Question

Ask for coefficients from Co to C4?

The Taylor series for `f(x)=e^x` at a=−4 is summation n=0 to infinity of cn(x+4)^n
(c n* lower case)

Answer 1

`c_0 =e^-4`
`c_1 = e^-4`
`c_2 = e^-4/2`
`c_3 = e^-4/6`
`c_4 = e^-4/24`

and in general:

`c_n = e^-4/(n!)`

Explanation:

By definition of the Taylor series of a function `f(x)` around `x=x_0`:

`f(x) = sum_(n=0)^oo (f^((n))(x_0))/(n!)(x-x_0)^n`

Now as:

`(d^nf)/dx^n e^x = e^x`

we have for `f(x) = e^x` and `x_0=-4`:

`f^((n))(-4) = e^(-4)`

and:

`e^x = sum_(n=0)^oo e^(-4)/(n!)(x+4)^n = e^-4sum_(n=0)^oo (x+4)^n/(n!)`

So:

`e^x = e^-4 (1+(x+4)+(x+4)^2/2+(x+4)^3/6+(x+4)^4/24)+ o(x^5)`