let say that from given function f(t), we want to construct new function which is given from existed function by this way
where T is some constant let say T=3; of course k can't be from -infinity to infinity in reality because we can't do infinity summation using computer,so it is first my afford
first let us define our function
function y=f(t);
y=-1/(t^2);
end
and second program
k=-1000:1:999;
F=zeros(1,length(k));
T=3;
for t=1:length(k)
F(t)=sum(f(t+k*T));
end
but when i am running second program ,i am getting
>> program
Error using ^
Inputs must be a scalar and a square matrix.
To compute elementwise POWER, use POWER (.^) instead.
Error in f (line 2)
y=-1/(t^2);
Error in program (line 5)
F(t)=sum(f(t+k*T));
so i have two question related to this program :
1.first what is error why it shows me mistake
how can i do it in excel? can i simplify it somehow? thanks in advance
EDITED :
i have changed my code by this way
k=-1000:1:999;
F=zeros(1,length(k));
T=3;
for t=1:length(k)
result=0;
for l=1:length(k)
result=result+f(t+k(l)*T);
end
F(t)=result;
end
is it ok?
To solve your problem in a vectorized way, you'll have to change the function f such that it can be called with vectors as input. This is, as #patrik suggested, achieved by using the element-wise operators .* ./ .^ (Afaik, no .+ .- exist). Unfortunately the comment of #rayryeng is not entirely correct, which may have lead to confusion. The correct way is to use the element-wise operators for both the division ./ and the square .^:
function y = f(t)
y = -1 ./ (t.^2);
end
Your existing code (first version)
k = -1000:1:999;
F = zeros(1,length(k));
T = 3;
for t=1:length(k)
F(t) = sum(f(t+k*T));
end
then works as expected (and is much faster then the version you posted in the edit).
You can even eliminate the for loop and use arrayfun instead. For simple functions f, you can also use function handles instead of creating a separate file. This gives
f = #(t) -1 ./ (t.^2);
k = -1000:1:999;
t = 1:2000;
T = 3;
F = arrayfun(#(x)sum(f(x+k*T)), t);
and is even faster and a simple one-liner. arrayfun takes any function handle as first input. We create a function handle which takes an argument x and does the sum over all k: #(x) sum(f(x+k*T). The second argument, the vector t, contains all values for which the function handle is evaluated.
As proposed by #Divakar in comments, you can also use the bsxfun function:
f = #(t) -1 ./ (t.^2);
k = -1000:1:999;
t = 1:2000;
T = 3;
F = sum(f(bsxfun(#plus,k*T,t.')),2);
where bsxfun creates a matrix containing all combinations between t and k*T, they are all evaluated using f(...) and last, the sum along the second dimension sums over all k's.
Benchmarking
Lets compare these solutions:
Combination of for loop and sum (original question):
Elapsed time is 0.043969 seconds.
Go through all combinations in 2 for loops (edited question):
Elapsed time is 1.367181 seconds.
Vectorized approach with arrayfun:
Elapsed time is 0.063748 seconds.
Vectorized approach with bsxfun as proposed by #Divakar:
Elapsed time is 0.099399 seconds.
So (sadly) the first solution including a for loop beats both vectorized approaches. For larger k vectors (-10000:1:9999), this behavior can be reproduced. The conclusion seems to be that MATLAB has indeed learned how to optimize for loops.
Related
I have a vector, v, of N positive integers whose values I do not know ahead of time. I would like to construct another vector, a, where the values in this new vector are determined by the values in v according to the following rules:
- The elements in a are all integers up to and including the value of each element in v
- 0 entries are included only once, but positive integers appear twice in a row
For example, if v is [1,0,2] then a should be: [0,1,1,0,0,1,1,2,2].
Is there a way to do this without just doing a for-loop with lots of if statements?
I've written the code in loop format but would like a vectorized function to handle it.
The classical version of your problem is to create a vector a with the concatenation of 1:n(i) where n(i) is the ith entry in a vector b, e.g.
b = [1,4,2];
gives a vector a
a = [1,1,2,3,4,1,2];
This problem is solved using cumsum on a vector ones(1,sum(b)) but resetting the sum at the points 1+cumsum(b(1:end-1)) corresponding to where the next sequence starts.
