`f(x)=2x^3+x^2-4` and `h = 0.5`, estimate `f^'(2.5) and f^('')(2.5)`
using Four point Forward difference, Backward difference, Central difference formula numerical differentiation
Also find exact value of f', f'' and error for each estimation
Solution:
Equation is `f(x) = 2x^3+x^2-4`.
`:. f^'(x) = 6x^2+2x`
`:. f^('')(x) = 12x+2`
The value of table for `x` and `y`
| x | 1 | 1.5 | 2 | 2.5 | 3 | 3.5 | 4 |
|---|
| y | -1 | 5 | 16 | 33.5 | 59 | 94 | 140 |
|---|
Four-point CDF (Central difference formula)
`f^'(x)=1/(12h)[f(x-2h)-8f(x-h)+8f(x+h)-f(x+2h)]`
`f^'(2.5)=1/(12*0.5)[f(2.5-2*0.5)-8f(2.5-0.5)+8f(2.5+0.5)-f(2.5+2*0.5)]`
`f^'(2.5)=1/6[f(1.5)-8f(2)+8f(3)-f(3.5)]`
`f^'(2.5)=1/6[5-8(16)+8(59)-94]`
`f^'(2.5)=42.5`
Absolute Error:`|"exact value of " f^'(2.5)-(42.5)|=|42.5 -42.5|=0`
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^('')(2.5)=(2f(2.5)-5f(2.5+0.5)+4f(2.5+2*0.5)-f(2.5+3*0.5))/((0.5)^2)`
`f^('')(2.5)=(2f(2.5)-5f(3)+4f(3.5)-f(4))/(0.25)`
`f^('')(2.5)=(2(33.5)-5(59)+4(94)-(140))/(0.25)`
`f^('')(2.5)=32`
Absolute Error:`|"exact value of " f^('')(2.5)-(32)|=|32 -32|=0`
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^('')(2.5)=(-f(2.5-3*0.5)+4f(2.5-2*0.5)-5f(2.5-0.5)+2f(2.5))/((0.5)^2)`
`f^('')(2.5)=(-f(1)+4f(1.5)-5f(2)+2f(2.5))/(0.25)`
`f^('')(2.5)=(--1+4(5)-5(16)+2(33.5))/(0.25)`
`f^('')(2.5)=32`
Absolute Error:`|"exact value of " f^('')(2.5)-(32)|=|32 -32|=0`
This material is intended as a summary. Use your textbook for detail explanation.
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