`f(x)=sinx` and `h = 0.1`, estimate `f^'(0.8) and f^('')(0.8)`
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) = sin(x)`.
`:. f^'(x) = cos(x)`
`:. f^('')(x) = -sin(x)`
The value of table for `x` and `y`
| x | 0.5 | 0.6 | 0.7 | 0.8 | 0.9 | 1 | 1.1 |
|---|
| y | 0.4794 | 0.5646 | 0.6442 | 0.7174 | 0.7833 | 0.8415 | 0.8912 |
|---|
Four-point CDF (Central difference formula)
`f^'(x)=1/(12h)[f(x-2h)-8f(x-h)+8f(x+h)-f(x+2h)]`
`f^'(0.8)=1/(12*0.1)[f(0.8-2*0.1)-8f(0.8-0.1)+8f(0.8+0.1)-f(0.8+2*0.1)]`
`f^'(0.8)=1/1.2[f(0.6)-8f(0.7)+8f(0.9)-f(1)]`
`f^'(0.8)=1/1.2[0.5646-8(0.6442)+8(0.7833)-0.8415]`
`f^'(0.8)=0.6967`
Absolute Error:`|"exact value of " f^'(0.8)-(0.6967)|=|0.6967 -0.6967|=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^('')(0.8)=(2f(0.8)-5f(0.8+0.1)+4f(0.8+2*0.1)-f(0.8+3*0.1))/((0.1)^2)`
`f^('')(0.8)=(2f(0.8)-5f(0.9)+4f(1)-f(1.1))/(0.01)`
`f^('')(0.8)=(2(0.7174)-5(0.7833)+4(0.8415)-(0.8912))/(0.01)`
`f^('')(0.8)=-0.7246`
Absolute Error:`|"exact value of " f^('')(0.8)-(-0.7246)|=|-0.7174 +0.7246|=0.0072`
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^('')(0.8)=(-f(0.8-3*0.1)+4f(0.8-2*0.1)-5f(0.8-0.1)+2f(0.8))/((0.1)^2)`
`f^('')(0.8)=(-f(0.5)+4f(0.6)-5f(0.7)+2f(0.8))/(0.01)`
`f^('')(0.8)=(-0.4794+4(0.5646)-5(0.6442)+2(0.7174))/(0.01)`
`f^('')(0.8)=-0.7232`
Absolute Error:`|"exact value of " f^('')(0.8)-(-0.7232)|=|-0.7174 +0.7232|=0.0058`
This material is intended as a summary. Use your textbook for detail explanation.
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