abstract algebra
group theory
inequalities
vector
linear algebra
matrix
algorithm

How do I find what group $(\mathbb{Z}\times \mathbb{Z})/<(m,n)>$ is isomorphic to, in the form $\prod \mathbb{Z}_{p_{i}^{n_{i}}}\times \mathbb{Z}^{n}$ from the fundamental theorem of finitely generated abelian groups?

If A is a subgroup of B then rank (B/A) <= rank B - rank A.

In our case B = ZxZ and A = Z(m,n) is a (free) subgroup of rank 1. Therefore rank (B/A) <= 2 - 1 = 1. If rank (B/A) = 0 then B/A is a torsion group, but it is easy to see that not all elements of ZxZ have torsion modulo A. So rank(B/A) = 1. That gives us the free component of B/A : Z^1.

Now the torsion component. Let's examine the elements of B that have torsion mod A. Let (x,y) be an arbitrary element of B. We must have k(x,y) = l(m,n) for some k and l, i.e. kx = lm and ky = ln. If n != 0 then we must have y != 0 and the condition becomes x/y = m/n , so x = m/n * y. If we put y = n / gcd(m,n) then x = m / gcd(m,n). That yields a torsion element of order gcd(m,n). More generally we could choose k-multiples of these values for x and y. For every k | gcd(m,n) we get  a torsion element of order gcd(m,n)/k = N. The torsion groups are all the Z/(N).

That completes the solution. (But check for multiplicities!)

See also my other posting today.

For $m=0$ or $n=0$, our group $A: = (\mathbb{Z}\times \mathbb{Z}) / \langle (m,n) \rangle$ is already cyclic. So let's assume $m \neq 0, n \neq 0$. As henrik pointed out, the torsion-free part of $A$ is $\mathbb{Z}$. To determine $Tor(A)$, we see that the following statements are equivalent:
$\exists z \in \mathbb{Z}\backslash \{0\}: z \overline{(a,b)}= 0$
$\exists z \in \mathbb{Z}\backslash \{0\}, u \in \mathbb{Z}: z(a,b)=u(m,n)$
$\exists z \in \mathbb{Z}\backslash \{0\}, m | za \wedge b = a n/m$
$b = a n/m$
$m | an, b = an/m$
$m/\gcd(m,n) | a \wedge b = an/m$
$\exists k \in \mathbb{Z}: a = k m/\gcd(m,n) \wedge b = k n/\gcd(m,n)$
Thus, $Tor(A)$ is the cyclic group generated by $x : = \overline{(m/\gcd(m,n),n/\gcd(m,n))}$. From the above and  $m|za \Leftrightarrow m/\gcd(m,a) | z$, it follows that the order of $\overline{(a,b)}$ is equal to $|m|/\gcd(m,a)$. Hence the order of $x$ is equal to $\gcd(m,n)$.

The result is $A \cong \mathbb{Z}\oplus \mathbb{Z}/\gcd(m,n)$.

@henrik: You used the inequality $rank(A/B)+rank(B) \leq rank(A)$. But of course, the equality $rank(A/B)+rank(B) = rank(A)$ is true (for arbitrary abelian groups). In our case, the subgroup has rank $1$ if both $m$ and $n$ are not zero.

You are right. If B is a finitely generated abelian group and A < B any subgroup, we have rank B/A = rank B - rank A.

But I didn't want to assume that formula which is somewhat hard. It is much easier to see that rank B/A <= rank B - rank A. Because if A = F + T is a decomposition in a free group and a torsion group, any set of elements in B that are linearly independent mod A can be extended to a set that is rank A larger by including a basis for F, and the extended set is linearly independent in B. So rank B >= rank B/A + rank A.

Here's a proof of equality: Consider first Zn = Zx...xZ (n times). Let e1,...,en be the standard basis. If x1,...,xm is a maximal linearly independent set in Zn, then x1,...,xm,ei is linearly dependent for each i. Hence there is a nontrivial linear combination a1*x1+...+am*xm+a*ei = 0 with a != 0. But then all the ei are spanned by x1,...,xm over Q so it is a basis in Qn. Therefore m = n.

We have found that every maximal linearly independent set in Zn has n elements. But beware, it need not be a basis!

Consider now an arbitrary finitely generated abelian group B. Decompose B = F + T. Let n = rank B = rank F. If x1,...,xm are linearly independent in B, so are their projections into F along T. If they are maximal in B they remain so in F after projection. Therefore m = n since F is free.

So every maximal linearly independent set in B has n = rank B elements.

Now let A < B be a subgroup and A = F + T the usual decomposition. Let x1,...,xm be a basis for F (rank F = rank A = m). Choose a maximal linearly independent set S in B containing these elements. They are x1,...,xm,y1,...,yk with m + k = n since S has n elements. If a1*y1+...+ak*yk = f + t is in A and N*t = 0 then by multiplying the equation with N we must conclude that a1=...=ak=0. So y1,...,yk are linearly independent  mod A. But then rank B/A >= k = n - m = rank B - rank A.

In general, the rank of an abelian group $A$ is defined to be the dimension of the $\mathbb{Q}$-vector space $A \otimes_\mathbb{Z}\mathbb{Q}$. Since it is isomorphic to the localisation $(\mathbb{Z}\backslash \{0\})^{-1}A$, the rank of $A$ coincides with the maximal cardinality of a linear independent subset.

Clearly, $A$ has rank $0$ iff $A$ is torsion, and the rank of the free abelian group $\mathbb{Z}^{(X)}$ equals $|X|$. Thus, the rank of a finitely-generated abelian group, which comes from the structure theorem, concides with the rank as defined above.

Finally, the rank is additive on short exact sequences, which follows from the corresponding property for the dimension of vector spaces and the flatness of $\mathbb{Q}$. In particular, $rank(B)+rank(A/B)=rank(A)$, where $B$ is a subgroup of $A$ and $A$ is an arbitrary abelian group.

It seems like a good idea to reduce the problem to vector spaces. But what do you mean by this localization? And is it obvious that Q is a flat Z-module? Maybe, I just haven't tried to check it.

flatness of $\mathbb{Q}$: http://www.mathlinks.ro/Forum/viewtopic.php?t=86267
localization: algebra-source of your choice, or http://en.wikipedia.org/wiki/Localization_of_a_module

Aha, now I see what you mean. You consider Q as the localization of Z at the 0 ideal and extend scalars for A with Z -> Q. Then it's clear.

The flatness of Q is actually fairly obvious. I used it in my proof without thinking about it as flatness.

Let 0 -> A' -> A -> A'' -> 0 be a short exact sequence of abelian groups. We want to show that 0 -> QxA' -> QxA -> QxA'' -> 0 is exact. (x means tensor product) Take the element a = m1/n1 x a1' + ... + mk/nk x ak' in QxA' and suppose that m1/n1 x a1 + ... + mk/nk x ak = 0 where ai' maps to ai. We must show that a = 0. But all we need to do is to multiply away the denominators and use the flatness of Z (as a Z-module) to conclude that N1 x a1' + ... + Nk x ak' = 0 where for instance N1 = m1n2...nk. Then use the fact that QxA' is a Q-vectorspace to multiply with 1/(n1...nk). That's all! We're through. I mean, we have shown the injectivity of QxA' -> QxA. The surjectivity of QxA -> QxA'' is easy.

Exactly the same proof can be used to show that S'A (ring of fractions) is a flat A-module for any commutative ring A with 1 and multiplicative subset S.

I don't understand anything being said in this thread - I have only been studying algebra for 6 weeks! However, I have another solution. Let $a = m/gcd(m,n), b= n/gcd(m,n)$. Then by choosing $(c,d)$ such that $ad-bc = 1$ we get $\mathbb{Z}\times \mathbb{Z}= \mathbb{Z}(a,b) \oplus \mathbb{Z}(c,d)$.

Now we have a natural automorphism of $\mathbb{Z}\times \mathbb{Z}$ where $(a,b)$ corresponds to $(1,0)$ and $(m,n)$ to $(gcd(m,n),0)$. Thus
$(\mathbb{Z}\times \mathbb{Z})/<(m,n)> \cong (\mathbb{Z}\times \mathbb{Z})/<(gcd(m,n),0)> \cong \mathbb{Z}\times \mathbb{Z}_{gcd(m,n)}$

[quote="henrik"]But all we need to do is to multiply away the denominators and use the flatness of Z (as a Z-module) to conclude that N1 x a1' + ... + Nk x ak' = 0 where for instance N1 = m1n2...nk.[/quote]

I don't understand what you do here ...

@Kalle: Wow, very nice (and simple) solution :-). We can generalize this:

Let $a_{1}, ... , a_{n}\in \mathbb{Z}$, not all $=0$. Then $d : = gcd(a_{1},...,a_{n}) \neq 0$, and $a_{1}/d , ... , a_{n}/d$ are coprime. It is well-known, that there's a $n \times n$ matrix of integers, whose first column is $s_{1}= (a_{1}/d , ... , a_{n}/d)$ and whose determinant is $1$. This yields an automorphism of $\mathbb{Z}^{n}$, which is given by $s_{i}\to e_{i}$. Hence $\langle (a_{1},...,a_{n}) \rangle$ corresponds to $\langle (d,0,...0) \rangle$, and

$\mathbb{Z}^{n}/ \langle (a_{1},...,a_{n}) \rangle \cong \mathbb{Z}^{n}/ \langle (d,0,..0) \rangle \cong \mathbb{Z}/d \times \mathbb{Z}^{n-1}$

Yeah, I guess I made you use heavy theorems because it wasn't clear that the problem came from an introductory course in algebra. We have not at all proven the fundamental theorem of finitely generated abelian groups, only stated it.

Can you point me to some information about the generalization of euclid's algorithm to find an integer matrix with determinant 1? I've never seen such a theorem.

We proved the fundamental theorem of finitely generated abelian groups in a more advanced class this week. I finally understand why the solution I proposed is to prove the theorem in a special case, and what it means. 1.5 years later. Whoo! (My choice of classes is not very algebra-oriented)

[quote="henrik"]Kalle and Germany: You're actually proving the structure theorem in a special case! Not necessary every time.  :)

Here's another fancy way to prove the result: Using homological algebra. To find the free component of the quotient we use the rank argument as before. Now the torsion component. There is a theorem in homological algebra saying that Tor1(A,Q/Z) is the torsion subgroup of the abelian group A for every A.

Tor1(-,Q/Z) is the first derived functor of  - x Q/Z (tensor product). The system of derived functors Torn is a universal homological delta-functor, so for every short exact sequence of abelian groups...

0 -> A' -> A -> A'' -> 0

, we have a long exact sequence...

...-> Tor1(A',Q/Z) -> Tor1(A,Q/Z) -> Tor1(A'',Q/Z) -> A' x Q/Z -> A x Q/Z -> A'' x Q/Z -> 0.

Consider the short exact sequence 0 -> Z -> Z2 -> B -> 0 where Z -> Z2 is k -> k(m,n). Tor1(Z2,Q/Z) = 0 because Z2 is torsion free. So we have the exact sequence:

0 -> Tor1(B,Q/Z) -> Z x Q/Z -> Z2 x Q/Z -> B x Q/Z -> 0.

Tor1(B,Q/Z) is the group we want to compute. Because of exactness at Z x Q/Z it is the kernel of Z x Q/Z -> Z2 x Q/Z.  Any element of Z x Q/Z = Q/Z has the form 1 x a/b with gcd(a,b) = 1. Its image in Z2 x Q/Z is (m,n) x a/b, which is 0 if and only if b | gcd(m,n). Such elements can be written uniquely 1 x a/b = 1 x a'/gcd(m,n) with a' in {0,...,gcd(m,n)-1}.

This proves that the kernel is Z/gcd(m,n) and we're through.

Ok, why so much fuss about such a simple problem? I really prefer my first proof (corrected and completed). But this provides more insight and is a general method that can be used in more complicated situations.[/quote]
Thank!  :rotfl:


[quote="nealth"]Lars Erikson got a perfect score a two day test. Each day he had 4.5 hours to solve 3 problems. If each problem was worth 7 points, then on average, how many minutes did he need to earn 1 point? Round your answer to nearest integer.[/quote]

[hide]$9\times 60\div 42\approx13$?[/hide]

[hide]If it takes 4.5 hours for 3 problems, then $\frac{4.5}{3}=1.5$ hours are spent on each problem on average.

