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17074216作业7

作者:17074216丁龙   发布时间:2019-05-26 21:43:40   浏览次数:143

3.4exercise

1. Decide whether the equations form a cubic spline

(b) S(x) =2x3 + x2 + 4x + 5 on [0,1]  

(x − 1)3 + 7(x − 1)2 + 12(x − 1) + 12 on [1,2]

Answer:

Property 1

There are n=3 data points. (0,5) (1,12) (2,32)

Property 2

S1(x)=6X^2+2X+4 ; S2(x)=3(x-1)^2+14(x-1)+12

S1(1)=6+2+4=12 ; S2(1)= 12

So, S1(1)= S2(1)

Property 3

S1’’(x)=12X+2 ; S2’’(x)=6(x-1) +14

S1’’(1)=14 ; S2’’(1)= 14

So, S1’’(1)= S2’’(1)

Therefore,S(x) is a cubic spline.

7题?

3.4 Computer Problems

1. Find the equations and plot the natural cubic spline that interpolates the data points (a) (0,3),(1,5), (2,4), (3,1) (b) (−1,3), (0,5), (3,1), (4,1), (5,1)

Answer:

Code:

%Program 3.5 Calculation of spline coefficients

%Calculates coefficients of cubic spline

%Input: x,y vectors of data points

% plus two optional extra data v1, vn

%Output: matrix of coefficients b1,c1,d1;b2,c2,d2;...

function coeff=splinecoeff(x,y)

n=length(x);v1=0;vn=0;

A=zeros(n,n); % matrix A is nxn

r=zeros(n,1);

for i=1:n-1 % define the deltas

   dx(i)= x(i+1)-x(i); dy(i)=y(i+1)-y(i);

end

for i=2:n-1 % load the A matrix

   A(i,i-1:i+1)=[dx(i-1) 2*(dx(i-1)+dx(i)) dx(i)];

   r(i)=3*(dy(i)/dx(i)-dy(i-1)/dx(i-1)); % right-hand side

end

% Set endpoint conditions

% Use only one of following 5 pairs:

A(1,1) = 1; % natural spline conditions

A(n,n) = 1;

%A(1,1)=2;r(1)=v1;

% curvature-adj conditions

%A(n,n)=2;r(n)=vn;

%A(1,1:2)=[2*dx(1) dx(1)];r(1)=3*(dy(1)/dx(1)-v1);

%clamped

%A(n,n-1:n)=[dx(n-1) 2*dx(n-1)];r(n)=3*(vn-dy(n-1)/dx(n-1));

%A(1,1:2)=[1 -1];

% parabol-term conditions, for n>=3

%A(n,n-1:n)=[1 -1];

%A(1,1:3)=[dx(2) -(dx(1)+dx(2)) dx(1)];

% not-a-knot, for n>=4

%A(n,n-2:n)=[dx(n-1) -(dx(n-2)+dx(n-1)) dx(n-2)];

coeff=zeros(n,3);

coeff(:,2)=Ar; % solve for c coefficients

for i=1:n-1 % solve for b and d

   coeff(i,3)=(coeff(i+1,2)-coeff(i,2))/(3*dx(i));

   coeff(i,1)=dy(i)/dx(i)-dx(i)*(2*coeff(i,2)+coeff(i+1,2))/3;

end

coeff=coeff(1:n-1,1:3);

(a):

>> splinecoeff([0;1;2;3],[3;5;4;1])

ans =

   2.6667         0   -0.6667

0.6667   -2.0000    0.3333

S(x)=

(b):

>> splinecoeff([-1;0;3;4;5],[3;5;1;1;1])

ans =

   2.5629         0   -0.5629

   0.8742   -1.6887    0.3176

  -0.6824    1.1698   -0.4874

0.1950   -0.2925    0.0975

S(x)=

4.2 Exercises

1. Fit data to the periodic model y = F3(t) = c1 + c2 cos2πt + c3 sin 2πt. Find the 2-norm error

and the RMSE.

Answer:

(a)

1=c1+c2+0*c3

3=c1+0*c2+c3

2=c1-c2+0*c3

0=c1+0*c2-c3

A=[1,1,0;1,0,1;1,-1,0;1,0,-1]

b=[1;3;2;0]

c =

   1.5000

  -0.5000

1.5000

r =

    0

    0

    0

    0

n=norm(r,2)

n =

    0

RMSE=0

(b)

1=c1+c2+0*c3

3=c1+0*c2+c3

2=c1-c2+0*c3

1=c1+0*c2-c3

A=[1,1,0;1,0,1;1,-1,0;1,0,-1]

b=[1;3;2;1]

c =

   1.7500

  -0.5000

1.0000

r =

  -0.2500

   0.2500

  -0.2500

0.2500

n=norm(r,2)

n =

0.5000

RMSE=0.2500

(c)

3=c1+c2+0*c3

1=c1-c2+0*c3

3=c1+c2+0*c3

2=c1-c2+0*c3

A=[1,1,0;1,-1,0;1,1,0;1,-1,0]

b=[3;1;3;2]

A为奇异矩阵,无解

3. Fit data to the exponential model by using linearization. Find the 2-norm of the difference

between the data points yi and the best model c1ec2ti.

Answer:

lnyi=lnc1+c2*ti

(a)

0=lnc1-2c2

ln2=lnc1

ln2=lnc1+c2

ln5=lnc1+2c2

A=[1,-2;1,0;1,1;1,2]

b=[0;ln2;ln2;ln5]

c =

0.6586

0.3615

  exp(0.658559070187365)

ans =

       1.9320

So,yi=1.9320*e0.3615*ti

(b)

0=lnc1

0=lnc1+c2

ln2=lnc1+c2

ln4=lnc1+2c2

A=[1,0;1,1;1,1;1,2]

b=[0;0;ln2;ln4]

c =

  -0.1733

   0.6931

exp(-0.173286795139986)

ans =

0.8409

So,yi=0.8409*e0.6931ti

4.2 Computer Problems

1. Fit the monthly data for Japan 2003 oil consumption, shown in the following table, with the

periodic model (4.9), and calculate the RMSE:

Answer:

yi=c1ec2ti.

A=[1,6.224;1,6.665;1,6.241;1,5.302;1,5.073;1,5.127;1,4.994;1,5.012;1,5.108;1,5.377;1, 5.510 ;1,6.372]

b=[ln1; ln2; ln3; ln4; ln5; ln6;ln7;ln8;ln9;ln10; ln11;ln12]

c=(A'*A)(A'*b)

c =

5.2284

  -0.6381

exp(5.228409900541486)

ans =

 186.4960

So,yi=186.4960*e-0.6381ti

r=b-A*c

r =

-1.2571

-0.2825

  -0.1476

  -0.4591

  -0.3821

  -0.1653

  -0.0960

   0.0490

   0.2281

   0.5051

   0.6852

   1.3223

>> norm(r,2)

ans =

2.1454

RMSE=(norm(r,2)^2/12)^(1/2)

RMSE =

0.6193

3.Consider the world population data of Computer Problem 3.1.1. Find the best exponential fifit ofthe data points by using linearization. Estimate the 1980 population, and fifind the estimationerror

Answer:








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