I am trying to code Horner’s Method for Polynomial Evaluation but for some reason its not working for me and I'm not sure where I am getting it wrong.
These are the data I have:
nodes = [-2, -1, 1]
x = 2
c (coefficients) = [-3, 3, -1]
The code I have so far is:
function y = horner(x, nodes, c)
n = length(c);
y = c(1);
for i = 2:n
y = y * ((x - nodes(i - 1)) + c(i));
end
end
I am supposed to end up with a polynomial such as (−1)·(x+2)(x+1)+3·(x+2)−3·1 and if x =2 then I am supposed to get -3. But for some reason I don't know where I am going wrong.
Edit:
So I changed my code. I think it works but I am not sure:
function y = horner(x, nodes, c)
n = length(c);
y = c(n);
for k = n-1:-1:1
y = c(k) + y * (x - nodes((n - k) + 1));
end
end
This works:
function y = horner(x, nodes, c)
n = length(c);
y = 0;
for i = 1:n % We iterate over `c`
tmp = c(i);
for j = 1:i-1 % We iterate over the relevant elements of `nodes`
tmp *= x - nodes(j); % We multiply `c(i) * (x - nodes(1)) * (x -nodes(2)) * (x- nodes(3)) * ... * (x - nodes(i -1))
end
y += tmp; % We added each product to y
end
% Here `y` is as following:
% c(1) + c(2) * (x - nodes(1)) + c(3) * (x - nodes(1)) * (x - nodes(2)) + ... + c(n) * (x - nodes(1)) * ... * (x - nodes(n - 1))
end
(I'm sorry this isn't python but I don't know python)
In the case where we didn't have nodes, horner's method works like this:
p = c[n]
for i=n-1 .. 1
p = x*p + c[i]
for example for a quadratic (with coeffs a,b,c) this is
p = x*(x*a+b)+c
Note that if your language supports fma
fma(x,y,x) = x*y+z
then horner's method can be written
p = c[n]
for i=n-1 .. 1
p = fma( x, p, c[i])
When you do have nodes, the change is simple:
p = c[n]
for i=n-1 .. 1
p = (x-nodes[i])*p + c[i]
Or, using fma
p = c[n]
for i=n-1 .. 1
p = fma( (x-nodes[i]), p, c[i])
For the quadratic above this leads to
p = (x-nodes[1]*((x-nodes[2])*a+b)+c
I want to solve a system of THREE differential equations with the Runge Kutta 4 method in Matlab (Ode45 is not permitted).
After a long time spent looking, all I have been able to find online are either unintelligible examples or general explanations that do not include examples at all. I would like a concrete example on how to implement my solution properly, or the solution to a comparable problem which I can build on.
I have come quite far; my current code spits out a matrix with 2 correct decimals on most of the components, which I am quite happy with.
However, when the step-size is decreased, the errors become enormous. I know the for-loop I have created is not entirely correct. I may have defined the functions incorrectly, but I am quite certain that the problem is solved if some minor changes are made to the for-loop because it appears to be solving the equation-system fairly well already in its current state.
clear all, close all, clc
%{
____________________TASK:______________________
Solve the system of differential equations below
in the interval 0<t<1, with stepsize h = 0.1.
x'= y x(0)=1
y'= -x-2e^t+1 y(0)=0 , where x=x(t), y=y(t), z=z(t)
z'= -x - e^t + 1 z(0)=1
THE EXACT SOLUTIONS for x y and z can be found in this pdf:
archives.math.utk.edu/ICTCM/VOL16/C029/paper.pdf
_______________________________________________
%}
h = 0.1;
t = 0:h:1
N = length(t);
%Defining the functions
x = zeros(N,1);%I am not entierly sure if x y z are supposed to be defined in this way.
y = zeros(N,1)
z = zeros(N,1)
f = #(t, x, y, z) -x-2*exp(t)+1;%Question: Do i need a function for x here as well??
g = #(t, x, y, z) -x - exp(t) + 1;
%Starting conditions
x(1) = 1;
y(1) = 0;
z(1) = 1;
for i = 1:(N-1)
K1 = h * ( y(i));%____I think z(i) is supposed to be here, but i dont know in what way.
L1 = h * f( t(i) , x(i) , y(i) , z(i));
M1 = h * g( t(i) , x(i) , y(i) , z(i));
K2 = h * (y(i) + 1/2*L1 + 1/2*M1);%____Again, z(i) should probably be here somewhere.
L2 = h * f(t(i) + 1/2*h, x(i)+1/2*K1 , y(i)+1/2*L1 , z(i)+1/2*M1);
M2 = h * g(t(i) + 1/2*h, x(i)+1/2*K1 , y(i)+1/2*L1 , z(i)+1/2*M1);
K3 = h * (y(i) + 1/2*L2 + 1/2*M2);%____z(i). Should it just be added, like "+z(i)" ?
