covert data as matrix

Question

hasan s 2021-2-23

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/754109-covert-data-as-matrix

评论： hasan s 2021-3-21

Hi

need help .... to CONVERT data 5*15 ,that I get it from the following program ....as MATRIX contain row i=5 and column j =15

this is the program:

Q =(unifrnd(0,1,5,15))
save Q

if there is alot of number ... that is difficult to write matrix by using ; in every row.

WHAT I WRITE in the same program???? IF ANY PROF. KNOW...thanks alot

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Answer 1

Walter Roberson 2021-2-23

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/754109-covert-data-as-matrix#answer_631374

在 MATLAB Online 中打开

Q =(unifrnd(0,1,5,15))
Q = 5×15
6639    0.4524    0.1932    0.7793    0.4019    0.0852    0.7317    0.5443    0.6400    0.9254    0.4015    0.3222    0.5093    0.5372    0.8024
1512    0.8782    0.8470    0.3862    0.3625    0.4341    0.9869    0.7286    0.4532    0.8665    0.5559    0.6165    0.0237    0.7311    0.2724
5689    0.8799    0.7683    0.6209    0.7952    0.0832    0.7453    0.1717    0.6402    0.4705    0.7399    0.6973    0.7990    0.9779    0.6402
6130    0.2077    0.8421    0.2649    0.0269    0.1636    0.5503    0.0439    0.2371    0.0030    0.7877    0.7767    0.4297    0.3760    0.0445
0602    0.1058    0.4839    0.0780    0.9417    0.1768    0.0868    0.8750    0.1341    0.3844    0.3747    0.5078    0.4752    0.1798    0.6737
save('Q.txt', 'Q', '-ascii', '-double')
!cat Q.txt
6392079079920463e-01   4.5235571596635848e-01   1.9316148221534890e-01   7.7929796267166485e-01   4.0187340004345129e-01   8.5151634550524102e-02   7.3169640609405273e-01   5.4429391635246049e-01   6.3998783353082334e-01   9.2536320087337998e-01   4.0149054447634047e-01   3.2223963473914463e-01   5.0930643031449951e-01   5.3716002761190185e-01   8.0235908232593778e-01
5121990903952298e-01   8.7824242681845399e-01   8.4704362397489430e-01   3.8621816305805401e-01   3.6246145933476215e-01   4.3414004021693398e-01   9.8688763144956315e-01   7.2858731426382739e-01   4.5318793660766010e-01   8.6646738130811141e-01   5.5589156556919395e-01   6.1649313001817796e-01   2.3700376853586835e-02   7.3108013038230801e-01   2.7240432119647917e-01
6888971954278122e-01   8.7987852225658303e-01   7.6834295200565705e-01   6.2087530894804055e-01   7.9523156025077957e-01   8.3189592572391668e-02   7.4527204071029118e-01   1.7168914014464565e-01   6.4024339383210793e-01   4.7051120988032114e-01   7.3991134708839257e-01   6.9727632429789299e-01   7.9901163482237803e-01   9.7786053731897582e-01   6.4020663985588955e-01
1299751870622943e-01   2.0769354912309579e-01   8.4212549598643649e-01   2.6494012680566281e-01   2.6923822357603600e-02   1.6360315270545678e-01   5.5026885098899259e-01   4.3926305010117850e-02   2.3709963300152437e-01   3.0286948911313338e-03   7.8774144114460520e-01   7.7669296695025258e-01   4.2969450904609452e-01   3.7597209698334411e-01   4.4524513343272121e-02
0167559639795587e-02   1.0578433110097485e-01   4.8392347456417129e-01   7.8019304296682912e-02   9.4166940478339789e-01   1.7681457887673979e-01   8.6762820992417611e-02   8.7495394112995439e-01   1.3413795755485414e-01   3.8441577935736448e-01   3.7470793006034464e-01   5.0775182098339389e-01   4.7524575666372515e-01   1.7981150219493136e-01   6.7370366631287792e-01

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Walter Roberson 2021-2-23

在 MATLAB Online 中打开

After the line

Q = unifrnd(0,1,5,15);

then Q is already the matrix of data. You can

disp(Q)
4793    0.8762    0.9996    0.9766    0.1466    0.5964    0.1777    0.9023    0.9839    0.7071    0.8686    0.4813    0.8447    0.1321    0.4197
8243    0.4757    0.0873    0.4524    0.3155    0.6068    0.1987    0.0429    0.4452    0.7281    0.2872    0.0883    0.7353    0.0966    0.9672
1782    0.7150    0.8373    0.8565    0.8762    0.6715    0.0130    0.8911    0.7683    0.0246    0.3806    0.5532    0.8728    0.1065    0.3088
3164    0.9974    0.8146    0.8010    0.7779    0.1290    0.6169    0.6383    0.7632    0.1078    0.5700    0.6541    0.3456    0.7388    0.9914
3927    0.3446    0.4765    0.3013    0.5837    0.6138    0.8402    0.1154    0.0497    0.2661    0.1674    0.3935    0.6073    0.8192    0.7342

if you need to, or pass it to something else, like

std(Q)
ans = 1×15
    0.2425    0.2714    0.3643    0.2868    0.3069    0.2224    0.3449    0.4151    0.3637    0.3319    0.2742    0.2157    0.2147    0.3668    0.3114

hasan s 2021-2-24

在 MATLAB Online 中打开

thank you very much for your reply.

