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 times a high enough power of 10 is approximately .
A commenter named Robert suggested looking at the difference between this approximation for and itself. He remarked that the error is also approximately too (when multiplied by a sufficiently high power of 10)!
If we shift the decimal point over we get an approximation of :
Now we look at the error
which if we shift again is
Wow! Now why is this true?
The key observation is that
or in general,
where denotes the -digit integer of all 5’s. With a little algebra we see that
As I mentioned last time, if the calculator is in degree mode and , then . So,
This implies that
Now, the error in this approximation is
We can rewrite this as
Thanks for providing the explanation!
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.
“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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