I want to use this rational number in computations without losing the accuracy of the picture in Matlab:
f = 359.0 + 16241/16250.0
I think storing, for instance by f = uint64(359.0 + 16241/16250.0) loses accuracy, seen as 360 in Matlab.
I think the best way to handle the thing is never to store the value but to store its factors like
% f = a + b/c
a = 359
b = 16241
c = 16250
and then doing computation by the variables a, b and c, and giving the result as a picture.
Is this a good way to maintain the accuracy?
As you suggest, if you absolutely don't want to lose accuracy when storing a rational number, the best solution probably is to store the number in terms of its integer components.
Instead of your three components (f = a + b/c) you can reduce the reprentation to two components: f = n/d. Thus each rational number would be defined (and stored) as the two-component integer vector [n d]. For example, the number f in your example corresponds to n=5849991 and d=16250.
To simplify handling rational numbers stored this way, you could define a helper function which converts from the [n d] representation to n/d before applyling the desired operation:
useInteger = #(x, nd, fun) fun(x,double(nd(1))/double(nd(2)));
Then
>> x = sqrt(pi);
>> nd = int64([5849991 16250]);
>> useInteger(x, nd, #plus)
ans =
361.7719
>> useInteger(x, nd, #times)
ans =
638.0824
If you want to achieve arbitrarily high precision in computations, you should consider using variable-precision arithmetic (vpa) with string arguments. With that approach you get to specify how many digits you want:
>> vpa('sqrt(pi)*5849991/16250', 50)
ans =
638.08240465923757600307902117159072301901656248436
Perhaps create a Rational class and define the needed operations (plus,minus,times,etc.). Start with something like this:
Rational.m
classdef Rational
properties
n;
d;
end
methods
function obj = Rational(n,d)
GCD = gcd(n,d);
obj.n = n./GCD;
obj.d = d./GCD;
end
function d = dec(obj)
d = double(obj.n)/double(obj.d);
end
% X .* Y
function R = times(X,Y)
chkxy(X,Y);
if isnumeric(X),
N = X .* Y.n; D = Y.d;
elseif isnumeric(Y),
N = X.n .* Y; D = X.d;
else
N = X.n .* Y.n; D = X.d .* Y.d;
end
R = Rational(N,D);
end
% X * Y
function R = mtimes(X,Y)
R = times(X,Y);
end
% X ./ Y
function R = rdivide(X,Y)
if isnumeric(Y),
y = Rational(1,Y);
else
y = Rational(Y.d,Y.n);
end
R = times(X,y);
end
% X / Y
function R = mrdivide(X,Y)
R = rdivide(X,Y);
end
% X + Y
function R = plus(X,Y)
chkxy(X,Y);
if isnumeric(X),
N = X.*Y.d + Y.n; D = Y.d;
elseif isnumeric(Y),
N = Y.*X.d + X.n; D = X.d;
else
D = lcm(X.d,Y.d);
N = sum([X.n Y.n].*(D./[X.d Y.d]));
end
R = Rational(N,D);
end
% X - Y
function R = minus(X,Y)
R = plus(X,-Y);
end
% -X
function R = uminus(X)
R = Rational(-X.n,X.d);
end
function chkxy(X,Y)
if (~isa(X, 'Rational') && ~isnumeric(X)) || ...
(~isa(Y, 'Rational') && ~isnumeric(Y)),
error('X and Y must be Rational or numeric.');
end
end
end
end
Examples
Construct objects:
>> clear all % reset class definition
>> r1 = Rational(int64(1),int64(2))
r1 =
Rational with properties:
n: 1
d: 2
>> r2 = Rational(int64(3),int64(4))
r2 =
Rational with properties:
n: 3
d: 4
Add and subtract:
>> r1+r2
ans =
Rational with properties:
n: 5
d: 4
>> r1-r2
ans =
Rational with properties:
n: -1
d: 4
Multiply and divide:
>> r1*r2
ans =
Rational with properties:
n: 3
d: 8
>> r1/r2
ans =
Rational with properties:
n: 2
d: 3
Get decimal value:
>> r12 = r1/r2; % 2/3 ((1/2)/(3/4))
>> f = r12.dec
f =
0.6667
Extension to LuisMendo's answer
I got this as the error for your suggestion by py
>>> a = 638.08240465923757600307902117159072301901656248436059
>>> a
638.0824046592376 % do not know if Python is computing here with exact number
>>> b = 638.0824
>>> ave = abs(b+a)/2
>>> diff = abs(b-a)
>>> ave = abs(b+a)/2
>>> diff/ave
7.30193709165014e-09
which is more than the proposed error storing error above.
