EXPANSION OF FUNCTIONS OF ONE & SEVERAL VARIABLES OBJECTIVES At the end of this session, you will be able to understand: Maclaurins Theorem Some Important Expression Taylors Theorems Function of Two Variables Total Differential Coefficients MACLAURINS THEOREM: If f(x) can be expanded in ascending powers of x, then 2 3( ) (0) '(0) ''(0) '''(0) ...... (0) .......2! 3! !nnx x xf x f xf f f fn= + + + + + + Proof. Suppose (1) 2 30 1 2 3( ) ........................ ................nnf x a a x a x a x a x = + + + + + Where are constant to be evaluated. 0 1 2 3, , , ,........................ na a a a a By successive differentiation (1) w.r.t. x, we get 2 3 11 2 3 42 22 3 43 4'( ) 2 3 4 ....................... ................(2)''( ) 2 3.2 4.3 ....................... ( 1) ...........(3)'''( ) 3.2 4.3.2 .......................nnnnf x a a x a x a x na xf x a a x a x n n a xf x a a x= + + + += + + += +3( 1)( 2) .......(4)nnn n n a x + + In general, ''( ) ( 1)( 2)............3.2.1 ...........(5)nf x n n n a term containing x = + Now putting x = 0 in (1) to (5), we get 0 1 2 3''(0) '''(0) (0)(0); '(0); ; ;.........; ,.......2! 3! !nnf f fa f a f a a an= = = = = 1Putting these values of constants in (1), we get 2 3( ) (0) '(0) ''(0) '''(0) ...... (0) .......2! 3! !nnx x xf x f xf f f fn= + + + + + + If we take f(x) = y; f(0) = (y)0; f(0) = (y1)0, f(0) = (y2)0, f(0) = (y3)0;..f(0) = (yn)0, then the above theorem takes the form as y = (y)0 + x (y1)0 + ...... ) (!......... ) (! 3) (! 20 0 330 22+ + + + nnynxyxyx or y = (y)0 + xy1(0) + ...... ) 0 (!......... ) 0 (! 3) 0 (! 23322+ + + + nnynxyxyx SOME IMPORTANT EXPANSIONS: 1. Expansion of ex. (Exponential series): Let f(x) = Then f(0) = = 1; xe0e ( )( ) ( )( )0 so that 0 1, where n = 1, 2, 3, ............n n xf x e f e = = = Substituting these values in Maclaurins series 2 3( ) (0) '(0) ''(0) '''(0) .......2! 3!x xf x f xf f f = + + + + We get ex = 1 + x + .....!......! 3 ! 23 2+ + + +nx x x n 2. Expansion of sin x. (Sine series): Let f(x) = sinx. Then f(0) = 0, f (x) = cosx so that f (0) = 1, f (x) = sinx so that f (0) = 0, f (x) = -cosx so that f (0) = -1, and so on. In general, f n (x) = sin 1 2x n s | |+ |\ . o that f n (x) = sin |.|

