I am struggling to convert a function, which I require for my civil engineering project, from symbolic expression. I need to use fzero to find the root of the function. Here H should be the variable and I need to find out the value of H. The function goes like
function x_c = f_x_c(s,H0,VA,Lo,qc,EAo,NF,Sj,Fj)
if (s < 0) || (s > Lo)
disp('The value of s is invalid')
disp(['s = ' num2str(s)]);
return
end
C1 = H/qc;
if NF == 0
n = 0;
sn = 0;
sum_Fj = 0;
end
if NF >= 1
Sj_Q = [0; Sj; Lo];
%Determine n and sn if 0 <= s < Lo:
if s < Lo
STOP = 0;
k = 0;
while STOP == 0
k = k + 1;
if (s >= Sj_Q(k,1)) && (s < Sj_Q((k + 1),1))
STOP = 1;
end
end
n = k - 1;
sn = Sj_Q(k,1);
end
%Determine n and sn if s = Lo:
if s == Lo
n = NF;
sn = Sj(NF,1);
end
sum_Fj = sum(Fj(1:n,1));
end
x_c = (H/EAo)*s;
x_c = x_c + C1*asinh((qc*s - VA + sum_Fj)/H) + ...
- C1*asinh((qc*sn - VA + sum_Fj)/H);
for j = 1:n
sk = Sj_Q((j + 1),1);
sk_1 = Sj_Q(j,1);
sum_Fj = sum(Fj(1:(j - 1)));
x_c = x_c + ...
+ C1*asinh((qc*sk - VA + sum_Fj)/H) + ...
- C1*asinh((qc*sk_1 - VA + sum_Fj)/H);
end
I want to use this f_x_c.m file in the main file where I will find the roots of this equation.
Could someone guide me how I can do that?
I have tried doing it using the following code but I wasn't successful.
if (s < 0) || (s > Lo)
disp('The value of s is invalid')
disp(['s = ' num2str(s)]);
return
end
C1 = #(H) (H/qc);
if NF == 0
n = 0;
sn = 0;
sum_Fj = 0;
end
if NF >= 1
Sj_Q = [0; Sj; Lo];
%Determine n and sn if 0 <= s < Lo:
if s < Lo
STOP = 0;
k = 0;
while STOP == 0
k = k + 1;
if (s >= Sj_Q(k,1)) && (s < Sj_Q((k + 1),1))
STOP = 1;
end
end
n = k - 1;
sn = Sj_Q(k,1);
end
%Determine n and sn if s = Lo:
if s == Lo
n = NF;
sn = Sj(NF,1);
end
sum_Fj = sum(Fj(1:n,1));
end
x_c =#(H) (H/EAo)*s;
x_c =#(H) (x_c(H) + (C1(H))*asinh((qc*s - VA + sum_Fj)/H) + ...
- (C1(H))*asinh((qc*sn - VA + sum_Fj)/H));
for j = 1:n
sk = Sj_Q((j + 1),1);
sk_1 = Sj_Q(j,1);
sum_Fj = sum(Fj(1:(j - 1)));
x_c =#(H) (x_c(H) + ...
+ C1(H)*asinh((qc*sk - VA + sum_Fj)/H) + ...
- C1(H)*asinh((qc*sk_1 - VA + sum_Fj)/H));
end
Edit:
I want to solve the following equation in the main file:
equation = f_x_c(inext_length, H0, vertical_reaction, inext_length, qc, EAo, NF, hanger_arc_length, point_hanger_force) + 1400;
% Whatever equation f_x_c returns, I have to add another number to it(like here it is 1400), then solve this equation using fzero.
So, in the main file, I wrote like:
equation = #(H) f_x_c(inext_length, H0, vertical_reaction, inext_length, qc, EAo, NF, hanger_arc_length, point_hanger_force);
equation = #(H) (equation(H) + 1400);
answer = fsolve(equation, H0);
A mock answer to your question probably looks like
function x_c = f_x_c(H,A,B,C,D)
x_c = H*A;
x_c = x_c + B*asinh(C/H) - B*asinh(D/H);
end
and a call to solver is
H = fzero(#(H)(f_x_c(H,1,1,1,1)+1400),1);
Related
I want the function P to look like this:
-1 + 0.6366*(x+pi/2) + (-0.000)*(x + pi/2)*(x)
and right now it looks like this
(5734161139222659*x)/9007199254740992 + (5734161139222659*pi)/18014398509481984 - (8131029572207409*x*(x + pi/2))/324518553658426726783156020576256 - 1.
