# Thread: Differentiation of Power Series

1. ## Differentiation of Power Series

Let $\displaystyle s(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - ...$and $\displaystyle c(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} -$ ... 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

2. Originally Posted by tbyou87
Let $\displaystyle s(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - ...$and $\displaystyle c(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} -$ ... 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 $\displaystyle (s^2+c^2)' = 0$

(c)Conclude that $\displaystyle s^2+c^2 = 1$.

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

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 $\displaystyle s(x) = \sin x$ and $\displaystyle c(x) = \cos x$

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

so, for instance:

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

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

$\displaystyle = c(x)$

4. 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

5. 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

6. ah! i remember the short way

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

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