In the last few months of 2023 and beginning of 2024, several banks in Italy have proposed deposit accounts with a favorable yield per year (around 4%) for a year or more. However some offers guarantee 5% $\tiny{for}$ $\tiny{6}$ $\tiny{months}$.
For computing the total interest for multiple years, or for a fraction of it, we need the following formula:
$$ \begin{aligned} I_t = I_0 \cdot (1+r)^t \end{aligned} $$
where $t$ is time, $I_t$ is the investment at time $t$, $r$ is the interest rate and $I_0$ is the initial investment ($t = 0$). The compound interest rate is $(1+r)^t-1$.
Therefore, the interest rate for 6 months with the yearly 5% is $(1+0.05)^{\frac{1}{2}}-1 = 1.02469-1=2.469 \%$.
It looks like that for a quick estimate, rather then try in your head to compute the square root of $1+somethingsmall$, is enough to divide by two the yearly interest.
Let’s try with other small values of interest rate: $1\%, 2\%, 4\%, 5\%, 10\%$
yearly rate (%) | divide by 2 estimate (%) | exact rate (%) |
---|---|---|
1 | 0.5 | 0.499 |
2 | 1 | 0.995 |
4 | 2 | 1.980 |
5 | 2.5 | 2.469 |
10 | 5 | 4.881 |
The quick formula is (of course) a good approximation. But for larger rates knowing the square root of something known close to it, it is more accurate and still easy to compute. That’s the case for $20\%, 40\%, 70\%, 90\%$ that translates in computing the square root of $1.2, 1.4, 1.7, 1.9$ that are close to $\sqrt{\frac{121}{100}}, \sqrt{\frac{144}{100}}, \sqrt{\frac{169}{100}}, \sqrt{\frac{196}{100}}$.
yearly rate (%) | divide by 2 estimate (%) | exact rate (%) | close square root estimate (%) |
---|---|---|---|
20 | 10 | 9.545 | 10 $(since \sqrt{1.2} \approx \sqrt{1.21} = \frac{\sqrt{121}}{\sqrt{100}} = 1.1)$ |
40 | 20 | 18.322 | 20 $(since \sqrt{1.4} \approx \sqrt{1.44} = \frac{\sqrt{144}}{\sqrt{100}} = 1.2)$ |
70 | 35 | 30.384 | 30 $(since \sqrt{1.7} \approx \sqrt{1.69} = \frac{\sqrt{169}}{\sqrt{100}} = 1.3)$ |
90 | 45 | 37.840 | 40 $(since\sqrt{1.90} \approx \sqrt{1.96} = \frac{\sqrt{196}}{\sqrt{100}} = 1.4)$ |
And for the case of extremely high return, that when summed to the initial investment are perfect squares, the close-square-root is the exact rate:
yearly rate (%) | divide by 2 estimate (%) | exact rate (%) | close square root estimate (%) |
---|---|---|---|
300 | 150 | 200 | 200 |
800 | 400 | 300 | 300 |
1500 | 750 | 400 | 400 |
At higher rates the estimate with the “close square root” is much more accurate. But what can we do when the perfect square is close but not too close?
Heron Method for Square Root
I remembered that I read a blog post that could help me and after digging in my browser history I found Estimating Square Roots in Your Head by Gregory Gundersen. This method is named after Heron of Alexandria, famous mathematician and engineer (inventor of the first vending machine, automatic door and wind-powered organ). It comes down to approximate the square root of a number adjusting a little the square root of the closest perfect square.
Suppose to have $n$ and you can decompose it $n = g^2 + x $. A rough estimate of $\sqrt{n} = \dfrac{g + \frac{n}{g}}{2}$.
