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H: The Broken Calculator Problem
So here is the Problem :-
Tom has a specific calculator . Unfortunately, all keys are broken except for one row$: 1,2,3,+,-$. Tom presses a sequence of $5$ random keys; where at each stroke, each key is equally likely to be pressed. The calculator then evaluates the entire expression, yielding a result of E. Find the Expected Value of E.
Before doing this we need to remember some facts :-
$(i)$ Excess Operators will be parsed as signs. For e.g. :- $-2-+3$ gives $E = -5$.and $-+-31$ gives $E = 31$
$(ii)$ Trailing Operators are discarded . For e.g. :- $2-+--$ gives $E = 2$
$(iii)$ Negative Sums are allowed . For e.g. :- $13 - 22$ give $E = -9$.
$(iv)$ A string consisting only of operators , gives $E$ as $0$ .
This Problem looks very interesting to me . First of all there can be many different types of sums for E and second, it's definitely not quite easy to get the expected value of it, and I don't know who to start doing it . Any ideas for this problem will be greatly appreciated !!
AI: Any string that starts with an operator is cancelled out by switching every operator in front of a number (if an operator is in front another operator, then leave it untouched). For example, if you have $−1234$, it is cancelled out by $+1234$. If you have $++123$, it is cancelled out by $+−123$. If you have $−1++3$, it is cancelled out by $+1+−3$. Therefore, we have to calculate the sum of all the results obtained from strings starting with a number.
Now, let $d$ be a number string of length at most $5$ with digits in $\{1,2,3\}$. Let $s(d)$ denote the sum of all values corresponding to strings of length $5$ starting with $d$ and the entry next to the end of $d$ is a sign (or if $d$ has length $5$ already, it is not followed by any sign). Show that
$$s(d)=\left\{\begin{array}{ll} \text{value}(d)\cdot\left(2\cdot 5^{4-\text{length}(d)}\right)&\text{if }\text{length}(d)<5\,,\\
\text{value}(d)\cdot 1&\text{if }\text{length}(d)=5
\,,\end{array}\right.$$
where $\text{length}(d)$ is the length of $d$, and $\text{value}(d)$ is the value of the string $d$ when interpreted as an integer.
If $S$ is the sum of all $s(d)$ where $d$ runs over all the number strings of length at most $5$ with digits in $\{1,2,3\}$, then show that
$$\begin{align}S&=3^0\cdot 6\cdot (2\cdot 5^3)+3^1\cdot 66\cdot (2\cdot 5^2)+3^2\cdot 666\cdot (2\cdot 5)\\&\phantom{abcde}+3^3\cdot 6666\cdot 2+3^4\cdot 66666\cdot 1=5831250\,.\end{align}$$
The expected value is then
$$\dfrac{S}{5^5}=\frac{5831250}{3125}=1866\,.$$
If the calculator calculates the results in base $b$, and there are $k$ available digits $t_1,t_2,\ldots,t_k$ (the available signs are still $+$ and $-$), then the expected value of the results from pressing the calculator $n$ times is
$$\frac{\sum\limits_{j=1}^k\,t_j}{(k+2)^n}\,\left(\sum_{r=1}^{n-1}\,k^{r-1}\,\frac{b^r-1}{b-1}\,\left(2\cdot (k+2)^{n-1-r}\right)+k^{n-1}\,\frac{b^n-1}{b-1}\right)\,.$$
I am leaving a proof and the simplification of the long expression above for the curious reader.
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H: What is value of this integral? $\int_{0}^{\infty}\frac{\log(1+4x^2)(1+9x^2)(9+x^2)+(9+x^2)\log(4+x^2)(10+10x^2)}{(9+x^2)^{2}(1+9x^{2})}dx$
What is value of this integral $$I=\int_{0}^{\infty}\frac{\log(1+4x^2)(1+9x^2)(9+x^2)+(9+x^2)\log(4+x^2)(10+10x^2)}{(9+x^2)^{2}(1+9x^2)}dx$$
My work :
\begin{align*}I&=\int_{0}^{\infty}\frac{\log(1+4x^2)}{9+x^2}dx+\int_{0}^{\infty}\frac{\log(4+x^2)(10+10x^2)}{(9+x^2)(1+9x^2)}dx\\
&=\int_{0}^{\infty}\frac{\log(1+4x^2)}{9+x^2}dx+\int_{0}^{\infty}\frac{\log(4+x^2)}{1+9x^2}dx+\int_{0}^{\infty}\frac{\log(4+x^2)}{9+x^2}dx\\
&=j_{1}+j_{2}+j_{3}\\
\end{align*}
$$j_{1}=\int_{0}^{\infty}\frac{\log(1+4x^2)}{9+x^2}dx=\sum_{n=0}^{\infty}\frac{(-1)^n(2^{2(n+1)})}{n+1}\int_{0}^{\infty}\frac{x^{2(n+1)}}{9+x^2}dx$$
$$j_{2}=\log(4)\int_{0}^{\infty}\frac{1}{1+(3x)^2}dx+\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(n+1)(2^{2(n+1)})}\int_{0}^{\infty}\frac{x^{2(n+1)}}{1+9x^2}dx=\frac{\log(4)\pi}{6}+\sum_{n=0}^{\infty}\frac{(-1)^n}{(n+1)2^{2(n+1)}}\int_{0}^{\infty}\frac{x^{2(n+1)}}{1+9x^2}dx$$
$$j_{3}=\int_{0}^{\infty}\frac{\log(4+x^2)}{9+x^2}dx=\frac{\log(4)\pi}{54}+\sum_{n=0}^{\infty}\frac{(-1)^n}{(n+1)2^{2(n+1)}}\int_{0}^{\infty}\frac{x^{2(n+1)}}{9+x^2}dx$$
Wait for a review to find solutions to this
AI: Consider using differentiation under the integral sign. Parameterize the integral as the following:
$$I(a)=\int_0^{\infty} \frac{\ln{\left(1+ax^2\right)}}{9+x^2} + \frac{\ln{\left(a+x^2\right)}}{1+9x^2}+ \frac{\ln{\left(a+x^2\right)}}{9+x^2} \; \mathrm{d}x$$
The integral in question is $I(4)$. First, differentiate $I(a)$ with respect to $a$: \begin{align*}
I'(a)&=\int_0^{\infty} \frac{x^2}{(9+x^2)(1+ax^2)}+\frac{1}{(1+9x^2)(a+x^2)}+\frac{1}{(9+x^2)(a+x^2)} \; \mathrm{d}x \\
&=\int_0^{\infty} -\frac{10 (a - 1)}{(a - 9) (9 a - 1) (a + x^2)} + \frac{2 (9 a - 41)}{(a - 9) (9 a - 1) (x^2 + 9)} + \frac{9}{(9 a - 1) (9 x^2 + 1)} + \frac{1}{(1 - 9 a) (a x^2 + 1)} \; \mathrm{d}x \\
&=\frac{\pi}{6} \left(\frac{\frac{19}{\sqrt{a}}+9}{3a+10\sqrt{a}+3}\right)\\
\end{align*}
Now, $I'(a)$ with respect to $a$ from $0$ to $4$:
\begin{align*}
I(4)&=\frac{\pi}{6} \int_0^4 \frac{\frac{19}{\sqrt{a}}+9}{3a+10\sqrt{a}+3} \; \mathrm{d}a \\
&=\frac{\pi}{6} \int_0^4 \frac{\frac{1}{\sqrt{a}}}{\sqrt{a}+3} + \frac{\frac{6}{\sqrt{a}}}{3\sqrt{a}+1} \; \mathrm{d} a \\
&=\frac{\pi}{6} \left(2 \int_0^4 \frac{\mathrm{d}\left(\sqrt{a}+3\right)}{\sqrt{a}+3} + 4 \int_0^4 \frac{\mathrm{d}\left(3\sqrt{a}+1\right)}{3\sqrt{a}+1}\right) \\
&=\frac{\pi}{3} \left(\ln{\left(\sqrt{a}+3\right)}+2\ln{\left(3\sqrt{a}+1\right)}\right) \bigg \rvert_0^4 \\
I(4) &= \int_{0}^{\infty}\frac{\log(1+4x^2)(1+9x^2)(9+x^2)+(9+x^2)\log(4+x^2)(10+10x^2)}{(9+x^2)^{2}(1+9x^2)} \mathrm{d}x =\boxed{\frac{\pi \ln{\left(\frac{245}{3}\right)}}{3}}\\
\end{align*}
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H: Is it always possible to get local basis elements of a vector field module on a manifold?
I'm studying about vector fields from Schuller's lectures on youtube. Given a smooth manifold $M$, if we have a smooth module $\Gamma(TM)$ consisting of all smooth vector fields on the manifold, it's clear to me that we can't get global "basis vector fields". In other words we can't come up with a global component representation of vector fields (in terms of the basis vector fields) on the entire manifold.
However, I'm unsure if it's possible to always come up with even local basis vector fields.
Given a vector field $X\in\Gamma(TM)$, is it always possible to come up with a local basis representation of the vector field $X$?
Would be grateful if anyone could clarify!
AI: Yes, any chart $(U,x)$ on the manifold $M$ gives rise to a local basis of vector fields defined on $U$,$\left\{\frac{\partial}{\partial x^1}, \dots, \frac{\partial}{\partial x^{m}}\right\}$, where $m= \dim M$; i.e for every $p\in U$, $\left\{\frac{\partial}{\partial x^1}(p), \dots, \frac{\partial}{\partial x^{m}}(p)\right\}$ is a basis for $T_pM$.
|
H: Show that the elements of the sequence are divisible by $2^n$
I am trying to prove the following:
Consider the sequence defined by $A_{n+2}=6A_{n+1}+2A_n, A_0=2, A_1=6$. Show that $2^n|A_{2n-1}$ but $2^{n+1}\nmid A_{2n-1}$.
The first terms of this sequence are 2, 6, 40, 252, 1592, 10056, 63520.
In fact, the maximal exponent of $2$ that divides $A_n$ seems to follow a pattern with a period of 4:
1, 1, 3, 2, 3, 3, 5, 4, 5, 5, 7, 6...
But I haven't been able to prove this.
AI: Using induction, it's easy to prove that $A_n=(3+\sqrt{11})^n+(3-\sqrt{11})^n$ (roots of the characteristic polynomial of the corresponding recurrence $t^2-6t-2$ are $3\pm\sqrt{11}$).
Define
$$
B_n:=A_{2n+1}~\text{for}~n\geq 0.
$$
Then,
$$
B_n=(3+\sqrt{11})^{2n+1}+(3-\sqrt{11})^{2n+1}.
$$
It's easy to check that $B_0=6$, $B_1=252$ and
$$
B_{n+2}=40B_{n+1}-4B_n
$$
(because $(3\pm\sqrt{11})^2=20\pm 6\sqrt{11}$ are roots of $t^2-40t+4$).
Now, define $C_n:=\frac{B_n}{2^{n+1}}$. Then, we can rewrite previous equality as
$$
2^{n+3}C_{n+2}=40\cdot 2^{n+2}C_{n+1}-4\cdot 2^{n+1} C_n.
$$
or
$$
C_{n+2}=20C_{n+1}-C_n.
$$
Also, $C_0=3$ and $C_1=63$.
It's clear from the recurrence relation for $\{C_n\}$ that all $C_n$ are integers. Moreover, each $C_n$ is odd (because $C_0,C_1$ are odd and from the recurrence relation: $C_{n+2}\equiv C_n\pmod 2$).
Therefore, $A_{2n+1}=2^{n+1}\cdot C_n$ for some odd $C_n$. Thus, $2^{n+1}\mid A_{2n+1}$ and $2^{n+2}\not\mid A_{2n+1}$, as desired.
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H: How is notation $\{f_n(x)\}\nearrow x$ interpreted?
In the context of continuity of probabilities we define $\{A_n\}\nearrow A$ to mean that $A_1 \subseteq A_2 \subseteq A_3 \subseteq ...$ and $\cup_n A_n=A$, where $A, A_1, A_2, A_3, ... \in \mathcal{F}$ for some probability triplet $(\Omega, \mathcal{F}, P)$. I have no problems with this notation.
But how is $\{f_n(x)\}\nearrow x$ interpreted, where $f_n:[0,\infty) \rightarrow [0,\infty)$ and $x$ is a random variable? I don't see how $f_1(x) \subseteq f_2(x) \subseteq f_3(x) \subseteq ...$ considering $f_n(x)$ is not a set but a real value. A possible interpretation might be that it's range of $f_n$ that is a subset of the range of $f_{n+1}$; but then it's unclear how $\cup_n f_n(x)=x$, i.e. how the union of ranges would correspond to a real value.
(Apparently $\{f_n(x)\}\nearrow x$ is useful when deriving the expectation of general non-negative random variables.)
AI: It means that $f_n(x)\le x$ for all $n,$ and that $\lim_{n\to\infty}f(x)=x.$ Verbally, we may say that $f_n(x)$ converges to $x$ from below. Often, it also indicates that $f_n(x)$ is monotonically increasing, but this varies from source to source.
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H: Proof that $\lim_{x\rightarrow \infty} f(x+a) =\lim_{x\rightarrow \infty} f(x) $ for constant $a$
It is easy to understand why the following statement is true for variable $x$ and constant $a$:
$\lim_{x\rightarrow \infty} f(x+a) =\lim_{x\rightarrow \infty} f(x) $
However, is there a proof for it, or is it true by definition?
AI: You're almost there. Let $l:= \lim\limits_{x\to \infty} f(x)$. Now, let $\epsilon>0$ be arbitrary. Then, by hypothesis, there exists a $c\in \Bbb{R}$ such that for all $x\in \Bbb{R}$, if $x>c$ then $|f(x)-l| < \epsilon$.
I'm not sure if this will help you or not, but let's just temporarily define $g(x):= f(x+a)$. Our goal is to show $\lim\limits_{x\to \infty}g(x) = l$ as well. Now, note that
\begin{align}
x>c-a & \implies x+a > c \\
&\implies |f(x+a) - l| < \epsilon \\
& \implies |g(x) - l| < \epsilon
\end{align}
So, what we have done is shown the existence of a $\tilde{c}$ (namely $c-a$) such that if $x>\tilde{c}$ then $|g(x)-l| < \epsilon$. Therefore, we have shown that $\lim\limits_{x\to \infty}g(x) = l$, exactly as we wanted to.
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H: Combination to find integers satisfying a condition
Let $n$ and $k$ be positive integers such that $n\ge\frac{k(k+1)}{2}$.
The number of solutions $(x_1,x_2,\dots,x_{k})$, with $x_1\ge1$, $x_2\ge2$,..., $x_{k}\ge k$ for all integers satisfying $x_1+x_2+\dots+x_{k}=n$ is?
I substituted the last equation in the first inequality.
$$x_1+x_2+\dots+x_{k}\ge\frac{k(k+1)}{2}.$$
Took $x_1$ as $1+ t_1$, $x_2$ as $2+t_2$...where $t_i\ge 0$. On simplifying by using sum of k numbers, I end with with
$t_1+t_2+\dots+t_{k} \ge0$. Since $x_1,x_2$... are in increasing order, and sum of all $t$ values is $0$, I conclude that this is only possible when $t=0$. Therefore only one solution is possible when $LHS = RHS$. The inequality is not valid.
But the answer is $\frac{1}{2}(2n-k^2+k-2)$. What am I missing here?
AI: No, the sequence $x_1,x_2,\dots,x_{k}$ is not necessarily increasing, so your conclusion is not correct.
Let $t_k=x_k-k\geq 0$, then we have to count the number of non-negative integer solutions of
$$t_1+t_2+\dots+t_k=n-\frac{k(k+1)}{2}$$
which is, by Stars-and-Bars, given by
$$\binom{n-\frac{k(k+1)}{2}+k-1}{k-1}=\binom{\frac{2n-k^2+k-2}{2}}{k-1}.$$
P.S. My answer is correct. Probably there is a typo in your book. The integer $\binom{\frac{2n-k^2+k-2}{2}}{k-1}$ is precisely the coefficient of $t^n$ in
$$ (t + t^2 +t^3+ \dots)(t^2 + t^3 +t^4+\dots)\dots (t^k + t^{k+1}+t^{k+2}+\dots),$$
that is
$$[t^n]\prod_{j=1}^k\frac{t^j}{1-t}=[t^{n-\frac{k(k+1)}{2}}](1-t)^{-k}
=(-1)^{{n-\frac{k(k+1)}{2}}}\binom{-k}{n-\frac{k(k+1)}{2}}=\binom{n-\frac{k(k+1)}{2}+k-1}{k-1}.$$
Numerical example. Take $n=8$ and $k=3$, then it is easy to see that
$$ (t + t^2 +t^3+ \dots)(t^2 + t^3+ t^4 +\dots)(t^3 + t^{4}+t^{5}+\dots)
=t^6+3t^7+6t^8+\dots$$
and the coefficient of $t^8$ is $6$:
$$\binom{\frac{16-9+3-2}{2}}{3-1}=\binom{4}{2}=6.$$
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H: Evaluate $\lim_{x\to -\infty} \frac{x^4\sin( \frac 1x )+ x^2}{1+|x|^3}$
Let $y=-x$
$$\lim_{y\to \infty} \frac{y^4 \sin (-\frac 1y) + y^2}{1+|y|^3}$$
$$=\lim_{y\to \infty} \frac{y^3 ((-y \sin \frac 1y) +\frac 1y)}{y^3(1+\frac{1}{y^3})}$$
At $y\to \infty$, $y\sin \frac 1y=0$ (What I think is happening)
Therefore the entire limit is $0$
I pointed out where I think I went wrong, but I am not able to correct it. What should be the right procedure?
AI: Let
$$L=\lim_{x\to -\infty} \frac{x^4\sin( \frac 1x )+ x^2}{1+|x|^3}$$
$$\implies L=\lim_{x \to -\infty} \frac{x^4/x+x^2}{1-x^3}=\lim_{x\to -\infty}\frac{ 1+1/x}{1/x^3-1}=-1.$$
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H: If entropy says the number of bits an information needs, why in this case it's less than one?
If,$3/4$ of the times, is raining in a city and $1/4$ is not, the entropy $(-\log(3/4))$ would say we need almost $0.415$ of a bit to say it's raining, and $2$ bits $(-\log(1/4) )$ to say it is not, right?? How can we make sense of this, how can we have less than one and more than $0$ bits?
AI: Having partial bits says that you can do more with the information than answer that one question. The number of bits given by an answer in your case is $-\frac 34 \log_2\left(\frac 34\right)-\frac 14 \log_2\left(\frac 14\right)\approx 0.811$. I should be able to find an encoding scheme that would represent the rain/no rain status of $1000$ days in $811$ bits.
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H: Can we estimate the profuct $(1+x_1)(1+x_2)\ldots(1+x_n)$ by the term $x_1\ldots x_n$?
Let $x_1, x_2, \ldots x_n$ be postive integers with $x_i\ge 2(1\le i\le n)$. Can we have the following inequality:
$$(1+x_1)(1+x_2)\cdots(1+x_n)\le f(n)x_1\cdots x_n,$$
where $f(n)$ is a term dependent on $n$ and I hope it is increasing in Polynomials or lower order. Does anyone can do it? Many thanks!
AI: Your constraint can be written
$$\left(\frac1{x_1}+1\right)\left(\frac1{x_2}+1\right)\cdots\left(\frac1{x_n}+1\right)\le\left(\frac32\right)^n$$ and this is tight.
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H: Local $\mathbb{k}$-algebra homomorphism
Let $(A,m)$ and $(B,n)$ be local commutative rings that are also $\mathbb{k}$-algebras. Let $\phi :B \rightarrow A$ be a local $\mathbb{k}$-algebra homomorphism.
Suppose that $A/m \cong \mathbb{k}$. I want to show that $B/n \cong \mathbb{k}$.
If we call the maps
$\tau : \mathbb{k} \rightarrow B \twoheadrightarrow B/n$
$\overline{\phi}: B/n \rightarrow A/m$,
$\rho : \mathbb{k} \rightarrow A \twoheadrightarrow A/m$
I have shown that $\overline{\phi}\circ\tau = \rho$. $\textbf{If we suppose further that $\rho$ is an isomorphism}$, as $\overline{\phi}$ is injective, it follows that $\tau$ must be surjective and hence an isomorphism (as it is already a nonzero ring homomorphism from a field).
But I don't see why $\rho$ has to be an isomorphism?
I don't think that the fact $A/m \cong \mathbb{k}$ implies that $\rho$ is an isomorphism, as the inclusion $k(t^2) \hookrightarrow k(t)$ gives a counterexample (see Local $k$-algebra with residue field $k$). So I am not sure that my approach is even on the right track!
This problem is related to Hartshorne Chapter 2, exercise 2.15.
Any help would be appreciated :).
AI: If you want $A/m\cong k$ as a $k$-algebra then that is saying that $\rho$
is an isomorphism, at least in my book.
But one can have $A/m\cong k$ just as a ring isomorphism and then $B/n$
need not be isomorphic to $k$. As an example, take a tower of field
extension $k\subset L\subset K$ with $k\cong K$ and $k\not\cong L$. Then let
$(A,m)=(K,0)$ and $(B,n)=(L,0)$.
Why do such fields exist? For example, take $k=\Bbb C$, $L=\Bbb C(t)$ and $K$
the algebraic closure of $L$.
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H: What is the proof that $(a+b)^2 >a^2 + b^2$?
I would like to know if there is a theorem that proves that
$$(a+b)^2>a^2+ b^2$$
where $ab>0$
I am also wondering whether there is a name associated with this inequality.
AI: By distributing out, you get that
$${(a+b)^2 = a^2 + 2ab + b^2 = (a^2 + b^2) + 2ab}$$
Now - since ${a,b>0}$, then ${2ab>0}$. In other words, ${(a+b)^2 = (a^2 + b^2) + \text{a little bit}}$. So
$${(a+b)^2 = (a^2 + b^2) + 2ab > a^2 + b^2}$$
Edit: Indeed as @KeithBackman has remarked - this argument still works even if both ${a,b<0}$ - the point is, we just need ${2ab>0}$ for this argument to work. If they are both negative, ${2ab}$ is still ${>0}$, so the argument holds still :)
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H: Triangle tangent to 3 Parabolas, finding the common area
Triangle $ABC$, $AB=4$, $BC=15$, $AC=13$.
Two sides are tangents to the respective Parabolas.
We have to find the area shaded.
My approach-
I tried finding the area of the quadratures(Archimedes) formed but it
doesn't help as- I have to find the area between the 2 intersecting
Parabolas also which I can't find.
AI: Hints based on the paper Properties of Parabolas Inscribed in a Triangle by J. A. Bullard, American Mathematical Monthly, volume 42, issue 10 (1935).
The vertices of the shaded area are along the medians. Such points divide each median in the ratio $1:8$. The three medians divide the shaded area in $6$ equivalent parts. The area of each part is $5/162$ of the whole area of the triangle $\triangle ABC$.
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H: Hilbert function is not eventually equal to a polynomial function
Consider the ring $S=k[x_1,\ldots,x_r]$, graded by $\mathbb N$ with each variable in degree $1$. Let $M$ be finitely generated graded $S$-module. The Hilbert function is defined by $H_M(d)=\dim_k M_d$. It is a well-known theorem of Hilbert that if $M$ is a finitely generated graded $S$-module, then $H_M(d)$ agrees with a polynomial for large $d$.
