This post updates my question in Solve finds wrong solution?
Given the "simple" function in x:
f = (x + 3/10)^(1/2) - (9*(x + 3/10)^5)/5 - (x + 3/10)^6 + (x - 1)/(2*(x + 3/10)^(1/2));
Find the zeros by the call
solve(f,x)
This yields 3 zeros:
ans =
0.42846617518653978966562924618638
0.15249587894102346284238111155954
0.12068186494007759990714181154349
A simple look at the plot shows that the third root is nonsense:
I have a serious problem because I need to retrieve the smallest zero from the above vector. Calling min(ans) returns a wrong zero. What can I do to workaround?
This is a non-polynomial equation, and it will probably fallback to a numeric solver (non-symbolic). So there might be numerical errors, or the numeric algorithm might get stuck and report false solutions, I'm not sure...
What you can do is substitute the solutions back into the equation, and reject ones that are above some specified threshold:
% define function
syms x real
syms f(x)
xx = x+3/10;
f(x) = sqrt(xx) - 9/5*xx^5 - xx^6 + (x - 1)/(2*sqrt(xx));
pretty(f)
% find roots
sol = solve(f==0, x, 'Real',true)
% filter bad solutions
err = subs(f, x, sol)
sol = sol(abs(err) < 1e-9); % this test removes the 2nd solution
% plot
h = ezplot(f, [0.1 0.5]);
line(xlim(), [0 0], 'Color','r', 'LineStyle',':')
xlabel('x'), ylabel('f(x)')
% programmatically insert data tooltips
xd = get(h, 'XData'); yd = get(h ,'YData');
[~,idx] = min(abs(bsxfun(#minus, xd, double(sol))), [], 2);
dcm = datacursormode(gcf);
pos = [xd(idx) ; yd(idx)].';
for i=1:numel(idx)
dtip = createDatatip(dcm, h);
set(get(dtip,'DataCursor'), 'DataIndex',idx(i), 'TargetPoint',pos(i,:))
set(dtip, 'Position',pos(i,:))
end
We are left only with the two desired solutions (one is rejected by our test):
/ 3 \5
9 | x + -- |
/ 3 \ \ 10 / / 3 \6 x - 1
sqrt| x + -- | - ------------- - | x + -- | + ----------------
\ 10 / 5 \ 10 / / 3 \
2 sqrt| x + -- |
\ 10 /
sol =
0.42846617518653978966562924618638
0.12068186494007759990714181154349 % <== this one is dropped
0.15249587894102346284238111155954
err(x) =
-9.1835496157991211560057541970488e-41
-0.058517436737550288309001512815475 % <==
1.8367099231598242312011508394098e-40
I also tried using MATLAB's numeric solvers, which were able to find the two solutions given reasonable starting points:
fzero function (a derivative-free root-finding algorithm)
fsolve function (from the Optimization Toolbox)
(See this related question)
fcn = matlabFunction(f); % convert symbolic f to a regular function handle
opts = optimset('Display','off', 'TolFun',1e-9, 'TolX',1e-6);
% FZERO
sol2(1) = fzero(fcn, 0.1, opts);
sol2(2) = fzero(fcn, 0.5, opts);
disp(sol2) % [0.1525, 0.4285]
% FSOLVE
sol3(1) = fsolve(fcn, 0.0, opts);
sol3(2) = fsolve(fcn, 1.0, opts);
disp(sol3) % [0.1525, 0.4285]
For comparison, I tried solving the equation directly into MuPAD, as well as in Mathematica.
Mathematica 9.0
MuPAD (R2014a)
Of course we can always directly call MuPAD from inside MATLAB:
% f is the same symbolic function we've built above
>> sol = evalin(symengine, ['numeric::solve(' char(f) ' = 0, x, AllRealRoots)'])
sol =
[ 0.15249587894102346284238111155954, 0.42846617518653978966562924618638]
The above call is equivalent to searching the entire range x = -infinity .. infinity (which can be slow!). We should assist numeric::solve by providing a specific search ranges when possible:
>> sol = feval(symengine, 'numeric::solve', f==0, 'x = 0 .. 1', 'AllRealRoots')
sol =
[ 0.15249587894102346284238111155954, 0.42846617518653978966562924618638]
One workaround could be taking min(ans), and then checking that f(min(ans))==0. If not, then use the next-smallest value.
Related
Assume we have three equations:
eq1 = x1 + (x1 - x2) * t - X == 0;
eq2 = z1 + (z1 - z2) * t - Z == 0;
eq3 = ((X-x1)/a)^2 + ((Z-z1)/b)^2 - 1 == 0;
while six of known variables are:
a = 42 ;
b = 12 ;
x1 = 316190;
z1 = 234070;
x2 = 316190;
z2 = 234070;
So we are looking for three unknown variables that are:
X , Z and t
I wrote two method to solve it. But, since I need to run these code for 5.7 million data, it become really slow.