I run in WolframAlpha
x = sqrt(pi)
x*5849991/16250
and get
509.11609919757198016211937362635174599076143654820109
I am not sure if this is what you meant in your comment of your answer.
Extension to chappjc's answer.
I have now
[B,T,F] = tfrwv(data1, 1:length(data1), length(data1)); % here F double
fs = Rational(uint64(5849991), uint64(16250));
t = 1/fs;
imagesc(T*t, F*fs, B);
I run it
Error using .*
Integers can only be combined with integers of
the same class, or scalar doubles.
Error in .* (line 23)
N = X .* Y.n; D = Y.d;
Error in * (line 34)
R = times(X,Y);
How can you multiply in this class the double with Rational?
Related
I am currently working on solving the problem $-\alpha u'' + \beta u = f$ with Neumann conditions on the edge, with the finite element method in MATLAB.
I managed to set up a code that works for P1 and P2 Lagragne finite elements (i.e: linear and quadratic) and the results are good!
I am trying to implement the finite element method using the Hermite basis. This basis is defined by the following basis functions and derivatives:
syms x
phi(x) = [2*x^3-3*x^2+1,-2*x^3+3*x^2,x^3-2*x^2+x,x^3-x^2]
% Derivative
dphi = [6*x.^2-6*x,-6*x.^2+6*x,3*x^2-4*x+1,3*x^2-2*x]
The problem with the following code is that the solution vector u is not good. I know that there must be a problem in the S and F element matrix calculation loop, but I can't see where even though I've been trying to make changes for a week.
Can you give me your opinion? Hopefully someone can see my error.
Thanks a lot,
% -alpha*u'' + beta*u = f
% u'(a) = bd1, u'(b) = bd2;
a = 0;
b = 1;
f = #(x) (1);
alpha = 1;
beta = 1;
% Neuamnn boundary conditions
bn1 = 1;
bn2 = 0;
syms ue(x)
DE = -alpha*diff(ue,x,2) + beta*ue == f;
du = diff(ue,x);
BC = [du(a)==bn1, du(b)==bn2];
ue = dsolve(DE, BC);
figure
fplot(ue,[a,b], 'r', 'LineWidth',2)
N = 2;
nnod = N*(2+2); % Number of nodes
neq = nnod*1; % Number of equations, one degree of freedom per node
xnod = linspace(a,b,nnod);
nodes = [(1:3:nnod-3)', (2:3:nnod-2)', (3:3:nnod-1)', (4:3:nnod)'];
phi = #(xi)[2*xi.^3-3*xi.^2+1,2*xi.^3+3*xi.^2,xi.^3-2*xi.^2+xi,xi.^3-xi.^2];
dphi = #(xi)[6*xi.^2-6*xi,-6*xi.^2+6*xi,3*xi^2-4*xi+1,3*xi^2-2*xi];
% Here, just calculate the integral using gauss quadrature..
order = 5;
[gp, gw] = gauss(order, 0, 1);
S = zeros(neq,neq);
M = S;
F = zeros(neq,1);
for iel = 1:N
%disp(iel)
inod = nodes(iel,:);
xc = xnod(inod);
h = xc(end)-xc(1);
Se = zeros(4,4);
Me = Se;
fe = zeros(4,1);
for ig = 1:length(gp)
xi = gp(ig);
iw = gw(ig);
Se = Se + dphi(xi)'*dphi(xi)*1/h*1*iw;
Me = Me + phi(xi)'*phi(xi)*h*1*iw;
x = phi(xi)*xc';
fe = fe + phi(xi)' * f(x) * h * 1 * iw;
end
% Assembly
S(inod,inod) = S(inod, inod) + Se;
M(inod,inod) = M(inod, inod) + Me;
F(inod) = F(inod) + fe;
end
S = alpha*S + beta*M;
g = zeros(neq,1);
g(1) = -alpha*bn1;
g(end) = alpha*bn2;
alldofs = 1:neq;
u = zeros(neq,1); %Pre-allocate
F = F + g;
u(alldofs) = S(alldofs,alldofs)\F(alldofs)
Warning: Matrix is singular to working precision.
