Author Topic: TT's Maths Thread (Read 147072 times) Tweet Share

humph · « **Reply #1170 on:** April 06, 2011, 12:31:36 am »

Let's first restrict ourselves to the open half-plane $\Re(s) > 1/2$ . Then for any $n \geq 1$ ,
$n^{-2s} \Gamma(s) \pi^{-s} = \int^{\infty}_{0}{n^{-2s} \pi^{-s} e^{-x} x^{s - 1} \, dx} = \int^{\infty}_{0}{e^{-\pi n^2 t} t^{ s - 1} \, dt}$
by making the substitution $x = \pi n^2 t$ . Summing over all $n \geq 1$ and interchanging the order of summation and integration, which is justified by absolute convergence (which relies on the crucial fact that $\Re(s) > 1/2$ ), we find that
$ \begin{align} \zeta(2s) \Gamma(s) \pi^{-s} & = \int^{\infty}_{0}{\sum^{\infty}_{n = 1}{e^{-\pi n^2 t}} t^{ s - 1} \, dt} \\ & = \frac{1}{2} \int^{1}_{0}{(\theta(it) - t^{-1/2}) t^{s - 1} \, dt} + \frac{1}{2} \int^{1}_{0}{(t^{-1/2} - 1) \, dt} + \frac{1}{2} \int^{\infty}_{1}{(\theta(it) - 1) t^{s - 1} \, dt} \\ & = \frac{1}{2} \int^{1}_{0}{(\theta(it) - t^{-1/2}) t^{s - 1} \, dt} + \frac{1}{2s - 1} - \frac{1}{2s} + \frac{1}{2} \int^{\infty}_{1}{(\theta(it) - 1) t^{s - 1} \, dt}, \end{align} $
where again the fact that $\Re(s) > 1/2$ was crucial in allowing us to evaluate the middle integral. The key point now is although everything is currently only well-defined on $\Re(s) > 1/2$ , the terms on the right-hand side actually define a function that is well-defined (in fact, holomorphic) on $\{s \in \mathbb{C} : s \neq 0, 1/2, 1\}$ (so that the right-hand side, and hence the left-hand side, extends to a meromorphic function on $\mathbb{C}$ with singularities at $s = 0, 1/2, 1$ ). The middle two terms clearly satisfy this. For the first term, we actually require the functional equation
$ \theta(it) = \frac{1}{\sqrt{t}} \theta\left(\frac{i}{t}\right) $
for $t > 0$ ; this is a standard application of the Poisson summation formula in Fourier analysis. Using this, we see that
$ \frac{1}{2} \int^{1}_{0}{(\theta(it) - t^{-1/2}) t^{s - 1} \, dt} = - \frac{1}{2} \int^{\infty}_{1}{(\theta(it) - 1) t^{-s - 1/2} \, dt}. $
So we must show that
$ - \frac{1}{2} \int^{\infty}_{1}{(\theta(it) - 1) t^{-s - 1/2} \, dt}, \quad \int^{\infty}_{1}{(\theta(it) - 1) t^{s - 1} \, dt} $
both converge for all $s \in \mathbb{C}$ . Let's look at the second integral. We already know this converges for $\Re(s) > 1/2$ . If $\Re(s) \leq 1/2$ , then it's easy to show that
$ \left|\int^{\infty}_{1}{(\theta(it) - 1) t^{s - 1} \, dt}\right| \leq \sum^{\infty}_{n = 1}{\int^{\infty}_{1}{e^{-\pi n^2 t} \, dt} = \frac{1}{\pi} \sum^{\infty}_{n = 1}{\frac{1}{n^2}} < \infty, $
and hence this integral converges. A similar argument proves this for the first integral.

Mao · « **Reply #1171 on:** April 06, 2011, 01:12:31 am »

You are quite right, I did make a typo (because I was expecting the integrals to cancel out and got lazy).

I also didn't care too much about convergence. I don't make a very good mathematician.