Thus, it takes $\frac{1.5}{7}\cdot60$ minutes to earn one point, which is approximately $13$ minutes.[/hide]

[hide]$\frac{1.5}{7}\times 60 \approx \boxed{13}$[/hide]

i bet it says the answer somewhere on the USAMO site =x

[hide]The answer is $\frac{9*42}{60}$ which is approximately $13$

$13$[/hide]

[hide]
Since he has 4.5 hours and 3 problems, on average, he does 1 problem in 1.5 hours.  It then takes him $\frac{1.5}{7}*60$ minutes to earn $1$ point. Since $\frac{1.5}{7}*60=13$, the answer is $\boxed{13}$[/hide]

[quote="12,21,11,5"]Does "*" just mean "times" :huh:[/quote]
yes :wink:

[hide]4.5 hours: 3 probs. =90 min/problem$\approx$ [b]13[/b] minutes/point[/hide]
[hide= [color=white]hide[/color]](what test was Lars taking?)[/hide]

$\text{all of that can be simplifed into}$ $\frac{270}{21}$ $\text{which is about}$ $\boxed{13}$

did you mean to say 270/21.

because 210/21 =10. :wink:


Given a pair of isogonal conjugates $ P$ and $ Q$ in a triangles $ ABC$, let $ D$ and $ E$ be the feet of the perpendiculars from $ P$ onto $ AB$ and $ AC$ respectively.

Prove that $ AQ$ is perpendicular to $ DE$.

[hide]

Let the intersection of AQ and PE be X, and the intersection of AQ and DE be Y. ADPE is cyclic, as $ \angle{ADP}\plus{}\angle{PEA}\equal{}180^\circ$. Therefore, $ \angle{APD}\equal{}\angle{AED}$. As $ \angle{PAD}\equal{}\angle{QAE}$ and $ \angle{AEP}\equal{}\angle{ADP}$, $ \angle{AXE}\equal{}\angle{APD}$. Therefore, $ AYE\sim AEX$, and $ \angle{AYE}\equal{}\angle{AEP}\equal{}90^\circ$.

[/hide]


suppose u have suppose u  have $ 1-chloro 2-methyl butane$ and $ sec-butyl chloride$ which would undergo $ S_{N}1$ faster

[hide="my thought"]at low temp. it should be $ sec-butyl chloride$ but at sufficiently high temp for substitution it should be $ 1-chloro 2-methyl butane$[/hide]

[hide="Opinion"]On ionization, 1-chloro-2-methylbutane forms a primary carbocation, while sec-butyl chloride forms a secondary one, which is more stable: so sec-butyl chloride should react faster by an SN1 mechanism. However, I don't see how the temperature can invert that.
[/hide]

[hide="may be"]my thought was that though the first has a higher activation energy than the secnd but then it rearranges to form a stable tertiary carbocation.but he first has no way but to remain as such .so if the temp  is suuficiently low then 1 is difficult but if temp is high then bot 1 and 2 take place but 2 can revert back as back reaction needs 0 activation energy but 1 will take time to do so[/hide]

[quote="Pardesi"]as back reaction needs 0 activation energy[/quote]

At least one the species (carbocation or chloride) must be solvated, and those solvent molecules must first be "draw back" in order for recombination to occur.

[hide="Also"]Don't also forget that 1-chloro-2-methylbutane is a primary halide, and even with the beta-methyl group, SN1 would be very hard to force because the SN2 path is much more easy. Even for secondary halides, for SN1 to occur the solvent should be considerably polar and the nucleophile should be soft: otherwise, SN2 or elimination is likely to occur instead.[/hide]


group theory
abstract algebra
superior algebra
superior algebra unsolved

1.Show that a finite group is isomorphic to a subgroup of A_n, n is a certain positive integer.
2.G is a non-Abelian simple group, show that Aut(G) is a complete group.
Thanks~!

Question 2 is standard. (See for example Rotman 1999, theor. 7.14, p. 162.)

For question 1, send each even element f of S_n onto the element of A_n+2 which is equal to f in [1, n] and to identity in [n+1, n+2]. Send each odd element f of S_n onto the element of A_n+2 which is equal to f in [1, n] and whose restriction to [n+1, n+2]  is the (only) transposition of [n+1, n+2].
So, S_n is isomorphic to a subgroup of A_n+2. By Cayley, each finite group is isomorphic to a subgroup of a S_n, thus 1 is demonstrated.


i don't know  about chemical web .
   i want to  have free book download .  can you help me ???

[quote="tranthanhnam"]...   i want to  have free book download .  can you help me ???[/quote]
Try

http://lib.org.by/_djvu/Ch_Chemistry/


superior algebra
superior algebra unsolved

Does there exist an Euclidean ring which is properly contained in the field $ V$ of real numbers, and whose quotient field is $ V$?


Does anyone have a clever way of quickly and efficiently balancing equations? The educated "trial and error" method seems simple enough, but it would be even better if there was a faster way of doing it.

Useful tips would also be nice.

Thanks

Usually the best way is simply inspection, especially in non-redox reactions.  In redox reactions, I usually balance by half-reaction.  If you really want, you can always set up systems of linear equations.

My wicked (not evil, mind you) Chem I teacher gave kids these equations to balance:

$\text{C} + \text{O}_2 \longrightarrow \text{CO} + \text{CO}_2$

$\text{NO}_2 \longrightarrow \text{N}_2 + \text{N}_2\text{O}_4$

$\text{H}^+ + \text{OH}^- \longrightarrow \text{H}_2\text{O} + \text{H}_2$

Sorry, that's not really relevant.  But it's fun.

You can balance equations using algebra. Search for balancing equations using algebra ( or something like that )


you can have more than one page per solution, right?

Yep. Just make sure you clearly label the order of the pages in a solution.


vector
function
linear algebra
linear algebra unsolved

1. R^2x2 is a vector space of 2 by 2 real matrices. Suppose V is the subset of R^2x2 consisting of lower triangular matrices
(a) show that V is a subspace of R^2x2 and prove that the dim(V) = 3
(b) consider R with addition only. Show that V with multiplication contains a copy of  R with addition by finding a function f so f : R --> V, so f(a+b) = f(a) f(b), in particular f(0) = I.

Magibosu, how did you try to solve these problems?


[color=purple] Let triangle $ ABC$ in plane, with any $ x,y,z > 0$ prove that
   $ \frac {y \plus{} z}{x}.\frac {1}{A} \plus{} \frac {z \plus{} x}{y}.\frac {1}{B} \plus{} \frac {x \plus{} y}{z}.\frac {1}{C}\ge\frac {18}{\pi}$
  Beautiful I think  :D [/color]

[quote]$ \frac {y + z}{x}.\frac {1}{A} + \frac {z + x}{y}.\frac {1}{B} + \frac {x + y}{z}.\frac {1}{C}\ge\frac {18}{A + B + C}$[/quote]

$ \| \begin{array}{cc} \\
By \ Am - Gm \\
 \\
\end{array} \| \begin{array}{cc} \frac {y + z}{x} \ge 2\sqrt {\frac {yz}{x^2}} \\
 \\
\frac {x + z}{y} \ge 2\sqrt {\frac {xz}{y^2}} \\
 \\
\frac {y + x}{z} \ge 2\sqrt {\frac {yx}{z^2}} \end{array} \| \Rightarrow \sum \frac {y + z}{x \cdot A} \ge 2 \sum \sqrt {\frac {yz}{x^2}} \cdot \frac {1}{A}$

$ 2 \sum \sqrt {\frac {yz}{x^2}} \cdot \frac {1}{A} \stackrel {Am - Gm}{\ge} 6 \sqrt [3]{\sqrt {\frac {yz}{x^2}} \cdot \sqrt {\frac {xz}{y^2}} \cdot \sqrt {\frac {yx}{z^2}} \cdot \frac {1}{A\cdot B \cdot C}} = \frac {6}{\sqrt [3]{A\cdot B\cdot C}}$

Then we should prove:

$ \frac {6}{\sqrt [3]{A\cdot B\cdot C}} \ge \frac {18}{A + B + C} \Leftrightarrow A + B + C \ge 3 \sqrt [3]{A\cdot B\cdot C}$, which is true by $ Am - Gm$.

[color=purple] You're right Ahiles it's similar as my proff. With this idea we can prove the following
      $ \frac {y \plus{} z}{x}.\frac {1}{a^2.A} \plus{} \frac {z \plus{} x}{y}.\frac {1}{b^2B} \plus{} \frac {x \plus{} y}{z}.\frac {1}{c^2.C}\ge\frac {18}{\pi Rr}$
 [/color]


analytic geometry
geometry
inequalities
congruent triangles

Both $ |x|\plus{}|y| \leq 2$ and $ \minus{}1 \leq x \leq 1$ are graphed in a coordinate plane. What is the number of square units in the area of the region bounded by these inequalities?

$ |x|\plus{}|y|\leq2$ graphs a diamond-like square with vertices $ (2,0)$, $ (0,\minus{}2)$, $ (\minus{}2,0)$, and $ (0,2)$. Restricting $ x$ to $ \minus{}1 \leq x \leq 1$ cuts off the left and right corners of the square. The area of the remaining hexagon is the area of the square minus the area of the two cut-off tiny triangular portions. The area of a square is $ s^2$, and here $ s\equal{}2\sqrt2$, so the area of the square is $ 8$. The two congruent triangles combined have an area of $ 2$, as each cover $ 1$ square unit ($ h\equal{}2$ and $ B\equal{}2$ in the basic triangle area formula of $ A\equal{}\frac{1}{2} Bh$). Thus, the area of the hexagon is $ 8\minus{}2\equal{}\fbox{6}$.

bruh this problem is too much work for ftw

just memorize

this isn't a hard problem if you draw it out

[quote=bingo2019]who has time to draw it out when some ftw games r 10 secs each problem tho[/quote]

yea...


Would you consider this as a proper paragraph? It's a conversation on the phone, and you are only given one side:

"Hello? Do I sound like the Green Lotus? What do you mean "more like the Green Dragon"? No, this is a slim white telephone in an upscale part of the community. Now good day to you."

Should I do that, or this:

"Hello?"
"Yes, is this the Green Lotus?"
"Do I sound like the Green Lotus?"
"More like the Green Dragon."
"What do you mean "more like the Green Dragon"?"
"Are you?"
"No, this is a slim white telephone in an upscale part of the community. Now good day to you."

Which one should I use?

Are you only writing one side of the conversation?  I would go with the first option if so.

The latter method is preferred. In the first one, the conversation between two people gets confusing, since you can be unsure of who is talking; this problem is even more pronounced in dialogoues with 3 or more people in it.

Quotes do count as as proper paragraph, I believe.

EDIT: Wow, even t0rajir0u beat me here.

Keeping Up Appearances?  Man, I love that show.   :D


Let  $A = x^{x^{... infinite...^{x}}}$  , where x is a positive real number .

Right side of the equation is limit of Fn(x) as n tends to infinity where

$F_{n}(x) = x^{x^{x^{x^{ ... n times ... }}}}$ ,

Please tell me when there is no x satisfy the equation.

$F_{n}(x) =x \uparrow \uparrow n$ (Knuth's up-arrow notation)


advanced fields
advanced fields unsolved

Hello. Could someone help me with this problem please?

I want to show that the (uncountable) set of all increasing sequences of natural numbers has not a countable cofinal set.

Thank you in advance.

For the future: you don't need to repost, you can edit your posts.

Let $ A$ be the set of all increasing integer sequences.

$ A$ is isomorph to the set of all integer sequences via
\[ (a_i)_i \mapsto (\Sigma_{k\equal{}0}^i a_k)_i .\]
Therefore $ |A|\equal{}2^\omega$ and
\[ cf(A) \equal{} cf(2^\omega) > \omega \]
(follows from a corollary of König's Theorem).


inequalities
algebra
polynomial
quadratics
function
special factorizations

I can't seem to get this one, I must not be seeing something really simple or at least something quite important  :(

Prove that for a real number $ a$ the following inequality is true: $ a^4\plus{}1\ge a^3\plus{}a$

$ a^4\minus{}a^3\minus{}a\plus{}1\ge 0$

Now, this is a polynomial in $ a$, so the first thing we do is to check for roots. Equality holds when $ a\equal{}1$ so factoring out $ a\minus{}1$ might be a good plave to start.

