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,
$\begin{aligned} & {{e}^{2x+3}}={{e}^{3}}\left( {{e}^{2x}} \right) \\\ & \Rightarrow {{e}^{2x+3}}={{e}^{3}}\times {{e}^{2x}}.........\left( 1 \right) \\\ \end{aligned}$
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,

& {{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{aligned}$$ Now, we will put a = 0 in above equation. Thus, we will get, $$\begin{aligned} & \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{aligned}$$ Thus, we have got the expansion of $ {{e}^{t}} $ . Now, we will find the expansion of $ {{e}^{2x+3}} $ . From (1) we know that $ {{e}^{2x+3}}={{e}^{2x}}\times {{e}^{3}} $ . So, we will put $ t=2x $ in (2). Thus, we will get, $$\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{aligned} & 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{aligned} $ **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.