I got this equation after I solved that problem here
I would like to solve it using Matlab.
The function roots returns polynomial roots. Simply pass the coefficients:
roots([1,-6,-36,216,-324])
Use solve from MATLAB's symbolic mathematics toolbox:
>> syms x;
>> y = solve('x^4-6*x^3-36*x^2+216*x-324==0')
y =
7.7446378738164683022795580182987
-6.3360292312480789716536487435108
2.2956956787158053346870453626061 - 1.1543655214730370697054534567177*i
2.2956956787158053346870453626061 + 1.1543655214730370697054534567177*i
The first line of code declares that x is a symbolic mathematics variable that we can use with the toolbox. Next, we use solve and we put in a string that describes the equation that is seen in your post. Bear in mind that when we multiply coefficients, you need to use the * operator, and for equality, we need to use double equals, or ==. The output should give you four roots, as dictated by the fundamental theorem of algebra. You'll see that you have two real roots, as well as two imaginary roots.
Alternatively, you can use roots in MATLAB by specifying a vector of coefficients starting from the highest order down to the lowest as per Daniel's answer.
Related
I'm using octave 3.8.1 which works like matlab.
I have an array of thousands of values I've only included three groupings as an example below:
(amp1=0.2; freq1=3; phase1=1; is an example of one grouping)
t=0;
amp1=0.2; freq1=3; phase1=1; %1st grouping
amp2=1.4; freq2=2; phase2=1.7; %2nd grouping
amp3=0.8; freq3=5; phase3=1.5; %3rd grouping
The Octave / Matlab code below solves for Y so I can plug it back into the equation to check values along with calculating values not located in the array.
clear all
t=0;
Y=0;
a1=[.2,3,1;1.4,2,1.7;.8,5,1.5]
for kk=1:1:length(a1)
Y=Y+a1(kk,1)*cos ((a1(kk,2))*t+a1(kk,3))
kk
end
Y
PS: I'm not trying to solve for Y since it's already solved for I'm trying to solve for Phase
The formulas located below are used to calculate Phase but I'm not sure how to put it into a for loop that will work in an array of n groupings:
How would I write the equation / for loop for finding the phase if I want to find freq=2.5 and amp=.23 and the phase is unknown I've looked online and it may require writing non linear equations which I'm not sure how to convert what I'm trying to do into such an equation.
phase1_test=acos(Y/amp1-amp3*cos(2*freq3*pi*t+phase3)/amp1-amp2*cos(2*freq2*pi*t+phase2)/amp1)-2*freq1*pi*t
phase2_test=acos(Y/amp2-amp3*cos(2*freq3*pi*t+phase3)/amp2-amp1*cos(2*freq1*pi*t+phase1)/amp2)-2*freq2*pi*t
phase3_test=acos(Y/amp3-amp2*cos(2*freq2*pi*t+phase2)/amp3-amp1*cos(2*freq1*pi*t+phase1)/amp3)-2*freq2*pi*t
Image of formula below:
I would like to do a check / calculate phases if given a freq and amp values.
I know I have to do a for loop but how do I convert the phase equation into a for loop so it will work on n groupings in an array and calculate different values not found in the array?
Basically I would be given an array of n groupings and freq=2.5 and amp=.23 and use the formula to calculate phase. Note: freq will not always be in the array hence why I'm trying to calculate the phase using a formula.
Ok, I think I finally understand your question:
you are trying to find a set of phase1, phase2,..., phaseN, such that equations like the ones you describe are satisfied
You know how to find y, and you supply values for freq and amp.
In Matlab, such a problem would be solved using, for example fsolve, but let's look at your problem step by step.
For simplicity, let me re-write your equations for phase1, phase2, and phase3. For example, your first equation, the one for phase1, would read
amp1*cos(phase1 + 2 freq1 pi t) + amp2*cos(2 freq2 pi t + phase2) + amp3*cos(2 freq3 pi t + phase3) - y = 0
Note that ampX (X is a placeholder for 1, 2, 3) are given, pi is a constant, t is given via Y (I think), freqX are given.
