i am trying to solve these 2 equations with 2 unknowns.
i want to find the angle of a1(alpha1), a2(alpha2) in radians.
however when i solve it in maple i get 0.112 for a2 and 0.089 for a1.
but some other guys used matlab and they get something different with plot and 2 angles..
can someone tell me if i am right or he is?
MY equations:
equation 1: 4/(2*pi)*(cos(2*a1)+cos(2*a2))=0
equation 2: 4/(10*pi)*(cos(10*a1)+cos(10*a2))=0
his matlab:
j=0;
for i=0:pi/100:1
j=j+1;
a2(j)=i;
a1_1(j)=acos(-cos(2*i))/2;
a1_2(j)=acos(-cos(10*i))/10;
end
plot(a2,a1_1,'-k',a2,a1_2,'-b','LineWidth',1.4);
I wanna plot like him..but i am not sure his Equations a1_1 is correct?
by the way the main equation it came from:
bn = 4/(npi)(cos(na1)+cos(na2))=0
bn = 0 and n is 2th and 10th harmonics to be eliminated
cos(A)+cos(B) = 0
happens for
A + B = pi + 2*k*pi or A - B = pi + 2*k*pi
Both of the two original equations could resolve the the same of the two variants or to different variants. Resolving to the same equation, say the first, will produce
7*(a1+a2) = pi + 2*k7*pi and 13*(a1+a2) = pi + 2*k13*pi
which can only hold if
0 = 6*pi + 2*(13*k7-7*k13) <=> 3 = 7*k13-13*k7
which implies a1+a2 = pi + 2*m*pi. .
In the case of different to different variant, one obtains (modula sign variations of a1, a2)
7*(a1+a2) = pi + 2*k7*pi and 13*(a1-a2) = pi + 2*k13*pi
and thus
a1 = (pi/7+pi/13)/2 + k7*pi/7 + k13*pi/13
a2 = (pi/7-pi/13)/2 + k7*pi/7 - k13*pi/13
and all possible sign variations, since cosine is an even function. However, simultaneous sign variations are also covered by variations of k7 and k13. Thus you obtain additionally (up to) 2*7*13=182 solutions to your problem.
Related
I have tried solving 4 equations of 4 unknowns in MATLAB and Mathematica.
I used vpasolve for finding the unknowns in MATLAB. Here is the MATLAB code.
Y1 = 0.02;
l1 = 0.0172;
syms Y2 Y3 l2 l3;
lambda = [0.0713 0.0688 0.0665];
b1 = 0.1170;
b3 = 0.1252;
t2_1 = (2*pi/lambda(1))*l2;
t3_1 = (2*pi/lambda(1))*l3;
t2_3 = (2*pi/lambda(3))*l2;
t3_3 = (2*pi/lambda(3))*l3;
t1_1 = (2*pi/lambda(1))*l1;
t1_3 = (2*pi/lambda(3))*l1;
eq1 = 2*Y1*tan(t1_1)+Y2*tan(t2_1)+Y3*tan(t3_1)==0;
eq2 = 2*Y1*tan(t1_3)+Y2*tan(t2_3)+Y3*tan(t3_3)==0;
eq3 = b1== (t1_1*Y1)+(t2_1*(Y2/2)*((sec(t2_1)^2)/(sec(t1_1)^2)))+(t3_1*(Y3/2)*((sec(t3_1)^2)/(sec(t1_1)^2)));
eq4 = b3== (t1_3*Y1)+(t2_3*(Y2/2)*((sec(t2_3)^2)/(sec(t1_3)^2)))+(t3_3*(Y3/2)*((sec(t3_3)^2)/(sec(t1_3)^2)));
E=[eq1 eq2 eq3 eq4];
S=vpasolve(E,[Y2,Y3,l2,l3]);
For the same equations, I wrote the below code in Mathematica.
eqns = {2*Y1*Tan[t11]+Y2*Tan[t21]+Y3*Tan[t31]==0,
2*Y1*Tan[t13]+Y2*Tan[t23]+Y3*Tan[t33]==0,
t11*Y1 + t21*(Y2/2)*(Sec[t21]^2/Sec[t11]^2) + t31*(Y3/2)*(Sec[t31]^2/Sec[t11]^2)-b1==0,
t13*Y1 + t23*(Y2/2)*(Sec[t23]^2/Sec[t13]^2) + t33*(Y3/2)*(Sec[t33]^2/Sec[t13]^2)-b3==0};
NMinimize[Norm[Map[First, eqns]], {Y2,Y3,l2,l3}]
But both are giving me different solutions and those are not the required solutions. I suppose I should use some other function for solving the equations. Can anyone help me finding out how to solve these equations? Your help is highly appreciated. Thank you.
