The Taylor series of a function is a power series, all of whose derivatives match their corresponding derivatives of the function.
Let us derive the Taylor series of a function #f(x)#, centered at #c#.
Let
#f(x)=sum_{n=0}^infty a_n(x-c)^n#
#=a_0+a_1(x-c)+a_2(x-c)^2+cdots#,
where coefficients #a_1, a_2, a_3,...# are to be determined.
By taking the derivatives,
#f'(x)=a_1+2a_2(x-c)+3a_3(x-c)^2+cdots#
#f''(x)=2a_2+3cdot2 a_3(x-c)+4cdot3 a_4(x-c)^2+cdots#
#f'''(x)=3cdot2 a_3+4cdot3cdot2a_4(x-c)+5cdot4cdot3a_5(x-c)^2+cdots#
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By plugging in #x=c#,
#f(c)=a_0=0! cdot a_0#
#f'(c)=a_1=1! cdot a_1#
#f''(c)=2a_2=2! cdot a_2#
#f'''(c)=3cdot2 a_3=3! cdot a_3#
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#f^{(n)}(c)=n! cdot a_n#
By dividing by #n!#,
#a_n={f^{(n)}(c)}/{n!}#
Hence, we have the Taylor series of #f(x)#, centered at #c#
#f(x)=sum_{n=0}^infty{f^{(n)}(c)}/{n!}(x-c)^n#.
I hope that this was helpful.