After your valuable information prof. Walter ,

I take

Q1 = unifrnd(0,1,10,1000);
RST(Q1)

but , take the data Q1, in the another program , alot of outputs is NAN. That is another program

sum1=0;sum2=0;sum3=0;sum4=0;sum5=0;sum6=0;sum7=0;sum8=0;sum9=0;
sum10=0;sum11=0;sum12=0;sum13=0;sum14=0;
err=10;
A=1;
B=2;
G=3;
Q=Q1;
while(err>[0.00001;0.00005;0.00002])
for j=1:1000
      for i=1:10  
        sum1=sum1+1/(A(j)+(G(j)*B(j)*((Q(i,j))^(G(j)-1))));
        sum2=sum2+(Q(i,j));
        sum3=sum3+(G(j)*((Q(i,j))^(G(j)-1)))/(A(j)+(G(j)*B(j)*((Q(i,j))^(G(j)-1))));
        sum4=sum4+(Q(i,j))^(G(j));
        sum5=sum5+(((G(j)*log(Q(i,j)))*((Q(i,j))^(G(j)-1)))+((Q(i,j))^(G(j)-1)))/(A(j)+(B(j)*G(j)*((Q(i,j))^(G(j)-1))));
        sum6=sum6+((Q(i,j))^(G(j)))*log(Q(i,j));
        sum7=sum7+1/(A(j)+(G(j)*B(j)*((Q(i,j))^(G(j)-1))))^2;
        sum8=sum8+(G(j)*((Q(i,j))^(G(j)-1)))/(A(j)+(B(j)*G(j)*((Q(i,j))^(G(j)-1))))^2;
        sum9=sum9+(((G(j)*log(Q(i,j)))*((Q(i,j))^(G(j)-1)))+((Q(i,j))^(G(j))))/(A(j)+(B(j)*G(j)*((Q(i,j))^(G(j)-1))))^2;
        sum10=sum10+(((G(j))^2)*((Q(i,j))^(2*G(j)-1)))/(A(j)+(B(j)*G(j)*((Q(i,j))^(G(j)-1))))^2;
        sum11=sum11+((((G(j))*log(Q(i,j)))*((Q(i,j))^(2*(G(j)-1))))+((Q(i,j))^(2*(G(j)-1))))/(A(j)+(B(j)*G(j)*((Q(i,j))^(G(j)-1))))^2;
        sum12=sum12+(((G(j)*(log(Q(i,j)))^2)*((Q(i,j))^(G(j)-1)))+((2*log(Q(i,j)))*((Q(i,j))^(G(j)-1))))/(A(j)+(B(j)*G(j)*((Q(i,j))^(G(j)-1))));
        sum13=sum13+(((G(j)*log(Q(i,j)))*((Q(i,j))^(G(j)-1)))+((Q(i,j))^(G(j)-1)))^2/(A(j)+(B(j)*G(j)*((Q(i,j))^(G(j)-1))))^2;
        sum14=sum14+(log(Q(i,j)))^2*((Q(i,j))^(G(j)));
    end
    x=[sum1-sum2;sum3-sum4;B(j)*sum5-B(j)*sum6];
    y1=(-sum1);
    y2=(-sum8);
    y3=(-B(j))*sum9;
    y4=(-sum8);
    y5=(-sum10);
    y6=((-B(j))*(G(j)))*sum11+sum5-sum6;
    y7=(-B(j))*sum9;
    y8=(-(B(j)*G(j)))*sum11+sum5-sum6;
    y9=(B(j))*sum12-((B(j))^2)*sum13-(B(j))*sum14;
    d=[y1,y2,y3;y4,y5,y6;y7,y8,y9];
    d=real(d);
    x=real(x);
    z0=[A(j);B(j);G(j)]
    p=inv(d)*x
    z=z0-p;
    err=abs(z-z0);
    A(j+1)=z(1);
    B(j+1)=z(2);
    G(j+1)=z(3);
    z0=z;
    end
end

A

I dont know the reason why ???what I change please to get all values of A are real, not NAN???

I must take column1, column 2,..., column 1000

I must take the first column of matrix to get first value of A

and take the second column of matix to find second value of A

and so on to get 1000 value of A

Can you help me ???

hasan s 2021-2-25

编辑：hasan s 2021-2-25

在 MATLAB Online 中打开

yes..prof. Walter

I need , as you say >> the value of sum1 for j = 2 only depends upon the information for this column, and does not include values calculated in previous column.

