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I need help computing

$$ \frac{d}{dx}\prod_{n=1}^{2014}\left(x+\frac{1}{n}\right)\biggr\rvert_{x=0} $$

The answer provided is $\frac{2015}{2\cdot 2013!}$, however, I do not know how to arrive at this answer. Does anyone have any suggestions?

  • 4
    It's not really an infinite product, is it? Just very large, right? The answer to the question is just the coefficient of $x$ in the expansion of the product. What I would suggest is that you try the problem for smaller values of $2014$ and see if you get the corresponding expression. That will give you an idea of how to generalize it and obtain the desired value. – 2017-02-16
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    @BrianTung I would give you correct answer but I can not since this is a comment. – 2017-02-16
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    Don't sweat it. As long as you figure out how to approach it, that's fine. – 2017-02-17

8 Answers 8

14

HINT:

Write $\prod_{n=1}^{2014}\left(x+\frac1n\right)=e^{\sum_{n=1}^{2014}\log\left(x+\frac1n\right)}$

Can you proceed now?

  • 0
    Dang, just beat me. +1 – 2017-02-16
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    Brevan, that has happened to me so many times here. – 2017-02-16
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    Hey, thanks for the response! How are the two answers equal to each other ? – 2017-02-16
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    @Ekesh *Copied from below my answer:* Note that $\log(ab) = \log(a)+\log(b)$. We can generalize this to get $\log\left(\prod_{k=a}^b f(k) \right) = \sum_{k=a}^b \log[f(k)]$. Then just note that $e^{\log(x)} = x$ – 2017-02-16
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    How does taking logs help? – 2017-02-16
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    @RobArthan It is quite a bit easier here, I believe, to work with the derivatives of a sum of logarithm functions that it is to take the derivative of a product of functions. – 2017-02-16
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    Yes, but this is just a polynomial in $x$ and all we need to do is to find the coefficient of $x$. – 2017-02-16
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    @RobArthan I understand. And inasmuch as the polynomials are affine, it is easier yet. But in my opinion, this way forward is just a bit easier and avoids the potential "fat finger" error in bookkeeping. – 2017-02-16
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    Logarithmic differentiation is the key when you need to differentiate an expression consisting of product/quotient of many sub-expressions. And it is almost necessary if one has to differentiate something like $\{f(x) \} ^{g(x)}$. +1 – 2017-02-17
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    @ParamanandSingh Thank you as always! – 2017-02-17
9

We can write the product as $$\exp\left(\sum_{n=1}^{2014}\log(x+1/n)\right)$$

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    Just like mine. ;-)) so (+1)! – 2017-02-16
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    Hey, how are the two answers equal ? – 2017-02-16
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    @Ekesh Note that $\log(ab) = \log(a)+\log(b)$. We can generalize this to get $\log\left(\prod_{k=a}^b f(k) \right) = \sum_{k=a}^b \log[f(k)]$. Then just note that $e^{\log(x)} = x$ – 2017-02-16
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    Fixed a typo and +1 – 2017-02-17
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    I appear to have just been downvoted, which is curious because Dr. MV wasn't. Would the downvoter please explain? – 2017-02-23
6

You are being asked to evaluate the derivative of a polynomial $f(x) = a_0 + a_1 x + \ldots $ when $x = 0$. Since $f'(x) = a_1 + 2a_2 x + \ldots$, the result will be $a_1$. In your case (putting $N = 2014$) $a_1$ is given by

$$ \sum_{n=1}^N\prod_{\begin{array}{c}m = 1\\m \neq n\end{array}}^N\frac{1}{m} = \sum_{n=1}^N \frac{n}{N!} = \frac{1}{N!}\sum_{n=1}^N n = \frac{1}{N!}\frac{1}{2}N(N+1) = \frac{N+1}{2\cdot(N-1)!} $$

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    Rob, this is fine, but in my humble opinion is a bit more work than the logarithm path you had questioned. (+1) for the solution. – 2017-02-16
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    @Dr.MV: perhaps you could complete your answer rather than leaving it as a hint to show us why it is less work to do it your way. – 2017-02-17
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    Rob, as I said in a comment following my post, I believe that the use of logarithms avoids pitfalls that come with the bookkeeping required in your approach. Moreover, I believe that it is conceptually easier. So, I'll leave it as a hint and others can pursue the hint toward a full solution. – 2017-02-17
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    @Dr.MV: so your way is better, but you're not prepared to show us why! – 2017-02-17
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    Rob, how many times have I qualified my statements as "in my opinion," or "I believe." Now, let's move on. And how on earth would I be able to show the virtue of avoiding fat-finger error? – 2017-02-17
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    I'm very happy to move on, but I am still wondering why it was too much work for you to show how the one line algebraic solution above is more work than the one you hinted at. – 2017-02-17
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    @RobArthan I definitely agree with Dr. MV here. The solution using e(log(blah)) is fairly straightforward to actually implement. It's fairly cut and dry to actually make it work. Your solution (while it is a good one!) has a lot more opportunities to make mistakes. I believe that is all that Dr. MV is saying. At some point during the derivation of the solution using your method it would be easier to make a small mistake or have slightly misguided reasoning and end up with the wrong answer. Honestly your answer might be less work but that doesn't mean it's "easier". – 2017-02-17
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    @Dason: I don't teach mathematics, so, if you do, I bow to your superior knowledge of the mistakes students make. This is really about a trade-off between algebra and analysis: my solution works over any ring but requires a bit of thought about how to determine a coefficient of a polynomial given as a product of affine polynomials; a solution using $\exp$ and $\log$ only works over certain rings such as $\Bbb{R}$ or $\Bbb{C}$ but makes the algebra a bit simpler provided you know the chain rule and the derivatives of $\exp$ and $\log$. – 2017-02-17
5

