differentiating series

[sambellamy]

i am asked to show that

f(x) = Ʃ_n=0^∞ (-1)ⁿx²ⁿ / (2n)!

is a solution to the differential equation f(x) + f''(x) = 0.

This is what I did:

f'(x) = Ʃ_n=1^∞ (-1)ⁿ(2n)x^2n-1 / (2n)!

f''(x) = Ʃ_n=2^∞ (-1)ⁿ(2n)(2n-1)x^2n-2 / (2n)!

f''(x) = Ʃ_n=0^∞ (-1)ⁿ⁺²(2n+2)(2n+1)x²ⁿ / (2n+2)!

f''(x) = Ʃ_n=0^∞ (-1)ⁿ⁺²(2n+2)(2n+1)x²ⁿ / (2n+2)(2n+1)(2n)!

f''(x) = Ʃ_n=0^∞ (-1)ⁿx²ⁿ / (2n)!

but now I have f''(x) = f(x), so adding them, I got

2(-1)ⁿx²ⁿ / (2n)!

Where did I go wrong? Are there more algebraic steps I can do to get this to equal zero?

 

[Ishuda]

i am asked to show that

f(x) = Ʃ_n=0^∞ (-1)ⁿx²ⁿ / (2n)!

is a solution to the differential equation f(x) + f''(x) = 0.

This is what I did:

f'(x) = Ʃ_n=1^∞ (-1)ⁿ(2n)x^2n-1 / (2n)!

f''(x) = Ʃ_n=1^∞ (-1)ⁿ(2n)(2n-1)x^2n-2 / (2n)! <<<<<<<<<
...

Where did I go wrong? Are there more algebraic steps I can do to get this to equal zero?

The index changes because usually the series is for xⁿ and the term n (n-1) x^n-2 would vanish for both 0 and 1. However, this series for x²ⁿ and 2n (2n-1) doesn't vanish for n=0 AND 1.

Also, two more things - (1) you want f'' = -f, not f'' = f and (2) Although you can show what you have is a solution, it is not necessarily the complete solution. The solution you have is the cosine function but the sine function is also a solution.

 

[sambellamy]

The index changes because usually the series is for xⁿ and the term n (n-1) x^n-2 would vanish for both 0 and 1. However, this series for x²ⁿ and 2n (2n-1) doesn't vanish for n=0 AND 1.

I think this is one point where I get confused. I assumed the index was increased by one integer for each derivative taken. Where is the term n (n-1) x^n-2 that we are basing the index change on? Do we need to raise the index only if it would completely erase a term?

 

[Ishuda]

I think this is one point where I get confused. I assumed the index was increased by one integer for each derivative taken. Where is the term n (n-1) x^n-2 that we are basing the index change on? Do we need to raise the index only if it would completely erase a term?

The increase in index doesn't happen as a matter of taking the derivative, it happens because the terms 'ignored' are zero. Lets take your function
$\displaystyle f(x) = \Sigma_0^\infty (-1)^n \frac {x^{2n}}{(2n)!}$
The first derivative is
$\displaystyle f'(x) = \Sigma_0^\infty (-1)^n (2 n) \frac {x^{2n-1}}{(2n)!}$
Now the value of the term for n=0 is zero, so, if we wish, we can just start the summation at n=1, i.e.
$\displaystyle f'(x) = \Sigma_1^\infty (-1)^n (2 n) \frac {x^{2n-1}}{(2n)!}$
However, when we take the second derivative
$\displaystyle f''(x) = \Sigma_1^\infty (-1)^n (2 n) (2n -1) \frac {x^{2n-2}}{(2n)!}$
that term for n = 1 is not zero so we can't drop the n=1 index.

A different example, suppose
$\displaystyle g(x) = \Sigma_0^\infty (-1)^n \frac {x^n}{n!}$
so
$\displaystyle g'(x) = \Sigma_0^\infty (-1)^n n \frac {x^{n-1}}{n!}$
and the term for n=0 is again zero and we can start the summation at 1, i.e.
$\displaystyle g'(x) = \Sigma_1^\infty (-1)^n n \frac {x^{n-1}}{n!}$
and
$\displaystyle g''(x) = \Sigma_1^\infty (-1)^n n (n-1) \frac {x^{n-2}}{n!}$
This time, though, the value for the term for n=1 is also zero, so we can start the summation at n=2, i.e.
$\displaystyle g''(x) = \Sigma_2^\infty (-1)^n n (n-1) \frac {x^{n-2}}{n!}$