Question

Find the taylor polynomial t3(x) for the function f centered at the number
a. f(x) = eâ4xsin(2x), a = 0

Answer 1

$e^{-4x}=\displaystyle\sum_{n=0}^\infty\frac{(-4x)^n}{n!}=1+(-4x)+\dfrac{(-4x)^2}2+\dfrac{(-4x)^3}6+\cdots$
$e^{-4x}=1-4x+8x^2-\dfrac{32x^3}3+\cdots$

$\sin2x=\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^k(2x)^{2k+1}}{(2k+1)!}=(2x)-\dfrac{(2x)^3}6+\cdots$
$\sin2x=2x-\dfrac{4x^3}3+\cdots$

$e^{-4x}\sin2x=\left(1-4x+8x^2-\dfrac{32x^3}3+\cdots\right)\left(2x-\dfrac{4x^3}3+\cdots\right)$
$e^{-4x}\sin2x=2x-8x^2+\dfrac{44x^3}3+\cdots$

$\implies T_3(x)=2x-8x^2+\dfrac{44x^3}3$

Answer 2

The Taylor polynomial $T_3(x)$ will be written as $2x-8x^2+\dfrac{44x^3}{3}+......$.

Given:

The given function is $f(x) = e^{-4x}sin(2x)$.

It is required to find the Tylor polynomial $t_3(x)$ centered at a=0.

Now, the expansion of the function $e^{-4x}$ can be written as,

$e^{-4x}=\sum\dfrac{(-4x)^n}{n!}\\e^{-4x}=1+(-4x)^1+\dfrac{(-4x)^2}{2!}+\dfrac{(-4x)^3}{3!}+.....\\e^{-4x}=1-4x+\dfrac{16x^2}{2}-\dfrac{64x^3}{6}+.....\\e^{-4x}=1-4x+8x^2-\dfrac{32x^3}{3}+.....$

Similarly, the expansion of the function $sin(2x)$ will be,

$sin(2x)=\sum\dfrac{(-1)^n(2x)^{2n+1}}{(2n+1)!}\\=\dfrac{2x}{1!}+\dfrac{-(2x)^3}{3!}+.....\\=2x-\dfrac{4x^3}{3}+......$

So, the function $f(x) = e^{-4x}sin(2x)$ will be written as,

$f(x) = e^{-4x}sin(2x)\\f(x)=(1-4x+8x^2-\dfrac{32x^3}{3}+.....)(2x-\dfrac{4x^3}{3}+......)\\f(x)=2x-8x^2+16x^3-\dfrac{4x^3}{3}+.......\\f(x)=2x-8x^2+\dfrac{(48-4)x^3}{3}+......\\f(x)=2x-8x^2+\dfrac{44x^3}{3}+......$

Therefore, the Taylor polynomial $T_3(x)$ will be written as $2x-8x^2+\dfrac{44x^3}{3}+......$.

For more details, refer to the llink:

https://brainly.com/question/15739221