Let $ a,b,c$ be nonnegative real numbers. Prove that
\[ 9\sum a^3 \plus{}75\sum ab^2 \ge 53\sum a^2b\plus{}93abc.\]
This inequality is not very nice and not very hard to prove, but from it, the following inequality follows

Let $ a,b,c$ be nonnegative real numbers, then
\[ \frac{a}{\sqrt{a\plus{}b}} \plus{}\frac{b}{\sqrt{b\plus{}c}}\plus{}\frac{c}{\sqrt{c\plus{}a}} \le \frac{5}{4}\sqrt{a\plus{}b\plus{}c}.\]

It is easy to see that the second one is Jack Garfunkel's Inequality. ;) :)

Regards

P/s: The first one helps us to find a very simple proof for Jack's Inequality.

The first one is really easy to solve, in fact we have:
\[ 9\sum a^3 \plus{} 75\sum ab^2 \ge 53\sum a^2b \plus{} 93abc
\]

\[ \iff \frac {49}{2}[ (a \minus{} b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2] \plus{} (2a \plus{} b \minus{} 3c)^2\left(\frac {9b\plus{}c}{a} \minus{} 10\right)\ge 0
\]
which is always correct if we assume that $ a \equal{} min\{a,b,c\}$
But I am curious what is the relationship between Jack's inequality and your. Let me think later, now I must go to bed  :D

[quote="zaizai-hoang"]The first one is really easy to solve, in fact we have:
\[ 9\sum a^3 \plus{} 75\sum ab^2 \ge 53\sum a^2b \plus{} 93abc
\]

\[ \iff \frac {49}{2}[ (a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2] \plus{} (2a \plus{} b \minus{} 3c)^2\left(\frac {9b \plus{} c}{a} \minus{} 10\right)\ge 0
\]
which is always correct if we assume that $ a \equal{} min\{a,b,c\}$
But I am curious what is the relationship between Jack's inequality and your. Let me think later, now I must go to bed  :D[/quote]

The identity is very nice but how about $ a \equal{} \max \left\{ {a,b,c} \right\}$.

I think your solution is not true, zaizai.

If this beatiful identity is true and i don't doubt about that you can write it three times for a,b and c. Surely one of them is non negative and all are equal.

[quote="can_hang2007"]

Let $ a,b,c$ be nonnegative real numbers, then
\[ \frac {a}{\sqrt {a \plus{} b}} \plus{} \frac {b}{\sqrt {b \plus{} c}} \plus{} \frac {c}{\sqrt {c \plus{} a}} \le \frac {5}{4}\sqrt {a \plus{} b \plus{} c}.
\]
It is easy to see that the second one is Jack Garfunkel's Inequality. ;) :)

Regards

P/s: The first one helps us to find a very simple proof for Jack's Inequality.[/quote]
I've just found the my 7th-proof for this inequality. :)

[quote="zaizai-hoang"]The first one is really easy to solve, in fact we have:
\[ 9\sum a^3 \plus{} 75\sum ab^2 \ge 53\sum a^2b \plus{} 93abc
\]

\[ \iff \frac {49}{2}[ (a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2] \plus{} (2a \plus{} b \minus{} 3c)^2\left(\frac {9b \plus{} c}{a} \minus{} 10\right)\ge 0
\]
which is always correct if we assume that $ a \equal{} min\{a,b,c\}$
But I am curious what is the relationship between Jack's inequality and your. Let me think later, now I must go to bed  :D[/quote]
This type of solution is not very new. It is just an extension of $ b \equal{} a \plus{} x,c \equal{} a \plus{} y$ where $ a \equal{} \min\{a,b,c\}$ ;)

Now, you can think about how my inequality imples Jack's one. ;) :) It will be very interesting. :)

Honey_S, are you sure my solution is wrong ?! I think we only need assume $ a \equal{} \min\{a,b,c\}$ because when $ a \equal{} \max\{a,b,c\}$ we can rewrite "three times for a,b and c" as AlexandrosG said. It is not an important thing in here. Canhang_2007 have explained why I could find out the identity :wink:
For the moment I don't have enought time to think about  them, nextweek I will take park in an important contest so I cann't think any more about this. See you later :)

[quote="zaizai-hoang"]The first one is really easy to solve, in fact we have:
\[ 9\sum a^3 \plus{} 75\sum ab^2 \ge 53\sum a^2b \plus{} 93abc
\]

\[ \iff \frac {49}{2}[ (a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2] \plus{} (2a \plus{} b \minus{} 3c)^2\left(\frac {9b \plus{} c}{a} \minus{} 10\right)\ge 0
\]
which is always correct if we assume that $ a \equal{} min\{a,b,c\}$
But I am curious what is the relationship between Jack's inequality and your. Let me think later, now I must go to bed  :D[/quote]
zaizai how can you find this solution

[quote="Inequalities Master"][quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers. Prove that
\[ 9\sum a^3 \plus{} 75\sum ab^2 \ge 53\sum a^2b \plus{} 93abc.
\]
This inequality is not very nice and not very hard to prove, but from it, the following inequality follows

Let $ a,b,c$ be nonnegative real numbers, then
\[ \frac {a}{\sqrt {a \plus{} b}} \plus{} \frac {b}{\sqrt {b \plus{} c}} \plus{} \frac {c}{\sqrt {c \plus{} a}} \le \frac {5}{4}\sqrt {a \plus{} b \plus{} c}.
\]
It is easy to see that the second one is Jack Garfunkel's Inequality. ;) :)

Regards

P/s: The first one helps us to find a very simple proof for Jack's Inequality.[/quote]

When are you planning to show us all how does this imply Jack's inequality?[/quote]

We are still waiting for an answer, Can.... Please help. :(

[quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers. Prove that
\[ 9\sum a^3 \plus{} 75\sum ab^2 \ge 53\sum a^2b \plus{} 93abc.
\]
This inequality is not very nice and not very hard to prove, but from it, the following inequality follows

Let $ a,b,c$ be nonnegative real numbers, then
\[ \frac {a}{\sqrt {a \plus{} b}} \plus{} \frac {b}{\sqrt {b \plus{} c}} \plus{} \frac {c}{\sqrt {c \plus{} a}} \le \frac {5}{4}\sqrt {a \plus{} b \plus{} c}.
\]
It is easy to see that the second one is Jack Garfunkel's Inequality. ;) :)

Regards

P/s: The first one helps us to find a very simple proof for Jack's Inequality.[/quote]
Now, let me explain why my inequality implies the Jack's one. :)

