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Solving simultaneous equations with complex coefficients using real methods My circuits analysis textbook teases that there's a way to convert a set of n complex equations into a set of 2n real equations, which can then be solved using any calculator that can solve real simultaneous equations. That is, no capability with complex numbers needed.
e.g.:
$(25 +j100)I_1 - (10+j80)I_2=100\angle0^\circ\\$ (1)
$-(10+j80)I_1+(30+j190)I_2=0$ (2)
I say "teases" because they point me to their website, where after a lengthy sign-up process, I find that the material isn't actually there.
Does anybody know what method they're referring to?
I know how to do this with Cramer's Rule, but that requires a matrix calculator that understands complex numbers (they exist but they're not common).
Here's the page:
https://books.google.com.au/books?id=VLbycoxwas8C&pg=PA959&lpg=PA959&dq=%22Solving+Simultaneous+Equations+with+Complex+Coefficients+Using+Any+Calculator%22&source=bl&ots=Bf9PJRGo3o&sig=B-ssojUUL4fnJXxOFu6VMw0vP_0&hl=en&sa=X&ei=rAekU5_xE5Tr8AWpqoEQ#v=onepage&q=%22Solving%20Simultaneous%20Equations%20with%20Complex%20Coefficients%20Using%20Any%20Calculator%22&f=false
|
The main idea is to split each equation into a real and a complex part.
To easily see how to do this take a look at complex multiplication as a linear transformation.
$(a + bi) * (c + di) = (ac - bd) + (bc + ad)i$ will become
$\left(\begin{array}{cc}c&-d\\d&c\end{array}\right)
\left(\begin{array}{c}a\\b\end{array}\right)=
\left(\begin{array}{c}ac-bd\\bc+ad\end{array}\right)$
You can use this pattern to rewrite your example as
$\left(\begin{array}{cccc|c}
25&-100&-10&80&100\\
100&25&-80&-10&0\\
-10&80&30&-190&0\\
-80&-10&190&30&0
\end{array}\right)$
|
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|
Derivative function went wrong I am trying to take the derivative of this function but I am facing some difficulties.
$$f(x)= e^{\ln(e^{7x^2+11})}$$
My answer was : $7e^{(7(x^2))}*14x$
I cancelled the $\ln$ with the $e$ first, then I downgrade the $7$ and keep the $\exp$. as it is, after that I took the derivative of the $7x^2$ and the result was the one on top.
|
$$f(x)= e^{\ln(e^{7x^2+11})}=\exp(\ln(\exp(7x^2+11)))$$
So
$$f(x)= \exp(7x^2+11), \implies f'(x)=14x\exp(7x^2+11)$$
|
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|
$ 7\mid x \text{ and } 7\mid y \Longleftrightarrow 7\mid x^2+y^2 $
Show that
$$ 7\mid x \text{ and } 7\mid y \Longleftrightarrow 7\mid x^2+y^2 $$
Indeed,
First let's show
$7\mid x \text{ and } 7\mid y \Longrightarrow 7\mid x^2+y^2 $
we've $7\mid x \implies 7\mid x^2$ the same for $7\mid y \implies 7\mid y^2$ then
$ 7\mid x^2+y^2 $
*
*Am i right and can we write $a\mid x \implies a\mid x^P ,\ \forall p\in \mathbb{N}^*$
Now let's show
$7\mid x^2+y^2 \Longrightarrow 7\mid x \text{ and } 7\mid y$
$7\mid x^2+y^2 \Longleftrightarrow x^2+y^2=0 \pmod 7 $
for
\begin{array}{|c|c|c|c|c|} \hline
x& 0 & 1 & 2& 3 & 4 & 5 & 6 \\ \hline
x^2& 0 & 1 & 4& 2 & 2 & 4 & 1 &\pmod 7\\ \hline
y& 0 & 1 & 2& 3 & 4 & 5 & 6 \\ \hline
y^2& 0 & 1 & 4& 2 & 2 & 4 & 1 & \pmod 7 \\ \hline
\end{array}
which means we have one possibility that $x=y= 0 \pmod 7 $
*
*Am I right and are there other ways?
|
$►$ If $x=7x_1$ and $y=7y_1$ then $x^2+y^2=7(x_1^2+y_1^2$).
$►$ If $x^2+y^2\equiv0\pmod7\iff x^2\equiv -y^2\pmod7$ then because of $\mathbb F_7^2=\{1,4,2,0\}$ and $(-1)\mathbb F_7^2=\{6,3,5,0\}$ the only possibility for
$x^2\equiv -y^2\pmod7$ is that both $x^2$ and $y^2$ are equal to $0$ modulo $7$ so $x$ and $y$ are equal to $0$ modulo $7$ (since $7$ is prime). Then $x\equiv y\equiv0\pmod7$.
|
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|
Soft sheaves adapted to $f_!$ I'm reading Gelfand-Manin, Homological Algebra. I understand that the class of soft sheaves is sufficiently large, because every injective sheaf is soft. Now to see that this class is adapted to $f_!$, I have to show that every acyclic complex of soft sheaves is mapped by the functor $f_!$ to an acyclic complex.
For this, I'd say it's enough to just show that if
$0\rightarrow \mathcal{F}\rightarrow \mathcal{G}\rightarrow\mathcal{H}\rightarrow 0$
is an exact sequence of soft sheaves, then $0\rightarrow \mathcal{f_!F}\rightarrow \mathcal{f_!G}\rightarrow\mathcal{f_!H}\rightarrow 0$ is also exact.
But in the book, they say the above is only enough because of an exercise from earlier which says that in an exact sequence of sheaves $0\rightarrow \mathcal{F}\rightarrow \mathcal{G}\rightarrow\mathcal{H}\rightarrow 0$ , if $\mathcal{F}$ and $\mathcal{G}$ are soft, then so is $\mathcal{H}$.
Can somebody tell me where this exercise is needed in the proof of the adaptedness of soft sheaves to $f_!$ ?
|
It is not needed in the proof, it just makes it easier. Because of that exercise you only need to prove that $f_!$ is right exact.
|
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If $f:\mathbb{R}^n \times \mathbb{R}^m \to \mathbb{R}^p$ is a bilinear function, then how to show that If $f:\mathbb{R}^n \times \mathbb{R}^m \to \mathbb{R}^p$ is a bilinear function, then how to show that $$\lim\limits_{(h,k) \to (0,0)} \dfrac{|f(h,k)|}{|(h,k)|} = 0$$.
|
For simplicity, I'll be using the euclidean norm for $\Bbb R^n$, $\Bbb R^m$ and $\Bbb R^n\times\Bbb R^m$, and the $\sup$ norm for the target space $\Bbb R^p$. The choice of norms is irrelevant to the result since all norms are equivalent in finite dimension.
There are coefficients $a_{ij;\,l}\in\Bbb R$, where $(i,j)$ ranges over $\lbrace 1,\dots,n\rbrace\times\lbrace 1,\dots,m\rbrace$ and $k\in\lbrace 1,\dots,p\rbrace$ such that for all $h=(h_1,\dots,h_n)\in\Bbb R^n,k=(k_1,\dots,k_m)\in\Bbb R^m$, the $l$-th coordinate of $f(h,k)$ is equal to
$$\left[f(h,k)\right]_l=\sum_{i,j}a_{ij;\,l}h_ik_j$$
If you let $A$ be the maximum of the absolute values $|a_{ij;\,l}|$, then for all $l$,
$$|\left[f(h,k)\right]_l|\leq A\sum_{i,j}|h_ik_j|\leq \frac12A\sum_{i,j}h_i^2+k_j^2=\frac12A(m|h|^2+n|k|^2)\leq\frac12A(m+n)|(h,k)|^2$$
Thus, for all $(h,k)\neq(0,0)$
$$\frac{|f(h,k)|}{|(h,k)|}\leq C|(h,k)|$$
for $C=\frac12A(m+n)$. The right hand side tends to zero as $(h,k)$ tends to zero.
|
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Proving $\limsup\frac 1 {a_n}=\frac 1 {\liminf a_n}$ and $\limsup a_n\cdot \limsup \frac 1 {a_n} \ge 1$
Let $a_n$ be a sequence such that $\forall n\in \mathbb n: 0<a\le a_n\le b <\infty.$
Prove:
*
*$\displaystyle\limsup_{n\to\infty}\frac 1 {a_n}=\frac 1 {\displaystyle\liminf_{n\to\infty}a_n}$
*$\displaystyle\limsup_{n\to\infty}a_n\cdot \limsup_{n\to\infty}\frac 1 {a_n} \ge 1$ and there's an equality iff $a_n$ is converging.
*
*Suppose there are two subsequences: $a_{n_l}, \ a_{n_k}$ such that $\lim a_{n_k} = k, \ \lim a_{n_l}=l$ and suppose $l\le k$, so $\lim \frac 1 {a_{n_k}}=\frac 1 k , \ \lim \frac 1 {a_{n_l}}=\frac 1 l$ so clearly: $\frac 1 k\le \frac 1 l\le l\le k$ so it's easy to see once the largest limit (supermum) is 'inverted' it has to become the smallest limit (infimum).
I realize this doesn't show equality, I don't know how to do the other way and I'm not even sure if what I did is good.
*If $a_n$ converges, suppose to $L$ as its limit then we have: $L\cdot \frac 1 L=1$.
If it does not converge then $a_n$ may tend to infinity or won't have a limit. From 1 we can change it to $\displaystyle\limsup_{n\to\infty}a_n\cdot \frac 1 {\displaystyle\liminf_{n\to\infty}a_n} \ge 1$ and from BW, every sequence has a converging subsequence, and since for converging subsequences: $\liminf a_n\le \limsup a_n$ we have $\frac {\limsup a_n} {\liminf a_n}=\limsup a_n\cdot \limsup\frac 1 {a_n} \ge 1$.
This should probably be in absolute value since one of those subsequnce limits can be negative, but it isn't in absolute value in the question.
|
To get the first one, you just need to notice that
$$\sup\{1/{a_k}; k\ge n\} = \frac1{\inf\{a_k; k\ge n\}}$$
and take limit for $n\to\infty$ to get
$$\lim_{n\to\infty}\sup\{1/a_k; k\ge n\} = \lim_{n\to\infty}\frac1{\inf\{a_k; k\ge n\}} = \frac1{\lim_{n\to\infty}\inf\{a_k; k\ge n\}}.$$
(Although you should also check the cases when the LHS is $+\infty$ and when you have zero in the denominator.)
Once you have shown the first part, you get
$$\limsup_{n\to\infty} \frac1{a_n} = \frac1{\liminf_{n\to\infty} a_n} \ge \frac1{\limsup_{n\to\infty} a_n}.$$
Then you can simply multiply this by $\limsup\limits_{n\to\infty} a_n$.
Again, you should check separately the cases where some of the above values is $+\infty$ or when you divide by zero.
|
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Möbius transformation: proving the image of the unit circle is a line Problem
1) Find the Möbius transformation which maps the points $0,i,-i$ to $0,1,\infty$ respectively.
2) Prove that the image of the circle centered at $0$, of radius $1$ is the line $\{Re(z)\}=1$.
In $1)$ I didn't have problems, the homographic transformation $T(z)$ which satisfies the conditions given is $T(z)=\dfrac{2z}{z+i}$.
I don't know how to solve $(2)$. If I denote the circle by $C$, I want to show that $T(C)=\{Re(z)=1\}$. I've tried to prove the two inclusions of these sets but I couldn't, I would appreciate some help.
|
An element of $C$ is $e^{it}$ with $t\in\mathbb{R}$:
$$T(e^{it})=\frac{2e^{it}}{e^{it}+i}=\frac{2e^{it}(e^{-it}-i)}{2+2\sin(t)}=\frac{2-2ie^{it}}{2+2\sin(t)}=\frac{1-i(\cos(t)+i\sin(t))}{1+\sin(t)}=$$
$$=\frac{1+\sin(t)-i\cos(t)}{1+\sin(t)}=1-i\frac{\cos(t)}{1+\sin(t)}\ .$$
This is a parametrization of the line $Re(z)=1$.
|
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Why do some sources call calculus, "the calculus"? No need to cite specific sources since I think it's a fairly common thing to see. What's up with that?
Thank you
Edit: I've seen it in several places. Here's where I'm currently looking at it at:
Calculus: An Intuitive and Physical Approach (Second Edition): Here on amazon. Read the sample provided by Amazon and you'll see it several times in the first chapter.
|
Because 'calculus' meant a set of rules for calculating things whereas 'the calculus' meant 'the infinitesimal calculus'. The qualifier was lost in the academic war over the foundations of the subject.
|
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Understanding trig interval I have kind of a random question I'm hoping someone could help me with.
So I was thinking about the interval $[-\pi, \pi]$ for a trig functions. Isn't this is the same interval as $[0, 2\pi]?$ The reason why I say that (and maybe this is where my confusion is) is because couldn't $[-\pi, \pi]$ be split as $[-\pi, 0] \cup [0, \pi],$ which when drawn the angle of rotation hits all the vital points on the unit circle just like $[0, 2\pi]?$
Thanks in advance!
|
I feel it would depend on the specific trig function only if you include hyperbolic.
|
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Related to the construction of $\Bbb C$ (generalisation) To construct $\Bbb C$, we consider $\Bbb R^2$ endowed with the operations: $$\begin{align} (a,b) + (c,d) &:= (a+c, b+d) \\ (a,b) \cdot (c,d) &:= (ac - bd, ad+bc)\end{align} $$
then write $(0,1_{\Bbb R}) = i$, go on writing $(a,b)$ as $a+ib$, etc. Maybe the question is silly and has a trivial explanation, but nevertheless, I'll ask: Has anyone ever tried to repeat the procedure with $\Bbb C^2$, "nesting imaginary units"?
More specifically, I mean, define on $\Bbb C^2$ the operations: $$\begin{align} (z_1,z_2) + (w_1,w_2) &:= (z_1 + w_1, z_2+w_2) \\ (z_1,z_2) \cdot (w_1,w_2) &:= (z_1w_1 - z_2w_2, z_1w_2+z_2w_1)\end{align}$$
then write $(0,1_{\Bbb C}) = j$, where $j$ is another "imaginary unit" such that $j^2 = -1_{\Bbb C}$, write $(z_1,z_2) = z_1 + jz_2$ and so on? (I'm just using this subscript $\Bbb C$ for emphasis). It seems to me that we would get $i^2 = j^2$, but that wouldn't mean necessarily that $i = j$, would it? Also, would be this related in any way to the quaternions?
This, way, if we can make $\Bbb C^2$ a field, we could make any $\Bbb R^{2n}, n \in \Bbb Z$ a field, by repeating the process, no? I don't know if any problem would appear if I expanded everything in terms of the real and imaginary parts of every component. We would get some mixed terms $ij$ and $ji$. Surely I could adventure myself in the calculations, but if someone already thought of this, and it ended up being meaningless, I won't keep hitting my head on the wall. Thank you for the attention, and I hope I managed to get my idea through.
|
Close. To get the complexes, use real matrices
$$
\left( \begin{array}{rr}
a & b \\
-b & a
\end{array}
\right) .
$$
To get the quaternions, use complex matrices
$$
\left( \begin{array}{rr}
\alpha & \beta \\
-\bar{\beta} & \bar{\alpha}
\end{array}
\right) .
$$
|
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How to show that no. of elements $x$ of group $G$ such that $x^3=e$ is odd? Let G be a finite group G. Then How can I show that no. of elements $x$ of group $G$ such that $x^3=e$ is odd ?
I read this question in an Algebra book. Since $e^3=e$, e must be one of those elements. But how to find for non trivial elements ?
|
We claim that the number of elements in $T = \{g\in G: g^3 = e, g\neq e\}$ is an even number $N$. The claim follows from recalling that $e^3 = e$, hence the number of elements with trivial cubes is $N+1$, an odd number.
In fact, if $g\in T$, then $\{g,g^2,g^4...\}\subset T$, so, since $T$ is finite, there is a minimal number $N = |T|$ such that $g^{2N} \in T$, and moreover $g^{2N} = g$ since $N$ is minimal. Suppose by contradiction that
$$
2^N-1 = 1 \bmod{3} \text{ or } 2^N-1 = 2 \bmod{3}.
$$
Then, respectively, $g^{2N} = g^2$ and $g^{2N} = e$, both which cannot be since $g \neq g^{-1} = g^2$ and $e\notin T$. Hence
$$
2^N-1 = 0\bmod{3}.
$$
By induction, this is true if $N$ is a non-zero even integer, since $2^2 -1 = 3$ and
$$
2^N = 1+3k \implies 2^{N+2} = 1 + 3(k+1).
$$
Moreover, it follows that $2^{2N+1}-1 = 2 \bmod{3}$, hence $N$ is even. This completes the proof.
|
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Why is $(\sec x)' = \tan x\sec x$ and not $\tan x$? As far as I understood, the Fundamental Theorem of Calculus states that the integral of a function is its anti-derivative. And yet, although the integral of $\tan x$ is $\sec x$, the derivative of $\sec x$ is $\tan x\sec x$.
