1. Using Three point Forward difference, Backward difference, Central difference formula numerical differentiation to find solution
| x | 1 | 1.05 | 1.10 | 1.15 | 1.20 | 1.25 | 1.30 |
| f(x) | 1 | 1.02470 | 1.04881 | 1.07238 | 1.09545 | 1.11803 | 1.14018 |
`f^'(1.15) and f^('')(1.15)`
Solution:
The value of table for `x` and `y`
| x | 1 | 1.05 | 1.1 | 1.15 | 1.2 | 1.25 | 1.3 |
|---|
| y | 1 | 1.0247 | 1.0488 | 1.0724 | 1.0954 | 1.118 | 1.1402 |
|---|
Three-point FDF (Forward difference formula)
`f^'(x)=1/(2h)[-3f(x)+4f(x+h)-f(x+2h)]`
`f^'(1.15)=1/(2*0.05)[-3f(1.15)+4f(1.15+0.05)-f(1.15+2*0.05)]`
`f^'(1.15)=1/0.1[-3f(1.15)+4f(1.2)-f(1.25)]`
`f^'(1.15)=1/0.1[-3(1.0724)+4(1.0954)-1.118]`
`f^'(1.15)=0.4663`
Three-point BDF (Backward difference formula)
`f^'(x)=1/(2h)[f(x-2h)-4f(x-h)+3f(x)]`
`f^'(1.15)=1/(2*0.05)[f(1.15-2*0.05)-4f(1.15-0.05)+3f(1.15)]`
`f^'(1.15)=1/0.1[f(1.05)-4f(1.1)+3f(1.15)]`
`f^'(1.15)=1/0.1[1.0247-4(1.0488)+3(1.0724)]`
`f^'(1.15)=0.466`
Three-point CDF (Central difference formula)
`f^'(x)=(f(x+h)-f(x-h))/(2h)`
`f^'(1.15)=(f(1.15+0.05)-f(1.15-0.05))/(2*0.05)`
`f^'(1.15)=(f(1.2)-f(1.1))/0.1`
`f^'(1.15)=(1.0954-1.0488)/0.1`
`f^'(1.15)=0.4664`
Three-point FDF (Forward difference formula) for second derivatives
`f^('')(x)=(f(x)-2f(x+h)+f(x+2h))/(h^2)`
`f^('')(1.15)=(f(1.15)-2f(1.15+0.05)+f(1.15+2*0.05))/((0.05)^2)`
`f^('')(1.15)=(f(1.15)-2f(1.2)+f(1.25))/(0.0025)`
`f^('')(1.15)=(1.0724-2(1.0954)+1.118)/(0.0025)`
`f^('')(1.15)=-0.196`
Three-point BDF (Backward difference formula) for second derivatives
`f^('')(x)=(f(x-2h)-2f(x-h)+f(x))/(h^2)`
`f^('')(1.15)=(f(1.15-2*0.05)-2f(1.15-0.05)+f(1.15))/((0.05)^2)`
`f^('')(1.15)=(f(1.05)-2f(1.1)+f(1.15))/(0.0025)`
`f^('')(1.15)=(1.0247-2(1.0488)+1.0724)/(0.0025)`
`f^('')(1.15)=-0.216`
Three-point CDF (Central difference formula) for second derivatives
`f^('')(x)=(f(x-h)-2f(x)+f(x+h))/(h^2)`
`f^('')(1.15)=(f(1.15-0.05)-2f(1.15)+f(1.15+0.05))/(0.05)^2`
`f^('')(1.15)=(f(1.1)-2f(1.15)+f(1.2))/(0.0025)`
`f^('')(1.15)=(1.0488-2(1.0724)+1.0954)/(0.0025)`
`f^('')(1.15)=-0.2`
This material is intended as a summary. Use your textbook for detail explanation.
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