This is the original entry. See the updated version here.

# How does a computer/calculator compute logarithms?

## The curious student’s frustration

There are many functions on a scientific or graphing calculator that we are introduced to as high school students that, we are told, just work. You select the function, put in the value that you need to calculate, hit “=” or “ENTER”, and SHABAM! You have the correct answer to some arbitrary number of digits that you are ensured are all 100% accurate. Pictured: The natural log of 2 to (an arbitrary) 32 digits on the built-in Microsoft calculator.

This is wonderful! Necessary, even, for if our calculators and computers calculated logarithms inaccurately, as well as exponentials, trig functions, and square roots, to name but a few, a lot of scientific and engineering work would be broken and end in catastrophe. But how do we know that the value on the calculator is, in fact, accurate? How did the calculator crunch the input number and give us the output? I remember this question being asked in high school, and the answer being no more than a handwave. To be fair to Algebra II teachers (which is where I was first introduced to logarithms, if memory serves), the answer is beyond the scope of algebra, and lies in calculus. But it is an answer worth knowing, and it’s an answer that is very knowable, even if you don’t have a background in calculus. Calculus is the method for deriving the formula used in calculators and computers, but the formula itself is a pretty simple polynomial.

## The Geometric Series

Most of us were introduced to polynomial equations in algebra. For a quick refresher, a polynomial is an expression involving at least one variable (usually $x$) and addition, subtraction, multiplication, division, and integer power operators. Here are a few examples:

Line:
$y = 5 + 2x$

Parabola:
$y = 1 - x - x^{2}$

Quartic:
$y = 4 - 2x - 5x^{2} + \frac{1}{2}x^{3} + x^{4}$

Note: The links lead to Desmos graphs where you can change the parameters to see how it changes the graph.

While these equations of polynomials contain a finite number of terms, we can have polynomials with an infinite number of terms. These are called series, and one of the simplest types of series is called a geometric series. A geometric series contains an initial term, usually denoted by the variable $a$, and each successive term is multiplied by a ratio, usually denoted by the variable $r$. Here are a couple of examples:

$\sum_{n=1}^{\infty} \frac{1}{2^n} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots$ $\sum_{n=0}^{\infty} 2 \cdot (-3)^n = 2 - 6 + 18 - 54 + \ldots$

In the first series, the initial value is $\frac{1}{2}$, and each term after it is the previous term multiplied by $\frac{1}{2}$. The second series begins with the number 2, and each term after it is the previous term multiplied by -3. Because the first series’ terms get smaller in absolute value for each successive term, the sum approaches, or converges, to a value. The second series’ terms get larger in absolute value with each successive term, so the series diverges without resolving to a defined value.

The general formula for a geometric series is

$\sum_{n=0}^{\infty} ar^n = a + ar + ar^2 + \ldots + ar^n + \ldots$

If we set this formula equal to $s$, we can do some rearranging to find a function that this series is equivalent to:

$s = a + ar + ar^2 + \ldots + ar^n + \ldots$

By pulling an $r$ out of all the terms after the first one, we get

$s = a + r(a + ar + ar^2 + \ldots + ar^{n-1} + \ldots)$

Despite the fact that $ar^n$ becomes $ar^{n-1}$, we have an infinite number of terms, so if we were to expand out the terms in the parentheses, we get the same series as $s$ in the first line; therefore, we can substitute the terms in the parentheses with $s$!

$s = a + rs$

Move $rs$ to the left side of the equation:

$s - rs = a,$

combine like-terms:

$s(1 - r) = a,$

and solve for $s$:

$s = \frac{a}{1 - r}.$

We can now set both $s$ expressions equal to each other:

$\frac{a}{1 - r} = \sum_{n=0}^{\infty} ar^n = a + ar + ar^2 + \ldots + ar^n + \ldots$

Note that this assumes the series converges. This formula does not hold if the series diverges.

