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Here is my summation:

$$\sum_{n=0}^{N/2-1}\frac1{25+nB}=A$$ where $0.01\le A\le2$, $10\le N\le2000$ and I need to find the $B$ for different values of $N$ I calculate this summation online (check here) But the Digamma function makes the output function complex and it is not easy to solve with a calculator or even by hand. I need to calculate the result of this summation with hand easily. Is there any way to convert my summation to simple formula? Something like this:

\begin{align*} \sum^n_{k=1}k&=\frac{n(n+1)}{2}\\[5pt] \sum^n_{k=1}k^2&=\frac{n(n+1)(2n+1)}{6}\\[5pt] \end{align*}

John Jin
  • 145

2 Answers2

1

Exact Solution:

Consider the function: $$ I'(x) = \sum_{k=0}^n\frac{1}{x+kb} $$

If we integrate both sides: $$ I(x) + C = \sum_{k=0}^n\log\left(x+kb\right) = \log\prod_{k=0}^n\left(x+kb\right) = \log\frac{(x+nb)!}{(x-1)!} $$

In other words: $$ I(x) + C = \log\Gamma(x+nb+1) - \log\Gamma(x) $$

If we take the derivative on both sides once more: $$ I'(x) = \psi(x+nb+1) - \psi(x),\quad\text{where}\quad \psi(x) = \frac{d}{dx}\log\Gamma(x) $$

Where $\psi(x)$ is known as digamma function. And that's why your summation depends on the digamma function, because its the derivative of $I(x)$ at $x=25$. Thus, there's no way to convert the sum to a simple formula, as that would be equivalent of saying there exists a simple formula for the digamma function. However, there's the possibility of using an approximation.

Approximate Solution:

Asymptotic expansion of the gamma function can be written like this:

$$\ln\Gamma(x) = x\ln x - x + \frac{1}{2}\log\frac{2\pi}{x} + O\left(\frac{1}{x}\right)$$ $$\psi(x) = \ln x - \frac{1}{2x} + O\left(\frac{1}{x^2}\right) $$

Thus, an approximate solution is: $$ I'(x) = \ln (x+nb+1) - \ln (x) + \frac{1}{2x} - \frac{1}{2(x+nb+1)} + O\left(\frac{1}{x^2}\right) $$ $$ I'(x) = \ln\left(1 + \frac{nb+1}{x}\right) + \frac{nb+1}{2x(x+1+nb)} + O\left(\frac{1}{x^2}\right) $$ $$ I'(x) = \ln\left(1 + \frac{nb+1}{x}\right) + O\left(\frac{1}{x^2}\right) $$

=]

To translate this to your situation, let $A = I'(25)$, $b=B$, $x=25$ and $n = N/2-1$, and thus you get: $$ A \approx \ln\left(1 + \frac{1-B}{25} + \frac{NB}{50}\right) $$

=]

kid
  • 103
1

Your sum is $$ A = \psi \left( {\left( {\tfrac{N}{2} - 1} \right)B + 26} \right) - \psi (25). $$ It is not possible to express your sum in terms of elementary functions. Nevertheless, one can obtain good upper and lower bounds for $B$. It is known that $$ \log \left( {x + \tfrac{1}{2}} \right) < \psi (x + 1) < \log (x+1) $$ for all $x>0$. Thus $$ \log \left( {\left( {\tfrac{N}{2} - 1} \right)B + \tfrac{{51}}{2}} \right) < A + \psi (25) < \log \left( {\left( {\tfrac{N}{2} - 1} \right)B + 26} \right), $$ i.e., $$ \frac{{2e^{A + \psi (25)} - 52}}{{N - 2}} < B < \frac{{2e^{A + \psi (25)} - 51}}{{N - 2}}, $$ with $\psi(25)=3.198742512\ldots$. For example, with $A=1$ and $N=500$, we get $$ 0.163 \ldots < B < 0.165 \ldots . $$ You can go beyond this, by using asymptotic approximations for the inverse of the digamma function (cf. https://doi.org/10.1007/s11139-014-9659-3). In particular, $$ B \approx \frac{{2e^{A + \psi (25)} - 51}}{{(N - 2)}} - \frac{1}{{12(N - 2)e^{A + \psi (25)} }}. $$ With $A=1$ and $N=500$, this gives $B=0.16506787525\ldots$, whereas the true value is $B = 0.16506787531\ldots$.

Gary
  • 31,845