anonymous 5 years ago P(x) = 1 - x + (x²/2!) - ... + (-1)^n (x^n/n!) RTP that P(x) has no double zeroes for n≥2

1. anonymous

I'm using proof by contradiction. P'(x) = -1 + x - (x²/2!) + ... + (-1)^2 [nx^(n-1) / n!] Assume that there is a multiple root A, ie P(A) = 0 and P'(A) = 0 ie 1 - A + (A²/2!) - ... + (-1)^n (A^n/n!) = 0 ...<1> and -1 + A - (A²/2!) + ... + (-1)^2 [nA^(n-1) / n!] ...<2> Now, <1> + <2>: 1 - A + (A²/2!) - ... + (-1)^n (A^n/n!) + -1 + A - (A²/2!) + ... + (-1)^2 [nA^(n-1) / n!] = 0 All except for the last terms cancel out. (-1)^n (A^n/n!) + + (-1)^2 [nA^(n-1) / n!] = ? I'm also not sure if the last term in <2> should be negative or not. Help please?

2. anonymous

to me, P(x) looks like the Taylor polynomial of e^(-x)... just to throw out a possibility

3. anonymous

but I will try this as well

4. anonymous

actually, the last term in P'(x) should cancel with one of the terms in P(x), the reason is this $P(x) = 1 + ... + (-1)^{n-1}{x^{n-1} \over (n-1)!}+ (-1)^n{x^n \over n!}$ and P'(x) is $P'(x) = -1+...+(-1)^n{x^{n-1} \over (n-1)!}$

5. anonymous

so the last term of P'(x) and the second to last term of P(x) have opposite signs no matter what n is since one has (-1)^n-1 and the other has (-1)^n so they will cancel each other letting us know that P(x)+P'(x) = $(-1)^n{x^n \over n!}$

6. anonymous

does that help ?

7. anonymous

Ah, awesome!! Thanks so much :D