matlab – error while loading shared libraries: file too short

I am trying to install MATLAB_R2020b on Ubuntu 20.04.

1]I extracted the .iso file

2]Then used the chmod -R 777 command because I didn’t have permission to execute the install

3]And finally I ran the install with sudo ./install inside the appropriate folder.

But then I get the following error:

/home/antonis/Downloads/MATLAB_R2020b_Linux64/bin/glnxa64/install_unix: error while loading shared libraries: /home/antonis/Downloads/MATLAB_R2020b_Linux64/bin/glnxa64/ file too short

After some searching, I found out that people having the same issue, unzipped the file using the following command:

unzip -X -K

However my version of Matlab is in a .iso form so I suppose I can’t use that. Also because I am using a free university license of Matlab, I am forced to download it from the university platform in an .iso form.

I am new to Linux so any help is much appreciated.


MathWorks MATLAB R2020b v9.9.0.1524771 Update 2 macOS | NulledTeam UnderGround

MathWorks MATLAB R2020b v9.9.0.1524771 Update 2 macOS | 1.85 GB

Company MathWorks most complete software for computational computer produces; the main program the company that actually Passport is software MATLAB (short for Mat rix Lab Oratory and means lab Matrix) is one of the most advanced software, algorithms and math and a programming language developed generation fourth is possible to visualize and plot functions and data is provided. Icon and bookmark MATLAB with the manufacturer’s logo is the same as the wave equation, L-shaped membrane and special functions have been extracted. MATLAB competitors such as Mathematica, Maple and Mathcad noted. Facilities and software features MathWorks MATLAB:

-perform a variety of complex mathematical calculations and heavy
-Development environment for managing code, files, and data
-explore ways to achieve this solution
-A variety of mathematical functions for linear algebra, statistics, Fourier analysis, optimization, filtering, numerical integration and …
-drawing two-dimensional and three-dimensional graphics functions for visualizing data as
-Design and construction of user interfaces under the programming languages C ++, C or Java
-Ability to test and measure the exact functions and graphs
-the possibility of communication signal processing, image and video
-There are various Jbhabzarhay engineering companies for specific applications such as
telecommunications, control, fuzzy, estimates, statistics, data collection, simulation systems,
neural networks, probability and …

System Requirements:
OS:Windows 10 (version 1803 or higher)/Windows 7 Service Pack 1/Windows Server 2016/Windows Server 2019

CPU:Minimum:Any Intel or AMD x64 processor/RecommendedAny Intel or AMD x64 processor with four logical cores and AVX2 instruction set

Disk:Minimum:2 GB of HDD space for MATLAB only, 4-6 GB for a typical installation / Recommended:An SSD is recommended A full installation of all MathWorks products may take up to 32
GB of disk space

RAM:Minimum:4 GB / Recommended:8 GB

-No specific graphics card is required.
-Hardware accelerated graphics card supporting OpenGL 3.3 with 1GB GPU memory is recommended.
-GPU acceleration using the Parallel Computing Toolbox requires a CUDA GPU
-For Polyspace, 4 GB per core is recommended

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Can someone please comment on the efficiency of Mathematica, Maple and Matlab in conditional plotting?

I use Mathematica to plot

f(x_) := -(1/
    8) (18 - 280 x^2 - 4 x^4 + 8 x^6 + 
     2 x^8 + (81 + 108 x^2 + 54 x^4 + 12 x^6 + x^8) Cos(
       4 π x) + (-64 + 128 x^2 - 64 x^4) Cos(
       2 (-2 + Sqrt(5)) π x) + (1 - 4 x^2 + 6 x^4 - 4 x^6 + 
        x^8) Cos(
       4 (-2 + Sqrt(5)) π x) + (-18 + 24 x^2 + 4 x^4 - 8 x^6 - 
        2 x^8) Cos(
       2 π x - 2 (-2 + Sqrt(5))π x) + (-18 + 24 x^2 + 
        4 x^4 - 8 x^6 - 2 x^8) Cos(
       2 π x + 2 (-2 + Sqrt(5)) π x)) Sin(π x)^2;
Plot({0, If(f(x) >= 0, 10^1.5)}, {x, 0, 200}, 
 PlotStyle -> {Directive(White, CapForm("Butt"), Opacity(-1), 
   Directive(Black, CapForm("Butt"), Opacity(10^10), Thickness(.05))},
  PlotPoints -> 5000000, MaxRecursion -> 6, WorkingPrecision -> 100, 
 AspectRatio -> 1/40, Axes -> {True, False})

