These results are relatively easy to prove (first year college-level calculus needed) and could be a good test to refresh your math skills. We posted another simple one (with a probabilistic / number theory flavor) a while back, see here.
Prove the following:
where g = 0.5772156649... is the Euler–Mascheroni constant.
Solution:
Let us introduce the function f(x), defined as follows:
The answer to the first question is simply -f(-1). How do we compute it? Here is the key:
An exact solution is available for this integral (I computed similar integrals as exercises during my last high school year), and the result can be found here. This WolframAlpha tool allows you to automatically make the cumbersome but simple, mechanical, boring computations. The result involves logarithms and Arctan( (2x+1) / SQRT(3) ) which has know values (a fraction of Pi) if x = 0, 1 or -1.
To answer the second question, one can use generalized harmonic numbers H(x) with x = 1/3 (see details here) together with the well known asymptotic expansion for the diverging harmonic series (this is where the Euler–Mascheroni constant appears.) The mechanics behind this are as follows. Consider A = B + C, with
Both A and B diverge, but C converges. We are interested in an asymptotic expansion, here, for A. The asymptotic expansion for B is well known: It involves the Euler–Mascheroni constant , and (log n) / 3. The computations for C is linked to the function f(x) as x tends to 1, and brings in the other numbers in the final result: log(3), and Pi / SQRT(3).
Note that if we define h(x) as
then C = h(1) and g(x) can be computed using a technique and integral very similar to that used for f(x).
Generalization:
The idea is to decompose a function g(x) that has a Taylor series, into three components. It generalizes the well known decomposition into two components: The even and odd part of g. It goes as follows:
where the a's are the Taylor coefficients.
Related article:
For related articles from the same author, click here or visit www.VincentGranville.com. Follow me on LinkedIn.
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Comment
It is worth mentioning that the identity $\Sigma_{k=0}{\infty} x^{3k} = \frac{1}{1-x^3}$ and the term by term integration hold for $x \in (0,1)$. To justify the evaluation of the series at $ x= - 1$ still equals the value of the integral, one appeals to Abel's theorem.
Posted 10 May 2021
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