a. Another Maclaurin series we examined is for sin x: a. Find the series for sin⁡(x2)x . Express the result in the summation notation, not just as a partial listing of the series terms. b. Another Maclaurin series we examined is for sin x: d. Find the series for integral sin⁡(x2)x

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a. Another Maclaurin series we examined is for sin x: a. Find the series for sin⁡(x2)x . Express the result in the summation notation, not just as a partial listing of the series terms.  b. Another Maclaurin series we examined is for sin x:  d. Find the series for integral sin⁡(x2)x

Benedict · Accepted Answer

The maclurin series is the Taylor series expansion about point a = 0.Find the value for f(x) at a = 0.f(x)=sin⁡xf(a)=sin⁡af(0)=sin⁡0So, f(0)=0Consider the formula for the Taylor series.Find the value for the derivatives required to find the Taylors series.f(x)=∑n=0∞fn(a)n!(x−a)n=f(a)0!(x−a)0+f1(a)1!(x−a)1+f2(a)2!(x−a)2+...=0+f1(a)1!(x−a)1+f2(a)2!(x−a)2+...=f1(a)1!(x−a)1+f2(a)2!(x−a)2+...Further, simplify.Now, f′(x)=ddx(sin⁡x)So, f1(x)=cos⁡xf1(0)=cos⁡0=1f″(x)=ddx(cos⁡x)So, f2(x)=−sin⁡xf2(0)=−sin⁡x=0f‴(x)=ddx(−sin⁡x)f3(x)=−cos⁡xf3(0)=−cos⁡0=−1Then, find the Taylor series centered at a = 0.Thus, find the Maclurins series.At a=0sin⁡(x2)x=x2x−x66x+x10120x−...x−16x5+1120x9−...x1!−13!x5+15!x9−...∑n=0∞(−1)nx4n+1(2n+1)!Hence, sin⁡(x2)x=∑n=0∞(−1)nx4n+1(2n+1)!

Jeffrey Jordon · Accepted Answer

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a. Another Maclaurin series we examined is for sin x: a. Find the series for frac{sin(x^{2})}{x} . Express the result in the summation notation, not j

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