Here are some examples:
yearly rate (%) | divide by 2 estimate (%) | exact rate (%) | Heron estimate (%) |
---|---|---|---|
30 | 15 | 14.018 | 14 $(since \frac{1.1+\frac{1.3}{1.1}}{2} \approx 1.14)$ |
40 | 20 | 18.321 | 18.5 $(since \frac{1.2+\frac{1.4}{1.2}}{2} \approx 1.185)$ |
50 | 25 | 22.464 | 22.5 $(since \frac{1.2+\frac{1.5}{1.2}}{2} = 1.225)$ |
80 | 40 | 34.164 | 35 $(since \frac{1.3+\frac{1.8}{1.3}}{2} \approx 1.35)$ |
100 | 50 | 41.421 | 40 $(since \frac{1.4+\frac{2}{1.4}}{2} \approx 1.4)$ |
200 | 100 | 73.205 | 72.5 $(since \frac{1.7+\frac{3}{1.7}}{2} \approx 1.725)$ |
500 | 250 | 144.949 | 145 $(since \frac{\frac{25+\frac{600}{25}}{2}}{10} = 24.5)$ |
1000 | 500 | 231.662 | 233 $(since \frac{3+\frac{11}{3}}{2} \approx 3.33)$ |
A really nice result! Before showing why it works, let’s consider (another) famous formula related to compound interest.
Rule of 70
The rule of 70 is used to determine approximately the number of years it takes for a variable to double by dividing the number 70 by the variable’s rate. It is used to compute how many years $(t)$ it would take for an investment to double given the annual rate of return $(r)$. In short:
$$ \begin{aligned} t = \frac{70}{r} \end{aligned} $$
For instance, take $r=2 (\%)$ to double your billion dollar you need roughly 35 years. How far is this estimation from the real value? Plugging in the value in compound interest formula $(1+r)^t$ we get $(1+0.02)^{35} = 1.02^{35} = 1.99988$.
Pretty sweet an error under $0.1\%$! Again, let’s try with different values of interest rates:
rate (%) | estimated years | investment after years |
---|---|---|
1 | 70 | 2.007 |
2 | 35 | 1.999 |
4 | 17.5 | 1.986 |
5 | 14 | 1.980 |
10 | 7 | 1.949 |
20 | 3.5 | 1.893 |
50 | 1.4 | 1.764 |
70 | 1 | 1.700 |
100 | 0.7 | 1.625 |
It was just not luck, but you can see how increasing the rate the formula gets less close to the right result.
Taylor Series
The fact that those approximations work when we are close enough to a perfect square for Heron Method and for low rates in Rule of 70 should be an hint that there is something to do with Taylor’s series. Indeed straight out of Wikipedia:
In mathematics, the Taylor series or Taylor expansion of a function is an infinite sum of terms that are expressed in terms of the function’s derivatives at a single point. For most common functions, the function and the sum of its Taylor series are equal near this point.
or if you prefer the formula:
$$ \begin{aligned} f(x) = \sum\limits_{n=0}^{\infty} \dfrac{{f^{(n)}(a)}}{{n!}}(x - a)^{n} \end{aligned} $$
where the notation $f^{(n)}$ denotes the $n^{th}$ derivative of $f$.
Taylor feat Heron
Please forgive me for this low quality meme, but I couldn’t help making (and sharing) it:
Back to math, the approximation at the second order of the Taylor series for the square root is $\sqrt{x} \approx \sqrt{a} + \dfrac{x-a}{2\sqrt{a}} = \dfrac{2a+x-a}{2\sqrt{a}} = \dfrac{a+x}{2\sqrt{a}} = \dfrac{\sqrt{a}+\frac{x}{\sqrt{a}}}{2}$
that is exactly the Heron formula.
