In the previous section,
we saw infinite geometric series. In this section,
we will see exponential series.
• Consider the series:
$1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} +~.~.~$
• The sum of this series is denoted by the letter e.
• e is very important for calculus and other areas of mathematics, economics, financial studies, science and engineering.
• Note that all terms in the series, are +ve. So e will be +ve.
Let us find the value of e. It can be done in 8 steps:
1. Two series are written below. They are written in such a way that, the terms in the second series come directly below the corresponding terms in the first series.
\begin{array}{ll}{} &{\frac{1}{3!}} & {~+~} &{\frac{1}{4!}} & {~+~} &{\frac{1}{5!}} & {~+~.~.~.~+~} &{\frac{1}{n!}} & {~+~.~.~.~\color{green}{\text{- - - (I)}}} &{} \\
{} &{\frac{1}{2^2}} & {~+~} &{\frac{1}{2^3}} & {~+~} &{\frac{1}{2^4}} & {~+~.~.~.~+~} &{\frac{1}{2^{n-1}}} & {~+~.~.~.~\color{green}{\text{- - - (II)}}} &{} \\
\end{array}
2. We see an interesting point. It can be written in 3 steps:
(i) Take any term in (II). Note down the power of 2 in that term.
(ii) Take the corresponding term in (I). Note down the denominator of that term.
(iii) The power will be one less than the denominator.
3. Another interesting point can also be observed. It can be written in 4 steps:
(i) Take the first terms: $\frac{1}{3!}~\text{and}~\frac{1}{2^2}$
• We have:
$\frac{1}{3!} = \frac{1}{6}~\text{and}~\frac{1}{2^2} = \frac{1}{4}$
• So the first term in (I) is less than the corresponding term in (II)
(ii) Take the second terms: $\frac{1}{4!}~\text{and}~\frac{1}{2^3}$
• We have:
$\frac{1}{4!} = \frac{1}{24}~\text{and}~\frac{1}{2^3} = \frac{1}{8}$
• So the second term in (I) is less than the corresponding term in (II)
(iii) Take the third terms: $\frac{1}{5!}~\text{and}~\frac{1}{2^4}$
• We have:
$\frac{1}{5!} = \frac{1}{102}~\text{and}~\frac{1}{2^4} = \frac{1}{16}$
• So the third term in (I) is less than the corresponding term in (II)
(iv) Based on the above 3 steps, we can write:
Whenever n is greater than 2,
$\frac{1}{n!}~<~\frac{1}{2^{n-1}}$
4. We saw that, each term in (I) is less than the corresponding term in (II). So we can write:
$\begin{array}{ll}{} &{\frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!}~ +~.~.~.~+~\frac{1}{n!}~+~.~.~.} & {~<~} &{\frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4}~ +~.~.~.~+~\frac{1}{2^{n-1}}~+~.~.~.} &{} \\
{\Rightarrow} &{\left(1 + \frac{1}{1!} + \frac{1}{2!} \right)~+~\left(\frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!}~ +~.~.~.~+~\frac{1}{n!}~+~.~.~.\right)} & {~<~} &{\left(1 + \frac{1}{1!} + \frac{1}{2!} \right)~+~\left(\frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4}~ +~.~.~.~+~\frac{1}{2^{n-1}}~+~.~.~.\right)~\color{green}{\text{- - - (A)}}} &{} \\
{\Rightarrow} &{\left(1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!}~ +~.~.~.~+~\frac{1}{n!}~+~.~.~.\right)} & {~<~} &{\left(1 + 1 + \frac{1}{2^1} \right)~+~\left(\frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4}~ +~.~.~.~+~\frac{1}{2^{n-1}}~+~.~.~.\right)} &{} \\
{\Rightarrow} &{1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!}~ +~.~.~.~+~\frac{1}{n!}~+~.~.~.} & {~<~} &{1 + \left(1 + \frac{1}{2^1} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4}~ +~.~.~.~+~\frac{1}{2^{n-1}}~+~.~.~.\right)~\color{green}{\text{- - - (B)}}} &{} \\
{\Rightarrow} &{1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!}~ +~.~.~.~+~\frac{1}{n!}~+~.~.~.} & {~<~} &{1 + \frac{1}{1 - \frac{1}{2}}} &{} \\
{\Rightarrow} &{1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!}~ +~.~.~.~+~\frac{1}{n!}~+~.~.~.} & {~<~} &{3} &{} \\
{\Rightarrow} &{e} & {~<~} &{3} &{} \\
\end{array}$
◼ Remarks:
• Line marked as A:
Here we add $\left(1 + \frac{1}{1!} + \frac{1}{2!} \right)$ on both sides.
• Line marked as B:
Consider the RHS of this line. We have a portion enclosed in brackets. This portion is a infinite geometric series. It’s sum can be easily calculated. It works out to 2.
