Power Series

[skyd92]

Again, don't even know where to start with this problem...

Problem: "Find a power series that converges to each function: n(x)= 3/(3-5x)"

Thanks!

 

[DrMike]

Do you know the power series for 1/(1-x) ?

Then, 3/(3-5x) = 1/(1-(5x/3))... write out the first poer series, but substitute 5x/3 in place of x everywhere.

 

[skyd92]

Ok, but then what? I don't know what that is supposed to result in.

 

[BigGlenntheHeavy]

You don't know much of anything, do you?

 

[skyd92]

If you don't want to help, then don't respond. I am taking this course by myself and I am trying the best that I can. So, yes, I do have a lot of questions, but that is the reason why I came to this site, to get my problems solved. I'm sorry if I may seem dumb to you, but I am just looking for someone that can actually help me.

 

[BigGlenntheHeavy]

Why don't you try helping yourself.

 

[skyd92]

I have tried, but when I say I have no clue what I am doing, I seriously have no clue what I am doing. I am just learning this concept and am having a hard time grasping it. So I'm sorry if I need a little hand-holding, but I am seriously new at this.

 

[skyd92]

can you maybe point me in the right direction? try to elaborate on what the previous guy was saying? help me help myself? Give me something more to work with or a better explanation?

 

[BigGlenntheHeavy]

DrMike pointed you in the right direction, can't you take it from there? If not, then you are over your head.

 

[Deleted member 4993]

Ok, but then what?

Did you do the expansion as suggested by the good_doctor?

If you did - please show us and we can explain the next step better.
I don't know what that is supposed to result in.

 

[galactus]

Rewrite $\displaystyle \frac{3}{3-5x}=\frac{1}{1-\frac{5}{3}x}$

and note that, generally, $\displaystyle \sum_{k=0}^{\infty}c^{k}x^{k}=\frac{1}{1-cx}$

That means you have $\displaystyle \sum_{k=0}^{\infty}(\frac{5}{3})^{k}x^{k}=\frac{1}{1-\frac{5}{3}x}=\frac{3}{3-5x}$

It is directly related to the famous geometric series $\displaystyle \sum_{k=0}^{\infty}x^{k}=\frac{1}{1-x}, \;\ x<1$. Only in this case, c=1.

See now?.

The idea is that we are finding a closed form for the series instead of writing them all out and adding one at a time.

As long as x<1, we can find the sum of the series by using the closed form.

Say we want to add them all up from 0 to 50 when x=1/2, arbitrarily.

$\displaystyle \sum_{k=0}^{50}(\frac{5}{3})^{k}(1/2)^{k}=1+\frac{5}{6}+\frac{25}{36}+................+\underbrace{.00010988}_{\text{50th term}}=5.99945057591$

But, instead of doing that, we just plug x=1/2 into the closed form and see that we get 6. As we add more and more terms and k heads off toward infinity, the series approaches 6. I just used 50 to show you how it gets closer and closer to 6 in this case. Try it with, say, x=1/10. It converges to 6/5.

$\displaystyle \frac{3}{3-5(\frac{1}{2})}=6$

That's the idea. See?. Does that help turn a light on a wee bit?.