# How do I Approximate $\log{2}\approx 0.693$ without using the Maclaurin series?

How do I approximate the value $\log{2}\approx 0.693$ without using the Maclaurin series?

The book gives the hint: consider $f(x)=e^x-e^{-x}-2x$.

• $f(\log 2) = \frac{3}{2}-2\log 2$. Also: $$f(x)=\frac{2x^3}{3!} + \frac{2x^5}{5!}\dots$$ – Thomas Andrews Sep 19 '15 at 15:35
• TBH isn't this still using Maclaurin series? – MrYouMath Sep 19 '15 at 15:39
• @MrYouMath Specifically, because the question says "without using the Maclaurin series." I took that to mean the specific, slow-converging sequence: $1-1/2+1/3-\cdots$. – Thomas Andrews Sep 19 '15 at 15:47
• As an aside, notice that $\ln2=0.7\color{red}-0.007$, while $\sqrt2=0.7\color{red}+0.007$. – Lucian Sep 19 '15 at 16:59
• See the selected answer here for a good way to approximate log function:math.stackexchange.com/questions/135368/… – NoChance Sep 19 '15 at 18:04

If you start from $\ln(2)=\int_1^2\frac{1}{x}\,dx$, then Simpson's Rule with $n=4$ gets you there fast:

$$\ln(2)\approx\frac{1}{12}\left(1+4\left(\frac45\right)+2\left(\frac23\right)+4\left(\frac47\right)+\frac12\right)\approx0.693\ldots$$

Surely not what the book's author had in mind, but still instructive:

On a calculator with a $\sqrt{\phantom2}$ button you could approximate $\log c$ by repeatedly hitting $\sqrt{\phantom2}$, getting a sequence of numbers each about twice as close to $1$ as the one before. After $n$ iterations we reach $c^{2^{-n}}$, which for large $n$ is about $1 + 2^{-n} \log c$. So subtract $1$ and multiply by $2^n$ to approximate $\log c$. On a 10-digit calculator you can get 5-digit accuracy (confirmed by comparison with the output of the "$\ln$" button...) by iterating enough times to get within about $10^{-5}$ of $1$, which is to say $n \cong 16.6 + \log_2 \ln c$ [with $16.6$ arising as $\log_2 (10^5)$].

For $c=2$ we get $n \cong 16$, and indeed calculation of $$2^{16} (2^{2^{-16}} - 1) = 2^{16} \left(\sqrt{\sqrt{\sqrt{\cdots\sqrt 2}}} - 1 \right)$$ yields a number between $0.6931$ and $0.6932$.

• This is Feynman's approach: feynmanlectures.caltech.edu/I_22.html – Michael Lugo Sep 28 '15 at 14:08
• Thank you for this reference. Feynman credits it to "Mr. Briggs of Halifax, in 1620" $-$ who didn't have a $\sqrt{\phantom2}$ button (let alone a $\ln$ button) and computed each square root by hand! – Noam D. Elkies Sep 28 '15 at 14:38

Motivated by the hint, $\ln(2)$ is a fixed point of $$f(x)=\frac{e^x-e^{-x}-2x-\frac32}{-2}$$ (This is a bit of a departure from the hint.)

It happens to be an attractor. Now, I'm assuming you can exponentiate an arbitrary decimal, since the hint gives you $e^x$. You have

\begin{align} a_0&=1\\ a_1&=f(a_0)=0.57\ldots\\ a_2&=f(a_1)=0.71\ldots\\ a_3&=f(a_2)=0.6867\ldots\\ a_4&=f(a_3)=0.6947\ldots\\ a_5&=f(a_4)=0.6927\ldots\\ a_6&=f(a_5)=0.6932\ldots\\ a_7&=f(a_6)=0.69312\ldots\\ a_8&=f(a_7)=0.69315\ldots\\ a_9&=f(a_8)=0.69314\ldots\\ \end{align} Continue for more accuracy.

• It looks as if $f(\log(2))=-\log(2)$. Did you mean to use $-f(x)$? Since $-f'(\log(2))=\frac34$ (which is less than $1$ in absolute value), iteration will converge if the initial point is close enough to $\log(2)$. – robjohn Sep 20 '15 at 8:46
• @robjohn I did mean that. I had a $-2$ in my original denominator. Thanks. – alex.jordan Sep 20 '15 at 15:51

A standard definition of $\log 2$ is $$\log 2:=\int_1^2{dx\over x}\ .$$ Consider the function $$h(x):=2-{4x\over3}+{8x^2\over27}\ .$$ Writing $x:={3\over2}+t$ $\>\bigl(-{1\over2}\leq t\leq{1\over2}\bigr)$, one computes $${1\over{3\over2}+t}-h\left({3\over2}+t\right)=-{16t^3\over 81+54 t}\tag{1}$$ (and this explains the choice of $h$). Using $(1)$ we obtain $$\log 2=\int_1^2 h(x)\>dx+R={56\over81}+R$$ with $$R=\int_{-1/2}^{1/2}{-16t^3\over 81+54t}\>dt=\int_0^{1/2}16t^3\left({1\over 81-54 t}-{1\over 81+54t}\right)\>dt\ .$$ From $$0\leq\left({1\over 81-54 t}-{1\over 81+54t}\right)\leq {108 t\over 81^2-27^2}\qquad\left(0\leq t\leq{1\over2}\right)$$ it then follows that $$0<R\leq{1\over 540}\ .$$ This proves the estimate $$0.691\leq{56\over81}\leq\log2\leq{1123\over1620}\leq0.69321\ .$$

