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Let $ a \in [0,1)$ and $ \phi_n(u) \equal{} \left( \frac {u \minus{} a}{u \plus{} a} \right)^n, I_n(a) \equal{} \int_0^1 |\phi_n(u)| \frac {du}{\sqrt {1 \minus{} u^2}}$.

I'm interested in the asymptotoics(behavious depending on $ n$) of $ \displaystyle \inf_{a \in [0,1)} I_n(a)$.

Do you have any ideas how to find this asymptotics.

Here is a "back of envelope" computation. Split the integral into $ \int_0^a$ and $ \int_a^1$. Note that the first one is an increasing function of $ a$ and the second is a decreasing one. So, basically we are interested in the meeting point. Let's make an assumption that they meet with $ na\gg 1$ and use the standard Laplace asymptotic method for both of them (this is exactly the assumption needed to justify Laplace for the second integral). The first integral in this range is about $ \frac a{2n}$. The second one is a bit trickier but it finally evaluates to something about $ \frac C{\sqrt {an}}\left(\frac {1\minus{}a}{1\plus{}a}\right)^n\approx  C\frac {e^{\minus{}2an}}{\sqrt {an}}$, provided that $ a\ll n^{\minus{}1/2}$, where $ C$ is some constant whose exact value is of no interest to us unless we want higher order terms in our asymptotic formula. Now, denoting $ t\equal{}an$, we see that our objective is to minimize $ \frac{t}{2n^2}\plus{}\frac{Ce^{\minus{}2t}}{\sqrt t}$. This is a piece of cake because the first term is slow and the second one is fast, so the answer for this minimization problem is $ \frac {\log n}{2n^2}$, which is the answer for the original problem as well. It remains to note that the optimal $ a$ is about $ \frac{\log n}n$, which is within the assumed range needed to justify our formulae. The end.

P.S. This argument is extremely sketchy, of course, but if you manage to understand what is done here yourself, next time you'll be able to do such problem without any outside help and in next to no time ;). If you see that it is too hard to recover the details from what I wrote, just let me know and I post the full argument when I have more time :).

Thanks a lot for you answer.
I've got the asymptotics you said for the first integral. But i don't see for the moment how to deal with the second integral, when we also have that at the point $ u\equal{}1$ where $ \phi_n(u)$ attains its maximum in $ [a,1]$, the value of $ \frac{1}{\sqrt{1\minus{}u^2}}$ tends to $ \infty.$

Ah, but Laplace asymptotic formula works for weights with weak power singularities at the point of maximum as well! The idea is the same: if you need the asymptotics for the integral $ \int_0^\text{whatever}x^\nu f(x)^n$ with $ \nu>\minus{}1$ and $ f$ attains its only maximum $ M$ at $ 0$ with $ f'(0)\equal{}\minus{}L<0$, you can replace $ f(x)$ by $ Me^{\minus{}(L/M)x}$, expand the integral to $ \plus{}\infty$ and get the answer involving the Gamma-function.

Ok, now it seems that i see how to get the asymptotics of the second integral as well.
But i have a few probably stupid question about the next steps you did :

[quote="fedja"] The second one is a bit trickier but it finally evaluates to something about $ \frac C{\sqrt {an}}\left(\frac {1 \minus{} a}{1 \plus{} a}\right)^n\approx C\frac {e^{ \minus{} 2an}}{\sqrt {an}}$, provided that $ a\ll n^{ \minus{} 1/2}$, where $ C$ is some constant whose exact value is of no interest to us unless we want higher order terms in our asymptotic formula. [/quote]

Where is this $ 2a$ in $ C\frac {e^{ \minus{} 2an}}{\sqrt {an}}$ coming from ? Shouldn't it just be $ C\frac {e^{ \minus{}n \ln{\frac{1\minus{}a}{1\plus{}a} } }}{\sqrt {an}}$ be ?

[quote="fedja"]
Now, denoting $ t \equal{} an$, we see that our objective is to minimize $ \frac {t}{2n^2} \plus{} \frac {Ce^{ \minus{} 2t}}{\sqrt t}$. This is a piece of cake because the first term is slow and the second one is fast, so the answer for this minimization problem is $ \frac {\log n}{2n^2}$, which is the answer for the original problem as well.[/quote]

Could you please explain how you got this value $ \frac {\log n}{2n^2}$.

[quote="eugene"]
Where is this $ 2a$ in $ C\frac {e^{ - 2an}}{\sqrt {an}}$ coming from ? Shouldn't it just be $ C\frac {e^{ - n \ln{\frac {1 - a}{1 + a} } }}{\sqrt {an}}$ ?
[/quote]
You mean $ e^{ n \ln{\frac {1 - a}{1 + a} } }$ in the numerator? (minus is an obvious misprint, I hope). Yes, that's correct but $ \ln{\frac {1 - a}{1 + a}=-2a+O(a^2)}$ and if $ a$ is not too large, $ O(a^2)$ can be neglected :)

[quote="eugene"][quote="fedja"]
Now, denoting $ t = an$, we see that our objective is to minimize $ \frac {t}{2n^2} + \frac {Ce^{ - 2t}}{\sqrt t}$.
[/quote]
Could you please explain how you got this value $ \frac {\log n}{2n^2}$.[/quote]
Sure. Take a look at the second term. It goes down exponentially and starts at some constant when $ t=1$. You want it to get comparable to the first term at the point of minimum. The first term grows quite slowly and starts at $ n^{-2}$. So, you want the second term to be also about (this "about" is really very lax) $ n^{-2}$. The main factor is, of course, $ e^{-2t}$, which tells you that $ t$ must be about $ \log n$. Now, once you got a clear idea of what you expect your optimal $ t$ to be, the formal proof is easy. Plugging $ t=\log n$, you get $ \frac{\log n}{2n^2}+\text{something much less}$. If $ t>(1-\delta)\log n$, you get at least $ (1-\delta)\frac{\log n}{2n^2}$ from the first term. If $ t<(1-\delta)\log n$, then the second term is at least $ Cn^{-2+2\delta}/\sqrt{\log n}$, which is much greater. So the minimum is, indeed, almost $ \frac{\log n}{2n^2}$. Generally speaking, if you have a turtle crawling up the mountain and a hare rushing down, the minimum of the sum of their altitudes (assuming that the hare cannot go below the sea level) is almost the same as the altitude of the turtle at the point of their meeting (actually you have to wait a short time after they meet to let the hare descend to a negligible altitude but the turtle won't be able to go far in that time anyway :).

Thanks a lot, Fedja, for nice explanation.

[quote="fedja"]
Now, once you got a clear idea of what you expect your optimal $ t$ to be, the formal proof is easy. Plugging $ t \equal{} \log n$, you get $ \frac {\log n}{2n^2} \plus{} \text{something much less}$. If $ t > (1 \minus{} \delta)\log n$, you get at least $ (1 \minus{} \delta)\frac {\log n}{2n^2}$ from the first term. If $ t < (1 \minus{} \delta)\log n$, then the second term is at least $ Cn^{ \minus{} 2 \plus{} 2\delta}/\sqrt {\log n}$, which is much greater. So the minimum is, indeed, almost $ \frac {\log n}{2n^2}$. [/quote]

You probably meant here if $ t > (1 \plus{} \delta)\log n$.

Nope, I meant exactly what I wrote (it is $ e^{\minus{}2t}$, remember?).

I meant $ (1 \plus{} \delta)$ in the first inequality: you took $ t \equal{} \log n$ and then want to consider values $ t > (1 \plus{} \delta)\log n$ and $ t < (1 \minus{} \delta)\log n$ and show that for both of these cases we get a bigger value.
But, anyway, it doesn't matter much.

Btw, would it be possible to solve the following generalization:

Let $ a_i \in [0,1)$ and $ \phi_{n_i}(u) \equal{} \left( \frac {u \minus{} a_i}{u \plus{} a_i} \right)^n, i \equal{} 1,2,\ldots ,k$ , $ a \equal{} (a_1,a_2, \ldots, a_k)$.
$ I_n(a) \equal{} \int_0^1 |\phi_{n_1}(u)| |\phi_{n_2}(u)| \ldots |\phi_{n_k}(u)| \frac {du}{\sqrt {1 \minus{} u^2}}$.

I'm interested in the asymptotoics(behavious depending on $ n$) of $ \displaystyle \inf_{a_i \in [0,1)} I_n(a)$.

Do you have any ideas how to find this asymptotics.

Uh-oh! The part near $ 1$ gives little trouble: under very mild assumptions, it is equivalent to $ C\frac{e^{\minus{}2n\sum a_j}}{\sqrt {n\sum a_j}}$, but the one near $ 0$ is a big headache. I'll see what I can do but I do not promise anything fast: multivariable optimization is always much harder than one variable one, you know... :) All I can say right away is that I expect a power law (with, perhaps, an occasional $ \log$ factor) with power growing linearly in $ k$. Do you need the exact asymptotics or just the order of magnitude?

By the way, do I understand right that $ n$ is the same in all $ \varphi$? (Your notation $ \varphi_{n_i}$ is a bit confusing because it makes one think that there are different $ n_i$ but the right hand side in the formula you wrote has no such things) :?

OK, it took me some time to figure out what is going on here and by now I believe that the answer is
\[ \frac{k\cdot (k\minus{}1)!^2\cdot (\log n)^{2k\minus{}1}}{2\cdot 4^{k\minus{}1}\cdot n^{2k}},\]
which more or less agrees with my intuition. The argument that led me to it is still quite shaky. I'll post it when I have more time unless you have a solid reason to believe that the answer cannot possibly be this one. By the way, where did such fancy minimization problem come from?

The problem comes from asymptotics of some another expression.
I was trying to find the asymptotic behaviour of it, and on the way run into the one posted here.

I knew before that with $ k$ numbers $ a_i$ the answer should asymptotically be $ \frac{(\log n)^{2k\minus{}1}}{n^{2k}}$ for other reasons for the initial expression, so i  would say that i have solid reason that the answer is the one you mentioned :). (i didn't thought about the constant though).

I would be very grateful if you could post you ideas how you came to this answer, when you have time.

[color=red]I edited a bit because trying to correct one mistake I made, I forgot to correct another one, so the initial answer turned out to be correct after all but it was only because I originally made an even number of mistakes. Now everything should be fine :)[/color]

Here is a brief ketch of what I did (I hope that it makes sense but there are a couple of things I didn't try to justify). Assume that $ a_1\ge a_2\ge\dots\ge a_k$. First of all, let's consider the case when $ k \equal{} 2$. If $ a_1$ is comparable to $ a_2$, it can easily be seen that you do not gain much compared with the case $ a_1 \equal{} a_2$, i.e., the order of magnitude is still about $ n^{ \minus{} 2}$, so to get less, one needs to consider $ a_1\gg a_2$. Then the contribution of the part near $ 1$ is about $ \frac {e^{ \minus{} 2a_1n}}{\sqrt {a_1n}}$ (with very lax "about", but, since this is a fast term, we can affors to be very sloppy in its estimate).

The term near $ 0$ splits into $ \int_0^{a_2} \plus{} \int_{a_2}^{a_1}$. The first integral is about $ \frac {a_2}{2n}$ (same as in the one-variable version) and the whole game is to estimate the second integral. Now, the function $ \frac {x \minus{} a_2}{x \plus{} a_2}\frac {a_1 \minus{} x}{a_1 \plus{} x}$ attains its maximum for $ a_2\ll x\ll a_1$, which (with certain reservations) allows to use exponential approximation $ \exp\left\{ \minus{} 2\frac {a_2}x \minus{} 2\frac {x}{a_1}\right\}$ and conclude that the maximum is about $ e^{ \minus{} 4\sqrt {\frac {a_2}{a_1}}}$ and it is attained somewhere near $ x \equal{} \sqrt {a_1a_2}$. Now, using Laplace asymptotic formula (again, with some reservations: the ratio $ a_2/a_1$ should be neither too small nor too large for this step), we see that this second integral is about $ a_1\sqrt [4]{\frac {a_2}{a_1}}n^{\minus{}1/2}e^{ \minus{} 4n\sqrt {\frac {a_2}{a_1}}}$.

If we believe in the approximate expression for this term, the sum is not hard to optimize (the same old hare and turtle story). We see that one needs to make $ a_1$ of order $ \log n/n$ and $ a_2/a_1$ of order $ (\log n/n)^2$ (I'll not try to figure out the constants for this particular case; we'll do it for the general case below). Now comes a big leap of faith: we shall assume that in general we have pretty much the same situation: the minimum is attained if $ a_1\gg a_2\gg\dots\gg a_k$ and, moreover, the ratios $ a_{j}/a_{j \minus{} 1}$ are about $ (\log n/n)^2$. How do we know that? There is a rational way to come to this conclusion but it is neither short, not completely rigorous, so we'll better see what follows from here and try to justify everything a posteriori.

Then $ a_j \equal{} C_j(\log n/n)^{2j \minus{} 1}$ and the final answer should be $ \frac {C_k(\log n)^{2k \minus{} 1}}{2n^{2k}}$. This is our "turtle". All "hares" (integrals from $ a_{j}$ to $ a_{j \minus{} 1}$; with such fast decay of $ a_j$ only 2 factors will really matter in each of them and we shall get the same approximation formula for each of them as in the $ k \equal{} 2$ case) should be below this level. This allows to find the relations for the constants: $ C_1\approx k$, $ 4\sqrt {\frac {C_j}{C_{j \minus{} 1}}}\approx 2k \minus{} (2j \minus{} 3) \minus{} 1$ leading to the answer
\[ C_k \equal{} k\cdot 4^{ \minus{} (k \minus{} 1)}\cdot[(k\minus{}1)!]^2\,.
\]
Now, how to justify everything. The first thing to do is to check the validity of the Laplace approximation for the second integral in the $ k \equal{} 2$ case when $ a_2/a_1\asymp (\log n/n)^2$. To be honest, I haven't done it but it should be a routine thing and even if it breaks (it should break really badly to change the answer because it is a hare estimate; for instance, if it turns out that you have an extra $ \log n$ factor in the right approximation formula, it won't move the final answer by an inch; only a missing power of $ n$ has enough strength to change the constant in the final asymptotics), it is clear what correction should be made in the argument. After that, you need to show that if $ a_{j}/a_{j \minus{} 1}\ll (\log n/n)^2$, then the integral from $ a_j$ to $ a_{j \minus{} 1}$ is much greater than, say, $ a_{j \minus{} 1}/n$, so we shall effectively reduce the number of variables in this case. Finally, once you know that each $ a_j/a_{j \minus{} 1} > c(\log n/n)^2$, you can say that to beat the assumed answer, you'll need to also have $ a_{j}/a_{j \minus{} 1} < C(\log n/n)^2$, which brings you right into the range where all the above estimates were done. And in that range we made an optimal choice.

