I have this equation: p = (1 + (q-1)*B*T) ^ (-1/q-1)
The values p, T are known and the diagram p - T makes curbe. I want to calculate q and B so that curbe be as close to straight line is posible.
Some values are:
T p
1 0,999147061
2 0,997121331
3 0,994562513
Is there any way to make matlab (or sth else) to give me the values of B and q ?
Related
I am new with Octave and I have a problem.I thought the following codes were the same, but they produces different results. What is the difference? Thanks
Octave/Matlab: Difference between e^(-1*z) and exp(-1*z)
g = 1./(1 + e^(-1*z));
g = 1./(1 + exp(-1*z));
Where z is a vector, element or matrix
In Octave
exp(1) equals e where e is Euler's number.
There are 4 operations/functions that are to be noted here:
e^x is same as expm(x) and e.^(x) is same as exp(x).
e^x and expm(m) represent e raise to the matrix x.
e.^(x) and exp(x) represent exponential ex for each element in matrix x.
If x is a scalar then all (e^x, expm(x), e.^x and exp(x)) are mathematically equal.
For your case, z is a matrix and hence you get different results.
In MATLAB,
e is not defined in MATLAB. exp(x) and expm(x) have same definitions in MATLAB as those that are described for Octave above.
PS: e or E are also used for E-notation in both MATLAB and Octave but that's a different thing.
In Octave, it is important to note that e^x and exp(x), where x is a double precision scalar variable, are not necessarily the same.
For instance:
>> a = e ^ 2
a = 7.3891
>> b = exp (2)
b = 7.3891
>> b - a
ans = 8.8818e-16
The reason is that exp (2) uses a dedicated algorithm to compute the exponential function, while e ^ 2 actually calls the function e () to get the value of e, and then squares it:
>> c = realpow (e (), 2)
c = 7.3891
>> c - a
ans = 0
Another reason why e ^ x and exp (x) differ is that they compute completely different things when x is a square matrix, but this has already been discussed in Sardar's answer.
I am working on MATLAB on my own, and was doing problem 9 on Project Euler
It states
" A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,
a2 + b2 = c2
For example, 32 + 42 = 9 + 16 = 25 = 52.
There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc."
Below is the code I wrote; however, it compiles, but does not produce and output. I was hoping to get some feedback on what's wrong, so I can fix it.
Thanks,
syms a;
syms b;
syms c;
d= 1000;
d= a + b + c ;
ab= a.^2 + b.^2;
ab= c.^2;
c
I propose a vectorized way (that is, without using loops) to solve the problem. It may seem relatively complicated, especially if you come from other programming languages; but for Matlab you should get used to this way of approaching problems.
Ingredients:
Vectorization;
Indexing;
Transpose;
Implicit singleton expansion;
hypot;
find.
Read up on these concepts if you are not familiar with them, and then try to solve the problem yourself (which of course is the whole point of Project Euler). As a hint, the code below proceeds along these lines:
Generate a 1×1000 vector containing all possible values for a and b.
Compute a 1000×1000 matrix with the values of c corresponding to each pair a, b
From that compute a new matrix such that each entry contains a+b+c
Find the row and column indices where that matrix equals 1000. Those indices are the desired a and b (why?).
You'll get more than one solution (why?). Pick one.
Compute the product of the obtained a and b and the corresponding c.
Once you have tried yourself, you may want to check the code (move the mouse over it):
ab = 1:1000; % step 1
cc = hypot(ab,ab.'); % step 2
sum_abc = ab+ab.'+cc; % step 3
[a, b] = find(sum_abc==1000); % step 4
a = a(1); b = b(1); % step 5
prod_abc = a*b*cc(a,b); % step 6
This is my first stab at machine learning, and I can implement the code anyway that I want. I have Matlab access, which I think will be simpler than Python, and I have pseudo code for implementing a PLA. The last part of the code, however, absolutely baffles me, though it is simpler than the code I have seen on here thus far. It seems to be calling for the use of variables not declared. Here's what I have. I'll point out the number line at which I get stuck.
1) w <- (n + 1) X m (matrix of small random nums)
2) I <- I augmented with col. of 1s
3) for 1 = 1 to 1000
4) delta_W = (N + 1) X m (matrix of zeros) // weight changes
5) for each pattern 1 <= j <= p
6) Oj = (Ij * w) > 0 // j's are subscript/vector matrix product w/ threshold
7) Dj = = Tj - Oj // diff. between target and actual
8) w = w + Ij(transpose)*Dj // the learning rule
Lines 1 thru 4 are coded.
My questions are on line 5: What does "for each pattern mean" (i.e., how does one say it in code). Also, which j are they interested in, I have a j in the observation matrix and a j in the target matrix. Also, where did "p" come from (I have i's, j's, m's and n's but no p's)? Any thoughts would be appreciated.
"for each pattern" refers to the inputs. All they are saying is to run that loop where Ij is the input to the perceptron.
To write this in MATLAB, it really depends on how your data is oriented. I would store your inputs as a mXn matrix, where m is the number of inputs and n is the size of each input.
