Geometric series headache.

What exactly are you after?

Let F(i, s) be the population after s steps with an initial population of i

Are you trying to calculate F(F(i, s), 2*F(i, s))?

As the numbers get large, F(i, s) is essentially i*5^s. However, this is too high for small numbers.

If you're after the size of the number and not the exact number, I would use a spreadsheet to calculate the population after 11 steps (call this N). The final population (after 2N more steps) will approximately equal
N*5^(2N) = N*10^(1.4N)

which has approximately
1.4N+1+log(N) digits

So, if you start with a population of 2, after 11 steps, you have N=61035157. After 2N more steps, you have a number that consists of approximately 85449228 digits.

That's interesting. I didn't know that way of approximating the number of digits.

What I'm after I suppose is if there is a way to define the end result in any way at all (preferably a way a layperson would understand). I appreciate the size of the number makes it difficult to even write down if I did know it.

I have got as far as a point where the population is 3.97*10^17. I now have 7.94*10^17 steps.

If we define x as 3.97*10^17 I can say at the end of those steps I have a population of 1.25(x*5^2x). And now I get to repeat it twice that many times. Gulp. I can say 1.5625(x*5^2x)*5^(2.5(x*5^2x)) but that isn't terrifically useful.


I did have an idea logarithms would help but my understanding of the applications of e is very shaky.

The log in formula is common log (base 10) not natural log (ln, which is base e).

Where are you getting the 1.25 from? You must be starting with 2.

I've taken a closer look. Here's the formula:
F(i, s) = (i - 0.75) * (5^s) + 0.75

I think it is exact.

Also, 5^x = 10^(log(5)*x)
Again, that's common log (approximately 0.69897)

I hope this helps.

Yes that's helpful thankyou.

The 1.25 is from the fact that, if x are created at step s, x/5 must have been made in step s-1, x/25 in step s-2 and so on. That is just from knowing that each step adds 5 times as many to the sum as the step before, which is something I simply observed occuring.

So x*1.25 is the sum or the total population at step s. It is actually a number that tends towards 1.25 as far as I can ascertain. 1 + 0.2 + 0.04 + 0.008 + 0.0016 etc.

Anyway, with 5^x, I can translate it into scientific notation the way you showed. That is very useful for making it a little more accessible.

Cheers.

For example, y=5^s is as follows..
(n-th step signifies n= s+1, s =0, ...,19 )

You can examine with "log", but exact numbers are available for more..

That's nice for "small" numbers. But what about 5^(10^17) as ZZM mentioned in a post? Even if you can compute each digit in only 1 nanosecond, you'll be at it for over two years.

And, if you wanted to print the number in a small font (say 1/20" x 1/20") it would cover (no margins) both sides of over 9*10^11 sheets of paper. That's a stack of paper about 59,000 miles high!

You can use logarithm, of course.

If you think about 5^(10^17), then take the common logarithm, i.e.

And hence 5^(10^17) is the number of 69,897,000,433,601,875 digits.

I meant exact numbers.

Impossible on a usual computer currently available.

Cheers for the tips both of you.

Markr - the point made about the stack of paper miles high is well made. I've tried to think about how to apply what you told me about logs to the rest of the progression but truthfully I think I'll just accept it's an extremely large number.

You see the point where I have a population of 3.97*10^17 is a very early stage. If we call that a, I now get to perform the basic step within this stage 2a times.

The population at the end of that can be b, and it is ~ 1.25(a*5^2a).

Then again, c = 1.25(b*5^2b)

Then a larger stage where I get 8c-6 steps instead of just 2c.
So d = 1.25(c*5^(8c-6)).

Again e = 1.25(d*5^(8d-6))

And f = 1.25(e*(5^(8e-6)).

That completes one grand cycle and there are 5 more. It just so happens that the remaining 5 have the same stages as above, but twice as many times each.

Something else I left out earlier for simplicity's sake was at the end of each grand cycle I square the population and add that many new members before beginning the next.

I'm not kidding - I knew it was astronomical but I was hoping perhaps there was a way to approximate the size of the end result. Now though I am happy to just accept it as being huge. Before we get a tiny bit along the first cycle there are less protons in the universe, I've heard! (Though I can't make any claims about the truth of that.)

Thanks very much for your help though. I am glad to know about the tricks with log.

My point exactly.

Isn't this essentially a problem of the limit of f(x) as x approaches infinity where f(x)=ab^x (because it's geometric) and b is greater than 1 (because it's increasing). I'm not seeing any other factors that would limit the growth so wouldn't the population just increase without bound?

If b is greater than 1, the function f(x) diverges as x tends to infinity, while it converges to zero if b is less than 1. If b is unity, the value remains constant.

Are you suggesting that b be complex? I don't really see how that would change anything... but I haven't thought about it much. And can't be 1 because then it would stay the same.

No, I used the wording of b being a positive real number. When b is complex, one should use another wording: "modulus of b" is grater than (smaller than, or equal to) 1.

It is bounded in this particular case because there are a finite number of steps to ZZM's algorithm.

I see. I missed the part where he said there were only 11 steps.