00:01
For the first part of this problem, we are to find a mclaurian series for the natural log of 1 plus x cubed.
00:07
Now we recall the mclaurian series for 1 over 1 minus x is equal to the summation from n equals 0 to infinity of x raise of power of n.
00:18
So 1 over 1 plus x, that's equal to the summation from n equals 0 to infinity of negative x raise of power of n.
00:29
That's summation from n equal 0 to infinity of negative 1 raise to n times x raise a power of n.
00:36
Taking the n to derivative both sides, you have integral of 1 over 1 plus x dx, that's equal to the integral of the summation from n equal 0 to infinity of negative 1 raise to n times x rase to n dx.
00:51
That's natural log of the absolute value of 1 plus x equal to integral of 1 minus x plus x squared minus x cubed plus x 3rd with the 4th power and so on d x integrating term by term we have natural log of can just get rid of the absolute value anyway since you just want natural log of 1 plus x cubed so natural log of 1 plus x, that's equal to x minus x squared all over 2 plus x cubed over 3 minus x -rays to the 4th power over 4 and so on, and then plus c.
01:42
Note that if x is 0, you have natural log of 1 equal to 0 plus c, meaning c is also 0.
01:53
So natural log of 1 plus x, that's equal to summation from n equals 1 to infinity of negative 1 raise of power of n plus 1 times x raise the power of n all over n.
02:11
So if x is x cubed, we have natural log of 1 plus x cubed.
02:17
That's equal to summation from n equals 1 to infinity of negative 1.
02:23
Raised to n plus 1 times x raised to the third power times n or that's just 3n all over n.
02:33
For the radius of convergence we have to apply ratio test.
02:38
Now our nth term here is equal to negative 1 raise to n plus 1 times x raised to 3n all over n...