Regarding the natural logarithm, the geometric series we are interested in is the one where $a = 1$ and $r = -x$:

$\frac{1}{1-(-x)} = 1 + (-x) + (-x)^2 + (-x)^3 + \ldots + (-x)^n + \ldots$ $\frac{1}{1+x} = 1 - x + x^2 - x^3 + x^4 + \ldots + (-x)^n + \ldots$

This series converges when $|x| < 1$. Now, you might think that some of the algebraic manipulation we did earlier was invalid, but we can check our work. Below, I have a Desmos graph with both the function $\frac{1}{1+x}$ and the infinite series. By moving the slider for $N$, you can add successive terms to the series and see that, as more and more terms are added, the series graph becomes a better and better approximation to the function graph for $|x| < 1$.

Desmos: Geometric Series Pictured: The 5-term approximation of the geometric series from above (black line) compared to the function $\frac{1}{1+x}$ (green line). Note how the series approximation is only accurate for $|x| < 1$.

## Getting a series for the natural logarithm

“This is all very interesting, but what does this have to do with computing logarithms?”

It turns out that the function $\frac{1}{1+x}$ and the natural logarithm are directly related. Specifically, the area under the curve $\frac{1}{1+t}$ from $t = 0$ to any value $x$ is the natural logarithm of $1+x$. In mathematical symbols, this is rendered as

$\int_{0}^{x} \frac{1}{1+t} dt = \ln{(1+x)}$

While the understanding of the calculus involved is beyond the scope of this article, I made a Desmos graph where you can play with values of $x$ ($a$ in the graph) and see the area under the curve, with its exact value shown and a 1x1 square area shown as a comparison. Pictured: The integral of $\frac{1}{1+x}$ from 0 to $e - 1$ (shaded in blue) and a square of Area = 1 (shaded in green). Both shaded areas are equal in size.

Taking integrals is an inverse problem and for some functions is very difficult, if not impossible, to get the exact solution. However, for polynomials, it’s actually very easy. For a given polynomial term, increase the degree by one, divide the term by the new degree, and take the difference of the function evaluated at the two integrand values (the values at the top and bottom of the $\int$ ).

We’ll start by setting the natural logarithm equal to the integral of our geometric series:

$\ln{(1+x)} = \int_{0}^{x} \frac{1}{1+t} dt = \int_{0}^{x} (1 - t + t^2 - t^3 + t^4 + \ldots + (-t)^n + \ldots) dt$

The integral of a sum is the sum of the integrals of each term, so we can take this one term at a time. Starting with the first term,

$\int_{0}^{x} 1 dt = \int_{0}^{x} 1t^0 dt = \frac{t^{0+1}}{0+1} \Big|_0^x = \frac{t^1}{1} \Big|_0^x = t \Big|_0^x = x - 0 = x$

Let me explain the symbols above. We start with taking the integral of 1, a constant, in terms of $t$ (which is a dummy variable to distinguish it from $x$ in the answer). A constant can be represented as a zero-degree polynomial term, so we can represent 1 as 1 times $t^0$. We add one to the degree, so 0 becomes 1, then we divide by the new degree, which is 1 in this case. Our result is $t$. We then need to evaluate this term at $t = x$ and $t = 0$ and subtract them (this is what the vertical bar represents). The result is $x - 0$, which reduces to $x$. So the first term of our infinite series for $\ln{(1+x)}$ is $x$! Let’s do the second term:

$\int_{0}^{x} -t dt = \int_{0}^{x} -t^1 dt = -\frac{t^{1+1}}{1+1} \Big|_0^x = -\frac{t^2}{2} \Big|_0^x = -\frac{x^2}{2} - \biggl(-\frac{0^2}{2} \biggr) = -\frac{x^2}{2} + 0 = -\frac{x^2}{2}$

Our second term is $-\frac{x^2}{2}$. You may start to see the overall pattern, but we’ll do one more:

$\int_{0}^{x} t^2 dt = \frac{t^{2+1}}{2+1} \Big|_0^x = \frac{t^3}{3} \Big|_0^x = \frac{x^3}{3} - \frac{0^3}{3} = \frac{x^3}{3}$

With this, the pattern starts to come into focus. We have a series where each term is a polynomial divided by its degree, where all of the odd degree polynomial terms are positive and all the even terms are negative. Written as a series,

$\ln{(1+x)} = \sum_{n=0}^{\infty} (-1)^n \frac{x^{n+1}}{n+1} = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots$

In fact, if we were to integrate the general term of the geometric series, this is exactly what we get:

$\int_0^x \frac{1}{1+t} dt = \int_0^x \sum_{n=0}^{\infty} (-t)^n dt$ $= \sum_{n=0}^\infty \biggl( \int_0^x (-t)^n dt \biggr)$ $= \sum_{n=0}^\infty \biggl( \int_0^x (-1)^n t^n dt \biggr)$ $= \Biggl( \sum_{n=0}^\infty (-1)^n \frac{t^{n+1}}{n+1} \Biggr) \Bigg|_0^x$ $= \sum_{n=0}^{\infty} (-1)^n \frac{x^{n+1}}{n+1}$

So, we did it! We have a polynomial series that calculates the natural logarithm! Granted, we need to shift our input by 1, as this series is $\ln{(1+x)}$, but that’s a simple hurdle to jump!

Except we’re limited. Very limited. Remember, the geometric series only applies to the function $\frac{1}{1+x}$ for $|x| < 1$, otherwise it diverges. The same limitation exists here. What’s worse, the series only converges quickly for $|x| < \frac{1}{2}$; outside this range, the number of terms becomes too much even for a computer to calculate in a reasonable amount of time. This is fine if all the values you need to calculate a logarithm for are between 0.5 and 1.5, but most problems involving logarithms tend to have very large values.

Well, we tried. Pack up, go home, see you next time.

## Properties of logarithms

Actually, as it turns out, this would be the end if we were dealing with almost any other function. But we are dealing with logarithms, and logarithms are special. There are two properties that we can use to reduce our inputs so that we can calculate the logarithm of any value we want!

The first property is that multiplication in the input is equivalent to addition of the outputs.

This means that if the input is the product of two factors, the logarithm of that product is the same as the logarithms of each factor taken individually and added together. For example, if we wanted to find the natural logarithm of the value 6, we can compute either $\ln(6)$ or $\ln(2) + \ln(3)$, because $6 = 2 \cdot 3$.

The general form of this property is

$\log_b{(ac)} = \log_b{(a)} + \log_b{(c)}$

This also applies to division in the inputs, but instead, you subtract the outputs:

$\log_b{ \biggl( \frac{a}{c} \biggr) } = \log_b{(a)} - \log_b{(c)}$

The second property is that the logarithm of a value raised to some power is that power times the logarithm of the (unraised) value.

This means that exponents in the input of the logarithm can hop out and become a coefficient in the output. For example, the number 8 can also be written as $2^3$ $(2 \cdot 2 \cdot 2 = 8)$, so $\ln(8) = \ln(2^3) = 3\ln(2)$.

The general form of this property is

$\log_b{(a^c)} = c \log_b{(a)}$

Each of these properties allows us to reduce the argument of the logarithm in different ways. Using the first property, if we have the natural logarithm of a known value (2 and 10 are common choices), we can reduce the argument by powers of that constant until we get a small enough input. Let’s use 15 as an example. If we have the natural logarithm of 2 calculated, we can keep dividing 15 by 2 until it is close to 1 while keeping track of the number of times we divided by 2. After dividing by 2 four times, we reduce 15 to 0.9375, so $0.9375 \cdot 2^4 = 15$. We can plug this value into our series like this:

$\ln(15) = \ln(2^4 \cdot 0.9375) = 4 \ln(2) + \ln(0.9375)$ $= 4 \ln(2) + \ln(1 + (0.9375 - 1)) = 4 \ln(2) + \ln(1 + (-0.0625))$ $= 4 \ln(2) + (-0.0625) - \frac{(-0.0625)^2}{2} + \frac{(-0.0625)^3}{3} + \ldots$ $= 2.772588722239781... - 0.0625 - 0.001953125 - 0.000081380208333... + \ldots$

We can see that as terms are added to the series, the approximation gets more and more precise as each term gets smaller and smaller in magnitude. In this particular example, the next term is less than one tenth the absolute value of the previous term. This means that the approximation gains about one digit of accuracy for each term we add to the series. To see this, here is a list of the first eight approximations for $\ln(15)$, as well as the accepted value:

0 terms: 2.772588722239781
1 term: 2.710088722239781
2 terms: 2.708135597239781
3 terms: 2.708054217031448
4 terms: 2.708050402334182
5 terms: 2.708050211599319
6 terms: 2.708050201665211
7 terms: 2.708050201133027
8 terms: 2.708050201103923

Accepted value: 2.708050201102210

So now we have a formula for finding the natural logarithm of any number to arbitrary precision based on the number of terms we use in the series. We did it! I even have a Python script that implements this formula to return the natural logarithm of any value within 64-bit floating-point precision using 48 terms.

Now, this formula is perfectly good to use. After all, it’s the formula used in Python’s decimal module to find the natural logarithm for base-10 number representation. But what if there were a better formula? One that converged much faster to find the natural logarithm with fewer terms?

## The better formula

How can an infinite series converge faster? Well, it turns out that for inputs less than 1, series that have terms that increase by two degrees will converge much faster than series that only increase by one. For example, if you look at the series for sine and cosine, they increase in degree by 2 for each term and converge very quickly (in fact, if you look at the C implementations of the sine kernel and cosine kernel, they only use six polynomial terms to get double floating-point precision, a much quicker convergence than our 48 terms!). So how do we get this series to skip a degree for each term? Surely, we can’t just get rid of all the even-degree terms or all of the odd-degree terms and still get an accurate answer, right?

We can if we do it in a clever way. Let’s look at the series again:

$\ln{(1+x)} = \sum_{n=0}^{\infty} (-1)^n \frac{x^{n+1}}{n+1} = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots$

This is the series for $\ln{(1+x)}$, but what if we could get a series for a slightly different input? For instance, what would the series be for $\ln{(1-x)}$? We can rewrite $\ln{(1-x)}$ as $\ln{(1 + (-x)}$. To get its series, let’s plug (-x) into our natural log series:

$\ln{(1-x)} = \ln{(1 + (-x))} = (-x) - \frac{(-x)^2}{2} + \frac{(-x)^3}{3} - \frac{(-x)^4}{4} - \ldots$ $\ln{(1-x)} = -x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} + \ldots = \sum_{n=0}^{\infty} -\frac{x^{n+1}}{n+1}$

The difference between this series and our original series is that all the odd terms have become negative while the even terms remain negative. This is perfect! If you don’t understand why, let me put each series in expanded form with all the terms lined up:

$\ln{(1+x)} = +x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \frac{x^6}{6} + \ldots$ $\ln{(1-x)} = -x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \frac{x^5}{5} - \frac{x^6}{6} - \ldots$

It should be apparent from this form that if we add the two series together, the odd terms will cancel out, and if we subtract the second series from the first series, the even terms will cancel out. So, should we add the series together, or subtract them? This must be determined by seeing how the logarithm function input changes when we add or subtract the two functions together. If we add the two functions together, we get

$\ln{(1+x)} + \ln{(1-x)} = \ln{((1+x)(1-x))} = \ln{(1-x^2)}$

And if we subtract the two functions, we get

$\ln{(1+x)} - \ln{(1-x)} = \ln{ \biggl( \frac{1+x}{1-x} \biggr)}$

using the first property of logarithms. Let’s remind ourselves that $x$ is the value we plug into the series. If we want the natural logarithm of some value—let’s call it $u$—then we have to find the value of $x$ to plug into our series that will get us the natural log of $u$. We can find this by setting $u$ equal to the input of each new, combined logarithm and solve for $x$. For the added logarithm, set $u$ equal to $1 - x^2$:

