If i remember I start with a triangle and end with a square.
I like the square with two arms by the side. Feels very rewarding to colour that shit full.
I just state the property to be demonstrated a second time.
i just draw a smiley face at the end like :smile: but not too smug because then you look like an ass
I have a little question. My math teacher told me Anglo-Saxons prefer to use l'Hôpital's rule instead of Asymptotic analysis to solve certain limits. I was wondering if both methods were similarly "powerful" ? i.e. can every limit that can be solved by A.A. be solved using the l'Hôpital method ? I assume the former could have its limits, but perhaps every A.A. rule is covered up by this property ?
[QUOTE=_Axel;40049568]I have a little question. My math teacher told me Anglo-Saxons prefer to use l'Hôpital's rule instead of Asymptotic analysis to solve certain limits. I was wondering if both methods were similarly "powerful" ? i.e. can every limit that can be solved by A.A. be solved using the l'Hôpital method ? I assume the former could have its limits, but perhaps every A.A. rule is covered up by this property ?[/QUOTE]
'Anglo-Saxons' lol :v:
They're kinda different things... L'Hôpital's rule is a specific rule for finding the limit of a ratio of functions both of which tend to zero or infinity. Asymptotic analysis (though I've not done much of it) is, as far as I know, about describing how a function behaves as the argument tends to infinity...
Can you give an example or two of the sort of problem you're thinking of?
i sign my proof [i]"Original Do Not Copy"[/i]
Hope you guys don't mind me asking here, but I have a bit of a problem and I can't for the life of me work out the solution (which I'm certain is fairly simple.. which is why it's bugging me so much.)
I'm trying to work out the minimum time a car will take to travel down a straight road of a certain length, with a certain required start and end speed, and a certain possible acceleration and deceleration and a certain top speed. It should be a simple problem but I can't get my head around it.
If the road is long enough for the car to reach it's top speed before it has to decelerate, it's simple as. You can just calculate the three parts (accelerating, cruising, decelerating) separately and add the results. My problem is, of course, when the road is too short for that so there is no cruising period and the car would need to begin to decelerate before it has reached top sp
Ultimately I'm trying to find time as a function of start speed, end speed, top possible speed, possible acceleration, possible deceleration, and road length. If that's not feasible, I need a way to calculate the time or distance along the road at which the car should stop accelerating and start braking.
This is for an F1 race simulator, by the way. I can solve this by iterating along the length of the road and comparing the distance it'd need to decelerate by the distance left on the road, but that's a really horrible, hacky method.
EDIT: Right, done some maths and uh, it isn't looking good.
If we assume the entire length of the road, S, is equal to the distance the car accelerates for and the distance it decelerates for we get.
[b]S = s1+s2[/b]
Where [b]s1=1/2 (u+v)*t1[/b] and [b]s2= 1/2 (v+v2)*t2[/b] (u is the starting velocity, V is the velocity it stops accelerating, and v2 is the end velocity)
So [b]2S =u*t1 + (t1+t2)v + v2*t2) [/b] (1)
Now, [b] t1 = (v-u)/a1[/b] (2) and [b]t2=(v2-v)/a2[/b] (3)
Substitue (2) and (3) into (1), we get:
[b]2S = u* (v-u)/a1 + [ (v-u)/a1 + (v2 -v)/a2] *v + v2 * (v2-v)/a2[/b]
Now if we solve for v (which should be the velocity at which the car stop accelerating and starts decelerating) according to wolfram alpha we get:
[b]V = root( a1*v2^2 + 2*a1*a2*S + v2*u^2) / root(v2-a1)[/b]
Great. Or not. for starters, we see that now if our acceleration (a1) value is greater than the end of the road speed (v2), we'll get an error. Similarly, if the top expression is negative the computer will throw a fit- and in fact, because a2 will always be negative (decelerating) it will usually be negative so find can't get a root.
Making the deceleration positive just to test it, I found the formula seems to produce fairly ok results - but it's not quite right I'm afraid. Any thoughts?
Does the car change accelerations instantly?
[QUOTE=JohnnyMo1;40058960]Does the car change accelerations instantly?[/QUOTE]
Yeah, and the acceleration and deceleration is linear (for the time being, anyway).
Well, it would always be linear optimally
[QUOTE=chaz13;40058282]Hope you guys don't mind me asking here, but I have a bit of a problem and I can't for the life of me work out the solution (which I'm certain is fairly simple.. which is why it's bugging me so much.)[/QUOTE]
What I would do, is I would take the kinematic equations, which I derived in Latex for you below.