What is "the" MacLaurin series you are referring to? To get $\log(2)$, for instance, we may evaluate the Taylor series of $\log(1-x)$ around $x=0$ at $x=-1$, or at $x=\frac{1}{2}$. In the second case we get the pretty fast-convergent series: $$\log(2) = \sum_{n\geq 1}\frac{1}{n 2^n} \tag{1}$$ that can be further "accelerated" through Euler's method, getting: $$\log(2) = 1-\sum_{n\geq 1}\frac{1}{n(n+1) 2^n}\tag{2}$$ or: $$\log(2) = \frac{1}{2}+2\sum_{n\geq 1}\frac{1}{n(n+1)(n+2)\,2^n}.\tag{3}$$ Another fast-convergent representation (a BBP-type series) that comes from the integral of a rational function over $[0,1]$ is: $$\log(2)=\frac{1}{3}\sum_{n\geq 0}\frac{(-1)^n}{(27)^n}\left(\frac{3}{6n+1}-\frac{2}{6n+3}-\frac{1}{6n+4}\right)\tag{4}$$ and the last representation requires just $2$ terms (!) to reach the wanted accuracy.

Another chance is given by the following technique (essentially due to Beuker). The integral: $$I = \int_{0}^{1}\frac{x^6(1-x)^6}{1+x}\,dx = -\frac{19519}{440}+64 \log(2)\tag{5}$$ is clearly positive but smaller than $\frac{1}{4^6}$. The approximation: $$\log(2)\approx \frac{19519}{28160}\tag{6}$$ hence meet the required accuracy.

If $y=\ln(x)$, then $$y'=\frac{1}{x}\quad\text{and}\quad y'=e^{-y}$$ and you can numerically solve either differential equation to $x=2$, using initial condition $y(1)=0$.

With the first differential equation, using the Runge-Kutta method with only two steps: $$\begin{array}{rrr|rrrr} n&x_n&y_n&k_{1,n}&k_{2,n}&k_{3,n}&k_{4,n}\\ 0&1&0\\ &&&\frac{1}{1}=1&\frac{1}{1+\frac{1}{4}}=\frac{4}{5}&\frac{1}{1+\frac{1}{4}}=\frac{4}{5}&\frac{1}{1+\frac{1}{2}}=\frac{2}{3}\\ 1&\frac32&0+\frac{\frac{1}{2}}{6}\left(1+2\cdot\frac{4}{5}+2\cdot\frac{4}{5}+\frac{2}{3}\right)=\frac{73}{180}\\ &&&\frac{1}{\frac{3}{2}}=\frac{2}{3}&\frac{1}{\frac{3}{2}+\frac{1}{4}}=\frac{4}{7}&\frac{1}{\frac{3}{2}+\frac{1}{4}}=\frac{4}{7}&\frac{1}{\frac{3}{2}+\frac{1}{2}}=\frac{1}{2}\\ 2&2&\frac{73}{180}+\frac{\frac{1}{2}}{6}\left(\frac{2}{3}+2\cdot\frac{4}{7}+2\cdot\frac{4}{7}+\frac{1}{2}\right)=\frac{1747}{2520} \end{array}$$ And $\frac{1747}{2520}=0.693\ldots$. For more accuracy, you have to start over with a smaller step size (and hence, more steps).

If you really don't want to use power series methods, here is one way: $e^{.6}\approx 1.822\ldots$ which is less than $2$. $e^{.7}\approx 2.013\ldots$ which is larger than $2$. So try $.65$: $e^{.65}\approx 1.915\ldots$. That's too small so try half way between $0.65$ and $0.7$, $0.675$: $e^{.675}\approx 1.964\ldots$. That's still too small so try half way between $0.675$ and $0.67$, $0.6725$: $e^{.6725}\approx 1.959\ldots$. Continue like that to desired accuracy.

• How do you calculate $e^x$? – YoTengoUnLCD Sep 19 '15 at 16:03
• This method is called binary search, and I don't agree with the down votes. The hint suggests that maybe it is acceptable to exponentiate arbitrary decimals. In any case OP did not rule that out. – alex.jordan Sep 20 '15 at 8:03

$\ln(2)$ is a zero to $f(x)=e^x-2$. Assuming you can exponentiate an arbitrary decimal, Newton's method converges fast.

\begin{align} a_0&=1\\ a_1&=a_0-\frac{e^{a_0}-2}{e^{a_0}}=a_0-1+2e^{-a_0}=0.7357\ldots\\ a_2&=a_1-1+2e^{-a_1}=0.6940\ldots\\ a_3&=a_2-1+2e^{-a_2}=0.6931\ldots\\ \end{align}

and continue to the desired precision.

Here's one for base 10, instead of base $e$:

$2^{10} = 1024 \approx 1000$, so $10 \log 2 \approx 3$ or $\log 2 \approx .3$.

• When going only for one decimal, this already follows form $2^3\approx 10$ – Hagen von Eitzen Sep 20 '15 at 7:46
• This is accurate to almost 3 places: log 2 =.301... – marty cohen Sep 20 '15 at 14:22
• @HagenvonEitzen 1024 for 1000 is an error of 2.4% while 8 for 10 is an error of 20%. – Did Sep 20 '15 at 16:02