How do you write the taylor series about 0 for the function #(1+x)^3#?

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Answer 1 · 2015-08-02T14:05:41

The Taylor series computes as #1+3x+3x^2+x^3# as expected...

Let #f(x) = (1+x)^3#

Then:

#f'(x) = 3(1+x)^2#

#f''(x) = 6(1+x)#

#f'''(x) = 6#

So #f(0) = 1#, #f'(0) = 3#, #f''(0) = 6# and #f'''(0) = 6#

So the Taylor series for #f(x)# about #0# gives us:

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

#=1/1+3/1x+6/2x^2+6/6x^3#

#=1+3x+3x^2+x^3#