Question

How do you find the Maclaurin Series for $\sin \left( {{x^2}} \right)?$

Answer 1

First find the Maclaurin Series for $\sin x$ then put ${x^2}$ in place of the argument that is $x$. Maclaurin Series of a function $f(x)$ is given as following:
$f(x) = f(0) + \dfrac{{f'(0)}}{{1!}}(x) + \dfrac{{f''(0)}}{{2!}}({x^2}) + \dfrac{{f'''(0)}}{{3!}}({x^3}) + ...$
Where $f'(0),\;f''(0)\;{\text{and}}\;f'''(0)$ are derivatives of first, second and third order at $x = 0$ of the given function $f(x)$

Complete step by step solution:
In order to find the Maclaurin Series for $\sin ({x^2})$ we need to first
to find the Maclaurin Series of $\sin x$ and then we will replace $x\;{\text{with}}\;{x^2}$ then we will get the required Maclaurin series for $\sin ({x^2})$
Maclaurin series of a function $f(x)$ is given by
$f(x) = f(0) + \dfrac{{f'(0)}}{{1!}}(x) + \dfrac{{f''(0)}}{{2!}}({x^2}) + \dfrac{{f'''(0)}}{{3!}}({x^3}) + ...$
So first finding the Maclaurin Series for $f(x) = \sin x$
Here $f(x) = \sin x$ so finding respective values at $x = 0$
$f(0) = \sin (0) = 0 \\\ f'(0) = \cos (0) = 1 \\\ f''(0) = - \sin (0) = 0 \\\ f'''(0) = - \cos (0) = - 1 \\\$
Putting these values above in order to find the Maclaurin Series expansion of $f(x) = \sin x$
$f(x) = f(0) + \dfrac{{f'(0)}}{{1!}}(x) + \dfrac{{f''(0)}}{{2!}}({x^2}) + \dfrac{{f'''(0)}}{{3!}}({x^3}) + ... \\\ \sin x = 0 + \dfrac{1}{{1!}}(x) + \dfrac{0}{{2!}}({x^2}) + \dfrac{{( - 1)}}{{3!}}({x^3}) + ... \\\ \sin x = \dfrac{x}{{1!}} - \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^5}}}{{5!}} - ... \\\$
This is the Maclaurin Series of a sine function
Hence for Maclaurin Series expansion for $f(x) = \sin ({x^2})$ we will replace $x\;{\text{with}}\;{x^2}$ we will get
$\sin x = \dfrac{x}{{1!}} - \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^5}}}{{5!}} - ... \\\ \sin \left( {{x^2}} \right) = \dfrac{{{x^2}}}{{1!}} - \dfrac{{{{\left( {{x^2}} \right)}^3}}}{{3!}} + \dfrac{{{{\left( {{x^2}} \right)}^5}}}{{5!}} - ... \\\$
Now we will use the law of indices for brackets to simplify it,
$\sin \left( {{x^2}} \right) = \dfrac{{{x^2}}}{{1!}} - \dfrac{{{x^{2 \times 3}}}}{{3!}} + \dfrac{{{x^{2 \times 5}}}}{{5!}} - ... \\\ \sin \left( {{x^2}} \right) = \dfrac{{{x^2}}}{{1!}} - \dfrac{{{x^6}}}{{3!}} + \dfrac{{{x^{10}}}}{{5!}} - ... \\\$
So $\sin \left( {{x^2}} \right) = \dfrac{{{x^2}}}{{1!}} - \dfrac{{{x^6}}}{{3!}} + \dfrac{{{x^{10}}}}{{5!}} - ...$ is
the required Maclaurin Series for the given trigonometric function $\sin ({x^2})$
Now we should more simplify and generalize this expansion in order to understand the expansion more easily
It can be generalized with the use of sigma summation notation as follows:
$\sin \left( {{x^2}} \right) = \dfrac{{{x^2}}}{{1!}} - \dfrac{{{x^6}}}{{3!}} + \dfrac{{{x^{10}}}}{{5!}} - ... \\\ \sin \left( {{x^2}} \right) = \sum\limits_{n = 0}^\infty {\dfrac{{{x^{4n + 2}}}}{{(2n + 1)!}} \times {{( - 1)}^n}} \\\$

Note: Maclaurin Series is a special case of Taylor Series, we obtain it by substituting ${x_0} = 0$ in the Taylor series. Maclaurin Series is very useful in finding the approximate values for any argument of a given function, this is to be noted that it gives approximate values instead of the absolute value, if you want absolute value then go for regular methods.