I have a code that needs to evaluate the arc length equation below:
syms x
a = 10; b = 10; c = 10; d = 10;
fun = 4*a*x^3+3*b*x^2+2*c*x+d
int((1+(fun)^2)^.5)
but all that returns is below:
ans = int(((40*x^3 + 30*x^2 + 20*x + 10)^2 + 1)^(1/2), x)
Why wont matlab evaluate this integral? I added a line under to check if it would evaulate int(x) and it returned the desired result.
Problems involving square roots of functions may be tricky to intgrate. I am not sure whether the integral exists or not, but it if you look up the integral of a second order polynomial you will see that this one is already quite a mess. What you would have, would you expand the function inside the square root, would be a ninth order polynomial. If this integral actually would exist it may be too complex to calculate.
Anyway, if you think about it, would anyone really become any wiser by finding the analytical solution of this? If that is not the case a numerical solution should be sufficient.
EDIT
As thewaywewalk said in the comment, a general rule to calculate these kinds of integrals would be valuable, but to know the primitive function to the particular integral would probably be overkill (if a solution could be found).
Instead define the function as an anonymous function
fun = #(x) sqrt((4*a*x.^3+3*b*x.^2+2*c*x+d).^2+1);
and use integral to evaluate the function between some range, eg
integral(fun,0,100);
for evaluating the function in the closed interval [0,100].
Related
I want to minimize a function like below:
Here, n can be 5,10,50 etc. I want to use Matlab and want to use Gradient Descent and Quasi-Newton Method with BFGS update to solve this problem along with backtracking line search. I am a novice in Matlab. Can anyone help, please? I can find a solution for a similar problem in that link: https://www.mathworks.com/help/optim/ug/unconstrained-nonlinear-optimization-algorithms.html .
But, I really don't know how to create a vector-valued function in Matlab (in my case input x can be an n-dimensional vector).
You will have to make quite a leap to get where you want to be -- may I suggest to go through some basic tutorial first in order to digest basic MATLAB syntax and concepts? Another useful read is the very basic example to unconstrained optimization in the documentation. However, the answer to your question touches only basic syntax, so we can go through it quickly nevertheless.
The absolute minimum to invoke the unconstraint nonlinear optimization algorithms of the Optimization Toolbox is the formulation of an objective function. That function is supposed to return the function value f of your function at any given point x, and in your case it reads
function f = objfun(x)
f = sum(100 * (x(2:end) - x(1:end-1).^2).^2 + (1 - x(1:end-1)).^2);
end
Notice that
we select the indiviual components of the x vector by matrix indexing, and that
the .^ notation effects that the operand is to be squared elementwise.
For simplicity, save this function to a file objfun.m in your current working directory, so that you have it available from the command window.
Now all you have to do is to call the appropriate optimization algorithm, say, the quasi Newton method, from the command window:
n = 10; % Use n variables
options = optimoptions(#fminunc,'Algorithm','quasi-newton'); % Use QM method
x0 = rand(n,1); % Random starting guess
[x,fval,exitflag] = fminunc(#objfun, x0, options); % Solve!
fprintf('Final objval=%.2e, exitflag=%d\n', fval, exitflag);
On my machine I see that the algorithm converges:
Local minimum found.
Optimization completed because the size of the gradient is less than
the default value of the optimality tolerance.
Final objval=5.57e-11, exitflag=1
Lets say, I have a function 'x' and a function '2sin(x)'
How do I output the intersects, i.e. the roots in MATLAB? I can easily plot the two functions and find them that way but surely there must exist an absolute way of doing this.
If you have two analytical (by which I mean symbolic) functions, you can define their difference and use fzero to find a zero, i.e. the root:
f = #(x) x; %defines a function f(x)
g = #(x) 2*sin(x); %defines a function g(x)
%solve f==g
xroot = fzero(#(x)f(x)-g(x),0.5); %starts search from x==0.5
For tricky functions you might have to set a good starting point, and it will only find one solution even if there are multiple ones.
The constructs seen above #(x) something-with-x are called anonymous functions, and they can be extended to multivariate cases as well, like #(x,y) 3*x.*y+c assuming that c is a variable that has been assigned a value earlier.
When writing the comments, I thought that
syms x; solve(x==2*sin(x))
would return the expected result. At least in Matlab 2013b solve fails to find a analytic solution for this problem, falling back to a numeric solver only returning one solution, 0.
An alternative is
s = feval(symengine,'numeric::solve',2*sin(x)==x,x,'AllRealRoots')
which is taken from this answer to a similar question. Besides using AllRealRoots you could use a numeric solver, manually setting starting points which roughly match the values you have read from the graph. This wa you get precise results:
[fzero(#(x)f(x)-g(x),-2),fzero(#(x)f(x)-g(x),0),fzero(#(x)f(x)-g(x),2)]
For a higher precision you could switch from fzero to vpasolve, but fzero is probably sufficient and faster.
