I am trying to write a line composed of two segments as a single equation in :
y = m1*x + c1 , for x<=x1
y = m2*x + c2 , for x>=x1
My questions are:
How can I write the function of this combined line as a single equation?
How can I write multiple functions (valid in separate regions of a linear parameter space) as a single equation?
Please explain both how to express this mathematically and how to program this in general and in Matlab specifically.
You can write this equation as a single line by using the Heaviside step function, https://en.wikipedia.org/wiki/Heaviside_step_function.
Combining two functions into one:
In fact, what you are trying to do is
f(x) = a(x) (for x < x1)
f(x) = q (for x = x1), where q = a(x1) = b(x1)
f(x) = b(x) (for x > x1)
The (half-maximum) Heaviside function is defined as
H(x) = 0 (for x < 0)
H(x) = 0.5 (for x = 0)
H(x) = 1 (for x > 0)
Hence, your function will be
f(x) = H(x1-x) * a(c) + H(x-x1) * b(x)
and, therefore,
f(x) = H(x1-x) * (m1*x+c1) + H(x-x1) * (m2x+c2)
If you want to implement this, note that many programming languages will allow you to write something like
f(x) = (x<x1)?a(x):b(x)
which means if x<x1, then return value a(x), else return b(x), or in your case:
f(x) = (x<x1)?(m1*x+c1):(m2x+c2)
Matlab implementation:
In Matlab, you can write simple functions such as
a = #(x) m1.*x+c1,
b = #(x) m2.*x+c2,
assuming that you have previously defined m1, m2, and c1, c2.
There are several ways to using/implementing the Heaviside function
If you have the Symbolic Math Toolbox for Matlab, you can directly use heaviside() as a function.
#AndrasDeak (see comments below) pointed out that you can write your own half-maximum Heaviside function H in Matlab by entering
iif = #(varargin) varargin{2 * find([varargin{1:2:end}], 1, 'first')}();
H = #(x) iif(x<0,0,x>0,1,true,0.5);
If you want a continuous function that approximates the Heaviside function, you can use a logistic function H defined as
H = #(x) 1./(1+exp(-100.*x));
Independently of your implementation of the Heaviside function H, you can, create a one-liner in the following way (I am using x1=0 for simplicity) :
a = #(x) 2.*x + 3;
b = #(x) -1.5.*x + 3;
Which allows you to write your original function as a one-liner:
f = #(x) H(-x).*a(x) + H(x).*b(x);
You can then plot this function, for example from -10 to 10 by writing plot(-10:10, f(-10:10)) you will get the plot below.
Generalization:
Imagine you have
f(x) = a(x) (for x < x1)
f(x) = q (for x = x1), where q = a(x1) = b(x1)
f(x) = b(x) (for x1 < x < x2)
f(x) = r (for x = x2), where r = b(x2) = c(x2)
f(x) = c(x) (for x2 < x < x3)
f(x) = s (for x = x2), where s = c(x3) = d(x3)
f(x) = d(x) (for x3 < x)
By multiplying Heaviside functions, you can now determine zones where specific functions will be computed.
f(x) = H(x1-x)*a(c) + H(x-x1)*H(x2-x)*b(x) + H(x-x2)*H(x3-x)*c(x) + H(x-x3)*d(x)
PS: just realized that one of the comments above talks about the Heaviside function, too. Kudos to #AndrasDeak .
Related
For example, FX = x ^ 2 + sin (x)
Just for curiosity, I don't want to use the CVX toolbox to do this.