To solve your specific problem, we can do something similar. As you need two entries per step, we use a vector 0.5 * ones(1,sum(b*2+1)) together with floor. As you in addition only want the entry 0 to occur once, we will just have to start each sequence at 0.5 instead of at 0 (which would yield floor([0,0.5,...]) = [0,0,...]).
So in total we have something like
% construct the list of 0.5s
a = 0.5*ones(1,sum(b*2+1))
% Reset the sum where a new sequence should start
a(cumsum(b(1:end-1)*2+1)+1) =a(cumsum(b(1:end-1)*2+1)+1)*2 -(b(1:end-1)+1)
% Cumulate it and find the floor
a = floor(cumsum(a))
Note that all operations here are vectorised!
Benchmark:
You can do a benchmark using the following code
function SO()
b =randi([0,100],[1,1000]);
t1 = timeit(#() Nicky(b));
t2 = timeit(#() Recursive(b));
t3 = timeit(#() oneliner(b));
if all(Nicky(b) == Recursive(b)) && all(Recursive(b) == oneliner(b))
disp("All methods give the same result")
else
disp("Something wrong!")
end
disp("Vectorised time: "+t1+"s")
disp("Recursive time: "+t2+"s")
disp("One-Liner time: "+t3+"s")
end
function [a] = Nicky(b)
a = 0.5*ones(1,sum(b*2+1));
a(cumsum(b(1:end-1)*2+1)+1) =a(cumsum(b(1:end-1)*2+1)+1)*2 -(b(1:end-1)+1);
a = floor(cumsum(a));
end
function out=Recursive(arr)
out=myfun(arr);
function local_out=myfun(arr)
if isscalar(arr)
if arr
local_out=sort([0,1:arr,1:arr]); % this is faster
else
local_out=0;
end
else
local_out=[myfun(arr(1:end-1)),myfun(arr(end))];
end
end
end
function b = oneliner(a)
b = cell2mat(arrayfun(#(x)sort([0,1:x,1:x]),a,'UniformOutput',false));
end
Which gives me
All methods give the same result
Vectorised time: 0.00083574s
Recursive time: 0.0074404s
One-Liner time: 0.0099933s
So the vectorised one is indeed the fastest, by a factor approximately 10.
This can be done with a one-liner using eval:
a = eval(['[' sprintf('sort([0 1:%i 1:%i]) ',[v(:) v(:)]') ']']);
Here is another solution that does not use eval. Not sure what is intended by "vectorized function" but the following code is compact and can be easily made into a function:
a = [];
for i = 1:numel(v)
a = [a sort([0 1:v(i) 1:v(i)])];
end
Is there a way to do this without just doing a for loop with lots of if statements?
Sure. How about recursion? Of course, there is no guarantee that Matlab has tail call optimization.
For example, in a file named filename.m
function out=filename(arr)
out=myfun(in);
function local_out=myfun(arr)
if isscalar(arr)
if arr
local_out=sort([0,1:arr,1:arr]); % this is faster
else
local_out=0;
end
else
local_out=[myfun(arr(1:end-1)),myfun(arr(end))];
end
end
end
in cmd, type
input=[1,0,2];
filename(input);
You can take off the parent function. I added it just hoping Matlab can spot the recursion within filename.m and optimize for it.
would like a vectorized function to handle it.
Sure. Although I don't see the point of vectorizing in such a unique puzzle that is not generalizable to other applications. I also don't foresee a performance boost.
For example, assuming input is 1-by-N. In cmd, type
input=[1,0,2];
cell2mat(arrayfun(#(x)sort([0,1:x,1:x]),input,'UniformOutput',false)
Benchmark
In R2018a
>> clear all
>> in=randi([0,100],[1,100]); N=10000;
>> T=zeros(N,1);tic; for i=1:N; filename(in) ;T(i)=toc;end; mean(T),
ans =
1.5647
>> T=zeros(N,1);tic; for i=1:N; cell2mat(arrayfun(#(x)sort([0,1:x,1:x]),in,'UniformOutput',false)); T(i)=toc;end; mean(T),
ans =
3.8699
Ofc, I tested with a few more different inputs. The 'vectorized' method is always about twice as long.
Conclusion: Recursion is faster.
