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.

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)

[Roolientha Denaud]

It says LIne: 10 Column:18

[Abhijeet Gadge]

Invalid expression. When calling a function or indexing a variable, use parentheses. Otherwise, check for mismatched delimiters.

[Walter Roberson]

@Abhijeet Gadge

That often means that you have a mismatch of ( or ), such as

sin(1/2))

Invalid expression. When calling a function or indexing a variable, use parentheses. Otherwise, check for mismatched delimiters.

[Leila Nait-Kaci]

Hello, im new to matlab and need some assistance: i keep getting this error even though i do not have any mismatch with my parentheses, Y= sin^2((2/3)*pi)+(1/3)*cos^2((2/3)*pi) If anyone could help me within the next two days (i have a test coming up) i would be very thankful.

[Rik]

sin^2 is a common mathematical notation, but not valid Matlab syntax. In Matlab you need to write it like this:

sin(x)^2  % for matrix multiplication (usually only for scalar inputs)
sin(x).^2 % for element-wise power (use for array input)

[John D'Errico]

@Leila Nait-Kaci

It is a common mathematical way of writing the power of a trig function, as for example:

sin^2(x)

HOWEVER, that does not make it valid MATLAB syntax. If you want to square the sin(x), then write:

sin(x)^2

[Leila Nait-Kaci]

It solved the problem Thank you very much!

[Adriana]

Can someone help me please? I don't know why it isn't working.

[Adriana]

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);

w = 108.28; %Velocidad Angular en rad/s

%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),no); %Regresar a poner no
vx(1,1) = vox;

aox = 0;
ax = zeros(length(t),no); %Regresar a poner no
ax(1,1) = aox;

int1=-0.055;
int2=0.055;

xo = int1+(int2-int1) *rand(1,no); %Aqui 
x = zeros(length(t),no); %Aqui
x(1,1) = xo;

voy = 0;
vy = zeros(length(t),no); %Aqui
vy(1,1) = voy;

int3=0.14;
int4=0;

yo = int3+(int4-int3) *rand(1,no); %Aqui
y = zeros(length(t),no); %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);

Invalid expression. When calling a function or indexing a variable, use parentheses. Otherwise, check for mismatched delimiters.

    x(i) = x(i-1) + vx(i-1,1)*dt;
    
    vy(i-1,1) = -g + (((po*g*V)-((A*g*V).*y(i-1,1)))/m) - ((6*pi*r*n*v(i-1,1))/m);
    y(i) = y(i-1) + vy(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
end

Can someone help me?

[Walter Roberson]

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);

You have x(i-1,1)(m-p.*V) . That code asks to take x, index it at (i-1,1), and then index the result at (m-p.*V) . However, MATLAB does not permit () indexing of calculated values. The calculated value would be a scalar, and it seems unlikely that m-p.*V will always happen to be exactly 1

Likewise you have ((m.*vx(i-1,1))/dt))(dt/m) which requests that the result of calculating ((m.*vx(i-1,1))/dt)) be indexed at dt/m which seems unlikely to be an exact integer.

MATLAB has absolutely no implicit multiplication, not anywhere that I have found. If you want to multiply, you must use the .* or * operators

[Adriana]

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);

w = 108.28; %Velocidad Angular en rad/s

%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]

I think you should open your own Question about this.

Answer 1

[Luna]

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

[Roolientha Denaud]

Still Didn't 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]

If it still didn't work, why did you mark this as accepted answer?

[Luna]

I am sorry, I am on my mobile right now and can’t copy paste the code directly.

Roolientha,

You have two sin functions they have brackets. When it says error in line 10 and column 51, even the error on that line, sometimes it might not be on that column.

also Rik is right you should check when you want to make multiplication like 3*5 or a*b or a*(pi*N) you should always use *.

if you miss one star it gives an error. Matlab does not get 3a you should be typing 3*a.

[Aman Kumar Sharma]

S = 1/(pf)(Pm(pf+1*i*(sqrt(1-(pf)^2))));

[Walter Roberson]

@Aman Kumar Sharma

MATLAB has absolutely no implied multiplication. 1/(pf)(something) is just invalid syntax in MATLAB, and Pm(something) is a request to either call a function named Pm or else to index an array named Pm

If you want to have multiplication in MATLAB, you need to use the multiplication operator, which is typicall .* (but sometimes you can get away with using the * operator instead -- the * operator means something else when the two sides are both non-scalar.)

[Merve Özkanat]

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]

MATLAB never uses [] for indexing, only for constructing lists and arrays. Indexing is always ()

Answer 2

[Rik]

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.

[Roolientha Denaud]

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]

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)

Answer 3

[Steven Lord]

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.

[Merve Özkanat]

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''.

Answer 4

[Aya Nabil]

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');

[Rik]

This is not an answer, but a question.

The standard Matlab syntax is to separate function calls by comma, semicolon, or newline. You did none of that on your last line.

