Taylors And Maclaurins Series

EXPANSION OF FUNCTIONS OF ONE & SEVERAL VARIABLES OBJECTIVES At the end of this session, you will be able to understand: Maclaurin’s Theorem Some Important Expression Taylor’s Theorems Function of Two Variables Total Differential Coefficients

MACLAURIN’S THEOREM: If f(x) can be expanded in ascending powers of x, then

x2 x3 xn n f ( x) = f (0) + xf '(0) + f ''(0) + f '''(0) + ...... + f (0) + ....... 2! 3! n! Proof. Suppose

f ( x) = a0 + a1 x + a2 x 2 + a3 x 3 + ........................an x n + ................ (1)

Where a0 , a1 , a2 , a3 ,........................an are constant to be evaluated. By successive differentiation (1) w.r.t. x, we get

f '( x) = a1 + 2a2 x + 3a3 x 2 + 4a4 x 3 .......................nan x n −1 + ................(2) f ''( x) = 2a2 + 3.2a3 x + 4.3a4 x 2 .......................n(n − 1)an x n − 2 + ...........(3) f '''( x) = 3.2a3 + 4.3.2a4 x....................... + n(n − 1)(n − 2)an x n −3 + .......(4) In general, f ''( x) = n(n − 1)(n − 2)............3.2.1an + term containing

x

...........(5)

Now putting x = 0 in (1) to (5), we get a0 = f (0); a1 = f '(0); a2 =

f ''(0) f '''(0) f n (0) ; a3 = ;.........; an = ,....... 2! 3! n!

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Putting these values of constants in (1), we get

x2 x3 xn n f ( x) = f (0) + xf '(0) + f ''(0) + f '''(0) + ...... + f (0) + ....... 2! 3! n! 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 x2 x3 xn y = (y)0 + x (y1)0 + ( y 2 ) 0 + ( y 3 ) 0 + ......... + ( y n ) 0 + ...... 2! 3! n! or

y = (y)0 + xy1(0) +

x2 x3 xn y 2 (0) + y 3 (0) + ......... + y n (0) + ...... 2! 3! n!

SOME IMPORTANT EXPANSIONS: 1. Expansion of ex. (Exponential series): Let f(x) = e x Then f(0) = e0 = 1;

f ( n ) ( x ) = e x so that f ( n ) ( 0 ) = e0 = 1, where n = 1, 2, 3, ............ Substituting these values in Maclaurin’s series

x2 x3 f ( x) = f (0) + xf '(0) + f ''(0) + f '''(0) + ....... 2! 3! x2 x3 xn We get e = 1 + x + + + ...... + + ..... 2! 3! n! x

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.

1  1   In general, f n (x) = sin  x + nπ  so that f n (x) = sin  nπ  2   2  When n = 2m, f n (0) = sin m π = 0 and when n = 2m + 1,

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1  1   1  f n (0) = sin  (2m + 1)π  = sin  mπ + π  = (−1) m sin  π  = (−1) m 2  2   2  Substituting these values in Maclaurin’s series, we get Sinx = 0 + x.1 + 0 +

or Sinx = x -

x3 x 2 m +1 (−1) + 0 + ...... + 0 + (−1) m + ...... 3 (2m + 1)!

x3 x5 x 2 m +1 + − ...... + (−1) m + ...... 3! 5! (2m + 1)!

Similarly, we may obtain the Cosine series: Cosx = 1 -

x2 x4 x5 x 2m + − + ....... + (−1) m + ..... 2! 4! 6! (2m)!

3. Expansion of log (1+x):

Let f (x) = log (1+x). Then f (0) = log 1 = 0; f n (x) =

(−1) n −1 (n − 1)! ( x + 1) n

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 Maclaurin’s series, we have f (x) = f(0) + x f ’(0) +

x2 xn n f ' ' (0) + ..... + f (0) + ...., 2! n!

we get log (1 + x) = 0 + x.1 -

x2 x3 x4 xn .1!+ .2!− .3!+..... + (−1) n −1 (n − 1)!+..... 2! 3! 4! n!

n x2 x3 x4 n −1 x or log (1 + x) = x + − + ...... + (−1) + ..... 2 3 4 n

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4. 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 Maclaurin’s series for (x), we get (1+x) n =1 + nx +

n(n − 1) 2 n(n − 1).....(n − m + 1) x m + x + ..... + …… 2! m!