I understand the calculation and you get $\tan x\sec x$ as the derivative, but how does it make sense in light of the fundamental theorem? What am I missing here?
|
The actual anti-derivative of $\tan{x}$ is:
$$\int\tan{x}\,\mathrm{d}x=\int\frac{\sin{x}}{\cos{x}}\,\mathrm{d}x=\int\frac{-\mathrm{d}(\cos{x})}{\cos{x}}=-\ln{(\cos{x})}+\text{constant}.$$
This gives us the definite integral,
$$\int_{0}^{x}\tan{u}\,\mathrm{d}u=-\ln{(\cos{x})}=\ln{\left(\frac{1}{\cos{x}}\right)}=\ln{(\sec{x})}.$$
Applying the fundamental theorem of calculus to this integral gives us:
$$\begin{align}
\frac{d}{dx}\int_{0}^{x}\tan{u}\,\mathrm{d}u&=\frac{d}{dx}\ln{(\sec{x})}\\
\implies \tan{x}&=\frac{\frac{d}{dx}(\sec{x})}{\sec{x}}.
\end{align}$$
Multiplying both sides by $\sec{x}$ yields:
$$\frac{d}{dx}(\sec{x})=\tan{x}\sec{x}.$$
|
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Two children paradox: where is my reasoning wrong? I hope here is the good place to be asking this. Apologies otherwise.
The statement is as follows:
"Ms Michu has two children. We know one of the two is a girl, we call that girl Ludivine. What is the probability that Ludivine has a brother, rather than a sister?"
This is with the usual assumptions that there is an equal chance for one person to be a boy or a girl, no cis or gemels, etc.
I read that the answer should be 2/3. Demonstration is :
For any given set of 2 children, there are 4 different equiprobable combinations:
GG BB BG GB
We get rid of the bb combination because there is no girl. That leaves us with 3 possibilities, of which 2 match the criteria of the statement, hence the probability of Ludivine having a brother is of 2/3.
But, if I follow the following reasoning, I obtain a probability of 1/2:
Ludivine(L) is either the elder(e) or the youngest(y), with an equal probability of 1/2.
If she is the elder(e), it gives 2 combinations, with again a probability of 1/2 for each:
B(y) G(L)(e)
G(y) G(L)(e)
If she is the youngest we again have 2 equiprobable combination :
G(L)(y) B(e)
G(L)(y) G(e)
By combining those probabilities, we have 2 cases out of 4 where she has brother. Therefore she has a probability of 2/4 or 1/2 to have a brother.
Now I feel like my reasoning is wrong somewhere, but I can't see where or why, even though i went through quite a few topics on the subject; hence why I'm re-discussing that old topic...
Thanks in advance!
|
The probability of the situation you have described as "B(y) G(L)(e)" is $\frac{1}{4}$, as you have said. But the probability of "G(y) G(L)(e)" is $\frac{1}{8}$: probability that elder child is a girl, $\frac{1}{2}$; probability that younger child is a girl, $\frac{1}{2}$; probability that you give the name Ludivine to the elder of the two girls, $\frac{1}{2}$. Therefore these two events are not equiprobable.
|
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every continuous map $S^1 \rightarrow S^1$ can be extended to continuous map $B^2 \rightarrow B^2$ Let $S^1$ denote the unit circle, and $B^2$ denote the closed unit disk. I came across this question and got stuck:
Q:) Every continuous map $f :S^1 \rightarrow S^1$ can be extended to continuous map $B^2 \rightarrow B^2$. If $f$ is a homeomorphism, then we can choose the extension to be a homeomorphism also.
I tried the obvious approach, namely for $x\in B^2-{0},$ define the extension by $x\rightarrow ||x||.f(x/||x||)$, but this does not seem to work,or I cannot figure out why this might work. What am I missing?
Thanks in advance.
|
Riccardo's comment is still useful you just need to be clever in how you apply it. Every continuous map $f\colon S^1\to S^1$ can be extended to a map $\tilde{f}\colon S^1\to B^2$ by composing with inclusion $i\colon S^1\to B^2$ of the boundary, so
$$\tilde{f}=f\circ i$$
and then because $B^2$ is contractible, $\tilde{f}$ must be nullhomotopic, therefore the map can be extended to a map $B^2\to B^2$ because of the theorem which says that any map from the circle to a space is nullhomotopic if and only if that map can be extended to the disc.
For the second part, just note that the identity on the circle can be extended to a homemorphism of the disc (also the identity) and so, using the fact that a homeomorphism of the circle is homotopic to the identity on the circle (or reflection), you can easily form you map in pieces (do the outer annulus first via the relevant homotopy and then 'fill in the hole' with a disc which corresponds to the identity.)
|
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Help! Totally stuck up with this limit.. I don't even know where to start with this
$$\displaystyle\lim_{x \rightarrow \frac{\pi}{6}} (2+\cos {6x})^{\ln |\sin {6x}|}$$
Please help me out (Hints in the right direction would be appreciated)
|
When $f(x)\to1$ and $g(x)\to\infty$ then in order to find $\lim f(x)^{g(x)}$, first find $\log\lim f(x)^{g(x)} = \lim\log(f(x)^{g(x)})= \lim(g(x)\log f(x))=L$ and conclude that $\lim f(x)^{g(x)}=\exp L$.
In this case you I'd try L'Hopital's rule applied to $\dfrac{\log f(x)}{1/g(x)}$.
|
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Limit points of $\cos n$.
Find the limit point of the sequence $\{s_n\}$ given by $s_n=\cos n $.
I know by this post Limit of sequence $s_n = \cos(n)$ that the sequence does not converge. But I don't know how to search those points.
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The set $A=\{n+2\pi k:n,k\in\mathbb{Z}\}$ is dense on $\mathbb{R}$. Given a $y\in[-1,1]$ there existe an $x\in\mathbb{R}$ such that $\cos x=y$. Since $A$ is dense on $\mathbb{R}$ there existe a sequence $s_m=n_m+2\pi k_m$ of elements in $A$ such that $\lim\limits_{m\rightarrow\infty}s_m=x$. Then $$\lim\limits_{m\rightarrow\infty}\cos {n_m}=\lim\limits_{m\rightarrow\infty}\cos(n_m+2\pi k_m)=\cos \left(\lim\limits_{m\rightarrow\infty}(n_m+2\pi k_m)\right)=\cos x=y.$$ This implies that the limits points of $\cos n$ is all point in $[-1,1]$.
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Finite Group with Nilpotent Subgroup of Prime Power Index is Solvable Let $G$ be a finite group, and assume that $H$ is a nilpotent subgroup whose index is a prime power. WLOG, we can say that the index of $H$ is the highest power of $p$ which divides the order of $G$.
I want to show that $G$ is solvable, and I'm allowed to use Burnside's $p^aq^b$ theorem, as well as Hall's Theorem (which states that if every Sylow-$p$ subgroup has a complement, then $G$ is solvable).
So say $|G|=p^aq_1^{\alpha_1}\cdots q_n^{\alpha_n}$. Then since $H$ is nilpotent, it's the direct product of its Sylow subgroups, so $H=Q_1\times\cdots\times Q_n$. $H$ is obviously a complement for $G$'s Sylow-$p$ subgroup $P$, so I'll be done if I can show there's a complement to $Q_i$, the Sylow-$q_i$. Now, if we let $H_1=H/Q_i$, my first idea is to consider $PH_i$, which has the desired order, but of course there's no guarantee that this is a subgroup, since neither of the two subgroups need to be normal.
I also tried assuming that $G$ is a minimal counterexample and thus simple and showing that some nontrivial proper subgroup is normal, but since $p$ does not have to be the largest prime, I can't argue that the action of $P$ on the $Q_i$'s by conjugation is trivial.
Are either of these avenues fruitful, or should I try something else?
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The main argument in the character theoretic proof of Burnside's $p^aq^b$ theorem actually proves that a finite simple group cannot have a conjugacy class of prime power order bigger than $1$. In your problem, $H$ is nilpotent, so it has nontrivial centre. Then, for $1 \ne h \in Z(H)$, either $h \in Z(G)$, in which case $G$ is not simple, or $|{\rm Cl}_G(h)|$ is a nontrivial power of $p$, so again $G$ is not simple. Hence, by induction, $G$ is solvable.
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$f'$ is bounded and isn't continuous on $(a,b)$, so there's a point $y\in(a,b)$ such that $\lim_{x\to y}f'$ does not exist
Prove/disprove: $f$ has a bounded derivative and $f'$ isn't continuous on $(a,b)$, so there's a point $y\in(a,b)$ such that $\displaystyle\lim_{x\to y}f'$ does not exist.
I think that if $f'$ isn't continuous on the interval, then maybe we could have two disjoint sub-intervals, like for example $(a,c), (d,b)$ such that $d-c=\dfrac {a+b} 3$ so there's a substantial gap in the interval $(a,b)$ where $f'$ isn't defined so it follows that it won't have a limit there, for example: on $\dfrac {c+d}2$.
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The very fundamental thing one needs to observe here is that a derivative can't have jump discontinuity. If $f'(x) \to L$ as $x \to c$ then $f'(c) = L$ and thus $f'$ is continuous at $c$. Hence it is not possible for a derivative to have a limit at point and not to be continuous at that point. It follows that there will be points where limit of derivative $f'(x)$ does not exist.
Now to the proof of the statement mentioned in bold above. One proof is already given by leshik in his answer. Here is another proof based on mean value theorem. Suppose that $f(x)$ is differentiable in neighborhood of $c$. Suppose $x \to c$ and $f'(x) \to L$ as $x \to c$. Then we have $f(x) - f(c) = (x - c)f'(d)$ where $d$ lies somewhere between $c$ and $x$. Thus $$\frac{f(x) - f(c)}{x - c}=f'(d)$$ When $x\to c$ then LHS tends to $f'(c)$ and RHS tends to $L$ because $d \to c$. It follows that $f'(c) = L = \lim_{x \to c}f'(x)$. Hence $f'(x)$ is continuous at $c$.
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Is the set discrete?
Is the set $S=\{(m+\frac{1}{2^{|p|}},n+\frac{1}{2^{|q|}}):m,n,p,q\in \mathbb{Z}$} discrete in $\mathbb{R}^2$?
I'm not getting how shall I check discrete here?
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Hint: Take a particular point $P=\left(m+\frac{1}{2^p}, n+\frac{1}{2^q}\right)$. We want to show that the point is an isolated point. To do this, show that our set contains no point other than $P$ within distance $\min\left(\frac{1}{2^{p+2}},\frac{1}{2^{q+2}}\right)$ of $P$.
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Show that $\dfrac{\sqrt{8-4\sqrt3}}{\sqrt[3]{12\sqrt3-20}} =2^\frac{1}{6}$ This was the result of evaluating an integral by two different methods. The RHS was obtained by making a substitution, the LHS was obtained using trigonometric identity's and partial fractions.
Now I know that these two are equal, but I just can't prove it. I tried writing the LHS in terms of powers of $2$, but can't get any further to the desired result.
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We have
$$8 - 4\sqrt{3} = 2(4-2\sqrt{3}) = 2(3-2\sqrt{3}+1) = 2(\sqrt{3}-1)^2,$$
so $\sqrt{8-4\sqrt{3}} = \sqrt{2}(\sqrt{3}-1)$.
Then we see that looking at $(\sqrt{3}-1)^3$ is a good idea:
$$(\sqrt{3}-1)^3 = (4-2\sqrt{3})(\sqrt{3}-1) = 6\sqrt{3} - 10,$$
so $12\sqrt{3}-20 = 2(\sqrt{3}-1)^3$ and $\sqrt[3]{12\sqrt{3}-20} = \sqrt[3]{2}(\sqrt{3}-1)$.
$\frac{\sqrt{2}}{\sqrt[3]{2}} = 2^{1/6}$ is then easy to see.
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prove that there infinitely many primes of the form $8k-1$ Using the fact that $$\left ( \frac{2}{p} \right )=(-1)^{\frac{p^2-1}{8}}$$
for each prime $p>2$,prove that there infinitely many primes of the form $8k-1$.
I thought that we could I assume that there is a finite number of primes of the form $8k-1$: $p_1,p_2 \dots ,p_k$
Could we maybe set $N=8p_1p_2 \cdots p_k-1 >1$
Then $N$ has a prime divisor $p$.$p$ can be of the form $8n+1,8n+3,8n+5 \text{ or } 8n+7$..
How could I continue?? Also...how can I use this: $\left ( \frac{2}{p} \right )=(-1)^{\frac{p^2-1}{8}}$ ?
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Let $p_1,p_2, \ldots, p_k$ be the list of ALL primes of the form $8s+7$. Let
$$N=(p_1p_2 \dotsb p_k)^2-2.$$
Note that $N \equiv 7 \pmod{8}$ and is odd. If $p$ is a prime that divides $N$, then
$$(p_1p_2 \dotsb p_k)^2 \equiv 2 \pmod{p}.$$ Thus
$$\left(\frac{2}{p}\right)=1.$$
Thus $p \equiv \pm 1 \pmod{8}$.
So all primes that divide $N$ must be of the from $8s+1$ or $8s+7$. But not all of them can be of the form $8s+1$ (ask why???)
So there must be one of the form $q=8s+7$. Now see if you can proceed from here.
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Using continuity to evaluate limits I hope you guys are enjoying your weekend. I have a question about limits.
This homework problem asks me to use continuity to evaluate this limit, I would like to double-check that I have following the right procedure.
The problem is as follows:
$$\lim_{x\to \pi}\sin(x + \sin x)$$
I break the problem up into two seperate limits:
$$\lim_{x\to \pi}\sin x + \lim_{x\to \pi}\sin(\sin x)$$
Because $\sin x$ is continuous in its domain and its domain includes all real numbers, both limits are continuous and have a domain that includes all real numbers.
I can therefore plug in $\pi$ and conclude that the limit is $0$.
Is my methodology correct or am I making a mistake?
Thanks!
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Using continuity to evaluate the problem means that you can use the following fact (assuming you proved it in class, not sure what else your teacher might have been asking) for a continuous function $f:\mathbb{R} \to \mathbb{R}$:
$$\lim_{x\to a} f(x)=f(\lim_{x\to 0} x).$$
So $\lim_{x\to \pi}\sin(x+\sin x)=\sin(\lim_{x\to \pi}x+\sin x)=\sin(\pi+0)=0$
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Is this correct and sufficient to show limit does not exist? Find limit or show that it does not exist:
$$\lim_{(x,y) \to (0,0)} \frac{ 2x^{2}y^{3/2} }{y^{2}+x^{8}}$$
using the path $x=m y^{1/4}$:
$$\lim_{(my^{1/4},y) \to (0,0)} \frac{ 2m^{2}y^{1/2}y^{3/2}}{y^{2}+m^{8}y^{2}}$$ $$\lim_{(my^{1/4},y) \to (0,0)} \frac{ 2m^{2}y^{2}}{y^{2}(1+m^{8})}$$ $$\lim_{(my^{1/4},y) \to (0,0)} \frac{ 2m^{2}}{1+m^{8}}$$
Does not exist, because limit is path dependent.
I am asking because I have a test in a couple days. I want to make sure I'm not committing some mortal math sin. Also, the Latex tutorial is great, I think I just wasted 20 minutes making that mess.
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Looks good to me! Good job. Intuitively you can guess that the limit won't exist because the overall power in the denominator dominates the power of the numerator. To realize this, you can look at straight-line paths to the origin by setting $y=ax$
and taking a limit that way. The numerator will have degree $\dfrac72$ but the denominator will have degree $8$.
– Cameron Williams
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Basic Trigonometry Question If $\cos{(A-B)}=\frac{3}{5}$ and $\sin{(A+B)}=\frac{12}{13}$, then find $\cos{(2B)}$.
Correct answer = 63/65.
I tried all identities I know but I have no idea how to proceed.
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$$\cos(2B)=\cos(A+B+\underbrace{A-B})=\cos(A+B)\cos(A-B)-\sin(A+B)\sin(A-B)$$
$$A-B=\arccos\frac3{15}$$
Using the principal value, $$\implies 0<A-B<\frac\pi2\implies\sin(A-B)>0$$
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True or False: The circumradius of a triangle is twice its inradius if and only if the triangle is equilateral. Let $R$ be the circumradius and $r$ be the inradius. The if part is clear to me. For an equilateral triangle, the circumcentre, the incentre and the centroid are the same point. So, by property of cebntroid $AG:GD=2:1\Rightarrow AG=2GD$. Thus $R=2r$.
But is the converse true? Whether $R=2r$ implies that the triangle must be equilateral ? We know some relations involving circumradius and inradius, like $R=\dfrac{abc}{4\Delta}, r=\dfrac{\Delta}{s}$, where $\Delta$ is the area of the triangle and $s$ is its semi-perimeter i.e. $s=\dfrac{a+b+c}{2}$. But then how to show that the triangle is equilateral if $R=2r$.
I would be thankful if anyone can help me.
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Let $a,b,c$ are the sides of a triangle, $A=$ area of the triangle, $s=$ semi-perimeter.
$R=\dfrac{abc}{4A}, r=\dfrac{A}{s}$
We have to show $R\geq 2r$.
The relation $\dfrac{abc}{4A}\geq \dfrac{2A}{s}$ holds
if $abc\geq \dfrac{8A^2}{s}$
if $abc\geq 8(s-a)(s-b)(s-c)$
if $abc\geq (b+c-a)(c+a-b)(a+b-c)$
This is true for all triangles.
When $a=b=c$, the equality holds.This is the case of an equilateral triangle.
So, $R=2r$.
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2-Frobenius Groups of order 25920 A group $G$ is called a 2-Frobenius group if $G=ABC$, where $A$ and $AB$ are normal subgroups of $G$, $AB$ is a Frobenius group with kernel $A$ and complement $B$ and $BC$ is a Frobenius group with kernel $B$ and complement $C$.
