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https://unimath.github.io/agda-unimath/foundation.mere-equivalences.html | 1,726,830,345,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652246.93/warc/CC-MAIN-20240920090502-20240920120502-00786.warc.gz | 534,377,967 | 41,379 | # Mere equivalences
Content created by Fredrik Bakke, Egbert Rijke, Jonathan Prieto-Cubides, Elisabeth Stenholm, Julian KG, fernabnor and louismntnu.
Created on 2022-02-12.
module foundation.mere-equivalences where
Imports
open import foundation.binary-relations
open import foundation.decidable-equality
open import foundation.functoriality-propositional-truncation
open import foundation.mere-equality
open import foundation.propositional-truncations
open import foundation.univalence
open import foundation.universe-levels
open import foundation-core.equivalences
open import foundation-core.propositions
open import foundation-core.sets
open import foundation-core.truncated-types
open import foundation-core.truncation-levels
## Idea
Two types X and Y are said to be merely equivalent, if there exists an equivalence from X to Y. Propositional truncations are used to express mere equivalence.
## Definition
mere-equiv-Prop :
{l1 l2 : Level} → UU l1 → UU l2 → Prop (l1 ⊔ l2)
mere-equiv-Prop X Y = trunc-Prop (X ≃ Y)
mere-equiv :
{l1 l2 : Level} → UU l1 → UU l2 → UU (l1 ⊔ l2)
mere-equiv X Y = type-Prop (mere-equiv-Prop X Y)
abstract
is-prop-mere-equiv :
{l1 l2 : Level} (X : UU l1) (Y : UU l2) → is-prop (mere-equiv X Y)
is-prop-mere-equiv X Y = is-prop-type-Prop (mere-equiv-Prop X Y)
## Properties
### Mere equivalence is reflexive
abstract
refl-mere-equiv : {l1 : Level} → is-reflexive (mere-equiv {l1})
refl-mere-equiv X = unit-trunc-Prop id-equiv
### Mere equivalence is symmetric
abstract
symmetric-mere-equiv :
{l1 l2 : Level} (X : UU l1) (Y : UU l2) → mere-equiv X Y → mere-equiv Y X
symmetric-mere-equiv X Y =
map-universal-property-trunc-Prop
( mere-equiv-Prop Y X)
( λ e → unit-trunc-Prop (inv-equiv e))
### Mere equivalence is transitive
abstract
transitive-mere-equiv :
{l1 l2 l3 : Level} (X : UU l1) (Y : UU l2) (Z : UU l3) →
mere-equiv Y Z → mere-equiv X Y → mere-equiv X Z
transitive-mere-equiv X Y Z f e =
apply-universal-property-trunc-Prop e
( mere-equiv-Prop X Z)
( λ e' →
apply-universal-property-trunc-Prop f
( mere-equiv-Prop X Z)
( λ f' → unit-trunc-Prop (f' ∘e e')))
### Truncated types are closed under mere equivalence
module _
{l1 l2 : Level} {X : UU l1} {Y : UU l2}
where
is-trunc-mere-equiv : (k : 𝕋) → mere-equiv X Y → is-trunc k Y → is-trunc k X
is-trunc-mere-equiv k e H =
apply-universal-property-trunc-Prop
( e)
( is-trunc-Prop k X)
( λ f → is-trunc-equiv k Y f H)
is-trunc-mere-equiv' : (k : 𝕋) → mere-equiv X Y → is-trunc k X → is-trunc k Y
is-trunc-mere-equiv' k e H =
apply-universal-property-trunc-Prop
( e)
( is-trunc-Prop k Y)
( λ f → is-trunc-equiv' k X f H)
### Sets are closed under mere equivalence
module _
{l1 l2 : Level} {X : UU l1} {Y : UU l2}
where
is-set-mere-equiv : mere-equiv X Y → is-set Y → is-set X
is-set-mere-equiv = is-trunc-mere-equiv zero-𝕋
is-set-mere-equiv' : mere-equiv X Y → is-set X → is-set Y
is-set-mere-equiv' = is-trunc-mere-equiv' zero-𝕋
### Types with decidable equality are closed under mere equivalences
module _
{l1 l2 : Level} {X : UU l1} {Y : UU l2}
where
has-decidable-equality-mere-equiv :
mere-equiv X Y → has-decidable-equality Y → has-decidable-equality X
has-decidable-equality-mere-equiv e d =
apply-universal-property-trunc-Prop e
( has-decidable-equality-Prop X)
( λ f → has-decidable-equality-equiv f d)
has-decidable-equality-mere-equiv' :
mere-equiv X Y → has-decidable-equality X → has-decidable-equality Y
has-decidable-equality-mere-equiv' e d =
apply-universal-property-trunc-Prop e
( has-decidable-equality-Prop Y)
( λ f → has-decidable-equality-equiv' f d)
### Mere equivalence implies mere equality
abstract
mere-eq-mere-equiv :
{l : Level} {A B : UU l} → mere-equiv A B → mere-eq A B
mere-eq-mere-equiv = map-trunc-Prop eq-equiv | 1,200 | 3,773 | {"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} | 2.953125 | 3 | CC-MAIN-2024-38 | latest | en | 0.419439 |
https://webot.org/info/en/?search=Fourier_series | 1,670,078,739,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710933.89/warc/CC-MAIN-20221203143925-20221203173925-00577.warc.gz | 658,100,168 | 88,407 | # Fourier series Information
https://en.wikipedia.org/wiki/Fourier_series
A Fourier series ( [1]) is a summation of harmonically related sinusoidal functions, also known as components or harmonics. The result of the summation is a periodic function whose functional form is determined by the choices of cycle length (or period), the number of components, and their amplitudes and phase parameters. With appropriate choices, one cycle (or period) of the summation can be made to approximate an arbitrary function in that interval (or the entire function if it too is periodic). The number of components is theoretically infinite, in which case the other parameters can be chosen to cause the series to converge to almost any well behaved periodic function (see Pathological and Dirichlet conditions). The components of a particular function are determined by analysis techniques described in this article. Sometimes the components are known first, and the unknown function is synthesized [A] by a Fourier series. Such is the case of a discrete-time Fourier transform.
Convergence of Fourier series means that as more and more components from the series are summed, each successive partial Fourier series sum will better approximate the function, and will equal the function with a potentially infinite number of components. The mathematical proofs for this may be collectively referred to as the Fourier Theorem (see § Convergence). The figures below illustrate some partial Fourier series results for the components of a square wave.
Another analysis technique (not covered here), suitable for both periodic and non-periodic functions, is the Fourier transform, which provides a frequency-continuum of component information. But when applied to a periodic function all components have zero amplitude, except at the harmonic frequencies. The inverse Fourier transform is a synthesis process (like the Fourier series), which converts the component information (often known as the frequency domain representation) back into its time domain representation.
Since Fourier's time, many different approaches to defining and understanding the concept of Fourier series have been discovered, all of which are consistent with one another, but each of which emphasizes different aspects of the topic. Some of the more powerful and elegant approaches are based on mathematical ideas and tools that were not available in Fourier's time. Fourier originally defined the Fourier series for real-valued functions of real arguments, and used the sine and cosine functions as the basis set for the decomposition. Many other Fourier-related transforms have since been defined, extending his initial idea to many applications and birthing an area of mathematics called Fourier analysis.
## Analysis process
This section describes the analysis process that derives the parameters of a Fourier series that approximates a known function, ${\displaystyle s(x).}$ An example of synthesizing an unknown function from known parameters is discrete-time Fourier transform.
### Common forms
The Fourier series can be represented in different forms. The amplitude-phase form, sine-cosine form, and exponential form are commonly used and are expressed here for a real-valued function ${\displaystyle s(x)}$. (See § Complex-valued functions and § Other common notations for alternative forms).
The number of terms summed, ${\displaystyle N}$, is a potentially infinite integer. Even so, the series might not converge or exactly equate to ${\displaystyle s(x)}$ at all values of ${\displaystyle x}$ ( such as a single-point discontinuity) in the analysis interval. For the well-behaved functions typical of physical processes, equality is customarily assumed, and the Dirichlet conditions provide sufficient conditions.
The integer index, ${\displaystyle n}$, is also the number of cycles the ${\displaystyle n^{\text{th}}}$ harmonic makes in the function's period ${\displaystyle P}$. [B] Therefore:
• The ${\displaystyle n^{\text{th}}}$ harmonic's wavelength is ${\displaystyle {\tfrac {P}{n}}}$ and in units of ${\displaystyle x}$.
• The ${\displaystyle n^{\text{th}}}$ harmonic's frequency is ${\displaystyle {\tfrac {n}{P}}}$ and in reciprocal units of ${\displaystyle x}$.
Fig 1. The top graph shows a non-periodic function s(x) in blue defined only over the red interval from 0 to P. The function can be analyzed over this interval to produce the Fourier series in the bottom graph. The Fourier series is always a periodic function, even if original function s(x) wasn't.
#### Amplitude-phase form
The Fourier series in amplitude- phase form is:
Fourier series, amplitude-phase form
${\displaystyle s_{\scriptscriptstyle N}(x)={\frac {A_{0}}{2}}+\sum _{n=1}^{N}A_{n}\cdot \cos \left({\tfrac {2\pi }{P}}nx-\varphi _{n}\right)}$
(Eq.1)
• Its ${\displaystyle n^{\text{th}}}$ harmonic is ${\displaystyle A_{n}\cdot \cos \left({\tfrac {2\pi }{P}}nx-\varphi _{n}\right)}$.
• ${\displaystyle A_{n}}$ is the ${\displaystyle n^{\text{th}}}$ harmonic's amplitude and ${\displaystyle \varphi _{n}}$ is its phase shift.
• The fundamental frequency of ${\displaystyle s_{\scriptscriptstyle N}(x)}$ is the term for when ${\displaystyle n}$ equals 1, and can be referred to as the ${\displaystyle 1^{\text{st}}}$ harmonic.
• ${\displaystyle {\tfrac {A_{o}}{2}}}$ is sometimes called the ${\displaystyle 0^{\text{th}}}$ harmonic or DC component. It is the mean value of ${\displaystyle s(x)}$.
Clearly Eq.1 can represent functions that are just a sum of one or more of the harmonic frequencies. The remarkable thing, for those not yet familiar with this concept, is that it can also represent the intermediate frequencies and/or non-sinusoidal functions because of the potentially infinite number of terms (${\displaystyle N}$).
Fig 2. The blue curve is the cross-correlation of a square wave and a cosine function, as the phase lag of the cosine varies over one cycle. The amplitude and phase lag at the maximum value are the polar coordinates of one harmonic in the Fourier series expansion of the square wave. The corresponding Cartesian coordinates can be determined by evaluating the cross-correlation at just two phase lags separated by 90º.
The coefficients ${\displaystyle A_{n}}$ and ${\displaystyle \varphi _{n}}$ can be understood and derived in terms of the cross-correlation between ${\displaystyle s(x)}$ and a sinusoid at frequency ${\displaystyle {\tfrac {n}{P}}}$. For a general frequency ${\displaystyle f,}$ and an analysis interval ${\displaystyle [x_{0},x_{0}+P],}$ the cross-correlation function:
${\displaystyle \mathrm {X} _{f}(\tau )={\tfrac {2}{P}}\int _{x_{0}}^{x_{0}+P}s(x)\cdot \cos \left(2\pi f(x-\tau )\right)\,dx;\quad \tau \in \left[0,{\tfrac {2\pi }{f}}\right]}$
(Eq.2)
is essentially a matched filter, with template ${\displaystyle \cos(2\pi fx)}$. [C] The maximum of ${\displaystyle \mathrm {X} _{f}(\tau )}$ is a measure of the amplitude ${\displaystyle (A)}$ of frequency ${\displaystyle f}$ in the function ${\displaystyle s(x)}$, and the value of ${\displaystyle \tau }$ at the maximum determines the phase ${\displaystyle (\varphi )}$ of that frequency. Figure 2 is an example, where ${\displaystyle s(x)}$ is a square wave (not shown), and frequency ${\displaystyle f}$ is the ${\displaystyle 4^{\text{th}}}$ harmonic. It is also an example of deriving the maximum from just two samples, instead of searching the entire function. That is made possible by a trigonometric identity:
Equivalence of polar and Cartesian forms
${\displaystyle \cos \left({\tfrac {2\pi }{P}}nx-\varphi _{n}\right)\ \equiv \ \cos(\varphi _{n})\cdot \cos \left({\tfrac {2\pi }{P}}nx\right)+\sin(\varphi _{n})\cdot \sin \left({\tfrac {2\pi }{P}}nx\right)}$
(Eq.3)
Substituting this into Eq.2 gives:
{\displaystyle {\begin{aligned}\mathrm {X} _{n}(\varphi )&={\tfrac {2}{P}}\int _{P}s(x)\cdot \cos \left({\tfrac {2\pi }{P}}nx-\varphi \right)\,dx;\quad \varphi \in [0,2\pi ]\\&=\cos(\varphi )\cdot \underbrace {{\tfrac {2}{P}}\int _{P}s(x)\cdot \cos \left({\tfrac {2\pi }{P}}nx\right)\,dx} _{\triangleq \ a_{n}}+\sin(\varphi )\cdot \underbrace {{\tfrac {2}{P}}\int _{P}s(x)\cdot \sin \left({\tfrac {2\pi }{P}}nx\right)\,dx} _{\triangleq \ b_{n}}\\&=\cos(\varphi )\cdot a_{n}+\sin(\varphi )\cdot b_{n}\end{aligned}}}
which introduces the definitions of ${\displaystyle a_{n}}$ and ${\displaystyle b_{n}}$. [2] And we note for later reference that ${\displaystyle a_{0}}$ and ${\displaystyle b_{0}}$ can be simplified:
${\displaystyle a_{0}={\tfrac {2}{P}}\int _{P}s(x)dx~,\quad b_{0}=0~.}$
The derivative of ${\displaystyle \mathrm {X} _{n}(\varphi )}$ is zero at the phase of maximum correlation.
${\displaystyle \mathrm {X} '_{n}(\varphi _{n})=\sin(\varphi _{n})\cdot a_{n}-\cos(\varphi _{n})\cdot b_{n}=0\quad \longrightarrow \quad \tan(\varphi _{n})={\frac {b_{n}}{a_{n}}}\quad \longrightarrow \quad \varphi _{n}=\arctan(b_{n},a_{n})}$
And the correlation peak value is:
{\displaystyle {\begin{aligned}A_{n}\triangleq \mathrm {X} _{n}(\varphi _{n})\ &=\cos(\varphi _{n})\cdot a_{n}+\sin(\varphi _{n})\cdot b_{n}\\&={\frac {a_{n}}{\sqrt {a_{n}^{2}+b_{n}^{2}}}}\cdot a_{n}+{\frac {b_{n}}{\sqrt {a_{n}^{2}+b_{n}^{2}}}}\cdot b_{n}={\frac {a_{n}^{2}+b_{n}^{2}}{\sqrt {a_{n}^{2}+b_{n}^{2}}}}&={\sqrt {a_{n}^{2}+b_{n}^{2}}}.\end{aligned}}}
Therefore ${\displaystyle a_{n}}$ and ${\displaystyle b_{n}}$ are the Cartesian coordinates of a vector with polar coordinates ${\displaystyle A_{n}}$ and ${\displaystyle \varphi _{n}.}$
#### Sine-cosine form
Substituting Eq.3 into Eq.1 gives:
${\displaystyle {\displaystyle s_{\scriptscriptstyle N}(x)={\frac {A_{0}}{2}}+\sum _{n=1}^{N}\left[A_{n}\cos(\varphi _{n})\cdot \cos \left({\tfrac {2\pi }{P}}nx\right)+A_{n}\sin(\varphi _{n})\cdot \sin \left({\tfrac {2\pi }{P}}nx\right)\right]}}$
In terms of the readily computed quantities, ${\displaystyle a_{n}}$ and ${\displaystyle b_{n}}$, recall that:
${\displaystyle \cos(\varphi _{n})=a_{n}/A_{n}}$
${\displaystyle \sin(\varphi _{n})=b_{n}/A_{n}}$
${\displaystyle A_{0}={\sqrt {a_{0}^{2}+b_{0}^{2}}}={\sqrt {a_{0}^{2}}}=a_{0}}$
Therefore an alternative form of the Fourier series, using the Cartesian coordinates, is the sine-cosine form: [D]
Fourier series, sine-cosine form
${\displaystyle s_{\scriptscriptstyle N}(x)={\frac {a_{0}}{2}}+\sum _{n=1}^{N}\left(a_{n}\cos \left({\tfrac {2\pi }{P}}nx\right)+b_{n}\sin \left({\tfrac {2\pi }{P}}nx\right)\right)}$
(Eq.4)
#### Exponential form
Another applicable identity is Euler's formula:
{\displaystyle {\begin{aligned}\cos \left({\tfrac {2\pi }{P}}nx-\varphi _{n}\right)&{}\equiv {\tfrac {1}{2}}e^{i\left(2\pi nx/P-\varphi _{n}\right)}+{\tfrac {1}{2}}e^{-i\left(2\pi nx/P-\varphi _{n}\right)}\\[6pt]&=\left({\tfrac {1}{2}}e^{-i\varphi _{n}}\right)\cdot e^{i2\pi (+n)x/P}+\left({\tfrac {1}{2}}e^{-i\varphi _{n}}\right)^{*}\cdot e^{i2\pi (-n)x/P}\end{aligned}}}
(Note: the ∗ denotes complex conjugation.)
Therefore, with definitions:
${\displaystyle c_{n}\triangleq \left\{{\begin{array}{lll}A_{0}/2&=a_{0}/2,\quad &n=0\\{\tfrac {A_{n}}{2}}e^{-i\varphi _{n}}&={\tfrac {1}{2}}(a_{n}-ib_{n}),\quad &n>0\\c_{|n|}^{*},\quad &&n<0\end{array}}\right\}={\tfrac {1}{P}}\int _{P}s(x)\cdot e^{-i2\pi nx/P}\,dx,}$
the final result is:
Fourier series, exponential form
${\displaystyle s_{_{N}}(x)=\sum _{n=-N}^{N}c_{n}\cdot e^{i2\pi nx/P}}$
(Eq.5)
This is the customary form for generalizing to § Complex-valued functions. Negative values of ${\displaystyle n}$ correspond to negative frequency (explained in Fourier transform § Use of complex sinusoids to represent real sinusoids).
### Example
Plot of the sawtooth wave, a periodic continuation of the linear function ${\displaystyle s(x)=x/\pi }$ on the interval ${\displaystyle (-\pi ,\pi ]}$
Animated plot of the first five successive partial Fourier series
Consider a sawtooth function:
${\displaystyle s(x)={\frac {x}{\pi }},\quad \mathrm {for} -\pi
${\displaystyle s(x+2\pi k)=s(x),\quad \mathrm {for} -\pi
In this case, the Fourier coefficients are given by
{\displaystyle {\begin{aligned}a_{n}&={\frac {1}{\pi }}\int _{-\pi }^{\pi }s(x)\cos(nx)\,dx=0,\quad n\geq 0.\\[4pt]b_{n}&={\frac {1}{\pi }}\int _{-\pi }^{\pi }s(x)\sin(nx)\,dx\\[4pt]&=-{\frac {2}{\pi n}}\cos(n\pi )+{\frac {2}{\pi ^{2}n^{2}}}\sin(n\pi )\\[4pt]&={\frac {2\,(-1)^{n+1}}{\pi n}},\quad n\geq 1.\end{aligned}}}
It can be shown that the Fourier series converges to ${\displaystyle s(x)}$ at every point ${\displaystyle x}$ where ${\displaystyle s}$ is differentiable, and therefore:
{\displaystyle {\begin{aligned}s(x)&={\frac {a_{0}}{2}}+\sum _{n=1}^{\infty }\left[a_{n}\cos \left(nx\right)+b_{n}\sin \left(nx\right)\right]\\[4pt]&={\frac {2}{\pi }}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n}}\sin(nx),\quad \mathrm {for} \quad x-\pi \notin 2\pi \mathbb {Z} .\end{aligned}}}
(Eq.6)
When ${\displaystyle x=\pi }$, the Fourier series converges to 0, which is the half-sum of the left- and right-limit of s at ${\displaystyle x=\pi }$. This is a particular instance of the Dirichlet theorem for Fourier series.
This example leads to a solution of the Basel problem.
### Convergence
A proof that a Fourier series is a valid representation of any periodic function (that satisfies the Dirichlet conditions) is overviewed in § Fourier theorem proving convergence of Fourier series.
In engineering applications, the Fourier series is generally presumed to converge almost everywhere (the exceptions being at discrete discontinuities) since the functions encountered in engineering are better-behaved than the functions that mathematicians can provide as counter-examples to this presumption. In particular, if ${\displaystyle s}$ is continuous and the derivative of ${\displaystyle s(x)}$ (which may not exist everywhere) is square integrable, then the Fourier series of ${\displaystyle s}$ converges absolutely and uniformly to ${\displaystyle s(x)}$. [3] If a function is square-integrable on the interval ${\displaystyle [x_{0},x_{0}+P]}$, then the Fourier series converges to the function at almost every point. It is possible to define Fourier coefficients for more general functions or distributions, in such cases convergence in norm or weak convergence is usually of interest.
### Complex-valued functions
If ${\displaystyle s(x)}$ is a complex-valued function of a real variable ${\displaystyle x,}$ both components (real and imaginary part) are real-valued functions that can be represented by a Fourier series. The two sets of coefficients and the partial sum are given by:
${\displaystyle c_{_{Rn}}={\frac {1}{P}}\int _{P}\operatorname {Re} \{s(x)\}\cdot e^{-i{\tfrac {2\pi }{P}}nx}\ dx}$ and ${\displaystyle c_{_{In}}={\frac {1}{P}}\int _{P}\operatorname {Im} \{s(x)\}\cdot e^{-i{\tfrac {2\pi }{P}}nx}\ dx}$
${\displaystyle s_{_{N}}(x)=\sum _{n=-N}^{N}c_{_{Rn}}\cdot e^{i{\tfrac {2\pi }{P}}nx}+i\cdot \sum _{n=-N}^{N}c_{_{In}}\cdot e^{i{\tfrac {2\pi }{P}}nx}=\sum _{n=-N}^{N}\left(c_{_{Rn}}+i\cdot c_{_{In}}\right)\cdot e^{i{\tfrac {2\pi }{P}}nx}.}$
Defining ${\displaystyle c_{n}\triangleq c_{_{Rn}}+i\cdot c_{_{In}}}$ yields: [4] [5]
${\displaystyle s_{_{N}}(x)=\sum _{n=-N}^{N}c_{n}\cdot e^{i{\tfrac {2\pi }{P}}nx}}$
(Eq.7)
This is identical to Eq.5 except ${\displaystyle c_{n}}$ and ${\displaystyle c_{-n}}$ are no longer complex conjugates. The formula for ${\displaystyle c_{n}}$ is also unchanged:
{\displaystyle {\begin{aligned}c_{n}&={\tfrac {1}{P}}\int _{P}\operatorname {Re} \{s(x)\}\cdot e^{-i{\frac {2\pi }{P}}nx}\ dx+i\cdot {\tfrac {1}{P}}\int _{P}\operatorname {Im} \{s(x)\}\cdot e^{-i{\tfrac {2\pi }{P}}nx}\ dx\\[4pt]&={\tfrac {1}{P}}\int _{P}\left(\operatorname {Re} \{s(x)\}+i\cdot \operatorname {Im} \{s(x)\}\right)\cdot e^{-i{\tfrac {2\pi }{P}}nx}\ dx\ =\ {\tfrac {1}{P}}\int _{P}s(x)\cdot e^{-i{\tfrac {2\pi }{P}}nx}\ dx.\end{aligned}}}
### Other common notations
The notation ${\displaystyle c_{n}}$ is inadequate for discussing the Fourier coefficients of several different functions. Therefore, it is customarily replaced by a modified form of the function (${\displaystyle s}$, in this case), such as ${\displaystyle {\hat {s}}[n]}$ or ${\displaystyle S[n]}$, and functional notation often replaces subscripting:
{\displaystyle {\begin{aligned}s_{\infty }(x)&=\sum _{n=-\infty }^{\infty }{\hat {s}}[n]\cdot e^{i\,2\pi nx/P}\\[6pt]&=\sum _{n=-\infty }^{\infty }S[n]\cdot e^{i\,2\pi nx/P}&&\scriptstyle {\mathsf {common\ engineering\ notation}}\end{aligned}}}
In engineering, particularly when the variable ${\displaystyle x}$ represents time, the coefficient sequence is called a frequency domain representation. Square brackets are often used to emphasize that the domain of this function is a discrete set of frequencies.
Another commonly used frequency domain representation uses the Fourier series coefficients to modulate a Dirac comb:
${\displaystyle S(f)\ \triangleq \ \sum _{n=-\infty }^{\infty }S[n]\cdot \delta \left(f-{\frac {n}{P}}\right),}$
where ${\displaystyle f}$ represents a continuous frequency domain. When variable ${\displaystyle x}$ has units of seconds, ${\displaystyle f}$ has units of hertz. The "teeth" of the comb are spaced at multiples (i.e. harmonics) of ${\displaystyle {\tfrac {1}{P}}}$, which is called the fundamental frequency. ${\displaystyle s_{\infty }(x)}$ can be recovered from this representation by an inverse Fourier transform:
{\displaystyle {\begin{aligned}{\mathcal {F}}^{-1}\{S(f)\}&=\int _{-\infty }^{\infty }\left(\sum _{n=-\infty }^{\infty }S[n]\cdot \delta \left(f-{\frac {n}{P}}\right)\right)e^{i2\pi fx}\,df,\\[6pt]&=\sum _{n=-\infty }^{\infty }S[n]\cdot \int _{-\infty }^{\infty }\delta \left(f-{\frac {n}{P}}\right)e^{i2\pi fx}\,df,\\[6pt]&=\sum _{n=-\infty }^{\infty }S[n]\cdot e^{i\,2\pi nx/P}\ \ \triangleq \ s_{\infty }(x).\end{aligned}}}
The constructed function ${\displaystyle S(f)}$ is therefore commonly referred to as a Fourier transform, even though the Fourier integral of a periodic function is not convergent at the harmonic frequencies. [E]
## History
The Fourier series is named in honor of Jean-Baptiste Joseph Fourier (1768–1830), who made important contributions to the study of trigonometric series, after preliminary investigations by Leonhard Euler, Jean le Rond d'Alembert, and Daniel Bernoulli. [F] Fourier introduced the series for the purpose of solving the heat equation in a metal plate, publishing his initial results in his 1807 Mémoire sur la propagation de la chaleur dans les corps solides (Treatise on the propagation of heat in solid bodies), and publishing his Théorie analytique de la chaleur (Analytical theory of heat) in 1822. The Mémoire introduced Fourier analysis, specifically Fourier series. Through Fourier's research the fact was established that an arbitrary (at first, continuous [6] and later generalized to any piecewise-smooth [7]) function can be represented by a trigonometric series. The first announcement of this great discovery was made by Fourier in 1807, before the French Academy. [8] Early ideas of decomposing a periodic function into the sum of simple oscillating functions date back to the 3rd century BC, when ancient astronomers proposed an empiric model of planetary motions, based on deferents and epicycles.
The heat equation is a partial differential equation. Prior to Fourier's work, no solution to the heat equation was known in the general case, although particular solutions were known if the heat source behaved in a simple way, in particular, if the heat source was a sine or cosine wave. These simple solutions are now sometimes called eigensolutions. Fourier's idea was to model a complicated heat source as a superposition (or linear combination) of simple sine and cosine waves, and to write the solution as a superposition of the corresponding eigensolutions. This superposition or linear combination is called the Fourier series.
From a modern point of view, Fourier's results are somewhat informal, due to the lack of a precise notion of function and integral in the early nineteenth century. Later, Peter Gustav Lejeune Dirichlet [9] and Bernhard Riemann [10] [11] [12] expressed Fourier's results with greater precision and formality.
Although the original motivation was to solve the heat equation, it later became obvious that the same techniques could be applied to a wide array of mathematical and physical problems, and especially those involving linear differential equations with constant coefficients, for which the eigensolutions are sinusoids. The Fourier series has many such applications in electrical engineering, vibration analysis, acoustics, optics, signal processing, image processing, quantum mechanics, econometrics, [13] shell theory, [14] etc.
### Beginnings
Joseph Fourier wrote:[ dubious ]
${\displaystyle \varphi (y)=a_{0}\cos {\frac {\pi y}{2}}+a_{1}\cos 3{\frac {\pi y}{2}}+a_{2}\cos 5{\frac {\pi y}{2}}+\cdots .}$
Multiplying both sides by ${\displaystyle \cos(2k+1){\frac {\pi y}{2}}}$, and then integrating from ${\displaystyle y=-1}$ to ${\displaystyle y=+1}$ yields:
${\displaystyle a_{k}=\int _{-1}^{1}\varphi (y)\cos(2k+1){\frac {\pi y}{2}}\,dy.}$
This immediately gives any coefficient ak of the trigonometrical series for φ(y) for any function which has such an expansion. It works because if φ has such an expansion, then (under suitable convergence assumptions) the integral
{\displaystyle {\begin{aligned}a_{k}&=\int _{-1}^{1}\varphi (y)\cos(2k+1){\frac {\pi y}{2}}\,dy\\&=\int _{-1}^{1}\left(a\cos {\frac {\pi y}{2}}\cos(2k+1){\frac {\pi y}{2}}+a'\cos 3{\frac {\pi y}{2}}\cos(2k+1){\frac {\pi y}{2}}+\cdots \right)\,dy\end{aligned}}}
can be carried out term-by-term. But all terms involving ${\displaystyle \cos(2j+1){\frac {\pi y}{2}}\cos(2k+1){\frac {\pi y}{2}}}$ for jk vanish when integrated from −1 to 1, leaving only the ${\displaystyle k^{\text{th}}}$ term.
In these few lines, which are close to the modern formalism used in Fourier series, Fourier revolutionized both mathematics and physics. Although similar trigonometric series were previously used by Euler, d'Alembert, Daniel Bernoulli and Gauss, Fourier believed that such trigonometric series could represent any arbitrary function. In what sense that is actually true is a somewhat subtle issue and the attempts over many years to clarify this idea have led to important discoveries in the theories of convergence, function spaces, and harmonic analysis.
When Fourier submitted a later competition essay in 1811, the committee (which included Lagrange, Laplace, Malus and Legendre, among others) concluded: ...the manner in which the author arrives at these equations is not exempt of difficulties and...his analysis to integrate them still leaves something to be desired on the score of generality and even rigour.[ citation needed]
### Fourier's motivation
Heat distribution in a metal plate, using Fourier's method
The Fourier series expansion of the sawtooth function (above) looks more complicated than the simple formula ${\displaystyle s(x)={\tfrac {x}{\pi }}}$, so it is not immediately apparent why one would need the Fourier series. While there are many applications, Fourier's motivation was in solving the heat equation. For example, consider a metal plate in the shape of a square whose sides measure ${\displaystyle \pi }$ meters, with coordinates ${\displaystyle (x,y)\in [0,\pi ]\times [0,\pi ]}$. If there is no heat source within the plate, and if three of the four sides are held at 0 degrees Celsius, while the fourth side, given by ${\displaystyle y=\pi }$, is maintained at the temperature gradient ${\displaystyle T(x,\pi )=x}$ degrees Celsius, for ${\displaystyle x}$ in ${\displaystyle (0,\pi )}$, then one can show that the stationary heat distribution (or the heat distribution after a long period of time has elapsed) is given by
${\displaystyle T(x,y)=2\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n}}\sin(nx){\sinh(ny) \over \sinh(n\pi )}.}$
Here, sinh is the hyperbolic sine function. This solution of the heat equation is obtained by multiplying each term of Eq.6 by ${\displaystyle \sinh(ny)/\sinh(n\pi )}$. While our example function ${\displaystyle s(x)}$ seems to have a needlessly complicated Fourier series, the heat distribution ${\displaystyle T(x,y)}$ is nontrivial. The function ${\displaystyle T}$ cannot be written as a closed-form expression. This method of solving the heat problem was made possible by Fourier's work.
### Complex Fourier series animation
Complex Fourier series tracing the letter 'e'. (The Julia source code that generates the frames of this animation is here [16] in Appendix B.)
An example of the ability of the complex Fourier series to trace any two dimensional closed figure is shown in the adjacent animation of the complex Fourier series tracing the letter 'e' (for exponential). Note that the animation uses the variable 't' to parameterize the letter 'e' in the complex plane, which is equivalent to using the parameter 'x' in this article's subsection on complex valued functions.
In the animation's back plane, the rotating vectors are aggregated in an order that alternates between a vector rotating in the positive (counter clockwise) direction and a vector rotating at the same frequency but in the negative (clockwise) direction, resulting in a single tracing arm with lots of zigzags. This perspective shows how the addition of each pair of rotating vectors (one rotating in the positive direction and one rotating in the negative direction) nudges the previous trace (shown as a light gray dotted line) closer to the shape of the letter 'e'.
In the animation's front plane, the rotating vectors are aggregated into two sets, the set of all the positive rotating vectors and the set of all the negative rotating vectors (the non-rotating component is evenly split between the two), resulting in two tracing arms rotating in opposite directions. The animation's small circle denotes the midpoint between the two arms and also the midpoint between the origin and the current tracing point denoted by '+'. This perspective shows how the complex Fourier series is an extension (the addition of an arm) of the complex geometric series which has just one arm. It also shows how the two arms coordinate with each other. For example, as the tracing point is rotating in the positive direction, the negative direction arm stays parked. Similarly, when the tracing point is rotating in the negative direction, the positive direction arm stays parked.
In between the animation's back and front planes are rotating trapezoids whose areas represent the values of the complex Fourier series terms. This perspective shows the amplitude, frequency, and phase of the individual terms of the complex Fourier series in relation to the series sum spatially converging to the letter 'e' in the back and front planes. The audio track's left and right channels correspond respectively to the real and imaginary components of the current tracing point '+' but increased in frequency by a factor of 3536 so that the animation's fundamental frequency (n=1) is a 220 Hz tone (A220).
### Other applications
The discrete-time Fourier transform is an example of a Fourier series.
Another application is to solve the Basel problem by using Parseval's theorem. The example generalizes and one may compute ζ(2n), for any positive integern.
## Table of common Fourier series
Some common pairs of periodic functions and their Fourier Series coefficients are shown in the table below.
• ${\displaystyle s(x)}$ designates a periodic function defined on ${\displaystyle 0.
• ${\displaystyle a_{0},a_{n},b_{n}}$ designate the Fourier Series coefficients (sine-cosine form) of the periodic function ${\displaystyle s(x)}$.
Time domain
${\displaystyle s(x)}$
Plot Frequency domain (sine-cosine form)
{\displaystyle {\begin{aligned}&a_{0}\\&a_{n}\quad {\text{for }}n\geq 1\\&b_{n}\quad {\text{for }}n\geq 1\end{aligned}}}
Remarks Reference
${\displaystyle s(x)=A\left|\sin \left({\frac {2\pi }{P}}x\right)\right|\quad {\text{for }}0\leq x
{\displaystyle {\begin{aligned}a_{0}=&{\frac {4A}{\pi }}\\a_{n}=&{\begin{cases}{\frac {-4A}{\pi }}{\frac {1}{n^{2}-1}}&\quad n{\text{ even}}\\0&\quad n{\text{ odd}}\end{cases}}\\b_{n}=&0\\\end{aligned}}} Full-wave rectified sine [17]: p. 193
${\displaystyle s(x)={\begin{cases}A\sin \left({\frac {2\pi }{P}}x\right)&\quad {\text{for }}0\leq x
{\displaystyle {\begin{aligned}a_{0}=&{\frac {2A}{\pi }}\\a_{n}=&{\begin{cases}{\frac {-2A}{\pi }}{\frac {1}{n^{2}-1}}&\quad n{\text{ even}}\\0&\quad n{\text{ odd}}\end{cases}}\\b_{n}=&{\begin{cases}{\frac {A}{2}}&\quad n=1\\0&\quad n>1\end{cases}}\\\end{aligned}}} Half-wave rectified sine [17]: p. 193
${\displaystyle s(x)={\begin{cases}A&\quad {\text{for }}0\leq x
{\displaystyle {\begin{aligned}a_{0}=&2AD\\a_{n}=&{\frac {A}{n\pi }}\sin \left(2\pi nD\right)\\b_{n}=&{\frac {2A}{n\pi }}\left(\sin \left(\pi nD\right)\right)^{2}\\\end{aligned}}} ${\displaystyle 0\leq D\leq 1}$
${\displaystyle s(x)={\frac {Ax}{P}}\quad {\text{for }}0\leq x
{\displaystyle {\begin{aligned}a_{0}=&A\\a_{n}=&0\\b_{n}=&{\frac {-A}{n\pi }}\\\end{aligned}}} [17]: p. 192
${\displaystyle s(x)=A-{\frac {Ax}{P}}\quad {\text{for }}0\leq x
{\displaystyle {\begin{aligned}a_{0}=&A\\a_{n}=&0\\b_{n}=&{\frac {A}{n\pi }}\\\end{aligned}}} [17]: p. 192
${\displaystyle s(x)={\frac {4A}{P^{2}}}\left(x-{\frac {P}{2}}\right)^{2}\quad {\text{for }}0\leq x
{\displaystyle {\begin{aligned}a_{0}=&{\frac {2A}{3}}\\a_{n}=&{\frac {4A}{\pi ^{2}n^{2}}}\\b_{n}=&0\\\end{aligned}}} [17]: p. 193
## Table of basic properties
This table shows some mathematical operations in the time domain and the corresponding effect in the Fourier series coefficients. Notation:
• Complex conjugation is denoted by an asterisk.
• ${\displaystyle s(x),r(x)}$ designate ${\displaystyle P}$-periodic functions or functions defined only for ${\displaystyle x\in [0,P].}$
• ${\displaystyle S[n],R[n]}$ designate the Fourier series coefficients (exponential form) of ${\displaystyle s}$ and ${\displaystyle r.}$
Property Time domain Frequency domain (exponential form) Remarks Reference
Linearity ${\displaystyle a\cdot s(x)+b\cdot r(x)}$ ${\displaystyle a\cdot S[n]+b\cdot R[n]}$ ${\displaystyle a,b\in \mathbb {C} }$
Time reversal / Frequency reversal ${\displaystyle s(-x)}$ ${\displaystyle S[-n]}$ [18]: p. 610
Time conjugation ${\displaystyle s^{*}(x)}$ ${\displaystyle S^{*}[-n]}$ [18]: p. 610
Time reversal & conjugation ${\displaystyle s^{*}(-x)}$ ${\displaystyle S^{*}[n]}$
Real part in time ${\displaystyle \operatorname {Re} {(s(x))}}$ ${\displaystyle {\frac {1}{2}}(S[n]+S^{*}[-n])}$
Imaginary part in time ${\displaystyle \operatorname {Im} {(s(x))}}$ ${\displaystyle {\frac {1}{2i}}(S[n]-S^{*}[-n])}$
Real part in frequency ${\displaystyle {\frac {1}{2}}(s(x)+s^{*}(-x))}$ ${\displaystyle \operatorname {Re} {(S[n])}}$
Imaginary part in frequency ${\displaystyle {\frac {1}{2i}}(s(x)-s^{*}(-x))}$ ${\displaystyle \operatorname {Im} {(S[n])}}$
Shift in time / Modulation in frequency ${\displaystyle s(x-x_{0})}$ ${\displaystyle S[n]\cdot e^{-i{\frac {2\pi x_{0}}{P}}n}}$ ${\displaystyle x_{0}\in \mathbb {R} }$ [18]: p. 610
Shift in frequency / Modulation in time ${\displaystyle s(x)\cdot e^{i{\frac {2\pi n_{0}}{P}}x}}$ ${\displaystyle S[n-n_{0}]\!}$ ${\displaystyle n_{0}\in \mathbb {Z} }$ [18]: p. 610
## Symmetry properties
When the real and imaginary parts of a complex function are decomposed into their even and odd parts, there are four components, denoted below by the subscripts RE, RO, IE, and IO. And there is a one-to-one mapping between the four components of a complex time function and the four components of its complex frequency transform: [19]
${\displaystyle {\begin{array}{rccccccccc}{\text{Time domain}}&s&=&s_{_{\text{RE}}}&+&s_{_{\text{RO}}}&+&is_{_{\text{IE}}}&+&\underbrace {i\ s_{_{\text{IO}}}} \\&{\Bigg \Updownarrow }{\mathcal {F}}&&{\Bigg \Updownarrow }{\mathcal {F}}&&\ \ {\Bigg \Updownarrow }{\mathcal {F}}&&\ \ {\Bigg \Updownarrow }{\mathcal {F}}&&\ \ {\Bigg \Updownarrow }{\mathcal {F}}\\{\text{Frequency domain}}&S&=&S_{\text{RE}}&+&\overbrace {\,i\ S_{\text{IO}}\,} &+&iS_{\text{IE}}&+&S_{\text{RO}}\end{array}}}$
From this, various relationships are apparent, for example:
• The transform of a real-valued function (sRE + sRO) is the even symmetric function SRE + i SIO. Conversely, an even-symmetric transform implies a real-valued time-domain.
• The transform of an imaginary-valued function (i sIE + i sIO) is the odd symmetric function SRO + i SIE, and the converse is true.
• The transform of an even-symmetric function (sRE + i sIO) is the real-valued function SRE + SRO, and the converse is true.
• The transform of an odd-symmetric function (sRO + i sIE) is the imaginary-valued function i SIE + i SIO, and the converse is true.
## Other properties
### Riemann–Lebesgue lemma
If ${\displaystyle S}$ is integrable, ${\textstyle \lim _{|n|\to \infty }S[n]=0}$, ${\textstyle \lim _{n\to +\infty }a_{n}=0}$ and ${\textstyle \lim _{n\to +\infty }b_{n}=0.}$ This result is known as the Riemann–Lebesgue lemma.
### Parseval's theorem
If ${\displaystyle s}$ belongs to ${\displaystyle L^{2}(P)}$ (periodic over an interval of length ${\displaystyle P}$) then: ${\textstyle {\frac {1}{P}}\int _{P}|s(x)|^{2}\,dx=\sum _{n=-\infty }^{\infty }{\Bigl |}S[n]{\Bigr |}^{2}}$
### Hesham's identity
If ${\displaystyle s}$ belongs to ${\displaystyle L^{4}(P)}$ (periodic over an interval of length ${\displaystyle P}$), and ${\displaystyle S[n]}$ is of a finite-length ${\displaystyle M}$ then: [20]
for ${\displaystyle S[n]\in \mathbb {C} }$, then ${\displaystyle {\frac {1}{P}}\int _{P}|s(x)|^{4}\,dx=\sum _{k=0}^{M-1}S[k]\sum _{l=0}^{M-1}S^{*}[l]{\Bigg [}{\underset {k\geq l}{\sum _{m=k-l}^{M-1}}}S^{*}[m]S[m-(k-l)]+{\underset {k
and for ${\displaystyle S[n]\in \mathbb {R} }$, then ${\displaystyle {\frac {1}{P}}\int _{P}|s(x)|^{4}\,dx=\sum _{k=0}^{M-1}S[k]\sum _{l=0}^{M-1}S[l]\sum _{m=|k-l|}^{M-1}S[m]S[m-|k-l|]}$
### Plancherel's theorem
If ${\displaystyle c_{0},\,c_{\pm 1},\,c_{\pm 2},\ldots }$ are coefficients and ${\textstyle \sum _{n=-\infty }^{\infty }|c_{n}|^{2}<\infty }$ then there is a unique function ${\displaystyle s\in L^{2}(P)}$ such that ${\displaystyle S[n]=c_{n}}$ for every ${\displaystyle n}$.
### Convolution theorems
Given ${\displaystyle P}$-periodic functions, ${\displaystyle s_{_{P}}}$ and ${\displaystyle r_{_{P}}}$ with Fourier series coefficients ${\displaystyle S[n]}$ and ${\displaystyle R[n],}$ ${\displaystyle n\in \mathbb {Z} ,}$
• The pointwise product:
${\displaystyle h_{_{P}}(x)\triangleq s_{_{P}}(x)\cdot r_{_{P}}(x)}$
is also ${\displaystyle P}$-periodic, and its Fourier series coefficients are given by the discrete convolution of the ${\displaystyle S}$ and ${\displaystyle R}$ sequences:
${\displaystyle H[n]=\{S*R\}[n].}$
• The periodic convolution:
${\displaystyle h_{_{P}}(x)\triangleq \int _{P}s_{_{P}}(\tau )\cdot r_{_{P}}(x-\tau )\,d\tau }$
is also ${\displaystyle P}$-periodic, with Fourier series coefficients:
${\displaystyle H[n]=P\cdot S[n]\cdot R[n].}$
• A doubly infinite sequence ${\displaystyle \left\{c_{n}\right\}_{n\in Z}}$ in ${\displaystyle c_{0}(\mathbb {Z} )}$ is the sequence of Fourier coefficients of a function in ${\displaystyle L^{1}([0,2\pi ])}$ if and only if it is a convolution of two sequences in ${\displaystyle \ell ^{2}(\mathbb {Z} )}$. See [21]
### Derivative property
We say that ${\displaystyle s}$ belongs to ${\displaystyle C^{k}(\mathbb {T} )}$ if ${\displaystyle s}$ is a 2π-periodic function on ${\displaystyle \mathbb {R} }$ which is ${\displaystyle k}$ times differentiable, and its ${\displaystyle k^{\text{th}}}$ derivative is continuous.
• If ${\displaystyle s\in C^{1}(\mathbb {T} )}$, then the Fourier coefficients ${\displaystyle {\widehat {s'}}[n]}$ of the derivative ${\displaystyle s'}$ can be expressed in terms of the Fourier coefficients ${\displaystyle {\widehat {s}}[n]}$ of the function ${\displaystyle s}$, via the formula ${\displaystyle {\widehat {s'}}[n]=in{\widehat {s}}[n]}$.
• If ${\displaystyle s\in C^{k}(\mathbb {T} )}$, then ${\displaystyle {\widehat {s^{(k)}}}[n]=(in)^{k}{\widehat {s}}[n]}$. In particular, since for a fixed ${\displaystyle k\geq 1}$ we have ${\displaystyle {\widehat {s^{(k)}}}[n]\to 0}$ as ${\displaystyle n\to \infty }$, it follows that ${\displaystyle |n|^{k}{\widehat {s}}[n]}$ tends to zero, which means that the Fourier coefficients converge to zero faster than the kth power of n for any ${\displaystyle k\geq 1}$.
### Compact groups
One of the interesting properties of the Fourier transform which we have mentioned, is that it carries convolutions to pointwise products. If that is the property which we seek to preserve, one can produce Fourier series on any compact group. Typical examples include those classical groups that are compact. This generalizes the Fourier transform to all spaces of the form L2(G), where G is a compact group, in such a way that the Fourier transform carries convolutions to pointwise products. The Fourier series exists and converges in similar ways to the [−π,π] case.
An alternative extension to compact groups is the Peter–Weyl theorem, which proves results about representations of compact groups analogous to those about finite groups.
The atomic orbitals of chemistry are partially described by spherical harmonics, which can be used to produce Fourier series on the sphere.
### Riemannian manifolds
If the domain is not a group, then there is no intrinsically defined convolution. However, if ${\displaystyle X}$ is a compact Riemannian manifold, it has a Laplace–Beltrami operator. The Laplace–Beltrami operator is the differential operator that corresponds to Laplace operator for the Riemannian manifold ${\displaystyle X}$. Then, by analogy, one can consider heat equations on ${\displaystyle X}$. Since Fourier arrived at his basis by attempting to solve the heat equation, the natural generalization is to use the eigensolutions of the Laplace–Beltrami operator as a basis. This generalizes Fourier series to spaces of the type ${\displaystyle L^{2}(X)}$, where ${\displaystyle X}$ is a Riemannian manifold. The Fourier series converges in ways similar to the ${\displaystyle [-\pi ,\pi ]}$ case. A typical example is to take ${\displaystyle X}$ to be the sphere with the usual metric, in which case the Fourier basis consists of spherical harmonics.
### Locally compact Abelian groups
The generalization to compact groups discussed above does not generalize to noncompact, nonabelian groups. However, there is a straightforward generalization to Locally Compact Abelian (LCA) groups.
This generalizes the Fourier transform to ${\displaystyle L^{1}(G)}$ or ${\displaystyle L^{2}(G)}$, where ${\displaystyle G}$ is an LCA group. If ${\displaystyle G}$ is compact, one also obtains a Fourier series, which converges similarly to the ${\displaystyle [-\pi ,\pi ]}$ case, but if ${\displaystyle G}$ is noncompact, one obtains instead a Fourier integral. This generalization yields the usual Fourier transform when the underlying locally compact Abelian group is ${\displaystyle \mathbb {R} }$.
## Extensions
### Fourier series on a square
We can also define the Fourier series for functions of two variables ${\displaystyle x}$ and ${\displaystyle y}$ in the square ${\displaystyle [-\pi ,\pi ]\times [-\pi ,\pi ]}$:
{\displaystyle {\begin{aligned}f(x,y)&=\sum _{j,k\in \mathbb {Z} }c_{j,k}e^{ijx}e^{iky},\\[5pt]c_{j,k}&={\frac {1}{4\pi ^{2}}}\int _{-\pi }^{\pi }\int _{-\pi }^{\pi }f(x,y)e^{-ijx}e^{-iky}\,dx\,dy.\end{aligned}}}
Aside from being useful for solving partial differential equations such as the heat equation, one notable application of Fourier series on the square is in image compression. In particular, the jpeg image compression standard uses the two-dimensional discrete cosine transform, a discrete form of the Fourier cosine transform, which uses only cosine as the basis function.
For two-dimensional arrays with a staggered appearance, half of the Fourier series coefficients disappear, due to additional symmetry. [22]
### Fourier series of Bravais-lattice-periodic-function
A three-dimensional Bravais lattice is defined as the set of vectors of the form:
${\displaystyle \mathbf {R} =n_{1}\mathbf {a} _{1}+n_{2}\mathbf {a} _{2}+n_{3}\mathbf {a} _{3}}$
where ${\displaystyle n_{i}}$ are integers and ${\displaystyle \mathbf {a} _{i}}$ are three linearly independent vectors. Assuming we have some function, ${\displaystyle f(\mathbf {r} )}$, such that it obeys the condition of periodicity for any Bravais lattice vector ${\displaystyle \mathbf {R} }$, ${\displaystyle f(\mathbf {r} )=f(\mathbf {R} +\mathbf {r} )}$, we could make a Fourier series of it. This kind of function can be, for example, the effective potential that one electron "feels" inside a periodic crystal. It is useful to make the Fourier series of the potential when applying Bloch's theorem. First, we may write any arbitrary position vector ${\displaystyle \mathbf {r} }$ in the coordinate-system of the lattice:
${\displaystyle \mathbf {r} =x_{1}{\frac {\mathbf {a} _{1}}{a_{1}}}+x_{2}{\frac {\mathbf {a} _{2}}{a_{2}}}+x_{3}{\frac {\mathbf {a} _{3}}{a_{3}}},}$
where ${\displaystyle a_{i}\triangleq |\mathbf {a} _{i}|,}$ meaning that ${\displaystyle a_{i}}$ is defined to be the magnitude of ${\displaystyle \mathbf {a} _{i}}$, so ${\displaystyle {\hat {\mathbf {a} _{i}}}={\frac {\mathbf {a} _{i}}{a_{i}}}}$ is the unit vector directed along ${\displaystyle \mathbf {a} _{i}}$.
Thus we can define a new function,
${\displaystyle g(x_{1},x_{2},x_{3})\triangleq f(\mathbf {r} )=f\left(x_{1}{\frac {\mathbf {a} _{1}}{a_{1}}}+x_{2}{\frac {\mathbf {a} _{2}}{a_{2}}}+x_{3}{\frac {\mathbf {a} _{3}}{a_{3}}}\right).}$
This new function, ${\displaystyle g(x_{1},x_{2},x_{3})}$, is now a function of three-variables, each of which has periodicity ${\displaystyle a_{1}}$, ${\displaystyle a_{2}}$, and ${\displaystyle a_{3}}$ respectively:
${\displaystyle g(x_{1},x_{2},x_{3})=g(x_{1}+a_{1},x_{2},x_{3})=g(x_{1},x_{2}+a_{2},x_{3})=g(x_{1},x_{2},x_{3}+a_{3}).}$
This enables us to build up a set of Fourier coefficients, each being indexed by three independent integers ${\displaystyle m_{1},m_{2},m_{3}}$. In what follows, we use function notation to denote these coefficients, where previously we used subscripts. If we write a series for ${\displaystyle g}$ on the interval ${\displaystyle \left[0,a_{1}\right]}$ for ${\displaystyle x_{1}}$, we can define the following:
${\displaystyle h^{\mathrm {one} }(m_{1},x_{2},x_{3})\triangleq {\frac {1}{a_{1}}}\int _{0}^{a_{1}}g(x_{1},x_{2},x_{3})\cdot e^{-i2\pi {\frac {m_{1}}{a_{1}}}x_{1}}\,dx_{1}}$
And then we can write:
${\displaystyle g(x_{1},x_{2},x_{3})=\sum _{m_{1}=-\infty }^{\infty }h^{\mathrm {one} }(m_{1},x_{2},x_{3})\cdot e^{i2\pi {\frac {m_{1}}{a_{1}}}x_{1}}}$
Further defining:
{\displaystyle {\begin{aligned}h^{\mathrm {two} }(m_{1},m_{2},x_{3})&\triangleq {\frac {1}{a_{2}}}\int _{0}^{a_{2}}h^{\mathrm {one} }(m_{1},x_{2},x_{3})\cdot e^{-i2\pi {\frac {m_{2}}{a_{2}}}x_{2}}\,dx_{2}\\[12pt]&={\frac {1}{a_{2}}}\int _{0}^{a_{2}}dx_{2}{\frac {1}{a_{1}}}\int _{0}^{a_{1}}dx_{1}g(x_{1},x_{2},x_{3})\cdot e^{-i2\pi \left({\frac {m_{1}}{a_{1}}}x_{1}+{\frac {m_{2}}{a_{2}}}x_{2}\right)}\end{aligned}}}
We can write ${\displaystyle g}$ once again as:
${\displaystyle g(x_{1},x_{2},x_{3})=\sum _{m_{1}=-\infty }^{\infty }\sum _{m_{2}=-\infty }^{\infty }h^{\mathrm {two} }(m_{1},m_{2},x_{3})\cdot e^{i2\pi {\frac {m_{1}}{a_{1}}}x_{1}}\cdot e^{i2\pi {\frac {m_{2}}{a_{2}}}x_{2}}}$
Finally applying the same for the third coordinate, we define:
{\displaystyle {\begin{aligned}h^{\mathrm {three} }(m_{1},m_{2},m_{3})&\triangleq {\frac {1}{a_{3}}}\int _{0}^{a_{3}}h^{\mathrm {two} }(m_{1},m_{2},x_{3})\cdot e^{-i2\pi {\frac {m_{3}}{a_{3}}}x_{3}}\,dx_{3}\\[12pt]&={\frac {1}{a_{3}}}\int _{0}^{a_{3}}dx_{3}{\frac {1}{a_{2}}}\int _{0}^{a_{2}}dx_{2}{\frac {1}{a_{1}}}\int _{0}^{a_{1}}dx_{1}g(x_{1},x_{2},x_{3})\cdot e^{-i2\pi \left({\frac {m_{1}}{a_{1}}}x_{1}+{\frac {m_{2}}{a_{2}}}x_{2}+{\frac {m_{3}}{a_{3}}}x_{3}\right)}\end{aligned}}}
We write ${\displaystyle g}$ as:
${\displaystyle g(x_{1},x_{2},x_{3})=\sum _{m_{1}=-\infty }^{\infty }\sum _{m_{2}=-\infty }^{\infty }\sum _{m_{3}=-\infty }^{\infty }h^{\mathrm {three} }(m_{1},m_{2},m_{3})\cdot e^{i2\pi {\frac {m_{1}}{a_{1}}}x_{1}}\cdot e^{i2\pi {\frac {m_{2}}{a_{2}}}x_{2}}\cdot e^{i2\pi {\frac {m_{3}}{a_{3}}}x_{3}}}$
Re-arranging:
${\displaystyle g(x_{1},x_{2},x_{3})=\sum _{m_{1},m_{2},m_{3}\in \mathbb {Z} }h^{\mathrm {three} }(m_{1},m_{2},m_{3})\cdot e^{i2\pi \left({\frac {m_{1}}{a_{1}}}x_{1}+{\frac {m_{2}}{a_{2}}}x_{2}+{\frac {m_{3}}{a_{3}}}x_{3}\right)}.}$
Now, every reciprocal lattice vector can be written (but does not mean that it is the only way of writing) as ${\displaystyle \mathbf {G} =m_{1}\mathbf {g} _{1}+m_{2}\mathbf {g} _{2}+m_{3}\mathbf {g} _{3}}$, where ${\displaystyle m_{i}}$ are integers and ${\displaystyle \mathbf {g} _{i}}$ are reciprocal lattice vectors to satisfy ${\displaystyle \mathbf {g_{i}} \cdot \mathbf {a_{j}} =2\pi \delta _{ij}}$ (${\displaystyle \delta _{ij}=1}$ for ${\displaystyle i=j}$, and ${\displaystyle \delta _{ij}=0}$ for ${\displaystyle i\neq j}$). Then for any arbitrary reciprocal lattice vector ${\displaystyle \mathbf {G} }$ and arbitrary position vector ${\displaystyle \mathbf {r} }$ in the original Bravais lattice space, their scalar product is:
${\displaystyle \mathbf {G} \cdot \mathbf {r} =\left(m_{1}\mathbf {g} _{1}+m_{2}\mathbf {g} _{2}+m_{3}\mathbf {g} _{3}\right)\cdot \left(x_{1}{\frac {\mathbf {a} _{1}}{a_{1}}}+x_{2}{\frac {\mathbf {a} _{2}}{a_{2}}}+x_{3}{\frac {\mathbf {a} _{3}}{a_{3}}}\right)=2\pi \left(x_{1}{\frac {m_{1}}{a_{1}}}+x_{2}{\frac {m_{2}}{a_{2}}}+x_{3}{\frac {m_{3}}{a_{3}}}\right).}$
So it is clear that in our expansion of ${\displaystyle g(x_{1},x_{2},x_{3})=f(\mathbf {r} )}$, the sum is actually over reciprocal lattice vectors:
${\displaystyle f(\mathbf {r} )=\sum _{\mathbf {G} }h(\mathbf {G} )\cdot e^{i\mathbf {G} \cdot \mathbf {r} },}$
where
${\displaystyle h(\mathbf {G} )={\frac {1}{a_{3}}}\int _{0}^{a_{3}}dx_{3}\,{\frac {1}{a_{2}}}\int _{0}^{a_{2}}dx_{2}\,{\frac {1}{a_{1}}}\int _{0}^{a_{1}}dx_{1}\,f\left(x_{1}{\frac {\mathbf {a} _{1}}{a_{1}}}+x_{2}{\frac {\mathbf {a} _{2}}{a_{2}}}+x_{3}{\frac {\mathbf {a} _{3}}{a_{3}}}\right)\cdot e^{-i\mathbf {G} \cdot \mathbf {r} }.}$
Assuming
${\displaystyle \mathbf {r} =(x,y,z)=x_{1}{\frac {\mathbf {a} _{1}}{a_{1}}}+x_{2}{\frac {\mathbf {a} _{2}}{a_{2}}}+x_{3}{\frac {\mathbf {a} _{3}}{a_{3}}},}$
we can solve this system of three linear equations for ${\displaystyle x}$, ${\displaystyle y}$, and ${\displaystyle z}$ in terms of ${\displaystyle x_{1}}$, ${\displaystyle x_{2}}$ and ${\displaystyle x_{3}}$ in order to calculate the volume element in the original cartesian coordinate system. Once we have ${\displaystyle x}$, ${\displaystyle y}$, and ${\displaystyle z}$ in terms of ${\displaystyle x_{1}}$, ${\displaystyle x_{2}}$ and ${\displaystyle x_{3}}$, we can calculate the Jacobian determinant:
${\displaystyle {\begin{vmatrix}{\dfrac {\partial x_{1}}{\partial x}}&{\dfrac {\partial x_{1}}{\partial y}}&{\dfrac {\partial x_{1}}{\partial z}}\\[12pt]{\dfrac {\partial x_{2}}{\partial x}}&{\dfrac {\partial x_{2}}{\partial y}}&{\dfrac {\partial x_{2}}{\partial z}}\\[12pt]{\dfrac {\partial x_{3}}{\partial x}}&{\dfrac {\partial x_{3}}{\partial y}}&{\dfrac {\partial x_{3}}{\partial z}}\end{vmatrix}}}$
which after some calculation and applying some non-trivial cross-product identities can be shown to be equal to:
${\displaystyle {\frac {a_{1}a_{2}a_{3}}{\mathbf {a} _{1}\cdot (\mathbf {a} _{2}\times \mathbf {a} _{3})}}}$
(it may be advantageous for the sake of simplifying calculations, to work in such a Cartesian coordinate system, in which it just so happens that ${\displaystyle \mathbf {a} _{1}}$ is parallel to the x axis, ${\displaystyle \mathbf {a} _{2}}$ lies in the xy-plane, and ${\displaystyle \mathbf {a} _{3}}$ has components of all three axes). The denominator is exactly the volume of the primitive unit cell which is enclosed by the three primitive-vectors ${\displaystyle \mathbf {a} _{1}}$, ${\displaystyle \mathbf {a} _{2}}$ and ${\displaystyle \mathbf {a} _{3}}$. In particular, we now know that
${\displaystyle dx_{1}\,dx_{2}\,dx_{3}={\frac {a_{1}a_{2}a_{3}}{\mathbf {a} _{1}\cdot (\mathbf {a} _{2}\times \mathbf {a} _{3})}}\cdot dx\,dy\,dz.}$
We can write now ${\displaystyle h(\mathbf {G} )}$ as an integral with the traditional coordinate system over the volume of the primitive cell, instead of with the ${\displaystyle x_{1}}$, ${\displaystyle x_{2}}$ and ${\displaystyle x_{3}}$ variables:
${\displaystyle h(\mathbf {G} )={\frac {1}{\mathbf {a} _{1}\cdot (\mathbf {a} _{2}\times \mathbf {a} _{3})}}\int _{C}d\mathbf {r} f(\mathbf {r} )\cdot e^{-i\mathbf {G} \cdot \mathbf {r} }}$
writing ${\displaystyle d\mathbf {r} }$ for the volume element ${\displaystyle dx\,dy\,dz}$; and where ${\displaystyle C}$ is the primitive unit cell, thus, ${\displaystyle \mathbf {a} _{1}\cdot (\mathbf {a} _{2}\times \mathbf {a} _{3})}$ is the volume of the primitive unit cell.
### Hilbert space interpretation
In the language of Hilbert spaces, the set of functions ${\displaystyle \left\{e_{n}=e^{inx}:n\in \mathbb {Z} \right\}}$ is an orthonormal basis for the space ${\displaystyle L^{2}([-\pi ,\pi ])}$ of square-integrable functions on ${\displaystyle [-\pi ,\pi ]}$. This space is actually a Hilbert space with an inner product given for any two elements ${\displaystyle f}$ and ${\displaystyle g}$ by:
${\displaystyle \langle f,\,g\rangle \;\triangleq \;{\frac {1}{2\pi }}\int _{-\pi }^{\pi }f(x)g^{*}(x)\,dx,}$ where ${\displaystyle g^{*}(x)}$ is the complex conjugate of ${\displaystyle g(x).}$
The basic Fourier series result for Hilbert spaces can be written as
${\displaystyle f=\sum _{n=-\infty }^{\infty }\langle f,e_{n}\rangle \,e_{n}.}$
Sines and cosines form an orthonormal set, as illustrated above. The integral of sine, cosine and their product is zero (green and red areas are equal, and cancel out) when ${\displaystyle m}$, ${\displaystyle n}$ or the functions are different, and π only if ${\displaystyle m}$ and ${\displaystyle n}$ are equal, and the function used is the same.
This corresponds exactly to the complex exponential formulation given above. The version with sines and cosines is also justified with the Hilbert space interpretation. Indeed, the sines and cosines form an orthogonal set:
${\displaystyle \int _{-\pi }^{\pi }\cos(mx)\,\cos(nx)\,dx={\frac {1}{2}}\int _{-\pi }^{\pi }\cos((n-m)x)+\cos((n+m)x)\,dx=\pi \delta _{mn},\quad m,n\geq 1,}$
${\displaystyle \int _{-\pi }^{\pi }\sin(mx)\,\sin(nx)\,dx={\frac {1}{2}}\int _{-\pi }^{\pi }\cos((n-m)x)-\cos((n+m)x)\,dx=\pi \delta _{mn},\quad m,n\geq 1}$
(where δmn is the Kronecker delta), and
${\displaystyle \int _{-\pi }^{\pi }\cos(mx)\,\sin(nx)\,dx={\frac {1}{2}}\int _{-\pi }^{\pi }\sin((n+m)x)+\sin((n-m)x)\,dx=0;}$
furthermore, the sines and cosines are orthogonal to the constant function ${\displaystyle 1}$. An orthonormal basis for ${\displaystyle L^{2}([-\pi ,\pi ])}$ consisting of real functions is formed by the functions ${\displaystyle 1}$ and ${\displaystyle {\sqrt {2}}\cos(nx)}$, ${\displaystyle {\sqrt {2}}\sin(nx)}$ with n= 1,2,.... The density of their span is a consequence of the Stone–Weierstrass theorem, but follows also from the properties of classical kernels like the Fejér kernel.
## Fourier theorem proving convergence of Fourier series
These theorems, and informal variations of them that don't specify the convergence conditions, are sometimes referred to generically as Fourier's theorem or the Fourier theorem. [23] [24] [25] [26]
The earlier Eq.7
${\displaystyle s_{_{N}}(x)=\sum _{n=-N}^{N}S[n]\ e^{i{\tfrac {2\pi }{P}}nx},}$
is a trigonometric polynomial of degree ${\displaystyle N}$ that can be generally expressed as:
${\displaystyle p_{_{N}}(x)=\sum _{n=-N}^{N}p[n]\ e^{i{\tfrac {2\pi }{P}}nx}.}$
### Least squares property
Parseval's theorem implies that:
Theorem — The trigonometric polynomial ${\displaystyle s_{_{N}}}$ is the unique best trigonometric polynomial of degree ${\displaystyle N}$ approximating ${\displaystyle s(x)}$, in the sense that, for any trigonometric polynomial ${\displaystyle p_{_{N}}\neq s_{_{N}}}$ of degree ${\displaystyle N}$, we have:
${\displaystyle \|s_{_{N}}-s\|_{2}<\|p_{_{N}}-s\|_{2},}$
where the Hilbert space norm is defined as:
${\displaystyle \|g\|_{2}={\sqrt {{1 \over P}\int _{P}|g(x)|^{2}\,dx}}.}$
### Convergence theorems
Because of the least squares property, and because of the completeness of the Fourier basis, we obtain an elementary convergence result.
Theorem — If ${\displaystyle s}$ belongs to ${\displaystyle L^{2}(P)}$ (an interval of length ${\displaystyle P}$), then ${\displaystyle s_{\infty }}$ converges to ${\displaystyle s}$ in ${\displaystyle L^{2}(P)}$, that is, ${\displaystyle \|s_{_{N}}-s\|_{2}}$ converges to 0 as ${\displaystyle N\to \infty }$.
We have already mentioned that if ${\displaystyle s}$ is continuously differentiable, then ${\displaystyle (i\cdot n)S[n]}$ is the ${\displaystyle n^{\text{th}}}$ Fourier coefficient of the derivative ${\displaystyle s'}$. It follows, essentially from the Cauchy–Schwarz inequality, that ${\displaystyle s_{\infty }}$ is absolutely summable. The sum of this series is a continuous function, equal to ${\displaystyle s}$, since the Fourier series converges in the mean to ${\displaystyle s}$:
Theorem — If ${\displaystyle s\in C^{1}(\mathbb {T} )}$, then ${\displaystyle s_{\infty }}$ converges to ${\displaystyle s}$ uniformly (and hence also pointwise.)
This result can be proven easily if ${\displaystyle s}$ is further assumed to be ${\displaystyle C^{2}}$, since in that case ${\displaystyle n^{2}S[n]}$ tends to zero as ${\displaystyle n\rightarrow \infty }$. More generally, the Fourier series is absolutely summable, thus converges uniformly to ${\displaystyle s}$, provided that ${\displaystyle s}$ satisfies a Hölder condition of order ${\displaystyle \alpha >1/2}$. In the absolutely summable case, the inequality:
${\displaystyle \sup _{x}|s(x)-s_{_{N}}(x)|\leq \sum _{|n|>N}|S[n]|}$
proves uniform convergence.
Many other results concerning the convergence of Fourier series are known, ranging from the moderately simple result that the series converges at ${\displaystyle x}$ if ${\displaystyle s}$ is differentiable at ${\displaystyle x}$, to Lennart Carleson's much more sophisticated result that the Fourier series of an ${\displaystyle L^{2}}$ function actually converges almost everywhere.
### Divergence
Since Fourier series have such good convergence properties, many are often surprised by some of the negative results. For example, the Fourier series of a continuous T-periodic function need not converge pointwise.[ citation needed] The uniform boundedness principle yields a simple non-constructive proof of this fact.
In 1922, Andrey Kolmogorov published an article titled Une série de Fourier-Lebesgue divergente presque partout in which he gave an example of a Lebesgue-integrable function whose Fourier series diverges almost everywhere. He later constructed an example of an integrable function whose Fourier series diverges everywhere ( Katznelson 1976).
## Notes
1. ^ In this context, synthesis is the creation of a complex waveform by summation of simpler waveforms.
2. ^ Some texts define P=2π to simplify the sinusoid's argument at the expense of generality.
3. ^ The scale factor ${\displaystyle {\tfrac {2}{P}},}$ which could be inserted later, results in a series that converges to ${\displaystyle s(x)}$ instead of ${\displaystyle {\tfrac {P}{2}}s(x).}$
4. ^ Some authors define ${\displaystyle a_{0}}$ differently than ${\displaystyle \left.a_{n}\right|_{n=0}.}$ Rather their scale factor is just ${\displaystyle {\tfrac {1}{P}},}$ and that of course changes Eq.4 accordingly.
5. ^ Since the integral defining the Fourier transform of a periodic function is not convergent, it is necessary to view the periodic function and its transform as distributions. In this sense ${\displaystyle {\mathcal {F}}\{e^{i{\frac {2\pi nx}{P}}}\}}$ is a Dirac delta function, which is an example of a distribution.
6. ^ These three did some important early work on the wave equation, especially D'Alembert. Euler's work in this area was mostly comtemporaneous/ in collaboration with Bernoulli, although the latter made some independent contributions to the theory of waves and vibrations. (See Fetter & Walecka 2003, pp. 209–210).
7. ^ These words are not strictly Fourier's. Whilst the cited article does list the author as Fourier, a footnote indicates that the article was actually written by Poisson (that it was not written by Fourier is also clear from the consistent use of the third person to refer to him) and that it is, "for reasons of historical interest", presented as though it were Fourier's original memoire.
## References
1. ^ "Fourier". Dictionary.com Unabridged (Online). n.d.
2. ^ Dorf, Richard C.; Tallarida, Ronald J. (1993). Pocket Book of Electrical Engineering Formulas (1st ed.). Boca Raton,FL: CRC Press. pp. 171–174. ISBN 0849344735.
3. ^ Tolstov, Georgi P. (1976). Fourier Series. Courier-Dover. ISBN 0-486-63317-9.
4. ^ Wolfram, Eric W. "Fourier Series (eq.30)". MathWorld--A Wolfram Web Resource. Retrieved 3 November 2021.
5. ^ Cheever, Erik. "Derivation of Fourier Series". lpsa.swarthmore.edu. Retrieved 3 November 2021.
6. ^ Stillwell, John (2013). "Logic and the philosophy of mathematics in the nineteenth century". In Ten, C. L. (ed.). Routledge History of Philosophy. Vol. VII: The Nineteenth Century. Routledge. p. 204. ISBN 978-1-134-92880-4.
7. ^ Fasshauer, Greg (2015). "Fourier Series and Boundary Value Problems" (PDF). Math 461 Course Notes, Ch 3. Department of Applied Mathematics, Illinois Institute of Technology. Retrieved 6 November 2020.
8. ^ Cajori, Florian (1893). A History of Mathematics. Macmillan. p. 283.
9. ^ Lejeune-Dirichlet, Peter Gustav (1829). "Sur la convergence des séries trigonométriques qui servent à représenter une fonction arbitraire entre des limites données" [On the convergence of trigonometric series which serve to represent an arbitrary function between two given limits]. Journal für die reine und angewandte Mathematik (in French). 4: 157–169. arXiv:.
10. ^ "Ueber die Darstellbarkeit einer Function durch eine trigonometrische Reihe" [About the representability of a function by a trigonometric series]. Habilitationsschrift, Göttingen; 1854. Abhandlungen der Königlichen Gesellschaft der Wissenschaften zu Göttingen, vol. 13, 1867. Published posthumously for Riemann by Richard Dedekind (in German). Archived from the original on 20 May 2008. Retrieved 19 May 2008.
11. ^ Mascre, D.; Riemann, Bernhard (1867), "Posthumous Thesis on the Representation of Functions by Trigonometric Series", in Grattan-Guinness, Ivor (ed.), Landmark Writings in Western Mathematics 1640–1940, Elsevier (published 2005), p. 49, ISBN 9780080457444
12. ^ Remmert, Reinhold (1991). Theory of Complex Functions: Readings in Mathematics. Springer. p. 29. ISBN 9780387971957.
13. ^ Nerlove, Marc; Grether, David M.; Carvalho, Jose L. (1995). . Elsevier. ISBN 0-12-515751-7.
14. ^ Wilhelm Flügge, Stresses in Shells (1973) 2nd edition. ISBN 978-3-642-88291-3. Originally published in German as Statik und Dynamik der Schalen (1937).
15. ^ Fourier, Jean-Baptiste-Joseph (1888). Gaston Darboux (ed.). Oeuvres de Fourier [The Works of Fourier] (in French). Paris: Gauthier-Villars et Fils. pp. 218–219 – via Gallica.
16. ^ Sepesi, G (13 February 2022). "Zeno's Enduring Example". Towards Data Science. pp. Appendix B.
17. Papula, Lothar (2009). Mathematische Formelsammlung: für Ingenieure und Naturwissenschaftler [Mathematical Functions for Engineers and Physicists] (in German). Vieweg+Teubner Verlag. ISBN 978-3834807571.
18. ^ a b c d Shmaliy, Y.S. (2007). Continuous-Time Signals. Springer. ISBN 978-1402062711.
19. ^ Proakis, John G.; Manolakis, Dimitris G. (1996). (3rd ed.). Prentice Hall. p. 291. ISBN 978-0-13-373762-2.
20. ^ Sharkas, Hesham (2022). "Solution of Integral of The Fourth Power of a Finite-Length Exponential Fourier Series". ResearchGate. doi: 10.13140/RG.2.2.31527.83368/2.
21. ^ "Characterizations of a linear subspace associated with Fourier series". MathOverflow. 2010-11-19. Retrieved 2014-08-08.
22. ^
23. ^ Siebert, William McC. (1985). Circuits, signals, and systems. MIT Press. p. 402. ISBN 978-0-262-19229-3.
24. ^ Marton, L.; Marton, Claire (1990). Advances in Electronics and Electron Physics. Academic Press. p. 369. ISBN 978-0-12-014650-5.
25. ^ Kuzmany, Hans (1998). Solid-state spectroscopy. Springer. p. 14. ISBN 978-3-540-63913-8.
26. ^ Pribram, Karl H.; Yasue, Kunio; Jibu, Mari (1991). Brain and perception. Lawrence Erlbaum Associates. p. 26. ISBN 978-0-89859-995-4.
### Further reading
• William E. Boyce; Richard C. DiPrima (2005). Elementary Differential Equations and Boundary Value Problems (8th ed.). New Jersey: John Wiley & Sons, Inc. ISBN 0-471-43338-1.
• Joseph Fourier, translated by Alexander Freeman (2003). The Analytical Theory of Heat. Dover Publications. ISBN 0-486-49531-0. 2003 unabridged republication of the 1878 English translation by Alexander Freeman of Fourier's work Théorie Analytique de la Chaleur, originally published in 1822.
• Enrique A. Gonzalez-Velasco (1992). "Connections in Mathematical Analysis: The Case of Fourier Series". American Mathematical Monthly. 99 (5): 427–441. doi: 10.2307/2325087. JSTOR 2325087.
• Fetter, Alexander L.; Walecka, John Dirk (2003). Theoretical Mechanics of Particles and Continua. Courier. ISBN 978-0-486-43261-8.
• Katznelson, Yitzhak (1976). An introduction to harmonic analysis (Second corrected ed.). New York: Dover Publications, Inc. ISBN 0-486-63331-4.
• Felix Klein, Development of mathematics in the 19th century. Mathsci Press Brookline, Mass, 1979. Translated by M. Ackerman from Vorlesungen über die Entwicklung der Mathematik im 19 Jahrhundert, Springer, Berlin, 1928.
• Walter Rudin (1976). (3rd ed.). New York: McGraw-Hill, Inc. ISBN 0-07-054235-X.
• A. Zygmund (2002). Trigonometric Series (third ed.). Cambridge: Cambridge University Press. ISBN 0-521-89053-5. The first edition was published in 1935. | 19,158 | 62,548 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 403, "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} | 3.890625 | 4 | CC-MAIN-2022-49 | latest | en | 0.909204 |
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### Section 4.7 : IVP's With Step Functions
In this section we will use Laplace transforms to solve IVP’s which contain Heaviside functions in the forcing function. This is where Laplace transform really starts to come into its own as a solution method.
To work these problems we’ll just need to remember the following two formulas,
\begin{align*}\mathcal{L}\left\{ {{u_c}\left( t \right)f\left( {t - c} \right)} \right\} & = {{\bf{e}}^{ - c\,s}}F\left( s \right) & & {\mbox{where }}F\left( s \right) = \mathcal{L}\left\{ {f\left( t \right)} \right\}\\ {\mathcal{L}^{\, - 1}}\left\{ {{{\bf{e}}^{ - c\,s}}F\left( s \right)} \right\} & = {u_c}\left( t \right)f\left( {t - c} \right) & & {\mbox{where }}f\left( t \right) = {\mathcal{L}^{\, - 1}}\left\{ {F\left( s \right)} \right\}\end{align*}
In other words, we will always need to remember that in order to take the transform of a function that involves a Heaviside we’ve got to make sure the function has been properly shifted.
Let’s work an example.
Example 1 Solve the following IVP. $y'' - y' + 5y = 4 + {u_2}\left( t \right){{\bf{e}}^{4 - 2t}},\hspace{0.25in}y\left( 0 \right) = 2\,\,\,\,\,\,\,\,y'\left( 0 \right) = - 1$
Show Solution
First let’s rewrite the forcing function to make sure that it’s being shifted correctly and to identify the function that is actually being shifted.
$y'' - y' + 5y = 4 + {u_2}\left( t \right){{\bf{e}}^{ - 2\left( {t - 2} \right)}}$
So, it is being shifted correctly and the function that is being shifted is $${{\bf{e}}^{ - 2t}}$$. Taking the Laplace transform of everything and plugging in the initial conditions gives,
\begin{align*}{s^2}Y\left( s \right) - sy\left( 0 \right) - y'\left( 0 \right) - \left( {sY\left( s \right) - y\left( 0 \right)} \right) + 5Y\left( s \right) & = \frac{4}{s} + \frac{{{{\bf{e}}^{ - 2s}}}}{{s + 2}}\\ \left( {{s^2} - s + 5} \right)Y\left( s \right) - 2s + 3 & = \frac{4}{s} + \frac{{{{\bf{e}}^{ - 2s}}}}{{s + 2}}\end{align*}
Now solve for $$Y(s)$$.
\begin{align*}\left( {{s^2} - s + 5} \right)Y\left( s \right) & = \frac{4}{s} + \frac{{{{\bf{e}}^{ - 2s}}}}{{s + 2}} + 2s - 3\\ \left( {{s^2} - s + 5} \right)Y\left( s \right) & = \frac{{2{s^2} - 3s + 4}}{s} + \frac{{{{\bf{e}}^{ - 2s}}}}{{s + 2}}\\ Y\left( s \right) & = \frac{{2{s^2} - 3s + 4}}{{s\left( {{s^2} - s + 5} \right)}} + {{\bf{e}}^{ - 2s}}\frac{1}{{\left( {s + 2} \right)\left( {{s^2} - s + 5} \right)}}\\ Y\left( s \right) & = F\left( s \right) + {{\bf{e}}^{ - 2s}}G\left( s \right)\end{align*}
Notice that we combined a couple of terms to simplify things a little. Now we need to partial fraction $$F(s)$$ and $$G(s)$$. We’ll leave it to you to check the details of the partial fractions.
\begin{align*}F\left( s \right) & = \frac{{2{s^2} - 3s + 4}}{{s\left( {{s^2} - s + 5} \right)}} = \frac{1}{5}\left( {\frac{4}{s} + \frac{{6s - 11}}{{{s^2} - s + 5}}} \right)\\ G\left( s \right) & = \frac{1}{{\left( {s + 2} \right)\left( {{s^2} - s + 5} \right)}} = \frac{1}{{11}}\left( {\frac{1}{{s + 2}} - \frac{{s - 3}}{{{s^2} - s + 5}}} \right)\end{align*}
We now need to do the inverse transforms on each of these. We’ll start with $$F(s)$$.
\begin{align*}F\left( s \right) & = \frac{1}{5}\left( {\frac{4}{s} + \frac{{6\left( {s - \frac{1}{2} + \frac{1}{2}} \right) - 11}}{{{{\left( {s - \frac{1}{2}} \right)}^2} + \frac{{19}}{4}}}} \right)\\ & = \frac{1}{5}\left( {\frac{4}{s} + \frac{{6\left( {s - \frac{1}{2}} \right)}}{{{{\left( {s - \frac{1}{2}} \right)}^2} + \frac{{19}}{4}}} - \frac{{8\frac{{\sqrt {19} }}{2}\frac{2}{{\sqrt {19} }}}}{{{{\left( {s - \frac{1}{2}} \right)}^2} + \frac{{19}}{4}}}} \right)\\ f\left( t \right) & = \frac{1}{5}\left( {4 + 6{{\bf{e}}^{\frac{t}{2}}}\cos \left( {\frac{{\sqrt {19} }}{2}t} \right) - \frac{{16}}{{\sqrt {19} }}{{\bf{e}}^{\frac{t}{2}}}\sin \left( {\frac{{\sqrt {19} }}{2}t} \right)} \right)\end{align*}
Now $$G(s)$$.
\begin{align*}G\left( s \right) & = \frac{1}{{11}}\left( {\frac{1}{{s + 2}} - \frac{{s - \frac{1}{2} + \frac{1}{2} - 3}}{{{{\left( {s - \frac{1}{2}} \right)}^2} + \frac{{19}}{4}}}} \right)\\ & = \frac{1}{{11}}\left( {\frac{1}{{s + 2}} - \frac{{s - \frac{1}{2}}}{{{{\left( {s - \frac{1}{2}} \right)}^2} + \frac{{19}}{4}}} + \frac{{\frac{5}{2}\frac{{\sqrt {19} }}{{\sqrt {19} }}}}{{{{\left( {s - \frac{1}{2}} \right)}^2} + \frac{{19}}{4}}}} \right)\\ g\left( t \right) & = \frac{1}{{11}}\left( {{{\bf{e}}^{ - 2t}} - {{\bf{e}}^{\frac{t}{2}}}\cos \left( {\frac{{\sqrt {19} }}{2}t} \right) + \frac{5}{{\sqrt {19} }}{{\bf{e}}^{\frac{t}{2}}}\sin \left( {\frac{{\sqrt {19} }}{2}t} \right)} \right)\end{align*}
Okay, we can now get the solution to the differential equation. Starting with the transform we get,
\begin{align*}Y\left( s \right) & = F\left( s \right) + {{\bf{e}}^{ - 2s}}G\left( s \right)\\ y\left( t \right) & = f\left( t \right) + {u_2}\left( t \right)g\left( {t - 2} \right)\end{align*}
where $$f(t)$$ and $$g(t)$$ are the functions shown above.
