|
On April 27 2010 19:16 space_yes wrote:Show nested quote +On April 27 2010 19:08 hacpee wrote:On April 27 2010 19:03 space_yes wrote:
The OP's 2nd graph was time to collect x number workers. You can use an exponential fit for that. I'm not saying you should, but I am saying you can.
Regardless, as I stated previously, you're only dealing with integer values for workers so you don't need the time for gas collection between worker values i.e. 1.5 workers so using an exponential model on such a small domain isn't necessary which others have already point out in their own words. Try to model it. Thats just my advice. y=e^-x is what you're saying the function is. What is y? What is x? How does that relate to dy/dx=-y(because that is the differential equation you need to solve for y=e^-x). Does it make sense? ![[image loading]](http://www.teamliquid.net/staff/Arrian/gasexpfunction.png) y = interval between returns x = number of workers Look at the fitting function at the bottom of the graph it is of the form y = be^(ax).
So that function will read.
So as the rate of change of the interval becomes larger(more positive) with respect to workers, the interval decreases. Thats how the differential equation reads. It doesn't make sense intuitively.
|
I didn't make the graph  but it does make sense. When you have 1 worker the time to collect gas is at its max and it decreases as you add workers b/c your collection rate is increasing. It is increasing at an increasingly slower rate! Because you are dealing with integer values for workers only, the exponential fit is unnecessary.
|
On April 27 2010 19:41 space_yes wrote:I didn't make the graph but it does make sense. When you have 1 worker the time to collect gas is at its max and it decreases as you add workers b/c your collection rate is increasing. It is increasing at an increasingly slower rate! Because you are dealing with integer values for workers only, the exponential fit is unnecessary.
Here is why it doesn't make sense. If there are no workers, if it follows an exponential model, what is the interval?
|
On April 27 2010 19:45 hacpee wrote:Show nested quote +On April 27 2010 19:41 space_yes wrote:I didn't make the graph but it does make sense. When you have 1 worker the time to collect gas is at its max and it decreases as you add workers b/c your collection rate is increasing. It is increasing at an increasingly slower rate! Because you are dealing with integer values for workers only, the exponential fit is unnecessary. Here is why it doesn't make sense. If there are no workers, if it follows an exponential model, what is the interval?
There can be no gas collection when there are 0 workers so there is no model  The domain is 1-3 or 1-6 ^_^
EDIT: you could get around the problem of exponential fitting at e^-x = 0 by using a system of differential equations and an initial condition but hey we're really talking about 1,2, or 3 workers on gas it's not necessary
|
I know it is disturbing to think that we can use e^x to model most kinds of growth or decay but eventually you will see how awesome of a function it is.
For example consider Euler's Identity:
e^(i * pi) + 1 = 0
pretty cool!
|
On April 27 2010 19:47 space_yes wrote:Show nested quote +On April 27 2010 19:45 hacpee wrote:On April 27 2010 19:41 space_yes wrote:I didn't make the graph but it does make sense. When you have 1 worker the time to collect gas is at its max and it decreases as you add workers b/c your collection rate is increasing. It is increasing at an increasingly slower rate! Because you are dealing with integer values for workers only, the exponential fit is unnecessary. Here is why it doesn't make sense. If there are no workers, if it follows an exponential model, what is the interval? There can be no gas collection when there are 0 workers so there is no model The domain is 1-3 or 1-6 ^_^ EDIT: you could get around the problem of exponential fitting at e^-x = 0 by using a system of differential equations and an initial condition but hey we're really talking about 1,2, or 3 workers on gas it's not necessary
using a system of differential equations? So you would have two or more solutions?
As I said, the key is to try to model it using your own intuition. Then you will see that it can't be exponential decay.Yes, you can play with the coefficients and try to force an exponential function onto the 1/x function. They look pretty similar. However the relationship isn't exponential decay.
|
Just curious: did you draw those graphs manually or you have some kind of tool ;P?
|
On April 27 2010 20:07 hacpee wrote:
using a system of differential equations? So you would have two or more solutions?
You would have an infinite number of solutions
As I said, the key is to try to model it using your own intuition. Then you will see that it can't be exponential decay.Yes, you can play with the coefficients and try to force an exponential function onto the 1/x function. They look pretty similar.
There is a reason why 1/x and 1/e^x = e^-x look similar ^_^
However the relationship isn't exponential decay.
I'm not claiming gas mining intervals are governed by exponential decay, I am only affirming that e^-x can be used as a fit (model) to the data and that it maintains necessary characteristics of proportionality. I've stated numerous times it is unnecessary (especially if you already have a function to describe that data that isn't some sort of approximation) b/c you have integer only values for the workers over a small domain.
The key to modeling isn't intuition. You couldn't be more wrong. People refused to believe the world was round b/c their intuition told them they'd fall off it if that were the case. No one truly believed until Foucault's Pendulum.
Anyways, I need to sleep ^_^ ~~
|
On April 27 2010 18:12 space_yes wrote:obviously you've not taken calculus
Speaking as a professional mathematician, you really need to keep your nonsense to yourself. You don't understand much of anything you've been saying in this thread yourself, some meaningless calculus newbie errors interspersed with "go read wikipedia if you don't understand".
On April 27 2010 16:24 space_yes wrote:
As you can see, the exponential function's value is proportional to its previous values. Considering e^-x it is inversely proportional to its previous values. It is this property of prior dependence that makes it particularly good for modeling growth and decay.
