Taylor Series Integration

1. Dec 2, 2009

vucollegeguy

1. The problem statement, all variables and given/known data

Use taylor series method to compute the integral from 1 to 2 of [sin(x2)] / (x2) with 10 -3 precision

2. Relevant equations

3. The attempt at a solution

I'm not sure where to start. Someone please help me.

Last edited: Dec 2, 2009
2. Dec 2, 2009

clamtrox

Do you know what a Taylor series is? You need to expand sin(x^2) as a Taylor series and then, given that the series satisfies some very broad conditions, you can integrate each term separately.

3. Dec 2, 2009

vucollegeguy

Yes.

sin(x2) after integration = (x3/3) -(x7)/(7*3!) + (x11)/(11*5!)-...
right?

can you help me with the rest of it?

4. Dec 2, 2009

clamtrox

$$\int_1^2 \frac{\sin(x^2)}{x^2} = \sum_{n=0}^{\infty} \int_1^2 \frac{(-1)^n (x^2)^{2n+1}}{(2n+1)! x^2} dx$$

Last edited: Dec 2, 2009
5. Dec 2, 2009

vucollegeguy

Thank you so much! So that stuff to that right of the equal sign, how do I use that solve the integral from 1 to 2 - do I just plug in the upper and lower limits and subtract the lower from the upper value? And the precision of 10 -3 - how do I get that?

I really appreciate your help.

6. Dec 2, 2009

clamtrox

The right hand side is easy to integrate now; all you have is a constant times x^(another constant).

7. Dec 2, 2009

vucollegeguy

Ok, integrating the right side gives:

(-1) n+1 * (x 2) 2n) / (2n!*x2)

right, or wrong?

8. Dec 2, 2009

clamtrox

Wrong.

$$\int_1^2 \frac{\sin(x^2)}{x^2} = \sum_{n=0}^{\infty} \int_1^2 \frac{(-1)^n (x^2)^{2n+1}}{(2n+1)! x^2} dx = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}\int_1^2 dx x^{4n}$$

9. Dec 2, 2009

vucollegeguy

I was totally wrong.
Ok, so now is where I plug into the upper and lower limits and subtract those values?

10. Dec 2, 2009

ideasrule

Yes. To get 10^-3 precision, you should keep on calculating terms until they go well below 10^-3.

11. Dec 2, 2009

vucollegeguy

Got it.
Thank you all for your help!