To prove Jack's Inequality, we will take squares both sides and rewrite it as
\[ \sum \frac {a^2}{a \plus{} b} \plus{} 2\sum \frac {ab}{\sqrt {(a \plus{} b)(b \plus{} c)}} \le \frac {25}{16} \sum a,
\]
or
\[ 2\sum \frac {ab}{\sqrt {(a \plus{} b)(b \plus{} c)}} \le \frac {9}{16} \sum a \plus{} \sum \frac {ab}{a \plus{} b}.
\]
Now, applying Cauchy Schwarz Inequality, we have
\[ \sum \frac {ab}{\sqrt {(a \plus{} b)(b \plus{} c)}} \le \sqrt {\left( \sum ab\right) \left( \sum \frac {ab}{(a \plus{} b)(b \plus{} c)}\right)} \equal{} \sqrt {\frac {\left( \sum ab \right) \left( \sum a^2b \plus{} 3abc\right)}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}}.
\]
Our inequality is now reduced from
\[ 2\sqrt {\frac {\left( \sum ab\right) \left( \sum a^2b \plus{} 3abc\right)}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}} \le \frac {9}{16}\sum a \plus{} \sum \frac {ab}{a \plus{} b}.
\]
Normalizing for $ a \plus{} b \plus{} c \equal{} 1$ and setting $ q \equal{} ab \plus{} bc \plus{} ca,r \equal{} abc$, this inequality becomes
\[ 2\sqrt {\frac {q\left( \sum a^2b \plus{} 3r\right) }{q \minus{} r} } \le \frac {9}{16} \plus{} \frac {q^2 \plus{} r}{q \minus{} r}.
\]
Now, using the my inequality above: $ 9\sum a^3 \plus{} 75\sum ab^2 \ge 53\sum a^2b \plus{} 93abc,$ we get
\[ 128\sum a^2b \le 9\sum a^3 \plus{} 75\sum ab(a \plus{} b) \minus{} 93abc \equal{} 3(3 \plus{} 16q \minus{} 97r).
\]
Therefore
\[ \sum a^2b \plus{} 3r \le \frac {3(3 \plus{} 16q \plus{} 31r)}{128}.
\]
And we can deduce our inequality to
\[ 2\sqrt {\frac {3q(3 \plus{} 16q \plus{} 31r)}{128(q \minus{} r)}} \le \frac {9}{16} \plus{} \frac {q^2 \plus{} r}{q \minus{} r},
\]
or equivalently,
\[ 7r \plus{} 9q \plus{} 16q^2 \ge 2\sqrt {6q(q \minus{} r)(3 \plus{} 16q \plus{} 31r)},
\]

\[ (7r \plus{} 9q \plus{} 16q^2)^2 \ge 24q(q \minus{} r)(3 \plus{} 16q \plus{} 31r).
\]
As $ (7r \plus{} 9q \plus{} 16q^2)^2 \ge 2\cdot 7r\cdot (9q \plus{} 16q^2) \plus{} (9q \plus{} 16q^2)^2 \ge 224q^2r\plus{}(9q\plus{}16q^2)^2$, and
\[ (q \minus{} r)(3 \plus{} 16q \plus{} 31r) \equal{} q(3 \plus{} 16q) \plus{} (15q \minus{} 3)r \minus{} 31r^2 \le q(3 \plus{} 16q) \plus{} 3qr,
\]
We have
\[ LHS \minus{} RHS \ge 224q^2r\plus{}(9q\plus{}16q^2)^2 \minus{} 24q[q(3 \plus{} 16q) \plus{} 3qr] \equal{} 152q^2r \plus{} q^2(3 \minus{} 16q)^2 \ge 0.
\]
This completes the proof of Jack's Inequality. :)

[quote="can_hang2007"][quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers. Prove that
\[ 9\sum a^3 \plus{} 75\sum ab^2 \ge 53\sum a^2b \plus{} 93abc.
\]
This inequality is not very nice and not very hard to prove, but from it, the following inequality follows

Let $ a,b,c$ be nonnegative real numbers, then
\[ \frac {a}{\sqrt {a \plus{} b}} \plus{} \frac {b}{\sqrt {b \plus{} c}} \plus{} \frac {c}{\sqrt {c \plus{} a}} \le \frac {5}{4}\sqrt {a \plus{} b \plus{} c}.
\]
It is easy to see that the second one is Jack Garfunkel's Inequality. ;) :)

Regards

P/s: The first one helps us to find a very simple proof for Jack's Inequality.[/quote]
Now, let me explain why my inequality implies the Jack's one. :)

To prove Jack's Inequality, we will take squares both sides and rewrite it as
\[ \sum \frac {a^2}{a \plus{} b} \plus{} 2\sum \frac {ab}{\sqrt {(a \plus{} b)(b \plus{} c)}} \le \frac {25}{16} \sum a,
\]
or
\[ 2\sum \frac {ab}{\sqrt {(a \plus{} b)(b \plus{} c)}} \le \frac {9}{16} \sum a \plus{} \sum \frac {ab}{a \plus{} b}.
\]
Now, applying Cauchy Schwarz Inequality, we have
\[ \sum \frac {ab}{\sqrt {(a \plus{} b)(b \plus{} c)}} \le \sqrt {\left( \sum ab\right) \left( \sum \frac {ab}{(a \plus{} b)(b \plus{} c)}\right)} \equal{} \sqrt {\frac {\left( \sum ab \right) \left( \sum a^2b \plus{} 3abc\right)}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}}.
\]
Our inequality is now reduced from
\[ 2\sqrt {\frac {\left( \sum ab\right) \left( \sum a^2b \plus{} 3abc\right)}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}} \le \frac {9}{16}\sum a \plus{} \sum \frac {ab}{a \plus{} b}.
\]
Normalizing for $ a \plus{} b \plus{} c \equal{} 1$ and setting $ q \equal{} ab \plus{} bc \plus{} ca,r \equal{} abc$, this inequality becomes
\[ 2\sqrt {\frac {q\left( \sum a^2b \plus{} 3r\right) }{q \minus{} r} } \le \frac {9}{16} \plus{} \frac {q^2 \plus{} r}{q \minus{} r}.
\]
Now, using the my inequality above: $ 9\sum a^3 \plus{} 75\sum ab^2 \ge 53\sum a^2b \plus{} 93abc,$ we get
\[ 128\sum a^2b \le 9\sum a^3 \plus{} 75\sum ab(a \plus{} b) \minus{} 93abc \equal{} 3(3 \plus{} 16q \minus{} 97r).
\]
Therefore
\[ \sum a^2b \plus{} 3r \le \frac {3(3 \plus{} 16q \plus{} 31r)}{128}.
\]
And we can deduce our inequality to
\[ 2\sqrt {\frac {3q(3 \plus{} 16q \plus{} 31r)}{128(q \minus{} r)}} \le \frac {9}{16} \plus{} \frac {q^2 \plus{} r}{q \minus{} r},
\]
or equivalently,
\[ 7r \plus{} 9q \plus{} 16q^2 \ge 2\sqrt {6q(q \minus{} r)(3 \plus{} 16q \plus{} 31r)},
\]

\[ (7r \plus{} 9q \plus{} 16q^2)^2 \ge 24q(q \minus{} r)(3 \plus{} 16q \plus{} 31r).
\]
As $ (7r \plus{} 9q \plus{} 16q^2)^2 \ge 2\cdot 7r\cdot (9q \plus{} 16q^2) \plus{} (9q \plus{} 16q^2)^2 \ge 224q^2r \plus{} (9q \plus{} 16q^2)^2$, and
\[ (q \minus{} r)(3 \plus{} 16q \plus{} 31r) \equal{} q(3 \plus{} 16q) \plus{} (15q \minus{} 3)r \minus{} 31r^2 \le q(3 \plus{} 16q) \plus{} 3qr,
\]
We have
\[ LHS \minus{} RHS \ge 224q^2r \plus{} (9q \plus{} 16q^2)^2 \minus{} 24q[q(3 \plus{} 16q) \plus{} 3qr] \equal{} 152q^2r \plus{} q^2(3 \minus{} 16q)^2 \ge 0.
\]
This completes the proof of Jack's Inequality. :)[/quote]

Beautiful proof for a very hard inequality, my dear friend. :)


geometry
3D geometry
analytic geometry
number theory proposed
number theory

Let $ X \equal{} \{A_1, A_2, ... , A_n \}$ be set of dinstinct 3-element subsets of the set $ \{1,2,3,...,36\}$ such that
(a) $ A_i$ and $ A_j$ have non empty intersections for every $ i$ and $ j$
(b) The intersections of all elements of $ X$ is a ampty set.
Show that $ n \ge 100$ . Determine the number of  such sets $ X$ when $ n\equal{}100$

I have an idea of using geometry :D

Take 6.n points inside the cube [1,36] ^3 whose coordinates are permutations of the elements of certain set Ai

....... :rotfl:


geometry
3D geometry
vector
geometric transformation
reflection
probability and stats

Randomly and independently chosen $n$ points in the inteval [0,1].Find the expectation of the least distance between two points of the$n$ points.
I guess the value may be $\frac{1}{n^{2}-1}$,but not having the ability to prove it. :blush:

We have $(y_{1},y_{2},\dots,y_{n})$ uniformly distributed in an $n$-dimensional hypercube. Let $Y=\min_{i\ne j}|y_{i}-y_{j}|.$ $Y$ is the quantity you are considering - the "least distance between two of the $n$ points."

