How can I write a matlab code for this iterative function.

x(n+1)=(1/(n+1)+(1-(1/(n+1)))*x(n)+5*(x(n)-1)/(exp(x(n))))

[Dyuman Joshi]

What is the value of the first element of the sequence? You have written the equation, you need to add a loop or you can use recursion.

[Walter Roberson]

are you sure it is 1/n+1 and not 1/(n+1) ?

[biratu lemessa]

the question that I need to iterate is this x(n+1)=1/(n+1)+(1-1/(n+1))*x(n)-5*(x(n)-1)/(exp(x(n))). But still it doen't run how can I do?

Answer 1

[Dyuman Joshi]

tolerance=0.000001;
%As it is not known at which iteration the sequence will converge, I have
%assumed 1000 iterations for preallocation of x 
k=1000;
x=zeros(1,k);
x(1)=0.5;
for n=1:k
    x(n+1)=1/(n+1)+(1-1/(n+1))*x(n)-5*(x(n)-1)/(exp(x(n)));
    %Check for tolerance, another option for checking tolerance is
    %abs(x(n+1)-x(n))/abs(x(n))<tol, note that the result might be
    %different in that case, it's upto you to choose which method to use
    if abs(x(n+1)-x(n))<tolerance
        fprintf('The sequence converges after %d iterations for tolerance = %f', n, tolerance)
        break;
    end
end

The sequence converges after 94 iterations for tolerance = 0.000001

x=x(1:n+1); %discard the zeros
format long
disp(x)

  Columns 1 through 9

   0.500000000000000   2.266326649281583   1.187675646656826   0.854617807067713   1.192955435194717   0.868156620583146   1.163680903337047   0.887604915198019   1.131425095116121

  Columns 10 through 18

   0.906311497659785   1.104084879025953   0.922883104252097   1.082035542164216   0.937164596743137   1.064428383729875   0.949286820328331   1.050404246023775   0.959447912093624

  Columns 19 through 27

   1.039260635834704   0.967862007062625   1.030437602503192   0.974745666029741   1.023484504028818   0.980311300399718   1.018034285987070   0.984761319457214   1.013786112572562

  Columns 28 through 36

   0.988282805885377   1.010493442171635   0.991043533899946   1.007955166924509   0.993189800022176   1.006008432754297   0.994846025193740   1.004522475351900   0.996115758361740

  Columns 37 through 45

   1.003393206646930   0.997083599564924   1.002538465503489   0.997817595201277   1.001893904290104   0.998371760512707   1.001409486226271   0.998788499089087   1.001046554104671

  Columns 46 through 54

   0.999100787825405   1.000775416334113   0.999334068735385   1.000573387255714   0.999507837165259   1.000423216016248   0.999636944276976   1.000311840561424   0.999732645922297

  Columns 55 through 63

   1.000229408743413   0.999803435150452   1.000168515568749   0.999855695291484   1.000123613117731   0.999894207380366   1.000090556922328   0.999922541260303   1.000066259190022

  Columns 64 through 72

   0.999943354996391   1.000048425021510   0.999958622773054   1.000035352648487   0.999969807492606   1.000025782755712   0.999977990924539   1.000018785234436   0.999983971469759

  Columns 73 through 81

   1.000013674344976   0.999988337348532   1.000009945349267   0.999991521223690   1.000007227307356   0.999993840856691   1.000005248001301   0.999995529293015   1.000003807929629

  Columns 82 through 90

   0.999996757223015   1.000002761066885   0.999997649512350   1.000002000655724   0.999998297399099   1.000001448733896   0.999998767477791   1.000001048427057   0.999999108306091

  Columns 91 through 95

   1.000000758285673   0.999999355255947   1.000000548129853   0.999999534070717   1.000000396004657

Answer 2

[Chunru]

% initial value of x(0)
x = 0;
for n=1:10
    x = (1/n)- (1-(1/n+1)) *x +5*exp(x);     % check the formula.  this one is diverging
    fprintf("i=%3d    x=%9.3f\n", i, x)
end

i=  0    x=    6.000
i=  0    x= 2020.644
i=  0    x=      Inf
i=  0    x=      Inf
i=  0    x=      Inf
i=  0    x=      Inf
i=  0    x=      Inf
i=  0    x=      Inf
i=  0    x=      Inf
i=  0    x=      Inf

[biratu lemessa]

thanks a lot