Let $A$ be a nilpotent group of order $2^4.3^4=1296$ and order $B$ equal to $5$ and order $C$ equal to $4$. So we know that $BC$ is the only Frobenius group of order $20$. Now, I would like to know that whether we can find all such 2-Frobenius group $G=ABC$ of order 25920?
Actually I am interested in finding $Irr(G)$ as well. (Irr(G) is the set of irreducible characters of group $G$).
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The smallest dimensional modules for a cyclic group of order $5$ over both ${\mathbb F}_2$ and ${\mathbb F}_3$ have dimension $4$, so $A$ must be a direct product of elementary abelian groups of orders $2^4$ and $3^4$. Also, $BC$ has a single faithful $4$-dimensional module over both ${\mathbb F}_2$ and ${\mathbb F}_3$, so there is a unique $2$-Frobenius group $ABC$ of the type you describe.
You get $2$-Frobenius groups $A_1BC$ and $A_2BC$ as the semidirect product of these two irreducible modules with $BC$, where $|A_1|=2^4$, $|A_2|=3^4$, and $ABC$ is a subdirect product $(A_1 \times A_2)BC$ of $A_1BC$ and $A_2BC$. Hence $ABC$ has an intransitive permutation representation of degree $97$.
You should probably use GAP (or Magma) to compute the character table of $ABC$. It has $111$ conjugacy classes.
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Axiom of Pairing Axiom of Pairing states that if $a,b$ are sets, $\exists$ a set $A$ such that $A=\{{a,b\}}$. My question is that why we can't use Axiom of specification to define $A=\{x|x=a \vee x=b\}$?
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User wrote:
My question is that why we can't use Axiom of specification to define $A=\{x|x=a \vee x=b\}$?
To apply Specification to construct $A=\{a,b\}$, $a$ and $b$ would have to have been assumed or proven to be elements of some other set $B$, i.e. $a\in B$ and $b\in B$. Then $A=\{x|x\in B \land [x=a \vee x=b]\}$. Pairing has no such requirement.
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how to solve this question of polynomials
Given the polynmial is exactly divided by $x+1$, when it is divided by
$3x-1$, the remainder is $4$. The polynomial leaves remainder $hx+k$ when divided by
$3x^2+2x-1$. Find $h$ and $k$.
This is the question which is confusing me.. i have done this question like this:
$p(x) = g1(x)(x+1) +0 $
$p(x) = g2(x)(3x-1) +4 $
$p(x) = g3(x)(3x+2x-1)+hx+k $
$=> p(x) = g3(x)(x+1)(3x-1) + hx+k $
now putting the value of $x$ in each $p(x) $
$p(-1) = 0 \tag{1}$
$p(1/3) = 4 \tag{2}$
$p(-1,1/3) = hx+k \tag{3}$
from equation $(2)$ and $(3)$
$4=hx+k\tag{4}$
now putting the value of $x$ in equation $(4)$.
$h(-1)+k=4 => -h+k=4 \tag{5}$
$h(1/3)+k=4 => h+3k=12 \tag{6}$
now adding equations $(5)$ and $(6)$
$ h+3k-h+k = 4+12 =>4k=16 => k=4$
now puting the value of of $k$ in equation $(5)$
$h=0$
therefore my answer is $h=0,k=4$ but the answers are$ h=3, k=3$. please help me to sort out my problem
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The error is that $\,p(-1) = 0\,\Rightarrow\, h(-1) + k = \color{#c00}0,\,$ not $\,\color{#c00}4.\,$ Fixing that yields the given answer.
Remark $\ $ This is a special case of the Chinese Remainder Theorem (CRT) or, equivalently, Lagrange interpolation. Either of these methods can be applied to solve the general case.
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The nonexistence of a polynomial I'm studying algebraic geometry. To illustrate a nonalgebraic set, it is given that a unit circle except for a point on it in cartesian product or whole plane except for one point. Why doesn't a polynomial whose zeros are a unit circle except for a point or whole plane except for one point exist? I'll be glad if one explains this situation.
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Polynomials are continuous functions. A function that is identically zero on the plane except at one point is not continuous. The same goes for the unit circle.
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Proving that mean KDR in a videogame is one This is not related to schoolwork. A friend of mine challenged me to prove that the mean KDR (assuming players can only die at the hands of other players) must always be equal to one. I have gotten through the logic part, and am now faced with the math part (which I am less capable of doing), and would like some help.
My logic is as follows: Every kill event is also a death event, and every death event is also a kill event. Therefore, the total number of kills must always equal the number of deaths. We know then that TOTAL kdr = 1. For a more explicit expression of this, we can represent the game as a directed graph whose nodes are players and whose edges represent kills. In such a graph, the in-degree of a node is its deaths and its out-degree is its kills.
I don't think that will be necessary. Instead I'm just sticking with that kills and deaths are distributed a certain way among players, and that each player's kdr = kills/deaths for that player. Proving what I have written in the following picture should finalize the whole proof. If it isn't possible to prove this, maybe the graph abstraction will help us.
EDIT: Forgot to mention earlier that if a player has 0 deaths, the denominator of their KDR is 1, not 0 (to prevent infinite KDR, widely used in games).
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This is clearly false,if there are two player: Alice and Bob and Bob kills alice 20 times and Bob kills alice once then Alice's ratio will be $\frac{1}{20}$ and Bob's ratio will be $\frac{20}{1}$, the mean of these two ratio's is $\frac{401}{40}>20$
Construction to increase mean kdr arbitrarily for any number of players:
Pick 1 player $v$ and make him kill all other players once. Then make all other players kill $v$ $k$ times. What is the sum of all the kdr now?
$\underbrace{\frac{n-1}{(n-1)k}}_{\text{player v}}+\underbrace{(n-1)\frac{k}{1}}_\text{the other n-1 players}\geq(n-1)k$.
So the arithmetic mean of the kdr's is at least $\frac{(n-1)k}{n}$. Increasing $k$ arbitrarily increases the kdr mean arbitrarily.
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Textbooks, lecture notes, and articles from arXiv for undergraduate students I have found some interesting textbooks and articles on arXiv, such as the following one, that are accessible to an undergraduate student:
Course of linear algebra and multidimensional geometry, by
Ruslan Sharipov.
My experience makes me believe that in arXiv there may be not only research papers, but also plenty of freely available interesting textbooks, lecture notes, and articles that can
*
*give insightful explanations of topics that are commonly found in an undergraduate degree program;
*but also enrich and broaden the mathematical culture of an undergraduate student.
Could you point out some good examples of such material?
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I found this one a while ago:
Euclidean plane and its relatives;
a minimalist introduction
Anton Petrunin
https://arxiv.org/abs/1302.1630
btw...
Here is a "hack" you can do to look for books on arxiv. Do an advanced search for "these lecture notes" OR "this book" in the abstract. Show 200 results per page. Use your browser search now for "these lecture notes" and then for "this book" ont he book. arxiv's search can do exact match for author name, but not for text in the abstract, it seems. That is why the second step is needed with your browser search.
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Proof that $\lim_{n\to\infty} n\left(\frac{1}{2}\right)^n = 0$ Please show how to prove that $$\lim_{n\to\infty} n\left(\frac{1}{2}\right)^n = 0$$
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Notice that
$$
\frac{(n+1)/2^{n+1}}{n/2^n}=\frac12\left(1+\frac1n\right)\tag{1}
$$
For $n\ge2$, the ratio in $(1)$ is at most $\frac34$. At $n=2$, $\dfrac{n}{2^n}=\dfrac12$. Therefore, $(1)$ implies
$$
\frac{n}{2^n}\le\frac12\left(\frac34\right)^{n-2}\tag{2}
$$
for $n\ge2$. Hopefully, it is clearer that
$$
\lim_{n\to\infty}\frac12\left(\frac34\right)^{n-2}=0\tag{3}
$$
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Question about a particular limit (at infinity) I have a question about limits, a problem specifically. I have been asked to solve the following limit in any way I see fit:
$$\lim_{x\to 2\pi^-}x\csc x$$
I know that the domain of $\csc$ is all numbers except for $n\pi$, and I know I could probably plug in numbers close to $2\pi$ to get the limit, and I'm pretty sure this is going to be a limit at infinity, but is there an easier way?
I feel like I should be able to do some limit-algebra and solve it that way, will I just have to look at the graph or plug in numbers? I was going to take out on of the x's and put it in front of the limit but because it is a variable I don't think I can do that. Any tips?
Thanks!
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Hint: The limit is
$$\lim_{x \rightarrow {2\pi}^{-}}\frac{x}{\sin x}$$
For values of $x$ close to but less than $2\pi,$ the values will be
$$ \frac{\text{numbers close to} \;\; 2\pi}{\text{negative numbers close to} \;\; 0} $$
which tells us the limit is $\ldots$
(Note that $\sin x$ is negative when $x$ is in the $4$th quadrant.)
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$x^{1+\epsilon}$ is not uniformly continuous on $[0,\infty)$ There are two questions.
First: is the proof underneath correct?
Let $\epsilon>0$ and let $f(x)=x^{1+\epsilon}$. I aim to show that $f$ is not uniformly continuous on $[0,\infty)$.
We will show that for any $\delta>0$ there exists $x$ and $y$ such that $|x-y|<\delta$ and $|f(x)-f(y)|\geq 1$.
Now let $\delta>0$ be given. Consider $f'(x)=(1+\epsilon)x^\epsilon$. We have that $f'\rightarrow\infty$ as $x\rightarrow \infty$ and that $f'$ is monotone increasing. Now choose an $x_0$ large enough so that $f'(x)$ is larger than $2/\delta$ for all $x\geq x_0-\delta$.
Now consider $|f(x_0)-f(y)|$ for some $y\in(x_0+\delta/2,x_0+\delta)$. The Mean Value Theorem guarantees us that there is a $\zeta\in(x_0, y)$ such that $|f(x_0)-f(y)|=|f'(\zeta)||x_0-y|$.
We thus have that
$$
\begin{array}{rl}
|f'(\zeta)||x_0-y|&=(1+\epsilon)\zeta^\epsilon|x_0-y|\\
&\geq(1+\epsilon)\zeta^\epsilon\delta/2\\
&\geq(1+\epsilon)x_0^\epsilon\delta/2\\
&\geq2/\delta\cdot\delta/2\\
&=1
\end{array}
$$
Thus for this choice of $x_0$ and $y$, we have that $|f(x_0)-f(y)|\geq 1$ and $f$ is not uniformly continuous.
Second: If all is well and good, what other ways are there to prove this claim? I feel like the approach I took was a little bulky.
|
Your proof is fine, and the same proof works for any $f$ with $f' \to \infty$ as $x \to \infty$. Personally, I think your approach is excellent in that it is extremely clear. Perhaps you could have been more terse, but I like your style.
Another approach, however, if you insist on being as terse as possible, is to note that $f$ is uniformly continuous if and only if it maps Cauchy pairs of cofinal sequences to pairs of cofinal sequences. See Pedro's answer for an explanation.
(in fact, this function takes Cauchy sequences to Cauchy sequences but is not uniformly continuous. The condition of taking Cauchy sequences to Cauchy sequences is sufficient, however, when considering functions over bounded sets).
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Solve the integeral equation (C.S.I.R) Let $\lambda_1, \lambda_2$ be the eigen value and $f_1 , f_2$ be the coressponding eigen functions for the homogeneous integeral equation
$$ \phi(x) - \lambda \int_0^1 (xt +2x^2) \phi(t) dt = 0 $$
Then
*
*$\lambda_1 = -18 - 6 \sqrt{10} , \lambda_2 = -18 + 6 \sqrt{10}$
*$\lambda_1 = -36 - 12 \sqrt{10} , \lambda_2 = -36 + 12 \sqrt{10}$
*$\int_0^1f_1(x)f_2(x) dx = 1$
*$\int_0^1f_1(x)f_2(x) dx = 0$
I have solved simply and get the values of $\lambda_1 \ \ and \ \ \lambda_2$ belong to (1).
Please tell me about (3) and (4) option.
Thank you.
|
Let's rewrite your equation in the following form:
\begin{align}
\frac{1}{\lambda}\phi(x) = x\int_{0}^{1}t\phi(t) \mathrm{d}t + 2x^{2}\int_{0}^{1} \phi(t) \mathrm{d}t = c_{1}x + c_{2}x^{2}
\end{align}
for $c_{1}: = \int_{0}^{1}t\phi(t) \, \mathrm{d}t$ and $c_{2}:= 2\int_{0}\phi(t)\, \mathrm{d}t$. From this we can see that $\phi$ is a quadratic polynomial with no constant term.
Substituting $\phi(x) = c_{1}x + c_{2}x^{2}$, we find that
\begin{align}
\frac{c_{1}}{\lambda} x + \frac{c_{2}}{\lambda}x^{2}
&=x \int_{0}^{1} t(c_{1}t+ c_{2}t^{2}) \, \mathrm{d}t + 2x^{2} \int_{0}^{1} c_{1}t + c_{2}t^{2} \, \mathrm{d}t \\
&=x(c_{1}/3 + c_{2}/4) + 2x^{2}(c_{1}/2 + c_{2}/3)
\end{align}
so that
\begin{align}
\frac{c_{1}}{\lambda} &= \frac{c_{1}}{3} + \frac{c_{2}}{4} \\
\frac{c_{2}}{\lambda} &= c_{1} + \frac{2c_{2}}{3} \\
\end{align}
It can be verified that if both $c_{1}$ and $c_{2}$ are nonzero, then $\lambda = -18 \pm 6\sqrt{10}$.
Since $c_{1} = \left(\frac{1}{\lambda} - \frac{2}{3} \right)c_{2}$
\begin{align}
f_{1}(x) &= -\left(\frac{1}{18+6\sqrt{10}} + \frac{2}{3}\right)c_{2}x + c_{2}x^{2}
= -\frac{1}{6}(1+\sqrt{10})c_{2}x + c_{2}x^{2} \\
f_{2}(x) &= -\left(\frac{1}{18-6\sqrt{10}} + \frac{2}{3}\right)c_{2}'x + c_{2}'x^{2} = -\frac{1}{6}(1-\sqrt{10})c_{2}'x + c_{2}'x^{2}
\end{align}
so that
\begin{align}
\int_{0}^{1} f_{1}(x)f_{2}(x) \, \mathrm{d}x
&=
c_{2}c_{2}' \int_{0}^{1} x^{4} -\frac{1}{3}x^{3} - \frac{1}{4}x^{4} \, \mathrm{d}x = \frac{c_{2}c_{2}'}{30}
\end{align}
So we can't claim that these eigenfunctions are orthogonal, or that their inner product is 1, unless we choose a scaling such that $c_{2}c_{2}' = 30$.
|
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|
Finding $\sin^{-1}(x)$ without using a calculator I don't understand how to compute $\sin^{-1} (0.6293)$, to figure out the angle without using a calculator. I understand how to find the answer if I use a calculator but I don't understand the necessary steps to solve the problem without a calculator.
Am I wrong to assume you have to know at least the value of the opposite or hypotenuse length of the given triangle? Can you find the angle if you only have the sine and nothing else and you don't have a calculator?
|
There are several approaches to this problem, all of them are a pain in the neck without a calculator. The first approach you can try is to construct a triangle which could give you a decent triangle or if you know some calculus, you can use the Taylor series of $\arcsin x$ and use that. I'll illustrate the Taylor series approach.
The Taylor series of $\arcsin x$ is
$$x + \frac{x^3}{6} + \frac{x^5}{40} + O(x^7)$$.
Using this information, if we plug in $.6293$ into the above then we'll get an "okay" approximation. The more terms you add, the better but it will get harder and harder to find the powers of $.6293$. So, we have
$$\arcsin(.6923) = .6293 + \frac{.6293^3}{6} + \frac{.6293^5}{40} + O(.6293^7) \approx .6782$$
The real answer is $\approx .6806$ so you be the judge.
|
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How to establish this inequality without using induction? Given the Fibonacci sequence $a_1 = 1$, $a_2 = 2$, $\ldots$, $a_{n+1} = a_n + a_{n-1} $ for $n \geq 2$, how to derive, without using induction, the inequality $$ a_n < (\frac{1+\sqrt{5}}{2})^n $$ for $n = 1, 2, 3, \ldots$?
I know how to establish the above inequality using induction.
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The fibbonacci numbers have a closed form: $a_n = \dfrac{1}{\sqrt{5}}\left[\left(\dfrac{1+\sqrt{5}}{2}\right)^{n+1} - \left(\dfrac{1-\sqrt{5}}{2}\right)^{n+1}\right]$.
Since $\left|\dfrac{1-\sqrt{5}}{2}\right| < 1$, we have $-1 < \left(\dfrac{1-\sqrt{5}}{2}\right)^{n+1} < 1$ for all $n \ge 1$.
Can you figure out what to do from here?
|
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Can $R[[x]]$ contain constants?
Consider the ring $R[[x]]$ of formal power series $\sum_{n=0}^\infty a_nx^n$ with coefficients in $R$. I was wondering whether $R[[x]]$ contains elements of $R$ (polynomials of degree $0$).
I'm trying to solve Commutative Algebra problems. I feel it is possible, as all of $\{a_1,a_2,\dots\}$ simply have to be equal to $0$, which I think is possible.
Thank you.
|
Yes, it can. ${}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}$
|
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Criticism on truth of Gödel sentence in standard interpretation Mendelson in his book mentioned the Gödel sentence and argued that in standard interpretation it is true. But Peter Milne in his article (On Godel Sentences and What They Say) criticized: "But we know that we cannot move from consistency of T to truth of γ
when no restriction has been put on the means of representation, the kind
of formulas used". But I cant understand what he means. Can you help me?
|
There is no real conflict.
