I want to integrate "\int_{0}^{1}(exp(-int_{0}^{y}f(x)dx))dy" with my basic trapezoid algorithm. I recieve an error declaration, but I should define g as a function. Do you have any idea how to do it?
Thanks a lot for any answer!
function y = trapapadbl(low1, up1,low2,up2,intstep1,intstep2,f)
g = 0;
step1 = (up1 - low1) / intstep1;
step2 = (up2 - low2) / intstep2;
for j = low1 : step1 : up1
g = g + feval(f,j);
end
g = #(y)(g - (feval(f, low1) + feval(f, up1))/2) * step1;
for i = low2 : step2 : up2
y= y + feval(g,i);
end
y= (y - (feval(g, low2) + feval(g, up2))/2) * step2;
>> trapapadbl(0,1,0.1,0,1,0.1,#sin)
??? Undefined function or variable "y".
Error in ==> trapapadbl at 12
y= y + feval(g,i);
Without working too hard to try to understand your code (!) the error is that y was never initialized. You can't add anything to y until it has a value. When I initialize y to 0, the code runs, but I get 0 as an output, which is not what happens when you integrate sin from 0 to 1. I may be calling the function wrong, but it's something to look out for!
Furthermore, your code is confusing, because you use the variable g as both a double (a number) and a function, even in the same line! The same problem happens as y is the input to your anonymous function, but also a double later on. It's syntactically correct, but a little hard to read. Consider using a different variable name, or including clear comments (or both!)
Related
SO basically it keeps saying error at line 14 which is the "else" code is at i dont get it why it is synta error please help
clear
clc
function f=f(x)
f = x^3 + 2*x^2 - 3*x -1
endfunction
disp ("sample input"): regulaFalsi (1,2,10^-4, 100)
function regulaFalsi(a, b, TOL, N)
i = 1
FA = f(a)
finalOutput =(i, a , b , a + (b-a)/2, f(a + (b-a) /2)
printf ("%-20s%-20s%-20s%-20s%-20s\n","n","a_n","b_n","p_n","f(p_n)")
while (i <= N),
p = (a*f(b)-b*f(a))/f(b) - f(a))
FP = f(p)
if (FP == 0 | aba (f(p)) < TOL) then
break
else
printf("%-20.8g %-20.8g %-20.8g %-20.8g %-20.8g\n", i, a, b, p, f(p))
end
i = i + 1
if (FA + FP > 0) then
a = p
else
b = p
end
end
I have been trying to fix this code for my assignment but i dont know why it keeps giving me syntax error
No indentation does not matter but you wrote
disp ("sample input"): regulaFalsi (1,2,10^-4, 100)
instead of
disp ("sample input"); regulaFalsi (1,2,10^-4, 100)
a colon : instead of a semi colon ;
Moreover an "end" is missing to close the regulaFalsi function definition
In addition to Serge's answer:
regulaFalsi() is defined ''after'' the first call to it. Sure that this first call will fail.
Although finalOutput is unused (and so likely useless), the definition finalOutput =(i, a , b , a + (b-a)/2, f(a + (b-a) /2) misses a closing ). It is likely the origin of the error.
f(): here it works, but in a more general way it is not a good idea to name the function's output with the same name as the function itself.
putting clear at the head of your script(s) is not really a good idea. It is most often useless, and most often violent enough to erase useful objects, like libraries loaded on the fly, etc.
When you report an error, please report the full actual error message. It is most often more useful than only comments or "personal translation" of the error.
How do I evaluate the function in only one of its variables, that is, I hope to obtain another function after evaluating the function. I have the following piece of code.
deff ('[F] = fun (x, y)', 'F = x ^ 2-3 * y ^ 2 + x * y ^ 3');
fun (4, y)
I hope to get 16-3y ^ 2 + 4y ^ 3
If what you want to do is to write x = f(4,y), and later just do x(2) to get -36, that is called partial application:
Intuitively, partial function application says "if you fix the first arguments of the function, you get a function of the remaining arguments".
This is a very useful feature, and very common Functional Programming Languages, such as Haskell, but even JS and Python now are able to do it. It is also possible to do this in MATLAB and GNU/Octave using anonymous functions (see this answer). In Scilab, however, this feature is not available.
