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I've to find a value for 's' were the infinit sum gives me the value 50. Is that possible and how do I've to calculate that value.

I've no idea how te begin so, help me!

Solve s for:

$$\sum\limits _{n=1}^{\infty}\frac{1}{n^s}=50$$

(and I know that it's the Riemann zeta function for real numbers bigger than 1)

Jan Eerland
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  • Where is this problem from? – MT_ Apr 20 '15 at 16:45
  • Approximative solution: We have $\lim_{s\to 1}(s-1)\zeta(s) = 1$ so $\zeta(s) \approx \frac{1}{s-1}$ for $s$ close to $1$. $s=1 + \frac{1}{50} = 1.02$ turns out to be pretty close to the true result. You want any better then you can try a higher order expansion or more simply: a numerical solution. – Winther Apr 20 '15 at 16:47
  • I am working on a project, where I've to use it! If it is possible – Jan Eerland Apr 20 '15 at 16:47
  • A quick plot in Mathematica shows me, that this is (probably) not possible. The values of that function are all below, at least, 5. – Stefan Perko Apr 20 '15 at 16:48
  • @StefanPerko : Not if $s$ goes to $1$ with $s < 1$. – pitchounet Apr 20 '15 at 16:49
  • @StefanPerko That is an artefact in the plot. For $s=1$ the sum is infinity and for $s=2$ the sum is $\pi^2/6 \approx 1.64$. The function is continuous inbetween so attains every value $> 1$ for some $s$. – Winther Apr 20 '15 at 16:50
  • @jibounet Oh yeah. Of course, where was my head... – Stefan Perko Apr 20 '15 at 16:50
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    There seems to be no motivation for this question. I mean you could at least admit that you don't care about the answer and that you only want to see methods to estimate it. That would at least make the question seem designed for people rather than computers.... – Zach466920 Apr 20 '15 at 17:02

5 Answers5

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We have that the function $\zeta(s)$ is decreasing on the interval $(1,2)$. Since $\lim_{s\to 1^+}\zeta(s)=+\infty$ while $\zeta(2)=\frac{\pi^2}{6}<50$ we have $\zeta(s_0)=50$ for some $s_0\in(1,2)$. Exploiting the asymptotic expansion in a right neighbourhood of $s=1$ (formula $(15)$ here, it follows from summation by parts) we have: $$ \zeta(s)\approx \frac{1}{s-1}+\gamma $$ so a reasonable estimate for $s_0$ is given by: $$ s_0\approx\frac{51-\gamma}{50-\gamma} $$ and we can increase the accuracy of such approximation by using Newton's method, for instance.

Jack D'Aurizio
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The Laurent series expansion of the Riemann Zeta function around $s=1$ is $$\zeta(s) = \dfrac1{s-1} + \sum_{n=0}^{\infty} \dfrac{(-1)^n}{n!} \gamma_n (s-1)^n$$ Further, recall that since $\zeta(2) = \dfrac{\pi^2}6 < 50$ and since $\zeta(s)$ is a rapidly decreasing function close to $1$, we have $s \in (1,2)$. Approximating $\zeta(s)$ with the first three terms gives us a quadratic in $s$, i.e., $$\dfrac1{s-1} + \gamma_0 - \gamma_1(s-1) \approx 50$$ Solving the quadratic gives us $s \approx 1.020234186$. Better approximations can be obtained by considering more terms and solving the resulting polynomial equation using numerical root finding techniques.

Adhvaitha
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$$s\simeq1.0202341852181139861332276132729412593842116437344617189788796190\ldots$$

Lucian
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  • This does not seem to address "I've no idea how to begin so, help me!" other that perhaps pulling out a CAS. – robjohn Apr 20 '15 at 16:53
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    @robjohn: other that perhaps pulling out a CAS - Precisely. There is no closed form for it. Numerical algorithms are the only way out. – Lucian Apr 20 '15 at 17:04
  • @robjohn But that's exactly the point. In the comments OP said that the value of $s$ is only needed for calculation. The derivation of the answer is therefore irrelevant to the question at hand, at least it seems that way to me. – MT_ Apr 20 '15 at 17:13
  • There are ways to get approximate answers without the black box of a CAS. Jack D'Aurizio's and user17762's answers give an idea of how to compute an approximation. – robjohn Apr 20 '15 at 17:43
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Mathematica yields $$s \approx 1.020234185218113986133227613272941259384211643734461718978879619075931...$$

I don't think you are going to get a good answer without just numerically approximating it.

MT_
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Since I have seen this graph for only real numbers (source) enter image description here - I have start thinking about to approximate the zeta function into a more elementary function: Hyperbola:

$\zeta(x)=\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^x} \rightarrow H(x)=\frac{\alpha x+\beta}{\lambda x-x_0}+y_0$ for x>1

To fit our hyperbola we gonna need particular values of the Riemann zeta function, like:

$(2;\frac{\pi^2}{6})$ and $(4;\frac{\pi^4}{90})$

One of the variables for $H(x)$ is $x_0=1$

Of course $y_0$ should be $1$ but we gonna set $y_0=0$ and $\alpha=0$ as well.

Hense: $h(x)=\frac{\beta}{\lambda x-1}$

(For X variables we need (at least) X points to find our formula)

Result: $h(x)=\frac{\pi^4}{6}\frac{1}{30-\pi^2+\frac{4(\pi^2-15)}{x}}$ for $1<x<5.2$

Solving for $h(x)=50$ we get $x=\frac{4(\pi^2-15)}{\pi^2(\frac{\pi^2}{300}+1)-30}\approx1.036$ not really close but by this simplification is good.

Epilog: A $H(x)$ can be fit for $x<1$ as well- here an example:

Via $(-1;-\frac{1}{12})$,$(0;-\frac{1}{2})$ and $(\frac{1}{2};\zeta(\frac{1}{2}))$ I have fit:

$h_2(x)=\frac{5(b+1)|x|}{12(b|x|+1)}-\frac{1}{2}$;where $b=\frac{17+24\zeta(\frac{1}{2})}{1+12\zeta(\frac{1}{2})}$;$x<1$

Krzysztof Myśliwiec
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