EPTall that comes to my mind is let f(x) = x, let g(x) = x (thus, all f(Q) = g(Q)), then f(x) = g(ax + b) for a = 1, b = 0.
Math Puzzle #3 - Page 2
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AlienAlias
United States324 Posts
all that comes to my mind is let f(x) = x, let g(x) = x (thus, all f(Q) = g(Q)), then f(x) = g(ax + b) for a = 1, b = 0. | ||
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mieda
United States85 Posts
On September 11 2010 06:35 AlienAlias wrote: I don't really understand the question ._. all that comes to my mind is let f(x) = x, let g(x) = x (thus, all f(Q) = g(Q)), then f(x) = g(ax + b) for a = 1, b = 0. The question is that for *any* two polynomials f,g with rational coefficients satisfying f(Q) = g(Q), there exists some rational constants a,b such that f(x) = g(ax + b). You've only considered one particular case of f(x) = x and g(x) = x. That's one out of infinitely many f,g satisfying the conditions of the problem. | ||
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mieda
United States85 Posts
On September 11 2010 06:29 naptiem wrote: Here's an attempt for any field, F: + Show Spoiler + Since f(F)=g(F), g(F) includes f(x) and there must be an element s of F where g(s)=f(x). Rewrite s in the form ax+b for some a,b in F: If x = 0, let b = s and a be any number in F. Then ax+b = a 0 + s = s. If x is not 0, let a = x^-1, the multiplicative inverse of x in F, and b = s - 1. Then ax+b = (x^-1) x + (s - 1) = 1 + (s - 1) = s. In both cases, ax+b = s and g(ax+b) = g(s) = f(x). Edit: Never mind, I think I misread the problem. K. In fact, you can easily come up with counterexamples over the complex or over the reals. The generalization I have in mind is to number fields with real embeddings, and the significant fact is the existence of a lattice L of F (F is finite extension of Q, with real embedding), i.e. L is free abelian group with rank = [F:Q], and the L sitting in F tensor_Q R, where the latter is given a topology as a vector space, is a discrete subgroup. | ||
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AlienAlias
United States324 Posts
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mieda
United States85 Posts
On September 11 2010 06:55 AlienAlias wrote: so basically the question is to prove that if f(x) and g(x) are polynomials (rational etc) of the same degree, there will be some a and b in which f(x) = g(ax + b)? or is there something else in this whole talk of sets and fields that I'm missing? No. The question is properly stated as it is. You're adding the assumption that deg f = deg g now, which will follow from the condition f(x) = g(ax + b) but that requires a bit of work still. | ||
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AlienAlias
United States324 Posts
On September 11 2010 06:57 mieda wrote: No. The question is properly stated as it is. You're adding the assumption that deg f = deg g now, which will follow from the condition f(x) = g(ax + b) but that requires a bit of work still. I'm sorry, I'm just trying to understand the original problem because I'm not exactly sure what Suppose f(Q) = g(Q) (i.e. the sets of values of f and g on the rationals are the same). means. In my first post, I thought this meant the same thing as f(x) = g(x), which means by law of identity they are the same function and thus the question is silly.However, apparently it involves things by the names of 'sets' and 'embedded fields' and other things I've not heard of before, so I'm trying to see if I can simplify it to terms that I would understand. I saw an example of f(x) = x^2 and g(x) = (2x)^2 for which f(Q) = g(Q), so I figured that whole thing meant the degree was equal. | ||
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mieda
United States85 Posts
On September 11 2010 07:03 AlienAlias wrote: I'm sorry, I'm just trying to understand the original problem because I'm not exactly sure what means. In my first post, I thought this meant the same thing as f(x) = g(x), which means by law of identity they are the same function and thus the question is silly. However, apparently it involves things by the names of 'sets' and 'embedded fields' and other things I've not heard of before, so I'm trying to see if I can simplify it to terms that I would understand. I saw an example of f(x) = x^2 and g(x) = (2x)^2 for which f(Q) = g(Q), so I figured that whole thing meant the degree was equal. I see. I'll try to clarify what f(Q) = g(Q) means. First, f(Q) means the set of all numbers in the range of f when you restrict the domain of f on the rationals. So for example, if f(x) = x^2 then f(Q) is the *set* or collection of all numbers y such that there exists some rational number x with y = x^2. For example, 25 is in this set because 25 = 5^2. 25/36 is also in the set because 25/36 = (5/6)^2 . You take the collection of all these numbers y such that there exists some x rational with y = f(x), and that's given a notation f(Q). Likewise g(Q) is the collection of all numbers y such that there exists (depending on y) some rational number x such that y = g(x). So the condition f(Q) = g(Q) just says the two sets are equal. In plain terms, it says that for any rational number x, you can find a rational number y (depending on x) such that f(x) = g(y). And vice versa. | ||
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Muirhead
United States557 Posts
On September 11 2010 06:29 mieda wrote: Need a lot more details, and I'm not convinced of this argument either. You're saying WLOG assume 0 as their constant terms? And then you're assuming deg f = deg g (this is true, but you didn't mention how to prove this at all). And then there's the next paragraph which doesn't seem to make much sense to me. Enjoy your Friday anyway!  My argument works. By "being divisible by p" k times I mean that the p-adic valuation of the rational number is blah blah... I think I assumed f,g have the same degree but it doesn't affect the argument if they have different agrees. | ||
