photo

Editor


Last seen: plus d'un an il y a |  Actif depuis 2018

Followers: 0   Following: 0

Message

Statistiques

  • Thankful Level 3
  • Thankful Level 2

Afficher les badges

Feeds

Afficher par

Question


Newton's method returns NaN when solving
I am trying to solve a 7D system of equations using the Newton's method. However, my code returns NaN after running. I have chec...

plus d'un an il y a | 1 réponse | 0

1

réponse

Question


How to computing confidence bounds of the regression model using the bootstrap method
I would like to compute and plot the bootstrap confidence bounds of the non-linear regression model. I have computed the bootstr...

presque 2 ans il y a | 1 réponse | 0

1

réponse

Question


Data fit problem, 'Data must be numeric, datetime, duration or an array convertible to double.'
Dear all. I am trying to fit my model to data and determine the confidence interval for the fitted parameters. However, when I r...

presque 2 ans il y a | 1 réponse | 0

1

réponse

Question


Curve fitting to data using fit
I have data x sampled at times t. I would like to fit my function to this data. Below is a code. clear; close all x=[100; 85.4...

presque 2 ans il y a | 1 réponse | 0

1

réponse

Question


Index exceeds the number of array elements (11)
I need help please. I am trying write a program with the following steps: (i) to obtain x(i) and y(i), i=1,2,...,m (ii) to cr...

presque 4 ans il y a | 1 réponse | 0

1

réponse

Question


Is there an easy way to make numerical simulations of the ODE of the form dx1/dt=x1(2-x1-x2), dx2/dt=x2(3-x1-x2-x3) for any xn?
I have tried doing for 'n=5' and here is my code; dx = @(t,x) [x(1); x(1)*(2-x(1)-x(2)); x(2); x(2)*(3-x(1)...

plus de 6 ans il y a | 1 réponse | 0

1

réponse

Question


Is there an easy way to make numerical simulations of the ODE of the form dx1/dt=x1(2-x1-x2), dx2/dt=x2(3-x1-x2-x3) for any xn?
I have tried doing for 'n=5' and here is my code; dx = @(t,x) [x(1); x(1)*(2-x(1)-x(2)); x(2); x(2)*(3-x(1)-x(2)-x(...

plus de 6 ans il y a | 1 réponse | 0

1

réponse