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Let $0<a_1<a_2< . . . <a_n$ be real numbers .I need to show that the equation $$ \frac{a_1}{a_1-x} + \cdots +\frac{a_n}{a_n-x}=2015 $$ has exactly $n$ real roots.

Please tell me the steps, do I need to simplify the equation? It seems painful to me. Please dont solve the question

Harry Peter
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The Learner
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  • My first guess would be to try an inductive argument. Have you done this? – Stan Tendijck May 02 '19 at 13:48
  • The question doesn't say it has no complex roots, only that there are $n$ real roots. – saulspatz May 02 '19 at 13:48
  • No I haven't tried out inductively although – The Learner May 02 '19 at 13:48
  • @saulspatz the equation is of degree n so at max it is having n real roots – The Learner May 02 '19 at 13:50
  • Have you thought about what the graph looks like? – saulspatz May 02 '19 at 13:50
  • You're right about the number of roots. I was just about to delete my comment, but you beat me to it by an eyelash. I think looking at the graph will be helpful, though. – saulspatz May 02 '19 at 13:53
  • @saulspatz can you please provide your statements in the form of an answer. It will definitely help me. Shall be thankful to you if you do the needful – The Learner May 02 '19 at 13:53
  • @Stan Tendijick how can I go about inductively if I am to prove it for $n+1$ . Please provide an answer to this part only . I have seen that for n=1 the linear equation is having root which is real – The Learner May 02 '19 at 13:56
  • I would show this via intermediate value theorem. You can find $n+1$ "flips", locations where the parity of the function changes (from <2015 to >2015). I won't detail how to do this as this is the point of the problem, but this is a nice hint. – Rushabh Mehta May 02 '19 at 13:59
  • The graph should be having n asymptotes, and since the derivative is positive the graph between these asymptotes should be strictly increasing . Now I can manage using IVT . Thanks for your cooperation guys – The Learner May 03 '19 at 11:06

3 Answers3

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The function $$f(x)=\sum_{i=1}^n{a_i\over a_i-x}$$ is continuous except at the points $a_1,a_2,\dots,a_n.$ When $x$ is close to $a_i$ but smaller than $a_i, \ f(x)$ is very large ("goes to $\infty.$") When $x$ is close to $a_i$ but larger than $a_i$ then $f(x)$ goes to $-\infty.$ Now you need to examine the behavior of $f(x)$ when $x$ is greater than all the $a_i$ and when $x$ is smaller than all the $a_i$ and you can sketch the graph. Then just count the number of times it crosses the line $y=2015.$

If you have difficulty seeing how to do this for a general $n$, start by doing it for $n=3$ say.

saulspatz
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Let $g(x)=f(x)-2015$.

(see illustration below).

Reducing $g(x)$ do a same denominator gives a rational fraction which is $0$ if and only if its numerator is $0$. And this numerator is clearly a $n$th degree polynomial.

We are going first to find $n-1$ roots...

Indeed, on each interval $(a_i,a_{i+1})$, one has

$$\lim_{x \to a_i+0} \ g(x)=-\infty+A=-\infty$$

where $A:=\sum_{j=1, j\neq i}^n\frac{a_j}{a_j-a_i}-2015,$

whereas :

$$\lim_{x \to a_{i+1}-0} \ g(x)=+\infty+B=+\infty$$

where $B:=\sum_{j=1, j\neq (i+1)}^n\frac{a_j}{a_j-a_{i+1}}-2015.$

Thus, by continuity of $g$ on interval $(a_i,a_{i+1})$, there is at least a real root of $f$ on this interval (intermediate values theorem). See remark below.

Thus we have found at least one root on each interval $(a_i,a_{i+1})$. It remains to find the last one (we are at Easter time : looking for the last egg...).

Is it "hidden" in $(-\infty,a_1)$ or $(a_n,+\infty)$ ?

Up to you (you have said you don't want a full solution...).

enter image description here

Fig. 1 : An illustration with $g(x):=\dfrac{2}{2-x}+\dfrac{3}{3-x}+\dfrac{5}{5-x}-10$ (we have taken $10$ instead of $2015$ because the latter is too big). Axes are black, asymptotes are red.

Remark : [in fact there is only one root by interval, because $g$ is increasing on each interval, but we don't need it].

Jean Marie
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A proof by contradiction.

This is a polynomial of degree n on x. So, there are n number of roots. We have to prove that n roots are real. There is no complex root of the equation. Let, there are two complex roots $s + ir$ and $ s – ir$ (As complex roots come in conjugate pair)

Now, $ s + ir $will satisfy the equation.

a1/(a1 – s – ir) + a2/(a2 – s – ir) + .... + an/(an – s – ir) = 2015

Similarly, a1/(a1 – s + ir) + a2/(a2 – s + ir) + ..... + an/(an – s + ir) = 2015

Now, subtracting the equations we get,

{a1/(a1 – s – ir) – a1/(a1 – s + ir)} + {a2/(a2 – s – ir) – a2/(a2 – s + ir)} + ..... + {an/(an – s – ir) – an/(an – s + ir)} = 2015 – 2015

a1(2ir)/{(a1 – s)2 + r2} + a2(2ir)/{(a2 – s)2 + r2} + ..... + an(2ir)/{(an – s)2 + r2} = 0

(2ir)[a1/{(a1 – s)2 + r2} + a2/(a2 – s)2 + r2}+......+ an/{(an – s)2 + r2}] = 0

Now, a1, a2, ..., an all are greater than 0. The expression inside the square bracket is greater than 0 as denominators are sum of square numbers and greater than $0$ . $r$ has to be $0$. The imaginary part of the roots are zero. The roots are no more complex. Our assumption was wrong. There is no complex root of the equation. All roots are real. There are $n $ number of real roots.

The Learner
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  • Can this be also a suitable method? – The Learner May 02 '19 at 15:09
  • Sorry I haven't used MathJax , I was in a hurry to show if my attempt is correct or not – The Learner May 02 '19 at 15:09
  • No, you don't need to argue about complex roots because expression $g(x)=f(x)-2015$ is the partial fraction decomposition of a rational fraction $N(x)/D(x)$ with deg(N)=deg(D)=n which is $0$ if and only if its denominator is zero ; as we are able to find $n$ real roots for $N(x)=0$, there is no place for complex roots (a polynomial with degree $n$ has $n$ roots, here all distinct). – Jean Marie May 02 '19 at 16:10