# Differentiation of Power Series

• Nov 3rd 2007, 08:56 PM
tbyou87
Differentiation of Power Series
Let $s(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - ...$and $c(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} -$ ... for x in R.

Prove that s' = c and c' = s.

I think you are supposed to differentiate term by term but i'm not sure exactly how that is done.

Thanks
• Nov 3rd 2007, 08:59 PM
ThePerfectHacker
Quote:

Originally Posted by tbyou87
Let $s(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - ...$and $c(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} -$ ... for x in R.

Prove that s' = c and c' = s.

I think you are supposed to differentiate term by term but i'm not sure exactly how that is done.

Thanks

Differenciate term by terms.

(b)Show that $(s^2+c^2)' = 0$

(c)Conclude that $s^2+c^2 = 1$. :D
• Nov 3rd 2007, 09:00 PM
Jhevon
Quote:

Originally Posted by tbyou87
Let $s(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - ... and c(x) = 1 - /frac{x^2}{2!} + /frac{x^4}{4!} - ... for x in R.$

Prove that s' = c and c' = s.

I think you are supposed to differentiate term by term but i'm not sure exactly how that is done.

Thanks

i suppose we are not to assume that $s(x) = \sin x$ and $c(x) = \cos x$

just differentiate term by term. meaning, differentiate each term separately

so, for instance:

$s(x) = x - \frac {x^3}{3!} + \frac {x^5}{5!} + \cdots$

$\Rightarrow s'(x) = 1 - \frac {x^2}{2!} + \frac {x^4}{4!} - \cdots$

$= c(x)$
• Nov 3rd 2007, 09:49 PM
tbyou87
So the first one makes sense to me.

I'm guessing for part b and c we are still using differentiation term by term. I guess i'm not seeing how to group them together so that they equal 0 for b and 1 for c.

Thanks
• Nov 3rd 2007, 09:52 PM
Jhevon
Quote:

Originally Posted by tbyou87
So the first one makes sense to me.

I'm guessing for part b and c we are still using differentiation term by term. I guess i'm not seeing how to group them together so that they equal 0 for b and 1 for c.

Thanks

for b you can do pretty easily by the chain rule. no term by term stuff.

for c, i forgot the short way, try doing it the loooonnnggg tteeedddiiiooouuusss way of actually multiplying out the power series and summing them
• Nov 3rd 2007, 10:04 PM
Jhevon
ah! i remember the short way

since you (should have, by now) proved that $\left( s^2(x) + c^2(x) \right)' = 0$ it means that $s^2(x) + c^2(x) = k$ for $k$ some constant.

indeed, since the series $s(x)$ and $c(x)$ work for all $x$, we have that the constant $k$ is the same for all $x$. in particular, try $x = 0$ and the conclusion is immediate