Mendelson is talking about a particular type of Gödel sentence, constructed in a standard way following the pattern of Gödel's original paper. [This is the kind of sentence people usually have in mind when they speak of a Gödel sentence for a theory $T$, without further qualification -- the sort of sentence constructed to "say", relative to a coding scheme, "I am unprovable in $T$".]
Milne on the other hand is talking about Gödel sentences in a more generic sense which is rather common nowadays, according to which any old fixed points $\gamma$ of the negation of the provability predicate for a theory $T$ counts as a Gödel sentence for the theory [i.e. any $\gamma$ such that $T \vdash \gamma \leftrightarrow \neg\mathsf{Prov}(\ulcorner\gamma\urcorner)$]. If $T$ is consistent but unsound, then the negation of the provability predicate can have false fixed points, so $T$ can have false Gödel sentences in the general sense.
Milne is complaining about people casually moving from a true claim about standardly constructed Gödel sentences (true if the theory is consistent) to false claim about generic Gödel sentences.
This is all explained in my Gödel book (pp. 182-183 of the second edition).
|
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Equation , powers of two I want to find the sum of the roots of the equation $$4(4^x + 4^{-x}) - 23(2^x + 2^{-x}) + 40 = 0 $$ in real numbers. I tried the substitution $ 2^x = t $ but then it turns into a quartic equation which I couldn't solve. I think its roots sum to zero so I want to prove it without actually finding the roots.
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$$4\left(a^2+\frac1{a^2}\right)-23\left(a+\frac1a\right)+40=0$$
$$\implies 4\left[\left(a+\frac1a\right)^2-2\right]-23\left(a+\frac1a\right)+40=0$$
$\displaystyle a+\frac1a=b\implies 4(b^2-2)-23b+40=0\iff 4b^2-23b+32=0$
|
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$\frac{\|\mathcal Ax\|_{n}}{\|x\|_n}$ is always bounded Given the matrix $A= (a_{i,j}) \in M_{n,n}$
$||A||=\sup\limits_{x\in X}\frac{\|\mathcal Ax\|_{n}}{\|x\|_n}$ where $|| $ . $|| _n$ is $ R^N$ norm
$\frac{\|\mathcal Ax\|_{n}}{\|x\|_n}$ is always bounded:
$||Ax||^2_{n}=(\sum\limits_{i,j=0}^n a_{i,j} x_j)^2\le\sum\limits_{i,j=0}^{n} a^2_{i,j} x^2_j
=(\sum\limits_{i,j=0}^n a^2_{i,j})(\sum\limits_{j=0}^n x^2_j)= M ||x||^2_n $
where $M=(\sum\limits_{i,j=0}^n a^2_{i,j})=||A||^2_n$
Is a correct proof?
Maybe:
$||Ax||^2_{n}=\sum\limits_{i}^n(\sum\limits_{j}^n a_{i,j} x_j)^2 \le\sum\limits_{i}^n(\sum\limits_{j}^n a^2_{i,j} x^2_j)
=\sum\limits_{i}^n(\sum\limits_{j}^n a^2_{i,j})(\sum\limits_{j=0}^n x^2_j)= M ||x||^2_n $
|
We have
$$
\left\{ \frac{\|\mathcal Ax\|_{n}}{\|x\|_n} : x \in \mathbb R^n \right\}
=
\left\{ \|\mathcal Ax\|_{n} : x \in \mathbb R^n, \|x\|_n=1\right\}
$$
The set on the right is compact because $\mathcal A$ and norm are continuous and the unit sphere is compact. Hence, the set on the left is bounded.
|
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Normal subgroup Suppose $N$ is a normal subgroup of $G$ and $H$ is a subgroup of $G$ that $|N|,[G:H]$ are finite and $(|N| , [G : H] ) =1$. Prove that $H\leq N$.
|
Hint: $$|HN|=\frac{|H|\cdot|N|}{|H \cap N|} $$ and $[HN:N]$ divides $[G:N]$. But $[HN:N]=[H:H \cap N]$, which in its turn divides $|H|$.
|
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Why there are irrational numbers? I do not quite get it. Why can't we represent all real numbers as a sum of rational numbers? Why do we need irrational numbers?
For example,
*
*$\pi=3.14159265358\cdots=3+10^{-1}+4*10^{-2}+10^{-3}+5*10^{-4}+\cdots$
*$e=2.71828182846\cdots=2+7*10^{-1}+10^{-2}+8*10^{-3}+2*10^{-4}+\cdots$
*And so on
|
Numbers that cannot be expressed rationally arise naturally as solutions of equations. The solution of $x^2=2$ & of $x(1-x)=1$, & any of an infinitude of specificable polynomial equations, cannot be expressed rationally. It can be proven directly, by a simple recipe, that no rational number can satisfy either of the two equations I have particularly cited; and the same can be proven similarly for the general case.
Moreover there is a class of numbers yet beyond this: the transcendental numbers, any of which can be shown to be not the solution of any polynomial equation comprising a finite number of terms having integer coefficients.
|
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World Cup probability question. What is the probability of none of the 32 teams of the World Cup bringing two consecutive draws in the first two games. Accept the fact that a win, draw or loss have the same probability to appear.
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Assume each game has a probability $p$ of ending in a draw, and that the results of each game are independent of each other. Lets analyze one group with teams W,X,Y,Z.
WLOG, the schedule for the first two games for each team is:
*
*W vs X
*Y vs Z
*W vs Y
*X vs Z
with the last two group games being W vs Z and X vs Y.
The ways for all four teams to avoid starting with two draws are:
All four games do not end in a draw - probability $(1-p)^4$
Only one game ends in a draw - probability $4p(1-p)^3$
Game 1&2 end in a draw while 3&4 do not, or vice-versa - probability $2p^2(1-p^2)$.
Thus, the probability that no one in this group starts with two draws is: $(1-p)^4 + 4p(1-p)^3 + 2p^2(1-p)^2 = (1-p)^2(1+2p-p^2)$
Therefore, the probability that no one across all $8$ groups starts with two draws is simply: $[(1-p)^2(1+2p-p^2)]^8$
If we assume $p = \dfrac{1}{3}$, then the desired probability is $\left(\dfrac{56}{81}\right)^8 \approx 0.05219$.
|
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Valuation associated to a non-zero prime ideal of the ring of integers I have a question from Frohlich & Taylor's book 'Algebraic Number Theory', p.64. I will keep the notation used there.
Let $K$ be a number field, $\mathcal o$ its ring of integers. Let $\mathfrak p$ be a non-zero prime ideal of $\mathcal o$ and $v=v_\mathfrak p$ the valuation of $K$ associated to $\mathfrak p$. Suppose that $\rho$ is a field automorphism of $K$ such that $v(x^\rho)=v(x)$ for all $x\in K$.
Why is this equivalent to the condition $\mathfrak p^\rho=\mathfrak p$?
Would it help to show that $(x\mathfrak o)^\rho=x^\rho\mathfrak o$ for all $x\in K$? - although I am not sure if this is even true.
Thanking you in advance.
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$v(x)$ is the number $n$ such that $x \in \mathfrak{p}^n-\mathfrak{p}^{n+1}$ now $v(x)\geq 1$ iff $x \in \mathfrak{p}$.
So we see that if $v(x^{\rho})=v(x)$ then
$$x^{\rho} \in \mathfrak{p} \Leftrightarrow x \in \mathfrak{p}$$
which means that $\mathfrak{p}^{\rho}=\mathfrak{p}$.
Conversely if $\mathfrak{p}^{\rho}=\mathfrak{p}$ then it is easy to see
$$x^{\rho} \in \mathfrak{p}^n-\mathfrak{p}^{n+1}\Leftrightarrow x \in \mathfrak{p}^n-\mathfrak{p}^{n+1}$$
and so $v(x^{\rho})=v(x)$.
|
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Show that some endomorphsm is not diagonalizable Given an endomorphism $f:V \rightarrow V$ on an $\mathbb{R}$-vector space, prove that if there is $v \in V-\{0\}$ such that $f^2(v)=-v$, then $f$ is not diagonalizable.
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Solved. I'm putting the solution. If A is a matrix for f in some basis, then $A^2v=-v$. If $A$ is in diagonal form, $A^2$ has only non-negative entries on the diagonal.
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How is $\mathbb{F}_4$ generated? I know $\mathbb{F}_4$ is a field while $\mathbb{Z}/(4)$ is just a ring. So how is $\mathbb{F}_4$ generated?
Complement: So what are the elements like in $\mathbb{F}_4$ like? Are they $\{0,1,x,x+1\}$? Is every field $\mathbb{F}_k$ has $k$ elements no matter whether $k$ is a prime?
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The answer of @MarceloBielsa is perfect, but I like an approach like what OP was working towards: the elements of $\mathbb F_4$ are $\{0,1,\omega,\omega+1\}$, where $\omega^2=\omega+1$, $(\omega+1)^2=\omega$, and $\omega(\omega+1)=1$. I think that that’s enough for filling in the multiplication table.
|
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help: isosceles triangle circumscribing a circle of radius r Please help me show that: the equilateral triangle of altitute $3r$ is the isosceles triangle of least area circumscribing a circle of radius $r$.
Iassumed the following:
base = $2a$
height = $h$
radius of circle = $r$
Area = $\frac{1}{2}(2a)h = ah$
$tan(2\theta) = \frac{h}{a}$ and $tan\theta = \frac{r}{a} $
Ii tried using double angle identities to represent h in terms of a and \theta but I have failed miserably.
This is what I ended up with:
$\displaystyle{%
h={2r\cos^{2}\left(\theta\right) \over 2\cos^{2}\left(\theta\right) - 1}}$
I know i need to differentiate the area and set to zero, but I can't get it. please direct me how to solve this problem. Thank you;
|
First, two corrections in the question
1) The height will be $3r/2$
2) This will maximize the area. Since the minimum area of an isosceles triangle will be of height $2r$ and zero base, hence zero area.
From the figure, $h=r+\sqrt{r^2-a^2}$, area $A=ah=a(r+\sqrt{r^2-a^2})$. Now equate $\frac{\partial A}{\partial a}=0$ to get the $a$ which maximizes the area. For your info,
$$
\frac{\partial A}{\partial a}= \frac{r\sqrt{r^2-a^2} + r^2-2a^2}{\sqrt{r^2-a^2}}; \\
\frac{\partial A}{\partial a}=0 \Rightarrow r\sqrt{r^2-a^2} + r^2-2a^2 = 0 \\
(2a^2-r^2)^2 = r^2(r^2-a^2) \\
a = \frac{\sqrt{3}}{2}r
$$
Hope this helps.
|
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1-form on Riemann Surface Good evening, I can not prove the following result:
Let $\omega $ be a meromorphic 1-form on $ \mathbb {C} _ {\infty} = \mathbb {C} \cup \infty $ such that $ \omega_{|\mathbb{C}} = f (z) dz $. Show that f is ratio of polynomial functions.
Any suggestions on how to develop the demonstration?
thanks.
|
As suggest in the comment, on another cover $\mathbb C$ (which corresponds to $\mathbb C\setminus\{0\} \cup \{\infty\}$ with coordinates $\tilde z$), write
$$\omega = \tilde f(\tilde z)$$
On the intersection of two coordinate $\mathbb C\setminus \{0\}$, we have $z = 1/\tilde z$, then
$$d\tilde z = d(1/ z) = -\frac{1}{ z^2} d z . $$
Write $f(z) = \sum_{n=k}^\infty a_n z^n$ and $\tilde f(\tilde z) = \sum_{n=m}^\infty b_n \tilde z^n$ for some $k, m\in \mathbb Z$, then
$$ \sum_{n=k}^\infty a_n z^n dz = f(z) dz = \tilde f(\tilde z) d\tilde z = \sum_{n=m}^\infty b_n \tilde z^n d\tilde z = \sum_{n=m}^\infty b_n z^{-n} \bigg(-\frac{1}{z^2}\bigg)dz$$
$$ \Rightarrow \sum_{n=k}^\infty a_n z^n = \sum_{n=m}^\infty (-b_n) z^{-n-2}$$
Then compare coefficients to see that $a_n = b_n = 0$ for large enough $n$.
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Calculation of the limit $\lim_{n \to +\infty} n^2x(1-x)^n, x \in [0,1]$ and the supremum How can I find this limit:
$$\lim_{n \to +\infty} n^2x(1-x)^n, x \in [0,1]$$
Do I have to use the L'Hospital rule? If so, do I have to differentiate with respect to $n$ or to $x$ ?
EDIT:
I also tried to find the supremum of $n^2x(1-x)^n$..I found that $f(x)=n^2x(1-x)^n$ achieves its maximum at the point $\frac{1}{n+1}$ and that $\sup_{x \in [0,1]} n^2x(1-x)^n=\frac{n}{e} \to +\infty$
Could you tell me if it is right?
|
This limit is always zero, there are two cases:
1)$x=0$-it's obvious.
2)$x \neq 0$, then the limit is zero, because $1-x <1$ and exponential function decrease faster than polynomial $n^2$
|
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Ways to sum to $n$ with $m$ integers that are $\le k$ Given three natural numbers $n$, $m$ and $k$, how many ways are there to write $n$ as the sum of $m$ natural numbers in the set $\{0, 1, \ldots, k\}$, where order does matter?
I've seen the "Ways to sum $n$ with $k$ numbers", but never where the different numbers are restricted by an upper bound.
I'm not even sure how to proceed with this question. I have a python script to calculate this (in essence, it tries the $(k+1)^m$ possible sums, computes them, and returns the number of sums whose result is $n$). I do have some recurrence equations, but I'm almost 100% sure they don't help that much. By fixing the first number of the sum to be $0, 1, \ldots, k$, and writing $P(n, m, k)$ as the solution of the problem:
P(n, m, k) = P(n, m - 1, k) + # If the first number is 0
P(n - 1, m - 1, k) + # If the first number is 1
...
P(n - k, m - 1, k) + # If the first number is k
and
P(n, 1, k) = 0 if n > k
1 if n <= k
Can this be solved in a more elegant way than brute force?
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It is not exactly what you want, but it is at least related:
OEIS A048887: Array T read by antidiagonals, where T(m,n) = number of compositions of n into parts <= m
What is missing is the condition that only those compositions are counted, which consist of exactly $m$ numbers, the above counts all.
So it could at least serve as an upper boundary.
BTW I wondered how the Online Encyclopedia of Integer Series handles multidimensional series, that is an answer: a series in two dimensions is given in (anti-) diagonal enumeration.
|
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|
Meaningful lower-bound of $\sqrt{a^2+b}-a$ when $a \gg b > 0$. I know that, for $|x|\leq 1$, $e^x$ can be bounded as follows:
\begin{equation*}
1+x \leq e^x \leq 1+x+x^2
\end{equation*}
Likewise, I want some meaningful lower-bound of $\sqrt{a^2+b}-a$ when $a \gg b > 0$.
The first thing that comes to my mind is $\sqrt{a^2}-\sqrt{b} < \sqrt{a^2+b}$, but plugging this in ends up with a non-sense lower-bound of $-\sqrt{b}$ even though the target number is positive.
\begin{equation*}
\big(\sqrt{a^2}-\sqrt{b} \big) - a < \sqrt{a^2+b}-a
\end{equation*}
How can I obtain some positive lower-bound?
|
The first three terms of $(1+x)^{\frac12}$ are $1 + \frac12 x - \frac18 x^2$. And you can check for yourself that $$\left(1 + \frac12 x - \frac18 x^2\right)^2 = 1 + x - \frac18 x^3 + \frac{1}{64}x^4$$
which is $\le 1+x$ whenever $\frac18 x^3 \ge \frac{1}{64}x^4$, i.e. for $0 \le x \le 8$.
Now just put $x = \frac{b}{a^2}$ to get
$$\sqrt{a^2+b} - a \ge \frac{b}{2a} - \frac{b^2}{8a^3}$$
whenever $0 \le b \le 8a^2$.
|
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If $4k^3+6k^2+3k+l+1$ and $4l^3+6l^2+3l+k+1$ are powers of two, how to conclude $k=1, l=2$ It is given that
$$4k^3+6k^2+3k+l+1=2^m$$
and
$$4l^3+6l^2+3l+k+1=2^n$$
where $k,l$ are integers such that $1\leq k\leq l$.
How do we conclude that the only solution is $k=1$, $l=2$?
I tried subtracting the two equations to get:
$$2(l-k)(2l^2+2lk+2k^2+3l+3k+1)=2^m(2^{n-m}-1)$$
But I am unable to proceed further.
Thanks for any help!
Updates:
An observation is that $k$ and $l$ have opposite parity.
Because if both $k$ and $l$ are even (or both odd), then $4k^3+6k^2+3k+l+1\equiv 1 \pmod 2$, which is a contradiction.
|
I have a simplification that I couldn't finish, but maybe someone better with polynomials can. It may also lead nowhere, obviously; but was too long as a comment.
