advanced computational fluid dynamics aa215a lecture 3...

Advanced Computational Fluid Dynamics

AA215A Lecture 3

Polynomial Interpolation: Numerical Differentiation

and Integration

Antony Jameson

Winter Quarter, 2016, Stanford, CALast revised on January 7, 2016

Contents

3 Polynomial Interpolation: Numerical Differentiation and Integration 23.1 Interpolation Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23.2 Error in the Interpolating Polynomial . . . . . . . . . . . . . . . . . . . . . . . . . . 33.3 Error of Derivative with Lagrange Interpolation . . . . . . . . . . . . . . . . . . . . . 33.4 Error of the kth Derivative of the Interpolation Polynomial . . . . . . . . . . . . . . 43.5 Interpolation with a Triangle of Polynomials . . . . . . . . . . . . . . . . . . . . . . . 53.6 Newton’s Form of the Interpolation Polynomial . . . . . . . . . . . . . . . . . . . . . 63.7 Hermite Interpolation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83.8 Integration Formulas using Polynomial Interpolation . . . . . . . . . . . . . . . . . . 103.9 Integration with a weight function . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103.10 Newton Cotes formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103.11 Gauss Quadrature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113.12 Formulas for Gauss integration with constant weight function . . . . . . . . . . . . . 133.13 Error bound for Gauss integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133.14 Discrete orthogonality of orthogonal polynomials . . . . . . . . . . . . . . . . . . . . 143.15 Equivalence of interpolation and least squares approximation using Gauss integration 143.16 Gauss Lobato Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153.17 Portraits of Lagrange, Newton and Gauss . . . . . . . . . . . . . . . . . . . . . . . . 16

1

Lecture 3

Polynomial Interpolation: NumericalDifferentiation and Integration

3.1 Interpolation Polynomials

So far we have not required Pn(x) to equal f(x) at any points in the interval. This permits somesmoothing! A natural approach is to make Pn(x) = a0 + a1x + ... + anx

n = f(x) at n + 1 pointsx0, x1, ..., xn. Then we have n+ 1 equations for the n+ 1 coefficients. Such a polynomial is calledan interpolation polynomial. A solution exists because the determinant of the equations is theVandemonde determinant

D =

∣∣∣∣∣∣∣∣∣

1 x0 ... xn0

1 x1 ... xn1

......

. . ....

1 xn ... xnn

∣∣∣∣∣∣∣∣∣=

∏

i>j

(xi − xj) 6= 0 (3.1)

This can be seen by noting that D vanishes if any xi equals any xj , but D is a polynomial of justthe same order as the product. The interpolation polynomial is unique since if there were 2 suchpolynomials Pn(x), Qn(x), then

Pn(xi)−Qn(xi) = 0 for i = 0, 1, ..., n (3.2)

But then Pn(x)−Qn(x) is an nth degree polynomial with n+ 1 roots, so it must be zero.In practice it is easiest to construct Pn(x) indirectly as a sum of specially chosen polynomials

rather than solve for the aj . To do this note that

φn,j(x) =(x− x0)(x− x1)...(x− xj−1)(x− xj+1)...(x− xn)

(xj − x0)(xj − x1)...(xj − xj−1)(xj − xj+1)...(xj − xn)= 1 when x = xj (3.3)

Then we can get

Pn(x) =n∑

j=0

f(xj)φn,j(x) (3.4)

which equals f(x) at x = x0, x1, ..., xn, and is the only such polynomial as has just been shown.The φn,j(x) are called Lagrange interpolation coefficients. We can write

φn,j(x) =wn(x)

(x− xj)w′n(xj)(3.5)

2

LECTURE 3. NUMERICAL DIFFERENTIATION AND INTEGRATION 3

wherewn(x) = (x− x0)(x− x1)...(x− xn) (3.6)

3.2 Error in the Interpolating Polynomial

We shall show that if f(x) has an (n + 1)th derivative and Pn(x) is an interpolating polynomialthen the remainder is

f(x)− Pn(x) =(x− x0)(x− x1)...(x− xn)

