## A Chat with a Calculator

### by Gregory Marton

I recently had the following chat with a particular calculator:

- e^(e^(e^(e^e))) = 10^(10^(10^6.219196780089781))
- e^(e^(e^(e^(e^e)))) = 10^(10^(10^(10^6.219196780089781)))
- e^(e^(e^(e^(e^(e^(e^(e^(e^e))))))) = 10^(10^(10^(10^(10^(10^(10^(10^6.219196780089781)))))))

It seems odd to me that putting a few more e’s down the bottom should result in it thinking there were the same number of extra 10s at the bottom. In fact, I’ve never seen a calculator answer in this form at all. I’m especially intrigued that the final power of ten seems to be the same in all three cases, so it can’t even just be estimating. Do you have any thoughts on what screwy counting could be behind these particular answers?

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## Lauren Ouellette:

For all of these examples, the calculator in question gives a “Computation timed out” error message following the output. However, it is able to correctly calculate e^(e^(e^(e))), which it represents as 2.33… × 10^1656520. Note that 10^1656520 = 10^(10^6.219196780089781), the power of ten seen in the above answers.

My guess is that in computing e^(e^(e^(e^e))), the calculator begins by finding the value for e^(e^(e^e)). It then tries to compute e^2.33… × e^(10^(10^6.219196780089781)). The computation of the latter term fails along the way. For some reason(?), it does not die, but actually returns a value of 10^(expt) rather than e^(expt). This gives 10^(10^(10^6.219196780089781)).

Presumably the computation of e^(e^(e^(e^(e^e)))) fails in the same way, returning 10 to the power of (the incorrect result for) e^(e^(e^(e^e))). This would explain why another power of 10 gets tacked on for each additional e.

I’ve got no idea what kind of bug might be causing the 10^(expt) error, though. I’d be curious to find out!

27 July 2011, 9:46 pm## Lauren Ouellette:

Oops, I meant to write (e^2.33…) ^ (10^(10^6.219196780089781)), not e^2.33… × e^(10^(10^6.219196780089781)). Also, I wonder whether the 10 tacked on each time is some default of the “power of 10 representation”, or whether it’s actually an initial approximation for e^2.33… .

27 July 2011, 10:13 pm## Austin:

I think your calculator is correct. 🙂

Any number has a power of 10 representation (and a power of e representation). See Wikipedia, “level-index arithmetic”. The counter-intuitive thing is “that putting a few more e’s down the bottom should result in it thinking there were the same number of extra 10s at the bottom.” However, try taking base-10 logarithms of a tall tower of e’s, and you’ll see why the resulting factors of 1/(ln 10) are vanishingly insignificant (until close to the end of the process).

(By the way, I can’t say I’m smart enough to figure this out on my own. I heard about it from Nick Trefethen, who made the same discovery as you and was equally startled by it.)

28 July 2011, 8:36 am## Philip Petrov:

Check this out:

http://www.wolframalpha.com/input/?i=e^%28e^%28e^%28e^%28e^%28e^%28e^%28e^%28e^e%29%29%29%29%29%29%29

28 July 2011, 12:21 pm## Philip Petrov:

Sorry, the link above got broken. Here it is:

wolfram alpha does the same thing

28 July 2011, 12:22 pm## Chris Chang:

Yes, the calculator is correct.

e^x = 10^(x / 2.302…), so, in particular, if we already have e^(e^(e^(e^e))) = 10^(10^(10^6.219…)), it follows that e^(e^(e^(e^(e^e)))) = 10^(10^(10^(10^6.219…)) / 2.302…). Division by 2.302… in the first exponent from the bottom of the tower is equivalent to subtracting 0.36… from the second exponent, which is equivalent to a microscopic change in the third exponent.

6 August 2011, 3:00 am