\| n21 When n = 2m, f n (0) = sin m = 0 and when n = 2m + 1, 2f n (0) = sin 1 1 1(2 1) sin ( 1) sin ( 1)2 2 2m mm m | | | |+ = + = = ` | | ) \ . \ . Substituting these values in Maclaurins series, we get Sinx = 0 + x.1 + 0 + ......)! 1 2 () 1 ( 0 ...... 0 ) 1 (31 2 3++ + + + + +mx x mm or Sinx = x - ......)! 1 2 () 1 ( ......! 5 ! 31 2 5 3++ + + +mx x x mm Similarly, we may obtain the Cosine series: Cosx = 1 - .....)! 2 () 1 ( .......! 6 ! 4 ! 22 5 4 2+ + + +mx x x x mm 3. Expansion of log (1+x): Let f (x) = log (1+x). Then f (0) = log 1 = 0; f n (x) = nnx n) 1 ()! 1 ( ) 1 (1+ So that f n (0) = (-1)n-1(n-1)!, where n = 1, 2, 3.. f (0) = (-1)1-1 (1-1)! = 1, f (0) = (-1)2-1(2-1)! = -1!, f (0) = (-1)3-1(3-1)! = 3!, f iv (0)= (-1)4-1(4-1)! = 3!, and so on. Substituting the values of f (0), f (0),f (0), etc. in Maclaurins series, we have f (x) = f(0) + x f (0) + ...., ) 0 (!..... ) 0 ( ' '! 22+ + + nnfnxfx we get log (1 + x) = 0 + x.1 - ..... )! 1 ( ) 1 (!..... ! 3 .! 4! 2 .! 3! 1 .! 214 3 2+ + + + nnx x x x nn or log (1 + x) = x - ..... ) 1 ( ......4 3 214 3 2+ + + + nx x x x nn 34. Expansion of (1 + x)n. (Binomial series): Let f(x) = (1 + x )n. Then f (0) = 1; f m (x) = n(n 1)(n 2).(n m + 1) (1 + x) n - m so that f m (0) = n(n - 1).(n - m + 1), where m = 1, 2, 3, f(0) = n, f (0) = n(n 1), f (0) = n (n-1) (n-2) and so on. Substituting the values of f (0), f (0), f (0) etc. in Maclaurins series for (x), we get (1+x) n =1 + nx +!) 1 ).....( 1 (.....! 2) 1 (2m x m n n nxn n m+ + + + Example: Expand the following by Maclaurins theorem: (i) tanx (ii) log secx Solution: (i) Let y = tanx. Then (y) 0 = tan0 = 0 y1 = sec2x = 1 + tan2x = 1 + y2 so that (y1)0 = 1 + (y)20 = 1 + 0 = 1, y2 = 2yy1, so that (y2)0 = 2 (y)0 (y1)0 = 2 1 0 = 0, y3 = 2y1y1 + 2yy2 = 2y12 + 2yy2 so that (y3)0 = 2 , 2 0 12= + y4 = 4y1y2 + 2y1y2 + 2yy3 = 6 y1y2 + 2yy3 so that (y4)0 = 61 2 0 2 0 + = 0, y5 = 6y22 + 6y1y3 + 2y1y3 + 2yy4 = 6y22 + 8y1y3 + 2yy4 so that (y5)0 = 0+8 16 0 2 1 = + and so on. Now by Maclaurins we get 2 3 4 50 1 0 2 0 3 0 4 0 5 02 3 4 535 ( ) ( ) ( ) ( ) ( ) ( ) ................2! 3! 4! 5! tan 0 .1 .0 .2 .0 .16 ...............2! 3! 4! 5!2 ........3 15x x x xy y x y y y y yx x x xx xxx x= + + + + + + = + + + + + += + + + 4(ii) Let y = log sec x. then (y) 0 = log sec 0 = log 1 = 0, 1 1 02 2 2 22 1 2 0 1 03 1 2 3 0 1 0 2 02 24 2 1 3 4 05 2 3 2 3 1 4 2 3 1 4 5 01.sec ( ) 0.secsec 1 tan 1 ( ) 1 ( ) 12 ( ) 2( ) ( ) 0,2 2 ( ) 2 1 0 24 2 2 6 2 ( ) 6 1 0 2y xtanx tanx so that yxy x x x y sothat y yy y y sothat y y yy y y y sothat yy y y y y y y y y y y so that y= = == = + = = + = + == = == + = + == + + = + = +2 26 3 2 4 2 4 1 5 3 2 4 1 56 00 2 06 6 2 2 6 8 2, ( ) 0 8 1 2 0 16, .y y y y y y y y y y y y yso that y and soon == + + + = + += + + = Now by Maclaurins theorem, we get . .......... ) (! 4) (! 