Assignment No:2.          MTH 603

Question:1

Using difference operator formulas (Δ and ) the values given in the table below,

X         0.3       0.5       0.7       0.9       1.1       1.3

Y         3.9118 3.8234 3.6773 3.4807 3.2408 2.9648

Estimate the value of

(a)     y^' (0.3)                                                 (b)     y''(1.3)

Solution:-

(a)  y^' (0.3)

0.3 appears at the beginning of the table then forward difference table is more appropriate for first derivative, so we have to calculate

Forward difference table

X             Y           ∆y      ∆^2y                ∆^3y                 ∆^4y                 ∆^5y

0.3       3.9118

0.5       3.8234 -0.0884

0.7       3.6773 -0.1461            -0.0577

0.9       3.4807 -0.1966            -0.0505            0.0072

1.1       3.2408 -0.2399            -0.0433            0.0072 0

1.3       2.9648 -0.276  -0.0361            0.0072 0          0

From the forward difference table

h = 0.2

Using the forward difference formula for (x) , we have

Df(x)   =  1/h  [∆y -(∆^2 y)/2  +(∆^3 y)/3-(∆^4 y)/4  +(∆^5 y)/5]

Putting the values

y^' (0.3)=1/0.2 [-0.0884 + 0.0577/2  +0.0072/3-0/4+0/5]

y^' (0.3)=1/0.2 [-0.0884 + 0.02885 +0.0024-0+0]

y^' (0.3)=1/0.2 [-0.05715]

y^' (0.3)= -0.28575

(b)     y''(1.3)

1.3 occurs at the end of the table it is appropriate to use backward difference table for derivative, so we have to calculate

Backward difference table

X               Y        y   ^2 y       ^3y                ^4y                ^5y

0.3       3.9118 -0.0884            -0.0577            0.0072    0          0

0.5       3.8234 -0.1461            -0.0505            0.0072    0

0.7       3.6773 -0.1966            -0.0433            0.0072

0.9       3.4807 -0.2399            -0.0361

1.1       3.2408 -0.276

1.3       2.9648

From the backward difference table

h = 0.2

using the backward difference formula for y''(x)

D^2 f(x)= y''(x) =  1/h^2   [∆^2 y +∆^3 y +11/12 ∆^4 y +5/6 ∆^5 y]

Putting the values

y^'' (1.3)=1/(0.2)^2  [-0.0361+0.0072+11/12 (0)+5/6  (0)]

y^'' (1.3)=1/0.04[ -0.0289]

y^'' (1.3)= -0.7225