Differentiation of Power Series

**tbyou87** · November 3rd 2007, 07:56 PM

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

**ThePerfectHacker** · November 3rd 2007, 07:59 PM

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 $s^2+c^2 = 1$ .

**Jhevon** · November 3rd 2007, 08:00 PM

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:

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

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

$= c(x)$

**tbyou87** · November 3rd 2007, 08:49 PM

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

**Jhevon** · November 3rd 2007, 08:52 PM

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

**Jhevon** · November 3rd 2007, 09:04 PM

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

Thread: Differentiation of Power Series

LinkBack

Thread Tools

Display

Differentiation of Power Series

Similar Math Help Forum Discussions

Exploiting geometric series with power series to find Taylors series

question about power series, taylor series, maclaurin series

Finding Power Series and Evaluating as a Power Series

Power Series Differentiation & Integration

differentiation and power series representation

Search Tags