About a year ago, I was reading the Go source for computing
$\log(x)$ and was very surprised by the
implementation.^{1}

Even three years into a PhD in Applied Math (i.e. numerics), I still managed to learn something by diving in and trying to understand what was going on.

### (Almost-)Correctness

Really I just wanted a chance to share a neat error plot:

The high-level:

- The computation $\log(x)$ can be reduced to computing a related function $R(s)$ (where the value $s$ can be obtained from $x$ on a computer in some straightforward way)
- $R(s)$ can be approximated by a function that we can actually compute with simple floating point operations: $f(s)$
- The error $R(s) - f(s)$ is "equioscillating", i.e. it minimizes the maximum error, a so-called minimax solution

If you're still interested in some mathematical details, stick around. If not, I hope you enjoyed the pretty picture.

### Doing the Obvious

The "obvious" idea that comes to mind is to use a Taylor series
approximation. But $\log(x)$ is not defined at
$x = 0,$ so a Taylor series of a related function is
typically used^{2}:

$\log\left(1 + x\right) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots$

But, the implementation doesn't do the obvious thing. Instead it breaks down the argument into an exponent and a fractional part

$x = 2^k \left(\frac{1 + s}{1 - s}\right)$

which transforms the computation to:

$\log(x) = k \log(2) + \left[\log\left(1 + s\right) - \log\left(1 - s\right)\right].$

The first part $k \log(2)$ can be handled in maximal precision for the limited set of integer values that $k$ can be.

### Finding a Function to Approximate

For the other part

$\begin{aligned}\log\left(1 + s\right) - \log\left(1 - s\right) &= 2 s + \frac{2}{3} s^3 + \frac{2}{5} s^5 + \cdots \\ &= 2 s + s R(s). \end{aligned}$

From here, I expected a polynomial approximation that just takes the first few terms in $R(s)$. However, coefficients that are close (but not equal) are given:

$R(z) \approx L_1 s^2 + L_2 s^4 + \cdots + L_6 s^{12} + L_7 s^{14}$

```
L1 = 6.666666666666735130e-01 /* 3FE55555 55555593 */
L2 = 3.999999999940941908e-01 /* 3FD99999 9997FA04 */
L3 = 2.857142874366239149e-01 /* 3FD24924 94229359 */
L4 = 2.222219843214978396e-01 /* 3FCC71C5 1D8E78AF */
L5 = 1.818357216161805012e-01 /* 3FC74664 96CB03DE */
L6 = 1.531383769920937332e-01 /* 3FC39A09 D078C69F */
L7 = 1.479819860511658591e-01 /* 3FC2F112 DF3E5244 */
```

For example:

$L_1 = 2^{-1} \cdot 1.5555555555593_{16} = \frac{2}{3} + 2^{-53} \cdot \frac{185}{3} \approx \frac{2}{3}.$

### Not Go, Sun

You may have noted my typo above: I used $R(z)$ when I should've used $R(s)$. This typo is directly from the source; not my copy-paste error, but a copy-paste error in the original source.

The original authors aren't the Golang devs:

```
// The original C code, the long comment, and the constants
// below are from FreeBSD's /usr/src/lib/msun/src/e_log.c
// and came with this notice. The go code is a simpler
// version of the original C.
```

It seems they copied **everything**, even the typos. Another
typo can be found when the description mentions
"a special Reme algorithm" used to compute the coefficients
$L_j$.

This typo is a bit more egregious: it actually uses a
**Remez** algorithm.

### Remez Algorithm

This algorithm takes a polynomial approximation like

$R(s) \approx \frac{2}{3} s^2 + \frac{2}{5} s^4 + \cdots + \frac{2}{15} s^{14}$

### Is That All You've Got?

I hope to follow this post up with some code to actually compute the coefficients that give an equioscillating error (and to show some coefficients that do slightly better).