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Sulaiman Khan
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Gauss Forward Interpolation Method

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

P(x)=yp=y0+p⋅Δy0+p(p−1)2!⋅Δ2y−1+(p+1)p(p−1)3!⋅Δ3y−1+(p+1)p(p−1)(p−2)4!⋅Δ4y−2+⋯ 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:

  • ypy_p is the interpolated value at xpx_p
  • y0y_0 is the initial value corresponding to x0 x_0
  • Δny \Delta^n y are the forward differences of the function values.
  • p=x−x0h,hp = \frac{x-x_0}{h}, h is the uniform difference between the xx values.
  • The termsp,(p−1),(p+1)…p,(p-1),(p+1) \dots are the factors involving the relative position of 𝑥 with respect to the data points.

Example of Gauss Forward Interpolation

Given the following data points:

xy
2.524.145
322.043
3.520.225
418.644
4.517.262
516.047

We are tasked with finding ypy_p where x=3.75x = 3.75.

Step 1: taking the closest to x as x0=3.5, x \text{ as } x_0 = 3.5 , h=x2−x1 h = x_2 - x_1

Step 2: Use the formula p=x−x0h p = \frac{x - x_0}{h}.

p=3.75−3.50.5=0.5 p = \frac{3.75 - 3.5}{0.5} = 0.5

Step 3: Calculate the central differences for the yy values.

xpyΔy\Delta yΔ2y\Delta^2 yΔ3y\Delta^3 yΔ4y\Delta^4 yΔ5y\Delta^5 y
2.5-124.145-2.102
3-0.522.043-1.8180.284-0.047
3.5020.225-1.5810.237-0.0380.009-0.003
40.518.644-1.3820.199-0.0320.006
4.5117.262-1.2150.167
51.516.047

Step 4: Apply the Gauss Forward Interpolation formula:

P(x)=yp=y0+p⋅Δy0+p(p−1)2!⋅Δ2y−1+(p+1)p(p−1)3!⋅Δ3y−1+(p+1)p(p−1)(p−2)4!⋅Δ4y−2+⋯ 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

Step 5: Substituting the values into the interpolation formula:

P(3.75)=20.225+(0.5)⋅−1.5810+(0.5∗(0.5−1))2!⋅0.2370+((0.5+1)∗0.5∗(0.5−1))3!⋅−0.0380+((0.5+1)∗0.5∗(0.5−1)∗(0.5−2))4!⋅0.0090+((0.5+1)∗(0.5+2)∗0.5∗(0.5−1)∗(0.5−2))5!⋅−0.0030P(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 + \frac{((0.5+1)∗0.5∗(0.5−1)∗(0.5−2))}{4!} \cdot 0.0090 + \frac{((0.5+1)∗(0.5+2)∗0.5∗(0.5−1)∗(0.5−2))}{5!} \cdot -0.0030

Step 6: After solving the equation, the interpolated value of yy at x=3.75x = 3.75 is approximately:

P(3.75)≈19.4074P(3.75) \approx 19.4074

This is the final result of the Gauss Forward Interpolation method.

Gauss Forward Interpolation Calculator

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