An almost elementary differentiation problem....

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SUMMARY

The discussion focuses on expressing the second derivative of the Eta function, defined as $$\eta(x) = \sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k^x}$$, at $$x=4$$ in terms of the Zeta function, $$\zeta(x) = \sum_{k=1}^{\infty}\frac{1}{k^x}$$. Participants concluded that the relationship between the derivatives of these functions can be established using known properties of the Zeta function. Specifically, the second derivative of the Eta function at $$x=4$$ can be expressed as a combination of the Zeta function and its derivatives at the same point.

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  • Understanding of the Zeta function and its properties
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Let $$\zeta(x)$$ be the Zeta function (where, for convenience, $$x$$ is assumed to be $$> 1$$).

$$\zeta(x) = \sum_{k=1}^{\infty}\frac{1}{k^x}$$Similarly, define the Eta function (alternating Zeta function) by the following series - where again, in this case, we assume $$x > 1$$:$$\eta(x) = \sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k^x}$$
Problem:Express the second derivative of the Eta function - at $$x=4$$ - in terms of the Zeta function (differentiated or otherwise), at $$x=4$$.
 
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Isn't it just a matter of applying the product rule on $ \displaystyle \displaystyle \eta(x) = \left(1-2^{1-x}\right) \zeta(x)$ twice?
 
Prometheus said:
Isn't it just a matter of applying the product rule on $ \displaystyle \displaystyle \eta(x) = \left(1-2^{1-x}\right) \zeta(x)$ twice?

Thanks for taking part, Prometheus! :D

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Yes, you're absolutely correct. The idea of the problem was to, potentially, encourage people who're less familiar with the functional relation you cited to develop it for themselves, then differentiate it.
 

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