$ (a\minus{}1)(a^3\minus{}1)\ge 0$

You should see here that $ a\minus{}1$ is also a factor of $ a^3\minus{}1$, so we factor it out again.

$ (a\minus{}1)^2(a^2\plus{}a\plus{}1)\ge 0$.

Now, $ (a\minus{}1)^2 \ge 0$ always holds since it is a square, so all we need to do now is show that $ a^2\plus{}a\plus{}1\ge 0$.

But this is a quadratic in a, and remembering the completing the square operation here,

$ a^2\plus{}a\plus{}1\equal{}(a\plus{}\frac{1}{2})^2\plus{}1\minus{}\frac{1}{4}\equal{}(a\plus{}\frac{1}{2})^2\plus{}\frac{3}{4}\ge \frac{3}{4} >0$, so we are done.

Unless I am remembering this theorem incorrectly, shifting everything to the RHS gives:
$ a^4 \minus{} a^3 \minus{} a \plus{} 1 \geq 0$
There are two sign changes, so by Descartes' Rule of Signs, there are either 2 or 0 real roots to the function $ f(x) \equal{} x^4 \minus{} x^3 \minus{} x \plus{} 1$. We see that the root -1 has a multiplicity of 2, and thus there are no other real roots and the function does not ever switch sign (unless you consider negative zeros :O).

EDIT: rofler did the same thing, except he didn't use wierd theorems

there are three cases
(1) - $ a$ is a positive real
there fore $ a(a^3 \minus{} 1) \ge a^3 \minus{}1$
which is true with equality when a=1
(2) - $ a$ is a negative real
the LHS is anyways positive but the RHS is still negative so done
(3) - $ a$ is 0
this is obvious

:?: How to prove Descartes' Rule of Signs? :?:

$ a^3*a \plus{} 1*1\ge a^3*1 \plus{} a*1$.
Equality holds when $ a \equal{} 1$.
If $ a\ne1$, we have the above inequality by rearrangement...

[quote="t0rajir0u"]As a general rule, any problem of this kind you can also do by weighted AM-GM.  Consider the inequalities

$ a^4 \plus{} 1 \plus{} 1 \plus{} 1 \ge 4a$
$ a^4 \plus{} a^4 \plus{} a^4 \plus{} 1 \ge 4a^3$.[/quote]

I think we need condition $ |a| \ge a$ for your solution because a is real number (we don't know that $ a \ge 0$ or turnaround)

$ a^4 \plus{} 1 \plus{} 1 \plus{} 1 \ge 4|a|\ge a$
$ a^4 \plus{} a^4 \plus{} a^4 \plus{} 1 \ge 4|a^3| \ge 4a^3$.

;)  :|   :|

Here is my proof:
By Cauchy, $ (a^4\plus{}1)(a^2\plus{}a^2) \ge (a^3\plus{}a)^2$
$ a^4\plus{}1 \ge (a^3\plus{}a) \cdot \frac{a^2\plus{}1}{2a}$
But by AM-GM, $ a^2\plus{}1 \ge 2a$
The problem is obvious if a is negative.

If you homogenize by letting $ b \equal{} 1$ and writing $ a^4 \plus{} b^4 \geq a^3b \plus{} ab^3$, it's true by Muirhead, so an AM-GM solution must exist. :D
[hide]$ \frac {a^4 \plus{} a^4 \plus{} a^4 \plus{} 1}{4} \geq a^3$ and $ \frac {a^4 \plus{} 1 \plus{} 1 \plus{} 1}{4} \geq a$ both by AM-GM. Add. (note this works even if $ a$ is negative since $ a^4 \geq 0$)[/hide]

Just wondering: why don't you directly use Muirhead to prove the ineq but rather use it to show an AM-GM solution exists?

Also does anyone know how to prove Descartes' Rule of Signs?

[quote="mathwizarddude"]Just wondering: why don't you directly use Muirhead to prove the ineq but rather use it to show an AM-GM solution exists?[/quote]
Muirhead is equivalent to AM-GM, but AM-GM is conceptually nicer and less blunt.  Muirhead on this problem is overkill.

[quote="mathwizarddude"]Also does anyone know how to prove Descartes' Rule of Signs?[/quote]
[url=http://www.cut-the-knot.org/fta/ROS2.shtml]Cut the Knot[/url] has a great discussion.

[quote="tonypr"]I can't seem to get this one, I must not be seeing something really simple or at least something quite important  :(

Prove that for a real number $ a$ the following inequality is true: $ a^4 \plus{} 1\ge a^3 \plus{} a$[/quote]

this is merely an applcation of the rearrangement inequality. :wink:


Hi,

It seems that the TODO package doesn't work inside theorem, itemize environment, etc...

Is this problem familiar?
[b]How can I solve it?[/b]

Thanks, Yochai.

[quote="Peter VDD"]what is the TODO package? :)[/quote]

Hi Peter,

The TODO package is a utility for generating a TODO list for your latex document. [b]Very recommended![/b]

Check it: [url]http://www.ctan.org/tex-archive/macros/latex/contrib/todo/?action=/tex-archive/macros/latex/contrib/[/url]
(take a look on the dvi for documentation)

By the way, I saw you post more then 800 messages, could you tell me a little about this forum users profile?

Good luck, Yochai.

p.s: if any familiar with this package and have a solution...[url][/url]

for the todo package, am I correct that this is mainly for bigger documents to keep structure? I think you will not need it here, latex was just implemented to make nice mathematical formulae, not whole documents on this site :)

Documents can be made on your computer (eg. in PDF format) and attached when posting.

[quote="twittoy"]could you tell me a little about this forum users profile? [/quote]
eum, what do you mean by "this forum users profile"? :)
You mean the profile on http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=3947 ?  Or the small one next to every post?

Welcome on the site anyways! :)

First, [color=blue]thank you [/color]very much for the welcome.
Greetings to those present.

[b]Regarding the TODO package[/b]:
It isn't for keeping document structure. It is for producing a TODO list (i.e. things you need to do/fix/add) for your latex work/thesis/article/book if you writing one.

[b]Regarding my question about this forum users profiles[/b]:
I just want to know who is around... This can help me know what questions I can ask (and expect to get an answer :)) -- that is to know what is this forum about since it is not clear from the site info. I thought the latex forum is for sharing knowledge about the "LaTeX" typesetting system.

Yochai.

I haven't used the todo package but a couple of things you could try (assuming you have already tried emailing the author)
1. post in the comp.text.tex newsgroup which is where the real [tex]\LaTeX [/tex]experts hang out
2. Try out the fixme package which [i]may[/i] do what you want, though again I haven't tried it

[quote="twittoy"]I just want to know who is around... This can help me know what questions I can ask (and expect to get an answer :)) -- that is to know what is this forum about since it is not clear from the site info. I thought the latex forum is for sharing knowledge about the "LaTeX" typesetting system.[/quote]

Welcome to our site!

Absolutely, this forum is the place to ask questions about LaTeX.

However, due to the nature of our site, we generally deal with questions regarding mathematical typesetting.  Which is not to say that you shouldn't post your question, just that it's somewhat less likely that you might receive an answer to your liking!

There are a lot of highly skilled TeX/LaTeX users here (myself, I have 15 years experience using LaTeX, although I would hesitate to call myself an "expert"), so usually someone will know the answer to your question.

Unfortunately, I've never used the [b]todo[/b] package, so I can't help you here.  I'll echo [b]stevem[/b]'s suggestion about comp.text.tex, or you can also go to [url]http://www.tug.org[/url]

Thank you both for your answers. I will try it ASAP.

I must say that this forum is very friendly, and I am happy to be here. (Thank you Google ;))

Hope that I will be able to contribute. For the moment, regarding this topic, I want to recommend you all about the [url=http://www.ctan.org/tex-archive/macros/latex/contrib/todo/?action=/tex-archive/macros/latex/contrib/]TODO package[/url] for managing the TODO task of your article/book/report.

Thanks again, Yochai.


[size=150]True or False Questions. If it is true, provide an explanation. If it is false, provide a counter-example.[/size]

[img]http://img135.imageshack.us/img135/9017/02lg9.jpg[/img]

18. for all x
19. If f and g are both increasing on (a,b), then f + g is increase on (a,b).
20. If f and g are both increasing on on (a,b) and f(x) >g(x) for all x, then f - g is alwys increasing on (a,b)
21. If f and g are both increasing on (a,b), then fg is increasing on (a,b)
22. If f `(x) exists and is non-zero for all x, then f(0) ? f(1).
23. lim as x approaches 0  for x/e^x  = 1

Hey this seems like homework to me! If it isnt dont mind this post  :)

this is not homework, it's final exam. and i'm confused on most of them.


[quote="dangvh"]Could someone tell me what is wrong with this proof?

Let S = 1 + 2 + 4 + 8 + 16+ 32...[/quote]There's your mistake. $S$ is not a number that you can do arithmetic on.

[quote="WindSlicer"]Hmm, for S-1, do you have to subtract 1 from each value?[/quote]

No, S-1 can be expressed like this,

S-1 = 1 - 1 + 2 + 4 + 8 +16 +32

which yields,

S-1= 2 + 4 + 8 +16 + 32

Well you can see that S is a diverging geometric series, so S does not have a definite number... That means that if the sequence keeps going, it will approach infinity. As I know, you cannot use infinity this way...

Or you were using the a_1/(1-r) formula, you'd still get -1, which is invalid since the sum diverges.

[quote="AntonioMainenti]There's your mistake. $S$ is not a number that you can do arithmetic on.[/quote]

Ah thanks

[quote="dangvh"]Could someone tell me what is wrong with this proof?

Let S = 1 + 2 + 4 + 8 + 16+ 32...

S is an infinite geometric series (just to make that clear lol)

Then,

2S = 2 + 4 + 8 + 16 + 32 + 64...

and.

S-1 = 2 + 4 + 8 + 16 + 32 + 64

So one can say,

2S = S-1
S=-1

This is obviously wrong but I just wanted to know where the mistake was[/quote]

$S=\infty$, it is not a convergent series, therefore, even when $2S=S-1$ is somehow true, it is also true that $2S-S$ is not defined!

Yes, keep in mind when doing those sorts of things that you cannot do conventional arithmatic with infinity.  Usually, it is best to think of infinity as a limit, not a number.


What are the equations of the asymptotes of y=tanx between 0 and 2 pi.


a,b,c are positive real numbers, what's the minimum of "(a+3c)/(a+2b+c)+4b/(a+b+2c)-8c/(a+b+3c)"

Problem:
If $a,b,c>0$ ,what's the minimum of $A=\frac{a+3c}{a+2b+c}+\frac{4b}{a+b+2c}-\frac{8c}{a+b+3c}$ ?

Answer:
$min(A)=-\frac{4+3\sqrt{2}}{140+99\sqrt{2}}$ with equality if $b=(1+\sqrt{2})a,c=(4+3\sqrt{2})a$

Solution:
$A=\frac{3x+1}{2y+x+1}+\frac{4y}{y+2x+1}-\frac{8x}{y+3x+1}$

,where $x=\frac{c}{a},y=\frac{b}{a}$.

After many calculations we find that for $x=4+3\sqrt{2}$ and $y=1+\sqrt{2}$ ,$A$ takes the minimum value

,but I haven't a good solution yet (just ideas).


Let $ x_0,x_1,b$ be the positive real numbers and $ 0<a<1$$ (a \in R)$.
The sequence $ (x_n)_{n\geq 0}$ is defined by $ x_{n\plus{}1}\equal{}ax_n\plus{}\frac{(1\minus{}a)x_{n\minus{}1}}{1\plus{}bx_{n\minus{}1}}$ $ \forall n \geq 1$.
Compute $ lim(nx_n)$ when $ n \to \plus{} \infty$.

