More about the neat calculator trick

Yesterday I wrote about a neat calculator trick that I had just learned.

We saw that if the calculator was set to degree mode, then {\displaystyle \sin(\frac{1}{555\cdots 5})} times a high enough power of 10 is approximately {\pi}.

A commenter named Robert suggested looking at the difference between this approximation for {\pi} and {\pi} itself. He remarked that the error is also approximately {\pi} too (when multiplied by a sufficiently high power of 10)!

For example,

\displaystyle \sin(1/5555555555)=3.14159265390395250385303346595\ldots\times 10^{-12}.

If we shift the decimal point over we get an approximation of {\pi}:

\displaystyle 10^{12}\sin(1/5555555555)=3.14159265390395250385303346595\ldots

Now we look at the error

\displaystyle 10^{12}\sin(1/5555555555)-\pi=3.1415926539039008267\ldots\times 10^{-10},

which if we shift again is

\displaystyle 10^{10}(10^{12}\sin(1/5555555555)-\pi)=3.1415926539039008267\ldots

Wow! Now why is this true?

The key observation is that

\displaystyle \frac{1}{180}=0.00\bar{5}=0.00555555+00000000\bar{5}=555555\cdot10^{-7}+10^{-5}\frac{1}{180},

or in general,

\displaystyle \frac{1}{180}=n_{k}\cdot10^{-k-2}+10^{-k}\frac{1}{180},

where {n_{k}} denotes the {k}-digit integer of all 5’s. With a little algebra we see that

\displaystyle \frac{1}{n_{k}}=10^{-k-2}\cdot\frac{180}{1-10^{-k}}.

As I mentioned last time, if the calculator is in degree mode and {x\approx 0}, then {\sin(x)\approx \pi x/180}. So,

\displaystyle \sin\Big(\frac{1}{n_{k}}\Big)\approx\frac{\pi}{180}\cdot 10^{-k-2}\cdot\frac{180}{1-10^{-k}}=\frac{\pi10^{-k-2}}{1-10^{-k}}.

This implies that

\displaystyle 10^{k+2}\sin\Big(\frac{1}{n_{k}}\Big)\approx\frac{\pi}{1-10^{-k}}\approx \pi.

Now, the error in this approximation is

\displaystyle 10^{k+2}\sin\Big(\frac{1}{n_{k}}\Big)-\pi\approx\frac{\pi}{1-10^{-k}}-\pi=\frac{10^{-k}\pi}{1-10^{-k}}.

We can rewrite this as

\displaystyle 10^{k}(10^{k+2}\sin\Big(\frac{1}{n_{k}}\Big)-\pi)\approx\frac{\pi}{1-10^{-k}}\approx \pi.


  1. Robert says:

    Thanks for providing the explanation!

  2. Nemo says:

    Instead of remembering the decimal expansion for 1/180, I usually just remember that .kkkkkk… is k/9 (where k is any single digit). .1111… is 1/9, from which it follows that .222… is 2/9, etc.

    The analogous statement is true in other numerical bases; for example, .222222… in base 3 is 2/3.

    All of which follows trivially from the formula for infinite geometric sums, but remembering it this way is sometimes useful.

    For example, 555555 is approximately 5/9 times 10^6.

  3. moioci says:

    “The analogous statement is true in other numerical bases; for example, .222222… in base 3 is 2/3.”

    Shouldn’t that be 2/3 for base _4_, ie, the denominator is base – 1? To state it a different way, .22222… in base 3 is like .99999… in base 10 = 1.

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