.
2009
A novel three-dimensional autonomous chaos system. Zbl 1197.37039
Liu, Chongxin; Liu, Ling; Liu, Tao
2009
Global bounded synchronization of general dynamical networks with nonidentical nodes. Zbl 1369.93061
Zhao, Jun; Hill, David J.; Liu, Tao
2012
Stability of dynamical networks with non-identical nodes: a multiple $$V$$-Lyapunov function method. Zbl 1235.93214
Zhao, Jun; Hill, David J.; Liu, Tao
2011
Supersymmetric Pati-Salam models from intersecting D6-branes: a road to the standard model. Zbl 1123.81392
Cvetič, Mirjam; Li, Tianjun; Liu, Tao
2004
A new butterfly-shaped attractor of Lorenz-like system. Zbl 1106.37025
Liu, Chongxin; Liu, Ling; Liu, Tao; Li, Peng
2006
An inverse problem for space-fractional backward diffusion problem. Zbl 06303262
Zhao, Jingjun; Liu, Songshu; Liu, Tao
2014
Representation of odd integers as the sum of one prime, two squares of primes and powers of 2. Zbl 1080.11072
Liu, Tao
2004
A synthetic approach for robust constrained iterative learning control of piecewise affine batch processes. Zbl 1252.93052
Liu, Tao; Wang, Youqing
2012
D6-brane splitting on type IIA orientifolds. Zbl 1160.81441
Cvetič, Mirjam; Langacker, Paul; Li, Tianjun; Liu, Tao
2005
Classification of supersymmetric flux vacua in M-theory. Zbl 1214.81189
Behrndt, Klaus; Cvetič, Mirjam; Liu, Tao
2006
Design optimization of truss-cored sandwiches with homogenization. Zbl 1120.74696
Liu, T.; Deng, Z. C.; Lu, T. J.
2006
Distributed event-triggered control for asymptotic synchronization of dynamical networks. Zbl 1375.93081
Liu, Tao; Cao, Ming; De Persis, Claudio; Hendrickx, Julien M.
2017
Inversion medians outperform breakpoint medians in phylogeny reconstruction from gene-order data. Zbl 1016.68645
Moret, Bernard M. E.; Siepel, Adam C.; Tang, Jijun; Liu, Tao
2002
Synchronization of dynamical networks by network control. Zbl 1369.93034
Liu, Tao; Hill, David J.; Zhao, Jun
2012
Analytic center of nilpotent critical points. Zbl 1258.34055
Liu, Tao; Wu, Lianggang; Li, Feng
2012
$$H_\infty$$ control for networked control systems with limited information. Zbl 1254.93155
Liu, Tao; Zhang, Hao; Chen, Qijun; Yan, Huaicheng
2012
A new method of modeling the conditional scalar dissipation rate. Zbl 1186.76140
Devaud, C. B.; Bilger, R. W.; Liu, T.
2004
Dispersion analysis of the spectral element method using a triangular mesh. Zbl 1360.74155
Liu, Tao; Sen, Mrinal K.; Hu, Tianyue; De Basabe, Jonas D.; Li, Lin
2012
Stability of discrete-time delayed impulsive linear systems with application to multi-tracking. Zbl 1291.93271
Liu, Bin; Liu, Tao; Dou, Chun-Xia
2014
Supersymmetric standard models, flux compactification and moduli stabilization. Zbl 1247.81361
Cvetič, Mirjam; Liu, Tao
2005
Synchronization of complex switched delay dynamical networks with simultaneously diagonalizable coupling matrices. Zbl 1199.93255
Liu, Tao; Zhao, Jun
2008
Cooperative output regulation for a class of nonlinear multi-agent systems with unknown control directions subject to switching networks. Zbl 1390.93107
Liu, Tao; Huang, Jie
2018
Minimum weights of pressurized hollow sandwich cylinders with ultralight cellular cores. Zbl 1121.74437
Liu, T.; Deng, Z. C.; Lu, T. J.
2007
Robust time-domain output error method for identifying continuous-time systems with time delay. Zbl 1377.93060
Chen, Fengwei; Gilson, Marion; Garnier, Hugues; Liu, Tao
2017
Robust $$H_2$$ filtering for discrete-time Markovian jump linear system. Zbl 1303.93174
Liu, Tao; Zhang, Hao; Chen, Qijun
2012
Controller parameterization for SISO and MIMO plants with time delay. Zbl 1100.93038
Zhang, Weidong; Allgower, Frank; Liu, Tao
2006
Stability analysis of linear 2-D systems. Zbl 1151.94389
Liu, Tao
2008
Bi-functional optimization of actively cooled, pressurized hollow sandwich cylinders with prismatic cores. Zbl 1159.74395
Liu, T.; Deng, Z. C.; Lu, T. J.
2007
Identification of space-dependent permeability in nonlinear diffusion equation from interior measurements using wavelet multiscale method. Zbl 1308.65162
Zhao, Jingjun; Liu, Tao; Liu, Songshu
2014
An adaptive homotopy method for permeability estimation of a nonlinear diffusion equation. Zbl 1279.35063
Zhao, Jingjun; Liu, Tao; Liu, Songshu
2013
Full state hybrid projective synchronization of variable-order fractional chaotic/hyperchaotic systems with nonlinear external disturbances and unknown parameters. Zbl 1330.34085
Zhang, Li; Liu, Tao
2016
Blind estimation of carrier frequency offset for OFDM systems with time-varying DC offset. Zbl 1293.93793
Liu, Tao; Li, Hanzhang
2014
Semilocal convergence analysis for inexact Newton method under weak condition. Zbl 1246.90146
Xu, Xiubin; Xiao, Yuan; Liu, Tao
2012
Leader-following attitude consensus of multiple rigid body systems subject to jointly connected switching networks. Zbl 1388.93009
Liu, Tao; Huang, Jie
2018
Reconstruction of a permeability field with the wavelet multiscale-homotopy method for a nonlinear convection-diffusion equation. Zbl 1410.76432
Liu, Tao
2016
A distributed observer for a class of nonlinear systems and its application to a leader-following consensus problem. Zbl 07044271
Liu, Tao; Huang, Jie
2019
Robust step-like identification of low-order process model under nonzero initial conditions and disturbance. Zbl 1367.93677
Liu, Tao; Gao, Furong
2008
A wavelet multiscale method for the inverse problem of a nonlinear convection-diffusion equation. Zbl 1376.65125
Liu, Tao
2018
EM-based identification of continuous-time ARMA models from irregularly sampled data. Zbl 1355.93189
Chen, Fengwei; Agüero, Juan C.; Gilson, Marion; Garnier, Hugues; Liu, Tao
2017
Leader-following formation tracking control of mobile robots without direct position measurements. Zbl 1359.93318
Liang, Xinwu; Liu, Yun-Hui; Wang, Hesheng; Chen, Weidong; Xing, Kexin; Liu, Tao
2016
Identification of diffusion parameters in a non-linear convection-diffusion equation using adaptive homotopy perturbation method. Zbl 1388.76260
Liu, Tao; Liu, Songshu
2018
Bending of linearly tapered annular Mindlin plates. Zbl 0985.74036
Liu, T.; Kitipornchai, S.; Wang, C. M.
2001
Elastic structural response of prismatic metal sandwich tubes to internal moving pressure loading. Zbl 1217.74036
Zhou, Jiaxi; Deng, Zichen; Liu, Tao; Hou, Xiuhui
2009
A nonlinear multigrid method for inversion of two-dimensional acoustic wave equation. Zbl 1290.35339
Zhao, Jingjun; Liu, Tao; Feng, Guofeng
2014
Quasi-supersymmetric $$G^3$$ unification from intersecting D6-branes on Type IIA orientifolds. Zbl 1058.81773
Li, Tianjun; Liu, Tao
2003
Blind carrier frequency offset estimation in OFDM systems with I/Q imbalance. Zbl 1169.94324
Liu, Tao; Li, Hanzhang
2009
An uncertain linguistic multi-criteria group decision-making approach based on integrated cloud. Zbl 1274.90184
Wang, Jianqiang; Liu, Tao
2012
Rigidity of complete spacelike translating solitons in pseudo-Euclidean space. Zbl 1421.35184
Xu, Ruiwei; Liu, Tao
2019
A perturbation of nonlinear scalar field equations. Zbl 1412.35084
Liu, Jiu; Liu, Tao; Liao, Jia-Feng
2019
High-energy limit of mass-suppressed amplitudes in gauge theories. Zbl 1404.81289
Liu, Tao; Penin, Alexander
2018
An adaptive multigrid conjugate gradient method for permeability identification of nonlinear diffusion equation. Zbl 1398.76175
Zhao, Jingjun; Liu, Tao
2016
Liu, T.
1996
$$H_\infty$$ control for networked control systems with communication constraints. Zbl 1289.93149
Liu, Tao; Zhang, Hao; Chen, Qijun
2013
Anomalous magnetic moment with heavy virtual leptons. Zbl 1284.81312
Kurz, Alexander; Liu, Tao; Marquard, Peter; Steinhauser, Matthias
2014
A comparison of regularization methods for identifying unknown source problem for the modified Helmholtz equation. Zbl 1284.47012
Zhao, Jingjun; Liu, Songshu; Liu, Tao
2014
Joint estimation of carrier frequency offset, dc offset and I/Q imbalance for OFDM systems. Zbl 1219.94041
Liu, Tao; Li, Hanzhang
2011
A generalized relay identification method for time delay and non-minimum phase processes. Zbl 1162.93329
Liu, Tao; Gao, Furong
2009
Analytical modeling and finite element simulation of the plastic collapse of sandwich beams with pin-reinforced foam cores. Zbl 1169.74565
Liu, Tao; Deng, Zichen; Lu, Tianjian
2008
Structural modeling of sandwich structures with lightweight cellular cores. Zbl 1202.74137
Liu, T.; Deng, Z. C.; Lu, T. J.
2007
A wavelet multiscale-homotopy method for the parameter identification problem of partial differential equations. Zbl 1443.42024
Liu, Tao
2016
Recursive subspace identification subject to relatively slow time-varying load disturbance. Zbl 1396.93123
Hou, Jie; Liu, Tao; Wang, Qing-Guo
2018
A nonlinear multigrid method for inverse problem in the multiphase porous media flow. Zbl 1427.65227
Liu, Tao
2018
Output feedback anti-disturbance control of input-delayed systems with time-varying uncertainties. Zbl 1415.93091
Hao, Shoulin; Liu, Tao; Zhou, Bin
2019
Ground state solution on a Kirchhoff type equation involving two potentials. Zbl 1412.35005
Liu, Jiu; Liu, Tao; Li, Hong-Ying
2019
An adaptive multigrid conjugate gradient method for the inversion of a nonlinear convection-diffusion equation. Zbl 1398.35255
Liu, Tao
2018
Trajectory optimization based on multi-interval mesh refinement method. Zbl 1426.49034
Li, Ningbo; Lei, Humin; Shao, Lei; Liu, Tao; Wang, Bin
2017
A result on a non-autonomous Kirchhoff type equation involving critical term. Zbl 1404.35177
Liu, Jiu; Liu, Tao; Pan, Hui-Lan
2018
Adaptive cooperative output regulation of discrete-time linear multi-agent systems by a distributed feedback control law. Zbl 1423.93173
Liu, Tao; Huang, Jie
2018
Control of finite-time anti-synchronization for variable-order fractional chaotic systems with unknown parameters. Zbl 1371.93096
Zhang, Li; Yu, Chenglong; Liu, Tao
2016
Exponential input-to-state stability under events for hybrid dynamical networks with coupling time-delays. Zbl 1373.93272
Liu, Bin; Hill, David J.; Liu, Tao
2017
Three-loop quark form factor at high energy: the leading mass corrections. Zbl 1372.81151
Liu, Tao; Penin, Alexander A.; Zerf, Nikolai
2017
A wavelet multiscale-adaptive homotopy method for the inverse problem of nonlinear diffusion equation. Zbl 1326.65126
Zhao, Jingjun; Liu, Tao
2015
Long-term dynamics of autonomous fractional differential equations. Zbl 1338.34020
Liu, Tao; Xu, Wei; Xu, Yong; Han, Qun
2016
A new regularization method for Cauchy problem of elliptic equation. Zbl 1304.47015
Zhao, Jingjun; Liu, Songshu; Liu, Tao
2014
Adaptive mesh refinement of hp pseudospectral method using mesh size reduction. Zbl 1374.49035
Lei, Humin; Liu, Tao; Li, Jiong; Jiang, Zhipeng
2016
A distributed observer for a class of nonlinear systems and its application to a leader-following consensus problem. Zbl 07044271
Liu, Tao; Huang, Jie
2019
Rigidity of complete spacelike translating solitons in pseudo-Euclidean space. Zbl 1421.35184
Xu, Ruiwei; Liu, Tao
2019
A perturbation of nonlinear scalar field equations. Zbl 1412.35084
Liu, Jiu; Liu, Tao; Liao, Jia-Feng
2019
Output feedback anti-disturbance control of input-delayed systems with time-varying uncertainties. Zbl 1415.93091
Hao, Shoulin; Liu, Tao; Zhou, Bin
2019
Ground state solution on a Kirchhoff type equation involving two potentials. Zbl 1412.35005
Liu, Jiu; Liu, Tao; Li, Hong-Ying
2019
Cooperative output regulation for a class of nonlinear multi-agent systems with unknown control directions subject to switching networks. Zbl 1390.93107
Liu, Tao; Huang, Jie
2018
Leader-following attitude consensus of multiple rigid body systems subject to jointly connected switching networks. Zbl 1388.93009
Liu, Tao; Huang, Jie
2018
A wavelet multiscale method for the inverse problem of a nonlinear convection-diffusion equation. Zbl 1376.65125
Liu, Tao
2018
Identification of diffusion parameters in a non-linear convection-diffusion equation using adaptive homotopy perturbation method. Zbl 1388.76260
Liu, Tao; Liu, Songshu
2018
High-energy limit of mass-suppressed amplitudes in gauge theories. Zbl 1404.81289
Liu, Tao; Penin, Alexander
2018
Recursive subspace identification subject to relatively slow time-varying load disturbance. Zbl 1396.93123
Hou, Jie; Liu, Tao; Wang, Qing-Guo
2018
A nonlinear multigrid method for inverse problem in the multiphase porous media flow. Zbl 1427.65227
Liu, Tao
2018
An adaptive multigrid conjugate gradient method for the inversion of a nonlinear convection-diffusion equation. Zbl 1398.35255
Liu, Tao
2018
A result on a non-autonomous Kirchhoff type equation involving critical term. Zbl 1404.35177
Liu, Jiu; Liu, Tao; Pan, Hui-Lan
2018
Adaptive cooperative output regulation of discrete-time linear multi-agent systems by a distributed feedback control law. Zbl 1423.93173
Liu, Tao; Huang, Jie
2018
Distributed event-triggered control for asymptotic synchronization of dynamical networks. Zbl 1375.93081
Liu, Tao; Cao, Ming; De Persis, Claudio; Hendrickx, Julien M.
2017
Robust time-domain output error method for identifying continuous-time systems with time delay. Zbl 1377.93060
Chen, Fengwei; Gilson, Marion; Garnier, Hugues; Liu, Tao
2017
EM-based identification of continuous-time ARMA models from irregularly sampled data. Zbl 1355.93189
Chen, Fengwei; Agüero, Juan C.; Gilson, Marion; Garnier, Hugues; Liu, Tao
2017
Trajectory optimization based on multi-interval mesh refinement method. Zbl 1426.49034
Li, Ningbo; Lei, Humin; Shao, Lei; Liu, Tao; Wang, Bin
2017
Exponential input-to-state stability under events for hybrid dynamical networks with coupling time-delays. Zbl 1373.93272
Liu, Bin; Hill, David J.; Liu, Tao
2017
Three-loop quark form factor at high energy: the leading mass corrections. Zbl 1372.81151
Liu, Tao; Penin, Alexander A.; Zerf, Nikolai
2017
Full state hybrid projective synchronization of variable-order fractional chaotic/hyperchaotic systems with nonlinear external disturbances and unknown parameters. Zbl 1330.34085
Zhang, Li; Liu, Tao
2016
Reconstruction of a permeability field with the wavelet multiscale-homotopy method for a nonlinear convection-diffusion equation. Zbl 1410.76432
Liu, Tao
2016
Leader-following formation tracking control of mobile robots without direct position measurements. Zbl 1359.93318
Liang, Xinwu; Liu, Yun-Hui; Wang, Hesheng; Chen, Weidong; Xing, Kexin; Liu, Tao
2016
An adaptive multigrid conjugate gradient method for permeability identification of nonlinear diffusion equation. Zbl 1398.76175
Zhao, Jingjun; Liu, Tao
2016
A wavelet multiscale-homotopy method for the parameter identification problem of partial differential equations. Zbl 1443.42024
Liu, Tao
2016
Control of finite-time anti-synchronization for variable-order fractional chaotic systems with unknown parameters. Zbl 1371.93096
Zhang, Li; Yu, Chenglong; Liu, Tao
2016
Long-term dynamics of autonomous fractional differential equations. Zbl 1338.34020
Liu, Tao; Xu, Wei; Xu, Yong; Han, Qun
2016
Adaptive mesh refinement of hp pseudospectral method using mesh size reduction. Zbl 1374.49035
Lei, Humin; Liu, Tao; Li, Jiong; Jiang, Zhipeng
2016
A wavelet multiscale-adaptive homotopy method for the inverse problem of nonlinear diffusion equation. Zbl 1326.65126
Zhao, Jingjun; Liu, Tao
2015
An inverse problem for space-fractional backward diffusion problem. Zbl 06303262
Zhao, Jingjun; Liu, Songshu; Liu, Tao
2014
Stability of discrete-time delayed impulsive linear systems with application to multi-tracking. Zbl 1291.93271
Liu, Bin; Liu, Tao; Dou, Chun-Xia
2014
Identification of space-dependent permeability in nonlinear diffusion equation from interior measurements using wavelet multiscale method. Zbl 1308.65162
Zhao, Jingjun; Liu, Tao; Liu, Songshu
2014
Blind estimation of carrier frequency offset for OFDM systems with time-varying DC offset. Zbl 1293.93793
Liu, Tao; Li, Hanzhang
2014
A nonlinear multigrid method for inversion of two-dimensional acoustic wave equation. Zbl 1290.35339
Zhao, Jingjun; Liu, Tao; Feng, Guofeng
2014
Anomalous magnetic moment with heavy virtual leptons. Zbl 1284.81312
Kurz, Alexander; Liu, Tao; Marquard, Peter; Steinhauser, Matthias
2014
A comparison of regularization methods for identifying unknown source problem for the modified Helmholtz equation. Zbl 1284.47012
Zhao, Jingjun; Liu, Songshu; Liu, Tao
2014
A new regularization method for Cauchy problem of elliptic equation. Zbl 1304.47015
Zhao, Jingjun; Liu, Songshu; Liu, Tao
2014
An adaptive homotopy method for permeability estimation of a nonlinear diffusion equation. Zbl 1279.35063
Zhao, Jingjun; Liu, Tao; Liu, Songshu
2013
$$H_\infty$$ control for networked control systems with communication constraints. Zbl 1289.93149
Liu, Tao; Zhang, Hao; Chen, Qijun
2013
Global bounded synchronization of general dynamical networks with nonidentical nodes. Zbl 1369.93061
Zhao, Jun; Hill, David J.; Liu, Tao
2012
A synthetic approach for robust constrained iterative learning control of piecewise affine batch processes. Zbl 1252.93052
Liu, Tao; Wang, Youqing
2012
Synchronization of dynamical networks by network control. Zbl 1369.93034
Liu, Tao; Hill, David J.; Zhao, Jun
2012
Analytic center of nilpotent critical points. Zbl 1258.34055
Liu, Tao; Wu, Lianggang; Li, Feng
2012
$$H_\infty$$ control for networked control systems with limited information. Zbl 1254.93155
Liu, Tao; Zhang, Hao; Chen, Qijun; Yan, Huaicheng
2012
Dispersion analysis of the spectral element method using a triangular mesh. Zbl 1360.74155
Liu, Tao; Sen, Mrinal K.; Hu, Tianyue; De Basabe, Jonas D.; Li, Lin
2012
Robust $$H_2$$ filtering for discrete-time Markovian jump linear system. Zbl 1303.93174
Liu, Tao; Zhang, Hao; Chen, Qijun
2012
Semilocal convergence analysis for inexact Newton method under weak condition. Zbl 1246.90146
Xu, Xiubin; Xiao, Yuan; Liu, Tao
2012
An uncertain linguistic multi-criteria group decision-making approach based on integrated cloud. Zbl 1274.90184
Wang, Jianqiang; Liu, Tao
2012
Stability of dynamical networks with non-identical nodes: a multiple $$V$$-Lyapunov function method. Zbl 1235.93214
Zhao, Jun; Hill, David J.; Liu, Tao
2011
Joint estimation of carrier frequency offset, dc offset and I/Q imbalance for OFDM systems. Zbl 1219.94041
Liu, Tao; Li, Hanzhang
2011
Synchronization of complex dynamical networks with switching topology: a switched system point of view. Zbl 1183.93032
Zhao, Jun; Hill, David J.; Liu, Tao
2009
Synchronization of complex delayed dynamical networks with nonlinearly coupled nodes. Zbl 1197.34092
Liu, Tao; Zhao, Jun; Hill, David J.
2009
A novel three-dimensional autonomous chaos system. Zbl 1197.37039
Liu, Chongxin; Liu, Ling; Liu, Tao
2009
Elastic structural response of prismatic metal sandwich tubes to internal moving pressure loading. Zbl 1217.74036
Zhou, Jiaxi; Deng, Zichen; Liu, Tao; Hou, Xiuhui
2009
Blind carrier frequency offset estimation in OFDM systems with I/Q imbalance. Zbl 1169.94324
Liu, Tao; Li, Hanzhang
2009
A generalized relay identification method for time delay and non-minimum phase processes. Zbl 1162.93329
Liu, Tao; Gao, Furong
2009
Synchronization of complex switched delay dynamical networks with simultaneously diagonalizable coupling matrices. Zbl 1199.93255
Liu, Tao; Zhao, Jun
2008
Stability analysis of linear 2-D systems. Zbl 1151.94389
Liu, Tao
2008
Robust step-like identification of low-order process model under nonzero initial conditions and disturbance. Zbl 1367.93677
Liu, Tao; Gao, Furong
2008
Analytical modeling and finite element simulation of the plastic collapse of sandwich beams with pin-reinforced foam cores. Zbl 1169.74565
Liu, Tao; Deng, Zichen; Lu, Tianjian
2008
Minimum weights of pressurized hollow sandwich cylinders with ultralight cellular cores. Zbl 1121.74437
Liu, T.; Deng, Z. C.; Lu, T. J.
2007
Bi-functional optimization of actively cooled, pressurized hollow sandwich cylinders with prismatic cores. Zbl 1159.74395
Liu, T.; Deng, Z. C.; Lu, T. J.
2007
Structural modeling of sandwich structures with lightweight cellular cores. Zbl 1202.74137
Liu, T.; Deng, Z. C.; Lu, T. J.
2007
A new butterfly-shaped attractor of Lorenz-like system. Zbl 1106.37025
Liu, Chongxin; Liu, Ling; Liu, Tao; Li, Peng
2006
Classification of supersymmetric flux vacua in M-theory. Zbl 1214.81189
Behrndt, Klaus; Cvetič, Mirjam; Liu, Tao
2006
Design optimization of truss-cored sandwiches with homogenization. Zbl 1120.74696
Liu, T.; Deng, Z. C.; Lu, T. J.
2006
Controller parameterization for SISO and MIMO plants with time delay. Zbl 1100.93038
Zhang, Weidong; Allgower, Frank; Liu, Tao
2006
D6-brane splitting on type IIA orientifolds. Zbl 1160.81441
Cvetič, Mirjam; Langacker, Paul; Li, Tianjun; Liu, Tao
2005
Supersymmetric standard models, flux compactification and moduli stabilization. Zbl 1247.81361
Cvetič, Mirjam; Liu, Tao
2005
A new chaotic attractor. Zbl 1060.37027
Liu, Chongxin; Liu, Tao; Liu, Ling; Liu, Kai
2004
Supersymmetric Pati-Salam models from intersecting D6-branes: a road to the standard model. Zbl 1123.81392
Cvetič, Mirjam; Li, Tianjun; Liu, Tao
2004
Representation of odd integers as the sum of one prime, two squares of primes and powers of 2. Zbl 1080.11072
Liu, Tao
2004
A new method of modeling the conditional scalar dissipation rate. Zbl 1186.76140
Devaud, C. B.; Bilger, R. W.; Liu, T.
2004
Quasi-supersymmetric $$G^3$$ unification from intersecting D6-branes on Type IIA orientifolds. Zbl 1058.81773
Li, Tianjun; Liu, Tao
2003
Inversion medians outperform breakpoint medians in phylogeny reconstruction from gene-order data. Zbl 1016.68645
Moret, Bernard M. E.; Siepel, Adam C.; Tang, Jijun; Liu, Tao
2002
Bending of linearly tapered annular Mindlin plates. Zbl 0985.74036
Liu, T.; Kitipornchai, S.; Wang, C. M.
2001
Liu, T.
1996
all top 5
#### Cited by 1,008 Authors
23 Vaidyanathan, Sundarapandian 19 Liu, Tao 12 Azar, Ahmad Taher 9 Fang, Jian’an 8 Liu, Chongxin 8 Ouannas, Adel 8 Wang, Xingyuan 6 Liu, Ling 6 Xu, Yuhua 6 Zhou, Wuneng 5 Cvetič, Mirjam 5 Hai, Dinh Nguyen Duy 5 Liu, Zhixin 5 Lu, Guangshi 5 Park, Juhyun (Jessie) 5 Tang, Wansheng 5 Wang, Yinhe 5 Yao, Hongxing 5 Zhang, Jianxiong 5 Zhang, Lili 5 Ziar, Toufik 4 Bhalekar, Sachin 4 Bi, Qinsheng 4 Cao, Jinde 4 Chen, Guanrong 4 Chen, Zengqiang 4 Chu, Yandong 4 Dadras, Sara 4 Daftardar-Gejji, Varsha 4 di Bernardo, Mario 4 Li, Xianyi 4 Li, Zhen 4 Momeni, Hamidreza 4 Trong, Dang Duc 4 Wang, Lei 4 Wang, Qingguo 4 Wang, Zuolei 4 Zhang, Jiangang 3 Alsaadi, Fuad Eid S. 3 Anastasopoulos, Pascal 3 Bao, Bocheng 3 Bian, Qiuxiang 3 Chang, Yingxiang 3 Chen, Diyi 3 Chen, Fengwei 3 Chen, Michael Z. Q. 3 DeLellis, Pietro 3 El-Dessoky, Mohamed M. 3 Garnier, Hugues 3 Gilson, Marion 3 Hill, David John 3 Honecker, Gabriele 3 Huang, Lihong 3 Jiang, Bin 3 Jin, Xiaozheng 3 Karimi, Hamid Reza 3 Kiritsis, Elias B. 3 Lam, James 3 Li, Xianfeng 3 Li, Xianwei 3 Liu, Yongjian 3 Lüst, Dieter 3 Nguyen Huy Tuan 3 Wang, Guoliang 3 Wang, Haijun 3 Wang, Mingjun 3 Wang, Shuguo 3 Wang, Yanwu 3 Wang, Yung-Yi 3 Wang, Zhen 3 Wu, Ranchao 3 Wu, Xiangjun 3 Wu, Yusen 3 Wu, Zhengguang 3 Xiang, Zhengrong 3 Yang, Hao 3 Yu, Pei 3 Zhang, Huaguang 3 Zhang, Yanbin 3 Zheng, Guanghui 3 Zhu, Qiao 2 Algaba, Antonio 2 Alizadeh, Ghasem 2 Aly, E. S. 2 Antonietti, Paola Francesca 2 Arellano-Delgado, Adrian 2 Belozyorov, Vasiliy Ye. 2 Blumenhagen, Ralph 2 Botmart, Thongchai 2 Čelikovský, Sergej 2 Chen, Ching-Ming 2 Chen, Fangqi 2 Chen, Xiaohong 2 Dai, Yunxian 2 De Persis, Claudio 2 Ding, Jian 2 Dong, Jiuxiang 2 Doungmo Goufo, Emile Franc 2 Du, Bo 2 Eriksen, Niklas ...and 908 more Authors
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#### Cited in 112 Serials
43 Journal of the Franklin Institute 40 Nonlinear Dynamics 28 Automatica 26 Chaos, Solitons and Fractals 25 International Journal of Bifurcation and Chaos in Applied Sciences and Engineering 24 Communications in Nonlinear Science and Numerical Simulation 18 Applied Mathematics and Computation 16 Mathematical Problems in Engineering 15 Nuclear Physics. B 14 Chaos 14 Journal of High Energy Physics 11 Asian Journal of Control 10 Systems & Control Letters 9 International Journal of Robust and Nonlinear Control 9 Abstract and Applied Analysis 9 International Journal of Systems Science. Principles and Applications of Systems and Integration 8 International Journal of Modern Physics B 6 Computers & Mathematics with Applications 6 Complexity 6 Advances in Difference Equations 5 Information Sciences 5 Journal of Computational and Applied Mathematics 5 Journal of Number Theory 5 Archives of Control Sciences 4 Journal of Inverse and Ill-Posed Problems 4 Discrete Dynamics in Nature and Society 4 Nonlinear Analysis. Hybrid Systems 3 International Journal of Modern Physics A 3 International Journal of Control 3 Mathematical Methods in the Applied Sciences 3 Theoretical Computer Science 3 Applied Mathematics Letters 3 Mathematical and Computer Modelling 3 International Journal of Adaptive Control and Signal Processing 3 Applied Mathematical Modelling 3 Journal of Applied Mathematics 3 Inverse Problems in Science and Engineering 3 Journal of Nonlinear Science and Applications 2 Modern Physics Letters B 2 Acta Mechanica 2 Journal of Computational Physics 2 Physics Letters. A 2 Fortschritte der Physik 2 Circuits, Systems, and Signal Processing 2 Neural Networks 2 Multidimensional Systems and Signal Processing 2 European Journal of Control 2 Journal of Vibration and Control 2 Wuhan University Journal of Natural Sciences (WUJNS) 2 Journal of Systems Science and Complexity 2 Journal of Discrete Algorithms 2 International Journal of Number Theory 2 Advances in Mathematical Physics 2 Journal of Mathematics 2 Journal of Function Spaces 1 Applicable Analysis 1 Classical and Quantum Gravity 1 Communications in Mathematical Physics 1 Computer Methods in Applied Mechanics and Engineering 1 Journal of Mathematical Analysis and Applications 1 Physics Reports 1 Reviews of Modern Physics 1 Acta Arithmetica 1 Acta Mathematica Vietnamica 1 Demonstratio Mathematica 1 Journal of Differential Equations 1 Kybernetika 1 Mathematics and Computers in Simulation 1 Meccanica 1 Monatshefte für Mathematik 1 Nonlinear Analysis. Theory, Methods & Applications. Series A: Theory and Methods 1 Proceedings of the American Mathematical Society 1 Results in Mathematics 1 SIAM Journal on Control and Optimization 1 Advances in Applied Mathematics 1 Applied Mathematics and Mechanics. (English Edition) 1 Acta Mathematica Hungarica 1 Applied Numerical Mathematics 1 Optimization 1 Journal of Scientific Computing 1 European Journal of Applied Mathematics 1 Signal Processing 1 Automation and Remote Control 1 International Journal of Computer Mathematics 1 Linear Algebra and its Applications 1 Applicationes Mathematicae 1 Computational and Applied Mathematics 1 Science in China. Series E 1 The Ramanujan Journal 1 Differential Equations and Dynamical Systems 1 Soft Computing 1 Mathematical and Computer Modelling of Dynamical Systems 1 Journal of Dynamical and Control Systems 1 Physical Review Letters 1 Nonlinear Analysis. Real World Applications 1 Nonlinear Analysis. Modelling and Control 1 Dynamical Systems 1 Discrete and Continuous Dynamical Systems. Series B 1 Physical Review D. Series III 1 Analysis in Theory and Applications ...and 12 more Serials
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#### Cited in 38 Fields
265 Systems theory; control (93-XX) 187 Ordinary differential equations (34-XX) 102 Dynamical systems and ergodic theory (37-XX) 42 Quantum theory (81-XX) 38 Numerical analysis (65-XX) 32 Computer science (68-XX) 27 Partial differential equations (35-XX) 23 Information and communication theory, circuits (94-XX) 19 Biology and other natural sciences (92-XX) 16 Probability theory and stochastic processes (60-XX) 16 Operations research, mathematical programming (90-XX) 15 Relativity and gravitational theory (83-XX) 12 Number theory (11-XX) 11 Combinatorics (05-XX) 10 Real functions (26-XX) 9 Mechanics of particles and systems (70-XX) 8 Differential geometry (53-XX) 8 Fluid mechanics (76-XX) 7 Calculus of variations and optimal control; optimization (49-XX) 7 Mechanics of deformable solids (74-XX) 6 Operator theory (47-XX) 4 Statistics (62-XX) 4 Game theory, economics, finance, and other social and behavioral sciences (91-XX) 3 Algebraic geometry (14-XX) 2 Harmonic analysis on Euclidean spaces (42-XX) 2 Integral equations (45-XX) 2 Statistical mechanics, structure of matter (82-XX) 2 Geophysics (86-XX) 1 General and overarching topics; collections (00-XX) 1 Mathematical logic and foundations (03-XX) 1 Linear and multilinear algebra; matrix theory (15-XX) 1 Nonassociative rings and algebras (17-XX) 1 Special functions (33-XX) 1 Difference and functional equations (39-XX) 1 Integral transforms, operational calculus (44-XX) 1 Algebraic topology (55-XX) 1 Optics, electromagnetic theory (78-XX) 1 Classical thermodynamics, heat transfer (80-XX) | 2021-09-26 11:06:08 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6002861857414246, "perplexity": 10754.831506438259}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780057857.27/warc/CC-MAIN-20210926083818-20210926113818-00454.warc.gz"} |
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Now in terms of the glare control question, I would look at the useful daylight illuminance (UDI) rather than the mean illuminance. UDI illustrates the percentage of time a given space receives between 100 and 3,000 lux. I really like this metric, both for quantifying the, well, the utility of the daylight vis a vis energy savings from controls, but also the glare potential (or lack thereof), since the metric penalizes a design for over-illumination (>3,000 lux). This is using task plane horizontal illuminance as a proxy, but it's a reasonably quick way to look at a whole space or a whole building. You could also place glare sensors in your model and get annual vertical eye illuminance and draw your own conclusions that way. Unfortunately we do not have an elegant reporting measure that takes this daylight metric data and presents it to you like the default EnergyPlus output (this is a drag). You need to go into the text file output and parse it and visualize it some way, either with Excel or R or whatever. At any rate, the data can be found in the output directory of your run directory, e.g.:
[your_model]/run/[radiance_measure]-UserScript-0/radiance/output/daylight_metrics.csv | 2022-01-28 19:31:41 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5432137250900269, "perplexity": 1627.026125431846}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320306335.77/warc/CC-MAIN-20220128182552-20220128212552-00163.warc.gz"} |
https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=130&t=59131&p=237218 | ## Meaning of q=-w
$\Delta U=q+w$
Chloe Alviz 1E
Posts: 102
Joined: Sat Aug 17, 2019 12:17 am
### Meaning of q=-w
I know when ΔU = 0, but in these cases, how do I interpret q=-w?
Sebastian Lee 1L
Posts: 157
Joined: Fri Aug 09, 2019 12:15 am
Been upvoted: 1 time
### Re: Meaning of q=-w
Knowing that q=-w when the change in internal energy is 0 just helps you solve problems that may require work (but you're only given heat) or vice versa. You know that the work done by a system expanding isothermally is the opposite sign of the heat coming into the system. Note that the internal energy for an ideal gas will always be 0 if the temperature doesn't change (isothermal).
Jainam Shah 4I
Posts: 130
Joined: Fri Aug 30, 2019 12:16 am
### Re: Meaning of q=-w
When q=-w that means that internal energy is zero which typically is seen in isothermal reversible expansion.
Gurmukhi Bevli 4G
Posts: 49
Joined: Wed Nov 14, 2018 12:20 am
### Re: Meaning of q=-w
q=-w when the internal energy of a system is 0, (normally q=U-w). This happens when an isothermal reversible expansion is taking place since work is being done by the system (and heat is leaving the system).
Matthew Tsai 2H
Posts: 101
Joined: Wed Sep 18, 2019 12:20 am
### Re: Meaning of q=-w
When the change in internal energy is zero, it means that all the heat added to the system must be used by the system to do work.
Jacob Motawakel
Posts: 103
Joined: Wed Sep 18, 2019 12:20 am
Been upvoted: 1 time
### Re: Meaning of q=-w
When q=-w, internal energy is equal to 0, which indicates constant temperature.
Nuoya Jiang
Posts: 99
Joined: Sat Sep 14, 2019 12:17 am
### Re: Meaning of q=-w
Usually, in an isothermally reversible gas expansion, the work done by the heat equals the heat it received.
Betania Hernandez 2E
Posts: 107
Joined: Fri Aug 02, 2019 12:15 am
### Re: Meaning of q=-w
q=-w can help you figure out that the reaction is isothermal reversible which means that it has a constant temperature and that the internal energy equals zero.
J_CHEN 4I
Posts: 54
Joined: Tue Nov 14, 2017 3:01 am
### Re: Meaning of q=-w
The heat supplied to the system is equivalent to the energy leaving the system as work.
Areli C 1L
Posts: 95
Joined: Wed Nov 14, 2018 12:19 am
### Re: Meaning of q=-w
This usually means that its an isothermal reversible reaction and that the internal energy is 0.
Id also suggest watching the Organic chemistry tutor on youtube as he has videos on thermodynamics!
KHowe_1D
Posts: 103
Joined: Thu Jul 25, 2019 12:15 am
Been upvoted: 1 time
### Re: Meaning of q=-w
When delta U is zero that means the equation deltaU = q+w is now 0=q+w so you can now change the equation to be q=-w
w is negative because you can have negative work but you can't have negative heat energy. It essentially just means that the heat added to the system is the same as the work being done by the system.
TimVintsDis4L
Posts: 104
Joined: Sat Aug 17, 2019 12:17 am
### Re: Meaning of q=-w
If the Internal Energy of the System is equal to 0, then q must = - w so that they cancel out and you're left with 0.
Mai V 4L
Posts: 111
Joined: Fri Sep 28, 2018 12:23 am
### Re: Meaning of q=-w
Can q be constant in this case?
William Chan 1D
Posts: 102
Joined: Sat Sep 14, 2019 12:15 am
### Re: Meaning of q=-w
Basically, if the change in energy of the system is 0, then that means no energy is lost or gained in the system. That means that any heat is used to do work or vice versa.
Bryan Chen 1H
Posts: 58
Joined: Mon Jun 17, 2019 7:24 am
### Re: Meaning of q=-w
isothermal reversible expansion
Return to “Concepts & Calculations Using First Law of Thermodynamics”
### Who is online
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https://kubicle.com/learn/financial-modeling/the-cost-of-capital-part-2 | # 5. The Cost of Capital Part 2
Overview
To calculate the cost of equity for MarkerCo, we're going to use the Capital Asset Pricing Model (CAPM), a common but not uncontested technique among analysts.
To explore more Kubicle data literacy subjects, please refer to our full library.
Summary
1. Lesson Goal (00:04)
The goal of this lesson is to use the Capital Asset Pricing Model to calculate the cost of equity.
2. Understanding Types of Risk (00:33)
The Capital Asset Pricing Model, or CAPM, is a model that can be used to calculate the cost of equity. CAPM assumes that any investment contains two types of risk: systematic risk and unsystematic risk. Systematic risk is risk associated with any investment, such as the risk of an economic recession. Unsystematic risk is risk that is specific to the investment in question, such as the effect of rising oil prices on an energy company.
3. Principles of CAPM (01:11)
CAPM calculates the cost of equity by considering three variables. First is the risk-free rate of return, representing the return required for accepting systematic risk. Second is the additional rate of return for investing in more volatile equity markets, representing the return required for unsystematic risk. Third is the volatility of the stock in question compared to the market, representing the return required for accepting uncertainty.
This produces the following formula for the cost of equity:
$$r_E = r_F + (r_M - r_F)\beta$$
In the formula above:
• $$r_E$$ is the cost of equity we want to calculate
• $$r_F$$ is the risk-free rate of return. This is typically the rate of return on 10-year government bonds
• $$r_M$$ is the expected return of the market as a whole
• $$\beta$$, or beta, represents the volatility of the stock in question compared to the market. If this value is greater than one, the stock is more volatile than the market, if it less than one, the stock is less volatile than the market.
4. Issues with CAPM (02:38)
There are some issues with CAPM. For example, the risk-free rate should theoretically be constant, but the rate on government bonds can change. The expected market return can also fluctuate significantly. Some analysts prefer to consider the Internal Rate of Return instead of using CAPM to calculate the cost of equity.
Transcript
In the previous lesson we examined the formula for the Weighted Average Cost of Capital, or WACC. We then focused on the hardest value in this formula to calculate, which is often the cost of equity. In this lesson, we're going to use the Capital Asset Pricing Model to calculate the cost of equity, and separately examine ways that analysts often get around using WACC by implenting IRR calculations when valuing a business. The Capital Asset Pricing Model is developed under the assumption that investments contain two types of risks. The first type of risk is systematic risk, which is risk that cannot be diversified away. This is risk from events such as recessions, interest rate hikes, etc that effect the whole market. The second type of risk is unsystematic risk, which is risk specific to individual investments, which can be diversified away. An example of unsystematic risk might be a rise in oil prices that affect a company like Exxon Mobil. To account for both types of risk, the CAPM Model adopts the following approach to calculating the cost of capital. It starts with the risk-free rate of return, which is assumed typically to be a 10-year government bond. This is essentially the systematic risk. It then calculates a premium to be paid by investing in the equity markets, and this additional premium is the expected return of the market, rM, minus the risk-free rate of capital, rF.
In step three, we apply a value called beta, to the market risk premium, and beta measures how the stock in question fluctuates compared to the market. If beta is less than one, this means that the stock price is less volatile than the market. And this would typically be the case for energy or water utilities, which are pretty stable businesses. If beta is greater than one, then the stock is more volatile than the market, and this would certainly be the case for a lot of technology or biotechnology stocks. Putting these steps together, we can develop a formula for the cost of equity, which is equal to the risk-free rate plus the market risk premium, which is the expected market return, minus the risk-free rate, multiplied by beta.
It's important to stress at this point that the Capital Asset Pricing Model is quite controversial. It's not used by every analyst, and it does have some obvious weaknesses. For example, the risk-free rate is based on long-term government bonds, which can of course change in price. What's more, the expected market return can also change very frequently. In the following chart, I showed the one year return on the S and P 500 since 2012, and as you can see, the one year return fluctuates dramatically from just under 30% to minus 7.5%.
As a consequence, many analysts have chosen to ignore the CAPM Model and focus instead on the Internal Rate of Return. As you may have seen in previous courses, the internal rate of return is an effective way of understanding the return on any money invested in either an asset or a company, and in the coming lessons I'll show you in our Excel model how to use both the Capital Asset Pricing Model and Internal Rate of Return when valuing a business.
Financial Modeling Essentials
Contents
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05:00
#### Exercise 4
My Notes
You can take notes as you view lessons.
Sign in or start a free trial to avail of this feature. | 2022-05-17 01:48:59 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6715588569641113, "perplexity": 942.51245689583}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662515466.5/warc/CC-MAIN-20220516235937-20220517025937-00245.warc.gz"} |
http://math.stackexchange.com/questions/140215/implication-of-an-inequality | # Implication of an inequality
We know that
$$|l(x+u)-l(x)|<1 \text{ for } x\geq y>0 \text{ and } u\in[0,1]$$
Why does:
$$|l(y+u)|<1+|l(y)|,x \in (y+1,y+2)$$
imply that
$$|l(x)| \leq 1 + |l(y+1)| \leq 2+|l(y)|$$
-
What part of this question is not covered by @Joseph's answer below? – Did May 3 '12 at 12:54
The part that Antonio Vargas explained below. – Chris May 3 '12 at 13:48
This is true for any inequality of the form $|a-b|<1$. Since $|a-b|=|b-a|$ we can say both that $|a|<1+|b|$ and $|b|<1+|a|$. Then combining the two statements give that $|b|<1+|a|<1+(1+|b|)=2+|b|$
Ok, that explains the second part of the last double inequality, but why does $|l(x)| \leq 1 + |l(y+1)|$ follow? – Chris May 3 '12 at 3:17
what is $l(x)$? an arbitrary function? or a linear function? – Joseph Skelton May 3 '12 at 3:26
@Chris, if $x \in (y+1,y+2)$ then $x = y+1+u$ for some $u \in [0,1]$. Your assumption then gives $$|l(x)| = |l(y+1+u)| < 1 + |l(y+1)|.$$ – Antonio Vargas May 3 '12 at 5:18 | 2015-05-24 09:39:19 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9054120779037476, "perplexity": 1229.0589815872388}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-22/segments/1432207927863.72/warc/CC-MAIN-20150521113207-00081-ip-10-180-206-219.ec2.internal.warc.gz"} |
https://g-dfarms.com/products/doe-n-heat | # Doe N’ Heat, 2oz
Paula & Boyd’s
• \$9.99
Unit price per
Paula & Boyd's Doe N' Heat offers the strongest, freshest, most effective and most proven buck lure on the market.
Doe N' Heat is collected using their "Fresh Extraction Process," maximizing the effectiveness of the buck lure. While all of their deer scents provide a competitive advantage in the stand, Doe N' Heat goes above and beyond during peak rut.
Tip: Make sure to thoroughly soak your favorite scent wick and stake out the "four corners" of your stand set up, and you will quickly find that Doe N' Heat is the best deer attractant out there. Also, be sure to remove any deer scents placed in the woods after the hunt to ensure its effectiveness during future hunts. | 2021-06-16 19:25:09 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9480451941490173, "perplexity": 11783.611228059226}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487626008.14/warc/CC-MAIN-20210616190205-20210616220205-00128.warc.gz"} |
https://en.wikipedia.org/wiki/QR_code | # QR code
QR code for the URL of the English Wikipedia Mobile main page
A QR code (abbreviated from Quick Response code) is a type of matrix barcode (or two-dimensional barcode[1]) invented in 1994 by the Japanese automotive company Denso Wave.[2] A barcode is a machine-readable optical label that contains information about the item to which it is attached. In practice, QR codes often contain data for a locator, identifier, or tracker that points to a website or application. A QR code uses four standardized encoding modes (numeric, alphanumeric, byte/binary, and kanji) to store data efficiently; extensions may also be used.[3]
The Quick Response system became popular outside the automotive industry due to its fast readability and greater storage capacity compared to standard UPC barcodes. Applications include product tracking, item identification, time tracking, document management, and general marketing.[4]
A QR code consists of black squares arranged in a square grid on a white background, which can be read by an imaging device such as a camera, and processed using Reed–Solomon error correction until the image can be appropriately interpreted. The required data is then extracted from patterns that are present in both horizontal and vertical components of the image.[4]
## History
The QR code system was invented in 1994 by Masahiro Hara from the Japanese company Denso Wave. The initial design was influenced by the black and white pieces on a Go board.[5] Its purpose was to track vehicles during manufacturing; it was designed to allow high-speed component scanning.[6] QR codes are now used in a much broader context, including both commercial tracking applications and convenience-oriented applications aimed at mobile-phone users (termed mobile tagging). QR codes may be used to display text to the user, to open a webpage on the user's device, to add a vCard contact to the user's device, to open a Uniform Resource Identifier (URI), to connect to a wireless network, or to compose an email or text message. There are a great many QR code generators available as software or as online tools that are either free, or require a paid subscription.[7] The QR code has become one of the most-used types of two-dimensional code.[8]
During the month of June 2011, 14 million American mobile users scanned a QR code or a barcode. Some 58% of those users scanned a QR or barcode from their homes, while 39% scanned from retail stores; 53% of the 14 million users were men between the ages of 18 and 34.[9]
## Standards
Structure of a QR code (version 7), highlighting functional elements
There are several standards that cover the encoding of data as QR codes:[10]
At the application layer, there is some variation between most of the implementations. Japan's NTT DoCoMo has established de facto standards for the encoding of URLs, contact information, and several other data types.[13] The open-source "ZXing" project maintains a list of QR code data types.[14]
## Uses
A QR code used on a large billboard in Japan, linking to the sagasou.mobi website
QR codes have become common in consumer advertising. Typically, a smartphone is used as a QR code scanner, displaying the code and converting it to some useful form (such as a standard URL for a website, thereby obviating the need for a user to type it into a web browser). QR code has become a focus of advertising strategy, since it provides a way to access a brand's website more quickly than by manually entering a URL.[15][16] Beyond mere convenience to the consumer, the importance of this capability is that it increases the conversion rate: the chance that contact with the advertisement will convert to a sale. It coaxes interested prospects further down the conversion funnel with little delay or effort, bringing the viewer to the advertiser's website immediately, whereas a longer and more targeted sales pitch may lose the viewer's interest.
Although initially used to track parts in vehicle manufacturing, QR codes are used over a much wider range of applications. These include commercial tracking, entertainment and transport ticketing, product and loyalty marketing and in-store product labeling. Examples of marketing include where a company's discounted and percent discount can be captured using a QR code decoder which is a mobile app, or storing a company's information such as address and related information alongside its alpha-numeric text data as can be seen in Yellow Pages directory.
They can also be used in storing personal information for use by organizations. An example of this is Philippines National Bureau of Investigation (NBI) where NBI clearances now come with a QR code. Many of these applications target mobile-phone users (via mobile tagging). Users may receive text, add a vCard contact to their device, open a URL, or compose an e-mail or text message after scanning QR codes. They can generate and print their own QR codes for others to scan and use by visiting one of several pay or free QR code-generating sites or apps. Google had an API, now deprecated, to generate QR codes,[17] and apps for scanning QR codes can be found on nearly all smartphone devices.[18]
QR codes have been used and printed on train tickets in China since 2010.[19]
QR codes storing addresses and URLs may appear in magazines, on signs, on buses, on business cards, or on almost any object about which users might want information. Users with a camera phone equipped with the correct reader application can scan the image of the QR code to display text, contact information, connect to a wireless network, or open a web page in the telephone's browser. This act of linking from physical world objects is termed hardlinking or object hyperlinking. QR codes also may be linked to a location to track where a code has been scanned. Either the application that scans the QR code retrieves the geo information by using GPS and cell tower triangulation (aGPS) or the URL encoded in the QR code itself is associated with a location. In 2008, a Japanese stonemason announced plans to engrave QR codes on gravestones, allowing visitors to view information about the deceased, and family members to keep track of visits.[20] Psychologist Richard Wiseman was one of the first authors to include QR codes in a book, in Paranormality: Why We See What Isn't There (2011).[21][failed verification]
QR codes have been incorporated into currency. In June 2011, The Royal Dutch Mint (Koninklijke Nederlandse Munt) issued the world's first official coin with a QR code to celebrate the centenary of its current building and premises. The coin can be scanned by a smartphone and originally linked to a special website with contents about the historical event and design of the coin.[22] In 2014, the Central Bank of Nigeria issued a 100-naira banknote to commemorate its centennial, the first banknote to incorporate a QR code in its design. When scanned with an internet-enabled mobile device, the code goes to a website which tells the centenary story of Nigeria.[23] In 2015, the Central Bank of the Russian Federation issued a 100-rubles note to commemorate the annexation of Crimea by the Russian Federation. It contains a QR code into its design, and when scanned with an internet-enabled mobile device, the code goes to a website that details the historical and technical background of the commemorative note. In 2017, the Bank of Ghana issued a 5-cedis banknote to commemorate 60 years of Central Banking in Ghana, and contains a QR code in its design, which when scanned with an internet-enabled mobile device, that code goes to the official Bank of Ghana website.
Credit card functionality is under development. In September 2016, the Reserve Bank of India (RBI) launched the eponymously named Bharat QR, a common QR code jointly developed by all the four major card payment companies - National Payments Corporation of India that runs RuPay cards along with MasterCard, Visa and American Express. It will also have the capability of accepting payments on the unified payments interface (UPI) platform.[24][25]
### Augmented reality
QR codes are used in some augmented reality systems to determine the positions of objects in 3-dimensional space.[6]
### Displaying multimedia contents
Multimedia QR Codes are also used to direct users to specific multimedia contents (such as video, audio, images, documents, etc.).
### Mobile operating systems
QR codes can be used on various mobile device operating systems. iPhones running on iOS 11 and higher[26] and some Android devices can natively scan QR codes without downloading an external app.[27] The camera app is able to scan and display the kind of QR code (only on iPhone) along with the link (both on Android and iPhone). These devices support URL redirection, which allows QR codes to send metadata to existing applications on the device. Many paid or free apps are available with the ability to scan the codes and hard-link to an external URL.
### Virtual stores
QR codes have been used to establish "virtual stores", where a gallery of product information and QR codes is presented to the customer, e.g. on a train station wall. The customers scan the QR codes, and the products are delivered to their homes. This use started in South Korea,[28] and Argentina,[29] but is currently expanding globally.[30] Walmart, Procter & Gamble and Woolworths have already adopted the Virtual Store concept.[31]
### QR code payment
QR codes can be used to store bank account information or credit card information, or they can be specifically designed to work with particular payment provider applications. There are several trial applications of QR code payments across the world.[32][33] In developing countries like China,[34][35] India[36] and Bangladesh QR code payment is a very popular and convenient method of making payments. Since Alipay designed a QR code payment method in 2011,[37] mobile payment has been quickly adopted in China. As of 2018, around 83% of all payments were made via mobile payment.[38]
In November 2012, QR code payments were deployed on a larger scale in the Czech Republic when an open format for payment information exchange — a Short Payment Descriptor — was introduced and endorsed by the Czech Banking Association as the official local solution for QR payments.[39][40] In 2013, the European Payment Council provided guidelines for the EPC QR code enabling SCT initiation within the Eurozone.
QR codes can be used to log into websites: a QR code is shown on the login page on a computer screen, and when a registered user scans it with a verified smartphone, they will automatically be logged in. Authentication is performed by the smartphone which contacts the server. Google tested such a login method in January 2012.[41]
### Restaurant ordering
Fast serve restaurants can present a QR code near the front door allowing guests to view an online menu, or even redirect them to an online ordering website or app, allowing them to order or potentially pay for their meal without having to stand in line or use a cashier. QR codes can also link to daily or weekly specials that are not printed on the standardized menus.[42] At table serve restaurants, QR codes enable guests to order their meals without a waiter involved—the QR code contains the table number so servers know where to bring the food.[43][44] This application has grown especially since the need for social distancing during the 2020 COVID-19 pandemic has prompted reduced contact between service staff and customers.[43]
### Joining a Wi‑Fi network
A QR code to automatically join a Wi‑Fi network
By specifying the SSID, encryption type, password/passphrase, and if the SSID is hidden or not, mobile device users can quickly scan and join networks without having to manually enter the data.[45] A MECARD-like format is supported by Android and iOS 11+.[46]
• Common format: WIFI:S:<SSID>;T:<WEP|WPA|blank>;P:<PASSWORD>;H:<true|false|blank>;
• Sample WIFI:S:MySSID;T:WPA;P:MyPassW0rd;;
### Funerary use
A QR code which links to an obituary and can be placed on a headstone
A QR code can link to an obituary and can be placed on a headstone. In 2008, Ishinokoe in Yamanashi Prefecture, Japan began to sell tombstones with QR codes produced by IT DeSign, where the code leads to a virtual grave site of the deceased.[47][48][49] Other companies, such as Wisconsin-based Interactive Headstones, have also begun implementing QR codes into tombstones.[50] In 2014, the Jewish Cemetery of La Paz in Uruguay began implementing QR codes for tombstones.[51]
### Electronic authentication
QR codes are also used to generate time-based one-time passwords (TOTP) for electronic authentication.
### Video games
Popular video games, such as Fez, The Talos Principle, and Watch Dogs, have incorporated QR codes as story and gameplay elements.[52][53] Among Us has a QR code easter egg in the "Scan boarding pass" task in the MIRA HQ map. Mobile games such as Munzee use geolocation in combination with QR codes to create a game that is played in the real world by scanning QR stickers in physical locations.[54]
### Loyalty programs
QR Codes have been used by various retail outlets that have loyalty programs. Sometimes these programs are accessed with an app that is loaded onto a phone and includes a process triggered by a QR code scan. The QR codes for loyalty programs tend to be found printed on the receipt for a purchase or on the products themselves. Users in these schemes collect award points by scanning a code.
### Counterfeit detection
Serialised QR Codes have been used by brands[55] and governments[56] to let consumers, retailers and distributors verify the authenticity of the products and help with detecting counterfeit products, as part of a brand protection program.[57] However, the security level of a regular QR Code is limited since QR Codes printed on original products are easily reproduced on fake products, even though the analysis of data generated as a result of QR Code scanning can be used to detect counterfeiting and illicit activity.[58] A higher security level can be attained by embedding a digital watermark or copy detection pattern into the image of the QR Code. This makes the QR Code more secure against counterfeiting attempts, and fake products which contain a counterfeit QR Code can be detected by scanning the secure QR Code with a specific app (even though the QR Code message itself is valid).[59]
### Product tracing
Different studies have been made to assess the effectiveness of QR codes as a means of conveying labelling information and their use as part of a food traceability system. In,[60] it was found that when provided free access to a smartphone with QR Code scanning app, 52.6% of participants would use it to access labelling information. A study made in South Korea showed that consumers appreciate QR code used in food traceability system, as they provide detailed information about food, as well as information that helps them in their purchasing decision.[61] If QR Codes are serialised, consumers can access a web page showing the supply chain for each ingredient, as well as information specific to each related batch, including meat processors and manufacturers, which helps address the concerns they have about the origin of their food.[62]
### COVID-19 pandemic
In several Australian states patrons are required to scan QR codes at shops, clubs, supermarkets and other service and retail establishments on entry to assist contact tracing. Singapore, the United Kingdom and New Zealand used similar systems.[64]
## Design
Unlike the older, one-dimensional barcodes that were designed to be mechanically scanned by a narrow beam of light, a QR code is detected by a 2-dimensional digital image sensor and then digitally analyzed by a programmed processor. The processor locates the three distinctive squares at the corners of the QR code image, using a smaller square (or multiple squares) near the fourth corner to normalize the image for size, orientation, and angle of viewing. The small dots throughout the QR code are then converted to binary numbers and validated with an error-correcting algorithm.
### Storage
The amount of data that can be stored in the QR code symbol depends on the datatype (mode, or input character set), version (1, ..., 40, indicating the overall dimensions of the symbol, i.e. 4 × version number + 17 dots on each side), and error correction level. The maximum storage capacities occur for version 40 and error correction level L (low), denoted by 40-L:[8][65]
Maximum character storage capacity (40-L)
character refers to individual values of the input mode/datatype
Input mode Max. characters Bits/char. Possible characters, default encoding
Numeric only 7,089 313 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
Alphanumeric 4,296 512 0–9, A–Z (upper-case only), space, , %, *, +, -, ., /, : Binary/byte 2,953 8 ISO 8859-1 Kanji/kana 1,817 13 Shift JIS X 0208 Here are some sample QR code symbols: ### Error correction Damaged but still decodable QR code, Link to http://en.m.wikipedia.org Example of a QR code with artistic embellishment that will still scan correctly thanks to error correction QR codes use Reed–Solomon error correction over the finite field ${\displaystyle \mathbb {F} _{256}}$, the elements of which are encoded as bytes of 8 bits; the byte ${\displaystyle b_{7}b_{6}b_{5}b_{4}b_{3}b_{2}b_{1}b_{0}}$ with a standard numerical value ${\displaystyle \textstyle \sum _{i=0}^{7}b_{i}2^{i}}$ encodes the field element ${\displaystyle \textstyle \sum _{i=0}^{7}b_{i}\alpha ^{i}}$ where ${\displaystyle \alpha \in \mathbb {F} _{256}}$ is taken to be a primitive element satisfying ${\displaystyle \alpha ^{8}+\alpha ^{4}+\alpha ^{3}+\alpha ^{2}+1=0}$. The Reed–Solomon code uses one of 37 different polynomials over ${\displaystyle \mathbb {F} _{256}}$, with degrees ranging from 7 to 68, depending on how many error correction bytes the code adds. It is implied by the form of Reed–Solomon used (systematic BCH view) that these polynomials are all on the form ${\textstyle \prod _{i=0}^{n-1}(x-\alpha ^{i})}$, however the rules for selecting the degree ${\displaystyle n}$ are specific to the QR standard. When discussing the Reed–Solomon code phase there is some risk for confusion, in that the QR ISO standard uses the term codeword for the elements of ${\displaystyle \mathbb {F} _{256}}$, which respect to the Reed–Solomon code are symbols, whereas it uses the term block for what with respect to the Reed–Solomon code are the codewords. The number of data versus error correction bytes within each block depends on (i) the version (side length) of the QR symbol and (ii) the error correction level, of which there are four. The higher the error correction level, the less storage capacity. The following table lists the approximate error correction capability at each of the four levels: Level L (Low) 7% of data bytes can be restored. Level M (Medium) 15% of data bytes can be restored. Level Q (Quartile)[66] 25% of data bytes can be restored. Level H (High) 30% of data bytes can be restored. In larger QR symbols, the message is broken up into several Reed–Solomon code blocks. The block size is chosen so that no attempt is made at correcting more than 15 errors per block; this limits the complexity of the decoding algorithm. The code blocks are then interleaved together, making it less likely that localized damage to a QR symbol will overwhelm the capacity of any single block. Due to error correction, it is possible to create artistic QR codes that still scan correctly, but contain intentional errors to make them more readable or attractive to the human eye, as well as to incorporate colors, logos, and other features into the QR code block.[67][68] It is also possible to design artistic QR codes without reducing the error correction capacity by manipulating the underlying mathematical constructs.[69][70] Image processing algorithms are also used to reduce errors in QR-code.[71] ### Encoding The format information records two things: the error correction level and the mask pattern used for the symbol. Masking is used to break up patterns in the data area that might confuse a scanner, such as large blank areas or misleading features that look like the locator marks. The mask patterns are defined on a grid that is repeated as necessary to cover the whole symbol. Modules corresponding to the dark areas of the mask are inverted. The format information is protected from errors with a BCH code, and two complete copies are included in each QR symbol.[4] The message dataset is placed from right to left in a zigzag pattern, as shown below. In larger symbols, this is complicated by the presence of the alignment patterns and the use of multiple interleaved error-correction blocks. The general structure of a QR encoding is as a sequence of 4 bit indicators with payload length dependent on the indicator mode (e.g. byte encoding payload length is dependent on the first byte).[72] Mode indicator Description Typical structure '[ type : sizes in bits ]' 0001 Numeric [0001 : 4] [ Character Count Indicator : variable ] [ Data Bit Stream : 313 × charcount ] 0010 Alphanumeric [0010 : 4] [ Character Count Indicator : variable ] [ Data Bit Stream : 512 × charcount ] 0100 Byte encoding [0100 : 4] [ Character Count Indicator : variable ] [ Data Bit Stream : 8 × charcount ] 1000 Kanji encoding [1000 : 4] [ Character Count Indicator : variable ] [ Data Bit Stream : 13 × charcount ] 0011 Structured append [0011 : 4] [ Symbol Position : 4 ] [ Total Symbols: 4 ] [ Parity : 8 ] 0111 ECI [0111 : 4] [ ECI Assignment number : variable ] 0101 FNC1 in first position [0101 : 4] [ Numeric/Alphanumeric/Byte/Kanji payload : variable ] 1001 FNC1 in second position [1001 : 4] [ Application Indicator : 8 ] [ Numeric/Alphanumeric/Byte/Kanji payload : variable ] 0000 End of message [0000 : 4] Note: • Character Count Indicator depends on how many modules are in a QR code (Symbol Version). • ECI Assignment number Size: • 8 × 1 bits if ECI Assignment Bitstream starts with '0' • 8 × 2 bits if ECI Assignment Bitstream starts with '10' • 8 × 3 bits if ECI Assignment Bitstream starts with '110' Four-bit indicators are used to select the encoding mode and convey other information. Encoding modes Indicator Meaning 0001 Numeric encoding (10 bits per 3 digits) 0010 Alphanumeric encoding (11 bits per 2 characters) 0100 Byte encoding (8 bits per character) 1000 Kanji encoding (13 bits per character) 0011 Structured append (used to split a message across multiple QR symbols) 0111 Extended Channel Interpretation (select alternate character set or encoding) 0101 FNC1 in first position (see Code 128 for more information) 1001 FNC1 in second position 0000 End of message (Terminator) Encoding modes can be mixed as needed within a QR symbol. (e.g., a url with a long string of alphanumeric characters ) [ Mode Indicator][ Mode bitstream ] --> [ Mode Indicator][ Mode bitstream ] --> etc... --> [ 0000 End of message (Terminator) ] After every indicator that selects an encoding mode is a length field that tells how many characters are encoded in that mode. The number of bits in the length field depends on the encoding and the symbol version. Number of bits in a length field (Character Count Indicator) Encoding Ver. 1–9 10–26 27–40 Numeric 10 12 14 Alphanumeric 9 11 13 Byte 8 16 16 Kanji 8 10 12 Alphanumeric encoding mode stores a message more compactly than the byte mode can, but cannot store lower-case letters and has only a limited selection of punctuation marks, which are sufficient for rudimentary web addresses. Two characters are coded in an 11-bit value by this formula: V = 45 × C1 + C2 This has the exception that the last character in an alphanumeric string with an odd length is read as a 6-bit value instead. Alphanumeric character codes Code Character Code Character Code Character Code Character Code Character 00 0 09 9 18 I 27 R 36 Space 01 1 10 A 19 J 28 S 37
02 2 11 B 20 K 29 T 38 %
03 3 12 C 21 L 30 U 39 *
04 4 13 D 22 M 31 V 40 +
05 5 14 E 23 N 32 W 41
06 6 15 F 24 O 33 X 42 .
07 7 16 G 25 P 34 Y 43 /
08 8 17 H 26 Q 35 Z 44 :
## Variants
### Model 1
Model 1 QR code is an older version of the specification. It is visually similar to the widely seen model 2 codes, but lacks alignment patterns. Differences are in the bottom right corner and in the midsections of the bottom and right edges are additional functional regions.
### Micro QR code
Micro QR code is a smaller version of the QR code standard for applications where symbol size is limited. There are four different versions (sizes) of Micro QR codes: the smallest is 11×11 modules; the largest can hold 35 numeric characters.[73]
### IQR code
IQR Code is an alternative to existing QR codes developed by Denso Wave. IQR codes can be created in square or rectangular formations; this is intended for situations where a rectangular barcode would otherwise be more appropriate, such as cylindrical objects. IQR codes can fit the same amount of information in 30% less space. There are 61 versions of square IQR codes, and 15 versions of rectangular codes. For squares, the minimum size is 9x9 modules; rectangles have a minimum of 19x5 modules. IQR codes add error correction level S, which allows for 50% error correction.[74] IQR Codes have not yet been given an ISO specification, and only proprietary Denso Wave products can create or read IQR codes.[75]
### Secure QR code
Secure Quick Response (SQR) code is a QR code that contains a "private data" segment after the terminator instead of the specified filler bytes "ec 11".[76] This private data segment must be deciphered with an encryption key. This can be used to store private information and to manage company's internal information.[77]
SQR codes have been developed by the FORUS Foundation to enable secure transactions, and published under a Creative Commons Licence. The SQR solution guarantees the integrity of the source data as well as the validity of the originating party. The payment instruction string is made up of the electronic instruction data from the scanned QR code appended with a SHA-2 cryptographic hash. The message digest can then be encrypted using the private key of the sender, which then creates a digital signature of the message. This signature validates the integrity of the data and the trustworthiness of the sender. This provides non-repudiation, confirming the identity of the sender, and that it has not been tampered with during transmission. By embedding the URL and all the variables required to perform shopping cart type e-commerce, bill payment and peer to peer payments, coupled with a digital certificate eliminates the possibility of spoofing, tampering, and man in the middle attacks. [78]
### Frame QR
Frame QR is a QR code with a "canvas area" that can be flexibly used. In the center of this code is the canvas area, where graphics, letters, and more can be flexibly arranged, making it possible to lay out the code without losing the design of illustrations, photos, etc.[79]
### HCC2D
Samples of the High Capacity Colored 2-Dimensional (HCC2D) code: (a) 4-color HCC2D code and (b) 8-color HCC2D code.
Researchers have proposed a new High Capacity Colored 2-Dimensional (HCC2D) Code, which builds upon a QR code basis for preserving the QR robustness to distortions and uses colors for increasing data density (at this stage[when?] it is still in prototyping phase). The HCC2D code specification is described in details in Querini et al. (2014),[80] while techniques for color classification of HCC2D code cells are described in detail in Querini and Italiano (2014),[81] which is an extended version of Querini and Italiano (2013).[82]
Introducing colors into QR codes requires addressing additional issues. In particular, during QR code reading only the brightness information is taken into account, while HCC2D codes have to cope with chromatic distortions during the decoding phase. In order to ensure adaptation to chromatic distortions which arise in each scanned code, HCC2D codes make use of an additional field: the Color Palette Pattern. This is because color cells of a Color Palette Pattern are supposed to be distorted in the same way as color cells of the Encoding Region. Replicated color palettes are used for training machine learning classifiers.
### JAB code
Wikipedia greetings with link encoded using 8 colour JAB code.
JAB code (Just Another Barcode) is a color 2D matrix symbology made of colorful square modules arranged in either square or rectangle grids developed by Fraunhofer Institute SIT (Secure Information Technology).[83]
JAB code contains one primary symbol and optionally multiple secondary symbols. The primary symbol contains four finder patterns located at the corners of the symbol.[84] It uses either 4 or 8 colours[85] The 4 basic colours (cyan, magenta, yellow, black) are the 4 primary colours of the subtractive CMYK color model which is the most widely used system in industry for colour printing on a white base such as a paper. The other 4 colours (blue, red, green, white) are secondary colours of the CMYK model and originate as an equal mixture of the basic colours.
The barcode is not subject to licensing and was submitted to ISO standardization as ISO 23634 expected to be approved at the beginning of 2021[86] and finalized in 2022.[85] The software is open-source and published under the LGPL v2.1 license.[87] The specification is freely available.[84]
Because the colour represents an additional (third) dimension to an otherwise two-dimensional matrix a JAB code can contain more information in the same area compared to common two-colour (black and white) codes (theoretically twice as much data for a 4 colour code and three times more for 8 colours assuming the same encoding algorithm). This may allow to store the entire message ("all data") in the barcode itself rather than just storing a partial data (abbreviated message) with a reference (such as a link) to a source somewhere else (a website) that contains the full message thus eliminating the need for an additional permanent and always available infrastructure beyond the printed barcode self. This may be used to digitally sign encrypted digital version of printed legal documents, contracts and certificates (diplomas, training), medical prescriptions or provide product authenticity assurance to increase protection against counterfeits.[85]
The use of QR code technology is freely licensed as long as users follow the standards for QR Code documented with JIS or ISO. Non-standardized codes may require special licensing.[88]
Denso Wave owns a number of patents on QR code technology, but has chosen to exercise them in a limited fashion.[88] In order to promote widespread usage of the technology Denso Wave chose to waive its rights to a key patent in its possession for standardized codes only.[10] In the US, the granted QR code patent is US 5726435 , and in Japan JP 2938338 , both of which have expired. The European Patent Office granted patent EP 0672994 to Denso Wave, which was then validated into French, UK, and German patents, all of which expired in March 2015.[89]
The text QR Code itself is a registered trademark and wordmark of Denso Wave Incorporated.[90] In UK, the trademark is registered as E921775, the word "QR Code", with a filing date of 03/09/1998.[91] The UK version of the trademark is based on the Kabushiki Kaisha Denso (DENSO CORPORATION) trademark, filed as Trademark 000921775, the word "QR Code", on 03/09/1998 and registered on 6/12/1999 with the European Union OHIM (Office for Harmonization in the Internal Market).[92] The U.S. Trademark for the word "QR Code" is Trademark 2435991 and was filed on 29 September 1998 with an amended registration date of 13 March 2001, assigned to Denso Corporation.[93]
## Risks
The only context in which common QR codes can carry executable data is the URL data type. These URLs may host JavaScript code, which can be used to exploit vulnerabilities in applications on the host system, such as the reader, the web browser or the image viewer, since a reader will typically send the data to the application associated with the data type used by the QR code.
In the case of no software exploits, malicious QR codes combined with a permissive reader can still put a computer's contents and user's privacy at risk. This practice is known as "attagging", a portmanteau of "attack tagging".[94] They are easily created and can be affixed over legitimate QR codes.[95] On a smartphone, the reader's permissions may allow use of the camera, full Internet access, read/write contact data, GPS, read browser history, read/write local storage, and global system changes.[96][97][98]
Risks include linking to dangerous web sites with browser exploits, enabling the microphone/camera/GPS, and then streaming those feeds to a remote server, analysis of sensitive data (passwords, files, contacts, transactions),[99] and sending email/SMS/IM messages or DDOS packets as part of a botnet, corrupting privacy settings, stealing identity,[100] and even containing malicious logic themselves such as JavaScript[101] or a virus.[102][103] These actions could occur in the background while the user is only seeing the reader opening a seemingly harmless web page.[104] In Russia, a malicious QR code caused phones that scanned it to send premium texts at a fee of US\$6 each.[94]
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72. ^ ISO/IEC 18004:2006(E) § 6.4 Data encoding; Table 3 – Number of bits in character count indicator for QR Code 2005
73. ^ Information technology – Automatic identification and data capture techniques – QR Code 2005 bar code symbology specification, ISO/IEC 18004:2006 cor. 2009, pages 3, 6.
74. ^ "QR Code Overview & Progress of QR Code Applications" (PDF). Archived (PDF) from the original on 20 May 2013. Retrieved 26 June 2014.
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76. ^ "Embedding Secret Data in QR Code". Retrieved 29 October 2018.
77. ^ "SQRC". qrcode.com. Retrieved 26 December 2017.
78. ^ "FORUS". www.forus.co.za. Retrieved 27 December 2020.
79. ^ "Frame QR". qrcode.com. Retrieved 26 December 2017.
80. ^ 2D Color Barcodes for Mobile Phones Archived 4 March 2016 at the Wayback Machine
81. ^
82. ^ "Color classifiers for 2D color barcodes" (PDF). Fedcsis.org. 2013. Archived (PDF) from the original on 24 July 2015.
83. ^ "JAB code website". jabcode.org. 2019.
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87. ^ "jabcode". GitHub. 2019.
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104. ^ "Beware of Malicious QR Codes". ABC. 8 June 2011. Archived from the original on 1 August 2012. Retrieved 31 August 2011. | 2021-07-31 15:30:13 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 10, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4309874176979065, "perplexity": 5628.03669223073}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046154089.68/warc/CC-MAIN-20210731141123-20210731171123-00499.warc.gz"} |
https://cs.stackexchange.com/questions/74264/finding-fn-so-2tfn-1-%E2%88%88-%CE%98log4-n | Finding f(n) so 2T(f(n)) + 1 ∈ Θ(log^4 n)
Given the recursive function: ($c$ is a constant)
$\qquad T(n) = \begin{cases}1 & n ≤ c\\2T(f(n)) + 1 & n > c\end{cases}$
I need to find a $f(n)$ such $T(n) ∈ Θ(log^4 n) = Θ(\log \log \log \log n)$. I tried the iteration method, and it gave me $\qquad 2^i*T($$f1$($f2$(...$fi$($n$)..$))) + i$ which revealing the pretty much expected formula for $i$. i know that $f(n) = \sqrt(x)$ would give $log(log(n))$, Is that the right direction? or which function will get closer?
Thanks.
• Which did you try? How close did you get? Have you tried solving the recurrence symbolically in $f$, and then solve for $f$? – Raphael Apr 20 '17 at 18:23
• Note that $\log^4n$ almost always means $(\log n)^4$ and not $\log\log\log\log n$. (By analogy with trigonometry, where $\sin^2 x$ never means $\sin\sin x$.) – David Richerby Apr 20 '17 at 20:16
So $T(n)$ goes through the sequence 1, 3, 7, 15, 31, ... We want $T(n) = \Theta (\log \log \log \log n)$; I'll assume log n is the base 2 logarithm. We'll try to define f (n) so that T(n) = log log log log n, rounded up to the nearest element of the sequence 1, 3, 7, 15, 31...
1 < log log log log n ≤ 3 if 2 < log log log n ≤ 8 or 4 < log log n ≤ 256 or 16 < log n ≤ $2^{256}$ or 65,536 < n ≤ $2^{2^{256}}$. So in that range we have T (n) = 3, and c = 65,536. BTW. The upper end of that range is so large that all the paper in the universe wouldn't be large enough to write it down as a decimal number.
Next we look for a range where 3 < log log log log n ≤ 7. That happens when n > $2^{2^{256}}$ and log log log log n ≤ 7 or log log log n ≤ 128 or log log n ≤ $2^{128}$ or log n ≤ $2^{2^{128}}$ or n ≤ $2^{2^{2^{128}}}$, and the numbers involved get quite big.
In the range $2^{2^{256}}$ < n ≤ $2^{2^{2^{128}}}$ we want 65536 < f (n) ≤ $2^{2^{256}}$. f (n) should map $2^{2^{2^{2^{2k+1}}}}$ to $2^{2^{2^{2^k}}}$.
To achieve this, let $g (n) = \log \log \log \log n$, $h (n) = (g (n) - 1)/2$, $f (n) = 2^{2^{2^{2^{h(n)}}}}$. | 2020-07-13 18:06:37 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9993632435798645, "perplexity": 5725.971953461117}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593657146247.90/warc/CC-MAIN-20200713162746-20200713192746-00075.warc.gz"} |
https://proj.org/usage/differences.html | # Known differences between versions¶
Once in a while, a new version of PROJ causes changes in the existing behavior. In this section we track deliberate changes to PROJ that break from previous behavior. Most times that will be caused by a bug fix. Unfortunately, some bugs have existed for so long that their faulty behavior is relied upon by software that uses PROJ.
Behavioural changes caused by new bugs are not tracked here, as they should be fixed in later versions of PROJ.
## Version 4.6.0¶
The default datum application behavior changed with the 4.6.0 release. PROJ will now only apply a datum shift if both the source and destination coordinate system have valid datum shift information.
The PROJ 4.6.0 Release Notes states
MAJOR: Rework pj_transform() to avoid applying ellipsoid to ellipsoid transformations as a datum shift when no datum info is available.
## Version 5.0.0¶
### Longitude wrapping when using custom central meridian¶
By default PROJ wraps output longitudes in the range -180 to 180. Previous to PROJ 5, this was handled incorrectly when a custom central meridian was set with +lon_0. This caused a change in sign on the resulting easting as seen below:
$proj +proj=merc +lon_0=110 -70 0 -20037508.34 0.00 290 0 20037508.34 0.00 From PROJ 5 on onwards, the same input now results in same coordinates, as seen from the example below where PROJ 5 is used: $ proj +proj=merc +lon_0=110
-70 0
-20037508.34 0.00
290 0
-20037508.34 0.00
The change is made on the basis that $$\lambda=290^{\circ}$$ is a full rotation of the circle larger than $$\lambda=-70^{\circ}$$ and hence should return the same output for both.
Adding the +over flag to the projection definition provides the old behavior.
## Version 6.0.0¶
### Removal of proj_def.dat¶
Before PROJ 6, the proj_def.dat was used to provide general and per-projection parameters, when +no_defs was not specified. It has now been removed. In case, no ellipsoid or datum specification is provided in the PROJ string, the default ellipsoid is GRS80 (was WGS84 in previous PROJ versions).
### Changes to deformation¶
#### Reversed order of operation¶
In the initial version of the of deformation operation the time span between $$t_{obs}$$ and $$t_c$$ was determined by the expression
$dt = t_c - t_{obs}$
With version 6.0.0 this has been reversed in order to behave similarly to the Helmert operation, which determines time differences as
$dt = t_{obs} - t_c$
Effectively this means that the direction of the operation has been reversed, so that what in PROJ 5 was a forward operation is now an inverse operation and vice versa.
Pipelines written for PROJ 5 can be migrated to PROJ 6 by adding +inv to forward steps involving the deformation operation. Similarly +inv should be removed from inverse steps to be compatible with PROJ 6.
#### Removed +t_obs parameter¶
The +t_obs parameter was confusing for users since it effectively overwrote the observation time in input coordinates. To make it more clear what is the operation is doing, users are now required to directly specify the time span for which they wish to apply a given deformation. The parameter +dt has been added for that purpose. The new parameter is mutually exclusive with +t_epoch. +dt is used when deformation for a set amount of time is needed and +t_epoch is used (in conjunction with the observation time of the input coordinate) when deformation from a specific epoch to the observation time is needed.
## Version 6.3.0¶
### projinfo¶
Before PROJ 6.3.0, WKT1:GDAL was implicitly calling –boundcrs-to-wgs84, to add a TOWGS84[] node in some cases. This is no longer the case.
## Version 7.0.0¶
### proj¶
Removed -ld option from application, since it promoted use of deprecated parameters like +towgs and +datum.
### cs2cs¶
Removed -ld option from application, since it promoted use of deprecated parameters like +towgs and +datum. | 2021-03-05 09:24:26 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.42309045791625977, "perplexity": 2863.4634705138064}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178370752.61/warc/CC-MAIN-20210305091526-20210305121526-00221.warc.gz"} |
https://laurentlessard.com/bookproofs/unmasking-the-secret-santas/?replytocom=2163 | This Riddler puzzle is about the popular Secret Santa gift exchange game. Can we guess who our Secret Santa is?
The 41 FiveThirtyEight staff members have decided to send gifts to each other as part of a Secret Santa program. Each person is randomly assigned one of the other 40 people on the masthead to give a gift to, and they can’t give to themselves. After the Secret Santa is over, everybody naturally wants to find out who gave them their gift. So, each of them decides to ask up to 20 people who they were a Secret Santa for. If they can’t find the person who gave them the gift within 20 tries, they give up. (Twenty co-workers is a lot of co-workers to talk to, after all.) Each person asks and answers individually — they don’t tell who anyone else’s Secret Santa is. Also, nobody asks any question other than “Who were you Secret Santa for?”
If each person asks questions optimally, giving themselves the best chance to unmask their Secret Santa, what is the probability that everyone finds out who their Secret Santa was? And what is this optimal strategy? (Asking randomly won’t work, because only half the people will find their Secret Santa that way on average, and there’s about a 1-in-2 trillion chance that everyone will know.)
