Another power series problem

1. Nov 30, 2004

kdinser

This one involves differentiation and integration of a power series.

I need to find the first derivative, second derivative, and integral of this.

$$f(x)=\sum(\frac{x}{2})^n$$

if $$\frac{d}{dx}f(x)=\sum na_n(x-c)^{n-1}$$

I guess I'm having problems figuring out what $$a_n$$ is.

Shouldn't it be 2^n and give

$$\sum\frac{n}{2^n}(x)^n$$ I know this is wrong, how is the book getting
$$\sum(\frac{n}{2})(\frac{x^n}{2})$$

I think if I can just figure out how to find $$a_n$$ I'll be all set, I hope

Last edited: Nov 30, 2004
2. Nov 30, 2004

Justin Lazear

Use the theorems

$$\int \left( \sum f \right) = \sum \left( \int f \right)$$

and

$$D \left( \sum f \right) = \sum \left( D f \right)$$

--J

3. Nov 30, 2004

kdinser

I totally screwed up the first function, it's correct now.

The bottom line for me is, if this is my problem
$$f(x)=\sum(\frac{x}{2})^n$$
what is $$a_n$$ and how do we know?

Thanks a lot for any help, this stuff is really driving me crazy. The book doesn't have nearly enough examples and the solutions manual just plain sucks.

4. Nov 30, 2004

Justin Lazear

You're going to need to start including the indices for the summations.

Regardless, I don't know why you're so worried about determining an $a_n$. It's not necessary and not particularly convenient, either. Just use the theorem I posted above, i.e.

$$\frac{d}{dx} \left( \sum_{n=0} \frac{x^n}{2^n} \right)= \sum_{n=0}\frac{d}{dx} \left( \frac{x^n}{2^n} \right)$$

--J

5. Nov 30, 2004

James R

What you have there is a general fact for the derivative of any power series of the form:

$$f(x) = \sum\limits_0^N a_n (x - c)^n$$

Because a power series can be differentiated term-by-term, the derivative is:

$$f'(x) = \sum\limits_0^N n a_n (x - c)^{n-1}$$

In your example, if you want to apply the formula mindlessly

$$a_n = \frac{1}{2^n}, \qquad c = 0$$

6. Nov 30, 2004

Justin Lazear

Just a note, the f' series could just as well start at n = 1, since the n = 0 term contributes nothing. Your book seems to use this fact.

--J

7. Nov 30, 2004

kdinser

Thanks, I think I see where I've been making most of my mistakes with power series in general, and why they have been giving me fits. In some situations I've been thinking of n as a variable and trying to treat it as such.

8. Nov 30, 2004

HallsofIvy

Staff Emeritus
$$f(x)=\sum(\frac{x}{2})^n= \sum\frac{1}{2^n}x^n$$

$$a_n= \frac{1}{2^n}$$

9. Nov 30, 2004

kdinser

one more thing is still bothering me though.

When I take the derivative of
$$\sum_{n=0}\frac{d}{dx} \left( \frac{x^n}{2^n} \right)$$

I come up with
$$\sum_{n=1}\left( \frac{n}{2^n} \right)\left( \frac{x}{2} \right) ^{n-1}$$

The book has
$$\sum_{n=1}\left( \frac{n}{2} \right)\left( \frac{x}{2} \right) ^{n-1}$$

I can see that they would be equivalent at n=1, but the terms after that are not going to match. I think I'm still missing something important here.

10. Nov 30, 2004

James R

$$\frac{d}{dx} \left( \frac{x^n}{2^n} \right) = \frac{n}{2^n}x^{n-1} = \frac{n}{2}\frac{x^{n-1}}{2^{n-1}} = \left(\frac{n}{2}\right)\left(\frac{x}{2}\right)^{n-1}$$

11. Nov 30, 2004

kdinser

Thanks James, makes sense now. I see my mistake now.