Workround
Nonetheless, Scilab itself uses a workarounds to carry a function with its arguments without fully evaluating. You see this being used in ode(), fsolve(), optim(), and others:
Create a list containing the function and the arguments to partial evaluation: list(f,arg1,arg2,...,argn)
Use another function to evaluate such list and the last argument: evalPartList(list(...),last_arg)
The implementation of evalPartList() can be something like this:
function y = evalPartList(fList,last_arg)
//fList: list in which the first element is a function
//last_arg: last argument to be applied to the function
func = fList(1); //extract function from the list
y = func(fList(2:$),last_arg); //each element of the list, from second
//to last, becomes an argument
endfunction
You can test it on Scilab's console:
--> deff ('[F] = fun (x, y)', 'F = x ^ 2-3 * y ^ 2 + x * y ^ 3');
--> x = list(fun,4)
x =
x(1)
[F]= x(1)(x,y)
x(2)
4.
--> evalPartList(x,2)
ans =
36.
This is a very simple implementation for evalPartList(), and you have to be careful not to exceed or be short on the number of arguments.
In the way you're asking, you can't.
What you're looking is called symbolic (or formal) computational mathematics, because you don't pass actual numerical values to functions.
Scilab is numerical software so it can't do such thing. But there is a toolbox scimax (installation guide) that rely on a the free formal software wxmaxima.
BUT
An ugly, stupid but still sort of working solution is to takes advantages of strings :
function F = fun (x, y) // Here we define a function that may return a constant or string depending on the input
fmt = '%10.3E'
if (type(x)==type('')) & (type(y)==type(0)) // x is string is
ys = msprintf(fmt,y)
F = x+'^2 - 3*'+ys+'^2 + '+x+'*'+ys+'^3'
end
if (type(y)==type('')) & (type(x)==type(0)) // y is string so is F
xs = msprintf(fmt,x)
F = xs+'^2 - 3*'+y+'^2 + '+xs+'*'+y+'^3'
end
if (type(y)==type('')) & (type(x)==type('')) // x&y are strings so is F
F = x+'^2 - 3*'+y+'^2 + '+x+'*'+y+'^3'
end
if (type(y)==type(0)) & (type(x)==type(0)) // x&y are constant so is F
F = x^2 - 3*y^2 + x*y^3
end
endfunction
// Then we can use this 'symbolic' function
deff('F2 = fun2(y)',' F2 = '+fun(4,'y'))
F2=fun2(2) // does compute fun(4,2)
disp(F2)
I asked a question a few days ago here and got an answer that seems like it would work- it involves using linsolve to find the solutions to a system of equations that are all modulo p, where p is a non-prime integer.
However, when I try to run the commands from the provided answer, or the linsolve help page, I get an error saying linsolve doesn't support arguments of type 'sym'. Is using linsolve with sym variables only possible in R2013b? I've also tried it with my school's copy, which is R2012b. Here is the code I'm attempting to execute (from the answer at the above link):
A = [0 5 4 1;1 7 0 2;8 1 0 2;10 5 1 0];
b = [2946321;5851213;2563617;10670279];
s = mod(linsolve(sym(A),sym(b)),8)
And the output is:
??? Undefined function or method linsolve' for input arguments of type 'sym'.
I've also tried to use the function solve for this, however even if I construct the equations represented by the matrices A and b above, I'm having issues. Here's what I'm attempting:
syms x y z q;
solve(5*y + 4*z + q == 2946321, x + 7*y + 2*q == 5851213, 8*x + y + 2*q == 2563617, 10*x + 5*y + z == 10670279,x,y,z,q)
And the output is:
??? Error using ==> char
Conversion to char from logical is not possible.
Error in ==> solve>getEqns at 169
vc = char(v);
Error in ==> solve at 67
[eqns,vars] = getEqns(varargin{:});
Am I using solve wrong? Should I just try to execute my code in R2013b to use linsolve with symbolic data types?
The Symbolic Math toolbox math toolbox has changed a lot (for the better) over the years. You might not have sym/linsolve, but does this work?:
s = mod(sym(A)\sym(b),8)
That will basically do the same thing. sym/linsolve just does some extra input checking and and rank calculation to mirror the capabilities of linsolve.
You're using solve correctly for current versions, but it looks like R2010b may not understand the == operator (sym/eq) in this context. You can use the old string format to specify your equations:
eqs = {'5*y + 4*z + q = 2946321',...
'x + 7*y + 2*q = 5851213',...
'8*x + y + 2*q = 2563617',...
