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Topic: Incorrect Quotient Rule (Read 750 times) |
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ThudnBlunder
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Incorrect Quotient Rule
« on: Dec 12th, 2006, 11:02am » |
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While building a bridge a lazy engineer with rusty maths used the equation
[f(x)/g(x)]' = f'(x)/g'(x)
Why did the bridge not collapse?
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| « Last Edit: Dec 12th, 2006, 7:10pm by ThudnBlunder » |
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koonie32
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Re: Incorrect Quotient Rule
« Reply #1 on: Dec 12th, 2006, 12:05pm » |
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My guess. F(X) = 0
G(X) = any function with a degree of at least 1
I don't know. That would work though.
0(g(X)) - (0)(g'(x)) / g(x)^2 = 0
f'(x) / g'(X) = 0 / 1
therefore [f(x)/g(x)]' done the correct way would equal f'(x) / g'(x) done the wrong way.
However those numbers are so way off. I doubt this is the answer.
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| « Last Edit: Dec 12th, 2006, 2:24pm by koonie32 » |
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Obob
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Re: Incorrect Quotient Rule
« Reply #2 on: Dec 12th, 2006, 1:54pm » |
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Straightforward algebra using the correct quotient rule (f/g)'=(f'g-fg')/g^2 shows that f'/f = (g')^2/(g(g'-g)).
I have not fully deduced what this implies f and g must be. It would make sense to let g be some arbitrary function, and then determine if the right hand side is of the form f'/f for some f. This amounts to determining the intersection of the ranges of the maps phi: f->f'/f and psi: g->(g')^2/(g(g'-g)), where f and g are suitably nice functions.
First assume f is meromorphic on some region U in the complex plane. All poles of f'/f are simple. Hence if g is meromorphic, a necessary condition for there to be some f with phi(f)=psi(g) is that psi(g) have only simple poles.
Now assume g vanishes to order m at z and g' vanishes to order n at z. We say g vanishes to a negative order at z if it has a pole of opposite that order at z. If m is not zero, then n=m-1. In this case, the order of vanishing of g'-g at z is also n. The order of vanishing of psi(g) at z is then 2n-m-n=n-m=m-1-m=-1. Thus psi(g) has a simple pole in this case.
On the other hand, assume the order of vanishing of g at z is 0, so g(z) is defined and nonzero. Then g' can have a zero of any nonnegative order n. If n>0, the order of vanishing of g'-g at z is 0; otherwise it can be any nonnegative number m. It follows that the order of vanishing of psi(g) at z is 2n if n>0, and -m if n=0.
This last calculation shows that the answer is not as nice as possible: it is not true that the images of phi and psi are equal. Indeed, if g'-g vanishes to order at least 2 and g, g' are both nonzero at z, then psi(g) has a pole of order at least 2. Since all poles of phi(f) are simple, there is no f with phi(f)=psi(g).
Does anybody have any other ideas for how to attack this?
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koonie32
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Re: Incorrect Quotient Rule
« Reply #3 on: Dec 12th, 2006, 2:06pm » |
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Lol my way is way off. Isn't it?
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towr
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Re: Incorrect Quotient Rule
« Reply #4 on: Dec 12th, 2006, 2:11pm » |
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If [f(x)/g(x)]' denotes the maximum tensile strength needed of steel cables in the bridge, and f(x)'/g(x)' happens to have a higher value, then the bridge would only get stronger.
However, that's ignoring the fact this is the Putnam section, and not the 'what happened' section 
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| « Last Edit: Dec 12th, 2006, 2:11pm by towr » |
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Obob
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Re: Incorrect Quotient Rule
« Reply #5 on: Dec 12th, 2006, 2:31pm » |
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Your answer is in no way incorrect, koonie, but a mathematician would say it is incomplete. Usually when a question asks someting like "how can (f/g)'=f'/g'," the desired solution is some sort of complete classification of all the possible choices f and g.
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koonie32
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Re: Incorrect Quotient Rule
« Reply #6 on: Dec 12th, 2006, 2:52pm » |
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Oh i see. Ya that would make much more sense. We have not gone far in calc yet but its a nice class.
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ThudnBlunder
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Re: Incorrect Quotient Rule
« Reply #7 on: Dec 12th, 2006, 7:47pm » |
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on Dec 12th, 2006, 1:54pm, Obob wrote:| Does anybody have any other ideas for how to attack this? |
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Well, from your f'/f = (g')2/(g(g' - g)) the solution I have integrates both sides to give
f(x) = eF(x,t)
where
F(x,t) = INT (from 0 to x) of
g'(t)2.dt
-------------------
g(t).[g'(t) - g(t)]
and then applies a little guesswork.
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| « Last Edit: Dec 13th, 2006, 6:03am by ThudnBlunder » |
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Grimbal
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Re: Incorrect Quotient Rule
« Reply #8 on: Dec 13th, 2006, 3:06am » |
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on Dec 12th, 2006, 11:02am, THUDandBLUNDER wrote:While building a bridge a lazy engineer with rusty maths used the equation
[f(x)/g(x)]' = f'(x)/g'(x)
Why did the bridge not collapse? |
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Maybe he used that particular equation in the estimation of how fast the paint will dry?
Or he was on the right side of wrong. He made the bridge stronger and more expensive than necessary, but nobody ever knew about that.
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| « Last Edit: Dec 13th, 2006, 3:06am by Grimbal » |
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ThudnBlunder
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Re: Incorrect Quotient Rule
« Reply #9 on: Dec 13th, 2006, 5:42am » |
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on Dec 13th, 2006, 3:06am, Grimbal wrote:
Maybe he used that particular equation in the estimation of how fast the paint will dry?
Or he was on the right side of wrong. He made the bridge stronger and more expensive than necessary, but nobody ever knew about that. |
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I heard that you were going into rehab, Grimbal. You must have been relieved to learn that help is available.
In future, I will dryly write something of the form
When is F[f(x),g(x)] = G[f'(x),g'(x)]?
But then you would probably speculate about the time. 
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| « Last Edit: Dec 13th, 2006, 5:56am by ThudnBlunder » |
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Grimbal
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Re: Incorrect Quotient Rule
« Reply #10 on: Dec 13th, 2006, 6:08am » |
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Oops, is this the putnam forum? Sorry.
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ThudnBlunder
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Re: Incorrect Quotient Rule
« Reply #11 on: Dec 13th, 2006, 6:17am » |
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on Dec 13th, 2006, 6:08am, Grimbal wrote:| Oops, is this the putnam forum? Sorry. |
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Out already? Haha, I think I understand now. You have been spending too much time in the 'WTF?' forums. 
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| « Last Edit: Dec 13th, 2006, 6:20am by ThudnBlunder » |
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Sameer
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Re: Incorrect Quotient Rule
« Reply #12 on: Dec 13th, 2006, 10:39am » |
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on Dec 13th, 2006, 6:17am, THUDandBLUNDER wrote:
Out already? Haha, I think I understand now. You have been spending too much time in the 'WTF?' forums. |
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seems like you have been too T&B?  As you clearly got influenced and put a math riddle in the wordings of "What happened?"
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ThudnBlunder
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Re: Incorrect Quotient Rule
« Reply #13 on: Dec 13th, 2006, 12:51pm » |
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on Dec 13th, 2006, 10:39am, Sameer wrote:
seems like you have been too T&B? As you clearly got influenced and put a math riddle in the wordings of "What happened?" |
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No, I don't normally bother with them.
But perhaps there have been too many such puzzles in Easy of late.
Anyway, the original equation is satisfied by
g(x) = eax
f(x) = exp[a2x/(a-1)]
for some constant a.
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