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.

3 years ago | 1 answer | 0

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...

3 years ago | 1 answer | 0

Answered

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 ...

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 ...

3 years ago | 0

| accepted

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...

3 years ago | 1 answer | 0

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 ...

3 years ago | 1 answer | 0

Answered

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...

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...

4 years ago | 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...

4 years ago | 2 answers | 0

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...

4 years ago | 1 answer | 0

Answered

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...

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...

4 years ago | 0

| accepted

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

4 years ago | 1 answer | 0