Uph-ph-ph. That was a long story! Hope it helps a bit :) Feel free to ask as many questions as you need to understand everything.

Thank you very much for you answer.

[quote="fedja"]

Then $ a_j \equal{} C_j(\log n/n)^{2j \minus{} 1}$ and the final answer should be $ \frac {C_k(\log n)^{2k \minus{} 1}}{2n^{2k}}$. This is our "turtle". All "hares" (integrals from $ a_{j}$ to $ a_{j \minus{} 1}$; with such fast decay of $ a_j$ only 2 factors will really matter in each of them and we shall get the same approximation formula for each of them as in the $ k \equal{} 2$ case) should be below this level. This allows to find the relations for the constants: $ C_1\approx k$, $ 4\sqrt {\frac {C_j}{C_{j \minus{} 1}}}\approx 2k \minus{} (2j \minus{} 3) \minus{} 1$ leading to the answer
\[ C_k \equal{} k\cdot 4^{ \minus{} (k \minus{} 1)}\cdot[(k \minus{} 1)!]^2\,.
\]
[/quote]

I'll write my understanding how you find the constants $ C_j$ just to be sure i understood it right.
In the whole expression $ \int_{0}^{a_n} \plus{} \int_{a_n}^{a_{n \minus{} 1}} \plus{} \ldots \int_{a_1}^{1}$, the last integral has asymptotics
$ \frac {e^{ \minus{} 2a_1n}}{\sqrt {a_1n}}$, and if $ a_1 \equal{} C_1 \frac {\log n}{n}$, then this is approximately $ \frac {1}{n^{2C_1} \sqrt {C_1 \log n}}$, from the other side the main asymptotic

Seems like your connection got broken when you were posting :? I can see that you, most likely, understood everything correctly but, to confirm it with certainty, I would prefer to see the entire text you intended to post :)

Sorry, indeed i didn't finish my post. I've made a correction above by now.

* Strange, but it disappeared once again. I'll try to post it separately here:

I'll write my understanding how you find the constants $ C_j$ just to be sure i understood it right.
In the whole expression $ \int_{0}^{a_n} \plus{} \int_{a_n}^{a_{n \minus{} 1}} \plus{} \ldots \int_{a_1}^{1}$,
the last integral has asymptotics $ \frac {e^{ \minus{} 2a_1n}}{\sqrt {a_1n}}$, and if $ a_1 \equal{} C_1 \frac {\log n}{n}$,
then this is approximately $ \frac {1}{n^{2C_1} \sqrt {C_1 \log n}}$, from the other side the
main asymptotic term $ \frac {a_k}{2k} \asymp \frac {1}{n^{2k}}$ , taking into account that
$ a_k \asymp C_k \left( \frac {log n}{n} \right)^{2k \minus{} 1}$.

Then requiring that the asymptotics of $ \int_{a_{j}}^{a_{j \minus{} 1}}$, which is
$ a_{j \minus{} 1}\sqrt [4]{\frac {a_j}{a_{j \minus{} 1}}}n^{ \minus{} 1/2}e^{ \minus{} 4n\sqrt {\frac {a_j}{a_{j \minus{} 1}}}}$
must be $ \frac {1}{n^{2k}}$ we find the ratios $ \frac {C_j}{C_{j \minus{} 1}}$.

Sorry for reviving an old thread for probably stupid reason.

Suppose i initially wanted to find the asymptotics of $ \displaystyle \inf_{a \in [0,1)} I_n(a)$, where
$ I_n(a) = \int_0^1 \frac{|\phi_n(u)|}{1+\phi_n(u)^2} \frac {du}{\sqrt {1 - u^2}}$ and
$ \phi_n(u)$ is still the same function  $ \phi_n(u)= \left( \frac {u - a}{u + a} \right)^n$

Well, $ \phi_n(u)$ definitely tends to $ 0$ at any point $ 0<u<1$, but unfortunately not uniformly on $ [0,1]$ so that we can't actually say that ${{ \int_0^1 \frac{|\phi_n(u)|}{1+\phi_n(u)^2} \frac {du}{\sqrt {1 - u^2}}=  \int_0^1 |\phi_n(u)|} \frac {du}{\sqrt {1 - u^2}} + o ( \int_0^1 |\phi_n(u)|} \frac {du}{\sqrt {1 - u^2}} )$ when $ n \to \infty$. Am i right ?

How can we find the asymptotics of $ \displaystyle \inf_{a \in [0,1)} I_n(a)$, at least using what Fedja has already shown ?

Thanks.

You have to go to my post #2 and reevaluate the first integral. Basically, now you get $ \int_0^\infty \frac {e^{ \minus{} x}}{1 \plus{} e^{ \minus{} 2x}}$ instead of just $ \int_0^\infty e^{ \minus{} x}$ as the factor in the Laplace asymptotic formula which changes the constant a bit. Just don't tell me you want $ k$ such new factors now... :roll: (if you want, the story is the same: investigate the corrections you need to make in the Laplace asymptotic formula first (you'll get the same sort of integral as above but with $ e^{ \minus{} x^2}$ instead of $ e^{ \minus{} x}$ for the coefficient) and recalculate the constants :))

Thanks a lot, Fedja. I just thought that  the asymptotics could somehow follow from what is already done. But fortunately a new recalculation here is not difficult.

Again related question:

Let $ a_i \in [0,1)$ and $ \phi_{n_i}(u) \equal{} \left( \frac {u \minus{} a_i}{u \plus{} a_i} \right)^n, i \equal{} 1,2,\ldots ,k$.
$ I_n(a) \equal{} \int_0^1 \phi_{n_1}(u) \phi_{n_2}(u) \ldots \phi_{n_k}(u) \frac {du}{\sqrt {1 \minus{} u^2}}$,

What about the asymptotics of  $ \displaystyle \inf_{a_i \in [0,1)} | I_n(a) |$ in this case? (i.e. instead of taking modules inside we put it in front of the integral.)

The asymptotics of each of the summand in $ I_n(a)$ is still the same, but now it has the form $ |I_n(a)|\equal{}|I_{1n}(a)\minus{}I_{2n}(a)\plus{}
\ldots \plus{}I_{kn}(a)|$, and it is not so clear to me whether the infimum would still be attained at the same point $ a$ as in the case considered by Fedja before.

I tried to do it in the first case with $ k\equal{}1$, there it worked and asymptotics is indeed attained at $ a\equal{}\log n$ (at least to my opinion). But in that case it there are only two summand and it is quite easy to investigate the expression like $ |I_{1n}(a)\minus{}I_{2n}(a)|$ when there are only two summands.

I would be grateful for your help :)

If $ n$ is even, it does not matter where to put the absolute value. When $ n$ is odd and $ k\equal{}1$, you obviously have $ I_n$ negative when $ a\approx 1$ and positive when $ a\approx 0$, so you can choose $ a\in(0,1)$ such that $ I_n\equal{}0$. So, I do not see much sense in the new question unless you meant something different from what you wrote... :?

[quote="fedja"]If $ n$ is even, it does not matter where to put the absolute value. When $ n$ is odd and $ k \equal{} 1$, you obviously have $ I_n$ negative when $ a\approx 1$ and positive when $ a\approx 0$, so you can choose $ a\in(0,1)$ such that $ I_n \equal{} 0$. So, I do not see much sense in the new question unless you meant something different from what you wrote... :?[/quote]

Thanks for your answer.
Sorry, my bad. What i wrote is indeed not quite what i meant.
I meant slightly different  question (i just didn't write it down first since i stupidly thought that it would be the same as i asked):

Let $ a_i \in [0,1)$ and $ \phi_{n_i}(u) \equal{} \left( \frac {u \minus{} a_i}{u \plus{} a_i} \right)^n, i \equal{} 1,2,\ldots ,k$. By $ I_n(a)$ i actually mean the following expression $ I_n(x,a) \equal{} \int_0^1 \frac{x^2}{x^2\plus{}(1\minus{}x^2)u^2}  \phi_{n_1}(u) \phi_{n_2}(u) \ldots \phi_{n_k}(u) \frac {du}{\sqrt {1 \minus{} u^2}}$ for each $ x \in [\minus{}1,1]$.

For the case $ k\equal{}1$ using Laplace method, the asymptotics of $ I_n(x,a)$ should be the following $ I_n(x,a) \asymp \frac{a}{2n} \minus{}x^2C\frac{\exp(\minus{}2an)}{\sqrt{an}}$ when $ n \to \infty$.
Now i actually am interested in the following thing: $ \epsilon_n \equal{} \displaystyle \inf_{a_i \in [0,1)} \sup_{x \in [\minus{}1,1]} | I_n(x,a) |$.

Why i thought that we again have here the answer $ \epsilon_n \equal{} \frac{\log n}{2n^2}$:
If $ a>\log n$ then $ |I_n(x,a)| \asymp \frac{a}{2n} \minus{}x^2C\frac{\exp(\minus{}2an)}{\sqrt{an}}$ and the supremum $ \sup_{x \in [\minus{}1,1]} | I_n(x,a) |$ is attained at $ x\equal{}0$ hence the infumum in this case is greater than $ \frac{\log n}{2n^2}$.
If $ a < \log n$ then $ |I_n(x,a)| \asymp x^2C\frac{\exp(\minus{}2an)}{\sqrt{an}} \minus{}\frac{a}{2n}$ the supremum $ \sup_{x \in [\minus{}1,1]} | I_n(x,a) |$ is attained at $ x\equal{}1$ hence the infumum in this case is again greater than $ \frac{\log n}{2n^2}$.

I'm terribly sorry for making you wait so long :blush:. Here is the (sketch of an) argument I had in mind.

First of all, the factor $ \frac{x^2}{x^2\plus{}(1\minus{}x^2)u^2}$, $ x\in[\minus{}1,1]$ can be written as $ \frac 1{1\plus{}Mu^2}$, $ M\equal{}x^{\minus{}2}\minus{}1\in[0,\plus{}\infty]$.

Secondly, the contribution of the intervals $ [a_i/\sqrt n,a_i\sqrt n]$ can be estimated by $ e^{\minus{}c\sqrt n}$ (the factor, corresponding to $ a_n$ is already small enough).

Let now $ J_j\equal{}[a_j\sqrt n, a_{j\plus{}1}/\sqrt n]$ with the convention that the first integral starts at $ 0$, the last one ends at $ 1$, and that $ J_j\equal{}\varnothing$ if the left end exceeds the right one. The total integral is then $ \sum_j \int_{J_j}\plus{}O(e^{\minus{}c\sqrt n})$.

The next observation is that, for each $ j$, we can choose $ M$ such that the factor $ \frac 1{1\plus{}Mu^2}$ is close to $ 1$ on $ J_k$ with $ k\le j$ and to $ 0$ on $ J_k$ with $ k>j$. Since the multiplicative gap between the intervals $ J_k$ is of order $ n$, this is possible.

Thus, for each $ j$, we can choose $ M$ such that the whole integral is
\[ \Bigl|\sum_0^j I_j\Bigr| \pm \Bigl[\delta\sum_k|I_k|\plus{}e^{\minus{}c\sqrt n}\Bigr]\]
with arbitrarily small $ \delta>0$ where $ I_j$ is the integral over $ J_j$ without the factor $ \frac 1{1\plus{}Mu^2}$.

Now the key observation is that, if
\[ \Bigl|\sum_0^j I_j\Bigr|\le A\plus{}\delta\sum_k|I_k|,\]
then
$ |I_0|\le C(\delta) A$, $ |I_j|\le 2C(\delta)A$ with $ C(\delta)\to 1$ as $ \delta\to 0\plus{}$. Indeed, we have $ \Bigl|I_j\Bigr|\le 2A\plus{}2\delta\sum_k|I_k|$ for all $ j\ge 1$. Adding these inequalities with the inequality for $ I_0$, we obtain $ \sum_j|I_j\Bigr|\le 2AK\plus{}2K\delta\sum_k|I_k|$ where $ K$ is the (fixed) number of intervals, which allows us to estimate the sum of $ |I_j|$ by $ \frac{2K}{1\minus{}2K\delta}A$. The rest should be clear.

The final step is to notice that, if $ A$ is less than the answer to the original problem, then the inequalities $ |I_j|\le 2A$ for $ j\ge 1$ imply the same estimates for the ratios $ a_{j}/a_{j\plus{}1}$ as before (they were at the level of power of $ n$, not of the constant in front of it), and, thereby, the estimate $ |I_0|>A$, which gives the desired contradiction.

As usual, feel free to ask questions and, please, accept my apologies for the delay.

Thanks a lot for your answer, Fedja. No need for apologies :)

I think i understand the idea and the structure except the last paragraph with estimates:

[quote="fedja"]
The final step is to notice that, if $ A$ is less than the answer to the original problem, then the inequalities $ |I_j|\le 2A$ for $ j\ge 1$ imply the same estimates for the ratios $ a_{j}/a_{j \plus{} 1}$ as before (they were at the level of power of $ n$, not of the constant in front of it), and, thereby, the estimate $ |I_0| > A$, which gives the desired contradiction.
[/quote]

As far as i understand the $ |I_j| \asymp a_j\sqrt [4]{\frac {a_j}{a_{j \plus{} 1}}\cdot n}\exp\left( \minus{} 4n\sqrt {\frac {a_j}{a_{j \plus{} 1}}\cdot n}\right)$ (*) Now if $ A \equal{} A_0 \equal{} C \frac {\log^{2k \minus{} 1} n}{n^{2k}}$, the constant obtained in the previous case, then requiring that $ |I_j|\leq 2A$ for all $ j \geq 1$ basically gives an estimate for $ a_j/a_{j \plus{} 1}$ of the following type:
$ a_j/a_{j \plus{} 1} \geq C_j \frac {\log^2n}{n^3}$ (unfortunately with $ n^3$ in the denominator because of the multiplier $ n$ in the square root in (*)) for appropriate constants $ C_j$ similarly to the previous case.
Multiplying such inequalities we get an estimate for $ a_0$, somewhat similar to
$ a_0 \geq C_0 \frac {\log^{2k}}{n^{3k}}$ (given that we also know the estimate from below for $ a_k$). But $ |I_0| \asymp \frac {a_0}{n\sqrt {n}}$ which as far as i understand doesn't give immediately that $ |I_0| > A$ ($ A$ is of order $ \frac {1}{n^{2k}}$)  ($ k$ is the number of numbers $ a_1,a_2,\ldots,a_k$).