[img]http://latex.codecogs.com/gif.latex?v_{f}%20=%20\int%20a%20dt%20=%20at+%C2%B4v_{0}[/img]
[img]http://latex.codecogs.com/gif.latex?x_{f}%20=%20\int%20v_{f}%20dt%20=%20\frac{1}{2}at^{2}+v_{0}t+x_{0}[/img]
I then took the last one and moved x-naught over to make the average change in x be 1/2at²/2 + v-naught*time.
[img]http://latex.codecogs.com/gif.latex?_{\Delta}x%20=%20\frac{1}{2}at^{2}+v_{0}t[/img]
So then, I set delta-x1 to be the accelerating portion and delta-x2 to be the decelerating portion of the track where delta-x1 + delta-x2 would be equal to the total length of the track necessary, s, where d is the rate of deceleration in the final portion of the track and assuming Vf to be the top speed. You would then get s(t) to be as shown below. (Note, I chose to keep the negative sign for the decelerating vector contained within d. You could easily apply the magnitude of the vector as d and make the whole term a negative between the two quantities.)
[img]http://latex.codecogs.com/gif.latex?s(t)%20=%20_{\Delta}x_{1}%20+%20_{\Delta}x_{2}%20=%20(\frac{1}{2}at^{2}+v_{0}t)+(%20\frac{1}{2}dt^{2}+v_{f}t)[/img]
[QUOTE=mastoner20;40060781]
stuff
[/QUOTE]
That was quite a nice derivation.
[QUOTE=mastoner20;40060781]What I would do, is I would take the kinematic equations, which I derived in Latex for you below.[/QUOTE]
Genius! Thanks for your help, I really appreciate it.
(If anyone has any insight on why my method above failed I'd be interested to know.. unless I slipped up in the math, I'm convinced it must provide some sort of physical result.. just not the correct one. I found the a1<v2 limit particularly interesting).
Thank you, sirs. And not a problem, Chaz. I'll go over your maths and see if anything jumps at me. Right now, the only thing I can see is if you set your parameters so that your velocity vectors are always greater than or equal to zero, you can mitigate the third velocity variable (though, I don't know if that will help with yours at all or not).
EDIT: I have to ask why you are searching for S (total average distance), but stop with the Final velocity (maximum)? Velocity, as I understood from your question, is going to be an element that is reliant upon the beginning conditions to reach, specifically initial velocity and acceleration rate, as well as a function of time, in which you are using velocity to substitute in in order to solve for S. Furthermore, your V_f is going to be the upper limit of your problem, as your acceleration will change to either null, or your decelerating quantity upon reaching that speed, so I would assume it to be an already known quantity.
[QUOTE=mastoner20;40069893]Thank you, sirs. And not a problem, Chaz. I'll go over your maths and see if anything jumps at me. Right now, the only thing I can see is if you set your parameters so that your velocity vectors are always greater than or equal to zero, you can mitigate the third velocity variable (though, I don't know if that will help with yours at all or not).
EDIT: I have to ask why you are searching for S (total average distance), but stop with the Final velocity (maximum)? Velocity, as I understood from your question, is going to be an element that is reliant upon the beginning conditions to reach, specifically initial velocity and acceleration rate, as well as a function of time, in which you are using velocity to substitute in in order to solve for S. Furthermore, your V_f is going to be the upper limit of your problem, as your acceleration will change to either null, or your decelerating quantity upon reaching that speed, so I would assume it to be an already known quantity.[/QUOTE]
The maximum velocity in my case was the only unknown in the equation (and time, but I removed that). You know the length of the road, the initial and final velocity, the maximum possible velocity (but not the true maximum velocity) and the maximum acceleration and deceleration. From that I intended to use t=(v-u)/a + (vf - v)/d to find the final time.
I got my SAT scores back. 680 in math.
I'm not so sure how I feel about that. Honestly, it didn't apply any of the math that I've learned since... well probably 8th grade Geometry. I don't understand how people are so pretentious about their scores when it's simply being good at word problems that are worded to make you think. It's more a critical thinking test than it is a math test.
[QUOTE=chaz13;40070240]The maximum velocity in my case was the only unknown in the equation (and time, but I removed that). You know the length of the road, the initial and final velocity, the maximum possible velocity (but not the true maximum velocity) and the maximum acceleration and deceleration. From that I intended to use t=(v-u)/a + (vf - v)/d to find the final time.[/QUOTE]
Ahhh, okay.
