A textbook of Computer Based Numerical and Statiscal Techniques part 16 ppsx

Horizontal lines between these values point out the incorrect functional value 46363 coefficient of first middle term in 1 – p5 is +10.. 10000 − = Let, given a set of equidistant values

Sol The difference table for the given table is:

−

−

−

−

−

−

5191

9529

Here, sum of fifth differences is small which may be neglected 0.270 and –270 are the adjacent values which are equal in magnitude and opposite in sign Horizontal lines between these values point out the incorrect functional value 46363 coefficient of first middle term in

(1 – p)5 is +10

∴ Error is given by 10e = 270 ⇒ e = 27

Hence, correct functional value 46363 27 4.6336

10000

− =

Let, given a set of equidistant values of arguments and its corresponding value of f(x) Suppose for n + 1 equidistant argument values x = a, a + h, a + 2h, , a + nh, are given.

y= f(x) = f(x0), f(x1), f(x2) , f(a + nh) i.e., f(x n).

Let one of the value of f(x) is missing Say it f(i) To determine this missing value of f(x), assume that f(x) can be represented by a polynomial of degree (n – 1) since n values of f(x) are

known

Hence,∆n−1f x( ) = constant and ∆n f x( ) 0=

Therefore, (E – I) n f(x) = 0 because ∆ = E – 1

[E n− n C E n− I+n C E n− I− ( 1)+ − n E n n− I n] ( ) 0f x =

( ) ( ) ( ) ( 1) ( ) 0

E f x − C E − f x + C E − f x − + − f x =

For first tabulated value of x, put x = 0

(0) (0) (0) ( 1) (0) 0

2

or ( ) ( 1) ( 1) ( 2) ( 1) (0) 0

2

n

n n

f n −nf n− + − f n− − + − f =

(1)

In equation (1), except missing term, each term is known and hence from this way missing term can be obtained

If two values of f(x) are mssing then in that case only (n – 1) values of f(x) can be given by

a polynomial of degree (n –2) i.e.,∆n−1f x( ) 0= or (E –1) n–1 f(x) = 0.

This gives for x = 0, (the first tabulated value) and for x = 1, (second tabulated value) and

by solving these two we get the two missing values for given function f(x) Similarly method proceeds to find three and more missing terms in given function f(x).

Example 6 Estimate the missing term in the following table:

=

x

y f x Explain why values differ from 3 3 or 27.

Sol Since we have given 4 values, therefore

4f x( ) 0, x

i.e., (E4−4E3+6E2−4E+1) ( ) 0,f x = ∀x

i.e., E f x4 ( ) 4− E f x3 ( ) 6+ E f x2 ( ) 4− Ef x( )+f x( ) 0, = ∀x

i.e., f x( + −4) 4 (f x+ +3) 6 (f x+ −2) 4 (f x+ +1) f x( ) 0, = ∀x

(on taking interval of differencing being 1)

On putting x = 0, we get

(4) 4 (3) 6 (2) 4 (1) (0) 0

Substituting the value of f(0), f(1), f(2), f(4) in (1), we get

81 4 (3) 6 9 4 3 1 0− f + × − × + =

i.e., 4f(3) = 124

(function values are 3n type and this is not a polynomial)

Example 7 Find the missing value of the data:

x

f x

Sol Since 4 values are known, let us assume the fourth order differences being zero Also since one value is unknown, we assume

4f x( ) 0, x

i.e (E4−4E3+6E2−4E+1) ( ) 0,f x = ∀x

i.e., E f x4 ( ) 4− E f x3 ( ) 6+ E f x2 ( ) 4− Ef x( )+ f x( ) 0, = ∀x

i.e., f x( + −4) 4 (f x+ +3) 6 (f x+ −2) 4 (f x+ +1) f x( )= ∀0, x

(on taking interval of differencing being 1)

On putting x = 0, we get

(4) 4 (3) 6 (2) 4 (1) (0) 0

Substituting the value of f(0), f(1), f(2), f(4) in (1), we get

37 4(21) 6(13) 4 (1) 7− + − f + =0

38 4 (1) 0f f(1) 9.5 Hence, the required missing value is 9.5

Example 8 Find the missing values in the table:

x

f x

Sol Difference table is as follows:

−

−

+ −

− −

− −

−

1

1 3

3

3

2.4

y

y y

y

as only three entries y0, y2, y4 are given, the function y can be represented by a second degree

polynomial

On solving these, we get

y1 = 2.925, y2 = 0.225

Example 9 Obtain the missing terms in the following table:

Sol Here we have six known values, therefore sixth differences being zero

i.e., ∆6 ( ) 0f x = For all values of x

i.e., (E6−6E5+15E4−20E3+15E2−6E+1) ( ) 0,f x = ∀x

i.e., E f x6 ( ) 6− E f x5 ( ) 15+ E f x4 ( ) 20− E f x3 ( ) 15+ E f x2 ( ) 6 ( )− Ef x + f x( ) 0,= ∀x i.e., f x( + −6) 6 (f x+ +5) 15 (f x+ −4) 20 (f x+ +3) 15 (f x+ −2) 6 (f x+ +1) f x( ) 0, = ∀x (1)

On putting x = 1 and x = 2 in equation (1), we get

(7) 6 (6) 15 (5) 20 (4) 15 (3) 6 (2) (1) 0

Putting the value of f(8), f(7), f(6), f(4), f(2), f(1) in equation (1) and (2), we get

343 – 6 × 216 + 15f(5) – 20 × 64 + 15f(3) – 6 × 8 + 1 = 0 Also 512 – 6 × 343 +15 × 216 – 20f (5) + 15 × 64 – 6f (3) + 8 = 0

On solving these two, we get f(3) = 27 and f(5) = 125.

Example 10 Assuming that the following values of y belong to a polynomial of degree 4, compute

the next three values:

Sol For the given data, difference table is:

−

−

−

−

−

−

∆

∆

∆

∆

∆

3 2 2

3

3

2

3

2

6 7

2

2

6

7

y y

y y

Since values of y belong to a polynomial of degree 4, fourth difference must be constant.

Therefore other fourth order differences will be 16

1 16

y

2 1 16

2 24

y

3 2 24

3 28

y

4 3 28

∆ − ∆ =

5 4 30

y −y =

Again, ∆4y2 = 16 then after solving, we get y6 = 129 and ∆4y3 = 16 gives y7 = 351

PROBLEM SET 3.2

1 Locate the error in the following table and correct them?

( ) 0.112046 0.120204 0.128350 0.136462 0.144600 0.152702 0.160788 0.168857 0.176908

x

f x

[Ans. f(3.63) = 0.136482]

2 Locate the error in the following: –1, 0, 7, 26, 65, 124, 215, 342, 511

[Ans correct value = 63, Error = 2]

3 Obtain the missing term in the following table:

x

f x

[Ans f(2.1) = 0.123, f(2.4) = 0.0900]

4 Estimate the production for the year 1964 and 1966 from the following data:

Year

Production

[Ans f(1964) = 306 f(1966 = 390)]

5 Given, log 100 = 2, log 101 = 2.0043, log 103 = 2.0128, log 104 = 2.0170 find log 102

[Ans log 102 = 2.0086]

6 Estimate the missing term in the following:

x y

7 Find the first term of the series whose second and subsequent terms are 8, 3, 0, –1, 0

[Ans First term is 15.]

8 Obtain the missing term in the following table:

x

f(x)

[Ans f(0.1)=0.123, f(0.4) = 0.090]

9 Evaluate the production of wool in the year 1935 from the given data:

Year x

Production (y)

[Ans 6.6]

The relation E = 1 + ∆ ⇒ E n = (1 + ∆)n has been used to express E n y x in terms of y x and its differences (1+∆)n has been expanded by binomial theorem without using y x in it Such methods

of operations are known as method of separation of symbols Point to be noted that the operations

on symbols has no meaning without operand y x i.e.,

yx+nh = E n y x = (1 + ∆)n y x

= (1 + n C1∆ + n C2∆2 + )y x

= y x + n C1∆yx + n C2∆2y x+

This type of operation in which we separate the operand from operator is called separation

of symbols

Example 1 Show that ∆r y k = ∇r y k+r

Sol We know ∇ = 1 – E–1

Therefore ∇r y k+r = (1 – E–1)r y k+r

r

E E

−

  y k+r

= (E – 1) r (E –r y k+r)

= (E – 1) r (E –r y k+r )

Example 2 Show that −

=

∆

∑

n 1 2

k 0

f k = ∆ f n – ∆f 0

Sol

1 2 0

n k

−

=

∆

∑ f k = 1( )2

0 1

n k E

−

=

−

∑ f k

2 0

n k

−

=

− +

∑ f k = 1( 2 1 )

0

2

n

k

−

=

∑

= f2 – 2 f1 + f0 + f3 – 2 f2 + f1 + f4 – 2 f3 + f2 + f5 – 2f4 + f3

+ f n–1 – 2f n–2 + f n–3

+ f n – 2f n–1 + f n–2 + f n+1 – 2f n + f n–1

= f n+1 – f n + f 0 – f1, on adding and canceling the diagonal terms

=(f n+1 – f n ) – (f 1 – f0)

= ∆f n – ∆f0.

