Question

Find the Taylor series for $f\left( x \right) = \cos x$.

Answer 1

Taylor’s series is an infinite sum of terms that are expressed in terms of the derivatives of the function at a single point.

Complete step by step solution:
The Taylor’s series for a function $f\left( x \right)$ at $x = a$(where $a$ is any real or complex number) can be found by the formula : $\sum\limits_{n = 0}^\infty {\dfrac{{{f^{\left( n \right)}}\left( a \right)}}{{n!}}} {\left( {x - a} \right)^n}$ (where ${f^{(n)}}(x)$ is the nth derivative of $f(x)$ at $x = a$).
Let us find the Taylor series for $f(x) = \cos x$ at $x = 0$. First we need to find the derivatives of $\cos x$ and their values for $x = 0$.
$f(x) = \cos x \Rightarrow f(0) = \cos (0) = 1$
$f'(x) = - \sin x \Rightarrow f'(0) = - \sin (0) = 0$

$f'''(x) = \sin x \Rightarrow f'''(0) = \sin (0) = 0$

Observe that after $f'''(x)$, the value of is equal to $f(x)$ , then ${f^{(5)}}(x)$ is equal to $f'(x)$ and the cycle repeats itself after every four values i.e. $1,0, - 1,0$.
Thus, the sum of the series is as follows:

Substituting the values :
= \cos (0) + \dfrac{{{x^1}}}{{1!}}\left\\{ { - \sin (0)} \right\\} + \dfrac{{{x^2}}}{{2!}}\left\\{ { - \cos (0)} \right\\} + \dfrac{{{x^3}}}{{3!}}\left\\{ {\sin (0)} \right\\} + \dfrac{{{x^4}}}{{4!}}\left\\{ {\cos (0)} \right\\} + .....
$= 1 + \dfrac{{{x^1}}}{{1!}}(0) + \dfrac{{{x^2}}}{{2!}}( - 1) + \dfrac{{{x^3}}}{{3!}}(0) + \dfrac{{{x^4}}}{{4!}}(1) + .....$
$= 1 + \dfrac{{{x^2}}}{{2!}}( - 1) + \dfrac{{{x^4}}}{{4!}}(1) - .....$

It is seen that only the even power of $x$ exists in the final series and there are alternate positive and negative signs, therefore the summation of the series can be written as:
$\sum\limits_{n = 0}^\infty {\dfrac{{{{\left( { - 1} \right)}^n}{x^{2n}}}}{{\left( {2n} \right)!}}}$ [ here $n \in W$ and we take $2n$ because only the even powers of $x$ exist and ${( - 1)^n}$ because of the alternate positive and negative signs.]

Hence, the Taylor’s series for $f(x) = \cos x$ is:
$1 - \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^4}}}{{4!}} - \dfrac{{{x^6}}}{{6!}} + ....$

Note:
The factorial of a number denoted by $n!$ is equal to the product of all numbers from $1$ to that number. For example $4! = 4 \times 3 \times 2 \times 1 = 24$ and $3! = 3 \times 2 \times 1 = 6$.