I have been working on solving some equation in a more complicated context. However, I want to illustrate my question through the following simple example.
Consider the following two functions:
function y=f1(x)
y=1-x;
end
function y=f2(x)
if x<0
y=0;
else
y=x;
end
end
I want to solve the following equation: f1(x)=f2(x). The code I used is:
syms x;
x=solve(f1(x)-f2(x));
And I got the following error:
??? Error using ==> sym.sym>notimplemented at 2621
Function 'lt' is not implemented for MuPAD symbolic objects.
Error in ==> sym.sym>sym.lt at 812
notimplemented('lt');
Error in ==> f2 at 3
if x<0
I know the error is because x is a symbolic variable and therefore I could not compare x with 0 in the piecewise function f2(x).
Is there a way to fix this and solve the equation?
First, make sure symbolic math is even the appropriate solution method for your problem. In many cases it isn't. Look at fzero and fsolve amongst many others. A symbolic method is only needed if, for example, you want a formula or if you need to ensure precision.
In such an old version of Matlab, you may want to break up your piecewise function into separate continuous functions and solve them separately:
syms x;
s1 = solve(1-x^2,x) % For x >= 0
s2 = solve(1-x,x) % For x < 0
Then you can either manually examine or numerically compare the outputs to determine if any or all of the solutions are valid for the chosen regime – something like this:
s = [s1(double(s1) >= 0);s2(double(s2) < 0)]
You can also take advantage of the heaviside function, which is available in much older versions.
syms x;
f1 = 1-x;
f2 = x*heaviside(x);
s = solve(f1-f2,x)
Yes, the Heaviside function is 0.5 at zero – this gives it the appropriate mathematical properties. You can shift it to compare values other than zero. This is a standard technique.
In Matlab R2012a+, you can take advantage of assumptions in addition to the normal relational operators. To add to #AlexB's comment, you should convert the output of any logical comparison to symbolic before using isAlways:
isAlways(sym(x<0))
In your case, x is obviously not "always" on one side or the other of zero, but you may still find this useful in other cases.
If you want to get deep into Matlab's symbolic math, you can create piecewise functions using MuPAD, which are accessible from Matlab – e.g., see my example here.
I'm trying to compute a rather ugly integral using MATLAB. What I'm having problem with though is a part where I multiply a very big number (>10^300) with a very small number (<10^-300). MATLAB returns 'inf' for this even though it should be in the range of 0-0.0005. This is what I have
besselFunction = #(u)besseli(qb,2*sqrt(lambda*(theta + mu)).*u);
exponentFuncion = #(u)exp(-u.*(lambda + theta + mu));
where qb = 5, lambda = 12, theta = 10, mu = 3. And what I want to find is
besselFunction(u)*exponentFunction(u)
for all real values of u. The problem is that whenever u>28 it will be evaluated as 'inf'. I've heared, and tried, to use MATLAB function 'vpa' but it doesn't seem to work well when I want to use functions...
Any tips will be appreciated at this point!
I'd use logarithms.
Let x = Bessel function of u and y = x*exp(-u) (simpler than your equation, but similar).
Since log(v*w) = log(v) + log(w), then log(y) = log(x) + log(exp(-u))
This simplifies to
log(y) = log(x) - u
This will be better behaved numerically.
The other key will be to not evaluate that Bessel function that turns into a large number and passing it to a math function to get the log. Better to write your own that returns the logarithm of the Bessel function directly. Look at a reference like Abramowitz and Stegun to try and find one.
If you are doing an integration, consider using Gauss–Laguerre quadrature instead. The basic idea is that for equations of the form exp(-x)*f(x), the integral from 0 to inf can be approximated as sum(w(X).*f(X)) where the values of X are the zeros of a Laguerre polynomial and W(X) are specific weights (see the Wikipedia article). Sort of like a very advanced Simpson's rule. Since your equation already has an exp(-x) part, it is particularly suited.
To find the roots of the polynomial, there is a function on MATLAB Central called LaguerrePoly, and from there it is pretty straightforward to compute the weights.
I have a function in MATLAB which takes another function as an argument. I would like to somehow define a piecewise inline function that can be passed in. Is this somehow possible in MATLAB?
Edit: The function I would like to represent is:
f(x) = { 1.0, 0.0 <= x <= 0.5,
-1.0, 0.5 < x <= 1.0
where 0.0 <= x <= 1.0
You really have defined a piecewise function with three break points, i.e., at [0, 0.5, 1]. However, you have not defined the value of the function outside of the breaks. (By the way, I've used the term "break" here, because we are really defining a simple form of spline, a piecewise constant spline. I might also have used the term knot, another common word in the world of splines.)
