Series expansion of exponential and logarithmic functions

Series expansion of exponential and logarithmic functions for entrance exams.

1. Series Expansion of Exponential and Logarithmic
Functions


Leonhard Euler, the great Swiss mathematician int roduced and named the
number e in his calculus text in 1748.
e
x
= 1+ ++.............
Putting x=1 in there, we get:
e
1
=1+1/1!+1/2!+1/3!+....
which gives
e=1+1+.5+.17+...
so e is approximately 2.67or 2< e< 3 and the value of e
2
= 7.3 by using exp. Series.
e
3x
= 1+ ++.............
which simplifies to:
e
3x
= 1+3x+9x
2
/2!+27x
3
/3!+81x
4
/4!+.....
To write e
x+2
as a series, it is best to rewrite e
x+2
as e
x
. e
2
so:
e
x+2
= e
2
(1+ ++.............)
This is true for any constant c in e
x+c
, try writing down the first four terms of
the series for e
x+4
,
The series for e
x+4

Here is start of the series for e
x
again:
e
x
= 1+ ++.............
Now recall that e
x+4
is the same as e
x
.e
4
, so we just need to multiply the series
for e
x
by the number e
4
. That gives us:

2. e
x+4
= e
4
(1+ ++.............)
e
x
= 1 + ++.............
e
-x
= 1 - - +............. [coefft. Of x
n
ix (-1)
n
/n!]
by adding & subtracting above two series , we get
= 1+ x
2
/2! +x
4
/4!+........... and =x+ x
3
/3! +x
5
/5!+...........
= = and = =
Some important points should be noticed
=e =
(ii) = e – 1 , = e – 2
(iii) = e – 1 , = e – 2 =
Example 1. Find the sum of the following to infinity
(i) 1/1!+2/2!+3/3!+..... (ii) x/1!+2x
2
/2!+3x
3
/3!+.....
Solution: (i) 1/1!+2/2!+3/3!+..... = = = e
(ii) = x ( ) = xe
x
.
Example2. Find the sum of the following to infinity:
(i) (i) (ii)
(ii)
(iii) Solution: (i) = = = = +
=e+e=2e.
(iv) (ii) = =x( ) + x
2
( ) = xe
x
+ x
2
e
x
.
(v)

3. (vi) Example 3. Find the sum of the following to infinity:
(vii) (i) (ii)
(viii)
(ix) Solution: (i) = = = = +
(x) = + + = +
+
(xi) = e + 3e + e = 5e.
(xii) (ii) = = + +x =
(x
3
+3x
2
+x)e
x

Some questions with hints :
Q.1 Find the sum of following
(i) x + + +........ (ii) 1 + + +........
Solution: (i) Tn = = = ½() ,
Sn = ½( (2xe
x
+ x
2
e
x
)
[ by example 1 & 2 we can find (ii) answer is ½(2e+e)= 3e/2]
Q.2 find the sum of the following series:
(i) (ii) (iii) + ++........
(iv) 12/2! + 28/3! + 50/4!+78/5!+.......... (v) 2/1! + 6/2! + 12/3!+20/4!+.........
(vi) 1+ 3/2! + 6/3! +10/4! +.......
Solution: (i) = = = - 2 +
[by above examples ] 2e-1-2e+2+e-2 = e-1.
= + = 5e +2e = 7e.
= = + 5 +4

4. = 2e5e +4e = 11e.
[ assume Tn = an
2
+bn+c , put n=2,3,4 and by solving eqns. We
will get a= 3, b=1 & c=-2] by above part (i) we get, 3(2e-1)+(e-1) -2(e-2) = 5e.
= [assume Tn = an
2
+bn+c , put n=1,2,3 and by solving eqns. We will get
a= 1, b=1 & c=0] by above part (ii) we get, 2e+e = 3e.
[assume Tn = an
2
+bn+c , put n=1,2,3 and by solving eqns. We will
get a= 1/2, b=1/2 & c=0] by above part (ii) we get, ½ (2e+e) = 3e/2.
3 find the sum of (i) (ii)
if S = , find 2S.
: (i) [ by using C(n,2) = n(n-1)/2! And by above results ]
−2)
= e
2
– 1. (iii) S = =1/2 = ½ e
3
.
4. (i) If = B0+B1x+B2x
2
+.....+Bnx
n
, then find the value of (Bn – Bn-1).
find the sum of 2/3!+ 4/5! +6/7!+..... (iii) 2/1! +4/3!+6/5!+......
find the sum of (x+y)(x-y) +1/2!(x+y)(x-y)(x
2
+y
2
)+1/3!(x+y)(x-y)(x
4
+y
4
+x
2
y
2
)+......
: (i) e
x
(1-x)
-1
= B0+B1x+B2x
2
+.....+Bnx
n