Say our inputs look like :
input = [1 5 -1;
2 3 2;
4 5 6;
... ]
First 'augment' this with a column of ones for the bias input:
[r c] = size(input);
input = [input ones(r,1)];
Then, your for loop will simply be:
for inputNumber = 1:r
pattern = input(inputNumber,:);
and you can continue from there.
So I'm trying to implement the Simpson method in Matlab, this is my code:
function q = simpson(x,f)
n = size(x);
%subtracting the last value of the x vector with the first one
ba = x(n) - x(1);
%adding all the values of the f vector which are in even places starting from f(2)
a = 2*f(2:2:end-1);
%adding all the values of the f vector which are in odd places starting from 1
b = 4*f(1:2:end-1);
%the result is the Simpson approximation of the values given
q = ((ba)/3*n)*(f(1) + f(n) + a + b);
This is the error I'm getting:
Error using ==> mtimes
Inner matrix dimensions must agree.
For some reason even if I set q to be
q = f(n)
As a result I get:
q =
0 1
Instead of
q =
0
When I set q to be
q = f(1)
I get:
q =
0
q =
0
I can't explain this behavior, that's probably why I get the error mentioned above. So why does q have two values instead of one?
edit: x = linspace(0,pi/2,12);
f = sin(x);
size(x) returns the size of the array. This will be a vector with all the dimensions of the matrix. There must be at least two dimensions.
In your case n=size(x) will give n=[N, 1], not just the length of the array as you desire. This will mean than ba will have 2 elements.
You can fix this be using length(x) which returns the longest dimension rather than size (or numel(x) or size(x, 1) or 2 depending on how x is defined which returns only the numbered dimension).
Also you want to sum over in a and b whereas now you just create an vector with these elements in. try changing it to a=2*sum(f(...)) and similar for b.
The error occurs because you are doing matrix multiplication of two vectors with different dimensions which isn't allowed. If you change the code all the values should be scalars so it should work.
To get the correct answer (3*n) should also be in brackets as matlab doesn't prefer between / and * (http://uk.mathworks.com/help/matlab/matlab_prog/operator-precedence.html). Your version does (ba/3)*n which is wrong.
I need to create a polynomial of the form:
P(x) = q(1,1) + q(2,2)(x-z(1)) + q(3,3)(x-z(1))(x-z(2)) + --- + q(2n, 2n)(x-z(1))(x-z(2))...(x-z(2n)) NOTE: The indices of the equation have been shifted to accomodate MATLAB.
in MATLAB. Consult this link here specifically slides 15 and 16.
I have the matrix Q filled, so I have the diagonal, and I also have z(1:2n) filled.
I'm having a hard time figuring out a way to create a polynomial that I can graph this polynomial. I've tried to use a for loop to append each term to P(x), but it doesn't operate the way I thought it would.
So far, my code will calculate the coefficients (presented as Q(0,0) -> Q(2n+1, 2n+1) in the problem above) without a problem.
I'm having an issue with the construction of a degree n polynomial of the form described above. Plotting makes more sense now, create a vector x with evaluative values, and then run them through the polynomial "function" and plot the x vector against the resulting vector.
So I just need to create this polynomial.
I would use diag and cumprod to help you accomplish this. First use diag to extract the diagonals of your matrix Q. After, use cumprod to generate a vector of cumulative products.
How cumprod works on a vector is that for each element in the vector, the i'th element collects products from 1 up to the i'th element. As an example, if we had a vector V = [1 2 3 4 5], cumprod(V) would produce [1 2 6 24 120]. The 4th element (as an example) would be 1*2*3*4, representing the products from the 1st to the 4th element.
As such, this is the code that I would do:
qdiag = diag(Q);
xMinusZ = x - z; % Takes z and does x - z for every element in z
cumProdRes = cumprod(xMinusZ);
P = sum(qdiag .* [1;cumProdRes(1:end-1)]);
P should give you P(x) that you desired. Make sure that z is a column vector to make it compatible with the diagonals extracted from Q.
NB: I believe there is a typo in your equation. The last term of your equation (going with your convention) should have (x-z(2n-1)) and not (x-z(2n)). This is because the first term in your equation does not have z.
Here's an example. Let's suppose Q is defined
Q = [1 2 3 4; 5 6 7 8; 9 10 11 12; 13 14 15 16];
The vector z is:
z = [4;3;2;1];
Let's evaluate the function at x = 2
Extracting the diagonals of Q should give us Q = [1;6;11;16]. Subtract x from every element of z should give us:
xMinusZ = [-2;-1;0;1];
Using the equation that you have above, we have:
P = 1 + 6*(-2) + 11*(-2)*(-1) + 16*(-2)*(-1)*(0) = 11
This is what the code should give.
What if we want to do this for more than one value of x?
As you have stated in your post, you want to evaluate this for a series of x values. As such, you need to modify the code so that it looks like this (make sure that x is a column vector):
qdiag = diag(Q);
xMinusZ = repmat(x,1,length(z)) - repmat(z',length(z),1);
cumProdRes = cumprod(xMinusZ,2);
P = sum(repmat(qdiag',length(z),1).*[ones(length(z),1) cumProdRes(:,1:end-1)],2);
P should now give you a vector of outputs, and so if you want to plot this, simply do plot(x,P);