(Derivatives at mid or near mid)
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(y0 + y1)
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(2u - 1)
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+
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[u(u - 1)]
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(D2y-1
+ D2y0)
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| yu = |
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+
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X (Dy0) |
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X
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+
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2
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2
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2!
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2
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{u[u - (1/2)](u - 1)}
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[u(u2 - 1)(u -
2)]
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(D4y-2
+ D4y-1)
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X
(D3y-1)
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+ |
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X
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3!
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4!
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2
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{u[u - (1/2)](u2
- 1)(u - 2)}
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[u(u2 - 1)(u2
- 2)(u - 3)]
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(D6y-3
+ D6y-2)
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+ ... |
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X
(D5y-2)
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+
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X
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5!
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6!
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2
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Where u = (x - a)/h
Therefore, du/dx = 1/h
(d/dx) [y(u)] = {(d/du)(du/dx) [y(u)]} = {(d/du)[y(u)](1/h)}
= [y'(u)]/h
Differentiating w.r.t. u
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(2u - 1)
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(D2y-1
+ D2y0)
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[3u2 - 3u + (1/2)]
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+
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Dy0
+
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X
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+
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X (D3y-1) |
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2!
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2
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3!
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(4u3 - 6u2
- 2u + 2)
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(D4y-2
+ D4y-1)
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X
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+.... |
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4!
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2
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Q. 12. A portion of a table of sines is given
below: (June 2002, June 99)
| Angle in Radians |
Sine |
| 0.25 |
0.2474 |
| 0.26 |
0.2571 |
| 0.27 |
0.2667 |
| 0.28 |
0.2764 |
| 0.29 |
0.2860 |
Find the derivative of this function at x = 0.27.
Solution. Here, we have to find out the derivative
near the middle of the table. So, we use Bessel's
formula.
The difference table is shown below:
| u |
x |
f(x) |
D |
D2 |
D3 |
D4 |
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-2
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0.25
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0.2474
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0.0097
0.0096
0.0097
0.0096
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-0.0001
0.0001
-0.0001
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-0.0002
-0.0002
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0
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-1
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0.26
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0.2571
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0
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0.27
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0.2667
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1
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0.28
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0.2764
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2
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0.29
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0.2860
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a = 0.27, h = 0.01
u = (x - 0.27)/h
Therefore, du/dx = 1/h
(d/dx) [y(u)] = [y'(u)]/h .....(i)
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(2u - 1)
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(D2y-1
+ D2y0)
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[3u2 - 3u + (1/2)]
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+
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| y'(u) = |
Dy0
+
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X
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+
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X (D3y-1) |
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2!
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2
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3!
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(4u3 - 6u2
- 2u + 2)
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(D4y-2
+ D4y-1)
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X
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4!
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2
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Putting u = 0
y'(0) = Dy0
- [(1/4) X (D2y-1
+ D2y0)]
+ [(1/12) X (D3y-1)]
+ [(1/24) X (D4y-2
+ D4y-1)]
or y'(0) = 0.0097 - {(1/4) X [0.0001 + (-0.0001)]}
+ [(1/12) X (-0.0002)] + [(1/24) X 0]
or y'(0) = 0.0096
From equation (i)
(d/dx) [y(u)] = [y'(0)]/h
or y'(0.27) = (0.0096)/0.01
or y'(0.27) = 0.96
Q. 13. A portion of a table of a function
f(x) is given below: (June 2000)
|
x
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0.71
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0.72
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0.73 |
0.74 |
0.75 |
0.76 |
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f(x)
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0.2474
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0.2571
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0.2667 |
0.2764 |
0.2860 |
0.2910 |
Find the derivative of this function at x = 0.74.
Solution. The difference table is shown below:
|
u
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x
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f(x)
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D
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D2
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D3
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D4
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D5
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-3
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0.71
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0.2474
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0.0097
0.0096
0.0097
0.0096
0.0050
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-0.0001
0.0001
-0.0001
-0.0046
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0.0002
-0.0002
-0.0045
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-0.0004
-0.0043
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-0.0039
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-2
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0.72
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0.2571
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-1
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0.73
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0.2667
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0
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0.74
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0.2764
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1
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0.75
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0.2860
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2
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0.76
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0.2910
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Differentiating w.r.t. u and putting u = 0, we get
y'(0) = Dy0
- [(1/4) X (D2y-1
+ D2y0)]
+ [(1/12) X (D3y-1)]
+ [(1/24) X (D4y-2
+ D4y-1)]
- [(1/120) X (D5y-2)]
or y'(0) = 0.0096 - {(1/4) X [(-0.0001)
+ (-0.0046)]} + [(1/12) X (-0.0045)] + [(1/24) X (-0.0043)]
- [(1/120) X (-0.0039)]
or y'(0) = 0.0103
Here, h = 0.01
Therefore, y'(0.74) = (0.0103)/0.01 = 1.03
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