N-Term Approximation for Matrices
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Hello,
Utilizing a function I created to calculate the N-term approximation for a matrix:
function[eA]= N_Approx( N,A,I )
sum=0;
for i=1:N
ai=(A^i)/(factorial(i));
sum=sum+ai;
end
eA=I+sum;
I would like to solve this function over a range of values of N however when I set my range for N (ie N=1:10), the approximation function above only outputs a matrix for the last number of the range. Is there a way to store the matrix created for every iteration?
Any help would be appreciated.
Thank you! CAAJ
1 Comment
Guillaume
on 30 Jan 2017
DO NOT use sum as a variable name. Soon enough, you will try to use the sum function in the same code and end up wondering why it comes up with some weird error. That will be because matlab is trying to index into your sum matrix instead of calling the sum function.
Accepted Answer
Jan
on 30 Jan 2017
Edited: Jan
on 30 Jan 2017
The power operation is expensive and you can avoid the repeated application:
function Result = N_Approx(N, A, I)
Result = zeros([size(A), N]);
Ai = eye(size(A));
SumAi = I; % If I should be added to each matrix
facti = 1;
for i = 1:N
facti = facti * i; % cumulative product instead of factorial
Ai = Ai * A; % cumulative product instead of power
SumAi = SumAi + Ai / facti;
Result(:, :, i) = SumAi;
end
Even if runtime does not matter in your case, it is a good programming practice to reduce the number of power operations, when possible.
If the variable "I" should not be added to each matrix, use: "SumAi = 0" instead of "SumAi = I" and add I to the final Result only.
You can access the intermediate matrices and the last value by:
Result(:, :, k)
with k goes from 1 to N.
As Walter has said already: There are many of questions in the forum caused by using "sum" as name of a variable, when later in the code it is tried to use as function. Therefore it is recommended to avoid this.
3 Comments
Jan
on 30 Jan 2017
The shown code replies all 10 matrices, but stored in a 3D-array: The first matrix is: Result(:, :, 1) and so on. Now you can inspect the differences between using 9 or 10 terms by comparing Result(:, :, 9) and Result(:, :, 10).
More Answers (2)
John BG
on 30 Jan 2017
1.
init sum to all zeros same size as resulting coefficients by replacing
sum=0;
with
sum=zeros([size(A),N]);
2.
in the loop, instead of accumulating in variable sum, stack the iterations indifferent layers of sum. If A is size 3x3, now sum is size 3x3xN
for i=1:N
ai=(A^i)/(factorial(i));
sum(:,:,i)=ai;
end
example
A=magic(3)
N=4
for i=1:N
ai=(A^i)/(factorial(i));
sum(:,:,i)=ai;
end
to read each coefficient use the following
sum(:,:,1)
=
8.00 1.00 6.00
3.00 5.00 7.00
4.00 9.00 2.00
sum(:,:,2)
ans =
45.50 33.50 33.50
33.50 45.50 33.50
33.50 33.50 45.50
sum(:,:,3)
=
199.50 171.50 191.50
179.50 187.50 195.50
183.50 203.50 175.50
.
CAAJ
if you find this answer useful would you please be so kind to mark my answer as Accepted Answer?
To any other reader, please if you find this answer of any help solving your question,
please click on the thumbs-up vote link,
thanks in advance
John BG
7 Comments
Walter Roberson
on 30 Jan 2017
John,
Here, we do not believe that our role is only to literally answer Questions; we believe that part of our role is to teach people how to write clear and efficient and bug-free MATLAB programs. Part of that involves pointing out to people where their programs are weak, and suggesting improvements. Your efforts here would be more effective if you were to do that as well.
CAAJ
on 30 Jan 2017
2 Comments
John BG
on 30 Jan 2017
Edited: John Kelly
on 3 Feb 2017
this is why I give you a layered 3D matrix as answer, you can then pull each term and do whatever you want, sum it or anything else you find convenient.
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