How to convert S array so that the values are not symbolic?
syms P x
f = sin(x);
f = matlabFunction(f);
X = [-pi/2, 0, pi/2];
Y = f(sym(X));
P = MetN(X,Y,x)
P = matlabFunction(P);
function [P] = MetN(X,Y,x)
n = length(X);
for i = 1:n
A(i,1) = 1;
end
for i = 2:n
for j = 2: n
if i >= j
produs = 1;
for k =1:j-1
produs = produs * (X(i) - X(k));
end
A(i,j) = produs;
end
end
end
S = SubsAsc(A, Y);
S = double(S);
disp(S);
sym produs
P = double(sym(S(1)));
for i = 2:n
produs = 1;
for j = 1:i-1
produs = produs * (x - sym(X(j)));
end
disp(produs);
P = P + double(S(i))*produs;
end
end
function [x] = SubsAsc(A,b)
n = length(b);
x(1) = (1/A(1,1))*b(1);
for k = 2:n
s = 0;
for j = 1:k-1
s = s + A(k,j)*x(j);
end
x(k) = (1/A(k,k))*(b(k)-s);
end
end
The output you currently have is because symbolic uses exact arithmetic, so it outputs it as a rational number (hence the ugly fraction).
To have it output P using decimals, use vpa(). For instance output P using decimals to 5 significant digits
>> vpa(P, 5)
ans =
0.63662*x - 2.5056e-17*x*(x + 1.5708)
This will, however, also round pi, so you can't really have the best of both worlds here.
I'm trying to write a matlab program to calculate an integral by means of trapezoidal and simpsons rule. The program for trapezoidal is as follows:
function [int, flag, stats] = trapComp(f, a, b, tol, hMin)
% Initialise variables
h = b - a;
n = 1;
int = h / 2 * (f(a) + f(b));
flag = 1;
if nargout == 3
stats = struct('totalErEst', [], 'totalNrIntervals', [], 'nodesList', []);
end
while h > hMin
h = h / 2;
n = 2 * n;
if h < eps % Check if h is not "zero"
break;
end
% Update the integral with the new nodes
intNew = int / 2;
for j = 1 : 2 : n
intNew = intNew + h * f(a + j * h);
end
% Estimate the error
errorEst = 1 / 3 * (int - intNew);
int = intNew;
if nargout == 3 % Update stats
stats.totalErEst = [stats.totalErEst; abs(errorEst)];
stats.totalNrIntervals = [stats.totalNrIntervals; n / 2];
end
if abs(errorEst) < tol
flag = 0;
break
end
end
end
Now simpsons rule I cant really quite get around. I know its very similar but I cant seem to figure it out.
This is my simpson code:
function [int, flag, stats] = simpComp(f, a, b, tol, hMin)
% Initialise variables
h = b - a;
n = 1;
int = h / 3 * (f(a) + 4 * f((a+b)/2) + f(b));
flag = 1;
if nargout == 3
stats = struct('totalErEst', [], 'totalNrIntervals', [], 'nodesList', []);
end
while h > hMin
h = h / 2;
n = 2 * n;
if h < eps % Check if h is not "zero"
break;
end
% Update the integral with the new nodes
intNew = int / 2;
for j = 1 : 2 : n
intNew = intNew + h * f(a + j * h);
end
% Estimate the error
errorEst = 1 / 3 * (int - intNew);
int = intNew;
if nargout == 3 % Update stats
stats.totalErEst = [stats.totalErEst; abs(errorEst)];
stats.totalNrIntervals = [stats.totalNrIntervals; n / 2];
end
if abs(errorEst) < tol
flag = 0;
break
end
end
end
Using this, however, gives an answer for an integral with a larger error than trapezoidal which i feel it shouldnt.
Any help would be appreciated
I'd like to simulate the following function:
l(t) = mu + Σ (1 + (t-t_i)/alpha)^(beta)
I wrote the function as follow:
function[l] = custom(m,t,H,par)
k = length(H);
n = length(t);
l = par.mu(m)*ones(n,1);
for i = 1:n
for j = 1:k
h = H{j};
h = h(h < t(i));
if ~isempty(h)
d = t(i) - h;
l(i) = l(i) + sum((1+ (d/par.alpha(m,j)))^ par.beta(m,j));
end
end
end
Now, how can I simulate this function for t=1000, par.mu= 0.5, par.alpha = 0.1, par.beta = 0.3?
