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)=x2cos⁡x2

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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)=x2cos⁡x2

Nola Robson · Accepted Answer

Here the objective is to find the Taylor’s series of f(x)=x2cos⁡x2 around a=0Taylor’s series of f(x) around x=a isf(x)=f(a)+f′(a)1!(x−a)+f″(a)2!(x−a)2+f‴(a)3!(x−a)3+...+fn(a)n!(x−a)n+R(z)Let f(x)=cos⁡x, and a=0f(x)=cos⁡x→f(0)=cos⁡0=1f′(x)=−sin⁡x→f′(0)=−sin⁡0=0f″(x)=−cos⁡x→f″(0)=−cos⁡0=−1f‴(x)=sin⁡x→f‴(0)=sin⁡0=0f4(x)=cos⁡x→f4(0)=cos⁡0=1Therefore the required Taylor’s series of cos⁡(x) isf(x)=f(a)+f′(a)1!(x−a)+f″(a)2!(x−a)2+f‴(a)3!(x−a)3+f4(a)4!(x−a)4+...cos⁡x=1+01!(x−0)+(−1)2!(x−0)2+03!(x−0)3+14!(x−0)4+05!(x−0)5+(−1)6!(x−0)6...cos⁡x=1−x22!+x44!−x66!+...cos⁡x=∑k=0∞(−1)k(2k)!x2kTo find the Taylor&#039;s series of cos⁡x2, replace x by x2 in Taylor&#039;s series of cos⁡x, socos⁡x=1−(x2)22!+(x4)24!−(x6)26!+...cos⁡x=1−x42!+x84!−x126!+...To find the Taylor&#039;s series of x2cos⁡x2, replace x by x2 in Taylor&#039;s series of cos⁡x2, sox2cos⁡x=x(1−x42!+x84!−x126!+...)

Jeffrey Jordon · Accepted Answer

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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

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