Doing a MacLaurin Series and more

hereiscassie
Messages
6
Reaction score
0
Doing a MacLaurin Series and more!

The function f is defined by f(x) = 1/(1+x^3). The MacLaurin series for f is given by
1 - x^3 + x^6 - x^9 +...+ (-1)^n(x^3n) +...
which converges to f(x) for -1 < x < 1.

a) Find the first three nonzero terms and the general term for the MacLaurin series for f '(x).

b) Use your results from part (a) to find the sum of the infinite series -3/2^2 + 6/2^5 - 9/2^8 +...+ (-1)^n[3n/ 2^(3n - 1)] +...

c) Find the first four nonzero terms and the general term for the MacLaurin series representing the intergral from 0 to x of f(t) dt.

d) Use the first three nonzero terns of the infinite series found in part (c) to approxiamate the integral from 0 to .5 of f(t) dt. What are the properties of the terms of the series representing the integral from 0 to .5 of f(t) dt that guarantee that this approxiamation is within 1/10,000 of the exact value of the integral.


I don't even know what to do in this one! I need some major help! especially in part b and d! DX

Please and Thank You! :)
 
Physics news on Phys.org


Could you rewrite the formula in LaTex?
It's terrible to read your 'formula'
 


What did you get for parts a and c?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

Similar threads

Replies
48
Views
5K
Replies
6
Views
2K
Replies
2
Views
5K
Replies
16
Views
3K
Replies
10
Views
2K
Replies
4
Views
3K
Back
Top