There can be a fair amount of work involved in solving differential equations that involve Heaviside functions.
Let’s take a look at another example or two.
Example 2 Solve the following IVP. $y'' - y' = \cos \left( {2t} \right) + \cos \left( {2t - 12} \right){u_6}\left( t \right)\quad \quad y\left( 0 \right) = - 4,\;y'\left( 0 \right) = 0$
Show Solution
Let’s rewrite the differential equation so we can identify the function that is actually being shifted.
$y'' - y' = \cos \left( {2t} \right) + \cos \left( {2\left( {t - 6} \right)} \right){u_6}\left( t \right)$
So, the function that is being shifted is $$\cos \left( {2t} \right)$$ and it is being shifted correctly. Taking the Laplace transform of everything and plugging in the initial conditions gives,
\begin{align*}{s^2}Y\left( s \right) - sy\left( 0 \right) - y'\left( 0 \right) - \left( {sY\left( s \right) - y\left( 0 \right)} \right) & = \frac{s}{{{s^2} + 4}} + \frac{{s{{\bf{e}}^{ - 6s}}}}{{{s^2} + 4}}\\ \left( {{s^2} - s} \right)Y\left( s \right) + 4s - 4 & = \frac{s}{{{s^2} + 4}} + \frac{{s{{\bf{e}}^{ - 6s}}}}{{{s^2} + 4}}\end{align*}
Now solve for $$Y(s)$$.
\begin{align*}\left( {{s^2} - s} \right)Y\left( s \right) & = \frac{{s + s{{\bf{e}}^{ - 6s}}}}{{{s^2} + 4}} - 4s + 4\\ Y\left( s \right) & = \frac{{s\left( {1 + {{\bf{e}}^{ - 6s}}} \right)}}{{s\left( {s - 1} \right)\left( {{s^2} + 4} \right)}} - 4\frac{{s - 1}}{{s\left( {s - 1} \right)}}\\ & = \frac{{1 + {{\bf{e}}^{ - 6s}}}}{{\left( {s - 1} \right)\left( {{s^2} + 4} \right)}} - \frac{4}{s}\\ Y\left( s \right) & = \left( {1 + {{\bf{e}}^{ - 6s}}} \right)F\left( s \right) - \frac{4}{s}\end{align*}
Notice that we combined the first two terms to simplify things a little. Also, there was some canceling going on in this one. Do not expect that to happen on a regular basis. We now need to partial fraction $$F(s)$$. We’ll leave the details to you to check.
\begin{align*}F\left( s \right) & = \frac{1}{{\left( {s - 1} \right)\left( {{s^2} + 4} \right)}} = \frac{1}{5}\left( {\frac{1}{{s - 1}} - \frac{{s + 1}}{{{s^2} + 4}}} \right)\\ f\left( t \right) & = \frac{1}{5}\left( {{{\bf{e}}^t} - \cos \left( {2t} \right) - \frac{1}{2}\sin \left( {2t} \right)} \right)\end{align*}
Okay, we can now get the solution to the differential equation. Starting with the transform we get,
\begin{align*}Y\left( s \right) & = F\left( s \right) + F\left( s \right){{\bf{e}}^{ - 6s}} - \frac{4}{s}\\ y\left( t \right) & = f\left( t \right) + {u_6}\left( t \right)f\left( {t - 6} \right) - 4\end{align*}
where $$f(t)$$ is given above.
Example 3 Solve the following IVP. $y'' - 5y' - 14y = 9 + {u_3}\left( t \right) + 4\left( {t - 1} \right){u_1}\left( t \right)\quad \quad y\left( 0 \right) = 0,\;y'\left( 0 \right) = 10$
Show Solution
Let’s take the Laplace transform of everything and note that in the third term we are shifting 4$$t$$.
\begin{align*}{s^2}Y\left( s \right) - sy\left( 0 \right) - y'\left( 0 \right) - 5\left( {sY\left( s \right) - y\left( 0 \right)} \right) - 14Y\left( s \right) & = \frac{9}{s} + \frac{{{{\bf{e}}^{ - 3s}}}}{s} + 4\frac{{{{\bf{e}}^{ - s}}}}{{{s^2}}}\\ \left( {{s^2} - 5s - 14} \right)Y\left( s \right) - 10 & = \frac{{9 + {{\bf{e}}^{ - 3s}}}}{s} + 4\frac{{{{\bf{e}}^{ - s}}}}{{{s^2}}}\end{align*}
Now solve for $$Y(s)$$.
\begin{align*}\left( {{s^2} - 5s - 14} \right)Y\left( s \right) - 10 & = \frac{{9 + {{\bf{e}}^{ - 3s}}}}{s} + 4\frac{{{{\bf{e}}^{ - s}}}}{{{s^2}}}\\ Y\left( s \right) & = \frac{{9 + {{\bf{e}}^{ - 3s}}}}{{s\left( {s - 7} \right)\left( {s + 2} \right)}} + \frac{{4{{\bf{e}}^{ - s}}}}{{{s^2}\left( {s - 7} \right)\left( {s + 2} \right)}} + \frac{{10}}{{\left( {s - 7} \right)\left( {s + 2} \right)}}\\ Y\left( s \right) & = \left( {9 + {{\bf{e}}^{ - 3s}}} \right)F\left( s \right) + 4{{\bf{e}}^{ - s}}G\left( s \right) + H\left( s \right)\end{align*}
So, we have three functions that we’ll need to partial fraction for this problem. I’ll leave it to you to check the details.
\begin{align*}F\left( s \right) & = \frac{1}{{s\left( {s - 7} \right)\left( {s + 2} \right)}} = - \frac{1}{{14}}\frac{1}{s} + \frac{1}{{63}}\frac{1}{{s - 7}} + \frac{1}{{18}}\frac{1}{{s + 2}}\\ f\left( t \right) & = - \frac{1}{{14}} + \frac{1}{{63}}{{\bf{e}}^{7t}} + \frac{1}{{18}}{{\bf{e}}^{ - 2t}}\end{align*} \begin{align*}G\left( s \right) & = \frac{1}{{{s^2}\left( {s - 7} \right)\left( {s + 2} \right)}} = \frac{5}{{196}}\frac{1}{s} - \frac{1}{{14}}\frac{1}{{{s^2}}} + \frac{1}{{441}}\frac{1}{{s - 7}} - \frac{1}{{36}}\frac{1}{{s + 2}}\\ g\left( t \right) & = \frac{5}{{196}} - \frac{1}{{14}}t + \frac{1}{{441}}{{\bf{e}}^{7t}} - \frac{1}{{36}}{{\bf{e}}^{ - 2t}}\end{align*} \begin{align*}H\left( s \right) & = \frac{{10}}{{\left( {s - 7} \right)\left( {s + 2} \right)}} = \frac{{10}}{9}\frac{1}{{s - 7}} - \frac{{10}}{9}\frac{1}{{s + 2}}\\ h\left( t \right) & = \frac{{10}}{9}{{\bf{e}}^{7t}} - \frac{{10}}{9}{{\bf{e}}^{ - 2t}}\end{align*}
Okay, we can now get the solution to the differential equation. Starting with the transform we get,
\begin{align*}Y\left( s \right) & = 9F\left( s \right) + {{\bf{e}}^{ - 3s}}F\left( s \right) + 4{{\bf{e}}^{ - s}}G\left( s \right) + H\left( s \right)\\ y\left( t \right) & = 9f\left( t \right) + {u_3}\left( t \right)f\left( {t - 3} \right) + 4{u_1}\left( t \right)g\left( {t - 1} \right) + h\left( t \right)\end{align*}
where $$f(t)$$, $$g(t)$$ and $$h(t)$$ are given above.
Let’s work one more example.
Example 4 Solve the following IVP. $y'' + 3y' + 2y = g\left( t \right),\hspace{0.25in}y\left( 0 \right) = 0\,\,\,\,\,\,\,\,y'\left( 0 \right) = - 2$
where,
$g\left( t \right) = \left\{ {\begin{array}{*{20}{l}}2&{\hspace{0.25in}t < 6}\\t&{\hspace{0.25in}6 \le t < 10}\\4&{\hspace{0.25in}t \ge 10}\end{array}} \right.$
Show Solution
The first step is to get $$g(t)$$ written in terms of Heaviside functions so that we can take the transform.
$g\left( t \right) = 2 + \left( {t - 2} \right){u_6}\left( t \right) + \left( {4 - t} \right){u_{10}}\left( t \right)$
Now, while this is $$g(t)$$ written in terms of Heaviside functions it is not yet in proper form for us to take the transform. Remember that each function must be shifted by a proper amount. So, getting things set up for the proper shifts gives us,
\begin{align*}g\left( t \right) & = 2 + \left( {t - 6 + 6 - 2} \right){u_6}\left( t \right) + \left( {4 - \left( {t - 10 + 10} \right)} \right){u_{10}}\left( t \right)\\ g\left( t \right) & = 2 + \left( {t - 6 + 4} \right){u_6}\left( t \right) + \left( { - 6 - \left( {t - 10} \right)} \right){u_{10}}\left( t \right)\end{align*}
So, for the first Heaviside it looks like $$f\left( t \right) = t + 4$$ is the function that is being shifted and for the second Heaviside it looks like $$f\left( t \right) = - 6 - t$$ is being shifted.
Now take the Laplace transform of everything and plug in the initial conditions.
\begin{align*}{s^2}Y\left( s \right) - sy\left( 0 \right) - y'\left( 0 \right) + 3\left( {sY\left( s \right) - y\left( 0 \right)} \right) + 2Y\left( s \right) & = \frac{2}{s} + {{\bf{e}}^{ - 6s}}\left( {\frac{1}{{{s^2}}} + \frac{4}{s}} \right) - {{\bf{e}}^{ - 10s}}\left( {\frac{1}{{{s^2}}} + \frac{6}{s}} \right)\\ \left( {{s^2} + 3s + 2} \right)Y\left( s \right) + 2 & = \frac{2}{s} + {{\bf{e}}^{ - 6s}}\left( {\frac{1}{{{s^2}}} + \frac{4}{s}} \right) - {{\bf{e}}^{ - 10s}}\left( {\frac{1}{{{s^2}}} + \frac{6}{s}} \right)\end{align*}
Solve for $$Y(s)$$.
\begin{align*}\left( {{s^2} + 3s + 2} \right)Y\left( s \right) & = \frac{2}{s} + {{\bf{e}}^{ - 6s}}\left( {\frac{1}{{{s^2}}} + \frac{4}{s}} \right) - {{\bf{e}}^{ - 10s}}\left( {\frac{1}{{{s^2}}} + \frac{6}{s}} \right) - 2\\ \left( {{s^2} + 3s + 2} \right)Y\left( s \right) & = \frac{{2 + 4{{\bf{e}}^{ - 6s}} - 6{{\bf{e}}^{ - 10s}}}}{s} + \frac{{{{\bf{e}}^{ - 6s}} - {{\bf{e}}^{ - 10s}}}}{{{s^2}}} - 2\\ Y\left( s \right) & = \frac{{2 + 4{{\bf{e}}^{ - 6s}} - 6{{\bf{e}}^{ - 10s}}}}{{s\left( {s + 1} \right)\left( {s + 2} \right)}} + \frac{{{{\bf{e}}^{ - 6s}} - {{\bf{e}}^{ - 10s}}}}{{{s^2}\left( {s + 1} \right)\left( {s + 2} \right)}} - \frac{2}{{\left( {s + 1} \right)\left( {s + 2} \right)}}\\ & \\ Y\left( s \right) & = \left( {2 + 4{{\bf{e}}^{ - 6s}} - 6{{\bf{e}}^{ - 10s}}} \right)F\left( s \right) + \left( {{{\bf{e}}^{ - 6s}} - {{\bf{e}}^{ - 10s}}} \right)G\left( s \right) - H\left( s \right)\end{align*}
Now, in the solving process we simplified things into as few terms as possible. Even doing this, it looks like we’ll still need to do three partial fractions.
We’ll leave the details of the partial fractioning to you to verify. The partial fraction form and inverse transform of each of these are.
\begin{align*}F\left( s \right) & = \frac{1}{{s\left( {s + 1} \right)\left( {s + 2} \right)}} = \frac{{\frac{1}{2}}}{s} - \frac{1}{{s + 1}} + \frac{{\frac{1}{2}}}{{s + 2}}\\ f\left( t \right) & = \frac{1}{2} - {{\bf{e}}^{ - t}} + \frac{1}{2}{{\bf{e}}^{ - 2t}}\end{align*} \begin{align*}G\left( s \right) & = \frac{1}{{{s^2}\left( {s + 1} \right)\left( {s + 2} \right)}} = - \frac{{\frac{3}{4}}}{s} + \frac{{\frac{1}{2}}}{{{s^2}}} + \frac{1}{{s + 1}} - \frac{{\frac{1}{4}}}{{s + 2}}\\ g\left( t \right) & = - \frac{3}{4} + \frac{1}{2}t + {{\bf{e}}^{ - t}} - \frac{1}{4}{{\bf{e}}^{ - 2t}}\end{align*} \begin{align*}H\left( s \right) & = \frac{2}{{\left( {s + 1} \right)\left( {s + 2} \right)}} = \frac{2}{{s + 1}} - \frac{2}{{s + 2}}\\ h\left( t \right) & = 2{{\bf{e}}^{ - t}} - 2{{\bf{e}}^{ - 2t}}\end{align*}
Putting this all back together is going to be a little messy. First rewrite the transform a little to make the inverse transform process possible.
$Y\left( s \right) = 2F\left( s \right) + {{\bf{e}}^{ - 6s}}\left( {4F\left( s \right) + G\left( s \right)} \right) - {{\bf{e}}^{ - 10s}}\left( {6F\left( s \right) + G\left( s \right)} \right) - H\left( s \right)$
Now, taking the inverse transform of all the pieces gives us the final solution to the IVP.
$y\left( t \right) = 2f\left( t \right) - h\left( t \right) + {u_6}\left( t \right)\left( {4f\left( {t - 6} \right) + g\left( {t - 6} \right)} \right) - {u_{10}}\left( t \right)\left( {6f\left( {t - 10} \right) + g\left( {t - 10} \right)} \right)$
where $$f(t)$$, $$g(t)$$, and $$h(t)$$ are defined above.
So, the answer to this example is a little messy to write down, but overall the work here wasn’t too terribly bad.
Before proceeding with the next section let’s see how we would have had to solve this IVP if we hadn’t had Laplace transforms. To solve this IVP we would have had to solve three separate IVP’s. One for each portion of $$g(t)$$. Here is a list of the IVP’s that we would have had to solve.
1. $$0 < t < 6$$ $y'' + 3y' + 2y = 2,\hspace{0.25in}y\left( 0 \right) = 0\,\,\,\,\,\,\,\,y'\left( 0 \right) = - 2$
The solution to this IVP, with some work, can be made to look like,
${y_1}\left( t \right) = 2f\left( t \right) - h\left( t \right)$
2. $$6 \le t < 10$$ $y'' + 3y' + 2y = t,\hspace{0.25in}y\left( 6 \right) = {y_1}\left( 6 \right)\,\,\,\,\,\,\,\,y'\left( 6 \right) = {y'_1}\left( 6 \right)$
where, $$y_{1}(t)$$ is the solution to the first IVP. The solution to this IVP, with some work, can be made to look like,
${y_2}\left( t \right) = 2f\left( t \right) - h\left( t \right) + 4f\left( {t - 6} \right) + g\left( {t - 6} \right)$
3. $$t \ge 10$$ $y'' + 3y' + 2y = 4,\hspace{0.25in}y\left( {10} \right) = {y_2}\left( {10} \right)\,\,\,\,\,\,\,\,y'\left( {10} \right) = {y'_2}\left( {10} \right)$
where, $$y_{2}(t)$$ is the solution to the second IVP. The solution to this IVP, with some work, can be made to look like,
${y_3}\left( t \right) = 2f\left( t \right) - h\left( t \right) + 4f\left( {t - 6} \right) + g\left( {t - 6} \right) - 6f\left( {t - 10} \right) - g\left( {t - 10} \right)$
There is a considerable amount of work required to solve all three of these and in each of these the forcing function is not that complicated. Using Laplace transforms saved us a fair amount of work. | 7,491 | 17,697 | {"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} | 4 | 4 | CC-MAIN-2024-33 | latest | en | 0.84212 |
http://www.carrawaydavie.com/liters-to-gallons-conversion-table/ | 1,571,406,200,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986682998.59/warc/CC-MAIN-20191018131050-20191018154550-00358.warc.gz | 232,163,517 | 8,669 | ## What is a liters to gallons conversion table
What is a liters to gallons conversion table as well as the US?
We will consider many other values needed in everyday life, such as liters to gallons conversion table, how many ounces in a gallon, gallons to an unce and many others.
There are 1000 milliliters in a liter (1000 mL in 1 liter).
Liters to fluid flows (fl oz) volume units conversion factors are listed below. To find out how much the liter is in the x liters, multiply the liter by the conversion factor.
1 Liter = 33.814022 US Fluid Ounces
1 Liter = 35.195079 Imperial Fluid Ounces
So, liters to gallons conversion table!
## What is a liquid ounce (fl oz) liters to gallons conversion table
A liquid ounce is an imperial as well as US customary measurement system volume unit. 1 US fluid ounce equals to 29.5735 mL and 1 royal (UK) fluid ounce equals to 28.4131 mL. So, liters to gallons conversion table?
There are 33.8140226 ounces in a liter because one liter (liter) is defined as the volume of 1 kilogram of water and there are 33.814 fluid ounces in the same volume of water.
There are 35.195079 imperial (UK) fl. ounces in a liter, because 1 fl. an ounce is 1.04084 imperial fl. ounce, and that makes 1.04084 * 33.814 (US fl oz in a liter)= 35.195079 imperial fl. oz in a liter. liters to gallons conversion table
## We need to know liters to gallons conversion table
1 gal is equal to 128 oz
Convert fluid formula to this formula:
pints = fluid ounces × 0.0625
1 Gallon (US, Fluid) = 128 Ounces (US, Fluid)
1 Gallon (UK, Fluid) = 160 Ounces (UK, Fluid)
liters to gallons conversion table
In Summary :.
1 gallon = 4 quarts = 8 pints = 16 mugs = 128 fluid ounces.
liters to gallons conversion table | 464 | 1,723 | {"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} | 3.171875 | 3 | CC-MAIN-2019-43 | latest | en | 0.850219 |
http://dimacs.rutgers.edu/Institute/96/course/lecture.6.html | 1,508,719,532,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825497.18/warc/CC-MAIN-20171023001732-20171023021732-00126.warc.gz | 102,055,223 | 3,434 | ## Lecture # 6
Instructor: Scot Drysdale July 1, 1996
Back to DIMACS Lectures
Homework Review - page 178 #3 & 5
Clarification on #5 In the Voronoi diagram the average number of edges is the same as the average degree in the Delaunay triangulation, which is 6. This means that on average you have 6 or fewer neighbors to look at to find the nearest neighbor.
There are 3n - 6 green edges in the Delaunay triangulation. There are 2n - 4 vertices. The average degree of a vertex in a Delaunay triangulation is the sum over all vertices in the Delaunay triangulation of degree of v multiplied by 1/n. This is at most 1/n times the sum over all edges, where each endpoint is counted separately. This is 1/n 2 (3n-6) = 6n/n - 12/n = 6 - 12/n.
Solution to # 3: Let B be the nearest neighbor of site A. Then there are no sites within the circle centered at A passing through B. The midpoint of segment AB is M. It is by definition equidistant from A and B, so will be on a boundary separating their Voronoi regions unless some point C lies closer to M than A and B.
But there can be no such C. The circle with diameter AB is strictly inside of the circle centered at A passing through B. There are no points inside the large circle, so there can be nothing inside the smaller circle.
Pizza Parlor location problem (discussion continued from last period)
The only places where a new pizza parlor can be placed that might maximize the distance to the nearest existing site are:
Voronoi vertices ( 2n - 4 = q(n) of them)
corners of boundary region. (Define this number to be b.)
Intersections between Voronoi edges and county boundary . q(n) if convex county boundary. If it is not convex then it is upper bounded by O(bn)
Software Demo
Voronoi - written at Dartmouth by Paul Chew.
Uses quick draw which is only for Mac, written in Pascal.
A Web version using Java applets is available at:
http://www.cs.cornell.edu/Info/People/chew/Delaunay.html
Incremental Algorithm basic idea is to add a point which creates a new region and update. First find what region the new point is in. Pick a region and start walking around the edges asking if you are on the correct side. When you find you are on the wrong side then cross over to a new region. If you go all the way around then you have found the correct region. This takes q( n) time on the average, but q(n) time in the worst case. Since the update is usually fast (expect to have at most 6 neighbors - see above), the search time is important. How can this be done?
If we could preprocess, we could do this search quickly.
Region search - Vertical lines placed through the Voronoi vertices (slide 1) - use a binary search to place the point in a slab. Then using the edges ask above or below. This method uses LOTS of memory, q(n2) space for n points. But the search can be done in 2log n time.
A method that works better for uniformly distributed points is: Grid bucketing- locate points in a grid. For a new point, find a point in the grid square where it lies or a nearby one, and use this point to start the search. This is actually a very fast method. (Expected time q(n) for the whole Delaunay diagram.) (Note - in the worst case the incremental method can take q(n2) time, because each new point may have to interact with all previous points. This can happen when the points are on a spiral.
Geometers Sketch Pad Demo
Find what triangle you are in, connect the point to the the three vertices of the triangle it is in. Put a circle through the 3 points that form the triangle to check the edge. If there any points inside the circle you will need to remove the edge and replace it with the other diagonal.
This process is known as the Guibas & Stolfi Algorithm.
Incremental in the worse case takes n log n time.
Divide and Conquer Method -
split the set in half, left and right
recursively check both sides
separate red and blue points
zipper method - red points stay on one side, blue on the other side of the line.
The line becomes part of the Voronoi diagram.
Proof by intimidation!
Take the convex hull of each side to find the place to start.
The time needed is n log n
The calculations needed are complex.
Guivas & Stolfi found an alternate way to do this using the dual of the Voronoi Diagram, which is the Delaunay triangulation. Think of an infinite bubble coming up the purple polygimal path, as it bumps into a Voronoi vertex it has an empty circle and then creates an edge.
Supplemental Notes:
Let's Make A Deal - The Monty Hall Problem can be found on the web at:
http://www.math.toronto.edu/mathnet/recreation.html
even if you are familiar with this problem you should visit this site, it has my personal stamp of approval.
MUDS discussion
Many of the words we use today in technology originated in the book:
Neuromancer by William Gibson, Ace Books, NY 1984
## Homework 5
1) 5.5.6[8] p. 192
2) In class we saw a divide-and-conquer algorithm for computing the Voronoi diagram. We computed a polygonal bisector curve that separated the points closer to some site in the left half from those closer to some point in the right half. Is it possible for this bisector curve to enter a Voronoi region through an edge and then leave the region through the SAME edge? Give an example or explain why it cannot happen.
Mailto: dimacs-www@dimacs.rutgers.edu | 1,296 | 5,329 | {"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} | 3.78125 | 4 | CC-MAIN-2017-43 | latest | en | 0.909996 |
https://encyclopediaofmath.org:443/index.php?title=Mathieu_functions | 1,670,432,084,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711200.6/warc/CC-MAIN-20221207153419-20221207183419-00196.warc.gz | 263,155,185 | 6,335 | # Mathieu functions
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The $2 \pi$- periodic solutions of the Mathieu equation
$$\frac{d ^ {2} u }{dz ^ {2} } + ( a + 16q \cos 2z) u = 0,\ \ z \in \mathbf R ,$$
which exist only when the point $( a, q)$ in the parameter plane lies on the boundary of the stability zones. A Mathieu function is even or odd, and is unique up to a factor; the second linearly-independent solution grows linearly in $z$ for $| z | \rightarrow \infty$, provided $q \neq 0$. The even Mathieu functions are the eigen functions of the integral equation
$$G( z) = \lambda \int\limits _ {- \pi } ^ \pi e ^ {k \cos z \cos t } G( t) dt,\ \ k = \sqrt 32q .$$
An analogous equation is satisfied by the odd Mathieu functions. The notation for Mathieu functions is:
$$ce _ {0} ( z, q), ce _ {1} ( z, q) , . . . ; \ se _ {1} ( z, q) , se _ {2} ( z, q) , . . . .$$
For $q \rightarrow 0$ these functions reduce to the trigonometric system
$$1 , \cos z , . . . ; \ \ \sin z, \sin 2z \dots$$
and they possess the same orthogonality properties on the interval $(- \pi , \pi )$. The Mathieu functions admit Fourier-series expansions which converge for small $| q | \leq r _ {n}$; the coefficients of these series are convergent power series in $q$, for example,
$$ce _ {0} ( z, q) = \ 1 +$$
$$+ \sum _ { n= } 1 ^ \infty \left [ 2 ^ {n+} 1 \frac{q ^ {n} }{( n!) ^ {2} } - \frac{n( 3n+ 4) 2 ^ {n+} 3 q ^ {n+} 2 }{(( n+ 1)!) ^ {2} } + O( q ^ {n+} 4 ) \right ] \cos 2nz.$$
#### References
[1] E.T. Whittaker, G.N. Watson, "A course of modern analysis" , Cambridge Univ. Press (1952) pp. Chapt. 2 [2] H. Bateman (ed.) A. Erdélyi (ed.) , Higher transcendental functions , 3. Automorphic functions , McGraw-Hill (1955) [3] G. Sansone, "Equazioni differenziali nel campo reale" , 1 , Zanichelli (1948) [4] M.J.O. Strett, "Lamésche-, Mathieusche- und verwandte Funktionen in Physik und Technik" , Springer (1932) [5] N.W. Mac-Lachlan, "Theory and application of Mathieu functions" , Clarendon Press (1947)
How to Cite This Entry:
Mathieu functions. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Mathieu_functions&oldid=47791
This article was adapted from an original article by M.V. Fedoryuk (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article | 777 | 2,316 | {"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} | 2.984375 | 3 | CC-MAIN-2022-49 | latest | en | 0.509009 |
afforess.github.io | 1,632,847,861,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060877.21/warc/CC-MAIN-20210928153533-20210928183533-00198.warc.gz | 2,244,794 | 11,037 | ## Area.Position module
Tools for working with `<x,y>` coordinates.
### Usage
``local Position = require('__stdlib__/stdlib/area/position')``
### Constructor Methods
new(pos) Returns a correctly formated position object. construct(x, y) Creates a table representing the position from x and y. update(pos, x, y) Update a position in place without returning a new position. load(pos) Load the metatable into the passed position without creating a new one. from_string(pos_string) Converts a position string to a position. from_key(pos_string) Converts a string key position to a position. from_chunk_position(pos) Gets the left top tile position of a chunk from the chunk position. from_pixels(pos) Convert position from pixels
### Metamethods
Metamethods Position tables are returned with these metamethods attached.
### Position Methods
add(pos1, ...) Addition of two positions. subtract(pos1, ...) Subtraction of two positions.. multiply(pos1, ...) Multiplication of two positions. divide(pos1, ...) Division of two positions. mod(pos1, ...) Modulo of two positions. closest(pos1, positions) Return the closest position to the first position. farthest(pos1, positions) Return the farthest position from the first position. between(pos1, pos2) The middle of two positions. projection(pos1, pos2) The projection point of two positions. reflection(pos1, pos2) The reflection point or two positions. store(pos) Stores the position for recall later, not deterministic. recall(pos) Recalls the stored position. normalize(pos) Normalizes a position by rounding it to 2 decimal places. abs(pos) Abs x, y values ceil(pos) Ceil x, y values. floor(pos) Floor x, y values. center(pos) The center position of the tile where the given position resides. round(pos) Rounds a positions points to the closest integer. perpendicular(pos) Perpendicular position. swap(pos) Swap the x and y coordinates. flip(pos) Flip the signs of the position. flip_x(pos) Flip the x sign. flip_y(pos) Flip the y sign. lerp(pos1, pos2, alpha) Lerp position of pos1 and pos2. trim(pos, max_len) Trim the position to a length. offset_along_line(pos1, pos2, distance_from_pos2) Returns the position along line between source and target, at the distance from target. translate(pos, direction, distance) Translates a position in the given direction. random(pos[, minimum=0][, maximum=1][, random_tile=false]) Return a random offset of a position. average(positions) Return the average position of the passed positions. min(positions) Return the minimum position of the passed positions. max(positions) Return the maximum position of the passed positions. min_xy(positions) Return a position created from the smallest x, y values in the passed positions. max_xy(positions) Return a position created from the largest x, y values in the passed positions. intersection(pos1_start, pos1_end, pos2_start, pos2_end) The intersection of 4 positions.
### Position Mutate Methods
normalized(pos) Normalizes a position by rounding it to 2 decimal places. absed(pos) Abs x, y values ceiled(pos) Ceil x, y values in place. floored(pos) Floor x, y values. centered(pos) The center position of the tile where the given position resides. rounded(pos) Rounds a positions points to the closest integer. swapped(pos) Swap the x and y coordinates. flipped(pos) Flip the signs of the position.
### Position Conversion Methods
to_pixels(pos) Convert to pixels from position to_chunk_position(pos) Gets the chunk position of a chunk where the specified position resides.
### Area Conversion Methods
expand_to_area(pos, radius) Expands a position to a square area. to_area(pos, width, height) Expands a position into an area by setting pos to left_top. to_tile_area(pos) Converts a tile position to the area of the tile it is in. to_chunk_area(pos) Get the chunk area the specified position is in. chunk_position_to_chunk_area(pos) Get the chunk area for the specified chunk position.
### Position Functions
len_squared(pos) Gets the squared length of a position len(pos) Gets the length of a position to_string(pos) Converts a position to a string. to_string_xy(pos) Converts a position to an x, y string. to_key(pos) Converts a position to a string suitable for using as a table index. unpack(pos) Unpack a position into a tuple. pack(pos) Packs a position into an array. is_zero(pos) Is this position {0, 0}. inside(pos, area) Is a position inside of an area. is_simple_position(pos) Is this a simple position. is_complex_position(pos) Is this a complex position. is_Position(pos) Does the position have the class attached is_position(pos) Is this any position atan2(pos1, pos2) Return the atan2 of 2 positions. angle(pos1, pos2) The angle between two positions cross(pos1, pos2) Return the cross product of two positions. equals(pos1, pos2) Tests whether or not the two given positions are equal. less_than(pos1, pos2) Is pos1 less than pos2. less_than_eq(pos1, pos2) Is pos1 less than or equal to pos2. distance_squared(pos1[, pos2]) Calculates the Euclidean distance squared between two positions, useful when sqrt is not needed. distance(pos1[, pos2={x=0]) Calculates the Euclidean distance between two positions. manhattan_distance(pos1[, pos2]) Calculates the manhatten distance between two positions. direction_to(pos1, pos2) Returms the direction to a position using simple delta comparisons. complex_direction_to(pos1, pos2, eight_way) Returns the direction to a position. increment(pos[, inc_x=0][, inc_y=0][, increment_initial=false]) Increment a position each time it is called. increment_closure([new_inc_x=0][, new_inc_y=0]) A closure which the increment function returns.
### Constructor Methods
# new(pos)
Returns a correctly formated position object.
Parameters:
• pos : (Position) the position table or array to convert
Returns: Usage:
``Position.new({0, 0}) -- returns {x = 0, y = 0}``
# construct(x, y)
Creates a table representing the position from x and y.
Parameters:
Returns:
# update(pos, x, y)
Update a position in place without returning a new position.
Parameters:
Returns:
• Concepts.Position the passed position updated.
Load the metatable into the passed position without creating a new one.
Always assumes a valid position is passed
Parameters:
• pos : (Position) the position to set the metatable onto
Returns:
• (Position) the position with metatable attached
# from_string(pos_string)
Converts a position string to a position.
Parameters:
• pos_string : (string) the position to convert
Returns:
# from_key(pos_string)
Converts a string key position to a position.
Parameters:
• pos_string : (string) the position to convert
Returns:
# from_chunk_position(pos)
Gets the left top tile position of a chunk from the chunk position.
Parameters:
Returns:
# from_pixels(pos)
Convert position from pixels
Parameters:
Returns:
### Metamethods
# Metamethods
Position tables are returned with these metamethods attached.
Methods that return a position will return a NEW position without modifying the passed positions.
Fields:
• __class
• __index : If key is not found, see if there is one availble in the Position module.
• __sub : Subtracts one position from another. Returns a new position.
• __mul : Multiply 2 positions. Returns a new position.
• __div : Divide 2 positions. Returns a new position.
• __mod : Modulo of 2 positions. Returns a new position.
• __unm : Unary Minus of a position. Returns a new position.
• __len : Length of a single position.
• __eq : Are two positions at the same spot.
• __lt : Is position1 less than position2.
• __le : Is position1 less than or equal to position2.
• __tostring : Returns a string representation of the position
• __concat : calls tostring on both sides of concact.
• __call : copy the position.
• __debugline
### Position Methods
Parameters:
• pos1 : (Position)
• ... : () position or x, y values.
Returns:
# subtract(pos1, ...)
Subtraction of two positions..
Parameters:
• pos1 : (Position)
• ... : () position or x, y values
Returns:
# multiply(pos1, ...)
Multiplication of two positions.
Parameters:
• pos1 : (Position)
• ... : () position or x, y values
Returns:
# divide(pos1, ...)
Division of two positions.
Parameters:
• pos1 : (Position)
• ... : () position or x, y values
Returns:
# mod(pos1, ...)
Modulo of two positions.
Parameters:
• pos1 : (Position)
• ... : () position or x, y values
Returns:
# closest(pos1, positions)
Return the closest position to the first position.
Parameters:
• pos1 : (Positions) The position to find the closest too
• positions : (array) array of Concepts.Position
Returns:
# farthest(pos1, positions)
Return the farthest position from the first position.
Parameters:
• pos1 : (Positions) The position to find the farthest from
• positions : (array) array of Concepts.Position
Returns:
# between(pos1, pos2)
The middle of two positions.
Parameters:
Returns:
• (Position) pos1 the middle of two positions
# projection(pos1, pos2)
The projection point of two positions.
Parameters:
Returns:
# reflection(pos1, pos2)
The reflection point or two positions.
Parameters:
Returns:
# store(pos)
Stores the position for recall later, not deterministic.
Only the last position stored is saved.
Parameters:
# recall(pos)
Recalls the stored position.
Parameters:
Returns:
# normalize(pos)
Normalizes a position by rounding it to 2 decimal places.
Parameters:
Returns:
# abs(pos)
Abs x, y values
Parameters:
Returns:
# ceil(pos)
Ceil x, y values.
Parameters:
Returns:
# floor(pos)
Floor x, y values.
Parameters:
Returns:
# center(pos)
The center position of the tile where the given position resides.
Parameters:
Returns:
• (Position) A new position at the center of the tile
# round(pos)
Rounds a positions points to the closest integer.
Parameters:
Returns:
# perpendicular(pos)
Perpendicular position.
Parameters:
Returns:
# swap(pos)
Swap the x and y coordinates.
Parameters:
Returns:
• (Position) A new position with x and y swapped
# flip(pos)
Flip the signs of the position.
Parameters:
Returns:
• Concepts.Position A new position with flipped signs
# flip_x(pos)
Flip the x sign.
Parameters:
Returns:
• Concepts.Position A new position with flipped sign on the x
# flip_y(pos)
Flip the y sign.
Parameters:
Returns:
• Concepts.Position A new position with flipped sign on the y
# lerp(pos1, pos2, alpha)
Lerp position of pos1 and pos2.
Parameters:
Returns:
# trim(pos, max_len)
Trim the position to a length.
Parameters:
# offset_along_line(pos1, pos2, distance_from_pos2)
Returns the position along line between source and target, at the distance from target.
Parameters:
• pos1 : (Position) where the line starts and extends from.
• pos2 : (Position) where the line ends and is offset back from.
• distance_from_pos2 : (number) backwards from pos1 for the new position.
Returns:
• (Position) a point along line between source and target, at requested offset back from target.
# translate(pos, direction, distance)
Translates a position in the given direction.
Parameters:
Returns:
# random(pos[, minimum=0][, maximum=1][, random_tile=false])
Return a random offset of a position.
Parameters:
• pos : (Position) the position to randomize
• minimum : (number) the minimum amount to offset (default: 0)
• maximum : (number) the maximum amount to offset (default: 1)
• random_tile : (boolean) randomize the location on the tile (default: false)
Returns:
• (Position) a new random offset position
# average(positions)
Return the average position of the passed positions.
Parameters:
• positions : (array) array of Concepts.Position
Returns:
# min(positions)
Return the minimum position of the passed positions.
Parameters:
• positions : (array) array of Concepts.Position
Returns:
# max(positions)
Return the maximum position of the passed positions.
Parameters:
• positions : (array) array of Concepts.Position
Returns:
# min_xy(positions)
Return a position created from the smallest x, y values in the passed positions.
Parameters:
• positions : (array) array of Concepts.Position
Returns:
# max_xy(positions)
Return a position created from the largest x, y values in the passed positions.
Parameters:
• positions : (array) array of Concepts.Position
Returns:
# intersection(pos1_start, pos1_end, pos2_start, pos2_end)
The intersection of 4 positions.
Parameters:
• pos1_start
• pos1_end
• pos2_start
• pos2_end
Returns:
### Position Mutate Methods
# normalized(pos)
Normalizes a position by rounding it to 2 decimal places.
Parameters:
Returns:
• (Position) the normalized position mutated
# absed(pos)
Abs x, y values
Parameters:
Returns:
• (Position) the absolute position mutated
# ceiled(pos)
Ceil x, y values in place.
Parameters:
Returns:
• (Position) the ceiled position mutated
# floored(pos)
Floor x, y values.
Parameters:
Returns:
• (Position) the floored position mutated
# centered(pos)
The center position of the tile where the given position resides.
Parameters:
Returns:
• (Position) the centered position mutated
# rounded(pos)
Rounds a positions points to the closest integer.
Parameters:
Returns:
• (Position) the rounded position mutated
# swapped(pos)
Swap the x and y coordinates.
Parameters:
Returns:
• (Position) the swapped position mutated
# flipped(pos)
Flip the signs of the position.
Parameters:
Returns:
• Concepts.Position the flipped position mutated
### Position Conversion Methods
# to_pixels(pos)
Convert to pixels from position
Parameters:
Returns:
# to_chunk_position(pos)
Gets the chunk position of a chunk where the specified position resides.
Parameters:
• pos : (Position) a position residing somewhere in a chunk
Returns:
Usage:
``local chunk_x = Position.chunk_position(pos).x``
### Area Conversion Methods
Expands a position to a square area.
Parameters:
• pos : (Position) the position to expand into an area
• radius : (number) half of the side length of the area
Returns:
# to_area(pos, width, height)
Expands a position into an area by setting pos to left_top.
Parameters:
Returns:
# to_tile_area(pos)
Converts a tile position to the area of the tile it is in.
Parameters:
Returns:
# to_chunk_area(pos)
Get the chunk area the specified position is in.
Parameters:
Returns:
# chunk_position_to_chunk_area(pos)
Get the chunk area for the specified chunk position.
Parameters:
Returns:
### Position Functions
# len_squared(pos)
Gets the squared length of a position
Parameters:
Returns:
# len(pos)
Gets the length of a position
Parameters:
Returns:
# to_string(pos)
Converts a position to a string.
Parameters:
• pos : (Position) the position to convert
Returns:
• (string) string representation of the position
# to_string_xy(pos)
Converts a position to an x, y string.
Parameters:
• pos : (Position) the position to convert
Returns:
# to_key(pos)
Converts a position to a string suitable for using as a table index.
Parameters:
• pos : (Position) the position to convert
Returns:
# unpack(pos)
Unpack a position into a tuple.
Parameters:
• pos : (Position) the position to unpack
Returns:
• (tuple) x, y
# pack(pos)
Packs a position into an array.
Parameters:
Returns:
# is_zero(pos)
Is this position {0, 0}.
Parameters:
Returns:
# inside(pos, area)
Is a position inside of an area.
Parameters:
Returns:
• (boolean) Is the position inside of the area.
# is_simple_position(pos)
Is this a simple position.
{num, num}
Parameters:
Returns:
# is_complex_position(pos)
Is this a complex position.
{x = number, y = number}
Parameters:
Returns:
# is_Position(pos)
Does the position have the class attached
Parameters:
Returns:
# is_position(pos)
Is this any position
Parameters:
Returns:
# atan2(pos1, pos2)
Return the atan2 of 2 positions.
Parameters:
Returns:
# angle(pos1, pos2)
The angle between two positions
Parameters:
Returns:
# cross(pos1, pos2)
Return the cross product of two positions.
Parameters:
Returns:
# equals(pos1, pos2)
Tests whether or not the two given positions are equal.
Parameters:
Returns:
• (boolean) true if positions are equal
# less_than(pos1, pos2)
Is pos1 less than pos2.
Parameters:
Returns:
# less_than_eq(pos1, pos2)
Is pos1 less than or equal to pos2.
Parameters:
Returns:
# distance_squared(pos1[, pos2])
Calculates the Euclidean distance squared between two positions, useful when sqrt is not needed.
Parameters:
Returns:
• (number) the square of the euclidean distance
# distance(pos1[, pos2={x=0])
Calculates the Euclidean distance between two positions.
Parameters:
Returns:
• (number) the euclidean distance
# manhattan_distance(pos1[, pos2])
Calculates the manhatten distance between two positions.
Parameters:
Returns:
• (number) the manhatten distance
# direction_to(pos1, pos2)
Returms the direction to a position using simple delta comparisons.
Parameters:
Returns:
# complex_direction_to(pos1, pos2, eight_way)
Returns the direction to a position.
Parameters:
Returns:
# increment(pos[, inc_x=0][, inc_y=0][, increment_initial=false])
Increment a position each time it is called.
This can be used to increment or even decrement a position quickly.
Do not store function closures in the global object; use them in the current tick.
Parameters:
• inc_x : (number) optional increment x by this amount (default: 0)
• inc_y : (number) optional increment y by this amount (default: 0)
• increment_initial : (boolean) Whether the first use should be incremented (default: false)
Returns:
Usage:
``````local next_pos = Position.increment({0,0})
for i = 1, 5 do next_pos(0,1) -- returns {x = 0, y = 1} {x = 0, y = 2} {x = 0, y = 3} {x = 0, y = 4} {x = 0, y = 5}``````
``````local next_pos = Position.increment({0, 0}, 1)
next_pos() -- returns {1, 0}
next_pos(0, 5) -- returns {1, 5}
next_pos(nil, 5) -- returns {2, 10}``````
``````local next_pos = Position.increment({0, 0}, 0, 1)
surface.create_entity{name = 'flying-text', text = 'text', position = next_pos()}
surface.create_entity{name = 'flying-text', text = 'text', position = next_pos()} -- creates two flying text entities 1 tile apart``````
# increment_closure([new_inc_x=0][, new_inc_y=0])
A closure which the increment function returns.
> Do not call this directly and do not store this in the global object.
Parameters:
• new_inc_x : (number) (default: 0)
• new_inc_y : (number) (default: 0)
Returns: | 4,341 | 18,306 | {"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} | 2.53125 | 3 | CC-MAIN-2021-39 | latest | en | 0.581844 |
https://brilliant.org/problems/joaos-functional-equation/ | 1,624,163,548,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487655418.58/warc/CC-MAIN-20210620024206-20210620054206-00175.warc.gz | 146,264,841 | 8,360 | # João's functional equation
Algebra Level 2
Let $f$ be a function from the positive integers to the positive integers that satisfies the property $f (x + y) = f (x) f (y)$
for all pairs of positive integers $(x,y)$. If we are given that $f(2) = 9$, what is the value of $f(5)$?
This problem is posed by João B.
× | 96 | 318 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 5, "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} | 2.75 | 3 | CC-MAIN-2021-25 | latest | en | 0.826898 |
https://web2.0calc.com/questions/how-to-solve_5 | 1,511,414,074,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806736.55/warc/CC-MAIN-20171123050243-20171123070243-00249.warc.gz | 689,158,057 | 6,357 | +0
# how to solve
0
283
2
tan x + 4 sec x = 4 , 0<x<360
Guest May 27, 2015
#2
+78741
+5
Ah....Mathcad......you did very well but you gave up a little too quickly, here........!!!!......notice the following factorization....
15sec^2(x)-32sec(x)+17=0
(15sec(x) - 17) (sec(x) -1) = 0
So either......
sec(x) = 1 which happens at 0 and 360.....but, both of these are out of the requested interval....or....
Here's a graph..........https://www.desmos.com/calculator/fpewbunwdj
Notice that the solution of 28.072° is "extraneous"......this will frequently happen when we square both sides of an equation......the only "good" solution in the requested interval occurs at about 331.928°
CPhill May 28, 2015
Sort:
#1
+122
+5
tan(x)+4sec(x)=4
tan(x)=4(1-sec(x)) Now square both sides to get
tan^2(x)=16(1-sec(x))^2 and using the identity tan^2(x)=sec^2(x)-1 on left hand side
sec^2(x)-1=16(1-sec(x))^2 Expanding brackets and collecting terms gives
15sec^2(x)-32sec(x)+17=0.
Now just solve the quadratic in sec(x) to get the 2 values of the solution. Solution has real roots as b^2-4ac=1024-1020=4,but formula will need to be used as this will not factorise. I am now going for beer.
#2
+78741
+5
Ah....Mathcad......you did very well but you gave up a little too quickly, here........!!!!......notice the following factorization....
15sec^2(x)-32sec(x)+17=0
(15sec(x) - 17) (sec(x) -1) = 0
So either......
sec(x) = 1 which happens at 0 and 360.....but, both of these are out of the requested interval....or....
Here's a graph..........https://www.desmos.com/calculator/fpewbunwdj
Notice that the solution of 28.072° is "extraneous"......this will frequently happen when we square both sides of an equation......the only "good" solution in the requested interval occurs at about 331.928°
CPhill May 28, 2015
### 8 Online Users
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners. See details | 645 | 2,094 | {"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} | 4.0625 | 4 | CC-MAIN-2017-47 | longest | en | 0.840283 |
https://aviation.stackexchange.com/questions/29340/does-the-dihedral-effect-happen-during-coordinated-flight?noredirect=1 | 1,709,260,677,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474893.90/warc/CC-MAIN-20240229234355-20240301024355-00167.warc.gz | 112,097,988 | 37,077 | # Does the dihedral effect happen during coordinated flight?
I have researched some about dihedral effect due to a question that I had, and I discovered that my question is essentially already answered, here and here, collectively.
This has led me to a new question because the information talks about slips.
If an airplane is flown in perfect coordination during a turn, does the dihedral effect still work to level the wings?
No.
The dihedral effect depends on a flow asymmetry between left and right wing. Coordinated flight means that both wings have the same flow conditions, and the only asymmetry between both is caused by control deflections.
You do have a flow asymmetry during a turn (coordinated or not) from the speed difference between the inner and the outer wing which grows with the turn rate and wing span. This does indeed cause more lift on the outer wing, but due to the difference in dynamic pressure, not due to dihedral. Since dihedral causes a side force component of lift, you can argue that the outer wing creates more side force which pushes the plane into the turn. But the pilot will compensate the difference in dynamic pressure with the ailerons to stop any rolling motion once the desired turn rate is stabilized. This will also equal the lift on both wings.
Another asymmetry is caused by the location of the vertical tail aft of the center of gravity. Essentially, the tail swings around during the turn, which causes a sideslip condition at the vertical tail. The result is a side force which grows with turn rate and tail lever arm, and due to the high location of most vertical tails also causes a small rolling moment. But again the wing's dihedral has nothing to do with that.
Dihedral only comes into play if you build up a sideslip, and then the turn is not coordinated any longer. @Zeus is right with his point that dihedral stabilizes the aircraft and makes it easier to keep the turn a coordinated one. Now we can split hairs all day if flying in turbulence makes a perfectly coordinated turn impossible. When it comes to dynamic stability, a moderate dihedral effect is highly recommended.
No but yes.
In a perfect turn (or level flight), everything will be symmetrical (except for the minor effects that Peter mentioned). So seemingly no effect.
However, the whole concept of stability requires that when a disturbance occurs, an opposing force or moment is generated. Even if you think that you are making a perfect turn, this perfection comes largely from inherent stability of the airplane, which automatically compensates small disturbances that are always happening. This includes the 'dihedral effect'.
Try to make a perfect turn in a severely unstable aircraft, say, with strong anhedral (perhaps by programming it in a simulator), and you'll see what I mean.
• Of course, my question has a hypothetical element for the purpose of isolating the specific phenomenon. Jun 15, 2016 at 15:47
• Uh, no: The disturbance will awaken the dihedral effect, but this will both cause a rolling motion and a yawing moment, so this will not just reduce the disturbance, but will awaken a lot of additional effects which together will make the aircraft roll and yaw in what is called a dutch roll. Jun 15, 2016 at 16:33
• @PeterKämpf, of course, there will be many effects (this is why I said 'includes'), but if we start talking about dynamic stability, it will be yet another topic. My point was just that in the real world, 'perfect' conditions don't exist without some sort of stability, which in turn involves the effects in question. It's worth keeping this in mind.
– Zeus
Jun 16, 2016 at 0:12
• Your point about flying with strong anhedral is very valid. +1 Jun 16, 2016 at 8:24
• Would dihedral have the opposite effect in a skid then? Jun 17, 2016 at 4:49 | 839 | 3,816 | {"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} | 3.15625 | 3 | CC-MAIN-2024-10 | latest | en | 0.968564 |
https://community.intel.com/t5/Intel-oneAPI-Math-Kernel-Library/A-large-Sparse-Matrix-and-Sparse-Matrix-multipy/td-p/1175111 | 1,627,427,549,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153491.18/warc/CC-MAIN-20210727202227-20210727232227-00404.warc.gz | 208,248,688 | 48,043 | Community
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Beginner
303 Views
## A large Sparse Matrix and Sparse Matrix multipy?
C=A*B
A is a large sparse matrix 10*5000000000, and B is a Sparse matrix with 5000000000*10.
A is created with csr, and B is create with csc.
each matrix may only have a few elements (NNZ=10)
How I can do this multipy? Did mkl_sparse_spmm support the multipy between csr and csc?
10 Replies
Black Belt
303 Views
I do not know of any routine to do exactly what you ask for.
The problem with multiplying sparse matrices is that, in general, their product is unlikely to be sparse, or at least not as sparse as the individual matrices and, what is more of a problem, the pattern of non-zero entries in the product matrix is neither known nor easy to compute. It is best to avoid forming the product of sparse matrices, if that is at all possible.
The CSR representation of A is the same as the CSC representation of AT. This property can be used in two ways. If the matrix is being formed from a COO representation or otherwise, interchanging the row/column coordinates is all that needs to be done to obtain the CSR representation instead of the CSC representation. Secondly, many BLAS level 2 and 3 routines give you an option to transpose one of the input matrices before performing an operation such as ?AXPY.
Employee
303 Views
Hi Burness,
I recalled, the computational routines required CSR or BSR model. But it is convenient to use COO format. for example,
first mkl_sparse_?_create_coo
Creates a handle for a matrix in COO format.
then mkl_sparse_convert_csr
then mkl_sparse_spmm, will do your sparse matrix multiply.
Computes the product of two sparse matrices and
stores the result as a sparse matrix.
MKL provide sample code under MKL install folder, for example /opt/intel/mkl/examples/examples_core_c.tgz . unzip them. in Spblas*
You can refer to them.
Best Regards,
Ying
Beginner
303 Views
Ying H. (Intel) wrote:
Hi Burness,
I recalled, the computational routines required CSR or BSR model. But it is convenient to use COO format. for example,
first mkl_sparse_?_create_coo
Creates a handle for a matrix in COO format.
then mkl_sparse_convert_csr
then mkl_sparse_spmm, will do your sparse matrix multiply.
Computes the product of two sparse matrices and
stores the result as a sparse matrix.
MKL provide sample code under MKL install folder, for example /opt/intel/mkl/examples/examples_core_c.tgz . unzip them. in Spblas*
You can refer to them.
Best Regards,
Ying
I'm sorry that I have not explain it clearly.
Suppose I have 2*5billion Matrix A and 5billion*2 Matrix B multipy:
first i will init these two matrix, for 2*5billion it is ok to new the A with csr, but for 5billion*2 if we new with csr, we must new a array to store 5billion+1 colIndex(OOM), but if we new with csc, it returns 'SPARSE_STATUS_NOT_SUPPORTED' when we use mkl_sparse_spmm.
Is there a method to solve this problem?
Employee
303 Views
Hello,
Hmm, if csc format, then how about first call mkl_sparse_s_create_csc based on your B array. , then call
mkl_sparse_convert_csr (you don't need create any new array, just sparse matrix * )
then call mkl_sparse_spmm
best Regards,
Ying
Beginner
303 Views
Ying H. (Intel) wrote:
Hello,
Hmm, if csc format, then how about first call mkl_sparse_s_create_csc based on your B array. , then call
mkl_sparse_convert_csr (you don't need create any new array, just sparse matrix * )
then call mkl_sparse_spmm
best Regards,
Ying
I will try first , but isn't it cause OOM?
Beginner
303 Views
duan, burness wrote:
Ying H. (Intel) 写道：
Hello,
Hmm, if csc format, then how about first call mkl_sparse_s_create_csc based on your B array. , then call
mkl_sparse_convert_csr (you don't need create any new array, just sparse matrix * )
then call mkl_sparse_spmm
best Regards,
Ying
I will try first , but isn't it cause OOM?
hi Ying:
I have try the method you share, but it seems that when I run the code before mkl_sparse_convert_csr. It happen OOM
maybe when I convert to csr, it will new a 5billion+1 array? Could you help me with this problem?
Employee
303 Views
## Hello,
As i see, here the key problem is that large matrix, so cause Out of memory, right? .
But you had mentioned, 2*5billion Matrix A and 5billion*2 Matrix B multiply, each matrix may only have a few elements (NNZ=10), then you don't need to call mkl function, thus avoid the big array malloc , just do multiply manually, is it ok?
Second consideration about the large matrix, 2*5billion Matrix A and 5billion*2 Matrix B multiply, how many memory on your machine and how do you store and load/ operate the matrix in your application?
If suppose 4 byte, one index array will take at least 1*5 *10^9 * 4 byte =20G memory. if you have to operation such big matrix, what i can recommend for such matrix maybe consider big data solution, like cluster sparse blas or pblas in MKL and anther sister library data analysis accelerate library, there are some algorithms to support distributed CSR sparse matrix, like SVD, PCA etc.
## Intel® Data Analytics Acceleration Library (Intel® DAAL …
https://software.intel.com/en-us/intel-daal
Best Regards,
Ying
Beginner
303 Views
Hi,
I had a similar issue where the CSR became more hindering than memory saving, in my case I didn't have billions of dimensions but still annoying because I had to perform the operation repeatedly and all the memory allocation and initialisation for no reason just costs a lot of time.
Since the number of nonzero entries is quite small in comparison to the dimension you'll most likely want to implement it by hand. In that case you'll be most efficient doing it in the coordinate representation in which case each of your matrices would be fully described by 20 indices, 10 values and the dimensions, i.e. 32 numbers rather than 5*10^9+10+10. Since your nnz is actually small you might even achieve some performance by sorting the entries of A and B appropriately apriori.
The SPARSE_STATUS_NOT_SUPPORTED is returned because you tried to use anything but CSR format for the matrix-matrix multiplication - at least that is what I made of it when I last attempted to use the COO representation of two matrices to compute a sparse product.
From what I can tell, the mkl_sparse_spmm function is a pretty interface to the mkl_?csrmultcsr sblas routine, i.e. it can compute the product of CSR-CSR (and potentially in extension CSC-CSC), if you try anything more exotic you get the not supported status. I believe, the analogue is true for the relationship between mkl_sparse_?_add and mkl_?csradd. This is based on my experience and I don't really know what happens behind the mkl_sparse_spmm call, perhaps someone could confirm or correct my statement?
Best,
Manuel
Employee
303 Views
duan, burness wrote:
Quote:
duan, burness wrote:
Ying H. (Intel) 写道：
Hello,
Hmm, if csc format, then how about first call mkl_sparse_s_create_csc based on your B array. , then call
mkl_sparse_convert_csr (you don't need create any new array, just sparse matrix * )
then call mkl_sparse_spmm
best Regards,
Ying
I will try first , but isn't it cause OOM?
hi Ying:
I have try the method you share, but it seems that when I run the code before mkl_sparse_convert_csr. It happen OOM
maybe when I convert to csr, it will new a 5billion+1 array? Could you help me with this problem?
Hi,
First of all for your matrices it is better to use spmmd function - it's look like your result is dense.
However, right now this function supported only csr format and transposition of first matrix only. So the only way i see is to divide both matrices on chanks, convert second matrix from csc to csr, call spmmd, add result in accum and remove temp data, something like it (pseudocode):
divide matrix A and B on several chunks (with 1 millions of rows, for example)
for (i in chunk) {
mkl_sparse_s_create_csc(B_i);
mkl_sparse_convert_csr(B_i, Bcsr_i);
mkl_sparse_d_spmmd(SPARSE_OPERATION_NON_TRANSPOSE,A_i, Bcsr_i, C_i);
C +=C_i;
mkl_sparse_delete(Bcsr_i);
}
Thanks,
Alex
Employee
303 Views
Manuel S. wrote:
Hi,
I had a similar issue where the CSR became more hindering than memory saving, in my case I didn't have billions of dimensions but still annoying because I had to perform the operation repeatedly and all the memory allocation and initialisation for no reason just costs a lot of time.
Best,
Manuel
Hi Manuel,
Can i ask which function you call repeatedly to understand how we can reduce number of memory allocation in your case?
Thanks,
Alex | 2,191 | 8,637 | {"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} | 2.828125 | 3 | CC-MAIN-2021-31 | longest | en | 0.877 |
http://www.wyzant.com/resources/answers/16045/need_some_quick_help_with_this_preferable_walk_through_instead_of_giving_me_an_answer_i_want_to_know_how_to_do_this | 1,394,209,816,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1393999645570/warc/CC-MAIN-20140305060725-00011-ip-10-183-142-35.ec2.internal.warc.gz | 604,931,417 | 10,152 | Search 73,104 tutors
0 0
## Need some quick help with this. ( Preferable walk through instead of giving me an answer I want to know how to do this)
Write an Equation of the perpendicular to y = 3x + 1 Passing though 0,4
Write an Equation of the perpendicular to y = 3x + 1 Passing though (0,4)
The line you want has a y-intercept already given, at x=0, the y-axis, y=4. Thank you Nataliya D.
Its slope must be -1/3 the negative of the reciprocal of the slope of 3 of the given line
y=-(1/3)x+4 | 148 | 496 | {"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} | 3.515625 | 4 | CC-MAIN-2014-10 | latest | en | 0.870472 |
https://qiita.com/niiyz/items/44df5fee4709880294e3 | 1,544,599,356,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823785.24/warc/CC-MAIN-20181212065445-20181212090945-00167.warc.gz | 732,928,778 | 12,228 | OCamlの練習 ソート
More than 1 year has passed since last update.
• OCamlの練習
• 野球成績型のリストを勝率順に並び替える
baseball.ml
```(* 野球成績型 *)
type baseball_seiseki = {
name: string; (* 名前 *)
won: int; (* 勝ち数 *)
drawn: int; (* 引けわけ数 *)
lost: int; (* 負け数 *)
}
(* 2016データ *)
let hiroshima = {name = "広島"; won = 89; drawn = 2; lost = 52}
let kyojin = {name = "巨人"; won = 71; drawn = 3; lost = 69}
let dena = {name = "DeNA"; won = 69; drawn = 3; lost = 71}
let hanshin = {name = "阪神"; won = 64; drawn = 3; lost = 76}
let yakuruto = {name = "ヤクルト"; won = 64; drawn = 1; lost = 78}
let chuunichi = {name = "中日"; won = 58; drawn = 3; lost = 82}
let npb2016 = [hiroshima; kyojin; dena; hanshin; yakuruto; chuunichi]
(* 勝率計算 勝率 ＝ 勝数 / (勝数 + 負数) *)
(* shouritsu : int -> int -> int *)
let shouritsu won lost = int_of_float(float_of_int(won) /. float_of_int(won + lost) *. 100.0)
(* 勝率TEST *)
let test1 = shouritsu 0 0 = 0
let test2 = shouritsu 1 0 = 100
let test3 = shouritsu 0 1 = 0
let test4 = shouritsu 100 50 = 66
let test5 = shouritsu 50 50 = 50
(* 勝率 降順位置に挿入 *)
(* baseball_seiseki_insert : baseball_seiseki list -> baseball_seiseki list *)
let rec baseball_seiseki_insert list target_team = match list with
| [] -> [target_team]
| ({name = loop_name; won = loop_won; drawn = loop_drawn; lost = loop_lost}) as loop_team :: rest ->
match target_team with {name = target_name; won = target_won; drawn = target_drawn; lost = target_lost} ->
if shouritsu target_won target_lost < shouritsu loop_won loop_lost then loop_team :: baseball_seiseki_insert rest target_team
else target_team :: list
(* 挿入TEST *)
let test_npb1 = [hiroshima; kyojin; dena; yakuruto; chuunichi]
let test_npb2 = [kyojin; dena; yakuruto]
let test1 = baseball_seiseki_insert [] hanshin = [hanshin]
let test2 = baseball_seiseki_insert test_npb1 hanshin = [hiroshima; kyojin; dena; hanshin; yakuruto; chuunichi]
let test3 = baseball_seiseki_insert test_npb2 chuunichi = [kyojin; dena; yakuruto; chuunichi]
(* 勝率降順ソート *)
(* baseball_seiseki_sort : baseball_seiseki list -> baseball_seiseki list *)
let rec baseball_seiseki_sort list = match list with
| [] -> []
| first :: rest -> baseball_seiseki_insert (baseball_seiseki_sort rest) first
(* 勝率降順ソートTEST *)
let test_npb1 = [hanshin; hiroshima; chuunichi; kyojin; dena; yakuruto;]
let test_npb2 = [chuunichi; kyojin; hanshin; hiroshima; dena; yakuruto]
let test1 = baseball_seiseki_sort [] = []
let test2 = baseball_seiseki_sort test_npb1 = npb2016
let test3 = baseball_seiseki_sort test_npb2 = npb2016
```
インサート Trace
```# #trace baseball_seiseki_insert;;
baseball_seiseki_insert is now traced.
# baseball_seiseki_insert [hiroshima; kyojin; dena; yakuruto; chuunichi] hanshin;;
baseball_seiseki_insert <--
[{name = "広島"; won = 89; drawn = 2; lost = 52};
{name = "巨人"; won = 71; drawn = 3; lost = 69};
{name = "DeNA"; won = 69; drawn = 3; lost = 71};
{name = "ヤクルト"; won = 64; drawn = 1; lost = 78};
{name = "中日"; won = 58; drawn = 3; lost = 82}]
baseball_seiseki_insert --> <fun>
baseball_seiseki_insert* <--
{name = "阪神"; won = 64; drawn = 3; lost = 76}
baseball_seiseki_insert <--
[{name = "巨人"; won = 71; drawn = 3; lost = 69};
{name = "DeNA"; won = 69; drawn = 3; lost = 71};
{name = "ヤクルト"; won = 64; drawn = 1; lost = 78};
{name = "中日"; won = 58; drawn = 3; lost = 82}]
baseball_seiseki_insert --> <fun>
baseball_seiseki_insert* <--
{name = "阪神"; won = 64; drawn = 3; lost = 76}
baseball_seiseki_insert <--
[{name = "DeNA"; won = 69; drawn = 3; lost = 71};
{name = "ヤクルト"; won = 64; drawn = 1; lost = 78};
{name = "中日"; won = 58; drawn = 3; lost = 82}]
baseball_seiseki_insert --> <fun>
baseball_seiseki_insert* <--
{name = "阪神"; won = 64; drawn = 3; lost = 76}
baseball_seiseki_insert <--
[{name = "ヤクルト"; won = 64; drawn = 1; lost = 78};
{name = "中日"; won = 58; drawn = 3; lost = 82}]
baseball_seiseki_insert --> <fun>
baseball_seiseki_insert* <--
{name = "阪神"; won = 64; drawn = 3; lost = 76}
baseball_seiseki_insert* -->
[{name = "阪神"; won = 64; drawn = 3; lost = 76};
{name = "ヤクルト"; won = 64; drawn = 1; lost = 78};
{name = "中日"; won = 58; drawn = 3; lost = 82}]
baseball_seiseki_insert* -->
[{name = "DeNA"; won = 69; drawn = 3; lost = 71};
{name = "阪神"; won = 64; drawn = 3; lost = 76};
{name = "ヤクルト"; won = 64; drawn = 1; lost = 78};
{name = "中日"; won = 58; drawn = 3; lost = 82}]
baseball_seiseki_insert* -->
[{name = "巨人"; won = 71; drawn = 3; lost = 69};
{name = "DeNA"; won = 69; drawn = 3; lost = 71};
{name = "阪神"; won = 64; drawn = 3; lost = 76};
{name = "ヤクルト"; won = 64; drawn = 1; lost = 78};
{name = "中日"; won = 58; drawn = 3; lost = 82}]
baseball_seiseki_insert* -->
[{name = "広島"; won = 89; drawn = 2; lost = 52};
{name = "巨人"; won = 71; drawn = 3; lost = 69};
{name = "DeNA"; won = 69; drawn = 3; lost = 71};
{name = "阪神"; won = 64; drawn = 3; lost = 76};
{name = "ヤクルト"; won = 64; drawn = 1; lost = 78};
{name = "中日"; won = 58; drawn = 3; lost = 82}]
- : baseball_seiseki list =
[{name = "広島"; won = 89; drawn = 2; lost = 52};
{name = "巨人"; won = 71; drawn = 3; lost = 69};
{name = "DeNA"; won = 69; drawn = 3; lost = 71};
{name = "阪神"; won = 64; drawn = 3; lost = 76};
{name = "ヤクルト"; won = 64; drawn = 1; lost = 78};
{name = "中日"; won = 58; drawn = 3; lost = 82}]
#
```
ソート Trace
```# #untrace baseball_seiseki_insert;;
baseball_seiseki_insert is no longer traced.
# #trace baseball_seiseki_sort;;
baseball_seiseki_sort is now traced.
# baseball_seiseki_sort [hanshin; hiroshima; chuunichi; kyojin; dena; yakuruto;];;
baseball_seiseki_sort <--
[{name = "阪神"; won = 64; drawn = 3; lost = 76};
{name = "広島"; won = 89; drawn = 2; lost = 52};
{name = "中日"; won = 58; drawn = 3; lost = 82};
{name = "巨人"; won = 71; drawn = 3; lost = 69};
{name = "DeNA"; won = 69; drawn = 3; lost = 71};
{name = "ヤクルト"; won = 64; drawn = 1; lost = 78}]
baseball_seiseki_sort <--
[{name = "広島"; won = 89; drawn = 2; lost = 52};
{name = "中日"; won = 58; drawn = 3; lost = 82};
{name = "巨人"; won = 71; drawn = 3; lost = 69};
{name = "DeNA"; won = 69; drawn = 3; lost = 71};
{name = "ヤクルト"; won = 64; drawn = 1; lost = 78}]
baseball_seiseki_sort <--
[{name = "中日"; won = 58; drawn = 3; lost = 82};
{name = "巨人"; won = 71; drawn = 3; lost = 69};
{name = "DeNA"; won = 69; drawn = 3; lost = 71};
{name = "ヤクルト"; won = 64; drawn = 1; lost = 78}]
baseball_seiseki_sort <--
[{name = "巨人"; won = 71; drawn = 3; lost = 69};
{name = "DeNA"; won = 69; drawn = 3; lost = 71};
{name = "ヤクルト"; won = 64; drawn = 1; lost = 78}]
baseball_seiseki_sort <--
[{name = "DeNA"; won = 69; drawn = 3; lost = 71};
{name = "ヤクルト"; won = 64; drawn = 1; lost = 78}]
baseball_seiseki_sort <--
[{name = "ヤクルト"; won = 64; drawn = 1; lost = 78}]
baseball_seiseki_sort <-- []
baseball_seiseki_sort --> []
baseball_seiseki_sort -->
[{name = "ヤクルト"; won = 64; drawn = 1; lost = 78}]
baseball_seiseki_sort -->
[{name = "DeNA"; won = 69; drawn = 3; lost = 71};
{name = "ヤクルト"; won = 64; drawn = 1; lost = 78}]
baseball_seiseki_sort -->
[{name = "巨人"; won = 71; drawn = 3; lost = 69};
{name = "DeNA"; won = 69; drawn = 3; lost = 71};
{name = "ヤクルト"; won = 64; drawn = 1; lost = 78}]
baseball_seiseki_sort -->
[{name = "巨人"; won = 71; drawn = 3; lost = 69};
{name = "DeNA"; won = 69; drawn = 3; lost = 71};
{name = "ヤクルト"; won = 64; drawn = 1; lost = 78};
{name = "中日"; won = 58; drawn = 3; lost = 82}]
baseball_seiseki_sort -->
[{name = "広島"; won = 89; drawn = 2; lost = 52};
{name = "巨人"; won = 71; drawn = 3; lost = 69};
{name = "DeNA"; won = 69; drawn = 3; lost = 71};
{name = "ヤクルト"; won = 64; drawn = 1; lost = 78};
{name = "中日"; won = 58; drawn = 3; lost = 82}]
baseball_seiseki_sort -->
[{name = "広島"; won = 89; drawn = 2; lost = 52};
{name = "巨人"; won = 71; drawn = 3; lost = 69};
{name = "DeNA"; won = 69; drawn = 3; lost = 71};
{name = "阪神"; won = 64; drawn = 3; lost = 76};
{name = "ヤクルト"; won = 64; drawn = 1; lost = 78};
{name = "中日"; won = 58; drawn = 3; lost = 82}]
- : baseball_seiseki list =
[{name = "広島"; won = 89; drawn = 2; lost = 52};
{name = "巨人"; won = 71; drawn = 3; lost = 69};
{name = "DeNA"; won = 69; drawn = 3; lost = 71};
{name = "阪神"; won = 64; drawn = 3; lost = 76};
{name = "ヤクルト"; won = 64; drawn = 1; lost = 78};
{name = "中日"; won = 58; drawn = 3; lost = 82}]
#
``` | 3,271 | 8,219 | {"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} | 2.59375 | 3 | CC-MAIN-2018-51 | longest | en | 0.679351 |
https://www.projectguru.in/conduct-generalized-least-squares-test/ | 1,721,324,685,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514848.78/warc/CC-MAIN-20240718161144-20240718191144-00433.warc.gz | 808,733,046 | 19,611 | # How to conduct generalized least squares test?
By Riya Jain & Priya Chetty on December 17, 2017
In statistics, Generalised Least Squares (GLS) is one of the most popular methods for estimating unknown coefficients of a linear regression model when the independent variable is correlating with the residuals. The Ordinary Least Squares (OLS) method only estimates the parameters in the linear regression model. Also, it seeks to minimize the sum of the squares of the differences between the observed responses in the given dataset and those predicted by a linear function. The main advantage of using OLS regression for estimating parameters is that it is easy to use. However, OLS gives robust results only if there are no missing values in the data and there are no major outliers in the data set. Moreover, the OLS regression model does not take into account unequal variance or ‘heteroskedastic errors’. Due to heteroscedastic errors, the results are not robust and also create bias.
Therefore, the generalized least squares test is crucial in tackling the problem of outliers, heteroskedasticity and bias in data. It is capable of producing estimators that are ‘Best Linear Unbiased Estimates’. Thus, the GLS estimator is unbiased, consistent, efficient and asymptotically normal.
## Major assumption for generalized least square regression analysis
The OLS model has main assumption of having estimators as BLUE. But as violation of BLUE assumption of Gauss Markov theorem result in having serial correlation and homoskedasticity assumption violation. This prevents model fitting and hence impact assessment or relationship development could not be done. Thus, GLS as an improvised model to deal with these violations has below stated assumptions
• The error variances are heteroskedastic
• Errors are correlated
• Normally distributed
• Absence of multicollinearity
In the absence of first two assumptions, the OLS estimators and the GLS estimators are the same. Thus, the difference between OLS and GLS is the assumptions of the error term of the model. There are 3 different perspectives from which one can understand the GLS estimator:
• A generalization of OLS
• Transforming the model equation to a new model whose errors are uncorrelated and have equal variances that is homoskedastic.
## Example of generalized least squares test
This section explains the process of applying GLS with the use of a case study. The sample dataset contains data from 30 students. The aim is to review the impact of self-efficiency and ability (independent variable) on achievement (dependent variable). For this case study first, a simple linear regression is performed and the results are compared with the generalized least squares test.
### Step 1: Linear regression
Since the dependent variable is continuous in nature, it is important to confirm if the dependent variable follows a normal distribution. The distribution of residuals of the dependent variable (achievement) is normal, with skewness -0.18 and kurtosis 1.95. As the table above shows, linear regression was performed to check the relationship between achievement and self-efficiency and ability. The parameter estimate was 0.003 with a p-value of 0.989. For another independent variable ability, the parameter estimate was -0.047 with a p-value of 0.823. This shows that none of the independent variables are statistically significant as the p-value is greater than 0.05.
The interpretation of coefficients of the independent variables is as follows:
• The independent variable ‘self-efficiency’ is positively related to dependent variable ‘achievement’. However other independent variable ‘ability’ is negatively attributed to the dependent variable.
• The estimates parameter and p value shows that the sample size was inadequate to demonstrate the true spectrum of relationship .
• Furthermore, for every unit of rise in self-efficiency, the dependent variable also increases by 1 unit, keeping all other factors same.
### Step 2: Weighted least squares regression
After performing the weighted analysis, self-efficiency was found to influence the achievement more, with a beta coefficient of 0.045 and a value of 0.021. This shows that the regression coefficient is statistically significant. Ability influenced the achievement less, with a beta coefficient of 0.014 with a value of 0.046. Both the p values are statistically significant which indicates that GLS is a better fit than simple regression done previously. Therefore there is the significant importance of ranking or relationship between dependent variable ‘achievement’ and independent variable ‘self- efficiency’ and ‘ability’.
## Application of generalized least squares
• GLS model is useful in regionalisation of hydrologic data.
• GLS is also useful in reducing autocorrelation by choosing an appropriate weighting matrix.
• It is one of the best methods to estimate regression models with auto correlate disturbances and test for serial correlation (Here Serial correlation and auto correlate are same things).
• One can also learn to use the maximum likelihood technique to estimate the regression models with auto correlated disturbances.
• The GLS procedure finds extensive use across various domains.The goal of GLS method to estimate the parameters of regional regression models of flood quantiles.
• GLS is widely popular in conducting market response model, econometrics and time series analysis.
A number of available software support the generalized least squares test, like R, MATLAB, SAS, SPSS, and STATA.
Priya is the co-founder and Managing Partner of Project Guru, a research and analytics firm based in Gurgaon. She is responsible for the human resource planning and operations functions. Her expertise in analytics has been used in a number of service-based industries like education and financial services.
Her foundational educational is from St. Xaviers High School (Mumbai). She also holds MBA degree in Marketing and Finance from the Indian Institute of Planning and Management, Delhi (2008).
Some of the notable projects she has worked on include:
• Using systems thinking to improve sustainability in operations: A study carried out in Malaysia in partnership with Universiti Kuala Lumpur.
• Assessing customer satisfaction with in-house doctors of Jiva Ayurveda (a project executed for the company)
• Predicting the potential impact of green hydrogen microgirds (A project executed for the Government of South Africa)
She is a key contributor to the in-house research platform Knowledge Tank.
She currently holds over 300 citations from her contributions to the platform.
She has also been a guest speaker at various institutes such as JIMS (Delhi), BPIT (Delhi), and SVU (Tirupati).
I am a master's in Economics from Amity University. Having a keen interest in Econometrics and data analysis, I was a part of the Innovation Project of Daulat Ram College, Delhi University. My core expertise and interest are in environment-related issues. Apart from academics, I love music and exploring new places. | 1,402 | 7,090 | {"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} | 2.984375 | 3 | CC-MAIN-2024-30 | longest | en | 0.88697 |
http://windowssecrets.com/forums/showthread.php/160147-Index%28-%29 | 1,503,061,903,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886104636.62/warc/CC-MAIN-20170818121545-20170818141545-00448.warc.gz | 458,903,089 | 15,150 | 1. ## Index(?)
Column A, Rows 1-10, Worksheet 1 are the letters A-J. Column A, Rows 1-10, Worksheet 2 are the letters A-J, Column B, Rows 1-10, names of clients. What I would like to do on Worksheet 1, column C, is "match" the client name on Worksheet 2 with the associated letters on Worksheet 1. For example, Column A, Row 1, Worksheet 1 contains A. Column A, Row 1, Worksheet 2 contains an A as well, and Column B, Row 1 contains the name "John". What I would like to do is insert a formula in Worksheet 1, Column C, Row 1 that matches the two As, and would result in the name "John" being inserted.
2. JL,
This should do the trick:
=INDEX(Sheet2!A1:B10,MATCH(\$A1,Sheet2!A1:A10,0),2)
Place the above formula in cell C1 of Sheet1 and then fill down.
indexmatch.JPG
Of course you can improve the formula by naming the range Sheet2!A1:A10 and Sheet2!A1:B10 and using the names in the formula.
HTH
JLKIRK-IndexMatch.xlsx
3. A shorter version of the index/match is to index on only 1 column:
=INDEX('Worksheet 2'!\$B\$1:\$B\$10,MATCH(A1,'Worksheet 2'!\$A\$1:\$A\$10,0))
But in this case the more direct way is to use a VLOOKUP:
=VLOOKUP(A1,'Worksheet 2'!\$A\$1:\$B\$10,2,0)
Steve
PS: RG, I think locking the cell ranges is preferred in this instance otherwise if the order is changed you may not find a match (if the appropriate match is on a row < low your lookup value is on)
4. Originally Posted by sdckapr
PS: RG, I think locking the cell ranges is preferred in this instance otherwise if the order is changed you may not find a match (if the appropriate match is on a row < low your lookup value is on)
Exactly! That's why I suggested using Named Ranges.
BTW: I didn't suggest VLookup because the OP didn't specify that the key letters would always be in alpha order.
5. Exactly! That's why I suggested using Named Ranges.
But even without named ranges, you can lock in the col and rows of the range so they range will not vary when the formula is copied. Your formula uses a different range for each row and if the value being looked is in a row > the row it is occupied in the table, it will not find the match...[for example if you the letter "A" in cell A2 of Sheet1, a match will not be found since it looks in Sheet2!A2:A11 so it misses the "A" in cell Sheet2!A1]
I didn't suggest VLookup because the OP didn't specify that the key letters would always be in alpha order.
I did not presume that they would be. Vlookup does not require them to be in Alphabetical order. Just like MATCH it can be used for an exact lookup with the optional parameter of zero ("0") at the end.
Steve
6. Steve,
You're correct the search ranges should be locked down. I was in a hurry to get out the door to pick up my wife's aunt at the airport an hour and a half away and wasn't thinking that filling down would change the references. Thanks for the reminder.
7. Hi RG
What speed were you doing to pick up your wife's aunt? Did you get there on time? Did you have to carry the luggage? How's your back? And let's get our priorities right here. You had a compelling choice between answering an Excel question, or obeying the boss.
I'm still in hospital. Day 10 I think. It must be the morphine.
zeddy
8. Zeddy.
In order:
60-70 MPH
Yes
Yes
Great!!!!!!
No choice! If Momma ain't happy Nobody's happy!
Got any extra Morphine?
Get well SOON!
#### Posting Permissions
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• | 939 | 3,494 | {"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} | 2.671875 | 3 | CC-MAIN-2017-34 | latest | en | 0.851047 |
http://mathhelpforum.com/geometry/208381-prove-similar-triangles-parallelogram.html | 1,529,550,296,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864019.29/warc/CC-MAIN-20180621020632-20180621040632-00194.warc.gz | 206,422,658 | 11,444 | # Thread: Prove similar triangles in parallelogram
1. ## Prove similar triangles in parallelogram
Given a parallelogram ABCD and 2 segments FH and EG intersecting the diagonal AC at point P and terminating on 2 opposite parallelogram sides respectively, plus segments HG and EF.
Pprove that triangles EFP and HGP are similar.
Could I get a hint on how to solve this problem?
If it could be proven that HG and EF are parallel or by the same token, that angle AFE=GHC, the proof would follow. I have been unable to make any headway.
2. ## Re: Prove similar triangles in parallelogram
first, show $\displaystyle \Delta HCG ~ \Delta FAE$ using angle-angle.
showing these two triangles are similar will show $\displaystyle m\angle{AFE} = m\angle{CHG}$
$\displaystyle m\angle{CHP} = m\angle{AFP}$ since they are alternate-interior angles
... now you can show $\displaystyle m\angle{GHP} = m\angle{EFP}$. All you need is one more pair of angles (which should be obvious from the diagram)
3. ## Re: Prove similar triangles in parallelogram
Thank you skeeter. I still don't see how triangles HCG and FAE are proven similar. Angle EAF=HCG. But to say that angle AEF=CFH or AFE=CHG, would assume that EF is parallel to HG which I still don't see how to prove.