I change it ,

B=2;
G=3;
Q=Q1;
for j=1:1000
    sum1=0;sum2=0;sum3=0;sum4=0;sum5=0;sum6=0;sum7=0;sum8=0;sum9=0;
    sum10=0;sum11=0;sum12=0;sum13=0;sum14=0;
      for i=1:10  
        sum1=sum1+1/(A(j)+(G(j)*B(j)*((Q(i,j))^(G(j)-1))));
        sum2=sum2+(Q(i,j));
        sum3=sum3+(G(j)*((Q(i,j))^(G(j)-1)))/(A(j)+(G(j)*B(j)*((Q(i,j))^(G(j)-1))));
        sum4=sum4+(Q(i,j))^(G(j));
        sum5=sum5+(((G(j)*log(Q(i,j)))*((Q(i,j))^(G(j)-1)))+((Q(i,j))^(G(j)-1)))/(A(j)+(B(j)*G(j)*((Q(i,j))^(G(j)-1))));
        sum6=sum6+((Q(i,j))^(G(j)))*log(Q(i,j));
        sum7=sum7+1/(A(j)+(G(j)*B(j)*((Q(i,j))^(G(j)-1))))^2;
        sum8=sum8+(G(j)*((Q(i,j))^(G(j)-1)))/(A(j)+(B(j)*G(j)*((Q(i,j))^(G(j)-1))))^2;
        sum9=sum9+(((G(j)*log(Q(i,j)))*((Q(i,j))^(G(j)-1)))+((Q(i,j))^(G(j))))/(A(j)+(B(j)*G(j)*((Q(i,j))^(G(j)-1))))^2;
        sum10=sum10+(((G(j))^2)*((Q(i,j))^(2*G(j)-1)))/(A(j)+(B(j)*G(j)*((Q(i,j))^(G(j)-1))))^2;
        sum11=sum11+((((G(j))*log(Q(i,j)))*((Q(i,j))^(2*(G(j)-1))))+((Q(i,j))^(2*(G(j)-1))))/(A(j)+(B(j)*G(j)*((Q(i,j))^(G(j)-1))))^2;
        sum12=sum12+(((G(j)*(log(Q(i,j)))^2)*((Q(i,j))^(G(j)-1)))+((2*log(Q(i,j)))*((Q(i,j))^(G(j)-1))))/(A(j)+(B(j)*G(j)*((Q(i,j))^(G(j)-1))));
        sum13=sum13+(((G(j)*log(Q(i,j)))*((Q(i,j))^(G(j)-1)))+((Q(i,j))^(G(j)-1)))^2/(A(j)+(B(j)*G(j)*((Q(i,j))^(G(j)-1))))^2;
        sum14=sum14+(log(Q(i,j)))^2*((Q(i,j))^(G(j)));
      end
    sum1
    sum2
    sum3
    sum4
    sum5
    sum6
    sum7
    sum8
    sum9
    sum10
    sum11
    sum12
    sum13
    sum14
    x=[sum1-sum2;sum3-sum4;B(j)*sum5-B(j)*sum6];
    y1=(-sum1);
    y2=(-sum8);
    y3=(-B(j))*sum9;
    y4=(-sum8);
    y5=(-sum10);
    y6=((-B(j))*(G(j)))*sum11+sum5-sum6;
    y7=(-B(j))*sum9;
    y8=(-(B(j)*G(j)))*sum11+sum5-sum6;
    y9=(B(j))*sum12-((B(j))^2)*sum13-(B(j))*sum14;
    d=[y1,y2,y3;y4,y5,y6;y7,y8,y9];
    d=real(d);
    x=real(x);
    z0=[A(j);B(j);G(j)]
    p=inv(d)*x
    z=z0-p;
    err=abs(z-z0);
    A(j+1)=z(1);
    B(j+1)=z(2);
    G(j+1)=z(3);
    z0=z;
    end
A;

When I run the program , all sum is NAN ...except sum 2...since it is the only which contain data(i,j) only

but other sums contain A(j), B(j), G(j),log(Q(i,j));

the NAN of sums give all the 1000 - value of A is NAN

What I do ??? please

hasan s 2021-2-26

在 MATLAB Online 中打开

I change it again to the following..but also most of values of A is NAN.

