Question

How do you find the Taylor series for $f(x) = x\,\cos ({x^2})$ ?

Answer 1

Hint : Here in this question we have to find the Taylor’s series. By using the formula of Taylors series $f(x) = f(a) + f'(a)(x - a) + \dfrac{{f''(a)}}{{2!}}{(x - a)^2} + \dfrac{{f'''(a)}}{{3!}}{(x - a)^3} + ...$ Substituting the value of a to the formula hence we can determine the solution for the given question.

Complete step-by-step answer :
We have to determine the nth Taylor’s series by using the Taylor’s series expansion. Here the function is $f(x) = x\,\cos ({x^2})$ . The formula for the Taylor’s series expansion is given by $f(x) = f(a) + f'(a)(x - a) + \dfrac{{f''(a)}}{{2!}}{(x - a)^2} + \dfrac{{f'''(a)}}{{3!}}{(x - a)^3} + ...$ the value of a is 0. When a is 0, the Taylor’s series is given by
$f(x) = f(0) + f'(0)(x) + \dfrac{{f''(0)}}{{2!}}{(x)^2} + \dfrac{{f'''(0)}}{{3!}}{(x)^3} + ...$ The general form of Taylor’s series expansion is written as $f(x) = \sum\limits_{n = 0}^\infty {\dfrac{{{f^{(n)}}(a)}}{{n!}}} {(x - a)^n}$
Now consider the function $f(x) = x\,\cos ({x^2})$ , the value of f at x=0 is $f(0) = 0$
We know that the Taylor’s series expansion for cosine trigonometry ratio when x = 0 is given by
$\cos (x) = 1 - \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^4}}}{{4!}} \mp ...$
Therefore the Taylor’s series for the $\cos ({x^2})$ for x = 0 and it is given by
$\Rightarrow \cos ({x^2}) = 1 - \dfrac{{{{\left( {{x^2}} \right)}^2}}}{{2!}} + \dfrac{{{{\left( {{x^4}} \right)}^2}}}{{4!}} \mp ...$
On simplifying we get
$\Rightarrow \cos ({x^2}) = 1 - \dfrac{{{x^4}}}{{2!}} + \dfrac{{{x^8}}}{{4!}} \mp ...$
When the series is multiplied by x and Taylor’s series is given by
$\Rightarrow x\left( {\cos ({x^2})} \right) = x\left( {1 - \dfrac{{{x^4}}}{{2!}} + \dfrac{{{x^8}}}{{4!}} \mp ...} \right)$
On simplifying we get
$\Rightarrow x\cos ({x^2}) = x - \dfrac{{{x^5}}}{{2!}} + \dfrac{{{x^9}}}{{4!}} \mp ...$
This can be written in general form but here it is very difficult to write in the general form. Therefore we write as it is.
Hence we have determined the Taylor series for the given function.
So, the correct answer is “ $x - \dfrac{{{x^5}}}{{2!}} + \dfrac{{{x^9}}}{{4!}} \mp ...$ ”.

Note : In Taylor’s series expansion we have the derivatives, so we have to apply the differentiation to the function. If the function is a logarithmic function then we use standard differentiation formula and then we determine the solution. While solving the above function we use simple arithmetic operations.