Numerical Methods in Engineering (ENGR20005)
Lecture 14 Dr. Leon Chan
lzhchan@unimelb.edu.au Department of Mechanical Engineering The University of Melbourne
Slides prepared by Prof.Andrew Ooi
L14.1:
The Derivative Matrix: Finite Difference
First Derivative
f0
x0 x1 x2 x3 x4 x5
f1
f2 f3 f4
Δ
Δ
Δ
f5
It is sometimes convenient to think of differentiation in terms of a matrix operation. To
ΔΔ
see this, let’s consider you have the 6 equally spaced data points shown above. The grid is your series of x values, x0, x1 ….. x5, and the corresponding values of fs are denoted by f0, f1, …. ,f5. For this example, we will use 2nd order central spatial discretization at the interior nodes. So at the interior points, x1 . . x4,
f 1′ = f 2 − f 0 2Δ
f 2′ = f 3 − f 1 2Δ
f 3′ = f 4 − f 2 2Δ
f 4′ = f 5 − f 3 2Δ
3
CDA
First Derivative
f0
f1
f2 f3 f4 ΔΔΔ
f5
ΔΔ
x0 x1 x2 x3 x4 x5
Since, the function f is only defined inside the domain, we do not have the value of the
function at x = x−1 and x = x6. Thus we cannot use the central difference formula at the end points. It is common to use a lower order scheme at the end points. For this
example, we will use 1st order one-sided scheme at the end points to approximate the spatial derivatives.
f 1′ = f 2 − f 0 2Δ
f 2′ = f 3 − f 1 2Δ
f 3′ = f 4 − f 2 2Δ
f 4′ = f 5 − f 3 2Δ
CDA
FDA
BDA
f 0′ = f 1 − f 0 Δ
f 5′ = f 5 − f 4 Δ
4
First Derivative
ΔΔ
x0 x1 x2 x3 x4 x5
f 1′ = f 2 − f 0 2Δ
f 2′ = f 3 − f 1 2Δ
f 3′ = f 4 − f 2 2Δ
f 4′ = f 5 − f 3 2Δ
CDA
FDA
BDA
f 0′ = f 1 − f 0 Δ
f 5′ = f 5 − f 4 Δ
f0′ f0
f1′ f1
f2′ =1 f2
f3′ Δ f3 f4′ f4
f5′ f5
−1 1 0 0 0 0
−1/2 0 1/2 0 0 0 0 −1/2 0 1/2 0 0 0 0 −1/2 0 1/2 0 0 0 0 −1/2 0 1/2
0 0 0 0 −1 1
5
f1
f2 f3 f4 ΔΔΔ
f0
f5
Second Derivative
f0
Δ
f1
f2 f3 f4
Δ
Δ
Δ
f5 Δ
x0 x1 x2 x3 x4 x5 ′′
Similarly, we can calculate the second derivatives, fi, using (higher order) central difference formula for the interior points and lower order one-sided scheme at the end points
f′′1 = f2 − 2f1 + f0 Δ2
f′′2 = f3 − 2f2 + f1 Δ2
f′′3 = f4 − 2f3 + f2 Δ2
f′′4 = f5 − 2f4 + f3 Δ2
f ′′ = f2 − 2f1 + f0 0 Δ2
f ′′ = f5 − 2f4 + f3 5 Δ2
6
f′′1 = f2 − 2f1 + f0 Δ2
f′′2 = f3 − 2f2 + f1 Δ2
f′′3 = f4 − 2f3 + f2 Δ2
f′′4 = f5 − 2f4 + f3 Δ2
f ′′ = f2 − 2f1 + f0 0 Δ2
f ′′ = f5 − 2f4 + f3 5 Δ2
Using all the above equations, we can define a second derivative matrix that operates on the
′′
vector f that will give us the second derivative f .
f′′ f
00
1 −2 1 0 0 0
1 −2 1 0 0 0 0 1 −2 1 0 0 0 0 1 −2 1 0 0 0 0 1 −2 1
0 0 0 1 −2 1
f′′ f 11
f′′1 f 2 =Δ f2
f′′2 3 3
f ′′ f4
4f
f′′ 5
5
7
L14.2:
The Derivative Matrix: Spectral Differentiation
Spectral Differentiation
Suppose we want to approximate the derivative of a function f(x). One way of doing so is to approximate the function f(x) using a Lagrange polynomial fn(x) of order n at the nodes x0, x1, …, xn and then take the derivative. Let’s consider for the moment the case where x ∈ [−1,1]
f (1)
f(−1)
−1 1
f (x) f1(x)
x
Consider the n = 1 case, where the interpolation nodes are x0 = − 1, x1 = 1, then the
interpolating polynomial through the points f(x0) and f(x1) is given by
f1(x)=( x−1 )f0+ f1 L0(x) −1 − 1
f1(x) = 12 (1 − x)f0 + 12 (1 + x)f1 9
Note that for brevity, I will L1(x) use the notation fi ≡ f(xi).
