Generating piecewise function and plotting it

Hi! I'm very new for Matlab at the moment. I tried the solve these piecewise equations by vectorized way. But for the condition one, I get error beacause of putting insufficient argument. I need help and waiting your solutions. here is my solution ;

function y=piecewise(x)

%ilk şart first cond

x1=x(x<-1)

y(x<-1)=-1

%ikinci şart second cond

x2=x(-1<=x & x<2);

y(-1<=x & x<2)=x2.^2+x;

%ücüncü şart third cond

x3=x(2<=x & x<4)

y(2<=x & x<4)=-x3+2

%dördüncü şart fourth cond

x4=x(4<=x & x<6)

y(4<=x & x<6)=x4.^3+2*x.^2-3

%besinci şart fifth cond

x5=x(6<=x)

y(6<=x)=2*exp(0.1*x5-0.6)

x=-5:0.1:10

y=piecewise(x)

plot(x,y)

Answer 1

[Rohit Sinha]

You could try this method

syms x %makes x a symbolic variable
f= piecewise(x<-1, -1, x>=-1 & x<2, x^2+x, x>=2 & x<4, -x+2, x>=4 & x<6, x^3+2*x^2-3, x>=6, 2*exp(0.3*(x-6))); %makes a piecewise function for given conditions
fplot(f) %plots the piecewise function

Answer 2

[Christopher Creutzig]

Your code mixes some x and x2 etc., as in x2.^2+x, which cannot work, as they have different sizes. Try this small modification:

function y=piecewise_fn(x)
    x1=x(x<-1);
    y(x<-1)=-1;
    
    x2=x(-1<=x & x<2);
    y(-1<=x & x<2)=x2.^2+x2;
    
    x3=x(2<=x & x<4);
    y(2<=x & x<4)=-x3+2;
    
    x4=x(4<=x & x<6);
    y(4<=x & x<6)=x4.^3+2*x4.^2-3;
    
    x5=x(6<=x);
    y(6<=x)=2*exp(0.1*x5-0.6);
end

Answer 3

[Amjad Green]

can you write the entire question in english

[ÖMER FARUK KURANLI]

Create the following piecewise function using relation operators and plot it in the range (-5,10).

[Walter Roberson]

Amjad Green: the policy here is that people can post in whatever natural language they feel most comfortable writing in. This resource is not restricted to English.

English happens to be the language understood by most of the people reading these Questions, so people are more likely to get a usable response if the Question happens to be in English, but it is decidedly not a requirement.

Mathworks implemented Japanese, Korean, and Chinese localizations for MATLAB Central, so they actively expect people to post in those languages.

[Samuel Armistead]

Why would someone answer a question asked in english, in a different langauge. May have well not answered at all. Let me go teach math in China using my English, see how far that gets them.

[Walter Roberson]

Sameul:

1. the question was originally asked in a different language.
2. The volunteers can choose to respond in whatever language they feel most comfortable in. People do not always feel comfortable expressing complex concepts in a different language. Sometimes the code that people post is enough to understand the response, and if not then there is always Google Translate to assist translation. If someone whose native language is Arabic posts in poor English, and the person who is trying to understand the code has a native language of Turkish, then Arabic structures expressed in weak English, and then translated by google to Turkish, is probably not going to be as accurate as if the Arabic poster had posted a confident and more detailed response in Arabic, and the original poster user Google Translate to convert to Turkish.

I emphasize: Everyone is welcome to post in whatever natural language they feel most comfortable in. Period.

[Tawsif Habib]

Let's focus on the main problem

[Leane Armada]

pungolay nlang bla

[john Snori]

We can implement piecewise functions by:

• Treating each function separately and merge and plot them on the same graph
• If-else statement along with for-loop
• Switch-case statement
• Using built-in function of Matlab which returns the piecewise function using a single-line command.

Source: https://www.entechin.com/how-to-plot-a-piecewise-function-in-matlab/

Answer 4

[baskar ayyakkannu]

function v = RocketVelocity(t)

if t(0<=t && t<=8)

v = 10*t^2-5*t;

elseif 8<=t && t<=16

v = 624-3*t;

elseif 16<=t && t<=26

v = 36*t +12*(t-16)^2;

elseif t>26

v = 2136*exp(-0.1*(t-26));

else

v = 0;

end

end

Answer 5

[baskar ayyakkannu]

solve the below error

Operands to the logical AND (&&) and OR (||) operators must be convertible to logical scalar

values. Use the ANY or ALL functions to reduce operands to logical scalar values.

Error in RocketVelocity (line 2)

if t(0<=t && t<=8)

code:

function v = RocketVelocity(t)
if t(0<=t && t<=8)
    v = 10*t^2-5*t;
elseif 8<=t && t<=16
    v = 624-3*t;
elseif 16<=t && t<=26
    v = 36*t +12*(t-16)^2;
elseif t>26
    v = 2136*exp(-0.1*(t-26));
else
    v = 0;
end
end

[Walter Roberson]

Your code has more than one problem.

Your code is only expecting scalar t. If t is a vector then using the && operator with it is wrong. && can only be used with scalars. && is the "short-circuit and", which executes the left side and only bothers to execute the right side of the && if the left side is (scalar) non-zero. For example,

if isfield(fp, 'usega') && usega > 2

does not execute the usega > 2 test unless the left side is true.

If you are expecting vector t then you need to change all of the && to & such as

elseif 16<=t & t<=26

Next problem: if t is a vector then 16<=t & t<=26 would be a vector. if you "if" a vector of values then the result is considered true only if all of the values are non-zero, equivalent to

elseif all(16 <= t(:) & t(:)<=26)

It would be uncommon for all elements in your vector to be within the same numeric range, so most of the time you would end up with the else branch that assigns v = 0

Note that your else branch assigns a scalar 0 even if t is non-scalar.

Next problem: if your t is not a square 2D array, then t^2 is an error. t^2 is t*t where the * is "matrix multiplication", also known as "inner product". For the * operator (and so the ^ operator) the number of columns of the left operand must be the same as the number of rows of the right operand; for the ^ operation that means that the matrix must be square (including scalar). If you want to square each element of a matrix individually you would use the .^2 operation instead of the ^2 operation.

Next: you have

if t(0<=t && t<=8)

after fixing that to

if t(0<=t & t<=8)

then you would be testing t against that range, constructing a logical array the same size as t, and you would be using that logical array to index t (because t(something) is a request to index t). So you would extract all of the elements of t that are in range 0 to 8, and you would be applying "if" to that list of elements, which is equivalent to testing whether they are non-zero; and the all() rule still applies. So you have effectively coded

if all( reshape(t(0<=t & t<=8), [], 1) ~= 0)

which is a weird way to test

if all(t(:) ~= 0)

Remember, your code is only going to pass one of the if/elseif/else tests, so you have problems if your t contains a range of values.

I recommend that you read about logical indexing:

https://www.mathworks.com/company/newsletters/articles/matrix-indexing-in-matlab.html

https://www.mathworks.com/help/matlab/math/array-indexing.html#MatrixIndexingExample-3