EDIT
If I use the below code for finding the roots, I'm able to find the solutions but I just want positive roots. I tried the code which you have mentioned for getting positive roots but its not working for this and I don't know why. Can you please check this once?
Y1=0.0125;l1=0.010563;lambda={0.0426,0.0401,0.0403,0.0423,0.0413}; b1 = 0.0804;
b3 = 0.0258;
t2_1 = (2*Pi/lambda[[1]])*l2;
t3_1 = (2*Pi/lambda[[1]])*l3;
t2_3 = (2*Pi/lambda[[5]])*l2;
t3_3 = (2*Pi/lambda[[5]])*l3;
t1_1 = (2*Pi/lambda[[1]])*l1;
t1_3 = (2*Pi/lambda[[5]])*l1;
eqns = {2*Y1*Tan[t11]+Y2*Tan[t21]+Y3*Tan[t31]==0,
2*Y1*Tan[t13]+Y2*Tan[t23]+Y3*Tan[t33]==0,
t11*Y1 + t21*(Y2/2)*(Sec[t21]^2/Sec[t11]^2) + t31*(Y3/2)*(Sec[t31]^2/Sec[t11]^2)-b1==0,
t13*Y1 + t23*(Y2/2)*(Sec[t23]^2/Sec[t13]^2) + t33*(Y3/2)*(Sec[t33]^2/Sec[t13]^2)-b3==0};
Try
Y1=0.02;l1=0.0172;lambda={0.0713,0.0688,0.0665};b1=0.1170;b3=0.1252;
t21=(2*Pi/lambda[[1]])*l2;t31=(2*Pi/lambda[[1]])*l3;t23=(2*Pi/lambda[[3]])*l2;
t33=(2*Pi/lambda[[3]])*l3;t11=(2*Pi/lambda[[1]])*l1;t13=(2*Pi/lambda[[3]])*l1;
eqns={2*Y1*Tan[t11]+Y2*Tan[t21]+Y3*Tan[t31]==0,
2*Y1*Tan[t13]+Y2*Tan[t23]+Y3*Tan[t33]==0,
t11*Y1+t21*(Y2/2)*(Sec[t21]^2/Sec[t11]^2)+t31*(Y3/2)*(Sec[t31]^2/Sec[t11]^2)-b1==0,
t13*Y1+t23*(Y2/2)*(Sec[t23]^2/Sec[t13]^2)+t33*(Y3/2)*(Sec[t33]^2/Sec[t13]^2)-b3==0};
tbl=Table[
{y2i,y3i,l2i,l3i}=RandomReal[{0,.2},4];
Quiet[root=Check[FindRoot[eqns,{{Y2,y2i,0,.2},{Y3,y3i,0,.2},{l2,l2i,0,.2},{l3,l3i,0,.2}}],False]];
If[root===False,Nothing,root]
,{256}];
roots=Sort[Map[{Y2,Y3,l2,l3}/.#&,tbl],Norm[#1]<Norm[#2]&]
If[roots=={},"It found no roots in that range using those coefficients",
Map[First,eqns]/.{Y2->roots[[1,1]],Y3->roots[[1,2]],l2->roots[[1,3]],l3->roots[[1,4]]}]
That will look for roots starting at 256 different random locations greater than, but near 0. The way I have written FindRoot this time it will stop if the search is outside the range 0,.2 You can change that range if needed.
Next your functions have many local minima that trap FindRoot. So it should discard all local minima found inside that range.
And I have hidden the warning messages, usually not a good idea to do.
Then it will Sort the roots to show those nearest 0 first.
If you want it to sort to show the results nearest some given value then I can modify the Sort to show those first, I just need to know where you want the roots nearest to.
And finally it substitutes the Y2,Y3,l2,l3 of the smallest root it found back into the four equations to demonstrate that the results are very very near zero and this is a real root instead of a local minima.
After trying that a few times I found one root at {0.0203704,0.0225972,0.0163842,0.0181147} and that looks very close to your required values if I swap Y2 and Y3 and swap l2 and l3. Is that perhaps the root that you are looking for?
If you need more then please tell me exactly what is needed.
Please check ALL this VERY carefully to make certain I have made no mistakes.
Finding m and c for an equation y = mx + c, with the help of math and plots.
y is data_model_1, x is time.
Avoid other MATLAB functions like fitlm as it defeats the purpose.
I am having trouble finding the constants m and c. I am trying to find both m and c by limiting them to a range (based on smart guess) and I need to deduce the m and c values based on the mean error range. The point where mean error range is closest to 0 should be my m and c values.
load(file)
figure
plot(time,data_model_1,'bo')
hold on
for a = 0.11:0.01:0.13
c = -13:0.1:-10
data_a = a * time + c ;
plot(time,data_a,'r');
end
figure
hold on
for a = 0.11:0.01:0.13
c = -13:0.1:-10
data_a = a * time + c ;
mean_range = mean(abs(data_a - data_model_1));
plot(a,mean_range,'b.')
end
A quick & dirty approach
You can quickly get m and c using fminsearch(). In the first example below, the error function is the sum of squared error (SSE). The second example uses the sum of absolute error. The key here is ensuring the error function is convex.