Consider only the case $y_{1}<y_{2}<\cdots<y_{n}.$ This is one of $n!$ equally likely orderings - geometrically, it's a simplex, and $n!$ of these simplices can be assembled to make the hypercube.

$P(Y>a)=n!P(0<y_{1}<y_{2}-a<y_{3}-2a<\cdots<y_{n}-(n-1)a<1-(n-1)a).$

This is $n!$ times the $n$-volume of a simplex, and we can assemble $n!$ such simplices into one hypercube with edge length $1-(n-1)a.$ Hence:

$P(Y>a)=[1-(n-1)a]^{n}$ for $0<a<\frac1{n-1}.$

This is the cumulative distribution from a classic "order statistics" problem. In fact, $Y$ has precisely the same distribution as the minimum of $n$ independent random variables each uniformly distributed on the interval $\left[0,\frac1{n-1}\right].$

This is a well-studied problem, and one of the results is that

$E(Y)=\frac1{n+1}\cdot\frac1{n-1}=\frac1{n^{2}-1}.$

Your guess was correct.

Hello Kent,
can you please elaborate on your solution? How do you go from [0,1] interval to $n-$dimensional hypercube? I also didn't understand the volume argument: how do you know how many such symplexes you can pack in such and such cube?
Thank you

Record the $n$ guesses $(y_1, \ldots, y_n)$.  This is a vector in $[0, 1]^n$, i.e., the $n$-dimensional hypercube.

For each permutation $w = w_1 w_2 \ldots w_n \in S_n$ (written in one-line notation, so the set of $w_i$ is equal to the set of integers between $1$ and $n$), we have an associated cone $x_{w_1} < x_{w_2} < \ldots < x_{w_n}$.  The intersection of these cones with the cube are minimal simplices, and they are congruent: reflections through the hyperplanes $x_i = x_j$ are the same as applying transpositions $(i, j)$ (written in cycle notation) and we can turn any permutation into any other by a sequence of transpositions.  Thus they all have the same volume, in particular $\frac{1}{n!}$.


calculus
integration
real analysis
real analysis unsolved

what can you say about integral like this :
$ \int_{\minus{}1}^{1}\frac{t^{2n}(1\minus{}t^{2})^{2n}}{(1\plus{}it)^{3n\plus{}1}}dt$ :huh:

I can say I would immediately put that into Mathematica. :lol:

I can say that it's same :=
$ K_{n} \equal{} \int_{0}^{1}\frac {t^{2n}(1 \minus{} t^{2})^{2n}( (1 \plus{} it)^{3n \plus{} 1} \plus{} (1 \minus{} it)^{3n \plus{} 1})}{(1 \plus{} t^{2})^{3n \plus{} 1}}dt$
The first few values read:
$ K_{1} \equal{} 14\pi \minus{} 4$
$ K_{2} \equal{} 968\pi \minus{} \frac {45616}{15}$
also note that $ K_{n} \minus{} > 0$ when $ n \minus{} > \infty$
you can also note that we have rational approximation to $ \pi$whsoe qualities exceed $ 1.0449$
1. $ \pi$ is irrational
2.   hence irrational measure ,upper bound will be equal to
$ 23.271..$


geometry
circumcircle
trigonometry
inequalities proposed
inequalities

Let $ a,b,c$ denote the sides of a triangle opposite the angles $ A,B$, and $ C$, respectively. Let R the circumradius of the triangle. If angle $ A>90$°, prove that

$ 2R^2\geq\frac{bc}{\sin A}$

[quote="Ligouras"]Let $ a,b,c$ denote the sides of a triangle opposite the angles $ A,B$, and $ C$, respectively. Let R the circumradius of the triangle. If angle $ A > 90$°, prove that

$ 2R^2\geq\frac {bc}{\sin A}$[/quote]
Using $ b \equal{} 2R \cdot \sin B, c \equal{} 2R \cdot \sin C$ we find that it is equivalent to:
$ \sin A \ge 2 \sin B \cdot \sin C$.
Now $ 2 \sin B \cdot \sin C \equal{} \cos B \minus{} C \minus{} \cos B \plus{} C \equal{} \cos B \minus{} C \plus{} \cos A \le 1 \plus{} \cos A$
So it suffices to prove $ \sin A \minus{} \cos A \ge 1$ when $ A > 90^{\circ}$:
Since $ \sin A, \minus{}\cos A \ge 0$ we see that $ (\sin A \minus{} \cos A)^2 \ge \sin^2 A \plus{} \cos^2 A \equal{} 1$, and we are done :)

[size=150]
More Strong:

$ 4R^2 \geq \frac {b^2 \plus{} c^2}{\sin{A}}$

$ < \equal{} > \frac {( \minus{} b^2 \minus{} c^2 \plus{} a^2)Q}{(a \plus{} b \plus{} c)( \minus{} a \plus{} b \plus{} c)(a \minus{} b \plus{} c)(a \plus{} b \minus{} c)} \geq 0$
$ Q \equal{} \minus{} b^6 \plus{} b^4a^2 \plus{} b^4c^2 \plus{} 6b^2c^2a^2 \plus{} b^2c^4 \plus{} c^4a^2 \minus{} c^6,$

only prove that:
$ Q \geq 0$

But $ Q \equal{} (b^4 \plus{} 6b^2c^2 \plus{} c^4)(a^2 \minus{} b^2 \minus{} c^2) \plus{} 8b^4c^2 \plus{} 8b^2c^4 > 0$.
proof end.
BQ[/size]


function
Gauss
vector
algebra
domain
linear algebra
linear algebra theorems

Let S be a set, F a field, and V(S;F) the space of all functions from S into F:

(f+g)(x) = f(x) + g(x)
(cf)(x) = cf(x)

Let W be any n-dimensional subspace of V(S;F). Show that there exist points $ x_{1}$,..,$ x_{n}$ in S and functions $ f_{1}$,..,$ f_{n}$ in W such that $ f_{i}$($ x_{j}$) = $ d{}_{i}{}_{j}$.

All I am able to figure out is that these functions must form a basis. Aside from that, I don't see how to pick the points in S, especially since S is just an arbitrary set and not even a linear space. Also, the problem seems to imply that S needs to contain n elements in order for V(S;F) to contain an n-dimensional subspace. Can someone be kind in solving this in detail?

If $ |S| = n$ then we might as well have $ S = \{1, 2, \ldots, n\}$ and the functions $ \{f_{i}\mid f_{i}(j) = \delta_{i,j}\}$ form a basis for $ V(S; F)$, so in particular $ V(S; F)$ is $ n$-dimensional and does not have any subspaces of dimension larger than $ n$.

For the original question, we have a subspace of dimension $ n$.  Choose $ n$ linearly independent functions in the subspace.  Then you just apply a variant of Gauss-Jordan elimination: take the first function, take a point at $ S$ on which it is not zero, and subtract multiples of it from your other selected functions so that on that point they become zero.  Now repeat with each of your functions in turn, and you're done.