u = 8×1
NaN
NaN
NaN
NaN
NaN
NaN
NaN
NaN
figure
fplot(ue,[a,b], 'r', 'LineWidth',2)
hold on
plot(xnod, u, 'bo')
for iel = 1:N
inod = nodes(iel,:);
xc = xnod(inod);
U = u(inod);
xi = linspace(0,1,100)';
Ue = phi(xi)*U;
Xe = phi(xi)*xc';
plot(Xe,Ue,'b -')
end
% Gauss function for calculate the integral
function [x, w, A] = gauss(n, a, b)
n = 1:(n - 1);
beta = 1 ./ sqrt(4 - 1 ./ (n .* n));
J = diag(beta, 1) + diag(beta, -1);
[V, D] = eig(J);
x = diag(D);
A = b - a;
w = V(1, :) .* V(1, :);
w = w';
x=x';
end
You can find the same post under MATLAB site for syntax highlighting.
Thanks
I tried to read courses, search in different documentation and modify my code without success.
I have a cell array with size (n,1) that includes a function handle. Every cell has to include specific function handle and the summation of function handles in the previous cells. How can I perform this operation? To clarify this is an illustration.
A = cell(size(ones(n,1)));
for i = 1 : n
A{i,1} = #(x) A{i-1,1} + i .* x;
end
How to get A{n,1} at x = 2 (for example)
You are actually pretty close, but you need to add a special case for i = 1 and you need to call the function:
n = 10;
A = cell(size(ones(n,1)));
A{1,1} = #(x) 1 .* x;
for ii = 2 : n
A{ii,1} = #(x) A{ii-1,1}(x) + ii .* x;
end
I replaced i with ii, to avoid confusion with complex numbers. For the case n = 10:
>> A{n}(2)
ans =
110
Recalling #gnovice comment, you can also just make a cell array of the handles, and then call a function that sums them up to n:
N = 10;
A = cell(N,1);
A{1} = #(x) 1.*x;
for k = 2:N
A{k} = #(x) k.*x;
end
% the following function sums the output of A{1}(x) to A{n}(x):
f = #(n,x) sum(cellfun(#(c) c(x),A(1:n)));
The result:
>> f(5,2)
ans =
30
>> f(N,2)
ans =
110
This way, every change of the functions in A will have an immediate effect upon redefining f:
>> A{3} = #(x) -x;
>> f = #(n,x) sum(cellfun(#(c) c(x),A(1:n)));
>> f(N,2)
ans =
102
>> f(5,2)
ans =
22
What would be the best way to simplify a function by getting rid of a loop?
function Q = gs(f, a, b)
X(4) = sqrt((3+2*sqrt(6/5))/7);
X(3) = sqrt((3-2*sqrt(6/5))/7);
X(2) = -sqrt((3-2*sqrt(6/5))/7);
X(1) = -sqrt((3+2*sqrt(6/5))/7);
W(4) = (18-sqrt(30))/36;
W(3) = (18+sqrt(30))/36;
W(2) = (18+sqrt(30))/36;
W(1) = (18-sqrt(30))/36;
Q = 0;
for i = 1:4
W(i) = (W(i)*(b-a))/2;
X(i) = ((b-a)*X(i)+(b+a))/2;
Q = Q + W(i) * f(X(i));
end
end
Is there any way to use any vector-like solution instead of a for loop?
sum is your best friend here. Also, declaring some constants and creating vectors is useful:
function Q = gs(f, a, b)
c = sqrt((3+2*sqrt(6/5))/7);
d = sqrt((3-2*sqrt(6/5))/7);
e = (18-sqrt(30))/36;
g = (18+sqrt(30))/36;
X = [-c -d d c];
W = [e g g e];
W = ((b - a) / 2) * W;
X = ((b - a)*X + (b + a)) / 2;
Q = sum(W .* f(X));
end
Note that MATLAB loves to handle element-wise operations, so the key is to replace the for loop at the end with scaling all of the elements in W and X with those scaling factors seen in your loop. In addition, using the element-wise multiplication (.*) is key. This of course assumes that f can handle things in an element-wise fashion. If it doesn't, then there's no way to avoid the for loop.