3) (! 2) ( ) (0 440 330 220 1 0 + + + + + = yxyxyxy x y y 2 3 4 5 62 4 6logsec 0 .0 .1 .0 .2 .0 .16............2! 3! 4! 4! 4! ........2 12 45x x x x xx xx x x = + + + + + += + + + Example 2: By Maclaurins theorems show that 3 2 4 2 5 3 722 2 2 2cos 1 ......... 2 cos . ...........3! 4! 5! 7! 4 !nnx x x x x n xe x xn= + + + + + Solution. ( )1001 02 2 03 3 04cos . then cos 0 1cos sin (cos sin ), ( ) 1,(cos sin ) ( sin cos ) 2 sin , ( ) 0,2 sin 2 cos 2 (sin cos ), ( ) 2,2 (sin cosxx x xx x xx x xxLet y e x y ey e x e x e x x sothat yy e x x e x x e x so that yy e x e x e x x sothat yy e x x= = == = == + = == = + = = +24 02 2 25 1 5 0 6 2 6 02 37 3 7 0) 2 (cos sin ) 4 cos 2 ( ) 22 , ( ) 2 , 2 ( ) 0,2 ( ) 2 .x xe x x e x y sothat yy y so that y y y so that yy y sothat y and soon = = = = = = == =, In general, 5( )12 220(1 1) cos( tan 1) (2) cos( / 4)1 so that (2) cos( )4n nnnny x n xy n= + + = +=n Now by Maclaurins theorem, we get 200 1 0 2 02 3 4 5 6 72 2 323 2 4 2 5 3 7( ) ( ) ( ) ............. ( ) ...................2! ! 1 .1 .0 ( 2) ( 2 ) ( 2 ) .0 2 ....2! 3! 4! 5! 6! 7!12 cos ......! 42 2 2 2 1 .....3! 4! 4! 7!nnnnx xy y x y y ynx x x x x xxxnnx x x xx= + + + + += + + + + + + + +| |+ + |\ .= + +21..2 cos ...........4 !nn xnn| | + |\ . TAYLORS THEOREM: If ) can be expanded in ascending powers of x, then. h x f + (2( ) ( ) '( ) "( ) ..... "( ) .......2! !nh hf x h f x hf x f x f xn+ = + + + + + Proof. 2 3 40 1 2 3 42 41 2 3 4( ) .......... .......... ......(1) (1) . . ., , '( ) 2 3 4 ................... ...........(2) "( )nnLet f x h A Ah A h A h A h A hBy successive differentiation of wr t h we havef x h A A h A h A hf x h+ = + + + + + + ++ = + + + ++ =22 3 43 42 3.2 4.3 ................... ...........(3) "'( ) 3.2.1. 4.3.2. ........................ ............(4)...........................................................................A A h A hf x h A A h+ + ++ = + +..... Putting h = 0 in (1), (2), (3) and (4), we get 0 1 2 30 1 2 3( ) : '( ) : ''( ) 2 : '"( ) 3.2.1.1 1( ); '( ); "( ); "'( ) .2! 3!f x A f x A f x A f x AA f x A f x A f x A f x and so on= = = = = = = = Substituting these values in (1) we get the Taylors theorem as ) 5 ...( ....... ) ( "!. .......... ) ( "! 2) ( ' ) ( ) (2+ + + + + = + x fnhx fhx hf x f h x f n 6Cor.1. Putting a = 0 in (5) we get ..... .......... .......... ) 0 ( "!........ ) ( "! 2) ( ' ) ( ) (2+ + + + + = + fnha fha hf a f h a f n Corl.2 Putting a = 0and h = x in Cor.1we get the Maclaurins theorem 2( ) (0) '(0) "(0) ........ "(0) ..................2! !nx hf x f xf f fn= + + + + + Cor.3 Putting x = h and h = a in (5), we get 2( ) ( ) '( ) "( ) ............ ( ) ......2! !nna xf a h f h af h f h f hn+ = + + + + + Cor.4 Putting h = x - a in Cor.1, we get ......... ) (!)" (.......... ) ( "! 