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 \(a\) is \(P_n(x)=\sum_{k = 0}^{n}\frac{f^{(k)}(a)}{k!}(x - a)^k\). Here \(a = 0\), so \(P_n(x)=\sum_{k=0}^{n}\frac{f^{(k)}(0)}{k!}x^k\)

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

Given \(f(x)=\ln(1 + 4x)\). Using the chain rule, if \(y=\ln(u)\) and \(u = 1+4x\), then \(y^\prime=\frac{u^\prime}{u}\). 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: Find \(P_1(x)\)

For \(n = 1\), \(P_1(x)=\frac{f(0)}{0!}x^0+\frac{f^\prime(0)}{1!}x^1\)
Since \(0!=1\), \(1!=1\), \(f(0) = 0\), \(f^\prime(0)=4\)
\(P_1(x)=0 + 4x\)

Answer:

\(P_1(x)=4x\)