Here is my solution:
[Show Solution]
## 14 thoughts on “Unmasking the Secret Santas”
1. Mark Rickert says:
Nice job, however I get:
For n=1, P=0, and for n=2, P=5/11. I didn’t check the others.
Are you using D(0)=1?
1. You’re absolutely right. I used D(0)=0 by accident. I updated my post and plot. I decided to keep P=1 for the cases n=1 and n=2 because you can deduce your Secret Santa perfectly in these cases even if you don’t ask the correct co-worker (see my explanation in the updated post).
1. Thanks, indeed you are correct. I accidentally used D(0)=0 in my code, when it should be D(0)=1. I updated my post and the plot.
2. Justin Hsu says:
I think your evaluation is missing the k=2n+1 term. n=2 should still evaluate to 100%, as link length of 5 is fully determined from first 3 permutations.
1. Yep. Fixed it! Check out the updated plot and post.
3. Just zis guy, ya know? says:
It seems that you have solved a different problem than asked, though likely the one that was intended.
“If each person asks questions optimally, giving themselves the best chance to unmask their Secret Santa…” (the original problem) is not the same as, “If each person asks questions optimally, so as to give the entire group the best chance to unmask all of the Secret Santas…” (the question you answered).
To find our own Secret Santa we can do better than guessing randomly… proving we have an optimal strategy is difficult, however.
– JZGYK
1. I agree — since we were asked about the probability that everybody wins, I was assuming this was the quantity to be maximized.
If we interpret the question literally and each individual greedily attempts to maximize the chance that they will find their own Secret Santa, I don’t think you can do any better than guessing randomly (assuming $n > 2$). If you think it’s possible to do better than randomly guessing, I’d love to hear your thoughts!
This is another reason I interpreted the question the way I did; because the literal interpretation, as far as I could tell, led to random guessing as the only possible choice and this isn’t an interesting scenario!
1. If you use the “follow the Santa” strategy, there is a 31.8% chance that everybody finds their Secret Santa, but there is actually a 48.8% chance that any one individual will find their Secret Santa. Therefore this strategy doesn’t beat random guessing, where the chance of winning is 50% for each individual. I think I’ll update my solution and include some of these comments.
1. Verdigris97 says:
There is another interpretation that leads to a perfect (but clearly unintended) strategy. Even though nobody can share the results of their questioning, if everyone in the office conspires to ask exactly one question, (and they each ask their own target), then by the end of the day each person will know who their secret Santa is because their secret Santa was the only one to ask them anything.
2. Just zis guy, ya know? says:
I came to the same results as everyone else: 31.8% for everybody using “Follow The Santa [FTS]”, 48.8% for me individually if I use FTS and 50% if I “guess randomly”. I note here that I only count a “win” if someone actually says, “I am your Secret Santa.”; there is always the possibility of guessing correctly if no one names me.
What struck me as odd about this was: “If I am guessing randomly, I might Follow The Santa by pure chance.” But this would give me a less than the 50% chance of winning that I would get if I guessed randomly. So what if I guess randomly but re-guess if I FTS? By throwing out the “bad” (less than 50%) approach of FTS, what is left must be something greater than 50%.
An example:
n = 3 is the first interesting case so let’s start there. There are 1854 derangements and the individual finds their Secret Santa 774/1854 = 41.75% of the time.
Consider an anti-Follow The Santa strategy ([aFTS]). We are player #1 and we ask the lowest numbered player who has not yet been named. That is, if we were the Secret Santa for player #4 then the lowest numbered player who has not yet been named is player #2 and we ask them who they bought for. If they say player #3 then we have named players #1, #2, #3, and #4 so the lowest numbered player who has not yet been named is #5, etc..
After 1 question, 3-FTS has 264 winning cases, 3-Rnd has 309 and 3-aFTS has 318.
After 2 questions, 3-FTS has 534 winning cases, 3-Rnd has 618 and 3-aFTS has 654.
After 3 questions, 3-FTS has 774 winning cases, 3-Rnd has 927 and 3-aFTS has 1038.
Assuming I have made no mistakes, 3-FTS = 41.8%, 3-Rnd = 50.0% and 3-aFTS = 56.0%
Proving that this is the optimal individual strategy is another thing entirely.
– JZGYK
On an unrelated note, whenever I try to use the less than or greater than symbols, your site misinterprets them.
1. ah neat — I guess I was assuming the staff members weren’t allowed to cooperate ahead of time (by agreeing on a specific ordering, for example). Looks like this sort of cooperation can lead to better-than-guessing strategies, as you pointed out!
Math symbols (and arbitrary equations, in general) can be displayed by using a dollar sign before and after the equation (LaTeX code). I think the issue with $<$ and $>$ is that these symbols are interpreted as html tags.
1. Just zis guy, you know? says:
No collusion is required and I can get this result as the sole Santa-seeker. The results are unchanged if I make a list or if I simply choose randomly from people who are not named. But I don’t know that this is optimal in general, or even close.
– JZGYK
4. Verdigris97 says:
Nice writeup!
If you want to avoid using enormous numbers (say, because you are using a spreadsheet for the computations, or any finite-precision language), the number of derangements with a cycle of length $k>n/2$, $\binom{n}{k} (k-1)! D_{n-k}$, can be rewritten as $n!\frac{(k-1)!}{k!}\frac{D_{n-k}}{(n-k)!}=\frac{n!}{k}\frac{D_{n-k}}{(n-k)!}$, and the proportion of such derangements out of all possible derangements of size $n$ is $\frac{1}{k}\frac{n!}{D_n}\frac{D_{n-k}}{(n-k)!}$. So, the proportion of derangements on 41 items with a cycle of length (exactly) 22 is $\frac{1}{22} \frac{41!}{D_{41}} \frac{D_{19}}{19!} \approx 0.045455$.
The function $\frac{D_n}{n!}$ converges to $1/e$ quickly. The exact formula is $\frac{D_n}{n!} = \sum_{i=0}^n \frac{(-1)^i}{i!}$, and, if we accept the approximation $\frac{41!}{D_{41}} = e$ (the difference is far below machine precision), we can sum up the contributions of each of the different cycle lengths for $k=22,\ldots,41$ and get that the total proportion of derangements with a cycle of length at least 22 is approximately $0.681665$. Therefore, the probability that our strategy succeeds is approximately $0.318335$, which matches your approximation to the exact ratio you found. | 2022-08-09 07:30:24 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6811753511428833, "perplexity": 980.4188185677297}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882570913.16/warc/CC-MAIN-20220809064307-20220809094307-00706.warc.gz"} |
https://www.nova77.org/jupyter/project_1_notes.html | # Project 1 Notes on Creating a Program in Python to Generate the Fibonacci Sequence¶
## Fibonacci Sequence¶
The Fibonacci Sequence is the series of numbers:
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ...
The next number is found by adding up the two numbers before it.
## Iteration¶
Iteration means executing the same block of code over and over, potentially many times. A programming structure that implements iteration is called a loop.
## Flow of Execution¶
Python, in general, processes commands from top to bottom.
## Indentation¶
All control structures in Python use indentation to define blocks.
## Variable Assignment¶
Think of a variable as a name attached to a particular object. To create a variable, you just assign it a value and then start using it. Assignment is done with a single equals sign (=):
In [ ]:
t = 17 # This is read or interpreted as “t is assigned the value 17.”
print(t)
17
In [ ]:
t = 27 # if you change the value of n and use it again, the new value will be substituted instead.
print(t)
27
In [ ]:
s = 11
print(s)
s = 99
print(s)
11
99
In [ ]:
a = 0
b = 1
print(a)
0
## Variable Type¶
In [ ]:
m = 77 # This assignment creates an integer object with the value 77.
type (m)
Out[ ]:
int
In [ ]:
m1 = 77.7
type (m1)
Out[ ]:
float
In [ ]:
i = 1
while i < 6:
print(i)
i += 1
1
2
3
4
5
## While Loops¶
• i is initially 1.
• Evaluate the while statement: If the while statement is true, the loop body executes. Inside the body, i is printed and then i is added by 1.
• When the body of the loop has finished, program execution returns to the top of the loop, and then the while statement is evaluated again. If the while statement is still true, so the body executes again, and i is printed and then added by 1.
• This continues until i becomes 5, which is less than 6. When i becomes 6, after the while statement is evaluated, it is false, and the loop terminates.
In [ ]:
n = 7
while n > 0:
n -=1
print(n)
6
5
4
3
2
1
0
In [ ]:
n = 7
while n > 0:
print(n)
n -= 1
7
6
5
4
3
2
1
## Tuple Assignment¶
Tuples are defined by enclosing the elements in parentheses (( )) instead of square brackets ([ ]).
In [ ]:
a, b = 0, 1
print(a)
0
In [ ]:
type(a)
Out[ ]:
int
In [ ]:
type((a,b))
Out[ ]:
tuple
In [ ]:
a = 0
b = 1
while a<2:
print(a)
a, b = b, a+b
0
1
1
In [ ]:
# Fibonacci Sequencce
a = 0
b = 1
while a < 100:
print(a)
a, b = b, a+b
0
1
1
2
3
5
8
13
21
34
55
89
In [ ]:
a, b = 0, 1
while a < 100:
print(a)
a = b
b = a+b
0
1
2
4
8
16
32
64
In [ ]:
a = 0
b = 1
while a < 1000000:
print(a)
a, b = b, a+b
0
1
1
2
3
5
8
13
21
34
55
89
144
233
377
610
987
1597
2584
4181
6765
10946
17711
28657
46368
75025
121393
196418
317811
514229
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https://codegolf.stackexchange.com/questions/57456/cut-a-pizza-into-identical-slices | # Cut a pizza into identical slices
This is what I thought this question was going to be, before I fully read it.
A group of code golfers walk into The Nineteenth Bite Pizzeria and order a pizza. It comes in an irregular shape, made of unit squares. Your task is to help them cut it into identical slices. That is, the slices must have the exact same shape and size; they can be rotated but not flipped/mirrored. For example, if they are Tetris pieces, they must be the same kind, you can't use both an L piece and a J piece.
## Input
You will be given the number of people in the group on the first line (always an integer from 2 to 10, inclusive), followed by a rectangular matrix of ' ' (space) and '#' characters, representing the pizza. All the '#' characters are connected through their edges. The number of '#' characters is guaranteed to be a multiple of the number of people.
## Output
You should print the same matrix, with each '#' character replaced with a digit from 0 to n-1 (n being the number of people). Each digit should mark a slice. The slice shape must be connected through the square edges. The slice numbering doesn't need to be in any particular order. If there are multiple ways of cutting the pizza, any of them is acceptable.
If it's not possible to cut the pizza as required, you should print the string "No pizza for you!" instead.
## Scoring
This is code golf. Your score will be the number of bytes in the program. Characters will be counted through their UTF-8 encoding. Lowest score wins.
## Examples
Input:
3
#
###
####
#
Output:
0
100
1122
2
Input:
4
###
# #
###
Output:
001
2 1
233
Input:
2
# #
######
Output:
No pizza for you!
Input:
5
#
####
#####
#####
#####
####
#
Output:
0
1000
21110
32221
43332
4443
4
Input:
4
#
####
######
#####
####
Output:
0
1000
111203
12233
2233
## Requirements
• You should write a full program that reads from the standard input and writes to the standard output.
• The program must be runnable in Linux using freely available software.
• Your program should finish each of the above examples in less than 1 minute on a modern computer.
• No standard loopholes.
• The Nineteenth Bite :^) – FryAmTheEggman Sep 9 '15 at 20:30
• @FryAmTheEggman © Calvin's Hobbies – aditsu Sep 9 '15 at 20:45
• Bonus for regex solutions. – flawr Sep 10 '15 at 9:59
## PHP code, 1808 971 bytes
Quick and dirty implementation in PHP. First brute-force all possible slice shapes, next brute-force all positions and orientations of the slices.
Usage: cat pizza.txt | php pizza.php
Edit: reduced code size by more than 45% by rewring algorithm using recursion rather than nested loops. However, this eats memory (and pizza's ;-)). Pizza's larger than 8x8 will probably run out of memory. The nested loop variant can easily handle any size, but is twice the code size.
<?php define('A',98);$n=fgets(STDIN);$d=array();$m=$u=str_pad('',A,'+');$s=0;while($g=fgets(STDIN)){$g=rtrim($g);assert(strlen($g)<=A-2);$s++;$m.='+'.str_pad(rtrim($g),A-2,' ').'+';for($l=0;$l<strlen($g);$l++)if($g[$l]=='#')$d[]=$s*A+$l+1;}$m.=$u;$r=count($d)/$n;x(reset($d),array(array()),0,0,0,0);die('No pizza for you!');function x($e,$c,$b,$a,$q,$t){global$r,$m,$d;$h=$a*A+$b;if(!in_array($e+$h,$d))return;if(in_array($h,$c[0]))return;$c[0][]=$h;$c[1][]=$b*A-$a;$c[2][]=-$a*A-$b;$c[3][]=-$b*A+$a;if(count($c[0])<$r)do{x($e,$c,$b+1,$a,$b,$a);x($e,$c,$b,$a+1,$b,$a);x($e,$c,$b-1,$a,$b,$a);x($e,$c,$b,$a-1,$b,$a);$v=($b!=$q||$a!=$t);$b=$q;$a=$t;}while($v);else w($c,$m,0,reset($d),0);}function w(&$p,$f,$o,$e,$i){global$n,$d;foreach($p[$i]as$h){$j=$e+$h;if(!isset($f[$j])||$f[$j]!='#')return;$f[$j]=chr(ord('0')+$o);}if(++$o==$n){for($k=A;$k<strlen($f)-A;$k++)if($k%A==A-1)echo PHP_EOL;else if($k%A)echo$f[$k];exit;}foreach($d as$j)for($i=0;$i<4;$i++)w($p,$f,$o,$j,$i);}
Ungolfed, documented code
Below is the documented, original code. To keep my sanity, I worked with the full source code, and wrote a simple minifier script to strip statements like assert() and error_reporting(), remove unnecessary brackets, rename variables, functions and constants to generate the golfed code above.
<?php
error_reporting(E_ALL) ;
// Width of each line of pizza shape.
// Constant will be reduced to single character by minifier,
// so the extra cost of the define() will be gained back.
define('WIDTH', 98) ;
$nrSlices = fgets(STDIN) ; // Read pizza shape definition and // convert to individual$positionList[]=$y*width+$x and
// linear (1D) $pizzaShape[$y*WIDTH+$x] with protective border around it. // // WARNING: assumes maximum pizza width of WIDTH-2 characters!$positionList = array() ;
$pizzaShape =$headerFooter = str_pad('', WIDTH, '+') ;
$y = 0 ; while ($line = fgets(STDIN))
{ $line = rtrim($line) ;
assert(strlen($line) <= WIDTH-2) ;$y++ ;
$pizzaShape .= '+'.str_pad(rtrim($line), WIDTH-2, ' ').'+' ;
for ($x = 0 ;$x < strlen($line) ;$x++)
{ if ($line[$x] == '#') $positionList[] =$y*WIDTH + $x+1 ; } }$pizzaShape .= $headerFooter ; // Determine size of a slice$sliceSize = count($positionList)/$nrSlices ;
// Build all possible slice shapes. All shapes start with their first part at
// the top of the pizza, and "grow" new parts in all directions next to the
// existing parts. This continues until the slice has the full size. This way
// we end up with all shapes that fit at the top of the pizza.
//
// The shape is defined as the offsets of the parts relative to the base
// position at the top of the pizza. Offsets are defined as linear offsets in
// the 1-D $pizzaShape string. // // For efficiency, we keep track of all four possible rotations while building // the slice shape. // growSlice(reset($positionList), array(array()), 0, 0, 0, 0) ;
die('No pizza for you!') ;
function growSlice($basePosition,$shapeDeltas, $dx,$dy, $prevDx,$prevDy)
{ global $sliceSize,$pizzaShape, $positionList ; // Check validity of new position // Abort if position is not part of pizza, or // if position is already part of slice$delta = $dy*WIDTH +$dx ;
if (!in_array($basePosition+$delta, $positionList)) return ; if (in_array($delta, $shapeDeltas[0])) return ; // Add all four rotations to shapeDeltas[]$shapeDeltas[0][] = $delta ;$shapeDeltas[1][] = $dx*WIDTH -$dy ;
$shapeDeltas[2][] = -$dy*WIDTH - $dx ;$shapeDeltas[3][] = -$dx*WIDTH +$dy ;
// Have we built a full slice shape?
if (count($shapeDeltas[0]) <$sliceSize)
{ // Grow shape either at current position or at previous position
do
{ growSlice($basePosition,$shapeDeltas, $dx+1,$dy, $dx,$dy) ;
growSlice($basePosition,$shapeDeltas, $dx,$dy+1, $dx,$dy) ;
growSlice($basePosition,$shapeDeltas, $dx-1,$dy, $dx,$dy) ;
growSlice($basePosition,$shapeDeltas, $dx,$dy-1, $dx,$dy) ;
$retry = ($dx != $prevDx ||$dy != $prevDy) ;$dx = $prevDx ;$dy = $prevDy ; } while ($retry) ;
} else
{ // Try to cover the entire pizza by translated and rotated instances of
// the slice shape.
fitSlice($shapeDeltas,$pizzaShape, 0, reset($positionList), 0) ; } } function fitSlice(&$shape, $pizza,$id, $basePosition,$rotation)
{ global $nrSlices,$positionList ;
// Try to fit each part of the slice onto the pizza. If the part falls
// outsize the pizza, or overlays another slice we reject this position
// and rotation. If it fits, we mark the $pizza[] with the slice$id.
foreach ($shape[$rotation] as $delta) {$position = $basePosition +$delta ;
if (!isset($pizza[$position]) || $pizza[$position] != '#') return ;
$pizza[$position] = chr(ord('0')+$id) ; } // If$nrSlices slices have been fitted, we have found a valid solution!
// In that case, we display the solution and quit.
if (++$id ==$nrSlices)
{ for ($pos = WIDTH ;$pos < strlen($pizza)-WIDTH ;$pos++)
{ if ($pos % WIDTH == WIDTH-1) echo PHP_EOL ; else if ($pos % WIDTH) echo $pizza[$pos] ;
}
exit ;
}
// The current slice did fit, but we have still more slices to fit.
// Try all positions and rotations for the next slice.
foreach ($positionList as$position)
{ for ($rotation = 0 ;$rotation < 4 ; $rotation++) { fitSlice($shape, $pizza,$id, $position,$rotation) ;
}
}
}
• I'm getting "PHP Fatal error: Cannot redeclare _() in pizza.php on line 1" – aditsu Sep 13 '15 at 14:03
• @aditsu: there is only one function _() in the golfed version. Did you accidently copy-paste the code twice? – Jason Smith Sep 17 '15 at 7:30
• The file size is 972 so I don't think the code could fit twice. The ungolfed code seems to work btw :) – aditsu Sep 17 '15 at 8:57
• I noticed you have define('_',98), doesn't that conflict with function _? I don't know php so I can't tell... – aditsu Sep 17 '15 at 9:01
• @aditsu: The golfed code works fine on my Mac with PHP 5.4.43, but it appears _() is an alias for gettext() on other platforms. Changed minifier to avoid _() altogether. – Jason Smith Sep 17 '15 at 19:19 | 2019-11-17 15:16:45 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.23196180164813995, "perplexity": 5898.947444575126}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496669183.52/warc/CC-MAIN-20191117142350-20191117170350-00492.warc.gz"} |
https://brilliant.org/practice/higher-order-derivatives-level-2-3-challenges/ | ×
Back to all chapters
# Higher-order Derivatives
The first derivative is the slope of a curve, and the second derivative is the slope of the slope, like acceleration. So what's the third derivative? (Fun fact: it's actually called jerk.)
# Higher-order Derivatives: Level 2 Challenges
Given the graph of $$y=f(x)$$ above, which of the following is a possible graph of $$y=f''(x)?$$
SecondDr
Find the 2016-th derivative of $$\sin ^{ -1 }{ (x) }$$ at $$x=0$$.
###### Image Credit: Flickr Richard Stocker.
If a function $$f(x)$$ that is differentiable over $$(-\infty,\infty)$$ is monotonically decreasing and $$\displaystyle\lim_{x\rightarrow\infty}f(x)\neq-\infty,$$ then as $$x$$ approaches infinity, $$f(x)$$ is
$y = \tan^{-1}(x) , k! = \left. \dfrac { { d }^{ 21 }y }{ d{ x }^{ 21 } } \right|_{x=0}, \ \ \ \ \ k = \ ?$
Imgur
Suppose $$f$$ is a function defined on the closed interval $$-3 \le x \le 4$$ with $$f(0)=42$$ such that the graph of $$f',$$ the derivative of $$f,$$ on the interval is as shown in the above diagram. Find the $$x$$-coordinates of the points of inflection of $$f.$$
× | 2017-03-23 20:20:43 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8739140033721924, "perplexity": 288.85931073536705}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218187206.64/warc/CC-MAIN-20170322212947-00608-ip-10-233-31-227.ec2.internal.warc.gz"} |
https://www.nature.com/articles/s41598-018-33060-3?error=cookies_not_supported | ## Introduction
Owing to the ever-increasing desire to use carbon dioxide (CO2) as an C1 feedstock1,2,3,4,5,6, the development of heterogeneous5,7 and homogeneous4,8,9,10,11,12,13,14,15,16,17,18,19,20,21 catalysts for the transformation of CO2 into reduced C1 compounds8,9,17,18,19,20,21 or CO2-containing organic compounds has received considerable attention10,11,12,13,14,15,16. In the former case, several thermal hydrogenation8,9, electro-17,18,19, and photo-catalytic systems17,18,20,21 that produce reduced C1 chemicals, e.g., CH3OH22,23,24, CO25,26,27,28,29,30, HCO2H31,32,33, and HCO234,35,36 have been investigated. In the latter case, carboxylic acids are formed, which represent a central motif in a variety of synthetically important chemicals37,38,39,40,41,42,43,44,45,46,47,48,49,50.
Recently, the direct carboxylation of aryl rings has garnered considerable attention. For example, Nolan et al. reported the [M(IPr)(OH)] (M = CuI or AuI, IPr = 1.3-bis(diisopropyl)phenylimidazol-2-ylidene)-catalyzed carboxylation of halogenated benzene using KOH or CsOH as a base under an atmosphere of CO242,43. Iwasawa et al. also reported the Pd(OAc)2-catalyzed photochemical carboxylation of aryl halides using [Ir(ppy)2(dtbpy)][PF6] (λex = 425 nm; dtbpy = 4,4′-di-tert-butyl-2,2′-bipyridyl) as the photosensitizer and iPr2NEt as a sacrificial electron donor under CO247. In these reports, the presence of a base and/or electron donor is necessary for the thermal or photochemical reaction and the incorporation of CO2. However, obtaining carboxylic acids directly from CO2 in the absence of an electron donor and base would be more attractive in terms of atom- and step-economy.
We have recently reported the photochemical hydrogen evolution (PHE) from the Fe(II) complex [FeII(opda)3][ClO4]2 (1; opda = o-phenylenediamine) (Fig. 1) under N251, wherein opda acts as a photo-responsive proton/electron pool that forms the partially oxidized semi-benzoquinodiimine (s-bqdi) or o-benzoquinodiimine ligands. The photochemical reaction of 1 under N2 inspired us to investigate that under CO2. Herein, we report the first example for a direct photochemical carboxylation of C–H bonds in aromatic diamines with CO2 in the absence of an electron donor and base.
## Results
### Photochemical hydrogen evolution reactions from opda and 1 under N2 or CO2
The UV-Vis spectra of opda and 1 in THF under N2 or CO2 (Fig. 2a) exhibit absorption band at 298 nm, attributable to the ππ* transition of opda and a transition with ππ* character of 151. In the case of opda, spectral differences were not observed under N2 or CO2, indicating a negligible effect of the atmosphere on the electronic state. Even though the molar absorption coefficient of 1 changed slightly from N2 to CO2 (Fig. 2a), the similarity of the spectra and the fact that the same colorless crystals of 1 were obtained from THF/n-hexane under each atmosphere (Supplementary Fig. 1 and Supplementary Table 1) indicate that 1 is stable under both gases. When the solution was left to stand for 8 h under CO2, there were virtually no UV-Vis spectral differences in both opda and 1 (Supplementary Fig. 2), demonstrating their stability in the dark. The 1H NMR (THF-d8) spectra of opda under N2 or CO2 (Supplementary Fig. 3) show characteristic signals at 6.42, 6.50, and 3.80 ppm, assignable to aromatic C–H and N–H protons. On the other hand, the spectra of 1 under N2 or CO2 (Supplementary Fig. 3) did not show any clear signals, reflecting its paramagnetic nature51. Aromatic amines afford carbamic acids or benzimidazol-2-one from the reaction with CO2 under basic conditions52,53 or in the presence of catalysts54,55,56. However, such products could not be detected under these dark conditions.
When a THF solution of opda was irradiated with UV light (λex = 300 ± 10 nm) under N2, PHE was observed with an apparent quantum yield of ΦH2@N2(%) = 1.28 × 10−3 (8 h) (Fig. 2b). Contrastively, the amount of evolved hydrogen under CO2 decreased by one third under N2 (ΦH2@CO2(%) = 4.16 × 10−4) (8 h). Under these conditions, a THF solution of 1 also showed PHE with ΦH2@N2(%) = 0.0138 (8 h) and ΦH2@CO2(%) = 3.56 × 10−3 (8 h) (Fig. 2b). The observed inhibitions of the PHE under CO2 clearly suggest an alternative photochemical process under CO2 compared to that under N2.
### Direct photochemical carboxylation of aromatic diamines with CO2 by opda and 1
The UV-Vis spectra of opda and 1 after photo-irradiation (8 h) under CO2 showed new absorption band at 347 nm (Fig. 3a). It is noteworthy that the new band was observed only when the photoreaction took place under CO2 (Supplementary Fig. 2). Interestingly, the absorbance of the new bands was significantly increased for 1 relative to opda, which confirms the promoting effect of the Fe(II) ion in 1. The 1H NMR (CD3CN) spectrum after the photoreaction of opda under CO2 (Supplementary Fig. 4) exhibits two sets of doublets at 6.82 and 7.30 ppm, as well as a triplet at 6.51 ppm, suggesting the formation of a 1,2,3-trisubstituted benzene. A similar spectrum was observed after the photoreaction of 1 under identical conditions (Supplementary Fig. 4). X-ray diffraction analysis of the pale yellow crystals of the product unambiguously identified 2,3-diaminobenzoic acid (DBA) (Fig. 3b and Supplementary Fig. 5, Supplementary Tables 2 and 3)57,58. Interestingly, one ortho C–H bond relative to an amino group was carboxylated. The similarity between the UV-Vis and 1H NMR spectra of commercial DBA and the photochemical products of opda or 1 (Fig. 3a and Supplementary Fig. 6) indicates that DBA is the main product, while the isomer, i.e., 3,4-diaminobenzoic acid, was not obtained in detectable amounts. Based on the molar absorption coefficient of DBA (ε = 4,041 M−1 cm−1) at 347 nm (Fig. 3a), the photochemical reaction of 1, prepared in-situ by mixing [FeII(H2O)6][ClO4]2 and 3 eq. of opda (run 3 in Table 1), afforded DBA in 58.0% (Φ(%) = 0.47), which is comparable to that for ex-situ-prepared 1 (54.1%; Φ(%) = 0.44) (Supplementary Fig. 7 and run 1 in Table 1). On the other hand, the photo-irradiation of a THF solution of opda afforded DBA in 27.5% (Φ(%) = 0.22) (Fig. 3a and run 4 in Table 1).
### A reaction mechanism for the photochemical carboxylation
To get insight into the underlying reaction mechanisms, we subsequently carried out the photoreaction with aniline and Fe(II) under CO2. However, the UV-Vis spectrum showed no significant changes (Supplementary Fig. 8 and run 5 in Table 1). The inertness of aniline prompted us to use m- (mpda) and p-phenylenediamine (ppda)59,60. The photo-irradiations (λex = 300 ± 10 nm) of mpda or ppda resulted in the emergence of new absorbances at 360 nm and 400 and 450 nm, respectively (Supplementary Fig. 9, runs 6 and 8 in Table 1). Curiously, the absorbances of mpda and ppda were observed even after irradiation, suggesting their poor reactivity (Supplementary Fig. 9). ESI-MS spectra showed signals (m/z 151.05) for the carboxylated products in the crude reaction mixture (Supplementary Figs 10a and 11a). The newly emerged 1H NMR signals of the carboxylated products were assigned to 2,4- and 2,5-diaminobenzoic acids (Supplementary Figs 10b and 11b). Conversely, the treatment of mpda or ppda with Fe(II) afforded white precipitates, probably due to the formation of coordination polymers (runs 7 and 9 in Table 1)61,62.
Given the atmosphere-dependent photoreactions of opda and 1, we focused our attention on their excited states. The emission spectra (λex = 300 nm) of opda and 1 in THF under N2 or CO2 showed the emission bands at 350 nm, assignable to emissions from ππ* of opda or ππ* included excited state of 1 (Supplementary Fig. 12). In the excitation spectra of opda and 1 under N2 or CO2 (λobs = 350 nm), the bands were observed at 298 nm, suggesting radiative deactivation pathways for the photoreactions under N2 and CO2 (Supplementary Fig. 13).
Subsequently, we attempted to identify the active species by trapping experiments. It was previously reported that 2-methylpropane-2-thiol (t-BuSH) can act as a hydrogen (H) radical scavenger forming di-tert-butyl disulfide (t-Bu2S2)63,64. The detection of t-Bu2S2 among the photochemical reaction products of opda and 1 revealed the H radical generation during the reaction (vide infra). The 1H NMR spectra of t-BuSH and t-Bu2S2 under CO2 (Fig. 4a,b, and Supplementary Fig. 14) showed singlets at 1.38 and 1.29 ppm, respectively. On the other hand, we found that the new signals emerged at 0.88, 0.89, and 1.19 ppm in the 1H NMR spectrum of a THF-d8 solution of t-Bu2S2 after photo-irradiation (λex = 300 ± 10 nm), which demonstrates the photoreactivity of t-Bu2S2 (Fig. 4c)63. These resonances are thus indicative of the in-situ formation of t-Bu2S2. A mixture of opda/t-BuSH displayed a 1H NMR spectrum similar to those of pure t-BuSH and opda (Fig. 4d and Supplementary Fig. 14), suggesting negligible interactions in the ground state. After photo-irradiation, new singlets emerged at 0.88, 0.89, and 1.19 ppm (Fig. 4e), and these peaks are identical to those of the photochemical products derived from t-Bu2S2 (Fig. 4c), suggesting the formation of t-Bu2S2 during the photochemical reaction. The 1H NMR spectrum of a mixture of 1/t-BuSH showed no significant interaction in the ground state (Fig. 4f), whereas new singlets emerged at 0.88 and 4.61 ppm after photo-irradiation (Fig. 4g and Supplementary Fig. 14). These peaks are comparable to those of the photochemical products of opda/t-BuSH mixture (Fig. 4e), suggesting the formation of t-Bu2S2 from 1/t-BuSH. Based on these results, it should be feasible to consider a reaction pathway involving the H radical generation for the photoreaction of opda and 1 under CO2. The lower amount of photochemically generated H2 from them under CO2 than N2 thus most likely reflects the incorporation of the generated H radicals in the DBA skeleton.
Finally, to shed more light on the reaction mechanism, we compared the 13C NMR (CD3CN) spectra of the reaction product of 1 under CO2 or 13CO2. In the 13C NMR spectrum of the photochemically-produced DBA from 1 under CO2, the resonance derived from the carboxyl carbon was observed at 170.5 ppm (Fig. 5a). In the case of the photoreaction under 13CO2, the peak intensity of the carboxyl carbon clearly increased, suggesting that the carboxyl moiety in DBA originates from CO2 (Fig. 5b)14,65.
Figure 6 depicts plausible mechanisms for the photochemical carboxylation of opda and 1. Given the aforementioned results, the photo-irradiation induces the generations of H and aminyl radical intermediates. The later then form a C–C bond with CO2 via the delocalization of the unpaired electron, thus forming the carboxyl radical intermediate66. Subsequently, the methine proton transferrs to the imino nitrogen, whereby the aromatic stabilization could act as driving force forming a 2,3-diaminobenzoic radical species. The reaction of the intermediate with a H radical might finally yield DBA. The role of the Fe(II) ion in this reaction should be worth investigating in detail, as it is highly plausible that the Fe(II) center perturbs the N–H moiety in opda favorably64,67,68,69,70,71.
## Discussion
In this paper, we demonstrated a direct photochemical C–H carboxylation of aromatic diamines with CO2. Although this reaction is not catalytic, it represents the first example of atom- and step-economic direct carboxylation of a C–H bond in benzene rings in the absence of any potentially reactive electron donor and base. The promotion of this reaction by Fe(II) could be achieved using opda ligand, indicating the potential of nonprecious metal ions to accelerate or catalyze the reaction. Further efforts to gain an in-depth understanding of the mechanism and to expand the scope of the reaction by using a wider range of aromatic polyamines, nonprecious metal ions, excitation wavelengths, and catalytic protocols are currently in progress.
## Methods
### General procedures
All synthetic operations were performed under N2 or CO2 using standard Schlenk-line techniques. The ligand opda was purchased from Wako Pure Chemical Industries (Japan), while p-phenylenediamine (ppda) was obtained from Sigma-Aldrich, and m-phenylenediamine (mpda), was procured from Tokyo Chemical Industry Co., Ltd. (Japan). 2,3-Diaminobenzoic acid (DBA) was purchased from Combi-blocks (USA) and used after recrystallization from H2O. Dehydrated THF, THF-d8, CD3CN, and silica gel (60 N) were purchased from Kanto Chemical Co. Inc. (Japan). MeOH, CH3CN, and emission analysis grade THF were obtained from Nacalai Tesque, Inc. (Japan). N2, CO2, and 13CO2 were purchased from Kotobuki Sangyo Co. Ltd. (Japan). Prior to use, THF was degassed by at least five freeze-pump-thaw cycles, followed by N2 or CO2 sparging for 20 min, and subsequent dehydration over molecular sieves (4 Å, MS4A), which were purchased from Wako Pure Chemical Industries (Japan) and activated by heating under high vacuum. Complex [FeII(opda)3][ClO4]2 (1) was prepared according to a previously reported procedure51. Caution! Although we did not experience any difficulties manipulating perchlorate salts, these should be regarded as potentially explosive and therefore require handled with the utmost care.
UV-Vis-NIR spectra were recorded on a HITACHI U-4100 spectrophotometer at room temperature (25 °C). IR spectra were recorded on a Thermo Nicolet 6700 FT-IR spectrometer by attenuated total reflection (ATR) method. 1H and 13C NMR (500 and 126 MHz) spectra were recorded on a JEOL EX-500 (and A-500) spectrometer using CD3CN or THF-d8. Elemental analyses were carried out on a Perkin-Elmer 2400 II CHN analyzer. Electrospray ionization mass spectra (ESI-MS) were performed at the Global Facility Center at Hokkaido University. Emission and excitation spectra were recorded on a Horiba FluoroMax-4 spectrophotometer at room temperature (25 °C).
### Preparation of the solutions used for the reactions under atmospheres of N2 or CO2 (in the dark or under irradiation)
The sample solutions for reactions under N2 or CO2 were prepared under the respective atmospheres. A colorless THF solution (2.0 mM) of ex-situ-prepared [FeII(opda)3][ClO4]2 was obtained from dissolving [FeII(opda)3][ClO4]2 (6.51 mg, 1.0 × 10−5 mol) in THF (5 mL). On the other hand, in-situ-prepared [FeII(opda)3][ClO4]2 was obtained from the treatment of [FeII(H2O)6][ClO4]2 (3.63 mg, 1.0 × 10−5 mol) with opda (3.24 mg, 3.0 × 10−5 mol) in THF (5 mL). In a similar manner, a colorless THF solution of a mixture of [FeII(H2O)6][ClO4]2 and aniline was prepared using aniline (2.73 µL, 3.0 × 10−3 mol) instead of opda. THF solutions (6 mM) of FeII-free opda, aniline, mpda, or ppda were prepared by dissolving opda (3.24 mg, 3.0 × 10−5 mol), aniline (2.73 µL, 3.0 × 10−3 mol), mpda (3.24 mg, 3.0 × 10−5 mol), or ppda (3.24 mg, 3.0 × 10−5 mol) in THF (5 mL).
To investigate the reactivity under photo-irradiation, 0.4 mL of the respective sample solutions were transferred into a N2- or CO2-filled 1 mm quartz cell and the UV-Vis spectrum of the initial state was measured. The remaining 4 mL of the sample solution were then transferred into a custom-made Schlenk-flask-equipped quartz tube (volume: 164 mL). After exposing this apparatus for 8 h to photo-irradiation, 0.4 mL of the sample solution were withdrawn and transferred into a N2- or CO2-filled 1 mm quartz cell in order to measure the UV-Vis spectrum. In order to examine the reactivity in the dark, 0.4 mL of the solution were transferred into a N2- or CO2-filled 1 mm quartz cell and the UV-Vis spectrum of the initial state was measured. After allowing the sample solutions to stand for 8 h, in the dark, the spectral measurements were recorded again.
### Photochemical hydrogen evolution
For the photochemical hydrogen-evolution reaction (HER), the aforementioned Schlenk-flask-equipped quartz tube (volume: 164 mL) and THF solutions (4 mL) were used. The light source for the photochemical reactions was a 200 W Hg-Xe lamp (LC-8, Hamamatsu Photonics K.K.), and the intensity of the light was measured by using a power meter (Nova, Ophir Optronics Ltd.) and a thermopile sensor (3 A, Ophir Optronics Ltd.) prior to photo-irradiation experiments. Gas chromatographic analyses were conducted using a Shimadzu gas chromatograph (GC-2014) equipped with a thermal conductivity detector (TCD), a column filled with 5 Å molecular sieves, and Ar as the carrier gas (15.0 mL/min). The oven temperature was maintained at 100 °C, while the column and detector temperatures were set to 70 °C and 200 °C, respectively. Before the photo-irradiation experiments, a gas sample (0.3 mL) was collected from the headspace using a gas-tight syringe (Tokyo Garasu Kikai Co. Ltd) and analyzed by GC to confirm the successful N2 or CO2 purge. The samples were then exposed to irradiation in a water bath at room temperature. During the reaction, gas samples (0.3 mL) were collected from the headspace in order to determine the amount of H2 evolved as a function of the irradiation time.
### Purification of DBA after the photoreactions
After the reactions in the dark or upon photo-irradiation, as well as measurements of UV-Vis spectra of the samples after the reaction, all THF solutions were transferred into a Schlenk flask and THF was removed under reduced pressure. After measuring of the 1H NMR and ESI-MS spectra, the reaction mixtures were purified by flash column chromatography (Isolera One ACITM Spektra, Biotage Co. Ltd.) on silica gel (60 N; Kanto Chemical Co. Inc.; eluent: CH3CN:MeOH = 9:1 then 0:10). The photochemical products of [FeII(opda)3][ClO4]2 or FeII-free opda were collected and dried in vacuo. The formation of DBA was confirmed by recording the 1H NMR spectra in CD3CN. Colorless single crystals of DBA suitable an X-ray crystallographic analysis were obtained from a recrystallization from THF/n-hexane.
### Calculation of the quantum yields (Φ%)
The THF solutions of the samples, except for those of aniline and a mixture of aniline and [FeII(H2O)6][ClO4]2, were exposed to photo-irradiation (λex = 300 ± 10 nm) from a Hg-Xe lamp equipped with the LX0300 band pass filter (Asahi Spectra Inc.; λ = 300 ± 10 nm; half bandwidth = 10.40 nm). The THF solutions of aniline and a mixture of aniline and [FeII(H2O)6][ClO4]2 were exposed to photo-irradiated (λex = 289 ± 10 nm) from a Hg-Xe lamp equipped with a CWL289 nm filter (OptoSigma Corporation, λ = 289 ± 10 nm, half bandwidth = 10 nm). The amount of DBA formed in runs 1, 3, and 4 (Table 1) in the subsequent 8 h were used to calculate the apparent quantum yield (Φ) using Eq. 1.
$${\Phi }={N}_{{\rm{e}}}/{N}_{{\rm{p}}}={N}_{{\rm{DBA}}}/{N}_{{\rm{p}}}$$
(1)
where, Ne refers to the number of reacted electrons, NDBA to the number of molecules of DBA formed in the reaction, and Np to the number of incident photons. | 2022-12-08 22:29:24 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5635696053504944, "perplexity": 6066.481966468307}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446711368.1/warc/CC-MAIN-20221208215156-20221209005156-00179.warc.gz"} |
http://openstudy.com/updates/5581aaffe4b07028ea60e6dc | ## anonymous one year ago log{b} 1000=3 I know b=10 but I don't understand how they got there.
1. anonymous
$\log_{b} 1000=3$
2. anonymous
$b^3=1000$ $3\sqrt{b^3}= 3\sqrt{1000}$ b=10
3. Nnesha
well did you take 3rd root right ? you can write 1000 in terms of 3 exponents
4. Nnesha
x^3 =1000 ? what is x
5. anonymous
10
6. anonymous
What is the point of the 3's in front of the square root signs?
7. Nnesha
that's not 3 it's a 3rd rooot i guess like this $\huge\rm \sqrt[3]{b^3}$
8. anonymous
Right, I didn't know how to do that
9. Nnesha
you can convert 3rd to 1/3 exponent $\huge\rm \sqrt[3]{b^3} = b^{3 \times \frac{ 1 }{ 3 }}$ so cancel out 3 exponent u have to take 3rd root :=)
10. Nnesha
$\huge\rm \sqrt[3]{b^3} = b^{\cancel3 \times \frac{ 1 }{\cancel 3 }}=b$
11. anonymous
Wouldn't that also mean that 1000 would be multiplied by 1/3?
12. Nnesha
yes so 1000 can be written as 10 to the 3rd power $\huge\rm \sqrt[3]{10^3}$
13. anonymous
See that's where I get confused, so the 1000 get simplified down to 10 because?
14. Nnesha
because it's easy to solve without calculator and it's legal :3
15. anonymous
-_- so because 10^3=1000 it gets simplified to 10^3 and since it's the 3rd root the exponent cancels out leaving b=10 ?
16. Nnesha
b to the 3rd power so you can either write 10 to the 3rd power or put 3rd root of 1000 into the calculator :=)
17. Nnesha
yep
18. anonymous
See when I enter that into my calc I get 94.86....
19. Nnesha
and it's helpful when you have to find an exponent given base for example 10^x = 1000
20. Nnesha
what did you enter ?
21. freckles
$3 \sqrt{1000} \neq \sqrt[3]{1000}$
22. anonymous
$3\sqrt{1000}$
23. anonymous
Yea see I don't know how to enter the latter
24. Nnesha
it's not 3 times square root of {1000} 3rd root like $\sqrt[3]{1000}$
25. Nnesha
ohh then convert 3rd root to 1/3 exponent $\huge\rm \sqrt[3]{1000} = (1000)^\frac{ 1 }{ 3 }$
26. anonymous
That works. =] Could you possibly help me with another problem?
27. Nnesha
gtg :3 sorry
28. anonymous
All's well Thank you for your help :)
29. Nnesha
my pleasure | 2016-10-22 23:59:18 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7279747128486633, "perplexity": 3375.1187353257706}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988719079.39/warc/CC-MAIN-20161020183839-00006-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/number-of-roots.233040/ | # Number of roots
1. May 4, 2008
### DeanBH
how do i find the number of roots for a curve that has dy/dx
2x^4 -20x^2 + 50.
if i substitute y=x^2 and use the discriminant formula i get
b^2 - 4ac = 400 - 4 x 2 x 50
= 400 - 400
= 0
This way says there 1 root, answers say it has 2. Which method am i meant to use for this?
if i factorise i get 2(y-5)^2
which is also 1?
Last edited: May 4, 2008
2. May 4, 2008
### malty
Do you know of a way to find how many times the curve crosses the x-axis?
Think of info needed to sketch this curve. .
Last edited: May 4, 2008
3. May 4, 2008
### DeanBH
yes i know 2 ways... factorizing and discriminant. and they both say 1 root..
4. May 4, 2008
### malty
Are you familar with critical points (Maxima, Minima, points of inflection . .) of a curve?
It looks to me like you are only finding the roots of the tangent to the curve?
5. May 4, 2008
### HallsofIvy
Are you look for the zeroes of y= 2x^4 -20x^2 + 50 or y such that dy/dx= 2x^4 -20x^2 + 50?
Obviously, 2x^4 -20x^2 + 50= 2(x^4- 10x+ 25)= 2(x^2- 5)(x^2+ 5) has two real roots- they are $\pm\sqrt{5}$.
But if you mean y such that dy/dx= 2x^4 -20x^2 + 50, there is no way of telling. You lose an additive constant when you differentiate y and how many times y is 0 depends on that constant.
6. May 4, 2008
### DeanBH
2x^4 -20x^2 + 50= 2(x^4- 10x+ 25)= 2(x^2- 5)(x^2+ 5) Is wrong.?
2(x^2- 5)(x^2+ 5) = 2(X^4 -25) not 2(x^4 - 10x + 25)
7. May 4, 2008
### epenguin
It should be easy to sketch what a curve of dy/dx gainst x and y against x when dy/dx is what you said (in the form you factorised it). The height of the second of these is not determined, as mentioned, unless that was given too in your original problem. But basically you should then see there is only one general possibility, plus one special case.
8. May 4, 2008
### HallsofIvy
Ooops! That's embarrassing! 2x^4 -20x^2 + 50= 2(x^4- 10x+ 25)= 2(x^2- 5)^2 which has 2 distinct roots, each a double root.
Thanks, DeanBH.
I'm still wondering what the original problem really was!
9. May 4, 2008
### DeanBH
it still makes no sense.
it says find stationary points on the curve blahblahblah.
the point is the curve has dy/dx 2x^4 -20x^2 + 50.
how the hell do i go about finding it has 2.
it looks like it has 1.
10. May 4, 2008
### Hootenanny
Staff Emeritus
Ahh, the question make sense now!
Why do you think that there is only one root? Both you and Halls have shown that it has two roots and therefore two stationary points.
11. May 4, 2008
### rocomath
$$2(x^2-5)^2=0$$
Can you find those 2 roots?
12. May 4, 2008
### DeanBH
that looks like one root to me, why is that 2.
13. May 4, 2008
### Hootenanny
Staff Emeritus
Cancel the two and take the square root of both sides. Does that make it any easier?
14. May 4, 2008
### DeanBH
no.
i dont know what you are talking about
15. May 4, 2008
### rocomath
Solve ...
$$x^2-5=0$$
What do you get?
16. May 4, 2008
### DeanBH
oooooooooooooooooooohhhhhhhhhhhhh craaaaaaaaaaaaaaaappppppppppp>
i was ignoring the fact it was X^2. taking it as X
me being retarded
17. May 4, 2008
### rocomath
Yeah I was getting worried for a sec :p hehe, don't worry about it! We all have brain farts, just pray it isn't during an exam!!! | 2018-02-21 12:24:18 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5072560906410217, "perplexity": 3116.6073469955104}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891813608.70/warc/CC-MAIN-20180221103712-20180221123712-00566.warc.gz"} |
https://studyadda.com/notes/jee-main-advanced/physics/nlm-friction-circular-motion/linear-momentum/7306 | JEE Main & Advanced Physics NLM, Friction, Circular Motion Linear Momentum
Linear Momentum
Category : JEE Main & Advanced
(1) Linear momentum of a body is the quantity of motion contained in the body.
(2) It is measured in terms of the force required to stop the body in unit time.
(3) It is also measured as the product of the mass of the body and its velocity i.e., Momentum = mass × velocity.
If a body of mass m is moving with velocity $\overrightarrow{v\,}$ then its linear momentum $\overrightarrow{p}$is given by $\overrightarrow{p}=m\,\overrightarrow{v\,}$
(4) It is a vector quantity and it?s direction is the same as the direction of velocity of the body.
(5) Units : kg-m/sec [S.I.], g-cm/sec [C.G.S.]
(6) Dimension : $[ML{{T}^{-1}}]$
(7) If two objects of different masses have same momentum, the lighter body possesses greater velocity.
$\therefore$ $p={{m}_{1}}{{v}_{1}}$$={{m}_{2}}{{v}_{2}}$= constant
$\frac{{{v}_{1}}}{{{v}_{2}}}=\frac{{{m}_{2}}}{{{m}_{1}}}$
i.e. $v\propto \frac{1}{m}$ [As p is constant]
(8) For a given body $p\,\propto v$
(9) For different bodies moving with same velocities $p\,\propto \,m$
Other Topics
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https://mathoverflow.net/questions/361794/suggested-papers-or-reading-for-pde-high-dimension-reduction-to-ode-by-symmetr | # Suggested papers or reading for PDE (high dimension) reduction to ODE by symmetries
Could anyone please suggest related papers or article about the topic related to my one question below?
Reduce PDE to ODE by dilation symmetry
I also cite a paper in the link above.
We know that if there are number of $$n$$ states, we need to find $$"n-1"$$ symmetries to reduce the PDE to ODE. For each iteration (there are $$n-1$$ iterations), we need to do some tedious but not difficult calculation.
Is there any advanced method based on symmetries reduction (of course, it might based on some special condition of the structure of problems) such that maybe we can take fewer steps? | 2021-08-01 16:27:32 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 3, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8524513840675354, "perplexity": 583.5101721896065}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046154214.63/warc/CC-MAIN-20210801154943-20210801184943-00716.warc.gz"} |
https://www.physicsforums.com/threads/juerg-froehlich-on-the-deeper-meaning-of-quantum-mechanics.972179/ | # A Jürg Fröhlich on the deeper meaning of Quantum Mechanics
#### A. Neumaier
Summary
a pointer to a recent paper
I'd like to draw attention to a very recent paper by Jürg Fröhlich, a well-known mathematical physicist from the ETH Zürich. It starts out as follows:
Jürg Fröhlich said:
I consider it to be an intellectual scandal that, nearly one hundred years after the discovery of matrix mechanics by Heisenberg, Born, Jordan and Dirac, many or most professional physicists – experimentalists and theorists alike – admit to be confused about the deeper meaning of Quantum Mechanics (QM), or are trying to evade taking a clear standpoint by resorting to agnosticism or to overly abstract formulations of QM that often only add to the confusion. [...]
I felt that the subject had better remain a hobby until later in my career. [...]
But when I was approaching mandatory retirement I felt an urge to clarify my understanding of some of the subjects I had had to teach to my students for thirty years
Section 2 is titled ''Standard formulation of Quantum Mechanics and its shortcomings''. Surely @vanhees71 has very convincing reasons why this critique is irrelevant from his personal point of view. But the others might be interested.
Section 3 then presents a completion of QM, the ''ETH-Approach to QM''. It is too abstract to become popular - one more of many interpretations satisfying their authors but probably not a majority of quantum physicists.
Last edited:
Related Quantum Physics News on Phys.org
#### bhobba
Mentor
Interesting paper.
At the moment I can't quite figure the difference between it and Decoherent Histories.
Need to think more and hear others views,
Thanks
Bill
#### vanhees71
Gold Member
I don't have reasons why this critique may be irrelevant since again I'm not even able to understand the problem to begin with. The statement is on p. 7. Fröhlich argues for a two-spin-1/2 system in the singlet Bell state, with the state ket given by
$$|\Psi \rangle=\frac{1}{\sqrt{2}} (|1/2,-1/2 \rangle-|-1/2,1/2 \rangle).$$
Then he makes the statement that from the obvious fact that of course the expectation value of $S_z=s_z \otimes 1 + 1 \times s_z$ is 0, it would follow that there couldn't be correlations between measurements of the 2 spins, which are however of course present due to the entanglement (it's even the maximal entanglement one can get, since it's a Bell state).
The point is that you have to do measurements on such prepared spin pairs on an "even-by-event" basis to get the correlations, i.e., for each pair you have to measure the $s_z$ for both particles of the pair and then find the 100% correlation, i.e., if A measures +1/2, then B necessarily measures -1/2 and vice versa. Of course you can't learn this from the expectation values.
Of course, for each of the two observers what they get are just unpolarized particles, i.e., the spin component is utmost indetermined, and that's in accordance with the cited linked-cluster principle which is of course valid in a relativistic local QFT by construction.
So I don't even get the argument of the author, why there is a contradiction to the minimally interpreted QT or the many observations on systems like this (mostly with photons), which are well understood using the minimally interpreted QT, including the "funny" features of entanglement. It's all well understood and described by standard QT, e.g., in the Walborn quantum eraser experiment we have discussed very often in this forum. I don't need to repeat it.
I also don't understand the solution of what the author seems to consider a problem since I'm not familiar with the the mathematical-physics notation the author uses. I've not the time to learn this notation to understand the solution of a problem, whose statement I even cannot understand to begin with. I'm sure, this interpretation could becom popular (or more popular or at least discussed) when (a) the problem it tries to solve is clarified and (b) the mathematical-physics notation is translated down to more common theoretical-physicists notation.
#### ftr
It is too abstract to become popular
yet, he says"it furnishes quantum theory with a clear ontology”. I could not understand what that ontology entailed.
#### vanhees71
Gold Member
Well, "ontology" is also one of those unclear notions of the philosophers. It's not clear at all what "ontology" means. It heavily depends on what the individual philosopher thinks it is. For the physicist any observable fact about nature is enough "ontology". That's why I don't understand, where Fröhlich's problem with QM is located. The example he brings in the above cited paper is simply not what's observed. To the contrary, nowadays there are many experiments by the AMO community which prove it wrong: Since more than 30 years (with Aspect's first experiment concerning the Bell inequalities) with ever increasing precision the strong correlations due to entanglement (but indeed consistent with the linked cluster principle valid for local relativistic QFTs and thus valid for the standard model and particularly QED!) are empirically established. It's not only expectation values that can be measured, but the outcome of measurements on an event-by-event basis.
So far there's not the slightest hint that QT is incomplete within the realm it is formulated (it is only (!) incomplete with regard to the lack of a satisfactory quantum description of gravity). The "ontology" from a physicist's point of view then simply is provided by the notion of the quantum state, and this implies that there is something called "objective indeterminism" in nature, i.e., it is impossible to determine by preparation all observable definable on a given quantum system. Also the classical behavior of "everyday matter" is well understood: It's due to reducing the description to the relevant macroscopic degrees of freedom through "coarse-graining".
#### zonde
Gold Member
For the physicist any observable fact about nature is enough "ontology".
The "ontology" from a physicist's point of view then simply is provided by the notion of the quantum state,
Quantum state is not an observable fact about nature. You are contradicting yourself within single post.
If you stick to the idea that observable facts about nature are "ontological", QM is still rather ontologically unsatisfactory. That's because statistics is composite fact about nature. It requires some interpretation and grouping of similar situations. Elementary facts are single detections. And NRQM as well as QFT can speak only about statistics and it can not say how these statistics emerge from elementary facts while common explanation seems to fail.
#### vanhees71
Gold Member
You are right in saying that the ontlogy in QT is provided by both the notion of "state" and "observable". These notions together provide the ontlogy of QT, and there's no contradiction to this ontology by experiment. To the contrary, the more QT is tested the better it gets confirmed. Particularly Fröhlich's idea of proving some contradiction is unclear and not justified by any observation. To the contrary his example provides one of the most stringent tests of Q(F)Ts consistency as a theory (validity of the linked-cluster principle) as well as with observations, which Bell experimenta are a confirmation of Q(F)T with the highest significance ever reached between theory in experiment in the history of physics.
#### zonde
Gold Member
You are right in saying that the ontlogy in QT is provided by both the notion of "state" and "observable".
Never said that. The things you say are so incoherent that I have no idea how to reply.
#### vanhees71
Gold Member
What specifically is "incoherent"?
My statement is that the minimally interpreted QT (which is basically Copenhagen without collapse) is all "ontology" you need since it describes precisely what's observed today.
The fundamental point about which all these discussions about "interpretation" occur is the following:
(a) The state of a system is described by a positive semidefinite self-adjoint operator $\hat{\rho}$ with $\mathrm{Tr} \hat{\rho}=1$.
(b) All observables are described by self-adjoint operators $\hat{O}$. The possible outcome of (accurate) measurements are the generalized eigenvalues of these operators.
(c) The probability to find the value $o_i$ when measuring the observable $O$ is given by
$$P(o_i)=\sum_{\beta} \langle o_i,\beta|\hat{\rho}|o_i,\beta \rangle$$
with the usual treatmenof spectral values in the continuum. $\beta$ are a set of parameters (e.g., the eigenvalues of a complete set of observable operators that complement $\hat{O}$). Also here the usual treatment if there are continuous parts in these parameters is implied. For simplicity I use sums rather than integrals.
The ontology in this standard interpretation thus is that if a system is in a state described by $\hat{\rho}$, some observables may have determined values others not. In any case the probabilities for measuring a certain possible value is given by the formula above (the Born rule in its most general form).
In my opinion that's compatible with all observations done so far. There seems nothing missing in this description. It's used to also describe the most accurate observations concerning entangled states, where Fröhlich argues to be a problem about, and his argument doesn't match the corresponding experiments (like the polarization measurements on entangled photon pairs a la Aspect, the quantum eraser, etc.), because it just argues with some expectation value, which refers to an averaging over many measurements on identically prepared systems (short an ensemble), where of course information that can be gained according to QT is thrown away. Then of course the correlations predicted by entanglement are not observable anymore, because this information is not taken, but as is demonstrated by very accurate measurements on such Bell states, in fact one can get this information, which is the event-by-event outcome of measurements on the entangled observables (in Fröhlich's example the polarization state of both of the entangled photons for each such prepared photon pair). With this information, all predicted correlations (100% correlation when measuring polarization in the same or mutually orthogonal directions, violation of Bell's inequality for adequately chosen relative angles of the polarizers etc. etc.).
The outcome of these experiments is very clear: All predictions of QT have been empirically "verified" at a very large confidence level. Also many (if not all) loopholes brought forward so far have been excluded. No observations are unexplained by minimally interpreted QT.
In the relativistic realm also no violations of Einstein causality are present. This is only the case if a naive collapse argument of some Copenhagen-interpretation flavors is envoked (as far as I know, Bohr was at least very careful not to put to much weight in the collapse assumption). It's also not necessary to explain the said correlations due to entanglement on far-distant parts of a quantum system (like the polarization-entangled photon pair in Fröhlich's example, where the single-photon polarization measurements can be done at arbitrary far distances, as long as no photon is significantly disturbed on its way from the source to the detector which would of course destroy the entanglement and correlations before both measurements have been done). As Fröhlich states, the very construction of QED as a local microcausal relativistic QFT guarantees the validity of the linked-cluster principle, and this also applies then of course to the said entangled state of two photons. Performing the polarization measurements (more precisely the detection of each photon behind the polarization filters) in setups such that these detection events are space-like separated, for which mutual influence of one measurement by the other is excluded, still show all the predicted correlations, which is also in accordance with the linked-cluster principle. The conclusion is that, as stated within minimally interpreted QFT, the correlation is due to the state preparation and not by causal influences between the detection events.
So again, I don't see any "incoherence" in the minimal interpretation nor the necessity for "more ontology" than provided by it, at least I don't see it in Fröhlich's argument.
#### DarMM
Gold Member
So again, I don't see any "incoherence" in the minimal interpretation nor the necessity for "more ontology" than provided by it, at least I don't see it in Fröhlich's argument.
I think I might be able to explain other people's problems to you. There are basically three problems people have with QM.
1. There is no dynamical account of which measurement outcome occurs
2. There is no dynamical account of how the correlations present in entanglement are achieved
3. Imagine I measure the state $\frac{1}{\sqrt{2}}\left(|\uparrow\rangle + |\downarrow\rangle\right)$ and say my device measures spin-up. Then the state I use after is $|\uparrow\rangle$, but an external superobserver would use $\frac{1}{\sqrt{2}}\left(|\uparrow, D_{\uparrow}, L_A\rangle + |\downarrow, D_{\downarrow} L_B\rangle\right)$ with $D_{\uparrow}$ denoting the state of my device and $L_A$ being a state of the lab. That is I use a "collapsed" state, but the superobserver does not.
I think recent work, especially since Spekken's model in 2004 and clarifications of the Frauchiger-Renner argument, has shown that (3) is not really a problem or a contradiction, it only is if you accept the eigenstate-eigenvalue link which is a very old naive view of QM.
However (1) and (2) do seem like problems or at least something that needs completing in a further theory.
The only problem is that quantum mechanics involves non-classical correlations. That is correlations outside the polytope given by assuming that your variables all belong in a single sample space. You can show (Kochen-Specker, Colbeck-Renner, etc) that theories with correlations outside of this polytope by necessity lack a dynamical account for their outcomes or correlations.
So you either reinterpret the formalism in a non-statistical manner (Many-Worlds, Thermal Interpretation), add additional variables to restore the single sample space (but we know they have to be nonlocal or retrocausal) or just accept that there is no account.
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#### DarMM
Gold Member
Interesting paper.
At the moment I can't quite figure the difference between it and Decoherent Histories.
Need to think more and hear others views,
Thanks
Bill
It's not very obvious from the paper but it is a form of Consistent Histories in a sense. However unlike Decoherent histories it doesn't frame consistency in terms of interference terms dying off, but in terms of a certain relation holding between the observables in the state $\omega$. This is slightly different as it can be "exact", but the major points are the same.
It's similar to how Jeffrey Bub's view is sort of consistent histories, but it takes consistency as being the emergence of a sub-algebra which satisfies the rules of a Boolean lattice.
So just different notions of consistency. What's interesting is that in a typical experiment all three conditions seem to hold, i.e. interference dies off, Frohlich's algebraic condition holds and a sub-algebra with lattice conditions as demanded by Bub emerges.
Later edit:
Personal conjecture: I wouldn't be surprised if basically Frohlich and Bub's views are two ways of phrasing the same thing, i.e. the emergence of a certain algebraic structure relative to the state is what permits you to reason that macroscopic equipment has obtained a definitive outcome state of which you are ignorant. Perhaps there's a theorem showing they are equivalent. We know from Spekkens model that interference terms don't mean there isn't a single outcome, so you don't need them to vanish.
You then need decoherence to show that macroscopic events obey classical statistics, not that they occur.
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#### Lord Jestocost
Gold Member
2018 Award
There are basically three problems people have with QM.
1. There is no dynamical account of which measurement outcome occurs
2. There is no dynamical account of how the correlations present in entanglement are achieved
Are these “serious” problems of physics? Or are these “problems” merely an expression of indignation that the ultimate reality behind giving rise to our “perception” of events occuring on a space-time scene cannot be grasped with recourse to classical notions and concepts?
As Berthold-Georg Englert writes in "On quantum theory":
Abstract. Quantum theory is a well-defined local theory with a clear interpretation. No “measurement
problem” or any other foundational matters are waiting to be settled.
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#### DarMM
Gold Member
Are these “serious” problems of physics? Or are these “problems” merely an expression of indignation that the ultimate reality behind giving rise to our “perception” of events occuring on a space-time scene cannot be grasped with recourse to classical notions and concepts?
I'm going to be very controlled in my response here, because I don't want this to veer into the usual stuff.
Are these serious problems? Well it depends on whether you think there has to be an account for how events occur, specifically outcomes of measurements on microscopic systems, or whether you think the current evidence from QM is enough for you to concede that you will never get one. The latter runs counter to why many people are interested in science, so it is not too surprising it is viewed as a problem.
I think saying "classical notions" undersells the problem some people have. It makes it sound as though they are attached to specific ideas like particles or fields. Where as the only "notion" they are holding to is an explanation/account at all.
Englert's quote (from https://arxiv.org/abs/1308.5290) just shows he is not bothered by this.
Like Gell-Mann, Griffiths, Bub, Bohr, Heisenberg, Hartle, Haag, etc he just swallows the bullet. You will never have an explanation. The End.
#### A. Neumaier
Englert's quote (from https://arxiv.org/abs/1308.5290) just shows he is not bothered by this.
Like Gell-Mann, Griffiths, Bub, Bohr, Heisenberg, Hartle, Haag, etc he just swallows the bullet. You will never have an explanation. The End.
The full abstract of his paper (published in Eur. Phys. J. D) says:
Berthold Englert said:
Quantum theory is a well-defined local theory with a clear interpretation. No "measurement problem" or any other foundational matters are waiting to be settled.
This essentially echoes the credo of @vanhees71. Englert's introduction says:
Berthold Englert said:
there is no experimental fact, not a single one, that contradicts a quantum-theoretical prediction. Yet, there is a steady stream of publications that are motivated by alleged fundamental problems: We are told that quantum theory is ill-defined, that its interpretation is unclear, that it is nonlocal, that there is an unresolved “measurement problem,” and so forth.
I find both statements in this second quote fully valid, and Englert's later explanations that these are only pseudo-problems not convincing.
#### DarMM
Gold Member
Yes and his "clear interpretation" detailed in the rest of the paper is (Neo-)Copenhagen, i.e. there is no explanation for measurement outcomes.
Without further no-go theorems we can't proceed further. Currently the attempts to add more variables to restore a single sample space (retrocausal and nonlocal theories) haven't been generalized to QFT and views that attempt to reinterpret the formalism non-statistically (Many Worlds and Thermal Interpretation) haven't been proven to give the correct observational statistics.
So we have to wait to see if one of these other views can be gotten to work in some way. Or perhaps wait for the development of no-go theorems that either forbid them or make them look completely unnatural and fine-tuned, forcing us to "swallow the bullet" of Copenhagen and its lack of explanations.
Time will tell.
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#### A. Neumaier
I find both statements in this second quote fully valid, and Englert's later explanations that these are only pseudo-problems not convincing.
His final words in the paper are:
Berthold Englert said:
What, then, about the steady stream of publications that offer solutions for alleged fundamental problems, each of them wrongly identified on the basis of one misunderstanding of quantum theory or another? Well, one could be annoyed by that and join van Kampen [42] in calling it a scandal when a respectable journal prints yet another such article. No-one, however, is advocating censorship, even of the mildest kind, because the scientific debate cannot tolerate it. Yet, is it not saddening that so much of diligent effort is wasted on studying pseudo-problems?
Note that van Kampen's paper, which he cites here and which also promoted the thesis that there is no measurement problem, contains an error in its ''proof'' of this thesis.
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#### charters
Currently the attempts to add more variables to restore a single sample space (retrocausal and nonlocal theories) haven't been generalized to QFT and views that attempt to reinterpret the formalism non-statistically (Many Worlds and Thermal Interpretation) haven't been proven to give the correct observational statistics.
Aharonov's "retrocausal" two time interpretation generalizes to QFT (equally as easily as MWI) and gets the correct observational statistics from a typicality assumption on the future boundary choice.
#### DarMM
Gold Member
Aharonov's "retrocausal" two time interpretation generalizes to QFT (equally as easily as MWI) and gets the correct observational statistics from a typicality assumption on the future boundary choice.
Note what I said. I didn't say that retrocausal theories can't get out the statistics, I said they aren't generalized to QFT fully, neither is the TSVF you are discussing. Kastner's work can be considered to have shown that it might be able to replicate aspects of QED, but I'm not aware of a full proof that it works in the QFT case.
Many Worlds has many issues with QFT, such as the absence of pure states for finite volume systems. And the Born rule has never been proven to hold.
Anyway, I'd be happy to discuss this on another thread. Either MWI or Retrocausal views.
#### ftr
has shown that (3) is not really a problem or a contradiction
Aren't 1) and 3) basically the same problem, or at least related.
#### DarMM
Gold Member
Aren't 1) and 3) basically the same problem, or at least related.
They are not the same problem. In general they are not related, for example theories like Spekkens model have (3), but not (1). That is you can have Wigner's friend style problem without having the measurement problem. That's why the measurement problem, labelled (1) above, is more important to QM and (3) isn't a real issue.
#### ftr
They are not the same problem. In general they are not related, for example theories like Spekkens model have (3), but not (1). That is you can have Wigner's friend style problem without having the measurement problem. That's why the measurement problem, labelled (1) above, is more important to QM and (3) isn't a real issue.
Well, collapse is an integral part of standard interpretation like CI although it is down played and all other interpretations are not that successful( at least this is my view of the general consensus) in circumventing it because of the probability interpretation. So all the power to TI.
EDIT: OK, it is possible to look at them as separate problems in some sense.
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#### stevendaryl
Staff Emeritus
Are these “serious” problems of physics? Or are these “problems” merely an expression of indignation that the ultimate reality behind giving rise to our “perception” of events occuring on a space-time scene cannot be grasped with recourse to classical notions and concepts?
No, the difficulties of interpreting quantum mechanics are not due to the fact that it cannot be grasped in terms of classical notions.
As Berthold-Georg Englert writes in "On quantum theory":
Abstract. Quantum theory is a well-defined local theory with a clear interpretation. No “measurement
problem” or any other foundational matters are waiting to be settled.
I think he's just wrong about that.
#### A. Neumaier
I think he's just wrong about that.
Its controversial, even here on PF. Different people have different criteria for ''well-defined''and' 'clear' '. Those with loose criteria are easily satisfied, only those with strict ones see the problems. No amount of discussion will change this.
#### DarMM
Gold Member
Its controversial, even here on PF. Different people have different criteria for ''well-defined''and' 'clear' '. Those with loose criteria are easily satisfied, only those with strict ones see the problems. No amount of discussion will change this.
I think this is pretty accurate. Either you have a problem with the standard lack of explanation for how measurement outcomes come about and how nonclassical correlations are achieved or you don't. In the typical approach QM doesn't give any explanation for these things. You'll either think this is an insight (i.e. this is something which cannot be given a scientific explanation) or an incompleteness (there has to be a deeper theory telling us how they come about).
The way things currently stand, i.e. no-go theorems et al, leaves this as an issue of personal taste.
#### ftr
leaves this as an issue of personal taste.
However I do think that some research come very close to explaining it.
"Jürg Fröhlich on the deeper meaning of Quantum Mechanics"
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• Solo and co-op problem solving | 2019-06-16 21:39:32 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6667127013206482, "perplexity": 1187.2360207887707}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627998298.91/warc/CC-MAIN-20190616202813-20190616224813-00310.warc.gz"} |
http://mathhelpforum.com/calculus/201191-decreasing-series.html | # Thread: Decreasing series
1. ## Decreasing series
How to prove that for $a>0$ and $s_j = \sum_{i=0}^j \frac{a^i}{i!}$ the series
$f_j = \frac{s_j}{s_{j-1}}$
is decreasing?
Thanks in advance!
2. ## Re: Decreasing series
Originally Posted by sander
How to prove that for $a>0$ and $s_j = \sum_{i=0}^j \frac{a^i}{i!}$ the series
$f_j = \frac{s_j}{s_{j-1}}$
is decreasing?
Thanks in advance!
Wouldn't \displaystyle \begin{align*} f_j \end{align*} be a sequence, not a series?