2) () ( ' ) ( ) ( )] ( [ ) (2++ ++ + = + = a fna xa fa xa f a x a f a x a f x f n Example: Expand log sin(x + h) in powers of h by Taylors Theorem Solution: 2 2( ) logsin( )1 ( ) logsin . '( ) cos cot ,sin" cos , '"( ) 2cos cot ,......................................................f x h x hf x x Futher f x xxf ec x f x ec x x+ = + = = == =x By Taylors Theorem, we get 2 32 32 2( ) ( ) '( ) "( ) "'( ) .............2! 3!2logsin( ) logsin cot cos cos cot ..................2! 3!h hf x h f x hf x f x f xh hx h x h x ec x ec x x+ = + + + + + = + + + Example: Show that ..... ..........3 2log ) log(3322 + + = +hxhxhxh h x Solution: Since we are to expand log ( x + h ) in power of x, therefore we are to use the from given in Cor3. putting x for a in Cor3, we get. 722 3 ( ) ( ) '( ) "( ) ................ (1)2!( ) log( )1 1 2 ( ) log ; '( ) ; "( ) ; ''( ) ,............. (1) xf x h f x xf h f hNow f x h x hf h h f h f h f hh h hPutting thesevaluein we get+ = + + ++ = + = = = =2 32 3 log( ) log ..............2 3x x xx h hh h h+ = + + + FUNCTION OF TWO VARIABLES: Taylors Theorem for function of two variables. To expand f ( x + h , y+ k ) in powers of h and k, in case f (x , y) and all its parietal derivatives are continuous in a certain domain of the point (x, y) Taking f (x+h ,y+k) as a function of one variable, say x i.e.x varies while y remains constant then expanding by Taylors theorem we have 22( , ) ( ,( , ) ( , ) ......(1)2!f x y k h f x y kf x h y k f x y k hv x + ++ + = + + + + Further, expanding each term on the right hand side of (1) by Taylors theorem taking y as variable and x as constant, we have 222 222 222( , ) ( , ) ( , )( , ) ( , ) ...... ( , ) .......2!( , )( , ) ....... ............2!1( , ) ( , ) 22!f x y k f x y f x yf x h y k f x y k h f x y ky x yh f x yf x y ky yf f f ff x h y k f x y h k h hk kx y x y xy + + = + + + + + + ` )| | + + + + | \ .| | + + = + + + + + | \ .222......fy| | + |\ . Or in symbolic from, we get "21 1( , ) ( , ) ..... ......2! !f ff x h y k f x y h k f h k f h k fx y x y n x y| || | | | | + + = + + + + + + + + | | | \ . . \\ .where f = f(x ,y) 1 2 21 1 2( 1).....2!n n n nn n n nn n n.nnf f n n fh k f h nh k h k kx y x x y x y | | + = + + + + | \ . fy [by Binomial Theorem] 8We now give an accurate statement of Taylors Theorem for function of two variables. If f (x, y) possesses continuous partial derivates upto nth order for all points (x,y) in the region then we have ( ) , a x a h b y b k + + ,211( , ) ( , ) ( , ) ( , ) ..........2!1 1.... ( , ) ( , )( 1)! ! 0 1n nf a h b k f a b h k f a b h k f a bx y x yh k f a b h k f a h b kn x y n x ywhere | | | | + + = + + + + + | | \ . \ .| | | | + + + + + + | | \ . \ .< < If u = f( y ); then to show that : dyyudxxudu += Give u = (y); then ). , ( y y x x f u u + + = + ) , ( ...... ) , ( ) , ( ) , ( y x f yyfxxfy x f y x f y y x x f u )`+||.|

