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)=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 \(P_n(x)=\sum_{k = 0}^{n}\frac{f^{(k)}(0)}{k!}x^{k}\), where \(f^{(k)}(x)\) is the \(k -\)th derivative of \(f(x)\)

Step2: Find the first - order derivative of \(f(x)\)

Given \(f(x)=\ln(1 + 4x)\). Using the chain rule \((\ln(u))^\prime=\frac{u^\prime}{u}\), where \(u = 1+4x\) and \(u^\prime=4\). So \(f^\prime(x)=\frac{4}{1 + 4x}\)

Step3: Evaluate \(f(x)\) and \(f^\prime(x)\) at \(x = 0\)

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

Step4: Construct \(P_1(x)\)

\(P_1(x)=\frac{f(0)}{0!}x^{0}+\frac{f^\prime(0)}{1!}x^{1}\). Since \(\frac{f(0)}{0!}=0\) and \(\frac{f^\prime(0)}{1!}=4\), then \(P_1(x)=4x\)

Answer:

\(P_1(x)=4x\)