Question

Find the expression that represent ${\left( {1 + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^4}}}{{4!}} + \dfrac{{{x^6}}}{{6!}} + ....} \right)^2}$ in the ascending power of $x$.
A. $1 + \dfrac{{{2^2}{x^2}}}{{2!}} + \dfrac{{{2^4}{x^4}}}{{4!}} + ...$
B. $1 + \dfrac{{{{\left( {2x} \right)}^2}}}{{2!}} + \dfrac{{{2^2}{x^4}}}{{4!}} + ...$
C. $1 + \dfrac{{{{\left( {2x} \right)}^2}}}{{2 \cdot 2!}} + \dfrac{{2{x^4}}}{{4!}} + ...$
D. $1 + \dfrac{{{{\left( {2x} \right)}^2}}}{{2 \cdot 2!}} + \dfrac{{{{\left( {2x} \right)}^4}}}{{2 \cdot 4!}} + ...$

Answer 1

Hint: Use the Taylor expansion for ${e^x}$, and ${e^{ - x}}$. To get the terms of the given expression, add the expansions of the two terms. After that, simplify the expression to arrive at the desired result.

Formula Used:
${e^x} = 1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^4}}}{{4!}} + \dfrac{{{x^5}}}{{5!}} + \dfrac{{{x^6}}}{{6!}} + ....$
${e^{ - x}} = 1 - \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} - \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^4}}}{{4!}} - \dfrac{{{x^5}}}{{5!}} + \dfrac{{{x^6}}}{{6!}} - ....$

Complete step by step solution:
The given expression is ${\left( {1 + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^4}}}{{4!}} + \dfrac{{{x^6}}}{{6!}} + ....} \right)^2}$.
Let’s consider the Taylor expansions of ${e^x}$, and ${e^{ - x}}$.
${e^x} = 1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^4}}}{{4!}} + \dfrac{{{x^5}}}{{5!}} + \dfrac{{{x^6}}}{{6!}} + ....$
${e^{ - x}} = 1 - \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} - \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^4}}}{{4!}} - \dfrac{{{x^5}}}{{5!}} + \dfrac{{{x^6}}}{{6!}} - ....$

Now add both equations.
${e^x} + {e^{ - x}} = \left( {1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^4}}}{{4!}} + \dfrac{{{x^5}}}{{5!}} + \dfrac{{{x^6}}}{{6!}} + ....} \right) + \left( {1 - \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} - \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^4}}}{{4!}} - \dfrac{{{x^5}}}{{5!}} + \dfrac{{{x^6}}}{{6!}} - ....} \right)$
Simplify the above equation.
${e^x} + {e^{ - x}} = \left( {1 + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^4}}}{{4!}} + \dfrac{{{x^6}}}{{6!}} + ....} \right) + \left( {1 + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^4}}}{{4!}} + \dfrac{{{x^6}}}{{6!}} + ....} \right)$
$ \Rightarrow {e^x} + {e^{ - x}} = 2\left( {1 + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^4}}}{{4!}} + \dfrac{{{x^6}}}{{6!}} + ....} \right)$ $.....\left( 1 \right)$
Divide both sides by 2.
$1 + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^4}}}{{4!}} + \dfrac{{{x^6}}}{{6!}} + .... = \dfrac{{{e^x} + {e^{ - x}}}}{2}$
Now take square on both sides.
${\left( {1 + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^4}}}{{4!}} + \dfrac{{{x^6}}}{{6!}} + ....} \right)^2} = {\left( {\dfrac{{{e^x} + {e^{ - x}}}}{2}} \right)^2}$
$ \Rightarrow {\left( {1 + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^4}}}{{4!}} + \dfrac{{{x^6}}}{{6!}} + ....} \right)^2} = \dfrac{{{e^{2x}} + {e^{ - 2x}} + 2{e^x}{e^{ - x}}}}{4}$ [Since ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$]
Simplify the above equation.
${\left( {1 + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^4}}}{{4!}} + \dfrac{{{x^6}}}{{6!}} + ....} \right)^2} = \dfrac{1}{4}\left( {{e^{2x}} + {e^{ - 2x}} + 2} \right)$
Solve the term ${e^{2x}} + {e^{ - 2x}}$ using equation $\left( 1 \right)$.
${\left( {1 + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^4}}}{{4!}} + \dfrac{{{x^6}}}{{6!}} + ....} \right)^2} = \dfrac{1}{4}\left( {2\left( {1 + \dfrac{{{{\left( {2x} \right)}^2}}}{{2!}} + \dfrac{{{{\left( {2x} \right)}^4}}}{{4!}} + \dfrac{{{{\left( {2x} \right)}^6}}}{{6!}} + ....} \right) + 2} \right)$
$ \Rightarrow {\left( {1 + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^4}}}{{4!}} + \dfrac{{{x^6}}}{{6!}} + ....} \right)^2} = \dfrac{1}{4}\left( {2 + 2\left( {\dfrac{{{{\left( {2x} \right)}^2}}}{{2!}} + \dfrac{{{{\left( {2x} \right)}^4}}}{{4!}} + \dfrac{{{{\left( {2x} \right)}^6}}}{{6!}} + ....} \right) + 2} \right)$
$ \Rightarrow {\left( {1 + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^4}}}{{4!}} + \dfrac{{{x^6}}}{{6!}} + ....} \right)^2} = \dfrac{1}{4}\left( {4 + 2\left( {\dfrac{{{{\left( {2x} \right)}^2}}}{{2!}} + \dfrac{{{{\left( {2x} \right)}^4}}}{{4!}} + \dfrac{{{{\left( {2x} \right)}^6}}}{{6!}} + ....} \right)} \right)$
$ \Rightarrow {\left( {1 + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^4}}}{{4!}} + \dfrac{{{x^6}}}{{6!}} + ....} \right)^2} = 1 + \dfrac{{{{\left( {2x} \right)}^2}}}{{2 \cdot 2!}} + \dfrac{{{{\left( {2x} \right)}^4}}}{{2 \cdot 4!}} + \dfrac{{{{\left( {2x} \right)}^6}}}{{2 \cdot 6!}} + ....$

Option ‘D’ is correct

Note: The Taylor expansion is the standard technique used to obtain a linear or a quadratic approximation of a function of one variable. It is used to calculate the value of a whole function at each point if the functional values and derivatives are identified at a single point.
The Taylor expansion of the term ${e^x}$ is: $\sum\limits_{i = 0}^n {\left( {\dfrac{{{x^i}}}{{i!}}} \right)} = 1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^4}}}{{4!}} + \dfrac{{{x^5}}}{{5!}} + \dfrac{{{x^6}}}{{6!}} + ....$