I read this line in the initial question: "Ultimately I'm trying to find time as a function of start speed, end speed, top possible speed, possible acceleration, possible deceleration, and road length. If that's not feasible, I need a way to calculate the time or distance along the road at which the car should stop accelerating and start braking." and got confused, sorry.
You should be able to take t = all that you just stated and plug into the equation I derived from a.
Also, just to make it easier on you, can't you simplify the case from three to simply two velocities? I'm assuming we're talking in one-dimension, and that you come to a complete stop (whether it's about a turn so that the ultimate final velocity goes to a different dimension, or altogether stops dead in its track at the end of a straight). This would hopefully make the t-substitution easier for you.
[QUOTE=Joey90;40051383]'Anglo-Saxons' lol :v:
They're kinda different things... L'Hôpital's rule is a specific rule for finding the limit of a ratio of functions both of which tend to zero or infinity. Asymptotic analysis (though I've not done much of it) is, as far as I know, about describing how a function behaves as the argument tends to infinity...
Can you give an example or two of the sort of problem you're thinking of?[/QUOTE]
What, you don't say Anglo-Saxons there ? :v:
I mean, perhaps the term Asymptotic analysis isn't the correct one. I merely searched the page for "Equivalence" on the French wikipedia and toggled the language to English. I'm not sure if it's really the same thing.
For example, to solve lim[x->0]:[(cosx-1)/(chx-1)], using l'Hôpital's rule you'd do:
lim[x->0]:[(cosx-1)/(chx-1)] = lim[x->0]:[(cosx-cos0)/(chx-ch0)] = lim[x->0]:[-sinx/shx] = lim[x->0]:[-(sinx-sin0)/(shx-sh0)] = lim[x->0]:[-cosx/chx] = -1
Using "Equivalence" you'd do:
cosx-1 ~(0) -x²/2
chx-1 ~(0) x²/2
=> (cosx-1)/(chx-1) ~(0) -(x²/2)/(x²/2)=-1 => (cosx-1)/(chx-1) -> -1
[QUOTE=_Axel;40074950]What, you don't say Anglo-Saxons there ? :v:
I mean, perhaps the term Asymptotic analysis isn't the correct one. I merely searched the page for "Equivalence" on the French wikipedia and toggled the language to English. I'm not sure if it's really the same thing.
For example, to solve lim[x->0]:[(cosx-1)/(chx-1)], using l'Hôpital's rule you'd do:
lim[x->0]:[(cosx-1)/(chx-1)] = lim[x->0]:[(cosx-cos0)/(chx-ch0)] = lim[x->0]:[-sinx/shx] = lim[x->0]:[-(sinx-sin0)/(shx-sh0)] = lim[x->0]:[-cosx/chx] = -1
Using "Equivalence" you'd do:
cosx-1 ~(0) -x²/2
chx-1 ~(0) x²/2
=> (cosx-1)/(chx-1) ~(0) -(x²/2)/(x²/2)=-1 => (cosx-1)/(chx-1) -> -1[/QUOTE]
OK, I think I see what you mean.
For functions tending to zero (and the other case can be done by thinking of x/y as (1/y)/(1/x)) you would probably work out the asymptotic behaviour using a Taylor expansion, right? Which is usually calculated by differentiating the function...
So as a general principle, both methods work by differentiating the function, which means that they probably work on the same things.
Here's a very non-rigorous way of seeing why it's the same:
If you write the Taylor expansions:
f(x) = f(0) + f'(0)x + ...
g(x) = g(0) + g'(0)x + ...
We assume f(0)=g(0)=0, so
f(x) = f'(0)x + ...
g(x) = g'(0)x + ...
then f(x)/g(x) = (f'(0)x+...)/(g'(0)x + ...)
so when you use your asymptotic method, you just throw away the higher order terms to get
f(x)/g(x) ~ (f'(0)x)/(g'(0)x)
when you use l'Hôpital, you just use f'(0) and g'(0) - which is the same as above, just cancelling the x! So you should get the same answer.
(In problems which you have to apply l'hopital multiple times, like your example does, you'll need to use more terms of the taylor expansion, but it's the same idea)
[QUOTE=Joey90;40075455]OK, I think I see what you mean.
For functions tending to zero (and the other case can be done by thinking of x/y as (1/y)/(1/x)) you would probably work out the asymptotic behaviour using a Taylor expansion, right? Which is usually calculated by differentiating the function...