4. ## Re: Prove similar triangles in parallelogram
... you are correct. I'll have to stare at it awhile and see if I can draw a conclusion.
5. ## Re: Prove similar triangles in parallelogram
Thanks. I have been at it several days. I drew it in a computer and the lines (EF and FG) are indeed parallel, just can't figure out how to prove it.
6. ## Re: Prove similar triangles in parallelogram
Triangles AEP and CGP are similar so mAP = PC, mEP = GP and mAE = CG for some constant m.
Triangles APF and CPH are similar so nAP = PC, nAF = CH and nFP = HP for some constant n.
Equating PC from both sets of equations shows that m = n, in which case, (replacing n by m), mEP = GP and mFP = HP.
Now feed that information into the triangles EPF and GPH together with the common angles EPF and GPH and we can show that the two triangles are similar. (There might be some well known theorem that, for the moment, I've forgotten, but failing that use of the cosine rule in the two triangles shows that mEF = HG).
7. ## Re: Prove similar triangles in parallelogram
Thank you very much, BobP
8. ## Re: Prove similar triangles in parallelogram
Originally Posted by BERMES39
Thank you very much, BobP
one simple construction consisting of two lines makes the solution easy
9. ## Re: Prove similar triangles in parallelogram
I was wrong in my assumption about construction lines.Using Bob's similar triangles
Triangles AEP-CGP AP/PC =PF/HP
Triangles APF-CPH AP/PC= EP/PG
PF/HP=EF/PG
mEPF =mHPG
Triangles EPF -HPG
SAS theorem
If two sides and the included angle of one are in same proportion as the correspondingsides and included angle ofother they are similar | 769 | 2,917 | {"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} | 3.859375 | 4 | CC-MAIN-2018-26 | latest | en | 0.914649 |
https://aakashsrv1.meritnation.com/ask-answer/question/how-can-we-derive-the-formula-of-karl-pearson-s-cefficient-o/correlation/3380421 | 1,669,707,524,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710690.85/warc/CC-MAIN-20221129064123-20221129094123-00250.warc.gz | 110,257,224 | 8,365 | # how can we derive the formula of karl pearson's cefficient of correlation in shortcut method?
The following steps are involved in the computation of Karl Pearson's coefficient of correlation using the short-cut method.
Step 1: Decide an assumed mean for the X-series and calculate deviations of each item of the series from it. Total the deviations and denote as ∑dx.
Step 2: Similarly, calculate the deviations in Y-series from an assumed mean and determine the summation as Σdy.
Step 3: Now, square the deviations of X-series and the Y-series and obtain their total as Σdx2 and Σdy2 respectively.
Step 4: Multiply the deviations obtained in the step 1 and step 2 and obtain the total as Σdx·dy.
Step 5: The following formula is applied to compute Karl Pearson's coefficient of correlation.
• 9
What are you looking for? | 188 | 830 | {"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} | 3.015625 | 3 | CC-MAIN-2022-49 | latest | en | 0.917267 |
https://grammarhow.com/euro-symbol-before-or-after-number/ | 1,723,410,266,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641008472.68/warc/CC-MAIN-20240811204204-20240811234204-00021.warc.gz | 224,952,122 | 27,462 | # Euro Symbol Before or After the Number? (20€ or €20)
Currency amounts and signs are interesting when you really think about them. Knowing where to place the euro symbol is something that changes based on which language you’re writing with. This article will explore how to write a number with the euro sign.
## Euro Symbol Before or After the Number?
In English, it is conventionally appropriate to place the euro symbol before the number. It is common to see €500 written. In non-English-speaking countries, you might find that the euro comes after the number. For example, in Swedish, it’s more common to write 500 €.
You might not have considered it before, but the language you use plays an important part in where the euro sign goes. To most English speakers, it should always go before the number. This is also true for other signs like £ and \$.
There are some countries in Europe that don’t use a sign at all. Instead, they would just write the cost of something (i.e. 400) without using currency, as it is already implied.
If you are not an English speaker, you might find that either 20 € or EUR 20 is used instead of the more conventional choice where the euro sign comes first.
## Euro Symbol Before the Number
You should place the euro symbol before the number when you are writing in conventional English. It’s common to see it used in this way because most major currencies are also written with the sign first. You can use “€30” when you want to show thirty euros, for example.
It might help to go through some examples of how the euro sign is used in some sentences:
• I thought she was going to lend me €30. I can see now that she was only joking about that.
• I don’t have €65 to give you right now. Please, come back a little later to see if that changes.
• She wanted it for €550, but I told her that it was over a 25% discount. I couldn’t go that low!
• They gave me €15 to pay him a present. I’m sure I can find something good with that.
## Euro Symbol After the Number
Certain European countries place the euro sign after the number when writing in their own language. It’s common for countries like Spain, Italy, France, and Germany to place the euro symbol after the number (or exclude it entirely) when writing monetary values.
You might also find that some people place the euro sign after the number because of how it’s used in spoken English. After all, most people would say “three euros,” which would be more logical to write as “3€” since the euros come second.
Here are some examples of how the euro sign could come after the number:
• I’m going to have to ask you for 5€ on top of what you’ve already given me.
• Wait, you’re selling this for 26€? That’s basically a steal! I’ll take it.
• It’s not going to sell for 100€. You’re going to have to lower that price.
• She wanted it for 6€, so I gave it to her because I felt generous.
## Pound Sign Before or After the Number?
When the pound sign is used, it should always come before the number. Since it’s common for written English to place all currency symbols before numbers, it makes sense that the UK would stick to these rules when using their own currency.
It is only ever appropriate to place the pound sign before a number. You won’t often find it come after.
The only time when it might come after is when a non-native speaker mistakes its placement. Again, it seems more logical to write “20£” since you would say “twenty pounds.” Nevertheless, this is very rare, and you should not use it.
• I thought that she was going to get me the £10 I needed to move on with this.
• It’s going to be over £3,000 when it’s all added up. Can you handle that?
• I’m not sure if £250 is going to cut it. Is there no way that you can increase that?
• We wanted £650,000. We didn’t even get half of that.
## Final Thoughts
When you are using the euro sign, convention dictates that it should come before the number. This is very common in written English. However, if you are not using English, but a European language to write euros, you will often find that the symbol comes after the number. | 932 | 4,099 | {"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} | 2.703125 | 3 | CC-MAIN-2024-33 | latest | en | 0.960445 |
https://tolstoy.newcastle.edu.au/R/e13/help/11/03/8616.html | 1,586,246,011,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371675859.64/warc/CC-MAIN-20200407054138-20200407084638-00043.warc.gz | 686,750,248 | 4,291 | # Re: [R] Help with plotting a line that is multicoloured based on levels of a factor
From: Jim Lemon <jim_at_bitwrit.com.au>
Date: Sat, 26 Mar 2011 14:20:19 +1100
On 03/26/2011 07:19 AM, Pam Allen wrote:
> Hello again,
>
> I wrote an example that better represents my data, since the coloured points
> are actually consecutive, but with variable lengths:
>
> date=as.Date(c(1:300))
> flow=sin(2*pi/53*c(1:300))
> levels=c(rep(c("high","med","low"),100))
> data=cbind.data.frame(date, flow, levels)
>
> library(zoo)
> z<- zoo(data\$flow, data\$date)
> zz=cbind.data.frame(date=as.Date(rownames(cbind.data.frame(rollapply(z, 2,
> align = "right", FUN="+")))),flow.change=(rollapply(z, 2, align =
> "right",FUN="+" )))
> names(zz)=c("date","todays.flow","next.day.flow")
> zzz=cbind.data.frame(zz[,1], (zz[,3]-zz[,2]))
> names(zzz)=c("date","change.flow")
> data2=merge(data, zzz)
> rate=zoo(data2\$change.flow,data2\$date)
> x=cbind.data.frame(date=as.Date(rownames(cbind.data.frame(rollapply(rate, 2,
> align="left", FUN="+")))), sign=rollapply(rate, 2, align="left",FUN="+"))
> names(x)=c("date","todays.change","next.day.change")
> xx=cbind.data.frame(x[,1],(x[,3]*x[,2]))
> names(xx)=c("date","sign")
> data2=merge(data2, xx)
>
> data3=cbind(data2,pass1=
> ifelse(data2\$flow<0, "extreme.low",
> ifelse(data2\$flow>=0.9, "extreme.high","NA")))
> data4=cbind(data3, pass2=
>
> ifelse(data3\$flow<0.8&data3\$flow>0&data3\$change.flow>=0&data3\$change>=0&data3\$pass1=="NA","medium"
> ,
>
> ifelse(data3\$flow<0.7&data3\$flow>0&data3\$change.flow<0&data3\$change<0&data3\$pass1=="NA","medium","NA")))
> data4\$pass1=paste(data4\$pass1, data4\$pass2)
> data4\$pass1=replace(data4\$pass1, data4\$pass1=="NA NA", "low")
>
> dat=cbind(data4[,1:5], class=
> ifelse(data4\$pass1=="extreme.high NA","1.Extreme.High",
> ifelse(data4\$pass1=="NA medium","2.Medium",
> ifelse(data4\$pass1=="low","3.Low",
> ifelse(data4\$pass1=="extreme.low NA","4.Extreme.Low",NA)))))
>
> colour=ifelse(dat\$class=="1.Extreme.High","red",
> ifelse(dat\$class=="2.Medium","green",
> ifelse(dat\$class=="3.Low","blue",
> ifelse(dat\$class=="4.Extreme.Low","purple",""))))
> plot(dat\$date, dat\$flow, col=colour)
>
>
> What I would like to do is to plot this using a line with the correct
> colours instead of points, i.e.:
>
> plot(dat\$date, dat\$flow, col=colour, type="l") ##Doesn't work, because the
> line is continuous
>
Hi Pam,
I couldn't get the above example to run, and I've arrived at this problem rather late. However, if I have guessed your original question correctly, this might help:
date=c(1:300)
flow=sin(2*pi/53*c(1:300))
levels=c(rep(c("high","med","low"),100)) data=cbind.data.frame(date, flow, levels) plot(data\$date,data\$flow,type="n")
library(plotrix)
color.scale.lines(data\$date,data\$flow,
col=color.scale(data\$flow,extremes=c("blue","red")))
The "clplot" function in plotrix is similar and might also be of interest.
Jim
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https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Received on Sat 26 Mar 2011 - 03:24:18 GMT
Archive maintained by Robert King, hosted by the discipline of statistics at the University of Newcastle, Australia.
Archive generated by hypermail 2.2.0, at Sat 26 Mar 2011 - 07:50:25 GMT.
Mailing list information is available at https://stat.ethz.ch/mailman/listinfo/r-help. Please read the posting guide before posting to the list. | 1,132 | 3,536 | {"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} | 2.640625 | 3 | CC-MAIN-2020-16 | latest | en | 0.458353 |
https://sailboat.guide/cape-dory/330 | 1,656,676,078,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103940327.51/warc/CC-MAIN-20220701095156-20220701125156-00173.warc.gz | 535,654,684 | 13,375 | # Cape Dory 330
1985 — 1988
Designer
Carl Alberg
Builder
Cape Dory Yachts
Association
Cape Dory Sailboat Owners Association
# Built
27
Hull
Monohull
Keel
Long
Rudder
?
Construction
FG
### Dimensions
Length Overall
33 0 / 10.1 m
Waterline Length
24 6 / 7.5 m
Beam
10 2 / 3.1 m
Draft
4 9 / 1.5 m
Displacement
13,300 lb / 6,033 kg
Ballast
5,500 lb / 2,495 kg
### Rig and Sails
Type
Cutter
Reported Sail Area
561′² / 52.1 m²
Total Sail Area
561′² / 52.1 m²
Sail Area
253′² / 23.6 m²
P
39 0 / 11.9 m
E
12 11 / 4 m
Air Draft
?
Sail Area
308′² / 28.6 m²
I
44 9 / 13.6 m
J
13 8 / 4.2 m
Forestay Length
46 9 / 14.3 m
Make
?
Model
?
HP
?
Fuel Type
?
Fuel Capacity
?
### Accomodations
Water Capacity
?
Holding Tank Capacity
?
?
Cabins
?
Hull Speed
6.3 kn
Classic: 6.63 kn
### Hull Speed
The theoretical maximum speed that a displacement hull can move efficiently through the water is determined by it's waterline length and displacement. It may be unable to reach this speed if the boat is underpowered or heavily loaded, though it may exceed this speed given enough power. Read more.
Formula
Classic hull speed formula:
Hull Speed = 1.34 x √LWL
A more accurate formula devised by Dave Gerr in The Propeller Handbook replaces the Speed/Length ratio constant of 1.34 with a calculation based on the Displacement/Length ratio.
Max Speed/Length ratio = 8.26 ÷ Displacement/Length ratio.311
Hull Speed = Max Speed/Length ratio x √LWL
6.33 knots
Classic formula: 6.63 knots
Sail Area/Displacement
16.0
<16: under powered
### Sail Area / Displacement Ratio
A measure of the power of the sails relative to the weight of the boat. The higher the number, the higher the performance, but the harder the boat will be to handle. This ratio is a "non-dimensional" value that facilitates comparisons between boats of different types and sizes. Read more.
Formula
SA/D = SA ÷ (D ÷ 64)2/3
• SA: Sail area in square feet, derived by adding the mainsail area to 100% of the foretriangle area (the lateral area above the deck between the mast and the forestay).
• D: Displacement in pounds.
15.99
<16: under powered
16-20: good performance
>20: high performance
Ballast/Displacement
41.4
>40: stiffer, more powerful
### Ballast / Displacement Ratio
A measure of the stability of a boat's hull that suggests how well a monohull will stand up to its sails. The ballast displacement ratio indicates how much of the weight of a boat is placed for maximum stability against capsizing and is an indicator of stiffness and resistance to capsize.
Formula
Ballast / Displacement * 100
41.36
<40: less stiff, less powerful
>40: stiffer, more powerful
Displacement/Length
403.4
>350: ultraheavy
### Displacement / Length Ratio
A measure of the weight of the boat relative to it's length at the waterline. The higher a boat’s D/L ratio, the more easily it will carry a load and the more comfortable its motion will be. The lower a boat's ratio is, the less power it takes to drive the boat to its nominal hull speed or beyond. Read more.
Formula
D/L = (D ÷ 2240) ÷ (0.01 x LWL)³
• D: Displacement of the boat in pounds.
• LWL: Waterline length in feet
403.37
<100: ultralight
100-200: light
200-300: moderate
300-400: heavy
>400: very heavy
Comfort Ratio
34.0
30-40: moderate bluewater cruising boat
### Comfort Ratio
This ratio assess how quickly and abruptly a boat’s hull reacts to waves in a significant seaway, these being the elements of a boat’s motion most likely to cause seasickness. Read more.
Formula
Comfort ratio = D ÷ (.65 x (.7 LWL + .3 LOA) x Beam1.33)
• D: Displacement of the boat in pounds
• LWL: Waterline length in feet
• LOA: Length overall in feet
• Beam: Width of boat at the widest point in feet
34.04
<20: lightweight racing boat
20-30: coastal cruiser
30-40: moderate bluewater cruising boat
40-50: heavy bluewater boat
>50: extremely heavy bluewater boat
Capsize Screening
1.7
<2.0: better suited for ocean passages
### Capsize Screening Formula
This formula attempts to indicate whether a given boat might be too wide and light to readily right itself after being overturned in extreme conditions. Read more.
Formula
CSV = Beam ÷ ³√(D / 64)
• Beam: Width of boat at the widest point in feet
• D: Displacement of the boat in pounds
1.73
<2: better suited for ocean passages
>2: better suited for coastal cruising
### Notes
An updated version of the CAPE CORY 33.
### For Sale
Have a sailboat to sell? | 1,289 | 4,433 | {"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} | 2.671875 | 3 | CC-MAIN-2022-27 | latest | en | 0.784833 |
http://csharphelper.com/blog/2014/07/determine-whether-a-point-is-inside-a-polygon-in-c/ | 1,618,945,146,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039490226.78/warc/CC-MAIN-20210420183658-20210420213658-00603.warc.gz | 25,769,211 | 15,897 | # Determine whether a point is inside a polygon in C#
One way to determine whether a point lies within a polygon is to add up the angles between the point and adjacent points on the polygon taken in order. For example, if the point in question is P and points A and B are adjacent on the polygon, then you look at the angle APB.
If the total of all the angles is 2π or -2π, then the point is inside the polygon. If the total is zero, the point is outside. You can verify this intuitively with some simple examples using squares or triangles.
To use the example program, draw a polygon and position the mouse over the point you want to check. Then press F3 to invoke the Point in Polygon test. Alternatively you can use the Alt key to open the program’s menu, use the arrow keys to select the test, and press Enter to perform the test. The program draw the point in green if it lies inside the polygon and red if it lies outside of the polygon.
The following code shows how the program’s Polygon class determines whether a point lies inside the polygon.
```// Return True if the point is in the polygon.
public bool PointInPolygon(float X, float Y)
{
// Get the angle between the point and the
// first and last vertices.
int max_point = Points.Length - 1;
float total_angle = GetAngle(
Points[max_point].X, Points[max_point].Y,
X, Y,
Points[0].X, Points[0].Y);
// Add the angles from the point
// to each other pair of vertices.
for (int i = 0; i < max_point; i++)
{
total_angle += GetAngle(
Points[i].X, Points[i].Y,
X, Y,
Points[i + 1].X, Points[i + 1].Y);
}
// The total angle should be 2 * PI or -2 * PI if
// the point is in the polygon and close to zero
// if the point is outside the polygon.
// The following statement was changed. See the comments.
//return (Math.Abs(total_angle) > 0.000001);
return (Math.Abs(total_angle) > 1);
}```
This code first gets the angle made by the polygon’s first and last points, plus the target point. It then loops through the polygon’s other points calculating their angles and adding them up. When it finishes, the method returns true if the sum of the angles is close to 2π or -2π.
The following code shows the GetAngle method that the PointInPolygon method uses to find angles.
```// Return the angle ABC.
// Return a value between PI and -PI.
// Note that the value is the opposite of what you might
// expect because Y coordinates increase downward.
public static float GetAngle(float Ax, float Ay,
float Bx, float By, float Cx, float Cy)
{
// Get the dot product.
float dot_product = DotProduct(Ax, Ay, Bx, By, Cx, Cy);
// Get the cross product.
float cross_product = CrossProductLength(Ax, Ay, Bx, By, Cx, Cy);
// Calculate the angle.
return (float)Math.Atan2(cross_product, dot_product);
}```
This code uses the DotProduct and CrossProductLength methods to get the angle’s dot product and length of the cross product. (If the two vectors are AB and BC, then the dot product is given by |AB| * |BC| * Cos(theta) and the length of the cross product is given by |AB| * |BC| * Sin(theta), where theta is the angle between the two vectors.)
If you divide the length of the cross product by the length of the dot product, you get:
```(|AB| * |BC| * Sin(theta)) / (|AB| * |BC| * Cos(theta)) =
Sin(theta)) / Cos(theta)) =
Tan(theta)```
The Math.ATan2 method takes opposite and adjacent side lengths for a right triangle and returns an angle with the corresponding tangent. That fact plus the previous equations means Math.Atan2(cross_product, dot_product) returns the angle between the two vectors.
To see the CrossProductLength method, download the example or see the post Determine whether a polygon is convex in C#. The following code shows the DotProduct method.
```// Return the dot product AB · BC.
// Note that AB · BC = |AB| * |BC| * Cos(theta).
private static float DotProduct(float Ax, float Ay,
float Bx, float By, float Cx, float Cy)
{
// Get the vectors' coordinates.
float BAx = Ax - Bx;
float BAy = Ay - By;
float BCx = Cx - Bx;
float BCy = Cy - By;
// Calculate the dot product.
return (BAx * BCx + BAy * BCy);
}```
If two vectors A and B have X and Y components <Ax, Ay> and <Bx, By>, then their dot product is simply:
`Ax * Bx + Ay * By`
The DotProduct method simply calculates and returns that value. For more information on dot products, see Wikipedia and Wolfram MathWorld.
This entry was posted in algorithms, geometry, graphics, mathematics and tagged , , , , , , , , , , , , . Bookmark the permalink.
### 17 Responses to Determine whether a point is inside a polygon in C#
1. Felipe Vaz says:
Great code man, worked like a charm for me. I was having a problem with the code from http://social.msdn.microsoft.com/forums/en-US/winforms/thread/95055cdc-60f8-4c22-8270-ab5f9870270a/ because its was saying that a point is inside when i used a very complex polygon, and with your code it worked very well, thank you so much.
2. Anonymous says:
Thank you!
Great applicaton :))
3. murat says:
great application thank you!
4. TFO says:
There are some great graphics tools in .NET
Regions
```Region region1 = new Region(new Rectangle(50, 0, 50, 150));
Point point = new Point(60, 10);
if (region1.IsVisible(point)
{
//Point is within region
}```
The region also has a transform. It can be used if the region happens to be a selection box or selection shape on your screen. You can apply the inverse transform (if any) used when some objects where drawn and test if a known point falls within your selection box.
5. Mike says:
THANKKKKK YOUUUU!!!! GREAT Solution
6. Himanshu Gupta says:
Thanks a Ton…
It was exactly what I needed.
Congratulations, very good job! I’ve used it with double data type
The PointInPolygon method fails for the below polygon:
https://imgur.com/QkXxkMM
Otherwise it is a great method and i am using it in my project where there are no such polygons!
• RodStephens says:
I think this is a rounding error. When I step through an example where it incorrectly thinks a point is inside the polygon, the total of the angles is 0.000001013279, which is slightly larger than the value 0.000001 that it uses for comparison.
There’s really no need to use such a small comparison value because the total should be either 2π, -2π, or 0.
I think it should work if you change the test in the PointInPolygon method to the following.
`return (Math.Abs(total_angle) > 1);`
I have made that change to the code.
Thank you!!!
Hi!
Also all the other polygon related operations!
If I were you, I would create a class library of polygon operations and also publish a nuget package…
Cheers,
Should this algorithm work also on spherical polygons? Especially near sphere poles?
• RodStephens says:
If you mean should it work in three-dimensional polygons, the answer is “sort of.” This code only works in two-dimensions, so you would need to project the three-dimensional points into two-dimensional space before using the method.
That’s one way a three-dimensional drawing tool can perform hit testing. They convert the points that make up the objects into the viewing projection and then do something similar to what’s shown here to see if the point lies inside the resulting transformed polygons. They also need to find the points of intersection so they can tell which intersection is closest to the camera.
If you’re trying to perform hit testing, then you might be better off just allowing the 3D tool to find the point of intersection for you.
If you’re asking whether it will work with a “polygon” that has points that lie on a sphere, then the answer is “not quite.” A triangle is still a triangle if its points lie on a sphere, but in general more complex polygons won’t actually be polygons because they won’t be flat. You can still project them with the viewing transformation to make them flat and then the previous discussion applies.
And if I’m completely missing your question, post a follow-up. 😉 | 1,920 | 7,931 | {"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} | 3.375 | 3 | CC-MAIN-2021-17 | longest | en | 0.737647 |
https://study.com/academy/topic/algebraic-expressions-equations.html | 1,579,563,832,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250601040.47/warc/CC-MAIN-20200120224950-20200121013950-00037.warc.gz | 685,136,736 | 24,436 | # Ch 3: Algebraic Expressions & Equations
### About This Chapter
Tackle algebraic equations and expressions with the help of the fun lessons in this engaging chapter. Our mobile-friendly video and text lessons cover all the topics you're learning about in class, while our quizzes help you get ready to take a test or catch up after missing class.
## Algebraic Expressions & Equations - Chapter Summary
Our informative chapter on algebraic expressions and equations explains variables, coefficients and like terms. Each short lesson is followed by a multiple-choice quiz you can use to test your understanding before you move on in the chapter. Some lessons include practice problems. All of our learning tools are available to you 24 hours a day on your mobile device or computer. If you have questions along the way, reach out to one of our instructors for help. This chapter is designed to help you:
• Translate words into algebraic expressions
• Simplify algebraic expressions in practice problems
• Express relationships as algebraic expressions
• Use exponents to simplify expressions
• Explain how algebraic expressions are simplified with negative signs
• Solve linear equations with substitution
• Describe what an inequality is
• Translate math sentences into inequalities
• Use open sentences and true/false statements to write equations with inequalities
13 Lessons in Chapter 3: Algebraic Expressions & Equations
Test your knowledge with a 30-question chapter practice test
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##### Summary:
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Support | 500 | 2,482 | {"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} | 2.703125 | 3 | CC-MAIN-2020-05 | longest | en | 0.895911 |
https://stat.ethz.ch/pipermail/r-sig-geo/2010-June/008500.html | 1,444,795,442,000,000,000 | text/html | crawl-data/CC-MAIN-2015-40/segments/1443738099622.98/warc/CC-MAIN-20151001222139-00221-ip-10-137-6-227.ec2.internal.warc.gz | 1,180,105,589 | 3,222 | # [R-sig-Geo] How to extract coordinates values from a shapefile?
Paul Hiemstra p.hiemstra at geo.uu.nl
Fri Jun 11 11:43:34 CEST 2010
```On 06/09/2010 11:01 PM, Dylan Beaudette wrote:
> On Wednesday 09 June 2010, Hadley Wickham wrote:
>
>> On Wed, Jun 9, 2010 at 3:24 PM, Edzer Pebesma
>>
>> <edzer.pebesma at uni-muenster.de> wrote:
>>
>>> The example provided by Matt assumes that each polygon consists of a
>>> single ring, and doesn't have islands, lakes etc. The function below
>>> pasts all coordinates to a single 2-column matrix. For clarity's sake, I
>>> avoided nested sapply's.
>>>
>>> library(maptools)
>>> package="maptools")[1], IDvar="FIPSNO", proj4string=CRS("+proj=longlat
>>> +ellps=clrk66")) allcoordinates = function(x) {
>>> ret = NULL
>>> polys = x at polygons
>>> for(i in 1:length(polys)) {
>>> pp = polys[[i]]@Polygons
>>> for (j in 1:length(pp))
>>> ret = rbind(ret, coordinates(pp[[j]]))
>>> }
>>> ret
>>> }
>>> q = allcoordinates(xx)
>>>
>>> # check:
>>> plot(xx)
>>> lines(q, col='blue')
>>>
>> And that's basically what fortify does, except it covers a few more cases.
>>
>>
> Not to be nit-picky, but wouldn't it be safer to pre-allocated "ret" instead
> of dynamically appending? This would allow the suggested function to scale to
> very large geometries.
>
In addition to my previous suggestion, I tried using lapply + do.call
instead of the for loop based implementation of edzer:
library(maptools)
package="maptools")[1], IDvar="FIPSNO", proj4string=CRS("+proj=longlat
+ellps=clrk66"))
allcoordinates = function(x) {
ret = NULL
polys = x at polygons
for(i in 1:length(polys)) {
pp = polys[[i]]@Polygons
for (j in 1:length(pp))
ret = rbind(ret, coordinates(pp[[j]]))
}
ret
}
allcoordinates_lapply = function(x) {
polys = x at polygons
return(do.call("rbind", lapply(polys, function(pp) {
do.call("rbind", lapply(pp at Polygons, coordinates))
})))
}
q = allcoordinates(xx)
z = allcoordinates_lapply(xx)
all.equal(q,z)
So far it produces the same output. And now comes the nice part, doing a
test with a large polygon set and timing the performance of both functions:
> system.time(x <- allcoordinates(nuts))
user system elapsed
22.781 8.572 31.422
> system.time(y <- allcoordinates_lapply(nuts))
user system elapsed
0.248 0.004 0.250
> all.equal(x,y)
[1] TRUE
So apart from the imo nicer looking function in terms of syntax, it is
also several orders of magnitude faster. This effect will only become
stronger if the polygon set becomes larger. So, lapply + do.call is your
friend :).
cheers,
Paul
> sessionInfo()
R version 2.11.0 (2010-04-22)
i486-pc-linux-gnu
locale:
[1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C
[3] LC_TIME=en_US.UTF-8 LC_COLLATE=en_US.UTF-8
[5] LC_MONETARY=C LC_MESSAGES=en_US.UTF-8
[7] LC_PAPER=en_US.UTF-8 LC_NAME=C
[11] LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] maptools_0.7-34 lattice_0.18-5 sp_0.9-62 foreign_0.8-40
loaded via a namespace (and not attached):
[1] grid_2.11.0
>
> Cheers,
> Dylan
>
>
>
>
--
Drs. Paul Hiemstra
Department of Physical Geography
Faculty of Geosciences
University of Utrecht
Heidelberglaan 2
P.O. Box 80.115
3508 TC Utrecht
Phone: +3130 274 3113 Mon-Tue
Phone: +3130 253 5773 Wed-Fri
http://intamap.geo.uu.nl/~paul | 1,054 | 3,438 | {"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} | 2.96875 | 3 | CC-MAIN-2015-40 | longest | en | 0.715492 |
https://forum.allaboutcircuits.com/threads/bootstrap-capacitor-increase-input-impedance-of-voltage-follower-how.115121/ | 1,627,117,383,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046150134.86/warc/CC-MAIN-20210724063259-20210724093259-00132.warc.gz | 267,440,163 | 23,031 | # Bootstrap capacitor increase input impedance of voltage follower ...how??
#### Himanshoo
Joined Apr 3, 2015
265
Hello guys ...
According to theory in order to increase the input impedance bootstrap capacitor is employed…which couples the same output voltage to point B ..and as the circuit is a voltage follower Vin = Vout ..thus the voltage difference across R1 is very small ..which means that no or very little current will flow through R1 ..which can be also visualised as a very high resistance present..
Analysis:
Without bootstrap capacitor Cb:
Zin = R1; figure 1..
With bootstrap capacitor:
Zin = (1+Aol)R1 ....where Aol = open loop gain of opamp.
As its clear from above equation that input impedance increase by a factor of (1+Aol)..
since from figure 1. we have
i1 = V1/R1 = 500mv / 68K = 7.35 mA.
from figure 2 ..we have
Zin = (999.996 - 999.998)mV / 7.35mA (forget Zin = (1+Aol) R1 for the time being)
= 2.72* 10^-4 k
But the value of Zin should be very 2 large ..but is not agreeing with the theory.. why?? please justify..
#### Attachments
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Last edited:
#### Alec_t
Joined Sep 17, 2013
12,148
from figure 2 ..we have Zin = (999.996 - 999.998)mV / 7.35k
How do you arrive at that?
#### Himanshoo
Joined Apr 3, 2015
265
How do you arrive at that?
the way iam thinking is that ...no matter how much resistance is connected in series... current i1(in this case wont change )
as it is very usual to think that Cb is making R1 look like a very high resistance as the voltage difference across R1 is very small i.e(999.996-999.998) ..its ok uptil now as a very small current would flow across R1..
since current i1 in both the cases are same..and the voltage difference is very low....hence it force me to think that now Zin might have increased..as Zin is literally equal to R1 (in absence of Cb)..simple ohm law..
#### Jony130
Joined Feb 17, 2009
5,241
Your equation is wrong. Please notice that for frequency larger than R1||R2 Cb act just like a short circuit.
So R1 resistor is connected between the input and the output of the amplifier. The voltage across R1 resistor is equal to the difference between Vin and Vout.
So the situation looks like this
Now let as try to find a input resistance.
Rin = Vin/Iin
In = (Vin - Vout)/R = (Vin - A*Vin)/R = Vin * (1 - A)/R
Rin = Vin/Iin = R/(1 - A)
So if Vin = 1V and A = 0.5V/V we get 0.5V at the output. So the input current is equal to:
Iin = (Vin - Vout)/R = 0.5V/10Ω = 50mA and therefore
Rin = 1V/50mA = 20Ω
If we increase the gain to 0.9V/V we have
Iin = (Vin - Vout)/R = 0.1V/10Ω = 10mA
So the input resistance is equal to
Rin = 1V/10mA = 100Ω
#### Himanshoo
Joined Apr 3, 2015
265
I spot it ...what i misunderstood was that Zin depends upon (Vin-Vout)..which isn't the situation actually it depends on Vin...hence you are right Jony ...instead Iin depends on R1
frequency larger than R1||R2 Cb act just like a short circuit.
what does this mean...are you talking in terms of impedance...please clarify...
#### Jony130
Joined Feb 17, 2009
5,241
I spot it ...what i misunderstood was that Zin depends upon (Vin-Vout)..which isn't the situation actually it depends on Vin...hence you are right Jony ...instead Iin depends on R1
Also notice that in fact Rin is equal to (R1/(1 - A))||Rint where Rint is a amplifier input resistance. Because only if Rint = ∞-->Rin = R1/(1 - A)
Also what will happens if the amplifier gain is negative (inverting amplifier)?
what does this mean...are you talking in terms of impedance...please clarify...
I was trying to say that for a frequency where Xcb << R1||R2 - the Rin is equal to R1/(1 - A). But for the lower frequency Rin = R1+R2.
#### Himanshoo
Joined Apr 3, 2015
265
#### Himanshoo
Joined Apr 3, 2015
265
Also what will happens if the amplifier gain is negative (inverting amplifier)?
The input impedance in that case would be ...Zin=[1+(Aol/Acl)]R1...since i am not completely sure of it...as i refer book named as"Operational Amplifier and Linear ICs"..by david bell...I dont know that the book requires some serious proof reading.....
Instead I urge you take a look over it....coz I fear that I am learning some kind of crap...
#### Jony130
Joined Feb 17, 2009
5,241
which one?
This one
Zin = (1+Aol)R1
The input impedance in that case would be ...Zin=[1+(Aol/Acl)]R1...since i am not completely sure of it...as i refer book named as"Operational Amplifier and Linear ICs"..by david bell...I dont know that the book requires some serious proof reading.....
Instead I urge you take a look over it....coz I fear that I am learning some kind of crap...
This time your equation ([1+(Aol/Acl)]R) only work for voltage-series (series shunt) feedback amplifiers.
And this has only a effect on a resistor inside a feedback loop (only on amplifier input resistance). | 1,390 | 4,852 | {"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} | 3.28125 | 3 | CC-MAIN-2021-31 | longest | en | 0.938532 |
http://oeis.org/A308471 | 1,571,800,998,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987828425.99/warc/CC-MAIN-20191023015841-20191023043341-00268.warc.gz | 148,168,062 | 3,998 | This site is supported by donations to The OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A308471 Lowest outliers for A057660. 1
1, 4, 6, 12, 24, 30, 60, 120, 180, 210, 360, 420, 840, 1260, 2520, 4620, 9240, 13860, 27720, 55440, 60060, 120120, 180180, 360360, 720720, 1441440, 1801800, 2042040, 3063060, 6126120, 12252240, 24504480, 30630600, 36756720, 38798760 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS A057660(n) is a multiplicative function bounded above by n*(n-1)+1, which is reached whenever n is 1 or prime. These numbers are the n such that the ratio between A057660(n) and the upper bound reaches a record low. Motivated by Daniel Forgues's conjecture that this sequence consists of 4 and A051451. A subsequence of A025487. LINKS Charlie Neder, Table of n, a(n) for n = 1..215 EXAMPLE A057660(60060)/(60060*60059+1) = 1211716737/3607143541 ~ 0.3359214, and every number less than 60060 has a ratio > 0.34, so 60060 is in this sequence. CROSSREFS Cf. A057660, A025487, A051451. Sequence in context: A054167 A303198 A120211 * A095416 A162688 A070232 Adjacent sequences: A308468 A308469 A308470 * A308472 A308473 A308474 KEYWORD nonn AUTHOR Charlie Neder, May 29 2019 STATUS approved
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Last modified October 22 22:34 EDT 2019. Contains 328335 sequences. (Running on oeis4.) | 535 | 1,606 | {"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} | 3.4375 | 3 | CC-MAIN-2019-43 | latest | en | 0.709176 |
https://finanzasmentor.com/problem-24 | 1,670,084,658,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710933.89/warc/CC-MAIN-20221203143925-20221203173925-00055.warc.gz | 292,885,726 | 6,711 | # Math help apps for iphone
This Math help apps for iphone supplies step-by-step instructions for solving all math troubles. We will give you answers to homework.
## The Best Math help apps for iphone
Best of all, Math help apps for iphone is free to use, so there's no reason not to give it a try! However, a better way is to subtract or add terms. This can be done using one of three strategies: If you have two numbers and one is bigger than the other, you can ignore the smaller one and just add or subtract that one’s value from both sides of the inequality. For example: 3x > 4 5 + x In this case, you would subtract 4 from both sides, leaving 3 > 5 6 – 4 , which is true because 6 > 5. This method can also be used to turn an inequality into a statement about addition or subtraction, as in “I am more than \$100 poorer than my friend.” If you have two numbers and one is less than the other, you can ignore the bigger one and just add or subtract that one’s value from both sides of the inequality. For example: 6 10 12 + 8 = ? = 15 20 In this case, you would add 8 to both sides, leaving 6 10 12 – 8 , which is true because 12 20 . This method can also be | 297 | 1,167 | {"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} | 4.15625 | 4 | CC-MAIN-2022-49 | latest | en | 0.944517 |
https://www.physicsforums.com/threads/dice-game.325130/ | 1,601,155,155,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400245109.69/warc/CC-MAIN-20200926200523-20200926230523-00529.warc.gz | 984,401,000 | 15,448 | Dice game
There is 5 players with 5 dices each = 25 throws per game
Whats the probability that there is X dices with the value Y(1-6) on the table.
for example. probability for 8 three's
this is calculated through the binomial theorem right?
http://en.wikipedia.org/wiki/Binomial_theorem
where n=25, k=8 och p=1/6
calculating the probability of having X dices with the same value could be calculated with the above times six right?
so far so good.
now it gets a bit more complicated.
how to calculate:
how many dices of the same value is the most probable result? (assuming all 25 dices has been thrown)
Also, conditional probability:
i know that i have got Z dices with the same value (1-6 it doesnt matter which value). What is now the most probable result of the number of dices with the same value on the table?(i know my own result, but i dont know the 4 other players result)
for instance: i get 4 three's. what is the most likely number of three's on the table totally of the 25dices?
please refer to probability theory and distributions if you know them.
Related Set Theory, Logic, Probability, Statistics News on Phys.org
HallsofIvy
Homework Helper
First "dices" is the third person singular present tense of a verb, what you do to vegetables; it is not the plural of the noun "dice". "dice" is itself the plural of the noun "die". I say this not as a criticism but because many people here do not have English as a first language- and it is a complicated language!
The binomial distribution you describe would be the probability of getting "x" dice to be a specific number. The probability of getting "x" dice the same (but any number between 1 and 6) would be a "multinomial" distribution because you might get, say "x" dice equal to 5, "y" dice equal to 3, etc.
The other way, determining the "most likely" number of dice that are the same (and again, there might be "x" dice that are equal to one number, "y" dice that are equal to another, etc.) requires calculating all the probability of each such number and seeing which is the largest.
haha thank you I did not know that about dices. English is not my first language.
anyway the multinomial formula being
(n!/(x1!...xk!))*p^x1...p^xk
in the above case:
how many dices of the same value is the most probable result? (assuming all 25 dices has been thrown)
n=25 p=1/6
but what exactely is x1 to xk? | 602 | 2,382 | {"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} | 3.953125 | 4 | CC-MAIN-2020-40 | latest | en | 0.937771 |
https://school.eckovation.com/337-prime-number/ | 1,600,869,255,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400210996.32/warc/CC-MAIN-20200923113029-20200923143029-00529.warc.gz | 576,038,750 | 30,387 | # Identifying a Prime Number – Simple method
###### METHOD TO CHECK WHETHER A NUMBER IS A PRIME OR NOT …. ?
PRIME NUMBER : A number is greater than 1 is called a prime number, if it has only two factors, namely 1 and the number itself.
Procedure to find out the prime number:
Suppose A is given number.
Step 1: Find a whole number nearly greater than the square root of A.
K > square root(A)
Step 2: Test whether A is divisible by any prime number less than K. If yes A is not a prime number. If not, A is prime number.
EXAMPLE :
Find out whether 337 is a prime number or not?
Step 1: 19 > square root (337) Prime numbers less than 19 are 2, 3, 5, 7, 11, 13, 17
(19*19=361 and 18*18=324.So,we need to find a whole number number greater than the square root of 337.As 324<337<361 we need to take 19)
Step 2: 337 is not divisible by any of them
Therefore 337 is a prime number
Correct! Wrong!
Correct! Wrong!
Correct! Wrong!
Correct! Wrong!
#### Is 599 a prime number ?
Correct! Wrong!
can you say which number is a prime and which is not .....?
Extraordinary....!
Good Job
Better luck next time
Don't miss out!
Learn new things. Get an article everyday. | 325 | 1,175 | {"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} | 4.1875 | 4 | CC-MAIN-2020-40 | latest | en | 0.828602 |
https://ar.scribd.com/document/82136632/Chem-373-Lecture-15-Angular-Momentum-III | 1,563,592,973,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526408.59/warc/CC-MAIN-20190720024812-20190720050812-00257.warc.gz | 309,877,136 | 63,714 | You are on page 1of 21
# Lecture 15: Angular Momentum-III. The material in this lecture covers the following in Atkins.
## Rotational Motion Section 12.7 Rotation in three dimensions
Lecture on-line Angular Momentum-III (3D. Part-2 (PDF) Angular Momentum-III (3D. Part-2) (Powerpoint) Handout for this lecture
Rotation..Quantum Schrdinger eq. for particle moving Mechanics 3D on sphere with radius a Consider now a particle moving on the surface of a sphere with the radius a Its Hamiltonian is given by
h2 2 h2 2 H= +V= 2m 2m sin ce the potential energy is uniform over the sphere and can be put to zero 2 2 2 2 = + + x 2 y 2 z2
We must solve h2 2 = E 2m
Rotation..Quantum Mechanics 3D
## Schrdinger eq. for particle moving on sphere with radius a
2 2 2 = + 2 L r 2 r r h 2r 2
L2 = h 2 [
d2
sin 2 d 2
+ sin
d d
sin
d d
1 d2 d2 d = - h 2 [ 2 + cot + ] 2 d 2 d d sin
We must solve h2 2 = E 2m
Rotation..Quantum Mechanics 3D
## h2 2 2 2 L2 2= = E; + We must solve 2 r r h 2 r 2 2m r h 2 2 (a, , ) 2 (a, , ) L2 (a, , )
2m
r 2
h 2r 2
] = E (a, , )
Wavefunction does not depend on r as a variable ( r is a constant r = a). Thus only last term on r.h.s. different from zero
L2 (a, , ) 2ma 2
[
] = E (a, , )
We must have H =
1 2ma
L2 2
2 ( , ) L
] = E (, ) 2ma 2
Rotation..Quantum Mechanics 3D
## We must solve H (, ) = E (, ) What is (, ) ?
We must have H =
Thus or L2 (, ) = 2ma 2 E (, )
Thus the eigenfunctions (, ) to H must be the eigenfunctions to L2 .
1 2ma
L2 2
1 2ma
L2 (, ) = E (, ) 2
(, ) = Yl,m ((, ) =
## 2l + 1 (l | m!| |m| Pl (cos ) exp[im] 4 (l + | m!|)
Rotation..Quantum Mechanics 3D
## Schrdinger eq. for particle moving on sphere with radius a
and what is E ?
Thus or L2 (, ) = 2ma 2 E (, ); (, ) = Ylm (, ) L2 Ylm (, ) = h 2 l(l + 1)Ylm (, ) = 2ma 2 EYlm (, ) 1) ( ma ) h 2 l(l + If = 2 , 2 E
= Ylm (, )
Finally
## h 2 l(l + 1) h 2 l(l + 1) = ; I = ma 2 E = 2I 2ma 2
Rotation..Quantum Mechanics 3D
## L2 Ylm (, ) = h 2 l(l + 1)Ylm (, )
Lz Ylm (, ) = hmYlm (, )
HYlm (, ) =
2 l(l + 1) h
2I
Ylm (, )
Schrdinger eq. for particle moving Rotation..Quantum on sphere with radius a Mechanics 3D H, L2 and Lz for a particle moving on a sphere have common eigenfunctions Ylm We already knew that Lz and L2 had commen eigenfunctions since [Lz , L2 ] = 0 = 1 L2 However H 2ma 2 Thus it is readily shown that [H, Lz ] = [H, L2 ] = 0
Lz , L2 and will give the outcomes (L2 )obs = h 2 l(l + 1); (Lz )obs = hm
2 2
## h l(l + 1) (L )obs (E)obs = = each time 2I 2I
Rotation..Quantum Mechanics 3D
l=0
Properties of solutions to Schrdinger eq. for particle moving on sphere with radius a
m=0
2l + 1 (l | m!| |m| (, ) = Yl,m ((, ) = Pl (cos ) exp[im] 4 (l + | m!|) L2Yl,m = h 2 l(l + 1)Yl,m; L zYl,m = mlhYl,m
## For l = 0 we have m = 0; 1 Yo,o = 4 (L2 )obs = 0; (L z )obs = 0
Value of Yoo is uniform over sphere
Rotation..Quantum Mechanics 3D l = 1 m = 0, 1, 1
Properties of solutions to Schrdinger eq. for particle moving on sphere with radius a 2l + 1 (l | m!| |m| (, ) = Yl,m ((, ) = Pl (cos ) exp[im] 4 (l + | m!|)
## L2Yl,m = h 2 l(l + 1)Yl,m; L zYl,m = mlhYl,m
l 1 m 0 Ylm 3 Y1,0 = cos 4 3 Y1,1 = sin exp[i] 8 3 Y1,-1 = sin exp[ i] 8 (L z )obs (L2 )obs 0h 1h - 1h 2h 2 2h 2 2h 2
1 1
1 -1
Rotation..Quantum Mechanics 3D
Properties of solutions to Schrdinger eq. for particle moving on sphere with radius a
(L2 )obs = 2h 2 and (Lz )obs = mh
## For l = 1 we have m = -1,0,1;
L
z L L L L L z L z z =
h
=0
= -
|L| = 2h
We have three states all with L2 = h 2 2 r That is the length | L | of L is in all three cases | L |= h 2 r However L is oriented differently in the three states with L z = h, 0, h For each of the states a
meassurement of L2 and L z will always give the same value e.g l = 1,m = 1 gives (L2 )obs = 2h 2 ;(Lz )obs = h
Rotation..Quantum Mechanics 3D
Properties of solutions to Schrdinger eq. for particle moving on sphere with radius a
## For l = 1 we have m = -1,0,1; and (Lz )obs = mh
L z L L L L L |L| = 2h z = L z z =
(L2 )obs = 2h 2
What about L x or L y ?
h
=0
We have (L x 2 + L y 2 )obs = (L2 )obs - (L2 )obs = 2h 2 h 2 = h 2 z However a meassurement of L x or L y can have as outcome -h,0, h
Rotation..Quantum Mechanics 3D
Properties of solutions to Schrdinger eq. for particle moving on sphere with radius a
## For l = 1 we have m = -1,0,1; and (Lz )obs = mh
(L2 )obs = 2h 2
For each of the states a meassurement of L2 and L z will always give the same value e.g l = 1,m = 1 gives (L2 )obs = 2h 2 ;(Lz )obs = h
L z L L L L L |L| = 2h z = L z z =
h
=0
What about the expectation values < L x > and < L y > representing the avarage value from many meassurements and what are the possible values for L x and L y for each meassurement ?
l = 2 Rotation..Quantum Mechanics 3D
m = 0, 1, 1, 2, 2
2l + 1 (l | m!| |m| (, ) = Yl,m ((, ) = Pl (cos ) exp[im ] 4 (l+ | m!|) < L2 >= h 2l(l + 1) < L z >= mlh
m 0
15 Y2,0 = ( 3 cos 1) 8
Ylm
(Lz )obs 0h 1 h 2h
(L2 )obs 6 h2 6 h2 6 h2
## 15 1 Y2,1 = m sin cos exp[ i] 8 15 2 Y2,2 = sin 2 exp[ 2i] 32
Rotation..Quantum Mechanics 3D 2l + 1 (l | m!| |m| (, ) = Yl,m ((, ) = Pl (cos ) exp[im ] 4 (l+ | m!|)
< L2 >= h 2l(l + 1) < L z >= mlh For l = 2 we have m = -2, -1, 0, 1, 2 < L2 >= 6h 2 L We have five states all with L2 = h 2 6 z r L =2 h That is the length | L | of L is in all z three cases | L |= h 6 L L =h r z However L is oriented differently L =0 in th three states with L z L z = 2h, h, 0, h, 2h L L = -h z For each of the states a L = - 2h z meassurement of L2 and L z
|L| = 6h
will always give the same value e.g l = 2, m = 1 gives < L2 >= 6h 2; < L z >= h
Rotation..Quantum Mechanics 3D
Properties of solutions to Schrdinger eq. for particle moving on sphere with radius a
(a) A summary of the Fig. 12.31. However, because the azimuthal angle of the vector around the z- axis is indeterminate, a better representation is as in (b), where each vector lies at an unspecified azimuthal angle on its cone.
Rotation..Quantum Mechanics 3D
2Ylm (, ) = h2l(l + 1) L
HYlm (, ) =
same energy
2l(l + 1) h 2 ma
Ylm (, )
L z Ylm (, ) = hm
Different orientation
The permitted orientations of angular momentum when l = 2. We shall see soon that this representation is too specific because the azimuthal orientation of the vector (its angle around z) is indeterminate.
Electron spin
Stern and Gerlach observed in 1921 that a beam of electrons (Ag - atoms) split in an inhomogeneous magnetic field into two beams
## They attributed to an internal spin angular momentum of the electron
1 l=s= 2
Electron spin m = ms = 1 2
r S
An electron spin (s = 1/2) can take only two orientations with respect to a specified axis.
An electron (top) is an electron with ms = +1/2; a electron (bottom) is an electron with ms = - 1/2.
The length of the spin - angular momentum is
r S
1 1 h 3 S | = h s(s + 1) = h ( + 1) = 2 2 2
## What you must know from this lecture
1. For a particle moving on a sphere we have that H, L2 and Lz commutes. Thus they must have common eigenfunctions. These eigenfunctions are the speherical harmonics
2. The corresponding eigenvalue equations are L2 Ylm (, ) = h 2 l(l + 1)Ylm (, ) Lz Ylm (, ) = hmYlm (, ) HYlm (, ) =
2 l(l + 1) h
2I
Ylm (, )
## What you must know from this lecture
2 and will give the outcomes Lz , L (L2 )obs = h 2 l(l + 1); (Lz )obs = hm h 2 l(l + 1) (E)obs = each time
## 3. For a state described by Ylm (, ) measurements of
2I
4. For a given l we have 2l + 1 eigenfunctions Ylm (, ) all with the same energy and the same length L2 = h l(l + 1) of the angular momentum. The all have different projections (orientations) Lz = mh of the angular momentum onto the z - axis. | 2,985 | 7,686 | {"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} | 3.765625 | 4 | CC-MAIN-2019-30 | latest | en | 0.751171 |
https://republicofsouthossetia.org/question/the-circle-has-an-area-of-12-56-square-feet-what-is-the-radius-of-the-circle-15919003-37/ | 1,643,313,748,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305288.57/warc/CC-MAIN-20220127193303-20220127223303-00167.warc.gz | 536,913,146 | 14,419 | ## The circle has an area of 12.56 square feet. What is the radius of the circle
Question
The circle has an area of 12.56 square feet. What is the radius of the circle
in progress 0
2 weeks 2022-01-14T11:58:10+00:00 2 Answers 0 views 0
## Answers ( )
Step-by-step explanation:
A
=
π
r
2
12.56
π
=
r
2
r
=
12.56
π
r=1.999
2. The answer would be 2ft if the area is 12.56 | 140 | 389 | {"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} | 3.640625 | 4 | CC-MAIN-2022-05 | latest | en | 0.85577 |
https://math.stackexchange.com/questions/1430386/use-the-definition-of-an-improper-integral-to-evaluate-the-given-integral-int/1430443 | 1,713,987,068,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819847.83/warc/CC-MAIN-20240424174709-20240424204709-00523.warc.gz | 340,053,846 | 36,535 | # use the definition of an improper integral to evaluate the given integral.$\int_3^\infty\frac{1}{(x+2)}dx$
I try to solve the first question
\begin{align} \lim \limits_{t \to \infty} {\int_{3}^{t} \frac{1}{x+2}}dx \end{align} and my answer for it is
$$\ln(\infty+2)-\ln(5)$$
However, now I'm stuck on question 2, on $a$, $b$ and don't know how to solve them
• One question at a time, please.
– Yes
Sep 11, 2015 at 2:58
• $$ln(\infty + 2) - ln(5) = ln(\infty)-ln(5) = ln(\frac{\infty}{5})=ln(\infty)=\infty$$ Or simply $$\ln(\infty+2)-\ln(5) = \infty$$ Sep 11, 2015 at 3:03
• yes, I solved this and I know its answer ,but my problem in question 2 Sep 11, 2015 at 3:15
The volume of the solid from rotating the curve $f(x)=e^{-x}$ around the $x$ axis, is given by
$$\pi\int_0^\infty (e^{-x})^2\,dx=\pi\int_0^\infty e^{-2x}\,dx$$
The volume of the solid from rotating the curve $f(x)=e^{-x}$ around the $y$ axis, is given by
$$\pi\int_0^1 (-\log y)^2\,dy=\pi\int_0^1 \log^2 (y)\,dy$$
Now, can you determine whether these integrals converge or diverge?
As I stated in the comments, the answer to the first question is $\infty$, which is what you would have gotten had you worked out the answer a bit more. (Nitpicking)
The second question involves a concept called finding the volume by rotation.
Finding the volume of a solid revolution is a method of calculating the volume of a 3D object formed by a rotated area of a 2D space. Finding the volume is much like finding the area, but with an added component of rotating the area around a line of symmetry - usually the x or y axis.
I will work out the answer to part a) of question 2, it's your responsibility to read up on this if you don't know what's going on.
This is the graph.
Now, what they want to know is what is the volume of a shape generated by rotating $e^{-x}$ around the $x$-axis.
Here's a visual of a similar shape being created,
Here's a visual of the completed shape.
Note that the equation is asking for the volume of these shapes, so just imagine them as being filled in instead of hollow.
So, let's walk through the equations required to do this, I'm only giving a basic overview because it is midnight and I'm exhausted (This is what the link is for).
$$\pi \int_0^\infty \text{Outer_Radius}^2-\text{Inner_Radius}^2dx$$ $$=\pi \int_0^\infty e^{-2x}-0^2dx$$ $$=\pi \int_0^\infty e^{-2x}dx$$
The question is does this integral converge or diverge?
Part b) asks you to do the exact same thing, except in terms of $y$, not $x$.
So,
$$y=e^{-x}$$ $$ln(\frac{1}{y}) = x$$
Now we just put everything in terms of $y$ and rotate around the $y$-axis.
Here's a visual.
Note that the $y$-value for the equation doesn't extend beyond 1, so that's the upper limit of the integral.
$$\pi \int_0^1 \text{Outer_Radius}^2-\text{Inner_Radius}^2dy$$ $$=\pi \int_0^1 ln(\frac{1}{y})^2-0^2dy$$ $$=\pi \int_0^1 (-ln(y))^2dy$$ $$=\pi \int_0^1 ln^2(y)dy$$
Does this integral converge or diverge? | 957 | 2,970 | {"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": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2024-18 | latest | en | 0.925813 |
https://www.unitconverters.net/weight-and-mass/tonne-to-pound-troy-or-apothecary.htm | 1,716,533,528,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058684.66/warc/CC-MAIN-20240524060414-20240524090414-00836.warc.gz | 924,008,041 | 3,798 | Home / Weight and Mass Conversion / Convert Tonne to Pound (troy Or Apothecary)
# Convert Tonne to Pound (troy Or Apothecary)
Please provide values below to convert tonne [t] to pound (troy or apothecary), or vice versa.
From: tonne To: pound (troy or apothecary)
### Tonne to Pound (troy Or Apothecary) Conversion Table
Tonne [t]Pound (troy Or Apothecary)
0.01 t26.7922888072 pound (troy or apothecary)
0.1 t267.9228880719 pound (troy or apothecary)
1 t2679.228880719 pound (troy or apothecary)
2 t5358.457761438 pound (troy or apothecary)
3 t8037.686642157 pound (troy or apothecary)
5 t13396.144403595 pound (troy or apothecary)
10 t26792.28880719 pound (troy or apothecary)
20 t53584.57761438 pound (troy or apothecary)
50 t133961.44403595 pound (troy or apothecary)
100 t267922.8880719 pound (troy or apothecary)
1000 t2679228.880719 pound (troy or apothecary)
### How to Convert Tonne to Pound (troy Or Apothecary)
1 t = 2679.228880719 pound (troy or apothecary)
1 pound (troy or apothecary) = 0.0003732417 t
Example: convert 15 t to pound (troy or apothecary):
15 t = 15 × 2679.228880719 pound (troy or apothecary) = 40188.433210785 pound (troy or apothecary) | 431 | 1,175 | {"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} | 2.59375 | 3 | CC-MAIN-2024-22 | latest | en | 0.571152 |
https://www.prcboard.com/2016/09/LET-Reviewer-General-Education-Math-Part-1.html?showComment=1496112392027 | 1,553,565,554,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912204768.52/warc/CC-MAIN-20190326014605-20190326040605-00244.warc.gz | 867,980,462 | 63,567 | # LET Reviewer General Education GenEd: Mathematics Part 1
In this website you will find the LET Reviewers in General Education (Gen Ed), Professional Education (Prof Ed) and Major Area of Specialization.
In efforts of this portal to attend on requests from PRCBoard Facebook fan page members, we consolidated some of the questions which may likely help out takers during the exams. RELATED ARTICLES:
Below is the LET Reviewer for General Education GENED: Mathematics Part 1. We encourage readers/ reviewees to use the comment boxes after the article for discussion.
1. Two buses leave the same station at 8:00 pm. One bus travels north at the rate of 30 kph and the other travels east at 40 kph. How many kilometers apart are the buses at 10 pm?
a. 140 km
b. 100 km
c. 70 km
d. 50 km
2. Calculate the mean absolute deviation of the following numbers: 60, 80, 100, 75 and 95
a. 12.4
b. 14.2
c. 16.1
d. 18.9
3. Which of the following is the factorization of the binomial x2 - 42?
a. (x + 4)(x + 2)
b. (x – 4)2
c. x(x + 2x + 2)
d. (x – 4)(x + 4)
4. What value of x will satisfy the equation: 0.4(5x - 1470) = x?
a. 490
b. 2,130
c. 1470
5. Which of the following has the greatest value:
a. 3 + 32 + (3 + 3)2
b. 33
c. [(3 + 3)2]2
d. (3 + 3 + 3)2
6. The average of 5 different counting numbers is 20. What is the highest possible value that one of the numbers can have?
a. 20
b. 40
c. 30
d. 90
7. Three brothers inherited a cash amount of P62,000 and they divided it among themselves in the ratio of 5:4:1. How much more is the largest share than the smallest share?
a. P75,000
b. P30,000
c. P24,800
8. What is the missing terms in the series 5, 20, 80, ___,1280, ___, 20, 480?
a. 50;210
b. 40;160
c. 35;135
d. 320;5120
9. At what rate per annum should P2400 be invested so that it will earn an interest of P800 in 8 years?
a. 6 ½ %
b. 5 ½ %
c. 4.17 %
d. 6 %
10. The area of a rectangle is (x2 + 2x - 8). If its length is x + 4, what is its width?
a. x + 2
b. x - 2
c. x + 1
d. x + 6
11. What is the value of 12⅙ - 3 ⅜ - 5 ⅔ + 20 ¾?
a. 21 1/8
b. 22
c. 23 7/8
d. 21
12. The vertex angle of an isosceles triangle is 20°. What is the measure of one of the base angles?
a. 150°
b. 60°
c. 75°
d. 80°
13. Ana and Beth do a job together in three hours. Working alone, Ana does the job in 5 hours. How long will it take Beth to do the job alone?
a. 3 and 1/3 hours
b. 2 and 1/3 hours
c. 3 hours
d. 7 and 1/2 hours
14. How much greater is the sum of the first 50 counting numbers greater than the sum of the first 100 counting numbers?
a. 110
b. 3,775
c. 3,155
d. 1200
15. Which of the following has the largest value?
a. 85
b. 39
c. 65
d. 94
16. A water tank contains 18 liters when it is 20% full. How many liters does it contain when 50% full?
a. 60
b. 30
c. 58
d. 45
17. The edges of a rectangular solid have these measures: 1.5 feet by 1½ feet by 3 inches. What is its volume in cubic inches?
a. 324
b. 225
c. 972
d. 27
18. In a certain school, the ratio of boys to girls is 5 is to 7. If there are 180 boys and girls in the school, how many boys are there?
a. 105
b. 90
c. 45
d. 75
19. Ruben’s grades in 6 subjects are 88, 90, 97, 90, 91 and 86? What is the least grade that he should aim for in the 7th subject if he has to have an average of 88?
a. 92
b. 74
c. 88
d. 85
20. On a certain day, three computer technicians took turns in manning a 24-hour internet shop. The number of hours Cesar, Bert, and Danny were on duty was in the ratio 3:4:5, respectively. The shop owner pays them P50 per hour. How much would Danny receive for that day?
a. P 230
b. P500
c. P160
d. P480
21. A retailer buys candies for P90.25. The pack has 35 pieces of candies. If she sells each candy for P2.25, how much profit does she make?
a. P11.50
b. P56.25
c. P37.50
d. P18.75
22. An online shop sells a certain calculator for P950 and charges P150 for shipping within Manila, regardless of the number of calculators ordered. Which of the following equations shows the total cost (y) of an order as a function of the number of calculators ordered (x)?
a. y = (950 + 150)x
b. y = 150x +950
c. x = 950y + 150
d. y = 950x + 150
23. One side of a 45° - 45° - 90° triangle measures x cm. What is the length of its hypotenuse?
a. X √3 cm
b. X cm
c. (X √3)/2 cm
d. X √2 cm
24. The legs of one right triangle are 9 and 12, while those of another right triangle are 12 and 16. How much longer is the perimeter of the larger triangle than the perimeter of the smaller triangle?
a. 84
b. 7
c. 12
d. 14
25. Determine the midpoint of the line segment joining the points (7, -3) and (-1, 6).
a. (2, 3/2)
b. (2, -3/2)
c. (3, 3/2)
d. (1, 5/2)
26. Which of these has the longest perimeter?
a. A square 21 cm on a side
b. A rectangle 19 cm long and 24 cm wide
c. An equilateral triangle whose side is 28 cm
d. A right triangle whose two legs are 24 and 32 cm
27. How many square inches are in 2 square yard?
a. 900
b. 144
c. 1296
d. 2,592
28. In a playground for Kindergarten kids, 18 children are riding tricycles or bicycles. If there are 43 wheels in all, how many tricycles are there?
a. 8
b. 9
c. 7
d. 11
29. Nelia takes ¾ hour to dress and get ready for school. It takes 4/5 hour to reach the school. If her class starts promptly at 8:00 am; what is the latest time she can jump out of bed in order not to be late for school?
a. 6:42 am
b. 6:27 am
c. 6:57 am
d. 7:02 am
30. Which common fraction is equivalent to 0.215?
a. 43/200
b. 27/125
c. 21/50
d. 108/375
Answers: 1B 2A 3D 4D 5C 6D 7C 8D 9C 10B 11C 12D 13D 14 15A 16D 17C 18D 19B 20B 21 22D 23D 24C 25C 26D 27D 28C 29B 30A
DISCLAIMER: This website believes that education is a right, not a privilege. This portal does not claim ownership to any materials posted. Likewise, questions are not influenced by PRC professional regulatory board. The main purpose of the reviewer is to assist examinees who chose to review online.
1. AnonymousJuly 27, 2017
nasa pinaka last po yung answers
2. wala po bang answer key?
4. Answer to number 14 is B. 3,775
Using Gauss formula... the sum of 1 to n = n(n+1)/ 2
1 to 50 = 50(50+1)/2 = 1275
1 to 100 = 100(100+1)/2 = 5050
So, 5050 - 1275 = 3,775
1. May God bless you!!!!!
5. Ano po answer dun sa number 14 at number 21?
1. AnonymousMay 16, 2017
14 B
21 A
2. hey, ask ko lang po .. san ba mkikita mga part 2 neto ?
3. 21 walang sagot error.
binili nya ang candy sa halagang P90.25
then 35 lahat ng candy sa isang pack. tapos binenta nya ng tig P2.25 each
so 35 * P2.25 = P78.75
P78.75 lang kita nya so wala siyang profit luge pa sya ng P11.25 kaya error.
ang tanong kasi magkano profit nya.
6. panu po nakuha ung number 1? bakit po B ang answer?
Thank you
1. Daisyrie VasquezMay 22, 2017
First, you have to get each other's distance after 2 hours:
North: d=st=30(2)=60
East: d=st=40(2)=80
Then, you have to apply Pythagorean Theorem in order to get the distance r^2 = a^2 + b^2 = 60^2 + 80^2 = 3600 + 6400=10 000
Get the square root of 10, 000 to get r= 100 km.
7. sa number 27 po, pano po nakuha?
thank you
1. 1 yard = 36 inches
1 yard^2 = 1296 inches^2
2 yard^2 = 2592 in^2
or,
2 yards = 72 inches
(2 yards = 72 in)^2
4 yards^2 = 5184 in^2
2 yards^2 = 2592
8. 4. What value of x will satisfy the equation: 0.4(5x - 1470) = x? a. 490 b. 2,130 c. 1470
ang naka-indicate po na answer ay D. pero wla po sa choices
x=588 tama po?
1. AnonymousJune 23, 2017
sa number 4 po ba ganito po ba process ng pag solve? distribution property, then transpose po.
kasi po wala pong nakaindicate na choices D po.
0.4(5x-1470)=x
0.4(5x)-0.4(1470)=x
2x-588=x
2x=588
2x/2=588/2
x=294
2. AnonymousJuly 11, 2017
588 po. di nyo na po na subtract ung isang x resulting to divide 588 by 2 which is wrong.
3. Hello po what is the right solution po sa # 4 ? salamat po sa maka ka sagot.
4. 0.4(5x-1470)=x
0.4(5x)-0.4(1470)=x
2x-588=x (TRANSFER X AND -500 AND CHANGE SIGNS)
2X-X=588
X=588
9. #21. A retailer buys candies for P90.25. The pack has 35 pieces of candies. If she sells each candy for P2.25, how much profit does she make?
2.25*35=78.75
90-78.75=11.25
But suppose to be the price of each candy is more than 2.50 or 2.60 because
90.25/35=2.5785714 or 2.58
10. Pano po nakuha yung sagot sa no. 1? Solution please thanks
1. using Pythagorean theorem
2. solution po sa number 1?
3. AnonymousMay 05, 2017
8:00 pm they start to part ways, and the question is how many kilometers apart after 10:00 pm, so the one that traveling north with a rate of 30kph (kilometer per Hour)using that rate we multiply it by two because the difference of 8:00 and 10:00 is 2hours. so the distance of the bus that traveling north from the starting point is 60km. the same thing with the bus that travels east, multiply it and it obtain 80km far from the starting point. so not using an illustration you form a right angle. to get the distance of the two, use the pythagorean theorem (c^2 = a^2 +b^2)
Let a = distance of bus traveling north from the starting point which is 60km
and b = distance of bus traveling east from the starting point which is 80km
c^2 = 60^2 + 80^2
then you'll obtain the distance of 100km
4. paano po basahin yun ^ na ito? anu po siya s math?
5. mga sir at mam,
pa help na po s mga number na ito
number - 3,4, 5, 8, 10, 11, 12, 15, 16, 18, 21, 22, 23, 25, 26, 29, 30.
11. AnonymousMay 01, 2017
At what rate per annum should P2400 be invested so that it will earn an interest of P800 in 8 years?ano po ba ang solution nito
1. AnonymousJune 09, 2017
800/8 years = 100 yearly interest
To get its interest rate,
100/2,400 = 4.17%
2. 9. At what rate per annum should P2400 be invested so that it will earn an interest of P800 in 8 years?
a. 6 ½ %
b. 5 ½ %
c. 4.17 %
d. 6 %
solution:
principal(P) = 2,400
time (t) = 8 years
interest = 800
rate (r) = ?
formula: I=Prt
800 = 2400(r) (8)
800 = 2400 (8) r
800 = 19200 r
800/19200 = r
0.041666 = r
0.041666 (100) = r
12. AnonymousMay 25, 2017
Paano po yung number 11. Na stuck ako e. Maraming salamat po sa sasagot!
1. AnonymousJuly 08, 2017
Try mong gawing decimal ung my fraction...halimbawa dun sa dulo...20 3/4 gawin mong 20.75......3 devide 4 is 0.75....gawin mo rin dun sa iba makukuha mo na sagot
13. AnonymousMay 30, 2017
ano po ang answer Ng no.14 and 21
15. AnonymousJuly 07, 2017
Pano po yung shortcut ngpagsolve sa number 14?
16. AnonymousJuly 11, 2017
Ano po ung solution sa 2 ..
1. Mean Absolute Deviation (MAD) is simply calculating the positive distance (absolute value) of the mean from each data point and divided by the number of data points.
In No. 2 , the data points are 60, 80, 100, 75, & 95
The number of data points is 5.
To get the mean, simple add all the data points and then divide by the number of data points. (Parang averaging lang)
Mean = 60+80+100+75+95 = 410/5 = 82
Then the positive (absolute distance of the mean from each data points:
| 60 - 82 | = 22
| 80 - 82 | = 2
| 100 - 82 | = 18
| 75 - 82 | = 7
| 95 - 82 | = 13
Then add all the distances and divide by the number of data points
MAD = 22+2+18+7+13 = 62/5 = 12.40 Letter A
17. AnonymousJuly 11, 2017
Ung easy way solution po ng mga deviation? Slamat po
18. AnonymousJuly 21, 2017
Parang mali po ung answer sa number 19???
1. 19. Ruben’s grades in 6 subjects are 88, 90, 97, 90, 91 and 86? What is the least grade that he should aim for in the 7th subject if he has to have an average of 88?
(88+90+97+90+91+86+x)/7 = 88
(542+x)/7 = 88
542+x=88*7
x=616-542
x=74
a. 92
b. 74
c. 88
d. 85
19. This comment has been removed by the author.
20. how po 17?
1. No. 17...Volume in cubic inches
Given: 1.5 feet by 1 1/2 feet by 3 inches
The formula of volume is side x side x side (or edge x edge x edge)
Since the volume is to be in the form cubic inches, we need to convert all the measurement in inches:
1.5 feet = 1.5 x 12 inches = 18 inches
1 1/2 feet is the same as 1.5 feet = 18 inches
3 inches is already in inches
Then just multiply the measurements = 18x18x 3 = 972 cubic inches
21. pwede nnyo e show youe solution lahay sa nakaka alam ng mga sagot. wala na po kasi akong idea pa help naman po
22. paano pa ung mean absolute deviation? sa no.2 po
23. 60+80+100+75+95=410/5=82
60 of 82 = 22
80 of 82 = 2
100 of 82 = 18
75 of 82 = 7
95 of 82 = 13
22+2+18+7+13=62/5=12.4
24. Sa #14 anu po yung sagot? Thanks
25. Pano ung solution sa #6?
1. # 6 ==> 5 different counting numbers, the average is 100, what is possible highest number of one number?
Trial and error po.
First determine the total of the 5 numbers.
Since the average is 20, it suggests that the total of the 5 numbers is 100 ( because 100/5 = 20)
If we try 90, the total of the remaining 4 numbers should equal 10 ( 100 - 90 = 10)
Find 4 different numbers that totals 10 ==> 1, 2, 3, & 4 = 10
So the 90 is the highest possible number of one number in the group.
2. sir ano solution ng # 28&29?
3. solution Number 28
3T+2B=43 assuming no kids have trainng wheels on bicycles.
T+B=18 assume each bike and trike has only one kid on it.
3T+2(18-T)=43
T=43-36=7
26. Paano kinuha yung 24
1. First, you have to solve for the hypotenuse of both triangles.
Let x be the hypotenuse of the smaller triangle
y be the hypotenuse of the larger triangle
by Pythagorean theorem:
x=√[(9^2)+(12^2)]
x=15
y=√[(12^2)+(16^2)]
y=20
Then subtract the perimeter of the smaller triangle from the larger triangle
(12+16+y)-(9+12+x)
(12+16+20)-(9+12+15)
48-36
12
27. Pano po yung number 20? Nahihirapan na po kasi ako pag yung ratio talo talo na xD Salamat :D
1. add mo yung numbers sa ratio. 3+4+5 = 12
then gawin mong fraction yung number sa ratio.
3/12 : 4/12 : 5/12
cesar = 3/12 bert = 4/12 danny = 5/12
multiply mo na yung kay danny sa 24 para makuha yung hours ni danny.
5/12 * 24 = 10
then multiply 10 sa rate which is P50.
P500 po yung sagot :)
2. anu po number yan?
28. Y number is letter d? Dva the average of the 5 numbers is 20?
29. can you show computations/solutions please
thanks
30. Number 1. The answer is 140kph. Therefore: the answer is A.
1. Hi! I think you just added 60kph + 80kph, tama ba? That only works if the two vehicles left in OPPOSITE directions. :) For this problem, you need to find the HYPOTENUSE of the triangle formed by the two buses. The distance traveled by each bus represents the legs of the triangle. Applying the PYTHAGOREAN THEOREM, you get 100kph. That's basically just the square root of (60^2 + 80^2). ^_^
31. The answer in number 4 is 588. Please check this item. Thanks
1. Hello po sa # 4 hindi po ba ganito ang pag solve?
0.4(5x-1470)=x
0.4x5x=2x
0.4x1470=588
2x588=x
2x+x=3x
3x/3 =remove
588/3 =196
Tama po ba?
32. Hi how do you solve number 13? Thanks!
1. Ana = 1/5 of the job (3 hrs)
Beth = x (3hrs)
Ana and Beth = 1 job
1/5(3) + 3x = 1
3/5 + 3x = 1
3x = 1 - 3/5
3x = 2/5
x = 2/5 /3
x= 7.5
2. paano kinuha yung 7.5 nah sagut?? bakit x=2/5/3 bakit ang sagut is 7.5??? ano ang solution pls
3. 1/5 = ana
1/n = beth
1/3 = both
1/5 + 1/n = 1/3
1/n = 1/3 - 1/5
1/n = 5/15 - 3/15
1/n = 2/15
2n = 15
n = 15/2
n = 7 1/2
4. AnonymousJune 24, 2018
pano po if unknown si ana? then 7.5 si beth? pano po?
33. solution of number 7 please. thanks
1. solution for number 7
5x + 4x + 1x = 62000
10x = 62000
10 10
x = 6200
5(6200) = 31000
4(6200) = 24800
1(6200) = 6200
31000-6200 = 24000
34. sa number 21 po anu sagot?
1. #21. A retailer buys candies for P90.25. The pack has 35 pieces of candies. If she sells each candy for P2.25, how much profit does she make?
a. P11.50
b. P56.25
c. P37.50
d. P18.75
Profit = Sales - Cost of goods
Profit = (35*2.25) - 90.25
Profit = 78.75 - 90.25
Profit = (-) 11.50
It is a negative profit..so it's a loss.
35. Hi po mga maam and sir ano po solution ng number18?
1. #18 is a ratio and proportion just take note of the total population which is 180.
boys + girls = 180.
the ratio is
(boys) / (boys+girls) = 5 / (5+7)
5/12 of the entire population are boys = 75.
(girls) / (boys+girls) = 7 / (5+7)
7/12 of the entire population are girls = 105.
36. pa check naman ng number 3 ...thanks
37. AnonymousMay 11, 2018
38. paano po yung 25? tama po ba yung [(x1 + x2)/2, (y1 + y2)/2]? Thanks
1. AnonymousMay 31, 2018
iadd mo lang yung dalawang x values diveded by 2 para magkaroon ka ng x midpoint. same process sa y values para naman sa y midpoint
39. AnonymousJune 12, 2018
paano po yung solution sa number 9, i'm so confused po i really need help
40. AnonymousJune 13, 2018
yung item # 4, answer is D. Bakit po ganun, eh choices is up to C lang?
41. AnonymousJune 15, 2018
bakit po ganun yung 3? ^_^
42. AnonymousJune 24, 2018
How to solve item no.3 ?? plss
1. 3. Which of the following is the factorization of the binomial x^2 - 4^2?
a. (x + 4)(x + 2)
b. (x – 4)^2
c. x(x + 2x + 2)
d. (x – 4)(x + 4)
The binomial is in the form of (a^2 – b^2), a difference of two squares.
Difference of two squares can be factored as (a+b) (a-b)
Let
a = x
b = 4
So.. (a+b) (a-b) = (x+4) (x-4)
43. in no. 21 why is it profit? suppose to be dba dapat lugi cya???
44. Wala po bang solutions?
45. opo walang solution?
46. nasaan po ang letter d sa no. 4?
47. Hindi po ba 140 yung sagot sa number 1? Dba po 30x2=60 then 40x2=80 2 dahil 2hrs na yung nakalipas tas 60+80=140?
1. paiwas po ang takbo nila mam... north ang isa.. west po yung isa forming a triangle kaya po pythagorean formula po
48. Anong tamang sagot sa nos. 14 and 21?
1. Answer to number 14 is B. 3,775
Using Gauss formula... the sum of 1 to n = n(n+1)/ 2
1 to 50 = 50(50+1)/2 = 1275
1 to 100 = 100(100+1)/2 = 5050
So, 5050 - 1275 = 3,775
49. Pano po ung series of number
50. solution for number 1 please | 6,488 | 17,595 | {"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} | 4.25 | 4 | CC-MAIN-2019-13 | longest | en | 0.831566 |
https://playtaptales.com/10-to-the-power-of-411/ | 1,656,488,591,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103624904.34/warc/CC-MAIN-20220629054527-20220629084527-00539.warc.gz | 507,049,921 | 3,879 | ## 10 to the power of 411
What is 10 to the power of 411 (10^411)? The answer is a Sestrigintacentillion. This is based on the Conway-Wechsler system for naming numbers.
## 10 to the power of 411 written out
10^411 is written out as:
1,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000 | 347 | 787 | {"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} | 2.625 | 3 | CC-MAIN-2022-27 | longest | en | 0.707344 |
https://groups.yahoo.com/neo/groups/new_ai_geostats/conversations/topics/117?l=1 | 1,510,960,744,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934804019.50/warc/CC-MAIN-20171117223659-20171118003659-00531.warc.gz | 637,023,615 | 22,833 | Sorry, an error occurred while loading the content.
AI-GEOSTATS: SUMMARY: Nscore transform & kriging of log normal data sets
Expand Messages
• Dear all, I’m sorry for being so late with the summary of the replies I got to the following question - What are the drawbacks of the normal score
Message 1 of 1 , Mar 4, 2001
Dear all,
I�m sorry for being so late with the summary of the replies I got to the
following question
- What are the drawbacks of the normal score transformation
- What are the latest developments that have been made to handle properly data
sets that have a log normal distribution.
I have cut and pasted here under bits and parts of the many replies I
received. Thanks a lot to:
Andrew, Joao Felipe, Isobel Clark, Paulo Justiniano Ribeiro Jr, Warr Benjamin,
Hirotaka Saito, Nelleke Swager, Syed Abdul Rahman Shibli, Raymond J. O'Connor
I also received a two pages long reply from Donald Myers. I have put the full
text in the archives of AI-GEOSTATS.
-----------------------------------------------------------------------
A. Comments on skewed data sets
The skewness of a data set can have many different origins and its
interpretation is of course highly subjective. Many assumptions have therefore
to be made.
Most of geostatistics is "distribution free", i.e., the derivation of the
simple kriging, ordinary kriging and universal kriging equations do not depend
on a distributional assumption (contrary to what is sometimes claimed).
However if a distributional assumption is to be useful it should be
multivariate rather than just univariate. Essentially none of the
transformations that are used in geostatistics can really preserve or produce
multivariate distributional properties, they are only univariate
transformations. For example, a histogram might appear lognormal and a log
transformation might then appear normal, this does not imply anything about
multivariate lognormality.
When handling skewed data sets, one can
1) remove the long tail and dismiss them as another population, i.e. work with
the main subset
2) dismiss the long tail as a set of "erroneous" data (this might be difficult
to justify)
3) use the data "as is" and use more robust measures, e.g. madogram, and do
not work with squared differences which are quite sensitive to long tails. The
choice of the sill becomes a problem in such a case. In the case of
multivariate lognormality, one can compute the relationship between the
variogram/covariance of the original and the variogram/covariance of the
transformed. This relationship is
essentially unknown in all other cases because it requires again, knowing the
multivariate distribution in analytic form (and being able to carrying out
certain complicated multiple integrations). The multivariate
transform must be known in analytic form and have a unique inverse. There are
examples in the literature of using power series approximations for the
transformation but too often the approximation is reduced to a linear one.
4) use a transformation and work in the transformed domain before
backtransforming (watch out for possible biases, where applicable).
5) use an indicator transform for different thresholds and regard the
connectivity of extreme values foremost on your agenda. This might be
difficult to implement in practice, particularly with sparse datasets
and the deterioration of the number of "pairs" at extreme thresholds where you
would normally want the best "resolution" anyway (median indicator kriging is
a possible workaround).
From the replies I have received, the last seems to be the most frequently
chosen option.
B. Problems with Normal Score Transformation (NST)
NST are useful to reveal the spatial correlation of highly skewed data sets.
Nevertheless, when a transformation is made prior to the estimation, several
problems will remain, First, one has introduced an element of ranking rather
than interval or ratio data for the original. Although one uses the NST data
as satisfying the requirements of normality, the back transformation process
can only recover the point estimates (e.g., for confidence limits) within the
resolution afforded by the original data at that point. If you have sparsely
distributed data there, the limit estimate has an uncertainty reflecting the
corresponding coarse steps (more a measurement error than an estimation
error).
Second, if one has ties in the original data, the NST assigns them to the
corresponding block of contiguous normal scores. Thus extra variance is
introduced as a result of handling the ties.
There are two types of nscore transformation:
1) a frequency based NST: data are transformed in order to get a histogram
showing a normal distribution.
Inconvenient: The ordering of the tied values introduces a bias when doing a
back-transformation, especially if there are many zero values
2) an empiricaly based NST: the transformation uses the cumulative
distribution and assigns the equivalent in the Gaussian space. When performing
a back-transformation, one get the original value.
Inconvenient: the histogram of the transformed data is often not normal.
Nevertheless, the results after kriging and simulation appear to be relatively
robust.
