Show that the following equation has exactly $n$ real roots.

Question

If $0<a_1< \cdots < a_n$, show that the following equation has exactly $n$ real roots. $$ \frac{a_1}{a_1-x}+\frac{a_2}{a_2-x}+ \frac{a_3}{a_3-x}+ \cdots + \frac {a_n}{a_n - x} = 2015$$

My working:
The LHS easily becomes $$a_1(a_2-x)...(a_n-x)+a_2(a_1-x)(a_3-x)...(a_nx)+.....+a_n(a_1-x)(a_2-x)...(a_{n-1} -x)$$ This shows it has $n$ roots

But what about exactly and real?

@stressed-out clearly, it is a polynomial with highest power of $x$=$n$ — ami_ba, Dec 07 '17 at 09:57
Hey! I see no reason of downvoting this beautiful Q...I also did show my working — ami_ba, Dec 07 '17 at 10:01
@amityas: Use the intermediate value theorem on the intervals $(a_{i-1},a_{i})$. It seems to work. :) (I didn't downvote your question. I think your question doesn't deserve to be downvoted. At least you've tried.) — stressed out, Dec 07 '17 at 10:02
My mind says that approx. 99% of stackexchange users are spam sadists ready for down voting a Q. only the holy 1% are answerers — ami_ba, Dec 07 '17 at 10:30
Also have a look at https://math.stackexchange.com/questions/1866687/number-of-real-roots-of-fraca-1a-1-x-fraca-2a-2-x-fraca-na-n and the linked posts. — StubbornAtom, Dec 07 '17 at 10:55

score 2 · Accepted Answer · answered Dec 07 '17 at 10:16

2

Hint:

Let $x=a+ib$ a root

Using Complex conjugate root theorem, $a-ib$ also be another root

$$2015-2015=\sum_{r=1}^n\left(\dfrac{a_r}{a_r-(a-ib)}-\dfrac{a_r}{a_r-(a+ib)}\right)=\sum_{r=1}^n?$$

Equate the imaginary parts to find $b=0$

answered Dec 07 '17 at 10:16

lab bhattacharjee

274,582

Ok...but how to prove that it has $n$ distinct? – ami_ba Dec 07 '17 at 10:22
Please ans if u are there – ami_ba Dec 07 '17 at 10:33
@amityas, Why do we need distinct ? – lab bhattacharjee Dec 07 '17 at 11:07

score 2 · Answer 2 · answered Dec 07 '17 at 10:17

Sketch the graph of the expression on the left hand side of the equation. Try cases n=1, 2 and 3 to get a feel for the general case.

Note that $f(x) = \dfrac{a_1}{a_1-x} + \dfrac{a_2}{a_2-x} + \dfrac{a_3}{a_3-x} + \dots + \dfrac{a_n}{a_n-x} $ has discontinuities/poles at $x=a_1, a_2, a_3, \dots, a_n$ and it is continuous everywhere else.

Also in each of the intervals $(-\infty, a_1), (a_1, a_2), (a_2, a_3), \dots, (a_n, \infty)$ the function $f(x)$ is a monotonically increasing function of $x$ because as $x$ increases each of the denominators decreases so each of the terms $\dfrac{a_i}{a_i-x}$ increases (because the $a_i$ are positive).

As $x$ approaches $-\infty$, each of the terms is small and positive and so $f(x)$ approaches 0 from above; as $x$ approaches $\infty$, each of the terms is small and negative and so $f(x)$ approaches 0 from below.

It should be clear from your sketch that the graph of $y=f(x)$ crosses the horizontal line $y=2015$ at exactly n points, one in each of the intervals $(-\infty, a_1), (a_1, a_2), (a_2, a_3), \dots, (a_{n-1}, a_n)$. So the equation $f(x) = 2015$ has exactly n real roots.

There is nothing special about the value 2015 - for any real value on the right hand side, the same argument shows that equation has exactly n real roots.

Show that the following equation has exactly $n$ real roots.

2 Answers2