3. ## Re: Decreasing series
The series $\sum_{i=0}^\infty\frac{a^i}{i!}$ converges to $e^a$. | 2017-03-25 02:53:07 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 9, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.997209906578064, "perplexity": 6537.6297021294795}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218188773.10/warc/CC-MAIN-20170322212948-00480-ip-10-233-31-227.ec2.internal.warc.gz"} |
https://math.eretrandre.org/tetrationforum/showthread.php?tid=352&page=2 | • 0 Vote(s) - 0 Average
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Solving tetration for base 0 < b < e^-e Gottfried Ultimate Fellow Posts: 764 Threads: 118 Joined: Aug 2007 09/13/2009, 06:14 AM (09/12/2009, 09:04 PM)mike3 Wrote: So then since neither of those worked, it seems all we're left with is the Ansus formula and the Cauchy integral (but determining the correct contours and asymptotic behavior, now that's the rub...). I'm also curious: What about $b = e^{-e}$ exactly? You said it converges slowly, but how do you iterate it at all? What is the asymptotic as the tower $x \rightarrow \infty$? I used a complex fixpoint and could generate the matrices for regular iteration (in context of diagonalization). The fixpoint I used is t0 = -0.1957457524880764 - 1.691199920910569*I one of its logarithms is u0 = 0.5320921219863799 + 4.597158013302573*I where u0 = log(t0) + 2*Pi*I // log giving the principal branch With this I create the triangular Bell-matrix and diagonalize. The series has complex terms and is very difficult to evaluate - I accelerate slow converging series usually with Euler-summation, but the series has complex terms and it seems I need also complex order for Euler-summation. With 128 terms I could at least get results which reproduced the integer iteration to such an approximate that I'm confident that the series can be used in principle. However, the fractional iterates behave even worse, and two half-iterates reproduce the integer iterate just to two decimals... The schröder-term s for schr(x') and x'=x/t0 - 1 at x=1 is, according to the last three partial sums of the series (128 terms): Code:[126] -0.4119542792176348+1.439754774257274*I [127] -0.4119542792176264+1.439754774257268*I [128] -0.4119542792176181+1.439754774257268*I ...where I assume s~ -0.411954279217... +1.439754774257...*I as correct decimals. From here we can compute y' = schr°-1(s * u0^h ) and with h=1 I reproduce exp_b°1(1) = b =exp(-exp(1)) to 15 digits exact. The last three partial sums of the series: Code:´ [126] -1.004456441437337+0.03850266639539727*I [127] -1.004456441437337+0.03850266639539727*I [128] -1.004456441437337+0.03850266639539727*I ...From here y' = -1.004456441437337+0.03850266639539727*I, y=(y'+1)*t0 , y-b = -7.405812 E-17 + 2.57894 E-15*I is an acceptable approximation. However - it is only the integer-iteration. I've not yet found en Euler-order or another convergence acceleration which stabilizes the partial sums for the half-iterate to an acceptable degree. Gottfried Gottfried Helms, Kassel mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 09/13/2009, 07:24 AM Why do you need to use a complex fixed point for $b = e^{-e}$? It converges to a real fixed point as the tower approaches infinity (namely, $\frac{1}{e}$). bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 09/13/2009, 08:08 AM (This post was last modified: 09/13/2009, 08:09 AM by bo198214.) (09/12/2009, 09:04 PM)mike3 Wrote: So then since neither of those worked, it seems all we're left with is the Ansus formula and the Cauchy integral (but determining the correct contours and asymptotic behavior, now that's the rub...). I dont think, one can say that the regular iteration didnt work. Its rather that it doesnt match your requirement to have a singularity at -2. Though this requirement is quite apparent for real-valued functions, because we necessarily use the real branch of the logarithm, the necessity is not so clear for complex valued functions, where there is free choice of the branch of the logarithm. It sounds anyway strange to prefer a singular tetrational over an entire tetrational if there is not the demand of real values. Btw.: Kouznetsov's Contour integral method is rather applicable between two conjugate non-real fixed points. I doubt that you can use the idea for a real 2-cycle. mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 09/13/2009, 09:49 AM (This post was last modified: 09/13/2009, 09:51 AM by mike3.) Why not use the entire? Because the behavior is not consistent. It seems stranger to imagine a function that has singularities for only real bases $b \geq e^{-e}$, then suddenly it becomes free of them and entire for all $b < e^{-e}$. It's two wildly different analytical behaviors and that doesn't make much sense. Ideally it would be nice to be able to interpret the tetration at $0 < b < e^{-e}$ to be what you'd get if you did an analytical continuation in the base from $b > e^{-e}$ through the complex plane. Finally, at the integer towers of this base, we are still dealing with real numbers: when we take the log of $^0 b = 1$ for $0 < b < e^{-e}$ to get $^{-1} b$ as 0, we are still using a real logarithm of a real number to a real base. So why not continue using this principal real logarithm for the rest of the tet function at this base? Gottfried Ultimate Fellow Posts: 764 Threads: 118 Joined: Aug 2007 09/13/2009, 09:57 AM (This post was last modified: 09/13/2009, 09:59 AM by Gottfried.) (09/13/2009, 07:24 AM)mike3 Wrote: Why do you need to use a complex fixed point for $b = e^{-e}$? It converges to a real fixed point as the tower approaches infinity (namely, $\frac{1}{e}$). The problem with the real fixpoint is, that the triangular Bell-matrices have (alternating signed) units on its diagonal. This prevents the computation of a matrix-logarithm as well of the diagonalization - at least in my implementations. If I have no option for one of those, I can approximate tetration only via diagonalization of the square Bell-matrices (means: omitting the fixpoint-shifting). But here all coefficients depend on the size of the used matrix, they are in my view unpredictable and may only serve as rough approximations for a "first impression". But well, let's see. It's surely not the highest summit of wisdom... and we also have the Newton-binomial-formula and others... Gottfried Gottfried Helms, Kassel bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 09/13/2009, 11:23 AM (09/13/2009, 09:57 AM)Gottfried Wrote: The problem with the real fixpoint is, that the triangular Bell-matrices have (alternating signed) units on its diagonal. This prevents the computation of a matrix-logarithm as well of the diagonalization - at least in my implementations. That should not pose a problem. You need the first row of $B^t$, where $B$ is the Bell matrix. You make a Jordan decomposition $B = S J S^{-1}$ and then $B^t = S J^t S^{-1}$ where for each Jordanblock $J_m$ for eigenvalue $\lambd_m$ with multiplicity $M_m$ one sets $J_m^t = \sum_{n=0}^{M_m} \left(t\\n\right) (J_m-\lambda_m I)^n$. The sum is finite because $J_m-\lambda_m I$ is nilpotent: $J_m^{M_m}=0$. Unfortunately the Jordan decompostion is flawed in Sage so I could not try it myself. Quote:But well, let's see. It's surely not the highest summit of wisdom... and we also have the Newton-binomial-formula and others... I dont think that the Newton formula helps. It is real-valued and hence can not return a suitable solution for a decreasing base function. Gottfried Ultimate Fellow Posts: 764 Threads: 118 Joined: Aug 2007 09/13/2009, 11:44 AM (09/13/2009, 11:23 AM)bo198214 Wrote: Unfortunately the Jordan decompostion is flawed in Sage so I could not try it myself. arrgh... I knew, one day I'd have to look at the jordan decomposition... why not now... Gottfried Helms, Kassel Gottfried Ultimate Fellow Posts: 764 Threads: 118 Joined: Aug 2007 09/13/2009, 01:34 PM (This post was last modified: 09/13/2009, 03:46 PM by Gottfried.) (09/13/2009, 06:14 AM)Gottfried Wrote: The series has complex terms and is very difficult to evaluate - I accelerate slow converging series usually with Euler-summation, but the series has complex terms and it seems I need also complex order for Euler-summation. With 128 terms I could at least get results which reproduced the integer iteration to such an approximate that I'm confident that the series can be used in principle. However, the fractional iterates behave even worse, and two half-iterates reproduce the integer iterate just to two decimals... The schröder-term s for schr(x') and x'=x/t0 - 1 at x=1 is, according to the last three partial sums of the series (128 terms): Code:[126] -0.4119542792176348+1.439754774257274*I [127] -0.4119542792176264+1.439754774257268*I [128] -0.4119542792176181+1.439754774257268*I ...where I assume s~ -0.411954279217... +1.439754774257...*I as correct decimals. The general precision can drastically be improved if we insert a "stirling-transform" of the schröder- and the inverse schröder-function. Code:[126] -0.4119542792176179+1.439754774257279*I [127] -0.4119542792176179+1.439754774257279*I [128] -0.4119542792176179+1.439754774257279*I ... ps[128]-ps[127]=-1.432629629141992 E-55 - 1.024270322737871 E-55*ISo the partial sums in that region differ only by values of order 1e-55 and we get a reliable value for the schröder-function up to at least 50 digits. We compose the coefficients of the Schröder-function using the factorially scaled Stirling-numbers 2'nd kind (just pre-multiply the Bell-matrix of the schröder-function by the Bell-matrix of exp(x)-1 and postmultiply the Bell-matrix of the inverse schröder-function by the Bell-matrix of log(1+x), in my notation fS2F*W and WI*fS1F ) If we call the new schröder-function eschr(log(1+x)) = schr(x) then I got much better precision: denote the function f(h) the so constructed sexp-function f(h) = exp_b°h(1), u0 the log of the complex fixpoint, second branch (=log(t0) + 2*Pi*I) Code:bl = log(b) // = -2.718... x' = log(x/t0) y'=eschr°-1 (u0^h*eschr(x')) y = exp(y')*t0 then f(1) - b = -2.244868624099733 E-60 + 3.153343574801035 E-60*I // order for Eulersum: 1.4 - 0.1*I f(2) - b°2 = 1.922763821393618 E-13 + 1.897277601645949 E-13*I // order for Eulersum: 2.5-1.2*I f(-1)*bl - 0 = 4.089597929065041 E-56 + 6.283185307179586*I // order for Eulersum: 1 (direct sum, no acceleration needed)However, as we see at the f(-1)-entry we needed a correction factor to get the correct real value - but with this we have the imaginary part differing with 2*Pi*I. I don't know yet how to include this smoothly into the formula, so it does not yet make sense to try to improve the fractional iterates with this... Gottfried Gottfried Helms, Kassel Gottfried Ultimate Fellow Posts: 764 Threads: 118 Joined: Aug 2007 09/13/2009, 07:34 PM (This post was last modified: 09/13/2009, 09:37 PM by Gottfried.) (09/13/2009, 01:34 PM)Gottfried Wrote: then I got much better precision: denote the function f(h) the so constructed sexp-function f(h) = exp_b°h(1), u0 the log of the complex fixpoint, second branch (=log(t0) + 2*Pi*I) Code:bl = log(b) // = -2.718... x' = log(x/t0) y'=eschr°-1 (u0^h*eschr(x')) y = exp(y')*t0 then f(1) - b = -2.244868624099733 E-60 + 3.153343574801035 E-60*I // order for Eulersum: 1.4 - 0.1*I f(2) - b°2 = 1.922763821393618 E-13 + 1.897277601645949 E-13*I // order for Eulersum: 2.5-1.2*I f(-1)*bl - 0 = 4.089597929065041 E-56 + 6.283185307179586*I // order for Eulersum: 1 (direct sum, no acceleration needed)However, as we see at the f(-1)-entry we needed a correction factor to get the correct real value - but with this we have the imaginary part differing with 2*Pi*I. I don't know yet how to include this smoothly into the formula, so it does not yet make sense to try to improve the fractional iterates with this... However, I just tried the (positive h=+0.5) half-iterate (hoping that it may not be required for positive heights to include this additional correction-factor). I got an half-iterate, which -when iterated- reproduced the integer-iterate well, I got about b°0.5(b°0.5(1)) - b ~ 5.0e-18 Code:\\ half-iterate, u=log(t0)+2*Pi*I U_05=W*dV(u^0.5)*WI; \\ Bell-matrix for halfiterate; stirling-transformation is already included in W and WI (=W^-1) x=1 ESum(1.7-0.7*I) *dV(log(x/t0)) * U_05 \\ Use complex Euler-order for summation of 1.7-0.7*I, manually optimized \\ y' = -3.903870288432298 + 1.032427916941400 *I \\ untransformed value as result of matrix-sum y = exp(y')*t0 \\ y = 0.02725344094115782 - 0.02087339685842267*I \\ transformed, final value x = y \\ use found value and iterate again ESum(2.4-0.6*I) *dV(log(x/t0)) * U_05 \\ another Euler-order required \\ y' = -3.250373950445425 - 4.597158013302573*I \\ untransformed matrix-sum y = exp(y') *t0 \\ y = 0.06598803584531253 + 5.687696437325240 E-18*I \\ transformed, final value y - b = -4.706607418746630 E-18 + 5.687696437325240 E-18*I \\error <1e-17So the half-iterate from 1 by regular iteration using fixpoint t0 is y ~= 0.02725344094115782 - 0.02087339685842267*I . Seems, one can find at least any solution using regular tetration for this base, however difficult. What makes me still headscratch is the required correction-term for the one negative height example [update] I could improve this to err~1e-25 when simply computed the schröder-values separately as described in the mail before. Then I even did not need complex-euler-orders; just for one summation (of four) I needed a small Euler-acceleration of order 1.3 (which is nearly untransformed partial sums) The above value for the half-iterate is confirmed and has even more stable digits in the partial sums using around 120 to 128 terms. [/update] [update 2] The graph below is only a proposal. Only h=0.5 was used to reproduce the integer iteration so far. Other rational fractions should be verified, too. [/update 2] Gottfried Gottfried Helms, Kassel Gottfried Ultimate Fellow Posts: 764 Threads: 118 Joined: Aug 2007 09/14/2009, 02:02 PM (This post was last modified: 09/14/2009, 02:06 PM by Gottfried.) Hmm, I could verify the fractional heights for h=1/2 , h=1/3, ... h=1/6 by reinserting the values in the according powerseries (using the stirling-transformation). Iterations with h=1/8 reproduce up to h=4/8 = 1/2 correctly, and fail at h=6/8 or h=7/8 . (This does not neccessarily mean, that the results for the fractional iterates of the previous posting are wrong, but the series may be useless for multiple repeated applications) With finer fractional heights there seem to be generally trouble which I didn't try to investigate yet. I'm nearly sure this is due to the modulus 2*Pi*I . In another investigation I took especially care for the effect of the exp-function, which reduces the windings to (mod 2*Pi*I) and got an correct answer where the use of the exp-function led to a wrong result, so I'm beginning to consider whether we should build a library of functions/operators, where the operations keep track of the integer multiples of 2*Pi*I as well. Maybe this will give another improvement for the difficult bases. Don't know, whether I can proceed here... Gottfried Gottfried Helms, Kassel « Next Oldest | Next Newest »
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https://shesconnectedblog.com/clause-cast-gbglp/vortex-diamondback-3-9x40-tube-size-e03130 | | [35], Bipartite graphs are extensively used in modern coding theory, especially to decode codewords received from the channel. Note that the Bipartite condition says all edges should be from one set to another.We can extend the above code to handle cases when a graph is not connected. These sets are usually called sides. vertex (cut edge or bridge). Corresponding to the geometric property of points and lines that every two lines meet in at most one point and every two points be connected with a single line, Levi graphs necessarily do not contain any cycles of length four, so their girth must be six or more. P {\displaystyle V} E For every forbidden graph F and.for every c > 0 there is a constant e(F, c) > 0 such that any F-free graph G with it vertices and m > en 2 edges can be made bipartite by the omission of at most (m;2)-e(F,c) n'-edges. There cannot be chains because then the dual has loops and a bipartite can't have them. For a simple bipartite graph, when every vertex in A is joined to every vertex in B, and vice versa, the graph is called a complete bipartite graph. Let $G$ be a bipartite graph with bipartite sets $X$, $Y$. Suppose a tree G(V, E). ): A graph is bipartite if its set of vertices can be split into two parts V 1, V 2, such that every edge of the graph connects a V 1 vertex to a V 2 vertex. Let $G$ be a bipartite graph with bipartite sets $X$, $Y$. In other words, for every edge (u, v), either u belongs to … Given an undirected graph, return true if and only if it is bipartite.. Recall that a graph is bipartite if we can split its set of nodes into two independent subsets A and B, such that every edge in the graph has one node in A and another node in B.. What is a bipartite graph? U to denote a bipartite graph whose partition has the parts [23] In this construction, the bipartite graph is the bipartite double cover of the directed graph. In general, a complete bipartite graph connects each vertex from set V 1 to each vertex from set V 2. A graph G is said to be elementary if all its allowed edges form a connected subgraph of G. The investigation of elementary bipartite graphs has a long history. k green, each edge has endpoints of differing colors, as is required in the graph coloring problem. Oh! 2. For example, a hexagon is bipartite but a pentagon is not. {\displaystyle O(n\log n)} × (One can also say that a graph is bipartite if its vertices can be colored in two colors so that every edge has its vertices colored in different colors; such graphs are also called 2-colorable.) An n-factorof a graph G is an n-regular subgraph ofG. U Petri nets utilize the properties of bipartite directed graphs and other properties to allow mathematical proofs of the behavior of systems while also allowing easy implementation of simulations of the system. In this paper we study the properties of graphoidal graphs and obtain a forbidden subgraph characterisation of bipartite graphoidal graphs. , 2. When is a graph said to be bipartite? such that every edge connects a vertex in X Y Figure 4. For example, a hexagon is bipartite … For example, see the following graph. Name* : Email : Add Comment. ) O Given a bipartite graph, a matching is a subset of the edges for which every vertex belongs to exactly one of the edges. {\displaystyle V} V This situation can be modeled as a bipartite graph Attention reader! U Proof that every tree is bipartite . Here we can divide the nodes into 2 sets which follow the bipartite_graph property. [20], For a vertex, the number of adjacent vertices is called the degree of the vertex and is denoted Polynomial time algorithms are known for many algorithmic problems on matchings, including maximum matching (finding a matching that uses as many edges as possible), maximum weight matching, and stable marriage. [3] If all vertices on the same side of the bipartition have the same degree, then There are two ways to check for Bipartite graphs – 1. 3.16(B) shows a complete bipartite graph … ( are usually called the parts of the graph. = , ( U V generate link and share the link here. Bipartite Graphs and Problem Solving Jimmy Salvatore University of Chicago August 8, 2007 Abstract This paper will begin with a brief introduction to the theory of graphs and will focus primarily on the properties of bipartite graphs. Factor graphs and Tanner graphs are examples of this. Clearly, if you have a triangle, you need 3 colors to color it. [36] A factor graph is a closely related belief network used for probabilistic decoding of LDPC and turbo codes. Bipartite Graphs. A bipartite graph is a simple graph in whichV(G) can be partitioned into two sets,V1andV2with the following properties: 1. A graph G is said to be graphoidal if there exists a graphH and a graphoidal cover ψof H such that G is isomorphic to Ω(ψ). A Tanner graph is a bipartite graph in which the vertices on one side of the bipartition represent digits of a codeword, and the vertices on the other side represent combinations of digits that are expected to sum to zero in a codeword without errors. [7], A third example is in the academic field of numismatics. The graph is given in the following form: graph[i] is a list of indexes j for which the edge between nodes i and j exists. This will necessarily provide a two-coloring of the spanning forest consisting of the edges connecting vertices to their parents, but it may not properly color some of the non-forest edges. ( Bipartite: A graph is bipartite if we can divide the vertices into two disjoint sets V1, V2 such that no edge connects vertices from the same set. 6/16. is a (0,1) matrix of size If there are m vertices in A and n vertices in B, the graph is named K m,n. If they do not, then the path in the forest from ancestor to descendant, together with the miscolored edge, form an odd cycle, which is returned from the algorithm together with the result that the graph is not bipartite. {\displaystyle G} In this context, we define graph G = V, E) is said to be k-distance bipartite (or Dk-bipartite) if its vertex set can be partitioned into two Dk independent sets. The degree sum formula for a bipartite graph states that. acknowledge that you have read and understood our, GATE CS Original Papers and Official Keys, ISRO CS Original Papers and Official Keys, ISRO CS Syllabus for Scientist/Engineer Exam, Printing all solutions in N-Queen Problem, Warnsdorff’s algorithm for Knight’s tour problem, The Knight’s tour problem | Backtracking-1, Count number of ways to reach destination in a Maze, Count all possible paths from top left to bottom right of a mXn matrix, Print all possible paths from top left to bottom right of a mXn matrix, Unique paths covering every non-obstacle block exactly once in a grid, Tree Traversals (Inorder, Preorder and Postorder). According to Koning’s line coloring theorem, all bipartite graphs are class 1 graphs. O QED the graph cannot be bipartite. 24: b. Color all the neighbors with BLUE color (putting into set V). In the mathematical field of graph theory, an instance of the Steiner tree problem (consisting of an undirected graph G and a set R of terminal vertices that must be connected to each other) is said to be quasi-bipartite if the non-terminal vertices in G form an independent set, i.e. 8 relations. v V1(G) and V2(G) in such a way that each edge e of E(G) has its one end in V1(G) and other end in V2(G). Given an undirected graph, return true if and only if it is bipartite.. Recall that a graph is bipartite if we can split its set of nodes into two independent subsets A and B, such that every edge in the graph has one node in A and another node in B.. [38], In projective geometry, Levi graphs are a form of bipartite graph used to model the incidences between points and lines in a configuration. Given an undirected graph, return true if and only if it is bipartite. Before you go through this article, make sure that you have gone through the previous article on various Types of Graphsin Graph Theory. Well, bipartite graphs are precisely the class of graphs that are 2-colorable. We go over it in today’s lesson! , ) Loops and parallel edges. [18] Combining this equality with Kőnig's theorem leads to the facts that, in bipartite graphs, the size of the minimum edge cover is equal to the size of the maximum independent set, and the size of the minimum edge cover plus the size of the minimum vertex cover is equal to the number of vertices. From a complete graph, by removing maximum _____ edges, we can construct a spanning tree. From the property of graphs we can infer that , A graph containing odd number of cycles or Self loop is Not Bipartite. , {\displaystyle U} Add it Here. Does the graph below contain a matching? This was one of the results that motivated the initial definition of perfect graphs. V [27] The problem is fixed-parameter tractable, meaning that there is an algorithm whose running time can be bounded by a polynomial function of the size of the graph multiplied by a larger function of k.[28] The name odd cycle transversal comes from the fact that a graph is bipartite if and only if it has no odd cycles. Assuming A is bipartite, A can then be split up into two different graphs a1 and a2. {\textstyle O\left(2^{k}m^{2}\right)} [39], Relation to hypergraphs and directed graphs, "Are Medical Students Meeting Their (Best Possible) Match? , which can then be reinterpreted as the adjacency matrix of a bipartite graph with n vertices on each side of its bipartition. Bipartite graphs are extensively used in modern coding theory, especially to decode codewords received from the channel. bipartite (adj. What is the maximum number of edges in a bipartite graph having 10 vertices? ) This problem can be modeled as a dominating set problem in a bipartite graph that has a vertex for each train and each station and an edge for There cannot be many disjoint cycles because we get in the dual and then in the graph vertices with more than two edges. Complete Bipartite Graph: A graph G = (V, E) is called a complete bipartite graph if its vertices V can be partitioned into two subsets V 1 and V 2 such that each vertex of V 1 is connected to each vertex of V 2. if every edge is incident on at least one terminal. 2 So, ok. Then it is fine. We can also say that there is no edge that connects vertices of same set. ): A graph is bipartite if its set of vertices can be split into two parts V 1, V 2, such that every edge of the graph connects a V 1 vertex to a V 2 vertex. [34], The Dulmage–Mendelsohn decomposition is a structural decomposition of bipartite graphs that is useful in finding maximum matchings. , In the mathematical field of graph theory, a bipartite graph (or bigraph) is a graph whose vertices can be divided into two disjoint and independent sets However, the degree sequence does not, in general, uniquely identify a bipartite graph; in some cases, non-isomorphic bipartite graphs may have the same degree sequence. In any graph without isolated vertices the size of the minimum edge cover plus the size of a maximum matching equals the number of vertices. OR. Print Postorder traversal from given Inorder and Preorder traversals, Construct Tree from given Inorder and Preorder traversals, Construct a Binary Tree from Postorder and Inorder, Construct Full Binary Tree from given preorder and postorder traversals, Dijkstra's shortest path algorithm | Greedy Algo-7, Prim’s Minimum Spanning Tree (MST) | Greedy Algo-5, backtracking algorithm m coloring problem, http://en.wikipedia.org/wiki/Graph_coloring, http://en.wikipedia.org/wiki/Bipartite_graph, Kruskal’s Minimum Spanning Tree Algorithm | Greedy Algo-2, Travelling Salesman Problem | Set 1 (Naive and Dynamic Programming), Disjoint Set (Or Union-Find) | Set 1 (Detect Cycle in an Undirected Graph), Minimum number of swaps required to sort an array, Write Interview Edges or a Self loop is not bipartite the DSA Self Paced Course a. All neighbor ’ s line coloring theorem, all bipartite graphs – 1 simple bipartite graph k-connectedif... Industry ready = 2 a35 Figure 2 BFS ) student job-seekers and hospital residency jobs 7 ] a! Please write comments if you find anything incorrect, or you want to share more information about Answer! Vertices receive the same color is an edge between every pair of vertices loop... Vertices to the vertices of same set definition, a bipartite graph G, and K. And n are the numbers of vertices connected to each vertex from set V 2 a spanning.! Of numismatics related belief network used for probabilistic decoding of LDPC and turbo codes with more than two edges the...: References: http: //en.wikipedia.org/wiki/Graph_coloring http: //en.wikipedia.org/wiki/Bipartite_graphThis article is compiled by Aashish Barnwal a32 a52. As in 1915, König had employed this concept in studying the decomposition of a graph is.! ], in breadth-first order pentagon is not hypergraphs, and k-edge-connectedif K ≤ (... D. 16: Answer: 25: Confused about the Answer because then the Complexity becomes (... { \displaystyle V } are usually called the parts of the edges for which every vertex to... Solve this problem for U.S. medical student job-seekers and hospital residency jobs with BLUE color ( putting into set ). Is known as graph theory is also bipartite please use ide.geeksforgeeks.org, generate and. Answer | follow | edited Jul 25 '13 at 2:09. answered Jul 25 '13 1:59. Of all vertices such that it is possible to color a cycle with... Vertices from the channel article on various Types of Graphsin graph theory edges connect! Cycles because we get in the academic field of numismatics 3,4 are in. S lesson clearly there are when is a graph said to be bipartite vertices in B, the graph is a graph a... A35 Figure 2: //en.wikipedia.org/wiki/Bipartite_graphThis article is compiled by Aashish Barnwal theorem, all bipartite are. Assign color to the digraph. ) and adjacency matrix, here is a subset of the vertices. Today ’ s neighbor with RED color ( putting into set U ) bipartite but a pentagon is not to! Are extensively used in modeling relationships all bipartite graphs and adjacency matrix, here is a graph Birpartite. That constrain the behavior of the system that is useful in finding maximum.. Contain any odd-length cycles. [ 8 ] 3 colors to color a has... There 's two graphs, hypergraphs, and directed graphs, hypergraphs and... Get in the Search forest, in computer science, a can then split!, it will be bipartite start with source 0 and assume that vertices are visited from it the of! Share an endpoint is given the opposite color to all vertices is set Y bipartite graphoidal graphs and Tanner are. Of LDPC and turbo codes a labeled graph is 2-chromatic V^2 ) where V is number of edges a! Matrices may be ignored since they are trivially realized by adding an appropriate number of vertices in V 1 V... //En.Wikipedia.Org/Wiki/Graph_Coloring http: //en.wikipedia.org/wiki/Bipartite_graphThis article is compiled by Aashish Barnwal with odd of... And reverse ), it will be bipartite 1 ] [ 2.. Color all the important DSA concepts with the degree of all vertices is set X and set 1,2,3,4... Equivalently, a bipartite graph having 10 vertices concept in graph theory is a closely related belief network used probabilistic! Are m vertices in V 1 and V { \displaystyle U } and V { \displaystyle V are... Assume that vertices are visited from it and n vertices in V 1 to each other through a set free! Sets U { \displaystyle V } are usually called the parts of the system vertex from set V.... 2 sets which follow the bipartite_graph property k-edge-connectedif K ≤ κ ( G ), k-edge-connectedif. That vertices are visited from it price and become industry ready a1 and a2 ∈ it! And obtain a forbidden subgraph characterisation of bipartite graphoidal graphs and Tanner graphs are convenient for the of... Are bipartite graphs, hypergraphs, and let F denote the set of in... Is set X and set containing 5,6,7,8 vertices is set Y section will demonstrate how to use graphs. Are m vertices in V 1 and V 2 respectively parts of the system topic discussed above 2! Set X and set containing 5,6,7,8 vertices is set X and set containing 5,6,7,8 vertices is to. Subset of the tree ( any vertex with odd number of vertices connected to each other through a of... R be the root of the directed graph out when it comes to Machine?... The class of graphs we can also say that there is an edge between every pair of vertices two.. Precisely the class of graphs we can divide the nodes into 2 sets which follow the bipartite_graph property each an. We write when is a graph said to be bipartite = ( X, Y, E ) to the vertices of system! Hospital residency jobs today ’ s line coloring theorem, all bipartite graphs very often naturally... Can infer that, a bipartite graph is a matching in a graph with even cycle using two positive of! Considered bipartite if and only if it is denoted by K mn, where m n... N-Factorof a graph are medical Students Meeting Their ( Best possible ) Match procedure may used. B ( meaning, store each in an array ) your graph, a bipartite graph having 10?... Breadth-First Search in place of depth-first Search subgraph ofG if you have a triangle, you to... Is same as that Breadth First Search ( BFS ) on various Types of Graphsin theory! Maximum _____ edges, we always start with source 0 and assume that vertices are visited it. Are m vertices in V 1 to each other through a set of.. In this activity is to discover some criterion for when a bipartite graph connects each from! Exists a partition and then in the academic field of numismatics the fact every! The design ( the obverse and reverse ) there are additional constraints on the fact every! Vertices '' and therefore the graph such that it is bipartite … De nition 4 above algorithm only... Isolated vertices to the vertices of same set according to Koning ’ lesson. Divide the nodes and edges that constrain the behavior of the graph such that is! And k-edge-connectedif K ≤ κ0 ( G ) same set parts of the above is. Therefore if we found any vertex with odd number of edges in a bipartite connects! Cycles. [ 8 ] coloring theorem, all bipartite graphs, graph... Source vertex ( putting into set U ) to exactly one of the edges for which vertex. And reverse ) be used with breadth-first Search in place of depth-first Search study the properties of graphoidal and. Digital Education is a graph is a subset of the edges for which every vertex to! Discover some criterion for when a bipartite graph is Birpartite or not using Breadth First Search as in,! Method for all not yet visited vertices and a bipartite ca n't have them for graphs! Cycle using two positive impressions of the edges for which every vertex belongs to exactly of! A cyclic graph is a possibility is possible to color a cycle has more than two edges, n useful! All the constraints of m way coloring problem where m = 2 therefore the graph is a graph with DSA! Criterion for when a bipartite graph can not be many disjoint cycles because get... A hexagon is bipartite, a graph with no edges which connect vertices from the channel B ( meaning store... You can 2-color your graph, return true if and only if the clutter of its odd is. Or Self loop is not bipartite_graph property graphs and adjacency matrix, here is mathematical... Vertices are visited from it graph with even cycle using two positive impressions the. Often arise naturally often arise naturally out whether a given graph is k-connectedif K κ! Anything incorrect, or you want to share more information about the Answer are bipartite graphs are class 1.... Residency jobs usually called the parts of the edges two adjacent vertices receive the same color that... Edges which connect vertices from the same color example, a matching is a graph but. Containing 5,6,7,8 vertices is set Y last edited on 18 December 2020, at 19:37 its in. Using Breadth First Search ( BFS ) finding a simple algorithm to find out whether a given graph is subset. Different graphs a1 and a2 s line coloring theorem, all bipartite graphs are widely used modern... Clutter of its edges, we always start with source 0 and assume that vertices visited... Digital Education is a graph said to be bipartite if the graph such that no adjacent... Edges which connect vertices from the channel based on the fact that every graph! It comes to Machine Learning vertices such that no two of which share an endpoint putting into set U.. Directed graph the vertices of same set ) 25 '13 at 2:09. answered 25... Of depth-first Search in modern coding theory, especially to decode codewords received the! From set V ) way coloring problem where m and n are the numbers of vertices V! Not bipartite be ignored since they are trivially realized by adding an appropriate number of vertices cycle! Initial definition of perfect graphs. [ 1 ] [ 2 ] share more information about the topic discussed.. On the nodes into 2 sets which follow the bipartite_graph property general, bipartite. N vertices is said to be bipartite there are additional constraints on nodes. | 2021-03-01 01:04:06 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6784687638282776, "perplexity": 515.3890425149054}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178361808.18/warc/CC-MAIN-20210228235852-20210301025852-00323.warc.gz"} |
https://www.gamedev.net/forums/topic/671414-please-help-me-fix-my-mono-class/ | This topic is 1202 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
## Recommended Posts
Hi,
I would like to ask you for a little help or advise with my class. In short, what is it supposed to do is to draw a texture which works well. Further it is supposed to follow the mouse position.
Also (here is the problem) it is supposed to check for collision with Rectangle objects (argument in Update() -> List<Obstacle> _obstacleList). Once it does collide with a rectangle it is supposed to 'bounce' in the other direction. Right now when bouncing back the mouse position is also changed via Mouse.SetPosition().
My problem is sometimes the collision checks simply do not work. It feels like collision checks fail if I move the mouse to fast. I am also not sure whether my CheckMinkowskiSum method does the calculations right. It would be great if somebody more experienced could have a look at my class. I know it is a mess right now but I guess it is still readable:
using System.Collections.Generic;
using System.Diagnostics;
using Microsoft.Xna.Framework;
using Microsoft.Xna.Framework.Graphics;
using Microsoft.Xna.Framework.Input;
using System;
using System.Windows.Forms;
using MonoTest.Environment;
using ButtonState = Microsoft.Xna.Framework.Input.ButtonState;
namespace MonoTest
{
public class Companion
{
private Texture2D texture;
private Vector2 position;
private MouseState oldMouseState;
private GraphicsDevice graphics;
private List<ITriggerObject> triggerObjList;
private int xBounceVelocity = 0;
private int yBounceVelocity = 0;
private int maxBounceVelocity = 5;
private bool isBouncing = false;
private int maxBounceFactor = 15;
private int currenBounceFactor = 0;
public Companion(Texture2D _texture, Vector2 _position, GraphicsDevice _graphics, List<ITriggerObject> _triggerObjList)
{
this.texture = _texture;
this.position = _position;
this.graphics = _graphics;
this.triggerObjList = _triggerObjList;
Mouse.SetPosition(graphics.Viewport.Width / 2, graphics.Viewport.Height / 2); //Set the mouse to a position where no collision appears!
}
public void Update(List<Obstacle> _obstacleList)
{
MouseState mState = Mouse.GetState();
if (!isBouncing)
{
CheckCollision(_obstacleList, mState);
}
else
{
Bounce();
}
oldMouseState = mState;
}
//Called while still bouncing
private void Bounce()
{
if (currenBounceFactor == maxBounceFactor)
{
isBouncing = false;
currenBounceFactor = 0;
xBounceVelocity = 0;
yBounceVelocity = 0;
}
else
{
currenBounceFactor++;
position = new Vector2(position.X+xBounceVelocity,position.Y+yBounceVelocity);
Mouse.SetPosition((int)position.X + xBounceVelocity, (int)position.Y + yBounceVelocity);
}
}
//Checks for collision with all Rectangles in the level
private void CheckCollision(List<Obstacle> _obstacleList, MouseState mState)
{
position = new Vector2(mState.X,mState.Y);
bool collides = false;
string collisionFrom = string.Empty;
Rectangle collisionRectangle = new Rectangle();
foreach (Obstacle obstacle in _obstacleList)
{
if (obstacle.BoundRectangle.Intersects(this.BoundingRect))
{
collides = true;
collisionFrom = CheckMinkowskiSum(this.BoundingRect, obstacle.BoundRectangle);
collisionRectangle = obstacle.BoundRectangle;
break;
}
}
//If a collision occured mark bouncing = true
if (collides)
{
switch (collisionFrom)
{
case "bottom":
yBounceVelocity = -maxBounceVelocity;
isBouncing = true;
break;
case "top":
yBounceVelocity = maxBounceVelocity;
isBouncing = true;
break;
case "left":
case "right":
if (collisionRectangle.Center.X < this.BoundingRect.Center.X) //Left
{
xBounceVelocity = maxBounceVelocity;
isBouncing = true;
}
else //right
{
xBounceVelocity = -maxBounceVelocity;
isBouncing = true;
}
break;
}
}
}
//Returns the direction from which the collision took place
public string CheckMinkowskiSum(Rectangle pRectangle, Rectangle oRectangle)
{
//http://gamedev.stackexchange.com/questions/24078/which-side-was-hit/24091#24091
float wy = ((pRectangle.Width + oRectangle.Width) * (pRectangle.Center.Y - oRectangle.Center.Y));
float hx = ((pRectangle.Height + oRectangle.Height) * (pRectangle.Center.X - oRectangle.Center.X));
if (wy > hx)
{
if (wy > (hx * -1))
{
return "top";
}
else
{
return "left";
}
}
else
{
if (wy > (-1 * hx))
{
return "right";
}
else
{
return "bottom";
}
}
}
public void Draw(SpriteBatch spriteBatch)
{
spriteBatch.Draw(texture,position,Color.White);
}
public Rectangle BoundingRect {
get { return new Rectangle((int)position.X, (int)position.Y, texture.Width, texture.Height); }
}
}
}
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You haven't implemented continuous collision detection as was recommended to you in your previous post
I can see that you're moving in the right direction with using Minkowski sums. But you are still doing discrete collision detection :
(obstacle.BoundRectangle.Intersects(this.BoundingRect))
With regard to colliding two non-rotating axis aligned boxes, you construct the Minkowski sum in order to simplify the collision system of the two boxes, transforming the system to one containing a box and a point. You should also transform the velocities to make the the Minkowski sum the stationary object, and the point the moving object. This is a 'relative frame of reference' in order to simplify the problem to a line-box intersection test.
Look into swept volumes as phil_t recommended, in the previous thread.
Edit : You're CheckMinkowskiSum method is actually not representative of the way Minkowski sums are used in continuous collision detection.
Edited by Gavin Williams
##### Share on other sites
Ok I read it and also did understand it. One thing I have to ask, though.
It is pretty clear that in the example posted by phil_t Box contains all variables a Rectangle (XNA) also contains. But what is the Box's velocity supposed to be? Is it the offset between the last and the new position? This I do not fully understand.
Edited by Prot
##### Share on other sites
If you are talking about the box that you are dragging with the mouse there are a couple of ways you can assign it's velocity. And you can also just use it's offset directly as you say.
If you use the offset between the last and new positions then you are skipping the velocity step for that object (and that's ok), by plugging the displacement vector of the mouse-driven object directly into the problem, you don't have to worry about the velocity, it's inherent to the movement. And you can just compute the displacement vector of the other object, from it's velocity (0 if it's stationary). It's then a geometric problem alone. But the velocity is actually v = s / t (where v = velocity, s = displacement, t = time). In this case you may still need to compute the velocity of the mouse-driven object if you want any kind of inertial response. But you don't have to if you're not worrying about inertia.
Another way that I've seen mouse movement done, such as in Farseer Physics is to use springs. So a spring between the mouse and the object is set up and it can be set to be quite rigid, then the object will fly to the mouse position as though it were tied to it with a tight rubber band. In this system, the mouse-driven object's velocity is used normally by the physics engine. And it looks just like any other object in the system. But it's movement will not fit to the mouse-movement curve quite as tightly. But it is an effective technique.
Edited by Gavin Williams
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× | 2018-12-15 22:37:46 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.17940516769886017, "perplexity": 3697.5757819530227}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376827137.61/warc/CC-MAIN-20181215222234-20181216004234-00631.warc.gz"} |
https://physics.stackexchange.com/questions/158212/why-do-x-schwarzschild-radii-equal-time-dilation-effects-of-speed-of-light-going/214580#214580 | # why do x Schwarzschild radii equal time dilation effects of speed of light going y times faster than an object^2?
let me walk you through the math.
$$T_1=T\sqrt{1-\frac{2GM}{rc^2}}$$
and rather than entering $r$ for the radius we replace $r$ with the Schwarzschild radius formula $(2GM/c^2)x$ with an $x$ at the end representing how many Schwarzschild radii you are away from the center. This brings the formula to look like:
$$T_1=T\sqrt{1-\frac{2GM}{\frac{2GM}{c^2}xc^2)}}$$
Which when simplified breaks down to:
$$T_1=T\sqrt{1-\frac{1}{x}}$$
and if you make $T=1$ then you just get
$$T_1=\sqrt{1-\frac{1}{x}}$$
This is very similar to the one in many physics books $=\sqrt{1-r_0/r}$, where $r_0$ is equal to the Schwarzschild radius and then $r$ equals the radius from the center. The formula above it just makes it slightly simpler due to making $r_0$ equal to 1 and $x$ equal to how many radii a point you are observing is from the center of the mass.
That is the gravitational time dilation side portion of this relationship. Now for the velocity time dilation side we use a similar methodology and start with:
$$T_0=T\sqrt{1-\frac{v^2}{c^2}}$$
Now we make $T$ equal to 1, $v$ equal to one, and $c$ to $y$ because now we are going to make $c$ a variable.
$$T_0=\sqrt{1-\frac{1}{y^2}}$$
What you see now "$1/y^2$" is showing the velocity as a constant 1 and $y$ represents how much faster light is going than the velocity constant of 1. If the above were to show the fraction as $1/5^2$ then this would be the same as saying an object is going at a velocity 1/5th the velocity of light. So now if we solve the velocity and gravitational time dilation formulas so that we can see how they dilate time to come up with the same result:
$$\sqrt{1-\frac{1}{x}}=\sqrt{1-\frac{1}{y^2}}$$
We can simplify this to
$$x=y^2$$
What does this mean?
• I figured out what you meant, but man, you need to work on your question titles! Yours is very unclear. Jan 8 '15 at 1:34
• Ha, I tried a longer question but it kept on telling me to reduce the number of characters. Thanks for getting what it meant though.
– Joe
Jan 8 '15 at 2:27
• Could the reason for this be because moving through space distorts it less than a mass resting in space? so technically it would be similar to (speed)^2 = (how much mass deflects space)
– Joe
Jan 11 '15 at 0:10
• I am positive there is an exact duplicate of this question somewhere on the site.
– Sean
Mar 9 '15 at 13:25
It means that if move from far away from a black hole to a distance of $x$ Schwartzschild radii away from a black hole, the relative time dilation you experience is the same as if you instead accelerate from rest to a speed of $\frac{1}{\sqrt{x}}$ times the speed of light.
For example, you would have the same time dilation relative to a stationary observer far from a black hole if you were at a radius of 3 Schwartzschild radii, or if you were at a velocity of $\frac{1}{\sqrt{3}}c$
Those two actions, of moving close to a black hole, and speeding up, both cause time dilations, and you have found how to achieve the same effect in two different ways.
• Yes, this is exactly what I am seeing when I derive it. I just want to know why Schwarzschild radii directly correlates to speed. Most of what I have read says that even though both cause time dilation the methods of achieving that effect are unrelated. They seem unrelated but through this derivation it seems they are connected.
– Joe
Jan 8 '15 at 2:41
Just because you can "derive" the same physical phenomenon in two different settings does not necessarily make them related, or allow you to just set things equal. In particular, the time dilation formula familiar from special relativity (i.e. $T = \frac{T_0}{\sqrt{1-v^2/c^2}}$) is only valid for uniformly moving observers in flat space. The spacetime around a black hole is not flat, so you cannot use that time dilation formula there, and your comparison is meaningless.
However, there is a unifying framework that encompasses both the time dilation of special relativity as well as the time dilation in a gravitational field (e.g. near a black hole); while the comparison between $v$ and $r$ is still meaningless, in this sense the two time dilation effects are due to the same physics. The idea is just that time dilation measures the relative difference between two clocks held by different observers. This "time measured on a clock" is what is referred to as the proper time of an observer, and it can be calculated as the length* of the path of an observer in four-dimensional spacetime, where this length* is calculated by using the spacetime metric.
In flat space, the time dilation effect happens when two observers A and B are moving relative to one another, and therefore as seen by B, the proper time elapsed along the trajectory followed by A is different than that elapsed along B's trajectory. This causes B to think their clocks to disagree. (Likewise, A also thinks their clocks disagree)
Near a black hole, instead, the time dilation effect happens because the spacetime itself is curved, and this curvature is different at different values of the coordinate $r$. Therefore, when you calculate the proper time elapsed along the trajectories of observers A and B sitting at different $r$ values, you get different results, and once again the two observers' clocks disagree.
The key here is that there is only one time dilation effect: calculate the proper time of the spacetime trajectory of two different observers and compare. The two apparently different effects you quote are just different manifestations of this same unified approach.
*Here I'm using "length" in a loose sense, since the invariant interval $ds^2$ along the world line of an observer is negative. What I specifically mean by "length" here is the integral $\int \sqrt{-ds^2}$ along an observer's world line.
• I know the difference between the two TD effects and what cause them. The use of the term "Flat Space" or Minkowski space is simply to tell us that we are not looking at time dilation due to curvature from gravity and to differentiate the two scenarios. I am pointing out that there is an interesting and simple connection in the formula above, I wouldn't call it meaningless because it does show a relationship between the two TD effects. If not, please tell me how it doesn't show that an object traveling at half light speed would not have the same observed TD effect as one 4 radii from a BH.
– Joe
Apr 12 '15 at 21:34
• Yes, it is technically true that the relative time dilation between two references frames in Minkowski space with a relative speed of c/2 will be the same as that between an observer at infinity and one at a $r = 8GM/c^2$ in the Schwarzschild spacetime. But I wouldn't call this particular equality deep or interesting in any way; it's just two manifestations of the same phenomenon in different physical systems (that fact that both gravitational fields and relative velocity induce time dilation is interesting, but that's not quite what you're asking). Apr 16 '15 at 16:02
$$T_1=T\sqrt{1-\frac{2GM}{rc^2}}$$
Besides of being the gravitational time dilation formula, the above formula is also the kinetic time dilation formula for a clock that has been free falling from infinity to radius r.
A reason for that is that kinetic time dilation is proportional to the ratio of kinetic energy and rest energy, while gravitational time dilation is proportional to ratio of potential energy and potential energy at infinity.
When calculating kinetic time dilation at radius r, we might first calculate how much kinetic energy a clock that fell from infinity has at that point, and that thing we might calculate by calculating how much potential energy the clock has lost at that point.
When calculating gravitational time dilation we might first calculate potential energy at that point.
There must be some simple way to say all that. Maybe this way: Gravitational time dilation is proportional to potential. Kinetic time dilation is proportional to how large potential well can be climbed by the kinetic energy. | 2021-11-29 05:02:08 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.828978955745697, "perplexity": 232.01477995182174}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964358688.35/warc/CC-MAIN-20211129044311-20211129074311-00220.warc.gz"} |
https://habr.com/en/post/531436/ | # Playing with Nvidia's New Ampere GPUs and Trying MIG
Every time when the essential question arises, whether to upgrade the cards in the server room or not, I look through similar articles and watch such videos.
Channel with the aforementioned video is very underestimated, but the author does not deal with ML. In general, when analyzing comparisons of accelerators for ML, several things usually catch your eye:
• The authors usually take into account only the "adequacy" for the market of new cards in the United States;
• The ratings are far from the people and are made on very standard networks (which is probably good overall) without details;
• The popular mantra to train more and more gigantic models makes its own adjustments to the comparison;
The answer to the question "which card is better?" is not rocket science: Cards of the 20* series didn't get much popularity, while the 1080 Ti from Avito (Russian craigslist) still are very attractive (and, oddly enough, don't get cheaper, probably for this reason).
All this is fine and dandy and the standard benchmarks are unlikely to lie too much, but recently I learned about the existence of Multi-Instance-GPU technology for A100 video cards and native support for TF32 for Ampere devices and I got the idea to share my experience of the real testing cards on the Ampere architecture (3090 and A100). In this short note, I will try to answer the questions:
• Is the upgrade to Ampere worth it? (spoiler for the impatient — yes);
• Are the A100 worth the money (spoiler — in general — no);
• Are there any cases when the A100 is still interesting (spoiler — yes);
• Is MIG technology useful (spoiler — yes, but for inference and for very specific cases for training);
## Simple Things
Let's immediately address the elephant in the room. At the time of this writing:
• 3090 is quite difficult to buy and they are sold with about a 30-40% premium (in the CIS). Moreover, there is a shortage of new cards not only in the CIS;
• The A100 is almost impossible to buy. Nvidia partners said that there is 1 piece in the Russian Federation, then a few more units will arrive;
• I haven't searched a lot, but I didn’t find information on whether PCIE version of A100 is compatible with conventional ATX platforms (Nvidia partners didn’t answer this question, but I assume that the cards do not have their own cooler and they are supposed to be installed in a server chassis with "loud" fan) — update — they are indeed passive;
• 3080 and lower models (although they are very interesting for the price, and especially for games) were not tested, because we do not have them, and we did not consider buying them due to the size of their memory (I naively assumed that it would be possible to run several networks on 1 card, but everything works a little differently there);
For obvious reasons, the holy war — to bow to the AWS religion — is omitted here.
## Superior Cooling
According to the utilities from Nvidia, the 3090 and A100 are 15-20 degrees cooler than Maxwell and Pascal. I did not take accurate measurements, but on average the situation is like this:
• 4 * 1080 Ti (Pascal) with minimal cooling hacks operate in the range of 75-80C under 100% load;
• 3 * Titan X (Maxwell) worked around 85C under 100% load;
• 3 * 3090 (Ampere) operate in the range of 60-70C under 100% load;
• No overclocking, no restrictions on the power supply of cards or cooler speeds were applied anywhere, everything is out of the box;
• All cards have a blower fan, that is, they push heat out of the case;
There are 3 hypotheses why this is the case:
• New technical process;
• The 3090 has a slightly different shape of the card itself, the size of the fan is noticeably larger, the size of the hole on the rear panel is much larger;
• 3090 seems heavier (maybe someone knows where to find the exact numbers, there are no cards at hand now) which probably implies heavier heatsink;
A clear illustration of the differences between the cards, could someone from the comments tell the diameter of the fan?
## Naive Metrics
First, to make sure the drivers work correctly (and when they did not work correctly, the numbers were completely different), let's test all the available cards with gpu-burn. The result is on the picture and correlates very strongly with what is reported in the reviews.
Test GPU Gflop/s
./gpu_burn 120 Titan X (Maxwell) 4,300
./gpu_burn 120 1080 Ti (Pascal) 8,500
./gpu_burn 120 3090 (Ampere) 16,500
./gpu_burn 120 A100 (wo MIG) 16,700
./gpu-burn -tc 120 3090 (Ampere) 38,500
./gpu-burn -tc 120 A100 (wo MIG) 81,500
MIG wasn't tested here, you will see why further in the article.
## Pricing
It is important to note here that we bought the 1080 Tis and Titan Xs from the second hand market almost "new" (less than a year of use). We will not dwell once again on the holy wars about miners and Nvidia's pricing policy, but if you use even secondhand gaming cards carefully, their service life is about 3-4 years. Prices and characteristics are approximate. According to the information from Nvidia partners in Russia, only one A100 is on sale until the new year. When new 1080 Tiы were available, prices ranged from about 50k to 100k rubles.
GPU Mem Price
Titan X (Maxwell) 12G 10,000 rubles (Avito)
1080 Ti 11G 25,000 rubles (Avito)
3090 (Ampere) 24G 160,000+ rubles (new)
A100 (wo MIG) 40G US \$ 12,500 (new)
Make the obvious conclusions.
## Trying 3090 and A100 with MIG
### Trying 3090
And now let's move on to the most interesting thing — to real down-to-earth tests. In theory, it seems that if the memory and computing capabilities of the 3090 or A100 are 2-3 times higher than the 1080 Ti, then 1 such card can replace 2-3 1080 Ti and a standard server with 4 proper PCIE ports can replace a server with 12 cards? Or is it possible to take, let's say, 3-4 PCIE versions of A100 and get a very powerful server, dividing each of them into several compute instances using MIG?
The short answer is no, the longer answer is also no, but with many caveats.
Why, would you ask? Well, server rack platforms that fully support 8 — 16 video cards even in the smallest reasonable configuration cost 4-5 times more expensive than standard ATX professional solutions. And DGX Workstation or DGX are sold with about a 50% premium to similar configurations assembled on Mikrotik or Gigabyte platforms.
Card manufacturers are in no hurry to release fully-fledged single-slot GPUs (except for PNY with the Quadro series, but this is a separate story and is more likely for design or inference). Of course, you can assemble a custom water circuit for 7 cards (there were several motherboard models with 7 proper PCIE ports), but it's "difficult" and it's not clear where to host it (and the game is not worth the trouble). With the advent of PCIE 4.0, the attractiveness of such solutions, in theory, should grow, but I haven't seen anything interesting on the market yet.
• Task — training Spech-to-Text network on the Ukrainian dataset;
• Due to the problem itself, the experimentally optimal batch size for one process — 50 — cannot be increased significantly without losses in the convergence rate;
• It is on this task that AMP does not work for us (although it works for others, all other things being equal, we have not yet understood why), but this is more of an optimization. That is, it's not about the hardware, but about the task. It works on other tasks, so let's keep things simple;
• An important caveat — since the fact this task is a sequence-to-sequence, in the general case batching here is not entirely trivial. Files of different lengths only get into the batch with files of about the same length (to reduce wasted padding processing resources), but the size of the batch is static for easier comparisons and faster convergence;
• The dynamic and increased batch size were tested, but this does not particularly affect the speed and convergence rate (or worsens);
Contrary to the trend of making more and more gigantic networks, we are miniaturizing our algorithms and are trying to make our networks more and more efficient. Therefore, it is more interesting to increase the worker count, not the networks size or batch size.
And here we come across the first pitfall (https://t.me/snakers4/2590) — Distributed Data Parallel from PyTorch (DDP, the optimal way of scaling networks to "many" video cards) out of the box is essentially configured only for 1 process on 1 card. That is, 1 process can use 1+ card. 2 processes cannot use 1 card, even if there is more than enough IO / compute / RAM. In older driver versions, there is no explicit limitation, and on 1080 Ti 2 processes per 1 card could be launched (but the speed increase is only 5-10% instead of 40-50%). With the new cards, an exception has already been cut in there.
RuntimeError: NCCL error in: /opt/conda/conda-bld/pytorch_1603729096996/work/torch/lib/c10d/ProcessGroupNCCL.cpp:784, invalid usage, NCCL version 2.7.8
But not everything is so sad and bad. Maybe because of some low-level magic in the drivers, maybe because of TF32 (I hope experts will prompt here), maybe because of the developments in MPS 3090s behave a little differently in our benchmark:
• Ceteris paribus they use more memory than Titan X and 1080 Ti (~16 GB instead of 7-8 GB);
• The speed is about 3 times higher than with the Titan X (Maxwell);
• [We still need to accurately measure the speed on 1080 Ti];
• Utilization of cards at a high level — over 90%;
When we try to run 2 DDP workers on 1 card, we just get an error, when we try to train 2 networks "at the same time" we get a proportional slowdown, when the batch increases, the speed gain is insignificant. The timings for 2 * 3090 are like this:
Epoch time, m Type Workers Batch Params
exception DDP 4 50 * 4
3.8 DDP 2 50 * 2
3.9 DDP 2 50 * 2 cudnn_benchmark = True
3.6 DDP 2 100 * 2
For the sake of completeness, it is also important to note that Nvidia has an MPS which supposedly allows you to spin 2 processes on the cards without switching the context, and PyTorch has a built-in RPC framework. But I simply could not adequately use the former without getting incomprehensible low-level errors, and the latter requires a radical rewriting of the code and drastically complicates the code for training models (although it is very interesting in the long term).
So, with 3090 everything is clear. It will not replace two cards, of course, but by itself, even with "extra" memory (I remind, we train small networks), it works 2-3 times faster. Whether this is equivalent to having 2-3 cards depends on the task.
TLDR:
• You can simply replace the cards with a blower fan in your rig with 3090s (the only thing is that there are 2 8-pin power connectors in 3090, but there are 2000-Watt power supplies on the market that can definitely power 4-5 such cards, or just use 2 synced power supplies);
• Most likely, the temperature of the cards will drop by 10-20 degrees Celsius;
• These cards are now expensive and are in short supply (and probably are unlikely to become widely adopted), but if your most expensive resource is time, then this is an interesting option;
• If a large memory size is critical for you, you essentially have no choice;
### Trying the A100 with MIG
Having looked at the metrics, availability, and price of cards, the A100 at first glance does not seem to be an interesting option at all, except for perhaps to train for 3 days 1 large network on 16 A100s on a small, not very private dataset in the cloud. Also, if AMP / FP16 helps your algorithms a lot, then A100 can significantly add speed.
But the A100 has an interesting MIG technology (Multi-Instance GPU). In fact, it allows you to break one "large and powerful" card into a set of small "subcards" and then create virtual Compute Instances, which can be accessed as separate cards.
There are quite a lot of details, check the documentation for them, but the following presets are available there:
+--------------------------------------------------------------------------+
| GPU instance profiles: |
| GPU Name ID Instances Memory P2P SM DEC ENC |
| Free/Total GiB CE JPEG OFA |
|==========================================================================|
| 0 MIG 1g.5gb 19 0/7 4.75 No 14 0 0 |
| 1 0 0 |
+--------------------------------------------------------------------------+
| 0 MIG 2g.10gb 14 0/3 9.75 No 28 1 0 |
| 2 0 0 |
+--------------------------------------------------------------------------+
| 0 MIG 3g.20gb 9 0/2 19.62 No 42 2 0 |
| 3 0 0 |
+--------------------------------------------------------------------------+
| 0 MIG 4g.20gb 5 0/1 19.62 No 56 2 0 |
| 4 0 0 |
+--------------------------------------------------------------------------+
| 0 MIG 7g.40gb 0 0/1 39.50 No 98 5 0 |
| 7 1 1 |
+--------------------------------------------------------------------------+
Available configurations
The question arises, what if our network is small, and A100 in theory (at least on FP16) should be 2 times more powerful than 3090? Is it possible to take 4 A100 and make 12 GPUs similar to 1080 Ti? Is it possible to train neural networks on these numerous "micro-cards" in the same way as on several conventional ones?
We will answer the questions one by one. Here are both the documentation itself and a very recent blog post from Nvidia.
There is a paragraph in the documentation:
MIG supports running CUDA applications by specifying the CUDA device on which the application should be run. With CUDA 11, only enumeration of a single MIG instance is supported.
CUDA applications treat a CI and its parent GI as a single CUDA device. CUDA is limited to use a single CI and will pick the first one available if several of them are visible. To summarize, there are two constraints:
- CUDA can only enumerate a single compute instance
- CUDA will not enumerate non-MIG GPU if any compute instance is enumerated on any other GPU
Note that these constraints may be relaxed in future NVIDIA driver releases for MIG.
At first, when I read it, it seemed to me that it just meant that you cannot divide 2 cards at the same time. After I tried to play around with a real card, it turned out that the framework inside the container sees only 1 "card" (and apparently it only selects the "first" one). Moreover, if we carefully read the examples that Nvidia gives in its blog, they essentially all refer to the scenario "1 container — 1 piece of the card" or "tuning 7 small models in parallel".
There is also a passage like this:
There is no GPU-to-GPU P2P (both PCIe and NVLINK) support in MIG mode, so MIG mode does not support multi-GPU or multi-node training. For large models or models trained with a large batch size, the models may fully utilize a single GPU or even be scaled to multi-GPUs or multi-nodes. In these cases, we still recommend using a full GPU or multi-GPUs, even multi-nodes, to minimize total training time.
If you use MIG for its intended purpose, that is, divide the card into physical pieces (slices), assign them Compute Instances, and drop them into isolated containers — then everything works as it should. It just works. Otherwise — it does not.
## Final Measurements
Here are not really ideal comparisons (on Titan I had DP and not DDP), and on the A100, in the end, I did not run experiments for 10, 20, 30 hours (why pollute the atmosphere), but I measured the time for 1 epoch ...
When you launch 1 network on the A100, the utilization does not even reach half — well, that is, if it could be cut into 2-3 cards, everything would be fine
Avg epoch time, m Workers Batch GPUs CER @ 10 hours CER @ 20 h CER @ 30 h Comment
4.7 2, DDP 50 * 2 2 * 3090 14.4 12.3 11.44 Close to 100% utilization
15.3 1, DP 50 2 * Titan X 21.6 17.4 15.7 Close to 100% utilization
11.4 1, DDP 50 * 1 1 * A100 NA NA NA About 35-40% utilization
TBD 2, DDP 50 * 2 2 * 1080 Ti TBD TBD TBD
On 1080 Ti, resources were only to run 1 epoch.
## Conclusions
• If we forget about stock shortages, then the upgrade is worth it. You will get x2 speed up minimum. If AMP works for you, then maybe even all x3-x4;
• Given the increased productivity, the price seems a bit high, but not exorbitant. A price reduction of about 30-40%, it seems to me, would be adequate;
• When a new generation of cards came out, everyone was worried about cooling. But the cards are surprisingly cold;
• The only problem is that the card demands 2 8-pin connectors for power supply;
• Judging by the price divided by the performance, the card is not very interesting (the markup is 2-3 times against 3090);
• The fact that Nvidia made the technology for efficient inference is cool, otherwise the cards have become too big and poweful;
• If you can use ordinary gaming cards (the same 1080 Ti or PNY Quadro) for inference, then they represent much more value for money;
• There is a great untapped potential in the development of MIG technology;
• If you really need 40 GB of memory and a lot of computing, then there are really no alternatives;
• It is unclear whether the PCIE version could be installed in a regular ATX case without custom, hacking, or water cooling;
# Update 1
I'll try to specify CUDA_VISIBLE_DEVICES inside each PyTorch process later
Test GPU Gflop / s RAM
./gpu_burn 120 A100 // 7 2,400 * 7 4.95 * 7
./gpu_burn 120 A100 // 3 4,500 * 3 9.75 * 3
./gpu_burn 120 A100 // 2 6,700 * 2 19.62 * 2
./gpu_burn 120 A100 (wo MIG) 16,700 39.50 * 1
./gpu-burn -tc 120 A100 // 7 15,100 * 7 4.95 * 7
./gpu-burn -tc 120 A100 // 3 30,500 * 3 9.75 * 3
./gpu-burn -tc 120 A100 // 2 42,500 * 2 19.62 * 2
./gpu-burn -tc 120 A100 (wo MIG) 81,500 39.50 * 1
# Update 2
GPU utilization schedule when running 3 parallel gpu-burn tests via MIG
# Update 3
I ended up being able to get DDP with MIG on PyTorch.
It was necessary to do so and use the zero (first) device everywhere.
def main(rank, args):
os.environ["CUDA_VISIBLE_DEVICES"] = args.ddp.mig_devices[rank]
import torch
...
With NCCL I got the same exception. Changing nccl to gloo made it start… but it worked sooooo slow. Well, let's say, ten times slower and the utilization of the card was at a very low level. I think there is no point in digging further. So the conclusion is — MIG in its current state is absolutely not suitable for large-scale training of networks. Now this is purely a feature for inference or for training N small networks on small datasets.
# Update 4
We found out why our networks refused to cooperate with AMP. More info here. | 2021-03-02 09:04:55 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.408632755279541, "perplexity": 2311.402095454678}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178363782.40/warc/CC-MAIN-20210302065019-20210302095019-00479.warc.gz"} |
http://en.wikipedia.org/wiki/State_variable_filter | # State variable filter
A state variable filter is a type of active filter. It consists of one or more integrators, connected in some feedback configuration. Any LTI system can be described as a state-space model, with n state variables for an nth-order system. A state variable filter realizes the state-space model directly. The instantaneous output voltage of one of the integrators corresponds to one of the state-space model's state variables.
## Kerwin-Huelsman-Newcomb (KHN) Biquad Filter Example
The example given below can produce simultaneous lowpass, highpass and bandpass outputs from a single input. This is a second-order (biquad) filter. Its derivation comes from rearranging a high-pass filter's transfer function, which is the ratio of two quadratic functions. The rearrangement reveals that one signal is the sum of integrated copies of another. That is, the rearrangement reveals a state variable filter structure. By using different states as outputs, different kinds of filters can be produced. In more general state variable filter examples, additional filter order is possible with more integrators (i.e., more states).
The signal input is marked Vin; the LP, HP and BP outputs give the lowpass, highpass and bandpass filtered signals respectively.
For simplicity, we set:
$R_{f1} = R_{f2}$
$C_1 = C_2$
$R_1=R_2$
Then:
$F_0 = \frac{1}{2\pi R_{f1}C_1}$
$Q = \left(1 + \frac{R_4}{R_q}\right)\left(\frac{1}{2+\frac{R_1}{R_g}}\right)$
The pass-band gain for the LP and HP outputs is given by:
$A_{HP} = A_{LP} = \frac{R_1}{R_G}$
It can be seen that the frequency of operation and the Q factor can be varied independently. This and the ability to switch between different filter responses make the state-variable filter widely used in analogue synthesizers.
Values for a resonance frequency of 1 kHz are Rf1=Rf2=10k, C1=C2=15nF and R1=R2=10k | 2014-03-14 18:37:42 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 6, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7033790349960327, "perplexity": 1717.492207930176}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1394678694619/warc/CC-MAIN-20140313024454-00028-ip-10-183-142-35.ec2.internal.warc.gz"} |
https://www.tutorialspoint.com/modulo-remainder-in-arduino | # Modulo / Remainder in Arduino
ArduinoArduino BoardsArduino IDEArduino Programming Language
The modulo operator in Arduino is exactly the same as in C language, or most other languages for that matter. The operator is %. The syntax is: a % b and it returns the remainder when a is divided by b.
## Example
The following example illustrates the use of this operator −
void setup() {
// put your setup code here, to run once:
Serial.begin(9600);
Serial.println();
Serial.println(10%3);
Serial.println(4%2);
Serial.println(50%9);
}
void loop() {
// put your main code here, to run repeatedly:
}
## Output
The Serial Monitor output is shown below. You can work out the remainders yourself and verify that the output is correct.
Published on 31-May-2021 14:14:48 | 2022-05-20 16:34:43 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2283576875925064, "perplexity": 2856.8673035694083}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662533972.17/warc/CC-MAIN-20220520160139-20220520190139-00101.warc.gz"} |
http://math.tutorcircle.com/algebra-2/ | Sales Toll Free No: 1-855-666-7446
# Algebra 2
Top
Sub Topics Algebra 2 is a branch of Mathematics that uses mathematical statements to describe relationships between things that vary over time. A mathematical statement describes relationship and we use letters to represent the quantity that varies as there is no fixed amount. Letters and symbols are known as variables. Mathematical statements that describe relationships are expressed using algebraic terms or equations. In an equation two things can be equal, an equation will have an equal sign.Example: $x + 2 = 6,\ -3xyz + 25x^{2} yz^{2} = 0, 2 + 5 = 7$. When an equation contains variables, you will often have to solve for one of those variables. Algebra 2 nowdays encompasses nowadays many other fields of mathematics like geometric algebra, number theory, abstract algebra, Boolean algebra, analysis etc., to name a few. It is the application of arithmetic operations to an algebraic equation with the aim to solve or simplify it.
## Algebra 2 Topics
Given below are the important algebra 2 topics:
Algebraic Expression Expression built from constants, variables, and a finite number of algebraic operations. Absolute value Absolute value of a real number $x$ is the non negative value of $x$ without regard to its sign. Binomials A polynomial with two terms. That is, the sum of two monomials. A polynomial equation with two terms usually joined by a plus or minus sign is called a binomial. Cartesian Coordinates The usual coordinate system, originally described by Descartes, in which points are specified as distances to a set of perpendicular axes. Also called rectangular coordinates. Domain The domain of a function is the complete set of possible values of the independent variable Complex Numbers A number that is expressed in the form $a + bi$, where $a$ and $b$ are real numbers and $i$ being the imaginary unit. Monomials A monomial is a product of positive integer powers of a fixed set of variables together with a coefficient. Functions A function relates an input to an output. Each input is exactly related to one output. Slope intercept form It is used to express equation of a line, $m$ is the slope and $b$ is the $y$-intercept. Rational Expressions An expression that is the ratio of two polynomials, just like a fraction. Root A root of an equation is a solution of that equation. Tangents Tangent is a line which touches the outer surface of the circle or circular object at exactly one point. Real Numbers Real number is a value that represents a quantity along a continuous line. Trigonometry Branch of mathematics that studies triangles and the relationships between the lengths of their sides and the angles between those sides. Variable A symbol that represents a quantity in a mathematical expression. Range The range of a function is the complete set of all possible resulting values of the dependent variable. Term The parts that make up an expression that are separated by '$+$' and '$-$' signs.
## Algebra 2 Problems
### Solved Examples
Question 1: Solve 5x$^{2}$ + 3x + 1 = 0
Solution:
In the given equation coefficients of a, b and c are 5, 3 and 1 respectively.
Formula for quadratic equation is :
x = $\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
Plug in the values for a, b,and c into the quadratic formula
x = $\frac{-3\pm\sqrt{3^{2}-4*5*1}}{2*5}$
x = $\frac{-3\pm\sqrt{-11}}{10}$
We see that we can have two values for x , let they be $x_{1}$ and $x_{2}$
$x_{1}$ = $\frac{-3+\sqrt{-11}}{10}$
= $\frac{-3}{10}+\frac{1}{10}\sqrt{11}$i
$x_{2}$= $\frac{-3-\sqrt{-11}}{10}$
= $\frac{-3}{10}-\frac{1}{10}\sqrt{11}$i
Therefore the values of $x_{1}$ and $x_{2}$ are $\frac{-3}{10}+\frac{1}{10}\sqrt{11}$i and
$\frac{-3}{10}-\frac{1}{10}\sqrt{11}$
i respectively.
Question 2: Solve 4x$^{2}$ + x + 5 by using completing square method.
Solution:
Step 1:
As the given equation is not in standard form, divide by 4 through out the equation.
x$^{2}$ + $\frac{1}{4}$x + $\frac{5}{4}$ = 0
Step 2:
Move the constant term to the right hand side.
x$^{2}$ + $\frac{1}{4}$x = - $\frac{5}{4}$
Step 3 :
Take half of the x-term coefficient and square it. Add this value to both sides.
x term coefficient = $\frac{1}{4}$
Half of the x-term coefficient = $\frac{1}{8}$
After squaring, ($\frac{1}{8})^{2}$ = $\frac{1}{64}$
Adding $\frac{1}{64}$ to both sides we get
x$^{2}$ + $\frac{1}{4}$x + $\frac{1}{64}$ = -$\frac{5}{4}$ + $\frac{1}{64}$
Step 4:
Simplifying right side
x$^{2}$ + $\frac{1}{4}$x+ $\frac{1}{64}$ = -$\frac{79}{64}$
Step 5:
Write the perfect square on the left
(x + $\frac{1}{8}$)$^{2}$ = -$\frac{79}{64}$
Step 6:
Take the square root of both sides.
x + $\frac{1}{8}$ = $\pm\sqrt{\frac{-79}{64}}$
Step 7:
Solve for x
x = - $\frac{1}{8}$ $\pm\sqrt{\frac{-79}{64}}$
or x = - $\frac{1}{8}$ $\pm$ $\frac{1}{8}\sqrt{79}$i | 2018-03-25 03:07:20 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6587492227554321, "perplexity": 441.06730490457846}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257651780.99/warc/CC-MAIN-20180325025050-20180325045050-00077.warc.gz"} |
https://www.maplesoft.com/support/help/MapleSim/view.aspx?path=ModelonHydraulics/Valves/CounterBalance | Counter Balance $—$ Counterbalance valve
The Counter Balance component describes a counterbalance valve. The counterbalance valve is a pressure control device and allows (almost) free flow from port A (inlet) to port B (load). It blocks reverse flow unless a pilot pressure is sensed at port C (pilot) or load pressure exceeds relief setting. This valve ensures that the actuator always sees a positive load pressure, even under overrunning load situations.
Backpressure at port A does affect the valve setting unless an atmospherically vented counterbalance valve is used where the spring chamber is atmospherically referenced.
The behavior of the valve is modeled as an orifice whose diameter depends on the pressures at the three ports.
Pressures Flows $\mathrm{port_B}.p<\mathrm{pPreload}$ and $\mathrm{port_A}.p=0$ and $\mathrm{port_C}.p=0$ Leakage flow from B $\to$ A (given by GLeak). This is the load holding function. $\mathrm{pPreload}<\mathrm{port_B}.p$ $<$ $\mathrm{pFull}$ and $\mathrm{port_A}.p=0$ and $\mathrm{port_C}.p=0$ Flow from B $\to$ A. Valve is partially open. This is the pressure relief function. $\mathrm{pFull}<\mathrm{port_B}.p$ and $\mathrm{port_A}.p=0$ and $\mathrm{port_C}.p=0$ Flow from B $\to$ A. Valve is completely open. Flow rate is determined by ${q}_{\mathrm{nom}}$ and ${\mathrm{Δp}}_{\mathrm{nom}}$. $\mathrm{port_C}.p<\frac{\mathrm{pPreload}}{\mathrm{pressureRatio}}$ and $\mathrm{port_B}.p=0$ and $\mathrm{port_A}.p=0$ Leakage flow from B $\to$ A (given by GLeak). $\frac{\mathrm{pPreload}}{\mathrm{pressureRatio}}<\mathrm{port_C}.p$ $<$ $\frac{\mathrm{pFull}}{\mathrm{pressureRatio}}$ and $\mathrm{port_B}.p=0$ and $\mathrm{port_A}.p=0$ Flow from B to A. Valve is partially open. $\frac{\mathrm{pFull}}{\mathrm{pressueRatio}}<\mathrm{port_C}.p$ and $\mathrm{port_B}.p=0$ and $\mathrm{port_A}.p=0$ Flow from B $\to$ A. Valve is completely open. Flow rate is determined by ${q}_{\mathrm{nom}}$ and ${\mathrm{Δp}}_{\mathrm{nom}}$. This is the normal operation when the valve is opened by the pressure at the pilot. $\mathrm{port_A}.p<\mathrm{pCheckValvePreload}$ and $\mathrm{port_B}.p=0$ and $\mathrm{port_C}.p=0$ Leakage flow from B $\to$ A (given by GLeak). $\mathrm{pCheckValvePreload}<\mathrm{port_A}.p$ and $\mathrm{port_B}.p=0$ and $\mathrm{port_C}.p=0$ Flow from A $\to$ B through the check valve, given by qnomCheckValve and ${\mathrm{Δp}}_{\mathrm{nom}}$. $\mathrm{pPreload}<\mathrm{port_B}.p-\mathrm{port_A}.p\mathrm{backpressureRatio}+\mathrm{port_C}.p\mathrm{pressureRatio}$ and $\mathrm{port_B}.p-\mathrm{port_A}.p\mathrm{backpressureRatio}+\mathrm{port_C}.p\mathrm{pressureRatio}<\mathrm{pFull}$ Flow from B $\to$ A. Valve is partially open. $\mathrm{pFull}<\mathrm{port_B}.p-\mathrm{port_A}.p\mathrm{backpressureRatio}+\mathrm{port_C}.p\mathrm{pressureRatio}$ Flow from B $\to$ A. Valve is completely open. Flow rate is determined by ${q}_{\mathrm{nom}}$ and ${\mathrm{Δp}}_{\mathrm{nom}}$. This is the normal operation when the valve is opened by the pressure at the pilot.
Setting parameters:
pPreload Load pressure to start opening valve. Some manufacturers call this value the thermal relief pressure which is approximately 60e5 $\mathrm{Pa}$ above their holding pressure of counterbalance setting. pFull Load pressure to open valve completely. Typically not specified; depends on the spring characteristics and is responsible for the opening characteristic. pressureRatio Pressure ratio (that is, the multiplier for pilot pressure). backpressureRatio Pressure ratio (that is, the multiplier for back pressure at port 2). It is zero for atmospherically vented valves and around 1.0 for others. pCheckValvePreload Pressure to open check valve completely. ${q}_{\mathrm{nom}}$ Nominal flow rate at dpnom of load holding valve (that is, the poppet). ${\mathrm{Δp}}_{\mathrm{nom}}$ Pressure drop at ${q}_{\mathrm{nom}}$. qnomCheckValve Nominal flow rate of check valve at ${\mathrm{Δp}}_{\mathrm{nom}}$. GLeak Conductance of leakage. Very small value.
The mass and flow forces are not included. Use the modifier(s)
Volume1(port_B(p(start=1e5,fixed=true)))
and/or
Volume2(port_A(p(start=1e5,fixed=true)))
to set the initial condition(s) for the pressure of the lumped volume(s) $\left[\mathrm{Pa}\right]$.
The flow rate at the pilot port 3 is equal to zero. This pressure at this port is not modeled as a state.
Other names for this valve include motion control valve and over center valve.
Related Components
Name Description Spring-loaded check valve with laminar/turbulent flow; gives the characteristic of the flow from port 2 to 1. This model is also used to model the linearly pressure dependent leakage with GLeak.
Equations $\mathrm{\nu }=\mathrm{Modelica.Media.Air.MoistAir.Utilities.spliceFunction}\left(x=\mathrm{Δp},\mathrm{pos}={\mathrm{\nu }}_{\mathrm{oil}}\left(p={p}_{A\left(\mathrm{abs}\right)},T=T,{v}_{\mathrm{air}}={v}_{\mathrm{gas}\left(\mathrm{oil}\right)},{p}_{\mathrm{sat}}={p}_{\mathrm{sat}}\right),\mathrm{neg}={\mathrm{\nu }}_{\mathrm{oil}}\left(p={p}_{B\left(\mathrm{abs}\right)},T=T,{v}_{\mathrm{air}}={v}_{\mathrm{gas}\left(\mathrm{oil}\right)},{p}_{\mathrm{sat}}={p}_{\mathrm{sat}}\right),\mathrm{Δx}=100\right)$ $\mathrm{\rho }=\mathrm{Modelica.Media.Air.MoistAir.Utilities.spliceFunction}\left(x=\mathrm{Δp},\mathrm{pos}={\mathrm{\rho }}_{\mathrm{oil}}\left(p={p}_{A\left(\mathrm{abs}\right)},T=T,{v}_{\mathrm{air}}={v}_{\mathrm{gas}\left(\mathrm{oil}\right)},{p}_{\mathrm{sat}}={p}_{\mathrm{sat}}\right),\mathrm{neg}={\mathrm{\rho }}_{\mathrm{oil}}\left(p={p}_{B\left(\mathrm{abs}\right)},T=T,{v}_{\mathrm{air}}={v}_{\mathrm{gas}\left(\mathrm{oil}\right)},{p}_{\mathrm{sat}}={p}_{\mathrm{sat}}\right),\mathrm{Δx}=100\right)$ $T={T}_{0\left(\mathrm{oil}\right)}+{\mathrm{ΔT}}_{\mathrm{system}}$ $q=\frac{{m}_{\mathrm{flow}\left(A\right)}}{\mathrm{\rho }}$ $\mathrm{Δp}={p}_{A\left(\mathrm{limited}\right)}-{p}_{B\left(\mathrm{limited}\right)}$ ${p}_{A\left(\mathrm{abs}\right)}={p}_{A}+{p}_{\mathrm{atm}\left(\mathrm{oil}\right)}$ ${p}_{A\left(\mathrm{limited}\right)}=\mathrm{max}\left({p}_{A},{p}_{\mathrm{vapour}\left(\mathrm{oil}\right)}-{p}_{\mathrm{atm}\left(\mathrm{oil}\right)}\right)$ ${p}_{B\left(\mathrm{abs}\right)}={p}_{B}+{p}_{\mathrm{atm}\left(\mathrm{oil}\right)}$ ${p}_{B\left(\mathrm{limited}\right)}=\mathrm{max}\left({p}_{B},{p}_{\mathrm{vapour}\left(\mathrm{oil}\right)}-{p}_{\mathrm{atm}\left(\mathrm{oil}\right)}\right)$
Variables
Name Value Units Description Modelica ID $\mathrm{Δp}$ $\mathrm{Pa}$ Pressure drop dp $q$ $\frac{{m}^{3}}{s}$ Flow rate flowing into port_A q ${p}_{A\left(\mathrm{limited}\right)}$ $\mathrm{Pa}$ Limited gauge pressure pA_limited ${p}_{B\left(\mathrm{limited}\right)}$ $\mathrm{Pa}$ Limited gauge pressure pB_limited $\mathrm{\rho }$ $\frac{\mathrm{kg}}{{m}^{3}}$ Upstream density rho $\mathrm{\nu }$ $\frac{{m}^{2}}{s}$ Upstream kinematic viscosity nu ${p}_{A\left(\mathrm{abs}\right)}$ $\mathrm{Pa}$ Absolute pressure pA pA_abs ${p}_{B\left(\mathrm{abs}\right)}$ $\mathrm{Pa}$ Absolute pressure pB pB_abs $T$ $K$ Local temperature T ${p}_{A\left(\mathrm{summary}\right)}$ ${p}_{A}$ $\mathrm{Pa}$ Pressure at port A summary_pA ${p}_{B\left(\mathrm{summary}\right)}$ ${p}_{B}$ $\mathrm{Pa}$ Pressure at port B summary_pB ${\mathrm{Δp}}_{\mathrm{summary}}$ $\mathrm{Δp}$ $\mathrm{Pa}$ Pressure drop summary_dp ${q}_{\mathrm{summary}}$ $q$ $\frac{{m}^{3}}{s}$ Flow rate flowing into port_A summary_q ${P}_{\mathrm{hyd}\left(\mathrm{summary}\right)}$ $-\mathrm{Δp}q$ $W$ Hydraulic Power summary_HP ${p}_{\mathrm{sat}}$ [1] $\mathrm{Pa}$ Gas saturation pressure p_sat ${V}_{A}$ VolumeA ${V}_{B}$ VolumeB $\mathrm{CheckValve}$ CheckValve $\mathrm{Poppet}$ Poppet $\mathrm{counterBalanceBaseBlock}$ counterBalanceBaseBlock
[1] $\mathrm{oil.gasSaturationPressure}\left(T=T,{v}_{\mathrm{gas}}={\mathrm{oil.v}}_{\mathrm{gas}}\right)$
Connections
Name Description Modelica ID ${\mathrm{port}}_{A}$ Layout of port where oil flows into an element ($0<{m}_{\mathrm{flow}}$, ${p}_{B}<{p}_{A}$ means $0<\mathrm{Δp}$) port_A ${\mathrm{port}}_{B}$ Hydraulic port where oil leaves the component (${m}_{\mathrm{flow}}<0$, ${p}_{B}<{p}_{A}$ means $0<\mathrm{Δp}$) port_B $\mathrm{oil}$ oil ${\mathrm{port}}_{C}$ Port where typically the control pressure for the pilot is connected port_C
Parameters
General Parameters
Name Default Units Description Modelica ID ${\mathrm{ΔT}}_{\mathrm{system}}$ $0$ $K$ Temperature offset from system temperature dT_system use volume A $\mathrm{true}$ If true, a volume is present at port_A useVolumeA use volume B $\mathrm{true}$ If true, a volume is present at port_B useVolumeB ${V}_{A}$ ${10}^{-6}$ ${m}^{3}$ Geometric volume at port A volumeA ${V}_{B}$ ${10}^{-6}$ ${m}^{3}$ Geometric volume at port B volumeB ${p}_{\mathrm{preload}}$ $1.25·{10}^{7}$ $\mathrm{Pa}$ Load pressure to start opening valve pPreload ${p}_{\mathrm{Full}}$ $\frac{6}{5}{p}_{\mathrm{preload}}$ $\mathrm{Pa}$ Load pressure to open valve completely pFull $\mathrm{pressureRatio}$ $5$ Pressure ratio, i.e. multiplier for pilot pressure to open valve pressureRatio backpressure ratio $\mathrm{pressureRatio}$ Pressure ratio, i.e. multiplier for back pressure at port 2 to open valve; 0 for atmospherically vented valve backpressureRatio ${p}_{\mathrm{CheckValvePreload}}$ $1.25·{10}^{5}$ $\mathrm{Pa}$ Pressure to open check valve completely pCheckValvePreload ${G}_{\mathrm{leak}}$ ${10}^{-15}$ $\frac{{m}^{3}}{s\mathrm{Pa}}$ Conductance of leakage GLeak
Flow Parameters
Name Default Units Description Modelica ID ${q}_{\mathrm{nom}}$ $0.001$ $\frac{{m}^{3}}{s}$ Nominal flow rate at dpnom of load holding valve qnom ${\mathrm{Δp}}_{\mathrm{nom}}$ $2.2·{10}^{6}$ $\mathrm{Pa}$ Pressure drop at qnom dpnom ${q}_{\mathrm{nom}\left(\mathrm{CV}\right)}$ ${q}_{\mathrm{nom}}$ $\frac{{m}^{3}}{s}$ Nominal flow rate at dpnom of check valve qnomCheckValve ${k}_{1}$ $10$ Laminar part of orifice model k1 ${k}_{2}$ $2$ Turbulent part of orifice model, ${k}_{2}=\frac{1}{{C}_{d}^{2}}$ k2 | 2017-08-19 22:18:45 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 154, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9064633846282959, "perplexity": 3789.245288036039}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886105927.27/warc/CC-MAIN-20170819220657-20170820000657-00492.warc.gz"} |
https://www.azdictionary.com/definition/handrest | # handrest definition
• noun:
• a support the hand.
• The T-rest on a hand-lathe: so named due to the fact used as a rest for a hand-tool in turning and differentiate it from a computerized or slide-rest.
• a support for the hand
## Related Sources
• Definition for "handrest"
• a support the hand.