Example: Expand the following by Maclaurin’s 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 × 0 × 1 = 0, y3 = 2y1y1 + 2yy2 = 2y12 + 2yy2 so that (y3)0 = 2 × 12 + 0 = 2 ,

y4 = 4y1y2 + 2y1y2 + 2yy3 = 6 y1y2 + 2yy3 so that (y4)0 = 6× 1 × 0 + 2 × 0 × 2 = 0, y5 = 6y22 + 6y1y3 + 2y1y3 + 2yy4 = 6y22 + 8y1y3 + 2yy4 so that (y5)0 = 0+8 × 1 × 2 + 0 = 16 and so on. Now by Maclaurin’s we get x2 x3 x4 x5 y = ( y )0 + x( y1 )0 + ( y2 )0 + ( y3 )0 + ( y4 )0 + ( y5 ) 0 +................ 2! 3! 4! 5! 2 3 4 5 x x x x ∴ tan x = 0 + x.1 + .0 + .2 + .0 + .16 + ............... 2! 3! 4! 5! 3 2 x = x + + x5 + ........ 3 15

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(ii)

Let y = log sec x. then (y) 0 = log sec 0 = log 1 = 0,

1 .sec xtanx = tanx so that ( y1 )0 = 0. sec x y2 = sec 2 x = 1 + tan 2 x = x = 1 + y 21 so that ( y2 )0 = 1 + ( y1 )02 = 1 y1 =

y3 = 2 y1 y2 so that ( y3 )0 = 2( y1 )0 ( y2 )0 = 0, y4 = 2 y22 + 2 y1 y3 so that ( y4 )0 = 2 × 12 + 0 = 2 y5 = 4 y2 y3 + 2 y2 y3 + 2 y1 y4 = 6 y2 y3 + 2 y1 y4 so that ( y5 )0 = 6 × 1 × 0 + 2 × 0 × 2 = 0 y6 = 6 y32 + 6 y2 y4 + 2 y2 y4 + 2 y1 y5 = 6 y32 + 8 y2 y4 + 2 y1 y5 so that , ( y6 )0 = 0 + 8 × 1 × 2 + 0 = 16, and so on. Now by Maclaurin’s theorem, we get

y = ( y ) 0 + x( y1 ) 0 +

∴

x2 x3 x4 ( y 2 ) 0 + ( y3 ) 0 + ( y 4 ) 0 + ........... 2! 3! 4!

x2 x3 x4 x5 x6 .1 + .0 + .2 + .0 + .16............ 2! 3! 4! 4! 4! 2 4 6 x x x = + + + ........ 2 12 45

log sec x = 0 + x.0 +

Example 2: By Maclaurin’s theorems show that

e x cos x = 1 + x −

n 2 x 3 2 2 x 4 2 2 x 5 23 x 7 nπ x n . + ........... − − + + ......... + 2 2 cos 3! 4! 5! 7! 4 n!

Solution.

Let y = e x cos x. then ( y )0 = e0 cos 0 = 1 y1 = e x cos x − e x sin x = e x (cos x − sin x), so that ( y1 )0 = 1, y2 = e x (cos x − sin x) + e x (− sin x − cos x) = −2e x sin x, so that ( y2 )0 = 0, y3 = −2e x sin x − 2e x cos x = −2e x (sin x + cos x), so that ( y3 )0 = −2, y4 = −2e x (sin x + cos x) − 2e x (cos x − sin x) = −4e x cos x = −22 y so that ( y 4 )0 = −2, y5 = −22 y1 , so that ( y5 )0 = −22 , y6 = −22 y2 so that ( y6 )0 = 0, y7 = −22 y3 so that ( y7 )0 = 23 and so on. In general,

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n

n

yn = (1 + 1) 2 cos( x + n tan −1 1) = (2) 2 cos( x + nπ / 4) n 1 so that ( yn )0 = (2) 2 cos( nπ ) 4

Now by Maclaurin’s theorem, we get

x2 xn ( y2 )0 + ............. + ( yn )0 + ................... 2! n! 2 3 4 x x x x5 x6 x7 = 1 + x.1 + .0 + (−2) + (−22 ) + (−22 ) + .0 + 23 + .... 2! 3! 4! 5! 6! 7! n x 1  + 2n 2 cos  π n  + ...... n! 4 

y = ( y )0 + x( y1 )0 +

n n 2 x 3 2 2 x 4 2 2 x 5 23 x 7 1 x 2 = 1+ x − − − − + .......2 cos  nπ  + ........... 3! 4! 4! 7!  4  n!

