I have this question from my exam where it's asked to find the sum: $$S=\sum_{k=1}^n \frac{1}{(1-r_k)^2}$$ where $r_k$ are the roots of $$f(x)=x^n-2x+2\quad,n\ge3$$ I recalled this relation $$\frac{f'(x)}{f(x)}=\sum_{k=1}^n \frac{1}{x-r_k}$$ and quickly realised that $$\frac{d}{dx}\frac{1}{x}=-\frac{1}{x^2}$$ and using derivative would easily produce my answer. Indeed $$\frac{d}{dx}\left( \frac{f'(x)}{f(x)} \right)=\frac{f''(x)f(x)-(f'(x))^2}{f(x)^2}=-\sum_{k=1}^n \frac{1}{(x-r_k)^2}$$ Evaluating at $x=1$ gives $$S=\frac{(n-2)^2-n(n-1)}{1^2}=4-3n$$ However after the exam I realised that I have no ideea why $$\frac{f'(x)}{f(x)}=\sum_{k=1}^n \frac{1}{x-r_k}$$ and I just memorized it, is there a nice way to show it? And of course maybe a more elegant solution to this exam question?

  • $\begingroup$ Are you aware of the Newton-Raphson Method? $\endgroup$ – Harry Alli Feb 21 '18 at 13:34
  • $\begingroup$ Never heard of, but thanks I will google it. $\endgroup$ – Zacky Feb 21 '18 at 13:36


Using logarithm,

$$\ln (f(x))=\ln (x-r_1)+\ln(x-r_2)+\cdots +\ln(x-r_n)$$

Taking derivative,

$$\frac{f'(x)}{f(x)}= \frac{1}{(x-r_1)}+\frac{1}{(x-r_2)} +\cdots +\frac{1}{(x-r_n)}$$


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    $\begingroup$ The issue with this post that Joonas Ilmavirta points out in his own answer can be fixed simply by replacing $\ln(\,)$ with $\ln |\,|$ everywhere. $\endgroup$ – Paul Sinclair Feb 21 '18 at 17:40
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    $\begingroup$ But then the function may not be differentiable any more. @Paul $\endgroup$ – Jaideep Khare Feb 21 '18 at 18:34
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    $\begingroup$ In that case can't we just split ln|x-a| into ln(x-a) when x>a and ln(a-x) when x<a. Derivating first gives $$\frac{d}{dx}ln(x-a)=\frac{1}{x-a}$$ and $$\frac{d}{dx}ln(a-x) = \frac{1}{a-x}(a-x)' =\frac{1}{x-a}$$ So $$\frac{d}{dx}ln|x-a|=\frac{1}{x-a}, x \neq a$$ $\endgroup$ – Zacky Feb 21 '18 at 18:53
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    $\begingroup$ @Joonas $\log(ab)=\log a+\log b$, doesn't hold true, I agree, but there is a constant factor of $2n\pi$ which vanishes on taking derivative. $\endgroup$ – Jaideep Khare Feb 22 '18 at 12:07
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    $\begingroup$ Actually, the logarithmic derivative is a linear operator in itself, with nice properties. One does not need to go through the problematic step of writing $\log(f(x))$. $\endgroup$ – Tom-Tom Feb 27 '18 at 19:53

While Jaideep Khare's computation with the logarithm gives a good quick way to memorize or derive the result, there is an issue: The logarithms of negative numbers are not defined if one works over the reals, and care must be taken with branches if one moves to the complex plane. When $r_k$ is fixed and $x$ ranges through all the reals, $x-r_k$ will inevitably get negative values.