A=1;
B=2;
G=3;
Q=Q1;
   for j=1:1000
      s1=0;sum1=0;
    for i=1:n 
       s1=s1+1
        sum1=sum1+1/(A(j)+(G(j)*B(j)*((Q(i,j))^(G(j)-1))));
    end
    sum1
    s2=0;sum2=0;
    for i=1:n 
       s2=s2+1;
        sum2=sum2+(Q(i,j));
        end
    sum2;
    s3=0;sum3=0;
    for i=1:n 
    s3=s3+1 ; 
        sum3=sum3+(G(j)*((Q(i,j))^(G(j)-1)))/(A(j)+(G(j)*B(j)*((Q(i,j))^(G(j)-1))));
        end 
    sum3;
    s4=0;sum4=0;
    for i=1:n
    s4=s4+1; 
        sum4=sum4+(Q(i,j))^(G(j));
        end
    sum4;
    s5=0;sum5=0;
    for i=1:n 
       s5=s5+1 ;   
        sum5=sum5+(((G(j)*log(Q(i,j)))*((Q(i,j))^(G(j)-1)))+((Q(i,j))^(G(j)-1)))/(A(j)+(B(j)*G(j)*((Q(i,j))^(G(j)-1))));
        end
    sum5;
    s6=0;sum6=0;
    for i=1:n 
     s6=s6+1 ;  
        sum6=sum6+((Q(i,j))^(G(j)))*log(Q(i,j));
        end
    sum6;
    s7=0;sum7=0;
       for i=1:n 
      s7=s7+1 ; 
        sum7=sum7+1/(A(j)+(G(j)*B(j)*((Q(i,j))^(G(j)-1))))^2;
        end
       sum7;
       s8=0;sum8=0;
       for i=1:n 
       s8=s8+1 ;   
        sum8=sum8+(G(j)*((Q(i,j))^(G(j)-1)))/(A(j)+(B(j)*G(j)*((Q(i,j))^(G(j)-1))))^2;
        end
       sum8;
       s9=0;sum9=0;
       for i=1:n
       s9=s9+1;    
        sum9=sum9+(((G(j)*log(Q(i,j)))*((Q(i,j))^(G(j)-1)))+((Q(i,j))^(G(j))))/(A(j)+(B(j)*G(j)*((Q(i,j))^(G(j)-1))))^2;
        end
       sum9;
       s10=0;sum10=0;
       for i=1:n 
       s10=s10+1; 
        sum10=sum10+(((G(j))^2)*((Q(i,j))^(2*G(j)-1)))/(A(j)+(B(j)*G(j)*((Q(i,j))^(G(j)-1))))^2;
        end
       sum10;
       s11=0;sum11=0;
       for i=1:n
       s11=s11+1 ;
        sum11=sum11+((((G(j))*log(Q(i,j)))*((Q(i,j))^(2*(G(j)-1))))+((Q(i,j))^(2*(G(j)-1))))/(A(j)+(B(j)*G(j)*((Q(i,j))^(G(j)-1))))^2;
        end
       sum11
       s12=0;sum12=0;
       for i=1:n
       s12=s12+1 ;
        sum12=sum12+(((G(j)*(log(Q(i,j)))^2)*((Q(i,j))^(G(j)-1)))+((2*log(Q(i,j)))*((Q(i,j))^(G(j)-1))))/(A(j)+(B(j)*G(j)*((Q(i,j))^(G(j)-1))));
       end
       sum12
      s13=0;sum13=0;
       for i=1:n
       s13=s13+1 ; 
        sum13=sum13+(((G(j)*log(Q(i,j)))*((Q(i,j))^(G(j)-1)))+((Q(i,j))^(G(j)-1)))^2/(A(j)+(B(j)*G(j)*((Q(i,j))^(G(j)-1))))^2;
       end
       sum13
       s14=0;sum14=0;
       for i=1:n
       s14=s14+1 ;
        sum14=sum14+(log(Q(i,j)))^2*((Q(i,j))^(G(j)));
      end
    sum14
    
    x=[sum1-sum2;sum3-sum4;B(j)*sum5-B(j)*sum6];
    y1=(-sum1);
    y2=(-sum8);
    y3=(-B(j))*sum9;
    y4=(-sum8);
    y5=(-sum10);
    y6=((-B(j))*(G(j)))*sum11+sum5-sum6;
    y7=(-B(j))*sum9;
    y8=(-(B(j)*G(j)))*sum11+sum5-sum6;
    y9=(B(j))*sum12-((B(j))^2)*sum13-(B(j))*sum14;
    d=[y1,y2,y3;y4,y5,y6;y7,y8,y9];
    d=real(d);
    x=real(x);
    z0=[A(j);B(j);G(j)]
    p=d\x;
    z=z0-p;
    err=abs(z-z0);
    A(j+1)=z(1);
    B(j+1)=z(2);
    G(j+1)=z(3);
    z0=z;
    end
A

what I change ???

