Constructing a Taylor Series
Key Questions

One of the benefits of a Taylor series is the ease of differentiation since it can be done term by term. So, the Talyor series
#f(x)=sum_{n=0}^infty{f^{(n)}(a)}/{n!}(xa)^n# can be differentiated as
#f'(x)=sum_{n=0}^infty{f^{(n)}(a)}/{n!}[(xa)^{n}]'# by Power Rule,
#=sum_{n=1}^infty{f^{(n)}(a)}/{n!}[n(xa)^{n1}]# (Note:
#n# starts from#1# since the term is zero when#n=0# )
by simplifying a little further,#=sum_{n=1}^infty{f^{(n)}(a)}/{(n1)!}(xa)^{n1}# I hope that this was helpful.

Let us solve
#y''+y=0# by Power Series Method.Let
#y=sum_{n=0}^inftyc_nx^n# , where#c_n# is to be determined.By taking derivatives,
#y'=sum_{n=1}^inftync_nx^{n1} Rightarrow y''=sum_{n=2}^inftyn(n1)c_nx^{n2}# We can rewrite
#y''+y=0# as#sum_{n=2}^inftyn(n1)c_nx^{n2}+sum_{n=0}^inftyc_nx^n=0# by shifting the indices of the first summation by 2,
#Rightarrow sum_{n=0}^infty(n+2)(n+1)c_{n+2}x^n+sum_{n=0}^inftyc_nx^n=0# by combining the summations,
#Rightarrow sum_{n=0}^infty[(n+2)(n+1)c_{n+2}+c_n]x^n=0# ,#Rightarrow (n+2)(n+1)c_{n+2}+c_n=0# #Rightarrow c_{n+2}={c_n}/{(n+2)(n+1)}# Let us look at even coefficients.
#c_2={c_0}/{2cdot1}=c_0/{2!}#
#c_4={c_2}/{4cdot3}={1}/{4cdot3}cdot{c_0}/{2!}=c_0/{4!}#
#c_6={c_4}/{6cdot5}={1}/{6cdot5}cdot c_0/{4!}={c_0}/{6!}#
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#c_{2n}=(1)^n{c_0}/{(2n)!}# Let us look at odd coefficients.
#c_3={c_1}/{3cdot2}={c_1}/{3!}#
#c_5={c_3}/{5cdot4}={1}/{5cdot4}cdot{c_1}/{3!}={c_1}/{5!}#
#c_7={c_5}/{7cdot6}={1}/{7cdot6}cdot{c_1}/{5!}={c_1}/{7!}#
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#c_{2n+1}=(1)^n{c_1}/{(2n+1)!}# Hence, the solution can be written as:
#y=sum_{n=0}^inftyc_nx^n# by splitting into even terms and odd terms,
#=sum_{n=0}^inftyc_{2n}x^{2n}+sum_{n=0}^inftyc_{2n+1}x^{2n+1}# by pluggin in the formulas for
#c_{2n}# and#c_{2n+1}# we found above,#=c_0sum_{n=0}^infty(1)^n{x^{2n}}/{(2n)!}+c_1 sum_{n=0}^infty(1)^n{x^{2n+1}}/{(2n+1)!}# by recognizing the power series,
#=c_0 cosx+c_1 sinx#
I hope that this was helpful.

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(xc)^n# #=a_0+a_1(xc)+a_2(xc)^2+cdots# ,where coefficients
#a_1, a_2, a_3,...# are to be determined.By taking the derivatives,
#f'(x)=a_1+2a_2(xc)+3a_3(xc)^2+cdots# #f''(x)=2a_2+3cdot2 a_3(xc)+4cdot3 a_4(xc)^2+cdots# #f'''(x)=3cdot2 a_3+4cdot3cdot2a_4(xc)+5cdot4cdot3a_5(xc)^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!}(xc)^n# .
I hope that this was helpful.
Questions
Power Series

Introduction to Power Series

Differentiating and Integrating Power Series

Constructing a Taylor Series

Constructing a Maclaurin Series

Lagrange Form of the Remainder Term in a Taylor Series

Determining the Radius and Interval of Convergence for a Power Series

Applications of Power Series

Power Series Representations of Functions

Power Series and Exact Values of Numerical Series

Power Series and Estimation of Integrals

Power Series and Limits

Product of Power Series

Binomial Series

Power Series Solutions of Differential Equations