L3 = h * f(t(i) + 1/2*h, x(i) + 1/2*K2 , y(i) + 1/2*L2 , z(i) + 1/2*M2);
M3 = h * g(t(i) + 1/2*h, x(i) + 1/2*K2 , y(i) + 1/2*L2 , z(i) + 1/2*M2);
K4 = h * (y(i) + L3 + M3);%_____z(i) ... ?
L4 = h * f( t(i)+h , x(i)+K3 , y(i)+L3, z(i)+M3);
M4 = h * g( t(i)+h , x(i)+K3 , y(i)+L3, z(i)+M3);
x(i+1) = x(i)+1/6*(K1+2*K2+2*K3+K4);
y(i+1) = y(i)+1/6*(L1+2*L2+2*L3+L4);
z(i+1) = z(i)+1/6*(M1+2*M2+2*M3+M4);
end
Answer_Matrix = [t' x y z]
So your main issue was not defining x properly. You were propagating its value using the Runge Kutta 4 (RK4) method, but never actually defined what its derivative was!
At the bottom of this answer is a function which can take any given number of equations and their initial conditions. This has been included to address your need for a clear example for three (or more) equations.
For reference, the equations can be directly lifted from the standard RK4 method described here.
Working Script
This is comparable to yours, but uses slightly clearer naming conventions and structure.
% Initialise step-size variables
h = 0.1;
t = (0:h:1)';
N = length(t);
% Initialise vectors
x = zeros(N,1); y = zeros(N,1); z = zeros(N,1);
% Starting conditions
x(1) = 1; y(1) = 0; z(1) = 1;
% Initialise derivative functions
dx = #(t, x, y, z) y; % dx = x' = dx/dt
dy = #(t, x, y, z) - x -2*exp(t) + 1; % dy = y' = dy/dt
dz = #(t, x, y, z) - x - exp(t) + 1; % dz = z' = dz/dt
% Initialise K vectors
kx = zeros(1,4); % to store K values for x
ky = zeros(1,4); % to store K values for y
kz = zeros(1,4); % to store K values for z
b = [1 2 2 1]; % RK4 coefficients
% Iterate, computing each K value in turn, then the i+1 step values
for i = 1:(N-1)
kx(1) = dx(t(i), x(i), y(i), z(i));
ky(1) = dy(t(i), x(i), y(i), z(i));
kz(1) = dz(t(i), x(i), y(i), z(i));
kx(2) = dx(t(i) + (h/2), x(i) + (h/2)*kx(1), y(i) + (h/2)*ky(1), z(i) + (h/2)*kz(1));
ky(2) = dy(t(i) + (h/2), x(i) + (h/2)*kx(1), y(i) + (h/2)*ky(1), z(i) + (h/2)*kz(1));
kz(2) = dz(t(i) + (h/2), x(i) + (h/2)*kx(1), y(i) + (h/2)*ky(1), z(i) + (h/2)*kz(1));
kx(3) = dx(t(i) + (h/2), x(i) + (h/2)*kx(2), y(i) + (h/2)*ky(2), z(i) + (h/2)*kz(2));
ky(3) = dy(t(i) + (h/2), x(i) + (h/2)*kx(2), y(i) + (h/2)*ky(2), z(i) + (h/2)*kz(2));
kz(3) = dz(t(i) + (h/2), x(i) + (h/2)*kx(2), y(i) + (h/2)*ky(2), z(i) + (h/2)*kz(2));
kx(4) = dx(t(i) + h, x(i) + h*kx(3), y(i) + h*ky(3), z(i) + h*kz(3));
ky(4) = dy(t(i) + h, x(i) + h*kx(3), y(i) + h*ky(3), z(i) + h*kz(3));
kz(4) = dz(t(i) + h, x(i) + h*kx(3), y(i) + h*ky(3), z(i) + h*kz(3));
x(i+1) = x(i) + (h/6)*sum(b.*kx);
y(i+1) = y(i) + (h/6)*sum(b.*ky);
z(i+1) = z(i) + (h/6)*sum(b.*kz);
end
% Group together in one solution matrix
txyz = [t,x,y,z];
Implemented as function
You wanted code which can "be applied to any equation system". To make your script more usable, let's take advantage of vector inputs, where each variable is on its own row, and then make it into a function. The result is something comparable (in how it is called) to Matlab's own ode45.
% setup
odefun = #(t, y) [y(2); -y(1) - 2*exp(t) + 1; -y(1) - exp(t) + 1];
y0 = [1;0;1];
% ODE45 solution
[T, Y] = ode45(odefun, [0,1], y0);
% Custom RK4 solution
t = 0:0.1:1;
y = RK4(odefun, t, y0);
% Compare results
figure; hold on;
plot(T, Y); plot(t, y, '--', 'linewidth', 2)
You can see that the RK4 function (below) gives the same result of the ode45 function.
