How do you uses a Taylor series to approximate integrals?

Question

astrophysics · Answer

You will need your power series representations

kkutie7 · Answer

Taylor series:
$$e^{x^{2}}=1+x^{2}+\frac{ x^{4} }{ 2! }+...$$
Integral:$$\int\limits_{0}^{1}e^{x^{2}}dx$$

tkhunny · Answer

Taylor Series turns your function into a polynomial.  Any 1st year calculus student can find the anti-derivative of a polynomial.  :-)

kkutie7 · Answer

Oh is that all I have to do? I thought it was more complicated than that =)

astrophysics · Answer

Right, so $$e^x = \sum_{n=0}^{\infty} \frac{ x^n }{ n! }$$ then we have $$e^{x^2} = \sum_{n=0}^{\infty} = \frac{ x^{2n} }{ n! }$$ so we can let $$e^{x^2} = \sum_{n=0}^{\infty} \frac{ x^{2n} }{ n! }$$ now we can integrate both sides! $$\int\limits_{0}^{1} e^{x^2} dx = \int\limits_{0}^{1} \sum_{n=0}^{\infty} \frac{ x^{2n} }{ n! }$$

astrophysics · Answer

Note we set them equal to each other, to show they equal so now you may expand the series and integrate as you would regularly!

tkhunny · Answer

Of course, equality holds only if you use infinitely many terms.  If you drop ANY of them, you have your approximation.

astrophysics · Answer

Right

kkutie7 · Answer

the answer is 1.433. I got it