Question

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

Answer 1

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?

Answer 2

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

Answer 3

but I will try this as well

Answer 4

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

\[P'(x) = -1+...+(-1)^n{x^{n-1} \over (n-1)!} \]

Answer 5

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) =

\[(-1)^n{x^n \over n!}\]

Answer 6

does that help ?

Answer 7

Ah, awesome!! Thanks so much :D