Q stringlengths 18 13.7k | A stringlengths 1 16.1k | meta dict |
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Find G. C. F. of $(8n^3 + 8n, 2n+1)$ I'm stuck with this problem, I divided $8n^3 + 8n$ by $2n+1$ and obtained $5$, so now my G. C. F is $\gcd(2n+1, -5)$.
What's next? I can't divide $2n+1$ by $-5$.
| Well, you figured out that $\gcd(8n^3 +8n, 2n+1) = \gcd(2n+1, -5)$.
And as $\gcd(\pm a, \pm b) = \gcd(a b)$ we know $\gcd(8n^3 + 8n, 2n+1) = \gcd(2n+1,5)$.
As $5$ is prime then $\gcd(2n+1, 5)$ is either $1$ or $5$.
It is $5$ if $5|2n+1$. ANd it is $1$ if $5\not\mid 2n+1$.
And $5|2n+1 \iff$
$2n+1 \equiv 0 \pmod 5 \iff$
$2n \equiv -1 \pmod 5 \iff$
$n \equiv 2 \pmod 5$.
So the answer is:
$\gcd(8n^3+8n, 2n+1) =\begin{cases} 5 &\text{if } n\equiv 2 \pmod 5\\1&\text{if } n\not\equiv 2 \pmod 5\end{cases}$
| {
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Show that if $x^p-x-1$ is reducible in $F[x]$ where $F$ has characteristic $p$ then it splits in $F[x]$ into monic distinct factors. So far, I have shown that the function is separable, after all, $D_f(x)=-1$ so they are relatively prime. I am having trouble showing that it splits. Our field $F$ contains $GF(F_p)$ thus it contains the splitting field of $x^n-1$. However, i am not sure how helpful that is.
| Hint: $x$ is a root means that $x^p = x+1$; but $(x+1)^p = x^p+1$, so $x+1$ is a root too.
| {
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Prove that $ST = TS$ if $S$ and $T$ have same eigenvectors
Suppose $V$ is finite-dimensional, $T \in \mathcal{L}(V)$ has dim $V$ distinct eigenvalues, and $S \in \mathcal{L}(V)$ has the same eigenvectors as $T$ (not necessarily with the same eigenvalues). Prove that $ST = TS$.
My general thinking is take $v \in V$ as a linear combination of the eigenvectors $v_1 + \dots + v_n$. Then
$$S(Tv) = S(\lambda_Tv) = \lambda_S\lambda_Tv = \lambda_T \lambda_S v = T(\lambda_S v) = TSv$$
Something feels incomplete and I think this is the right direction, but it doesn't feel like I am using everything that is given to me. Not sure where to go from here or how to troubleshoot this myself
| Let $\lambda_1,\dots,\lambda_n$ be the distinct eigenvalues of $\textsf{T}$. Let $v_1,\dots,v_n$ be the corresponding eigenvectors. Also, let $\mu_1,\dots,\mu_n$ be the eigenvalues of $\textsf S$.
So, we have
$$\textsf{T}(v_i) = \lambda_iv_i \qquad (i=1,2,\dots,n)$$
and
$$\textsf{S}(v_i) = \mu_iv_i \qquad (i=1,2,\dots,n)$$
Now, since $\{v_1,\dots,v_n\}$ is a basis for $\textsf V$, it is sufficient to show that $(\textsf{ST})(v_i)=(\textsf{TS})(v_i)$ for all $i$. But this is easy, just see :
$$\begin{align}
(\textsf{ST})(v_i) &= \textsf{S}(\textsf{T}(v_i)) = \textsf{S}(\lambda_iv_i) \\
&= \lambda_i\textsf{S}(v_i) = \lambda_i(\mu_iv_i) = \mu_i(\lambda_iv_i) \\
&= \mu_i\textsf{T}(v_i) = \textsf{T}(\mu_iv_i) \\
&= \textsf{T}(\textsf{S}(v_i)) = (\textsf{TS})(v_i)
\end{align}$$
| {
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Artin's theorem and Brauer's theorem on Characters
Artin: Let $\chi$ be rational valued complex character of (finite) group $G$. Then $\chi$ can be written as $\mathbb{Q}$-linear combination of characters $1_H^G$ for some cyclic subgroups $H$ of $G$.
In the theorem of Brauer, the subgroups $H$ are allowed to be elementary subgroups it is quite general than Artin's theorem in following sense:
Brauer: Every irreducible complex character $\chi$ of $G$ can be written as $\mathbb{Z}$-linear combinations of characters $\lambda_H^G$ for some subgroups $H$ of $G$, which are elementary subgroups, and $\lambda$ is a linear character of $H$.
Q. In the theorem of Brauer, comparing with Artin,
1) subgroups $H$ are allowed to be elementary (which include cyclic subgroups of $G$ also).
2) For elementary subgroups $H$, we consider $\lambda_H^G$, with $\lambda$ a linear character of $H$, not necessarily $1$.
I was wondering, what will happen if we consider subgroups $H$ to be elementary but the characters of $G$ to be $1_H^G$ (instead of $\lambda_H^G$); these induced characters are rational valued; so by integral combination of such characters of $G$, can we get all the rational valued characters of $G$?
A (sub)group $K$ is said to be $p$-elementary, for a prime $p$, if $K=C_m\times P$ where $C_m$ is cyclic group of order coprime to $p$ (may be trivial) and $P$ is a $p$-group (may be trivial). We say that $K$ is elementary (sub)group if it is $p$-elementary for some prime $p$.
| The answer seems to be negative:
If we take the characters $\{ (1_H)^G \,\, | \,\, H \mbox{ is elementary subgroup of $G$} \}$, then an integer valued character of $G$ may not be integral combination of these specific induced characters.
Consider the quaternion group $G=Q_8=\langle x,y\rangle$. All subgroups of $Q_8$ are elementary subgroups:
$$1, \,\, \langle x^2\rangle,\,\, \langle x\rangle,\,\, \langle y\rangle,\,\, \langle xy\rangle,\,\, G.$$
Consider trivial characters of these subgroups, and induce to $G$, and denote them, respectively by $\psi_1, \ldots, \psi_6.$
Consider $\chi$ the degree $2$ complex irreducible character of $G$, it is integer values.
Suppose $\chi=a_1 \psi_1 + \cdots + a_6\psi_6$ with $a_i\in\mathbb{Z}$. Then
$$(*) \,\,\,\,\,\,\,2=\chi(1)=a_1 \psi_1(1)+\sum_{i>1} a_i\psi_i(1) \mbox{ and } -2=\chi(x^2)=a_1 \psi_1(x^2)+\sum_{i>1} a_i\psi_i(x^2)$$
Note for $2\le i\le 6$, $\psi_i(1)=\psi_i(x^2)$; whereas, $\psi_1(1)=8$ and $\psi_1(x^2)=0$.
Subtracting equations in (*) and noting that the five-term sums in both are same, we get
$4=8a_1$, so $a_1\notin \mathbb{Z}$.
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Is a function absolutely continuous if and only if its derivative is in $L^1$? So I was reading about the Fundamental Theorem of Calculus for Lebesgue Integrals. It said $F$ is absolutely continuous on $[a,b]$ iff $F'$ exists $a.e.$, $F'\in L^1$ and $$F(x)-F(a)=\int_a^x F'dm.$$
Well, but the Fundamental Theorem of Calculus for Riemann Integrals we know that if $F$ is differentiable everywhere, then $$F(x)-F(a)=\int_a^xF'(x)dx.$$
Given that the Riemann Integral, when it exists, equals the Lebesgue integral for $L^1$ functions, would this mean that if $F$ is not absolutely continuous, would I be correct in concluding that $F'\not\in L^1.$ In particular, at least one of
$$\int_a^b (F')^+dm,$$
$$\int_a^x(F')^-dm$$
was infinite.
| Correction
If a differentiable function $f$ is continuously differentiable except a measure-zero set of points on a closed interval, then Newton-Leibniz formula holds for $f$.
There is a function called Volterra's function differentiable everywhere, having a bounded derivative but its derivative is not Riemann integrable.
Answer
If you assume the differentibility of $f$ on the whole closed interval, then the answer is positive.
In fact, we have the following conclusion.
If $f$ is a continuous function of differentibility except countable points on closed interval $[a,b]$, then following conditions are equivalent:
(1) Newton-Leibniz formula holds for $f$ and $f'$ on every subinterval of $[a,b]$ in the sense of Lebesgue integral.
(2) $f$ is of absolute continuity.
(3) $f$ is of bounded variation.
(4) $f'$ is Lebesgue integrable.
Reference: J. Yeh, Real Analysis Theory of Measure and Integration, pp. 272-274.
Puzzlement
If we denote the set of absolutely continuous functions by $AC[a,b]$, the set of functions of bounded variation by $BV[a,b]$, the set of continuous functions which are differentiable except countable points on $[a,b]$ by $ D^{ec}[a,b] $, and the set of functions a.e. differentiable whose derivatives are included in $L^1[a,b]$ by $L^{1,1}[a,b]$.
It is clear that $BV[a,b]\cap D^{ec}[a,b]=L^{1,1}[a,b]\cap D^{ec}[a,b]=AC[a,b]\cap D^{ec}[a,b]$.
I wonder if $AC[a,b]\subseteq D^{ec}[a,b]$. If true, we have $L^{1,1}[a,b]\cap D^{ec}[a,b]=AC[a,b]$, that is amazing. Or could we construct an absolutely continuous function undifferentiable on an uncountable set?
| {
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Let $a_n$ and $b_n$ be sequences . if ... $\lim _{n\to \infty } |a_n - b_n| = 1$ prove that $b_n$ is Convergent Let $a_n$ and $b_n$ two sequences .
if $a_n$ is Convergent
and $\lim _{n\to \infty } |a_n - b_n| = 1$
prove that $b_n$ is also Convergent
I know that it is not true but I need to find example I think
$a_n=\frac{1-n}{n}$
$b_n = \left(-\frac{1}{2}\right)^n$
Is this correct and if not how to prove it ?
thanks
| Hint: Try $a_n=0$ and $b_n=(-1)^n$.
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A subset of a set with a specified property Consider the set $A=\{1,2,3,...,2000\}$. We are asked to find a subset of maximum size so that the difference of any two members of this subset is not a prime number. I think one subset of maximum size can be $\{1,5,..., 1397\}$. i.e., numbers of the form $4k-3$ for natural $k$. It is easy to show that one can not add more elements to this set as its difference with some element would be $ 2,3$ or $5$. Hence the maximum size of such subsets is $500$. Now we have to prove that in any set of $501$ elements chosen from $A$, there are always two elements whose difference is a prime number. To do so, we apply the pigeonhole principle to the holes $\{1,2,3,4\}, \{5,6,7,8\}, \dots \{1997,1998,1999,2000\}$ and $501$ numbers as pigeons. Please help me if the proof is complete or not. If not, how to show that $500$ is the maximum size? What are the suitable holes?
Thank you.
| Here is an argument for not more than $500$ elements of $S$.
We will use your pigeonholes of $\{1,2,3,4\}, \{5,6,7,8\},...\{1997,1998,1999,2000\}$.
That is, the pigeonhole sets are of the form $\{4k+1,4k+2,4k+3,4k+4\}$ for $k=0,1,...,499$.
Let $S$ be the set with no prime differences that we are constructing.
Suppose there are two elements in $S$ from any of the pigeonhole sets, they must differ by $1$ (since the other options are to differ by $2$ or $3$, both prime). I claim that (1) there are then no numbers from the next (greater) pigeonhole set in $S$; and (2) there are no numbers from the previous pigeonhole set.
For (1) there are three cases:
Case 1: $4k+1$ and $4k+2$ are in $S$, but then each element of the next pigeonhole set ($\{4k+5, 4k+6, 4k+7, 4k+8\}$ cannot be in $S$. [We have $(4k+5)-(4k+2)$ is prime; $(4k+6)-(4k+1)$ is prime, $(4k+7)-(4k+2)$ is prime, and $(4k+8)-(4k+1)$ is prime.]
Cases 2 and 3: Where respectively $4k+2$ and $4k+3$ are in $S$ and $4k+3$ and $4k+4$ are in $S$ are similar.
For (2) (nothing in the previous pigeonhole set), the cases are similar.
We conclude that if there are two elements from one pigeonhole, then this is balanced out by no elements from the next or previous pigeonhole set. So again there are at most $2$ out of any $8$ consecutive numbers in $S$. So $S$ cannot contain more than $500$ elements.
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Solving circular arrangement problem with k identical and m distinct positions. There are 11 chairs around a circular table. In how many ways can we arrange 10 people in these seats?
If
(i) There are 11 identical chairs placed equally apart around the table
My solution is:
First fix the one person in any of the chair.
Then permute other 9 people in remaining 10 seats.
i.e $10P9$ which is same as $10!$
(ii) If there are 11 distinctly coloured chairs placed equally apart around the table.
My solution is:
All the chairs are distinct.
So arrangements like
$A_1A_2A_3A_4A_5A_6A_7A_8A_9A_{10}$ and $A_{10}A_1A_2A_3A_4A_5A_6A_7A_8A_9$ would be different.
So it is like linear arrangement.
Arrangement of 10 people in 11 chairs can be done in $11P10$, ways which is same as $11!$
(iii) If there are 10 identically coloured chair and 1 chair is distinctly coloured.
My solution is:
First fix one person in the coloured seat.
Then permute remaining 9 people in remaining 10 seats.
i.e $10P9$ which is same as $10!$
The answer key says
(i)$9!$
(ii)$11!$
(iii)$11!$
Consider that I am a semi-beginner and learning this subject.
What am I doing wrong?
| In part I, Fix the empty chair and $10$ people can be seated in 10 identical chairs in $10P10= 10!$ ways.
In part II, by fixing one chair, you have made the chairs in a linear fashion. Then there are 11 ways a person can be seated, 10 ways the second person can be seated and so on until the last person can be seated in 2 ways to give $(11*10*9...2) = 11!$ ways.
In part III, similar reasoning, by fixing the coloured chair, you have made the chairs in a linear fashion. Then there are 10 ways a person can be seated in the colored chair and the rest of 9 could be seated in the 10 identical chairs in $10P9$ ways + You can keep the colored chair empty and seat the 10 in 10 identical chairs in $10P10$ ways to a total of $(10.10!+10!) = 11!$ ways.
| {
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Findind Hamiltonian functions for flow Consider the symplectic space $(\mathbb{R}^2, \omega_{st}=dx \wedge dy$. I want to find a Hamiltonian function, such that its time-1-flow is of the form
$\varphi(x,y)= (2x, \dfrac{1}{2} y)$
My ansatz:
Consider $\varphi_t(x,y):=(2xt, \dfrac{1}{2} y t)$. Now
$\dfrac{d}{dt} \varphi_t(x,y)= (2x, \dfrac{1}{2} y)$
The corresponding vector field would be $X(x,y)= 2x \ dx + \dfrac{1}{2} y \ dy $.
Let $v= a \ dx + b \ dy$
Then $\omega_{st}(X,v)=2xb-\dfrac{1}{2} y a$
Is that correct so far or is it complete nonsense?
| Let's look for a flow of the form $$\varphi_t(x,y) = \left(2xf(t), \dfrac{y}{2f(t)}\right)$$ for some function $f(t)$. No matter what happens, this'll always be a symplectomorphism. Since $\varphi_0 = {\rm Id}_{\Bbb R^2}$, we need $f(0) = 1/2$ and $\varphi_1= \varphi$ gives $f(1) = 1$. Moreover, the group-property of the flow now implies that $f$ is affine. So we necessarily have $f(t) = (t+1)/2$. With this we have that $$X_{(x,y)} = \frac{{\rm d}}{{\rm d}t}\bigg|_{t=0} \varphi_t(x,y) = \frac{{\rm d}}{{\rm d}t}\bigg|_{t=0} \left(x(t+1), \frac{y}{t+1}\right) = x\partial_x\big|_{(x,y)}-y\partial_y\big|_{(x,y)}.$$Now we have that $$\omega(X, \cdot) = ({\rm d}x\wedge {\rm d}y)(x\partial_x-y\partial_y, \cdot) = \begin{vmatrix} x & {\rm d}x \\ -y & {\rm d}y\end{vmatrix} = x\,{\rm d}y + y\,{\rm d}x = {\rm d}(xy). $$Alright, so there you have it: $H(x,y) = xy$.
| {
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Counting the number of partitions of $\mathbb{R}$ into countable subsets I'm trying to determine the number of partitions of $\mathbb{R}$ into countable subsets. Obviously each partition will contain uncountably many sets and the cardinality of the set of all such partitions is bounded above by the number of relations on $\mathbb{R}$ (which is $2^c$) but I haven't gotten any further than that. Thanks for your help!
| The number of such partitions is $2^{\mathfrak c}$.
First, each partition $\pi$ is a collection of disjoint subsets of $\mathbb R$, so it has size at most $\mathfrak c$, and so there are at most $|[\mathcal P(\mathbb R)]^{\le\mathfrak c}|=2^{\mathfrak c}$ partitions, where $[A]^{\le\kappa}$ denotes the collection of subsets of $A$ of size at most $\kappa$.
Second, split $\mathbb R$ as the disjoint union $A\cup B\cup C$, each set of size $\mathfrak c$. Fix a partition $\{D_x:x\in C\}$ of $A$, and a partition $\{E_x:x\in C\}$ of $B$, where each $D_x$ and each $E_x$ is countably infinite. Given $F\subseteq C$, consider the partition $\rho_F$ of $\mathbb R$ into countably infinite sets given by the sets $D_x$ for $x\notin F$, $D_x\cup\{x\}$ for $x\in F$, $E_x\cup\{x\}$ for $x\notin F$, and $E_x$ for $x\in F$. The assignment $F\mapsto \rho_F$ is injective, and therefore there are at least $2^{\mathfrak c}$ partitions of $\mathbb R$ into countably infinite sets.
| {
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How can I express that 2 is the only prime number that is even using predicate logic I defined some predicates below:
$B(a)$: a is a prime number
$C(a, b)$: b is divisible by $a$
Let $a, b$ be integers greater than $1$
My attempt is below, but I am not sure whether it is correct.
$$\forall a\in \mathbb{Z}, a = 2 \Leftrightarrow B(a) \land C(a, 2) $$
| This is also known as a definite descriptions in Bertrand Russell's theory of descriptions,
There is only one $x$ satisify $P$:
$$\exists x_0, P(x_0)\wedge (\forall x_1,P(x_1)\rightarrow x_0=x_1)\tag{1}$$
There is only one $x$ satisify $P$, and that $x$ satisify $Q:$
$$\exists x_0, P(x_0)\wedge (\forall x_1,P(x_1)\rightarrow x_0=x_1)\land Q(x_0)\tag{2}$$
In some sense, this also read as $Q$ is the only thing satisify $P$.
So we want the property $P(x)$ be $x$ is prime and even, $Q$ be $x$ equal to $2$.
(Note that statement didn't claime the uniqueness of "been a $2$".)
Define$P(x):(\forall m,m\mid x\to(m=1\lor m=p))\land 2\mid x\tag{$x$ is prime, also even}$
Also $Q(x):x=2$, then In math notation, we can write:
\begin{align}
&\exists x_0, (\forall m,m\mid x_0\to(m=1\lor m=p))\land 2\mid x_0\\
&\wedge (\forall x_1,(\exists x_1, (\forall m,m\mid x_1\to(m=1\lor m=p))\land 2\mid x_1)\rightarrow x_0=x_1)\land x_0=2\\
\end{align}
Use predicates, that is same as $(2)$.
| {
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Field extension of isomorphism of elliptic curves "If $A,B$ are elliptic curves over a field $k$ of characteristic not 2 or 3 and if $A,$ isomorphic over an extension of $k$, then they become isomorphic over an extension of $k$ of degree $\leq 6$. "
$\textbf{Q:}$ Since $A,B$ are exactly described by $g_2$ and $g_3$ or equivalently their $j$ invariants, suppose $g_2^A=c^4g_2^B$ and $g_3^A=c^6g_3^B$ with $c^4,c^6\in k$. The statement is saying there is an isomorphism over an extension of $k$. The only possibility is degree 2 extension or degree 6 extension. What does the book mean isomorphic over an extension of $k$? I can only see application of such conclusion for degree 2 or degree 6 extension and application of such conclusion for extension beyond 6 is redundant. Should not $j$ invariant automatically conclude isomorphism over arbitrary extension?