\|++ = + + = yyuxxu +=to first order of approximation (replacing f by u) Thanking limits, we have . dyyudxxudu+= TOTAL DIFFERENTIAL COEFFICIENTS: ( , ) ( ) ( ) , .. . , .. .........(1)If u f x y where x t and y t then we know thatu udu dx dyx ydu dx dyut du dt dx dt dy dtdt dt dtdu u dx u dydt x dt y dt = = = = + = = = = + B Again if u = f =(x, y) where x = (t1 ,t2)then 1 12 2 u u x u yt x t y tu u x u yandt x t y t = + ` = + )12 9Example: Expand in power of (x - 1) and ( y + 2 ) by Taylors theorem. 2( , ) 3 2 f x y x y y = + Solution: By Taylors theorem 22 32 21( , ) ( , ) {( ) ( , ) ( ) ( , )} {( ) ( , )2!1 2( )( ) ( , ) ( ) ( , )} ( ) ( , )3! 3( ) ( ) ( , ) 3( )( ) ( , ) x y xxxy xy xxxxxy xyzf x y f a b x a f a b y b f a b x a f a bx a y b f a b y b f a b x a f a bx a y b f a b x a y b f a b= + + + + + + + + }3 ( ) ( , ) ............... .......(1)yyyy b f a b + + Given a = 1, b = -2 22( , ) 3 2 (1, 2) 10( , ) 2 (1, 2) 2( 1)( 2) 4( , ) 3 (1, 2) 4( , ) 2 (1, 2) 4( , ) 2 (1, 2) 2( , ) 0 (1, 2) 0( , ) 0 (1, 2) 0( , ) 2 (1, 2)x xy yxx xxxy xyyy yyxxx xxxxxy xxyf x y x y y ff x y xy ff x y x ff x y y ff x y y ff x y ff x y ff x y f= + = = = = = + = += = = == == == = 2( , ) 0 (1, 2) 0( , ) 0 ( , ) 0xyy xyyyyy yyyf x y ff x y f x y= == = Putting a = 1, b = -2 and values in (1), we get 2 22 32 2 3213 2 10[( 1)( 4) ( 2)(4)] [ 1) ( 4)2!1 2( 1)( 2)(2) ( 2) (0)] [( 1) (0)3! 3( 1) ( 2)(2) 3( 1)( 2) (0) ( 2) (0)].3 2 10 4( 1) 4( 2)x y y x y xx y y xx y x y yor x y y x y+ = + + + + + + + + + + + + + ++ = + + 2 22( 1) 2( 1)( 2) ( 1) ( 2). x x y x y + + + + Example: Expand eax by Maclaurins theorem. Solution: y = e ax, them (y)0 = e0 = 1 (by putting x = 0), 1001 1 02 2 2 0 22 000, ( ) ,( )............... ................................... .....................( )axaxn ax n nn ny ae y ae ay a e y a e ay a e y a e a= = == = == = . = By Maclaurins theorem, we get 2 30 1 0 2 0 3 0 02 32 3 ( ) ( ) ( ) ( ) ....... ( ) .............2! 3! !1 ....... ...........2! 3! !nnnax nx x xy y x y y y ynx x xe xa a a an= + + + + + + = + + + + + + Note: If a = 1, then 2 31 ......... ....2! 3! !nx x x xe xn= + + + + + This is known as Exponential series. Example: Expand ex sex x by Maclaurins theorem. Solution : Let y = ex sec x then (y)0 = e0 sec (0) = 1 11 0 02 22 1 2 0 0 1 02 2 23 1 2 12 21sec sec tan sec (1 tan ) (1 tan ), ( ) ( ) [1 tan(0)] 1sec (1 tan ), ( ) ( ) sec (0) ( ) (1 tan 0) 1 1 2,2 sec tan (1 tan ) sec =2 sec 2 sec tx xxy e x e x xe x x y x y yy y x y x y y yy y sex x y x x y x y xy x y x= += + = + = + == + + = + + = += + + + ++23 0 1 0 0 2 0an (1 tan )( ) 2( ) 1 2( ) (0) ( ) .1 4x y xy y y y+ += + + == By Maclaurins theorem, we get 2 20 1 0 2 0 3 032 3sec ( ) ( ) ( ) ( ) ............2! 3!2 1 .1 .2 .4 ........2! 3!2sec 1 ..........3xxx xe x y x y y yx xxe x x x x= + + + += + + + + = + + + + 11 Example: Use Maclaurins theorem to find the expansion of in ascending powers of x to the containing ) log( xe l +4x . Solution: Let then ( ). 1 log( xe y + = 2 log ) log( )00 = + = e l y 01 1 0022 21 12 1 1 2 0 1 0 1 03 21, ( )2 1(1 ) . [1 ] 1.( ) (1 ) ( ) (1 )1 11 11 1(1 ), ( ) ( ) [1 ( ) ] 12 2 (1xxx x x x x x x xx x x xxx xe ey ye l ee e e e e e e eyl e e l e eey ye ey y y y y yy y= = =+ ++ + = = =+ + + + = = + + = = = = = 121 1 2 3 0 2 0 1 0 2 02 1 22 24 3 3 1 2 4 0 3 0 3 0 1 0 2 0) ( ), ( ) ( ) 2( ) ( )1 1 1 2 2. . 02 2 4 2 2 , ( ) ( ) 2( ) ( ) 2( )1 10 0 2.16 8y y y y y y yy y yy y y y y y y y y y+ = = = == = = = By Maclaurins theorem, we get ........ ) (! 3) (! 2) ( ) ( ) 1 log(0 320 220 1 0 + + + + = + yxyxy x y ex .. ..........81.! 40 .! 3 41.! 2 21. 2 log4 2 2+|.|