You probably meant some another way to get better estimates or i am just wrong somewhere.

Where did you get that extra $ n$ under the square root in the exponent from? :? The approximation for the integral from $ a_j$ to $ a_{j\plus{}1}$ is [b]exactly[/b] the same as before (up to $ \pm$, of course). All that I said was that the integrals from $ a_j$ to $ a_j\sqrt n$ and from $ a_{j\plus{}1}/\sqrt n$ to $ a_{j\plus{}1}$ can be safely ignored, not that we need to replace $ a_j$ by $ a_{j}\sqrt n$ in the previous estimate for one hump or anything like that. So, there is no $ n^3$, just the same good old $ n^2$.

Thanks for the answer, Fedja. Now i see what you mean.
Below i write my understanding  of your proof. Sorry that it is rather long. If you happen to have time,please tell me whether i got your point right. Thanks :)

In the beginning fix the numbers $ a_1,a_2,\ldots,a_q$. Just to be consistent with the previous case i let's consider the opposite ordering of $ a_i$'s, i.e. assume $ a_q < \ldots < a_2 < a_1$. The integrals
$ I_j \equal{} \int_{a_j}^{a_j\sqrt n} \plus{} \int_{a_j\sqrt n}^{a_{j \plus{} 1}/\sqrt n} \plus{} \int_{a_{j \plus{} 1}/ \sqrt n}^{a_{j \plus{} 1}} \equal{} O(e^{ \minus{} c\sqrt n}) \plus{} \int_{J_j}$ with $ J_j \equal{} [a_j \sqrt n,a_{j \plus{} 1}/ \sqrt n]$. Therefore the whole integral $ I \equal{} \sum_j \int_{J_j} \plus{} O(e^{ \minus{} c\sqrt n})$.

Let's fix $ \epsilon > 0$ and find such $ M$ that $ 1 \minus{} \frac {1}{1 \plus{} Mu^2} < \epsilon$ on $ J_k$ with $ k \leq j$ and $ \frac {1}{1 \plus{} Mu^2} < \epsilon$ on $ J_k$ with $ k > j$ (*). Indeed due to
the inequality $ \frac {1 \minus{} \epsilon}{\epsilon a^2_{j \plus{} 1}n} < \frac {\epsilon n}{(1 \minus{} \epsilon) a^2_{j \plus{} 1}}$ (which is true say for $ \epsilon \equal{} 1/n$) one has that there exists $ M$ with
$ \frac {1 \minus{} \epsilon}{\epsilon a^2_{j \plus{} 1}n} < M < \frac {\epsilon n}{(1 \minus{} \epsilon) a^2_{j \plus{} 1}}$ and for such $ M$ one indeed has (*).

Thus, for each $ j$, we can choose $ M$ such that
\[ |I| \equal{} \Bigl|\sum_0^j I_i\Bigr| \pm \Bigl[\delta\sum_{k \equal{} 1}^K |I_k| \plus{} e^{ \minus{} c\sqrt n}\Bigr]
\]
Let's prove that there exists $ j$ such that
\[ \Bigl|\sum_0^j I_i\Bigr| \geq A_0 \plus{} \delta\sum_{k \equal{} 1}^K |I_k| \quad \quad \quad (1)
\]
where $ A_0$ is the asymptotic with constant obtained in the previous case.

Indeed suppose that on the contrary for all $ j$:
\[ \Bigl|\sum_0^j I_i\Bigr| < A_0 \plus{} \delta\sum_{k \equal{} 1}^K |I_k|
\]
Then for $ j \equal{} 0$, $ |I_0|\leq C(\delta)A_0$. For $ j \geq 1$
\[ \Bigr|I_j \Bigr| \minus{} \Bigr| \sum_{i \equal{} 0}^{j \minus{} 1} I_i \Bigr| \leq A_0 \plus{} \delta\sum_{k \equal{} 1}^K |I_k|.
\]
Hence the estimate $ \Bigl|I_j\Bigr|\le 2A_0 \plus{} 2\delta\sum_k|I_k|$. Summing such inequalities for all $ j$ obtain $ \sum_{k \equal{} 1}^K |I_k| \leq \frac {2K}{1 \minus{} 2K\delta}A_0$ and plugging into the estimate for $ |I_j|$ we get that $ |I_j|\le 2C(\delta)A_0$.

Now since we know that $ |I_j| \asymp a_{j \minus{} 1}\sqrt [4]{\frac {a_{j \minus{} 1}}{a_j} } \exp \left( \minus{} 4n\sqrt {\frac {a_{j \minus{} 1}}{a_j}} \right)$ (we can forget about the summand $ O(e^{ \minus{} c\sqrt n})$ which is $ o(A_0)$ we can estimate $ a_{j \minus{} 1}/a_j$. First we note that there exist constants $ B_j$ such that $ a_{j \minus{} 1}/a_j \geq B_j \frac {\log^2 n}{n^2}$ therefore $ a_j \geq C_j \left(\frac {\log n}{n} \right)^{2j}$ for some constants $ C_j$.

If $ a_{j \minus{} 1}/a_j \equal{} (2q \minus{} 2j \plus{} 2 \minus{} \tau)^2/16 \frac {\log^2 n}{n^2}$ for some positive $ \tau$ then we obtain the contradiction. Therefore for all $ j$ : $ a_{j \minus{} 1}/a_j \geq (2q \minus{} 2j \plus{} 2)^2/16 \frac {\log^2 n}{n^2}$ and $ a_0 \geq D_0$ where $ D_0 \equal{} \frac {q\cdot (q \minus{} 1)!^2\cdot (\log n)^{2q \minus{} 1}}{2\cdot 4^{q \minus{} 1}\cdot n^{2q}}$ is the asymptotic in $ A_0$ obtained in the previous case and $ |I_0|\geq A_0$.

This contradiction proves (1) which implies that for $ \sup_M |I| \geq A_0$ for any set $ a \equal{} (a_1,a_2,\ldots,a_q)$ and hence $ \inf_a \sup_M |I| \geq A_0$.

I glanced at it briefly and nothing stroke my eye as "obviously wrong" except that the ordering of $ a_j$ should be direct here: $ a_1<\dots<a_q$. (at least, all you wrote makes sense if (and only if)  this ordering is chosen).

Reviving an old thread once again).

[quote="eugene"]
Btw, would it be possible to solve the following generalization:

Let $ a_i \in [0,1)$ and $ \phi_{n_i}(u) \equal{} \left( \frac {u \minus{} a_i}{u \plus{} a_i} \right)^n, i \equal{} 1,2,\ldots ,k$ , $ a \equal{} (a_1,a_2, \ldots, a_k)$.
$ I_n(a) \equal{} \int_0^1 |\phi_{n_1}(u)| |\phi_{n_2}(u)| \ldots |\phi_{n_k}(u)| \frac {du}{\sqrt {1 \minus{} u^2}}$.
I'm interested in the asymptotoics(behavious depending on $ n$) of $ \displaystyle \inf_{a_i \in [0,1)} I_n(a)$.
[/quote]

Now i am interested in the following similar question: in the question above all the fractions $ \frac {u \minus{} a_i}{u \plus{} a_i}$ come with the same multilicity $ n$, there are $ k$ of them in total. Denote $ N \equal{} nk$. Suppose one doesn't require necessary that all these fractions $ \frac {u \minus{} a_i}{u \plus{} a_i}$ come with the same multilicity, but rather $ \phi_{n_i}(u) \equal{} \left( \frac {u \minus{} a_i}{u \plus{} a_i} \right)^{n_i}, i \equal{} 1,2,\ldots ,k$ and $ \sum n_i \equal{} N$.
And then the same question about the asymptotics in terms of $ N$ and $ k$ of
\[ \displaystyle \inf_{a_i \in [0,1), n_i, \sum n_i \equal{} N} I_n(a)\]
(The infimum is also taken over the possible multiplicities $ n_i$). Would it be true that the infimum is attained when all the multiplicities are the same $ n_i \equal{} n$ ?

Thanks in advance!

This post is mainly to confirm that I didn't overlook your question. I'm not able to answer it fully yet. Seems like it is true but I need some time to become completely sure about that and in the next two weeks I'll have next to none.

I understand that you want a slick argument that woud just reduce the new problem to the one already solved but what I have now is just a back of envelope computation that shows that if $ n_j\equal{}p_jN$ with $ p_j>0,\sum_j p_j\equal{}1$, then equal $ p_j$ give you the optimal choice for the asymptotic constant in the end (you should be able to easily do that yourself by now, just go over the proof and introduce $ p_j$ everywhere; you'll end up getting their product squared in the denominator in the final asymptotic expression).

Thank you, Fedja, for your reply. Yes, indeed I see the way you meant via recalculating all the expressions introducing $ p_j$ and having their product squared in $ C_k \equal{} \frac{k [(k\minus{}1)!]^2 }{2^{2k\minus{}2} (p_1p_2 \ldots p_k)^2}$. If we want to minimize this expression we must have all $ p_j$'s equal.
As you corrrectly assumed i was indeed hoping for some a bit different reason, as you said "slick argument".  So, if you happen to have one some time in future when you have time, i would be glad if you posted it.
Thanks again :)


advanced fields
advanced fields unsolved

Let $ \langle\cdot,\cdot\rangle$ be the Lorentz metric on $ \mathbb R^{n\plus{}1}$ defined by $ \langle x,y\rangle\equal{}\minus{}x_0y_0\plus{}x_1y_1\plus{}\ldots\plus{}x_ny_n$, where $ x\equal{}(x_0,\ldots,x_n)$ and $ y\equal{}(y_0,\ldots,y_n)$.
Let $ H^n\equal{}\{x\in\mathbb R^{n\plus{}1}\mid\langle x,y\rangle\equal{}\minus{}1,x_0>0\}$.
Prove that $ \langle\cdot,\cdot\rangle$ induces a complete Riemannian metric on $ H^n$.

To show that $ \langle\cdot,\cdot\rangle$ induces a Riemannian metric on $ H^n$, do I need to show that, given a fixed $ x\in H^n$, for $ v\in T_xH^n$, $ \langle v,v\rangle>0$?

If so, how do I show that $ v_0^2<v_1^2\plus{}\ldots\plus{}v_n^2$? We know $ T_xH^n: \equal{}\{v\in\mathbb R^{n\plus{}1}\mid\langle v,x\rangle\equal{}0\}$.

How do I show the induced metric is complete? By definition, I must show that for all $ x\in H^n$, any geodesic $ \alpha(t)$ with $ \alpha(0)\equal{}x$ is defined for all $ t\in\mathbb R$, or that the exponential map is defined on all of $ T_xH^n$.

To show it's positive definite, it's probably easiest to use the fact that it's isometric to a different model of hyperbolic space, where the metric is positive definite (do you know that?), i.e. if you can prove that the Lorentz metric is the pullback of the Poincare metric on the hyperbolic disk, you're done.

For the second part, it's relatively easy to solve the geodesic equation on the hyperbola.  Let your base point be the south pole (i.e. (0,0,...,1)), solve the equation there, and then use the fact that hyperbolic space is homogeneous and isotropic to shift your base point.

You have a little typo in your definition of $ H^n$.

For your first question, try performing a Lorentz transformation that moves $ x$ to the point $ (1,0,\ldots,0)$.

For the second question, by Hopf-Rinow, you only need to show that you can continue any geodesic starting from the point $ (1,0,\ldots,0)$.  These geodesics are easy to compute explicitly, by symmetry.  (Actually, Hopf-Rinow is overkill, since you can use Lorentz transformations.)


logarithms
induction
algebra solved
algebra

Take logarithms you get $\ln(u_{n+1})=2\ln(u_n)-\ln(2)$. Which is a classic recurence.

It would have been better to put $u_1=a$ in order to avoid any confusions. ;)

Very easy one...are you sure that's an olympiad-level problem??  ;)   Assuming that $u_1=u$, the answer is (for $k\geq 2$)
\[ u_k=\frac{1}{2^{(k-1)^2-1}} u^{2^{k-1}} \] easy proof follows by induction

$u_{n+1}=\frac{u_n^b}a \Longrightarrow u_n = a^{\frac 1{b-1}}(\frac{u_0}{a^\frac 1{b-1}})^{b^n}$
 :cool: Try induction :cool:


Hi,

Has someone dates for IPHO 2010 which will be held in Croatia ?

Thank you very much.

It doesn't seem like it's known yet.  The IPhO website seems pretty far behind as they have the "next" competition listed as this year's, which already occurred.


looking for an example of a transformation that does preserves addition but does NOT preserve scalar
multipliclation

ANYONE???????

[I assume you're talking about a map between vector spaces over some field $ K$]

You can't do it if the field is $ \mathbb{Q}$, or if the field has $ p$ elements with $ p$ prime. If $ K$ is anything else, we can make it a vector space over one of these base fields, and then mess with that.
For $ \mathbb{R}$, this does not lend itself well to examples- look up "Hamel basis" to see some of what's going on. Any "construction" invokes the Axiom of Choice or an equivalent.

An explicit example:
Consider $ \mathbb{C}$ as a vector space over itself. The transformation $ f(x\plus{}iy)\equal{} x\plus{}2iy$ preserves addition but not scalar multiplication.

well..I thought I saw it, but I don't..f(x+iy)=x+2iy, seems to do
f(kx+kiy)=kx+2kiy=kf(x)?? What am I missing?????

What you're missing is complex scalars. The transformation jmerry proposed is real-linear, but not complex linear. If $ k\in\mathbb{R},$ then indeed $ f(kx)\equal{}kf(x).$ But in general, $ f(ix)\ne if(x).$


(a) Find a set $ A$ of ten positive integers such that no six distinct elements of $ A$ have a sum which is divisible by $ 6$.

(b) Is it possible to find such a set if "ten" is replaced by "eleven"?