Ref. Lang, Elliptic Functions, Chpt 1, Sec 4. pg 18
|
Should not $j$-invariant automatically conclude isomorphism over arbitrary extension?
No, the $j$-invariant only classifies elliptic curves over an algebraically closed field. Curves with the same $j$-invariant may not be isomorphic over $k$ as they could be twists. That is, a curve and its twist might not be isomorphic over $k$, but there is an isomorphism defined over some extension of $k$.
For instance, consider the elliptic curves $E_1 : y^2 = x^3 - 1$ and $E_2: y^2 = x^3 + 1$. One can show that these elliptic curves are not isomorphic over $\mathbb{Q}$. However, there is an easy isomorphism over $\mathbb{Q}(i)$:
\begin{align*}
E_1 &\to E_2\\
(x,y) &\mapsto (-x,iy) \, .
\end{align*}
(Indeed, I cooked up this example by forming the quadratic twist of $E_1$ by $-1$.)
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Continuity of the stochastic process $X_t=\int_0^t(a+b\frac{u}{t}) \, dW_u$ I am wondering about the continuity of the stochastic process
$$X_t=\int_0^t \left(a+b\frac{u}{t}\right)\,dW_u$$
which has variance $t$ and normally distributed for
$a^2+\frac{b^2}{3}+ab=1$
The process seems to be discontinuous at $t=0$ except for $b=0.$
Is this stochastic process really discontinuous or does the limit $t\to 0$ have a finite value
| We have $$\begin{align*}X_t&=\int_0^t \left(a+b\frac{u}{t}\right)\,dW_u \\\\ &=aW_t + \frac{b}{t}\int_0^t u\, dW_u\end{align*}$$
Using Itô's formula we get:
$$\int_0^t u\, dW_u = tW_t - \int_0^t W_s\, ds$$
and so
$$X_t = (a+b)W_t - \frac{b}{t}\int_0^t W_s \, ds$$
We know: $W_s$ is continous a.s. hence from the Fundamental theorem of calculus it follows that $$\int_0^t W_s \, ds$$ is differentiable a.s. in $t=0$ which means nothing else that $$\lim_{t\to 0} \frac{1}{t}\int_0^t W_s \, ds$$ exists a.s.
All together we get: $$\lim_{t \to 0} X_t = \lim_{t \to 0}\frac{b}{t}\int_0^t W_s \, ds$$ exists a.s. and so $X_t$ is continous in $0$.
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Find a triangle such that : area are natural number and sides is prime numbers Find a triangle such that :
$area=S\in\mathbb{N^{*}}$ and the sides $a,b,c$ prime numbers
I need find this triangle not by imagine I need by a prof or something
I try $2,3,5$ , $3,5,11$ , $13,11,7$ but I tried
I would like a explain to find this sides
I have already see your hints and ideas
| One possibility is that the area is $0$, such as your $2,3,5$ attempt. In fact, $2$ together with any twin prime works.
If we don't want the area to be $0$, then I think the most helpful tool we have at our disposal is Heron's formula for the area:
$$
S=\sqrt{s(s-a)(s-b)(s-c)}\\
s=\frac{a+b+c}2
$$
We see that if $s$ is not an integer (and thus is half of some odd integer), then there is no hope for the area to be an integer either (as the radicand necessarily becomes an odd number divided by $16$). And since most primes are odd, that means one of the sides must be $2$. And since we don't want area $0$, we must have an isosceles triangle. Say $a=2$ and $b=c$.
So, with this, we get $s=b+1$, and Heron's formula becomes
$$
S=\sqrt{(b+1)(b-1)\cdot1^2}\\
=\sqrt{b^2-1}
$$
For this to be an integer, we need $b=1$, which 1) isn't a prime, and 2) gives us area $0$. So there are no positive integer area triangles with prime sides.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to nondimensionalize a second order differential equation
A body of mass $m$ is thrown upwards in a vertical direction from the
earth's surface with a velocity $v$. The air resistance is supposed to
be taken into account by Stoke's law $F_R= - cv$ for the flow
resistance in viscous fluids, which is reasonable for small
velocities. Here $c$ is a coefficient depending on the shape and size
of the body. The motion is supposed to depend on the mass $m$, the
velocity $v$, the gravitaional acceleration $g$ and the friction
coefficient $c$ with dimension $[c]= \frac{M}{T}$.
The initial value problem for the height is assumed to take the form
$$ mx''+cx'=-mg$$
$$x(0)=0$$
$$x'(0)=v$$
Non dimensionalize the differential equation.
First I rewrote it to
$$ x'' = -g-\frac{c}{m} x'$$
$$x(0)=0$$
$$x'(0)=v$$
Than I wrote down the fundamental dimensions, mass $M$, time $T$ and height $H$.
After that I determined the dimensions of the involved quantities
$[x] = H$
$[g]=\frac{H}{T^2}$
$[v]=\frac{H}{T}$
$[c]=\frac{M}{T}$
$[m]=M$
$[t]= T$
I thought to nondimensionalize I to define $\tau = \frac{t}{\bar{t}}$, $y=\frac{x}{\bar{x}}$ and $z= \frac{m}{\bar{m}}$ where $\bar{t},\bar{x} and \bar{m}$ are characteristic quantities.
Now I wanted to do a change of variables, and write $x''$ as $y''(\tau)$,but that is where it went wrong.
$$\frac{d^2x}{dt^2}=\frac{d}{dt}(\bar{x}\frac{dy}{dt}) = \bar{x}\frac{d^2y}{dt^2}=0$$
I'm pretty new to this and I tried to look up similar problems and try the methods they used, but I just don't know what to do. Any nod in the right direction is much appreciated :)
| Calling
$$
\cases{
\tau = \frac{t}{t_o}\\
\eta = \frac{x}{x_o}
}
$$
we have
$$
\frac{d^n}{dt^n} = \frac{1}{t_o^n}\frac{d^n}{d\tau^n}
$$
and after substitution
$$
m\frac{x_o}{t_o^2}\eta''+ c\frac{x_o}{t_o}\eta' + m g = 0
$$
or
$$
\eta'' + \frac{c t_o}{m}\eta'+\frac{t_o^2 g}{x_o} = 0
$$
now determining $x_o,t_o$ such that
$$
\cases{
\frac{c t_o}{m} = 1\\
\frac{t_o^2 g}{x_o} = 1
}
$$
we have
$$
t_o = \frac mc,\ \ \ x_o = \frac{m^2}{c^2}g
$$
and consequently
$$
\eta''+\eta'+1=0
$$
etc.
| {
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Reference for Derivation of Higher order Runge Kutta I have a problem about determining $a_1, a_2, k_1, k_2, \ldots a_n, k_n,\ldots$
In the general form of the Higher Order Runge-Kutta below :
$$y_{r+1}=y_r+a_1k_1+a_2k_2+\cdots+a_nk_n$$
For the convenient, i'll write it down The Runge Kutta $2^{\text{nd}}$ in my book just in case if the formula is different from your views:
$$
\begin{align}
k_1&=hf(x_r,y_r)\\
k_2&=hf(x_r+p_1h,y_r+q_{11}k_1)\\
y_{r+1}&=y_r+(a_1k_1+a_2k_2)
\end{align}
$$
I'm not really sure, but some sources talk about Butcher Tableau, slope for each $k_n$, Taylor expansion, and rooted trees, for determining the $a_n$. And i don't really understand what their relation is, especially for Butcher Tableau, and rooted trees.
Could you explain all of these for me? Or just give me a good reference that discusses about derivation of the higher order Runge-Kutta method in detail, please?
Because, some books just skip over the derivation part.
| A good source on Butcher tableaus and B-trees would, almost canonically, be the book by Butcher himself. For a short overview, see the three sets of slides with an introduction to B-trees, demonstration of the method for up to order 4 and outlook to implicit methods, or a historical overview
You could of course also read the most original source, the introduction of Runge-Kutta methods by M. Wilhelm Kutta (1901), "Beitrag zur näherungsweisen Lösung totaler Differentialgleichungen", where the order conditions are established and solved up to order 4, with an order 5 method that turned out not to be correct for systems.
| {
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How to arrive from $x^2y^2－x^2－y^2－6xy＋4$ to $(xy＋x＋y－2）(xy－x－y－2）$? How do we arrive from
$x^2y^2－x^2－y^2－6xy＋4$
to
$(xy＋x＋y－2）(xy－x－y－2）$
?
According to my book these two are equal, but I can't understand how to transform one to another.
| $$x^2y^2-x^2-y^2\color{red}{-6xy}+4=x^2y^2\color{red}{-4xy}+4-x^2\color{red}{-2xy}-y^2\\=(xy-2)^2-(x+y)^2=(xy+x+y-2)(xy-x-y-2)$$
| {
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A symmetric, diagonally dominant matrix A with real positive diagonal entries is positive definite Suppose $A \in \mathbb{R}^{n\times n}$ is symmetric and diagonally dominant with positive diagonal entries. I have to prove that $A$ is positive definite but without using theorems, just algebraically.
I´ve started with: $$x^T A x = \sum_{i=1}^n a_{ii} x_i^2 + \sum_{i=j} a_{ij} x_i x_j > \sum_{i=1}^n \sum_{i\neq j} |a_{ij}| x_i^2 + \sum_{i\neq j} a_{ij} x_i x_j$$ but I could much further. I was thiking about: $$ \sum_{1=1}^n \sum_{j>i} |a_{ij}| (x_i^2 + x_j^2) + 2 a_{ij} x_i x_j$$ but I´m not sure how to continue.
| I guess you write $x^TAx=\sum_{i=1}^n a_{i,i}x_i^2+\sum_{i\neq j}a_{i,j}x_ix_j\ge\sum_{i=1}^n (\sum_{i\neq j}|a_{i,j}|)x_i^2-\sum_{i\neq j}|a_{i,j}||x_i||x_j| =\sum_{j>i }(|a_{i,j}|(x_i^2+x_j^2-2|x_i||x_j|))\ge 0$
| {
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How to find a substitution that transforms an equation into a particular equation. (Roots of polynomials). Find the value of $c$ so that the substitution $x=u+c$ transforms the equation $x^3-12x^2+45x-54=0$ into the equation $u^3-3u^2=0$
My first idea is to find the roots of $u^3-3u^2=0$ And getting $u=0$ and $u=3$, however I am not sure how to continue.
If someone could explain how to tackle this sort of problem, I would greatly appreciate it. Thanks.
| Here's a relatively fast approach that avoids factoring a cubic.
Notice that $u^3 - 3 u^2 = u^2 (u - 3)$ has a double root at $u = 0$, so if $x = u + c$, the given polynomial, $$p(x) := x^3 − 12 x^2 + 45 x − 54$$ has a double root at $x = (0) + c = c$.
Since $c$ is a double root, it is also a root of $$p'(x) = 3 (x^2 - 8 x + 15) = 3 (x - 3) (x - 5) ,$$ so $c = 3$ or $c = 5$. But $5 \not\mid 54$, so the Rational Root Theorem implies that $p(5) \neq 0$, leaving
$$\color{#df0000}{\boxed{c = 3}} .$$
| {
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Why do we get rid of the $-$ sign in this equation I am finding interquartile ranges for the exponential function cdf as part of a project for university:
$$Q\left(\frac{1}{4}\right) = -\frac{1}{\lambda}\ln\left(1-\frac{1}{4}\right)= -\frac{1}{\lambda}\ln\left(\frac{3}{4}\right)= \frac{1}{\lambda}\left(\ln(4) - \ln(3)\right)$$ I do understand the quotient rule of logarithm but the rule states that if I have $\ln(\frac{3}{4})$ this should be the same as $\ln(3) - \ln(4)$ but in the above case is $\ln(4)-\ln(3)$, furthermore, where did the $(-)$ sign go at the end of the equation for $\frac{1}{\lambda}$? what am I missing?
| A minus sign switches the order of subtraction: ${-}(a-b) = b-a$. You can see this by writing ${-}(a-b)$ as ${-}(a + {-}{-}b)$ and distributing the negative: ${-}(a-b) = {-}(a + {-}b) = -a + {-}{-}b = b -a$.
Alternatively: Use the identity ${-}\ln(x) = \ln(1/x)$.
| {
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Riemann upper sum of Riemann integral I read this lecture note https://www.math.ucdavis.edu/~hunter/m125b/ch1.pdf. For all partitions $P$ of finite closed interval $[a,b]$, define the upper Riemann sum $f$ w.r.t. partition $P$ by $U(f; P)$. If $f$ is bounded, then $m(b-a)\leq U(f; P)$, where $m\leq f.$
why $\inf_{P} U(f; P)$ exists? We do not need to claim $U(f; P)$ is decreasing?
And how about the $\lim_{P} U(f; P)$? Here is the limit of the net.
| The collection of partitions $\mathcal{P}$ is directed set with preorder $\supseteq$ where $P' \supseteq P$ indicates that $P'$ is a refinement of $P$, that is the set of endpoints of subintervals of $P$ is contained in the set of endpoints of subintervals of $P'$.
The net $\{U(f;P) \}$ defined on $\mathcal{P}$ is non-increasing in the sense that $P' \supseteq P$ implies that $U(P';f) \leqslant U(P;f)$. In other words, upper Darboux sums decrease as partitions are refined.
Since the net $\{U(f;P) \}$ is bounded below and non-increasing it is convergent.
| {
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Solving system of Congruences with Two Variables (x and y) I know a bit about the Chinese Remainder theorem but what do I do in the case I was asked to solve a system of congruences such as this with two variables:
$3x + y = 7$ (mod 8)
$4x + 3y = 1$ (mod 8)
| The first equation is in a form suggesting substitution method as you have $y$ with coefficient $1$:
*
*$y \equiv 7-3x \pmod 8$
Plug this into the second equation and solve for $x$:
$$4x + 3(7-3x) \equiv 4x +21 -9x \stackrel{21=16+5}{\equiv} -5x +5 \equiv 1 \pmod 8$$
Now, note that the $5^2 \equiv (24+1) \equiv 1 \pmod 8$. Using this you get
$$\Leftrightarrow 5x \equiv 4 \pmod 8 \stackrel{\cdot 5}{\Leftrightarrow} \boxed{x =} 20 \equiv \boxed{4 \pmod 8}$$
Plugging this back into the equation $y \equiv 7-3x\pmod 8$ gives
$$\boxed{y=} 7-3\cdot 4 \equiv -5 \equiv \boxed{3 \pmod 8}$$
| {
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write the given third order linear equation as an equivalent system of first order equations with initial values. Write the given third order linear equation as an equivalent system of first-order equations with initial values:
$$-(‴+3\sin(t)′) = 2t$$
with $$y(3)=−2, y'(3)=3, y''(3)=0$$
Use $_1=$, $_2=′$, and $_3=″$.
I was able to get $_3'$=$−(3\sin()_2)−2$. Does this mean that $x_1 = 0, x_3 =0$ but what is $x_2$ and the constant? What is the $t$?
| You have that :
$$\begin {cases} x_1=y \\ x_2=y'=x'_1 \\x_3=y''=x'_2 \end{cases} \implies \begin {cases} x'_1=x_2 \\ x'_2=x_3 \end{cases} $$
And as you noted you have also that
$$x'_3=-3\sin(t)x_2-2t$$
Therefore you can write the DE equation as :
$$\begin{pmatrix} x_1 \\ x_2\\x_3 \end{pmatrix}'= \begin{pmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & -3\sin(t) & 0 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}+ \begin{pmatrix} 0 \\ 0 \\ -2t \end{pmatrix}$$
The "$t$" is the inhomogeneous part of the equation.
| {
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Intersection of subgroups of given orders and normality Quite confused about how to solve this question..As it would have been easier to solve if the group was given to be cyclic, but no such case here
Let $H$ and $K$ be subgroups of a group $G$ of orders $14$ and $21$ respectively. If $H \cap K \neq\{e\}$, here $e$ is the identity of the group $G$, then show that $H \cap K$ is a normal subgroup of $G$.
| This is not true. The issue/trick is that there is no information about the group $G$ so I can pick it to have very few normal subgroups.
For example, we can take $G=S_n$ for some $n\geq 7$ (why?). Then $G$ has very few normal subgroups (see here), and in particular none of order $7$ (why is $7$ relevant?).
For a concrete counter-example, take $H=\langle (1,2,\ldots,7)(8,\ldots,14)\rangle$ and $K=\langle (1,2,\ldots,7)(8,\ldots,21)\rangle$.
| {
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HCF $(x,y) = 16$ and LCM $(x,y) = 48000$. Then the possible number of pairs $(x,y)$ Let $S$ be the set of all ordered pairs $(x,y)$ of positive integers, with HCF $(x,y) = 16$ and LCM $(x,y) = 48000$. The
number of elements in $S$ is
My Attempt : $48000= 2^7. 3 . 5^3$ As the L.c.m contains $2^7$ as a factor and G.c.d contains $2^4$ as a factor , we can assure that one element contains $2^7$ and the other one contains $2^4$. And $3$ and $5^3$ will be divided between two numbers in a manner so that the L.C.M and G.C.D remain as given.
So the possible pairs are $(2^7 .3 . 5^3 , 2^4)$ , $(2^7 .3 , 5^3.2^4)$ , $(2^7 , 3.5^3.2^4)$ , $(2^7.5^3 , 3.2^4)$.
So I think the number of elements in the set $S$ is $4$. Have I gone wrong anywhere? Can anyone please help me ?
| WLOG let $\dfrac xX=\dfrac yY=16$ so that $(X,Y)=1$
We have $48000\cdot16=xy=16^2XY\iff XY=3000=3\cdot2^3\cdot5^3$
So, the possible values of $X$ can be
take none of the factors $$1$$
take one of the factors $$3;2^3;5^3$$
take two of the factors $$2^3\cdot5^3;3\cdot5^3;2^3\cdot3$$
take all three of the factors $$3\cdot2^3\cdot5^3$$
| {
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Asking about $R$-submodules which are not finitely generated.
Let $R=\mathbb{Z}[X_{1},X_{2},...,X_{n},...]$ the polynomial ring in infinitely many variables with coefficients in $\mathbb{Z}$. Find (and explain why) a $R^{R}$-submodule (being $R^R$ the regular module over $R$) which is not finitely generated.
I need help about this issue, thank you so much.
| This is, I believe, the standard example of the fact that a submodule of a finitely generated module needs not be finitely generated.
Hint: The regular module is finitely generated (as it always is for a unital ring) by $1$. We are looking for a submodule (an ideal of $R$, basically) which is not finitely generated. $R$ is a polynomial ring. What kinds of ideals do you have in a polynomial ring? Can you think of one that's not finitely generated?
| {
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$\int_{0}^{\infty}e^{-x}p_n(x)p_m(x)=0$ when $p_n(x)=(-1)^nn!\sum_{k=0}^{n}\binom{n}{k}\frac{(-x)^k}{k!}$ Let $n \in \mathbb N_0$. Consider the polynomials $p_n$ defined by
$$p_n(x)=(-1)^nn!\sum_{k=0}^{n}\binom{n}{k}\frac{(-x)^k}{k!}$$
I want to show that $\int_{0}^{\infty}e^{-x}p_n(x)p_m(x)=0$ for $n \neq m$.