If you absolutely know that you will never evaluate the function outside of [0,1], then there is no problem. So then just define a piecewise function with ONE break point, at x = 0.5. The simple way to define a piecewise constant function like yours is to use a logical operator. Thus the test (x > 0.5) returns a constant, either 0 or 1. By scaling and translating that result, it is easy to generate a function that does what you wish.
constfun = #(x) (x > 0.5)*2 - 1;
An inline function does a similar thing, but inline functions are VERY slow compared to an anonymous function. I would strongly recommend use of the anonymous form. As a test, try this:
infun = inline('(x > 0.5)*2 - 1','x');
x = 0:.001:1;
tic,y = constfun(x);toc
Elapsed time is 0.002192 seconds.
tic,y = infun(x);toc
Elapsed time is 0.136311 seconds.
Yes, the inline function took wildly more time to execute than did the anonymous form.
A problem with the simple piecewise constant form I've used here is it is difficult to expand to when you have more break points. For example, suppose you wished to define a function that took on three different values depending on what interval the point fell in? While this can be done too with creative use of tests, carefully shifting and scaling them, it can get nasty. For example, how might one define the piecewise function that returns
-1 when x < 0,
2 when 0 <= x < 1,
1 when 1 <= x
One solution is to use a unit Heaviside function. So first, define a basic unit Heaviside function.
H = #(x) (x >= 0);
Our piecewise function is now derived from H(x).
P = #(x) -1 + H(x)*3 + H(x-1)*(-1);
See that there are three pieces to P(x). The first term is what happens for x below the first break point. Then we add in a piece that takes effect above zero. Finally, the third piece adds in another offset in above x == 1. It is easily enough plotted.
ezplot(P,[-3,3])
More sophisticated splines are easily generated from this beginning. Se that I've called this construct a spline again. Really, this is where we might be leading. In fact, this is where this leads. A spline is a piecewise function, carefully tied together at a list of knots or break points. Splines in particular often have specified orders of continuity, so for example, a cubic spline will be twice differentiable (C2) across the breaks. There are also piecewise cubic functions that are only C1 functions. My point in all of this is I've described a simple beginning point to form any piecewise function. It works quite well for polynomial splines, although there may be a wee bit of mathematics required to choose the coefficients of these functions.
Another way to create this function is as an explicit piecewise polynomial. In MATLAB, we have the little known function mkpp. Try this out...
pp = mkpp([0 .5 1],[1;-1]);
Had you the splines toolbox, then fnplt will plot this directly for you. Assuming that you don't have that TB, do this:
ppfun = #(x) ppval(pp,x);
ezplot(ppfun,[0 1])
Looking back at the mkpp call, it is rather simple after all. The first argument is the list of break points in the curve (as a ROW vector). The second argument is a COLUMN vector, with the piecewise constant values the curve will take on in these two defined intervals between the breaks.
Several years ago I posted another option, piecewise_eval. It can be downloaded from the MATLAB Central file exchange. This is a function that will allow a user to specify a piecewise function purely as a list of break points, along with functional pieces between those breaks. Thus, for a function with a single break at x = 0.5, we would do this:
fun = #(x) piecewise_eval(x,0.5,{1,-1});
See that the third argument provides the value used in each segment, although those pieces need not be purely constant functions. If you wish the function to return perhaps a NaN outside of the interval of interest, this too is easily accomplished.
fun = #(x) piecewise_eval(x,[0 0.5 1],{NaN,1,-1,NaN});
My point in all of this rather lengthy excursion is to understand what a piecewise function is, and several ways to build one in MATLAB.
Unfortunately, MATLAB doesn't have a ternary operator which would make this sort of thing easier, but to expand slightly on gnovice's approach, you could create an anonymous function like so:
fh = #(x) ( 2 .* ( x <= 0.5 ) - 1 )
In general, anonymous functions are more powerful than inline function objects, and allow you to create closures etc.
If you really want to make an inline function (as opposed to an anonymous function), then the following would probably be the simplest way:
f = inline('2.*(x <= 0.5)-1');
However, as pointed out in the other answers, anonymous functions are more commonly used and are more efficient:
f = #(x) (2.*(x <= 0.5)-1);
I just had to solve that problem, and I think the easiest thing to do is use anonymous functions. Say that you have a piecewise function:
when x<0 : x^2 + 3x
when 0<=x<=4: e^x
when x>4 : log(x)
I'd first define logical masks for each piecewise region:
PIECE1 = #(x) x<0
PIECE2 = #(x) x>=0 & x<=4
PIECE3 = #(x) x>4
Then I'd put them all together:
f = #(x) PIECE1(x).*(x.^2+3*x) + PIECE2(x).*exp(x) + PIECE3(x).*log(x)
x = -10:.1:10
figure;
plot(x,f(x))