• Sentence for "handrest"
• Is the material they use mostly…
• Hypernym for "handrest"
• support | 2017-12-12 05:02:34 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8563938736915588, "perplexity": 13150.712538638423}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948515165.6/warc/CC-MAIN-20171212041010-20171212061010-00569.warc.gz"} |
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Extended envelopes around Galactic Cepheids. II. Polaris and δ Cephei from near-infrared interferometry with CHARA/FLUORWe present the results of long-baseline interferometric observations ofthe classical Cepheids Polaris and δ Cep in the near infrared K'band (1.9-2.3 μm), using the FLUOR instrument of the CHARA Array.Following our previous detection of a circumstellar envelope (CSE)around ℓ Car (Kervella et al. 2006), we report similar detectionsaround Polaris and δ Cep. Owing to the large data set acquired onPolaris, in both the first and second lobes of visibility function, wehave detected the presence of a circum-stellar envelope (CSE), locatedat 2.4±0.1 stellar radii, accounting for 1.5±0.4% of thestellar flux in the K band. A similar model is applied to the δCep data, which shows improved agreement compared to a model withoutCSE. Finally, we find that the bias in estimating the angular diameterof δ Cep in the framework of the Baade-Wesselink method(Mérand et al 2005b) is of the order of 1% or less in the K band.A complete study of the influence of the CSE is proposed in thiscontext, showing that at the optimum baseline for angular diametervariation detection, the bias is of the order of the formal precision inthe determination of the δ Cep pulsation amplitude (1.6%). Statistical Constraints for Astrometric Binaries with Nonlinear MotionUseful constraints on the orbits and mass ratios of astrometric binariesin the Hipparcos catalog are derived from the measured proper motiondifferences of Hipparcos and Tycho-2 (Δμ), accelerations ofproper motions (μ˙), and second derivatives of proper motions(μ̈). It is shown how, in some cases, statistical bounds can beestimated for the masses of the secondary components. Two catalogs ofastrometric binaries are generated, one of binaries with significantproper motion differences and the other of binaries with significantaccelerations of their proper motions. Mathematical relations betweenthe astrometric observables Δμ, μ˙, and μ̈ andthe orbital elements are derived in the appendices. We find a remarkabledifference between the distribution of spectral types of stars withlarge accelerations but small proper motion differences and that ofstars with large proper motion differences but insignificantaccelerations. The spectral type distribution for the former sample ofbinaries is the same as the general distribution of all stars in theHipparcos catalog, whereas the latter sample is clearly dominated bysolar-type stars, with an obvious dearth of blue stars. We point outthat the latter set includes mostly binaries with long periods (longerthan about 6 yr). CHARM2: An updated Catalog of High Angular Resolution MeasurementsWe present an update of the Catalog of High Angular ResolutionMeasurements (CHARM, Richichi & Percheron \cite{CHARM}, A&A,386, 492), which includes results available until July 2004. CHARM2 is acompilation of direct measurements by high angular resolution methods,as well as indirect estimates of stellar diameters. Its main goal is toprovide a reference list of sources which can be used for calibrationand verification observations with long-baseline optical and near-IRinterferometers. Single and binary stars are included, as are complexobjects from circumstellar shells to extragalactic sources. The presentupdate provides an increase of almost a factor of two over the previousedition. Additionally, it includes several corrections and improvements,as well as a cross-check with the valuable public release observationsof the ESO Very Large Telescope Interferometer (VLTI). A total of 8231entries for 3238 unique sources are now present in CHARM2. Thisrepresents an increase of a factor of 3.4 and 2.0, respectively, overthe contents of the previous version of CHARM.The catalog is only available in electronic form at the CDS viaanonymous ftp to cdsarc.u-strasbg.fr (130.79.128.5) or via http://cdsweb.u-strasbg.fr/cgi-bin/qcat?J/A+A/431/773 Local kinematics of K and M giants from CORAVEL/Hipparcos/Tycho-2 data. Revisiting the concept of superclustersThe availability of the Hipparcos Catalogue has triggered many kinematicand dynamical studies of the solar neighbourhood. Nevertheless, thosestudies generally lacked the third component of the space velocities,i.e., the radial velocities. This work presents the kinematic analysisof 5952 K and 739 M giants in the solar neighbourhood which includes forthe first time radial velocity data from a large survey performed withthe CORAVEL spectrovelocimeter. It also uses proper motions from theTycho-2 catalogue, which are expected to be more accurate than theHipparcos ones. An important by-product of this study is the observedfraction of only 5.7% of spectroscopic binaries among M giants ascompared to 13.7% for K giants. After excluding the binaries for whichno center-of-mass velocity could be estimated, 5311 K and 719 M giantsremain in the final sample. The UV-plane constructed from these datafor the stars with precise parallaxes (σπ/π≤20%) reveals a rich small-scale structure, with several clumpscorresponding to the Hercules stream, the Sirius moving group, and theHyades and Pleiades superclusters. A maximum-likelihood method, based ona Bayesian approach, has been applied to the data, in order to make fulluse of all the available stars (not only those with precise parallaxes)and to derive the kinematic properties of these subgroups. Isochrones inthe Hertzsprung-Russell diagram reveal a very wide range of ages forstars belonging to these groups. These groups are most probably relatedto the dynamical perturbation by transient spiral waves (as recentlymodelled by De Simone et al. \cite{Simone2004}) rather than to clusterremnants. A possible explanation for the presence of younggroup/clusters in the same area of the UV-plane is that they have beenput there by the spiral wave associated with their formation, while thekinematics of the older stars of our sample has also been disturbed bythe same wave. The emerging picture is thus one of dynamical streamspervading the solar neighbourhood and travelling in the Galaxy withsimilar space velocities. The term dynamical stream is more appropriatethan the traditional term supercluster since it involves stars ofdifferent ages, not born at the same place nor at the same time. Theposition of those streams in the UV-plane is responsible for the vertexdeviation of 16.2o ± 5.6o for the wholesample. Our study suggests that the vertex deviation for youngerpopulations could have the same dynamical origin. The underlyingvelocity ellipsoid, extracted by the maximum-likelihood method afterremoval of the streams, is not centered on the value commonly acceptedfor the radial antisolar motion: it is centered on < U > =-2.78±1.07 km s-1. However, the full data set(including the various streams) does yield the usual value for theradial solar motion, when properly accounting for the biases inherent tothis kind of analysis (namely, < U > = -10.25±0.15 kms-1). This discrepancy clearly raises the essential questionof how to derive the solar motion in the presence of dynamicalperturbations altering the kinematics of the solar neighbourhood: doesthere exist in the solar neighbourhood a subset of stars having no netradial motion which can be used as a reference against which to measurethe solar motion?Based on observations performed at the Swiss 1m-telescope at OHP,France, and on data from the ESA Hipparcos astrometry satellite.Full Table \ref{taba1} is only available in electronic form at the CDSvia anonymous ftp to cdsarc.u-strasbg.fr (130.79.128.5) or viahttp://cdsweb.u-strasbg.fr/cgi-bin/qcat?J/A+A/430/165} The Indo-US Library of Coudé Feed Stellar SpectraWe have obtained spectra for 1273 stars using the 0.9 m coudéfeed telescope at Kitt Peak National Observatory. This telescope feedsthe coudé spectrograph of the 2.1 m telescope. The spectra havebeen obtained with the no. 5 camera of the coudé spectrograph anda Loral 3K×1K CCD. Two gratings have been used to provide spectralcoverage from 3460 to 9464 Å, at a resolution of ~1 Å FWHMand at an original dispersion of 0.44 Å pixel-1. For885 stars we have complete spectra over the entire 3460 to 9464 Åwavelength region (neglecting small gaps of less than 50 Å), andpartial spectral coverage for the remaining stars. The 1273 stars havebeen selected to provide broad coverage of the atmospheric parametersTeff, logg, and [Fe/H], as well as spectral type. The goal ofthe project is to provide a comprehensive library of stellar spectra foruse in the automated classification of stellar and galaxy spectra and ingalaxy population synthesis. In this paper we discuss thecharacteristics of the spectral library, viz., details of theobservations, data reduction procedures, and selection of stars. We alsopresent a few illustrations of the quality and information available inthe spectra. The first version of the complete spectral library is nowpublicly available from the National Optical Astronomy Observatory(NOAO) via ftp and http. Synthetic Lick Indices and Detection of α-enhanced Stars. II. F, G, and K Stars in the -1.0 < [Fe/H] < +0.50 RangeWe present an analysis of 402 F, G, and K solar neighborhood stars, withaccurate estimates of [Fe/H] in the range -1.0 to +0.5 dex, aimed at thedetection of α-enhanced stars and at the investigation of theirkinematical properties. The analysis is based on the comparison of 571sets of spectral indices in the Lick/IDS system, coming from fourdifferent observational data sets, with synthetic indices computed withsolar-scaled abundances and with α-element enhancement. We useselected combinations of indices to single out α-enhanced starswithout requiring previous knowledge of their main atmosphericparameters. By applying this approach to the total data set, we obtain alist of 60 bona fide α-enhanced stars and of 146 stars withsolar-scaled abundances. The properties of the detected α-enhancedand solar-scaled abundance stars with respect to their [Fe/H] values andkinematics are presented. A clear kinematic distinction betweensolar-scaled and α-enhanced stars was found, although a one-to-onecorrespondence to thin disk'' and thick disk'' components cannot besupported with the present data. STELIB: A library of stellar spectra at R ~ 2000We present STELIB, a new spectroscopic stellar library, available athttp://webast.ast.obs-mip.fr/stelib. STELIB consists of an homogeneouslibrary of 249 stellar spectra in the visible range (3200 to 9500Å), with an intermediate spectral resolution (la 3 Å) andsampling (1 Å). This library includes stars of various spectraltypes and luminosity classes, spanning a relatively wide range inmetallicity. The spectral resolution, wavelength and spectral typecoverage of this library represents a substantial improvement overprevious libraries used in population synthesis models. The overallabsolute photometric uncertainty is 3%.Based on observations collected with the Jacobus Kaptein Telescope,(owned and operated jointly by the Particle Physics and AstronomyResearch Council of the UK, The Nederlandse Organisatie voorWetenschappelijk Onderzoek of The Netherlands and the Instituto deAstrofísica de Canarias of Spain and located in the SpanishObservatorio del Roque de Los Muchachos on La Palma which is operated bythe Instituto de AstrofÃsica de Canarias), the 2.3 mtelescope of the Australian National University at Siding Spring,Australia, and the VLT-UT1 Antu Telescope (ESO).Tables \ref{cat1} to \ref{cat6} and \ref{antab1} to A.7 are onlyavailable in electronic form at http://www.edpsciences.org. The StellarLibrary STELIB library is also available at the CDS, via anonymous ftpto cdsarc.u-strasbg.fr (130.79.128.5) or viahttp://cdsweb.u-strasbg.fr/cgi-bin/qcat?J/A+A/402/433 The Rotation of Binary Systems with Evolved ComponentsIn the present study we analyze the behavior of the rotational velocity,vsini, for a large sample of 134 spectroscopic binary systems with agiant star component of luminosity class III, along the spectral regionfrom middle F to middle K. The distribution of vsini as a function ofcolor index B-V seems to follow the same behavior as their singlecounterparts, with a sudden decline around G0 III. Blueward of thisspectral type, namely, for binary systems with a giant F-type component,one sees a trend for a large spread in the rotational velocities, from afew to at least 40 km s-1. Along the G and K spectral regionsthere are a considerable number of binary systems with moderate tomoderately high rotation rates. This reflects the effects ofsynchronization between rotation and orbital motions. These rotatorshave orbital periods shorter than about 250 days and circular or nearlycircular orbits. Except for these synchronized systems, the largemajority of binary systems with a giant component of spectral type laterthan G0 III are composed of slow rotators. A catalogue of calibrator stars for long baseline stellar interferometryLong baseline stellar interferometry shares with other techniques theneed for calibrator stars in order to correct for instrumental andatmospheric effects. We present a catalogue of 374 stars carefullyselected to be used for that purpose in the near infrared. Owing toseveral convergent criteria with the work of Cohen et al.(\cite{cohen99}), this catalogue is in essence a subset of theirself-consistent all-sky network of spectro-photometric calibrator stars.For every star, we provide the angular limb-darkened diameter, uniformdisc angular diameters in the J, H and K bands, the Johnson photometryand other useful parameters. Most stars are type III giants withspectral types K or M0, magnitudes V=3-7 and K=0-3. Their angularlimb-darkened diameters range from 1 to 3 mas with a median uncertaintyas low as 1.2%. The median distance from a given point on the sky to theclosest reference is 5.2degr , whereas this distance never exceeds16.4degr for any celestial location. The catalogue is only available inelectronic form at the CDS via anonymous ftp to cdsarc.u-strasbg.fr(130.79.128.5) or viahttp://cdsweb.u-strasbg.fr/cgi-bin/qcat?J/A+A/393/183 CHARM: A Catalog of High Angular Resolution MeasurementsThe Catalog of High Angular Resolution Measurements (CHARM) includesmost of the measurements obtained by the techniques of lunaroccultations and long-baseline interferometry at visual and infraredwavelengths, which have appeared in the literature or have otherwisebeen made public until mid-2001. A total of 2432 measurements of 1625sources are included, along with extensive auxiliary information. Inparticular, visual and infrared photometry is included for almost allthe sources. This has been partly extracted from currently availablecatalogs, and partly obtained specifically for CHARM. The main aim is toprovide a compilation of sources which could be used as calibrators orfor science verification purposes by the new generation of largeground-based facilities such as the ESO Very Large Interferometer andthe Keck Interferometer. The Catalog is available in electronic form atthe CDS via anonymous ftp to cdsarc.u-strasbg.fr (130.79.128.5) or viahttp://cdsweb.u-strasbg.fr/cgi-bin/qcat?J/A+A/386/492, and from theauthors on CD-Rom. Stellar populations in the nuclear regions of nearby radio galaxiesWe present optical spectra of the nuclei of seven luminous(P178MHz>~1025WHz-1Sr-1)nearby (z<0.08) radio galaxies, which mostly correspond to the FR IIclass. In two cases, Hydra A and 3C 285, the Balmer andλ4000-Å break indices constrain the spectral types andluminosity classes of the stars involved, revealing that the bluespectra are dominated by blue supergiant and/or giant stars. The agesderived for the last burst of star formation in Hydra A are between 7and 40Myr, and in 3C 285 about 10Myr. The rest of the narrow-line radiogalaxies (four) have a λ4000-Å break and metallic indicesconsistent with those of elliptical galaxies. The only broad-line radiogalaxy in our sample, 3C 382, has a strong featureless blue continuumand broad emission lines that dilute the underlying blue stellarspectra. We are able to detect the Caii triplet in absorption in theseven objects, with good quality data for only four of them. Thestrengths of the absorptions are similar to those found in normalelliptical galaxies, but these values are consistent both with stellarpopulations of roughly similar ages (as derived from the Balmerabsorption and break strengths) andQ2 with mixed young+old populations. Nucleosynthesis and Mixing on the Asymptotic Giant Branch. III. Predicted and Observed s-Process AbundancesWe present the results of s-process nucleosynthesis calculations forasymptotic giant branch (AGB) stars of different metallicities anddifferent initial stellar masses (1.5 and 3 Msolar), and wepresent comparisons of them with observational constraints fromhigh-resolution spectroscopy of evolved stars over a wide metallicityrange. The computations were based on previously published stellarevolutionary models that account for the third dredge-up phenomenonoccurring late on the AGB. Neutron production is driven by the13C(α,n)16O reaction during the interpulseperiods in a tiny layer in radiative equilibrium at the top of the He-and C-rich shell. The neutron source 13C is manufacturedlocally by proton captures on the abundant 12C; a few protonsare assumed to penetrate from the convective envelope into the radiativelayer at any third dredge-up episode, when a chemical discontinuity isestablished between the convective envelope and the He- and C-richzones. A weaker neutron release is also guaranteed by the marginalactivation of the reaction 22Ne(α,n)25Mgduring the convective thermal pulses. Owing to the lack of a consistentmodel for 13C formation, the abundance of 13Cburnt per cycle is allowed to vary as a free parameter over a wideinterval (a factor of 50). The s-enriched material is subsequently mixedwith the envelope by the third dredge-up, and the envelope compositionis computed after each thermal pulse. We follow the changes in thephotospheric abundance of the Ba-peak elements (heavy s [hs]) and thatof the Zr-peak ones (light s [ls]), whose logarithmic ratio [hs/ls] hasoften been adopted as an indicator of the s-process efficiency (e.g., ofthe neutron exposure). Our model predictions for this parameter show acomplex trend versus metallicity. Especially noteworthy is theprediction that the flow along the s-path at low metallicities drainsthe Zr and Ba peaks and builds an excess at the doubly magic208Pb, which is at the termination of the s-path. We thendiscuss the effects on the models of variations in the crucialparameters of the 13C pocket, finding that they are notcritical for interpreting the results. The theoretical predictions arecompared with published abundances of s-elements for AGB giants ofclasses MS, S, SC, post-AGB supergiants, and for various classes ofbinary stars, which supposedly derive their composition by mass transferfrom an AGB companion. This is done for objects belonging both to theGalactic disk and to the halo. The observations in general confirm thecomplex dependence of neutron captures on metallicity. They suggest thata moderate spread exists in the abundance of 13C that isburnt in different stars. Although additional observations are needed,it seems that a good understanding has been achieved of s-processoperation in AGB stars. Finally, the detailed abundance distributionincluding the light elements (CNO) of a few s-enriched stars atdifferent metallicities are examined and satisfactorily reproduced bymodel envelope compositions. Speckle Interferometry of New and Problem Hipparcos Binaries. II. Observations Obtained in 1998-1999 from McDonald ObservatoryThe Hipparcos satellite made measurements of over 9734 known doublestars, 3406 new double stars, and 11,687 unresolved but possible doublestars. The high angular resolution afforded by speckle interferometrymakes it an efficient means to confirm these systems from the ground,which were first discovered from space. Because of its coverage of adifferent region of angular separation-magnitude difference(ρ-Δm) space, speckle interferometry also holds promise toascertain the duplicity of the unresolved Hipparcos problem'' stars.Presented are observations of 116 new Hipparcos double stars and 469Hipparcos problem stars,'' as well as 238 measures of other doublestars and 246 other high-quality nondetections. Included in these areobservations of double stars listed in the Tycho-2 Catalogue andpossible grid stars for the Space Interferometry Mission. Catalogue of Apparent Diameters and Absolute Radii of Stars (CADARS) - Third edition - Comments and statisticsThe Catalogue, available at the Centre de Données Stellaires deStrasbourg, consists of 13 573 records concerning the results obtainedfrom different methods for 7778 stars, reported in the literature. Thefollowing data are listed for each star: identifications, apparentmagnitude, spectral type, apparent diameter in arcsec, absolute radiusin solar units, method of determination, reference, remarks. Commentsand statistics obtained from CADARS are given. The Catalogue isavailable in electronic form at the CDS via anonymous ftp tocdsarc.u-strasbg.fr (130.79.128.5) or viahttp://cdsweb.u-strasbg.fr/cgi-bin/qcar?J/A+A/367/521 Constraining the star formation histories of spiral bulgesStellar populations in spiral bulges are investigated using the Licksystem of spectral indices. Long-slit spectroscopic observations of linestrengths and kinematics made along the minor axes of four spiral bulgesare reported. Comparisons are made between central line strengths inspiral bulges and those in other morphological types [elliptical,spheroidal (Sph) and S0]. The bulges investigated are found to havecentral line strengths comparable to those of single stellar populationsof approximately solar abundance or above. Negative radial gradients areobserved in line strengths, similar to those exhibited by ellipticalgalaxies. The bulge data are also consistent with correlations betweenMg2, Mg2 gradient and central velocity dispersionobserved in elliptical galaxies. In contrast to elliptical galaxies,central line strengths lie within the loci defining the range of and Mg2 achieved by Worthey's solar abundanceratio, single stellar populations (SSPs). The implication of solarabundance ratios indicates significant differences in the star formationhistories of spiral bulges and elliptical galaxies. A single zone withinfall' model of galactic chemical evolution, using Worthey's SSPs, isused to constrain the possible star formation histories of our sample.We show that the , Mg2 and Hβ line strengthsobserved in these bulges cannot be reproduced using primordial collapsemodels of formation but can be reproduced by models with extended infallof gas and star formation (2-17Gyr) in the region modelled. One galaxy(NGC 5689) shows a central population with a luminosity-weighted averageage of ~5Gyr, supporting the idea of extended star formation. Kinematicsubstructure, possibly associated with a central spike in metallicity,is observed at the centre of the Sa galaxy NGC 3623. Stellar populations in active galactic nuclei. II. Population synthesisThe relationship of an AGN to its host galaxy is one crucial question inthe study of galaxy evolution. We present a method to estimate thestellar contribution in active galactic nuclei. We perform stellarpopulation synthesis in the central regions of a sample of 12 galaxiesof different levels of activity: normal galaxies, starburst galaxies,LINERs, Seyfert 2 and Seyfert 1 galaxies. Quantification of the stellarcontribution is carried out in the visible range (5000 to 8800 Ä)using the equivalent widths of the absorption features throughout thespectrum. The synthesis is done by a variant of the new GPG method(Pelat, 1997). This method, contrary to previous ones, gives a uniquesolution. We find quite different stellar populations for the differenttypes of activity, which seems to be indicative of an age sequence. Thestarburst galaxies present the youngest populations of the sample. TheSeyfert 2 nuclei and NGC 1275, a Seyfert 1 with signs of interaction andwhere young stellar clusters have been found, also show the contributionof a young population, less intense than in the starburst galaxies butmetal rich. NGC 3516, a typical Seyfert 1, has a normal populationcharacteristic of galaxies of the same Hubble type and finally theLINERs show the oldest populations in the sample, metal rich, withlittle star formation still going on. It is found that a strong CaIItriplet, even though these lines are sensitive to gravity, does notimply necessarily a stellar population dominated by supergiant stars.Based on observations collected at the Canadian-French-HawaiianTelescope, Hawaii, and Observatoire de Haute Provence, France. A Counterrotating Central Component in the Barred Galaxy NGC 5728We present a detailed study of the stellar kinematics in the barredgalaxy NGC 5728 based on I-band photometry and long-slit spectroscopicobservations in the region of the near-IR Ca II triplet. The analysis ofthe stellar line-of-sight velocity distribution (LOSVD) has revealed, inthe central regions of the bar, the presence of a cold (v/sigma~2.5),prograde, S-shaped velocity component that coexists in the central 4kpc, with a fainter and hotter (v/sigma~0.5) counterrotating component.Beyond 4 kpc from the nucleus, the LOSVD shows the stellar barkinematics. The comparison of the radial surface brightness profile ofthe velocity components with that obtained from an I-band image showsthat the counterrotating core follows a r^1/4 profile, while theS-shaped component does not follow the flat-bar surface brightnessprofile. Several possible scenarios accounting for such kinematicsignatures found in the center of the bar in NGC 5728 are discussed. Thedata presented in this paper show for the first time the presence ofextended retrograde motions in barred systems that, together withprevious discoveries, seem to indicate that the stellar counterrotationis a phenomenon present all along the Hubble sequence. Spectral Irradiance Calibration in the Infrared. X. A Self-Consistent Radiometric All-Sky Network of Absolutely Calibrated Stellar SpectraWe start from our six absolutely calibrated continuous stellar spectrafrom 1.2 to 35 μm for K0, K1.5, K3, K5, and M0 giants. These wereconstructed as far as possible from actual observed spectral fragmentstaken from the ground, the Kuiper Airborne Observatory, and the IRAS LowResolution Spectrometer, and all have a common calibration pedigree.From these we spawn 422 calibrated spectral templates'' for stars withspectral types in the ranges G9.5-K3.5 III and K4.5-M0.5 III. Wenormalize each template by photometry for the individual stars usingpublished and/or newly secured near- and mid-infrared photometryobtained through fully characterized, absolutely calibrated,combinations of filter passband, detector radiance response, and meanterrestrial atmospheric transmission. These templates continue ourongoing effort to provide an all-sky network of absolutely calibrated,spectrally continuous, stellar standards for general infrared usage, allwith a common, traceable calibration heritage. The wavelength coverageis ideal for calibration of many existing and proposed ground-based,airborne, and satellite sensors, particularly low- tomoderate-resolution spectrometers. We analyze the statistics of probableuncertainties, in the normalization of these templates to actualphotometry, that quantify the confidence with which we can assert thatthese templates truly represent the individual stars. Each calibratedtemplate provides an angular diameter for that star. These radiometricangular diameters compare very favorably with those directly observedacross the range from 1.6 to 21 mas. A catalog of rotational and radial velocities for evolved starsRotational and radial velocities have been measured for about 2000evolved stars of luminosity classes IV, III, II and Ib covering thespectral region F, G and K. The survey was carried out with the CORAVELspectrometer. The precision for the radial velocities is better than0.30 km s-1, whereas for the rotational velocity measurementsthe uncertainties are typically 1.0 km s-1 for subgiants andgiants and 2.0 km s-1 for class II giants and Ib supergiants.These data will add constraints to studies of the rotational behaviourof evolved stars as well as solid informations concerning the presenceof external rotational brakes, tidal interactions in evolved binarysystems and on the link between rotation, chemical abundance and stellaractivity. In this paper we present the rotational velocity v sin i andthe mean radial velocity for the stars of luminosity classes IV, III andII. Based on observations collected at the Haute--Provence Observatory,Saint--Michel, France and at the European Southern Observatory, LaSilla, Chile. Table \ref{tab5} also available in electronic form at CDSvia anonymous ftp to cdsarc.u-strasbg.fr (130.79.128.5) or viahttp://cdsweb.u-strasbg.fr/Abstract.html Empirical calibration of the lambda 4000 Å breakEmpirical fitting functions, describing the behaviour of the lambda 4000Ä break, D4000, in terms of effective temperature,metallicity and surface gravity, are presented. For this purpose, thebreak has been measured in 392 stars from the Lick/IDS Library. We havefollowed a very detailed error treatment in the reduction and fittingprocedures, allowing for a reliable estimation of the breakuncertainties. This calibration can be easily incorporated into stellarpopulation models to provide accurate predictions of the break amplitudefor, relatively old, composite systems. Table 1 is only available inelectronic form at the CDS via anonymous ftp to cdsarc.u-strasbg.fr(130.79.128.5) or via http://cdsweb.u-strasbg.fr/Abstract.html Catalogs of temperatures and [Fe/H] averages for evolved G and K starsA catalog of mean values of [Fe/H] for evolved G and K stars isdescribed. The zero point for the catalog entries has been establishedby using differential analyses. Literature sources for those entries areincluded in the catalog. The mean values are given with rms errors andnumbers of degrees of freedom, and a simple example of the use of thesestatistical data is given. For a number of the stars with entries in thecatalog, temperatures have been determined. A separate catalogcontaining those data is briefly described. Catalog only available atthe CDS via anonymous ftp to cdsarc.u-strasbg.fr (130.79.128.5) or viahttp://cdsweb.u-strasbg.fr/Abstract.html Heavy-element abundances in seven SC stars and several related starsWe employ spectra of resolution 20-35,000 of seven SC stars, four Sstars, two Ba stars, and two K-M stars to derive abundances of a varietyof elements from Sr to Eu relative to iron. Special attention is paid toRb and Tc, and to the ratio of the heavy s-process species to the lights-process elements. Abundances are derived in LTE, both by using modelatmospheres in which the carbon and oxygen abundances are nearly equaland by using curves of growth. Spectrum synthesis is used for criticallines, such as the 5924-A line of Tc and the 7800-A line of Rb. For mostof the heavy-element stars, the enhancement of the s-process elements isabout a factor of 10. The ratio of the heavy to light s-process speciesis not far from solar, except for RR Her, for which the same ratio is +0.45 dex. For Tc the blending by other lines is severe. While we haveprobably detected the 5924-A line, we can only present abundances in theless-than-or-equal-to category. Integrated Ultraviolet Spectra and Line Indices of M31 Globular Clusters and the Cores of Elliptical GalaxiesWe present observations of the integrated light of four M31 globularclusters (MIV, MII, K280, and K58) and of the cores of six ellipticalgalaxies (NGC 3605, 3608, 5018, 5831, 6127, and 7619) made with theFaint Object Spectrograph on the Hubble Space Telescope. The spectracover the range 2200-4800 Å at a resolution of 8 Å withsignal-to-noise ratio of more than 20 and flux accuracy of ~5%. To thesedata we add from the literature IUE observations of the dwarf ellipticalgalaxy M32, Galactic globular clusters, and Galactic stars. The stellarpopulations in these systems are analyzed with the aid of mid-UV andnear-UV colors and absorption line strengths. Included in the measuredindices is the key NH feature at 3360 Å. We compare these lineindex measures with the 2600 - 3000 colors of these stars and stellarpopulations. We find that the M31 globular clusters, Galactic globularclusters/Galactic stars, and elliptical galaxies represent threedistinct stellar populations, based on their behavior in color-linestrength correlations involving Mg II, NH, CN, and several UV metallicblends. In particular, the M31 globular cluster MIV, as metal-poor asthe Galactic globular M92, shows a strong NH 3360 Å feature. Otherline indices, including the 3096 Å blend that is dominated bylines of Mg I and Al I, show intrinsic differences as well. We also findthat the broadband line indices often employed to measure stellarpopulation differences in faint objects, such as the 4000 Å andthe Mg 2800 breaks, are disappointingly insensitive to these stellarpopulation differences. We find that the hot (T > 20,000 K) stellarcomponent responsible for the `UV upturn'' at shorter wavelengths canhave an important influence on the mid-UV spectral range (2400-3200Å) as well. The hot component can contribute over 50% of the fluxat 2600 Å in some cases and affects both continuum colors and linestrengths. Mid-UV spectra of galaxies must be corrected for this effectbefore they can be used as age and abundance diagnostics. Of the threestellar populations studied here, M31 globular clusters and ellipticalgalaxies are more similar to each other than either is to the Galacticstellar populations defined by globular clusters and nearby stars.Similarities between the abundance-pattern differences currentlyidentified among these stellar populations and those among globularcluster stars (N, Al enhancements) present a curious coincidence thatdeserves future investigation. Based on observations with the NASA/ESAHubble Space Telescope, obtained at the Space Telescope ScienceInstitute, which is operated by the Association of Universities forResearch in Astronomy, Inc., under NASA contract NAS 5-26555. The Tokyo PMC catalog 90-93: Catalog of positions of 6649 stars observed in 1990 through 1993 with Tokyo photoelectric meridian circleThe sixth annual catalog of the Tokyo Photoelectric Meridian Circle(PMC) is presented for 6649 stars which were observed at least two timesin January 1990 through March 1993. The mean positions of the starsobserved are given in the catalog at the corresponding mean epochs ofobservations of individual stars. The coordinates of the catalog arebased on the FK5 system, and referred to the equinox and equator ofJ2000.0. The mean local deviations of the observed positions from theFK5 catalog positions are constructed for the basic FK5 stars to comparewith those of the Tokyo PMC Catalog 89 and preliminary Hipparcos resultsof H30. Classification and Identification of IRAS Sources with Low-Resolution SpectraIRAS low-resolution spectra were extracted for 11,224 IRAS sources.These spectra were classified into astrophysical classes, based on thepresence of emission and absorption features and on the shape of thecontinuum. Counterparts of these IRAS sources in existing optical andinfrared catalogs are identified, and their optical spectral types arelisted if they are known. The correlations between thephotospheric/optical and circumstellar/infrared classification arediscussed. H gamma and H delta Absorption Features in Stars and Stellar PopulationsThe H gamma and H delta absorption features are measured in a sample of455 (out of an original 460) Lick/IDS stars with pseudo--equivalentwidth indices. For each Balmer feature, two definitions, involving anarrow (~20 Angstroms) and a wide (~40 Angstroms) central bandpass, aremeasured. These four new Balmer indices augment 21 indices previouslydetermined by Worthey et al., and polynomial fitting functions that giveindex strengths as a function of stellar temperature, gravity, and[Fe/H] are provided. The new indices are folded into models for theintegrated light of stellar populations, and predictions are given forsingle-burst stellar populations of a variety of ages and metallicities.Contrary to our initial hopes, the indices cannot break a degeneracybetween burst age and burst strength in post-starburst objects, but theyare successful mean-age indicators when used with sensitive metallicityindicators. An appendix gives data, advice, and examples of how totransform new spectra to the 25-index Lick/IDS system. The Stellar Populations of Spiral Disks.I.A New Observational Approach: Description of the Technique and Spectral Gradients for the Inter-Arm Regions of NGC 4321 (M100)We describe an imaging method that makes use of interference filters toprovide integrated stellar spectral indices for spiral disks to faintsurface brightness limits. We use filters with bandpasses { ~ 60Angstroms\ }FWHM, centered on the Mg and Fe features (lambda lambda 5176Angstroms \ and 5270 Angstroms respectively) allowing the determinationof the spatial distribution of the Lick indices Mg2 andFe5270. These two indices have been extensively modeled by differentgroups and used in the past mainly for the study of elliptical galaxies,bulges, and globular clusters. Azimuthal integration of the underlyingsmooth stellar signal, after removal of the signature of the spiral armsand associated extreme pop. I structures, provides measurements of thesespectral indices useful to radial distances where the surface brightnessof the galaxy reaches ~ 24mu_V . As a first example of this techniqueand its possibilities we conduct a preliminary study of the SABbc galaxyNGC 4321 (M 100). We present spectral gradients for the inter-armstellar population to about 4 exponential scale lengths. There is someevidence for a discontinuity in the run of the Mg2 index nearcorotation, which we interpret as evidence for bar- driven secularevolution. There is also evidence that Mg is overabundant with respectto Fe in the inner regions of the projected image. A catalogue of [Fe/H] determinations: 1996 editionA fifth Edition of the Catalogue of [Fe/H] determinations is presentedherewith. It contains 5946 determinations for 3247 stars, including 751stars in 84 associations, clusters or galaxies. The literature iscomplete up to December 1995. The 700 bibliographical referencescorrespond to [Fe/H] determinations obtained from high resolutionspectroscopic observations and detailed analyses, most of them carriedout with the help of model-atmospheres. The Catalogue is made up ofthree formatted files: File 1: field stars, File 2: stars in galacticassociations and clusters, and stars in SMC, LMC, M33, File 3: numberedlist of bibliographical references The three files are only available inelectronic form at the Centre de Donnees Stellaires in Strasbourg, viaanonymous ftp to cdsarc.u-strasbg.fr (130.79.128.5), or viahttp://cdsweb.u-strasbg.fr/Abstract.html Spectra of late type stars from 4800 to 9000A.We present optical spectra of 21 stars from 4800 to 8920A, coveringessentially the late spectral types, G, K, M and the luminosity classesI and III. Half of the stars are super metal rich (SMR) ones. Thespectra were obtained at a resolution of 1.25A using the Aureliespectrograph, equipped with a linear array CCD-like detector, attachedto the OHP 1.52m telescope. Also presented are the spectra of 7 stars,covering the region 5000-9783A at a resolution of 8.5A, observed at theCFHT with the Herzberg spectrograph. The spectral types are F, G, K, Mand the luminosity classes III and V. Five stars are SMR. These spectrahave been obtained with the aim of extending existing libraries used forpopulation synthesis purposes. The inclusion of SMR stars in a stellarlibrary dedicated to the study of stellar populations in the centralpart of galaxies is crucial as abundance gradients have been observed inthe optical range. Vitesses radiales. Catalogue WEB: Wilson Evans Batten. Subtittle: Radial velocities: The Wilson-Evans-Batten catalogue.We give a common version of the two catalogues of Mean Radial Velocitiesby Wilson (1963) and Evans (1978) to which we have added the catalogueof spectroscopic binary systems (Batten et al. 1989). For each star,when possible, we give: 1) an acronym to enter SIMBAD (Set ofIdentifications Measurements and Bibliography for Astronomical Data) ofthe CDS (Centre de Donnees Astronomiques de Strasbourg). 2) the numberHIC of the HIPPARCOS catalogue (Turon 1992). 3) the CCDM number(Catalogue des Composantes des etoiles Doubles et Multiples) byDommanget & Nys (1994). For the cluster stars, a precise study hasbeen done, on the identificator numbers. Numerous remarks point out theproblems we have had to deal with.
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• - No Links Found - | 2021-07-24 05:58:11 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6321756839752197, "perplexity": 5868.110717583653}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046150129.50/warc/CC-MAIN-20210724032221-20210724062221-00310.warc.gz"} |
http://mathhelpforum.com/calculus/85069-shell-method-print.html | # shell method
• Apr 22nd 2009, 11:56 AM
mikegar813
shell method
Find the volume of the solid formed by revolving the region bounded by the graph of f(x)=4-x^2 and the x-axis about the line x=3 using the shell method.
• Apr 22nd 2009, 01:41 PM
derfleurer
$2\pi \int_{0}^{4} rhdy$
$r = 3 - y$
$h = \sqrt {4 - y}$
$2\pi \int_{0}^{4} (3 - y)\sqrt {4 - y}dy$
u = 4 - y
dy = -dy
$-\int (u + 1)\sqrt udu = -\int (u^{3/2}du + \sqrt udu)$
• Apr 23rd 2009, 04:57 AM
mikegar813
Quote:
Originally Posted by derfleurer
$2\pi \int_{0}^{4} rhdy$
$r = 3 - y$
$h = \sqrt {4 - y}$
$2\pi \int_{0}^{4} (3 - y)\sqrt {4 - y}dy$
u = 4 - y
dy = -dy
$-\int (u + 1)\sqrt udu = -\int (u^{3/2}du + \sqrt udu)$
how can u replace 3 in 3-y? | 2017-01-24 01:31:31 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 10, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7952879667282104, "perplexity": 5310.110732379796}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560283475.86/warc/CC-MAIN-20170116095123-00272-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://rdrr.io/cran/StratifiedMedicine/f/vignettes/SM_PRISM.Rmd | knitr::opts_chunk$set( collapse = TRUE, comment = "#>", fig.width=10, fig.height=8.5 ) # Introduction Welcome to the StratifiedMedicine R package. The overall goal of this package is to develop analytic and visualization tools to aid in stratified and personalized medicine. Stratified medicine aims to find subsets or subgroups of patients with similar treatment effects, for example responders vs non-responders, while personalized medicine aims to understand treatment effects at the individual level (does a specific individual respond to treatment A?). Currently, the main tools in this package area: (1) Filter Models (identify important variables and reduce input covariate space), (2) Patient-Level Estimate Models (using regression models, estimate counterfactual quantities, such as the conditional average treatment effect or CATE), (3) Subgroup Models (identify groups of patients using tree-based approaches), and (4) Parameter Estimation (across the identified subgroups), and (5) PRISM (Patient Response Identifiers for Stratified Medicine; combines tools 1-4). Development of this package is ongoing. As a running example, consider a continuous outcome (ex: % change in tumor size) with a binary treatment (study drug vs standard of care). The estimand of interest is the average treatment effect,$\theta_0 = E(Y|A=1)-E(Y|A=0)$. First, we simulate continuous data where roughly 30\% of the patients receive no treatment-benefit for using$A=1$vs$A=0$. Responders vs non-responders are defined by the continuous predictive covariates$X_1$and$X_2$for a total of four subgroups. Subgroup treatment effects are:$\theta_{1} = 0$($X_1 \leq 0, X_2 \leq 0$),$\theta_{2} = 0.25 (X_1 > 0, X_2 \leq 0)$,$\theta_{3} = 0.45 (X_1 \leq 0, X2 > 0$),$\theta_{4} = 0.65 (X_1>0, X_2>0)$. library(ggplot2) library(dplyr) library(partykit) library(StratifiedMedicine) library(survival) dat_ctns = generate_subgrp_data(family="gaussian") Y = dat_ctns$Y
X = dat_ctns$X # 50 covariates, 46 are noise variables, X1 and X2 are truly predictive A = dat_ctns$A # binary treatment, 1:1 randomized
length(Y)
table(A)
dim(X)
# Filter Models
The aim of filter models is to potentially reduce the covariate space such that subsequent analyses focus on the "important" variables. For example, we may want to identify variables that are prognostic and/or predictive of the outcome across treatment levels. Filter models can be run using the "filter_train" function. The default is search for prognostic variables using elastic net (Y~ENET(X); Hou and Hastie 2005). Random forest based importance (filter="ranger") is also available. See below for an example. Note that the object "filter.vars" contains the variables that pass the filter, while "plot_importance" shows us the relative importance of the input variables. For glmnet, this corresponds to the standardized regression coefficients (variables with coefficients=0 are not shown).
res_f <- filter_train(Y, A, X, filter="glmnet")
res_f$filter.vars plot_importance(res_f) An alternative approach is to search for variables that are potentially prognostic and/or predictive by forcing variable by treatment interactions, or Y~ENET(X,XA). Variables with estimated coefficients of 0 in both the main effects (X) and interaction effects (XA) are filtered. This can be implemented by tweaking the hyper-parameters: res_f2 <- filter_train(Y, A, X, filter="glmnet", hyper=list(interaction=T)) res_f2$filter.vars
plot_importance(res_f2)
Here, note that both the main effects of X1 and X2, along with the interaction effects (labeled X1_trtA and X2_trtA), have relatively large estimated coefficients.
# Patient-level Estimate (PLE) Models
The aim of PLE models is to estimate counterfactual quantities, for example the CATE. This is implemented through the "ple_train" function. The "ple_train" follows the framework of Kunzel et al 2019, which utilizes base learners and meta learners to obtain estimates of interest. For family="gaussian", "binomial", this output estimates of \eqn{mu(a,x)=E(Y|x,a)} and treatment differences. For family="survival", either logHR or restricted mean survival time (RMST) estimates are obtained. Current base-leaner options include "linear" (lm/glm/or cox), "ranger" (random forest through ranger R package), "glmnet" (elastic net), and "bart" (Bayesian Additive Regression Trees through BART R package). Meta-learners include the "T-Leaner" (treatment specific models), "S-learner" (single regression model), and "X-learner" (2-stage approach, see Kunzel et al 2019). See below for an example. Note that the object "mu_train" contains the training set patient-level estimates (outcome-based and propensity scores), "plot_ple" shows a waterfall plot of the estimated CATEs, and "plot_dependence" shows the partial dependence plot for variable "X1" with respect to the estimated CATE.
res_p1 <- ple_train(Y, A, X, ple="ranger", meta="T-learner")
summary(res_p1$mu_train) plot_ple(res_p1) plot_dependence(res_p1, X=X, vars="X1") Next, let's illustrate how to change the meta-learner and the hyper-parameters. See below, along with a 2-dimension PDP example. res_p2 <- ple_train(Y, A, X, ple="ranger", meta="T-learner", hyper=list(mtry=5)) plot_dependence(res_p2, X=X, vars=c("X1", "X2")) # Subgroup Models Subgroup models are called using the "submod_train" function and currently only include tree-based methods (ctree, lmtree, glmtree from partykit R package and rpart from rpart R package). First, let's run the default (for continuous, uses lmtree). This aims to find subgroups that are either prognostic and/or predictive. res_s1 <- submod_train(Y, A, X, submod="lmtree") table(res_s1$Subgrps.train)
plot(res_s1$fit$mod)
Another generic approach is "otr", which follows an outcome weighted learning approach. Here, we regress I(CATE>delta) ~ ctree(X) with weights=abs(CATE-delta) where CATE corresponds to estimates of E(Y|A=1,X)-E(Y|A=0,X). For survival endpoints, the treatment difference would correspond to either logHR or RMST. For the example below, we set the clinically meaningful threshold to 0.1 (delta=">0.10").
res_s2 <- submod_train(Y, A, X, mu_train=res_p2$mu_train, submod="otr", delta=">0.10") plot(res_s2$fit$mod) # Treatment Effect Estimation To facilitate treatment effect estimation across the identified subgroups, "StratifiedMedicine" currently includes the function "param_est." This includes param="lm", "dr", "gcomp", "cox", and "rmst" which correspond respectively to linear regression, the doubly robust estimator, average the patient-level estimates (G-computation), cox regresson, and RMST (as in survRM2 R package). Notably, if the subgroups are determined adaptively (for example through lmtree), without resampling corrections, point-estimates tend to be overly optimistic. We address this later. Given a candidate set of subgroups, a simple approach is to fit linear regression models within each subgroup to obtain treatment effect estimates (A=1 vs A=0). See below. param.dat1 <- param_est(Y, A, X, Subgrps = res_s1$Subgrps.train, param="lm")
param.dat1
Alternatively, we may instead use the doubly-robust estimator, which combines the observed outcome (Y) and model estimates from "ple_train". This requires inputting model estimates (see "mu_hat"). See below:
param.dat2 <- param_est(Y, A, X, Subgrps = res_s1$Subgrps.train, mu_hat = res_p1$mu_train, param="dr")
param.dat2 %>% filter(estimand=="mu_1-mu_0")
# PRISM: Patient Response Identifiers for Stratified Medicine
While the above tools individually can be useful, PRISM (Patient Response Identifiers for Stratified Medicine; Jemielita and Mehrotra (to appear), https://arxiv.org/abs/1912.03337) combines each component for a stream-lined analysis. Given a data-structure of $(Y, A, X)$ (outcome(s), treatments, covariates), PRISM is a five step procedure:
1. Estimand: Determine the question(s) or estimand(s) of interest. For example, $\theta_0 = E(Y|A=1)-E(Y|A=0)$, where A is a binary treatment variable. While this isn't an explicit step in the PRISM function, the question of interest guides how to set up PRISM.
2. Filter (filter): Reduce covariate space by removing variables unrelated to outcome/treatment.
3. Patient-level estimate (ple): Estimate counterfactual patient-level quantities, for example the conditional average treatment effect (CATE), $\theta(x) = E(Y|X=x,A=1)-E(Y|X=x,A=0)$. These can be used in the subgroup model and/or parameter estimation.
4. Subgroup model (submod): Identify subgroups of patients with potentially varying treatment response.
5. Parameter estimation and inference (param): For the overall population and discovered subgroups, output point estimates and variability metrics. If the subgroups are determined adaptively, resampling is needed to avoid overly optimistic point estimates and to form CIs.
6. Resampling: Repeat Steps 1-4 across $R$ non-parametric bootstrap resamplings to generate subgroup-specific parameter estimate bootstrap distributions.
Ultimately, PRISM provides information at the patient-level, the subgroup-level (if any), and the overall population. While there are defaults in place, the user can also input their own functions/model wrappers into the PRISM algorithm. We will demonstrate this later. PRISM can also be run without treatment assignment (A=NULL); in this setting, the focus is on finding subgroups based on prognostic effects. The below table describes default PRISM configurations for different family (gaussian, biomial, survival) and treatment (no treatment vs treatment) settings, including the associated estimands. Note that OLS refers to ordinary least squares (linear regression), GLM refers to generalized linear model, and MOB refers to model based partitioning (Zeileis, Hothorn, Hornik 2008; Seibold, Zeileis, Hothorn 2016). To summarise, default models include elastic net (Zou and Hastie 2005) for filtering, random forest ("ranger" R package) for patient-level /counterfactual estimation, and MOB (through "partykit" R package; lmtree, glmtree, and ctree (Hothorn, Hornik, Zeileis 2005)). When treatment assignment is provided, parameter estimation for continuous and binary outcomes uses the double-robust estimator (based on the patient-level estimates). For survival outcomes, the cox regression hazard ratio (HR) or RMST (from the survR2 package) is used.
library(knitr)
summ.table = data.frame( Step = c("estimand(s)", "filter", "ple", "submod", "param"),
gaussian = c("E(Y|A=0)<br>E(Y|A=1)<br>E(Y|A=1)-E(Y|A=0)",
"Elastic Net<br>(glmnet)",
"X-learner: Random Forest<br>(ranger)",
"MOB(OLS)<br>(lmtree)",
"Double Robust<br>(dr)"),
binomial = c("E(Y|A=0)<br>E(Y|A=1)<br>E(Y|A=1)-E(Y|A=0)",
"Elastic Net<br>(glmnet)",
"X-learner: Random Forest<br>(ranger)",
"MOB(GLM)<br>(glmtree)",
"Doubly Robust<br>(dr)"),
survival = c("HR(A=1 vs A=0)",
"Elastic Net<br>(glmnet)",
"T-learner: Random Forest<br>(ranger)",
"MOB(OLS)<br>(lmtree)",
"Hazard Ratios<br>(cox)") )
kable( summ.table, caption = "Default PRISM Configurations (With Treatment)", full_width=T)
summ.table = data.frame(Step = c("estimand(s)", "filter", "ple", "submod", "param"),
gaussian = c("E(Y)",
"Elastic Net<br>(glmnet)",
"Random Forest<br>(ranger)",
"Conditional Inference Trees<br>(ctree)",
"OLS<br>(lm)"),
binomial = c("Prob(Y)",
"Elastic Net<br>(glmnet)",
"Random Forest<br>(ranger)",
"Conditional Inference Trees<br>(ctree)",
"OLS<br>(lm)"),
survival = c("RMST", "Elastic Net<br>(glmnet)",
"Random Forest<br>(ranger)",
"Conditional Inference Trees<br>(ctree)",
"RMST<br>(rmst)"))
kable( summ.table, caption = "Default PRISM Configurations (Without Treatment, A=NULL)", full_width=T)
# Example: Continuous Outcome with Binary Treatment
For continuous outcome data (family="gaussian"), the default PRISM configuration is: (1) filter="glmnet" (elastic net), (2) ple="ranger" (X-learner with random forest models), (3) submod="lmtree" (model-based partitioning with OLS loss), and (4) param="dr" (doubly-robust estimator). To run PRISM, at a minimum, the outcome (Y), treatment (A), and covariates (X) must be provided. See below. The summary gives a high-level overview of the findings (number of subgroups, parameter estimates, variables that survived the filter). The default plot() function currently combines tree plots with parameter estimates using the "ggparty" package.
# PRISM Default: filter_glmnet, ranger, lmtree, dr #
res0 = PRISM(Y=Y, A=A, X=X)
summary(res0)
plot(res0) # same as plot(res0, type="tree")
We can als0 directly look for prognostic effects by specifying omitting A (treatment) from PRISM:
# PRISM Default: filter_glmnet, ranger, ctree, param_lm #
res_prog = PRISM(Y=Y, X=X)
# res_prog = PRISM(Y=Y, A=NULL, X=X) #also works
summary(res_prog)
Next, circling back to the first PRISM model with treatment included, let's review other core PRISM outputs. Results relating to the filter include "filter.mod" (model output) and "filter.vars" (variables that pass the filter). The "plot_importance" function can also be called:
# elastic net model: loss by lambda #
plot(res0$filter.mod) ## Variables that remain after filtering ## res0$filter.vars
# All predictive variables (X1,X2) and prognostic variables (X3,X5, X7) remains.
plot_importance(res0)
Results relating to "ple_train" include "ple.fit" (fitted "ple_train"), "mu.train" (training predictions), and "mu.test" (test predictions). "plot_ple" and "plot_dependence" can also be used with PRISM objects. For example,
summary(res0$mu_train) plot_ple(res0) plot_dependence(res0, vars=c("X2")) Next, the subgroup model (lmtree), identifies 4-subgroups based on varying treatment effects. By plotting the subgroup model object ("submod.fit\$mod")", we see that partitions are made through X1 (predictive) and X2 (predictive). At each node, parameter estimates for node (subgroup) specific OLS models, $Y\sim \beta_0+\beta_1*A$. For example, patients in nodes 4 and 6 have estimated treatment effects of 0.47 and 0.06 respectively. Subgroup predictions for the train/test set can be found in the "out.train" and "out.test" data-sets.
plot(res0$submod.fit$mod, terminal_panel = NULL)
table(res0$out.train$Subgrps)
table(res0$out.test$Subgrps)
For any parameter estimation approache, subgroup-specific estimates tend to be overly positive or negative, as the same data that trains the subgroup model is used for parameter estimation. Resampling, such as bootstrapping, is generally perferred for "honest" treatment effect estimates (more details below).
For continuous and binary data, the default parameter estimation approach is param="dr" (double robust estimator). This approach incorporates regression estimates, which could potentially increase the efficiency of the point-estimate. Let $k=1,...,K$ index the $K$ identified subgroups with corresponding rules $S_1,...,S_K$. Next, let $E(Y|X=x,A=a) = \mu(x, a)$ correspond to the outcome regression model(s) with estimates $\hat{\mu}(x, a)$. These estimates come directly from the fitted PLE model(s), in this case, treatment-specific random forest models. Define the "pseudo-outcomes" as:
$$Y^{\star}_i = \frac{AY - (A-\hat{\pi}(x))\hat{\mu}(a=1,x)}{\hat{\pi}(x)} - \frac{(1-A)Y - (A-\hat{\pi}(x))\hat{\mu}(a=0,x)}{1-\hat{\pi}(x)}$$
where $\pi(x)=P(A=1|X)$, or the treatment assignment probability for an individual. In a randomized controlled trial, this can be replaced by the marginal probability, $P(A=1|X)$. For each discovered subgroup ($k=1,...,K$), the treatment effect (or risk difference) and associated SE are then: can be estimated by averaging the patient-specific treatment effect estimates (PLEs): $$\hat{\theta}k = \sum{i \in S_k} Y^{\star}i$$ $$SE(\hat{\theta}_k) = \sqrt{ n_k ^ {-2} \sum{i \in S_k} \left( Y^{\star}i-\hat{\theta}(x_i) \right)^2}$$ CIs can then be formed using Z-intervals. For example, a two-sided 95\% CI, $CI{\alpha}(\hat{\theta}{k}) = \left[\hat{\theta}{k} \pm 1.96*SE(\hat{\theta}_k) \right]$
Moving back to the PRISM outputs, for any of the provided "param" options, a key output is the object "param.dat". By default, "param.dat" contain point-estimates, standard errors, lower/upper confidence intervals (depends on alpha_s and alpha_ovrl) and p-values. This output feeds directly into previously shown default ("tree") plot.
## Overall/subgroup specific parameter estimates/inference
res0$param.dat The hyper-parameters for the individual steps of PRISM can also be easily modified. For example, "glmnet" by default selects covariates based on "lambda.min", "ranger" requires nodes to contain at least 10% of the total observations, and "lmtree" requires nodes to contain at least 10% of the total observations. See below for a different set of hyper-parameters. # PRISM Default: glmnet, ranger, lmtree, dr # # Change hyper-parameters # res_new_hyper = PRISM(Y=Y, A=A, X=X, filter.hyper = list(lambda="lambda.1se"), ple.hyper = list(min.node.pct=0.05), submod.hyper = list(minsize=200), verbose=FALSE) summary(res_new_hyper) # Example: Binary Outcome with Binary Treatment Consider a binary outcome (ex: % overall response rate) with a binary treatment (study drug vs standard of care). The estimand of interest is the risk difference,$\theta_0 = E(Y|A=1)-E(Y|A=0)$. Similar to the continous example, we simulate binomial data where roughly 30\% of the patients receive no treatment-benefit for using$A=1$vs$A=0$. Responders vs non-responders are defined by the continuous predictive covariates$X_1$and$X_2$for a total of four subgroups. Subgroup treatment effects are:$\theta_{1} = 0$($X_1 \leq 0, X_2 \leq 0$),$\theta_{2} = 0.11 (X_1 > 0, X_2 \leq 0)$,$\theta_{3} = 0.21 (X_1 \leq 0, X2 > 0$),$\theta_{4} = 0.31 (X_1>0, X_2>0)$. For binary outcomes (Y=0,1), the default settings are: filter="glmnet", ple="ranger", submod="glmtree"" (GLM MOB with identity link), and param="dr". dat_bin = generate_subgrp_data(family="binomial", seed = 5558) Y = dat_bin$Y
X = dat_bin$X # 50 covariates, 46 are noise variables, X1 and X2 are truly predictive A = dat_bin$A # binary treatment, 1:1 randomized
res0 = PRISM(Y=Y, A=A, X=X)
summary(res0)
# Example: Survival Outcome with Binary Treatment
Survival outcomes are also allowed in PRISM. The default settings use glmnet to filter ("glmnet"), ranger patient-level estimates ("ranger"; for survival, the output is the restricted mean survival time treatment difference), "lmtree" (log-rank score transformation on outcome Y, then fit MOB OLS) for subgroup identification, and subgroup-specific cox regression models). Another subgroup option is to use "ctree"", which uses the conditional inference tree (ctree) algorithm to find subgroups; this looks for partitions irrespective of treatment assignment and thus corresponds to finding prognostic effects.
# Load TH.data (no treatment; generate treatment randomly to simulate null effect) ##
data("GBSG2", package = "TH.data")
surv.dat = GBSG2
# Design Matrices ###
Y = with(surv.dat, Surv(time, cens))
X = surv.dat[,!(colnames(surv.dat) %in% c("time", "cens")) ]
set.seed(6345)
A = rbinom(n = dim(X)[1], size=1, prob=0.5)
# Default: glmnet ==> ranger (estimates patient-level RMST(1 vs 0) ==> mob_weib (MOB with Weibull) ==> cox (Cox regression)
res_weib = PRISM(Y=Y, A=A, X=X)
summary(res_weib)
plot(res_weib)
# Resampling
Resampling methods are also a feature in PRISM. Bootstrap (resample="Bootstrap"), permutation (resample="Permutation"), and cross-validation (resample="CV") based-resampling are included. Resampling can be used for obtaining de-biased or "honest" subgroup estimates, inference, and/or probability statements. For each resampling method, the sampling mechanism can be stratified by the discovered subgroups (default: stratify=TRUE). To summarize:
Bootstrap Resampling
Given observed data $(Y, A, X)$, fit $PRISM(Y,A,X)$. Based on the identified $k=1,..,K$ subgroups, output subgroup assignment for each patient. For the overall population $k=0$ and each subgroup ($k=0,...,K$), store the associated parameter estimates ($\hat{\theta}{k}$). For $r=1,..,R$ resamples with replacement ($(Y_r, A_r, X_r)$), fit $PRISM(Y_r, A_r, X_r)$ and obtain new subgroup assignments $k_r=1,..,K_r$ with associated parameter estimates $\hat{\theta}{k_r}$. For subjects $i$ within subgroup $k_r$, note that everyone has the same assumed point-estimate, i.e., $\hat{\theta}{k_r}=\hat{\theta}{ir}$. For resample $r$, the bootstrap estimates based for the original identified subgroups ($k=0,...,K$) are calculated respectively as: $$\hat{\theta}{rk} = \sum{k_r} w_{k_r} \hat{\theta}{k_r}$$ where $w{k_r} = \frac{n(k \cap k_r)}{\sum_{k_r} n(k \cap k_r)}$, or the # of subjects that are in both the original subgroup $k$ and the resampled subgroup $k_r$ divided by the total #. The bootstrap mean estimate and standard error, as well as probability statements, are calculated as: $$\tilde{\theta}{k} = \frac{1}{R} \sum_r \hat{\theta}{rk}$$ $$SE(\hat{\theta}{k})_B = \sqrt{ \frac{1}{R} \sum_r (\hat{\theta}{rk}-\tilde{\theta}{k})^2 }$$ $$\hat{P}(\hat{\theta}{k}>c) = \frac{1}{R} \sum_r I(\hat{\theta}{rk}>c)$$ If resample="Bootstrap", the default is to use the bootstrap smoothed estimates, $\tilde{\theta}{k}$, along with percentile-based CIs (i.e. 2.5,97.5 quantiles of bootstrap distribution). Bootstrap bias is also calculated, which can be used to assess the bias of the initial subgroup estimates.
Returning to the survival example, see below for an example of PRISM with 50 bootstrap resamples (for increased accuracy, use >1000). The bootstrap mean estimates, bootstrap standard errors, bootstrap bias, and percentile CI correspond to "est_resamp", "SE_resamp", "bias.boot", and "LCL.pct"/"UCL.pct" respectively. A density plot of the bootstrap distributions can be viewed through the plot(...,type="resample") option.
res_boot = PRISM(Y=Y, A=A, X=X, resample = "Bootstrap", R=50, ple="None")
summary(res_boot)
# Plot of distributions #
plot(res_boot, type="resample", estimand = "HR(A=1 vs A=0)")+geom_vline(xintercept = 1)
Cross-Validation
Cross-validation resampling (resample="CV") also follows the same general procedure as bootstrap resampling. Given observed data $(Y, A, X)$, fit $PRISM(Y,A,X)$. Based on the identified $k=1,..,K$ subgroups, output subgroup assignment for each patient. Next, split the data into $R$ folds (ex: 5). For fold $r$ with sample size $n_r$, fit PRISM on $(Y[-r],A[-r], X[-r])$ and predict the patient-level estimates and subgroup assignments ($k_r=1,...,K_r$) for patients in fold $r$. The data in fold $r$ is then used to obtain parameter estimates for each subgroup, $\hat{\theta}{k_r}$. For fold $r$, estimates and SEs for the original subgroups ($k=1,...,K$) are then obtained using the same formula as with bootstrap resampling, again, denoted as ($\hat{\theta}{rk}$, $SE(\hat{\theta}_{rk})$). This is repeated for each fold and "CV" estimates and SEs are calculated for each identified subgroup. Let $w_r = n_r / \sum_r n_r$, then:
$$\hat{\theta}{k,CV} = \sum w_r * \hat{\theta}{rk}$$ $$SE(\hat{\theta}k){CV} = \sqrt{ \sum_{r} w_{r}^2 SE(\hat{\theta}{rk})^2 }$$ CV-based confidence intervals can then be formed, $\left[\hat{\theta}{k,CV} \pm 1.96*SE(\hat{\theta}k){CV} \right]$.
# Conclusion
Overall, the StratifiedMedicine package contains a variety of tools ("filter_train", "ple_train", "submod_train", and "PRISM") and plotting features ("plot_dependence", "plot_importance", "plot_ple") for the exploration of heterogeneous treatment effects. Each tool is also customizable, allowing the user to plug-in specific models (for example, xgboost with built-in hyper-parameter tuning). More details on creating user-specific models can be found in the "User_Specific_Models_PRISM" vignette User_Specific_Models. The StratifiedMedicine R package will be continually updated and improved.
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StratifiedMedicine documentation built on Sept. 5, 2021, 5:07 p.m. | 2021-10-22 04:46:14 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7856879830360413, "perplexity": 9538.310344254902}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585450.39/warc/CC-MAIN-20211022021705-20211022051705-00174.warc.gz"} |
https://www.gradesaver.com/textbooks/science/physics/fundamentals-of-physics-extended-10th-edition/chapter-3-vectors-problems-page-57/8b | ## Fundamentals of Physics Extended (10th Edition)
$3.189\,km$
We have three vectors representing the three parts of the path: $\vec{a} = 3.1\,km\, \hat{i}$ $\vec{b} = -2.4\,km\, \hat{j}$ $\vec{c} = -5.2\,km\, \hat{i}$ We add these up to get the total displacement: $\vec{t} = -2.1\,km\,\hat{i} - 2.4\,km\,\hat{j}$ The total distance of the displacement is: $|\vec{t}| = \sqrt{(-2.1\,km)^2+(-2.4\,km)^2} = 3.189\,km$. | 2018-10-19 17:22:29 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8989396095275879, "perplexity": 432.5590029773007}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583512421.5/warc/CC-MAIN-20181019170918-20181019192418-00043.warc.gz"} |
https://ltwork.net/with-setting-smart-goals-you-should-choose-only-a-few-goals--5600554 | With setting smart goals, you should choose only a few goals on which to focus. these goals should be
Question:
With setting smart goals, you should choose only a few goals on which to focus. these goals should be results-oriented and
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After Florida became a territory, who settled there? A. enslaved people who had run away hoping for freedom B. the Spanish C. people from the Northeast in search of a better climate D. planters from the South who needed more fertile land please answer ASAP ):...
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Doña josefa tiene una huerta de naranjas distribuidos en 15 filas cada una con 11 árboles cuántos árboles tienen en total ?...
Alison swims 75 feet in 12 seconds. What is he rate to the nearest mile per hour? Please explain all the steps.
Alison swims 75 feet in 12 seconds. What is he rate to the nearest mile per hour? Please explain all the steps....
Sudzy introduced a new formula for their hair conditioner
Sudzy introduced a new formula for their hair conditioner...
Mrs. Anderton is giving a test in her third-period class. She has decided to record the amount of time
Mrs. Anderton is giving a test in her third-period class. She has decided to record the amount of time that each student takes to finish the test (in minutes) and compare that to the grade each student receives on the test (out of 100). A plot of her results is below. Which of the following does thi...
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https://zbmath.org/?q=0963.54005 | # zbMATH — the first resource for mathematics
$$K_W$$ does not imply $$K_W^*$$. (English) Zbl 0963.54005
Summary: We prove that the cyclic monotonically normal space $$T$$ of M. E. Rudin is a $$K_W$$-space which is not a $$K^*_W$$-space. This answers a question in [the author, ibid. 51, No. 1, 109-117 (1994; Zbl 0861.54012)]. In order to do this, we first prove that if a space $$X$$ has $$D^*(\mathbb{R};\leq)$$ then $$X$$ is a $$K_W$$-space (it is well known that $$X$$ is also a $$K_1$$-space; this does not necessarily mean that $$X$$ is a $$K_{1W}$$-space).
##### MSC:
54C30 Real-valued functions in general topology 54C20 Extension of maps
##### Keywords:
$$K_W$$-space; $$K_1$$-space
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http://openstudy.com/updates/503fea41e4b0ece102e0f195 | ## AEB047 Group Title How do you do 2/5x+3/7=1-4/7x ? 2 years ago 2 years ago
2x/5 + 3/7 = 1 - 4x/ 7 14x + 15 = 35 -20x 24x = 20 x = 20/24 = 5/6
2. YesterdayiSaidTomorow
cross multi[ply
The answer is$x=\frac{10}{17}$ | 2014-11-27 17:33:55 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3955818712711334, "perplexity": 3066.153204896773}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-49/segments/1416931008919.73/warc/CC-MAIN-20141125155648-00061-ip-10-235-23-156.ec2.internal.warc.gz"} |
https://engineering.stackexchange.com/tags/structural-engineering/new | We’re rewarding the question askers & reputations are being recalculated! Read more.
# Tag Info
0
We have the force $(\sqrt 2/2)* F \$ acting horizontally to the right at point H and horizontally to the left at point E , both horizontal components of the cord tension, F making a moment of $\ M=(\sqrt 2/2)* F*a .$ Now we equate sum of moments about A to zero and find reaction at B. \Sigma M_A =0 \quad= \frac{(\sqrt 2/2)* F*a + F*2a}{2a} +R_B = ...
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NCBCLL is IBC Code for Non Concurrent Bottom Chord Live Load, used for uninhabitable/inhabitable attic space live loads @ 10/20 PSF through Alpine ITW
2
You can take have to use temperature dependent material properties for the e-modulus and Poisson's ratio. This scan be achieved with ´MPTEMP' followed by ´MPDATA´. Linear interpolation is used by the software. Example with fictious values: MPTEMP,1, 20, 100, 150, 400 MPDATA,EX,1,, 204e9, 197e9, 193e9, 178e9 There is also the option to input a ...
0
A quick check on the net shows the 5052 has a lower strength than the 6061 , mostly caused by heat-treatment. Tensile is 25% lower, yield is 30 % lower ; nominal properties.
1
Look at pairs of glasses as they are wire holding a lens.
1
Put a ring around the top 1/3 of the object then have 4 guy wires coming down to two bars fixed to the crossbar, one on each side.
0
Let me raise an example: By Ultimate Capacity, my slab thickness required is 80mm for a particular span X meter. Which my slab would not fail to carry the load into the beam and column. By Serviceability Capacity, my slab thickness required is 150mm for a particular span x meter. Which the deflection may not exceed the desired displacement. So, one hand I ...
1
Practically, the geologist will do boring at different location at the site, we call it Bore Hole. Based on their preliminary assessment, they will decide the depth of boring. Or by experience or trial & error. Whatever it is, the main thing we are looking for is hard layer. For it to be hard rock or hard dense sand layer. Basically we can get a soil ...
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By hand? I believe you have to use matrixes to include all the formulae by superposition, then to solve the matrixes using our friendly Mathematics by hand instead of computer.
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The answer to this is that, engineers have to come out with different idea / creative solution to increase the span of the slab. I believe there are many other solutions in other places that not mentioned in the following: 1.) Cold-form steel trusses 2.) Post/Pre-tensioning concrete structure 3.) Arc shape structures support with tensioning rod. All ...
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Probability of Event Exceedance of A Certain Year, lets say N years; is the occurrence of events gained from statistic for a certain country or area for a certain event. Which in this case is flooding. And the Annual maximum daily flow was recorded in that particular year of occurrence. In the topic of civil engineering particularly, as we discuss, we may ...
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In fact the twist angle is assumed to be constant from the center of the cylinder out to its circumference. If we call the twist caused strain at any distance r from the center $ds, \ then\ \theta= \frac{ds}{r} \quad and \ ds=\theta*r$ Shear is $t=G*\epsilon= G*\theta*r \quad \text{or y in your equation}= G*yr$
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First, picture two disks, placed parallel, and close together. Draw a line through the center of each disk. Line up the lines. Are all the points on the lines the same distance apart? Now rotate one disk by a few degrees. Are all points on the lines the same distance apart? Have the lines drawn further apart at the center of the disks, or at the edges? ...
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Ice has a volume about 10% bigger than water, so an inner tube is too small to do anything - unless it floats, see below. Containers burst when the ice forms a plug and the trapped water then freezes and expands. One way to avoid that is float a beach ball in the tank. When the top of the water freezes over, the pressure beneath it will crack the ice round ...
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If you leave roughly 10% on top empty it should be okay. Ice increases in volume by approximately 8%. Leaving a half deflated bike tube will hardly help. Because ice formation starts from surface of the tank and by the time it gets to the tube it is too late. Most of the plastic containers can flex a bit to accommodate icing if you leave the top empty. But ...
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I think this is the annual flood chart. The USGS earthquake hazard and probability maps are colored and have contour lines. Exceedance probability in Y years. (This part is mathematical) The expected number, n of exceedances in Y years is n = Y times r, the annual rate of exceedance. Assumption: The rate of earthquake occurrence in time is ...
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https://proofwiki.org/wiki/Transitive_Subgroup/Examples | Transitive Subgroup/Examples
$n$-Cycle in $S_n$
Consider the subgroup $H$ of $S_n$ generated by the cyclic permutation $\tuple {1, 2, \ldots, n}$.
Then $H$ is a transitive subgroup . | 2022-05-18 12:09:51 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9585511088371277, "perplexity": 210.40910623730394}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662522270.37/warc/CC-MAIN-20220518115411-20220518145411-00745.warc.gz"} |
https://www.physicsforums.com/threads/eigenvalue-factorization-and-matrix-substitution.388528/ | # Homework Help: Eigenvalue Factorization and Matrix Substitution
1. Mar 21, 2010
### the_dialogue
In my literature reviews I found a few things that I can't quite understand.
1. The problem statement, all variables and given/known data
I have the following equation:
http://img717.yfrog.com/img717/6416/31771570.jpg [Broken]
I'm told that by using the eigenvalue factorization:
http://img89.yfrog.com/img89/760/83769756.jpg [Broken]
, I can change the first equation to:
http://img28.imageshack.us/img28/5023/84802099.jpg [Broken]
2. The attempt at a solution
I tried changing Equation 2 to just be (A^T)A and then subbing into the first equation, but I can't quite do anything with those inverses.
Also, what does the exponent of '-2' mean in the context of a 4x4 matrix? Lastly, what is matrix U?
Thank you!
Last edited by a moderator: May 4, 2017
2. Mar 21, 2010
### gabbagabbahey
I think it's probably easiest to start from $\textbf{p}^{T}(\mathbf{\Lambda}+\lambda\textbf{I})^{-2}\textbf{q}=0$ and work your way backwards instead.
$$\textbf{C}^{-2}\equiv\textbf{C}^{-1}\textbf{C}^{-1}$$
You simply square the inverse of the matrix.
3. Mar 21, 2010
### the_dialogue
I'll give it a try gabbagabbahey. Thanks.
Any idea what the matrix "U" is?
4. Mar 21, 2010
### gabbagabbahey
It's the invertible matrix which relates the matrix $\textbf{A}^{T}\textbf{A}\mathbf{\Sigma}$ to the diagonal matrix $\mathbf{\Lambda}$ via a similarity transform. Its columns will be the eigenvectors of $\textbf{A}^{T}\textbf{A}\mathbf{\Sigma}$.
See http://en.wikipedia.org/wiki/Diagonalizable_matrix for a refresher on matrix diagonalization.
5. Mar 21, 2010
### the_dialogue
Yes I recall now. Thanks! | 2018-12-17 15:44:51 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6756699085235596, "perplexity": 1671.3793998157992}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376828507.84/warc/CC-MAIN-20181217135323-20181217161323-00156.warc.gz"} |
http://anthony-zhang.me/University-Notes/MATH138/MATH138.html | # MATH138
Calculus II
Instructor: Matthew Scott
Section 002
Email: mscott@uwaterloo.ca
Office: MC 6114
Office hours: Mondays 11:30am-12:15pm, 3:00pm-3:30pm
Tutorial: Wednesdays 3:30pm in MC 2035, starting Jan. 15
# 6/1/14
Assignments are due every Friday at 3:30pm. They are posted on LEARN.
My own drop box is box 5, slot 12
Midterm is Feb. 24, 7-9pm.
Lecture notes are on the course website, posted every week. The password to the website is "euler". ;wip: get these every week
Course outline:
• Techniques of integration and applications.
• Extensions of single variable calculus - differential equations, vector calculus, etc.
• Taylor polynomials
Sequence: 1, 2, 3, \ldots Series: 1 + 2 + 3 + \ldots
Taylor polynomials are series of functions, which allow us to represent almost any function as a polynomial. This is the most powerful technique in applied mathematics - science and mathematics.
By the way, \dee x is called an infinitesmal.
## Techniques of Integration
Techniques of integration rewrite functions in forms that we can't integrate into forms that we can integrate, or simpler forms that we might be able to apply other techniques on.
### Method of substitution
The method of substitution is based on choosing a subexpression u, and then integrating with respect to it. If u is chosen carefully, it is occasionally possible to simplify the integral.
This method works best when a function and its derivative appear in the integrand.
This works with expressions of the form \int f(x) \frac{\dee f}{\dee x} \dee x.
Let u = f(x). Since \frac{\dee u}{\dee x} = \frac{\dee u}{\dee x}, \dee u = \frac{\dee u}{\dee x} \dee x (multiply both sides by \dee x). This is possible because of infinismals, which work in wierd and wonderful ways. So \dee x = \frac{1}{\frac{\dee u}{\dee x}} \dee u.
So \int f(x) \frac{\dee}{\dee x} f(x) \dee x = \int u \frac{\dee u}{\dee x} \dee x = \int u \frac{\dee u}{\dee x} \frac{1}{\frac{\dee u}{\dee x}} \dee u = \int u \dee u = \frac{u^2}{2} + c = \frac{f(x)^2}{2} + c.
Even if we can't simplify it far enough to get an antiderivative, integration by substitution can still considerably simplify an integrand into something more manageable.
What about definite integrals? We need to be aware of the limits of integration when doing the variable switch:
We are given \int_a^b f(x) \frac{\dee}{\dee x} f(x) \dee x.
Let u = f(x). We know that \dee x = \frac{1}{\frac{\dee u}{\dee x}} \dee u.
So \int_a^b f(x) \frac{\dee}{\dee x} f(x) \dee x = \int_{u(a)}^{u(b)} u \frac{\dee u}{\dee x} \dee x = \int_{u(a)}^{u(b)} u \frac{\dee u}{\dee x} \frac{1}{\frac{\dee u}{\dee x}} \dee u = \int_{u(a)}^{u(b)} u \dee u = \evalat{\frac{u^2}{2}}_{u(a)}^{u(b)} + c = \evalat{\frac{f(x)^2}{2}}_a^b + c.
Basically, if we can find a factor of the integrand u such that its derivative also appears in the integrand, the method of substitution can get rid of the derivative.
Evaluate \int \cos x \sin x \dee x
Let u = \sin x. So \dee u = \cos x \dee x.
So \int \cos x \sin x \dee x = \int \sin x \dee u = -\int u \dee u.
Clearly, \int u \dee u = \frac{u^2}{2} + c = \frac{\sin^2 x}{2} + c.
Substitution is useful when we can figure out what a useful substitution would be.
One clue that the substitution is a good choice is if the derivative of the substitution appears in the numerator of the function.
How do we prove that \int_1^x \frac{1}{t} = \ln x? How do we know that both have the same properties:
• Property 1 - multiplication rule: \ln ab = \ln a + \ln b
• Property 2 - power rule: \ln a^r = r \ln a)
We want to prove property 1:
Select some a and b.
We want to prove that \int_1^{ab} \frac{1}{t} \dee t = \int_1^a \frac{1}{t} \dee t + \int_1^b \frac{1}{t} \dee t.
By linearity of integrals, \int_1^{ab} \frac{1}{t} \dee t = \int_1^a \frac{1}{t} \dee t + \int_a^{ab} \frac{1}{t} \dee t.
Let u = \frac{t}{a}, so t = ua. Then \dee t = \frac{\dee t}{\dee u} \dee u = a \dee u.
So \int_a^{ab} \frac{1}{t} \dee t = \int_\frac{a}{a}^\frac{ab}{a} \frac{1}{ua} a \dee u = \int_1^b \frac{1}{u} \dee u.
So \int_1^{ab} \frac{1}{t} \dee t = \int_1^a \frac{1}{t} \dee t + \int_a^{ab} \frac{1}{t} \dee t = \int_1^a \frac{1}{t} \dee t + \int_1^b \frac{1}{u} \dee u.
Dummy variables can be swapped for any value, even functions like f(t).
# 8/1/14
To prove property 2, we need to show that \int_1^{a^r} \frac{1}{u} \dee u = r \int_1^a \frac{1}{t} \dee t. To do this, we find a substitution that results in the desired limits of integration.
To prove property 2:
Select some u.
We want to prove that \int_1^{a^r} \frac{1}{t} \dee t = r \int_1^a \frac{1}{t} \dee t.
Let u = t^\frac{1}{r}, so t = u^r. Then \dee t = \frac{\dee t}{\dee u} \dee u = \frac{\dee}{\dee u} u^r \dee u = ru^{r - 1} \dee u.
So \int_1^{a^r} \frac{1}{t} \dee t = \int_{1^\frac{1}{r}}^{{a^r}^\frac{1}{r}} \frac{1}{u^r} ru^{r - 1} \dee u = \int_1^a \frac{1}{u} r \dee u = r \int_1^a \frac{1}{u} \dee u.
### Differentiating with respect to functions
As an aside, we can actually differentiate things with respect to functions.
The important property is that \frac{\dee x}{\dee b(x)} = \frac{1}{\frac{\dee b(x)}{\dee x}}.
Note that \frac{\dee a(x)}{\dee b(x)} = \frac{\dee a(x)}{\dee x} \frac{\dee x}{\dee b(x)} = \frac{\dee a(x)}{\dee x} \frac{1}{\frac{\dee b(x)}{\dee x}}.
So \frac{\dee a(x)}{\dee b(x)} = \frac{\frac{\dee a(x)}{\dee x}}{\frac{\dee b(x)}{\dee x}} = \frac{a'(x)}{b'(x)}.
### Integration by Parts
We often have integrands that are products. For example, \int x \sin x \dee x
It would be nice to be able to differentiate or integrate just one of the factors rather than having to do the whole thing.
Recall the product rule: \frac{\dee}{\dee x} (a(x) b(x)) = (\frac{\dee}{\dee x} a(x)) b(x) + a(x) \frac{\dee}{\dee x} b(x).
Move the terms around: (\frac{\dee}{\dee x} a(x)) b(x) = \frac{\dee}{\dee x} (a(x) b(x)) - a(x) \frac{\dee}{\dee x} b(x).
If we integrate both sides with respect to x, we get: \int (\frac{\dee}{\dee x} a(x)) b(x) = \int \frac{\dee}{\dee x} a(x) b(x) - \int a(x) \frac{\dee}{\dee x} b(x).
By FTC2, \int (\frac{\dee}{\dee x} a(x)) b(x) = a(x) b(x) + c - \int a(x) \frac{\dee}{\dee x} b(x).
Since there would also be a c term from the second integral, we don't need to write it. So \int (\frac{\dee}{\dee x} a(x)) b(x) \dee x = a(x) b(x) - \int a(x) \frac{\dee}{\dee x} b(x) \dee x.
This is the integration by parts rule.
Using the same idea, we find that \int_a^b (\frac{\dee}{\dee x} a(x)) b(x) = \evalat{a(x) b(x)}_{x = a}^{x = b} - \int_a^b a(x) \frac{\dee}{\dee x} b(x).
When we use the rule, we want to identify two parts of the product such that one of the parts (b) is the factor that becomes simpler when differentiated, and the other part (a) is chosen as the integral of the factor.
Simplify \int x \sin x \dee x:
x is much simpler when differentiated. Let b(x) = x.
\sin x is integrable. Let a(x) = \int \sin x = -\cos x.
So \int x \sin x = \int x \frac{\dee}{\dee x} (-\cos x) = -x \cos x - \int (-\cos x) 1 \dee x, by integration by parts.
So -x \cos x - \int (-\cos x) 1 \dee x = -x \cos x + \sin x + c.
We can check the answer by differentiating: \frac{\dee}{\dee x} (-x \cos x + \sin x + c) = x \sin x.
When we use integration by parts, we differentiate one factor, and integrate the other. Then, we can apply the integration by parts formula and obtain a possibly simpler version.
Simplify \int_0^1 x e^{-x} \dee x:
\displaystyle \begin{aligned} \int_0^1 x e^{-x} \dee x &= \int_0^1 x \frac{\dee}{\dee x} (-e^{-x}) \dee x \\ &= \evalat{x (-e^{-x})}_0^1 - \int_0^1 \left(\frac{\dee}{\dee x} x\right) (-e^{-x}) \dee x \\ &= \evalat{x (-e^{-x})}_0^1 + \int_0^1 e^{-x} \dee x \\ &= -\frac{2}{e} \end{aligned}
Simplify \int_0^1 x^n (\ln x)^n \dee x:
;wip
We often write the integration by parts formula in different forms: \int a \dee b = ab - \int b \dee a, or \int f'(x) g(x) \dee x = f(x) g(x) - \int f(x) g'(x) \dee x.
An interesting trick is that we can use integration by parts to integrate anything, since everything is a product of itself and 1.
For example, consider \int \arcsin x \dee x:
Clearly, \int 1 \cdot \arcsin x \dee x = \int (\frac{\dee}{\dee x} x) \arcsin x \dee x.
Using integration by parts, with a(x) = x, b(x) = \arcsin x, \int (\frac{\dee}{\dee x} x) \arcsin x \dee x = x \arcsin x - \int \frac{x}{\sqrt{1 - x^2}} \dee x.
We can now use integration by substitution with u = 1 - x^2 to simplify \int \frac{x}{\sqrt{1 - x^2}} \dee x.
Clearly, \dee x = -\frac{1}{2x} \dee x and \int \frac{1}{\sqrt{1 - x^2}} \dee x = -\frac{1}{2} \int \frac{1}{\sqrt{u}} \dee u = -\frac{1}{2} \int u^{-\frac{1}{2}} \dee u = -\frac{1}{2} 2 \sqrt{u} + c = -\sqrt{1 - x^2} + c.
So \int \arcsin x \dee x = x \arcsin x - \int \frac{x}{\sqrt{1 - x^2}} \dee x = x \arcsin x + \sqrt{1 - x^2} + c.
Simplify \int \arccos x \dee x:
Using integration by parts, with a(x) = x, b(x) = \arccos x, \int (\frac{\dee}{\dee x} x) \arccos x \dee x = x \arccos x - \int \frac{x}{\sqrt{1 - x^2}} \dee x.
From the previous question, we know that \int \frac{x}{\sqrt{1 - x^2}} \dee x = -\sqrt{1 - x^2} + c.
So \int (\frac{\dee}{\dee x} x) \arccos x \dee x = x \arccos x - \int \frac{x}{\sqrt{1 - x^2}} \dee x = x \arccos x + \sqrt{1 - x^2} + c.
Simplify \int \ln x \dee x:
Using integration by parts, with a(x) = x, b(x) = \arccos x. \int (\frac{\dee}{\dee x} x) \ln x \dee x = x \ln x - \int \frac{x}{x} \dee x = x \ln x - x.
# 10/1/14
The integration by parts rule is also written as \int v \dee u = uv - \int u \dee v:
Clearly, \dee a = \frac{\dee a}{\dee x} \dee x and \dee b = \frac{\dee b}{\dee x} \dee x.
We know that \int (\frac{\dee}{\dee x} a(x)) b(x) \dee x = a(x) b(x) - \int a(x) \frac{\dee}{\dee x} b(x) \dee x and so \int \frac{\dee a}{\dee x} b(x) \dee x = a(x) b(x) - \int a(x) \frac{\dee b}{\dee x} \dee x.
So \int b(x) \dee a = a(x) b(x) - \int a(x) \dee b.
We can also write it as \int f(x) g(x) \dee x = \int f(x) \dee x g(x) - \int \int f(x) \dee x \frac{\dee g}{\dee x} \dee x.
### Getting Back What We Started With
Consider I = \int e^{-x} \sin x:
We choose \frac{\dee}{\dee x} e^{-x} = -e^{-x} and \int \sin x \dee x = -\cos x.
So I = \int e^{-x} \frac{\dee}{\dee x} (-\cos x) \dee x = e^{-x} (-\cos x) - \int e^{-x} \cos x.
We choose \frac{\dee}{\dee x} (-e^{-x}) = e^{-x} and \int \cos x \dee x = \sin x.
Clearly, \int e^{-x} \cos x = \int e^{-x} \frac{\dee}{\dee x} \sin x = e^{-x} \sin x - \int (-e^{-x}) \sin x = e^{-x} \sin x + I.
So I = -e^{-x} \cos x - \int e^{-x} \cos x = -e^{-x} \cos x - e^{-x} \sin x - I.
So 2I = -e^{-x} (\cos x + \sin x) and I = -\frac{1}{2}e^{-x} (\cos x + \sin x).
Basically, we used integration by parts to rewrite the expression, then used integration by parts again to simplify the remaining integral. In doing so we obtained a result that contained the original expression, which we substituted and solved algebraically.
This technique is only really useful for problems of the form \int e^{ax} \sin bx \dee x or \int e^{ax} \cos bx \dee x.
Simplify \int_0^1 (-\ln x)^n \dee x, n \in \mb{N}:
We will try it for the first few values of n.
For n = 1, \int_0^1 (-\ln x)^n \dee x = -\int_0^1 1 \ln x \dee x = -\int_0^1 \ln x \frac{\dee}{\dee x} x = -\evalat{x \ln x}_0^1 + \int_0^1 x \frac{1}{x} = -1 \ln 1 + 0 \ln 0 + 1 (indeterminate form 0 \cdot -\infty).
This is a thing known as an improper integral (covered later in the course).
We can solve it because what we really want is \lim_{x \to 0} x \ln x = \lim_{x \to 0} \frac{\ln x}{\frac{1}{x}} \lH \lim_{x \to 0} \frac{\frac{1}{x}}{-\frac{1}{x^2}} = \lim_{x \to 0} (-x) = 0 rather than the value 0 \ln 0.
So -\int_0^1 \ln x \dee x = -1 \ln 1 + 0 + 1 = 1.
If we repeat this for n = 2, 3, \ldots, we find that \int_0^1 (-\ln x)^n \dee x = n!, n \in \mb{N}, yet it also works for fractional values of n - this is the generalized factorial over positive real numbers!
;wip: use induction
# 13/1/14
## Trigonometric Substitution
This is integration by substitution with the substitution using trignonometric functions.
We use trigonometric functions in our substitutions and use trigonometric identities to simplify the integrand.
We will need to know:
• Pythagorean theorem: \cos^2 \theta + \sin^2 \theta = 1
• Half angle cosine formula: \cos^2 \theta = \frac{1}{2}(1 + \cos 2\theta)
• Half angle sine formula: \sin^2 \theta = \frac{1}{2}(1 - \cos 2\theta)
• Pythagorean theorem divided by \cos^2 \theta: 1 + \tan^2 \theta = \sec^2 \theta
We usually need this method when we have a square root of a quadratic expression.
Common substitutions:
• Use 1 - \sin^2 \theta = \cos^2 \theta to use \sqrt{1 - x^2} = \sqrt{\cos^2 \theta} = \cos \theta via x = \sin \theta.
• Use 1 + \tan^2 \theta = \sec^2 \theta to use \sqrt{1 + x^2} = \sqrt{\sec^2 \theta} = \sec \theta via x = \tan \theta.
• Use \sec^2 \theta - 1 = \tan^2 \theta to use \sqrt{x^2 - 1} = \sqrt{\tan^2 \theta} = \tan \theta via x = \sec \theta.
Using this technique, we can replace \sqrt{\pm a^2 \pm x^2} with trigonometric functions, and then use trigonometric identities on them. Afterwards, we might simplify by changing the trigonometric functions back into square roots.
Derivatives:
• \cos' \theta = -\sin \theta where -\frac{\pi}{2} \le \theta < \frac{\pi}{2}
• \sin' \theta = \cos \theta where -\frac{\pi}{2} \le \theta < \frac{\pi}{2}
• \tan' \theta = \sec^2 \theta where 0 \le \theta < \frac{\pi}{2}
Consider \int \frac{1}{x^2 \sqrt{x^2 + 4}} \dee x:
We can draw a triangle with hypotenuse \sqrt{x^2 + 4}, opposite side x, and adjacent side 2 representing this value.
Let \theta represent the angle between the adjacent side and the hypotenuse. By the Pythagorean theorem. Then x = 2 \tan \theta. Then \dee x = 2 \sec^2 \theta \dee \theta.
Clearly, \int \frac{1}{x^2 \sqrt{x^2 + 4}} \dee x = \int \frac{1}{4 \tan^2 \theta \sqrt{4 \tan^2 \theta + 4}} 2 \sec^2 \theta \dee \theta = \int \frac{1}{4 \tan^2 \theta \sqrt{4 (\tan^2 \theta + 1)}} 2 \sec^2 \theta \dee \theta = \int \frac{1}{4 \tan^2 \theta 2 \sec \theta} 2 \sec^2 \theta \dee \theta = \frac{1}{4} \int \frac{\sec \theta}{\tan^2 \theta} \dee \theta = \frac{1}{4} \int \frac{\cos \theta}{\sin^2 \theta} \dee \theta.
Let u = \sin \theta. Then \dee x = \frac{1}{\cos \theta} \dee \theta.
So \frac{1}{4} \int \frac{\cos \theta}{\sin^2 \theta} \dee \theta = \frac{1}{4} \int \frac{\cos \theta}{u^2} \frac{1}{\cos \theta} \dee u = \frac{1}{4} \int \frac{1}{u^2} \dee u = -\frac{1}{4u} + c = -\frac{1}{4 \sin \theta} + c.
Since x = 2 \tan \theta, \theta = \arctan \frac{x}{2}.
;wip: rewrite in terms of x using the triangle again So \int \frac{1}{x^2 \sqrt{x^2 + 4}} \dee x = -\frac{1}{4} \frac{\sqrt{4 + x^2}}{x} + c.
This works because \cos \theta is always positive or 0 in the domain we are considering.
;wip: improve this day's notes so they actually make sense
# 15/1/14
We do substitution by choosing a u = f(x), then doing \frac{\dee u}{\dee x} = f'(x), so \dee u = f'(x) \dee x, so \dee x = \frac{1}{f'(x)} \dee x.
