I figured out I can write a function to calculate the probability of at least two share the same birthday:
function retval = bday (total, n)
retval= 1- prod((total - n + 1: total)/total)
endfunction
octave:62> bday(365, 23)
retval = 0.5073
ans = 0.5073
Now if I hope to plot a 3d chart using mesh, where total is a vector 1000:100:2000, and n is another vector 20:5:40. How should I do it?
Related
Write an octave function to implement f(x) = sin(3x)/(0.4+(x-2)^2).
Write an octave script to interpolate between the values of f(x) = sin(3x)/(0.4+(x-2)^2) sampled uniformly at up to 9 points in the interval x = [0,4].
I'm confused as to what this question is asking. I interpreted the 1st part as defining a function fx that can be called from anywhere to return the values of f(x) for a given x, but I'm not sure if the x's have to be inputs.
For the 2nd part, am I correct in using the interpl function?
My attempt:
Function file fx.m
function fx
x=(0:0.25:4);
y = sin(3*x)/(0.4+(x-2))^2
endfunction
But this only returns 1 value for y. I need to return 9 uniformly spaced samples. I feel as though I need to use a for loop somehow...
Script intpl.m
1;
yi=interpl(x,y,0.4:0.4:3.6)
I think your teacher wants something like:
function y = f(x)
y = ....x..... (fill your formula here but use elementwise operations [1])
endfunction
and then use this function for the given range:
x = linspace (0, 4, 9);
y = f(x)
if you want to have this in one file foo.m be sure to not start the file with the function definition. I normally use "1;" so your script foo.m becomes:
1;
function y = f(x)
x = ....;
endfunction
x = linspace (...);
y = f(x)
plot (x, y) # if you want to plot it
[1] https://www.gnu.org/software/octave/doc/interpreter/Arithmetic-Ops.html
I am doing a simple task on ocatve.
I have to administer drug dose at day 1,7,21 and 28..
i wrote a function like that:
function xdot = f (x,t)
a=[1;7;21;28]
drug=0; ### initially drug is zero
if (t==a)
drug=drug+57.947;
else
drug=drug+0;
endif
xdot=(-0.4077)*(x)+ drug; passing the value of drug to differential equation
endfunction
In the main file i called this function in lsode:
t=linspace(0,30,30);
x0=0;
y=lsode(#ex,x0,t); ### ex is the file name where function is written
plot(t,y,'o')
This program doesn't work.. it displays all the time zero value for drug. Can anybody help me that how to administer dug dose with certain time step by manipulating linspace function.
It looks like you have a simple clearance model, and at each time in a, you want the dose to be delivered instantaneously. That is, at each time in a, the amount of drug in the subject increases by 57.947.
If that is the model that you have in mind, implementing it in the formula for xdot will not work well. You would actually need to implement it as a "delta function", and lsode won't work with that.
Instead, I recommend solving the problem in stages, corresponding to the time intervals [0, 1], [1, 7], [7, 21], and [21, 28]. During one stage, the differental equation is simply xdot = -0.4077*x. In the first stage, the initial condition is 0. In the next stage, the initial condition is the final value of the previous stage plus the dosage amount 57.947.
Here's a script:
dose = 57.947;
a = [1 7 21 28 30];
x0 = 0.0;
t0 = 0;
t = [];
y = [];
for t1 = a
tinterval = linspace(t0, t1, 2*(t1 - t0) + 1);
yinterval = lsode(#(x, t) -0.4077*x, x0, tinterval);
t = [t tinterval];
y = [y yinterval'];
t0 = t1;
x0 = yinterval(end) + dose;
end
plot(t, y, "linewidth", 2);
The script creates this plot:
Note that the differential equation xdot = -k*x has the solution x0*exp(-k*(t-t0)), so the call to lsode could be replaced with
yinterval = x0*exp(-0.4077*(tinterval - t0));
If you do this, also remove the transpose from yinterval two lines below:
y = [y yinterval];
If you want to keep the implementation of the administration of the drug dose within the formula for xdot, you'll need to distribute it over a small time interval. It could be implemented as a short rectangular pulse of width w and height 57.974/w. You'll also need to ensure that lsode takes sufficiently small internal time steps (less than w) so that it "sees" the drug dose.
You probably want replace t==a by ismember (t, a). Also, you should erase the else clause as it has no effect on the answer.
UPDATE
Consider rewriting your function as:
function xdot = f (x,t)
xdot = -0.4077*x + 57.947*ismember (t, [1 7 21 28])
endfunction
I am a first time MATLAB user. I have a 2d array of t vs f in MATLAB. This 2d array correponds to a function, say f(t). I am using ode45 to solve a set of differential equations and f(t) is one of the coefficients i.e. I have a set of equations of the form x'(t)=f(t)x(t) or variations thereof. How do I proceed?
I need a way to convert my array of t vs f into a function that can be used in the ode45 method. Presumably, the conversion will do some linear interpolation and give me the equation of the best fit curve.
Or if there is another approach, I'm all ears!
Thank you!