Now, consider the graded ring $S=k[z_1,z_2]$, with $\deg z_1=2,\deg z_2=3$. I have to show that the Hilbert function $H_S(d)$ is not eventually equal to a polynomial function of $d$.
The homogeneous component $S_d$ is the $k$-submodule of $S$ generated by the monomials of degree $d$. Since $\deg z_1=2$ and $\deg z_2=3$, a monomial $z_1^{\alpha_1}z_2^{\alpha_2}$ has degree $2\alpha_1+3\alpha_2$ (I'm not quite sure about this statement. Can someone verify please?). So given $d$, I have to find the total number of non-negative integer solutions to $2\alpha_1+3\alpha_2=d$. So this reduces to a problem in combinatorics. The generating function associated with the term $2\alpha_1$ is $\frac1{1-t^2}$ and with $3\alpha_2$ is $\frac1{1-t^3}$. After some tedious calculations(which I'm not including here), I got the final answer as
$$H_S(d)=\begin{cases}
\dfrac{(-1)^d}{4}+\dfrac14+\dfrac{d+1}6+\dfrac13,& d\equiv 0\pmod 3\\
\dfrac{(-1)^d}{4}+\dfrac14+\dfrac{d+1}6-\dfrac13,& d\equiv 1\pmod 3\\
\dfrac{(-1)^d}{4}+\dfrac14+\dfrac{d+1}6,& d\equiv 2\pmod 3
\end{cases},$$
and it seems to agree with the coefficients given here. But even after all these calculations, I'm struggling to find a solid argument to conclude the proof.
AI: The generating function of $H_S(d)$ is
$$f(t)=\frac1{(1-t^2)(1-t^3)}.$$
The generating function of a sequence $(a_n)_{n=0}^\infty$ of a function which is
eventually polynomial always has the form
$$g(t)=\frac{G(t)}{(1-t)^N}$$
where $G$ is a polynomial. In particular, $g(t)$ has poles in the complex plane
only at $t=1$. But $f(t)$ has a pole at $t=-1$.
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H: Confusion about probability space associated with infinite coin flips
Let $\Omega = \{ \omega = (\omega_1, \omega_2, \ldots) : \omega_j = 1 \text{ or } 0 \}$. For each positive integer $n$, let $\Omega_n = \{ \omega = (\omega_1, \ldots, \omega_n) : \omega_j = 1 \text{ or } 0 \}$. We can consider $\Omega_n$ as a probability space with $\sigma$-algebra $ 2^{\Omega_n}$ and probability induced by $\mathbb{P}_n(\omega) = 2^{-n}$.
We define $F_n$ to be the collection of all subsets $A$ of $\Omega$ such that there is an
$E \in 2^{\Omega_n}$ with \begin{equation}
A = \{(\omega_1, \omega_2, \ldots) : (\omega_1, \ldots, \omega_n) \in E\}. \tag 1
\end{equation}
$F_n$ is a finite $\sigma$-algebra (containing $2^{2^n}$ subsets) and $F_1 \subset F_2 \subset F_3 \subset \cdots$ (i.e., an ascending sequence of $\sigma$-algebras). If $A$ is of the form $(1),$ we let $\mathbb{P}(A) = \mathbb{P}_n(E_n)$. This gives a function $\mathbb{P}$ on on \begin{equation}
F^{0} = \bigcup_{j=1}^{\infty} F_j
\end{equation}
This is followed by the proposition that $F^{0}$ is an algebra but not a $\sigma$-algebra.
All of the above and most of the proof makes sense to me. But in the proof for the proposition I just mentioned states: $ \Omega \in F_0$ since $\Omega \in F^1$. I get a bit confused because $F_1$ should just consist of the first flip being tails and the first flip being heads. Does this mean that $F_1$ consists of each as the first element in two infinite sequences where such as $(1, \omega_2, \ldots)$ and $(0, \omega_2, \ldots)$, where $\omega_j, j > 1$ are all just not given? But then doesn't $F_1$ trivially contain all possible sequences? Maybe it's because I'm a bit rusty on measure theory or I'm missing something but I'm a bit confused.
AI: Note that $F_n$ is not a subset of $\Omega$, it is a set of subsets of $\Omega$. So it doesn't make sense to ask about $F_1$ containing all possible infinite sequences.
By definition, $F_1$ consists of four subsets of $\Omega$ (corresponding to the four subsets of $\Omega_1$). These four subsets are:
$\Omega$ (corresponding to $E=\Omega_1$)
$\emptyset$ (corresponding to $E=\emptyset$)
all sequences in $\Omega$ that start with $0$ (corresponding to $E=\{(0)\}$)
all sequences in $\Omega$ that start with $1$ (corresponding to $E=\{(1)\}$)
In general, $\Omega_n$ has size $2^n$ hence has $2^{2^n}$ subsets. These $2^{2^n}$ subsets of $\Omega_n$ correspond to $2^{2^n}$ subsets of $\Omega$, which together make up the collection $F_n$. The correspondence is: start with a subset $E$ of $\Omega_n$ (so this is some collection of sequences of $0$'s and $1$'s of length $n$). Then define the subset $A_E$ of $\Omega$ to consist of all infinite sequences whose first $n$ entries are a sequence in $E$. Now $F_n$ is the collection $\{A_E:E\subseteq \Omega_n\}$.
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H: How to solve $\int_0^1dx\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}\,dy$
The original question is:
Prove that:$$\begin{aligned}\\
\int_0^1dx\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}\,dy\neq\int_0^1dy&\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}\,dx\\
\end{aligned}\\$$
But I can't evaluate the integral $$\int_0^1dx\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}\,dy$$
At first, I assumed $x^2+y^2=z^2$. But, it is so complicated. Then, I assumed $x=r\cos\theta$ and $y=r\sin\theta$. But, I can't calculate the limits. Solving the equations I got three values of $\theta$ i.e. $\theta=0$, $\theta=\frac{\pi}{4}$ and $\theta=\frac{\pi}{2}$. I am just confused. Please help.
AI: Hint:
$$\int_0^1\frac {x^2-y^2}{(x^2+y^2)^2}\, dy = \int_0^1\frac{\partial}{\partial y} \left(\frac{y}{x^2 + y^2}\right)\, dy = \frac{1}{1+x^2} $$
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H: If the largest positive integer is n such that $\sqrt{n - 100} + \sqrt{n + 100}$ is a rational no. , find the value of $\sqrt{n - 1}$ .
So here is the Problem :-
If the largest positive integer is n such that $\sqrt{n - 100} + \sqrt{n + 100}$ is a rational no. , find the value of $\sqrt{n - 1}$ .
What I tried :- I think that for $\sqrt{n - 100} + \sqrt{n + 100}$ to be a rational no. , both $(n - 100)$ and $(n + 100)$ have to be squares. Suppose :- $(n - 100)$ = $k^2$ and $(n + 100)$ = $m^2$ for some positive integers $k,m$ , and in the end I could only deduce that $(m + 10)(m - 10) = k^2 + 100$ , but then I couldn't proceed .
Also by guesswork, I could deduce that for $n = 125$, both nos. do become squares, although I don't know whether $n = 125$ is the highest or not.
Any hints or explanations to this problem will be greatly appreciated !
AI: Let $\sqrt{n-100} + \sqrt{n+100} = p$, where $p$ is rational.
$$\implies 2n + 2\sqrt{n^2 - 10000} = p^2$$
But that must mean that $2\sqrt{n^2 - 10000}$ is rational.
Which must mean that $\sqrt{n^2 - 10000}$ is rational.
$$\implies n^2 - 10000 = k^2$$
$$\implies (n+k)(n-k) = 10000$$ The problem requires us to maximize $n$, notice that we'll get the maximum value of $n$ if we split $10000 = 5000 \times 2$ and set $n+k = 5000$ and $n-k = 2$ to get $n = 2501$.
Hence, $\boxed{\sqrt{n-1} = 50}$
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H: How to prove that zeros of two polynomials interlace implies the Wronskian is nonnegative or nonpositive?
Given two real polynomials $f,g$ with the roots are all real. We say the zeros of $f,g$ interlace if
$$\alpha_1\leq\beta_1\leq\alpha_2\leq\beta_2\leq\ldots$$ or
$$\beta_1\leq\alpha_2\leq\beta_2\leq\alpha_2\leq\ldots$$
where $\alpha_i,\beta_j$ are the roots of $f,g$, respectively. Define the Wroskian $W[f,g]=f^{'}g-fg^{'}$.
How to prove the following result?
If the zeros of $f,g$ interlace, then $W[f,g]$ is either nonnegative or nonpositive on the whole real axis
$\mathbb{R}$.
AI: I'll give you a hint. Try to prove the following.
Note that we necessarily have the roots of $f,g$ are distinct, and $\lvert\deg f-\deg g\rvert\leq 1$. Furthermore we can write (assuming WLOG $\deg f\leq\deg g$) $f=ag+\sum_{j=1}^{\deg g}b_jg_j$ where $g_j(x):=g(x)/(x-\beta_j)$, unique $(a,b_1,\dots,b_{\deg g})\in\mathbb{R}^{1+\deg g}$ since the $\beta_j$ are distinct. The interlacing condition is equivalent to: $b_j$, for all $j$, are of the same sign. So
$$
\frac{W[f,g]}{g^2}=\left(\frac{f}g\right)'=\sum_j\frac{-b_j}{(x-\beta_j)^2}
$$
or equivalently, $W[f,g]=-\sum_j b_jg_j^2$.
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H: Finding unknown constants
As $\lim_{x\rightarrow 0}\frac{\tan 2x - 2\sin ax}{x-x\cos 2x} = b$ where $a$ and $b$ are real constants. Then what is the value of $a$ and $b.$ Can someone help me with idea on how to approach this problem.
AI: Hint:
If $a\ne 1$, we have $\tan 2x-2\sin ax\sim_0 2(1-a)x$, whereas $1-\cos 2x\sim_0 2x^2$, so
$$\frac{\tan 2x-2\sin ax}{x(1-\cos 2x)}\sim_0\frac{1-a}{x^2}\to \pm\infty.$$
Therefore, if the quotient has a (finite) limit, necessarily $a=1$. Proceed using Taylor's formula at least up to order $3$ to determine the value of this limit.
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H: Questions on the proof of the strong law of large numbers
Here is the proof of the strong law of large numbers presented in the textbook I'm currently using.
Let $X_1,X_2,...$ be a sequence of independent identically distributed random variables and assume that $E[(X_i)^{4}] < \infty$. Prove the strong law of large numbers.
We note that the assumption $E[(X_i)^{4}] < \infty$ implies that the expected value of the $X_i$ is finite. Indeed, using the inequality $|x| \leq 1 + x^{4}$, we have $E[|X_i|] \leq 1 + E[(X_i)^{4}] < \infty$. Let us assume first that $E[X_i]=0$. We will show that $$E[\sum_{n=1}^{\infty} \frac{(X_1+X_2+ \cdot \cdot \cdot X_n)^{4}}{n^{4}}] < \infty$$
We have $$E[\frac{(X_1+X_2+ \cdot \cdot \cdot X_n)^{4}}{n^{4}}] = \frac{1}{n^{4}} \sum_{i_1=1}^{n} \sum_{i_2=1}^{n} \sum_{i_3=1}^{n} \sum_{i_4=1}^{n} E[X_{i_1}X_{i_2}X_{i_3}X_{i_4}]$$
Let us consider the various terms in this sum. If one of the indices is different from all of the other indices, the corresponding term is equal to zero. For example, if $i_1$ is different from $i_2,i_3,i_4$, the assumption of $E[X_i]=0$ yields, $$E[X_{i_1}X_{i_2}X_{i_3}X_{i_4}] = E[X_{i_1}]E[X_{i_2}X_{i_3}X_{i_4}]=0$$ Therefore, the nonzero terms in the above sum are either of the form E[(X_i)^{4}] (there are n such terms), or of the form $E[(X_i)^{2}(X_j)^{2}]$, with $i \neq j$. We conclude that there are $n$ terms of the first type and $3n(n - 1)$ terms of the second type. Thus $$E[(X_1+X_2+ \cdot \cdot \cdot X_n)^{4}]=nE(X_1)^{4}+3n(n-1)E[(X_1)^{2}(X_2)^{2}]$$ Using the inequality $xy \leq (x^{2}+y^{2})/2$, we obtain $E[(X_1)^{2}(X_2)^{2} \leq E[(X_1)^{4}]$ and $E[(X_1+X_2+ \cdot \cdot \cdot X_n)^{4}] \leq 3n^{2}E[(X_1)^{4}]$.
It follows that $$E[\sum_{n=1}^{\infty} \frac{(X_1+X_2+ \cdot \cdot \cdot X_n)^{4}}{n^{4}}] = \sum_{n=1}^{\infty} \frac{1}{n^{4}} E[(X_1+X_2+ \cdot \cdot \cdot X_n)^{4}] \leq \sum_{n=1}^{\infty}\frac{3}{n^{2}}E[(X_1)^{4}] < \infty$$
This implies that $(X_1+ \cdot \cdot \cdot +X_n)^{4}/n^{4}$ converges to zero with probability 1 and therefore, $(X_1+ \cdot \cdot \cdot+ X_n)/n$ also converges to zero with probability 1 , which is the strong law of large numbers.
Here are my questions about this proof :
1.Where does the assumption of $E[(X_i)^{4}] < \infty$ comes from ? It just feels so out of the blue.
For the part where they used the inequality $xy \leq (x^{2}+y^{2})/2$ to obtain $E[(X_1)^{2}(X_2)^{2} \leq E[(X_1)^{4}]$, is it that they first squared both sides of the inequality then because the r.vs are identical the right side of the inequality could be written as $\frac{(2x^{2})^{2}}{4}=x^{4}$ ?
Could someone explain how the monotone convergence theorem allows for such an equality $E[\sum_{n=1}^{\infty} \frac{(X_1+X_2+ \cdot \cdot \cdot X_n)^{4}}{n^{4}}] = \sum_{n=1}^{\infty} \frac{1}{n^{4}} E[(X_1+X_2+ \cdot \cdot \cdot X_n)^{4}]$. Reason being that the book said the theorem is beyond the scope of the book, so there isn't much about it.
How is this implication established ? What allows this to be concluded ? $(X_1+ \cdot \cdot \cdot +X_n)^{4}/n^{4}$ converges to zero with probability 1 and therefore, $(X_1+ \cdot \cdot \cdot+ X_n)/n$ also converges to zero with probability 1.
Thank you and I really appreciate any help given ! :)
AI: The whole argument here depends on $E(X^4)<\infty$. The conclusion is true
with only $E(|X|)<\infty$ but much harder to prove.
Applying $xy\le\frac12(x^2+y^2)$ to $x=X_1^2$ and $y=X_2^2$ and taking expectation gives
$$E(X_1^2X_2^2)\le\frac{E(X_1^4)+E(X_2^4)}2.$$
As the variables are identically distributed, $E(X_2^4)=E(X_1^4)$.
MCT implies that for variables $Y_1,Y_2,\ldots$ with each $Y_n\ge0$ then
$$E\left(\sum_{n=1}^\infty Y_n\right)=\sum_{n=1}^\infty E(Y_n).$$
It's like integrating a sum of non-negative functions.
From the sum of the expectations being finite, one gets via a Borel-Cantelli argument
that $n^{-4}(X_1+\cdots+X_n)^4\to0$ almost surely, and that is exactly the same as
$n^{-1}(X_1+\cdots+X_n)\to0$.
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H: Question regarding the proof that every non-negative measurable function is the limit of simple functions
$f$ be a non-negative measurable function. For each $x$ define $k_n(x)$ to be the unique integer satisfying $$k_n(x)\leq 2^nf(x)<k_n(x)+1$$ Define $$s_n(x)=\begin{cases}\frac{k_n(x)}{2^n},\text{ if }f(x)<n
\\n,\text{ if }f(x)\geq n\end{cases}$$
Then $(s_n)$ is the required sequence of simple functions. This is done in every proof I've seen. Now is there any reason to assign them the value $n$ when $f(x)\geq n$? I understand that $k_n=2^nn$ when $f(x)=n$ so it seems like the natural choice. But the proof still works if we define $s_n(x)$ to be $0$ if $f(x)\geq n$ right?
AI: For $s_n$ to be a simple function it needs to take on only finitely many values. A natural choice is to just use the dyadic numbers. We first fix $n$ and consider all numbers of the form $\{\frac{j}{2^n}\leq n: j\in\mathbb{N}\}$, this is a finite set. So we choose $s_n$ to only take values from this set and require that $s_n(x)\leq f(x)$. In addition to choosing $n$ so that $s_n$ only takes a finite number of values, there is an obvious connection between $s_n$ and $n$ that makes it easy to recognize.
The reason why you do not set $s_n(x) = 0$ if $f(x) > n$ is because $f$ can take values at $\infty$. Consider the measurable function $f =\infty$. Your idea would imply that $s_n(x) = 0$ for all $x$. But then $s_n(x)\not\to f(x)$ anywhere.
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H: Inequality involving integral and second derivative
UCLA basic exam spring 2017 problem 8
Show that there is a constant $C$ so that $$\left| \frac{f(0)+f(1)}{2}-\int_0^1 f(x)\, \mathrm dx \right| \leq C \int_0^1 \lvert f''(x)\rvert \, \mathrm dx$$ for every $C^2$ function $f:\mathbb{R} \rightarrow \mathbb{R}$
The domain of $f$ is just $[0, 1]$. For $f''=0$, problem is easy. So the problem is:
Prove that there is a constant $C$ so that $$ \left\{ \frac{ \left| \frac{f(0)+f(1)}{2}- \int_0^1 f(x) \, \mathrm dx \right|}{ \int_0^1 \lvert f"(x)\rvert \, \mathrm dx} \mid f:[0, 1] \rightarrow \mathbb{R}, \,f \text{ is } C^2, \, f'' \neq 0 \right\} $$ is bounded
Please give a hint. Thanks!
AI: Here's a hint for proving from scratch.
Hint:
Observe we have that
\begin{align}
\frac{f(1)+f(0)}{2}-\int^1_0 f(x)\ dx =&\ f(0)+\frac{1}{2}\int^1_0 f'(x)\ dx-\int^1_0\left[f(0)+\int^x_0 f'(\tau)\ d\tau\right] dx\\
=&\ \frac{1}{2}\int^1_0 f'(x)\ dx-\int^1_0 f'(\tau)(1-\tau)\ d\tau\\
=&\ \frac{1}{2}\left[f'(x)x\big|^1_0 -\int^1_0 xf''(x)\ dx \right]-\left[\frac{1}{2}f'(\tau)(1-\tau)^2\big|^1_0-\frac{1}{2}\int^1_0 f''(\tau)(1-\tau)^2 \right]\\
=&\ \frac{1}{2}[f'(1)-f'(0)]+\frac{1}{2}\int^1_0 f''(x)((1-x)^2-x)\ dx
\end{align}
In short, use integration by parts.
|
H: How to solve the ODE $y' = \frac{x+y-2}{y-x-4}$?
I am trying to solve the ODE $$y' = \frac{x+y-2}{y-x-4} \tag1 $$
This is a homogeneous special form ODE.
Let $x = u -1$ and $y=v+3$ in order to transform it to a homogeneous ODE. Hence,
$$ (1) \iff v'(u) = \frac{u+v(u)}{v-u} \tag 2$$
At last let $v(u) = z(u)u \iff v'(u) = z'(u)u+z(u)$ therefore,
$$ (2) \iff \frac{z'(u)}{z(u)+1} = \frac1u \tag 3$$
$(3)$ is a seperate variable ODE. Therefore we integrate both sides.
$$ \int \frac{z'(u)}{z(u)+1} \,du = \int \frac1u \,du $$
This integral seems to be easy to evaluate
$$ \int \frac{z'(u)}{z(u)+1} \, du $$
but I can't get my head around it. Any ideas?
AI: Your integral is just:
$$I=\int \dfrac {dz}{z+1}=\ln |z+1|+C$$
$$y' = \frac{x+y-2}{y-x-4}$$
Is easy to integrate:
$$(y-x-4)dy-(x+y-2)dx=0$$
$$(y-4)dy-(x-2)dx-(xdy+ydx)=0$$
$$(y-4)dy-(x-2)dx-dxy=0$$
Integration gives us :
$$\dfrac {y^2}2-4y-\dfrac {x^2}2+2x-xy=C$$
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H: find bound on condition number of matrix given matrix norm
Suppose $A$ is a $202 \times 202$ matrix with $\|A\|_2 = 100$ and $\|A\|_f = 101$. Give the sharpest lower bound on the 2-norm condition number $k(A)$.
I know $k(A) = \|A\|\cdot\|A^{-1}\| = 100 \|A^{-1}\|$
I also know that $\|A\| = \sup_x \frac{\|Ax\|}{\|x\|} = 100$
I'm also not sure, but I think that since the norm of $x$ is a scalar, that I may be able to bring it into the fraction so that $\|A\| = \sup_x \left\|A\frac{x}{\|x\|}\right\| = 100$, if this is true then I can restrict $x$ to be a unit vector. Not sure if this would help.
Can anyone explain how to do this problem?
This question is from Numerical Linear Algebra - Trefethen & Bau
AI: $100=\big\Vert A \big\Vert_2 = \sigma_1$
$101^2 = \big\Vert A \big\Vert_F^2 = \sigma_1^2 + \sigma_2^2 + ....+ \sigma_{201}^2 + \sigma_{202}^2$
$k(A) =\frac{\sigma_1}{\sigma_{n}} =\frac{\sigma_1}{\sigma_{202}}= \frac{100}{\sigma_{202}}$
To get a lower bound on $k(A)$ we want to maximize the denominator. So what's the largest value that $\sigma_{202}$ may take?
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H: Show that $\left \langle A^tx, y \right \rangle = \left \langle x, Ay \right \rangle$
Let $V = \mathbb{R}^n$ and $A\in\mathbb{M}_{n\times n}(\mathbb{R})$ show that $\left \langle A^tx, y \right \rangle = \left \langle x, Ay \right \rangle$.
I managed to demonstrate the exercise in the case of the standard inner product but I don't know how to do it for any inner product.
Is there an easy way to do it without using adjoint transformations?
AI: This fails in general if you don't take the standard inner product.
For example, take $\langle x,y\rangle := x_1y_1 + 2x_2y_2$ on $\Bbb R^2$ and take
$$
A := \begin{bmatrix} 0 & 2 \\ 1 & 0\end{bmatrix}
$$
then $\langle A^Te_1,e_2\rangle = \langle 2e_2,e_2\rangle = 4$ but $\langle e_1,Ae_2\rangle = \langle e_1,2e_1\rangle = 2$.
Edit: Just to clarify what you might have been looking for instead, given a Hilbert space $H$ (which for these purposes are nice inner product spaces such as $\Bbb R^n$ with the standard inner product) and a bounded linear operator $T:H\to H$ (which in the finite-dimensional setting, this is any linear transformation), its adjoint operator is the linear operator $T^*:H\to H$ characterised by the property that
$$
\langle Tx,y\rangle = \langle x,T^*y\rangle
$$
which is then just the definition of the adjoint operator (since in the case where $T$ is bounded, an adjoint always exists and is unique). In $\Bbb R^n$ with the standard inner product, what you have shown is that the adjoint operator is given by taking the transpose matrix, but as the above example shows, the adjoint operator necessarily depends also on the inner product in question.
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H: The set of all [group/ring/module/etc.] structures on a cardinal number
Given a cardinal number $\kappa$, how can one construct the set of the (algebraic) structures of a certain kind (group, ring, module, etc.) whose base set is a subset of $\kappa$?
