Significant digits using a logarithm and a floating point number

Asked Mar 13 '15 at 09:23

Active Mar 13 '15 at 10:44

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I want to print the significant digits of a number n represented using a double d as n = 2^d. Is there a function of d which computes the significant digits of n? A maximum number (larger than 0) for the significant digits is also fine, like DBL_DIG in float.h.

Edit: I will be doing many additions to d such that 2^d >> MAX_DOUBLE. I suspect therefore the number of significant digits in n will be less than the significant digits of d. Is this true? By addition I mean (hopefully this makes sense):

Let n = n1 + n2, n1 = 2^d1, n2 = 2^d2 then I will perform d = d1 + d2, such that n = 2^d.

The reason I am using this is because it is nearly impossible to let d overflow and I am fine with loosing lots of precision in n.

I will be using a function described here: http://buddy.sourceforge.net/manual/group__info_ge551a6cc544c7d50a3d7c7fb7f5f9374.html#ge551a6cc544c7d50a3d7c7fb7f5f9374.

edited Mar 13 '15 at 10:44

asked Mar 13 '15 at 09:23

meijuh

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I'm not sure what you mean by "compute the significant digits of `n`" as distinct from "compute `n`". Do you want to know how many significant digits there are? (It's the same as the number of significant digits of `d`.) – Sneftel Mar 13 '15 at 09:30
@Sneftel: I added relevant usage details to the question – meijuh Mar 13 '15 at 10:24
Do you mean `n = n1 * n2`? – Simon Byrne Mar 13 '15 at 11:28
@SimonByrne I don't think so – meijuh Mar 13 '15 at 14:00
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What you have doesn't make sense: addition in log space corresponds to multiplication: `2^(d1+d2) = 2^d1 * 2^d2` – Simon Byrne Mar 13 '15 at 14:20

Significant digits using a logarithm and a floating point number

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