\| + + + + = x x xx ..........192181212 log ) 1 log(4 2+ + + = + x x x ex Example: Expand by Maclaurins theorem ) sin 1 log( x + Solution: Let then ), sin 1 log( x y + = . 0 1 log ) (0 = = y .(1) 2 212 21 1cos sin1 cos2 2.cos1 1 1 1 sin 1 sincos sin 2sin cos2 2 2x xxy xx x 12x x x= = =+ + | |+ + |\ . x 12 xxx xx xx xx x21tan 121tan 121sin21cos21sin21cos21sin21cos21sin21cos22 2+=+=|.|

\| += (by dividing Num. and Den. By cos x1) 2 11 1tan4 2y x | | =

\ .| 141tan ) (0 1 = = y (2) ,21.4141sec22 |.|

\||.|

\| = x y 141sec21) (20 2 = = y (3) 2321 1 1 1 1 12sec tan2 4 2 2 4 21 1 1 1 1sec tan2 4 2 4 2y xx x | || | | ||= | | | \ .\ . \ .\ | | | | | |= | | | \ . \ . \ . 12x ||.1 = , ,1 2 y y 3 0 2 0 1 0( ) ( ) ( ) y y y = = + .(4) 24 4 1 2[ ] y y y y = + , 2 1 1 ) ( ) ( ) ( ) (022 1 0 3 0 4 = = = y y y y ..(5) . . .. .. . . .. .. By Maclaurins theorem, we get. ....... ) (! 4) (! 3) (! 2) ( ) ( ) sin 1 log(0 440 330 220 1 0 + + + + + = + yxyxyxy x y x .......... ) 2 (! 4 ! 3 ! 204 3 2+ + + = x x xx .......12 3 24 3 2+ + = x x xx Example: Expand by Maclaurins theorem. x1sin Solution: Let y = ..(1) x1sin Differentiating it with respect to x, we get 131 ) 1 () 1 (12 2121 = = x yxy ..(2) Differentiating it again, we have 0 ) 1 (0 ) 1 ( 0 ) 1 ( 2 ) 2 (12222 122 121= = + = + xy x y x y xy x y y x y (3) Differentiating (3) n times by Leibnitzs theorem, we have ( )22 1 1 2 1 11 ( 2 ) ( 2) .n n nn n n ny x c y x c y y x c y+ + + + + + = .1 0n ( ) 0 . ) 2 (! 2) 1 () 2 ( 11 122 = + + + + n n n n n ny x y yn nx ny x y 0 ) 1 2 ( ) 1 (21 22= + + + n n n y n xy n y x ..(4) Putting x = 0 in (1), (2), (3), we get ( yn+2 ) 0 - n2 ( yn)0 = 0 Putting n = 1, 2, 3 , we have 1 ) ( ) ( ) ( 0 ) ( 1 ) (0 3 0 1 0 3 0 1 0 3 = = = y y y y y . 0 ) ( ) ( 4 ) ( 0 ) ( 4 ) (0 4 0 2 0 4 0 2 0 4 = = = y y y y y 20 5 0 3 0 5 0 3 0 53 9 ) ( ) ( 9 ) ( 0 ) ( 9 ) ( = = = = y y y y y 0 ) ( ) ( 16 ) ( 0 ) ( 16 ) (0 6 0 4 0 6 0 4 0 6 = = = y y y y y 2 20 7 0 5 0 7 0 5 0 75 . 3 9 25 ) ( ) ( 25 ) ( 0 ) ( 25 ) ( = = = = y y y y y Hence by Malarias theorem, we have ... .......... .! 7. 5 . 3! 5. 3! 3......... ) 5 3 .(! 70 .! 6) 3 (! 50 .! 41 .! 30 .! 21 . 0....... ) (! 33) (! 2) ( ) ( sin7 2 2 5 2 32 27 625 4 3 20 3 0 220 1 01+ + + =+ + + + + + + + =+ + + + =x x xxx x x x x xxyxyxy x y x 14.......71.25.43.2151.43.2131.21sin7 5 3 1+ + + + = x x x x x Example: Expand 1(sin )2x in ascending powers of x. Solution: Let (y2 1) (sin= y1) 0 = 0 ( )( )11212 sin1y xx= ( )( )( ) 2 ) ( 0 2 14 2 2 10 ) ( 4 14 ) (sin 4 ) 1 (0 2 1 22121 2 120 12122 1 212= = = = = = = y xy y x y xy y y x y y y x y x y x Differentiating n times by Leibnitzs theorem, we get | | 0 ) 2 (! 2) 1 () 2 ( ) 1 (1 122 = +

