Estimation quiz!

[EdS2]

This is from an old computer magazine: which is larger, the number of millimetres in a mile or the number of square yards in a square kilometre?

Any other quick numeric or numeracy quizzes to share?

(Of course I liked this one because I got it right...)

[Paul Dale]

Interesting problem. Being confident in both systems of units makes it easier.

Which is larger \( e^{\pi} \) or \( \pi ^ e \)?


Pauli

[cruff]

(07-30-2020 11:27 AM)Paul Dale Wrote:  Which is larger \( e^{\pi} \) or \( \pi ^ e \)?

Neither? It appears, to 9 significant digits, these are the same number. Too early in the morning to work out why at the moment.

[Paul Dale]

They differ in their second digit...

Might the arguments to y^x have been backwards?


Pauli

[KeithB]

Which is larger eπ or πe?

isn't an estimation problem. 22 or 23 are the same for an estimate.

[Albert Chan]

(07-30-2020 10:54 AM)EdS2 Wrote:  which is larger, the number of millimetres in a mile or the number of square yards in a square kilometre?

We can estimate relative size without multiplying

1 yard = (36 in) (25.4 mm/in) = 914.4 mm → 1 yard ≈ 1 m

1 mile = 1760 yard ≈ 1.760e6 mm              // over-estimated 10%
1 km^2 = (1e3 m)^2 ≈ 1e6 yard^2           // under-estimated 20%

Since 1.76 > 1.30, we have mile/mm > km^2/yard^2

(07-30-2020 11:27 AM)Paul Dale Wrote:  Which is larger \( e^{\pi} \) or \( \pi ^ e \)?

e^x = 1 + x + x^2/2! + x^3/3! + ... ≥ 1+x
x ≥ ln(1+x)

→ if x≠0, x > ln(1+x)

ln(e^y) = y
ln(y^e) = e*(1 + ln(1 + (y/e-1))) ≤ e*(1 + (y/e-1)) = y

→ if y≠e, ln(y^e) < y ⇒ e^y > y^e
→ e^pi > pi^e

[Werner]

Or, even simpler:

Since e^x = 1 + x + x^2/2 +…, e^x > 1 + x if x>0
Take x = pi/e - 1, x>0 because pi>e
e^(pi/e-1) > 1 + pi/e - 1
e^(pi/e) / e > pi/e
e^(pi/e) > pi
(e^(pi/e))^(e) > pi^(e)
e^pi > pi^e

Cheers, Werner

[Massimo Gnerucci]

(07-30-2020 02:24 PM)Werner Wrote:  Or, even simpler:

Since e^x = 1 + x + x^2/2 +…, e^x > 1 + x if x>0
Take x = pi/e - 1, x>0 because pi>e
e^(pi/e-1) > 1 + pi/e - 1
e^(pi/e) / e > pi/e
e^(pi/e) > pi
(e^(pi/e))^(e) > pi^(e)
e^pi > pi^e

Cheers, Werner

Clever! ;)

[Albert Chan]

Another way, let f = ln(e^x/x^e) = x - e*ln(x)

f' = 1 - e/x = 0 ⇒ x=e           // locate extremum
f'' = e/x^2 > 0                     // 2nd derivative test, f(e) is minimum

if f x ≠ e: f(x) > f(e) = 0 ⇒ e^x > x^e

→ e^pi > pi^e

[rprosperi]

Albert and Werner: It's at times like this that I like to remind people this is exactly why we have calculators - when presented with a problem like this, we can carefully select the weapon of choice, perhaps an HP model not recently used, or an unusual competitive product, and bang it out.

With all due respect to your ever-obvious extreme math skills, I simply say: "Calculators gentlemen! The tool of discerning geeks everywhere"

As for the expected question of how to reply if someone puts a gun to my head and demands I solve the problem without a calculator; my reply: "shoot"

[Werner]

Two remarks: the proof I presented is not mine, I found it somewhere.
And: you win, Albert!
Cheers, Werner

[Albert Chan]

Third way, using the fact that continuous compouding maximize capital growth.
For finite periods n, r>0: (1 + r/n)^n < e^r

Let r = pi - e, n = e:

pi^e = (e + r)^e = e^e * (1 + r/n)^n < e^e * e^r = e^pi

→ e^pi > pi^e

[rprosperi]

(07-30-2020 06:10 PM)Albert Chan Wrote:  Third way, using the fact that continuous compouding maximize capital growth.

Thanks Albert, that page is actually quite good. It's a clear and easy to follow derivation of the various interest formulae. [bookmarked]

[Paul Dale]

(07-30-2020 01:56 PM)KeithB Wrote:  isn't an estimation problem. 22 or 23 are the same for an estimate.

They differ in the second significant digit just like the original problem.
Why would it be an estimation problem?

Pauli

[cruff]

(07-30-2020 01:04 PM)Paul Dale Wrote:  They differ in their second digit...
Might the arguments to y^x have been backwards?

Indeed, pilot error. That's what I get for not looking at the calculator key legend closely.

[EdS2]

Very nice view of the e^pi question!

For the headline (metric and imperial) I got it first by reckoning that both numbers were the same order of magnitude, and then by reckoning that miles are a bit on the large side for metric units, as are square yards, and both of those push in the same direction.

The puzzle is from PCW magazine, in a puzzle feature called Leisure Lines. There's a corresponding book, but almost all the puzzles are not estimations. Here's the one I spotted which is:

Quote:If you could fold a sheet of rice paper, a thousandth of an inch thick, successively doubling the thickness, 50 times over, how thick would the result be?

[Content warning for that book: many cultural references have not dated well. Math and Logic Puzzles for PC Enthusiasts, by J. J. Clessa]

[Paul Dale]

\( 1000 \approx 2^{10} \), so \( \approx 2^{40} \) inches thick -- thicker than the sun's diameter by a factor of 5 or 6.


Pauli

[Valentin Albillo]

(07-31-2020 01:40 PM)Paul Dale Wrote:  \( 1000 \approx 2^{10} \), so \( \approx 2^{40} \) inches thick -- thicker than the sun's diameter by a factor of 5 or 6.

Well, no.

2^40 inches are some 27.9275 million kilometers, and the sun's diameter is about 1.3927 million kilometers, so the factor is ~20.

Your estimation was wrong by some 300-400%.

V.

[Paul Dale]

Thanks for the correction.
Regardless, it is a huge number.

Pauli

[EdS2]

Very good. I thought Pauli's estimate interesting - evidently each of us have our favourite yardsticks. For me, the distance to the Moon is the biggest one(*) in miles, at a quarter of a million. I don't (didn't) even have a figure in mind for the size of the Sun. I think in future I'll count it as nearly a million miles. (I'm from the UK but old enough to be mostly unmetricated.)

A bit of mental arithmetic, reckoning 30 inches to the yard and 1500 yards to the mile, and 30*30 about 1000, got me to the answer of 20 million miles, which isn't too far off.

(I'd like to make a joke comparing a Pauli estimate to a Fermi estimate, but I can't do it.)

(*) Oh, no, I do have the figure 93 million miles for the distance to the Sun. Perhaps the earliest large number I came across. And two digits of precision!