Method one (using "solve"):
tic
S = solve( eq1 , eq2 , eq3 , X , Z , t ,...
'ReturnConditions', true, 'Real', true);
toc
X = double(S.X(1))
Z = double(S.Z(1))
t = double(S.t(1))
results of method one:
X = 316190;
Z = 234060;
t = -2.9280;
Elapsed time is 0.770429 seconds.
Method two (using "fsolve"):
coeffs = [a,b,x1,x2,z1,z2]; % Known parameters
x0 = [ x2 ; z2 ; 1 ].'; % Initial values for iterations
f_d = #(x0) myfunc(x0,coeffs); % f_d considers x0 as variables
options = optimoptions('fsolve','Display','none');
tic
M = fsolve(f_d,x0,options);
toc
results of method two:
X = 316190; % X = M(1)
Z = 234060; % Z = M(2)
t = -2.9280; % t = M(3)
Elapsed time is 0.014 seconds.
Although, the second method is faster, but it still needs to be improved. Please let me know if you have a better solution for that. Thanks
* extra information:
if you are interested to know what those 3 equations are, the first two are equations of a line in 2D and the third equation is an ellipse equation. I need to find the intersection of the line with the ellipse. Obviously, we have two points as result. But, let's forget about the second answer for simplicity.
My suggestion it's to use the second approce,which it's the recommended by matlab for nonlinear equation system.
Declare a M-function
function Y=mysistem(X)
%X(1) = X
%X(2) = t
%X(3) = Z
a = 42 ;
b = 12 ;
x1 = 316190;
z1 = 234070;
x2 = 316190;
z2 = 234070;
Y(1,1) = x1 + (x1 - x2) * X(2) - X(1);
Y(2,1) = z1 + (z1 - z2) * X(2) - X(3);
Y(3,1) = ((X-x1)/a)^2 + ((Z-z1)/b)^2 - 1;
end
Then for solving use
x0 = [ x2 , z2 , 1 ];
M = fsolve(#mysistem,x0,options);
If you may want to reduce the default precision by changing StepTolerance (default 1e-6).
Also for more increare you may want to use the jacobian matrix for greater efficencies.
For more reference take a look in official documentation:
fsolve Nonlinear Equations with Analytic Jacobian
Basically giving the solver the Jacobian matrix of the system(and special options) you can increase method efficency.
I was given a piece of Matlab code by a lecturer recently for a way to solve simultaneous equations using the Newton-Raphson method with a jacobian matrix (I've also left in his comments). However, although he's provided me with the basic code I cannot seem to get it working no matter how hard I try. I've spent many hours trying to introduce the 'func' function but to no avail, frequently getting the message that there aren't enough inputs. Any help would be greatly appreciated, especially with how to write the 'func' function.
function root = newtonRaphson2(func,x,tol)
% Newton-Raphson method of finding a root of simultaneous
% equations fi(x1,x2,...,xn) = 0, i = 1,2,...,n.
% USAGE: root = newtonRaphson2(func,x,tol)
% INPUT:
% func = handle of function that returns[f1,f2,...,fn].
% x = starting solution vector [x1,x2,...,xn].
% tol = error tolerance (default is 1.0e4*eps).
% OUTPUT:
% root = solution vector.
if size(x,1) == 1; x = x'; end % x must be column vector
for i = 1:30
[jac,f0] = jacobian(func,x);
if sqrt(dot(f0,f0)/length(x)) < tol
root = x; return
end
dx = jac\(-f0);
x = x + dx;
if sqrt(dot(dx,dx)/length(x)) < tol
root = x; return
end
end
error('Too many iterations')
function [jac,f0] = jacobian(func,x)
% Returns the Jacobian matrix and f(x).
h = 1.0e-4;
n = length(x);
jac = zeros(n);
f0 = feval(func,x);
for i =1:n
temp = x(i);
x(i) = temp + h;
f1 = feval(func,x);
x(i) = temp;
jac(:,i) = (f1 - f0)/h;
end
The simultaneous equations to be solved are:
sin(x)+y^2+ln(z)-7=0
3x+2^y-z^3+1=0
x+y+Z-=0
with the starting point (1,1,1).
However, these are arbitrary and can be replaced with anything, I mainly just need to know the general format.
Many thanks, I know this may be a very simple task but I've only recently started teaching myself Matlab.
