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)=\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 \(f(x)\):
The first - derivative \(f^{\prime}(x)=4e^{4x}\), the second - derivative \(f^{\prime\prime}(x)=16e^{4x}\).

Step2: Evaluate the derivatives at \(a = 0\)

When \(x = 0\), \(f(0)=e^{0}=1\), \(f^{\prime}(0)=4e^{0}=4\), \(f^{\prime\prime}(0)=16e^{0}=16\).

Step3: Calculate \(P_{2}(x)\)

Using the formula \(P_{2}(x)=\frac{f(0)}{0!}x^{0}+\frac{f^{\prime}(0)}{1!}x^{1}+\frac{f^{\prime\prime}(0)}{2!}x^{2}\).
Since \(0!=1\), \(1!=1\), \(2! = 2\), we have \(P_{2}(x)=1 + 4x+\frac{16}{2}x^{2}\).

Answer:

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