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+pΔy−1+2!(p+1)p⋅Δ2y−1+3!(p+1)p(p−1)⋅Δ3y−2+4!(p+2)(p+1)p(p−1)⋅Δ4y−2+ Where:
- yp is the interpolated value at xp
- y0is the initial value corresponding to x0 its the First value of xp greater than x given, to interpolate.
- Δny are the Backward differences of the function values taken from the central difference table.
- p=hx−x0,h is the uniform difference between the x values.
- The termsp,(p−1),(p+1)… 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:
x | y |
---|
1931 | 15 |
1941 | 20 |
1951 | 27 |
1961 | 39 |
1971 | 52 |
We are tasked with finding yp where x=1946.
Step 1: taking the closest to x as x0=1951,h=x2−x1=1931−1941
Step 2: Use the formula p=hx−x0.
p=101946−1951=−0.5 Step 3: Calculate the central differences for the y values.
x | p | y | Δy | Δ2y | Δ3y | Δ4y |
---|
1931 | -20 | 15 | | | | |
1941 | -10 | 20 | 5 | 2 | | |
1951 | 0 | 27 | 7 | 5 | 3 | 7 |
1961 | 10 | 39 | 12 | 1 | -4 | |
1971 | 20 | 52 | 13 | | | |
Step 4: Apply the Gauss Backward Interpolation formula:
P=y0+pΔy−1+2!(p+1)p⋅Δ2y−1+3!(p+1)p(p−1)⋅Δ3y−2+4!(p+2)(p+1)p(p−1)⋅Δ4y−2+⋯ Step 5: Substituting the values into the interpolation formula:
P(1946)=27+(−0.5)⋅7+2!(−0.5+1)∗(−0.5)⋅5+3!(−0.5+1)∗(−0.5)∗(−0.5−1)⋅3+4!(−0.5+2)(0.5+1)∗(−0.5)∗(0.5−1)⋅−7 Step 6: After solving the equation, the interpolated value of y at x=1946 is approximately:
P(1946)≈22.8984375 This is the final result of the Gauss Backward Interpolation method.