How Do You Apply Power Series to Evaluate Functions and Estimate Errors?

[kreil]

I'm having trouble with a few homework problems, so here are the problems and my thoughts.

Homework Statement

Use the power series to evaluate the function
f(x)= \frac{1}{\sqrt{1+x^4}}-cos(x^2)
at x=0.01. Use the first two terms in the series to approximate the function, but estimate the error introduced by truncating the series.

The attempt at a solution
My main problem with this question is that it appears the function is equal to zero at, and in the neighborhood of, x=0. Also, the values of all the derivatives of the function are zero at x=0, so my power series expansion looks like this: f(x)=0. Am I missing something here, or is this really a "trick" question?

2. Homework Statement
Find a two term approximation and an error bound for the integral
\int_0^t e^{-x^2}dx
in the interval 0<t<0.1

The attempt at a solution
I'm not sure how to start this one...should I treat the integrand as the function or is the integral included? If the function is just the integrand, I don't see any problems. However, if the integral is included in the function then how would I proceed?

Any thoughts or hints you all could provide would be most appreciated. This HW is due tomorrow so quick replies are welcome!

Josh

 

[michalll]

The first function is in first two aproximations always zero.
For the second i would suggest
\int_0^t e^{-x^2}dx = 0.5\int e^{-s}s^{-1/2}ds
then approximate e^{-s} which would lead to
0.5\int \sum_0^\infty((-1)^n (1/n!)s^{n})s^{-1/2}ds
which can be easily integrated

 

[kreil]

The first function is in first two aproximations always zero.
For the second i would suggest
\int_0^t e^{-x^2}dx = 0.5\int e^{-s}s^{-1/2}ds
then approximate e^{-s} which would lead to
0.5\int \sum_0^\infty((-1)^n (1/n!)s^{n})s^{-1/2}ds
which can be easily integrated

What formula did you use to convert the integral? Also, are the limits of integration the same? Where does the two term approximation come in? Is estimating the error in the expansion of the e-fcn enough?

 

[michalll]

I used McLaurin formula for e^x. The limits aren't the same but you can get back to x after integrating. The two term approximations means you use only two first terms in the sum. The sum is alternating so the error won't be greater than the absolute value of the third term in this case.

 

[kreil]

how would i go about switching back to x after the following:

0.5\int (1-s)s^{-1/2}ds=-\frac{\sqrt{s}}{3}(s-3)

 

[michalll]

The substitution used was s=x^{2}