# I have a code but it keep saying "invalid expression. when calling a function or indexing a variable, use parentheses. otherwise check for mismatched delimiters.

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Roolientha Denaud il 15 Nov 2018
Commentato: DGM il 28 Ott 2023
F=[];
A=[];
for f=0:200:1800
F =[F,f]; %in HZ
f0 = 500;% in Hz
f1= 1500;% in Hz
N = 20;
a = 0:5:25
a= abs((a*sin[pi(f-f0)*N]))/ (sin[pi(f-f0)]))+ abs((a*sin[pi(f-f1)*N]))/ (sin[pi(f-f1)]));
A= [A, a];
end
plot (F, A)
##### 12 CommentiMostra 10 commenti meno recentiNascondi 10 commenti meno recenti
clear all; close all; clf; clc;
g = 9.81; %m/s2 Valor de Aceleración Gravitacional
p = 960; %Densidad del Fluido en kg/m^3;
V = 5.08e-10; %Volumen de la Partícula en m^3
m = 1.3e-6; %Masa de la Partícula en kg
r = 4.95e-4; %Radio de la Partícula en m
po = 2559.055; %Densidad de la Partícula en kg/m^3
n = 0.046; %Viscosidad en kg/m/s
b = (-p*g*V)+(m*g)/(2*r^2*(po-p)*g)/9*n; %Constante de Amortiguamiento
v = (2*r^2)*((po-p)*g/9*n);
%no = 20; %Número de partículas
dt = 0.01;
to = 0;
tf = 10;
t= (to:dt:tf)';
T = numel(t);
vox = 0;
vx = zeros(length(t),1); %Regresar a poner no
vx(1,1) = vox;
aox = 0;
ax = zeros(length(t),1); %Regresar a poner no
ax(1,1) = aox;
int1=-0.055;
int2=0.055;
xo = int1+(int2-int1) *rand(1,1); %Aqui
x = zeros(length(t),1); %Aqui
x(1,1) = xo;
voy = 0;
vy = zeros(length(t),1); %Aqui
vy(1,1) = voy;
int3=0.14;
int4=0;
yo = int3+(int4-int3) *rand(1,1); %Aqui
y = zeros(length(t),1); %Aqui
y(1,1) = yo;
for i=2:1:T
vx(i-1,1) = ((w^2*x(i-1,1)*(m-p.*V)) - (b.*vx(i-1,1)) + ((m.*vx(i-1,1))/dt))*(dt/m);
x(i) = x(i-1) + vx(i-1,1)*dt;
figure(1)
subplot(1,2,1)
plot(t(i,1),vx(i,1),'r-o')
grid on;
xlabel('Tiempo [segundos]');
ylabel('Posición Vertical y [metros]');
hold on
subplot(1,2,2)
plot(t(i,1),x(i,1),'r-o')
grid on;
xlabel('Tiempo [segundos]');
ylabel('Posición Vertical y [metros]');
hold on
en
We have these two graphs but we need to have an oscilation type of graph. Can you help me?
Walter Roberson il 29 Nov 2022

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### Risposta accettata

Luna il 15 Nov 2018
Modificato: Luna il 15 Nov 2018
Hi Roolientha,
Your expression is invalid because you are calling a function sin with brackets not paranthesis.
sin[..] %check and correct them with paranthesis
sin(..) %because sin is a function like abs,plot,etc
##### 7 CommentiMostra 5 commenti meno recentiNascondi 5 commenti meno recenti
Merve Özkanat il 3 Ott 2022
Hi I get the same error, could you help me please?
I want to create a yxoordinat vector [11x1] with values from a vector d.
d is a vector with 363 values.
d(afg)=dr; % d is a [363,1] vector with values.
ycoordinat=zeros(11,1);
for i=56:66
for s=1:11
ycoordinat[s,1]=d((i-1)*3+1) % d((i-1)*3+1) is the value that should be put into a vector.
end
end
But it says ''Invalid expression. When calling a function or indexing a variable, use parentheses. Otherwise, check for mismatched delimiters''.
Walter Roberson il 4 Ott 2022
MATLAB never uses [] for indexing, only for constructing lists and arrays. Indexing is always ()

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### Più risposte (8)

