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7. Gauss Backward formula (Numerical Interpolation) example ( Enter your problem )
  1. Formula & Example-1
  2. Example-2
  3. Example-3
Other related methods
  1. Newton's Forward Difference formula
  2. Newton's Backward Difference formula
  3. Newton's Divided Difference Interpolation formula
  4. Lagrange's Interpolation formula
  5. Lagrange's Inverse Interpolation formula
  6. Gauss Forward formula
  7. Gauss Backward formula
  8. Stirling's formula
  9. Bessel's formula
  10. Everett's formula
  11. Hermite's formula
  12. Missing terms in interpolation table

1. Formula & Example-1
(Previous example)
3. Example-3
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2. Example-2





2. Find Solution using Gauss Backward formula
xf(x)
193112
194115
195120
196127
197139
198152

x = 1966


Solution:
The value of table for `x` and `y`

x193119411951196119711981
y121520273952

Gauss's backward difference interpolation method to find solution

`h=1941-1931=10`

Taking `x_0=1961` then `p=(x-x_0)/h=(x-1961)/10`

Now the central difference table is
`x``p=(x-1961)/10``y``Deltay``Delta^2y``Delta^3y``Delta^4y``Delta^5y`
1931-312
3
1941-2152
50
1951-12023
73-10
19610275-7
12-4
19711391
13
1981252


`x = 1966`

`p = (x - x_0)/h = (1966 - 1961)/10 = 0.5`

`y_0=27, Delta y_(-1)=7,Delta^2y_(-1)=5,Delta^3y_(-2)=3,Delta^4y_(-2)=-7,Delta^5y_(-3)=-10`

Gauss's backward interpolation formula is
`y_p=y_0+p Delta y_(-1) + ((p + 1)p)/(2!) * Delta^2y_(-1) + ((p + 1)p(p - 1))/(3!) * Delta^3y_(-2) + ((p + 2)(p + 1)p(p - 1))/(4!) * Delta^4y_(-2) + ((p + 2)(p + 1)p(p - 1)(p - 2))/(5!) * Delta^5y_(-3)`

`y_(0.5) = 27 + (0.5)(7) + ((0.5 + 1)(0.5))/(2) * (5) + ((0.5 + 1)(0.5)(0.5 - 1))/(6) * (3) + ((0.5 + 2)(0.5 + 1)(0.5)(0.5 - 1))/(24) * (-7) + ((0.5 + 2)(0.5 + 1)(0.5)(0.5 - 1)(0.5 - 2))/(120) * (-10)`

`y_(0.5)=27 +3.5 +1.875 -0.1875 +0.2734375 -0.1171875`

`y_(0.5)=32.34375`


Solution of Gauss's backward interpolation is `y(1966) = 32.34375`


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