Hence, you are, in fact, dealing with a non-linear vector equation of the form
F(phase) = 0
where F is a multi-dimensional (vector) function taking a multi-dimensional (vector) input variable phase (comprised of phase1, phase2,..., phaseN). And you are looking for the set of phaseX, where all of the components of your vector function F are zero. N.B. F is a shorthand for your functions. Therefore, the first component of F, called f1, for example, is
f1 = amp1*cos(phase1+...) + amp2*cos(phase2+...) + amp3*cos(phase3+...) - y = 0.
Hence, f1 is a one-dimensional function of phase1, phase2, and phase3.
The technical term for what you are trying to do is find a zero of a non-linear vector function, or find a solution of a non-linear vector function. In Matlab, there are different approaches.
For a one-dimensional function, you can use fzero, which is explained at http://www.mathworks.com/help/matlab/ref/fzero.html?refresh=true
For a multi-dimensional (vector) function as yours, I would look into using fsolve, which is part of Matlab's optimization toolbox (which means I don't know how to do this in Octave). The function fsolve is explained at http://www.mathworks.com/help/optim/ug/fsolve.html
If you know an approximate solution for your phases, you may also look into iterative, local methods.
In particular, I would recommend you look into the Newton's Method, which allows you to find a solution to your system of equations F. Wikipedia has a good explanation of Newton's Method at https://en.wikipedia.org/wiki/Newton%27s_method . Newton iterations are very simple to implement and you should find a lot of resources online. You will have to compute the derivative of your function F with respect to each of your variables phaseX, which is very simple to compute since you're only dealing with cos() functions. For starters, have a look at the one-dimensional Newton iteration method in Matlab at http://www.math.colostate.edu/~gerhard/classes/331/lab/newton.html .
Finally, if you want to dig deeper, I found a textbook on this topic from the society for industrial and applied math: https://www.siam.org/books/textbooks/fr16_book.pdf .
As you can see, this is a very large field; Newton's method should be able to help you out, though.
Good luck!
I am doing a project where i find an approximation of the Sine function, using the Least Squares method. Also i can use 12 values of my own choice.Since i couldn't figure out how to solve it i thought of using Taylor's series for Sine and then solving it as a polynomial of order 5. Here is my code :
%% Find the sine of the 12 known values
x=[0,pi/8,pi/4,7*pi/2,3*pi/4,pi,4*pi/11,3*pi/2,2*pi,5*pi/4,3*pi/8,12*pi/20];
y=zeros(12,1);
for i=1:12
y=sin(x);
end
n=12;
j=5;
%% Find the sums to populate the matrix A and matrix B
s1=sum(x);s2=sum(x.^2);
s3=sum(x.^3);s4=sum(x.^4);
s5=sum(x.^5);s6=sum(x.^6);
s7=sum(x.^7);s8=sum(x.^8);
s9=sum(x.^9);s10=sum(x.^10);
sy=sum(y);
sxy=sum(x.*y);
sxy2=sum( (x.^2).*y);
sxy3=sum( (x.^3).*y);
sxy4=sum( (x.^4).*y);
sxy5=sum( (x.^5).*y);
A=[n,s1,s2,s3,s4,s5;s1,s2,s3,s4,s5,s6;s2,s3,s4,s5,s6,s7;
s3,s4,s5,s6,s7,s8;s4,s5,s6,s7,s8,s9;s5,s6,s7,s8,s9,s10];
B=[sy;sxy;sxy2;sxy3;sxy4;sxy5];
Then at matlab i get this result
>> a=A^-1*B
a =
-0.0248
1.2203
-0.2351
-0.1408
0.0364
-0.0021
However when i try to replace the values of a in the taylor series and solve f.e t=pi/2 i get wrong results
>> t=pi/2;
fun=t-t^3*a(4)+a(6)*t^5
fun =
2.0967
I am doing something wrong when i replace the values of a matrix in the Taylor series or is my initial thought flawed ?
Note: i can't use any built-in function
If you need a least-squares approximation, simply decide on a fixed interval that you want to approximate on and generate some x abscissae on that interval (possibly equally spaced abscissae using linspace - or non-uniformly spaced as you have in your example). Then evaluate your sine function at each point such that you have
y = sin(x)
Then simply use the polyfit function (documented here) to obtain least squares parameters
b = polyfit(x,y,n)
where n is the degree of the polynomial you want to approximate. You can then use polyval (documented here) to obtain the values of your approximation at other values of x.