5. Earlier we said that:
$1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} +~.~.~=~e$
• Consider the LHS. The first two terms are 1 and 1.
• So we can write:
e will be always greater than 2.
6. Let us compare the results:
♦ From (4) we have: e < 3
♦ From (5) we have: e > 2
• So we can write: 2 < e < 3
7. The approximate value of e is 2.71828. The accuracy will increase when we increase the number of terms in the series.
• e is an irrational number like π.
8. Consider the equation that we use to evaluate e:
$e~=~1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} +~.~.~.$
♦ In the LHS, the power of e is 1.
♦ In the RHS, all numerators are 1.
✰ These ones are actually, powers of 1.
• So we can write:
$e^1~=~1 + \frac{1^1}{1!} + \frac{1^2}{2!} + \frac{1^3}{3!} + \frac{1^4}{4!} +~.~.~.$
• If the power of e is 2, then we can write:
$e^2~=~1 + \frac{2^1}{1!} + \frac{2^2}{2!} + \frac{2^3}{3!} + \frac{2^4}{4!} +~.~.~.$
• If the power of e is 7, then we can write:
$e^7~=~1 + \frac{7^1}{1!} + \frac{7^2}{2!} + \frac{7^3}{3!} + \frac{7^4}{4!} +~.~.~.$
◼ We can write the general form:
• If the power of e is x, then:
$e^x~=~1 + \frac{x^1}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} +~.~.~.~+ \frac{x^n}{n!}~+~.~.~.$
Now we will see a solved example:
Solved example A.3
Find the coefficient of x2 in the expansion of e2x+3 as a series in the powers of x.
Solution:
1. We have the general form:
$e^x~=~1 + \frac{x^1}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} +~.~.~.~+ \frac{x^n}{n!}~+~.~.~.$
2. In our present case, we have (2x+3) in the place of x. So we can write:
$e^{2x+3}~=~1 + \frac{(2x+3)^1}{1!} + \frac{(2x+3)^2}{2!} + \frac{(2x+3)^3}{3!} + \frac{(2x+3)^4}{4!} +~.~.~.~+ \frac{(2x+3)^n}{n!}~+~.~.~.$
3. The general term is: $\frac{(2x+3)^n}{n!}$
The numerator of this general term can be expanded using binomial theorem. We get:
$\begin{array}{ll}{} &{\frac{(2x+3)^n}{n!}} & {~=~} &{\frac{(3+2x)^n}{n!}} &{} \\
{} &{} & {~=~} &{\frac{1}{n!}\left[3^n + \rm{{}^n C_1}3^{n-1}(2x) +\rm{{}^n C_2}3^{n-2}(2x)^2 + \rm{{}^n C_3}3^{n-3}(2x)^3~+~.~.~.~+~(2x)^n \right]} &{} \\
\end{array}$
4. Note the third term of the above expansion. It is the term with "x2".
• So we can write:
General form of the coefficient of x2 is $\frac{\rm{{}^n C_2}3^{n-2}(2x)^2}{n!}$
5. In (3), we considered the general term. For each term, there will be a term with x2. We want the sum of coefficients of all such terms. That sum will be:
$\sum_{n=2}^{n=\infty}{\frac{\rm{{}^n C_2}3^{n-2}2^2}{n!}}$
(We saw similar problems in section 9.5)
• This can be simplified as shown below:
$\begin{array}{ll}{} &{\sum_{n=2}^{n=\infty}{\frac{\rm{{}^n C_2}3^{n-2}2^2}{n!}}} & {~=~} &{\sum_{n=2}^{n=\infty}{\frac{\frac{n!}{2! \times (n-2)!}\times 3^{n-2} \times 4}{n!}}} &{} \\
{} &{} & {~=~} &{\sum_{n=2}^{n=\infty}{\frac{\frac{1}{1 \times (n-2)!}\times 3^{n-2} \times 2}{1}}} &{} \\
{} &{} & {~=~} &{2\sum_{n=2}^{n=\infty}{\frac{3^{n-2}}{(n-2)!}}} &{} \\
{} &{} & {~=~} &{2 \left[\frac{3^{2-2}}{(2-2)!} + \frac{3^{3-2}}{(3-2)!} + \frac{3^{4-2}}{(4-2)!} + \frac{3^{5-2}}{(5-2)!}~+~.~.~. \right]} &{} \\
{} &{} & {~=~} &{2 \left[\frac{3^{0}}{0!} + \frac{3^{1}}{1!} + \frac{3^{2}}{2!} + \frac{3^{3}}{3!}~+~.~.~. \right]} &{} \\
{} &{} & {~=~} &{2 \left[\frac{1}{1} + \frac{3^{1}}{1!} + \frac{3^{2}}{2!} + \frac{3^{3}}{3!}~+~.~.~. \right]} &{} \\
{} &{} & {~=~} &{2 \left[1 + \frac{3^{1}}{1!} + \frac{3^{2}}{2!} + \frac{3^{3}}{3!}~+~.~.~. \right]} &{} \\