[quote="AndrewTom"]
(b) Is it possible to find such a set if "ten" is replaced by "eleven"?[/quote]

Among any $ 2n\minus{}1$ integer numbers , we can find $ n$ numbers such that have a sum which is divisible by $ n$.  :wink:

[quote="AndrewTom"]How can we do that?[/quote]
 First we should prove this lemma for $ p$ ,when $ p$ is a prime number, which can be proved by Chevalley`s theorem in polyminals.
Then we can prove this for all natural number.(By induction)


Hi,
The problem states that:
Prove that $f(n) = 1-n$, is the only integer valued function defined on integers such that
$f(f(n))=n$; and
$f(f(n+2)+2)=n,R$ for all integers $'n'$.

I have proved it by Induction. Is that right?

The statement is wrong. Any function $f(n)=b-n$ where $b$ is an integer constant satisfies the conditions.

[hide="Proof"]From the definition of a function, we have $f(n_{1})=f(n_{2})\implies n_{1}=n_{2}$. Now assume $n_{1}=n_{2}$. Then the first equation gives $f(f(n_{1}))=f(f(n_{2}))\implies f(n_{1})=f(n_{2})$. Therefore $n_{1}=n_{2}\iff f(n_{1})=f(n_{2})$ which shows that $f$ is bijection. Hence the two equations give $f(n)=f(n+2)+2$, which means that $f(n)$ is linear, giving $an+b=a(n+2)+b+2\iff a=-1\iff f(n)=b-n$. Checking proves that $b$ can have any integer value.[/hide]

And generally, induction is not the method of choice in such problems, because it doesn't prove the uniqueness.

If I'm not mistaken, the original proplem stated $f(0) = 1$, as well.

This is the 1992 Putnam A1, which I proved [url=http://wangsblog.com/jeffrey/?p=227]here[/url].


Let $a$ and $b$ be given real numbers such that $1<a^3<b$. Find all continuous functions $ f: (a,b)\to \mathbb R$ satisfying \[ f(xyz)\equal{}f(x)\plus{}f(y)\plus{}f(z),\] for all $x,y,z \in  (a,b)$ such that $xyz < b$.

[quote="Xaenir"]Find all continuous functions $ f: (a,b)\to R$ satisfying $ f(xyz) = f(x) + f(y) + f(z)$,$ \forall x,y,z$ in $ (a,b)$
and $ 1 < a^3 < b$[/quote]

Rather interesting problem requiring a lot of rigor. :)

Ok, so the problem is :
Find all continuous functions $ f : (a,b)\to\mathbb R$ such that $ f(xyz)=f(x)+f(y)+f(z)$ $ \forall x,y,z\in(a,b)$ such that $ xyz\in(a,b)$ with $ 1<a<a^3<b$

I prefer working with additions rather than multiplications, so I suggest to use $ g(x)=f(e^x)$
$ g(x)$ is continuous and defined in $ (c,d)$ with $ c=\ln(a)$ and $ d=\ln(b)$ and $ 0<c<3c<d$

So the new problem is :
Find all continuous functions $ g : (c,d)\to\mathbb R$ such that $ g(x+y+z)=g(x)+g(y)+g(z)$ $ \forall x,y,z\in(c,d)$ such that $ x+y+z\in(c,d)$ with $ 0<c<3c<d$

Let $ x_0\in(c,\frac{d-c}{2})$
Let $ u\in(0,d-c-2x_0)$
Let $ p,q\in\mathbb N$ such that $ x_0+\frac{p}{q}u<d-c-x_0$

Let $ n\in\mathbb N_0$ such that $ x_0+(n+1)\frac{u}{q} < d-c-x_0$ ($ n$ exists since $ x_0+\frac{u}{q}\leq$ $ x_0+u<x_0+(d-c-2x_0)=d-c-x_0)$
With such $ n$, we have $ d>d-2x_0-(n+1)\frac{u}{q} > c$. Then : Let $ z\in(c,d-2x_0-(n+1)\frac{u}{q})$

$ g(x_0+(x_0+(n+1)\frac{u}{q})+z)=g(x_0)+g(x_0+(n+1)\frac{u}{q})+g(z)$
$ g((x_0+\frac{u}{q})+(x_0+n\frac{u}{q})+z)=g(x_0+\frac{u}{q})+g(x_0+n\frac{u}{q})+g(z)$

And so $ g(x_0+(n+1)\frac{u}{q})=g(x_0+n\frac{u}{q})+(g(x_0+\frac{u}{q})-g(x_0))$

And so $ g(x_0+k\frac{u}{q})=g(x_0)+k(g(x_0+\frac{u}{q})-g(x_0))$ $ \forall k\in\mathbb N_0$ such that $ x_0+k\frac{u}{q} < d-c-x_0$

So :
$ k=p$ $ \implies$ $ g(x_0+p\frac{u}{q})=g(x_0)+p(g(x_0+\frac{u}{q})-g(x_0))$
$ k=q$ $ \implies$ $ g(x_0+q\frac{u}{q})=g(x_0)+q(g(x_0+\frac{u}{q})-g(x_0))$

$ \implies$ $ g(x_0+\frac{p}{q}u)=g(x_0)+\frac{p}{q}(g(x_0+u)-g(x_0))$

$ \implies$ $ g(x_0+r\cdot u)=g(x_0)+r(g(x_0+u)-g(x_0))$ $ \forall r\in\mathbb Q^+$ such that $ x_0+r\cdot u<d-c-x_0$

Using continuity, we get : $ g(x_0+y u)=g(x_0)+y(g(x_0+u)-g(x_0))$ $ \forall y\in\mathbb R^+$ such that $ x_0+y\cdot u<d-c-x_0$

$ \implies$ $ g(x_0+y)=g(x_0)+y\frac{g(x_0+u)-g(x_0)}{u}$ $ \forall y\in\mathbb R^+$ such that $ x_0+y<d-c-x_0$

$ \implies$ $ g(x)=g(x_0)+(x-x_0)\frac{g(x_0+u)-g(x_0)}{u}$ $ \forall x\in [x_0,d-c-x_0)$

$ \implies$ $ g(x)=\alpha x + \beta$ $ \forall x\in [x_0,d-c-x_0)$ (and $ \alpha$ and $ \beta$ dont depend on $ u$)

$ \implies$ $ \boxed{g(x)=\alpha x + \beta \forall x\in (c,d-2c)}$

Since $ d>3c$, we have $ c<\frac{d}{3}<d-2c$ and so :
$ \forall x\in(c,\frac{d}{3})$, we know that ${ x\in (c,d-2c)}$ and that $ 3x\in(c,d)$. So : $ g(3x)=3g(x)=3\alpha x+3\beta$

So, $ \boxed{g(x)=\alpha x + 3\beta \forall x\in (3c,d)}$ and we have three cases :

1) $ d-2c > 3c$
In this case, the two intervals $ (c,d-2c)$ and $ (3c,d)$ have a common part and so $ \beta=0$ and $ g(x)=\alpha x$ $ \forall x\in (c,d)$
And these conditions are obviously enough.

2) $ d-2c < 3c$
In this case, the two intervals $ (c,d-2c)$ and $ (3c,d)$ have no common part and so :
$ g(x)=\alpha x + \beta \forall x\in (c,d-2c)$
$ g(x)=\alpha x + 3\beta \forall x\in (3c,d)$
$ g(x)$ may have any value for $ x\in[d-2c,3c]$ (in respect with continuity)
And these conditions are enough since $ x+y+z\in(c,d)$ $ \implies$ $ x,y,z\in(c,d-2c)$

3) $ d-2c=3c$
We are in the same situation than in point 2 but the continuity at $ d-2c=3c$ implies $ \beta=0$ and so we are in the case 1).

Final solution :
==============
1) If $ 1 < a < a^3 < a^5\leq b$ : $ f(x)=\alpha\ln(x)$ ($ \alpha\in\mathbb R$)

2) If $ 1 < a < a^3 < b < a^5$ : Let $ \alpha,\beta\in\mathbb R$

$ \forall x\in(a,\frac{b}{a^2})$ : $ f(x)=\alpha\ln(x)+\beta$

$ \forall x\in(a^3,b)$ : $ f(x)=\alpha\ln(x)+3\beta$

$ \forall x\in[\frac{b}{a^2},a^3]$ : $ f(x)$ may have any value in respect with continuity


Solutions to the Round 4 problems are now posted on the [url=http://www.usamts.org]USAMTS website[/url].

Another typo:

"If the particle follows the blue arrow ending at $ (3, 4)$, then the probability of winning from there is $ \frac{1}{2}$, by symmetry."

Seems it should be $ (4, 3)$.

When will we know our Round 4 scores/feedback and prize choices?


geometry
geometric transformation
homothety
parallelogram
AwesomeMath
summer program
ratio

how are we supposed to do the problem if the only description you give is the ppicture.  its not even high resolution

ya its a real bad diagram too I cannot tell which sides are supposed to be equal in length :o

Well the diagram is pretty clear if you zoom in, but maybe this should be moved to the Pre-Olympiad forum.

so its three different sized equilateral triangles joined at their vertices to form a scalene triangle in the center..

hm

[quote="rnwang2"]I think the midpoints suggest a homothety, or whatever.
But this is definitely not HSB![/quote]
Aha... at least you remembered that... now if you find the right centre of homothety the problem is solved almost immediately.

[quote="probability1.01"]Aha... at least you remembered that... now if you find the right centre of homothety the problem is solved almost immediately.[/quote]
So you're supposed to draw in the parallelograms, right? I don't see the center...:help:

NB: I have 666 posts.

hmmm...this problem is very similar to the number 10 on the awesomemath test B.  the only difference is that t was the centers of the triangles. ill move it

please transfer it to the its appropriate forum, if needed.

Let AR be a median of the original triangle. Consider a homothety (with ratio 2:1) about A  taking P to D, Q to E, and R to F. Prove DEF is equilateral with congruent triangles.

can some1 please explain what is homothety? I EARLIER HEARD IT FROM MOONMATH.... :blush:


This is Problem 8 on the 1993 AIME:
Let $ S\,$ be a set with six elements.  In how many different ways can one select two not necessarily distinct subsets of $ S\,$ so that the union of the two subsets is $ S\,$?  The order of selection does not matter; for example, the pair of subsets $ \{a, c\}\,$, $ \{b, c, d, e, f\}\,$ represents the same selection as the pair $ \{b, c, d, e, f\}\,$, $ \{a, c\}\,$.

The solution in the wiki was:
Call the two subsets $ m$ and $ n$. For each of the elements in $ S$, we can assign it to either $ m$, $ n$, or both. This gives us $ 3^6$ possible methods of selection. However, because the order of the subsets does not matter, each possible selection is double counted, except the case where both $ m$ and $ n$ contain all $ 6$ elements of $ s$. So our final answer is then $ \frac {3^6 \minus{} 1}{2} \plus{} 1 \equal{} \boxed{365}$

Then:
I tried to solve it, and I found that we first partition the 6 elements into two sets, and there should be $ 2^6/2\equal{}32$ ways to do so. Then, for each element, we can choose if we want to duplicate it by also putting it in the other subset, and there are $ 2^6\minus{}1\equal{}63$ ways to do so. My final answer was then $ 32 \cdot 63\equal{}2016$. Obviously, this is the wrong answer, but where did I go wrong?

The duplication step overcounts.  Putting an element in set $ A$ and choosing to duplicate it is the same as putting an element in set $ B$ and choosing to duplicate it, which is why the official solution does the duplication along with the assignment.

OK. Thanks. Can you explain how much it overcounted by? It overcounted by a large amount.

You're basically doing $ 4^6$ instead of the $ 3^6$ in the solution you gave.


[quote="eagerkill"][img]http://img291.imageshack.us/img291/1562/untitledly7.jpg[/img][/quote]

Why can you set c = 0 ?

And please don't use Imageshack or similar image storage sites. As far as I know, images uploaded there will be deleted eventually. You can attach images here on MathLinks by using the "Add an Attachment" feature when posting.

  Darij

[quote="eagerkill"]Because after homogenization, I thinks I can forget the constraint a^2+b^2+c^2=1 and establish another constraint. And new constraint I establish is c=0. So, is it wrong ??[/quote]

Sorry, I didn't see your reply until now.
You cannot establish every constraint you would like to. For instance, I hope it is clear to you that you cannot establish the constraint $a^{2}+b^{2}+c^{2}=0$ ;) . Before establishing a constraint you have to make sure that it is justified, i. e. that proving the inequality under the constraint will easily yield a proof for the general case. For instance, you can add a constraint $a+b+c=1$ to an inequality for three numbers $a$, $b$, $c$ if you are sure that proving the inequality for the case $a+b+c=1$ will yield a proof of this inequality for all $a$, $b$, $c$. Basically, establishing the constraint $a+b+c=1$ is the same as saying "WLOG we assume that $a+b+c=1$", and every time you say "WLOG" you have to make sure that there is really no loss of generality. See http://www.mathlinks.ro/Forum/viewtopic.php?t=20553 for more explanation.

Actually, dehomogenization is a pretty useless tool except for some special cases (mostly those constraints that allow a trigonometric substitution).

  Darij


abstract algebra
advanced fields
advanced fields unsolved

If $ f: S^n \to S^n$ is continuous and $ \Sigma f$ is the suspension of $ f$, how to show that $ f$ and $ \Sigma f$ have the same (Brouwer) degree?

(using some Mayer-Vietoris technique)

Sorry, you'll have to explain that to me...

The isomorphism $ H^n(S^n) \equal{} H^{n\plus{}1}(\Sigma S^n)$ is natural, so if $ f$ is multiplication by $ n$, so is $ \Sigma f$.

Thanks. Why is this an isomorphism? How does one construct it?

It is the boundary map of Mayer-Vietoris. Its kernel and cokernel are the homology groups of hemispheres, which are contractible, hence $ 0$ for $ i > 0$.


Let the medians of the triangle $ABC$ intersect at point $M$. A line $d$ through $M$ intersects the circumcircle $ABC$ at $X$ and $Y$ so that $A$ and $C$ lie on the same side of $d$. Prove that $BX \cdot BY = AX \cdot AY+CX \cdot CY$.

