# Chapter 11 - Section 11.10 - Taylor and Maclaurin Series - 11.10 Exercises - Page 771: 5

$x,x^{2},\frac{x^{3}}{2}, \frac{x^{4}}{6}$

#### Work Step by Step

$f(x)=xe^{x}$ $f'(x)=(x+1)e^{x}; f'(0)=1$ $f''(x)=(x+2)e^{x}; f''(0)=2$ $f'''(x)=(x+3)e^{x}; f'''(0)=3$ $f^{(4)}(x)=(x+4)e^{x}; f^{(4)}(0)=4$ n-th term of Taylor series is $\dfrac{f^{n-1}(0)x^{n-1}}{(n-1)!}$ Ist term of Taylor series is $\frac{f(0)x^{0}}{(0)!}=0$ Second term of Taylor series is $\frac{f'(0)x^{1}}{1!}=x$ Third term of Taylor series is $\frac{f''(0)x^{2}}{2!}=x^{2}$ Fourth term of Taylor series is $\frac{f'''(0)x^{3}}{3!}=\frac{x^{3}}{2}$ Fifth term of Taylor series is $\frac{f^{(4)}(0)x^{4}}{4!}=\frac{x^{4}}{6}$ Hence the first four non-zero terms of the Taylor's series are: $x, x^{2}, \frac{x^{3}}{2}, \frac{x^{4}}{6}$

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