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Muirhead
United States557 Posts
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mieda
United States85 Posts
On September 11 2010 07:14 Muirhead wrote: My argument works. By "being divisible by p" k times I mean that the p-adic valuation of the rational number is blah blah... I think I assumed f,g have the same degree but it doesn't affect the argument if they have different agrees. p-adic valuation of what? How did you reduce to the relatively prime case? And what is relatively prime? | ||
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Muirhead
United States557 Posts
On September 11 2010 07:16 mieda wrote: p-adic valuation of what? How did you reduce to the relatively prime case? And what is relatively prime? I'll write it up in more detail later if nobody else gets it by tonight :D. My Friday is filled with the all important and super sexy Starcraft LAN at MIT :D | ||
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mieda
United States85 Posts
On September 11 2010 07:18 Muirhead wrote: I'll write it up in more detail later if nobody else gets it by tonight :D. My Friday is filled with the all important and super sexy Starcraft LAN at MIT :D Becareful there, Starcraft can take up all your time even on weekdays if you're not careful  I quit Harvard CSL since I was spending way too much time on ICCUP :p | ||
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Muirhead
United States557 Posts
On September 11 2010 07:19 mieda wrote: Becareful there, Starcraft can take up all your time even on weekdays if you're not careful I quit Harvard CSL since I was spending way too much time on ICCUP :pHm you wouldn't happen to be Arnav Tripathy, Alex Zhai, Zachary Abel, or Yi Sun would you? I imagine you at least know those people  | ||
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mieda
United States85 Posts
On September 11 2010 07:21 Muirhead wrote: Hm you wouldn't happen to be Arnav Tripathy, Alex Zhai, Zachary Abel, or Yi Sun would you? I imagine you at least know those people I do know Zachary, met him at the math lounge. I don't know the rest. In fact, are you taking a course at Harvard by any chance? | ||
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category
United States85 Posts
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mieda
United States85 Posts
On September 11 2010 06:25 Muirhead wrote: f(Q) clearly equals g(Q) if f(x)=g(ax+b) Thus WLOG f and g have integer coefficients that are relatively prime and 0 as their constant terms. Suppose f(x)=a_nx^n+...a_1*x. Suppose g(x)=b_nx^n+...+b_1*x. Let p^r be a prime power dividing a_1 but not b_1, if such a prime power exists. Then f(x) will always be divisible by p either less than or equal to 0 times or more than r times, while g(p) is divisible by p at least once and at most r times. So this means that f(Q)=g(Q) then a_1= plus/minus b_1 One can continue inductively through all the coefficients This is wrong and you should be careful there, it's a common error that a lot of students seem to make. Take f(x) = x^3 + 4x and g(x) = x^3 + x^2 - 2x. the 2-adic valuation (additive) v(4) = 2 and v(2) = 1. Here your r = 2. But v(g(2)) = 3 > 2 = r now. Also, you seem to be proving that if f(x) = a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x and g(x) = b_n x^n +b_{n-1} x^{n-1} + ... + b_1 x are of those form (no constant terms and contents of f,g are 1) and f(Q) = g(Q), then all the corresponding coefficients are +/- of each other. Here's a counterexample: Take g(x) = x^3 + x^2 - 2x as above. Then g(x+1) = x^3 + 4x^2 + 3x. But -2 and 3 have different 2-adic and 3-adic valuations, even though the images of g(x) and g(x+1) on the rationals are the same. There are some other road blocks that I can immediately think of in just doing simple valuation calculations, so you'd better write up a detailed attempt. It most likely won't work. As Richard likes to say, devil's always in the detail :p . Speaking of which, I remember I was trying to show him some calculations I did with etale cohomology of some rigid analytic spaces (the method I used, as you might know was to compute combinatorially by finding some open affinoids with good reduction - well you need a semistable model also - and you can imagine how complicated it can get with spectral sequences of rapoport-zink) and I was very sure some parts were standard calculations, little did I know he caught one error and proceeded to tell me a lengthy story with "devil's in the detail" as his point. Of course this problem is nowhere as complicated as computing spectral sequences, and does admit elementary solutions, but he taught me to always be careful Valuation calculations is the first thing one would try for this problem. I think you probably need to do a little more than simple valuation calculations, but who knows. Edit: Rapoport is visiting Harvard at the moment. If you do algebraic geometry you should probably talk to him while he's still here ^^ | ||
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KristianJS
2107 Posts
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mieda
United States85 Posts
On September 11 2010 09:56 KristianJS wrote: Looks interesting, I'll have a go. Thinking geometrically seems to give some intuition and I have some vague ideas as to how the specific behaviour of polynomials can come into play, but nothing crystallized yet. Bedtime too anyway so will try again tomorrow Great! | ||
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mieda
United States85 Posts
Edit: I'll post it tomorrow or day after tomorrow probably. | ||
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FiBsTeR
United States415 Posts
Remember me mieda? We played on iccup and east before. Do you play sc2 now? Oh and btw it's not being able to solve problems like these that push me further and further into the less pure land of CS.  | ||
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