Question

Help with sequences and series:
Show that the function f(x) = sum from 0 to infinity ((-1)^n * x^(2n))/((2n)!) is a solution of the differential equation f''(x) + f(x) = 0.

Answer 1

I'll try to make to write f(x) in the equation editor....one sec.

Answer 2

\[\sum_{0}^{\infty} ((-1)^{n} x^{2n})/((2n)!)\]

Answer 3

That's f(x), and I need to show that it's a solution to the differential equation f''(x) + f(x) = 0.

Answer 4

take the second derivative of the sumation and add it to the sumation itself to see if it goes to 0

Answer 5

the summation is jsut a polynomial

Answer 6

in teh eq editor, if you type in \frac{}{} you can fill in the {}s with your top and bottom arguments

Answer 7

Oh I see, f(x) is just cos(x)! Thanks a lot!

Answer 8

\[\sum_{0}^{\infty} \frac{(-1)^{n} x^{2n}}{(2n)!}\]
\[D_x[\sum ]=\sum_{1}^{\infty} \frac{(-1)^{n} 2n\ x^{(2n-1)}}{(2n)!}\]
\[D_{xx}[\sum ]=\sum_{2}^{\infty} \frac{(-1)^{n} 2n(2n-1)\ x^{(2n-2)}}{(2n)!}\]

Answer 9

lol, that too

Answer 10

i think i need more practice on my summation derivatives :) but your way is sufficient i believe