If you add the two equations, you get
$$4(k^3 + l^3)+ 6(k^2 + l^2) + 4 (k + l) + 2 = 2^m + 2^n$$
If you look at this modulo (k+l), it implies (mod k+l):
$$4(k+l)^{3} + 6(k+l)^{2} + 2 = 2^m + 2^n$$
Substituting $x:= k+l$, we get
$$4x^{3} + 6x^{2} + 2 = 2^m + 2^n$$
That's where I get stuck.
|
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|
What is the domain of the function $F(x)=\int_{0}^{x}\frac{\operatorname{arctan}(t)}{t}dt$? What is the domain of the following function?
$$F(x)=\int_{0}^{x}\frac{\operatorname{arctan}(t)}{t}dt$$
On the one hand, the internal function is not defined at 0, but on the other hand, it's defined on every other point except zero, so the region can be calculated.
|
The function can be extended in $x=0$ to obtain a continuous function defined on the whole real line. Hence such extended function is Riemann integrable and hence its integral on $[0,x]$ is well defined for all $x$. Now remember that if you modify a function in a finite number of points, its integral does not change. Hence you can extended your function with any value in $x=0$ and obtain an Riemann integrable function.
However, strictly speaking, Riemann integrals on the interval $[a,b]$ is defined only for functions which are defined on the whole interval $[a,b]$. If the function is only defined on the interval $(a,b]$ this should be considered an improper integral i.e.:
$$
F(x) = \int_0^x f(t)\, dt = \begin{cases}\lim_{a\to 0^+} \int_a^x f(t)\, dt & \text{if } x>0 \\ \lim_{a\to 0^-}\int_a^x f(t)\, dt & \text {if $x<0$ }\\ 0 &\text{if } x=0\end{cases}.
$$
But, again, since the function $f$ can be extended with continuity in $x=0$ the limit exists and is equal to the integral on $[0,x]$ of the extended function.
|
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|
Permutation, Combinatorics Stuck here : there are 100 objects labeled 1, 2,...100. They are arranged in all possible ways. How many arrangements are there in which object 28 comes before object 29.
My approach : Consider object 28 & object 29 , a single object. Now we have a total of 99 objects which can be per mutated in 99! ways .
But the answer is 4950*98! .
What's wrong with my approach?
|
You can take object 28 and 29 as alike. So the permutation order i.e. 28 before 29 is not disturbed. This gives no. of arrangements as 100!/2! = 98!*99*100/2=98!*4950
|
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Intuition Behind Compactification I'm heading into my second semester of analysis, and I still don't have a good intuition of when a set is compact. I know two definitions, covering compact and sequentially compact, but both of those seem very difficult to apply "real time". In $\mathbb{R}^n$ I know we have Heine-Borel, which is a very easy way to check things, but I would like to know a different way to easily check for compactness, if there is one.
For instance, can you compactify any space? Compactifying Euclidean Space is easy to understand because of Heine-Borel, but can you compactify, for instance, $\mathbb{R}^2$ with the discrete metric?
|
There's a few different questions going on here, but I'll focus on the last one:
Yes, it is possible to compactify any space. An easy way to do so is to take your space $X$ and add a point called "$\infty$", and we say that a set $G$ containing $\infty$ is open if and only if $(X \cup \{\infty\}) \setminus G$ is a compact set in $X$ (and therefore in $X \cup \{\infty\}$). This is called the "one-point compactification".
So, taking your example: the only compact sets in $\mathbb{R}^2$ under the discrete metric are those with finitely many points. So, we can take the compactification $\mathbb{R}^2 \cup \{\infty\}$ by saying that we only call a set containing $\infty$ open if it contains all but finitely many of the elements of $\mathbb{R}^2$. Notice that this is very much distinct from the discrete topology on $\mathbb{R}^2 \cup \{\infty\}$.
|
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The limits of the sum of functions whose limits do not exist I have a homework problem that I'm not sure how to start. I tried Google for similar examples it didn't turn up anything. Could someone tell me the name of the concept to look into? The problem is as follows:
Show by example that $\lim_{x\to c}f(x) + g(x)$ can exist even if both $\lim_{x\to c}f(x)$ and $\lim_{x\to c}g(x)$ do not exist.
|
What about $f(x)=x$ and $g(x)=-x$. Neither limit as $x\to\infty$ exists as a real number, yet the limit of the sum is $0$. Or $f(x)=sinx$ and $g(x)=-sinx$. This time neither limit exists (as a real number OR +/-$\infty$, but the limit of the sum is $0$.
To get this phenomenon as $x$ approaches some real number $c$, think of a function $f$ with a vertical asymptote at $x=c$, and then take that function's negation as $g$.
|
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Sequences where each number is a divisor of one less than the next Let $N,k$ be fixed. Call a sequence of positive integers $a_1,a_2,\dots,a_k$ good if for each $i$, $a_i$ is a divisor of $a_{i-1}-1$. Consider the set
$$S = \{x : \text{$x$ occurs in some good sequence of length $k$ that ends in $N$}\}$$
of numbers that appear in some good sequence of length $k$ ending in $N$.
Is it possible to get an estimate for $|S|$, as a function of $k,N$? Is it possible to get a reasonable upper bound on this? Is there any reason to expect that $|S|$ might be asymptotically much smaller than $N$, say $O((\log N)^c)$ or something like that?
Example: for $k=3$, $N=27$, we have $S=\{1,2,3,4,5,6,12,13,25,26,27\}$, so $|S|=11$. The set of good sequences of length 3 that end in 27 are:
1,2,27
1,13,27
1,26,27
2,13,27
3,13,27
4,13,27
5,26,27
6,13,27
12,13,27
25,26,27
So there does appear to be some structure here, but I'm not sure if there's anything that allows clean reasoning about it or about such sequences.
|
I'm sure sharper things can be said, but here are some estimates to calibrate thinking.
Let $f_k(N)$ be the function you describe. Note that $f_1$ is identically $1$, while $$f_k(N) = \sum_{d\mid(N-1)} f_{k-1}(d)$$ for all $k\ge2$. So for example, $f_2(N) = \tau(N-1)$ where $\tau$ is the number-of-divisors function.
Already this prohibits the possibility that $f_k(N) \ll (\log N)^c$ for any $c$, since there are integers $N-1$ (the primorials, for example) that have at least $\exp((\log2+o(1))\log N/\log\log N)$ divisors.
On the other hand, $\tau(N) \ll_\epsilon N^\epsilon$ for every $\epsilon>0$. From this fact and the recusrive formula for $f_k(N)$, it's easy to deduce that $f_k(N) \ll_{k,\epsilon} N^\epsilon$ by induction on $k$.
|
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Drawing previously undrawn cards from a deck Suppose you have a deck of $y$ cards. First, randomly select $y-x$ distinct cards and sign the face of each, then shuffle all the cards back in to the deck. Proceed as follows:
Draw a card. If it is already signed, replace the card and shuffle the deck. If it is not yet signed, sign it, then replace the card and shuffle the deck. My question is what is the probability that you will draw an unsigned card as a function of time? For instance, when $t=1$, the probability is $x/y$. When $t=2$, the probability is $\frac{x(x-1)}{y^2}+\frac{(y-x)(x)}{y^2}$ and so on. I have written a Python program that will compute the probabilities for small values of $t$, but the run time is $O(2^t)$ and was wondering if there is a simpler way to solve this problem
My solution:
The number of summands for time $t$ is $2^{t-1}$, and each is of the form $\frac{a_1a_2\dots a_t}{y^t}$. The set of $t$-tuples $a_1a_2\dots a_t$ appearing in the sum can be put in bijection with the set of odd binary strings of length $t$ as follows: If the $i$-th digit of the string is 1, then $a_i$ is $x$ minus the sum of the previous digits of the string; otherwise, $a_i$ is equal to $y-x$ plus the sum of the previous digits. For instance, $1101$ corresponds with $x(x-1)(y-x+2)(x-2)$ and $1011$ corresponds with $x(y-x+1)(x-1)(x-2)$. It is pretty simple to write an algorithm that will loop over all such binary strings to find the probability.
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From the description of the problem, we can set up the following recurrence:
\begin{align*}
p(n,x) &= \left(1-\frac{x}{y}\right)p(n-1,x)+\frac{x}{y}q(n-1,x-1) \\
q(n,x) &= \left(1-\frac{x}{y}\right)p(n-1,x)+\frac{x}{y}q(n-1,x-1) \\
p(0,x) &= 0 \\
q(0,x) &= 1 \\
p(n,0) &= q(n,0) = 0
\end{align*}
and the required answer is $p(n,x)$ where $x$ is the number of unsigned cards.
The functions $p$ and $q$ are to know whether the last pick is a signed card or not.
$p(2,x)$ matches with your answer.
Expanding and simplifying the recurrence gives
\begin{align*}
p(n,x) &= \frac{x}{y}\left(\frac{y-1}{y}\right)^{n-1}
\end{align*}
|
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What is the significance of the slope of the tangent line of a function? Why is the derivative so important? As I finished calc 1. I can use the product rule and chain rule and resolve integrals. But I feel like its too mechanical for my taste. I know the procedure and I execute on paper without really understanding or experiencing the "ahaa moment".
For example, when I was learning geometry in elementary school, the "ahaa moment" for me was when I had to move furniture in my room and needed to find areas of stuff. I'm trying to find the equivalent application of the derivative and integral as I learn calculus. Could someone demystify this?
Thank you in advance.
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The differential is the slope or the rate of change.
On a roller-coaster:
*
*Your velocity is the rate of change of position.
*Your acceleration is the rate of change of your velocity.
*Your jerk is the rate of change of your acceleration.
*Your jounce is the rate of change of your jerk.
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Lp space and sequence For what value of $p$ the sequence $\displaystyle x_{n}=\frac{1}{n}$ is on $l^p$ (where $\displaystyle l^p = \lbrace (x_1,x_2,...)| x_{i}\in\mathbb{C}\hspace{0.1cm}\text{and}\hspace{0.1cm} \left(\sum_{i=1}^{\infty}|x_{i}|^p\right)^{1/p}<\infty\rbrace$).
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If p=1, the series diverge. If p>2 it converges, so $x_n \in l^p$ for all $p\ge 2$ (since $l^p \subset l^{p+1}$)
|
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Can zero divisors be in the denominator when we localize rings? Can we localize rings with zero divisors? Can those zero divisors be in the denominator?
I thought defining $$\frac{a}{b}=\frac{c}{d} \text{ iff }t(ad-bc)=0 \text{ where $b,d,t$ belong to the same multiplicative system}$$
accommodated for that little detail. But my professor thinks not. I am confused.
Any help would be greatly appreciated.
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Your definition is correct and even necessary: if you don't include the factor $t$ into the definition, then you will in general not get an equivalence relation between pairs (a,b) of ring elements. This however is necessary to define the notion of a fraction a/b.
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Derivative of a function is the equation of the tangent line? So what exactly is a derivative? Is that the EQUATION of the line tangent to any point on a curve? So there are 2 equations? One for the actual curve, the other for the line tangent to some point on the curve? How can the equation of the tangent line be the same equation throughout the curve?
|
So what exactly is a derivative?
The derivative is instantaneous (i.e. at any given precise moment in time) the rate of change of a dependent variable (usually $y$) with respect to the independent variable (usually $x$).
For straight lines, the derivative is simply the slope or gradient of the line.
For curves, whose gradient (slope) is constantly changing, the derivative gives the instantaneous rate of change at a given point; i.e. the slope of the tangent at that point.
There is one and only one tangent (that's what tangent means in Latin- "touching"to a curve at any given point.
The tangent is a straight line that just touches the curve.
So there are 2 equations? One for the actual curve, the other for the line tangent to some point on the curve?
In a sense, yes. There's the equation of the curve, $y=f(x)$, which could be any function ($x^2, e^x, \ln(x), \tan(x),$ etc.), and then there's the equation of the tangent line at a given point.
If you choose a different point on the curve, not only will you change the gradient of the tangent at the point, but also the equation of the tangent.
How can the equation of the tangent line be the same equation throughout the curve?
This is not true.
I'll show you with a sketch of an arbitrary function $y=f(x).$
I've just picked 3 random points on the curve (denoted by purple dots).
At each of these points, I've drawn the tangents to the curve at that specific point. As you can see, these are clearly different lines, so they can't possibly have the same equation.
|
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Parametric form of curve $\vert z+i\vert = 1$ I need to integrate a complex function through the curve $\vert z+i\vert = 1$.
As far as I know I need the parametric form of this curve.
I know that when I have $\vert z\vert = 1$, the parametric form is something like $\cos(t) + i\sin(t)$.
But what's different when I have that "$+i$"?
|
$$
|z-z_0| = r
$$
is the equation of a circle centered in $z_0$ with radius r. Its parametric form is
$$
z = z_0 + re^{it} = z_0 + r(\cos t + i \sin t)
$$
|
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How were 'old-school' mathematics graphics created? I really enjoy the style of technical diagrams in many mathematics books published in the mid-to-late 20th century. For example, and as a starting point, here is a picture that I just saw today:
Does anybody know how this graphic was created? Were equations used for the surfaces and then a plotting program used? How was the line-hatching achieved? Here is a another gorgeous picture from David Hilbert's "Geometry and the Imagination":
Again, how was this created? Was it done by hand, then scanned in?
More pressingly: how do I create these kinds of images? Certainly, most of us are familiar with Matlab, Geogebra, gnuplot, or other software for creating mathematical figures, as we are also familiar with vector-based programs like Inkscape and Adobe Illustrator. I've looked at 'old-school' programs like IPE (a little bit like XFig), but still, I don't find them as attractive as the examples above. There is then LaTeX solutions like TikZ. I guess they must surely be hand-drawn, but I would like to know about the process for how these were drawn (and the equipment used).
By way of note, there is an article here about trying to use 3d modeling programs and shaders to duplicate hand-drawn figures.
|
Often the illustrations were drawn by hand, by the mathematicians themselves. The book A Topological Picturebook by George K. Francis (Springer, 1987) describes how one learns to do this:
This book is about how to draw mathematical pictures. … Theirs [the geometers of the 19th century] was a wonderfully straightforward way of looking at rather complex things, notably Riemann surfaces and geometrical constructions over the complex numbers. They drew pictures, built models and wrote manuals on how to do this. … I resolved to try to do the same for the mathematics of my contemporaries.
The first example is how to draw a hyperbolic paraboloid on the blackboard:
No software is used, but there are techniques one can learn.
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How to derive the formula to calculate the amount of cubes in a pyramid? The pyramid looks like:
For which I managed to derive the formula for the count of cube sides (ignoring the top). This was easy by simply thinking about it as a triangle:
If we have 4 squares wide pyramid, then the total sides represented graphically looks like:
1*4 xxxx
+2*4 xxxxxxxx
+3*4 xxxxxxxxxxxx
+4*4 xxxxxxxxxxxxxxxx
= 40 sides total.
Which is obviously a triangular area.
So I calculated half of the rectangle + remaining halves of the squares diagonally:
n*(4*n) (4*n)
------- + ----- = 2n^2+2n
2 2
Hurray!
Now to the real problem:
How could I visualize the count of cubes in the same pyramid in a similar way?
My brain just doesn't work in 3d. I had an idea about calculating the volume of pyramid, but I couldn't really get anywhere from there.
The answer is:
2n^3+3n^2+n
-----------
6
But I have now forgotten where I found this piece of magic. It works perfectly, but I don't know why.
So I would like to understand how someone came up with that formula? Preferrably in visual representation which would be easy to understand.
It's really easy to visualize as a loop (n = 4):
1*1 x
+2*2 xxxx
+3*3 xxxxxxxxx
+4*4 xxxxxxxxxxxxxxxx
= 30 cubes total.
But I don't know how to start compressing this loop into one single formula.
Edit: I added one mandatory tag that I could think of somehow relating to this question.
|
Consider each level separately. On the $k$-th level ($k$ starting at $1$ and counting from the top of the pyramid), there are $k^2$ blocks. Hence we just have to compute $1^2+2^2+\cdots+n^2$.
There is a standard result that
$$\sum\limits_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{6}=\frac{2n^3+3n^2+n}{6},$$
which is provable by induction or various counting methods. Here is the wikipedia article on square pyramidal numbers as they are called.
Visual proof (Taken from https://math.stackexchange.com/a/48152/160289), due to Man-Keung Siu. It appeared in the March 1984 issue of Mathematics Magazine:
(n+½)*(n+1)*n
------------- = Amount of cubes.
3
|
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How to tell if a Fibonacci number has an even or odd index Given only $F_n$, that is the $n$th term of the Fibonacci sequence, how can you tell if $n \equiv 1 \mod 2$ or $n \equiv 0 \mod 2$?
I know you can use the Pisano period, however if $n \equiv 1$ or $n \equiv 2$ $\mod \pi(k)$, it can never be found, where $k$ is in $\pi(k)$ (The Pisano period).
Also there is the fact that if $\sqrt{5F_n^2+ 4}$ is an integer then $n
\equiv 0 \mod 2$, but is there a faster way?
Lastly, because $F_1 = F_2 = 1$, that would have to be an exception for whatever rule/formula that would apply.
|
Assuming that $F_n\geq 2$, you can check the parity of $n$ depending on the sign of the difference between $\frac{1+\sqrt{5}}{2}F_n$ and the closest integer. If negative, then $n$ is even, if positive, then $n$ is odd.
|
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Compact $\omega$-limit set $\Rightarrow$ connected Consider the flow $\varphi: \mathbb{R} \times \mathbb{R}^n \to \mathbb{R}^n$ and $L_{\omega}(x)$ the $\omega$-limit set of a point $x \in \mathbb{R}^n$. How can I show that if $L_{\omega}(x)$ is compact, then it is connected? I think one should assume it is connected and then get an absurd. Some help here would be nice.
Also, how can a limit set not be compact or connected? All the (common) examples I can think are compact and connected. Can someone give an exemple?
|
So... After some more thinking I got a proof.