TAYLOR’S THEOREM: If f ( x + h ) can be expanded in ascending powers of x, then.

f ( x + h) = f ( x) + hf '( x) +

h2 hn f "( x) + ..... + f "( x) + ....... n! 2!

Proof.

Let f ( x + h) = A0 + A1 h + A2 h 2 + A3 h3 + A4 h 4 + .......... + An h n + ..........

......(1)

By successive differentiation of (1) w.r.t., h, we have f '( x + h) = A1 + 2 A2 h + 3 A3 h 2 + 4 A4 h 4 + ...................

...........(2)

f "( x + h) = 2 A2 + 3.2 A3 h + 4.3 A4 h 2 + ...................

...........(3)

f "'( x + h) = 3.2.1. A3 + 4.3.2. A4 h + ........................

............(4)

........................................ ........................................ Putting h = 0 in (1), (2), (3) and (4), we get

f ( x) = A0 : f '( x) = A1 : f ''( x) = 2 A2 : f '"( x) = 3.2.1. A3 ⇒ A0 = f ( x); A1 = f '( x); A2 =

1 1 f "( x); A3 = f "'( x) and so on. 2! 3!

Substituting these values in (1) we get the Taylor’s theorem as

f ( x + h) = f ( x) + hf ' ( x) +

h2 hn f " ( x) + ........... + f " ( x) + ....... 2! n!

...(5)

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Cor.1. Putting a = 0 in (5) we get

f (a + h) = f (a) + hf ' (a ) +

h2 hn f " (a ) + ........ + f " (0) + ......................... n! 2!

Corl.2 Putting a = 0and h = x in Cor.1we get the Maclaurin’s theorem x2 hn f ( x) = f (0) + xf '(0) + f "(0) + ........ + f "(0) + .................. n! 2! Cor.3 Putting x = h and h = a in (5), we get a2 xn n f (a + h) = f (h) + af '( h) + f "(h) + ............ + f (h) + ...... n! 2! Cor.4 Putting h = x - a in Cor.1, we get

f ( x) = f [a + ( x − a )] = f (a) + ( x − a) f ' (a ) +

( x − a) 2 ( x − a)" n f " (a) + .......... + f (a) + ......... n! 2!

Example: Expand log sin(x + h) in powers of h by Taylor’s Theorem Solution:

f ( x + h) = log sin( x + h)

∴

1 cos x = cot x, sin x f " = − cos ec 2 x, f '"( x) = 2 cos ec 2 x cot x,

f ( x) = log sin x. Futher f '( x) =

............................. ......................... By Taylor’s Theorem, we get h2 h3 f "( x) + f "'( x) + ............. 2! 3! h2 2h 3 2 cos ec 2 x cot x + .................. ⇒ log sin( x + h) = log sin x + h cot x − cos ec x + 2! 3! f ( x + h) = f ( x) + hf '( x) +

x x2 x3 + − .......... ..... Example: Show that log( x + h) = log h + − h 2h 2 3h 3 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.

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f ( x + h) = f ( x) + xf '(h) +

x2 f "(h) + ................ 2!

(1)

Now f ( x + h) = log( x + h) 1 1 2 ∴ f (h) = log h; f '(h) = ; f "(h) = − 2 ; f ''(h) = 3 ,............. h h h Putting these value in (1) we get log( x + h) = log h +

x x2 x3 − 2 + 3 + .............. h 2h 3h

FUNCTION OF TWO VARIABLES: Taylor’s 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 Taylor’s theorem we have

f ( x + h, y + k ) = f ( x , y + k ) + h

∂f ( x, y + k ) h 2 ∂f ( x, y + k + + ......(1) ∂v 2! ∂x 2

Further, expanding each term on the right hand side of (1) by Taylor’s theorem taking y as variable and x as constant, we have  ∂f ( x, y ) k 2 f ( x, y ) ∂  ∂f ( x, y ) f ( x + h, y + k ) = f ( x , y ) + k + + ...... + h  f ( x, y ) + k + ....... 2 ∂y ∂x  ∂y 2! ∂y 

