question find the first five non - zero terms...
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 now, the first five non - zero terms of the maclaurin series for cos(4x) are
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: Identify non - zero terms
From the derivatives and their values at $x = 0$:
- $f(0)=1$, so the first term is $1$.
- $f^{\prime\prime}(0)=- 16$, and the second term is $\frac{-16}{2!}x^{2}=-8x^{2}$.
- $f^{(4)}(0)=256$, and the third term is $\frac{256}{4!}x^{4}=\frac{256}{24}x^{4}=\frac{32}{3}x^{4}$.
- $f^{(6)}(0)=-4096$, and the fourth term is $\frac{-4096}{6!}x^{6}=\frac{-4096}{720}x^{6}=-\frac{256}{45}x^{6}$.
- $f^{(8)}(0)=65536$, and the fifth term is $\frac{65536}{8!}x^{8}=\frac{65536}{40320}x^{8}=\frac{256}{1575}x^{8}$.
## Step3: Write in sigma notation
We observe that the general term of the Maclaurin series for $\cos(4x)$ has the form $\frac{(-1)^{n}(4x)^{2n}}{(2n)!}$. So, $\cos(4x)=\sum_{n = 0}^{\infty}\frac{(-1)^{n}(4x)^{2n}}{(2n)!}$.
# Answer:
The first five non - zero terms are $1 - 8x^{2}+\frac{32}{3}x^{4}-\frac{256}{45}x^{6}+\frac{256}{1575}x^{8}$, and the Maclaurin series in sigma notation is $\sum_{n = 0}^{\infty}\frac{(-1)^{n}(4x)^{2n}}{(2n)!}$