Assume $L_{\omega}(x)$ is not connected. So there are disjoint open sets $A,B\in \mathbb{R}^n$ such that $L_{\omega} \subset A \cup B$ and $A \cap L_{\omega},B\cap L_{\omega}$ are non empty. Therefore, there are sequences $\{t_n\},\{s_n\}$ such that $\displaystyle\lim_{n \to \infty} t_n=\lim_{n \to \infty} s_n=\infty$, $t_n<s_n<t_{n+1}$ and $\varphi_{t_n}(x)\in A, \varphi_{s_n}(x)\in B$ for all $n$.
Now $\{\varphi_t(x); t\in (t_n,s_n)\}$ is a curve joining a point in $A$ to a point in $B$. Then there is a $r_n\in (t_n,s_n)$ such that $\varphi_{r_n}(x) \notin A\cup B$. Note that $\displaystyle \lim_{n \to \infty}r_n=\infty$.
We know that $\{\varphi_{r_n}(x)\}\subset L_{\omega}(x)$. Since it is compact, there is a subsequence of $\{\varphi_{r_n}(x)\}$ converging to a point $y \notin A \cup B$. But that means that $y \in L_{\omega}(x)$. This contradiction completes the proof.
|
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Algebraically, What Does $\Bbb R$ get us? In terms of the basic algebraic operations -- addition, negation, multiplication, division, and exponentiation -- is there any gain moving from $\Bbb Q$ to $\Bbb R$?
Say we start with $\Bbb N$: $\Bbb N$ is closed under addition and multiplication. But then we decide we'd like a number system that's closed under negation as well, so we construct $\Bbb Z$. Great. But then we decide we'd like to extend this number system further to be closed under division and so we construct $\Bbb Q$. The next step is closure under exponentiation - but when we construct that number system, we don't get $\Bbb R$, we get a subset of $\Bbb C$ which I'll call $\Bbb Q_{\exp}$.
Now clearly when constructing $\Bbb R$ from $\Bbb Q$ we do gain completeness, but our gain is then analytic, not necessarily algebraic. Do we gain any algebraic advantage in constructing $\Bbb R$ from $\Bbb Q$ similar to what we get at each of the other constructions I mentioned?
|
I thinks the motivation for $\mathbb{R}$ is not algebraic, but rather it corresponds to our geometric intuition about which numbers are possible. Also it was founded at a time when we viewed the universe as continuous space.
|
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|
Minimize Sum a_i / Sum b_i over subsets I have two positive finite sequences $a_i$ and $b_i$, with $0 \leqslant i \leqslant n$. The problem is to find the subset $I$ of $\{0, ..., n\}$ that minimizes:
$$\frac{\sum_{i \in I} a_i}{\sum_{i \in I} b_i}$$
in an efficient way from the algorithmic point of view. Have you any ideas please?
[EDIT] SORRY: the cardinal of $I$ is a given value in the problem... let's call it $m$.
|
Hint: suppose you have some subset $I$ with a ratio $\frac ab$ and add one more element to make a subset $I'$. Does the ratio increase or decrease?
|
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|
Ways to study mathematics while commuting I spend approximately 3 to 4 hours on public transport everyday. I try to maximize the usage of this time by checking email etc on my phone.
Are there any tips to study mathematics while commuting?
Thanks for sharing! Just want to make full use of the time spent commuting!
|
Go to Amazon and search for whatever branch of mathematics you're looking for. Then change the order of the books to "average customer review" and you'll have a massively curated list of the best books in that topic.
|
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|
Integration of exponential and square root function I need to solve this $$\int_{-\pi}^{\pi} \frac{e^{ixn}}{\sqrt{x^2+a^2}}\,dx,$$
where $i^2=-1$ and $a$ is a constant.
|
By definition, $\displaystyle\int_0^\infty\frac{\cos x}{\sqrt{x^2+a^2}}dx=K_0\big(|a|\big)$, where K is a Bessel function. Letting $x=nt$, we
have $\displaystyle\int_0^\infty\frac{\cos(nt)}{\sqrt{t^2+a^2}}dt=K_0\big(|an|\big)$. Unfortunately, there are no “incomplete” Bessel functions,
so your integral does not possess a closed from even in terms of such special expressions, unless, of
course, $n=0$, in which case the answer is simply $\text{arcsinh}\dfrac\pi{|a|}$ . As an aside, for positive values of a
and n, we have $\displaystyle\int_0^\infty\frac{\sin(nx)}{\sqrt{x^2+a^2}}dx=\frac\pi2\Big[I_0(an)-L_0(an)\Big]$, where I is another Bessel function,
and L is a Struve function.
|
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Distribution related to brownian bridge Let $B(t)$ be a Brownian Bridge and $U$ is uniformly distributed on $(0,1)$. I wish to know the distribution function $B(U)$. Is it possible? As we know, $B(t)\sim N(0,t(1-t))$. But, I haven't a clue when $t$ is replaced by random variable $U$. Could anyone help me?
|
Let $p_t$ denote the PDF of $B(t)$ and assume that $U$ is independent of $B$ with PDF $f_U$, then the distribution of $B(U)$ has PDF
$$
q(\ )=\int p_t(\ )f_U(t)\mathrm dt.
$$
In the present case, $U$ is uniform on $(0,1)$ and, for every $t$ in $(0,1)$,
$$
p_t(x)=\frac1{\sqrt{2\pi t(1-t)}}\mathrm e^{-x^2/(2t(1-t))},
$$
hence
$$
q(x)=\int_0^1\frac1{\sqrt{2\pi t(1-t)}}\mathrm e^{-x^2/(2t(1-t))}\mathrm dt.
$$
The change of variable $4t(1-t)=1/u^2$ yields
$$
q(x)=\frac2{\sqrt{2\pi}}\int_0^\infty\mathrm e^{-2x^2(1+u^2)}\frac{\mathrm du}{1+u^2}.
$$
Differentiating this and identifying $q'(x)$ yields finally
$$
q(x)=2\int_{|x|}^\infty\mathrm e^{-2u^2}\mathrm du.
$$
Note that, to compute some characteristics of the distribution of $B(U)$, one may find more convenient to bypass the PDF $q$ and to go back to the definition of $B(U)$, for example,
$$
E(B(U))=\int_0^1E(B(t))\mathrm dt=0,
$$
and
$$
E(B(U)^2)=\int_0^1E(B(t)^2)\mathrm dt=\int_0^1t(1-t)\mathrm dt=\frac16.
$$
Likewise, for every suitable measurable function $A$,
$$
E(A(B(U)))=\int_0^1E(A(B(t)))\mathrm dt=\int_0^1\int_\mathbb RA(x)p_t(x)\mathrm dx\mathrm dt.
$$
|
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Diameter of a circle using 3 nonlinear points I am trying to find the diameter of a circle using 3 points on its circumference. 2 of the points are 5 feet from eachother while the third point is centered between the other 2. The ceter point is 1 foot from a line drawn between the other 2 points.
|
Hint: If two chords $AB$ and $CD$ of a circle intersect at $P$, then $AP\cdot PB=CP\cdot PD$. Draw the diameter joining the red and green lines in your diagram. So, $AP=PB=\dfrac{5}{2}$ and $CP=1$, and hence $DP=\dfrac{25}{4}$. Now you can compute the diameter.
|
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Finding the basis of an intersection of subspaces
We have subspaces in $\mathbb R^4: $
$w_1= \operatorname{sp} \left\{
\begin{pmatrix} 1\\ 1 \\ 0 \\1 \end{pmatrix} ,
\begin{pmatrix} 1\\ 0 \\ 2 \\0 \end{pmatrix},
\begin{pmatrix} 0\\ 2 \\ 1 \\1 \end{pmatrix} \right\}$,
$w_2= \operatorname{sp} \left\{
\begin{pmatrix} 1\\ 1 \\ 1 \\1 \end{pmatrix} ,
\begin{pmatrix} 3\\ 2 \\ 3 \\2 \end{pmatrix},
\begin{pmatrix} 2\\ -1 \\ 2 \\0 \end{pmatrix} \right\}$
Find the basis of $w_1+w_2$ and the basis of $w_1\cap w_2$.
So in order to find the basis for $w_1+w_2$, I need to make a $4\times 6$ matrix of all the six vectors, bring it to RREF and see which vector is LD and the basis would be the LI vectors.
But the intersection of these 2 spans seems empty, or are they the LD vectors that I should've found before ?
In general, how is the intersection of subspaces defined ?
|
Hint: the intersection of these two spans is NOT empty. What you need to do is find a new spanning set for $w_2$ that contains some of the vectors from the spanning set for $w_1$. The common vectors will span the intersection. Now that you have a basis for $w_1\cap w_2$, you can extend it to a basis of $w_1+w_2$ by adding on the vectors that are in $w_1$ or $w_2$ but not in $w_1\cap w_2$.
|
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Periodic continuous function which is integrable on $\mathbb{R}$ Let $f:\mathbb{R}\to\mathbb{R}$ be a $T$-periodic function, that is $f(t+T)=f(t)$ for all $t\in \mathbb{R}$. Assume that
$$\int_0^{+\infty}|f(s)|ds<+\infty.$$
Now if we assume in addition that $f$ is continuous, my intuition tells me that we must have necessarily $f=0$, is this correct ?
|
This is correct. The way you can see this is by considering the maximum of $|f|$, call it $L$. For any $x$ such that $|f(x)|=L$, we have that $|f(y)| > \frac{L}{2}$ for all $|x-y| < \delta$ (for a sufficient choice of $\delta$). Can you see how to argue it from here?
|
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|
Areas of contemporary Mathematical Physics I have often heard that some developments in Physics such as Gauge Theory, String Theory, Twistor Theory, Loop Quantum Gravity etc. have had a significant impact on pure mathematics especially geometry and conversely.
I am interested in knowing a list of areas of Mathematical Physics which have important and interesting open research problems. Please mention such areas and some references where one can get started in each of them.
Just to give an idea of what I have in mind following areas come to my mind for example when I say Mathematical Physics : Knot Theory, Mirror Symmetry, Atiyah-Singer Index Theorem & Dirac Operators, Topological Quantum Field Theory etc. I believe that such listing could be useful to other members of the m.se community as well.
Is there an article/website/blog where I can find such listing ?
I had earlier asked a question about the existence of a website similar to string wiki, but unfortunately it does not seem to exist. Unfortunately this does not have a very systematic classification of sub areas of Mathematical Physics though it does provide some references. Another very useful website exists for Physics but I am unaware of a similar one for Maths.
Please note that my question deals with interactions between Pure Mathematics and Fundamental Theoretical Physics. There are interesting and valuable aspects like applications of mathematics in statistical mechanics or fluid mechanics but for the purposes of this question, let us exclude them.
Edit : If it is not possible to give a complete listing, please mention some main areas along with canonical references. To give a better idea of what kind of things I am looking for here are two examples Advanced CFT and Differetntial Topology and QFT though suggestions do not have to be in these directions.
|
Here is the 2010 Mathematics Subject Classification list. It has 6500 entries, working out the mathematical physics projection operator is left as an exercise to the reader.
|
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Lagrange Bürmann Inversion Series Example I am trying to understand how one applies Lagrange Bürmann Inversion to solve an implicit equation in real variables(given that the equation satisfies the needed conditions). I have tried looking for examples of this, but all I have found is the wikipedia article for the topic and the examples there were too rushed(or requiring too much knowledge of an outside topic) for me to understand.
Could someone please walk me through an example of how to use this (beautiful) theorem so that I may use it for myself?
|
An example of solution of a transcendental equation by means of Lagrange inversion can be the following.
Consider the transcendental equation:
$$(x-a)(x-b) = l e^ x $$
You can rewrite as:
$$x = a+ \frac{l e^ x}{(x-b)} $$
Applying Lagrange inversion:
$$x = a+ \sum_{n=1}\frac{l^n}{n!}\left[\left(\frac{d}{dx}\right)^{n-1}\frac{e^{nx}}{(x-b)^n} \right]_{x=a}$$
and developing the derivative, you can find an explicit solution:
$$x=a+\sum_{n=1}^{n=\infty} \frac{1}{n n!}\left( \frac{nle^a}{a-b}\right)^n B_{n-1}\left( \frac{-2}{n(a-b)}\right)$$
where $B_n(x)$ are the Bessel polynomials are defined as:
$$B_n(x)=\sum_{k=0}^n\frac{(n+k)!}{(n-k)!k!}\left(\frac{x}{2}\right)^k$$
See: http://en.wikipedia.org/wiki/Bessel_polynomials
Another solution can be obtained swapping $a$ with $b$:
$$x=b+\sum_{n=1}^{n=\infty} \frac{1}{n n!}\left( \frac{nle^b}{b-a}\right)^n B_{n-1}\left( \frac{-2}{n(b-a)}\right)$$
A numerical example:
$$x^2-1=e^{x-1}$$
gives as solutions:
$$x=1+\sum_{n=1}^{n=\infty} \frac{1}{n n!}\left( \frac{n}{2}\right)^n B_{n-1}\left( \frac{-1}{n}\right) = 1.78947...$$
$$x=-1+\sum_{n=1}^{n=\infty} \frac{1}{n n!}\left( \frac{-ne^{-2}}{2}\right)^n B_{n-1}\left( \frac{1}{n}\right) = -1.0617135...$$
For details, see :
"Generalization of Lambert W-function, Bessel polynomials and transcendental equations"
http://arxiv.org/abs/1501.00138
|
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Area of a spherical cap formed by the plane containing any side of an inscribed regular tetrahedron I was trying to think about this problem today and realized that practically all of my high school geometry has deserted me, so "how to find it" answers would be greatly appreciated. As to the actual problem: Imagine that a unit sphere has a regular tetrahedron inscribed in it. The plane containing any arbitrary side of the tetrahedron cuts the sphere's surface into two parts with different areas. What are the areas of the two parts, and how would I go about finding this sort of thing on my own?
|
(Another hint) Find the distance from the center of the inscribed tetrahedron to the center of one of its faces, call that $r$ [Once one can get the coordinates of some regular tetrahedron's vertices and center, after rescaling one can get this $r$.]
Once that $r$ is known, there is likely an available formula on-line for the area of a spherical cap given the distance of the cutting plane from the sphere origin. Alternately maybe this can be set up as a surface integral and done directly.
|
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Prove that $\dfrac{0.5x^2 + x + 1}{x^2 + x + 1}$ is a strictly decreasing function. This is part of an actuarial science problem. Unfortunately, the official solution of this problem takes the derivative of
$$\dfrac{0.5x^2 + x + 1}{x^2 + x + 1}\text{, } \quad x \geq 0\text{.}$$
and shows that it is always $\leq 0$. However, this does not at all show that the function is strictly decreasing.
I'm trying to prove this myself. If I assume $x > y$, I want to show that
$$\dfrac{0.5x^2 + x + 1}{x^2 + x + 1} < \dfrac{0.5y^2 + y + 1}{y^2 + y + 1}\text{.}$$
Needless to say, this does not look clean if I were to "work backwards."
Any suggestions?
|
Differentiating is a clumsy way of solving the problem. However, let's look at the derivative. It is equal to
$$-\frac{x(0.5x+1)}{(x^2+x+1)^2}.$$
The denominator is bounded away from $0$. The numerator is negative for $x\gt 0$. Thus (Mean Value Theorem) our function is strictly decreasing in the interval $(0,\infty)$, indeed in the interval $[0,\infty)$.
|
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What am I supposed to do here now? $\tan(\pi/8) = \sqrt{2} -1$ complex analysis
Find $\sqrt{1+i}$, and hence show $\tan(\pi/8) = \sqrt{2}-1$
Okay so I know that $\sqrt{1+i} = 2^{1/4}e^{i\pi/8}$ and I know $\sin x = \frac{e^{ix} - e^{-ix}}{2i}$ and $\cos x = \frac{e^{ix} + e^{-ix}}{2}$
If i directly substitute those definitions of sine and cosine into $$\tan(x) = \frac{\sin(x)}{\cos(x)}$$
then I am going to end up with a complex number in the form of $x + iy$. My key says
$$\tan \pi/8 = \sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}} = \sqrt{2}-1.$$
Did they just use $\Re(\tan \pi/ 8) = \frac{\Re (\sin )}{\Re \cos}$?
|
You can work it out by rationalizing *
You can use a half-angle formula for Tan, i.e., a formula for Tan(B/2)
$tan(B/2) = (1 − cos B) / sin B = sin B / (1 + cos B)$
For CosB=SinB =$\sqrt \frac{2}{2}$, then , $Tan \pi/8= \frac{\frac{\sqrt{2}}{2}}{1+\frac{\sqrt{2}}{2}}=\frac{\sqrt2}{2+\sqrt2}=\frac{2\sqrt2-2}{2}=\sqrt{2}-1$
If you want to arrive at the actual values of $\sqrt{2}-1, \sqrt{2}+1$ for sin, cos, you can use DeMoivre's theorem:
$(Cos\theta+iSin\theta)^{1/2}=(Cos\theta/2+ iSin \theta/2)$, and then you can use half-angle formulas for each of sine and cosine:
$cos(B/2) = ± \sqrt{([1 + cos B] / 2)}$
$Sin(B/2) = ±\sqrt{([1 - cos B] / 2})$
And in this case, $SinB=CosB= \frac{\sqrt{2}}{2}$
|
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Induced Lie algebra homomorphism from Lie group homomorphism: left translation A general result of Lie Theory is that every Lie group homomorphism $\Phi: G\rightarrow H$ induces a Lie algebra homomorphism $\phi: \frak{g} \rightarrow \frak{h}$.
