# Approximating the impact of inflation

The other day someone mentioned to me a rule of thumb that he was using to estimate the number of years $n$ it would take for inflation to destroy half of the purchasing power of today’s money:
$$n = \frac{70}{p}$$
Here $p$ is the inflation in percent, e.g. if the inflation rate is $2\%$ then today’s money would buy only half of today’s goods and services in 35 years. You can also think of a saving account with an interest rate of $2\%$ that would double your money in 35 years.

It is not difficult to derive this formula. The starting point is:
$$2K = K (1 + \frac{p}{100})^n$$
This is equivalent to:
$$2 = (1 + \frac{p}{100})^n$$
Taking the log gives:
$$\log(2) = n \log(1 + \frac{p}{100})$$
The first term of the Taylor series approximation of $\log(1+x)$ for small $x$ is $x$. Hence for small $p$ I can set:
$$\log(2) \doteq n \, \frac{p}{100}$$
Next I have to estimate the value for $\log(2)$. Writing it as an integral leads to:
$$\log(2) = \int_1^2 \frac{1}{x} \,dx$$
Using Simpson’s rule I can approximate the integral with:
$$\int_1^2 \frac{1}{x} \,dx \doteq \frac{2-1}{6} (1+4\frac{2}{1+2}+\frac{1}{2} ) = \frac{25}{36} \doteq 0.7$$
Thus,
$$n \doteq \frac{70}{p}$$
Plotting the two formulas against each other reveals that the approximation works pretty well, even for inflation rates up to 10%.

### R Code

Here is the R code to reproduce the plot.
curve(70/x, from=1, to=10,       xlab=“Inflation rate p%”,       ylab=“Number of years for purchaing power to half”,       main=“Impact of inflation on purchasing power”,      col=“blue”,       type=“p”, pch=16, cex=0.5)curve(log(2)/(log(1+x/100)),       from=1, to=10, add=TRUE,       col=“red”)legend(“topright”,        legend=c(“70/p”,“log(2)/log(1+p/100)”),        bty=“n”,       col=c(“blue”, “red”),        pch=c(16,16), pt.cex=c(1,1))