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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
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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm using proof by contradiction. P'(x) = 1 + x  (x²/2!) + ... + (1)^2 [nx^(n1) / 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^(n1) / n!] ...<2> Now, <1> + <2>: 1  A + (A²/2!)  ... + (1)^n (A^n/n!) + 1 + A  (A²/2!) + ... + (1)^2 [nA^(n1) / n!] = 0 All except for the last terms cancel out. (1)^n (A^n/n!) + + (1)^2 [nA^(n1) / n!] = ? I'm also not sure if the last term in <2> should be negative or not. Help please?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0to me, P(x) looks like the Taylor polynomial of e^(x)... just to throw out a possibility

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but I will try this as well

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0actually, the last term in P'(x) should cancel with one of the terms in P(x), the reason is this \[P(x) = 1 + ... + (1)^{n1}{x^{n1} \over (n1)!}+ (1)^n{x^n \over n!}\] and P'(x) is \[P'(x) = 1+...+(1)^n{x^{n1} \over (n1)!} \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so 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)^n1 and the other has (1)^n so they will cancel each other letting us know that P(x)+P'(x) = \[(1)^n{x^n \over n!}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ah, awesome!! Thanks so much :D
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