- #1
Bazzinga
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[PLAIN]http://img600.imageshack.us/img600/1210/11096142.png
Hey I was wondering if you guys could help me out with this question...
I think I have the right power series:
[tex]= \frac{1}{1-x} + \frac{x}{1-x} [/tex]
[tex]= (1+x+x^{2}+x^{3}+...)(x+x^{2}+x^{3}+x^{4}+...) [/tex]
[tex]= 1+2x+2x^{2}+2x^{3}+2x^{4}+... [/tex]
[tex]= 1 + \sum^{infinity}_{n=1} 2x^{n} [/tex]
Then I integrate to get:
[tex] \int^{1/10}_{0} 1 + \sum^{infinity}_{n=1} 2x^{n} dx [/tex]
[tex]=x + \sum^{infinity}_{n=1} \frac{2x^{n+1}}{n+1}|^{1/10}_{0} [/tex]
[tex]= \frac{1}{10} + \sum^{infinity}_{n=1} \frac{(1/5)^{n+1}}{n+1} [/tex]
So I have the integral, I'm just not sure how to find an answer such that the error is less than 10-5 (I've never been any good with errors...)
We did learn in class today something where the error is less than
[tex] \frac{M}{(n+2)!} (b-a)^{n+2} [/tex]
Am I supposed to rearrange this formula? And what's M?
Hey I was wondering if you guys could help me out with this question...
I think I have the right power series:
[tex]= \frac{1}{1-x} + \frac{x}{1-x} [/tex]
[tex]= (1+x+x^{2}+x^{3}+...)(x+x^{2}+x^{3}+x^{4}+...) [/tex]
[tex]= 1+2x+2x^{2}+2x^{3}+2x^{4}+... [/tex]
[tex]= 1 + \sum^{infinity}_{n=1} 2x^{n} [/tex]
Then I integrate to get:
[tex] \int^{1/10}_{0} 1 + \sum^{infinity}_{n=1} 2x^{n} dx [/tex]
[tex]=x + \sum^{infinity}_{n=1} \frac{2x^{n+1}}{n+1}|^{1/10}_{0} [/tex]
[tex]= \frac{1}{10} + \sum^{infinity}_{n=1} \frac{(1/5)^{n+1}}{n+1} [/tex]
So I have the integral, I'm just not sure how to find an answer such that the error is less than 10-5 (I've never been any good with errors...)
We did learn in class today something where the error is less than
[tex] \frac{M}{(n+2)!} (b-a)^{n+2} [/tex]
Am I supposed to rearrange this formula? And what's M?
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