Exponential and Logarithmic Series
2.0 Deduction of other important series from the exponential $(e^x)$ series
2.0 Deduction of other important series from the exponential $(e^x)$ series
- ${e^{ax}} = 1 + \frac{{ax}}{{1!}} + \frac{{{a^2}{x^2}}}{{2!}} + \frac{{{a^3}{x^3}}}{{3!}} + ... + \frac{{{a^r}{x^r}}}{{r!}} + ...\infty $
Proof: We know, $${e^x} = 1 + \frac{x}{{1!}} + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + ... + \frac{{{x^r}}}{{r!}} + ...\infty $$ Put $x=ax$ in the above series.
Therefore, we get, $${e^{ax}} = 1 + \frac{{ax}}{{1!}} + \frac{{{a^2}{x^2}}}{{2!}} + \frac{{{a^3}{x^3}}}{{3!}} + ... + \frac{{{a^r}{x^r}}}{{r!}} + ...\infty $$
- ${e^{ - x}} = 1 - \frac{x}{{1!}} + \frac{{{x^2}}}{{2!}} - \frac{{{x^3}}}{{3!}} + ... + {( - 1)^r}\frac{{{x^r}}}{{r!}} + ...\infty $
Proof: We know, $${e^{ax}} = 1 + \frac{{ax}}{{1!}} + \frac{{{a^2}{x^2}}}{{2!}} + \frac{{{a^3}{x^3}}}{{3!}} + ... + \frac{{{a^r}{x^r}}}{{r!}} + ...\infty $$ Put $a=-1$,
So, the above equation becomes, $${e^{ - x}} = 1 - \frac{x}{{1!}} + \frac{{{x^2}}}{{2!}} - \frac{{{x^3}}}{{3!}} + ... + {( - 1)^r}\frac{{{x^r}}}{{r!}} + ...\infty $$
- ${a^x} = 1 + \frac{{\left( {{{\log }_e}a} \right)}}{{1!}}x + \frac{{{{\left( {{{\log }_e}a} \right)}^2}}}{{2!}}{x^2} + \frac{{{{\left( {{{\log }_e}a} \right)}^3}}}{{3!}}{x^3} + ... + \frac{{{{\left( {{{\log }_e}a} \right)}^r}}}{{r!}}{x^r} + ...\infty $
Proof: We know, $${e^x} = 1 + \frac{x}{{1!}} + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + ... + \frac{{{x^r}}}{{r!}} + ...\infty $$ We can write, $$\begin{equation} \begin{aligned} {a^x} = {e^{{{\log }_e}{a^x}}}\quad \left( {as,\;{e^{{{\log }_e}a}} = a} \right) \\ or \\ {a^x} = {e^{x{{\log }_e}a}} \\ {a^x} = {e^{\alpha x}}\quad \left( {where\;\alpha = {{\log }_e}a} \right) \\\end{aligned} \end{equation} $$ ${where\;\alpha = {{\log }_e}a}$
So, expanison of ${e^{\alpha x}}$ is, $${e^{\alpha x}} = 1 + \frac{{\alpha x}}{{1!}} + \frac{{{\alpha ^2}{x^2}}}{{2!}} + \frac{{{\alpha ^3}{x^3}}}{{3!}} + ... + \frac{{{\alpha ^r}{x^r}}}{{r!}} + ...\infty $$ Therefore, $${a^x} = 1 + \frac{{\left( {{{\log }_e}a} \right)}}{{1!}}x + \frac{{{{\left( {{{\log }_e}a} \right)}^2}}}{{2!}}{x^2} + \frac{{{{\left( {{{\log }_e}a} \right)}^3}}}{{3!}}{x^3} + ... + \frac{{{{\left( {{{\log }_e}a} \right)}^r}}}{{r!}}{x^r} + ...\infty $$
Question 1. If $x<1$, find the coefficient of $x^n$ in the expansion of $\left( {\frac{{{e^x}}}{{1 - x}}} \right)$.