[Merve Özkanat]

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]

You're using square brackets instead of parentheses.

Answer 5

[Tahlia Jones]

[2*3^12[cos140+sin140]/(5+3i)^3(2+3i);-(3-5i)3/25[cos-60sin-60]]

[Walter Roberson]

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)

Answer 6

[ali sh]

mu=mu0*(1+omega*(3-(n2)^2);

Invalid expression. When calling a function or indexing a variable, use parentheses. Otherwise, check for mismatched

delimiters.

[Steven Lord]

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.

Answer 7

[Jobin Geevarghese Thampi]

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.

[Jobin Geevarghese Thampi]

can anyone help me on this

[Walter Roberson]

 if n(ii,jj)>=5 && n(ii,jj)<=18

Answer 8

[Kamilu Sanusi]

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];
answer=inputdlg(prompt,dlg_title,num_lines,defaultans);

% Reading all inputs and change string to numeric value
% ****************************%
v1 = str2num(answer(1));      % Read power input V1
Z_1 = str2num(answer(2));     % Read impedance input z1
Z_12 = str2num(answer(3));     % Read impedance input z12
L_B2 = str2num(answer(4));     % Read load bus input s2
v2 = str2num(answer(5));     % Read voltage 2 v2
pubase = str2num(answer(6));  % PU MVA base

% Variales declared
% Convert impedances to admittances
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('Bus admittance matrix:')
disp('   yi1              yi2')
disp(y)
disp('V2 value=')
disp(v2)
disp('s1 value=')
disp(s1)
disp('slose_12 value=')
disp(slose_12)

[Kamilu Sanusi]

It keeps pointing to propmt row

[Walter Roberson]

In MATLAB, () can occur in a few different contexts:

• (one_expression) indicates that one_expression is to be grouped together for priority. For example, 3*(5+7) . Here, one_expression is intended to indicate that you have a single expression not separated by commas
• name(expression1, expression2, expression3, ...) might indicate that the given name is an array that is being indexed at the given locations. Each comma marks the end of a dimension
• name(expression1, expression2, expression3, ...) might indicate that the given name is the name of variable containing a function handle, and that the function is to be executed passing in the values as parameters. Each comma marks the end of a parameter.
• name(expression1, expression2, expression3, ...) might indicate that the given name is the name of a function, and that the function is to be executed passing in the values as parameters. Each comma marks the end of a parameter.
• There are some cases in which extra () will be silently ignored, such as some ways of writing for loops
• () can sometimes occur after keywords in order to give options for the keywords; see for example passing in a pool to parfor

In MATLAB, there are no cases in which you can have (expression1, expression2, expression3, ...) as an expression to create a list of any kind. For example your

defaultans =('1.06+0i','6i','0.2i','1.6+0.20i','1+0i','100');

is simply invalid in MATLAB, not a way to create a list of character vectors.

In MATLAB, to create a list of character vectors, use { } instead of ( ) . So for example,

defaultans = {'1.06+0i','6i','0.2i','1.6+0.20i','1+0i','100'};

If you were trying to create a list of numeric values, then using { } would not necessarily be wrong, but typically you would instead use [ ] to create a numeric vector. For example,

answer_values = [1.06+0i, 6i, 0.2i, 1.6+0.20i, 1+0i, 100];

[Kamilu Sanusi]

@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]

https://www.mathworks.com/help/matlab/ref/inputdlg.html#d123e751295

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.

Answer 9

[michael]

function [f]=fad(0.01,0.1,0.15,0.5); 
%
% Factor de Amplificación Dinámica
%
% Por: Roberto Aguiar Falconi 
%          CEINCI-ESPE
% ---------------------------------------
% [f]=fad(z1,z2,z3,z4)
% ---------------------------------------
% 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---

[michael]

>> newtonraphson

File: newtonraphson.m Line: 1 Column: 28

Invalid expression. When calling a function or indexing

a variable, use parentheses. Otherwise, check for

mismatched delimiters.

[Walter Roberson]

function [f]=fad(0.01,0.1,0.15,0.5);

What is the function going to do with the 0.01?

Where does the function get values for z1 z2 z3 z4?

% z1: Factor de amortiguamiento 1

comments are not executable. MATLAB is not going to look at the comment and deduce that z1 is a parameter to the function.

[michael]

I should do this example, but I don't know how to complement it

[Walter Roberson]

@michael

Suppose you define three functions:

function cost = fad1(3, 2)
   cost = slices_of_pizza * 2.50 + pieces_of_chicken * 1.99;
end
function cost = fad2(3, 2)
   cost = pieces_of_chicken * 1.99 + slices_of_pizza * 2.50;
end
function cost = fad3(3, 2)
   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

function cost = fad3(3, 2)
   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

function cost = fad4(3,2)
   %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]

See

https://www.mathworks.com/help/matlab/ref/function.html

https://www.mathworks.com/help/matlab/matlab_prog/local-functions-in-scripts.html

For further reading, look at the bottom of those pages.