photo

Oddur Bjarnason


Labyrinth

Last seen: presque 3 ans il y a Actif depuis 2016

Followers: 0   Following: 0

Message

Statistiques

  • First Review
  • Thankful Level 1

Afficher les badges

Feeds

Afficher par

Question


Why does the function call "function [hdr, record] = edfReadOddur( 'tinnaprime.edf',varargin )" give an error?
This is the error message: Error: File: edfReadOddur.m Line: 1 Column: 40 Unexpected MATLAB expression.

plus de 7 ans il y a | 1 réponse | 0

1

réponse

Question


edf to Matlab conversion?
I need to convert edf to Matlab. I have tried to use edfRead. The function call is function [hdr, record] = edfread('aesa1.ed...

plus de 7 ans il y a | 1 réponse | 0

1

réponse

Réponse apportée
Distinguishing matrices in for loops
Combining the suggestions of David and KSSV I believe that I have found a solution to my problem. I have been trying to write a ...

plus de 7 ans il y a | 0

| A accepté

Question


Distinguishing matrices in for loops
I want to distinguish matrices in a for loop. The following does not work: T=[0,1,-0.1,0.8,0;0,0,0,1,0;-0.2,-1,0,-0.2,0;0...

plus de 7 ans il y a | 1 réponse | 0

1

réponse

Question


Logistic Function Transform of vector values
I need to transform the elements of a vector by a logistic function into a vector with elements with values between 0 and 1. I ...

plus de 7 ans il y a | 1 réponse | 0

1

réponse

Réponse apportée
Test based on inequality of two vectors does not succeed.
function stationarystates(S0,T) %This function is a simple model of a Markov chain % S0 is the initial state % T is the t...

environ 8 ans il y a | 0

Question


Test based on inequality of two vectors does not succeed.
stationarystates([1,0],[0.8,0.2;0.6,0.4]) function stationarystates( S0,T ) %This function is a simple model of a Mar...

environ 8 ans il y a | 2 réponses | 0

2

réponses

Question


Correct and incorrect answer from linsolve
The following gives a correct answer: A=[-2/5,1/5;2/5,-1/5;1,1] B=[0;0;1] linsolve(A,B) ans =0.3333 0.6667 The...

environ 8 ans il y a | 1 réponse | 0

1

réponse

Réponse apportée
I have a system of 3 linear equations with 2 variables. I can solve it by substitution but not with linsolve.
I found the answer to my question: 3*x/5+y/5=x 2*x/5+0.8*y=y x+y=1 3*x/5-5/5*x+y/5=0 2*x/5+4/5*y-5/5*y=0 x+y=1 -2/5...

environ 8 ans il y a | 0

| A accepté

Question


I have a system of 3 linear equations with 2 variables. I can solve it by substitution but not with linsolve.
3*x/5+y/5=x 2*x/5+0.8*y=y x+y=1

environ 8 ans il y a | 1 réponse | 0

1

réponse