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 2020-11-22 Answered
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.
\(f(x)=x^2\cos x^2\)

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Nola Robson
Answered 2020-11-23 Author has 6098 answers
Here the objective is to find the Taylor’s series of \(f(x)=x^2\cos x^2\) around a=0
Taylor’s series of f(x) around x=a is
\(f(x)=f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f'''(a)}{3!}(x-a)^3+...+\frac{f^n(a)}{n!}(x-a)^n+R(z)\)
Let \(f(x)=\cos x,\) and a=0
\(f(x)=\cos x\to f(0)=\cos0=1\)
\(f'(x)=-\sin x\to f'(0)=-\sin0=0\)
\(f''(x)=-\cos x\to f''(0)=-\cos0=-1\)
\(f'''(x)=\sin x\to f'''(0)=\sin0=0\)
\(f^4(x)=\cos x\to f^4(0)=\cos0=1\)
Therefore the required Taylor’s series of \(\cos(x)\) is
\(f(x)=f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f'''(a)}{3!}(x-a)^3+\frac{f^4(a)}{4!}(x-a)^4+...\)
\(\cos x=1+\frac{0}{1!}(x-0)+\frac{(-1)}{2!}(x-0)^2+\frac{0}{3!}(x-0)^3+\frac{1}{4!}(x-0)^4+\frac{0}{5!}(x-0)^5+\frac{(-1)}{6!}(x-0)^6...\)
\(\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+...\)
\(\cos x=\sum_{k=0}^\infty\frac{(-1)^k}{(2k)!}x^{2k}\)
To find the Taylor's series of \(\cos x^2\), replace x by \(x^2\) in Taylor's series of \(\cos x\), so
\(\cos x=1-\frac{(x^2)^2}{2!}+\frac{(x^4)^2}{4!}-\frac{(x^6)^2}{6!}+...\)
\(\cos x=1-\frac{x^4}{2!}+\frac{x^8}{4!}-\frac{x^{12}}{6!}+...\)
To find the Taylor's series of \(x^2\cos x^2\), replace x by \(x^2\) in Taylor's series of \(\cos x^2\), so
\(x^2\cos x=x(1-\frac{x^4}{2!}+\frac{x^8}{4!}-\frac{x^{12}}{6!}+...)\)
\(x^2\cos x=x^2-\frac{x^6}{2!}+\frac{x^{10}}{4!}-\frac{x^{14}}{6!}+...\)
\(x^2\cos x=\sum_{k=0}^\infty\frac{(-1)^k}{(2k)!}x^{4k+2}\)
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