How do you find the taylor series of #sin (x^2)#?

1 Answer
Bill K.
Jun 11, 2015

The answer (about #x=0#) is #x^2-x^6/(3!)+x^10/(5!)-x^14/(7!)+\cdots#

Explanation:

You can get this answer in two ways:

1) Replace #x# with #x^2# in the well-known Taylor series for #sin(x)=x-x^3/(3!)+x^5/(5!)-x^7/(7!)+\cdots# (about #x=0#).

2) Let #f(x)=sin(x^2)# and use the formula #f(0)+f'(0)x+(f''(0))/(2!)x^2+(f'''(0))/(3!)x^3+\cdots#. Note that #f'(x)=2x cos(x^2)#, #f''(x)=2cos(x^2)-4x^2sin(x^2)#, etc... #f(0)=0#, #f'(0)=0#, #f''(0)=2#, etc...

Method 1 is clearly superior for this example.

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