A partial solution:
Let $ u_n\equal{}1/x_n$. $ u_{n\plus{}1}\minus{}u_n\equal{}\dfrac{(1\minus{}a)(\minus{}u_n^2\plus{}u_nu_{n\minus{}1}\plus{}bu_n)}{u_n(1\minus{}a)\plus{}au_{n\minus{}1}\plus{}ab}$.
We assume (that is numerically true) (*): $ u_n\equal{}c{n}\plus{}d\log(n)\plus{}e\plus{}o(1)$ where $ c,d,e$ are real numbers. Thus $ u_n\sim{cn}$ and
$ u_{n\plus{}1}\minus{}u_n\sim(1\minus{}a)(\minus{}u_n^2\plus{}u_nu_{n\minus{}1}\plus{}bu_n)/(cn)\sim(1\minus{}a)(\minus{}u_n\plus{}u_{n\minus{}1}\plus{}b)\sim(1\minus{}a)(\minus{}c\plus{}b)$
Thus $ c\equal{}(1\minus{}a)(\minus{}c\plus{}b)$ and $ c\equal{}\dfrac{(1\minus{}a)b}{2\minus{}a}$.
Conclusion: $ lim(nx_n)\equal{}\dfrac{2\minus{}a}{(1\minus{}a)b}$.
It remains to prove (*).

Here we prove that $ x_n\rightarrow{0}$:
$ x_n\geq{0}$ but $ x_n$ is not decreasing ! Let $ v_n\equal{}\sup\{x_n,x_{n\minus{}1}\}$. $ x_{n\plus{}1}\leq{a}x_n\plus{}(1\minus{}a)x_{n\minus{}1}\leq{v}_n$ then $ v_{n\plus{}1}\leq{v}_n$ and $ v_n\rightarrow{l}\geq{0}$. Moreover $ \limsup{x_n}\equal{}{l}$. We extract a subsequence $ x_{n_k\plus{}1}$ s.t. $ x_{n_k\plus{}1}\rightarrow{l}$. The function $ x\in]0,\infty[\rightarrow\dfrac{(1\minus{}a)x}{1\plus{}bx}$ is increasing; then $ x_{n_k\plus{}1}\equal{}ax_{n_k}\plus{}\frac{(1\minus{}a)x_{n_k\minus{}1}}{1\plus{}bx_{n_k\minus{}1}}$ implies $ l\leq{a}l\plus{}\dfrac{(1\minus{}a)l}{1\plus{}bl}$; we obtain $ bl\leq{0}$ and $ l\equal{}0$.   QED

With my last post #3 we know that $ u_n$ tends to $ \infty$. It's not sufficient; we must prove (for instance) that $ u_{n\plus{}1}\minus{}u_n\rightarrow{c}\not\equal{}{0}$.
Indeed then $ u_n\sim{c}n$ and $ u_n\sim{u}_{n\minus{}1}$. $ u_{n\plus{}1}\minus{}u_{n}\equal{}\frac{(1\minus{}a)(\minus{}u_{n}^{2}\plus{}u_{n}u_{n\minus{}1}\plus{}bu_{n})}{u_{n}(1\minus{}a)\plus{}au_{n\minus{}1}\plus{}ab}$ implies $ c\sim(1\minus{}a)(\minus{}u_{n}^{2}\plus{}u_{n}u_{n\minus{}1}\plus{}bu_{n})/u_n\equal{}(1\minus{}a)(\minus{}u_{n}\plus{}u_{n\minus{}1}\plus{}b)\sim(1\minus{}a)(\minus{}c\plus{}b)$. We conclude as in post #2.
It remains to prove that $ u_{n\plus{}1}\minus{}u_n\rightarrow{c}\not\equal{}{0}$.


calculus
integration
trigonometry
inequalities
calculus computations

Let $ V(a)$ be the volume of the solid obtained by revolvong the region bounded by the curve $ y \equal{} x\sqrt {x\sin ax}\ (0\leq x\leq \frac {\pi}{a})$ and $ x$ axis around the $ x$ axis.

For $ a > 0$, find the minimum value of $ V(a) \plus{} V\left(\frac{1}{a}\right)$.

$ V(a)\equal{} \pi \int_0^{\frac{\pi}{a}}x^3\sin ax dx$

$ \equal{}\frac{\pi}{a^4} \int_0^{\pi}u^3\sin u du$

so $ V(a)\equal{}\frac{1}{a^4}V(1)$  so assuming V(1) is not zero (which is obviously isnt intuitively),

we know that the range of V(a) must contain 1, which means the minimum value is obtained there and is 2

We have $ V(a) \equal{} \pi\int_{0}^{\pi/a}x^3\sin axdx$. This integral can be solved by integration by parts. After some routine computations we arrive at

$ V(a) \equal{} \pi\int_{0}^{\pi/a}x^3\sin axdx \equal{} \frac {\pi^2(\pi^2 \minus{} 6)}{a^4}.$

I just determined $ V(a)$ as an exercise, because I think it's clear that $ V(a) \plus{} 1/V(a) \geq 2$ as the inequality $ x \plus{} \frac {1}{x} \geq 2$ for $ x > 0$ is well known.Anyway, since it's clear that $ V(a) > 0$ the minimal value is 2, isn't it? :maybe:

kunny先生へ：正解はなんですか、先生？前言ったように、答えは２だと思います。もっとも、任意の正の実数$ x$に対して、 $ x \plus{} \dfrac{1}{x}\geq 2$ですけど。つまり、$ V(a)$は正の実数である限り、どのように定義されたにかかわらず$ V(a) \plus{} \frac {1}{V(a)}\geq 2$が成り立ちますね。計算の間違いがあれば、ぜひ言ってくださいね。

[quote="kunny"]
@Jutaro The calculation of $ V(a)$ is wrong. :(
[/quote]

I want to check my computation before trying to solve the rest of the problem. I got that

$ \int x^3\sin axdx \equal{} \frac {3(a^2x^2 \minus{} 2)\sin ax}{a^4} \minus{} \frac {(a^2x^2 \minus{} 6)x\cos ax}{a^3}\plus{}C$

and at least this seems to be correct since I checked it with Mathematica. So, when we evaluate this expression for the integration limits 0 and $ \pi/a$ don't we get $ \frac {\pi(\pi^2 \minus{} 6)}{a^4}$? I'm sorry if I'm making some really stupid mistake. :blush:

hello, i have also got $ V(a) \equal{} \frac {\pi^2(\pi^2 \minus{} 6)}{a^4}$ and so we have
$ V(a) \plus{} V(\frac {1}{a}) \equal{} \pi^2(6 \minus{} \pi^2)(a^4 \plus{} \frac {1}{a^4})\geq2\pi(6 \minus{} \pi^2)$.
Sonnhard.

Dr Sonnhard Graubner, we are looking for a volume of a solid, so the integral is:

$ V \equal{} \pi \int f^2 \,dx$

You are missing the $ \pi$ in front of your integral..

hello, yes, you have right, i have on my sheet $ V(a)\equal{}\pi\int_{0}^{\frac{\pi}{a}}x^3\sin(ax)\,dx$. So we get $ V(a)\plus{}V(\frac{1}{a})\equal{}\pi^2(\pi^2\minus{}6)(a^4\plus{}\frac{1}{a^4})$.
Sonnhard.

[quote="kunny"]Oops! I made a typo. :oops:

@Jutaro The calculation of $ V(a)$ is wrong. :(

P.S. You are not allowed to post in Japanese in this forum.[/quote]

Can kunny show us his solution?? :D


conics
ellipse
analytic geometry
graphing lines
slope
number theory proposed
number theory

Let $a,b \in \mathbb{Q}$ and $p \in \mathbb{P}$ such that $a^2+b^2p=1$. Prove that there exist $c,d \in \mathbb{Q}$ such that $a=\frac{c^2-d^2p}{c^2+d^2p}$ and $b=\frac{2cd}{c^2+d^2p}$.

P.S. I know that using Hilbert's Theorem 90 it's obvious (and that's my solution as well) but can anyone solve it using only elementary means ?  ;)

Note that $x^2+py^2=1$ is the equation of an ellipse and that we are looking for the rational points on it.
There is obviosly the point $P(-1,0)$ on it. Now take any line $p$ through $P$ with rational slope and this line will intersect the ellipse at a rational point (it's easy to see, otherwise just solve the equations). In contrary, every rational point except $(1,0)$ gives such a line, thus we have a bijection.
Now calculate the intersection for a line with given slope and your characterising equations should come out (I haven't checked, but thats the geometrical way to find such theorems).

Edit: I think this gives  $a=\frac{c^2-pd^2}{c^2+pd^2}, b=\frac{2cd}{c^2+pd^2}$, and your way should give that too (otherwise $a>1$).


factorial
number theory solved
number theory

find all a,b,c positive integers such that
1-)$a!+b!+c! =a!b!$
2-)$a!b! =a!+b!+2^{c}$
3-)$a!b! =a!+b!+c^{2}$

You're right. I've corrected them, even if they seem useless now.

No, they are not useless now, in fact the third solves the third problem (it implies that $a!-1$ and $b!-1$ are sum of two squares, thus that $a,b \leq 3$).

Ok, so I only solve the first two :

a) Wlog, we may assume that $a\leq b.$ Then $a!$ divides $c!$ so that $a\leq c.$
If $a<b$ then $b!$ must not divide $c!$ so that $b>c.$ Thus :
$a!b! =a!+b!+c! \leq b!+2[(b-1)!]$.
This leads to $(a!-1)b \leq 2$ which forces $a=1$ or $a=b=2.$ It is easy to verify that these two possibilities give no solution to the problem.

Thus $a=b$ and the equation reduces to $(b!(b!-2) = c!.$
Let $c=b+d$ so that $b!-2 = (b+d)(b+d-1)\cdots(b+1).$
For $b\geq 4$ the LHS is not divisible by $4$ thus the RHS has at most three factors.
For $c=b+1$ we easily get th solution $a=b=3,c=4,$ which has to be refused according to our conditions.
It is straightforward to verify that there is no solution in the cases $c=b+2$ and $c=b+3.$

For $b\leq 3$ we easily verify that $a=b=3,c=4$ is a solution.

Pierre.

b) Wlog we may assume that $a\leq b.$
Ths $a!$ divides $2^{c}$ so that $a!$ is a power of $2$, which forces $a=1$ or $2.$
It is easy to verify that $a=1$ gives no solution.
For $a=2$ the equation reduces to $b! = 2+2^{c}.$
If $c\geq 2$ then we must have $b!$ not divisible by $4$ wich leads to $b \leq 3.$
Conversely, we see that the only solution in that case is $a=c=2,b=3.$

For $c\leq 2$ it is straightforward to verify that it gives no solution.

Pierre.

thank you everyone, but i couldn't understand this
[quote="ZetaX"]it implies that $a!-1$ and $b!-1$ are sum of two squares, thus that $a,b \leq 3$.[/quote]

Let $ p \equiv \minus{} 1 \mod 4$ be prime. Then $ c^{2}\equiv \minus{} 1 \mod p$ has no solution since otherwise we would, using that $ \frac {p \minus{} 1}{2}$ is odd, get $ 1\equiv c^{p \minus{} 1}\equiv (c^{2})^{\frac {p \minus{} 1}{2}}\equiv ( \minus{} 1)^{\frac {p \minus{} 1}{2}}\equiv \minus{} 1 \mod p$.

But that gives us that there is no prime $ p \equiv \minus{} 1 \mod 4$ that divides $ c! \plus{} 1$, thus no such prime divides $ a! \minus{} 1$. Now all prime divisors are $ \equiv 1 \mod 4$ (it's clear that the number is odd), thus also any of their products is $ \equiv 1 \mod 4$, but one of this products is $ a! \minus{} 1$, giving $ a! \minus{} 1 \equiv 1 \mod 4$.
But for $ a > 3$ the latter is clearly wrong, and the same argument works to show that $ b < 4$.


I'm from Turkey. I'm preparing for national olympiads.
I need some document. Can you send me some documents?
-in english please- I will be grateful if you do it.
I'm sorry for my poor english.
 :)

I would go to contests link up at the top of this website, and select a contest from past years.

Of course I downloaded all documents in this site.
But I need recent documents besides this site.
Speak with you later...

Hojoo Lee has written a couple of very good documents on inequalities and number theory: you can find these on [url=http://myhome.naver.com/hojoolee/orange.html]his website[/url].

However, when preparing for olympiads, there is no substitute for just solving problems, and there are enough in the resources section on this website to keep you going for years.


[quote="anirudh"]How many three-digit positive integers are there?[/quote]

[hide]
999-100+1 = 900
[/hide]

[i]oops.added another digit[/i]

[hide]Instead of counting.. you can just find out how many possibilities there are...