For example, using the axioms of ZFC, how can one justify the existence of the set of all group structures whose base set is a subset of the cardinal $\kappa$? What would this set look like with the set-builder notation?
AI: I'll address the case of groups first, since I imagine you'll probably be able to extend it readily to the other cases.
At the end of the day, a group is a set $G$ equipped with a choice of an element $e\in G$ and a choice of a function $\cdot:G\times G\to G$ satisfying some properties.
Therefore, for any set $G$, the set of all group structures on $G$ is a subset of $G\times(G^{G\times G})$.
Now, for a cardinal $\kappa$ (or any set, really), we have a set of subsets $G\subseteq\kappa$, for which we get a set of group structures contained in $G\times(G^{G\times G})$, so the set of all groups whose base set sits inside $\kappa$ is a subset of
$$
\bigcup_{G\subseteq\kappa}G\times(G^{G\times G})
$$
which, being a set-indexed union of sets, is a set.
If you wanted to describe this set explicitly using set-builder notation, you can just define it set $\{(G,e,\cdot) \mid G\subseteq\kappa; e\in G; (\cdot):G\times G\to G; \text{group axioms}\}$ or even
$$
\left\{(G,e,\cdot)\in\bigcup_{G\in2^\kappa}\{G\}\times G\times(G^{G\times G}) \middle| \text{group axioms}\right\}
$$
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H: Cantor's theorem and countable additivity of Lebesgue measure
In this book, with more than 300 pages, the author claims that Cantor was wrong:
Transfinity, Wolfgang Mückenheim
https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf
In particular, the author lists in more than 100 pages authors that had concerns about Cantor's argument. He even cites Sharon Shelah who wrote
Of course, the most severe skeptics will even deny the mathematical
content of Cantor’s theorem ($2^{\aleph_0} > \aleph_0$). To these we
have nothing to say at all, beyond a reasonable request that they
refrain from using the countable additivity of Lebesgue measure.
CARDINAL ARITHMETIC FOR SKEPTICS, Shelah
https://arxiv.org/pdf/math/9201251.pdf
Is there any chance that indeed Cantor's theory is inconsistent? In the paper above, Shelah refers to concerns about "consistency" in Cantor's theory - how far may these concerns eventually reach?
AI: Well, the answer to the question "What is wrong in that book?" is "practically everything." Moreover, the mathematical errors (as opposed to situations where he misunderstands quoted arguments of others) are not at all original to Mückenheim, but rather are the general cranky arguments against the infinite in mathematics - with seemingly one exception, which may be instructive and which is the reason I'm writing this answer.
(To clarify: the prospects for inconsistency in various set theories is actually an interesting topic, but Mückenheim's book does not form a serious contribution to it. If you're interested, one relevant term is "consistency strength.")
First, let me briefly summarize what's not original. The bulk of Mückenheim's book is a reiteration of the standard arguments-from-incredulity, that set theory displays "bad" features and is therefore clearly inconsistent (although Mückenheim either misunderstands or deliberately misuses the technical term "inconsistent" - he conflates formal inconsistency and inconsistency with physical reality). For example, there is the "paradox" of the banker who at day $n\in\mathbb{N}$ gains $10$ dollars but spends $1$ dollar, and yet winds up "at the end of the day" completely broke based on which dollars they chose to spend. The "paradoxes" of this general flavor are completely resolved once we uncover the implicit assumptions that the relevant set-theoretic operations are well-defined and continuous in the appropriate senses, which they aren't; basically, the justification for the arguments against these situations boils down to trying to lift results about finite sets to infinite sets without justifying their continued validity.
The following error, however, does seem original to Mückenheim. (See here if you can view deleted posts.) Consider two different set-theoretic implementations of the natural numbers: as the von Neumann numerals $$0_V=\{\}, 1_V=\{\{\}\}, 2_V=\{\{\}, \{\{\}\}\}, 3_V=\{\{\}, \{\{\}\}, \{\{\}, \{\{\}\}\}\}, ..., (i+1)_V=i\cup\{i_V\}, ...$$ versus the Zermelo numerals $$0_Z=\{\}, 1_Z=\{\{\}\}, 2_Z=\{\{\{\}\}\}, 3_Z=\{\{\{\{\}\}\}\}, ..., (i+1)_Z=\{i_Z\}, ...$$ Now take an appropriate "set-theoretic limit of the natural numbers" in each sense: we have $$\limsup_{i\in\mathbb{N}}i_V=\{i_V: i\in\mathbb{N}\}\not=\emptyset$$ but $$\limsup_{i\in\mathbb{N}}i_Z=\emptyset.$$ Aha! says Mückenheim, we have here a contradiction. Well, no, we don't - what we have is two different implementations which behave differently with respect to a set-theoretic operation. But that set-theoretic operation is not meaningful at the level of the structure being implemented itself! This is basically the same error as looking at two programs which compute the same function and being confused about how one is longer than the other: "the length of the program" is not a property of a bare function.
So this mistake reveals the need to distinguish between the thing being implemented and the choice of implementation, and more importantly between operations/relations defined on the level of the thing being implemented vs. the implementation framework. There are indeed interesting things to say about this (the relevant logical term is "interpretation") ... but Mückenheim doesn't. However, since this does appear to be an original confusion and is vaguely related to something interesting it seems worth mentioning.
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H: Let $f$ be a differentiable function with $f ′(x) = 4$. If $f(0) = 2$, find $f(3)$.
So obviously this is some anti derivative stuff, and I the anti derivative would've been $2x^2$, with $f(x)=4x$ and then just plug in $x=3$ to get the answer. That's not the case tough, because with $f(x)=4x$, I see now that $f(0)$ does not equal $2$.
AI: As $ f(x) = \int f'(x)dx + C = 4x + C$, we have that $f(0) = C = 2$, hence $f(x) = 4x + 2$.
Hence, $f(3) = 12 + 2 = 14.$
|
H: Finding the Last Eigenvalue for a Matrix
$K$ is a $3 \times 3$ real symmetric matrix such that $K = K^3$. Furthermore, we are given that:
\begin{align*}
K(1, 1, 1) \ \ & = \ \ (0, 0, 0) \\
K(1, 2, -3) \ \ & = \ \ (1, 2, -3)
\end{align*}
So we know that $0, 1$ are two of the eigenvalues of $K$. What can I do to ascertain that the last eigenvalue is $0, 1$ or something else?
\begin{align*}
\det(K - \lambda I ) \ \ & = \ \ -\lambda(1 - \lambda)(c - \lambda) \\
& = \ \ -\lambda(\lambda - 1)(\lambda - c) \\
& = \ \ -\lambda^3 +(c + 1)\lambda^2 - c \lambda \\ \\
K^3 - K \ \ & = \ \ K(K - 1)(K + 1)
\end{align*}
If I know for sure that $K$ is not idempotent, then I know that $c = -1$:
\begin{align*}
\det (K - \lambda I) - K^3 + K \ \ = \ \ (c+1)K^2 - (c+1)K \ \ = \ \ 0
\end{align*}
But since it was not given explicitly, I don't think this will work.
AI: Each of $0,1$ and $-1$ is a possible value of the last eigenvalue. In fact, if you complete $u=(1,1,1)^T/\sqrt{3}$ and $v=(1,2,-3)^T/\sqrt{14}\}$ to an orthonormal basis $\{u,v,w\}$ of $\mathbb R^3$, then $K=vv^T+\lambda ww^T$ will satisfy $K^3=K$ for every $\lambda\in\{0,1,-1\}$ and $(0,u),(1,v)$ and $(\lambda,w)$ are three eigenpairs of $K$.
However, as the eigenvalues sum to the trace, one can at least ascertain that the remaining eigenvalue is $\operatorname{tr}(K)-1$.
|
H: projective space minus a closed point
Let $k$ be an algebraically closed field and let $\mathbb P^n_k=\text{Proj}(k[x_0,x_1,...,x_n])$ .
If $n\ge 2$, and $p\in \mathbb P^n_k$ is a closed point, then can $\mathbb P^n_k\setminus \{p\}$ be a Projective variety ?
Considering global section ring doesn't give any contradiction because since $\{p\}$ is a closed subset of codimension $\ge 2$, so $\mathcal O(\mathbb P^n_k\setminus \{p\})\cong \mathcal O(\mathbb P^n_k)\cong k$. Similarly, the Picard group or class group also doesn't give any contradiction.
Please help.
AI: No. As any map $f:X\to Y$ with $X$ a projective variety is closed, if $\Bbb P^n_k\setminus\{p\}$ were projective, then it's image under the natural embedding in $\Bbb P^n_k$ would be closed. But it's not (if it were, $\Bbb P^n_k$ would be reducible, which is obviously nonsense).
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H: Practical algorithm to calculate power subgroup of a polycyclic group
I am looking for a practical algorithm to calculate the power subgroup $G^n := \langle g^n \mid g \in G \rangle$ of a (possibly infinite) polycyclic group $G$. A theoretical algorithm is given in [1], but it does not appear to be practical, as it involves enumerating all finite index normal subgroups with index less than a certain integer.
I came up with the following algorithm (given in GAP-code), which reduces the problem to finite polycyclic groups.
PowerSubgroup := function( G, n )
local L, N, p, Q, Qn;
L := List( Pcp( G ), g -> g^n );
N := NormalClosure( G, SubgroupNC( G, L ) );
p := NaturalHomomorphismByNormalSubgroupNC( G, N );
Q := Image( p );
Qn := SubgroupNC( Q, List( Q, q -> q^n ) );
return PreImage( p, Qn );
end;
Unfortunately, for this finite polycyclic group Q, I couldn't think of anything other than just taking the set of all $n$-th powers. As one may expect, this creates pretty terrible bottleneck. I feel like there might be a more efficient approach, perhaps by exploiting the polycyclic presentation or the derived series somehow...
[1] Baumslag, Gilbert; Cannonito, Frank B.; Robinson, Derek J. S.; Segal, Dan, The algorithmic theory of polycyclic-by-finite groups, J. Algebra 142, No. 1, 118-149 (1991). ZBL0774.20019.
AI: As an improvement on what you have already, you could compute representatives of the conjugacy classes of $Q$, and then $Q^n$ is the normal closure of the subgroup generated by their $n$-th powers.
I think what I would try is to choose a collection of random elements of the group, let $N$ be the normal closure of the subgroup generated by their $n$-th powers, and then construct $Q/N$. You can check using the Exponent function whether $Q/N$ has exponent dividing $n$. If not, repeat with $Q$ replaced by $Q/N$.
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H: You can prove that compacts in the weak topology are limited without the Eberlein-Šmulian theorem
You can prove that compacts in the weak topology are bounded without the Eberlein-Šmulian theorem?
With Smuliam's Theorem this is immediate, because if you assume by contradiction that $K$ is unborded in the norm topology, exists $\{x_n\}_k \subset K$ such that $||x_k||>k$ for all $k$, due to the compactness in the weak topology, being relatively compact, there is a limit point, so the sequence cannot explode in norm. I would like to know if there is any way to do this without using this Theorem?
AI: Correct me if I'm missing something subtle, but I think this follows directly from the uniform boundedness principle. Let $X$ be a Banach space and let $K \subset X$ be weakly compact. Consider $K$ as a subset of $X^{**}$. Each $f \in X^*$ is continuous with respect to the weak topology on $X$, so the set $\{f(x) : x \in K\} = f(K)$ is a compact subset of $\mathbb{R}$ or $\mathbb{C}$ and hence bounded. Thus the set $K \subset X^{**}$ is pointwise bounded, and so by the uniform boundedness principle it is bounded in norm. Since the norm on $X^{**}$ agrees with the norm on $X$ (the canonical embedding is an isometry) we have shown that $K$ is bounded in norm.
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H: Prove that the diagonals of a rhombus are orthogonal.
I'm trying to solve some of the problems in Ahlfors' Complex Analysis book. On the section about analytic geometry, the following problem is stated:
Prove that the diagonals of a rhombus are orthogonal.
Since the idea was to use complex analysis tools to solve it, I came up with the following.
I take any arbitrary rhombus and draw it on the complex plane. After this, I re-orient it such that one of the corners is lying on the intersection of the real and imaginary axis and one of the diagonals is on the imaginary axis.
Relating one of the sides of the rhombus with the imaginary number $z$, I obtain the following scenario:
Recalling that the reflection about the imaginary axis is given by $z \to -\overline{z}$.
Using this construction, if indeed the diagonals are orthogonal this would mean that $\frac{z -\overline{z}}{z +\overline{z}}$ is purely imaginary (since multiplying by $i$ results in a $90^\circ$ rotation in the complex plane). To show this, I do
$$
\overline{\left(\frac{z -\overline{z}}{z +\overline{z}}\right)} = \frac{\overline{z} -\overline{\overline{z}}}{\overline{z} +\overline{\overline{z}}} = \frac{\overline{z} -z}{\overline{z} + z} = - \left(\frac{z -\overline{z}}{z +\overline{z}}\right)
$$
And since
\begin{align}
z = a+ib \text{ is purely imaginary } \iff a=0 \iff a =-a \iff a -ib = -a-ib \iff \overline{z} = -z
\end{align}
this concludes the solution.
Is my solution correct? Thank you very much!
AI: This is essentially right. There's one problem with
if indeed the diagonals are orthogonal this would mean that ... is
purely imaginary
Here what you want is the converse:
if ... is purely imaginary then the diagonals are orthogonal
You might try an alternative proof using $z$ and $z+e^{i\theta}z$ for the first two edges of the rhombus.
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H: Second order derivative of a chain rule (regarding reduction to canonical form)
I've been stuck on this for a couple of days. So this is from this book ("Partial Differential Equations in Mechanics 1", page 125).
Section 4.2 Reduction to canonical forms, which leads to the development of the Laplace equation.
In this section, I don't understand how they expand the second-order partial derivative:
Where,
Here is what I got so far. When I do it, I only get to have 4 terms, and not 5 like what's in the book. Here I apply product rule first and then the chain rule (Note, I'm using square brackets to indicate that I am taking the partial derivative of whatever is in them. Just to keep it organized).
$$\begin{align}
\frac{\partial}{\partial x}\frac{\partial u}{\partial x} &= \\
&= \frac{\partial}{\partial x} \biggl( \frac{\partial u}{\partial \xi} \frac{\partial \xi}{\partial x} + \frac{\partial u}{\partial \eta} \frac{\partial \eta}{\partial x} \biggr) \\
&=\frac{\partial}{\partial x} \biggl( \frac{\partial u}{\partial \xi} \frac{\partial \xi}{\partial x}\biggr) + \frac{\partial}{\partial x} \biggl(\frac{\partial u}{\partial \eta} \frac{\partial \eta}{\partial x} \biggr) \\
&= \frac{\partial}{\partial x} \biggl[ \frac{\partial u}{\partial \xi} \biggr] \frac{\partial \xi}{\partial x} + \frac{\partial u}{\partial \xi} \frac{\partial}{\partial x} \biggl[ \frac{\partial \xi}{\partial x} \biggr] +
\frac{\partial}{\partial x} \biggl[ \frac{\partial u}{\partial \eta} \biggr] \frac{\partial \eta}{\partial x} + \frac{\partial u}{\partial \eta} \frac{\partial}{\partial x} \biggl[ \frac{\partial \eta}{\partial x} \biggr] \\
\text{Now the chain rule:}\\
&= \frac{\partial}{\partial \xi}\biggl[\frac{\partial u}{\partial \xi}\biggr] \frac{\partial \xi}{\partial x} \frac{\partial \xi}{\partial x}
+ \frac{\partial u}{\partial \xi} \frac{\partial^2 \xi}{\partial x^2}
+ \frac{\partial}{\partial \eta}\biggl[\frac{\partial u}{\partial \eta}\biggr] \frac{\partial \eta}{\partial x} \frac{\partial \eta}{\partial x}
+ \frac{\partial u}{\partial \eta} \frac{\partial^2 \eta}{\partial x^2} \\
&=\frac{\partial^2 u}{\partial \xi^2} \biggl(\frac{\partial \xi}{\partial x} \biggr)^2
+ \frac{\partial u}{\partial \xi} \frac{\partial^2 \xi}{\partial x^2}
+ \frac{\partial^2 u}{\partial \eta^2} \biggl(\frac{\partial \eta}{\partial x} \biggr)^2
+ \frac{\partial u}{\partial \eta} \frac{\partial^2 \eta}{\partial x^2}
\end{align}
$$
My tree of the chain rule looks like this (is it correct?)
In addition, if someone could explain why this chain rule is valid? Granted, this may be a whole topic on its own, so if you could just point to some resource or what this particular operation is called, that would do.
$$
\frac{\partial}{\partial x}\biggl[ \frac{\partial u}{\partial \xi} \biggr] = \frac{\partial}{\partial \xi} \biggl[\frac{\partial u}{\partial \xi}\biggr]\frac{\partial \xi}{\partial x}
$$
Thank you in advance.
UPDATE:
(as per answer by @peek-a-boo)
P.S. Corrections or edits are welcomed.
AI: You have a mistake when calculating $\dfrac{\partial}{\partial x}\left[\dfrac{\partial u}{\partial \xi}\right]$ and $\dfrac{\partial}{\partial x}\left[\dfrac{\partial u}{\partial \eta}\right]$ in the middle where you're missing a extra step in the chain rule. For simplicity, just call $v:= \dfrac{\partial u}{\partial \xi}$. Then by equation (4.11), we have
\begin{align}
\dfrac{\partial v}{\partial x} &= \dfrac{\partial v}{\partial \xi} \dfrac{\partial \xi}{\partial x} + \dfrac{\partial v}{\partial \eta} \dfrac{\partial \eta}{\partial x}.
\end{align}
So, if we plug in the definition of $v$, we get
\begin{align}
\dfrac{\partial}{\partial x}\left[\dfrac{\partial u}{\partial \xi}\right] &= \dfrac{\partial^2 u}{\partial \xi^2} \dfrac{\partial \xi}{\partial x} + \dfrac{\partial^2 u}{\partial \eta \partial \xi} \dfrac{\partial \eta}{\partial x}.
\end{align}
Similarly,
\begin{align}
\dfrac{\partial}{\partial x}\left[\dfrac{\partial u}{\partial \eta}\right] &= \dfrac{\partial^2 u}{\partial \xi \partial \eta} \dfrac{\partial \xi}{\partial x} + \dfrac{\partial^2 u}{\partial \eta^2} \dfrac{\partial \eta}{\partial x}.
\end{align}
Finally, when you put all of this together, just remember that mixed partial derivatives are equal: $\dfrac{\partial^2 u}{\partial \xi \partial \eta} = \dfrac{\partial^2 u}{\partial \eta \partial \xi}$ (that's how the factor of $2$ comes up in equation $4.13$)
And yes, your chain rule tree looks right (that's how you can get 4.11 and 4.12). You can also create similar chain-rule trees for $\frac{\partial u}{\partial \xi}$ and $\frac{\partial u}{\partial \eta}$. As for why the chain rule is valid... well that's a completely different issue, which you should probably ask in a separate question if this answer isn't sufficient.
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H: how to calculate this problem please help me with steps
Person A can finish one task in 10 days. What if he work 10% faster for first day how much time it takes to finish the work now?
AI: Since he finishes $1$ task in $10$ days, he finishes $\frac{1}{10}$ of a task in $1$ day. $10%$ more than $\frac{1}{10}$ is just $\frac{11}{100}$ on the first day. So, we want to solve $\frac{11}{100}+\frac{d-1}{10}=1$ where $d$ is the amount of days now needed. We have $11+10d-10=100$, so $d=\frac{99}{10}$, so $\frac{99}{10}$ days or $14256$ minutes.
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H: Nilpotents of $\Bbb{Z}_n$
I am trying to find the nilpotents of the ring $\Bbb{Z}_n$.
Let $\bar{a}$ $\in$ $\Bbb{Z}_n$ be a nilpotent.
Then by definition, $\exists$ $m \in \Bbb{N}$ such that $\bar{a}^m$ = $\bar{0}$.
From here, we get that $a^m$ $\in$ $n\Bbb{Z}$.
So we have that $n|a^m$.
Let $p$ be any prime divisor of $n$.
Clearly $p|n$.
Now $p|n$ and $n|a^m$.
So we have that $p|a^m$.
So from here, $p|a$.
Let $n = p_1^{r_1}p_2^{r_2}..p_k^{r_k}$ i.e. the prime factorization of $n$.
By the above observation, $p_i|a$ for all $i = 1,2,...k$.
Let $r$ = $max$$({r_1,r_2..r_k})$ .
Then first of all, I want to know that how can we claim that $a^r =0$?
And from here, how could I find the nilpotents of the ring?
AI: $p_i|a\Rightarrow p_i^r|a^r\Rightarrow p_1^r\cdots p_k^r|a^r$, then we get $n|p_1^r\cdots p_k^r|a^r$ (because $r_i\le r$). Thus $\tilde{a}^r=0$
And the last bit of your argument shows that, if $n|a^r$ then $a$ is nilpotent. So you have the following: $\tilde{a}$ is nilpotent in $\mathbb{Z}/n\mathbb{Z}$ if and only if $n|a^m$ for some $m\in\mathbb{N}$
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H: Identities with respect to composition
In my abstract algebra textbook, when introducing category it says that morphisms should satisfy several properties and two of them are:
For every object $A$ of $C$, there exists (at least) one morphism $1_A \in \text{Hom}_C(A,A)$, the ‘identity’ on A.
and
The identity morphisms are identities with respect to composition: that is, for all $f \in\text{Hom}_C(A,B)$ we have $$f1_A = f,\quad 1_Bf = f.$$
Why is it important to point out the second property? Could there be an identity that doesn't satisfy the second property?
AI: A morphism whose domain coincides with the codomain is called an 'endomorphism'.
The first statement in itself only asserts that for every object $A$ there is an endomorphism $A\to A$. (How we call or denote this distinguished endomorphism has no strict meaning until the second statement.)
Together with the second statement, they assert that there exists such an endomorphism which acts like an identity from both sides. Hence the name and notation.
For example, in the category of sets, exactly the identity functions satisfy the second statement, and thus, any set with at least two elements has other endomorphisms, too.
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H: If $f: \mathbb{Z} × \mathbb{N} \to \mathbb{Z}, f(x,y)=x^2+y$. Prove if is an injection and prove if is an surjection.
If $f:\mathbb{Z} × \mathbb{N} \to \mathbb{Z}$.
$f(x,y)=x^2+y$.
$f:\mathbb{Z} × \mathbb{N}$ is a relation
Prove if is an injection and prove if is an surjection.
Help, I do not know how to proof it.
AI: The comments above have been rather clear. Let me give a more explicit answer.
Let us recall the definitions of injectivity and surjectivity. Let $A$ and $B$ be sets. Then, a function $f: A \to B$ is said to be injective iff:
$$\forall x,y \in A: f(x) = f(y) \implies x = y$$
and it is said to be surjective iff:
$$f(A) = B$$
where $f(A)$ is the image set of $A$ under the map $f$.
Now, we will note that for your specific case, $(x,y) \in \mathbb{Z} \times \mathbb{N}$. So, $x \in \mathbb{Z}$ and $y \in \mathbb{N}$. If we claim that the function $f: \mathbb{Z} \times \mathbb{N} \to \mathbb{Z}$ is surjective, then consider the integer $-1$. Since it is the case that:
$$\forall x \in \mathbb{Z}: x^2 \geq 0$$
$$\forall y \in \mathbb{N}: y \geq 0$$
So, their sum $x^2+y = f(x,y) \geq 0$ for any $(x,y)$. But that just means that there does not exist a pair $(x,y)$ such that $f(x,y) = -1$. Hence, this function cannot be surjective.
