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$$f(x)=\sum_{i=1}^x\frac1i$$ $$f'(x)=?$$ NOTE: I am defining f(x) for all natural number and considering the function as you zoom out

Is it even possible to take this derivative? Why or why not? If so, what meaning does the derivative convey? What function does f(x) approach as you zoom out? These are all questions that popped into my head when considering this function and now I look to you to answer them because my small brain can not.

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    $\begingroup$ What does it even mean to have $x$ as the end of a summation, if $x$ isn't a natural number? $\endgroup$
    – Alan
    Sep 22 at 22:26
  • $\begingroup$ Well, for this I am assuming x is a natural number and I am considering the function as you zoom out $\endgroup$ Sep 22 at 22:27
  • $\begingroup$ It doesn't make sense. I wouldn't put more thought on it. $\endgroup$
    – jjagmath
    Sep 22 at 22:28
  • $\begingroup$ Then what does the derivative mean on a function that inputs natural numbers? $\endgroup$
    – Alan
    Sep 22 at 22:31
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    $\begingroup$ @jjagmath THIS shows that it does make sense as others have "put more thought into it." ;-)) $\endgroup$
    – Mark Viola
    Sep 22 at 22:40

2 Answers 2

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$f(x)$ has a natural extension to complex $x$ as the harmonic number function $$H(x)=\psi(x+1)+\gamma$$ Here $\psi$ is the digamma function. It follows that $$H'(x)=\psi^{(1)}(x+1)$$ with a trigamma function.

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  • $\begingroup$ What is exact definition of the digamma function? I have seen many people use it,but they just talk about the properties of it $\endgroup$ Sep 22 at 22:43
  • $\begingroup$ @SamuelMorrison Look it up on Wikipedia, won't you? $\endgroup$ Sep 22 at 22:45
  • $\begingroup$ sry for asking that question $\endgroup$ Sep 22 at 22:46
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It might interest you to know that the premise for what you are attempting to do, for a general $f(i)$ not just for the particular case of $f(i)=\frac{1}{i}$, goes under the name of Indefinite_Sum or "Anti-Delta"..

In your particular case, since the Digamma function satisfies the functional identity $$ \Delta _x \psi (x) = \psi (x + 1) - \psi (x) = \frac{1}{x} $$ then $$ \sum\limits_x {\frac{1}{x}} = \Delta _x ^{ - 1} \left( {\frac{1}{x}} \right) = \psi (x) $$

Therefore $$ f(x) = \sum\limits_{k = 1}^x {\frac{1}{k}} = \sum\nolimits_{k = 1}^{x + 1} {\frac{1}{k}} = \psi (x + 1) - \psi (1) $$

note that the second sum is the correct translation of the first under the definition of the indefinite sum, when this is made "definite".

Consequently $$ f'(x) = \psi ^{\left( 1 \right)} (x + 1) $$

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