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Power Series
- Thread starter skyd92
- Start date
BigGlenntheHeavy
Senior Member
- Joined
- Mar 8, 2009
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- 1,577
You don't know much of anything, do you?
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
Senior Member
- Joined
- Mar 8, 2009
- Messages
- 1,577
Why don't you try helping yourself.
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.
BigGlenntheHeavy
Senior Member
- Joined
- Mar 8, 2009
- Messages
- 1,577
DrMike pointed you in the right direction, can't you take it from there? If not, then you are over your head.
D
Deleted member 4993
Guest
skyd92 said: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.
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?.
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?.