Note that c = Beta(1) and m = Beta(2).
Reproducible example (MATLAB code[1]):
% Generate some example data
N = 50;
X = 2 + 13*random(makedist('Beta',.7,.8),N,1);
Y = 5 + 1.5.*X + randn(N,1);
% Example 1
SSEh =#(Beta) sum((Y - (Beta(1) + (Beta(2).*X))).^2);
Beta0 = [0.5 0.5]; % Initial Guess
[Beta SSE] = fminsearch(SSEh,Beta0)
% Example 2
SAEh =#(Beta) sum(abs(Y-(Beta(1) + Beta(2).*X)));
[Beta SumAbsErr] = fminsearch(SAEh,Beta0)
This is a quick & dirty approach that can work for many applications.
#Wolfie's comment directs you to the analytical approach to solve a system of linear equations with the \ operator or mldivide(). This is the more correct approach (though it will get a similar answer). One caveat is this approach gets the SSE answer.
[1] Tested with MATLAB R2018a
I'm given the a(k) matrix and e(n) and I need to compute y(n) from the following eq:
y(n) = sum(k = 1 to 10 )( a(k)*y(n-k) ) + e(n).
I have the a matrix(filter coefficients) and the e matrix (the residual), therefore there is only 1 unknown: y, which is built by the previous 10 samples of y.
An example to this equation:
say e(0) (my first residual sample) = 3
and y(-10) to y(-1) = 0
then y(0), my first sample in the signal y, would just be e(0) = 3:
y(0) = a(1)*y(-1) + a(2)*y(-2) + .... + a(10)*y(-10) + e(0) = e(0) = 3
and if e(1) = 4, and a(1) = 5, then
y(1) = a(1)*y(0) + a(2)*y(-1) + a(3)&y(-2) + ... + a(10)*y(-9) + e(1) = 19
The problem is
I don't know how to do this without loops because, say, y(n) needs y(n-1), so I need to immediately append y(n-1) into my matrix in order to get y(n).
If n (the number of samples) = say, 10,000,000, then using a loop is not ideal.
What I've done so far
I have not implemented anything. The only thing I've done for this particular problem is research on what kind of Matlab functions I could use.
What I need
A Matlab function that, given an equation and or input matrix, computes the next y(n) and automatically appends that to the input matrix, and then computes the next y(n+1), and automatically append that to the input matrix and so on.
If there is anything regarding my approach
That seems like it's on the wrong track, or my question isn't clear enough, of if there is no such matlab function that exists, then I apologize in advance. Thank you for your time.
I would like to solve the following equation: tan(x) = 1/x
What I did:
syms x
eq = tan(x) == 1/x;
sol = solve(eq,x)
But this gives me only one numerical approximation of the solution. After that I read about the following:
[sol, params, conds] = solve(eq, x, 'ReturnConditions', true)
But this tells me that it can't find an explicit solution.
How can I find numerical solutions to this equation within some given range?
I've never liked using solvers "blindly", that is, without some sort of decent initial value selection scheme. In my experience, the values you will find when doing things blindly, will be without context as well. Meaning, you'll often miss solutions, think something is a solution while in reality the solver exploded, etc.
For this particular case, it is important to realize that fzero uses numerical derivatives to find increasingly better approximations. But, derivatives for f(x) = x · tan(x) - 1 get increasingly difficult to accurately compute for increasing x:
As you can see, the larger x becomes, the better f(x) approximates a vertical line; fzero will simply explode! Therefore it is imperative to get an estimate as closely to the solution as possible before even entering fzero.
So, here's a way to get good initial values.
Consider the function
f(x) = x · tan(x) - 1
Knowing that tan(x) has Taylor expansion:
tan(x) ≈ x + (1/3)·x³ + (2/15)·x⁵ + (7/315)·x⁷ + ...
