Question

Find the coefficient of $ {{x}^{2}} $ in the expansion of $ {{e}^{2x+3}} $ as a series in powers of x.

Answer 1

Hint: To solve the question given above, we will first write $ {{e}^{2x+3}} $ in the form of $ {{e}^{t}} $ using the identity: $ {{e}^{a+b}}={{e}^{a}}.{{e}^{b}} $ . Then we will first find out what is Taylor’s expression for any $ f\left( x \right) $ . Then we will see the formula of expansion of $ f\left( x \right) $ . To find the expansion of $ {{e}^{x}} $ , we will take $ f\left( x \right) $ as $ {{e}^{x}} $ and then we will expand the function in the form of summation of powers of x. Then, we will find the expansion of $ {{e}^{2x+3}} $ by putting 2x in place of x and multiplying by $ {{e}^{2}} $ . Then, in the expansion obtained, we will find the coefficient of $ {{x}^{2}} $ . This will be the required answer.

Complete step-by-step answer:
To start with, we will first try to write $ {{e}^{2x+3}} $ in the form of $ {{e}^{t}} $ . We know that $ {{e}^{a+b}}={{e}^{a}}.{{e}^{b}} $ . Thus we will get,
 $ \begin{align}
  & {{e}^{2x+3}}={{e}^{3}}\left( {{e}^{2x}} \right) \\
 & \Rightarrow {{e}^{2x+3}}={{e}^{3}}\times {{e}^{2x}}.........\left( 1 \right) \\
\end{align} $
Now, we will try to expand $ {{e}^{t}} $ using Taylor’s expansions. Taylor expansion of a function $ f\left( t \right) $ is an infinite sum of terms that are expressed in terms of the function’s derivatives at a single point. The Taylor expansion of $ f\left( t \right) $ is given by:
 $ f\left( t \right)=f\left( a \right)+\dfrac{f'\left( a \right)}{1!}\left( t-a \right)+\dfrac{f''\left( a \right)}{2!}{{\left( t-a \right)}^{2}}+......+\dfrac{{{f}^{\left( n \right)}}\left( a \right)}{n!}{{\left( t-a \right)}^{n}} $
Now, let us assume $ f\left( t \right)={{e}^{t}} $
\[{{e}^{t}}={{e}^{a}}+\dfrac{{}_{t=a}\left| \dfrac{d}{dt}\left( {{e}^{t}} \right) \right.}{1!}\left( t-a \right)+\dfrac{{}_{t=a}\left| \dfrac{{{d}^{2}}}{d{{t}^{2}}}\left( {{e}^{t}} \right) \right.}{2!}{{\left( t-a \right)}^{2}}+......+\dfrac{{}_{t=a}\left| f\left( {{e}^{t}} \right) \right.}{n!}{{\left( t-a \right)}^{n}}\]
Now, we know that no matter how many times we differentiate $ {{e}^{t}} $ , it will still remain $ {{e}^{t}} $ . Thus, we will get,
\[\begin{align}
  & {{e}^{t}}={{e}^{a}}+\dfrac{{}_{t=a}\left| {{e}^{t}} \right.}{1!}\left( t-a \right)+\dfrac{{}_{t=a}\left| {{e}^{t}} \right.}{2!}{{\left( t-a \right)}^{2}}+......+\dfrac{{}_{t=a}\left| {{e}^{t}} \right.}{n!}{{\left( t-a \right)}^{n}} \\
 & \Rightarrow {{e}^{t}}={{e}^{a}}+\dfrac{{{e}^{a}}}{1!}\left( t-a \right)+\dfrac{{{e}^{a}}}{2!}{{\left( t-a \right)}^{2}}+......+\dfrac{{{e}^{a}}}{n!}{{\left( t-a \right)}^{n}} \\
\end{align}\]
Now, we will put a = 0 in above equation. Thus, we will get,
\[\begin{align}
  & \Rightarrow {{e}^{t}}={{e}^{0}}+\dfrac{{{e}^{0}}}{1!}\left( t-0 \right)+\dfrac{{{e}^{0}}}{2!}{{\left( t-0 \right)}^{2}}+......+\dfrac{{{e}^{0}}}{n!}{{\left( t-0 \right)}^{n}} \\
 & \Rightarrow {{e}^{t}}=1+\dfrac{t}{1!}+\dfrac{{{t}^{2}}}{2!}+......+\dfrac{{{t}^{n}}}{n!}.......\left( 2 \right) \\
\end{align}\]
Thus, we have got the expansion of $ {{e}^{t}} $ . Now, we will find the expansion of $ {{e}^{2x+3}} $ . From (1) we know that $ {{e}^{2x+3}}={{e}^{2x}}\times {{e}^{3}} $ . So, we will put $ t=2x $ in (2). Thus, we will get,
\[\Rightarrow {{e}^{t}}=1+\dfrac{2x}{1!}+\dfrac{{{\left( 2x \right)}^{2}}}{2!}+......+\dfrac{{{\left( 2x \right)}^{n}}}{n!}.......\left( 3 \right)\]
Now, we will substitute the value of $ {{e}^{2x}} $ from (3) to (1). Thus, we will get,
 $ {{e}^{2x+3}}={{e}^{3}}\times \left[ 1+\dfrac{2x}{1!}+\dfrac{{{\left( 2x \right)}^{2}}}{2!}+......+\dfrac{{{\left( 2x \right)}^{n}}}{n!} \right] $
Now, we have to find the coefficient of $ {{x}^{2}} $ in the above series. The coefficient of $ {{x}^{2}} $ in above series is given by,
 $ \begin{align}
  & coefficient\ of\ {{x}^{2}}={{e}^{3}}\times \dfrac{{{\left( 2 \right)}^{2}}}{2!} \\
 & \Rightarrow coefficient\ of\ {{x}^{2}}={{e}^{3}}\times \dfrac{4}{2} \\
 & coefficient\ of\ {{x}^{2}}=2{{e}^{3}} \\
\end{align} $

Note: The Taylor’s expansion which we have applied to a function $ f\left( t \right) $ has certain limitations. To apply this Taylor expansion to any function $ f\left( t \right) $ , $ f\left( t \right) $ should be continuous and it should be infinitely differentiable at “a”. The function $ f\left( t \right)={{e}^{t}} $ is continuous and infinitely differentiable at all points so we have been able to find its expansion.