(n+ 1)!f (n+1)(ξ) (3.7)

where ξ is in the interval defined by x0, x1, ..., xn.Let Sn(x) be defined by

f(x)− Pn(x) = wn(x)Sn(x) (3.8)

wherewn(x) = (x− x0)(x− x1)...(x− xn) (3.9)

DefineF (z) = f(z)− Pn(z)− wn(z)Sn(x) (3.10)

This is continuous in z and vanishes at n+2 points x0, x1, ..., xn, x. Thus by Rolle’s theorem F ′(z)vanishes at n+1 points, F ′′(z) vanishes at n points, ..., and finally F (n+1)(z) vanishes at one pointξ. But

dn+1

dzn+1Pn(z) = 0 (3.11)

dn+1

dzn+1wn(z)Sn(x) = (n+ 1)!Sn(x) (3.12)

ThusF (n+1)(z) = fn+1(z)− (n+ 1)!Sn(x) (3.13)

and setting z = ξ

Sn(x) =1

(n+ 1)!f (n+1)(ξ) (3.14)

f(x)− Pn(x) =wn(x)

(n+ 1)!f (n+1)(ξ) (3.15)

Since the error depends on wn(x) we naturally ask how the xi should be distributed to minimizewn(x).

3.3 Error of Derivative with Lagrange Interpolation

Suppose thatf(x) = Pn(x) +

wn

(n+ 1)!f (n+1)(ξ) (3.16)

Then

f ′(x) = P ′n(x) +w′n

(n+ 1)!f (n+1)(ξ) +

wn

(n+ 1)!f (n+2)(ξ)

dξ

dx(3.17)


where

w′n =n∑

j=0

n∏

k 6=j

(x− xk) (3.18)

Then at an interpolation point,

f ′(xj) = P ′n(xj) +n∏

k 6=j

(xj − xk)f (n+1)

(n+ 1)!(ξ) (3.19)

with equal intervals h

f ′(x0) = P ′(x0) +hn

n+ 1f (n+1)(ξ) (3.20)

since(x1 − x0)(x2 − x0)...(xn − x0) = n!hn (3.21)

With one interval,

f ′(x0) =f(x1)− f(x0)

h+h

2f ′′(ξ) (3.22)

3.4 Error of the kth Derivative of the Interpolation Polynomial

Let fn+1(x) be continuous and let

Rn(x) = f(x)− Pn(x) (3.23)

Then

R(k)n (x) =

n−k∏

j=0

(x− ξj)f (n+1)(ξ)

(n+ 1− k)!(3.24)

where the distinct points ξj are independent of x and lie in the intervals xj < ξj < xj+k for j =0, 1, ..., n− k and ξ(x) is some point in the interval containing x and the ξj .Proof:Rn = f−Pn has n+1 continuous derivatives and vanishes at n+1 points x = xj , for j = 0, 1, ..., n.Apply Rolle’s theorem k ≤ n times. The zeros are then distributed as illustrated.


This leads to the following table for the distribution of the n− k + 1 zeros ξj of R(k)n .

R R(1) R(2) · · · R(k)

x0

x1 (x0, x1)x2 (x1, x2) (x0, x2)...xk (xk−1, xk) (xk−2, xk) · · · (x0, xk)...xn (xn−1, xn) (xn−2, xn) · · · (xn−k, xn)

Define

F (z) = R(k)n (z)− α

n−k∏

j=0

(z − ξj) (3.25)

For any x distinct from the ξj choose α such that

F (x) = 0 (3.26)

Then F (z) has n − k + 2 zeros and by Rolle’s theorem F (n−k+1)(z) has one zero, η, say, in theinterval containing x and the ξj .Thus

0 = F (n−k+1)(η) (3.27)

= Rn+1n (η)− α(n− k + 1)! (3.28)

= fn+1(η)− α(n− k + 1)! (3.29)

or

α =fn+1(η)

(n− k + 1)!(3.30)

It follows that for a mesh width bounded by h

R(k)n ≤Mhn−k+1 (3.31)

where M is proportional to sup∣∣fn+1(x)

∣∣ in the interval.