You need to create a new file called myfunc.m (or whatever name you like) which takes a single input parameter - a column vector x - and returns a single output vector - a column vector y such that y = f(x).
For example,
function y = myfunc(x)
y = zeros(3, 1);
y(1) = sin(x(1)) + x(2)^2 + log(x(3)) - 7;
y(2) = 3*x(1) + 2^x(2) - x(3)^3 + 1;
y(3) = x(1) + x(2) + x(3);
end
You can then refer to this function as #myfunc as in
>> newtonRaphson2(#myfunc, [1;1;1], 1e-6);
The reason for the special notation is that Matlab allows you to call a function with no parameters by omitting the parens () that follow it. So for example, Matlab interprets myfunc as you calling the function with no arguments (so it tries to replace it with its result) whereas #myfunc refers to the function itself, rather than its result.
Alternatively you can write a function directly using the # notation, as in
>> newtonRaphson2(#(x) exp(x) - 3*x, 2, 1e-2)
ans =
1.5315
>> newtonRaphson2(#(x) exp(x) - 3*x, 1, 1e-2)
ans =
0.6190
which are the two roots of the equation exp(x) - 3 * x = 0.
Edit - as an aside, your professor has terrible coding style (if the code in your question is a direct copy-paste of what he gave you, and you haven't mangled it along the way). It would be better to write the code like this, with indentation making it clear what the structure of the code is.
function root = newtonRaphson2(func, x, tol)
% Newton-Raphson method of finding a root of simultaneous
% equations fi(x1,x2,...,xn) = 0, i = 1,2,...,n.
%
% USAGE: root = newtonRaphson2(func,x,tol)
%
% INPUT:
% func = handle of function that returns[f1,f2,...,fn].
% x = starting solution vector [x1,x2,...,xn].
% tol = error tolerance (default is 1.0e4*eps).
%
% OUTPUT:
% root = solution vector.
if size(x, 1) == 1; % x must be column vector
x = x';
end
for i = 1:30
[jac, f0] = jacobian(func, x);
if sqrt(dot(f0, f0) / length(x)) < tol
root = x; return
end
dx = jac \ (-f0);
x = x + dx;
if sqrt(dot(dx, dx) / length(x)) < tol
root = x; return
end
end
error('Too many iterations')
end
function [jac, f0] = jacobian(func,x)
% Returns the Jacobian matrix and f(x).
h = 1.0e-4;
n = length(x);
jac = zeros(n);
f0 = feval(func,x);
for i = 1:n
temp = x(i);
x(i) = temp + h;
f1 = feval(func,x);
x(i) = temp;
jac(:,i) = (f1 - f0)/h;
end
end
I'm basicaly trying to find the product of an expression that goes like this:
(x-(N-1)/2).....(x+(N-1)/2) for even value of N
x is a value that I will set at the beginning that changes too but that is a different problem...
let's say for the sake of argument that for now x is a constant (ex x=1)
example for N=6
(x-5/2)(x-3/2)(x-1/2)(x+1/2)(x+3/2)*(x+5/2)
the idea was to create a row vector every element of which is each individual term (P(1)=x-5/2) (P(2)=x-3/2)...etc and then calculate its product
N=6;
x=1;
P=ones(1,N);
for k=(-N-1)/2:(N-1)/2
for n=1:N
P(n)=(x-k);
end
end
y=prod(P);
instead this creates a vector that takes only the first value of the epxression and then
repeats the same value at each cell.
there is obviously a fundamental problem with my loop but I just can't see it.
So if anyone can help with that OR suggest a better way to calculate the product I would be grateful.
Use vectorized commands
Why use a loop when you can use vectorized commands like prod?
y = prod(2 * x + [-N + 1 : 2 : N - 1]) / 2;
For convenience, you may want to define an anonymous function for it:
f = #(N,x) reshape(prod(bsxfun(#plus, 2 * x(:), -N + 1 : 2 : N - 1) / 2, 2), size(x));
Note that the function is compatible with a (row or column) vector input x.
Tests in MATLAB's Command Window
>> f(6, [2,2]')
ans =
-14.7656
4.9219
-3.5156
4.9219
-14.7656
>> f(6, [2,2])
ans =
-14.7656 4.9219 -3.5156 4.9219 -14.7656
Benchmark
Here is a comparison of rayreng's approach versus mine. The former emerges as the clear winner... :'( ...at least as N increases.
