Question

use substitution to find the taylor series at x = 0 of the function \\( \ln \left(1+3 x^{4}\
ight) \\).
what is the taylor series for \\( \ln \left(1+3 x^{4}\
ight) \\) at x = 0?
a. \\( 3 x^{4}-\frac{3^{2} x^{8}}{2}+\frac{3^{3} x^{12}}{3}-\frac{3^{4} x^{16}}{4}+\cdots \\)
b. \\( 3 x^{4}-\frac{3 x^{8}}{2}+\frac{3 x^{12}}{3}-\frac{3 x^{16}}{4}+\cdots \\)
c. \\( 1-3 x^{4}+\frac{3 x^{8}}{2}-\frac{3 x^{12}}{3}+\cdots \\)
d. \\( -3 x^{4}+\frac{3^{2} x^{8}}{2}-\frac{3^{3} x^{12}}{3}+\frac{3^{4} x^{16}}{4}-\cdots \\)

Explanation:

Step1: Recall the Taylor series of \(\ln(1 + t)\)

The Taylor series of \(\ln(1 + t)\) at \(t = 0\) is \(\sum_{n = 1}^{\infty}\frac{(- 1)^{n+1}t^{n}}{n}=t-\frac{t^{2}}{2}+\frac{t^{3}}{3}-\frac{t^{4}}{4}+\cdots\)

Step2: Substitute \(t = 3x^{4}\)

Substitute \(t = 3x^{4}\) into the series \(\ln(1 + t)\).
When \(t = 3x^{4}\), we have:

$$
LATEXBLOCK0
$$

For \(n = 1\), the term is \(\frac{(-1)^{1 + 1}3^{1}x^{4\times1}}{1}=3x^{4}\)
For \(n = 2\), the term is \(\frac{(-1)^{2+1}3^{2}x^{4\times2}}{2}=-\frac{3^{2}x^{8}}{2}\)
For \(n = 3\), the term is \(\frac{(-1)^{3 + 1}3^{3}x^{4\times3}}{3}=\frac{3^{3}x^{12}}{3}\)
For \(n = 4\), the term is \(\frac{(-1)^{4+1}3^{4}x^{4\times4}}{4}=-\frac{3^{4}x^{16}}{4}\)

Answer:

A. \(3x^{4}-\frac{3^{2}x^{8}}{2}+\frac{3^{3}x^{12}}{3}-\frac{3^{4}x^{16}}{4}+\cdots\)