I am trying to use the fixed point iteration method with initial approximation x(1)=0 to obtain an approximation to the root of the equation f(x)=3x+sin(x)e^x=0.
The stopping criterion is
|x(k+1)-x(k)|<0.0001
x(1) = 0;
n = 100;
for k = 1:n
f(k) = 3*x(k) +sin(x(k))-exp(x(k));
if (abs(f(k))<0.0001)
break;
end
syms x
diff(f(k));
x(k+1) = x(1)- (f(k))/(diff(f(k)));
end
[x' f']
This is the error I am getting: Error using / Matrix dimensions must
agree. Error in prac2Q2 (line 15)
x(k+1) = x(1)- (f(k))/(diff(f(k)));
I would suggest to calculate the derivative by hand and use that term as denominator or to save the derivative in another variable and use this as the denominator.
Derivative as Variable
f(k) = ...;
df(k) = diff(f(k));
x(k+1) = x(k) - f(k) / df(k);
PS: I cannot test this, because I do not have access to the Symbolic Toolbox right now.
If you're looking for the root of 3*x +sin(x)-exp(x) you want to resolve this equation:
3*x + sin(x) - exp(x) = 0
The easiest way will be to isolate x in one side of the equation:
x = (exp(x) - sin(x))/3 % now iterate until x = (exp(x) - sin(x))/3
Now I would recommand to use an easier fixed point method: x(k+1) = (x(k)+f(x(k)))/2
x = 1 % x0
while 1
y = (exp(x)-sin(x))/3; % we are looking for the root not for a fixed point !!! y = f(x)
x = (x+y)/2 % after a few iterations x == y, so x = (x+y)/2 or x = 2x/2
if abs(x-y) < 1e-10
break
end
end
And you obtain the correct result:
x = 0.36042
No need of symbolic math.
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
I'm having some issues getting my RK2 algorithm to work for a certain second-order linear differential equation. I have posted my current code (with the provided parameters) below. For some reason, the value of y1 deviates from the true value by a wider margin each iteration. Any input would be greatly appreciated. Thanks!
Code:
f = #(x,y1,y2) [y2; (1+y2)/x];
a = 1;
b = 2;
alpha = 0;
beta = 1;
n = 21;
h = (b-a)/(n-1);
yexact = #(x) 2*log(x)/log(2) - x +1;
ye = yexact((a:h:b)');
s = (beta - alpha)/(b - a);
y0 = [alpha;s];
[y1, y2] = RungeKuttaTwo2D(f, a, b, h, y0);
error = abs(ye - y1);
function [y1, y2] = RungeKuttaTwo2D(f, a, b, h, y0)
n = floor((b-a)/h);
y1 = zeros(n+1,1); y2 = y1;
y1(1) = y0(1); y2(1) = y0(2);
for i=1:n-1
ti = a+(i-1)*h;
fvalue1 = f(ti,y1(i),y2(i));
k1 = h*fvalue1;
fvalue2 = f(ti+h/2,y1(i)+k1(1)/2,y2(i)+k1(2)/2);
k2 = h*fvalue2;
y1(i+1) = y1(i) + k2(1);
y2(i+1) = y2(i) + k2(2);
end
end
Your exact solution is wrong. It is possible that your differential equation is missing a minus sign.
y2'=(1+y2)/x has as its solution y2(x)=C*x-1 and as y1'=y2 then y1(x)=0.5*C*x^2-x+D.
If the sign in the y2 equation were flipped, y2'=-(1+y2)/x, one would get y2(x)=C/x-1 with integral y1(x)=C*log(x)-x+D, which contains the given exact solution.
0=y1(1) = -1+D ==> D=1
1=y1(2) = C*log(2)-1 == C=1/log(2)
Additionally, the arrays in the integration loop have length n+1, so that the loop has to be from i=1 to n. Else the last element remains zero, which gives wrong residuals for the second boundary condition.
Correcting that and enlarging the computation to one secant step finds the correct solution for the discretization, as the ODE is linear. The error to the exact solution is bounded by 0.000285, which is reasonable for a second order method with step size 0.05.
I want to write a program that makes use of Newtons Method:
To estimate the x of this integral:
Where X is the total distance.
I have functions to calculate the Time it takes to arrive at a certain distance by using the trapezoid method for numerical integration. Without using trapz.
function T = time_to_destination(x, route, n)
h=(x-0)/n;
dx = 0:h:x;
y = (1./(velocity(dx,route)));
Xk = dx(2:end)-dx(1:end-1);
Yk = y(2:end)+y(1:end-1);
T = 0.5*sum(Xk.*Yk);
end
and it fetches its values for velocity, through ppval of a cubic spline interpolation between a set of data points. Where extrapolated values should not be fetcheable.
function [v] = velocity(x, route)
load(route);
if all(x >= distance_km(1))==1 & all(x <= distance_km(end))==1
estimation = spline(distance_km, speed_kmph);
v = ppval(estimation, x);
else
error('Bad input, please choose a new value')
end
end
Plot of the velocity spline if that's interesting to you evaluated at:
dx= 1:0.1:65
Now I want to write a function that can solve for distance travelled after a certain given time, using newton's method without fzero / fsolve . But I have no idea how to solve for the upper bound of a integral.
