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A couple of maths-related questions..

In our lives, time doesn't have an end.

At a person's end of lives (or at the end of this life if there's no reincarnation), there's a well-deserved peaceful timeless sleep.

But of course you never experience a time when you don't experience.

Only your survivors experience the time after your complete shutdown.

So a person, even when they die, never reaches oblivion. They just go into sleep, but never experience a time when they don't experience.

And, even at the beginning of the arrival of that sleep, the person has no idea that there ever was, or could have been such things as worldly life, body, identity, time, or events--or trouble, misfortune, need, want or lack.

Michael829

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Michael829

Exactly. I and a group of other 12 year old boys where talking about that back in junior high. One guy said "I cant imagine that the Universe could go on forever!". And then a second guy said "its just as hard to imagine that the Universe stops at some kind of brick wall". I silently agreed with the second guy. I thought to myself that "even if there were some kind of brick wall at 'the end of the universe' then....what's just beyond that brick wall...and whats beyond that...?"

So it cuts both ways. The Universe being either finite, or infinite, is hard to grasp.

supposing you had a list of every person in the world, how many orders can you list that?

First I want to emphasize how big these numbers are. Of course when one says a trillion trillion, that means a trillion trillions. And likewise, when one says a trillion trillion trillion, that means a trillion of those trillion trillions. So, with each "trillion" added to the name of that number, the number is multiplied by a trillion.

Now, suppose you're using a type-size such that each letter occupies a square millimeter. That's actually a fairly typical type-size.

Then, when you write the number of orders in which a list of all the world's people could be written, in the form of trillion trillion trillion trillion trillion trillion trillion... , then you'd need a piece of land about 1000 feet by 1000 feet, to write that number in that form. ...to write a trillion that many times in a row. That 1000 feet by 1000 feet piece of land would be needed to write that long a string of the word "trillion".

If you wanted to instead write out all the zeros, it would take about 50% more space, and so the side dimension of the square would be about 25% longer.

The number could also be written as 1E(1E11). Ten to the hundred-billionth power.

I like to express a number in the form of trillion trillion trillion trillion...(and so on), because it gives an especially clear description of how big the number is.

A factorial of that size isn't convenient to calculate directly, but that answer, above, is for a lower-bound for the factorial, a number such that the factorial will be somewhat larger than the stated number. But it's also approximate and rounded-off.

Michael829

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Michael829

This Ted video explains factorials in general, and the deck of card problem in particular quite well.

One new point not already mentioned is the way that you calculate the factorial of any number is simply by multiplying each interger up to and including that number. The factorial of four is 4 times three times two times one (which equals 24).

For the deck of 52 cards its 52 times 51 etc. That yields that staggering 8 followed by 67 zeroes number.

Here is a poser.

The pack of 52 is only the basic regular number cards and face cards of each of the four suites.

What if you add in the joker to make a pack of 53?

How many possible card shuffles would there by if you include the one joker?

The way to calculate that would be simply multiply the factorial of 52 (52!), which is our above (8 time ten to the 67th) number, by the number 53.

That would equal 424 times ten to the 67th power (or you could round it to 4.24 times ten to the 69th).

Last edited by naturalplastic on 18 Dec 2017, 2:40 pm, edited 1 time in total.

I've encountered two estimates for the number of atoms in the observable universe: 10^78 and 10^85

Those numbers can be written as 1E78 and 1E85

So the number of ways that a list of 10 billion people could be arranged, around 10 to the hundred-billionth power, or IE(1E11), is incomparably larger than the number of atoms in the observable universe.

And, referring the QR code, that little rectangular pattern of smaller black or white squares, which you scan, to reach a website....The number of possible configurations of black and white small squares, around 6E368, if I remember correctly, is also a lot more than the number of atoms in the observable universe.

Michael Ossipoff

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Michael829

Invite all seven billion members of the human race to one place for a family get together?

You wanna know how many possible ways that you could seat all of the guests?

That would be the factorial of seven billion.