Varying N, fixed x
Fixed N (= 10), vector x of varying length
Fixed N (= 100), vector x of varying length
Benchmark code
function benchmark
% varying N, fixed x
clear all
n = logspace(2,4,20)';
x = rand(1000,1);
tr = zeros(size(n));
tj = tr;
for k = 1 : numel(n)
% rayreng's approach (poly/polyval)
fr = #() rayreng(n(k), x);
tr(k) = timeit(fr);
% Jubobs's approach (prod/reshape/bsxfun)
fj = #() jubobs(n(k), x);
tj(k) = timeit(fj);
end
figure
hold on
plot(n, tr, 'bo')
plot(n, tj, 'ro')
hold off
xlabel('N')
ylabel('time (s)')
legend('rayreng', 'jubobs')
end
function y = jubobs(N,x)
y = reshape(prod(bsxfun(#plus,...
2 * x(:),...
-N + 1 : 2 : N - 1) / 2,...
2),...
size(x));
end
function y = rayreng(N, x)
p = poly(linspace(-(N-1)/2, (N-1)/2, N));
y = polyval(p, x);
end
function benchmark2
% fixed N, varying x
clear all
n = 100;
nx = round(logspace(2,4,20));
tr = zeros(size(n));
tj = tr;
for k = 1 : numel(nx)
disp(k)
x = rand(nx(k), 1);
% rayreng's approach (poly/polyval)
fr = #() rayreng(n, x);
tr(k) = timeit(fr);
% Jubobs's approach (prod/reshape/bsxfun)
fj = #() jubobs(n, x);
tj(k) = timeit(fj);
end
figure
hold on
plot(nx, tr, 'bo')
plot(nx, tj, 'ro')
hold off
xlabel('number of elements in vector x')
ylabel('time (s)')
legend('rayreng', 'jubobs')
title(['n = ' num2str(n)])
end
function y = jubobs(N,x)
y = reshape(prod(bsxfun(#plus,...
2 * x(:),...
-N + 1 : 2 : N - 1) / 2,...
2),...
size(x));
end
function y = rayreng(N, x)
p = poly(linspace(-(N-1)/2, (N-1)/2, N));
y = polyval(p, x);
end
An alternative
Alternatively, because the terms in your product form an arithmetic progression (each term is greater than the previous one by 1/2), you can use the formula for the product of an arithmetic progression.
I agree with #Jubobs in that you should avoid using for loops for this kind of computation. There are cases where for loops perform fast, but for something as simple as this, avoid using loops if possible.
An alternative approach to what Jubobs has suggested is that you can consider that polynomial equation to be in factored form where each factor denotes a root located at that particular location. You can use poly to convert these factors into a polynomial equation, then use polyval to evaluate the expression at the point you want. First, generate your roots by linspace where the points vary from -(N-1)/2 to (N-1)/2 and there are N of them, then plug this into poly. Finally, for any values of x, put this into polyval with the output of poly. The advantage of this approach is that you can evaluate multiple points of x in a single sweep.
Going with what you have, you would simply do this:
p = poly(linspace(-(N-1)/2, (N-1)/2, N));
out = polyval(p, x);
With your example, supposing that N = 6, this would be the output of the first line:
p =
1.0000 0 -8.7500 0 16.1875 0 -3.5156
As such, this is saying that when we expand out (x-5/2)(x-3/2)(x-1/2)(x+1/2)(x+3/2)(x+5/2), we get:
x^6 - 8.75x^4 + 16.1875x^2 - 3.5156
If we take a look at the roots of this equation, this is what we get:
r = roots(p)
r =
-2.5000
2.5000
-1.5000
1.5000
-0.5000
0.5000
As you can see, each term corresponds to one factor in your polynomial equation, so we do have the right mindset here. Now, all you have to do is use p with your values of x into polyval to obtain your results. For example, if I wanted to evaluate that polynomial from -2 <= x <= 2 where x is an integer, this is the result I get:
polyval(p, -2:2)
ans =
-14.7656 4.9219 -3.5156 4.9219 -14.7656
Therefore, when x = -2, the result is -14.7656 and so on.
Though I would recommend the solution by #Jubobs, it is also good to check what the issue is with your loop.
The first indication that something is wrong, is that you have a nested loop over 2 variables, and only index with one of them to store the result. Probably you just need a single loop.
Here is a loop that you may be interested in that should do roughly what you need:
N=6;
x=1;
k=(-N-1)/2:(N-1)/2
P = ones(size(k));
for n=1:numel(k)
P(n)=(x-k(n));
end
y=prod(P);
I tried to keep the code close to the original, so hopefully it is easy to understand.
I am having difficulty in finding roots of a nonlinear equation. I have tried Matlab and Maple both, and both give me the same error which is
Error, (in RootFinding:-NextZero) can only handle isolated zeros
The equation goes like
-100 + 0.1335600000e-5*H + (1/20)*H*arcsinh(2003.40/H)
The variable is H in the equation.