Walter Roberson 2021-2-26

在 MATLAB Online 中打开

rng(123456)

digits(32)

N = 1000;

Q1 = rand(10,N);

A = zeros(1,N+1,'sym');

B = zeros(1,N+1,'sym');

G = zeros(1,N+1,'sym');

A(1) = 1;

B(1) = 2;

G(1) = 3;

Q = vpa(Q1);

for j=1:1000

sum1=0;sum2=0;sum3=0;sum4=0;sum5=0;sum6=0;sum7=0;sum8=0;sum9=0;

sum10=0;sum11=0;sum12=0;sum13=0;sum14=0;

for i=1:10

sum1=sum1+1/(A(j)+(G(j)*B(j)*((Q(i,j))^(G(j)-1))));

sum2=sum2+(Q(i,j));

sum3=sum3+(G(j)*((Q(i,j))^(G(j)-1)))/(A(j)+(G(j)*B(j)*((Q(i,j))^(G(j)-1))));

sum4=sum4+(Q(i,j))^(G(j));

sum5=sum5+(((G(j)*log(Q(i,j)))*((Q(i,j))^(G(j)-1)))+((Q(i,j))^(G(j)-1)))/(A(j)+(B(j)*G(j)*((Q(i,j))^(G(j)-1))));

sum6=sum6+((Q(i,j))^(G(j)))*log(Q(i,j));

sum7=sum7+1/(A(j)+(G(j)*B(j)*((Q(i,j))^(G(j)-1))))^2;

sum8=sum8+(G(j)*((Q(i,j))^(G(j)-1)))/(A(j)+(B(j)*G(j)*((Q(i,j))^(G(j)-1))))^2;

sum9=sum9+(((G(j)*log(Q(i,j)))*((Q(i,j))^(G(j)-1)))+((Q(i,j))^(G(j))))/(A(j)+(B(j)*G(j)*((Q(i,j))^(G(j)-1))))^2;

sum10=sum10+(((G(j))^2)*((Q(i,j))^(2*G(j)-1)))/(A(j)+(B(j)*G(j)*((Q(i,j))^(G(j)-1))))^2;

sum11=sum11+((((G(j))*log(Q(i,j)))*((Q(i,j))^(2*(G(j)-1))))+((Q(i,j))^(2*(G(j)-1))))/(A(j)+(B(j)*G(j)*((Q(i,j))^(G(j)-1))))^2;

sum12=sum12+(((G(j)*(log(Q(i,j)))^2)*((Q(i,j))^(G(j)-1)))+((2*log(Q(i,j)))*((Q(i,j))^(G(j)-1))))/(A(j)+(B(j)*G(j)*((Q(i,j))^(G(j)-1))));

sum13=sum13+(((G(j)*log(Q(i,j)))*((Q(i,j))^(G(j)-1)))+((Q(i,j))^(G(j)-1)))^2/(A(j)+(B(j)*G(j)*((Q(i,j))^(G(j)-1))))^2;

sum14=sum14+(log(Q(i,j)))^2*((Q(i,j))^(G(j)));

end

% disp(sum1)

% disp(sum2)

% disp(sum3)

% disp(sum4)

% disp(sum5)

% disp(sum6)

% disp(sum7)

% disp(sum8)

% disp(sum9)

% disp(sum10)

% disp( sum11)

% disp(sum12)

% disp(sum13)

% disp(sum14)

x=[sum1-sum2;sum3-sum4;B(j)*sum5-B(j)*sum6];

y1=(-sum1);

y2=(-sum8);

y3=(-B(j))*sum9;

y4=(-sum8);

y5=(-sum10);

y6=((-B(j))*(G(j)))*sum11+sum5-sum6;

y7=(-B(j))*sum9;

y8=(-(B(j)*G(j)))*sum11+sum5-sum6;

y9=(B(j))*sum12-((B(j))^2)*sum13-(B(j))*sum14;

d=[y1,y2,y3;y4,y5,y6;y7,y8,y9];

d=real(d);

x=real(x);

z0=[A(j);B(j);G(j)];

if rank(d) < size(d,1)

fprintf('Early termination at j = %d due to rank failure\n', j);

break;

end

p=d\x;

z=z0-p;

err=abs(z-z0);

A(j+1)=z(1);

B(j+1)=z(2);

G(j+1)=z(3);

z0=z;

end

Early termination at j = 20 due to rank failure

disp(vpa(A(1:j),5))

disp(vpa(B(1:j),5))

disp(vpa(G(1:j),5))

With this particular rng seed and digits(50), the output will be

Early termination at j = 29 due to rank failure

[1.0, 1.0349, 0.86581, 0.99576, 1.0175, -0.18171, -5.6105, -5.8035, -6.7836, -29.428, -2.7838, -0.10434, 0.44927, 1.2216, 1.5661, 1.4736, 1.3823, 1.7725, 1.6324, 1.4034, 1.7778, 2.0341, 2.0359, 1.7352, 1.9228, 2.0852, 2.1461, 1.7618, 2.0146]

[2.0, 1.801, 2.8963, 1.3459, 2.3176, 5.4269, 22.873, 29.808, 46.481, 116.8, 340.77, 1660.9, 1538.6, 1379.8, 1382.8, 1427.6, 1745.9, 1747.5, 1736.8, 1706.3, 1707.5, 1704.6, 1701.4, 1685.4, 1684.3, 1684.5, 1673.8, 1665.6, 1664.2]

[3.0, 2.9351, 2.8183, 2.4457, 3.3412, 0.52283, 1.9538, 2.1644, 2.4915, 6.3134, 15.635, 55.825, 69.911, 79.92, 82.984, 112.25, 194.11, 197.78, 248.55, 291.33, 293.3, 306.1, 319.32, 387.32, 398.36, 401.89, 435.72, 550.72, 567.52]

What is happening? Well, once G gets large enough then since Q < 1, Q^G will be small and most of your sums get small but the first one stays comparatively large. This leads to small x(2) and especially small x(3) which leads to large responses on the p\x and that leads to even larger G values . Eventually everything in d except the first element vanishes compared to the first element, giving you a matrix of rank 1 (or 2, depending on how much arithmetic error you allowed for.)