$9*10*10=900$[/hide]

[hide]
there are 9 possible digits for the hundreds place value, 10 for the tens, and 10 for the ones so you would do 9*10*10 which is 900[/hide]

[hide="my solution"] i got it using 2 ways...i got the highest 3 digit integer (999) and subtracted 99 from it because from 1 to 99, the integers are not 3 digit integers. so...$999-99=900$ :D
also, i noticed the number of different possibilities for each place value, and got 9 for the hundreds value, 10 for tens, and 10 for the ones. $9*10*10=900$ integers. :rotfl:   [/hide]


Can someone prove this formula for me?

$\sqrt{a+b\sqrt{c}}=\sqrt{\frac{a+q}{2}}+\sqrt{\frac{a-q}{2}}$
where $q^2=a^2-b^2c$

Also, please tell me if the formula is incorrect  :huh:

well this is usually how you do it.  You assume $\sqrt{a + b\sqrt{c}} = \sqrt{x} + \sqrt{y}$ for certain $x$ and $y$ (hopefully rational), and you square to get $a+b\sqrt{c} = x + y + 2\sqrt{xy}$.  Then you solve the system $x + y = a$, and $2\sqrt{xy} = b\sqrt{c}$.

[quote="eryaman"]Well it seems theres a problem...either $a>q$, $a=q$, or $a<q$.  For the 1st and 3rd cases...you get a nonreal answer by taking the squareroot of a negative.[/quote]

I guess in the 3rd case you take the negative square root.  Because the way to obtain the formula is to square both my equation, subtract the second from the first, and then square root it, then you have $x-y = blah$, and then you get the formula by solving this one with the first equation.  But since you took the square root, you have to take both the plus and minus sign into account.

Well, just square both sides and substitute:

$a + b\sqrt{c} = \frac{a+q}{2} + \frac{a - q}{2} + 2\sqrt{\frac{a^2 - q^2}{4}}$

$a + b\sqrt{c} = a + \sqrt{a^2 - q^2}$

$b\sqrt{c} = \sqrt{a^2 - (a^2 - b^2 c)}$

$b\sqrt{c} = \sqrt{b^2 c}$

$b\sqrt{c} = b\sqrt{c}$


Beginning with the second day of a seven-day week, Potter ate 4 more biscuits than he ate the day before. He ate a total of 105 biscuits during the week. How many biscuits did he eat on the fifth day?

akhil0422

Ok, you can make an equation.

$ x \plus{} (x\plus{}4) \plus{} (x\plus{}4\plus{}4) \plus{} (x\plus{}4\plus{}4\plus{}4) \plus{} (x\plus{}4\plus{}4\plus{}4\plus{}4) \plus{} (x\plus{}4\plus{}4\plus{}4\plus{}4\plus{}4) \plus{} (x \plus{} 4\plus{}4\plus{}4\plus{}4\plus{}4\plus{}4) \equal{} 105$

$ 7x \plus{} 84 \equal{} 105$

$ 7x \equal{} 21$

$ x \equal{} 3$

Thus, he ate $ 3 \plus{} 4(4)$ or $ \boxed{19}$ biscuits.  Whoa!

Try not to click ask on forum and then type out a new post. The GameBot does that for you.


geometry
trapezoid
cyclic quadrilateral
geometry proposed

let $ABCD$ be a square inscribed in the circle $(O)$ .On the  minor are $\hat{BC}$ ,take an arbitrary point $M$ distinct from $B;C$ .The  lines $CM$ and $BD$ intersect at a point $E$,the lines $CM$ and $BA$ intersect at a point $F$.Prove that the triangles $ABE$ and $DOF$ have equal areas

Is there anything wrong with this question? what if D,O,F are collinear?

Sorry,my problem is:

let $ABCD$ be a square inscribed in the circle $(O)$ .On the  minor arc $BC$ ,take an arbitrary point $M$ distinct from $B;C$ .The  lines $CM$ and $BD$ intersect at a point $E$,the lines $CM$ and $BA$ intersect at a point $F$.Prove that the triangles $ABE$ and $DOF$ have equal areas

I think it's not true ...
We have $S_{DOF}= S_{BOF}$
So $S_{DOF}= S_{ABE}\Leftrightarrow S_{BOF}= S_{ABE}$
Moreover, $\angle OBF = \angle ABE$ so  $S_{BOF}= S_{ABE}\Leftrightarrow BO.BF =BA.BE \Leftrightarrow BF = \sqrt{2}BE$
Note that $\angle FBE = 45^{o}$ so $BF = \sqrt{2}BE \Leftrightarrow FE \perp BE$. It's wrong
Could you check it for me, please ? Thank you

[quote="toan hoc"]No I think my problem is true![/quote]

Yes it is true but not for the point $F\equiv CM\cap BA.$ The point $F,$ must be as the intersection point of $DM,$ $BA.$

So, we can easy to prove that the triangles $\bigtriangleup ABE,$ $DOF,$ have equal areas, because of $BE\perp EF,$ from the cyclic quadrilateral $MBEF.$

Hence, from the trapezium $AOFE,$ the wanted conclusion it follows $($ Because of $(DOF) = (BOF)$ $).$

Kostas Vittas.

PS. The [b][size=109]PDatK40sp's[/size][/b] remark is correct for the problem's statement you presented $($ $F\equiv CM\cap BA$ $).$


function
trigonometry
ratio
complex analysis
complex analysis unsolved

this question has no context, just caught myself thinking about it a little and didnt get too far.

for what real numbers a,b does $ \frac{\zeta (a)}{\zeta (b)}$ have nice closed forms..?

Well if $ a,b$ are even positive integers you have an easy one.  Another one is if $ a \plus{} b \equal{} 1$, in which case
\[ \frac {\zeta(a)}{\zeta(b)} \equal{} 2(2\pi)^{ \minus{} b}\Gamma\left(b\right)\cos{\frac {b\pi}{2}}
\]
In fact that equality holds for complex $ a,b$ as well.  No other nice closed form expressions for that ratio I could think of.


geometry
trapezoid
geometric transformation
homothety
rectangle
ratio
similar triangles

1. Trapezoid with base lengths a and b is divided with Line segment EF, which is parallel with base lengths a and b, on two equal pieces. Determine |EF| .

2. Fresh mushrooms contain 90% water and dried mushrooms contain 12% water. How many kilograms of dried mushrooms we can get from 12 kilograms of fresh ones.

Side note: In problem number 2 I constantly get 1.6 kg and solution is supposed to be 1.36 kg...

2.

$ 12$ kg of fresh mushrooms $ \equal{}$ $ 10.8$ kg of water and $ 1.2$ kg of mushroom.

This is reduced to $ 1.2$ kg of mushroom and $ w$ kg of water where $ w \equal{} 0.12(1.2 \plus{} w)$

Therefore $ w \equal{} 0.1636$.

Adding this water to the $ 1.2$ kg of mushroom, we have $ 1.3636$ kg of dried mushrooms.

I hope I got it right.

1.

Here I think you need to know more than just $ a$ and $ b$. For example we can draw the side of length $ a$ and then below and parallel to it have any number of sides whose length is $ b$.

Letting the height be $ h$ and the distance from side $ b$ to $ EF$ be $ h_{1}$, I get $ a \plus{} \frac{h_{1}(a\minus{}b)}{h}$ but I guess I've done something silly.

For the mushroom problem, instead of having a seperate volume of water, you can simply put $ 1.2\equal{}0.88X$ was 1.2 is the mass of the mushrooms without water, which equals to 88% of the total mass because 12% is water, you can simply divide it out instead of using a whole new variable.

Thank you a lot for the mushroom problem :) I have just realised how wrong my approach to the problem was.

As for the trapezoid problem, solution should be:

$ EF=\sqrt{\frac{a^2+b^2}{2}}$

but I can't get to that solution :S

For 1, if I'm understanding your question correctly, you just extend the sides of the trapezoid into a triangle, and use the fact that you have now created two similar triangles.

For the first one, you can also drop perpendiculars and then abuse homothecy to find the small segments of the triangles, then set up a small system and simplify (easier than it sounds).

Gotta run, but I'll post it later.

Edit: Ok, here's what I got... [hide="Solution"]
Let the trapezoid be $ ABCD$ such that $ AB \equal{} a$ and $ CD \equal{} b$. Let the projections of $ A$ and $ B$ onto $ EF$ be $ A'$ and $ B'$. Let $ EA' \equal{} x$ and $ FB' \equal{} y$.

Extending the lines $ AA'$ and $ BB'$ we form a rectangle with one length $ a$. The two right triangles formed by these extensions are also split into two parts of equal area by $ EF$.

The homothecy that takes $ \triangle BB'F$ to the larger mapping with twice the area therefore has a ratio of $ \sqrt {2}$, as does the homothecy that maps $ \triangle AA'E$ to the larger triangle. Therefore, $ \sqrt {2}(x \plus{} y) \plus{} a \equal{} b\implies x \plus{} y \equal{} \frac {b \minus{} a}{\sqrt {2}}$.

That means that $ EF \equal{} x \plus{} y \plus{} a \equal{} \boxed{a \plus{} \frac {b \minus{} a}{\sqrt {2}}}$.
[/hide]

Anyone see the mistake?

Edit: ...I see what I did there. Thanks AndrewTom

If the area of the triangles are in the ratio $ 1: 2$ then the ratios of the rectangles must be as well. But the two rectangles $ A'B'GH$ (where $ G$ and $ H$ are below $ A'$ and $ B'$ respectively) and $ ABB'A'$ are equal in area and so have the same height. This contradicts $ AG$ being $ \sqrt{2}(AA')$.

For the first problem, we can let $ EF \equal{} x$, the height of the biggest trapezoid $ h$, and the height of the trapezoid with base $ a,x$ be $ y$, and it follows that the height of the trapezoid with base $ x,b$ is $ h \minus{} y$.

From similar triangles, $ \frac h{(b \minus{} a)/2} \equal{} \frac y{(x \minus{} a)/2}$.

Equating areas, $ (a \plus{} x)y \equal{} (x \plus{} b)(h \minus{} y)$.
Substituting $ y \equal{} h(x \minus{} a)/(b \minus{} a)$ obtained from the first equation, the h cancels, and we are left with

$ \frac{(x\minus{}a)(a\plus{}2x\plus{}b)}{b\minus{}a}\equal{}x\plus{}b$.

Solving for $ x$ we get $ x \equal{} \sqrt {\frac {a^2 \plus{} b^2}2}$.

Hi mathwizarddude. You seem to have assumed that the original trapezium is isosceles. I suspect that is what Nexolas neglected to tell us. If so, you are (of couse) correct.

Thanks for the solution, mathwizarddude. :D I will ahve to look more into this :)

AndrewTom, I haven't neglected anything. Nowhere in the problem is written that trapezium is isosceles. And I reallydo not see why you think he, mathwizarddude, assumed that trapezium is isosceles... Well, if I knew how to solve this problem I would've known what is assumed, and what is not. :-S

In my solution I have to admit that I indeed assumed the special case of an isosceles trapezoid because of $ \frac {h}{(b \minus{} a)/2} \equal{} \frac {y}{(x \minus{} a)/2}$

However, I think this can be easily compensated by considering $ \frac {h}{k(b \minus{} a)} \equal{} \frac {y}{k(x \minus{} a)}$ for $ 0 < k\le1$. :)
BTW, where did you get this problem? Also, is there any geometric significance and meanings behind the length being a square root mean?

I was wrong, Nexolas.  :blush:

mathwizarddude, I've just worked through your amendment ($ \frac{h}{k(b\minus{}a)} \equal{} \frac{y}{k(x\minus{}a)}$ for $ 0<k\le 1$) and, of course, it works. Well done!

I'd also like to know where the question came from as it is quite a nice one.

[quote="AndrewTom"]I was wrong, Nexolas.  :blush:

mathwizarddude, I've just worked through your amendment ($ \frac {h}{k(b \minus{} a)} \equal{} \frac {y}{k(x \minus{} a)}$ for $ 0 < k\le 1$) and, of course, it works. Well done!

I'd also like to know where the question came from as it is quite a nice one.[/quote]

Not a problem, AndrewTom.

