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
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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