 ∂f ( x, y ) h2 ∂ 2  + .......  + ............ f ( x, y ) + k 2  ∂y 2! ∂y    ∂f ∂2 f  ∂f  1  ∂ 2 f ∂2 f ⇒ f ( x + h, y + k ) = f ( x, y ) +  h + k  +  h 2 2 + 2hk + k 2 2  + ...... ∂y  2!  ∂x ∂x∂y ∂y   ∂x Or in symbolic from, we get " 2  ∂f ∂f  1 ∂ ∂  1 ∂ ∂  f ( x + h, y + k ) = f ( x, y ) +  h + k  f +  h + k  f + ..... +  h + k  f + ...... where ∂y  ∂y  ∂y  2!  ∂x n !  ∂x  ∂x f = f(x ,y) +

n

n n  ∂ n(n − 1) n − 2 2 ∂ n f ∂  ∂n f n ∂ f n −1 n ∂ f. h + k  f = h h k + nh k n −1 + + ..... + k 2! ∂y  ∂x n ∂x ∂y ∂x n −1∂ 2 y ∂y n  ∂x [by Binomial Theorem]

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We now give an accurate statement of Taylor’s Theorem for function of two variables. If f (x, y) th possesses continuous partial derivates upto n order for all points (x,y) in the region ( a ≤ x ≤ a + h, b ≤ y ≤ b + k ) , then we have 2

 ∂ ∂  ∂  1 ∂ f (a + h, b + k ) = f (a, b) +  h + k  f (a, b) +  h + k  f (a, b) + .......... ∂y  ∂y  2!  ∂x  ∂x ∂  1  ∂ .... + h + k  ∂y  (n − 1)!  ∂x

n −1

n

∂  1 ∂ f (a, b) +  h + k  f (a + θ h, b + θ k ) n !  ∂x ∂y  where 0 < θ < 1

If u = f( y ); then to show that du =

∂u ∂u dx + dy : ∂x ∂y

Give u = (y); then u + δ u = f ( x + δ x, y + δy ).

 ∂ f ∂f ∴ δu = f ( x + δx, y + δy ) − f ( x, y ) =  f ( x, y ) +  δ x+ δ ∂y  ∂x 

=

  y  + ...... − f ( x, y )  

∂u ∂u δ x + δ y to first order of approximation (replacing f by u) ∂y ∂x

Thanking limits, we have du =

∂u ∂u dx + dy. ∂x ∂y

TOTAL DIFFERENTIAL COEFFICIENTS:

If u = f ( x, y ) where x = φ (t ) and y = φ (t ) , then we know that ∂u ∂u du = dx + dy. ∂x ∂y du dx dy But .dt du = dx = .dt , dy = .dt dt dt dt du ∂u dx ∂u dy ∴ = . + dt ∂x dt ∂y dt

.........(1)

Again if u = f =(x, y) where x = φ (t1 ,t2)then

and

∂u ∂u ∂x ∂u ∂y  = + ∂t1 ∂x ∂t1 ∂y ∂t1   ∂u ∂u ∂x ∂u ∂y  = + ∂t2 ∂x ∂t2 ∂y ∂t2 

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Example: Expand f ( x, y ) = x y + 3 y − 2 in power of (x - 1) and ( y + 2 ) by Taylor’s theorem. 2

Solution: By Taylor’s theorem

1 {( x − a ) 2 f xx (a, b) 2! 1 + 2( x − a )( y − b) f xy (a, b) + ( y − b) 2 f xy (a, b)} + ( x − a )3 f xxx (a, b) 3! 2 2 + 3( x − a ) ( y − b) f xxy (a, b) + 3( x − a )( y − b) f xyz (a, b)

f ( x, y ) = f (a, b) + {( x − a ) f x (a, b) + ( y − b) f y (a, b)} +

+ ( y − b)3 f yyy (a, b)

} + ...............

