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How to differentiate series $\sum\limits_{n =1}^{100} n(101 - n) \times \log(x - n)$?

I was solving a problem which is mentioned below:

If $f(x) = \prod\limits_{n=1}^{100} (x-n)^{n(101 - n)}$ then find $\dfrac{f(101)}{f'(101)}$.

I took log on both sides of the given function and I am unable to find the derivative of the series.

Can anyone just give me a hint how to differentiate the series as I'm just a beginner to this topic. I would be greatly thankful.

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    Since the sum is finite (up to $100$), you can differentiate each term independantly. – TheSilverDoe Dec 16 '21 at 17:33
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    The derivative of $\ln (x-n)$ is $\frac{1}{x-n}$ – Alan Abraham Dec 16 '21 at 17:34
  • @AlanAbraham but it's inside the summation. Doesn't that matter? –  Dec 16 '21 at 17:34
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    Look at what TheSilverDoe said about differentiating the summation. – Alan Abraham Dec 16 '21 at 17:35
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    Derivative of a (finite number of terms) sum = sum of the derivatives of each term, whether there are 2 or 120340123 terms in the sum. – MasB Dec 16 '21 at 17:39
  • To emphasize, the number of terms must be some static number not related to the variable you are differentiating by. There is a popular fake proof where people try to perform $\frac{d}{dx}[x^2]$ by saying $\frac{d}{dx}\left[\sum\limits_{i=1}^x x\right] = \sum\limits_{i=1}^x 1 = x$. That simply is not allowed. – JMoravitz Dec 16 '21 at 17:58

1 Answers1

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Kudos to the OP for being wary of assuming that $F(x)=\sum f_n(x)$ implies $F'(x)=\sum f_n'(x)$, but for finite sums that assumption is indeed justified. Consequently, the derivative of $F(x)=\sum_{n=1}^{100}n(101-n)\log(x-n)$ is

$$F'(x)=\sum_{n=1}^{100}{n(101-n)\over x-n}$$

and so

$$F'(101)=\sum_{n=1}^{100}n={100\cdot101\over2}=5050\quad\text{(Ligget se!)}^*$$

The justification for bringing the derivative inside a finite sum is essentially just mathematical induction on the linearity formula $(f+g)'=f'+g'$. It's only when you have an infinite sum that things get dicey and you have to be careful.

$^*$See Versions of the Gauss Schoolroom Anecdote collected by Brian Hayes.

Barry Cipra
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