1 + x/1! + x
2
/2! + x
3
/3!+......+x
n-1
/(n-1)!+x
n
/n!+....)(1+x+x2+x3+....+x
n-
1
+x
n
+...)
B0+B1x+B2x
2
+.....+Bnx
n

Coeffts. Of x
n
and x
n-1
on the both sides, we get 1/n!+1/(n-1)!+....+1/2!+1/1!=1=Bn
1(n-1)!+1/(n-2)!+....+1/2!+1/1!+1=Bn-1. Therefore Bn – Bn-1 = 1/n!.
For (ii) & (iii) ,we will take help of above results and , then add &
subtract 1 in Nr. Accordingly .[answer of (ii) e
-1
(iii) e]

5. For (iv) we get (x
2
-y
2
)+1/2!(x
4
-y
4
)+1/3!(x
6
-y
6
)+..... =( - 1)- ( – 1) = .
e = [by using binomial
theorem = 1+n. + +.....]
= 1+1+1/2!+1/3!+..... as 1/n, 2/n, 3/n ...=0 as n .

Let a >0, then for all real values of x,
a
x
= 1+ x(logea) + x
2
/2! (logea)
2
+..........
how we will you get this series, put x=cx in exp. Series , then
e
cx
= 1+cx +(cx)
2
/2! + .......... = 1+ x(logea) + x
2
/2! (logea)
2
+..........
[let e
c
=a it implies c= logea & e
cx
= (e
c
)
x
= a
x
.]

6. The graph of y = e
x
is upward-sloping, and increases faster as x increases. The graph always
lies above the x-axis but can get arbitrarily close to it for negative x; thus, the x-axis is a
horizontal asymptote. The slope of the tangent to the graph at each point is equal to
its y coordinate at that point. The inverse function is the natural logarithm ln(x); because of
this, refer to the exponential function as the antilogarithm.

The natural logarithm function, if considered as a real-valued function of a real
variable, is the inverse function of the exponential function, leading to the identities:
,
,put x=1 in loge(1+x) then loge2= 1-1/2+1/3-1/4+..... is valid.
Natural Log Series

7. Loge3 = 2[1/2+(1/2)
3
/3+(1/2)
5
/5+........] , by putting (1+x)/(1-x)=3 it gives
x=1/2, similarly we can do for log4,log5 and so on......
We know that = 2(x + .............)=
1+0.083+0.012+...=1.098
0.61< loge2 < o.76 [as we can write loge2= (1-1/2)+(1/3-1/4)+..... >0.61 and
loge2= 1-(1/2-1/3)-(1/4-1/5)-.....<0.76 we can write log2 < 1 < log3
Some examples:
1. If y = x- x
2
/2+x
3
/3-x
4
/4+.....and if |x|<1, prove that x=y+y
2
/2!+y
3
/3!+....
2. Prove that ...........has the same sum as the series
...........
prove that 2 logx – log(x+1) – log(x-1) = ...........
. If y > 0 , then prove that logy = 2 { +.........}
And hence find the value of log2 to three places of decimals.
5. prove that for |x| <1,
= x - ..........
6. if are the roots of x
2
– px +q=0, prove that
log(1+px+qx
2
) = ()x-(
Solution: 1. y=log(1+x) it implies e
y
= 1+x , write exponential series.
2.Put x = 1/(n+1) it gives x++..... = - (- x- +.....)= - log(1-x)
Put x=1/(1+n), we get - = = ...........
3. L.H.S. = = - = ...........[by using laws of
logarithmics]
4. Put x = , then R.H.S. = 2(x + .............)=
= logy, then put y=2 in question it gives 0.693= log2.