C. Performing kriging with log normal data sets
Most of the replies underlined the frequent use of an indicator approach. If
Lognormal kriging seems to be the solution for log normal data sets, it is
based on the strict assumption that the data set is log normal, assumption
which is almost impossible to verify unless one has an extensive knowledge of
the data set.
If one is willing to assume multi-variate lognormality (univariate is not
really sufficient) then the transformation is theoretically known and has a
unique inverse that is also known. Even in this case there
is the problem of a bias in the re-transformed estimates. A number of authors
have written on this, Journel, Dowd being two of them (see various papers in
Math. Geology). As pointed out in those papers the correction in the case of
Simple Kriging (punctual) is essentially solved, a good approximation is
available in the case of Ordinary Kriging (punctual). There are some
theoretical problems in the case of block kriging that are usually handled in
an almost ad-hoc way, e.g., if the point values are multi-variate lognormal
then the block values theoretically should not be either univariate or
multivariate lognormal. There seems to be little in the literature pertaining
to a mixing of lognormality and non-constant drift(mean). If the non-constant
mean is not first removed then the complications resulting from a non-linear
transformation are much worse since the non-constant mean and the mean zero
random component are not separately transformed.
For other non-linear transforms (other than the log in the case of
multivariate lognormality), even knowing the inverse transform in analytic
form is not sufficient to allow computing the bias adjustment unless
one also knows the MULTIVARIATE distribution in analytic form. Even then, the
actual mechanics of doing so can be very tedious or complicated. That is,
while there is a nice theorem on change of variables in a multiple integral,
the actual step of applying it to a specific problem can be very tedious and
complicated. Moreover the theorem has moderately strong assumptions which are
not always satisfied.
In the case of multivariate lognormality, one can also determine the
adjustment needed in the kriging variances. This aspect seems to have
attracted little attention in the case of other non-linear transforms and it
is at least as difficult a problem.
Apparently, lognormal kriging and indicator kriging produce very similar
results.
D. Recent developments:
The litterature seems to be quite poor in publications on non-parametric
geostatistics.
The Box-Cox family of transformations which has the log-normal as a particular
case has been recently proposed.
SUGGESTED READING
CHRISTENSEN, O.F., DIGGLE, P.J. AND RIBEIRO JR, P.J. (2001). Analysing
positive-valued spatial data: the transformed Gaussian model. In GeoENV III -
Geostatistics for environmental applications, Quantitative Geology and
Geostatistics, Kluwer Series (to appear)
CLARK I. 1996 "Lognormal kriging applied to non-lognormal deposits: two case
studies",
5th International Geostatistics Congress, Wollongong Australia, 22--27
September
CLARK I. 1997. Geostatistics applied to skewed data", Conference of the
International Section on Mathematical Methods in Geology (Mining P��bram
Symposia) of the International Association for Mathematical Geology, Prague,
6--10 October, Matematicke Metody V Geologii: P��bram Scientiae Rerum
Montanarum
CLARK I. 1998. Geostatistical estimation and the lognormal distribution
Geocongress, Pretoria RSA, June
SAITO, H. and P. GOOVAERTS. 2000. Geostatistical interpolation of positively
skewed and censored data in a dioxin contaminated site. Environmental Science
& Technology, vol.34, No.19: 4228-4235.
Gregoire Dubois
Institute of Mineralogy and Petrography
Dept. of Earth Sciences
University of Lausanne
Switzerland
http://www.ai-geostats.org
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Your message has been successfully submitted and would be delivered to recipients shortly. | 2,165 | 9,803 | {"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} | 2.96875 | 3 | CC-MAIN-2017-47 | latest | en | 0.910572 |
http://nrich.maths.org/public/leg.php?code=-415&cl=3&cldcmpid=7324 | 1,477,143,248,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988718987.23/warc/CC-MAIN-20161020183838-00406-ip-10-171-6-4.ec2.internal.warc.gz | 193,356,182 | 7,731 | # Search by Topic
#### Resources tagged with STEM - General similar to The Olympic Torch Tour:
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### What's the Point of Squash?
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### Shaping the Universe II - the Solar System
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The second in a series of articles on visualising and modelling shapes in the history of astronomy.
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##### Stage: 3, 4 and 5 Challenge Level:
This part introduces the use of Logo for number work. Learn how to use Logo to generate sequences of numbers.
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Design and test a paper helicopter. What is the best design? | 1,283 | 5,521 | {"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} | 3.75 | 4 | CC-MAIN-2016-44 | longest | en | 0.85892 |
http://slideplayer.com/slide/3846102/ | 1,527,368,042,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867859.88/warc/CC-MAIN-20180526190648-20180526210648-00636.warc.gz | 274,818,739 | 23,981 | # CHAPTER 9 Thermal & Moisture Protection
## Presentation on theme: "CHAPTER 9 Thermal & Moisture Protection"— Presentation transcript:
CHAPTER 9 Thermal & Moisture Protection
CON 4003 Construction Estimating Prof R. V. Locurcio
Presentations Today… CH 9 – Thermal & Moisture Protection
Thursday… HW#5 – Jose Vega Thursday… China Presentation
Waterproofing & Dampproofing
Consists of a water resistent coat that is sprayed or brushed onto the exterior of the foundation to prevent moisture intrusion into the building
Waterproofing Sprayed or rolled-on waterproofing
…bid by the square ft. Waterproof membranes and building paper …bid as a sheet or roll good
Building Paper and Vapor Barriers
A vapor barrier is often used to refer to any material for damp proofing, typically a plastic or foil sheet, that resists penetration of moisture through the wall, ceiling or floor assemblies of buildings
Moisture & Vapor Barriers
Building Paper and Vapor Barriers
Where: Behind masonry or siding Under shingles Between insulation and inside finish of exterior walls Must be lapped to be effective
Waterproof membrain or building paper used
to establish the waterproofing system Drain board: added to minimize the chance of water penetration of the waterproof membrane There must be a drain system at the bottom of the drain board to prevent the accumulation of water Building paper Fiberglass Drain board Drain system
The area of waterproofing = 148 ft x 8.33ft = 1232.84 SF
Problem 9-1 How much waterproofing is needed to cover the foundation walls of a 24-foot by 50-foot basement? The wall is 8 feet high. The wall sits on a footing that is 12 inches high and extends 16 inches beyond the wall on the exterior side. Assume that the waterproofing starts 12 inches below the top of the foundation wall and covers the top surface of the footings. If one gallon of waterproofing material covers 75 square feet, how many gallons of waterproofing material are needed ? Solution: L= 50ft, W= 24ft, Perimeter = ( ) = 148 LF The height of the waterproofing = 8ft -1ft+ (16in)/(12 in/ft) = 8.33 LF Water proofing The area of waterproofing = 148 ft x 8.33ft = SF Add four 16in by 16in corners on the top of the footing Total area = x((16/12) x(16/12)) = SF 1240 SF Quantity = = 16.6 gallons A-A Cross section of foundation 75 gal/SF
Insulation Fiberglass Batt Insulation
paperbacked so to be installed with the paper on the interior side of wall Blown Insulation fiberglass shredded newspaper spun rock Exterior Insulation Finish System Consists of a rigid insulation layer covered with a synthetic stucco system. The stucco system includes a base coat a reinforcing layer, a brown coat, and a colored finish coat.
Insulation Fiberglass Batt insulation Paperbacked or Unfaced
Common size precut 15” & 23” Blown Insulation Made from Fiberglass shredded newspaper, and blown rock. Spray Foam Insulation Made from Isocyanate & Polyol Resin
The length of wall = 25ft, height = 8ft
Problem 9-6 How many rolls of insulation are needed for Problem 5 if the insulation comes in rolls 15 inches wide and 32 feet long? Don’t include the waste factor. Solution: The length of wall = 25ft, height = 8ft The number of rows is one row 8 feet high. The number of columns is calculated using Eq. (4-17) as follows: The stud spacing is 16” Number columns = (25 ft) (12 in/ft) =18.75 Columns (15in + 1in) 19 number of sheets of 8-foot by 15 in-wide insulation. The length of roll is 32ft, so 4 sheets from each roll: No. of rolls = 19 = 4.75 ~ 5 rolls 4
25LF 16” 8LF 16” 16” 16” 25LF X 12”/LF = 300 inches Panels = 300/16 = or 19 panels Roll = 32 LF / 8 = 4 panels /roll Rolls = 19 / 4 = = 5 rolls
Exterior Insulation Finish System
“Consists of a rigid insulation layer covered with a synthetic stucco system.” “The stucco system includes a base coat a reinforcing layer, a brown coat, and a colored finish coat.”
Shingle Roofs Asphalt Shingles specified by expected life
20, 25, 30- year three tab or architectural Flashing required at all penetrations vent pipes intersections with wall or chimney Drip Edge required at all edges including gable end Starter Shingles required for all roof edges that are parallel to the rows of shingles. placed on the roof edges then completely covered up by first layer of shingles
Shingle Installation
Hip and Valley Roof Hip Roof is a type of roof where all sides slope downwards to the walls, usually with a fairly gentle slope
Asphalt Impregnated Felt on Roofs
Slope greater then 2:12 and less than 4:12 The felt must be lapped 19 in Slope greater than or equal to 4:12 The felt must be lapped 2 in Length sloped = Lenght plan view (1+ Slope^2)^0.5 Area surface = Area plan view (1+ Slope^2)^0.5 Hips and Valleys increase the waste To do a proper takeoff for a roof, each roof surface must be taken off individually
Area of Triangular Roof Section
***Length of Slope*** = same length from trapezoidal section Bilding Paper Building Paper Building Paper Average Length of a row for triangle areas is the average of the length of the ridge and the eave plus the width of the felt plus the end lap. ( (Ridge + Eave) / 2 ) + felt width + end lap Area of Triangular Section = (Average Length) x (no. Rows) x (felt width)
Roof Drip Edge
Siding Soffit and Fascia - 1
Siding, Soffit, and Fascia -2
Vinyl Soffit Aluminum Fascia
Siding Soffit and Fascia - 3
Problem 9-12: Determine the quantity of insulation , mesh reinforcement , and stucco needed to cover the end of the building in Fig the insulation comes in 4ft by 8ft sheets. The reinforcement comes in rolls 100 ft long by 3ft wide and needed to be lapped 3 inches. One cubic foot of base or brown coat will cover 60 square feet of wall, and one bucket of finish coat will cover 120 square feet of wall. Solution:
4.67’ 12 4 Part B 8’ W 4’X4’ D 6’X6’ 8” 4’
Stucco Wall Assembly
EIFS Wall Assembly
Roof Assembly
Roof Coverings
The End | 1,539 | 5,925 | {"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} | 3.203125 | 3 | CC-MAIN-2018-22 | latest | en | 0.821323 |
https://socialsci.libretexts.org/Bookshelves/Economics/Book%3A_Introduction_to_Economic_Analysis/03%3A_Quantification/3.01%3A_Elasticity | 1,642,371,627,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300244.42/warc/CC-MAIN-20220116210734-20220117000734-00591.warc.gz | 587,370,126 | 25,173 | # 3.1: Elasticity
• Anonymous by request
Learning Objectives
• What is the best way of measuring the responsiveness of demand?
• What is the best way of measuring the responsiveness of supply?
Let $$x (p)$$ represent the quantity purchased when the price is p, so that the function x represents demand. How responsive is demand to price changes? One might be tempted to use the derivative, $$x′$$, to measure the responsiveness of demand, since it measures how much the quantity demanded changes in response to a small change in price. However, this measure has two problems. First, it is sensitive to a change in units. If I measure the quantity of candy in kilograms rather than in pounds, the derivative of demand for candy with respect to price changes even if the demand itself is unchanged. Second, if I change price units, converting from one currency to another, again the derivative of demand will change. So the derivative is unsatisfactory as a measure of responsiveness because it depends on units of measure. A common way of establishing a unit-free measure is to use percentages, and that suggests considering the responsiveness of demand to a small percentage change in price in percentage terms. This is the notion of elasticity of demand. The concept of elasticity was invented by Alfred Marshall (1842–1924) in 1881 while sitting on his roof. The elasticity of demand is the percentage decrease in quantity that results from a small percentage increase in price. Formally, the elasticity of demand, which is generally denoted with the Greek letter epsilon, ε, (chosen mnemonically to indicate elasticity) is
$$\varepsilon=-d x x d p p=-p x d x d p=-p x^{\prime}(p) x(p)$$
The minus sign is included in the expression to make the elasticity a positive number, since demand is decreasing. First, let’s verify that the elasticity is, in fact, unit free. A change in the measurement of x doesn’t affect elasticity because the proportionality factor appears in both the numerator and denominator. Similarly, a change in the measure of price so that p is replaced by r = ap, does not change the elasticity, since as demonstrated below,
$$\varepsilon=-r d d r x(r / a) x(r / a)=-r x^{\prime}(r / a) 1 a x(r / a)=-p x^{\prime}(p) x(p)$$
the measure of elasticity is independent of a, and therefore not affected by the change in units.
How does a consumer’s expenditure, also known as (individual) total revenue, react to a change in price? The consumer buys x(p) at a price of p, and thus total expenditure, or total revenue, is TR = px(p). Thus,
$$d \text { dp } p x(p)=x(p)+p x^{\prime}(p)=x(p)\left(1+p x^{\prime}(p) x(p)\right)=x(p)(1-\varepsilon)$$
Therefore,
$d dp TR 1 p TR =1−ε.$
In other words, the percentage change of total revenue resulting from a 1% change in price is one minus the elasticity of demand. Thus, a 1% increase in price will increase total revenue when the elasticity of demand is less than one, which is defined as an inelastic demand. A price increase will decrease total revenue when the elasticity of demand is greater than one, which is defined as an elastic demand. The case of elasticity equal to one is called unitary elasticity, and total revenue is unchanged by a small price change. Moreover, that percentage increase in price will increase revenue by approximately 1 – ε percent. Because it is often possible to estimate the elasticity of demand, the formulae can be readily used in practice. Table 3.1 provides estimates on demand elasticities for a variety of products.
Table 3.1 Various Demand Elasticities
Product Product ε
Salt Movies 0.9
Matches Shellfish, consumed at home 0.9
Toothpicks Tires, short-run 0.9
Airline travel, short-run Oysters, consumed at home 1.1
Residential natural gas, short-run Private education 1.1
Gasoline, short-run Housing, owner occupied, long-run 1.2
Automobiles, long-run Tires, long-run 1.2
Legal services, short-run Automobiles, short-run 1.2-1.5
Tobacco products, short-run Restaurant meals 2.3
Residential natural gas, long-run Airline travel, long-run 2.4
Fish (cod) consumed at home Fresh green peas 2.8
Physician services Foreign travel, long-run 4.0
Taxi, short-run Chevrolet automobiles 4.0
Gasoline, long-run Fresh tomatoes 4.6
From http://www.mackinac.org/archives/1997/s1997-04.pdf; cited sources: James D. Gwartney and Richard L. Stroup,, Economics: Private and Public Choice, 7th ed., 1995; 8th ed., 1997; Hendrick S. Houthakker and Lester D. Taylor, Consumer Demand in the United States, 1929–1970 (1966; Cambridge: Harvard University Press, 1970); Douglas R. Bohi, Analyzing Demand Behavior (Baltimore: Johns Hopkins University Press, 1981); Hsaing-tai Cheng and Oral Capps, Jr., "Demand for Fish," American Journal of Agricultural Economics, August 1988; and U.S. Department of Agriculture.
When demand is linear, $$$$x(p)=a-b p$$$$, the elasticity of demand has the form
$$\varepsilon=\mathrm{bp} \mathrm{a}-\mathrm{bp}=\mathrm{p} \text { a } \mathrm{b}-\mathrm{p}$$
This case is illustrated in Figure 3.1.
Figure 3.1 Elasticities for linear demand
If demand takes the form x(p) = a * p−ε, then demand has constant elasticity, and the elasticity is equal to ε. In other words, the elasticity remains at the same level while the underlying variables (such as price and quantity) change.
The elasticity of supply is analogous to the elasticity of demand in that it is a unit-free measure of the responsiveness of supply to a price change, and is defined as the percentage increase in quantity supplied resulting from a small percentage increase in price. Formally, if s(p) gives the quantity supplied for each price p, the elasticity of supply, denoted by η (the Greek letter “eta,” chosen because epsilon was already taken) is
$$\eta=\operatorname{ds} \mathrm{s} \mathrm{dp} \mathrm{p}=\mathrm{p} \mathrm{s} \mathrm{ds} \mathrm{dp}=\mathrm{ps}^{\prime}(\mathrm{p}) \mathrm{s}(\mathrm{p})$$
Again, similar to demand, if supply takes the form s(p) = a * pη, then supply has constant elasticity, and the elasticity is equal to η. A special case of this form is linear supply, which occurs when the elasticity equals one.
## Key Takeaways
• The elasticity of demand is the percentage decrease in quantity that results from a small percentage increase in price, which is generally denoted with the Greek letter epsilon, ε.
• The percentage change of total revenue resulting from a 1% change in price is one minus the elasticity of demand.
• An elasticity of demand that is less than one is defined as an inelastic demand. In this case, increasing price increases total revenue.
• A price increase will decrease total revenue when the elasticity of demand is greater than one, which is defined as an elastic demand.
• The case of elasticity equal to one is called unitary elasticity, and total revenue is unchanged by a small price change.
• If demand takes the form x (p) = a * p−ε , then demand has constant elasticity, and the elasticity is equal to ε.
• The elasticity of supply is defined as the percentage increase in quantity supplied resulting from a small percentage increase in price.
• If supply takes the form s (p) = a * p η, then supply has constant elasticity, and the elasticity is equal to η.
## EXERCISES
1. Suppose a consumer has a constant elasticity of demand ε, and demand is elastic $$$$(ε > 1)$$$$. Show that expenditure increases as price decreases.
2. Suppose a consumer has a constant elasticity of demand ε, and demand is inelastic $$$$(ε < 1)$$$$. What price makes expenditure the greatest?
3. For a consumer with constant elasticity of demand $$$$(ε > 1)$$$$, compute the consumer surplus.
4. For a producer with constant elasticity of supply, compute the producer profits. | 1,877 | 7,727 | {"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": 9, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2022-05 | latest | en | 0.948589 |
https://becalculator.com/22-8-cm-to-inch-converter.html | 1,679,515,401,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296944452.74/warc/CC-MAIN-20230322180852-20230322210852-00178.warc.gz | 171,396,934 | 16,426 | # 22.8 cm to inch converter
## FAQs on 22.8 cm to inch
### How many inches in a cm?
If you are looking to convert 22.8 centimeters into an inch-length number, first you need to know how many inches 1 cm equals.
Here I can give you a direct indication that one centimeter is equivalent to 0.3937 inches.
### How do you convert 1 cm into inches?
For 1cm to inches conversion, multiply 1cm using a conversion factor 0.3937.
This makes it much easier to convert 22.8 cm to inches.
So 1 cm to inches = 1 times 0.3937 = 0.3937 inches.
This will allow you to answer the following question easily and quickly.
• What is one centimeter into inches?
• What is cm into inches conversion?
• What is the equivalent of 1 cm in inches?
• What does 1 cm equal to in inches?
### Definition:Centimeter
Centimeter is an International Standard Unit of Length. It is equal to one hundredth of a meter. It’s approximately equivalent to 39.37 inches.
### Meaning of Inch
The length units of the Anglo-American continent are measured in inches. 12 inches equals 1 foot, and 36 inches equals 1 yard. According to modern standards, one inch equals 2.54 centimeters.
### How do I convert 22.8 cm to inches?
By the above, you have fully grasped cm to inches.
The following is the specific algorithm:
Value in inches = value in cm × 0.3937
So, 22.8 cm to inches = 22.8 cm × 0.3937 = 0.897636 inches
This formula can be used to answer the related questions:
• What is 22.8 cm in inches?
• How do you convert cm to inches?
• How to translate cm to inches?
• What is standard measurement for cm to inches?
• What size are 22.8 cm into inches?
cm inch 22.4 cm 0.881888 inch 22.45 cm 0.8838565 inch 22.5 cm 0.885825 inch 22.55 cm 0.8877935 inch 22.6 cm 0.889762 inch 22.65 cm 0.8917305 inch 22.7 cm 0.893699 inch 22.75 cm 0.8956675 inch 22.8 cm 0.897636 inch 22.85 cm 0.8996045 inch 22.9 cm 0.901573 inch 22.95 cm 0.9035415 inch 23 cm 0.90551 inch 23.05 cm 0.9074785 inch 23.1 cm 0.909447 inch 23.15 cm 0.9114155 inch | 616 | 2,006 | {"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} | 3.53125 | 4 | CC-MAIN-2023-14 | longest | en | 0.83289 |
https://anygamble.com/testing-roulette-systems-the-five-number-system | 1,685,791,494,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649193.79/warc/CC-MAIN-20230603101032-20230603131032-00193.warc.gz | 125,209,706 | 16,958 | # AnyGamble
Mar 20, 2017
Testing roulette system
# Testing Roulette Systems - The Five Number System
This system called System of 5 numbers is played in the relatively simple way. You choose any 5 numbers and you keep betting on them. The basic series consists of 17 spins with the following progression: in the first 7 spins bet by one unit on each number, in further five spins your bet by 6 units on each number and then you bet 3 times by 11 units and finally 2 times by 16 units. Total bankroll is 510 units altogether.
## What do the Statistics say?
On the following table you can see the system from the viewpoint of the probability and the expected profit:
Spin Probability Profit in Series Expected Profit 1 13.51% 31 4,1892 2 11.69% 26 3.0387 3 10.11% 21 2.1227 4 8.74% 16 1.3987 5 7.56% 11 0.8317 6 6.54% 6 0.3923 7 5.66% 1 0.0566 8 4.89% 151 7.3856 9 4.23% 121 5.1185 10 3.66% 91 3.3292 11 3.16% 61 1.9301 12 2.74% 31 0.8483 13 2.37% 156 3.6921 14 2.05% 101 2.0674 15 1.77% 46 0.8143 16 1.53% 146 2.2353 17 1.32% 66 0.8739 18 8,47% -510 -43.2206 Total 100% -2.8956
Unfortunately, the expected profit resulted in a loss, therefore in the long term, the system should go below zero. Let’s have a look at the progression of the basic system on the following graph:
Our expectations became true and in spite of the short-term growth, the system sloped downward, (it had a downward trend). There were several options on how to modify the system but we were not able to apply all of them.
There was no point in altering the series length. It follows (from the table above) that as long as the series was, always the expected profit results in a loss (we had tested a series of 5 spins and a series of 25 spins). At the first sight, you would assume that there is a chance that you hit at least one number within 17 spins, but empirically based research had confirmed that there were also series when the roulette did not hit any number of a certain pentad 50 times in a row.
Applying the wait strategy would not be also supposed to have a success. We have attempted to apply at least a liner series progression; it means that after each unsuccessful series, we increased the roulette bet size on each number by the value of initial bet and after making a profit we reset the bet size back to initial bet:
As you could see, the system appeared to go to a plus and had an upward trend but the local slumps were drastically huge and reached values of hundreds of thousands below zero, therefore an idea of series progression was not successful. We have tried to apply some modification by employing a spin progression of the bet in one series in order that a win always reached a minimum value.
Unfortunately, we did not observe any amelioration and the progression practically remained the same as before. We can conclude that this system in this form is not feasible in the long term. | 797 | 2,902 | {"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} | 4.0625 | 4 | CC-MAIN-2023-23 | latest | en | 0.952116 |
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# John Thompson
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Connect the blocks in the model to produce the following signal: <<http://blogs.mathworks.com/images/seth/cody/connect-blocks...
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8年以上 前 | ダウンロード 4 件 | | 432 | 1,621 | {"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} | 2.90625 | 3 | CC-MAIN-2022-21 | latest | en | 0.712818 |
https://math.stackexchange.com/questions/282898/integral-help-here-please | 1,571,219,191,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986666959.47/warc/CC-MAIN-20191016090425-20191016113925-00001.warc.gz | 597,461,729 | 34,752 | I have to solve this
$$\int \frac{dx}{(x+1)\sqrt{x^2+2x}}$$
First make the substitution $u=x+1$ and get
$$\int \frac{ds}{u\sqrt{u^2-1}}$$
Next I substitute $s=\sqrt{u^2-1}$ and I get $ds=u\,du/\sqrt{u^2-1}$.
I have integral of $1/s^2+1\dots$ I replace for $s$ and I get the integral as
$$\arctan\sqrt{u^2-1} \, .$$
The answer on my textbook is $-\arcsin[1/(x+1)]$
• U n r e a d a b l e. – Git Gud Jan 20 '13 at 18:27
• try putting in latex – Santosh Linkha Jan 20 '13 at 18:33
• @GitGud U n h e l p f u l – Fly by Night Jan 20 '13 at 18:40
• @GitGud I see... – Fly by Night Jan 20 '13 at 18:42
• @GitGud: for the new user, your comment is not at all helpful. experimentX at least suggested a link that leads to helpful information. – robjohn Jan 20 '13 at 19:26
You wanted $\;s=\sqrt{u^2-1}$, so $\;s^2=u^2-1$, $\;2s\,ds=2u\,du$, and $\;s\,ds=u\,du$.
Substituting gives us: \begin{align}\int\frac{du}{u\sqrt{u^2-1}} &= \int\frac{u\,du}{u^2 \sqrt{u^2-1}} \\&= \int\frac{s\,ds}{(s^2+1)s}\\&= \int\frac{ds}{s^2+1} \\ \\&= \arctan s + C. \end{align}
Now don't forget to "back substitute": $s = \sqrt{u^2-1} = \sqrt{(x+1)^2 - 1} = \sqrt{x^2 + 2x}$
So our answer is: $$\quad\arctan(\sqrt{x^2 + 2x}) + C$$
Note that with trig substitution, and trigonometric solutions, there can be any number of solutions (constants may vary), as is evident by the number of trigonometric identities! E.g.:
$$\arctan\sqrt{x^2+2x}-\left(\arcsin\left(\frac1{x+1}\right)-\frac\pi2\right) = 0$$
$$-\arcsin\frac1{x+1}=\arctan\sqrt{x^2+2x}-\text{ constant} = \arctan \sqrt{x^2 + 2x} + C$$
To check your integration answers, you can always take the the derivative of your solutions, and if you can derive the original integrand, you are "good to go." Try differentiating each solution (your's and the text's) and both will give you the equivalent of your original integrand.
• – Yves Jan 20 '13 at 23:43
• @Yves - It you expand those factors $(x+1)$ etc, and simplify, I'm willing to bet that d/dx of each function will be equivalent, through identities. It just amounts to algebraic manipulation and trig identities. – Namaste Jan 20 '13 at 23:48
• yes it's exactly the same.... however if I plot the two functions in Wolfram|Alpha (not their derivatives) they seem quite different... – Yves Jan 21 '13 at 0:42
• Yves Because they differ by constants, that can "throw off" the graphs. Integrals yield "families of functions/solutions"... – Namaste Jan 21 '13 at 0:46
• Here they should differ only by one real constant (assuming we are in R), but it seems more complicated, maybe because the two functions are not defined for the same values ? wolframalpha.com/input/… – Yves Jan 21 '13 at 1:05
If $s=\sqrt{u^2-1}$ then $s^2=u^2-1$ so $2s\,ds=2u\,du$, and $s\,ds=u\,du$.
$$\int\frac{du}{u\sqrt{u^2-1}} = \int\frac{u\,du}{u^2 \sqrt{u^2-1}} = \int\frac{s\,ds}{(s^2+1)s} = \int\frac{ds}{s^2+1} = \arctan s + \text{constant}.$$
Now draw a right triangle in which the "opposite" side has length $\sqrt{u^2-1}$ and the "adjacent" side has length $1$. By the Pythagorean theorem, the hypotenuse has length $u$, and you see the trigonometric identity $$\arctan\sqrt{u^2-1} = \arcsin\frac{\sqrt{u^2-1}}{u}$$
After your work, there are several possibilities. For example, we can let $v=\frac{1}{u}$. Then $u=\frac{1}{v}$ and $du=-\frac{1}{v^2}\,dv$. Do the rest of the substituting, noting that $\frac{1}{\sqrt{u^2-1}}=\frac{v}{\sqrt{1-v^2}}$. Quickly we arrive at $$\int -\frac{dv}{\sqrt{1-v^2}},$$ which is familiar.
A more standard substitution is $u=\sec\theta$. Then $du=\sec\theta\tan\theta$, and $\sqrt{u^2-1}=\tan\theta$. Everything cancels!
Let $u=1/t$, and get
$$\int \frac{du}{u\sqrt{u^2-1}}=-\int \frac{dt}{\sqrt{1-t^2}}=-\arcsin t +C.$$
Michael Hardy's on the right track here, but I think the answer is incomplete. First of all, it's sloppy to leave that u in there. Your answer in terms of $x$ is $\tan^{-1}\sqrt{x^2+2x}$. As Michael Hardy suggested, create the right triangle, marking one angle. Let the length of the side opposite that angle be $\sqrt{x^2+2x}$ and the adjacent side have a length of $1$. That gives the hypoteneuse a length of $x+1$.
Now, you will see that the angle opposite the angle you marked has a sine of $\frac1{x+1}$. These 2 angles add up to $\frac\pi2$ radians, so you have
$$\tan^{-1}\sqrt{x^2+2x}+\sin^{-1}\frac1{x+1}=\frac\pi2$$
$$-\sin^{-1}\frac1{x+1}=\tan^{-1}\sqrt{x^2+2x}-\frac\pi2$$
Including the constant of integration in your answer, $\tan^{-1}\sqrt{x^2+2x}+C$, we see that the 2 answers are equivalent. Both forms are correct. | 1,666 | 4,572 | {"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": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2019-43 | latest | en | 0.6984 |
turutaemprendedora.com | 1,555,812,937,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578530161.4/warc/CC-MAIN-20190421020506-20190421042506-00396.warc.gz | 570,937,749 | 4,321 | # Ross elementary analysis homework solutions
Math 104. Introduction to Analysis Department of Mathematics University of California, Berkeley Homework will be assigned once a week and will be due the following week (except possibly in the weeks when there is a midterm). Kenneth Ross, Elementary Analysis: The Theory of Calculus, Springer, 2003. Math 312: Real Analysis I am Stephen G. Simpson, a Professor of Mathematics at Penn State University. Math 312 is an introductory course on real analysis. I taught it in in Spring 2009.
As stated in the Penn State catalog of courses, the only prerequisite for Math 312 is Math 141. Spring 2015, Math 500, Real Analysis Course Information. Course Coverage: This course intends to cover some basic topics of real analysis at the undergraduate level: the real number system, limits and sequences, sequences and series of functions, continuity and differentiability of functions, Riemann we will cover Chapter 1 to Get instant access to our stepbystep Elementary Analysis solutions manual.
Our solution manuals are written by Chegg experts so you can be assured of the highest quality! Author: Kenneth A Ross. 0 solutions. Frequently asked questions. Understanding Elementary Analysis homework has never been easier than with 48: For each homework, the TA will type up the solutions of three problems for you (including 2 graded problems).
Problems 4. 11 and 4. 15 are omitted from HW2. Textbook: Kenneth A. Ross, Elementary Analysis: Ross elementary analysis homework solutions theory of Calculus, Second edition, Springer Undergraduate Texts in Mathematics. Course description: This course is an introduction to rigorous mathematical proofs involving properties of real numbers, continuity, differentiation, integration, and infinite sequences and series.
Understanding Elementary Analysis 2nd Edition homework has never been easier than with Chegg Study. Why is Chegg Study better than downloaded Elementary Analysis 2nd Edition PDF solution manuals? It's easier Elementary Analysis: The Theory of Calculus, by Kenneth A.
Ross Additional reading Principles of Mathematical Analysis, by Walter Rudin Real Mathematical Analysis, by Charles Pugh Basic Analysis: Introduction to Real Analysis A random selection of the assigned homework will be graded. Homework solutions will be published on this Elementary Real Analysis (Second Edition, 2008) by Brian S. Thomson, Judith B. Bruckner, and Andrew M. Bruckner Other useful textbooks: Elementary Analysis.
The Theory of Calculus by Kenneth A. Ross and Understanding Analysis by Stephen Abbott. Syllabus If you have a problem with the Homework will be assigned each Friday Download homework solutions to ross elementary analysis math (PDF, ePub, Mobi) Books homework solutions to ross elementary analysis math (PDF, ePub, Mobi) Page 1
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Email: [email protected] | 589 | 2,886 | {"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} | 2.90625 | 3 | CC-MAIN-2019-18 | latest | en | 0.901246 |
https://www.chegg.com/homework-help/intersecting-normal-line-normal-curve-x2-2xy-3y2-0-1-1-inter-chapter-3.6-problem-45e-solution-9780321388506-exc | 1,553,558,425,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912204736.6/warc/CC-MAIN-20190325234449-20190326020449-00533.warc.gz | 740,914,284 | 27,662 | Student Solutions Manual Part 2 for University Calculus (1st Edition) Edit edition Problem 45E from Chapter 3.6: Intersecting normal The line that is normal to the curve x2 ...
We have solutions for your book!
Chapter: Problem:
Intersecting normal The line that is normal to the curve x2 + 2xy - 3y2 = 0 at (1, 1) intersects the curve at what other point?
Step-by-step solution:
Chapter: Problem:
• Step 1 of 4
Differentiating both sides of the equation of the curve, we have
• Chapter , Problem is solved.
Corresponding Textbook
Student Solutions Manual Part 2 for University Calculus | 1st Edition
9780321388506ISBN-13: 032138850XISBN: Authors:
This is an alternate ISBN. View the primary ISBN for: University Calculus: Alternate 1st Edition Textbook Solutions | 198 | 768 | {"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} | 2.734375 | 3 | CC-MAIN-2019-13 | latest | en | 0.829517 |
https://nrich.maths.org/867/clue | 1,576,213,439,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540548544.83/warc/CC-MAIN-20191213043650-20191213071650-00449.warc.gz | 478,452,266 | 4,654 | ### Odd Differences
The diagram illustrates the formula: 1 + 3 + 5 + ... + (2n - 1) = n² Use the diagram to show that any odd number is the difference of two squares.
### Factorial
How many zeros are there at the end of the number which is the product of first hundred positive integers?
### Rachel's Problem
Is it true that $99^n$ has 2n digits and $999^n$ has 3n digits? Investigate!
# Really Mr. Bond
##### Age 14 to 16 Challenge Level:
Is it always the case that when you square a number whose last digit is 5 you always end with 25?
By breaking the number down into a form (x + 5) it may then be possible to see what is happening and why. | 172 | 652 | {"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} | 3.578125 | 4 | CC-MAIN-2019-51 | latest | en | 0.891044 |
http://metamath.tirix.org/mpeascii/mdslle1i | 1,721,533,449,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517550.76/warc/CC-MAIN-20240721030106-20240721060106-00140.warc.gz | 21,200,410 | 2,608 | # Metamath Proof Explorer
## Theorem mdslle1i
Description: Order preservation of the one-to-one onto mapping between the two sublattices in Lemma 1.3 of MaedaMaeda p. 2. (Contributed by NM, 27-Apr-2006) (New usage is discouraged.)
Ref Expression
Hypotheses mdslle1.1
`|- A e. CH`
mdslle1.2
`|- B e. CH`
mdslle1.3
`|- C e. CH`
mdslle1.4
`|- D e. CH`
Assertion mdslle1i
`|- ( ( B MH* A /\ A C_ ( C i^i D ) /\ ( C vH D ) C_ ( A vH B ) ) -> ( C C_ D <-> ( C i^i B ) C_ ( D i^i B ) ) )`
### Proof
Step Hyp Ref Expression
1 mdslle1.1
` |- A e. CH`
2 mdslle1.2
` |- B e. CH`
3 mdslle1.3
` |- C e. CH`
4 mdslle1.4
` |- D e. CH`
5 ssrin
` |- ( C C_ D -> ( C i^i B ) C_ ( D i^i B ) )`
6 3 2 chincli
` |- ( C i^i B ) e. CH`
7 4 2 chincli
` |- ( D i^i B ) e. CH`
8 6 7 1 chlej1i
` |- ( ( C i^i B ) C_ ( D i^i B ) -> ( ( C i^i B ) vH A ) C_ ( ( D i^i B ) vH A ) )`
9 id
` |- ( B MH* A -> B MH* A )`
10 ssin
` |- ( ( A C_ C /\ A C_ D ) <-> A C_ ( C i^i D ) )`
11 10 bicomi
` |- ( A C_ ( C i^i D ) <-> ( A C_ C /\ A C_ D ) )`
12 11 simplbi
` |- ( A C_ ( C i^i D ) -> A C_ C )`
13 1 2 chjcli
` |- ( A vH B ) e. CH`
14 3 4 13 chlubi
` |- ( ( C C_ ( A vH B ) /\ D C_ ( A vH B ) ) <-> ( C vH D ) C_ ( A vH B ) )`
15 14 bicomi
` |- ( ( C vH D ) C_ ( A vH B ) <-> ( C C_ ( A vH B ) /\ D C_ ( A vH B ) ) )`
16 15 simplbi
` |- ( ( C vH D ) C_ ( A vH B ) -> C C_ ( A vH B ) )`
17 1 2 3 3pm3.2i
` |- ( A e. CH /\ B e. CH /\ C e. CH )`
18 dmdsl3
` |- ( ( ( A e. CH /\ B e. CH /\ C e. CH ) /\ ( B MH* A /\ A C_ C /\ C C_ ( A vH B ) ) ) -> ( ( C i^i B ) vH A ) = C )`
19 17 18 mpan
` |- ( ( B MH* A /\ A C_ C /\ C C_ ( A vH B ) ) -> ( ( C i^i B ) vH A ) = C )`
20 9 12 16 19 syl3an
` |- ( ( B MH* A /\ A C_ ( C i^i D ) /\ ( C vH D ) C_ ( A vH B ) ) -> ( ( C i^i B ) vH A ) = C )`
21 11 simprbi
` |- ( A C_ ( C i^i D ) -> A C_ D )`
22 15 simprbi
` |- ( ( C vH D ) C_ ( A vH B ) -> D C_ ( A vH B ) )`
23 1 2 4 3pm3.2i
` |- ( A e. CH /\ B e. CH /\ D e. CH )`
24 dmdsl3
` |- ( ( ( A e. CH /\ B e. CH /\ D e. CH ) /\ ( B MH* A /\ A C_ D /\ D C_ ( A vH B ) ) ) -> ( ( D i^i B ) vH A ) = D )`
25 23 24 mpan
` |- ( ( B MH* A /\ A C_ D /\ D C_ ( A vH B ) ) -> ( ( D i^i B ) vH A ) = D )`
26 9 21 22 25 syl3an
` |- ( ( B MH* A /\ A C_ ( C i^i D ) /\ ( C vH D ) C_ ( A vH B ) ) -> ( ( D i^i B ) vH A ) = D )`
27 20 26 sseq12d
` |- ( ( B MH* A /\ A C_ ( C i^i D ) /\ ( C vH D ) C_ ( A vH B ) ) -> ( ( ( C i^i B ) vH A ) C_ ( ( D i^i B ) vH A ) <-> C C_ D ) )`
28 8 27 syl5ib
` |- ( ( B MH* A /\ A C_ ( C i^i D ) /\ ( C vH D ) C_ ( A vH B ) ) -> ( ( C i^i B ) C_ ( D i^i B ) -> C C_ D ) )`
29 5 28 impbid2
` |- ( ( B MH* A /\ A C_ ( C i^i D ) /\ ( C vH D ) C_ ( A vH B ) ) -> ( C C_ D <-> ( C i^i B ) C_ ( D i^i B ) ) )` | 1,406 | 2,713 | {"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} | 3.0625 | 3 | CC-MAIN-2024-30 | latest | en | 0.296564 |
https://primepuzzles.net/puzzles/puzz_676.htm | 1,718,529,263,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861657.69/warc/CC-MAIN-20240616074847-20240616104847-00226.warc.gz | 414,032,657 | 4,232 | Problems & Puzzles: Puzzles
Puzzle 676 primes in a row such that... Bernardo Recaman Santos and Carlos Rivera, as a result of a short & swift email interchange, composed the following puzzle: Q1. Find the smallest prime P (greater than 11) such that it is possible to split and arrange all the primes, 2, 3, 5, 7, 11, 13, ..., P, in two rows in each of which the difference between adjacent primes is always a perfect square. ( A single row is not possible because primes of the form 4n+1 cannot mix with primes of the form 4n+3.). Q2. Find the smallest prime P such that it is possible to split and arrange all the primes, 2, 3, 5, 7, ..., P, in two circles (or necklaces) in each of which the difference between adjacent primes is always a perfect square.
Contributions came from Giovanni Resta, Jean Brette, Hakan Summakoglu & Emmanuel Vantieghem.
***
All of them found the same solutions for both questions, except Jean Brette who only found solution for Q1 "working at hand".
The smallest partition in two rows is
(5, 41, 37, 53, 17, 13, 29), and
(31, 47, 11, 2, 3, 19, 23, 7, 43).
The smallest range of primes that can partitioned into two cycles
is 2,3,...,353 (71 primes):
Two examples of these two cycles are:
{5, 41, 37, 101, 137, 281, 181, 197, 97, 61, 317, 313, 349, 353, 29,
173, 157, 257, 241, 277, 293, 229, 193, 337, 13, 17, 53, 89, 233,
269, 73, 109, 113, 149, 5}
and
{2, 3, 103, 139, 283, 347, 331, 7, 11, 47, 31, 131, 167, 151,
251, 107, 71, 67, 211, 311, 307, 271, 127, 191, 227, 263, 199, 163,
179, 79, 223, 239, 43, 59, 23, 19, 83, 2}
or:
{5,41,37,101,137,281,181,197,97,241,277,293,229,193,337,353,349,
313,317,61,257,157,173,29,13,17,53,89,233,269,73,109,113,149,5}
and
{2,11,47,191,227,211,311,307,271,127,131,31,67,71,107,251,151,
7,331,347,283,139,103,167,23,59,43,239,223,79,179,163,263,199,
3,19,83,2}
***
Records | Conjectures | Problems | Puzzles
Home | Melancholia | Problems & Puzzles | References | News | Personal Page | Puzzlers | Search | Bulletin Board | Chat | Random Link Copyright © 1999-2012 primepuzzles.net. All rights reserved. | 791 | 2,105 | {"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} | 3.96875 | 4 | CC-MAIN-2024-26 | latest | en | 0.822978 |
http://www.transtutors.com/questions/finance-4250.htm | 1,511,348,325,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806569.66/warc/CC-MAIN-20171122103526-20171122123526-00672.warc.gz | 507,852,562 | 13,837 | # Finance
Nelson and cooper Inc. has two different plans for financing a $6,000,000 expansion project which are Currently under consideration: Plan 1 : is financing the entire project by the issuance of common stock,$10 par. Plan 2:call for issuance of 12% bond for $3,000,000 with the remaining capitalization for the common stock,$10 par. A) compute the earning per share on common stock.
Related Questions in Employee and Labor Relations
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Solution Preview :
The book value per common share denotes the dollar value available for common shareholders after all assets are liquidated and all debtors are paid
• accounting assignment February 11, 2011
,000 shares of $10 par value common stock and 20,000 shares of 6%,$100 par value preferred stock . (No dividends are in arrears.) The book value per share of common stock is: A. \$39. B.... | 511 | 2,198 | {"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} | 2.53125 | 3 | CC-MAIN-2017-47 | latest | en | 0.901598 |
https://github.com/erichson/DMDpack | 1,660,644,139,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572286.44/warc/CC-MAIN-20220816090541-20220816120541-00095.warc.gz | 268,581,304 | 38,193 | {{ message }}
/ DMDpack Public
Dynamic Mode Decomposition
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Dynamic Mode Decomposition
The dynamic mode decomposition (DMD) is an equation-free, data-driven matrix decomposition that is capable of providing accurate reconstructions of spatio-temporal coherent structures arising in nonlinear dynamical systems, or short-time future estimates of such systems. The DMD method provides a regression technique for least-square fitting of video snapshots to a linear dynamical system. The method integrates two of the leading data analysis methods in use today:
Fourier transforms and Principal Components. Originally introduced in the fluid mechanics community, DMD traces its origins to Bernard Koopman in 1931 and can be seen as a special case of Koopman theory. Meanwhile DMD has emerged as a powerful tool for analyzing dynamics of nonlinear systems and in the last few years alone, DMD has seen tremendous development in both theory and application. In theory, DMD has seen innovations around compressive architectures, multi-resolution analysis and de-noising algorithms. In addition to continued progress in fluid dynamics, DMD has been applied to new domains, including neuroscience, epidemiology, robotics, and the current application of video processing and computer vision.
The DMDpack includes the following implementations
• Exact DMD (dmd) facilitating truncated, partial or randomized SVD
• Compressed DMD (cdmd)
• Robust DMD (tdmd) using total least squares
Installation
To build and install DMDpack, run from within the main directory in the release: python setup.py install
After successfully installing DMDpack, the unit tests can be run by: python setup.py test
See the documentation for more details.
Example
Get started:
import numpy as np
from dmd import dmd, cdmd
from dmd import tools
First, lets create some (noise-free) toy data:
# Define time and space discretizations
x=np.linspace( -9, 9, 200)
t=np.linspace(0, 3*np.pi , 80)
dt=t[2]-t[1]
X, T = np.meshgrid(x,t)
# Create two patio-temporal patterns
S1 = 0.9* np.cos(X)*(1.+0.* T)
S2 = ( (1./np.cosh(X)) * np.tanh(X)) *(2.*np.exp(1j*2.8*T))
S= S1+S2
The high-dimensional signal S is superimposed as a time-independent signal S1 and time-dependent signal S2, shown in the following figure:
DMD is a data processing algorithm which decomposes a matrix S in space and time. The matrix S is decomposed as S = FBV, where the columns of F contain the dynamic modes. The modes are ordered according to the amplitudes stored in the diagonal matrix B. V is a Vandermonde matrix describing the temporal evolution. Hence, using the dynamic mode decomposition, we aim to separate S into its underlying components, as follows
Fmodes, b, V, omega = dmd(F.T, dt=dt, k=k, return_vandermonde=True, return_amplitudes=True)
Plotting the continuous-time eigenvalues: we see that S is indeed superimposed from two underlying signals, i.e., mode 1 and 2. All the other modes are bounded far away from the origin and can be considered as unstable, i.e, not relevant for recovering the underlying signals. Hence, we can approximate S, S1 and S2 as follows:
Sre = (Fmodes[:,0:2]*b[0:2]).dot(V[0:2,:]).T
S1re = (Fmodes[:,0:1]*b[0]).dot(V[0:1,:]).T
S2re = (Fmodes[:,1:2]*b[1]).dot(V[1:2,:]).T
Plotting the approximated signals, we indeed capture the underlying signals faithfully
The visual evidence is convincing, let's compute the approximation error in addition:
print(np.linalg.norm(S-Sre, 'fro')/np.linalg.norm(S, 'fro')*100)
print(np.linalg.norm(S1-S1re, 'fro')/np.linalg.norm(S1, 'fro')*100)
print(np.linalg.norm(S2-S2re, 'fro')/np.linalg.norm(S2, 'fro')*100)
which supports the visual evidence
2.82741507892e-12
1.60385599574e-12
4.72877417589e-12
Dynamic Mode Decomposition
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No packages published | 1,004 | 3,949 | {"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} | 2.671875 | 3 | CC-MAIN-2022-33 | longest | en | 0.857054 |
http://www.independent.co.uk/news/education/education-news/games-1152866.html | 1,498,585,561,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128321497.77/warc/CC-MAIN-20170627170831-20170627190831-00433.warc.gz | 558,179,425 | 17,859 | # games
Goats and cars revisited
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The Independent Online
Last week's piece about goats, cars and the laws of probability has attracted a considerable postbag. The paradox began by putting you in the position of a game show contestant who has to pick one of three doors. All you know is that one of the doors has this week's star prize of a car behind it; the other two offer only a mangy goat. When you have picked one door, the host (who knows which door conceals the star prize) opens one of the doors you did not choose and shows a goat behind it. He then offers to let you change your mind. Most people then stick to their original choice. In fact, as explained last week, you will double your winning chance if you change your mind.
Antonio Carvalho writes from Henley on Thames: "Let's freeze the game show at the point when the host asks the contestant if he would like to change his mind. To avoid clutter, let's remove the door already shown to have a goat behind it. As far as the contestant is concerned, the circumstances remain unchanged, ie the door initially chosen by him has only one-third of a chance of having the star prize behind it. Now let's bring in a passer- by from the street, show him the remaining two doors, and give him a go at the star prize. The odds for the passer-by cannot but be 50-50! But this is impossible, because the odds must be the same for the passer-by and the initial contestant. Unless probability is a subjective concept, in which case a whole chapter of maths goes down the plughole. Can you help? (I'd like to get some sleep!)"
That, I think, sums up the psychological aspect of the paradox rather better than I did last week. Ian Bellamy identifies the paradox as one cited by Frederick Mosteller of Harvard 30 years ago. Mosteller's explanation, he says, is to draw up a table of probabilities of the various possible outcomes when the contestant picks door A from three doors A, B and C. Four things can then happen:
1: A and B are goats; host opens B;
2: A and C are goats; host opens C;
3: B and C are goats; host opens B;
4: B and C are goats; host opens C.
Since the probability that the prize is behind any particular door is one in three, the probability of 1 and 2 are each one in three, while 3 and 4 each have only one in six chances of happening. If the contestant sticks to his original choice, he wins in cases 3 and 4, but loses in 1 and 2. The total probability of a win is therefore 1/6 + 1/6 = 1/3. If he changes his mind, he wins in cases 1 and 2, and loses in 3 and 4: overall chance of a win 1/3 + 1/3 = 2/3 . Mr Bellamy concludes: "The Paradox, Mosteller sternly insists, is apparent only to those who fail to employ the proper sample space, which has to include the actions of the game show host."
But I am sure that will not totally cure Mr Carvalho's sleeplessness. For the real paradox lies not in the calculation of the probabilities, but in the nagging feeling that the host's action in opening one door after you have made your choice is completely irrelevant.
Jim Bragg, however, has given us a very simple example that should cure any lingering example:
"I had difficulty intuitively (not logically) understanding why you should always opt to change your door, until someone explained it to me thus: The game show host offers a million doors to you. After you choose one, the host closes all doors except yours and one other. The odds on the other door having the star prize now seem a lot better, don't they?"
And that explains everything. A chap coming in from outside would still see only two apparently equal doors. You, however, would know that one door (your original choice) was picked at random from a million, the other was selected from 999,999 as the only one that might have a goat behind it. That's a very special door. Change your mind! | 898 | 3,860 | {"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} | 3.5625 | 4 | CC-MAIN-2017-26 | latest | en | 0.967071 |
https://synapses.polytechnique.fr/catalogue/2021-2022/ue/812/MAP535-regression?from=D4 | 1,718,323,488,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861517.98/warc/CC-MAIN-20240613221354-20240614011354-00674.warc.gz | 517,003,781 | 13,642 | v2.11.0 (5518)
# PA - C8 - MAP535 : Regression
## Descriptif
Objectives
The objective of this course is to introduce linear. Regression plays a key role in many problems and it is absolutely essential for a datascientist to understand the theory and the practice of regression analysis. It is also an important vehicle to address the statistical challenges in statistical learning : model selection, penalisation, resampling (bootstrap, cross-validation) robustness, detection of outliers, and also methods to detect deviations from an assumed model. The course will also serve as a motivation to sharpen the understanding of statistical techniques, covering both estimation and tests.
Syllabus
1. Introduction to statistical learning
Regression: Learning objectives and applications
Linear models : interpretation, examples
Least-Square estimators properties (bias, variance)
Case study: univariate and multivariate regression
Multivariate Linear Regression: Parametric casee
Construction of least-square estimators
2. Parametric true model
Distribution of least-squares estimatorsAsymptotic properties
Gaussian case (distribution of the parameters, confidence regions)
Confidence intervals and tests
Classical regression diagnostic (leverage points)
Case studyAlgo: understanding multiple linear regression with R (lm summary, detecting outliers, understanding classical regression diagnosis)
3. Residual analysis (homoscedasticity, non-linear dependence)
Outlier detection (leverage effects, influence, introduction to robsut statistics)
Functional modelintroduction to non-parametric regression : from parameters to functions
Multiple models for a single problemFunction classes, model selection
Variable choice / Basis / Spline
Bias / Variance (Approximation error / Estimation Error)
Case study : Spline regression
4. Model Selection and Resampling
Approximation Error / Estimation Error
Learning Error / Generalization Error
Resampling based method: jacknife, bootstrap, and Cross Validation
Case study: model selection with CV
5. Model Selection and Unbiased Risk Estimation
Unbiased Risk Estimation
AIC/BIC Penalization and Exhaustive Exploration
Forward / Backward and Stochastic Exploration
Multiple tests
6. Model Selection and Penalization
Restricted Model and Penalization
Ridge and Lasso
Numerical algorithm: Gradient Descent and Coordinate Descent
Case study: Coordinate Descent and Lasso
Langue du cours : Anglais
Numérique sur 20
## Pour les étudiants du diplôme Data Science for Business
Le rattrapage est autorisé (Note de rattrapage conservée)
L'UE est acquise si note finale transposée >= C
• Crédits ECTS acquis : 4 ECTS
La note obtenue rentre dans le calcul de votre GPA.
## Pour les étudiants du diplôme Echanges PEI
Le rattrapage est autorisé (Note de rattrapage conservée)
## Pour les étudiants du diplôme Titre d’Ingénieur diplômé de l’École polytechnique
Le rattrapage est autorisé (Note de rattrapage conservée)
L'UE est acquise si note finale transposée >= C
• Crédits ECTS acquis : 5 ECTS | 694 | 3,058 | {"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} | 2.625 | 3 | CC-MAIN-2024-26 | latest | en | 0.732175 |
https://privatewriting.us/blog/2020/11/16/ | 1,611,150,123,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703520883.15/warc/CC-MAIN-20210120120242-20210120150242-00397.warc.gz | 508,054,982 | 10,536 | # Daily Archives: November 16, 2020
## Create a plan for interpreting and judging the data, and making recommendations.
1. How and where will data be recorded?
2. Create a plan for interpreting and judging the data, and making recommendations.
3. How will a record of the evaluation efforts and subsequent curriculum alterations be maintained?
4. When, how, and to whom will data, evaluation results, and recommendations be reported?
5. Develop a plan for review of the evaluation process
6. Describe any faculty development needed to support curriculum evaluation. What is the basis of this assessment?
## Calculate the price of each bond in the following scenarios
Price sensitivity and convexity. It is recommended to use a spreadsheet to solve this problem.
Suppose the US zero-coupon rate curve is given as:
Consider the following three bonds with face value \$100:
(a) Calculate the arbitrage price of each bond.
(b) The price sensitivity of a bond (also known as the ‘dollar value of one basis point’ or DV01) is defined
as the change in price when all rates go up 1 basis point (i.e. a +0.01%
parallel shift of the entire zero-coupon rate curve). Compute the price sensitivity of each bond.
(c) Calculate the price of each bond in the following scenarios:
(i) 10 basis point rate increase (+0.10%);
(ii) 1 point rate increase (+1%).
….
Zero-coupon rate curve and expectations. The short-term zero-coupon rate curve of the euro zone is given as:
The current refinancing rate of the European Central Bank (ECB) is at 2.75%. This is the rate at which banks can borrow from the ECB for 2 weeks. The ECB Board of Governors will meet in 2 weeks and potentially decide on a new refinancing rate R.
(a) Without making any calculation can you guess if the market expects the ECB to:
(i) Lower its rate by 25 bps (i.e. R = 2.5%)?
(ii) Leave its rate unchanged (i.e. R = 2.75%)?
(iii) Raise its rate by 25 bps (i.e. R = 3%)?
(iv) Raise its rate by 50 bps (i.e. R = 3.25%)?
(v) Any other scenario?
(b) The….
## True or False? The three questions are independent.
True or False? The three questions are independent.
(a) “The average monthly return of Kroger Co. in 2009-10 was 0.28% (including dividends). Therefore, its
annual return was (1 + 0.28%)12 − 1 ≈ 3.4%.”
(b) “To calculate the annual volatility of a series of monthly returns I may either compute their standard
deviation and multiply it by or equivalently I may annualize each monthly return and then compute the corresponding standard deviation.”
(c) “The return of my portfolio is 15% per year and its risk is 25% per year. The stock of MeToo.Com has a 15% return and 30% risk. Hence, adding MeToo.Com to my portfolio would increase its risk but not its return.”
## What is the realized risk-return profile of Richky Corp.?
a.Risk-free rate and Sharpe ratio Using the data for the Treasury bond in determine the theoretical risk-free rate rf so that the Sharpe ratio of the T-Bond be equal to 1.
b: Risk and return of Richky Corp. The table below gives the stock price of Richky Corp. at the end of
each month over the past year. The risk-free rate was constant at 5%.
(a) Given a \$134 initial stock price at the end of the previous year and a \$13 dividend per share distributed on 30 June, calculate the monthly returns of Richky Corp. Assume that the dividend is reinvested in the stock.
(b) What is the realized risk-return profile of Richky Corp.?
(c) You are the Chief Financial Officer of Richky Corp. At a….
## Which portfolio would you choose to obtain an expected return around 5.25%?
Currency portfolio. It is recommended that you solve this problem using a spreadsheet. You are a euro-zone investor with 1 billion euros to be invested in dollars (USD), yen (JPY), or pounds sterling (GBP). You are given the following market data and forecasts:
(a) Plot the three currencies on a risk-return chart, taking the interest produced by each currency into account.
(b) Draw the risk-return evolution of a portfolio which gradually switches from dollars to yen (i.e. 100% in dollars initially, then 90% in dollars and 10% in yen, etc.). Repeat this question for a portfolio which gradually switches from yen to pounds, and then from pounds to dollars.
(c) Plot the risk-return profiles of all possible portfolios made of the three currencies, considering only long investment….
## What is the payoff at maturity of a portfolio long a zero-coupon bond with face value \$90 and an in-the-money European call option struck at \$90 on an underlying stock S currently trading at \$100?
Option payoffs. The three questions are independent.
(a) In each of the three examples in identify the underlying assets, the maturity date, and the payoff formula.
(b) Find a portfolio of European options on an underlying asset S with maturity T whose payoff matches the figure below:
(c) What is the payoff at maturity of a portfolio long a zero-coupon bond with face value \$90 and an in-the-money European call option struck at \$90 on an underlying stock S currently trading at \$100? Assuming no dividends, no arbitrage, infinite liquidity, and the ability to short-sell, show that this portfolio must be worth more than \$100.
## Do you think barrier options should be more expensive than plain vanilla options of same characteristics?
Barrier option A ‘knock-out barrier option’ is a call or put option which may only be exercised at maturity if the price of the underlying never hits a pre-agreed barrier price H throughout the life of the option. Symmetrically, a ‘knock-in barrier option’ may only be exercised at maturity if the price of the underlying hits the barrier price H.
(a) Do you think barrier options should be:
(i) more expensive than plain vanilla options of same characteristics?
(ii) less expensive than plain vanilla options of same characteristics?
(iii) more expensive in some cases and less expensive in other cases? (Please specify.)
(b) In this question we consider calls with strike 100 and 1-month maturity. The underlying spot price is
## What is the 1-year forward exchange rate?
Forward exchange rate. This problem is about forward contracts in foreign exchange and goes beyond the scope of equity derivatives. The spot exchange rate of the euro is S dollars, i.e. to buy one euro one must pay S dollars. The euro zone yield curve is flat at rEU while the American yield curve is flat at rUS. A forward contract on the euro-dollar is an agreement to receive euros and pay dollars at a pre-agreed date T and exchange rate F.
(a) Starting with €1, find two ways to have dollars in a year’s time by investing or borrowing in either currency and exchanging between currencies through the spot and forward markets. Using an arbitrage argument, establish that the 1-year forward exchange rate of the euro must….
## Find an upper bound for z(T) as a function of T, τ and z(T + τ ).
Forward interest rate. This problem is about forward interest rates and goes beyond the scope of equity derivatives. Consider a market with no arbitrage and infinite liquidity where investors can lend and borrow money for any maturity T at the zero-coupon rate z(T). Let z(T, τ ) denote the forward rate agreed today on a loan beginning at time t = T and ending at time t = T + τ . For instance, the ‘6-month × 1-year’ forward rate z(0.5, 1) is the rate agreed today on a loan starting in six months and ending in eighteen months.
(a) In this question the zero-coupon rate curve is given as: z(T) = 5% + T × 0.5%.
(i) Draw z(T) as a function of maturity T.
(ii) Suppose…. | 1,742 | 7,635 | {"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} | 3.484375 | 3 | CC-MAIN-2021-04 | latest | en | 0.892658 |
https://www.mathworks.com/matlabcentral/cody/problems/106-weighted-average/solutions/260514 | 1,508,735,204,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825575.93/warc/CC-MAIN-20171023035656-20171023055656-00574.warc.gz | 940,548,234 | 11,479 | Cody
# Problem 106. Weighted average
Solution 260514
Submitted on 13 Jun 2013 by Mandeep Singh
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% x = [1 2 3]; w = [10 15 20]; y_correct = 100/3; assert(isequal(weighted_average(x,w),y_correct)) x = [0 -2 3]; w = [10 0 10]; y_correct = 10; assert(isequal(weighted_average(x,w),y_correct))
ans = 33.3333 ans = 10 | 158 | 483 | {"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} | 2.59375 | 3 | CC-MAIN-2017-43 | latest | en | 0.675027 |
http://www.rajibroy.com/?p=20534 | 1,585,936,457,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370515113.54/warc/CC-MAIN-20200403154746-20200403184746-00280.warc.gz | 286,080,537 | 9,001 | 23 February 2020
Sunday morning logic puzzle (a trifle easier than before)
Three smart men were trying to find out who is the smartest of them all. Or was at least most logical and quickest thinker. To resolve that, they went to a woman who had the reputation of being the smartest of them all. She devised a quick problem for them. She blindfolded them and painted a dot on each of their foreheads.
She then told them “I have painted a dot on each of your foreheads. The dots are either white or black in color. I am going to soon open your blindfolds. You will be able to see other’s dots but not your own. I am then going to ask you to raise your hand if you see a white dot (on either of the two friend’s foreheads). Then I am going to ask you to guess the color of your own dot”.
Saying so, she opened their blindfolds.
“You may raise your hand now if you see a white dot”
All three men raised their hands.
The three men started pondering and soon one of them – who was indeed a little smarter than the others – said “I have a white dot on my forehead”.
Question: How did the man figure out what color he had?
Posted February 23, 2020 by rajibroy in category "Puzzles | 283 | 1,181 | {"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} | 3.40625 | 3 | CC-MAIN-2020-16 | latest | en | 0.983552 |
https://buy-essay.com/please-help-explain-why-its-saying-its-wrong-where-it-shows-the-red-xs-also-please-help-me-answer-the-rest-of-the/ | 1,670,199,203,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711001.28/warc/CC-MAIN-20221205000525-20221205030525-00333.warc.gz | 182,656,741 | 14,197 | On February 1, 2018, Arrow Construction Company entered into a three-year construction contract to build a bridge for a price of \$8,570,000. During 2018, costs of \$2,190,000 were incurred with estimated costs of \$4,190,000 yet to be incurred. Billings of \$2,690,000 were sent, and cash collected was \$2,440,000. In 2019, costs incurred were \$2,690,000 with remaining costs estimated to be \$3,885,000. 2019 billings were \$2,940,000 and \$2,665,000 cash was collected. The project was completed in 2020 after additional costs of \$3,990,000 were incurred. The company’s fiscal year-end is December 31. Arrow recognizes revenue over time according to percentage of completion. Requlrec: 1. Compute the amount of revenue and gross profit or loss to be recognized in 2018, 2019, and 2020 using the percentage of completion method? 2a. Prepare journal entries for 2018 to record the transactions described (credit “various accounts” for construction costs incurred) 2b. Prepare journal entries for 2019 to record the transactions described (credit “various accounts” for construction costs incurred). 3a. Prepare a partial balance sheet to show the presentation of the project as of December 31, 2018. 3b. Prepare a partial balance sheet to show the presentation of the project as of December 31, 2019. Answer is not complete. Complete this question by entering your answers in the tabs below Req 1 Req 2A Req 2B Req 3A 2q Compute the amount of revenue and gross profit or loss to be recognized in 2018, 2019, and 2020 using the percentage of completion method? (Do not round intermediate calculations. Loss amounts should be indicated with a minus sign. Round your final answers to the nearest whole dollar.) Choose denominator % complete to dste Choose numerator Actusl costs to date | ÷ Estimated total costs。 2018 2019 2020 8,380,000 8,785,000 34.3260% 55.8760% 100.0000% 2.190.000 ÷ 4.880.000 2018 Recognized in prior To date Recognized in 2018 2,941,224 2.190,000 751,224 Construction 2941,224 2.190,000 751,224 Gross profit (loss) 2019 Recognized in prior To date Recognized in 2019 Construction 2.941,224 2.190,000 4.770.919 1.829,895 2.190,000 Gross profit (loss) 2020 Recognized in prior To date Recognized in 2020 Construction
1.
2018
To Date Recognised in prior years Recognised in 2018 Construction Revenue \$2,941,740* \$0 \$2,941,740 Construction Expense \$2,190,000 \$0 \$2,190,000 Gross Profit (loss) \$751,740** \$0 \$751,740
*\$8,570,000 x 34.3260% = \$2,941,740
**\$2,941,740 – \$2,190,000 = \$751,740
2019
To date Recognised in prior years Recognised in 2019 Construction Revenue \$4,771,433 \$2,941,740 \$1,829,693 Construction Expense \$4,880,000 \$2,190,000 \$2,690,000 Gross Profit (Loss) (\$108,567) \$751,740 (\$860,307)
2020
To date Recognised in prior years Recognised in 2020 Construction Revenue \$8,570,000 \$4,771,433 \$3,798,567 Construction Expense \$8,870,000 \$4,880,000 \$3,990,000 Gross Profit (Loss) (\$300,000) (\$108,567) (\$191,433)
2a Year General Journal Debit Credit 2018 Construction in Progress \$2,190,000 Various Accounts \$2,190,000 2018 Accounts Receivable \$2,690,000 Billings on Construction Contract \$2,690,000 2018 Cash \$2,440,000 Accounts Receivable \$2,440,000 2018 Construction in Progress \$751,740 Construction Expense \$2,190,000 Construction Revenue \$2,941,740
2b Year General Journal Debit Credit 2019 Construction in Progress \$2,690,000 Various Accounts \$2,690,000 2019 Accounts Receivable \$2,940,000 Billings on Construction Contract \$2,940,000 2019 Cash \$2,665,000 Accounts Receivable \$2,665,000 2019 Construction Expense \$2,690,000 Construction in Progress \$860,307 Construction Revenue \$1,829,693
3a Balance sheet 2018 Current Assets: Account Receivable \$250,000 Costs and profit in excess of billings \$251,740
3b Balance sheet 2019 Current Assets: Account Receivable \$525,000 Current Liabilities: Billings in Excess of costs \$858,567
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#### Resources tagged with Divisibility similar to Have You Got It?:
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### There are 40 results
Broad Topics > Numbers and the Number System > Divisibility
### Peaches Today, Peaches Tomorrow....
##### Stage: 3 Challenge Level:
Whenever a monkey has peaches, he always keeps a fraction of them each day, gives the rest away, and then eats one. How long could he make his peaches last for?
### Ben's Game
##### Stage: 3 Challenge Level:
Ben passed a third of his counters to Jack, Jack passed a quarter of his counters to Emma and Emma passed a fifth of her counters to Ben. After this they all had the same number of counters.
### Three Times Seven
##### Stage: 3 Challenge Level:
A three digit number abc is always divisible by 7 when 2a+3b+c is divisible by 7. Why?
### Repeaters
##### Stage: 3 Challenge Level:
Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (e.g. 594594). Explain why whatever digits you choose the number will always be divisible by 7, 11 and 13.
##### Stage: 3 Challenge Level:
List any 3 numbers. It is always possible to find a subset of adjacent numbers that add up to a multiple of 3. Can you explain why and prove it?
### American Billions
##### Stage: 3 Challenge Level:
Play the divisibility game to create numbers in which the first two digits make a number divisible by 2, the first three digits make a number divisible by 3...
### What Numbers Can We Make Now?
##### Stage: 3 Challenge Level:
Imagine we have four bags containing numbers from a sequence. What numbers can we make now?
### Neighbours
##### Stage: 2 Challenge Level:
In a square in which the houses are evenly spaced, numbers 3 and 10 are opposite each other. What is the smallest and what is the largest possible number of houses in the square?
### Division Rules
##### Stage: 2 Challenge Level:
This challenge encourages you to explore dividing a three-digit number by a single-digit number.
### Dozens
##### Stage: 3 Challenge Level:
Do you know a quick way to check if a number is a multiple of two? How about three, four or six?
### Counting Factors
##### Stage: 3 Challenge Level:
Is there an efficient way to work out how many factors a large number has?
### What Numbers Can We Make?
##### Stage: 3 Challenge Level:
Imagine we have four bags containing a large number of 1s, 4s, 7s and 10s. What numbers can we make?
### Curious Number
##### Stage: 2 Challenge Level:
Can you order the digits from 1-3 to make a number which is divisible by 3 so when the last digit is removed it becomes a 2-figure number divisible by 2, and so on?
### The Remainders Game
##### Stage: 2 and 3 Challenge Level:
A game that tests your understanding of remainders.
### What an Odd Fact(or)
##### Stage: 3 Challenge Level:
Can you show that 1^99 + 2^99 + 3^99 + 4^99 + 5^99 is divisible by 5?
### Eminit
##### Stage: 3 Challenge Level:
The number 8888...88M9999...99 is divisible by 7 and it starts with the digit 8 repeated 50 times and ends with the digit 9 repeated 50 times. What is the value of the digit M?
### Skeleton
##### Stage: 3 Challenge Level:
Amazing as it may seem the three fives remaining in the following `skeleton' are sufficient to reconstruct the entire long division sum.
### Oh! Hidden Inside?
##### Stage: 3 Challenge Level:
Find the number which has 8 divisors, such that the product of the divisors is 331776.
### Going Round in Circles
##### Stage: 3 Challenge Level:
Mathematicians are always looking for efficient methods for solving problems. How efficient can you be?
### Remainders
##### Stage: 3 Challenge Level:
I'm thinking of a number. When my number is divided by 5 the remainder is 4. When my number is divided by 3 the remainder is 2. Can you find my number?
### Remainder
##### Stage: 3 Challenge Level:
What is the remainder when 2^2002 is divided by 7? What happens with different powers of 2?
### Book Codes
##### Stage: 2 Challenge Level:
Look on the back of any modern book and you will find an ISBN code. Take this code and calculate this sum in the way shown. Can you see what the answers always have in common?
### Legs Eleven
##### Stage: 3 Challenge Level:
Take any four digit number. Move the first digit to the 'back of the queue' and move the rest along. Now add your two numbers. What properties do your answers always have?
### Differences
##### Stage: 3 Challenge Level:
Can you guarantee that, for any three numbers you choose, the product of their differences will always be an even number?
### Just Repeat
##### Stage: 3 Challenge Level:
Think of any three-digit number. Repeat the digits. The 6-digit number that you end up with is divisible by 91. Is this a coincidence?
##### Stage: 3 Challenge Level:
Powers of numbers behave in surprising ways. Take a look at some of these and try to explain why they are true.
### Elevenses
##### Stage: 3 Challenge Level:
How many pairs of numbers can you find that add up to a multiple of 11? Do you notice anything interesting about your results?
### AB Search
##### Stage: 3 Challenge Level:
The five digit number A679B, in base ten, is divisible by 72. What are the values of A and B?
### Digat
##### Stage: 3 Challenge Level:
What is the value of the digit A in the sum below: [3(230 + A)]^2 = 49280A
### Gran, How Old Are You?
##### Stage: 2 Challenge Level:
When Charlie asked his grandmother how old she is, he didn't get a straightforward reply! Can you work out how old she is?
### Ewa's Eggs
##### Stage: 3 Challenge Level:
I put eggs into a basket in groups of 7 and noticed that I could easily have divided them into piles of 2, 3, 4, 5 or 6 and always have one left over. How many eggs were in the basket?
### Divisively So
##### Stage: 3 Challenge Level:
How many numbers less than 1000 are NOT divisible by either: a) 2 or 5; or b) 2, 5 or 7?
### Factoring Factorials
##### Stage: 3 Challenge Level:
Find the highest power of 11 that will divide into 1000! exactly.
### Powerful Factorial
##### Stage: 3 Challenge Level:
6! = 6 x 5 x 4 x 3 x 2 x 1. The highest power of 2 that divides exactly into 6! is 4 since (6!) / (2^4 ) = 45. What is the highest power of two that divides exactly into 100!?
### Gaxinta
##### Stage: 3 Challenge Level:
A number N is divisible by 10, 90, 98 and 882 but it is NOT divisible by 50 or 270 or 686 or 1764. It is also known that N is a factor of 9261000. What is N?
### Square Routes
##### Stage: 3 Challenge Level:
How many four digit square numbers are composed of even numerals? What four digit square numbers can be reversed and become the square of another number?
### Digital Roots
##### Stage: 2 and 3
In this article for teachers, Bernard Bagnall describes how to find digital roots and suggests that they can be worth exploring when confronted by a sequence of numbers.
### Flow Chart
##### Stage: 3 Challenge Level:
The flow chart requires two numbers, M and N. Select several values for M and try to establish what the flow chart does. | 1,765 | 7,106 | {"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} | 4.15625 | 4 | CC-MAIN-2017-34 | latest | en | 0.889163 |
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# Phys documents - Chapter 3 Relativity 17 3.2.4 The trajectory of a photon For the trajectory of a photon(and for each particle with zero restmass
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Chapter 3: Relativity 17 3.2.4 The trajectory of a photon For the trajectory of a photon (and for each particle with zero restmass) holds ds 2 =0 . Substituting the external Schwarzschild metric results in the following orbital equation: du d ϕ ± d 2 u d ϕ 2 + u - 3 mu ² =0 3.2.5 Gravitational waves Starting with the approximation g μ ν = η μ ν + h μ ν for weak gravitational ±elds and the de±nition h ± μ ν = h μ ν - 1 2 η μ ν h α α it follows that ± h ± μ ν =0 if the gauge condition h ± μ ν / x ν =0 is satis±ed. From this, it follows that the loss of energy of a mechanical system, if the occurring velocities are ± c and for wavelengths ² the size of the system, is given by: dE dt = - G 5 c 5 ³ i,j ± d 3 Q ij dt 3 ² 2 with Q ij = ´ ± ( x i x j - 1 3 δ ij r 2 ) d 3 x the mass quadrupole moment. 3.2.6 Cosmology If for the universe as a whole is assumed: 1. There exists a global time coordinate which acts as x 0 of a Gaussian coordinate system, 2. The 3-dimensional spaces are isotrope for a certain value of
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## This note was uploaded on 11/16/2010 for the course ENGR 201 taught by Professor Elder during the Spring '10 term at Blinn College.
Ask a homework question - tutors are online | 445 | 1,515 | {"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} | 2.84375 | 3 | CC-MAIN-2017-13 | latest | en | 0.862703 |
http://math.stackexchange.com/questions/279111/do-these-axioms-fully-describe-the-integers?answertab=oldest | 1,441,233,531,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440645293619.80/warc/CC-MAIN-20150827031453-00184-ip-10-171-96-226.ec2.internal.warc.gz | 154,768,159 | 22,259 | # Do these axioms fully describe the integers?
Here, I use Peano-like axioms to describe the set of integers $Z$. They are based on two successor functions, each starting with a common point of $0$, and a principle of induction for the integers.
Let $Z$, $Pos$, $Neg$, $s$, $s'$ and $0$ be such that:
$Pos\subset Z$
$Neg\subset Z$
$Z=Pos\cup Neg$ (edit)
$\forall x (x\in Pos \wedge x\in Neg \leftrightarrow x=0)$
$s:Pos\rightarrow Pos$
$s$ is injective
$s':Neg\rightarrow Neg$
$s'$ is injective
$\forall x\in Pos (s(x)\neq 0)$
$\forall x\in Neg (s'(x)\neq 0)$
$\forall m ((0\in m\wedge \forall x\in Pos (x\in m\rightarrow s(x)\in m) \wedge \forall x\in Neg (x\in m\rightarrow s'(x)\in m))\rightarrow \forall x\in Z (x\in m))$
Note that, contrary to the usual convention, I have had to include $0$ in both sets $Pos$ and $Neg$.
Lemma: $0\in Z, Pos, Neg$
See my follow-up below
-
The usual construction seems much simpler to me. It defines an integer as a pair $(m, n)$ of natural numbers, with $(a,b) = (c,d)$ whenever $a+d = c+b$. This is simpler because it handles 0 uniformly, instead of making it a strange special case, and because it makes positive and negative the same kind of thing. One can define $(a,b) + (c,d)$ easily as $(a+c, b+d)$, without having to do separate definitions for positive and negative integers. In your system, you'll have to handle several separate cases depending on whether you are adding positive to positive, positive to negative, etc. – MJD Jan 15 '13 at 18:00
@MJD Your construction is simpler only when you describe informally like this. Note that, for the equality of order pairs, we usually have $(a,b)=(c,d)\leftrightarrow a=c \wedge b=d$. – Dan Christensen Jan 15 '13 at 18:17
Have fun defining addition. – MJD Jan 15 '13 at 18:38
@MJD I have been able to construct an add function for the natural numbers starting with a version of Peano's Axioms and a single successor function s. It should be only a bit more complicated with two successor functions s and s' in this case. – Dan Christensen Jan 15 '13 at 20:31
Yes, this characterizes the integers as long as the quantifiers on subsets range over all subsets. If you just take the axioms you have for Pos, these give Peano's axioms, which uniquely capture the natural numbers up to isomorphism in full second-order semantics. The same it true for Neg. Thus the overall structure for these axioms will be the integers, up to isomorphism.
Contrary to some claims, it is not very hard to define addition. First, there is a canonical semigroup isomorphism between (Pos, $s$) and (Neg, $s'$) preserving $0$. So this gives a notion of $-x$ for each $x$. Now we only have to define addition for positive numbers, which is described on the Wikipedia article, and then we use the negation operation to define addition for arbitrary integers.
-
Could you please elaborate a little on addition? I agree that defining addition of negative integers will look very similar to defining it for positive integers. But with the structure that OP has, I think mixing the two will yield a confusing mess. How does one calculate ${}^-3 + 5$ in this system? – MJD Jan 15 '13 at 19:32
@MJD: we can emulate the ordered pair construction to define addition without mentioning pairs. To add $x$ and $y$, imagine $x$ is the form $(a,0)$ or $(0,a)$, and $y$ is in the form $(b,0)$ or $(0,b)$. Then the sum will be of the form $(c,0)$, $(0,d)$, or $(c,d)$. In the first case the answer is $c$, in second case it is $-d$, and in the third case the answer is $c - d$, which can be defined in two cases. If $c > d$ then $c - d$ is the unique positive $x$ with $x + d = c$, otherwise it is negative of the unique $y$ with $c + y = d$, and these only refer to addition of positive numbers. – Carl Mummert Jan 15 '13 at 20:38
That is significantly more complicated than the definition with pairs, and it depends on defining $>$ first, which the pairs definition doesn't. I only claimed that a definition in this system would require multiple cases to treat addition of positive and negative number as separate cases, and that is exactly what you did. – MJD Jan 15 '13 at 23:50
@MJD: the same problem will occur with pairs as soon as we eliminate the equivalence relation by putting each pair in standard form $(c,0)$ or $(0,d)$. In other words the pairs version avoids cases only by avoiding the actual equality relation. – Carl Mummert Jan 16 '13 at 3:01
It seems that you are attempting to describe a second order theory such that, up to isomorphism, all models are the integers.
While your formulation is not entirely clear I think it can, with a bit more work, be turned into a finite list of second order axioms that will do the job.
-
There are only 11 axioms here. (I added one. See edit.) – Dan Christensen Jan 15 '13 at 16:24
Now 12. (See edit) – Dan Christensen Jan 15 '13 at 18:19
Back to 11! $0\in Z$ is derivable from the other axioms. – Dan Christensen Jan 15 '13 at 18:50
This has a model in the naturals. Take $Z$ to be the naturals, $Pos$ to be the evens, $Neg$ to be the odds plus 0. Define $s$ to be $+2$ and $s'$ to be $+2$ except $s'0=1$.
-
Having a model in the naturals is not a problem. – André Nicolas Jan 15 '13 at 7:21
@AndréNicolas.Oops, that last comment of mine, deleted, shows it is before first-espresso-of-the-day. Yes it is second order, which was indeed what I was originally assuming. Anyway, what's missing then is something that distinguishes the order structure of the integers. – Peter Smith Jan 15 '13 at 8:02
@PeterSmith I was trying to avoid an order structure at this level. Of course, you should be able construct an add function and define $$a\leq b\leftrightarrow \exists c\in Pos (a+c=b)$$ – Dan Christensen Jan 15 '13 at 16:20
@DanChristensen But you haven't (yet) defined addition ... – Peter Smith Jan 15 '13 at 16:36
I am working on it. Again, I think you should be able to construct the add function from these axioms (using Cartesian product, power set and subset rules). Just added another axiom (see edit). 12 now. – Dan Christensen Jan 15 '13 at 17:53
Follow-up
After much tinkering, I have settled on the following Peano-like axioms for the integers:
Let $Z, L, R, 0, s$ be such that:
$R\subset Z$, the non-negative integers (right)
$L\subset Z$, the non-positive integers (left)
$Z=R\cup L$
$\forall x (x\in R \wedge x\in L \leftrightarrow x=0)$
$s: Z\rightarrow Z$, a bijection
$\forall x (x\in R \rightarrow s(x)\in R)$
$\forall x (x\in L \rightarrow s^{-1}(x)\in L)$
$\forall x (x\in R \rightarrow s(x)\neq 0)$
$\forall x (x\in L \rightarrow s^{-1}(x)\neq 0)$
$\forall P ((P\subset Z \wedge 0\in P\wedge \forall x (x\in P\rightarrow s(x)\in P) \wedge \forall x (x\in P\rightarrow s^{-1}(x)\in P) ) \rightarrow Z\subset P)$
- | 1,937 | 6,800 | {"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} | 3.890625 | 4 | CC-MAIN-2015-35 | longest | en | 0.797652 |
https://www.enotes.com/topics/science/questions/solution-0-365g-hcl-per-l-water-prepared-10-0ml-336288 | 1,718,332,648,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861520.44/warc/CC-MAIN-20240614012527-20240614042527-00737.warc.gz | 698,983,650 | 17,225 | # A solution of 0.365g of HCl per L of water is prepared and 10.0ml of this is the diluted to 100ml. what is the pH of the solution? (HCl=36.5g/mole)
The mols of HCl present in 0.365g is,
n = 0.365 g / 36.5 g/mol
n = 0.01 mol.
