I want to ask for the normalization coefficient of the solution f(x).
The implementation method in Mathematica is as follows:
f[x] := Sin[(n Pi x)/(2 a)];
norm = FullSimplify[1/Integrate[Abs[f[x]]^2, {x, 0, a}], n \[Element] Integers]
The final results are as follows:
2/a
I want to implement the same function in Maple, and have tried the following：
assume(n::posint);
psi := x -> sin(1/2*n*pi*x/a);
c := 1/int(abs(psi(x))^2, x = 0 .. a);
simplify(solve(c = 1, a));
Is there a good way to deal with it if it is implemented with Maple?
restart;
kernelopts(version);
Maple 2022.0, X86 64 LINUX, Mar 8 2022, Build ID 1599809
assume(n::posint);
f := abs(sin(1/2*n*Pi*x/a))^2:
simplify(combine(1/int(convert(f,Heaviside,x),
x=0..a)));
2
-
a
simplify(combine(1/int(f,x=0..a)))
assuming a::real;
2
-
a
Related
Im trying to evaluate this function in maple
but I keep getting this answer, why isn't maple integrating properly. I tried numerically integrating it and it works but I need the analytical solution too.
restart;
sig := x->(exp((x-t)/a)-exp((-x-t)/a))
/(exp((x-t)/a)+exp((-x-t)/a)):
new := convert(simplify(convert(expand(sig(x)),trigh)),tanh);
new := tanh(x/a)
simplify(expand(convert(sig(x) - new, exp)));
0
Now, you originally wrote int(f*sig(x)/x,x).
You didn't indicate that f was a function of x, and as a mere constant it's not really important and could simply be pulled out in front of the integral as a constant factor. If f is some function of x then you really need to state what it is!
Let's consider int(sig(x)/x,x=c..d). Using the simplification new, that is just,
Q := Int( new/x, x=c..d );
Q := Int(tanh(x/a)/x, x = c .. d)
QQ := IntegrationTools:-Change(Q, y=x/a, y);
QQ := Int(tanh(y)/y, y = c/a .. d/a)
You said that you wanted an "analytical solution" by which I take it you mean an explicit formula for the symbolic integration result. But what do you want if the integral does not (mathematically) have a closed form exact, symbolic result?
Would you be content with an (exact, symbolic) series approximation?
H := (a,ord,c,d)
-> int(convert(series(eval(new/x,:-a=a),x,ord),
polynom),x=c..d):
# order 5
H(a, 5, c, d);
3 3 / 5 5\
d - c -c + d 2 \-c + d /
----- - -------- + ------------
a 3 5
9 a 75 a
For a specific example, taking a=2 and an (exact) series approximation of order 25, then the integral from x=0 to x=1 gets evaluated as an exact rational.
evalf(H(2, 25, 0, 1));
0.4868885956
Here's the numeric integration for those same values,
evalf(Int( eval(new/x,a=2), x=0..1 ));
0.4868885956
Specialized numeric quadrature could be as good as a series approximation for a variety of applications, but of course that would depend on what you intend on doing with the result.
This raises the question: what do you hope to do with some supposed "analytical result" that you cannot do with a black-box function that generates the floating-point numeric approximation? Why do you "need" an "analytic result"?
BTW, another way to simplify it (in case the construction above of new does not succeed in your Maple version):
new := convert(simplify(expand( numer(sig(x))/exp(-t/a) ))
/simplify(expand( denom(sig(x))/exp(-t/a) )),
compose,trigh,tanh);
/x\
new := tanh|-|
\a/
I have a program that does some work to get a matrix w, which is 3(n+1) by 3(n+1). I have a vector fbar that is 3(n+1) by 1. I want to get the matrix that, when w is multiplied by it, gives fbar.
In mathematical notation, w * A = fbar. I have w and fbar, and I want A.
I tried to solve it with this command:
fsolve({seq(multiply(w, A)[i, 1] = fbar[i, 1], i = 1 .. 3*(n+1))})
but I don't understand the response Maple gave:
fsolve({2.025881905 A1[2,1]+7.814009150 A1[3,1]+...
-7.071067816 10^(-13) A1[3,1]-0.0004999999990
A1[4,1]-0.0007071067294 A1[5,1]-0.0004999999990 A1[6,1])
A3[6,1]=0},{A1[1,1],A1[2,1],A1[3,1],A1[4,1],A1[5,1],A1[6,1],A\
2[1,1],A2[2,1],A2[3,1],A2[4,1],A2[5,1],A2[6,1],A3[1,1],A3[2,1]\
,A3[3,1],A3[4,1],A3[5,1],A3[6,1]})
What does this mean, and how can I get a more meaningful answer?
