I'm trying to make Maple solve a complex equation, but it produces an incorrect result.
The following images tells it all :
At (3) I would expect to get something close to 1 (as (2) shows), yet it gives me something that doesn't make any sense. Is it that the || (to express the complex number modulus) operator has another significance in the solve() function?
The more appropriate function here is fsolve.
Example 1
restart:
G:=(w,L)->(5+I*L*2*Pi*w)/(150+I*L*2*Pi*w);
evalf(5*abs(G(10,1)));
fsolve(5*abs(G(10,L))=%,L=0..10)
Example 2
As above, you need to specify the interval L=0..1 where the solution might be.
G:=(f,L)->(256.4+I*L*2*Pi*f)/(256.4+9845+I*L*2*Pi*f);
evalf(5*abs(G(20000,0.03602197444)));
fsolve(5*abs(G(20000,L))=%,L=0..1);
If you are facing difficulties to specify the interval then you should plot it first, it will give you an idea about it?
plot(5*abs(G(20000,L)),L=0..1)
Restrict the values of L in the solve command with the assuming command.
Example 1:
G:= (w,L) -> (50+I*L*2*Pi*w)/(150+I*L*2*Pi*w);
result := evalf(5*abs(G(10,1)));
solve({5*abs(G(10,L)) = result},L) assuming L::real;
{L = 1.000000000}, {L = -1.000000000}
Example 2:
G:=(f,L) -> (256.4+I*2*Pi*L*f)/(256.4+9845+I*2*Pi*L*f);
result := 5*abs(G(20000,0.03602197444));
solve({5*abs(G(20000,L)) = result},L) assuming L::real;
{L = 0.03602197445}, {L = -0.03602197445}
Related
From the documentation, we see the following example:
g = gallery('integerdata',3,[15,1],1);
x = gallery('uniformdata',[15,1],9);
y = gallery('uniformdata',[15,1],2);
A = table(g,x,y)
func = #(x, y) (x - y);
B = rowfun(func,A,...
'GroupingVariable','g',...
'OutputVariableName','MeanDiff')
When the function func is applied to A in rowfun how does it know that there are variables in A called x and y?
EDIT: I feel that my last statement must not be true, as you do not get the same result if you did A = table(g, y, x).
I am still very confused by how rowfun can use a function that does not actually use any variables defined within the calling environment.
Unless you specify the rows (and their order) with the Name/Value argument InputVariables, Matlab will simply take column 1 as first input, column 2 as second input etc, ignoring eventual grouping columns.
Consequently, for better readability and maintainability of your code, I consider it good practice to always specify InputVariables explicitly.
I'm trying to use Newton's method to approximate where the gradient of a function = 0 But I'm guessing error messages when I try to put in guesses for an iteration of Newton's, and I don't know what the problem is.
(For reference, Profit = 144TVa - 0.07TVb TVb - 0.01 TVa^2 + 174TVb -0.01TVb^2 - 4E5)
with(LinearAlgebra):
with(VectorCalculus):
DProfit := Gradient(Profit, [TVa, TVb])
F := unapply(DProfit, TVa, TVb)
J := Jacobian(DProfit)
Guess := V->evalf(V-Multiply(Jinv,F(V))):
But when I try to evaluate Guess at any points, it gives me an error:
Guess(3000,3000)
Error, (in Guess) invalid input: F uses a 2nd argument, TVb, which is missing
Guess(<3000,3000>
Error, (in Guess) invalid input: F uses a 2nd argument, TVb, which is missing
even though F(3000,3000) returns [-66,-36]
Thanks for the help.
Figured out how to fix it!
Instead of:
Guess := V->evalf(V-Multiply(Jinv,F(V))):
I did:
Guess := V->evalf(V-Multiply(Jinv,F(V[1],V[2]))):
I have been using MATLAB fminunc function to solve my optimization problem. I want to try the minFunc package :
http://www.di.ens.fr/~mschmidt/Software/minFunc.html
When using fminunc, I defined a function funObj.m which gives me the objective value and the gradient at any point 'x'. It also takes in several external inputs say, {a,b,c} which are matrices. So the function prototype looks like :
function [objVal,G] = funObj(x,a,b,c)
I want to use the same setup in the minFunc package. From the examples, I figured this should work :
options.Method='lbfgs';
f = #(x)funObj(x,a,b,c);
x = minFunc(f,x_init,options);
But when I call this way, I get an error as:
Error using funObj
Too many output arguments.