Solution: We have, $$\frac{{{e^x}}}{{1 - x}} = {e^x}{(1 - x)^{ - 1}}$$ We know the expansion of $(1+x)^n$ as, $${(1 + x)^n} = 1 + \frac{{nx}}{{1!}} + \frac{{n(n - 1){x^2}}}{{2!}} + \frac{{n(n - 1)(n - 2){x^3}}}{{3!}} + ...\infty $$ So, $$\begin{equation} \begin{aligned} {(1 - x)^{ - 1}} = 1 + \frac{{( - 1)( - x)}}{{1!}} + \frac{{( - 1)( - 2){{( - x)}^2}}}{{2!}} + \frac{{( - 1)( - 2)( - 3){{( - x)}^3}}}{{3!}} + ...\infty \\ {(1 - x)^{ - 1}} = 1 + x + {x^2} + {x^3} + ...\infty \\\end{aligned} \end{equation} $$ Also, $${e^x} = 1 + \frac{x}{{1!}} + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + ... + \frac{{{x^r}}}{{r!}} + ...\infty $$ So, $${e^x}{(1 - x)^{ - 1}} = \left( {1 + \frac{x}{{1!}} + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + ... + \frac{{{x^r}}}{{r!}} + ...\infty } \right)\left( {1 + x + {x^2} + {x^3} + ... + {x^n} + ...\infty } \right)$$ For coefficient of $x^n$, $$\begin{equation} \begin{aligned} 1 \times {x^n} = {x^n} \\ \left( {\frac{x}{{1!}}} \right) \times {x^{n - 1}} = \left( {\frac{{{x^n}}}{{1!}}} \right) \\ \left( {\frac{{{x^2}}}{{2!}}} \right) \times {x^{n - 2}} = \left( {\frac{{{x^n}}}{{2!}}} \right) \\ \left( {\frac{{{x^3}}}{{3!}}} \right) \times {x^{n - 3}} = \left( {\frac{{{x^n}}}{{3!}}} \right) \\ ..... \\ ..... \\ ..... \\ \left( {\frac{{{x^n}}}{{n!}}} \right) \times 1 = \left( {\frac{{{x^n}}}{{n!}}} \right) \\\end{aligned} \end{equation} $$ So, adding the above terms we get, $${x^n}\left[ {1 + \frac{1}{{1!}} + \frac{1}{{2!}} + \frac{1}{{3!}} + ... + \frac{1}{{n!}}} \right]$$ Therefore the coffiecient of $x^n$ is, $$\left[ {1 + \frac{1}{{1!}} + \frac{1}{{2!}} + \frac{1}{{3!}} + ... + \frac{1}{{n!}}} \right]$$
Question 2. If $x<1$, find the coefficient of $x^n$ in the expansion of $\frac{{{e^x}}}{{{{\left( {1 - x} \right)}^2}}}$
Solution: We have, $$\frac{{{e^x}}}{{{{\left( {1 - x} \right)}^2}}} = {e^x}{(1 - x)^{ - 2}}$$ We know the expansion of $(1+x)^n$ as, $${(1 + x)^n} = 1 + \frac{{nx}}{{1!}} + \frac{{n(n - 1){x^2}}}{{2!}} + \frac{{n(n - 1)(n - 2){x^3}}}{{3!}} + ...\infty $$ So, $$\begin{equation} \begin{aligned} {(1 - x)^{ - 2}} = 1 + \frac{{( - 2)( - x)}}{{1!}} + \frac{{( - 2)( - 3){{( - x)}^2}}}{{2!}} + \frac{{( - 2)( - 3)( - 4){{( - x)}^3}}}{{3!}} + ...\infty \\ {(1 - x)^{ - 2}} = 1 + 2x + 3{x^2} + 4{x^3} + ...+(n+1){x^n}+...\infty \\\end{aligned} \end{equation} $$ Also, $${e^x} = 1 + \frac{x}{{1!}} + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + ... + \frac{{{x^r}}}{{r!}} + ...\infty $$ So, $${e^x}{(1 - x)^{ - 2}} = \left( {1 + \frac{x}{{1!}} + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + ... + \frac{{{x^r}}}{{r!}} + ...\infty } \right)\left( {1 + 2x + 3{x^2} + 4{x^3} + ...+(n+1){x^n}+...\infty } \right)$$ For coefficient of $x^n$, $$\begin{equation} \begin{aligned} 1 \times (n+1){x^n} = (n+1){x^n} \\ \left( {\frac{x}{{1!}}} \right) \times n{x^{n - 1}} = \left( {\frac{{n{x^n}}}{{1!}}} \right) \\ \left( {\frac{{{x^2}}}{{2!}}} \right) \times (n-1){x^{n - 2}} = \left( {\frac{{(n-1){x^n}}}{{2!}}} \right) \\ \left( {\frac{{{x^3}}}{{3!}}} \right) \times (n-2){x^{n - 3}} = \left( {\frac{{(n-2){x^n}}}{{3!}}} \right) \\ ..... \\ ..... \\ ..... \\ \left( {\frac{{{x^n}}}{{n!}}} \right) \times 1 = \left( {\frac{{{x^n}}}{{n!}}} \right) \\\end{aligned} \end{equation} $$ So, adding the above terms we get, $${x^n}\left[ {(n+1) + \frac{n}{{1!}} + \frac{(n-1)}{{2!}} + \frac{(n-2)}{{3!}} + ... + \frac{1}{{n!}}} \right]$$ Therefore the coffiecient of $x^n$ is, $$\left[ {(n+1) + \frac{n}{{1!}} + \frac{(n-1)}{{2!}} + \frac{(n-2)}{{3!}} + ... + \frac{1}{{n!}}} \right]$$
Question 3. Prove that the coefficient of $x^n$ in the expansion, $$=\left[ {1 + \frac{{\left( {p + qx} \right)}}{{1!}} + \frac{{{{\left( {p + qx} \right)}^2}}}{{2!}} + \frac{{{{\left( {p + qx} \right)}^3}}}{{3!}} + ...\infty } \right] = \frac{{{e^a}{b^n}}}{{n!}}$$
Solution: We know, $$={e^y} = 1 + \frac{y}{{1!}} + \frac{{{y^2}}}{{2!}} + \frac{{{y^3}}}{{3!}} + ... + \frac{{{y^r}}}{{r!}} + ...\infty $$ On comparing both the series we get, $$=y=p+qx$$ So, $$=\left[ {1 + \frac{{\left( {p + qx} \right)}}{{1!}} + \frac{{{{\left( {p + qx} \right)}^2}}}{{2!}} + \frac{{{{\left( {p + qx} \right)}^3}}}{{3!}} + ...\infty } \right] = {e^{(p + qx)}}$$ Therefore, $$={e^{(p + qx)}} = {e^p}.\;{e^{qx}}$$ On expanding $e^{qx}$ we get, $$={e^{(p + qx)}} = {e^p}\left( {1 + \frac{{qx}}{{1!}} + \frac{{{q^2}{x^2}}}{{2!}} + \frac{{{q^3}{x^3}}}{{3!}} + ... + \frac{{{q^n}{x^n}}}{{n!}} + ...\infty } \right)$$ So, the coefficient of $x^n$ is, $$=\frac{{{e^p}{q^n}}}{{n!}}$$
Question 4. Find the coefficient of $y^n$ in the expansion of ${e^{{e^y}}}$
Solution: Let, ${e^y} = x$
So, ${e^{{e^y}}}=e^x$
We know the expansion of $e^x$, $$={e^x} = 1 + \frac{x}{{1!}} + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + ... + \frac{{{x^r}}}{{r!}} + ...\infty $$ or $$ = {e^{{e^y}}} = 1 + \frac{{{e^y}}}{{1!}} + \frac{{{e^{2y}}}}{{2!}} + \frac{{{e^{3y}}}}{{3!}} + ... + \frac{{{e^{ry}}}}{{r!}} + ...\infty $$ After further expansion, we get, $$ = {e^{{e^y}}} = 1 + \frac{{\left( {1 + \frac{y}{{1!}} + \frac{{{y^2}}}{{2!