Question

Use the definition of a Taylor series to find the first three non zero terms of the Taylor series for the given function centered at a=1. Write the Taylor series in summation notation

Answer 1

`e^(4x)=e^4+4e^4(x-1)+8e^4(x-1)^2+...`

`\displaystyle e^(4x)=\sum^(\infty)_(n=0) (4^ne^4)/(n!)(x-1)^n`

Explanation:

Taylor series expansions of f(x) at the point x = a

`\text{f}(x)=\text{f}(a)+\text{f}\:'(a)(x-a)+\frac{\text{f}\:''(a)}{2!}(x-a)^2+\frac{\text{f}\:'''(a)}{3!}(x-a)^3+...+\frac{\text{f}\:^((r))(a)}{r!}(x-a)^r+...`

This expansion is valid only if
`\text{f}\:^((n))(a)` exists and is finite for all
`n \in \mathbb{N}`, and for values of x for which the infinite series converges.

`\textsf{Let }\text{f}(x)=e^(4x) \textsf{ and }a=1`

`\text{f}(x)=\text{f}(1)+\text{f}\:'(1)(x-1)+\frac{\text{f}\:''(1)}{2!}(x-1)^2+...`

`\boxed{\begin{minipage}{5.5 cm}\underline{Differentiating $e^(f(x))$}\\\\If $y=e^(f(x))$, then $\frac{\text{d}y}{\text{d}x}=f\:'(x)e^(f(x))$\\\end{minipage}}`

`\text{f}(x)=e^(4x) \implies \text{f}(1)=e^4`

`\text{f}\:'(x)=4e^(4x) \implies \text{f}\:'(1)=4e^4`

`\text{f}\:''(x)=16e^(4x) \implies \text{f}\:''(1)=16e^4`

Substituting the values in the series expansion gives:

`e^(4x)=e^4+4e^4(x-1)+(16e^4)/(2)(x-1)^2+...`

Factoring out e⁴:

`e^(4x)=e^4\left[1+4(x-1)+8}(x-1)^2+...\right]`

Taylor Series summation notation:

`\displaystyle \text{f}(x)=\sum^(\infty)_(n=0) \frac{\text{f}\:^((n))(a)}{n!}(x-a)^n`

Therefore:

`\displaystyle e^(4x)=\sum^(\infty)_(n=0) (4^ne^4)/(n!)(x-1)^n`