(x+1) 1+1
f 1 ( x ) = 12 ( 1 − x ) f 0 + 12 ( 1 + x ) f 1
Taking the derivative of f1(x) gives f 1′ ( x ) = − 12 f 0 + 12 f 1
f1 = f(1)
f0 = f(−1)
In matrix form, we can write
x1 = 1 x Typically, we want to evaluate the derivatives at the interpolation nodes, x0 and x1.
f 1′ ( x ) = [ − 1 1 ] { f 0 } 2 2 f1
Writing the output in a matrix we have
{f1′(x0)} = {f0} f1′(x1) f1
Note that for this case, f1′(x0) = f1′(x1) = − 12 f0 + 12 f1 10
Differentiation matrix [D]1
−12 12 −12 12
x0 = − 1
f1(x)
[D]1
L0′(x) − 12
− 12
L1′(x) 12
12
{f′(x )} −1 1 0 = 2
f 1′ ( x 1 ) − − 1 1 2
1{f} 2 0
1 1f1 2
x0 = − 1
x1 = 1 The rows of [D]1correspond to the different nodes and the columns the derivative
of the Lagrange basis functions.
Hence, the derivative at the interpolation nodes, x0 and x1 is given by
−12 12 −12 12
{ df } = [D]1{f} where [D]1 = dx
11
f1 = f(0) f2 = f(1)
f (x) f2(x)
f0 = f(−1)
As for the n = 2 case, with the interpolation nodes x0 = − 1, x1 = 0, and x2 = 1, the
x1 = 0 x2 = 1 f2(x) = f0 + (x − 1)(x + 1) f1 + f2
x0 = − 1
interpolating polynomial through the points f0 ≡ f(x0), f1 ≡ f(x1) and f2 ≡ f(x2) is given by
x
(x−0)(x−1) (−1−0)(−1−1)
(x − 0)(x + 1) (1−0)(1+1)
L0(x)
(0−1)(0+1)
f 2 ( x ) = 12 x ( x − 1 ) f 0 + ( 1 + x ) ( 1 − x ) f 1 + 12 x ( x + 1 ) f 2
12
L1(x) L2(x)
f 2 ( x ) = 12 x ( x − 1 ) f 0 + ( 1 + x ) ( 1 − x ) f 1 + 12 x ( x + 1 ) f 2 Taking the derivative of f2(x) gives
f 2′ ( x ) = ( x − 12 ) f 0 − 2 x f 1 + ( x + 12 ) f 2 which in matrix form is given by
[(1) (1)]f0 f 2′ ( x ) = x − 2 − 2 x x + 2 f 1
f1 = f(0)
f2(x) f2 = f(1)
x1 = 0
f (x)
f0 = f(−1) f2
x0 = − 1
x2 = 1 x=−1
x =0 1
x2 = 1
And again, if we evaluate f2′(x) at the interpolation nodes, x0, x1 and x2
−3 2−1 f2′(x0) 2 2
f0 f2′(x1) = −1 0 1 f1
0
2 2
f′(x) 1 3 f 22 1−23 2
2
2
[D]2
13
L0′(x)
− 32 − 12
12
L1′(x)
2
0 −2
L2′(x)
− 12 12
32
which gives the differentiation matrix
−32 2 −12 −12 0 12 12 −2 32
[D]2 =
for equally spaced grids with 3 interpolation nodes and x ∈ [−1,1]
In general, suppose that the function f(x) is represented by the nth order Lagrange polynomial (you have sampled the function at n + 1 points)
f(x) ≈ fn(x) = ∑n fjLj(x) j=0
we will assume that x ∈ [−1,1] but the interpolation nodes need not be equally spaced. The derivative of f(x) can be approximated by just
differentiating the equation above n
f′(x) ≈ fn′(x) = ∑ fjLj′(x)
j=0
If we now evaluate the derivative at the interpolation nodes x = xi,
f′(xi) ≈ fn′(xi) = ∑n fjLj′(xi) j=0
then the above equation can be expressed in matrix form as
15
fn′(x0) fn′(x1) =
fn′(x2) .
fn′(xn)
f0
f1
f2 .
fn
f′(xi) ≈ fn′(xi) = ∑n fjLj′(xi) j=0
matrix form
L0′(x0) L1′(x0) L2′(x0)
L0′(x1) L1′(x1) L2′(x1)
. . . . . . . . .
