# 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

a. Another Maclaurin series we examined is for sin x: a. Find the series for $\frac{\mathrm{sin}\left({x}^{2}\right)}{x}$ . 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 $\frac{\mathrm{sin}\left({x}^{2}\right)}{x}$
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Benedict

The maclurin series is the Taylor series expansion about point a = 0.
Find the value for f(x) at a = 0.
$f\left(x\right)=\mathrm{sin}x$
$f\left(a\right)=\mathrm{sin}a$
$f\left(0\right)=\mathrm{sin}0$
So, $f\left(0\right)=0$
Consider the formula for the Taylor series.
Find the value for the derivatives required to find the Taylors series.
$f\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{{f}^{n}\left(a\right)}{n!}\left(x-a{\right)}^{n}$
$=\frac{f\left(a\right)}{0!}\left(x-a{\right)}^{0}+\frac{{f}^{1}\left(a\right)}{1!}\left(x-a{\right)}^{1}+\frac{{f}^{2}\left(a\right)}{2!}\left(x-a{\right)}^{2}+...$
$=0+\frac{{f}^{1}\left(a\right)}{1!}\left(x-a{\right)}^{1}+\frac{{f}^{2}\left(a\right)}{2!}\left(x-a{\right)}^{2}+...$
$=\frac{{f}^{1}\left(a\right)}{1!}\left(x-a{\right)}^{1}+\frac{{f}^{2}\left(a\right)}{2!}\left(x-a{\right)}^{2}+...$
Further, simplify.
Now, ${f}^{\prime }\left(x\right)=\frac{d}{dx}\left(\mathrm{sin}x\right)$
So, ${f}^{1}\left(x\right)=\mathrm{cos}x$
${f}^{1}\left(0\right)=\mathrm{cos}0$
$=1$
${f}^{″}\left(x\right)=\frac{d}{dx}\left(\mathrm{cos}x\right)$
So, ${f}^{2}\left(x\right)=-\mathrm{sin}x$
${f}^{2}\left(0\right)=-\mathrm{sin}x$
$=0$
${f}^{‴}\left(x\right)=\frac{d}{dx}\left(-\mathrm{sin}x\right)$
${f}^{3}\left(x\right)=-\mathrm{cos}x$
${f}^{3}\left(0\right)=-\mathrm{cos}0$
$=-1$
Then, find the Taylor series centered at a = 0.
Thus, find the Maclurins series.
At a=0
$\frac{\mathrm{sin}\left({x}^{2}\right)}{x}=\frac{{x}^{2}}{x}-\frac{{x}^{6}}{6x}+\frac{{x}^{10}}{120x}-...$
$x-\frac{1}{6}{x}^{5}+\frac{1}{120}{x}^{9}-...$
$\frac{x}{1!}-\frac{1}{3!}{x}^{5}+\frac{1}{5!}{x}^{9}-...$
$\sum _{n=0}^{\mathrm{\infty }}\left(-1{\right)}^{n}\frac{{x}^{4n+1}}{\left(2n+1\right)!}$
Hence, $\frac{\mathrm{sin}\left({x}^{2}\right)}{x}=\sum _{n=0}^{\mathrm{\infty }}\left(-1{\right)}^{n}\frac{{x}^{4n+1}}{\left(2n+1\right)!}$

Jeffrey Jordon