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How do you find the Maclaurin Series for #F(x) = x*sin(x)#?

Answer 1

Isaiah Allen

Maclaurin series for #xsinx# is

= #x^2-x^4/(3!)+x^6/(5!)-x^8/(7!)+x^10/(9!).......#

A Maclaurin series is a Taylor series expansion of a function about #0#. For example,

#f(x)=f(0)+f'(0)x+(f''(0))/(2!)x^2+(f^(3)(0))/(3!)x^3+(f^4(0))/(4!)x^4+.......#

Thus Maclaurin series for #sinx# would be

#sinx=sin0+cos0*x-sin0/(2!)x^2-cos0/(3!)x^3+sin0/(4!)x^4+.......#

= #0+1*x-0*x^2-1/(3!)x^3+1/(5!)x^5.......#

= #x-1/(3!)x^3+1/(5!)x^5-1/(7!)x^7+1/(9!)x^9.......#

Hence, Maclaurin series for #xsinx# is

= #x^2-x^4/(3!)+x^6/(5!)-x^8/(7!)+x^10/(9!).......#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.