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Nonlinear Systems and Polynomial
Interpolation
Adrian Down
February 14, 2006
1 Review: Nonlinear systems of equations
1.1 Linear approximation
1.1.1 Taylor series
Recall the Taylor series approximation of a system of equations at a point x
near a point a,
f (x) = f (a) + J(a) (x − a) + O
|x − a|
2
where J is the n × n Jacobian matrix,
J ≡
∂ 1 f 1 (a)... ∂nf 1 (a)
.. .
∂ 1 fn(a)... ∂nfn(a)
1.1.2 Newton’s method
Newton’s method finds the zeros of a function by finding the successive in-
tercepts of the tangents to the curve with the x axis,
f (xν ) + f
′ (xν )(xν+1 − xν ) = 0
By analogy, the iterative solution to the system of functions can be found
from
f (xν ) + J(xν )(xν+1 − xν )
1.2 Discretized boundary value problem
1.2.1 Matrix representation
Recall the heat flow boundary value problem considered previously. We found
that
T 1
Tn
2
r(x 1 , T 1 )
. . .
. . .
r(xn, Tn)
Returning to the previous notation,
f (T) =
T 1
Tn
2
r(x 1 , T 1 )
. . .
. . .
r(xn, Tn)
1.2.2 Jacobian
We now need to compute the Jacobian.
Note. If the system of equations is linear, the Jacobian matrix is the same
as the original coefficient matrix.
J(T) =
−2 + h
2 ∂r ∂T
(x 1 , T 1 ) 1
1 − 2 + h
2 ∂r ∂T
(xn, Tn)
Since the element of r in the jth row depends only on Tj , the partial deriva-
tives of r contribute only diagonal elements to J.
The iteration equation is
J(Tν )(Tν+1 − Tν ) = −f (Tν )
As long as the Jacobian matrix is non-singular, the iterative process will
converge.
Proof. Assume for the sake of contradiction that rankV < n + 1. Then V
has a nontrivial null space. In particular, ∃ a set of coefficients { αk } such
that
V
α 0
α 1
. . .
αn
Writing the polynomial,
p(x) = α 0 + α 1 x +... + αnx
n
Because αnx
n may be 0, the degree m of p(x) is less than or equal to n.
By the definition of p(x), p vanishes at the n+1 distinct points x 0 ,... , xn.
However, by the fundamental theorem of algebra, a polynomial of degree m
has at most m distinct roots. This is a contradiction, since m < n + 1. The
contradiction shows that rankV = n + 1.
Note. rankV = n + 1 ⇔ V is nonsingular.
2.3 Alternative polynomial methods
2.3.1 Computational limitations
Although V is nonsingular, the problem of finding { ak } is not always easy.
As more polynomials are included in the interpolation, the determinant of
V approaches 0, which implies that V becomes very nearly singular and the
polynomial interpolation method breaks down.
det V =
i>j
(xi − xj )
xi − xj → 0 as n → ∞
Thus we must seek other ways to accomplish the interpolation.
Example. Consider the interval [0, 1] with n = 4. The determinant is
(x 3 − x 0 ) (x 2 − x 0 ) (x 1 − x 0 ) (x 3 − x 1 ) (x 2 − x 1 ) (x 3 − x 2 )
Thus the V matrix is very nearly singular, even for a simple example with
small n
2.3.2 Lagrange form
We seek a polynomial of the form
p(x) =
∑^ n
k=
ykLk(x)
Suppose we choose Lk such that
Lk(xl) =
0 if k 6 = l
1 if k = l
≡ δkl
With this substitution,
p(xl) =
∑^ n
k=
ykδkl = yl
Hence the interpolation achieves the desired values of y at each point { xk }.
Our task is then to choose an L such that Lk(xl) = δkl. We take a product
of the roots and normalize,
Lk(x) =
j 6 =k
(x − xj )
(xk − xj )
There are n monomial factors in the numerator, and so the degree of the
polynomial is n, as desired.
Although this expression is explicit, it is not optimum because of the
computations required to compute L.
2.3.3 Newton form
p(x) = c 0 + c 1 (x − x 0 ) + c 2 (x − x 0 )(x − x 1 ) +...
- cn (x − x 0 )(x − x 1 )... (x − xn− 1 ) ︸ ︷︷ ︸ n factors
