Exponential and Logarithmic Series

Have you ever noticed the use of the word exponential in our daily life?

It is used to express something which is rising or increasing at a steady and a usual rapid rate or something which increases quickly by large amounts.

Examples:

(i) Prices have increased exponentially i.e. at an exponential rate.

(ii) The budget is increasing at an exponential rate.

(iii) We all want to progress exponentially and achieve success in our lives.

Something is said to increase or decrease exponentially if its rate of change must be expressed using exponents.
A graph of such a rate would appear not as a straight line, but as a curve that continually becomes steeper or shallower.
In Math, it refers to a Mathematical (function, curve, series, or equation) containing, or involving one or more numbers or quantities raised to an exponent.

So, here we try to identify what does Exponential series mean in mathematics.

Sequence

A sequence is a list or arrangement of numbers in a definite order written according to some rule or pattern.

Series

If the n terms of the sequence are a₁, a₂, a₃, a₄, ……., a_n then the expression a₁ + a₂ + a₃ + a₄ + ……. + a_n is called the series associated with the given sequence.

Exponential Series

1. A series of the form

is called the Exponential series, x ∈ R. Here e^x is called the exponential function.

Replace x by –x in the above series given by (1), we obtain

$e^{-x}$ = 1 + $\frac{(-x)}{1!}+\frac{(-x)^2}{2!}+\frac{(-x)^3}{3!}+\frac{(-x)^4}{4!}+..............+\frac{(-x)^r}{r!}+......$ ∞ =

2. $e^{-x}$ = 1 – $\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+..............+\frac{(-x)^r}{r!}+......$ ∞ =

On adding (1) and (2) we have

$e^x$ + $e^{-x}$ = 2

3. $\frac{e^x +e^{-x}}{2}$ = 1 + $\frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!}+.........$ ∞

Subtracting (2) from (1),

$e^x - e^{-x}$ = 2

4. $\frac{e^x - e^{-x}}{2}$ = x + $\frac{x^3}{3!}+\frac{x^5}{5!}+.......$ ∞ =

Particular cases:

Put x = 1 in (1) and (2), we have

5. $e^1 = 1 + \frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+........$ ∞ =

Here e is an irrational number such that 2 < e < 3.

The value of e = 2.7182818284

6. $e^{-1}=1+\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+.......$ ∞ =

Put x = 1 in (3) and (4) we have

7. $\frac{e+e^{-1}}{2}=1+\frac{1}{2!}+\frac{1}{4!}+........$ ∞ =

8. $\frac{e-e^{-1}}{2}=1+\frac{1}{3!}+\frac{1}{5!}+........$ ∞ =

Logarithmic Series

The logarithmic series is defined for |x| < 1 and is given by

$log_{e}$ (1+x) = x – $\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+..........$ ∞

Replace x by –x in the above series, we obtain

$log_{e}$ (1 – x) = – x – $\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}-..........$ ∞ or

– $log_{e}$ (1-x) = x + $\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+..........$ ∞

On adding (1) and (2) we have

$log_{e}$ (1+x) + $log_{e}$ (1-x) = $log_{e}$ (1- $x^2$ ) = -2 , |x|< 1

Subtracting (2) from (1), we have

$log_{e}$ (1+x)- $log_{e}$ (1-x) = $log_{e}$ $\left ( \frac{1+x}{1-x} \right )$ = 2

Put x = 1 in (1), we have

$log_{e}$ 2=1 – $\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+..........$ ∞ = $\frac{1}{1.2}+\frac{1}{3.4}-\frac{1}{5.6}+..........$ ∞

Note: log_e(1 + x) is defined for x = 1.

Examples

Now let’s consider some examples on exponential and logarithmic series.

Example 1: Find the value of 1+ $\frac{5}{1!}+\frac{5^2}{2!}+\frac{5^3}{3!}+\frac{5^4}{4!}+................$ ∞ .

Since, $e^x$ = 1 + $\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+................+\frac{x^r}{r!}+.......$ ∞.

Hence, put x = 5 in above expression, we have

$e^5$ = 1 + $\frac{5}{1!}+\frac{5^2}{2!}+\frac{5^3}{3!}+\frac{5^4}{4!}+................$ ∞

Example 2: Find the value of 1 – $\frac{(2.5)}{1!}+\frac{(2.5)^{2}}{2!}-\frac{(2.5)^{3}}{3!}+\frac{(2.5)^{4}}{4!}-................$ ∞

Since, $e^{-x}$ = 1 – $\frac{x}{1!}+\frac{x^2}{2!}-\frac{x^3}{3!}+\frac{x^4}{4!}+.....................$ ∞

Hence, put x = 2.5 in above expression, we have

$e^{(-2.5)}$ = $e^{(-\frac{5}{2})}$ = 1 – $\frac{(2.5)}{1!}+\frac{(2.5)^{2}}{2!}-\frac{(2.5)^{3}}{3!}+\frac{(2.5)^{4}}{4!}-................$ ∞

Example 3: Find the value of $\frac{2}{1!}+\frac{4}{3!}+\frac{6}{5!}+\frac{8}{7!}+............$ ∞ .

$\frac{2}{1!}+\frac{4}{3!}+\frac{6}{5!}+\frac{8}{7!}+............$ ∞ = $\frac{(1+1)}{1!}+\frac{(1+3)}{3!}+\frac{(1+5)}{5!}+\frac{(1+7)}{7!}+............$ ∞

= $\frac{e-e^{-1}}{2}$ + $\frac{e+e^{-1}}{2}$ =e

Example 4: Find the value of (0.5) – $\frac{(0.5)^2}{2}$ + $\frac{(0.5)^3}{2}$ – $\frac{(0.5)^4}{4}$ +……….∞

Since, $log_{e}$ (1+x) = x – $\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+..........$ ∞

Hence, put x = 0.5 we have

$log_{e}$ (1+0.5) = (0.5) – $\frac{(0.5)^2}{2}+\frac{(0.5)^3}{3}-\frac{(0.5)^4}{4}+..........$ ∞

log $\left ( 1+\frac{1}{2} \right )$ = log $\left (\frac{3}{2} \right )$ = (0.5) – $\frac{(0.5)^2}{2}+\frac{(0.5)^3}{3}-\frac{(0.5)^4}{4}+..........$ ∞

Example 5: Find the value of $\frac{1}{3}+\frac{1}{2.3^2}+\frac{1}{3.3^3}+\frac{1}{4.3^4}+......$ .∞.

Since, $log_{e}$ (1-x) = -x – $\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}-.........$ ∞

– $log_{e}$ (1-x) = x + $\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+.........$ ∞

Hence, put x= $\frac{1}{3}$ we have

Check Point

Find the value of $1+ \frac{3}{1!}++ \frac{3^2}{2!}+ \frac{3^3}{3!}+ \frac{3^4}{4!}+...............$ ∞
Find the value of $1+ \frac{(0.5)}{1!}+ \frac{(0.5)^2}{2!}+ \frac{(0.5)^3}{3!}+ \frac{(0.5)^4}{4!}+...............$ ∞
Find the value of $\frac{4}{1!}+\frac{4^3}{3!}+\frac{4^5}{5!}+.............$ ∞
Find the value of $-(0.5)-\frac{(0.5)^2}{2}-\frac{(0.5)^3}{3}-\frac{(0.5)^4}{4}-............$ ∞
Find the value of $\frac{1}{n}-\frac{1}{2n^2}+\frac{1}{3n^3}-\frac{1}{4n^4}+............$ ∞