Rik il 15 Nov 2018
Modificato: Rik il 15 Nov 2018
You are using square brackets for the sin function, instead of parentheses. Also, you were missing a lot of * signs. You should check if the four constituents are indeed what you need in the line below.
a= abs((a*sin(pi*(f-f0)*N))) / sin(pi*(f-f0)) + abs((a*sin(pi*(f-f1)*N))) / sin(pi*(f-f1));
You have another problem here: you are trying to plot A against F, but A will have 6 values for each value of F, because of that first assignment of a.
You could speed this up by first fixing the dynamic growth of these vectors by pre-allocating them, and it should also be possible to remove the loop entirely once you have code that does what you intend to do.
##### 2 CommentiMostra NessunoNascondi Nessuno
Roolientha Denaud il 15 Nov 2018
Still didnt work it says " Error: File: S.m Line: 10 Column: 51
Invalid expression. When calling a function or indexing a variable, use parentheses. Otherwise, check for
mismatched delimiters."
Rik il 15 Nov 2018
Have you copied the line exactly as I posted it? Because for me it doesn't error at that line. I do have an error when plotting, because there you haven't explained how you want to extend F to have enough values. The code below assumes you want to repeat the values of F.
F=0:200:1800;%in HZ
f0 = 500;% in Hz
f1= 1500;% in Hz
N = 20;
a = 0:5:25;
A=zeros(numel(a),numel(F));
for n=1:numel(F)
f=F(n);
a= abs((a*sin(pi*(f-f0)*N))) / sin(pi*(f-f0)) + abs((a*sin(pi*(f-f1)*N))) / sin(pi*(f-f1));
A(:,n)=a;
end
plot (F, A)

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Steven Lord il 15 Nov 2018
If you're editing this code in an editor other than the MATLAB Editor, I advise you to copy and paste it into the MATLAB Editor. If you have Code Analyzer enabled in the Preferences (it is enabled by default) you should see a red square in the upper-right corner of the Editor window and several colored lines (three orange, one red) below it. Orange lines indicate places where your code should work, but Code Analyzer has a suggestion that may improve its robustness and/or performance. Red lines indicate lines of code where its analysis indicates errors that will prevent your code from running at all. Clicking on the red line brings the cursor to the line in question, where the suspected error should be underlined.
If you hover over the underlined suspected error (or warning), you will receive some additional information that may help you understand the suggestion or error. In this case, once I fixed the sin[...] sections there is still an error. Counting, you have mismatched parentheses. Here I've written numbers below the locations of the parentheses in your code. Everywhere I see a ( I add 1 to the number, and everywhere I see a ) I subtract 1.
a= abs((a*sin(pi(f-f0)*N)))/ (sin(pi(f-f0))))+ abs((a*sin(pi(f-f1)*N)))/ (sin(pi(f-f1))));
12 3 4 3 210 1 2 3 210*
The * indicates -1, where I saw more closing parentheses than opening parentheses. Delete that ) and continue counting on that line of code.
##### 1 CommentoMostra -1 commenti meno recentiNascondi -1 commenti meno recenti
Merve Özkanat il 3 Ott 2022
Hi I get the same error, could you help me please?
I want to create a yxoordinat vector [11x1] with values from a vector d.
d is a vector with 363 values.
d(afg)=dr; % d is a [363,1] vector with values.
ycoordinat=zeros(11,1);
for i=56:66
for s=1:11
ycoordinat[s,1]=d((i-1)*3+1) % d((i-1)*3+1) is the value that should be put into a vector.
end
end
But it says ''Invalid expression. When calling a function or indexing a variable, use parentheses. Otherwise, check for mismatched delimiters''.

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Aya Nabil il 19 Mag 2020
Modificato: Aya Nabil il 19 Mag 2020
Invalid expression. When calling a function or indexing a variable, use parentheses. Otherwise, check for mismatched delimiters
what's is the problem?
it always shows me this comment for this code
Line: 1 Column: 23
[t,c]=ode45('prob5_16',[0,6],[0.2]);
clinear=0.2+0.111*(1-exp(-t/1.11));
plot(t,c,t,clinear)gridxlabel('time')ylabel('concentration')>>title('Solution for Problem 5.16');
##### 3 CommentiMostra 1 commento meno recenteNascondi 1 commento meno recente
Merve Özkanat il 3 Ott 2022
Hi I get the same error, could you help me please?
I want to create a yxoordinat vector [11x1] with values from a vector d.
d is a vector with 363 values.
d(afg)=dr; % d is a [363,1] vector with values.
ycoordinat=zeros(11,1);
for i=56:66
for s=1:11
ycoordinat[s,1]=d((i-1)*3+1) % d((i-1)*3+1) is the value that should be put into a vector.
end
end
But it says ''Invalid expression. When calling a function or indexing a variable, use parentheses. Otherwise, check for mismatched delimiters''.
Rik il 4 Ott 2022
You're using square brackets instead of parentheses.

