Why 2^52 is equal to 2^52+1 in Matlab ? And how to fix it?

Question

For some reason 2^52 is equal to 2^52+1 in Matlab, but why ? And how can i fix this ? For more info please run the code below and check the results.

Here it is the outputs(with vpa):

>> format long
>> digits(500)
>> vpa(2^52)

ans =

4503599627370496.0

>> vpa(2^52+1)

ans =

4503599627370496.0

>> isequal(vpa(2^52), vpa(2^52+1))

ans =

     1

>> vpa(2^52+1)

ans =

4503599627370496.0

>> ans+1

ans =

4503599627370497.0

>> vpa(2^52+1000)

ans =

4503599627371496.0

https://ibb.co/iDDAwF

(the outputs without vpa)

>> 2^52

ans =

    4.503599627370496e+015

>> 2^52+1

ans =

    4.503599627370497e+015

>> isequal(2^52, 2^52+1)

ans =

     0

>> 2^52+1

ans =

    4.503599627370497e+015

>> ans+1

ans =

    4.503599627370498e+015

>> 2^52+1000

ans =

    4.503599627371496e+015

Edit: This isn't a duplicate and has nothing to do with floating point errors.

Thats odd. I just ran all the commands in matlab you have in your image and got all the expected answers — RSon1234, Feb 22 '17 at 07:19
Please, don't post pictures to code. It is difficult to copy. Then, what is `vpa`? My Matlab does not have it. — Bernhard, Feb 22 '17 at 07:22
vpa = Variable-precision arithmetic (https://www.mathworks.com/help/symbolic/vpa.html) — Kitiara, Feb 22 '17 at 07:24
Possible duplicate of [Why is 24.0000 not equal to 24.0000 in MATLAB?](http://stackoverflow.com/questions/686439/why-is-24-0000-not-equal-to-24-0000-in-matlab) — Adriaan, Feb 22 '17 at 07:27
You have a basic floating point error; you reached the end of the largest numbers MATLAB can represent without loosing precision. — Adriaan, Feb 22 '17 at 07:27
@Adriaan I ran these exact commands on my machine in matlab and got the correct results — RSon1234, Feb 22 '17 at 07:30
@Adriaan It's an integer. Besides Matlab can differenciate between 2^52 and 2^53. — Kitiara, Feb 22 '17 at 07:31
can you check if the results are correct without using vpa? tell me what you get. Because even without vpa, I show the difference — RSon1234, Feb 22 '17 at 07:35
@RSon1234 Looks like vpa is the source of the problem. Check my post. — Kitiara, Feb 22 '17 at 07:42
@RSon1234: this behavior depends on the Matlab version used. Newer versions will sometimes evaluate the expression inside `vpa` as if it were already symbolic rather than in floating point. However, that behavior shouldn't be depended on. Something like `vpa(2)^52+1` should always be used. — horchler, Feb 22 '17 at 17:20

Mikhail · Accepted Answer · 2017-02-25T03:59:19.817

7

vpa_item=vpa('2^52');
vpa_item2=vpa('1+2^52');
disp(isequal(vpa_item, vpa_item2));

results in 0

You can use "symbolic expressions" to bypass limitations of finite or floating point arithmetic on the input end.

--edit--

The linked page says vpa(1+sym(2)^52) is the paradigmatic expression, although both methods should work.

What will not work (in general) is

value_affected_by_imprecise_arithemetic = 1+2^52;
vpa(value_affected_by_imprecise_arithemetic)

edited Feb 25 '17 at 03:59

answered Feb 22 '17 at 07:39

Mikhail

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So do you know why I am getting the correct result doing this as the OP did and the OP is getting incorrect results? – RSon1234 Feb 22 '17 at 07:41
@RSon1234 So, in R2015a `isequal(vpa(2^52), vpa(2^52+1))` results in `0` which is the correct behavior. Maybe due to a version mismatch? idk... – Mikhail Feb 22 '17 at 07:42
@Mikhail I just tested `isequal(vpa(num2str(2^52)), vpa(num2str(2^52+1)))` and it worked, thanks. – Kitiara Feb 22 '17 at 08:33
@Mikhail Which begs the question why `isequal(vpa(num2str(2^57)), vpa('2^57'))` = `0` ? and `isequal(vpa(str2num(num2str(2^57))), vpa('2^57'))` = `1` ? That is absolutely weird. – Kitiara Feb 22 '17 at 09:02
1

Don't use strings with symbolic math – this is advised against at the top of the documentation for `vpa` – and this functionality won't even work in future versions of Matlab. Use `vpa(2)^52+1` or `2^vpa(52)+1`, etc. Expression like `vpa(2^52+1)` may not work as `2^52+1` may be evaluated first in floating point (behavior in older Matlab versions may differ). – horchler Feb 22 '17 at 17:14
@Mikhail then why vpa(2^52+1000) is working but not vpa(2^52+1) ? Anyone know the reason ? Btw i have R2011a. – Kitiara Feb 23 '17 at 09:21
@horchler same question goes for you too – Kitiara Feb 23 '17 at 09:21
@Kitiara One reason is that `vpa(some_value)` requires that `some_value` is correct on the input end. When you write a complicated express, it is evaluated before it is transformed into a `vpa ` object, which defeats the purpose. – Mikhail Feb 23 '17 at 19:38
@Mikhail But the same thing happens for 2^52+1000 case too. – Kitiara Feb 23 '17 at 21:29
You need to convert to a vpa type before you do the math. ie, vpa(2)^52 – Mikhail Feb 23 '17 at 23:46

Why 2^52 is equal to 2^52+1 in Matlab ? And how to fix it?

1 Answers1