}} + ... + \frac{{{y^n}}}{{n!}} + ...\infty } \right)}}{{1!}} + \frac{{\left( {1 + \frac{{2y}}{{1!}} + \frac{{{{\left( {2y} \right)}^2}}}{{2!}} + ... + \frac{{{{\left( {2y} \right)}^n}}}{{n!}} + ...\infty } \right)}}{{2!}} + \frac{{\left( {1 + \frac{{3y}}{{1!}} + \frac{{{{\left( {3y} \right)}^2}}}{{2!}} + ... + \frac{{{{\left( {3y} \right)}^n}}}{{n!}} + ...\infty } \right)}}{{3!}} + ...\infty $$ So, the coefficient of $y^n$ is, $$\begin{equation} \begin{aligned} = \frac{1}{{1!n!}} + \frac{{{2^n}}}{{2!n!}} + \frac{{{3^n}}}{{3!n!}} + ...\infty \\ = \frac{1}{{n!}}\left( {1 + \frac{{{2^n}}}{{2!}} + \frac{{{3^n}}}{{3!}} + ...\infty } \right) \\\end{aligned} \end{equation} $$
Question 5. Prove that, $${e^2} - 1 = \frac{{1 + \frac{{{2^2}}}{{2!}} + \frac{{{2^4}}}{{3!}} + \frac{{{2^6}}}{{4!}} + ...}}{{1 + \frac{1}{{2!}} + \frac{2}{{3!}} + \frac{{{2^2}}}{{4!}} + ...}}$$
Solution: We will solve numerator and denominator separately.
For Numerator, $$ = 1 + \frac{{{2^2}}}{{2!}} + \frac{{{2^4}}}{{3!}} + \frac{{{2^6}}}{{4!}} + ...$$ The above equation can be written as, $$\begin{equation} \begin{aligned} = \frac{1}{{{2^2}}}\left\{ {{2^2} + \frac{{{2^4}}}{{2!}} + \frac{{{2^6}}}{{3!}} + \frac{{{2^8}}}{{4!}} + ...} \right\} \\ = \frac{1}{{{2^2}}}\left\{ {\frac{{\left( {{2^2}} \right)}}{1} + \frac{{{{\left( {{2^2}} \right)}^2}}}{{2!}} + \frac{{{{\left( {{2^2}} \right)}^3}}}{{3!}} + \frac{{{{\left( {{2^2}} \right)}^4}}}{{4!}} + ...} \right\} \\ = \frac{1}{{{2^2}}}\left\{ { - 1 + \left[ {1 + \frac{{\left( {{2^2}} \right)}}{1} + \frac{{{{\left( {{2^2}} \right)}^2}}}{{2!}} + \frac{{{{\left( {{2^2}} \right)}^3}}}{{3!}} + \frac{{{{\left( {{2^2}} \right)}^4}}}{{4!}} + ...} \right]} \right\} \\ = \frac{{{e^{{2^2}}} - 1}}{{{2^2}}} \\ = \frac{{{e^4} - 1}}{4} \\\end{aligned} \end{equation} $$ For Denominator, $$ = 1 + \frac{1}{{2!}} + \frac{2}{{3!}} + \frac{{{2^2}}}{{4!}} + ...\infty $$ The above equation can be written as, $$\begin{equation} \begin{aligned} = \frac{1}{{{2^2}}}\left( {{2^2} + \frac{{{2^2}}}{{2!}} + \frac{{{2^3}}}{{3!}} + \frac{{{2^4}}}{{4!}} + ...\infty } \right) \\ = \frac{1}{{{2^2}}}\left( {1 + 1 + \frac{2}{{1!}} + \frac{{{2^3}}}{{3!}} + \frac{{{2^4}}}{{4!}} + ...\infty } \right)\quad \quad \left( {{2^2} = 1 + 1 + \frac{2}{{1!}}} \right) \\ = \frac{{{e^2} + 1}}{{{2^2}}} \\ = \frac{{{e^2} + 1}}{4} \\\end{aligned} \end{equation} $$ On solving numerator & denominator, $$\begin{equation} \begin{aligned} = \frac{{\frac{{{e^4} - 1}}{4}}}{{\frac{{{e^2} + 1}}{4}}} \\ = {e^2} - 1 \\\end{aligned} \end{equation} $$