For injectivity, consider the integer $2$ and let $y = 1$. Then, we can see that:
$$f(-1,1)= (-1)^2+1 = 2 = 1^2+1 = f(1,1)$$
Since each image has multiple preimages under $f$, it follows that this is not an injective map.
If you do not understand the terminology that I have used above, then you are free to ask questions about it. On the other hand, if a large portion of it does look foreign, then I'd suggest that you have a look at your notes and review the relevant material before attempting such questions.
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H: Personal proof to the notion that the set of rational numbers is countable
I am not a student of core mathematics and hence as a result, all of my educational (engg.physics) background is based on the notion of applied mathematics rather than core mathematics, but I am attending a lecture in probability theory where the professor proved that the set of rational numbers is bijective to the set of natural numbers and hence the set of rational numbers $\mathbb{Q}$ is countable.
I am trying to prove it in a different way and would like the "core" mathematicians to check if I am making any logical fallacy in my arguments. Here it goes
Let me define $\mathbb{N}^2$ as
$$\mathbb{N}^2=\left\{(x,y)|x,y\in\mathbb{N}\right\}$$
Hence, I claim that $\mathbb{N}^2$ is countable, since $f:\mathbb{N}\to\mathbb{N}^2$ is clearly bijective.
Now, I define
$$\mathbb{Q}=\left\{\frac{p}{q}|(p,q)\in\mathbb{N}^2\right\}$$
Since, $f:\mathbb{N}^2 \to \mathbb{Q}$ is clearly surjective and $\mathbb{N}^2$ is countable, that implies that $\mathbb{Q}$ is countable
Is this proof of mine logically correct? If not, please let me know if I have assumed something somewhere where I should not have assumed stuff. Your help is really appreciated.
AI: No, it is not correct, because:
you did not define any of the $f$'s;
the set $\left\{\frac pq\,\middle|\,(p,q)\in\Bbb N^2\right\}$ is not equal to $\Bbb Q$; it is equal to $\{q\in\Bbb Q\mid q>0\}$.
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H: Find all functions $f:\Bbb R^+\to\Bbb R^+$ s.t. for all $x\in \Bbb R^+$ the following is valid: $f\bigg(\frac{1}{f(x)}\bigg)=\frac{1}{x}$
Find all functions $f:\Bbb R^+\to\Bbb R^+$ s.t. for all $x\in \Bbb R^+$ the following is valid:
$$f\bigg(\frac{1}{f(x)}\bigg)=\frac{1}{x}$$
I tried to substitute $\frac{1}{x}$ for $x$ and compare the equations:
$$f\bigg(\frac{1}{f(\frac{1}{x})}\bigg)=x $$
From this I found one solution $f(x)=x.$
AI: Using the hint of @Somos we can substitute
$$g(x)=-\log{(f(\exp{(x)}))}$$
so that the equation becomes
$$g(g(x))=x$$
So we just need $g(x)$ to be an involution (of which there are infinitely many). Choosing any involution $g(x)$ defined over $\mathbb{R}$ gives a solution
$$f(x)=\exp{(-g(\log{(x)}))}$$
Some solutions include
$$(g(x),f(x))\in\{(x,1/x),(-x,x)\}$$
I believe these are the only continuous solutions but there should be infinitely many discontinuous ones.
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H: Prove that if $F$ is a family of sets and $A\in F$, then $\bigcap F\subseteq A$
Question is from "How to Prove it" by Vellenman.
I am struggling to even understand how this theorem could ever be true?
I get that the first step is to assume x is an arbitrary element of $\bigcap F$, as the definition of $\bigcap F\subseteq A$ breaks down to $\forall x(x\in\bigcap F \rightarrow x\in A)$. Suppose $x\in \bigcap F$.
From here I am now confused about the definition of $\bigcap F$ and how $\bigcap F$ could even be a proper subset of $A$?
AI: A proof in Velleman's style would possibly look like this:
Let $x$ be an arbitrary element of $\bigcap\mathcal F$. From $x\in\bigcap\mathcal F$ and $A\in\mathcal F$ we obtain $x\in A$ $($by universal instantiation$)$. Since $x$ is arbitrary, $\forall x(x\in\bigcap\mathcal F\rightarrow x\in A)$ and so $\bigcap\mathcal F\subseteq A$. $Q.E.D.$
Universal instantiation is well explained in the book.
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H: Inverse Laplace transform of $\exp(-s)F(s)$
$$y(s)=1/s - 1/s^2 + \exp(-s)\cdot(1/s^2)$$
I'm struggling with $$\exp(-s)\cdot(1/s^2)$$
formulas: $$f(t-1)u(t-1)\to \exp(-s)F(s)$$
But $$f(t)=t\cdot u(t)$$
is this true:
inverse Laplace transform of $$\exp(-s)F(s)=(t-1)u(t-1)u(t-1)$$
AI: It's simply:
$$\mathcal{L^{-1}} \left (e^{-s}\dfrac 1 {s^2} \right)=u(t-1)(t-1)$$
Since you have that:
$$\mathcal{L^{-1}} \left (e^{-cs} F(s)\right)=u(t-c)f(t-c)$$
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H: Proof of Morera's Theorem for Triangular Contours
Sorry if this has been proven previously on MSE but I cannot find an obvious duplicate. I am attempting to prove the stronger version of Morera's theorem namely:
If $f:U\mapsto\mathbb{C}$ is a continuous function on an open set $U$ such
that $\int_\gamma f(z)\,\mathrm{d}z=0$ for all triangular contours $\gamma$ contained in $U$, then $f$ is holomorphic on $U$.
Proof (attempt):
Let $a\in U$. Since $U$ is open, $\exists\,r\gt0$ such that $B(a,r)=\{z\in\mathbb{C}:|z-a|\lt r\}\subseteq U$. Now consider $f$ restricted to the domain $B(a,r)$. Using the given assumptions, this restriction of $f$ is continuous and satisfies $\int_\gamma f(z)\,\mathrm{d}z=0$ for all triangular contours $\gamma$ contained in $B(a,r)$. Then we can define $F:B(a,r)\mapsto\mathbb{C}$ by
$$F(z)=\int_{[a,z]}f(w)\,\mathrm{d}w$$
where $[a,z]$ is the line segment from $a$ to $z$ in $\mathbb{C}$. This function is now well-defined as $B(a,r)$ is connected. Next we can calculate
\begin{align}
F'(z)
&=\lim_{h\to0}\frac{F(z+h)-F(z)}h\\
&=\lim_{h\to0}\frac{\int_{[a,z+h]}f(w)\,\mathrm{d}w-\int_{[a,z]}f(w)\,\mathrm{d}w}h\\
&=\lim_{h\to0}\frac{\overbrace{\int_{[a,z+h]}f(w)\,\mathrm{d}w+\int_{[z+h,z]}f(w)\,\mathrm{d}w+\int_{[z,a]}f(w)\,\mathrm{d}w}^{=\int_\gamma f(w)\,\mathrm{d}w=0}+\int_{[z,z+h]}f(w)\,\mathrm{d}w}h\\
&=\lim_{h\to0}\frac{\int_{[z,z+h]}f(w)\,\mathrm{d}w}h\\
&=\lim_{h\to0}\frac1h\int_0^1f(z+ht)\cdot h\,\mathrm{d}t\\
&=\lim_{h\to0}\int_0^1f(z+ht)\,\mathrm{d}t\\
&=\int_0^1\lim_{h\to0}f(z+ht)\,\mathrm{d}t\\
&=\int_0^1f(z)\,\mathrm{d}t\qquad(f\text{ continuous})\\
&=f(z)\\
\end{align}
Thus $F$ is holomorphic on $B(a,r)$ with derivative $f$. So, in particular, we can apply Cauchy's differentiation formula to give
$$f'(a)=F''(a)=\frac1{\pi i}\int_\gamma\frac{F(z)}{(z-a)^3}\mathrm{d}z$$
for a suitable contour $\gamma$. But $a\in U$ was chosen arbitrarily and hence $f$ is holomorphic on $U$.
AI: It looks good. Hope the following adds something.
Regarding the computation of the derivative $F'$, an alternative using triangle inequality: We have from your formulas for $h\neq 0$ $$\dfrac{F(z+h)-F(z)}{h}-f(z)=\frac{1}{h}\int_{[z,z+h]}\{f(\zeta)-f(z)\}d\zeta$$ Now by continuity of $f$, given $\epsilon>0$ there exists $\delta>0$ such that $|\zeta-z|<\delta\Rightarrow |f(\zeta)-f(z)|<\epsilon$ so for $0<|z+h-z|=|h|<\delta$ we get using triangle inequality for contours $$|\dfrac{F(z+h)-F(z)}{h}-f(z)|\le\frac{1}{h} \max{|f(\zeta)-f(z)|}_{\zeta\in [z,z+h]}.h<\epsilon$$
In the end instead of using Cauchy's formula, we can use the fact that derivative of holomorphic is holomorphic (from what I have seen the Cauchy formulas follow from this in composition with the usual formula): We see that $F\in Hol(B(a,r))$ as a result. And so $F'=f\in Hol(B(a,r))$. Now since $a$ was arbitrary we are done.
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H: Find $a,b,c$ if ${(1+3+5+.....a)}+{(1+3+5+....b)}={(1+3+5+.....c)}$
If $${(1+3+5+.....a)}+{(1+3+5+....b)}={(1+3+5+.....c)}$$
and $$(a+b+c)=21, a\gt6$$
We have to find $a,b,c$
My attempt
I use a little fact that the sum of first $n$ odd numbers is $n^2$
From that i get $$a^2+b^2=c^2$$
Which means that the solutions are pythagorean triples, and found no such soultion.
While if i try to do this by hand, $$a=7,b=5,c=9$$ satisfies the question but not the pythaoarean equation that i set up above
Why is this so?
AI: If $a=2n-1$ so $$1^2+3^2+...+a^2=n^2=\left(\frac{a+1}{2}\right)^2.$$
Id est, $$(a+1)^2+(b+1)^2=(c+1)^2$$ or
$$a^2+b^2+2a+2b+2=(22-a-b)^2$$ or
$$ab-23(a+b)=-241$$ or
$$(a-23)(b-23)=288.$$
Also we have $a\geq7$, $b+c\leq14$, $c>b$ and $c>a$.
Can you end it now?
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H: Number of non-zero solutions of an equation in $F \times F$ where $F$ is a field
Let $F$ be a field of order $32$.
Then I need to find the number of non-zero solutions $(a,b)$ $\in$ $F\times F$ of the equation $x^2+xy+y^2 = 0$.
I know that characteristic of a ring with unity $1$ is the order of $1$ in the group $(R,+).
Since $F$ has unity being a field, the order of unity i.e. $1$ divides $2^5$.
Since characteristic of a field is either $0$ or prime.
Since we have a finite field, hence characteristic can't be $0$ and hence it is $2$.
So the characteristic of $F \times F$ is $lcm(2,2)$ which is $2$.
Since order of $F \times F$ is $2^{32}$. So from here, can I say that $F \times F$ is a field?
If I can, then if $x = 0$ then $y$ will be $0$ and also the other way round.
So to get the required solutions, $x$ and $y$ both should be non-zero.
From here, how could I proceed?
AI: You are correct that the characteristic of $F$ is $2$.
I think you misunderstood the role of $F \times F$ here, though. $(a,b) \in F \times F$ is just saying that $(a,b)$ is a pair of elements, each in $F$. That is, $a \in F$ and $b \in F$, just that we want to count the number of such pairs, where order matters. $F \times F$ is being used here simply as a set of pairs, not really as a field or even as a ring.
We're looking for solutions of $x^2 + xy + y^2 = 0$ where $x$ and $y$ are elements of $F$.
The answer is, there are no solutions (other than $(0,0)$).
As you noticed, the field properties give that if $x=0$ then $y=0$, and vice versa. So suppose $x$ and $y$ are elements in $F$ with $x \neq 0$, $y \neq 0$, and $x^2+xy+y^2=0$.
In any field, polynomial multiplication gives
$$ (x-y)(x^2 + xy + y^2) = x^3 - y^3 $$
So we must also have
$$x^3 - y^3 = (x-y) \cdot 0 $$
$$x^3 = y^3$$
$$(xy^{-1})^3 = 1$$
The multiplicative group of $F$ has $31$ elements ($0$ is not a member). Since $31$ is prime, that group is cyclic, so the multiplicative order of every element other than identity $1$ is $31$. So the only way the cube of $xy^{-1}$ can be $1$ is if $xy^{-1} = 1$. Then we must have $x=y$, so
$$0 = x^2+xy+y^2 = 3x^2 = x^2$$
But this would require $x=0$, contradicting the assumption. There cannot be any nonzero solutions.
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H: Finding coefficients of a quadratic with roots having certain intervals
If the roots of the quadratic equation $$(4p−p^2 −5)x^2 −(2p−1)x+3p=0$$ lie on either side of unity, then the number of integral values of $p$ is?
Okay so I'm having a hard time what the question means by both sides of unity. Does it mean one root is on $ x>1 $ and other in $ x<1$? After that, how do I adjust the coefficent so that one root is greater tha one?
AI: If the leading coefficient is positive, then the condition $f(1)\lt 0$ suffices. If it is negative, then we want $f(1)\gt 0$. These conditions can be condensed into $$(4p-p^2-5) f(1)\lt 0 $$ Can you finish?
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H: Integral in $3D$
Let $R = [0,1]\times [0,1]$ and $f(x,y) = x.$ Note that $f$ is uniformly continuous on $R$. Use the definition of an integral to show that $\displaystyle\int_R f(v)dv = 0.5.$
Drawing this out, it is easy to see that this integral is the volume of a triangular prism with base area $0.5$ and height $1$, which is $0.5$. However, I'm more concerned about how to prove why this is indeed $0.5$.
By the definition of integration, $\displaystyle\int_R f(v)dv = L\displaystyle\int_R f(v)dv = \sup \{L(f, G): \text{ G grid on $R$}\} = U\displaystyle\int_R f(v)dv = \inf\{U(f, G) : \text{G grid on $R$}\}.$ For any grid $G = \{R_1,\cdots, R_p\},$ we have that $L(f, G) = \sum_{R_i\cap R\neq \emptyset} m_i|R_i| = \sum_{R_i} m_i|R_i|$ and $U(f, G) = \sum_{R_i} M_i|R_i|,$ where $M_i = \mathrm{sup}\{f(x) : x \in R_i\}$ and $m_i = \mathrm{inf}\{f(x) : x \in R_i\}.$ To show that $\sup\{L(f, G) : \text{ G grid on $R$}\} = 0.5,$ it suffices to show that for any $\epsilon > 0,$ we can find a grid $G$ so that $0.5 - \epsilon < L(f, G)$ and that for all grids $G,L(f, G)\leq 0.5$. Since $\sup\{L(f, G) : \text{ G grid on $R$}\} = \inf\{U(f, G) : \text{ G grid on $R$}\}$, it suffices to show the latter case, that is, for any $\epsilon > 0, \exists$ a grid $G = \{R_1,\cdots, R_k\}$ so that $0.5 - \epsilon < \displaystyle\sum_{R_i}m_i|R_i|$ and for all grids $G,L(f, G)\leq 0.5$. Let $G$ be a grid of $R$. Then $G$ is a collection of rectangles $R_k$ of the form $[x_{k-1}^1, x_k^1]\times [x_{k-1}^2\times x_k^2],$ where $P_j(G)=\{x_k^j : 1\leq k\leq l, x_{k-1}^j < x_k^j, x_0^j = x_j, x_l^j = b_j\}$ is a partition of $[a_j, b_j]$, and $R = [a_1,b_1]\times [a_2,b_2].$ We have that each $[x_{k-1}^j, x_k^j]\subseteq [0,1].$ Also, $m_k = \inf \{f(x) : x \in R_k\} = x_{k-1}^1.$ Hence $\displaystyle\sum_{R_k}m_k|R_k| = \displaystyle\sum_{R_k}x_{k-1}^1(x_k^1 - x_{k-1}^1)(x_k^2 - x_{k-1}^2).$ However, I'm not quite sure how to show that $\displaystyle\sum_{R_k}x_{k-1}^1(x_k^1 - x_{k-1}^1)(x_k^2 - x_{k-1}^2) \leq 0.5$ and that we may find a set of $R_k's$ so that $\displaystyle\sum_{R_k}x_{k-1}^1(x_k^1 - x_{k-1}^1)(x_k^2 - x_{k-1}^2)>0.5-\epsilon.$
I know that $\displaystyle\sum_{R_k}x_{k-1}^1(x_k^1 - x_{k-1}^1)(x_k^2 - x_{k-1}^2) \leq \displaystyle\sum_{R_k} x_{k-1}^1$ as $[x_{k-1}^j, x_k^j]\subseteq [0,1]$ for $1\leq j\leq 2$
AI: For a general grid (partition) $G$ of $[0,1]\times[0,1]$ we have subrectangles of the form $[x_{j-1},x_j] \times [y_{k-1},y_k]$ for $j = 1, \ldots, n$ and $k = 1, \ldots,m.$
The upper sum is given by
$$U(f,G) = \sum_{j=1}^n\sum_{k=1}^m x_{j}(x_j-x_{j-1})(y_k - y_{k-1}) = \sum_{j=1}^nx_{j}(x_j-x_{j-1})\sum_{k=1}^m (y_k - y_{k-1}) = \sum_{j=1}^nx_{j}(x_j-x_{j-1}), $$
and, similarly, the lower sum is
$$L(f,G) = \sum_{j=1}^nx_{j-1}(x_j-x_{j-1}), $$
Since $x_{j-1} \leqslant \frac{1}{2}(x_{j-1} + x_j)\leqslant x_j$, we have
$$L(f,G)\leqslant \underbrace{\sum_{j=1}^n \frac{1}{2}(x_{j-1} + x_j)(x_j - x_{j-1})}_{ = \frac{1}{2}\sum_{j=1}^n (x_j^2 - x_{j-1}^2) = \frac{1}{2}} \leqslant U(f,G)$$
We also have
$$U(f,G) - L(f,G) = \sum_{j=1}^n (x_j - x_{j-1})^2 \leqslant \max_{j=1,\ldots,n}(x_j - x_{j-1})$$
Thus,
$$\tag{1}L(f,G) \leqslant \frac{1}{2} \leqslant U(f,G) \leqslant L(f,G) + \max_{j=1,\ldots,n}(x_j - x_{j-1})$$
Choosing a sufficiently fine grid $G$ where $\max_{j=1,\ldots,n}(x_j - x_{j-1})< \epsilon$, we get
$$\tag{2}\frac{1}{2} - \epsilon \leqslant L(f,G) \leqslant \frac{1}{2}$$
(1) and (2) together show that for any $\epsilon > 0$ there is a grid $G$ such that
$$\frac{1}{2} - \epsilon \leqslant L(f,G) \leqslant \frac{1}{2} \leqslant U(f,G) \leqslant \frac{1}{2} + \epsilon$$
This proves Riemann integrability and since the integral lies between lower and upper sums we must have
$$\int_R f = \frac{1}{2}$$
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H: Prove that $m(\{x\in[0,1]:\lim \sup_{j\rightarrow\infty}f_j(x)\geq\frac{1}{2}\})\geq\frac{1}{2}$ under these conditions...
Question: Suppose for each $j\in\mathbb{N}, f_j:[0,1]\rightarrow\mathbb{R}$ is Lebesgue measurable such that $0\leq f_j\leq\frac{3}{2}$ and $\int_0^1 f_j dm=1$. Prove that $m(\{x\in[0,1]:\lim \sup_{j\rightarrow\infty}f_j(x)\geq\frac{1}{2}\})\geq\frac{1}{2}$.
Thoughts/Attempt: Let $A=\{x\in[0,1]:\lim \sup_{j\rightarrow\infty}f_j(x)\geq\frac{1}{2}\}$, and $B=\{x\in[0,1]:\lim \sup_{j\rightarrow\infty}f_j(x)<\frac{1}{2}\}$. Suppose, by contradiction, that $m(A)<\frac{1}{2}$. So, we can split up the integral as $$\int_0^1f_jdm=\int_Af_jdm+\int_Bf_jdm$$ where we get equality by the integral in the assumption. Now, $\int_Af_jdm<\frac{1}{2}$, by our (contradiction) assumption. And, $\int_Bf_jdm<\frac{1}{2}$, using our set $B$. Therefore, $$\int_0^1f_jdm<\frac{1}{2}+\frac{1}{2}=1$$ a contradiction, since this integral must equal $1$ from our assumption. Hence, we contradict that $m(A)<\frac{1}{2}$.
However, I am not quite sure if this works, because our sets are dealing with the $\lim\sup f_j(x)$ as $x\in[0,1]$, but wouldn't I have to compensate in the integral since the image of $f_j$ is all of $\mathbb{R}$?
AI: To handle the $\limsup$ for your integrals over $B$, I suggest using Fatou's lemma.
Fatou's lemma only applies to non-negative functions, and $\tfrac 3 2 - f_n(x)$ is a non-negative function. Applying Fatou to $\tfrac 3 2 - f_n(x)$ on $B$, we have
$$ \int_B \liminf_{n \to \infty}\left(\tfrac 3 2 - f_n (x)\right) dm \leq \liminf_{n \to \infty} \int_B \left(\tfrac 3 2 - f_n (x)\right) dm,$$
or equivalently,
$$ \limsup_{n \to \infty} \int_B f_n(x) dm \leq \int_B \limsup_{n \to \infty} f_n(x) \leq \tfrac 1 2 m(B).$$
Meanwhile, the correct inequality for the integrals over $A$ is
$$ \int_A f_n(x) dm \leq \tfrac 3 2 m(A)$$
for each $n \in \mathbb N$. (It seems like you've missed the factor of $\tfrac 3 2$, coming from the upper bound on $f_n$.)
Thus
$$ 1 = \limsup_{n \to \infty}\int_0^1 f_n(x) dx \leq \tfrac 3 2 m(A) + \tfrac 1 2 m(B) = m(A) + \tfrac 1 2,$$
which clearly implies that $m(A) \geq \tfrac 1 2$.
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H: Probability two uniform distribution(0,1) = 2/9
Two numbers are independently and uniformly chosen from the interval (0,1). What is the probability that the sum of the numbers is less than 1 and the product of the numbers is less than 2/9? (Note that both conditions hold simultaneously.)
Given than $n_1\sim\operatorname{uniform}(0,1)$ and $n_2\sim\operatorname{uniform}(0,1)$
$$Z = n_1\cdot n_2 \sim \operatorname{uniform}(0,1)$$
If $X\sim \operatorname U(0,1)$ then
$$P(X < x) = \frac{x-a}{b-a}$$
$$P\left(z < \frac{2}{9}\right) = \frac{\frac{2}{9}-0}{1-0} = \frac{2}{9}$$
Is this correct or am i making an error somewhere?
AI: In other words, what is the area of the region in the square $0\le x\le 1,\,0\le y\le 1$ satisfying $x+y<1,\,xy<\tfrac29$? It's the double integral$$\begin{align}\int_0^1dx\int_0^{\min\left\{1-x,\,\tfrac{2}{9x}\right\}}dy&=\int_0^1dx\min\left\{1-x,\,\tfrac{2}{9x}\right\}\\&=\int_0^{1/3}(1-x)dx+\int_{1/3}^{2/3}\tfrac{2}{9x}dx+\int_{2/3}^1(1-x)dx\\&=\tfrac13+\tfrac29\ln2\\&\approx0.4874.\end{align}$$You'll want to double-check my arithmetic, but Monte Carlo agrees.