3.5 Interpolation with a Triangle of Polynomials

Set

w0(x) = (x− x0)w1(x) = (x− x0)(x− x1)wk(x) = (x− x0)(x− x1)...(x− xk)

Now to interpolate f(x) with values fi at xi we set

p(x) = a0 + a1(x− x0) + a2(x− x0)(x− x1)... (3.32)


Then

p0(x0) = f0 = a0

p1(x1) = f1 = a0 + a1(x1 − x0)p2(x2) = f2 = a0 + a1(x2 − x0) + a2(x2 − x0)(x2 − x1)

...

so we can solve for the ai in succession

ak =fk − a0 − a1(xk − x0)− ...− ak−1(xk − x0)...(xk − xk−2)

(xk − x0)(xk − x1)...(xk − xk−1)=fk − pk−1(xk)wk−1(xk)

(3.33)

3.6 Newton’s Form of the Interpolation Polynomial

The Lagrangian form of the interpolation polynomial has the disadvantage that the coefficientshave to be recomputed if a new interpolation point is added. To avoid this let

Pn(x) =n∑

k=0

akwk−1(x) (3.34)

wherewk(x) = (x− x0)(x− x1)...(x− xk) and w−1(x) = 1 (3.35)

We can determine the ak so that

Pn(xj) = f(xj) for j = 0, 1, ..., n (3.36)

Suppose this has been done for Pk−1, and we now add xk. Then

Pk(x) = Pk−1(x) + akwk−1(x) (3.37)

Pk(xj) = Pk−1(xj) = f(xj) for j = 0, 1, ..., k − 1 (3.38)

Pk(xk) = Pk−1(xk) + akwk−1(xk) = f(xk) (3.39)

where ak = f(xk)−Pk−1(xk)wk−1(xk) .

On the other hand the coefficient of the highest order term in the Lagrange expression is

n∑

j=0

f(xj)w′n(xj)

(3.40)

so by comparison

an =n∑

j=0

f(xj)w′n(xj)

(3.41)

Thus an is a linear combination of f(xj) for j = 0, 1, ..., n. Multiply by

xn − x0 = xn − xj + xj − x0 (3.42)


Then

an(xn − x0) = −n∑

j=0

f(xj)w′n(xj)

(xj − xn) +n∑

j=1

f(xj)w′n(xj)

(xj − x0)

= −n∑

j=0

f(xj)∏k 6=j(xj − xk)

+n∑

j=1


(3.43)

These are just the expressions for an−1 in the intervals x0 to xn−1 and x1 to xn. Thus if we define

f [x0, ..., xn] = an =n∑

j=0


(3.44)

we find thatf [x0, ..., xn](x0 − xn) = f [x1, ..., xn]− f [x0, ..., xn−1] (3.45)

and starting fromf [x0] = f(x0) (3.46)

we havef [x0, x1] =

f [x1]− f [x0]x1 − x0

(3.47)

f [x0, x1, x2] =f [x1, x2]− f [x0, x1]

x2 − x0(3.48)

giving Newton’s form of the interpolation polynomial

Pn(x) = f [x0] + (x− x0)f [x0, x1] + ...+ (x− x0)...(x− xn−1)f [x0, ..., xn] (3.49)

The square bracketed expressions are Newton’s divided differences, and we now have Newton’s formfor the interpolation polynomial:

Pn(x) = f [x0] + (x− x0)f [x0, x1] + ...+ (x− x0)...(x− xn−1)f [x0, ..., xn] (3.50)

where each new term is independent of the previous terms.To estimate the magnitude of the higher divided differences we can use the result already

obtained for the remainder.Theorem: Let x0, x1, ..., xk−1 be distinct points and let f be continuous in interval containing allthese points. Then for some point ξ in this interval

f [x0, ..., xk−1, x] =f (k)(ξ)k!