The hint is that for $m,n \in \mathbb N_0$ we have:
$$\sum_{k=0}^{n}\sum_{j=0}^{m}(-1)^{j+k}\binom{j+k}{k}\binom{n}{k}\binom{m}{j}=\delta_{m,n}$$
$p_n(x)p_m(x)=(-1)^{n+m}n!m!\sum_{k=0}^{n}\sum_{j=0}^{m}(-1)^{j+k}\binom{n}{k}\binom{m}{j}\frac{x^{j+k}}{j!k!}$. But I don't know how to continue or how I can bring this into a form to use the given hint
| In order to show the orthogonality of $p_n$ and $p_m$ (with respect to the inner product $\langle f,g\rangle = \int_{0}^{+\infty}f(x)g(x)e^{-x}\,dx$) for $n\neq m$ it is enough to prove the orthogonality of $p_n(x)$ and $x^j$ for $j<\deg(p_n)=n$. Now
$$\langle p_n(x), x^j \rangle = (-1)^n n!\sum_{k=0}^{n}\binom{n}{k}\frac{(-1)^k}{k!}\int_{0}^{+\infty}x^{k+j}e^{-x}\,dx $$
equals
$$ (-1)^n n!j!\sum_{k=0}^{n}\binom{n}{k}(-1)^k \binom{k+j}{j} $$
where the sum is the $n$-th power of the forward difference operator $\delta:q(x)\mapsto q(x+1)-q(x)$ applied to the polynomial $q(x)=\binom{x+j}{j}$. Since $\deg q = j<n$, the RHS is zero.
| {
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Formula for sum of combinations I am trying to find a closed-form formula or, at least, a different and more useful representation for the following sum of combinations:
\begin{equation}
\sum_{i=1}^{n}\frac{n!}{i!(n-i)!}\times\frac{(-1)^i}{i}
\end{equation}
| We have that
$$\sum_{i=1}^{n}\frac{n!}{i!(n-i)!}\times\frac{(-1)^i}{i}=\sum_{i=1}^{n}\frac{(-1)^i}{i}\binom{n}{i}$$
then refer to
*
*Sum of Pascal's triangle column
*Let $n$ be a positive integer, Prove that $\sum_{k=1}^n\frac{ (-1)^{k-1}}{k}{n \choose k} = H_n$
| {
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Area triangle (possible) The area of a right triangle is an integer greater than $ 85 $. If the hypotenuse measures $ 20 $, what is the area of this triangle?
It’s just a 3-4-5 right triangle, so the area is $\boxed{96}$.
This is true? If so, how to prove it?
But he said the area is integer, not the sides
Can you use Weitzenbock inequality in this problem?
| We need
$$a^2+b^2=20^2\implies a^2=400-b^2$$
which by inspection, taking $a$ and $b$ integers, leads to the unique solution $a=12$ and $b=16$ such that $\frac12 ab\ge 85$ and the area in that case is equal to $96$.
For $a$ and $b$ reals we have
$$\frac12ab=\frac12a\sqrt{400-a^2}=86 \implies a\sqrt{400-a^2}=172 $$$$\implies a=\sqrt{186}\pm\sqrt{14},\quad b=\sqrt{186}\mp\sqrt{14}$$
which is the unique solution for the area equal to $86$.
In a similar way we can obtain integer solutions up to $100$ which is the largest area we can obtain for $a=b=10 \sqrt 2$.
| {
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Show that the smallest eigenvalue is strictly positive. I have a $2\times 2$ real, symmetric and positive definite matrix $B_{x,n}$ which depends on a point $x\in[0,1]$ and $n\in\mathbb{R}$. I want to show that for sufficiently large $n$, the smallest eigenvalue of $B_{x,n}$ is strictly positive uniformly on $x$.
To do so, I managed to show that $B_{x,n}\to B$ as $n\to\infty$ with $B$ positive definite (and so, has only strict positive eigenvalues).
It is known that the smallest eigenvalue equals the solution $\inf_{\lVert z \rVert=1}z'B_{x,n}z$. Then for each nonzero $z\in\mathbb{R}^2$, $z'B_{x,n}z\to z'Bz>0$. However, I believe that the desired result can be shown only if $z'B_{x,n}z\to z'Bz>0$ uniformly on $\{z:\lVert z\rVert=1\}$. Is it right?
Can you give me suggestions?
This is Lemma 1.5 of Tsybakov's Introduction to nonparametric estimation, by the way.
| Since $B$ is positive definite then $\det(B)>0$ and $\operatorname{tr}(B)>0$.
Since the determinant and the trace are continuous functions and $\displaystyle\lim_{n\rightarrow \infty}B_{x,n}=B$ then $\displaystyle\lim_{n\rightarrow \infty}\det(B_{x,n})=\det(B)$ and $\displaystyle\lim_{n\rightarrow \infty}\operatorname{tr}(B_{x,n})=\operatorname{tr}(B)$.
So there is $N$ such that if $n>N$ then $\det(B_{x,n})>\frac{1}{2}\det(B)>0$ and $\operatorname{tr}(B_{x,n})>\frac{1}{2}\operatorname{tr}(B)>0$.
Therefore the product and the sum of the two eigenvaues of $B_{x,n}$ are positive for $n>N$. So the eigenvalues are positive for $n>N$.
| {
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Why is $X^*$ with the weak-* topology not a locally compact vector space? I know that locally compact Hausdorff topological vector space must be isomorphic to $\mathbb{C}^d$ or $\mathbb{R}^d$ for some $d\in \mathbb{N}$.
However the Banach Alaouglu theorem says that the closed unit ball of $X^*$ is compact with respect to the weak-* topology, which seemed to me at first sight to mean that it is locally compact. But I know that this space with the weak-* topology is also Hausdorff, and allegedly should mean that $X^*$ with the weak-* topology is finite dimensional.
My question is then why is the closed unit ball in $X^*$ not a compact neighbourhood of $0$?
I found this thread saying that exactly, but I do not understand why. Is the unit ball of $X^*$ not a neighbourhood of $0\in X^*$ with the weak-* topology, which is contained in the aforementioned set?
| The open unit ball in the strong topology is NOT open in the weak$^*$ topology. The weak$^*$ topology is way coarser than the strong one (less open sets).
To get an idea, consider $\ell^\infty (\mathbb N)$. A local basis of open neighbourhoods in the strong topology are the sets
$$
U_n=\left\{(a_k)_{k\in\mathbb N}: |a_k|< 1/n,\,\,\text{for all $k\in\mathbb N$} \right\}, \,\,n\in\mathbb N.
$$
In the case of the weak$^*$ topology, a local basis of open neighbourhoods are finite intersections of the sets
$$
W_{m,n}=\left\{(a_k)_{k\in\mathbb N}: |a_m|\le 1/n\right\}, \,\,m,n\in\mathbb N.
$$
Each $\overline U_n=\left\{(a_k)_{k\in\mathbb N}: |a_k|\le 1/n,\,\,\text{for all $k\in\mathbb N$} \right\}$ is compact in the weak$^*$ sense, but its interior is NOT open in the strong sense.
| {
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Computing a sum of fractions using sigma notation Evaluate the following using sigma notation
$$\frac34 + \frac65 + \frac96 + \frac{12}{7} + \frac{15}{8}$$
For the denominator I get $\sum_{i=4}^8 i$ but how about the numerator?
| What about
$$\sum_{i=4}^8 \frac{3(i-3)}{i}=3\sum_{i=4}^81-9\sum_{i=4}^8\frac1i$$
| {
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$f : X\longrightarrow Y$ isomorphism $ \Rightarrow (?)\, \, f^*:Div(Y)\longrightarrow Div(X)$ isomorphism. Let $X$ be a complex manifold. Denote by Div$(X)$ the Weil divisors group of $X$.
We have to:
Let $f : X \longrightarrow Y$ be a holomorphic map of connected
complex manifolds and suppose that $f$ is dominant, i.e. $f(X)$ is dense in $Y$.
Then the pull-back defines a group homomorphism
$$
f^* :Div(Y) \longrightarrow Div(X).
$$
Question: If $f : X \longrightarrow Y$ is an analytic isomorphism ( bi-holomorphic map), then is it true that $f^*:Div(Y)\longrightarrow Div(X)$ is a group isomorphism?
Thanks
| The inverse map gives rise to a map on Div's in the opposite direction. Since $\text{Div}(\cdot)$ is functorial, and composition of $f$ with its inverse gives the identity map, this means that the map on Div's is inverse to the original one.
| {
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Proving sequence $(\frac{e^n}{n^{100}})_{n\in\mathbb{N}}$convergence I can see that since $e^n$ grows asymptotically faster than the polynomial $n^{100}$ as n approaches to $\infty$, $e^n>0$ and $n^{100}>0$, so $\lim_{n\to\infty} \frac{e^n}{n^{100}}=\infty$.
But I'm stuck on proving the convergence of the sequence $(\frac{e^n}{n^{100}})_{n\in\mathbb{N}}$. Could anyone guides me? Thanks in advance.
| Let $k$ be any fixed positive integer (e.g. 100, in your example).
$$\dfrac{e^n}{n^k}=\dfrac{\sum_{i=0}^\infty n^i/i!}{n^k} > \dfrac{n^{k+1}/(k+1)!}{n^k} = \dfrac{n}{(k+1)!}\to\infty $$
A single term of the series for $e^n$ dominates the power function $n^k$
| {
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How to find the second derivative of y in $y^2 = x^2 + 2x$? I have a problem to solve:
use implicit differentiation to find $\frac{dy}{dx}$ and then $\frac{d^2y}{dx^2}$. Write the solutions in terms of x and y only
It means that I need to differentiate the equation one time to find $y'$ and then once more to find $y''$.
The correct answer from the textbook is $y' = \frac{x + 1}{y}$ and $y'' = \frac{x^2 + 2x}{y^3}$. I got the first derivative right, but I can't understand how did they get the second one, or is it a typo (unlikely), since I have $y'' = \frac{1}{y} - \frac{(x + 1)^2}{y^3}$
I did this:
$$
y^2 = x^2 + 2x\\
2yy' = 2x + 2\\
yy' = x + 1\\
y' = \frac{x + 1}{y}\\
$$
I tried to get to the second derivative from both $yy' = x + 1$, $y' = \frac{x + 1}{y}$ and $2yy' = 2x + 2$. But every time I had that dangling constant (1 or 2), which lead to the dangling $\frac{1}{y}$ in my answer.
Like here:
$$
yy' = x + 1\\
y'y' + yy'' = 1\\
yy'' = 1 - (y')^2\\
y'' = \frac{1 - (y')^2}{y}\\
y'' = \frac{1}{y} - \frac{(y')^2}{y}\\
y'' = \frac{1}{y} - \frac{(\frac{x + 1}{y})^2}{y}\\
y'' = \frac{1}{y} - \frac{(x + 1)^2}{y^3}
$$
I don't see any way to get from my answer to the textbook's one with a transformation, no way to get rid from y in the numerator. And the correct answer doesn't have a "y" there.
Could someone either point to an error in my solution, or corroborate the suspicion that it indeed may be a typo.
| From $y'y'+yy''=1$ multiply by $y^2$.
Then $(yy')^2+y^3y''=(x+1)^2+y^3y''=y^2=x^2+2x \iff y^3y''=x^2+2x-x^2-2x-1=-1$
If we continue your calculation
$y''=\dfrac 1y-\dfrac{(x+1)^2}{y^3}=\dfrac{y^2-(x+1)^2}{y^3}=\dfrac{(x^2+2x)-(x^2+2x+1)}{y^3}=\dfrac{-1}{y^3}$
Gives the same result, so I guess the textbook result is erroneous (i.e. it gives $yy''=1$ which does not agree with derivatives of $\pm\sqrt{x^2+2x}$)
| {
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Prove that $4\tan^{-1}\left(\frac{1}{5}\right) - \tan^{-1}\left(\frac{1}{239}\right)= \frac{\pi}{4}$
Prove that $4\tan^{-1} \left(\dfrac{1}{5}\right) - \tan^{-1}\left(\dfrac{1}{239}\right)=\dfrac{\pi}{4}.$
I was wondering if there was a shorter solution than the method below?
Below is my attempt using what I would call the standard approach to these kinds of problems.
The expression on the left hand side is equivalent to $$\tan^{-1}\left[\tan \left(4\tan^{-1}\left(\dfrac{1}{5}\right)\right)-\tan^{-1}\left(\dfrac{1}{239}\right)\right]\\
=\tan^{-1}\left(\dfrac{\tan(4\tan^{-1}(\frac{1}{5}))-\frac{1}{239}}{1+\frac{1}{239}\tan(4\tan^{-1}(\frac{1}{5}))}\right)\tag{1}.$$
We have that $$\tan\left(4\tan^{-1}\left(\frac{1}{5}\right)\right)=\dfrac{2\tan(2\tan^{-1}(\frac{1}{5}))}{1-\tan^2(2\tan^{-1}(\frac{1}{5})}\tag{2}$$
and that
$$\tan\left(2\tan^{-1}\left(\frac{1}{5}\right)\right)=\dfrac{2\cdot \frac{1}{5}}{1-(\frac{1}{5})^2}=\dfrac{5}{12}\tag{3}.$$
Plugging in the result of $(3)$ into $(2)$ gives
$$\tan\left(4\tan^{-1}\left(\frac{1}{5}\right)\right) = \dfrac{2\cdot \frac{5}{12}}{1-(\frac{5}{12})^2}=\dfrac{120}{119}\tag{4}.$$
Pluggin in the result of $(4)$ into $(1)$ gives that the original expression is equivalent to
$$\tan^{-1}\left(\dfrac{\frac{120}{119}-\frac{1}{239}}{1+\frac{1}{239}\cdot\frac{120}{119}}\right)=\tan^{-1}\left(\dfrac{\frac{119\cdot 239 + 239-119}{239\cdot 119}}{\frac{119\cdot 239+120}{119\cdot 239}}\right)=\tan^{-1}(1)=\dfrac\pi4,$$
as desired.
| As advised by Maximilian Janisch, you should use the $\tan x$ formula rather $\tan^{-1}x$:
$$\tan\left[4\tan^{-1} \left(\dfrac{1}{5}\right) - \tan^{-1}\left(\dfrac{1}{239}\right)\right]=\tan\left[\dfrac{\pi}{4}\right] \iff \\
\frac{\tan\left[4\tan^{-1} \left(\dfrac{1}{5}\right)\right]-\frac1{239}}{1+\tan\left[4\tan^{-1} \left(\dfrac{1}{5}\right)\right]\cdot \frac1{239}}=1 \iff \\
\tan\left[4\tan^{-1} \left(\dfrac{1}{5}\right)\right]=\frac{120}{119} \iff \\
\frac{2\tan\left[2\tan^{-1} \left(\dfrac{1}{5}\right)\right]}{1-\tan^2\left[2\tan^{-1} \left(\dfrac{1}{5}\right)\right]}=\frac{120}{119} \iff \\
\frac{2\cdot \frac{2\cdot \frac15}{1-\frac1{5^2}}}{1-\left[\frac{2\cdot \frac15}{1-\frac1{5^2}}\right]^2}=\frac{120}{119} \iff \\
\frac{\frac5{6}}{1-\frac{25}{144}}=\frac{120}{119} \ \checkmark$$
| {
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Find :$ \lim_{n\rightarrow \infty} \int_{0}^{1} f(x^n) dx $?
Given $f$ is continuous on $[0,1]$ , Then $$ \lim_{n\rightarrow \infty} \int_{0}^{1} f(x^n) dx $$ is
*
*$ f(1)$
*$f(0)$
*$1$
*$0$
I thinks it will be $0$ because $ \lim_{n\rightarrow \infty} \int_{0}^{1} x^n dx = \lim_{n\rightarrow \infty}\frac{1}{n+1} = 0 $
Is its true ?
| The limit is $f(0)$ by DCT. A dominating integrable function is the supremum of $|f|$.
| {
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What is the maximum possible number of edges of a graph with n vertices and k components? Following is my attempt:
For maximum possible edges $->$ all k components must be connected sub-graphs
Maximum possible edges in a graph with n vertices = ${n \choose 2}$, I thought of removing k-1 edges, but only to realise that doing so doesn't divide the graph into k components.
How should I proceed now?
| I am assuming your question is the following:
What is the maximum number of edges in a graph with $n$ vertices and $k$ connected components?
This is equivalent to maximizing the function
$$
f(x_1,...,x_k) = \sum_{i} \binom{x_i}{2} =\frac{1}{2} \sum_{i} x_i^2-x_i
$$
subject to the constraints
$$
\sum_{i}x_i = n
$$
and $x_i \in \mathbb{N}^{>0}$.
Why? Because of none of the components $C_i$ can have an edge between them and $\binom{|C_i|}{2}$ is the maximum edges subject to this constraint; the $x_i \in \mathbb{N}^{>0}$ is necessary to ensure that there are $k$ components.
From here we have two possible approaches:
*
*combinatorial optimization
*Lagrange multipliers
but the first is much nicer.
Combinatorial approach Notice that if we take a "ball" (i.e. vertex) out of the $x_j$ "box" (i.e. component) and place it in the "$x_i$" box then we get that $\Delta f = x_i-x_j+1$. Therefore it seems that the solution is to just place all of the balls in one box because $x_i>x_j$ implies $f'=f+\Delta f = f+ x_i-x_j+1$ is a better solution. This is not allowed because of the condition $x_i \in \mathbb{N}^{>0}$. But we can set
$x_1=n-k+1$ and $x_2=...=x_k=1$.
This is the optimal configuration. Indeed suppose you had any other solution $z_1,...,z_k$ for a contradiction. Order the $z_i$ in decreasing size. Suppose that $z_i >1$ for some $i>2$ then we get that setting $z_i$ equal to $z_i-1$ and $z_1$ equal to $z_1+1$ gives us a better solution contradicting the optimality of the $z_i$.
| {
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A Theorem from Jacobson: $U^G\otimes_F V\cong (U\otimes_F V_H)^G$
Theorem: Let $G$ be a finite group, $H\le G$, $F$ a field. Let $U$ be an $F[H]$-module and $V$ an $F[G]$-module; $U^G$ is induced $F[G]$-module. Then,
$$U^G\otimes_F V\cong (U\otimes_F V_H)^G \,\,\,\, \mbox{(isomorphic as $F[G]$-modules)}$$
This theorem is mentioned in Jacobson's volume 2 (see $\S 5.10$,Theorem 5.17(3), p. 292-293). But I didn't get clue for the proof by Jacobson. He points out the following in beginning of proof of this theorem: we have two $F[G]$-modules $(F[G]\otimes_{F[H]} U)\otimes_F V$ and $F[G]\otimes_{F[H]} (U\otimes_F V)$, and the natural $F$-vector space isomorphism of these spaces, $(a\otimes x)\otimes y\mapsto a\otimes (x\otimes y)$ is not $F[G]$-module homomorphism. (Then he constructs new $F[G]$-modue isomorphism of them.)
Q. I tried to write alternate proof in following way; I wanted to know, is it correct?
(i) Let $t=[G:H]$ and $G=\cup_{i=1}^t g_iH$ ($g_1=1$). Then W.L.G, we can take $U^G=\oplus_i g_iU$.
(ii) So $U^G\otimes_F V\cong (U\otimes_F V) \oplus \cdots \oplus (g_tU\otimes_F V)$ as $F[G]$-module.
(iii) The decomposition of $F[G]$-module in (ii) satisfies following properties:
*
*$G$ permutes the $t$ many components (subspaces) transitively.
*stabilizer of subspace $U\otimes_F V$ is precisely $H$ (so $U\otimes_F V=U\otimes_F V_H$).
*So the $F[G]$-module in RHS of $(ii)$ is isomorphic to $(U\otimes_F V_H)^G$ as $F[G]$-module. q.e.d.
| Yes, your proof is correct. This characterisation of induction, namely that an $F[G]$-module $W$ having an $F[H]$-submodule $U$ such that $\dim W = [G:H]\dim U$ and $U$ generates $W$ as an $F[G]$-module is very powerful and gives many other elegant proofs.
| {
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how many $4\times 4$ matrices with specific entries Suppose we construct $4\times 4$ matrix with the following requirements:
Entry $1,1$ is $a$. We have to insert three elements $b$, one element $c$ and one element $d$. Other entries are $0$. How many such matrices are there?