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H: Upper bound for amount of intervals in intersection of interval sets
I have two sets of numbers which are unions of disjoint intervals, and I have to find an upper bound for how many of such intervals can there be in the intersection of the two sets.
Here's a diagram of how that would look like. After trying for a couple of examples it seems to me that this upper bound is |A| + |B| - 1, where |A| would be the amount of intervals in set A. However, I'm not sure this is the best upper bound, and I'm also not sure of how one would go about proving it
AI: Let the two families of intervals be $\mathscr{A}$ and $\mathscr{B}$. Let
$$\mathscr{A}_0=\left\{A\in\mathscr{A}:A\cap\bigcup\mathscr{B}=\varnothing\right\}$$
and
$$\mathscr{B}_0=\left\{B\in\mathscr{B}:B\cap\bigcup\mathscr{A}=\varnothing\right\}\;.$$
For each $A\in\mathscr{A}\setminus\mathscr{A}_0$ let
$$\mathscr{B}_A=\{B\in\mathscr{B}:A\cap B\ne\varnothing\}\;,$$
and let
$$\mathscr{I}_A=\{A\cap B:B\in\mathscr{B}_A\}\;.$$
It’s not hard to see that $|\mathscr{I}_A|=|\mathscr{B}_A|$. It’s also clear that if $A_0,A_1\in\mathscr{A}$, $B_0\in\mathscr{B}_{A_0}$, and $B_1\in\mathscr{B}_{A_1}$, then $A_0\cap B_0\ne A_1\cap B_1$, so $\mathscr{I}_{A_0}\cap\mathscr{I}_{A_1}=\varnothing$, and it follows that the total number of intervals is
$$|\mathscr{A}_0|+|\mathscr{B}_0|+\sum_{A\in\mathscr{A}\setminus\mathscr{A}_0}|\mathscr{I_A}|=|\mathscr{A}_0|+|\mathscr{B}_0|+\sum_{A\in\mathscr{A}\setminus\mathscr{A}_0}|\mathscr{B}_A|\;.\tag{1}$$
Next, note that no member of $\mathscr{B}$ belongs to more than two of the sets $\mathscr{B}_A$ for $A\in\mathscr{A}$, and this happens only when $B\in\mathscr{B}_{A_0}\cap\mathscr{B}_{A_1}$ for adjacent intervals $A_0,A_1\in\mathscr{A}$. Thus, at most $|\mathscr{A}|-1$ members of $\mathscr{B}$ can be counted twice in the last term of $(1)$, all are counted at least once, and none can be counted more than twice. It follows that
$$\begin{align*}
|\mathscr{A}_0|+|\mathscr{B}_0|+\sum_{A\in\mathscr{A}\setminus\mathscr{A}_0}|\mathscr{B}_A|&\le|\mathscr{A}_0|+|\mathscr{B}|+|\mathscr{A}\setminus\mathscr{A}_0|-1\\
&=|\mathscr{A}|+|\mathscr{B}|-1\;.
\end{align*}$$
It’s easy to show that this upper bound can be attained.
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H: Normality is transitive along a chain of index 2 subgroups
Let $G$ be a finite group. Let $G_1$ be a subgroup of $G$ such that $[G:G_1]=2$. Now suppose that $H$ is a normal subgroup of $G_1$ with odd order. Prove that $H$ is a normal subgroup of $G$.
Some remarks/partial progress:
I have shown that if $G$ is a group, $G_1$ is a subgroup of index $2$, and $H$ is a subgroup of odd order, then $H\subset G_1$. I don't think this is relevant to the problem at hand though.
If $H$ is a group of odd order, squaring is injective.
Since $G_1$ in the original problem statement has index $2$, any element $g\in G$ satisfies $g^2 \in G_1$.
If $g\in G$ is in $G_1$, then clearly $gH=Hg$, so it suffices to show our claim for $g\in G\setminus G_1$. Note further than $G\setminus G_1$ is the set of all elements of even order.
AI: The answer is "no". Take the wreath product $G=C_3\wr C_2$ of order $18$. It has a subgroup $G_1=C_3\times C_3$ of index $2$. The subgroup $G_1$ has several subgroups $H$ of order 3, each of which is normal in the Abelian group $G_1$ but only one of which is normal in $G$.
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H: Dense Subspace of Extremally Disconnected Space is Extremally Disconnected
Problem 15G of Willard is -
Every dense subspace and every open subspace of an extremally disconnected space is extremally disconnected.
I've been able to prove the 'open subspace' part of the problem, but the result for 'dense subspace' has been evading me. Any help would be appreciated!
AI: $\newcommand{\cl}{\operatorname{cl}}$
Suppose that $D$ is dense in an extremally disconnected space $X$, and let $U$ be (relatively) open in $D$; then $U=D\cap V$ for some open $V$ in $X$. I claim that $\cl_DU=D\cap\cl_XV$. If this is true, then $\cl_DU$ is clearly open in $D$, and the desired result follows.
Proof of claim: Clearly $\cl_XV$ is a closed set in $X$ containing $U$, so $D\cap\cl_XV$ is a closed set in $D$ containing $U$, and therefore $\cl_DU\subseteq D\cap\cl_XV$.
Now suppose that $x\in D\cap\cl_XV$, and let $W$ be any open nbhd of $x$ in $D$; $W=D\cap G$ for some open $G$ in $X$. Clearly $G\cap V\ne\varnothing$, so $$W\cap U=(D\cap G)\cap(D\cap V)=D\cap(G\cap V)\ne\varnothing\;,$$ since $D$ is dense in $X$. Thus, $x\in\cl_DU$, and since $x$ was arbitrary, $D\cap\cl_XV\subseteq\cl_DU$ and hence $\cl_DU=D\cap\cl_XV$. $\dashv$
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H: Given $N, b$ find the largest value of $k$ such that $N=a \times b^k$. $a,b,k,N \in \mathbb{Z}$
I'm trying to find a function that produces $k$ in this problem using $N$ and $b$ as inputs.
Given $N, b$ find the largest value of $k$ such that $N=a \times b^k$. $a,b,k,N \in \mathbb{Z}$
My instinct is that this is not possible.
E.g. If $N=300$ and $b=10$ then $k=2$ since $300=3\times10^2$
Is there a name for this concept? I've struggled to google anything useful / relevant.
AI: $k$ is the $b$-adic order of $N$.
It would be difficult to find a function that delivers $k$ from a given $N, b$, however $k$ is easily calculated by repeadtedly dividing $N$ by $b$ until it would give a fraction. In your example, we see that $\frac{300}{10^2}=3$, but $\frac{300}{10^3}=0.3\not\in\Bbb Z$, and so $k=2$
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H: Surface integrals: why do we use two integrals to find the area of a surface?
Why is it a double sum of rectangles and not a single sum?
Recently I found another way to calculate the area of a surface on my course of differential geometry and really confused me because its one.
$r_u=$ the partial with respect of $u$
$r_v=$ the partial with respect of $v$
and we take the modulus of the cross product
$$\iint |r_u(u,v)\:\times\:r_v(u,v)| \, du \, dv$$
AI: You can always think of the double sum as a single sum where you add up all of the rectangles. But since the rectangles come in rows and columns, it's convenient to think of this as a double sum. This then naturally leads to the practical method of calculating a double integral as an iterated integral. (But there's still a theorem to prove to show that this actually works under reasonable conditions, and the pathological counterexamples where it doesn't.)
There are other definitions of the area of a surface involving a sum that is not naturally thought of as a double sum. Instead of dividing the surface into rectangles, divide it into triangles. Of course, any way of dividing into rectangles gives a triangulation (a way of dividing into triangles), by simply dividing each rectangle in half, but most triangulations won't come this way. Then you add up the areas of all of the triangles (a single sum) and take the limit as the largest length of any side of any triangle goes to zero. If you tag your triangulations (by picking a point in each triangle, or without loss of generality by picking a vertex of each triangle), then you can also define more general surface integrals in this way. (And when relevant, you can orient a triangle by picking a cyclic order of its vertices, define when adjacent triangles have compatible orientations, and thereby do integrals that require a surface to be oriented.)
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H: Alteration in Binary Strings Question
Question:
An alteration in a binary string is said to occur when the string encounters one of the following two patterns -
“01” or “10”.
For example, the string 1101001 has exactly 4 alterations in it - which occur at positions 3,4,5
and 7. Count the total number of n bit strings that have at least 2 alterations.
I'm completely stumped on this question. After spending some time looking at it, I can determine that a combination will need to be used to try and count the total number of possibilities. However, I'm not sure how you would be able to set this up using some sort of counting method.
Any sort of solution/guidance is appreciated.
AI: Let's calculate the answer by finding number of n-bit strings (no restriction) - ... strings with 0 alteration - ... with 1 alteration.
When n=2 there are no strings with 2 alterations or more, so the answer is 0. Assume $n\geq 2$
Number of n-bit strings is $\fbox{$2^n$}$, as each bit can either be 0 or 1.
Strings with 0 alteration means there are no "switch" in the string, which implies the string is either 0....0 or 1....1. There are $\fbox{$2$}$ cases here.
Finally, the number of strings with 1 alteration must be in the form 0...01....1 or 1...10....0 with at least one 0 and one 1. Consider the form 0...01...1. There can be 1 0 as the prefix, 2 0, ..., (n-1) 0, giving us (n-1) cases. The same applies for 1...10...0, giving us $\fbox{$2(n-1)$}$ cases in total.
Therefore, the number of strings with 2 or more alterations is $\fbox{$2^n-2n$}$.
Bonus: Number of strings with exactly k alterations is 2*(n-1 choose k).
Proof: Let's consider the number strings with exactly k alterations
that start with a `0` and double the answer.
The n-bit string looks like this: 0 _ _ _ _ _ _ _ _ _ _
Now let's insert a "pipe" between each bit: 0|_|_|_|_|_|_|_|_|_|_
For example one string might look like this: 0000000|1111|0000000 (where k=2)
Each pipe represents an alteration, and we must choose k of them out of the (n-1)
Thus there are 2*(n-1 choose k) where choose is the binomial coefficient
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H: Prove that the axiom of choice is necessary in order to prove something else.
My mathematical background is perhaps a little lacking on this topic, but I've been searching and haven't come up with a satisfactory answer to this question. I have no idea how to approach the problem or if it has been answered.
I have seen numerous cases where it is asserted that "the axiom of choice is necessary" to complete a particular proof but I've yet to see a case where the necessity of the axiom of choice is actually itself proven.
For example consider the following question: Additive function $:\mathbb{R} \rightarrow \mathbb{R}$ that is not linear.
The answers and comments note that you can't construct such a function without the axiom of choice. Then it seems like this would imply, for example, that any function with a closed form
which is additive is necessarily linear. Is this correct? How would one go about proving the necessity of the axiom of choice?
I apologize in advance that I don't have my own attempt to share since I'm not even sure how to approach such a proof.
AI: Let's take the axioms of a group. We have a binary operator, $*$, and the axioms state that there is a neutral element, $e$, and that $*$ is associative, and for every $x$ there is $y$ such that $x*y=e$.
Question. Is it true that for every $x,y$ it holds that $x*y=y*x$?
Well, there are infinitely many proofs from these finitely many axioms. So how can we tell? Checking them one by one is futile.
The answer is that if this was provable, then every model of the axioms would also satisfy the above property. In other words, every group would be commutative. So if we can find a group which is not commutative, then we effectively proved that the axioms of a group are not sufficient for proving that for every $x$ and $y$, $x*y=y*x$.
And indeed, it is not hard to find non-commutative groups.
So, going back to $\sf ZF$ and $\Bbb Q$-linear operators on $\Bbb R$. How would you prove that $\sf ZF$ is not sufficient for proving the existence of such discontinuous operators? Well, you'd show that there are models of $\sf ZF$ in which there are no such operators.
Now we know that any $\Bbb Q$-linear operator which is also Baire measurable is continuous (one can also use Lebesgue measurability, for example). So if we can find a model of $\sf ZF$ in which every such linear function is Baire measurable, then every such function is also continuous.
And indeed, this was shown possible by Solovay, and later improved upon by Shelah. In other words, they exhibited models of $\sf ZF$ in which every function $f\colon\Bbb{R\to R}$ is Baire measurable, and in particular any $\Bbb Q$-linear operator. So every such operator is continuous.
These constructions are extremely technical utilising not only the technique of forcing, but also extended techniques of symmetric extensions, and often relying on theorems in analysis as well. But with time, one can learn them.
TL;DR: To prove that some axioms don't prove a statement, it is usually easier to prove that there is a model of the axioms where the statement is false. This is true for $\sf ZF$ as well.
To show that the axiom of choice is necessary for proving something we need to show that:
$\sf ZFC$ implies this something, and
there is a model of $\sf ZF$ where this something is false.
The difficult part—conceptually—is getting your head around models of $\sf ZF$, because it's the foundation of mathematics. But once you've gone through that step, the rest is just a technicality.
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H: Proof of equivalence of two statements about relationship between two generating functions
I am trying to prove the equivalence of the following two statement:
$(a_i)_{0}^{\infty}$ and $(b_i)_{0}^{\infty}$ are infinite sequences of numbers, such that their elements are related in the following manner: $b_n=\sum_{k=0}^{n}{{n}\choose{k}}a_k$.
$(a_i)_{0}^{\infty}$ and $(b_i)_{0}^{\infty}$ are infinite sequences of numbers, such that their generating functions are related in the following manner: $B(x)=e^{x}A(x)$
The way I tried to prove it. The j-th derivative of function of given power series is:
$G(x)=\sum_{k=0}^{\infty}g_{i}x^i \iff \frac{d^j}{dx^j}G(x)=\sum_{k=0}^{\infty}g_{i}\frac{d^j}{dx^j}x^i=\sum_{k=0}^{\infty}g_{i}\frac{i!}{(i-j)!}x^{i-j}=\sum_{k=j}^{\infty}g_{i}\frac{i!}{(i-j)!}x^{i-j}$
Therefore:
$H(x)=\frac{d^j}{dx^j}G(x) \iff h_k=g_{k+j}\frac{(k+j)!}{k!}$
Now consider j-th derivative of $e^{x}A(x)$, by the product rule:
$\frac{d^j}{dx^j}(e^xA(x))=\sum_{l=0}^{j}{{j}\choose{l}}A^{(l)}(x)(e^x)^{(j-l)}=\sum_{l=0}^{j}{{j}\choose{l}}A^{(l)}(x)e^x=e^x\sum_{l=0}^{j}{{j}\choose{l}}A^{(l)}(x)$
Therefore:
$\frac{d^j}{dx^j}B(x)=e^x\sum_{l=0}^{j}{{j}\choose{l}}A^{(l)}(x)$
By pluggin in $x=0$:
$\frac{d^j}{dx^j}B(0)=e^0\sum_{l=0}^{j}{{j}\choose{l}}A^{(l)}(0)$
$\frac{d^j}{dx^j}B(0)=\sum_{l=0}^{j}{{j}\choose{l}}A^{(l)}(0)$
When zero is plugged in, only the coefficient of the zeroth power remains, thus:
$j!b_j=\sum_{l=0}^{j}{{j}\choose{l}}l!a_l$
$b_j={\frac{1}{j!}}\sum_{l=0}^{j}{\frac{j!}{(j-l)!l!}}l!a_l$
$b_j=\sum_{l=0}^{j}{\frac{a_l}{(j-l)!}}$
$b_n=\sum_{l=0}^{n}{\frac{a_k}{(n-k)!}}$
Fromula is obviously wrong, what was my mistake?
AI: It appears to me that the problem is talking about exponential generating functions. In that case
$$A(x)=\sum_{n\ge 0}a_n\frac{x^n}{n!}\;,$$
and
$$e^xA(x)=\left(\sum_{n\ge 0}\frac{x^n}{n!}\right)\left(\sum_{n\ge 0}a_n\frac{x^n}{n!}\right)\;.$$
The coefficient of $x^n$ in this product is
$$\sum_{k=0}^n\left(\frac{a_k}{k!}\cdot\frac1{(n-k)!}\right)=\sum_{k=0}^n\frac{a_k}{k!(n-k)!}\;,$$
so
$$\begin{align*}
e^xA(x)&=\sum_{n\ge 0}\sum_{k=0}^n\frac{a_k}{k!(n-k)!}x^n\\
&=\sum_{n\ge 0}\sum_{k=0}^n\frac{n!a_k}{k!(n-k)!}x^n\\
&=\sum_{n\ge 0}\sum_{k=0}^n\binom{n}ka_kx^n\\
&=\sum_{n\ge 0}b_nx^n\;.
\end{align*}$$
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H: 3 new Points lying on Jerabek Hyperbola?
R is the circumcenter, H is the orthocenter of the triangle ABC.
Then points F,G,E are the points of intersection of the altitudes with the sides of the triangle ABC. U, V, W are the intersection points of cevians AR, BR, CR with the sides of the triangle ABC.
X is defined as the intersection of lines GU, EF. Y is defined as the intersection of GU, WV and Z is the intersection point of lines WV, EF. Then it can be shown and proven that lines AX, BY, CZ always intersect at the point RH.
UPDATE: Actually, 4 'RH' points exist. **3 of them always belong to Jerabek hyperbola that goes through A,B,C,R,H and the fourth RH point lies on that hyperbola only in a special case. All the points of this configuration are shown here.
I wonder what is known about these RH points, K points, S point and related conics?
In fact, 3 RH points can be any of those or completely new ones:
54 (Kosnita point), 64 isogonal conjugate of the de Longchamps point), 65 (orthocenter of the contact triangle), 66 (isogonal conjugate of the Exeter point), 67 (isogonal conjugate of the far-out point), 68 (Prasolov point), 69, 70, 71, 72, 73, 74, 248, 265, 290, 695, 879, 895, 1173, 1175, 1176, 1177, 1242, 1243, 1244, 1245, 1246, 1439, 1798, 1903, 1942, 1987, 2213, 2435, 2574, 2575, 2992, and 2993.
AI: Referring back to my previous answer to a related question, we'll consider things in a bit more generality, and handle an ambiguity in the formulation of this property.
Given $\triangle ABC$ we define points $P_+$ and $P_-$ (which correspond to the orthocenter and circumcenter in OP's current question) via barycentric coordinates
$$P_\pm = \frac{\alpha_\pm A+\beta_\pm B+\gamma_\pm C}{\alpha_\pm+\beta_\pm+\gamma_\pm} \tag{1}$$
Cevians through $P_+$ and $P_-$ meet the sides of the triangle in six points, $D_\pm$, $E_\pm$, $F_\pm$ (OP's $F$, $G$, $E$, $U$, $V$, $W$), and pairs of these points determine lines that meet pairwise at three more points, $D$, $E$, $F$ (OP's $X$, $Y$, $Z$). Finally, although there's ambiguity in choosing pairs of points that ultimately determine $D$, $E$, $F$, the cevians $\overleftrightarrow{AD}$, $\overleftrightarrow{BE}$, $\overleftrightarrow{CF}$ always concur. As per my other answer, the points of concurrency (which I denote with sub-scripted (or not) $K$) have barycentric coordinates
$$\begin{align}
K_A &= \left(\frac12:\frac{1}{\dfrac{\alpha_+}{\beta_+}+\dfrac{\alpha_-}{\beta_-}}:\frac{1}{\dfrac{\alpha_+}{\gamma_+}+\dfrac{\alpha_-}{\gamma_-}}\right) \tag2\\[4pt]
K_B &= \left(\frac{1}{\dfrac{\beta_+}{\alpha_+}+\dfrac{\beta_-}{\alpha_-}}:\frac12:\frac{1}{\dfrac{\beta_+}{\gamma_+}+\dfrac{\beta_-}{\gamma_-}}\right) \tag3 \\[4pt]
K_C &= \left(\frac{1}{\dfrac{\gamma_+}{\alpha_+}+\dfrac{\gamma_-}{\alpha_-}}:\frac{1}{\dfrac{\gamma_+}{\beta_+}+\dfrac{\gamma_-}{\beta_-}}:\frac12\right) \tag4 \\[4pt]
K\phantom{_X} &= \left(\frac1{\beta_+\gamma_- + \beta_-\gamma_+}:
\frac1{\gamma_+\alpha_-+\gamma_-\alpha_+}:\frac1{\alpha_+\beta_-+\alpha_- \beta_+}\right) \tag5
\end{align}$$
It's not clear which of these points OP intends by "$RH$", but $75\%$ of the time, it doesn't matter. :) To see why, note that the five-point conic through $A$, $B$, $C$, $P_+$, $P_-$ has barycentric equation
$$\frac{\alpha_+}{x} \left(\frac{\beta_+}{\beta_-} - \frac{\gamma_+}{\gamma_-}\right)
+ \frac{\beta_+}{y}\left( \frac{\gamma_+}{\gamma_-} - \frac{\alpha_+}{\alpha_-}\right)
+\frac{\gamma_+}{z}\left( \frac{\alpha_+}{\alpha_-} -\frac{\beta_+}{\beta_-}\right)= 0 \tag{6}$$
where $x:y:z$ are the barycentric coordinates of any point on the conic. The reader can verify that $K_A$, $K_B$, $K_C$ (one of which is what I suspect OP intends to be point "$RH$") automatically satisfy $(6)$, and that $K$ satisfies it only if
$$\left( \frac{\alpha_+}{\alpha_-}-\frac{\beta_+}{\beta_-} \right)\left( \frac{\beta_+}{\beta_-}-\frac{\gamma_+}{\gamma_-}\right)\left(\frac{\gamma_+}{\gamma_-}-\frac{\alpha_+}{\alpha_-}\right)=0 \tag{7}$$
(In a comment to the original question, I included squares in a formula corresponding to the above. My second pass at the problem seems to make the squares unnecessary. I'm going with that.)
Observe that if, say, the first factor of $(7)$ vanishes, equivalently, if $\alpha_+/\beta_+=\alpha_-/\beta_-$ then (via my previous answer) cevian points $F_+$ and $F_-$ coincide. Consequently we can interpret $(7)$ in general as the condition that $\overleftrightarrow{P_+P_-}$ passes through a vertex of $\triangle ABC$.
In OP's specific case using the circumcenter and orthocenter,
$$\begin{align}
R &= (\alpha_+:\beta_+:\gamma_+) = (\sin 2 A: \sin 2 B : \sin 2C ) \\
H &= (\alpha_-:\beta_-:\gamma_-) = (\tan A:\tan B:\tan C)
\end{align} \tag8$$
condition $(7)$ transforms to
$$(\cos^2A-\cos^2B)(\cos^2B-\cos^2C)(\cos^2C-\cos^2A)=0 \tag{9}$$
so that (non-degenerate) $\triangle ABC$ must be isosceles for the point $K$ to lie on the conic.
Regarding the "triangle center-ness" of these $K$-points, we can note that, even in general, $K_A$, $K_B$, $K_C$ are typically not triangle centers in the Kimberling sense, as they aren't symmetrically-defined by the parent triangle. As formulas $(2)$, $(3)$, $(4)$ show, the point you get depends upon how the vertices are labeled. (That's why the subscripts make sense!)
On the other hand, point $K$ (via $(5)$) is symmetrically-defined, and it therefore generally corresponds to a triangle center. In fact, it turns out that this point is known as the Ceva Point for points $P_+$ and $P_-$. In particular, the Ceva Point for the orthocenter and circumcenter is designated $X(1105)$ in Kimberling's Encyclopedia of Triangle Centers.