Ln′(x0) Ln′(x1) Ln′(x2)
L0′(x2) L1′(x2) L2′(x2)
. . …..
. . ….. L0′(xn) L1′(xn) L2′(xn) . . . Ln′(xn)
[D]n =
L0′(x0) L1′(x0) L2′(x0)
L0′(x1) L1′(x1) L2′(x1)
. . . . . . . . .
Ln′(x0) Ln′(x1) Ln′(x2)
L0′(x2) L1′(x2) L2′(x2)
. . …..
. . ….. L0′(xn) L1′(xn) L2′(xn) . . . Ln′(xn)
Hence, in order to find the derivative matrix, we need to find an expression for L′(x ) where
Lj(x) = k=0,k≠j
x−xk xj − xk
ji ∏n
D =L′(x) ij ji
we need to differentiate Lj(x) and evaluate the derivative at x = xi. 16
D =L′(x) ij ji
fn′(x0)
fn′(x1)
=
L0′(x0) L1′(x0) L2′(x0)
L0′(x1) L1′(x1) L2′(x1)
L0′(x2) L1′(x2) L2′(x2)
. . . . . . . . .
Ln′(x0) f0
Ln′(x1) f1
fn′(x2)
Ln′(x2) f2
. …….. . ……..
L0′(xn) L1′(xn) L2′(xn) . . . Ln′(xn) fn (xi −x0)…(xi −xi−1)(xi −xi+1)….(xi −xn)
fn′(xn) It can be shown that
D =L′(x)= ij ji
How do we get the equation for Dij above? Proof on the next slide. 17
i≠j 1+..+1+1+…+1i=j
(xi −xj)(xj −x0)…(xj −xj−1)(xj −xj+1)….(xj −xn)
xi −x0 xi −xi−1 xi −xi+1 xi −xn
To find
L j ( x ) = ∏n x − x k k=0,k≠j xj − xk
Lj′(xi)
Consider the case where n = 3 L1(x) = (x − x0)(x − x2)(x − x3)
(x1 − x0)(x1 − x2)(x1 − x3) Φ3(x) = (x − x0)(x − x1)(x − x2)(x − x3)
L1(x) = Φ3(x)
(x − x1)Φ′3(x1)
L1(x) = (x − x0)(x − x1)(x − x2)(x − x3) (x − x1)Φ′3(x1)
L1(x) = (x − x0)(x − x2)(x − x3) Φ′3(x1)
So we need to show that
Φ′3(x1) = (x1 − x0)(x1 − x2)(x1 − x3)
If we take the log of both sides of the equation
log Φ3(x) = log((x − x0)(x − x1)(x − x2)(x − x3))
= log(x − x0) + log(x − x1) + log(x − x2) + log(x − x3)
Differentiate both sides with respect to x gives Φ′3(x)= 1 + 1 + 1 + 1
Φ3(x) x − x0 x − x1 x − x2 Φ ′3 ( x ) = ( x − x 0 ) ( x − x 1 ) ( x − x 2 ) ( x − x 3 ) 1
x − x3
+ 1 + 1 +
x−x )
For x = x0,
Φ′3(x0) = (x0 − x1)(x0 − x2)(x0 − x3)
For x = x1,
Φ′3(x0) = (x1 − x0)(x1 − x2)(x1 − x3)
Hence, in general
Φ ′n ( x i ) = ( x i − x 0 ) . . . ( x i − x i − 1 ) ( x i − x i + 1 ) . . . . ( x i − x n )
(x−x
1
x−x x−x 0123
ln(Lj(x)) = ln
i=j
∏n x−xk k=0,k≠j xj − xk
n
respect to x
L′(x) n x−x
∑ ln k=0,k≠j
x−xk (x−x)
ln(Lj(x)) =
Differentiate both sides with
j=∑jk1 L(x) (x−x x−x)
j k=0,k≠j kjk
n Lj′(x) = Lj(x) ∑
k=0,k≠j
1 (x−x)
Hence
n Lj′(xi) = Lj(xi) ∑
1 (x−x)
k=0,k≠j
Note that the term in the RHS goes to
infinity if i ≠ j. Here we are only considering the case where i = j. So
Li(xi) = 1. Thus n
Li′(xi)= ∑ k=0,k≠j
jk
k
ik
(x−x) ik
1
