290 posts
Location
St.Louis, MO
Posted 22 February 2015 - 09:40 PM
I think solving this will be pretty advanced….
Anyways
I wrote the following code:
function limit(zz)
aa=0
for bb=2,zz do
aa=aa+(1/(bb-1))
end
return aa
end
And I was wondering,
What is the limit for aa?
I spent 30 mins of machine time to compute limit(10000000) and it was 16.695311265857.
I was wondering what limit(infinity) would be?
I know any machine would take a hecka long time to figure that out (infinity to be exact) but humans can solve equations and limits.
So is anyone up to the task?
This isn't my homework, I'm just bored.
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Posted 23 February 2015 - 12:37 AM
theoretically, from my glance, the limit is infinity.
although, what you've done kinda breaks math… 1 over infinity… xD
Edited on 22 February 2015 - 11:38 PM
1080 posts
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In the Matrix
Posted 23 February 2015 - 12:54 AM
I'm confused how it took you 30 minutes to compute that, unless you were running it in CC. This code takes ~10 seconds to complete at 10 million, the same you did for 30 minutes.
290 posts
Location
St.Louis, MO
Posted 23 February 2015 - 01:02 AM
I was making it display the percentage of computations that are done every calculation
and i didnt use locals.
After I posted the OP I then made it display percentages every 10000 operations
theoretically, from my glance, the limit is infinity.
although, what you've done kinda breaks math… 1 over infinity… xD
How does this Equal infinity+1? Heck, from what I see, Its not over 100.
1220 posts
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Earth orbit
Posted 23 February 2015 - 01:27 AM
FWIW, I *think* GeforceFan meant 1 over infinity as in '1 / infinity', not 'infinity + 1'.
EDIT: spelling correction
Edited on 23 February 2015 - 12:33 AM
1080 posts
Location
In the Matrix
Posted 23 February 2015 - 01:28 AM
To be entirely technical, your calculation is actually to infinity - 1 since you start at 2. The best bet to get to the actual end with the number would be zz + 1
7083 posts
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Tasmania (AU)
Posted 23 February 2015 - 02:30 AM
theoretically, from my glance, the limit is infinity.
although, what you've done kinda breaks math… 1 over infinity… xD
As x nears infinity, 1 over x nears 0. Granted, no matter how many times the loop iterates, you'll never
reach a number of infinite size, so you'll always be adding
something - but the numbers get so small that the actual
limit of the function will be lower than infinity, because you reach the point where you're no longer
ever adding enough to reach to the next whole number.
Generally, this type of problem is solved by running the function a few times with various variables, then plotting the results on a graph. Examining the slope of the line gives you a pretty good indication as to where the output would stop getting any higher.
290 posts
Location
St.Louis, MO
Posted 23 February 2015 - 03:37 AM
I excluded 1 because the output would be infinity
8543 posts
Posted 23 February 2015 - 05:58 AM
Well, if someone wants to play around with the plotting thing, I did happen to leave it running for a while, here are two data points:
> print(limit(100000000))
18.997896403853
> print(limit(100000000000))
25.905651686526
524 posts
Location
Cambridge, England
Posted 23 February 2015 - 07:48 AM
This is a pretty classical calculus problem.
The question can be restated as: "What is the Sum from 2 to Infinity of '1/(X-1)'?"
Which can be simplified to: "What is the Sum from 1 to Infinity of '1/X'?", which we can solve by Integrating 1/X.
The integral of 1/X is Ln(x), which has no limit. So there's your answer.
Some googling revealed that this series has a name: the
Harmonic Series, and has a few interesting properties.
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Posted 26 February 2015 - 10:55 PM
theoretically, from my glance, the limit is infinity.
although, what you've done kinda breaks math… 1 over infinity… xD
This is a pretty classical calculus problem.
The question can be restated as: "What is the Sum from 2 to Infinity of '1/(X-1)'?"
Which can be simplified to: "What is the Sum from 1 to Infinity of '1/X'?", which we can solve by Integrating 1/X.
The integral of 1/X is Ln(x), which has no limit. So there's your answer.
Some googling revealed that this series has a name: the
Harmonic Series, and has a few interesting properties.
Yes! I'm right!