Find Missing terms in interpolation table
| x | f(x) |
| 1961 | 200 |
| 1962 | 220 |
| 1963 | 260 |
| 1964 | ? |
| 1965 | 350 |
| 1966 | ? |
| 1967 | 430 |
Solution:
The value of table for `x` and `y`
| x | 1961 | 1962 | 1963 | 1964 | 1965 | 1966 | 1967 |
|---|
| y | 200 | 220 | 260 | ? | 350 | ? | 430 |
|---|
Here we are given 5 values of y. We assume that y is polynomial of degree 4.
Hence `Delta^5 y_k=0`
`(E-1)^5 y_k=0`
`(E^5-5E^4+10E^3-10E^2+5E-1) y_k=0`
taking `k=0`
`y_5-5y_4+10y_3-10y_2+5y_1-y_0=0`
`:.b-5(350)+10(a)-10(260)+5(220)-200=0`
`:.b+10a-3450=0`
`:.b+10a=3450`
taking `k=1`
`y_6-5y_5+10y_4-10y_3+5y_2-y_1=0`
`:.430-5(b)+10(350)-10(a)+5(260)-220=0`
`:.-5b-10a+5010=0`
`:.-5b-10a=-5010`
Now solving this 2 equations using substitution method
`b+10a=3450`
and `-5b-10a=-5010`
`-5(b+2a)=-5 * 1002`
`b+2a=1002`
Suppose,
`10a+b=3450 ->(1)`
and `2a+b=1002 ->(2)`
Taking equation `(1)`, we have
`10a+b=3450`
`=>b=-10a+3450 ->(3)`
Putting `b=-10a+3450` in equation `(2)`, we get
`2a+b=1002`
`2a+(-10a+3450)=1002`
`=>2a-10a+3450=1002`
`=>-8a+3450=1002`
`=>-8a=1002-3450`
`=>-8a=-2448`
`=>a=306 ->(4)`
Now, Putting `a=306` in equation `(3)`, we get
`b=-10a+3450`
`=>b=-10(306)+3450`
`=>b=-3060+3450`
`=>b=390`
`:.b=390" and "a=306`
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then
