Gauss Backward Interpolation Method

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

P=y0+pΔy1+(p+1)p2!Δ2y1+(p+1)p(p1)3!Δ3y2+(p+2)(p+1)p(p1)4!Δ4y2+ P = y_0 + p \Delta y_{-1} + \frac{(p+1)p}{2!} \cdot \Delta^2 y_{-1} + \frac{(p+1)p(p-1)}{3!} \cdot \Delta^3 y_{-2} + \frac{(p+2)(p+1)p(p-1)}{4!} \cdot \Delta^4 y_{-2} +

Where:

  • ypy_p is the interpolated value at xpx_p
  • y0y_0 is the initial value corresponding to x0 x_0 its the First value of xpx_p greater than xx given, to interpolate.
  • Δny \Delta^n y are the Backward differences of the function values taken from the central difference table.
  • p=xx0h,hp = \frac{x-x_0}{h}, h is the uniform difference between the xx values.
  • The termsp,(p1),(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 Backward Interpolation

Given the following data points:

xy
193115
194120
195127
196139
197152

We are tasked with finding ypy_p where x=1946x = 1946.

Step 1: taking the closest to x as x0=1951, x \text{ as } x_0 = 1951 , h=x2x1=19311941 h = x_2 - x_1 = 1931 - 1941

Step 2: Use the formula p=xx0h p = \frac{x - x_0}{h}.

p=1946195110=0.5 p = \frac{1946 - 1951}{10} = -0.5

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

xpyΔy\Delta yΔ2y\Delta^2 yΔ3y\Delta^3 yΔ4y\Delta^4 y
1931-2015
1941-102052
19510277537
19611039121-4
1971205213

Step 4: Apply the Gauss Backward Interpolation formula:

P=y0+pΔy1+(p+1)p2!Δ2y1+(p+1)p(p1)3!Δ3y2+(p+2)(p+1)p(p1)4!Δ4y2+ P = y_0 + p \Delta y_{-1} + \frac{(p+1)p}{2!} \cdot \Delta^2 y_{-1} + \frac{(p+1)p(p-1)}{3!} \cdot \Delta^3 y_{-2} + \frac{(p+2)(p+1)p(p-1)}{4!} \cdot \Delta^4 y_{-2} +\cdots

Step 5: Substituting the values into the interpolation formula:

P(1946)=27+(0.5)7+(0.5+1)(0.5)2!5+(0.5+1)(0.5)(0.51)3!3+(0.5+2)(0.5+1)(0.5)(0.51)4!7P(1946) = 27 + (-0.5) \cdot 7 + \frac{(-0.5+1)∗(-0.5)}{2!} \cdot 5 + \frac{(-0.5+1)∗(-0.5)∗(-0.5−1)}{3!} \cdot 3 + \frac{(-0.5+2)(0.5+1)∗(-0.5)∗(0.5−1)}{4!} \cdot -7

Step 6: After solving the equation, the interpolated value of yy at x=1946x = 1946 is approximately:

P(1946)22.8984375P(1946) \approx 22.8984375

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

Gauss Backward Interpolation Calculator

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