{} &{} & {~=~} &{2e^3} &{} \\
\end{array}$
6. So the coefficient of x2 is 2e3.
Alternate method:
1. $\begin{array}{ll}{} &{e^{2x+3}} & {~=~} &{e^3 \times e^{2x}} &{} \\
{} &{} & {~=~} &{e^3 \left[1 + \frac{2x}{1!} + \frac{(2x)^{2}}{2!} + \frac{(2x)^{3}}{3!}~+~.~.~. \right]} &{} \\
\end{array}$
2. Consider the third term in the above result. It is the term with x2.
• This term can be written as: $\frac{4x^2 e^3}{2!}~=~2x^2 e^3$.
3. So we can write:
The coefficient of x2 is $2e^3$
Solved example A.4
Find the value of e2 rounded off to one decimal place.
Solution:
1. Let us expand e2:
$\begin{array}{ll}{} &{e^2} & {~=~} &{1 + \frac{2^1}{1!} + \frac{2^2}{2!} + \frac{2^3}{3!} + \frac{2^4}{4!} + \frac{2^5}{5!} + \frac{2^6}{6!} + \frac{2^7}{7!} + \frac{2^8}{8!}~+~.~.~.} &{} \\
{} &{} & {~=~} &{\left(1 + \frac{2^1}{1!} + \frac{2^2}{2!} + \frac{2^3}{3!} + \frac{2^4}{4!} \right) + \left(\frac{2^5}{5!} + \frac{2^6}{6!} + \frac{2^7}{7!} + \frac{2^8}{8!}~+~.~.~.\right)} &{} \\
{} &{} & {~=~} &{\left(1 + 2 + 2 + \frac{8}{6} + \frac{16}{24} \right) + \left(\frac{2^5}{5!} + \frac{2^6}{6!} + \frac{2^7}{7!} + \frac{2^8}{8!}~+~.~.~.\right)} &{} \\
{} &{} & {~=~} &{\left(5 + \frac{4}{3} + \frac{2}{3} \right) + \left(\frac{2^5}{5!} + \frac{2^6}{6!} + \frac{2^7}{7!} + \frac{2^8}{8!}~+~.~.~.\right)} &{} \\
{} &{} & {~=~} &{7 + \left(\frac{2^5}{5!} + \frac{2^6}{6!} + \frac{2^7}{7!} + \frac{2^8}{8!}~+~.~.~.\right)} &{} \\
\end{array}$
2. In the last line of the above result, consider the portion inside brackets. We will modify that portion as shown below:
$\begin{array}{ll}{} &{\frac{2^5}{5!} + \frac{2^6}{6!} + \frac{2^7}{7!} + \frac{2^8}{8!}~+~.~.~.} & {~=~} &{\frac{2^5}{5!} + \frac{2^6}{6 \times 5!} + \frac{2^7}{7 \times 6 \times 5!} + \frac{2^8}{8 \times 7 \times 6 \times 5!}~+~.~.~.} &{} \\
{\Rightarrow} &{\frac{2^5}{5!} + \frac{2^6}{6!} + \frac{2^7}{7!} + \frac{2^8}{8!}~+~.~.~.} & {~=~} &{\frac{1}{5!}\left(\frac{2^5}{1} + \frac{2^6}{6} + \frac{2^7}{7 \times 6} + \frac{2^8}{8 \times 7 \times 6 }~+~.~.~. \right)} &{} \\
{\Rightarrow} &{\frac{2^5}{5!} + \frac{2^6}{6!} + \frac{2^7}{7!} + \frac{2^8}{8!}~+~.~.~.} & {~<~} &{\frac{1}{5!}\left(\frac{2^5}{1} + \frac{2^6}{6} + \frac{2^7}{6 \times 6} + \frac{2^8}{6 \times 6 \times 6 }~+~.~.~. \right)~\color{green}{{\text{- - - (A)}}}} &{} \\
{\Rightarrow} &{\frac{2^5}{5!} + \frac{2^6}{6!} + \frac{2^7}{7!} + \frac{2^8}{8!}~+~.~.~.} & {~<~} &{\frac{1}{5!}\left(\frac{2^5}{1-\frac{2}{6}} \right)~\color{green}{\text{- - - (B)}}} &{} \\
{\Rightarrow} &{\frac{2^5}{5!} + \frac{2^6}{6!} + \frac{2^7}{7!} + \frac{2^8}{8!}~+~.~.~.} & {~<~} &{\frac{1}{5!}\left(\frac{2^5}{\frac{2}{3}} \right)} &{} \\
{\Rightarrow} &{\frac{2^5}{5!} + \frac{2^6}{6!} + \frac{2^7}{7!} + \frac{2^8}{8!}~+~.~.~.} & {~<~} &{\frac{1}{5 \times 4 \times 3 \times 2 \times 1}\left(\frac{2^5 \times 3}{2} \right)} &{} \\
{\Rightarrow} &{\frac{2^5}{5!} + \frac{2^6}{6!} + \frac{2^7}{7!} + \frac{2^8}{8!}~+~.~.~.} & {~<~} &{\frac{1}{5 \times 1 \times 1 \times 1 \times 1}\left(\frac{2 \times 1}{1} \right)} &{} \\
{\Rightarrow} &{\frac{2^5}{5!} + \frac{2^6}{6!} + \frac{2^7}{7!} + \frac{2^8}{8!}~+~.~.~.} & {~<~} &{\frac{2}{5}} &{} \\
{\Rightarrow} &{\frac{2^5}{5!} + \frac{2^6}{6!} + \frac{2^7}{7!} + \frac{2^8}{8!}~+~.~.~.} & {~<~} &{0.4} &{} \\
\end{array}$
◼ Remarks:
•
Line marked as (A):
In this line, consider the denominators of the RHS.