You can increase digits() to postpone the exact place the rank becomes a problem, but you don't get far.

hasan s 2021-2-27

编辑：hasan s 2021-2-27

在 MATLAB Online 中打开

I change it : Is that right?? is the algorithm right for newton raphson of non linear equations??

N = 1000;
Q1 = rand(10,N);
Q=reshape(Q1,1,[]); % convert matrix to row vector
A(1) = 1;
B(1) = 2;
G(1) = 3;
for i=1:2
    sum1=0;sum2=0;sum3=0;sum4=0;sum5=0;sum6=0;sum7=0;sum8=0;sum9=0;
    sum10=0;sum11=0;sum12=0;sum13=0;sum14=0;
    for j=1:10000
        sum1=sum1+1/(A(i)+(G(i)*B(i)*((Q(j))^(G(i)-1))));
        sum2=sum2+(Q(j));
        sum3=sum3+(G(i)*((Q(j))^(G(i)-1)))/(A(i)+(G(i)*B(i)*((Q(j))^(G(i)-1))));
        sum4=sum4+(Q(j))^(G(i));
        sum5=sum5+(((G(i)*log(Q(j)))*((Q(j))^(G(i)-1)))+((Q(j))^(G(i)-1)))/(A(i)+(B(i)*G(i)*((Q(j))^(G(i)-1))));
        sum6=sum6+((Q(j))^(G(i)))*log(Q(j));
        sum7=sum7+1/(A(i)+(G(i)*B(i)*((Q(j))^(G(i)-1))))^2;
        sum8=sum8+(G(i)*((Q(j))^(G(i)-1)))/(A(i)+(B(i)*G(i)*((Q(j))^(G(i)-1))))^2;
        sum9=sum9+(((G(i)*log(Q(j)))*((Q(j))^(G(i)-1)))+((Q(j))^(G(i))))/(A(i)+(B(i)*G(i)*((Q(j))^(G(i)-1))))^2;
        sum10=sum10+(((G(i))^2)*((Q(j))^(2*G(i)-1)))/(A(i)+(B(i)*G(i)*((Q(j))^(G(i)-1))))^2;
        sum11=sum11+((((G(i))*log(Q(j)))*((Q(j))^(2*(G(i)-1))))+((Q(j))^(2*(G(i)-1))))/(A(i)+(B(i)*G(i)*((Q(j))^(G(i)-1))))^2;
        sum12=sum12+(((G(i)*(log(Q(j)))^2)*((Q(j))^(G(i)-1)))+((2*log(Q(j)))*((Q(j))^(G(i)-1))))/(A(i)+(B(i)*G(i)*((Q(j))^(G(i)-1))));
        sum13=sum13+(((G(i)*log(Q(j)))*((Q(j))^(G(i)-1)))+((Q(j))^(G(i)-1)))^2/(A(i)+(B(i)*G(i)*((Q(j))^(G(i)-1))))^2;
        sum14=sum14+(log(Q(j)))^2*((Q(j))^(G(i)));
    end
    %     disp(sum1)
%     disp(sum2)
%     disp(sum3)
%     disp(sum4)
%     disp(sum5)
%     disp(sum6)
%     disp(sum7)
%     disp(sum8)
%     disp(sum9)
%     disp(sum10)
%     disp( sum11)
%     disp(sum12)
%     disp(sum13)
%     disp(sum14)
    x=[sum1-sum2;sum3-sum4;B(i)*sum5-B(i)*sum6];
    y1=(-sum1);
    y2=(-sum8);
    y3=(-B(i))*sum9;
    y4=(-sum8);
    y5=(-sum10);
    y6=((-B(i))*(G(i)))*sum11+sum5-sum6;
    y7=(-B(i))*sum9;
    y8=(-(B(i)*G(i)))*sum11+sum5-sum6;
    y9=(B(i))*sum12-((B(i))^2)*sum13-(B(i))*sum14;
    d=[y1,y2,y3;y4,y5,y6;y7,y8,y9];
    d=real(d);
    x=real(x);
    z0=[A(i);B(i);G(i)]
    p=d\x;
    z=z0-p;
    err=abs(z-z0);
    A(i+1)=z(1);
    B(i+1)=z(2);
    G(i+1)=z(3);
    z0=z;
end
A
B
G

when run ,I get

A =

1.0000 1.2030 1.2808 1.3323 1.3486 1.3537 1.3557 1.3565 1.3569 1.3570 1.3571

B =

2.0000 2.6133 2.7633 2.7282 2.7355 2.7333 2.7333 2.7331 2.7331 2.7330 2.7330

G =

3.0000 4.7106 6.1480 6.9308 7.2657 7.3095 7.3186 7.3223 7.3240 7.3247 7.3250

but ...why the out put of A,B,G have ten output ,while I determinant i=2 , ??

thanks a lot for your help prof. Walter

hasan s 2021-2-27

编辑：hasan s 2021-2-27

在 MATLAB Online 中打开

ok prof. thanks for reply

what about the program??is right?