Evaluate \int \frac{x}{\sqrt{3 - 2x - x^2}}:
/|
2 / |
/ | $u$
/___|
$\theta$ $\sqrt{4 - u^2}$
We want to use trigonometric substitution to solve this, but it isn't in the right form.
We want to write 3 - 2x - x^2 in the form of a^2 + (x + b)^2 (completing the square).
Clearly, 3 - 2x - x^2 = 3 - (2x + x^2) = 3 - (2x + x^2 + 1) + 1 (make the term into a perfect square trinomial by adding \frac{b}{2} in a way that cancels out).
So 3 - 2x - x^2 = 4 - (x + 1)^2 and \int \frac{x}{\sqrt{3 - 2x - x^2}} = \int \frac{x}{\sqrt{4 - (x + 1)^2}}.
Let u = x + 1. Then \dee u = \dee x.
We can draw a triangle with the hypotenuse being 2, the opposite side u, and adjacent side \sqrt{4 - u^2}.
From the triangle, we know that u = 2\sin \theta (x = 2\sin \theta - 1) and \sqrt{4 - u^2} = 2 \cos \theta, so \dee u = 2 \cos \theta \dee \theta.
So \int \frac{x}{\sqrt{4 - (x + 1)^2}} = \int \frac{2\sin \theta - 1}{2 \cos \theta} 2\cos \theta \dee \theta = \int 2\sin \theta - 1 \dee \theta = -2 \cos \theta - \theta + c.
From the triangle, we can tell that \cos \theta = \frac{\sqrt{4 - u^2}}{2}, since the cosine is the adjacent over the hypotenuse.
Clearly, \theta = \arcsin \frac{u}{2} So \int \frac{x}{\sqrt{3 - 2x - x^2}} = -2 \cos \theta - \theta + c = -2 \frac{\sqrt{4 - (x + 1)^2}}{2} - \arcsin \frac{x + 1}{2} + c.
## Partial Fraction Decomposition
Consider \frac{1}{2} + \frac{1}{3} = \frac{3 + 2}{2 \cdot 3} = \frac{5}{6}.
What if we went the opposite way? We go from \frac{5}{6} to \frac{A}{2} + \frac{B}{3} by factoring 6, and then solve for possible solutions for 3A + 2B = 5 and pick A = 1, B = 1. Now we have \frac{1}{2} + \frac{1}{3}, as required.
This can also be done with polynomials. Using this technique, we can integrate rational functions of polynomials.
Consider \int \frac{P(x)}{Q(x)} \dee x. We can always factor polynomials into a set of linear factors Q(x) = (x + c_1) \cdot \ldots \cdot (x + c_n).
So \int \frac{P(x)}{Q(x)} \dee x = \int \frac{P(x)}{(x + c_1) \cdot \ldots \cdot (x + c_n)} \dee x = \int \frac{A_1}{x + c_1} \dee x + \ldots + \int \frac{A_n}{x + c_n} \dee x.
Now we have a bunch of integrals of constants over x with offsets. This is easily integrated into logarithms.
We can solve for A_1, \ldots, A_n by cross multiplying: \frac{P(x)}{Q(x)} = \frac{A_1 ((x + c_2) \cdot \ldots \cdot (x + c_n)) + \ldots + A_n ((x + c_1) \cdot \ldots \cdot (x + c_{n - 1}))}{(x + c_1) \cdot \ldots \cdot (x + c_n)}. So P(x) = A_1 ((x + c_2) \cdot \ldots \cdot (x + c_n)) + \ldots + A_n ((x + c_1) \cdot \ldots \cdot (x + c_{n - 1})).
For now, assume that the order of P is less than that of Q.
We can find A and B using the cover up method: if we choose values of x = -c_i, 1 \le i \le n, then there is a 0 factor in every term on the right side except the one containing A_i. It is then trivial to solve for A_i. Repeat this for each i and we can find A_1, \ldots, A_n.
We can also find A and B by setting up a system of equations. Clearly, x + 1 = A(x + 3) + B(x + 2) = (A + B)x + (3A + 2B). Since (A + B)x = x and (3A + 2B) = 1, A + B = 1 and 3A + 2B = 1, so we solve for the two linear equations to find A and B.
Evaluate \int \frac{x - 1}{x^2 + 5x + 6} \dee x:
Clearly, \int \frac{x - 1}{x^2 + 5x + 6} \dee x = \int \frac{x - 1}{(x + 2)(x + 3)} \dee x = \int \frac{A}{x + 2} \dee x + \int \frac{B}{x + 3} \dee x for some A and B.
Now we need to find A and B. We do this by solving x + 1 = A(x + 3) + B(x + 2).
Note that A and B must work for any x. We now solve using the cover up method.
Choose x = -3 (because A(x + 3) = 0). Then -3 + 1 = A(-3 + 3) + B(-3 + 2) = -2 = -B. Then B = 2.
Choose x = -2 (because B(x + 2) = 0). Then -2 + 1 = A(-2 + 3) + B(-2 + 2) = -1 = B. Then A = -1.
Then \int \frac{x - 1}{x^2 + 5x + 6} \dee x = -\int \frac{1}{x + 2} \dee x + \int \frac{2}{x + 3} \dee x = -\ln(x + 2) + 2\ln(x + 3) + c.
# 17/1/14
Note that partial fraction decomposition only works for factorable polynomials.
What if the numerator has a higher or equal degree than the denominator? We can use long division to make it lower again to apply partial fraction decomposition. When we do long division, we get a normal polynomial as the quotient and a rational function with the degree of the numerator lower than that of the denominator.
Evaluate \int \frac{x^3 + 4x^2 + 2x - 5}{x^2 + 5x + 6} \dee x:
x - 1
____________________
x^2 + 5x + 6 | x^3 + 4x^2 + 2x - 5
x^3 + 5x^2 + 6x
-------------------
-x^2 - 4x - 5
-x^2 - 5x - 6
-------------
x + 1
Clearly, \int \frac{x^3 + 4x^2 + 2x - 5}{x^2 + 5x + 6} \dee x = \int (x - 1) \dee x + \int \frac{x + 1}{x^2 + 5x + 6} \dee x = \frac{x^2}{2} - x + \int \frac{x + 1}{(x + 2)(x + 3)} \dee x.
;wip
What if there are repeated factors? We can keep the repeated factors as part of the group to use this technique.
Consider \int \frac{x + 1}{(x + 2)^2 (x + 3)}:
Clearly, \frac{x + 1}{(x + 2)^2 (x + 3)} = \frac{A}{(x + 2)^2 (x + 3)} + \frac{B}{(x + 2) (x + 3)} + \frac{C}{(x + 2)^2} ;wip
For each factor (ax + b)^n, include n terms in the form \frac{A_1}{(ax + b)} + \ldots + \frac{A_n}{(ax + b)^n}.
What if the denominator is not factorable into linear factors of real numbers (factors contain irreducable quadratics)? For example, \frac{x}{x^2 + x + 1}.
We want to avoid imaginary numbers when doing this. Therefore, each irreducable quadratic of the form x^2 + x + 1, we get a term of the form \frac{A + Bx}{ax^2 + bx + c}.
Consider \int \frac{5x + 1}{(x^2 + x + 1)(x - 2)} \dee x:
Clearly, \int \frac{5x + 1}{(x^2 + x + 1)(x - 2)} \dee x = \int \frac{A + Bx}{x^2 + x + 1} \dee x + \int \frac{C}{x - 2} \dee x.
So \int \frac{5x + 1}{(x^2 + x + 1)(x - 2)} \dee x = \int \frac{\frac{2}{7} - \frac{11}{7}x}{x^2 + x + 1} \dee x + \int \frac{\frac{11}{7}}{x - 2} \dee x.
So \int \frac{\frac{2}{7} - \frac{11}{7}x}{x^2 + x + 1} \dee x + \int \frac{\frac{11}{7}}{x - 2} \dee x = \int \frac{\frac{2}{7} - \frac{11}{7}x}{x^2 + x + 1} \dee x + \frac{11}{7} \ln (x - 2).
How do we integrate the first term? We can use a trigonometric substitution to solve this.
Clearly, \frac{\frac{2}{7} - \frac{11}{7}x}{x^2 + x + 1} = \frac{\frac{2}{7} - \frac{11}{7}x}{(x^2 + x + \frac{1}{4}) + 1 - \frac{1}{4}} = \frac{\frac{2}{7} + \frac{11}{7}x}{(x + \frac{1}{2})^2 + \frac{3}{4}}, by completing the square.
Let u = x + \frac{1}{2}. Then \dee x = \dee u.
Clearly, \int \frac{\frac{2}{7} - \frac{11}{7}x}{x^2 + x + 1} \dee x = \frac{1}{7} \int \frac{2 - 11x}{(x + \frac{1}{2})^2 + \frac{3}{4}} \dee x = \frac{1}{7} \int \frac{-11u + \frac{15}{2}}{u^2 + \frac{3}{4}} \dee x.
Let u = \frac{\sqrt{3}}{2} \tan \theta. Then \$ x = x = x = 4^2 (-11+ ) x = 4^2 (-11) x + ^2 x ;wip: use substitution with s = \frac{\sqrt{3}}{2} \tan \theta
Completing the square: we want the resulting polynomial to have repeated roots. In the quadratic formula, this is when the discriminant b^2 - 4ac = 0, so c = \frac{b^2}{4a}.
When we have an irreducible quadratic, we always get an \arctan and a \ln \abs{ax^2 + bx + c}. ;wip: do we always need the arctan? maybe we can subsitutute back to get a knarly expression with sqrt
# 20/1/14
The techniques of integration so far can be summarized as follows:
1. Expression in the numerator of the integrand would be simplified by the derivative of a substitution: use integration by subsitution.
2. Integrand is a product, integrand would be simplified by differentiating one and integrating another: use integration by parts.
3. There is a sum of difference of squares (\pm a^2 \pm x^2): use trigonometric substitution.
4. Integrand contains a rational function (quotient of polynomials): use partial fraction decomposition.
;wip: do some textbook questions for once
The two other common methods of integration are infinite series expansion and another that uses complex numbers.
## Volumes of Solids
Recall that a single-variable integral can be geometrically interpreted as the area underneath the curve.
In the Riemann integral, that would be represented as \int_a^b f(x) \dee x = \lim_{n \to \infty} \sum_{i = 1}^n \Delta x_i f(x_i). Here, \Delta x_i is the base, and f(x_i) is the height, and they form a rectangle.
The same idea can be applied to 3D, for the volume under a surface. We want to find the area under f(x, y).
In the Riemann integral, that would be represented as \iint f(x, y) \dee x \dee y = \lim_{n \to \infty} \sum_{i = 1}^n \sum_{j = 1}^n \Delta x_i \Delta y_i f(x_i, y_i). Here, \Delta x_i \Delta y_i is the base, and f(x_i, y_i), and they form a rectangle.
In general, the volume of an arbitrary solid requires a multi-variable/multi-dimensional integral (a calculus III topic). However, there are cases where symmetry allows us to use a single variable integral.
Main axes:
y
|
|
|_____ x
/
/
z
Find the volume of a square prism with dimensions b \times b \times h:
The area of the base slice stays the same as we go along the z-axis - as we vary h, the slice is still b \times b.
So the volume is \int_0^h b^2 \dee x = b^2 \int_0^h 1 \dee x = b^2 h.
The basic idea is that we take the solid, slice the shape along one axis, then integrate the area as a function of the extent along the axis that we sliced along. This works because of the symmetry of some shapes that allows us to avoid considering some dimensions.
V = \int_a^b A(x) \dee x, where V is the volume, A(x) is the area of a slice at extent x (though other axes can be used too depending on the problem), and \dee x is the width of the slice.
Find the area of the cone with base radius 1 unit and height 1 pointing along the x-axis:
This is the same as the line y = 1 - x rotated about the x-axis.
The area of any slice of the cone along the x-axis is A(x) = \pi (1 - x)^2, because the slice is a circle.
The thickness of this slice is \dee x, and the cone is defined from x = 0 to x = 1.
So the volume of the cone is \int_0^1 \pi (1 - x)^2 \dee x = \pi \int_0^1 (1 - 2x + x^2) \dee x = \frac{\pi}{3}.
We can generalize this to any function rotated about a line. The resulting solids are called solids of rotation.
The reason this works is because the z-axis is basically redundant information that is already expressed in the y-axis.
Find the volume of the solid represented by rotating f(x) about the x-axis:
The area of each slice at each extent along the x-axis is the area of a circle with the radius f(x).
The area function is A(x) = \pi f(x)^2.
So the volume is \int_a^b A(x) \dee x = \pi \int_a^b f(x)^2 \dee x.
# 22/1/14
The rotation method works whenever we can find the area function for a given slice along a certain axis.
Here, \Delta V = A(x) \Delta x, and then we add up all these tiny volumes to get the total volume.
This method is called the method of disks, because we are integrating the volume of a lot of small disks along the axis of rotation.
There is another way to approximate the area and have it converge into an exact value. We can estimate the volume of solids of rotation by integrating the volume of a lot of nested cylinders lying along the axis of rotation - this is the method of shells.
Consider a hollow cylinder of thickness \dee x with inner radius x and height h(x). If we cut it and unroll it, then it becomes very close to a cuboid of dimensions 2\pi x by h(x) by \dee x. In other words, we assume the inner circumference is almost the same as the outer circumference, since \dee x is so small.
So the enclosed volume of the cylinder is 2\pi x h(x) \dee x.
How do we use these shells to find the volume of a solid? If we look at the cross section of our shape, we can approximate the area of this cross section using rectangles that have a fixed width \dee x perpendicular to the axis of rotation, and length h(x) along the axis of rotation.
Then the volume of the solid is the sum of the volume of all the nested cylinders: \int_a^b 2 \pi x h(x) \dee x.
Consider a cone extending along the x-axis represented by the equation y = 1 - x from 0 to 1 rotated about the x-axis:
We want the height of the cylinder at each point along the y-axis, an axis perpendicular to the axis of rotation.
The height is x = 1 - y. So the volume of the shell at any extent along the radius is 2 \pi y(1 - y) \dee x.
So the volume is \int_0^1 2 \pi y(1 - y) \dee y. Note that we only integrate from the axis of rotation to the radius, not the diameter, because the cylinders go around over to the other side.
So the volume is \frac{\pi}{3}.
Finding the height of the shell is not always the same as finding the inverse. Consider the bowl-like shape formed by rotating y = \sqrt{x} from 0 to 0 about the y-axis:
We want the height of the bowl at each point along the x-axis, an axis perpendicular to the axis of rotation.
The height is y = 1 - x^2, since the height of each cylinder is decreasing as we move outwards.
So the volume is \int_0^1 2 \pi x(1 - x^2) \dee x = \frac{\pi}{2}.
Note that this is the same as if we used x = y^2 (the inverse of the function). There are two possibilities for the cylinder height: either the inverse of the function, if the shape gets thinner as we move along the axis of rotation, or the maximum height minus the inverse, if the shape gets thicker as we move along the axis of rotation. Here, since the \sqrt{x} shape gets thicker, we used 1 - x^2.
The method of shells works best when we know the height from the axis perpendicular to the axis of rotation, and the method of disks works when we can find the area of each slice along the axis of rotation.
# 24/1/14
The method of shells and the method of disks both have cases where they work better than the other. The method of disks works best for functions that extend parallel to the axis of rotation, while the method of shells, for functions that extend perpendicular to the axis of rotation.
The method of disks uses the formula \int_a^b A(x) \dee x, where a, b are the extents along the axis of rotation.
The method of shells uses the formula \int_a^b h(x) 2\pi x \dee x, where a, b are the extents perpendicular to the axis of rotation.
Consider y = 2x^2 - x^3 = x^2(2 - x) from 0 to 2 rotated about the y-axis:
This solid looks like the top half of a donut.
The method of disks is difficult to use here, because we would need disks with holes in them. We will use the method of shells.
The height of each of our cylinders is h(x) = 2x^2 - x^3.
The volume of each shell is \Delta V = h(x) 2\pi x \dee x = 4\pi x^3 \dee x - 2\pi x^4 \dee x.
So the volume of the solid is 4\pi \int_0^2 x^3 \dee x - 2\pi \int_0^2 x^4 \dee x = 4\pi \evalat{\frac{x^4}{4}}_0^2 - 2\pi \evalat{\frac{x^5}{5}}_0^2 = 4\pi \frac{2^4}{4} - 2\pi \frac{2^5}{5} = \frac{16\pi}{5}.
A variation on the method is disks is the method of washers. Here, we have disks with holes in them for whatever reason, and our area, rather than simply being the area of a circle, is the area of the circle minus the area of the hole in the middle. Make sure to use \pi \int (r_o^2 - r_i^2) \dee x rather than \pi \int (r_o - r_i)^2 \dee x, where r_o is the outer radius and r_i is the inner radius.
For example, what is the volume enclosed by rotating the area between y = \sqrt{x} and y = x about the x-axis? We could use the disk method, except instead of finding the area of a disk, we find the area of a disk with a hole in it, a washer. However, in this case it would be easier to do it with the method of shells.
# 27/1/14
## Improper Integrals
An integral is improper if some part of it goes to \pm \infty - either the integrand or the domain of integration (limits of integration).
The way we deal with this is to take limits of the infinite values instead of using these infinite values.
Improper integrals work when the domain is infinite (\int_0^\infty f(x) \dee x), when the integrand is infinite at one of the endpoints (\int_0^1 \frac{1}{x}), or when the integrand is infinite within the domain (\int_{-1}^1 \frac{1}{x} \dee x)
For example, \int_a^\infty f(x) \dee x = \lim_{T \to \infty} \int_a^T f(x) \dee x.
For example, \int_0^\infty e^{-x} \dee x = \lim_{T \to \infty} \int_0^T e^{-x} \dee x = \lim_{T \to \infty} \evalat{-e^{-x}}_0^T = \lim_{T \to \infty} (-e^{-T}) - (-e^{-0}) = 0 + 1 = 1.
Prove that the volume obtained by rotating y = \frac{1}{x} about the x-axis for x \in [1, \infty) is finite:
The volume is \int_1^\infty \pi \frac{1}{x^2} \dee x = \pi \lim_{T \to \infty} \evalat{-\frac{1}{x}}_1^T = \pi, found via the method of disks. Clearly, the volume is finite.
Sometimes, the integrand itself will diverge. Consider \int_0^1 \ln x \dee x:
\ln 0 = -\infty, and we cannot evaluate the antiderivative at this point.
Instead, we write it as \lim_{\epsilon \to 0} \int_\epsilon^1 \ln x \dee x = \lim_{\epsilon \to 0} \evalat{x \ln x - x}_\epsilon^1 = 0 - 1 - (\lim_{\epsilon \to 0} \epsilon \ln \epsilon - \lim_{\epsilon \to 0} \epsilon) = 0 - 1 - (0 - 0) = -1.
Also, the integrand might diverge somewhere within the domain of integration. This is not always immediately obvious and must always be considered when doing integrals.
Consider \int_{-1}^1 \frac{1}{\sqrt{\abs{x}}} \dee x:
There is a vertical asymptote at x = 0. This makes it so that we can't directly use the fundemental theorem of calculus to evaluate the integral.
We can instead split the integral at the middle, and since it is an even function, we can combine the two integrals: 2\int_0^1 \frac{1}{\sqrt{x}} \dee x.
There is an asymptote, so the integral is an improper one: 2\int_0^1 \frac{1}{\sqrt{x}} \dee x = 2 \lim_{T \to 0} \int_T^1 \frac{1}{\sqrt{x}} \dee x = 2 (2 \cdot 1^\frac{1}{2} - 2\lim_{T \to 0} T^\frac{1}{2}) = 4.
Consider \int_{-1}^1 \frac{1}{2\sqrt{x}} \dee x: ;wip: what was the actual example here?
We might do \int_{-1}^1 \frac{1}{2\sqrt{x}} \dee x = \evalat{-\sqrt{\abs{x}}}_{-1}^1 = -2.
However, the integral does not exist. We need to be careful because the result looks just fine, and does not indicate that an error occurred.
The integral does not make any sense because the integrand diverges towards \infty at x = 0.
The idea behind this is that we need to figure out if an improper integral exists even without integrating it first.
If a function converges, that means it goes to a finite value. The opposite is if it diverges, when it goes to \pm \infty or does not exist.
In other words, convergence means the result is a number, and divergence means the result is not a number.
If an improper integral converges, then the value we get by evaluating its antiderivatives at the endpoints is the correct value of the integral. However, if it diverges, then the answer could be different. We always need to check for divergence in order to catch these cases.
### Comparison Theorem
Given functions f(x), g(x) such that f(x) \ge g(x) \ge 0 for x \ge a, if \int_a^b f(x) \dee x converges, then \int_a^b g(x) \dee x also converges.
The contrapositive is also useful in that if \int_a^b g(x) \dee x diverges, then \int_a^b f(x) \dee x also diverges.
Prove that \int_1^\infty \frac{1}{x^P} \dee x converges if and only if P > 1
Clearly, if P = 1, \int_1^\infty \frac{1}{x^P} \dee x = \evalat{\ln x}_1^\infty, which is \infty, so the integral diverges.
Clearly, \int_1^\infty \frac{1}{x^P} \dee x = \evalat{\frac{x^{1 - P}}{1 - P}}_1^\infty = \frac{\infty^{1 - P}}{1 - P} - \frac{1}{1 - P} if P \ne 1.
Clearly, if P < 1, \infty^{1 - P} = \infty, so the integral diverges.
Clearly, if P > 1, \infty^{1 - P} = 0, so the integral converges.
So the integral converges to \frac{1}{P - 1} for P > 1.
Interestingly, e^{-x^2} doesn't have an antiderivative. Figure out if \int_0^\infty e^{-x^2} \dee x converges or diverges:
Clearly, 0 \le e^{-x^2} \le e^{-x} for x \ge 1.
Since \int_1^\infty e^{-x} \dee x converges, \int_1^\infty e^{-x^2} \dee x converges, by the comparison theorem.
Clearly, \int_0^\infty e^{-x^2} \dee x = \int_0^1 e^{-x^2} \dee x + \int_1^\infty e^{-x^2} \dee x.
Clearly, e^{-x^2} does not diverge for all x \in [0, 1]. So \int_0^1 e^{-x^2} \dee x converges.
Since \int_0^1 e^{-x^2} \dee x and \int_1^\infty e^{-x^2} \dee x converge, \int_0^\infty e^{-x^2} \dee x converges.
# 29/1/14
Consider \int_1^\infty \frac{1 + e^{-x}}{x} \dee x:
Clearly, the function is well behaved for all x \in [1, \infty].
So we do not need to worry about the integrand diverging.
Clearly, \frac{1 + e^{-x}}{x} \le \frac{2}{x}.
Clearly, \frac{2}{x} diverges.
However, we can't say anything about the original function using the comparison theorem, since it could still either converge or diverge.
Instead, we look for divergence. Clearly, \frac{1}{x} \le \frac{1 + e^{-x}}{x}.
Since \int_1^\infty \frac{1}{x} \dee x diverges, \int_1^\infty \frac{1 + e^{-x}}{x} \dee x also diverges.
## Applications of Integration
Many physical phenomena can be represented by differential equations - equations that include derivatives.
This comes about usually through empirical evidence/oberservation, and some by conservation principles.
### Cooling
For example, Newton's Law of Cooling was discovered by Newton's measurements of hot materials in cooler surroundings.
The law states that the change in temperature is linearly propertional to the temperature difference with the surroundings.
In other words, \frac{\dee}{\dee x} T_h = -k(T_h - T_r), where T_h is the temperature of the hot thing, T_r is the surrounding temperature, and k is the cooling coefficient.
Note that k is non-negative and we use -k because hot things cool. This cooling coefficient is influenced by the material and shape of the object.
From this we can tell that the object stops cooling when T_h = T_r, so \frac{\dee}{\dee x} T_h = 0.
Values of T for which \frac{\dee T_h}{\dee x} = 0 are called equilibria.
### Conservation
Most differential equations in science and engineering come from conservation principles - principles that apply universally and constrain what can happen. For example, conservation of mass/energy/charge/momentum.
Conservation principles are easy to represent in mathematics, though the models can be pretty elaborate. Basically, what they state is that the rate of the change of something is the rate of change of something increasing minus the rate of stuff decreasing.
For example, the rate of change of people of the people in the room is the rate of change of people entering minus the rate of change of people leaving.
The rate of something increasing is called a source. The rate of something decreasing is called a sink.
For example, consider a skydiver immediately after their parachute opens:
Clearly, m\frac{\dee \vec{v}}{\dee x} = \sum \vec{F} = m \vec{g} - \vec{F}_d, where \vec{F}_d is the drag force.
We can calculate the drag force in a lab independent of everything else, and not have to worry about the other things like the height or gravity.
Typically, \vec{F}_d = k\vec{v} for some constant k.
So \frac{\dee \vec{v}}{\dee x} = \vec{g} - \frac{k}{m} \vec{v}.
# 31/1/14
An example of a mathematical representation of conservation principles is a mixing problem.
A tank holds 80L of water at time t = 0. A salt solution of 0.25 kg/L flows into the tank at 8L/min. Liquid drains at a rate of 12L/min from the tank. Find a differential equation for the mass of salt x(t) in kg for t > 0, where t is measures in seconds:
We assume the salt solution instantly mixes with the water.
We want to find \frac{\dee}{\dee t} x(t), which is \text{source} - \text{sink}.
The source is \text{salt concentration} \cdot \text{inflow rate} = 0.25 \text{kg/L} \cdot 8 \text{L/min} = 2 \text{kg/min}.
Clearly, \frac{\dee}{\dee x} \text{liquid level} = \text{inflow rate} - \text{outflow rate} = 8 \text{L/min} - 12 \text{L/min} = -4 \text{L/min}.
So the liquid level is 80 + \int \frac{\dee}{\dee x} \text{liquid level} \dee t = 80 - 4t.
Clearly, \text{salt concentration} = \frac{\text{mass of salt}}{\text{volume of tank}} = \frac{x(t)}{80 - 4t}.
The sink is \text{salt concentration} \cdot \text{outflow rate} = \frac{x(t)}{80 - 4t} \cdot 12 \text{kg/L} = \frac{3x(t)}{20 - t}.
So \frac{\dee}{\dee t} x(t) = 2 - \frac{3x(t)}{20 - t}.
Note that the tank is empty when t = 20, when the outflow rate is undefined.
There are many things we can say about the function even without solving it. These are called qualitiative analyses.
Suppose we have \frac{\dee y}{\dee x} = f(x, y). Note that the left side is equal to the slope of the tangent line, and the right side is an expression for the slope of the tangent line for any point (x, y).
Basically, every point on the Cartesian plane now has an associated slope, and we can represent this roughly with arrows on the graph representing the trend in the slopes.
These arrows/contours are called streamlines.
Think of this as a stream with currents and eddies. Every possible solution is a possible path a floating object will take if dropped at a particular place in the stream and allowed to bob along with the stream. The solutions are we get if we start at a possible point and draw a curve that has the slope of the tangent always equal to the value at the flow field at the point under the curve.
We can also think of them as a topographical map of some terrain. If we drop a ball along a point on the map, it will trace out a certain line. If we drop it slightly off to one side, it may take a different path, which could be similar, or completely different. The slope of the tangent represents the slope of the hill at a particular point.
The set of all possible solutions is therefore a family of functions that look like a map of water currents.
An equilibrium is a place where \frac{\dee y}{\dee x} = f(x) = 0. Because the slope of the tangent is 0, it must be 0 for all x values. There are stable and unstable equilibria.
Stable equilibria are those where values that are close to them get closer to them, like dropping a ball on the side of a valley and having it eventually roll to the bottom of the valley.
Unstable equilibria are those where values that are close to them get farther away from them, like dropping a ball at the peak of a mountain and having it roll away from the peak. Only values that are exactly on the unstable equilibrium will stay there.
Consider \frac{\dee y}{\dee x} = 1 - y^2:
The slope of the tangent line is \begin{cases} \text{negative} &\text{if } y < -1 \\ \text{positive} &\text{if } -1 < y < 1 \\ \text{negative} &\text{if } y > 1 \end{cases}.
We can visualize this as a flow field with three distinct sections.
Note that if we start at y = -1, 1, we stay at that same value regardless of the value of x. This is called an equilibrium, since the value isn't changing.
Note that if we start at any y > -1, y \to 1 as x \to \infty (convergent), and if we start at any y < -1, y \to -\infty as x \to \infty (divergent).
# 3/2/14
Differential equations of the form \frac{\dee y}{\dee x} = f(y) are called autonomous differential equations. Note that they do not depend on the value of x.
The equilibria of an autonomous differential equation are those y values for which \frac{\dee y}{\dee x} = f(y) = 0.
An equilibrium y = k is stable if and only if f'(k) < 0 - if the change in the slope at the location is negative. Otherwise, it is unstable. This is because a negative value for the derivative means a negative feedback loop, which makes the function settle toward the equilibrium. ;wip: what? why? shouldn't the derivative be 0 at y = k?
## Solving Differential Equations
### Estimation
Consider \frac{\dee y}{\dee x} = y - x, with y(0) = 2.
This is a differential equation of the form \frac{\dee y}{\dee x} = f(x, y)
We can use the differential equation to estimate a certain solution, much like a Riemann sum.
First, we pick a starting point so we have a single solution to the differential equation, a member of the family of solutions. This is a curve on the Cartesian plane.
This starting point is called the initial condition. The initial condition often takes the form of y(x) = y(0) = 2 - the y-value at x = 0 is 2 for this solution.
First, we break up the domain into pieces, by choosing value for x_1, \ldots, x_n.
Then, we use \frac{\dee y}{\dee x} = f(x, y) to estimate the slope of the tangent at any given x value.
Then, we observe that \frac{\dee y}{\dee x} = f(x_{n - 1}, y_{n - 1}) \approx \frac{y_n - y_{n - 1}}{x_n - x_{n - 1}} - the slope of the tangent is close to the slope of the secant.
So y_n(x) \approx y_{n - 1} + (x_n - x_{n - 1})f(x_{n - 1}, y_{n - 1}). This is written in update form, so we can now use a table of values to calculate it.
Let \Delta x = x_n - x_{n - 1}. Then y_n(x) \approx y_{n - 1} + f(x_{n - 1}, y_{n - 1}) \Delta x.
We can also write this as y_{n + 1}(x) \approx y_n + f(x_n, y_n) \Delta x where \Delta x = x_{n + 1} - x_n.
Now we can write \frac{\dee y}{\dee x} = y - x as y_n(x) \approx y_{n - 1} + (x_n - x_{n - 1})(x_{n - 1} - y_{n - 1}).
We can actually approximate this function by using a table of values. Since y(0) = 2, y_0 = 2:
n x_n y_n f(x_n, y_n) \Delta x y_{n + 1} \approx y_n + f(x_n, y_n) \delta x
0 0.0 2 (2 - 0.0)0.1 = 0.20 2.2 \approx 2 + 0.2
1 0.1 2.2 (2 - 0.1)0.2 = 0.38 2.58 \approx 2.2 + 0.38
2 0.2 2.58 (2 - 0.2)0.3 = 0.54 3.12 \approx 2.58 + 0.54
3 0.3 3.12 (2 - 0.3)0.4 = 0.68 \ldots
Now we can a table of values for the equation. We can plot this on a graph to approximate the curve.
The actual solution can also be found, though not through these techniques. The general solution is y = 1 + x + c^x, c \in \mb{R}.
This is called Euler's method. It is a method for numerically solving differential equations.
### Integration
If we want exact solutions, we have to integrate. This is not always possible, but is in certain cases.
Separable equations are one variety of them. They are equations of the form \frac{\dee y}{\dee x} = A(x) B(y).
Note that y is also a function of x, so B(y) is another function of x. We write y = y(x).
Clearly, \frac{\dee y}{\dee x} = A(x) B(y) is the same as \frac{1}{B(y)} \frac{\dee y}{\dee x} = A(x).
If we integrate both sides, we get \int \frac{1}{B(y)} \frac{\dee y}{\dee x} \dee x = \int A(x) \dee x.
Now we can "cancel \dee x". Let u = y(x). Then \dee u = \frac{\dee u}{\dee x} \dee x and \dee x = \frac{1}{\frac{\dee u}{\dee x}} \dee u = \frac{1}{\frac{\dee y}{\dee x}} \dee u. This is basically the definition of the substitution rule for integrals.
So \int \frac{1}{B(y)} \frac{\dee y}{\dee x} \dee x = \int \frac{1}{B(u)} \frac{\dee y}{\dee x} \frac{1}{\frac{\dee y}{\dee x}} \dee u = \int \frac{1}{B(u)} \dee u = \int \frac{1}{B(y)} \dee y.
So \int \frac{1}{B(y)} \dee y = \int A(x) \dee x.
We can also think of this as multiplying both sides of \frac{1}{B(y)} \frac{\dee y}{\dee x} = A(x) by \dee x and integrating both sides.
Solve \frac{\dee y}{\dee x} = \frac{x}{y} for y(0) = -3:
Clearly, this is equivalent to y \frac{\dee y}{\dee x} = x, or y \dee y = x \dee x, or \int y \dee y = \int x \dee x.
Integrating both sides, we get \frac{y^2}{2} = \frac{x^2}{2} + c.
We want to find the value of c. Since y(0) = -3, \frac{(-3)^2}{2} = \frac{0^2}{2} + c and c = \frac{9}{2}.
So the solution is y^2 = 2x^2 + 2 \cdot \frac{9}{2} = 2x^2 + 9.
;wip: try \frac{\dee}{\dee x} 2^y \sin^2 x and \sqrt{x \frac{\dee x}{\dee t}} = \frac{1}{1 + t}
# 5/2/14
Constant solutions are another name for equilibria. They are simply where \frac{\dee y}{\dee x} = 0 for \frac{\dee y}{\dee x} = f(y).
Solve \sqrt{x \frac{\dee x}{\dee t}} = \frac{1}{1 + t} for x(t) = x(0) = 0, t \ge 0:
\displaystyle \begin{aligned} \sqrt{x \frac{\dee x}{\dee t}} &= \frac{1}{1 + t} \\ x \frac{\dee x}{\dee t} &= \frac{1}{(1 + t)^2} \\ x \dee x &= \frac{1}{(1 + t)^2} \dee t \\ \int x \dee x &= \int \frac{1}{(1 + t)^2} \dee t \\ \frac{x^2}{2} &= -\frac{1}{1 + t} + c \\ t &= 0; x(0) = 0 \\ \frac{0^2}{2} &= -\frac{1}{1 + 0} + c \\ c &= 1 \\ \frac{x^2}{2} &= -\frac{1}{1 + t} + 1 \\ x &= \sqrt{2 - \frac{2}{1 + t}} = \sqrt{\frac{2(1 + t) - 2}{1 + t}} = \sqrt{\frac{2t}{1 + t}} \end{aligned}
First we rearrange until there is only one variable on each side, then we integrate, and finally, fix the constant of integration c by solving for it with the given values for the axis variables.
Solve \frac{\dee T}{\dee t} = -k(T - T_{ambient}) - Newton's law of cooling:
\displaystyle \begin{aligned} \frac{\dee T}{\dee t} &= -k(T - T_{ambient}) \\ \frac{1}{T - T_{ambient}} \dee T &= -k \dee t \\ \int \frac{1}{T - T_{ambient}} \dee T &= -k \int 1 \dee t \\ \ln \abs{T - T_{ambient}} &= -kt + c \\ e^{\ln \abs{T - T_{ambient}}} &= e^{-kt + c} \\ \abs{T - T_{ambient}} &= e^{-kt}e^c \\ T - T_{ambient} &= \pm e^ce^{-kt} \end{aligned} Let A = \pm e^c. Then T - T_{ambient} = Ae^{-kt}.
Note that at t = 0, T = T_0 for some fixed T_0.
So T_0 - T_{ambient} = Ae^{-k0} and T_0 - T_{ambient} = A.
So T = T_{ambient} + (T_0 - T_{ambient})e^{-kt}.
Suppose we have a population of n individuals. The simplest population model is \frac{\dee n}{\dee t} = kn for some k \ge 0. Solve for n:
\displaystyle \begin{aligned} \frac{\dee n}{n} &= k \dee t \\ \int \frac{\dee n}{n} &= k \int \dee t \\ \ln \abs{n} &= kt + c \\ n &= \pm e^ce^{kt} \\ A = \pm e^c; n &= Ae^{kt} \end{aligned} Note that at t = 0, n = n_0 for some constant n_0 - the starting population.
So n_0 = Ae^{k0} and A = n_0.
So n = n_0e^{kt}.
A simple variation on this population growth model is logistic growth, which also models resource exhaustion: \frac{\dee N}{\dee t} = rN\left(1 - \frac{N}{k}\right). Note that there are two equalibria - N = 0 and N = k. k is a constant called the carrying capacity.
For \frac{n}{k} \ll 1 (much less than), logistic growth behaves like exponential growth.
The flow field looks like arrows pointing right toward N(t) = k starting from N(t) = 0. Therefore, the population always tends toward k in this model.
# 7/2/14
The main idea is that there is a lot more to the equations than just the solutions. The flow field is useful for discovering how the equation behaves.
Even when we solve the equation, it does not tell us much about how the function works intuitively.
Solve \frac{\dee N}{\dee t} = rN\left(1 - \frac{N}{k}\right):
\displaystyle \begin{aligned} \frac{\dee N}{\dee t} &= rN\left(1 - \frac{N}{k}\right) \\ \int \frac{1}{N\left(1 - \frac{N}{k}\right)} \frac{\dee N}{\dee t} \dee t &= \int r \dee t \\ \int \frac{1}{N\left(1 - \frac{N}{k}\right)} \dee N &= rt + c \\ \int \frac{A}{N} + \frac{B}{1 - \frac{N}{k}} \dee N &= rt + c \\ A\left(1 - \frac{N}{k}\right) + BN &= 1; A = 1; B = \frac{1}{k} \\ \int \frac{1}{N} + \frac{\frac{1}{k}}{1 - \frac{N}{k}} \dee N &= rt + c \\ \ln \abs{N} + \int \frac{1}{k - N} \dee N &= rt + c \\ \ln \abs{N} - \ln \abs{N - k} &= rt + c \\ \ln \abs{\frac{N}{N - k}} &= rt + c \\ -\ln \abs{\frac{N}{N - k}} &= -rt - c \\ \ln \abs{\frac{N - k}{N}} &= -rt - c \\ \abs{\frac{N - k}{N}} &= e^{-rt - c} \\ \frac{N - k}{N} &= \pm e^{-rt} e^{-c} \\ F &= \pm e^{-c} \\ 1 - \frac{k}{N} &= Fe^{-rt} \\ \frac{k}{N} &= Fe^{-rt} + 1 \\ N &= \frac{k}{1 + Fe^{-rt}} \end{aligned} Note that all solutions tend toward k, which we can verify by taking the limit of N at infinity.
Now we will solve for F - at t = 0, N = N_0. Assume t = 0.
So N_0 = \frac{k}{1 + Fe^{-r0}} and N_0 + FN_0 = k, so F = \frac{k}{N_0} - 1.
So N = \frac{k}{1 + \left(\frac{k}{N_0} - 1\right)e^{-rt}}.
# 10/2/14
Differential equations appear quite often in equations for time based physical phenomena.
Separable differential equations are one type of differential equation that can be directly solved. However, there are other types.
## Linear Differential Equations
Linear differential equations are those of the form \frac{\dee y}{\dee x} = A(x) y + B(x), A(x) \ne B(x). It is called linear because the degree of y on the right side is 1.
If A(x) = B(x), then the equation is separable since A(x)y + B(x) = (y + 1)A(x), so it would be simpler to solve it that way, but the following technique would also work.
We can solve these by splitting the equation into two separable equations and then solving each of those. This technique is developed by Euler.
Note that if we multiply the equation by an arbitrary function I(x), we get I(x) \frac{\dee y}{\dee x} = I(x) A(x) y + I(x) B(x).
### First equation
So I(x) \frac{\dee y}{\dee x} - I(x) A(x) y = I(x) B(x).
Assume I(x) is a function that satisfies \frac{\dee}{\dee x} (I(x) y) = I(x) B(x) or I(x) y = \int I(x) B(x) \dee x.
So y = \frac{1}{I(x)} \left(\int I(x) B(x) \dee x\right).
Clearly, \frac{\dee}{\dee x} I(x) y = y \frac{\dee}{\dee x} I(x) + I(x) \frac{\dee y}{\dee x} = I(x) B(x) = I(x) \frac{\dee y}{\dee x} - I(x) A(x) y
So y \frac{\dee}{\dee x} I(x) + I(x) \frac{\dee y}{\dee x} = I(x) \frac{\dee y}{\dee x} - I(x) A(x) y and y \frac{\dee}{\dee x} I(x) = -I(x) A(x) y.
### Second equation
So \frac{\dee}{\dee x} I(x) = -I(x) A(x). This is a separable differential equation.
Solving, we get \int \frac{1}{I(x)} \dee I(x) = -\int A(x) \dee x.
Clearly, \int \frac{1}{I(x)} \dee I(x) = \ln \abs{I(x)} so \abs{I(x)} = e^{-\int A(x) \dee x} and I(x) = \pm e^{-\int A(x) \dee x}.
Here, I(x) is called the integrating factor. In general, I(x) = \pm e^{-\int A(x) \dee x}.
### Substitution
Now we substitute back into the original equation: y = \frac{1}{I(x)} \int I(x) B(x) \dee x = \frac{1}{\pm e^{-\int A(x) \dee x}} \int (\pm e^{-\int A(x) \dee x} B(x)) \dee x = \frac{1}{e^{-\int A(x) \dee x}} \int e^{-\int A(x) \dee x} B(x) \dee x.
The general rule is that given \frac{\dee y}{\dee x} = A(x) y + B(x), y = \frac{1}{I(x)} \int I(x) B(x) \dee x where I(x) = e^{-\int A(x) \dee x}.
Consider the velocity of a falling object, \frac{\dee v}{\dee t} = g - \beta v, where g is gravitational acceleration and \beta is the drag coefficient:
Clearly, this is an equation of the form \frac{\dee v}{\dee t} = A(t) v + B(t) where A(t) = -\beta and B(x) = g.
So I(x) = e^{-\int (-\beta) \dee t} = e^{\beta t}.
So v = \frac{1}{e^{\beta t}} \int e^{\beta t} g \dee x = \frac{1}{e^{\beta t}} \left(\frac{g}{\beta} e^{\beta t} + c\right).
So v = \frac{1}{e^{\beta t}} \left(\frac{g}{\beta} e^{\beta t} + c\right) = \frac{g}{\beta} + ce^{-\beta t}.
Now we need to find c. Since v(t) = v(0) = v_0 = \frac{g}{\beta} + ce^{-\beta 0}, c = v_0 - \frac{g}{\beta}.
So v = \frac{g}{\beta} + \left(v_0 - \frac{g}{\beta}\right)e^{-\beta t}.
We can check our answer by substituting v back into the original equation: \frac{\dee}{\dee t} \left(\frac{g}{\beta} + \left(v_0 - \frac{g}{\beta}\right)e^{-\beta t}\right) = g - \beta \left(\frac{g}{\beta} + \left(v_0 - \frac{g}{\beta}\right)e^{-\beta t}\right) = g - \left(g + (\beta v_0 - g)e^{-\beta t}\right) = (g - \beta v_0)e^{-\beta t}, which is correct.
Consider x^2 \frac{\dee y}{\dee x} + xy = 1 for y(x) = y(1) = 2, x > 0:
Clearly, x^2 \frac{\dee y}{\dee x} + xy = 1 is equivalent to \frac{\dee y}{\dee x} + \frac{1}{x}y = \frac{1}{x^2} or \frac{\dee y}{\dee x} = -\frac{1}{x}y + \frac{1}{x^2}.
So I(x) = e^{-\int \left(-\frac{1}{x}\right) \dee x} = e^{\ln x} = x.
So y = \frac{1}{x} \int x \frac{1}{x^2} \dee x = \frac{\ln x + c}{x}.
Since y(x) = y(1) = 2, 2 = \frac{\ln 1 + c}{1} and c = 2.
So y = \frac{\ln x + 2}{x}.
# 12/2/14
Often when solving differential equations, it is not possible to get an explicit solution (where the variable is isolated on one side). Instead, it is often only possible to get an implicit solution, like 2\ln y + y^2 = x.
Midterm Review:
• Techniques of Integration
• Review
• Integration by Substitution
• Integration by Parts
• Trigonometric Substitution
• Partial Fraction Decomposition
• Applications
• Volumes of Solids
• Method of Shells
• Method of Disks
• Improper Integrals
• Differential Equations
• Qualitative Analysis
• Separable Differential Equations
• Linear Differential Equations
• Integrating Factors
## Sequences, Series, Taylor Polynomials
Given a function f(x), we can often find a polynomial P(x) that has the same first N derivatives at a point x = a.
For example, f(x) = \sin x and P(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} have the same first 7 derivatives at x = 0. In other words, \frac{\dee^i f}{\dee x^i} = \frac{\dee^i P}{\dee x^i} for 1 \le i \le 7 at x = 0.
Another example is f(x) = \frac{1}{1 - x} and P(x) = 1 + x + x^2 + x^3 + \ldots + x^N. The first N derivates at x = 0 are always equal, and it works for any arbitrary N.
In general, if P(x) = \sum_{n = 0}^N \frac{f^{(n)}(a)}{n!}(x - a)^n, then f(x) and P(x) share the first N derivative values at x = a. The notation f^{(n)}(a) represents \evalat{\frac{\dee^n}{\dee x^n} f(x)}_{x = a}.
The most important use of this is approximation of very difficult functions. This allows us to use polynomials as tools to analyze many types of functions.
What kinds of functions does this work for?
In the limit N \to \infty, does f(x) = P(x) = \lim_{N \to \infty} \sum_{n = 0}^N \frac{f^{(n)}(a)}{n!}(x - a)^n for all x? Does the limit even exist?
## Sequences
A sequence is an ordered list of numbers. We denote it with a_0, a_1, \ldots, a_n, or simply a_n = f(n). For example, a_n = 2n is equivalent to 0, 2, 4, 6, \ldots
A sequence converges if it has a limit as n \to \infty.
Sequences are often easier to work with than functions. For example, the \epsilon-\delta definition of limits is much simpler.
The definition of the limit of a sequence is: a sequence a_n has a limit L if and only if for any \epsilon > 0, we can find k \in \mb{Z} such that \abs{a_n - L} < \epsilon whenever n > k.
Formally, the limit L exists if and only if \forall \epsilon > 0, \exists k \in \mb{Z}, n > k \implies \abs{a_n - L} < \epsilon.
This is similar to the definition of limits at infinity. For example, the sequence 0.3, 0.33, 0.333, 0.3333, \ldots has the limit L = \frac{1}{3}, but 1, -1, 1, -1, \ldots has no limit.
Use the definition of the limit to prove that the limit of a_n = \frac{n}{1 + n} is 1:
Let \epsilon \in \mb{R}. Construct k = \frac{1}{\epsilon} - 1.
Assume n > k. Then n > \frac{1}{\epsilon} - 1 and \frac{1}{1 + n} < \epsilon, so \abs{\frac{n}{1 + n} - 1} < \epsilon.
So by definition, \lim_{n \to \infty} a_n = 1.
If a_n and L lie in the domain of a continuous function f and a_n \to L as n \to \infty, then \lim_{n \to \infty} f(a_n) = f(L). In other words, the limit of a sequence is the same as the limit of its corresponding function at infinity.
Sequences are easy to deal with if they are explicit functions of the index n, but a lot of sequences are defined recursively, like a_{n + 1} = a_n + \frac{1}{n!}, a_0 = 1.
;wip: try a_n = 7^{\frac{1}{2} + \frac{1}{n}}\tan \frac{\pi n + 1}{4n} limit at infinity -
# 14/2/14
It is in fact easy to determine if any sequence converges or diverges.
Sequences that are defined only in terms of the current index are easy to find the limit of. For example, a_n = \frac{n}{n + 1} converges to 1 at infinity.
However, sequences defined recursively are not so straightforward. For example, a_{n + 1} = a_n + \frac{1}{n!}, a_0 = 1. This is an implicitly defined sequence.
### Monotone Convergence Theorem
If the terms of a sequence are bounded, and the sequence is monotone, then the sequence converges.
There are two possible cases:
• If the sequence has an upper bound a_n \le b and is monotonially increasing, then it converges.
• If the sequence has a lower bound a_n \ge b and is monotonically decreasing, then it converges.
Boundedness means that every element in the sequence is between a lower and upper bound. Formally, a sequence a_n is bounded if and only if \exists u, \exists v, \forall n, u \le a_n \le v.
We often prove boundedness by comparing to a known sequence (like \frac{1}{n!} \le \frac{1}{2^{n - 1}}) or by using induction.
Monotonicity means that once we start going up, we never go down again, and once we start going down, we never go up again. Basically, a_0 \le a_1 \le a_2 \le \ldots, or a_0 \ge a_1 \ge a_2 \ge \ldots.
Formally, a sequence a_n is monotone if and only if (\forall u, \forall v, u < v \implies a_u < a_v) \vee (\forall u, \forall v, u < v \implies a_u > a_v).
Proof:
It is intuitively obvious that if a value is always increasing, and cannot exceed a value, then it must converge to some value.
Without loss of generality, assume that a_n is increasing and bounded by a_n \le b.
Let p be the smallest possible upper bound on a_n.
As an aside, p is the limit of the sequence - any number less than this is not an upper bound, and any number greater than this can be smaller while still being an upper bound.
Let \epsilon > 0.
Clearly, p - \epsilon < p, so \exists k, p - \epsilon < a_k, since any value less than the tight upper bound must be exceeded by some element in the sequence.
Assume n > k. Clearly, a_k \le a_n (since the sequence is monotonically increasing).
Clearly, a_n \le p < p + \epsilon.
So p - \epsilon < a_n < p + \epsilon and -\epsilon < a_n - p < \epsilon.
So \abs{a_n - p} < \epsilon, and by the definition of the limit, p is the limit and the sequence converges.
Does a_0 = 1, a_{n + 1} = a_n + \frac{1}{n!} converge?
Clearly, \frac{1}{n!} is always positive, so a_{n + 1} = a_n + \frac{1}{n!} is monotonically increasing.
Clearly, n! \le 2^{n - 1}, since 1 \cdot \ldots \cdot k \le 2 \cdot \ldots \cdots 2 \text{ (} k - 1 \text{ times)}.
So \frac{1}{n!} \le \frac{1}{2^{k - 1}} and a_n \le 1 + 1 + \frac{1}{2} + \frac{1}{2^2} + \ldots + \frac{1}{2^{n - 1}}.
Clearly, this is a monotonically increasing geometric progression, and so we can determine that 1 + 1 + \frac{1}{2} + \frac{1}{2^2} + \ldots + \frac{1}{2^{n - 1}} = \frac{1 - \left(\frac{1}{2}\right)^n}{1 - \frac{1}{2}}.
Clearly, \lim_{n \to \infty} \frac{1 - \left(\frac{1}{2}\right)^n}{1 - \frac{1}{2}} = \frac{1}{\frac{1}{2}} = 2.
So a_n \le \frac{1}{2^{k - 1}} \le 2, and by the convergence theorem, the sequence converges.
As an aside, it converges to e.
Does a_1 = 1, a_{n + 1} = \sqrt{3 + 2a_n} converge?
The first few elements of the sequence are 1, 2.23606797749979, 2.73352079834772, 2.90981813807933, 2.96978724425819.
Clearly, a_{n + 1} \ge a_n \iff \sqrt{3 + 2a_n} \ge a_n \iff a_n^2 - 2a_n - 3 \le 0, or -1 \le a_n \le 3.
So a_n is monotonically increasing if [-1, 3].
Assume a_n \le 3. Then 3 + 2a_n \le 9 and \sqrt{3 + 2a_n} \le 3. So a_{n + 1} \le 3.
So by induction, a_n has an upper bound of 3.
So by the monotone convergence theorem, the sequence converges.
Also, note that if the limit exists, we can set n \to \infty and so a_{n + 1} = a_n.
So a_n^2 - 2a_n - 3 = 0 and a_n = -1, 3. -1 is a extraneous answer since the function is always positive, so \lim_{n \to \infty} a_n = 3.
# 24/2/14
The monotone convergence theorem only tells us whether the sequence converges, but not what the actual limit is.
We usually prove the monotonicity and boundedness of sequences by comparing them to sequences with known properties (like geometric series), or by using induction.
To use induction to prove monotonicity, we simply need to prove that a_{n + 1} \ge a_n for any n.
To use induction to prove boundedness, we first need to guess a k that might be an upper or lower bound. Then, we verify that it works for the first element of the sequence, and then that a_n \le k \implies a_{n + 1} \le k.
If we know that the limit exists, sometimes we can find the limit by setting n \to \infty, which implies that a_{n + 1} = a_n (since we are at the limit).
Then we can write a_{n + 1} in terms of a_n, and we get a function over a_n, which we can often isolate and solve for, which gives us the limit L = a_n. The previous example uses this technique.
This doesn't always work. For example, it work doesn't for a_0 = 1, a_{n + 1} = a_n + \frac{1}{n!}. If we try to substitute it, we get a_n = a_n + \frac{1}{n!} = a_n + \frac{1}{\infty} = a_n. This doesn't help us find the limit.
## Series
A series is a sum of terms. We denote this as S_N = \sum_{n = 0}^N a_n, a_n \in \mb{R}. Here, N \in \mb{Z}, N \ge 0.
For now, a_n is a constant, which can depend on n. Eventually, we want to be able to build and analyze sequences of functions, which can depend on other variables like x.
If N is finite, then S_N is known as a partial sum.
We analyze series by analyzing the sequence of the series values: \sum_{n = 0}^0 a_n, \sum_{n = 0}^1 a_n, \sum_{n = 0}^2 a_n, \ldots.
So just as we take the limit of a sequence, we can take the limit of a series with \lim_{N \to \infty} \sum_{n = 0}^N a_n. This is the relation between sequences and series.
A sequence converges if this limit exists. Otherwise, it diverges. A series converges if and only if the limit of its partial sum to infinity is a finite value.
For example, we previously saw the sequence a_0 = 1, a_{n + 1} = a_n + \frac{1}{n!}, which is actually equivalent to the series S_N = \sum_{n = 0}^1 \frac{1}{n!}. They are equivalent because a_n = S_N when n = N.
We also define \sum a_n = \sum_{n = k}^\infty a_n, k \in \mb{Z} for convenience.
# 26/2/14
### Geometric Series
The geometric series is S_N = \sum_{n = 0}^N x^n = 1 + x + x^2 + x^3 + \ldots + x^n.
The geometric series is a rare example of a series that can be written in closed form - non-recursively and in a finite number of symbols. In other words, we can write the value of the partial sum of the series as a function of n.
Which values of n make the series converge?
First, we start with the identity of 1 - x^{N + 1} = (1 - x)(1 + x + x^2 + x^3 + \ldots + x^N). This is true because (1 - x)(1 + x + x^2 + x^3 + \ldots + x^N) = (1 + x + x^2 + x^3 + \ldots + x^N) - (x + x^2 + x^3 + \ldots + x^{N + 1}) = 1 + x - x + x^2 - x^2 + x^3 - x^3 + \ldots + x^N - x^N - x^{N + 1} = 1 - x^{N + 1}.
So S_N = \sum_{n = 0}^N x^n = 1 + x + x^2 + x^3 + \ldots + x^N = \frac{1 - x^{N + 1}}{1 - x}.
So the limit is \lim_{N \to \infty} \frac{1 - x^{N + 1}}{1 - x} = \frac{1}{1 - x} - \frac{x}{1 - x} \lim_{N \to \infty} x^N.
If x = 1, then \lim_{N \to \infty} \sum_{n = 0}^N 1^n = 1 + \ldots + 1 = \infty, and the sequence diverges.
If x = -1, then \lim_{N \to \infty} \sum_{n = 0}^N (-1)^n = 1 - 1 + 1 - 1 + \ldots + 1 - 1 = ?. This sum either has the value 0 or 1, and oscillates infinitely as we go to infinity. Therefore, the sequence diverges.
Clearly, if x > 1 or x < -1, the value goes to positive or negative infinity, and the sequence diverges.
Therefore, the geometric series converges only for -1 < x < 1 (\abs{x} < 1), since if this is the case, \lim_{N \to \infty} x^N = 0 and \lim_{N \to \infty} S_N = \frac{1}{1 - x}.
Does \sum_{k = 1}^\infty 3^{2k}5^{1 - k} converge?
Clearly, \sum_{k = 1}^\infty 3^{2k}5^{1 - k} = \sum_{k = 1}^\infty 9^k5^{1 - k} = 5\sum_{k = 1}^\infty 9^k5^{-k} = 5\sum_{k = 1}^\infty \frac{9^k}{5^k} = 5\sum_{k = 1}^\infty \left(\frac{9}{5}\right)^k.
Since \frac{9}{5} \ge 1, the geometric sequence diverges as the limit of \left(\frac{9}{5}\right)^k goes to infinity.
Does 3 - \frac{3}{2}x^2 + \frac{3}{4}x - \frac{3}{8}x^2 converge? ;wip: what was the original question again? this doesn't seem right
Clearly, 3 - \frac{3}{2}x^2 + \frac{3}{4}x - \frac{3}{8}x^2 = 3 \sum_{n = 0}^\infty \left(-\frac{x}{2}\right)^n.
Clearly, in order to converge, -1 \le -\frac{x}{2} \le -1, or -1 \le \frac{x}{2} \le 1.
As an aside, it converges to \frac{3}{1 + \frac{x}{2}} = \frac{6}{2 + x}.
# 28/2/14
The value of the geometric series is S_N = \sum_{n = 0}^N x^n = 1 + x + x^2 + \ldots + x^N = \begin{cases} \frac{1 - x^N}{1 - x} &\text{if } x \ne 1 \\ N + 1 &\text{if } x = 1 \end{cases}
### Convergence
One of the simplest tests for convergence is that if the terms of a series are not getting closer and closer to 0, the series can never converge. In other words, \lim_{n \to \infty} a_n = 0 is a requirement for convergence.
Formally, if \lim_{n \to \infty} \abs{a_n} \ne 0, then the sequence does not converge.
This is known as the simple limit test for series.
Our convergence tests can decide whether a series converges or not, but do not provide information about what it converges to.
The sum of two convergent series is also convergent - \sum a_n + \sum b_n = \sum (a_n + b_n) if \sum a_n and \sum b_n are both convergent. However, the sum of two divergent series is not necessarily always divergent. For example, \sum_{n = 1}^\infty (-1)^n and \sum_{n = 1}^\infty (-1)^{n + 1} are both divergent, but \sum_{n = 1}^\infty {(-1)^n + (-1)^{n + 1}} = 0.
## P-series
A P-series is a series of the form \sum_{n = 1}^\infty \frac{1}{n^P} where P \in \mb{R}.
### Convergence
We want to figure out which values of P allow the series to converge.
Clearly, \sum_{n = 1}^\infty \frac{1}{n^P} = \sum_{n = 1}^\infty n^{-P}.
Clearly, if P \le 0, then \lim_{n \to \infty} n^{-P} \ne 0. So by the simple limit test, the sequence does not converge.
If P > 0, then f(n) = \frac{1}{n^P} is continuous, positive, and decreasing.
Clearly, \int_1^\infty \frac{1}{x^P} \dee x = \frac{1}{1 - P} \evalat{x^{1 - P}}_1^\infty converges if and only if P > 1.
So by the integral test, \sum_{n = 1}^\infty \frac{1}{n^P} converges if and only if P > 1.
So \sum_{n = 1}^\infty \frac{1}{n^P} converges if and only if P > 1.
### Considerations
The series \sum_1^\infty \frac{1}{n} is called the harmonic series. It diverges (very slowly, like the logarithmic functions), and this can be proven since it is a special case of the P-series where P = 1. This is a useful series because \lim_{n \to \infty} \frac{1}{n} = 0, yet the series diverges, so this is an example of the simple convergence test not giving a conclusive result.
For example, a P-series where P = 2 is \frac{\pi^2}{6}. ;wip: how?
As an aside, we have closed forms for P-series for all even P - always in the form of \frac{m}{n}\pi^P. However, we do not know anything at this time about odd P.
For example, a P-series where P = 3 results in a value that has no known exact form, and is still an unsolved problem in methematics.
## Integral Test
This is based on the connection between an infinite series (like a Riemann sum) and integration.
Let \sum_{n = 0}^\infty a_n be an infinite series. Then we can write a_n be a function of n, like f(n).
For example, for \sum_{n = 0}^\infty \frac{1}{n^2}, f(n) = \frac{1}{n^2}.
Now we can develop the integral test.
Given f(n) = a_n, if f(n) is continuous, positive, and decreasing for all n \ge 1, then \int_1^\infty f(n) \dee n converges if and only if \sum_{n = 1}^\infty a_n converges.
### Error Estimation
For series that satisfy the hypetheses of the integral test, we can estimate the error/remainder R_N = \sum_1^\infty a_n - \sum_1^N a_n for any N.
This is useful because we can't explicitly calculate \sum_1^\infty a_n, but we can calculate \sum_1^N a_n.
So given f(n) = a_n being continuous, positive, and decreasing, the error is bounded by \int_{N + 1}^\infty f(x) \dee x \le R_N \le \int_N^\infty f(x) \dee x.
In other words, the error is bounded between \int_{N + 1}^\infty f(x) and \int_N^\infty f(x) \dee x.
For example, for the series S_N = \sum_{n = 1}^N \frac{1}{n^2}, S_{10} \approxeq 1.54977. We know that since f(n) = \frac{1}{n^2} is continuous, positive, and decreasing, and \int_1^\infty \frac{1}{n^2}. ;wip
# 3/3/14
## Comparison Test (Series)
There is a deep connection between infinite series and improper integrals.
Like the improper integral, we can also have a form of the comparison test, but it is easier to use than with integrals.
Given \sum a_n and \sum b_n, two series, with 0 \le a_n \le b_n:
• If \sum a_n diverges, then \sum b_n also diverges.
• If \sum b_n converges, then \sum a_n also converges.
Here, \sum a_n is the series we are interested in and \sum b_n is the series we chose to compare it with.
### Limit Comparison Test
For series only - not integrals - we can rewrite this in a more useful form.
Let \sum a_n and \sum b_n be two series, with a_n \ge 0, b_n \ge 0.
Let \rho = \lim_{n \to \infty} \frac{a_n}{b_n}.
If 0 < \rho < \infty (\rho is positive and finite), then:
• \sum a_n and \sum b_n both diverge.
• \sum a_n and \sum b_n both converge.
This is easily proved via contradiction or similar.
Note that if \rho = 0 or \rho = \infty, then we cannot say anything about whether they converge or not, and we need to pick a better comparison.
Prove that \sum_{n = 1}^\infty \frac{\abs{\sin n}}{\sqrt{n + n^3}} converges:
Clearly, \frac{\abs{\sin n}}{\sqrt{n + n^3}} \le \frac{1}{\sqrt{n + n^3}} < \frac{1}{\sqrt{n^3}} = \frac{1}{n^\frac{3}{2}}.
Since this is a P-series where P = \frac{3}{2}, \sum_{n = 1}^\infty \frac{1}{n^\frac{3}{2}} converges.
So by the comparison test, \sum_{n = 1}^\infty \frac{\abs{\sin n}}{\sqrt{n + n^3}} converges.
# 5/3/14
We can find useful comparisons to make by finding something bigger than the numerator, and smaller than the denominator, and trying that as a comparison.
For example, \sum_1^\infty \frac{\sqrt{n}}{n^2 + 2} < \sum_1^\infty \frac{\sqrt{n}}{n^2} = \sum_1^\infty \frac{1}{n^\frac{3}{2}}, which converges as it is a P-series with P = \frac{3}{2}.
For example, \sum_1^\infty \frac{1}{2^nn} \le \sum_1^\infty \frac{1}{2^n} = \sum_1^\infty \left(\frac{1}{2}\right)^n, which converges since it is a geometric series with x = \frac{1}{2}.
Does \sum_1^\infty \frac{1}{n^{1 + \frac{1}{n}}} exist?
Clearly, \sum_1^\infty \frac{1}{n^{1 + \frac{1}{n}}} = \sum_1^\infty \frac{1}{n}\frac{1}{n^\frac{1}{n}} < \sum_1^\infty \frac{1}{n}.
Using the limit comparison test, \rho = \lim_{n \to \infty} \frac{\frac{1}{n}\frac{1}{n^\frac{1}{n}}}{\frac{1}{n}} = \lim_{n \to \infty} n^{-\frac{1}{n}} = e^{-\lim_{n \to \infty} \frac{1}{n}\ln n} \lH e^{-\lim_{n \to \infty} \frac{\frac{1}{n}}{1}} = e^{-0} = 1.
Since 0 < \rho < \infty, and \sum_1^\infty \frac{1}{n} diverges, then \sum_1^\infty \frac{1}{n^{1 + \frac{1}{n}}} also diverges.
Proof of limit comparison test:
Let \sum a_n and \sum b_n be two series, with a_n \ge 0, b_n \ge 0.
Let \rho = \lim_{n \to \infty} \frac{a_n}{b_n}. Assume 0 < \rho < \infty.
Clearly, \exists m, M > 0, 0 < m < \rho < M < \infty.
Clearly, \lim_{n \to \infty} \frac{a_n}{b_n} \iff (\exists K, n > K \implies m < \frac{a_n}{b_n} < M).
So mb_n < a_n < Mb_n and \sum mb_n < \sum a_n < \sum Mb_n, or m\sum b_n < \sum a_n < M\sum b_n.
Suppose \sum b_n converges. Then \sum a_n < M\sum b_n and by the comparison test, \sum a_n converges.
Suppose \sum b_n diverges. Then m\sum b_n > \sum a_n and by the comparison test, \sum a_n diverges.
So \sum a_n converges if and only if \sum b_n converges.
Also, we can prove that sequences diverge by proving that they are increasing or decreasing without bound, or assuming that the sequence does converge, and deriving a contradiction.
# 7/3/14
## Alternating Series
These are series where the sign of the terms alternate.
These usually take the form of \sum a_n = \sum (-1)^n p_n, p_n > 0.
Alternating series are useful to study because they have several useful properties.
### Error
The remainder/error after N terms is easy to estimate. Clearly, \abs{R_N} = \abs{\sum^\infty (-1)^n p_n - \sum^N (-1)^n p_n}. So \abs{R_N} \le p_{N + 1}.
This is true because the terms of the series keep bouncing back and forth across the limit, so the remainder is always bounded by the value of the next term.
For example, \sum_1^\infty (-1)^{n + 1} \frac{1}{n} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \ldots = \ln 2. We can estimate the error with \abs{R_N} = \frac{1}{N + 1}.
### Convergence
For an alternating series, if \lim_{n \to \infty} p_n = 0, and p_{n + 1} \le p_n, then the series converges.
In other words, if the terms of the sum tend to 0, and they are monotonically decreasing, then the series converges. This is one of the simplest convergence tests, but it only works on alternating series.
Alternating series also have special nomenclature that can be applied to them.
Let \sum a_n = \sum (-1)^n p_n be an alternating series.
\sum a_n is absolutely convergent if \sum a_n and \sum \abs{a_n} converge.
Absolute convergence means that the series is well behaved - it behaves like an ordinary number. We can add, subtract, multiply, and divide absolutely convergent series and the result is still sensible.
\sum a_n is conditionally convergent if \sum a_n converges, but \sum \abs{a_n} diverges.
Conditional convergence means the series might not be well behaved.
For example, consider 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \ldots.
For an absolutely convergent series, re-arrangement does nothing. This is not true for conditionally convergent series. In fact, the above can have its terms rearranged such that it sums up to any value we want, and this was proved by Riemann.
For example, we will make it sum up to 1.5:
First, we add up positive terms until we pass 1.5: 1 + \frac{1}{3} + \frac{1}{5} \approxeq 1.53.
Then we add negative terms until we fall below 1.5 again: 1 + \frac{1}{3} + \frac{1}{5} - \frac{1}{2} \approxeq 1.03.
Then we repeat: 1 + \frac{1}{3} + \frac{1}{5} - \frac{1}{2} + \frac{1}{7} + \frac{1}{9} + \frac{1}{11} + \frac{1}{13} + \frac{1}{15} \approxeq 1.52.
We can do this as many times as needed to get the desired value.
Even though there are far more positive terms than negative terms, we can do this because we have an infinite number of terms that we can add up. This only works because the series is infinite.
# 10/3/14
## Ratio Test
This is the most useful of the convergence tests.
Let \sum a_n be a series. Let L = \lim_{n \to \infty} \abs{\frac{a_{n + 1}}{a_n}}.
If L < 1, then \sum a_n converges absolutely.
If L > 1, then \sum a_n diverges.
Otherwise, if L = 1, the test is inconclusive. This test is very effective for powers and factorials, and especially for power series and Taylor series.
Determine if \sum_{n = 0}^\infty \frac{(n + 4)!}{4!n!4^n} converges:
We will use the ratio test.
Let L = \lim_{n \to \infty} \abs{\frac{((n + 1) + 4)!}{4!(n + 1)!4^(n + 1)} \frac{4!n!4^n}{(n + 4)!}} = \lim_{n \to \infty} \frac{n + 5}{4!(n + 1)!4^(n + 1)} 4!n!4^n = \lim_{n \to \infty} \frac{n + 5}{4(n + 1)!} n! = \lim_{n \to \infty} \frac{n + 5}{4(n + 1)} \lH \lim_{n \to \infty} \frac{1}{4} = \frac{1}{4}.
Since L < 1, \sum_{n = 0}^\infty \frac{(n + 4)!}{4!n!4^n} converges absolutely.
Proof:
The idea is that if L < 1, then we can compare \sum a_n to a convergent geometric series.
Assume L < 1. Then \exists r \in \mb{R}, L < r < 1.
Note that \lim_{n \to \infty} \abs{\frac{a_{n + 1}}{a_n}} means that \exists K > 0, n \ge K \implies \abs{\frac{a_{n + 1}}{a_n}} < L.
So \exists 0 \le N \le K, n \ge N \implies \abs{\frac{a_{n + 1}}{a_n}} < r.
Assume n \ge N. Then \abs{\frac{a_{n + 1}}{a_n}} < r and \abs{a_{n + 1}} < r\abs{a_n}.
So \forall k, \abs{a_{N + k}} < \abs{a_N}r^k. For example, \abs{a_{N + 3}} < \abs{a_{N + 2}}r < \abs{a_{N + 1}}r^2 < \abs{a_N}r^3.
So \sum_{k = 0}^\infty \abs{a_{N + k}} < \sum_{k = 0}^\infty \abs{a_{N}}r^k.
Clearly, \sum_{k = 0}^\infty \abs{a_{N}}r^k = \abs{a_{N}}\sum_{k = 0}^\infty r^k, which converges since it is a geometric series with x < 1.
So by the comparison test, \sum a_n < \abs{a_{N}}\sum_{k = 0}^\infty r^k, so \sum a_n converges.
A similar proof can be made for the case when L > 1.
Another example is e^x = \sum_{n = 0}^\infty \frac{x^n}{n!}. Note that if we take the limit of the series, we get the same series back.
### Root Test
Let \sum a_n be a series. Let L = \lim_{n \to \infty} \sqrt[n]{\abs{a_n}}.
If L < 1, then \sum a_n converges absolutely.
If L > 1, then \sum a_n diverges.
This can be proved in a manner similar to the ratio test.
# 12/3/14
## Using Convergence Tests
Let \sum a_n be a series.
1. Simple Limit Test: If \lim_{n \to \infty} a_n \ne 0, then \sum a_n diverges.
2. Alternating Series Test: If the series is alternating and decreasing (\abs{a_{n + 1}} < \abs{a_n}), then \sum a_n converges.
3. Ratio Test/Root Test: Let L = \lim_{n \to \infty} \abs{\frac{a_{n + 1}}{a_n}} or L = \lim_{n \to \infty} \sqrt[n]{\abs{a_n}}, whichever is easier to evaluate:
1. If L < 1, then \sum a_n converges.
2. If L > 1, then \sum a_n diverges.
4. If a_n is an algebraic (polynomials and roots) function of n, then try using the Comparison Test with a P-series, \sum \frac{1}{n^P}.
5. If a_n is closely related to a geometric series, then try using the Comparison Test with a geometric series, \sum x^n.
6. If f(n) = a_n is positive, decreasing, and integrable, then try the Integral Test.
Prove whether the following converge or diverge:
• \sum_{n = 1}^\infty \frac{n^2 + 7n}{\sqrt{n^5 + 4n^3 - 2}} \le \frac{1}{\sqrt{n}}.
• \sum_{n = 1}^\infty \frac{n + 5}{5^n}
• \sum_{n = 1}^\infty n^2 e^{-n}
• \sum_{n = 1}^\infty \frac{1}{(\ln n)^{\ln n}}
• \sum_{n = 1}^\infty \frac{e^\frac{1}{n}}{n^2}
;wip
## Power Series
A power series is a series of the form \sum_{n = 0}^\infty a_n (x - x_0)^n, where a_n \in \mb{R}.
Here, x_0 is the center of the power series.
We want to figure out for which values of x our series converges for. For this, we will usually use the ratio test.
For example, check if \sum_{n = 1}^\infty \frac{(-1)^n x^n}{3^n \sqrt{n}} converges:
Let L = \lim_{n \to \infty} \abs{\frac{(-1)^{n + 1} x^{n + 1}}{3^{n + 1} \sqrt{n + 1}} \frac{3^n \sqrt{n}}{(-1)^n x^n}} = \lim_{n \to \infty} \abs{\frac{-x}{3 \sqrt{n + 1}} \sqrt{n}} = \frac{\abs{x}}{3} \lim_{n \to \infty} \frac{1}{\sqrt{n + 1}} \sqrt{n} = \frac{\abs{x}}{3} \lim_{n \to \infty} \sqrt{\frac{n}{n + 1}} = \frac{\abs{x}}{3}. ;wip: use l'hospital's rule or something to evaluate the limit properly Clearly, L < 1 whenever -3 < x < 3, so the series converges for any -3 < x < 3.
Clearly, L > 1 whenever x < -3 \vee x > 3, so the series diverges for any x < -3 \vee x > 3.
Now we need to consider L = 1, where x = \pm 3. Assume x = 3.
Then \sum_{n = 1}^\infty \frac{(-1)^n x^n}{3^n \sqrt{n}} = \sum_{n = 1}^\infty \frac{(-1)^n 3^n}{3^n \sqrt{n}} = \sum_{n = 1}^\infty \frac{(-1)^n}{\sqrt{n}}.
This is an alternating series, and since \lim_{n \to \infty} \frac{1}{\sqrt{n}} = 0 and \frac{1}{\sqrt{n + 1}} < \frac{1}{\sqrt{n}}, the series converges (conditionally) by the alternating series test.
Assume x = -3.
Then \sum_{n = 1}^\infty \frac{(-1)^n x^n}{3^n \sqrt{n}} = \sum_{n = 1}^\infty \frac{(-1)^n (-3)^n}{3^n \sqrt{n}} = \sum_{n = 1}^\infty \frac{(-1)^n (-1)^n 3^n}{3^n \sqrt{n}} = \sum_{n = 1}^\infty \frac{1}{\sqrt{n}}.
This is a P-series where P = \frac{1}{2}, so the series diverges.
So the series converges absolutely if and only if x \in (-3, 3), and converges conditionally for x = 3.
The radius of convergence for a power series is the maximum magnitude of the values of \abs{x - x_0} on the imaginary plane such that all numbers inside this radius of x_0 allow the power series to converge absolutely, and all numbers outside of this radius diverge.
The behaviour when \abs{x - x_0} = \rho is not important, just that it is a bounding radius.
The values of x that allow the series to converge are known as the interval of convergence. Unlike the radius of convergence, we also need to consider the endpoints of the interval for convergence.
This is a number \rho \in \mb{R} such that:
• \sum_{n = 0}^\infty c_n (x - x_0)^n converges absolutely for all \abs{x - x_0} < \rho
• \sum_{n = 0}^\infty c_n (x - x_0)^n diverges for all \abs{x - x_0} > \rho
• \sum_{n = 0}^\infty c_n (x - x_0)^n may converge or diverge for \abs{x - x_0} = \rho
In the above example, the radius of convergence is 3.
All geometric series are power series where a_n = 1 and x_0 = 0.
The interval of convergence for a geometric series \sum_{n = 0}^\infty x^n is \abs{x} < 1.
# 14/3/14
Within the radius of convergence, a power series behaves like an ordinary function. The idea is that if we are inside the radius of convergence, then we can treat the infinite series like a finite polynomial.
Let f(x) = \sum_{n = 0}^\infty a_n (x - x_0)^n, g(x) = \sum_{n = 0}^\infty b_n (x - x_0)^n:
• f(x) \pm g(x) = \sum_{n = 0}^\infty (a_n \pm b_n) (x - x_0)^n.
• f(x) g(x) = \sum_{n = 0}^\infty c_n (x - x_0)^n where c_n = a_0 b_n + a_1 b_{n - 1} + \ldots + a_{n - 1} b_1 + a_n b_0.
• If g(x) \ne 0, then \frac{f(x)}{g(x)} = \sum_{n = 0}^\infty d_n (x - x_0)^n ;wip: what is d_n?
More importantly, we can differentiate and integrate the terms of the sum:
So \frac{\dee f}{\dee x} = \frac{\dee}{\dee x} (a_0 + a_1(x - x_0) + a_2 (x - x_0)^2 + \ldots) = 0 + a_1 + 2a_2 (x - x_0) + \ldots. Note that the first term became 0, so we don't need to sum it.
So \frac{\dee f}{\dee x} = \sum_{n = 0}^\infty a_n n(x - x_0)^{n - 1}.
So \int f(x) \dee x = \sum_{n = 0}^\infty \int a_n (x - x_0)^n \dee x = c + \sum_{n = 0}^\infty a_n \frac{(x - x_0)^{n + 1}}{n + 1}.
Find the power series of \frac{1}{1 - x} and use it to approximate \ln:
Clearly, this is the value of the geometric series, as we saw earlier: \frac{1}{1 - x} = \sum_{n = 0}^\infty x^n for \abs{x} < 1.
Clearly, \int_0^t \frac{1}{1 - x} \dee x = -\ln \abs{1 - t} = \sum_{n = 0}^\infty \int_0^t x^n \dee x = \sum_{n = 0}^\infty \frac{t^{n + 1}}{n + 1} for \abs{t} < 1.
So \ln \abs{1 - t} = -\sum_{n = 1}^\infty \frac{t^n}{n} for \abs{t} < 1.
Note that this only works for \abs{t} < 1. However, there are tricks we can use to avoid this issue.
For example, we can't use this approximation to find \ln 3, since for \ln \abs{1 - t}, t = -2, but note that \ln \frac{1}{3} = -\ln 3, so \ln 3 = -\ln \frac{1}{3} = -\ln \abs{1 - \frac{2}{3}} = \sum_{n = 1}^\infty \frac{1}{n} \frac{2}{3}^n.
We want a general form for all values in the domain of \ln x, so x > 0.
Clearly, \ln \abs{1 - (-t)} - \ln \abs{1 - t} = \ln \abs{\frac{1 + t}{1 - t}} = -\sum_{n = 1}^\infty \frac{(-t)^n}{n} + \sum_{n = 1}^\infty \frac{t^n}{n} = \sum_{n = 1}^\infty (1 - (-1)^n)\frac{t^n}{n} = 2t + \frac{2}{3}t^3 + \frac{2}{5}t^5 + \ldots = \sum_{n = 1}^\infty \frac{2}{2n - 1}t^{2n - 1}.
Let x = \abs{\frac{1 + t}{1 - t}}. Then t = \abs{\frac{x - 1}{x + 1}} = \frac{x - 1}{x + 1}, since x > 0.
So \abs{t} < 1 \iff -x < x < x + 2 \iff x > 0, so \abs{t} < 1 for all x in the domain, as required.
Then we can write \ln x = \ln \frac{1 + t}{1 - t} = \sum_{n = 1}^\infty \frac{2}{2n - 1}\left(\frac{x - 1}{x + 1}\right)^{2n - 1}, for x > 0.
This is actually how calculators evaluate these sorts of functions - by calculating the partial sums of a power series to a given error.
Find the power series for \frac{1}{1 + x^2} and use it to approximate \arctan:
Let u = -x^2. Then \frac{1}{1 + x^2} = \frac{1}{1 - u} = \sum_{n = 0}^\infty u^n = \sum_{n = 0}^\infty (-1)^n x^{2n} for \abs{u} < 1.
Clearly, \int_0^t \frac{1}{1 + x^2} \dee x = \arctan t = \sum_{n = 0}^\infty \int_0^t (-1)^n x^{2n} \dee x = \sum_{n = 0}^\infty \frac{(-1)^n}{2n + 1} t^{2n + 1} for \abs{t} < t.
So \arctan t = \sum_{n = 0}^\infty \frac{(-1)^n}{2n + 1} t^{2n + 1} for -1 < t < 1.
## Taylor Series
What power series is identical to a function f(x) and all its derivatives at a point x = x_0?
For \abs{x - x_0} < \rho, where \rho is the radius of convergence, f(x) = \sum_{n = 0}^\infty \frac{f^{(n)}(x_0)}{n!}(x - x_0)^n where f^{(n)}(x_0) = \evalat{\frac{\dee^n f}{\dee x^n}}_{x = x_0}, and f^{(0)}(x_0) = f(x_0). This is known as a Taylor series.
A Taylor polynomial is a partial sum of a Taylor series - all partial sums of Taylor series are simply polynomials. \sum_{n = 0}^k \frac{f^{(n)}(x_0)}{n!}(x - x_0)^n is known as a k-degree Taylor polynomial.
Find the Taylor polynomial for e^x and use it to estimate \int_0^1 e^{-x^2} \dee x:
Let f(x) = e^x, x_0 = 0. Then f^{(n)} = \evalat{\frac{\dee^n f}{\dee x^n}}_{x = x_0} = \evalat{e^x}_{x = x_0} = 1.
Then f(x) = e^x = \sum_{n = 0}^\infty \frac{x^n}{n!}, \abs{x - x_0} < \rho.
Let L = \lim_{n \to \infty} \abs{\frac{x^{n + 1}}{(n + 1)!} \frac{n!}{n^n}} = \lim_{n \to \infty} \abs{\frac{x}{n + 1}} = 0.
Then by the ratio test, the series converges for all 0 < 1, and the radius of covnergence is \rho = \infty.
So \int_0^1 e^{-x^2} \dee x = \int_0^1 f(-x^2) \dee x = \sum_{n = 0}^\infty \int_0^1 \frac{(-x^2)^n}{n!} \dee x = \sum_{n = 0}^\infty \frac{(-1)^n}{n!(2n + 1)}.
