00:01
For this problem, we want to approximate the value of the natural log of 1 .2 using taylor series for f of x, which is equal to the natural log of 1 plus x, and evaluating this tailor series at x equals 0 .2.
00:14
Now, let's recall, 1 over 1 minus x is equal to 1 plus x squared plus x cubed, x2 the 4th, and so on.
00:26
Or in summation format, that's just summation from n equals 0 to infinity of x raised a power n.
00:35
So 1 over 1 plus x, which is the same as 1 over 1 minus negative x, that's just equal to 1 plus negative x plus negative x squared plus negative x cubed plus negative x through the fourth and so on.
00:54
Or that's just the same as 1 minus x plus x squared minus x cubed plus x to the fourth and so on, which in summation format is equal to summation from n equals 0 to infinity of negative x raised of power n.
01:13
So if i take the antiderivative of 1 over 1 plus x and it's equivalent taylor series, i will have natural log of 1 plus x.
01:26
That's equal to, we have x minus x squared over 2 plus x cubed over 3 minus x to the 4th over 4 plus x to the 5th over 5 and so on.
01:41
Or that's the same as the summation from n equals 0 to infinity of negative 1 raised to power n times x raised to n plus 1 over n plus 1.
01:55
That means you will use this periseries here.
01:59
To approximate we will evaluate ln of 1 plus x using its tailor series at x equals 0 .2...