Since there are three $b$ elements, out of $4\cdot 4 -1$ ($a$ takes up the first entry) we can choose $3$ random entries, but we do not distinguish between $b$ elements so the choices are $$\frac{{15 \choose 3}}{3!}.$$ Then we proceed to choose any of the $12$ entries for element $c$ and then $11$ choices for element $d$. Thus there are $$\frac{{15 \choose 3}}{3!}\cdot 12\cdot 11$$ such matrices. Is this correct?
| You do not need to divide by $3!$, the binomial counts unordered groups. $${15 \choose 3}\cdot 12\cdot 11$$
| {
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Random walk on Non-negative integers (Invariant Dist.) I wanted to confirm my answer to the following:
Given a random walk on the non-negative integers, beginning at 0 and provided transition probabilities of $\frac{1}{n+1}$ to the right and $\frac{n}{n+1}$ to the left at some arbitrary $\ n$ we have a probability transition matrix as follows.
$$PTM=\begin{bmatrix}
0 & 1 & 0 & 0 & 0 & \cdots \\
1/2 & 0 & 1/2 & 0 & 0 & \cdots\\
0 & 2/3 & 0 & 1/3 & 0 & \cdots\\
0 & 0 & 3/4 & 0 & 1/4 & \cdots\\
0 & 0 & 0 & 4/5 & 0 & \cdots \\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots
\end{bmatrix}$$
To determine the invariant distribution, we must find some $\mu$ such that $\mu P=\mu$. Provided that we assume our initial step for the walk is distributed according to $W_0 \sim \mu$, we have that the invariant distribution can be calculated as follows:
$$\frac{1}{2} \mu_1 + \sum_{n=1}^{\infty} \mu_{n-1}\frac{1}{n} + \mu_{n+1}\frac{n+1}{n+2}=1$$
When calculating the above expression per term I create the following expressions:
$$\frac{1}{2} \mu_1 = \mu_0$$
$$\mu_0 + \frac{2}{3} \mu_2 = \mu_1$$
$$\vdots$$
$$\mu_{n-1}\frac{1}{n} + \mu_{n+1}\frac{n+1}{n+2} = \mu_n$$
Solving the above expression I have the following invariant distribution:
$$\mu = \left \langle \mu_0 \ 2\mu_0 \ \frac{3}{2}\mu_0 \ \frac{2}{3}\mu_0 \ \frac{5}{24}\mu_0 \ \frac{1}{20}\mu_0 \ 0 \cdots \right \rangle$$
| This is the right approach and the answer (up to the 0 in the last component in your last equation, which seems to be a typo) is also correct. The general form of the distribution is $\mu_n = \frac{n+1}{n!} \mu_0$, which you can prove by induction.
To make this a probability distribution, choose $\mu_0 = \frac{1}{2e}$. Then all terms sum to 1.
| {
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Show that if $n$ and $m$ are both positive, then $nm$ is positive. This problem is on Terrence Tao's Analysis I, page 36.
Let $n,m$ be natural numbers. Then $n \times m=0$ if and only if atleast one of $n,m$ is equal to zero. In particular, if $n$ and $m$ are both positive, then $nm$ is also positive.
I know that multiplication is defined as recursive addition. $1 \times m := m$ and $(n+1) \times m= (n \times m) + m$. But, that's all that comes to my mind. I am not sure, how to find and organize my arguments to prove the above result. Should I use induction? Any hints, tips that lead me to prove this result would be helpful.
As a second question, what's a reasonable amount of time, I should spend on thinking about a theorem and sketching a proof, before I move on, perhaps try another question and then come back later? And should I search for a solution on the Internet, if I still don't get it.
| Proof(Rough sketch).
Let $n,m$ be positive natural numbers. We induct on $n$ keeping $m$ fixed.
(I) Claim. $1 \times m$ is positive.
$1 \times m := m$ by the definition of multiplication and $m$ is positive.
(II) We inductively assume that $n \times m$ is positive.
(III) We would like to show that $(n+1)\times m$ is positive.
By the definition of multiplication:
$\begin{aligned}
(n+1) \times m &= (n \times m) + m
\end{aligned}$
$n \times m$ is positive from the inductive assumption. $m$ is a positive natural number. The sum of two positive natural numbers is a positive number. This closes the induction.
| {
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Is $f(x)=\begin{cases}x \ \text{if } x\in [0,1),\\3-x \ \text {if} \ x \in [1,2]\end{cases}$ continuous from $[0,2]$ to $[0,2]$? Yes/No Consider the map
$f : [0,2] \rightarrow [0,2] $ defined by $ f(x) = \begin{cases} x \ \text{if
x} \in [0,1),\\3-x \ \text {if} \ x \in [1,2]\end{cases}$.
Is $f$ is continuous?
My attempt: Yes
Only one - sided limits can be taken at $x=0,x=2$ and for the other points, the double sided limit can be argued.
By this we can say that $f$ is continuous
Is it true?
| Notice that:
$$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} x = 1,$$
and
$$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} 3-x = 2.$$
Since
$$\lim_{x \to 1^-} f(x) \neq \lim_{x \to 1^+} f(x),$$
we can conclude that this function is not continuous $[0, 2]$.
| {
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Contour integral of rational function to some power I am interested in the following integral $$\int_{0}^{2\pi}d\theta \left(\frac{1-|x|^2}{|1-x e^{-i\theta}|^2}\right)^{y}$$
Does it has a nice closed form expression?
I know that $$\int_{0}^{2\pi}d\theta \left(\frac{1-|x|^2}{|1-x e^{-i\theta}|^2}\right)=2\pi$$
by contour integration.
Any hints are welcome.
| Assuming $|x|=r<1$, the integral is $$\int_0^{2\pi}\left(\frac{1-r^2}{1-2r\cos\theta+r^2}\right)^y d\theta.$$ It reduces to the hypergeometric function. A contour integration approach is possible here too; for $0<\Re y<1$, we integrate $\big(z(1-rz)^y(1-r/z)^y\big)^{-1}$ along the circle $|z|=1$, and squeeze the contour down to just encircling the segment $(0,r)$ of the real line (the cut for the integrand); the result is analytically continued w.r.t. $y$. The same result is obtained directly from the binomial series (let $y=1+\alpha$) $$(1-z)^{-1-\alpha}=\sum_{n=0}^\infty\binom{n+\alpha}{n}z^n\qquad(|z|<1)$$ writing $1-2r\cos\theta+r^2=(1-re^{i\theta})(1-re^{-i\theta})$: $$\int_0^{2\pi}\frac{d\theta}{(1-2r\cos\theta+r^2)^{1+\alpha}}=2\pi\sum_{n=0}^\infty\binom{n+\alpha}{n}^2 r^{2n}.$$ This has a (true) closed form only when $\alpha$ (i.e. $y$) is an integer (and when it is half an integer, the result is a sort of complete elliptic integrals).
| {
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Series expansion for $\arctan(1-x)$ Series expansion for $\arctan(1-x)$
I try to expand this function into its Taylor series by means of differentiating it and then integrating it terms by terms but I fail to obtain the correct result.
The derivative of $\arctan(1-x)=-\dfrac{1}{x^2-2x+2}$. By using long division, I can obtain the series expansion for $-\dfrac{1}{x^2-2x+2}$ and it is
$-\dfrac{1}{2}-\dfrac{x}{2}-\dfrac{x^2}{4}+\dfrac{x^4}{8}+\dfrac{x^5}{8}+\dfrac{x^6}{16}...$
I integrate this series terms by terms and I obtain:
$0-\dfrac{x}{2}-\dfrac{x^2}{4}-\dfrac{x^3}{12}+\dfrac{x^5}{40}+\dfrac{x^6}{48}-\dfrac{x^7}{122}...$
The series according to Wolfram is: $\dfrac{\pi}{4}-\dfrac{x}{2}-\dfrac{x^2}{4}-\dfrac{x^3}{12}+\dfrac{x^5}{40}+\dfrac{x^6}{48}-\dfrac{x^7}{122}...$.
I notice that if I use the indefinite integral, there is going to be an arbitrary constant left after the integrating process. How do I obtain this $\dfrac{\pi}{4}$? Without it, is my series wrong?
Is there any other methods to expand this function into Maclaurin series without directly employing the Maclaurin formula
| When you integrate it, you get a $+C$. To get it's value, you can substitute in $x=0$ to get that $C=\arctan(1)$.
| {
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Prove that $( R \circ S ) \cap T = \varnothing$ iff $(\mathrm{R}^{-1} \circ T) \cap S= \varnothing$. I am having a bit of a hard time proving the following statement:
Show that
$( R \circ S ) \cap T = \varnothing$ iff $(\mathrm{R}^{-1} \circ T) \cap S= \varnothing$.
I sort of understand composition of functions, inverse functions, and set theory individually, but when put together I seem to get confused.
It would be wonderful if anyone could be of help, as I'm not even sure how to begin.
Thank you!
EDIT: See attempt at the question in the answers
| $( R \circ S ) \cap T = \varnothing$ iff $(\mathrm{R}^{-1} \circ T) \cap S= \varnothing \iff ( R \circ S ) \cap T \ne \varnothing$ iff $(\mathrm{R}^{-1} \circ T) \cap S \ne \varnothing$
Let R, S, T be relations on the same set A
*
*$( R \circ S ) \cap T \ne \varnothing \to (\mathrm{R}^{-1} \circ T) \cap S \ne \varnothing$
Assuming $( R \circ S ) \cap T \ne \varnothing$,
Then, $\exists x, y \in A$ such that $(x,y) \in R \circ S$ and $(x,y) \in T$
$(x,y) \in R \circ S \implies \exists z$ such that $(x,z) \in R$
and $(z,y) \in S$
$(x,z) \in R \implies (z,x) \in \mathrm{R}^{-1}$
From $(z,x) \in \mathrm{R}^{-1}$ and $(x,y) \in T$, we can say that $(z,y) \in \mathrm{R}^{-1} \circ T$
Now, from $(z,y) \in \mathrm{R}^{-1} \circ T$ and $(z,y) \in S$, we can conclude that $(z,y) \in (\mathrm{R}^{-1} \circ T) \cap S$.
*$( R \circ S ) \cap T \ne \varnothing \leftarrow (\mathrm{R}^{-1} \circ T) \cap S \ne \varnothing$
Assuming $(\mathrm{R}^{-1} \circ T) \cap S \ne \varnothing$,
Then, $\exists x, y \in A$ such that $(x,y) \in \mathrm{R}^{-1} \circ T$ and $(x,y) \in S$
$(x,y) \in \mathrm{R}^{-1} \circ T \implies \exists z$ such that $(x,z) \in \mathrm{R}^{-1}$ and $(z,y) \in T$
$(x,z) \in \mathrm{R}^{-1} \implies (z,x) \in R$
From $(z,x) \in R$ and $(x,y) \in S$, we can say that $(z,y) \in R \circ S$
Now, from $(z,y) \in R \circ S$ and $(z,y) \in T$, we can conclude that $(z,y) \in ( R \circ S ) \cap T$.
| {
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Let $\Omega$ be a half-plane and $f_n$ a sequence of holomorphic functions on $\Omega$. I want to solve the second part of this 2-part problem. The first part, which I solved, states the following:
Let $\Omega$ be a bounded region and $f_n$ a sequence of holomorphic functions on $\Omega$ and continuous on $\overline\Omega$. Prove that if $f_n$ converges uniformly in $\partial \Omega$ then it converges in $H(\Omega)$.
The second part asks whether this is true if $\Omega$ is a half plane.
My intuition tells me that this does not hold, but I can't seem to find an "easy" counterexample, or a proof to show that it does hold.
Any help is more than appreciated :)
| $e^{-n(1+e^{z})} \to 0$ uniformly on the real line but it does not converge in $H(\Omega)$ where $\Omega$ is the upper half plane. [Put $z=2+i\pi$].
| {
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How do I differentiate $\sin(x)^{\cos(x^3)}$? I'm taking a calc 1 class and I have come across a function that I'm having difficulty finding answers on the web.
$y=\sin(x)^{\cos(x^3)}$
I know there's some chain rule to apply, but what do I do with the cos(x)?
I am assuming this:
$y' = \cos(x^3).(\sin(x)')^{\cos(x^3) - 1}$
Is my thinking correct?
| Use that
$$y=e^{\cos (x^3)\log(\sin x)}$$
and by chain rule we have
$$y=e^{f(x)}\implies y'=f'(x)e^{f(x)}$$
| {
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Smallest $m$ such that, for any $m$ unit squares in an $n\times n$ grid, the centers of some four of them are vertices of a parallelogram
Consider an $n\times n$ grid formed by $n^2$ unit squares. We define the center of a unit square as the intersection of its diagonals.
Find the smallest integer $m$ such that, choosing and $m$ unit squares in the grid, we always get four unit squares among them whose centers are vertices of a parallelogram
I was trying to derive a formula that represented the number of squares that can’t be selected after $k$ squares have already been chosen but I couldn’t because it doesn’t take into account straight lines and point outside the grid
Solutions would be appreciated
Taken from the 2016 Pan African Maths Olympiad
http://pamo-official.org/problemes/PAMO_2016_Problems_En.pdf
| A partial solution
First note that if we choose the $2n-1$ squares along two adjacent edges of the grid then no parallelograms are formed and so $m$ has to be at least $2n$.
Note also that there are precisely $(2n-1)^2$ different vectors between centres of squares. One of these is the zero vector and the other vectors occur in pairs as v and -v.
For any set of $2n$ squares there are $\begin{pmatrix}2n\\2\\\end{pmatrix}$ choices of pairs of squares. Comparing this with the number of vectors given above we see that there have to be two pairs where the centres differ by the same vector. However, this does not complete the proof because the two vectors might be in the same straight line and therefore not form opposite edges of a parallelogram.
I hope these ideas help you get a full proof.
| {
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If we have an equilateral triangle with a square inscribed in it, could we prove that the triangles we get are congruent?
(I forgot to add a last point. Let $X$ be the midpoint of $\overline{GF}$.)
Could we prove that triangles $CGX$, $CFX$, $GAS$, and $FBE$ are all congruent?
| Let $M$ be the midpoint of $\overline{AB}$. As $C,G$ and $A$ are collinear, $C,X$ and $M$ are collinear and $\overline{GX}$ is parallel to $\overline{AM}$, $\triangle CXG$ must be similar to $\triangle CMA$. Hence, $\frac{GX}{XC}=\frac{AM}{MC}$. Then, observe that $XC=MC-2GX$ and, by Pythagoras', $MC=\sqrt{(2AM)^2-AM^2}=\sqrt{3}AM$.
Hence, $\frac{GX}{\sqrt{3}AM-2GX}=\frac{AM}{\sqrt{3}AM}$ and $GX=\frac{\sqrt{3}}{\sqrt{3}+2}AM$. Given that $\triangle CXG$ and $\triangle GDA$ are oriented the same way, if they were congruent, we would have $GX=AD=\frac{AM}{2}$ but this is not the case as $\frac{\sqrt{3}}{\sqrt{3}+2}\neq \frac12$. Hence the triangles are not congruent and furthermore, $G$ is not the midpoint of $AC$.
| {
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Contours and Closed Path
Sorry, but I thought I am very good in complex analysis, but then I saw this question which makes me question myself "Is closed path different from contours?"
Now I know the difference between them by the logic that contours have orientation.
This problem is from Chapter 9 of Complex Analysis by Ian Stewart and David Tall.
But still to be frank, I honestly dont know how to prove this, as I can see that there is no orientation mentioned of $\gamma$, this is not a contour.
But still I don't think so it is the correct way to prove it.
I plotted the sketch and yes it is closed, but I feel there must be some analytical way to prove this is closed but not a contour?
Please let me know or solve it please?
| It seems that a contour is defined as "made up of a finite number of smooth paths which have non-zero continuous derivatives". See p.91.
Your path is continuous (note that $\lim_{t \to 0} \gamma(t) = \lim_{t \to 1} \gamma(t) = 0$). However it is not differentiable at $t = 0, 1$ since
$$\dfrac{t + it\sin(\pi/t)}{t-0} = 1 +i\sin(\pi/t)$$
does not have a limit as $t \to 0$.
| {
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How does Variational AutoEncoder (VAE) get mean and variance? Could somebody explain to me how VAE works in this tutorial (look only at cell which starts with class Sampling...)?
input=(batch_size=64, flatten_pixels=784) goes to Dense(64, 'relu') layer. The result goes through Dense(32) twice in parallel. One output of it they call z_mean, another z_log_var. Why do they think that output of two identical Dense(32) layers with the same input should return mean and log variance? Reading about this topic I found that z_mean becomes a mean and z_log_var becomes a log variance during the loss minimization process.
Also, as far as I know:
KL Divergence = CrossEntropy - Entropy = -∑plog(q) - (-∑plog(p))
I know what KL Divergence is, I understand binary and cathegorical crossentropy, but I still can't understand why the loss calculated by the formula below forces one output of Dense(32) to be mean and another output to be log variance:
kl_loss = - 0.5 * tf.reduce_mean(
z_log_var - tf.square(z_mean) - tf.exp(z_log_var) + 1)
| The $n=1$ case of this answer obtains the KL loss as $-\frac12(1+\ln\sigma^2-\sigma^2-\mu^2)$. This is your formula with z_mean $\mu$ and z_log_var $\ln\sigma^2$.
| {
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prove that $5^{2n+1} - 3^{2n+1} - 2^{2n+1}$ is divisible by 30 for all integers n ≥ 0. Prove that $5^{2n+1} - 3^{2n+1} - 2^{2n+1}$ is divisible by 30 for all integers n ≥ 0.
I have tried induction as follows.
Step 1:
Try n = 0, we get: $5 - 3 - 2 = 0$, which is divisible by 30.
Try n = 1, we get: $5^{3} - 3^{3} - 2^{3} = 90$, which is also divisible by 30.
Step 2:
Assume it is true for n = k.
So we are assuming the following equality is true: $5^{2k+1} - 3^{2k+1} - 2^{2k+1} = 30M$, for some integer M.
Step 3:
Now we look at the next case: n = k + 1.
$5^{2(k+1)+1} - 3^{2(k+1)+1} - 2^{2(k+1)+1}$
= $5^{2k+3} - 3^{2k+3} - 2^{2k+3}$
= $25\times5^{2k+1} - 9\times3^{2k+1} - 4\times2^{2k+1}$
= $21\times5^{2k+1} + 4\times5^{2k+1} - 5\times3^{2k+1} - 4\times3^{2k+1} - 4\times2^{2k+1}$
= $21\times5^{2k+1} - 5\times3^{2k+1} + 4\times[5^{2k+1} - 3^{2k+1} - 2^{2k+1}]$
= $21\times5^{2k+1} - 5\times3^{2k+1} + 4\times30M$. (Assumed in step 2)
The last term is divisible by 30. But I cannot get a factor 30 out of the first two terms. I can show divisibility by 15 as follows:
= $7\times3\times5\times5^{2k} - 5\times3\times3^{2k} + 4\times30M$
= $7\times15\times5^{2k} - 15\times3^{2k} + 4\times15\times2M$
But how do I show divisibility by 30?
| Let $a_n = 5^{2n+1} - 3^{2n+1} - 2^{2n+1} = 5 \cdot 25^n - 3 \cdot 9^n - 2 \cdot 4^n$. Then $a_{n+3} = 38 a_{n+2} - 361 a_{n+1} + 900 a_n$ (*). Therefore, you only need to check the claim for $n=0,1,2$, which is immediate.
(*) Because $(x-25)(x-9)(x-4) = x^3 - 38 x^2 + 361 x - 900$. The coefficients are not important here. The important point is that $a_n$ satisfies a linear recurrence with integer coefficients.
| {
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$\lambda^2$ is an eigenvalue of $A^2$. Prove that $\lambda$ or $-\lambda$ is an eigenvalue of the matrix $A$ Let $A$ be a $n\times n$ real matrix. Let $\lambda \in \mathbb{R}$ such that $\lambda^2$ is an eigenvalue of the matrix $A^2$. Prove that $\lambda$ or $-\lambda$ is an eigenvalue of the matrix $A$.
