MATLAB: 1e301-10^301 ~= 0

big numbersMATLAB

Why does 1e301-10^301 not equal zero in matlab, but 1e300-10^300 does?

1e301-10^301 is 1.18961352678223e+285.

Best Answer

1e301 is computed entirely at parse time, as a double precision number.

10^301 involves computing 10 and 301 at parse time, each as double precision numbers, and then computing the one to the power of the other at run time, resulting in a double precision number. As this is a floating point calculation, it is subject to standard floating point roundoff .

1e301 comes out as the first representable double precision number greater than 10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 and 10^301 is one bit different, coming out as the last representable double precision number less than 10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 . 10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 itself is not exactly representable in double precision.

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MATLAB: How to compute 3^301 as an integer

You can never represent 3^301 as an integer in double precision, because it exceeds 2^53-1. It also clearly exceeds the capacity of UINT64.

This is true for any integer type in MATLAB, except for syms (the symbolic toolbox) or my own VPI toolbox, found on the file exchange.

X = sym(3)^301
X =
410674437175765127973978082146264947899391086876012309414440570235106991532497229781400618467066824164751453321793982128440538198297087323698003
X = vpi(3)^301
X =
    41067443717576512797397808214626494789939108687601230941444057023510
699153249722978140061846706682416475145332179398212844053819829708732369
8003

What matters is you need to use a tool that can store the full number. Doubles, or uint64 are unable to do so.

If you only want to store it as a floating point number though, a double has no problem with doing so, as long as the result iswithin the dynamic range of a double. It is easily so.

3^301
ans =
     4.10674437175765e+143

although if you pushed things too far,

3^1001
ans =
   Inf

it will overflow a double. So you just need to understand the limits of a double.

If you truly intend to work woth very large numbers in floating point, then your options reduce to syms again, or to my HPF toolbox, also found on the file exchange.

So, here 40 dibits, using the symbolic tools:

X = sym(3)^301;
vpa(X,40)
ans =
4.106744371757651279739780821462649478994e143

Or in HPF, using 500 digits of precision as a floating point number:

X = hpf(3,500)^1001
X =
3966212458442419910671365779256433097896266098256444502994761104680485792040114698622334941551824185891729333090617747067852863746301856735596028175773861022082860675594962411634044322686894523823171200352765854123386843446869283314184698519162260023471518534376441033216641163191116687452396104764748408662490805691335746870685249194923663690755669212254696812828119726515539079816086550295056821764327181278269014316592308443051156946241497496031843098396519308306708565660003

Best Answer

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MATLAB: How to compute 3^301 as an integer

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