What is the correct way to call minFunc for my case?
**EDIT : Alright, here is a sample function that I want to use with minFunc. Lets say I want to find the minimum of a*(b-x)^2, where a,b are scalar parameters and x being a scalar too. The MATLAB objective function will then look like :
function obj = testFunc(x,a,b)
obj = a*(b-x)^2;
The function call to minimize this using fminunc (in MATLAB ) is simply:
f = #(x)testFunc(x,a,b);
x = fminunc(f,x_init);
This gives me the minimum of x = 10. Now, How do I do the same using minFunc ?
"Note that by default minFunc assumes that the gradient is supplied, unless the 'numDiff' option is set to 1 (for forward-differencing) or 2 (for central-differencing)."
The error is because only one argument is returned by the function. You can either return the gradient as a second argument or turn on numerical differencing.
Agree with Mark. I think the simplest way to solve it is
minFunc(#testFunc, x_init, a, b, c)
In MATLAB temporary function can only have one return value. So f = #(x)testFunc(x,a,b); let your method drop gradient part every time. Because minFunc can accept extra paramters, you can pass a, b and c after x_init. I think this would work.
temp(i,1) = rand(1)*(pb(1,num).pos(i,1) - pw(1,num).pos(i,1));
This line gives the following error:
Error using ==> minus
Not enough input arguments.
The following are the definitions of pb and pw.
pw=struct('fitness',[],'pos',{});
pb=struct('fitness',[],'pos',{});
pos is a 2 x 1 array.
When tracking down errors like this, I break the problem up into smaller bits. Especially when the logic isn't readily apparent. Not only does it provide a path that can be used to step through your function using the debugger, but it also makes it more readable.
I've taken liberty with the intermediate variable names.
thisPb = pb(1,num);
thisPw = pw(1,num);
initialPos= pw.pos(i,1);
finalPos = pb.pos(i,1);
whos initialPos finalPos
temp(i,1) = rand(1) * (finalPos - initialPos);
The line with whos will print out the values. Make sure that finalPos and initialPos are both numbers.
One way that you can get this error is when num is an empty matrix.
The expression
>> s(x).a
can return a variable number of outputs, depending on the size of x.
If x = [1,2,3] for example, it will return three values (as long as s has at least three elements).
If x = [] on the other hand, then s(x).a will return no outputs, so the expression
>> disp(s(x).a)
will give you a Not enough input arguments error, which is almost certainly what you're seeing. You should check that num is not empty.
Are you sure, that all values are really initialised? Try to check this before your codeline.
disp(pb(1,num).pos(i,1))
disp(pw(1,num).pos(i,1))
temp(i,1) = rand(1)*(pb(1,num).pos(i,1) - pw(1,num).pos(i,1));
I've been given the task of translating a piece of MATLAB code into IDL and have
hit a roadblock when I came across the MATLAB function accumarry(). The
function, described here
is used to sum elements in one array, based on indices given in another. Example
1 perhaps explains this better than the actual function description at the top
of the page. In trying to reproduce Example 1 in IDL, I haven't been able to avoid a for loop, but I'm confident that it is possible. My best attempt is the following:
vals = [101,102,103,104,105]
subs = [0,1,3,1,3]
n = max(subs)+1
accum = make_array(n)
for i = 0, n-1 do begin
wVals = where(subs eq i,count)
accum[i] = count eq 0 ? 0 : total(vals[wVals])
endfor
print,accum
; 101.000 206.000 0.00000 208.000
Any advice on improving this would be greatly appreciated! I expected IDL to have a similar built-in function, but haven't been able to track one down. Perhaps some magic with histogram binning?
I found a number of possible solutions to this problem on Coyote's IDL site (not surprisingly.)
http://www.idlcoyote.com/code_tips/drizzling.html
I ended up using the following, as a compromise between performance and readability:
function accumarray,data,subs
mx = max(subs)
accum = fltarr(mx+1)
h = histogram(subs,reverse_indices=ri,OMIN=om)
for j=0L,n_elements(h)-1 do if ri[j+1] gt ri[j] then $
accum[j+om] = total(vals[ri[ri[j]:ri[j+1]-1]])
return,accum