I know how to prove the converse (and there are multiple threads regarding it), but I'm not sure how to show the other direction
I have:$$$$
$$\begin{gather}
A^2\bar{x}=AA\bar{x}=\lambda^2\bar{x} \\
\text{since if}\:A \:\text{and}\:B\: \text{share eigenvectors, so does} \: AB,\: \text{we let}\: A\bar{x}=\mu\bar{x}\Rightarrow A^{-1}\bar{x}=\frac{1}{\mu}\bar{x}, \\
A^{-1}AA\bar{x}=A\bar{x}=\frac{1}{\mu}\lambda^2\bar{x} \\
\Rightarrow \frac{1}{\mu}\lambda^2=\mu \Rightarrow \lambda^2=\mu^2 \\
\therefore\mu=\pm\lambda
\end{gather}$$
Here I'm using the fact that if $A\bar{x}=k\bar{x},\:\text{then}\: A^2\bar{x}=c\bar{x}$ but I feel like I will need the converse of that statement to make my proof valid
| Let $x$ be the eigenvector associated to $\lambda^2$, consider $V=Vect(x,A,(x))$ it is stable by $A$. The matrix of $A$ in $\{x,A(x)\}$ is:
$\pmatrix{0&\lambda^2\cr 1&0}$ this implies that the characteristic polynomial of the restriction of $A$ to $V$ is $X^2-\lambda^2$.
| {
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Algebraic degree of $\cos\left(\frac{p\pi}{q}\right)$ How can we find the algebraic degree of $\cos\left(\frac{p\pi}{q}\right)$ for $p$, $q$ coprime integers?
I know that the algebraic degree of $e^{\frac{p\pi i}{q}}$ is $\phi(q)$, since cyclotomic polynomials are irreducible. I also know that $$\cos(x)=\frac{e^{ix}+e^{-ix}}{2}.$$ But my knowledge of abstract algebra is quite slim, and I don't know how to relate these two facts.
I've checked on Google, and I haven't been able to find anything related. Any result or reference would be appreciated.
| Hint: Assume that $|q|\neq 1$. There is a quadratic polynomial $f(x)$ with coefficients in $\mathbb{K}:=\mathbb{Q}\Biggl(\cos\left(\frac{p\pi}{q}\right)\Biggr)$ such that $\exp\left(\frac{p\pi\text{i}}{q}\right)$ is a root of $f(x)$. Prove that $f(x)$ is irreducible over $\mathbb{K}$ by showing that $$\left[\mathbb{Q}\Bigg(\exp\left(\frac{p\pi\text{i}}{q}\right)\Bigg)_{\vphantom{\frac{1}{\frac{1}{\frac{1}{1}}}}}^{\vphantom{\frac{1}{\frac{1}{\frac{1}{1}}}}}:\mathbb{K}\right]>1\,.$$
| {
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The solution for $a+b+c+d = 4$ and $\left( \frac{1}{a^{12}} + ... + \frac{1}{d^{12}} \right) \left( 1 + 3abcd \right) = 16$ $a,b,c,d > 0$, then find the solution for
$$a+b+c+d = 4$$
and
$$ \left( \frac{1}{a^{12}} + \frac{1}{b^{12}} + \frac{1}{c^{12}} + \frac{1}{d^{12}} \right) \left( 1 + 3abcd \right) = 16$$
Attempt:
We know that $a=b=c=d=1$ is a solution. Are there any other solutions? By AM-GM,
$$ \frac{4}{4} \ge (abcd)^{1/4} \implies abcd \le 1 $$
Also by HM-GM we have
$$ \frac{4}{ \frac{1}{a^{12}} +\frac{1}{b^{12}} + \frac{1}{c^{12}} + \frac{1}{d^{12}} } \le \left( a^{12} b^{12} c^{12} d^{12} \right)^{1/4} $$
$$ \frac{4}{(abcd)^{3}} \le \frac{1}{a^{12}} +\frac{1}{b^{12}} + \frac{1}{c^{12}} + \frac{1}{d^{12}} $$
then
$$ 4 \le \frac{4}{(abcd)^{3}} \le \frac{1}{a^{12}} +\frac{1}{b^{12}} + \frac{1}{c^{12}} + \frac{1}{d^{12}} $$
and of course $1 + 3abcd \le 4$.
Next,
$$ \frac{1}{a^{12}} +\frac{1}{b^{12}} + \frac{1}{c^{12}} + \frac{1}{d^{12}} =
\frac{ (bcd)^{12} + (acd)^{12} + (abd)^{12} + (abc)^{12} }{(abcd)^{12}} $$
so we have
$$\left( \frac{1}{a^{12}} + \frac{1}{b^{12}} + \frac{1}{c^{12}} + \frac{1}{d^{12}} \right) \left( 1 + 3abcd \right)
= \frac{ (bcd)^{12} + (acd)^{12} + (abd)^{12} + (abc)^{12} }{(abcd)^{12}}
+ \frac{ 3(bcd)^{12} + 3(acd)^{12} + 3(abd)^{12} + 3(abc)^{12} }{(abcd)^{11}}
= 16 $$
| Note that
\begin{align*}
\left(\frac{1}{a^{12}} + \frac{1}{b^{12}} + \frac{1}{c^{12}} + \frac{1}{d^{12}}\right)(1 + 3abcd)
&\geq \frac{4}{(abcd)^3}(1 + 3abcd) \\
&= \frac{4}{(abcd)^3} + \frac{12}{(abcd)^2} \\
&\geq 4 + 12 \\
&= 16
\end{align*}
where the first inequality holds by AM-GM, and the last inequality holds since $abcd \leq 1$ (again established by AM-GM). But by the equality conditions of AM-GM, equality is attained in each of these inequalities only when $a, b, c, d$ are equal.
| {
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Boolean algebra: simplify (A+B)$'C'$(C+D) I need to solve / simplify the output of a logic circuit. The output is $(A+B)'C'(C+D)$
I wrote the truth table and it returns 0 in all outputs. I tried to solve it using laws of boolean algebra:
$(A+B)' = A'B'$
And $C′(C+D) = (C′C + C′D)$
$C′C = 0$ [Boolean law $A.A′ = 0$]
Does that mean $(C′C + C′D) = 0$ as well?
If that's the case, the result of the entire output would be zero. Am I right or is there any mistakes? Is there any other law applied here?
Update: I made a mistake in the truth table. It doesn't return zero in all outputs.
| $C'C=0$ will not imply $C'C + C'D=0$, and $C'C + C'D=1$ when $C=0,D=1.$
Here is how to simplify the expression:
\begin{align}
&(A+B)'C'(C+D)\\
&\equiv(A'B')C'(C+D)\tag*{de Morgan’s Theorem}\\
&\equiv(A'B')(C'C+C'D)\tag*{Distributive law}\\
&\equiv(A'B')(0+C'D)\tag*{Complement}\\
&\equiv(A'B')(C'D)\tag*{Identity}\\
\end{align}
Hence it has A'B'C'D as it's minimal CNF&DNF form.
| {
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Why does the boundary of mobius strip wrap twice around core circle but not any other line? A space $X$ deformation retracts onto a subspace $A$ if there exists continuous map $F:X\times [0,1]\rightarrow X$ such that $F(x,0)=x,F(x,1)\in A,F(a,t)=a$ $\forall a\in A$.
The mobius strip deformation retracts onto its core circle. But I don't understand how, under this deformation retraction,
*
*The boundary circle wraps twice around the core circle
*Any other line wraps only once around the core circle
I think 1 happens because at each step of the deformation retraction, the boundary goes around twice and so in the final step (i.e. $t=1$) it just becomes twice the circle. But I see this reason working equally well for any other line. After all, none of the lines actually coincide with the central line until the last step.
Any help would be appreciated.
| The Möbius strip $M$ can be obtained by gluing two opposite edges of a rectangle by an orientation reversing homeomorpism. Explictly, we may define
$$M = [0,1 ] \times [-1,1]/\sim$$
where $(0,t) \sim (1,-t)$. Let $p : [0,1 ] \times [-1,1] \to M$ denote the quotient map.
Then you get various types of embedded circles.
Type 1: The core circle $C = p([0,1] \times \{0\})$. It can be parameterized by $c : [0,1] \to M, c(x) = [x,0]$. The map $r : M \to C, r([x,t]) = [x,0]$, is a well-defined strong deformation retraction.
Type 2: Circles $C_t = p([0,1] \times \{-t,t\})$ for $t \in (0,1]$, especially the boundary circle $C_1$. These can be parameterized by $c_t(x) = [2x,-t]$ for $x \le 1/2$ and $c_t(x) = [2x-1,t]$ for $x \ge 1/2$. These circles do no intersect the core circle. The maps $r_t = r \mid_{C_t} : C_t \to C$ are $2$-$1$. In fact, they are covering maps with two sheets.
Type 3. Circles $C'_t = p(D_t)$ for $t \in [0,1]$, where $D_t = \{(x,t(2x-1)) \mid x \in [0,1]\}$. These can be parameterized by $c'_t(x) = [x,t(2x-1)]$. We have $C'_0 = C$ and for $t > 0$ these circles intersect the core circle in the single point $[1/2,0]$. The maps $r'_t = r \mid_{C'_t} : C'_t \to C$ are homeomorphisms.
There are of course many more embedded circles, but let us confine to the above.
It is now obvious that the circles $C_t$ wrap twice around the core circle, and that the circles $C'_t$ wrap once around the core circle.
| {
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Nimbers for Kayles I have a problem with calculating the nimbers for this game.
I know that for Kayles the following is true:
\begin{array}{c|c}
\hline
Hight & Nimber \\ \hline
12 & ^*4 \\ \hline
11 & ^*6 \\ \hline
10 & ^*2 \\ \hline
9 & ^*4 \\ \hline
8 & ^*1 \\ \hline
7 & ^*2 \\ \hline
6 & ^*3 \\ \hline
5 & ^*4 \\ \hline
4 & ^*1 \\ \hline
3 & ^*3 \\ \hline
2 & ^*2 \\ \hline
1 & ^*1 \\ \hline
0 & ^*0 \\ \hline
\end{array}
My attempt to get to these numbers:
n(0) = mex{} = 0
n(1) = mex{0} = 1
n(2) = mex{0, 1} = 2
However,
n(3) = mex{1, 2} = 0
which is obviously the wrong result.
What I am trying to figure out is how to calculate those nimber values. Any help is appreciated.
| Community wiki answer so the question can be marked as answered:
A third option for the game of length $3$, in addition to leaving a game of length $1$ or a game of length $2$, is to leave two games of length $1$, with value $1\oplus1=0$, so $n(3)=\mathrm{mex}\{0,1,2\}=3$.
| {
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Verification of a basic Riemann summation problem: $f(x) = 1+x$ where $x\in [-1, 2]$ This is the very first time I solve a problem involving Riemann's summation so I would like to verify whether I get it correctly. Problem statement (hopefully, I have translated the problem statement in a comprehensible way.):
Find an integral sum $\sigma_n$ for a function $f(x) = 1+x$ on the interval $x\in [-1, 2]$. Choose points so that they are in the middle of each partition.
I've started by considering the interval. Suppose that we split the interval into $n$ parts. That means that each interval is of length:
$$
\Delta x = {3\over n}
$$
The interval begins at $x = -1$, so we might find the points $\zeta_k$ lying in the middle of each interval. So that means:
$$
\zeta_k = -1 + \frac{3k + 3(k-1)}{2n}
$$
Here $-1$ is the beginning of the range of $x$, and $\frac{3k + 3(k-1)}{2n}$ is chosen because we need a center point in each split. Here is a sandbox for the above.
Now by we want to find a sum:
$$
\begin{align}
\sigma_n &= \sum_{k=1}^nf(\zeta_k)\Delta x \\
&= \sum_{k=1}^n \left(1 + \frac{3k + 3(k-1)}{2n} -1\right)\Delta x \\
&= \sum_{k=1}^n \frac{3k + 3(k-1)}{2n}\Delta x \\
&= \sum_{k=1}^n \frac{3k + 3(k-1)}{2n}\cdot {3\over n} \\
&= {9\over 2n^2}\sum_{k=1}^n (2k-1) \\
&= {9\over 2n^2}\left({2n(n+1)\over 2}-n\right)\\
&= {9\over 2n^2}n^2 = {9\over 2}
\end{align}
$$
Here the sum yields a value. But as far as I understood if $\sigma_n$ is a function of $n$ then we have to take the limit of $\sigma_n$ as $n\to\infty$.
Does my writing make sense at all?
| Your work looks correct. In your particular example, $\sigma_n$ is independent of $n$, but if you had taken points on the left (or on the right) side of each partition interval, instead of in the middle, or if you had a non-linear function such as $f(x)=1+x^2$, then only in the limit as $n\to\infty$ would the Riemann sum give the exact value for the integral. It is worth trying those as other exercises to see what I am talking about.
| {
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Derivative of a generic polynomial function Let $P:M_n(\mathbb{C})\to\mathbb{C}$ be a polynomial function, $A=(A_{ij}),E_{ji}\in M_n(\mathbb{C})$,where $E_{ji}$ is defined as matrix filled with zeros except $1$ in j-th row and i-th column and $t\in \mathbb{R}$.
I want to take derivative of $P((I+tE_{ij})A)$ with respect to $t$ and evaluate it in $t=0$ afterwards.
My textbook suggests it is $$\frac{d}{dt}P((I+tE_{ji})A)|_{t=0}=\sum_{k,l}\frac{\partial P}{\partial A_{lk}}(A)\cdot (E_{ji}A)_{lk}.$$
My intuition says it makes use of chain rule, but I am failing to see where comes the sum from.
| Hint
You have
$$A= \sum\limits_{k,l}A_{kl}E_{kl}$$
| {
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Prove that $\mathbb{Z}$ is a UFD while $\mathbb{Z}[\sqrt{-5}]$ is not. An integral domain $R$ is called a unique factorization domain (UFD) if every nonzero, nonunit element of $R$ can be uniquely written as a product of irreducible elements, up to reordering the factorization and taking associates of the irreducible factors (e.g. $10 = (2)(5) = (-5)(-2)\in\mathbb{Z}$).
$1.$ Prove that $\mathbb{Z}$ is a UFD.
$2.$ Prove that $\mathbb{Z}[\sqrt{-5}]$ is not a UFD.
I think $1$ is equivalent to the proving the uniqueness part of the Fundamental Theorem of Arithmetic.
As for $2$, $5 = 1\cdot 5$, where $1$ and $5$ are irreducible, and $5 = (-1)\cdot (\sqrt{-5})^2$, where $(-1)$ and $(\sqrt{-5})$ are also irreducible, so it has two distinct factorizations in $\mathbb{Z}[\sqrt{-5}].$ Thus, $\mathbb{Z}[\sqrt{-5}]$ is not a UFD. Do I need to prove that $-1,1,5,\sqrt{-5}$ are irreducible? And if not, is this proof still correct?
| For 1: the definition says "can be uniquely written", so you essentially have to prove the Fundamental Theorem of Artithmetic (not just the "uniqueness part).
For 2: are really 1,-1 and 5 irreducible?
Instead, note that $2\cdot 3=6=(1+\sqrt{-5})\cdot(1-\sqrt{-5})$
PS: Remember that irreducible elements are not units by definition
| {
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How is the subspace identified? In Linear Algebra class, I was learning about vector spaces and subspaces but can't quite grasp the concepts. We were given the following:
*
*$w = \{(a,b,c) \in R^3 : a = 1\}$
*$w = \{(a,b,c) \in R^3 : a = 0\}$
We were told that (2) is a subspace, while (1) is not because (1) is "not closed under addition". What does that mean?
| Set 1 is "not closed under addition" because vectors in this set are of the form $\left [ \begin{array}{c}
1 \\
y\\
z\\
\end{array} \right ]$. If I add any two of these together I get the result
$$
\left [ \begin{array}{c}
1 \\
a\\
b\\
\end{array} \right ] + \left [ \begin{array}{c}
1 \\
c\\
d\\
\end{array} \right ]\;\; =\;\; \left [ \begin{array}{c}
2 \\
a+c\\
b+d\\
\end{array} \right ]
$$
and this doesn't belong to the set since our first entry isn't 1. The second set is "closed under addition" since vectors in this set are of the form $\left [ \begin{array}{c}
0 \\
y\\
z\\
\end{array} \right ]$ and adding any two of them yields:
$$
\left [ \begin{array}{c}
0 \\
a\\
b\\
\end{array} \right ] + \left [ \begin{array}{c}
0 \\
c\\
d\\
\end{array} \right ] \;\; =\;\; \left [ \begin{array}{c}
0 \\
a+c\\
b+d\\
\end{array} \right ].
$$
We see that addition in this set preserves the first entry remaining zero, guaranteeing that adding two vectors in this set keeps you in the set.
| {
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How is a relation defined? I've read multiple ways of defining relations and was wondering what is generally accepted. One of these ways is as follows.
$xRy=C\iff C\subseteq A\times B$
and the cartesian product is defined as
$A\times B=D\iff (\forall x)(\forall y)((x,y)\in D\iff x\in A \land y\in B)$
Is this definition generally accepted or is there a more formal definition?
| In set theory, relations are sets; thus "to be a relation" is a property of sets.
We define the pair $(x,y)$, for example with Kuratowski's definition, and then we define the cartesian product of two sets $A$ and $B$ :
$A \times B = \{ (x,y) \mid x \in A \text { and } y \in B \}$.
Finally, we define when a set is a (binary) relation :
$\text {Rel}(A) \text { iff } \forall x \ [x \in A \to \exists y \exists z (x=(y,z))]$.
The symbol $xRy$ is an abbreviation for $(x,y) \in R$.
Due to the fact that $A \times B$ is the set of all pairs $(x,y)$ with $x \in A$ and $y \in B$, we have that :
for every $C$, if $C \subseteq A \times B$, then $\text {Rel}(C)$.
| {
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Find the value of $x$ in this equation. $2^x -x=5$
I can't solve this equation, but I can see that $x=3$ (through trial and error) but I don't know how to attain the value $3$.
| $$2^x - x=5 \implies 2^x = 5+x$$
It is trivial to see that for $x>3$ ,
$$2^x > 5+x$$
Hence we are only left with $3$ possibilities for $x\in \mathbb {N}$ , which yields $3$ as a solution to the equation.
For $x \in \mathbb {Z}$ , we observe that for $x<-5$,
$$2^x > 5+x$$
Hence the only useful range for the solution is $[-5,3]$ which can be found using trial and error.
| {
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Edge colouring number of a k-regular graph and relation to bridges I am researching graph theory, and am confused about this problem. I am considering a $k$-regular connected graph $G$, where $k\geq2$ and where $\chi'(G)=k$ (the edge colouring number.) I then want to be able to show that $G$ does not contain any bridges.
However, I do not know how to solve this. I can show that as, $\chi'(G)=k$, the edge colouring number is the same as the maximum degree of a vertex, there must be an even number of vertices in $G$. But this does not tell me whether $k$ must be odd or even. I feel like I want to remove the bridge, and then use the handshaking lemma to show that for one of the new connected subgraphs formed by removing the bridge the equality does not match, as one side is odd and the other even, but I cannot seem to gather the right information to do this. Or perhaps this is the wrong method.
Any help appreciated.
| Note with an edge colouring $k$, you can find $k$ distinct perfect matchings, one for each colour class.
*
*Suppose a bridge $uv$ exist, then the removal of this bridge gives two distinct components, say $C_1$ and $C_2$.
*Next, we note that the union of two disjoint perfect matchings form a disjoint union of cycle that spans the graph.
*Now, pick two colour classes, one of which contains $uv$, and there must be a cycle between $u$ and $v$.
*However, this meant that either
*
*$uv$ has a loop (contradicting $uv$ bridge as the other edge still connects the two components)
*Or that some vertex in $C_1$ must be connected to some other vertex in $C_2$ to close the loop.
