Gauss Forward Interpolation Method

Gauss Forward Interpolation is used to interpolate a value close to the beginning or middle of the data set. The method uses central differences to create an interpolation polynomial.

Formula for Gauss Forward Interpolation:

P(x)=y_p = y_0 + p \cdot \Delta y_0 + \frac{p(p-1)}{2!} \cdot \Delta^2 y_{-1} + \frac{(p+1) p (p-1)}{3!} \cdot \Delta^3 y_{-1} + \frac{(p+1) p (p-1)(p-2)}{4!} \cdot \Delta^4 y_{-2} + \cdots

Where:

$y_p$ is the interpolated value at $x_p$
$y_0$ is the initial value corresponding to $x_0$
$\Delta^n y$ are the forward differences of the function values.
$p = \frac{x-x_0}{h}$ , where $h$ is the uniform difference between $x$ values.

Example of Gauss Forward Interpolation

Given the following data points:

$x$	$y$
2.5	24.145
3	22.043
3.5	20.225
4	18.644
4.5	17.262
5	16.047

We are tasked with finding $y_p$ where $x = 3.75$ .

Step 1: Choose Initial Point

Take the point closest to $x$ as $x_0 = 3.5$ , with $h = 0.5$ .

Step 2: Calculate p

p = \frac{3.75 - 3.5}{0.5} = 0.5

Step 3: Calculate Central Differences

$x$	$p$	$y$	$\Delta y$	$\Delta^2 y$	$\Delta^3 y$	$\Delta^4 y$	$\Delta^5 y$
2.5	-1	24.145	-2.102
3	-0.5	22.043	-1.818	0.284	-0.047
3.5	0	20.225	-1.581	0.237	-0.038	0.009	-0.003
4	0.5	18.644	-1.382	0.199	-0.032	0.006
4.5	1	17.262	-1.215	0.167
5	1.5	16.047

Step 4 & 5: Substitute into Formula

P(3.75) = 20.225 + (0.5) \cdot (-1.5810) + \frac{0.5(0.5-1)}{2!} \cdot 0.2370 + \frac{(0.5+1)(0.5)(0.5-1)}{3!} \cdot (-0.0380)

Conclusion

P(3.75) \approx 19.397

Therefore, using Gauss Forward Interpolation, the estimated value of $y$ at $x = 3.75$ is approximately 19.397.

Gauss Forward Interpolation Calculator

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Gauss Forward Interpolation Method

Gauss Forward Interpolation is used to interpolate a value close to the beginning or middle of the data set. The method uses central differences to create an interpolation polynomial.

Formula for Gauss Forward Interpolation:

P(x)=y_p = y_0 + p \cdot \Delta y_0 + \frac{p(p-1)}{2!} \cdot \Delta^2 y_{-1} + \frac{(p+1) p (p-1)}{3!} \cdot \Delta^3 y_{-1} + \frac{(p+1) p (p-1)(p-2)}{4!} \cdot \Delta^4 y_{-2} + \cdots

Where:

$y_p$ is the interpolated value at $x_p$
$y_0$ is the initial value corresponding to $x_0$
$\Delta^n y$ are the forward differences of the function values.
$p = \frac{x-x_0}{h}$ , where $h$ is the uniform difference between $x$ values.