+ + + + + n n n n n ny x y yn nx ny x y 0 ) 1 2 ( ) 1 (21 22= + + + n n n y n xy n y x 0 2 2) ( ) ( n n y n y = + (1) Putting n = 1,2,3,.., in (1), we get ( ) ( ) ( ) ( )( ) ( ) ( ) ( )2 23 1 4 20 0 0 02 25 3 6 40 0 0 01. 0 2 . 2 .2;3 . 0 4 . 2 .2 .2;y y y yy y y y= = = == = = =2 4 Hence 2 31 20 1 0 2 0 3 02 2 4 2 2 6(sin ) ( ) ( ) ( ) ( ) ......2! 3!2. 2 .2. 4 .2 .2................. ......(2)2! 4! 6!x xx y x y y yx x x= + + + += + + + Deductions: 1. If we put x = sin in the above result, we get. 15.......! 6sin 2 . 2. 4! 4sin2 . 2! 2sin 26 224222+ + + = 2. If we differentiate both sides of (2) w.r.t x, we get ( ).........! 52 . 2 . 4! 32 . 2 21sin 252 23221+ + + =x xxx x ( ).........5 . 3. 4 . 2321sin5321+ + + = xx xxx Example: Expand sin-1 (x + h) in power of x as far as the term in x3. Solution: First we observe that we are to expand sin-1 (x + h) in ascending powers of x. so let f(h) = sin -1h. Then f(h + x) = sin-1 ( h + x) Thus we are to expand f (h + x) in power of x. So by Taylors theorem, we have f( h + x) = f (h) + xf(h)+ .... ) ( ' "! 3) ( "! 23 2+ + h fxh fx .(1) Now f(h) = sin 1h.. Therefore f (h) = 2 1/ 221(1 )1 hh = f (h) = 2 3/ 2(1 ) h h 2 3/ 2 2 5/ 22 3/ 2 2 2 5/ 2 2 5/ 2 2 22 5/ 2 2'''( ) (1 ) ( 3/ 2)(1 ) ( 2 ) (1 ) 3 (1 ) (1 ) [(1 ) 3 ) =(1 ) (1 2 ), etc. f h h h h hh h h h hh h = + = + = + + h Substituting these values in (1), we have Sin-1 (h + x ) = Sin-1 h + (1-h2)1/2 x + (x2/2!) h (1-h2)-3/2 + (x3/3!) (1-h2)-5/2 (1-2h2)+ 16 Example: Use Taylors theorem to prove that tan-1(x + h) = tan-1x +(hsin )2 3sin sin 2 sin3( sin ) ( sin ) .....1 2 3h h + .+(-1)n-1(hsin )n.....,sin+nn where 0 = cot-1x Solution: Let f(x) = tan-1x. then f (x + h) = tan-1 (x + h ). Expanding f (x + h) in power of h by Taylors theorem we have f ( x + h) = f (x) + 2'( ) ''( ) ..... ''( ) ....1! 2! !nh h hf x f x f xn+ + + + .(1) Now f ( x ) = tan-1x. Therefore f (x) = Dn tan-1x = (-1) n-1 (n-1)! sinn sin n . Where = cot-1x Putting n= 1, 2, 3, ..in it , we get f (x)= sin sin , f (x) = -1! sin2 sin2 f (x) = 2! Sin3 sin3 .etc. Substituting these values in (1), we have tan-1 (x + y) = tan-1 x +hsin sin - 2 32 3sin sin 2 2!sin sin3 .......2! 3!h h + 1( 1) ( 1)!sin sin ....!nn nhn nn + + = tan-1 x + hsin2 3 1sin sin 2 sin3 sin( sin ) ( sin ) ..... ( 1) ( sin ) .......1 2 3 n n nh h hn + + + Example: Expand ex cosy near the point 1,4 |