I have 100 equations with 5 variables. Is there a function in Matlab which I can use to find the optimal solution of these equations?
My problem is to find argmin ||(a-ic)^2 + (b-jd)^2 + e - h(i,j)|| over all i, j from -10 to 10. ie.
%% Note: not Matlab code. Just showing the Math.
for i = -10:10
for j = -10:10
(a-ic)^2 + (b-jd)^2 + e = h(i,j)
known: h(i,j) is a 10*10 matrix,and i,j are indexes
expected: the optimal result of a,b,c,d,e
You can try using lsqnonlin as follows.
%% define a helper function in your .m file
function f = fun(x)
a=x(1); b=x(2); c=x(3); d=x(4); e=x(5); % Using variable names from your question. In other situations, be careful when overwriting e.
f=zeros(21*21,0); % size(f) is taken from your question. You should make this a variable for good practice.
for i = -10:10
for j = -10:10
f(10*(i+10+1)+(j+10+1)) = (a-i*c)^2 + (b-j*d)^2 + e - h(i,j); % 10 is taken from your question.
end
end
end
(Aside, why is your h(i,j) taking negative indices??)
In your main function you can simply write
function out=myproblem(x0)
out=lsqnonlin(#fun,x0);
end
In your cmd, you can call with specific initial try such as
myproblem([0,0,0,0,0])
Helper function over anonymous because in my experience helpers get sped up by JIT while anonymous do not. I also opted to reshape in the loops as an opposed to actually call reshape after because I expect reshape to cost significant extra time. Remember that O(1) in fun is not O(1) in lsqnonlin.
(As always, a solution to a nonlinear problem is not guaranteed.)
I'm trying to implement the forward Euler method using matlab, but don't understand the error I'm getting. This is what I have written:
function y = ForwardEulerMethod(f,y0,T,N)
h=T/N;
t=zeros(N+1,1);
for i=0:N
t(i)=i.*h; %line 5
end
y=zeros(N+1,1);
y(0)=y0;
for i=1:N
y(i)=y(i-1)+h.*f(t(i-1),y(i-1));
end
end
My error is with line 5 and says, "Subscript indices must either be real positive integers or logicals." I am familiar with this rule, but don't see how I'm breaking it. I'm just trying to replace a zero at each location in t with a numerical value. What am I missing?
Agree with #vijoc above. You are indexing with 0 at multiple places. You could either change how you are indexing the values or get rid of the for loop altogether, like below:
function y = ForwardEulerMethod(f,y0,T,N)
h=T/N;
t=0:N .* h; % this takes the place of the first for-loop
y=zeros(N+1,1);
y(1)=y0;
for i=2:N+1
y(i)=y(i-1)+h.*f(t(i-1),y(i-1));
end
end
You could even replace the second loop if the function f takes vector inputs like so:
y(1) = y0;
y(2:end) = y(1:end-1) + h .* f(t(1:end-1), y(1:end-1));
You're iterating over i = 0:N and using it as t(i)=i.*h, so you're trying to access t(0) during the first iteration. Matlab indexing starts from 1, hence the error.
You also have other lines which will cause the same error, once the execution gets that far.
I have the equation 1 = ((π r2)n) / n! ∙ e(-π r2)
I want to solve it using MATLAB. Is the following the correct code for doing this? The answer isn't clear to me.
n= 500;
A= 1000000;
d= n / A;
f= factorial( n );
solve (' 1 = ( d * pi * r^2 )^n / f . exp(- d * pi * r^2) ' , 'r')
The answer I get is:
Warning: The solutions are parametrized by the symbols:
k = Z_ intersect Dom::Interval([-(PI/2 -
Im(log(`fexp(-PI*d*r^2)`)/n)/2)/(PI*Re(1/n))], (PI/2 +
Im(log(`fexp(-PI*d*r^2)`)/n)/2)/(PI*Re(1/n)))
> In solve at 190
ans =
(fexp(-PI*d*r^2)^(1/n))^(1/2)/(pi^(1/2)*d^(1/2)*exp((pi*k*(2*i))/n)^(1/2))
-(fexp(-PI*d*r^2)^(1/n))^(1/2)/(pi^(1/2)*d^(1/2)*exp((pi*k*(2*i))/n)^(1/2))
You have several issues with your code.