I have the following code for simulation of this function but it's not efficient...How can I make it better?
function[h] = simcustom(t,par)
h = -log(rand)/par.mu;
if h < t
do5 = 1;
i = 0;
j = 0;
k = 0;
n = 1;
while true
if do5;
k = k + 1;
Lstar(k) = custom(1,h(n),{h},par); %#ok
end
j = j + 1;
U(j) = rand; %#ok
i = i + 1;
u(i) = -log(U(j))/Lstar(k); %#ok
if i == 1
Tstar(i) = u(i); %#ok
else
Tstar(i) = Tstar(i-1) + u(i);
end
if Tstar(i) > t
h = {h};
break
end
j = j + 1;
U(j) = rand;
if U(j) <= custom(1,Tstar(i),{h},par)/Lstar(k)
n = n + 1;
h = [h Tstar(i)];
do5 = true;
else
k = k + 1;
Lstar(k) = custom(1,Tstar(i),{h},par);
do5 = false;
end
end
else
h = {[]};
end
And here is the application:
par2.mu = 0.5;
par2.alpha = 0.1;
par2.beta = 0.3;
t = 1000;
h2 = simcustom(t,par2);
I'm trying to implement an inversion counter in MATLAB using MergeSort, but for some reason, some of the answers are way off. For example, the number of inversions in [3, 4, 8, 1] is 3, but I'm getting 2. However, the array is being sorted correctly, so I think that the way that I'm counting the split inversions is the problem.
Here's my code:
function [count, sorted] = mergesort(A)
% count is the number of inversions; sorted is the sorted array.
n = length(A);
if n == 1
count = 0;
sorted = A;
else
m = ceil(n/2);
[count1, sorted1] = mergesort(A(1:m));
[count2, sorted2] = mergesort(A(m+1:n));
[crosscount, sorted] = merge(sorted1, sorted2);
count = count1 + count2 + crosscount;
end
end
function [crosscount, z] = merge(x, y)
n = length(x); m = length(y); z = zeros(1, n+m);
ix = 1;
iy = 1;
crosscount = 0;
for iz = 1:(n+m);
if ix > n
z(iz) = y(iy);
iy = iy + 1;
elseif iy > m
z(iz) = x(ix);
ix = ix + 1;
crosscount = crosscount + (n + 1 - ix); %this might be wrong
elseif x(ix) <= y(iy)
z(iz) = x(ix);
ix = ix + 1;
elseif x(ix) > y(iy)
z(iz) = y(iy);
iy = iy + 1;
crosscount = crosscount + 1; %im pretty sure this is right
end
end
end
Alright, so a friend helped me figure it out. My intuition was correct, but I needed help from an actual programmer to understand where I went wrong:
elseif iy > m
z(iz) = x(ix);
ix = ix + 1;
crosscount = crosscount + (n + 1 - ix); **%this might be wrong - this is actually wrong, since you don't want to count if you're traversing beyond the edge of the right array, and since the actual counting is happening in the last if case**
elseif x(ix) <= y(iy)
z(iz) = x(ix);
ix = ix + 1;
elseif x(ix) > y(iy)
z(iz) = y(iy);
iy = iy + 1;
crosscount = crosscount + **(n + 1 - ix)** **this is right because you're counting the remaining numbers in your left array that are also greater than y(iy) and just adding that to your count**
end
I'm trying to make a B-Spline Function
first i set the variables and made the Knot vector
# cmpp.m
% Set variables
k = 3; % (8 mod 2 + 2 + 1)
p = k - 1; % Order = 2
n = 2*k - 1; % Control points = 5
l = n + p + 1; % Vector size n + p + 1 = 8
% Create the Knot vector
% Kv = [0 0 0 1 2 3 3 3] size = 8
knoten = 0; % set all knots to 0
Kv = [];
for j=1:1:l
Kv = [ Kv 0 ];
end
for i=1:1:l
if (i > n)
if (i <= n)
Kv(i) = knoten + 1;
knoten = knoten + 1;
else
Kv(i) = knoten;
end
else
Kv(i) = knoten;
end
end
then i worte a function to create the basic function
# f.m
function N = f(N,t,i,k,u,x,s)
if (u < x)
N(i,k) = ((((u-t(i)).*f(t,i,k-1,u+s,x,s)) / (t(i+k-1) - t(i))) + (((t(i+k)-u).*f(t,i+1,k-1,u+s,x,s)) / (t(i+k) - t(i+1))));
if ((u >= t(i)) && (u < t(i+1)))
N(i,1) = 1;
else
N(i,1) = 0;
end
end
end
and called it in cmpp.m
# cmpp.m
...
...
...
N = zeros(l,k);
x = (n+1) - (k-1);
s = 1;
N = [N f(N,Kv,1,k,0,x,s)];
but i get always this error in Matlab
>> cmpp
Subscripted assignment dimension mismatch.
Error in f (line 3)
N(i,k) = ((((u-t(i)).*f(t,i,k-1,u+s,x,s)) / (t(i+k-1) - t(i))) +
(((t(i+k)-u).*f(t,i+1,k-1,u+s,x,s)) / (t(i+k) - t(i+1))));
Error in cmpp (line 32)
N = [N f(N,Kv,1,k,0,x,s)]