TrueTears · « **Reply #1172 on:** April 18, 2011, 01:58:30 am »

hey guys, just some questions regarding this problem, I've done most of the essential working but just some details i want to clarification over:

"Let $*$ be a binary relation on $\mathbb{R} \backslash \{0\}$ defined by $a * b = a|b|$ . Is $<\mathbb{R}, *>$ a group? Explain"

My first question is that it says at the start that $*$ is a binary operation ON $\mathbb{R} \backslash \{0\}$ but then the binary structure $<\mathbb{R}, *>$ means that $*$ is now a binary operation on $\mathbb{R}$ instead of $\mathbb{R} \backslash \{0\}$

so thus in showing whether $<\mathbb{R}, *>$ is a group or not, do we take the binary operation $*$ on $\mathbb{R}$ or $\mathbb{R} \backslash \{0\}$ as this is actually an essential part of showing whether $<\mathbb{R}, *>$ is a group or not

eg, if we use $\mathbb{R}$ then when we are trying to show that for all $x \in \mathbb{R}$ there is an identity element $e$ such that $e * x = x * e = x$

thus $e*x = e|x$ |

and assume $e|x| = x$ then we have $e = \frac{x}{|x|}$

but if $x = 0$ , then $e$ is undefined and thus $x = 0$ does not have an identity element in $\mathbb{R}$

however if we use $\mathbb{R} \backslash \{0\}$ we can clearly see that $e = \frac{x}{|x|}$ is the identity element since $x$ can not be 0 and thus $e$ is defined

cheers

Also another question:

[IMG]http://img850.imageshack.us/img850/1990/matrixgroup.jpg[/img]

First, what does it mean by "all operations performed modulo 2"? My understanding is that say we multiply 2 matrices that are 2x2 and we get something like

$\begin{bmatrix} 3 & 2 \\ 4 & 5 \end{bmatrix}$

then actually the matrix should be:

$\begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}$

in other words every entry in the matrix has to be modulo 2?

Also what does "let S be the set of 2x2 matrices A with entries in $\{0, 1\}$ such..." mean?

does that mean all the entries in these matrices are either 0 or 1?

and lastly, how are we meant to find the elements of the group? do we actually have to list all 2^4 possible matrices with entries either 0 or 1 and pick the ones where $det(A) \not\equiv 0 \pmod{2}$ ? or am i interpreting it wrong?

Also how are we meant to find the identity and inverse element?

because by the definition of the identity element, it is an element e in S such that for all x in S, $e * x = x * e = x$

how can we find the matrix e in S that satisfies the above for every x in S?

thanks

/0 · « **Reply #1173 on:** April 18, 2011, 02:46:07 am »

With the 1st question, I assume they really mean $\langle \mathbb{R}\setminus \lbrace 0\rbrace,*\rangle$ ... I might be wrong, but sometimes they get lazy with notation.
[Actually don't take my word on this one, since i'm really not sure]

With the second question, all operations performed modulo 2 means every entry in the resulting matrix must be modulo two. In the example you gave the matrix would actually be the identity.

And yep, if all the matrices' entries are in $\lbrace 0,1\rbrace$ then they are only 0 or 1, and furthermore it is possible to list out all the elements of $S$ by simply constructing all invertible 2x2 matrices with 0s and 1s. From a quick glance there should only be ~~2 elements~~ (lol or not). (but check)

TrueTears · « **Reply #1174 on:** April 18, 2011, 02:53:53 am »

Cheers /0 !

I found 6 elements in S though lol

$\begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}$

$\begin{bmatrix} 0 & 1 \\ 1 & 0\end{bmatrix}$

$\begin{bmatrix} 1 & 1 \\ 1 & 0\end{bmatrix}$

$\begin{bmatrix} 0 & 1 \\ 1 & 1\end{bmatrix}$

$\begin{bmatrix} 1 & 0 \\ 1 & 1\end{bmatrix}$

$\begin{bmatrix} 1 & 1 \\ 0 & 1\end{bmatrix}$

i listed out the entire 2^4 possibilities and calculated each det, those above satisfies the condition for S [namely, the set of all 2x2 matrices A with entires in $\{0, 1\}$ such that $det(A) \not\equiv 0 \pmod{2}$ ]

assuming those 6 elements are right... how do we find the inverse and inverse for the group?

actually is my reasoning right for finding the inverse?

we require an element e in S such that for all x in S, $e * x = x * e = x$

thus say we let $e = \begin{bmatrix} a & b \\ c & d\end{bmatrix}$ and pick a specific x from S, ie, pick $x = \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}$

thus we find that $e = \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}$ satisfies $e * x = x * e = x$

Now since we ALREADY know that $<S, *>$ is a group and that the identity element in a group is unique, then $e = \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}$ must be the identity element for the group!