Which Lie algebra homomorphism is induced by left (or right)-translations:
$L_g h = gh$ for $h,g \in G$, which is a map $G \rightarrow G$ ?
A first idea would be looking at the corresponding pushforward: $L_{ g \star} $ is a map between tangent vectors at $g$ and $h$ respectively. For $ X \in T_g G$ the pushforward is
$L_{ g \star} X = X' \in T_{gh} G$ and therefore this is a map between different tangent spaces which does not help me, because A Lie algebra homomorphism has to be a map from $T_e G$ to $T_e G$.
Any help or ideas finding the corresponding Lie algebra homomorphism for left translations would be much appreciated.
|
Left and right translations are not Lie group homomorphisms; they don't even preserve the identity, and the induced map on Lie algebras is obtained by looking at derivatives at the identity. However, conjugation by a fixed element $g \in G$ is, and the induced map on $\mathfrak{g}$ gives a representation $G \to \text{Aut}(\mathfrak{g})$ called the adjoint representation. This is itself a Lie group homomorphism, and differentiating it gives the adjoint representation
$$\mathfrak{g} \ni x \mapsto (y \mapsto [x, y]) \in \text{Der}(\mathfrak{g})$$
of $\mathfrak{g}$ (and this is one way to define the Lie bracket).
|
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Group homomorphism $f$ is surjective iff $g$ is Let $G$ be an additive group, and let $u, v:G\to G$ to be two endomorphisms. Define $f(x) = x- v(u(x))$ and $g(x) = x-u(v(x))$. The question is to show that $f$ is surjective iff $g$ is.
I'm only able to show that $u:\ker f\cong \ker g$, but unable to show the statement needed. Observe that $u(f(x)) = g(u(x))$ and $v(g(x)) = f(v(x))$.
|
We have the following general fact: In a ring $R$ (not assumed to be commutative) with elements $u,v$, if $1-uv$ is invertible, then $1-vu$ is invertible. In fact, one checks that
$(1-vu)^{-1} = 1 + v (1-uv)^{-1} u$
(This formula has a nice motivation using the geometric series.)
We may apply this to the ring $R=\mathrm{End}(G)$ for an abelian group $G$ (with $1=\mathrm{id}_G$) and see that $1-uv$ is an isomorphism if and only if $1- v u$ is an isomorphism.
But we can also use the formula above to prove that $1-vu$ is surjective when $1-uv$ is surjective:
Let $g \in G$. Choose some $h \in G$ with $(1-uv)(h) = u(g)$, i.e. $h - u(v(h))=u(g)$. Then $g + v(h) \in G$ satisfies
$$(1-vu)(g + v(h))=g + v(h) - v(u(g+v(h)))=g + v(h - u(v(h)) - u(g))=g.$$
This proof also works when $G$ is not abelian (where $x-y$ abbreviates $x+(-y)$ and we use $-(x+y)=(-y)+(-x)$ in the calculation).
|
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Show that this matrix is invertible I have the following exercise:
Show that the matrix $A=(a_{ij})$ where $a_{ij}=i^{j-1}$, $i,j=1, \dots n$ is invertible.
Do I have to show that the determinant is equal to $0$?
|
From the wiki article on Vandermonde matrices, the determinant of the mentioned matrix would be
$$det = \prod_{1 \le i < j \le n}(j - i)$$
But since $j > i$, the above product will always be positive, i.e. not equal to $0$.
|
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Please check my solution of $\int \sin^6(x)\cos^3(x) dx$ $$\int \sin^6(x)\cos^3(x) dx = \int \sin^6(x)(1-\sin^2(x))\cos(x)dx$$
$$\int \sin^6(x)\cos(x)dx - \int\sin^8x\cos xdx$$
Now, $\cos xdx = d(\sin x)$
$$\int u^6du - \int u^8du = \frac{1}{7}u^7 - \frac{1}{9}u^9 + C$$
$$\frac{1}{7}\sin^7(x) - \frac{1}{9}\sin^9(x) + C$$
However, WolframAlpha says it's:
Can anyone tell if those expressions are equal?
|
$11 + 7 \cos 2x = 11 + 7 - 14\sin^2 x = 2(9 - 7\cos^2 x) \Rightarrow\\
\dfrac {1}{126} (11 + 7 \cos 2x) = \dfrac{1}{63}(9 - 7\sin^2x) = \boxed{\dfrac{1}{7} - \dfrac{\sin^2x}{9}}\Rightarrow\\
\\
\text{The expressions are equal.}$
|
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"url": "https://math.stackexchange.com/questions/850618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
The Cesàro Mean Theorem in the infinite case. I am trying to prove the Cesàro Mean Theorem in the infinite case. Let me state my problem more precisely.
Problem. Let $ (a_{n})_{n \in \mathbb{N}} $ be a sequence in $ \mathbb{R} $ such that $ \displaystyle \lim_{n \to \infty} a_{n} = \infty $. Then prove that
$$
\lim_{n \to \infty} \frac{1}{n} \sum_{k = 1}^{n} a_{k} = \infty.
$$
I tried to use only the definition of convergence, but I got stuck. I know that the result is an immediate consequence of the Cesàro-Stolz Theorem, but I am looking for a solution using the definition of convergence alone. Any suggestions? Thanks to all helpers!
|
Since $a_{n}\to\infty$ as $n\to\infty$ then there exists $N$ such that if $n\ge N$ then $a_{n}\ge M>0$. This means at most finitely many terms are negative. By choosing $N$ large enough we may also assume that $\sum_{k=1}^{n}a_{k}>0$ for $n\ge N$. By perhaps choosing $n$ even larger we may assume that $\lvert\frac{\sum_{k=1}^{N}a_{k}}{n}\rvert<1$ So:
$\frac{\sum_{k=1}^{n}a_{k}}{n}=\lvert\frac{\sum_{k=1}^{n}a_{k}}{n}\rvert\ge\frac{\sum_{k=N+1}^{n}a_{k}}{n}-\frac{\sum_{k=1}^{N}a_{k}}{n}>\frac{n-N-1}{n}M-1=(M-1)-\frac{N+1}{n}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
how to derive the fact that the integral of $1/\sin^2(x) = -\cot (x)$ I know how that the integral of $\dfrac{1}{\sin^2(x)} = -\cot (x)$, but how does derive this fact? Can you use half-angle formula to do this integral?
|
When you asked how to "derive" the fact that $\int \frac{1}{\sin^2 x} dx = -\cot x$, I thought maybe someone had suggested that this fact might be true, and asked if you could prove it. If that were so, then you would merely need to differentiate $-\cot x$.
In fact I would not be at all surprised to learn that the first person to discover how to
integrate $\frac{1}{\sin^2 x}$ was not actually trying to solve that problem specifically, but rather was simply interested in finding the derivative of $f(x) = \cot x$. Once they had done that, they knew the integral of $f'(x)$, which means they also knew how to integrate $-f'(x) = \frac{1}{\sin^2 x}$.
Most of the other methods described here are useful things to know about. But it all comes down to remembering a formula. I remember hating this part of first-year calculus, because it seemed like too much rote memorization of facts, especially the ones that said "the integral of ... is ...". I didn't want to have to remember all that. I probably would have been happier if I had just decided to make the best of it and had made myself a set of flash cards.
What you might do is to remember that $\int \sec^2 x dx = \tan x$ and that
$\sec^2 x = \frac{1}{\cos^2 x}$, and you might ask yourself whether the integral of $\frac{1}{\cos^2 x}$ might not be closely related to the integral of $\frac{1}{\sin^2 x}$.
In fact the derivatives and integrals of trigonometric functions are closely related to those of their cofunctions, since $\sin x = \cos\left(\frac{\pi}{2} - x\right)$ and so forth. Well, then, given that $\tan x$ is the answer to $\int\frac{1}{\cos^2 x} dx$,
will $\cot x$ solve the new problem? Almost; it turns out that
$$\frac{d}{dx} \cot x = -\frac{1}{\sin^2 x},$$
so the solution to your problem is $-\cot x$.
Another method that seems slightly less like guesswork:
knowing that $\sin x = \cos\left(\frac{\pi}{2} - x\right)$ and that we can integrate
$\frac{1}{\cos^2 x} = \sec^2 x$,
substitute $u = \frac{\pi}{2} - x$.
Then $x = \frac{\pi}{2} - u$ and $dx = -du$, so
$$
\begin{eqnarray*}
\int \frac{1}{\sin^2 x} dx &=& \int -\frac{1}{\sin^2 \left(\frac{\pi}{2} - u\right)} du \\
&=& -\int \frac{1}{\cos^2 u} du \\
&=& -\tan u \\
&=& -\tan \left(\frac{\pi}{2} - x\right) \\
&=& -\cot x.
\end{eqnarray*}
$$
Or you might use the tangent substitution method explained by @Felix Marin, especially if it's on an exam and you think of that method first.
I was happier in second-year calculus and much happier in third-year real analysis, by the way, because it became more about "big ideas" and not so much about the details of how to compute this or that function. (And along the way, a lot of the techniques I had to memorize in high school came to make a lot more sense in light of other patterns I was learning.)
|
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|
Matlab Code to simulate trajectories of Ito process. I need some help to generate a Matlab code in order to do the following question. Can somebody help me in this regard. Any sort of hint that could be helpful will surely be appreciated..
Q: "Simulate $N=25$ trajectories of the Ito Process X satisfying the following SDE
$dXt = \mu X_tdt + \sigma X_tdB_t.$
with $X_0=1$, $\mu=1.5$, $\sigma=1.0$ and their Euler approximations with equidistant time steps of size $\Delta=2^{-4}$ corresponding to the same sample paths of the Wiener process on the time interval $[0,T]$ for $T=1$. Evaluate the absolute error by the statistic defined below
m=$\frac1N$$\sum_{k=0}^N $|$X_{T,k}$-$Y_{T,k}$|
where $X_{T,k}$ and $Y_{T,k}$ respectively are the $k$-th simulated trajectories of Ito process and their Euler approximation corresponding to same sample paths of the Wiener process"
I have created the following code on Matlab for the above question. Can somebody correct me if I'm wrong somewhere.
randn('state',100)
mu=1.5; sigma=1; Xzero=1;
T=1; N=25; dt=T/N;
dW=sqrt(dt)*randn(1,N);
W=cumsum(dW);
Xtrue=Xzero*exp((mu-0.5*sigma^2)*([dt:dt:T])+sigma*W);
Xem=zeros(1,N);
Xem(1)=Xzero+dt*mu*Xzero+mu*Xzero*dW(1);
for j=2:N
Xem(j)=Xem(j-1)+dt*mu*Xem(j-1)+sigma*Xem(j)*dW(j);
end
|
The increment of Brownian motion $B_{t+ \Delta }- B_t$ is normally distributed with mean $0$ and standard deviation $\sqrt{\Delta}.$
Generate a sample path using the discrete Euler approximation:
$$X_{k+1}=X_{k} + \mu X_k \Delta + \sigma X_k\sqrt{\Delta}\xi\,\,(k=1,2,...),$$
where $\xi$ is a random number with a standard normal distribution.
To generate random samples for $\xi$, first generate a uniformly distributed random number $r \sim$ U(0,1) and take $\xi = N^{-1}(r)$ where $N$ is the standard normal distribution function.
|
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|
Volume of the cooking pot
A cooking pot has a spherical bottom, while the upper part is a truncated cone. Its vertical cross-section is shown in the figure.If the volume of food increases by 15% during cooking, what is the maximum initial volume of food that can be cooked without spoiling ?
It is clear that we have to evaluate the volume of the cooking pot, first. One important observation that I had made was that that whole cooking pot is a part of the spherical sector(of the sphere of which the bottom is a part). So if we can evaluate the volume of the spherical sector and subtract the volume of the truncated cone problem will be solved. However I am unable to evaluate the volume of the spherical sector.
Any other method of solving the sum will also be accepted.
As a twelfth standard student, a well-explained solution (preferably with diagram) will be necessary for my clear understanding.
|
So first consider the bottom spherical part. We know that the distance in the y (vertical) direction from the top of the spherical part to the bottom is 20cm and that the distance across is 40. It should be easy to see that this implies that the radius is 20cm and that we are dealing with a half-sphere. Thus, we can use the formula for the the volume of a sphere $\frac{4}{3}*\pi*r^3$ and divide by 2 where r = 20cm.
Next we need the conical part. We can think of this as two cones, one is missing however, which is the top part. So, we can calculate the big cone and subtract the small cone to get the almost-cone's volume. We can use the standard formulae here too. $V=\pi*r^2*h$ So we have a height of 20 (explained later) and r =20. Finally we need the small cone, which has a radius of merely 10. But we don't have the height? However, we do have that the cone is the same cone as before right? So the slope of this missing cone must be the same. So it must have half the height (of the total) because it has half the radius. So we have $V=\pi*r^2*h$ with r = 10 and h =10.
As a result we have $\frac{4}{6}*\pi*20^3 + \pi*20^2*20 - \pi*10^2*10$
|
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|
How to find $\int_0^{\pi}\frac{\sin n\theta}{\cos\theta-\cos\alpha}d\theta$ I was doing some work in physics and I came up with a definite integral. I tried everything I could but couldn't solve the integral. The integral is
$$
\int_0^\pi {\sin\left(n\theta\right)\over
\cos\left(\theta\right) - \cos\left(\alpha\right)}\,{\rm d}\theta\,,
\qquad\qquad n\ \in\ {\mathbb Z}\,,\qquad 0\ \leq\ \alpha\ \leq\ 2\pi
$$
There is singularity at $\alpha=\theta$ which increases its difficulty.
I tried complex analysis but couldn't solve it.please help me with method,also provide an answer with proof if you like, I would appreciate it.
|
Thanks to complex analysis, it is rather easy to obtain
$$
\int_0^{\pi}{\cos\left(n\theta\right)\over
\cos\left(\theta\right) - \cos\left(\alpha\right)}\,{\rm d}\theta=\frac{\pi cos(n\alpha)}{sin(\alpha)}
$$
By the way, this result is also obtained in attachment, but with a method much more complicated than usual. In fact, this complicated method is proposed for the much more difficult integral :
$$
\int_0^{\pi}{\sin\left(n\theta\right)\over
\cos\left(\theta\right) - \cos\left(\alpha\right)}\,{\rm d}\theta\
$$
The closed form obtained involves the Incomplete Beta function in the complex range. Nevertheless, one of the parameters is nul, which makes think that further simplification might be possible (may be involving polylogarithms)
Several numerical checking were done. Of course, for the numerical computation of the integral (noted $I$ in attachement), the Cauchy principal value is considered.
|
{
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|
Help to evaluate this limit $\lim_{x \to \infty}x^{\frac{1}{x}}$ What is the value of this limit?
$$
\lim_{x \to \infty}x^{\frac{1}{x}}
$$
I have never encountered such a limit before, so any help or advice would be much appreciated.
|
An approach similar to G Tony Jacobs: use the continuity of logarithm (i.e. $\log \lim f(x) = \lim \log f(x)$) to log the expression to get
$$
L f(x) = \frac{\log x}{x}
$$
then show it converges to $0$ by L'Hospital's rule, then exponentiate back to get 1.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to prove $\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{m^2+n^2}=+\infty$ How to prove $$\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{m^2+n^2}=+\infty.$$
I try to do like $$\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{m^2+n^2}=\sum_{N=1}^\infty \sum_{n+m=N}^\infty \frac{1}{m^2+n^2}=\sum_{N=1}^\infty \sum_{m=1}^{N-1} \frac{1}{m^2+(N-m)^2}$$
$$\frac{1}{m^2+(N-m)^2}\leq \frac{2}{N^2}$$
but it doesn't work.
|
$$
\begin{align}
\sum_{m=1}^\infty\sum_{n=1}^\infty\frac1{m^2+n^2}
&\ge\sum_{m=1}^\infty\sum_{n=1}^\infty\frac1{m^2+n^2+2mn+m+n}\\
&=\sum_{m=1}^\infty\sum_{n=1}^\infty\left(\frac1{m+n}-\frac1{m+n+1}\right)\\
&=\sum_{m=1}^\infty\frac1{m+1}\\[6pt]
&=\infty
\end{align}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Exercise with matrix A) For which $a,b$ is the matrix $A=\begin{bmatrix}
a & 0\\
b & b
\end{bmatrix}$ invertible?
B) Calculate $A^{1000}$ where $A$ is the above matrix with $a=1$ and $b=2$.
$$$$
I have done the following:
A) $$\begin{vmatrix}
a & 0\\
b & b
\end{vmatrix} \neq 0 \Rightarrow ab \neq 0 \Rightarrow a \neq 0 \text{ and } b \neq 0$$
B) $$A=\begin{bmatrix}
1 & 0\\
2 & 2
\end{bmatrix}$$
How can I calculate $A^{1000}$??
Is it maybe $$A^n=\begin{bmatrix}
1 & 0\\
2 \cdot n & 2^n
\end{bmatrix}$$
|
Your matrix is diagonalizable, because u have 2 different eigenvalues $\lambda_1=1$ and $\lambda_2=2$. Then you find an invertible matrix $S$, such that $SAS^{-1}=D$ with a diagonal matrix D, which contains your eingenvalues:
Now we obtain the following:
$A^{1000}=(S^{-1}DS)^{1000}=S^{-1}DS*S^{-1}DS*....*S^{-1}DS=S^{-1}D^{1000}S$ (Because $S^{-1}S=I$).
