Question

How do you write the Taylor series for $f\left( x \right) = \cosh x$?

Answer 1

This problem deals with expansion of Taylor series of a given function. In mathematics, the Taylor series of a function is an infinite sum of terms that are expressed in terms of the function’s derivatives at a single point. For most common functions, the function and the sum of its Taylor series are equal near this point.
The Taylor series of a function is defined as:
$\Rightarrow \sum\limits_{n = 0}^\infty {\dfrac{{{f^n}\left( {{x_0}} \right)}}{{n!}}} {\left( {x - {x_0}} \right)^n}$

Complete step-by-step answer:
we know that the Taylor series of a function is defined, as given below:
$\Rightarrow \sum\limits_{n = 0}^\infty {\dfrac{{{f^n}\left( {{x_0}} \right)}}{{n!}}} {\left( {x - {x_0}} \right)^n}$
Here the $n$ in ${f^n}\left( {{x_0}} \right)$ denotes the nth derivative of $f(x)$, and the $n$ in ${\left( {x - {x_0}} \right)^n}$ is the exponent or the power.
Here $f(x) = \cosh x$.
We are going to find the Taylor series of $\cosh x$, but around $x = 0$, then we set ${x_0} = 0$, and use the above definition.
Now finding the derivatives as shown below:
$\Rightarrow f\left( x \right) = \cosh x$, here when $x = 0$, $f(0) = \cosh (0) = 1$
$\therefore f(0) = 1$
Now the first derivative is:
$\Rightarrow f'\left( x \right) = \sinh x$, here when $x = 0$, $f'(0) = \sinh (0) = 0$
$\therefore f'(0) = 0$
Now the second derivative is :
$\Rightarrow f''\left( x \right) = \cosh x$, here when $x = 0$, $f''(0) = \cosh (0) = 1$
$\therefore f''(0) = 1$
Now the third derivative is:
$\Rightarrow f'''\left( x \right)\sinh x$, here when $x = 0$, $f'''(0) = \sinh (0) = 0$
$\therefore f'''(0) = 0$
Now the fourth derivative is :
$\Rightarrow {f^{IV}}\left( x \right) = \cosh x$, here when $x = 0$, ${f^{IV}}(0) = \cosh (0) = 1$
$\therefore {f^{IV}}(0) = 1$
Now the fifth derivative is:
$\Rightarrow {f^V}\left( x \right)\sinh x$, here when $x = 0$, ${f^V}(0) = \sinh (0) = 0$
$\therefore {f^V}(0) = 0$
Now the sixth derivative is :
$\Rightarrow {f^{VI}}\left( x \right) = \cosh x$, here when $x = 0$, ${f^{VI}}(0) = \cosh (0) = 1$
$\therefore {f^{VI}}(0) = 1$
And so on.
Now substituting these in the Taylor expansion formula, as shown:
$\Rightarrow \sum\limits_{n = 0}^\infty {\dfrac{{{f^n}\left( 0 \right)}}{{n!}}} {\left( {x - 0} \right)^n}$
$\Rightarrow \dfrac{{f\left( 0 \right)}}{{0!}}{\left( {x - 0} \right)^1} + \dfrac{{f'\left( 0 \right)}}{{1!}}\left( {x - 0} \right) + \dfrac{{f''\left( 0 \right)}}{{2!}}{\left( {x - 0} \right)^2} + \dfrac{{f'''\left( 0 \right)}}{{3!}}{\left( {x - 0} \right)^3} + \dfrac{{{f^{IV}}\left( 0 \right)}}{{4!}}{\left( {x - 0} \right)^4} + \dfrac{{{f^V}\left( 0 \right)}}{{5!}}{\left( {x - 0} \right)^5} + \dfrac{{{f^{VI}}\left( 0 \right)}}{{6!}}{\left( {x - 0} \right)^6} + .....$
Substituting the values of the derivatives, as shown:
$\Rightarrow \dfrac{1}{1}{\left( {x - 0} \right)^0} + \dfrac{0}{{1!}}{\left( {x - 0} \right)^1} + \dfrac{1}{{2!}}{\left( {x - 0} \right)^2} + \dfrac{0}{{3!}}{\left( {x - 0} \right)^3} + \dfrac{1}{{4!}}{\left( {x - 0} \right)^4} + \dfrac{0}{{5!}}{\left( {x - 0} \right)^5} + \dfrac{1}{{6!}}{\left( {x - 0} \right)^6} + ......$
$\Rightarrow 1 + 0 + \dfrac{{{x^2}}}{2} + 0 + \dfrac{{{x^4}}}{{24}} + 0 + \dfrac{{{x^6}}}{{720}} + ........$
The terms multiplied with zero, will be zero.
$\Rightarrow 1 + \dfrac{{{x^2}}}{2} + \dfrac{{{x^4}}}{{24}} + \dfrac{{{x^6}}}{{720}} + ........$
$\therefore \cosh x = 1 + \dfrac{{{x^2}}}{2} + \dfrac{{{x^4}}}{{24}} + \dfrac{{{x^6}}}{{720}} + ........$
For more accuracy we can find more derivatives and continue building up the series.

Final Answer: The Taylor series expansion of $\cosh x = 1 + \dfrac{{{x^2}}}{2} + \dfrac{{{x^4}}}{{24}} + \dfrac{{{x^6}}}{{720}} + ........$

Note:
Please note that a Taylor series is a series expansion of a function about a point. A one-dimensional Taylor series is an expansion of a real function about a point. This is also known as a Maclaurin series.