This is a simple example with passing function as a parameter to right hand side of the ODE. Create files in the same directory and run main
main.m
t = 0:0.2:10; f = sin(t);
fun = #(xc) interp1(t, f, xc);
x0=0.5
% solve diff(x, t)=sin(t)
% pass function as parameter
[tsol, xsol] = ode45(#(t, x) diff_rhs(t, x, fun),[0.0 8.0], 0.5);
% we solve equation dx/dt = sin(x), x(0)=x0
% exact solution is x(t) = - cos(t) + x(0) + 1
plot(tsol, xsol, 'x');
hold on
plot(tsol, -cos(tsol) + x0 + 1, '-');
legend('numerical', 'exact');
diff_rhs.m
function dx = diff_rhs(t, x, fun)
dx = fun(t);
end
References in the documentation: interp1 and anonymous Functions.
I am learning MatLab on my own, and I have this assignment in my book which I don't quite understand. Basically I am writing a function that will calculate sine through the use of Taylor series. My code is as follows so far:
function y = sine_series(x,n);
%SINE_SERIES: computes sin(x) from series expansion
% x may be entered as a vector to allow for multiple calculations simultaneously
if n <= 0
error('Input must be positive')
end
j = length(x);
k = [1:n];
y = ones(j,1);
for i = 1:j
y(i) = sum((-1).^(k-1).*(x(i).^(2*k -1))./(factorial(2*k-1)));
end
The book is now asking me to include an optional output err which will calculate the difference between sin(x) and y. The book hints that I may use nargout to accomplish this, but there are no examples in the book on how to use this, and reading the MatLab help on the subject did not make my any wiser.
If anyone can please help me understand this, I would really appreciate it!
The call to nargout checks for the number of output arguments a function is called with. Depending on the size of nargout you can assign entries to the output argument varargout. For your code this would look like:
function [y varargout]= sine_series(x,n);
%SINE_SERIES: computes sin(x) from series expansion
% x may be entered as a vector to allow for multiple calculations simultaneously
if n <= 0
error('Input must be positive')
end
j = length(x);
k = [1:n];
y = ones(j,1);
for i = 1:j
y(i) = sum((-1).^(k-1).*(x(i).^(2*k -1))./(factorial(2*k-1)));
end
if nargout ==2
varargout{1} = sin(x)'-y;
end
Compare the output of
[y] = sine_series(rand(1,10),3)
and
[y err] = sine_series(rand(1,10),3)
to see the difference.
Hi
I am trying to sum two function handles, but it doesn't work.
for example:
y1=#(x)(x*x);
y2=#(x)(x*x+3*x);
y3=y1+y2
The error I receive is "??? Undefined function or method 'plus' for input arguments of type 'function_handle'."
This is just a small example, in reality I actually need to iteratively sum about 500 functions that are dependent on each other.
EDIT
The solution by Clement J. indeed works but I couldn't manage to generalize this into a loop and ran into a problem. I have the function s=#(x,y,z)((1-exp(-x*y)-z)*exp(-x*y)); And I have a vector v that contains 536 data points and another vector w that also contains 536 data points. My goal is to sum up s(v(i),y,w(i)) for i=1...536 Thus getting one function in the variable y which is the sum of 536 functions. The syntax I tried in order to do this is:
sum=#(y)(s(v(1),y,z2(1)));
for i=2:536
sum=#(y)(sum+s(v(i),y,z2(i)))
end
The solution proposed by Fyodor Soikin works.
>> y3=#(x)(y1(x) + y2(x))
y3 =
#(x) (y1 (x) + y2 (x))
If you want to do it on multiple functions you can use intermediate variables :
>> f1 = y1;
>> f2 = y2;
>> y3=#(x)(f1(x) + f2(x))
EDIT after the comment:
I'm not sure to understand the problem. Can you define your vectors v and w like that outside the function :
v = [5 4]; % your 536 data
w = [4 5];
y = 8;
s=#(y)((1-exp(-v*y)-w).*exp(-v*y))
s_sum = sum(s(y))
Note the dot in the multiplication to do it element-wise.
I think the most succinct solution is given in the comment by Mikhail. I'll flesh it out in more detail...
First, you will want to modify your anonymous function s so that it can operate on vector inputs of the same size as well as scalar inputs (as suggested by Clement J.) by using element-wise arithmetic operators as follows:
s = #(x,y,z) (1-exp(-x.*y)-z).*exp(-x.*y); %# Note the periods
Then, assuming that you have vectors v and w defined in the given workspace, you can create a new function sy that, for a given scalar value of y, will sum across s evaluated at each set of values in v and w:
sy = #(y) sum(s(v,y,w));
If you want to evaluate this function using an array of values for y, you can add a call to the function ARRAYFUN like so:
sy = #(y) arrayfun(#(yi) sum(s(v,yi,w)),y);
Note that the values for v and w that will be used in the function sy will be fixed to what they were when the function was created. In other words, changing v and w in the workspace will not change the values used by sy. Note also that I didn't name the new anonymous function sum, since there is already a built-in function with that name.