You can do this directly with the LinearSolve functional from the LinearAlgebra package if w and fbar are defined as a matrix and vector respectively. The below code makes a reproducible example. Note that the solution of LinearSolve should be equal to x.
w := Matrix(<<1,2,3>|<4,5,6>|<7,8,10>>);
LinearAlgebra[ReducedRowEchelonForm](%); ## Full rank => 1 solution)
x := <1,2,3>;
fbar := w.x;
## Solve the equation w.x = fbar
LinearAlgebra[LinearSolve](w,fbar);
I am trying to create a custom function for pearson's correlation coefficient with this code in matlab 2010
function [p] = customcorr(o)
x := a
y := b
x_mean := mean(a)
y_mean := mean(b)
x_std := std(a)
y_std := std(b)
n := length(o)
r := (1/(n-1))*((x-x_mean)*(y-y_mean))/(x_std*y_std)
end
But i get an error when trying to execute it
Error in ==> customcorr at 2
x := a
Anybody might know what the problem is? Thank you
First, check the correct MATLAB syntax: a "normal" assignment is done by =, not by :=.
Second, you use a and b but these are not defined as parameters of the function. Replace the function head by function p = customcorr(a,b).
Third, I am not really sure what o should be, I assume it can be replaced by length(a) or length(b).
The estimator for an unbiased correlation coefficient is given by
(from wikipedia)
Thus you need to sum all the (a-a_mean).*(b-b_mean) up with sum. Note that it is required to write .* to get the element-wise multiplication. That way you subtract the mean from each element of the vectors, then multiply the corresponding a's and b's and sum up the results of these multiplications.
Together this is
function p = customcorr(a,b)
a_mean = mean(a);
b_mean = mean(b);
a_std = std(a);
b_std = std(b);
n = length(a);
p = (1/(n-1)) * sum((a-a_mean).*(b-b_mean)) / (a_std*b_std);
end
What MATLAB does in their corr function (besides many other interesting things) is, they check the number of arguments (nargin variable) to see if a and b were supplied or not. You can do that by adding the following code to the function (at the beginning)
if nargin < 2
b = a;
end
I'm working on some Matlab code to perform something called the Index Calculus attack on a given cryptosystem (this involves calculating discrete log values), and I've gotten it all done except for one small thing. I cant figure out (in Matlab) how to solve a linear system of congruences mod p, where p is not prime. Also, this system has more than one variable, so, unless I'm missing something, the Chinese remainder theorem wont work.
I asked a question on the mathematics stackexchange with more detail/formatted mathjax here. I solved the issue in my question at that link, and now I'm attempting to find a utility that will allow me to solve the system of congruences modulo a non-prime. I did find a suite that includes a solver supporting modular arithmetic, but the modulus must be prime (here). I also tried stepping through to modify it to work with non-primes, but whatever method is used doesn't work, because it requires all elements of the system have inverses modulo p.
I've looked into using the ability in Matlab to call MuPAD functions, but from my testing, the MuPAD function linsolve (which seemed to be the best candidate) doesn't support non-prime modulus values either. Additionally, I've verified with Maple that this system is solvable modulo my integer of interest (8), so it can be done.
To be more specific, this is the exact command I'm trying to run in MuPAD:
linsolve([0*x + 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], Domain = Dom::IntegerMod(8))
Error: expecting 'Domain=R', where R is a domain of category 'Cat::Field' [linsolve]
The same command returns correct values if I change the domain to IntegerMod(23) and IntegerMod(59407), so I believe 8 is unsuitable because it's not prime. Here is the output when I try the above command with each 23 and 59407 as my domain:
[x = 1 mod 23, y = 1 mod 23, z = 12 mod 23, q = 14 mod 23]
[x = 14087 mod 59407, y = 1 mod 59407, z = 14365 mod 59407, q = 37320 mod 59407]
These answers are correct- x, y, z, and q correspond to L1, L2, L3, and L4 in the system of congruences located at my Math.StackExchange link above.
I'm wondering if you tried to use sym/linsolve and sym/solve previously, but may have passed in numeric rather than symbolic values. For example, this returns nonsense in terms of what you're looking for:
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(A,b),8)
But if you convert the numeric values to symbolic integers, sym/linsolve will keep everything in terms of rational fractions. Then
s = mod(linsolve(sym(A),sym(b)),8)
returns the expected answer
s =
6
1
6
4
This just solves the system linear system using symbolic math as if it were a normal matrix. For large systems this can be expensive, but I'd imagine no more than using MuPAD's numeric::linsolve or linalg::matlinsolve. sym/mod should return the modulus of the numerator of each solution component. I believe that you will get an error if the modulus and the denominator are not at least coprime.
sym/solve can also be used to solve this in a similar manner:
L = sym('L',[4,1]);
[L1,L2,L3,L4] = solve(A*L==b);
s = mod([L1;L2;L3;L4],8)
A possible issue with using either sym/solve or sym/linsolve is that if there are multiple solutions to the linear congruence problem (as opposed to the linear system), this approach may not return all of them.
