Question

Can someone explain how I would do this? a and b please...determine the degree of the Maclaurin polynomial...ect.

Answer 1

anyone have a clue? << just an explanation on how i can perform this calculation would suffice

Answer 2

1+x+x^2+x^3+\cdots\!

so the Taylor series for x−1 at a = 1 is

1-(x-1)+(x-1)^2-(x-1)^3+\cdots.\!

By integrating the above Maclaurin series we find the Maclaurin series for log(1 − x), where log denotes the natural logarithm:

-x-\frac{1}{2}x^2-\frac{1}{3}x^3-\frac{1}{4}x^4-\cdots\!

and the corresponding Taylor series for log(x) at a = 1 is

(x-1)-\frac{1}{2}(x-1)^2+\frac{1}{3}(x-1)^3-\frac{1}{4}(x-1)^4+\cdots,\!

and more generally, the corresponding Taylor series for log(x) at some a = x0 is:

\log ( x_0 ) + \frac{1}{x_0} ( x - x_0 ) - \frac{1}{x_0^2}\frac{( x - x_0 )^2}{2} + \cdots.

The Taylor series for the exponential function ex at a = 0 is

1 + \frac{x^1}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!}+ \cdots = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \frac{x^5}{120} + \cdots\! = \sum_{n=0}^\infty \frac{x^n}{n!}.

Answer 3

Hello, I hope that Assists you