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A = [A-x(1) B-x(2) C-x(3);D-x(4) E-x(5) F-x(6); G-x(7) H-x(8) I-x(9)]
I have to obtain x(1)...x(9) for det(A) = 0.
Given a 3x3 matrix A
its determinant is
therefore you need to solve |A| = 0. For your case we are given
The easiest solution for x so that |A| = 0 is when
a - x(1) = 0
b - x(2) = 0
c - x(3) = 0
which leads to
x(1) = a
x(2) = b
x(3) = c
so
x = A
is the most trivial solution. There exists an infinite number of solutions to this problem, this is just one. You could choose another solution where
a - x(1) != 0
b - x(2) != 0
c - x(3) != 0
and then you would have to set
ei - fh = 0
di - fg = 0
dh - eg = 0
which would involve simultaneous equations.
I suggest before trying to code up a solution you work through one by hand like I've done here.
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Suppose H(xy) = H(x) * H(y) . Clearly preimage properties are violated. How can we find x, y such that H(x) = H(y) mod (2^k)
2^k is a very special modulus. You can prove the following strengthened version of Euler's theorem:
Suppose that x is any postive integer. Then x^phi(2^k) (mod 2^k) is equal to either 0 or 1, with 1 if and only if x is odd.
Proof:
If x is odd, then gcd(x,2^k) = 1, hence x^phi(2^k) (mod 2^k) = 1 by Euler's theorem.
Suppose that x is even. The result is trivially true if x = 0, so suppose that x > 0. Write x = (2^s)*y where y is odd and s > 0. Note that
phi(2^k)` = 2^(k-1)
But then,
x^phi(2^k) = (2^s*y)^phi(2^k)
= (2^s)^phi(2^k) * y^phi(2^k)
= 2^(s*phi(2^k)) * y^phi(2^k)
= 2^(s*2^(k-1)) * y^phi(2^k)
= 0 * y^phi(2^k) = 0 (mod 2^k)
the last line follows from the fact that s*2^(k-1) >= k hence 2^(s*2^(k-1)) is a multiple of 2^k.
Note that if x is even then you actually have x^k = 0 (mod 2^k), so raising x to the power phi(2^k) is overkill for any but the smallest k.
Given this lemma, it is now trivial to see that there exists distinct x,y with either H(x) = H(y) = 0 or H(x) = H(y) = 1. Since it seems to be homework, I'll leave the details to you.
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I'm learning about signal processing and currently I have to do an speech synthesizer in Matlab. For emulate the resonator system of the mouth I've this transfer function:
R(z) = 1 - z ^(-1)
Can I implement this system with filter function in Matlab? I don't know how to extract the coeficients "a" and "b"...
Note: y = filter(b, a, x), where x is the input signal that we have to filter.
Thank you all!
Consulting the documentation for filter, you represent a transfer function as a rational function of coefficients such that:
The desired transfer function you want, Y(z) / X(z) = R(z) is equal to:
R(z) = 1 - z^{-1}
Here a(1) is implicitly equal to 1. Therefore, b(1) = 1 and b(2) = -1 referring to the above equation. All of the coefficients in the denominator are 0 except for a(1) which is equal to 1.
As such, a = 1; b = [1 -1]; and so filtering your signal is simply:
a = 1; b = [1 -1];
y = filter(b, a, x);
x is the signal of interest you want to filter.
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I am new to Matlab and for the values of a, l and w i need to find all the values for l in the data set and the corresponding w values.
a=10;
l=(0:10)
w=(0:10)
for l,d
if a == l.*w
disp(l)
disp(w)
end
end
Not sure what you want to do, but I think your code could be put as follows:
a = 10;
l = 0:a; %// actually, it would suffice to check numbers up to floor(a/2)
ind = rem(a,l)==0; %// logical index that tells if remainder is zero or not
disp([l(ind); a./l(ind)])
Result:
1 2 5 10
10 5 2 1
You could do it more directly with Matlab's factor function:
f = factor(a);
disp([f; a./f])
Result:
2 5
5 2
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I have multiple arrays that I need to identify and interpolate to a set number. the set number will be the 'length?' of the biggest array. I need to How could I identify each array length and create a loop to interpolate each array to that specific length? Sorry if I am not providing enough detail.
A = rand(10,2);
B = rand(20,2);
C = rand(5,2);
%find max length, for you cell array you want: max(cellfun(#(x) length(x), MyCellArray))
n = max([length(A), length(B), length(C));
%repeat for each, i.e. loop through the cell array
x = A(:,1);
y = A(:,2);
m = min(x);
M = max(x);
d = (M - m) / n;
xi = m:d:M;
Ai = interp1(x, y, xi);
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I'm trying to understand why we're iterating through Mandelbrot points until |z| < 4.
why 4? is there somekind of a law? or is it based on statistical measurements?
thanks, igal
Consider the Mandelbrot set with along y=0, which would would be z(i) = z(i-1)^2 + c.
Consider when c = (x=-2, y=0)
z(0) = 0
z(1) = 0^2 + -2 = -2
z(2) = (-2)^2 + -2 = 4 - 2 = 2
z(3) = 2^2 + -2 = 4 - 2 = 2
z(...) = 2^2 + -2 = 4 - 2 = 2
This example (x=-2,y=0) is the point with the greatest magnitude that will never blow up. Thus when z^2 > 4, there is no point in further iteration since you already know it will blow up.
All other points where the magnitude of the point >= 2 will blow up.