How do you find the taylor series of #1 1x^2#?

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Answer 1 · 2015-08-09T20:31:35

The Taylor series basically hands us back #11x^2# ...

The Taylor series for #f(x)# about #0# is:

#f(0)+(f'(0))/(1!)x+(f''(0))/(2!)x^2+...#

For #f(x) = 11x^2#, we have

#f'(x) = 22x, f''(x) = 22, f'''(x) = 0,...#

So

#f(0) = 0#, #f'(0) = 0#, #f''(0) = 22#, #f'''(0) = 0#, ...

and the Taylor series is:

#f(0)+(f'(0))/(1!)x+(f''(0))/(2!)x^2+(f'''(0))/(3!)x^3+...#

#=0+0/(1!)x+22/(2!)x^2+0/(3!)x^3+...#

#=11x^2#