Finding the number of roots which are real

Question

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

Answer 1

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.

Answer 2

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...).

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].

Answer 3

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.