Constructing a Taylor Series
Key Questions
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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!}(x-a)^n# can be differentiated as
#f'(x)=sum_{n=0}^infty{f^{(n)}(a)}/{n!}[(x-a)^{n}]'# by Power Rule,
#=sum_{n=1}^infty{f^{(n)}(a)}/{n!}[n(x-a)^{n-1}]# (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)}/{(n-1)!}(x-a)^{n-1}# I hope that this was helpful.
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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^{n-1} Rightarrow y''=sum_{n=2}^inftyn(n-1)c_nx^{n-2}# We can rewrite
#y''+y=0# as#sum_{n=2}^inftyn(n-1)c_nx^{n-2}+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.
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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.
Questions
Power Series
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Introduction to Power Series
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Differentiating and Integrating Power Series
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Constructing a Taylor Series
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Constructing a Maclaurin Series
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Lagrange Form of the Remainder Term in a Taylor Series
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Determining the Radius and Interval of Convergence for a Power Series
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Applications of Power Series
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Power Series Representations of Functions
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Power Series and Exact Values of Numerical Series
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Power Series and Estimation of Integrals
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Power Series and Limits
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Product of Power Series
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Binomial Series
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Power Series Solutions of Differential Equations