The point where $\overleftrightarrow{AK_A}$, $\overleftrightarrow{BK_B}$, $\overleftrightarrow{CK_C}$ concur —which I call $K_\star$ in my previous answer— is also symmetrically-defined, so it counts as a "triangle center" of some kind. In fact, it is known as the crosspoint of $P_+$ and $P_-$. Its orthocenter-circumcenter incarnation is Kimberling's $X(185)$. (Interestingly, the ETC notes that this point is the orthocenter of the tangential triangle of the Jerabek hyperbola.)
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H: Integrate ${\sin(x)\cos(x)}$ by parts, by letting ${u=\cos(x),dv=\sin(x)dx}$
I was able to integrate by parts using $u=\sin(x)$ but I'm trying to do it the other way.
With $\int \sin\left(x\right) \cos\left(x\right) dx$
$u = \cos\left(x\right)$
$dv = \sin\left(x\right) dx$
$v = -\cos\left(x\right)$
Then,
$I = uv - \int v \ du = -\cos\left(x\right)\cos\left(x\right) + \int \cos\left(x\right)\sin\left(x\right) \ dx$
but then I get $I = -\cos^2\left(x\right) + I \, $ and those $I$'s would cancel out and I would get zero? Assuredly, I'm missing a negative sign somewhere
AI: You are indeed missing a minus sign. ${\frac{d}{dx}(\cos(x))=-\sin(x)}$. So the ${-\int vdu}$ part actually is
$${-\int (-\cos(x))(-\sin(x))dx=-\int\sin(x)\cos(x)=-I}$$
So
$${I = \cos^2(x) - I}$$
Which implies that
$${I = \frac{\cos^2(x)}{2}}$$
(obviously add the +c at the end). As required
Edit: This is a working solution (and nothing wrong with it). However, it's worth noting that it's also doable just by standard substitution. Notice that it is of the form
$${\int f(x)f'(x)dx}$$
where ${f(x) = \sin(x)}$. Letting ${u=\sin(x)}$ you get that ${du = \cos(x)dx\Rightarrow dx=\frac{du}{\cos(x)}}$, hence
$${\Rightarrow \int u\cos(x)\frac{du}{\cos(x)}=\int udu=\frac{1}{2}u^2 + c}$$
but we know ${u=\sin(x)}$ so
$${\int \sin(x)\cos(x)dx = \frac{1}{2}\cos^2(x) + c}$$
So both solutions give the same answer, but indeed substitution is easier :)
Edit Edit: @MarkViola gave yet another way to integrate this by using the identity ${\sin(2x)=2\sin(x)\cos(x)}$. Notice that means your function, ${\sin(x)\cos(x)}$ is nothing but ${\frac{\sin(2x)}{2}}$, and hence
$${\int \sin(x)\cos(x)dx = \int \frac{\sin(2x)}{2}dx=-\frac{1}{4}\cos(2x)+c}$$
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H: Inverse of matrix plus identity
I have a matrix $A$ which is symmetric and positive definite, and I am curious about the properties of $(I+A)^{-1}$.
I can tell that the matrix will exist (that is, $I+A$ will be symmetric and invertible), and thus that $(I+A)^{-1}$ will also be symmetric. I am curious if there are any other known properties of this inverse.
For example, is there anything we can say about the products $A(I+A)^{-1}$ or $(I+A)^{-1}A$?
AI: For example, one can say
$A(I + A)^{-1} = (A^{-1})^{-1}(I + A)^{-1}$
$= ((I + A)A^{-1})^{-1} = (A^{-1} + AA^{-1})^{-1} = (I + A^{-1})^{-1}. \tag 1$
It is also easy to see that
$(I + A)^{-1}A = (I + A^{-1})^{-1}. \tag 2$
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H: Help with Independent Probability Problem
There are n dice of different colors being rolled simultaneously. The numbers that show up on the faces are
added up to compute X. How many different die rolls result in X being divisible by 3? (basically how many
of the n tuples with each number being between 1 and 6 are such that the sum of these n numbers is divisible
by 3)
I was having some problem figuring out how exactly to answer this question. We discussed how to solve this problem using a known number of dice, but I can't figure out how I would accomplish this using a variable number of dice, as it is hard to calculate the possibility of all numbers that the dice can sum to and be divisible by 3. Any help would be appreciated.
AI: Hints:
How many $n$-tuples are possible in total?
What proportion have a sum which is a multiple of $3$? (Easier than you might think at first - consider the effect of the last dice to roll)
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H: Can inflection points be determined from the local extrema of the first derivative?
The local extrema of the first derivative determining the inflection points makes sense; for $x = c$ to be a local extremum for the first derivative, the first derivative's derivative (i.e., the second derivative) would need to change signs as it passes through $x = c$ and $f''(c)=0$. Wikipedia seems to agree:
the graph of the differentiable function has an inflection point at (x, f(x)) if and only if its first derivative, f′, has an isolated extremum at x
The reason I ask is because of a multiple choices question from my highschool math book:
Which statement verifies that the function $f$ has a point of inflection at $x = c$?
The two answers I was stuck on were
The derivative $f'$ has a local maximum at $x = c$
The sign of $f''$ changes at $x=c$
The answer is 2, but I don't really understand why 1 isn't correct. Maybe because inflection points can also happen if the first derivative has local minima? Though with the way the question is worded this shouldn't matter.
AI: I would think that statement 1. is incorrect for the reason you suggested: since inflection points also occur at local minimums of the first derivative, statement 1. does not wholly verify whether a point is an inflection point.
I will agree that this is not necessarily the best way to word the question, but statement 2. encompasses more cases (both local minimum and maximum), which makes it the better answer—and that's just how it goes with multiple-choice questions.
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H: Prove that there exists a sequence $(a_n)_{n=1}^{\infty}$
I have the following question that,
Let $A$ be a nonempty set of real numbers with a lower bound. Prove that there exists a sequence $(a_n)_{n=1}^{\infty}$ such that $a_n \in A $ for all $n$ and $\lim_{n \to \infty} a_n = \inf(A)$.
Adopting solution from the post
For any $\varepsilon>0$ there is $x \in A$ (that depends of $\varepsilon$) for which $x < \inf(A) + \varepsilon$ (why?). Thus, for any $n \in \mathbb Z^+$ there exists $a_n \in A$ such that $$a_n < \inf(A) + \frac1n.$$
Accepting any editing recommendation for this question.
AI: For any $\varepsilon>0$ there is $x \in A$ (that depends of $\varepsilon$) for which $x < \inf(A) + \varepsilon$ (why?). Thus, for any $n \in \mathbb Z^+$ there exists $a_n \in A$ such that $$a_n < \inf(A) + \frac1n.$$ Can you continue from here?
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H: Chess Board counting problem
Consider a n x n chess board. Count the number of shortest paths from the position (0,0) to the position (100,100) if each move can either be a horizontal step or a vertical step. You may assume that n is larger than 100.
AI: Each move can be either $(x,y)\to(x+1,y)$ (we call it "h") or $(x,y)\to(x,y+1)$ (we call it "v") so a shortest path is sequence of $100$ "h" and $100$ "v" mixed up in
$$\frac{200!}{100!\cdot 100!}={200\choose 100} \hbox{ ways.}$$
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H: $2^x$ is irrational if $x$ is irrational?
Prove/Disprove that if $x$ is irrational, then $2^x$ is also irrational.
My attempt for the proof:
Suppose $2^x>0$ is a rational number, then $2^x=\frac{a}{b}$ for some natural numbers $a$ and $b$. Taking logarithm with base $2$ on both sides to get, $x=\log_2 \frac{a}{b}$. Here I stuck! how to reach at $x$ is rational?
AI: This is false in general, take $x=\frac{\ln 3}{\ln 2}$, then $2^x=e^{x\ln 2}=e^{\ln 3}=3\in\mathbb{Q}$ but $x\notin\mathbb{Q}$. Otherwise, there would exist $p,q\geqslant 1$ coprimes such that $x=\frac{p}{q}$, that is to say $q\ln 3=p\ln 2$ and thus $3^q=2^p$ which is not because $2$ and $3$ are coprimes.
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H: Relationship between convexity of geometric figures and functions.
This is a rather simple question that made me curious while studying basic geometry. In Hardy's number theory book, a region is convex if it is possible, through every point $P$ in the boundary - to draw at least one line $l$ such that the whole of the region lies on one side of $l$. On the other hand, a function $f:\mathbb{R}\rightarrow\mathbb{R}$ is convex if $\forall x_{1},x_{2}\in X,\forall t\in [0,1]: f(tx_{1}+(1-t)x_{2})\leq tf(x_{1})+(1-t)f(x_{2})$ according to Wikipedia.
Now, $f(x)=-x^2$ is not convex in the second definition, but I think the region bounded by the function is convex is the first definition. I want to ask how the two definitions of convexity are related (if at all) and where the seeming disparity of the observation above stems from. Any insight into the topic of convexity would be much appreciated.
AI: There are two slightly different ideas here. Just sticking to intuitive ideas:
Convex set or region: in a sense the points on a line segment joining any two points in the set lies in the set. I think Hardy's version that you quote is equivalent in two dimensions
Convex function: in a sense the points on or above the graph of a function form a convex set. I think your definition from Wikipedia is equivalent
In your example of $f(x)=-x^2$, the points on or below the graph of the function form a convex set, and that is in fact the definition of a concave function
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H: Question about the definition of chain homotopy
I recently learned about the definition of chain homotopy.
If $f^\bullet, g^\bullet\colon C^\bullet\to D^\bullet$ are chain maps, then the definition is the following.
A chain homotopy between $f^\bullet$ and $g^\bullet$ is a family of
morphisms $T^n\colon C^n\to D^{n-1}$ such that for all $n\in \mathbb{Z}$ holds
$$f^n - g^n = \partial^{n-1}_D \circ T^n + T^{n+1}\circ \partial^n_C.$$
My question: What i do not understand is, what exactly the difference or the sum of maps in $$f^n - g^n = \partial^{n-1}_D \circ T^n + T^{n+1}\circ \partial^n_C.$$
are supposed to mean. What does $f^n-g^n$ mean in this diagram? What does the sum on the right hand side of the equation then tell me? Or generally, what does the above equation exactly describe?
Thanks for any help!
AI: Both $f^n$ and $g^n$ are maps $C^n \to D^n$, thus their difference $f^n-g^n$ is defined by $$(f^n-g^n)(x)=f^n(x)-g^n(x)$$ for any $x \in C^n$. Similarly, the right hand side of the equation is defined by $$(\partial_D^{n-1} \circ T^n + T^{n+1} \circ \partial_C^n)(x) =\partial_D^{n-1} (T^n (x)) + T^{n+1} (\partial_C^n (x))$$ for any $x \in C^n$.
The utility of two maps being chain homotopic is that they induce the same map on cohomology. Since $f^{\bullet}$ and $g^{\bullet}$ are chain maps, they induce well defined maps on cohomology $$\tilde{f}^{\bullet}, \tilde{g}^{\bullet} : H^{\bullet}(C) \to H^{\bullet}(D)$$ defined by $$\tilde{f}^{n}([x])=[f^n(x)]$$ for any $[x] \in H^n(C)$, and similarly for $\tilde{g}^{\bullet}$. Now, since we know $f$ and $g$ are homotopic, for any $[x] \in H^n(C)$ and any representative $x$ of the class $[x]$, we have $$[(f^n-g^n)(x)]=[\partial_D^{n-1} (T^n (x))] + [T^{n+1} (\partial_C^n (x))]$$
which simplifies to $$[f^n(x)-g^n(x)]=[\partial_D^{n-1} (T^n(x))],$$ as $\partial_C^n(x)=0$, since $x$ represents a cohomology class. Therefore $$[f^n(x)]=[g^n(x)] +[\partial_D^{n-1}(T^n(x))]$$ and since $[\partial_D^{n-1}(T^n(x))]$ is the trivial cohomology class, we have $$[f^n(x)]=\tilde{f}^n([x])=\tilde{g}^n([x])=[g^n(x)].$$ That is, $\tilde{f}^{\bullet}=\tilde{g}^\bullet$ as maps $ H^n(C) \to H^n(D)$.
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H: Deck of Cards combinatorics
{A,2,3,4,5,6,7,8,9,10,J ,Q ,K} and four suits - {Hearts, Diamonds, Spades, Clubs} . A Hand is a set of 5 cards picked up from the standard deck. How many different hands contain at least one of the following two cards : {K of Hearts, Q of Diamonds} ?
I'm having a little trouble finding the answer to this problem. I know that since two cards out of 5 have to be certain cards, that the choices for the other 3 cards is C(50,3). I also know that there are C(52,5) choices of possible hands for a deck of cards. However, I'm not sure what I would do if I needed to count the total number of hands containing the specified cards, but I'm thinking it would contain some sort of indirect counting method
AI: Hint
You already know that there are $C(52,5)$ choices of possible hands for a deck of cards in total
How many possible hands include neither the King of Hearts nor the Queen of Diamonds?
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H: Let $E$ a normed vector space and $L \neq E$ a vector subspace of $E$. Can $L$ contain any open ball of $E$?
I was trying to solve a problem of the book Elementos de Topologia Geral by Elon Lages and I found an exercise that I don't have success. The exercise is: prove that any vector subspace $L≠E$ of a normed vector space $E$ couldn't contain any open ball of $E$.
Could someone give me a hint? Thank you.
AI: Think about a ball for a moment. There is a point in the centre, and then there are vectors (possibly short ones) from that point in all directions. If a ball is contained in a subspace, then all of those directions are contained in the subspace. Since the subspace is closed under scalar multiplication, we can stretch these direction vectors as much as we like, until we get every point in the space.
So, let's prove this. Let's suppose that $L$ is a subspace of $E$, and that $B[y; r] \subseteq L$. I aim to show that $L = E$. We already know $L \subseteq E$, so let's show $E \subseteq L$.
Suppose $x \in E$. I want to shrink down the vector from $0$ to $x$ until it is length $r$. The resulting vector is $r\frac{x}{\|x\|}$. Consequently, $y + r\frac{x}{\|x\|} \in B[y; r] \subseteq L$. Further, $y \in B[y; r] \subseteq L$, so
$$r\frac{x}{\|x\|} = \left(y + r\frac{x}{\|x\|}\right) - y \in L,$$
since $L$ is closed under subtraction. Since it is also closed under scalar multiplication,
$$x = \frac{\|x\|}{r}\left(r\frac{x}{\|x\|}\right) \in L$$
as well. Therefore, $E = L$ as required.
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H: Percentage question on playing a game
I was doing some past papers of AMC when I got stuck on this question:
Two people were playing a video game. One morning each of them won 70% of their games. That afternoon, they played the same number of games as each other and each won them all. One of the player's winning percentage for the day rose to 85%, while the other player's winning percentage rose to 90%. What is the minimum possible total number of games that they could've played that morning?
I couldn't find an efficient method of working out the solution. By trial and error I found out that they could have played a minimum of 30 games in total, (10 for one and 20 for the other) and by each winning 20 more games, the required result is achieved. I'm sure there's another much more efficient method other than trial and error
AI: They clearly each played a multiple of $10$ games in the morning and a multiple of $10$ (or of $20$ but that is a multiple of $10$) games in total, so a multiple of $10$ games in the afternoon.
Let's suppose the first player played $10a$ games in the morning and $10b$ in the afternoon. You know $7a+10b=8.5(a+b)$ so $a=b$ and the smallest positive integer solution is $a=1,b=1$. As a check this gives $7/10$ games won in the morning, $10/10$ in the afternoon and $17/20$ overall
Let's suppose the second player played $10c$ games in the morning and $10c$ in the afternoon. You know $7c+10d=9(c+d)$ so $2c=d$ and the smallest positive integer solution is $c=1,d=2$. As a check this gives $7/10$ games won in the morning, $20/20$ in the afternoon and $27/30$ overall
To me this suggests $10+10$ games played in the morning and $20+30=50$ overall
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H: Mapping from $Re(z)\leq0$ onto the disk $|z|\leq1$
How am I supposed to do it? I really have no idea at all.
AI: Hint: Recall the Möbius transformation
$$
f\colon z\mapsto\frac{z+1}{z-1}
$$
maps the open left half plane onto the open unit disc. This almost does what you want: $f(\{z\in\mathbb{C}\mid\operatorname{Re}z\leq 0\})=\overline{D}(0,1)-\{1\}$. Can you think of some $g\colon\mathbb{C}\to\mathbb{C}$ such that $g\circ f$ would do?
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H: Proving the function is always positive in the interval $(0,1)$
Let $f_0(x)=\frac{1}{1-x}$ and define $f_{n+1}(x)=xf'_n(x)$ prove that
$f_{n+1}(x)>0$ for $0<x<1$ and $n\in \mathbb{N}$
my attempt was as following. Since the expression invloves the natural numbers, induction will be a good method of proof. After evaluating few terms, I found that
$f_n(x)=\frac{g(x)}{{(1-x)}^{2n}}$
So the proof is equivelent to prove $g(x)>0$
Any hint will be appreciated
AI: Let $\epsilon > 0$ and consider the compact set $[\epsilon, 1-\epsilon]$. Then on this set we have that the power series $$\sum_{k=0}^\infty x^k = \frac{1}{1-x}$$
and that the convergence is occurring inside the radius of convergence so we know that
$$ f_1(x) = x\frac{d}{dx}\sum_{k=0}^\infty x^k = x\sum_{k=0}^\infty \frac{d(x^k)}{dx} = \sum_{k=1}^\infty kx^k$$
By induction you can show that for any $n$ that $$\sum_{k=1}^\infty k^n x^k$$
has the same radius of convergence as the original power series by using the root test. Hence we can continue to switch the order of the series and derivative. By doing this we can calculate by induction that $$f_n(x) = \sum_{k=1}^\infty k^n x^k$$
for any $x\in[\epsilon, 1-\epsilon]$. Then since $x > 0$, we can see that $f_n(x)$ is always equal to a convergent series of positive numbers. Hence $f_n(x) > 0$ for all $x\in[\epsilon, 1-\epsilon]$. You can show this holds for all $\epsilon >0$, hence the same conclusion holds on $(0,1)$ as desired.
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H: What is the value of $L$ and the arcs $AG$ and $BH$?
What is the value of $L$ and the arcs $AG$ and $BH$, as a function of $d, R$ and $r$?
Note that $G, H, C, D$ are collinear and $AB$ tangent to both circles.
The problem was inspired by a question from the ACT exam.
AI: Hint: Please draw a line from center $C$ to line $AD$, parallel to $AB$. It will be perpendicular to line $AD$.
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H: Analogue of Turing recognizable languages
A language $S$ is called Turing recognizable if for some Turing machine $S$ is exactly the set of inputs when the machine halts. How can we call the language which is the set of outputs for some Turing machine? How are these two classes related?
AI: Any language "generated" by a Turing machine that always halts the way you mentioned (i.e., $s\in S$ iff $T(x)=s$ for some input $x$ to the fixed Turing machine $T$) will be Turing recognisable.
The way to recognise the language $S$ is to use the Turing machine which does the following:
Machine A: Turing machine to recognise the language $S=\{s\mid \exists x : T(x)=s\}$, where $T$ always halts
given an input $s$ (to check whether $s\in S$)
for all possible input strings $x$:
simulate $T(x)$
if $T(x)=s$, then accept
If $s\in S$, then the above machine will eventually halt and accept when it comes to a string $x$ such that $T(x)=s$; if $s\notin S$ then this machine will run forever.
Nonempty Turing recognisable languages can also be "generated" by a Turing machine in this way too.
Suppose $S$ is a language recognised by the Turing machine $T$ that is nonempty, so that we have some known element $s_0\in S$, then consider the following algorithm:
Machine B: Turing machine which always halts to generate the nonempty language $S\ni s_0$ recognised by $T$
take as input a string $s$ and a number $n$
simulate $T(s)$
if $T(s)$ halts in $n$ steps and accepts, then print $s$
otherwise, print $s_0$
The language generated by the above machine will be $S$: for any $s\in S$, suppose $T(s)$ accepts $s$ in $n$ steps, then the above machine will print $s$ after input $(s,n)$.
If $s\notin S$, then no possible input will make the above machine print $s$.
However, the Turing recognisable language $\varnothing$ cannot really be generated by a Turing machine that always halts.
If we become more flexible and consider languages generated by a Turing machine which is allowed to run forever, then we can modify the machines to fix this
Machine A': Turing machine to recognise the language $S = \{s \mid \exists x:T(x)=s\}$ where $T$ may not halt
given an input $s$
for some enumeration of pairs $(x,n)$ where $x$ is a string and $n$ is a number
simulate $T(x)$ for $n$ steps
if $T(x)$ halts and outputs $s$, then accept
Machine B': Turing machine which may not always halt that generates a language $S$ recognised by $T$
takes as input a string $s$
simulate $T(s)$
if $T(s)$ eventually halts and accepts, print $s$
if $T(s)$ halts and rejects, loop forever
Therefore, in summary:
Nonempty Turing recognisable languages are equivalently sets of outputs generated by some Turing machine that always halts
Turing recognisable languages are equivalently sets of outputs generated by some Turing machine that may not halt at every input
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H: Why is the general solution to linear homogeneous differential equation with constant coefficients different if roots are distinct or repeating?
Consider the second order linear homogeneous differential equation:
$$ax'' + bx' + c = 0$$
If we start from the assumption that the solution has the form
$$ x(t) = e^{rt}$$
Then, if we solve the characteristic polynomial, then we will have 2 values for $r$, so the following solutions are valid:
$$x(t) = e^{r_1t}$$
and
$$x(t) = e^{r_2t}$$
So, I would expect the general solution to look like this:
$$x(t) = c_1e^{r_1t} + c_2e^{r_2t}$$
But, if the roots are repeating (e.g. $r_1=r_2$), we write the solution as:
$$x(t) = c_1e^{r_1t} + c_2te^{r_1t}$$
But, why? I know if we have repeating roots, then, we can factor out the constant, and we end up with a single constant multiplied by an exponential e.g., $(c_1 + c_2)e^{r_1t}$, but isn't that form still valid? I mean, can we just use that form? Why use $x(t) = c_1e^{r_1t} + c_2te^{r_1t}$ instead of $(c_1 + c_2)e^{r_1t}$.
AI: You are looking for the general solution. That is why only using solutions $ce^{rt}$, when $r$ is a double root, is insufficient. (Writing $(c_1 + c_2)e^{rt}$ conveys no knowledge beyond solutions of the form $ce^{rt}$.) If “solve” only meant find some solution then just give the solution $0$. Do you agree that is useless?
Remember that you want the general solution in order to find the solution fitting some initial conditions. A solution where $y(0) = 0$ and $y’(0) = 1$ can’t be $ce^{rt}$. But $te^{rt}$ is a solution fitting those initial conditions. If you refuse to consider solutions like $te^{rt}$ when $r$ is a double root then you'll never be able to solve that ODE when $y(0) = 0$ and $y'(0) = 1$.
There are many situations in math where multiple roots behave differently than distinct roots. An example in basic calculus is partial fraction decompositions. If $a \not= b$ then
$$
\frac{1}{(x-a)(x-b)} = \frac{c}{x-a} - \frac{c}{x-b}
$$
where $c = 1/(a-b)$, but this is not valid when $a = b$. The partial fraction decomposition for $1/(x-a)^2$ is, well, itself. There is nothing to do in that case.
