63 is not equal to 2^(log(63)/log(2)) in Matlab

Question

Because of the floating point error 2^(log(63)/log(2)) isn't equal to 63. Check the results below:

format long;
>> 2^(log(63)/log(2))

ans =

  63.000000000000014

And unfortunatelly i can't use vpa on a logarithm according to the Matlab documents:

Unlike exact symbolic values, double-precision values inherently contain round-off errors. When you call vpa on a double-precision input, vpa cannot restore the lost precision, even though it returns more digits than the double-precision value. However, vpa can recognize and restore the precision of expressions of the form p/q, pπ/q, (p/q)1/2, 2q, and 10q, where p and q are modest-sized integers.

So how can i solve this issue ? I have very big numbers like 2^200 and i get very big errors.

Edit: I'm not asking why it is happening. I'm asking how to make this work as 100% accurate so this isn't a duplicate.

The best solution so far:

Unfortunatelly the solution that is suggested by @Sardar_Usama isn't always working as intended. Check the results below:

>> sym(2^(log(2251799813685247)/log(2)))

ans =

2251799813685259

On the other hand

>> 2^(log(vpa(2251799813685247))/log(vpa(2)))

ans =

2.2517998136852470000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000e0*10^0*10^15

is much much more closer to 2251799813685247 = 2^51. It's error is around ~9.491*10^-494 which makes this the best solution so far but there is still some error.

What's the specific issue? Is it how to avoid floating point inaccuracy in general? Or is it how to evaluate 2^(log(k) / log(2)) without inaccuracy? — Paul Hankin, Feb 23 '17 at 17:24
What's wrong with it not being exactly 63? It's only a `2*eps(63)` difference. In general, numerical calculations shouldn't rely on exact values. — horchler, Feb 23 '17 at 17:56
@horcler It gets significantly higher when you work with big numbers. — Kitiara, Feb 23 '17 at 17:59

score 4 · Answer 1 · answered Feb 23 '17 at 17:32

4

If you cannot use round or vpa, there is a slower way of dealing this, if you have Symbolic Math Toolbox, by creating symbolic numbers . i.e.

a = sym(2^(log(63)/log(2)))

This will give you sym class 63 which you can later convert to double using:

double(a)

This is what you'll get:

>> format long
>> a = sym(2^(log(63)/log(2)))

a =

63

>> double(a)

ans =

    63

answered Feb 23 '17 at 17:32

Sardar Usama

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Unfortunatelly this isn't always working as intended. Check this out `sym(2^(log(2251799813685247)/log(2)))` = `2251799813685259`. (btw 2251799813685247 is 2^51) On the other hand, `2^(log(vpa(2251799813685247))/log(vpa(2)))` is much much more closer to `2251799813685247`. It's error is around `~2.252*10^-192` which is not causing any trouble for my calculations but there is still some error. – Kitiara Feb 24 '17 at 00:19
@Kitiara There are only two options: *symbolic arithmetic* and *numeric arithmetic* (which includes *variable precision* and *double precision*). There *is* no other option. Getting 100% accuracy is not always possible. Read this article: https://www.mathworks.com/help/symbolic/recognize-and-avoid-roundoff-errors.html – Sardar Usama Feb 24 '17 at 06:22

63 is not equal to 2^(log(63)/log(2)) in Matlab

1 Answers1