Your comment that "$ S$ isn't linear" suggests that you don't yet have a feel for what the space $ V(S; F)$ looks like.  In particular, you're thinking of the elements of $ S$ as the "numbers," but they're not: it's the functions $ S \to F$ that are the "numbers" you want.

can you be specific on what you mean by take away multiples? i will work on it today and see if i know what you mean

Let's suppose $ S = \{a, b, c\}, F = \bf Q$.  Then $ V(S; F)$ is a 3-dimensional vector space over $ \bf Q$.  Consider any two-dimensional subspace of this space, e.g. the space generated by the functions $ f_{1}$ and $ f_{2}$ where $ f_{1}(a) = 3$, $ f_{1}(b) = 0$ and $ f_{1}(c) =-3$ while $ f_{2}(a) = 2$, $ f_{2}(b) = 3$ and $ f_{2}(c) =-2$.  (These are linearly independent.)  Pick a point in the domain of $ f_{1}$ where $ f_{1}$ is non-zero.  (Such a point exists by independence.)  I like $ a$.  Now, Gauss-Jordan eliminate: replace $ f_{2}$ with $ g = f_{2}-\frac{f_{2}(a)}{f_{1}(a)}f_{1}$, which has the property that $ g(a) = 0$.  However, by linear independence of $ f_{1}$ and $ f_{2}$, $ g$ takes some nonzero value at some point.  Then, use this value to perform a similar transformation and replace $ f_{1}$ with something that is 0 at that point.  Now, rescale both functions so that they are 1 at the respective points we've chosen.

This is almost exactly the same as Gauss-Jordan elimination.  If you pretend that the set $ S$ is always finite, you can write each of your functions out in one-line notation and then you can even do your operations on matrices.


Is there a polynomial P(x) with integer coefficients such that P(10)=400,
P(14)=440 and P(18)=520.

[quote="Rust"]No, there is not, because $(18-10)\not |(P(18)-P(10))=120.$[/quote]

$8\cdot15=120$ so sorry but this is not a proof.

A more important mistake out of your post you have is

"P(+a)."  This should have been typed as "P(x+a)" otherwise why have the plus sign included with the "a?"

(I can't speak to whether or not you have properly developed a proof, for instance, without informing the reader the name of what formula or discoverer's process you are calling upon, before you show it being worked on.)


The SC Community is pretty active.  We just added a bunch of new users, and I encourage all South Carolinians to join!

NJ, California, Massachusetts, or NY probably.

Dense population. Lots of smart people. yea. :wink:

connecticut, definitely.(althogh 80% of the posts there are nonmath related)

yes. I would tend to think one of the more north-eastern states for some reason. maybe it is false stereotypes created. i don't mean to offend anyone.

apparently, i went to the NJ one and i was the first person to post for ages.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=113717[/url]


geometry
geometric transformation
reflection
analytic geometry
graphing lines
slope
vector

Find the shortest total distance from $(-21, 33)$ to $(0,63)$ if one must go from $(-21, 33)$ to the line whose equation is $y=3x-7$ and then from that line to $(0,63)$.
[hide]
Point-Line equation
Line= $Ax+By+c = 0$
Point= $(r,s)$
Distance:
$\frac{|Ar+Bs+C|}{\sqrt{A^{2}+B^{2}}}$
I tried that but I did not find out at which point the distance line from $(-21,33)$ intersect the given line $(y=3x-7)$.[/hide]

There's a neat method for these types of problems. If the points are A and B, reflect A across the line to A' and find A'B.

[hide="General solution outline for this problem"]Find the image of $(0,63)$ when it is reflected over the line $y = 3x-7$. The shortest distance will be the distance from $(-21,33)$ to this image.[/hide]

I saw this method in Mathematical Circles! That book is cool.
Anyways, how do you find the reflection image?

I don't know if this is the most efficient method but I would:

1) find the distance from it to the line.
2) find the slope of a line perpendicular to the given line.
3) Turn the magnatude and slope given in steps 1 and 2 into a vector
4) Double the vector
5) Translate the point through that vector.


Let $a, b, c>0$ such that $ab+bc+ca=3.$ Prove that

  \[ a^{3}+b^{3}+c^{3} +6abc\geq 9.  \]

I'm expecting nice solutions for this nice inequality. ;)

Cezar - I needed to cut off your solution as this round of Polish olympiad is still in progress

Locked till 7th December (or smth around this - don't remember exact date)

Cezar, you actually KNEW that this problem shouldn't be posted yet  :mad:

It appeared a few days ago, and I locked that topic, remember?!

And I got PM'ed by [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=13617]jerry144[/url] asking me please to solve it for him... this is really getting annoying with the polish olympiad... I think we should do something about this. Somehow. :?

I'm pretty sure there is only one guy (having many different accounts or something) responsible for this.

[quote="Cezar Lupu"]Let $ a, b, c\geq0$ such that $ bc \plus{} ca \plus{} ab \equal{} 3.$ Prove that $ a^3 \plus{} b^3 \plus{} c^3 \plus{} 6abc\geq 9.$[/quote]$ a^3 \plus{} b^3 \plus{} c^3 \plus{} 6abc \minus{} 9 \equal{} 3(a \plus{} b \plus{} c \minus{} 3) \plus{} \sum{a(a \minus{} b)(a \minus{} c)}\geq0,$

which is true by $ (a \plus{} b \plus{} c)^2\geq3(bc \plus{} ca \plus{} ab)$ and [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=1166]Schur inequality[/url].

By the way, the stronger inequality $ a^3 \plus{} b^3 \plus{} c^3 \minus{}3abc\geq 6\sqrt{3}(1\minus{}abc)$ holds for the same conditions.

[quote="Cezar Lupu"]Let $ a, b, c > 0$ such that $ ab \plus{} bc \plus{} ca \equal{} 3.$ Prove that
\[ a^{3} \plus{} b^{3} \plus{} c^{3} \plus{} 6abc\geq 9.
\]
I'm expecting nice solutions for this nice inequality. ;)[/quote]

Let $ a\plus{}b\plus{}c\equal{}p;ab\plus{}bc\plus{}ca\equal{}q\equal{}3;abc\equal{}r$. We have:

$ a^{3} \plus{} b^{3} \plus{} c^{3} \plus{} 6abc\geq 9.$

$ \leftrightarrow p^3\minus{}3pq\plus{}3r\plus{}6r\ge\ 9$

$ \leftrightarrow p^3\minus{}9p\plus{}9r \ge\ 9$

Follow Schur $ 9r \ge\ p(4q\minus{}p^2) \equal{} 12p \minus{} p^3$

$ \rightarrow LHS \ge\ p^3\minus{}9p\plus{} 12p \minus{} p^3 \ge\ 9$

$ \leftrightarrow p \ge\ 3$ which is obvious.

We have done! :)


I can only prove that :
tan -1x + tan -1y
= tan -1 (x + y)/(1-xy) when : -:pi:/2 < tan -1x + tan -1y < :pi:/2
= tan -1 (x + y)/(1-xy) + :pi: when : :pi:/2 < tan -1x + tan -1y  < :pi:
= tan -1 (x + y)/(1-xy) - :pi: when : -:pi: < tan -1x + tan -1y < -:pi:/2

(Does it 's correct. ?)

but I see from somewhere that
tan -1x + tan -1y
= tan -1 (x + y)/(1-xy) when : xy < 1
= tan -1 (x + y)/(1-xy) + :pi: when : xy > 1 and x > 0
= tan -1 (x + y)/(1-xy) - :pi: when : xy > 1 and x < 0

Can anyone prove this. ? Please help me.  :D

Try taking the tangent of the inequality.

