I would like to write a Matlab code to calculate the following:
\sum_{k=0}^{N-1} \frac{1}{k!} \sum_{i=0}^{k} {k \choose i}(a-1)^{k-i} a^k
and my code is:
N = 3;
a = [3 4];
for k = 0:N-1
f = 0;
for i = 0:k
f = f + nchoosek(k,i).* a.^k .* (a-1).^(k-i);
end
sumoff = sum(f);
all = (( 1./ (factorial(k))).*sumoff);
end
overall= sum(all);
'all' variable gives different value when it is inside the for loop rather than outside. But I want it to calculate when k = 0:N-1. What am I doing wrong?
Thank you.
The issue is your current code overwrites all on every iteration. Moving it outside the loop also doesn't work because you'll only save the result of the last iteration.
To save the all of every iteration, define all as a vector and then assign each intermediate result into that vector:
N = 3;
a = [3 4];
% preallocate a vector for `all`
all = nan(N-1, 1);
for k = 0:N-1
f = 0;
for i = 0:k
f = f + nchoosek(k,i) .* a.^k .* (a-1).^(k-i);
end
sumoff = sum(f);
% assign your intermediate result into the `all` vector
all(k+1) = ((1./(factorial(k))) .* sumoff);
end
overall = sum(all);
Related
I'm trying to call a numerical integration function (namely one that uses the trapazoidal method) to compute a definite integral. However, I want to pass more than one value of 'n' to the following function,
function I = traprule(f, a, b, n)
if ~isa(f, 'function_handle')
error('Your first argument was not a function handle')
end
h = (b-a)./ n;
x = a:h:b;
S = 0;
for j = 2:n
S = S + f(x(j));
end
I = (h/2)*(f(a) + 2*S + f(b)); %computes indefinite integral
end
I'm using; f = #(x) 1/x, a = 1 and b = 2. I'm trying to pass n = 10.^(1:10) too, however, I get the following output for I when I do so,
I =
Columns 1 through 3
0.693771403175428 0.069377140317543 0.006937714031754
Columns 4 through 6
0.000693771403175 0.000069377140318 0.000006937714032
Columns 7 through 9
0.000000693771403 0.000000069377140 0.000000006937714
Column 10
0.000000000693771
Any ideas on how to get the function to take n = 10.^(1:10) so I get an output something like,
I = 0.693771403175428, 0.693153430481824, 0.693147243059937 ... and so on for increasing powers of 10?
In the script where you are calling this from, simply iterate over n
k = 3;
f = #(x)1./x;
a = 1; b = 2;
I = zeros(k,1);
for n = 1:k
I(n) = traprule(f, a, b, 10^n);
end
% output: I = 0.693771403175428
% 0.693153430481824
% 0.693147243059937
Then I will contain all of the outputs. Alternatively you can adapt your function to use the same logic to loop over the elements of n if it is passed
as a vector.
Note, you can improve the efficiency of your traprule code by removing the for loop:
% This loop operates on every element of x individually, and is inefficient
S = 0;
for j = 2:n
S = S + f(x(j));
end
% If you ensure you use element-wise equations like f=#(x)1./x instead of f=#(x)1/x
% Then you can use this alternative:
S = sum(f(x(2:n)));
I'm trying to figure out a way to make a plot of a function in Matlab that accepts k parameters and returns a 3D point. Currently I've got this working for two variables m and n. How can I expand this process to any number of parameters?
K = zeros(360*360, number);
for m = 0:5:359
for n = 1:5:360
K(m*360 + n, 1) = cosd(m)+cosd(m+n);
K(m*360 + n, 2) = sind(m)+sind(m+n);
K(m*360 + n, 3) = cosd(m)+sind(m+n);
end
end
K(all(K==0,2),:)=[];
plot3(K(:,1),K(:,2),K(:,3),'.');
end
The code you see above is for a similar problem but not exactly the same.
Most of the time you can do this in a vectorized manner by using ndgrid.
[M, N] = ndgrid(0:5:359, 1:5:360);
X = cosd(M)+cosd(M+N);
Y = sind(M)+sind(M+N);
Z = cosd(M)+sind(M+N);
allZero = (X==0)&(Y==0)&(Z==0); % This ...
X(allZero) = []; % does not ...
Y(allZero) = []; % do ...
Z(allZero) = []; % anything.
plot3(X,Y,Z,'b.');
A little explanation:
The call [M, N] = ndgrid(0:5:359, 1:5:360); generates all combinations, where M is an element of 0:5:359 and N is an element of 1:5:360. This will be in the form of two matrices M and N. If you want you can reshape these matrices to vectors by using M = M(:); N = N(:);, but this isn't needed here.
