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Forgive me if my question is slightly incorrect.

I once saw a note, where it was stated that to compute $\tan\left(10^{100}\right)$ (by hand, without calculator) we need to know the $100$-th digit of $\pi$.

Is it true, and if it indeed is, how is this $100$-th number related to the task?

  • Does the digit of pi include $3$ befor the decimal point, or does it start with $1415...$? Because it is important. – KKZiomek Nov 25 '16 at 19:37

3 Answers3

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Suppose you want to find $\tan(1000)$ by hand. Since you're Feynman and you're really smart you can do this fairly accurately for numbers in the interval $(-\frac \pi 2, \frac \pi 2)$ but since the tangent function is periodic you can reduce it to that case by using that $1000-318\pi=0.973 \dotsb$ but if your approximation of $\pi$ were of by $10^{-3}$, say $3.141$ you would get $1000-318\times 3.141 = 1.161$ and since the error in the ratio of $\tan(1.161)$ and $\tan(0.973)$ is more than 10% you lost your bet.

Sophie
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A user in the comments answers this in the following question here.

To quote

because the tangent function is periodic of period π, to know what the value is you need to know 10100modπ; this is effectively the same as knowing π to a hundred places`

edit: link to the answer is here

Community
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KyleKW
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The digits needed are actually the 101st digit and 102nd digit after the decimal point of $1\over\pi$. These two digits happen to be "12" (courtesy of Wolframalpha) so $\tan\left(10^{100}\right)\approx\tan(0.12\pi)$.

To see why this works, let's follow the entire calculation:

Fetching the first 102 digits of $1\over\pi$ from Wolframalpha, we get:

0.318309886183790671537767526745028724068919291480912897495334688117793595268453070180227605532506171912... (note the "12" in the end)

So $10^{100}\over\pi$ is:

3183098861837906715377675267450287240689192914809128974953346881177935952684530701802276055325061719.12...

Which we will write as $N+0.12...$ (where $N$ is an integer).

Now we can write:

$\tan\left(10^{100}\right)=\tan\left[\left(10^{100}\over\pi\right)\cdot\pi\right]\approx\tan\left[\left(N+0.12\right)\cdot\pi\right]=\tan\left(N\pi+0.12\pi\right)$

And since $\tan(x)$ has a period of $\pi$, we get:

$\tan(N\pi+0.12\pi)=\tan(0.12\pi)$

Which can further be approximated as:

$\tan(0.12\pi)\approx\tan(0.38)\approx0.38$ (the last step uses the approximation of $\tan(x)\approx{x}$ for small $x$)

KobMiller
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    As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. – Community Nov 11 '21 at 00:54