Can you hint me.? :maybe: I try to use the fact that:
tan x < 0  if and only if x < 0
tan x > 0  if and only if x > 0
But it 's not the same as I told you.

hi phongthong,

are you familiar with the identity:

tan(u+v)=(tan u + tan v)/(1-tan u tan v)

this can be derived readily from

[hide]

sin (u+v)=sin u cos v + cos u sin v

cos (u+v)=cos u cos v - sin u sin v

[/hide]

??

try setting [tex]u=tan^{-1} x, v=tan^{-1} y[/tex]

The tangent sum formula is
$\tan{a+b} = \frac{\tan a + \tan b}{1 - \tan a \tan b}.$

If we let $a = \tan^{-1} x$ and $b = \tan^{-1} y$, then we have
$\tan{\tan^{-1} x+\tan^{-1} y} = \frac{\tan \tan^{-1} x + \tan \tan^{-1} y}{1 - \tan \tan^{-1} x \tan \tan^{-1} y}$
$\tan{\tan^{-1} x+\tan^{-1} y} = \frac{x + y}{1 - xy}$
Taking the inverse tangent of both sides, we have
$\tan^{-1} x+\tan^{-1} y = \tan^{-1} \left( \frac{x + y}{1 - xy} \right)$

Note that $\tan^{-1} x$ represents all angles $\theta$ for which $\tan \theta = x$, so that $\tan^{-1} x$ does not have a single, fixed value.  (You may be thinking of Tan<sup>-1</sup> x, the [i]principal[/i] inverse tangent, which is always between -pi/2 and pi/2.)  Thus, the $\pm \pi$ things don't matter, and what you have said is essentially correct.

Hi. Everyone  :D
Thanks for answer. But I think it 's not my question.
I mean that how do you know about the condition

xy < 1 , xy > 1 and x > 0, xy < 1 and x < 0

Can anyone explain to me.?

Sorry if my question is too silly.  :cool:

May be it 's too hard for intermediate topics.?

You wish to prove that the conditions

$-{{\pi}\over{2}} < \tan^{-1}x + \tan^{-1}y < {{\pi}\over{2}}$

and

$xy < 1$

are equivalent.

Set $a = \tan^{-1}x$, $b = \tan^{-1}y$. Then $x = \tan a$ and $y = \tan b$.

Recall the formula

$\tan (a + b) = {{\tan a + \tan b}\over{1 - \tan a \tan b}}$.

Rearrange to get

$\tan a \tan b = 1 - {{\tan a + \tan b}\over{\tan (a + b)}}$. We need to use the condition $-{{\pi}\over{2}} < a + b < {{\pi}\over{2}}$ to prove that $1 - {{\tan a + \tan b}\over{\tan (a + b)}} < 1 \Rightarrow {{\tan a + \tan b}\over{\tan (a + b)}} > 0$, i.e. both the numerator and the denominator have the same sign.

When $-{{\pi}\over{2}} < a + b < 0$, $\tan a + \tan b < 0$, $tan (a + b) < 0$. The inequality holds.

When ${{\pi}\over{2}} > a + b > 0$, $\tan a + \tan b > 0$, $tan (a + b) > 0$. The inequality holds.

(the inequality is indeterminate for a + b = 0, but it obviously holds for the original condition).

QED.

The issue (maybe) is that if we think of $\tan^{-1}$ as having a range of $(-\pi/2,\pi/2)$, then it is not necessarily true that $\tan^{-1}(\tan x)=x$.

I don't believe that there is a consistent usage of $\tan^{-1}$ versus $\mathrm{Tan}^{-1}$.


Thanks.  Another question:

While looking at the Shoelace theorem, I also found a theorem called "Pick's theorem."  Are both of them worth knowing?  What can Pick's theorem do that the Shoelace theorem cannot?

Aren't they essentially the same thing?

It does say "It is similar to the shoestring formula, and though it is less powerful it is a good tool to have in solving problems."
Pick's Theorem can only be used when all of the vertices lie on lattice points, while the Shoelace Theorem can be used whenever.

[quote="AdithyaGanesh"]What's the easiest way to find the area of an $ n$-gon given $ n$ arbitrary vertices?  I get these problems
in FTW a lot.[/quote]

Rephrased:

"Given $ n$ arbitrary vertices, what is the easiest way to find the area of
a polygon with $ n$ sides?"

Arbitrary vertices for a polygon having four or more sides do not
define a unique polygon, and for a given number of sides, their
areas are not necessarily equal to each other.  --> EDIT

For instance, if the vertices are $ (0, 0), (4, 0), (2, \sqrt {12}),$ and $ (1, 1)$,
then you could connect the four vertices at different times to form three
different quadrilaterals, all having areas different from one another.

[quote="Arrange your tan"][quote="AdithyaGanesh"]What's the easiest way to find the area of an $ n$-gon given $ n$ arbitrary vertices?  I get these problems
in FTW a lot.[/quote]

Rephrased:

"Given $ n$ arbitrary vertices, what is the easiest way to find the area of
a polygon with $ n$ sides?"

Arbitrary vertices for a polygon having four or more sides do not
give the same area necessarily.

For instance, if the vertices are $ (0, 0), (4, 0), (2, \sqrt {12}),$ and $ (1, 1)$,
then you could connect the four vertices at different times to form three
different quadrilaterals, all having areas different from each other.[/quote]

I realize that.  I was asking for a simple method to calculate the area of a polygon when you are given the vertices.  Previously, I was drawing a small graph and dividing the polygon into triangles, squares and rectangles.  However, this took a long time.

No, you claim you realize that, but you *didn't* mention
particulars in your question.  You have got to be more
specific with your polygon conditions.  I answered the
question as given.

Make more limitations in your question, then, such as the
vertices must be of a convex polygon where there is only
one polygon that can connect those vertices.


floor function
ceiling function
combinatorics unsolved
combinatorics

show that
\[
\sum\limits_{k = 0}^{\left[ {\frac{n}{2}} \right]} {\left( {\left( {\begin{array}{*{20}c}
   n  \\
   k  \\
\end{array}} \right)} \right.}  - \left. {\left( {\begin{array}{*{20}c}
   n  \\
   {k - 1}  \\
\end{array}} \right)} \right)^2  = \frac{1}{{n + 1}}\left( {\begin{array}{*{20}c}
   n  \\
   {2n}  \\
\end{array}} \right)
\]

I guess you meant $\frac{1}{n + 1} \cdot {{2n} \choose n}$. :)

Which 'french material'?  :huh:

Pierre.

Indeed, the assertion should be that

$\sum_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor}\left(\binom{n}{k}-\binom{n}{k-1}\right)^2=\frac{1}{n+1}\cdot\binom{2n}{n}$

holds for every integer $n\geq 0$.

The [i]proof[/i] is not difficult. We start with some purely technical steps to get a nicer upper limit for the sum than $\left\lfloor\frac{n}{2}\right\rfloor$:
We have to prove that $S=\frac{1}{n+1}\cdot\binom{2n}{n}$, where $S=\sum_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor}\left(\binom{n}{k}-\binom{n}{k-1}\right)^2$. Now, consider the sum $T=\sum_{k=0}^{n+1}\left(\binom{n}{k}-\binom{n}{k-1}\right)^2$. This sum T consists of
- all the terms of the sum S - they form the "left half" of the sum T, corresponding to values of k in the range $0\leq k\leq\left\lfloor\frac{n}{2}\right\rfloor$;
- all the terms of the sum S being repeated again in the "right half" of the sum T, corresponding to values of k in the range $\left\lceil\frac{n}{2}\right\rceil+1\leq k\leq n+1$ (in fact, for every k, we have

$\left(\binom{n}{k}-\binom{n}{k-1}\right)^2=\left(\binom{n}{n-k}-\binom{n}{n-\left(k-1\right)}\right)^2=\left(\binom{n}{\left(n+1-k\right)-1}-\binom{n}{n+1-k}\right)^2$
$=\left(\binom{n}{n+1-k}-\binom{n}{\left(n+1-k\right)-1}\right)^2$,

so that each term from the "left half" of the sum T has an equal "mirror" term in the "right half" of the sum);
- additionally, only if n is odd, the term $\left(\binom{n}{\left\lceil\frac{n}{2}\right\rceil}-\binom{n}{\left\lfloor\frac{n}{2}\right\rfloor}\right)^2$; but this term can be ignored, since it equals 0, because $\binom{n}{\left\lceil\frac{n}{2}\right\rceil}=\binom{n}{\left\lfloor\frac{n}{2}\right\rfloor}$ (in fact, $\left\lceil\frac{n}{2}\right\rceil=n-\left\lfloor\frac{n}{2}\right\rfloor$).
Hence, the sum T consists of all the terms of the sum S, each of them taken twice, and nothing more (or a zero term, which doesn't matter). So we have $T=2\cdot S$. Thus, instead of proving that $S=\frac{1}{n+1}\cdot\binom{2n}{n}$, it will be enough to show that $T=\frac{2}{n+1}\cdot\binom{2n}{n}$. Now,