The function RK4 is simply a "condensed" version of the above script, it will work for however many equations you want to use. For broad use, you would want to include input-checking in the function. I have left this out for clarity.
function y = RK4(odefun, tspan, y0)
% ODEFUN contains the ode functions of the system
% TSPAN is a 1D vector of equally spaced t values
% Y0 contains the intial conditions for the system variables
% Initialise step-size variables
t = tspan(:); % ensure column vector = (0:h:1)';
h = t(2)-t(1);% define h from t
N = length(t);
% Initialise y vector, with a column for each equation in odefun
y = zeros(N, numel(y0));
% Starting conditions
y(1, :) = y0(:)'; % Set intial conditions using row vector of y0
k = zeros(4, numel(y0)); % Initialise K vectors
b = repmat([1 2 2 1]', 1, numel(y0)); % RK4 coefficients
% Iterate, computing each K value in turn, then the i+1 step values
for i = 1:(N-1)
k(1, :) = odefun(t(i), y(i,:));
k(2, :) = odefun(t(i) + (h/2), y(i,:) + (h/2)*k(1,:));
k(3, :) = odefun(t(i) + (h/2), y(i,:) + (h/2)*k(2,:));
k(4, :) = odefun(t(i) + h, y(i,:) + h*k(3,:));
y(i+1, :) = y(i, :) + (h/6)*sum(b.*k);
end
end
Ok, turns out it was just a minor mistake where the x-variable was not defined as a function of y (as x'(t)=y according to the problem.
So: Below is a concrete example on how to solve a differential equation system using Runge Kutta 4 in matlab:
clear all, close all, clc
%{
____________________TASK:______________________
Solve the system of differential equations below
in the interval 0<t<1, with stepsize h = 0.1.
x'= y x(0)=1
y'= -x-2e^t+1 y(0)=0 , where x=x(t), y=y(t), z=z(t)
z'= -x - e^t + 1 z(0)=1
THE EXACT SOLUTIONS for x y and z can be found in this pdf:
archives.math.utk.edu/ICTCM/VOL16/C029/paper.pdf
_______________________________________________
%}
%Step-size
h = 0.1;
t = 0:h:1
N = length(t);
%Defining the vectors where the answer is stored.
x = zeros(N,1);
y = zeros(N,1)
z = zeros(N,1)
%Defining the functions
e = #(t, x, y, z) y;
f = #(t, x, y, z) -x-2*exp(t)+1;
g = #(t, x, y, z) -x - exp(t) + 1;
%Starting/initial conditions
x(1) = 1;
y(1) = 0;
z(1) = 1;
for i = 1:(N-1)
K1 = h * e( t(i) , x(i) , y(i) , z(i));
L1 = h * f( t(i) , x(i) , y(i) , z(i));
M1 = h * g( t(i) , x(i) , y(i) , z(i));
K2 = h * e(t(i) + 1/2*h, x(i)+1/2*K1 , y(i)+1/2*L1 , z(i)+1/2*M1);
L2 = h * f(t(i) + 1/2*h, x(i)+1/2*K1 , y(i)+1/2*L1 , z(i)+1/2*M1);
M2 = h * g(t(i) + 1/2*h, x(i)+1/2*K1 , y(i)+1/2*L1 , z(i)+1/2*M1);
K3 = h * e(t(i) + 1/2*h, x(i) + 1/2*K2 , y(i) + 1/2*L2 , z(i) + 1/2*M2);
L3 = h * f(t(i) + 1/2*h, x(i) + 1/2*K2 , y(i) + 1/2*L2 , z(i) + 1/2*M2);
M3 = h * g(t(i) + 1/2*h, x(i) + 1/2*K2 , y(i) + 1/2*L2 , z(i) + 1/2*M2);
K4 = h * e( t(i)+h , x(i)+K3 , y(i)+L3, z(i)+M3);
L4 = h * f( t(i)+h , x(i)+K3 , y(i)+L3, z(i)+M3);
M4 = h * g( t(i)+h , x(i)+K3 , y(i)+L3, z(i)+M3);
x(i+1) = x(i)+1/6*(K1+2*K2+2*K3+K4);
y(i+1) = y(i)+1/6*(L1+2*L2+2*L3+L4);
z(i+1) = z(i)+1/6*(M1+2*M2+2*M3+M4);
end
Answer_Matrix = [t' x y z]
I want to calculate the integral of the following function using int in MATLAB. I tried the following way, but it always shows the error Input arguments must be 'double':
syms x
int(exp(x) .* 1000 .* square(1000 .* (x -1)))
or
syms x
int(exp(x) * 1000 * square(1000 * (x -1)), x, -1000, 1000)
or
syms x
int(exp(x) * 1000 * square(1000 * (x -1)), x, -1000.0, 1000.0)
My function definitions are:
function y = step(t)
y = (1 + sign(t))/2;
end
function y = square(t)
y = step(t + 0.5) - step(t - 0.5);
end
You need to declare your functions to be symbolic:
syms square
syms step2
Why step2? Because step is an already defined function of a common toolbox.