Here is a way to prove the desired result without resorting to logarithms: By the fundamental theorem of algebra a polynomial can be written in terms of its roots and a coefficient $a\in\mathbb C$ as $$ f(x) = a(x-r_1)\cdots(x-r_n). $$ The coefficient of $x^n$ is $a$ and in your case you know it to be $1$, but it is unimportant. The derivative can be computed by differentiating each term in the product in turn: $$ f'(x) = a(x-r_2)\cdots(x-r_n) + a(x-r_1)(x-r_3)\cdots(x-r_n) + \dots + a(x-r_1)\cdots(x-r_{n-1}), $$ where in the $n$th term of the sum the $n$th term of the product has been dropped. When $f(x)\neq0$, it is easy to divide each term of $f'(x)$ by $f(x)$, and we get $$ \frac{f'(x)}{f(x)} = \frac{1}{x-r_1} + \frac{1}{x-r_2} +\dots+ \frac{1}{x-r_n}. $$

The result and the proof are also applicable to complex polynomials, whereas with logarithms $f(z)$ will always hit the branching set of the logarithm, no matter how you choose it. Also, $\log(ab)=\log(a)+\log(b)$ is not true in general for complex $a$ and $b$.

Finally, let me emphasize that this result holds for a polynomial function $f$ (with roots counted with multiplicity), not for all functions in general. In your question you are working with polynomials but you speak about "functions", so beware of generalizing the result beyond what's true.

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    $\begingroup$ The trick with the log is slick, but I have to give the point to this answer, since it's more general. $\endgroup$ – David Stanley Feb 21 '18 at 15:05

Another approach, which is longer but would also work for more general symmetric expressions in the roots, is to use Vieta's formulae.

If a polynomial $f(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$ has roots $r_1, r_2, \dots, r_k$, then the ratio $-\frac{a_{n-1}}{a_n}$ gives us the sum $r_1 + r_2 + \dots + r_k$.

To turn this into the sum we want, we should modify $f(x)$ slightly.

First, $f(t-x)$ has roots $t - r_1, t - r_2, \dots, t - r_k$.

Second, $f(t-1/x)$, though it is not a polynomial, is still zero when $x = 1/(t-r_1), \dots, 1/(t-r_k)$.

However, multiplying through by $x^n$ gives the polynomial $x^n f(t-1/x)$ which has roots at $1/(t-r_1), \dots, 1/(t-r_k)$.

To apply the Vieta's formula we want to use, it remains to figure out the coefficients of $x^n$ and $x^{n-1}$ in $$ x^n f(t-1/x) = x^n a_n (t-1/x)^n + x^n a_{n-1} (t - 1/x)^{n-1} + \dots + x^n a_1 (t-1/x) + x^n a_0. $$ To get a multiple of $x^n$ from a term $x^n a_k (t-1/x)^k$, we have to choose the $t^k$ term of the binomial expansion of $(t-1/x)^k$. This tells us that the coefficient of $x^n$ in this polynomial is $$ a_n t^n + a_{n-1} t^{n-1} + \dots + a_0 = f(t). $$ To get a multiple of $x^{n-1}$ from a term $x^n a_k (t-1/x)^k$, we have to choose the $\binom k1 t^{k-1}(-1/x)$ term of the binomial expansion of $(t-1/x)^k$. This tells us that the coefficient of $x^{n-1}$ in this polynomial is $$ - a_n \cdot n t^{n-1} - a_{n-1} \cdot (n-1)t^{n-2} - \dots - a_1 = -f'(t). $$ Putting these together, we get that the sum of the roots of $x^n f(t-1/x)$ is $$ \sum_{i=1}^n \frac{1}{t-r_i} = \frac{f(t)}{f'(t)}, $$ as desired.

(For the specific case of the polynomial $f(x) = x^n - 2x + 2$, this would go faster since we wouldn't have to derive the general formula.)


In general, if $$f(x)=\prod_{k=1}^{n}f_k(x)$$ Then $$\ln f(x)=\sum_{k=1}^{n}\ln f_k(x)$$ and thus $$\frac{f'(x)}{f(x)}=\sum_{k=1}^{n}\frac{f_k'(x)}{f_k(x)}$$


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