Question: Given $\lambda_1<\lambda_2<\lambda_3 \in \Bbb R, a_1,a_2,a_3>0 $

Show that this equation has exactly 2 solutions:

$\frac {a_1}{x-\lambda_1}+\frac {a_2}{x-\lambda_2}+\frac {a_3}{x-\lambda_3}=0$

What I did: Try 1: Since we are studying continuity, Weirstrass theorem and Mean value theorem at the moment, I think one of these theorems should somehow be involved in the proof... Assuming that x is not equal to any of the lambdas I multiplied by the denumerators and got a quadratic equation- From this I figure that there must be at most 2 solutions. I thought about using the discriminant to prove that that there are exactly 2, but the calculation is very long, and I don't think this is what the lecturer intended for us to do, considering what we are studying now. Try 2: Thought of using the Intermediate value theorem. So I defined a function $f(x)=\frac {a_1}{x-\lambda_1}+\frac {a_2}{x-\lambda_2}+\frac {a_3}{x-\lambda_3}$ and I want to find when it equals 0. but Intermediate value theorem only involves two lambdas, and also I don't know if the lambdas are negative or positive (some/all of them)- Should I divide to cases?

Hints are welcome at first :-)

  • 1
    $\begingroup$ Hint: What sign is $\lim_{x\to \lambda_1^+} f(x)$ and $\lim_{x\to \lambda_2^-} f(x)$? $\endgroup$ – Arthur May 22 '13 at 15:17
  • $\begingroup$ The first one equals $\infty$ and the second equals $-\infty$, so now I should find a point in $(\lambda_1,\lambda_2)$, and repeat the process for limits of $(\lambda_2,\lambda_3)$ and find a point there? $\endgroup$ – jreing May 22 '13 at 15:34
  • $\begingroup$ You don't need to find them, just point out that they exist. But yeah, that's the idea. $\endgroup$ – Arthur May 22 '13 at 15:40
  • 1
    $\begingroup$ You also have to point out that there is only one point. For this, besides the continuity, you should use the fact that f(x) is monotone on those intervals. $\endgroup$ – gvo May 22 '13 at 15:48
  • $\begingroup$ @gvo He has already figured out that there cannot be more than two total, since the function $(x-\lambda_1)(x-\lambda_2)(x-\lambda_3)f(x)$ is a quadratic polynomial, it cannot have more than 2 zeroes. Since you cannot lose zeroes by multiplying with linear factors, $f(x)$ has a maximum of 2 zeroes. Since by the intermediate value theorem there is a zero between each of the asymptotes, there are at least 2. Therefore it has to be exactly 2. $\endgroup$ – Arthur May 22 '13 at 22:21

Yes, try to divide into cases.

More complete answer :

For $x \in ]- \infty, \lambda_1[$, all your terms are negative -> No solutions

For $x \in ]\lambda_1, \lambda_2[$ -> use the continuity of the function as you said on the try 2

For $x \in ]\lambda_2, \lambda_3[$ -> use the continuity of the function as you said on the try 2

For $x \in ]- \infty, \lambda_3[$, all your terms are positive -> No solutions


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