Let $N$ be the midpoint of $AC$ and $A',B',C',N'$ the orthogonal projections of $A,B,C,N$ on the line $d.$  Segment $NN'$ becomes the median of the trapezoid $ACC'A'$  $\Longrightarrow$ $NN' \equal{} \frac {_1}{^2}(AA' \plus{} CC').$ But from  $\triangle MBB' \sim \triangle MNN',$ we get the proportion $\frac {_{BB'}}{^{NN'}} \equal{} \frac {_{BM}}{^{NM}} \equal{} 2.$  Hence, it follows that $BB' \equal{} AA' \plus{} CC' \ (\star).$

On the other hand, if $R$ denotes the circumradius of $\triangle ABC,$ we have the relations

$BX \cdot BY \equal{} 2R \cdot BB' \ , \ AX \cdot AY \equal{} 2R \cdot AA' \ , \ CX \cdot CY \equal{} 2R \cdot CC'.$

Combining these expressions with $(\star)$ yields

$\frac {BX \cdot BY}{2R} \equal{} \frac {AX \cdot AY}{2R} \plus{} \frac {CX \cdot CY}{2R}  \Longrightarrow  BX \cdot BY \equal{} AX \cdot AY \plus{} CX \cdot CY.$


怎么下载、安装、使用LATEX？

单击最上面一排的第三了: Latex Help
如果你只想在这个论坛内使用的话, 就没必要下载. 只需在code两边加上dollar sign就可以了.
例如: a+b \geq 2 \sqrt{ab} 加上之后,就变成了:
$  a+b \geq 2 \sqrt{ab} $

[quote="honglei"]怎样写CODE?请详细告诉我.万分感谢! :blush:[/quote]
还是在那里, 有一个联接好象叫: symbols and signs.


Problem 23 on the 1986 AHSME:

Let $ N\equal{}69^5\plus{}5 \cdot 59^4\plus{}10 \cdot 69^3\plus{}10 \cdot 69^2\plus{}5 \cdot 69\plus{}1$. How many positive integers are factors of $ N$?

(A) 3
(B) 5
(C) 69
(D) 125
(E) 216

Let x=69

Therefore, the expression becomes:

x^5 + 5x^4 + 10x^3 + 10x^2 + 5x + 1

notice the binomial coefficients, we factor and get

(x+1)^5

replacing x with 69

70^5 = 2^5 * 5^5 * 7^5

Thus, the number of factors is

(5+1)*(5+1)*(5+1) = [b]216[/b]

Sorry I don't know any LaTeX   :blush:

[quote="Eulers_Apprentice"]$ x^{5} \plus{} 5x^{4} \plus{} 10x^{3} \plus{} 10x^{2} \plus{} 5x \plus{} 1 \equal{} (x \plus{} 1)^{5} \Longleftrightarrow$
$ N \equal{} 70^{5} \equal{} 2^{5}\cdot 5^{5} \cdot7^{5}$

$ \therefore \; d(N) \equal{} (5 \plus{} 1)^{3} \equal{} \boxed{216}\ \ \ \ \blacksquare$
[/quote]
[hide="Code"]
[code]$x^{5} + 5x^{4} + 10x^{3} + 10x^{2} + 5x + 1 = (x + 1)^{5} \Longleftrightarrow$
$N = 70^{5} = 2^{5}\cdot 5^{5} \cdot7^{5}$

$\therefore \; d(N) = (5 + 1)^{3} = \boxed{216}\ \ \ \  \blacksquare$[/code][/hide][url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:About]LaTeX Wiki[/url]

The problem in the original post says,  :P
[quote="ThetaPi"]Problem 23 on the 1986 AHSME:

Let $ N \equal{} 69^5 \plus{} 5 \cdot \color{blue} 59 \color{black} ^4 \plus{} 10 \cdot 69^3 \plus{} 10 \cdot 69^2 \plus{} 5 \cdot 69 \plus{} 1$. How many positive integers are factors of $ N$?
[/quote]

Probably just a typographical error.


calculus
integration
geometry unsolved
geometry

What is the average thickness of a slab sawed at random from a round log ?

Orl, are you sure this is not "Advanced Fields" or "Calculus"?

  darij

[quote="darij grinberg"]Orl, are you sure this is not "Advanced Fields" or "Calculus"?
[/quote]

Yes, the solution involves in $\int$ or even $\int \int$ depending on which solution you prefer. But yet the problems sounds so elementary.  :lol:


MATHCOUNTS
AMC
USA(J)MO
USAMO
AIME
FTW
number theory

I'm gonna make a tournament, with difficulty ranging from hard Classroom Math to National Mathcounts. The tests will not be in any particular form(a.k.a you might have a short answer problem right in-between two multiple-choice problems. You may not use calculators at all, and I have a system of sign-ups:

Anyone who has not completed middle-school can compete, with a few exceptions:

If you have gotten in the top 3rd at National Mathcounts, or have made USAMO at least once, you may not compete, so other middle-schoolers have a chance :wink:
The first test should be up in a few days.

[quote="pythag011"]I join. Hi. Put USAMO problems in it.[/quote]

You made usamo...
can i join?

[quote="Urc"]can i join?[/quote]

Sure! If you want to join, just say so!

[quote="1=2"]
If you have gotten in the top 3rd at National Mathcounts, or have made USAMO at least once, you may not compete, so other middle-schoolers have a chance :wink:[/quote]

I have to comment on this. 75th is BARELY in the top 3rd :rotfl:
Why didn't you say top 57?

Now I feel honored. :D

But I finished middle school... :(

I wanna join!  I made AIME, but not USAMO.  Also, I've never done MC.

But my FTW rating is 1881 from memming, so don't use FTW problems :D

You got it.

Post here if you want to be in!

[quote="1=2"]And from the lack of popularity I have decided to withdraw from middleschool tournaments altogether. Bye!  :bye:[/quote]

Wickedestjr, have you seen the last post of 1=2?

[quote="pythag011"]...Since the rules do not allow me to participate, I am taking over the tournament, if that's ok with you, 1=2?[/quote]

hm dont you have ur own tourney?

naw, just let this die. You have your own USAMO tournament.

yeah, but it died...maybe I made the problems a little hard...

[quote="pythag011"]Due to the death of this tourney, I am hosting it now, and this time I will make easy equestions. But they still might be sort of hard.[/quote]

Hmm...easy but hard?  Like:

1. (insert easy problem here, like 1+1)
2. (insert hard problem here, like IMO)
3. (insert easy)
4. (hard)

like that?

Like this is the first round.

1. There is a Venn diagram. All of the regions have the same area. If the radius of each of teh circles is 1, and the centers from an equilateral triangle, find the side length.

2.How many ways are tehre to choose 5 people from 10 people standing in a line if none of them can be standing next to each otehr.

3. Find all primes p such that $ p^2\plus{}2$ is NOT a multiple of 3.

1. how many circles are in the venn diagram???

the cetners form an equilateral triangle. Quote.

Wondering:  would we need proofs?  or just answers?

NOOOOOOOO Proofs. If you want proofs, go turn in solutions for my other tourney.

Was that the real first round, or just an example of a possible first round?

Okay, well I am going to have to quit. Sorry. I really want to compete but I don't think that I will have enough time, because I am already hosting 5 math tournaments, and 3 other games.

THE BONUS ROUND (for USAMO qualifiers)

if you want to make USAMO this year, or already have, solve the following quesitons.

The first problem. PM me if you solve it with SOLUTIONS to get the second one.

This will be good practice for writing proofs.

The fisrt problem is pretty easy.

Define the fibonacci sequence as $ F_1 \equal{} 1,$ $ F_2 \equal{} 1,$, and $ F_{n\plus{}1} \equal{} F_n \plus{} F_{n\minus{}1}.$

If m and n are positive integers, prove taht gcd($ F_m, F_n$) = $ F_{gcd(m,n)}$


algebra
polynomial
analytic geometry
inequalities
floor function
combinatorics unsolved
combinatorics

Let $ A$ be set of all $ n$-tuples $ (a_1, a_2, \ldots, a_n)$, where $ 0 \leqslant a_i \leqslant k$.

We say that element $ a \equal{} (a_1, a_2, \ldots, a_n)$ dominate element $ b \equal{} (b_1, b_2, \ldots, b_n)$ iff $ a_i >\equal{} b_i$ for every $ i \equal{} 1, 2, \ldots n$.

Determinate maximal subset $ F$ of $ A$, such that there are no two elements $ a, b \in F$ with $ a$ dominating $ b$.

For $ k \equal{} 1$ we have original Sperner's lemma, and domination is simple subset relation.
I think that final result is family with all $ n$-tuples with equal sum, or equivalently maximal coefficient in polynomial
\[ (1 \plus{} x \plus{} x^2 \plus{} \ldots \plus{} x^k)^n\]

Thank you for this beautiful problem. Did you ask to determine a maximal subset $ F$ or its maximal length? I can provide with the maximum length and that equals to $ (k + 1)^{n - 1}$.

Proof: The set if all n-tuples $ T(n)$ with each co ordinate $ 0\leq a_i\leq k,1\leq i\leq n$ forms a poset under the relation $ a < b\quad iff \quad a_i\leq b_i\quad \forall i$.

For $ n > 1$,let $ \{(i,e_{n - 1})|e_{n - 1}\in T(n - 1)\}$ denote all n-tuples whose first co ordinate is $ i$ and $ j$th co ordinate is $ (j - 1)$th coordinate of $ e_{n - 1}%Error. "QUAD" is a bad command.
\forall 2\leq j\leq n$

Let $ C_{e_{n - 1}} = \{(i,e_{n - 1})|\forall 1\leq i\leq n\}$.
Observe that,
$ \{C_e|\forall e\in T(n - 1)\}$ forms $ (k + 1)^{n - 1}$ disjoint minimum chains.By Dilworth's theorem, we have a antichain of desired length.(qed)

[quote="alexilic"]We say that element $ a = (a_1, a_2, \ldots, a_n)$ dominate element $ b = (b_1, b_2, \ldots, b_n)$ iff $ a_i > = b_i$ for every $ i = 1, 2, \ldots n$. [/quote]
We can get to another variant of the problem by changing the inequality in the line to ...$ a_i>b_i$...

You are wrong!
For $ k \equal{} 1$, maximal number of elements is $ \binom{2n}{n}$ which is greater than $ 2^{n \minus{} 1}$, as you claim.

Keep thinking...

yes, I am wrong but not in the sense you said. I disagree on the fact that if $ k\equal{}1$, the maximal antichain has length $ 2n\choose n$. $ T(n,k\equal{}1)$ contains $ 2^n$ elements. Let me list all of them for n=3,

{(000),(001),(010),(011),(100),(101),(110),(111)}

$ {6\choose 3}\equal{}20>|T(n,k\equal{}1)|$. Here you are wrong.

In my previous post I suggested to break $ T(n)$ into chains like this,
$ C_1\equal{}\{(000),(100)\}$
$ C_2\equal{}\{(001),(101)\}$
$ C_3\equal{}\{(010),(110)\}$
$ C_4\equal{}\{(011),(111)\}$

If this was a minimum number of chains as I claimed, we would have got $ 2^2$ length antichain. But the longest antichains present here are $ \{(100),(010),(001)\}$ and $ \{(011),(101),(110)\}$. Here , I went wrong.

Here is a minimum configuration of chains
$ C_1\equal{}\{(000),(001),(101)\}$
$ C_2\equal{}\{(100),(110)\}$
$ C_3\equal{}\{(010),(011),(111)\}$

I am thinking about the possible closed form ie., the minimum number of chains for a general n and k. Let me know about new developments.

I mistyped bound... The correct answer for $ k \equal{} 1$ is $ \binom{n}{\lfloor \frac{n}{2} \rfloor}$.

For $ n \equal{} 3$ we get $ 3$ sets and one such example is $ \{001, 010, 100\}$.

We shall pick out the pairs in the following fashion: (0,k) (1,k-1),..
For a specific pair (m,n) we construct a n-tuple using exactly $ \lfloor \frac {n}{2} \rfloor$ m's. This can be done in $ n\choose {\lfloor \frac {n}{2} \rfloor}$ ways ensuring maximality in this process. Any two tuples here are not comparable. Build such $ n$ tuples for every pair  and we observe that any two n-tuples across two pairs are not comparable. We thus obtain a antichain of length ${ \lfloor \frac {k+1}{2} \rfloor {n\choose {\lfloor \frac {n}{2} \rfloor} }}$. Observe that we cannot add any other n-tuple to this antichain. How do we prove or disprove that this is the longest antichain ?

Imitating proof of a Sperner lemma we can obtain following inequality $ \sum_{(a_1,a_2,...,a_n)\in F}{\frac {{k \choose a_1}{k\choose a_2}...{k \choose a_n}}{{nk \choose a_1 + a_2 + ... + a_n}}}\leq 1$, where $ \mathcal l{F}$ denotes set with required properties.


Group $ \beta$
[b]Theoretical Round. Problems to solve.[/b]

[b]1. Galaxy.[/b] The bolometric (total) stellar magnitudeof a galaxy visible from a distance of $ L_1\equal{}3\text{Mpc}$ is $ m_1\equal{}6^m.88$. Find the bolometric stellar magnitude $ m_2$ of this galaxy visible from a distance of $ L_2\equal{}3\text{Gpc}$

[b]2. Space Sail.[/b] A space probe for research of the asteriod belt is sent to objects of researches by using a solar sail. The probe-system of mass $ m\approx 1\text{ton}$ was moving around the Sun in a circular orbit of radius $ 1\text{A.U.}$ Then such a sail (that was earlier a part of the probe-system) was opened and after half a revolution around the Sun the probe has reached Ceres. Estimate the area $ S$ and thickness $ d$ of this sail. Consider the sail is of mirror type. The typical distances of the asteroid belt bodies from the Sun are $ 2.8\text{A.U.}$. The constant of total solar irradiation is $ A\approx 1.37 \text{kW/m}^2$

[b]3. Alcohol in Universe.[/b] Astronomers based at Britain's Jodrell Bank Observatory say they have spotted a cloud of alcohol in deep space that measure 288 billion miles (463 billion kilometers) across, a finding that could shed light on how giant stars are formed from primordial gas. Some nations even decided to send expeditions to this area to taste the cloud but... enthusiasm disappeared after learning that the cloud is so remote that it will take us many million years to reach it on the modern techniques. The density of molecules in the cloud is large in comparison with interstellar gas density but very small from our custom point of view, only about $ 10\text{atoms/mm}^3$. Estimate the temperature such a cloud should have to be stable and not get dispersed in future (before the international expedition will reach the cloud). Suppose the alcohol is ethyl ($ C_2H_5OH$)

[b]5. Visit.[/b] Extraterrestrial animals (animaloids) decided to invade the Earth. The navigation devices of the spacecrafts of vituloids (lat. vitulus=seal) needs that sometimes after the landing the Polar star ($ \alpha\equal{}2^h32^m$, $ \delta\equal{}89^o16'$ at the epoch of starting the colonization) nust be exactly in Zenith and the devices of crocodiloids needs that sometime the Polar star must be exactly on the Horizon. The devices are placed on the top of the spacecrafts and these can land on the Earth only in vertical positions. Also every device has to be invisible directly for the device of any other spacecraft. Estimate, how many spacecraft of every type of animaloids can land on the Earth. Consider the Earth to be spherical, the spacecrafts' bottom to be on its surface (may stand either on ground or on the water) and height of every spacecraft is $ h\equal{}10m$

Enjoy!  :lol:

P.S.: I have ommited $ 4$ as it involves a photograph. :)

Anyone trying?
No. $ 2$ and $ 3$ are my favorites. I think there is a very elegant method(though not tried yet) for $ 2$, using the fact that luminosity follows inverse square law.

hey  i have done most of these and yes that elegant method was tremendous but the thing is i dont know latex !@#$ 98$

No. 2 is a very nice problem. I couldn't solve it at all. Shreyas Patankar, can you please post your solution?