Every linear second-order ODE with constant coefficients has a $2$-dimensional solution space. That property is true whether or not the roots of the quadratic polynomial are equal or distinct. But concrete formulas for a basis of the solution space are different in the cases of distinct roots and repeated roots.
Perhaps you don't understand how someone could discover the extra solution $te^{rt}$ when $r$ is a double root. Here is some motivation. In the case of distinct roots $r_1$ and $r_2$, you have solutions $c_1e^{r_1t} + c_2e^{r_2t}$. In particular, $(e^{r_1t} - e^{r_2t})/(r_1-r_2)$ is a solution. Now let $r_2 \to r_1$. By L'Hopital's rule,
$$
\lim_{r_2 \to r_1} \frac{e^{r_1t} - e^{r_2t}}{r_1-r_2} = te^{r_1t}.
$$
That suggests that when $r_2 = r_1$ we should check if $te^{r_1t}$ fits the ODE, and you can check it really does. Another way to think about this is that when $r_1 \not= r_2$, the function $y(t) = (e^{r_1t} - e^{r_2t})/(r_1-r_2)$ satisfies $y(0) = 0$ and $y'(0) = 1$. For a single $r_1$, $y(0) = te^{r_1t}$ also satisfies $y(0) = 0$ and $y'(0) = 1$.
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H: Finding values using the equation of $x$ that satisfies $\left\{ x \right\} + \left\{ {\frac{1}{x}} \right\} = 1$
Let the real number $x$ satisfies $\left\{ x \right\} + \left\{ {\frac{1}{x}} \right\} = 1$ and $k$ denotes the value of $\left\{ {{x^3}} \right\} + \left\{ {\frac{1}{{{x^3}}}} \right\} = k$ (where $\{\cdot\}$ denotes
fractional part function). Find the value of $k$.
My approach is as follows: $x$ cannot be an integer,
$x = n + a,a \in \left( 0,1 \right)$
$$a + \left\{ {\frac{1}{{n + a}}} \right\} = 1 \Rightarrow \left\{ {\frac{1}{{n + a}}} \right\} = 1 - a = t,t \in \left( {0,1} \right)$$
Not able to proceed from here.
AI: Hint: Since $\{x\}+\{x^{-1}\}$ is an integer, that means $x+x^{-1}$ is also an integer. So $x^3+x^{-3}=(x+x^{-1})^3-3(x+x^{-1})$ is also an integer. Now show $x^3$ is irrational to conclude $\{x^3\}+\{x^{-3}\}=1$.
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H: Determining the leading coefficient of Vandermonde's Determinant.
On Shilov's Book "Linear Algebra", when calculating the Vandermonde's Determinant,
the author concludes the leading coefficient of the product of the roots of the determinant (seen as a polynomial) is
$$ W(x_1, ..., x_{n-1}) $$
Shilov proceded to see the solution of the determinant as a polynomial of degree $(n - 1)$ in $x_n$, noting that from the definition of Vandermonde's determinant, we have vanishing cases when $x_n$ is equal to any other parameter in $W$ (any determinant with two equal columns vanishes). It makes sense to propose the solution as (polynomial remainder theorem)
$$ W(x_1, .., x_{n}) = k\prod_{k = 1}^{n - 1}(x_n - x_k)$$
with
$$ W(x_1, \ldots, x_n) = \left| \begin{array}{cccc} 1 & 1 & \cdots & 1\\ x_1 & x_2 & \cdots & x_n \\ x_1^2 & x_2^2 & \cdots & x_n^2\\ \cdot & \cdot & \cdots & \cdot \\ x_1^{n-1} & x_2^{n-1} & \cdots & x_n^{n-1}\\ \end{array} \right| $$
Then the author states $$ k = W(x_1,...,x_{n−1})$$ can be found when expanding the las column of the Vandermonde's determinant.
My question is, how did he conclude that?
When expanding $W$ with respect to the last column we do have
$$ W = 1 \cdot A_{1n} - x_nA_{2n} + ... + x_n^{n - 1}W(x_1, ..., x_{n - 1})$$
I believe he then compares this two expressions:
$$ k\prod_{k = 1}^{n - 1}(x_n - x_k) = 1 \cdot A_{1n} + x_nA_{2n} + ... + x_n^{n - 1}W(x_1, ..., x_{n - 1})$$
What's the reason he only takes the last co-factor to determine the value of $k$?
AI: To make it a bit more clear, let's consider $$W(x_1,\ldots, x_{n-1},X) $$
which is indeed a polynomial in indeterminate $X$ of degree $n-1$ with roots $x_1,\ldots, x_{n-1}$.
By polynomial factor theorem, there is a constant $k$ such that
$$ W(x_1,\ldots, x_{n-1},X) =k \prod_{i=1}^{n-1} (X-x_i). $$
We note that $k$ is the leading coefficient of this polynomial.
On the other hand, by performing expansion by the last column in determinant $W(x_1,\ldots, x_{n-1},X)$ (just like you did, but with $X$ instead of $x_n$), we see that the leading coefficient of polynomial $W(x_1,\ldots, x_{n-1},X) $ is exactly
$W(x_1,\ldots, x_{n-1})$ (all expansion cofactors are of course functions of $x_1,\ldots, x_{n-1}$ only and they are the coefficients of the polynomial in $X$).
As we are referring to the leading coefficient of the same polynomial, this implies:
$$k= W(x_1,\ldots, x_{n-1}).$$
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H: Polygon Diagonal Combinatorics
A diagonal for a polygon is defined as the line segment joining two non-adjacent points. Given an n-sided polygon, how many different diagonals can be drawn for this polygon?
I know that the number of diagonals is C(n,2). However, I don't know how to account for the fact that you can't draw a diagonal for two points that are adjacent to each other.
AI: There are two easy ways to take that restriction into account. You can start with the number of pairs of vertices, $\binom{n}2$, and subtract the number of pairs of adjacent vertices, $n$. Or you can observe that at each vertex there are $n-3$ other vertices that are not adjacent. That means that each vertex is an end of $n-3$ diagonals, so there are $n(n-3)$ ends of diagonals and therefore $\frac{n(n-3)}2$ diagonals.
As a check,
$$\binom{n}2-n=\frac{n(n-1)}2-n=\frac{n(n-3)}2\;.$$
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H: Isomorphism Definition - Error in Hoffman Kunze?
I am Referring to the book Linear Algebra by Hoffman and Kunze $2$e, page $84$ section $3.3$.
It defines Isomorphism as follows:
If $V$ and $W$ are vector spaces over a field $\mathbb{F}$, then any one-one $T$ $\in$ $\mathcal{L}(V, W)$ of $V$ onto $W$ is called an Isomorphism from $V$ to $W$.
Standing alone, the definition seems correct, but the authors go on to say that:
if $V$ is Isomorphic to $W$ via an Isomorphism $T$, then $W$ is Isomorphic to $V$ because $T^{-1}$ is an Isomorphism from $W$ to $V$.
I feel this is erroneous, because as per the definition given $T$ is one-one, which doesn't imply the existence of an inverse on its own.
To correct it, either, the term "bijective" must be used instead of one-one, or there should be an additional condition that $~\dim(V) = \dim(W).$
Am I correct in this interpretation? Or am I missing something out?
I tried looking up in the various errata for this book, but couldn't find anything related to this.
Thanks.
AI: Note that in the definition of isomorphism given by the author, $T$ is onto. So $T$ is bijective by definition.
Additional information:
In general, people use the term into when they describe a map from a set $A$ into $B$. So onto is used for a specific purpose.
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H: What are some ways to prove that a k-partite graph is nonplanar?
I am reading papers on graph theory and I encountered one work that talked about the planarity of a certain graph. In one of the proofs of a theorem, the author stated that the graph $K_{1,2,3}$ is nonplanar. Is this a known result? How can one check that this is planar? And does it hold for other graphs of the form $K_{1,2,n}$?
AI: $K_{1,2,3}$ contains a copy of $K_{3,3}$, which is well known not to be planar.
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H: Prove the inequality $u^{\alpha}v^{1-\alpha} \leq \alpha u + (1-\alpha)v $
I'm trying to prove the inequality highlighted by the red line in the picture, but I do not know how. Please provide detail as much as you can, thank you!
For a minimal family, the sufficient statistic $\mathbf{T}$ is also minimal sufficient. For a proof, see Lehmann:
Theory of Point Estimation, Example 5.9, pp. 43-44
If we parametrize the family using $\eta=c(\theta),$ this is called the natural parametrization (or the canonical parametrization). We then write
$$
f(\mathbf{x}, \mathbf{\eta})=\exp \left(\mathbf{\eta}^{T} \mathbf{T}(\mathbf{x})-A(\mathbf{\eta})\right) h(\mathbf{x})
$$
where
$$
A(\mathbf{\eta})=\log \int_{E} \exp \left(\mathbf{\eta}^{T} \mathbf{T}(\mathbf{x})\right) h(\mathbf{x}) d \mathbf{x}
$$
The natural parameter space is $\mathbf{H}=\{\boldsymbol{\eta}: A(\boldsymbol{\eta})<\infty\} .$
Theorem 1 : H is a convex set.
Proof: Let $0<\alpha<1$ and take $\eta$ and $\eta_{1}$ in $\mathbf{H}$. Write
$$
A\left(\alpha \eta+(1-\alpha) \eta_{1}\right)=\log \int_{E}\left(\exp \left(\eta^{T} \mathbf{T}(\mathbf{x})\right) h((\mathbf{x}))\right)^{\alpha}\left(\exp \left(\mathbf{\eta}_{1}^{T} \mathbf{T}(\mathbf{x})\right) h((\mathbf{x}))\right)^{1-\alpha} d \mathbf{x}$$
But $\underline{u^{\alpha} v^{1-\alpha} \leq \alpha u+(1-\alpha) v}$ (take logarithms of both sides and use the fact that the logarithm is a concave function), whence
$$A\left(\alpha \eta+(1-\alpha) \eta_{1}\right) \leq \alpha A(\eta)+(1-\alpha) A\left(\eta_{1}\right)<\infty$$
AI: For $u$ and $v$ are positives and $0\leq \alpha\leq1$ it's just AM-GM because $\alpha+1-\alpha=1.$
The AM-GM it's the following.
Let $\alpha_i\geq0$ such that $\sum\limits_{cyc}\alpha_i=1$ and $x_i>0$. Prove that:
$$\alpha_1x_1+\alpha_2x_2+...+\alpha_nx_n\geq x_1^{\alpha_1}x_2^{\alpha_2}...x_n^{\alpha_n},$$
which is just Jensen for a $\ln$ function:
$$\ln(\alpha_1x_1+\alpha_2x_2+...+\alpha_nx_n)\geq\alpha_1\ln{x_1}+\alpha_2\ln{x_2}+...+\alpha_n\ln{x_n}.$$
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H: Independence between fractional parts of consecutive sums of independent uniforms
Let $X_1,X_2$ be independent $\text{Uniform}(0,1)$ random variables. Define $U_1 = X_1 - \lfloor X_1 \rfloor$ and $U_2 = X_1 + X_2 - \lfloor X_1 + X_2 \rfloor$ where $\lfloor a \rfloor$ is the largest integer less or equal to $a \in \mathbb{R}$. We can take as given that $U_1$ and $U_2$ are also $\text{Uniform}(0,1)$.
We want to show that $U_1$ and $U_2$ are independent. We can do so by arguing in cases that $P(U_1 \leq u_1, U_2 \leq u_2) = P(U_1 \leq u_1) P(U_2 \leq u_2) = u_1 u_2$ for (1) $u_1 > u_2$ and (2) $u_1 \leq u_2$. I'm struggling with arguing two lines in case (2). The solution given is as follows:
\begin{align} P(U_1 \leq u_1, U_2 \leq u_2) &= P(X_1 - \lfloor X_1 \rfloor \leq u_1, X_1 + X_2 - \lfloor X_1 + X_2 \rfloor \leq u_2) \\\\
&= P\left(X_1 - \lfloor X_1 \rfloor \leq u_1, 0 \leq X_1 + X_2 \leq u_2\right) + P(X_1 - \lfloor X_1 \rfloor \leq u_1, 1 \leq X_1 + X_2 \leq 1 + u_2) \\\\
&= E\left[P(X_1 - \lfloor X_1 \rfloor \leq u_1, 0 \leq X_1 + X_2 \leq u_2 | X_1 = x)\right] \\\\
&\ \ \ \ + E\left[P(X_1 - \lfloor X_1 \rfloor \leq u_1, 1 \leq X_1 + X_2 \leq 1 + u_2 | X_2 = y)\right] \\\\
&= \int_0^{u_1} P(0 \leq X_1 + X_2 \leq u_2 | X_1 = x)\ dx + \int_0^{u_1} P(1 \leq X_1 + X_2 \leq 1 + u_2 | X_2 = y)\ dy \\\\
&= \int_0^{u_1} P(X_2 \leq u_2 - x)\ dx + \int_0^{u_1} P(1 - y \leq X_2 \leq 1)\ dy \\\\
&= \int_0^{u_1} u_2 - x\ dx + \int_0^{u_1} y\ dy \\\\
&= u_1 u_2 \end{align}
as desired. Between the third and fourth lines, I'm unclear why it must be the case that
$$
E\left[P(X_1 - \lfloor X_1 \rfloor \leq u_1, 1 \leq X_1 + X_2 \leq 1 + u_2 | X_2 = y)\right] = \int_0^{u_1} P(1 \leq X_1 + X_2 \leq 1 + u_2 | X_2 = y)\ dy.
$$
"Pulling" the indicator $I(X_1 \leq u_1)$ conditioned on $X_1 = x$ makes sense for the first expectation, but it doesn't seem justified conditioned on $X_2 = y$. That is, $X_1$ and $X_1 + X_2$ aren't conditionally independent given $X_2 = y$. Additionally, assuming this holds, I'm struggling with arguing that
$$\int_0^{u_1} P(1 \leq X_1 + X_2 \leq 1 + u_2 | X_2 = y)\ dy = \int_0^{u_1} P(1 - y \leq X_2 \leq 1)\ dy$$
between the fourth and fifth lines. It's possible that there's a typo in the solution where the second expectation in the third line should actually be conditioned on $X_1 = x$, say. While this would rectify my first issue, I'm still unclear why the equality holds between the fourth and fifth lines.
AI: Yes, there is a typo between the third and fourth lines. It is supposed to be
$$
E\left[P(X_1 - \lfloor X_1 \rfloor \leq u_1, 1 \leq X_1 + X_2 \leq 1 + u_2 | X_1 = y)\right] = \int_0^{u_1} P(1 \leq X_1 + X_2 \leq 1 + u_2 | X_1 = y)\ dy.
$$
As for your second question we have $y\leq u_1$ on this interval which by the intial assumption implies $y\leq u_2\longrightarrow 1\leq 1+u_2-y$.
So the right side of the inequality $P(1-y\leq X_2\leq 1+u_2-y)$ vanishes to $P(1-y\leq X_2\leq 1)$ because $X_2\leq 1$.
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H: Evaluating $\int _0^1\frac{\ln \left(x^3+1\right)}{x+1}\:dx$
What methods would work best to find $\displaystyle \int _0^1\frac{\ln \left(x^3+1\right)}{x+1}\:dx$
As usual with this kind of integral i tried to differentiate with the respect of a parameter
$$\int _0^1\frac{\ln \left(ax^3+1\right)}{x+1}\:dx$$
$$\int _0^1\frac{x^3}{\left(x+1\right)\left(ax^3+1\right)}\:dx=\frac{1}{a-1}\int _0^1\left(\frac{ax^2-ax+a}{ax^3+1}-\frac{1}{x+1}\right)\:dx$$
in the end $2$ of these integrals are nice but the other $2$ are not
$\displaystyle \int _0^1\frac{-ax}{ax^3+1}\:dx$ and $\displaystyle \int _0^1\frac{a}{ax^3+1}\:dx$
Is there a better approach to this?
AI: Instead of using Feynman's trick at once use the following substitution first
$$\underbrace{\int _0^1\frac{\ln \left(1+x^3\right)}{1+x}\:dx}_{x=\frac{1-t}{1+t}}$$
$$=\ln \left(2\right)\underbrace{\int _0^1\frac{1}{1+x}\:dx}_{\ln \left(2\right)}+\int _0^1\frac{\ln \left(1+3x^2\right)}{1+x}\:dx-3\underbrace{\int _0^1\frac{\ln \left(1+x\right)}{1+x}\:dx}_{\frac{1}{2}\ln ^2\left(2\right)}$$
Now make use of the general result proved here.
$$\int _0^1\frac{\ln \left(b+ax^2\right)}{1+x}\:dx$$
$$=-\frac{\ln ^2\left(b\right)}{4}-\frac{\text{Li}_2\left(-\frac{a}{b}\right)}{2}-\frac{\ln ^2\left(a+b\right)}{4}+\frac{\ln \left(b\right)\ln \left(a+b\right)}{2}-\arctan ^2\left(\sqrt{\frac{a}{b}}\right)+\ln \left(2\right)\ln \left(a+b\right)$$
So by setting $a=3$ and $b=1$ we have
$$-\frac{\text{Li}_2\left(-3\right)}{2}-\arctan ^2\left(\sqrt{3}\right)+\ln ^2\left(2\right)-\frac{1}{2}\ln ^2\left(2\right)$$
Thus,
$$\int _0^1\frac{\ln \left(1+x^3\right)}{1+x}\:dx=-\frac{\text{Li}_2\left(-3\right)}{2}-\frac{\pi ^2}{9}+\frac{1}{2}\ln ^2\left(2\right)$$
Where $\text{Li}_2\left(z\right)$ is the Dilogarithm function.
As suggested by others one could also turn the integral into
$$\int _0^1\frac{\ln \left(1+x^3\right)}{1+x}\:dx=\int _0^1\frac{\ln \left(1-x+x^2\right)}{1+x}\:dx+\int _0^1\frac{\ln \left(1+x\right)}{1+x}\:dx$$
From the result above we obtain the value of that other integral which is
$$\int _0^1\frac{\ln \left(1-x+x^2\right)}{1+x}\:dx=-\frac{\text{Li}_2\left(-3\right)}{2}-\frac{\pi ^2}{9}$$
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H: How to express a Python loop and a condition mathematically?
I have the following Python code that I want to express mathematically.
W = 0
for i in range(5):
if (A > i_1) and ( (A < i_2) or (A < i_3) ):
W = W + 1
This is a generic problem that I am trying to achieve; looping over some values and checking some condition. If the condition is satisfied, I want to add 1 to the counter (W in this case); otherwise, skip to the next value. How can I express this using a mathematical formula?
To be more specific:
for row in DF1.iterrows():
pTime = row['A1_time']
W = 0
for row2 in DF1.iterrows():
a1 = row2['A1_time']
a2 = row2['A2_time']
a3 = row2['A3_time']
if (pTime > a1) and ( (pTime < a2) or (pTime < a3) ):
W = W + 1
AI: To me it looks like you are just computing the cardinality of the set of elements meeting your conditions.
So something like $|\{x : P(x)\}|$ where $P(x)$ means $x $ satisfies your condition.
I can’t be totally sure how to elaborate until you clarify what the index has to do with the inner conditions.
Borrowing a bit of python syntax and using it with set notation, what I'm saying holds above for your latest example:
$|\{(x,y)\in DF1\times DF1: x.a1time > y.a1time \text{ and } x.a1time < max(y.a2time, y.a3time)\}|$
I was able to simplify the OR condition slightly by using max. In fact, this is far closer to how I would do it with real python:
len([(x,y) for x in DF1.iterrows() for y in DF1.iterrows()
if x['a1_time'] > y['a1_time'] and x['a1_time'] < max(y['a2_time'],y['a3_time']))])
It can also be compressed a little bit more by using things from the itertools library, but this is more elementary python.
Since it looks like you don't care about the actual entries, just their count, another trick is to just create a list of $1$'s indicating each successful hit, then using sum. This might be a good alternative because you can do it all with generator expressions, conserving memory.
from itertools import product
sum(1 for x,y in product(DF1.iterrows(), DF1.iterrows())
if x['a1_time'] > y['a1_time'] and x['a1_time'] < max(y['a2_time'],y['a3_time'])))
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H: Doubts on a game that two players take turns to take an element and xor it to their sums.
This is originally a question from https://codeforces.com/contest/1383/problem/B, but this is more of a math problem than an algorithmic one, so I decided to post here.
There is a game where two players take turns to take an element from an array $A$ containing non-negative integers. Say Alice is the one who go first and Bob is the second. Both Alice and Bob has an initial score of $0$. For each turn, a player can pick an element $v$ from $A$ and remove it), and his score becomes $\text{his original score} \oplus v$, where $\oplus$ denotes $\text{xor}$. When $A$ becomes empty, the game ends, and the one with larger sum wins. If both players play optimally, who will be the one winning, or the game will end in a draw.
Solution can be found here: https://codeforces.com/blog/entry/80562
Basically the strategy is simple. For example $A = \{110_2, 100_2, 010_2, 010_2\} $. Consider the most significant bit, If the number of ones is odd, then one of them will win immediately (depending on the parity of the number of zeros and number of ones $mod$ 4).
However, the solution mentions that when the number of ones of the most significant bit is even, we move on to the next bit and do the same.
I do not understand why we can do the same on the next bit because the choice of $1^{st}$ most significant bit somehow limit the choice of the $2^{nd}$ most significant bit. Using the example mentioned above, when we are considering the $2^{nd}$ bit, we know that $110_2$ and $100_2$ must go to different players, otherwise it will not end in a tie in the $1^{st}$ bit. Furthermore, the choice of $1^{st}$ and $2^{nd}$ bit will limit the choice of $3^{rd}$ bit.
However, the solution in the editorial seems to imply these won't affect the result but there is no explanation. Can anyone kindly explain why this is the case? Thanks in advance.
AI: With an even number of $1^{st}$ significant bits, the winner cannot be decided on the $1^{st}$ significant bit no matter how Alice and Bob play.
In order for it to affect a game’s outcome. One side must have a move history containing an odd number of $1^{st}$ significant bits. But the total number is even, so the other player will also have made an odd number of such moves. Both players’ scores have a $1$ in front, and the winner is decided based on the remainder.
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H: If $X$ is a Banach space any capable conditions to make sure that $X$ is also Hilbert?
Let $X$ be a Banach space and suppose that $X$ is isometric with a Hilbert space.Is it true to say that $X$ is also a Hilbert space?If I assume that $X$ is isomorphic with a Hilbert space is it still true?
AI: YES for the first question, and NO for the second. If $T$ is an isometry from $X$ and $Y$ then $ \langle Tx, Ty \rangle=\langle x, y \rangle$ gives an inner product on $Y$ which makes it a Hilbert space. $\mathbb R^{2}$ with any norm is isomorphic to $\mathbb R^{2}$ with the Euclidean norm . Take the norm $\|(x,y)\|=|x|+|y|$ for a counter-example to the second statement. I leave it to you to verify that this norm does not satisfy Parallelogram Law which means that the space is not an inner product space. For Parallelogram Law see https://en.wikipedia.org/wiki/Parallelogram_law
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H: Given $k, a \in \mathbb{R}$, find a polynomial $P$ such that $P(k) = a$
You are given two real numbers $k,a \in \mathbb{R}$, and you are promised that there is a polynomial with integer coefficients $P \in \mathbb{Z}[X]$ such that evaluating it on $k$ yields $a$, i.e.
