# Answer to Question #229342 in Electrical Engineering for Alock kumar

Question #229342

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.

1
2021-08-26T01:04:57-0400

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+pyn

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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