Claim: There exists such that
Proof:
The interpolation polynomial may be expressed using dividing difference coefficients i.e.
which implies
In general, the divided difference coefficients may be expressed as a factorial weighted point of a derivative of i.e.
Hence,
which implies
From the hint we know that
which implies
Since is quadratic, is constant i.e.
for all
Therefore,
By the fundamental theorem of calculus,
Therefore,
Now suppose and has 4 continuous derivatives. In this case show
where . What is in terms of the derivatives of ?
|
We know that , because . Now, by we can conclude that there exists such that .
Find such that is a polynomial of degree and this set is orthogonal on with respect to the weight function . (Note: , )
|
To find orthogonal use the Gram Schmidt method.
Let be a basis of .
Choose .
From Gram Schmidt, we have
, where
Therefore
Proceeding with Gram Schmidt, we have
where
Therefore
Derive the 2-point Gaussian formula
i.e. find the weights and nodes
|
The nodes and are the roots of the th orthogonal polynomial i.e.
Applying the quadratic formula yields the roots:
The approximation is exact for polynomials at most of degree . Hence, we have the following system of equations
Solving the solving the system of equation by substitution yields the weights:
Write down the Jacobi iteration for solving in a way that it would be programmed on a computer
|
Choose
for
- <convergence condition>
end
Where , is diagonal, are lower and upper triangular, respectively.
The diagonal entries of are non-zero since otherwise would not exist.
Therefore contains off-diagonal non-zero entries.
The computation during each iteration is given by
Therefore there are multiplies in each iteration.
Assume that is strictly diagonally dominant: for
Show that the Jacobi iteration converges for any guess . (Hint: You may use Gerschgorin's theorem without proving it.)
|
Let .
Theorem 8.2.1 [SB] states that if and only if the Jacobi iteration converges.
Matrix multiplication and the definitions of gives the explicit entrywise value of
for and
Then, using Gerschgorin's Theorem and diagonal dominance, we have the result.