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Tahlia Jones il 20 Mar 2021
[2*3^12[cos140+sin140]/(5+3i)^3(2+3i);-(3-5i)3/25[cos-60sin-60]]
##### 1 CommentoMostra -1 commenti meno recentiNascondi -1 commenti meno recenti
Walter Roberson il 20 Mar 2021
MATLAB has no implied multiplication. You have to put in every multiplication operator.
Also in MATLAB, function calls have to have their parameters in () such as sin(pi/3)
The MATLAB sin and cos operations are in radians but there is sind() and cosd()
Remember to put in () around denominators that involve expressions A/B*C is (A/B)*C not A/(B*C)

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ali sh il 31 Gen 2022
mu=mu0*(1+omega*(3-(n2)^2);
Invalid expression. When calling a function or indexing a variable, use parentheses. Otherwise, check for mismatched
delimiters.
##### 1 CommentoMostra -1 commenti meno recentiNascondi -1 commenti meno recenti
Steven Lord il 31 Gen 2022
You have three left parentheses ( but only two right parentheses ). Thus you have a mismatch in your parentheses. You need one more ) but where to put it depends on what you're trying to compute.

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Jobin Geevarghese Thampi il 17 Mar 2022
Modificato: Walter Roberson il 17 Mar 2022
functiom k=summa(n)
[row col]=size(n);
k=0;
for ii=1:row
for jj=1:col
if (ii,jj)>=5 && (ii,jj)<=18
k=k+n(ii,jj);
end
end
end
File: summa.m Line: 6 Column: 15
Invalid expression. When calling a function or indexing a variable, use parentheses. Otherwise, check
for mismatched delimiters.
##### 2 CommentiMostra NessunoNascondi Nessuno
Jobin Geevarghese Thampi il 17 Mar 2022
can anyone help me on this
Walter Roberson il 17 Mar 2022
if n(ii,jj)>=5 && n(ii,jj)<=18

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Kamilu Sanusi il 2 Mag 2022
Modificato: Walter Roberson il 2 Mag 2022
Hello please I am having a similar problem, i need assistance please. What I keep getting is "Invalid expression. When calling a function or indexing a variable, use parentheses. Otherwise, check for mismatched delimiters"
and is pointing to Line 2 column 35
Here is the code:
clear all
clc
% User Interface pop-up window
prompt=('Power Input_Bus1 [p.u.]:','Z_1 [p.u.]:','Z_12 [p.u.]:','Load_Bus2:','V2:','P.U. MVA Base:');
dlg_title='Power Flow Analysis';
defaultans =('1.06+0i','6i','0.2i','1.6+0.20i','1+0i','100');
num_lines = [ones(size(defaultans')) ones(size(defaultans'))*50];
% Reading all inputs and change string to numeric value
% ****************************%
pubase = str2num(answer(6)); % PU MVA base
% Variales declared
G1= 1/Z_1 % Self reactance
y12 = 1/Z_12
% Identify load P (real) and Q (imag) value
s2 = -1*L_B2,
p2 = real(s2);
q2 = imag(s2);
% Create the bus admittance matrix
y = [(G1+y12) -(y12);-(y12) (y12)];
% Load flow calculation using Gauss siedel
n = 1 %n y loop declaration
while n > 0.00001
v2e = (((p2-(q2)*i)/conj(v2))+y12*v1)/y12;
e2 = abs(v2e-v2);
v2 = v2e;
em = [e2];
n = max(em);
end
% s1 calculation
s1 = v1*(v1*(G1+y12)-(y12*v2)); % slack bus apparaent power
P1 = real(s1);
Q1 = imag(s1);
% Determine line flow and line losses
I12 = y12*(v1-v2);
I21 = -I12;
% Line flow
s12 = v1*conj(I12);
s21 = v2*conj(I21);
% Line losses
slose_12 = s12 + s21;
%***********************************************************************%
% Displar results
% **********************************************************************%
disp(' yi1 yi2')
disp(y)
disp('V2 value=')
disp(v2)
disp('s1 value=')
disp(s1)
disp('slose_12 value=')
disp(slose_12)
##### 4 CommentiMostra 2 commenti meno recentiNascondi 2 commenti meno recenti
Kamilu Sanusi il 2 Mag 2022
@Walter Roberson, thank you so much, your really inputs really helped.
Please I am a beginner still learning from original codes written by people. Kindly, can you explain what the
num_lines = [ones(size(defaultans')) ones(size(defaultans'))*50]; and answer=inputdlg(prompt,dlg_title,num_lines,defaultans); are meant to implement? and also that specific number 50 in the num_line.
Thank you so much
Walter Roberson il 2 Mag 2022
defaultans is a cell row vector. defaultans' is its transpose, and so is a cell column vector. size(defaultans') would be a vector with the number of entries in defaultans as its first value, and 1 (only one column for the transpose) as the second value. ones() of that would be an array that is all ones and has as many ones in a column as there were entries in defaultans .
Likewise the second part is similar but the *50 multiplies it by 50.
Those are together in [] so what you get is an array that has as many rows as there are entries in defaultans, and the first column is all 1 and the second column is all 50.
When that array is passed in to inputdlg() in the position it is in, it gives information about the dimensions of each input area. The leading 1 means "reserve one line of text for each input area" and the trailing 50 means "reserve 50 characters for each input area".
Why 50? Well, that appears to be an arbitrary design decision. It looked good to whoever wrote the code.