$T=\sum_{k=0}^{n+1}\left(\binom{n}{k}-\binom{n}{k-1}\right)^2=\sum_{k=0}^{n+1}\left(\binom{n}{k}^2+\binom{n}{k-1}^2-2\cdot\binom{n}{k}\cdot\binom{n}{k-1}\right)$
$=\sum_{k=0}^{n+1}\binom{n}{k}^2+\sum_{k=0}^{n+1}\binom{n}{k-1}^2-2\sum_{k=0}^{n+1}\binom{n}{k}\cdot\binom{n}{k-1}$.

Since $\sum_{k=0}^{n+1}\binom{n}{k}^2=\sum_{k=0}^{n+1}\binom{n}{k-1}^2=\sum_{k=0}^n\binom{n}{k}^2$ (in fact, these sums differ only in terms like $\binom{n}{-1}^2$ and $\binom{n}{n+1}^2$ which are zero) and $\sum_{k=0}^{n+1}\binom{n}{k}\cdot\binom{n}{k-1}=\sum_{k=0}^{n}\binom{n}{k}\cdot\binom{n}{k-1}$ (since these sums only differ in the zero term $\binom{n}{n+1}\cdot\binom{n}{n}$), this becomes

$T=\sum_{k=0}^n\binom{n}{k}^2+\sum_{k=0}^n\binom{n}{k}^2-2\sum_{k=0}^n\binom{n}{k}\cdot\binom{n}{k-1}$
$=2\left(\sum_{k=0}^n\binom{n}{k}^2-\sum_{k=0}^n\binom{n}{k}\cdot\binom{n}{k-1}\right)$
$=2\left(\sum_{k=0}^n\binom{n}{k}\cdot\binom{n}{k}-\sum_{k=0}^n\binom{n}{k}\cdot\binom{n}{n-\left(k-1\right)}\right)$
$=2\left(\sum_{k=0}^n\binom{n}{k}\cdot\binom{n}{n-k}-\sum_{k=0}^n\binom{n}{k}\cdot\binom{n}{n+1-k}\right)$.

Now, by the Vandermonde identity,

$\sum_{k=0}^n\binom{n}{k}\cdot\binom{n}{n-k}=\binom{n+n}{n}=\binom{2n}{n}$;
$\sum_{k=0}^n\binom{n}{k}\cdot\binom{n}{n+1-k}=\binom{n+n}{n+1}=\binom{2n}{n+1}=\frac{\left(2n\right)!}{\left(n+1\right)!\cdot\left(n-1\right)!}$
$=\frac{\left(2n\right)!}{n!\cdot\left(n+1\right)\cdot\frac{n!}{n}}=\frac{n}{n+1}\cdot\frac{\left(2n\right)!}{n!\cdot n!}=\frac{n}{n+1}\cdot\binom{2n}{n}$,

so that

$T=2\left(\binom{2n}{n}-\frac{n}{n+1}\cdot\binom{2n}{n}\right)=\frac{2}{n+1}\cdot\binom{2n}{n}$,

qed.

  Darij


How many of you have participated in this before?

I'm in AP Chemistry now, so I'm interested in this.

if you want more info on the olympiad, see this site: http://membership.acs.org/W/WashDC/olympiad.html
or, you can try searching for the USNCO on the internet. i've never participated in this olympiad, so, this is as much as i know about it.

There's also some information in the Chemistry forum in AoPS too, at [url=http://artofproblemsolving.com/Forum/index.php?f=330&sid=bbc1cc8f2dcf7a3e7e0dc05c61639647]artofproblemsolving.com/Forum/index.php?f=330&sid=bbc1cc8f2dcf7a3e7e0dc05c61639647[/url]

I went to an Olympiad this April......The result was disgusting. One point below the cut-off!


modular arithmetic
LaTeX
logarithms
number theory proposed
number theory

$ 2^y \pmod n = b$

Let us call $ m$ the least common power $ 2$ of $ (n,b)$ and $ r_1, r_2$ the remains.

$ n=2^m + r_1$
$ b=2^m + r_2$

Example

$ 2^y \pmod {91} = 74$

$ n=91$
$ b=74$

$ m=64=2^6$

$ n=2^6 + 27$
$ b=2^6 + 10$

$ 2^y \pmod {91} = 74$ is equivalent for some integer $ k$ to $ 2^y = kn+b$

So we can write

$ 2^y = k(2^m + r_1) + 2^m + r_2  = 2^m\cdot(k+1) + k*r_1 + r_2$

then $ k\cdot r_1 + r_2$ has to be equal to $ 2^m$ so the equation can hold

we have then to solve

$ k\cdot r_1 + r_2 = 2^m$

$ k$ is unknown

$ k= (2^m - r_2)/r_1$

if we replace $ k$ by its value in the equation

$ 2^y = \frac{2^m - r_2}{r_1}\cdot(2^m + r_1) + 2^m + r_2$

By simplifying

$ 2^y = 2^m\cdot (k+2)$

It works only if $ k=2^t - 2$ (edit mistake)
It is easy to test $ k$ by computing $ m,r_1$ and $ r_2$
$ k= (2^m - r_2)/r_1$

Example: $ n=91$, $ m=6$, $ r_1=27$, $ r_2=10$

$ k= (64 - 10)/27 = 2$

$ 2^y \pmod {91}=74$
$ 2^y =(2^6)\cdot(2+2)=2^8$
$ 2^8 \pmod {91} =74$

If we have the first solution then it is done

$ 2^{92} \pmod {91} = 74$
$ 2^8 \pmod {91} = 74$

Any comment or any idea to improve my solution?
Thank you any useful comment.
I'm working on another idea to solve any case.
My question is : how many cases can be solved this way? few? lot?

[color=red][b]Moderator edit: $ \text{\LaTeX}$-ed[/b][/color]

If we use [u]negative numbers [/u]for $ r_1$ and $ r_2$ with $ 2^m$ bigger than $ n$ and $ b$ we can find the appropriate $ k$ such as :
$ 1- k +$ or $ - 2 = 2^t$ (edit correction)
$ 2- k$ integer ($ r_1$ dividing $ 2^m - r_2$)

But I do not know if we can find the solution for any $ n$.
If it is poosible than we can solve a lot of cases maybe all the cases.
Can we find quickly $ 2^m$ such as the 2 conditions above hold?

[b]It is solvable! I was wrong on my calculations, but my idea works! [/b]

It is more complicated but we can $ (m,k,r_1,r_2)$ such as $ 2^y$ will be equal to $ (2^m)\cdot f(k)$ where $ f(k)$ is equal to some $ 2^t$
$ f(k)$ is a linear congruence
We start from the equation

$ 2^y = kn + b$

we have to split $ n$ and $ b$ such as
$ n=2^s + r_1$
$ b= 2^l + r_2$

then we will have

$ 2^y = kn + b = k\cdot(2^s + r_1) + 2^l + r_2 = 2^s (k+ 2^{l-s}) + kr_1 + r_2$

we have to solve $ 2^s = kr_1 + r_2$

such as $ 2^y = 2^sf(k)$ where $ f(k)$ is power of 2.