This problem came from problem collection of tasks, problems if you like, for practicing to solve test for acceptance to university (I doubt this was grammatically correct, but I could not express myself better). And this one was not the hardest one, now when I have seen the idea for solving this problem :)

I will sure post some more problems if I could not solve them :)


vector
integration
calculus
calculus computations

Use the vectorfield
\[\mathbf{F}(x,y,z)=\frac{-y\mathbf{i}+x\mathbf{j}}{x^{2}+y^{2}}\]
to together with Stokes thm, show that every orientable surface in $\mathbb{R}^{3}$ with
the unit circle $\{(x,y,z),\, | \, x^{2}+y^{2}=1, \, z=0\}$ as its boundary must intersect the $z$-axis.

The statement almost contains the solution: suppose that $S$ is an orientable surface in $\mathbb{R}^{3}$ with the unit circle $C=\{(x,y,z)\colon x^{2}+y^{2}=1, \, z=0\}$ as its boundary, and $S$ does not intersect the $z$-axis. Since $\nabla\times\vec F=0$ outside of the $z$-axis, we can use the Stokes theorem to obtain $\int_{C}\vec F\cdot d\vec r=0$. The latter is shown to be false by a direct computation.


inequalities
complex analysis
combinatorics proposed
combinatorics

Let $S$ be an 8-element subset of $\{1,2,...,100\}$. Show that there are two disjoint subsets $A,B\subset S$ with the same number of elements and the same sum of their elements.

I can't believe myself this is true, but my proof seems to work... notice that $S=\{1,2,4,8,16,32,64\}$ is a 7-element set where no two subsets have the same sum of their elements; the problem claims that with just one more element, it is always assured that there are two subsets with the same sum of their elements that additionally have the same number of elements!

Isn't $\{1,2,3,6,12,24,48,96\}$ a counterexample? Every number greater than $2$ in this $8$-element set is the sum of the numbers in the set which are smaller than it, so if we assume, for instance, that $x\in A$ is the largest number in $A\cup B$, then $x$ alone is greater than the sum of the elements in $B$ ($A,B$ must each have at least two elements, of course, so $x\ge 3$).

Sorry if I'm missing something obvious.

Your statement is false. An counter example is

$S \: =\: \{1, 2, 4, 8, 16, 32, 64, 96\}.$

Assume your statement is true in this case. Let $A$ and $B$ be two disjoint subsets of $S$ such that $|A| \,=\, |B| \,=\,n$.

* $n = 4.$ Impossible since the sum of the elements in $S$ are odd.

* $n = 3.$ The number 96 has to be in $A \cap B.$ Let say $96 \in A.$ Then both the numbers 32 and 64 must be in $B$. If not, the sum of the 3 elements in $B$

$S(B) \; \leq \; 64 \:+\: 16 \:+\: 8 \;=\; 88 \; < \; 96 \; < \; S(A).$

Hence

$a_{1}\:+\: a_{2}\:+\: 96 \;=\; b_{1}\:+\: 32 \:+\: 64$

where $a_{1}, \, a_{2}\in A$ and $b_{1}\in B$. But this is equivalent to

$a_{1}\:+\: a_{2}\;=\; b_{1}$

which is impossible.

* $n = 2$. This is obviously not possible since the sum of the third and second largest element in $S$ equals the largest element in $S$ (i.e. $32 \:+\: 64 \: =\: 96$).

Indeed, I found the same counterexample yesterday. Spending some more thoughts on it, I could then find the mistake in my argument.

The problem given to me by mathmanman was the same problem for a 10-element subset. I couldn't find a proof for that one...

[quote="Peter Scholze"]Let $S$ be an 8-element subset of $\{1,2,...,100\}$. Show that there are two disjoint subsets $A,B\subset S$ with the same number of elements and the same sum of their elements.  [/quote]

It is a somewhat notorious problem to prove that 9 is the optimal number of elements for this problem.  That 10 works is a famous, easy and seminal problem.

[quote="fleeting_guest"]
It is a somewhat notorious problem to prove that 9 is the optimal number of elements for this problem.  That 10 works is a famous, easy and seminal problem.[/quote]
Hmm, then could you please post your solution?
I hope you're not confusing with the same problem, but without the assumption that the two subsets have the same number of elements. (since this problem is indeed famous and easy; but with that assumption..)

[quote="mathmanman"]I hope you're not confusing with the same problem, but without the assumption that the two subsets have the same number of elements. (since this problem is indeed famous and easy; but with that assumption..)[/quote]

Unfortunately, I indeed confused it with the unrestricted problem: given a 9-element set with elements up to 100 (the exact bound is slightly higher than 100) there are two subsets with the same sum but possibly different number of elements.    There are two proofs, one a somewhat complicated analysis of various inequalities specific to 9 and another using complex analysis in a very clever way that gives bounds for larger $n > 9$.   It would take a long time for me to find or (maybe) reconstruct either argument, sorry -- I meant that the problem for $n=9$ is famously difficult.


let O is the center of circumcircle of ABC triangle.M is a point on AB arc.K & N are 2 point on AB such that :
$ MK\bot OA$ and $ MN\bot OB$
(MK,AC)=L and (MN,BC)=P.
what is <MLP ?
 say it by <A,<B,<C

Did not you forget some condition ?
[hide]KL=MN
(see [url]http://www.mathlinks.ro/viewtopic.php?t=145391[/url])[/hide]


I forgot to mention. I only have 5 dollars, meaning only the person with the best answer gets 5 dollars. Sorry! I wish I were wealthier too!  :lol:

Best answer to what? I don't think I was there when you told us.

i don't think you were there, ask the others.


algebra
polynomial
inequalities
function
absolute value
number theory
prime numbers

Your original statement did not include that modulus. Anyway, what about $ X-1$?

sorry i edited it now , i was missing some of the conditions, now i have edited it.but still i think it is missing something. i need time to express it in maths language,hard for me.
$ p(x)=x-1$
$ |1|=|-1|$

[quote="binomial_4eva"]i have discovered something new my self about polynomials :lol:
can someone prove me wrong ?or prove me right?

any polynomial with all the coefficients being different,  i mean like $ |a_{n}|\neq{|a_{n-1}|}\neq{|a_{n-2}|}\neq{|a_{n-3}|}\cdots{|a_{1}|}\neq{|a_{0}|}$
then $ |x|\geq deg(P(x))+1$[/quote]

Still trivially false.  Let $ P(x) = 2x-1$.

Actually, let $ P(x) = x$.

The next time you think you've discovered something new about polynomials, at least try to make it true for linear ones.  You just tried to prove that every linear polynomial with distinct coefficients has a root with absolute value greater than or equal to $ 2$, which is absurd.

If by $ |x|$ he means the greatest value of $ |a_{k}|$, then maybe he has a point.

The statement doesn't make any sense, just by plugging in $ x = 0$. The RHS is always positive, so the inequality doesn't hold... you know there's something wrong when the "intent" of the problem isn't even clear. I don't see any specification that $ x$ is a root of $ P$, either.

i will write down the correct theorem soon, i am working on it at the moment, it is abit confusing when turning the idea from the thing i am doing to polynomials, i agree with all of you, i should have atleast tried it for linear polys to see if it works.but neway soon i will write the theorem here for all to prove me wrong.

[quote="binomial_4eva"]i will write the theorem here for all to prove me wrong.[/quote]

This may be somewhat foreign to you, but in mathematics there is no equivalent of "innocent until proven guilty."  If you want to call it a theorem, you must prove it and present a correct proof.  It is not up to others to show it is false.

Maybe he will provide us with Binomial_4eva's Hypothesis or Conjecture .

[quote="binomial_4eva"]ya lol spix is right.[/quote]

You're funny  :P We should rename Mathlinks like uhm "The Great Theorem Making Machine". Really, there are like thousands of new theorems on this forum, it's great.

Please read what I wrote for you on another recent thread of yours.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=156242&start=21]It's here[/url]

[quote="binomial_4eva"]i have discovered something new my self about polynomials :lol:
can someone prove me wrong ?or prove me right?

any polynomial with all the coefficients being different,  i mean like $ |a_{n}|\neq{|a_{n-1}|}\neq{|a_{n-2}|}\neq{|a_{n-3}|}\cdots{|a_{1}|}\neq{|a_{0}|}$
then $ |x|\geq deg(P(x))+1$

Edited:coffiecients being different, always positive[/quote]

You probably found that after looking at only one example and trying to generalize. In mathematics, you can't do that. You have to prove that your theorem always holds.

For example, I'll make the claim that the prime numbers go on forever. Now, I can't just say that based on looking at the first few primes. And I also can't check every possible case. What I need is something like this: Assume for the sake of argument that there are only a finite number of primes. Let's multiply them all together and call the number $ x$. Now consider the number $ x+1$. It's not divisible by 2, because it'll leave a remainder of 1. It's not divisible by 3, because it'll leave a remainder of 1. In fact, it's not divisible for any prime number, because it'll always leave a remainder of 1 (you have to prove this step too, but it's not that hard--try it). So either $ x+1$ is a new prime number, or else $ x+1$ is divisible by another prime number that wasn't on our list. In any case, our original assumption led to a logical contradiction, so it must have been wrong. Therefore, there are an infinite number of primes.

(Sorry if that was a little off-topic...)

[quote="mathnerd314"]You probably found that after looking at only one example and trying to generalize. In mathematics, you can't do that. You have to prove that your theorem always holds.[/quote]

In lieu of a proof, at least try more examples.  Try, for example, to construct examples that will cause your conjecture to fail.

Another thing that happens a lot is that you forget about the trivial case  :P

Especially in set theory ... "what about the empty set?  :rotfl: "

Catching the trivial case and actually being able to spot it in your proof takes much practice. I realize that many people, though good in math (not as good on AoPs   :wink: ) make mistakes on trivial cases (especially division by zero). So it is very possible that you somehow overlooked the trivial cases. No?

I know this because I go to seen other math forums, and I go to tell you, AoPs is really really good at doing proofs and covering all cases.

Off topic story: This is really funny if you were there. In my analysis class when we were doing stuff on limits of functions, my professor kept overlooking the cases involving 0 and 1, I kept on correcting him over and over again, got real funny. Soon he looked at me and said "what will I do without you".


we have a Circle $C$ and 2 tangent lines $\ell_1,\ell_2$ to the circle and point $P$, construct segment $AB\ ,A\in \ell_1\ ,B\in \ell_2$ such that the line $AB$ is tangent to the circle and $\angle APB=\alpha=\text{constant}$.

there is a very useful lemma in harmonic wich can solve it very easy...

asheg I know U solve it but plz don't give any hint. ok??? ;)

Please don' pay attention to Tavakolli words.

this is solvable problem mens.

Try to solve  it's so nice.

[quote="lomos_lupin"]:o  hey man are you INSANE, ?? Wow you generalized every part of the problem up to this, you should have given , it in the OPEN Problems .Cause its really unsolvable via the elementry tools of geometry.

I know we all can solve this in a jetty ( Indeed its a magical solution ) but its impossible to be found ,i will just give the much more SOLVABLE versian of the problem :(if i may amir  :) )

[color=blue]problem[/color]Let $l_1$ and $l_2$ be two parallal line and tangetn to circle $C$ at $R,S$ .Let $P$ be a point on the $RS$.Find points $A \in l_1,B \in l_2$ such that

1.$\angle APB=$min possible value
2.$\angle APB=K$ where a is a given angle.[/quote]

[quote="lomos_lupin"]:(if i may amir  :) ).[/quote]

lomos lupin consider taht ure not amir saeedi...


geometry conics ellipse calculus MATHCOUNTS

Area of an ellipse = ab*Pi, where a and b are the major and minor axis of the ellipse.

[quote]sorry, when i said elliptical i meant looks like an ellipse but not defined as one.[/quote]

uh...please explain. Generally, when something looks like an elliipse, it is an ellipse.  :D

Yeah, unless it's squiggly or is obviously not an ellipses, it's pretty safe to assume it is one, I think, yet even though I'm the moderator here, I've never actually participated in MC, so I'm not sure. :)

[quote="tetrahedr0n"]You can always use calculus to find the under any strange curves. (Even 3-d curves like toruses and such.) I guess this would not be mc though.[/quote]
you never know. i used calculus on a mc problem last year and got it right (3D)....easier than thinking about what exactly the shape was.

[color=cyan]Are you perhaps refering to the region that you get when two circles partially overlap each other?[/color]

i do understand what you are saying. there was a post on this topic some time ago on the mathcounts forum.