So as a general principle, both methods work by differentiating the function, which means that they probably work on the same things.
Here's a very non-rigorous way of seeing why it's the same:
If you write the Taylor expansions:
f(x) = f(0) + f'(x)x + ...
g(x) = g(0) + g'(0)x + ...
We assume f(0)=g(0)=0, so
f(x) = f'(x)x + ...
g(x) = g'(0)x + ...
then f(x)/g(x) = (f'(x)x+...)/(g'(x)x + ...)
so when you use your asymptotic method, you just throw away the higher order terms to get
f(x)/g(x) ~ (f'(x)x)/(g'(x)x)
when you use l'Hôpital, you just use f'(x) and g'(x) - which is the same as above, just cancelling the x! So you should get the same answer.
(In problems which you have to apply l'hopital multiple times, like your example does, you'll need to use more terms of the taylor expansion, but it's the same idea)[/QUOTE]
Thank you ! I should have seen the link between differentiation and Taylor expansions. The thing is we don't systematically use Taylor-Young to calculate Taylor expansions, as successive differentiation can sometimes be tiresome, and rather use general formulas we found with T-L (for example the expansion of exp,cos,ch...) and applying composition rules to work out asymptotic behaviour. Also shouldn't it be "f(x)/g(x) ~ (f'(0)x)/(g'(0)x)" instead of "f(x)/g(x) ~ (f'(x)x)/(g'(x)x)" ? Taylor expansions are polynomial approximations, unless we aren't talking about the same thing. I guess it's just a typo.
[t]http://i.imgur.com/nQiNjxz.png[/t]
Progress on Khan Academy..
Pretty bad :P
[QUOTE=_Axel;40082075]Thank you ! I should have seen the link between differentiation and Taylor expansions. The thing is we don't systematically use Taylor-Young to calculate Taylor expansions, as successive differentiation can sometimes be tiresome, and rather use general formulas we found with T-L (for example the expansion of exp,cos,ch...) and applying composition rules to work out asymptotic behaviour. Also shouldn't it be "f(x)/g(x) ~ (f'(0)x)/(g'(0)x)" instead of "f(x)/g(x) ~ (f'(x)x)/(g'(x)x)" ? Taylor expansions are polynomial approximations, unless we aren't talking about the same thing. I guess it's just a typo.[/QUOTE]
You're right, it was just a typo. I think I've corrected it now.
I guess using composition laws to work out asymptotic behaviour just corresponds to similar laws that you could work out for Taylor series...
For a limit f(x)/g(x) we're only interested in the leading non-zero term of the Taylor Expansion. Asymptotic Analysis is basically just stating what that is, l'hopital is (indirectly) finding it by differentiation.
Using any composition rules for asymptotic analysis comes down to saying the same thing about the leading term of the Taylor expansion.
This is all assuming that your asymptotic approximation is a polynomial one (and thus must just be a truncation of the Taylor expansion)... If it isn't then possibly something different might happen, not really sure.
I've starting reading Categories for the Working Mathematician.
Shit's abstract, yo.
Is there a continuous function in which rational points have the same period as their denominator? (e.g. f(f(f(2/3))) = 2/3)
What do you mean by period?
I mean the smallest nonzero number n which satisfies:
[img]http://upload.wikimedia.org/math/6/3/1/631627146282e93f27440d74611b3d13.png[/img]
For a specific x.
Okay so how about this:
f can't be x at the rationals so for it to be continuous only the integers and numbers outside a cycle can be fixed points. But since f isn't x there are hyperbolic fixed points which precludes periodic points in some interval.
[QUOTE=Krinkels;40135794]Okay so how about this:
f can't be x at the rationals so for it to be continuous only the integers and numbers outside a cycle can be fixed points. But since f isn't x there are hyperbolic fixed points which precludes periodic points in some interval.[/QUOTE]
I don't know many ways to analyse things to do with composing a function with itself (pretty much only the contraction mapping theorem)... I'd usually interpret periodic to mean something like f(x+c) = f(x).
I had a look at the problem and couldn't really make much progress. It might be worth trying it for dyadic rationals first (ones of the form n/2[sup]k[/sup]) since it at least simplifies the problem of fractions in lowest form etc.
Certainly if there is such a function, it can't be differentiable (look at f(1/n!)) so I don't know if you can really talk about hyperbolic fixed points and stuff...
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