.......(1)

Given a = 1, b = -2

f ( x, y ) = x 2 y + 3 y − 2 ⇒ f (1, −2) = −10 f x ( x, y ) = 2 xy ⇒ f x (1, −2) = 2(−1)(−2) = −4 f y ( x, y ) = x 2 + 3 ⇒ f y (1, −2) = +4 f xx ( x, y ) = 2 y ⇒ f xx (1, −2) = −4 f xy ( x, y ) = 2 y ⇒ f xy (1, −2) = 2 f yy ( x, y ) = 0 ⇒ f yy (1, −2) = 0 f xxx ( x, y ) = 0 ⇒ f xxx (1, −2) = 0 f xxy ( x, y ) = 2 ⇒ f xxy (1, −2) = 2 f xyy ( x, y ) = 0 ⇒ f xyy (1, −2) = 0 f yyy ( x, y ) = 0 ⇒ f yyy ( x, y ) = 0 Putting a = 1, b = -2 and values in (1), we get

1 [ x − 1) 2 (−4) 2! 1 + 2( x − 1)( y + 2)(2) + ( y + 2) 2 (0)] + [( x − 1)3 (0) 3! 2 + 3( x − 1) ( y + 2)(2) + 3( x − 1)( y + 2) 2 (0) + ( y + 2)3 (0)].

x 2 y + 3 y − 2 = 10[( x − 1)(−4) + ( y + 2)(4)] +

or x 2 y + 3 y − 2 = −10 − 4( x − 1) + 4( y + 2) − 2( x − 1) 2 + 2( x − 1)( y + 2) + ( x − 1) 2 ( y + 2). Example: Expand eax by Maclaurin’s theorem. Solution: y = e ax, them (y)0 = e0 = 1 (by putting x = 0),

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y1 = ae ax , ⇒

( y1 )0 = ae0 = a,

y 2 = a 2 e ax ⇒ ( y2 )0 = a 2 e0 = a 2

∴

............... ...............

.................... .....................

yn = a n e ax

⇒ ( yn ) 0 = a n e 0 = a n .

By Maclaurin’s theorem, we get

x2 x3 xn y = ( y )0 + x( y1 )0 + ( y2 )0 + ( y3 )0 + ....... + ( yn )0 + ............. n! 2! 3! 2 3 n x x x ⇒ e ax = 1 + xa + a 2 + a 3 + ....... + a n + ........... n! 2! 3! Note: If a = 1, then

ex = 1 + x +

x 2 x3 xn + + ......... + .... 2! 3! n!

This is known as Exponential series. Example: Expand ex sex x by Maclaurin’s theorem. Solution : Let y = ex sec x then (y)0 = e0 sec (0) = 1

y1 = e x sec x + e x sec x tan x = e x sec x(1 + tan x) = y (1 + tan x), ( y1 )0 = ( y )0 [1 + tan(0)] = 1 y2 = y sec 2 x + y1 (1 + tan x), ( y2 )0 = ( y )0 sec 2 (0) + ( y1 )0 (1 + tan 0) = 1 + 1 = 2, y3 = y1 sex 2 x + 2 y sec 2 x tan x + y2 (1 + tan x) + y1 sec2 x =2 y1 sec 2 x + 2 y sec 2 x tan x + y2 (1 + tan x) ( y3 )0 = 2( y1 )0 1 + 2( y )0 (0) + ( y2 )0 .1 = 4 ∴By Maclaurin’s theorem, we get

x2 x2 e sec x = ( y )0 + x( y1 )0 + ( y2 )0 + ( y3 )0 + ............ 2! 3! 3 x2 x = 1 + x.1 + .2 + .4 + ........ 2! 3! 2 ⇒ e x sec x = 1 + x + x 2 + x3 + .......... 3 x

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Example: Use Maclaurin’s theorem to find the expansion of log(l + e ) in ascending powers of x to the containing x 4 . x

0 x Solution: Let y = log(1 + e ). then ( y ) 0 = log(l + e ) = log 2

ex e0 1 = = , ( y ) 1 0 l + e0 2 1 + ex ex (1 + e x )e x − e x .e x e x [1 + e x − e x ] 1 = = y2 = . (l + e x ) 2 (1 + e x ) 2 (l + e x ) (1 + e x )

y1 =

= y1

 ex  1 = − y 1 1  x  1 + ex  1+ e  1  1 1 = 1− 2  2  2 ( y3 )0 = ( y2 )0 − 2( y1 )0 ( y2 )0

⇒ y 2 = y1 (1 − y1 ),

( y2 )0 = ( y1 )0 [1 − ( y1 )0 ] =

y3 = y2 (1 − y1 ) + y1 (− y2 ),

1 1 1 − 2. . = 0 2 2 4 ( y4 )0 = ( y3 )0 − 2( y3 ) 0 ( y1 ) 0 − 2( y2 2 ) 0

= y2 − 2 y1 y2

=

y4 = y3 − 2 y3 y1 − 2 y2 2 ,

= 0 − 0 − 2. ∴By Maclaurin’s theorem, we get

log(1 + e x ) = ( y ) 0 + x ( y1 ) 0 +

1 1 =− 16 8

x2 x2 ( y2 ) 0 + ( y 3 ) 0 + ........ 2! 3!