(3.51)

Proof: From Newton’s formula

f(x)− Pk−1(x) = (x− x0)...(x− xk−1)f [x0, ..., xk−1, x] = (x− x0)...(x− xk−1)f (k)(ξ)k!

(3.52)

by the remainder theorem. But x is distinct from x0, x1, ..., xk−1 so the theorem follows on dividingout the factors.


3.7 Hermite Interpolation

We can generalize interpolation by matching derivatives as well as values at the interpolationpoints. Such a polynomial is called the osculating polynomial and the procedure is called Hermiteinterpolation.

Let H2n+1(x) be a polynomial of degree 2n+ 1 such that

H2n+1(x) = f(xj) for j = 0, 1, ..., n (3.53)

H ′2n+1(x) = f ′(xj) for j = 0, 1, ..., n (3.54)

The 2n + 2 coefficients are needed to satisfy 2n + 2 conditions. H2n+1(x) can be found indirectlyas in Lagrange interpolation.

Let ψn,j(x) and γn,j(x) be polynomials of degree 2n+ 1 such that

ψn,j(xi) = δij , ψ′n,j(xi) = 0, for i = 0, 1, ..., n

γn,j(xi) = 0 , γ′n,j(xi) = δij , for i = 0, 1, ..., n

Then the Hermite interpolation polynomial is

H2n+1(x) =n∑

j=0

(f(xj)ψn,j(x) + f ′(xj)γn,j(x)

)(3.55)

It can be directly verified by differentiation that the required polynomials are

ψn,j(x) = (1− 2φ′n,j(x)(x− xj))φ2n,j(x) (3.56)

andγn,j(x) = (x− xj)φ2

n,j(x) (3.57)

where φn,j(x) are the Lagrange polynomials satisfying

φn,j(xi) = δij for i = 0, 1, ..., n (3.58)

The error in Hermite interpolation is

f(x)−H2n+1(x) = ω2n(x)

f2n+2(ξ)(2n+ 2)!

(3.59)

whereωn(x) = (x− x0)(x− x1)...(x− xn) (3.60)

This can be proved in the same way as the error estimate for Lagrange interpolation where now

F (ξ) = f(ξ)−H2n+1(ξ)− ω2n(ξ)Sn(x) (3.61)

and on the first application of Rolle’s theorem F ′(ξ) has 2n+ 2 distinct zeros.The Hermite polynomial can be obtained via a passage to the limit from the Lagrange polyno-

mial P2n+1(x) which interpolates f(x) at the 2n+ 2 points

x0, x0 + h, x1, x1 + h, ..., xn, xn + h (3.62)


as h→ 0 to produce n+ 1 pairs of coincident points. Define

N(h) =∏

k=1,n, k 6=j

(x− xk + h) (3.63)

andD(h) =

∏

k=1,n, k 6=j

(xj − xk + h) (3.64)

Then the contribution of the points xj and xj + h to P2n+1(x) is

Q(x) = f(xj)N(0)D(0)

N(−h)D(−h)

x− (xj + h)xj − (xj + h)

+ f(xj + h)N(0)D(h)

N(−h)D(0)

x− xj

h

Expanding f(xj + h) in a Taylor series

Q(x) = f(xj)N(0)D(0)

N(−h)D(−h)

x− (xj + h)xj − (xj + h)

+(f(xj) + hf ′(xj) +O(h2)

) N(0)D(h)

N(−h)D(0)

x− xj

h

= f(xj)N(0)D(0)

[N(−h)D(−h) −

x− xj

hN(−h)

(1

D(−h) −1

D(h)

)]

+ f ′(xj)(x− xj)N(0)D(0)

N(−h)D(h)

+O(h)

