Thread Subject: Rounding a nuber after 4th digit

what is a function that can do the following operation?

?(1.23456)=1.2346
?(1.23451)=1.2345

It means that I want to round the number just after the
fourth digit.

"Hosein " <Kalaeimh@yahoo.com> wrote in message <g4tmoj$sl8
$1@fred.mathworks.com>...
> what is a function that can do the following operation?
>
> ?(1.23456)=1.2346
> ?(1.23451)=1.2345
>
> It means that I want to round the number just after the
> fourth digit.
>

% If you mean fourth DECIMAL digit, then:

x=[1.23456 1.23451]
y = round(10000*x)/10000

In article <g4tmoj$sl8$1@fred.mathworks.com>,
Hosein <Kalaeimh@yahoo.com> wrote:
>what is a function that can do the following operation?
>
>?(1.23456)=1.2346
>?(1.23451)=1.2345
>
>It means that I want to round the number just after the
>fourth digit.

You can't do it in any binary floating point arithmetic machine.
There is no exact binary floating point representation
possible for 1.2345 or 1.2346

>> sprintf('%.250g',1.2345)

ans =

1.2344999999999999307220832633902318775653839111328125

>> sprintf('%.250g',1.2346)

ans =

1.234599999999999919708670859108678996562957763671875


You can only do this for "presentation purposes", that is as
character strings:

>> sprintf('%.4f',1.23456)

ans =

1.2346

>> sprintf('%.4f',1.23451)

ans =

1.2345


--
  "Whenever there is a hard job to be done I assign it to a lazy
  man; he is sure to find an easy way of doing it."
                                              -- Walter Chrysler

here is a simple function tht i wrote for repetitive use..



% Author: Ravi Chilumula, Last Modified: Mon, jul 07, 2008, 15:20PM EST
% Function Usage: % truncate after specified number of digits
% x= 1.345678;, truncate(x,4) rounds it after 4 decimal points

function [outp] = truncate(inp, noOfDigits) % do polyfit using the logarithm of the x-input

outp = ((10^noOfDigits).*inp)/(10^noOfDigits);

In article <31452591.1215458531550.JavaMail.jakarta@nitrogen.mathforum.org>,
ravi.chilumula@gmail.com <ravi.chilumula@gmail.com> wrote:
>here is a simple function tht i wrote for repetitive use..
>
>
>
>% Author: Ravi Chilumula, Last Modified: Mon, jul 07, 2008, 15:20PM EST
>% Function Usage: % truncate after specified number of digits
>% x= 1.345678;, truncate(x,4) rounds it after 4 decimal points
>
>function [outp] = truncate(inp, noOfDigits) % do polyfit using the logarithm of the x-input
>
>outp = ((10^noOfDigits).*inp)/(10^noOfDigits);


Your function does not do what it claims to do. It *cannot* do
what it claims to do. 1.3456 and 1.3457 are not *possible* to
represent exactly in ANY binary floating point number system.

1/1000 is 1/(2^3 * 5^3). The 1/2^3 part is representable exactly
in a binary floating point number systems, but 1/125 is NOT
exactly representable in binary floating point. 1/5 has a repeating
binary representation, binary 0.100110011001(repeated infinitely)
just the same way as in decimal, 1/7 is 0.1428571428571(repeated infinitely)
1/10 and powers thereof are not representable exactly in
IEEE binary floating point.

Your routine does not truncate after the given number of decimal
places: it just produces a number that is -close to- the truncated value.

Actually it doesn't even do that, if double precision numbers
are input: you do not have a fix() or round() or ceil() or floor()
so the output is going to be the same as the input except in
cases of overflow.
--
  "The study of error is not only in the highest degree
  prophylatic, but it serves as a stimulating introduction to the
  study of truth." -- Walter Lipmann

"Hosein " <Kalaeimh@yahoo.com> wrote in message
news:g4tmoj$sl8$1@fred.mathworks.com...
> what is a function that can do the following operation?
>
> ?(1.23456)=1.2346
> ?(1.23451)=1.2345
>
> It means that I want to round the number just after the
> fourth digit.

For clarification, why do you want to do this? Do you want to try to
perform computions using these truncated numbers, or do you just want to
display them?

In the former case, you'll probably need to use something like Fixed-Point
Toolbox:

http://www.mathworks.com/products/fixed/

In the latter, try to do this using FPRINTF:


x = [1.23456 1.23451];
fprintf('%1.4f\n', x)


--
Steve Lord
slord@mathworks.com