You can check this within some interval [a,b] by checking if the second derivative is nonnegative. For this you have to define a vector of x-values, find the numerical second derivative and check whether it is not too negative:
a = 0;
b = 1;
margin = 1e-5;
point_count = 100;
f=#(x) x.^2 + sin(x);
x = linspace(a, b, point_count)
is_convex = all(diff(x, 2) > -margin);
Since this is a numerical test, you need to adjust the parameter to the properties of the function, that is if the function does wild things on a small scale we might not be able to pick it up. E.g. with the parameters above the test will falsely report the function f=#(x)sin(99.5*2*pi*x-3) as convex.
clear
syms x real
syms f(x) d(x) d1(x)
f = x^2 + sin(x)
d = diff(f,x,2)==0
d1 = diff(f,x,2)
expSolution = solve(d, x)
if size(expSolution,1) == 0
if eval(subs(d1,x,0))>0
disp("condition 1- the graph is concave upward");
else
disp("condition 2 - the graph is concave download");
end
else
disp("condition 3 -- not certain")
end
So I have three seperate function in MATLAB where each have its designated purpose.
The first one calculates the partial derivative
The second finds the roots for a system of two equations and two variables.
The third is supposed to find the critical point.
I have tried doing everything separately and it worked. However when I do this through my CriticalPoint function it just keeps running(it says "busy" in the lower left corner).
I have tried solving this manually. Meaning that I've used each function as intended, saved the values in my workspace and used for the next function.
The only thing different is how I treated my partials.
I have a function which takes evaluates the partial derivative at a point [x,y].
derivative = Pderiv(f, a, b, i)
%The i denotes if the partial derivative is take with respect to x or y
%where if i == 1, then it means that partial derivative is with respect to
%x and if i == 2 then it is with respect to y
The thing I did differently was that I found the partial derivative for f(x,y) manually using pen and paper.
I.e
df/dx and df/dy (these are partials, sorry for not using the proper symbol)
And then I inserted it into my function:
[x0,y0] = MyNewton(f, g, a, b)
Where df/dx is the argument for "f" and df/dy is the argument for "g".
This gives me the correct x & y values when df/dx=df/dy=0.
What I want to have it do is for a given function f(x,y)
I want to find the partial derivative where the input values are still represented by x and y.
I was told by my teacher that the following expression:
g=#(x,y)NPderiv(f,x,y,1);
would be sufficent. Is it though? Since it works when I do it manually but the program runs forever when I define my function g as above.
I'm not sure if it is relevant but here is my code for the partial derivative:
function derivative = NPderiv(f, a, b, i)
h = 0.0000001;
fn=zeros(1,2);
if i == 1
fn(i) = (f(a+h,b)-f(a,b))/h;
elseif i==2
fn(i) = (f(a,b+h)-f(a,b))/h;
end
derivative = fn(i);
end
function for my critical point function:
function [x,y] = CriticalPoint(f, a, b)
g=#(x,y)NPderiv(f,x,y,1); %partial derivative for f(x,y) with respect to x
z=#(x,y)NPderiv(f,x,y,2);%partial derivative for f(x,y) with respect to y
[x,y]=MyNewton(g,z,a,b);
end
MyNewton function:
function [x0,y0] = MyNewton(f, g, a, b)
y0 = b;
x0 = a;
tol = 1e-18;
resf = tol + 1;
resg = tol + 1;
while resf > tol && resg > tol
dfdx = NPderiv(f, x0, y0, 1);
dfdy = NPderiv(f, x0, y0, 2);
dgdx = NPderiv(g, x0, y0, 1);
dgdy = NPderiv(g, x0, y0, 2);
jf1 = [f(x0,y0) dfdy; ...
g(x0,y0) dgdy];
jg1 = [dfdx f(x0,y0); ...
dgdx g(x0,y0)];
j2 = [dfdx dfdy; ...
dgdx dgdy];
jx = det(jf1)./det(j2);
jy = det(jg1)./det(j2);
x0 = x0 - jx;
y0 = y0 - jy;
resf = abs(f(x0,y0));
resg = abs(g(x0,y0));
end
end
I am creating a function that takes in data (x,y,z) and an anonymous function (M) as inputs. M's inputs are data (x,y,z) and a parameter (theta).
