The Riemann Zeta Function

davidmccormick · #1 $Old$ November 27th, 2008, 01:06 PM

Consider the function

$f(s) = \sum_{r=1} ^ {\infty} \frac{1}{r^s}$ .
I have so far managed to show that the series converges for each $s\in\ (1,{\infty})$ and that this series defines a continuous function $f : (1,{\infty}) \rightarrow\mathbb{R}$ . I am however struggling to show that:

(i) $f$ is differentiable and that $f'(s) < 0$ for all $s \in\ (1,{\infty})$ .
(ii) $f$ is differentiable and that $f''(s) > 0$ for all $s \in\ (1,{\infty})$ .

Any help would be greatly appreciated.
thanks

Mathstud28 · #2 $Old$ November 27th, 2008, 03:29 PM

Quote:

Originally Posted by davidmccormick $View Post$

Consider the function

$f(s) = \sum_{r=1} ^ {\infty} \frac{1}{r^s}$ .
I have so far managed to show that the series converges for each $s\in\ (1,{\infty})$ and that this series defines a continuous function $f : (1,{\infty}) \rightarrow\mathbb{R}$ . I am however struggling to show that:

(i) $f$ is differentiable and that $f'(s) < 0$ for all $s \in\ (1,{\infty})$ .
(ii) $f$ is differentiable and that $f''(s) > 0$ for all $s \in\ (1,{\infty})$ .

Any help would be greatly appreciated.
thanks

Note that on the specified interval that $\zeta(x)$ converges uniformly on $(1,\infty)$ since those points are on the interior of its interval of convergence. Now note that $\frac{1}{r^x}$ is differentiable to $\frac{-x}{r^{x+1}}$ and that by the Weirstrass test or using the ratio/root test that this is uniformly convergent on $(0,\infty)$ we can conclude that

$\forall{x}\in(1,\infty)~\zeta'(x)=\sum_{n=1}^{\infty}\frac{-x}{r^{x+1}}$

Now it is obvious that $\forall{x}\in(1,\infty)~\frac{-x}{r^{x+1}}<0$ , so there is part i and for part two repeat a similar process of establishing, uniform convergence of $\sum_{n=1}^{\infty}\frac{-x}{r^{x+1}}$ on $(1,\infty)$ , uniform convergence of $\sum_{n=1}^{\infty}\frac{x(x+1)}{r^{x+2}}$ on $(1,\infty)$ , and the differentiability of $\frac{-x}{r^{x+1}}$ on $(1,\infty)$

davidmccormick · #3 $Old$ November 27th, 2008, 06:18 PM

I feel really stupid because your explanation is very clear but I don't follow....can you please explain why the derivative of $\frac{1}{r^x}$ is $\frac{-x}{r^{x+1}}$ because I make it out to be $-r^{-x}ln(r)$ . The second thing is the ratio test on $\frac{-x}{r^{x+1}}$ gives convergence, does that necessarily imply uniform convergence.
Sorry about this and thanks a lot

Mathstud28 · #4 $Old$ November 27th, 2008, 08:05 PM

Quote:

Originally Posted by davidmccormick $View Post$

I feel really stupid because your explanation is very clear but I don't follow....can you please explain why the derivative of $\frac{1}{r^x}$ is $\frac{-x}{r^{x+1}}$ because I make it out to be $-r^{-x}ln(r)$ . The second thing is the ratio test on $\frac{-x}{r^{x+1}}$ gives convergence, does that necessarily imply uniform convergence.
Sorry about this and thanks a lot

Because $\zeta(x)$ is a function of x so $\zeta'(x)=\frac{d}{dx}\sum_{n=1}^{\infty}\frac{1}{r^x}$ now since the appropriate conditions were met as I showed you $\frac{d}{dx}\sum_{n=1}^{\infty}\frac{1}{r^x}=\sum_{n=1}^{\infty}\frac{d}{dx}\frac{1}{r^x}=\sum_{n=1}^{\infty}\frac{-x}{r^{x+1}}$ . And to answer your other question, I mistyped, forgive me. Disregard the ratio/root test comment and stick the Weirstrass M-test.

Moo · #5 $Old$ November 28th, 2008, 12:37 AM

Quote:

Originally Posted by Mathstud28 $View Post$

Because $\zeta(x)$ is a function of x so $\zeta'(x)=\frac{d}{dx}\sum_{n=1}^{\infty}\frac{1}{r^x}$ now since the appropriate conditions were met as I showed you $\frac{d}{dx}\sum_{n=1}^{\infty}\frac{1}{r^x}=\sum_{n=1}^{\infty}\frac{d}{dx}\frac{1}{r^x}=\sum_{n=1}^{\infty}\frac{-x}{r^{x+1}}$ . And to answer your other question, I mistyped, forgive me. Disregard the ratio/root test comment and stick the Weirstrass M-test.

As you said, it is a function of x.

