w 2025 ww 9: problem 1 (1 point) question fin...
w 2025 ww 9: problem 1 (1 point) question find the first five non - zero terms of the maclaurin series for cos(4x) and then use that to write cos(4x) as a maclaurin series using sigma notation. solution we have f(x)=cos(4x) f(0)=1 f(x)= - 4sin(4x) f(0)=0 f(x)= - 16cos(4x) f(0)= - 16 f(x)=64sin(4x) f(0)=0 f^(4)(x)=256cos(4x) f^(4)(0)=256 f^(5)(x)= - 1024sin(4x) f^(5)(0)=0 f^(6)(x)= - 4096cos(4x) f^(6)(0)= - 4096 f^(7)(x)=16384sin(4x) f^(7)(0)=0 f^(8)(x)=65536cos(4x) f^(8)(0)=65536
Answer
# Explanation:
## Step1: Recall Maclaurin series formula
The Maclaurin series of a function $f(x)$ is given by $f(x)=\sum_{n = 0}^{\infty}\frac{f^{(n)}(0)}{n!}x^{n}=f(0)+f^{\prime}(0)x+\frac{f^{\prime\prime}(0)}{2!}x^{2}+\frac{f^{\prime\prime\prime}(0)}{3!}x^{3}+\frac{f^{(4)}(0)}{4!}x^{4}+\cdots$
## Step2: Find first - five non - zero terms
We know that $f(0) = 1$, $\frac{f^{\prime\prime}(0)}{2!}=\frac{- 16}{2}=-8$, $\frac{f^{(4)}(0)}{4!}=\frac{256}{24}=\frac{32}{3}$, $\frac{f^{(6)}(0)}{6!}=\frac{-4096}{720}=-\frac{256}{45}$, $\frac{f^{(8)}(0)}{8!}=\frac{65536}{40320}=\frac{256}{1575}$
So the first five non - zero terms are $1 - 8x^{2}+\frac{32}{3}x^{4}-\frac{256}{45}x^{6}+\frac{256}{1575}x^{8}$
## Step3: Write in sigma notation
Notice that for $n = 0,1,2,\cdots$, the general term of the Maclaurin series of $\cos(4x)$ is $(-1)^{n}\frac{(4x)^{2n}}{(2n)!}$. So $\cos(4x)=\sum_{n = 0}^{\infty}(-1)^{n}\frac{(4x)^{2n}}{(2n)!}$
# Answer:
First five non - zero terms: $1 - 8x^{2}+\frac{32}{3}x^{4}-\frac{256}{45}x^{6}+\frac{256}{1575}x^{8}$; Sigma notation: $\sum_{n = 0}^{\infty}(-1)^{n}\frac{(4x)^{2n}}{(2n)!}$