1. Formula & Example-1 (table data)
Formula
1. For `x=x_0`
`[(dy)/(dx)]_(x=x_0) = 1/h * ( Delta Y_0 - 1/2 * Delta^2 Y_0 + 1/3 * Delta^3 Y_0 - 1/4 * Delta^4 Y_0 + ...)`
`[(d^2y)/(dx^2)]_(x=x_0) = 1/h^2 * ( Delta^2 Y_0 - Delta^3 Y_0 + 11/12 * Delta^4 Y_0 + ...)`
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2. For any value of `x`
`[(dy)/(dx)] = 1/h * ( Delta Y_0 + (2t-1)/(2!) * Delta^2 Y_0 + (3t^2-6t+2)/(3!) * Delta^3 Y_0 + (4t^3-18t^2+22t-6)/(4!) * Delta^4 Y_0 + ...)`
`[(d^2y)/(dx^2)] = 1/h^2 * ( Delta^2 Y_0 + (t-1) * Delta^3 Y_0 + (12t^2-36t+22)/24 * Delta^4 Y_0 + ...)`
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Examples
1. Using Newton's forward/backward differentiation method to find solution
| x | f(x) | | 0.0 | 1.0000 | | 0.1 | 0.9975 | | 0.2 | 0.9900 | | 0.3 | 0.9776 | | 0.4 | 0.8604 |
x = 0
Solution:
Numerical differentiation method to find solution.
The value of table for `X` and `Y`
| X | 0 `x_0=0` | 0.1 `x_1=0.1` | 0.2 `x_2=0.2` | 0.3 `x_3=0.3` | 0.4 `x_4=0.4` |
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| Y | 1 `y_0=1` | 0.9975 `y_1=0.9975` | 0.99 `y_2=0.99` | 0.9776 `y_3=0.9776` | 0.8604 `y_4=0.8604` |
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Newton's forward differentiation table is as follows.
| X | Y(X) | `DeltaY` | `Delta^2Y` | `Delta^3Y` | `Delta^4Y` | | 0 `X_0=0` | 1 `Y_0=1` | | | | | | | -0.0025 `-0.0025=0.9975-1`
`DeltaY_0=Y_1-Y_0` | | | | | 0.1 `X_1=0.1` | 0.9975 `Y_1=0.9975` | | -0.005 `-0.005=-0.0075--0.0025`
`Delta^2Y_0=DeltaY_1-DeltaY_0` | | | | | -0.0075 `-0.0075=0.99-0.9975`
`DeltaY_1=Y_2-Y_1` | | 0.0001 `0.0001=-0.0049--0.005`
`Delta^3Y_0=Delta^2Y_1-Delta^2Y_0` | | | 0.2 `X_2=0.2` | 0.99 `Y_2=0.99` | | -0.0049 `-0.0049=-0.0124--0.0075`
`Delta^2Y_1=DeltaY_2-DeltaY_1` | | -0.1 `-0.1=-0.0999-0.0001`
`Delta^4Y_0=Delta^3Y_1-Delta^3Y_0` | | | -0.0124 `-0.0124=0.9776-0.99`
`DeltaY_2=Y_3-Y_2` | | -0.0999 `-0.0999=-0.1048--0.0049`
`Delta^3Y_1=Delta^2Y_2-Delta^2Y_1` | | | 0.3 `X_3=0.3` | 0.9776 `Y_3=0.9776` | | -0.1048 `-0.1048=-0.1172--0.0124`
`Delta^2Y_2=DeltaY_3-DeltaY_2` | | | | | -0.1172 `-0.1172=0.8604-0.9776`
`DeltaY_3=Y_4-Y_3` | | | | | 0.4 `X_4=0.4` | 0.8604 `Y_4=0.8604` | | | | |
The value of `x` at you want to find `f(x) : x_0 = 0`
`h = x_1 - x_0 = 0.1 - 0 = 0.1`
`[(dy)/(dx)]_(x=x_0) = 1/h * ( Delta Y_0 - 1/2 * Delta^2 Y_0 + 1/3 * Delta^3 Y_0 - 1/4 * Delta^4 Y_0)`
`:.[(dy)/(dx)]_(x=0) = 1/0.1 * (-0.0025 - 1/2 xx -0.005 + 1/3 xx 0.0001 - 1/4 xx -0.1)`
`:.[(dy)/(dx)]_(x=0) = 0.25033`
`[(d^2y)/(dx^2)]_(x=x_0) = 1/h^2 * ( Delta^2 Y_0 - Delta^3 Y_0 + 11/12 * Delta^4 Y_0 )`
`:.[(d^2y)/(dx^2)]_(x=0) = 1/0.01 * (-0.005 - 0.0001 + 11/12 xx -0.1)`
`:.[(d^2y)/(dx^2)]_(x=0) = -9.67667`
Solution for `Pn'(0) = 0.25033`
Solution for `Pn''(0) = -9.67667`
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then
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