Also1

D(−h) −1

D(h)= 2h

D′(0)D2(0)

+O(h2) (3.65)

Thus in the limit as h→ 0

Q(x) = f(xj)N2(0)D2(0)

(1− 2(x− xj)

D′(0)D(0)

)

+ f ′(xj)(x− xj)N2(0)D2(0)

whereN(0)D(0)

= φn,j(x) (3.66)

and D′(0) equals ddxN(0) evaluated at xj , so that

D′(0)D(0)

= φ′n,j(xj) (3.67)


3.8 Integration Formulas using Polynomial Interpolation

Let Pn−1(x) interpolate f(x) at x0, x1, ..., xn. Then

Pn(x) =n∑

j=0

f(xj)φn,j(x) (3.68)

where

φn,j(x) =

∏k 6=j(x− xk)∏k 6=j(xj − xk)

(3.69)

so thatφn,j(xk) = δjk (3.70)

Now approximate∫ ba f(x)dx by

∫ ba Pn(x)dx where typically a = x0, b = xn. Then

∫ b

af(x)dx =

n∑

j=0

Ajf(xj) + E(f) (3.71)

where E(f) denotes the error, and

Aj =∫ b

aφn,j(x)dx (3.72)

The result is exact if f(x) is a polynomial of degree ≤ n, since then Pn(x) = f(x). This is calledprecision of degree n.

3.9 Integration with a weight function

We may include a weight function w(x) ≥ 0 in the integral. Then

∫ b

af(x)w(x)dx =

n∑

j=n

Ajf(xj) + E(f) (3.73)

where now

Aj =∫ b

aφn,j(x)w(x)dx (3.74)

Depending on w(x), analytical expressions are not necessarily available for evaluating the coeffi-cients Aj .

3.10 Newton Cotes formulas

These are derived by approximating f(x) by an interpolation formula using n + 1 equally spacedpoints in [a, b] including the end points. The first 3 such formulas are:Trapezoidal rule ∫ b

af(x)dx =

h

2(f(a) + f(b)) + E(f), h = b− a (3.75)


Simpson’s rule∫ b

af(x)dx =

h

3f(a) +

4h3f(a+ h) +

h

3f(b) + E(f), h =

b− a

2(3.76)

and Simpson’s 38 formula

∫ b

af(x)dx =

3h8f(a) +

9h8f(a+ h) +

9h8f(a+ 2h) +

3h8f(b) + E(f), h =

b− a

3(3.77)

3.11 Gauss Quadrature

When the interpolation points x are fixed the integration formula has n + 1 degrees of freedomcorresponding to the coefficients Aj for j = 0, ..., n. Accordingly we can find values of Aj whichyield exact values of the integral of all polynomials of degree n,

Pn(x) = a0 + a1x+ ...+ anxn, (3.78)

since these also have n + 1 degrees of freedom corresponding to the coefficients aj for j = 0, ..., n.If we are also free to choose the integration points xj , then we have 2n + 2 degrees of freedomcorresponding to xj and Aj , for j = 0, ..., n. Now it is possible to find values of xj and Aj whichenable exact integration of polynomials of degree ≤ 2n + 1. One could try to find the requiredvalues by solving 2n+ 2 nonlinear equations

n∑

j=0

Ajf(xj) =∫ b

af(x)dx (3.79)

where f(x) = 1, x, ..., x2n+1 in turn. However the required values can be found indirectly as follows.Let φi(x) be orthogonal polynomials for the weight function w(x), so that

∫ b

aφj(x)φk(x)w(x) = 0 for j 6= k (3.80)

Choose xj as the zeros of φn+1(x). Then integration using polynomial interpolation is exact forpolynomials up to degree 2n+ 1.Proof:If f(x) is a polynomial of degree ≤ 2n+ 1 it can be uniquely expressed as

f(x) = q(x)φn+1(x) +R(x) (3.81)

where q(x) and R(x) are polynomials of degree ≤ n. Then

∫ b

af(x)w(x)dx =

∫ b

aq(x)φn+1(x)w(x)dx+

∫ b

aR(x)w(x)dx =

∫ b

aR(x)w(x)dx (3.82)

since φn+1(x) is orthogonal to all polynomials of degree < n+ 1.