I need to determine the dimension of the parameter theta inside FUNC
EDIT: (To add some context)
I have data that follows a known data generating process (DGP). For example, I can generate data using a linear instrumental variable DGP with 1 endogenous variable (hence theta will be dimension 1):
n = 100; q = 10;
theta0 = 1; % true param value
e = randn(n, 1); % 2nd stage error
u = randn(n, 1); % 1st stage error
z = randn(n, q); % instrument
x = z * ones(q, 1) + u; % endog variable
y = x * theta0 + e; % dependent variable
Then I want to estimate theta0 using my own variation of generalized linear methods (FUNC)
M = #(x,y,z,theta) z' * (y - x * theta); % moment condition
thetahat = FUNC(M, x, y, z); % estimate theta0
and FUNC.m is
function out = FUNC(M, x, y, z)
k = ; % (!!!) <-- this is what I need to find out!
objFunc = #(theta) M(x, y, z, theta)' * M(x, y, z, theta);
out = fminunc(objFunc, ones(1, k)); % <-- this is where its used
end
In the above example, the DGP is a linear IV model. However, I should be able to use my function for any other DGP.
Other DGPs could, for example, define M as follows:
% E.g. 1) theta is dimension 1
M=#(x,y,z,theta) z' * (y - x * theta);
% E.g. 2) theta is dimension 2
M=#(x,y,z,theta) z' * (y - (x * theta(1))^theta(2));
% E.g. 3) theta is dimension 3
M=#(x,y,z,theta) z' * (y - (theta(1) + x * theta(2))^theta(3));
The (super bad) hack that I am currently using for (!!!) is:
for ktest = [3,2,1] % the dimension of theta will never be higher than 3
try
M(x, y, z, ones(1, ktest);
k = ktest;
end
end
Since you know already what the form and requirements of your function M will be when you pass it to FUNC, it doesn't make much sense to then require FUNC to determine it based only on M. It would make much more sense to pass flag values or needed information to FUNC when you pass it M. I would write FUNC in one of two ways:
function out = FUNC(M, x, y, z, k) % Accept k as an argument
...
end
function out = FUNC(M, x, y, z, theta0) % Pass the initial guess, of the correct size
...
end
If you really want to let FUNC do the extra work, then the answer from excaza is how I would do it.
Old answer below. not really valid since the question was clarified, but I'm leaving it temporarily...
I think you have two better options here...
Make M a cell array of anonymous functions:
You could make your input M a cell array of possible anonymous functions and use the number of values in theta as the index. You would pass this M to FUNC:
M = {#(x,y,z,theta) z' * (y - x * theta), ...
#(x,y,z,theta) z' * (y - (x * theta(1))^theta(2)), ...
#(x,y,z,theta) z' * (y - (theta(1) + x * theta(2))^theta(3))};
Then somewhere inside FUNC:
out = M{numel(theta)}(x, y, z, theta);
Make M a normal function instead of an anonymous one:
An anonymous function is good for quick, simple formulas. Add in conditional logic and you should probably just make it a fully-fledged function. Here's an example with a switch statement (good for if you have a number of different formulas):
function out = M(x, y, x, theta)
switch numel(theta)
case 1
out = z' * (y - x * theta);
case 2
out = z' * (y - (x * theta(1))^theta(2));
case 3
out = z' * (y - (theta(1) + x * theta(2))^theta(3));
end
end
And here's an example that sets some defaults for parameters (good for if you have one formula with different ways to set its parameters, like you seem to have):
function out = M(x, y, x, theta)
switch numel(theta)
case 1
p1 = 0;
p2 = theta;
p3 = 1;
case 2
p1 = 0;
p2 = theta(1);
p3 = theta(2);
case 3
p1 = theta(1);
p2 = theta(2);
p3 = theta(3);
end
out = z' * (y - (p1 + x * p2)^p3);
end
MATLAB doesn't store any information about the size of the inputs to an anonymous function. While a better idea would be to modify your code so you don't have to do these kinds of gymnastics, if your function definition is known to fit a narrow band of possibilities you could use a regular expression to parse the function definition itself. You can get this string from the return of functions.