we can use that to approximate the function f(x). Truncating after the second term, we can write:
f(x) ≈ x · (x + (1/3)·x³) - 1
Now, key to realize is that tan(x) repeats with period π. Therefore, it is most useful to consider the family of functions:
fₙ(x) ≈ x · ( (x - n·π) + (1/3)·(x - n·π)³) - 1
Evaluating this for a couple of multiples and collecting terms gives the following generalization:
f₀(x) = x⁴/3 - 0π·x³ + ( 0π² + 1)x² - (0π + (0π³)/3)·x - 1
f₁(x) = x⁴/3 - 1π·x³ + ( 1π² + 1)x² - (1π + (1π³)/3)·x - 1
f₂(x) = x⁴/3 - 2π·x³ + ( 4π² + 1)x² - (2π + (8π³)/3)·x - 1
f₃(x) = x⁴/3 - 3π·x³ + ( 9π² + 1)x² - (3π + (27π³)/3)·x - 1
f₄(x) = x⁴/3 - 4π·x³ + (16π² + 1)x² - (4π + (64π³)/3)·x - 1
⋮
fₙ(x) = x⁴/3 - nπ·x³ + (n²π² + 1)x² - (nπ + (n³π³)/3)·x - 1
Implementing all this in a simple MATLAB test:
% Replace this with the whole number of pi's you want to
% use as offset
n = 5;
% The coefficients of the approximating polynomial for this offset
C = #(npi) [1/3
-npi
npi^2 + 1
-npi - npi^3/3
-1];
% Find the real, positive polynomial roots
R = roots(C(n*pi));
R = R(imag(R)==0);
R = R(R > 0);
% And use these as initial values for fzero()
x_npi = fzero(#(x) x.*tan(x) - 1, R)
In a loop, this can produce the following table:
% Estimate (polynomial) Solution (fzero)
0.889543617524132 0.860333589019380 0·π
3.425836967935954 3.425618459481728 1·π
6.437309348195653 6.437298179171947 2·π
9.529336042900365 9.529334405361963 3·π
12.645287627956868 12.645287223856643
15.771285009691695 15.771284874815882
18.902410011613000 18.902409956860023
22.036496753426441 22.036496727938566 ⋮
25.172446339768143 25.172446326646664
28.309642861751708 28.309642854452012
31.447714641852869 31.447714637546234
34.586424217960058 34.586424215288922 11·π
As you can see, the approximant is basically equal to the solution. Corresponding plot:
To find a numerical solution to a function within some range, you can use fzero like this:
fun = #(x)x*tan(x)-1; % Multiplied by x so fzero has no issue evaluating it at x=0.
range = [0 pi/2];
sol = fzero(fun,range);
The above would return just one solution (0.8603). If you want additional solutions, you will have to call fzero more times. This can be done, for example, in a loop:
fun = #(x)tan(x)-1/x;
RANGE_START = 0;
RANGE_END = 3*pi;
RANGE_STEP = pi/2;
intervals = repelem(RANGE_START:RANGE_STEP:RANGE_END,2);
intervals = reshape(intervals(2:end-1),2,[]).';
sol = NaN(size(intervals,1),1);
for ind1 = 1:numel(sol)
sol(ind1) = fzero(fun, mean(intervals(ind1,:)));
end
sol = sol(~isnan(sol)); % In case you specified more intervals than solutions.
Which gives:
[0.86033358901938;
1.57079632679490; % Wrong
3.42561845948173;
4.71238898038469; % Wrong
6.43729817917195;
7.85398163397449] % Wrong
Note that:
The function is symmetric, and so are its roots. This means you can solve on just the positive interval (for example) and get the negative roots "for free".
Every other entry in sol is wrong because this is where we have asymptotic discontinuities (tan transitions from +Inf to -Inf), which is mistakenly recognized by MATLAB as a solution. So you can just ignore them (i.e. sol = sol(1:2:end);.
Multiply the equation by x and cos(x) to avoid any denominators that can have the value 0,
f(x)=x*sin(x)-cos(x)==0
Consider the normalized function
h(x)=(x*sin(x)-cos(x)) / (abs(x)+1)
For large x this will be increasingly close to sin(x) (or -sin(x) for large negative x). Indeed, plotting this this is already visually true, up to an amplitude factor, for x>pi.
For the first root in [0,pi/2] use the Taylor approximation at x=0 of second degree x^2-(1-0.5x^2)==0 to get x[0]=sqrt(2.0/3) as root approximation, for the higher ones take the sine roots x[n]=n*pi, n=1,2,3,... as initial approximations in the Newton iteration xnext = x - f(x)/f'(x) to get
n initial 1. Newton limit of Newton
0 0.816496580927726 0.863034004302817 0.860333589019380
1 3.141592653589793 3.336084918413964 3.425618459480901
2 6.283185307179586 6.403911810682199 6.437298179171945
3 9.424777960769379 9.512307014150883 9.529334405361963
4 12.566370614359172 12.635021895208379 12.645287223856643
5 15.707963267948966 15.764435036320542 15.771284874815882
6 18.849555921538759 18.897518573777646 18.902409956860023
7 21.991148575128552 22.032830614521892 22.036496727938566
8 25.132741228718345 25.169597069842926 25.172446326646664
9 28.274333882308138 28.307365162331923 28.309642854452012
10 31.415926535897931 31.445852385744583 31.447714637546234
11 34.557519189487721 34.584873343220551 34.586424215288922
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.