Also the approximate integral is

I =n∑

j=0

Ajf(xj)

=n∑

j=0

Ajq(xj)φn+1(xj) +n∑

j=0

AjR(xj)

=n∑

j=0

AjR(xj)

(3.83)

since φn+1(xj) is zero for j = 0, 1, ..., n. But then

I =∫ b

aR(x)w(x)dx (3.84)

exactly, since R(x) is a polynomial of degree ≤ n.The degree of precision 2n+1 is the maximum attainable. Consider the polynomial of φ2

n+1(x)of degree 2n+ 2. Then

I = 0 (3.85)

since φn+1(xj) = 0 for j = 0, 1, ..., n. But

∫ b

aφ2

n+1(x)w(x)dx > 0 (3.86)

because φ2n+1(x) > 0 except at the zeros x0, x1, ..., xn.

When applied to an arbitrary smooth function f(x) which can be expanded as

f(a) + (x− a)f ′(a) + ...+(x− a)2n+2

(2n+ 2)!f (2n+2)(ξ) (3.87)

the error is proportional to a bound on |f (2n+2)(x)| because the preceding terms in the Taylor seriesare integrated exactly.


3.12 Formulas for Gauss integration with constant weight function

Number of Points xj Aj Precision Order

1 0 2 1 2

2 ± 1√3

1 3 4

30 8

9 5 6±√

155

59

4±r

3−2q

65

718+

√30

36 7 8

±r

3+2q

65

718−√30

36

5

0 128225

9 10±√

5− 2√

107

322+13√

70900

±√

5 + 2√

107

322−13√

70900

Table 3.1: Formulas for Gauss integration with constant weight function

3.13 Error bound for Gauss integration

Let H2n+1(x) be the Hermite interpolation polynomial to f(x) at the integration points. The Gaussintegration formula is exact for H2n+1(x). Hence

∫ b

aH2n+1(x)w(x)dx =

n∑

j=0

AjQ(xj) =n∑

j=0

Ajfj (3.88)

Thus ∫ b

af(x)w(x)dx−

n∑

i=0

Ajfj =∫ b

a(f(x)−H2n+1(x))w(x)dx (3.89)

According to the error formula for Hermite interpolation

f(x)−H2n+1(x) = P 2n(x)

f (2n+2)(ξ)

(2n+ 2)!(3.90)

wherePn(x) = (x− x0)(x− x1)...(x− xn) (3.91)

is the orthogonal polynomial normalized so that its leading coefficient is unity.Then the error is

E =1

(2n+ 2)!

∫ b

aw(x)P 2

n(x)f (2n+2)(ξ)dx (3.92)

Since w(x)P 2n(x) ≥ 0 the error lies in the range between

min f (2n+2)(ξ)A and max f (2n+2)(ξ)A (3.93)


where

A =1

(2n+ 2)!

∫ b

aw(x)P 2

n(x)dx (3.94)

and henceE = Af (2n+2)(η) (3.95)

for some value of η in [a, b].

3.14 Discrete orthogonality of orthogonal polynomials

Let φj(x) be orthogonal polynomials satisfying

(φj , φk) =∫ b

aφj(x)φk(x)w(x)dx = 0 for j 6= k (3.96)

Let xi be the zeros of φn+1(x).Then if j ≤ n, k ≤ n, φj(x)φk(x) is a polynomial of degree ≤ 2n. Accordingly (φj , φk) is

evaluated exactly by Gauss integration. This implies that the φj satisfy the discrete orthogonalitycondition

n∑

i=0

Aiφj(xi)φk(xi) = 0, given j 6= k, j ≤ n, k ≤ n (3.97)

where Ai are the coefficients for Gauss integration at the zeros of φn+1.