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michael il 28 Ott 2023
Modificato: DGM il 28 Ott 2023
%
% Factor de Amplificación Dinámica
%
% Por: Roberto Aguiar Falconi
% CEINCI-ESPE
% ---------------------------------------
% ---------------------------------------
% z1: Factor de amortiguamiento 1
% z2: Factor de amortiguamiento 2
% z3: Factor de amortiguamiento 3
% z4: Factor de amortiguamiento 4
% r : Relación entre la frecuencia excitación a frecuencia natural
% f : Factor de amplificación dinámica
hold off
dr=0.02;r=0;
for i=1:150
r=r+dr;
f(i)=1.0/(sqrt((1-r^2)^2+(2*z1*r)^2));f1(i)=1.0/(sqrt((1-r^2)^2+(2*z2*r)^2));
f2(i)=1.0/(sqrt((1-r^2)^2+(2*z3*r)^2));f3(i)=1.0/(sqrt((1-r^2)^2+(2*z4*r)^2));
rr(i)=r;
end
plot (rr,f); hold on
plot (rr,f1,'--'); plot (rr,f2,':'); plot (rr,f3,'-.')
xlabel('r'); ylabel('Factor de amplificacion');
axis([0,3,0,5]);
text (2.0,4.5,'z1 ---- ','Fontname','symbol'); text (2.0,4.0,'z2 - - -','Fontname','symbol')
text (2.0,3.5,'z3 .......','Fontname','symbol'); text (2.0,3.0,'z4 .-.-.-','Fontname','symbol')
hold off
% ---fin---
##### 5 CommentiMostra 3 commenti meno recentiNascondi 3 commenti meno recenti
Walter Roberson il 28 Ott 2023
Suppose you define three functions:
cost = slices_of_pizza * 2.50 + pieces_of_chicken * 1.99;
end
cost = pieces_of_chicken * 1.99 + slices_of_pizza * 2.50;
end
if rand() < 0.5
cost = slices_of_pizza * 2.50 + pieces_of_chicken * 1.99;
else
cost = pieces_of_chicken * 1.99 + slices_of_pizza * 2.50;
end
end
What would be your expected output for each of the two functions?
Are you assuming that the first-mentioned variable gets assigned the first value (3) and the second-mentioned variable gets assigned the second value (2) ? If so then is "first mentioned" according to parsing or according to execution?
If it is according to parsing (fad3 mentions slices_of_pizza first even though that statement might not be executed) then suppose you have
if false
cost = bottles_of_2L_softdrink * 3.99 + cartons_of_500ml_ice_cream * 6.99;
elseif rand() < 0.5
cost = slices_of_pizza * 2.50 + pieces_of_chicken * 1.99;
else
cost = pieces_of_chicken * 1.99 + slices_of_pizza * 2.50;
end
end
then is it the implication that 3 would be associated with bottles_of_2L_softdrink and 2 would be associated with cartons_of_500ml_ice_cream because those are the first two variables mentioned in the code (in a statement never executed)? If so then how many slices_of_pizza and how many pieces_of_chicken ?
Is it the algorithm that you when encounter a variable that has not been assigned a value yet, that the variable will be assigned the value that is next on the parameter list? So if you had
%each piece of chicken is accompanied by a finger cleaning packet
cost = pieces_of_chicken * 1.99 + peices_of_chicken * 0.03 + slices_of_pizza * 2.50;
end
then should pieces_of_chicken be associated with 3, and peices_of_chicken which is a distinct variable name to MATLAB should be associated with 2, and slices_of_pizza should be associated with ... what? MATLAB doesn't know that peices_of_chicken is a typing mistake for pieces_of_chicken ...
Can you come up with any sustainable principle for associating numeric values in a function declaration with particular variable names in the function body?
DGM il 28 Ott 2023

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