Now this is in 1 L. If you take 10ml out of this and dilute it upto 100 ml, the concentration has been diluted 10 times.
Therefore final concentration is, C,
C = (0.01 mol/l)/10
C = 0.001 mol per liter.
HCl disassociates according to following reaction,
HCl ------> H+ + Cl-
Therefore the [H+] concentrtion is 0.001 mol per litre.
The definition of pH is,
pH = -log[H+]
pH = -log[0.001]
pH = -(-3)
pH = 3
The pH of the solution is 3. | 237 | 678 | {"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} | 3.640625 | 4 | CC-MAIN-2024-26 | latest | en | 0.901875 |
https://disap.io/dynamics | 1,590,516,610,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347391277.13/warc/CC-MAIN-20200526160400-20200526190400-00132.warc.gz | 317,037,803 | 14,071 | Dynamics is the branch of physics that addresses the motion of material objects with regards to the factors that affect them (momentum, force, mass, energy).
Dynamics is a very essential unit to the study of physics, as it makes up the foundation of Newtonian physics. Newton’s three laws are tendencies of nature that Sir Isaac Newton discovered that begin to describe and explain basic physical principles.
Law #1: An object at rest or in motion will remain at rest or in motion unless acted upon by a net external force. This may seem a little confusing at first because if you (for example) roll a tennis ball on a floor, it will eventually stop. While it may seem like no net external force is acting on it, there is actually friction, air resistance, etc. acting on the ball, making it come to a stop despite the lack of any obvious “net external forces”.
Law #2: Force is equal to the mass of an object times the object’s acceleration (Fnet = ma). This formula is extremely essential to the study of physics, as it is used in many different instances to solve a large variety of problems.
Law #3: Every reaction has an equal and opposite reaction. In order for two forces to be an action-reaction pair, they must exert the forces on each other; for example, the force of you on earth (Fye) = the force of earth on you (Fey). Two forces that would NOT be an action-reaction pair would not act on each other.
#### Example Problem #1
Which of Newton's three laws explains why, when you're cruising in an airplane, you don't feel any forces acting on you?
Answer: 2nd law
Explanation: Because you are travelling at a constant velocity in the airplane, a = 0m/s^2. If a = 0, then, according to Fnet = ma, Fnet must equal zero. This is why you don't feel like you're moving when you really are (except for instances of acceleration, where a does have a value and Fnet does not equal zero).
#### Solving More Complex Dynamics Problems
Free body diagrams are a very helpful tool in visualizing complex problems. A free body diagram features a dot (the object being analyzed) and arrows either pointing to or from the object that represent the forces acting on the objects.
Friction will often find its way into tricky physics problems. The two types of friction essential to a basic understanding of physics are static friction (FS) and kinetic friction (FK). Static friction is the friction an object experiences while it is stationary, whereas kinetic friction is the friction that an object experiences while it is moving. The formula for calculating the force of friction (FF) is that Ff = μ(FN). The letter mu (μ) is a ratio that we use to help us solve for the force of friction or the normal force.
Inclined planes are another way that dynamics problems can get more complicated. In order to work with inclined planes, you must find the components of forces to help you solve the problem. For example, with a force like gravity, you’d have to break the force into its x-component (working parallel to the ramp) and its y-component (working normal (perpendicular) to the ramp). So, Fgy = Fgcosθ = mgcosθ and Fgx = Fgsinθ = mgsinθ. Remember, on an inclined plane, the normal (perpendicular) force is going to be the force perpendicular to the plane the object is on, NOT necessarily vertical. Similarly, for forces that are being applied at an angle, you would use trigonometry to find the components of the force, which you can then use to solve the problem.
#### Example Problem #2
If you have a block (5kg) on a plane inclined at an angle of 45 degrees with a coefficient of kinetic friction of .35 and you release it, what will the block’s acceleration be?
Answer: 4.60m/s^2
Explanation: First, solve for a using Newton’s 2nd Law: Fnet = ma. Once you have done this, you should have a = Fnet/m. Since we know m is 5kg, we just need to solve for Fnet. The net force will be Fgx (mgsinθ or 5 x 10 x sin45 = 35.36N) - Ff (μk(FN) = .35(35.36) = 12.37N) = 22.99. 22.99/5 = 4.60 m/s^2. | 996 | 4,026 | {"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} | 4.0625 | 4 | CC-MAIN-2020-24 | latest | en | 0.940395 |
https://infinitylearn.com/surge/ncert-solutions/class-8/maths/ncert-solutions-for-class-8-maths-chapter-4-practical-geometry-ex-4-3/ | 1,696,147,302,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510810.46/warc/CC-MAIN-20231001073649-20231001103649-00712.warc.gz | 346,759,429 | 20,838 | NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.3
# NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.3
## NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.3
NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Exercise 4.3
Ex 4.3 Class 8 Maths Question 1.
MO = 6 cm, ∠R = 105°, OR = 4.5 cm, ∠M = 60°, ∠O = 105°
PL = 4 cm, LA = 6.5 cm, ∠P = 90°, ∠A = 110°, ∠N = 85°
(iii) Parallelogram HEAR
HE = 5 cm, EA = 6 cm, ∠R = 85°
(iv) Rectangle OKAY
OK = 7 cm, KA = 5 cm
Solution:
(i) Construction:
Step I: Draw OR = 4.5 cm
Step II: Draw two angles of 105° each at O and R with the help of protactor.
Step III: Cut OM = 6 cm.
Step IV: Draw an angle of 60° at M to meet the angle line through R at E.
Thus, MORE is the required quadrilateral.
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(ii) Construction:
Step I: Draw LA = 6.5 cm
Step II: Draw an angle of 75° at L and 110° at A with the help of a protractor.
[∵ 360° – (110° + 90° + 85°) = 75°] Step III: Cut LP = 4 cm.
Step IV: Draw an angle of 90° at P which meets the angle line through A at N.
Thus PLAN is the required quadrilateral.
(iii) Construction: (Opposite sides of a parallelogram are equal)
Step I: Draw HE = 5 cm.
Step II: Draw an angle of 85° at E and cut EA = 6 cm.
Step III: Draw an arc with centre A and radius 5 cm.
Step IV: Draw another arc with centre H and radius 6 cm to meet the previous arc at R.
Step V: Join HR and AR
Thus, HEAR is the required parallelogram.
(iv) Construction:
(Each angle of a rectangle is 90° and opposite sides are equal.)
Step I: Draw OK = 7 cm.
Step II: Draw the angle of 90° at K and cut KA = 5 cm.
Step III: Draw an arc with centre O and radius 5 cm.
Step IV: Draw another arc with centre A and radius 7 cm to meet the previous arc at Y.
Step V: Join OY and AY.
Thus OKAY is the required rectangle.
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Verify OTP Code (required) | 821 | 2,753 | {"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} | 4.8125 | 5 | CC-MAIN-2023-40 | latest | en | 0.823466 |
https://www.techwhiff.com/issue/3x-8-12-answer-this-problem--71858 | 1,656,239,589,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103205617.12/warc/CC-MAIN-20220626101442-20220626131442-00648.warc.gz | 1,077,756,632 | 12,456 | # -3x + 8 = 12 answer this problem
-3x + 8 = 12
### 1. Which equation is equivalent to the equation 6x + 9 = 12? AX+9 = 6 B. 2x + 3 = 4 C. 3x + 9 = 6 D. OX + 12 = 9
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### Kyle's car average 380 miles on a 16-gallon tank of gas. How far can he drive on 6 gallons of gas? A. 160 1/2 miles B. 168 1/2 miles C. 142 1/2 miles D. 140 3/4 miles E. 149 1/4 miles
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### Which word in this excerpt from act I, scene I, of Richard III means “to listen”? Yea, Richard, when I know; for I protest As yet I do not: but, as I can learn, He hearkens after prophecies and dreams; And from the cross-row plucks the letter G. And says a wizard told him that by G His issue disinherited should be; And, for my name of George begins with G, It follows in his thought that I am he. These, as I learn, and such like toys as these Have moved his highness to commit me now.
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### Liquid octane CH3CH26CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose 2.3 g of octane is mixed with 12.4 g of oxygen. Calculate the minimum mass of octane that could be left over by the chemical reaction. Round your answer to 2 significant digits.
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### A car starts from rest at a stop sign. It accelerates at 4.6 m/s^2 for 6.2 s , coasts for 2.1s , and then slows down at a rate of 3.3 m/s^2 for the next stop sign. How far apart the are the stop signs?
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### During the early 1800s and late 1900s, the United States underwent huge economic and technological changes. The development of a society that relied on free enterprise and innovation led to new inventions and increasingly efficient businesses. These changes helped make the United States one of the world’s strongest economies and industrial centers. From Thomas Edison’s light bulb to Henry Ford’s affordable automobiles to Frederick Taylor’s time-study analysis, US innovations influenced business,
During the early 1800s and late 1900s, the United States underwent huge economic and technological changes. The development of a society that relied on free enterprise and innovation led to new inventions and increasingly efficient businesses. These changes helped make the United States one of the w... | 1,771 | 6,675 | {"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} | 3.21875 | 3 | CC-MAIN-2022-27 | latest | en | 0.843907 |
https://www.storyboardthat.com/storyboards/rustypotato/pudddddddinggg-scenario | 1,548,132,667,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583826240.93/warc/CC-MAIN-20190122034213-20190122060213-00605.warc.gz | 937,325,239 | 16,138 | # PUDDDDDDDINGGG scenario
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#### Storyboard Description
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#### Storyboard Text
• Puddings at this factory must be 120 grams in weight, or 5 grams away. If they don't meet the requirement, they will get thrown away.
• |x+5|=120 -(x+5)=120 -5 -5 -x-5=120 |x|=115 +5 +5 -x=125 x=-125
• WHAT ARE YOU DOING??? <---x------x------x---> 115 120 125
• It's 6 grams away.
• It's alright, want to go eat lunch together?
• Ok!
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Video Marketing | 213 | 967 | {"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} | 2.578125 | 3 | CC-MAIN-2019-04 | latest | en | 0.59946 |
https://answerprime.com/increasing-and-decreasing-intervals-calculator/ | 1,708,788,663,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474541.96/warc/CC-MAIN-20240224144416-20240224174416-00617.warc.gz | 103,593,105 | 40,820 | # Increasing and decreasing intervals calculator
Use a graphing calculator to find the intervals on which the function is increasing or decreasing f(x)-x/25 2 , for-5sxs5 Determine the interval(s) on which the function is increasing. Select the correct choice below and fil in any answer boxes in your choi The furpction is increasing on the intervals) (Type your answer in interval notation. Round to three decimal places as needed. Use a comma to separate answers as need O B. The function is never increasing Determine the interval(s) on which the function is decreasing. Select the correct choice below and fill in any answer boxes in your cho O A. The function is decreasing on the interval(s) O B. The function is never decreasing. (Type your answer in interval notation. Round to three decimal places as needed. Use a comma to separate answers as need Click to select and 24 MacBook Air
f(x) = x sqrt ( 25 – x^2 )
graph shown below
function is increasing on
the interval ( –
3.536 , 3.536 )
function is decreasing on
the interval [ – 5
, – 3.536 ) U ( 3.536 , 5 ]
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Scroll to Top | 266 | 1,127 | {"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} | 2.65625 | 3 | CC-MAIN-2024-10 | longest | en | 0.88791 |
https://awrcorp.com/download/faq/english/docs/Analyst_User_Guide/ch07s01s06.html | 1,590,744,264,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347402885.41/warc/CC-MAIN-20200529085930-20200529115930-00113.warc.gz | 260,925,379 | 2,996 | ### 7.1.6. Determining Units
Each expression variable evaluates to a numeric value with an associated unit, for example 5 cm. See “Setting Working Units” for more information. Some examples of units are distance, mass, charge and electric field. If a variable's expression is '1' then it is said to have None unit. You can add and subtract terms of the same unit. Parameters have a unit as well and they accept expressions that resolve to the same unit as them. A parameter's unit is constant. The Value column of the “Variables Control” shows the expression variable's evaluated value in working units. The tooltip of the cells in this column show the name of the unit if the unit is one of the standard types listed in the units dialog box.
In the previous figure, variable v1 has expression 5 m which evaluates to 500 cm because cm is the current length working unit. v2 has expression 10[m^2]. The square brackets allow you to define more complex units in this case such as area. You could also define v2 as 10m*m or 10*(m^2). The square bracket notation allows you to put the unit specifier right after the numeric value. The working unit for area in the above figure is m². | 266 | 1,181 | {"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} | 2.8125 | 3 | CC-MAIN-2020-24 | latest | en | 0.881095 |
https://softwarequtech.github.io/SimulaQron/html/_sources/NativeModeCorrRng.rst.txt | 1,618,561,448,000,000,000 | text/plain | crawl-data/CC-MAIN-2021-17/segments/1618038088731.42/warc/CC-MAIN-20210416065116-20210416095116-00057.warc.gz | 627,808,952 | 3,728 | Generate correlated randomness ============================== Having started the virtual quantum nodes, let us now run a simple test application, which already illustrates some of the aspects in realizing protocols. Our objective will be to realize the following protocol which will generate 1 shared random bit between Alice and Bob. Evidently, there would be classical means to achieve this trivial task chosen for illustration. * Alice generates 1 EPR pair, that is, two maximally entangled qubits :math:A and :math:B of the form :math:|\Psi\rangle_{AB} = \frac{1}{\sqrt{2}} \left(|0\rangle_A |0\rangle_B + |1\rangle_A |1\rangle_B\right) * She sends qubit :math:B to Bob. * Both Alice and Bob measure their respective qubits to obtain a classical random number :math:x \in \{0,1\}. Before seeing how this example works, let us again simply run the code:: cd examples/nativeMode/corrRNG sh doNew.sh Next to a considerable about of debugging information, you should be seeing the following two lines:: ALICE: My Random Number is 0/1 BOB: My Random Number is 0/1 Note that the order of these two lines may differ, as it does not matter who measures first. So what is actually going on here ? Let us first look at how we will realize the example by making an additional step (3) explicit: * Alice generates 1 EPR pair, that is, two maximally entangled qubits :math:A and :math:B of the form :math:|\Psi\rangle_{AB} = \frac{1}{\sqrt{2}} \left(|0\rangle_A |0\rangle_B + |1\rangle_A |1\rangle_B\right) * She sends qubit :math:B to Bob. * Bob is informed of the identifier of the qubit and is informed it has arrived. * Both Alice and Bob measure their respective qubits to obtain a classical random number :math:x \in \{0,1\}. While the task we want to realize here is completely trivial, the addition of step 3 does however already highlight a range of choices on how to realize step 3 and the need to find good abstractions to allow easy application development. One way to realize step 3 would be to hardwire Bobs measurement: if the hardware can identify the correct qubit from Alice, then we could instruct it to measure it immediately without asking for a notification from Alice. It is clear that in a network that is a bit larger than our tiny three node setup, identifying the right setup requires a link between the underlying qubits and classical control information: this is the objective of the classical/quantum combiner, for which we will provide code in version 0.2 of SimulaQron. This version simply allows a completely barebones access to the virtual nodes without implementing such convenient abstractions in order to allow you to explore such possibilities. To this end, we will here actually implement the following protocol for mere illustration purposes. We emphasize that this would be inefficient on a real quantum network since it requires Bob to store his qubit until Alice's control message arrives, which can be a significant delay causing the qubit to decohere in the meantime. * Alice generates 1 EPR pair, that is, two maximally entangled qubits :math:A and :math:B of the form :math:|\Psi\rangle_{AB} = \frac{1}{\sqrt{2}} \left(|0\rangle_A |0\rangle_B + |1\rangle_A |1\rangle_B\right) * She sends qubit :math:B to Bob. * Alice sends Bob the correct identifier of the qubit, and tells him to measure it. * Both Alice and Bob measure their respective qubits to obtain a classical random number :math:x \in \{0,1\}. To realize this, we thus need not only the connection to the virtual quantum node servers, but Alice and Bob themselves need to run a client/server to exchange classical control information. Before looking at the code, we node that the setup of these servers is again determined by a configuration file, namely config/classicalNet.cfg. This file defines which nodes act as servers in the classical communication network listening for control information to execute the protocol. You want to copy this to whatever example you are running. It takes the same format as above, where in our example only Bob will act run a server:: # Configuration file for servers on the classical communication network # # For each host its informal name, as well as its location in the network must # be listed. # # [name], [hostname], [port number] # Bob, localhost, 8812 The first thing that happens if we execute the script doNew.sh is that after some setting up it will call run.sh, executing:: #!/bin/sh python3 bobTest.py & python3 aliceTest.py Let us now look at the programs for Alice and Bob. Alice will merely run a client on the classical communication network that connects to Bob to be found in aliceTest.py. Using the template (see general Examples section) which establishes the connections to the local virtual nodes, we thus need to provide client code for Alice to implement the protocol above. The function runClientNode will automatically be executed once Alice connected to her local virtual quantum node simulating the underlying hardware, and to Bob's server:: ##################################################################################################### # # runClientNode # # This will be run on the local node if all communication links are set up (to the virtual node # quantum backend, as well as the nodes in the classical communication network), and the local classical # communication server is running (if applicable). # @inlineCallbacks def runClientNode(qReg, virtRoot, myName, classicalNet): """ Code to execute for the local client node. Called if all connections are established. Arguments qReg quantum register (twisted object supporting remote method calls) virtRoot virtual quantum ndoe (twisted object supporting remote method calls) myName name of this node (string) classicalNet servers in the classical communication network (dictionary of hosts) """ logging.debug("LOCAL %s: Runing client side program.",myName) # Create 2 qubits qA = yield virtRoot.callRemote("new_qubit_inreg",qReg) qB = yield virtRoot.callRemote("new_qubit_inreg",qReg) # Put qubits A and B in a maximally entangled state yield qA.callRemote("apply_H") yield qA.callRemote("cnot_onto",qB) # Send qubit B to Bob # Instruct the virtual node to transfer the qubit remoteNum = yield virtRoot.callRemote("send_qubit",qB, "Bob") # Tell Bob the ID of the qubit, and ask him to measure bob = classicalNet.hostDict["Bob"] yield bob.root.callRemote("process_qubit", remoteNum) # Measure qubit A to obtain a random number x = yield qA.callRemote("measure") print("ALICE: My Random Number is ",x,"\n") reactor.stop() Let us now look at Bob's server program to be found in bobTest.py. Observe that Alice will call process_qubit above. Not included in the code below are several standard methods that require no change to be used in examples.:: ##################################################################################################### # # localNode # # This will be run if the local node acts as a server on the classical communication network, # accepting remote method calls from the other nodes. class localNode(pb.Root): # This can be called by Alice to tell Bob to process the qubit @inlineCallbacks def remote_process_qubit(self, virtualNum): """ Recover the qubit and measure it to get a random number. Arguments virtualNum number of the virtual qubit corresponding to the EPR pair received """ qB = yield self.virtRoot.callRemote("get_virtual_ref",virtualNum) # Measure x = yield qB.callRemote("measure") print("BOB: My Random Number is ", x, "\n") | 1,680 | 7,499 | {"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} | 2.953125 | 3 | CC-MAIN-2021-17 | latest | en | 0.876412 |
http://www.ks.uiuc.edu/Research/namd/mailing_list/namd-l.2011-2012/3960.html | 1,721,304,435,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514828.10/warc/CC-MAIN-20240718095603-20240718125603-00228.warc.gz | 51,731,121 | 4,800 | # Re: restarting umbrella sampling simulation.
From: ipsita basu (ibasu788_at_gmail.com)
Date: Mon Sep 10 2012 - 01:48:44 CDT
Thank you sir for your help as I've learnt a lot from you.
In the harmonic part regarding center, targetcenter etc I was wrong,
you are absolutely correct that I've to set center at 70 and target at
20. It is very much clear from the output.
Now when I did some analysis from the output dcd file, I found that
all the segments are moving equally from +z to -z direction. But I
only one segment to move which I had set in the group1 portion. This
part in the configuration file is like:
colvar {
width 1.0
name distanceZ
distance {
group1 {
atomNumbersRange 94313 - 97288
}
group2 {
dummyAtom ( 00.99, 00.39, -61.89 )
}
}
}
where group1 is defined above. I want the segment in group1 is moving
towards group2 keeping group2 fixed and as there is a large no in
group2 segment, I defined it like above.
And my desire is to decrease the distance between the two above
defined groups from the value set at centers to targetcenters only by
moving group1 without affecting the rest of the system.
But as the whole system is moving, the desired distance is not
decreasing,it initially decreases then increases. My questions are:
1.Is the selection mentioned above correct?
2.Should I put some position restraints to the system except group1?
3.Is there anything else I missed or I misunderstood?
On 9/5/12, Jérôme Hénin <jhenin_at_ifr88.cnrs-mrs.fr> wrote:
> On 5 September 2012 09:34, ipsita basu <ibasu788_at_gmail.com> wrote:
>> Sorry to bother you again. But I have again some doubts which I want
>> to make clear.
>> One point is:
>> If I set centers at 70.0, the traj file shows that the distance is
>> increasing from 68.3 and if I set centers at 60.0 then the distance is
>> decreasing from 68.3, from this observation I think that the initial
>> distance is 68.3 and it tries to reach to the value set at centers.
>
> Yes, this sounds right. Remember that the current value of the
> restraint centers are always in the NAMD output.
>
>> Then what is the importance of targetcenters,
>> targetnumsteps,targetnumstages etc.?
>
> If everything is set up right, they should matter.
>
>> Second point is:
>> From namd list I get an instruction “give a fixed center for the
>> harmonic restraint, and run the simulation for some time, then you
>> give a different center and repeat.” and according to your suggestion
>> not to change the configuration file when restarting the job.
>
> That is only if you want all your windows to be run sequentially, each
> one starting from the end of the previous one.
>
>> So can I
>> run the whole umbrella sampling simulation for just one time setting
>> centers at 20.0 (my desire is to change the distance from 68.3 to 20)
>> ?
>
> My guess is, you should keep centers at 70 and set targetCenters at 20.
>
>> Can't I analyze it with wham with only one traj file?
>
> Everything is possible.
>
>> And if the
>> above things are possible then what will be the value of targetcenters
>> in my configuration file when centers at 20.0?
>
> I don't think the value of centers should be 20 here.
>
> Jerome
>
>
>
>> On 9/4/12, Jérôme Hénin <jhenin_at_ifr88.cnrs-mrs.fr> wrote:
>>> Ipsita,
>>> Please keep the list in CC so that others can follow this conversation.
>>>
>>> The config file should not be changed for restarting (i.e. "centers
>>> 70" should stay that way), as NAMD should get from the restart file
>>> information about the current stage of the transformation.
>>>
>>> The very first stage should be numbered zero. Please look for output
>>> lines mentioning changes in restraint centers in the log file of the
>>> first NAMD run.
>>>
>>> Finally, be aware that targetNumSteps gives the number of steps per
>>> stage, not the total. The total simulation time should be
>>> (targetNumStages + 1) * targetNumSteps.
>>> The +1 is for sampling the last point: for 100 stages, there are 101
>>> points to be sampled.
>>>
>>> to improve it as we realize what's needed, thanks to user feedback
>>> like yours.
>>>
>>> Jerome
>>>
>>>
>>> On 4 September 2012 10:10, ipsita basu <ibasu788_at_gmail.com> wrote:
>>>> Whatever I choose the value of centers, keeping fixed targetCenters
>>>> 50.00 , the NAMD output is :
>>>> colvars: Moving restraint stage 1 : setting centers to { 59.802 }.
>>>> Then the center will move to 59.802 in the next 1000000 steps? Then
>>>> what is the meaning of setting centers, and targetCenters?
>>>> Yes I include the colvarsInput parameter as :
>>>> colvarsInput out-us/da-us-1.out.res.colvars.state
>>>>
>>>> On 9/4/12, Jérôme Hénin <jhenin_at_ifr88.cnrs-mrs.fr> wrote:
>>>>> Ipsita,
>>>>>
>>>>> Better than looking at the colvars trajectory, you can look in the
>>>>> NAMD output for lines saying :
>>>>> Moving restraint stage n : setting centers to x
>>>>>
>>>>> When restarting the job, do you add the colvarsInput parameter in the
>>>>> NAMD config file?
>>>>>
>>>>> Cheers,
>>>>> Jerome
>>>>>
>>>>>
>>>>>
>>>>> On 4 September 2012 07:42, ipsita basu <ibasu788_at_gmail.com> wrote:
>>>>>> Sir,
>>>>>> I have a query regarding restarting umbrella sampling simulation of
>>>>>> membrane protein system.
>>>>>> When I start it one portion of my configuration is:
>>>>>> harmonic {
>>>>>> name lenpot
>>>>>> colvars distanceZ
>>>>>> centers 70.00
>>>>>> forceConstant 5.0
>>>>>> targetCenters 60.00
>>>>>> targetNumSteps 1000000
>>>>>> targetNumStages 100
>>>>>> where the distance between the two selected group is 70 which I want
>>>>>> to decrease to 60 in 1000000 steps. The output traj is like:
>>>>>> 999800 6.04585123909563e+01
>>>>>> # step distanceZ
>>>>>> 1000000 6.04704550193212e+01
>>>>>> Now when I restart the job I set the above portion as :
>>>>>> harmonic {
>>>>>> name lenpot
>>>>>> colvars distanceZ
>>>>>> centers 60.00
>>>>>> forceConstant 5.0
>>>>>> targetCenters 50.00
>>>>>> targetNumSteps 1000000
>>>>>> targetNumStages 100
>>>>>> But doing this the output traj file is like:
>>>>>> 1403600 6.03582445087823e+01
>>>>>> 1403800 6.03622946744465e+01
>>>>>> which means the distance is not decreasing anymore, so I omit the
>>>>>> option centers 60.00 and then I get the error: Error: must define the
>>>>>> initial centers of the restraints.
>>>>>> So should I set the value for centers and target centers, then what
>>>>>> is
>>>>>> the difference between the two?
>>>>>> Or should I set off the options:
>>>>>> targetCenters 50.00
>>>>>> targetNumSteps 1000000
>>>>>> targetNumStages 100
>>>>>> --
>>>>>> Ipsita Basu
>>>>>> Research Fellow
>>>>>> c/o : Dr. Chaitali Mukhopadhyay
>>>>>> Rajabazar Science College
>>>>>> Kolkata - 700009
>>>>>>
>>>>>
>>>>
>>>>
>>>> --
>>>> Ipsita Basu
>>>> Research Fellow
>>>> c/o : Dr. Chaitali Mukhopadhyay
>>>> Rajabazar Science College
>>>> Kolkata - 700009
>>>
>>
>>
>> --
>> Ipsita Basu
>> Research Fellow
>> c/o : Dr. Chaitali Mukhopadhyay
>> Rajabazar Science College
>> Kolkata - 700009
>>
>
```--
Ipsita Basu
Research Fellow | 1,919 | 6,992 | {"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} | 2.53125 | 3 | CC-MAIN-2024-30 | latest | en | 0.943154 |
https://kubicle.com/learn/excel/correlation | 1,550,371,046,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247481428.19/warc/CC-MAIN-20190217010854-20190217032854-00542.warc.gz | 574,758,032 | 17,416 | # 7. Correlation
Subtitles Enabled
# Sign up for a free trial to continue watching this lesson.
Overview
Correlation measures how two variables move in relation to each other, a useful metric for finding out what's driving revenue and profits.
Lesson Notes
Correlation
- Correlation is a statistical measure of how two variables move in relation to each other
- The correlation coefficent is a number between -1 and +1 that indicates the strength of a correlation
Measuring correlation:
--- -1 Perfect negative correlation
--- +1 Perfect positive correlation
--- 0 to +/- 0.3 is a weak correlation
--- +/- 0.3 to +/- 0.7 is a moderate correlation
--- +/- 0.7 to +/-1 is a strong correlation
New functions
=CORREL() Calculate the correlation between two continuous variables
Note: This formula can only be used for continuous variables. See this link to learn more about the different correlation coefficients that are used for other variable types.
Keyboard shortcuts
CTRL + F3: Open name manager
ALT + N: Create new name
CTRL + →: Move to end of data region
CTRL + SHIFT + →: Select all cells within data region
SHIFT + →: Select adjacent cell
ALT + E, S, F: Paste formulas
ALT + O, C, A: Autofit column width
CTRL + Page Up/Page Down: Move to next sheet
Transcript
In the previous lesson, we saw that there appeared to be a relationship between the day of the week and our daily revenue. In truth, there's often a number of variables that have relationships with the revenue, and it will be great to somehow quantify the strength of these relationships. Step forward correlation. The correlation formula accepts two columns of data and generates a correlation coefficient with the value between one and minus one. A correlation of one, shown on the example on the left, is known as perfect positive correlation. Whenever one value increases, the other increases at the exact same rate. For a correlation of minus one, one value goes up when another goes down, and in exactly the same proportion. In real life, you're very unlikely to find a correlation of one or minus one, but it's useful to know what the extremes look like. The final chart on the right shows a correlation of zero, where there is absolutely no relationship between the two variables. In this lesson, we're going to check two correlations for Virtuoso Sports Bar. First we're going to check if a correlation exists between the revenue and the total number of transactions, and then we're going to check if a correlation exists between daily revenue and the external temperature. As a rule of thumb, correlations between zero and plus or minus 0.3 are considered weak, between 0.3 and 0.7, moderate, and correlations of 0.7 and above are considered strong. Let's start off by finding the correlation between the number of transactions on each day and the total daily revenue. Our relationship would reasonably expect to have a strong correlation. I'll start off by creating a named range with control F3 and alt N, and this will be called transactions, and in the refers to box, I'll simply select my column.
I'll then escape, and go to my output cell, and to calculate a correlation, we'll write equals correl, open a bracket, and the first array will be revenues, and the second array will be transactions.
I'll close the bracket and press enter, and this gives me a correlation 0.98, which we can see is a very strong correlation. Let's now see if a correlation exists between revenue and daily temperature. The sports bar has quite a big beer garden, so we might expect that on warmer days, we get more customers, and hence, more revenues. Let's start by creating a named range for temperature, so I'll press control F3, alt N for a new name, and I'll call it temperature.
In the refers to box, I'll simply select my temperature column.
Then press OK and escape.
I'll now go to my output cell and write equals correl, revenues and then temperature.
Close the bracket and press enter, and here we find that the correlation between temperature and revenue is actually quite weak. This means that the company might not be getting any additional customers on hotter days, which is what the beer garden is actually for. Given this result, the company might want to consider using the beer garden space for a different purpose.
Here we calculated two correlations in two separate steps, but the analysis tool pack allows us to perform multiple correlations at the same time. Let's go to the data tab in the ribbon and click on data analysis.
And we'll select correlation and press OK.
We'll then scroll up to the top of the page and select our input range, and we'll include the labels in the first row, and the output will be in a new worksheet. I'll press OK to complete. If I adjust the column widths very quickly with Alt+O+C+A, you can see the correlations between the three variables. First we have the correlation between revenue and revenue, which is obviously one.
Here is the correlation between revenue and transactions, which we saw earlier was 0.98, and here is the correlation between temperature and revenue, and temperature and the number of transactions. When you need to complete multiple correlations on a single data set, you can see how the analysis tool pack can save you quite a lot of time. Overall, the correlation function in Excel is quite useful when quantifying the relationship between two variables; however, it does come with a lot of limitations, and if you're not aware of these, it's very easy to make mistakes. The primary constraint regards the type of data the correlation function can accept. There are different types of correlation coefficients for different types of data, and the correl function represents only one of these. Unless both data columns are continuous variables, IE, they are numerical, and they can be measured along a continuum, then you shouldn't use the correl function, so for example, we couldn't use the correl function when finding the correlation between the day of the week and revenue. We'd need to calculate a different correlation coefficient for this data. In the show notes, I'll link to some articles that discuss the other types of correlation coefficients, and I'll show you how to calculate some of these in a later lesson.
Contents
3 mins
4 mins
4 mins
3 mins
4 mins
4 mins
#### 7. Correlation
5 mins
My Notes
You can take notes as you view lessons.
Sign in or start a free trial to avail of this feature. | 1,372 | 6,488 | {"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} | 4.28125 | 4 | CC-MAIN-2019-09 | longest | en | 0.907057 |
https://books.openbookpublishers.com/10.11647/obp.0168/Section0001.xhtml | 1,638,474,785,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2021-49/segments/1637964362287.26/warc/CC-MAIN-20211202175510-20211202205510-00551.warc.gz | 202,450,601 | 49,348 | # I. Mental Skills
Even a superficial glance at history shows … great innovators … did vast amounts of computation and gained much of their insight in this way. I deplore the fact that contemporary mathematical education tends to give students the idea that computation is demeaning drudgery to be avoided at all costs.
Harold M. Edwards (1936– )
Fermat’s Last Theorem
We start our journey in a way that should be accessible to everyone – with a quick romp through important ideas from secondary school mathematics. The content is at times very elementary; but the problems often hint at something more challenging. The items included in this first chapter also highlight selected facts, techniques and ideas. Some of this early material is included to introduce certain ideas and techniques that later chapters will assume to be “known”. A few problems appeal to more advanced ideas (such as complex numbers), and are included here to indicate that “mental skills” are not restricted to elementary material.
Pencil and paper will be needed, but the items tend to focus on things which a student of mathematics should know by heart, or should learn to see at a glance, or should be able to calculate inside the head. In later problems (e.g. from Problem 18 onwards) the emphasis on mental skills should be interpreted as “ways of thinking”, rather than being taken to mean that everything should be done in your head. This is especially true where extended calculations or proofs are required.
Some of the items in this chapter (such as Problems 1 and 2) should be thoroughly familiar, and are included to underline this fact, rather than because we anticipate that they will need much active attention. Most of the early items in this first chapter are either core or auxiliary. However, there are also some real gems, which may even warrant a place in the the standard core.
The chapter is largely devoted to underlining the need for mastery of a repertoire of instantly available techniques, that can be used mentally, quickly, and flexibly to analyse less familiar problems at sight. But it also seeks to emphasise connections. Hence readers should be prepared to challenge their previous experience, in case it may have led to methods and results being perceived too narrowly.
We repeat the comment made in the section About this book. The “essence of mathematics”, which is referred to in the title, is largely implicit in the problems, and is there for the reader to extract. There is some discussion of this essence in the text interspersed between the problems. But, to avoid spilling the beans prematurely, and hence spoiling the problems, many important observations are buried away in the solutions, or in the Notes which follow many of the solutions.
## 1.1. Mental arithmetic and algebra
### 1.1.1 Times tables.
Problem 1 Using only mental arithmetic:
(a) Compute for yourself, and learn by heart, the times tables up to 9 × 9.
(b) Calculate instantly:
(i) 0.004 × 0.02
(ii) 0.0008 × 0.07
(iii) 0.007 × 0.12
(iv) 1.08 ÷ 1.2
(v) (0.08)2
Multiplication tables are important for many reasons. They allow us to appreciate directly, at first hand, the efficiency of our miraculous place value system – in which representing any number, and implementing any operation, are reduced to a combined mastery of
(i) the arithmetical behaviour of the ten digits 0–9, and
(ii) the index laws for powers of 10.
Fluency in mental and written arithmetic then leaves the mind free to notice, and to appreciate, the deeper patterns and structures which may be lurking just beneath the surface.
### 1.1.2 Squares, cubes, and powers of 2.
Algebra begins in earnest when we start to calculate with expressions involving powers. As one sees in the language we use for squares and cubes (i.e. 2nd and 3rd powers), these powers were interpreted geometrically for hundreds and thousands of years – so that higher powers, beyond the third power, were seen as being somehow unreal (like the 4th dimension). Our uniform algebraic notation covering all powers emerged in the 17th century (with Descartes (1596–1650)). But before one begins to work with algebraic powers, one should first aim to achieve complete fluency in working with numerical powers.
Problem 2
(a) Compute by mental arithmetic (using pencil only to record results), then learn by heart:
(i) the squares of positive integers: first up to 122; then to 312
(ii) the cubes of positive integers up to 113
(iii) the powers of 2 up to 210 .
(b) How many squares are there: (i) < 1000? (ii) < 10 000? (iii) < 100 000?
(c) How many cubes are there: (i) < 1000? (ii) < 10 000? (iii) < 1 000 000?
(d) (i) Which powers of 2 are squares? (ii) Which powers of 2 are cubes?
(e) Find the smallest square greater than 1 that is also a cube. Find the next smallest.
Evaluating powers, and the associated index laws, constitute an example of a direct operation. For each direct operation, we need to think carefully about the corresponding inverse operation – here “extracting roots”. In particular, we need to be clear about the distinction between the fact that the equation x2 = 4 has two different solutions, while $\sqrt{4}$ has just one value (namely 2).
Problem 3
(a) The operation of “squaring” is a function: it takes a single real number x as input, and delivers a definite real number x2 as output.
– Every positive number arises as an output (“is the square of something”).
– Since x2 = (−x)2, each output (other than 0) arises from at least two different inputs.
– If a2 = b2, then 0 = a2 b2 = (ab)(a+b), so either a = b, or a = −b. Hence no two positive inputs have the same square, so each output (other than 0) arises from exactly two inputs (one positive and one negative).
Hence each positive output y corresponds to just one positive input, called $\sqrt{\mathit{y}}$.
Find:
(i) $\sqrt{49}$ (ii) $\sqrt{144}$ (iii) $\sqrt{441}$ (iv) $\sqrt{169}$
(v) $\sqrt{196}$ (vi) $\sqrt{961}$ (vii) $\sqrt{96 100}$
(b) Let a > 0 and b > 0. Then $\sqrt{\mathit{ab}}>0$, and $\sqrt{\mathit{a}}×\sqrt{\mathit{b}}>0$, so both expressions are positive.
Moreover, they have the same square, since
$\left(\sqrt{\mathit{ab}}{\right)}^{2}=\mathit{ab}=\left(\sqrt{\mathit{a}}{\right)}^{2}·\left(\sqrt{\mathit{b}}{\right)}^{2}=\left(\sqrt{\mathit{a}}×\sqrt{\mathit{b}}{\right)}^{2}.$
$\therefore \sqrt{\mathit{a}×\mathit{b}}=\sqrt{\mathit{a}}×\sqrt{\mathit{b}}.$
Use this fact to simplify the following:
(i) $\sqrt{8}$ (ii) $\sqrt{12}$ (iii) $\sqrt{50}$
(iv) $\sqrt{147}$ (v) $\sqrt{288}$ (vi) $\sqrt{882}$
(c) [This part requires some written calculation.] Exact expressions involving square roots occur in many parts of elementary mathematics. We focus here on just one example – namely the regular pentagon.
Suppose that a regular pentagon ABCDE has sides of length 1.
(i) Prove that the diagonal AC is parallel to the side ED.
(ii) If AC and BD meet at X, explain why AXDE is a rhombus.
(iii) Prove that triangles ADX and CBX are similar.
(iv) If AC has length x, set up an equation and find the exact value of x.
(v) Find the exact length of BX.
(vi) Prove that triangles ABD and BXA are similar.
(vii) Find the exact values of cos 36°, cos 72°.
(viii) Find the exact values of sin 36°, sin 72°.
Every calculation with square roots depends on the fact that “ $\sqrt{\phantom{\rule{0ex}{0ex}}}$ is a function”. That is: given y > 0,
$\sqrt{\mathit{y}}$ denotes a single value – the positive number whose square is y.
The equation x2 = y has two roots, namely $\mathit{x}=±\sqrt{\mathit{y}}$; however, $\sqrt{\mathit{y}}$ has just one value (which is positive).
The mathematics of the regular pentagon is important – and generally neglected. It is included here to underline the way exact expressions involving square roots arise naturally.
In Problem 3(c), parts (iii) and (vi) require one to identify similar triangles using angles. The fact that “corresponding sides are then proportional” leads to a quadratic equation – and hence to square roots.
Parts (vii) and (viii) illustrate the fact that basic tools, such as
• the trigonometric identity cos2 θ + sin2 θ =1,
• the Cosine Rule, and
• the Sine Rule
should be part of one’s stock-in-trade. Notice that the exact values for
cos 36°, cos 72°, sin 36°, and sin 72°
also determine the exact values of
sin54° = cos36°, sin 18° = cos72°, cos54° = sin36°, and cos 18° = sin72°.
### 1.1.3 Primes
Problem 4
(a) Factorise 12 345 as a product of primes.
(b) Using only mental arithmetic, make a list of all prime numbers up to 100.
(c)(i) Find a prime number which is one less than a square.
(ii) Find another such prime.
There are 4 prime numbers less than 10; 25 prime numbers less than 100; and 168 prime numbers less than 1000.
Problem 4(c) is included to emphasise a frequently neglected message:
Words and images are part of the way we communicate.
But most of us cannot calculate with words and images.
To make use of mathematics, we must routinely translate words into symbols. For example, unknown numbers need to be represented by symbols, and points in a geometric diagram need to be properly labelled, before we can begin to calculate, and to reason, effectively.
### 1.1.4 Common factors and common multiples
To add two fractions we need to find a common multiple, or the LCM, of the two given denominators. To cancel fractions, or to simplify ratios, we need to be able to spot common factors and to find HCFs. Two positive integers a, b which have no (positive) common factors other than 1 (that is, with HCF (a,b) = 1) are said to be relatively prime, or coprime.
Problem 5 [This problem requires a mixture of serious thought and written proof.]
(a) I choose six integers between 10 and 19 (inclusive).
(i) Prove that some pair of integers among my chosen six must be relatively prime.
(ii) Is it also true that some pair must have a common factor?
(b) I choose six integers in the nineties (from 90–99 inclusive).
(i) Prove that some pair among my chosen integers must be relatively prime.
(ii) Is it also true that some pair must have a common factor?
(c) I choose n + 1 integers from a run of 2n consecutive integers.
(i) Prove that some pair among the chosen integers must be relatively prime.
(ii) Is it also true that some pair must have a common factor?
### 1.1.5 The Euclidean algorithm
School mathematics gives the impression that to find the HCF of two integers m and n, one must first obtain the prime power factorisations of m and of n, and can then extract the HCF from these two expressions. This is fine for beginners. But arithmetic involves unexpected subtleties. It turns out that, as the numbers get larger, factorising integers quickly becomes extremely difficult - a difficulty that is exploited in modern encryption systems. (The limitations of any method that depends on first finding the prime power factorisation of an integer should have become clear in Problem 4(b), where it is all too easy to imagine that 91 is prime, and in Problem 4(c)(ii), where students regularly think that 143, or that 323 are prime.)
Hence we would like to have a simple way of finding the HCF of two integers without having to factorise each of them first. That is what the Euclidean algorithm provides. We will look at this in more detail later. Meanwhile here is a first taste.
Problem 6
(a) (i) Explain why any integer that is a factor (or a divisor) of both m and n must also be a factor of their difference mn, and of their sum m + n.
(ii) Prove that
$\begin{array}{c}HCF\left(m,n\right)=HCF\left(m-n,n\right).\hfill \end{array}$
(iii) Use this to calculate in your head HCF (1001, 91) without factorising either number.
(b) (i) Prove that: $HCF\left(m,m+1\right)=1$.
(ii) Find $HCF\left(m,2m+1\right)$.
(iii) Find $HCF\left({m}^{2}+1,m-1\right)$.
### 1.1.6 Fractions and ratio
Problem 7 Which is bigger: 17% of nineteen million, or 19% of seventeen million?
Problem 8
(a) Evaluate
$\begin{array}{c}\left(1+\genfrac{}{}{0.1ex}{}{1}{2}\right)\left(1+\genfrac{}{}{0.1ex}{}{1}{3}\right)\left(1+\genfrac{}{}{0.1ex}{}{1}{4}\right)\left(1+\genfrac{}{}{0.1ex}{}{1}{5}\right).\hfill \end{array}$
(b) Evaluate
$\begin{array}{c}\sqrt{1+\genfrac{}{}{0.1ex}{}{1}{2}}×\sqrt{1+\genfrac{}{}{0.1ex}{}{1}{3}}×\sqrt{1+\genfrac{}{}{0.1ex}{}{1}{4}}×\sqrt{1+\genfrac{}{}{0.1ex}{}{1}{5}}×\sqrt{1+\genfrac{}{}{0.1ex}{}{1}{6}}×\sqrt{1+\genfrac{}{}{0.1ex}{}{1}{7}}.\hfill \end{array}$
(c) We write the product “$4×3×2×1$ ” as “4!” (and we read this as “4 factorial”). Using only pencil and paper, how quickly can you work out the number of weeks in 10! seconds?
Problem 9 The “DIN A” series of paper sizes is determined by two conditions. The basic requirement is that all the DIN A rectangles are similar ; the second condition is that when we fold a given size exactly in half, we get the next smaller size. Hence
• a sheet of paper of size A3 folds in half to give a sheet of size A4 - which is similar to A3; and
• a sheet of size A4 folds in half to give a sheet of size A5; etc..
(a) Find the constant ratio
r = “(longer side length) : (shorter side length)”
for all DIN A paper sizes.
(b) (i) To enlarge A4 size to A3 size (e.g. on a photocopier), each length is enlarged by a factor of r. What is the “enlargement factor” to get from A3 size back to A4 size?
(ii) To “enlarge” A4 size to A5 size (e.g. on a photocopier), each length is “enlarged” by a factor of $1r$. What is the enlargement factor to get from A5 size back to A4 size?
Problem 10
(a) In a sale which offers “15% discount on all marked prices” I buy three articles: a pair of trainers priced at £ 57.74, a T-shirt priced at £ 17.28, and a yo-yo priced at £ 4.98. Using only mental arithmetic, work out how much I should expect to pay altogether.
(b) Some retailers display prices without adding VAT — or “sales tax” — at 20% (because their main customers need to know the pre-VAT price). Suppose the prices in part (a) are the prices before adding VAT. Each price then needs to be adjusted in two ways — adding VAT and subtracting the discount. Should I add the VAT first and then work out the discount? Or should I apply the discount first and then add the VAT?
(c) Suppose the discount in part (b) is no longer 15%. What level of discount would exactly cancel out the addition of VAT at 20%?
Problem 11
(a) Using only mental arithmetic:
(i) Determine which is bigger:
$12+15 or 13+14?$
(ii) How is this question related to the observation that 10 < 12?
(b) [This part will require some written calculation and analysis.]
(i) For positive real numbers x, compare
$1x+2+1x+5 and 1x+3+1x+4.$
(ii) What happens in part (i) if x is negative?
### 1.1.7 Surds
Problem 12
(i) $(2+1)2$ (ii) $(2+1)2$ (iii) $(1+2)3$
(b) Simplify:
(i) $10+46$
(ii) $5+26$
(iii) $3+52$
(iv) $10−25$
The expressions which occur in exercises to develop fluency in working with surds often appear arbitrary. But they may not be. The arithmetic of surd arises naturally: for example, some of the expressions in the previous problem have already featured in Problem 3(c). In particular, surds will feature whenever Pythagoras’ Theorem is used to calculate lengths in geometry, or when a proportion arising from similar triangles requires us to solve a quadratic equation. So surd arithmetic is important. For example:
• A regular octagon with side length 1 can be surrounded by a square of side $2+1$ (which is also the diameter of its incircle); so the area of the regular octagon equals $(2+1)2−1$ (the square minus the four corners).
$2−1$ features repeatedly in the attempt to apply the Euclidean algorithm, or anthyphairesis, to express $2$ as a “continued fraction”.
$10−25$ may look like an arbitrary, uninteresting repeated surd, but is in fact very interesting, and has already featured as 4 sin 36° in Problem 3(c).
• One of the simplest ruler and compasses constructions for a regular pentagon ABCDE (see Problem 185) starts with a circle of radius 2, centre O, and a point A on the circle, and in three steps constructs the next point B on the circle, where AB is an edge of the inscribed regular pentagon, and
$AB¯=10−25.$
## 1.2. Direct and inverse procedures
We all learn to calculate — with numbers, with symbols, with functions, etc. But we may not notice that most calculating procedures come in pairs.
• First we learn a direct, deterministic, handle-turning technique, where answers are easy to churn out (as with addition, or multiplication, or working out powers, or multiplying brackets in algebra, or differentiating).
• Then we try to work backwards, or to “undo” this direct operation (as with subtraction, or division, or finding roots, or factorising, or integrating). This inverse procedure requires one to be completely fluent in the corresponding direct procedure; but it is much more demanding, in that one has to juggle possibilities as one goes, in order to home in on the required answer.
To master inverse procedures requires a surprising amount of time and effort. And because they are harder to master, they can easily get neglected. Even where they receive a lot of time, there are aspects of inverse procedures which tend to go unnoticed.
Problem 13 In how many different ways can the missing digits in this short multiplication be completed?
$\frac{\begin{array}{c}\phantom{\rule{0.33em}{0ex}}\phantom{\rule{0.33em}{0ex}}\phantom{\rule{0.33em}{0ex}}\phantom{\rule{0.33em}{0ex}}\square 6\hfill \\ ×\phantom{\rule{0.33em}{0ex}}\phantom{\rule{0.33em}{0ex}}\phantom{\rule{0.33em}{0ex}}\square \hfill \end{array}}{\underset{}{\square \phantom{\rule{0.33em}{0ex}}2\phantom{\rule{0.33em}{0ex}}\phantom{\rule{0.33em}{0ex}}8}}$
One would like students not only to master the direct operation of multiplying digits effectively, but also to notice that the inverse procedure of
“identifying the multiples of a given integer that give rise to a specified output”
depends on
the HCF of the multiplier and the base (10) of the numeral system.
• Multiplying by 1, 3, 7, or 9 induces a one-to-one mapping on the set of ten digits 0—9; so an inverse problem such as “7 × □ ends in 6” has just one digit-solution.
• Multiplying by 2, 4, 6, or 8 induces a two-to-one mapping onto the set of even digits (multiples of 2); so an inverse problem such as “6 × □ ends in 4” has two digit-solutions, and an inverse problem such as “6 × □ ends in 3” has no digit-solutions.
• Multiplying by 5 induces a five-to-one mapping onto the multiples (0 and 5) of 5, so an inverse problem such as “5 × □ ends in 0” has five digit-solutions and an inverse problem such as “5 × □ ends in 3” has no digit-solutions at all.
• Multiplying by 0 induces a ten-to-one mapping onto the multiples of 0(namely 0); so an inverse problem such as “0 × □ ends in 0” has ten digit-solutions and an inverse problem such as “0 × □ ends in 3 (or any digit other than 0)” has no digit-solutions at all.
The next problem shows — in a very simple setting — how elusive inverse problems can be. Here, instead of being asked to perform a direct calculation, the rules and the answer are given, and we are simply asked to invent a calculation that gives the specified output.
Problem 14
(a) In the “24 game” you are given four numbers. Your job is to use each number once, and to combine the four numbers using any three of the four basic arithmetical operations — using the same operation more than once if you wish, and as many brackets as you like (but never concatenating different numbers, such as “3” and “4” to make “34”). If the given numbers are 3, 3, 4, 4, then one immediately sees $3×4+3×4=24$. With 3, 3, 5, 5 it may take a little longer, but is still fairly straightforward. However, you may find it more challenging to make 24 in this way:
(i) using the four numbers 3, 3, 6, 6
(ii) using the four numbers 3, 3, 7, 7
(iii) using the four numbers 3, 3, 8, 8.
(b) Suppose we restrict the numbers to be used each time to “four 4s” (4, 4, 4, 4), and change the goal from “make 24”, to “make each answer from 0—10 using exactly four 4s”.
(i) Which of the numbers 0—10 cannot be made?
(ii) What if one is allowed to use squaring and square roots as well as the four basic operations? What is the first inaccessible integer?
Calculating by turning the handle deterministically (as with addition, or multiplication, or multiplying out brackets, or differentiating) is a valuable skill. But such direct procedures are usually only the beginning. Using mathematics and solving problems generally depend on the corresponding inverse procedures — where a certain amount of juggling and insight is needed in order to work backwards (as with subtraction, or division, or factorisation, or integration). For example, in applications of calculus, the main challenge is to solve differential equations (an inverse problem) rather than to differentiate known functions.
Problem 14 captures the spirit of this idea in the simplest possible context of arithmetic: the required answer is given, and we have to find how (or whether) that answer can be generated. We will meet more interesting examples of this kind throughout the rest of the collection.
### 1.2.1. Factorisation
Problem 15
(a) (i) Expand ($a+b)2$ and $(a+b)3$.
(ii) Without doing any more work, write out the expanded forms of $(a−b)2$ and $(a−b)3$.
(b) Factorise (i) $x2+2x+1$ (ii) $x4−2x2+1$ (iii) $x6−3x4+3x2−1$.
(c)(i) Expand (a - b)(a + b).
(ii) Use (c)(i) and (a)(i) to write down (with no extra work) the expanded form of
$(a−b−c)(a+b+c)$
and of
$(a−b+c)(a+b−c).$
(d) Factorise $3x2+2x−1$.
## 1.3. Structural arithmetic
Whenever the answer to a question turns out to be unexpectedly nice, one should ask oneself whether this is an accident, or whether there is some explanation which should perhaps have led one to expect such a result. For example:
• Exactly 25 of the integers up to 100 are prime numbers — and 25 is exactly one quarter of 100. This is certainly a beautifully memorable fact. But it is a numerical fluke, with no hidden mathematical explanation.
• 11 and 101 are prime numbers. Is this perhaps a way of generating lots of prime numbers:
$11,101,1001,10001,100001,…?$
It may at first be tempting to think so — until, that is, you remember what you found in Problem 6(a)(iii).
Problem 16 Write out the first 12 or so powers of 4:
$4, 16, 64, 256, 1024, 4096, 16384, 65536, …$
Now create two sequences:
the sequence of final digits: 4, 6, 4, 6, 4, 6, …
the sequence of leading digits: 4, 1, 6, 2, 1, 4, 1, 6, …
Both sequences seem to consist of a single “block”, which repeats over and over for ever.
(a) How long is the apparent repeating block for the first sequence? How long is the apparent repeating block for the second sequence?
(b) It may not be immediately clear whether either of these sequences really repeats forever. Nor may it be clear whether the two sequences are alike, or whether one is quite different from the other. Can you give a simple proof that one of these sequences recurs, that is, repeats forever?
(c) Can you explain why the other sequence seems to recur, and decide whether it really does recur forever?
Problem 17 The 4 by 4 “multiplication table” below is completely familiar.
$1234246836912481216$
What is the total of all the numbers in the 4 by 4 square? How should one write this answer in a way that makes the total obvious?
## 1.4. Pythagoras’ Theorem
From here on the idea of “mental skills” tends to refer to ways of thinking rather than to doing everything in your head.
Pythagoras’ Theorem is one of the first truly surprising results in school mathematics: it is hard to see why anyone would think of “adding the squares of the two shorter sides”. Despite the apparent attribution to a named person (Pythagoras), the origin of the theorem, and its proof, are unclear. There certainly was someone called Pythagoras (around 500 BC). But the main ancient references to him were written many hundreds of years after he died, and are not very reliable. The truth is that we know very little about him, or his theorem. The proof in Problem 18 below appeared in Book I of Euclid's thirteen books of Elements (written around 300 BC — two hundred years after Pythagoras). Much that is said (wrongly) to stem from Pythagoras is attributed in some sources to the Pythagoreans — a loose term which refers to any philosopher in what is seen as a tradition going back to Pythagoras. (This is a bit like interpreting anything called Christian in the last 2000 years as stemming directly from Christ himself.)
Clay tablets from around 1700 BC suggest that some Babylonians must have known “Pythagoras’ Theorem”; and it is hard to see how one could know the result without having some kind of justification. But we have no evidence of either a clear statement, or a proof, at that time. There are also Chinese texts that refer to Pythagoras’ Theorem (or as they call it, “Gougu”), which are thought to have originated BC — though the earliest surviving edition is from the 13th century AD. There is even an interesting little book by Frank Swetz, with the tongue-in-cheek title Was Pythagoras Chinese?.
The history may be confused, but the result — and its Euclidean proof — embodies something of the surprise and elegance of the very best mathematics. The Euclidean proof is included here partly because it is one that can, and should, be remembered (or rather, reconstructed — once one realises that there is really only one possible way to split the “square on the hypotenuse” in the required way). But, as we shall see, the result also links to exact mental calculation with surds, to trigonometry, to the familiar mnemonic “CAST”, to the idea of a “converse”, to sums of two squares, and to Pythagorean triples.
### 1.4.1 Pythagoras’ Theorem, trig for special angles, and CAST
Problem 18 (Pythagoras’ Theorem) Let ΔABC be a right angled triangle, with a right angle at C. Draw the squares ACQP, CBSR, and BAUT on the three sides, external to ΔABC. Use the resulting diagram to prove in your head that the square BAUT on BA is equal to the sum of the other two squares by:
• drawing the line through C perpendicular to AB, to meet AB at X and UT at Y
• observing that PA is parallel to QCB, so that ΔACP (half of the square ACQP, with base AP and perpendicular height AC) is equal in area to ΔABP (with base AP and the same perpendicular height)
• noting that ΔABP is SAS-congruent to ΔAUC, and that ΔAUC is equal in area to ΔAUX (half of rectangle AUYX, with base AU and height AX).
• whence ACQP is equal in area to rectangle AUYX
• similarly BCRS is equal in area to BTYX.
The proof in Problem 18 is the proof to be found in Euclid's Elements Book 1, Proposition 47. Unlike many proofs,
• it is clear what the proof depends on (namely SAS triangle congruence, and the area of a triangle), and
• it reveals exactly how the square on the hypotenuse AB divides into two summands — one equal to the square on AC and one equal to the square on BC.
Problem 19
(a) Use Pythagoras’ Theorem in a square ABCD of side 1 to show that the diagonal AC has length $2$. Use this to work out in your head the exact values of sin 45°, cos 45°, tan 45°.
(b) In an equilateral triangle ΔABC with sides of length 2, join A to the midpoint M of the base BC. Apply Pythagoras’ Theorem to find AM. Hence work out in your head the exact values of sin 30°, cos 30°, tan 30°, sin 60°, cos 60°, tan 60°.
(c)(i) On the unit circle with centre at the origin O:(0, 0), mark the point P so that P lies in the first quadrant, and so that OP makes an angle θ with the positive x-axis (measured anticlockwise from the positive x-axis). Explain why P has coordinates (cos θ, sin θ).
(ii) Extend the definitions of cos θ and sin θ to apply to angles beyond the first quadrant, so that for any point P on the unit circle, where OP makes an angle θ measured anticlockwise from the positive x-axis, the coordinates of P are (cos θ, sin θ). Check that the resulting functions sin and cos satisfy:
* and cos are both positive in the first quadrant,
* is positive and cos is negative in the second quadrant,
* and cos are both negative in the third quadrant, and
* is negative and cos is positive in the fourth quadrant.
(iii) Use (a), (b) to calculate the exact values of cos 315°, sin 225°, tan 210°, cos 120°, sin 960°, tan (-135°).
(d) Given a circle of radius 1, work out the exact area of a regular n-gon inscribed in the circle:
(i) when n = 3
(ii) when n = 4
(iii) when n = 6
(iv) when n = 8
(v) when n = 12.
(e) Given a circle of radius 1, work out the area of a regular n-gon circumscribed around the circle:
(i) when n = 3
(ii) when n = 4
(iii) when n = 6
(iv) when n = 8
(v) when n = 12.
Knowing the exact values of sin, cos and tan for the special angles 0°, 30°, 45°, 60°, 90° is like knowing one's tables. In particular, it allows one to evaluate trigonometric functions mentally for related angles in all four quadrants (using the CAST mnemonic — C being in the SE of the unit circle, A in the NE quadrant, S in the NW quadrant, and T in the SW quadrant — to remind us which functions are positive in each quadrant). These special angles arise over and over again in connection with equilateral triangles, squares, regular hexagons, regular octagons, regular dodecagons, etc., where one can use what one knows to calculate exactly in geometry.
Problem 20
(a) Use Pythagoras’ Theorem to calculate the exact length of the diagonal AC in a square ABCD of side length 2.
(b) Let X be the centre of the square ABCD in part (a). Draw lines through X parallel to the sides of ABCD and so divide the large square into four smaller squares, each of side 1. Find the length of the diagonals AX and XC.
Pythagoras’ Theorem holds the key to calculating exact distances in the plane. To calculate distances on the Earth's surface one needs a version of Pythagoras for “right angled triangles” on the sphere. We address this in Chapter 5.
### 1.4.2 Converses and Pythagoras’ Theorem
Each mathematical statement of the form
“if … (Hypothesis H),
then … (Consequence C)”
has a converse statement — namely
“if C,
then H”.
If the first statement is true, there is no a priori reason to expect its converse to be true. For example, part (c) of Problem 25 below proves that
“if an integer has the form 4k + 3,
then it cannot be written as the sum of two squares”.
However, the converse of this statement
“if an integer cannot be written as a sum of two squares,
then it has the form 4k + 3”
is false — since 6 cannot be written as the sum of two squares.
Despite this counterexample, whenever we prove a standard result, it makes sense to ask whether the converse is also true. For example,
“if PQRS is a parallelogram, then opposite angles are equal:
∠P = ∠R, and ∠Q = ∠S” (see Problem 157(ii)).
However you may not have considered the truth (or otherwise) of the converse statement:
If ABCD is a quadrilateral in which opposite angles are equal (∠A = ∠C and ∠B = ∠D), is it true that ABCD has to be a parallelogram?
The next problem invites you to prove the converse of Pythagoras’ Theorem. You should not use the Cosine Rule, since this is a generalisation of both Pythagoras’ Theorem and its converse.
Problem 21 Let ABC be a triangle. We use the standard labelling convention, whereby the side BC opposite A has length a, the side CA opposite B has length b, and the side AB opposite C has length c. Prove that, if c2 = a2 + b2, then ∠BCA is a right angle.
### 1.4.3 Pythagorean triples
The simplest example of a right angled triangle with integer length sides is given by the familiar triple 3, 4, 5:
$32+42=52.$
Any such integer triple is called a Pythagorean triple.
The classification of all Pythagorean triples is a delightful piece of elementary number theory, which is included in this chapter both because the result deserves to be memorised, and because (like Pythagoras’ Theorem itself) the proof only requires one to juggle a few simple ideas that should be part of one's armoury.
Pythagorean triples arise in many contexts (e.g. see the text after Problem 180). The classification given here shows that Pythagorean triples form a family depending on three parameters p, q, s (in which s is simply a “scaling” parameter, so the most important parameters are p, q). As a warm-up we consider two “one-parameter subfamilies” related to the triple 3, 4, 5.
Problem 22 Suppose a2 + b2 = c2 and that b, c are consecutive integers.
(a) Prove that a must be odd — so we can write it as a = 2m + 1 for some integer m.
(b) Prove that c must be odd — so we can write it as c = 2n + 1 for some integer n. Find an expression for n in terms of m.
Problem 22 reveals the triple (3, 4, 5) as the first instance (m = 1) of a one-parameter infinite family of triples, which continues
$(5,12,13) (m=2), (7,24,25) (m=3), (9,40,41) (m=4),…,$
whose general term is
$(2m+1, 2m(m+1), 2m(m+1)+1).$
The triple (3, 4, 5) is also the first member of a quite different “one-parameter infinite family” of triples, which continues
$(6,8,10), (9,12,15), ….$
Here the triples are scaled-up versions of the first triple (3, 4, 5).
In general, common factors simply get in the way:
If a2 + b2 = c2 and HCF(a, b) = s, then s2 divides a2 + b2, and a2 + b2 = c2; so s divides c.
And if a2 + b2 = c2 and HCF(b, c) = s, then s2 divides c2 - b2 = a2, so s divides a.
Hence a typical Pythagorean triple has the form (sa, sb, sc) for some scale factor s, where (a, b, c) is a triple of integers, no two of which have a common factor: any such triple is said to be primitive (that is, basic — like prime numbers). Every Pythagorean triple is an integer multiple of some primitive Pythagorean triple. The next problem invites you to find a simple formula for all primitive Pythagorean triples.
Problem 23 Let (a, b, c) be a primitive Pythagorean triple.
(a) Show that a and b have opposite parity (i.e. one is odd, the other even) — so we may assume that a is odd and b is even.
(b) Show that
$b22=c−a2c+a2,$
where
$HCFc−a2,c+a2=1$
and $c−a2,c+a2$ have opposite parity.
(c) Conclude that
$c+a2=p2 and c−a2=q2,$
where HCF(p, q) = 1 and p and q have opposite parity, so that c = p2 + q2, a = p2 - q2, b = 2pq.
(d) Check that any pair p, q having opposite parity and with HCF(p, q) = 1 gives rise to a primitive Pythagorean triple
$c=p2+q2,a=p2−q2,b=2pq$
satisfying a2 + b2 = c2.
Problem 24 The three integers a = 3, b = 4, c = 5 in the Pythagorean triple (3, 4, 5) form an arithmetic progression: that is, c - b = b - a. Find all Pythagorean triples (a, b, c) which form an arithmetic progression — that is, for which c - b = b - a.
### 1.4.4 Sums of two squares
The classification of Pythagorean triples tells us precisely which squares can be written as the sum of two squares. We now turn to the wider question: “Which integers are equal to the sum of two squares?”
Problem 25
(a) Which of the prime numbers < 100 can be written as the sum of two squares?
(b) Find an easy way to immediately write (a2 + b2)(c2 + d2) in the form (x2 + y2). (This shows that the set of integers which can be written as the sum of two squares is “closed” under multiplication.)
(c) Prove that no integer (and hence no prime number) of the form 4k + 3 can be written as the sum of two squares.
(d) The only even prime number can clearly be written as a sum of two squares: 2 = 12 + 12. Euler (1707—1783) proved that every odd prime number of the form 4k + 1 can be written as the sum of two squares in exactly one way. Find all integers < 100 that can be written as a sum of two squares.
(e) For which integers N < 100 is it possible to construct a square of area N, with vertices having integer coordinates?
In Problem 25 parts (a) and (d) you had to decide which integers < 100 can be written as a sum of two squares as an exercise in mental arithmetic. In part (b) the fact that this set of integers is closed under multiplication turned out to be an application of the arithmetic of norms for complex numbers. Part (e) then interpreted sums of two squares geometrically by using Pythagoras’ Theorem on the square lattice. These exercises are worth engaging in for their own sake. But it may also be of interest to know that writing an integer as a sum of two squares is a serious mathematical question — and in more than one sense.
Gauss (1777—1855), in his book Disquisitiones arithmeticae (1801) gave a complete analysis of when an integer can be represented by a ‘quadratic form’, such as x2 + y2 (as in Problem 25) or x2 - 2y2 (as in Problem 54(c) in Chapter 2).
A completely separate question (often attributed to Edward Waring (1736—1798)) concerns which integers can be expressed as a kth power, or as a sum of n such powers. If we restrict to the case k = 2 (i.e. squares), then:
• When n = 2, Euler (1707—1783) proved that the integers that can be written as a sum of two squares are precisely those of the form
$m2×p0×p1×p2×…×ps,$
where p0 = 1 or 2, and p1 < p2< … < ps are odd primes of the form 4l + 1.
• When n = 3, Legendre (1752—1833) and Gauss proved between them that the integers which can be written as a sum of three squares are precisely those that are not of the form 4m × (8l + 7).
• When n = 4, Lagrange (1736—1813) had previously proved that every positive integer can be written as a sum of four squares.
## 1.5. Visualisation
Problem 26 (Pages of a newspaper) I found a (double) sheet from an old newspaper, with pages 14 and 27 next to each other. How many pages were there in the original newspaper?
Problem 27 (Overlapping squares) A square ABCD of side 2 sits on top of a square PQRS of side 1, with vertex A at the centre O of the small square, side AB cutting the side PQ at the point X, and ∠AXQ = θ.
(a) Calculate the area of the overlapping region.
(b) Replace the two squares in part (a) with two equilateral triangles. Can you find the area of overlap in that case? What if we replace the squares (i.e. regular 4-gons) in part (a) with regular 2n-gons?
Problem 28 (A folded triangle) The equilateral triangle ΔABC has sides of length 1cm. D and E are points on the sides AB and AC respectively, such that folding ΔADE along DE folds the point A onto A' which lies outside ΔABC.
What is the total perimeter of the region formed by the three single layered parts of the folded triangle (i.e. excluding the quadrilateral with a folded layer on top)?
Problem 29 (A + B = C) The 3 by 1 rectangle ADEH consists of three adjacent unit squares: ABGH, BCFG, CDEF left to right, with A in the top left corner. Prove that
$∠DAE+∠DBE=∠DCE.$
Problem 30 (Dissections)
(a) Joining the midpoints of the edges of an equilateral triangle ABC cuts the triangle into four identical smaller equilateral triangles. Removing one of the three outer small triangles (say AMN, with M on AC) leaves three-quarters of the original shape in the form of an isosceles trapezium MNBC. Show how to cut this isosceles trapezium into four congruent pieces.
(b) Joining the midpoints of opposite sides of a square cuts the square into four congruent smaller squares. If we remove one of these squares, we are left with three-quarters of the original square in the form of an L-shape. Show how to cut this L-shape into four congruent pieces.
Problem 31 (Yin and Yang) The shaded region in Figure 1, shaped like a large comma, is bounded by three semicircles — two of radius 1 and one of radius 2.
Cut each region (the shaded region and the unshaded one) into two ‘halves’, so that all four parts are congruent (i.e. of identical size and shape, but with possibly different orientations).
Figure 1: Yin and Yang
In Problem 31 your first thought may have been that this is impossible. However, since the wording indicated that you are expected to succeed, it was clear that you must be missing something — so you tried again. The problem then tests both flexibility of thinking, and powers of visualisation.
### 1.6.1. Sine Rule
School textbooks tend to state the Sine Rule for a triangle ABC without worrying why it is true. So they often fail to give the result in its full form:
Theorem If R is the radius of the circumcircle of the triangle ABC, then
$asinA=bsinB=csinC=2R.$
This full form explains that the three ratios
$asinA,bsinB,csinC$
are all equal because they are all equal to the diameter 2R of the circumcircle of ΔABC — an additional observation which may well suggest how to prove the result (see Problem 32).
Problem 32 Given any triangle ABC, construct the perpendicular bisectors of the two sides AB and BC. Let these two perpendicular bisectors meet at O.
(a) Explain why OA = OB = OC.
(b) Draw the circle with centre O and with radius OA. There are three possibilities:
(i) The centre O lies on one of the sides of triangle ABC.
(ii) The centre O lies inside triangle ABC.
(iii) The centre O lies outside triangle ABC.
Case (i) leads directly to the Sine Rule for a right angled triangle ABC (remembering that sin 90° = 1). We address case (ii), and leave case (iii) to the reader.
(ii) Extend the line BO to meet the circle again at the point A'. Explain why ∠BA'C = ∠BAC = ∠A, and why ∠A'CB is a right angle. Conclude that
$sinA=BCA′B=a2R,$
and hence that
$asinA=2R=bsinB=csinC.$
Problem 33 Let Δ = area(ΔABC).
(a) Prove that
$Δ=12⋅ab⋅sinC.$
(b) Prove that $4RΔ=abc$.
### 1.6.2 Radians and spherical triangles
There is no God-given unit for measuring distance; different choices of unit give rise to answers that are related by scaling. However the situation is different for angles. In primary and secondary school we measure turn in degrees — where a half turn is 180°, a right angle is 90°, and a complete turn is 360°. This angle unit dates from the ancient Babylonians (~ 2000 BC). We are not sure why they chose 360 units in a full turn, but it seems to be related to the approximate number of days in a year (the time required for the heavens to make a complete rotation in the night sky), and to the fact that they wrote their numbers in “base 60”. However the choice is no more objectively mathematical than measuring distance in inches or in centimetres.
After growing up with the idea that angles are measured in degrees, we discover towards the end of secondary school that:
there is another unit of measure for angles — namely radians.
It may not at first be clear that this is an entirely natural, God-given unit. The size of, or amount of turn in, an angle at the point A can be captured in an absolute way by drawing a circle of radius r centred at the point A, and measuring the arc length which the angle cuts off on this circle. The angle size (in radians) of the angle at A is then defined to be the ratio
That is,
size of angle at the point A = arc length cut off on a circle of radius 1, centred at the apex A.
Hence a right angle is of size $π2$ radians; a half turn is equal to $π$ (radians); a full turn is equal to $2π$ (radians); each angle in an equilateral triangle is equal to $π3$ (radians); the three angles of a triangle have sum $π$; and the angles of a polygon with n sides have sum $(n−2)π$ (see Problem 230 in Chapter 6).
For a while after the introduction of radians we continue to emphasise the word radians each time we give the measure of an angle in order to stress that we are no longer using degrees. But this is not really a switch to a new unit: this new way of measuring angles is in some sense objective — so we soon drop all mention of the word “radians” and simply refer to the size of an angle (in radians) as if it were a pure number.
This switch affects the meaning of the familiar trigonometric functions. And though we continue to use the same names (sin, cos, tan, etc.), they become slightly different as functions, since the inputs are now always assumed to be in radians.
The real payoff for making this change stems from the way it recognizes the connection between angles and circles. This certainly makes calculating circular arc lengths and areas of sectors easy (an arc with angle θ on a circle of radius r now has length θ r; and a circular sector with angle 2θ now has area θr2). But the main benefit — which one hopes all students appreciate eventually — is that this change of perspective highlights the fundamental link between sin x, cos x, and ex:
• “cos x” becomes the derivative of sin x
• “-sin x” becomes the derivative of cos x, and
• the three functions are related by the totally unexpected identity
$\begin{array}{c}{e}^{i\theta }=\mathrm{cos}\theta +i\mathrm{sin}\theta .\hfill \end{array}$
The next problem draws attention to a beautiful result which reveals, in a pre-calculus, pre-complex number setting, a beautiful consequence of thinking about angles in terms of radians. The goal is to discover a formula for the area of a spherical triangle in terms of its angles and $π$, which links the formula for the circumference of a circle with that for the surface area of a sphere.
Suppose we wish to do geometry on the sphere. There is no problem deciding how to make sense of points. But it is less clear what we mean by (straight) lines, or line segments.
Before making due allowance for the winds and the tides, an airline pilot and a ship's Captain both need to know how to find the shortest path joining two given points A, B on a sphere. If the two points both lie on the equator, it is plausible (and correct) that the shortest route is to travel from A to Balong the equator. If we think of the equator as being in a horizontal plane through the centre O of the sphere, then we may notice that we can change the equator into a circle of longitude by rotating the sphere so that the “horizontal” plane (through O) becomes a “vertical” plane (through O). So we may view two points A and B which both lie on the same circle of longitude as lying on a “vertical equator” passing through A, B and the North and South poles: the shortest distance from A to B must therefore lie along that circle of longitude.
If we now rotate the sphere through some other angle, we get a “tilted equator” passing through the images of the (suitably tilted) points A and B: these “tilted equators” are called great circles. Each great circle is the intersection of the sphere with a plane through the centre O of the sphere. So
to find the shortest path from A to B:
• take the plane determined by the points A, B and the centre of the sphere O;
• find the great circle where this plane cuts the sphere;
• then follow the arc from A to B along this great circle.
Once we have points and line segments (i.e. arcs of great circles) on the sphere, we can think about triangles, and about the angles in such a triangle. In a triangle ABC on the sphere, the sides AB and AC are arcs of great circles meeting at A. By rotating the sphere we can imagine A as being at the North pole; so the two sides AB and AC behave just like arcs of two circles of longitude emanating from the North pole. In particular, we can measure the angle between them (this is exactly how we measure longitude): the two arcs AB, AC of circles of longitude set off from the North pole A in different horizontal directions before curving southwards, and the angle between them is the angle between these two initial horizontal directions (that is, the angle between the plane determined by O, A, B and the plane determined by O, A, C).
Figure 2: Angles on a sphere
Problem 34 Imagine a triangle ABC on the unit sphere (with radius r = 1), with angle α between AB and AC, angle β between BC and BA, and angle γ between CA and CB. You are now in a position to derive the remarkable formula for the area of such a spherical triangle.
(a) Let the two great circles containing the sides AB and AC meet again at A'. If we imagine A as being at the North pole, then A' will be at the South pole, and the angle between the two great circles at A' will also be α. The slice contained between these two great circles is called a lune with angle α.
(i) What fraction of the surface area of the whole sphere is contained in this lune of angle α? Write an expression for the actual area of this lune.
(ii) If the sides AB and AC are extended backwards through A, these backward extensions define another lune with the same angle α, and the same surface area. Write down the total area of these two lunes with angle α.
(b)(i) Repeat part (a) for the two sides BA, BC meeting at the vertex B, to find the total area of the two lunes meeting at B and B' with angle β.
(ii) Do the same for the two sides CA, CB meeting at the vertex C, to find the total area of the two lunes meeting at C and C' with angle γ.
(c)(i) Add up the areas of these six lunes (two with angle α, two with angle β, and two with angle γ). Check that this total includes every part of the sphere at least once.
(ii) Which parts of the sphere have been covered more than once? How many times have you covered the area of the original triangle ABC? And how many times have you covered the area of its sister triangle A'B'C'?
(iii) Hence find a formula for the area of the triangle ABC in terms of its angles — α at A, β at B, and γ at C.
### 1.6.3. Polar form and sin(A+B)
The next problem is less elementary than most of Chapter 1, but is included here to draw attention to the ease with which the addition formulae in trigonometry can be reconstructed once one knows about the polar form representation of a complex number. Those who are as yet unfamiliar with this material may skip the problem — but should perhaps remember the underlying message (namely that, once one is familiar with this material, there is no need ever again to get confused about the trig addition formulae).
Problem 35
(a) You may know that any complex number z = cosθ + isinθ of modulus 1 (that is, which lies on the unit circle centred at the origin) can be written in the modulus form z = e. Use this fact to reconstruct in your head the trigonometric identities for sin(A + B) and for cos(A + B). Use these to derive the identity for tan(A + B).
(b) By choosing X, Y so that $A=X+Y2$, and $B=X−Y2$, use part (a) to reconstruct the standard trigonometric identities for
$sinX+sinY, sinX−sinY, cosX+cosY, cosX−cosY.$
(c) (i) Check your answer to (a) for sin(A + B) by substituting A = 30°, and B = 60°.
(ii) Check your answer to (b) for cos X - cos Y by substituting X = 60°, and Y = 0°.
(d) (i) If $A+B+C+D=π$, prove that
$sinAsinB+sinCsinD=sin(B+C)sin(B+D).$
(ii) Given a cyclic quadrilateral WXYZ, with ∠XWY = A, ∠WXZ = B, ∠YXZ = C, ∠WYX = D, deduce Ptolemy's Theorem:
$WX×YZ+WZ×XY=WY×XZ.$
## 1.7. Regular polygons and regular polyhedra
Regular polygons have already featured rather often (e.g. in Problems 3, 12, 19, 27, 28, 29). This is a general feature of elementary mathematics; so the neglect of the geometry of regular polygons, and their 3D companions, the regular polyhedra, is all the more unfortunate. We end this first chapter with a first brief look at polygons and polyhedra.
### 1.7.1. Regular polygons are cyclic
Problem 36 A polygon ABCDE… consists of n vertices A, B, C, D, E, …, and n sides AB, BC, CD, DE… which are disjoint except that successive pairs meet at their common endpoint (as when AB, BC meet at B). A polygon is regular if any two sides are congruent (or equal), and any two angles are congruent (or equal). Can a regular n-gon ABCDE… always be inscribed in a circle? In other words, does a regular polygon automatically have a “centre”, which is equidistant from all n vertices?
### 1.7.2. Regular polyhedra
Problem 37 (Wrapping)
(a) You are given a regular tetrahedron with edges of length 2. Is it possible to choose positive real numbers a and b so that an a by b rectangular sheet of paper can be used to “wrap”, or cover, the regular tetrahedron without leaving any gaps or overlaps?
(b) Given a cube with edges of length 2, what is the smallest sized rectangle that can be used to wrap the cube in the same way without cutting the paper? (In other words, if we want to completely cover the cube, what is the smallest area of overlap needed? How small a fraction of the paper do we have to waste?)
Problem 38 (Cross-sections) Can a cross-section of a cube be:
(i) an equilateral triangle?
(ii) a square?
(iii) a polygon with more than six sides?
(iv) a regular hexagon?
(v) a regular pentagon?
Problem 39 (Shadows) Can one use the Sun's rays to produce a plane shadow of a cube:
(i) in the form of an equilateral triangle?
(ii) in the form of a square?
(iii) in the form of a pentagon?
(iv) in the form of a regular hexagon?
(v) in the form of a polygon with more than six sides?
The imparting of factual knowledge is for us a secondary consideration. Above all we aim to promote in the reader a correct attitude, a certain discipline of thought, which would appear to be of even more essential importance in mathematics than in other scientific disciplines. …
General rules which could prescribe in detail the most useful discipline of thought are not known to us. Even if such rules could be formulated, they would not be very useful. Rather than knowing the correct rules of thought theoretically, one must have assimilated them into one's flesh and blood ready for instant and instinctive use. Therefore for the schooling of one's powers of thought only the practice of thinking is really useful.
G. Pólya (1887–1985) and G. Szegö (1895–1985)
## 1.8. Chapter 1: Comments and solutions
1.
(a) Assuming that the 2×, 3×, 4×, and 5× tables are known, and that one has understood that the order of the factors does not matter, all that remains to be learned is 6× 6, 6× 7, 6× 8, 6× 9; 7× 7, 7× 8, 7× 9; 8× 8, 8× 9; and 9× 9.
(b) (i) $(4×10−3)×(2×10−2)=8×10−5=0.00008$
(ii) $(8×10−4)×(7×10−2)=56×10−6=0.000056$
(iii) $(7×10−3)×(12×10−2)=84×10−5=0.00084$
(iv) $1.08÷1.2=10.8÷12=108÷(12×10)=0.9$
(v) $(8×10−2)2=64×10−4=0.0064$
2.
(a) (i) 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144; 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900, 961
(ii) 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331
(iii) $1=20$, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024
(b) (i) 31 ($312=961$, $322=210=1024$)
(ii) 99 ($1002=104=10000$)
(iii) 316 ($3102=96100<100000<3202$; so look more carefully between 310 and 320)
(c) (i) 9 ($93=729<103=1000$)
(ii) 21 ($203=8000$, $223=10648$)
(iii) 99 ($1003=106=1000000$)
(d) (i) Those powers $2e$ of the form $22n$ for which the exponent e is a multiple of 2: i.e. $e≡0 mod2$.
(ii) Those powers 2e of the form $23n$ for which the exponent e is a multiple of 3: i.e. $e≡0 mod3$.
(e) $64=26=43=82$. $729=36=93=272$.
3.