If you were to have yet another variable, you would use: [M, N, P] = ndgrid(0:5:359, 1:5:360, 10:5:1000).
By the way: The code part where you delete the entry [0,0,0] doesn't do anything here, because this value doesn't appear. I see you only needed it, because you were allocating a lot more memory than you actually needed. Here are two versions of your original code, that are not as good as the ndgrid version, but preferable to your original one:
m = 0:5:359;
n = 1:5:360;
K = zeros(length(m)*length(n), 3);
for i = 1:length(m)
for j = 1:length(n)
nextRow = (i-1)*length(n) + j;
K(nextRow, 1) = cosd(m(i)) + cosd(m(i)+n(j));
K(nextRow, 2) = sind(m(i)) + sind(m(i)+n(j));
K(nextRow, 3) = cosd(m(i)) + sind(m(i)+n(j));
end
end
Or simpler, but a bit slower:
K = [];
for m = 0:5:359
for n = 1:5:360
K(end+1,1:3) = 0;
K(end, 1) = cosd(m)+cosd(m+n);
K(end, 2) = sind(m)+sind(m+n);
K(end, 3) = cosd(m)+sind(m+n);
end
end
I am currently trying to run a script that calls a particular function, but want to call the function inside a loop that halfs one of the input variables for roughly 4 iterations.
in the code below the function has been replaced for another for loop and the inputs stated above.
the for loop is running an Euler method on the function, and works fine, its just trying to run it with the repeated smaller step size im having trouble with.
any help is welcomed.
f = '3*exp(-x)-0.4*y';
xa = 0;
xb = 3;
ya = 5;
n = 2;
h=(xb-xa)/n;
x = xa:h:xb;
% h = zeros(1,4);
y = zeros(1,length(x));
F = inline(f);
y(1) = ya;
for j = 1:4
hOld = h;
hNew = hOld*0.5;
hOld = subs(y(1),'h',hNew);
for i = 1:(length(x)-1)
k1 = F(x(i),y(i));
y(i+1,j+1) = y(i) + h*k1;
end
end
disp(h)
after your comment, something like this
for j = 1:4
h=h/2;
x = xa:h:xb;
y = zeros(1,length(x));
y(1) = ya;
for i = 1:(length(x)-1)
k1 = F(x(i),y(i));
y(i+1,j+1) = y(i) + h*k1;
end
end
I wrote a code in Matlab which I predefine the variable "a" and then set up a for loop of 5 iterations where the variable "a" goes through some basic operations. However, the for loop output only saves the fifth iteration of "a." How do I save all 5 iterations in a 1x5 array?
The code is as follows:
a = 10;
k = 0.5;
n = 2;
for m = 1:5
a = a + (a*k) + n;
end
Edit:
I just found it that I have to create a new variable.
a = 10;
k = 0.5;
n = 2;
a_n = zeros(1,5);
for m = 1:5
a = a + (a*k) + n;
a_n(m) = a;
end
You may need to store value of a after each iteration into an another variable x
a = 10;
k = 0.5;
n = 2;
for m = 1:5
a = a + (a*k) + n;
x(m) = a;
end
x
Output:
x =
17.000 27.500 43.250 66.875 102.312
You would need to use a different variable to store the 5 iterations as an array.
Code would look something like this:
a = 10;
k = 0.5;
n = 2;
b = [];
for m = 1:5
a = (a + (a*k) + n)
b = [b a];
end
You can now print b for all 5 iteration values.
Here is an alternate way to update values into the 1-D matrix.
I need to evaluate following expression (in pseudo-math notation):
∑ipi⋅n
where p is a matrix of three-element vectors and n is a three-element vector. I can do this with for loops as follows but I can't figure out
how to vectorize this:
p = [1 1 1; 2 2 2];
n = [3 3 3];
s = 0;
for i = 1:size(p, 1)
s = s + dot(p(i, :), n)
end
Why complicate things? How about simple matrix multiplication:
s = sum(p * n(:))
where p is assumed to be an M-by-3 matrix.
I think you can do it with bsxfun:
sum(sum(bsxfun(#times,p,n)))
----------
% Is it the same for this case?
----------
n = 200; % depending on the computer it might be
m = 1000*n; % that n needs to be chosen differently
A = randn(n,m);
x = randn(n,1);
p = zeros(m,1);
q = zeros(1,m);
tic;
for i = 1:m
p(i) = sum(x.*A(:,i));
q(i) = sum(x.*A(:,i));
end
time = toc; disp(['time = ',num2str(time)]);