Even when I use PlotPoints-> 5*10^6 and MaxRecursion-> 15 it does not give me the whole result, some parts of the solutions are missing (I notice this when I plot it for a short domain like $90<x<100$). Then, can I be hoping to get the whole result by other software like Matlab or Maple? If someone has had experience in working with all these programs, please tell me which program is more efficient in detailed plotting? I am not familiar with those programs, and I want to know if they are more accurate in this sense, try them out.

Another question, if I increase the number of PlotPoints to $10^{10}$, will it be safe for my laptop (core i7)? Since it takes much time and fan starts working loudly.

P.S. Is there a forum like this for Maple?

Solving a multi-objective optimization problem as a mono-objective problem in MATLAB

Suppose the following multi-objective minimization problem:

min f1(x)
min f2(x)
subject to some constraints.

We all know that searching for pareto optimal solutions in a multi-objective minimization problem can be done by using the paretosearch function in MATLAB. For example:

(x,f) = paretosearch(FUN,NVARS,A,b,Aeq,beq,lb,ub,NONLCON,options);

Suppose now that we transform the above multi-objective minimization problem into a mono-objective problem by aggregating the two objectives f1 and f2 as F_w (x) = f1(x) + w*f2(x), where w designates the weight value. For example:

min F_w (x)= f1(x) + w*f2(x)
subject to some constraints.

The variable x tends to the optimal value as the weight w grows and tends to infinity.

My question: What function in MATLAB can replace paretosearch to find the pareto solutions of the above mono-objective minimization problem? I want a function that can take the weight w as an argument. Can someone provide me a detailed example in MATLAB?

Any help will be very appreciated!

Is my Matlab to Python translation correct?

MATLAB (original code):

function p=wfpt(t,v,a,z,err)

  tt=t/(a^2); % use normalized time
  w=z/a; % convert to relative start point

  % calculate number of terms needed for large t
  if pi*tt*err<1 % if error threshold is set low enough
      kl=sqrt(-2*log(pi*tt*err)./(pi^2*tt)); % bound
      kl=max(kl,1/(pi*sqrt(tt))); % ensure boundary conditions met
  else % if error threshold set too high
      kl=1/(pi*sqrt(tt)); % set to boundary condition

  % calculate number of terms needed for small t
  if 2*sqrt(2*pi*tt)*err<1 % if error threshold is set low enough
      ks=2+sqrt(-2*tt.*log(2*sqrt(2*pi*tt)*err)); % bound
      ks=max(ks,sqrt(tt)+1); % ensure boundary conditions are met
  else % if error threshold was set too high
      ks=2; % minimal kappa for that case
  % compute f(tt|0,1,w)
  p=0; %initialize density
  if ks<kl % if small t is better...
      K=ceil(ks); % round to smallest integer meeting error
      for k=-floor((K-1)/2):ceil((K-1)/2) % loop over k
          p=p+(w+2*k)*exp(-((w+2*k)^2)/2/tt); % increment sum
      p=p/sqrt(2*pi*tt^3); % add constant term
  else % if large t is better...
      K=ceil(kl); % round to smallest integer meeting error
      for k=1:K
          p=p+k*exp(-(k^2)*(pi^2)*tt/2)*sin(k*pi*w); % increment sum
      p=p*pi; % add constant term

