Let $0<a_1<a_2<\cdots<a_n$ be real numbers. Show that the equation has exactly $n$ real roots.

$\displaystyle \sum_{i=1}^n \dfrac{a_i} {a_i-x} =2015$

I multiplied out denominators and that was very long, but I don't think descarte's rule will be helpful. Please any help is appreciated.

  • $\begingroup$ i suppose u meant x in the denominator $\endgroup$ – user379195 May 11 '17 at 4:57
  • $\begingroup$ Yeah corrected that! $\endgroup$ – Aditya Narayan Sharma May 11 '17 at 4:59

$\displaystyle f(x)=\bigg(\sum\limits_{i=1}^n\frac{a_i}{a_i-x}\bigg)-2015\quad$ with all $a_i>0$.

$\displaystyle f'(x)=\sum\limits_{i=1}^n\underbrace{\frac{a_i}{(a_i-x)^2}}_{>0}>0\quad$ so $f$ is increasing on each interval where it is defined.

It is also clear that $f$ is continuous except in its poles, thus on $\mathbb R\setminus\{a_1,a_2,...,a_n\}$.

$\begin{cases} \displaystyle \lim\limits_{x\to a_i^-}f(x)=+\infty\\ \displaystyle \lim\limits_{x\to a_i^+}f(x)=-\infty\\ \end{cases}$

By applying intermediate value theorem and using that $f\ \nearrow$ strictly, there is exactly $1$ real root in $]a_i,a_{i+1}[$, making this $n-1$ roots for $i=1..n$.

We have also $\lim\limits_{x\to-\infty}f(x)=-2015<0$ so there is $1$ root in $]-\infty,a_1[$.

We have also $\lim\limits_{x\to+\infty}f(x)=-2015<0$ so there is no root in $]a_n,+\infty[$.

Combining the results, gives exactly $n$ roots in $\mathbb R$.

  • $\begingroup$ The interval method gives n-2 roots $\endgroup$ – user379195 May 11 '17 at 6:26
  • $\begingroup$ Why n-2 ? There are $n-1$ intervals. $\endgroup$ – zwim May 11 '17 at 6:56

another way is assume it has a complex root which is non real (it has roots by fundamental theorem of algebra). Let $x+iy$ be the complex root. Then we have $\sum_{i=1}^n a_i/(a_i-x-iy)=2015$ Multiply the lower side by conjugate to conclude $\sum_{i=1}^n (a_i^2-a_ix+ia_iy)/((a_i-x)^2+y^2)=2015$ Hence $\sum_{i=1}^na_iy/((a_i-x)^2+y^2)=0$ by comparing the imaginary part since each term has the same sign so $y=0$ . The fundamental theorem of algebra is applicable because the eqn reduces to solving a polynomial equation.

  • $\begingroup$ This method is great, thanks +1 $\endgroup$ – Aditya Narayan Sharma May 11 '17 at 5:18
  • $\begingroup$ i think the second method uses less information (but heavier machinery) $\endgroup$ – user379195 May 11 '17 at 5:21
  • $\begingroup$ Disproving the existence of any complex root what so ever is great, that's why I preferred this over IVT $\endgroup$ – Aditya Narayan Sharma May 11 '17 at 5:25
  • $\begingroup$ Guess there are no bugs now $\endgroup$ – user379195 May 11 '17 at 6:24
  • $\begingroup$ Surely the problem with this solution is that it assumes something (The Fundamental Theorem of Algebra) which is FAR more difficult to prove than what is asked in the question. I am guessing the OP has just learned the IVT, which is needed one way or another to do the question. $\endgroup$ – ancientmathematician May 11 '17 at 6:34

define $b_i=(a_i+a_{i+1})/2$ then considering the function $\prod_{i=1}^n(a_i-x)-\sum_{i=1}^n \prod_{j\neq i}(a_j-x)$ show that $f(b_i)$ is alternatively positive and negative. So there exists a real root in the intervals $(a_i,a_{i+1})$. Now multiply without expanding and put $a_1$ and $a_n$ respectively and compare with the behaviour of the polynomial you get by multiplying at $\infty,-\infty$ to get the remaining two roots which lie in $(-\infty,a_1)$ and $(a_n,\infty)$


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