According to the fundamental theorem of calculus I suppose the derivative of the integral is the function inside the integral, which is what I've tried to recreate as Time_to_destination / (1/velocity)
I added the constant I want to solve for to time to destination so its
(Time_to_destination - (input time)) / (1/velocity)
Not sure if I'm doing that right.
EDIT: Rewrote my code, works better now but my stopcondition for Newton Raphson doesnt seem to converge to zero. I also tried to implement the error from the trapezoid integration ( ET ) but not sure if I should bother implementing that yet. Also find the route file in the bottom.
Stop condition and error calculation of Newton's Method:
Error estimation of trapezoid:
Function x = distance(T, route)
n=180
route='test.mat'
dGuess1 = 50;
dDistance = T;
i = 1;
condition = inf;
while condition >= 1e-4 && 300 >= i
i = i + 1 ;
dGuess2 = dGuess1 - (((time_to_destination(dGuess1, route,n))-dDistance)/(1/(velocity(dGuess1, route))))
if i >= 2
ET =(time_to_destination(dGuess1, route, n/2) - time_to_destination(dGuess1, route, n))/3;
condition = abs(dGuess2 - dGuess1)+ abs(ET);
end
dGuess1 = dGuess2;
end
x = dGuess2
Route file: https://drive.google.com/open?id=18GBhlkh5ZND1Ejh0Muyt1aMyK4E2XL3C
Observe that the Newton-Raphson method determines the roots of the function. I.e. you need to have a function f(x) such that f(x)=0 at the desired solution.
In this case you can define f as
f(x) = Time(x) - t
where t is the desired time. Then by the second fundamental theorem of calculus
f'(x) = 1/Velocity(x)
With these functions defined the implementation becomes quite straightforward!
First, we define a simple Newton-Raphson function which takes anonymous functions as arguments (f and f') as well as an initial guess x0.
function x = newton_method(f, df, x0)
MAX_ITER = 100;
EPSILON = 1e-5;
x = x0;
fx = f(x);
iter = 0;
while abs(fx) > EPSILON && iter <= MAX_ITER
x = x - fx / df(x);
fx = f(x);
iter = iter + 1;
end
end
Then we can invoke our function as follows
t_given = 0.3; % e.g. we want to determine distance after 0.3 hours.
n = 180;
route = 'test.mat';
f = #(x) time_to_destination(x, route, n) - t_given;
df = #(x) 1/velocity(x, route);
distance_guess = 50;
distance = newton_method(f, df, distance_guess);
Result
>> distance
distance = 25.5877
Also, I rewrote your time_to_destination and velocity functions as follows. This version of time_to_destination uses all the available data to make a more accurate estimate of the integral. Using these functions the method seems to converge faster.
function t = time_to_destination(x, d, v)
% x is scalar value of destination distance
% d and v are arrays containing measured distance and velocity
% Assumes d is strictly increasing and d(1) <= x <= d(end)
idx = d < x;
if ~any(idx)
t = 0;
return;
end
v1 = interp1(d, v, x);
t = trapz([d(idx); x], 1./[v(idx); v1]);
end
function v = velocity(x, d, v)
v = interp1(d, v, x);
end
Using these new functions requires that the definitions of the anonymous functions are changed slightly.
t_given = 0.3; % e.g. we want to determine distance after 0.3 hours.
load('test.mat');
f = #(x) time_to_destination(x, distance_km, speed_kmph) - t_given;
df = #(x) 1/velocity(x, distance_km, speed_kmph);
distance_guess = 50;
distance = newton_method(f, df, distance_guess);
Because the integral is estimated more accurately the solution is slightly different
>> distance
distance = 25.7771
Edit
The updated stopping condition can be implemented as a slight modification to the newton_method function. We shouldn't expect the trapezoid rule error to go to zero so I omit that.
function x = newton_method(f, df, x0)
MAX_ITER = 100;
TOL = 1e-5;
x = x0;
iter = 0;
dx = inf;
while dx > TOL && iter <= MAX_ITER
x_prev = x;
x = x - f(x) / df(x);
dx = abs(x - x_prev);
iter = iter + 1;
end
end
To check our answer we can plot the time vs. distance and make sure our estimate falls on the curve.
...
distance = newton_method(f, df, distance_guess);
load('test.mat');
t = zeros(size(distance_km));
for idx = 1:numel(distance_km)
t(idx) = time_to_destination(distance_km(idx), distance_km, speed_kmph);
end
plot(t, distance_km); hold on;
plot([t(1) t(end)], [distance distance], 'r');
plot([t_given t_given], [distance_km(1) distance_km(end)], 'r');
xlabel('time');
ylabel('distance');
axis tight;
One of the main issues with my code was that n was too low, the error of the trapezoidal sum, estimation of my integral, was too high for the newton raphson method to converge to a very small number.
Here was my final code for this problem:
function x = distance(T, route)
load(route)
n=10e6;
x = mean(distance_km);
i = 1;
maxiter=100;
tol= 5e-4;
condition=inf
fx = #(x) time_to_destination(x, route,n);
dfx = #(x) 1./velocity(x, route);
while condition > tol && i <= maxiter
i = i + 1 ;
Guess2 = x - ((fx(x) - T)/(dfx(x)))
condition = abs(Guess2 - x)
x = Guess2;
end
end
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 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);