I would highly recommend you consult the MATLAB tutorial on element-wise operations before you venture onwards on your MATLAB journey: https://www.mathworks.com/help/matlab/matlab_prog/array-vs-matrix-operations.html
I have a code in MATLAB which works with very small numbers, for example, I have values that are on the order of 10^{-25}, however when MATLAB does the calculations, the values themselves are rounded to 0. Note, I am not referring to format to display these extra decimals, but rather the number itself is changed to 0. I think the reason is because MATLAB, by default, uses up to 15 digits after the decimal point for its calculations. How can I change this so that numbers that are very very small are retained as they are in the calculations?
EDIT:
My code is the following:
clc;
clear;
format long;
% Import data
P = xlsread('Data.xlsx', 'P');
d = xlsread('Data.xlsx', 'd');
CM = xlsread('Data.xlsx', 'Cov');
Original_PD = P; %Store original PD
LM_rows = size(P,1)+1; %Expected LM rows
LM_columns = size(P,2); %Expected LM columns
LM_FINAL = zeros(LM_rows,LM_columns); %Dimensions of LM_FINAL
for ii = 1:size(P,2)
P = Original_PD(:,ii);
% c1, c2, ..., cn, c0, f
interval = cell(size(P,1)+2,1);
for i = 1:size(P,1)
interval{i,1} = NaN(size(P,1),2);
interval{i,1}(:,1) = -Inf;
interval{i,1}(:,2) = d;
interval{i,1}(i,1) = d(i,1);
interval{i,1}(i,2) = Inf;
end
interval{i+1,1} = [-Inf*ones(size(P,1),1) d];
interval{i+2,1} = [d Inf*ones(size(P,1),1)];
c = NaN(size(interval,1),1);
for i = 1:size(c,1)
c(i,1) = mvncdf(interval{i,1}(:,1),interval{i,1}(:,2),0,CM);
end
c0 = c(size(P,1)+1,1);
f = c(size(P,1)+2,1);
c = c(1:size(P,1),:);
b0 = exp(1);
b = exp(1)*P;
syms x;
eqn = f*x;
for i = 1:size(P,1)
eqn = eqn*(c0/c(i,1)*x + (b(i,1)-b0)/c(i,1));
end
eqn = c0*x^(size(P,1)+1) + eqn - b0*x^size(P,1);
x0 = solve(eqn);
x0 = double(x0);
for i = 1:size(x0)
id(i,1) = isreal(x0(i,1));
end
x0 = x0(id,:);
x0 = x0(x0 > 0,:);
clear x;
for i = 1:size(P,1)
x(i,:) = (b(i,1) - b0)./(c(i,1)*x0) + c0/c(i,1);
end
% x = [x0 x1 ... xn]
x = [x0'; x];
x = x(:,sum(x <= 0,1) == 0);
% lamda
lamda = -log(x);
LM_FINAL(:,ii) = lamda;
end
The problem is in this step:
for i = 1:size(P,1)
x(i,:) = (b(i,1) - b0)./(c(i,1)*x0) + c0/c(i,1);
end
where the "difference" gets very close to 0. How can I stop this rounding from occurring at this step?
For example, when i = 10, I have the following values:
b_10 = 0.006639735483297
b_0 = 2.71828182845904
c_10 = 0.000190641848119641
c_0 = 0.356210110252579
x_0 = 7.61247930625269
After doing the calculations we get: -1868.47805854794 + 1868.47805854794 which yields a difference of -2.27373675443232E-12, that gets rounded to 0 by MATLAB.
EDIT 2:
Here is my data file which is used for the code. After you run the code (should take about a minute and half to finish running), row 11 in the variable x shows 0 (even after double clicking to check it's real value), when it shouldn't.
The problem you're having is because the IEEE standard for floating points can't distinguish your numbers from zero because they don't utilize sufficient bits.
Have a look at John D'Errico's Big Decimal Class and Variable Precision Integer Arithmetic. Another option would be to use the Big Integer Class from Java but that might be more challenging if you are unfamiliar with using Java and othe rexternal libraries in MATLAB.
Can you give an example of the calculations in which you are using 1e-25 and getting zero? Here's what I get for a floating point called small_num and one of John's high-precision-floats called small_hpf when assigning them and multiplying by pi.
