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

Answer:

If you center the series at x=1

$T_3(x) = e^2 + 4e^2 (x-1)+10(x-1)^2 + \frac{56}{3} e^2(x-1)^3 + R(x)$

Where $R(x)$ is the error.

Step-by-step 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

$T_3(x) = e^2 + 4e^2 (x-1)+10(x-1)^2 + \frac{56}{3} e^2(x-1)^3 + R(x)$

Where $R(x)$ is the error.