another power series problem

[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

[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

[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.

[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

[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

[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

[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.

[HallsofIvy]

To answer your original question, in
f(x)=\sum(\frac{x}{2})^n= \sum\frac{1}{2^n}x^n

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

[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.

[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}

[kdinser]

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