N = 1000;
Q1 = rand(10,N);
Q=reshape(Q1,1,[]); % convert matrix to row vector
A(1) = 1;
B(1) = 2;
G(1) = 3;
for i=1:2
    sum1=0;sum2=0;sum3=0;sum4=0;sum5=0;sum6=0;sum7=0;sum8=0;sum9=0;
    sum10=0;sum11=0;sum12=0;sum13=0;sum14=0;
    for j=1:10000
        sum1=sum1+1/(A(i)+(G(i)*B(i)*((Q(j))^(G(i)-1))));
        sum2=sum2+(Q(j));
        sum3=sum3+(G(i)*((Q(j))^(G(i)-1)))/(A(i)+(G(i)*B(i)*((Q(j))^(G(i)-1))));
        sum4=sum4+(Q(j))^(G(i));
        sum5=sum5+(((G(i)*log(Q(j)))*((Q(j))^(G(i)-1)))+((Q(j))^(G(i)-1)))/(A(i)+(B(i)*G(i)*((Q(j))^(G(i)-1))));
        sum6=sum6+((Q(j))^(G(i)))*log(Q(j));
        sum7=sum7+1/(A(i)+(G(i)*B(i)*((Q(j))^(G(i)-1))))^2;
        sum8=sum8+(G(i)*((Q(j))^(G(i)-1)))/(A(i)+(B(i)*G(i)*((Q(j))^(G(i)-1))))^2;
        sum9=sum9+(((G(i)*log(Q(j)))*((Q(j))^(G(i)-1)))+((Q(j))^(G(i))))/(A(i)+(B(i)*G(i)*((Q(j))^(G(i)-1))))^2;
        sum10=sum10+(((G(i))^2)*((Q(j))^(2*G(i)-1)))/(A(i)+(B(i)*G(i)*((Q(j))^(G(i)-1))))^2;
        sum11=sum11+((((G(i))*log(Q(j)))*((Q(j))^(2*(G(i)-1))))+((Q(j))^(2*(G(i)-1))))/(A(i)+(B(i)*G(i)*((Q(j))^(G(i)-1))))^2;
        sum12=sum12+(((G(i)*(log(Q(j)))^2)*((Q(j))^(G(i)-1)))+((2*log(Q(j)))*((Q(j))^(G(i)-1))))/(A(i)+(B(i)*G(i)*((Q(j))^(G(i)-1))));
        sum13=sum13+(((G(i)*log(Q(j)))*((Q(j))^(G(i)-1)))+((Q(j))^(G(i)-1)))^2/(A(i)+(B(i)*G(i)*((Q(j))^(G(i)-1))))^2;
        sum14=sum14+(log(Q(j)))^2*((Q(j))^(G(i)));
    end
    %     disp(sum1)
%     disp(sum2)
%     disp(sum3)
%     disp(sum4)
%     disp(sum5)
%     disp(sum6)
%     disp(sum7)
%     disp(sum8)
%     disp(sum9)
%     disp(sum10)
%     disp( sum11)
%     disp(sum12)
%     disp(sum13)
%     disp(sum14)
    x=[sum1-sum2;sum3-sum4;B(i)*sum5-B(i)*sum6];
    y1=(-sum1);
    y2=(-sum8);
    y3=(-B(i))*sum9;
    y4=(-sum8);
    y5=(-sum10);
    y6=((-B(i))*(G(i)))*sum11+sum5-sum6;
    y7=(-B(i))*sum9;
    y8=(-(B(i)*G(i)))*sum11+sum5-sum6;
    y9=(B(i))*sum12-((B(i))^2)*sum13-(B(i))*sum14;
    d=[y1,y2,y3;y4,y5,y6;y7,y8,y9];
    d=real(d);
    x=real(x);
    z0=[A(i);B(i);G(i)]
    p=d\x;
    z=z0-p;
    err=abs(z-z0);
    A(i+1)=z(1);
    B(i+1)=z(2);
    G(i+1)=z(3);
    z0=z;
end
A
B
G

when run ,I get

A =

1.0000 1.2030 1.2808 1.3323 1.3486 1.3537 1.3557 1.3565 1.3569 1.3570 1.3571

B =

2.0000 2.6133 2.7633 2.7282 2.7355 2.7333 2.7333 2.7331 2.7331 2.7330 2.7330

G =

3.0000 4.7106 6.1480 6.9308 7.2657 7.3095 7.3186 7.3223 7.3240 7.3247 7.3250

why the out put of A,B,G have ten output ,while I determinant i=2 , ??

