Question

How to do Taylor series expansion of ${e^{\dfrac{{^{ - {x^2}}}}{2}}}$ ?

Answer 1

In the above given question, we are given an exponential function as ${e^{\dfrac{{^{ - {x^2}}}}{2}}}$ . We have to expand the given exponential function into its Taylor series expansion. In order approach the Taylor series expansion of the given exponential function, first we have to find some of the initial derivatives of ${e^{\dfrac{{^{ - {x^2}}}}{2}}}$ . After that, we can substitute them in the Taylor series which is given by the formula $\sum {\dfrac{{{f^{\left( n \right)}}\left( a \right)}}{{n!}}} {\left( {x - a} \right)^n}$ .

Complete answer:
Given expression is ${e^{\dfrac{{^{ - {x^2}}}}{2}}}$ .
We have to find the Taylor series expansion of the given expression ${e^{\dfrac{{^{ - {x^2}}}}{2}}}$ .
Now, let us find some initial derivatives of ${e^{\dfrac{{^{ - {x^2}}}}{2}}}$ .
Here we have,
$\Rightarrow {f^{\left( 0 \right)}}\left( x \right) = f\left( x \right) = {e^{ - {{\dfrac{x}{2}}^2}}}$
And,
$\Rightarrow f'\left( x \right) = {e^{ - {{\dfrac{x}{2}}^2}}} \cdot x = - x{e^{ - {{\dfrac{x}{2}}^2}}}$
Also,

i.e.

Similarly,
$\Rightarrow f'''\left( x \right) = \left( {{x^2} - 1} \right)\left( { - x{e^{ - {{\dfrac{x}{2}}^2}}}} \right) + {e^{ - {{\dfrac{x}{2}}^2}}}\left( {2x} \right)$
That gives us,
$\Rightarrow f'''\left( x \right) = \left( { - {x^3} + x} \right){e^{ - {{\dfrac{x}{2}}^2}}} + 2x{e^{ - {{\dfrac{x}{2}}^2}}}$
That is,
$\Rightarrow f'''\left( x \right) = \left( { - {x^3} + 3x} \right){e^{ - {{\dfrac{x}{2}}^2}}}$
Also,
$\Rightarrow f''''\left( x \right) = \left( { - {x^3} + 3x} \right)\left( { - x{e^{ - {{\dfrac{x}{2}}^2}}}} \right) + 2x{e^{ - {{\dfrac{x}{2}}^2}}}$
That gives us,
$\Rightarrow f''''\left( x \right) = \left( {{x^4} + 3{x^2}} \right){e^{ - {{\dfrac{x}{2}}^2}}} - 3{x^2}{e^{ - {{\dfrac{x}{2}}^2}}} + 3{e^{ - {{\dfrac{x}{2}}^2}}}$
That is,
$\Rightarrow f''''\left( x \right) = \left( {{x^4} - 6{x^2} + 3} \right){e^{ - {{\dfrac{x}{2}}^2}}}$
Now, we have the values up to the fourth derivative of ${e^{\dfrac{{^{ - {x^2}}}}{2}}}$ .
Now putting these values in the Taylor series expansion, we can write the expansion of ${e^{\dfrac{{^{ - {x^2}}}}{2}}}$ as,

$$\Rightarrow \sum {\dfrac{{{f^{\left( n \right)}}\left( a \right)}}{{n!}}} {\left( {x - a} \right)^n} = \dfrac{{{f^{\left( 0 \right)}}\left( a \right)}}{{0!}}{\left( {x - a} \right)^0} + \dfrac{{{f^{\left( 1 \right)}}\left( a \right)}}{{1!}}{\left( {x - a} \right)^1} + \\\ \dfrac{{{f^{\left( 2 \right)}}\left( a \right)}}{{2!}}{\left( {x - a} \right)^2} + \dfrac{{{f^{\left( 3 \right)}}\left( a \right)}}{{3!}}{\left( {x - a} \right)^3} + \dfrac{{{f^{\left( 4 \right)}}\left( a \right)}}{{4!}}{\left( {x - a} \right)^4} + ... \\\$$

Substituting the above values of derivatives, we get

$$\Rightarrow \sum {\dfrac{{{f^{\left( n \right)}}\left( a \right)}}{{n!}}} {\left( {x - a} \right)^n} = {e^{ - {{\dfrac{a}{2}}^2}}} + - a{e^{ - {{\dfrac{a}{2}}^2}}}\left( {x - a} \right) + \\\ \dfrac{{\left( {{a^2} - 1} \right){e^{ - {{\dfrac{a}{2}}^2}}}}}{{2!}}{\left( {x - a} \right)^2} + \dfrac{{\left( { - {a^3} + 3a} \right){e^{ - {{\dfrac{a}{2}}^2}}}}}{{3!}}{\left( {x - a} \right)^3} + \dfrac{{\left( {{a^4} - 6{a^2} + 3} \right){e^{ - {{\dfrac{a}{2}}^2}}}}}{{4!}}{\left( {x - a} \right)^4} + ... \\\$$

Now, since the value of $a$ is not given, so if $a = 0$ then we have the above equation as,
$\Rightarrow \sum {\dfrac{{{f^{\left( n \right)}}\left( a \right)}}{{n!}}} {\left( {x - a} \right)^n} = 1 - \dfrac{{{x^2}}}{2} + \dfrac{{{x^4}}}{8} - \dfrac{{{x^6}}}{{48}} + ...$
Watching the pattern of the above series, we can write the next steps of series in the following manner as,
$\Rightarrow \sum {\dfrac{{{f^{\left( n \right)}}\left( a \right)}}{{n!}}} {\left( {x - a} \right)^n} = 1 - \dfrac{{{x^2}}}{2} + \dfrac{{{x^4}}}{8} - \dfrac{{{x^6}}}{{48}} + \dfrac{{{x^8}}}{{384}} - \dfrac{{{x^{10}}}}{{3840}} + ...$
Here, the nth term of the series is given by,
$\Rightarrow {{\rm T}_n} = \dfrac{{{f^{\left( n \right)}}\left( a \right)}}{{n!}}{\left( {x - a} \right)^n} = {\left( { - 1} \right)^n}\dfrac{{{x^{2n}}}}{{{2^n}n!}}$
Therefore, the Taylor series expansion can be written as,
$\Rightarrow \sum {\dfrac{{{f^{\left( n \right)}}\left( a \right)}}{{n!}}} {\left( {x - a} \right)^n} = \sum {{{\left( { - 1} \right)}^n}\dfrac{{{x^{2n}}}}{{{2^n}n!}}}$
Hence, the required Taylor series expansion of the expression ${e^{\dfrac{{^{ - {x^2}}}}{2}}}$ can be written as,
$\Rightarrow {e^{\dfrac{{^{ - {x^2}}}}{2}}} = \sum {\dfrac{{{f^{\left( n \right)}}\left( a \right)}}{{n!}}} {\left( {x - a} \right)^n} = \sum {{{\left( { - 1} \right)}^n}\dfrac{{{x^{2n}}}}{{{2^n}n!}}}$
Therefore, the Taylor series expansion of the expression ${e^{\dfrac{{^{ - {x^2}}}}{2}}}$ is $\sum {{{\left( { - 1} \right)}^n}\dfrac{{{x^{2n}}}}{{{2^n}n!}}}$ .

Note: The Taylor Series is also known as the Taylor Polynomial. It is the representation of a function as an infinite sum of terms calculated from the values of its derivatives at a single point $a$ . A Maclaurin series or Maclaurin Polynomial is a special case of the Taylor Series. It has one difference and that is, it uses zero as the single point i.e. $a = 0$ .