Let $ABC$ be an acute triangle, let $H$ be the foot of its altitude from $A$, and let $D$, $E$, $Q$ be the feet of the perpendiculars from an arbitrary point $P$ inside the triangle onto the lines $AB$, $AC$, $AH$, respectively. Prove that $\left|AB\cdot AD-AC\cdot AE\right|=BC\cdot PQ$.

Let $M=AP\cap BC$, and let $B',C'$ be the feet pf the perpendiculars from $B,C$ to $AP$ respectively.

We have $AB\cdot AD=AP\cdot AB',\ AC\cdot AE=AP\cdot AC'$, and we get $|AB\cdot AD-AC\cdot AE|=AP\cdot B'C'$. What we must show now is $AP\cdot B'C'=BC\cdot PQ$, which is equivalent to $\frac{AP}{PQ}=\frac{BC}{B'C'}$.

Since $\frac{BM}{MB'}=\frac{CM}{MC'}$, we get $\frac{BC}{B'C'}=\frac{BM}{MB'}$, so we have to prove $\frac{AP}{PQ}=\frac{BM}{MB'}$. This follows from the fact that $APQ,BMB'$ are similar right triangles (simple angle chase).

Another solution:

[b]Lemma 1.[/b] If $k_1$ and $k_2$ are two circles with centers $O_1$ and $O_2$, and if P and Q are two points such that $PQ\perp O_1O_2$, then the difference between the powers of the point P with respect to the circles $k_1$ and $k_2$ equals the difference between the powers of the point Q with respect to the circles $k_1$ and $k_2$.

[i]Proof.[/i] Since $PQ\perp O_1O_2$, we have $\angle PTO_1 = 90^{\circ}$, where T is the point of intersection of the lines PQ and $O_1O_2$. Thus, by the Pythagoras theorem, $O_1P^2=O_1T^2+TP^2$. Similarly, $O_2P^2=O_2T^2+TP^2$.

If $r_1$ and $r_2$ are the radii of the circles $k_1$ and $k_2$, then, since the centers of these two circles are $O_1$ and $O_2$, the power of the point P with respect to the circle $k_1$ equals $O_1P^2-r_1\ ^2$, and the power of the point P with respect to the circle $k_2$ equals $O_2P^2-r_2\ ^2$. Therefore, the difference between these two powers is

$\left(O_1P^2-r_1\ ^2\right)-\left(O_2P^2-r_2\ ^2\right)=O_1P^2-O_2P^2-r_1\ ^2+r_2\ ^2$
$=\left(O_1T^2+TP^2\right)-\left(O_2T^2+TP^2\right)-r_1\ ^2+r_2\ ^2=O_1T^2-O_2T^2-r_1\ ^2+r_2\ ^2$.

So this is the difference between the powers of the point P with respect to the circles $k_1$ and $k_2$. But similarly, we get the same expression for the difference between the powers of the point Q with respect to the circles $k_1$ and $k_2$. Thus, the two powers are equal, and Lemma 1 is proven.

Now, to the solution of the problem:

We will work with directed segments.

If X is the orthogonal projection of the point P on the line BC, we have < PXH = 90°; since we also have < XHQ = 90° and < PQH = 90°, it follows that the quadrilateral PQHX has three right angles, so that it is a rectangle. Thus, PQ = XH.

Since < PXB = 90° and < PDB = 90°, the points X and D lie on the circle with diameter PB. Denote this circle by y; the center of this circle is the midpoint Y of the segment PB. Similarly, the points X and E lie on the circle z with diameter PC, and the center of this circle is the midpoint Z of the segment PC. Now, since the points Y and Z are the midpoints of the sides PB and PC of triangle BPC, respectively, we have YZ || BC. On the other hand, since the line AH is an altitude in triangle ABC, we have $AH\perp BC$. Thus, $AH\perp YZ$. Since the points Y and Z are the centers of the circles y and z, we can thus apply Lemma 1 and conclude that the difference between the powers of the point A with respect to the circles y and z equals the difference between the powers of the point H with respect to the circles y and z. Now, the power of the point A with respect to the circle y is $AB\cdot AD$, the power of the point A with respect to the circle z is $AC\cdot AE$, the power of the point H with respect to the circle y is $HB\cdot HX$, and the power of the point H with respect to the circle z is $HC\cdot HX$. Thus, we have

$AB\cdot AD-AC\cdot AE = HB\cdot HX-HC\cdot HX$.

In other words,

$AB\cdot AD-AC\cdot AE = \left(HB-HC\right)\cdot HX = CB\cdot HX = BC\cdot XH$.

Since PQ = XH, we thus get

$AB\cdot AD-AC\cdot AE = BC\cdot PQ$.

For non-directed segments, this rewrites as

$\left|AB\cdot AD-AC\cdot AE\right| = BC\cdot PQ$.

$\blacksquare$

  Darij


I have been trying to prove that $ 12\cdot{x^3}\minus{}3\equal{}n^2$ has no solutions in naturals numbers with $ x > 1$

I first thought of finding a mod in which it makes no sense, but I just couldn't find one (after some attempts).
Thanks in advance.

If we are allowed to use Fermat's Last Theorem, I can solve this.

Clearly, $ 3 \mid n$ and $ n > 0$ since $ x > 1$. Let $ n \equal{} 3m$ for some integer $ m > 0$. Then \[ 12x^3 \minus{} 3 \equal{} 9m^2 \Rightarrow 4x^3 \minus{} 1 \equal{} 3m^2. \] Since the left-hand side is an odd number, $ m$ is odd. Let $ m \equal{} 2k\plus{}1$ for some integer $ k \geq 0$. By expanding, we get that \[ 4x^3 \equal{} 3(2k\plus{}1)^2 \equal{} 12k^2 \plus{} 12k \plus{} 4. \] Therefore, by dividing by four on both sides, \[ x^3 \equal{} 3k^2 \plus{} 3k \plus{}1 \Rightarrow x^3 \plus{} k^3 \equal{} (k\plus{}1)^3. \] If $ k \equal{} 0$, then $ x \equal{} 1$, which is impossible since $ x > 1$. Otherwise, $ k > 0$ and by Fermat's Last Theorem, there are no positive integer solutions to this equation.

(PS. I am sure that the equation $ x^3 \plus{} k^3 \equal{} (k\plus{}1)^3$ not having positive integer solutions can be proved using elementary means and using FLT is probably overkill.)


Let p>2 be a prime, and a, b, c, d be integers not divisible by p, such that:
{ra/p} + {rb/p} + {rc/p} + {rd/p} = 2
for any integer r not divisible by p.

Prove that at least two of the numbers a+b, a+c, a+d, b+c, b+d, c+d are divisible by p.

{} is the fractional part function.

An initial thought:
We probably have to have p | (a + b) and p | (c + d), or some permutation thereof.  If that is the case, we definitely have a working solution.  So we have to prove that in any other arrangement, there will be some r such that the sum doesn't work.  Also, we can take a, b, c, d and r in [1, p - 1].  Hmm...

Okay, further thoughts:  letting r = 1, we get
{a/p} + {b/p} + {c/p} + {d/p} = 2 which gives us
a + b + c + d = 2p.
Now, we know that p/2 is not integral, so each of a, b, c and d is strictly greater or less than p/2.  Furthermore, we know that at least one of the four must be greater and at least one of the four must be less.  WLOG assume a > p/2, b < p/2.  Then letting r = 2, we get
{2a/p} + {2b/p} + {2c/p} + {2d/p} = 2
(2a - p) + 2b + (2c + 2d - np) = 2p where n = 0, 1, or 2:  0 if c, d < p/2, 2 if c, d > p/2, 1 if they are split (one above, one below).
This gives us 2(a + b + c + d) = (3 + n)p from which we get n = 1.  So assume WLOG  a, c > p/2 > b, d.
Now we just have to show the sums are what we want.

I wonder if I used a similar argument for r = p - 1, would that finish it off?  Hmmm ... will continue thinking on it.

bump

I think you can also assume that say a = 1 <= b < p/2 <=c <=d, and try to prove that d = p-1

*note, you can assume a = 1, because if a != 1, there will be some k s.t. ak=1 mod p, and then a' = 1, b' = bk, c' = ck, d'=dk and relabel


I wanted to post a problem to advanced forum, but I don't know to which category (I was thinking about inequalities or geometry). Maybe there will be more such problems. What should I do with them? Maybe there's forum named other problems somewhere but I haven't found it.

You can put it in either Geometry or Inequalities. If it is mostly unappropriate moderators will move it from one to another. :)


prove that there are no  positive integer solutions $ x,y,z$ such that $ 2xz\equal{}y^{2}$ and $ x \plus{} z\equal{}1994$

[b]PLEASE HIDE YOUR SOLUTIONS[/b]

well here  is my solution .i am not sure whether i am right

[hide="Unsure solution"] the given equation we have $ 2xz$ to be a perfect square.$ 2$ is not a perfect square ,so we have $ xz = 2^{2n + 1}$ where $ n %Error. "belongsto" is a bad command.
\ N$
since a power of 2 will be divisible by numbers which are multiples of 2 alone,we have $ x = 2^{k}$ and $ z = 2^{l}$
lets assume without loss of generality that $ l > k$
we have $ 2^{k} + 2^{l} = 1994$
$ \implies$
$ 2^{k}(1 + 2^{l - k}) = 1994$
$ 2^{k - 1}(1 + 2^{l - k}) = 997$
this is not possible because the [b]L.H.S[/b] is even and the [b]R.H.S[/b] is odd.
so........... there are no solutions.

Am i correct in my solution[/hide]

well i think it can be solved [hide="using"]some infinite descent or something like that case.but i dont know wat it is  :( .[/hide]

this is wat i have
[hide]we have $ 1994 \equal{} x \plus{} z$
note that $ 1994 \equal{} 2 \cdot 997$(997 is a prime)
square both sides to get-
$ 4 \cdot (997)^2 \equal{} x^2 \plus{} z^2 \plus{} 2xz$
or,
$ 4 \cdot (997)^2 \equal{} x^2 \plus{} z^2 \plus{} y^2$
we have too cases,
since $ y^2$ is a perfect square and even .
$ y^2\equiv 0\mod 4$
since $ y^2 \equal{} 2xz$,
we have two cases,
     [u]Case 1[/u] - $ x,z \equal{} 4k$
     this means $ (x \plus{} z)^2 \equal{} (8a)^2 \equal{} 64\cdot a^2$
     also $ 64\cdot a^2 \equal{} 4 \cdot (997)^2$
     $ 997 \equal{} 4 \cdot 249 \plus{} 1$
     therefore, $ 997 \equal{} 4b \plus{} 1$
     or, $ 16 \cdot a^2 \equal{} 4b \plus{} 1$
     which is not  possible since the LHS is a multiple of 4 but the RHS isnt
     [u]Case 2[/u]- $ x,z\equiv 2\mod 4$
     therefore either $ x,z\equiv 2\mod 4$
     or $ z,x\equiv 1\ mod 4$ respectively
     but this makes $ x^2 \plus{} y^2 \plus{} z^2\equiv 1\mod 4$
     which is a contradiction
hence no solutions in integers.[/hide]

ok i guess my solution was long!
anyway can anyone tell what infinte descent is?

geniusbliss,
two cases need not be taken
$ y^2$ is always  congruent to 0 mod4 as it is even

ok you have edited it. :lol:

srinath.r: your solution is ocmpletely wrong.

geniusbliss:good solution but you have made a mistake though still the solution is correct.
$ x,z$ are even {whihc can be proved by divisibility}.
In Case 1: $ x\equal{}4k, z\equal{} 4k'$ is how you should have written because it is not necessary that $ x$ and $ z$ are same.
Still it comes to $ 16(k\plus{}k')^2 \equal{} 4*997$ whihc is not possible clearly.
Otherwise, a nice one.

Well, infinite descent cannot solve this one as for the second time u cannot say the same  even divisibilty when you get $ 997$.

well ok but i dint mean to write it as $ x,z \equal{} 4k$ i was thinking one thing but typed another :)


algebra
polynomial
geometry
angle bisector

[hide="A Start for #2"]\begin{align*}(1000a+100b+10c+d)+(a+b+c+d)&=2005\\ & \Downarrow\\ 1001a+101b+11c+2d&=2005\end{align*}
$ a=0$ results in a 3-digit number; $ a=2$ is impossible unless $ d=1.5$, but $ d$ must be an integer. $ \therefore a=1$.

\begin{align*}1001+101b+11c+2d&=2005\\ & \Downarrow\\
101b+11c+2d&=1004\end{align*}[/hide]

[quote="watcharapon"]1. let f(x) = x^6+x^5+x^4+x^3+x^2+x + 1 determine remnants for  f(x^7)/f(x)
 2. the sum of a four-digit number ands its four digits is 2005. what is this four-digit number?
 3. In triangle ABC, AB=10 and AC=18. Mis the midpoint of of BC, and the line through M parrallel to the bisector of CAB cuts AC at D. find the length of AD.
 4. Let x,y and z be positive numbers such that x+y+xy=8 y+z+yz=15 z+x+zx=35 find the value of x+y+z+xy
 5. Let A be the larger root of (2004x)^2 -(2003)(2005)x -1=0 and B be the smaller root of x^2 + 2003x - 2004 =0. Determine the value of A-B
 6.Determine all integers n such that n^4 - 4n^3 +15n^2-30n + 27 is a prime number
 7.Let a,b and c be real numbers such that a+bc = b+ca = c+ab =501. If M is  the maximum value of a+b+c and m is the minimum value of a+b+c . Determine the value of M + 2m
 8.the positive integers a,b and care such that a+b+c =20=ab + bc -ca-b^2 Determine all positive value of abc.[/quote]

Brief answers.

