The hyperbolic cosine function is defined as follows: $f(x)=\cosh (x)=\frac{e^{x}+e^{-x}}{2}$ .Use the Taylor polynomial for $e^{x}$ near 0 to find the Taylor polynomial of degree 4 for $f(x)=4 \cosh (x)$ .
A) $4-\frac{4 x^{2}}{2 !}+\frac{4 x^{3}}{3 !}-\frac{4 x^{4}}{4 !}$
B) $4\left(1+\frac{x^{2}}{2}+\frac{x^{4}}{24}\right)$
C) $4+\frac{3 x}{1 !}-\frac{3 x^{2}}{2 !}+\frac{x^{4}}{24}$
D) $4+4 x+\frac{4 x^{2}}{2}+\frac{4 x^{3}}{3 !}+\frac{4 x^{4}}{4 !}$

Question

Quizplus · Accepted Answer

The Taylor polynomial of $e^x$ near 0 up to degree 4 is $1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}$. For $e^{-x}$, it is $1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!}$. Adding these and dividing by 2 gives $\cosh(x) = 1 + \frac{x^2}{2!} + \frac{x^4}{4!}$. Multiplying by 4 gives $4\left(1 + \frac{x^2}{2!} + \frac{x^4}{4!}\right)$, which simplifies to $4\left(1 + \frac{x^2}{2} + \frac{x^4}{24}\right)$.

The Hyperbolic Cosine Function Is Defined as Follows f(x)=cosh⁡(x)=ex+e−x2f(x)=\cosh (x)=\frac{e^{x}+e^{-x}}{2}f(x)=cosh(x)=2ex+e−x​ Use the Taylor Polynomial For Defined

The Hyperbolic Cosine Function Is Defined as Follows $f(x)=\cosh (x)=\frac{e^{x}+e^{-x}}{2}$ Use the Taylor Polynomial For Defined