♦ We put 6 in the place of 7.
♦ We put 6 in the place of 8 also.
✰ So the denominators decrease. As a result, the fractions increase in values.
✰ Then the whole RHS becomes larger than the LHS. We change the "=" sign to "<" sign.
•
Line marked as (B):
In the line line marked as (A), in the RHS, the portion inside brackets is an infinite geometric series. It's sum can be easily calculated.
3. Now consider the last line of the result in (1).
Imagine that we replace $\frac{2^5}{5!} + \frac{2^6}{6!} + \frac{2^7}{7!} + \frac{2^8}{8!}~+~.~.~.$
by $\frac{1}{5!}\left(\frac{2^5}{1} + \frac{2^6}{6} + \frac{2^7}{6 \times 6} + \frac{2^8}{6 \times 6 \times 6 }~+~.~.~. \right)$
•
Then we can write: e2 < 7.4
4. Consider again the expansion of e2. We want the first seven terms:
$\begin{array}{ll}{} &{e^2} & {~=~} &{1 + \frac{2^1}{1!} + \frac{2^2}{2!} + \frac{2^3}{3!} + \frac{2^4}{4!} + \frac{2^5}{5!} + \frac{2^6}{6!} + \frac{2^7}{7!} + \frac{2^8}{8!}~+~.~.~.} &{} \\
{} &{} & {~=~} &{\left(1 + \frac{2^1}{1!} + \frac{2^2}{2!} + \frac{2^3}{3!} + \frac{2^4}{4!} + \frac{2^5}{5!} + \frac{2^6}{6!} \right) + \frac{2^7}{7!} + \frac{2^8}{8!} ~+~.~.~.} &{} \\
{} &{} & {~=~} &{\left(1 + 2 + 2 + \frac{2^3}{3 \times 2 \times 1} + \frac{2^4}{4 \times 3 \times 2 \times 1} + \frac{2^5}{5 \times 4 \times 3 \times 2 \times 1} + \frac{2^6}{6 \times 5 \times 4 \times 3 \times 2 \times 1} \right) + \frac{2^7}{7!} + \frac{2^8}{8!} ~+~.~.~.} &{} \\
{} &{} & {~=~} &{\left(1 + 2 + 2 + \frac{4}{3} + \frac{2}{3} + \frac{4}{15} + \frac{4}{45} \right) + \frac{2^7}{7!} + \frac{2^8}{8!} ~+~.~.~.} &{} \\
{} &{} & {~=~} &{\left(1 + 2 + 2 + \frac{60 + 30 + 12 + 4}{45} \right) + \frac{2^7}{7!} + \frac{2^8}{8!} ~+~.~.~.} &{} \\
{} &{} & {~=~} &{\left(1 + 2 + 2 + \frac{106}{45} \right) + \frac{2^7}{7!} + \frac{2^8}{8!} ~+~.~.~.} &{} \\
{} &{} & {~=~} &{\left(1 + 2 + 2 + 2.3555556 \right) + \frac{2^7}{7!} + \frac{2^8}{8!} ~+~.~.~.} &{} \\
{} &{} & {~=~} &{\left(7.355 \right) + \frac{2^7}{7!} + \frac{2^8}{8!} ~+~.~.~.} &{} \\
\end{array}$
•
It is clear that, e2 is greater than 7.355556.
•
This value when rounded off to one decimal place is: 7.4.
5. Let us compare the results:
♦ From (3), we have: e2 < 7.4
♦ From (4), we have: e2 > 7.4
6. So the value of e2 rounded off to one decimal place is 7.4.
In the next section, we will see mathematical modelling.
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