i≠j Rewrite Lj(x) as
Using the quotient rule, the derivative
of the Lagrange basis polynomials Li′(x) are given by
Lj′(x) = Φ′3(x)(x − xj) − Φ3(x) (x − xj)2Φ′3(xj)
at x = xi
Lj′(x) = Φ′3(x)(x − xj) − Φ3(xi)
So for i ≠ j Φ′n(xi) Lj′(xi) = (xi − xj)Φ′n(xj)
=0 (x − xj)2Φ′3(xj)
18
Lj(x) = (x − x0)⋯(x − xj−1)(x − xj+1)⋯(x − xn) (xj − x0)⋯(xj − xj−1)(xj − xj+1)⋯(xj − xn)
= Φn(x)
(x − xj)Φ′n(xj)
where the function Φn(x) is given by Φn(x) = (x − x0)(x − x1)⋯(x − xn−1)(x − xn)
(xi − xj)
(xi −x0)…(xi −xi−1)(xi −xi+1)….(xi −xn)
(xj −x0)…(xj −xj−1)(xj −xj+1)….(xj −xn)
1+..+1+1+…+1i=j xi −x0 xi −xi−1 xi −xi+1 xi −xn
D =L′(x)= ij ji
i≠j (14.1)
n Li′(xi)= ∑
k=0,k≠j
(x−x) ik
i=j
1
19
i≠j L′(x ) =
ji
Φ′n(xi)
(xi − xj)Φ′n(xj)
Example L14.1: Write a MATLAB program that implements Eq. (14.1) to approximate the derivative of
Compare with the exact answer
f(x) = ex sin(x)
df =ex(sin(5x)+5cos(5x))
Verify your program using an equally spaced grid with 11 nodal points.
dx
20
Lecture14M01.m
yreal=exp(xint).*sin(5*xint);
dyreal=exp(xint).*(sin(5*xint)+5*cos(5*xint));
figure(1)
hold off plot(xpoints,ypoints,’ko’,’MarkerSize’,20,’MarkerFaceColor’,’r’) hold on
plot(xint,yint,’b-‘,’Linewidth’,4)
plot(xint,yreal,’k-‘,’Linewidth’,4)
xlabel(‘x’);ylabel(‘f(x)’)
D=DerivMatrix(xpoints,n);
derivypoints=D*ypoints;
fn′(x1) hold off fn′(x2)
figure(2) plot(xint,dyreal,’k-‘,’Linewidth’,4) .
=
L0′(x2) L1′(x2) L2′(x2) … . . . …
Ln′(x2) .
holdon . fn(xn)
. . . … L0′(xn) L1′(xn) L2′(xn) …
. Ln′(xn)
Dij = Lj′(xi)
L0′(x0) L1′(x0) L2′(x0) …
L0′(x1) L1′(x1) L2′(x1) …
function D=DerivMatrix(x,n) D=zeros(n+1,n+1);
for i=1:n+1
num(i)=1.0; for k=1:n+1
if k ~= i num(i)=num(i)*(x(i)-x(k));
end end
end
for j=1:n+1
Ln′(x0)
Ln′(x1)
fn′(x0)
plot(xpoints,derivypoints,’ko’,’MarkerSize’,20,’MarkerFaceColor’,’b’) xlabel(‘x’);ylabel(‘df(x)/dx’)
f(x) = ex sin(x)
df =ex(sin(5x)+5cos(5x)) dx
Dij
f0
f1
f2 .
fn
21
clear all
xpoints=linspace(-1,1,11)’; ypoints=exp(xpoints).*sin(5*xpoints); n=length(xpoints)-1; %order of polynomial xint=-1:0.01:1;
yint=LagrangePolynomial(xpoints,ypoints,n,xint);
den(j)=1.0;
figure(2)
hold off
plot(xint,dyreal,’k-‘,’Linewidth’,4)
hold on plot(xpoints,derivypoints,’ko’,’MarkerSize’,20,’MarkerFaceColor’,’b’) xlabel(‘x’);ylabel(‘df(x)/dx’)
Dij
fn′(x0)
fn′(x1)
fn′(x2) = .
fn(xn)
L0′(x0) L1′(x0) L2′(x0)
L0′(x1) L1′(x1) L2′(x1)
. . . . . . . . .