(a) (i) 7; (ii) 12; (iii) 21; (iv) 13; (v) 14; (vi) 31; (vii) $10×31=310$
(b) (i) $22$; (ii) $23$; (iii) $52$; (iv) $73$; (v) $122$; (vi) $212$
(c) (i) ∠ABC = 108°. ΔBAC is isosceles ($BA=BC$), so ∠BAC = ∠BCA = 36°.
$∴∠CAE=∠BAE−∠BAC=72°=180°−∠AED.$
So AC is parallel to ED (since corresponding angles add to 180°).
(ii) AX is parallel to ED; similarly DX is parallel to EA. Hence AXDE is a parallelogram, with EA = ED.
(iii) The two triangles are both isosceles and ∠AXD = ∠CXB = 108° (vertically opposite angles). Hence ∠XAD = ∠XCB = 36°, and ∠XDA = ∠XBC = 36°.
(iv) AD: CB = DX: BX, so x: 1 = 1: (x - 1); hence x2 - x - 1 = 0 and $x=1+52$ — the Golden Ratio, usually denoted by the Greek letter $τ$ (tau), with approximate value $1.6180339887…$.
(v) $BX=x−1=5−12$ ($=1τ=τ−1$), with approximate value $0.6180339887…$.
(vi) We may either check that corresponding angles are equal in pairs (36°, 72°, 72°), or that corresponding sides are in the same ratio $x:1=1:(x−1$).
(vii) $cos36°=5+14$; $cos72°=5−14$ (drop perpendiculars from D to AB and from X to BC; or use the Cosine Rule).
(viii) Use $sin236°+cos236°=1$: $sin36°=10−254$; $sin72°=10+254$.
Note: The Golden Ratio crops up in many unexpected places (including the regular pentagon, and the Fibonacci numbers). Unfortunately much that is written about its ubiquity is pure invention. One of the better popular treatments, that highlights the number's significance, while taking a sober view of spurious claims, is the book The Golden Ratio by Mario Livio.
4.
(a) $12345=5×2469=3×5×823$. But is 823 a prime number?
It is easy to check that 823 is not divisible by 2, or 3, or 5, or 7, or 11. The Square Root Test (displayed below) tells us that it is only necessary to check four more potential prime factors.
Square Root Test: If $N=a×b$ with $a≤b$, then $a×a≤a×b=N$, so the smaller factor $a≤N$.
Hence, if N ($=823$ say) is not prime, its smallest factor > 1 is at most equal to $N$ ($=823<29$). Checking a = 13, 17, 19, 23 shows that the required prime factorisation is $12345=3×5×823$.
(b) There are 25 (2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97).
Notes:
(i) For small primes, mental arithmetic should suffice. But one should also be aware of the general Sieve of Eratosthenes (a Greek polymath from the $3rd$ century BC). Start with the integers 1—100 arranged in ten columns, and proceed as follows:
$\begin{array}{c}\begin{array}{cccccccccc}\hfill 1& \hfill 2& \hfill 3& \hfill 4& \hfill 5& \hfill 6& \hfill 7& \hfill 8& \hfill 9& \hfill 10\\ \hfill 11& \hfill 12& \hfill 13& \hfill 14& \hfill 15& \hfill 16& \hfill 17& \hfill 18& \hfill 19& \hfill 20\\ \hfill 21& \hfill 22& \hfill 23& \hfill 24& \hfill 25& \hfill 26& \hfill 27& \hfill 28& \hfill 29& \hfill 30\\ \hfill 31& \hfill 32& \hfill 33& \hfill 34& \hfill 35& \hfill 36& \hfill 37& \hfill 38& \hfill 39& \hfill 40\\ \hfill 41& \hfill 42& \hfill 43& \hfill 44& \hfill 45& \hfill 46& \hfill 47& \hfill 48& \hfill 49& \hfill 50\\ \hfill 61& \hfill 62& \hfill 63& \hfill 64& \hfill 65& \hfill 66& \hfill 67& \hfill 68& \hfill 69& \hfill 70\\ \hfill 71& \hfill 72& \hfill 73& \hfill 74& \hfill 75& \hfill 76& \hfill 77& \hfill 78& \hfill 79& \hfill 80\\ \hfill 81& \hfill 82& \hfill 83& \hfill 84& \hfill 85& \hfill 86& \hfill 87& \hfill 88& \hfill 89& \hfill 90\\ \hfill 91& \hfill 92& \hfill 93& \hfill 94& \hfill 95& \hfill 96& \hfill 97& \hfill 98& \hfill 99& \hfill 100\end{array}\hfill \end{array}$
Delete 1 (which is not a prime: see (ii) below).
Circle the first undeleted integer; remove all other multiples of 2.
Circle the first undeleted integer; remove all other multiples of 3.
Circle the first undeleted integer; remove all other multiples of 5.
Circle the first undeleted integer; remove all other multiples of 7.
All remaining undeleted integers < 100 must be prime (by the Square Root Test (see part (a)).
(ii) The multiplicative structure of integers is surprisingly subtle. The first thing to notice is that 1 has a special role, in that it is the multiplicative identity: for each integer n, we have $1×n=n$. Hence 1 is “multiplicatively neutral” — it has no effect.
The “multiplicative building blocks” for integers are the prime numbers: every integer > 1 can be broken down, or factorised as a product of prime numbers, in exactly one way. The integer 1 has no proper factors, and has no role to play in breaking down larger integers by factorisation. So 1 is not a prime. (If we made the mistake of counting 1 as a prime number, then we would have to make all sorts of silly exceptions — for example, to allow for the fact that $2=2×1=2×1×1=…$, so 2 could then be factorised in infinitely many ways.)
(iii) Notice that 91 = 7 × 13 is not a prime; so there is exactly one prime in the 90s — namely 97.
How many primes are there in the next run of 10 (from 100—109)?
How many primes are there from 190—199? How many from 200—210?
(c)(i) $3=22−1$.
(ii) Many students struggle with this, and may suggest 143, or 323, or even 63. The problem conceals a (very thinly) disguised message:
One cannot calculate with words.
To make use of mathematics, we must routinely translate words into symbols.
As soon as “one less than a square” is translated into symbols, bells should begin to ring. For you know that $x2−1=(x−1)(x+1)$, so $x2−1$ can only be prime if the smaller factor ($x−1$) is equal to 1.
5.
(a) (i) If we try to avoid such a “relatively prime pair”, then we must not choose any of 11, 13, 17, 19 (since they are prime, and have no multiples in the given range). So we are forced to choose the other six integers: 10, 12, 14, 15, 16, 18 — and there are then exactly two pairs which are relatively prime, namely 14, 15 and 15, 16.
(ii) If we try to avoid such a pair, then we can choose at most one even integer. So we are then forced to choose all five available odd integers, and our list will be: “unknown even, 11, 13, 15, 17, 19”. If the even integer is chosen to be 14, or 16, then every pair in my list has LCM = 1. So the answer is “No”.
(b) (i) If we try to avoid such a pair, then we must not choose 97 (the only prime number in the nineties). And we must not choose $95=5×19$ (which is relatively prime to all other integers in the given range — except for 90); and we must not choose $91=7×13$ (which is relatively prime to all other integers in the given range — except for 98). So we are forced to choose six integers from 90, 92, 93, 94, 96, 98, 99. Whichever integer we then omit leaves a pair which is relatively prime.
(ii) If we try to avoid such a pair, then we can choose at most one even integer. So we are then forced to choose all five available odd integers, and our list will be: “unknown even, 91, 93, 95, 97, 99”, and so must include the pair 93, 99 — with common factor 3. So the answer is “Yes”.
(c) In parts (a) and (b), the possible integers are limited (in (a) to the “teens”, and in (b) to the “nineties”); so it is natural to reach for ad hoc arguments as we did above. But in part (c) you know nothing about the numbers chosen.
Note: The question says that “I choose”, and asks whether “you” can be sure. So you have to find either a general argument that works for any n, or a counterexample. And the theme of this chapter indicates that it should not require any extended calculation.
The relevant “general idea” is the Pigeon Hole Principle which we may meet in the second part of this collection. So this problem may be viewed as a gentle introduction.
(i) Group the 2n consecutive integers
$a+1,a+2,…,a+2n$
into n pairs of consecutive integers
${a+1,a+2},{a+3,a+4},…,{a+(2n−1),a+2n}.$
– If we choose at most one integer from each pair, then we never get more than n integers.
– So as soon as we choose n+1 integers from 2n consecutive integers, we are forced to choose both integers in some pair k, k+1, and this pair of consecutive integers is always relatively prime (see Problem 6(b)(i)).
(ii) We saw in part (a)(ii) that, if n = 5 and the 2n integers start at 10, then we can choose six integers (either 11, 13, 14, 15, 17, 19, or 11, 13, 15, 16, 17, 19), and in each case every pair has LCM = 1. So for n = 5 the answer is “No” (because there is at least one case where one cannot be sure).
However, as soon as n is at least 6, we show that the argument in part (a)(ii) breaks down. As before, if we try to choose a subset in which no pair has a common factor, then we can choose at most one even integer. So we are forced to choose all the odd integers. But any run of at least six consecutive odd integers includes two multiples of 3. So for $n≥6$, the answer is “Yes”.
6.
(a)(i) Suppose k is a factor of m and n. Then we can write m = kp and n = kq for some integers p, q. Hence m - n = k(p - q), so k is a factor of m - n. Also m + n = k(p + q), so k is a factor of m + n.
(ii) Any factor of m and n is also a factor of their difference m - n; so the set of common factors of m and n is a subset of the set of common factors of m - n and n.
And any factor of m - n and n is also a factor of their sum m; so the set of common factors of m - n and n is a subset of the set of common factors of m and n.
Hence the two sets of common factors are identical. In particular, the two “highest common factors” are equal.
(iii) Subtract 91 from 1001 ten times to see that
$HCF(1001,91)=HCF(1001−910,91)=91.$
(b) (i) Subtract m from m + 1 once to see that
$HCF(m+1,m)=HCF(1,m)=1.$
(ii) Subtract m from 2m + 1 twice to see that
$HCF(2m+1,m)=HCF(m+1,m)=HCF(1,m)=1.$
(iii) Subtract m - 1 from m2 + 1 “m + 1 times” to see that
$HCF(m2+1,m−1)=HCF((m2+1)−(m2−1),m−1)=HCF(2,m−1).$
Hence, if m is odd, the HCF = 2; if m is even, the HCF = 1.
7. They are equal. (The first is
$17100×19000000,$
the second is
$19100×17000000,$
which are equal since multiplication is commutative and associative.)
8.
(a)
$32×43×54×65=62=3$
(b)
$32×43×54×65×76×87=82=2$
(c)
$\begin{array}{ccc}\hfill & \hfill & 10×9×8×7××5×4×3×2×1\phantom{\rule{0.222em}{0ex}}\text{seconds}\hfill \\ \hfill & \hfill & \phantom{\rule{2.00em}{0ex}}\phantom{\rule{2.00em}{0ex}}=\genfrac{}{}{0.1ex}{}{10×9×8×7×6×5×4×3×2×1}{60}\phantom{\rule{0.222em}{0ex}}\text{minutes}\hfill \\ \hfill & \hfill & \phantom{\rule{2.00em}{0ex}}\phantom{\rule{2.00em}{0ex}}=\genfrac{}{}{0.1ex}{}{10×9×8×7×6×5×4×3×2×1}{60×60}\phantom{\rule{0.222em}{0ex}}\text{hours}\hfill \\ \hfill & \hfill & \phantom{\rule{2.00em}{0ex}}\phantom{\rule{2.00em}{0ex}}=\genfrac{}{}{0.1ex}{}{10×9×8×7×6×5×4×3×2×1}{60×60×24}\phantom{\rule{0.222em}{0ex}}\text{days}\hfill \\ \hfill & \hfill & \phantom{\rule{2.00em}{0ex}}\phantom{\rule{2.00em}{0ex}}=\genfrac{}{}{0.1ex}{}{10×9×8×7×6×5×4×3×2×1}{60×60×24×7}\phantom{\rule{0.222em}{0ex}}\text{weeks}\hfill \\ \hfill & \hfill & \phantom{\rule{2.00em}{0ex}}\phantom{\rule{2.00em}{0ex}}=6\phantom{\rule{0.222em}{0ex}}\text{weeks (after cancelling)}.\hfill \end{array}$
Note: These three questions underline what we mean by structural arithmetic. Fractions should never be handled by evaluating numerators and denominators. Instead one should always be on the lookout for structural features which simplify calculation — such as cancellation.
9. (a) Suppose a rectangle in the “DIN A” series has dimensions a by b, with a < b. Folding in half produces a rectangle of size $b2$ by a. Hence $b:a=a:b2$, so $b2=2a2$, and $b:a=2:1$.
(b) (i) $1r$. (ii) r.
10.
(a) “15% discount” means the price actually charged is “85% of the marked price”. Hence each marked price needs to be multiplied by 0.85.
The distributive law says we may add the marked prices first and then multiply the total (exactly £ 80) by 0.85 to get
Note: The context (shopping, sales tax, and discount) is mathematically uninteresting. What matters here is the underlying multiplicative structure of the solution, which arises in many different contexts.
(b) “Add 20% VAT” means multiplying the discounted pre-VAT total (£68) by 1.2, or $65$. Hence the final price, with VAT added, is £ (1.2 × 0.85 × 80).
If the VAT were added first, the price before discount would be £ (1.2 × 80), and the final price after allowing for the discount would be £ (0.85 × 1.2 × 80).
Since multiplication is commutative, the two calculations have the same result, so the order does not matter (just as the final result in Problem 9 is the same whether one first enlarges A4 to A3 and then reduces A3 to A4, or first reduces A4 to A5 and then enlarges A5 to A4).
Note: Notice that we did not evaluate the two answers to see that they gave the same output £81.60. If we had, then the equality might have been a fluke due to the particular numbers chosen. Instead we left the answer unevaluated, in structured form, which showed that the equality would hold for any input.
(c) To cancel out multiplying by $65$ we need to multiply by $56$ — a discount of $16$, or $1623%$.
Note: This question has nothing to do with financial applications. It is included to underline the fact that although percentage change questions use the language of addition and subtraction (“increase”, or “decrease”), the mathematics suggests they should be handled multiplicatively.
11.
(a) (i) $2×5<3×4$, so
$72×5>73×4.$
Hence
$12+15>13+14.$
(ii) At first sight, “$10≤12$” may not seem related to “$12+15>13+14$”. Yet the crucial fact we started from in part (i) was “$2×5=10≤12=3×4$”.
(b) 10 < 12, so
$(x+2)(x+5)=x2+7x+10
(i) If all four brackets are positive (i.e. if x > − 2), then we also have 2x + 7 > 0, and it follows that
$\begin{array}{ccc}\genfrac{}{}{0.1ex}{}{1}{x+2}+\genfrac{}{}{0.1ex}{}{1}{x+5}\hfill & =\hfill & \genfrac{}{}{0.1ex}{}{2x+7}{\left(x+2\right)\left(x+5\right)}\hfill \\ \hfill & >\hfill & \genfrac{}{}{0.1ex}{}{2x+7}{\left(x+3\right)\left(x+4\right)}\hfill \\ \hfill & =\hfill & \genfrac{}{}{0.1ex}{}{1}{x+3}+\genfrac{}{}{0.1ex}{}{1}{x+4}.\hfill \end{array}$
(ii) When calculating with the given algebraic expression, the values
$x=−2,−3,−4,−5$
are “forbidden values”.
If x > − 2, then (as in part (i)) we have
$\begin{array}{ccc}\genfrac{}{}{0.1ex}{}{1}{x+2}+\genfrac{}{}{0.1ex}{}{1}{x+5}\hfill & =\hfill & \genfrac{}{}{0.1ex}{}{2x+7}{\left(x+2\right)\left(x+5\right)}\hfill \\ \hfill & >\hfill & \genfrac{}{}{0.1ex}{}{2x+7}{\left(x+3\right)\left(x+4\right)}\hfill \\ \hfill & =\hfill & \genfrac{}{}{0.1ex}{}{1}{x+3}+\genfrac{}{}{0.1ex}{}{1}{x+4}.\hfill \end{array}$
For permitted values of x < -2, one or more of the brackets (x + 2), (x + 5), (x + 3), (x + 4) will be negative. However, one can still carry out the algebra to simplify
$1x+2+1x+5=2x+7(x+2)(x+5)and$ $1x+3+1x+4=2x+7(x+3)(x+4)$
When $x=−72$ both expressions are equal, and equal to 0. The simplified numerators are both positive if $x>−72$, and both negative if $x<−72$; and the sign of the denominators changes as one moves through the four intervals −3 > x > −2, −4 > x > −3, −5 > x > −4, x > −5, with the inequality switching
$\begin{array}{c}\hfill \mathrm{from}“>”\left(\mathrm{for}x>-2\right)\phantom{\rule{2.00em}{0ex}}\mathrm{to}\phantom{\rule{1.00em}{0ex}}“<”\left(\mathrm{for}-3”\left(\mathrm{for}-3.5”\left(\mathrm{for}-5
12.
(a) (i) $3+22$; (ii) $3−22$; (iii) $7+52$.
Note: Notice that you can write down the answer to (ii) as soon as you have finished (i), without doing any further calculation.
(b) (i) $2+6$; (ii) $2+3$; (iii) $1+52$ (the Golden Ratio $1+52=τ$ is the larger root of the quadratic equation $x2−x−1=0$. Hence $3+52=τ+1=τ2$); (iv) $10−25$: this does not simplify further.
Note: Some readers may think an apology is in order for part (iv). The lesson here is that, while one should always try to simplify, there is no way of knowing in advance whether a simplification is possible. And there is no way out of this dilemma. So one is reduced to thinking: any simplification would involve $5$, and if one tries to solve $(a+b5)2=10−25$, then the solutions for a and b do not lead to anything “simpler”. (This repeated surd should perhaps have rung bells, as it was equal to the exact expression for 4sin 36° in Problem 3(c). It was included here partly because the question of its simplification should already have arisen when it featured in that context.)
13. In reconstructing the missing digits the number of possible solutions is determined by the highest common factor of the multiplier and 10. At the first step (in the units column):
because HCF(6, 10) = 2, $□×6=8 mod10$ has two solutions which differ by 5 — namely 3 and 8.
The first possibility then requires us to solve $(□×3)+1=2 mod10$: because $HCF(3,10)=1$, this has just one solution — namely 7. This gives rise to the solution $76×3=228$.
The second possibility requires us to solve $(□×8)+4=2 mod10$: because HCF(8, 10) = 2, this has two solutions which differ by 5 — namely 1 and 6. This gives rise to two further solutions: $16×8=128$, and $66×8=528$.
14.
(a) The solutions are entirely elementary, with no trickery. But they can be surprisingly elusive. And since this elusiveness is the only reason for including the problem, we hesitate to relieve any frustration by giving the solution.
The whole thrust of the “24 game” is to underline the scope for “getting to know” the many faces of a number like 24: for example, as 24=12 + 12 ($=3×4+3×4$ for 3, 3, 4, 4); as 24=25 −1 ($=5×5−3÷3$ for 3, 3, 5, 5); and as 24=27 - 3 ($=3×3×3−3$ for 3, 3, 3, 3). So one should be looking for ways of exploiting other important arithmetical aspects of 24— in particular, as $4×6$ and as $3×8$.
(b) (i) 0 = (4 − 4) + (4 − 4); $1=(4÷4)×(4÷4)$; $2=(4÷4)+(4÷4)$; $3=(4+4+4)÷4$; $4=((4−4)×4)+4$; $5=((4×4)+4)÷4$; $6=4+((4+4)÷4)$; $7=4+4−(4÷4)$; 8 = $(4+4)×(4÷4)$; . The output 10 seems to be impossible with the given restrictions.
(ii) With squaring and allowed we can manage $10=4+4+4−4$. Indeed, one can make everything up to 40 except (perhaps) 39.
15.
(a) (i) a2 + 2ab + b2; $a3+3a2b+3ab2+b3$
(ii) Replace b by (−b): $a2−2ab+b2$; $a3−3a2b+3ab2−b3$
(b) (i) $(x+1)2$; (ii) $(x2−1)2$; (iii) $(x2−1)3$
(c) (i) $a2−b2$
(ii) Replace “b” by “b + c”: $a2−(b+c)2=a2−b2−c2−2bc$
Replace “b” by “b − c”: $a2−(b−c)2=a2−b2−c2+2bc$
(d) One way is to rewrite this expression as a difference of two squares:
$\begin{array}{ccc}{\left(2x\right)}^{2}-\left({x}^{2}-2x+1\right)\hfill & =\hfill & {\left(2x\right)}^{2}-{\left(x-1\right)}^{2}\hfill \\ \hfill & =\hfill & \left(2x-\left(x-1\right)\right)\left(2x+\left(x-1\right)\right)\hfill \\ \hfill & =\hfill & \left(x+1\right)\left(3x-1\right)\hfill \end{array}$
Note: As so often, the messages here are largely implicit. In part (a)(ii) we\linebreak explicitly highlight the intention to use what you already know (by simply substituting “-b” in place of “b”. In part (b), you are expected to recognise (i), and then to see (ii) and (iii) as mild variations on the expansions of $(a−b)2$ and $(a−b)3$ in part (a). Part (c) repeats (in silence) the message of (a)(ii): think — don't slog it out. And part (d) encourages you to keep an eye out for thinly disguised instances of “a difference of two squares”.
16.
(a) Final digits: ‘block’ 4, 6 of length 2;
leading digits: “block” 4, 1, 6, 2, 1 of length 5.
(b) Claim The sequence of “units digits” really does recur.
Proof Given a power of 4 that has units digit 4, the usual multiplication algorithm for multiplying by 4 produces a number with units digit 6.
Given this new power of 4 with units digit 6, the usual multiplication algorithm for multiplying by 4 produces a number with units digit 4.
At this stage the sequence of units digits begins a new cycle.
[Alternatively: The units digit is simply equal to the relevant power of $4 mod10$. Multiplying by 4 changes 4 to $6 mod10$; multiplying by 4 changes 6 to $4 mod10$; — and the cycle repeats.]
(c) The sequence of leading digits seems to recur every 5 terms, because $45=210=1024$ is almost exactly equal to 1000. Each time we move on 5 steps in the sequence, we multiply by $45=1024$. As far as the leading digit is concerned, this has the same effect as multiplying the initial term (4) by slightly more than 1.024 (then adding any ‘carries'), which is very like multiplying by 1— and so does not change the leading digit (yet).
However, each time we move on 10 steps in the sequence, we multiply by $410=10242=1048576$. As far as the leading digit is concerned, this has the same effect as multiplying by slightly more than 1.048576.
When we move on 25 steps, we multiply by $425=1125899906842624$. And as far as the leading digit is concerned, this has the same effect as multiplying by slightly more than 1.12599906842624. And so on.
Eventually, the multiplier becomes large enough to change one of the leading digits.
17. The total is 100.
Having found this by direct calculation, we should think indirectly and notice that $100=102$.
And we should then ask: “Why 10? What has 10 got to do with the $4×$ multiplication table?”
A quick check of the $1×$ multiplication table (total = 1), the $2×$ multiplication table (total = 9), etc. may suggest what we should have seen immediately.
The first row has sum: $\left(1+2+3+4\right)$.
The second row has total $2×(1+2+3+4)$.
The third row has total $3×(1+2+3+4)$.
The fourth row has total $4×(1+2+3+4)$.
$∴$ The total is $(1+2+3+4)×(1+2+3+4)=(1+2+3+4)2$.
19.
(a) $sin45°=12=22=cos45°$; $tan45°=1$
(b) $AM=3$; $sin30°=12$, $cos30°=32$, $tan30°=13=33$; $sin60°=32$, $cos60°=12$, $tan60°=3$.
(c) (iii) $cos315°=cos45°=22$; $sin225°=−sin45°=−22$; $tan210°=tan30°=33$; $cos120°=−cos60°=−12$; $sin960°=sin240°=−sin60°=−32$; $tan(−135°)=tan45°=1$.
(d) Cut the n-gon into n“cake slices”, and use the formula “$12absinC$” for each slice.
(i) $334$; (ii) 2; (iii) $332$; (iv) $22$; (v) 3
(e) Work out the side length of the n-gon, then cut the n-gon into n“slices”, and use the formula “$12(base×height$)” for each slice.
(i) $33$; (ii) 4; (iii) $23$; (iv) $8(2−1)$; (v) $12(2−3)$
Note: There is no hidden trig here: all you need is Pythagoras’ Theorem. For example, in part (e)(iv) we can extend alternate sides of the regular octagon to form the circumscribed 2 by 2 square. The four corner triangles are isosceles right angled triangles with hypotenuse of length s (the side of the octagon). Hence each side of the square is equal to $s+2⋅s2=2$, whence $s=2(2−1)$.
20. (a) $8$; (b) $2$, $2$; (c) $8=4×2=22$
21. Construct the perpendicular from A to BC (possibly extended); let this meet the line BC at X. There are four possibilities:
(i) either X = C, in which case ∠BCA is a right angle as required; or X = B, in which case $b2=a2+c2$, contradicting $a2+b2=c2$;
(ii) $X≠B,C$, and C lies between B and X;
(iii) $X≠B,C$, and X lies between B and C;
(iv) $X≠B,C$, and B lies between X and C.
We analyse case (ii) and leave cases (iii) and (iv) to the reader.
(ii) ΔAXC and ΔAXB are both right angled triangles; so by Pythagoras’ Theorem we know that
$\begin{array}{ccc}A{C}^{2}\hfill & =\hfill & A{X}^{2}+X{C}^{2},\phantom{\rule{0.278em}{0ex}}\text{and}\hfill \\ A{B}^{2}\hfill & =\hfill & A{X}^{2}+X{B}^{2}\hfill \\ \hfill & =\hfill & A{X}^{2}+{\left(XC+CB\right)}^{2}\hfill \\ \hfill & =\hfill & A{X}^{2}+X{C}^{2}+C{B}^{2}+2XC\cdot CB\hfill \\ \hfill & =\hfill & A{C}^{2}+C{B}^{2}+2XC\cdot CB.\hfill \end{array}$
Since we are told that $AC2+CB2=AB2$, it follows that $2XC⋅CB=0$, contrary to $X≠C$.
Note: Notice that the proof of the converse of Pythagoras’ Theorem makes use of Pythagoras’ Theorem itself.
22.
(a) c = b + 1, so $a2=c2−b2=2b+1$. Hence a is odd, and we can write $a=2m+1$.
(b) Suppose $b=2n−1$ is also odd. Then $c2=4n2$ is divisible by 4— which contradicts the fact that $b2=4(n2−n)+1$, and $a2=4(m2+m)+1$, so $a2+b2$ leaves remainder 2 on division by 4.
Hence b = 2n is even and c = 2n + 1 is odd. But then
$(2m+1)2+(2n)2=a2+b2=c2=(2n+1)2,$
so $4(m2+m)=4n$, and $n=m(m+1)$.
23. (a) If a and b are both even, then $HCF(a,b)≠1$, so the triple would not be primitive.
If a and b are both odd, we use the idea from Problem 22(b). Suppose a = 2m + 1, b = 2n + 1; then $a2=4(m2+m)+1$, and $b2=4(n2+n)+1$, so $a2+b2=2×(2(m2+m+n2+n)+1)$. But this is “twice an odd number”, so cannot be equal to $c2$ (since c would have to be even, and any even square must be a multiple of 4).
Hence we may assume that a is odd and b is even: so c is is odd.
(b) Then $a2+b2=c2$ yields $b2=c2−a2=(c−a)(c+a)$, so
$b22=c−a2c+a2.$
Any common factor of $c+a2$ and $c−a2$ divides their sum c and their difference a, so $HCF(c−a2,c+a2)=1$. Since the difference of these two factors is a, which is odd, they have opposite parity.
(c) If two integers are relatively prime, and their product is a square, then each of the factors has to be a square (consider their prime factorisations). Hence $c+a2=p2$ and $c−a2=q2$, where $HCF(p,q)=1$ and p and q have opposite parity. Therefore
$c=p2+q2,a=p2−q2,b=2pq.$
(d) It is easy to check that any triple of the given form is (i) primitive, and (ii) satisfies $a2+b2=c2$.
24. Claim The only such triples are those of the form (3s, 4s, 5s).
Proof We show that the only primitive Pythagorean triple which forms an arithmetic progression is the familiar triple (3, 4, 5).
By Problem 23, one of the numbers in any primitive Pythagorean triple is even (namely 2pq) and two are odd (p and q are of opposite parity, so $p2−q2$ and $p2+q2$ are both odd).
∴ 2pq is the “middle term”, and the smallest and largest terms differ by $2q2$.
∴ the common difference d = c - b = b - a is equal to $q2$.
$2pq=p2$, so p = 2q.
Finally, since p and q are relatively prime, we must have q = 1, p = 2. QED
Note: Alternatively, let (a, b, c) be any Pythagorean triple (not necessarily primitive), which forms an arithmetic progression. Then
$a2=c2−b2=(c−b)(c+b)=(b−a)(c+b).$
So b(c + b) = a(a + b + c). Hence $a⋅3b=a(a+b+c)=b(c+b)$. It then follows that $3b2=b(a+b+c)=4ba$, so 3b = 4a and a: b = 3: 4.
25.
(a) $2=12+12$, $5=22+12$, $13=32+22$, $17=42+12$, $29=52+22$, $37=62+12$, $41=52+42$, $53=72+22$, $61=62+52$, $73=82+32$, $89=82+52$, $97=92+42$.
(b) $(a2+b2)(c2+d2)=(ac−bd)2+(ad+bc)2$.
Note: It is easy to check this identity once it is given, but most of us are not so fluent in algebra as to spot this handy identity without help! However, Chapter 1 is about “Mental skills”, and one such technique (once you have mastered it) arises from the arithmetic of complex numbers. If you have met complex numbers, then this identity can be written down immediately. Let us explain briefly how.
Every complex number $w=a+bi$ (where $i2=−1$) can be represented as a point in the complex plane with coordinates (a, b). The “size”, or modulus, of w is its length $|w|$ (the distance of (a, b) from the origin (0, 0)); and the square of this length $a2+b2$ is referred to as the norm of the complex number w=a+bi. The required identity is an immediate consequence of the two facts:
• the modulus of a product is equal to the product of the two moduli: $|wz|=|w|⋅|z|$, and
• the norm $a2+b2$ can be expressed algebraically as $a2+b2=(a+bi)(a−bi)$.
Once we know these facts:
• $a2+b2$ can be interpreted as the norm of w=a+bi, and
• $c2+d2$ as the norm of z = c + di;
the product of the two norms $(a2+b2)(c2+d2)$ is then equal to the norm of the product $w⋅z=(ac−bd)+(ad+bc)i$.
Note: If we choose z = c − di, then wz = (ac +bd) + (bc - ad)i, and we get a second identity: $(a2+b2)(c2+d2)=(ac+bd)2+(bc−ad)2$.
(c) The square $(2n)2$ of any even number 2n is a multiple of $22=4$.
Any odd number has the form 2n + 1; its square $(2n+1)2=4n2+4n+1$ is 1 more than a multiple of 4. So in the sum of two squares, we have
(i) both squares are even and their sum is a multiple of 4, or
(ii) one square is even and one is odd and their sum is of the form 4k + 1, or
(iii) both squares are odd and their sum is of the form 4k + 2.
Hence no number of the form 4k + 3 can be written as a sum of two squares.
(d) We are told that $2=12+12$, and that Euler showed every prime of the form 4k + 1 can be written as the sum of two squares. Part (b) then shows that any product of such primes can be written as the sum of two squares. And if we multiply a sum of two squares by a square, the result can again be written as the sum of two squares. This allows us to construct the list of forty six integers N < 100 which can be so written. These are precisely the integers of the form
“(a square) × (a product of distinct primes p, where p = 2 or p = 4k + 1)”:
0, 1, 2, 4, 5, 8, 9, 10, 13, 16, 17, 18, 20, 25, 26, 29, 32, 34, 36, 37, 40, 41, 45, 49,
50, 52, 53, 58, 60, 61, 64, 65, 68, 72, 73, 74, 80, 81, 82, 83, 85, 87, 89, 90, 97, 98.
(e) The side of such a square is the hypotenuse of a right angled triangle whose legs run in the x- and y- directions, and have integer lengths (because their vertices are at points with integer coordinates). Hence the answer is exactly the same as for part (d) (provided one does not quibble about the idea of a square with side 0 and area 0).
26. Most sheets in a newspaper are double sheets with four pages. If we assume that all sheets are double sheets, then the 13 pages before page 14 match up with the 13 pages after page 27, so there are 27 + 13 = 40 pages in all. (If the paper included inserted ‘single sheets’ — with just two pages, then there is no solution.)
27.
(a) If θ = 90°, then the overlap is clearly one quarter of the small square. In general, the continuations of the sides BA and DA cut the small square into four congruent quadrilaterals, one of which is the area of overlap. So the overlap is always one quarter of the lower square.
(b) The area of overlap for “a large equilateral ΔABC on top of a small equilateral ΔPQR” is not constant, but depends on the angle of orientation of the large triangle. However, if viewed in the right way, something similar works for a large regular 2n-gon on top of a small regular 2n-gon with one corner of the large polygon at the centre of the small one.
The key is to realise how the fraction “one quarter” arises for a regular 4-gon. There 2n = 4, so n = 2, and each vertex angle is equal to $360°2n(n−1)=90°$, which is exactly $n−12n=14$ of $360°$. For a regular 2n-gon, the large polygon always covers a fraction equal to exactly $n−12n$ of the small polygon.
28. 3 cm, the same as the perimeter of triangle ABC.2
What if A were folded to some point A'' on BC?
29. In extremis one may reach for trigonometry: if we denote the three angles by α (at A), β (at B), and γ (at C), then the arrangement of squares implies that $tanα=13$, $tanβ=12$, and $tanγ=1$, so we can use the standard identity
$tan(α+β)=tanα+tanβ1−tanαtanβ$
to see that $tan(α+β)=1=tanγ$.
2 From: Y. Wu, The examination system in China: the case of zhongkao mathematics. 12th International Congress on Mathematical Education. 8 July – 15 July, 2012, COEX, Seoul, Korea
However, it is worth looking for a more elementary explanation than ‘brute force calculation'. If we embed the horizontal 3 by 1 rectangle ADEH in the top right hand corner of a 4 by 4 square ZDXY, (with Z labelling the top left hand corner), then we can complete the square AEPQ, which has AE as one side, with P on side XY and Q on side YZ.
Then ∠AEH = ∠DAE, and ∠AEQ = ∠DCE. So all we need to explain is why ∠HEQ = ∠DBE — and this follows from the fact that EQ passes through the centre of the 4 by 4 square ZDXY.
30.
(a) Construct points P and Q inside the trapezium so that MNPQ is similar to BCMN. If the line through P parallel to AB meets BC at X, and the line through Q parallel to AC meets BC at Y, then MNPQ, NBXP, XYQP, MCYQ are the required pieces.
(b) Each of the four pieces must be three-quarters of one of the small squares. So we have to lose one quarter of each small square. There are various ways to do this, but most create non-congruent parts. Cut each of the smaller squares into quarters as for the original square. If O is the centre of the original square, lump together the three mini-squares which have O as a vertex to form an L-shape. Each of the three remaining small squares has lost a quarter and forms an identical L-shape.
31. To divide the shaded region in 2 congruent parts, rotate the lower small semicircle through the angle $π2$ anticlockwise about the centre of the large circle.
Note: The same idea allows one to divide the shaded region into n congruent parts: rotate the lower small semicircle successively through the angle $πn$ anticlockwise about the centre of the large circle.3
32.
(a) The point O lies on the perpendicular bisector of AB, and so is equidistant from the two endpoints A and B, so OA = OB. The point O also lies on the perpendicular bisector of BC, and so is equidistant from the two endpoints B and C, so OB = OC.
(b) (ii) ∠BA'C and ∠BAC are angles subtended in the same segment of the circle by the same chord BC, so are equal (“angles in the same segment”).
∠A'CB is the angle subtended on the circumference (at C) by the diameter A'B, and so must be a right angle. In ΔA'BC we then see that
$sinA=sinA′=a2R.$
If we now switch attention from the angle at A to the angle at B, and then to the angle at C, we can show that $sinB=b2R$, and that $sinC=c2R$.
3 From: Introductory Assignment, Gelfand Correspondence Program in Mathematics.
33.
(a) Drop a perpendicular from A to meet the line BC at X. Then $AX=b⋅sinC$, so
$Δ=12⋅(a×bsinC).$
(b) Substitute “$sinC=c2R$” (from the Sine Rule) into the formula in part (a).
34.
(a) (i) $α2π$; .
(ii) $4α$
(b) (i) $4β$; (ii) $4γ$
(c)(i) $4(α+β+γ)$
(ii) Triangle ABC and its sister triangle $A′B′C′$ are congruent, and each is covered 3 times.
(iii)
$\begin{array}{ccc}4\left(\alpha +\beta +\gamma \right)\hfill & =\hfill & \left(\text{total surface area of the unit sphere}\right)\hfill \\ \hfill & \hfill & \phantom{\rule{2.00em}{0ex}}+\left(4×\left(\text{area of the spherical triangle }ABC\right)\right)\hfill \\ \therefore area\left(△ABC\right)\hfill & =\hfill & \left(\alpha +\beta +\gamma \right)-\pi .\hfill \end{array}$
Note: In particular, the formula for the area of a spherical triangle implies:
• the angle sum $α+β+γ$ in any spherical triangle is always greater than $π$, and
• the larger the triangle ABC, the more its angle sum must exceed $π$.
35.
(a)
$\begin{array}{ccc}\mathrm{cos}\left(A+B\right)+i\mathrm{sin}\left(A+B\right)\hfill & =\hfill & {e}^{i\left(A+B\right)}\hfill \\ \hfill & =\hfill & {e}^{iA}\cdot {e}^{iB}\hfill \\ \hfill & =\hfill & \left(\mathrm{cos}A+i\mathrm{sin}A\right)\left(\mathrm{cos}B+i\mathrm{sin}B\right).\hfill \end{array}$
Hence
$sin(A+B)=sinAcosB+cosAsinB,$
and
$cos(A+B)=cosAcosB−sinAsinB.$
To reconstruct tan(A + B), divide these two expressions, and then divide numerator and denominator by “cos A cos B” to get
$tan(A+B)=sin(A+B)cos(A+B)=tanA+tanB1−tanAtanB.$
(b)
$\begin{array}{ccc}\mathrm{sin}X\hfill & =\hfill & \mathrm{sin}\left(A+B\right)\hfill \\ \hfill & =\hfill & \mathrm{sin}A\mathrm{cos}B+\mathrm{cos}A\mathrm{sin}B\hfill \\ \hfill & =\hfill & \mathrm{sin}\left(\genfrac{}{}{0.1ex}{}{X+Y}{2}\right)\mathrm{cos}\left(\genfrac{}{}{0.1ex}{}{X-Y}{2}\right)+\mathrm{cos}\left(\genfrac{}{}{0.1ex}{}{X+Y}{2}\right)\mathrm{sin}\left(\genfrac{}{}{0.1ex}{}{X-Y}{2}\right),\hfill \end{array}$
and
$\begin{array}{ccc}\mathrm{sin}Y\hfill & =\hfill & \mathrm{sin}\left(A-B\right)\hfill \\ \hfill & =\hfill & \mathrm{sin}A\mathrm{cos}B-\mathrm{cos}A\mathrm{sin}B\hfill \\ \hfill & =\hfill & \mathrm{sin}\left(\genfrac{}{}{0.1ex}{}{X+Y}{2}\right)\mathrm{cos}\left(\genfrac{}{}{0.1ex}{}{X-Y}{2}\right)-\mathrm{cos}\left(\genfrac{}{}{0.1ex}{}{X+Y}{2}\right)\mathrm{sin}\left(\genfrac{}{}{0.1ex}{}{X-Y}{2}\right),\hfill \end{array}$ $∴sinX+sinY=2sinX+Y2cosX−Y2.$
For sin X - sin Y, substitute “-Y” in place of Y to get:
$sinX−sinY=2sinX−Y2cosX+Y2.$
Similarly
$\begin{array}{ccc}\mathrm{cos}X+\mathrm{cos}Y\hfill & =\hfill & \mathrm{cos}\left(A+B\right)+\mathrm{cos}\left(A-B\right)\hfill \\ \hfill & =\hfill & \left(\mathrm{cos}A\mathrm{cos}B-\mathrm{sin}A\mathrm{sin}B\right)+\left(\mathrm{cos}A\mathrm{cos}B+\mathrm{sin}A\mathrm{sin}B\right)\hfill \\ \hfill & =\hfill & 2\mathrm{cos}A\mathrm{cos}B\hfill \\ \hfill & =\hfill & 2\mathrm{cos}\left(\genfrac{}{}{0.1ex}{}{X+Y}{2}\right)\mathrm{cos}\left(\genfrac{}{}{0.1ex}{}{X-Y}{2}\right)\hfill \end{array}$ $\begin{array}{ccc}\mathrm{cos}X-\mathrm{cos}Y\hfill & =\hfill & \mathrm{cos}\left(A+B\right)-\mathrm{cos}\left(A-B\right)\hfill \\ \hfill & =\hfill & \left(\mathrm{cos}A\mathrm{cos}B-\mathrm{sin}A\mathrm{sin}B\right)-\left(\mathrm{cos}A\mathrm{cos}B+\mathrm{sin}A\mathrm{sin}B\right)\hfill \\ \hfill & =\hfill & -2\mathrm{sin}A\mathrm{sin}B\hfill \\ \hfill & =\hfill & -2\mathrm{sin}\left(\genfrac{}{}{0.1ex}{}{X+Y}{2}\right)\mathrm{sin}\left(\genfrac{}{}{0.1ex}{}{X-Y}{2}\right).\hfill \end{array}$
(c)
(i) $sin(A+B)=sin90°=1$;
$sinAcosB+cosAsinB=122+322=1.$
(ii) $cosX−cosY=12−1=−12$;
$−2sinX+Y2sinX−Y2=−2sin230°=−12.$
(d) (i) $2sinAsinB=cos(A−B)−cos(A+B)$.
$\begin{array}{cc}\hfill & \therefore 2\mathrm{sin}A\mathrm{sin}B+2\mathrm{sin}C\mathrm{sin}D\hfill \\ \hfill & \phantom{\rule{2.00em}{0ex}}\phantom{\rule{2.00em}{0ex}}\phantom{\rule{2.00em}{0ex}}\phantom{\rule{2.00em}{0ex}}=\left[\mathrm{cos}\left(A-B\right)-\mathrm{cos}\left(A+B\right)\right]\hfill \\ \hfill & \phantom{\rule{2.00em}{0ex}}\phantom{\rule{2.00em}{0ex}}\phantom{\rule{2.00em}{0ex}}\phantom{\rule{2.00em}{0ex}}\phantom{\rule{2.00em}{0ex}}\phantom{\rule{2.00em}{0ex}}\phantom{\rule{2.00em}{0ex}}+\phantom{\rule{0.167em}{0ex}}\left[\mathrm{cos}\left(C-D\right)-\mathrm{cos}\left(C+D\right)\right]\hfill \\ \hfill & \phantom{\rule{2.00em}{0ex}}\phantom{\rule{2.00em}{0ex}}\phantom{\rule{2.00em}{0ex}}\phantom{\rule{2.00em}{0ex}}=\mathrm{cos}\left(A-B\right)+\mathrm{cos}\left(C-D\right)\hfill \\ \hfill & \phantom{\rule{2.00em}{0ex}}\phantom{\rule{2.00em}{0ex}}\phantom{\rule{2.00em}{0ex}}\phantom{\rule{2.00em}{0ex}}\phantom{\rule{2.00em}{0ex}}\phantom{\rule{2.00em}{0ex}}\phantom{\rule{2.00em}{0ex}}\left(\text{since}C+D=\pi -\left(A+B\right)\right)\hfill \\ \hfill & \phantom{\rule{2.00em}{0ex}}\phantom{\rule{2.00em}{0ex}}\phantom{\rule{2.00em}{0ex}}\phantom{\rule{2.00em}{0ex}}=2\mathrm{cos}\left(\genfrac{}{}{0.1ex}{}{A+C-\left(B+D\right)}{2}\right)\mathrm{cos}\left(\genfrac{}{}{0.1ex}{}{A+D-\left(B+C\right)}{2}\right)\hfill \\ \hfill & \phantom{\rule{2.00em}{0ex}}\phantom{\rule{2.00em}{0ex}}\phantom{\rule{2.00em}{0ex}}\phantom{\rule{2.00em}{0ex}}=2\mathrm{cos}\left(\genfrac{}{}{0.1ex}{}{\pi }{2}-\left(B+D\right)\right)\mathrm{cos}\left(\genfrac{}{}{0.1ex}{}{\pi }{2}-\left(B+C\right)\right)\hfill \\ \hfill & \phantom{\rule{2.00em}{0ex}}\phantom{\rule{2.00em}{0ex}}\phantom{\rule{2.00em}{0ex}}\phantom{\rule{2.00em}{0ex}}=2\mathrm{sin}\left(B+D\right)\mathrm{sin}\left(B+C\right).\hfill \end{array}$
Note: We can swap A and B without changing the expression “$sinAsinB+sinCsinD$”. Hence the same should be true of the RHS “sin(B + C)sin(B + D)”. Fortunately, since $A+B+C+D=π$, we know that sin(A + C) = sin(B + D), and sin(A + D) = sin(B + C).
(ii) In triangle WXY we see that $A+B+C+D=π$. Hence
$sinAsinB+sinCsinD=sin(A+D)sin(B+D).$
Now let R be the radius of the circle. Use “equality of angles in the same segment” and the Sine Rule (in its full form: see Problem 32) to write:
2Rsin A = XY, 2Rsin B = WZ, 2Rsin C = YZ, 2Rsin D = WX,
2Rsin(A+D) = WY, 2Rsin(B+D) = XZ.
$∴WX×YZ+WZ×XY=WY×XZ.$
36. Yes.
Let the perpendicular bisectors of AB and BC meet at the point O.
Then OA = OB and OB = OC, so the circle with centre O passing through A also goes through B and C.
We have to prove that this circle also passes through D, E, etc..
To do this we prove that ΔOBC ≡ ΔOCD.
We know that ΔOAB ≡ ΔOBC (by SSS-congruence: OA = OB and OB = OC by the construction of O; and AB = BC since both are sides of a regular n-gon). Moreover
$\begin{array}{ccc}\angle OAB\hfill & =\hfill & \angle OBA\text{(base angles of the isosceles triangle}△OAB\right)\hfill \\ \hfill & =\hfill & \angle OCB\text{(since}△OAB\equiv △OBC\right)\hfill \\ \hfill & =\hfill & \angle OBC\text{(base angles of the isosceles triangle}△OBC\right).\hfill \end{array}$
And ∠ABC = ∠BCD (angles of the same regular n-gon).
$∴$ ∠OCD = ∠BCD - ∠OCB = ∠ABC – ∠OBA = ∠OBC.
$∴$ ΔOBC ≡ △ OCD (by the SAS-congruence criterion).
Hence OC = OD.
Continuing in this way we can prove that OA = OD = OE, etc..
37.
(a) Yes. There are two nets for a regular tetrahedron. One of these consists of four equilateral triangles in a row (alternately right side up and right side down). In making the tetrahedron, the two sloping ends are glued together. So if we cut half a triangle from one end and stick it on the other end, we get a 4 by $3$ rectangle which folds round the tetrahedron exactly without any gaps or overlaps.
(b) The usual way to wrap a cube with edges of length 2 is to take a 4 by 8 rectangular piece of wrapping paper, to position the cube centrally on an edge of length 4(1 unit from each edge), and to wrap the paper to cover a circuit of four faces. The overlapping residue can then be folded down to cover each side face, with overlaps. Hence the ratio
“area of paper” : “surface area of cube” = 8 : 6.
The same ‘wastage rate’ can be achieved with a square $42$ by $42$ piece of paper. Position the cube centrally on the paper, but turned through an angle of 45°. Then fold the four corners of the paper up each of the four side faces (with folds to tuck in four ‘wasted’ isosceles right angled triangles — one in the middle of each edge of the paper, with total wasted area 8). Finally, the four isosceles right angled triangles at the four corners of the paper can be folded in to exactly cover the top face without further overlaps.
However, one can do significantly better if the paper can be folded back on itself. Take a 2 by 14 rectangle, and think of this as being marked into seven 2 by 2 squares. Place the cube to cover the central 2 by 2 square — leaving three 2 by 2 squares sticking out each side. Fold one 2 by 6 strip up to cover the top square, before folding back along a diagonal of the top square to reveal the inside of the paper and to cover half of the top square twice before folding down to cover one side square. Do the same with the other 2 by 6 strip, with the reverse fold along the diagonal of the top square resulting in the other half of the top square being covered twice, with the tail folding down to cover the other side square. Hence the ratio
“area of paper” : “surface area of cube” = 7 : 6.
(This lovely solution was provided by Julia Gog. We do not know whether one can do better.)
38.
(i) Yes. (Cut off a corner A say with a plane passing through the three neighbours of A.)
(ii) Yes. (Cut the cube parallel to a face.)
(iii) No. (Any cross-section of the cube is a polygon. Each edge of this polygonal cross-section is the line segment formed by the intersection of the cutting plane with one of the faces of the cube. Since the cube has just six faces, the cross-section can have at most six sides.)
(iv) Yes. (Let A and G be two opposite corners — so that AG passes through the centre O of the cube. Then the plane through O which is perpendicular to AG cuts the surface of the cube in a regular hexagon.)
(v) No. (It may not be clear how to prove this easily. It is perfectly possible to obtain a pentagonal cross-section by cutting with a plane that misses exactly one face. But if the cutting plane misses exactly one face, we can be sure that it must cut both faces belonging to some “opposite pair”; and these two faces are parallel, so the resulting edges of the cross-sectional pentagon are parallel. Hence the pentagon can never be regular.)
39.
(i) No; (ii) Yes; (iii) No; (iv) Yes; (v) No.
The 12 edges of the cube come in three groups of four — namely the four parallel edges in each of three directions.
Consider the four edges in one of these parallel groups. If the Sun's rays are parallel to these four edges, then each of these edges projects to a single vertex of the outline of the shadow — which is a projection of a square.
In all other cases two of the four parallel edges in the group give rise to shadows that form part of the boundary of the shadow polygon, while the other two edges project to the inside of the shadow (and so do not feature in the boundary of the shadow). Hence each of the three groups provides two edges to the boundary of the shadow polygon, and we obtain a hexagon.
To obtain a regular hexagon, align the Sun's rays parallel to the line AG joining two opposite corners A and G of the cube, and position the shadow plane perpendicular to this direction. The three edges at these two corners A and G then project to the inside of the shadow, while the six remaining edges project to a regular hexagon. (Since there are four body diagonals like AG, there are four ways to make such a projection. In each case, the six edges of the cube that project to the regular hexagon form a non-planar hexagon on the surface of the cube, that zig-zags its way round the polyhedron like a ‘wobbly equator’, turning alternately left and right each time it reaches a vertex. Such a closed circuit on a regular polyhedron is called a Petrie polygon — named after John Flinders Petrie (1907—1972), son of the famous Egyptologist Flinders Petrie). | 30,177 | 100,129 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 411, "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} | 2.65625 | 3 | CC-MAIN-2021-49 | longest | en | 0.961816 |
https://qsartoolbox.freshdesk.com/support/solutions/articles/16000069834-what-does-the-p-value-in-the-report-means- | 1,717,075,181,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971667627.93/warc/CC-MAIN-20240530114606-20240530144606-00214.warc.gz | 423,773,061 | 9,005 | When you perform a hypothesis test in statistics, a p-value helps you determine the significance of your results [D.J. Rumsey]. Hypothesis tests are used to test the validity of a claim that is made about a population. This claim that’s on trial, in essence, is called the null hypothesis.
The alternative hypothesis is the one you would believe if the null hypothesis is concluded to be untrue. The evidence in the trial is your data and the statistics that go along with it. All hypothesis tests ultimately use a p-value to weigh the strength of the evidence (what the data are telling you about the population). The p-value is a number between 0 and 1 and interpreted in the following way:
• A small p-value (typically ≤ 0.05) indicates strong evidence against the null hypothesis, so you reject the null hypothesis and state confidently that the alternative hypothesis is true.
• A large p-value (> 0.05) indicates weak evidence against the null hypothesis, so you fail to reject the null hypothesis. p-value bigger than the threshold does not mean that alternative hypothesis is wrong, it just shows that we cannot be sure about the conclusions we draw from the data.
• p-values very close to the cutoff (0.05) are considered to be marginal (could go either way).
For example, let we have 2 analogues similar to our chemical of interest and both of them show negative effect. We make the conclusion that all chemicals from this group (including our target) have some properties which explain the negative effect, so we infer that the target chemical is also negative. The null hypothesis states that our assumption is wrong and there is no relationship between the negative effect and the properties of chemicals in our group.
P-value is the probability for having 2 chemicals both with negative values due to pure chance. In this case it is 0.25 which is bigger than the typical threshold and because of this we cannot be sure if the picture we see is due to some relationship between the effect and the properties of chemicals in the group.
Now let’s consider another example, in which we have 8 analogues all of them having negative effect. In this case p-value is approx. 0.004, which is far below the threshold, so we are confident that the negative effect of chemicals is due to some property they possess, but not due to chance. | 483 | 2,345 | {"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} | 3.546875 | 4 | CC-MAIN-2024-22 | latest | en | 0.957527 |
https://mathematica.stackexchange.com/questions/222037/contracting-indices-in-lists | 1,713,345,997,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817146.37/warc/CC-MAIN-20240417075330-20240417105330-00880.warc.gz | 338,820,736 | 42,923 | # Contracting indices in lists
Using an answer to this question (Working with Dirac Gamma Matrices using FeynCalc - A simple problem), I defined
Clear[γ]
SetAttributes[γ, Listable]
γ[μ_] := If[μ == 0, KroneckerProduct @@ PauliMatrix[{1, μ}],
I KroneckerProduct @@ PauliMatrix[{2, μ}]
]
So γ is an object labelled by μ, which is an index whose value runs from 0 to 3 (0, 1, 2, 3). I would like to contract γ with γ, i.e.
,
where the summation over repeated indices is understood. (For physicists: I also defined γ with lower indices, which has the spatial components with a different sign, but this is not the point now).
The question is: how can I do this? I don't think I can use TensorProduct and/or TensorContract, since γ is not a tensor.
And what if I want to contract one or more indices of a γ-like object which is labelled by two indices or more?
P.S.: I do not want to use FeynCalc.
• Well, what a coincidence that the "answer" you mentioned was just given by me years ago :). May 18, 2020 at 5:01
index[_[x__]] := x (* <-- Extracts arguments from an expression. *)
muteIndexSum[list_] := Module[
{indices, repeatedIndices, result},
(*ALL indices: *)
indices = Table[index[gamma], {gamma, list}];
(*REPEATED indices: *)
repeatedIndices = {};
Do[If[Count[indices, i] == 2 && ContainsNone[repeatedIndices, {i}],
AppendTo[repeatedIndices, i]], {i, indices}];
(*Dot product of gamma matrices: *)
result = Apply[Dot, list];
Do[result = Sum[result /. i -> k, {k, 0, 3}], {i,
repeatedIndices}];
(* If there are no repeated indices,
this will just return the dot product of what you passed. *)
Return[result];
]
Since you have already used γ, I am using γm as a notation (using γ would mess things up):
muteIndexSum[{γm[μ], γm[ν], γm[μ]}]
Output:
γm[0].γm[ν].γm[0] + γm[1].γm[ν].γm[1] + γm[2].γm[ν].γm[2] + γm[3].γm[ν].γm[3]
If you contract all indices,
muteIndexSum[{γm[μ], γm[ν], γm[μ], γm[ν]}]
The output will be:
γm[0].γm[0].γm[0].γm[0] + γm[0].γm[1].γm[0].γm[1] + γm[0].γm[2].γm[0].γm[2] + γm[0].γm[3].γm[0].γm[3] + γm[1].γm[0].γm[1].γm[0] + γm[1].γm[1].γm[1].γm[1] + γm[1].γm[2].γm[1].γm[2] + γm[1].γm[3].γm[1].γm[3] + γm[2].γm[0].γm[2].γm[0] + γm[2].γm[1].γm[2].γm[1] + γm[2].γm[2].γm[2].γm[2] + γm[2].γm[3].γm[2].γm[3] + γm[3].γm[0].γm[3].γm[0] + γm[3].γm[1].γm[3].γm[1] + γm[3].γm[2].γm[3].γm[2] + γm[3].γm[3].γm[3].γm[3]
If you want to use your pre-defined γ matrices, just use the rule to substitute:
muteIndexSum[{γm[μ], γm[ν], γm[μ], γm[ν]}]/.{γm -> γ}
Output:
{{4, 0, 0, 0}, {0, 4, 0, 0}, {0, 0, 4, 0}, {0, 0, 0, 4}}
• Thank you, it works perfectly! I have a silly question although: if I define Clear[[Gamma]alti] SetAttributes[[Gamma]alti, Listable] [Gamma]alti[[Mu]_] := If[[Mu] == 0, KroneckerProduct @@ PauliMatrix[{1, [Mu]}], I KroneckerProduct @@ PauliMatrix[{2, [Mu]}]] and use for example muteIndexSum[{γm[μ], γm[ν], γm[μ], γm[ν]}]/.{γm -> γalti}, it does not work... How can the name be a problem?
– FB20
May 17, 2020 at 15:02
• Renaming should not cause problems. But sometimes Mathematica misbehaves. Try saving the .nb file, restarting Mathematica and evaluating the notebook. If changing the name is all you did, it should work just fine. May 17, 2020 at 15:10
• P.s.: If this solved your problem, you can mark it as the correct answer. May 17, 2020 at 15:31
• You were right, I restarted and it's all fine now. But I just noticed that your code doesn't work if I want to contract two different objects, like γalti and γbassi. I became aware of that since γ_mu γ^mu should be equal to 4I, when I is the 4x4 identity matrix. Any solution for that? In physics, when the metric is not euclidean, tensors with upper indices are not equal to tensors with lower indices. I'm not able to evaluate if for this other problem a new question is necessary.
– FB20
May 17, 2020 at 17:17
• *where I is the 4x4 identity matrix.
– FB20
May 17, 2020 at 17:28
What the shape of $$\gamma_\mu \gamma^\mu$$ do you expect? I am not sure whether the code below works to your satisfaction or not:
Total @ MapThread[Dot, {γ[Range[0, 3]], γ[Range[0, 3]]}]
Update
OK, let me make it in a more formal and natural way. First let some points made clear:
• Contraction of a pair of two same Greek indexes is accompanied with Minkowski metric signatured $$\mathrm{diag} g_{\mu\nu} = \{+, -, -, -\}$$ (g = DiagonalMatrix[SparseArray @ {1, -1, -1, -1}];);
• Contraction of a pair of two same Latin indexes is the same as that of a matrix dot product, or say the metric is just the identity matrix;
• Einstein convention is employed.
Then the quantity of interest in fact is
$$\gamma^\mu \gamma_\mu = (\gamma^\mu)^{mn} (\gamma_\mu)_{nl} = (\gamma^{^{\substack{\color{red}{1} \\ \mu}}})^{^{\substack{2\; \color{blue}{3} \\ mn}}} g_{_{\substack{\mu\nu \\ \color{red}{4}\,\color{cyan}{5}}}} (\gamma^{^{\substack{\color{cyan}{6} \\ \nu}}})_{_{\substack{nl \\ \color{blue}{7}8}}}$$
So it should work by using TensorProduct with TensorContract (pay attention to which two indexes are paired to contract):
γμ = γ[Range[0, 3]];
TensorContract[TensorProduct[γμ, g, γμ], {{1, 4}, {5, 6}, {3, 7}}]
In the same spirit $$\sigma^{\mu\nu}\sigma_{\nu\rho} = (\sigma^{\mu\nu})^{mn}\ g_{\nu\alpha}\ (\sigma^{\alpha\beta})_{nl}\ g_{\beta\rho} \\ \sigma^{\mu\nu}\sigma_{\mu\nu} = (\sigma^{\mu\nu})^{mn}\ g_{\mu\alpha}\ (\sigma^{\alpha\beta})_{nl}\ g_{\beta\nu}$$ should correspond to, respectively,
Clear[σ]
σ[μ_, ν_] := I (γ[μ].γ[ν] - γ[ν].γ[μ]) / 4
σμν = Outer[σ, Range[0, 3], Range[0, 3]] // SparseArray;
TensorContract[TensorProduct[σμν, g, σμν, g], {{2, 5}, {4, 9}, {6, 7}, {8, 11}}]
TensorContract[TensorProduct[σμν, g, σμν, g], {{1, 5}, {2, 12}, {4, 9}, {6, 7}, {8, 11}}]
Other contractions can be done similarly.
• Thank you for your answer! It works, but I have problem with two-indices objects. In my code, I also use: Clear[[Sigma]] SetAttributes[[Sigma], Listable] [Sigma][[Mu]_, [Nu]_] := I ([Gamma][[Mu]].[Gamma][[Nu]] - [Gamma][[Nu]].\ [Gamma][[Mu]])/4 Now what if I want to contract two indices of four, or every indices? It would be fantastic to use such a short and quick command like yours.
– FB20
May 17, 2020 at 14:55
• @FB20 Plz see my update. May 18, 2020 at 4:44
• Thanks a lot! I'm sorry to bother you again, but I have another doubt; take as an example your TensorContract[TensorProduct[σμν, g, σμν, g], {{2, 5}, {4, 9}, {6, 7}, {8, 11}}]. This is an object with two free (spacetime) indices: mu and rho. I would like to define a two-indices tensor which is precisely this quantity. I tried with the standard definition ":=" but I failed...
– FB20
May 18, 2020 at 20:39
• @FB20 Maybe you can use Part to extract from what is obtained there. May 19, 2020 at 6:30
• Ok I solved with a little trick but I have another problem ahaha. I made another question.
– FB20
May 19, 2020 at 10:22 | 2,454 | 6,820 | {"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": 4, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2024-18 | latest | en | 0.795475 |
readexiled.blogspot.com | 1,601,091,259,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400232211.54/warc/CC-MAIN-20200926004805-20200926034805-00174.warc.gz | 111,114,439 | 13,424 | ## Thursday, April 21, 2005
Fun With Math
Just like last year, my (would-be?) scientific calculations have determined the Top Ten College Hockey Programs Of All Time:
1 - Michigan (4034 points)
2 - North Dakota (3480)
3 - Denver (3122)
4 - Minnesota (3078)
5 - Boston University (2541)
6 - Wisconsin (2283)
7 - Boston College (1644)
8 - Michigan Tech (1585)
9 - Maine (1473)
10 - Lake Superior State (1329)
The only change in the rankings from last year is that Denver moves past Minnesota into third.
The calculations used were as follows:
1) Making it only to the NCAA semifinals and not to the title game gives points as follows: 1 point for 1948 semifinalists, 2 points for 1949 semifinalist, 3 points for 1950 semifinalists, and so on through 2005. More points are assigned the closer we get to the present under the What Have You Done For Me Lately Theorem.
2) Finishing second in the NCAAs gives points as follows: 116 points for 1948 runners-up, 117 points for 1949 runners-up, 118 points for 1950 runners-up, and so on through 2005. 116 points is given to the 1948 runners-up under the reasoning that finishing second is twice as great as being a semifinalist in 2005 (58 points.) The What Have You Done For Me Lately Theorem also applies here.
3) Winning the NCAA championship gives points as follows: 346 points for 1948 champions, 347 points for 1949 champions, 348 points for 1950 champions, and so on through 2005. 346 points is given to the 1948 champions under the reasoning that winning it all is twice as great as finishing second in 2005 (173 points.) The What Have You Done For Me Lately Theorem also applies here.
The Excel file I used to make these calculations is available for download here. Please feel free to comment or email me with comments or criticisms on my calculation process. | 458 | 1,819 | {"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} | 2.734375 | 3 | CC-MAIN-2020-40 | longest | en | 0.928048 |
https://math.stackexchange.com/questions/3366335/example-of-f-mathbbs-rightarrow-mathbbs-subset-mathbbr-mathbbs | 1,713,628,906,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817670.11/warc/CC-MAIN-20240420153103-20240420183103-00494.warc.gz | 345,215,257 | 36,879 | # Example of $f: \mathbb{S}\rightarrow \mathbb{S}\subset \mathbb{R},$ $\mathbb{S}$ closed with no fixed point
This is a textbook question.
"A function $$f: \mathbb{S}\rightarrow \mathbb{S}\subset \mathbb{R}$$ with $$\mathbb{S}$$ closed is called weakly contractive if \begin{align*} d(f(x), f(y)) Given an example of a weakly contractive mapping with no fixed point. Show that if $$\mathbb{S}$$ is compact, then $$f$$ has a fixed point. (Hint: Define $$g(x)=|f(x)-x|$$ and check its extreme value property on $$\mathbb{S}$$)"
Question?
Is the definition of weakly contractive correct/standard or should it be a $$\leq$$ symbol instead of $$<$$? (textbook, although good, does have some math typos)
IF it was $$\leq$$ then if $$\mathbb{S}=\mathbb{R}_+$$ and $$f(x)=x+c, c>0$$ could be a solution.
If it is indeed $$<$$ then I have no idea how to proceed. Any suggestions?
EDIT: perhaps $$f(x)=-1/x$$ could do it even with the strict inequality? (the idea here being exploiting $$\mathbb{S}$$ closed but not compact) Is that right? EDIT2: Just realized this does not work. $$f(x)=-1/x$$ is not defined at $$x=0$$. Making $$\mathbb{S}$$ not closed. EDIT3: There was a hint for the question, just added now.
No, the definition is correct. The point is that for the contraction mapping theorem to hold, it's not enough to say "points get closer when you map them under $$f$$". You need them to get closer by some factor $$\lambda<1$$ that does not depend on the points you plug in.
By the contraction mapping theorem, $$f$$ will have a fixed point if there exists $$\lambda<1$$ such that $$d(f(x),f(y))<\lambda d(x,y)$$ for all $$x,y \in S$$. So this tells you that no such $$\lambda$$ can exist for your $$f$$. In other words, for any $$\lambda<1$$, there exist $$x,y \in S$$ such that $$d(f(x),f(y)) \geq \lambda d(x,y)$$.
I think you'll find that Ross Millikan's function does the trick, since as $$x,y \to \infty$$, $$d(f(x),f(y)) \approx d(x,y)$$. To show that the inequality holds, suppose $$x, so $$e^{-x}>e^{-y}$$. Also note that by the choice of $$S$$, $$f$$ is increasing on $$S$$, so $$x+e^{-x}. Then: $$d(f(x),f(y)) = (y+e^{-y})-(x+e^{-x}) = (y-x)+ (e^{-y}-e^{-x})<(y-x)=d(x,y).$$
The adjective weakly would suggest that $$\le$$ is appropriate. I do not know what is standard. Your answer would be a good one in that case. Without the $$\le$$ you need to push things off to infinity like your $$+c$$ does, but still have a contraction. How about $$x \to x+e^{-x}$$ on $$x \ge 1$$?
• The "weak" refers to the fact that the RHS is just $d(x,y)$ and not $\lambda d(x,y)$ for some constant $\lambda<1$.
– kccu
Sep 23, 2019 at 2:59
• I was thinking about some sort of asymptotic function approaching f(x)=x from above. But that invalidates the inequality for small values of x, no? Sep 23, 2019 at 2:59
• @LucasMation: Not necessarily. That is why I chose $x \ge 1$ for $S$, so the derivative would be strictly less than $1$ in absolute value. Sep 23, 2019 at 3:00
• @LucasMation: No, at $x=1$ we have $f'(x)=1-\frac 1e$ Sep 23, 2019 at 3:11 | 971 | 3,055 | {"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": 43, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2024-18 | latest | en | 0.898732 |
http://www.slideserve.com/meira/more-with-the-normal-curves | 1,493,438,609,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917123270.78/warc/CC-MAIN-20170423031203-00465-ip-10-145-167-34.ec2.internal.warc.gz | 691,931,689 | 16,086 | More with the normal curves… - PowerPoint PPT Presentation
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More with the normal curves…. 6.3 Approximate Binomial Distribution & Test Hypotheses. about 68% of a data set lies in the range about 95% of a data set lies in the range almost all of a data set lies in the range. 34% of your data. 34% of your data. 13.5% of your data.
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
More with the normal curves…
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More with the normal curves…
6.3 Approximate Binomial Distribution & Test Hypotheses
about 68% of a data set lies in the range
about 95% of a data set lies in the range
almost all of a data set lies in the range
Normal Distribution:
A bell-shaped, normal curve that is symmetric about the mean; total area under the curve is 1
Bubye Penny?
23% of adults in the U.S. favor abolishing the penny and making the nickel the lowest denomination coin.
You are conducting a random survey of 500 adults. What is the probability that you will find at most 106 adults who favor abolishing the penny?
23% of adults in the U.S. favor abolishing the penny and making the nickel the lowest denomination coin.
You are conducting a random survey of 500 adults. What is the probability that you will find at most 106 adults who favor abolishing the penny?
WHOA! Way too much work!
• 23% of adults in the U.S. favor abolishing the penny and making the nickel the lowest denomination coin.
• You are conducting a random survey of 500 adults. What is the probability that you will find at most 106 adults who favor abolishing the penny?
n = trials p = probability
Normal Distribution
You try:
• Hyperopia is a condition that affects approximately 25% of the adult population in the U.S. Consider a random sample of 280 U.S. adults.
a.) What is the probability that 63 or more people are farsighted?
b.) What is the probability that fewer than 84 people are farsighted?
c.) What is the probability that at most 65 people are farsighted?
Accel math 2 period 6
Rejection?
• If you test a claim (a hypothesis) by collecting your own data, and if the probability of the event occurring is less than 5% then you should reject the original claim.
Do you trust everything you read?
• A recent Harris Poll claimed that 44% of adults cut back on their spending in order to pay for increased gas prices.
• To test this, you surveyed (hypothetically) 60 adults and found 21 of them cut back on spending in order to pay for gas.
• Should you reject the Harris Poll’s findings?
You try…
• Rework the example if 18 people out of the 60 adults cut back on their spending. | 751 | 3,095 | {"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} | 4.0625 | 4 | CC-MAIN-2017-17 | longest | en | 0.898525 |
https://www.instructables.com/Asymmetrical-Capacitor-Thrusters-the-Biefeld-Brow/ | 1,680,388,462,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296950363.89/warc/CC-MAIN-20230401221921-20230402011921-00657.warc.gz | 927,678,409 | 45,328 | # Asymmetrical Capacitor Thrusters: the Biefeld-Brown Effect
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## Introduction: Asymmetrical Capacitor Thrusters: the Biefeld-Brown Effect
Back in the 1920s, a young man named Thomas Townsend Brown discovered that if he charged a capacitor with high voltage direct current, the capacitor would exhibit thrust in the direction of the + positive electrode. Off and on for the rest of his life (he died in 1985) he worked on these devices, trying to make a practical propulsion device using this principle. While in college, Brown worked with physicist Paul Biefeld, and the professor's name was added to Brown's when the phenomenon was named "the Biefeld-Brown Effect" (hereafter "B-B").
Basically, a B-B device consists of a small or sharp electrode separated by dielectric material (an insulator, in other words) from a large or blunt electrode. In this form the device is a type of capacitor. When fed high voltage (20,000 volts) DC current, the capacitor will develop thrust leading from the small/sharp electrode and away from the large/blunt electrode. Brown maintained the device thrust toward the positive + pole and away from the negative - pole, but modern experiments have shown the B-B effect works in either polarity. What is important is the electrodes must be differently sized--hence an "asymmetrical capacitor."
A lot of people maintain the B-B effect is electrogravitic; that is, the electrical current acting on the capacitor somehow counters, or interferes with normal gravity. I do not believe this. The effect is clearly the result of excess ions being thrown off the small/sharp electrode, flowing toward the oppositely charged large/blunt electrode. In normal air, the ions pick and charge air ions, adding to the cascade and increasing the thrust. In a vacuum, or under insulating oil, asymmetrical capacitors can still exhibit thrust, though it is usually much weaker. This is because only the ions coming off the forward electrode are present without charged air ions to help push.
This Instructable will describe how to build twin rotary thrusters as a demonstration of the Biefeld-Brown effect. If you prefer to believe this is a form of anti-gravity, you are welcome to do so. My purpose is show you how to build the device, not debate Brown's electrogravitic theories.
New video as of 8/3/2012, showing new type rotors powered by relatively low voltage from a microwave oven transformer:
New video of the thrusters with a different power supply:
## Step 1: Parts You Will Need
Here is a list of the materials you will need to build the thrusters:
1 polyethylene cutting board, 14 x 17 inches by 3/8ths inch thick
other sizes and materials will work; wood will do if you coat it with insulating varnish
5 feet of 1/2 inch PVC tubing
ordinary hardware store plumbing pipe, white
2 1/2 inch PVC "T" joints
1 1/2 inch PVC 90 degree elbow
5 1/2 inch PVC end caps with flat ends
A lot of pipe caps come with rounded ends. Try to find squared-off, flat caps.
See http://www.creativeshelters.com/fittings/Display-All-PVCFittings.aspx for example
1 1/2 inch PVC plug
10 inches of 1.25 inch (about 40 mm) cardboard tubing
I got this from a roll of holiday wrapping paper. It should be as light -weight as possible without
compromising strength
2 12 oz. aluminum soft drink cans, any brand
2 aluminum discs, 40 mm diameter
Again, these should be as thin as possible to save weight, but be rigid too.
a few square inches of rubber sheet about the thickness of a coin
Foam rubber will do if it's dense
4 rubber appliance "feet" with 1 screw each
about 14 inches of 1/4 inch hardwood dowel
Poplar, basswood, etc.
6 feet of 22 gauge bell wire
6 8/32 brass bolts, each about 1 inch long
1 8/32 brass bolt, about 1.5 inches long
1 8/32 washer (brass, aluminum, or steel is OK)
3 knurled brass 8/32 nuts
3 8/32 brass hex nuts
shellac or polyurethane varnish
Super glue
aluminum HVAC tape
tube cutter, scissors, sharp knife, paper hole punch, a drill with 1/4, 1/16" and 8/32 bits
power supply--see separate page for info
## Step 2: The Capacitors
Asymmetrical capacitors are simple devices. First cut two lengths of 40mm cardboard tubing, 3.5 inches long. With a common hole punch, punch one hole in each tube as close to the mid-point as your punch can reach. It may not be at the balance point, but it won't matter.
Next coat the tubes thoroughly inside and out with shellac or polyurethane varnish. Let dry until completely hard.
Using the end of the tube closest to the punched hole, trace a circle around the tube on a piece of thin rubber or foam rubber. Cut this out carefully and glue over the open end of the tube nearest the punched hole.
Take an empty, clean, and dry soft drink can. With a sharp pair of scissors or an X-acto knife, cut off both ends. Cut the resulting tube of aluminum in a straight line along the long axis, making a single rectangle of aluminum. Flatten this out gently and draw two parallel lines 2.25 inches apart. Cut the aluminum along both lines. You'll end up with a rectangle of curled aluminum 2.25 inches long. Trim to 4.5 inches wide.
Wrap the aluminum around the cardboard tube snugly. Apply Super Glue to the inside edge of the metal cylinder. You can use clothes pins or other small clamps to keep the aluminum in place till the glue sets. When the glue is dry, cut a piece of aluminum duct tape 2.25 inches long and 1/2 an inch wide. Put this over the outside of the glued joint and smooth it down.
Repeat these steps to make a second thruster.
Take the two 40mm aluminum disks (I used ID tags, no hole). Glue these over the rubber end pieces with Super Glue. Center the disks carefully and allow to dry.
If your aluminum cylinders are carefully made, they should slip on over the cardboard tubes, yet be tight enough to stay in place.
The capacitors are done. Set them aside until later.
## Step 3: The Thruster Base and Frame
At the corners of your cutting board drill short pilot holes and install the four rubber appliance "feet". Be sure not to drill all the way through the board. If you do, fill the hole from the top with clear silicone sealant, otherwise you may get arcing to the exposed screw points.
Turn the board over. Using a ruler, find the exact center of the board. The simplest way to do this is to measure from corner to corner. Where the two lines intersect is the center of the board. Drill a 8/32 hole all the way through the board.
Chose two corners on opposite diagonals. Just inside from the rubber feet on each corner drill a 8/32 hole through the board. Take two 1/2 PVC flat headed pipe caps. Drill a 8/32 holes in the center of the cap (the center is often indicated by a small dot left over from the casting process). Put a 1 inch long 8/32 bolt upward through each hole in the cutting board. Push the bolts through the holes in the caps. The open end of the cap should face upward. Put a 8/32 nut on each bolt and tighten. Don't over tighten and crack the PVC.
Put a 1 inch 8/32 brass bolt through the hole in the center of the cutting board, pointing up. Put a washer over the bolt and screw on one inch-long threaded aluminum spacer. Tighten finger tight.
Cut two pieces of 1/2 inch PVC 9.5 inches long. Insert these in the pipe caps using a twisting motion. Don't use PVC cement! Next take your 90 degree elbow and two T joints. Cut two pieces of 1/2 inch tubing, each 8 inches long. Put the elbow on one, a T in the middle, and the other T at the other end. Take your 1/2 PVC plug and drill a 8/32 hole in the center of the plug. Insert a brass 1 inch 8/32 bolt down through the cap.
Cut about 15 inches of single strand 22 gauge bell wire. Strip about an inch of insulation off each end. Wrap one end tightly around the bolt inside the plug. Add a washer on the outside of the cap, then screw on your other aluminum spacer. The wire should be firmly attached to the bolt at this point. Feed the wire through the PVC tubing toward the second T joint. Insert the plug with metal parts into the center T joint. Press in until it fits tightly.
To the outside T joint fit a short length (about 1.25 inches) of 1/2 PVC tubing. Get a flat 1/2 inch pipe cap and drill the center for another 8/32 bolt. Thread a knurled 8/32 nut onto the 1.5 inch brass 8/32 bolts. Put the 8/32 bolt into the cap hole from the outside. Wrap the stripped end of the 22 gauge bell wire around the bolt. Fit a 8/32 nut and tighten the nut and knurled nut together, leaving about an inch of bolt protruding from the cap.
(The efficiency of the thruster can be increased by using better insulated, high voltage rated wire to connect the upper electrode to the top spindle connection. Where can you get HV wire? Salvage it from old microwave ovens--that's what I do.)
Set the frame and top tubing aside and proceed to the next step.
## Step 4: The Spindle
This is trickiest part of the assembly. I will give measurements from my model, but you may have to fiddle with them to get yours to fit and spin freely.
Cut a piece of 1/2 inch PVC tubing 6 inches long. By hand or with a drill press, drill a 1/4 hole at the exact middle of the tube. Be sure the hole is centered horizontally as well as longitudinally! The thrusters must be balanced in order to spin easily.
Drill 8/32 holes into two flat PVC 1/2 inch pipe caps. Take care these are centered as best as you can. Put 8/32 brass bolts through the holes from the inside and thread on a knurled brass nut on each. Leave loose for now.
These bolts will not spin in the 8/32 aluminum spacers as is. Using a file, or emery cloth, try to wear down the exposed threads on the bolts. You don't have to smooth them out completely, just ease them enough they will turn in the spacers without binding.
Cut your 1/4 hardwood dowel to 13 inches. Slip this through the hole in the spindle. Make sure it's centered on the spindle tube. It should be a snug slip-fit. If it's too loose, you may have to put a drop of Superglue on the joint after you center the dowel on the spindle.
Just below the top and bottom caps drill small (1/16th inch) holes on both sides of the spindle aligned with the dowel. Cut two lengths of 22 gauge bell wire 16 to 17 inches long. Strip all four ends 3/4 of an inch. In the very center of the wires, carefully strip off 1 inch of insulation, leaving bare wire in the middle. Make a loop in this bare section. Fit the loop over the head of one of the bolt in the pipe cap. Pull tight, then snug the knurled nut on the outside, trapping the wire against the cap. Repeat this process on the second cap. Feed the wire into the spindle and out again through the 1/16th holes, top and bottom. Press the caps in place.
Put a drop of glue or silicone cement on each end of the dowel. Carefully slip the dowels into the 1/4 holes punched in the thrusters. MAKE SURE THE THRUSTERS FACE THE RIGHT WAY--the flat discs of BOTH capacitors must be placed to "chase" each other around the spindle. If you put them one facing the other the thrust will cancel out and the device will not move. It doesn't matter if the device turns clockwise or counter-clockwise.
Push the dowels all the way through to the opposite side of the thruster and hold there until the glue sets. Make sure the thruster units are level! Thrust will be wasted if the capacitors are canted up or down.
When the spindle assembly is dry, insert the bottom smoothed bolt into the aluminum spacer mounted on the cutting board. Fit the top assembly to the two upright tubes, taking care to fit the top spacer over the smoothed brass bolt sticking out of the top of the spindle. Press the top assemble int place. Check that the spindle turns freely. If it doesn't, apply graphite (not oil!) to the spacers and adjust how much the top assembly presses down on the spindle.
Cut 3/4 inch squares of aluminum duct tape. The upper wire should reach the aluminum disk glued to the front of each capacitor. Tape the bare ends of the wire to the disks. The lower wire must be taped to the long aluminum cylinders at the rear of the thruster. Smooth the tape down firmly.
## Step 5: Power Supplies
The Biefeld-Brown Effect requires direct current. You may find sites on the Net or books that say you can use AC, but trust me, AC won't work. Neon sign transformers, oil burner ignition modules, etc. won't work unless rectified into DC. There is an easy DC power supply you can use.
Your best and easiest choice is an old TV set or computer monitor, the heavy, blocky kind with a cathode ray picture tube. These are usually called "CRT monitors." Computer monitors are often available for free as more and more people convert to flat LCD type screens. (I once called a person in my town who wanted to give away a free CRT monitor. She was so happy to get rid of it she delivered it to my house!)
CRTs contain high voltage capacitors that can hold a powerful charge for a long time! Be careful! If you pick up an old CRT, leave it unplugged for several days before you try to tinker with it. Even then, watch out! It takes very little effort to convert a CRT into a high voltage DC power supply, but BE CAREFUL!
You'll need a couple of cables, maybe 24 inches or more long. They must be able to handle high voltage. I salvage HV cable from microwave ovens, etc. but you can make your own by using heavy duty stranded 10 or 12 gauge house wire and slipping it into vinyl or silicone tubing. I pushed 10 gauge house wire into vinyl aquarium tubing.
Attach small alligator clips to each end of your HV cables.
Next, turn the CRT screen-down. You don't want to scratch or crack the picture tube (it may explode), so inverting the monitor on a pad or old towel is a good idea. Remove the CRT's rear cabinet. You might have to cut off the data cable first, as some monitors have the cable molded in so they can't be detached. You can also discard the pedestal if it is in the way.