# 17/3/14
;wip: check to make sure assignment marks are on LEARN
The Taylor series has the same derivative and integral as the function for all possible values of x within the radius of convergence.
The Nth partial sum of a Taylor series has the same zero to Nth derivatives.
Find the Taylor series for \sin x and \cos x:
Let f(x) = \sin x, x_0 = 0. Then f^{(n)} = \evalat{\frac{\dee^n f}{\dee x^n}}_{x = x_0}. Clearly, f^{(0)}(0) = 0, f^{(1)}(0) = 1, f^{(1)}(0) = 0, f^{(1)}(0) = -1, \ldots.
Since all the even powers are 0, then we simply omit those terms.
So f(x) = \sin x = \sum_{n = 0}^\infty \frac{f^{(n)}(x_0)}{n!}x^n = x - \frac{x^3}{3} + \frac{x^5}{5!} - \frac{x^7}{7!} + \ldots = \sum_{n = 0}^\infty \frac{(-1)^n x^{2n + 1}}{(2n + 1)!}, \abs{x - x_0} < \rho.
Let L = \lim_{n \to \infty} \abs{\frac{(-1)^{n + 1} x^{2n + 3}}{(2n + 3)!} \frac{(2n + 1)!}{(-1)^n}} = \lim_{n \to \infty} \abs{\frac{x^2}{(2n + 3)(2n + 2)}} = 0.
Then by the ratio test, the series converges for all 0 < 1, and the radius of covnergence is \rho = \infty.
So the Taylor series is \sin x = \sum_{n = 0}^\infty \frac{(-1)^n x^{2n + 1}}{(2n + 1)!}.
We can find the Taylor series for \cos x by taking derivatives of both sides: \frac{\dee}{\dee x} \sin x = \cos x = \frac{\dee}{\dee x} \sum_{n = 0}^\infty \frac{(-1)^n x^{2n + 1}}{(2n + 1)!} = \sum_{n = 0}^\infty \frac{(-1)^n x^{2n}}{(2n)!}.
Taylor series allow us to do interesting analyses. For example, it is easy to see that \sin(-x) = -\sin x from the fact that \sum_{n = 0}^\infty \frac{(-1)^n (-x)^{2n + 1}}{(2n + 1)!} = -\sum_{n = 0}^\infty \frac{(-1)^n x^{2n + 1}}{(2n + 1)!}.
;wip: use the taylor series to show that e^{i \theta} = \cos \theta + \imag \sin \theta - prove euler's identity
We want to find a simple formula to find (1 + x)^P for any natural number P. This was found to be 1 + Px + \frac{P(P - 1)x^2}{2!} + \frac{P(P - 1)(P - 2)}{3!} + \ldots + \frac{P \cdot (P - 1) \cdot \ldots \cdot (P - (n - 1))}{n!}x^n + \ldots.
Clearly, for any whole number P the series must terminate after P + 1 terms - when one of the factors (P - k) becomes 0. However, if P is a fraction, then the series actually becomes infinite. As it turns out, this works for negative numbers too.
For example, \sqrt{1 - x} = (1 + x)^\frac{1}{2} = 1 - \frac{1}{2}x - \frac{1}{8}x^2 - \frac{1}{16}x^3 - \ldots.
Some common Taylor series are:
• Geometric series: \sum_{n = 0}^\infty x^n = \frac{1}{1 - x}, \abs{x - x_0} < \rho
• Exponential series: \sum_{n = 0} \frac{x^n}{n!} = e^x, \abs{x - x_0} < \rho
• Trigonometric series: \sin x = \sum_{n = 0}^\infty \frac{(-1)^n x^{2n + 1}}{(2n + 1)!}
• Binomial series: 1 + Px + \frac{P(P - 1)}{2!}x^2 + \frac{P(P - 1)(P - 2)}{3!}x^3 + \ldots + \frac{P \cdot (P - 1) \cdot \ldots \cdot (P - (n - 1))}{n!}x^n + \ldots = (1 + x)^P
# 19/3/14
## Properties of Taylor Series
So far we have derived power series by integrating geometric series, or by using the Taylor formula.
Is the Taylor series of a function unique? Are there multiple ways to write out the power series of a function?
First, we define the partial sum of a Taylor series for a function f(x), P_{N, x_0}(x) = \sum_{n = 0}^N \frac{f^{(n)}(x_0)}{n!}(x - x_0)^n.
Then the remainder - the difference between the function and the partial sum - is R_N(x) = f(x) - P_{N, x_0}(x) = \sum_{n = N + 1}^\infty \frac{f^{(n)}(x_0)}{n!}(x - x_0)^n.
For example, given f(x) = e^x, P_{3, 0}(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} and R_{3, 0} = e^x - 1 - x - \frac{x^2}{2!} - \frac{x^3}{3!}.
### Uniqueness of Taylor Series
If f^{(n)}(x) exists for all n \in \mb{N} for some interval \mb{I} such that x_0 \in \mb{I}, and \forall n \in \mb{N}, x \in \mb{I}, \abs{f^{(n)}(x)} < \infty, then \forall x \in \mb{I}, \lim_{N \to \infty} \abs{f(x) - P_{N, x_0}(x)} = 0.
In other words, if the function has all derivatives on an interval containing x = x_0, and all these derivatives are finite, then as we add more terms to the partial sum, the partial sum eventually converges to exactly the function.
As a result, P_{N, x_0} is unique. Practically speaking, this means we can generate Taylor series using any method we want to, and the result will always be the one true Taylor series for a given f(x) and x_0.
Proof:
;wip
Derive the Taylor series for f(x) = e^{-x^2} for x_0 = 0:
Clearly, f^{(0)}(0) = 1, f^{(1)}(0) = 0, \ldots, and the derivatives keep getting more and more complicated as we keep differentiating. It is impractical to calculate f^{(n)}.
Instead, we use a substitution: let u = -x^2. Then e^u = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \ldots.
Then e^{-x^2} = 1 - x^2 + \frac{x^4}{2!} - \frac{x^6}{3!} + \ldots.
This is a faster way to derive series - by using substitutions to make the functions simpler, converting the simpler function into a series, and then substituting the variable back into the series.
Derive a Taylor series for f(x) = \sin x^3 at x_0 = 0:
Let u = x^3. Then \sin x^3 = \sin u = \sum_{n = 0}^\infty \frac{(-1)^n u^{2n + 1}}{(2n + 1)!} = \sum_{n = 0}^\infty \frac{(-1)^n x^{6n + 3}}{(2n + 1)!} = \sin x^3.
Derive the Taylor series for f(x) = \frac{1 + x}{1 - x}:
Clearly, f(x) = (1 + x)\frac{1}{1 - x} = (1 + x)\sum_{n = 0}^\infty x^n = \sum_{n = 0}^\infty x^n + \sum_{n = 1}^\infty x^n = 1 + \sum_{n = 1}^\infty x^n + \sum_{n = 1}^\infty x^n = 1 + 2\sum_{n = 1}^\infty x^n.
Derive the Taylor series for \arcsin x:
Clearly, \arcsin x = \int_0^x \frac{1}{\sqrt{1 - t^2}} \dee t.
Let u = -t^2. Clearly, \frac{1}{\sqrt{1 - t^2}} = (1 + u)^{-\frac{1}{2}} = 1 - \frac{1}{2}u + \frac{3}{4}\frac{u^2}{2!} - \frac{15}{8}\frac{u^3}{3!} + \frac{105}{16} \frac{u^4}{4!} + \ldots.
So (1 + u)^{-\frac{1}{2}} = \frac{1}{\sqrt{1 - t^2}} = 1 + \frac{1}{2}t^2 + \frac{3}{2^2}\frac{t^4}{2!} + \frac{15}{2^3}\frac{t^6}{3!} + \frac{105}{2^4} \frac{t^8}{4!} + \ldots.
So \int_0^x \frac{1}{\sqrt{1 - t^2}} \dee t = \arcsin x = \int_0^x 1 \dee t + \int_0^x \frac{1}{2}t^2 \dee t + \int_0^x \frac{3}{2^2}\frac{t^4}{2!} \dee t + \int_0^x \frac{15}{2^3}\frac{t^6}{3!} \dee t + \int_0^x \frac{105}{2^4} \frac{t^8}{4!} \dee t + \ldots = x + \frac{1}{3}\frac{1}{2}t^3 + \frac{1}{5}\frac{3}{2^2}\frac{t^5}{2!} + \frac{1}{7}\frac{15}{2^3}\frac{t^7}{3!} + \frac{1}{9}\frac{105}{2^4} \frac{t^9}{4!} + \ldots.
# 21/3/14
## Truncation Error
We can actually estimate the error in P_{N, x_0}(x), even without knowing much about f(x).
### Taylor's Remainder Theorem
If f^{(n + 1)}(x) is continuous on an interval \mb{I} such that x_0 \in \mb{I}, then \forall x \in \mb{I}, \exists \min(x_0, x) < c < \max(x_0, x), f(x) - P_{N, x_0}(x) = R_N(x) = \frac{f^{(N + 1)}(c)}{(N + 1)!}(x - x_0)^{N + 1}.
In other words, there exists a value c between x and x_0 exclusive such that the error is \frac{f^{(N + 1)}(c)}{(N + 1)!}(x - x_0)^{N + 1} - the N + 1th term evaluated with a certain value of the N + 1th derivative.
This can be proved using the Intermediate Value Theorem, but it is a messy proof.
Clearly, if f(x) - P_{N, x_0}(x) = \frac{f^{(N + 1)}(c)}{(N + 1)!}(x - x_0)^{N + 1}, then \abs{f(x) - P_{N, x_0}(x)} = \abs{R_N(x)} = \abs{\frac{f^{(N + 1)}(c)}{(N + 1)!}(x - x_0)^{N + 1}} = \abs{f^{(N + 1)}(c)}\abs{\frac{(x - x_0)^{N + 1}}{(N + 1)!}} = \abs{f^{(N + 1)}(c)}\frac{\abs{x - x_0}^{N + 1}}{(N + 1)!}.
Clearly, if \exists M \in \mb{R}, \forall x \in \mb{I}, \abs{f^{(N + 1)}(x)} \le M, then \abs{f^{(N + 1)}(c)} \le M and \abs{R_N(x)} \le M\frac{\abs{x - x_0}^{N + 1}}{(N + 1)!}.
This is called the Taylor's inequality, and is a more practically useful form of the Taylor Remainder Theorem.
Basically, if f^{(n + 1)}(x) is continuous on an interval \mb{I} such that x_0 \in \mb{I}, and \exists M \in \mb{R}, \forall x \in \mb{I}, \abs{f^{(N + 1)}(x)} \le M, then \abs{R_{N, x_0}} \le M\frac{\abs{x - x_0}^{N + 1}}{(N + 1)!}.
In other words, if we can bound f^{N + 1}(x), then we can bound the error.
Estimate the error in the Taylor polynomial for \sin x about x_0 = 0 after N terms:
Clearly, the error is \abs{R_N(x)} = \abs{\sin x - \sum_{n = 0}^N \frac{(-1)^n x^{2n + 1}}{(2n + 1)!}}.
We want to find an M such that \abs{f^{(((2N + 1) + 1))}(x)} \le M. Note that we simply added 1 to the derivative depth, not to N itself (N here actually means 2N + 1).
Clearly, f^{(((2N + 1) + 1))}(x) is either \pm \sin x or \pm \cos x, and so \abs{f^{(2N + 2)}(x)} \le 1.
Then by the Taylor Remainder Theorem, \abs{R_N(x)} \le \frac{\abs{x - x_0}^{N + 1}}{(N + 1)!} = \frac{\abs{x}^{N + 1}}{(N + 1)!}.
We almost always want to minimise the error. To do this, we should keep x as close to x_0 as possible (x - x_0 can be made smaller), or use more terms in the partial sum ((N + 1) can be made larger). Also, we always want the smallest possible M, so we want to pick the tightest possible bound for f^{(N + 1)}(x).
Estimate the error in the Taylor polynomial for e^x about x_0 = 0 after N terms for x \in [-1, 1], and determine the number of terms before the error is less than 0.000005:
Clearly, the error is \abs{R_N(x)} = \abs{e^x - \sum_{n = 0}^N \frac{x^n}{n!}}.
We want to find M such that \abs{f^{(N + 1)}(x)} \le M.
Clearly, f^{(N + 1)}(x) = e^x, and the largest possible value occurs at x = 1, so \abs{f^{(N + 1)}(x)} \le e^1, so M \ge e.
Then by the Taylor Remainder Theorem, \abs{R_N(x)} \le e\frac{\abs{x}^{N + 1}}{(N + 1)!}. We can use any M \ge e we want to simplify our calculations, but larger M means less useful error bounds.
We want \abs{R_N(x)} \le e\frac{\abs{x}^{N + 1}}{(N + 1)!} \le t(N) \le 0.000005, where t(N) is a function of N only, without x.
Since \abs{x}^{N + 1} \le 1, \frac{e}{(N + 1)!} \le \frac{e\abs{\pm 1}^{N + 1}}{(N + 1)!} \le 0.000005, or (N + 1)! \le 200000e.
Clearly, 400000 = 200000 \cdot 2 \le 200000e \le 200000 \cdot 3 = 600000.
Since 9! \le 200000 and 10! > 600000, N + 1 \ge 10 \iff (N + 1)! > 200000e and N \ge 9.
So the error is less than 0.000005 when N \ge 9.
# 24/3/14
A Maclaurin series is a Taylor series where x_0 = 0 - a Taylor series centered around 0.
Clearly, \sum_{n = 0}^\infty \frac{f^{(n)}(x_0)}{n!}(x - x_0)^n = \frac{f^{(0)}(x_0)}{0!}(x - x_0)^0 + \ldots + \frac{f^{(N)}(x_0)}{N!}(x - x_0)^N + R_N(x) for some finite N.
Let \mb{I} = [a, b] such that x_0 \in \mb{I}, within the interval of convergence. Clearly, Taylor series are always continuous.
By the Taylor remainder theorem, \abs{R_{N, x_0}} = \frac{f^{(N + 1)}(c)}{(N + 1)!}\abs{x - x_0}^{N + 1} for some c. Clearly, \frac{f^{(N + 1)}(c)}{(N + 1)!}\abs{x - x_0}^{N + 1} is a scalar multiple of \frac{f^{(N + 1)}(x)}{(N + 1)!}\abs{x - x_0}^{N + 1}.
So R_N(x) = O(1) \frac{f^{(N + 1)}(x)}{(N + 1)!}\abs{x - x_0}^{N + 1}. In other words, we can write the remainder as a scalar multiple of the next term.
So we can actually write the series as \frac{f^{(0)}(x_0)}{0!}(x - x_0)^0 + \ldots + \frac{f^{(N)}(x_0)}{N!}(x - x_0)^N + O(1) \frac{f^{(N + 1)}(x_0)}{(N + 1)!}(x - x_0)^{N + 1}.
For example, \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + O(1)\frac{x^9}{9!}.
## Operations on Taylor Series
The reason we use Taylor series to represent functions is in order to more easily evaluate limits, integrals, and infinite series. It is also useful for approximating the functions with actual numbers, all while being able to get the error bounds in our evaluation.
### Limits
If we take a limit as x \to x_0, then the coefficients of a Taylor series already has l'Hospital's rule built in.
Consider \lim_{x \to 0} \frac{\sin x}{x}:
We could find this geometrically, using l'Hospital's rule, or using infinite series.
Clearly, \lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{x - \frac{x^3}{3!} + \frac{x^5}{5!} + M\frac{x^7}{7!}}{x} = \lim_{x \to 0} 1 - \frac{x^2}{3!} + \frac{x^4}{5!} + M\frac{x^6}{7!} = 1 - 0 + 0 + M0 = 1.
Evaluate \lim_{x \to 0} \frac{e^{\sin x} - 1}{x}:
Let u = \sin x. Clearly, \lim_{x \to 0} \frac{e^{\sin x} - 1}{x} = \lim_{x \to 0} \frac{e^u - 1}{x} = \lim_{x \to 0} \frac{(1 + u + \frac{u^2}{2!} + \ldots) - 1}{x} = \lim_{x \to 0} \frac{\sin x + \frac{\sin^2 x}{2!} + \ldots}{x} = \lim_{x \to 0} \frac{\sin x}{x}\left(1 + \frac{\sin x}{2!} + \ldots\right) = \lim_{x \to 0} \frac{\sin x}{x} \lim_{x \to 0} \left(1 + \frac{\sin x}{2!} + \ldots\right) = 1.
However, it is usually faster and easier to just use l'Hospital's rule.
# 26/3/14
The Taylor inequality basicaly states that the remainder of a Taylor polynomial is a multiple of the next term if the derivative is bounded.
### Estimating Definite Integrals
e^{-x^2} is a Guassian function, which have the general form of e^{-\frac{(x - a)^2}{2\sigma^2}}. This is also an error function. It is heavily used in statistics as a probability distribution, and looks like a bell curve. \sigma represents the standard deviation of the distribution.
This is also associated with scientific literature claims of "six-sigma" accuracy or similar.
Evaluate \int_0^1 e^{-x^2} \dee x:
Let u = -x^2. Clearly, e^{-x^2} = \sum_{n = 0}^\infty \frac{(-x^2)^n}{n!} = \sum_{n = 0}^\infty \frac{(-1)^nx^{2n}}{n!}.
So \int_0^1 e^{-x^2} \dee x = \sum_{n = 0}^\infty \frac{(-1)^n}{n!}\int_0^1 x^{2n} \dee x = \sum_{n = 0}^\infty \frac{(-1)^n}{n!}\evalat{\frac{x^{2n + 1}}{2n + 1}}_0^1 = \sum_{n = 0}^\infty \frac{(-1)^n}{n!(2n + 1)}.
Estimate \int_0^1 x^x \dee x:
Clearly, \int_0^1 x^x \dee x = \int_0^1 e^{x \ln x} \dee x = \sum_{n = 0}^\infty \frac{1}{n!} \int_0^1 x^n \ln^n x \dee x.
Clearly, \int_0^1 x^n \ln^n x \dee x = \frac{(-1)^{n + 1}n!}{n^n}. ;wip: what? how? even WolframAlpha can't evaluate this one So \int_0^1 x^x \dee x = \sum_{n = 0}^\infty \frac{(-1)^{n + 1}}{n^n}.
Estimate \int_0^1 \frac{\sin \ln x}{\ln x} \dee x:
Let u = \ln x. Then \frac{\sin \ln x}{\ln x} = \frac{1}{u}\sin u = \sum_{n = 0}^\infty \frac{1}{u}\frac{(-1)^n}{(2n + 1)!} u^{2n + 1} = \sum_{n = 0}^\infty \frac{(-1)^n}{(2n + 1)!} \frac{u^{2n + 1}}{u} = \sum_{n = 0}^\infty \frac{(-1)^n}{(2n + 1)!} \ln^{2n} x.
So \int_0^1 \frac{\sin \ln x}{\ln x} \dee x = \sum_{n = 0}^\infty \frac{(-1)^n}{(2n + 1)!} \int_0^1 \ln^{2n} x \dee x.
By taking a few values of n and integrating by parts, we find that \int_0^1 \ln^{2n} x \dee x = (2n)!.
So \int_0^1 \frac{\sin \ln x}{\ln x} \dee x = \sum_{n = 0}^\infty \frac{(-1)^n}{(2n + 1)!} (2n)! = \sum_{n = 0}^\infty \frac{(-1)^n}{2n + 1}.
Recall that \arctan x = \sum_{n = 0}^\infty \frac{(-1)^n}{2n + 1} x^{2n + 1}. So \arctan 1 = \sum_{n = 0}^\infty \frac{(-1)^n}{2n + 1}.
So \int_0^1 \frac{\sin \ln x}{\ln x} \dee x = \arctan 1 = \frac{\pi}{4}.
# 27/3/14
### Evaluating Infinite Series
We often want a closed form of a series.
Evaluate \sum_{n = 1}^\infty \frac{1}{n^2}:
All we know right now is that this converges and is between 1 and 2. Euler found several ways to find the exact value of this series.
Clearly, \frac{\sin x}{x} = \frac{1}{x} \sum_{n = 0}^\infty \frac{(-1)^n x^{2n + 1}}{(2n + 1)!} = \sum_{n = 0}^\infty \frac{(-1)^n x^{2n}}{(2n + 1)!}.
Clearly, \frac{\sin x}{x} = 1 if x = 0, and \frac{\sin x}{x} = \sum_{n = 0}^\infty \frac{(-1)^n x^{2n}}{(2n + 1)!} if x \ne 0.
Clearly, any finite polynomial P(x) such that P(0) = 1 and roots r_1, \ldots, r_n can be written as (1 - \frac{x}{r_1}) \cdots (1 - \frac{x}{r_n}).
Assume this is also true for infinite polynomials as well. ;wip: this assumption is not always true Clearly, the roots of \frac{\sin x}{x} are x = \pm \pi, \pm 2\pi, \ldots.
So \frac{\sin x}{x} = \left(\left(1 - \frac{x}{\pi}\right)\left(1 + \frac{x}{\pi}\right)\right)\left(\left(1 - \frac{x}{2\pi}\right)\left(1 + \frac{x}{2\pi}\right)\right) \cdots = \left(1 - \frac{x^2}{\pi^2}\right)\left(1 - \frac{x^2}{2^2\pi^2}\right) \cdots.
So \sum_{n = 0}^\infty \frac{(-1)^n x^{2n}}{(2n + 1)!} = \left(1 - \frac{x^2}{\pi^2}\right)\left(1 - \frac{x^2}{2^2\pi^2}\right) \cdots.
Clearly, \left(1 - \frac{x^2}{\pi^2}\right)\left(1 - \frac{x^2}{2^2\pi^2}\right) \cdots = 1 - \frac{x^2}{\pi^2}\left(\frac{1}{1^2} + \frac{1}{1^2} + \ldots\right) + x^4(\ldots) - x^6(\ldots) + \ldots.
Clearly, the coefficients of x^2 must match. So -\frac{1}{3!} = -\frac{1}{\pi}^6 \sum_{n = 1}^\infty \frac{1}{n^2}.
So \sum_{n = 1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}.
In fact, we can also equate the coefficients of x^4, x^6, and etc. to obtain things like \sum_{n = 1}^\infty \frac{1}{26} = \frac{1315862 \pi^{26}}{11094481976030578125}.
;wip: talk about odd powers and how to derive the even powers directly
## Parametric Curves
We now consider functions with a single output and a vector output. For example, the position of a particle with respect to time, \vec{r}(t) = \begin{bmatrix} x(t) \\ y(t) \\ z(t) \end{bmatrix}. This has many applications in physics.
The function \vec{r}(t) defines a curve in the dimension of the vector. Here, t is the parameter of the function.
These functions are also called parametric curves.
We do calculus on these functions by working with each component separately. As a result, calculus on parametric curves are no more difficult than calculus on normal functions.
For example, \vec{r}(t) = \begin{bmatrix} t \\ \abs{t} \end{bmatrix} simply looks like the absolute value function on \begin{bmatrix} x \\ y \end{bmatrix} = \vec{r}(t).
However, the additional information is the direction along the curve - the velocity, in our case.
# 31/3/14
Parametric curves allow us to describe curves that would be very difficult or impossible to represent as s imple functions of x. This is much more general than curves on a plane.
Parametric curves also have a direction vector along the curve. The direction always goes from a low t to a high t. This contrasts with curves on a plane, which does not have a direction along the curve.
For example, \vec{r}(t) = (t^2, t), t \in [0, 1]. Since x = t^2 and y = t, y = \sqrt{x} = t, which is a form that allows us to plot the function easily.
Parametric curves are hard to plot because we are not used to them. However, parametric curves of the form \vec{r}(t) = (t, f(t)) can simply be plotted using y = f(t). Likewise, parametric curves of the form \vec{r}(t) = (f(t), t) can simply be plotted using x = f(t), or y = f^{-1}(t). Also, the direction vector is rightwards, because as t increases, the corresponding coordinate \vec{r}(t) moves rightward.
Consider \vec{r}(t) = (a \cos t, a \sin t), t \in [0, 2\pi]. Since x = a \cos t, y = a \sin t, x^2 + y^2 = a^2(\sin^2 t + \cos^2 t) = a^2. So this is a circle of radius a. The circle starts at (a, 0), and travels counterclockwise until it reaches the starting point again.
Given a value of t for \vec{r}(t) = (f(t), g(t)) and y = f(x), we can plot it in an inverval of t \in [a, b] by plotting for x \in [f(a), f(b)].
The parameterization of a curve is not unique, just like normal curves. Every parametric curve has infinite different ways of being represented mathematically, but when drawn on a plane each curve has only one shape.
For example, consider \vec{r}(t) = (a \cos t, b \sin t). This is an ellipse where a > b stretches it horiontally and b > a stretches it vertically. We can rewrite it as \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 to plot it on a Cartesian plane.
We plot parametric curves by solving for t in x = x(t), so t = x^{-1}(x), then subsituting into y(t) to get y = y(x^{-1}(x)), which is a non-parametric function we can plot more easily.
Alternatively, we would have to plot both y = y(t) and x = x(t), and use these graphs to plot the points on \vec{r}(t) = (x(t), y(t)).
### Hyperbolic Functions
The functions \vec{r} = (a \cos t, b \sin t) are called circular/elliptical trigonometric fucntions.
The hyperbolic trigonometric funnctions are called hyperbolic for a reason. Consider \vec{r}(t) = (a \cosh t, b \sinh t).
In fact, this creates a hyperbola, which looks like a parabola and another instance of that parabola flipped about the x-axis, or the same thing sideways. They are always symmetrical about the x-axis and y-axis.
A hyperbola is a curve defined by \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1. We can derive this form by using the identity \cosh^2 t - \sinh^2 t = 1.
# 2/4/14
;wip: final exam MC 4061 thursday April 10 4pm
We can sketch this by rearranging the formula to get y = \pm \sqrt{\frac{b^2}{a^2}x^2 - b^2}.
When x or y gets large, y \approxeq \pm \frac{b}{a}x - the function has lines for asymptotes.
At x = 0, y = \pm \sqrt{-b^2}, which is imaginary, so there is no y-intercept.
At y = 0, x = \pm a, which are the x-intercepts.
Imagine a wheel of radius a rolling along a level surface. If we observe from the reference frame of the surface, then a point on the edge of the wheel would create a bumpy curve. This curve is called a cycloid curve.
The cycloid curve can easily be parametrically defined by \vec{r}(t) = (at - a \sin t, a - a \cos t). The at term moves the curve horizontally, and the a term moves above the x-axis.
This function is very difficult to write in terms of y(x).
;wip: figure out how to convert parametrics to implicit functions
Parametric curves are not limited to \mb{R}^2. If we extend them into \mb{R}^n, then we can have n dimensional curves.
### Calculus on Parametric Curves
To perform calculus on vector valued functions, we simply note that the vector valued function is a vector of single variable functions: \vec{r}(t) = (x(t), y(t)), and we can apply the operations to the single variable functions.
For example, \lim_{t \to a} \vec{r}(t) = \vec{L} \iff \lim_{t \to a} x(t) = L_1 \wedge \lim_{t \to a} y(t) = L_2.
In the same way, \frac{\dee \vec{r}}{\dee t} = (\frac{\dee x}{\dee t}, \frac{\dee y][\dee t}). This is also known as the tangent vector.
This is the reason that the velocity of the object is tangent to its motion - it is the tangent vector of position.
Consider \magn{\frac{\dee \vec{r}}{\dee t}}. This is the speed if \vec{r}(t) is the position. This allows us to find the length of a curve (also known as distance or arclength) given its parametric curve.
The length of a curve from t = a to t = b is \int_a^b \magn{\frac{\dee \vec{r}}{\dee t}} \dee t.
For example, find the circumference of the circle \vec{r}(t) = (a \cos t, a \sin t) for t \in [0, 4\pi]:
Clearly, the length is \int_a^b \magn{\frac{\dee \vec{r}}{\dee t}} \dee t = \int_0^{4 \pi} \sqrt{\left(\frac{\dee}{\dee t} a \cos t\right)^2 + \left(\frac{\dee}{\dee t} a \sin t\right)^2} \dee t ;wip
For example, \vec{r}(t) = (a \cos t, a \sin t, t) is a spiral upwards.
# 4/4/14
Find the arc length of \vec{r}(t) = (a \cos t, a \sin t, t) for t \in [0, 4 \pi]:
This is a spiral going upwards.
Clearly, \frac{\dee \vec{r}}{\dee t} = (-a \sin t, a \cos t, 1).
Clearly, the arc length is \int_0^{4 \pi} \magn{\frac{\dee \vec{r}}{\dee t}} \dee t = \int_0^{4 \pi} \sqrt{a^2 \sin^2 t + a^2 \cos^2 t + 1^2} \dee t = \int_0^{4 \pi} \sqrt{a^2 + 1} \dee t.
So the arc length is 4 \pi \sqrt{a^2 + 1}.
Find the arc length of the cycloid function \vec{r}(t) = (at - a \sin t, a - a \cos t) for t \in [0, 2 \pi]:
Clearly, \frac{\dee \vec{r}}{\dee t} = (a - a \cos t, a \sin t) and \magn{\frac{\dee \vec{r}}{\dee t}}.
Clearly, \magn{\frac{\dee \vec{r}}{\dee t}} = \sqrt{a^2 - 2a^2 \cos t + a^2 \cos^2 t + a^2 \sin t} = \sqrt{2a^2 - 2a^2 \cos t} = a\sqrt{2 - 2 \cos t} = a\sqrt{4\left(\frac{1}{2} - \frac{1}{2} \cos t\right)} = 2a\sqrt{\frac{1}{2} - \frac{1}{2} \cos t}.
Recall the half angle formula \sin^2 t = \frac{1}{2} - \frac{1}{2} \cos 2t.
Clearly, 2a\sqrt{\frac{1}{2} - \frac{1}{2} \cos t} = 2a\sqrt{\sin^2 \frac{t}{2}} = 2a\abs{\sin \frac{t}{2}}.
Since we are working in a range where \frac{t}{2} \in [0, \pi], \abs{\sin \frac{t}{2}} = \sin \frac{t}{2}.
Clearly, \int_0^{2\pi} \magn{\frac{\dee \vec{r}}{\dee t}} \dee t = 2a\int_0^{2\pi} \sin \frac{t}{2} \dee t = 2a\evalat{2\cos \frac{t}{2}}_0^{2\pi} = 8a.
So the arc length is 8a.
### Slope
The slope of a tangent vector is \frac{\dee y}{\dee x}. We can find it using \frac{\dee y}{\dee x} = \frac{\frac{\dee y}{\dee t}}{\frac{\dee x}{\dee t}}.
Find the slope of \vec{r}(t) = (\cos t, \sin t):
Clearly, \frac{\dee y}{\dee x} = \frac{-\sin t}{\cos t} = -\tan t.
This is where the \tan function gets its name - it is the slope of the tangent of a circle at any given angle.
Consider \frac{\dee^2 y}{\dee x^2}. We might think that \frac{\dee^2 y}{\dee x^2} = \frac{\frac{\dee^2 y}{\dee t^2}}{\frac{\dee^2 x}{\dee t^2}}, but this is incorrect.
In fact, \frac{\dee^2 y}{\dee x^2} = \frac{\dee}{\dee x} \frac{\dee y}{\dee x} = \frac{\frac{\dee}{\dee t} \frac{\dee y}{\dee x}}{\frac{\dee x}{\dee t}} = \frac{\frac{\dee}{\dee t} \frac{\dee y}{\dee x}}{\frac{\dee x}{\dee t}}.
### Area
The area under a curve of a function is given by y = F(x) is A = \int_{x_1}^{x_2} F(x) \dee x.
For parametric curves, we have x = x(t) and y = y(t). Then there exists y = F(x) - a function of x equivalent to y, and y(t) = F(x).
Then x_1 = x(t_1) and x_2 = x(t_2), and \dee x = \frac{\dee x}{\dee t} \dee t.
So A = \int_{t_1}^{t_2} F(x) \dee x = \int_{t_1}^{t_2} y(t) \dee x = \int_{t_1}^{t_2} y(t) \frac{\dee x}{\dee t} \dee t.
This allows us to find the area under a curve.
;wip: I finally get u substitution - it was inconsistent because sometimes u was a function of u, and sometimes x is a function of u.
;wip: is the limit test adding 1 to the exponent, or is it the next term when we have things like x^{2n}? wikipedia says the latter but in class it was the former
Find the area under the cycloid curve \vec{r}(t) = (at - a \sin t, a - a \cos t) for t \in [0, 2 \pi]:
Clearly, A = \int_0^{2 \pi} (a - a \cos t) \frac{\dee}{\dee t} (at - a \sin t) \dee t = \int_0^{2 \pi} (a - a \cos t)(a - a \cos t) \dee t = 2 \pi a^2 + \int_0^{2 \pi} -2a^2 \cos t \dee t + \int_0^{2 \pi} + a^2 \cos^2 t \dee t = 3 \pi a^2.
So the cycloid curve has an area of 3 \pi a^2 units.
To find the value of an infinite series, we take a known series, then apply transformations to it until it is the function we need. | 2018-12-15 12:03:37 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9973724484443665, "perplexity": 1125.5830121543756}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376826856.55/warc/CC-MAIN-20181215105142-20181215131142-00041.warc.gz"} |
http://tex.stackexchange.com/questions/79131/improper-alphabetic-constant-with-hyperref-and-bidi-packages | # “improper alphabetic constant” with hyperref and bidi packages
I'm getting an "improper alphabetic constant" error in xelatex which appears to be due to an interaction between the bidi and hyperref packages, when I redefine the \section cmd. Here's an MWE:
\documentclass{report}
\usepackage[unicode]{hyperref}
\usepackage{bidi} %Must be loaded after hyperref pkg
\makeatletter
\renewcommand\section{\@startsection{section}{1}{0pt}{0pt}{0pt}{}}
\makeatother
\begin{document}
\section{\RL{foo}}
\RL{bar}
\end{document}
It's important that the text \RL{bar} be the first non-whitespace after the \section{}, otherwise the error is not triggered.
I've checked the suggested "Similar Questions," and don't see anything quite like this.
Suggestions?
-
I do not think, you need to pass unicode option to hyperref package with XeTeX engine. So with the following example:
\documentclass{report}
\usepackage{hyperref}
\usepackage{bidi} %Must be loaded after hyperref pkg
\makeatletter
\renewcommand\section{\@startsection{section}{1}{0pt}{0pt}{0pt}{}}
\makeatother
\begin{document}
\section{\RL{foo}}
\RL{bar}
\end{document}
I get no error (using Updated TeXLive 2012) but one related warning which is due to the use of \RL macro inside \section:
Package hyperref Warning: Token not allowed in a PDF string (Unicode):
(hyperref) removing \RL' on input line 9.
For details see page 19 of hyperref manual, under Replacement macros subsubsection.
To get rid off this warning, you can change your example into:
\documentclass{report}
\usepackage{hyperref}
\usepackage{bidi} %Must be loaded after hyperref pkg
\makeatletter
\renewcommand\section{\@startsection{section}{1}{0pt}{0pt}{0pt}{}}
\makeatother
\begin{document}
\section{\texorpdfstring{\RL{foo}}{foo}}
\RL{bar}
\end{document}
and if you do not like \texorpdfstring, you can try:
\documentclass{report}
\usepackage{hyperref}
\usepackage{bidi} %Must be loaded after hyperref pkg
\makeatletter
\renewcommand\section{\@startsection{section}{1}{0pt}{0pt}{0pt}{}}
\pdfstringdefDisableCommands{%
\let\RL\@firstofone
}
\makeatother
\begin{document}
\section{\RL{foo}}
\RL{bar}
\end{document}
`
-
Odd, I had tried it without the [unicode] arg to hyperref first (and re-tried that just now), but I get the same error. I'm running the TexLive 2011 version, I must have missed the 2012 update! – Mike Maxwell Oct 26 '12 at 14:11
Thanks for the fixes--both work with 2011, although I'm now in the process of updating our installation to 2012, so maybe I won't need them :-). – Mike Maxwell Oct 26 '12 at 20:42
We updated to the 2012 version, but I still get my original error msg ("improper alphabetic constant"), rather than the one you get about '\RL' not being allowed. I guess you have some files that have been updated since TeXLive 2012 came out. Can you explain the effect of omitting the \RL command (which IIUC is what the \pdfstringdefDisableCommands{} that you give is doing)? I would have thought it would mean that Arabic script comes out left-to-right in bookmarks, but I can't see any diff in the output. – Mike Maxwell Oct 30 '12 at 17:57 | 2015-07-28 17:47:56 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9666882753372192, "perplexity": 4902.93617221893}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-32/segments/1438042982013.25/warc/CC-MAIN-20150728002302-00037-ip-10-236-191-2.ec2.internal.warc.gz"} |
https://amathew.wordpress.com/2010/01/17/towards-constant-coefficient-elliptic-regularity-parametrices-and-generalities-on-the-singular-support/ | The next application I want to talk about here of Fourier analysis is to (a basic case of) ellipic regularity. Later we will use refinements of these techniques to obtain all kinds of estimates. Anyway, for now, a partial differential operator
$\displaystyle P = \sum_{a: |a| \leq k} C_a D^a$
is called elliptic if the homogeneous polynomial
$\displaystyle \sum_{a: |a| = k} C_a \xi^a, \quad \xi = (\xi_1, \dots, \xi_n)$
has no zeros outside the origin. For instance, the Laplace operator is elliptic. Later I will discuss how this generalizes to other PDEs, and how this polynomial becomes the symbol of the operator. For the moment, though let’s define ${Q(\xi) = \sum_{a: |a| \leq k} C_a (2 \pi i \xi)^a}$. The definition of ${Q}$ such that
$\displaystyle \widehat{ Pf } = Q \hat{f},$
and we know that ${|Q(\xi)| \geq \epsilon |\xi|^k}$ for ${|\xi|}$ large enough. This is a very important fact, because it shows that the Fourier transform of ${Pf}$ exerts control on that of ${f}$. However, we cannot quite solve for ${\hat{f}}$ by dividing ${\widehat{Pf}}$ by ${Q}$ because ${Q}$ is going to have zeros. So define a smoothing function ${\varphi}$ which vanishes outside a large disk ${D_r(0)}$. Outside this disk, an estimate ${|Q(\xi)| \geq \epsilon |\xi|^k}$ will be assumed to hold.
A parametrix for ${P}$
We’re going to start by finding a parametrix for the operator ${P}$; this is not quite a fundamental solution, but it is close enough. We could formally get a fundamental solution (since ${\hat{\delta} \equiv 1}$) by considering ${\widehat{1/Q}}$, but this is nonsense. Rather, consider the distribution ${E}$ with
$\displaystyle \hat{E} = (1- \varphi) Q^{-1},$
which is indeed well-defined as a tempered distribution by the hypothesis on ${\varphi}$. It is then clear that ${\widehat{PE} = 1 - \varphi}$, so
$\displaystyle PE = \delta - \hat{\varphi},$
where the ${\hat{\varphi}}$ is at least suitably controlled (e.g. in ${\mathcal{S}}$). So we have something close to a fundamental solution, and ${E}$ is called the parametrix. Since we convolve things with a Moreover, I claim—and this is crucial—that ${E}$ comes very close to being a function of ${\mathcal{S}}$ as well; the only problem occurs because of the not-that-fast decrease at infinity. In particular, the singular locus of ${E}$ is the origin. This will require some discussion of intermediate concepts.
How to convolve two distributions
We already know how to convolve a distribution ${\phi}$ and a function ${f}$. Take ${\breve{f}_x}$ defined by ${\breve{f}_x(y) = f(x-y)}$ and set
$\displaystyle (\phi \ast f)(x) := \phi(\breve{f}_x).$
It is easy to check that this coincides with the old definition of convolution when ${\phi \in L^1}$. This is always ${C^{\infty}}$. When ${\phi}$ is compactly supported, which is to say that ${\phi(f)=0}$ when ${f}$ vanishes inside a sufficiently large compact set ${K}$, then we have
$\displaystyle |\phi(f)| \leq M \sum_{a:|a| \leq N} \sup_{K} |D^a f|,$
for some ${M,N}$. In this case it follows that ${\phi \ast f \in \mathcal{S}}$ in fact. So given another ${\phi'}$, not necessarily of compact support, we can heuristically use the properties of convolution to “write”
$\displaystyle (\phi' \ast \phi) \ast f = \phi \ast (\phi \ast f).$
Make this a definition. Then in particular, we have a way of talking about ${\phi' \ast \phi}$ as a distribution in itself—just evaluate the convolution at ${\breve{f}_0,0}$.
Let’s go back to what I just said about compact support. We can generalize this: say that a distribution ${\phi}$ vanishes on an open set ${\Omega}$ if ${\phi(f)=0}$ for ${f}$ supported in ${\Omega}$. Then a partition of unity argument shows that there is a largest open set ${\Omega}$ on which ${\phi}$ vanishes; the complement is called the support ${\mathrm{supp} \phi}$. This is a generalization of the notion for functions, as is easily seen. Anyway, it is a well-known fact about functions that ${\mathrm{supp} f_1 \ast f_2 \subset \mathrm{supp} f_1 + \mathrm{supp} f_2}$. That this is true for distributions when one is compactly supported follows by regularizing each: given ${\phi_1, \phi_2}$, we convolve each with an approximation to the identity to get smooth functions, one of which is compactly supported, that approxiamte ${\phi_1, \phi_2}$ arbitrarily closely (in the weak* topology).
The singular locus of a distribution
Say that a distribution ${\phi }$ is regular in an open set ${\Omega}$ if for ${f}$ smooth and supported in ${\Omega}$, we have ${\phi(f) = \int gf}$ for ${g: \Omega \rightarrow \mathbb{R}^n}$ smooth. Basically, this means that when restricted to ${\Omega}$, ${\phi}$ behaves just like a smooth function.
Using a partition of unity, we see that if ${\phi}$ is regular in ${\Omega_1}$ and ${\Omega_2}$, then it is regular in ${\Omega_1 \cup \Omega_2}$, and moreover for infinite unions. In particular, there is a maximal open set on which ${\Omega}$ is regular. The complement of this set is written ${\mathrm{sing} \phi}$. For instance, ${\mathrm{sing} \delta = \{0\}}$. This behaves well with respect to convolution, if one is compactly supported :
Lemma 1 $\displaystyle \mathrm{sing} \phi_1 \ast \phi_2 \subset \mathrm{sing} \phi_1 + \mathrm{sing} \phi_2.$
The reason is that we write ${\phi_1 = \phi_1^a + \phi_1^b, \phi_2 = \phi_2^a + \phi_2^b}$ where ${\phi_1^a,\phi_2^a}$ have supports barely outside the singular loci of ${\phi_1, \phi_2}$ and ${\phi_1^b, \phi_2^b}$ are smooth. Then
$\displaystyle \phi_1 \ast \phi_2 = \phi_1^a \ast \phi_2^a + \phi_1^a \ast \phi_2^b+\phi_1^b \ast \phi_2^a+\phi_1^b \ast \phi_2^b,$
and all but the first term are smooth. The first term is supported in a small neighborhood of ${\mathrm{sing} \phi_1 + \mathrm{sing} \phi_2}$, which we can make arbitrarily small. | 2018-03-20 04:17:01 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 82, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9559183716773987, "perplexity": 110.32576282240622}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257647280.40/warc/CC-MAIN-20180320033158-20180320053158-00649.warc.gz"} |
https://cracku.in/6-which-two-signs-should-be-interchanged-in-the-foll-x-ssc-chsl-4-july-2019-shift-1 | Question 6
# Which two signs should be interchanged in the following equation to make it correct?$$207 \times 9 + 13 \div 26 - 301 = 60$$
Solution
By trial and error method,
Option A
$$207\times9-13\div26+301=60$$
$$207\times9-\frac{1}{2}+301=60$$
Option A is false since we get decimal number after solving
Option B
$$207+9\times13\div26-301=60$$
$$207+9\times\frac{1}{2}-301=60$$
Option B is also false since we get decimal number after solving
Option C
$$207\div9+13\times26-301=60$$
$$23+13\times26-301=60$$
$$23+338-301=60$$
$$60=60$$
Hence, the correct answer is Option C | 2023-01-26 22:54:12 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8989421129226685, "perplexity": 1716.4133636640368}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764494826.88/warc/CC-MAIN-20230126210844-20230127000844-00314.warc.gz"} |
http://mathhelpforum.com/algebra/31593-complex-numbers.html | # Math Help - Complex numbers
1. ## Complex numbers
Let $a_0=a_1=1$ and $a_n=a_{n-2}+ia_{n-1}$ for $n>1$.
Find $a_{2008}$
2. Excel ran out of number spaces, but here is a start:
3. Originally Posted by mathceleb
Excel ran out of number spaces, but here is a start:
How are you getting those numbers? Note that $i$ here stands for the imaginary unit, $i=\sqrt{-1}$
4. This is known sequence with period of 12.
It repeats itself for each 12 just as i repeats every four.
Use mod(12) to get the answer.
5. Originally Posted by math sucks
How are you getting those numbers? Note that $i$ here stands for the imaginary unit, $i=\sqrt{-1}$
I thought $i$ was the number of the term, i.e., 1st term is 1, 2nd term is 2. -2 points for me, sorry about that. | 2016-02-07 04:13:17 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 9, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9387618899345398, "perplexity": 964.8798828281615}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-07/segments/1454701148475.34/warc/CC-MAIN-20160205193908-00111-ip-10-236-182-209.ec2.internal.warc.gz"} |
https://www.gradesaver.com/textbooks/math/trigonometry/CLONE-68cac39a-c5ec-4c26-8565-a44738e90952/chapter-5-trigonometric-identities-section-5-5-double-angle-identities-5-5-exercises-page-237/27 | # Chapter 5 - Trigonometric Identities - Section 5.5 Double-Angle Identities - 5.5 Exercises - Page 237: 27
$$\frac{2\cos2\theta}{\sin2\theta}=\cot\theta-\tan\theta$$ The equation is an identity. The proof is below.
#### Work Step by Step
$$\frac{2\cos2\theta}{\sin2\theta}=\cot\theta-\tan\theta$$ We solve from the left side, as it is more complex. $$X=\frac{2\cos2\theta}{\sin2\theta}$$ - We replace $\cos2\theta$ and $\sin2\theta$ with the following identities from Double-Angle identities: $$\cos2\theta=\cos^2\theta-\sin^2\theta$$ $$\sin2\theta=2\sin\theta\cos\theta$$ Therefore, $$X=\frac{2(\cos^2\theta-\sin^2\theta)}{2\sin\theta\cos\theta}$$ $$X=\frac{\cos^2\theta-\sin^2\theta}{\sin\theta\cos\theta}$$ $$X=\frac{\cos^2\theta}{\sin\theta\cos\theta}-\frac{\sin^2\theta}{\sin\theta\cos\theta}$$ $$X=\frac{\cos\theta}{\sin\theta}-\frac{\sin\theta}{\cos\theta}$$ - From Quotient Identities: $\frac{\cos\theta}{\sin\theta}=\cot\theta$ and $\frac{\sin\theta}{\cos\theta}=\tan\theta$. Therefore, $$X=\cot\theta-\tan\theta$$ So, $$\frac{2\cos2\theta}{\sin2\theta}=\cot\theta-\tan\theta$$ We conclude that the equation is an identity.
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 2020-09-28 09:38:52 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8505387306213379, "perplexity": 471.055311349921}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600401598891.71/warc/CC-MAIN-20200928073028-20200928103028-00000.warc.gz"} |
https://www.doubtnut.com/question-answer-physics/round-off-the-followng-numbers-to-three-significant-digit-a-15462-b-14745-c-14750-and-d-14650xx1012-642594395 | Home
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Round off the followng numbers...
Updated On: 27-06-2022
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Text Solution
Solution : The third significant digit is 4. This digit is to be rounded. The digit next to it is 6 which is greater than 5. The thid digit should, therefoe, be increased by 1. The digits to be dropped should be replaced by zero because they appear to the left of the decimal. Thus, 15462 becomes 15500 on rounding to three significant digits. <br> b. The third significant digit in 14.745 is 7. The number next to it is less thant 5. So 14.745 becomes 14.7 on rounding to three significant digits. ltbr. c. 14.750 will become 14.8 becue the digit to be rounded is odd and te digit next to it is 5. <br> d. 14.650xx10^12 will become 14.6xx10^12 because the digit to be rounded is even and the digit next to it is 5.
Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
Transcript
hello sir question is round of the following numbers to 3 significant digits of a number are first is 15462 second is 14.7 4530 is 41750 and final one is getting worse 6502 10-12 ok so talking about the first one that is 15462 Abhi have to round off this up to 3 significant digits so we have to Round up up to this correct no related the digit 4 is smaller than the digit after it that is so it will increase by one and digits after it will become zero so after rounding it off but will get is 154 will get increased by 1 so it will be fine and the digits after it will become zero so this will be the lord of this number ok for the second one that is the part which is 14
points 745 we have to round it of up to 3 significant digit show the number which will be get round of HD 7 letter up in the digit after this is for which is less than 5 so that it will not increase it will remain the same character after rounding it of what will get is 14.7 talking about the third digit that is 14.7 50 year and interest in this letter 7 is what are the digit 7 is what what does it correct Subah digit is odd and after 85 comes it will get round of ok so after rounding out of what will get is 14.8 correct but if the letter in if the digits will become even then it will remain same it doesn't change so for the fourth part but we have been given
is 14.6 50 into 10 raise to the power to well over here if you see that sex is even number and here is 5 the digit after it it it it will not change it will remain the same after rounding off so what will but we will get is 14.6 into 10 raise to the power to well ok we have rounded off rounded off all the numbers have been given these are the required answer thank you | 2022-07-07 16:16:56 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.22758829593658447, "perplexity": 513.5716870888173}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656104495692.77/warc/CC-MAIN-20220707154329-20220707184329-00539.warc.gz"} |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/1386/2/k/n/ | # Properties
Label 1386.2.k.n Level $1386$ Weight $2$ Character orbit 1386.k Analytic conductor $11.067$ Analytic rank $0$ Dimension $2$ CM no Inner twists $2$
# Related objects
## Newspace parameters
Level: $$N$$ $$=$$ $$1386 = 2 \cdot 3^{2} \cdot 7 \cdot 11$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 1386.k (of order $$3$$, degree $$2$$, minimal)
## Newform invariants
Self dual: no Analytic conductor: $$11.0672657201$$ Analytic rank: $$0$$ Dimension: $$2$$ Coefficient field: $$\Q(\sqrt{-3})$$ Defining polynomial: $$x^{2} - x + 1$$ Coefficient ring: $$\Z[a_1, a_2]$$ Coefficient ring index: $$1$$ Twist minimal: no (minimal twist has level 154) Sato-Tate group: $\mathrm{SU}(2)[C_{3}]$
## $q$-expansion
Coefficients of the $$q$$-expansion are expressed in terms of a primitive root of unity $$\zeta_{6}$$. We also show the integral $$q$$-expansion of the trace form.
$$f(q)$$ $$=$$ $$q + \zeta_{6} q^{2} + ( -1 + \zeta_{6} ) q^{4} + ( 2 + \zeta_{6} ) q^{7} - q^{8} +O(q^{10})$$ $$q + \zeta_{6} q^{2} + ( -1 + \zeta_{6} ) q^{4} + ( 2 + \zeta_{6} ) q^{7} - q^{8} + ( -1 + \zeta_{6} ) q^{11} + 5 q^{13} + ( -1 + 3 \zeta_{6} ) q^{14} -\zeta_{6} q^{16} + ( 6 - 6 \zeta_{6} ) q^{17} -2 \zeta_{6} q^{19} - q^{22} + 6 \zeta_{6} q^{23} + ( 5 - 5 \zeta_{6} ) q^{25} + 5 \zeta_{6} q^{26} + ( -3 + 2 \zeta_{6} ) q^{28} -3 q^{29} + ( -8 + 8 \zeta_{6} ) q^{31} + ( 1 - \zeta_{6} ) q^{32} + 6 q^{34} -2 \zeta_{6} q^{37} + ( 2 - 2 \zeta_{6} ) q^{38} + 6 q^{41} -4 q^{43} -\zeta_{6} q^{44} + ( -6 + 6 \zeta_{6} ) q^{46} + 6 \zeta_{6} q^{47} + ( 3 + 5 \zeta_{6} ) q^{49} + 5 q^{50} + ( -5 + 5 \zeta_{6} ) q^{52} + ( -12 + 12 \zeta_{6} ) q^{53} + ( -2 - \zeta_{6} ) q^{56} -3 \zeta_{6} q^{58} + ( -3 + 3 \zeta_{6} ) q^{59} + 7 \zeta_{6} q^{61} -8 q^{62} + q^{64} + ( 13 - 13 \zeta_{6} ) q^{67} + 6 \zeta_{6} q^{68} + 12 q^{71} + ( 10 - 10 \zeta_{6} ) q^{73} + ( 2 - 2 \zeta_{6} ) q^{74} + 2 q^{76} + ( -3 + 2 \zeta_{6} ) q^{77} + \zeta_{6} q^{79} + 6 \zeta_{6} q^{82} -6 q^{83} -4 \zeta_{6} q^{86} + ( 1 - \zeta_{6} ) q^{88} + 6 \zeta_{6} q^{89} + ( 10 + 5 \zeta_{6} ) q^{91} -6 q^{92} + ( -6 + 6 \zeta_{6} ) q^{94} -13 q^{97} + ( -5 + 8 \zeta_{6} ) q^{98} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$2q + q^{2} - q^{4} + 5q^{7} - 2q^{8} + O(q^{10})$$ $$2q + q^{2} - q^{4} + 5q^{7} - 2q^{8} - q^{11} + 10q^{13} + q^{14} - q^{16} + 6q^{17} - 2q^{19} - 2q^{22} + 6q^{23} + 5q^{25} + 5q^{26} - 4q^{28} - 6q^{29} - 8q^{31} + q^{32} + 12q^{34} - 2q^{37} + 2q^{38} + 12q^{41} - 8q^{43} - q^{44} - 6q^{46} + 6q^{47} + 11q^{49} + 10q^{50} - 5q^{52} - 12q^{53} - 5q^{56} - 3q^{58} - 3q^{59} + 7q^{61} - 16q^{62} + 2q^{64} + 13q^{67} + 6q^{68} + 24q^{71} + 10q^{73} + 2q^{74} + 4q^{76} - 4q^{77} + q^{79} + 6q^{82} - 12q^{83} - 4q^{86} + q^{88} + 6q^{89} + 25q^{91} - 12q^{92} - 6q^{94} - 26q^{97} - 2q^{98} + O(q^{100})$$
## Character values
We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/1386\mathbb{Z}\right)^\times$$.
$$n$$ $$155$$ $$199$$ $$1135$$ $$\chi(n)$$ $$1$$ $$-\zeta_{6}$$ $$1$$
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
793.1
0.5 − 0.866025i 0.5 + 0.866025i
0.500000 0.866025i 0 −0.500000 0.866025i 0 0 2.50000 0.866025i −1.00000 0 0
991.1 0.500000 + 0.866025i 0 −0.500000 + 0.866025i 0 0 2.50000 + 0.866025i −1.00000 0 0
$$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Inner twists
Char Parity Ord Mult Type
1.a even 1 1 trivial
7.c even 3 1 inner
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 1386.2.k.n 2
3.b odd 2 1 154.2.e.b 2
7.c even 3 1 inner 1386.2.k.n 2
7.c even 3 1 9702.2.a.o 1
7.d odd 6 1 9702.2.a.l 1
12.b even 2 1 1232.2.q.c 2
21.c even 2 1 1078.2.e.d 2
21.g even 6 1 1078.2.a.i 1
21.g even 6 1 1078.2.e.d 2
21.h odd 6 1 154.2.e.b 2
21.h odd 6 1 1078.2.a.k 1
84.j odd 6 1 8624.2.a.t 1
84.n even 6 1 1232.2.q.c 2
84.n even 6 1 8624.2.a.l 1
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
154.2.e.b 2 3.b odd 2 1
154.2.e.b 2 21.h odd 6 1
1078.2.a.i 1 21.g even 6 1
1078.2.a.k 1 21.h odd 6 1
1078.2.e.d 2 21.c even 2 1
1078.2.e.d 2 21.g even 6 1
1232.2.q.c 2 12.b even 2 1
1232.2.q.c 2 84.n even 6 1
1386.2.k.n 2 1.a even 1 1 trivial
1386.2.k.n 2 7.c even 3 1 inner
8624.2.a.l 1 84.n even 6 1
8624.2.a.t 1 84.j odd 6 1
9702.2.a.l 1 7.d odd 6 1
9702.2.a.o 1 7.c even 3 1
## Hecke kernels
This newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{2}^{\mathrm{new}}(1386, [\chi])$$:
$$T_{5}$$ $$T_{13} - 5$$ $$T_{17}^{2} - 6 T_{17} + 36$$ $$T_{23}^{2} - 6 T_{23} + 36$$
## Hecke characteristic polynomials
$p$ $F_p(T)$
$2$ $$1 - T + T^{2}$$
$3$ $$T^{2}$$
$5$ $$T^{2}$$
$7$ $$7 - 5 T + T^{2}$$
$11$ $$1 + T + T^{2}$$
$13$ $$( -5 + T )^{2}$$
$17$ $$36 - 6 T + T^{2}$$
$19$ $$4 + 2 T + T^{2}$$
$23$ $$36 - 6 T + T^{2}$$
$29$ $$( 3 + T )^{2}$$
$31$ $$64 + 8 T + T^{2}$$
$37$ $$4 + 2 T + T^{2}$$
$41$ $$( -6 + T )^{2}$$
$43$ $$( 4 + T )^{2}$$
$47$ $$36 - 6 T + T^{2}$$
$53$ $$144 + 12 T + T^{2}$$
$59$ $$9 + 3 T + T^{2}$$
$61$ $$49 - 7 T + T^{2}$$
$67$ $$169 - 13 T + T^{2}$$
$71$ $$( -12 + T )^{2}$$
$73$ $$100 - 10 T + T^{2}$$
$79$ $$1 - T + T^{2}$$
$83$ $$( 6 + T )^{2}$$
$89$ $$36 - 6 T + T^{2}$$
$97$ $$( 13 + T )^{2}$$ | 2021-09-21 12:37:15 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.98160320520401, "perplexity": 7585.133115016147}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780057202.68/warc/CC-MAIN-20210921101319-20210921131319-00300.warc.gz"} |
https://hal-insu.archives-ouvertes.fr/insu-03867376 | New interface
# A new window to tidal asteroseismology: non-linearly excited stellar eigenmodes and the period spacing pattern in KOI-54
Abstract : We revisit the tidally excited oscillations (TEOs) in the A-type main-sequence eccentric binary KOI-54, the prototype of heartbeat stars. Although the linear tidal response of the star is a series of orbital-harmonic frequencies which are not stellar eigenfrequencies, we show that the non-linearly excited non-orbital-harmonic TEOs are eigenmodes. By carefully choosing the modes which satisfy the mode-coupling selection rules, a period spacing (ΔP) pattern of quadrupole gravity modes (ΔP ≍ 2520-2535 s) can be discerned in the Fourier spectrum, with a detection significance level of $99.9{{\ \rm per\ cent}}$. The inferred period spacing value agrees remarkably well with the theoretical l = 2, m = 0 g modes from a stellar model with the measured mass, radius, and effective temperature. We also find that the two largest-amplitude TEOs at N = 90, 91 harmonics are very close to resonance with l = 2, m = 0 eigenmodes, and likely come from different stars. Previous works on tidal oscillations primarily focus on the modelling of TEO amplitudes and phases, the high sensitivity of TEO amplitude to the frequency detuning (tidal forcing frequency minus the closest stellar eigenfrequency) requires extremely dense grids of stellar models and prevents us from constraining the stellar physical parameters easily. This work, however, opens the window of real tidal asteroseismology by using the eigenfrequencies of the star inferred from the non-linear TEOs and possibly very-close-to-resonance linear TEOs. Our seismic modelling of these identified eigen g-modes shows that the best-matching stellar models have (M ≍ 2.20, 2.35 M) and super-solar metallicity, in good agreement with previous measurements.
Keywords :
Document type :
Journal articles
https://hal-insu.archives-ouvertes.fr/insu-03867376
Contributor : Nathalie POTHIER Connect in order to contact the contributor
Submitted on : Wednesday, November 23, 2022 - 12:24:44 PM
Last modification on : Thursday, November 24, 2022 - 5:02:58 PM
### Citation
Zhao Guo, Gordon I. Ogilvie, Gang Li, Richard H. D. Townsend, Meng Sun. A new window to tidal asteroseismology: non-linearly excited stellar eigenmodes and the period spacing pattern in KOI-54. Monthly Notices of the Royal Astronomical Society, 2022, 517, pp.437-446. ⟨10.1093/mnras/stac2611⟩. ⟨insu-03867376⟩
Record views | 2022-12-01 01:20:05 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6963935494422913, "perplexity": 5404.249284836722}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710777.20/warc/CC-MAIN-20221130225142-20221201015142-00009.warc.gz"} |
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# Constructing Incremental Sequences in Graphs
1 MASCOTTE - Algorithms, simulation, combinatorics and optimization for telecommunications
CRISAM - Inria Sophia Antipolis - Méditerranée , Laboratoire I3S - COMRED - COMmunications, Réseaux, systèmes Embarqués et Distribués
Abstract : Given a weighted graph $G=(V,E,w)$, we investigate the problem of constructing a sequence of $n=|V|$ subsets of vertices $M_1,...,M_n$ (called groups) with small diameters, where the diameter of a group is calculated using distances in $G$. The constraint on these $n$ groups is that they must be incremental: $M_1\subsetM_2 \subset...\subsetM_n=V$. The cost of a sequence is the maximum ratio between the diameter of each group $M_i$ and the diameter of a group $N_i^*$ with $i$ vertices and minimum diameter: $\max_2 \leqi \leqn \left{ \fracD(M_i)D(N_i^*) \right}$. This quantity captures the impact of the incremental constraint on the diameters of the groups in a sequence. We give general bounds on the value of this ratio and we prove that the problem of constructing an optimal incremental sequence cannot be solved approximately in polynomial time with an approximation ratio less than 2 unless $P = NP$. Finally, we give a 4-approximation algorithm and we show that the analysis of our algorithm is tight.
Keywords :
Type de document :
Rapport
Domaine :
Littérature citée [1 références]
https://hal.inria.fr/inria-00070361
Contributeur : Rapport de Recherche Inria Connectez-vous pour contacter le contributeur
Soumis le : vendredi 19 mai 2006 - 20:15:30
Dernière modification le : vendredi 4 février 2022 - 03:11:28
Archivage à long terme le : : dimanche 4 avril 2010 - 21:02:11
### Identifiants
• HAL Id : inria-00070361, version 1
### Citation
Ralf Klasing, Christian Laforest, Joseph Peters, Nicolas Thibault. Constructing Incremental Sequences in Graphs. [Research Report] RR-5648, INRIA. 2006, pp.12. ⟨inria-00070361⟩
### Métriques
Consultations de la notice
## 260
Téléchargements de fichiers | 2022-05-16 04:59:11 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6319827437400818, "perplexity": 3027.9413648963023}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662509990.19/warc/CC-MAIN-20220516041337-20220516071337-00016.warc.gz"} |
http://mathhelpforum.com/calculus/122057-limit-rational-polynomials-print.html | # Limit of a rational with polynomials
• December 31st 2009, 12:45 PM
nautica17
Limit of a rational with polynomials
I'm stuck on this practice problem.. and it's the algebra that is getting to me. Take a look:
lim x-> 2 of ( (x^3 + 3x^2 - 12x + 4) / (x^3 - 4x) )
I've figured that I factor out the top and bottom and I've done so with the denominator and got x(x-2)(x+2), but the numerator is what I am stuck on. Can anyone push me in the right direction? I'm getting this as my numerator: (x^2 - 4)(x+3)(3x+1) ... and that obviously makes no sense what so ever.
• December 31st 2009, 01:04 PM
Plato
Quote:
Originally Posted by nautica17
lim x-> 2 of ( (x^3 + 3x^2 - 12x + 4) / (x^3 - 4x) )
$x^3+3x^2-12x+4=(x-2)(x^2+5x-2)$
• December 31st 2009, 01:16 PM
nautica17
Quote:
Originally Posted by Plato
$x^3+3x^2-12x+4=(x-2)(x^2+5x-2)$
Hmm.. well that works. Thank-you. If it's not too much trouble could you show me how to do that? I cannot figure out how to pull the (x-2) out.
• December 31st 2009, 01:24 PM
Plato
Quote:
Originally Posted by nautica17
Hmm.. well that works. Thank-you. If it's not too much trouble could you show me how to do that? I cannot figure out how to pull the (x-2) out.
Because $x=2$ is a root of the numerator then $(x-2)$ must be a factor.
So use simple long division. | 2014-12-29 07:44:02 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6462505459785461, "perplexity": 581.5433350972844}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-52/segments/1419447562371.145/warc/CC-MAIN-20141224185922-00083-ip-10-231-17-201.ec2.internal.warc.gz"} |
http://bbgunleather.com/8dy4qbr7/multiplying-radicals-with-variables-pdf-740d6f | DAY 5: RADICALS WITH VARIABLES Radicals and Variables Example(s) a. a2= b. x6= c. y12= d. 100d2= Simplifying Radicals with Variables 1. Break the radical into two (one that is a perfect root). You have remained in right site to begin getting this info. We know from the commutative property of multiplication that the order doesn't really matter when you're multiplying. To multiply $$4x⋅3y$$ we multiply the coefficients together and then the variables. 80 Simplifying Radicals with Coefficients When we put a coefficient in front of the radical, we are multiplying it by our answer after we simplify. In order to simplify a radical, all we need to do is take the … This can be divided which leaves the radical in the denominator. Answers to Adding and Subtracting Radicals of Index 2: With Variable Factors 1) −6 6 x 2) − 3ab 3) 5wz 4) − 2np 5) 4 5x 6) −4 6y 7) −2 6m 8) −12 3k We want to simplify the expression $$\sqrt 3 \left( {4\sqrt {10} + 4} \right)$$ Again, we want to use the typical rules of multiplying expressions, but we will additionally use our property of radicals, … 2. Multiplying radicals with coefficients is much like multiplying variables with coefficients. Then simplify and combine all like radicals. A b Arl 8ll arUizgRhyt js E 7rSe Us4ebrwvceTdG.K E KMBazdde 8 Xw0iVtghv zI Lnxf 7iunhi OtmeC IA hlXgSePb 7rxa Z m1A.I Worksheet by Kuta Software LLC This resource works well as independent practice, homework, extra credit or even as an assignment to leave for the substitute (includes answer The result is $$12xy$$. Multiplying Radical Expressions . Divide each exponent by the index (root). d) √1. (1) calculator Simplifying Radicals: Finding hidden perfect squares and taking their root. 2 18 2. We can add and subtract expressions with variables like this: $5x+3y - 4x+7y=x+10y$ There are two keys to combining radicals by addition or subtraction: look at the index, and look at the radicand. Once we multiply the radicals, we then look for factors that are a power of the index and simplify the radical whenever possible. Fol-lowing is a definition of radicals. Remember that you can multiply numbers outside the radical with numbers outside the radical and numbers inside the radical with numbers inside the radical, assuming the radicals have the same index. Radicals 7.1 Name _____Per_____ (DN) ON BACK OF PACKET LO: I can simplify radical expressions including adding, subtracting, multiplying, dividing and rationalizing denominators. A radical is an expression or a number under the root symbol. Adding Subtracting Multiplying Radicals Worksheets Dividing Radicals Worksheets Algebra 1 Algebra 2 Square Roots Radical Expressions Introduction Topics: Simplifying radical expressions Simplifying radical expressions with variables Adding radical expressions Multiplying radical expressions Removing radicals from the denominator Math Topics Find the perfect power that divides evenly into the coefficient. Jan 22, 2017 - Resources for radical expressions, equations, and functions. 18 multiplying radical expressions problems with variables including monomial x monomial, monomial x binomial and binomial x binomial. Radicals Simplifying Radicals With Mult Div Workshee … (1) calculator Simplifying Radicals: Finding hidden perfect squares and taking their root. Answers to Multiplying Radicals of Index 2: No Variable Factors 1) 6 2) 4 3) The Multiplication Property of Square Roots. Multiplying Radical Expressions. This worksheet correlates with the 1 2 day 2 simplifying radicals with variables power point it contains 12 questions where students are asked to simplify radicals that contain variables. If we take Warm up question #1 and put a 6 in front of it, it looks like this 6 125 6• 25 5 6•5 5 30 5 1. 3. See more ideas about middle school math, teaching math, math lessons. Displaying top 8 worksheets found for - Simplifying Radicals With Variables. ©c I2x0 X1U1Z xKeu4t SaC VSQoPfkt 9w1aArkeo BLzLjC8. 5. 3) When variables are involved, we use the same process. Here are the steps required for Multiplying Radicals With More Than One Term: Step 1: Distribute (or FOIL) to remove the parenthesis. It is common practice to write radical expressions without radicals in the denominator. A. Multiplying a two-term radical expression involving square roots by its conjugate results in a rational expression. Download File PDF Algebra 2 Multiplying And Dividing Radicals Answer Algebra 2 Multiplying And Dividing Radicals Answer Recognizing the exaggeration ways to get this ebook algebra 2 multiplying and dividing radicals answer is additionally useful. D. SIMPLIFY RADICALS WITH PERFECT PRINCIPAL ROOT USING EXPONENT RULE . 42 cannot be simplified, so we are finished. We do not leave radicals in … can be simplified to Since the 5 and 3 are both on the outside, we finish by multiplying them together. The 2 and the 7 are just constants that being multiplied by the radical expressions. Plus model problems explained step by step Multiplying radicals, though seemingly intimidating, is an incredibly simple process! Square-root expressions with the same radicand are examples of like radicals. Radicals - Higher Roots Objective: Simplify radicals with an index greater than two. Be looking for powers of 4 in each radicand. Multiplying Radicals. 6 72 The property states that whenever you are multiplying radicals together, you take the product of the radicands and … Let’s try one more example. Doing algebra expressions with exponents and variables ; trig identity word problems ; 5th grade math adding and subtracting expressions worksheet ; adding and multiplying fractions in same equation ; how to do newton raphson on casio graphics calculator ; ks3 science worksheets ; how to solve circle graphs problem ; graph circle on calculator The order does n't really matter When you 're multiplying ( one that is a perfect root ) root! Radicals Answer appropriately simple not leave radicals in the denominator an incredibly simple process that relatively. Coefficients together and then the variables ( one that is a perfect )... 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Taking their root pull out powers of 4, using the fact that that divides evenly into coefficient..., though seemingly intimidating, is an expression or a number under the root symbol into (. Simplify radicals with Mult Div Workshee … adding and subtracting radicals is much like multiplying variables coefficients. To find perfect squares and then the variables we then look for Factors are! Middle school math, math lessons a two-term radical expression involving square roots are.! Are finished same radicand are examples of like radicals by adding or...! 7 are just constants that being multiplied by the radical in the denominator and pull out of! We get into multiplying radicals with perfect PRINCIPAL root using EXPONENT RULE property! Does n't really matter When you 're multiplying, using the fact.. By multiplying them together EXPONENT by the index and simplify the radical whenever possible radicals,! | 2021-02-25 13:52:25 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8690734505653381, "perplexity": 1863.5296995078772}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178351134.11/warc/CC-MAIN-20210225124124-20210225154124-00388.warc.gz"} |
https://code.tutsplus.com/courses/perfect-workflow-in-sublime-text-2/lessons/configuring-and-mastering-split-windows | FREELessons: 34Length: 2.5 hours
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• Transcript
5.7 Configuring and Mastering Split Windows
As a former Vim user, I depend heavily on split windows. Unfortunately, it can be a bit difficult to remember the various keybindings for managing these windows in Sublime. We'll fix that in this video! | 2023-02-03 22:50:00 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8824800252914429, "perplexity": 14681.935493737463}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500076.87/warc/CC-MAIN-20230203221113-20230204011113-00364.warc.gz"} |
https://uniontestprep.com/pert/practice-test/math/pages/1 | # Question 1 Math Practice Test for the PERT
If the original price of an item was $30.00 and Joan only paid$24.00 for it, what percentage discount did Joan receive on her purchase? | 2018-01-20 07:11:22 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2192310094833374, "perplexity": 3236.2042474248497}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084889473.61/warc/CC-MAIN-20180120063253-20180120083253-00100.warc.gz"} |
https://codeforces.com/problemset/problem/755/E | E. PolandBall and White-Red graph
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
PolandBall has an undirected simple graph consisting of n vertices. Unfortunately, it has no edges. The graph is very sad because of that. PolandBall wanted to make it happier, adding some red edges. Then, he will add white edges in every remaining place. Therefore, the final graph will be a clique in two colors: white and red.
Colorfulness of the graph is a value min(dr, dw), where dr is the diameter of the red subgraph and dw is the diameter of white subgraph. The diameter of a graph is a largest value d such that shortest path between some pair of vertices in it is equal to d. If the graph is not connected, we consider its diameter to be -1.
PolandBall wants the final graph to be as neat as possible. He wants the final colorfulness to be equal to k. Can you help him and find any graph which satisfies PolandBall's requests?
Input
The only one input line contains two integers n and k (2 ≤ n ≤ 1000, 1 ≤ k ≤ 1000), representing graph's size and sought colorfulness.
Output
If it's impossible to find a suitable graph, print -1.
Otherwise, you can output any graph which fulfills PolandBall's requirements. First, output m — the number of red edges in your graph. Then, you should output m lines, each containing two integers ai and bi, (1 ≤ ai, bi ≤ n, ai ≠ bi) which means that there is an undirected red edge between vertices ai and bi. Every red edge should be printed exactly once, you can print the edges and the vertices of every edge in arbitrary order.
Remember that PolandBall's graph should remain simple, so no loops or multiple edges are allowed.
Examples
Input
4 1
Output
-1
Input
5 2
Output
41 22 33 44 5
Note
In the first sample case, no graph can fulfill PolandBall's requirements.
In the second sample case, red graph is a path from 1 to 5. Its diameter is 4. However, white graph has diameter 2, because it consists of edges 1-3, 1-4, 1-5, 2-4, 2-5, 3-5. | 2022-07-03 20:38:25 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3846197724342346, "perplexity": 705.701888598958}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656104249664.70/warc/CC-MAIN-20220703195118-20220703225118-00584.warc.gz"} |
https://www.greencarcongress.com/geothermal/ | ## DOE to issue funding opportunity on lithium extraction and conversion from geothermal brines
##### 05 October 2022
The US Department of Energy (DOE) announced its intent to issue a funding opportunity (DE-FOA-0002823) that will support research on the extraction and conversion of lithium from geothermal brines for use of lithium batteries in electric vehicles and clean energy storage. Geothermal brines—a byproduct of geothermal power—have become a proven,... Read more →
## Chevron and MOECO to collaborate on Advanced Closed Loop geothermal technology
##### 26 September 2022
Chevron New Energies, and Mitsui Oil Exploration (MOECO) signed a Joint Collaboration Agreement to explore the technical and commercial feasibility of advanced geothermal power generation in Japan. Building on Chevron and MOECO’s long-standing relationship, the new collaboration will study geothermal resource potential across Japan and will evaluate the effectiveness of... Read more →
Lars Carlstrom, the founder-CEO of Italvolt, announced the launch of a new company, Statevolt, which will construct a 54GWh Gigafactory in Imperial Valley, southern California with an expected CapEx of $4 billion. To launch the new facility, Statevolt has also signed a Letter of Intent (LOI) with Controlled Thermal Resources... Read more → The Salton Sea geothermal field in California potentially holds enough lithium to meet all of the US’ domestic battery needs, with even enough left over to export some of it. (Earlier post.) There are currently 11 commercial plants at the Salton Sea field producing geothermal energy, a process in which... Read more → ## DOE awards$8.4M for accessing geothermal potential from abandoned oil and gas wells
## DOE awards $5.5M to apply machine learning to geothermal exploration ##### 06 May 2019 The US Department of Energy (DOE) announced up to$5.5 million for 10 new projects to apply machine learning techniques to geothermal exploration and production. Machine learning—the use of advanced algorithms to identify patterns in and make inferences from data—could assist in finding and developing new geothermal resources. If applied... Read more →
The cost of generating power from renewable energy sources has reached parity or dropped below the cost of fossil fuels for many technologies in many parts of the world, according to a new report released by the International Renewable Energy Agency (IRENA). The report, “Renewable Power Generation Costs in 2014”,... Read more →
## California approves another $18M for clean energy projects;$12M for alternative vehicles
##### 26 October 2009
The Department of Energy (DOE) has selected 37 energy research projects for $151 million in funding through the recently formed Advanced Research Projects Agency-Energy (ARPA-E). This is the first round of projects funded under ARPA-E, which is receiving total of$400 million under the American Recovery and Reinvestment Act. Among... Read more →
PNNL’s metal-organic heat carrier (MOHC) in the biphasic fluid may help improve thermodynamic efficiency of the heat recovery process. This image represents the molecular makeup of one of several MOHCs. Source: PNNL. Click to enlarge. Scientists at the US Department of Energy’s Pacific Northwest National Laboratory (PNNL) have developed a... Read more →
## Potter Drilling to Test Oxford Catalysts’ Instant Steam Technology in Drilling Geothermal Wells
##### 21 May 2009
Oxford Catalysts Group PLC has entered into a memorandum of understanding (MOU) with Potter Drilling, Inc., a google.org funded company, to explore the incorporation of Oxford Catalysts’ Instant Steam technology (earlier post) into Potter Drilling’s hydrothermal spallation technology for drilling geothermal wells. Geothermal wells can be slow and expensive to... Read more →
## Johnson Controls Expands Into Solar, Geothermal, Wind, Biomass and Other Renewables
##### 20 February 2007
Johnson Controls, Inc. is expanding its business into designing, installing and servicing geothermal, solar, biomass, wind and other renewable sources as energy supply options for customers. Johnson Controls is a global leader in interior experience, building efficiency and power solutions. Johnson Controls provides batteries for automobiles and hybrid electric vehicles,... Read more →
## Ormat and DOE to Validate Electricity Generation from Oilfield Heat
##### 25 January 2007
Ormat binary geothermal power generation system. Click to enlarge. Ormat Technologies, Inc., a geothermal and recovered energy business, has signed a shared-cost Cooperative Research and Development Agreement (CRADA) with the US Department of Energy (DOE) to validate the feasibility of using geothermal power generation technology for the production of commercial... Read more →
## MIT-Led Study: Geothermal Could Supply Substantial Portion of Future US Power Need
##### 22 January 2007
Schematic of a conceptual two-well Enhanced Geothermal System in hot rock in a low-permeability crystalline basement formation. Click to enlarge. A comprehensive new MIT-led study of the potential for geothermal energy within the United States has found that Enhanced Geothermal System (EGS) technology could supply a substantial portion of US... Read more → | 2022-12-08 19:14:36 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5352137088775635, "perplexity": 9834.33095301738}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446711360.27/warc/CC-MAIN-20221208183130-20221208213130-00125.warc.gz"} |
http://aas.org/archives/BAAS/v26n4/aas185/abs/S11106.html | On the measurement of a Compton-Getting dipole in the photon number counts of $\gamma$-ray bursts
Session 111 -- Gamma Ray Bursts
Oral presentation, Thursday, 12, 1995, 10:00am - 11:30am
## [111.06] On the measurement of a Compton-Getting dipole in the photon number counts of $\gamma$-ray bursts
Caleb A. Scharf, Keith Jahoda, Elihu Boldt (NASA/GSFC)
One of the strongest pieces of evidence for the cosmological origin of $\gamma$-ray bursts would be the observation of a photon number count dipole (a Compton-Getting effect) due to our motion with respect to a distant burst parent population. Such an effect is the product of a dipole anisotropy in the distribution of bursts (Maoz 1994, ApJ 428, 454) and a dipole anisotropy due to counting photons in a finite energy bandwidth. The form of the Compton-Getting effect is discussed and Monte-Carlo simulations are performed to estimate the likelihood of its measurement under different circumstances. Using a catalogue of 410 bursts observed by the COMPTON/BATSE instrument (in the energy band 20-50$keV$), and the dipole determined from the Cosmic Microwave Background, we find that the observed dipole-aligned component has only a $\sim 10$\% chance of occuring for the null hypothesis of no Compton-Getting effect. However at the 90\% confidence level we can only constrain the Compton-Getting effect to be $> 0$\% and $< 40$\% in magnitude. Of the order of $10^{4}$ bursts would be necessary for a robust confirmation of the expected 1-2\% Compton-Getting effect. In addition the observed correlation with the direction of the Local Group motion has only a $\sim 5$\% probability of occuring by chance. These results are intriguing and suggest that future analyses of the angular distribution of $\gamma$-ray bursts should include photon number weighting.
C.A.S. acknowledges the NRC for their support through a Research Associateship. | 2014-10-01 08:19:54 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8102772235870361, "perplexity": 1411.27630318227}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-41/segments/1412037663372.35/warc/CC-MAIN-20140930004103-00422-ip-10-234-18-248.ec2.internal.warc.gz"} |
https://www.intmath.com/blog/mathematics/funny-graphs-from-graphjam-5611 | # Funny graphs - from GraphJam
By Murray Bourne, 31 Jan 2011
GraphJam uses pie charts, histograms and Venn diagrams to give us some really funny insights into the human condition.
Here is a sampling of their more family-friendly graphs:
What GraphJam is doing in these graphs could be a great concept for a class project in statistics - to encourage students to be creative (almost always missing in math classes), have fun making others laugh, and learn something about expressing statistics in a visual way.
I'll feature some more of these graphs in a later article.
Disclaimer: GraphJam is not always a family-friendly site. With that in mind, here is their link: Graphjam.
### 3 Comments on “Funny graphs - from GraphJam”
1. ritesh says:
interesting graphs.good way to love math
2. Anil says:
So simple yet true.Did make it look interesting for even those who hate Maths.
3. Funny graphs « garysmathsblog says:
[...] out some funny graphs at GraphJam. Go and check it [...]
### Comment Preview
HTML: You can use simple tags like <b>, <a href="...">, etc.
To enter math, you can can either:
1. Use simple calculator-like input in the following format (surround your math in backticks, or qq on tablet or phone):
a^2 = sqrt(b^2 + c^2)
(See more on ASCIIMath syntax); or
2. Use simple LaTeX in the following format. Surround your math with $$ and $$.
$$\int g dx = \sqrt{\frac{a}{b}}$$
(This is standard simple LaTeX.)
NOTE: You can mix both types of math entry in your comment.
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From Math Blogs | 2020-06-01 17:37:20 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6792309880256653, "perplexity": 9500.428954144661}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347419056.73/warc/CC-MAIN-20200601145025-20200601175025-00334.warc.gz"} |
https://acroz.dev/2016/02/16/neural-networks/ | # Neural Networks
This post is part of the series Neural Networks.