1. First, you're evaluating some parts in floating-point. This isn't always bad as long as you know the solution will be exact. However, factorial(500) overflows to Inf. In fact, for factorial, anything bigger than 170 will overflow and any input bigger than 21 is potentially inexact because the result will be larger than flintmax. This calculation should be preformed symbolically via sym/factorial:
n = sym(500);
f = factorial(n);
which returns an integer approximately equal to 1.22e1134 for f.
2. You're using a period ('.') to specify multiplication. In MuPAD, upon which most of the symbolic math functions are based, a period is shorthand for concatenation.
Additionally, as is stated in the R2015a documentation (and possibly earlier):
String inputs will be removed in a future release. Use syms to declare the variables instead, and pass them as a comma-separated list or vector.
If you had not used a string, I don't think that it would have been possible for your command to get misinterpreted and return such a confusing result. Here is how you could use solve with symbolic variables:
syms r;
n = sym(500);
A = sym(1000000);
d = n/A;
s = solve(1==(d*sym(pi)*r^2)^n/factorial(n)*exp(-d*sym(pi)*r^2),r)
which, after several minutes, returns a 1,000-by-1 vector of solutions, all of which are complex. As #BenVoigt suggests, you can try the 'Real' option for solve. However, in R2015a at least, the four solutions returned in terms of lambertw don't appear to actually be real.
A couple things to note:
MATLAB is not using the values of A, d, and f from your workspace.
f . exp is not doing at all what you wanted, which was multiplication. It's instead becoming an unknown function fexp
Passing additional options of 'Real', true to solve gets rid of most of these extraneous conditions.
You probably should avoid calling the version of solve which accepts a string, and use the Symbolic Toolbox instead (syms 'r')
We have an equation similar to the Fredholm integral equation of second kind.
To solve this equation we have been given an iterative solution that is guaranteed to converge for our specific equation. Now our only problem consists in implementing this iterative prodedure in MATLAB.
For now, the problematic part of our code looks like this:
function delta = delta(x,a,P,H,E,c,c0,w)
delt = #(x)delta_a(x,a,P,H,E,c0,w);
for i=1:500
delt = #(x)delt(x) - 1/E.*integral(#(xi)((c(1)-c(2)*delt(xi))*ms(xi,x,a,P,H,w)),0,a-0.001);
end
delta=delt;
end
delta_a is a function of x, and represent the initial value of the iteration. ms is a function of x and xi.
As you might see we want delt to depend on both x (before the integral) and xi (inside of the integral) in the iteration. Unfortunately this way of writing the code (with the function handle) does not give us a numerical value, as we wish. We can't either write delt as two different functions, one of x and one of xi, since xi is not defined (until integral defines it). So, how can we make sure that delt depends on xi inside of the integral, and still get a numerical value out of the iteration?
Do any of you have any suggestions to how we might solve this?
Using numerical integration
Explanation of the input parameters: x is a vector of numerical values, all the rest are constants. A problem with my code is that the input parameter x is not being used (I guess this means that x is being treated as a symbol).
It looks like you can do a nesting of anonymous functions in MATLAB:
f =
#(x)2*x
>> ff = #(x) f(f(x))
ff =
#(x)f(f(x))
>> ff(2)
ans =
8
>> f = ff;
>> f(2)
ans =
8
Also it is possible to rebind the pointers to the functions.
Thus, you can set up your iteration like
delta_old = #(x) delta_a(x)
for i=1:500
delta_new = #(x) delta_old(x) - integral(#(xi),delta_old(xi))
delta_old = delta_new
end
plus the inclusion of your parameters...
You may want to consider to solve a discretized version of your problem.
Let K be the matrix which discretizes your Fredholm kernel k(t,s), e.g.
K(i,j) = int_a^b K(x_i, s) l_j(s) ds
where l_j(s) is, for instance, the j-th lagrange interpolant associated to the interpolation nodes (x_i) = x_1,x_2,...,x_n.
Then, solving your Picard iterations is as simple as doing
phi_n+1 = f + K*phi_n
i.e.
for i = 1:N
phi = f + K*phi
end
where phi_n and f are the nodal values of phi and f on the (x_i).