explicit formula for divided differences
We will proceed by recursion on . When , the formula to be proven reduces to
which agrees with the definition of .
To prove that this is correct when , one needs to check that it the recurrence relation for divided differences.
Thus, we see that, if
Hence, by induction, the formula holds for all . ∎
This formula may be phrased another way by introducing the polynomials defined as
We may write
|Title||explicit formula for divided differences|
|Date of creation||2013-03-22 14:41:16|
|Last modified on||2013-03-22 14:41:16|
|Last modified by||rspuzio (6075)|