For example:
function [nelements] = findsizetheta(fh)
defstr = func2str(fh);
test = regexp(defstr, 'theta\((\d+)\)', 'tokens');
if isempty(test)
% Assume we have theta instead of theta(1)
nelements = 1;
else
nelements = max(str2double([test{:}]));
end
end
Which returns 1, 2, and 3 for your example definitions of M.
This assumes that theta is present in you anonymous function and that it is defined as a vector.
Also note that MATLAB cautions against utilizing functions in a programmatic manner, as its behavior may change in future releases. This was tested to function in R2017b.
This question is connected to this one. Suppose again the following code:
syms x
f = 1/(x^2+4*x+9)
Now taylor allows the function f to be expanded about infinity:
ts = taylor(f,x,inf,'Order',100)
But the following code
c = coeffs(ts)
produces errors, because the series does not contain positive powers of x (it contains negative powers of x).
In such a case, what code should be used?
Since the Taylor Expansion around infinity was likely performed with the substitution y = 1/x and expanded around 0, I would explicitly make that substitution to make the power positive for use on coeffs:
syms x y
f = 1/(x^2+4x+9);
ts = taylor(f,x,inf,'Order',100);
[c,ty] = coeffs(subs(ts,x,1/y),y);
tx = subs(ty,y,1/x);
The output from taylor is not a multivariate polynomial, so coeffs won't work in this case. One thing you can try is using collect (you may get the same or similar result from using simplify):
syms x
f = 1/(x^2 + 4*x + 9);
ts = series(f,x,Inf,'Order',5) % 4-th order Puiseux series of f about 0
c = collect(ts)
which returns
ts =
1/x^2 - 4/x^3 + 7/x^4 + 8/x^5 - 95/x^6
c =
(x^4 - 4*x^3 + 7*x^2 + 8*x - 95)/x^6
Then you can use numden to extract the numerator and denominator from either c or ts:
[n,d] = numden(ts)
which returns the following polynomials:
n =
x^4 - 4*x^3 + 7*x^2 + 8*x - 95
d =
x^6
coeffs can then be used on the numerator. You may find other functions listed here helpful as well.
I'm working on a problem that applies a luminosity equation:
E = 64.77* T^−4 ∫ x^−5( e^(1.432/Tx) -1 )^−1 dx
Where T = 3500;
to simp son's rule which is a few sums and such.
problem 17.8 here: http://my.safaribooksonline.com/book/computer-aided-engineering/9780123748836/-introduction-to-numerical-methods/ch17lev1sec10
What I've done is made a function simpson(fn, a, b, h) that runs simp son's rule correctly.
however, the problem is making that integral equation into a function handle that works. I've gotten it to work for simple function handles like
f = #x x.^2
but when I try and make the integral into a function:
fn = #(x)(64.77/T^4).*integral((x.^(-5)).*((exp(((1.432)./(3500.*x)))).^(-1)), 4e-5, 7e-5);
simp(fn, 5, 15, 1)
function s = simp(fn, a, b, h)
x1 = a + 2*h:2*h:b-2*h;
sum1 = sum(feval(fn, x1));
x2 = a + h:22*h:b-h;
sum2 = sum(feval(fn, x2));
s = h/3*(feval(fn, a) + feval(fn, b) + 4*sum2 + 2*sum1);
it doesn't work. error message is Integral: first input must be function handle.
Any help appreciated.
You're supposed to be evaluating the integral using Simpsons rule, whereas you are using integral to calculate the integral, then fn is not a function of x. You want to do this:
fn = #(x)(x.^(-5)).*((exp(((1.432)./(3500.*x)))).^(-1));
I=simp(fn,a,b,h);
E=(64.77/T^4)*I;