$\frac{d}{dx} \left(\frac{1}{r^x} \right)=\frac{d}{dx} \left(e^{-x \ln(r)}\right)=-\ln(r) \cdot \frac{1}{r^x}$

So davidmccormick, you are correct.

davidmccormick · #6 $Old$ November 28th, 2008, 03:32 AM

But all I've done is differentiating the first term of the original series. I am pretty sure that the derivative exist and is equal to $\zeta'(s) = -\sum_{r=1}^{\infty}\frac{lnr}{r^s} = -\sum_{r=2}^{\infty}\frac{lnr}{r^s}$ . But I have not actually shown that $\zeta$ is differentiable.
Any thoughts.
thanks

Opalg · #7 $Old$ November 28th, 2008, 04:19 AM

Quote:

Originally Posted by davidmccormick $View Post$

I am pretty sure that the derivative exist and is equal to $\zeta'(s) = -\sum_{r=1}^{\infty}\frac{\ln r}{r^s} = -\sum_{r=2}^{\infty}\frac{\ln r}{r^s}$ .

The series $-\sum_{r=1}^{\infty}\frac{\ln r}{r^s}$ converges uniformly in any interval [a,∞), where a>1 (by the Weierstrass M-test). So you can integrate this series term by term (getting $\zeta(s)$ ), and by the fundamental theorem of calculus the integrated series will have the required derivative, which is clearly negative.

The same argument, repeated, will show that the second derivative is $\sum_{r=1}^{\infty}\frac{(\ln r)^2}{r^s}$ , which is clearly positive.

CaptainBlack · #8 $Old$ November 28th, 2008, 04:21 AM

Quote:

Originally Posted by davidmccormick $View Post$

But all I've done is differentiating the first term of the original series. I am pretty sure that the derivative exist and is equal to $\zeta'(s) = -\sum_{r=1}^{\infty}\frac{lnr}{r^s} = -\sum_{r=2}^{\infty}\frac{lnr}{r^s}$ . But I have not actually shown that $\zeta$ is differentiable.
Any thoughts.
thanks

It is sufficient to show that the sequence of parital sums:

$S_N(x)=\sum_{r=1}^N \frac{1}{r^x}$

in a neighbourhood of $x$ is uniformly convergent, and that $S'_N(x)$ is also uniformly convergent to conclude that:

$\frac{d \zeta}{dx}=\frac{d}{dx}\left[ \lim_{N \to \infty} S_N(x) \right] = \lim _{N \to \infty} \frac{dS_N(x)}{dx}$

The uniform convergence on the sequence of partial sums for the zeta function can be demonstrated fairly easily on any closed interval $[a,b],\ 1<a<b$ , and with a bit more trouble for the sequence of derivatives of the partial sums. Together these prove that the derivative of the zeta function can be found by term by term differentiation of the series for the zeta function on $(1,\infty).$

CB

CaptainBlack · #9 $Old$ November 28th, 2008, 05:17 AM

Quote:

Originally Posted by Mathstud28 $View Post$

Note that on the specified interval that $\zeta(x)$ converges uniformly on $(1,\infty)$ since those points are on the interior of its interval of convergence.

This needs clarifying. If this is saying that the series is uniformly convergent on $(0, \infty)$ then I suspect its wrong.

CB

davidmccormick · #10 $Old$ November 28th, 2008, 05:55 AM

For the Weierstrass M-test what series do we use to show that $-\sum_{r=1}^{\infty}\frac{\ln r}{r^s}$ and $\sum_{r=1}^{\infty}\frac{(\ln r)^2}{r^s}$ respectively are uniformly convergent.

Opalg · #11 $Old$ November 28th, 2008, 07:30 AM

Quote:

Originally Posted by davidmccormick $View Post$

For the Weierstrass M-test what series do we use to show that $-\sum_{r=1}^{\infty}\frac{\ln r}{r^s}$ and $\sum_{r=1}^{\infty}\frac{(\ln r)^2}{r^s}$ respectively are uniformly convergent.

These series are uniformly convergent on the interval [a,∞), for any given a>1. In fact, $\frac{(\ln r)^k}{r^s}$ is decreasing as a function of s, because its derivative is $-\frac{(\ln r)^{k+1}}{r^s}$ . So its maximum value on the interval [a,∞) is its value at the left endpoint s=a, and we take $M_r = \frac{(\ln r)^k}{r^a}$ (where k = 1 or 2 as appropriate) in the M-test. The fact that $\sum_{r=1}^\infty \frac{(\ln r)^k}{r^a}$ converges (for a>1) then tells you that these series converge uniformly on that interval.

But note that uniform convergence on the interval [a,∞) for every a>1 does not imply uniform convergence on the interval (1,∞); and in fact the Riemann zeta series is not uniformly convergent on (1,∞).

davidmccormick · #12 $Old$ November 28th, 2008, 09:06 AM

you've been very helpful and thorough. thank you