I think that we can find a solution in less $ O(\log n)$

[color=red][b]Moderator edit: $ \text{\LaTeX}$-ed[/b][/color]

If someone can put my explanations in latex with more explicit formulas it will be helpful.

Thank you!


algebra
polynomial
abstract algebra
Ring Theory
superior algebra
superior algebra solved

If $\mathcal F$ is a finite field, then $\mathcal F [X]$ contains irreducible polynomials of every degree $>1$.

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=30473]here[/url] for a proof by counting.

Let a be a generator of the multiplicative group F-{0} and p the characteristic of F. The ring homomorphism Z_p[x]->F that sends a polynimial f(x) to f(a) is onto. The kernel is generated by a monic irreducible polynomial. From the isomorphism that is induced on the quotient, it follows that the degree of this polynomial is n, where p^n is the cardinality of F.

Sorry,
As soon as I posted the above I realised the question was for F not Z_p.

Let $|F|=p^m, m\ge 1$.
Let $n\ge 2$.
Denote by $F_d(X)=\displaystyle\prod_{deg(g)=d, g\in F[X] monic, irred}g(X)$.

We want to prove that $u(X) = X^{p^{mn}}-X = \displaystyle\prod_{d|n}F_d$. (*)

That is if $f$ is a monic irreducible polynomial over $F[X]$ of degree $d$, $f|u$ iff $d|n$.

'=>'
$f$ is an irreducible polynomial of degree $d$ so that $f|u$. We want $d|n$.

Let $a$ be a root of $f$. Then $[F[a]: F]=d\Rightarrow |F[a]|=p^{md}$,
so every element of $F[a]$ is a root of the polynomial $X^{p^{md}}-X$ (and these are all) . Now because $f|u$ it follows that $a^{p^{mn}}=a$.

Let $u_1a^{d-1}+u_2a^{d-2}+\ldots+u_{d-1}a+u_d$ be an element of $F[a]$.

We have that
$\left(u_1a^{d-1}+u_2a^{d-2}+\ldots+u_{d-1}a+u_d\right)^{p^{mn}}=u_1(a^{p^{mn}})^{d-1}+u_2(a^{p^{mn}})^{d-2}+\ldots+u_{d-1}a^{p^{mn}}+u_d$
$=u_1a^{d-1}+u_2a^{d-2}+\ldots+u_{d-1}a+u_d$. So every element of $F[a]$ verifies $x^{p^{mn}}-x=0$, so $x^{p^{md}}-x|x^{p^{mn}}-x$ so $md|mn\Rightarrow d|n$.

'<='
$f$ is an irreducible polynomial of degree $d$ so that $d|n$. We want $f|u$.

If $a$ is a root of $f$ then $|F[a]| = p^{md}$ so $a^{p^{md}}=a$. It follows that $f|X^{p^{md}}-X$. Because $d|n\Rightarrow$ $X^{p^{md}}-X|X^{p^{mn}}-X$
(it is easy to check this fact). Results that $f|u$.

We have proved (*). Denote by $A(k) = \left|\left\{g\in F[X]: g \textmd{ irreducible }, \deg g=k\right\}\right|$.

It is easy to check now that $\deg F_d=dA(d)$. From (*) we have that $(p^m)^n=\displaystyle\sum_{d|n}A(d)d$. By the Mobius inversion formula we get that
$A(n)=\displaystyle\frac1n\sum_{d|n}\mu(n/d)p^{md}$ which is obviously $\ge 1$ because it's a sum of powers of $p$ with coeffs in {-1/n,1/n}.
So we have proved that there are irreducible polynomials of any order over a finite field $F$.


Naperville Central HS 841
Naperville North HS 839
IL Math/Science Academy 785
Walter Payton College Prep 726
Glenbrook North HS 712
Adlai E. Stevenson HS 697
Evanston Twp. HS 689
Neuqua Valley HS 669
Oak Park River Forest 660
New Trier HS 653
Lyons Twp. HS 649
Whitney Young Magnet HS 640
Fremd HS 640
Hinsdale Central HS 639
Glenbrook South HS 618

wow, NCHS  :jump:

 :first:

Well, we're going to state.  I can believe I came in first in the regional for the freshman written with a 32 (after making stupid mistakes on 7 and 8!)  But we did manage to win our regional as a team, probably because our fresh/soph 2-person team (which I wasn't on) got triple everyone else's score.

I won't even talk about 8-person team... :blush:

Hmmm, is ICTM that fun of a competition as ARML or MATHCOUNTS?

1.  MathCounts is not "fun". Fun usually refers to creativity and enjoyability of problems.
2. ARML is fun.
3. ICTM is not fun, unless neither North nor IMSA gets 1st.

[quote="avec_une_h"]But we did manage to win our regional as a team, probably because our fresh/soph 2-person team (which I wasn't on) got triple everyone else's score.[/quote]
Even though we only got half of the 2 person questions right...

Wooo Cary-Grove.

We won our regionals. I came in 5 minutes late to my written competition because of orals, was all "F these problems, they suck,"  and still managed to get the second-highest score with 5 stupid mistakes. In two-person I more-than-doubled second place's score.

Our regional is trivial :D


induction
extremal principle
real analysis
real analysis unsolved

On the plane given n points $A_1,A_2,...,A_n$ s.t:
1)all of them are not collinear
2)the line passing through $A_i$ and $A_j$ passes some $A_k$ $\forall i,j \text{ and } i\ne j$
Show that such points are infinitely many.(n→∞)

to view all the Some problems from the Uzbekistan universities olympiads visit
 [url]http://www.mathlinks.ro/Forum/topic-81099.html[/url]

There's a well known problem (Sylvester): given a finite set of points such that the line passing through any two of them contains at least one other point of the set then all the points lie on a line.
The shortest proof uses the extremal principle.
From this, your problem follows easily.

What do you call the "extremal principle"?

I have slution.But it is very long.(The solution of Sylvesters problem)

Although I still don't know what was meant by the "extremal principle", I have found a straightforward solution to this problem (it's really obvious):

[color=blue][b]Theorem.[/b] Let $\left\{ A_{1},A_{2},\ldots,A_{n}\right\}$, where $n \in \mathbb N^\ast$, be a finite set of points such that any line passing through two of the points contains at least one other point from the set.

Then, all the points lie on a line.[/color]

(© Darij)

[i]Solution.[/i] We will prove the statement of the problem by induction over $n$.

For $n=1,2,3$, it is trivial. The condition given is actually the same as the conclusion.

Now let's suppose that we have proved it for $1,2,3,4,\ldots,n-1$.

WRONG :blush: [color=red]We are given the set $\left\{ A_{1},A_{2},\ldots,A_{n}\right\}$. Applying the statement of the theorem for $n-1$ and $\left\{ A_{1},A_{2},\ldots,A_{n-1}\right\}$, we see that $A_{1},A_{2},\ldots,A_{n-1}$ all lie on a line. Doing the same thing for $n=3$ and $A_{1},A_{n-1},A_{n}$, we see that $A_{1},A_{n-1},A_{n}$ are on the same  line.[/color]

Thus, $A_{1},A_{2},\ldots,A_{n}$ are collinear and the proof of the induction step is over.

$\blacksquare$

Now, the problem given at the "Uzbekistan University Olympiad" is just the contrapositive of the above theorem, so we're done.

[quote="perfect_radio"]Although I still don't know what was meant by the "extremal principle", I have found a straightforward solution to this problem (it's really obvious):

[color=blue][b]Theorem.[/b] Let $\left\{ A_{1},A_{2},\ldots,A_{n}\right\}$, where $n \in \mathbb N^\ast$, be a finite set of points such that any line passing through two of the points contains at least one other point from the set.

Then, all the points lie on a line.[/color]

(© Darij)

[i]Solution.[/i] We will prove the statement of the problem by induction over $n$.

