Question

tan 31 × 40 + x = tan 37.x
Given tan 31 = 0.6
And tan 37 = 0.75​

Answer 1

Answer:

Step-by-step explanation:

Let the function y  f(x) take the values y0

, y1

, , yn corresponding to

the values x0, x1, , xn of x. Let these values of x be equispaced such that

xi  x0  ih (i  0, 1, ). Assuming y(x) to be a polynomial of the nth degree

in x such that 0 01 1 () () , ,, . ( ) n n yx y yx y yx y    We can write

INTERPOLATION • 275

yx a a x x a x x x x a x x x x x x ( ) ( ) ( )( )( )    01 0 2 0 1 3 0 1 2 – ( – )( – ) – – –

– – 0 1 –1 ( )( ) ( – ) n n     ax x x x x x (1)

Putting x  x0

, x1

, , xn successively in (1), we get

0 01 0 11 0 2 0 12 0 22 0 2 1 y ay a ax x y a ax x ax x x x     , – , (–) ( ) ( )( ) – –

and so on.

From these, we find that 0 0 0 1 0 11 0 1 a y y y y a x x ah     , () – –

 1 0

1 a y h  

Also 1 2 1 12 1 22 0 2 1

2

12 0 2

( ) ( )( )

2

y y y ax x ax x x x

a h a hh y h a

     

   

   2

2 10 0 2 2

1 1

2 2! a yy y h h

    

Similarly 3

3 0 3

1

3! a y h

  and so on.

Substituting these values in (1), we obtain

2 3

00 0

0 0 01 012 2 3 ( ) ( ) ( )( ) ( )( )( ) 2! 3!

yy y yx y x x x x x x x x x x x x h h h

 

         

(2)

Now if it is required to evaluate y for x x0  ph, then

0 10 0 ( ) , ( ) ( 1) , x x ph x x x x x x ph h p h         

000 ( ) ( ) ( 1) ( 2) xx xx xx p hh p h         etc.

Hence, writting y(x) = y(x0

+ ph) = yp

, (2) becomes

2 3

00 0 0

( 1) ( 1)( 2)

2! 3! p

pp pp p y y py y y        

 

0

( 1) -1

3!

n pp p n

y       (3)

It is called Newton’s forward interpolation formula as (3) contains y0

and the forward differences of y0

Otherwise: Let the function y  f(x) take the values y0

, y1

, y2

, corresponding to the values x0

, x0  h, x0  2h,  of x. Suppose it is required to

evaluate f(x) for x  x0  ph, where p is any real number.

276 • NUMERICAL METHODS IN ENGINEERING AND SCIENCE

For any real number p, we have defined E such that

() ( ) p E f x f x ph  

0 00 ( ) ( ) (1 ) p p

p y f x ph E f x y      [ E  1  ]

2 3

0 0

( 1) ( 1)( 2) 1 2! 3!

pp pp p p yy

                (4)

[Using binomial theorem]

i.e., 2 3

00 0 0

( 1) ( 1)( 2)

2! 3! p

pp pp p y y py y y         

If y  f(x) is a polynomial of the nth degree, then n1

y0

and higher differences will be zero.

Hence (4) will become

 

2 3

00 0 0

0

( 1) ( 1)( 2)

2! 3!

( 1) 1

3!

p

n

pp pp p y y py y y

pp p n

y

        

   





Which is same as (3)

Obs. 1. This formula is used for interpolating the values of y

near the beginning of a set of tabulated values and extrapolating

values of y a little backward (i.e., to the left) of y0

.

Obs. 2. The first two terms of this formula give the linear interpolation while the first three terms give a parabolic interpolation and so on.