MWE
function test
syms x z
syms square
syms step2
z1 = int(exp(x) .* 1000 .* square(1000 .* (x -1)))
z2 = int(exp(x) * 1000 * square(1000 * (x -1)), x, -1000, 1000)
z3 = int(exp(x) * 1000 * square(1000 * (x -1)), x, -1000.0, 1000.0)
end
function y=step2(t)
y = (1 + sign(t))/2;
end
function y = square(t)
y = step2(t + 0.5) - step2(t - 0.5);
end
z1 = 500*exp(x)*(sign(1000*x - 1999/2) - sign(1000*x - 2001/2))
z2 = 1000*exp(1999/2000)*(exp(1/1000) - 1)
z3 = 1000*exp(1999/2000)*(exp(1/1000) - 1)
I have a symbolic function that looks like this
syms x y(x) h
fn(x) = y + (h^2*(diff(y(x), x) + 2))/2 + (h^5*diff(y(x), x, x, x, x))/120 + (h^3*diff(y(x), x, x))/6 + (h^4*diff(y(x), x, x, x))/24 + h*(2*x + y(x) - 1)
I'd like to replace all instances of derivatives of y with its first derivative, i.e.
subs(fn, sym('diff(y(x), x)'), dy)
where dy is already defined as
dy(x) = 2*x + y(x) - 1
The result is the following:
ans(x) =
y + (h^2*(2*x + y(x) + 1))/2 + (h^5*diff(y(x), x, x, x, x))/120 + (h^3*diff(y(x), x, x))/6 + (h^4*diff(y(x), x, x, x))/24 + h*(2*x + y(x) - 1)
It replaces the first derivative, but not the higher derivatives. What I want is for the h^5 term to have (h^5*diff(dy(x), x, x, x). Is there any way to do that?
My current method is pretty hackish and involves converting the sym to a string, replacing first derivatives with dy, then converting back to a sym and evaluating to reduce the order of each term of the series by one, but it has to be recursive because at each stage derivatives of dy are then replaced by something containing diff(y, ...). I was hoping there would be a cleaner way to deal with this.
You need to keep in mind that Matlab treats things like diff(y,x) and diff(y,x,2) as distinct variables. It doesn't know how to substitute diff(y,x) into diff(y,x,2) because such a general operation for an abstract function (one with out an explicit definition, e.g., y(x)) is ill-defined.
How about something like this that performs substitution from the opposite end, starting with the highest order derivatives:
syms y(x) h
dy(x) = 2*x + y - 1
fn(x) = y + (h^2*(diff(y, x) + 2))/2 + (h^5*diff(y, x, 4))/120 + (h^3*diff(y, x, 2))/6 + (h^4*diff(y, x, 3))/24 + h*(2*x + y - 1)
fn2 = subs(fn, diff(y, x, 4), diff(dy, x, 3));
fn2 = subs(fn2, diff(y, x, 3), diff(dy, x, 2));
fn2 = subs(fn2, diff(y, x, 2), diff(dy, x));
fn2 = subs(fn2, diff(y, x), dy);
This returns
fn2(x) =
y(x) + (h^2*(2*x + y(x) + 1))/2 + (h^3*(2*x + y(x) + 1))/6 + (h^4*(2*x + y(x) + 1))/24 + (h^5*(2*x + y(x) + 1))/120 + h*(2*x + y(x) - 1)
Or you can leave dy(x) as an abstract symbolic expression initially:
syms y(x) dy(x) h
fn(x) = y + (h^2*(diff(y, x) + 2))/2 + (h^5*diff(y, x, 4))/120 + (h^3*diff(y, x, 2))/6 + (h^4*diff(y, x, 3))/24 + h*(2*x + y - 1)
fn2 = subs(fn, diff(y, x, 4), diff(dy, x, 3));
fn2 = subs(fn2, diff(y, x, 3), diff(dy, x, 2));
fn2 = subs(fn2, diff(y, x, 2), diff(dy, x));
fn2 = subs(fn2, diff(y, x), dy)
which returns
fn2(x) =
y(x) + (h^4*diff(dy(x), x, x))/24 + (h^2*(dy(x) + 2))/2 + (h^5*diff(dy(x), x, x, x))/120 + (h^3*diff(dy(x), x))/6 + h*(2*x + y(x) - 1)