Question

Approximate f by a Taylor polynomial with degree n at the number a. Step 1 The Taylor polynomial with degree n = 3 is T3(x) = f(a) + f '(a)(x − a) + f ''(a) 2! (x − a)2 + f '''(a) 3! (x − a)3. The function f(x) = e2x2 has derivatives f '(x) = $$4x e2x2, f ''(x) = $$16x2+4 e2x2, and f '''(x) = $$48x+64x3 e2x2. Step 3 Therefore, T3(x) = . Submit Skip (you cannot come back)

Answer 1

If you center the series at x=1

`T_3(x) = e^2 + 4e^2 (x-1)+10(x-1)^2 + (56)/(3) e^2(x-1)^3 + R(x)`

Where
`R(x)` is the error.

Explanation:

From the information given we know that

`f(x) = e^(2x^2)`

`f'(x) = 4x e^(2x^2)` (This comes from the chain rule )

`f^((2))(x) = 4e^(2x^2) (4x^2+1)` (This comes from the chain rule and the product rule)

`f^((3))(x) = 16xe^(2x^2)(4x^2 + 3)` (This comes from the chain rule and the product rule)

If you center the series at x=1 then

`T_3(x) = e^2 + 4e^2 (x-1)+10(x-1)^2 + (56)/(3) e^2(x-1)^3 + R(x)`

Where
`R(x)` is the error.