Question

for the function $f(x)=\ln(1 + 4x)$, find the taylor polynomials of orders 0, 1, 2, and 3 generated by $f$ at $a = 0$.
$p_0(x)=0$
$p_1(x)=4x$
$p_2(x)=\square$

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 \(P_n(x)=\sum_{k=0}^{n}\frac{f^{(k)}(0)}{k!}x^k\).

First, find the derivatives of \(f(x)=\ln(1 + 4x)\).
The first - derivative: \(f^{\prime}(x)=\frac{4}{1 + 4x}\) (using the chain rule \((\ln(u))^\prime=\frac{u^\prime}{u}\), where \(u = 1+4x\) and \(u^\prime=4\)).
The second - derivative: \(f^{\prime\prime}(x)=-\frac{16}{(1 + 4x)^2}\) (using the quotient rule \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^2}\), here \(u = 4\), \(v=1 + 4x\), \(u^\prime=0\), \(v^\prime=4\)).

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

\(f(0)=\ln(1+4\times0)=0\)
\(f^{\prime}(0)=\frac{4}{1+4\times0}=4\)
\(f^{\prime\prime}(0)=-\frac{16}{(1 + 4\times0)^2}=-16\)

Step3: Construct \(P_2(x)\)

Using the Taylor polynomial 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 \(\frac{f(0)}{0!}=0\), \(\frac{f^{\prime}(0)}{1!}x = 4x\), and \(\frac{f^{\prime\prime}(0)}{2!}x^2=\frac{-16}{2}x^2=-8x^2\)

Answer:

\(P_2(x)=4x-8x^2\)