Not even coherent. An exponential function is proportional to it's derivative.
On April 27 2010 18:36 space_yes wrote:
Consider the Taylor series where a = 0 of degree 1.
If you do not know what this is do not respond to my post.
Possibly the most transparent attempt to win an argument by obfuscation. Totally irrelevant fluff.
Real calculus is defined on the reals, it works because the reals have a significant amount of useful properties which allow calculus. The reals are a field, a Hilbert space, a locally compact topological group, a continuum Using real calculus to analyse this is entirely meaningless. No algebraic, analytic or geometric representations have even been considered for this model yet, just some poor straw grabbing attempt to apply real variable calculus onto a problem for which it is completely unsuited.
|
The whole discussion here is rather amusing, since when do we need differential equations, taylor expansions, the limit and series expansion definitions of e^x to discuss three bro's mining gas.
I thought you wanted to keep your post understandable for non-mathematicians. Now why would you then define how to compute the real number e? Which also doesn't have anything to do with the problem we are looking at here, since the simple inverse proportionality is enough to model it.
Are you trying to say that e^-x is decreasing over time and therefore can be used to model things? Well sure, but fitting a few datapoints to the function doesn't make things prettier, you could use just any function to do that.
I think you should tuck that mathematical schlong back into your pants.
|
What are you ppl talking about?! ...Jezus Christ !
|
On April 27 2010 21:29 freakclub wrote:
I think you should tuck that mathematical schlong back into your pants.
I tried, but it's four dimensional and noncompact.
|
QuantumPenguin, I didn't mean yours, you have my permission to keep waving it around  We posted at the same time. I meant the one of space_yes, who has, as you pointed out, filled this thread with some serious mathematical bullshit.
|
The key to modeling isn't intuition. You couldn't be more wrong. People refused to believe the world was round b/c their intuition told them they'd fall off it if that were the case. No one truly believed until Foucault's Pendulum.
The pendulum proves the earth rotates, not that it is round.
The proof for the earth being round is more in the lines of Galileo who argued that when ships arrives you'll first see the top and then gradually more and more of the ship.
Just thought I would give my incredibly relevant input to the totally overboard math discussion... IT'S 1 vs 6 workers on gas for christs sake - the minor faults in the OP aren't relevant at all...
|
On April 27 2010 21:18 QuantumPenguin wrote:Speaking as a professional mathematician, you really need to keep your nonsense to yourself. You don't understand much of anything you've been saying in this thread yourself, some meaningless calculus newbie errors interspersed with "go read wikipedia if you don't understand". Show nested quote +On April 27 2010 16:24 space_yes wrote:
As you can see, the exponential function's value is proportional to its previous values. Considering e^-x it is inversely proportional to its previous values. It is this property of prior dependence that makes it particularly good for modeling growth and decay.
Not even coherent. An exponential function is proportional to it's derivative. Show nested quote +On April 27 2010 18:36 space_yes wrote:
Consider the Taylor series where a = 0 of degree 1.
If you do not know what this is do not respond to my post.
Possibly the most transparent attempt to win an argument by obfuscation. Totally irrelevant fluff. Real calculus is defined on the reals, it works because the reals have a significant amount of useful properties which allow calculus. The reals are a field, a Hilbert space, a locally compact topological group, a continuum Using real calculus to analyse this is entirely meaningless. No algebraic, analytic or geometric representations have even been considered for this model yet, just some poor straw grabbing attempt to apply real variable calculus onto a problem for which it is completely unsuited.
My first TL troll! Cool! ^_^ At least you took the time to take my posts out of context and do some googling. A professional mathematician too? Wow @_@ I'm so intimidated !! You can specifically state that you're not trying to come across as a math expert but people still try and make you out to be some evil asshole by first claiming you're spouting nonsense, and then when prove you're not, claim you're obviously just trying to show off! lolz
Not even coherent. An exponential function is proportional to it's derivative.
Did you intend this as a joke? All elementary functions are 'proportional to their derivatives' and d/dx of e^x is e^x... Anyways I don't want to feed the troll any more than I already have -_-~~
|
They're talking about bad maths, 3xist.
Stuff interpreted wrong, bad vocabulary transmitting the wrong idea, unnecessary complications for a topic that doesn't require them, mistakes, and show offs.
To sum it up: using a continuous graph (of an interpolated function) for this was a bad idea.
Most of the stuff up here in this page, you'll understand if you ever read take a Calculus I course / read a book on the subject / read wikipedia and research a bit.
It won't help you on getting more gas efficiency in SCII though.
|
Great arcticle, I love it!
|
SO, gas issue still exist on SC2?
|
On April 28 2010 01:42 brocoli wrote:
They're talking about bad maths, 3xist.
Stuff interpreted wrong, bad vocabulary transmitting the wrong idea, unnecessary complications for a topic that doesn't require them, mistakes, and show offs.
To sum it up: using a continuous graph (of an interpolated function) for this was a bad idea.
Most of the stuff up here in this page, you'll understand if you ever read take a Calculus I course / read a book on the subject / read wikipedia and research a bit.
It won't help you on getting more gas efficiency in SCII though.
So can you explain why it is or isn't 1/x? Because you seem to know how to explain it better. No I am not being sarcastic.
|
hacpee, you got it right in you earlier posts. space_yes doesn't know anything about maths, just don't listed to that rubbish.
|
|
|
|
|
|
EPT