[quote="ComplexZeta"]This isn't really relevant, but I would like to point out that intuitively, you don't need calculus to find the volume of a torus. What you do is to cut the torus into little wedge-shaped piece and then put them back the opposite way so that you end up with a cylinder. The radius of the cylinder is the same as the radius of the circular cross-section of the torus cut from a plane that goes through the central axis. (That doesn't make sense, does it? I hope you understand.) The height is the length of the center of the round bit. It's hard to explain without a picture, but I hope you understand what I mean.[/quote]
er. yeah, isn't that Pappus's theorem intuitively speaking?


Let x,y,z >0 and xyz=1
   Prove that (x+y+z)^5 >71(xy+yz+xz)

i will show that actually
$(x+y+z)^5\geq 81(xy+xz+yz)xyz$
for all $x,y,z>0$. but this immediately follows from muirhead's inequality since terms of the form $x^2y^2z$ sum to the smallest terms and the number of terms is right(equality holds for x=y=z). if you want to check, you can write $(x+y+z)^5$ explicitly down and then use muirhead :D

If you don't like Muirhead then Cauchy is a good choice.
Let see:
$(a+b+c)^2 \geq 3(ab+ac+bc)$
$(a+b+c)^3 \geq 27abc=27$
Hence $(a+b+c)^5\geq 81(ab+ac+bc)$ :)


Find the solution set of the inequality

6tgx/[1+(tgx) 2 ] > sin(1980-x)-sin(270-x)+tg20.tg40.tg80/ :sqrt: 3


search
combinatorics unsolved
combinatorics

calculate the quantity of odd numbers in n'th row of pascal's triangel.

I don't know what kind of foormula you are expecting...

But, if the binary expansion of $n$ has exactly $k$ digits $1$ then the number of odd terms in the $n^{th}$^is $2^k$.
This follows from Lucas' theorem : $C_n^k$ is odd if and only if each digit of $n$ in its binary expansion is not less than the corresponding digit in the binary expansion of $k$.
Thus, each of the $0$ digits of $n$ forces the corresponding digits of $k$ to be $0$. And for eah digit $1$ you have two possibilities...

Pierre.

thank you but I want a proof for it and I have known the answer. :P

Search here with Lucas theorem.
I gave a proof in another thread.

Pierre.


What is the smallest integer multiple of 9997, other than 9997 itself, which contains only odd digits?

The smallest is 33339995 which is 9997*3335. This answer can be seen from noticing that 9997 is very close to 10000, in fact it is (10000-3). This quantity multiplied by an odd number <3334 will yeild
(odd)*(10000)-(quantity<10000) which is also known as (even)*(10000)+(0<quantity<10000) which has an even first digit, and this quantity multiplied by any even number will end in an even number so an even last digit. Therefore n>3334 and we test 3335 as it is the first odd > 3334 and we see it works.


geometry
circumcircle
quadratics
inequalities
absolute value
algebra
geometry solved

Given a triangle $ABC$, find a point $M$ on the circumcircle of triangle $ABC$ such that $MA^{2}=MB\cdot MC$.

[color=blue][b]Problem.[/b] Given a triangle ABC, construct all points P on the circumcircle of triangle ABC satisfying $PA^{2}=PB\cdot PC$.[/color]

[i]Solution.[/i] Let r be an arbitrary positive number, and let k be the circle with center A and radius r. Let B', C' and P' be the images of the points B, C and P under the inversion with respect to this circle k. Of course, in order to construct the point P, it will be enough to construct the point P', since the point P can be obtained from the point P' by inverting in our circle k.

So we will concentrate on constructing the point P'. Well, first note that the inversion with respect to the circle k maps the circumcircle of triangle ABC to a line (since the circumcircle of triangle ABC passes through the center of inversion), and this line passes through the points B' and C' (since these points are the images of the points B and C in our inversion). Thus, the inversion with respect to the circle k maps the circumcircle of triangle ABC to the line B'C'. Since the point P lies on the circumcircle of triangle ABC, its image P' under the inversion with respect to the circle k must therefore lie on the line B'C'.

Since the points P' and B' are the images of the points P and B under the inversion with respect to the circle k (which has center A and radius r), we have $AP\cdot AP^{\prime}=r^{2}$ and $AB\cdot AB^{\prime}=r^{2}$. Thus, $AP\cdot AP^{\prime}=AB\cdot AB^{\prime}$, and $\frac{AP}{AB}=\frac{AB^{\prime}}{AP^{\prime}}$. Together with < PAB = < B'AP', this yields that the triangles PAB and B'AP' are similar; hence, $\frac{PA}{PB}=\frac{B^{\prime}A}{B^{\prime}P^{\prime}}$. Similarly, $\frac{PA}{PC}=\frac{C^{\prime}A}{C^{\prime}P^{\prime}}$.

Thus, we have the following chain of equivalent equations:

$PA^{2}=PB\cdot PC$
$\Longleftrightarrow$ $\frac{PA}{PB}\cdot\frac{PA}{PC}=1$
$\Longleftrightarrow$ $\frac{B^{\prime}A}{B^{\prime}P^{\prime}}\cdot\frac{C^{\prime}A}{C^{\prime}P^{\prime}}=1$
$\Longleftrightarrow$ $B^{\prime}A\cdot C^{\prime}A=B^{\prime}P^{\prime}\cdot C^{\prime}P^{\prime}$.

Now, what does this equation actually mean? Well, $B^{\prime}A\cdot C^{\prime}A$ is a fixed (and known) value, and $B^{\prime}P^{\prime}\cdot C^{\prime}P^{\prime}$ can be interpreted as the absolute value of the power of the point P' with respect to the circle with diameter B'C'. Thus, the equation $B^{\prime}A\cdot C^{\prime}A=B^{\prime}P^{\prime}\cdot C^{\prime}P^{\prime}$ signifies that the power of the point P' with respect to the circle with diameter B'C' is either $B^{\prime}A\cdot C^{\prime}A$ or $-B^{\prime}A\cdot C^{\prime}A$.

The locus of all points that have a given power with respect to a given circle $k_{1}$ is a circle concentric with the circle $k_{1}$, and this circle can be easily constructed. Hence,

- the locus of all points whose power with respect to the circle with diameter B'C' is $B^{\prime}A\cdot C^{\prime}A$ is a circle which can be easily constructed - denote this circle by $k_{+}$;
- the locus of all points whose power with respect to the circle with diameter B'C' is $-B^{\prime}A\cdot C^{\prime}A$ is a circle which can be easily constructed - denote this circle by $k_{-}$.

Now, since the equation $B^{\prime}A\cdot C^{\prime}A=B^{\prime}P^{\prime}\cdot C^{\prime}P^{\prime}$ signifies that the power of the point P' with respect to the circle with diameter B'C' is either $B^{\prime}A\cdot C^{\prime}A$ or $-B^{\prime}A\cdot C^{\prime}A$, this equation puts the point P' on one of the two circles $k_{+}$ and $k_{-}$. Since we also know that the point P' lies on the line B'C', we can thus construct the point P' as a common point of one of the two circles $k_{+}$ and $k_{-}$ with the line B'C'. We get 4, 3 or 2 different solutions for the point P', depending on whether both circles $k_{+}$ and $k_{-}$ intersect the line B'C' (this case yields 4 solutions), or the circle $k_{-}$ intersects the line B'C' and the circle $k_{+}$ is tangent to this line (this case yields 3 solutions), or the circle $k_{-}$ intersects the line B'C' and the circle $k_{+}$ doesn't have common points with B'C' (this case yields 2 solutions). (It is easy to show that the circle $k_{-}$ always intersects the line B'C'.) Consequently, inverting with respect to the circle k, we get 4, 3 or 2 solutions for the point P.

And the problem is solved.

The reader can easily show using the above that
- we get 4 solutions for the point P if $BC^{2}>4\cdot CA\cdot AB$;
- we get 3 solutions for the point P if $BC^{2}=4\cdot CA\cdot AB$;
- we get 2 solutions for the point P if $BC^{2}<4\cdot CA\cdot AB$.

  darij

:mad:  :mad:  darij , why always , always your solutions are better than the others :P

Solution: let us name the circle $C_1$. for every point on the $C_1$ like $R$ we give a translated point  with the name of $R^,$ such that
$R^,$ be on the $CR$ such that $RR^,=RB$ . the locus of the $R^, $ will be a circle
cause $2\angle BR^,C =\angle BRC$ . let us name the second circle $C_2$ and it center with $O$ and its radis with $r$.
 now for every point like $S$ on $C_1$ ,$SB.SC=$its power to $C_2$ wich means
$SO^2 - r^2$.if $SB.SC=SO^2 - r^2=SA^2 $ then we have that
$SO^2 -SA^2=r^2$, we know that $O, A , r $ are constant so we have a good lucus for such points like $S$(it will be line prependicular to $OA$ ,  but we have another locus for $S$ and it is the circle $C_1$ , so their common point will be the desired point.

so $M$ is found , darij didnt let me to berief so let me enjoy my solution :lol:

We want to find a point like M such that MA^2=MB.MC,it means
MA:MC=MB:MA(1).
According to Ptolemys theorem AB*MC+BC*AM=AC*MB [b]=>[/b]:MA[b]=>[/b][/b][/b]
BC+AB*(MC:MA)=AC*(MB:MA) and from (1)[b]=>[/b]
BC+AB(MA:MB)=AC*(MB:MA) and put MB:MA=x then we have
BC+AB:x=AC*x [b]=>[/b]*x [b]=>[/b] BC*x+AB=AC*(x^2) and BC,AB,AC are fixed so we can find x from this quadratic equation;so we can find MB:MA(it is fixed) [b]=>[/b]we have a [color=red]geometric locus[/color] for M(Apollunious circle). If we draw this Apollunious circle it will intersects circumcircle of triangle ABC in point M.   :D

My friends! in your solutions there is no attention to M when AB=AB or  $\angle A=90$  :idea:
What will you do with that? :?:

is any body exept jensen who can solve the problem with out using inversion.
unfortunately my solution is also by using inversion. :roll:

I have just corrected my solution, it contained some technical mistakes (I confused the power with its absolute value). However, I don't understand this:

[quote="jensen"]There is something else..
If AB=AC or angel BAC=90 => there is no M such that MA^2=MB*MC.(?)[/quote]

Why not?

  darij


the equation $ x^4\plus{}Ax^3\plus{}Bx^2\plus{}Cx\plus{}D\equal{}0$ with integers A,B,C,D has a two of its roots $ \minus{}2i\plus{}5 and 3\plus{}i$. what is the value of A?

The other roots are

$ 5\plus{}2i$ and $ 3\minus{}i$

Then, $ x_1\plus{}x_2\plus{}x_3\plus{}x_4\equal{}\minus{}A$

[quote="red_dog"]The other roots are

$ 5 \plus{} 2i$ and $ 3 \minus{} i$[/quote]

Why?

Because A, B, C, D are real coefficients. And if an equation with real coefficients has the root $ a\plus{}bi$ then it has also the root $ a\minus{}bi$.

As red_dog mentioned, two other roots are $ 5\plus{}2i$ and $ 3\minus{}i$.

By Vieta's, we know that $ \minus{}A$ is sum of four roots.

Thus, $ \minus{}A\equal{}5\plus{}2i\plus{}5\minus{}2i\plus{}3\plus{}i\plus{}3\minus{}i\equal{}16$

Hence, $ A\equal{}\minus{}16$

[quote="red_dog"]Because A, B, C, D are real coefficients. And if an equation with real coefficients has the root $ a \plus{} bi$ then it has also the root $ a \minus{} bi$.[/quote]

Right. You should reffer to that.

[quote="Zed_darkx"]Even the square root of a complex no.s gives you conjugate roots.[/quote]

Wrong.


limit
number theory unsolved
number theory

Suppose that $(a_n)$ is a sequence of positive integers such that $\lim\limits_{n\to \infty} \dfrac{n}{a_n}=0$
Prove that there exists $k$ such that there are at least $1990$ perfect squares between $a_1 + a_2 + ... + a_k$ and $a_1 + a_2 + ... + a_{k+1}$.

Let $b_n = \sum \limits_{i=1}^n a_n$

Claim:
$\lim\limits_{n\to \infty} \dfrac{n^2}{b_n}=0$
Proof:
By the condition of the problem, there exists $N$ such that for all $n>N$, $a_n>kn$, where k is an arbitrary constant.
Then $b_{2N}>N(kN)=kN^2$
$\dfrac{(2N)^2}{b_{2n}}=\dfrac{4}{k}$
which we can make arbitrarily small by increasing $k$.

