Taylor polynomials in the 1st, 2nd, 3rd and 4th order and Sigma expression?

It is easier to do part b first. A Taylor series is defined as:

f(x) = sigma( f^('j)(x0)*(x-x0)^j/j!,0,infinity)

where f^('j)(x0) means the j'th derivative of j evaluated at x0. In this case f(x)=e^(-2x), so in our sigma notation, we can just write:

f(x) = sigma(d^j/dx^j (e^(-2x)) | (x=x0) *(x-x0)^j/j!,0,infinity)

a) Use the formula we just created. We should create each term and then add them together to get each order of approximation.

0 order term = ln(2)

1 order term = f ' (x0)*(x-x0)^1/1! = -2*e^(-2x0)*(x-x0)^1/1 = -.5*(x-ln(2)

2 order term = f ' ' (x0)*(x-x0)^2/2! = f ' ' (x0)*(x-x0)^2/2

3 order term = f ' ' ' (x0))*(x-x0)^3/3! = f ' ' ' (x0))*(x-x0)^3/6

4 order term = f ' ' ' ' (x0))*(x-x0)^4/4! = f ' ' ' ' (x0))*(x-x0)^12

Now for each order we only add that many terms together. For the first order, its ln(2) -.5*(x-ln(2). For the third its ln(2) -.5*(x-ln(2) + f ' ' (x0)*(x-x0)^2/2, etc.

I'm sure you can figure out how to plug in the rest.