Example 3 Prove that +1

x 2

y = 1

2 (y x + y x+1 ) – 1

16 (

2

x

y

∆ + ∆2y x+1 ); Assuming that, ∆3y x = 0.

2

x

y

+ = E1/2 y x = (1 + ∆)1/2 y x = 1+ ∆ − ∆12 18 2

x

y

1

1

x

y +

x

y

∆ and ∆y x = y x+1 – y x

Therefore from equation (1), we have

1 2

x y

1

2

y + y + −y –

1 1

+

2 y x+y x+ −16 ∆ y x+ ∆ y x+ .

Example 4 Using the method of separation of symbol, show that

∆

u u u u u u u u

Sol On taking R H S of given identity

0

− ∆ + ∆ − ∆ +

     

=

1

1 1

2

−

(2+ ∆)− u = +(1 E)− u

= (1− +E E2−E3+ )u0

= u0−u1+u2−u3+ Hence proved

Example 5 Prove by the method of separation of symbols, that

2

u

Sol L.H.S of given identity

u

u + x+ x + x + x +

=

u + Eu + E u + E u +

=

0

e u =e +∆u =e ⋅e u∆

=

0

+ ∆ + ∆ + ∆ +

=

= R.H.S

Example 6 Prove that: u x = u x–1 + ∆u x−2 + ∆2u x−3+ + ∆n−1u x n− + ∆n u x n−

Sol To prove u x = u x–1 + ∆ux–2 + ∆2u x−3 + + ∆n−1u x n− + ∆n u x n− , shift the last term of right hand side and then solve left hand side

u x – n

x n

u −

∆ = (1 – ∆n E –n )u x

= 1

n

E

  ∆ 

−

   

  u x =

1

n

E (E

n –∆n ) u x = 1n

E

n n

E E

 − ∆ 

 − ∆ 

 u x 3 1 + ∆ = E ⇒ E – ∆ = 1

= 1n

E [E

n–1 + ∆E n−2 + ∆2E n−3+ +∆n− 1]u x

= (E–1 + ∆E−2 + ∆2E−3 + +∆n−1E−n )u x

= u x–1 + ∆u x−2 + ∆2u x−3+ +∆n−1u x n−

Example 7 Show that: u 2n –n C 1 2u 2n–1 + n C 2 2 2 u 2n–2 – + (–2) n u n = (–1) n (c – 2an)

Where u n = an 2 + bn + c

Sol L.H.S = u 2n –n C12u 2n–1 + n C2 22 u 2n–2 – + (–2)n u n

= E n u n –n C 1 ·2E n–1 u n + n C 2 2 2 E n–2 u n – + (–2) n u n

= (E n–n n C 1 2E n–1 + n C 2 2 2 E n–2 – +(–2) n u n

= (E – 2) n u n = (∆ – 1)n u n = (–1)n (1 – ∆)n u n

= (–1) n

( 1) 2 1

n n

 − ∆ + ∆ 

(On neglecting higher order differences as u n is a polynomial of second degree)

1

2

n

n n

1

2

1

2

= ( ) ( 2 ) { ( )2 2} ( )

1n  an bn c n a n 1 an bn n 1 n

2 2 1 2

n n

−



2

2

= ( )−1n (an2 +bn+c)−n(2an+ + +a b) a n( 2 −n)

= (–1)n [c –2an] = R.H.S.

Example 8 Using the method of separation of symbols, show that:

n

x n

u −

∆ = u x – nu x–1 + n n 1( − )

2 u x–2 – + (–1)

n u x–n

Sol R.H.S = u x – nu x–1 + ( 1)

2

n n−

u x–2 – + (–1)n u x–n

( 1) 2

n n

n n

= 1 1 ( 2 1) 2 ( 1) n x x

n n

= (1−E−1)n u x

E

 −  =  =

= ∆n E u−n x = ∆n u x−1=L.H.S

Example 9 Use the method of separation of symbols to prove the following identities:

1. + x ∆ + x ∆ + = + x ∆2 − +x ∆4 − +

0 1 1 2 2 x 1 x 1 2 x 2

2 u x−u x 1+ +u x 2+ −u x 3+ +

x (1/ 2) x (3 / 2) x ( 5 / 2) x (7 / 2)

3 u 0 + n C 1 u 1 x + n C 2 u 2 x 2 + n C 3 u 3 x 3 + = (1 + x) n u 0 + n C 1 (1 + x) n–1 x∆u 0

+ n C 2 (1 + x) n–2 x 2∆2 u 0 +

Sol

= + ∆2 −1 + ∆4 −2 +

= + ∆2 −1+ ∆4 −2+

[1 x C E x C E ]u x