  % convert to f(t|v,a,w)
  p=p*exp(-v*a*w -(v^2)*t/2)/(a^2);

Python Translation:

import numpy as np
def wfpt(t,v,a,z,err):

tt = t / (a**2) # use normalised time
w=z/a # convert to relative starting point

# calculate number of terms needed for large t
if ((math.pi)*tt*err) < 1:
    kl = math.sqrt(-2*log(pi*tt*err) ./ ((math.pi)**2)*tt)
    kl = max(kl, 1/((math.pi)*math.sqrt(tt)))
    kl = 1 / ((math.pi)*math.sqrt(tt))

# calculate number of terms needed for small t
if (2*(math.sqrt(2*math.pi*tt)*err) < 1):
    ks = 2 + math.sqrt(-2*tt .* log(2*sqrt(2*math.pi * tt) * err))
    ks = max(ks, math.sqrt(tt) + 1)
    ks = 2

# compute f(tt|0,1,w)
p = 0 # initialize the density
if ks < kl:
    K = math.ceil(ks)
    for k = np.arange(-math.floor((K-1)/2), math.ceil((K-1)/2))
        p = p + (w+2*k)*math.exp(-((w+2*k)**2)/2/tt)
    p = p/math.sqrt(2*pi*tt**3)
    K = math.ceil(kl)
    for k = np.arange(1, k):
        p = p+k*math.exp(-(k**2)*(math.pi**2)*tt/2)*sin(k*math.pi*w)
    p = p*math.pi

# convert to f(t|v,a,w)
p = p*exp(-v*a*w - (v**2)*t/2)/(a**2)

Specifically I am unsure if I translated the MATLAB operators / and ./ correctly.

programming languages – Random numbers with condition in matlab

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mathematics – This bisection method code in matlab is giving me an error

I am coding how to get the exact root using the bisection method in MATLAB.
Dichotomie means bisection method.
p is the function, a0 and b0 are the 2 points, nmax is the maximum iterations and tolx is the error.


function (r,n,errx) = Dichotomie(p,a0,b0,nmax,tolx)
%This function approximates the solution of p(x) using the method of
a(i) = a0;
b(i) = b0;
c(i) = (a0 + b0)/2;

while((i<=nmax) && (tx>tolx))   

      if (p(a0) * p(c0)) < 0        
        c(i+1) = (a(i+1) + b(i+1)) / 2;
         tx=abs(c(i+1) - c(i));      
         i = i + 1;
      elseif (p(a0) * p(c0)) > 0
          c(i+1) = (a(i+1) + b(i+1)) / 2;
          tx=abs(c(i+1) - c(i));
          i = i + 1;
      elseif (p(a0) * p(c0)) == 0
          a(i+1) = a(i);
          b(i+1) = b(i);
          c(i+1) = (a(i+1) + b(i+1)) / 2;
          tx=abs(c(i+1) - c(i));
          i = i + 1;




clc;clear all;close all;  
%f(x) = x^2 +3*x - 1  
p=(1 3 -1);  
(r,n,errx) = Dichotomie(p,a0,b0,nmax,tolx);  


Error using inline (line 50)  
Input must be a string.  

Error in Dichotomie (line 11)  

Função .format() do Python no Matlab

Há algum tempo que programa Python e nas aulas começámos a programar Matlab.

Descobri que existe a função fprintf() com a qual eu posso fazer uma formatação numa string, utilizando caracteres de controlo, por exemplo, n ou %s, %c, entre outros.

fprintf('Pi = %3.2fn DezPi = %3.2f', pi, 10*pi, 100*pi)

A minha pergunta é se existe alguma função mais simples que fprintf de alguma forma parecida com a função .format no Python, onde eu não necessito de utilizar caracteres de controlo.

Idade = 18
print ("Hello, I am {} years old !".format(Idade))