N-Term Approximation for Matrices

9 visualizaciones (últimos 30 días)
CAAJ
CAAJ el 30 de En. de 2017
Editada: John Kelly el 3 de Feb. de 2017
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 comentario
Guillaume
Guillaume el 30 de En. de 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.

Iniciar sesión para comentar.

Respuesta aceptada

Jan
Jan el 30 de En. de 2017
Editada: Jan el 30 de En. de 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 comentarios
Jan
Jan el 30 de En. de 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).
CAAJ
CAAJ el 30 de En. de 2017
Thank you very much Jan!

Iniciar sesión para comentar.

Más respuestas (2)

John BG
John BG el 30 de En. de 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 comentarios
John BG
John BG el 30 de En. de 2017
Guillon,
It's CAAJ that chose that particular name of variable
Walter Roberson
Walter Roberson el 30 de En. de 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.

Iniciar sesión para comentar.


CAAJ
CAAJ el 30 de En. de 2017
Just to clarify a bit, with inputting a range for my function beforehand, I would like to see how using 10 terms changes the matrix verses using 9 terms, and etc. I don't really understand why, but when I do set my range before hand, the function outputs only one matrix rather than the 10 different matrices created for the range of N values.
  2 comentarios
John BG
John BG el 30 de En. de 2017
Editada: John Kelly el 3 de Feb. de 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.
Walter Roberson
Walter Roberson el 3 de Feb. de 2017
Jan's answer does give a 3D matrix with all of the results.

Iniciar sesión para comentar.

Categorías

Más información sobre Creating and Concatenating Matrices en Help Center y File Exchange.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by