Error using fsurf (too many functions) but success using ezsurf

When using the fsurf command to plot this function, I recieve the message: "error using fsurf, too many functions."

By simply changing "fsurf" to "ezsurf" I successfully get my 4 figures (intended to model tumors).

I understand there is a difference in the plotting methods of fsurf and ezsurf, and I assume that fsurf, if correctly used, would give me more detailed figures. What could be causing the discrepancy? I appreciate any and all help!

syms u v 
for a = 3:4 
for b = 5:6 
    figure
fsurf((1+1/5*sin(a*u)*sin(b*v))*sin(v)*cos(u),(1+1/5*sin(a*u)*sin(b*v))*sin(v)*sin(u),(1+1/5*sin(a*u)*sin(b*v))*cos(v),[0 2*pi 0 pi])
end
end

Answer 1

[madhan ravi]

Just vectorize the function and use function handles. To learn the proper usage of the functions that get you into a dilemma is to read the documentation page of the function.

fsurf(@(u,v)(1+1/5*sin(a*u).*sin(b*v)).*sin(v).*cos(u),(1+1/5*sin(a*u).*sin(b*v)).*sin(v).*sin(u),(1+1/5*sin(a*u).*sin(b*v)).*cos(v),[0 2*pi 0 pi]) % just alter your line to this line
doc fsurf % to open the documentation of the function

However I'm not getting the same error message that you're getting and it's not familiar.

[Austin Vander Linden]

I appreciate the help, however I'm still getting the same error.

[Austin Vander Linden]

Update- I opened a fresh instance of matlab and the command works perfectly. I'm unsure what was causing the error message, but I'm fine now. Thank you !

Answer 2

[Balu]

Convert your interval argument to a double

fsurf(f, double(plot_range))

MATLAB doesn't always evaluate the final values of expressions. For example, you used 2*pi in your code. This has to be evaluated to 6.28. But if MATLAB finds it suitable, it can decide to just leave it as 2*pi and evaluate later when needed. But when your expressions are not evaluated, I think your plot range is considered a vector function. By doing double(plot_range), you get a scalar array, which is what fsurf needs.

[Walter Roberson]

I think this needs some clarification.

Suppose you have a code expression such as

sum([1 2*pi])

ans = 7.2832

then at some point 2*pi has to be calculated. The question is when it gets calculated. Does it get calculated as soon as the line is parsed, so that the internal tokenized code becomes

sum(horzcat(1, 6.28318530717959))

? Or does the calculation get held off at parsing time but the expression is calculated before sum() is called, similar to

temporary1 = prod(2, 3.14159265358979);

temporary2 = horzcat(1, temporary1);

sum(temporary2);

? Or does an unevaluated expression get passed in to sum(), to be calculated only when the result is actually needed ("lazy evaluation"), something like

node1 = LAZY(@prod, 2, pi)

node2 = LAZY(@horzcat, 1, node1)

sum( node2 )

The first answer to that is that the model of MATLAB is left to right evaluation, as modified by () and [] and other precedence (for example, multiplication has higher priority). During execution, as soon as both arguments of a binary operator are resolved, the operator is invoked, and the resulting value becomes the left operand for any further operators. Thus the execution of sum([1 2*pi]) is most similar to the second flow I show with temporary values.

Suppose you have

[sin(pi) myfunction() sin(pi)]

then a question would be whether that would get executed like

temporary1 = sin(pi);

temporary2 = myfunction();

horzcat(temporary1, temporary2, temporary1)

or like

temporary1 = myfunction();

temporary2 = sin(pi);

horzcat(temporary2, temporary1, temporary2)

or like

temporary1 = sin(pi);

temporary2 = myfunction();

temporary3 = sin(pi);

horzcat(temporary1, temporary2, temporary3)

The left-to-right rule I mentioned above eliminates the possibility that the middle value might be evaluated first, but leaves open the possibility that optimization might take place to save the value of common sub-expressions instead of recalculating them.

It turns out that the answer is that there might be deep optimizations but they have to operate as if they were not present. In particular, there is the possibility that myfunction() might alter the execution environment, sin(pi) might mean one thing on the left, and then after myfunction() is called possibly sin or pi might mean different things and so there might be a different result. This is not hypothetical; I have successfully tested such cases. (You can get some strange results if calling a function has the effect of replacing a function call with a reference to a variable.)

By the way: formally speaking, pi in MATLAB is a call to a function.

[Walter Roberson]

But when your expressions are not evaluated, I think your plot range is considered a vector function.

No. At the syntax level, MATLAB just knows that there is an expression at that point. The expression is fully evaluated; if it included any function calls, then the function would be invoked, and the result of the function would be what was passed as the parameter to fsurf().

[0 2*pi 0 pi] involves only calculations with double precision values, so the result of [0 2*pi 0 pi] is already a double precision vector. Calling double([0 2*pi 0 pi]) is not necessary at all and would have no effect in this case.

fsurf() and fplot() accept any numeric datatype for the range specification. You can

fsurf(@(x,y)x.^2-y.^2, uint8([8 11]))

so saying that you need to double() the plot range is incorrect advice.

[Balu]

@WalterRoberson I'd expect that to be the case for any programming language, but it doesn't appear that MATLAB is rigorous enough to do that.

When I have a range matrix of unevaluated expressions, I find that the double function is necessary to avoid the too many functions error from fsurf.

>> plot_range
[-pi/20, (11*pi)/20, -5, 5]
>> f
f =
sin(x)^20
>> fsurf(f, plot_range)
Error using fsurf
Too many functions. 
>> fsurf(f, double(plot_range))
>>

MATLAB version: 9.13.0.2145394 (R2022b) Update 3

[Walter Roberson]

@Balu

Ah, what is going on there is that you have defined your plot_range symbolically (you are getting symbolic π there), and your f is symbolic as well. Your fsurf(f,plot_range) is then passing two symbolic expressions to fsurf(), which is only expecting one symbolic expression. The range does need to be numeric (not symbolic) for fsurf(). But this has nothing to do with "vector function" or "scalar array".