Subroutine using if statement

What I'm trying to do is to supply different value of constant k at different x intervals. However, it seems that my y(x==3) is not equal to y_actual.

x=0:20;
k=X(x); %Subroutine func
y=k*x;

y_actual = (5+2)*3 % what i wanted
y(x==3)

function k=X(x)
if (x>=2)  
    w=5;
else 
    w=4;
end

k=w+2;

end

[SALAH ALRABEEI]

when u send X(x) which already conatins values <2, so the if will always go to the else (w=4)

[Anna Lin]

I see. So I shouldn't pass x into the function X?

Answer 1

[Star Strider]

I generally favour this sort of approach to such prolblems —

X = @(x) ((x>=2).*5 + (x<2).*4) + 2;

x = 0:20;
k = X(x);
y = k.*x;

figure
plot(x, y, '+-')
grid

xy = table(x(:),y(:), 'VariableNames',{'x','y'})

xy = 21×2 table
    x      y 
    __    ___

     0      0
     1      6
     2     14
     3     21
     4     28
     5     35
     6     42
     7     49
     8     56
     9     63
    10     70
    11     77
    12     84
    13     91
    14     98
    15    105

.

[Anna Lin]

Thank you so much! My actual assignment is much more complex with 2d heat equations, so i make this simple question trying to understand the concept. Anyways, thank you once again.

[Anna Lin]

what if I have another y interval to be considered? like a box area when (x<=a & y<=b)?

[Star Strider]

As always, my pleasure!

To include another condition, simply add it (providing the other conditions do not overlap oir conflict with existing conditions):

X = @(x) ((x>=2).*5 + (x<2).*4) + 2 + ((x<=a) & (y<=b)).*something

However, ‘X’ then needs to be a function of both ‘x’ and ‘y’ with appropriate additional constants and arguments, as necessary. It will be necessary to experiment with to get the desired result.

This approach generally works for most such situations, however more complicated situations may require additional such functions, or a completely different approach altogether.

.

[Anna Lin]

Alright, I'll work around with it. Thank you very much!

[Star Strider]

The image was not present earlier.

Example —

F = @(x,y) ((x<=3) & (y<=5)).*(-2) + ((x>=20) & (y<=5)).*2;

xv = linspace(0,40,40);
yv = linspace(0,10,10);
[Xm,Ym] = ndgrid(xv,yv);

Zm = F(Xm,Ym);

figure
surfc(Xm, Ym, Zm)
grid on
axis('equal')

.

[Anna Lin]

Thank you very very much!

[Star Strider]

As always, my pleasure!

Answer 2

[Image Analyst]

Lots of stuff wrong with this. For starters, you're passing in the whole x vector into the (poorly-named) X function yet your function seems to be expecting only a single value of x, not a whole multi-element vector. Take the semicolons off the ends of the lines and you'll see exactly what it's doing, which is exactly what you told it. You said x==3 which gives a 21 element logical vector with false everywhere except at index 4 (where x has the value 3) and it's true there.

ans =

1×21 logical array

0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

So then y(4) = 18 which is why it's giving you 18.

I'm not sure if you expect the k to be a vector or a scalar so I'm not sure how to tell you to fix it.

x=0:20
k=X(x) % Subroutine function
y=k*x

y_actual = (5+2)*3 % what i wanted
y(x==3)

function k=X(x)
if (x>=2)
	w=5;
else
	w=4;
end
k=w+2;
end

x =

Columns 1 through 20

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

Column 21

20

k =

6

y =

Columns 1 through 20

0 6 12 18 24 30 36 42 48 54 60 66 72 78 84 90 96 102 108 114

Column 21

120

y_actual =

21

ans =

18

[Anna Lin]

The k should be a vector.