Answer:
[tex]e^{4x}=e^4+4e^4(x-1)+8e^4(x-1)^2+...[/tex]
[tex]\displaystyle e^{4x}=\sum^{\infty}_{n=0} \dfrac{4^ne^4}{n!}(x-1)^n[/tex]
Step-by-step explanation:
Taylor series expansions of f(x) at the point x = a
[tex]\text{f}(x)=\text{f}(a)+\text{f}\:'(a)(x-a)+\dfrac{\text{f}\:''(a)}{2!}(x-a)^2+\dfrac{\text{f}\:'''(a)}{3!}(x-a)^3+...+\dfrac{\text{f}\:^{(r)}(a)}{r!}(x-a)^r+...[/tex]
This expansion is valid only if [tex]\text{f}\:^{(n)}(a)[/tex] exists and is finite for all [tex]n \in \mathbb{N}[/tex], and for values of x for which the infinite series converges.
[tex]\textsf{Let }\text{f}(x)=e^{4x} \textsf{ and }a=1[/tex]
[tex]\text{f}(x)=\text{f}(1)+\text{f}\:'(1)(x-1)+\dfrac{\text{f}\:''(1)}{2!}(x-1)^2+...[/tex]
[tex]\boxed{\begin{minipage}{5.5 cm}\underline{Differentiating $e^{f(x)}$}\\\\If $y=e^{f(x)}$, then $\dfrac{\text{d}y}{\text{d}x}=f\:'(x)e^{f(x)}$\\\end{minipage}}[/tex]
[tex]\text{f}(x)=e^{4x} \implies \text{f}(1)=e^4[/tex]
[tex]\text{f}\:'(x)=4e^{4x} \implies \text{f}\:'(1)=4e^4[/tex]
[tex]\text{f}\:''(x)=16e^{4x} \implies \text{f}\:''(1)=16e^4[/tex]
Substituting the values in the series expansion gives:
[tex]e^{4x}=e^4+4e^4(x-1)+\dfrac{16e^4}{2}(x-1)^2+...[/tex]
Factoring out e⁴:
[tex]e^{4x}=e^4\left[1+4(x-1)+8}(x-1)^2+...\right][/tex]
Taylor Series summation notation:
[tex]\displaystyle \text{f}(x)=\sum^{\infty}_{n=0} \dfrac{\text{f}\:^{(n)}(a)}{n!}(x-a)^n[/tex]
Therefore:
[tex]\displaystyle e^{4x}=\sum^{\infty}_{n=0} \dfrac{4^ne^4}{n!}(x-1)^n[/tex]
