It is supposed that Matlab uses binary floating point arithmetic, so we should expect that it could recognise exact binary numbers with no error. Look at the example:
>> format rat;i=[0:1:11];y=100+(96+2.*i)/(2^10);y'
ans =
3203/32
9409/94
4104/41
26126/261
12813/128
2903/29
25627/256
12113/121
6407/64
51257/512
5306/53
26931/269
The correct numbers can be found at Table 2 of page 15 from the attached Ebook (it is in greek, but tables are in english), where you can also compare the absolute accuracy of Octave in contrast to the errors of Matlab.
Book Σημειώσεις Εργαστηρίου Γραμμικών Μαθηματικών
This is not an error in MATLAB calculation, but accuracy of displaying.
Check the output of this code:
i=0:1:11;
y=100+(96+2*i)/(2^10);
rats(y',30)
looks better?
The command:
format rat
uses the same algorithm like rats function to approximate a number , but with default argument LEN = 13, which is not sufficient to properly display the numbers from this example.
@ (Grzegorz knor) well done and thanks for explanation ,,,, but i need to know the difference between rat & rats
rat (pi)
ans =
3 + 1/(7 + 1/(16))
=======================
rats(pi)
ans =
355/113
===========================
also if you explain the parameters of rats ,,,,, rats(y',30) whatt 30 means ???
thanks Advance
There is no diference, because RATS calls RAT function. Functions differ in way to display the result.
From MATLAB documentation:
rats(X,LEN) uses RAT to display rational approximations to
the elements of X. The string length for each element is LEN.
The default is LEN = 13, which allows 6 elements in 78 spaces.
Asterisks are used for elements which can't be printed in the
allotted space, but which are not negligible compared to the other
elements in X.
See also Algorithm section at bottom of the page:
http://www.mathworks.com/help/matlab/ref/rat.html
The relation between LEN argument in RATS and TOL argument in RAT is as follows:
tol = min(10^(-(LEN-1)/2) * norm(X(isfinite(X)),1),.1);
RATS function is not a time efficient function.
Look:
***
MATLAB:
>> y3=100+(96+2*3)/(2^10)
y3 =
26126/261
>> tic;rats(y3,30);toc
Elapsed time is 0.053786 seconds.
***
OCTAVE:
octave:1> format rat
octave:2> y3=100+(96+2*3)/(2^10)
y3 = 51251/512
octave:3> tic;rats(y3,30);toc
Elapsed time is 0 seconds.
***
From an expensive program like Matlab we want best and efficient accuracy at the default level.
@ Demetris - be careful: in your Octave code you've used 'format rat' and in the consequence time is 0 seconds. But I agree, in this case MATLAB is slower than Octave.
But let's run more complicated computations:
clc
!octave --eval 'y = 0:pi/3:100000;tic;rats(y,30);toc'
y = 0:pi/3:100000;
tic
rats(y,30);
toc
On my computer result is as follows:
Octave: Elapsed time is 14.7556 seconds.
MATLAB: Elapsed time is 11.652769 seconds.
There is no comparison at all, at least for my computer. Look:
octave:1> y3=100+(96+2*3)/(2^10)
y3 = 100.10
octave:2> tic;rats(y3,30);toc
Elapsed time is 0.001 seconds.
(less than 0.053786 seconds of Matlab)
***
octave:3> y = 0:pi/3:100000;
octave:4> tic
octave:5> rats(y,30);
octave:6> toc
Elapsed time is 33 seconds.
***
>> y = 0:pi/3:100000;
tic
rats(y,30);
toc
Elapsed time is 191.975279 seconds.
***
So Octave is approximately 6 times faster than Matlab at your example.
Versions:
GNU Octave, version 3.6.0
MATLAB 7.11.0 (R2010b)
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