$u = 1 - x^2$ $x^2 = 1 - u$ $x = \sqrt{1-u}$

For the subtracted logarithm, set $u$ equal to $\frac{1+x}{1-x}$:

$u = \frac{1+x}{1-x}$ $u(1-x) = 1 + x$ $u - ux = 1 + x$ $u - 1 = ux + x$ $ux + x = u - 1$ $x(u + 1) = u - 1$ $x = \frac{u-1}{u+1}$

The addition of the logarithms requires taking a square root—a complicated function in and of itself—to convert our main input into the series input, while the subtraction of the logarithms only requires the basic operations of addition, subtraction, and division. Subtracting the logarithms is the clear winner. Now we can take our two series and combine them in this manner:

$\ln{(1+x)} = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \frac{x^6}{6} + \ldots$ $-\ln{(1-x)} = -(-x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \frac{x^5}{5} - \frac{x^6}{6} - \ldots)$

Distributing the negative sign on the second series, we get

$\ln{(1+x)} = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \frac{x^6}{6} + \ldots$ $-\ln{(1-x)} = x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \frac{x^5}{5} + \frac{x^6}{6} + \ldots ,$

and then combining them gives us

$\ln{ \biggl( \frac{1+x}{1-x} \biggr)} = 2x + \frac{2}{3}x^3 + \frac{2}{5}x^5 + \frac{2}{7}x^7 + \ldots$ $\ln{ \biggl( \frac{1+x}{1-x} \biggr)} = \sum_{n=0}^{\infty} \frac{2}{2n+1}x^{2n+1} .$

Substituting our initial value, $u$, into the input, we get

$\ln{(u)} = \sum_{n=0}^{\infty} \frac{2}{2n+1} {\biggl( \frac{u-1}{u+1} \biggr)}^{2n+1} = 2{\biggl( \frac{u-1}{u+1} \biggr)} + \frac{2}{3} {\biggl( \frac{u-1}{u+1} \biggr)}^3 + \frac{2}{5} {\biggl( \frac{u-1}{u+1} \biggr)}^5 + \frac{2}{7} {\biggl( \frac{u-1}{u+1} \biggr)}^7 + \ldots$

We now have a series that increases by two degrees for each new term added! But how much quicker does this series converge compared to our old series? I have made a Desmos graph where you can compare the two series. The series for $\ln{(1+x)}$ is in red and our new series is in green. Slide the value for $N$ to add terms to the series approximations, and note how they converge both by looking at the graph and by looking at the calculation for the natural log of 2. (Also, notice the difference in the radius of convergence.) The old series takes 36 terms to reach Desmos’ calculator accuracy (11 decimal places), but the new series only takes 12! We reduced the number of terms we need by one third! And when you have the option to reduce the argument like we did with the example to find the natural log of 15, the number of terms reduces further. In my Python script for the new series, the code calculates the natural logarithm for any number with 15 terms, less than one third of the 48 terms needed for my script for the other series! This series is what is used in the logarithm function used in the C library, which was made during a time where memory was much more limited, so optimization was much more necessary than it is today; it only uses 8 terms of the series after optimizing the argument reduction algorithm to get an answer accurate to double floating-point precision. Pictured: A comparison of the 3-term approximations for the series $\ln{(1+x)}$ (red) and $\ln{\Bigl( \frac{x-1}{x+1} \Bigr)}$ (green) to $\ln{(x)}$ (dotted black). Even after only three terms, the latter series is a much better approximation than the former.

But regardless of whichever series is used under the hood, you now understand how your computer and your calculator compute and give you an accurate value to the natural logarithm function. And with the change of base formula, you can calculate the logarithm of any base.