EDIT: As you can't use polyfit you can generate the Vandermonde matrix for the least-squares approximation directly (the below assumes x is a row vector).
A = ones(length(x),1);
x = x';
for i=1:n
A = [A x.^i];
end
then simply obtain the least squares parameters using
b = A\y;
You can clearly optimise the clumsy Vandermonde generation loop above I have just written to illustrate the concept. For better numerical stability you would also be better to use a nice orthogonal polynomial system like Chebyshev polynomials of the first kind. If you are not even allowed to use the matrix divide \ function then you will need to code up your own implementation of a QR factorisation and solve the system that way (or some other numerically stable method).
I have to analyze 802.11 saturation throughput using matlab, and here is my problem. I'm trying to solve parametric equations below (parameters are m,W,a) using solve function and i get
Warning: Explicit solution could not be found
How could I solve above equations using matlab?
I guess you were trying to find an analytical solution for tau and p using symbolic math. Unless you're really lucky with your parameters (e.g. m=1), there won't be an analytical solution.
If you're interested in numerical values for tau and p, I suggest you manually substitue p in the first equation, and then solve an equation of the form tau-bigFraction=0 using, e.g. fzero.
Here's how you'd use fzero to solve a simple equation kx=exp(-x), with k being a parameter.
k = 5; %# set a value for k
x = fzero(#(x)k*x-exp(-x),0); %# initial guess: x=0
Hello i have one question. When I calculate a division in matlab: x/(pi.^2)
syms x
x/(pi.^2)
ans =
(281474976710656*v)/2778046668940015
the correct answer is x/9.8696, so why is matlab giving me this result?
Is it a bug?
You have to use the vpa() command "Variable-precision arithmetic". Check this code:
syms x real; % define x as a real symbolic variable (not a complex variable)
vpa( x/(pi.^2), 5) % second argument define number of significant digits
For trigonometric expressions involving pi, it is sometimes good to define sym('pi'):
syms x real;
pi_s = sym('pi');
expr = x/pi_s^2
I try to always use the 'real' tag when using the symbolic toolbox. If you do not use it you are going to see a lot of complex conjugates and other things that are not important for your problem, because x is probably real variable.
Hope this helps,
No it is not a bug:
2778046668940015/281474976710656 = 9.8696
I'm trying to solve a system of equations in the s-domain. So set up this system of equations in matrix form:
a=[.4*s+s+5 -5; -5 .5*s+5]
c=[3/s; 3/(2*s)]
(1/s)*a*b=c
I just get the error that s is undefined.
How can I solve for b in terms of s?
Matlab does not (naturally) do symbolic calculations --- which is what your code is trying to do. Matlab's variables need to be concrete numbers, or arrays, or structures, etc. They cannot just be placeholders for arbitrary numbers.
(UNLESS: You use the symbolic computing toolbox for Matlab. I haven't really used this because I prefer to do symbolic computing in environments such as Maple or Mathematica. You could even solve your problem on the Wolfram Alpha website)
But if you pick a specific value of s, computing what you want is easy:
s = 5;
a=[.4*s+s+5 -5; -5 .5*s+5];
c=[3/s; 3/(2*s)];
b = s*(a\c);
Where I have used the backslash operator for doing linear inversion.
You should now have that
(1/s)*a*b-c
is the zero vector.
EDIT: I looked into the symbolic toolbox. It looks like this is what you want (but you need to have the symbolic toolbox licensed and installed for it to work):
syms s;
a=[.4*s+s+5 -5; -5 .5*s+5];
c=[3/s; 3/(2*s)];
b = simple(s*(a\c))
The code to perform your calculation using symbolic operators is:
syms s; %This defines 's' as a symbolic token
a=[.4*s+s+5 -5; -5 .5*s+5]; %a and c inherit the symbolic properties from s
c=[3/s; 3/(2*s)];
result = solve('(1/s)*a*b=c','b') %Solve is the general symbolic toolbox algebraic solver.
This produces
result =
(c*s)/a
Generally speaking, Matlab performs best as a numerical toolbox. So depending on your application I would go with another approach, such as that demonstrated by Ian Hincks in another answer. But sometimes the situation demands a symbolic solution.