Lets assume we have an optimization problem like this:
x = zeros(1,10);
x(1,1) = 2;
for k = 1: 9
x(k+1) = 10 x(k);
end
Is it possible to write the equation without the for loop?

 Akzeptierte Antwort

John BG
John BG am 14 Feb. 2017

0 Stimmen

Sakil
ok for a matrix, look:
A=[1 2;3 4];n=[1:1:10];L=A(:);B=reshape(bsxfun(@power,L,n),[size(A),10])
the resulting exponential matrices are stored in the layers of B
B(:,:,1) =
1 2
3 4
B(:,:,2) =
1 4
9 16
B(:,:,3) =
1 8
27 64
B(:,:,4) =
1 16
81 256
B(:,:,5) =
1 32
243 1024
B(:,:,6) =
1 64
729 4096
B(:,:,7) =
1 128
2187 16384
B(:,:,8) =
1 256
6561 65536
B(:,:,9) =
1 512
19683 262144
B(:,:,10) =
1 1024
59049 1048576
.
Sakil
would you please be so kind to mark my answer as Accepted Answer,
thanks in advance
regards
John BG

Weitere Antworten (1)

John BG
John BG am 13 Feb. 2017

1 Stimme

this
x = zeros(1,10);
x(1) = 2;
for k = 1: 9
x(k+1) = 10 *x(k);
end
x
=
Columns 1 through 3
2.00 20.00 200.00
Columns 4 through 6
2000.00 20000.00 200000.00
Columns 7 through 9
2000000.00 20000000.00 200000000.00
Column 10
2000000000.00
is the same as
n=[1:1:10];x=.2*10.^n
x =
Columns 1 through 3
2.00 20.00 200.00
Columns 4 through 6
2000.00 20000.00 200000.00
Columns 7 through 9
2000000.00 20000000.00 200000000.00
Column 10
2000000000.00
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

3 Kommentare

Hi. What if the multiplier is not 10? say the multiplier is a 2 by 2 matrix A. Then the statement will be:
x = x(1) * (A.^(0:9));
but this seems to be not working. Can you please check that?
More compactly,
x = 2 * 10.^(0:9);
Jan
Jan am 14 Feb. 2017
@Sakil: The operation A.^(0:9) is not valid, therefore the readers have to guess, what you want as result. Which dimension does x have after this calculation?

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am 13 Feb. 2017

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