Now this only works because we already know that $<S, *>$ is a group beforehand right? since if we didn't know $<S,*>$ was a group then just solving $e * x = x * e = x$ for a specific $x$ doesn't mean that $e$ could be the identity element for the rest of the elements, in other words, there might not exist an identity element!

TrueTears · « **Reply #1175 on:** April 18, 2011, 05:06:25 am »

[IMG]http://img140.imageshack.us/img140/3783/rankofa.jpg[/img]

Also I don't get this question, I've reduced A down to its row echelon form of:

$\begin{bmatrix} 1 & 1 & t \\ 0 & 1 & -1 \\ 0 & 0 & 1\end{bmatrix}$ and $t \neq \{1,-2\}$

We see that the basis for the row space would be the vectors (1,1,t), (0,1, -1) and (0,0,1) so the rank(A) = 3

but the question says how does the rank of A vary with t? well even if t changes (as long as $t \neq \{1,-2\}$ ) then the rank of A is independent of t?

I don't get what the question is trying to ask... any clarifications?

evaever · « **Reply #1176 on:** April 18, 2011, 08:44:20 am »

rank of A is not a continuous function of t

humph · « **Reply #1177 on:** April 18, 2011, 02:22:46 pm »

For your first question, I'd imagine that it's a typo: it's definitely supposed to be $\mathbb{R} \setminus \{0\}$ throughout, not $\mathbb{R}$ .

Also note that your calculation shows that should the identity $e$ exist, it must satisfy $e = x/|x|$ for every $x \in \mathbb{R} \setminus \{0\}$ , but this is impossible, as $x/|x| = 1$ if $x > 0$ but $x/|x| = -1$ if $x < 0$ .

For your second question, I think /0 has explained this pretty well. So basically you're studying 2x2 matrices whose entries are either 1 or 0, whose determinant is either 1 or -1. You can shortcut here a bit by remembering that $\det(A) = \det(A^T)$ (which tells you that you only need to study 2^3 possibilities), and similarly that $\det(\mathbf{a}_1 \; \mathbf{a}_2) = - \det(\mathbf{a}_2 \; \mathbf{a}_1)$ for a matrix made of two column vectors $\det(\mathbf{a}_1 \; \mathbf{a}_2) = - \det(\mathbf{a}_2 \; \mathbf{a}_1)$ .
Once you've found them all, it's pretty obvious which one is the identity (same matrix as usual) - this is because it's ALWAYS the identity for any group of matrices (for obvious reasons). A little matrix multiplication tells you which matrices are inverses of each other. There are shortcuts to avoid all this calculation, but on the whole it shouldn't be too much work anyway.

For your third question, evaever has the right idea. Think of the rank as a function of $t$ ; that is, consider the function $\mathrm{rank}(A)(t)$ . As evaever says, it's a piecewise defined function that isn't continuous at two points.

TrueTears · « **Reply #1178 on:** April 18, 2011, 04:12:51 pm »

thanks guys

and also @humph, I've found $S = \{ \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 1 & 0\end{bmatrix}, \begin{bmatrix} 1 & 1 \\ 1 & 0\end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 1 & 1\end{bmatrix}, \begin{bmatrix} 1 & 0 \\ 1 & 1\end{bmatrix}, \begin{bmatrix} 1 & 1 \\ 0 & 1\end{bmatrix}\}$

and by my argument in the previous post (is that right?) I found $e = \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}$ to be the identity element.