'10*x + 5*y + z = 10670279'};
vars = {'x','y','z','q'};
[x,y,z,q] = solve(eqs{:},vars{:})
So I have a simple example of what I want to do:
restart;
assume(can, real);
f := {g = x+can*x*y, t = x+x*y};
assign(f[1]); g;
can := 2;
plot3d(g, x = 0 .. 100, y = 0 .. 100);
while this works:
restart;
f := {g = x+can*x*y, t = x+x*y};
assign(f[1]);
can := 2;
plot3d(g, x = 0 .. 100, y = 0 .. 100);
But that assumptions are really important for my real life case (for some optimisations with complex numbers) so I cant just leve can not preassumed.
Why it plots nothuing for me and how to make it plot?
The expression (or procedure) to be plotted must evaluate to a numeric, floating-point quantity. And so, for your expression g, the name can must have a specific numeric value at the time any plot of g is generated.
But you can produce a sequence of 3D plots, for various values of can, and display them. You can display them all at once, overlaid. Or you can display them in an animated sequence. And you can color or shade them each differently, to give a visual cue that can is changing and different for each.
restart;
f := {g = x+can*x*y, t = x+x*y};
eval(g,f);
N:=50:
Pseq := seq(
plot3d(eval(g,f),
x=0..10,y=0..10,
color=RGB(0.5,0.5,can/(2*N)),
transparency=0.5*(can/(N+1))),
can=1 .. N):
plots:-display(Pseq, axes=box);
plots:-display([Pseq],insequence=true,axes=box);
By the way, you don't have to assign to g just for the sake of using the equation for g that appears inside f. Doing that assignment (using assign, say, like you did) makes it more awkward for you subsequently to create other equations in terms of the pure name g unless you first unassign the name g. Some people find it easier to not make the assignment to g at all for such tasks, and to simply use eval as I've done above.
Now on to your deeper problem. You create an expression containing a local, assumed name. and then later on you want to use the same expression but with the global, unassumed version of that name. You can create the expression, with it containing the global, unassumed name instead of the local, assumed name, buy performing a substitution.
restart;
assume(can, real);
f := {g = x+can*x*y, t = x+x*y};
{g = x + can~ x y, t = x + x y}
assign(f[1]);
g;
x + can~ x y
can := 2:
g;
x + can~ x y
# This fails, because g contains the local name can~
plot3d(g, x=0..100, y=0..100);
# A procedure to make the desired substitution
revert:=proc(nm::name)
local len, snm;
snm:=convert(nm,string);
len:=length(snm);
if snm[-1]="~" then
return parse(snm[1..-2]);
else return parse(nm);
end if;
end proc:
# This is the version of the expression, but with global name can
subsindets(g,`local`,revert);
x + can x y
# This should work
plot3d(subsindets(g,`local`,revert),
x=0..100,y=0..100);
I'm having some problems with my code. Here it is:
lambdaz = 1.2;
n = 24;
mu = 0.00055e9;
lambda = sym('lambda','clear');
W = (((2.*mu)./n.^2)).*((lambda.^n)+(lambdaz.^n)+((lambda.^-n).*(lambdaz.^-n))-3);
dW_dlambda = diff(W, lambda);
W2=(((2.*mu)./n.^2).*(lambda.^n))+(((2.*mu)./n.^2).*(lambdaz.^n))+(((2.*mu)./n.^2).*((lambda.^-n).*(lambdaz.^-n)))-(3.*((2.*mu)./n.^2))
dW2_dlambda=diff(W2,lambda)
x=((((lambda.^2).*(lambdaz))-1).^-1).*(dW_dlambda);
x2=((((lambda.^2).*(lambdaz))-1).^-1).*(dW2_dlambda)
P2 = int(x2,lambda)
P=int(x,lambda);
P=(0:1000:26700)
plot(lambda,P)
When I try to plot lambda against P I get the "conversion to double from sym is not possible" error message. I'm not particularly fantastic at Matlab so any help would be gratefully received!
The plot function only works for numeric inputs. Both lambda and P are symbolic expressions (at least before overwrote P by setting it equal to a vector after the integration) that cannot be directly converted to to floating point. You get the same error if try to something like double(sym('exp(x)')). You have two options. The first is the ezplot function in the Symbolic Toolbox:
...
P = int(x,lambda);
ezplot(P,[-5 5]); % Plot's P from lambda = -5 to lambda = 5
Or you can use the subs function:
...
P = int(x,lambda);
lambda = -5:0.01:5;
plot(lambda,real(subs(P,'lambda',lambda)))
axis([lambda(1) lambda(end) -1e15 1e15])
I used real to suppress a warning for negative values of lambda.