# HOW TO FIND EXPANSION OF EXPONENTIAL FUNCTION

The series Σ(xn/n!), where n ∈ [1, +∞) is called an exponential series. It can be proved that this series converges for all values of x.

For any real number x,

Σ(xn/n!) where n ∈ [1, +∞)  =  ex, where

Some more results :

## Solved Questions

Write the first 6 terms of each exponential series :

Question 1 :

e5x

Solution :

ex  =  1 + (x/1!) + (x2/2!) + (x3/3!) + ..................

In the formula above, substitute x = 5x.

e5x :

=  1 + (5x/1!) + ((5x)2/2!) + ((5x)3/3!) + ((5x)4/4!) +  ((5x)5/5!) + ..................

=  1 + (5x/1!) + (25x2/2) + (125 x3/6) + (625 x4/24) +  (625 x5/24) + ..................

Question 2 :

e-2x

Solution :

e-x  =  1 - (x/1!) + (x2/2!) - (x3/3!) + ..................

In the formula above, substitute x = -2x.

e-2x :

= 1 - (2x/1!) + ((2x)2/2!) - ((2x)3/3!) + ((2x)4/4!) - (2x)5/5!) + ..................

=  1 - (2x/1!) + (4x2/2) + (8x3/6) + (16x4/24) +(32x5/120) + ..................

=  1 - 2x + 2x2 + (4x3/3) + (2x4/3) + (4x5/15) + .........

Question 3 :

e(1/2)x

Solution :

ex  =  1 + (x/1!) + (x2/2!) + (x3/3!) + ..................

In the formula above, substitute x = x/2.

e(1/2)x :

= 1 - (x/2/1!) + ((x/2)2/2!) - ((x/2)3/3!) + ((x/2)4/4!) -  ((x/2)5/5!) + ..................

=  1 - (x/2) + (x2/8) + (x3/48) + (x4/384) + (x5/3840) + .....

## Logarithmic Expansion

The series Σ(−1)n+1xn /n, where ∈ [1, +∞) is called a logarithmic series. This series converges for all values of x satisfying |x| < 1. This series converges when x = 1 also. For all values of x satisfying |x| < 1, the sum of the series is log(1 + x). Thus

log(1 + x)  =  x - (x2/2) + (x3/3) - (x4/4) + .......

for all values of x satisfying |x| < 1.

By taking -x in place of x, we get

log(1 - x)  =  -x - x2/2 - x3/3 - x4/4 .......

for all values of x satisfying |x| < 1.

Now,

log [(1+x) / (1-x)]  =  log(1 + x) - log(1 - x)

Using this we get

log [ (1 + x)/(1 - x)]  = 2 [x + x3/3  + x5/5 +............]

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