$\frac{d}{dx}\prod_{n=1}^{2014}\left(x+\frac{1}{n}\right) = \sum_\limits{i=1}^{2014} \prod_\limits{n\ne i}(x+\frac 1n)$

evaluating at $x=0$ gives us

$\sum_\limits{i=1}^{2014} \prod_\limits{n\ne i}(\frac 1n)\\\prod_\limits{n\ne i}(\frac 1n) = \frac {i}{2014!}\\ \frac 1{2014!}\sum_\limits{i=1}^{2014}i = \frac {(2015)(2014)}{2(2014!)} = \frac {2015}{2(2013!)}$

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    Really? First line: product rule $(x+a)(x+b) = (x+a) + (x+b)$ Next line: $\prod_\limits {n\ne i} \frac 1n = \frac {1}{1\cdot2\cdot3\cdots (i-1)(i+1)\cdots 2013\cdot 2014}$ multiply top and bottom by $i$ and you get $\frac {i}{2014!}$ last line is the sum of an arithmetic progression. – 2017-02-16
4

This big product is hard to work with, but we can turn it into an easier-to-work-with summation by taking advantage of logarithms. Let $\displaystyle f(x) = \prod_{n=1}^{2014} \left(x + \frac{1}{n} \right)$.

Chain rule gives $\displaystyle \Big(\ln(f(x)) \Big)' = \frac{f'(x)}{f(x)}$, which means $f'(x) = f(x) \Big( \ln(f(x)) \Big)'$.

A property of logarithms tells us that $\ln(f(x)) = \displaystyle \sum_{n=1}^{2014} \ln \left( x + \frac{1}{n} \right)$. The derivative of this is $\displaystyle \sum_{n=1}^{2014} \frac{1}{x + 1/n}$. Now let's evaluate $f'(x)$ at zero:

$$\displaystyle f'(0) = \left( \prod_{n=1}^{2014} \frac{1}{n} \right) \left( \sum_{n=1}^{2014} n \right)$$

Apply the famous formula for the sum of the first $n$ natural numbers and you arrive at what you're looking for.

2

Logarithmic differentiation is straightforward: note that for a differentiable function $f$ whose logarithm is well-defined, $$\frac{d}{dx}\left[\log f(x)\right] = \frac{f'(x)}{f(x)},$$ so we have with the choice $f(x) = \prod_{n=1}^m (x+1/n)$ $$\frac{df}{dx} = \frac{d}{dx}\left[\prod_{n=1}^m \left( x + \frac{1}{n} \right)\right] = \left(\prod_{n=1}^m \left(x + \frac{1}{n}\right)\right)\left( \frac{d}{dx}\left[\sum_{n=1}^m \log \left(x + \frac{1}{n}\right) \right] \right),$$ thus converting the derivative of a product into a sum of derivatives. Continuing, we find $$\frac{df}{dx} = f(x) \sum_{n=1}^m \frac{1}{x+1/n}.$$ Evaluating at $x = 0$ gives $$f'(0) = \frac{m(m+1)/2}{m!} = \frac{m+1}{2(m-1)!}.$$

  • 0
    Your first expression, $(\log(f))'=f'/f$, is correct, but appears to have been implemented incorrectly. You've written $f'=(log(f))'/f$ subsequently. – 2017-02-17
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    @Dr.MV Thanks for catching that error. Not sure why I did that! – 2017-02-17
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    You're quite welcome. My pleasure - pleased I could help. And I've made more mistakes on MSE than I can count. -Mark – 2017-02-17
0

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\left.\totald{}{x}\prod_{n = 1}^{2014}\pars{x + {1 \over n}} \right\vert_{\ x\ =\ 0} = 1 + {1 \over 2} + {1 \over 3} + \cdots + {1 \over 2014} = \bbx{\ds{H_{2014}}} \approx \ln\pars{2015} + \gamma \approx 8.1856 \end{align}

0

If expanded:

$$\frac{d}{dx} \prod_{n=1}^{2014} \left(x+\frac{1}{n}\right) \big\vert_{x=0}=\frac{d}{dx} \prod_{n=1}^{2014} \left(\frac{nx+1}{n}\right) \big\vert_{x=0}= \frac{\frac{d}{dx}\prod_{n=1}^{2014} (nx+1) \big\vert_{x=0}}{\prod_{n=1}^{2014} n}= \frac{\frac{d}{dx} \left[(x+1)(2x+1)(3x+1)\cdots(2014x+1)\right] \big\vert_{x=0}}{2014!}=\frac{\frac{d}{dx} \left[(C_{2014}x^{2014}+C_{2013}x^{2013}+\cdots+C_{2}x^2+ (1+2+\cdots+2014)x+1)\right] \big\vert_{x=0}}{2014!}=\frac{1+2+\cdots+2014}{2014!}=\frac{2015}{2\cdot2013!}$$