by Taylors Theorem. ||\ .Solution: By Taylors theorem F(x + h, y + k) = F (x, y) + ) 1 .....(! 31! 213 2+||.|

\|++||.|

\|++||.|\ + Fykxh Fykxh Fykxh

| Again ex cosy = F(x, y) =F ,4 4). 1 ( 1

|.|

\| + + y x where h = x -1, k = y - 4 = F |.|\ + + k h41

| 17 F(x, y) = ex cosy F1.42e | | =\ . | y eyF xcos = 24. 1 eyF=|.|

\| y exF xsin = 24. 122exF=|.|

\| y exF xcos22= 24. 122exF=|.|

\| y eyF xcos22 = 24. 122eyF =|.|

\| y ey xF xcos2 = 24. 12ey xF =|.|

\| Substituting these values in Taylors theorem, we get ex cosy =

|.|

\| + +242) 1 (2 eyexe +221( 1) 2( 1) ..........2! 4 42 2 2e e ex x y y | | | | | | | | + + + | | | |\ . \ . \ . \ . 18ADDITIONAL PROBLEMS: 1. Show that by Maclaurins theorem, 2 3 41log(1 ) ........... ( 1)2 3 4nn!x x x xx xn+ = + + + 2. By Maclaurins theorem, prove that ( )/ 22 22 32 3 13sin .......... sin( tan ) ...3! !nax na ba b b be bx bx abx x x nn a+= + + + + + 3. Apply Maclaurins theorem to obtain the term upto 4x in the expansion of l2og(1 sin ) x + 4. If y e, show that ( )1sin a = x( )2 22 11 (2 1)n nx y n xy n a y+ + + +20n = Hence by Maclaurins theorem show that 12 2 2 2sin 32 3(1 )1 ........2! 3!1 1Also deduce that 1 sin sin sin .....2! 3!a x a x a ae ax xe += + + + += + + + + 5. By Maclaurins theorem, show that 12 2 2 2 2cos 3 42(1 ) (2 )1 ........2! 3! 4!aa x a x a a a ae ax x x e | | + += + + |\ . 19