Taylor feat Rule of 70
From the formula of compound interest we have $I_t = I_0 \cdot (1+r)^t$. Under the assumption that $I_t = 2 \cdot I_0$ it simplifies into $ 2 = (1+r)^t$. Solving the equation for t:
$$ \begin{aligned} \ln{(1+r)^t} &= \ln{2} \\\ t\ln{(1+r)} &= \ln{2} \\\ t &= \dfrac{\ln{(2)}}{\ln{(1+r)}} \\\ \end{aligned} $$
Considering small values of $r$, the second order approximation of $\ln{(1+r)} \approx \ln(1) + r = r$. Therefore we obtain $t \approx \dfrac{\ln{2}}{r} \approx \dfrac {0.693}{r}$. Using the percentage value of $r$ we get (with a little round up) the rule of 70 (actually this rule has the name 72, 70 or 69).
Bonus Track: Rule of $\approx 100 \cdot \log{n}$
For the doubling of your investment you can use the rule of 70. But why stop here? Why don’t look for higher return (💎🙌)? Extrapolating from the above formula, and rounding a little bit for ease of memorization we get
$n$ | $\log(n)$ | Rule of $\approx 100 \cdot \log{n}$ |
---|---|---|
2 | 0.69 | 70 |
3 | 1.10 | 110 |
5 | 1.61 | 160 |
Thanks to the logarithmic properties $\log(a \cdot b) = \log(a) + \log(b)$ and $\log(a^k) = k \cdot \log(a)$, we can compute the rule for a bunch of multipliers just remembering 70, 110 and 160.
$n$ | Rule of $\approx 100 \cdot \log{n}$ |
---|---|
4 | 2$\cdot$70 = 140 |
6 | 70+110 = 180 |
8 | 3$\cdot$70 = 210 |
9 | 2$\cdot$110 = 220 |
10 | 70+160 = 230 |
12 | 70+180 = 250 |
Furthermore with a little of weighting we can estimate the rule for numbers in between. For instance, since $6 < 7 < 8 \rightarrow \log(6) < \log(7) < \log(8)$ thus for 7 we have the rule of $\dfrac{180+210}{2}=195$ (it’s a rough estimate). Following the same idea we can compute the rule for other prime numbers like 11, and 13.
$n$ | Rule of $\approx 100 \cdot \log{n}$ |
---|---|
7 | $\dfrac{180+210}{2}=195$ |
11 | $\dfrac{230+250}{2}=240$ |
13 | $\dfrac{250+(195+70)}{2} \approx 255$ |
Let’s compute the errors for small rates: 1%, 2% and 5%:
$n$ | $\approx 100 \cdot \log{n}$ | $t$ for $r=1\%$ | estimated $n$ with $r=1\%$ | $t$ for $r=2\%$ | estimated $n$ with $r=2\%$ | $t$ for $r=5\%$ | estimated $n$ with $r=5\%$ |
---|---|---|---|---|---|---|---|
2 | 70 | 70 | 2.007 | 35 | 2.000 | 14 | 1.980 |
3 | 110 | 110 | 2.988 | 55 | 2.972 | 22 | 2.925 |
4 | 140 | 140 | 4.027 | 70 | 4.000 | 28 | 3.920 |
5 | 160 | 160 | 4.914 | 80 | 4.875 | 32 | 4.765 |
6 | 180 | 180 | 5.996 | 90 | 5.943 | 36 | 5.792 |
7 | 195 | 195 | 6.961 | 97.5 | 6.895 | 39 | 6.705 |
8 | 210 | 210 | 8.081 | 105 | 7.999 | 42 | 7.762 |
9 | 220 | 220 | 8.927 | 110 | 8.831 | 44 | 8.557 |
10 | 230 | 230 | 9.861 | 115 | 9.750 | 46 | 9.434 |
11 | 240 | 240 | 10.893 | 120 | 10.765 | 48 | 10.401 |
12 | 250 | 250 | 12.032 | 125 | 11.886 | 50 | 11.467 |
13 | 255 | 255 | 12.646 | 127.5 | 12.489 | 51 | 12.040 |
Impressive (not knowing the math behind) that the error is even in the worst cases below 10%. Now you are ready to impress your financial advisor with these tricks!
Caveat: if the guaranteed return is really high is probably a scam and this is not financial advice!