But $D^{1000}=\begin{pmatrix} 1 & 0 \\ 0 & 2^{1000} \\ \end{pmatrix}$. Then it follows that : $A^{1000}=S^{-1}\begin{pmatrix} 1 & 0 \\ 0 & 2^{1000} \\ \end{pmatrix} S$
|
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|
Finding the range and domain of $f(x)=\tan (x)$ I am attempting to find the range and domain of $f(x)=\tan(x)$ and show why this is the case. I can seem to find the domain relatively well, however I run into problems with the range. Here's what I have done so far.
Finding the domain of $f(x)=\tan(x)$
Consider $f(x)=\tan(x)$ is defined as $f(x)=\tan(x)=\frac{\sin(x)}{\cos(x)}$, it is clear the domain of $f(x)$ is undefined when $\cos(x)=0$. $\cos(x)=0$ whenever $x=\frac{\pi}{2}+\pi k$ for integers $k$, so the domain of $f(x) =\tan (x)$ can be stated as $x\in\mathbb{R}, x \ne \frac{\pi}{2}+\pi k\text{ for integers k}$
Finding the range of $f(x)=\tan(x)$
To find the range of $f(x)=\tan(x)$ we must refer to the definition of $f(x)=\tan(x)=\frac{\sin(x)}{\cos(x)}$. From this we can see that $f(x)$ is undefined when $\cos(x)=0$, as the interval of $\cos(x)$ is $[-1,1]$ we will now need to split this into two cases: $-1\leqslant\cos(x)<0$ and $0<\cos(x)\leqslant1$.
Considering the first interval; $-1\leqslant\cos(x)<0$, as $\cos(x)\to0^-$: $\sin(x)\to1$ and $\tan(x)=\frac{\sin(x)}{\cos(x)}\approx\frac{1}{\text{very small(negative)}}\approx\text{very big(negative)}$.
In other words as $\cos(x)\to0^-$, $\tan(x)\to-\infty$
Considering the second interval; $0<\cos(x)\leqslant1$, as $\cos(x)\to0^+$: $\sin(x)\to1$ and $\tan(x)=\frac{\sin(x)}{\cos(x)}\approx\frac{1}{\text{very small(positive)}}\approx\text{very big(positive)}$.
In other words as $\cos(x)\to0^+$, $\tan(x)\to+\infty$.
This is where I am up to. What I want to know is how I can show definitively that $tan(x)$ can take all the values within the interval $[-\infty,\infty]$. Some people would call it a day here, and say that this shows that the range of $f(x)$ is $-\infty<\tan(x)<\infty$. However I nearly ran into a similar error when I was finding the range of $\sec(x)$, only to discover that although it does tend to positive and negative infinity it doesnt take any values in the interval $(-1,1)$. Where do I proceed from here?
EDIT: I am not looking for answers that use differentiation. Answers should be pre-calculus level.
|
It is easiest to use the intermediate value theorem when finding the range : You know that
$$
\lim_{x\to \pi/2} \tan(x) = +\infty \text{ and } \lim_{x\to -\pi/2} \tan(x) = -\infty
$$
So the image of the interval $(-\pi/2, \pi/2)$ must be $\mathbb{R}$
|
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|
Power iteration If $A$ is a matrix you can calculate its largest eigenvalue $\lambda_1$. What are the exact conditions under which the power iteration converges? Power iteration
Especially, I often see that we demand that the matrix is symmetric? Is this necessary?
What seems to be indespensable is that there is a largest eigenvalue (absolute value is large). But what about the structure of the eigenspace of this eigenvalue?
Apparently, many times it is not considered that the eigenspace to this largest eigenvalue does not need to be one-dimensional. So what happens if the eigenspace is two-dimensional. Can we still use this algorithm?
|
Let $$M:=\begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix},\ \text{then}\quad M \begin{pmatrix}1\\ 1 \end{pmatrix}=1\cdot \begin{pmatrix}1\\ 1 \end{pmatrix},\ \text{and} \quad M \begin{pmatrix}1\\ -1 \end{pmatrix}=-1\cdot \begin{pmatrix}1\\ -1 \end{pmatrix} $$
Now let $x^0 =(x^0_1,x^0_2) \neq (0,0)$, then the sequence $x^{k+1} = Mx^k$ will never converge since $M$ just switch the coordinates of $x^k$. If you want a matrix with maximal eigenvalue (without absolute value) is not simple and the algorithm don't converge, consider $$A = \begin{pmatrix}0 & 1 & 0 \\ 1 &0 & 0 \\ 0& 0 &1\end{pmatrix}, \quad A\begin{pmatrix}1 \\ 1 \\ 0 \end{pmatrix}=1\cdot \begin{pmatrix}1 \\ 1 \\ 0 \end{pmatrix},\quad A\begin{pmatrix}1 \\ -1 \\ 0 \end{pmatrix}=-1\cdot \begin{pmatrix}1 \\ -1 \\ 0 \end{pmatrix},\quad A\begin{pmatrix}0 \\ 0 \\ 1 \end{pmatrix}=1\cdot \begin{pmatrix}0 \\ 0\\ 1 \end{pmatrix}, $$
and the sequence $x^{k+1} = Ax^k$ will never converge except if you start directly with a eigenvector.
|
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|
Why set of natural numbers is infinite, while each natural number is finite? In his book Analysis Vol. 1, author Terence Tao argues that while each natural number is finite, the set of natural numbers is infinite (though has not defined what infinite means yet). Using Peano Axiom, if a property holds for P(0) and whenever P(n) is true, P(n+1) is also true, then it is true for all natural numbers. [See image attached at the end.]
However, he has not provided an argument/proof why the set of natural numbers in infinite.
If we go by the same argument, the set of natural numbers should also be finite.
Just like finiteness let’s say P is property called count associated with each natural number. Count can be defined as P(n) = n+1. (Intuitively count means number of elements in the set till n, or the number of elements in the set till n). Now P(0) = 1, which is finite. If P(n) is finite (i.e. n+1), then P(n+1) will also be finite. Hence, the number of elements in the set of natural numbers should also be finite.
|
The fact that the infinitude of the natural numbers is effectively taken as an axiom in set theory illustrates to some degree the difficulty in proving that there are infinitely many natural numbers. And you are correct to observe that "infinite" needs to be correctly defined before it makes sense to claim the natural numbers are infinite in number.
What we can quickly and easily do, is to prove that your claim there are finitely many natural numbers, is contradictory.
If as you state, $P(n)$ counts every natural number then since $P(n)=n+1$ it follows by the law of induction that $P(n)$ is itself a natural number and also by induction that $P(n)+1$ is a natural number. Then the contradiction immediately follows that for every $P(n)$ there is some integer greater than it so every $P(n)$ fails to enumerate the natural numbers.
It does not take a huge leap of logic to add to our system the axiom that some number exists called $\infty$ which is greater than all natural numbers.
Intuitively, my preference is to a) aspire to only use infinity in supposition with a view to demonstrating a contradiction and b) also I prefer the notion that infinity is algebraically independent of addition rather than that it exists and is greater than every integer, but this is VERY FAR from standard.
I don't think it's wise to criticise a Fields medal winner for not providing a proof of an Informal Remark. You can rest assured he is extremely adept at constructing rigorous formal arguments when required to do so.
|
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|
An Integration Calculation I'm just having a bit of difficulty understanding the last couple of steps made in the paper Horowitz & Hubeny - Quasinormal Modes of AdS Black Holes and the Approach to Thermal Equilibrium (p.8) which can be found at this link where the following is stated
$$\int_{r_+}^{\infty}dr[f|\psi'|^2+2i\omega\bar{\psi}\psi'+V|\psi|^2]=0$$
and taking the imaginary part gives
$$\int_{r_+}^{\infty}dr[\omega \bar{\psi}\psi'+\bar{\omega}\psi\bar{\psi'}]=0.$$
Integration by parts of the second term yields
$$(\omega - \bar{\omega})\int_{r_+}^{\infty}dr \bar{\psi} \psi'$$
$$=\bar{\omega}|\psi(r_+)|^2$$
given that $\psi(\infty)=0$ and $\psi'$ denotes differentiation w.r.t $r$. Substituting this final result back into the first equation we obtain
$$\int_{r_+}^{\infty}dr[f|\psi'|^2+V|\psi|^2]=\frac{|\omega^2|\psi(r_+)|^2}{Im\omega}.$$
My problem lies in showing these last two results; namely finding $\bar{\omega}|\psi(r_+)|^2$ from the previous equation and then showing the final substitution.
I've been trying for a very long time with integration by parts, using some complex identities involving the conjugate etc., but I can't arrive at the final result.
This is simply a case of me trying to fully understand a paper I'm interested in. Any help would be greatly appreciated.
|
Adding more intermediate steps:
\begin{align*}
\int_{r_+}^{\infty}dr[\omega\bar{\psi}\psi'+\bar{\omega}\bar{\psi}'\psi]&=
\int_{r_+}^{\infty}dr[\omega\bar{\psi}\psi'-\bar{\omega}\bar{\psi}\psi'+\bar{\omega}\bar{\psi}\psi'+\bar{\omega}\bar{\psi}'\psi]=\\
&=\int_{r_+}^{\infty}dr[\color{red}{\omega\bar{\psi}\psi'-\bar{\omega}\bar{\psi}\psi'}+\color{blue}{\bar{\omega}(\bar{\psi}\psi)'}]=\\
&=\color{red}{(\omega-\bar{\omega})\int_{r_+}^{\infty}dr\,\bar{\psi}\psi'}+
\color{blue}{\bar{\omega}|\psi|^2\biggl|_{r_+}^{\infty}}=\\
&=(\omega-\bar{\omega})\int_{r_+}^{\infty}dr\,\bar{\psi}\psi'-\bar{\omega}|\psi(r_+)|^2.
\end{align*}
Since the initial expression was previosly shown to be zero, we obtain the first result.
To show the second result, we substitute $\int_{r_+}^{\infty}dr\,\bar{\psi}\psi'$ by
$\displaystyle \frac{\bar{\omega}|\psi(r_+)|^2}{\omega-\bar{\omega}}$ in the 2nd term of your first equation:
$$2i\omega \int_{r_+}^{\infty}dr\,\bar{\psi}\psi'=2i\omega\cdot \frac{\bar{\omega}|\psi(r_+)|^2}{\omega-\bar{\omega}}=\frac{2i}{\omega-\bar{\omega}}|\omega|^2|\psi(r_+)|^2
=\frac{1}{\Im\omega}|\omega|^2|\psi(r_+)|^2=\frac{|\omega^2||\psi(r_+)|^2}{\Im\omega}.$$
|
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|
How to get $(\frac{x^2}{2}+\frac{1}{2x^2})^2$ from $1+(\frac{x^2}{2}-\frac{1}{2x^2})^2$? How can I get $(\frac{x^2}{2}+\frac{1}{2x^2})^2$ from $1+(\frac{x^2}{2}-\frac{1}{2x^2})^2$?
The book lists the former as the solution to that step.
This is part of an arc length problem, and I think I'm just hitting a mental roadblock on solving this.
|
$$\begin{eqnarray*}
1 + \left(\frac{x^2}{2} - \frac{1}{2x^2}\right)^2
&=& 1 + \left(\frac{x^2}{2}\right)^2 - 2\left(\frac{x^2}{2} \cdot \frac{1}{2x^2}\right) + \left(\frac{1}{2x^2}\right)^2 \\
&=& 1 + \left(\frac{x^2}{2}\right)^2 - \frac{1}{2} + \left(\frac{1}{2x^2}\right)^2 \\
&=& \left(\frac{x^2}{2}\right)^2 + \frac{1}{2} + \left(\frac{1}{2x^2}\right)^2 \\
&=& \left(\frac{x^2}{2}\right)^2 + 2\left(\frac{x^2}{2} \cdot \frac{1}{2x^2}\right) + \left(\frac{1}{2x^2}\right)^2 \\
&=& \left(\frac{x^2}{2} + \frac{1}{2x^2}\right)^2
\end{eqnarray*}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/851743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Evaluation of $ \lim_{x\rightarrow \infty}\left\{2x-\left(\sqrt[3]{x^3+x^2+1}+\sqrt[3]{x^3-x^2+1}\right)\right\}$
Evaluate the limit
$$
\lim_{x\rightarrow \infty}\left(2x-\left(\sqrt[3]{x^3+x^2+1}+\sqrt[3]{x^3-x^2+1}\right)\right)
$$
My Attempt:
To simplify notation, let $A = \left(\sqrt[3]{x^3+x^2+1}\right)$ and $B = \left(\sqrt[3]{x^3-x^2+1}\right)$. Now
$$
\begin{align}
2x^2 &= A^3-B^3\\
x &= \sqrt{\frac{A^3-B^3}{2}}
\end{align}
$$
So the limit becomes
$$\lim_{x\rightarrow \infty}\left(\sqrt{\frac{A^3-B^3}{2}}-A-B\right)$$
How can I complete the solution from this point?
|
$a+b=\dfrac{a^3+b^3}{a^2-ab+b^2}$
I think this identity can be used to simplify your expression.
Let $a=\sqrt[3]{x^3+x^2+1}$ and $b=\sqrt[3]{x^3-x^2+1}.$
Then
$a+b=\dfrac{(x^3+x^2+1)+(x^3-x^2+1)}{(x^3+x^2+1)^{2/3}-(x^3+x^2+1)^{1/3}(x^3+x^2+1)^{1/3}+(x^3+x^2+1)^{2/3}}\\=\dfrac{2(x+1/x^2)}{(1+1/x+1/x^3)^{2/3}-(1+1/x+1/x^3)^{1/3}(1+1/x+1/x^3)^{1/3}+(1+1/x+1/x^3)^{2/3}}$
As $x\rightarrow \infty$ we have $a+b\rightarrow 2x.$ Hence $$\lim_{x\rightarrow \infty}\left\{2x-\left(\sqrt[3]{x^3+x^2+1}+\sqrt[3]{x^3-x^2+1}\right)\right\}\\= \lim_{x\rightarrow \infty} {2x-(a+b)}=0$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/851849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 3
}
|
Finding tight upper/lower bounds for $\mathbb{E}[\frac{1}{1+X^{2}}]$ where $X$ is a RV with $\mathbb{E}[X]=0$ and $\mbox{Var}(X)=\nu<\infty $ The question is pretty much in the title.
My first thought was using Jensen's inquality to get some sort of lower bound. Since $\frac{1}{1+x^{2}}$
is convex on $\mathbb{R}\backslash\left[-\frac{1}{2},\frac{1}{2}\right]$
I think by Jensen's inequality I get a lower bound of the form: $$\frac{1}{1+\nu}\leq\frac{1}{1+\mathbb{E}\left[\left(X\cdot1_{\left\{ X>\frac{1}{2}\right\} }\right)^{2}\right]}\leq\mathbb{E}\left[\frac{1}{1+\left(X\cdot1_{\left\{ X>\frac{1}{2}\right\} }\right)^{2}}\right]\leq\mathbb{E}\left[\frac{1}{1+X^{2}}\right]
$$
I'm not sure if it's even correct to use Jensen's inequality in this manner, the formulation of the inequality I'm familiar with is for an integrable RV and a convex function over an interval of the form $\left(a,b\right)$
with $a,b$
possibly being $\pm\infty$.
EDIT: like a comment suggested it's not hard to apply Jensen's Inequality correctly to obtain the second inequality. Unfortunately I just noticed the third inequality is incorrect since $$\frac{1}{1+\left(X\cdot1_{\left\{ X\geq\frac{1}{2}\right\} }\right)^{2}}\geq\frac{1}{1+X^{2}} $$
Yet the bound itself still feels like it should be correct. I also verified you can approach it as desired with a uniform RV defined on the interval $\left(-\varepsilon,\varepsilon\right)$ as $\varepsilon$ tends to $0$. As can be seen:
$$\mathbb{E}\left[\frac{1}{1+X^{2}}\right]=\int\limits _{-\varepsilon}^{\varepsilon}\frac{1}{2\varepsilon}\cdot\frac{1}{1+x^{2}}dx=\frac{\tan^{-1}\left(\varepsilon\right)}{\varepsilon}\overset{\varepsilon\to0}{\longrightarrow}1$$
$$\frac{1}{1+\mathbb{E}\left[X^{2}\right]}=\frac{1}{1+\frac{\varepsilon^{2}}{3}}\overset{\varepsilon\to0}{\longrightarrow}1$$
I'd appreciate help showing a bound of the form $\frac{1}{1+\nu}\leq\mathbb{E}\left[\frac{1}{1+X^{2}}\right]$ indeed holds (or a contradicting example and an alternative bound).
|
The function $t\mapsto\frac1{1+t}$ is convex on $t\geqslant0$ hence
$$
E\left(\frac1{1+X^2}\right)\geqslant\frac1{1+E(X^2)}=\frac1{1+\nu}.
$$
The lower bound is attained when $P(X=\sqrt\nu)=P(X=-\sqrt\nu)=\frac12$.
On the other hand, if $P(X=0)=1-\frac\nu{x^2}$ and $P(X=x)=P(X=-x)=\frac\nu{2x^2}$ for some $|x|\geqslant\sqrt\nu$, then $E(X)=0$ and $E(X^2)=\nu$ while $$E\left(\frac1{1+X^2}\right)=1-\frac\nu{1+x^2},$$ hence the best universal upper bound valid for every $X$ such that $E(X)=0$ and $E(X^2)=\nu$ with $\nu\gt0$ is
$$
E\left(\frac1{1+X^2}\right)\lt1.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/851932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
}
|
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