# 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

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

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

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.

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