How do I find the roots (or the approximate roots) of this equation?
Matlab Code:
The function file:
function hor_force = horizontal(XY, XZ, Lo, EAo, qc, VA)
syms H
equation = (-1*ZZ) + (H/qc)*(cosh((qc/H)*(XZ- XB))) - H/qc + ZB;
hor_force = `solve(equation);`
The main file:
EAo = 7.5*10^7;
Lo = 100.17;
VA = 2002;
XY = 0;
ZY = 0;
XB = 50;
ZB = -2;
XZ = 100;
ZZ = 0;
ql = 40;
Error which Matlab shows:
Error using sym/solve (line 22)
Error using maplemex
Error, (in RootFinding:-NextZero) can only handle isolated zeros
Error in horizontal (line 8)
hor_force = solve(equation);
Error in main (line 34)
h = horizontal(XY, XZ, Lo, EAo, ql, VA)
http://postimg.org/image/gm93z3b7z/
You don't need the symbolic toolbox for this:
First, create an anonymous function that can take vectors at input (use .* and ./:
equation = #(H) ((-1*ZZ) + (H./qc).*(cosh((qc./H).*(XZ- XB))) - H./qc + ZB);
Second, create a vector that you afterwards insert into the equation to find approximately when the sign of the function changes. In the end, use fzero with x0 as the second input parameter.
H = linspace(1,1e6,1e4);
x0 = H(find(diff(sign(equation(H))))); %// Approximation of when the line crosses zero
x = fzero(equation, x0) %// Use fzero to find the crossing point, using the initial guess x0
x =
2.5013e+04
equation(x)
ans =
0
To verify:
You might want to check out this question for more information about how to find roots of non-polynomials.
In Maple, using the expression from your question,
restart:
ee := -100 + 0.1335600000e-5*H + (1/20)*H*arcsinh(2003.40/H):
Student:-Calculus1:-Roots(ee, -1e6..1e6);
[ 5 ]
[-1.240222868 10 , -21763.54830, 18502.23816]
#plot(ee, H=-1e6..1e6, view=-1..1);
I'm trying build a matlab function that will evaluate a function and vector that are sent in as parameters. I'm having a hard time trying to figure out how to send in the function so that it can be evaluated in the matlab function. I figured out how to do it without the function but I'm a little lost trying to evaluate it within a matlab function. Below is my code...
This is what I'm trying to do...
x = [x1 x2]';
f = x(x1)^2 + 2 * (x2)^2
x = [5 10];
f = (5)^2 + 2 * (10)^2 % which I would like to return 225, not a column vector
This is what I have and what I have tried...
x = [5 10]';
% without using a matlab function
% k = 1
% f = x(k)^2 + 2 * x(k + 1)^2; % returns the correct answer of 225
f = x^2 + 2 * x^2 % complains about the scalar 2
f = x.^2 + 2 * x.^2 % returns a column vector [75; 300]
function [value] = evalFunction(f,x)
value = f(x);
I've tried...
f = #(x) x.^2 + 2 * (x+1).^2;
value = evalFunction(#f,x) %Error: "f" was previously used as a variable
So I tried...
f = #(x) x.^2 + 2 * (x+1).^2;
value = evalFunction(f,x) %value = [97;342]
I'm new to matlab so any help is appreciated. I've been doing some research and found some stuff here on stackoverflow but can't seem to get it to work. I've seen there are other ways to do this, but I will eventually be adding more code to the matlab evalFunction function so I'd like to do it this way. Thanks!
Anonymous functions and function handles plus array indexing. Taking x as a 2-element vector, define and use your function like:
f = #(x) x(1).^2 + 2 * x(2).^2;
value = evalFunction(f,x) % but you can just do f(x) if that is all you need
However, if evalFunction does nothing other than evaluate f at x, then you don't need it at all. Just do f(x).
Alternately,
f = #(x1,x2) x1.^2 + 2*x2.^2;
value = evalFunction(f,x1,x2); % here your function will call it by f(x1,x2)
You are probably coming at this from a C background - in Matlab, x+1 is the entire vector x with 1 added - not the element offset by 1.
The function you need is
f = #(x)x(1).^2 + 2 * (x(2)).^2;
or, to be a little more "matlab-like":
f = #(x) [1 2] * x(1:2)'.^2;
Which performs the element-wise square of the first two elements of x as a column vector, and then does the matrix multiplication with [1 2], resulting in
1 * x(1) .^2 + 2 * x(2) .^2;
Which seems to be what you were asking for.
caveat: did not have opportunity to test this...