Once the cabinet is off, BE CAREFUL. Look for a thick cable ending in a black suction cup. This will be attached to the picture tube somewhere. Gently pry the suction cup off. Inside will be connectors used to channel high voltage from the monitor power supply to the picture tube. This will be your positive (+) HV connection.
Look around the other side of the picture tube. You should see a strand of bare wire running around the front edge of the tube. This is your negative (-) or ground connection. Clip one alligator clip to the + suction cup connection. Take the other HV cable and clip one end to the - ground wire.
You can cut small holes in the plastic cabinet that allow you to feed your HV cables out. Once you have your cables through the cabinet, put it back on the monitor and screw it back in place.
To operate the asymmetrical capacitor thrusters, set them on a large table or cleared floor area. Put your upside down CRT as far away as your cables will allow. Connect the positive CRT cable to the brass bolt on the top of the thruster frame. Clip the negative cable to the spacer on the bottom of the spindle. Be sure the cables don't touch or interfere with the turning of the thrusters.
Before you power up, make sure the CRT switch is on. It's best if you can plug the CRT into a variac or other isolation transformer. This gives you control of the input voltage and a safe on/off switch. If you don't have a variac, try to use a wall plug controlled by a wall switch.
KEEP WELL BACK from the thrusters and monitor while it's turned on. A CRT power supply can produce 20-30,000 volts, depending on the size of the monitor. That much DC can really hurt you! It can also jump a surprisingly long way from exposed metal connections to you. So stay back!
OK, so you're plugged in. You're well back from the thrusters. Hit the switch! Now what happens?
## Step 6: Troubleshooting
If you've led a clean life and followed every direction in this Instructable, your Biefeld-Brown thrusters should work right off. When you power up, there's a slight delay, then you'll hear a hissing coming from the capacitors. The thrusters will start to turn, slowly at first, then with increasing speed until they reach the maximum potential of your device. The thrusters I built reach a top speed of 46 RPM when powered by an old Dell CRT I estimate is putting out 27,000 volts. Thrusters work best on a level surface, in dry, cool air.
Suppose you flip your switch and nothing happens? Switch off and check your connections. Most likely your CRT is not working. They can short out, and when they do, they're dead. Keep your cables and connections tight and widely separated at all times.
Suppose you get hissing, but no rotation? Your spindle is probably binding. Switch off the power and check the pressure of the frame top against the spindles. Try the brass spindle posts. Do they turn freely, or do the threads still bind? You may have to buff the threads down a little more. Lubricate the spacers with a little powdered graphite (usually sold as lock lubricant). Don't use oil! Oil is an insulator. It will interfere with the transfer of power through the brass screws.
Suppose you get rotation, but it is very slow? Your power supply may be weak. It takes better than 20,000 volts to get the thrusters moving. Also, if your power cables are too close to the thruster (or too close to each other) the spindle may be hampered by the escaping high voltage corona.
When I first built my thrusters, they would turn a half revolution and stop, still hissing. This happened because I ran the + connection from the base to the top commutator inside the PVC pipe, and the bell wire was leaking enough high voltage to create a corona that 'caught' the capacitors as they passed. To fix this, I removed the bell wire from the PVC upright and made my + connection at the top of the frame, as described in the current Instructable.
You may experience arcing between the front and back thruster electrodes. This kills thrust and may have weird side effects around your house. Every time I get a high voltage DC spark, it drives a digital clock/thermometer in my upstairs bathroom crazy. To cure arcing, try sliding the cylinder electrodes a little farther away from the disk electrodes. Don't separate them too much or you'll kill the B-B effect. If that doesn't help, get a large electronic resistor in the low megaohm range and attach it to the + connection at the top of the thruster frame. Clip your + lead from the CRT to the resistor. That ought to cure most arcing problems.
Here's the NASA report on asymmetrical capacitors: http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20040171929_2004178266.pdf
Good luck, and be safe.
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interesting this project, i will analyze it a little, i am not a scientist, i am a lover of technology,
The original Thomas Brown devices work on the basis of applying an oscillating high voltage/charge (not high DC voltage). While some ionisation of the air can occur and movement of the ions can contribute slightly to the movement, the aim here is to get the surface of the electrodes of the asymmetric capacitor to emit radiation. it is the radiation that we should be looking at and to determine exactly how much force is exerted on the charged electrodes. This radiation is independent of any ionised air as Mr Brown confirmed in a vacuum chamber in France and a non-ionising oil bath. The secret to more rapid acceleration of a practical nature is achieved through the surface area of the electrodes facing away, whereas the surfaces facing inwards and relatively close have their radiation emissions cancelled out through destructive interference (because of the opposite polarity of the electrodes). What matters here is the radiation being emitted on the outer surfaces facing away into space. The reason for having different sized electrodes for the capacitor is to change the surface area such that when you apply the same oscillating voltage on both electrodes, the charge density is different on those surfaces. A higher charge density on the "smaller surface area" electrode is directly controlling the energy density of the radiation, and when the energy density is higher, the radiation exerts a recoiling force that is stronger on that electrode compared to the opposite "larger" electrode.
There is supporting evidence for this interpretation. Mr Brown noted an unusual form of acceleration at a fixed high voltage and frequency of the oscillating voltage. He claimed he had to reduce the voltage to prevent his devices from flying away. In other words, his devices can increase in acceleration and do it in a non-linear way. To understand this observation, you must imagine yourself situated far away from the capacitor when in operation and see it as a point charge emitting radiation predominantly in one direction. Then you have a situation explained by the Abraham-Lorentx formula. This is the formula first developed in 1905 by Max Abraham to describe the radiation reaction force on the point charge by the emitting radiation. And according to that controversial formula of classical electrodynamics, the acceleration is "exponential" as the mathematical solution to the formula shows for both velocity and acceleration.
In other words, you must correctly apply the right voltage and at the right frequency to emit the radiation for the device to properly accelerate it on a practical level and where we should expect to apply a new technology.
It is a pity he (and many others online) have never realised this fact. But fear not, an experiment will be performed very soon to show this mathematical solution is in action within Mr Brown's electrokinetic device.
Sorry to give a negative dampening view ! But this effect has been studied for over a 100 years and turned out to be a dead end. Its just a leaky high voltage capacitor thats trying to 'equalize the charges on its plates' by utilizing air molecules (or charged ubipresent solar wind particles in space ) to discharge itself. There are far better ways to generate thrust. The effect is miniscule from thrust point of view to make it useful. I suspect NASA is working not on this effect but on thrusters using solar wind in particular.
Please don't misunderstand my point of view on this device. It's just a primitive ion motor, a reaction driven pinwheel. I've never thought it had anything to do with anti-gravity, etc. It's an interesting experiment; that's why I wrote this, so that others could try it themselves.
Hello, thanks for the fine instructable.
Following the announcement of MIT's success:
https://www.insidescience.org/news/no-propellers-n...
I read the Paper with interest but it does not go far enough.
I have become interested in the idea of increasing the thrust of the motor as a stand-alone entity but have not found any follow-up research on-line. My initial thoughts came from the development on vacuum tubes and by-pass jet motors: Would a concentric inner tube double the ion flow? Maybe an outer annular ring would help focus/constrict the ion drift.
Have you any thought on these ideas or further references I could use?
Regards.
Hi! I was curious as to how much current goes into the asymmetrical capacitor?
I can only guess 0.00A current because capacitors block current.
Capacitors can block current, but if there is current leakage then there can appear input current going into the capacitor which would just be current replacing the current leaking out of the capacitor.
I guess what I am trying to ask is, is there any current leakage happening and if so how much? A lot?
I have one more question if you don't mind. I see that a high voltage pulsed DC is utilized, will a regular HV DC that is not pulsed be work as well? Thank you in advance.
Hi! I was curious as to how much current goes into the asymmetrical capacitor?
I can only guess 0.00A current because capacitors block current.
Capacitors can block current, but if there is current leakage then there can appear input current going into the capacitor which would just be current replacing the current leaking out of the capacitor.
I guess what I am trying to ask is, is there any current leakage happening and if so how much? A lot?
I have one more question if you don't mind. I see that a high voltage pulsed DC is utilized, will a regular HV DC that is not pulsed work as well? Thank you in advance.
I'm not sure how to express what's happening in terms of current. ACTs use high voltage DC--pulsed or not--In the thousands of volts range. For example, if you search for some of my other videos, you'll see I'm using as voltage source everything from a hand cranked Wimshurst machine to a microwave oven transformer (with voltage multiplier). A Wimshurst probably generates current in the microamp range, but a MOT can put out 15, 20 amps? Also, because ACTs are designed to leak ions (that's where the thrust comes from) they're poor capacitors. Pretty much as soon as the power is shut off the power leaks from the sharp surfaces of the ACT and dissipates. You can get a shock if you touch one too soon, but with a discharge rod, such as are commonly, made to discharge Leyden jars, shocks from an ACT can be avoided.
Also.. http://www.gizmag.com/cannae-reactionless-drive-space-propulsion/33210/
Propulsion of the ACTs is certainly the result of ion wind. It's the result of ions being emitted from the lead electrode and being accelerated by the trailing one. If you watch my other video on Instructables (https://www.instructables.com/id/Biefeld-Brown-Elec... you will see the effect of streaming ions even when the electricity applied is insufficient to move the thrusters. That's what's happening in your video; the ACT vibrates a bit because of the high voltage DC coursing through it. This phenomenon has been studied over the decades by everyone from NASA to a French aerospace company. In a vacuum (or near vacuum, since perfect vacuums are hard to obtain) a thruster can still twitch a bit because it is throwing off ions still. The ACT in your video doesn't really move much because there are no air ions to provide 'wind' to assist pushing. I once showed by rotary ACT to a physics professor I know. He was puzzled how something as heavy as the thrusters could be propelled so fast by mere ions. He even did calculations to figure out how much thrust was needed, and the figure he got seemed far too high for my simple apparatus to achieve. What he forgot was the cascading effect of ions coming off the electrodes, ionizing the air molecules around them and being drawn back by the rear electrode. This is an important distinction. A simple device with a single electrode will not produce as much thrust as an ACT, even if the single electrode spews tons of ions. The ions from the device alone cannot propel it; it takes air to complete the effect. I learned this from a Brazilian expert, Antonio Carlos M. de Queiroz. He has a wonderful site here (http://www.coe.ufrj.br/~acmq/links.html).
There's no antigravity going on here. The B-B effect can be surprising, but it's not inexplicable.
@Mr.Apol; The video was good to watch; I am curious why you turned lights off for part of it (I was looking for green glows or something). The instructions were even better, especially learning what can be done with old CRT's. My favorite line from your Instructable was, "May have weird side effects around your house." Nothling like playing around with 20,000-30,000 volts at home :)
Cheers! :)
Site
I turned the lights out to see any corona, etc. from the thrusters. In fact there is quiet but noticeable discharge between the electrodes. Since sparking is bad for thrust, I'd like to eliminate it, but I haven't found away to do so entirely, yet.
Paul
Sparking can be eliminated by using resistors (rated for high voltage).
@Mr. A; Oh! And I forgot, there is a wonderful SciFi story from years ago where the government showed a scientist an alien antigravity machine and, after that, he struggled to finally make one, which was 1,000 times bigger and heavier. After he showed them that it worked, they admitted to him that they had faked the evidence. Because he (by fiat) believed that it could be done, he came up with a working prototype. Ah, the SciFi of a few decades ago...
Cheers! :)
Site
@Mr. Apol; Hi! OK I will look again to see the discharge. However, as you wrote, discharge between the electrode does create .. interesting effects.
*now is thinking of Bride of Frankenstein but that's for another post*
Cheers! :)
Site
There is some evidence thrusters work in vacuum, otherwise NASA would not have been interested in them. They apparently don't work as well as they do in air, since they lack the cascading effect caused by air ions joining the flow.
PBT
Contrary to popular belief NASA also puts quite some research efforts into conventional propulsion inside the atmosphere. And there is no cascading effect, there is simply the effect of ionizing a gas. Put one of these under a bell jar with no other metal parts in the neighbourhood with whatever system you wish to use to measure force. You'll find nothing. These things work by ionising a gas and accelerating it. If there is no gas in the immediate area and it still works that means you are either ionising your electrodes (very bad as they won't last long in that situation) or you're working with an electron beam but that's impossible or extremely unlikely for reasons I won't even get into explaining. If you don't believe me you might want to read up a bit more on how electrohydrodynamic thrusters actually work.
The French apparently had some success with thrust in a vacuum, but ultimately their tests came to nothing:
http://projetmontgolfier.info/INTRODUCTION.html
I believe the Bahnson series of experiments also claimed some thrust in a vacuum, but I'd have to check.
Certainly under the right conditions (not in this device!) ionising the electrodes could occur. Isn't this the basic principle of the ion enginem a substance like caesium as the ion fuel?
I use the term cascade in its ordinary meaning, "a connected series, as of amplifiers for an increase in output." I suppose a better term might be an avalanche, wherein a small disturbance grows larger by taking new portions with it. The language is merely meant to be descriptive, not definitive.
Don't get me wrong. This effect is caused by ions. As I said, I don't believe in anti-gravity.
PBT | 6,937 | 30,556 | {"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} | 2.625 | 3 | CC-MAIN-2023-14 | longest | en | 0.930062 |
http://slideplayer.com/slide/1557703/ | 1,516,662,415,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891543.65/warc/CC-MAIN-20180122213051-20180122233051-00628.warc.gz | 316,080,238 | 18,084 | # Here is an interesting and lovely way to look at the beauty of mathematics, and of God, the sum of all wonders. The Beauty of Mathematics Wonderful World.
## Presentation on theme: "Here is an interesting and lovely way to look at the beauty of mathematics, and of God, the sum of all wonders. The Beauty of Mathematics Wonderful World."— Presentation transcript:
Here is an interesting and lovely way to look at the beauty of mathematics, and of God, the sum of all wonders. The Beauty of Mathematics Wonderful World Press the spacebar to continue
1 x 8 + 1 = 9 12 x 8 + 2 = 98 123 x 8 + 3 = 987 1234 x 8 + 4 = 9876 12345 x 8 + 5 = 987 65 123456 x 8 + 6 = 987654 1234567 x 8 + 7 = 9876543 12345678 x 8 + 8 = 98765432 123456789 x 8 + 9 = 987654321
1 x 9 + 2 = 11 12 x 9 + 3 = 111 123 x 9 + 4 = 1111 1234 x 9 + 5 = 11111 12345 x 9 + 6 = 111111 123456 x 9 + 7 = 1111111 1234567 x 9 + 8 = 11111111 12345678 x 9 + 9 = 111111111 123456789 x 9 +10= 1111111111
9 x 9 + 7 = 88 98 x 9 + 6 = 888 987 x 9 + 5 = 8888 9876 x 9 + 4 = 88888 98765 x 9 + 3 = 888888 987654 x 9 + 2 = 8888888 9876543 x 9 + 1 = 88888888 98765432 x 9 + 0 = 888888888 Brilliant, isnt it?
1 x 1 = 1 11 x 11 = 121 111 x 111 = 12321 1111 x 1111 = 1234321 11111 x 11111 = 123454321 111111 x 111111 = 12345654321 1111111 x 1111111 = 1234567654321 11111111 x 11111111 = 123456787654321 111111111 x 111111111 = 12345678987654321 And look at this symmetry:
101% From a strictly mathematical viewpoint: What Equals 100%? What does it mean to give MORE than 100%? Ever wonder about those people who say they are giving more than 100%? We have all been in situations where someone wants you to GIVE OVER 100% How about ACHIEVING 101%? What equals 100% in life? Now, take a look at this…
Heres a little mathematical formula that might help Answer these questions:
If: A B C D E F G H I J K L M N O P Q R S T U V W X Y Z Is represented as: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26.
If: H-A-R-D-W-O-R- K 8+1+18+4+23+15+18+11 = 98% And: K-N-O-W-L-E-D-G-E 11+14+15+23+12+5+4+7+5 = 96%
But: A-T-T-I-T-U-D-E 1+20+20+9+20+21+4+5 = 100% THEN, look how far the love of God will take you: L-O-V-E-O-F-G-O-D 12+15+22+5+15+6+7+15+4 = 101%
Therefore, one can conclude with mathematical certainty that: While Hard Work and Knowledge will get you close, and Attitude will Get you there, Its the Love of God that will put you over the top!
Have a great day… and God bless you!
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https://bioinformatics.stackexchange.com/questions/3348/knnimputation-for-missing-data/3353 | 1,702,313,065,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679515260.97/warc/CC-MAIN-20231211143258-20231211173258-00357.warc.gz | 162,566,132 | 40,846 | # knnImputation for missing data
I have a dataframe containing missing values. I want to impute them with knnImputation as shown below.
Now my question is: are data imputed row wise or column wise and is there a way to give this option in knnImputation?
A <-c(1,2,3,4,5, NA, NA)
B <-c(2,3,NA, 2,4, 7, 6)
C <-c(2,3,4, NA, 5, 1,7)
D<-21:27
data<-data.frame(A,B, C, D)
new_data<-knnImputation(data, k = 3, scale = T, meth = "weighAvg", distData = NULL)
• According to the code it seems like by row scaling by column but I am not sure. But you can change the orientation of your matrix by transposing it with t . Another option is to test it with a symmetric (or quasi-symmetric) matrix
– llrs
Jan 20, 2018 at 10:36
## 1 Answer
The data are imputed using values from the same column
Here's a simplified example to show what's going on:
A <-c(1,2,3,4,5)
B <-c(2,3,NA, 2,4)
C <- c(2,3,4, 6, 1)
data<-data.frame(A,B,C)
data
#> A B C
#>1 1 2 2
#>2 2 3 3
#>3 3 NA 4
#>4 4 2 6
#>5 5 4 1
# I'm using scale = F to make things simpler
knnImputation(data, k = 3, scale = F, meth = "weighAvg", distData = NULL)
#> A B C
#>1 1 2.000000 2
#>2 2 3.000000 3
#>3 3 2.594272 4
#>4 4 2.000000 6
#>5 5 4.000000 1
# calculate euclidean distances, dropping column of interest
dist(data[,-2])
#>1 2 3 4
#>2 1.414214
#>3 2.828427 1.414214
#>4 5.000000 3.605551 2.236068
#>5 4.123106 3.605551 3.605551 5.099020
# use k rows with smallest distance to row of interest to calculate weighted average
# using exp(-dist) as weights
weight_sum <- weight_sum = exp(-2.828427) + exp(-1.414214) + exp(-2.236068)
2*exp(-2.828427)/weight_sum + 3*exp(-1.414214)/weight_sum + 2*exp(-2.236068)/weight_sum
#>2.594271
If you want to impute using values from the same row, all you have to do is transpose the matrix:
t(knnImputation(t(data), k = 2, scale = F, meth = "weighAvg", distData = NULL))
#> A B C
#>[1,] 1 2.000000 2
#>[2,] 2 3.000000 3
#>[3,] 3 3.086729 4
#>[4,] 4 2.000000 6
#>[5,] 5 4.000000 1
(I had to lower k because there are only 2 extra columns) | 836 | 2,084 | {"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} | 2.734375 | 3 | CC-MAIN-2023-50 | latest | en | 0.691597 |
https://www.education.com/worksheets/third-grade/arrays-and-the-commutative-property/ | 1,669,611,610,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710473.38/warc/CC-MAIN-20221128034307-20221128064307-00591.warc.gz | 798,905,705 | 48,514 | # Search Printable 3rd Grade Worksheets
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Practicing Commutative Property with Arrays
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Practicing Commutative Property with Arrays
Children explore the concept of commutative properties with the visual aid of arrays.
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Glossary: Foundations of the Commutative Property
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Glossary: Foundations of the Commutative Property
Use this glossary with the EL Support Lesson Plan: Foundations of the Commutative Property.
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Vocabulary Cards: Foundations of the Commutative Property
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Vocabulary Cards: Foundations of the Commutative Property
Use these vocabulary cards with the EL Support Lesson Plan: Foundations of the Commutative Property.
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Multiplication and the Commutative Property
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Multiplication and the Commutative Property
This worksheets teaches one of three multiplication properties, commutative, which means that you can multiply numbers in any order and get the same product.
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Properties of Multiplication: Commutative
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Properties of Multiplication: Commutative
This worksheet introduces children to the commutative property, then uses a series of prompts to help them practice it.
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Practice Problems: Parking Lot Multiplication
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Practice Problems: Parking Lot Multiplication
Children will enjoy using arrays to learn about the commutative property of multiplication in this playful hands-on worksheet.
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Parking Lot Multiplication Word Problems #1
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Parking Lot Multiplication Word Problems #1
Give your students a real-world visual of cars and parking lots as they get comfortable with arrays and the commutative property of multiplication.
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Multiplication and the Associative Property
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Multiplication and the Associative Property
The associative property means you can rearrange the numbers in a multiplication problem and still get the same answer. Students will practice applying this property in various forms and thinking critically about the important math concept.
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Comparing Arrays
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Comparing Arrays
Use this worksheet to get your students looking at arrays and noticing the similarities and differences.
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Multiplication and the Distributive Property
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Multiplication and the Distributive Property
This worksheet helps third graders practice their understanding of the distributive property of multiplication.
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Matching Arrays to Equations
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Matching Arrays to Equations
This worksheet gives your 3rd grader practice translating arrays into equations and vice versa. Have him match the equations to the visual arrays.
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Describing Arrays with Cars
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Describing Arrays with Cars
Understanding arrays can make multiplication much easier for your child. Practice identifying rows and columns in arrays with this multiplication worksheet.
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Multiple Choice Arrays
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Multiple Choice Arrays
In this worksheet, your 3rd grader will match arrays to the corresponding equation.
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Multiplication: Star Arrays
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Multiplication: Star Arrays
Children will use star arrays to solve 11 single-digit multiplication problems.
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Multiplication: Finding the Total with Arrays
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Multiplication: Finding the Total with Arrays
Familiarize yourself with arrays and practice using them to write multiplication sentences in this worksheet.
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Multiplication and the Distributive Property
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Multiplication and the Distributive Property
The distributive property means you can multiply a sum or difference by multiplying each number separately and then adding or subtracting the products. This worksheet will help your kids apply this important concept and set them up for algebraic success!
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Associative Property Multiplication
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Associative Property Multiplication
Your 3rd grader will see the associative property represented visually with this colorful area model multiplication practice worksheet.
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Distributive Property of Multiplication
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Distributive Property of Multiplication
In this worksheet, children learn what the distributive property is, and how to use it to solve multiplication problems.
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Break it Up: Distributive Property 1
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Break it Up: Distributive Property 1
The distributive property is a great tool to help with mental math and simplifying larger multiplication problems. Use this scaffolded resource with your students as an introduction to the distributive property.
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Glossary: Explore the Associative Property of Multiplication
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Glossary: Explore the Associative Property of Multiplication
Use this glossary with the EL Support Lesson Plan: Explore the Associative Property of Multiplication.
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Break it Up: Distributive Property 2
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Break it Up: Distributive Property 2
Support your students as they use the distributive property to help with mental math and simplifying larger multiplication problems. Use this resource with your young mathematicians as they gain proficiency with the distributive property.
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Distributive Property Drawings
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Distributive Property Drawings
This great resource will help your students see the visual side of multiplication! With this exercise, your young mathematicians will draw area models and use the distributive property to solve tough multiplication problems.
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Vocabulary Cards: Explore the Associative Property of Multiplication
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Vocabulary Cards: Explore the Associative Property of Multiplication
Use these vocabulary cards with the EL Support Lesson Plan: Explore the Associative Property of Multiplication.
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Drawing Small Arrays with Grids
Worksheet
Drawing Small Arrays with Grids
Drawing arrays in grids makes it easier to visualize the numbers in a multiplication fact. Give this worksheet to your 3rd grader to practice this tricky skill. | 1,179 | 6,089 | {"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} | 3.609375 | 4 | CC-MAIN-2022-49 | latest | en | 0.812449 |
https://www.netlib.org/scalapack/explore-html/da/d4d/_b_i__cvvamn_8c_source.html | 1,642,766,140,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303356.40/warc/CC-MAIN-20220121101528-20220121131528-00384.warc.gz | 943,781,047 | 3,297 | ScaLAPACK 2.1 2.1 ScaLAPACK: Scalable Linear Algebra PACKage
BI_cvvamn.c
Go to the documentation of this file.
1 #include "Bdef.h"
2 void BI_cvvamn(int N, char *vec1, char *vec2)
3 {
4 SCOMPLEX *v1=(SCOMPLEX*)vec1, *v2=(SCOMPLEX*)vec2;
5 float diff;
6 BI_DistType *dist1, *dist2;
7 int i, k;
8
9 k = N * sizeof(SCOMPLEX);
10 i = k % sizeof(BI_DistType);
11 if (i) k += sizeof(BI_DistType) - i;
12 dist1 = (BI_DistType *) &vec1[k];
13 dist2 = (BI_DistType *) &vec2[k];
14
15 for (k=0; k < N; k++)
16 {
17 diff = Cabs(v1[k]) - Cabs(v2[k]);
18 if (diff > 0)
19 {
20 v1[k].r = v2[k].r;
21 v1[k].i = v2[k].i;
22 dist1[k] = dist2[k];
23 }
24 else if (diff == 0)
25 {
26 if (dist1[k] > dist2[k])
27 {
28 v1[k].r = v2[k].r;
29 v1[k].i = v2[k].i;
30 dist1[k] = dist2[k];
31 }
32 }
33 }
34 }
SCOMPLEX::r
float r
Definition: Bdef.h:38
BI_cvvamn
void BI_cvvamn(int N, char *vec1, char *vec2)
Definition: BI_cvvamn.c:2
SCOMPLEX
Definition: Bdef.h:38
Bdef.h
Cabs
#define Cabs(z)
Definition: Bdef.h:270
BI_DistType
#define BI_DistType
Definition: Bdef.h:72
SCOMPLEX::i
float i
Definition: Bdef.h:38 | 494 | 1,112 | {"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} | 2.734375 | 3 | CC-MAIN-2022-05 | longest | en | 0.148958 |
https://www.perlmonks.org/?displaytype=print;node_id=1182083;replies=1 | 1,575,958,201,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540525821.56/warc/CC-MAIN-20191210041836-20191210065836-00250.warc.gz | 810,187,514 | 3,047 | http://www.perlmonks.org?node_id=1182083
Wasn't sure if this belongs here, or in Obfuscations.
Having once run:
```use strict;
use warnings;
use Storable;
my (%table);
foreach my \$i (0..9) {
foreach my \$j (\$i..9) {
\$table{\$i}{\$j} = \$table{\$j}{\$i} = \$i + \$j
}
}
```use strict;
use warnings;
use Storable;
my @problem = @ARGV;
my (%matrix);
foreach my \$number (@problem) {
my \$log = 0;
push @{\$matrix{\$log++}}, \$_ for reverse (split //, \$number);
}
my \$col = 0;
while (exists \$matrix{\$col}) {
my @column = @{\$matrix{\$col}};
my \$first = shift @column;
while(scalar @column > 0) {
my \$second = shift @column;
\$first = \$table{\$first}{\$second};
if (length(\$first) > 1) {
\$first = substr(\$first,-1,1);
push @{\$matrix{\$col + 1}}, 1;
}
}
\$matrix{\$col++} = \$first;
}
printf "%s",\$matrix{\$col - \$_} for (1..\$col);
print "\n";
```H:\perl>perl adder.pl 1
1
H:\perl>perl adder.pl 21 14 99 6 12
152
1000000000000000000000000000000000000000000
Addition tables for other number systems are left as an exercise for the (extremely bored) reader. Vaguely apropos of Multiply Hex values. I started to write a program to do multiplication and realized I needed to figure out how to add first.
But God demonstrates His own love toward us, in that while we were yet sinners, Christ died for us. Romans 5:8 (NASB)
Replies are listed 'Best First'.
by pryrt (Prior) on Feb 15, 2017 at 18:21 UTC
I had started down a similar path inspired by the same node, but focusing on hex -- and stopping abruptly when I got busy on more pressing (\$work-related) issues :-). I built the times and addition tables, and had started on my addition subroutine... but this is as far as I got. Posted for entertainment value only. :-)
```package HexMath;
my (%times, %sums);
BEGIN { # create tables
for my \$x ( 0 .. 15 ) {
my \$hx = sprintf '%x', \$x;
for my \$y ( 0 .. 15 ) {
my \$hy = sprintf '%x', \$y;
next if exists \$times{\$hx}{\$hy};
\$times{\$hx}{\$hy} = \$times{\$hy}{\$hx} = sprintf '%x', \$x*\$y;
\$sums{\$hx}{\$hy} = \$sums{\$hy}{\$hx} = sprintf '%x', \$x+\$y;
}
}
printf "\n%4s: ", 'x ';
printf "%4x ", \$_ for 0..15;
for my \$x ( 0 .. 15 ) {
my \$hx = sprintf '%x', \$x;
printf "\n%4.4s: ", \$hx;
for my \$y ( 0 .. 15 ) {
my \$hy = sprintf '%x', \$y;
printf '%4.4s ', \$times{\$hx}{\$hy};
}
}
print "\n";
printf "\n%4s: ", '+ ';
printf "%4x ", \$_ for 0..15;
for my \$x ( 0 .. 15 ) {
my \$hx = sprintf '%x', \$x;
printf "\n%4.4s: ", \$hx;
for my \$y ( 0 .. 15 ) {
my \$hy = sprintf '%x', \$y;
printf '%4.4s ', \$sums{\$hx}{\$hy};
}
}
print "\n";
}
while( defined(my \$addend = shift)) {
my \$c = 0;
for my \$hexit ( 1 .. length(\$addend) ) {
my \$x = lc substr(\$answer, -\$hexit, 1);
my \$y = lc substr(\$addend, -\$hexit, 1);
my \$s = \$sums{\$x}{\$y};
#my \$u =
printf "\n\t#%d: %s + %s = %s\n", \$hexit, \$x, \$y, \$s;
} | 1,006 | 2,863 | {"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} | 2.59375 | 3 | CC-MAIN-2019-51 | latest | en | 0.746327 |
https://www.nagwa.com/en/videos/402128401747/ | 1,721,725,943,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518029.81/warc/CC-MAIN-20240723072353-20240723102353-00215.warc.gz | 752,579,795 | 33,975 | Question Video: Using Trigonometric Values of Special Angles to Evaluate Trigonometric Expressions | Nagwa Question Video: Using Trigonometric Values of Special Angles to Evaluate Trigonometric Expressions | Nagwa
# Question Video: Using Trigonometric Values of Special Angles to Evaluate Trigonometric Expressions
Find the value of sin² 45 + cos² 45.
01:21
### Video Transcript
Find the value of sine squared 45 plus cos squared 45.
To help us find the value of sine squared 45 plus cos squared 45, we’re actually gonna use one of our Pythagorean trigonometric identities. I’m just gonna show you three of them and then I’ll show you the one that we’re actually gonna be using.
Okay, so the example of our Pythagorean trig identities are sine 𝜃 plus cos squared 𝜃 equals one, tan squared 𝜃 plus one equals sec squared 𝜃, and one plus cot squared 𝜃 is equal to cosec squared 𝜃.
And now to actually solve this problem, find the values of sine squared 45 plus cos squared 45, we’re actually gonna be using this top one, which is sine squared 𝜃 plus cos squared 𝜃 is equal to one. And this is actually sometimes known as the fundamental Pythagorean trigonometric identity.
So if you see how we’re going to use it, well actually the way we gonna use it is that our 𝜃 is gonna be equal to 45. Cause as you can see in the identity, we’ve got sine squared 𝜃 plus cos squared 𝜃 equals one, so still the same 𝜃 in each part, where in ours we’ve got 45 in each part.
So therefore, we can say that sine squared 45 plus cos squared 45 must be equal to one. And we’ve actually achieved that because we’ve used our top Pythagorean trigonometric identity, which is sin squared 𝜃 plus cos squared 𝜃 equals one.
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• Realistic Exam Questions | 467 | 1,908 | {"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} | 4.125 | 4 | CC-MAIN-2024-30 | latest | en | 0.819349 |
https://nrich.maths.org/public/leg.php?code=5039&cl=2&cldcmpid=1010 | 1,558,651,667,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257432.60/warc/CC-MAIN-20190523224154-20190524010154-00557.warc.gz | 573,083,449 | 9,426 | # Search by Topic
#### Resources tagged with Interactivities similar to Bike Ride:
Filter by: Content type:
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Challenge level:
### There are 215 results
Broad Topics > Information and Communications Technology > Interactivities
### Take the Right Angle
##### Age 7 to 11 Challenge Level:
How many times in twelve hours do the hands of a clock form a right angle? Use the interactivity to check your answers.
### Take Your Dog for a Walk
##### Age 7 to 11 Challenge Level:
Use the interactivity to move Mr Pearson and his dog. Can you move him so that the graph shows a curve?
### An Unhappy End
##### Age 11 to 14 Challenge Level:
Two engines, at opposite ends of a single track railway line, set off towards one another just as a fly, sitting on the front of one of the engines, sets off flying along the railway line...
### Do You Measure Up?
##### Age 7 to 11 Challenge Level:
A game for two or more players that uses a knowledge of measuring tools. Spin the spinner and identify which jobs can be done with the measuring tool shown.
### Up and Across
##### Age 11 to 14 Challenge Level:
Experiment with the interactivity of "rolling" regular polygons, and explore how the different positions of the red dot affects its vertical and horizontal movement at each stage.
### Speeding Up, Slowing Down
##### Age 11 to 14 Challenge Level:
Experiment with the interactivity of "rolling" regular polygons, and explore how the different positions of the red dot affects its speed at each stage.
### How Far Does it Move?
##### Age 11 to 14 Challenge Level:
Experiment with the interactivity of "rolling" regular polygons, and explore how the different positions of the red dot affects the distance it travels at each stage.
### You Tell the Story
##### Age 7 to 11 Challenge Level:
Can you create a story that would describe the movement of the man shown on these graphs? Use the interactivity to try out our ideas.
### Subtended Angles
##### Age 11 to 14 Challenge Level:
What is the relationship between the angle at the centre and the angles at the circumference, for angles which stand on the same arc? Can you prove it?
### How Random!
##### Age 7 to 11 Challenge Level:
Explore this interactivity and see if you can work out what it does. Could you use it to estimate the area of a shape?
### Dotty Circle
##### Age 7 to 11 Challenge Level:
Watch this film carefully. Can you find a general rule for explaining when the dot will be this same distance from the horizontal axis?
### More Transformations on a Pegboard
##### Age 7 to 11 Challenge Level:
Use the interactivity to find all the different right-angled triangles you can make by just moving one corner of the starting triangle.
### Add the Weights
##### Age 7 to 11 Challenge Level:
If you have only four weights, where could you place them in order to balance this equaliser?
### Right Angles
##### Age 11 to 14 Challenge Level:
Can you make a right-angled triangle on this peg-board by joining up three points round the edge?
### Triangles in Circles
##### Age 11 to 14 Challenge Level:
Can you find triangles on a 9-point circle? Can you work out their angles?
### Estimating Angles
##### Age 7 to 14 Challenge Level:
How good are you at estimating angles?
### World of Tan 25 - Pentominoes
##### Age 7 to 11 Challenge Level:
Can you fit the tangram pieces into the outlines of these people?
### World of Tan 21 - Almost There Now
##### Age 7 to 11 Challenge Level:
Can you fit the tangram pieces into the outlines of the lobster, yacht and cyclist?
### World of Tan 22 - an Appealing Stroll
##### Age 7 to 11 Challenge Level:
Can you fit the tangram pieces into the outline of the child walking home from school?
### World of Tan 24 - Clocks
##### Age 7 to 11 Challenge Level:
Can you fit the tangram pieces into the outlines of these clocks?
### Rolling Around
##### Age 11 to 14 Challenge Level:
A circle rolls around the outside edge of a square so that its circumference always touches the edge of the square. Can you describe the locus of the centre of the circle?
### Overlapping Circles
##### Age 7 to 11 Challenge Level:
What shaped overlaps can you make with two circles which are the same size? What shapes are 'left over'? What shapes can you make when the circles are different sizes?
### World of Tan 26 - Old Chestnut
##### Age 7 to 11 Challenge Level:
Can you fit the tangram pieces into the outline of this brazier for roasting chestnuts?
### Shapely Tiling
##### Age 7 to 11 Challenge Level:
Use the interactivity to make this Islamic star and cross design. Can you produce a tessellation of regular octagons with two different types of triangle?
### Colour Wheels
##### Age 7 to 11 Challenge Level:
Imagine a wheel with different markings painted on it at regular intervals. Can you predict the colour of the 18th mark? The 100th mark?
### World of Tan 27 - Sharing
##### Age 7 to 11 Challenge Level:
Can you fit the tangram pieces into the outline of Little Fung at the table?
##### Age 7 to 11 Challenge Level:
Three beads are threaded on a circular wire and are coloured either red or blue. Can you find all four different combinations?
### World of Tan 29 - the Telephone
##### Age 7 to 11 Challenge Level:
Can you fit the tangram pieces into the outline of this telephone?
### World of Tan 30 - Logical Thinking
##### Age 7 to 11 Challenge Level:
Can you logically construct these silhouettes using the tangram pieces?
### Counters
##### Age 7 to 11 Challenge Level:
Hover your mouse over the counters to see which ones will be removed. Click to remove them. The winner is the last one to remove a counter. How you can make sure you win?
### Chocolate Bars
##### Age 7 to 11 Challenge Level:
An interactive game to be played on your own or with friends. Imagine you are having a party. Each person takes it in turns to stand behind the chair where they will get the most chocolate.
### Carroll Diagrams
##### Age 5 to 11 Challenge Level:
Use the interactivities to fill in these Carroll diagrams. How do you know where to place the numbers?
### World of Tan 20 - Fractions
##### Age 7 to 11 Challenge Level:
Can you fit the tangram pieces into the outlines of the chairs?
### 100 Percent
##### Age 7 to 11 Challenge Level:
An interactive game for 1 person. You are given a rectangle with 50 squares on it. Roll the dice to get a percentage between 2 and 100. How many squares is this? Keep going until you get 100. . . .
### Coordinate Tan
##### Age 7 to 11 Challenge Level:
What are the coordinates of the coloured dots that mark out the tangram? Try changing the position of the origin. What happens to the coordinates now?
### Factor Lines
##### Age 7 to 14 Challenge Level:
Arrange the four number cards on the grid, according to the rules, to make a diagonal, vertical or horizontal line.
### Code Breaker
##### Age 7 to 11 Challenge Level:
This problem is based on a code using two different prime numbers less than 10. You'll need to multiply them together and shift the alphabet forwards by the result. Can you decipher the code?
### Noughts and Crosses
##### Age 7 to 11 Challenge Level:
A game for 2 people that everybody knows. You can play with a friend or online. If you play correctly you never lose!
### Makeover
##### Age 5 to 11 Challenge Level:
Exchange the positions of the two sets of counters in the least possible number of moves
### Building Stars
##### Age 7 to 11 Challenge Level:
An interactive activity for one to experiment with a tricky tessellation
### The Path of the Dice
##### Age 7 to 11 Challenge Level:
A game for 1 person. Can you work out how the dice must be rolled from the start position to the finish? Play on line.
### L-ateral Thinking
##### Age 5 to 11 Challenge Level:
Try this interactive strategy game for 2
### Diagonal Dodge
##### Age 7 to 14 Challenge Level:
A game for 2 players. Can be played online. One player has 1 red counter, the other has 4 blue. The red counter needs to reach the other side, and the blue needs to trap the red.
### Icosian Game
##### Age 11 to 14 Challenge Level:
This problem is about investigating whether it is possible to start at one vertex of a platonic solid and visit every other vertex once only returning to the vertex you started at.
### Calculator Bingo
##### Age 7 to 11 Challenge Level:
A game to be played against the computer, or in groups. Pick a 7-digit number. A random digit is generated. What must you subract to remove the digit from your number? the first to zero wins.
### Coordinate Cunning
##### Age 7 to 11 Challenge Level:
A game for two people that can be played with pencils and paper. Combine your knowledge of coordinates with some strategic thinking.
### Counter Roundup
##### Age 7 to 11 Challenge Level:
A game for 1 or 2 people. Use the interactive version, or play with friends. Try to round up as many counters as possible.
### Khun Phaen Escapes to Freedom
##### Age 11 to 14 Challenge Level:
Slide the pieces to move Khun Phaen past all the guards into the position on the right from which he can escape to freedom.
### Penta Play
##### Age 7 to 11 Challenge Level:
A shape and space game for 2,3 or 4 players. Be the last person to be able to place a pentomino piece on the playing board. Play with card, or on the computer.
### Online
##### Age 7 to 14 Challenge Level:
A game for 2 players that can be played online. Players take it in turns to select a word from the 9 words given. The aim is to select all the occurrences of the same letter. | 2,214 | 9,613 | {"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} | 3.5 | 4 | CC-MAIN-2019-22 | latest | en | 0.866854 |
https://soimg.org/which-statement-is-true-about-line-h/ | 1,638,801,272,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363301.3/warc/CC-MAIN-20211206133552-20211206163552-00244.warc.gz | 609,803,966 | 5,266 | 9
01:4The diagram mirrors a number of planes, lines, and points.Which statement is true about line h?Line h intersects line fat 2 points, A and B.TLine h is the intersection of planes R and T.PLine h intersects plane P at allude C.Line h has points on planes R, P, and T.BhAIK09с C
The correct option is
(d) Line h has points on planes R, P, and also T.
You are watching: Which statement is true about line h?
Step-by-action explanation: A diagram is given via a number of planes, lines and points.
We are to choose the statement that is TRUE around the line h.
Option (a) is
"Line h intersects line f at 2 points, A and B".
From the diagram, we deserve to see that line h intersects line f at only one point, point B.
So, the provided statement is FALSE.
Option (b) is
"Line h is the intersection of planes R and also T".
We have the right to watch from the diagram that the intersection of the planes R and T is line g, not line h.
So, the offered statement is FALSE.
Option (c) is
"Line h intersects aircraft P at suggest C".
We deserve to watch that the line h intersects the aircraft P at allude B, not at allude C.
So, the provided statement is FALSE.
Option (d) is
Line h has actually points on planes R, P, and also T".
We have the right to watch that line h has point B, which lies on both planes R and P. Also, line h has actually a allude A that lies on airplane T.
So, line h has actually points that lie on the planes R, P and also T.
Thus, the given statement is TRUE around line h.
Option (d) is TRUE.
According via the diagram and utilizing the meanings of planes, lines, and also points, the statement true about line h is:
d. Line h has points on planes R, P, and also T.
Solution:
a. Line h intersects line f at point B, the the statement a. is false.
b. Line h is on airplane R, however is not in the intearea of 2 planes. Line g is in the interarea of planes R and T,, then statement b. is false.
c. Line l intersects airplane P at point C, then statement c. is false.
d. Line h has actually points on planes R, P, and also T, then statement d. is true.
According with the diagram and also making use of the definitions of planes, lines, and also points, the statement true about line h is:
d. Line h has points on planes R, P, and also T.
Solution:
a. Line h intersects line f at point B, the the statement a. is false.
b. Line h is on airplane R, yet is not in the interarea of two planes. Line g is in the interarea of planes R and T,, then statement b. is false.
c. Line l intersects plane P at suggest C, then statement c. is false.
d. Line h has points on planes R, P, and also T, then statement d. is true.
Step-by-step explanation:
answer: line h is the interarea of planes rand t.
i dont know it
step-by-step explanation:
Is D
Step-by-step explanation:
Ed 2020
line h is with a ascoceoles triangle
line h is with a ascoceoles triangle
Step-by-step explanation:
you watch that line h hregarding do with evrything so if you have actually a kind of perent simply multiply
The true statement is Line h has actually points on planes R, P, and T
Step-by-action explanation:
The remainder of the question is the attached figure
According to the graph, we will certainly check which option is true.
a. Line h intersects line f at two points, A and B (Wrong)
Because: h intersects line f at B only.
b. Line h is the intersection of planes R and also T (Wrong)
Because: g is the interarea of planes R and also T
c. Line h intersects aircraft P at allude C (Wrong)
Because: h intersects plane P at suggest B
d. Line h has points on planes R, P, and T (True)
Due to the fact that h has the allude B on the airplane P, h has the point A on the plane T
and also the points of h on the plane R
I think the answer is C...
See more: Einstein The Life Of A Genius By Jack Steinberg, Einstein: The Life Of A Genius: Isaacson, Walter
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Contents
## How does percentile dice work 5e?
There are actually two ten sided dice in a standard 7-dice D&D set. The faces of one are numbered 0 to 9, and the faces of the other go from 00 to 90 by tens. To make a ‘percentile’ roll, you roll both die and add them together, with one exception: a double zero roll is 100.
## Can you roll a 100 on percentile dice?
Yes, a d100 is the same as 2d10 with one as the percentile. A d100 goes 1–100, a d10 goes 0-9. Neither allows you to roll a 0, because of the way you count a percentile dice. (00 on the percentile and a 6 on the other dice forms 6, 00 on one and 0 on the other is 100, no option will result in 0.)
## What is percentile dice DND?
When rolling a D100 percentile dice, when the D10 results in a 10 (zero side up), DNDBeyond counts it as a zero instead of a 10. That means a dice roll of 00 and 0 would result in a “0” when the lowest number possible should be 01 and highest 100 (a 90 plus 10), as indicated by every D100 chart.
## Does 0 on d10 mean 10?
The die is a d10, and when such a die is rolled, the 0 is usually read as a 10. Some RPGs use it as a 0-9, but that’s more of the exception rather than the rule. The only time a 0 on a d10 is read as a 0 is when it’s rolled as a pair of d10s to get a d100, or percentile dice, where a “00” is 100.
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## How do you read a dice percentage?
Before rolling, choose one of the two dice to be your “tens” digit. Roll them, then arrange them in order (just like you were writing it down – tens to the left of ones). Read the number, that’s your answer. If you get a 0 in the tens spot, it means you rolled under 10.
## Why does a d10 have a 0?
The one with two digits is just telling you the tens digit, and the one with one digit is the ones digit. If you’re rolling only that die with 0, 10, 20, etc., then a 10 = a 1 and the 00 = a 10. 0 or 00 on a d10 just means the highest number.
## What is 00 0 on a d100?
Most people would say a 00 on the percentile and a 0 on the d10. BUT, given the facts I’ve just carefully laid out, that violates the rules you determine the rest of your rolls by. A 00-0 roll would, technically, be a straight up zero. Instead of a 1-100 scale, you’ve made yourself a 0-99 scale.
## How do you roll d%?
The numbers on the percent die represent themselves. 00 = 0 %, 90 = 90%, 60 = 60% and so on so on. We roll this die in addition to a D10, We take the number on the D10 and combine it with the percentage roll and that equals your percent total. Example: D% is rilled and a 30 comes up.
## When would you roll a d100?
Most uses of a d100 are for a DM
When a character mixes two potions together, you can roll on the Potion Miscibility table. If more than two are combined, roll again for each subsequent potion, combining the results.
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## What is 100 sided dice used for?
Zocchihedron is the trademark of a 100-sided die invented by Lou Zocchi, which debuted in 1985. Rather than being a polyhedron, it is more like a ball with 100 flattened planes. It is sometimes called “Zocchi’s Golfball”. Zocchihedra are designed to handle percentage rolls in games, particularly in role-playing games.
## How do you read dice rolls?
The first die represents the tens digit, and the second die the ones digit. For example, a roll of 1 followed by a roll of 5 will give a total of 15, while a roll of 3 followed by a roll of 6 will give a total of 36. The average result of the D66 is 38.5, and the standard deviation about 17.16. | 1,016 | 3,653 | {"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} | 3.90625 | 4 | CC-MAIN-2023-06 | latest | en | 0.913811 |
http://www.enotes.com/homework-help/what-m-1-m-2-1-m-1-2-226407 | 1,477,353,296,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719815.3/warc/CC-MAIN-20161020183839-00009-ip-10-171-6-4.ec2.internal.warc.gz | 434,003,707 | 10,282 | # What is m is 1/m^2 = 1/m - 1/2
hala718 | High School Teacher | (Level 1) Educator Emeritus
Posted on
Given the equation 1/m^2 = 1/m - 1/2.
We need to find the values of m that satisfies the equation.
Let us get rid of the denominator by multiplying by 2m^2.
==> 2m^2(1/m^2) = 2m^2(1/m) - 2m^2(1/2)
==> 2 = 2m - m^2
Now we will combine all terms on the left side.
==> m^2 - 2m + 2 = 0.
Now we will use the formula to solve for m.
==> m1= ( 2 + sqrt(4-8) / 2
= (2+sqrt-4)/2
= (2 + 2i)/2
= 1+ i
==> m2= 1-i
Then there are no real solution for m.
However, the equation has a complex solution.
==> m = { 1+i, 1-i}
giorgiana1976 | College Teacher | (Level 3) Valedictorian
Posted on
The values of m that verify the equation arethe roots of the equation.
1/m^2 = 1/m - 1/2
Since the LCD is 2m^2, we'll multiply by 2m^2 both sides.
2m^2/m^2 = 2m^2/m - 2m^2/2
We'll simplify and we'll get:
2 = 2m - m^2
We'll move all terms to one side:
m^2 - 2m + 2 = 0.
We'll re-write the equation:
m^2 - 2m + 1 + 1 = 0
(m-1)^2 = -1
We'll have square root both sides:
sqrt (m-1)^2 = sqrt -1
m-1 = -i
m1 = 1 - i
m-1 = +i
m2 = 1 + i
The solutions of the equation are :{1 - i ; 1 + i}. | 494 | 1,199 | {"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} | 4.375 | 4 | CC-MAIN-2016-44 | latest | en | 0.848433 |
https://www.assignmentexpert.com/homework-answers/physics/mechanics-relativity/question-23311 | 1,571,011,118,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986648343.8/warc/CC-MAIN-20191013221144-20191014004144-00086.warc.gz | 906,964,188 | 12,718 | 79 444
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# Answer to Question #23311 in Mechanics | Relativity for Gabriel
Question #23311
An elevator is moving upward at 1.20 m/s when it experiences an acceleration 0.33 m/s2 downward, over a distance of 0.69 m. What will its final velocity be?
The distance equal:
S = v*t - a*t^2/2
v - initial velocity
a - acceleration
t - time
t^2 - 2*v/a *t + 2*S/a = 0
t^2 - 7.2 t + 4.14 = 0
Solutions of this equation are:
t1& = 0.63 s &
t2 = 6.64 s
t1 - this is a time when elevator was moving upward
t2 - this is a time when elevator was moving downward
We need a time when elevator was moving upward, so t=t1 ->
v_f = v - a*t
v_f - final velocity
v_f = 1.2 - 0.63*0.33 = 1 m/s
Answer: final velocity will be 1 m/s
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https://hoursfinder.com/0-9-hours/350-watt-hours-in-amp-hours.html | 1,709,289,952,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475238.84/warc/CC-MAIN-20240301093751-20240301123751-00397.warc.gz | 307,782,270 | 5,422 | We collected information about 350 Watt Hours In Amp Hours for you. Follow the liks to find out everything about 350 Watt Hours In Amp Hours.
### Watt Hour to Amp Hour Conversion Calculator ...
https://electricscooterparts.com/watt-hour-to-amp-hour-conversion-calculator.html
Watt Hour to Amp Hour Conversion Calculator. This calculator converts a battery or battery packs Watt hours (Wh) rating into its Amp Hours rating. The Amp hour rating can be useful to know when purchasing new batteries that are rated on Amp hours. For individual batteries enter the Volts rating of the battery and the Watt hours rating of the ...
### Convert Amp hour to Watt hour (Ah to Wh)
https://www.convert-formula.com/ah-wh/
Convert Watt hour to Amp hour (Wh to Ah) Insert Watts-hour (Wh) and voltage (V) below and click on Calculate to obtain Amp-hours (Ah). Formula is (Wh)/ (V) = (Ah). For example, if you have a 10Wh battery rated at 5V, the current is 10Wh / 5V = 2Ah.
### Calculating watt-hours, amp-hours and battery sizing
https://www.alternate-energy.net/calculateamp02.html
Calculating watt-hours, amp-hours and battery sizing ... This is equal to 30 percent of the battery banks rated amp-hour capacity as expressed in amperes. Solar(pv) systems will usually not deliver more than a 10-20 C rate of charge. ... television 350 watts microwave oven 1200 watts air conditioner (12,000 btu) 3250 watts
### Battery Capacity Calculator
https://www.omnicalculator.com/other/battery-capacity
Jun 14, 2018 · If you want to convert between amp hours and watt-hours or find the C-rate of a battery, give this battery capacity calculator a try. It is a handy tool that helps you understand how much energy is stored in the battery that your smartphone or a drone runs on. Additionally, it provides you with a step-by-step instruction on how to calculate amp hours and watt-hours, so you will be able to ...Estimated Reading Time: 3 mins
### How to Calculate Battery Amp Hours - Deep Cycle Battery …
https://deepcyclebatterystore.com/calculate-battery-amp-hours/
Watt-hours = watts * hours / efficiency = 1250 / 0.85 = 1470 watt-hours. Since watts = amps * volts divide the watt hours by the voltage of the battery to get amp-hours of battery storage. Amp-hours (at 12 volts) = watt-hours / 12 volts = 1470 / 12 = 122.5 amp-hours. If you are using a different voltage battery the amp-hours will change by dividing it by the battery voltage you are using.Estimated Reading Time: 4 mins
### How Much Power Does a 350 Watt Solar Panel Produce ...
https://www.portablesolarexpert.com/how-much-power-does-a-350-watt-solar-panel-produce/
Jul 30, 2021 · If the solar panel reaches 350 watts two hours a day, we can assume an average of 297 to 315 watts for 5 hours, or whatever number of sunlight hours are available in your location. So let us say the solar panel has an average output of 315 watts or 90% of its peak rating. There are 6 hours …
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# NCERT solutions for class 10 Mathematics
## NCERT solutions for class 10 Mathematics
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## NCERT Solutions Chapter 13 Surface Areas and Volumes
From Class IX, you are familiar with some of the solids like cuboid, cone, cylinder, and
sphere (see Fig. 13.1). You have also learnt how to find their surface areas and volumes. You must have seen a truck with a container fitted on its back (see Fig. 13.2), carrying oil or water from one place to another. Is it in the shape of any of the four basic solids mentioned above? You may guess that it is made of a cylinder with two hemispheres as its ends.
Again, you may have seen an object like the one in Can you name it? A test tube, right! You would have used one in your science laboratory. This tube is also a combination of a cylinder and a hemisphere. Similarly, while travelling, you may have seen some big and beautiful buildings or monuments made up of a combination of solids mentioned above. If for some reason you wanted to find the surface areas, or volumes, or capacities of such objects, how would you do it? We cannot classify these under any of the solids you have already studied. In this chapter, you will see how to find surface areas and volumes of such objects.
## NCERT Solutions for Class 10th Mathematics
NCERT Solutions Class 10 Mathematics PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 10 Mathematics includes text book solutions from both part 1. NCERT Solutions for CBSE Class 10 Mathematics have total 15 chapters. Class 10 Mathematics ncert Solutions in pdf for free Download are given in this website. Ncert Mathematics class 10 solutions PDF and Mathematics ncert class 10 PDF solutions with latest modifications and as per the latest CBSE syllabus are only available in myCBSEguide.
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http://mathhelpforum.com/algebra/15161-strange-square-root-problem-print.html | 1,503,484,847,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886118195.43/warc/CC-MAIN-20170823094122-20170823114122-00631.warc.gz | 260,531,598 | 3,155 | # Strange Square Root Problem
• May 19th 2007, 06:19 PM
Scorpion
Strange Square Root Problem
Ok, I need some kind of algebraic method to:
- Determine which values of r give a real answer for t, and which values of r give a complex answer for t
http://i190.photobucket.com/albums/z...athscthing.jpg
Now i have been solving this easy enough using the quadratic formula
(let x= etc
then x^2=r+http://i190.photobucket.com/albums/z...athscthing.jpg(without the t=)
x^2-x=r
then x^2-x-r=0, solve using quadratic formula).
Now my teacher said that my method of trialling different values works, but isn't quite what he is after, he wants some kind of algebraic method to show which values produce a complex or real answer, like just shove in a value and it spits out an answer.
Thanks any help is appreciated
By the way - This is rather urgent.:)
• May 19th 2007, 10:26 PM
Scorpion
• May 20th 2007, 12:17 AM
CaptainBlack
Quote:
Originally Posted by Scorpion
Ok, I need some kind of algebraic method to:
- Determine which values of r give a real answer for t, and which values of r give a complex answer for t
http://i190.photobucket.com/albums/z...athscthing.jpg
Now i have been solving this easy enough using the quadratic formula
(let x= etc
then x^2=r+http://i190.photobucket.com/albums/z...athscthing.jpg(without the t=)
x^2-x=r
then x^2-x-r=0, solve using quadratic formula).
Now my teacher said that my method of trialling different values works, but isn't quite what he is after, he wants some kind of algebraic method to show which values produce a complex or real answer, like just shove in a value and it spits out an answer.
Thanks any help is appreciated
By the way - This is rather urgent.:)
I think this is the same as your method:
t = sqrt(r+ sqrt(r + ...))
so:
t^2 = r + sqrt(r + sqrt(r+..)) = r + t
so we have:
t^2 - t + r = 0
This has real roots for t if (and only if) the discriminant of this quadratic
greater or equal to zero.
In this case the discriminant is 1-4r, so we need:
1-4r>=0
or:
1>=4r
so we need:
r<=1/4.
RonL | 586 | 2,069 | {"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} | 3.890625 | 4 | CC-MAIN-2017-34 | longest | en | 0.894356 |
https://www.enotes.com/homework-help/solve-following-polynomial-inequalities-x3-5x2-2x-446850 | 1,521,818,852,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257648313.85/warc/CC-MAIN-20180323141827-20180323161827-00671.warc.gz | 783,592,532 | 9,999 | # Solve the following polynomial inequalities.x3 - 5x2 + 2x ≥ -8
ishpiro | Certified Educator
First, rewrite this inequality so that the polynomial function is compared to zero:
`x^3 - 5x^2 + 2x + 8 >=0`
To solve this ineqality, factor the polynomial on the left side. One can notice that -1 is the root of the polynomial because `(-1)^3 - 5*(-1)^2+2*(-1) + 8 = 0` , so x+ 1 is a factor of the polynomial. The rest of the factors can be found by synthetic division:
-1 | 1 - 5 2 8
-1 6 -8
------------------------
1 -6 8 0
The first three numbers of the last row are the coefficients of the second factor of the polynomial, `x^2 - 6x + 8` .
The polynomial on the left side is factored as
`(x+1)(x^2 - 6x + 8) = (x+1)(x-2)(x-4)`
To determine whether this expression is greater than 0, note that it will change from positive to negative (of vice versa) at the points -1, 2, and 4.
At the point x = 0, which is on interval (-1, 2), the expression is positive: (0+1)(0 - 2)(0 - 4) = 8
So the polynomial is
negative when x < - 1
positive on (-1, 2)
negative on (2, 4)
positive when x >4
Therefore the solution set of the given inequality is
(-1, 2), x > 4
or in interval notation
`(-1, 2)uu (4, oo).`
` ` | 416 | 1,237 | {"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} | 4.4375 | 4 | CC-MAIN-2018-13 | latest | en | 0.792128 |
https://regmsif.cf/page/2/ | 1,624,086,711,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487643703.56/warc/CC-MAIN-20210619051239-20210619081239-00088.warc.gz | 432,098,852 | 28,304 | ## Digital Image Processing Assignment III
• Gamma correction;
• Visibility enhancement;
• Histogram equalization.
### Step One: Define Some Functions
We add a function that generate a blank color image from a given image,
// generate a blank color image
void gen_color(BITMAP *bmImg, BITMAP *bmColor) {
bmColor->bmInfoSize = bmImg->bmInfoSize;
bmColor->bmInfo = (BITMAPINFO *) malloc(bmColor->bmInfoSize);
init_bmp(bmColor);
}
### Step Two: Gamma Correction
Gamma encoding of images is used to optimize the usage of bits when encoding an image, or bandwidth used to transport an image, by taking advantage of the non-linear manner in which humans perceive light and color. Gamma correction is a nonlinear operation used to encode and decode luminance values in video or still image systems. Gamma correction is, in the simplest cases, defined by the following power-law expression.
$$L_d=L_w^{\frac{1}{\gamma}}$$
void gamma_correct(BITMAP *bmImg, BITMAP *bmGamma, double gamma) {
gen_color(bmImg, bmGamma);
uint8_t *p = (uint8_t *) malloc(1 << 11);
for (int16_t i = 0; i < 256; ++i) {
p[i] = adjust(255 * pow(i / 255., 1 / gamma));
}
for (uint32_t h = 0; h < bmiHeader->biHeight; ++h) {
for (uint32_t w = 0; w < bmiHeader->biWidth; ++w) {
uint32_t pos = h * bmImg->bmBytesPerRow + w * bmImg->bmBytesPerPel;
bmGamma->bmData[pos] = p[bmImg->bmData[pos]];
bmGamma->bmData[pos + 1] = p[bmImg->bmData[pos + 1]];
bmGamma->bmData[pos + 2] = p[bmImg->bmData[pos + 2]];
}
}
free(p);
}
### Step Three: Visibility Enhancement
We use a logarithmic operator to adjust the pixel value,
$$L_d=\frac{\log(L_w+1)}{\log(L_{max}+1)}$$
where $L_d$ refers to display luminance, $L_w$ refers to original luminance, and $L_{max}$ is the maximal luminance in the original image.
This mapping function make sure that, no matter the dynamic range of the scene, the maximal luminance value will be 1 (white), and the others change smoothly.
void enhance(BITMAP *bmImg, BITMAP *bmEnhance) {
gen_color(bmImg, bmEnhance);
double *bmYUV = (double *) malloc(bmImg->bmBytesPerRow * bmiHeader->biHeight << 3);
double minD = 1, maxD = 0;
for (uint32_t h = 0; h < bmiHeader->biHeight; ++h) {
for (uint32_t w = 0; w < bmiHeader->biWidth; ++w) {
uint32_t pos = h * bmImg->bmBytesPerRow + w * bmImg->bmBytesPerPel;
uint8_t *B = &bmImg->bmData[pos];
uint8_t *G = &bmImg->bmData[pos + 1];
uint8_t *R = &bmImg->bmData[pos + 2];
// transform RGB into YUV
bmYUV[pos] = 0.299 * *R + 0.587 * *G + 0.114 * *B;
bmYUV[pos + 1] = -0.147 * *R - 0.289 * *G + 0.436 * *B;
bmYUV[pos + 2] = 0.615 * *R - 0.515 * *G - 0.100 * *B;
double D = log(bmYUV[pos] + 1) / log(256);
if (D < minD) {
minD = D;
}
if (D > maxD) {
maxD = D;
}
}
}
if (minD < maxD) {
for (uint32_t h = 0; h < bmiHeader->biHeight; ++h) {
for (uint32_t w = 0; w < bmiHeader->biWidth; ++w) {
uint32_t pos = h * bmImg->bmBytesPerRow + w * bmImg->bmBytesPerPel;
double *Y = &bmYUV[pos];
double *U = &bmYUV[pos + 1];
double *V = &bmYUV[pos + 2];
double D = log(*Y + 1) / log(256);
// histogram stretching
*Y = 255 * (D - minD) / (maxD - minD);
// transform YUV into RGB
bmEnhance->bmData[pos] = adjust(*Y + 2.032 * *U);
bmEnhance->bmData[pos + 1] = adjust(*Y - 0.395 * *U - 0.581 * *V);
bmEnhance->bmData[pos + 2] = adjust(*Y + 1.140 * *V);
}
}
}
free(bmYUV);
}
### Step Four: Histogram Equalization
Histogram equalization is a method in image processing of contrast adjustment using the image’s histogram. Let $p_i$ be the probability of an occurrence of a pixel of level $i$ in the image, then its new level $s_i=\sum_{k=0}^i p_k$. The image should be converted to Lab color space or HSL/HSV color space before equalization, so that the algorithm can be applied to the luminance or value channel without resulting in changes to the hue and saturation of the image.
void histeq(BITMAP *bmImg, BITMAP *bmHistEq) {
gen_color(bmImg, bmHistEq);
double *bmHSL = (double *) malloc(bmImg->bmBytesPerRow * bmiHeader->biHeight << 3);
for (uint32_t h = 0; h < bmiHeader->biHeight; ++h) {
for (uint32_t w = 0; w < bmiHeader->biWidth; ++w) {
uint32_t pos = h * bmImg->bmBytesPerRow + w * bmImg->bmBytesPerPel;
double B = bmImg->bmData[pos] / 255.;
double G = bmImg->bmData[pos + 1] / 255.;
double R = bmImg->bmData[pos + 2] / 255.;
double *H = &bmHSL[pos];
double *S = &bmHSL[pos + 1];
double *L = &bmHSL[pos + 2];
// transform RGB into HSL
double min = B < G ? B < R ? B : R : G < R ? G : R;
double max = B > G ? B > R ? B : R : G > R ? G : R;
if (min == max) {
*H = 0;
} else if (max == R) {
*H = 60 * (G - B) / (max - min);
} else if (max == G) {
*H = 60 * (2 + (B - R) / (max - min));
} else if (max == B) {
*H = 60 * (4 + (R - G) / (max - min));
}
if (*H < 0) {
*H += 360;
}
*L = (min + max) / 2;
if (min == 1 || max == 0) {
*S = 0;
} else {
*S = (max - *L) / (*L < 0.5 ? *L : 1 - *L);
}
}
}
// histogram equalization
double *p = (double *) malloc(1 << 11);
memset(p, 0, 1 << 11);
for (uint32_t h = 0; h < bmiHeader->biHeight; ++h) {
for (uint32_t w = 0; w < bmiHeader->biWidth; ++w) {
uint32_t pos = h * bmImg->bmBytesPerRow + w * bmImg->bmBytesPerPel;
}
}
for (int16_t i = 0; i < 256; ++i) {
if (i > 0) {
p[i] += p[i - 1];
}
}
for (uint32_t h = 0; h < bmiHeader->biHeight; ++h) {
for (uint32_t w = 0; w < bmiHeader->biWidth; ++w) {
uint32_t pos = h * bmImg->bmBytesPerRow + w * bmImg->bmBytesPerPel;
double *H = &bmHSL[pos];
double *S = &bmHSL[pos + 1];
double *L = &bmHSL[pos + 2];
// transform HSL into RGB
double C = (1 - fabs(*L * 2 - 1)) * *S;
double Hapos = *H / 60;
double X = C * (1 - fabs(fmod(Hapos, 2) - 1));
double R, G, B;
if (Hapos <= 1) {
R = C, G = X, B = 0;
} else if (Hapos <= 2) {
R = X, G = C, B = 0;
} else if (Hapos <= 3) {
R = 0, G = C, B = X;
} else if (Hapos <= 4) {
R = 0, G = X, B = C;
} else if (Hapos <= 5) {
R = X, G = 0, B = C;
} else {
R = C, G = 0, B = X;
}
double m = *L - C / 2;
bmHistEq->bmData[pos] = adjust((B + m) * 255);
bmHistEq->bmData[pos + 1] = adjust((G + m) * 255);
bmHistEq->bmData[pos + 2] = adjust((R + m) * 255);
}
}
free(p);
free(bmHSL);
}
## Digital Image Processing Assignment II
• Image binarization;
• Binary image erosion;
• Binary image dilation;
• Binary image opening;
• Binary image closing.
### Step One: Define Some Functions
Here we define two functions that generate a blank grayscale image from a given image,
// given bmHeader and bmInfo, initialize others
void init_bmp(BITMAP *bmImg) {
bmImg->bmData = (uint8_t *) malloc(bmImg->bmBytesPerRow * bmiHeader->biHeight);
}
// generate a blank grayscale image
void gen_gray(BITMAP *bmImg, BITMAP *bmGray) {
bmGray->bmInfo = (BITMAPINFO *) malloc(bmGray->bmInfoSize);
bmGray->bmInfo->bmiHeader.biBitCount = 8; // one byte per pixel
// initilize palette
for (int16_t i = 0; i < 256; ++i) {
rgb->rgbBlue = i;
rgb->rgbGreen = i;
rgb->rgbRed = i;
rgb->rgbReserved = 0;
}
init_bmp(bmGray);
// IMPORTANT: MODIFY byOffBits AND bfSize
}
and a function that change a color image into grayscale.
void color2gray(BITMAP *bmImg, BITMAP *bmGray) {
gen_gray(bmImg, bmGray);
uint8_t minY = 255, maxY = 0;
for (uint32_t h = 0; h < bmiHeader->biHeight; ++h) {
for (uint32_t w = 0; w < bmiHeader->biWidth; ++w) {
uint32_t pos = h * bmImg->bmBytesPerRow + w * bmImg->bmBytesPerPel;
uint8_t *B = &bmImg->bmData[pos];
uint8_t *G = &bmImg->bmData[pos + 1];
uint8_t *R = &bmImg->bmData[pos + 2];
uint8_t Y = adjust(0.299 * *R + 0.587 * *G + 0.114 * *B);
if (Y < minY) {
minY = Y;
}
if (Y > maxY) {
maxY = Y;
}
}
}
// rearrange gray indensity
for (uint32_t h = 0; h < bmiHeader->biHeight; ++h) {
for (uint32_t w = 0; w < bmiHeader->biWidth; ++w) {
uint32_t pos = h * bmImg->bmBytesPerRow + w * bmImg->bmBytesPerPel;
uint8_t *B = &bmImg->bmData[pos];
uint8_t *G = &bmImg->bmData[pos + 1];
uint8_t *R = &bmImg->bmData[pos + 2];
uint8_t Y = adjust(0.299 * *R + 0.587 * *G + 0.114 * *B);
uint32_t _pos = h * bmGray->bmBytesPerRow + w;
bmGray->bmData[_pos] = adjust(255. * (Y - minY) / (maxY - minY));
}
}
}
### Step Two: Change Gray to Binary
In order to binarize the image, we need determine a threshold, and change pixels whose grayscale is below the threshold to black, and the others to white. But how to find the optimal threshold? There’s an excellent algorithm called Otsu’s method, which, in brief, maximize inter-class variance.
void otsu_gray2binary(BITMAP *bmGray, BITMAP *bmBinary) {
gen_gray(bmGray, bmBinary);
uint8_t minG = 255, maxG = 0;
double *p = (double *) malloc(1 << 11);
memset(p, 0, 1 << 11);
for (uint32_t h = 0; h < bmiHeader->biHeight; ++h) {
for (uint32_t w = 0; w < bmiHeader->biWidth; ++w) {
uint32_t pos = h * bmGray->bmBytesPerRow + w;
++p[bmGray->bmData[pos]];
if (bmGray->bmData[pos] < minG) {
minG = bmGray->bmData[pos];
}
if (bmGray->bmData[pos] > maxG) {
maxG = bmGray->bmData[pos];
}
}
}
double muT = 0;
for (int16_t k = minG; k <= maxG; ++k) {
muT += k * p[k];
}
uint8_t threshold = 0;
double omegaK = 0, muK = 0, maxB = 0;
for (int16_t k = minG; k < maxG; ++k) {
omegaK += p[k];
muK += k * p[k];
double sigmaK = (muT * omegaK - muK) * (muT * omegaK - muK) / omegaK / (1 - omegaK);
if (maxB < sigmaK) {
maxB = sigmaK;
threshold = k;
}
}
free(p);
for (uint32_t h = 0; h < bmiHeader->biHeight; ++h) {
for (uint32_t w = 0; w < bmiHeader->biWidth; ++w) {
uint32_t pos = h * bmGray->bmBytesPerRow + w;
bmBinary->bmData[pos] = bmGray->bmData[pos] < threshold ? 0 : 255;
}
}
}
Otsu’s method sets the global threshold, but for some special image with different luminance, adaptive thresholding may produce better result. There are at least two ways to apply adaptive thresholding.
void pixel_adaptive_gray2binary(BITMAP *bmGray, BITMAP *bmBinary, uint32_t border, uint32_t offset) {
gen_gray(bmGray, bmBinary);
border >>= 1;
uint32_t *integral = (uint32_t *) malloc(bmGray->bmBytesPerRow * bmiHeader->biHeight << 2);
memset(integral, 0, bmGray->bmBytesPerRow * bmiHeader->biHeight * 4);
// get integral image
for (uint32_t h = 0; h < bmiHeader->biHeight; ++h) {
for (uint32_t w = 0; w < bmiHeader->biWidth; ++w) {
uint32_t pos1 = h * bmGray->bmBytesPerRow + w;
uint32_t pos2 = (h - 1) * bmGray->bmBytesPerRow + w;
uint32_t pos3 = h * bmGray->bmBytesPerRow + w - 1;
uint32_t pos4 = (h - 1) * bmGray->bmBytesPerRow + w - 1;
integral[pos1] = bmGray->bmData[pos1]
+ (h > 0 ? integral[pos2] : 0)
+ (w > 0 ? integral[pos3] : 0)
- (h > 0 && w > 0 ? integral[pos4] : 0);
}
}
for (uint32_t h = 0; h < bmiHeader->biHeight; ++h) {
for (uint32_t w = 0; w < bmiHeader->biWidth; ++w) {
uint32_t pos = h * bmGray->bmBytesPerRow + w;
// range checking
uint32_t x1 = h > border ? h - border : 0;
uint32_t x2 = h + border + 1 < bmiHeader->biHeight ? h + border + 1 : bmiHeader->biHeight;
uint32_t y1 = w > border ? w - border : 0;
uint32_t y2 = w + border + 1 < bmiHeader->biWidth ? w + border + 1 : bmiHeader->biWidth;
uint32_t pos1 = (x2 - 1) * bmGray->bmBytesPerRow + y2 - 1;
uint32_t pos2 = (x1 - 1) * bmGray->bmBytesPerRow + y2 - 1;
uint32_t pos3 = (x2 - 1) * bmGray->bmBytesPerRow + y1 - 1;
uint32_t pos4 = (x1 - 1) * bmGray->bmBytesPerRow + y1 - 1;
uint64_t cnt = (x2 - x1) * (y2 - y1);
uint64_t sum = integral[pos1]
- (x1 > 0 ? integral[pos2] : 0)
- (y1 > 0 ? integral[pos3] : 0)
+ (x1 > 0 && y1 > 0 ? integral[pos4] : 0);
bmBinary->bmData[pos] = bmGray->bmData[pos] * cnt * 100 < sum * (100 - offset) ? 0 : 255;
}
}
}
void block_adaptive_gray2binary(BITMAP *bmGray, BITMAP *bmBinary, uint32_t border, double step, uint32_t offset) {
gen_gray(bmGray, bmBinary);
uint32_t *integral = (uint32_t *) malloc(bmGray->bmBytesPerRow * bmiHeader->biHeight << 2);
memset(integral, 0, bmGray->bmBytesPerRow * bmiHeader->biHeight * 4);
// get integral image
for (uint32_t h = 0; h < bmiHeader->biHeight; ++h) {
for (uint32_t w = 0; w < bmiHeader->biWidth; ++w) {
uint32_t pos1 = h * bmGray->bmBytesPerRow + w;
uint32_t pos2 = (h - 1) * bmGray->bmBytesPerRow + w;
uint32_t pos3 = h * bmGray->bmBytesPerRow + w - 1;
uint32_t pos4 = (h - 1) * bmGray->bmBytesPerRow + w - 1;
integral[pos1] = bmGray->bmData[pos1]
+ (h > 0 ? integral[pos2] : 0)
+ (w > 0 ? integral[pos3] : 0)
- (h > 0 && w > 0 ? integral[pos4] : 0);
}
}
for (uint32_t h = 0; h < bmiHeader->biHeight; h += border * step) {
for (uint32_t w = 0; w < bmiHeader->biWidth; w += border * step) {
uint32_t pos = h * bmGray->bmBytesPerRow + w;
// range checking
uint32_t x1 = h;
uint32_t x2 = h + border + 1 < bmiHeader->biHeight ? h + border + 1 : bmiHeader->biHeight;
uint32_t y1 = w;
uint32_t y2 = w + border + 1 < bmiHeader->biWidth ? w + border + 1 : bmiHeader->biWidth;
uint32_t pos1 = (x2 - 1) * bmGray->bmBytesPerRow + y2 - 1;
uint32_t pos2 = (x1 - 1) * bmGray->bmBytesPerRow + y2 - 1;
uint32_t pos3 = (x2 - 1) * bmGray->bmBytesPerRow + y1 - 1;
uint32_t pos4 = (x1 - 1) * bmGray->bmBytesPerRow + y1 - 1;
uint64_t cnt = (x2 - x1) * (y2 - y1);
uint64_t sum = integral[pos1]
- (x1 > 0 ? integral[pos2] : 0)
- (y1 > 0 ? integral[pos3] : 0)
+ (x1 > 0 && y1 > 0 ? integral[pos4] : 0);
for (uint32_t _h = x1; _h < x2; ++_h) {
for (uint32_t _w = y1; _w < y2; ++_w) {
uint32_t _pos = _h * bmGray->bmBytesPerRow + _w;
// when overlapping, black has the priority
bmBinary->bmData[_pos] &= bmGray->bmData[_pos] * cnt * 100 < sum * (100 - offset) ? 0 : 255;
}
}
}
}
}
### Step Three: Erosion and Dilation
Let’s say that white is the foreground and black is the background. Assume you have a solid circle, and you can only place it on white pixels, which means the circle shouldn’t cover the black pixels. We make the available area where the center can be put white and the others black, and get a new image—the result of erosion.
void erode(BITMAP *bmBinary, BITMAP *bmErosion, uint32_t border) {
gen_gray(bmBinary, bmErosion);
border >>= 1;
for (uint32_t h = 0; h < bmiHeader->biHeight; ++h) {
for (uint32_t w = 0; w < bmiHeader->biWidth; ++w) {
uint32_t pos = h * bmBinary->bmBytesPerRow + w;
// range checking
uint32_t x1 = h > border ? h - border : 0;
uint32_t x2 = h + border + 1 < bmiHeader->biHeight ? h + border + 1 : bmiHeader->biHeight;
uint32_t y1 = w > border ? w - border : 0;
uint32_t y2 = w + border + 1 < bmiHeader->biWidth ? w + border + 1 : bmiHeader->biWidth;
uint8_t flag = 255;
for (uint32_t _h = x1; _h < x2; ++_h) {
for (uint32_t _w = y1; _w < y2; ++_w) {
uint32_t _pos = _h * bmBinary->bmBytesPerRow + _w;
flag &= bmBinary->bmData[_pos];
}
}
bmErosion->bmData[pos] = flag;
}
}
}
This time, you want to put the center of the circle on white pixels regardless of whether other points lie on black pixels. We make the area that can be covered by the circle white and the others black. This is what dilation do. Surprisingly, the code is very very similar with erosion.
void dilate(BITMAP *bmBinary, BITMAP *bmDilation, uint32_t border) {
gen_gray(bmBinary, bmDilation);
border >>= 1;
for (uint32_t h = 0; h < bmiHeader->biHeight; ++h) {
for (uint32_t w = 0; w < bmiHeader->biWidth; ++w) {
uint32_t pos = h * bmBinary->bmBytesPerRow + w;
// range checking
uint32_t x1 = h > border ? h - border : 0;
uint32_t x2 = h + border + 1 < bmiHeader->biHeight ? h + border + 1 : bmiHeader->biHeight;
uint32_t y1 = w > border ? w - border : 0;
uint32_t y2 = w + border + 1 < bmiHeader->biWidth ? w + border + 1 : bmiHeader->biWidth;
uint8_t flag = 0;
for (uint32_t _h = x1; _h < x2; ++_h) {
for (uint32_t _w = y1; _w < y2; ++_w) {
uint32_t _pos = _h * bmBinary->bmBytesPerRow + _w;
flag |= bmBinary->bmData[_pos];
}
}
bmDilation->bmData[pos] = flag;
}
}
}
### Step Four: Opening and Closing
In fact, opening operation and closing operation are just the combination of the two operations we mentioned above, the only difference is the order. Opening operation is an erosion followed by a dilation, while closing operation is the reverse.
Visually speaking, opening operation makes the gaps between black pixels disappeared, and closing operation makes the gaps between white pixels disappeared.
Note that these two operations are both idempotent, which means performing one of them multiple times is equivalent to performing it one time.
## Give Up Sometimes
The expression “Never, never give up” means to keep trying and never stop working for your goals. Do you agree or disagree with this statement? Use specific reasons and examples to support your answer.
Topic
Many people believe in the saying that we should never give up casually, which tells them to work incessantly for their goals. Admittedly, this spirit-lifting saying has given countless people a pat on the back, inspired them to strive for their long-held dreams and acquire prosperity in the end. But generally speaking, I suppose we don’t have to shy away from giving up. Sometimes we need to put aside our persistence and accept the abnegation.
First of all, the initial orientation matters. We can’t guarantee that the first step we take is towards the right direction, which means we can be heading the wrong without perceiving it. If so, it would be better to just turn around than to press on. There’s no need to sacrifice your success for simply senseless insistence. One example of this is an ancient idiom from China, narrating a man who intended to go south steered his carriage to the north regardless of the prompting from a passer-by. He had not realized the sheer meaninglessness of his action. Of course, he has not arrived his destination yet. Had he listened to others’ advice and turned back, he would have reached it.
Besides, we should pay heed to the progress. Not only are we likely to move forward to where we do not expect at the beginning, but also during the process, and thus adjusting ourselves dynamically should be attached significance to. In addition, even though we know clearly what we are doing and why we are doing it, we may lack the capacity to do it well, and the progress can fall short of our expectation. When we’re patently unsuitable for tackling certain type of things, insisting may not be the optimal option. Some classmates in my high school will be rather representative instances. As soon as they entered the school, each of them picked an Olympiad in a certain subject. Yet before long, they found themselves not very good at competition, thereby opting out, focusing on regular curriculum. It turned out ultimately that they obtained fairly high scores in the college entrance examination. We can derive from it the fact that giving up does not necessarily mean losing, but somehow means wining.
What’s more, the outcome may be perplexing. When it come out, it’s time to judge whether it is right or not. We can scarcely ensure that the result will always fulfil our anticipation, so when we receive an unexpected result which is against the basic facts, probably it should not be calculated in statistics. At this time, alleging that the result has no problem seems a bit stubborn. On the other hand, if the result does not correspond with what we used to deem it should be, the chances are what deceives you is not the result, but the acknowledgement. If we look back upon history of science and technology, there are innumerable examples, such as Galileo who doubted the previous law of free fall, Bohr who suspected the previous nuclear model, etc. All of these indicate that even the most seeming truth has the possibility to be broken someday. The interesting thing is that, the result and the truth were dropped alternately, which set the pace for the development of human civilization. We wouldn’t have today’s advancement if there were no giving up in the world.