x4  1  1 x2 1 x2 = log 2 + x. + . + .0 + . −  + ............ 4!  8  2 2! 4 3! 1 1 1 4 ⇒ log(1 + e x ) = log 2 + x + x 2 − x + .......... 2 8 192 Example: Expand log(1 + sin x ) by Maclaurin’s theorem Solution: Let y = log(1 + sin x), then

( y) 0 = log1 = 0.

….(1)

1 1 x − sin 2 x 1 cos x 2 2 y1 = .cos x = = 1 1 1 1 1 + sin x 1 + sin x   2 2  cos x + sin x  + 2sin x cos x 2 2  2 2  cos 2

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1 1 1 1 1 x − sin 2 x cos x − sin x 1 − tan x 2 2 = 2 2 = 2 = 2 1 1 1 1 1   cos x + sin x 1 + tan x  cos x + sin x  2 2 2 2 2   cos 2

1 (by dividing Num. and Den. By cos x ) 2

1  1 ⇒ y1 = tan  π − x  2  4 1  1 1 y 2 = sec 2  π − x . − , 4  2 4

1 ∴ ( y1 ) 0 = tan π = 1 4 1 1 ∴( y 2 ) 0 = − sec 2 π = −1 2 4

……(2) ……(3)

1 1  1  1   1  1 1 y3 = −  2sec 2  π − x   −  × tan  π − x   −   2 2  2  2   2  4 4  1  1  1  1 1 = −  −  sec 2  π − x  × tan  π − x   2  2  4 4  2 

∴ ( y3 )0 = −( y2 )0 ( y1 )0 = +1

= − y 2 , y1 ,

y4 = −[ y4 y1 + y2 2 ], …. …. ….. ….. …. …. ….. …..

…….(4)

2

∴ ( y 4 ) 0 = −( y 3 ) 0 ( y1 ) − ( y 2 ) 0 = −1 − 1 = −2 …..(5)

∴

By Maclaurin’s theorem, we get. x2 x3 x4 log(1 + sin x ) = ( y ) 0 + x ( y1 ) 0 + ( y 2 ) 0 + ( y3 ) 0 + ( y 4 ) 0 + ....... 2! 3! 4!

x2 x3 x4 = 0+ x− + − (−2) + .......... 2! 3! 4! x2 x3 x4 = x− + − + ....... 2 3 12 Example: Expand sin Solution: Let y =

−1

x by Maclaurin’s theorem.

sin −1 x

…..(1)

Differentiating it with respect to x, we get

13

y1 =

1

2

(1 − x ) 2

⇒ y1 (1 − x 2 ) = 1

…..(2)

Differentiating it again, we have 2

y1 (−2 x) + 2 y1 y 2 (1 − x 2 ) = 0 ⇒ − xy1 + y 2 (1 − x 2 ) = 0 ⇒ y 2 (1 − x 2 ) − xy1 = 0

……(3)

Differentiating (3) n times by Leibnitz’s theorem, we have

 yn + 2 (1 − x 2 ) + n c1 yn +1 (−2 x) + n c2 yn (−2)  −  yn +1 .x + n c1 yn .1 = 0    

(

)

⇒ y n + 2 1 − x 2 − ny n +1 (−2 x) +

n(n − 1) y n (−2) − y n +1 .x − ny n = 0 2!

⇒ (1 − x 2 ) y n+2 − (2n + 1) xyn+1 − n 2 y n = 0

…..(4)

Putting x = 0 in (1), (2), (3), we get

( yn+2 ) 0 - n2 ( yn)0 = 0 Putting n = 1, 2, 3…… , we have

( y3 ) 0 − 1( y1 ) 0 = 0 ⇒ ( y3 ) 0 = ( y1 ) 0 ⇒ ( y3 ) 0 = 1 ( y 4 ) 0 − 4( y 2 ) 0 = 0 ⇒ ( y 4 ) 0 = 4( y 2 ) 0 ⇒ ( y 4 ) 0 = 0. ( y 5 ) 0 − 9( y 3 ) 0 = 0 ⇒ ( y 5 ) 0 = 9( y 3 ) 0 ⇒ ( y 5 ) 0 = 9 = 3 2 ( y6 ) 0 − 16( y 4 ) 0 = 0 ⇒ ( y6 ) 0 = 16( y 4 ) 0 ⇒ ( y 6 ) 0 = 0