[hide="1"]
$ f(x) = (x^6 + x^5 + x^4 + x^3 + x^2 + x + 1)Q(x) + R(x)$, where $ R(x)$ is at most a fifth degree polynomial.

Let $ \omega = e^{2i\pi/7}$, then $ f(\omega) = f(\omega^2) = \cdots = f(\omega^6) = 0$ and $ f(\omega^7) = f((\omega^2)^7) = \cdots = f((\omega^6)^7) = 7$, so $ R(\omega) = R(\omega^2) = \cdots = R(\omega^6) = 7$. But thats six values, and $ R(x)$ is fifth degree polynomial, so $ \boxed{R(x) = 7}$.
[/hide]

[hide="2"]
Let $ s(n)$ be sum of digits of $ n$. Since $ s(n) \le 1 + 9 + 9 + 9 = 28$, nothing $ \le 1976$ works; trivial casework shows $ n \ge 2000$ doesn't work either. So $ n + s(n) = 2005 \Longrightarrow \overline{19ab} + (1 + 9 + a + b) = 2005 \Longrightarrow 11a + 2b = 95$. Since $ a, b$ are digits, by checking parities we see that $ \boxed{1979}$ is the unique answer.
[/hide]

[hide="3"]
Let $ N$ be foot of angle bisector. By [[angle bisector theorem]] we have $ BN = 5x, CN = 9x$ for some $ x$, and thus $ MC = \frac {BC}{2} = \frac {BN + CN}{2} = 7x$. Since $ \triangle CDM \sim \triangle CAN$ by corresponding angles etc,
\[ \frac {CD}{18} = \frac {7x}{7x + 2x} \Longrightarrow CD = 14
\]
And so AD = $ 18 - CD = \boxed{4}$.
[/hide]

[hide="4"]
Rewrite $ x + y + xy + 1 = 9$ etc, and factoring,
\begin{align} (x + 1)(y + 1) & = 9 \\
(y + 1)(z + 1) & = 16 \\
(x + 1)(z + 1) & = 36 \end{align}
Their product is
\begin{align}(x + 1)^2(y + 1)^2(z + 1)^2 = 9 \cdot 16 \cdot 36 \Longrightarrow (x + 1)(y + 1)(z + 1) = 72\tag{4}\end{align}
Divide each of $ (1),(2),(3)$ from $ (4)$ to get
\[ z + 1 = 8,\ x + 1 = 4.5,\ y + 1 = 2
\]
, and so $ x + y + z + xy = 3.5 + 1 + 7 + 3.5 = \boxed{15}$.
[/hide]

[hide="5"]
$ (2004^2x + 1)(x - 1) = 0$ and $ (x + 2004)(x - 1) = 0$, so $ A - B = 1 - ( - 2004) = \boxed{2005}$.
[/hide]

[hide="6"]
Rewrite as $ [(n - 2)n(n + 2)]n^2 + 3(5n^2 - 10n + 9)$; modulo 3, we see that at least one of $ n - 2, n, n + 2$ must be divisible by $ 3$ so the entire expression is vidisible by $ 3$. So the only time the answer can be prime is if
\[ n^4 - 4n^3 + 15n^2 - 30n + 27 = 3
\]
which doesn't seem to occur.
[/hide]

7

8

5.
from the RRT we know that A=1 and B=-2004 so $ A \minus{} B \equal{} 2005$
7
.$ a \plus{} bc \equal{} b \plus{} ca < \equal{} > a \minus{} ca \plus{} bc \minus{} b \equal{} 0 < \equal{} > a \minus{} b \minus{} c(a \minus{} b) \equal{} 0 < \equal{} > (a \minus{} b)(1 \minus{} c) \equal{} 0$
like this we get $ (1 \minus{} a)(b \minus{} c) \equal{} 0$ and $ (1 \minus{} b)(a \minus{} c) \equal{} 0$
because if $ a \equal{} b \equal{} c \equal{} 1 \equal{} > a \plus{} bc \equal{} 2$ and not $ 501$ we just know that $ a \equal{} b \equal{} c$
and
$ a^2 \plus{} a \minus{} 501 \equal{} 0$
$ D \equal{} 2008$
$ a1 \equal{} (1 \plus{} \sqrt{2008})/2$ and $ a2 \equal{} (1 \minus{} \sqrt{2008})/2$
so
$ M \equal{} 3*a1 \equal{} (3 \plus{} 3*(\sqrt{2008}))/2$
 and
 $ 2m \equal{} 6*a2 \equal{} 3 \minus{} 3*(\sqrt{2008})$
so
 [u]$ M \plus{} 2m \equal{} 15/2 \minus{} 3/2(\sqrt{2008})$[/u]


inequalities
number theory unsolved
number theory

prove that the equation dose not have any answer in Fibonacci numbers with a1=a2=1

 Fn = p^a    and p=4t+3      t>0

i searched for the condition(s) that $ n$ must have if such equation holds and i got this:
$ p$ is prime ,no?
if so,let's assume $ n \equal{} q_1^{\alpha_1}...q_k^{\alpha_k}$ and let $ k > 1$.
if $ n \equal{} 2^s.\beta$ and $ \beta > 1$ then obviously we must have $ F_{\beta} \equal{} p^b$ where $ b > 0$.but we know that for all $ m > 0$ we have $ F_{2m \minus{} 1}^2 \plus{} F_{2m \plus{} 1}^2 \plus{} 1 \equal{} 3F_{2m \minus{} 1}F_{2m \plus{} 1}$(see PEN L5).and if we put $ m \equal{} \frac {\beta \minus{} 1}{2}$ we get that $ F_{\beta}$ doesn't have any prime divisors equivalent to $ \minus{} 1$ modulo $ 4$.
so we get that $ n \equal{} 2^s$.but in this case since $ s$ must be greater than $ 1$ we get that $ F_4 | F_n \implies 3 |F_n \implies p \equal{} 3$ and it is a contradiction with $ p \equal{} 4t \plus{} 3,t > 0$.u can see that if $ s>2$ then it doesn't work at all!
i tried another way to solve this nice problem(without using what was mentioned in PEN L5) and i got only this: :(
we'll try to prove that $ n$ is a prime(i assume that $ n$ is odd if not then take the odd part of $ n$ and replace it by $ n$).
note that this odd part which is $ > 1$ exists (if not then $ n \equal{} 2^s \implies p \equal{} 3$).
we know that $ F_l | F_{lt}$.for all $ l,t \in N$.
it's obvious that in this case we must have

$ F_{q_1} \equal{} p^{t_1}$ and also $ F_{q_k} \equal{} p^{t_k}$

but since we had assumed $ k > 1$ and also

$ \gcd(F_{q_1},F_{q_k}) \equal{} F_1 \equal{} 1$.

so we have a clear contradiction by assuming that $ k > 1$.

so we must have $ n \equal{} q^b$ for some odd prime $ q$ and integer $ b$.

obviously we have $ b < a \plus{} 1$.

to prove it just see that $ F_q < F_{q^2} < ... < F_{q^{b}}$ and all $ F_{q^{i}},0 < i < b \plus{} 1$ divide $ p^a$ which implies that

$ F_q \equal{} p^{a_1} < F_{q^2} \equal{} p^{a_2} < ... < F_{q^{b}} \equal{} p^{a_b}$

and

$ 0 < a_1 < a_2 < ... < a_{b} \equal{} a$.

so it's obvious that $ b < a \plus{} 1$.

let's assume $ b > 1$ and $ q \neq p$.

we have proven in PEN L4 this property :

$ F_t^3 | F_{tl} \minus{} F_{t \plus{} 1}^l \plus{} F_{t \minus{} 1}^l$.now we put $ t \equal{} l \equal{} q$ which implies that

$ F_q^3 | F_{q^2} \minus{} F_{q \plus{} 1}^q \plus{} F_{q \minus{} 1}^q$ but we have $ F_q \equal{} p^{a_1}$ and

 $ p^{a_1 \plus{} 1} | F_q^3 | F_{q^2} \minus{} F_{q \plus{} 1}^q \plus{} F_{q \minus{} 1}^q$ and also since

$ F_{q^2} \equal{} p^{a_2},a_2 > a_1$
we get that
$ p^{a_1 \plus{} 1} | F_{q \plus{} 1}^q \minus{} F_{q \minus{} 1}^q \equal{} (F_{q \plus{} 1} \minus{} F_{q \minus{} 1}).(\frac {F_{q \plus{} 1}^q \minus{} F_{q \minus{} 1}^q}{F_{q \plus{} 1} \minus{} F_{q \minus{} 1}}) \equal{} (p^{a_1}).(\frac {F_{q \plus{} 1}^q \minus{} F_{q \minus{} 1}^q}{p^{a_1}}) \implies p | \frac {F_{q \plus{} 1}^q \minus{} F_{q \minus{} 1}^q}{p^{a_1}}$
which is a clear contradiction since
 $ \gcd(F_{q \plus{} 1} \minus{} F_{q \minus{} 1},\frac {F_{q \plus{} 1}^q \minus{} F_{q \minus{} 1}^q}{F_{q \plus{} 1} \minus{} F_{q \minus{} 1}}) \equal{} (F_q,q) \equal{} 1 \implies p \nmid \frac {F_{q \plus{} 1}^q \minus{} F_{q \minus{} 1}^q}{p^{a_1}}$
it means that we have $ b \equal{} 1$ or $ q \equal{} p$.
the first case seems difficult.(for me :D )it is $ F_q \equal{} p^a$.
for the second one
we get that $ F_q \equal{} q^{a_1},F_{q^2} \equal{} q^{a_2},...,F_{q^{b}} \equal{} q^{a_b} \equal{} q^a$.
by using the same theorem mentioned above this time for $ p \equal{} q$ we get that if $ b > 1$ then
$ F_{q}^{3}| F_{q^{2}} \minus{} F_{q \plus{} 1}^{q} \plus{} F_{q \minus{} 1}^{q} \implies q^{a_1 \plus{} 2} | F_{q \plus{} 1}^{q} \minus{} F_{q \minus{} 1}^{q} \equal{} \frac {F_{q \plus{} 1}^q \minus{} F_{q \minus{} 1}^q}{F_{q \plus{} 1} \minus{} F_{q \minus{} 1}}.F_q \implies q^2 | \frac {F_{q \plus{} 1}^q \minus{} F_{q \minus{} 1}^q}{F_{q \plus{} 1} \minus{} F_{q \minus{} 1}}$
note that i have used the fact that $ q^{a_1 \plus{} 2} | F_{q^2}$.to prove it just use inequality and prove that $ q.F_q < F_{q^2}$.
but u can easily see that (by the well-known theorem of lifting the exponent)
$ q \parallel{} \frac {F_{q \plus{} 1}^q \minus{} F_{q \minus{} 1}^q}{F_{q \plus{} 1} \minus{} F_{q \minus{} 1}}$.so we have a contradiction with $ b > 1$.
it means we have $ b \equal{} 1$.
and we have now $ F_q \equal{} q^a$.
so i have proven that if there exists $ n$ s.t. $ F_n \equal{} p^a$ and $ p$ and $ n$ are odd then $ n \equal{} q$ where $ q$ is prime.
hope i didn't make any mistake! :D

In fact, if we look for solution with $ a>1$ and $ p$ arbitrary, there are only $ 4$ such solutions: $ 1,1,8,144$.

[quote="zhoraster"]In fact, if we look for solution with $ a > 1$ and $ p$ arbitrary, there are only $ 4$ such solutions: $ 1,1,8,144$.[/quote]
i didn't try on it but i'm curious to know how u got this ?

It's not me who proved this, but [url=http://arxiv.org/abs/math/0403046]Bugeaud et al[/url]


I need some help figuring out this complicated formula.

An event with 5 possibilities.
The chance of each event is based on X. X can range from 0 to 1. We are given the following information.

The chance of event 1 occurring is:
x = 1/5, chance = 0%
x = 2/5 chance = 1%
x = 3/5 chance = 1%
x = 4/5 chance = 11%
x = 1 chance = 25%

The chance of event 2 occurring is:
x = 1/5, chance = 2%
x = 2/5 chance = 6%
x = 3/5 chance = 18%
x = 4/5 chance = 34%
x = 1 chance = 55%

The chance of event 3 occurring is:
x = 1/5, chance = 11%
x = 2/5 chance = 24%
x = 3/5 chance = 48%
x = 4/5 chance = 40%
x = 1 chance = 16%

The chance of event 4 occurring is:
x = 1/5, chance = 33%
x = 2/5 chance = 49%
x = 3/5 chance = 26%
x = 4/5 chance = 14%
x = 1 chance = 3%

The chance of event 5 occurring is:
x = 1/5, chance = 54%
x = 2/5 chance = 20%
x = 3/5 chance = 7%
x = 4/5 chance = 1%
x = 1 chance = 1%

Each event is mutually exclusive (only 1 event can occur at a time). The total chance over the 5 events must equal 100% for any one X value.

How do I figure out the chance of any event when x is any number?

i dont get it. could u explain why 2/5 chance=10% for example?

No, he's saying that first you pick an X value:

Say I pick 3/5.

