Elleanor Mckenzie

2021-09-01

For the function defined as follows, find the Taylor polynomials of degree 4 at 0.
$f\left(x\right)={e}^{x+1}$

nitruraviX

The function is, $f\left(x\right)={e}^{x+1}$
The Taylor polynomials of degree 4 at 0 is,
${P}_{4}\left(x\right)=f\left(0\right)+\frac{{f}^{\prime }\left(0\right)}{1!}x+\frac{f{}^{″}\left(0\right)}{2!}{x}^{2}+\frac{{f}^{\left(3\right)}\left(0\right)}{3!}{x}^{3}=\frac{{f}^{\left(4\right)}\left(0\right)}{4!}{x}^{4}$
Find the value of $f\left(x\right)$ at $x=0$ as,
$f\left(0\right)={e}^{0+1}$
$={e}^{1}$
$=e$
Find the first, second, third and fourth derivatives of $f\left(x\right)$ and find its corresponding values at $x=0$ as shown in below table.

The Taylor polynomials of degree 4 at 0 is calculated as follows.
${P}_{4}\left(x\right)=f\left(0\right)=\frac{{f}^{\prime }\left(0\right)}{1!}x+\frac{f{}^{″}\left(0\right)}{2!}{x}^{2}+\frac{{f}^{\left(3\right)}\left(0\right)}{3!}{x}^{3}+\frac{{f}^{\left(4\right)}\left(0\right)}{4!}{x}^{4}$
$=e+\frac{e}{1}x+\frac{e}{2}{x}^{2}+\frac{e}{6}{x}^{3}+\frac{e}{24}{x}^{4}$

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