11
$\begingroup$

Prove that all roots of $$\displaystyle \sum_{r=1}^{70} \dfrac{1}{x-r} =\dfrac{5}{4} $$

are real

I encountered this question in my weekly test. I tried setting $\displaystyle \sum_{r=1}^{70} \dfrac{1}{x-r} = \dfrac{P'(x)}{P(x)} $ ; where $\displaystyle P(x) = \prod_{r=1}^{70} (x-r)$ and tried to use some inequality, but to no avail.

Can we also generalize this?

Find the condition such that

$$\displaystyle \sum_{r=1}^{n} \dfrac{1}{x-r} = a $$

has all real roots.

$n\in \mathbb{Z^+} \ ; \ a\in \mathbb{R}$

$\endgroup$
0
20
$\begingroup$

Alternatively: suppose that the function has a complex root; call it $x+iy$; $y\neq 0$.

Then $$\sum_{r=1}^{70}\frac{1}{x-r+iy}=\frac{5}{4}$$

multiply by the conjugate to get

$$\sum_{r=1}^{70}\frac{x-r-iy}{(x-r)^2+y^2}=\frac{5}{4}$$

which implies

$$\operatorname{Im}\left( \sum_{r=1}^{70}\frac{x-r-iy}{(x-r)^2+y^2} \right)=0$$ or $$\sum_{r=1}^{70}\frac{-y}{(x-r)^2+y^2}=0$$ which is a contradiction since each term has the same sign and is nonzero.

$\endgroup$
3
  • $\begingroup$ One might mention that this is essentially the proof of the Gauss-Lucas Theorem $\endgroup$ – Martin R Mar 21 '16 at 14:41
  • $\begingroup$ +1, but -- why do you structure it as a proof by contradiction? (If you remove the unnecessary assumption that $y \neq 0$, isn't it a perfectly valid direct proof that the imaginary part of any solution is 0?) $\endgroup$ – ruakh Mar 25 '16 at 1:14
  • $\begingroup$ Good point. I guess my mind jumps to contradiction. $\endgroup$ – Elliot G Mar 25 '16 at 1:20
13
$\begingroup$

Hint: Note that the derivative of the left-hand side, when it exists, is negative. So our function is decreasing in any interval $(k,k+1)$ where $1\le k\le 70-1$.

In this interval, our function is very large positive when $x$ is a tiny bit larger than $k$, and very large negative when $x$ is a tiny bit smaller than $k+1$. So by the Intermediate Value Theorem our equation has a root between $k$ and $k+1$.

$\endgroup$
3
  • 1
    $\begingroup$ This yields 69 real roots of the 70th degree polynomial but then the 70th must also be real because the co-efficients are real so non-real roots occur in conjugate pairs. $\endgroup$ – DanielWainfleet Mar 21 '16 at 6:53
  • 1
    $\begingroup$ @user254665: I had left that out to leave something to do. Another way of identifying the $70$-th is to use the same argument on $(70,\infty)$. Just to the right of $70$ the function is very large, and it decreases to $0$, passing $5/4$ along the way. $\endgroup$ – André Nicolas Mar 21 '16 at 7:30
  • $\begingroup$ Ok. Good presentation. $\endgroup$ – DanielWainfleet Mar 21 '16 at 7:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.