Then the chance of event one happening is 1%.
The chance of event two is 18%
The chance of event three is 48%
The chance of event four is 26%
The chance of event five is 7%

That only applies when X = 3/5.


logarithms
superior algebra
superior algebra unsolved

Prove that if $ a,x,y$ are $ p$-adic integers different from $ 0$ and $ p | x, pa | xy$, then \[ \frac 1y \frac{(1\plus{}x)^y\minus{}1}{x} \equiv \frac{\log (1\plus{}x)}{x} \;\;\;\; ( \textrm{mod} \; a\ ) \\\\  .\]

[i]L. Redei[/i]

Using the power series expansions for the $p$-adic [url=https://en.wikipedia.org/wiki/P-adic_exponential_function]exponential and logarithm[/url], we have for $\log(1+x)=z$ that
\[
\frac{(1+x)^y-1}{y} = \frac{\exp\big(y\log(1+x)\big)-1}{y}=z+\frac{yz^2}{2!}+\frac{y^2z^3}{3!}+\dots
\]
Therefore, the problem reduces to
\[
S=\frac{yz^2}{2!}+\frac{y^2z^3}{3!}+\dots\equiv 0\pmod{ax},
\]
where it suffices to show $pS\equiv 0\pmod{x^2y}$ because $pa|xy$. However, this follows from the fact that $x^2|p\cdot \frac{z^k}{k!}$, where
\[
\frac{z}{x}=\frac{\log(1+x)}{x}=1-\frac{x}{2}+\frac{x^2}{3}\mp\dots
\]
is a $p$-adic integer (due to $v_p(x^{k-1})\ge k-1\ge v_p(k)$), so our divisibility in turn reduces to $p\cdot \frac{z^{k-2}}{k!}$ being a $p$-adic integer, which is clear, because by [url=http://artofproblemsolving.com/wiki/index.php?title=Legendre%27s_Formula]Legendre's formula[/url], $v_p(k!)\le \frac{k-1}{p-1}\le (k-2)v_p(z)+1$. $\blacksquare$


I'm not sure if this belongs in this forum, but here goes:
Using only algebra, simplify the below:
1.$\sqrt[3]{i}$

2. $\frac{1}{\sqrt{i}+i}$

Yes, probably this forum... by using only algebra do you mean complex trig stuff allowed?
[hide="1"]$i=e^{i\frac{\pi}{2}}, i^{\frac13}=e^{i\frac{\pi}{6}},e^{i\frac{5\pi}{6}},e^{i\frac{9\pi}{6}}$... that's pretty messy stuff...[/hide]
[hide="2"]rationalize twice: first multiply thru by $\sqrt{i}=i$, then by $i-1$. Now you just calculate the top...[/hide]

If "using only algebra" means that De Moivre is shunned, then we should proceed thus: $x^{3}=i\implies x^{6}=-1$, hence we should solve $x^{6}+1=0$ and then take only the three solutions that satisfy $x^{3}=i$ (the remaining three will give $x^{3}=-i$, of course).

Similar for the second problem: find the two values of $\sqrt{i}$ by solving $x^{4}+1=0$, and then substitute them and simplify.

Hint for both equations: Factorize


ok. Well, basically this topic is to right about funny things that happen....and to have fun!! JOKES, FUNNY STUFF, RIDDLES, and other.
Ok. now that's all I have to say..:P
hm.Do you want to have a riddle for the day? I will do it, but if I don't do it one day, someone else do it!
WE CAN!!! Then the next person below this post, posts a riddle, joke, but the riddle a day, I do....ok anyway LETS START!
[b]
Riddle 1 (Sunday riddle):What do you call a fish without an eye?[/b]
[hide="Answer, figure riddle out before looking!"]FSH!Get it, the i...?[/hide]

haha? that was just sooo funny k00lperson
but

[hide="What did one Smuckers bottle say to the other?"]frankly scarlett, i dont give a jam[/hide]

what is this?
<3~
mce mce mce

[hide="hint"]not <:3~[/hide]

..idk?

[b]Monday Riddle: Why was the baby ant confused?[/b]
[hide="Answer, solve riddle before looking!"]All his uncles were ants![/hide]

@k00lperson... 3 BLIND MICE!

kinda funny to give an IDK after your fish one.

[hide="i found this amazing joke online"]Fresh out of business school, the young man answered a want ad for an accountant. Now he was being interviewed by a very nervous man who ran a small business that he had started himself. "I need someone with an accounting degree," the man said. "But mainly, I'm looking for someone to do my worrying for me."

"Excuse me?" the accountant said.

"I worry about a lot of things," the man said. "But I don't want to have to worry about money.
Your job will be to take all the money worries off my back."

"I see," the accountant said. "And how much does the job pay?"

"I'll start you at eighty thousand."

"Eighty thousand dollars!" the accountant exclaimed. "How can such a small business afford a sum like that?"

"That," the owner said, "is your first worry."[/hide]

[b]Tuesday Riddle:Why was six afraid of seven? (easy one!)[/b]
[hide="Answer"]Because seven ate nine![/hide]

[hide=" clicky-->:rotfl: "]An inmate on death row was scheduled to be put to death by firing squad the following morning. One of the prison guards asked the inmate if he wanted something special for his last meal. The inmate declined the offer. Later, the prison guard asked the inmate if there was something special he wanted to do on his final day. Again, the inmate declined the offer. The following morning, as he inmate was being put before the firing squad, the guard asked him if he wanted a cigarette and a blindfold. "No," the inmate said, "just get it over with."

"Well, is there anything that I can do for you before you go?" asked the guard. The inmate thought for a moment, then replied, "Actually, music is my life. One thing I would really like would be to sing my favorite song, from beginning to end, without any interruptions." The guard nodded and agreed. "Go ahead," said the guard. The inmate started, "One billion bottles of beer on the wall..."

[/hide]

lol, but after he sang down to 1 ..did he die??lol

k00lperson, the point of the song was to delay his death. it takes a life time to count to 1 billion. so, he probably died halfway through it from old age[hide="."]what did the webkinz say at Christmas( P.S. you have to know about webkinz to answer this question.)[/hide]
there are two hide tags. one is miniminiminiminiminimini. who cares? [hide="Cheater!"] Answer:
I hope Santakinz brought me a lot of loot!!!! :blush: [/hide]

[hide="another one"]When Albert Einstein was making the rounds of the speaker's circuit, he usually found himself eagerly longing to get back to his laboratory work. One night as they were driving to yet another rubber-chicken dinner, Einstein mentioned to his chauffeur (a man who somewhat resembled Einstein in looks & manner) that he was tired of speechmaking.

"I have an idea, boss," his chauffeur said. "I've heard you give this speech so many times. I'll bet I could give it for you." Einstein laughed loudly and said, "Why not? Let's do it!" When they arrive at the dinner, Einstein donned the chauffeur's cap and jacket and sat in the back of the room. The chauffeur gave a beautiful rendition of Einstein's speech and even answered a few questions expertly.

Then a supremely pompous professor asked an extremely esoteric question about anti-matter formation, digressing here and there to let everyone in the audience know that he was nobody's fool. Without missing a beat, the chauffeur fixed the professor with a steely stare and said, "Sir, the answer to that question is so simple that I will let my chauffeur, who is sitting in the back, answer it for me."[/hide]

[b]Wednesday Riddle:What can fill a room but takes up no space?[/b]
[hide="Answer!"]Light! :P :wink: [/hide]

your wrong k00l. think about shadows. a good answer would be matter.

what do you call this?

fce

[hide="look behind you!"]when a blind fish meets a blind mouse.[/hide]

[b]Thursday Riddle:What walks all day on its head?[/b]
[hide="Answer"]A nail in a horse shoe![/hide]

A champion jumps over a castle and lands on a priest. The priest is then dragged of the landscape. What is happening here?

[hide="Answer"]
A chess game!! champion-knight, castle-rook, priest-bishop.[/hide]

Happy New Year!
[b]Friday Riddle:What did the boy squirrel say to the girl squirrel on Valentine's day?[/b]
[hide="Answer.."]I'm nuts about you! :P[/hide]


Three given circles have the same radius and pass through a common point $ P$. Their other points of pairwise intersections are $ A$, $ B$, $ C$. We define triangle $ A'B'C'$, each of whose sides is tangent to two of the three circles. The three circles are contained in $ \triangle A'B'C'$. Prove that the area of $ \triangle A'B'C'$ is at least nine times the area of $ \triangle ABC$


Maria boarded a plane from the airport on the 10th of august . The airport clock showed that it was Saturday shortly after 6 pm
when Maria returned the hand of the day clock had moved 135 degree from its position when Maria left the airport.
How many hours have passed since Maria boarded the plane on Saturday.
ans 63.

Well since the day clock moved $ 135$ degrees, it has moved $ \frac{135}{360}$ of the total week. Thus the number of days that have moved on is $ \frac{135}{360} \cdot 7 \equal{} 2.625$.

This is two whole days and $ \frac{5}{8}$ths of a day, so this is $ 2 \cdot 24 \plus{} \frac{5}{8} \cdot 24 \equal{} 48\plus{}15\equal{}63$ hours. (This is assuming that less than a week passed since she borded the plane.)

Hope this helped.

PS - Notice that the time Maria boarded the plane ins't important at all - all we need to know about is the day clock.


I have 3 points A, B, and C located in a 3D plan.
The x, y, and z coordinates of point A and B are known and also the distance between points A, B, and C is known as well (81 feet).
What I want to do is to determine the x, y, and z coordinates of point C.

So you mean $d\left(A,B\right)=d\left(A,C\right)=d\left(B,C\right)=81$, right?  In that case, you have $A$ and $B$ fixed, but there are an infinite amount of possibilities for $C$ which satisfy this.  The possibilities for $C$ actually turn out to form a circle.

A circle centered at the midpoint of $AB$ and having a radius of $\frac{81\sqrt3}{2}$ as a matter of fact...


The Dwarfs in the Land-under-the-Mountain have just adopted a completely new decimal currency system based on the Pippin, with gold coins to the value of $ 1$ Pippin, $ 10$ Pippins, $ 100$ Pippins and $ 1000$ Pippins.

In how many ways is it possible for a Dwarf to pay, in exact coinage, a bill of $ 1997$ Pippins?

Isn't this more of a informatics problem, more particular dynamic programming?

Hi Andrew;

This is a standard partition problem, easy to set up but hard to compute.

$ \frac {1}{(1 \minus{} x)(1 \minus{} x^{10})(1 \minus{} x^{100})(1 \minus{} x^{1000})}$

and your looking for the coefficient of x^1997, or

$ (1 \plus{} x \plus{} x^2 \plus{} x^3 \plus{} ... \plus{} )(1 \plus{} x^{10} \plus{} x^{20} \plus{} ... \plus{} )(1 \plus{} x^{100} \plus{} x^{200} \plus{} x^{300} \plus{} ... \plus{} )(1 \plus{} x^{1000})$

and your looking for the coefficient of x^1997. I expanded it with a computer to get that coefficient and it is 2650. So there 2650 ways to make 1997 in pippen currency.


abstract algebra
group theory
superior algebra
superior algebra unsolved

Let $ N$ and $ H$ two groups, $ \varphi$ and $ \psi$ two morphisms $ H \to \mathrm{Aut}(N)$.
If there exist $ u \in  \mathrm{Aut}(N)$ such that $ \psi(h) \equal{} u \circ \varphi(h) \circ u^{\minus{}1}$, or if there exist $ v \in  \mathrm{Aut}(H)$ such that $ \varphi \equal{} \psi \circ v$, then $ N \rtimes_{\varphi} H \simeq N \rtimes_{\psi} H$.
So these are two suffisiant conditions for semidirect products to be isomorphic.

I am looking for necessary conditions. In particular, I am interested to find conditions so that if $ N$ and $ H$ are two [b]finite[/b] groups there is a unique (up to isomorphism) semidirect product $ N \rtimes H$ that is not direct.

This question is interesting but I think it is too difficult (or too general).
Even for $ N\equal{}\mathbb{Z}/p^2\mathbb{Z}\times \mathbb{Z}/p\mathbb{Z}$ and $ H\equal{}\mathbb{Z}/p\mathbb{Z}$, with $ p>2$ prime, I think there are 3 non isomorphic, non abelian groups of the form $ N\rtimes H$.
Showing that two semidirect products are non isomorphic is often a difficult task.

[quote="JC_math"]This question is interesting but I think it is too difficult (or too general).
Even for $ N \equal{} \mathbb{Z}/p^2\mathbb{Z}\times \mathbb{Z}/p\mathbb{Z}$ and $ H \equal{} \mathbb{Z}/p\mathbb{Z}$, with $ p > 2$ prime, I think there are 3 non isomorphic, non abelian groups of the form $ N\rtimes H$.
Showing that two semidirect products are non isomorphic is often a difficult task.[/quote]

For what it is worth:

For p=2, there are three non-abelian groups of order 2^4 with abelian normal subgroups of type 4 × 2 that are complemented.  For p≥3, there are five non-abelian groups of order p^4 with abelian normal subgroups of type p^2 × p that are complemented.

In other words, your claim is true for p=2 too, and for p>2 it is even more true than you claimed.  The proof is just consulting the list of groups of order p^4.

Indeed JackSchmidt.
Do you think the initial question from Portland is "solvable"?

[quote="JC_math"]Do you think the initial question from Portland is "solvable"?[/quote]

I don't normally look at solving random questions like this.  First I look to see if there are any interesting special cases or motivation.  I don't see any interesting special cases or motivation for this question.

Semidirect products can be isomorphic for random reasons. Even listing the ones where some isomorphism takes N to N does not seem too interesting.  When H is simple, and N is elementary abelian, this might be a question about the representation theory of H, but even that question seems contrived (some sort of postage stamp problem where the denominations available are the dimensions of the non-principal irreducible representations).

Another aspect of the problem that strikes me as contrived is trying to ignore direct products as if they are not semidirect.  For instance, what about H=A5, N=A5?  There are up to conjugacy exactly two actions of H on N, the trivial action giving the direct product, and the conjugation action, in which H definitely does not centralize N.  I would not think that one would call the semidirect product formed from the second action the direct product, but on the other hand there is an isomorphism (centralizing N) between it and H × N.  How many semidirect products that are not direct does the pair (H,N) have?

Thank you JackSchmidt and JC_Math.

In relation to
[quote="JackSchmidt"]First I look to see if there are any interesting special cases or motivation. I don't see any interesting special cases or motivation for this question...[/quote]
The  motivation was simple, when you find one or several "sufficient conditions", it is fairly natrual to ask wheter or not we know some "necessary conditions", don't you think?
I thought about the question before posting, and because it seemed fairly difficult (at least to me) in the generic case, I reduced it to finite groups, in hope that it would be simpler. The example given by JC_Math convinced me that even for finite (even abelian) groups this is probably too difficult a question.

JackSchmidt,

Like you I normally don't answer random questions.
I asked myself a question not far from the one at the beginning of that thread a while back, which is why I knew about the characterization of groups of order $ p^4$.
Indeed, "this might be a question about the representation theory of H", or about wreath product of cyclic groups (to be used in representation theory:)), which was my case.