1. Originally Posted by Mathman
I thought I killed the thread with my last post.
You beat me with that for sure. When I tried to use my calculator and it wouldn't take in 15 decimals, I automatically gave up, not to speak of the MacLaurin series, which I've never heard of before.

2. Originally Posted by Daniel5555
Mathman

Let c = 6.28318530717956 m, R = radius of the Earth, r = 1 m.

We do know from formula that c = 2*Pi*R*sin(r/R).

So R*sin(r/R) = c/2*Pi.

Let x = c/2*Pi.

Using McLauren series, we can determine that sin(r/R) is approximately r/R - r^3/(6*R^3) if we use only two first terms.

So R*(r/R - r^3/6*R^3) = x.

Which trivially means that R = square_root(r^3/6*(r - x)).
I got R = sqrt( r^3 / ( 6r - 3C/pi ) ) , only I got 76.7 kilometers. ( It's a small world. )
It could be solved with a calculator with enough digits.

3. Well, Daniel and PolymerBob both got it algebraically. Yet when they put the numbers in their calculators the numerical answer was either 2.9 km or 76.7 km. This seems like a big difference, considering that the algebraic answers are identical.

Let me try it without a calculator. The circumference of the circle was measured with an accuracy of 10-15 meters, so let’s estimate pi to the same accuracy in our calculations:

Pi = 3.141592653589793

2Pi = 6,283185307179586

c = 6.283185307179560

So the difference between the Euclidean circumference 2 pi r and the spherical circumference 2 pi R sin (r/R), for r = 1, is

2 Pi – c = 2.6 x 10-14

(Note that we have at best only two significant figures in this problem, so the answer can turn out to be all over the place, but at least should have the right order of magnitude however we compute it.)

Now solve for R:

c = 2 pi R sin(r/R) = 2 pi R (r/R—r3/6 R3) = 2 pi r – (pi/3)(1/R2), so

1/R2 = (3/pi)(2pi – c) = (3/pi)x(2.6x10-14) = 2.5 X 10-14

the last approximation holding because pi is just a little bigger than 3.

Thus R2 = (1/2.5) x 1014 = .4 x 1014

So R = .63 x 107 = 6,300,000 meters. (The square root of 40 is somewhere between 6 and 7 – let’s guess 6.3 )

So the radius of the earth is 6,300 kilometers, which isn’t too bad an estimate. (Correct order of magnitude and at least one significant digit is right. If you do it algebraically and put the numbers in at the end, without using scientific notation, then you must divide by a number that is very close to zero.)

By the way, the error in using the polynomial approximation for the sine is vanishingly small in this example, because the earth is very much bigger than the circle on the ice.

------------------

Here is another one in the same spirit.

According to the Bible (Second Chronicles 4:2),

Also King Solomon made a molten sea of ten cubits from brim to brim, round in compass, and five cubits the height thereof; and a line of thirty cubits did compass it round about.
In other words, the circumference of Solomon’s bowl was 30 cubits and the diameter was 10 cubits. (A cubit is the length of your forearm from elbow to fingertips. L. cubitum = elbow).

What was the circumference of the earth back in Solomon’s time?

Solution: We have c = C sin (pi d/C). Put in c = 30, d = 10, and solve for C by inspection.

4. Originally Posted by skatinginbc
When I tried to use my calculator and it wouldn't take in 15 decimals, I automatically gave up, not to speak of the MacLaurin series, which I've never heard of before.
Actually, I found a better way. The calculator function that comes with Microsoft Windows goes out to 32 digits when in scientific mode. So using the nuclear weapon of calculation .......

3C / PI = 5.9999999999999747163987764178222

6r ( 6 x 1 ) minus this is 2.5283661223582177753345612998071 x 10E-14

The reciprocal of this is 39551727801646 ( rounded off )

The square root is 6,288,585 meters, or 6,289 kilometers. ( The correct value is 6,371 kilometers. )

5. Originally Posted by PolymerBob
6r ( 6 x 1 ) minus this is 2.5283661223582177753345612998071 x 10E-14
This is the key. If you use a small calculator that stores only a few digits of pi, this number will be off by many orders of magnitude, resulting in an answer that is way too small.

6. Mathman
PolymerBob
I don't know what happened in my case, because I actually used scientific mode of Windows Calculator. I calculated it once again, and it gave me the correct answer (6,288,984.something... m). I guess I just pressed wrong button or there were some other silly errors somewhere since it was late at night.

Well, the solutions we had: R = square_root(r^3/6*(r - x)) and R = sqrt( r^3 / ( 6r - 3C/pi ) ) are equivalent to what Mathman presented, so I think it's generally correct, 5 out of 10, I guess

7. I can't contribute to the discussion at hand. But I just read the following joke online. I think it will play well to this crowd:

A physicist and a mathematician watch a man walk into a phone booth. Shortly afterward, two men walk out of the booth.

The physicist says, "There must be something wrong with our measurements."

The mathematician says, "If one more person walks into the booth, it will be empty."

8. Originally Posted by Olympia
The mathematician says, "If one more person walks into the booth, it will be empty."
That sounds like something Sheldon (The Big Bang Theory) would say.

9. Originally Posted by Olympia
A physicist and a mathematician watch a man walk into a phone booth. Shortly afterward, two men walk out of the booth.

The physicist says, "There must be something wrong with our measurements."

The mathematician says, "If one more person walks into the booth, it will be empty."

Originally Posted by skatinginbc
That sounds like something Sheldon (The Big Bang Theory) would say.
My hero!

Sheldon's version's of the circle problem goes like this. (And this really is how extra-galactic astronomers attempt to measure the curvature of the universe. So far all measurements are consistent with the Euclidean model -- curvature = 0.)

Suppose we could draw a circle of radius 2 meters in empty space, far from any gravitating matter. Suppose that we could measure the circumference so accurately that we could compute pi = C/D out to 52 decimal places. Suppose that the 52nd decimal place of pi computed by this method turned out to be 7 instead of 8.

What is the curvature of the universe?

Hint: c = [2 (textbook pi) / k ] sin (rk). Solve for k as above. (Answer: .07 per billion light years. )

Sheldon told me that.

10. I told the joke to my math supervisor and he was laughing uproariously. There's something so satisfying about it as an illustration of negative numbers.

11. Mathman
Sheldon's version's of the circle problem goes like this. (And this really is how extra-galactic astronomers attempt to measure the curvature of the universe. So far all measurements are consistent with the Euclidean model -- curvature = 0.)

Suppose we could draw a circle of radius 2 meters in empty space, far from any gravitating matter. Suppose that we could measure the circumference so accurately that we could compute pi = C/D out to 52 decimal places. Suppose that the 52nd decimal place of pi computed by this method turned out to be 7 instead of 8.

What is the curvature of the universe?

Hint: c = [2 (textbook pi) / k ] sin (rk). Solve for k as above. (Answer: .07 per billion light years. )

Sheldon told me that.
The only thing that I can say is that this is truly fascinating.

Maybe I should have chosen purely math-related career, at least all this stuff is much funnier than I what I have to do those days...

12. Originally Posted by Mathman
Suppose we could draw a circle of radius 2 meters in empty space, far from any gravitating matter. Suppose that we could measure the circumference so accurately that we could compute pi = C/D out to 52 decimal places. Suppose that the 52nd decimal place of pi computed by this method turned out to be 7 instead of 8. What is the curvature of the universe?
When I read that paragraph, I thought of "Colorless green ideas sleep furiously", a famous sentence of Noam Chomsky.

13. I dont have anything to contribute but they showed me this pic at work and I was laughing all morning non stop, maybe it is not that funny afterall but I found it hilarious

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