now if we let $\begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}=a , \begin{bmatrix} 0 & 1 \\ 1 & 0\end{bmatrix}=b, \begin{bmatrix} 1 & 1 \\ 1 & 0\end{bmatrix} =c, \begin{bmatrix} 0 & 1 \\ 1 & 1\end{bmatrix}=d, \begin{bmatrix} 1 & 0 \\ 1 & 1\end{bmatrix} = f, \begin{bmatrix} 1 & 1 \\ 0 & 1\end{bmatrix}=g$

now the table of operations looks like:

[IMG]http://img19.imageshack.us/img19/3672/tableoperations.jpg[/img]

however isn't the inverse element in a group always unique? ie, there is only one just like the identity element? As my book says:

[IMG]http://img863.imageshack.us/img863/9718/uniqueness.jpg[/img]

however from the table we can clearly see that the inverse element for a is a since $a*a = a$

also $c*d = d*c = a$ so c's inverse element is d

but now we have 2 inverse elements, one being a and the other being d, contradicting the fact that the group can only have one unique inverse for all of its elements?

humph · « **Reply #1179 on:** April 18, 2011, 05:46:13 pm »

No, for EACH $a$ , there is a unique inverse $a'$ . Your statement is really quite silly, if you think about it: you're saying that there is a unique, fixed $a'$ such that for every $a$ , $a'$ is the inverse of $a$ . What about the most natural definition of a group, $\mathbb{R}$ (or $\mathbb{Q}$ , or $\mathbb{Z}$ ) under addition? Each element $a$ has a unique inverse $-a$ , but of course it's impossible for there to be one single "global" inverse for every point.

TrueTears · « **Reply #1180 on:** April 18, 2011, 05:48:18 pm »

oh wait i think i misinterpreted it, every inverse is unique for that specific element but there doesn't have to be 1 inverse for the whole group right?

so for each $y \in S$ , the inverse of y, given by y' has to satisfy $y * y' = y' * y = e$

so $y*y' = yy' = e$

treating y and y' as matrices

we can solve for y': $y^{-1}yy'=y^{-1}e$

so $y' = y^{-1}$

this also satisfies $y'*y = e$ since $y^{-1}y = e$

thus for every element y in S, the inverse of y is given by $y^{-1}$ , this is unique for EACH element not for ALL elements

is that right?

lol opps yeah i realised my stupid mistake right after you posted humph haha thanks again

Mao · « **Reply #1181 on:** April 18, 2011, 11:27:37 pm »

Quote from: TrueTears on April 18, 2011, 05:06:25 am

[IMG]http://img140.imageshack.us/img140/3783/rankofa.jpg[/img]

Also I don't get this question, I've reduced A down to its row echelon form of:

$\begin{bmatrix} 1 & 1 & t \\ 0 & 1 & -1 \\ 0 & 0 & 1\end{bmatrix}$ and $t \neq \{1,-2\}$

We see that the basis for the row space would be the vectors (1,1,t), (0,1, -1) and (0,0,1) so the rank(A) = 3

but the question says how does the rank of A vary with t? well even if t changes (as long as $t \neq \{1,-2\}$ ) then the rank of A is independent of t?

I don't get what the question is trying to ask... any clarifications?

During row operations, it seems you would need to divide by functions of t. The zeroes of these functions are values of t you should investigate.

You could also try to compute the determinant to see what values would play up.

TrueTears · « **Reply #1182 on:** April 19, 2011, 01:17:44 am »

yup thanks mao, i got it now

just have another Q

[IMG]http://img263.imageshack.us/img263/7585/solutionsforagroup.jpg[/img]

(e is the identity element)

I played around with examples and for $a^m=e$ to be true for each $a \in G$ then we need to have $a * (a*a*...*a) = e$ where $(a*a*...*a) = a'$ [the inverse of a]

however i am still a bit unsure on how to go about proving what's required... any help would be greatly appreciated!

kamil9876 · « **Reply #1183 on:** April 19, 2011, 10:40:02 pm »

So you want to show that one of $a^1,a^2...a^n$ is the identity, what if none of them were? Then what does the pigeonhole principle say?

TrueTears · « **Reply #1184 on:** April 20, 2011, 05:22:06 am »

cheers kamil

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