Devising a mathematical model of a system requires careful thought so that it accurately represents the system. In relatively simple applications, for example in calculating the trajectory of a ball launched with a certain velocity, the underlying physics can be represented directly by the model. In such an example, the equations of Newtonian mechanics can be integrated to calculate the path of the ball.
As the underlying system becomes more complex, for example when modelling a biological system, simplifications and approximations must be made, as the system becomes too large to model explicitly.
In machine learning, the physics or relationships underlying the data may not be known at all - in fact, it is usually our goal to discover these relationships through our analysis of the data with no or limited a priori knowledge. Deciding on a suitable mathematical model to fit to the available data can then be a difficult task - the choice of an unsuitable model may unfairly bias the results of the analysis.
## Inspiration from Neurology
Originally devised to model and study the mechanisms of the brain, computational neural networks have increasingly found utility as a powerful machine learning technique. The brain is modelled as a large network of neurons, each of which is considered to be a simple computational unit taking a number of inputs $$\{x_i\}$$ and providing a single output (or “activation”) $$a$$. The activation is calculated by applying weights $$\{w_i\}$$ and a bias term $$b$$ and passing to an activation function $$g$$:
$a = g \left( \sum_i w_i x_i + b \right)$
A common choice for $$g$$ is the sigmoid function:
$g(x) = \sigma(x) = \frac{1}{1 + e^{-x}}$
Neurons are often depicted as below, receiving a number of inputs from the left and providing an output to the right:
A more complex model can then be constructed by assembling a network of such neurons, where multiple layers of neurons have their inputs and outputs connected together:
More complex structures with reentrant positive feedback loops are also possible and may offer additional power, however here we concentrate on the simpler unidirectional, layer-arranged structure as above. Note that while some neurons in the above network appear to provide different outputs, the different arrows leaving a neuron in fact correspond to the same output being sent to multiple neurons in the next layer of the network.
## Intuition
By combining together a network of simple computational networks, it is possible to form a network capable of producing a wide range of mathematical representations. A set of inputs $$\{x_i\}$$ can be provided to the input layer, processed through the neural network, and a useful output or outputs $$\{a_i\}$$ then generated. This is how real neural networks in the brain are thought to work, and the application of the concept to machine learning turns out to be a powerful technique.
## Further Detail
For further detail on the implementation and application of neural networks to a handwritten digit recognition system, see the list of other posts in this series below. | 2021-03-06 01:28:02 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5976141691207886, "perplexity": 363.0604109954424}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178374217.78/warc/CC-MAIN-20210306004859-20210306034859-00328.warc.gz"} |
https://tspace.library.utoronto.ca/handle/1807/24701 | Home
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Please use this identifier to cite or link to this item: http://hdl.handle.net/1807/24701
Title: A Longitudinal Investigation into the Association of Smoking and Depression among Adolescents: Exposures, Outcomes, and Auxiliary Hypotheses Authors: Chaiton, Michael Advisor: Cohen, Joanna Department: Dalla Lana School of Public Health Keywords: DepressionSmoking Issue Date: 5-Aug-2010 Abstract: Introduction: The association between smoking and depression has been well established, but the nature of the relationship has not been determined. A synthesis of longitudinal studies examining the onset of smoking and depression among adolescents demonstrated consistent evidence of both smoking predicting depression and depression predicting smoking in multiple populations; however, more work is needed to develop and test the mechanisms associated with the onset of the co-occurrence of smoking and depression. This thesis examines the role of a broad range of potential confounders on the relationship between smoking and depression, and investigates a potential mechanism of effect. Method: Analyses were conducted using the Nicotine Dependence in Teens (NDIT) cohort which included 1293 students initially aged 12-13 years recruited from all grade seven classes in a convenience sample of ten secondary schools in Montreal, Canada surveyed twenty times over five years. Multiple regressions were performed to examine the temporal relationship of potential confounders on the relationship between smoking and depression and to empirically observe variables that could be intermediate on pathways between smoking and depression. A growth curve model was developed to test the effect of perceived self medication on changes in depression scores over time. Results: A concept map of the smoking and depression relationship in the NDIT cohort was developed according to the results of proportional hazard and fixed effect regressions in which friend smoking, stress, and anxiety-associated variables were identified as intermediate variables. Perceived self-medication was associated with decelerated rates of change of depressive symptoms over times, suggesting that smoking may increase mean levels of stress and depressive symptoms, but may offer the perception of control. Conclusion: In concert, this thesis suggests a model in which stress and the perceived control of psychobiological function using cigarettes lead to the development of increased depressive symptoms and increased cigarette use. URI: http://hdl.handle.net/1807/24701 Appears in Collections: DoctoralDalla Lana School of Public Health - Doctoral theses
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https://access.openupresources.org/curricula/our-hs-math/integrated/math-3/unit-3/lesson-1/ready_set_go.html | # Lesson 1Scott’s March MotivationDevelop Understanding
For each problem, place the appropriate inequality symbol between the two expressions to make the statement true.
If , then:
If , then:
If , then:
If , then:
If , then:
If , then:
If , then:
If , then:
If , then:
## Set
For problems 10–12, the recursive rule for a polynomial function is given in the form .
• Use the recursive rule to fill in the first 5 values of the function in the table of values.
• Identify the rate of change.
• Classify the type of polynomial function as linear, quadratic, or cubic.
### 10.
rate of change:
type of polynomial:
Input
Output
### 11.
rate of change:
type of polynomial:
Input
Output
### 12.
rate of change:
type of polynomial:
Input
Output
For problems 13–16,
• Find the rate of change for each table of values.
• Write the recursive form of each function. If the rate of change is constant, write in the constant value. If the rate of change is a function, write the equation of the function that describes the rate of change.
• Identify the type of function.
### 13.
Input
Output
rate of change:
recursive rule:
type of function:
### 14.
Input
Output
rate of change:
recursive rule:
type of function:
### 15.
Input
Output
rate of change:
recursive rule:
type of function:
### 16.
Input
Output
rate of change:
recursive rule:
type of function:
## Go
Find the quotient without using a calculator. If you have a remainder, write the remainder as a whole number. Example: r
### 18.
Is a factor of ? How do you know?
Find .
### 20.
Is a factor of ? How do you know?
### 22.
Is a factor of ? How do you know? | 2022-06-30 11:11:32 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 98, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.926508367061615, "perplexity": 3358.7954993960902}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103671290.43/warc/CC-MAIN-20220630092604-20220630122604-00059.warc.gz"} |
https://bytepawn.com/investigating-information-storage-in-quantized-autoencoders-with-pytorch-and-mnist.html | Investigating information storage in quantized Autoencoders with Pytorch and MNIST
Marton Trencseni - Sun 04 April 2021 - Machine Learning
Introduction
In this experiment I wanted to understand the compression ratio of Autoencoders: how much information (how many bits) does an Autoencoder encode in the encoding dimensions? Let's say an autoencoder is able to encode a 28x28 grayscale MNIST image (28x28x8 bits = 6272 bits) in a 32 dimensional encoding space with acceptable reconstruction loss. What is the compression ratio? With a CUDA/GPU, those 32 dimensions are actually 32 float32's, so it's 32x32 = 1024 bits, which corresponds to 6.1x (lossy) compression. But are all those 1024 bits really needed? Intuitively the entire float32 space is probably not used. A related question is, what is the "right" number of encoding dimensions to pick for Autoencoders?
The notebook is up on Github.
Experiment setup
To answer these questions, I took a simple Autoencoder neural network with a Linear+ReLu encoder and a Linear+Sigmoid decoder layer. Since I will want to quantize the bits between the encoder and a decoder, I use the sigmoid() function to get the encoder's output to be between 0 and 1, and then the inverse, the logit() function before feeding back to the decoder.
Code
The code is a straightforward Autoencoder neural network implemented in Pytorch, with some additional transformations in the forward() function to implement quantization. The arrows mark the departure from a vanilla Autoencoder:
class Autoencoder(nn.Module):
def __init__(self, encoding_dims):
super(Autoencoder,self).__init__()
self.encoder = nn.Sequential(
nn.Flatten(),
nn.Linear(img_dims*img_dims, encoding_dims),
nn.ReLU(),
)
self.decoder = nn.Sequential(
nn.Linear(encoding_dims, img_dims*img_dims),
nn.Unflatten(1, (1, img_dims, img_dims)),
nn.Sigmoid(),
)
def forward(self, x, quantize_bits=None):
x = self.encoder(x)
x = torch.sigmoid(x) <--- sigmoid
if quantize_bits is not None: <--- if not training
x = round_bits(x, quantize_bits) <--- .. then quantize the encoding
x = torch.logit(x, eps=0.001) <--- logit = inverse sigmoid
x = self.decoder(x)
return x
The function round_bits() quantizes the input number to 2**quantize_bits levels between 0 and 1:
def round_bits(x, quantize_bits):
mul = 2**quantize_bits
x = x * mul
x = torch.floor(x)
x = x / mul
return x
The main training loop trains the Autoencoder for different encoding_dims, and then tests the reconstruction loss for various values of quantize_bits:
for encoding_dims in [4, 8, 16, 32, 64, 128, 256]:
# train
autoencoder = Autoencoder(encoding_dims=encoding_dims).to(device)
distance = nn.BCELoss()
num_epochs = 50
for epoch in range(num_epochs):
imgs = Variable(imgs).to(device)
output = autoencoder(imgs)
loss = distance(output, imgs)
loss.backward()
optimizer.step()
# test
distance = nn.MSELoss()
for quantize_bits in [2, 4, 8, 16, 32]:
loss = 0
imgs = Variable(imgs).to(device)
output = autoencoder(imgs, quantize_bits=quantize_bits)
loss += distance(output, imgs)
Results
The results can be plotted to show the loss per encoding_dims, per quantize_bits:
The plot shows that:
• each float32 in the encoding stores around 8 bits of useful information (out of 32), since all of the curves flatten out after 8 bits
• 128 dimensions is the maximum required, since the next jump to 256 yield no significant decrease in loss
• overall, based on these curves, encodim_dims = 64 and quantize_bits = 8 appears to be a good trade-off (total_bits = 64*8 = 512 bits)
Alternatively we can plot total_bits = encoding_dims * quantize_bits on the x-axis:
This re-affirms that 512 bits --- which corresponds to 12x (lossy) compression --- is a good trade-off, or 1024 bits for 10% less loss. Loss does not decrease significantly after 1024 bits, that appears to be best the Autoencoder can accomplish. For reference, the entire MNIST training dataset, uncompressed is 28*28*8 * 60*1000 / 8 = 47,040,000 bytes. After gzip compression, the file size is 9,912,422 bytes, for a lossless compression ratio of 4.7x.
In the next post, I will explore what we lose with the Autoencoder's lossy compression in terms of recognizability of the digits. | 2021-06-24 20:33:11 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.513505756855011, "perplexity": 4820.096895584666}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623488559139.95/warc/CC-MAIN-20210624202437-20210624232437-00610.warc.gz"} |
https://www.physicsforums.com/threads/a-question-about-quadratic-residues.588015/ | 1. Mar 17, 2012
### yeland404
I need to prove that a be a odd integer that congruence X^2$\equiv$a mod 2
is always solvable with exactly one incongruent solution modulo 2.
this question is linked with (b) let a be an odd integer. Prove that the congruence X^2$\equiv$a mod 4 is solvable iff a$\equiv$1 mod 4. in this case ,prove that X^2$\equiv$a mod 4solutions has exactly two incongruent
solutions modulo 4.
these two seem to link with each other. And the proposition I learn is X^2$\equiv$a mod p has either no solution or two solutions, but p there is an odd prime number. HOw to apply to the queations above?
2. Mar 17, 2012
### Office_Shredder
Staff Emeritus
You should be able to just check these by hand. For example does x^2=2 (mod 4) have any solutions? Just plug in 0,1,2,3 for x and see what you get | 2018-06-19 03:29:10 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7820899486541748, "perplexity": 1219.9442326314427}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267861752.19/warc/CC-MAIN-20180619021643-20180619041643-00586.warc.gz"} |
https://stats.stackexchange.com/questions/475555/probit-model-with-gaussian-noise | # Probit model with Gaussian noise
Assume we have the following model setup $$\Phi^{-1}(D)=\alpha+\beta X+\epsilon$$ where $$\epsilon\sim N(0,\sigma^{2})$$ and $$D_{i}=\{0,1\}$$. This implies that $$\text{Pr}(D_{i}=1\,|\,X,\epsilon)=\Phi(\alpha+\beta X+\epsilon)$$ but we need to integrate out the $$\epsilon$$ \begin{align} \text{Pr}(D_{i}=1\,|\,X)&=\int_{\epsilon}\text{Pr}(D_{i}=1\,|\,X,\epsilon)\,f_{\epsilon}(\epsilon)\,d\epsilon\\ &=\Phi\bigg(\frac{\alpha+\beta x}{\sqrt{1+\sigma^{2}}}\bigg) \end{align}
So if we were to estimate the parameters of this model $$(\hat{\alpha},\hat{\beta},\hat{\sigma})$$ we could go about it by changing the likelihood function of the simple probit model from \begin{align} L=\sum_{i=1}^{n}D_{i}\log(\Phi(\alpha+\beta X+\epsilon))+(1-D_{i})\log(\Phi(\alpha+\beta X+\epsilon)) \end{align} to \begin{align} L^{*}=\sum_{i=1}^{n}D_{i}\log\bigg(\Phi\bigg(\frac{\alpha+\beta X}{\sqrt{1+\sigma^{2}}}\bigg)\bigg)+(1-D_{i})\log\bigg(1-\Phi\bigg(\frac{\alpha+\beta X}{\sqrt{1+\sigma^{2}}}\bigg)\bigg) \end{align} However, I've noticed that performing reliable optimisation of this likelihood is difficult. Given the toy example
n = 10000
a = -2
b = 0.01
x = runif(n, min = 1, max = 5) + rnorm(n, 0, 0.15)
p = pnorm(a + b*x)
d = rbinom(n, size = 1, prob = p)
y = tibble::as_tibble(data.frame(x, p, d))
and the likelihood defined as
fn = function(par, x, d) {
return(-sum(
d*log(pmax(10^-23, pnorm((par[1] + par[2]*x)/sqrt(1 + par[3]^2)))) +
(1-d)*log(pmax(10^-23, 1 - pnorm((par[1] + par[2]*x)/sqrt(1 + par[3]^2))))
))
}
and using quasi-Newton methods
optim(par = c(0, 0, 0.5),
fn = fn,
x = y$$x, d = y$$d,
method = "L-BFGS-B",
lower = c(-Inf, -Inf, 0),
upper = c(Inf, Inf, Inf),
hessian = TRUE)
typically doesn't behave very well. In fact, the $$\hat{\sigma}$$ usually just converges to a point near the starting value.
Are there any obvious changes (choice of algorithm, approximations to the likelihood function, better choice of starting values) that can be made to make the estimation of $$(\hat{\alpha},\hat{\beta},\hat{\sigma})$$ more reliable?
The model is not identified, meaning there is no unique solution to the optimization problem. There are infinite values of the parameters that will yield the same likelihood. For example, $$\alpha = .5$$, $$\beta = 1$$, and $$\sigma = 2$$ will yield the exact same likelihood as $$\alpha = 1$$, $$\beta = 2$$, and $$\sigma = \sqrt{19}$$.
More generally, consider the maximum of the likelihood, $$L^*$$, which is found when $$(\alpha, \beta, \sigma) = (\alpha^*, \beta^*, \sigma^*)$$. For any $$k$$, $$\left(k\alpha^*, k\beta^*, \sqrt{k^2+1+(k\sigma^*)^2}\right)$$ will yield the exact same likelihood. Therefore, there is no unique value of the parameters that maximizes the likelihood. This is why the optimization is unstable; any specific solution it arrives at will be due purely to numerical instability.
Note that in typical probit regression, we assume $$\sigma=0$$, i.e., that there is no latent variable $$\epsilon$$ that is unaccounted for. This is different from the latent variable formulation of probit regression, where we assume $$Y^*=X\beta+\epsilon$$ where $$\epsilon \sim N(0, 1)$$, and $$P(D=1|X) = P(Y^*>0|X)=P(X\beta + \epsilon > 0)$$ which implies $$P(D=1|X) = \Phi(X\beta)$$.
• Thanks @Noah. Yes I'm aware of the latent variable motivation for the probit model. I was trying to extend the specification of the usual probit model $\text{Pr}(D=1\,|\,X)=\Phi(\beta X)$ to include some other source of randomness. Unfortunately it seems this leads to problems with estimation. – epp Jul 5 '20 at 7:07 | 2021-07-28 01:36:35 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 28, "wp-katex-eq": 0, "align": 3, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9554196000099182, "perplexity": 795.201547868016}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046153515.0/warc/CC-MAIN-20210727233849-20210728023849-00456.warc.gz"} |
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Zbl 0834.26013
Alzer, Horst
The inequality of Ky Fan and related results.
(English)
[J] Acta Appl. Math. 38, No.3, 305-354 (1995). ISSN 0167-8019; ISSN 1572-9036/e
This survey paper presents refinements, extensions, and variants of the well-known Ky Fan inequality $$\prod^n_{i = 1} \bigl( y_i/(1 - y_i) \bigr)^{1/n} < \sum^n_{i = 1} y_i \left/ \sum^n_{i = 1} \right. (1 - y_i),$$ valid for all real numbers $y_i \in (0,1/2]$ $(i = 1, \ldots, n)$ which are not all equal. In the list of 54 references, there are 24 of the author of this paper.
[J.E.Pečarić (Zagreb)]
MSC 2000:
*26D15 Inequalities for sums, series and integrals of real functions
Keywords: arithmetic mean; geometric mean; Ky Fan inequality
Cited in: Zbl 0991.26013 Zbl 0872.26007
Highlights
Master Server | 2013-05-23 03:12:54 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5233567357063293, "perplexity": 4732.738588371297}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368702762497/warc/CC-MAIN-20130516111242-00098-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://de.maplesoft.com/support/help/view.aspx?path=StudyGuides%2FMultivariateCalculus%2FChapter4%2FExamples%2FSection4-10%2FExample4-10-3 | Example 4-10-3 - Maple Help
# Online Help
###### All Products Maple MapleSim
Chapter 4: Partial Differentiation
Section 4.10: Optimization on Closed Domains
Example 4.10.3
Find the extreme values of the function $f\left(x,y\right)={x}^{2}-xy+{y}^{2}+1$ on the domain $R$ consisting of the triangular region whose edges are the $y$-axis, and the lines $y=x$ and $y=4$.
© Maplesoft, a division of Waterloo Maple Inc., 2021. All rights reserved. This product is protected by copyright and distributed under licenses restricting its use, copying, distribution, and decompilation.
For more information on Maplesoft products and services, visit www.maplesoft.com | 2022-10-01 01:56:14 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 147, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2564938962459564, "perplexity": 6252.983568566943}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030335514.65/warc/CC-MAIN-20221001003954-20221001033954-00151.warc.gz"} |
https://stacks.math.columbia.edu/tag/0DXN | Lemma 52.9.3. In Lemma 52.9.2 if instead of the empty condition (2) we assume
1. if $\mathfrak p \in V(I)$ and $\mathfrak p \not= \mathfrak m$, then $\text{depth}_{A_\mathfrak p}(M_\mathfrak p) + \dim (A/\mathfrak p) > s$,
then the conditions also imply that $H^ i_{J_0}(M)$ is a finite $A$-module for $i \leq s$.
Proof. This is a special case of Lemma 52.8.3. $\square$
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar). | 2022-06-27 15:19:57 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 2, "x-ck12": 0, "texerror": 0, "math_score": 0.9851688742637634, "perplexity": 372.9419211107305}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103334753.21/warc/CC-MAIN-20220627134424-20220627164424-00177.warc.gz"} |
https://www.numerade.com/questions/suppose-the-series-sum-a_n-is-conditionally-convergent-a-prove-that-the-series-sum-n2-a_n-is-diverge/ | 💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!
JH
# Suppose the series $\sum a_n$ is conditionally convergent. (a) Prove that the series $\sum n^2 a_n$ is divergent.(b) Conditional convergence of $\sum a_n$ is not enough to determine whether $\sum na_n$ is convergent. show this by giving an example of conditionally convergent series such that $\sum na_n$ converges and an example where $\sum na_n$ diverges.
## a. This contradiction shows that $\sum n^{2} a_{m}$ diverges. b. In both cases, $\sum n a_{n}$ diverges by the Test for Divergence.
Sequences
Series
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##### Kristen K.
University of Michigan - Ann Arbor
Lectures
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### Video Transcript
we were given that the syriza vahan is conditionally conversion. Yeah, so in part A, we'd like to show and squared And the sum of this diverges so away to prove this is to go by contradiction. So here, Lis just point that out there. So let's go ahead and suppose the opposite. That would be the sum of and square a n converges Now by the test for diversions. This would imply that the end's term here goes to zero in the limit. But we can rewrite this in a way that will be convenient for us. So let's go ahead and rewrite the and square as a fraction. And if this lemon goes to zero, then I could go ahead and take absolute value. Here is well, so let me just go ahead and throw the absolute value there. Now we go to the limit comparison test. So here's another Theron and we have a fraction here. We know that the Siri's one over and square converges. You can use the pizzas to see this P equals two. It's bigger than one. And that implies that too. Some convergence. So in this case, limit comparison test would say so does the Siri's here. However, this cannot happen because in the given information we have that this syriza's conditionally conversion that it means part of the definition that if you take the absolute value of a n, the Siri's will diverge. So is it recap. We suppose that it converges because we're using proof by contradiction. And under this assumption in the proof, we end up showing that the Siri's an absolute value, converges. But that contradicts. They've given information, so that completes the proof. A party. Okay, now let's go to the next page for part B. So just remember still the given information I'll put that here they and some this was conditionally conversion. So I'll just abbreviate that. And now we want to show that this is not enough enough information to determine if the Siri's and and commercials So we'll show this by giving an example one in which it converges, and then one of which diversions. So let's look at an example here. Par I. So in this case, let's go ahead and define the sum of a M the following way. Now the Siri's is conditionally conversion. You can see that it conversions by the alternating Siri's test. But if you take absolute value, the negative one just becomes a one. And this diverges because you can use the pee test here. P equals one. Yeah, so that guarantees that this example were using, in part one here that it is indeed conditionally conversion. We have to check that because that is the assumption here in the problem. Now I look at the Siri's and a N that just becomes negative, one to the end and this diverges and to see that you can use the test for divergence. So here's one example in which the sum of nn divergence now will provide an example in which it converges this time well defined. The sum of an let's go ahead and do it negative one to the end over and let's do and natural log and in the denominator. So this is conditionally comm urgent. You can use alternating Siri's test that all apply convergence. But if you look at the absolute value, this diverges and to see that you could use the integral test so that once again guarantees that we're under this hypothesis up here, conditionally convergent. And then now go ahead and look at the sun stand there that becomes negative one to the end, over natural log of end in the Siri's those converge by the alternating serious tests. So we gave two examples, one in which the sum of many and diverse and that another example in which it converges and that solves proud part B of this problem.
JH
#### Topics
Sequences
Series
##### Kristen K.
University of Michigan - Ann Arbor
Lectures
Join Bootcamp | 2021-10-27 09:38:25 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9110020995140076, "perplexity": 442.4123269997138}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323588113.25/warc/CC-MAIN-20211027084718-20211027114718-00337.warc.gz"} |
https://www.khanacademy.org/math/multivariable-calculus/double_triple_integrals/double_integrals | Double integrals
6 videos
A single definite integral can be used to find the area under a curve. with double integrals, we can start thinking about the volume under a surface!
Double integral 1
VIDEO 10:29 minutes
Introduction to the double integral
Double integrals 2
VIDEO 9:50 minutes
Figuring out the volume under z=xy^2
Double integrals 3
VIDEO 8:03 minutes
Let's integrate dy first!
Double integrals 4
VIDEO 9:24 minutes
Another way to conceptualize the double integral.
Double integrals 5
VIDEO 9:51 minutes
Finding the volume when we have variable boundaries.
Double integrals 6
VIDEO 9:58 minutes
Let's evaluate the double integrals with y=x^2 as one of the boundaries. | 2015-07-04 01:51:30 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9720638394355774, "perplexity": 8123.093650977682}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-27/segments/1435375096293.72/warc/CC-MAIN-20150627031816-00186-ip-10-179-60-89.ec2.internal.warc.gz"} |
https://gcs-group.ro/air-flow-uujun/count-and-say-sequence-in-python-cbf7ce | 0. ANALYSIS. 'o, world!' ... Count¶ A sequence can also be queried for the number of times a particular item appears: In [52]: all_letters. I don't know why the sequences were not appended to … The rest of the numbers are obtained by the sum of the previous two numbers in the series. 0. So if you have the value 111221, these groups are 111, 22, and 1. If number is odd, then collatz() should print and return 3 * number + 1. Counting bases in a sequence Sorting DNA sequences by length 29 common beginner errors on one page When to use aggregate/filter/transform in Pandas Inventing new animals with Python Python tutorial. lintcode: Count and Say; Problem Statement. >>>del spam[1] ... in accession capital to say that I acquire in fact enjoyed account your weblog posts. The count-and-say sequence is a sequence of digit strings defined by the recursive formula:. 'lo, wo' The look-and-say sequence was introduced and analyzed by John Conway. Applications of LCS: Forms the basis for data comparison which will be used in the field of bioinformatics. Assign the value of … Tuples are created by the comma operator, but they are not within square brackets. Among operations that are supported by most sequence types, “in” and “not in” operations have equal priority as the comparison operations, and “+” and “*” operations have equal priority as the corresponding numeric operations. Print the Fibonacci sequence. This statement can be negated with either not (x in NewSeq) or x not in NewSeq. String Processing in Python: Look-and-Say Sequence - Duration: 11:34. Then that number is assigned to the Number variable. In Python, Sequences are the general term for ordered sets. Defining a list and indexing and appending it. 111221 1 is read off as "one 1" or 11. This time, we’ll work on the Count-and-Say problem from Leetcode. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. You could use a proper test framework or just write simple code: Python has a style guide called PEP 8. >>>spam[2] The lines show the growth of the numbers of digits in the look-and-say sequences with starting points 23 (red), 1 (blue), 13 (violet), 312 (green). NewSeq[-i] returns the i’th element from the end of NewSeq, so NewSeq [-1] will the last element of NewSeq, NewSeq [-2] will be the second -last element. >>>spam.insert(1, 'and') The look-and-say sequence is the sequence of below integers: 1, 11, 21, 1211, 111221, 312211, 13112221, 1113213211, … How is above sequence generated? 1 2. A tuple with single item ends with a trailing comma. In this way we can simply call that function inside the while loop. ['bacon', 'chicken', 42, 10] It could be printing double quotes or indicating a new line. Python len() is a built-in function in python. change how cnt works by being a bit more natural: set it to 0 and increment it in all cases. Python Algorithm; Introduction Basic Date Structure String ... Count and Say Question. In this tutorial, we will learn how to count the total number of digits in a number using python. NewSeq[i] returns the i’th character of NewSeq. 21 is read off as "one 2, then one 1" or 1211. 21 4. Decorators are another elegant representative of Python's expressive and minimalistic syntax. Note: The sequence of integers will be represented as a string. Introduction to Sequences in Python. My friend says that the story of my novel sounds too similar to Harry Potter. >>>print var I need 30 amps in a single room to run vegetable grow lighting. "[4] (Notice: first dimension has prime value 5, then it’s 7, 11, and following prime values.) 11 is read off as "two 1s" or 21. Out of these seven, three are the most popular. 'Hello, world!'. • isalnum( ) Do US presidential pardons include the cancellation of financial punishments? THE CERTIFICATION NAMES ARE THE TRADEMARKS OF THEIR RESPECTIVE OWNERS. Can I use Spell Mastery, Expert Divination, and Mind Spike to regain infinite 1st level slots? 21 is read off as "one 2, then one 1" or 1211. is sometimes referred to as the Cuckoo's Egg, from a description of Morris in … Use MathJax to format equations. This challenge requires us to write a function that gives the n th term of the count-and-say sequence in Python. The count-and-say sequence is the sequence of integers beginning as follows: 1, 11, 21, 1211, 111221, ... 21 is read off as one 2, then one 1 or 1211. 42 Slicing and dicing and indexing a string. The count-and-say sequence is the sequence of integers with the first five terms as following: 1. Code Review Stack Exchange is a question and answer site for peer programmer code reviews. >>>"Hello, world! Now I want to create a separate function that calculates the value of the nth term of the sequence. What is the standard practice for animating motion -- move character or not move character? If I'm the CEO and largest shareholder of a public company, would taking anything from my office be considered as a theft? "[3] 1. Enclosing parentheses are optional in tuples. The count-and-say sequence is the sequence of integers with the first five terms as following: 1. That means every integer (repeated continuously) is read off with its count value. Using python, count the number of digits in a number. >>>spam[1] Let’s see what escape sequences are and how to use them in Escape Sequences in Python. Whereas mutable objects are easy to change. So the first character of a string is at index 0, the second character at index 1 and so on. This is a sequence whose few terms are like below − 1; 11; 21; 1211; 111221; The string will be read like. Mutable Sequence Types in Python. • expandtabs( [tabsize]) >>>spam[1] Following are the operations that can be performed on a sequence: –. >>>spam 11 is read off as two 1s or 21. • isupper( ) 7. c++. In these Sequences in Python article, we shall talk about each of these sequence types in detail, show how these are used in python programming and provide relevant examples. Is it natural to use "difficult" about a person? LeetCode 153. ['bacon', 'and', 'chicken', 42, 10] • join( seq) Problem : Given an integer n, generate the nth count and say sequence. Taking this chance to have cnt initialised at the beginning of the loop, we'd have something like: I highly recommend Ned Batchelder's talk "Loop like a native". It’s more than a sequence. 题目翻译 Among other things, it gives guidelines about indentation. But in Python, there are number of types that all fit this description, each with special customization. The idea of the look-and-say sequence is similar to that of run-length encoding. rev 2021.1.21.38376, The best answers are voted up and rise to the top, Code Review Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us, Given an integer n, generate the nth count and say sequence, Ned Batchelder's talk "Loop like a native", Episode 306: Gaming PCs to heat your home, oceans to cool your data centers, Given a sequence of words, print all anagrams together, Optimize Cython code with np.ndarray contained, Return a rank for string of letters based on alphabetical permutations. By closing this banner, scrolling this page, clicking a link or continuing to browse otherwise, you agree to our Privacy Policy, New Year Offer - Python Training Program (36 Courses, 13+ Projects) Learn More, 36 Online Courses | 13 Hands-on Projects | 189+ Hours | Verifiable Certificate of Completion | Lifetime Access, Programming Languages Training (41 Courses, 13+ Projects, 4 Quizzes), Angular JS Training Program (9 Courses, 7 Projects), Practical Python Programming for Non-Engineers, Python Programming for the Absolute Beginner, Software Development Course - All in One Bundle. • Center(width[, fillchar]) Python: Tips of the Day. It means to say the nth digit is the sum ... 0, a = 0, b = 1 and count = 1 Step 3: while (count <= n) Step 4: print sum Step 5: Increment ... 4 to 7 Implementing the Fibonacci Series program in python. * operator repeats a sequence a defined number of times. >>>spam[2] Note: The sequence of integers will be represented as a string. Were the Beacons of Gondor real or animated? Python has a built-in function itertools.groupby for finding groups in an iterator, and using this function, the look-and-say step becomes: which can be rearranged into a single expression: and so the whole sequence can be generated like this: To select only the \$n\$th value of the sequence, use itertools.islice: For this type of exercice, it is easy to write simple tests so that you can have a quick feedback when you break something as you write/rewrite your function. countAndSay(1) = "1" countAndSay(n) is the way you would "say" the digit string from countAndSay(n-1), which is then converted into a different digit string. If the number of terms is more than 2, we use a while loop to find the next term in the sequence by adding the preceding two terms. In each step of the "count-and-say sequence" (which is more usually called the "look-and-say sequence") you have to find the groups of consecutive runs of identical digits. For any d other than 1, the sequence starts as follows: end (Optional) - ending index within the string where search ends. "[3:9] site design / logo © 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. • startswith( prefix[, start[, end]]) For new python developers and learners, this article should create essential learning objectives, for established programmers, this could be a revision module. The way to call count is we say the name of the sequence whether it's a string, integer, or tuple that we want to count on. >>>"Hello, world! How can I defeat a Minecraft zombie that picked up my weapon and armor? The look-and-say sequence is a concealed and mysterious topic of mathematics. Apart from these, there are many other methods and functions are available that can be implemented on strings, lists, and tuple etc. And here is our program. QGIS outer glow effect without self-reinforcement, Analysis of this sentence and the "through via" usage within. + operator combines two sequences in a process. A "LookAndSay" sequence generator in python. Introducing 1 more language to a trilingual baby at home. ['bacon', 'and', 'chicken', 42, 10] In this video, we will be considering the so-called "Look-and-Say" sequence. Details about these functions will be provided in subsequent articles. >>>string[-6:-1] 'chicken' 题解1 - 迭代. Syntax : list_name.count(object) Parameters : Object is the things whose count is to be returned. When we run the above code with count value 12, the else block is not gets executed because the while loop is terminated with a break statement. (I am not sure if the 8 space indent was intended or not in your question). 11 is read off as "two 1s" or 21. Counting Letters in DNA Strings¶. November 17, 2020 . he count-and-say sequence is the sequence of integers with the first five terms as following: 1 is read off as "one 1" or 11. Given an integer n, generate the _n_th sequence. jason1244 created at: 2 days ago | No replies yet. This blog is dedicated to a revision of the Python sequence and collections. Note: The sequence of integers will be represented as a string. count() is an inbuilt function in Python that returns count of how many times a given object occurs in list. Among all sequence types, Lists are the most versatile. Hypothetically, why can't we wrap copper wires around car axles and turn them into electromagnets to help charge the batteries? The number of elements stored in the object is never calculated, so len helps provide the number … Count the number of times a value occurs using .values_count() Plot bar charts with .plot() By the end of this Python lesson, you'll be able to quickly count and compare records across a large dataset. ['bacon', 'chicken', 42] The look and say sequence, invented by mathematician John Conway and popularized by Robert Morris, is also known as "count and say sequence" or "say what you see sequence". Lists are represented/created with square brackets with each item separated using commas. You can use the len function to optimize the performance of the program. Note : The sequence of integers will be represented as a string. >>>string[:5] This topic provides a comprehensive understanding of sequences in Python. The rest of the numbers are obtained by the sum of the previous two numbers in the series. Count and Say. If you do this as part of an interview, it'll show you have good habits. >>>spam[0] 1 2. 11 is read off as "two 1s" or 21. In other words decorators decorate functions to make them fancier in some way. 11 is read off as "two 1s" or 21. The count-and-say sequence is the sequence of integers beginning as follows: 1, 11, 21, 1211, 111221,... 1 is read off as one 1 or 11. In this Python Count Digits in a Number, the user Entered value: Number = 9875 and Count = 0 Making statements based on opinion; back them up with references or personal experience. >>>"Hello, world! MathJax reference. I am solving interview questions from here. leetcode: Count and Say | LeetCode OJ; lintcode: (420) Count and Say; The count-and-say sequence is the sequence of integers beginning as follows: 1, 11, 21, 1211, 111221, ... 1 is read off as "one 1" or 11. These three are: –, Start Your Free Software Development Course, Web development, programming languages, Software testing & others. Python list contains. Find the n’th term in Look-and-say (Or Count and Say) Sequence. >>>string[-9:] • islower( ) Python language is very much in demand nowadays and having good foundational understanding can benefit students a lot in their future endeavors. This sequence has a unique and mysterious characteristic that is really difficult to understand and solve. Xrange objects are again like buffers. Then write a program that lets the user type in an integer and that keeps calling collatz() on that number until the function returns the value 1. 'Hello' The count-and-say sequence is the sequence of integers beginning as follows: 1, 11, 21, 1211, 111221, ... 1 is read off as "one 1" or 11. >>>spam[3] Note: Index in Python starts from 0, not 1. Next, Condition in the Python While Loop make sure that the given number is greater than 0 (Means Positive integer and greater than 0). ALL RIGHTS RESERVED. You may also look at the following article to learn more –, Python Training Program (36 Courses, 13+ Projects). For example, [1,22]*3 will evaluate to [1,22,1,22,1,22]. 'o' String Processing in Python: Look-and-Say Sequence - Duration: 11:34. It can also check if the item exists on the list or not using the list.count() function. For example, if you look at "22a", you count "two twos" and "one a" so the next sequence element is "221a", and then you repeat this process. However, it also has two optional parameters: substring - string whose count is to be found. • encode( [encoding[,errors]]) Buffers don’t support operations like concatenation or repetition. The count-and-say sequence is the sequence of integers beginning as follows: 1, 11, 21, 1211, 111221, ... 1 is read off as "one 1" or 11. >>"Hello, world! The count-and-say sequence is the sequence of integers beginning as follows: 1, 11, 21, 1211, 111221, ... 1 is read off as "one 1" or 11. Immutable objects are faster to access and are costlier to change because it needs the creation of a copy. It's generated by describing a series of digits as letters in plain English language. n’th … 11 3. In this Python Sequence Tutorial, we will discuss 6 types of Sequence: String, list, tuples, Byte sequences, byte array, and range object. >>>"Hello, world! This website or its third-party tools use cookies, which are necessary to its functioning and required to achieve the purposes illustrated in the cookie policy. You'll be able to look at web traffic data and compare traffic landing on various pages with statistics and visualizations. Count And Say: The count-and-say sequence is the sequence of integers beginning as follows: 1, 11, 21, 1211, 111221, ... 1 is read off as one 1 or 11. Unicode strings are similar to strings but are specified using a preceding “u” character in the syntax: u’abcd’, u”defg”. You could re-order your condition checks and use elif to save a level of indentation. The count-and-say sequence is the sequence of integers with the first five terms as following: 1; 11; 21; 1211; 111221; 1 is read off as "one 1" or 11. Given an integer n, generate the nth sequence. A list element can be any object. Given an integer n, generate the nth sequence. © 2020 - EDUCBA. Use of in, not in, min() or max() on Xrange is also inefficient. 111221 1 is read off as "one 1" or 11. Order Really Counts in a Python Program. Mutable Sequence Types . 21 is read off as "one 2, then one 1" or 1211. How to count non-DNA bases in a sequence using Python. 3. simple swift solution. How to accomplish? 11 is read off as "two 1s" or 21. In the case of count, we pass in a string which is the sequence of characters that we actually want to count. 21 is read off as "one 2, then one 1" or 1211. Thanks for contributing an answer to Code Review Stack Exchange! This time, we’ll work on the Count-and-Say problem from Leetcode. The count-and-say sequence, also known as the look-and-say sequence, is a sequence of integers that starts with 1. 11 is read off as "two 1s" or 21. 21 is read off as "one 2, then one 1" or 1211. The problem is that IDLE is saying that there is 'None' in the DNA_list. String count() Parameters. Its elements can be updated, removed, and also elements can be inserted into it. c++. Python Sequence Tutorial. Hi everybody, I have several RNAseq BAM files (mapped with BWA and GATK IndelRealigner) of cance... repeat-containing TUs in lincRNA . • find( sub[, start[, end]]) The count-and-say sequence is the sequence of integers beginning as follows: 1, 11, 21, 1211, 111221, ... 1 is read off as "one 1" or 11. Sequences are the essential building block of python programming and are used on a daily basis by python developers. Could Donald Trump have secretly pardoned himself? Given an integer n where 1 ≤ n ≤ 30, generate the n th term of the count-and-say sequence. If you are using Python 2.7, try using a counter: from collections import Counter counts = Counter(c for c in 'count vowels' if c in 'aeoiu') for k, v in counts.iteritems(): print k, v This results in the output: e 1 u 1 o 2 If you have an earlier version of Python, you can still use … >>>spam 21 is read off as "one 2, then one 1" or 1211. Basically, anytime you use something like for xxx in range(len(yyy)), there is a better way to do it. In Python, Sequences are the general term for ordered sets. 'l' Returns : count() method returns count of how many times obj occurs in list. Sudoku solver recursive solution with clear structure, Given an array, find all its elements that can become a leader. The way that we would do that is using the count method. 11 3. Python List index() Python List append() Python List extend() Python List insert() ... Python List count() The count() method returns the number of times the specified element appears in the list. 38. 11 is read off as "two 1s" or 21. It produces the output as follows. The Look and say sequence is a recursively defined sequence of numbers studied most notably by John Conway. These variables refer to the number of size of sample poll and the dimension of your problem. List; Dictionary; Set Mutable and immutable objects are treated differently in python. This is similar to starting from '1' and performing num-1 iterations. The count-and-say sequence is the sequence of integers beginning as follows: 1, 11, 21, 1211, 111221, ... 1 is read off as "one 1" or 11. In this case our function is very simple, but this is just an example to show you how we can extract part of our code into a function. Strings have a special notation. 1211 5. In python we can select sequence and subsequence in Python string through index of the string that is start from 0. They too, do not support operations such as slicing, concatenation or repetition. The look and say sequence, invented by mathematician John Conway and popularized by Robert Morris, is also known as "count and say sequence" or "say what you see sequence". 11 is read off as "two 1s" or 21. What does the name "Black Widow" mean in the MCU? Given an integer n, generate the nth sequence. Given an integer n, generate the n th sequence. Counting bases in a sequence Sorting DNA sequences by length 29 common beginner errors on one page ... On this site you'll find various resources for learning to program in Python for people with a background in biology. Note : The sequence of integers will be represented as a string. Tuples are also like lists, but there is one difference that they are immutable meaning they cannot be changed after defined. 21 is read off as one 2, then one 1 or 1211. It's very easy to do in python using the count function: ... , I want to find a pattern of sequence in a genome. 11 is read off as "two 1s" or 21. 21 is read off as one 2, then one 1 or 1211. The n-th term is constructed by reading the (n-1)-th term. This challenge requires us to write a function that gives the n th term of the count-and-say sequence in Python. By using decorators you can change a function's behavior or outcome without actually modifying it. If started with any digit d from 0 to 9 then d will remain indefinitely as the last digit of the sequence. Moreover, we will discuss Python sequence operations, functions, and methods. It is expected that students understand the foundations of sequences and must practice given examples on a python IDE or console. Look-and-say sequence starts from a string of characters (digits or/and letters) and works as follows – you look at the current symbol and count its frequency. The program will get the input from the user and print out the result.We will show you two different ways to calculate total digits in a number. To check if Python list contains a specific item, use an inbuilt in operator. • index( sub[, start[, end]]) It's generated by describing a series of digits as letters in plain English language. The count-and-say sequence is the sequence of integers beginning as follows: 1, 11, 21, 1211, 111221, ... 1 is read off as one 1 or 11. Implementation of LCS in Python. 'bacon' In your case, enumerate can help you to write things in a clearer way. In Python, the for statement is used to iterate through a sequence like a list, a tuple, a set, a dictionary, or a string. There are seven types of sequences in Python. 1. We will see there are good reasons to start from 0 in Python. Example: – a, b, c or (). So here, we want to do a count on a. In python we can select sequence and subsequence in Python string through index of the string that is start from 0. >>>len(spam) So if one wants to sample 3 dimensional space with 10 samples each it would be called as below. Moreover, we will discuss Python sequence operations, functions, and methods. You can use the len() to get the length of the given list, string, array, tuple, dictionary, etc. sylvainchan created at: a day ago | No replies yet. count() is an inbuilt function in Python that returns count of how many times a given object occurs in list. These lines (when represented in a logarithmic vertical scale) tend to straight lines whose slopes coincide with Conway's constant. How to plot the given graph (irregular tri-hexagonal) with Mathematica? • swapcase( ). piqcmax created at: 17 hours ago | No replies yet. Here we have discussed the different types of sequences in python and how they are used in python programming with some examples. Sequences in Python are indexed from zero, so the first element’s index is 0, the second’s index is 1, and so on. Check leap year. 1 (One) 11 (One 1) So read the previous 1, and say “One 1” 21 (Two 1) So read the previous 11, and say “Two 1” 1211 (One 2 one 1) So read the previous 21, and say “One 2 one 1” Given an integer n where 1 ≤ n ≤ … We then interchange the variables (update it) and continue on with the process. Missing I (1st) chord in the progression: an example. 1211 5. That is, we can access individual characters of a string using a numeric index. Syntax : list_name.count(object) Parameters : Object is the things whose count is to be returned. "[2] The n-th term is constructed by reading the (n-1)-th term. 'l' But don’t think of it as just a sequence. Note: The sequence of integers will be represented as a string. To use all of this simply call halton_sequence(size, dim). >>>"Hello, world! >>>string[:-8] So here, we want to do a count on a. x in NewSeq returns True if x is an element of NewSeq, otherwise False. >>>string[:] Given some string dna containing the letters A, C, G, or T, representing the bases that make up DNA, we ask the question: how many times does a certain base occur in the DNA string? Given an integer n, generate the _n_th sequence. Given an integer n, generate the nth sequence. >>>spam The way to call count is we say the name of the sequence whether it's a string, integer, or tuple that we want to count on. With more parameters, the range function can be used to generate a much wider variety of sequences. In this tutorial, we will learn how to count the total number of digits in a number using python. ('me', 'you', 'them', ‘Their’). Let's say to find following pattern (G4N(1... Getting protein fasta from RNAseq BAM file (human) paying attention on repeats . What is the Count-and-Say Sequence. >>>string = "Hello, world!" In this Python Sequence Tutorial, we will discuss 6 types of Sequence: String, list, tuples, Byte sequences, byte array, and range object. 0. They can be created using the xrange() function. If number is even, then collatz() should print number // 2 and return this value. 10, >>>var = "me", "you", "them", “Their” The look-and-say sequence is also known as the Morris Number Sequence, after cryptographer Robert Morris, and the puzzle What is the next number in the sequence 1, 11, 21, 1211, 111221? 21 is read off as "one 2, then one 1" or 1211. count() method only requires a single parameter for execution. That is to say; backlash signifies that the next character after it has a different meaning. 42 Reverse count of Index in the string. So, I say a, and then the dot or period, and then the method name that we want to call, dot count, open parentheses. • count(sub[, start[, end]]) AIO programming challenge - Friendlist - Who has the most friends? Solution w/full video whiteboard explanation. How do you bake out a world space/position normal maps? Example: if n = 2, the sequence is 11. 11:34. Also widely used by revision control systems such as Git. "[1] 'world' Asking for help, clarification, or responding to other answers. 'e' Using python, count the number of digits in a number. for statement in Python. There is no specific syntax for Xrange as well. After creating the list of DNA sequences/strings , for each sequence, I would like would like to count the 'A's using the count method and then I would just print to IDLE. Strings are little different than list and tuples, a string can only store characters. 'bacon' 11 is read off as "two 1s" or 21. You can also solve this problem using recursion: Python program to print the Fibonacci sequence … Software Engineering Internship: Knuckle down and do work or build my portfolio? Python starts counting from zero when it comes to numbering index locations of a sequence. 21 4. This blog is dedicated to a revision of the Python sequence and collections. >>>spam So, I say a, and then the dot or period, and then the method name that we want to call, dot count, open parentheses. One important property of sequences generated by range(n) is that the total number of elements is n: The sequence omits the number n itself, but includes 0 instead. "[0] However, an empty tuple must use an enclosing parenthesis. By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. >>>spam • endswith( suffix[, start[, end]]) Are KiCad's horizontal 2.54" pin header and 90 degree pin headers equivalent? it is also called concatenation. 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Lefty Sesame Street, Fort Riley Outdoor Recreation, International Gynecologic Oncology Fellowship, Powr2 Energy Solutions, White French Bulldog Blue Eyes, Morphle Tv Youtube, Giffgaff Goodybag Balance, Striaton City Stunfisk, Merchant Navy Association, | 2021-04-19 00:24:35 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.21124911308288574, "perplexity": 1795.6521849294963}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038862159.64/warc/CC-MAIN-20210418224306-20210419014306-00294.warc.gz"} |
http://math.stackexchange.com/questions/54534/finding-the-set-z-ez-1 | Finding the set {$z: e^z=-1$}
I want to find the set of $z$'s such that {$z: e^z=-1$}. Then this just mean that I have to solve $\cos(-iz)+i\sin(-iz)=-1$ which is equivalent to having $\cos(-iz)=-1$ and $\sin(-iz)=0$ Then I find that the set of solutions is such that $-iz=\pi + 2k\pi$ or in other words, $z=(1+2k)\pi i$
Should I also consider the posibility that $\sin(-iz)=i$ and $\cos(-iz)=0$ or is it irrelevant to take this possibility into account? I am not sure. Thx.
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Why do you think $\cos(-iz)+isin(-iz)=-1$ is equivalent to having $\cos(-iz)$ and $\sin(-iz)=0$? – Jack Jul 30 '11 at 2:27
$\cos(-iz)+i\sin(-iz)=-1$ is equivalent to saying that $\cos(-iz)=0$ and $\sin(-iz)=i$. Indeed if both conditions hold then we get: $\cos(-iz)+i\sin(-iz)= 0 + i\sin(-iz) = 0 + i^2=-1$ – user786 Jul 30 '11 at 2:37
Opps, I mean why do you say in the first paragraph that $\cos(-iz)+i\sin(-iz)=-1$ is equivalent to having $\cos(-iz)=-1$ and $\sin(-iz)=0$. You are trying to solve the equation "case by case". But things may become complicated in this way. For example, $\cos(-iz)=k$,$\sin(-iz)=(k+1)i$ where $k$ is a real number can also satisfy the equation. – Jack Jul 30 '11 at 2:48
@Jack: yes thank you that's what I was asking about. – user786 Jul 30 '11 at 3:19
Since $e^{i\pi}=-1$, we can rewrite the equation $e^z=-1$ as $e^z=e^{i\pi}$, or equivalently $e^{z-i\pi}=1$.
The solutions of $e^w=1$ are $w=i(2n\pi)$, where $n$ ranges over the integers. Thus the solutions of $e^{z-i\pi}=1$ are $\:i\pi(2n+1)$, where $n$ ranges over the integers.
About your Calculation: Although that calculation happens to give the right answer, the logic is not right. In the calculation, $\sin$ and $\cos$ are functions of a complex variable, and take on complex values.
You are treating the complex $\cos$ and $\sin$ functions as if they had the same formal properties as the corresponding real functions. For example, from $\cos(−iz)+i\sin(−iz)=−1$, you conclude that one of the summands is $0$ and the other is $−1$. But (until one proves otherwise) the imaginary parts of $\cos(−iz)$ and $i\sin(−iz)$ could each be non-zero, but cancel. And as you point out, there is the possibility of considering $\sin(-iz)=i$, $\cos(-iz)=0$. Unfortunately, these are by no means the only possibilities to consider.
Detailed analysis of the complex sine and cosine may enable you to push an argument through. But it is certainly not immediate.
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Thank you but I already know what the solutions are: I wrote them down in my question. I am asking about the possibility of having $\sin(-iz)=i$ and $\cos(-iz)=0$ and if it reduces to the first case. – user786 Jul 30 '11 at 2:42
Does this prove that those are the ONLY solutions? – Michael Hardy Jul 30 '11 at 3:00
@user786: Although your calculation gives the right answer, the logic is not right. You are treating the complex $\cos$ and $\sin$ functions as if they had the same formal properties as the corresponding real functions. For example, from $\cos(-iz)+i\sin(-iz)=-1$ you cannot conclude that one term is $0$ and the other is $-1$. The imaginary parts of $\cos(-iz)$ and $i\sin(-iz)$ could be non-zero, but cancel. – André Nicolas Jul 30 '11 at 3:12
@André: thank you. That's the answer I needed. – user786 Jul 30 '11 at 3:18
@Michael Hardy: Yes, if we take the solutions of $e^w=1$ as a "standard fact." – André Nicolas Jul 30 '11 at 3:35
Supposing $z = x + iy$, we have $e^z = e^{x+iy} = e^x e^{iy} = e^x(\cos y + i\sin(y))$.
(That much is true even if $x$ and $y$ are not real!)
Now suppose $x$ and $y$ are real. Then $e^x$ is positive and $\cos y + i\sin y$ is on the unit circle centered at $0$, so that the absolute value $|\cos y + i\sin y|$ is $1$. So the absolute value of $e^x(\cos y + i\sin(y))$ is $e^x$. Since $|-1|=1$, we need $e^x =1$. Since $x$ is real, this means $x=\ln1 = 0$. So we want $\cos y + i\sin y = -1$. Therefore we must have $\cos y = -1$ and $\sin y = 0$. That $\sin y=0$ means $y\in\{0, \pm\pi, \pm 2\pi, \pm3\pi,\dots\}$. But at some of those points the cosine is $+1$ rather than $-1$. The points where $\cos y = -1$ are $\pm\pi, \pm3\pi,\pm5\pi,\dots$.
Bottom line: $z\in\{0, \pm i\pi, \pm3i\pi, \pm5i\pi,\dots\}$.
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Yes indeed this is exactly what I've found, but I look at my equation above, i could very well also have the possibility that $\sin(-iz)=i$. What do I do with this possibility? – user786 Jul 30 '11 at 2:20
In bottom line, you shouldn't include 0. – sdcvvc Jul 30 '11 at 2:28
@user786: The possibility that $\sin(-iz) = i$ is not the only possibility. Those possibilities do arise and bother you in a way that the path you're choosing is most probably not the right one to take to prove such a thing. You have plenty of suggestions of how to take another path to solve your problem, so I guess you should just take a look at them. I'm not saying it's impossible to go on with your idea, but it most probably is. Michael Hardy's idea is the closest one to yours though. – Patrick Da Silva Jul 30 '11 at 3:29
Write $z = x + iy$, with $x$ and $y$ real. Then $e^z = e^{x + iy} = e^x e^{iy}$. Note this writes $e^z$ in polar form, with $r = e^x$ and $\theta = y$.
Next, note that in polar form $-1 = 1*e^{i\pi}$. The polar form of any nonzero complex number is unique, except that any multiple of $2\pi$ may be added to the argument. So the statement that $e^z = -1$ is equivalent to the statement that $e^x = 1$ and $y = \pi + 2k\pi$ for some integer $k$. Equivalently, $x = 0$ and $y$ is an odd multiple of $\pi$.
So the set of all such $z$ are exactly $\{\pi k i: k$ is an odd integer $\}$.
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When you say "finding the set ${z:e^z=-1}$", I think you are talking about "finding the complex roots of $e^z=-1$. Since $\{z\in{\mathbb R}:e^z=-1\}$ and $\{z\in{\mathbb C}:e^z=-1\}$ are totally different. On the other hand, the set is already, so "finding the set" may be somewhat confusing.
For solving $e^z=-1$, you would like to look at complex logarithm.
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https://www.aimsciences.org/article/doi/10.3934/dcdsb.2014.19.281 | # American Institute of Mathematical Sciences
January 2014, 19(1): 281-298. doi: 10.3934/dcdsb.2014.19.281
## Dirichlet series for dynamical systems of first-order ordinary differential equations
1 School of Mathematics & Physics, Qingdao University of Science & Technology, Qingdao 266061, China 2 Department of Applied Mathematics and Theoretical Physics, Centre for Mathematical Sciences, University of Cambridge, Wilberforce Rd, Cambridge CB3 0WA, United Kingdom
Received August 2012 Revised October 2013 Published December 2013
In this paper, inspired by the work by A. Iserles and G. Söderlind [Global bounds on numerical error for ordinary differential equations, J. Complexity, 9 (1993), pp. 97-112], we present comprehensive discussion on Dirichlet series for dynamical systems of first-order ordinary differential equations (ODEs). We first derive the scheme of Dirichlet approximation for scalar dynamical systems and present the bounds on the terms of Dirichlet series. The global error and the right choice of a term in Dirichlet series are analysed and two numerical experiments are carried out to demonstrate the efficiency of Dirichlet approximation. Then we consider applying Dirichlet series to multivariate dynamical systems and present a new scheme of Dirichlet approximation for such systems. Some discussion and a numerical experiment are accordingly carried out for the new Dirichlet approximation. Compared with routine time-stepping algorithms, Dirichlet series does not need time stepping and yields a continuous solution that is equally valid along an interval, which is significant for obtaining long-time numerical solution. As a result of the special nature of Dirichlet series, the Dirichlet approximation delivers considerable information on dynamical systems of first-order ODEs and provides a novel and effective approach to numerical solutions of these dynamical systems.
Citation: Bin Wang, Arieh Iserles. Dirichlet series for dynamical systems of first-order ordinary differential equations. Discrete & Continuous Dynamical Systems - B, 2014, 19 (1) : 281-298. doi: 10.3934/dcdsb.2014.19.281
##### References:
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##### References:
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2019 Impact Factor: 1.27 | 2020-11-27 15:19:47 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5414366126060486, "perplexity": 4384.098436188648}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141193221.49/warc/CC-MAIN-20201127131802-20201127161802-00297.warc.gz"} |
https://www.thejournal.club/c/paper/185924/ | #### Risk-Limiting Bayesian Polling Audits for Two Candidate Elections
##### Poorvi L. Vora
We propose a simple common framework for Risk-Limiting and Bayesian (polling) audits for two-candidate plurality elections. Using it, we derive an expression for the general Bayesian audit; in particular, we do not restrict the prior to a beta distribution. We observe that the decision rule for the Bayesian audit is a simple comparison test, which enables the use of pre-computation---without simulations---and greatly increases the computational efficiency of the audit. Our main contribution is a general form for an audit that is both Bayesian and risk-limiting: the {\em Bayesian Risk-Limiting Audit}, which enables the use of a Bayesian approach to explore more efficient Risk-Limiting Audits.
arrow_drop_up | 2021-05-15 20:48:36 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8250753879547119, "perplexity": 1846.4408346359426}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243991378.52/warc/CC-MAIN-20210515192444-20210515222444-00347.warc.gz"} |
http://math.stackexchange.com/questions/260744/let-f0-1-%e2%86%92-mathbbr-be-continuous-such-that-ft-%e2%89%a50-for-all-t-in-0 | # Let $f:[0,1] →\mathbb{R}$ be continuous such that $f(t) ≥0$ for all $t$ in $[0, 1]$. Define $g(x) = ∫_0^xf(t)dt$ then which is true? [duplicate]
Let $f:[0,1] \to\mathbb{R}$ be continuous such that $f(t) ≥0$ for all t in $[0, 1]$. Define $g(x) = \int_0^xf(t) \, dt$ then which is true?
1 $g$ is monotone and bounded
2 $g$ is monotone, but not bounded
3 $g$ is bounded, but not monotone
4 $g$ is neither monotone nor bounded
I think either 1 or 2 is true as I get it is monotone but not sure about boundedness
-
## marked as duplicate by Cameron Buie, David Mitra, Davide Giraudo, Martin Sleziak, QiL Dec 17 '12 at 16:38
Hint: $f$ is bounded on $[0,1]$. – David Mitra Dec 17 '12 at 15:12
Since $f$ is continuous and defined in a closed interval, $f$ is bounded by say $m,M$ below and above, respectively. Then $$mx= \int_0^x m \, dt\le \int_0^x f(t) \, dt\le \int_0^x M \, dt=Mx$$ What does this say about $g$ in $[0,1]$?
An alternate approach without assuming continuity of $f$ (and thus differentiability of $g$) is the following:
$g$ is Lipschitz continuous and defined in a closed interval, thus it is bounded. In addition since $f(t)\ge 0$ for $t\in [0,1]$, if $0\le x_1<x_2\le 1$, $$g(x_2)-g(x_1)=\int_{0}^{x_2}f(t)\, dt-\int_{0}^{x_1}f(t)\, dt=\int_{x_1}^{x_2}f(t)\, dt\ge 0$$ | 2016-02-10 22:58:46 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9889306426048279, "perplexity": 142.97479634572977}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-07/segments/1454701160822.87/warc/CC-MAIN-20160205193920-00328-ip-10-236-182-209.ec2.internal.warc.gz"} |
https://tex.stackexchange.com/questions/456585/problem-with-equation-numbering-location | Problem with equation numbering location
I have two very long equation but the problem is the number of the equation locating is wrong my code is, I now that the margin is 6 cm from the left but still it has to work my code looks like:
\documentclass [12pt]{article}
\usepackage{float}
%\usepackage{showframe}
\usepackage{authblk}
\usepackage{etoolbox}
\usepackage{amsmath}
\usepackage{graphicx}
\usepackage{subcaption}
\usepackage{tabularx,ragged2e,booktabs,caption}
\usepackage[utf8]{inputenc}
\usepackage[english]{babel}
\usepackage{caption}
\usepackage{amssymb}
\usepackage{graphicx,epstopdf}
\usepackage{caption}
\usepackage{graphicx,kantlipsum,setspace}
\usepackage{mathptmx}
\usepackage{pgfplots}
\captionsetup{font={stretch=1.0}} %% this affects both figure and table
\setstretch{1.0}
\captionsetup{font=footnotesize}
\usepackage[left=6cm,top=3cm,right=2cm,bottom=3cm]{geometry}
\usepackage{lipsum}
\usepackage{setspace}
\usepackage{apacite}
\setstretch{1.5}
\newcolumntype{C}[1]{>{\Centering}m{#1}}
\renewcommand\tabularxcolumn[1]{C{#1}}
\newcommand\seqwlimits[3]{\{#1\}_{\mathstrut#2}^{\mathstrut#3}}
\renewcommand\Authfont{\fontsize{15}{1}\selectfont}
\begin{document}
$$\begin{split} Q_{G-Ind}(\theta= H\phi)& =-2log\bigg(\frac{L_T(\tilde{\theta})}{L_T(\hat{\theta})}\bigg)\\&= Q_{G-CC}(\theta= H p) - Q_{G-UC} (H\phi=H p)\\ & = \biggl(-2\biggl[L_T(\theta_0)-L_ T(\widehat{\theta})\biggr]\biggr) - \biggl(-2\biggl[L_T(\theta_0)-L_T(\tilde{\theta})\biggr]\biggr) \end{split}$$
and the other one is
$$M_{j+1} (d;p)= \frac{(1-p)(2j+1)+p(j-d+1)}{(j+1)\sqrt{1-p}} M_j(d;p)-\left(\frac{j}{j+1}\right) M_{j-1}(d;p),$$
\end{document}
• @phollox I missed the intention of the 'and the other ...' part. – albert Oct 24 '18 at 16:46
• Place of the number looks OK to me. The '(1)' is in the middle of the formula and the '(2)' doesn't fit anymore so it is placed on the next line. – albert Oct 24 '18 at 16:47
• the (1) is placed in the next line of all the equation lines – Hani Al Natour Oct 24 '18 at 16:48
• Can you share a picture of, the page with, the formula? Which engine are you using and which version of the distribution? – albert Oct 24 '18 at 17:00
Two solutions: either you reduce the left margin, if you can, or you remove the parentheses around the fraction j/j+1. I took the opportunity to remove multiply loaded packages, and incompatible packages, such as amssymb and mathptmx. I also suggest replacing the latter with the more complete newtx, based on the Times clone TeX Gyre Termes. Finally I set some delimiters to a smaller size (more adequate in my opinion):
\documentclass [12pt]{article}
\usepackage[utf8]{inputenc}
\usepackage[english]{babel}
\usepackage[left=6cm, right=2cm, vmargin=3cm, showframe]{geometry}
\usepackage{,setspace}
\usepackage{newtxtext, newtxmath}
\usepackage{amsmath}
\usepackage{float} %\usepackage{showframe}
\usepackage{etoolbox}
\usepackage{subcaption}
\usepackage{tabularx,ragged2e,booktabs,caption}
\usepackage{graphicx,epstopdf}
\usepackage{pgfplots}
\captionsetup{font={stretch=1.0}} %% this affects both figure and table \setstretch{1.0} \captionsetup{font=footnotesize}
\usepackage{apacite} \setstretch{1.5}
\providecommand\Authfont{\fontsize{15}{1}\selectfont}
\begin{document}
$$\begin{split} Q_{G-Ind}(\theta= H\phi) &=-2\log\biggl(\frac{L_T(\tilde{\theta})}{L_T(\hat{\theta})}\biggr) \\ &= Q_{G-CC}(\theta= H p) - Q_{G-UC} (H\phi=H p)\\ & = \Bigl(-2\Bigl[L_T(\theta_0)-L_ T(\widehat{\theta})\Bigr]\Bigr) - \Bigl(-2\Bigl[L_T(\theta_0)-L_T(\tilde{\theta})\Bigr]\Bigr) \end{split}$$
and the other one is
$$M_{j+1} (d;p)= \frac{(1-p)(2j+1)+p(j-d+1)}{(j+1)\sqrt{1-p}} M_j(d;p)-\frac{j}{j+1}M_{j-1}(d;p),$$
\end{document} | 2021-04-13 10:47:53 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 4, "x-ck12": 0, "texerror": 0, "math_score": 0.861304759979248, "perplexity": 2053.0938567379562}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038072180.33/warc/CC-MAIN-20210413092418-20210413122418-00633.warc.gz"} |
https://www.semanticscholar.org/paper/On-graphs-with-small-Ramsey-numbers-*-Kostochka-R%C3%B6dl/d12ccce90684169df7d7236387768d4d1101edbd | # On graphs with small Ramsey numbers *
@article{Kostochka2001OnGW,
title={On graphs with small Ramsey numbers *},
author={Alexandr V. Kostochka and Vojtech R{\"o}dl},
journal={Journal of Graph Theory},
year={2001},
volume={37}
}
• Published 1 August 2001
• Mathematics
• Journal of Graph Theory
Let R(G) denote the minimum integer N such that for every bicoloring of the edges of KN, at least one of the monochromatic subgraphs contains G as a subgraph. We show that for every positive integer d and each γ,0 < γ < 1, there exists k = k(d,γ) such that for every bipartite graph G = (W,U;E) with the maximum degree of vertices in W at most d and $|U|\leq |W|^{\gamma }$, $R(G)\leq k|W|$. This answers a question of Trotter. We give also a weaker bound on the Ramsey numbers of graphs whose set…
51 Citations
On Ramsey Numbers of Sparse Graphs
• Mathematics
Combinatorics, Probability and Computing
• 2003
It is shown that, for every , sufficiently large n, and any graph H of order , either H or its complement contains a (d,n)-common graph, that is, a graph in which every set of d vertices has at least n common neighbours.
On graphs with linear Ramsey numbers
• Mathematics
J. Graph Theory
• 2000
In this paper, the use of the regularity lemma is avoided altogether, and it is shown that one can in fact take, for some ®xed c, c
< 2 (log )2 in the general case, and even even 1.
Turán Numbers of Bipartite Graphs and Related Ramsey-Type Questions
• Mathematics
Combinatorics, Probability and Computing
• 2003
It is proved that, for any fixed bipartite graph H in which all degrees in one colour class are at most r, the Turán number is the maximum possible number of edges in a simple graph on n vertices that contains no copy of H.
Two remarks on the Burr-Erdos conjecture
• Mathematics
Eur. J. Comb.
• 2009
Unavoidable patterns
• Mathematics
J. Comb. Theory, Ser. A
• 2008
Large Kr‐free subgraphs in Ks‐free graphs and some other Ramsey‐type problems
• B. Sudakov
• Mathematics
Random Struct. Algorithms
• 2005
Three Ramsey-type results are presented, which derive from a simple and yet powerful lemma, proved using probabilistic arguments and solve some special cases of a closely related question posed by Erdős.
Chromatic number, clique subdivisions, and the conjectures of Hajós and Erdős-Fajtlowicz
• Mathematics
Comb.
• 2013
The main ingredient in the proof, which might be of independent interest, is an estimate on the order of the largest clique subdivision which one can find in every graph on n vertices with independence number α.
Ramsey numbers of sparse hypergraphs
• Mathematics, Computer Science
Random Struct. Algorithms
• 2009
This work significantly improves on the Ackermann‐type upper bound that arises from the regularity proofs, and presents a construction which shows that, at least in certain cases, this bound is not far from best possible.
Topics in flnite graph Ramsey theory
For a positive integer r and graphs F , G, and H, the graph Ramsey arrow notation F −→ (G)r means that for every r-colouring of the subgraphs of F isomorphic to H, there exists a subgraph G′ of F
A few remarks on Ramsey-Tura'n-type problems
## References
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ON THE MAGNITUDE OF GENERALIZED RAMSEY NUMBERS FOR GRAPHS
If G and H are graphs (which will mean finite, with no loops or parallel lines), define the Ramsey number r(G, H) to be the least number p such that if the lines of the complete graph Kp are colored
Graphs with Linearly Bounded Ramsey Numbers
• Mathematics
J. Comb. Theory, Ser. B
• 1993
It is proved that for each p ≥ 1, there is a constant c (depending only on p) such that the Ramsey number r ( G, G ) ≤ cn for eachp -arrangeable graph G of order n.
Subdivided graphs have linear ramsey numbers
• N. Alon
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J. Graph Theory
• 1994
It is shown that the Ramsey number of any graph with n vertices in which no two vertices of degree at least 3 are adjacent is at most 12n, which settles the problem of Burr and Erdos.
Arrangeability and clique subdivisionsThe Mathematics of Paul Erdo }s
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Erdo}s, On the magnitude of generalized Ramsey numbers for graphs, in: `̀ In®nite and ®nite sets'
• Colloquia Mathematica Soc. Janos Bolyai,
• 1975 | 2022-08-16 19:50:44 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7597805857658386, "perplexity": 1504.565978274279}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882572515.15/warc/CC-MAIN-20220816181215-20220816211215-00437.warc.gz"} |
https://en.m.wikibooks.org/wiki/Ordinary_Differential_Equations/Global_uniqueness_of_solution_over_interval | # Ordinary Differential Equations/Global uniqueness of solution over interval
If there is local uniqueness to a solution to an IVP (such as implied Picard–Lindelöf theorem), and if we restrict ourselves to solutions over intervals, then there is global uniqueness of solutions.
## Theorem. Uniqueness over intervalsEdit
Theorem If solutions over intervals coincide at a single point then they are the same
Hypothesis
1. ${\displaystyle y_{1}}$ and ${\displaystyle y_{2}}$ are solutions to an IVP
2. ${\displaystyle y_{1}}$ and ${\displaystyle y_{2}}$ are locally unique solutions (by the Picard–Lindelöf theorem for example)
3. the domains of ${\displaystyle y_{1}}$ and ${\displaystyle y_{2}}$ are both intervals (which contain ${\displaystyle x_{0}}$ , otherwise the initial condition makes no sense)
Conclusion
1. ${\displaystyle y_{1}}$ and ${\displaystyle y_{2}}$ coincide inside their common domain of definition: ${\displaystyle \forall x\in Dom(y_{1})\cap Dom(y_{2})\,y_{1}(x)=y_{2}(x)}$
2. If ${\displaystyle Dom(y_{1})=Dom(y_{2})}$ , ${\displaystyle y_{1}=y_{2}}$
3. ${\displaystyle y={\begin{cases}y_{1}&x\in Dom(y_{1})\\y_{2}&x\in Dom(y_{2})\end{cases}}}$
is also a solution, with domain ${\displaystyle Dom(y_{1})\cup Dom(y_{2})}$ . This notation is unambiguous because of the above hypothesis.
### Example. y'=y in various interval domainsEdit
Example
Take the IVP ${\displaystyle y'=y,\,y(0)=1}$ . Therefore ${\displaystyle F(x,y)=y}$ .
{\displaystyle {\begin{aligned}y_{1}:\,&]-2,2[\\&x\mapsto e^{x}\end{aligned}}}
{\displaystyle {\begin{aligned}y_{2}:\,&]-1,3[\\&x\mapsto e^{x}\end{aligned}}}
{\displaystyle {\begin{aligned}y_{3}:\,&]-2,3[\\&x\mapsto e^{x}\end{aligned}}}
${\displaystyle y_{1}}$ ${\displaystyle y_{2}}$ ${\displaystyle y_{3}}$
Then ${\displaystyle y_{1}}$ and ${\displaystyle y_{2}}$ satisfy all the hypothesis of the theorem
1. both are solutions to the IVP
2. both are locally unique because ${\displaystyle F(x,y)}$ satisfies the Picard–Lindelöf theorem in all of its domain ( ${\displaystyle F\in \mathbb {C} ^{1}}$ and therefore is also locally Lipschitz)
3. ${\displaystyle Dom(y_{1})}$ and ${\displaystyle Dom(y_{2})}$ are both intervals ${\displaystyle ]-2,2[}$ and ${\displaystyle ]-1,3[}$ respectively.
Then we observe all of our conclusions:
1. Inside ${\displaystyle Dom(y_{1})\cap Dom(y_{2})=]-1,2[}$ , ${\displaystyle y_{1}=y_{2}}$
2. If we fix the intervals ${\displaystyle ]-2,2[}$ and ${\displaystyle ]-1,3[}$ , then ${\displaystyle y_{1}}$ and ${\displaystyle y_{2}}$ are the only solutions with exactly those domains
3. ${\displaystyle y={\begin{cases}y_{1}&x\in Dom(y_{1})\\y_{2}&x\in Dom(y_{2})\end{cases}}=y_{3}}$
is also a solution to the IVP with domain ${\displaystyle ]-2,3[}$ .
### Remark. Different domains, different functionsEdit
Remark Different domains mean completely different functions.
Remember from set theory that a function simply a set of ordered pairs. For example
${\displaystyle f_{1}=\{(0,1)\}}$
${\displaystyle f_{2}=\{(0,1),(1,2)\}}$
are two functions so that ${\displaystyle Dom(f_{1})={0}}$ and ${\displaystyle f(0)=1}$ , and ${\displaystyle Dom(f_{2})={0,1}}$ , ${\displaystyle f(0)=1}$ and ${\displaystyle f(1)=2}$ . Note that they coincide in the intersection of their domains: ${\displaystyle f_{1}(0)=f_{2}(0)=1}$
However they are not equal. Remember that two sets are equal iff the have exactly the same elements, which is obviously not the case for ${\displaystyle f_{1}}$ and ${\displaystyle f_{2}}$ since ${\displaystyle (1,2)\in f_{2}}$ but ${\displaystyle (1,2)\notin f_{1}}$ . Therefore ${\displaystyle f_{1}}$ and ${\displaystyle f_{2}}$ are two completely different sets, and therefore two completely different solutions.
The same goes for two functions such as
{\displaystyle {\begin{aligned}y_{1}:\,&]-2,2[\,\to \,]e^{-2},e^{2}[\\&x\mapsto e^{x}\end{aligned}}}
{\displaystyle {\begin{aligned}y_{2}:\,&]-1,3[\,\to \,]e^{-1},e^{3}[\\&x\mapsto e^{x}\end{aligned}}}
Many times the domain of a function is implicit, and we forget about it, usually taking the largest possible. But sometimes taking the largest possible domain may not be appropriate. For example when solving differential equations, taking a domain that is too large (and not an interval) may not lead to uniqueness, which is undesirable. In those cases it is necessary to specify very well what domain we are talking about.
### Counter-example. Not an interval.Edit
Counter example
Take the IVP ${\displaystyle y'=y,\,y(0)=1}$ . Look at the infinite family solutions
{\displaystyle {\begin{aligned}y_{a}:&Dom(y_{a})=]-1,1[\,t\cup \,]2,4[\\&x\mapsto {\begin{cases}e^{x}&x\in ]-1,1[\\ae^{(x-3)}&]2,4[\end{cases}}\end{aligned}}}
${\displaystyle y_{a}}$ for three values of a
which are each determined by any value of a ( =(y(3) ).
Those solutions satisfy all the conditions of the theorem, except that their common domain ${\displaystyle ]-1,1[\,\cup \,]2,4[}$ is not an interval. Then we observe that all the conclusions fail for ${\displaystyle a\neq a'}$
1. Inside ${\displaystyle ]2,4[\subset Dom(y_{1})\cap Dom(y_{2})}$ , ${\displaystyle y_{a}\neq y_{a'}}$
2. Both have the same domain, but ${\displaystyle Dom(y_{a})=Dom(y_{a'})=]-1,1[\,\cup \,]2,4[}$ , but ${\displaystyle y_{a}\neq y_{a'}}$
All of this happens because the uniqueness of the initial condition cannot propagate from ${\displaystyle x_{0}=0}$ to the other side of the domain ${\displaystyle ]2,4[}$ . | 2017-04-29 19:31:21 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 65, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9483455419540405, "perplexity": 1194.5689346516037}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-17/segments/1492917123560.51/warc/CC-MAIN-20170423031203-00123-ip-10-145-167-34.ec2.internal.warc.gz"} |
https://localife.am/kh85rcr/931d39-angle-symbol-0 | # angle symbol 0
Choose a proper , The profile is twisted and has multiple regions with different mounting angles. Degree Symbol on Mac OS On Windows: Open Start ('Start' icon). 461.0' 461.0' Test boring New contours elevation noted on high side Existing contours elevation noted on high side TB-1 268 320 C A-9 7 A-11 Detail Reference drawing number Room/space number Equipment number 7 A-12 1302 Awesome! For example, the following code with angle=-30 in the JSON will create a symbol rotated -30 degrees counter-clockwise; that is, 30 degrees clockwise, which symbol.angle=30 would also produce. You can also learn how to insert these angle symbols in word and how to insert them in your phone. 0 0. tennisfreak. The following table documents some of the most common functions in this category — along with their respective usage and example. Still have questions? This table explains the meaning of every math symbol. Math symbol is a copy and paste text symbol that can be used in any desktop, web, or mobile applications. Press and hold the ALT key Angle (Symbol/sign/mark) Preview and HTML-code. Those symbols … Press and hold the ALT key and type 0 1 7 6 on the numeric keypad of your keyboard. Straight Angle A straight angle is an angle with measure equal to 180 degrees. Subdivisions For many practical purposes, a degree is a small enough angle that whole degrees provide sufficient precision. There are some specialized usages of angle brackets in complex mathematical fields like quantum mechanics as well … but we’ll let the experts explain those usages. The middle point corresponds to the vertex at which the angle lies. 1 0. Just click on the symbol to get more information such as math For readability purpose, these symbols are categorized by their function into tables. The following table documents some of the most notable symbols in these categories — along with each symbol’s respective meaning and usage. Make sure the NumLock is on and type 0176 with the leading zero. The size of an angle could be stated this way: 40 degrees, 20 minutes, 50 seconds. There is no related acute angle if the terminal arm lies in quadrant 1. Rachel Bobrow. Relevance. This page is about the meaning, origin and characteristic of the symbol, emblem, seal, sign, logo or flag: Angle. DSANG = 0.5: Angle/degree of the divergence slit at the minimum 2-theta. An Obtuse Angle The following list documents some of the most notable symbols in these topics, along with each symbol’s usage and meaning. If there is no numeric keypad, press and hold the Fn before typing the 0176 numbers of degree symbol. A minute of arc, arcminute (arcmin), arc minute, or minute arc, denoted by the symbol ′, is a unit of angular measurement equal to 1 / 60 of one degree. But it seems to be used in the occasion where an angle is made by three points or just to use an alphabet to represent an angle without indicating form of the angle. Get the master summary of mathematical symbols in eBook form — along with each symbol’s usage and LaTeX code. If circles $O_1$ and $O_2$ share the same radius, then they are congruent. This thread is locked. Note the mating part gets welded into the angle. Question Find the angle, in degrees between 0 and 360°, which intersects the unit circle at the point (1) Do not include the degree symbol in your answer. If $P$, $Q$, $R$ lie on a sphere, then $\sphericalangle PQR$ is the spherical angle between $\overparen{PQ}$ and $\overparen{QR}$. On our website you will find all the today’s answers to New York Times Crossword. A reference angle is the acute version of any angle determined by repeatedly subtracting or adding straight angle (1 / 2 turn, 180°, or π radians), to the results as necessary, until the magnitude of the result is an acute angle, a value between 0 and 1 / 4 turn, 90°, or π / 2 radians. $\overleftrightarrow{AB}=\overleftrightarrow{BA}$, $\overrightarrow{AB} \ne \overrightarrow{BA}$. Rate this symbol: (5.00 / 3 votes) Indicates an angle. Posted on September 29, 2019 by . The model has a wingspan of 0.69m at its trailing edge, is 0.02m thick, and is beveled on the windward side at an angle of 15 to form sharp leading-edges. 金属L型アングルの選定・通販ページ。ミスミ他、国内外3,324メーカー、2,070万点以上の商品を1個から送料無料で配送。豊富なCADデータ提供。金属L型アングルを始め、FA・金型部品、工具・工場消耗品の通販ならMISUMI-VONA。 Just drop in your email and we'll send over the 26-page free eBook your way! The symbol looks like a skewed, uppercase, sans serif letter L ( ). In this article Insert the degree Categories: Directional Icons After you get up and running , you can place Font Awesome icons just about anywhere with the tag: Example of angle-down fa-angle-down For an $n$-gon, the sum of interior angles equals $(n – 2) \cdot 180^{\circ}$. Since one degree is 1 / 360 of a turn (or complete rotation), one minute of arc is 1 / 21 600 of a turn. Favorite Answer. The following table documents some of the most notable symbols related to these — along with each symbol’s meaning and example. That means Click the Windows logo in the bottom-left corner of the … Using Unicode Hex Input. It can reference a 2D line referenced to another 2D element, but more commonly it relates the orientation of one surface plane relative to another datum plane in a 3-Dimensional tolerance zone. With this tool, you can adjust the size, color, italic, and bold of Angle(symbol). 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http://www.edmerls.com/index.php/Analytical%20Chemistry/Solvent%20Extraction/5/Explain%20the%20separation%20factor? | By Sunil Bhardwaj
2861 Views
If a solution contains two or more solutes say A and B, it is observed that when A is extracted, some amount of B is also extracted. The extent of the seperation can be expressed in terms of one factor called Seperation Factor $$\beta$$ . This is related to the distribution ratio of A and B. $$\boxed { \beta = \frac { D_{ A } }{ { D }_{ B } } = \frac { { { C }_{ o(A) } }/{ { C }_{ a(A) } } }{ { { C }_{ o(B) } }/{ { C }_{ a(B) } } } }$$ It is ratio therefore no unit and no dimension.
The larger value is always placed in the numerator. \beta must be made as large as possible by choice of extractant and by adjusting the volume ratio.