3.15 Equivalence of interpolation and least squares approximationusing Gauss integration

Let f∗(x) be the least squares fit to f(x) using orthogonal polynomials φj(x) for the interval [a, b]with weight function w(x) ≥ 0. Then

f∗(x) =n∑

j=0

c∗jφj(x) (3.98)

where

c∗j =(f, φj)(φj , φj)

=

∫ ba f(x)φj(x)w(x)dx∫ b

a φ2j (x)w(x)dx

(3.99)

Let xi be the zeros of φn+1(x), and let Pn(x) be the interpolation polynomial to f(x) satisfying

Pn(xi) = f(xi) for i = 0, 1, ..., n (3.100)

Pn(x) can be expanded as

Pn(x) =n∑

k=0

ckφk(x) (3.101)

Then the ck are determined by the conditions

n∑

k=0

ckφk(xi) = f(xi) for i = 0, 1, ..., n (3.102)


Also the polynomials φj(x) satisfy the discrete orthogonality condition

n∑

i=0

Aiφj(xi)φk(xi) = 0 given j 6= k, j ≤ n, k ≤ n (3.103)

where Ai are the coefficients for Gauss integration.Now multiplying (3.102) by Aiφj(xi) and summing over i, it follows that

cj =∑n

i=0Aif(xi)φj(xi)∑ni=0Aiφ2

j (xi)(3.104)

This is exactly the formula (3.99) for evaluating the least squares coefficient c∗j by Gauss integration.Alternative Proof:The coefficients cj are actually the least squares coefficients which minimize

J = ||f −n∑

j=0

cjφj ||2 (3.105)

in the discrete semi-norm in which

∫ f(x)−

n∑

j=0

cjφj(x)

2

w(x)dx (3.106)

is evaluated by Gauss integration at the zeros xi of φn+1(x) as

J =n∑

i=0

Ai

f(xi)−

n∑

j=0

cjφj(xi)

2

w(x0) (3.107)

But as a consequence of the interpolation condition (3.99) the coefficients cj yield the value J = 0,thus minimizing J .

3.16 Gauss Lobato Integration

To integrate f(x) over [-1,1] choose integration points at −1, 1, and the zeros of Qn−1, which arethe zeros of

(1 + x)(1− x)φn−1(x) (3.108)

Then letf(x) = q(x)(1− x2)φn−1(x) + r(x) (3.109)

where if f(x) is a polynomial of degree ≤ 2n− 1, q(x) and r(x) are polynomial of degree ≤ n− 2.Now let the polynomials φj be orthogonal for the weight function

w(x) = 1− x2 (3.110)

Then

I =∫ 1

−1f(x)dx =

∫ 1

−1q(x)φn−1(x)w(x)dx+

∫ 1

−1r(x)dx =

∫ 1

−1r(x)dx (3.111)


since φn−1 is orthogonal to all polynomials of lower degree. Also the approximate integral is

I =n∑

j=0

Ajf(xj) (3.112)

where

Aj =∫ 1

−1φn,j(x)dx (3.113)

Then

I =n∑

j=0

Ajq(xj)(1− x2j )φn−1(xj) +

n∑

j=0

Ajr(xj) =n∑

j=0

Ajr(xj) (3.114)

since (1− x2j )φn−1(xj) = 0 for j = 0, 1, ..., n. But then

I = I (3.115)

since r(x) is a polynomial of degree < n. Here φj are the Jacobi polynomials P (1,1)j , where P (α,β)

j

is orthogonal for the weight(1− x)α(1 + x)β (3.116)

3.17 Portraits of Lagrange, Newton and Gauss

Figure 3.1: Joseph-Louis Lagrange (1736-1813)


Figure 3.2: Sir Isaac Newton (1642-1726)

Figure 3.3: Carl Friedrich Gauss (1777-1855)

advanced computational fluid dynamics aa215a lecture 3...

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