Ln′(x0) Ln′(x1) Ln′(x2)
L0′(x2) L1′(x2) L2′(x2)
. . …..
. . ….. L0′(xn) L1′(xn) L2′(xn) . . . Ln′(xn)
f0
f1
f2 .
fn
i≠j
(xi −x0)…(xi −xi−1)(xi −xi+1)….(xi −xn)
(xi −xj)(xj −x0)…(xj −xj−1)(xj −xj+1)….(xj −xn)
1+..+1+1+…+1i=j xi −x0 xi −xi−1 xi −xi+1 xi −xn
df dx
dfn (approximation) dx
for i=1:n+1 num(i)=1.0; for k=1:n+1
if k ~= i num(i)=num(i)*(x(i)-x(k));
end end
end
for j=1:n+1 den(j)=1.0; for k=1:n+1
if k ~= j den(j)=den(j)*(x(j)-x(k));
end end
end
*(x(i)-x(j)));
for i=1:n+1 for k=1:n+1
if i~=k D(i,i)=D(i,i)+1./(x(i)-x(k));
end end
end end
function D=DerivMatrix(x,n) D=zeros(n+1,n+1);
for i=1:n+1 for j=1:n+1
if i ~= j D(i,j)=num(i)/(den(j)
end end
end
function yint=LagrangePolynomial(x,y,n,xint)
Dij = Lj′(xi) =
n=4 f(x) f4(x) n=4
df4 (xi) at equally spaced points
n = 10 f(x)
dx
n = 10
f10(x)
23
df10 (xi) at equally spaced points dx
Almost
to machine precision!
error= 1 ∑|f′(xi)−fn′(xi)| n+1
Error decreases when you increase the number of data points, n
24
n
L14.3:
Spectral Differentiation: Legendre and Lobatto Polynomials
In Lecture 12 (topic L12.2) you were showed that integral of f(x) can be approximated very accurately by using the following formula
I=∫1 f(x)dx=∑wif(xi) −1 i
where xi are the roots of the Legendre Polynomials, Pn(x) shown below. xi do not include the end points x = − 1 and x = 1. The roots for P4(x) is indicated by the red dots below.
P4(x)
26
In Workshop 7, you will see that if you want to include the endpoints in the evaluation
of the integration, you have to evaluate the function at the roots of the completed Lobatto polynomials
Lo (x)=(1−x ) dx =(n+1) Pn(x)−xPn+1(x)
̂ n+1 2 dPn ( )
̂
The explicit expressions for Lon(x) for small values of n are shown below
̂2 Lo1(x) = 1 − x
̂2 Lo2(x) = (1 − x )3x
̂122 Lo3(x)=2(1−x)(15x −3)
̂123
Lo4(x) = 8(1 − x )(140x − 60x)
27
28
̂2 Lo1(x) = 1 − x
̂2 Lo2(x) = (1 − x )3x
29
30
̂122 Lo3(x)=2(1−x)(15x −3)
31
̂123
Lo4(x) = 8(1 − x )(140x − 60x)
In Workshop 7, you will see that if you want to include the endpoints in the evaluation
of the integration, you have to evaluate the function at the roots of the completed Lobatto polynomials
Lo (x)=(1−x ) dx =(n+1) Pn(x)−xPn+1(x)
̂ n+1 2 dPn ( )
̂
The explicit expressions for Lon(x) for small values of n are shown below
̂2 Lo1(x) = 1 − x
̂2 Lo2(x) = (1 − x )3x
̂122 Lo3(x)=2(1−x)(15x −3)
̂123
Lo4(x) = 8(1 − x )(140x − 60x)
̂̂
L o3(x)
L o1(x)
̂
L o2(x)
̂
L o4(x)
32
I will now calculate derivatives using the derivative matrix consisting of derivatives of the Lagrange basis polynomials Lj(x) evaluated at the Gauss-Legendre (GLe) and the Gauss-
f0
f1
f2 .
fn
Lobatto (GLo) points which corresponds to the roots of the Legendre, Pn(x), and Lobatto, ̂
Lon+1(x), Polynomials respectively. fn′(x0)
L0′(x0) L1′(x0) L2′(x0)
L0′(x1) L1′(x1) L2′(x1)
. . . . . . . . .
Ln′(x0) Ln′(x1) Ln′(x2)
L0′(x2) L1′(x2) L2′(x2)
. . …..