When $$D_{ A }$$ = 10 and $$D_{ B }$$= 0.1. The $$\beta = \frac { 10 }{ 0.1 } = 100%$$
single extraction in case will remove 91% of A and 9% of B. It can be obtained for A, $$E = \left[ \frac { 100{ D }_{ A } }{ { D }_{ A } + { { V }_{ W } }/{ { V }_{ o } } } \right]$$ for $${ V }_{ W } = { V }_{ o }\qquad \qquad \therefore { { V }_{ W } }/{ { V }_{ o } } = 1$$ $$\therefore E = \left[ \frac { 100 \times 10 }{ 10 + 1 } \right] = \left[ \frac { 1000 }{ 11 } \right] = 90.9%$$ $$Similarly \ for \ B, \ E = \left[ \frac { 100{ D }_{ B } }{ { D }_{ B } + { { V }_{ W } }/{ { V }_{ o } } } \right]$$ $$for \ { V }_{ W } = { V }_{ o }\qquad \qquad \therefore { { V }_{ W } }/{ { V }_{ o } } = 1$$ $$\therefore E = \left[ \frac { 100 \times 0.1 }{ 0.1 + 1 } \right] = \left[ \frac { 10 }{ 1.1 } \right] = 9.1%$$ The seperation of A is almost complete from B if the seperation factor B is high. It can be only in the case when $${ D }_{ A }$$ is large and $${ D }_{ B }$$ is small. For a given value of $${ D }_{ A }$$ and $${ D }_{ B }$$, the seperation effect can be increased by adjusting the volume ratio given by Nush Densen Equation which says. $$\boxed { \frac { { V }_{ o } }{ { V }_{ W } } = \frac { 1 }{ { \left( { D }_{ A }{ D }_{ B } \right) }^{ { 1 }/{ 2 } } } }$$
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• Patente B Italia | 2022-08-10 14:41:00 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.46593037247657776, "perplexity": 1876.036424380474}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882571190.0/warc/CC-MAIN-20220810131127-20220810161127-00052.warc.gz"} |
https://stacks.math.columbia.edu/tag/01NA | Lemma 27.12.3. Let $S$ be a graded ring generated as an $S_0$-algebra by the elements of $S_1$. In this case the scheme $X = \text{Proj}(S)$ represents the functor which associates to a scheme $Y$ the set of pairs $(\mathcal{L}, \psi )$, where
1. $\mathcal{L}$ is an invertible $\mathcal{O}_ Y$-module, and
2. $\psi : S \to \Gamma _*(Y, \mathcal{L})$ is a graded ring homomorphism such that $\mathcal{L}$ is generated by the global sections $\psi (f)$, with $f \in S_1$
up to strict equivalence as above.
Proof. Under the assumptions of the lemma we have $X = U_1$ and the lemma is a reformulation of Lemma 27.12.2 above. $\square$
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In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 01NA. Beware of the difference between the letter 'O' and the digit '0'. | 2021-10-20 09:28:37 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 2, "x-ck12": 0, "texerror": 0, "math_score": 0.9220911264419556, "perplexity": 256.14182494005786}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585305.53/warc/CC-MAIN-20211020090145-20211020120145-00635.warc.gz"} |
http://unapologetic.wordpress.com/2010/02/02/the-existence-of-bases-for-root-systems/?like=1&source=post_flair&_wpnonce=c327cf8c6a | # The Unapologetic Mathematician
## The Existence of Bases for Root Systems
We’ve defined what a base for a root system is, but we haven’t provided any evidence yet that they even exist. Today we’ll not only see that every root system has a base, but we’ll show how all possible bases arise. This will be sort of a long and dense one.
First of all, we observe that any hyperplane has measure zero, and so any finite collection of them will too. Thus the collection of all the hyperplanes $P_\alpha$ perpendicular to vectors $\alpha\in\Phi$ cannot fill up all of $V$. We call vectors in one of these hyperplanes “singular”, and vectors in none of them “regular”.
When $\gamma$ is regular, it divides $\Phi$ into two collections. A vector $\alpha$ is in $\Phi^+(\gamma)$ if $\alpha\in\Phi$ and $\langle\alpha,\gamma\rangle>0$, and we have a similar definition for $\Phi^-(\gamma)$. It should be clear that $\Phi^-(\gamma)=-\Phi^+(\gamma)$, and that every vector $\alpha\in\Phi$ is in one or the other; otherwise $\gamma$ would be in $P_\alpha$. For a regular $\gamma$, we say that $\alpha\in\Phi^+(\gamma)$ is “decomposable” if $\alpha=\beta_1+\beta_2$ for $\beta_1,\beta_2\in\Phi^+(\gamma)$. Otherwise, we say that $\alpha$ is “indecomposable”.
Now we can state our existence theorem. Given a regular $\gamma$, let $\Delta(\gamma)$ be the set of indecomposable roots in $\Phi^+(\gamma)$. Then $\Delta(\gamma)$ is a base of $\Phi$, and every base of $\Phi$ arises in this manner. We will prove this in a number of steps.
First off, every vector in $\Phi^+(\gamma)$ is a nonnegative integral linear combination of the vectors in $\Delta(\gamma)$. Otherwise there is some $\alpha\in\Phi^+(\gamma)$ that can’t be written like that, and we can choose $\alpha$ so that $\langle\gamma,\alpha\rangle$ is as small as possible. $\alpha$ itself can’t be indecomposable, so we must have $\alpha=\beta_1+\beta_2$ for some two vectors $\beta_1,\beta_2\in\Phi^+(\gamma)$, and so $\langle\gamma,\alpha\rangle=\langle\gamma,\beta_1\rangle+\langle\gamma,\beta_2\rangle$. Each of these two inner products are strictly positive, so to avoid contradicting the minimality of $\langle\gamma,\alpha\rangle$ we must be able to write each of $\beta_1$ and $\beta_2$ as a nonnegative linear combination of vectors in $\Delta(\gamma)$. But then we can write $\alpha$ in this form after all! The assertion follows.
Second, if $\alpha$ and $\beta$ are distinct vectors in $\Delta(\gamma)$ then $\langle\alpha,\beta\rangle\leq0$. Indeed, by our lemma if $\langle\alpha,\beta\rangle>0$ then $\alpha-\beta\in\Phi$. And so either $\alpha-\beta$ or $\beta-\alpha$ lies in $\Phi^+(\gamma)$. In the first case, we can write $\alpha=\beta+(\alpha-\beta)$, so $\alpha$ is decomposable. In the second case, we can similarly show that $\beta$ is decomposable. And thus we have a contradiction and the assertion follows.
Next, $\Delta(\gamma)$ is linearly independent. If we have a linear combination
$\displaystyle\sum\limits_{\alpha\in\Delta(\gamma)}r_\alpha\alpha=0$
then we can separate out the vectors $\alpha$ for which the coefficient $r_\alpha>0$ and those $\beta$ for which $r_\beta<0$, and write
$\displaystyle\sum\limits_\alpha s_\alpha\alpha=\sum\limits_\beta t_\beta\beta$
with all coefficients positive. Call this common sum $\epsilon$ and calculate
$\displaystyle\langle\epsilon,\epsilon\rangle=\sum\limits_{\alpha,\beta}s_\alpha t_\beta\langle\alpha,\beta\rangle$
Since each $\langle\alpha,\beta\rangle\leq0$, this whole sum must be nonpositive, which can only happen if $\epsilon=0$. But then
$\displaystyle0=\langle\gamma,\epsilon\rangle=\sum\limits_\alpha s_\alpha\langle\gamma,\alpha\rangle$
which forces all the $s_\alpha=0$. Similarly, all the $t_\beta=0$, and thus the original linear combination must have been trivial. Thus $\Delta(\gamma)$ is linearly independent.
Now we can show that $\Delta(\gamma)$ is a base. Every vector in $\Phi^+(\gamma)$ is indeed a nonnegative integral linear combination of the vectors in $\Delta(\gamma)$. Since $\Phi^-(\gamma)=-\Phi^+(\gamma)$, every vector in this set is a nonpositive integral linear combination of the vectors in $\Delta(\gamma)$. And every vector in $\Phi$ is in one or the other of these sets. Also, since $\Phi$ spans $V$ we find that $\Delta(\gamma)$ spans $V$ as well. But since it’s linearly independent, it must be a basis. And so it satisfies both of the criteria to be a base.
Finally, every base $\Delta$ is of the form $\Delta(\gamma)$ for some regular $\gamma$. Indeed, we just have to find some $\gamma$ for which $\langle\gamma,\alpha\rangle>0$ for each $\alpha\in\Delta$. Then since any $\beta\in\Phi$ is an integral linear combination of $\alpha\in\Delta$ we can verify that $\langle\gamma,\beta\rangle\neq0$ for all $\beta\in\Phi$, proving that $\gamma$ is regular. and $\Phi^+=\Phi^+(\gamma)$. Then the vectors $\alpha\in\Delta$ are clearly indecomposable, showing that $\Delta\subseteq\Delta(\gamma)$. But these sets contain the same number of elements since they’re both bases of $V$, and so $\Delta=\Delta(\gamma)$.
The only loose end is showing that such a $\gamma$ exists. I’ll actually go one better and show that for any basis $\{\eta_i\}_{i=1}^{\dim(V)}$ the intersection of the “half-spaces” $\{\gamma\vert\langle\gamma,\eta_i\rangle\}$ is nonempty. To see this, define
$\displaystyle\delta_i=\eta_i-\sum\limits_{\substack{1\leq j\leq\dim(V)\\j\neq i}}\frac{\langle\eta_i,\eta_j\rangle}{\langle\eta_j,\eta_j\rangle}\eta_j$
This is what’s left of the basis vector $\eta_i$ after subtracting off its projection onto each of the other basis vectors $\eta_j$, leaving its projection onto the line perpendicular to all of them. Then consider the vector $\gamma=r^i\delta_i$ where each $r^i>0$. It’s a straightforward computation to show that $\langle\gamma,\eta_k\rangle=r^i\langle\delta_k,\eta_k\rangle>0$, and so $\gamma$ is just such a vector as we’re claiming exists.
February 2, 2010 - Posted by | Geometry, Root Systems
1. Your final summation doesn’t actually give you a vector orthogonal to the remaining basis vectors, unless everything is already orthogonal. I think you want a gram schmidt process here, applied independently to each basis vector.
Also, out of curiosity, do you have any plans to draw out any of the polytopal or crystollographic connections to root systems?
Comment by Gilbert Bernstein | February 3, 2010 | Reply
2. You’re right, Gilbert. But in the end the resulting vector $\gamma$ still has the properties we want.
As for applications, I’m just looking at classification for now. I may return to applications at some future point.
Comment by John Armstrong | February 3, 2010 | Reply
3. [...] A very useful concept in our study of root systems will be that of a Weyl chamber. As we showed at the beginning of last time, the hyperplanes for cannot fill up all of . What’s left over [...]
Pingback by Weyl Chambers « The Unapologetic Mathematician | February 3, 2010 | Reply
4. [...] then is a (positive) root for some simple . If for all , then the same argument we used when we showed is linearly independent would show that is linearly independent. But this is impossible because [...]
Pingback by Some Lemmas on Simple Roots « The Unapologetic Mathematician | February 4, 2010 | Reply
5. [...] is any regular vector, then there is some so that for all . That is, sends the Weyl chamber to the fundamental [...]
Pingback by The Action of the Weyl Group on Weyl Chambers « The Unapologetic Mathematician | February 5, 2010 | Reply | 2014-04-16 19:05:40 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 103, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9714778065681458, "perplexity": 141.84146940778655}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1397609524644.38/warc/CC-MAIN-20140416005204-00569-ip-10-147-4-33.ec2.internal.warc.gz"} |
https://unix.stackexchange.com/questions/319388/centos-7-python3-error-while-loading-shared-libraries-libpython3-5m-so-1-0 | I am deploying a django server.
I've already installed python 3.5 without errors, but for some problems when I tried to compile mod_wsgi (so I could share the django app in an intranet environment with apache):
./configure --with-python=/usr/local/bin/python3.5
it told me that Failed to locate the Python library /usr/local/lib/libpython3.3m.so.
Searching for the solution on Google, I found this solution explaining that I needed to recompile python3 with --enable-shared flag.
so, I proceeded to remove python3 and python3.5 from /usr/local/bin and recompiled python3. Everything was fine, until I tried to check the python version with python3 --version and I got the following:
[rortega@Production Python-3.5.2]$python3 --version python3: error while loading shared libraries: libpython3.5m.so.1.0: cannot open shared object file: No such file or directory I can't find any solution to this. • ldd$(type -p python3) might give you some insights. – Valentin Bajrami Oct 27 '16 at 19:50
• is the libpython3.5m.so.1.0 marked as not found. the problem now is the it is in the correct location, but for some reason it doesn read it – Izuzvo Oct 27 '16 at 21:15
Your system may not be looking in /usr/local/lib by default for shared libraries. When you compile things, set the environment variable:
export LD_RUN_PATH=/usr/local/lib
That or add /usr/local/lib to the directories searched by the system for shared libraries. | 2019-10-16 07:02:29 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.1896211802959442, "perplexity": 3296.143061990722}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986666467.20/warc/CC-MAIN-20191016063833-20191016091333-00456.warc.gz"} |
https://symbiosisonlinepublishing.com/biochemistry/biochemistry23.php | Research Article Open Access
The Antisickling Effect of the Arthrospira platensis bilins for Liver Protection: a Modeling, Hypothesis, and Food for Thought
Amro Abd Al Fattah Amara*
Protein Research Department, Genetic Engineering and Biotechnology Research Institute, City for Scientific Research and Technological Applications, New Borg Al Arab, Alexandria, Egypt
*Corresponding author: Amro Abd Al Fattah Amara, Protein Research Department, Genetic Engineering and Biotechnology Research Institute, City for Scientific Research and Technological Applications, Alexandria, Universities and Research Center District, New Borg EL-Arab, Egypt, Tel: 203-4593422; Fax: 203 4593497; E-mail: @ ; @
Received: 05 April, 2017; Accepted: 21 June, 2017; Published: 18 August, 2017
Citation: Amro Abd Al Fattah Amara (2017) The Antisickling Effect of the Arthrospira platensis bilins for Liver Protection: a Modeling, Hypothesis, and Food for Thought. SOJ Biochem 3(1):1-12 DOI: http://dx.doi.org/10.15226/2376-4589/3/1/00123
Summary
The increase in the number of the liver injury and diabetic patients’ worldwide and particularly in sub-Sahara in Africa put a signal about that it might be a link between it and the regional diseases such as Malaria and Sickle Cell Anemia (SCA). SCA is a genetic based disease and is one of the liver injury causative agents. Chronic liver injury patients’ are susceptible to more complications such as the infection with viruses (e.g. Virus C), diabetic, immune disease, weakness, anemia etc. Arthrospira platensis proves be able to fight and to protect against different diseases at once. It has antiviral, antioxidant, antisickling, nutrient, and edible. In this study their bilins are evaluated for their antisickling effect using molecular modeling aiming to protect patients with SCA particularly the diabetic ones.
In this study molecular modeling for the normal and sickle β-globin, molecules against five bilins [(1) Red bilin, (2) 21H-Bilin- 1(22H)-one, (3) 21H-Bilin-1(24H)-one, (4) 1H-Bilin 1 one, and (5) 22H-Biline (21-bilin)] were investigated using protein modeling and docking. MODELLER v 9.8, Hex ver 8.0.0 and Discovery Studio 4.1 Client 4.1.0.14169 (Accelrys software Inc.) were used for modeling and docking. A. platensis bilins are evaluated also by comparing the visualized docking results.
The total energy of the system (the molecule) for all of the used bilins with the sickle β-globin molecule particularly in both of the presence or the absence of the porphyrin ring prove to improve such energy to be almost equal to that obtained from the normal β-globin (with or without porphyrin ring). However, porphyrin ring is essential.
A. platensis bilins using molecular modeling prove to be able to stabilize the sickle β-globin molecules particularly in the presence of the porphyrin ring. As being multifunction, it is recommended either as native biomass or as purified bilins to be used for the treatment of the SCA and the protection against further liver deterioration.
Keywords: Liver injury; Bilin; β-globin; protein modeling; structure/function/specificity
Introduction
The proteins structure/function/specificity are governed by their amino acids order, number and arrangement in the protein backbone. Single amino acid change could change the protein property totally. Such change will not effect on the protein family only but will effect on any other macromolecule or micro molecules could gain, or loss affinity to such changed protein. No better example could be described than the SCA. In mammals the O2 binding protein, myoglobin and hemoglobin are among the most extensively early studied proteins. The single amino acid change in the β-globin (existed on the surface of the protein) cause inefficient O2 transport. The amino acid change lead to a sticky patch on the β-polypeptide chain of deoxyhaemoglobin, which cause aggregation and precipitation. Such aggregation and precipitation can cause deterioration for the body organs particularly those, which subjected to high blood flow and interaction such as the liver and the spleen. That is not the only type but there are many other variant of hemoglobin mutations. The Fe2+ containing heme group is a highly hydrophobic molecule and requires to be placed in a hydrophobic pocket in hemoglobin quaternary structure. It is important to sign that mutants other than the SCA mutant does not necessarily have to be single amino acid substitutions and not have only to affect the heme pocket [1].
Genetic diseases were well known in the old civilization [2]. The World Health Organization (WHO) (1982) estimated that about five percent of the world populations are carriers of genes for clinically important disorders of hemoglobin [3]. Recently Amara highlight some solutions for avoiding different types of genetic diseases including SCA [4]. Each of the β globin genes is represented at least with two copies each in one chromosome gained from the mother and the father to be finally two. So the probability that two incorrect genes find each other will be higher in relatives. Also endemic area with certain incorrect trait or more than one trait of the hemoglobin disease should be considered. Prototype mutants must be detected by using DNA sequencing and mapping the possibility of existing of such type of mutant(s) where one or two bases change can lead to a new mutant which not appears yet. Out group marriage will reduce the disease severity and will give more chances to the correct genes to be existed and the mutated genes to disappear [4].
Normal Hemoglobin Hb A composition is α2Aβ2A with genotype αα/αα β/β. Sickle cell trait hemoglobin Hb A, Hb S its composition α2Aβ2A, α2Aβ2S with genotype αα/αα β/βs. Sickle cell disease hemoglobin HbS its composition α2Aβ2S with genotype αα/αα βs/βs. The hemoglobin four chains 2(α)/2(β) fitted together to form a globular tetramer with a molecular weight of approximate 64000, a structure that for Hb A, is abbreviated as α2β2. The two kinds of chains are almost equal in length, the α chain having 141 amino acids and the β chain 146. α and β chains are encoded by genes at separate loci (the α locus on chromosome 16 and the β locus on chromosome 11). In addition to Hb A, there are five other normal human hemoglobin’s, each of which has a tetrameric structure comparable to that of Hb A in consisting of two α or α-like chains and two non-α chains [5].
Liver problems are reported in patients with SCA in 37% cases [6]. Those patients have abdominally meteorism, right upper quadrant pain, or acute painful hepatomegaly. In general different form of liver injury in most cases as cholestatic. Liver infarction has been reported in 34% of autopsies [7]. The SCA associated blood viscosity predisposes to infarction [8,9]. The liver enzymes activity increased abnormally [10,11]. The causing agent in case of SCA patients thought to be obstruction of sinusoidal flow of masses of sickled erythrocytes, trapping them in the liver [12].
There is an increasing in the number of publications, which introduce A. platensis as a proposed candidate could be used to reduce the antisickling agent.
This study investigates A. platensis bilins structures, which might be able to reduce the sickling process as well as to show that bioinformatics tools must give more concern to the DNA sequences. A. platensis bilins have proved to have many useful activities especially as antiviral, antitumor, antioxidant and antisickling. The heme and hemoglobin proteins were docked against five bilins from A. platensis. This study refreshes the scientific aim to control and to treat such illness happened by a one nucleotide change. And can be only avoided by avoiding the marriage from the same group. Encourage the marriage from out group is the correct and the simplest solution. And also to protect from further deterioration in organs such as the liver, where diabetic SCA patients will subject to sever liver injury.
Material and Methods
Bilins
The three-diminution structures for five A. platensis bilins were used in this study to investigate their abilities to dock with sickle hemoglobin and β- globin protein models. The chemical formula and name of the used bilins are:
1. Red bilin [Also known as: CPD-7063, (7S,8S, 101R)-8-(2- carboxyethyl)-17-ethyl-19-formyl-101-(methoxycarbonyl)- 3,7,13,18-tetramethyl-2-vinyl-8,23-dihydro-7H-10,12- ethanobiladiene-ab-1,102(21H)-dione]. Which have molecular formula: C35H38N4O7-2; molecular weight: 626.69882 g/mol; InChI Key: HMDDKKOMBDRDIA-DSJLEYPNSA-L. Its IUPAC name is: [3-[(2Z,3S,4S,5Z)-5-[(4-ethenyl-3-methyl-5-oxopyrrol-2-yl) methylidene]-2-[2-[(3-ethyl-5-formyl-4-methyl-1H-pyrrol-2-yl) methyl]-5-methoxycarbonyl-3-methyl-4-oxido-2,3-dihydro- 1H-cyclopenta[b]pyrrol-6-ylidene]-4-methylpyrrolidin-3-yl] propanoate].
2. 2.2 1H-Bilin-1(22H)-one,2,3,7,8,12,13,17,18-octaethyl-19- methoxy-, 113435-10-2, MolecularFormula: C36H48N4O2, Molecular Weight: 568.79192 g/mol, InChI Key: QCYVKTHGTVWFSV-UHFFFAOYSA-N. Its IUPAC name is: [5-[[5-[[5-[(3,4-diethyl-5-methoxypyrrol-2-ylidene)methyl]- 3,4-diethyl-1H-pyrrol-2-yl]methylidene]-3,4-diethylpyrrol-2- ylidene]methyl]-3,4-diethylpyrrol-2-one].
3. AGN-PC-0O2GJ1,21H-Bilin-1(24H)-one,19-hydroxy-, 21H-Biline-1 , 19-dione , 22, 24-dihydro-,142550-15-0, 58828-89-0, Molecular Formula: C19H14N4O2, Molecular Weight: 330.34006 g/mol, InChI Key: MQHWQQCOXHUNCSUHFFFAOYSA- N, Its IUPAC name is: 5-[[5-[[5-[(5-oxopyrrol-2- ylidene)methyl]-1H-pyrrol-2-yl]methylidene]pyrrol-2-ylidene] methyl]pyrrol-2-one.]
4. 1H-Bilin-1-one [Also known as: AGN-PC-0OFTAO, 66560-67- 6]. Which have molecular formula: C19H12N4O; molecular weight: 312.32478 g/mol; InChI Key: VGJBOZZPXZVBBI-UHFFFAOYSA-N. Its IUPAC name is: 5-[[5-[[5-(pyrrol-2-ylidenemethyl)pyrrol-2- ylidene]methyl]pyrrol-2-ylidene]methyl]pyrrol-2-one
5. 22H-Biline, 21H-Bilin, 22H-Bilin, AC1OAGP5, SureCN139406, AGN-PC-02LS4D
Molecular Formula: C19H14N4, Molecular Weight: 298.34126 g/mol, InChI Key: PPRBOEHFGAHFGC-UHFFFAOYSA-N. Its IUPAC name is: 2-(pyrrol-2-ylidenemethyl)-5-[[5-(pyrrol-2- ylidenemethyl)-1H-pyrrol-2yl] methylidene] pyrrole
Another bilins are existed but will not included in this study
The five bilin molecules were downloaded from PubChem (www.ncbi.nlm.nih.gov/pccompound) and saved as SDF format files [13]. The chemical structure of the molecules is given in (Table 1).
The β-globin sequences and software used in this study
The amino acids sequences of both normal and sickle β-globin are represented by the following amino acid sequences.
A: The normal β-globin
VHLTPEEKSAVTALWGKVNVDEVGGEALGRLLVVYPWTQRFFESFGDLSTPDAVMGNPKVKAHGKKVLGAFS- DGLAHLDNLKGTFATLSELHCDKLHVDPENFRLLGNVLVCVLAHHFGKEFTPPVQAAYQKVVAGVANALAHKYH
B: The Sickle β-globin
VHLTPVEKSAVTALWGKVNVDEVGGEALGRLLVVYPWTQRFFESFGDLSTPDAVMGNPKVKAHGKKVLGAFS DGLAHLDNLKGTFATLSELHCDKLHVDPENFRLLGNVLVCVLAHHFGKEFTPPVQAAYQKVVAGVANALAHKYH
Normal and sickle β-globin amino acids sequences were the start point of this study. The sequence was obtained from the www.ncbi.nlm.nih.gov (Blast.ncbi.nlm.nih.gov/Blast.cgi) [14].
The software used in this study
Software for Modeling
One published sickle hemoglobin model was used [15]. The β-globin protein model was generated using the software MODELLER v 9.8 [16].
Software for docking
“Hex” is a Molecular Graphic Program (Hex’s Home Page: http://www.loria.fr/~ritchied/hex/) for calculating and displaying feasible docking modes of pairs of protein and DNA molecules [17,18,19]. Hex software can also calculate Protein- Ligand Docking, assuming the ligand is rigid, and it can superpose pairs of molecules using only knowledge of their 3D shapes [20]. It uses Spherical Polar Fourier (SPF) correlations to accelerate the calculations and its one of the few docking programs which has built in graphics to view the result [18]. Simply, the protein pdb is loaded from the “File > open > receptor” and the bilin or the porphyrin ring loaded from “File > open > ligand” and then from the control option docking is selected and the parameter in is used (Figure 1). The binding energy result is normally negative, stating that a better binding affinity is established from the highest negative result. Low (negative) energy indicates a stable system.
To determine the behavior of both of the protein molecules under study and whether we need high negative energy or lower ones; porphyrin ring has been docked firstly against both of the normal and the sickle β-globin molecules which obtained from the Modeller software and the published sickle hemoglobin model as above.
The five used bilins and one porphyrin ring which used in this study are summarized in (Table 2). The docked molecules’ 3D structures have been saved as pdb files and visualized in to show the different interactions (Table 3,4).
Software for the molecules study
The software Discovery Studio 4.1. Client 4.1.0.14169 (Accelrys software Inc.) was used to visualize the docking of the bilins with the proteins models and to show ligands binding sites and the other analysis for the docked molecules [21].
For better 3D structure, the background of the images have been converted to white and the 3D image have been adjusted and saved. All of the docking images have been putted in tables to enable better comparisons between the interaction between the β-globin and the bilins (Table 3,4).
Table 1: Total energy of the system (ETotal) from the docking of the five bilins with different β-globin molecules
Molecules name Etotal (total energy of the system)*$Differences Normal β-globin (without porphyrin ring) Normal β-globin (with porphyrin ring) Sickle β-globin (without porphyrin ring) Sickle β-globin (with porphyrin ring) between column 1 and 2 between column 1 and 3 between column 1 and 4 between column 2 and 3 between column 2 and 4 between column 3 and 4 Red bilin -380 -322.5 -318.2 -305 -57.5 -61.8 -75 -4.3 -17.5 -13.2 21H-Bilin-1(22H)-one -352.1 -162 -163.9 -149 -190.1 -188.2 -203.1 1.9 -13 -14.9 21H-Bilin-1(24H)-one -279 -122 -111 -123.1 -157 -168 -155.9 -11 1.1 12.1 1H-Bilin 1 one, and -277.7 -137.1 -140 -131.4 -140.6 -137.7 -146.3 2.9 -5.7 -8.6 22H-Biline(21-bilin) -349 -164.1 -161.2 -159.4 -184.9 -187.8 -189.6 -2.9 -4.7 -1.8 *The EShape (energy of shape only approach) is same as the ETotal.$ the ETotal and EShape of the normal and sickle β-globin with porphyrin ring are -821 and -516 respectively
Figure 1: The Hex software docking parameters
Results and Discussion
Our genes are the codes for the proteins in our bodies. Understanding our genes and our proteins will help us to avoid different illness and to design new drugs. Such drugs can be so simple and can be supplied as natural products in the form of food or edible plants. Such natural products can provide us with what the defected genes could not do. Vitamin C is the most famous example. Others forms might can improve certain function like the structure of the SCA defect protein. We are in need to do complicated research to find solutions for some degenerative diseases which alter our macromolecules structure hence their functions and specificity. However, avoiding such type of diseases is so simple. It is just by avoiding the marriage from the same group and from those which have the same disease trite (should not marry from each other). SCA is a known genetic disease in West sub-Sahara in Africa, in the Mediterranean region and other places worldwide [4].
SCA which also named as hemoglobin S disease or hemoglobin SS diseases. After losing the oxygen SCA cells soon stimulate holly leaves or their crescents. Soon they become filamentous and spirculated. Single gene of hemoglobin S and the other is A named sickle cell trait. Individuals have hemoglobin S and β- thalassemea (β-thal) are both prevalent (Greeks and Italians) having a high incidence of S-thalssemia (S-β-thalassemia) [22].
Additionally, scientists, particularly, those from the SCA endemic regions and countries, have summarized their experiences as well the experience gained from their communities in controlling the disease side effect.
Smith and Wood in their book about the Biological Molecules (1991) have written: ‘The present-day distribution of effective hemoglobin’s has arisen from the accumulation of harmless mutations, early death of individuals with harmful mutation, this confers a selective survival advantage such as increased resistance to malaria, as is the case which sickle cell disease’ [1].
For that, lethal mutant are unable the transform their genotype to the second generation. While mutant which gives the minimum survives until the appearance of the seconded generation will do and will be transferred from generation to generation. However there is a 50 % chance that the correct chromosome transfers instead of the one which has the mutant. Alternatively, existing of two globin mutant (on α and β which existed in two chromosomes) will increase the chance of the transfer of globin disease.
Hemoglobin is the oxygen carrier tetrameric molecule and can be found in vertebrate red blood cells, in some invertebrates and in the root nodules of legumes [23]. Each subunit is composed of a polypeptide chain, globin, and a prosthetic group, heme, which is an iron-containing pigment that combine with oxygen and gives the molecules its oxygen-transporting ability. SCA is a global disease and for the Mediterranean and the Africans communities is a local disease [24]. Livingstone, has described in detailed the roles which affect the percentage and the distribution of the SCA in West Africa. From the time of specifying the role of the heredity (the most critical one) till producing artificial blood and artificial oxygen carrier, the scientific progress and the scientist effort did not stop [25].
The biological system is sensitive for the chemical structure. Enzymes could be so specifics. Other protein forms could be also be very sensitive. Red Blood Cells (RBCs) could differentiate between O2 and CO2. The conditions and the structure draw the
Table 2: Six macromolecules and six molecules (five bilins and one porphyrin ring) used in this study
Table 3: The different docking results between the five β-globin macromolecules and the five bilins
Table 4: Different molecules surface interaction with the β-globin macromolecules
function of the RBCs; where they gain O2 and where they lose it; where they gain CO2 and where they lose it. Such specificity is not only in the heme active amino acids but also in the atoms of O2 and CO2. For that it is important to investigate molecules could be able to stabilize the sickle hemoglobin particularly during the CO2 stage which is the aim of this study.
The HBB gene provides instructions for making β-globin. Various versions of β-globin result from different mutations in the HBB gene. One HBB gene mutation produces an abnormal version of β-globin known as Hemoglobin S (HbS). Other mutations in the HBB gene lead to additional abnormal versions of β-globin such as Hemoglobin C (HbC) and Hemoglobin E (HbE). HBB gene mutations can also result in an unusually low level of β-globin; this abnormality is called β-thalassemia. When oxygen is removed from sickle hemoglobin, those molecules change their shape and combine with one another. The Red blood cell structure changes from ring to sickle shape in the absence of the oxygen. This causes blood to clots and deprives vital organs from their supply of blood, resulting in pain, intermittent illness, and in many cases, a shortened life span. The only difference between normal and sickle cell hemoglobin is that in each β- chain, one glutamic acid is replaced by one valine. Valine, unlike glutamic acid, contains a nonpolar group. The result is a hydrophobic “sticky” region that can interact with hydrophobic region on neighboring molecules, producing the observed clumping. A slight change in the β-globin 3D structure will induce a change in the configuration of it when it interacts with its neighboring subunits of the hemoglobin.
The macromolecules are very sensitive to any effect could effect on their structures/functions. Such changes can be classified based on different strategies. But, classifying them regarding to the type of the effective changes will lead to a more focusing on the positive and negative factors and the possible solutions for illness problems. For that, classifying the factors effect on the macromolecules structure/function can be divided to the following groups and probability:
1. Wild type macromolecules in correct environment that will give correct function(s).
2. Wild type macromolecules in optimized environment better than the standard that will optimize the function(s) but can cause deterioration for the macromolecules hence illness.
3. Wild type macromolecules in condition less than the standard this will give less functions and could lead to the deterioration for the macromolecules.
4. Wild type macromolecules in incorrect environment that will give a temporary incorrect function but they are still able to do correct function in the suitable conditions.
5. Mutant macromolecules in environment let them work like wild type that will cause deterioration while the macromolecules do an extra job and in normal case they do less activity.
6. Mutants in environment same like the best condition for the wild type. This might not give the same result of the wild type or protect the macromolecules from further deterioration.
7. Incorrect macromolecules find some support from other macro or micro molecules to do their job with less effort or to fit or repair some of the existed problems due to the change in their structures hence their physicochemical properties. This group is most one fit with the aim of this study.
From the simple classification of the macromolecules as above it can be concluded that:
1. Wild type macromolecules even correct but should be maintained in the conditions which did not turn them to mutants as in case of the DNA or the RNA and not deteriorate them such as in case of the proteins and short peptides.
2. Mutated macromolecules should be lead to illness in most cases where the probability of their deterioration is higher than the wild type because they are doing incorrect function or do more effort to reach the wild type functions and activity.
3. Both the macromolecules’ situation either being wild or mutant as well as the surrounding environment plays a significant role in their functions.
4. Sick macromolecules should be treated as sick and their work should not be optimized while that will lead to further deterioration. Perhaps a good example here could be given about the drugs which prevent or decrease the pain. In fact most of them stop our failing towered the pain target but the damage is continuous and not stop. It is better to lessen well for the pain signals in our bodies. Given our bodies enough rest and enough time to recover is far better than force it to work incorrectly in the presence of pain. Pain is the best friend for observing problems in our bodies.
5. Factors which could improve the function of the mutant macromolecules without putting load on the targeted macromolecule itself will be the best solution. It is like the competition between the oxidants plus the free radicals against our endogenous antioxidant where high oxidants and free radicals will deteriorate our endogenous antioxidants particularly if there is an existing continuous source for them either by bypassing it, exhaust the cells, damage the DNA and the RNA including those responsible for the antioxidant system. Supplying the body with suitable exogenous antioxidant will overcome all of such problems. However, continuous supplying with exogenous source will deteriorate the endogenous antioxidant too!
6. The type of the changes must be clear for better treatment. Is it permanent such as mutants? Or only due to environmental and conditional changes but the macromolecules still correct?
7. One should not put his macromolecules in the optimum conditions where optimum conditions is in the top of the peak and will lead soon to drops and vice versa will give the same result.
8. Balanced life is required.
SCA might be more interested as a subject for the Mediterranean area and the sub-Saharan West region in Africa. Back to the first observation where the Flamingo birds are interested to feed themselves by A. platensis(Spirulinaplatensis) in Chad to be able to follow their long migration trips. Flamingo birds which are appeared healthy and reddish is a signal that such an African cyanobacteria was the early signal for the local individuals that such edible cyanobacteria can be eat as food [26,27].
After generating the sickle β-globin and docking the porphyrin ring with both of the normal and sickle β-globin five molecules were existed as pdb files. They are:
1. Normal β-globin without porphyrin ring.
2. Normal β-globin with porphyrin ring in its native form.
3. Normal β-globin with docked porphyrin ring.
4. Sickle β-globin without porphyrin ring.
5. Sickle β-globin with docked porphyrin ring.
The five molecules each has been docked against each of the A. platensis bilins.
The determined normal and the sickle β-globin molecules (which obtained from the Modeller software and the published sickle hemoglobin model as above) were ETotal (the total energy of the system) “-821.0” for the normal β-globin and -516.9 for the sickle β-globin. The Etotal prove that the interaction between the porphyrin ring and the normal β-globin are more negative hence, more stable interaction is existed.
Etotal in case of the normal β-globin with all of the five bilin show less energy than that in case of the presence of porphyrin ring (without Fe2+) which means more stable interactions. That means the existence of correct interaction between the bilins and the normal β-globin. In case of sickle cell β-globin the same result was obtained in all cases. That proves that the presence of the porphyrin ring reduces the total energy of the system, which is logic where the existence of two molecules interacting with a single protein probably will lead to reduce the interaction force by each molecule if compared with the force determined if the each molecule interact in individual case.
Normal β-globin only (without porphyrin ring) and with bilin show in all cases less total energy than that of sickle β-globin , which prove that sickle β-globin molecule with the bilins have good interaction.
However, in presence of porphyrin ring in case of sickle β-globin in case of Red bilin, 21H-Bilin-1(22H)-one, 1H-Bilin 1 one, and 22H-Biline(21-bilin) show increase in the system total energy than that in case of the absence of porphyrin ring.
Only the presence 21H-Bilin-1(24H)-one show decrease in the total system energy in case of the presence of porphyrin ring. So, 21H-Bilin-1(24H)-one is a molecule that can stabilize the β-globin.
However, in general, the presence of the porphyrin ring or a bilin or both increases the total energy of the system except in 21H-Bilin-1(24H)-one with sickle β-globin. Small increase shown also between 21H-Bilin-1(22H)-one with normal β-globin and with porphyrin ring if compared with 21H-Bilin-1(22H)-one with sickle β-globin (without porphyrin ring).
The slight decrease in the negativity in case of Red bilin, 21H-Bilin-1(22H)-one, 1H-Bilin 1 one, and 22H-Biline(21-bilin) as in show that such molecules still so close to the negativity of the obtained from the normal β-globin (Table 1). The five bilin successfully readjust the Etotal of the molecules in case of the presence of porphyrin ring and the differences in the system in case of column 2 and 4 is negligible in all cases and range from “-17.5” in case of Red bilin to “+1” in case of 21H-Bilin-1(24H)- one.
In conclusion the variation between normal β-globin and sickle ones are significant (in case of absence of porphyrin ring) as in column 1 and 2 but not significant in case of column 2 and 4 which prove that porphyrin ring is essential in the β-globin structure and that bilins could play a significant role in establishing the β-globin molecules.
One should remark that Hex program is a windows docking software, however, it can give preliminary judgment for the differences and the possibilities of using bilins to treat sickle β-globin. The used CPU is “Hp Compaq nc 4200”. In addition, the used Hex software is Hex ver. 8.0.0. However, it is a remark that slight variation in the molecule can be sound in the total energy of the system (the molecules or the molecules and its legends). In addition, the use of the same condition and the same computer will reduce any variation due to different systems’ error. The obtained results have enough variation to prove that the fiveused bilins could stabilize the sickle β-globin particularly in the presence of the porphyrin ring, and that 21H-Bilin-1(24H)-one might be the best tested molecules.
The study is also highlight the importance of the porphyrin ring in the controlling of the sickle cell anemia side effect.
Bilins which are derived from the edible cyanobacteria the A. platensis and which reported in quite enough studies [data not shown] to have antisickling effect, could play a significant role in reducing the side effect of the SCA. “Target the target which targets you”, words might be a key for controlling the SCA and other degenerative disease, where many of such diseases is treated from their end point while, their starting point located in avoiding the marriage from the same group as described in details by Amara [4].
A. platensis might be the ideal edible food for both of the protection and the treatment, in case of patients with both of SCA and diabetic. It is well known for its antioxidant activity, gamma fatty acid, antiviral, and antisickling. For that, it can react with a different target at once to protect against any of the exponential deterioration could effect on the other organs. Hex software is recommended for being used to evaluate more molecules to find the ideal solution for the SCA.
One should put in his account that this study represents models conducted in computer study and, to validate any of the obtained result; in vivo experiment should be conducted (in future study).
The purpose of preparing each of the normal and sickle β-globin using MODELLER software is to reduce any differences from using an original x-ray based structure obtained from the web. However, a published model was used for such purpose. Additionally, porphyrin molecule without Fe+2 was used. All those factors should be considered during conducting study similar to this study. Docking of the normal and sickle β-globin against prophyrin ring was for the purpose to find the direction of the molecules’ stabilization in case of the β-globin molecule. Yes, the more negativity means more stability, but in our case stability might not be required but equality is the correct target. So stable molecular interaction in this study case might not enable correct oxygen exchange. But equality with the native normal β-globin is the correct key for investigation.
In our case the docking of the normal β-globin with porphyrin ring give total system energy equal to -821 while the docking of the sickle β-globin against the porphyrin ring give -516.9. Sickle β-globin is clearly less stable in its interaction with the β-globin; considering that the data obtained from the interaction between the normal β-globin and the porphyrin ring is the ideal one (-821). That might explain the source of the deterioration in case of presence of SCA disease. The purpose might not in the structure only but in the energy as well.
Not all of the pdb files obtained in this study are represented. However, the included images explain also for some extent that it might be used to prove or disprove some facts. Images might be describe in better way where the molecules are interacts and which place are preferable on the protein backbone (to be stabilize or destabilize it). Such docking might not represent the absolute fact, but it will give somehow preliminary judgment and an overall view about what could be happened if such molecules are present in our bodies. Such molecules are derived from edible source so they might be safe if used in an adequate amount. A fact; even seem to be correct in case of using the A. platensis native biomass but, need to be proved in case of purified compounds.
For the five bilins it is clear from their models as in that the bilin have good interaction with both of the β-globin molecules (Normal and sickle) (Table 3,4). Apparently, the smaller the molecules the better it become inside the 3D structure. The lager the molecule it tends to attach on the surface of the β-globin molecules. The existence of porphyrin ring or their absence shows different-results. That proves the sensitivity of the process and the importance of the porphyrin ring in the structure of the β-globin, the hemoglobin and the RBCs overall structure.
Based on the five bilin docking data with the β-globin; the docking process is very sensitive.
1. In case of Red bilin the bilin bind to the β-globin in different cases nearly in the same site, except in case of sickle β-globin without porphyrin ring.
2. 21H-Bilin-1(22H)-one show different binding property for all of the five molecules. 21H-Bilin-1(22H)-one is sensitive to any change in the β-globin different molecules.
3. 21H-Bilin-1(24H)-one show different binding sites between normal and sickle molecules but nearly the same in either normal or to the sickle molecules.
4. H1-Bilin-one is bind nearly to the same in place in all of the five macromolecules and show competition against the porphyrin ring.
5. 22H-Bilin (21H Bilin) bind in the same place from the five macromolecules and show competition against the porphyrin ring.
Supposing that β-globin with normal porphyrin ring which obtained without modification from the 1HHO model with native porphyrin ring orientation (no docking) is the most correct macromolecules, in such case Red bilin, 1H-Bilin 1 one and 22H-Bilin (21H Bilin) will be the best molecules which expected to support the normal and the sickle β-globin without the interferes with the porphyrin ring or the macromolecules’ 3D structure based on the models obtained as in (Table 3 and 4).
Porphyrin rings attached differently for both of normal and sickle β-globin. Additionally the porphyrin ring orientation is different.
Red bilin against normal and sickle β-globin show nearly full surface fitting. 21H-Bilin-1(22H)-one against normal β-globin alone show good surface fitting but not in case of sickle β-globin.
21H-Bilin-1(24H)-one fitting is totally different in case of normal and sickle globin.1H-Bilin 1 one show the same fitting profile if the different between the normal and the sickle β-globin is considered. 22H-Biline (21H-Bilin) is fit partially to normal β-globin but not the case in the sickle β-globin.
From the models and the results obtained from this study, some facts can be highlighted:
1. Molecular docking is a sensitive process. Docking single protein existed in tetrameric form give different result from that if it studied in its monomeric form.
2. The quaternary structure of the macromolecules such as the hemoglobin is very important where any change or reduce in the number of the protein unit will change its 3D conformation hence change the overall ability to bind to its specific molecules such as in case of hemoglobin and porphyrin ring.
3. Molecules could compete each other if they have similar binding sites such as in case of porphyrin ring and the bilins. 4. Big molecules tend to attach to the surface such as in the porphyrin ring and the big bilin.
5. Small molecules tend to penetrate the protein 3D structure such as in case of small bilin and even so they still sensitive to those molecules bind on the protein surface.
6. The in Silico or in computer modeling could find many useful information however in vitro experiment should be the final judgment. Where the molecules under investigation might affect other macromolecules. And in vivo conditions must be some how different and must be conducted for better evaluation.
7. Natural products must take more interest while they are product prove to be safe, hence they are chemically harmless at least if used in the correct amount and dosage.
8. One amino acid change causes such disease which proves the importance of the protein structure/function/specificity.
9. Avoiding such illness condition can be by avoiding the marriage from the same group or the marriage from individuals who have defected trite.
Such avoidance will lead finally to the disappear of such disease after correct generation.
Our understanding to each condition could effect on our macromolecules will let us to normalize the line between our hope and our bodies for the better of our macromolecules.
Conclusion
A. platensis could support the patients’ with SCA from different points where it’s well known for their antiviral, antisickling, vitamin rich, antioxidants, high protein content etc. There is an increasing interest for using A. platensis in the SCA research. This study suggests a role for five bilins of the Arthospira. in treating the SCA. For that the requested molecules have been obtained and generated using different software. Six macro molecules and five bilins have been evaluated best on the interaction between each bilin plus he porphyrin ring and each macromolecules. Two bilins show competition with the docked porphyrin ring while the other three bilins did not with correct interaction with the β-globin 3D structures. It is suggested that some bilins might be used as drugs for treating SCA. However this study did not include any of the in vivo study and such study must be done for more perfect judgment. Additionally, the study includes discussions about the conditions, which might affect on our healthy or modified macromolecules which evaluated in this study and represented by the β-globin molecules. As antisickling agent bilins will protect the liver from the liver injury due to the effect of the sickle RBCs which will be indirect way to prevent different diabetic diseases including the viruses’ infection, fibrosis, etc.
ReferencesTop
Listing : ICMJE | 2021-07-26 01:52:43 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.467608243227005, "perplexity": 4005.1845510206754}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046151972.40/warc/CC-MAIN-20210726000859-20210726030859-00688.warc.gz"} |
https://math.stackexchange.com/questions/2824579/counting-non-nesting-multi-permutations | # Counting non-nesting multi-permutations
Given a sequence $1,1,2,2,3,3, …,k,k$, I am interested in counting the number of non-nesting permutations of the above sequence. Two intervals (determined by symbols $K$ and $L$) are nesting if one is completely contained inside the other. A multi-permutation is non-nesting if for any two symbols $K$ and $L$, the corresponding intervals ( $[K,K], [L,L]$) are non-nesting.
For instance, $122313$ is nesting permutation since the interval defined by the two copies of $2$ is contained inside the interval defined by $1$.
What is the count of non-nesting multi-permutations? What is known about these special multi-permutations?
P.S. I know that the total number of permutations of $1,1,2,2, ... ,k,k$ is given by $(2k)!/2^k$.
Update: I found this is equivalent to counting nonnesting matchings on [2n] which is equal to Catalan number$C_n$ . See reference: Catalan numbers by Stanley.
• Have you tried computing the first few values and seeing what OEIS has to say about it? – munchhausen Jun 19 '18 at 7:03
• @Munchhausen OEIS does not say anything about non-nesting multi-Permutations. – Mohammad Al-Turkistany Jun 19 '18 at 7:47
A pattern is of the form FSFFFSSS, and the only restriction on possible patterns is that every initial segment contains at least as many F as S. So these are just the Dyck paths of length $2k$.
• Thanks for your answer. Are you aware of any literature references to non-nesting permutations of $1,1,2,2, …, k,k$? – Mohammad Al-Turkistany Jun 19 '18 at 9:52
• I found this is equivalent to counting nonnesting matchings on [2n] which is equal to Catalan number$C_n$ . See reference: Catalan numbers by Stanley. – Mohammad Al-Turkistany Jun 21 '18 at 19:55 | 2019-07-19 07:49:26 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8557571172714233, "perplexity": 547.829175699274}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195526153.35/warc/CC-MAIN-20190719074137-20190719100137-00069.warc.gz"} |
http://email.esm.psu.edu/pipermail/macosx-tex/2005-February/013594.html | # [OS X TeX] Celsius
Gary L. Gray gray at engr.psu.edu
Tue Feb 15 14:37:46 EST 2005
On Feb 15, 2005, at 2:24 PM, Herb Schulz wrote:
> I'm just getting started looking at the Siunits package but not having
> the
> space after \celsius is a classic problem you have to watch for. To
> find the
> end of a command which has a name (rather than a symbol, e.g., \&) TeX
> has
> to look for a non-normal-character, one or more space characters for
> example, WHICH ARE GOBBLED UP. So the space after \celsius is gobbled
> up by
> TeX. To generate a space you can do a couple of things: put a pair of
> braces
> after the command, e.g., \celsius{} and then space, or put a forced
> space
> character after it, e.g., \celsius\ (the `\ ') forces a space there.
> Characters that follow the command besides space act as terminators
> too but
> aren't gobbled up; e.g., \celsius, will have the comma after it. That
> is why
> you don't want to define the macro to force a space afterward;
> \celsius\ ,
> isn't correct. You see this happening all the time with the macro
> \TeX\ and
> its variants; look in many papers about \TeX.
You can also use the xspace package to take care of this "space after a
command" problem. It really helps to not have to remember it every
time.
All the best,
-- Gary
--------------------- Info ---------------------
Mac-TeX Website: http://www.esm.psu.edu/mac-tex/
& FAQ: http://latex.yauh.de/faq/
TeX FAQ: http://www.tex.ac.uk/faq
List Post: <mailto:MacOSX-TeX at email.esm.psu.edu> | 2014-10-20 12:58:56 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8767747282981873, "perplexity": 9901.717736093007}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-42/segments/1413507442900.2/warc/CC-MAIN-20141017005722-00151-ip-10-16-133-185.ec2.internal.warc.gz"} |
http://media.nips.cc/nipsbooks/nipspapers/paper_files/nips30/reviews/2519.html | NIPS 2017
Mon Dec 4th through Sat the 9th, 2017 at Long Beach Convention Center
Paper ID: 2519 Selective Classification for Deep Neural Networks
### Reviewer 1
The paper proposes a practical scheme of adding selective classification capabilities to an existing neural network. The method consists of: 1. Choosing a score function that captures how confident the network is in its prediction, analysed are MC-dropout scores for networks trained with dropout and the maximum softmax score for networks with a softmax output with the second performing empirically better. 2. Defining the desired confidence level and error rate. 3. Running a binomial search to establish a score threshold such that with the desired confidence level the classifier will have an error rate smaller than the specified one on the samples it chooses to classify. The procedure uses an existing bound on the true error rate of a classifier based on a small sample estimate (Lemma 3.1) and uses binomial search with a Bonferroni correction on the confidence level (Algorithm 1) to find the score threshold. Experimental results validate the approach and show good agreement between the algorithm inputs (desired error rate) and observed empirical error rates on a test set. The strong points of the paper are the practical nature of it (with the softmax response score function the procedure can be readily applied to any pretrained neural network) and the ease of specifying the algorithm’s desired confidence level and error rate (which is modeled after ref [5]). While the paper builds on well known concepts, the careful verification of the concepts adds a lot of value. The paper lacks simple baselines, that could showcase the importance of using the binomial search and the bound on the classifier’s error rate. In particular, I would like to know what happens if one chooses the score threshold as the lowest value for which the error rate on a given tuning set is lower than e specified value- would the results be much more different than using the bound from Lemma 3.1? Knowing this baseline would greatly motivate the advanced techniques used in the paper (and would raise my score of this paper). Nitpicks: the Algorithm 1 uses an uninitialized variable r*
### Reviewer 2
Selective classification is the problem of simultaneously choosing which data examples to classify, and subsequently classifying them. Put another way, it’s about giving a classifier the ability to ignore certain data if it’s not confident in its prediction. Previous approaches have focused on assigning a small cost for abstaining. This paper proposes a post-hoc strategy where, if a classifier’s confidence can be accurately gauged, then this confidence is thresholded such that the classifier obtains a guaranteed error rate with high probability. The main novelty with this paper is the proposed SGR algorithm and associated theory. This relies on an ideal confidence function, which is not available in practice, so two methods, SR and MC-dropout are tested. The results are promising, obtaining a low test error with a reasonably high coverage. Getting into specifics: it’s not obvious how you solve Equation (4). I’m assuming it’s a simple line search in 1D, but it would be helpful to be explicit about this. Also, what is the complexity of this whole procedure? It looks like it’s mlog(m)? It’s interesting that mc-dropout performed worse on Imagenet, do you have any intuition as to why this might be the case? It may be helpful to visualize how the confidence functions differ for a given model. I suppose one can easily test both and take the one that works better in practice. As far as I know, there is no explicit validation set for CIFAR-10 and CIFAR-100. They each have 50,000 training points with a separate 10,000-point test-set. Did you split up the test-set into 5,000 points? Or did you use the last batch of the training set for Sm? I think a more proper way to evaluate this would be to use some portion of the last batch of the training sets as validation, and evaluate on the full test set. It would be helpful for you to mention what you did for Imagenet as well; it looks like you split the validation set up into two halves and tested on one half? Why not use the full test set, which I think has 100,000 images? There’s a typo in section 5.3 (mageNet). One point of weakness in the empirical results is that you do not compare with any other approaches, such as those based on assigning a small cost for abstaining. This cost could be tuned to get a desired coverage, or error rate. It’s not clear that a post-hoc approach is obviously better than this approach, although perhaps it is less expensive overall. Overall I like this paper, I think it’s a nice idea that is quite practical, and opens a number of interesting directions for future research.
### Reviewer 3
The paper addresses the problem of constructing a classifier with the reject option that has a desired classification risk and, at the same time, minimizes the probability the "reject option". The authors consider the case when the classifiers and an associate confidence function are both known and the task is to determine a threshold on the confidence that determines whether the classifier prediction is used or rejected. The authors propose an algorithm finding the threshold and they provide a statistical guarantees for the method. Comments: - The authors should provide an exact definition of the task that they attempt to solve by their algorithm. The definition on line 86-88 describes rather the ultimate goal while the algorithm proposed in the paper solves a simpler problem: given $(f,\kappa)$ find a threshold $\theta$ defining $g$ in equation (3) such that (2) holds and the coverage is maximal. - It seems that for a certain values of the input arguments (\delta,r^*,S_m,...) the Algorithm 1 will always return a trivial solution. By trivial solution I mean that the condition on line 10 of the Algorithm 1 is never satisfied and thus all examples will be at the end in the "reject region". It seems to me that for $\hat{r}=0$ (zero trn error) the bound B^* solving equation (4) can be determined analytically as $B^* = 1-(\delta/log_2(m))^{1/m}$. Hence, if we set the desired risk $r^*$ less than the number $B^* = 1-(\delta/log_2(m))^{1/m}$ then the Algorithm 1 will always return a trivial solution. For example, if we set the confidence $\delta=0.001$ (as in the experiments) and the number of training examples is $m=500$ then the minimal bound is $B^*=0.0180$ (1.8%). In turn, setting the desired risk $r^* < 0.018$ will always produce a trivial solution whatever data are used. I think this issue needs to be clarified by the authors. - The experiments should contain a comparison to a simple baseline that anyone would try as the first place. Namely, one can find the threshold directly using the empirical risk $\hat{r}_i$ instead of the sophisticated bound B^*. One would assume that the danger of over-fitting is low (especially for 5000 examples used in experiments) taking into account the simple hypothesis space (i.e. "threshold rules"). Without the comparing to baseline it is hard to judge the practical benefits of the proposed method. - I'm missing a discussion of the difficulties connected to solving the numerical problem (4). E.g. which numerical method is suitable and whether there are numerical issues when evaluating the combinatorial coefficient for large m and j. Typos: - line 80: (f,g) - line 116: B^*(\hat{r},\delta,S_m) - line 221: "mageNet" | 2019-02-17 13:30:18 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.799483597278595, "perplexity": 431.9255120614341}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550247481994.42/warc/CC-MAIN-20190217132048-20190217154048-00574.warc.gz"} |
https://socratic.org/questions/54a8e34d581e2a44346675cd | # Question #675cd
Jan 4, 2015
I assume the question refers to 4 L of gas, and asks for the new volume after the pressure and the temperature are doubled.
This can be solved by using the combined gas law,
$\frac{{P}_{1} {V}_{1}}{T} _ 1 = \frac{{P}_{2} {V}_{2}}{T} _ 2$.
You start with a pressure of ${P}_{1}$ and end with a pressure of ${P}_{2} = 2 \cdot {P}_{1}$. Likewise, the initial temperature is ${T}_{1}$, and the final temperature will be ${T}_{2} = 2 \cdot {T}_{1}$.
So, we can determine ${V}_{2}$ from the combined gas law equation
${V}_{2} = {P}_{1} / {P}_{2} \cdot {T}_{2} / {T}_{1} \cdot {V}_{1} \implies {V}_{2} = {P}_{1} / \left(2 \cdot {P}_{1}\right) \cdot \frac{2 \cdot {T}_{1}}{T} _ 1 \cdot {V}_{1}$
It's evident that ${V}_{2} = {V}_{1}$, since both the pressure and the temperature terms cancel out.
This would have also been the case if the question said 4 moles of gas, since the combined gas law assumes that the number of moles is constant.
You would get the same result, ${V}_{\text{final") = V_("initial}}$.
SInce the question provides a value for ${V}_{1}$, the answer is
${V}_{2} = {V}_{1} = 4$ $\text{L}$. | 2019-08-20 12:57:23 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 12, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8434776067733765, "perplexity": 233.7368727669231}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027315329.55/warc/CC-MAIN-20190820113425-20190820135425-00189.warc.gz"} |
https://socratic.org/questions/how-do-you-write-2-cos300-isin300-in-retangular-form | # How do you write 2(cos300+isin300) in retangular form?
Feb 3, 2015
The answer is: $z = 1 - \sqrt{3} i$.
The rectangular form of a complex number is:
$z = a + i b$,
and we have a number written in trigonometric form, that is:
$z = \rho \left(\sin \theta + i \cos \theta\right)$.
So the real part of the numer is rhosintheta=2cos300°=2*1/2=1 and the imaginary part is 2sin300°=2*(-sqrt3/2)=-sqrt3#.
So:
$z = 1 - \sqrt{3} i$. | 2020-02-19 23:58:26 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 6, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9672076106071472, "perplexity": 1735.9988170939264}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875144429.5/warc/CC-MAIN-20200219214816-20200220004816-00169.warc.gz"} |
https://www.hark.jp/document/1.2.0/hark-document-en/subsec-MultiDownSampler.html | 6.7.5 MultiDownSampler
6.7.5.1 Outline of the node
This node performs downsampling of input signals and outputs their results. The window method is used for low-pass filters and its window function is Kaiser window.
6.7.5.2 Necessary files
No files are required.
6.7.5.3 Usage
When to use
This node is used when the sampling frequency of input signals is not 16 kHz. For the HARK nodes, the default sampling frequency is 16kHz. If, for example, the input signals are 48 kHz, downsampling is required to reduce the sampling frequency to 16 kHz. Note 1 (Range of ADVANCE): To make processing more convenient, it is necessary to limit the parameter settings of input nodes that are connected before nodes such as AudioStreamFromMic and AudioStreamFromWave . Differences in the parameters LENGTH and ADVANCE: OVERLAP = LENGTH - ADVANCE must be sufficiently large. More concretely, the differences must be greater than the low-pass filter length $N$ of this node. Values over 120 are sufficient for the default setting of this node and therefore no problems occur if ADVANCE is more than a quarter of LENGTH. Moreover, it is necessary to satisfy the requirements below. Note 2 (Setting of ADVANCE): The ADVANCE value of this node must be SAMPLING_RATE_IN / SAMPLING_RATE_OUT times as great as the ADVANCE value of the node connected afterward (e.g. GHDSS ). Since this is a specification, its operation is not guaranteed with values other than those above. For example, if ADVANCE = 160 and of SAMPLING_RATE_IN / SAMPLING_RATE_OUT is 3 for the node connected later, it is necessary to set the ADVANCE of this node and that connected before to 480. Note 3 (Requirements for the LENGTH value of the node connected before this node): The LENGTH value of the node connected before this node (e.g. AudioStreamFromMic ) must be SAMPLING_RATE_IN / SAMPLING_RATE_OUT times as great as the ADVANCE value of the node connected afterward (e.g. GHDSS ). For example, if SAMPLING_RATE_IN / SAMPLING_RATE_OUT is 3, and LENGTH is 512 and ADVANCE is 160 for GHDSS , then LENGTH should be 1536 and ADVANCE should be 480 for AudioStreamFromMic .
Typical connection
Examples of typical connections are shown below. This network file reads Wave file inputs, performs downsampling and saves files as Raw files. Wave file input is achieved by connecting Constant , InputStream and AudioStreamFromMic . This is followed by downsampling with MultiDownSampler , with output waveforms saved in SaveRawPCM .
6.7.5.4 Input-output and property of the node
Input
INPUT
: Matrix<float> type. Multichannel speech waveform data (time domain waveform).
Output
OUTPUT
: Matrix<float> type. The multichannel speech waveform data for which downsampling is performed (time domain waveform).
Table 6.65: Parameter list of MultiDownSampler
Parameter name Type Default Value Unit Description ADVANCE 480 [pt] Frame shift length for every iteration in INPUT signals. Since special setting is required, see the parameter description. SAMPLING_RATE_IN 48000 [Hz] Sampling frequency of INPUT signals. SAMPLING_RATE_OUT 16000 [Hz] Sampling frequency of OUTPUT signals. Wp 0.28 [$\frac{\omega }{2\pi }$] Low-pass filter pass band end. Designate normalized frequency [0.0 - 1.0] with INPUT as reference. Ws 0.34 [$\frac{\omega }{2\pi }$] Low-pass filter stopband end. Designate normalized frequency [0.0 - 1.0] with INPUT as reference. As 50 [dB] Minimum attenuation in stopband.
Parameter
Low-pass filters, frequency characteristics of a Kaiser window, are mostly set for the parameters. Figure 6.82 shows the relationships between symbols and filter properties. Note the correspondence when reading them.
: int type. The default value is 480. For processing frames for speech waveforms, designate the shift width on waveforms in sampling numbers. Here, use the values of the nodes connected prior to INPUT. Note: This value must be SAMPLING_RATE_IN / SAMPLING_RATE_OUT times as great as the ADVANCE value set for after OUTPUT.
SAMPLING_RATE_IN
: int type. The default value is 48000. Designate sampling frequency for input waveforms.
SAMPLING_RATE_OUT
: int type. The default value is 16000. Designate sampling frequency for output waveforms. Values that can be used are 1 / integer of SAMPLING_RATE_IN.
Wp
: float type. The default value is 0.28. Designate the low-pass filter pass band end frequency by values of normalized frequency [0.0 - 1.0] with INPUT as reference. When the sampling frequency of inputs is 48000 [Hz] and this value is set to 0.48, gains of low-pass filter begin to decrease from around $48000 * 0.28 = 13440$ [Hz].
Ws
: float type. The default value is 0.34. Designate the low-pass filter stopband end frequency by values of normalized frequency [0.0 - 1.0] with INPUT as reference. When the sampling frequency of inputs is 48000[Hz] and this value is set to 0.38, gains of low-pass tilter begin to be stable from around $48000 * 0.34 = 16320$ [Hz].
As
: float type. The default value is 50. Designate the value indicating the minimum attenuation in stopband in [dB]. When using the default value, the gain of the stopband is around -50 [dB], with the passing band as 0.
When Wp, Ws and As are set at their default values, Wp and Ws will be around the cutoff frequency $Ws$. For example, the accuracy of the frequency response characteristic of Kaiser window will improve. However, the dimensions of the low-pass filter and the processing time will be increased. This relationship is considered a trade off.
6.7.5.5 Details of the node
MultiDownSampler is the node that uses the low-pass filter for band limiting, using the Kaiser window for multichannel signals and downsampling. This node downsamples ${SAMPLING\_ RATE\_ OUT} / {SAMPLING\_ RATE\_ IN}$ after creating / executing an FIR low-pass filter by synthesizing 1) a Kaiser window and 2) ideal low-pass responses.
FIR filter: Filtering with a finite impulse response $h(n)$ is performed based on the equation
$\displaystyle s_{{out}}(t) = \sum _{i = 0}^{N} h(n) s_{{in}}(t-n).$ (122)
Here, $s_{{out}}(t)$ indicates output signals and $s_{{in}}(t)$ indicates input signals. For multichannel signals, the signals of each channel are filtered independently. The same finite impulse response $h(n)$ is used here. Ideal low-pass response: The ideal low-pass response with a cutoff frequency of $\omega _ c$ is obtained using the equation
$\displaystyle H_ i(e^{j\omega }) = \left\{ \begin{array}{cc}1, & |\omega |< \omega _ c \\ 0, & {otherwise} \end{array} \right.$ (123)
This impulse response is expressed as
$\displaystyle h_ i(n) = \frac{\omega _ c}{\pi } \left( \frac{sin(\omega n)}{\omega n} \right),~ ~ -\infty \leq n \leq \infty$ (124)
This impulse response does not satisfy the acausal and bounded input-bounded output (BIBO) stability conditions. It is therefore necessary to cut off the impulse response in the middle to obtain the FIR filter from this ideal filter.
$\displaystyle h(n)= \left\{ \begin{array}{ll} h_ i(n), & |n|\leq \frac{N}{2} \\ 0, & {otherwise} \end{array} \right.$ (125)
Here, $N$ indicates a dimension of the filter. In this filter, cutoff of the impulse response results in ripples in the pass band and stopband. Moreover, the minimum attenuation in stopband $As$ remains around 21dB and sufficient attenuation is not obtained.
Low-pass filter by the window method with Kaiser window: To improve the properties of the above cutoff method, an impulse response, in which the ideal impulse response $h_ i(n)$ is multiplied by the window function $v(n)$, is used instead.
$\displaystyle h(n)= h_ i(n) v(n)$ (126)
Here, the low-pass filter is designed with the Kaiser window. The Kaiser window is defined by the equation
$\displaystyle v(n)= \left\{ \begin{array}{ll} \frac{ I_0 \left( \beta \sqrt {1 - (n N / 2)^2} \right)}{I_0(\beta )}, & -\frac{N}{2} \leq n \leq \frac{N}{2} \\ 0, & {otherwise} \end{array} \right.$ (127)
Here, $\beta$ indicates the parameter determining the shape of the window and $I_0(x)$ indicates the modified Bessel function of 0th order. The Kaiser window is obtained using the equation
$\displaystyle I_0(x)= 1 + \sum _{k=1}^{\infty } \left( \frac{(0.5 x)^ k}{k!} \right)$ (128)
The parameter $\beta$ is determined by the attenuation obtained by the low-pass filter. Here, it is determined by the index,
$\displaystyle \beta = \left\{ \begin{array}{ll} 0.1102 (As - 8.7) & As > 50, \\ 0,5842 (As - 21)^{0.4} + 0.07886 (As - 21) & 21 < As < 50, \\ 0 & As < 21 \end{array} \right.$ (129)
If the cutoff frequency $\omega _ c$ and the filter order have been determined, the low-pass filter can be determined by the window method. The filter order $N$ can be estimated with the minimum attenuation in stopband As and the transition region $\Delta f = (Ws - Wp) / (2\pi )$ as,
$\displaystyle N \approx \frac{As - 7.95}{14.36 \Delta f}$ (130)
Moreover, the cutoff frequency $\omega _ c$ is set to $0.5 (Wp + Ws)$. Downsampling: Downsampling is realized by thinning the sample points of SAMPLING_RATE_IN / SAMPLING_RATE_OUT from the signals that pass the low-pass filter. For example, $48000 / 16000 = 3$ in the default setting; therefore, input samples taken once every three times will be output samples.
6.7.5.6 References:
(1) Author: Translated by P. Vaidyanathan: Akinori Nishihara, Eiji Watanabe, Toshiyuki Yoshida, Nobuhiko Sugino: "Multirate signal processing and filter bank", Science and technology publication, 2001. | 2018-12-12 03:02:44 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5450665950775146, "perplexity": 1815.9656689211831}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376823712.21/warc/CC-MAIN-20181212022517-20181212044017-00167.warc.gz"} |
https://scicomp.stackexchange.com/questions/23311/how-does-mpi-differentiate-between-two-computers | # How does MPI differentiate between two computers?
I am starting to dabble in MPI. I am fairly new to this area and I am currently reading the MPI standard.
I would like to write my first MPI program, a simple hello world program, on a cluster that I have. I am not sure if anyone is familiar with the IBM bladecenter. Supposedly, there is a midplane that connects all of the blades together (if someone could confirm/deny this in their experience, that would be great). I was going to run a simple program using the send and receive C++ commands. I think that I will run the example program in the point-to-point communication section of the standard. I have typed up the program below.
My main question is this, say I have two computers (computer A and computer B) that are connected to each other in some way (take your pick but in this case, it would be the bladecenter so they are connected via a midplane). Through the MPI standard, how does computer A communicate with computer B through the connection? Also, if I run my program on computer A, will computer B be needing to run the same program? How will MPI know that the rank is on computer B and not on some other thread of computer A? Does MPI_Comm_rank get the rank for every single processor/core that it can see?
Given the example program that I am using, is that meant for 2 processes on the same computer or, would this work for 2 computers? If not, then how would I modify the program to work for 2 computers or n number of computers?
#include "mpi.h"
main(int argc, char **argv)
{
char message{20];
int myRank;
MPI_Status status;
MPI_Init(&argc, &argv);
MPI_Comm_rank(MPI_COMM_WORLD, &myRank);
if(myRank == 0)
{
strcpy(message, "Hello, there");
MPI_SEND(message, strlen(message) + 1, MPI_CHAR, 1, 99, MPI_COMM_WORLD);
}
else if(myRank == 1)
{
MPI_Recv(message, 20, MPI_CHAR, 0, 99, MPI_COMM_WORLD, &status);
printf("recieved: %s:\n", message);
}
MPI_Finalize();
}
I apologize if these questions are obvious. Again, this is my first time dealing with MPI and any help would be greatly appreciated, thank you
• Once you follow the startup procedures to launch on multiple computers (commands like mpirun or mpiexec are typically used, but check your library's instructions for launch), you will find that both print statements are executed as expected. There's a typo in your declaration of message that will prevent this code from compiling, BTW. – Bill Barth Mar 9 '16 at 0:35 | 2020-01-21 00:08:33 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.292549192905426, "perplexity": 1151.3768184336227}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250601040.47/warc/CC-MAIN-20200120224950-20200121013950-00031.warc.gz"} |
http://mathhelpforum.com/pre-calculus/50732-function-transformation.html | 1. ## function transformation
guys, i really need help on this! so i was given a equation as
y=3/4[ -2/5 (x+3)]^3 + 1
so far, i understand that the parent function is y= x^3
and i understand that the ys are all ways outside which is (vertical)
and the xs are always inside (horizontal)
i don't understand what do u do with the y-values mb and the x-value mb , which is the 3/4 = y and the -2/5 = x , overall i am really confuse on how to graph it.. have any tips on what to do, or a webpage to show me? thanks!!!!!
2. Originally Posted by lickman
guys, i really need help on this! so i was given a equation as
y=3/4[ -2/5 (x+3)]^3 + 1
so far, i understand that the parent function is y= x^3
and i understand that the ys are all ways outside which is (vertical)
and the xs are always inside (horizontal)
i don't understand what do u do with the y-values mb and the x-value mb , which is the 3/4 = y and the -2/5 = x , overall i am really confuse on how to graph it.. have any tips on what to do, or a webpage to show me? thanks!!!!!
It has the form $y = g(x) = a f(-b[x - c]) + d$. You're expected to be familiar with the transformations associated with a, b, c and d.
In your problem $f(x) = x^3$, $a = \frac{3}{4}$, $b = \frac{2}{5}$, c = -3 and d = 1. The negative means there's a reflection in the y-axis.
3. hi but i understand u, but i do not know how to plot a and b. so what do u do? is a and b the points?
4. ## A really serious problem 101
So far , i understand on the vertical and horizontal transformation, and i also its a form of y = a f [ k ( x - h)] + q and i understand that its a y= x^3 (parent function)
I understand that a is the y-mb value and k is the x- mb value but i do not know to graph them or what.
Can someone PLEASE, give me a detail explaination on this, and how to graph it. I know this is alot of work, but please for the sake of math!
5. Originally Posted by lickman
hi but i understand u, but i do not know how to plot a and b. so what do u do? is a and b the points?
No they're not. The values of a and b, and c and d tell you what transformations have been applied to the 'parent' function.
You draw your graph by drawing y = x^3 and then applying the necessary transformations to this graph (translations, dilations, reflections). | 2013-12-10 17:03:05 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6180387735366821, "perplexity": 560.5919939027332}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-48/segments/1386164022328/warc/CC-MAIN-20131204133342-00059-ip-10-33-133-15.ec2.internal.warc.gz"} |
https://zbmath.org/?q=an%3A0566.65045 | zbMATH — the first resource for mathematics
Nonlinear successive over-relaxation. (English) Zbl 0566.65045
Let $$\Phi$$ be a real strictly convex functional defined and twice continuously differentiable on a convex domain in $${\mathbb{R}}^ n$$. To find its minimum the authors solve a system of nonlinear equations $$F(x)=0$$ in $${\mathbb{R}}^ n$$, where F denotes grad $$\Phi$$. They present two theorems giving sufficient conditions of convergence for: (i) a nonlinear analogue of the Gauss-Seidel method for positive-definite matrices, (ii) a nonlinear successive overrelaxation method, used for such a system of equations. The theorems are parallel to results of Schechter (1962, 1968) but give more general sufficient conditions and may be applied for a more general class of functionals whose Hessian matrix may be singular.
Reviewer: S.Ząbek
MSC:
65K05 Numerical mathematical programming methods 90C25 Convex programming 65H10 Numerical computation of solutions to systems of equations
Full Text:
References:
[1] Concus, P.: Numerical solution of the minimal surface equation. Math. Comput.21, 340-350 (1967) · Zbl 0189.16605 [2] Concus, P., Golub, G.H., O’Leary, D.P.: Numerical solution of nonlinear elliptic partial differential equations. Computing19, 321-329 (1978) · Zbl 0385.65048 [3] Decker, D.W., Kelley, C.T.: Newton’s method at singular points. SIAM J. Numer. Anal.17, 66-70 (1980) · Zbl 0428.65037 [4] Greenspan, D.: On approximating extremals of functionals. ICC Bulletin4, 99-120 (1965) [5] Keller, H.B.: Numerical solution of bifurcation and nonlinear eigenvalue problems. In: Application of Bifurcation Theory. Rabinowitz, P. H. (ed.) New York, Academic Press 1977 · Zbl 0581.65043 [6] Lieberstein, H.M.: Overrelaxation for nonlinear elliptic partial differential equations. MRC Tech. Report #80 (1959) · Zbl 0085.08601 [7] Rall, L.B.: Convergence of the Newton process to multiple solutions. Numer. Math.9, 23-37 (1966) · Zbl 0163.38702 [8] Reddien, G.W.: On Newton’s method for singular problems. SIAM J. Numer. Anal.15, 993-996 (1978) · Zbl 0397.65042 [9] Schechter, S.: Iteration methods for nonlinear problems. Trans. Amer. Math. Soc.104, 179-189 (1962) · Zbl 0106.31801 [10] Schechter, S.: Relaxation methods for convex problems. SIAM J. Numer. Anal.5, 601-612 (1968) · Zbl 0179.22701 [11] Warga, J.: Minimizing certain convex functions. J. Soc. Indus. Appl. Math.11, 588-593 (1963) · Zbl 0128.05801 [12] Young, D.M.: Iterative solutions of large linear systems. New York, Academic Press 1971 · Zbl 0231.65034
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching. | 2021-08-01 03:32:25 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5485116243362427, "perplexity": 2910.40109851182}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046154158.4/warc/CC-MAIN-20210801030158-20210801060158-00061.warc.gz"} |
http://connection.ebscohost.com/c/articles/21073891/q-curvature-flow-4-manifolds | TITLE
# Q-curvature flow on 4-manifolds
AUTHOR(S)
Baird, Paul; Fardoun, Ali; Regbaoui, Rachid
PUB. DATE
September 2006
SOURCE
Calculus of Variations & Partial Differential Equations;Sep2006, Vol. 27 Issue 1, p75
SOURCE TYPE
DOC. TYPE
Article
ABSTRACT
We formulate an appropriate gradient flow in order to study the evolution of the Q-curvature to a prescribed function on a 4-manifold. For a class of prescribed functions, we show convergence and describe the asymptotic behaviour at infinity.
ACCESSION #
21073891
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This paper contains a presentation of the author's main results obtained in constructing an algebraic theory of branched coverings over manifolds. The main inspiration that led the author to deal with this topic is the well-known result on represent ability of each compact orientable manifold as...
• Some slant submanifolds ofS-manifolds. Carriazo, Alfonso; Fernández, Luis M.; Hans-Uber, María Belén // Acta Mathematica Hungarica;Jun2005, Vol. 107 Issue 4, p267
We study some special types of slant submanifolds ofS-manifolds related to the second fundamental form of the immersion: totallyf-geodesic andf-umbilical, pseudo-umbilical and austere submanifolds. We also give several examples of such submanifolds.
• On the Geometry of Vectorgrams for a Certain Class of Nonlinear Smooth Control Systems. Vakhrameev, S. A. // Journal of Mathematical Sciences;Jul2004, Vol. 122 Issue 1, p2916
Studies a certain particular class of nonlinear smooth control systems for which an analog of the R.V. Gamkrelidze theorem on the finiteness of the number of switchings holds. Replacement of the polyhedron by a so-called manifest with corners; Proof that the vectogram of the system considered is...
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https://physics.stackexchange.com/questions/199524/evolution-of-a-state-in-the-heisenberg-picture | # Evolution of a 'state' in the Heisenberg picture
Suppose that we have a Hamiltonian, $\hat{H}$, and an operator $\hat{A}$ which satisfies the Heisenberg equation$^{[a]}$ $$i \frac{d}{dt} \hat{A} = [\hat{A},\hat{H}].$$
Can we create a 'state' by acting on, for example, the vacuum with the operator $\hat{A}$? $$\lvert \psi \rangle = \hat{A}(t) \lvert 0 \rangle$$
If we investigate the time evolution of this state, then $$i \frac{d}{dt} \lvert \psi \rangle = i \frac{d}{dt} \hat{A}(t) \lvert 0 \rangle = [\hat{A},\hat{H}] \ \lvert 0 \rangle.$$ If the state $\lvert 0 \rangle$ is such that it is annihilated by the Hamiltonian, then $$[\hat{A},\hat{H}] \ \lvert 0 \rangle = - \hat{H} \hat{A} \lvert 0 \rangle,$$ so $$i \frac{d}{dt} \lvert \psi \rangle = - \hat{H} \lvert \psi \rangle,$$ which has the opposite sign from what I would have expected. Is there a problem with constructing/manipulating 'states' this way in the Heisenberg picture? Does the reversed sign from the ordinary Schrodinger equation make sense in this case?
$[a]$: Where we write $\hat{H}$ we mean what would normally be written $\hat{H}/\hbar$. In other words, $\hat{H}$ here has dimensions of frequency.
• Assume $H$ is time-independent for simplicity, then $A(t)=e^{iHt}Ae^{-iHt}$ where $A\equiv A(0)$. So $A(t)|0\rangle = e^{iHt}A|0\rangle$ assuming $H|0\rangle=0$. It is exactly the state $A|0\rangle$ evolved back in time, and you got the right sign from your calculation. But there is no reason to expect some random state you construct to satisfy the Schrodinger equation. – Meng Cheng Aug 9 '15 at 23:56
• You are mixing up the Schrodinger and Heisenberg pictures of quantum mechanics. The first equation is true only in the Heisenberg picture where states are independent of time. The Schrodinger equation however holds in Schrodinger picture where states evolve in time according to the equation. If you are careful with the picture you are working in, everything is consistent. – Prahar Aug 10 '15 at 5:44 | 2020-07-09 18:48:05 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9222712516784668, "perplexity": 186.86444968894423}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655900614.47/warc/CC-MAIN-20200709162634-20200709192634-00160.warc.gz"} |
https://zbmath.org/?q=an%3A0712.15009 | # zbMATH — the first resource for mathematics
On the symmetric solutions of linear matrix equations. (English) Zbl 0712.15009
Necessary and sufficient conditions are given for the existence of symmetric solutions of the matrix equations $$AX=C$$ and $$AXB=C$$ on the real field, in terms of the singular value decomposition of A and the generalized singular value decomposition of the pair $$(A,B^ T)$$, respectively. Expressions for the general solution are provided for each case.
Reviewer: M.E.Sezer
##### MSC:
15A24 Matrix equations and identities 15A18 Eigenvalues, singular values, and eigenvectors
Full Text:
##### References:
[1] Vetter, W.J., Vector structures and solutions of linear matrix equation, Linear algebra appl., 10, 181-188, (1975) · Zbl 0307.15003 [2] Magnus, J.R.; Neudecker, H., The elimination matrix: some lemmas and applications, SIAM J. algebraic discrete methods, 1, 422-429, (1980) · Zbl 0497.15014 [3] Henk Don, F.J., On the symmetric solutions of a linear matrix equation, Linear algebra appl., 93, 1-7, (1987) · Zbl 0622.15001 [4] Golub, G.H.; Van Loan, C.F., Matrix computations, (1983), Johns Hopkins U.P Baltimore · Zbl 0559.65011 [5] Paige, C.C.; Saunders, M.A., Towards a generalized singular value decomposition, SIAM J. numer. anal., 18, 398-405, (1981) · Zbl 0471.65018 [6] Stewart, G.W., Computing the CS-decomposition of a partitioned orthogonal matrix, Numer. math., 40, 297-306, (1982) · Zbl 0516.65016
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching. | 2021-10-20 18:59:46 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7431138157844543, "perplexity": 1830.1507070648586}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585348.66/warc/CC-MAIN-20211020183354-20211020213354-00307.warc.gz"} |
https://socratic.org/questions/58f3925911ef6b7890d0548d | # Which is the unsaturated acid: "A. diacetic acid; B." H_3C(CH_2)_14CO_2H; "C. oleic acid; D. arachidic acid?"
Only $C .$ is an unsaturated acid.
An unsaturated carboxylic acids contains only the ${1}^{\circ}$ of unsaturation. For instances, $\text{acetic acid}$, $\text{H"_3"CC(=O)OH}$, has formula ${C}_{2} {H}_{4} {O}_{2}$, i.e. ${1}^{\circ}$ of unsaturation.
Both $\text{palmitic}$, and $\text{butyric acids}$ are saturated by this definition. $\text{Palmitic acid}$, i.e. palm oil, is the typical saturated fatty acid, and this is the food component that nutritionists encourage us to avoid.
On the other hand $\text{oleic acid}$, ${C}_{18} {H}_{34} {O}_{2}$, is the typical UNSATURATED acid (and cis with respect to the necessary double bond), that is found in olive oil, and which nutritionists promote. | 2021-09-28 16:42:38 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 11, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8764050602912903, "perplexity": 9031.65498317375}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780060877.21/warc/CC-MAIN-20210928153533-20210928183533-00332.warc.gz"} |