8. 5. log(1+x).(1+x)
-1
=( x - +.......)(1-x+x
2
-x
3
+x
4
-x
5
+.....)
6. R.H.S. = log(1+ = log(1+( ) = log(1+px+qx
2
)
Some important questions with hints:
Q.1 Prove that = 2( ++....)
Q.2 prove that for |x| < ½ , log(1+3x+2x
2
) = 3x - .....
Q.3 prove that log(x+a) – log(x-a) = 2( +.......)
Q.4 prove that = logba.
Q.5 prove that = [1 - ...........]
Q.6 if x
2
y = (2x-y) and |x| <1, prove that
(y+ y
3
/3+y
5
/5+.....) = 2( x+x
3
/3+x
5
/5+.....)
Q.7 prove that (i) +........= ½ log2.
(ii) +.......= log2.
(iii) +........= 2-2log2.
Q.8 Find the sum of (i) +...........
(ii) +.........
(iii) +.............
(iv) - .........
Answers with hints
1. = (1+x)log(1+x)+(1-x)log(1-x)
= [log(1+x)+log(1-x)]+ x[log(1+x)+log(1-x)]
= -2(x
2
/2+x
4
/4+x
6
/6+....) + 2x (x+x
3
/3+x
5
/5+....) =2( ++....)

9. 2. L.H.S. = log[(1+x)(1+2x)] = log(1+x)+log(1+2x) = R.H.S.
3. L.H.S. = log{x(1+a/x)}-log{x(1-x/a)} = log(1+x/a) – log(1-x/a) [Use use series of
log(1+x) & log(1-x)]
4. Let x=a-1 & y=b-1, then L.H.S. becomes loge(1+x)/loge(1+y)=logea/logeb= logba
5. Put x = 1/(n+1), we get ( .......) =[1-(1-1/2)x – (1/2-1/3)x
2
-(1/3-
1/4)x
3
-.....] =[( 1 + .......) - ( x + +.......)]
[1/x( x + .......) - ( x + ......)]=-1/xlog(1-x)+log(1-x), put
the value of x ,it becomes nlog(1+1/n)=R.H.S.
6. L.H.S. =y(x
2
+1) =2x it implies y= 2x/(x
2
+1) , we get = by taking log on
the both sides & use log (1+x)/(1-x).
7. (i) put x=1/3 ,L.H.S. & use this result 2(x + .............)=
(ii)nth term = , n=1,2,3.... ( by partial fraction which is used for
integration) it can be written as 1= A(2n) + B(2n-1), find A & B by putting n=o &
n=1/2,so we get A=1,B =-1, then sum to n terms becomes =1-1/2+1/3-
1/4+.......=log2.
(iii) nth term = , n=1,2,3.... ( by partial fraction which is used for
integration) it can be written as 1= A(2n+1) + B(n), find A & B by putting
n=o & n=-1/2,so we get A=1,B =-2, then sum to n terms becomes
=2[1/2-1/3+1/4-1/5+.......]=-2 [-1/2+1/3-1/4+1/5+.......] =-2
[log2-1]= R.H.S.
Q.8 (i) nth term = , n=1,2,3....same as above
method we get A= 2, B=-3 &C=1 sum = =
= 2(1-1/2+1/3-1/4+....)+(-1/2+1/3-1/4+.......)=3log2-1. (ii) nth term
= , n=1,2,3....same as above method we get A= 1/2,
B=-1 &C=1/2 sum = =
= ½ {(1-1/2+1/3-1/4+....)+(-1/2+1/3-1/4+.......)}=log2-1/2.(iii)put x= 1/5 , sum =
= = - (1+x) – = – ) –
1= (1-x)
-1
+ log(1-x) – 1= by putting x=1/5.
(iv)L.H.S. = }+ }=1/2{log(1+1/2)+log(1+1/3)}
=1/2log(3/2X4/3)=1/2log2.