proof by contradiction? (1 Viewer)

[fantasia]

show that p(x) = 1 + x + x^2/2! + ... x^n/n! has no double root

i'm guessing this is a proof by contradiction, although i've forgotten what you do.

 

[Rorix]

P'(x) = ....
when P'(x) = 0, P(x) =/= 0 because...

 

[Estel]

show that p(x) = 1 + x + x^2/2! + ... x^n/n! has no double root
p'(x) = 1 + x + x^2/2! + ...x^(n-1)/(n-1)!
p(x)-p'(x) = x^n/n

Suppose that a is a double root of p(x)
p(a) = p'(a) = 0
i.e. p(a) - p'(a) = 0.
a^n/n = 0, a = 0
but then p(a) = 1
:. no double root.

 

[fantasia]

ah i get it now. thanks estel and rorix.