( y 7 ) 0 − 25( y 5 ) 0 = 0 ⇒ ( y 7 ) 0 = 25( y 5 ) 0 ⇒ ( y 7 ) 0 = 25 × 9 = 3 2.5 2 Hence by Malaria’s theorem, we have

x2 x3 sin x = ( y ) 0 + x( y1 ) 0 + ( y 2 ) 0 + ( y 3 ) 0 + ....... 2! 3! 2 3 4 x x x x5 x6 x7 2 2 2 = 0 + x.1 + .0 + .1 + .0 + (3) + .0 + .(3 5 ) + ......... 2! 3! 4! 5! 6! 7! 3 2 5 2 2 7 x 3 .x 3 .5 .x = x− + + . + ............. 3! 5! 7! −1

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1 1 1 3 1 1 3 5 1 ⇒ sin −1 x = x + . x 3 + . . x 5 + . . . x 7 + ....... 2 3 2 4 5 2 4 2 7 Example: Expand (sin

−1

x)2 in ascending powers of x.

−1 2

Solution: Let y = (sin )

y1 = 2 ( sin −1 x )

(y1) 0 = 0

1

(1 − x ) 2

2

⇒ (1 − x 2 ) y1 = 4(sin −1 x) 2 = 4 y

( ) ∴(y ) = 0 ⇒ (1 − x )2 y y − 2 xy = 4 y ⇒ (1 − x )y − xy − 2 = 0 ∴(y ) = 2 2

⇒ 1 − x 2 y1 = 4 y

1 0

2

2

1

2

1

1

2

2

1

2 0

Differentiating n times by Leibnitz’s theorem, we get

n(n − 1)   2 − + − + ( 1 ) ( 2 ) y x ny x y n (2) − [ y n +1 x + ny n ] = 0 n + 2 n + 1  2!  

⇒ (1 − x 2 ) y n + 2 − (2n + 1) xy n +1 − n 2 y n = 0

∴ ( y n + 2 ) = n2 ( y n ) 0

…(1)

Putting n = 1,2,3,….., in (1), we get

( y3 )0 = 1. ( y1 )0 = 0 ( y5 )0 = 32. ( y3 )0 = 0

( y4 )0 = 22. ( y2 )0 = 22.2; ( y6 )0 = 42. ( y4 )0 = 22.24.2;

Hence

x2 x3 (sin x) = ( y0 ) + x( y1 )0 + ( y2 )0 + ( y3 )0 + ...... 2! 3! 2 2 4 2 2 6 2.x 2 .2.x 4 .2 .2.x = + + + ................ 2! 4! 6! −1

2

......(2)

Deductions:

1. If we put x = sin θ in the above result, we get.

15

θ2 =

2 sin 2 θ sin 4 θ 2 2.2 sin 6 θ + 2 2 .2 + 4 2. + ....... 2! 4! 6!

2. If we differentiate both sides of (2) w.r.t x, we get

2 sin −1 x

(1 − x ) 2

⇒

x3 x5 2 2 = 2 x + 2 .2 + 4 .2 .2 + ......... 3! 5! 2

sin −1 x

(1 − x ) 2

2 3 2.4.x 5 = x+ x + + ......... 3 3.5

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 Taylor’s theorem, we have x2 x3 f " ( h) + f " ' (h) + .... …….(1) f( h + x) = f (h) + xf’(h)+ 2! 3! Now f(h) = sin –1h.. Therefore f ’(h) =

f ’’ (h) =

1 1− h

2

= (1 − h2 )−1/ 2

h(1 − h2 )−3 / 2

f '''(h) = (1 − h2 )−3 / 2 + h(−3/ 2)(1 − h2 )−5 / 2 (−2h) = (1 − h2 )−3 / 2 + 3h2 (1 − h2 )−5 / 2 = (1 − h2 )−5 / 2 [(1 − h2 ) + 3h2 ) =(1 − h2 )−5 / 2 (1 + 2h2 ), etc.