The number of perfect squares between $b_n$ and $b_{n+1}$ is $\sqrt{b_{n+1}}-\sqrt{b_{n}}$
There exists N such that for all n>N, $b_n > kn^2$

Let $u$ be the smallest integer such that
$\dfrac{b_{u+1}}{(n+1)^2} > 1990^2$
Then,
$\dfrac{b_u}{n^2} < 1990^2$
Taking the square root of both sides of the equation, then subracting, we get
$\sqrt{b_{u+1}}-\sqrt{b_u}>1990$, as desired.

Nice, there is a different  solution using Stolz theorem.

Let's define sequences $\{u_n\},\{b_n\}$ such that $u_n=1+2+...+n$, $b_n=a_1+a_2+...+a_n$.
Thus from the Stolz theorem because $\lim\limits_{n\to \infty}\dfrac{u_{n+1}-u_n}{b_{n+1}-b_n}=\lim\limits_{n\to \infty}\dfrac{n+1}{a_{n+1}}=0$ we have $\lim\limits_{n\to \infty}\dfrac{u_n}{b_n}=0$ so $\lim\limits_{n\to \infty}\dfrac{\sqrt{u_n}}{\sqrt{b_n}}=\lim\limits_{n\to \infty}\dfrac{\sqrt{\dfrac{n(n+1)}{2}}}{\sqrt{a_1+a_2+...+a_n}}=0$ so $\lim\limits_{n\to \infty}\dfrac{n}{\sqrt{a_1+a_2+...+a_n}}=0$

Suppose that for every $k \in Z_{+}$ $\sqrt{a_1+a_2+...+a_{k+1}}-\sqrt{a_1+a_2+...+a_{k}}<m$ where $m$ is a given positive integer.

That implies $\sqrt{a_1+...+a_{k+1}}<\sqrt{a_1}+km$ so $\dfrac{n}{\sqrt{a_1+a_2+...+a_n}}>\dfrac{n}{\sqrt{a_1}+nm}$ but obviously $\lim\limits_{n\to \infty}\dfrac{n}{\sqrt{a_1}+nm}=\dfrac{1}{m}$ - we have a contradiction cause the limit should be $0$.

See also the post:

http://www.mathlinks.ro/Forum/viewtopic.php?p=353420#353420


[hide]
The number of factors is the exponent + 1.
$14\cdot3\cdot3\cdot2=252$
$25\cdot25\cdot2=1250$
$(x+1)(y+1)(z+1)=xyz+xy+xz+yz+x+y+z+1$
$(x^2+1)(2x)=2x^3+2x$[/hide]

[quote="Drunken_Math"]Find the number of factors for the following numbers:

$2^{13}*5^2*19^2*37$

$2^{24}*13^{24}*19$

$2^x*3^y*5^z$ where $x$, $y$, and $z$ are positive integers.

$2^{x^2}*3^{2x-1}$[/quote]

[hide=" :)"]
$14 \times 3 \times 3 \times 2=\boxed{252}\\\\25 \times 25 \times 2=\boxed{1250}$
thats all i got know.[/hide]click me

eh???? click me? [color=green]click the smilie[/color]


1. I'm looking for an thorough (but understandable by stupid people like me) explanation on how one can define one's own environments and commands, or redefine existing ones.

2. Can someone detail [b]everything[/b] there is to know about numbering for me? (Sorry, search engines have failed, or I'm too inexperienced to understand the stuff on the web)

3. How can one make references to things other than section numbers, page numbers, figure numbers, and equation numbers?

Actually, I don't really want to do anything fancy, nothing that much more complicated than a newtheorem environment. All I want to do is to be able to type something like

[code]
\begin{problem} blablablablablablablablablablablablablablablablablablablablablablablablablablablablablabla blablablablablablablablablablablablablablabla

\begin{solution} blablablablablablablablablablablablablablablablablablablablablablablablablablablablablabla blablablablablablablablablablablablablablabla blablablablabla. blablablablablablablablablabla. blablablablablablablablablablablablablablabla. blablablablabla

\begin{lemma} blablablablablablablablablabla
\label{strange}
\begin{proof} blablablablablablablablablablablablablablabla blablablablablablablablablablablablablablablablablablablabla
blablablablablablablablablablablablablablablablablablablablablablablablablablablablablabla
blablablablablablablablablablablablablablabla \qed
\end{proof}

\end{lemma}

\begin{lemma} blablablablablablablablablabla
\label{diophantine}
\begin{proof}

\begin{case}{blablablablablablablablablablablabla, n > 0} \label{awesome}blablablablablablablablablablablablablablablablablablablablablablablablablablablablablabla blablablablablablablablablablablablablablablablablablablablablablablabla blablablablablablablablablablablablablablablablablabla blablablablablablablablablablablablablablablablablabla
\end{case}

\begin{case}{blablablablablablablablablablablabla, n < 0} \label{confusing}blablablablablablablablablablablablablablablablablablablablablablablablablablablablablabla blablablablablablablablablablablablablablablablablablablablablablablabla blablablablablablablablablablablablablablablablablabla blablablablablablablablablablablablablablablablablabla
\end{case}

blablablablablabla blablablablablablablablablablablablablablablablablabla

blablablablablablablablablablablablablablablablablablablablablablablablablablablablablabla blablablablablablablablablablablabla blablablablablablablablablablablablablablablablablabla blablablablablablablablablablablabla blablablablablabla blablablablablabla \ref{awesome} blablablablablablablablablablablablablablablablablabla \ref{confusing} blablablablablabla \ref{strange}

blablablablablablablablablablablablablablablablablabla \qed
\end{proof}
\end{lemma}

\ref{diophantine} blabla

\end{solution}

\end{problem}

%------------------------------------------------------------------------------

\begin{problem} blablablablablabla blablablablablablablablablablablabla blablablablablablablablablablablabla

\begin{solution} blablablablablablablablablablablablablablablablablabla
blablablablablablablablablablablablablablablablablablablablablablablabla
blablablablablablablablablablablablablablablablablablablablablablablablablablablablablablablablablablablabla
blablablablablabla \ref{diophantine}[problem] blabla

\begin{lemma} blablablablablablablablablabla
\begin{proof} blablablablablablablablablablablablablablabla blablablablablablablablablablablablablablablablablablablabla
blablablablablablablablablablablablablablablablablablablablablablablablablablablablablabla
blablablablablablablablablablablablablablabla \qed
\end{proof}

\end{solution}

\end{problem}
[/code]

and produce something like

[quote]
[b]Problem 1.[/b] blablablablablablablablablablablablablablablablablablablablablablablablablablablablablabla blablablablablablablablablablablablablablabla

[i]Solution.[/i] blablablablablablablablablablablablablablablablablablablablablablablablablablablablablabla blablablablablablablablablablablablablablabla blablablablabla. blablablablablablablablablabla. blablablablablablablablablablablablablablabla. blablablablabla

__[b]Lemma 1.[/b] blablablablablablablablablabla

__[i]Proof.[/i] blablablablablablablablablablablablablablabla
__blablablablablablablablablablablablablablablablablablablabla
__blablablablablablablablablablablablablablablablablablablablablablablablablablablablablabla
__blablablablablablablablablablablablablablabla [QED SYMBOL]

__[b]Lemma 2.[/b] blablablablablablablablablabla

__[i]Proof.[/i]

____[i]Case I.[/i] blablablablablablablablablablablabla, n > 0
____blablablablablablablablablablablablablablablablablablablablablablablablablablablablablabla
____blablablablablablablablablablablablablablablablablablablablablablablabla
____blablablablablablablablablablablablablablablablablabla blablablablablablablablablablablablablablablablablabla

____[i]Case II.[/i] blablablablablablablablablablablabla, n < 0
____blablablablablablablablablablablablablablablablablablablablablablablablablablablablablabla
____blablablablablablablablablablablablablablablablablablablablablablablabla
____blablablablablablablablablablablablablablablablablabla blablablablablablablablablablablablablablablablablabla

__blablablablablabla blablablablablablablablablablablablablablablablablabla

__blablablablablablablablablablablablablablablablablablablablablablablablablablablablablabla
__blablablablablablablablablablablabla blablablablablablablablablablablablablablablablablabla
__blablablablablablablablablablablabla blablablablablabla blablablablablabla Case 1
__blablablablablablablablablablablablablablablablablabla Case 2 blablablablablabla Lemma 1

__blablablablablablablablablablablablablablablablablabla  [QED SYMBOL]

Lemma 2 blabla

[b]Problem 2.[/b] blablablablablabla blablablablablablablablablablablabla blablablablablablablablablablablabla

[i]Solution.[/i] blablablablablablablablablablablablablablablablablabla
blablablablablablablablablablablablablablablablablablablablablablablabla
blablablablablablablablablablablablablablablablablablablablablablablablablablablablablablablablablablablabla
blablablablablabla Lemma 2 of Problem 1 blabla

__[b]Lemma 1.[/b]  blablablablablablablablablabla

__[i]Proof.[/i] blablablablablablablablablablablablablablabla
__blablablablablablablablablablablablablablablablablablablabla
__blablablablablablablablablablablablablablablablablablablablablablablablablablablablablabla
__blablablablablablablablablablablablablablabla [QED SYMBOL]

[/quote]

(The underscores are supposed to be indents.) I hope I didn't make a mistake... In particular, note how the numbering of Lemmas starts over with each problem, and the number of each Case starts over with each Lemma, but the numbering does not appear as "Lemma 1.1", but rather, just "Lemma 1". The same thing happens when these things are referenced. Also, note how the each Lemma and Case and their respective proofs/text are indented.

Basically what you want is to customise the theorem environment. The easiest way to do that is use the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=ntheorem]ntheorem package[/url] which will be included in your LaTeX distribution. Have a read of the documentation, look at the examples they give and commands like \theoremindent, \theoremnumbering and see how you get on.

If you still have trouble, come back here for help showing us how far you got.

I suppose it's a bit unfair expecting you to start from scratch, so here is a sample to start you off
[code]\usepackage[thmmarks]{ntheorem} %thmmarks ensures you get a qed symbol
\begin{document}

\newtheorem{problem}{Problem}
\renewcommand{\theproblem}{\arabic{problem}.} % add a . after the number
\theoremindent1cm % start lemma with 1cm indent
\newtheorem{lemma}{Lemma}[problem] % numbering starts again in each problem
\renewcommand{\thelemma}{\arabic{lemma}.}
\newtheorem{solution}{Solution}
\theoremindent1cm
\theoremnumbering{Roman} % change numbering to I,II etc
\newtheorem{case}{Case}
\renewcommand{\thecase}{\Roman{case}.}
\qedsymbol{$\blacksquare$} % use whatever symbol you want for qed
\theoremstyle{break} % proof to start on next line
\newtheorem*{proof}{Proof.} % * means don't number proofs[/code]


The integers 2 through 9 are each placed in the figure with one integer in each of the eight smallest triangles. The integers are placed so that the pairs of integers in each of the four smallest squares have the same sum. What is that sum?

[asy]size(101);
draw(unitsquare);
draw(shift(up)*shift(right)*unitsquare);
filldraw(shift(up)*unitsquare,gray(.6));
filldraw(shift(right)*unitsquare,gray(.6));
draw((0,0)--(2,2)^^(0,2)--(2,0));[/asy]

In order to have the same sum, we have to pair up the smallest number with the biggest, then the next smallest with the next biggest, and so on, until the two middle numbers are added together. So $ 2 \plus{} 9 \equal{} \boxed{11}$.

First way I saw isn't quite as brief, but it's probably worth stating for similar puzzles!

The sum of all the integers is $ \frac{9(10)}{2} \minus{}1\equal{}44$, and that will be divided equally between four small squares: $ \boxed{11}$.


algebra
polynomial
real analysis
real analysis unsolved

Prove that for every $\epsilon >0$ there exists a positive integer $n$ and there
are positive numbers $a_1,...,a_n$ such that for every $\epsilon < x < 2\pi - \epsilon$
we have $\sum_{k=1}^{n}a_k\cos(kx) < \frac{-1}{\epsilon}|\sum_{k=1}^{n}a_k\sin(kx)|$

Maybe Schweitzer competition, I am not sure.