Question

ex – 2 Use the following Taylor series to find the first four nonzero terms of the Taylor series for the function centered at 0. ta eX = 1 + x + — + ... + = 2! k=0 The first nonzero term is .

Answer 1

Using only first four nonzero terms of Taylor series of $e^{x}$ centered at .$x=0$

$e^{x}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}=1+x+\frac{x^{2}}{2}+\frac{x^{3}}{6}$

$e^{x}-2=1+x+\frac{x^{2}}{2}+\frac{x^{3}}{6}-2=-1+x+\frac{x^{2}}{2}+\frac{x^{3}}{6}$

$\frac{e^{x}-2}{x^{5}}=\frac{-1+x+\frac{x^{2}}{2}+\frac{x^{3}}{6}}{x^{5}}$

$=\frac{-1}{x^{5}}+\frac{x}{x^{5}}+\frac{\frac{x^{2}}{2}}{x^{5}}+\frac{\frac{x^{3}}{6}}{x^{5}}$

$=\frac{-1}{x^{5}}+\frac{1}{x^{4}}+\frac{1}{2x^{3}}+\frac{1}{6x^{2}}$

That is

$\boldsymbol{\frac{e^{x}-2}{x^{5}}=\frac{-1}{x^{5}}+\frac{1}{x^{4}}+\frac{1}{2x^{3}}+\frac{1}{6x^{2}}}$

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