The old saying “Never, never give up” will continue to exert its positive attitude and lift people up. However, I take the viewpoint that people sometimes need to give up what they think, have or believe. This is not persuading you to give up everything in your life, but encouraging you to give up in a more sensible way, in which you will better encounter what you love, accomplish what you dream, and make what you desire to be.
## Digital Image Processing Assignment I
• RGB->YUV;
• Color to gray: gray=Y in YUV color space;
• Rearrange gray intensity to lie between [0,255];
• Write a grayscale bmp;
• Change the luminance value Y;
• YUV->RGB;
• Write a color bmp.
### Step One: BMP File Structure
A common BMP file is comprised of four part: image file header, image information header, palette and image data.
The image file header is a struct whose length is 14 bytes. Here gives its definition,
typedef struct tagBITMAPFILEHEADER {
WORD bfType;
DWORD bfSize;
WORD bfReserved1;
WORD bfReserved2;
DWORD bfOffBits;
} BITMAPFILEHEADER;
and the explanation for every variable.
• bfType: must always be set to ‘BM’ to declare that this is a .bmp-file;
• bfSize: specifies the size of the file in bytes;
• bfReserved1: must always be set to zero;
• bfReserved2: must always be set to zero;
• bfOffBits: specifies the offset from the beginning of the file to the bitmap data.
The image information header is also a struct, while its length is 40 bytes. The definition
typedef struct tagBITMAPINFOHEADER {
DWORD biSize;
LONG biWidth;
LONG biHeight;
WORD biPlanes;
WORD biBitCount
DWORD biCompression;
DWORD biSizeImage;
LONG biXPelsPerMeter;
LONG biYPelsPerMeter;
DWORD biClrUsed;
DWORD biClrImportant;
} BITMAPINFOHEADER;
The explanation
• biSize: number of bytes to define BITMAPINFORHEADER structure;
• biWidth: image width (number of pixels);
• biHeight: image height (number of pixels), note that if it’s a positive number, the image is inverted, otherwise upright;
• biPlanes: number of planes, should always be 1;
• biBitCount: bits per pixel, which may be 1, 4, 8, 16, 24, 32;
• biCompression: compression type, only non-compression(BI_RGB) is discussed here;
• biSizeImage: image size with bytes, when biCompression is BI_RGB, biSizeImage is 0;
• biXPelsPerMeter: horizontal resolution, pixels per meter;
• biYPelsPerMeter: vertical resolution, pixels per meter;
• biClrUsed: number of color indices used in the bitmap, when it’s 0, all the palette items are used;
• biClrImportant: number of important color indices for image display, when it’s 0, all items are important.
The palette has a series of RGBQUADs, which is defined like this.
typedef struct tagRGBQUAD {
uint8_t rgbBlue;
uint8_t rgbGreen;
uint8_t rgbRed;
uint8_t rgbReserved;
} RGBQUAD;
Note that the order of the color is blue, green, and red, not the reverse. The number of RGBQUADs is decided by biBitCount and biClrUsed.
Next we need to define BITMAPINFO.
typedef struct tagBITMAPINFO {
} BITMAPINFO;
As we know nothing about the number of RGBQUADs an image uses, the BITMAPINFO should be defined as a pointer so that right amount of memory can be allocated to it.
The image data contains color of all pixels, and every biBitCount bit(s) represents a pixel.
At last, we define a struct to storage a full BMP image.
typedef struct tagBITMAP {
BITMAPINFO *bmInfo;
uint32_t bmInfoSize;
uint32_t bmBytesPerRow;
uint8_t bmBytesPerPel;
uint8_t *bmData;
} BITMAP;
When defining the two structs, we need to add a line #pragma pack(push, 1) to avoid struct padding.
### Step Two: Read/Write a BMP File
A BMP file is a binary file, so we need to add "b" to the second parameter when using fopen.
Another important thing is that the number of bytes in one row must always be adjusted to fit into the border of a multiple of four, and we need to calculate how many bytes are there in one row.
For convenience, we define an initialize function, which receives bmHeader and bmInfo, and initializes other variables.
// given bmHeader and bmInfo, initialize others
void init_bmp(BITMAP *bmImg) {
bmImg->bmData = (uint8_t *) malloc(bmImg->bmBytesPerRow * bmiHeader->biHeight);
}
The read function is showed below.
// read a BMP from file
void read_bmp(BITMAP *bmImg, char *filepath) {
FILE *fiInImg = fopen(filepath, "rb");
// if biBitCount is less than 16, use all the palette, otherwise do not use palette
} else {
}
bmImg->bmInfo = (BITMAPINFO *) malloc(bmImg->bmInfoSize);
}
init_bmp(bmImg);
fclose(fiInImg);
}
The write function looks similar with the read function.
// write a BMP to file
void write_bmp(BITMAP *bmImg, char *filepath) {
FILE *fiOutImg = fopen(filepath, "wb");
fwrite(bmImg->bmInfo, bmImg->bmInfoSize, 1, fiOutImg);
fclose(fiOutImg);
}
### Step Three: Change Color to Gray
To make things easier, we define a function to duplicate a BMP file,
// duplicate a BMP
void copy_bmp(BITMAP *bmDes, BITMAP *bmSrc) {
memcpy(bmDes, bmSrc, sizeof(BITMAP));
bmDes->bmInfo = (BITMAPINFO *) malloc(bmSrc->bmInfoSize);
memcpy(bmDes->bmInfo, bmSrc->bmInfo, bmSrc->bmInfoSize);
bmDes->bmData = (uint8_t *) malloc(bmSrc->bmBytesPerRow * bmiHeader->biHeight);
}
and a function to make sure the RGB value of a pixel lie between [0, 255].
// make RGB value legal
int16_t ret = (int16_t) (val + 0.5);
return ret < 0 ? 0 : ret > 255 ? 255 : ret;
}
We need to change RGB to YUV first, as the grayscale is determined by Y value. The formula is
$$\begin{bmatrix} 0.299 & 0.587 & 0.114 \\ -0.147 & -0.289 & 0.436 \\ 0.615 & -0.515 & -0.100 \end{bmatrix} \times \begin{bmatrix} R \\ G \\ B \end{bmatrix} = \begin{bmatrix} Y \\ U \\ V \end{bmatrix}$$
// calculate YUV value
double *bmYUV = (double *) malloc(sizeof(double) * bmImg.bmBytesPerRow * bmiHeader->biHeight);
for (uint32_t h = 0; h < bmiHeader->biHeight; ++h) {
for (uint32_t w = 0; w < bmiHeader->biWidth; ++w) {
uint32_t pos = h * bmImg.bmBytesPerRow + w * bmImg.bmBytesPerPel;
uint8_t *B = &bmImg.bmData[pos];
uint8_t *G = &bmImg.bmData[pos + 1];
uint8_t *R = &bmImg.bmData[pos + 2];
bmYUV[pos] = 0.299 * *R + 0.587 * *G + 0.114 * *B;
bmYUV[pos + 1] = -0.147 * *R - 0.289 * *G + 0.436 * *B;
bmYUV[pos + 2] = 0.615 * *R - 0.515 * *G - 0.100 * *B;
}
}
To change color to gray, we can simply make the R, G and B value of a pixel equal to its Y value. It would be more complex if we want to change it into an image with biBitCount equal to 8, because we need to set the palette manually.
One more step, rearrange gray intensity to lie between [0,255]. It’s just a math problem. So the code
// color to gray
BITMAP bmGray;
bmGray.bmInfo = (BITMAPINFO *) malloc(bmGray.bmInfoSize);
for (int i = 0; i < 256; ++i) {
rgb->rgbBlue = i;
rgb->rgbGreen = i;
rgb->rgbRed = i;
rgb->rgbReserved = 0;
}
init_bmp(&bmGray);
uint8_t min = 255, max = 0;
for (uint32_t h = 0; h < bmiHeader->biHeight; ++h) {
for (uint32_t w = 0; w < bmiHeader->biWidth; ++w) {
uint32_t pos = h * bmImg.bmBytesPerRow + w * bmImg.bmBytesPerPel;
double *Y = &bmYUV[pos];
if (*Y < min) {
min = *Y;
}
if (*Y > max) {
max = *Y;
}
}
}
// rearrange gray indensity
for (uint32_t h = 0; h < bmiHeader->biHeight; ++h) {
for (uint32_t w = 0; w < bmiHeader->biWidth; ++w) {
uint32_t pos = h * bmImg.bmBytesPerRow + w * bmImg.bmBytesPerPel;
double *Y = &bmYUV[pos];
uint32_t _pos = h * bmGray.bmBytesPerRow + w * bmGray.bmBytesPerPel;
bmGray.bmData[_pos] = adjust(255 * (*Y - min) / (max - min));
}
}
### Step Four: Change the Luminance
The luminance is depend on Y value, too. What we need to do is just changing the Y value and applying the inverse formula below.
$$\begin{bmatrix} 1.000 & 0.000 & 1.140 \\ 1.000 & -0.3946 & -0.5805 \\ 1.000 & 2.032 & -0.0005 \end{bmatrix} \times \begin{bmatrix} Y \\ U \\ V \end{bmatrix} = \begin{bmatrix} R \\ G \\ B \end{bmatrix}$$
And the code is simple.
// change luminance
BITMAP bmLight, bmDark;
copy_bmp(&bmLight, &bmImg);
copy_bmp(&bmDark, &bmImg);
for (uint32_t h = 0; h < bmiHeader->biHeight; ++h) {
for (uint32_t w = 0; w < bmiHeader->biWidth; ++w) {
uint32_t pos = h * bmImg.bmBytesPerRow + w * bmImg.bmBytesPerPel;
double *Y = &bmYUV[pos];
double *U = &bmYUV[pos + 1];
double *V = &bmYUV[pos + 2];
bmLight.bmData[pos] = adjust(*Y + 25 + 2.032 * *U - 0.0005 * *V);
bmLight.bmData[pos + 1] = adjust(*Y + 25 - 0.3946 * *U - 0.5805 * *V);
bmLight.bmData[pos + 2] = adjust(*Y + 25 + 1.140 * *V);
bmDark.bmData[pos] = adjust(*Y - 50 + 2.032 * *U - 0.0005 * *V);
bmDark.bmData[pos + 1] = adjust(*Y - 50 - 0.3946 * *U - 0.5805 * *V);
bmDark.bmData[pos + 2] = adjust(*Y - 50 + 1.140 * *V);
}
}
### Step Five: Complete Source Code
#### bmp.h
#ifndef _BMP_H_
#define _BMP_H_
#include <stdint.h>
#pragma pack(push, 1) // avoid struct padding
uint16_t bfType;
uint32_t bfSize;
uint16_t bfReserved1;
uint16_t bfReserved2;
uint32_t bfOffBits;
uint32_t biSize;
int32_t biWidth;
int32_t biHeight;
uint16_t biPlanes;
uint16_t biBitCount;
uint32_t biCompression;
uint32_t biSizeImage;
int32_t biXPelsPerMeter;
int32_t biYPelsPerMeter;
uint32_t biClrUsed;
uint32_t biClrImportant;
uint8_t rgbBlue;
uint8_t rgbGreen;
uint8_t rgbRed;
uint8_t rgbReserved;
typedef struct tagBITMAPINFO {
} BITMAPINFO;
typedef struct tagBITMAP {
BITMAPINFO *bmInfo;
uint32_t bmInfoSize;
uint32_t bmBytesPerRow;
uint8_t bmBytesPerPel;
uint8_t *bmData;
} BITMAP;
#pragma pack(pop)
#endif
#### bmp.c
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include "bmp.h"
// given bmHeader and bmInfo, initialize others
void init_bmp(BITMAP *bmImg) {
bmImg->bmData = (uint8_t *) malloc(bmImg->bmBytesPerRow * bmiHeader->biHeight);
}
// read a BMP from file
void read_bmp(BITMAP *bmImg, char *filepath) {
FILE *fiInImg = fopen(filepath, "rb");
// if biBitCount is less than 16, use all the palette, otherwise do not use palette
} else {
}
bmImg->bmInfo = (BITMAPINFO *) malloc(bmImg->bmInfoSize);
}
init_bmp(bmImg);
fclose(fiInImg);
}
// duplicate a BMP
void copy_bmp(BITMAP *bmDes, BITMAP *bmSrc) {
memcpy(bmDes, bmSrc, sizeof(BITMAP));
bmDes->bmInfo = (BITMAPINFO *) malloc(bmSrc->bmInfoSize);
memcpy(bmDes->bmInfo, bmSrc->bmInfo, bmSrc->bmInfoSize);
bmDes->bmData = (uint8_t *) malloc(bmSrc->bmBytesPerRow * bmiHeader->biHeight);
}
// write a BMP to file
void write_bmp(BITMAP *bmImg, char *filepath) {
FILE *fiOutImg = fopen(filepath, "wb");
fwrite(bmImg->bmInfo, bmImg->bmInfoSize, 1, fiOutImg);
fclose(fiOutImg);
}
// make RGB value legal
int16_t ret = (int16_t) (val + 0.5);
return ret < 0 ? 0 : ret > 255 ? 255 : ret;
}
int main(int argc, char *argv[]) {
BITMAP bmImg;
// calculate YUV value
double *bmYUV = (double *) malloc(sizeof(double) * bmImg.bmBytesPerRow * bmiHeader->biHeight);
for (uint32_t h = 0; h < bmiHeader->biHeight; ++h) {
for (uint32_t w = 0; w < bmiHeader->biWidth; ++w) {
uint32_t pos = h * bmImg.bmBytesPerRow + w * bmImg.bmBytesPerPel;
uint8_t *B = &bmImg.bmData[pos];
uint8_t *G = &bmImg.bmData[pos + 1];
uint8_t *R = &bmImg.bmData[pos + 2];
bmYUV[pos] = 0.299 * *R + 0.587 * *G + 0.114 * *B;
bmYUV[pos + 1] = -0.147 * *R - 0.289 * *G + 0.436 * *B;
bmYUV[pos + 2] = 0.615 * *R - 0.515 * *G - 0.100 * *B;
}
}
// color to gray
BITMAP bmGray;
bmGray.bmInfo = (BITMAPINFO *) malloc(bmGray.bmInfoSize);
for (int i = 0; i < 256; ++i) {
// to prove that the palette works well, play a small trick
rgb->rgbBlue = (i >> 4) << 4;
rgb->rgbGreen = (i >> 4) << 4;
rgb->rgbRed = (i >> 4) << 4;
rgb->rgbReserved = 0;
}
init_bmp(&bmGray);
uint8_t min = 255, max = 0;
for (uint32_t h = 0; h < bmiHeader->biHeight; ++h) {
for (uint32_t w = 0; w < bmiHeader->biWidth; ++w) {
uint32_t pos = h * bmImg.bmBytesPerRow + w * bmImg.bmBytesPerPel;
double *Y = &bmYUV[pos];
if (*Y < min) {
min = *Y;
}
if (*Y > max) {
max = *Y;
}
}
}
// rearrange gray indensity
for (uint32_t h = 0; h < bmiHeader->biHeight; ++h) {
for (uint32_t w = 0; w < bmiHeader->biWidth; ++w) {
uint32_t pos = h * bmImg.bmBytesPerRow + w * bmImg.bmBytesPerPel;
double *Y = &bmYUV[pos];
uint32_t _pos = h * bmGray.bmBytesPerRow + w * bmGray.bmBytesPerPel;
bmGray.bmData[_pos] = adjust(255 * (*Y - min) / (max - min));
}
}
write_bmp(&bmGray, "gray.bmp");
// change luminance
BITMAP bmLight, bmDark;
copy_bmp(&bmLight, &bmImg);
copy_bmp(&bmDark, &bmImg);
for (uint32_t h = 0; h < bmiHeader->biHeight; ++h) {
for (uint32_t w = 0; w < bmiHeader->biWidth; ++w) {
uint32_t pos = h * bmImg.bmBytesPerRow + w * bmImg.bmBytesPerPel;
double *Y = &bmYUV[pos];
double *U = &bmYUV[pos + 1];
double *V = &bmYUV[pos + 2];
bmLight.bmData[pos] = adjust(*Y + 25 + 2.032 * *U - 0.0005 * *V);
bmLight.bmData[pos + 1] = adjust(*Y + 25 - 0.3946 * *U - 0.5805 * *V);
bmLight.bmData[pos + 2] = adjust(*Y + 25 + 1.140 * *V);
bmDark.bmData[pos] = adjust(*Y - 50 + 2.032 * *U - 0.0005 * *V);
bmDark.bmData[pos + 1] = adjust(*Y - 50 - 0.3946 * *U - 0.5805 * *V);
bmDark.bmData[pos + 2] = adjust(*Y - 50 + 1.140 * *V);
}
}
write_bmp(&bmLight, "light.bmp");
write_bmp(&bmDark, "dark.bmp");
return 0;
}
## 题目描述
This is an interactive problem.
For two positive integers $x, y$, we define $\pi(x, y)$ to be the number of distinct primes that divide both $x$ and $y$. For example $\pi(2, 3) = 0, \; \pi(8, 16) = 1$ and $\pi(30, 105) = 2$.
For two positive integers $a, b$, where $a \le b$, we define $S(a, b)$ to be the sum of values $\pi(x, y)$ over all pairs of integers $(x, y)$ satisfying $a \le x \lt y \le b$.
Your task is to compute the values $S(a, b)$ for many query pairs $(a, b)$. To make your task more challenging, all the queries have to be answered online.
## 代码
#include <cstdio>
#include <cstring>
#include <algorithm>
int const N = 1000005;
int tp, prime[N], rec[N], vis[N], sum[N];
void init() {
for (int i = 2; i < N; ++i) {
if (!vis[i]) {
prime[tp++] = i;
}
for (int j = 0; j < tp && i * prime[j] < N; ++j) {
vis[i * prime[j]] = 1;
if (i % prime[j] == 0) {
break;
}
}
}
for (int i = 2; i < N; ++i) {
sum[i] = sum[i - 1] + !vis[i];
}
}
int main() {
init();
int q;
scanf("%d", &q);
for (; q--;) {
int a, b;
scanf("%d%d", &a, &b);
long long ans = 0;
for (int i = 0; i < tp && prime[i] < b;) {
int cnt = b / prime[i] - (a - 1) / prime[i];
int nxt = 1e9;
if (b / prime[i]) {
nxt = std::min(nxt, b / (b / prime[i]));
}
if ((a - 1) / prime[i]) {
nxt = std::min(nxt, (a - 1) / ((a - 1) / prime[i]));
}
ans += 1ll * cnt * (cnt - 1) / 2 * (sum[nxt] - i);
i = sum[nxt];
}
printf("%lld\n", ans);
fflush(stdout);
}
return 0;
}
## Spring II
It was a surprise to discover a new nest opposite the window.
Through the telescope, I scrupulously watched the doves in the nest, who were reposing restfully. Narrowly I discerned that they were a family comprised of a mother and two babies. Sighing with relief, I thought, “Finally the dove has returned, and luckily I can still see the family.” I even expected to witness the first flight of the baby doves.
I started wondering where the doves came from, as it was odd that they were inclined to construct their home here between the buildings, and after a period of cogitating, I ultimately deduced that once they might live in the decrepit houses nearby which were to be pulled down owing to the modernization and urbanization.
From this viewpoint, the doves seemed to be pitiable. “It can be tough for them to find a congenial shelter in the city.” I uttered in my heart, as I observed them fluttering their gray wings. It looked as if they were enjoying the serene life. Yet my brain did not cease contemplating.
Living in the era with rapid technological advancements which largely hasten our pace of life, we ineluctably forgo the joyous time being close to nature. And not until the demise of the erstwhile happiness we attained from nature comes do we realize that it is deplorable to turn ourselves into a hardened machine.
The blood-red sun was sinking and a livid cloud in the east received its rays. A twinge of woe surged through me. If we humanity kept pursuing interests by damaging the environment, what would the attendant consequence be? Would it only be driving the doves homeless?
I couldn’t help recalling the days when the city was shrouded in thick smog, and people could scarcely distinguish the things in the distance. And the rivers in the city, which were limpid before, had become turbid in recent decades. The foregoing instances were attributed to human activities. Not to mention the global warming, the ozone depletion, and the extinction of multitudinous species. Urbanization was just all of these in miniature.
The sky was tinted with black inch by inch, and the doves quieted down again. I shifted my eyes from them.
I leaned on the windowsill, looking up at the firmament, which used to be spangled with innumerable stars that were now unseen to us. Embedded aloft in the inky night sky, the pale yellow full moon cast its glow over the city, like an infinite piece of cloth mantling, yet faraway as the neon lights were, their harsh beams penetrated the textile that the moon carefully sewed without any difficulty, as if it was nonexistent. The dazzling rays pricked not only my eyes, but also my heart. “Perhaps the doves dread the blinding light, too.”
There was a time when all the doves could flit from tree to tree jovially in the dense and emerald woodlands, where they interacted with other creatures complying with the principles of nature. It was not until human beings arose that changes occurred. Trees were felled at first, and then sewage and fumes were disgorged by pipes. As the time elapsed our compunction was attenuated unawares by the profits we gained at the price of the environment we lived. We averted our gaze from what was prime, focusing on what was subordinate. The cupidity for gain and the scant heed we paid to conservation incurred the mire we confront today.
We really should pause, ruminate for a while, and inquire ourselves, what is our original orientation, and does what we endeavor to do diverge from that aim. Is our purpose to make every hue around us grey? Or to fetter ourselves in the forest of reinforced concrete? Or, to alter the earth to meet human’s multifarious demands regardless of the sustainability?
The onus is on us to maintain an agreeable environment. We need introspection, though it may seems arduous to atone for the devastation.
Fortunately, we gradually awake to the pernicious effects, and we are making a change. Eco-civilization construction has been put forward, and the departments concerned are sparing no effort to ameliorate the imbalanced ecology.
It was late spring, and every plant was flourishing. I heard the birds warbling a mellifluous night chorus, my heart mollified.
I entertain the hope that one day, we can thoroughly relish the felicity from nature. I envision an unspoiled environment. If we esteem nature, make optimal decision, and take prompt action.
I’m looking forward to the arrival of that very day.
## Spring I
I heard something twittering on the balcony.
Walking on my tiptoe, I reached the window, through which I spotted a turtle dove standing on the stainless-steel-made railing, with a sprig in its beak. Barely had I observed it attentively when it hastily fled away, leaving behind a fuzzy silhouette.
I was leaning on the windowsill, looking down, when a ball of white captivated me, which I instantly recognized as peach blossoms. Bathed in radiant sunlight, they stretched their exquisite bodies in all directions. The balmy breeze gently swayed the flowers, carrying the scent they exuded wherever it passed. Eventually here came the spring. As I was about to return to my room, my eyes rested on a half-finished bird nest that lay on the board put on the railings.
It was unbelievable that the dove was planning to build a nest on our balcony. Yet this was the case. What’s more, I witnessed two doves erecting their lovely home together. Before long, a complete nest came out.
I could see in the mild twilight, a tiny nest lying steadily behind the huge flower pots, and an adorable dove perching on the nest. The setting sun cast its silky beams on this little cute creature through the verdant plants in the pots, resembling a warm hand patting tenderly. This tranquil scene was set off by the distant golden horizon, which formed the most enchanting landscape in the world. I could picture that under its little body, eggs were being warmed up, and new life was being brewed.
But we all know that good times do not last long, which I grasped on that drizzly evening when the dove had gone away and had not returned.
I stared at the empty nest vacantly, a sense of loss welling up in my heart. There lay the sole egg placidly, reflecting the cold light which the moon shed on the earth. Why didn’t the dove return? Maybe, I unintentionally frightened it, or, it had found a superior habitat. But, did it desert its cherished egg? On this fairly chilly rainy night, where could it go? Would it be shivering all over in the freezing wind? As the moonlight suffused the ground, I perceived that, never would I know the answer.
I leaned on the windowsill, overlooking the swaying flowers. I could hear raindrops knocking the canopy, composing a rumpled melody agitating my thoughts. I could see them trickling down the eaves, descending, sinking into puddles, feebly splattering a spray of droplets. Also I could hear the moaning wind dashing through the alley, wildly rattling loose windows and doors. I could see it wobbling the leaves furiously, severing them from branches and tugging them down. But, in the meantime, I could see the blossoms that the tree threw up in the air, being hit by the rain and being swung by the wind, while they kept emanating aroma that wafted all around. I could see the myriad scattered petals which fell on the moist tarmac, all colors draining rapidly, yet somehow, they were peculiarly appealing with the lamplight flickering in dewdrops.
It suddenly occurred to me that, unlike what I considered before, what was called spring, was not a picture in a particular style. It not only meant arrival and growth, but also meant departure and wilting to some extent.
So the spring is not all about a couple of turtle doves who are busy constructing a cozy neat twiggy nest where the new life will be brought, nor about the pale pink blossoms that bloom wherever you behold, adorning the corner of the bright blue sky. The essence of the spring lies in the persistence that the doves possess, even if they get so close to human beings and lose one of the only two eggs. It lies in the patience with which the dove squats in its nest unceasingly, awaiting the arrival of the baby, no matter the icy drizzle and the roaring gale. It lies in the optimism which encourages the blossoms to bloom vigorously in the frigid sprinkle. It lies in the devotion of the blossoms as they make the world transfigured despite the adversity they are confronted with.
I lingered on the balcony, hearkening wind and rain, dismay dissipating gradually. And here came the spring. I could sense that, distinctly.
## The Beginning
The bell is about to ring. The new year is in sight.
Time flies as we grow and mature, leaving naivety behind. No sooner have we relived the past year in detail than we face the coming year hurriedly. Lay down the heavy burden, repack the travelling bag, take our initial dream, step forward readily and firmly.
It’s said those who have received compliments should be rather careful, for other’s acclaim may block the consciousness of weakness, which surely leads to failure. Yet by no means shall we belittle ourselves, since nobody will have confidence in us unless we are confident. Owning a distinct understanding of oneself, cultivating the better part, remedying what have been done that turns out a mistake, are ways to adjust our mentality.
When we gaze at the vast, dark night sky, sparsely-arranged stars hang high, whispering about what they have just witnessed on this tiny, lonely, but vigorous cerulean planet, such as a large amount of teenagers burning the midnight oil, waiting for that extraordinary moment, after which they may become an adult, or may bare his or her heart to someone else, or, may just sigh with the feeling that it seems something in life, which is hard to be aware of, has vanished silently and drastically. Whatever they see, the sparkling stars keep extending their best wishes to whom they have not known and may never know. The stars, the glinting stars, the stars that split the profound night sky, for how long will we still be able to look up at you?
When we overlook the pure white snowfield, accumulated snowflakes lie serenely, mirroring the warm, golden light that streetlamps cast, illumining the bottom of passers-by’s hearts, which originally filled with some annoying things, like working overtime without gaining anything, or getting into a conflict with someone he or she considers dearest, or, biting, frigid wind blowing right into cheek, as if thousands of sharp knives are pricking the skin. But at the time, the light disperses the dust, and pushes the haze in the heart aside, bringing peace and quiet, awakening hope for the future, for the year approaching. The snow, the crystal snow, the snow that decorates the empty city streets, will you still be here years later?
No matter how we get through the year just past, it doesn’t make any sense to immerse in the memory. We shall always look forward, but don’t forget, up in the sky, down on the ground, there must be someone else, caring you, supporting you, giving you hope whenever it seems impossible.
Anyhow, it’s a brand new beginning.
Goodbye, 2018. Hello, 2019.
## Light III
He could feel drizzle patting his face.
To the best of his ability, he stood up. Anew, darkness enveloped him.
He pondered about what he had just experienced, which was patently a dreadful nightmare. Yet everything was so real, that he couldn’t persuade himself not to recall that dream.
In a moment, he caught sight of something in the distance, something that was gigantic enough to be seen. It was hard for him to put down the desire to approach it. He started padding, but he sensed something preternatural. As if he was walking through a long narrow corridor, in which there was solely enormously faint light.
When he finally reached the thing, he was astonished to find that he was facing a boundless wall which was clearly insurmountable. Scarcely had he had time to figure out what was happening when he was sucked into an exceedingly strong whirlpool, losing his consciousness.
He was astounded again when he opened his eyes.
In his sight there turned seasons. He could see fresh green seedlings protruding, carefree swallows singing and dancing freely in spring, and the brilliant sun creeping up gradually, tiny trees slowly growing to a huge proportion during summer, and with autumn coming the golden foliage fell, embellishing the azure blue sky, and before long clean white snowflakes began to appear, which indicated that winter had reigned the ground. It was quite amazed that he was able to recognize the various colors in no time, as well as the weathers he used to feel and the planets passing over head. What he once could only heard of was just in front of him, like an ancient but exquisite picture scroll opening before his face by degrees. Wherever his vision went, the superb landscapes appealed to him.
He was immersing himself in this fabulous world when in an instant, as if a heavy gate was opened, recollections flooded his brain.
He attempted to cast his mind back, and something mysterious occurred.
He could see his father holding him in his arms, while his mother cradled his brother, staring at two little lives, humming a familiar song he had heard for many times. He could see he and his brother climbing up and down at home, and his brother frequently lending him a hand without which he might have fallen off the table that was still higher than him. He could see the family outing, during which his relatives did not get weary of guiding him, telling him what was going on around, such as fragrant flowers dancing with colorful butterflies. So joyful was he at that time that as he thought, in his heart a knot was untied, and the ice melted.
Never had he realized that so many fantastic memories had been missing since his third birthday, for which he would have lived delightedly, appreciating the love and hope on earth, summoning up the courage and confidence, to strive for an entirely different life.
But it was said there was no sense crying over every mistake, for what had been done was unchangeable. He smiled, blandly, as he floated in the air, higher and higher.
Looking down, he saw himself, just lying, peacefully, on the snow-covered ground.
“There lies a wall, up to the heaven, down to the hell, left to infinity, right to where you cannot see.”
“This wall is called death.”
“Only when a person is confronted with this wall will he look back upon his entire life.” | 14,458 | 48,301 | {"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} | 2.578125 | 3 | CC-MAIN-2021-25 | latest | en | 0.526011 |
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Chapter 7
Solution to Problems
1. Alphanumeric characters are transmitted as 7-bit ASCII words, with a single parity bit added, over a link with a transmission rate of 9.6 kbps.
a. How many characters are transmitted each second?
Answer: Each character (a letter or number, or symbol) is seven data bits plus one parity bit,
giving eight bits per character. Bit rate R b = 9.6 kbps, hence the link transmits 9.6 / 8 = 1200 characters per second.
b. If a typical page of text contains 500 words with an average of five char acters per word and a space between words, how long does it take to transmit a page?
Answer: One average page is 500 (written) words averaging five characters, with a space between words. Hence one page contains 500 ¥ 5 + 500 = 3000 characters = 24,000 bits.
Time to send one page is 24,000 / 9,600 = 2.5
seconds.
c. If the bit error rate on the link is 10 -5 , how many characters per page are detected as having
errors?
How many undetected errors are there?
We can detect one bit error in an 8 bit character,
and also 3, 5, or 7 bit errors. We cannot detect 3, 4, 6 or 8 bit errors. The probability of k errors in a word of n bits is found from the binomial equation (equation 7.4).
Answer: The probability of a bit error is 10 -5 .
P e (k)
=
(n, k) p k (1 – p) n k
where (n, k) =
n! / [k! (n – k)!]
For n = 8 and k = 1, 2, 3, we find for p = 10 -5
P e (1) = (8, 1) ¥ 10 -5 ¥ (0.99999) 7 where (8, 1) = 8! / [ 1 ¥ 7!] = 8 P e (1) = 8 ¥ 10 -5 P e (2) = 2.8 ¥ 10 -9 P e (3) = 5.6 ¥ 10 -14
All higher order terms are negligible, and P e (3) is negligible in comparison with P e (1). Hence the probability of a detected error in a character is 8 ¥ 10 -5 . With 3000 characters per
page, the probability that a character error is detected on any page is 3000 ¥ 8 ¥ 10 -5 = 0.24.
d.
On average how many pages can be transmitted before
(i) a detected error occurs (ii) an undetected error occurs? Answer: (i) With a bit error rate of 10 -5 , there are 0.24 errors per page, so we should expect to find one character error every four pages, on average.
(ii) The probability of an undetected error is P e (2) = 2.8 ¥ 10 -9 . With 3000 characters per
page, the probability of a character error on any page is 3000 ¥ 2.8 ¥ 10 -9 = 8.4 ¥ 10 -6 . We
would have to send 119,000 pages before we would expect to find an undetected error – equivalent to transmitting this text book 222 times. Since it takes 2.5 seconds to send one page at 9.6 kbps, it takes 22 minutes to transmit the whole text book, and 82 hours elapse before the undetected error occurs.
e. If the BER increases to 10 -3 , how many detected and undetected errors are there in a page of the text? Answer: With p = 10 -3 the new probabilities for 1, 2 and 3 errors per character are
P e (1) = 8 ¥ 10 -3 P e (2) = 2.8 ¥ 10 -5 P e (3) = 5.6 ¥ 10 -8
We can ignore the probability of three errors in the eight bit character.
(i) The probability of a detected character error is now P e (1) = 8 ¥ 10 -3 . With 3000
characters per page, there are 24 detected errors per page on average.
(ii) The probability of an undetected error is P e (2) = 2.8 ¥ 10 -5 , giving a probability of 0.084
errors per page, or one undetected character on every 12 pages, on average. A BER of 10 -3 is at the lower limit for transmission of text using single parity error checking.
2. A (6, 3) block code has a minimum distance of two.
a. How many errors can be detected in a codeword?
Answer: The minimum distance, d min , of a codeword is the smallest difference between two non-zero codewords in the set. The number of errors that can be detected is (d min – 1). Hence a code set with minimum distance two can detect one error.
b.
How many errors can be corrected in a codeword? The number of errors that can be corrected by a code set is given by (d min – 1) / 2,
rounded down to the next integer value. If d min = 2, (d min – 1) / 2 = ½ and the code cannot correct errors. The smallest value of d min in any code that can correct errors is d min = 3, when the code can correct one error.
3. A QPSK data link carries a bit stream at 1.544 Mbps and has an overall (C/N) o ratio of 16 dB
in the receiver at the VSAT earth station in clear air. The QPSK demodulator at the VSAT
station has an implementation margin of 1.0 dB.
3.0 dB reduction in the receiver (C/N) o ratio. For 0.01% of the year rain attenuation causes 6.0 dB reduction in the receiver (C/N) o ratio.
For 0.1% of the year rain attenuation causes
a. Calculate the BER in clear air, and the BER exceeded for 0.1% and 0.01% of the year. Answer: The symbol error rate a QPSK link is given by
P e = Q[÷(C/N) eff ] where (C/N) eff is the effective C/N ratio at the input to the QPSK demodulator, taking account of implementation margin. When Gray coding of the data bits is
employed (as is always the case) BER ª P e . power ratio of 31.62
Hence for (C/N) eff = 16.0 – 1.0 = 15.0 dB, a
BER ª Q[÷31.62] = 5.62
Using the Q[z] table of Appendix C, BER ª 10 -8 .
For 0.1% of the year rain attenuation causes 3.0 dB reduction in the receiver (C/N) o ratio.
Hence (C/N) eff
= 15.0 – 3.0 = 12.0 dB or a ratio of 15.85.
BER ª Q[÷15.85] = 3.98, BER ª 3.5 ¥ 10 -5 .
b. Repeat the calculation when data are transmitted using half rate FEC with a coding gain of 5.5 dB (at all BERs). The bit rate on the link remains at 1.544 Mbps when coding is
What is the data rate with half rate FEC applied to the data?
Answer: With half rate error correction coding applied to the bit stream, the effective (C/N) eff ratio is increased by 5.5 dB. Hence for clear air conditions, (C/N) eff = 16.0 – 1.0 + 5.5 = 20.5 dB, a power ratio of 112.2.
BER ª Q[÷112.2] = Q[10.59]
The Q[z] table of Appendix C gives values for Q[z] only to Q[8] because BER ª 0 for any value
of z larger than eight, indicating that there are no errors occurring on the link. For 0.1% of the year rain attenuation causes 3.0 dB reduction in the receiver (C/N) o ratio.
Hence (C/N) eff
= 15.0 – 3.0 + 5.5 = 17.5 dB, a ratio of 56.23.
BER ª
Q[÷56.23] = Q[7.50],
BER ª
3.2 ¥ 10 -14 .
At a bit rate of 1.544 Mbps, there is one error every 1 / (1.544 ¥ 3.2 ¥ 10 -14 ) = 4.94 ¥ 10 8 s, or
one error every 15.7 years. 99.9% of the year.
The link is operating error free with 3 dB of C/N reduction, for
For 0.01% of the year rain attenuation causes 6.0 dB reduction in the receiver (C/N) o ratio.
Hence (C/N) eff
BER ª
= 15.0 – 6.0 + 5.5 = 14.5 dB, a ratio of 28.18.
Q[÷28.18] = Q[5.31],
BER ª
5.5 ¥ 10 -8 .
There is now an error every 11.8 seconds on the link. Only half of the bits in the transmitted bit stream are data bits; the other half are parity bits used in the EC decoder to correct errors. Hence the data rate at the output of the link is 1.544 Mbps / 2 = 772 kbps.
c. Repeat the calculation of part (b) above when the bit rate on the link is increased to 3.088 Mbps with no increase in transmitter power. Answer: If the bit rate on the link is increases to 3.088 Mbps and half rate forward error correction is applied, the data rate at the link output is 1.544 Mbps. However, the bandwidth of the link must be doubled to allow the higher bit rate signal to be transmitted, and this doubles the noise power in the receiver and reduces overall C/N by 3.0 dB. In this case, overall C/N in clear air is 13.0 dB, so the BER in clear air corresponds to (C/N) eff = 13.0 - 1.0 + 5.5 = 17.5 dB, a ratio of 56.23.
BER ª Q[÷56.23] = Q[7.50], BER ª 3.2 ¥ 10 -14 and there are no errors on the link. For 0.1% of the year rain attenuation causes 3.0 dB reduction in the receiver (C/N) o ratio and (C/N) eff = 12.0 – 3.0 + 5.5 = 14.5 dB, a ratio of 28.18.
BER ª Q[÷28.18] = Q[5.31], BER ª 5.5 ¥ 10 -8 . There is now an error every 11.8 seconds on the link. For 0.01% of the year with 6.0 dB reduction in the receiver (C/N) o ratio,
(C/N) eff = 12.0 – 6.0 + 5.5 = 11.5 dB, a ratio of 14.13. BER ª Q[÷13.13] = Q[3.76], BER ª 8.4 ¥ 10 -5 .
There are now 262 errors every second on the link. This is towards the lower margin for operation of the link with speech signals (S/N ª 35 dB for 8-bit digital speech) and below the
threshold of BER = 10 -6 that is usually applied for data transmission.
4. The Analysis of a 56-kbps data link shows that it suffers burst errors that corrupt several adjacent bits. The statistics for burst errors on this link are given in Table P.4. Table P.4 Statistics for Burst Errors on a Link in Problem 4.
No. Adjacent Bits Corrupted Probability of Occurrence 2 4 ¥ 10 -2 3 2 ¥ 10 -3 4 3 ¥ 10 -4 5 1 ¥ 10 -6 6 2 ¥ 10 -9 7 5 ¥ 10 -11 8 1 ¥ 10 -12 9 3 ¥ 10 -14 10 2 ¥ 10 -17
a. Using Table P7.4, select a burst error correcting code that will reduce the probability of an uncorrected burst error below 10 -10 . Answer: We want to ensure that the probability of an uncorrected error is below 10 -10 . From Table P7.4, the probability of an uncorrected burst error is below 10 -10 for codes with burst error correction capability of 7 and more adjacent bits. The (103,88) code in Table P7.4 meets the requirements, with a code rate 88/103.
b. Calculate the data rate for messages sent over the link using the code you selected.
Answer: For a transmitted bit rate of R b = 56 kbps, the output of the FEC decoder is a message data rate of
R b
=
88/103 ¥ 56 = 47.84 kbps.
c. Estimate the average bit error rate for the coded transmission.
The probability of an uncorrected burst error is the sum of the probabilities that 8, 9,
10 … adjacent bits are in corrupted in the received bit stream, and the errors cannot be corrected
by the FEC decoder.
From Table P 7.4
P(8) = 1 ¥ 10 -12 - P(9) = 3 ¥ 10 -14 P(10) = 1 ¥ 10 -17
Higher order terms are negligible. Hence the probability of uncorrected error bits reaching the
decoder output is approximately
P e
ª
8 ¥ P(8)
+ 9 ¥ P(9)
=
8 ¥ 10 -12
+
27 ¥ 10 -14
= 8.27 ¥ 10 -12
At a transmission rate of 56 kbps, bursts of bit errors are occurring at the output of the FEC
decoder with a probability of approximately 10 -12 . The time between burst errors is 1 / (56 ¥ 10 3
¥ 10 -12 )
or 207 days. The link is essentially error free.
1.786 ¥ 10 7 s
=
5. A satellite link carries packet data at a rate of 256 kbps. The data are sent in 255-bit blocks
using a (255, 247) code that can detect three errors. The probability of a single bit error, p,
varies from 10 -6 under good conditions to 10 -3 under poor conditions. The one way link delay is 250 ms.
a. If no error detection is used, what is the message data rate for the link?
Message data rate without error detection is 247/255 ¥ 256 = 247.969 kbps.
b. For a link BER of 10 -6 , find the probability of detecting an error in a block of 255 bits when
error detection is applied
Hence find how often an error is detected.
presence of 1, 2, or 3 errors in the received bit stream. Hence the probability that we detect an error in any block is P(error detected) = P(1) + P(2) + P(3) Using equation 7.4
The error detection capability of the (247, 255) code is three bits, so we can detect the
P e (k)
=
(n, k) p k ( 1- p ) n-k
where (n, k) =
n! / [ k! ¥ (n-k)! ]
For k = 1, 2, and 3 with p = 10 -6 and a block length n = 255 bits
P
P e (2)
P e (3) = (255 ¥ 254 ¥ 253 / 6) ¥ 10 -18 ¥ (1 – 10 -6 ) 252 = 2.73 ¥ 10 -12
e (1)
= 255 ¥ 10 -6 ¥ (1 – 10 -6 ) 254 = 255 ¥ 10 -6 ¥ 0.9948 = 2.549 ¥ 10 -4 = (255 ¥ 254/2) ¥ 10 -12 ¥ (1 – 10 -6 ) 253 = 3.237 ¥ 10 -8
Thus the probability of a detected error is P(detected) = P(1) + P(2) + P(3) ª P(1) = 2.549 ¥ 10 -4
The link transmits bits at 256 kbps, and a block has 255 bits, so the block transmission rate is R block = 256 ¥ 10 3 / 255 = 1003.92 blocks per second.
The average time between block errors is T error = 1 / (R block ¥ P(error)) = 1 / (1003.92 ¥ 2.549 ¥ 10 -4 ) = 3.908 seconds
c. Estimate the probability that a block of 255 bits contains an undetected error when the link BER is 10 -3 and error detection is applied. Answer: Only three errors can be detected in any one block, so four or more errors are undetected. Hence, the probability that errors in the block are not detected is P(undetected error) = P(4) + P(5) + P(6) + … With a probability of bit error p = 10 -3 , the probability of undetected errors in the block is
P e (4) ª 172 ¥ 10 6 ¥ 10 -12 ¥ (1 – 10 -3 ) 251 = 1.337 ¥ 10 -4 P e (5) = 8.637 ¥ 10 9 ¥ 10 -15 ¥ (1 – 10 -3 ) 250 = 6.762 ¥ 10 -6 P e (6) = 3.599 ¥ 10 11 ¥ 10 -18 ¥ (1 – 10 -3 ) 249 = 2.805 ¥ 10 -7
Hence P undetected ª P(4) + P(5) + P(6) = 1.404 ¥ 10 -4
d. Find the message data throughput when the link BER is 10 -6 and a stop and-wait ARQ
system is used, assuming one retransmission always corrects the block. Answer: With stop-and-wait ARQ protocol, the transmitter must stop and wait until it receives an ACK or a NAK from the receive end of the link before it sends the next block. The transmission rate is dominated by the 500 ms waiting time for the block to reach the receiver and the receiver to respond with an ACK or NAK. The time to transmit one block is approximately
1 ms, so a single block transmission plus waiting time is 2 ¥ 250 + 1 = 501 ms, and throughput
averages 509 bits per second, and the message data rate is 247/255 ¥ 509 = 493 message bits per second. After 1003.92 blocks have been transmitted, we encounter a detected error and must retransmit the corrupted block. That occurs after 9.88 hours, so the extra time to resend blocks is not significant. Stop-and-wait is the wrong protocol for GEO satellite links, because of the long round trip time. Go-back-N or continuous transmission should be used. As a supplement to this question, let’s look at throughput with go-back-N and continuous transmission ARQ protocols. Go-back-N. With a go-back-N protocol, transmission is continuous until a NAK is returned by the receiver indicating that a corrupted block has been received. At that point, 501 ms worth of blocks has been transmitted since the now corrupted block was sent, and must be restransmitted. At 256 kbps bit rate, this corresponds to 502.96 blocks, so we must resend 503 blocks. On average, we can send 1003.92 blocks before an error occurs, and then we must resend 503 blocks, so the average message bit delivery rate is
R b = 247/255 ¥ 1003.92 / 1506.92 ¥ 256 kbps = 165.198 kbps. This result ignores any time associated with decoding a received packet at the transmitter to recover the NAK, so the message rate will be slightly lower in practice. Continuous transmission. With continuous transmission, we resend only the corrupted block when a NAK is received. Thus, on average, once every 1003.92 blocks we must add an extra block into the bit stream. The message data rate is then
R b = 247/255 ¥ 1003.92/1004.92 ¥ 256 kbps = 247.472 kbps.
6.
Repeat Problem 5 using a block length of 1024 bits and a (1024, 923) code that can detect 22
errors in a block. The (1024, 923) code can correct 10 errors. Find the average number of blocks that can be transmitted before an uncorrected error occurs when the BER is 10 -3 . Repeat the analysis for a BER of 10 -2 .
Note:
P b = 10 -3 in this case) can be calculated from the Poisson distribution more easily than from the
binomial distribution.
The probability of an unlikely event (11 or more errors in a block of 1024 data bits with
The Poisson distribution is given by
where
l
=
l
k
e
-
l
P(x = k)
=
k! N is the block length, and k is the number of bits in error.
N P b .
occurs when there are 11 through 22 bit errors in the 1024 bit block. We use the Poisson formula because it works well for a small number of errors in a large number of bits.
a. If we take no action regarding errors, the message data throughput is
The (1024, 923) code can correct 10 errors, so a detected but uncorrected block error
R b = 923/1024 ¥ 256 kbps = 230.75 kbps
b. Errors are corrected when there are 1, 2, …10 errors present in the block of 1024 bits. Hence
the probability that all the errors in the block are corrected is
P(corrected) = P(1) + P(2) + P(3) + …. From the Poisson formula
P e (k)
=
l k e -l / k!
with l
=
N ¥ p
and N = 1024, p = 10 -3 giving l
For k = 1, 2, 3 4, 5 P e (1) = 1.024 ¥ e -1.024 = 0.3677
P e (2) = 1.024 2 / 2 ¥ e -1.024 = 0.1928 P e (3) = 1.024 3 / 6 ¥ e -1.024 = 0.0643
P e (4) = 1.024 4 / 24 ¥ e -1.024 = 0.0165
= 1,024.
P e (5) = 1.024 5 / 120 ¥ e -1.024 = 0.0033 Higher order terms can be neglected. Thus the probability that the block contains a corrected error is approximately P(corrected) ª P(1) + P(2) + P(3) + P(4) + P(5) = 0.665
We can correct 10 errors in the block, so a retransmission is required when 11 or more errors are detected. Hence the probability of a retransmission is P(retransmission) = P(11) + P(12) + P(13) + P(14) + …
P e (11) = 1.024 11 / 11! ¥ e -1.024 = 1.168 ¥ 10 -8 P e (12) = 1.024 12 / 12! ¥ e -1.024 = 9.967 ¥ 10 -10 P e (13) = 1.024 13 / 13! ¥ e -1.024 = 7.851 ¥ 10 -11
Hence the probability or a retransmission is approximately
P(retransmission) ª 1.168 ¥ 10 -8 + 9.967 ¥ 10 -10 + 7.851 ¥ 10 -11 = 1.269 ¥ 10 -8
Block length is 1024 bits and transmission rate 256 kbps, so there are 250 blocks transmitted every second, one every 4 ms. The time between uncorrected block errors is
T error
=
1 / (250 ¥ 1.269 ¥ 10 -8 ) s
= 315,208 s = 87.6 hours.
This is a sufficiently long time that the 504 ms required to retransmit one or 125 corrupted blocks does not significantly reduce throughput in go back N and continuous transmission ARQ protocols. Stop and wait is still just as inefficient as in the previous example, of course. What is surprising in this example is that the system can operate successfully sending long blocks at a bad error rate. The ability to correct 10 bits in the 1024 block renders the link close to error free even when the BER is 10 -3 . The probability of eleven or more errors occurring in a block of 1024 bits is sufficiently low that almost all blocks are delivered correctly after forward error correction.
Bit error rate 10 -2
When the bit error rate is 10 -2 ,
N ¥ l
= 1024 ¥ 10 -2
= 10.24.
a. Message data throughput with no error detection leading to ARQ is unchanged at 230.75 kbps.
b. The probability of a retransmission increases considerably from the case where p = 10 -3 ,
because, on average, there will now be 10 errors per block. When 11 … 22 errors occur, a
retransmission is needed. Hence the probability of retransmission is P(retransmission) = P(11) + P(12) + P(13) …. P(22) Individual probabilities are shown below.
P e (11) = 10.24 11 / 11! ¥ e -10.24
= 0.1161
P e (12) = 10.24 12 / 12! ¥ e -10.24 = 0.0991 P e (13) = 10.24 13 / 13! ¥ e -10.24 = 0.0781 P e (14) = 10.24 14 / 14! ¥ e -10.24 = 0.0571 P e (15) = 10.24 15 / 15! ¥ e -10.24 = 0.390 P e (16) = 10.24 16 / 16! ¥ e -10.24 = 0.0249 P e (17) = 10.24 17 / 17! ¥ e -10.24 = 0.0150 P e (18) = 10.24 18 / 18! ¥ e 10.24 = 0.0085
Ignoring higher order terms
P(retransmission) ª 0.478 On average, one in every 2.28 blocks contains more than 10 errors and must be retransmitted.
We can transmit 2.28 ¥ 923 = 2108 message bits, on average, before we must repeat a block. The time to transmit 2.28 blocks is 9.12 ms, so a go-back-N ARQ protocol would, on average, send 2.28 blocks and then have to repeat 504 ms of transmissions, so throughput would average about 2 ¥ 2.28 = 4.56 blocks per second, or 4407 message bits per second.
With continuous transmission, we need to introduce a repeat block into the bit stream after every 2.28 blocks, so throughput would be 2.28/3.28 ¥ 230.75 = 160.4 kbps for message bits.
When the BER increases to 10 -2 there are too many errors for the go-back-N ARQ protocol to work efficiently. Continuous transmission ARQ slows down the bit stream, but is able to maintain a reasonable throughput. The answers here assumed that a single retransmission always correctly replaced a corrupted block. With p = 10 -2 there is a significant probability that a retransmitted block is corrupted, requiring yet another retransmission, which may also be corrupted. A BER of 10 -3 is usually regarded as the lower limit for most communication systems. With Turbo coding, it is possible to operate at higher bit error rates because of the powerful correction capabilities of turbo codes. | 6,401 | 20,351 | {"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} | 4.15625 | 4 | CC-MAIN-2020-10 | latest | en | 0.876422 |
http://conversion.org/area/board/hide | 1,643,396,514,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320306335.77/warc/CC-MAIN-20220128182552-20220128212552-00459.warc.gz | 12,805,130 | 6,888 | # board to hide conversion
Conversion number between board [bd] and hide is 1.5942250790736 × 10-8. This means, that board is smaller unit than hide.
### Contents [show][hide]
Switch to reverse conversion:
from hide to board conversion
### Enter the number in board:
Decimal Fraction Exponential Expression
[bd]
eg.: 10.12345 or 1.123e5
Result in hide
?
precision 0 1 2 3 4 5 6 7 8 9 [info] Decimal: Exponential:
### Calculation process of conversion value
• 1 board = (0.00774192) / (4.85622770688*10^05) = 1.5942250790736 × 10-8 hide
• 1 hide = (4.85622770688*10^05) / (0.00774192) = 62726400 board
• ? board × (0.00774192 ("m²"/"board")) / (4.85622770688*10^05 ("m²"/"hide")) = ? hide
### High precision conversion
If conversion between board to square-metre and square-metre to hide is exactly definied, high precision conversion from board to hide is enabled.
Since definition contain rounded number(s) too, there is no sense for high precision calculation, but if you want, you can enable it. Keep in mind, that converted number will be inaccurate due this rounding error!
### board to hide conversion chart
Start value: [board] Step size [board] How many lines? (max 100)
visual:
boardhide
00
101.5942250790736 × 10-7
203.1884501581471 × 10-7
304.7826752372207 × 10-7
406.3769003162943 × 10-7
507.9711253953678 × 10-7
609.5653504744414 × 10-7
701.1159575553515 × 10-6
801.2753800632589 × 10-6
901.4348025711662 × 10-6
1001.5942250790736 × 10-6
1101.7536475869809 × 10-6
Copy to Excel
## Multiple conversion
Enter numbers in board and click convert button.
One number per line.
Converted numbers in hide:
Click to select all
## Details about board and hide units:
Convert Board to other unit:
### board
Definition of board unit: ≡ 1 in × 1 ft. = 0.3048 × 0.0254 = 7.74192×10-3 m²
Convert Hide to other unit:
### hide
Definition of hide unit: ≈ 120 ac (variable). The hide was an English unit of land measurement. Definition variable, ≈ 12 × ac ≈ 120 × 4046.8564224 m²
← Back to Area units | 650 | 2,025 | {"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} | 3.40625 | 3 | CC-MAIN-2022-05 | latest | en | 0.678462 |
https://businesspost.ng/general/target-market-analysis-mean-median-and-mode/ | 1,713,664,424,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817699.6/warc/CC-MAIN-20240421005612-20240421035612-00206.warc.gz | 131,185,516 | 46,943 | # Target Market Analysis: Mean, Median and Mode
August 23, 2023
Averages are commonly used in the business and analyzing of the target market. First, you need to understand the 4 types of averages: the Mean, Mode, Median and Range. The Mean is best to find the average number of clients visiting your department store. The online mean calculator can be good for free online help for businesses. Mode is used to find the common number of products purchased on a particular day. The Median can be used to find the average number of products sold in a month.
The Range is used to find the number of clients in a particular market and the distortion in the marketplace. You can say that averages are one of the most used statistical tools, and their application is vast in the field of business and commerce. The average value calculator is best for finding the Mean, Mode, Median and Range of statistical data. This is used to find the nature of a target market and to determine the dynamics of the marketplace.
Now, We learn the applications of the various types of the average in businesses and commerce.
## What is the mean in algebra?
The Mean can be used to find the average number of clients in a particular Month. For example, if you have data for 12 months of the year and want t extract the average number of consumers each month. The mean generator can be used to find the mean values.
Jan Feb Mar April May June July Aug Sep Oct Nov Dec 4000 4100 4400 4300 4500 4400 4300 4000 4200 4000 4100 4200
Now add all the numbers of visited consumers in the 12 months as follows:
Mean Average Consumer per month = [Total consumers visited/Total Number of months] =[(4000+4100+4400+4300+4500+4400+4300+4000+4200+4000+4100+4200)/12]
= [50500/12]
Mean Average Consumer per month = 4208.333
The average numbers calculator commutes all the values of the Means in a matter of seconds.
## The Utilization of the Mode:
The Mode is used to find the most repeating values in our statistical data, and the simple online mean calculator assists in finding the mode values.
Jan Feb Mar April May June July Aug Sep Oct Nov Dec 4000 4100 4400 4300 4500 4400 4300 4000 4200 4000 4100 4200
Consider the statistical data:
Value Frequency 4000 3 (25%) 4100 2 (16.666666666667%) 4400 2 (16.666666666667%) 4300 2 (16.666666666667%) 4500 1 (8.3333333333333%) 4200 2 (16.666666666667%)
The online mean calculator indicates that 4000 appears 3 times in the data, and it is the mode or the most repeating value of our dataset. You can find the Mode to find the most repeating value in the marketplace; this can be quite useful in scanning the marketplace.The analysis of the target market is essential to make our planning up to mark and enable us to beat our competitors.
## The Median value finds pinpoint segregation of the data; the online mean calculator would be used to find the Median of our dataset values.
Jan Feb Mar April May June July Aug Sep Oct Nov Dec 4000 4100 4400 4300 4500 4400 4300 4000 4200 4000 4100 4200
Median = 4000, 4000, 4000, 4100, 4100, 4200, 4200, 4300, 4300, 4400, 4400, 4500
Median = 4200
## What is the Range?
The range of the dataset values is the dispersion of the data and the size of the target market. The online mean calculator displays the following Range of the dataset values:
Jan Feb Mar April May June July Aug Sep Oct Nov Dec 4000 4100 4400 4300 4500 4400 4300 4000 4200 4000 4100 4200
Range of the Consumer= (Biggest – Smallest)
Range of the Consumer=(4500-4000)=500
## Conclusion:
The average values of the Mean, Median, Mode and Range are great estimations to find the nature of the Target Market. The analysis of the Target Market is essential to formulate the decision-making. You can determine the various alternatives to decision-making if you have extracted the result from the average calculator. This result would clear the picture and the scenario of the target market and competitors analysis. You can determine the SWOT(Strengths, Weaknesses, Opportunities, and Threats) of the marketplace.
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By Modupe Gbadeyanka The former Governor of Kogi State, Mr Yahaya Bello, has been declared wanted by the Economic and Financial Crimes Commission (EFCC). | 1,301 | 5,231 | {"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} | 3.75 | 4 | CC-MAIN-2024-18 | latest | en | 0.895566 |
https://wrfranklin.org/nikola/Teaching/graphics-f2018/posts/finalexam-sol/ | 1,726,068,741,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651390.33/warc/CC-MAIN-20240911152031-20240911182031-00255.warc.gz | 582,536,694 | 7,488 | # CG ECSE-4750 Computer Graphics Final Exam Solution, RPI, Tues 2018-12-18
Name, RCSID: WRF solutions
## 1 Grade statistics
1. Number of final submissions: 27
2. Highest score: 74
3. Lowest score: 44
4. Average score: 61.78
5. Median score: 64
## 3 Rules
1. You have 180 minutes.
2. You may bring in two 2-sided 8.5"x11" papers with notes.
3. You may not share material with each other during the exam.
4. No collaboration or communication (except with the staff) is allowed.
5. There are thirty six questions. Check that your copy of this test has all the pages.
6. Full score is 76.
## 4 Questions
1. _____/2 Put these graphics pipeline components in the correct order:
2. primitive-assembly 2nd
3. rasterizer 3rd
2. _____/2 Which of the following 3D rotation methods is vulnerable to gimbal lock?
1. Euler angles *
2. matrix
3. quaternion
4. vector
3. _____/2 In the OpenGL pipeline, the Primitive Assembler does what?
1. fits together pieces of ancient Sumerian pottery.
2. rotates vertices as their coordinate systems change.
3. creates lines and polygons from vertices. *
4. finds the pixels for each polygon.
5. reports whether the keyboard and mouse are plugged in correctly.
4. _____/2 When rotating an object, what can happen to an object?
1. Straight lines might turn into curves.
2. Straight lines stay straight, but angles might change.
3. Straight lines stay straight, and angles don't change, but distances may change, either longer or shorter.
4. Straight lines stay straight, and angles don't change, but distances might get longer.
5. Straight lines stay straight, and angles and distances don't change. *
5. _____/2 If i and j are quaternions, what is i+j?
1. -k.
2. 0.
3. 1.
4. i+j, there is no simpler representation. *
5. k.
6. _____/2 If i and j are quaternions, what is ij?
1. -k.
2. 0.
3. 1.
4. i+j, there is no simpler representation.
5. k. *
7. _____/2 The quaternion i represents what rotation?
1. 180 degrees about the x-axis. *
2. 90 degrees about the x-axis.
3. 180 degrees about the y-axis.
4. 90 degrees about the y-axis.
5. no change, i.e., 0 degrees about anything.
8. _____/2 The parametric equation of a line through the points (1,1) and (2,3) is:
1. P = (1,1) + t(1,0) + u(0,1)
2. P = (1,1) + t(1,1)
3. P = (1,1) + t(1,2) *
4. P = (1,1) + t(2,3)
5. P = t(1,1) + u(2,3)
9. _____/2 The normal vector of length one to the plane through the points (1,0,0), (0,1,0), (0,0,1) is
1. (1,0,0)
2. $\left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right)$ *
3. (-1, 0, 0)
4. 3
5. (1/3, 1/3, 1/3)
10. You call gl.BufferSubData to do what?
1. to add or replace part of the buffer in the GPU. *
2. to define a submarine object.
3. to subtract some data in the buffer.
4. to tell the GPU to look for a pattern and substitute any occurrences,
5. to tell the GPU to use a subroutine instead of the main program.
11. _____/2 Translating the 2D homogeneous point by (1,2,3) by (in Cartesian terms) dx=1, dy=2 gives which new homogeneous point?
1. (1,2,3)
2. (1,2,3,4)
3. (2,4)
4. (2,4,3)
5. (4,8,3) *
as Cart: (1/3,2/3)+(1,2) = (4/3,8/3)
12. _____/2 Rotating the 2D Cartesian point (0,1) by 90 degrees gives what:
1. (1,0)
2. (-1,0) * (corrected)
3. (0,1)
4. (0,-1)
5. (-.7,.7)
13. _____/2 Give me a matrix M that has this property: for all vectors p, $Mp = \begin{pmatrix}2&3&1\end{pmatrix} \times p$. The operator is the vector cross product.
$\begin{pmatrix}0 & -1 & 3 \\ 1 & 0 & -2\\ -3 & 2 & 0 \end{pmatrix}$
14. _____/2 In a fragment shader, a sampler2D does something more than a simple array lookup like you have in most programming languages. What?
If the subscripts are fractions, this blends the adjacent texels.
15. _____/2 Tell me about view normalization. What is it? What is good about it?
Transforms the scene and the projection together to make the projection orthogonal down one axis and the clip region a 2x2x2 cube centered on the origin. However a cube in the scene becomes a truncated pyramid. The simpler projection and clip polyhedron are a win. E.g., the hardware is simpler.
16. _____/4 Compute the projection equations when the viewpoint is at (0,0,0) and the projection plane is 2x+y+z=4. Use Cartesian coordinates. You do not need to put the result into a matrix form, but simplify your result as much as possible.
The projection will be a scaling since the center is at the origin. Point (x,y,z) will scale to (ax,ay,az) where 2ax+ay+az=4, so a=4/(2x+y+z). So, the projection is:
x' = 4x/(2x+y+z), y' = 4y/(2x+y+z), z' = 4z/(2x+y+z),
You could start with the matrix mentioned in class and end with this:
$\left( \begin{array}{cccc} 4 &0&0&0\\ 0&4&0&0\\ 0&0&4&0 \\ 2&1&1&0 \end{array} \right)$
17. _____/2 Name the algorithm that was designed to rasterize a circle using only integer additions, subtractions, and shifts.
Bresenham's circle algorithm.
18. _____/2 In a shader, what is the difference between a uniform variable and a varying variable?
A uniform variable is a global constant that is the same for all vertices. A varying variable is computed in the vertex shader and then interpolated across the polygon to give a value at each fragment that is input to the fragment shader.
19. _____/2 Give 2 reasons why we usually use a sequence of low-degree curves instead of one high degree curve.
No local control: changing one control point can change the whole curve. Badly conditioned matrix: a small change in a control point can cause a big change in the curve. Sensitivity to numerical roundoff errors.
20. _____/2 What does the word rational mean in NURBS?
NURBS use homogeneous coordinates, so their Cartesian X, Y, and Z are ratios.
21. _____/2 What is the difference between Phong shading and Gouraud shading?
Gouraud shading computes a color for each vertex and then interpolates it across the polygon (face). Phong shading starts with a surface normal at each vertex of the polygon and interpolates an approximate normal for each pixel of the polygon. Then, in the fragment shader, this approximate normal is normalized an input into the lighting equation that is the subject of the following question.
22. _____/2 In the Phong lighting model, which computes the color or brightness of each pixel, how do you make a surface material more metallic or shiny?
The shininess is the exponent of the dot product of the view vector and reflected vector. Higher means shinier.
23. _____/2 Where is the 3D homogeneous point (1,0,0,0)?
At infinity in the Cartesian direction (1,0,0). Informally, at the end of the positive X axis. The limit of the Cartesian point sequence (x,0,0) as x approaches infinity. The center of projection for a parallel projection down the X axis.
24. _____/2 Consider a 2D cubic Bezier curve with Cartesian control points (0,0), (1,1), (2,2), (3,3). What is the point at t=0.5?
1/8 (0,0) + 3/8 (1,1) + 3/8 (2,2) + 1/8 (3,3) = (3/2, 3/2)
25. _____/2 Describe what the curve in the previous question looks like.
A straight line from (0,0) to (3,3).
26. _____/2 What is Constructive Solid Geometry?
Constructive solid geometry allows a modeler to create a complex surface or object by using Boolean operators to combine simpler objects - Wikipedia.
27. _____/2 What texture mapping problem does the mipmap solve?
The projected size of the texels are usually not close to the size of the pixels. This is a problem because: If the texels are smaller than the pixels, aliasing issues arise. If larger, then several adjacent pixels will be the same color, making the image look posterized. Mipmaps solve this by precalculating a pyramid of texture maps that are coarser and coarser.
28. _____/2 Mathematically, the aliasing problem in CG
1. happens when high frequency signals are not sampled often enough.
2. happens when low frequency signals are sampled too often.
3. is a new problem that did not occur with the old displays.
4. can be reduced by subsampling and averaging
5. both (a) and (d). *
29. _____/2 About polygon clipping:
1. A polygon might gain vertices. *
2. A polygon might gain area.
3. If a polygon's vertices are all outside the clip region, then it may be deleted.
4. Using a 6-stage pipeline keeps the latency the same.
5. Concave polygons become convex.
30. _____/2 A small light source that is twice as far away is 1/4 as bright. That is, there is an inverse square fall off for brightness. However, when modeling light in graphics, we usually don't do that. Why?
Various reasons. The dynamic range of brightness in the scene would then be too great. If the light is bigger than a point and we're not too far away, or if there is ambient light, then the brightness falls off more slowly.
31. _____/2 What tool maps spectral colors into a human perceptual coordinate system? Your use it to determine the visual result of mixing colors.
The CIE Chromaticity diagram.
32. _____/2 With a B-spline, the joints are usually invisible because the two segments meet with C2 continuity. Sometimes a designer may want there to be a visible kink at a joint, i.e., wants only C0 continuity. How can that be achieved?
To reduce C2 continuity to C0 continuity at a joint, make 3 control points coincide.
33. _____/2 What's the hardest part of implementing the painters algorithm?
Sorting the polygons back to front. This is always hard if they overlap in Z and sometimes impossible. Then you have to split a polygon.
34. _____/2 Consider a cube. It has both geometry and topology. What is the geometry and what is the topology?
Geometry: the positions of the vertices. Topology: which vertices (i.e., their id numbers) make up the edges and faces.
35. _____/2 Name the javascript WebGL function that sets the current buffer for future buffer operations.
bindBuffer
36. _____/2 Name the javascript WebGL function that declares that the current buffer contains 3 floats per vertex (or generally some number of some type per vertex).
vertexAttribPointer
End of final exam. Total: 76 points. | 2,753 | 9,965 | {"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} | 3.140625 | 3 | CC-MAIN-2024-38 | latest | en | 0.834285 |
https://www.studyadda.com/index.php?/notes/jee-main-advanced/mathematics/trigonometric-equations/cyclic-quadrilateral/9431 | 1,532,207,170,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592778.82/warc/CC-MAIN-20180721203722-20180721223722-00221.warc.gz | 975,779,794 | 16,550 | Category : JEE Main & Advanced
A quadrilateral PQRS is said to be cyclic quadrilateral if there exists a circle passing through all its four vertices P, Q, R and S.
Let a cyclic quadrilateral be such that
$PQ=a,\,QR=b,\,RS=c$ and $SP=d$.
Then$\angle Q+\angle S={{180}^{o}}$,$\angle A+\angle C={{180}^{o}}$
Let $2s=a+b+c+d$
Area of cyclic quadrilateral = $\frac{1}{2}(ab+cd)\sin Q$
Also, area of cyclic quadrilateral = $\sqrt{(s-a)(s-b)(s-c)(s-d)}$, where $2s=a+b+c+d$ and $\cos Q=\frac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}-{{d}^{2}}}{2(ab+cd)}$.
(1) Circumradius of cyclic quadrilateral : Circum circle of quadrilateral PQRS is also the circumcircle of $\Delta PQR$. $R=\frac{1}{4\Delta }\sqrt{(ac+bd)(ad+bc)(ab+cd)}=\frac{1}{4}\sqrt{\frac{(ac+bd)(ad+bc)(ab+cd)}{(s-a)(s-b)(s-c)(s-d)}}$.
(2) Ptolemy's theorem : In a cyclic quadrilateral PQRS, the product of diagonals is equal to the sum of the products of the length of the opposite sides i.e., According to Ptolemy's theorem, for a cyclic quadrilateral PQRS $PR.QS=PQ.RS+RQ.PS.$
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# When A Projectile Is At The Highest Point Of Its Trajectory, The Directions Of Its Velocity And Acceleration Are
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## Question
### Solution
Correct option is
#### SIMILAR QUESTIONS
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Puzzle 926. pandigital and prime numbers
In the table below I have organized some of the facts that I have just collected or that I have computed about pandigitals and prime numbers.
Pandigitals are one one the two classes: a) composed of the ten decimal digits, 0 to 9, or b) composed of the decimal digits 1 to 9. The first are named just "pandigitals" and the later are named "zero-less pandigitals".
Q1. I invite you to work over the three unknown facts corresponding to the last three rows of the table.
Q2. Do you know some other interesting fact about pandigitals and prime numbers?
Concept Zero-less pandigitals Pandigitals Smallest 123456789 1023456789 Largest 987654321 9876543210 Quantity 9! = 362880 10! = 3628800 Smallest prime factor 2 2 The only prime that divides all the members of the set 3 3 Largest prime factor 109739359 1097393447 Smallest prime that is not factor of any member of the set 44449 111119 The largest prime that only divides just one member of the set 109739359 1097393447 Members of the set with the largest quantity of primes divisors 589234176 =2^16*3^5*37, (22) 6398410752=2^21*3^3*113, (25) Members of the set with the largest quantity of distinct primes divisors 725638914= *3^2*7*11*13*17*23*103 and three more with 8 distinct prime factors 2148736590 = *3^2*5*7*11*13*17*23*61 and 16 more with 9 distinct prime factors Sum from 2 to P giving a pandigital P=92857, 110921 & 112997 (394521678, 547128369 & 572469138, respectively) P=155853, … 441461 (1063254978, …, 7803615924) (13 primes and pandigital values) The smallest prime that only divides just one member of the set of pandigitals Unknown Unknown The quantity of the primes that that only divides just one member of the set of pandigitals Unknown (Please send the whole list of primes and the corresponding pandigital) Unknown (Please send the whole list of primes and the corresponding pandigital) The quantity of the distinct prime factors involved in all the set of pandigitals Unknown Unknown
Contributions came from Emmanuel Vantieghem
***
Emmanuel wrote:
First of all, the number of pandigitals is not 10! but 9*9! = 3265920
Further :
The set of zero-less pandigitals :
* Number of distinct primes involved in the set : 144319
* Number of primes that divide just one number of the set : 109076
smallest such element : 11909 (dividing 142675893)
The set of pandigitals :
* Number of distinct primes involved in the set : 1102173
*Number of primes that divide just one number of the set : 834218
smallest such element : 293339 (dividing 1795234680)
As you can see, the number of primes that divide just only pandigital is quite big, so I send you the results in annex 1annex 2.
There you will find the prime couples {p,m} where p is the prime in question and m the associated pandigital
***
Records | Conjectures | Problems | Puzzles | 805 | 2,903 | {"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} | 3.796875 | 4 | CC-MAIN-2018-26 | latest | en | 0.819283 |
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How long is the distance between Earth and Pluto?
At its most distant, when the two bodies are on the opposite sides of the sun from one another, Pluto lies 4.67 billion miles (7.5 billion kilometers) from Earth. At their closest, the two are only 2.66 billion miles (4.28 billion km) apart.Feb 20, 2016
How long would it take to travel to Pluto?
With a maximum velocity of 590 miles per hour, the trip to Pluto will only take about 680 years. Which really puts things into perspective when considering just how wild it is that we have a spacecraft about to reach Destination Pluto. Launched in January of 2006, it now travels at more than 50,000 miles per hour.
Can you see Earth from Pluto?
As you get farther away, the sun's glare begins to overwhelm the view of Earth. From Pluto – even though Earth would be bright enough to see – you probably couldn't see it in the sun's glare. So that is the answer to the question of how far you could be from Earth, and still see it with your own eyes.Feb 5, 2021
What is the minimum distance between Earth and Pluto?
With due consideration of the high inclination 17.2o of Pluto's orbital plane to our ecliptic, the least lower bound for the minimum Earth-Pluto distance is 4.29 B km = 2.66 B miles, nearly.Mar 30, 2016
How long is a day on Pluto?
On approach in July 2015, the cameras on NASA's New Horizons spacecraft captured Pluto rotating over the course of a full “Pluto day.” The best available images of each side of Pluto taken during approach have been combined to create this view of a full rotation. Pluto's day is 6.4 Earth days long.Nov 20, 2015
Why is Pluto no longer a planet NASA?
Answer. The International Astronomical Union (IAU) downgraded the status of Pluto to that of a dwarf planet because it did not meet the three criteria the IAU uses to define a full-sized planet. Essentially Pluto meets all the criteria except one—it “has not cleared its neighboring region of other objects.”Nov 19, 2019
Can a human survive on Pluto?
No. The average surface temperature there is -229 C (that's -380 F for us in the United States), and there's no breathable oxygen (it's all frozen), and nothing to eat or any way to produce something to eat. Long-term survival at or near Pluto would require technology that hasn't been invented yet.
How long does it take reach Mars?
The trip to Mars will take about seven months and about 300 million miles (480 million kilometers). During that journey, engineers have several opportunities to adjust the spacecraft's flight path, to make sure its speed and direction are best for arrival at Jezero Crater on Mars.
Where is Voyager 1 now?
Voyager 1 is currently in the constellation of Ophiucus. The current Right Ascension of Voyager 1 is 17h 13m 23s and the Declination is +12° 02' 11” (topocentric coordinates computed for the selected location: Greenwich, United Kingdom [change]).
How far is Pluto from Sun?
From an average distance of 3.7 billion miles (5.9 billion kilometers), Pluto is 39 astronomical units away from the Sun. One astronomical unit (abbreviated as AU), is the distance from the Sun to Earth. From this distance, it takes sunlight 5.5 hours to travel from the Sun to Pluto.Aug 6, 2021
Is Pluto visible from Earth without a telescope?
Can you see Pluto without a telescope? No, the dwarf planet is too far and too small to reflect enough of the Sun's light to be visible from planet Earth without a telescope. Because of the distance away from Earth, Pluto requires the strongest of telescopes to have the best possible view.
Can you see the sun in space?
Our sun is one object in the sky that everyone can see, but no one can look at. Far and away the brightest object in the sky, the sun is easy to find, but it's so bright that one can't look directly at it without vision damage. ... SOHO stands for Solar and Heliospheric Observatory.Apr 10, 2009
How big is the universe?
The proper distance—the distance as would be measured at a specific time, including the present—between Earth and the edge of the observable universe is 46 billion light-years (14 billion parsecs), making the diameter of the observable universe about 93 billion light-years (28 billion parsecs).
What is the closest planet to Earth?
Although Venus is the planet that comes closest to Earth as it sweeps by on its orbit, Mercury stays the closest to Earth the longest, according to a commentary published Tuesday (March 12) in the magazine Physics Today.Mar 15, 2019
What planet is the longest distance from the Earth?
Our solar system is so big it is almost impossible to imagine its size if you use ordinary units like feet or miles. The distance from Earth to the Sun is 93 million miles (149 million kilometers), but the distance to the farthest planet Neptune is nearly 3 billion miles (4.5 billion kilometers).
How far to Pluto from Earth?
• The closest distance between the Earth and Pluto occurs when Earth is at its most distant from the Sun, and Pluto is at its closest. And the Sun, Earth and Pluto are lined up in a perfect line. The distance would be about 4.2 billion km.
How is Pluto compared to Earth?
• Pluto is NOT bigger than Earth. Pluto is smaller than Earth at just 1400 miles wide, which translates to 2300km or half the width of the United States. Compared to Earth as an entire planet, it’s only about 18% of Earth’s diameter. When you look at surface area alone, the difference is even greater. Pluto has just...
How is Pluto different from Earth?
• Pluto is very different from all the other planets in the Solar System, especially Earth. Pluto is a cold and dark place, as it is very far from the sun. It is so cold that no life could ever exist on the lonely planet.
Why is Pluto no longer a planet?
• Pluto follows the first two rules: It is round, and it orbits the sun. It does not, however, follow the third rule. It has not yet cleared the neighborhood of its orbit in space. Because it does not follow this rule, Pluto is no longer considered a planet. | 1,385 | 6,023 | {"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} | 3.109375 | 3 | CC-MAIN-2023-14 | latest | en | 0.938482 |
http://www.nag.com/numeric/cl/nagdoc_cl23/examples/source/g02hlce.c | 1,438,105,911,000,000,000 | text/plain | crawl-data/CC-MAIN-2015-32/segments/1438042982013.25/warc/CC-MAIN-20150728002302-00022-ip-10-236-191-2.ec2.internal.warc.gz | 597,398,410 | 2,153 | /* nag_robust_m_corr_user_fn (g02hlc) Example Program. * * Copyright 2002 Numerical Algorithms Group. * * Mark 7, 2002. * Mark 7b revised, 2004. */ #include #include #include #include static void NAG_CALL ucv(double t, double *u, double *ud, double *w, double *wd, Nag_Comm *comm); int main(void) { /* Scalars */ double bd, bl, tol; Integer exit_status, i__, indm, j, k, l1, l2, m, maxit, mm, n, nit, nitmon; Integer pdx; NagError fail; Nag_OrderType order; Nag_Comm comm; /* Arrays */ double *a = 0, *cov = 0, *theta = 0, *userp = 0, *wt = 0, *x = 0; #ifdef NAG_COLUMN_MAJOR #define X(I, J) x[(J-1)*pdx + I - 1] order = Nag_ColMajor; #else #define X(I, J) x[(I-1)*pdx + J - 1] order = Nag_RowMajor; #endif INIT_FAIL(fail); exit_status = 0; printf( "nag_robust_m_corr_user_fn (g02hlc) Example Program Results\n"); /* Skip heading in data file */ scanf("%*[^\n] "); /* Read in the dimensions of X */ scanf("%ld%ld%*[^\n] ", &n, &m); /* Allocate memory */ if (!(a = NAG_ALLOC(m*(m+1)/2, double)) || !(cov = NAG_ALLOC(m*(m+1)/2, double)) || !(theta = NAG_ALLOC(m, double)) || !(userp = NAG_ALLOC(2, double)) || !(wt = NAG_ALLOC(n, double)) || !(x = NAG_ALLOC(n * m, double))) { printf("Allocation failure\n"); exit_status = -1; goto END; } #ifdef NAG_COLUMN_MAJOR pdx = n; #else pdx = m; #endif /* Read in the X matrix */ for (i__ = 1; i__ <= n; ++i__) { for (j = 1; j <= m; ++j) scanf("%lf", &X(i__, j)); scanf("%*[^\n] "); } /* Read in the initial value of A */ mm = (m + 1) * m / 2; for (j = 1; j <= mm; ++j) scanf("%lf", &a[j - 1]); scanf("%*[^\n] "); /* Read in the initial value of theta */ for (j = 1; j <= m; ++j) scanf("%lf", &theta[j - 1]); scanf("%*[^\n] "); /* Read in the values of the parameters of the ucv functions */ scanf("%lf%lf%*[^\n] ", &userp[0], &userp[1]); /* Set the values of remaining parameters */ indm = 1; bl = 0.9; bd = 0.9; maxit = 50; tol = 5e-5; /* Change nitmon to a positive value if monitoring information * is required */ nitmon = 0; comm.p = (void *) userp; /* nag_robust_m_corr_user_fn (g02hlc). * Calculates a robust estimation of a correlation matrix, * user-supplied weight function plus derivatives */ nag_robust_m_corr_user_fn(order, ucv, indm, n, m, x, pdx, cov, a, wt, theta, bl, bd, maxit, nitmon, 0, tol, &nit, &comm, &fail); if (fail.code != NE_NOERROR) { printf("Error from nag_robust_m_corr_user_fn (g02hlc).\n%s\n", fail.message); exit_status = 1; goto END; } printf("\n"); printf("nag_robust_m_corr_user_fn (g02hlc) required %4ld " "iterations to converge\n\n", nit); printf("Robust covariance matrix\n"); l2 = 0; for (j = 1; j <= m; ++j) { l1 = l2 + 1; l2 += j; for (k = l1; k <= l2; ++k) printf("%10.3f%s", cov[k - 1], k%6 == 0 || k == l2?"\n":" "); } printf("\n"); printf("Robust estimates of theta\n"); for (j = 1; j <= m; ++j) printf(" %10.3f\n", theta[j - 1]); END: if (a) NAG_FREE(a); if (cov) NAG_FREE(cov); if (theta) NAG_FREE(theta); if (userp) NAG_FREE(userp); if (wt) NAG_FREE(wt); if (x) NAG_FREE(x); return exit_status; } void NAG_CALL ucv(double t, double *u, double *ud, double *w, double *wd, Nag_Comm *comm) { double t2, cu, cw; double *userp = (double *) comm->p; /* Function Body */ cu = userp[0]; *u = 1.0; *ud = 0.0; if (t != 0.0) { t2 = t * t; if (t2 > cu) { *u = cu / t2; *ud = *u * -2.0 / t; } } /* w function and derivative */ cw = userp[1]; if (t > cw) { *w = cw / t; *wd = -(*w) / t; } else { *w = 1.0; *wd = 0.0; } return; } | 1,272 | 3,404 | {"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} | 2.734375 | 3 | CC-MAIN-2015-32 | latest | en | 0.446301 |
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