a. Find the Taylor Series of $f(x) = x^{2}*sin(\frac{x^{2}}{3})$ about a = 0.
b. Find the Taylor Series of $f(x) = x^{3}*cos(x^{2})$ about a = 0
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For the first one, I recalled the sin(x) series expansion:
$sin(x) = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} + \frac{x^{9}}{9!} + ....$
Since the one I am working with is in the Maclaurin form I can simply replace all the x's in the sine expansion with what is within the sine function I am given, but first pulling out the $x^{2}$, like so:
$x^{2}*sin(\frac{x^2}{3}) = \frac{x^2}{3} - \frac{(\frac{x^2}{3})^{3}}{3!} + \frac{(\frac{x^2}{3})^{5}}{5!} - \frac{(\frac{x^2}{3})^{7}}{7!} + \frac{(\frac{x^2}{3})^{9}}{9!} + ....$
Then accounting for the pulled out $x^{2}$ I multiply it all by $x^{2}$
$x^{2}*sin(\frac{x^2}{3}) = x^{2}\bigg(\frac{x^2}{3} - \frac{(\frac{x^2}{3})^{3}}{3!} + \frac{(\frac{x^2}{3})^{5}}{5!} - \frac{(\frac{x^2}{3})^{7}}{7!} + \frac{(\frac{x^2}{3})^{9}}{9!} + ....\bigg)$
And that I think would be the Taylor Series in its expanded form.
For the second one, I recall the series expansion for f(x) = cos(x) about a = 0
$cos(x) = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \frac{x^{6}}{6!} + \frac{x^{8}}{8!} - ....$
Same process as above Replace all the x's of the cosine expansion with what's in the current cosine function:
$x^3*cos(x^2) = 1 - \frac{(x^2)^{2}}{2!} + \frac{(x^2)^{4}}{4!} - \frac{(x^2)^{6}}{6!} + \frac{(x^2)^{8}}{8!} - ....$
Then multiply it all by $x^3$
like so:
$x^3*cos(x^2) = x^3\bigg(1 - \frac{(x^2)^{2}}{2!} + \frac{(x^2)^{4}}{4!} - \frac{(x^2)^{6}}{6!} + \frac{(x^2)^{8}}{8!} - ....\bigg)$
Is this incorrect?
Please help
Thank you