Any method a. Use any analytical method to find the first four nonzero terms of the Taylor series centered at 0 for the following functions. You do not need to use the definition of the Taylor series coefficients. f(x)=x^2cos x^2

cistG

cistG

Answered question

2020-11-22

Any method
a. Use any analytical method to find the first four nonzero terms of the Taylor series centered at 0 for the following functions. You do not need to use the definition of the Taylor series coefficients.
f(x)=x2cosx2

Answer & Explanation

Nola Robson

Nola Robson

Skilled2020-11-23Added 94 answers

Here the objective is to find the Taylor’s series of f(x)=x2cosx2 around a=0
Taylor’s series of f(x) around x=a is
f(x)=f(a)+f(a)1!(xa)+f(a)2!(xa)2+f(a)3!(xa)3+...+fn(a)n!(xa)n+R(z)
Let f(x)=cosx, and a=0
f(x)=cosxf(0)=cos0=1
f(x)=sinxf(0)=sin0=0
f(x)=cosxf(0)=cos0=1
f(x)=sinxf(0)=sin0=0
f4(x)=cosxf4(0)=cos0=1
Therefore the required Taylor’s series of cos(x) is
f(x)=f(a)+f(a)1!(xa)+f(a)2!(xa)2+f(a)3!(xa)3+f4(a)4!(xa)4+...
cosx=1+01!(x0)+(1)2!(x0)2+03!(x0)3+14!(x0)4+05!(x0)5+(1)6!(x0)6...
cosx=1x22!+x44!x66!+...
cosx=k=0(1)k(2k)!x2k
To find the Taylor's series of cosx2, replace x by x2 in Taylor's series of cosx, so
cosx=1(x2)22!+(x4)24!(x6)26!+...
cosx=1x42!+x84!x126!+...
To find the Taylor's series of x2cosx2, replace x by x2 in Taylor's series of cosx2, so
x2cosx=x(1x42!+x84!x126!+...)
Jeffrey Jordon

Jeffrey Jordon

Expert2021-12-27Added 2605 answers

Answer is given below (on video)

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