Maclaurin Series(approximation)

• Dec 10th 2009, 03:03 AM
nameck
Maclaurin Series(approximation)
The Maclaurin series for sin x is

$\frac {\sin x}x = \sum\limits_{k = 0}^\infty {\frac{{( - 1)^k x^{2k} }}
{{(2k + 1)!}}} = 1 - \frac {x^2}{3!} + \frac {x^4}{5!} - \frac {x^6}{7!} + ..$

use the corresponding Maclaurin polynomial of degree 5 to approximate

$\int\frac{\sin x}{x} dx$

integration from -1 to 1

i dont have any idea to solve it..
• Dec 10th 2009, 03:35 AM
HallsofIvy
Quote:

Originally Posted by nameck
The Maclaurin series for sin x is

$\frac {\sin x}x = \sum\limits_{k = 0}^\infty {\frac{{( - 1)^k x^{2k} }}
{{(2k + 1)!}}} = 1 - \frac {x^2}{3!} + \frac {x^4}{5!} - \frac {x^6}{7!} + ..$

use the corresponding Maclaurin polynomial of degree 5 to approximate

$\int\frac{\sin x}{x} dx$

integration from -1 to 1

i dont have any idea to solve it..

Integrate "term by term"!

Since $\frac {\sin x}x = \sum\limits_{k = 0}^\infty {\frac{{( - 1)^k x^{2k} }}
{{(2k + 1)!}}} = 1 - \frac {x^2}{3!} + \frac {x^4}{5!} - \frac {x^6}{7!} + ..$

$\int_{-1}^1 \frac{sin x}{x}= \int_{-1}^1 dx- \frac{1}{3!}\int_{-1}^1 x^2 dx+ \frac{1}{5!}\int_{-1}^1 x^4 dx$

Since the problem said "polynomial of degree 5" and integrating $x^4$ will give degree 5, you can stop there.
• Dec 10th 2009, 04:04 AM
nameck
got it!! =)
• Dec 10th 2009, 04:19 AM
chisigma
The McLaurin expasion of the sin function is...

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

Only the 'odd' powers of x give contribution, so that is...

$\int_{-a}^{a} \sin x\cdot dx =0$ (2)

... no matter which is a...

http://digilander.libero.it/luposabatini/notte1.jpg

Merry Christmas from Italy

$\chi$ $\sigma$
• Dec 10th 2009, 04:44 AM
HallsofIvy
Yes, but this was sin(x)/x which is an even function.

Quote:

Originally Posted by chisigma
The McLaurin expasion of the sin function is...

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

Only the 'odd' powers of x give contribution, so that is...

$\int_{-a}^{a} \sin x\cdot dx =0$ (2)

... no matter which is a...

http://digilander.libero.it/luposabatini/notte1.jpg

Merry Christmas from Italy

$\chi$ $\sigma$

• Dec 10th 2009, 04:57 AM
chisigma
I have posted that in order to clarify the misundestanding caused by the fact that the function to be integrated is $\frac{\sin x}{x}$ and not $\sin x$ as written is the first post by nameck: the Maclaurin series for sin x is...

http://digilander.libero.it/luposabatini/notte1.jpg

Merry Christmas from Italy

$\chi$ $\sigma$
• Dec 10th 2009, 05:00 AM
nameck
i got 1.8922222..
is my answer correct?