In December 2024
Answer to Question #229342 in Electrical Engineering for Alock kumar
For the following function, find the numerical differentiation of first order using forward, backward and center finite divided-difference formula at a point, 1, =3.2s with step size, h =0.i s, where i= 72,
i(r) = 40sint -10t^2
Evaluate the values using the conventional differentiation formula also and hence comment which numerical method produce better result based on the percentage of true error.
Newton’s forward:
Equation is f(x)=i(r)=40sin(x)-10x2.
The value of table for x and y
x
2
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9
4
y
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
Newton's forward difference interpolation method to find solution
Newton's forward difference table is
x
y
Δy
2
0
0
2.1
0
0
2.2
0
0
2.3
0
0
2.4
0
0
2.5
0
0
2.6
0
0
2.7
0
0
2.8
0
0
2.9
0
0
3
0
0
3.1
0
0
3.2
0
0
3.3
0
0
3.4
0
0
3.5
0
0
3.6
0
0
3.7
0
0
3.8
0
0
3.9
0
0
4
0
The value of x at you want to find the f(x):x=3.8
h=x1-x0=2.1-2=0.1
p=x-x0h=3.8-20.1=18
Newton's forward difference interpolation formula is
y(x)=y0+pΔy0
y(3.8)=0+18×0
y(3.8)=0+0
y(3.8)=0
Therefore: Newton's method gives y(3.8)=0
Backwards:
The value of table for x and y
x
2
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9
4
y
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
Newton's backward difference interpolation method to find solution
Newton's backward difference table is
x
y
∇y
2
0
0
2.1
0
0
2.2
0
0
2.3
0
0
2.4
0
0
2.5
0
0
2.6
0
0
2.7
0
0
2.8
0
0
2.9
0
0
3
0
0
3.1
0
0
3.2
0
0
3.3
0
0
3.4
0
0
3.5
0
0
3.6
0
0
3.7
0
0
3.8
0
0
3.9
0
0
4
0
The value of x at you want to find the f(x):x=2.1
h=x1-x0=2.1-2=0.1
p=x-xnh=2.1-40.1=-19
Newton's backward difference formula is
y(x)=yn+p∇yn
y(2.1)=0+(-19)×0
y(2.1)=0+0
y(2.1)=0
Newton's backward interpolation method y(2.1)=0
Comment: The backward method produces a better percentage of true error
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