Promise: $\ \exists P \in \mathbb{Z}[X] \ \ \ s.t.\ \ P(k) = a$
Problem: find such $P$. In general, I'd expect more than one solution to exist; then I'd be interested in obtaining the one with lowest degree.
Note that $k$ may be any real number, and a particular case of interest (due to context, see below) is $k = -\frac{\sqrt{2}}{2}$.
Some context. I'm a PhD student in computer science and this problem came up in my research when trying to reverse engineer a black box. Essentially, $k$ is a parameter set a priori and $a$ is the output the black box produces; finding $P$ would let me know valuable information about the internal workings of the black box.
I am aware that, if it were the case that $k \in \mathbb{N}$ and $a > 0$, then I could obtain each coefficient of the polynomial by calculating remainders:
$p_0 = a \bmod k$, gives the coefficient of the 0-degree term;
$p_1 = \frac{a - p_0}{k} \bmod k$, gives the coefficient of the 1-degree term, and so on...
$p_2 = \dots$
However the fact that $k$ may be any real number prevents me from using this kind of approach from discrete maths.
AI: If we look just at $k=-\frac{\sqrt2}2$, suppose $P(k)=a$, then $2^Na=p\sqrt2+q$ for $N=\deg P$ and $p,q\in\Bbb Z$, so we can recover some polynomial $P$ just by doing the following:
keep doubling $a$ until it has the form $2^na=p\sqrt2+q$. Take $n$ to be minimal with this property.
since $k^2=\frac12$, we get that $a=p\cdot\frac{\sqrt2}2\cdot\frac1{2^{n-1}} + \frac q{2^n} = -p\cdot k\cdot k^{2(n-1)} + qk^{2n}$ so that the polynomial $P(x)=qx^{2n}-px^{2n-1}$ does the trick and has degree either $2n$ or $2n-1$ (...well, it could have degree $-\infty$ if we started with $a=0$ but I digress)
As for minimality, write $P(x) = \sum_{i=0}^{2n}=c_ix^i$ and note that $a=P(k) = \sum_{i=0}^n\frac{c_{2i}}{2^i} - \sum_{i=0}^{n-1}\frac{c_{2i+1}}{2^i}\sqrt2$ fits into the form $p\sqrt2+q$ after scaling by $2^n$. This is minimal if $c_{2n}$ is odd, and we can ensure that $c_{2n}$ is odd by the minimality of $P$ with $P(k)=a$: indeed, if $c_{2n}=2c_{2n}'$, then $c_{2n}k^{2n} = c_{2n}'k^{2(n-1)}$ using that $k^{-2}=2$, allowing us to reduce $P$ by one degree.
Therefore, the degree of minimal $P$ is twice the minimum $n$ such that $2^na=p\sqrt2+q$, showing that the above algorithm succeeds in finding a minimum-degree $P$ in the case that $a$ arises from an even-degree polynomial. The analysis for odd degree is pretty much the same.
However, this algorithm is fairly ad hoc and is difficult to generalise once $k\neq-\frac{\sqrt2}2$.
I'm not sure how ugly $k$ is allowed to be. For example, if $k$ is transcendental, then $P(k)=a$ will immediately tell you what $P$ is just from trying to represent $a$ in a computable way (relative to $k$, I suppose). In general, I'm not sure how $k$ and $a$ would be represented, if I'm assuming you are trying to computationally solve for $P$ somehow. For instance, in my above algorithm, I worked under the assumption that the computer was dealing with elements of $\Bbb Q(\sqrt2)$ or something (to ensure the computation is exact).
I'm sure a similar technique to the one above can be used for other $k$ that look "similar to" your case of interest $k=-\frac{\sqrt2}2$, though showing minimality might be more difficult.
Also, if $k\in\Bbb N$, then for any $P(x)\in\Bbb Z[x]$, we would get $P(k)\in\Bbb Z$, so the only $a$ that can be written as $a=P(k)$ would be integers, so the minimal-degree polynomial in this situation is really just the constant polynomial $P(x)\equiv a$. This continues to be the case if $k\in\Bbb Z$, so there's no need to iteratively use modular arithmetic to find $P(x)$.
|
H: Solving method for fractions having Prime numbers as Denominator.
I have a question , like
$30 = 1.18x$ , and $30 = 0.82 y$ , find $x+y$.
when solving, $x=1500/59$, $y=1500/41$. Since the denominator is involving prime numbers I was wondering is there any method or some logic is there to solve this question under 30 seconds.?
The actual question goes like this :
A man sold two gifts at $\$30$ each. On one gift he gained $18\%$ and on the other gift he lost $18\%$. What is his overall gain/loss(in $)
AI: Continuing from an4s's answer:
$$\frac{60}{1 - 0.18^2} = \frac{60}{1 - \left(\frac{9}{50} \right)^2} = \frac{60}{1 - \frac{81}{2500}} = \frac{60}{\frac{2500 - 81}{2500}} = \frac{60 \cdot 2500}{2500 - 81} = \frac{150 \ 000}{2419}.$$
Then using the options in the multiple-choice question:
$$\frac{150 \ 000}{2500} < \frac{150 \ 000}{2419} < \frac{150 \ 000}{2400}$$
$$\frac{150}{25} < \frac{150 \ 000}{2419} < \frac{500}{8}$$
$$60 < \frac{150 \ 000}{2419} < 62.5$$
Since the gifts are between $60$ and $62.5$ dollars, he has therefore made a loss. Then the answer must be between options $B$ and $C$. However, since his loss is strictly less than $60 - 62.5 = -2.5$ dollars, then the answer is option $B$.
|
H: (Verification) If $g \circ f$ is injective, then $f$ must also be injective.
Prove that if $g \circ f$ is injective, then $f$ must also be injective.
$g(f(x))$ is injective.
Then, $x\neq x’ \implies (g\circ f)(x) \neq (g\circ f)(x’)$.
Suppose for the sake of contradiction: $f(x)=f(x’) \nRightarrow x=x’$
Then this implies that it is logically possible that for some $x$ and $x’$:
$f(x)=f(x’) \implies (g\circ f)(x) \neq (g\circ f)(x’)
\implies g(f(x))\neq g(f(x’))$
But if $f(x)=f(x’)$, then $f(x)=f(x’) \implies g(f(x))\neq g(f(x))$ which is a contradiction.
Therefore $f(x)=f(x’) \implies x=x’$, that is, $f$ is also injective.
I’ve numbered the lines to make it easier to reference a mistake if I’ve made one. My concern is (3.). I think it should be correct to say that if (2.) holds, then that suggests that it is logically possible that $x\neq x’$ could hold under $f(x)=f(x’)$.
AI: The proof is, in principle, correct, but all the use of logic symbols are wrong. Proof: Suppose that $f$ is not injective. Then for some $x\ne x'$ we have $f(x)=f(x')$. But then $g(f(x))=g(f(x'))$, so $g\circ f$ is not injective, a contradiction.
|
H: Why is my integrand wrong?
I am given the following integral. $$\iiint\limits_E \sqrt{x^2+z^2} \ dV$$ bound by the paraboloid $y=x^2+z^2$ and the plane $y=9$.
Half way through the problem, it says the given integral can be simplified to $$\int_{-3}^3\int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} {\left(9-x^2-z^2\right)\sqrt{x^2+z^2}} \ dz \ dx$$. I understand how the integral has made it to this point.
However, it says that it is easier to solve by converting to polar coordinates in the xz-plane: $x=r\cos{\theta}$ and $z=r\sin{\theta}$.
This is where my understanding ends. This is the end part of the question.
$$\iiint \sqrt{x^2+z^2} \ dV = \iint (9-x^2-z^2)\sqrt{x^2+z^2} \ dA
\\ = \int_0^{2\pi}\int_0^3 {\mathrm{integrand \ here}} \ dr \ d\theta$$
This is the part where I am not able to get the question right. Substituting $x=r\cos{\theta}$ and $z=r\sin{\theta}$ in the integral, I get the integrated: $\left(9-r^2\right)\cdot\sqrt{r^2} = r\cdot \left( 9-r^2\right)$ as $r\geq 0$.
Where am I wrong in getting $r\cdot \left( 9-r^2\right)$ for the integrand?
AI: You need to multiply r to the integrand because $\mathrm{d}A=r\mathrm{d}r\mathrm{d}\theta$, which would make your integrand to $-\left(r^3-9r\right)\cdot r$.
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H: When is a Lie group action is a local diffeomorphism?
My whole question is actually in the title. Suppose a Lie group $G$ acting smoothly on a smooth manifold $M.$ I am looking for the conditions on the group action or the Lie group $G$ that makes $M$ and $g(M)$ are locally diffeomorphic for all $g\in G.$ Perhaps this is the case for all Lie group actions?
I was pondering about this question sometimes but couldn't get in to any conclusion. This might be obvious for the experts and people with sufficient background.
AI: If I'm interpreting your (edited) question correctly, the answer is yes: For any element $g\in G$, $\theta_g:M\to M$ is smooth, since the action itself is smooth, and has a smooth inverse $\theta_{g^{-1}}:M\to M$, since $\theta_{g^{-1}}\circ\theta_g=\theta_{g^{-1}g}=\theta_e=\text{id}_M$. Thus each $\theta_g$ is a global diffeomorphism.
|
H: Given the position vectors of the vertices of a triangle, prove that another point is the orthocentre of the triangle.
The original problem is:
If a, b, c, d are the position vectors of points A, B, C, D respectively such that $$(\vec{a}-\vec{d}). (\vec{b}-\vec{c})= (\vec{b}-\vec{d}). (\vec{c}-\vec{a})= 0$$then prove that D is the orthocentre of ${\Delta}$ ABC.
How do we go about proving that a point is the orthocentre of a triangle? I've tried expanding the dot product but I don't seem to get anywhere.
AI: That formula states that $AD\perp BC$ (so that $D$ is on the altitude from $A$ to $BC$)
and that $AC\perp BD$ (so that $D$ is on the altitude from $B$ to $AC$).
As $D$ is on two altitudes of the triangle, it is its orthocentre.
|
H: The region of convergence of the series $\sum_{n=1}^{\infty} \frac{(-1)^n n (x-1)^n}{(n^3+1)(3^n+1)}$.
I am trying to find the region of convergence of the series
$$
\sum_{n=1}^{\infty} \frac{(-1)^n n (x-1)^n}{(n^3+1)(3^n+1)}.
$$
I know that, by the limit comparison test,
$$
\sum_{n=1}^{\infty} \frac{n (x-1)^n}{(n^3+1)(3^n+1)}
$$
converges if and only if
$$
\sum_{n=1}^{\infty} \frac{(x-1)^n}{n^2 3^n}
$$
converges, and, by the Leibniz criterion,
$$
\sum_{n=1}^{\infty} \frac{(-1)^n (x-1)^n}{n^2 3^n}
$$
converges if $|x-1|\le 3$. But I do not know how the convergence of the original series follows from these two facts.
AI: https://en.wikipedia.org/wiki/Power_series check "Radius of convergence" here your $a_n=\frac{n(-1)^n}{\left(n^3+1\right)(3^n+1)}$. You just find $r=3$ but notice that your center here is $x_0=1$(and not $x_0=0$) so you demand $|x-1|<3$ and you check in the case $|x-1|=3$ if the series converges.
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H: Find the limit: $\lim_{x\to 3^-} \dfrac{(e^{(x+3)\ln 27})^{\frac{x}{27}} -9}{3^x-27}$
The expression simplifies to
$$\lim_{x\to 3^-}\dfrac{(27^{x+3})^{x/27} -9}{3^x-27}=\lim_{h\to 0} \frac{27^{(6-h)(3-h)/27} -9}{3^{3-h}-27}.$$
I simplified the denominator as follows: $\;27(3^{-h}-1)=-27h \ln 3$.
How should I simplify the numerator now?
AI: $$\lim_{h\to 0}\frac{27^{\frac{(6-h)(3-h)}{27}}-9}{3^{3-h}-27}=\lim_{h\to 0}\frac{27^{\frac{18-h^2-9h}{27}}-9}{27(3^{-h}-1)}$$
$$=\lim_{h\to 0}\frac{\left(9\cdot 27^{\frac{-h^2-9h}{27}}-9\right)}{27(3^{-h}-1)}$$
$$=\frac13\lim_{h\to 0}\frac{\left( 27^{\frac{-h^2-9h}{27}}-1\right)}{(3^{-h}-1)}$$
$$=\frac13\lim_{h\to 0}\frac{\left( 27^{\frac{-h^2-9h}{27}}-1\right)}{\frac{-h^2-9h}{27}}\cdot \frac{(-h)}{3^{-h}-1}\cdot \frac{\frac{h^2+9h}{27}}{h}$$
$$=\frac13\cdot \ln 27\cdot \frac{1}{\ln 3}\cdot \frac13$$
$$=\frac{3\ln3 }{9\ln 3}=\color{blue}{\frac13}$$
|
H: Continuity of a map between a product space equipped with product topology
Consider $[n]=\{1, 2, \cdots, n\}$ with the discrete topology and let $X=\prod_{n\geq 1}[n]$ be the product space with the product topology. For $x=(a_1, a_2, \cdots)$, define $T(x)=(1, a_1, a_2, \cdots)$. Then how to show the following?
The map $T:X\to X$ has a unique fixed point
The map $T:X\to X$ is continuous
For part 1, let $T(x)=x$, then $(a_1, a_2, \cdots)=T(x)=(1, a_1, a_2, \cdots)$ which means $a_1=1$, $a_2=a_1=1$, $a_3=a_2=1, \cdots$. Therefore, $x=(1, 1, 1, \cdots)$ is the unique fixed point.
But I don't know how to prove the second part. Please help how to initiate? Thanks in advance.
AI: Consider $U$ a basic open subset of the product space. That means that $U$ is the product of open subsets of the spaces $[n]$, from which though, only finite of them are not equal to the respective whole space $[n]$.
Observe that a product $[1]=\{1\}$ so every open set in the product topology is of the form $\{1\}\times U_2 \times U_3 \times \cdots$, where $U_n$ is on open set in $[n]$.
Now also observe that $T(a_1, a_2,\ldots)=(1,1,a_2,...)$. Thus if $U_2$ is an open set that doesn't contain $1$, i.e. $U_2=\{2\}$, then the inverse image of $U$ is the empty set.
If $U_2$ contains $1$, the inverse image of $U$ is the set $V=\{1\} \times U_3 \times U_4\times \cdots$.
In both cases the inverse image is an open set.
|
H: How to find the critical index $a$ of $x^af(x)$?
Let $f\in C^1(0,+\infty)$, $f(x)> 0$ in $(0,+\infty)$, $f(0+)=+\infty$, and $f$ is decrased in $(0,+\infty)$. Consider $g(x)=x^af(x)$, $(a>1)$. Now one can show that $\liminf_\limits{x\rightarrow0^+}g(x)=\limsup_\limits{x\rightarrow0^+}g(x)$, so we write $\lim_\limits{x\rightarrow0^+}g(x)>0$ to describe both $\lim_\limits{x\rightarrow0^+}g(x)=c>0$ and $\lim_\limits{x\rightarrow0^+}g(x)=+\infty$. Attention, for we only care about what happen near $x=0$，so we can weaken $(0,+\infty)$ to be $(0,\delta)$ and can smoothen $f$ with a smooth function when $x>\delta$.
Problem: Given the $f$ as above, define $g_a(x)=x^af(x)$,$ (a>1)$. If
(1) there exists $\alpha>1$ such that $\lim_\limits{x\rightarrow0^+}g_\alpha(x)>0$,
(2) there exists $\beta>\alpha$ such that $\lim_\limits{x\rightarrow0^+}g_\alpha(x)=0$.
Please show that there exists $\gamma\in[\alpha,\beta)$ such that
(i) $\lim_\limits{x\rightarrow0^+}g_\gamma(x)>0$,
(ii) for any $\epsilon>0$, $\lim_\limits{x\rightarrow0^+}g_{\gamma+\epsilon}(x)=0$.
That is, can one find the "critical index"?
Attempt, example and some information: There are many examples for the proposition:
(a) $f(x)=\frac{1}{x^2}\ln(1+\frac{1}{x})$, then $\lim_\limits{x\rightarrow0^+}g_2(x)>0$ and $\lim_\limits{x\rightarrow0^+}g_3(x)=0$, we can find the "critical index" to be $a=2\in[2,3)$
(b) $f(x)=x^{-2+e^x}\ln(1+\frac{1}{x})$ satisfies our conditions in $(0,\text{small}~\delta)$. One can easily find out the "critical index" is $a=1$.
I think the difficulty of this problem is that "if $a$ make $\lim_\limits{x\rightarrow0^+}g_a(x)=0$, then can one show there exist a small $\epsilon>0$ such that $\lim_\limits{x\rightarrow0^+}g_{a-\epsilon}(x)=0$?". I can't show this can you help me? Or this proposition is wrong and you can find the counterexamples? But I want this proposition to be true, does it need to add any extra conditions for $f(x)$?
If you need you can add the condition $|f'(x)|\leq C \frac{f(x)}{x}$. In fact these conditions for $f$ is given by other propositions and theorem. I just summarized it.
2020/7/26 Addition (you can choose to answer or just have a look): One gives the counterexample in the answer, so if I still want this proposition to be correct I must add some conditions to $f$. In fact, this is relative to this problem: estimate a integral with parameter
The problem in the link is one example of a kind of integrals with parameter, if one puts $u=1/r$, then it become the problem here. So the $f$ here should be like $\frac{1}{x^m}(\text{log term, arctan term})$ (the example (a)(b) above), and $\frac{-1}{x^2\log x}$ will not appear. So what can I give the conditions to $f$ such that the proposition can be ture? (I discussed with my friends and we turns out that, it's difficult to describe straight $f$ have such type $\frac{1}{x^m}(\text{log term, arctan term})$, so we want a more general lemma to due with it. )
AI: (I don't know if this answers your question, but it can be an hint.)
You can define the function $\lambda\colon [1,+\infty)\to [0,+\infty]$ by
$$
\lambda(a) := \limsup_{x\to 0+} x^a f(x).
$$
Since $g_a > g_b$ if $a < b$, we clearly have that $\lambda$ is monotone non-increasing.
By assumption, $\lambda(\alpha) > 0$ and $\lambda(\beta) = 0$ for some
$1 < \alpha < \beta$.
Let
$$
\gamma := \inf\{a\geq 1:\ \lambda(a) = 0\}.
$$
This "critical" value $\gamma$ has the following properties:
(1) $\lambda(a) = +\infty$ for every $a < \gamma$;
(2) $\lambda(a) = 0$ for every $a > \gamma$.
On the other hand, I think that $\lambda(\gamma)$ can be every element of $[0,+\infty]$ (depending on the choice of $f$).
Some examples:
(1) If you take $f(x) = - \log(x) / x^2$, you have that $\lambda(a) = 0$ if $a > 2$ and $\lambda(a) = +\infty$ if $a \leq 2$.
In this case, $\gamma = 2$ and $\lambda(\gamma) = +\infty$.
(2) If you take $f(x) = - \frac{1}{x^2 \log x}$, then again $\gamma = 2$, but in this case $\lambda(\gamma) = 0$.
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H: What is the supremum, infimum, maximum and minimum of the set $B:=\{x \in\mathbb{R}, \left||x-1|-|x-2|\right|<1\}$?
What is the supremum, infimum, maximum and minimum of the set
$$B:=\{x \in\mathbb{R}, \left|\left|x-1\right|-\left|x-2\right|\right|<1 \}?$$
I am not sure how to find any of them. I thought I have to find for which values is the equation right and then from the values I could have some interval for which I could say what is the min, max, sup and inf.. But whatever I put for $x$ I get that $1 < 1$.
AI: If $x \le 1$ then $$|1-x + x - 2|\lt 1 \implies 1 \lt 1$$Which is a contradiction. If $1\lt x \le 2$ then $$|x-1 + x - 2| \lt 1 \implies 1 \lt x \lt 2$$And if $x\gt 2$ then $$|x - 1 - x + 2| \lt 1 \implies 1 \lt 1$$
So the answer is $A = \{x \in \mathbb{R} | 1\lt x \lt2\}$.
I think you can find the values that you've mentioned then.
|
H: Derive $\mathbf b$ from $\mathbf a = \mathbf b× \mathbf c$
I have an equation:
$$\mathbf a = \mathbf b × \mathbf c,$$
where $\mathbf a$ $\mathbf b$ and $\mathbf c$ are 3-vectors.
How could I derive $b$ from the equation and express it in terms of $\mathbf a$ and $\mathbf c$?
AI: You can't. Given $c$ and a vector $a$ perpendicular perpendicular to $c$ choose $b$ in the plane perpendicular to $a$. Then $b\times c$ lies in the direction of $a$ and you want $|a|=|b||c|\sin \theta$ where $\theta$ is the angle between $b$ and $c$ taken in the correct direction.
So then all you need is $|b|\sin \theta = \frac {|a|}{|c|}$. If you choose $\theta$ then the length of $b$ is fixed, but you can choose any non-zero $\theta$, so $b$ is not unique and can't be determined.
|
H: Convergence of a Real number sequence
I want to prove if the following statement is true:
Let a sequence of real numbers $x_{m}\left(j\right)\to x\left(j\right)$ as $m \to \infty$, $ \forall j\in \mathbb{N}$. Then:
\begin{equation*}
\sum_{j \in \mathbb{N}} |x_{m}\left(j\right) - x\left(j\right)| \to 0 \text{ as } m\to \infty
\end{equation*}
Proof:
Since $x_{m}\left(j\right) \to x\left(j\right)$ as $m\to \infty$, this means that $\forall \epsilon>0$ $\exists m_{0} \in \mathbb{N}$ such that $|x_{m}\left(j\right) - x\left(j\right)|<\epsilon$ for all $m\ge m_{0}$.
For each $j\in \mathbb{N}$ we chose $\frac{\epsilon}{2^{j}}$. Then:
\begin{equation*}
0 \le \sum_{j \in \mathbb{N}} |x_{m}\left(j\right) - x\left(j\right)| < \sum_{j \in \mathbb{N}} \frac{\epsilon}{2^{j}} = \epsilon
\end{equation*}
which is the thesis.
I have big doubs about the last line: I don't know if it is logically correct or not. I quite sure that I can bound each term of the sum for a $j$ fixed and for each of them there exists an $m_{0}$ such that the bound holds. Is this still true when the term becomes infinite? Could anyone explain me the concept? Does it follow from the definition of convergence of series of real numbers?
AI: You are trying to prove something that is not even true. Here is a counter-example: Let $x_m(j)=1$ if $m=j$ and $0$ if $m \neq j$. Let $x_j=0$ for all $j$. Can you verify that this is a counter-example?
|
H: $H$ and $K$ are normal subgroups of a group $G$ with $K \le H$, then $aK =H \iff a \in H$
Let $H$ and $K$ are normal subgroups of a group $G$ with $K \le H$.
How can I prove that $aK =H \iff a \in H$?
I‘m stuck with the part when $a \in H$ then $aK$ contains $H$.
Given $u$ in $H$, how can I show that $u \in aK$ ？
Thanks for helping.
AI: This should work as a a counterexample:
Let $G = \mathbb Z$, $H = 2\mathbb Z$ and $K = 4\mathbb Z$. Since $G$ is abelian, $K, H \trianglelefteq G$, and $K\le H$. Now we can choose $a \in H$, say $a = 2$, and consider $aK = 2 + 4\mathbb Z$. This doesn't contain $H$, as for example $4 \in H$ but $4 \notin aK$.
|
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