Tutorial–6, B. Tech. Sem III, 24 July, 2016
1. Prove the following operator relations:
2
(i). E = 1 + ∆ = (1 − 5)−1 , (ii) ∆ = µδ + δ2 ,
(x)
(iii). ∆ log f (x) = log(1 + ∆f
)
f (x)
(iv). ∆[f (x)g(x)] = f (x)∆g(x) + g(x + h)∆f (x),
(v). 45 = 54 =q4 − 5 = δ 2 ,
2
I
(Difference Operators and Interpolation)
ON
(vi). ∆ = 21 δ 2 + δ 1 + δ4 , (vii). δ 3 y 1 = y2 − 3y1 + 3y0 − y−1 ,
2
(viii). δ(fk gk ) = µfk δ(gk ) + µgk δ(fk ),
(ix). √2+∆
= √2−5
, (x). ∆ = ehD − 1, (xi). µδ = sinh(hD),
1−5
1+∆
(xii). e−hD = 1 − 5,
1
(xiii). f [x, y, z] = x + y + z, f = x3 , (xiv). f [a, b, c] = abc
, f = a1 ,
n
(xv). f [a, b, c, ...(n symbols)...l] = 4n!hf (a)
n
4n ( x1 ), (v).
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2. Find for (h = 1)
2 3
(i). 42 (abex ), (ii). 4Exx3 , (iii). 4 tan−1 (ax), (iv).
(vi). 4(x + cosx).
2 x
E e
3. Prove that: ex = 4E ex . 4
and
2 ex
f0 + xf1 +
x2
f
2! 2
+ ... = ex (f0 + x 4 f0 +
x2
2!
42
E
x3 ,
4 f0 + ...)
4. Let Pn (x) = (x − x0 )(x − x1 )(x − x2 ).......(x − xn−1 ), where xi = x0 + ih, i being integer;
n!
show that 4Pn = nhPn−1 , hence show that 4r Pn = (n−r)!
hr Pn−r , r = 1 (1) n − 1, and
n
n
4 Pn = n!h .
5. Find the third divided difference with argument 2, 4, 9, 10 of the function f (x) = x3 − 2x.
6. Form a divided difference table for f (x) = x4 + 6x2 + x − 2 for values of x = −3 (1) 3,
show that 5th order diferences are zero.
7. Given that f (0) = 8, f (1) = 68, and f (5) = 123 determine f (2); calculate the error also.
8. Tabulate sin x for x0 = 30 (2) 40 and interpolate sin 310 and sin 330 . Compare with exact
values.
9. Tabulate ex for x = 1.7 (0.1) 2.2 and interpolate at x = 1.71, 2.15.
10. Find log10 1152.5 and log10 1161.3 using the following data:
x
1150
1155
1160
1165
1170
1175
log10 (x) 3.06069 3.06258 3.06445 3.06632 3.06818 3.07003
and error in the result.
11. Use Lagrange’s formula to interpolate the values of f (5) from
x
1 2 3 4
7
f (x) 2 4 9 16 128
How much it deviates from 25 .?
1
1180
3.07188
12. A third degree polynomial passes through the points (0, −1), (1, 1), (2, 1), (3, −2). Find
the polynomial. Ans: − 16 x3 − 12 x2 + 83 x − 1
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13. From the following data, find the number of students who obtained less than 45 marks:
Marks
30 − 40 40 − 50 50 − 60 60 − 70 70 − 80
No. of students
31
42
51
35
31
Ans: approx. 48
ON
14. Find the form of f (x), given that f (0) = 8, f (1) = 11, f (4) = 68 and f (5) = 123 also
determine f (2). Ans: x3 − x2 + 3x + 8, 18
15. Given that f (0) = −18, f (1) = 0 = f (3) = f (6), f (5) = −248 and f (9) = 13104 find the
form of f (x) assuming it to be a polynomial of degree 5th. {Hint: f (x) = (x − 1)(x −
3)(x − 6)φ(x) , φ(x) is a polynomial of degree 2 with φ(0) = 1, φ(5) = 31, φ(9) = 91 }
Ans: x5 − 9x4 + 18x3 − x2 + 9x − 18
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16. Find log10 301 using Newton’s Divided difference interpolation formula from the data:
x
300
304
305
307
Ans: 2.4786
log10 (x) 2.4771 2.4829 2.4843 2.4871
√
17. The values of y = x are listed below:
x 4
6
7
10
y 2 2.449 2.646 3.162
Compute x corresponding to y = 2.5. Ans: 6.25148
2