00:01
For this problem, we want to find the degree n of the mclaurian polynomial that is used such that the error in approximating f of 0 .5 for f of x, which is equal to sine of x, is less than 0 .001.
00:16
Now, we recall the truncation error when using a mclaurian polynomial, this is given by the absolute value of r sub n of x, which is less than equal to mcm, the absolute value of x raised n plus 1 all over n plus 1 factorial, where m here is the upper bound for the derivative of f at x.
00:50
Now if our f of x is equal to sine of x, since this is periodic, we have f sub n plus 1 of x equal to, we have cosine x, we can have negative sign of x, and then you can have negative cosine of x, and then lastly we will have sign of x, and then you will go back to these four, because this is periodic.
01:24
If you notice, is the values for the derivative, its maximum value is just 1 because the range of values for cosine and sine is from negative 1 to 1.
01:37
So the absolute value of the n plus 1 derivative of f at x will be less than or equal to 1.
01:46
So let's call this our m...