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 Taylor’s theorem to prove that sin θ sin 2θ sin 3θ tan-1(x + h) = tan-1x +(hsin θ ) − (h sin θ ) 2 + (h sin θ )3 − ..... 1 2 3 sin nθ ….+(-1)n-1(hsin θ )n + ....., where 0 = cot-1x n Solution: Let f(x) = tan-1x. then

f (x + h) = tan-1 (x + h ). Expanding f (x + h) in power of h by Taylor’s theorem we have f ( x + h) = f (x) +

h h2 hn f '( x) + f ''( x) + ..... + f ''( x) + .... n! 1! 2!

…….(1)

Now f ( x ) = tan-1x. Therefore f ’’(x) = Dn tan-1x = (-1) n-1 (n-1)! sinn θ sin n θ . Where θ = cot-1x Putting n= 1, 2, 3, ……..in it , we get f ’(x)= sin θ sin θ , f ’’(x) = -1! sin2 θ sin2 θ f ’’’(x) = 2! Sin3 θ sin3 θ .etc. Substituting these values in (1), we have h2 h3 sin 2 θ sin 2θ + 2!sin 3 θ sin 3θ − ....... 2! 3! n h + (−1) n −1 (n − 1)!sin n θ sin nθ + .... n! sin θ sin 2 θ sin 3θ sin nθ = tan-1 x + hsin θ − ( h sin θ ) 2 + ( h sin θ )3 + ....... ..... + ( −1) n −1 ( h sin θ ) n 1 2 3 n tan-1 (x + y) = tan-1 x +hsin θ sin θ -

 π Example: Expand ex cosy near the point 1,  by Taylor’s Theorem.  4 Solution: By Taylor’s theorem 2

3

 ∂ 1 ∂ 1 ∂ ∂  ∂  ∂  F(x + h, y + k) = F (x, y) +  h + k  F +  h + k  F +  h + k  F + .....(1) 2!  ∂x 3!  ∂x ∂y  ∂y  ∂y   ∂x

Again ex cosy = F(x, y)  π π  π  π   = F 1 + h + k  =F 1 + ( x − 1). +  y − , where h = x -1, k = y 4 4  4  4   

17

F(x, y) = ex cosy

∂F = e x cos y ∂y ∂F = −e x sin y ∂x ∂2F = e x cos y 2 ∂x

∂2F = −e x cos y 2 ∂y

∂2F = −e x cos y ∂x∂y

e  π = 2  4 ∂F  π  e 1.  = ⇒ ∂y  4  2 2 e ∂ F π

⇒ F 1.

1.  = ∂x 2  4  2 2 e ∂ F π 1.  = ⇒ 2  ∂x  4  2 2 ∂ F  π  −e  1.  = ⇒ ∂y 2  4  2 ⇒

∂2F  π  − e ⇒ 1.  = ∂x∂y  4  2

Substituting these values in Taylor’s theorem, we get

e  e  π  − e + ( x − 1) +y−  4  2  2  2  2 1 π   −e   π   −e    2 e + 2( x − 1)  y −   +y−   + ( x − 1)  + .......... 2!  4   2   4   2   2 

ex cosy =

18

ADDITIONAL PROBLEMS:

1. Show that by Maclaurin’s theorem, n x 2 x3 x 4 n −1 x log(1 + x) = x − + − + ........... + (−1) 2 3 4 n!

2. By Maclaurin’s theorem, prove that

a 2 + b2 ) ( 3a 2 b − b3 3 2 ax e sin bx = bx + abx + x + .......... + 3! n!

n/2

b x n sin(n tan −1 ) + ... a

3. Apply Maclaurin’s theorem to obtain the term upto x 4 in the expansion of log(1 + sin 2 x) 4. If

y = ea sin

−1

x

, show that (1 − x 2 ) yn + 2 − (2n + 1) xyn +1 − ( n 2 + a 2 ) yn = 0

Hence by Maclaurin’s theorem show that

a 2 x 2 a(12 + a 2 ) 3 e = 1 + ax + + x + ........ 2! 3! 1 1 Also deduce that eθ = 1 + sin θ + sin 2 θ + sin 3 θ + ..... 2! 3! a sin −1 x

5. By Maclaurin’s theorem, show that

e

a cos −1 x

  a2π a 2 x 2 a(1 + a 2 ) 3 a(22 + a 2 ) 4 = 1 − ax + − x + x ........  e 2! 3! 4!  

19