No idea what number that would add up to. Exactly....a lot.

If it were 10 billion it would be ten to the hundred billionth power.

No. Not "a 100 billion" . But a number written with a one followed by....a 100 billion zeroes.

The number of all atoms in the universe can be written by a one followed by 78 to 85 zeroes.

My girlfriend looked up the world's population for me, and said that it's 7.6 billion.

The number of orders in which a list of 7.6 billion people can be arranged is about 10 to the 75-billionth power

My result for 10 billion people, about 10 to the 100-billionth power, was correct for that number of people.

(I posted about this earlier tonight, but made some sort of error in what I said in that post.)

Michael829

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Michael829

Correction: For 7.6 billion people, the number of orders they can be listed in is ten to the 72-billionth power.

That's also that result that is given by Stirling's formula for approximating large factorials.

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

Previously, I'd said ten to the 75-billionth power.

That was because I made a small error when integrating the natural logarithm function.

It's an integration-by-parts. I've done integration by parts many times, but somehow I didn't use it right for the natural logarithm function, that time.

But the error resulted in only a small difference in the answer, as described above. I still don't know what my error was.

When I use the right integral for the natural logarithm function, my result is the same as that of Stirling's formula, because the simpler method that I use is perfectly accurate for very large factorials.

Michael Ossipoff

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Michael829

I'd said:

**Quote:**

But the error resulted in only a small difference in the answer, as described above. I still don't know what my error was.

Found the error. I'd left out one of the operations, when doing that particular problem. Without that omission, i get the right answer for the integral of the natural logarithm function. ...which gives me the same answer as Stirling's formula, for the factorial of 7.6 billion. That factorial is the number of ways that a list of all the world's people can be arranged:

Ten to the 72-billionth power.

Michael829

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Michael829

Sorry, but way I see it, there are 90 billion billion trillion trillion trillion trillion trillion trillion zillion kazillion blades of grass on this planet, and ten times that of grains of sand on this planet, and a billion times that of all the stars in space, and a zillion trillion atoms in the whole universe. Actually a zillion trillion is probably a teeny tiny part of ALL the atoms that make up every single existing thing in the entire universe.

That is how a person with poor math knowledge numerically sees the world. The world is so vast, there must be a lot more than a few billion grains of sand on this entire planet.

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That is how a person with poor math knowledge numerically sees the world. The world is so vast, there must be a lot more than a few billion grains of sand on this entire planet.

Not sure what youre saying here.

None of us on this thread said that there ONLY a "few billion grains of sand on the entire planet". If that's what you thought we said you misunderstood every one of us. I suspect that you don't understand exponential math.

The number of whole human beings alive at the moment is "a few billion" (a little more than seven billion as a matter of fact). The number of sand grains on Earth is way more than that by many orders of magnitude. I forget what he said above about sand grains, but the number atoms in the universe is between 10^78 to 10^85.

A billion is "a thousand million" (in American lingo). So a billion is 10^9.

So going by the lowest figure (10 to the 78th power) for the number of atoms that would mean that the number of atoms in the Universe is a million billion billion billion billion billion billion. At the high end of 10^85 it would be ten thousand , billion, billion, billion, billion, billion, billion, billion,billion, billion. In other words - kinda like what youre saying above.

(I've deleted my previous post, not because it was incorrect--everything i said in it was accurate. I deleted it because I felt that that talk about tiny and gigantic volumes was too extremely drastically humongous a topic to dump on people at this forum. But the number-comparison in this post seemed tolerable to post.)

One more thing:

Let's use O(N) to mean the number of orders in which N objects could be arranged. ( It could also be written as "N factorial", or "N!")

So the number of orders in which could be arranged all of the world's people could be called O(World's population).

Suppose that you listed and numbered all of the orders in which could be arranged all of the world's people.

In how many orders could *that* list be arranged?

Using the notation that I suggested above, we could call that number O(O(World population)).

That number would be about 10 to the power of (10 to the 72-billionth power).

