2. Find Missing terms in interpolation table
Solution:
The value of table for `x` and `y`
Here we are given 4 values of y. We assume that y is polynomial of degree 3.
Hence `Delta^4 y_k=0`
`(E-1)^4 y_k=0`
`(E^4-4E^3+6E^2-4E+1) y_k=0`
taking `k=0`
`y_4-4y_3+6y_2-4y_1+y_0=0`
`:.b-4(15)+6(8)-4(a)=0`
`:.b-4a-12=0`
`:.b-4a=12`
taking `k=1`
`y_5-4y_4+6y_3-4y_2+y_1=0`
`:.35-4(b)+6(15)-4(8)=0`
`:.-4b+a+93=0`
`:.-4b+a=-93`
Now solving this 2 equations using substitution method
`b-4a=12`
and `-4b+a=-93`
`4b-a=93`
Suppose,
`-4a+b=12 ->(1)`
and `-a+4b=93 ->(2)`
Taking equation `(1)`, we have
`-4a+b=12`
`=>b=4a+12 ->(3)`
Putting `b=4a+12` in equation `(2)`, we get
`-a+4b=93`
`-a+4(4a+12)=93`
`=>-a+16a+48=93`
`=>15a+48=93`
`=>15a=93-48`
`=>15a=45`
`=>a=3 ->(4)`
Now, Putting `a=3` in equation `(3)`, we get
`b=4a+12`
`=>b=4(3)+12`
`=>b=12+12`
`=>b=24`
`:.b=24" and "a=3`
This material is intended as a summary. Use your textbook for detail explanation.
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