. . ….. L0′(xn) L1′(xn) L2′(xn) . . . Ln′(xn)
fn′(x1) = fn′(x2)
. fn′(xn)
(xi −x0)…(xi −xi−1)(xi −xi+1)….(xi −xn)
(xi −xj)(xj −x0)…(xj −xj−1)(xj −xj+1)….(xj −xn)
i≠j
Dij = Lj′(xi) =
xi −x0 xi −xi−1 xi −xi+1 xi −xn
1+..+1+1+…+1i=j
Derivative matrix
Can evaluate derivative matrix using xi from
• Equally spaced points •
•
Roots of GLe
Roots of GLo
33
Consider the function
The exact derivative is given by
f(x) = exp(x)sin(5x)
= exp(x)[sin(5x) + 5 cos(5x)]
df dx
Using the derivative matrix associated with the equally spaced nodes of order n, the results for n = 4 and n = 10 are shown in the next slide.
34
n = 4
Equally spaced points
f(x) n = 4 df (x) f4(x) dx
f(x) = exp(x)sin(5x)
df4 (xi) at equally spaced points dx
n = 10
f (x)
df (x)
f10(x)
n = 10
dx
35
df10 (xi) at equally spaced points dx
Equally spaced points
n = 4 f(x) n = 4 df (x) f4(x) dx
df4 (xi) at equally spaced points dx
n = 10
f (x)
df (x)
f10(x)
n = 10
dx
36
df10 (xi) at equally spaced points dx
GLo points
n=4 f(x)
n=4
df (x) dx
f4(x)
df4 (xi) at GLo points dx
n = 10 f(x)
n = 10
df (x) dx
f10(x)
37
df10 (xi) at GLo points dx
GLo points
n=4 f(x)
n=4
df (x) dx
f4(x)
df4 (xi) at GLo points dx
n = 10 f(x)
n = 10
df (x) dx
f10(x)
38
df10 (xi) at GLo points dx
Almost
to machine precision!
error= 1 ∑|f′(xi)−fn′(xi)| n+1
Equally spaced points
Error decreases when you increase
the number of data points, n
39
n
Equally spaced points
GLe points
error= 1 ∑|f′(xi)−fn′(xi)| n+1
40
n
Usually, GLo and GLe points are
much more accurate than equally spaced points!!
GLo points
error= 1 ∑|f′(xi)−fn′(xi)| n+1
Equally spaced points
41
n
GLe points
Now consider the Runge function
over the interval [−1,1]. The simplest way of approximating f(x) is using an Nth order interpolating polynomial and sample f(x) at evenly spaced points, given by
xi = 2i − 1 n
for i = 0,…, n.
However, this does lead to large errors near the end points which may be seen in the figures below.
42
f(x) = 1 1+25×2
f(x) = 1 1+25×2
43
f2(x)
Equally spaced points
f(x) = 1 1+25×2
44
f6(x)
45
46
f6(x)
Equally Spaced Equally Spaced f10(x) 11 points f40(x) 41 points
GLo
11 points
f10(x)
GLo
41 points
f40(x)
47
Equally spaced 11 points
f40(x)
Equally Spaced 41 points
df10 (xi) at GLo points dx
GLo
11 points
f10(x)
GLo
41 points
f40(x)
48
Equally spaced 11 points
Equally spaced 41 points
df10 (xi) at GLo points dx
df40 (xi) at Equally spaced points dx
GLo
11 points
f10(x)
GLo
41 points
f40(x)
49
Equally spaced 11 points
Equally spaced 41 points
df10 (xi) at GLo points dx
df40 (xi) at Equally spaced points dx
GLo
11 points
f10(x)
GLo
41 points
f40(x)
50
Equally spaced 11 points
Equally spaced 41 points
df10 (xi) at GLo points dx
df40 (xi) at Equally spaced points dx
df10 (xi) at GLo points dx
GLo
11 points
51
GLo
41 points
f40(x)
Equally spaced 11 points
Equally spaced 41 points
df10 (xi) at GLo points dx
df40 (xi) at Equally spaced points dx
df10 (xi) at GLo points dx
df40 (xi) at GLo points dx
GLo
11 points
GLo
41 points
52
Equally spaced 11 points
Equally spaced 41 points
df10 (xi) at GLo points dx
GLo
11 points
df40 (xi) at Equally spaced points dx
GLo
41 points
df10 (xi) at GLo points dx
df40 (xi) at GLo points dx
53