Series expansion of exponential and logarithmic functions for entrance exams.

1. Series Expansion of Exponential and Logarithmic
Functions
Leonhard Euler, the great Swiss mathematician introduced and named the
number e in his calculus text in 1748.
ex = 1+ + +.............
Putting x=1 in there, we get:
e1=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 e2 = 7.3 by using exp. Series.
e3x = 1+ + +.............
which simplifies to:
e3x = 1+3x+9x2/2!+27x3/3!+81x4/4!+.....
To write ex+2 as a series, it is best to rewrite ex+2 as ex. e2 so:
ex+2 = e2(1+ + +.............)
This is true for any constant c in ex+c, try writing down the first four terms of
the series for ex+4,
The series for ex+4
Here is start of the series for ex again:
ex = 1+ + +.............
Now recall that ex+4 is the same as ex.e4, so we just need to multiply the series
for ex by the number e4. That gives us:

2. ex+4 = e4(1+ + +.............)
ex = 1 + + +.............
e-x = 1 - - +............. [coefft. Of xn ix (-1)n/n!]
by adding & subtracting above two series , we get
= 1+ x2/2! +x4/4!+........... and =x+ x3/3! +x5/5!+...........
= = 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!+2x2/2!+3x3/3!+.....
Solution: (i) 1/1!+2/2!+3/3!+..... = = =e
(ii) = x( ) = xex .
Example2. Find the sum of the following to infinity:
(i) (i) (ii)
(ii)
(iii) Solution: (i) = = = = +
=e+e=2e.
(iv) (ii) = =x( ) + x2 ( ) = xex + x2 ex .
(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 =
(x3+3x2+x)ex
Some questions with hints :
Q.1 Find the sum of following
(i) x + + +........ (ii) 1 + + +........
Solution: (i) Tn = = = ½( ) ,
Sn = ½( (2xex + x2ex)
[ 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 = an2+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 = an2+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 = an2+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)
= e2 – 1. (iii) S = =1/2 = ½ e3 .
4. (i) If = B0+B1x+B2x2+.....+Bnxn , 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)(x2+y2)+1/3!(x+y)(x-y)(x4+y4+x2y2)+......
: (i) ex(1-x)-1 = B0+B1x+B2x2+.....+Bnxn
1 + x/1! + x2/2! + x3/3!+......+xn-1/(n-1)!+xn/n!+....)(1+x+x2+x3+....+xn-
1 n
+x +...)
B0+B1x+B2x2+.....+Bnxn
Coeffts. Of xn and xn-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 &
-1
subtract 1 in Nr. Accordingly .[answer of (ii) e (iii) e]

5. For (iv) we get (x2-y2)+1/2!(x4-y4)+1/3!(x6-y6)+..... =( - 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,
ax = 1+ x(logea) + x2/2! (logea)2 +..........
how we will you get this series, put x=cx in exp. Series , then
ecx = 1+cx +(cx)2/2! + .......... = 1+ x(logea) + x2/2! (logea)2 +..........
[let ec =a it implies c= logea & ecx = (ec)x = ax.]

6. x
The graph of y = e 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- x2/2+x3/3-x4/4+.....and if |x|<1, prove that x=y+y2/2!+y3/3!+....
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 x2 – px +q=0, prove that
log(1+px+qx2) = ( )x-(
Solution: 1. y=log(1+x) it implies ey = 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+x2-x3+x4-x5+.....)
6. R.H.S. = log(1+ = log(1+( ) = log(1+px+qx2)
Some important questions with hints:
Q.1 Prove that = 2( + +....)
Q.2 prove that for |x| < ½ , log(1+3x+2x2) = 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 x2y = (2x-y) and |x| <1, prove that
(y+ y3/3+y5/5+.....) = 2( x+x3/3+x5/5+.....)
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(x2/2+x4/4+x6/6+....) + 2x (x+x3/3+x5/5+....) =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)x2-(1/3-
1/4)x3-.....] =[( 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(x2+1) =2x it implies y= 2x/(x2+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.