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Series expansion of exponential and logarithmic functions for entrance exams.

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- 1. Series Expansion of Exponential and LogarithmicFunctionsLeonhard Euler, the great Swiss mathematician introduced and named thenumber 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 givese=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 ofthe series for ex+4,The series for ex+4Here 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 seriesfor 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 & -1subtract 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 binomialtheorem = 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 , thenecx = 1+cx +(cx)2/2! + .......... = 1+ x(logea) + x2/2! (logea)2 +..........[let ec =a it implies c= logea & ecx = (ec)x = ax.]
- 6. xThe graph of y = e is upward-sloping, and increases faster as x increases. The graph alwayslies above the x-axis but can get arbitrarily close to it for negative x; thus, the x-axis is ahorizontal asymptote. The slope of the tangent to the graph at each point is equal toits y coordinate at that point. The inverse function is the natural logarithm ln(x); because ofthis, refer to the exponential function as the antilogarithm.The natural logarithm function, if considered as a real-valued function of a realvariable, 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 givesx=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 andloge2= 1-(1/2-1/3)-(1/4-1/5)-.....<0.76 we can write log2 < 1 < log3Some 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 thatlog(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 forintegration) 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.

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