Question

find the taylor polynomials of orders 0, 1, 2, and 3 generated by f at a.

f(x)=e^{4x}, a = 0

p_{0}(x)=1 (simplify your answer)

p_{1}(x)=1 + 4x

p_{2}(x)=1 + 4x + 8x^{2}

p_{3}(x)=\square

Explanation:

Step1: Recall the Taylor polynomial formula

The Taylor polynomial of order \(n\) for a function \(f(x)\) about \(a\) is \(P_{n}(x)=\sum_{k = 0}^{n}\frac{f^{(k)}(a)}{k!}(x - a)^{k}\). Here \(a = 0\), so \(P_{n}(x)=\sum_{k=0}^{n}\frac{f^{(k)}(0)}{k!}x^{k}\), and \(f(x)=e^{4x}\).
First, find the derivatives of \(f(x)\):

• \(f(x)=e^{4x}\), then \(f(0)=e^{4\times0}=1\)
• \(f^{\prime}(x)=4e^{4x}\), so \(f^{\prime}(0)=4e^{0}=4\)
• \(f^{\prime\prime}(x)=4^{2}e^{4x}\), then \(f^{\prime\prime}(0)=4^{2}e^{0}=16\)
• \(f^{(3)}(x)=4^{3}e^{4x}\), thus \(f^{(3)}(0)=4^{3}e^{0}=64\)

Step2: Calculate \(P_{3}(x)\)

Using the formula \(P_{3}(x)=\frac{f(0)}{0!}x^{0}+\frac{f^{\prime}(0)}{1!}x^{1}+\frac{f^{\prime\prime}(0)}{2!}x^{2}+\frac{f^{(3)}(0)}{3!}x^{3}\)

• \(\frac{f(0)}{0!}=1\) (since \(0!=1\) and \(f(0) = 1\))
• \(\frac{f^{\prime}(0)}{1!}x=\frac{4}{1!}x = 4x\)
• \(\frac{f^{\prime\prime}(0)}{2!}x^{2}=\frac{16}{2}x^{2}=8x^{2}\)
• \(\frac{f^{(3)}(0)}{3!}x^{3}=\frac{64}{6}x^{3}=\frac{32}{3}x^{3}\)

Answer:

\(P_{3}(x)=1 + 4x+8x^{2}+\frac{32}{3}x^{3}\)