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 \(a = 0\) (Maclaurin polynomial) is \(P_n(x)=\sum_{k = 0}^{n}\frac{f^{(k)}(0)}{k!}x^{k}=f(0)+f^{\prime}(0)x+\frac{f^{\prime\prime}(0)}{2!}x^{2}+\cdots+\frac{f^{(n)}(0)}{n!}x^{n}\)

Step2: Find the first - and second - derivatives of \(y = f(x)=\ln(1 + 4x)\)

• First derivative: Using the chain rule, if \(y=\ln(u)\) and \(u = 1+4x\), then \(y^{\prime}=\frac{dy}{du}\cdot\frac{du}{dx}\). Since \(\frac{d}{du}\ln(u)=\frac{1}{u}\) and \(\frac{d}{dx}(1 + 4x)=4\), \(f^{\prime}(x)=\frac{4}{1 + 4x}\)
• Second derivative: Using the quotient rule \((\frac{u}{v})^{\prime}=\frac{u^{\prime}v - uv^{\prime}}{v^{2}}\), where \(u = 4\) and \(v=1 + 4x\). \(u^{\prime}=0\) and \(v^{\prime}=4\). So \(f^{\prime\prime}(x)=\frac{0\times(1 + 4x)-4\times4}{(1 + 4x)^{2}}=-\frac{16}{(1 + 4x)^{2}}\)

Step3: Evaluate the function and its derivatives at \(x = 0\)

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

Step4: Calculate \(P_2(x)\)

\(P_2(x)=f(0)+f^{\prime}(0)x+\frac{f^{\prime\prime}(0)}{2!}x^{2}\)
Substitute \(f(0) = 0\), \(f^{\prime}(0)=4\), \(f^{\prime\prime}(0)=-16\) into the formula:
\(P_2(x)=0 + 4x+\frac{-16}{2}x^{2}=4x-8x^{2}\)

Answer:

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