Walter Roberson 2021-2-27

在 MATLAB Online 中打开

N = 1000;
Q1 = rand(10,N);
Q=reshape(Q1,1,[]); % convert matrix to row vector
A(1) = 1;
B(1) = 2;
G(1) = 3;
for i=1:2
    sum1=0;sum2=0;sum3=0;sum4=0;sum5=0;sum6=0;sum7=0;sum8=0;sum9=0;
    sum10=0;sum11=0;sum12=0;sum13=0;sum14=0;
    for j=1:10000
        sum1=sum1+1/(A(i)+(G(i)*B(i)*((Q(j))^(G(i)-1))));
        sum2=sum2+(Q(j));
        sum3=sum3+(G(i)*((Q(j))^(G(i)-1)))/(A(i)+(G(i)*B(i)*((Q(j))^(G(i)-1))));
        sum4=sum4+(Q(j))^(G(i));
        sum5=sum5+(((G(i)*log(Q(j)))*((Q(j))^(G(i)-1)))+((Q(j))^(G(i)-1)))/(A(i)+(B(i)*G(i)*((Q(j))^(G(i)-1))));
        sum6=sum6+((Q(j))^(G(i)))*log(Q(j));
        sum7=sum7+1/(A(i)+(G(i)*B(i)*((Q(j))^(G(i)-1))))^2;
        sum8=sum8+(G(i)*((Q(j))^(G(i)-1)))/(A(i)+(B(i)*G(i)*((Q(j))^(G(i)-1))))^2;
        sum9=sum9+(((G(i)*log(Q(j)))*((Q(j))^(G(i)-1)))+((Q(j))^(G(i))))/(A(i)+(B(i)*G(i)*((Q(j))^(G(i)-1))))^2;
        sum10=sum10+(((G(i))^2)*((Q(j))^(2*G(i)-1)))/(A(i)+(B(i)*G(i)*((Q(j))^(G(i)-1))))^2;
        sum11=sum11+((((G(i))*log(Q(j)))*((Q(j))^(2*(G(i)-1))))+((Q(j))^(2*(G(i)-1))))/(A(i)+(B(i)*G(i)*((Q(j))^(G(i)-1))))^2;
        sum12=sum12+(((G(i)*(log(Q(j)))^2)*((Q(j))^(G(i)-1)))+((2*log(Q(j)))*((Q(j))^(G(i)-1))))/(A(i)+(B(i)*G(i)*((Q(j))^(G(i)-1))));
        sum13=sum13+(((G(i)*log(Q(j)))*((Q(j))^(G(i)-1)))+((Q(j))^(G(i)-1)))^2/(A(i)+(B(i)*G(i)*((Q(j))^(G(i)-1))))^2;
        sum14=sum14+(log(Q(j)))^2*((Q(j))^(G(i)));
    end
    %     disp(sum1)
%     disp(sum2)
%     disp(sum3)
%     disp(sum4)
%     disp(sum5)
%     disp(sum6)
%     disp(sum7)
%     disp(sum8)
%     disp(sum9)
%     disp(sum10)
%     disp( sum11)
%     disp(sum12)
%     disp(sum13)
%     disp(sum14)
    x=[sum1-sum2;sum3-sum4;B(i)*sum5-B(i)*sum6];
    y1=(-sum1);
    y2=(-sum8);
    y3=(-B(i))*sum9;
    y4=(-sum8);
    y5=(-sum10);
    y6=((-B(i))*(G(i)))*sum11+sum5-sum6;
    y7=(-B(i))*sum9;
    y8=(-(B(i)*G(i)))*sum11+sum5-sum6;
    y9=(B(i))*sum12-((B(i))^2)*sum13-(B(i))*sum14;
    d=[y1,y2,y3;y4,y5,y6;y7,y8,y9];
    d=real(d);
    x=real(x);
    z0=[A(i);B(i);G(i)]
    p=d\x;
    z=z0-p;
    err=abs(z-z0);
    A(i+1)=z(1);
    B(i+1)=z(2);
    G(i+1)=z(3);
    z0=z;
end
z0 = 3×1
     1
     2
     3
z0 = 3×1
    1.1665
    2.6140
    4.6185
A
A = 1×3
    1.0000    1.1665    1.2490
B
B = 1×3
    2.0000    2.6140    2.7217
G
G = 1×3
    3.0000    4.6185    5.9808

Looks like 2+1 = 3 outputs to me.

As you do not initialize all of A, you are counting on A not existing already, but your A did exist, from a previous run in which your for i limit was 10 instead of 2.

Walter Roberson 2021-3-19

Sorry, I do not have the resources to assist you with any additional questions for a while.

hasan s 2021-3-21

thanks alot

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