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*suspiciously*large number of the digits of pi. Perhaps you have met one of these people. They can typically be found hiding behind bushes and under the counters at pastry shops, just...

*waiting*.

At the slightest hint of a mention of pi, they will jump out and start reciting the digits like there's a prize at the end. After rattling off numbers for a few minutes they abruptly come to an end, grin like an idiot, and walk away. It is an unpleasant encounter.

The sheer uselessness of this kind of thing has always bothered me, so I'd like to set a preliminary upper bound on the number of digits of pi that could ever possibly potentially kind of be useful (maybe). For those following along at home, now would be a good time to put on your numerology hats.

Alright, so I hear this thing pi is fairly useful when dealing with circles. Let's say we want to make a really big circle and have its diameter only deviate by a very small amount from the correct value. To do this successfully, we will have to know pi fairly well.

Let's take this to extremes now. Suppose I want to put a circle around the

*entire visible universe*such that the uncertainty in the diameter is the size of a

*single proton*. What would be the fractional uncertainty in the circumference in this case?

If we know pi exactly, then we have that

\[\delta C = \frac{\partial C}{\partial d} \delta d = \pi \delta d = C \frac{\delta d}{d}, \]

where d is the diameter and C is the circumference. In other words, the fractional uncertainty in the circumference is just

\[\frac{\delta C}{C} = \frac{\delta d}{d}. \]

Using a femtometer for the size of a proton and 90 billion light years for the size of the Universe [1], we get

\[\frac{\delta C}{C} = \frac{\delta d}{d} = \frac{10^{-15}\mbox{m}}{(90\times10^9)(3\times10^7\mbox{s})(3\times10^8\mbox{m s}^{-1})} \sim\frac{10^{-15}}{10^{27}}\sim10^{-42}.\]

Alright, so how well do we need to know pi to get a similar fractional uncertainty? Well, we have that

\[\frac{\delta \pi}{\pi} = \frac{\delta C}{C} = 10^{-42}, \]

so we can afford an uncertainty in pi of

\[ \delta \pi = \pi \times 10^{-42}\]

and thus we'll need to know pi to about 42 digits. How's that for an answer?

So if we have a giant circle the size of the

*entire visible universe*, we can find its diameter to within the size of a

*single proton*using pi to 42 digits. Therefore, I adopt this as the maximal number of digits that could ever prove useful in a physical sense (albeit under a somewhat bizzarre set of circumstances).

If reciting hundreds of digits is what makes you happy, go for it. But 42 digits is more than enough pi for me.

[1] "But I thought the Universe was only 13.7 billion years! What voodoo is this!?" Yeah, I know. See here for a nice explanation. [back]

Forgive me the trivial question, but is it 3.{42 digits}, or 3.{41 digits}?

ReplyDeleteThe uncertainty would be in the 42nd digit after the decimal point, so 3.{42 digits}. Although, it's all just for fun so if you like 41 better than 42, go with it.

ReplyDelete42 is a well known answer!

ReplyDelete"At the slightest hint of a mention of pi, they will jump out and start reciting the digits like there's a prize at the end."

ReplyDeleteThere is no "end" of pi.

But, there's an end to the digits they know (owing to finite brain size).

DeleteAnd what happens when we measure g-2 to more than 42 significant digits? I'm pretty sure pi figures into the theoretical calculation of its value under the Standard Model.

ReplyDeleteAre we ever in any danger of doing that?

Delete