*This contradicts $C_1$ and $C_2$ being two distinct components, and we are done.
| {
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$T_{a}(\text{ball}(\ell^p))$ is compact, where $T_a\colon\ell^p\to\ell^p$, $(T_a x)_{n}:=a_{n}x_{n}$ and $\mathbb{C}\ni a_n\to0$ as $n\to\infty$ Let $a\in\ell^{\infty}=\ell_{\mathbb{C}}^{\infty}$ and define $T_a\colon\ell^p\to\ell^p$, $(T_a x)_{n}:=a_{n}x_{n}$ for $1\leq p<\infty$. Suppose that $a_n\to0$. How do I prove that $T_{a}(\text{ball}(\ell^p))$ is compact? I have shown that $T_{a}$ is a compact operator. So $\overline{T_{a}(\text{ball}(\ell^p))}$ is compact. If $p\neq1$, then I think I know how to prove that $T_{a}(\text{ball}(\ell^p))$ is compact (by using reflexivity of $\ell^{p}$ and weak topology arguments). But I'm stuck on the case $p=1$. Any suggestions are greatly appreciated! Maybe there is a method that proves it for all $1\leq p<\infty$ at once?
| We can use the following criterion of compacity in a Banach space.
A subset $A$ is precompact (namely $\bar A$ is compact) if (and only if) it is bounded, and for every $\epsilon$ there is a finite dimensional vector subspace $V$ such that for every $y\in A$ $d(y,V)\leq \epsilon$
Let $ V_n$ the subspace of sequences such that $\forall k\geq n, x_k=0$ .
Then, for every sequence $x$ in the unit ball, $d(T_ax, V_n)\leq (\Sigma_{k>n} \vert a_k u_k \vert ^p )^{1/p}\leq Sup_{k>n} \vert a_k\vert $. This proves the result.
To prove the criterion, note that if a set $A$ satisfy it it is at a the distance $\epsilon $ of a closed ball of this finite dim. vector space, a set wich is covered by a finite number of ball of radius $\epsilon$. So $A$ is contained in a finite union of balls of radius 2$\epsilon$.
| {
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spectrum of elements in $C^*$ algebra Suppose $x,y$ are two invertible positive elements in a $C^*$ algebra $A$,if $\|x\|=\|y\|$,can we compute the spectrum $\sigma(x^{-1}y)$ of $x^{-1}y$?Does there exist a relationship between the spectrum of the multiplication of two elements and the norm of elements?
| You can't expect a relation. For instance consider
$$
x=\begin{bmatrix} 1&0\\0&\tfrac1n\end{bmatrix} ,\ \ \ y=\begin{bmatrix} 1&0\\0&1\end{bmatrix} .
$$
Then $\|x\|=\|y\|=1$, and $\|x^{-1}y\|=n$. The norm only sees the maximum of the spectrum, but nothing else.
For a more dramatic example consider the block matrices
$$\tag1
x=\begin{bmatrix} 1&0\\0& z\end{bmatrix} ,\ \ \ y=\begin{bmatrix} 1&0\\0&w\end{bmatrix} .
$$
We can take $z,w$ to be any two contractions, and we will still have $\|x\|=\|y\|=1$, while $\sigma(x^{-1}y)=\{1\}\cup \sigma(z^{-1}w)$. Now let $X\subset (1,\infty)$ be any compact set; let $v$ be an operator with $\sigma(v)=X$, and let $w=\tfrac1{\|v\|}v$. Then $w$ is a contraction. If we now take $z=\tfrac1{\|v\|}\,I$, we still have $\|x\|=\|y\|=1$, while $\sigma(z^{-1}w)=X$.
| {
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Show that a function is negative over its domain I would like to demonstrate that the following function is negative
\begin{equation}
f(x)=-\frac{t}{4\sqrt{x}^3}\bigg[1-\bigg(1+\sqrt{x}\bigg)^{\frac{1}{t-1}}\bigg]+\bigg(\frac{1}{1-t}\bigg)\bigg(\frac{t}{2\sqrt{x}}\bigg)^2\bigg(1+\sqrt{x}\bigg)^{\frac{2-t}{t-1}}
\end{equation}
for $x>0$ and $1>t>0$.
I have graphed this in Desmos and confirmed that the function is negative given my restrictions on $x$ and $t$.
Any help to solve this analytically would be most appreciated!
Many thanks.
| You can show fairly simply that the first term will be negative and the second term positive with positive x and 1 > t > 0.
Because they are added, showing that the overall function is negative becomes a question of showing that the first (negative) term dominates, which you can do by setting up and solving an inequality between them and show that it holds given the restrictions on x and t.
| {
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Problematic inequality & hint I would like to ask for hint for proving following inequality:
$$x^3(1+x)+y^3(1+y)+z^3(1+z)\geq \frac{3}{4}(1+x)(1+y)(1+z)$$
for all $x>0$, $y>0$, $z>0$ such that $xyz=1$.
Generally, I tried to find some elementary solution, but even with some calculus I didn't solve it.
Edit. My attempt:
Let $f:(0,\infty)\times(0,\infty)\rightarrow \mathbb{R}$ be function defined by equality:
$$ f(x,y)=x^3(1+x)+y^3(1+y)+\frac{1}{x^3y^3}(1+\frac{1}{xy})-\frac{3}{4}(1+x)(1+y)(1+\frac{1}{xy})$$
and I calculated partial derivatives:
$$\frac{\partial f}{\partial x }(x,y)=-\frac{4}{x^5y^4}-\frac{3}{x^4y^3}+4x^3+\frac{3(y+1)}{4x^2y}+3x^2-\frac{3(y+1)}{4}$$
$$\frac{\partial f}{\partial y }(x,y)=-\frac{4}{y^5x^4}-\frac{3}{y^4x^3}+4y^3+\frac{3(x+1)}{4y^2x}+3y^2-\frac{3(x+1)}{4}$$
I noticed that:
$$\frac{\partial f}{\partial x }(1,1)=\frac{\partial f}{\partial x }(1,1)=0.$$
I began to wonder if the function is convex (i.e. by demonstrating that $g(w)=w^3(1+w)$ is convex for positive values and $h(x,y)=(1+x)(1+y)(1+\frac{1}{xy})$ is also concave for positive x and y, with latter I had some calculation trouble).
| Since $x^3$ is convex for $x\ge 0$, we have that $x^3 \ge 1 + 3(x-1)$, using the tangent at $x=1$.(*) With this observation, it is enough to prove that
$$
\sum_{cyc} (3x-2)(1+x)\geq \frac{3}{4}(1+x)(1+y)(1+z)
$$
Expanding the terms, and using $x y z = 1$, gives the equivalent
$$
-30 + \sum_{cyc} x + 12 \sum_{cyc} x^2 - 3 \sum_{cyc} xy \ge 0
$$
Now note
$$
3 \sum_{cyc} x^2 - 3 \sum_{cyc} xy = 3(\frac12 \sum_{cyc} x^2 +\frac12 \sum_{cyc} y^2 - \sum_{cyc} xy ) = \frac{3}{2}(\sum_{cyc} (x-y)^2) \ge 0
$$
So it is enough to show
$$
-30 + \sum_{cyc} x + 9 \sum_{cyc} x^2 \ge 0
$$
For the two sums in here, we have by AM-GM:
$\sum_{cyc} x \ge 3 \sqrt[3]{xyz} = 3$
and by Titu's Lemma (Cauchy - Schwarz)
$\sum_{cyc} x^2 \ge \frac13 (\sum_{cyc} x )^2 \ge \frac93= 3$
So we have
$$
-30 + \sum_{cyc} x + 9 \sum_{cyc} x^2 \ge -30 + 3 +9 \cdot 3 = 0
$$
which proves the claim. $\qquad \Box$
(*) Alternatively, if you do not want calculus arguments, $x^3 \ge 1 + 3(x-1)$ can be shown elementary. Let $y = x-1$. Then we have $x^3 = (1+y)^3 = 1 + 3y + y^2(3+y)$
and this is $\ge 1 + 3y$ as long as $y \ge -3$ or $x \ge -2$, which is given since we are interested in $x \ge 0$.
| {
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Investigating the geometric patterns of $x^k+y^k=r^k$ for $k \in N$ I was investigating the graphs of equations of the form $x^k+y^k=r^k$.
I am not sure how to ask this so I will try to simplify the problem first.
For simplicity sake lets let $r=2$, now
For $k=1$, I get a line.
$x+y=2$.
For $k=2$, I get a circle with radius 2.
$x^2 + y^2= 4$.
Is there a name for the image I get when k=3?
$x^3+y^3=8$?
Are there more names of the geometric figures for higher powers of $k \in N$?
I have included a picture of these two cases where $k=3$ and $k=4$
| For even $k$, these are Lamé Curves also known as hyperellipses.
For odd $k$, you have part of a Lamé Curve in quadrant 1 and the curve asymptotically approaches the line $y = -x$ from above in quadrants two and four. I do not know of a name for the full curves for odd $k$. (They are examples of superelliptic curves, but that is almost useless as a label.)
| {
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The mean of $f$ minimizes $||f-\xi||_p$ over all $\xi\in\Bbb R$. [Reference request] I would like to know where I can find a reference for the following fact:
Let $\Omega$ be a set of finite measure, $1<p<\infty$ and $f\in L^p(\Omega,\mu)$. Denote $\bar f:=\frac 1{|\Omega|} \int_\Omega f d\mu$. Then
$$
\bar f = \arg\min \{ ||f-\xi||_p : \xi\in\Bbb R \}.
$$
For $p=2$ the result is extremely well-known whereas for $p=1$ the above doesn't hold.
I am surprise that I couldn't find a reference for the above result for a general $p>1$ since I assumed it to be somewhat well-known as well. Does anyone know a book where this theorem is stated?
| This result only holds for $p = 2$. Indeed, if we denote
$$J(\xi) = \int_\Omega |f - \xi|^p \, \mathrm d\mu,$$
we get (at least formally)
$$J'(\xi) = \int_\Omega p \, |f - \xi|^{p-2} \, (f - \xi) \, \mathrm d\mu.$$
For $p = 2$, this is $0$ iff $\xi$ is the mean of $f$. But this is not true for $p \ne 2$.
| {
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is log 0 is or is not undefined? per Figure 1.7 in pattern recognition and machine learning (free)
Plots of M = 9 polynomials fitted to the data set shown in Figure 1.2 using the regularized error
function (1.4) for two values of the regularization parameter λ corresponding to ln λ = −18 and ln λ = 0. The
case of no regularizer, i.e., λ = 0, corresponding to ln $λ = −∞$
ln $0 = −∞$
per this post
log 0 is undefined.
so, is log 0 is or is not undefined?
|
The logarithmic function $~\log_b(x)~$ (for any positive real number $b\ne 1$) is defined only for $~x>0~$.
If possible let $~\log_b(0)=k~$, where $~k~$ is any real number. Then by the definition of logarithm, $~b^k=0~$.
Now, any ‘real’ quantity (may it be positive or negative) raised to another real quantity can never result in $~0~$.
You may think about the case of $(\text{any number})^{0}$ or more specifically, $0^0$. But the interesting fact is that any number raised to the power $0$ results in $1$ and not $0$.
So, the value of the expression $~\log_b(0)~$ makes no sense and more specifically, we cannot determine any value for it or we cannot define it.
Thus $~\log_b(0)~$ is Indeterminate or Undefined or Does not exist.
For various values of the base $~b~$ $($let us consider $~b=10,~b=e,~b=2$ $)$, at $~x=0$, we have vertical asymptotes (see in the figure given below).
| {
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Dividing a polygon into 6 equal regions You are given a convex polygon, ie all its internal angles are less than 180 degrees. Prove that you can always draw three straight lines through a specific point inside this polygon, such that they divide it into 6 equal (by area) regions? Bonus questions:
*
*Can you prove that this can be achieved for non-convex polygons too? This time the point may not lie inside the polygon.
*Does this result extend to 4 lines dividing a convex polygon into 8 equal regions?
This question was inspired by the hard problem in the recent TopCoder Open Algorithm final, written by Michal Forisek (misof).
| Let's have a polygon. Let's draw a line at some angle $\theta_1$ to some fixed direction. By moving the line parallel to itself, we can make the whole polygon to lie either on one side of the line or the other. Thus, since the function “area on one side minus area on the another” is continuous, it should go through zero. Thus, for every $\theta_1$ there is a line $AD = L(\theta_1)$ that splits the polygon in halves.
Using the same mean value theorem we can show, that for every $\theta_1$, there is $\theta_2$ such that lines $L_1=AD=L(\theta_1)$ and $L_2=BE=L(\theta_2)$ split the area in a proportion $k=[APB]/[BPD]=1/2$. Indeed, if $\theta_2=\theta_1$, then $L_1=L_2$ and $k=0$. However, if $\theta_2=\theta_1+\pi$ (we rotated the line 2 by $\pi$ and $L_1=L_2$ again), then $k=\infty$. Given $k(\theta_2)$ is continuous, there should be some $\theta_2$, so $k=1/2$
By the same argument, for every $\theta_1$ there is always a $\theta_3$ and line $L_3=CF=L(\theta_3)$, so $k=[AQC]/[CQD]=2$.
Now consider point $R$, the intersection of $L_2$ and $L_3$. As we traverse from $\theta_1\to\theta_1+\pi$, lines $L2\leftrightarrow L3$ and $P\leftrightarrow Q$, but $R\to R$. However, now $R$ lies on the other side of line $L_1$. That means that during its journey, it crossed the line $L_1$. At this moment all three lines pass through one point.
By the way, we didn't use neither property of convexity, nor that it is a polygon. What mattered is that the area of the intersection of this shape with half-plane is continuous in respect to movements of the half-plane.
| {
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"url": "https://math.stackexchange.com/questions/3444763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving that an improper Riemann integral involving $f(x)$ exists given that $f(x)$ is Riemann integrable and periodic with period $1$
Given that $f(x)$ is Riemann integrable and periodic with period $1$
and $\int_{0}^{1}f\left(x\right)dx=0$, prove that
$$\int_{1}^{\infty}\frac{f\left(x\right)}{x^{s}}dx$$ exists for $s>0$
Here's my attempt to prove it for $s>1$, Define a sequence {$a_n$} by $$ {a_n}=\int_{1}^{n}\left|\frac{f\left(x\right)}{x^{s}}\right|dx$$
Using the fact that $f(x)$ is bounded, we have $$\int_{1}^{n}\left|\frac{f\left(x\right)}{x^{s}}\right|dx\ \le\ M\ \int_{1}^{n}\frac{1}{x^{s}}dx=M\left(\frac{1}{s-1}-\frac{1}{\left(s-1\right)n^{s-1}}\right)<\frac{M}{s-1}$$ for all $n$. So {$a_n$} is an increasing sequence bounded above and therefore converges which implies that {$b_n$}= $\int_{1}^{n}\frac{f\left(x\right)}{x^{s}}dx$ also converges. Is this correct? and even if it is, this argument can't be applied when $0<s\le1$.
| Hint:Apply Dirichlet's integral test.
| {
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Dimension of affine variety mod $p$ can only increase Let $X$ be an affine variety defined by polynomials over $\mathbb{Z}$.
Reducing the polynomials modulo $p$, we obtain a variety $X_p$ defined over the finite field $\mathbb{F}_p$.
Is it true that the dimension of $X_p$ is larger or equal to the dimension of $X$?
Note: It seems likely to me that the dimension stays the same for all but finitely many primes $p$, and this answers says that it is the case for projective varieties. My question is different: I'm asking about affine varieties, I'm asking about all primes, and I'm only asking for an inequality.
EDIT: To make the second part of the question explicit:
In addition, is it also true that $\dim X_p=\dim X$ for all but finitely many primes $p$?
| Let me state the following result (Lemma 05F7 in the Stacks Project):
Let $f:X\rightarrow S$ be a morphism of finite type where $S$ is an
irreducible scheme (with generic point $\eta$). If $n=\mathrm{dim}X_\eta$
then there exists a nonempty open $U\subseteq S$ such that for
all $s\in U$ one has $\mathrm{dim} X_s=n$.
Hence, whenever you have a variety $X$ over $\mathbf{Z}$ (which means a $\mathrm{Spec}(\mathbf{Z})$-scheme of finite type), the dimension of $X\otimes_\mathbf{Z} \mathbf{F}_p$ (the fiber over $(p)$) is equal to that of $X\otimes_\mathbf{Z}\mathbf{Q}$ (the fiber over the generic point of $\mathrm{Spec}(\mathbf{Z})$) for all but finitely many primes $p$.
Finally, your question amounts to knowing if $\mathrm{dim}X\leq\mathrm{dim}(X\otimes_\mathbf{Z}\mathbf{Q})$, but this is not always true: consider $X=\mathrm{Spec}(\mathbf{Z}\left[ X\right] /(2))=\mathrm{Spec}(\mathbf{F}_2\left[ X\right] )$ so that $\mathrm{dim}(X)=1$ while $\mathrm{dim}(X\otimes_\mathbf{Z}\mathbf{Q})=-\infty$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integrating function $h:=\lambda(A \cap (B-x))$ for $\lambda(A),\lambda(B)<\infty$ Given two sets $A,B$ with finite measure let $h:=\lambda(A \cap (B-x))$. Show this function is integrable and calculate its integral.
I thought about using the following identity, $\lambda(A)+\lambda(B)=\lambda(A \cup B)+\lambda(A \cap B)$, approximating both $A$ and $B$ from below with intervals, and using the monotone convergence theorem.
Thanks in advance for any help.
| $\lambda (A\cap (B-x)=\int 1_{A}(y)1_B(x+y)dy$
Use Tonelli's theorem to show that $\int_{\Bbb{R}} \int_{\Bbb{R}}F(x,y)dydx<\infty$ where $F(x,y)=1_{A}(y)1_B(x+y)$ is non-negative.
The integral is equal to $m(A)m(B)$ by translation invarianve of the Lebesgue measure.
| {
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Riemann sum of $\int_1^2 {1\over x^2} dx$. I've spent quite a time solving the following problem:
Evaluate using Riemann's sum:
$$
I = \int_1^2{1\over x^2} dx
$$
I was first trying the following approach, which didn't work since the summation seems undoable to me:
$$
\Delta x = {1\over n}\\
I = \lim_{n\to\infty}\sum_{k=1}^nf\left(1+{k\over n}\right)\Delta x \\
= \lim_{n\to\infty}\sum_{k=1}^n{n^2\over (k+n)^2} {1\over n} \\
= \lim_{n\to\infty}\sum_{k=1}^n{n\over (k+n)^2}
$$
Wolfram evaluates this sum in terms of digamma function which is too advanced.
Several hours has passed before I decided to reconsider the point to choose in each partition. Let:
$$
\Delta x = {1\over n}\\
x_k = 1 + {k\over n}\\
\begin{align}
I &= \lim_{n\to\infty}\sum_{k=1}^nf\left(\sqrt{x_k x_{k-1}}\right)\Delta x \\
&= \lim_{n\to\infty}\sum_{k=1}^n{1 \over x_k x_{k-1}}\Delta x \\
&= \lim_{n\to\infty}\sum_{k=1}^n{1 \over \left(1+{k\over n}\right)\left(1+{k-1\over n}\right)}\Delta x \\
&= \lim_{n\to\infty}\sum_{k=1}^n{n^2 \over (n+k)(n+k-1)}{1\over n}\\
&=\lim_{n\to\infty}\sum_{k=1}^n{n \over (n+k)(n+k-1)} \\
&=\lim_{n\to\infty}\sum_{k=1}^n\left({n \over (n+k-1)} - {n \over (n+k)}\right)\\
&= {n\over n} - {n\over 2n}\\
&= \boxed{{1\over 2}}
\end{align}
$$
This sum telescopes nicely. Now I'm wondering whether the first approach is even doable. I've met some other questions but the first one lists a hint I don't really understand and the second one is closed as a duplicate.