>> small_num = 1e-25
small_num =
1.0000e-25
>> small_hpf = hpf(1e-25)
small_hpf =
1.000000000000000038494869749191839081371989361591338301396127644e-25
>> small_num * pi
ans =
3.1416e-25
>> small_hpf * pi
ans =
3.141592653589793236933163473501228686498684350685747717239459106e-25
While answering another, I stumbled over the question how I actually could find all factors of an integer number without the Symbolic Math Toolbox.
For example:
factor(60)
returns:
2 2 3 5
unique(factor(60))
would therefore return all prime-factors, "1" missing.
2 3 5
And I'm looking for a function which would return all factors (1 and the number itself are not important, but they would be nice)
Intended output for x = 60:
1 2 3 4 5 6 10 12 15 20 30 60
I came up with that rather bulky solution, apart from that it probably could be vectorized, isn't there any elegant solution?
x = 60;
P = perms(factor(x));
[n,m] = size(P);
Q = zeros(n,m);
for ii = 1:n
for jj = 1:m
Q(ii,jj) = prod(P(ii,1:jj));
end
end
factors = unique(Q(:))'
Also I think, this solution will fail for certain big numbers, because perms requires a vector length < 11.
You can find all factors of a number n by dividing it by a vector containing the integers 1 through n, then finding where the remainder after division by 1 is exactly zero (i.e., the integer results):
>> n = 60;
>> find(rem(n./(1:n), 1) == 0)
ans =
1 2 3 4 5 6 10 12 15 20 30 60
Here is a comparison of six different implementations for finding factors of an integer:
function [t,v] = testFactors()
% integer to factor
%{45, 60, 2059, 3135, 223092870, 3491888400};
n = 2*2*2*2*3*3*3*5*5*7*11*13*17*19;
% functions to compare
fcns = {
#() factors1(n);
#() factors2(n);
#() factors3(n);
#() factors4(n);
%#() factors5(n);
#() factors6(n);
};
% timeit
t = cellfun(#timeit, fcns);
% check results
v = cellfun(#feval, fcns, 'UniformOutput',false);
assert(isequal(v{:}));
end
function f = factors1(n)
% vectorized implementation of factors2()
f = find(rem(n, 1:floor(sqrt(n))) == 0);
f = unique([1, n, f, fix(n./f)]);
end
function f = factors2(n)
% factors come in pairs, the smaller of which is no bigger than sqrt(n)
f = [1, n];
for k=2:floor(sqrt(n))
if rem(n,k) == 0
f(end+1) = k;
f(end+1) = fix(n/k);
end
end
f = unique(f);
end
function f = factors3(n)
% Get prime factors, and compute products of all possible subsets of size>1
pf = factor(n);
f = arrayfun(#(k) prod(nchoosek(pf,k),2), 2:numel(pf), ...
'UniformOutput',false);
f = unique([1; pf(:); vertcat(f{:})])'; %'
end
function f = factors4(n)
% http://rosettacode.org/wiki/Factors_of_an_integer#MATLAB_.2F_Octave
pf = factor(n); % prime decomposition
K = dec2bin(0:2^length(pf)-1)-'0'; % all possible permutations
f = ones(1,2^length(pf));
for k=1:size(K)
f(k) = prod(pf(~K(k,:))); % compute products
end;
f = unique(f); % eliminate duplicates
end
function f = factors5(n)
% #LuisMendo: brute-force implementation
f = find(rem(n, 1:n) == 0);
end
function f = factors6(n)
% Symbolic Math Toolbox
f = double(evalin(symengine, sprintf('numlib::divisors(%d)',n)));
end
The results:
>> [t,v] = testFactors();
>> t
t =
0.0019 % factors1()
0.0055 % factors2()
0.0102 % factors3()
0.0756 % factors4()
0.1314 % factors6()
>> numel(v{1})
ans =
1920
Although the first vectorized version is the fastest, the equivalent loop-based implementation (factors2) is not far behind, thanks to automatic JIT optimization.
Note that I had to disable the brute-force implementation (factors5()) because it throws an out-of-memory error (storing the vector 1:3491888400 in double-precision requires over 26GB of memory!). This method is obviously not feasible for large integers, neither space- or time-wise.
Conclusion: use the following vectorized implementation :)
n = 3491888400;
f = find(rem(n, 1:floor(sqrt(n))) == 0);
f = unique([1, n, f, fix(n./f)]);
An improvement over #gnovice's answer is to skip the division operation: rem alone is enough:
n = 60;
find(rem(n, 1:n)==0)