Using Four 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 |
|---|
Four-point CDF (Central difference formula)
`f^'(x)=1/(12h)[f(x-2h)-8f(x-h)+8f(x+h)-f(x+2h)]`
`f^'(1.15)=1/(12*0.05)[f(1.15-2*0.05)-8f(1.15-0.05)+8f(1.15+0.05)-f(1.15+2*0.05)]`
`f^'(1.15)=1/0.6[f(1.05)-8f(1.1)+8f(1.2)-f(1.25)]`
`f^'(1.15)=1/0.6[1.0247-8(1.0488)+8(1.0954)-1.118]`
`f^'(1.15)=0.4663`
Four-point FDF (Forward difference formula) for second derivatives
`f^('')(x)=(2f(x)-5f(x+h)+4f(x+2h)-f(x+3h))/(h^2)`
`f^('')(1.15)=(2f(1.15)-5f(1.15+0.05)+4f(1.15+2*0.05)-f(1.15+3*0.05))/((0.05)^2)`
`f^('')(1.15)=(2f(1.15)-5f(1.2)+4f(1.25)-f(1.3))/(0.0025)`
`f^('')(1.15)=(2(1.0724)-5(1.0954)+4(1.118)-(1.1402))/(0.0025)`
`f^('')(1.15)=-0.22`
Four-point BDF (Backward difference formula) for second derivatives
`f^('')(x)=(-f(x-3h)+4f(x-2h)-5f(x-h)+2f(x))/(h^2)`
`f^('')(1.15)=(-f(1.15-3*0.05)+4f(1.15-2*0.05)-5f(1.15-0.05)+2f(1.15))/((0.05)^2)`
`f^('')(1.15)=(-f(1)+4f(1.05)-5f(1.1)+2f(1.15))/(0.0025)`
`f^('')(1.15)=(-1+4(1.0247)-5(1.0488)+2(1.0724))/(0.0025)`
`f^('')(1.15)=-0.196`
This material is intended as a summary. Use your textbook for detail explanation.
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