Using a Power Series to approximate a definite integral

[Bazzinga]

[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:

$$= \frac{1}{1-x} + \frac{x}{1-x}$$
$$= (1+x+x^{2}+x^{3}+...)(x+x^{2}+x^{3}+x^{4}+...)$$
$$= 1+2x+2x^{2}+2x^{3}+2x^{4}+...$$
$$= 1 + \sum^{infinity}_{n=1} 2x^{n}$$

Then I integrate to get:

$$\int^{1/10}_{0} 1 + \sum^{infinity}_{n=1} 2x^{n} dx$$
$$=x + \sum^{infinity}_{n=1} \frac{2x^{n+1}}{n+1}|^{1/10}_{0}$$
$$= \frac{1}{10} + \sum^{infinity}_{n=1} \frac{(1/5)^{n+1}}{n+1}$$

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
$$\frac{M}{(n+2)!} (b-a)^{n+2}$$

Am I supposed to rearrange this formula? And what's M?

 

[grey_earl]

If you keep n+1 terms in your sum and neglect the rest, this formula gives you the (maximal) error you can have. M is the maximum of the (n+1)-st derivative of your function over the interval [a,b]=[0,1/10], which is in your case given by the next term in the power series, the first one you neglected, times (n+1)!, since you use a power series and not (directly) the derivative.
Beware: you may have to change n+1 for n+2 above, I'm not quite sure there, usually one truncates a series at the termn and then in the formulae there are n+1's insteadof the n+2 you gave us ...

 

[SteamKing]

One other note: When evaluating the limits inside the Sigma, exponentiation takes precedence over multiplication by coefficients. For 2 * x^(n+1) evaluated at x = 1/10, the correct evaluation is expressed as 2*(1/10)^(n+1), so that (1/10) is raised to the power (n+1) BEFORE multiplying by 2.

 

[Bazzinga]

I must have copied down the formula wrong, woops!

Im still not understanding.. I can find the error when finding an approximation for the series, but I don't understand how to find an approximation of the series when I have an error...

 

[grey_earl]

Instead of summing all terms in your result up to n=infinity, you just sum until a finite n=N. This is your approximation. Of course it isn't correct, but there comes the error formula: it tells you how big your mistake can be at most.
Therefore, you have to find a N such that the error, evaluated for that N, is smaller than 10^(-5), and use this N. (You should get N=4).

 

[Bazzinga]

I ended up using trial and error and I ended up with N=4. I'm not sure if there's a better way to find M, but I made a list of derivatives and found it that way and that worked! Thanks!