What would be the way to finish the initial approach? In the first approach, the problem is actually reduced to finding the limit which I couldn't handle. Also is there some intuition in choosing the "right" points in the partitions?
| I've just tried one more technique while solving a similar problem and it seems to work fine. Let's split the interval $[1, 2]$ with points $x_0, x_1, \dots, x_n$ so that they form a geometric progression. Let $q$ denote the denominator of geometric progression. So the interval is split by the points: $q, q^2, \dots, q^n$, therefore:
$$
\Delta x_1 = q - 1\\
\Delta x_2 = q^2 - q\\
\cdots\\
\Delta x_n = q^n - q^{n-1}\\
$$
Now pick the points $\zeta_k$ from the rightmost point of each subsegment. Calculate the value of the function in each point $\zeta_k$:
$$
f(\zeta_k) = \left\{{1\over q^2}, {1\over q^4}, \dots, {1\over q^{2k}}\right\}
$$
Now write the sum:
$$
\begin{align}
S_n &= \sum_{k=1}^n f(\zeta_k)\Delta x_k \\
&= \sum_{k=1}^n {1\over q^{2k}} (q^k - q^{k-1}) \\
&= {1\over q^2}(q-1) + {1\over q^4}(q^2-q) + \cdots + {1\over q^{2n}}(q^n-q^{n-1}) \\
&= {1\over q^2}(q-1) + {1\over q^3}(q-1) + \cdots + {1\over q^{2n-1}}(q-1) \\
&= (q-1)\left({1\over q^2} + {1\over q^3} + {1\over q^4} + \cdots + {1\over q^{n+1}}\right)
\end{align}
$$
By geometric sum:
$$
S_n = {q^n - 1 \over q^{n + 1}}
$$
Remember $q = \sqrt[n]{2}$. Now the only part left is taking the limit:
$$
\begin{align}
I &= \lim_{n\to\infty} S_n \\
&= \lim_{n\to\infty} {q^n - 1 \over q^{n + 1}}\\
&= \lim_{n\to\infty} {(\sqrt[n]{2})^n - 1 \over (\sqrt[n]{2})^{n + 1}}\\
&= \boxed{{1\over 2}}
\end{align}
$$
It looks like this can be generalized for:
$$
\int_a^b {1\over x^k}\mathop{dx}
$$
| {
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Proof of closed-form solution of the difference of two factorial series Context
I'm working on a problem tangentially related to the Kepler Equation 1. The details are very much in the weeds, and I'm not in a position to explain at this time exactly how I have arrived at Equation 1. Yet, I believe that the following holds true:
$$ \lim_{k\rightarrow \infty} \sum\limits_{s=0}^{ k- 1 } \,
\dfrac{
\left[ 2^{ 2\,(k- s) } \left[
( k- s) !\right]^4
- \pi\,2^{2(s-k) -1 }
\,
\left[ 2\,(k - s)]!\right]^2 [ 2\,(k -s) ]
\right]
}{[2(k- s)]![2\,(k- s) ]\,\left[ ( k - s) ! \right]^2}
=
\dfrac{ \pi }{2 } - 1
~\text{Eq}.~1.$$
I've plotted Equation 1 for various $k$. My results seem to indicate that the expression above is plausibly true. Beyond $k=50$ I run into floating point issues in the numerical calculation, and series is not computable.
I cannot figure out how to determine the veracity of the equation 1. I've seen some closed-form solutions to factorial series (e.g., [2]).[] Yet, I have not seen such an expression elsewhere.
Questions
*
*Does anyone have one or more references to a book that has many factorial series?
*Can anyone prove Equation 1 true or false?
*Can anyone illustrate the results for $k >> 50$?
Bibliography
1 Find the inverse of an equation reminiscent of Kepler's equation
[2] http://mathworld.wolfram.com/FactorialSums.html
| Starting from @marty cohen's answer and fimplifying, we have
$$f_1(k)=\sqrt{\pi }\,\frac{ \Gamma (k+1)}{\Gamma \left(k+\frac{1}{2}\right)}-1$$
$$f_2(k)=\frac{\Gamma \left(k+\frac{3}{2}\right)}{\sqrt{\pi }\, \Gamma (k+1)}-\frac{1}{2}$$
$$f_1(k)-\pi f_2(k)=\frac \pi 2-1+\sqrt \pi\left(\frac{\Gamma (k+1)}{\Gamma \left(k+\frac{1}{2}\right)}-\frac{\Gamma
\left(k+\frac{3}{2}\right)}{\Gamma (k+1)} \right)$$
Now, using Stirling approximation and continuing with Taylor series for large values of $k$
$$\log \left(\frac{\Gamma (k+1)}{\Gamma \left(k+\frac{1}{2}\right)}\right)=\frac{1}{2} \log \left({k}\right)+\frac{1}{8 k}-\frac{1}{192
k^3}+O\left(\frac{1}{k^5}\right)$$
$$\frac{\Gamma (k+1)}{\Gamma \left(k+\frac{1}{2}\right)}=t+\frac{1}{8 t}+\frac{1}{128 t^3}-\frac{5}{1024
t^5}+O\left(\frac{1}{t^7}\right)$$ where $\color{red}{t=\sqrt k}$.
Similarly
$$\log \left(\frac{\Gamma \left(k+\frac{3}{2}\right)}{\Gamma (k+1)}\right)=\frac{1}{2} \log \left({k}\right)+\frac{3}{8 k}-\frac{1}{8
k^2}+\frac{3}{64 k^3}-\frac{1}{64k^4}+O\left(\frac{1}{k^5}\right)$$
$$\frac{\Gamma
\left(k+\frac{3}{2}\right)}{\Gamma (k+1)}=t+\frac{3}{8 t}-\frac{7}{128 t^3}+\frac{9}{1024 t^5}+O\left(\frac{1}{t^7}\right)$$
$$f_1(k)-\pi f_2(k)=\frac \pi 2-1+\sqrt \pi\left(-\frac{1}{4 t}+\frac{1}{16 t^3}-\frac{7}{512t^5}+O\left(\frac{1}{t^7}\right)\right)$$
Computing for $k=10$, the exact value is
$$f_1(10)-\pi f_2(10)=\frac \pi 2-1+\frac{215955}{46189}-\frac{707825 \pi }{524288}\approx 0.4340968$$ while the above expansion gives
$$\frac{\pi }{2}-1-\frac{12487 }{51200}\sqrt{\frac{\pi }{10}}\approx 0.4340980$$
With regard to @user3113647's plots, we then have
$$S(k)-S(k-1)=\frac{1}{8} \sqrt{\pi }
\left(\frac{1}{k}\right)^{3/2}+O\left(\frac{1}{k^{7/2}}\right)$$ and then a slope of $-\frac 32$ in the logarithmic scale.
| {
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Prove that $\sum_{i=1}^{n}\frac{1}{\left(n+i\right)^{2}}\sim\frac1{2n}$ I would like a proof of the asymptotic relationship $$\sum_{i=1}^{n}\frac{1}{\left(n+i\right)^{2}}\sim\frac1{2n}$$
without assuming that the sum is a Riemann sum.
This problem arose from Question 1909556, which asks about the Riemann sum of $\int_1^2\frac1{x^2}\ \mathrm{d}x=\lim_{n\to\infty}\frac1n\sum_{i=0}^n\frac{n^2}{(n+i)^2}=\frac12$. It features a nebulous clue that $\lim_{n\to\infty}\frac{\sum_{i=1}^{2n}\frac1{i^2}-\sum_{i=1}^{n}\frac1{i^2}}{1/n}=\frac12$. I can't figure out how this clue works but one way of showing it could be with the asymptotic relationship is true, and from calculations, it seems to work. But I can't find any feasible way of proving it without assuming the value of the integral.
I would like to avoid assuming that the sum is simply the integral so that I can prove the integral from the sum. It also seems like a fairly simple relationship, so I would imagine there could be a nice proof.
| A more elementary approach.
$$\frac{1}{2n}=\sum_{i=1}^{n}\frac{1}{(n+i)(n+i-1)}$$ because the sum telescopes to $\frac{1}{n}-\frac{1}{2n}.$
So:
$$\begin{align}\frac{1}{2n}-\sum_{i=1}^{n}\frac{1}{\left(n+i\right)^{2}}&=\sum_{i=1}^{n}\left(\frac{1}{(n+i)(n+i-1)}-\frac{1}{(n+i)^2}\right)\\
&=\sum_{i=1}^{n}\frac{1}{(n+i)^2(n+i-1)}\tag{1}\\&<\sum_{i=1}^{n}\frac{1}{n^3}\\&=\frac{1}{n^2}
\end{align}$$
Also, the value at (1) is positive. So we have:
$$0<\frac{1}{2}-n\sum_{i=1}^{n}\frac{1}{(n+i)^2}<\frac{1}{n}$$ and hence$$n\sum_{i=1}^{n}\frac{1}{(n+i)^2}\to\frac{1}{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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how to calculate number of states for this logistics problem? Consider a logistics problem with 3 cities, 5 trucks and 3 packages. Each truck can be at any of the locations. A package can either be at one of the locations or in one of the trucks. What would the number of states be if an atomic representation is used?
would the answer be 5 (num of trucks) x 3 (num of cities) x 3 ^ 8 (packages can be in a city or a truck) x 8?
I feel like this is completely wrong. Can anyone give me a breakdown on how to solve these types of problems?
Thank you
| For Atomic Presentation, the total number of states would at least be $n^{p+t}$, where $n$ is the number of cities, $p$ is the number of packages, and $t$ is the number of trucks. In this case, the least states are: $3^{5+3} = 6561$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the significance of Topological Entropy? I'm taking a course on Dynamical systems and we learnt about topological entropy. I get that topological entropy is a quantity which measures how quickly distinct points separate asymptotically, and that it is a topological invariant (which makes it good). But how does knowing the topological entropy of a dynamical system help us understand the properties of a dynamical system better? For example, since positive entropy doesn't imply chaos and neither does chaos imply positive entropy as one would naively expect initially, entropy does not seem helpful in understanding chaotic behaviour.
So the question is, what insight does knowing the entropy give into a dynamical system?
| Perhaps the primary value is that it is related to other invariants outside of just topological dynamics, as expressed in the so-called Variational Principle, which says:
If $T : X \to X$ is a continuous self-map of a compact Hausdorff space $X$ then its topological entropy $h(T)$ is the equal to the supremum of the measure theoretic entropies $h_\mu(T)$, as $\mu$ varies over all probability measures on $X$ that are invariant under $T$ (meaning that $T^*(\mu)=\mu$).
Here's a Scholarpedia link which discusses these and many other matters regarding topological entropy.
Another nice thing about topological entropy is that in some restricted cases it gives very nice information about the number $P_n$ of periodic orbits of cardinality $n$ (see this other link on the same Scholarpedia page):
$$h(T) = \limsup_{n \to \infty} \frac{\log P_n}{n}
$$
| {
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Finding the angles and area of unusual shapes My friend sent me the following geometry problems. I think I have the first one, but I think the 2nd and 3rd are unsolvable, although I could be missing something.
My attempt:
*
*I'm pretty sure this one is 1080. I'm having a hard time writing out my explanation here, but I can justify it on paper. The idea is that the outer shape (if we ignore the triangles) is an octagon, and the sum of its angles is 1080, and I can show that the sum of the marked angles is also 1080.
*I don't think this one gives us enough information. If we moved the point D left or right, the angle of $x$ would change and the constraints would still be satisfied.
*I can't figure out the answer to this one. I think it's unsolvable as well, but I can't prove it like #2. We can't find the area of anything in this picture. The shaded region and the whole shape are both close to being a trapezoid, but they aren't.
| 2) Note $ \angle ABD = 90 -25 =65$, which yields $x=65 -25 =40$ due to isosceles triangle.
3) The shaded area is equal to the area of the quadrilateral minus the areas of the two small right triangles. Thus,
$$A= \frac 12\cdot 4(x+4) + \frac12 \cdot 6(2+y) -\frac12 \cdot 4x -\frac12\cdot 6y=14$$
where $x$ and $y$ are the unknown segment lengths that cancel out in the difference.
| {
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"timestamp": "2023-03-29T00:00:00",
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Difference between domain and co-domain in sets? Let's say I have a subset of the Cartesian plane, for example:
$\{(x, y) \in R \times R: 2x+3 > 5\}$.
If I am asked to find the co-domain of the following set, how would I do so? I know how to find the domain, which is done by finding all possible $(x,y)$ ordered pairs and then placing all the $x$ values in a set. But how is this different from the co-domain?
Thank you very much!
| Normally, domain and co-domain are defined on functions, but we'll use your textbook's definition for domain and co-domain on relations.
In your relation, note that your set describes only the $x$-values, $2x+3>5$, i.e. $x>1$. So the $x$-values that satisfy this condition is your domain, as defined in your textbook. However, since there is no condition on your $y$-values, all of $\mathbb{R}$ is your co-domain.
| {
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Find the largest divisor of 6006006006 that does not exceed 60,000 factoring it, we have
$6 * 1001*1000001 = 2*3*7*11*13* (100^3+1) = 2*3*7*11*13*101*9901$
with this prime factorization, how do you check the largest divisor without too much guessing and checking?
| One way is to look for targets to get close to and just use your own numerical nous. Here, the obvious targets are the given $60000$, and $6006006006/60000 \approx 100100$.
$2\cdot3\cdot9901=59406$.
For this instance, start with $60000=10000\cdot6$ and notice that $9901$ is a little less than $10000$, then look for $6$.
With a more complicated factorization it's worth checking quite a few options.
| {
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$x^4 + 1020x^3 - 4x^2 + 2039x+1$ is divisible by 1019. Give all integers x in the range [1,2018] such that $x^4 + 1020x^3 - 4x^2 + 2039x+1$
is divisible by 1019.
Applying mod 1019 to the coefficients, we have
$x^4 + x^3 - 4x^2 + x + 1$ which is equal to $(x-1)^2(x^2+3x+1)$
I tried making 1 of the factors equal to a multiple of 1019, but it looks tedious and i have not found a solution other than 1020.
answer given : 1, 492, 524, 1010, 1511, 1543
| As $1019$ is prime, one of the factors must be $\equiv 0\pmod{1019}$. For $(x-1)$ this obviously means that $x=1$ or $x=1020$.
For the factor $x^2+3x+1$, we'd expect two solutions $x=\frac{-3\pm\sqrt{5}}2$ - but what is $\sqrt 5$ in modular arithmetic? Any $y$ with $y^2\equiv 5\pmod{1019}$. Fortunately, just playing with $1019+5=1024$, we see that $32^2\equiv5\pmod{1019}$. Hence $\frac{-3+32}{2}\equiv\frac{1016+32}{2}=524$ and $\frac{-3-32}{2}\equiv\frac{1016-32}{2}=492$ are remainders mod $1019$ that solve too (i.e., $492$, $524$, $1511$,$1543$)
| {
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Can anyone explain to me why $(34)(123) = (124)$? $$\begin{align}
(34)H&= \{(34)(1),(34)(123),(34)(132)\}\\
& = \{(34),(124),(1432)\}.
\end{align} $$
Can anyone explain to me why $(34)(123) = (124)$?
I don't understand coset multiplying can you help me with this
| If $\sigma=(123)$, $\tau=(34)$, and we compose right-to-left (as is consistent with your other compositions), then
$$\begin{align}
1 &\stackrel{\sigma}{\mapsto} 2 \stackrel{\tau}{\mapsto}2, \\
2&\stackrel{\sigma}{\mapsto} 3 \stackrel{\tau}{\mapsto} 4, \\
4 &\stackrel{\sigma}{\mapsto} 4 \stackrel{\tau}{\mapsto} 3, \\
3 &\stackrel{\sigma}{\mapsto} 1 \stackrel{\tau}{\mapsto} 1,
\end{align}$$
so $\tau\sigma=(34)(123)=(1243)$.
| {
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Prove that exist $x\in \left\{ 1,...,14 \right\}$ such that $\sigma(x)=x$, where $\sigma\in S_{14}$ and $|\sigma|=28$?
Let $\sigma\in S_{14}$ which is an even permutation of the order of $28$. Prove that exist $x\in \left\{ 1,...,14 \right\}$ such that $\sigma(x)=x$.
My try:
We know that the permutation order is equal to the least common multiple of cycles that make up a given permutation and $28=2\cdot2\cdot7$.
So $\sigma$ must be character $(a_1 a_2 a_3 a_4)(b_1 b_2 ... b_7)$ - composition of row cycle $4$ and row cycle $7$
because if $\sigma$ would be a character $(a_1
,a_2)(b_1b_2)(c_1...c_7)$ then $|\sigma|=2\cdot7=14$ which is contrary to the assumption.
That's why $4+7=11$ elements elements undergo nontrivial permutations and $14-11=3$ elements pass on to each other.
So $\sigma$ has a character:
$$\sigma=\begin{pmatrix} a_1 & a_2 & a_3 & a_4 & b_1 & b_2 & b_3 & b_4 & b_5 & b_6 & b_7 & c_1 & c_2 & c_3 \\ a_2 & a_3 & a_4 & a_1 & b_2 & b_3 & b_4 & b_5 & b_6 & b_7 & b_1 & c_1 & c_2 & c_3\end{pmatrix}$$
Moreover we have information that $\sigma=(a_1 a_2 a_3 a_4)(b_1 b_2 ... b_7)$ is composition an even number of transpositions.
However these are my only thoughts and I don't know what to do next to come to the thesis.
EDIT:
According to the remark from @EricTowers $\sigma$ can still have a character $(a_1 a_2 a_3 a_4)(b_1 b_2 ... b_7)(c_1c_2)$ then $$\sigma=\begin{pmatrix} a_1 & a_2 & a_3 & a_4 & b_1 & b_2 & b_3 & b_4 & b_5 & b_6 & b_7 & c_1 & c_2 & c_3 \\ a_2 & a_3 & a_4 & a_1 & b_2 & b_3 & b_4 & b_5 & b_6 & b_7 & b_1 & c_2 & c_1 & c_3\end{pmatrix}$$
| You already have the prime factorisation of $28$. To get an element of order $28$, you need to partition $14$ into divisors of $28$ (namely, $1$, $2$, $4$, $7$, and $14$) so that their LCM is $28$.${}^\dagger$ So, what are the partitions of $14$ into those divisors, potentially including $1$, $4$, and $14$, such that the disjoint cycles of elements of $S_{14}$ form elements of order $28$ with cyclic decompositions composed of those divisors?
You'll find that you'll always need a $1$ in the cyclic decomposition. What does that imply?
You need to have at least one term of $7$ or $14$ in the partition. It should be obvious why you can't have a term $14$; can you have two terms of $7$? If the number of $7$s in the partition is odd, what does that say about the number of $1$s in the partition?${}^\dagger$
$\dagger$: I am thankful to @StevenStadnicki for the clarifying sentences provided in the comments.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\sqrt{2a^2+b+1}+\sqrt{2 b^2+a+1}\geq 4$ when $a+b=2$. Suppose that $a,b$ are non-negative numbers such that $a+b=2$. I want to prove $$\sqrt{2 a^2+b+1}+\sqrt{2 b^2+a+1}\geq 4.$$
My attempt: I tried proving $$2a^2+b+1\geq 4$$ but this seems wrong (try $a=0,b=1$).
How can I prove the above result?
| By Minkowski (triangle inequality) we obtain:
$$\sum_{cyc}\sqrt{2a^2+b+1}=\sum_{cyc}\sqrt{2a^2+\frac{b(a+b)}{2}+\frac{(a+b)^2}{4}}=$$
$$=\frac{1}{2}\sum_{cyc}\sqrt{9a^2+4ab+3b^2}=\frac{1}{2}\sum_{cyc}\sqrt{7a^2+b^2+8}\geq$$
$$\geq\frac{1}{2}\sqrt{7(a+b)^2+(b+a)^2+8(1+1)^2}=4.$$
We see that our inequality is true for all reals $a$ and $b$ such that $a+b=2$.
| {
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Finding the amount of possibilities to split two parties in pairs I have two parties. Each party contains $3$ people. I'm trying to figure out how many possibilities there are to divide those two parties into pairs so each pair contains a person from party A and a person from party B.
I think the answer is $6$. But I'm trying to prove it. First of all, I put the three people from party A in a row in $3!$ options. I also put the three people from party B in a separate row in $3!$ options. Each pair will be in the people by indexes (first pair contains the first person from the first row and the first person from the second row and so on).
But now we need to divide by something. How should I continue this idea?
| You are correct that there are six ways to match people from party $A$ with people from party $B$ in pairs.
Line up the people in party $A$ in some order, say alphabetically. There are three ways to match a person from party $B$ with the first person in line, two ways to match one of the remaining people from party $B$ with the second person in line, and one way to match the remaining person from party $B$ with the third person in line. Hence, there are $3! = 6$ ways to pair people from party $A$ with people from party $B$.
