# 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)),
pch=c(16,16), pt.cex=c(1,1)) 