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.

Expert's answer

**Newton’s forward:**

Equation is *f*(*x*)=*i*(*r*)=40sin(*x*)-10*x*2.

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**

**Δ***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*=3.8

*h*=*x*1-*x*0=2.1-2=0.1

*p*=*x*-*x*0*h*=3.8-20.1=18

Newton's forward difference interpolation formula is

*y*(*x*)=*y*0+*p*Δ*y*0

*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*=*x*1-*x*0=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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