To continue with your idea, match the person from party $A$ with the corresponding person in party $B$. Notice that doing so counts each possible pairing $3!$ times, once for each of the $3!$ ways we could arrange the same set of pairs in the two lines. Hence, there are
$$\frac{3!3!}{3!} = 3!$$
ways to pair people from party $A$ with people from party $B$.
| {
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If $D\in [AB$ s.t. $AD=BC$ and $\angle{ADC}=\frac{3}{4}\cdot \angle{ABC}$ find $\angle {A}$. Let $\triangle {ABC}$ s.t. $AB=AC$ and $\angle{A}>90$. If $D\in [AB$ s.t. $AD=BC$ and $\angle{ADC}=\frac{3}{4}\cdot \angle{ABC}$ find $\angle {A}$.
My idea: I denote $\angle B=4x$, then I apply "Sine theorem" in $\triangle {ABC}$ and $ \triangle {ADC}$.
I obtain $sin(5x)=2sin(3x)\cdot cos(4x)$. Now I am stuck.
I try to construct and to solve it with an elementary construction but I didn't succeed. Can I apply here "The pants theorem"?
|
Let $E$ on $BC$ such that $CE\cong AC\cong AB$, and denote $\angle ABC = \alpha$.
*
*$\triangle BDE$ is isosceles, therefore $\angle BDE =\angle BED= \frac{\alpha}2$.
*Since $\angle ADC = \frac34\alpha$, we have $\angle EDC = \frac14 \alpha$.
*Note that $\angle DEC = 180^\circ -\frac12\alpha$, so that $\triangle DEC$ is isoceles, too.
*Construct then the rhombus $DGCE$ and connect $A$ with $G$. Observe that $\triangle AGC$ is equilateral, $\triangle ADG$ is isosceles, and $DG\parallel BC$.
*We can write therefore $\angle BAC$ in two ways and obtain the equation $$180^\circ - 2\alpha = 60^\circ + \alpha.$$
*Thus $\alpha = 40^\circ$ and $\angle BAC = 100^\circ$.
| {
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Determine if the fuction satisfy a local or a uniform Lipschitz condition The question: Determine if $f(t,y) = \frac{t^2y}{1+y^2}$ satisfies a local or a uniform Lipschitz condition.
My thoughts: Well from my work, I have
$$ |f(t,x) - f(t,y)| \leq K |x-y|. $$
I can determine what $K$ is by the mean value theorem. For any $z \in [x,y]$,
$$ \Big|\frac{f(t,x)-f(t,y)}{x -y}\Big| = \Big|\frac{\partial f}{\partial y} \Big |. $$
Therefore $$\frac{\partial f}{\partial y} = \frac{t^2 (y^2-1)}{(y^2 + 1)^2}$$. My question is how can we show this function is either locally Lipschitz and not globally Lipschitz. I am guessing this function is not uniformly Lipschitz because of the $t^2$ term.
Thank you very much!
| $Df(t,y)=\begin{pmatrix}
\frac{2ty}{1+y^{2}} &\frac{1-y^2}{(y^2+1)^2}
\end{pmatrix}$ so $\|Df(x,y)\|\le \sqrt{4t^2+1}.$
Let $(t,y),(t_1,y_1),(t_2,y_2)\in \overline J\times \overline U$, where $J,U$ is some bounded open sets in $\mathbb R$
Then, the MVT (in two variables) and compactness show that
$|f(t_1,y_1)-f(t_2,y_2)| \le \sup_{(t,y)\in\overline J\times \overline U}\{\sqrt{4t^2+1}\}\cdot\|(t_1,y_1)-(t_2,y_2)\|\le$
$ K\|(t_1,y_1)-(t_2,y_2)\|$, where $K$ depends on $J$ and $U$. Thus, $f$ is locally Lipschitz.
Now, $f$ is differentiable eveywhere, so
$f(t+h,y+k)-f(t,y)=Df(t,y)(h,k)+r(h,k)$ where $\frac{r(h,k)}{|(h,k)|}\to 0$ as $(h,k)\to (0,0)$.
Set $y=1$ and consider the points $(t,1).$ We have then,
$\frac{|Df(t,1)(h,k)|}{|(h,k)|}=\frac{|th|}{|h|+|k|}$ so $\frac{|f(t+h,1+k)-f(t,1)|}{|(h,k)|}=\frac{|th|}{|h|+|k|}+\frac{r(h,k)}{|(h,k)|}\to |t|$ as $(h,k)\to (0,0)$
which implies that $f$ is not globally Lipschitz.
| {
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Show that function $E$ is non-multiplicative. Definition 1:
A function $f$ is said to be non-multiplicative if
$$f(ab)\ne f(a)f(b)$$
for all coprime integers $a,b>1$.
Definition 2:
We define the function $E$ as
$$E(n)= n+1-\tau (n)- \phi(n)$$
$$=\sum_{(n,d)\notin\{1,d\} \\ \ \ \ \ 1<d<n}1.$$
Here, $(n,d)$ denotes gcd$(n,d)$
The values of $E(n)$ A045763
Question:
Show that the function $E$ is non-multiplicative.
$\\$
Example:
Let $a=10$, $b=7$, $ab=n=70$.
We have $E(70)=39$, $E(10)=3$, $E(7)=0$, and
$E(70)\ne E(10)E(7)$.
| First recall that $n = \sum_{d | n} \phi(d)$, so using this we have
$$E(n) = \sum_{d | n} \phi(d) - \sum_{d | n} 1 - (\phi(n) - 1) = \sum_{d | n,\, d < n} (\phi(d) - 1)$$
Now note that (i) for coprime $a, b$, and $d | ab$, we have a unique decomposition $d = d_1 d_2$ for $d_1 | a$ and $d_2 | b$, and that (ii) for coprime $d_1, d_2$ we have
$$\phi(d_1d_2) - 1 = \phi(d_1)\phi(d_2) - 1 \geq (\phi(d_1) - 1)(\phi(d_2) - 1),$$
since $\phi(d_1), \phi(d_2) \geq 1$. Using these facts, it follows that for coprime $a, b > 1$, we have
\begin{align*}
E(ab)
&= \sum_{d | ab,\, d < ab} (\phi(d) - 1) \\
&= \sum_{\substack{d_1 | a,\, d_2 | b \\ (d_1, d_2) \neq (a, b)}}(\phi(d_1d_2) - 1) \\
&> \sum_{\substack{d_1 | a,\, d_2 | b \\ d_1 < a,\, d_2 < b}}(\phi(d_1d_2) - 1) \\
&\geq \sum_{\substack{d_1 | a,\, d_2 | b \\ d_1 < a,\, d_2 < b}}(\phi(d_1) - 1)(\phi(d_2) - 1) \\
&= E(a)E(b)
\end{align*}
where the first inequality is strict because the difference is
$$\sum_{d_1 | a,\,d_1 < a} (\phi(d_1b) - 1) + \sum_{d_2 | b,\,d_2 < b} (\phi(ad_2) - 1) \geq \phi(b) + \phi(a) - 2 > 0$$
since at least one of $a, b$ is divisible by a prime $\geq 3$. Thus $E(ab) > E(a)E(b)$ for coprime $a, b > 1$, so $E$ is non-multiplicative.
| {
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Are trigonometry functions Ratios or Distance? Consider a point, say $S(2,3)$. Now here $3$ indicate that $S$ is $3$ units away from x axis. Right?
Now consider what Wikipedia says:
The trigonometric functions cos and sin are defined, respectively, as
the x- and y-coordinate values of point A.
This definition of $sin$ and $cos$ is based on unit circle. In this definition sin is defined as Y-coordinate of point $A$ on unit circle. But now what do we mean by Y-coordinate?
Y-coordinate is distance between point $A$ to $x$-axis (Right? ).
How could sin or for that matter any trigonometric function can be a distance? Trigonometry functions, for acute angle, are defined as ratios of sides. How could they be "distance"(with unit) in one definition and "ratio" (unitless) in other?
| This is a good question. I think it best always to regard the values of $\sin$, $\cos$, and so on as ratios. What allows these values seemingly to be defined as distances in your quotation from Wikipedia is that that definition refers to the unit circle—a circle whose radius is $1$. A similar definition that works for circles of arbitrary radius would be
The trigonometric functions $\cos$ and $\sin$ are defined, respectively, as the $x$- and $y$-coordinate values of point $A$ divided by the radius of the circle.
In this definition, the values are again ratios.
Added: There is a sense in which lengths stated within some system of measurement are ratios too. To say that a tree is $3$ meters high is to say that the ratio of its height to that of the fundamental unit of measurement—whether that's defined by an actual meter stick somewhere, or something else—is $3$. When we quote units with our lengths, we are implicitly carrying along a physical length to be used for comparison. So mathematically, lengths quoted with units are pairs, where the two elements of the pair are
*
*the ratio of that which is being measured to the fundamental measuring unit, and
*the fundamental measuring unit itself.
When we talk about the unit circle, we are abstracting away from that a bit by using the circle's radius itself as a measure, rather than something external. All quoted lengths are now really ratios of lengths defined within the figure itself since the unit of measure is within the figure.
| {
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Is my understanding of cardinality of sets of strings correct? Is my understanding of cardinality of sets of strings correct?
1) A finite set of symbols (containing $x$ symbols) and strings of finite length $n$ gives us $x^n$ finitely many strings.
2) A finite set of symbols (containing $x$ symbols) and strings of countably infinite length $\aleph_0$ gives us $x^{\aleph_0}$ uncountably many strings.
3) A countably infinite set of symbols (containing $\aleph_0$ symbols) and strings of finite length $n$ gives us $\aleph_0^n$ countably infinitely many strings.
4) A countably infinite set of symbols (containing $\aleph_0$ symbols) and strings of countably infinite length $\aleph_0$ gives us ${\aleph_0}^{\aleph_0}$ uncountably many strings.
And I've learned that there are countably infinite number of strings of finite lengths given a set of finitely many symbols, which is different from saying $x^n$, since it's talking about all values of the finite number $n$. Is there a formula to capture this notion in relation to $x^n$?
| Apart from the issue about $x$ pointed out in the comments, your assertions are correct.
For the last one, if $X$ is a finite set of cardinality $x$, let $X^n$ denote the set of words of length $n$ on the alphabet $X$ (a finite set of cardinality $x^n$, as you observed). Note that, for $n = 0$, $X^0$ is the singleton containing the empty word.
Then the set of all finite words, usually denoted $X^*$, is the union $\bigcup_{n \geqslant 0} X^n$. As a countable union of finite sets, it is countable, and hence its cardinality is $1$ if $x = 0$ and $\aleph_0$ otherwise.
| {
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Line $11x+3y-48 = 0$ tangent a graph $f(x) = \frac{4x + 3}{3x - 6}$ at $(a,b)$ Line $11x+3y-48 = 0$ tangent a graph $f(x) = \frac{4x + 3}{3x - 6}$ at (a,b) when $a<b$
$a-b = ...$
Find gradient of the line, df(x) / dx
$\frac{(4x+3)3 - (3x-6)(4)} {(3x-6)^2}$
Which the same as $ -11/3$
$(12x+9 - 12x + 24 ) 3= -11(3x-6)^2$
If i solve x, literally i solve $a $ right?
Is there less complicated way to solve it?
| Rewrite
$$f(x)=\frac43+\frac{11}{3x-6}.$$
Now
$$f'(x)=-\frac{11}{(3x-6)^2}\cdot3$$
and $f'(x)=-11/3$ gives $(3x-6)=\pm1$, that is $x=7/3$ and $x=5/3$.
| {
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In how many ways can four men and four women be seated at a round table if no two men are to be in adjacent seats? In how many ways can four men and four women be seated at a round table if no two men are to be in adjacent seats?
Please use principle of inclusion exclusion to solve.
My approach:
Let $S_i$ represent the number of arrangements where $i$ men are adjacent. Thus, $S_2$, for instance, would be ${4 \choose 2}(2!)(6!)$, treating the two adjacent men as one person in the circular arrangement (leaving seven people so $6!$ ways), and there are $2!$ ways to arrange these two adjacent men.
The whole equation would look like:
$$7! - S_2 + S_3 - S_4 = 7! - {4 \choose 2}(2!)(6!) + {4 \choose 3}(3!)(5!) - {4 \choose 4}(4!)(4!) = -1296$$
I am not sure what I did wrong as the result should be positive.
| The error you made was counting the number of consecutive men rather than the number of pairs of adjacent men. Let $|A_i|$ be the number of pairs of adjacent men.
Let Angela be one of the women. Seat Angela. We will use her as our reference point. Relative to her, we can arrange the other seven people in $7!$ ways as we proceed clockwise around the table. From these, we must subtract those arrangements in which there are one or more pairs of adjacent men.
A pair of adjacent men: Seat Angela. Choose which two of the four men will sit in adjacent seats. We now have six objects to arrange in the remaining seats, the block of two men and the other five people. Those objects can be arranged in $6!$ ways as we proceed around the table. The two men can be arranged within the block in $2!$ ways. Hence, there are
$$|A_1| = \binom{4}{2}6!2!$$
such seating arrangements, which agrees with your count for two consecutive men.
Two pairs of adjacent men: This can occur in two ways. Either the pairs overlap, in which case there are three consecutive men or they are disjoint, so that there are two separate pairs of adjacent men. You did not consider the second of these possibilities.
Two overlapping pairs of adjacent men: Seat Angela. Choose which three of the four men will sit in consecutive seats. We have five objects to arrange in the remaining seats, the block of three men and the other four people. The objects can be arranged in $5!$ ways as we proceed clockwise around the table. Within the block, the three men can be arranged in $3!$ ways. Hence, there are
$$\binom{4}{3}5!3!$$
such seating arrangements, which agrees with your count for three consecutive men.
Two disjoint pairs of adjacent men: Seat Angela. Choose which of the other three men is paired with the youngest man. The other two men must form the other pair of adjacent men. We have five objects to arrange in the remaining seats, the two blocks of adjacent pairs of men and the other three women. The objects can be arranged in $5!$ ways as we proceed clockwise around the table. Within each block of two men, the men can be arranged in $2!$ ways. Hence, there are
$$\binom{3}{1}5!2!2!$$
such seating arrangements.
Thus,
$$|A_2| = \binom{4}{3}5!3! + \binom{3}{1}5!2!2!$$
Three pairs of adjacent men: Since there are only four men, this can only occur if the three pairs are overlapping, that is, if the four men sit in consecutive seats. Seat Angela. We have four objects to arrange in the remaining seats, the block of four men and the other three women. The objects can be arranged in $4!$ ways. The four men can be arranged within the block in $4!$ ways. Hence,
$$|A_3| = \binom{4}{4}4!4!$$
which agrees with your count for four consecutive men.
Thus, by the Inclusion-Exclusion Principle, the number of seating arrangements of four men and four women at a round table in which no two men sit in adjacent seats is
$$7! - \binom{4}{2}6!2! + \binom{4}{3}5!3! + \binom{3}{1}5!2!2! - \binom{4}{4}4!4! = 144$$
Check: Since there are four men and four women, as Bram28 points out, the only way for the men not to sit in adjacent seats is if the men and women alternate seats. Seat Angela. Doing so determines which of the remaining seats will be filled by women and which will be filled by men. The three seats that are available to the other three women can be filled in $3!$ ways as we proceed clockwise around the table relative to Angela. Once those seats have been filled, the remaining four seats can be filled by the four men in $4!$ ways as we proceed clockwise around the table relative to Angela. Hence, there are
$$3!4! = 144$$
seating arrangements of four men and four men at a round table in which no two of the men are adjacent, which agrees with the result we obtained by using the Inclusion-Exclusion Principle.
| {
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Characterizing convex hull by set of affinely independent vectors While the statement seems intuitively plausible, I am currently struggling to see a proof that any convex hull "spanned" by a set of $n+1$ affinely independent $n$-vectors, $\{x_1,x_2,...,x_{n+1}\}$, is uniquely characterized by this set of vectors.
That is to say, how can I see that $$\left\{\sum_{j=1}^{n+1}\lambda_j\,x_j:\sum_{j=1}^{n+1}\lambda_j=1\text{ and }\lambda_j\geq0\,\,\forall\,\,j\right\}=\left\{\sum_{j=1}^{k+1}\gamma_j\,z_j:\sum_{j=1}^{k+1}\gamma_j=1\text{ and }\gamma_j\geq0\,\,\forall\,\,j\right\}$$implies that $\{x_1,x_2,...,x_{n+1}\}=\{z_1,z_2,...,z_{k+1}\}.$
Thank you very much.
| Let use now use the barycentric coordinates associated with the simplex $\Delta$. For simplicity, let us write $s_j,t_j$ instead of $x_j-O,z_j-O$ for the vectors. Note that, thanks to the affine linear independence of the two sets, these new sets of vectors are linearly independent.
Since the two set have the same convex hull we already know that every element of one set can be written as a linear combination of elements from the other. Thus our problem is reduced to prove that such a linear combination consists of only one vector.
As a proof by contradiction, let
$$t_l=\sum \lambda_j s_j \\
\exists_{p\neq q}\lambda_p,\lambda_q≠0$$
This implies (together with the fact that the two set have the same convex hull)
$$s_p=\sum_{j=0}^{k+1}\gamma_jt_j\ (1)
\\
s_q=\sum_{j=0}^{k+1}\hat{\gamma}_jt_j\ (2)$$
Combining with the precedent requirement, we get
$$s_l=\lambda_p\sum_{j=0}^{k+1}\gamma_js_j+\lambda_q\sum_{j=0}^{k+1}\hat{\gamma}_js_j+...$$
By linear independence (and the fact that $\lambda\ge 0$) we obtain
$$\gamma_{j\neq l}=0\\ \hat{\gamma}_{j\neq l}=0$$
This yeld (thanks to (1),(2) )
$$s_p=\gamma_l t_l\\
s_q=\hat{\gamma}_l t_l$$
This contradicts the linear independence of $s_p, s_q$.
| {
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Prove that $x^2+px+p^2$ is a factor $(x+p)^n-x^n-p^n$, if $n$ be odd and not divisible by $3$. Question:
Prove that $x^2+px+p^2$ is a factor of $(x+p)^n-x^n-p^n$, if $n$ is odd and is not divisible by $3$.
My approach:
$$(x+p)^n-x^n-p^n=\sum_{r=0}^n\limits {n\choose r} x^{n-r}p^r-x^n-p^n$$ What can I do after that?
| You need to show every root of $x^2 + px + p^2$ is a root of $(x + p)^n - x^n - p^n$. The quadratic formula gives that $x = p\omega$ or $p\bar{\omega}$
where $w = -{1 \over 2} + {\sqrt{3} \over 2}i$ is a complex third root of unity. So what you need to show is that for the values of $n$ in question
that
$$(p\omega + p)^n - (p\omega)^n - p^n = 0$$
You also need this for $\bar{\omega}$ in place of $\omega$ but that will follow by taking complex conjugates of this equation.
Hence you need to show for your values of $n$ that
$$(\omega + 1)^n = \omega^n + 1$$
I won't finish this off but you can reduce this to checking a small list of $n$.
| {
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Combinatorics and Expected Value There are 15 candidates running for a given Senate seat, comprised of 10 men and 5 women. There are 32 polls, in which any of the candidates are equally likely to be ranked first, independently of the other polls (meaning the position of each candidate on any of the polls is purely random).
Find the expected value of the number of times that a woman will rank 1st in any of the polls.
I understand I need to find the probability of that any given woman will rank 1st in at least 1 of the polls, which I need direction on how exactly to approach.
| Looking at this problem, we can identify a discrete random variable.
Let $X$ be the number of polls where a woman ranks first. $X$ is a binomial variable with the distribution $X$~$Bin(32,\frac{1}{3})$, since the chance of a woman being ranked first is $\frac{5}{15}=\frac{1}{3}$.
Then the expected value of a binomial random variable is
\begin{align}
E(X)&=np\\
&=32\cdot\frac{1}{3}\\
&\approx 10.6667
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3448628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
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