Or, in text-symbols: 10^(10^72 Billion)

Now, a physicist named Max Tegmark estimated that, if this universe is infinite, and if the constants of physics are unchanged out to a great enough distance, and if matter-density doesn't change systematically, out to a great enough distance, then, if you go out far enough, you'll probably find a Hubble-volume of space in which everything, the things, the people, the events, are exactly the same as they are in the Hubble volume of space whose center we're at.

For example, there'd even be an exact copy of you, experiencing exactly the same events that you're experiencing here. Every flea on every cat would be exactly the same too. ...etc.

A "Hubble volume" is a volume of space amounting to a sphere roughly 20 billion lightyears across. That size has a cosmological significance, but that doesn't matter here. ...only that its diameter of 20 billion lightyears or so is a little less than a quarter of the diameter of the observable universe.

Anyway, Tegmark estimated that, if the universe is infinite, and the other conditions I mentioned are met, then, if you go out far enough, you'll encounter a Hubble volume of space in which everything and every event is identical to what it is in the Hubble volume that we're in the center of.

He said that the most likely distance for encountering that identical Hubble volume, expressed in meters, is ten to the power of (10 to the 118th power).

That is, 10^(10^118) meters.

Tegmark gives that distance in meters, because meters is the standard distance-unit in physics.

Actually, that number is so large that, if we divided it by the number that would convert it from meters to lightyears, and if that "118" is only accurate enough to be somewhere between 117.9 and 118.1, then the conversion from meters to lightyears wouldn't change how that number is written: It would still be about 10^(10^118).

So then, let's call it 10^(10^118) lightyears, instead of meters, because lightyears are the usual unit for interstellar distances.

Now, the question is, if there's probably an identical Hubble volume 10^(10^118) lightyears out, then, then how many Hubble volumes would there probably be, that aren't farther from here than 10^(10^72 billion) lightyears.

To get that answer, you'd divide 10^(10^72 billion) by 10^(10^118), and cube the result (multiply it by itself 3 times. N^3 = NXNXN).

But, like what i was saying before, about the smaller number, if the "72" is only accurate enough to mean "somewhere between 71.9 and 72.1", then dividing it by 10^(10^118), and then cubing the result, wouldn't change how the number is written. It would still be about 10^(10^72 billion).

So: Within a spherical volume of space, from here out to a number of lightyears equal to O(O(World Population)), there are probably about 10^(10^72 billion) Hubble volumes scattered about, that are identical to the Hubble volume that we're at the center of.

That's assuming that 1) Tegmark's estimate for the most likely distance to the closest identical Hubble volume is correct; and 2) That, even at that greater distance, the physics constants are unchanged; and 3) That, even out to that distance, there isn't a systematic change in matter-density.

Michael Ossipoff

Michael Ossipoff

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Michael829

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Nonsense.

To anyone who hasn't seen the factorial-notation, the factorial symbol "!" can easily be mistaken for a punctuation-mark that indicates strong emotional emphasis. Perhaps you hadn't heard about that punctuation-mark. It's called an "exclamation-point".

Yes, in a different mathematical topic, "O" has a different meaning. So what.

I made it clear that I was referring to the number of orders in which N objects could be arranged.. It's reasonable to define (for the purpose of this discussion) O(N) as the number of orders in which N objects can be arranged.

All that is required of definitions is they be clearly specified, and consistently used.

There's nothing unusual about the same letter being used with different meanings in mathematics and physics.

The letter "g" is often used to refer to the acceleration due to gravity at the Earth's surface (some standard value of it). But, in the notation of super-large numbers, gN or g(N), represents a certain function of N that is a very large number.It's a procedure-based notation for naming a very large number.

Some people instead use a capital "G" for that purpose, though "G" also represents the gravitational-constant.

Those are just a few of many examples of a letter being used with more than one meaning.

Sorry, but there aren't enough letters in Latin and Greek, to go around. Re-use of letters is common.

Michael Ossipoff

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Michael829

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