# 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))