Finally, using the MuPAD function numlib::ichrem (chinese remainder theorem for integers), here's some code that attempts to obtain the complete solution:
A = [0 5 4 1;1 7 0 2;8 1 0 2;10 5 1 0];
b = [2946321;5851213;2563617;10670279];
m = 10930888;
mf = str2num(strrep(char(factor(sym(m))),'*',' '));
A = sym(A);
b = sym(b);
s = sym(zeros(length(b),length(mf)));
for i = 1:length(mf)
s(:,i) = mod(linsolve(A,b),mf(i));
end
mstr = ['[' sprintf('%d,',mf)];
mstr(end) = ']';
r = sym(zeros(length(b),1));
for i = 1:length(b)
sstr = char(s(i,:));
r(i) = feval(symengine,'numlib::ichrem',sstr(9:end-2),mstr);
end
check = isequal(mod(A*r,m),b)
I'm not sure if any of this is what you're looking for, but hopefully it might be helpful. I think that it might be a good idea to put in a enhancement/service request with the MathWorks so that MuPAD and the other solvers can handle systems better in the future.
I am trying to calculate the convolution of
x(t) = 1, -1<=t<=1
x(t) = 0, outside
with itself using the definition.
http://en.wikipedia.org/wiki/Convolution
I know how to do using the Matlab function conv, but I want to use the integral definition. My knowledge of Matlab and WolframAlpha is very limited.
I am still learning Mathematica myself, but here is what I came up with..
First we define the piecewise function (I am using the example from the Wikipedia page)
f[x_] := Piecewise[{{1, -0.5 <= x <= 0.5}}, 0]
Lets plot the function:
Plot[f[x], {x, -2, 2}, PlotStyle -> Thick, Exclusions -> None]
Then we write the function that defines the convolution of f with itself:
g[t_] = Integrate[f[x]*f[t - x], {x, -Infinity, Infinity}]
and the plot:
Plot[g[t], {t, -2, 2}, PlotStyle -> Thick]
EDIT
I tried to do the same in MATLAB/MuPad, I wasn't as successful:
f := x -> piecewise([x < -0.5 or x > 0.5, 0], [x >= -0.5 and x <= 0.5, 1])
plot(f, x = -2..2)
However when I try to compute the integral, it took almost a minute to return this:
g := t -> int(f(x)*f(t-x), x = -infinity..infinity)
the plot (also took too long)
plot(g, t = -2..2)
Note the same could have been done from inside MATLAB with the syntax:
evalin(symengine,'<MUPAD EXPRESSIONS HERE>')
Mathematica actually has a convolve function. The documentation on it has a number of different examples:
http://reference.wolfram.com/mathematica/ref/Convolve.html?q=Convolve&lang=en
I don't know much about Mathematica so I can only help you (partially) about the Matlab part.
To do the convolution with the Matlab conv functions means you do it numerically. What you mean with the integral definition is that you want to do it symbolically. For this you need the Matlab Symbolic Toolbox. This is essentially a Maple plugin for Matlab. So what you want to know is how it works in Maple.
You might find this link useful (wikibooks) for an introduction on the MuPad in Matlab.
I tried to implement your box function as a function in Matlab as
t = sym('t')
f = (t > -1) + (t < 1) - 1
However this does not work when t is ob symbolic type Function 'gt' is not implemented for MuPAD symbolic objects.
You could declare f it as a piecewise function see (matlab online help), this did not work in my Matlab though. The examples are all Maple syntax, so they would work in Maple straight away.
To circumvent this, I used
t = sym('t')
s = sym('s')
f = #(t) heaviside(t + 1) - heaviside(t - 1)
Unfortunately this is not successful
I = int(f(s)*f(t-s),s, 0, t)
gives
Warning: Explicit integral could not be found.
and does not provide an explicit result. You can still evaluate this expression, e.g.
>> subs(I, t, 1.5)
ans =
1/2
But Matlab/MuPad failed to give you and explicit expression in terms of t. This is not really unexpected as the function f is discontinuous. This is not so easy for symbolic computations.
Now I would go and help the computer, fortunately for the example that you asked the answer is very easy. The convolution in your example is simply the int_0^t BoxFunction(s) * BoxFunction(t - s) ds. The integrant BoxFunction(s) * BoxFunction(t - s) is again a box function, just not one that goes from [-1,1] but to a smaller interval (that depends on t). Then you only need to integrate the function f(x)=1 over this smaller interval.
Some of those steps you would have to to by hand first, then you could try to re-enter them to Matlab. You don't even need a computer algebra program at all to get to the answer.
Maybe Matematica or Maple could actually solve this problem, remember that the MuPad shipped with Matlab is only a stripped down version of Maple. Maybe the above might still help you, it should give you the idea how things work together. Try to put in a nicer function for f, e.g. a polynomial and you should get it working.