In case you don’t want to return a BigInt, then the following might work for you too: function log10(bigint) { if (bigint < 0) return NaN; const s = bigint.toString(10); return s.length + Math.log10(“0.” + s.substring(0, 15)) } function log(bigint) { return log10(bigint) * Math.log(10); } function natlog(bigint) { if (bigint < 0) return NaN; … Read more
This really depends on the domain of what values you want to compute a logarithm of. For IEEE doubles, many processors can take logarithms in a single assembly instruction; x86 has the FYL2X and FYL2XP1 instructions, for example. Although typically instructions like these will only take the logarithm in some fixed base, they can be … Read more
logsumexp works by evaluating the right-hand side of the equation log(∑ exp[a]) = max(a) + log(∑ exp[a – max(a)]) I.e., it pulls out the max before starting to sum, to prevent overflow in exp. The same can be applied before doing vector dot products: log(exp[a] ⋅ exp[b]) = log(∑ exp[a] × exp[b]) = log(∑ exp[a … Read more
I think it does not matter what is the base of the log as the relative complexity is the same irrespective of the base used. So you can think of it as O(log2X) = O(log10X) Also to mention that logarithms are related by some constant. So log₁₀(x) = log₂(x) / log₂(10) So most of the … Read more
Use this identity: logb(n) = loge(n) / loge(b) Where log can be a logarithm function in any base, n is the number and b is the base. For example, in Java this will find the base-2 logarithm of 256: Math.log(256) / Math.log(2) => 8.0 Math.log() uses base e, by the way. And there’s also Math.log10(), … Read more
Java Number Cruncher: The Java Programmer’s Guide to Numerical Computing provides a solution using Newton’s Method. Source code from the book is available here. The following has been taken from chapter 12.5 Big Decimal Functions (p330 & p331): /** * Compute the natural logarithm of x to a given scale, x > 0. */ public … Read more
Notation As is the convention both in mathematics and programming, the “log” function is taken to be base-e. The “exp” function is the exponential function. Remember that these functions are inverses we take the functions as: exp : ℝ → ℝ+, and log : ℝ+ → ℝ. Solution You’re just solving a simple equation here: … Read more
If you are on a recent-ish x86 or x86-64 platform (and you probably are), use the bsr instruction which will return the position of the highest set bit in an unsigned integer. It turns out that this is exactly the same as log2(). Here is a short C or C++ function that invokes bsr using … Read more
In general, one doesn’t expand out log(a + b); you just deal with it as is. That said, there are occasionally circumstances where it makes sense to use the following identity: log(a + b) = log(a * (1 + b/a)) = log a + log(1 + b/a) (In fact, this identity is often used when … Read more
try plt.yscale(‘log’, nonposy=’clip’) http://matplotlib.org/api/pyplot_api.html#matplotlib.pyplot.yscale The issue is with the bottom of bars being at y=0 and the default is to mask out in-valid points (log(0) -> undefined) when doing the log transformation (there was discussion of changing this, but I don’t remember which way it went) so when it tries to draw the rectangles for … Read more