# Thread: Maclaurin series, dont know how to solve it

1. ## Maclaurin series, dont know how to solve it

Assume that equals its Maclaurin series for all x.
Use the Maclaurin series for to evaluate the integral

I get .334 for an answer, but evidently I'm doing something wrong, any insight would be outstanding, thanks

2. Originally Posted by N736RA
Assume that equals its Maclaurin series for all x.
Use the Maclaurin series for to evaluate the integral

I get .334 for an answer, but evidently I'm doing something wrong, any insight would be outstanding, thanks
A similar problem was dealt with here not so long ago. see if it helps

3. Originally Posted by N736RA
Assume that equals its Maclaurin series for all x.

Use the Maclaurin series for to evaluate the integral

I get .334 for an answer, but evidently I'm doing something wrong, any insight would be outstanding, thanks

Since we know that $sin(x)=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+1}}{(2n+1)!}$...we adapt this to $\sin(7x^2)$

$\int\sin(7x^2)dx=\int\sum_{n=0}^{\infty}\frac{(-1)^{n}(7x^2)^{2n+1}}{(2n+1)!}dx=\int\sum_{n=0}^{\i nfty}\frac{(-1)^{n}7^{2n+1}x^{4n+2}}{(2n+1)!}dx$

Integrating we get $\int_0^{.64}\sin(7x^2)dx=\sum_{n=0}^{\infty}\frac{ (-1)^{n}7^{2n+1}x^{4n+3}}{(2n+1)!\cdot(4n+3)}\bigg|_ 0^{.64}=.334$

4. so you agree that .334 should be the answer? thats what I got initially but evidently its wrong, any ideas? thanks a bunch for your help

5. Originally Posted by N736RA
so you agree that .334 should be the answer? thats what I got initially but evidently its wrong, any ideas? thanks a bunch for your help
I evaluated both the integral and the series on Mathcad(the infallable math program) and both yielded .334...so I would assume your source for answers has a typo

6. thanks for the help, I'll email the professor and see if theres a problem with the system, thanks again

7. Originally Posted by N736RA
thanks for the help, I'll email the professor and see if theres a problem with the system, thanks again
No problem!

8. Ok, so he got back to me, and evidently rather than the final answer, its looking for the first two terms in the series, so how would one go about solving for that? (wondering how to take both the x and the n into account) thanks

9. Originally Posted by N736RA
Ok, so he got back to me, and evidently rather than the final answer, its looking for the first two terms in the series, so how would one go about solving for that? (wondering how to take both the x and the n into account) thanks
$\sum_{n=0}^{1}\frac{(-1)^n7^{2n+1}x^{4n+3}}{(4n+3)(2n+1)!}\bigg|_0^{.63} =.262$

10. hmm no luck with that, are there any more decimals after that, for whatever stupid reason the program likes 10 decimals. And then how did you go about solving that sum (like how do you account for both variables?) thanks again for your help, and sorry if these are dumb questions haha

11. Originally Posted by N736RA
hmm no luck with that, are there any more decimals after that, for whatever stupid reason the program likes 10 decimals. And then how did you go about solving that sum (like how do you account for both variables?) thanks again for your help, and sorry if these are dumb questions haha
0.261759581135

12. no luck, sorry, any other ideas?

13. Originally Posted by N736RA
no luck, sorry, any other ideas?
Well maybe he wants the $a_1,a_2$ terms which are the second and third temrs...not the first and second...their sum would be a negative number so I dont think its them....I would have to say either your proffessor is making a mistake or I am misunderstanding something ...sorry

14. haha no problem, I would agree theres prlly an error, the professor is pretty useless when it comes to helping (its an online course), thanks for all of your help though! and just for kicks and giggles, what was the negative number you got? thanks again

15. Originally Posted by N736RA
haha no problem, I would agree theres prlly an error, the professor is pretty useless when it comes to helping (its an online course), thanks for all of your help though! and just for kicks and giggles, what was the negative number you got? thanks again
-0.242677011654

And no problem...thank me if you like

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