# Maclaurin series

• Apr 28th 2010, 08:50 PM
nhatie
Maclaurin series
3) Assume that http://webwork.umemat.maine.edu/webw...7f1bf381b1.png equals its Maclaurin series for all x.
Use the Maclaurin series for http://webwork.umemat.maine.edu/webw...5f4771bad1.png to evaluate the integral . Your answer will be an infinite series. Use the first two terms to estimate its value.

I am not sure how to do this!
• Apr 28th 2010, 09:00 PM
Failure
Quote:

Originally Posted by nhatie
3) Assume that http://webwork.umemat.maine.edu/webw...7f1bf381b1.png equals its Maclaurin series for all x.

Use the Maclaurin series for http://webwork.umemat.maine.edu/webw...5f4771bad1.png to evaluate the integral

. Your answer will be an infinite series. Use the first two terms to estimate its value.

I am not sure how to do this!

If you believe that $\sin(x)=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1}$ then surely you have that $\sin(5x^2)=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}(5x^2)^{2n+1}$.

Plugging this into your integral you get:

$\int_0^{0.64}\sin(5x^2)\, dx=\int_0^{0.64}\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}(5x^2)^{2n+1}\, dx$
Next thing to do is to be bold enough to exchange $\int$ and $\sum$, like this:

$=\sum_{n=0}^\infty \int_0^{0.64}\frac{(-1)^n}{(2n+1)!}(5x^2)^{2n+1}\, dx\approx \sum_{n=0}^1 \int_0^{0.64}\frac{(-1)^n}{(2n+1)!}(5x^2)^{2n+1}\, dx=\ldots$
• Apr 28th 2010, 09:08 PM
nhatie
Quote:

Originally Posted by Failure
If you believe that $\sin(x)=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1}$ then surely you have that $\sin(5x^2)=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}(5x^2)^{2n+1}$.

Plugging this into your integral you get:

$\int_0^{0.64}\sin(5x^2)\, dx=\int_0^{0.64}\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}(5x^2)^{2n+1}\, dx$
Next thing to do is to be bold enough to exhange $\int$ and $\sum$, like this:

$=\sum_{n=0}^\infty \int_0^{0.64}\frac{(-1)^n}{(2n+1)!}(5x^2)^{2n+1}\, dx\approx \sum_{n=0}^1 \int_0^{0.64}\frac{(-1)^n}{(2n+1)!}(5x^2)^{2n+1}\, dx=\ldots$

I kind of understand how to do that step but I can't seem to know what will next take place:D Thank you
• Apr 28th 2010, 09:12 PM
Failure
Quote:

Originally Posted by nhatie
I kind of understand how to do that step but I can't seem to know what will next take place:D Thank you

So let me elaborate this a bit more:

$\ldots=\int_0^{0.64} 5x^2\, dx+\int_0^{0.64}\frac{-1}{3!}(5x^2)^3\, dx=5\int_0^{0.64}x^2\, dx-\frac{125}{6}\int_0^{0.64}x^6\,dx=\ldots$
• Apr 29th 2010, 05:17 AM
nhatie
so it would be
(5)(x) from 0 to .64 - (125/6)(1/5)(x^5) from .0 to .64?
my answer of it is : 3.2 - .44739 = 2.7526
but they said it was wrong? did i do something wrong??
• Apr 29th 2010, 08:09 AM
Failure
Quote:

Originally Posted by nhatie
so it would be
(5)(x) from 0 to .64 - (125/6)(1/5)(x^5) from .0 to .64?
my answer of it is : 3.2 - .44739 = 2.7526
but they said it was wrong? did i do something wrong??

Apparently you must have made some mistake(s) when evaluating the remaning two integrals. To continue my own calculation:

$\ldots =5 \int_0^{0.64}x^2\, dx-\frac{125}{6}\int_0^{0.64}x^6\,dx= 5\cdot\Big[\frac{1}{3}x^3\Big]_{x=0}^{0.64}-\frac{125}{6}\cdot\Big[\frac{1}{7}x^7\Big]_{x=0}^{0.64}$
$=5\cdot\frac{1}{3}\cdot 0.64^3-\frac{125}{6}\cdot\frac{1}{7}\cdot 0.64^7\approx 0.306$
• Apr 29th 2010, 10:35 AM
nhatie
i try it but they said it was wrong :(
• Apr 29th 2010, 11:01 AM
Failure
Quote:

Originally Posted by nhatie
i try it but they said it was wrong :(

I'm sorry about that, but I cannot find a serious mistake in my calculation. If I ask a CAS to integrate $\int_0^{0.64}\sin(5*x^2)\, dx$ numerically I get the answer $\approx 0.32226$.
Now this does not fit exactly what I have found, namely $\approx 0.306$, but that's not particularly surprising because the next term in the series gives $\approx 0.0174$.
• Apr 29th 2010, 11:07 AM
nhatie
it is so weird!
• Apr 29th 2010, 11:12 AM
Failure
Quote:

Originally Posted by nhatie
it is so weird!

Are you sure you have given us the correct problem statement?
• Apr 29th 2010, 11:19 AM
nhatie
Quote:

Originally Posted by Failure
Are you sure you have given us the correct problem statement?

here
Assume that http://webwork.umemat.maine.edu/webw...7f1bf381b1.png equals its Maclaurin series for all x.
Use the Maclaurin series for http://webwork.umemat.maine.edu/webw...5f4771bad1.png to evaluate the integral . Your answer will be an infinite series. Use the first two terms to estimate its value.
• Apr 29th 2010, 11:41 AM
Failure
Quote:

Originally Posted by nhatie
here
Assume that http://webwork.umemat.maine.edu/webw...7f1bf381b1.png equals its Maclaurin series for all x.

Use the Maclaurin series for http://webwork.umemat.maine.edu/webw...5f4771bad1.png to evaluate the integral

. Your answer will be an infinite series. Use the first two terms to estimate its value.

If that's the case, then I take it that your teacher must be mistaken, or, another possibility, not telling you the truth. Why could that be? - Well, I remember that during my studies I once was told that my solution of a somewhat more difficult exercise was wrong. I immediately startet arguing with the teacher. After a few exchanges as to why I thought his objections were not valid, he began to smile and told me that he now really did believe that I was right.
So at that time, my teacher was simply testing whether I had found the solution to this exercise on my own (or at least understood the solution so well that I was able to argue for it convincingly). If a student gets the solution from someone else however, he or she typically will not immediately try to argue for its correctness, but instead rather quickly withdraw...