For $n=1,2,3$, it is trivial. The condition given is actually the same as the conclusion.

Now let's suppose that we have proved it for $1,2,3,4,\ldots,n-1$.

We are given the set $\left\{ A_{1},A_{2},\ldots,A_{n}\right\}$. Applying the statement of the theorem for $n-1$ and $\left\{ A_{1},A_{2},\ldots,A_{n-1}\right\}$, we see that $A_{1},A_{2},\ldots,A_{n-1}$ all lie on a line. Doing the same thing for $n=3$ and $A_{1},A_{n-1},A_{n}$, we see that $A_{1},A_{n-1},A_{n}$ are on the same  line.

Thus, $A_{1},A_{2},\ldots,A_{n}$ are collinear and the proof of the induction step is over.

$\blacksquare$

[/quote]
I am very much tempted to use this: :rotfl:

I see that nobody else care, so I have to bring the bad news: That proof is incorrect!!!!:rotfl: When you are watching the set $\{A_{1}, A_{2}, ... ,A_{n-1}\}$ they would be all collinear if on every line determined by some two points from this set, the third point is also from this set. It doesn't have to be that way, on line $A_{1}A_{2}$ third point can be $A_{n}$ which isn't in this set, so your proof is not good, here is the correct one:

Suppose opposite, that they aren't all on a line, and look at line $A_{1}A_{2}A_{3}$ (point $A_{2}$ is between points $A_{1}$ and $A_{3}$). Then it has to be some point which isn't on that line and look at one of them which distance from the line is the smallest. WLOG let that be $A_{4}$ and its projection on line $A_{1}A_{2}A_{3}$ is $A_{4}'$. At least two of points $A_{1},A_{2},A_{3}$ must be on the same side of the line with respect to the $A_{4}'$. Suppose that those points are $A_{1}$ and $A_{2}$ and that $A_{2}\in [A_{1}A_{4}']$. Then distance from $A_{2}$ to line $A_{1}A_{4}$ is smaller that $A_{4}A_{4}'$, so contradiction and the end of the proof!

Difference between the case with finite and infinite number of points is when you have finite number of points you can choose the smallest distance between some point and a line, but when you have infinite points, you can not!

Bye!


inequalities
trigonometry
geometry
inequalities unsolved

If $ a,b,c$ are sides of a triangle such that $ abc\equal{}a\plus{}b\plus{}c,$ show that
\[ (ab\minus{}2)(bc\minus{}2)(ca\minus{}2)\ >\ \frac4{(a\plus{}b\plus{}c)^2}\]

Notice that :

$ a\plus{}b\minus{}c\equal{}a\plus{}b\plus{}c\minus{}2c\equal{}abc\minus{}2c\equal{}c(ab\minus{}2)$

Therefore our inequality may be written as:

$ c(ab\minus{}2).b(ac\minus{}2).c(ab\minus{}2) > \frac{4}{a\plus{}b\plus{}c}$
$ \Leftrightarrow (a\plus{}b\plus{}c)(a\plus{}b\minus{}c)(b\plus{}c\minus{}a)(c\plus{}a\minus{}b) > 4$
$ \Leftrightarrow 16S^2 >4$
$ \Leftrightarrow abc >2R$
$ \Leftrightarrow a\plus{}b\plus{}c >2R$

which is obvious  :)

$ ab \equal{} \frac {a \plus{} b \plus{} c}{c}$

$ \implies$  $ (a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(a \plus{} c \minus{} b)(a \plus{} b \plus{} c) > 4$

$ \sqrt {(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(a \plus{} c \minus{} b)(a \plus{} b \plus{} c)} \equal{} 4A(ABC)$

$ a \equal{} 2R.sinA,b \equal{} 2R.sinB,c \equal{} 2R.sinC$

$ \implies$  $ 8R^{3}(sinAsinBsinC) \equal{} 2R(sinA \plus{} sinB \plus{} sinC)$  $ \implies$

$ 4R^{2}.(sinAsinBsinC) \equal{} (sinA \plus{} sinB \plus{} sinC)$

$ A(ABC) \equal{} 2R^{2}sinAsinBsinC$

$ \implies$  $ sinA \plus{} sinB \plus{} sinC > 1$

$ a \plus{} b \plus{} c > 2R$,

[quote="cnyd"]$ sinA \plus{} sinB \plus{} sinC > 1$ ? is it obvious?[/quote]

oh sorry it's not obvious I just see it evident in any triangle But anyway i'll prove it :
in any triangle there exists two angle  which are less than $ \frac {\pi}{2}$

Let's suppose that $ B$ and $ C$ are these angles  and $ B>C$

Then$ \frac {B \minus{} C}{2} < \frac {\pi}{3} \implies cos(\frac {B \minus{} C}{2}) > \frac {1}{2}\implies 2cosA.cos\frac {B \minus{} C}{2} > cosA > cosA^2$

and hence $ sinA \plus{} 2cosA.cos\frac {B \minus{} C}{2} > sin^2A \plus{} cos^2A \equal{} 1 \Leftrightarrow sinA \plus{} sinB \plus{} sinC > 1$

[quote="Abdek"][quote="cnyd"]$ sinA \plus{} sinB \plus{} sinC > 1$ ? is it obvious?[/quote]

oh sorry it's not obvious I just see it evident in any triangle But anyway i'll prove it :
in any triangle there exists two angle  which are less than $ \frac {\pi}{2}$

Let's suppose that $ B$ and $ C$ are these angles  and $ B > C$

Then$ \frac {B \minus{} C}{2} < \frac {\pi}{3} \implies cos(\frac {B \minus{} C}{2}) > \frac {1}{2}\implies 2cosA.cos\frac {B \minus{} C}{2} > cosA > cosA^2$

and hence $ sinA \plus{} 2cosA.cos\frac {B \minus{} C}{2} > sin^2A \plus{} cos^2A \equal{} 1 \Leftrightarrow sinA \plus{} sinB \plus{} sinC > 1$[/quote]
I don't understand the last equivalence.. But if $ A \equal{} \pi,B\equal{}C\equal{}0$ then $ \sin A \plus{} \sin B \plus{} \sin C \equal{} 0$, so that is probably the problem.
My counterexample once again: $ a\equal{}b\equal{}\sqrt{2},c\equal{}2\sqrt{2}$

[quote="Abdek"]But if $ a,b,c$ are the side of triangle then $ a \plus{} b > c$ and in your contre example we have the equality  :)  which in not possible because The $ LHS \equal{} 0 \implies \frac {4}{a \plus{} b \plus{} c} \equal{} 0$ which is absurd

in The last equivalence we have
 $ sinA \plus{} sinB \plus{} sinC \equal{} sinA \plus{} 2sin(\frac {B \plus{} C}{2})cos(\frac {B \minus{} C}{2}) \equal{} sinA \plus{} 2cos(A)cos(\frac {B \minus{} C}{2})$[/quote]

I think you are wrong because Mathias_DK meant $ (a,b,c)\rightarrow (\sqrt{2},\sqrt{2},2\sqrt{2})$

[quote="Abdek"]But if $ a,b,c$ are the side of triangle then $ a \plus{} b > c$ and in your contre example we have the equality    which in not possible because The $ LHS \equal{} 0 \implies \frac {4}{a \plus{} b \plus{} c} < 0$ which is absurd :)

in The last equivalence we have
 $ sinA \plus{} sinB \plus{} sinC \equal{} sinA \plus{} 2sin(\frac {B \plus{} C}{2})cos(\frac {B \minus{} C}{2}) \equal{} sinA \plus{} 2cos(A)cos(\frac {B \minus{} C}{2})$[/quote]
I find $ \sin \left ( \frac {B \plus{} C}{2} \right ) \equal{} \cos \left ( \frac {A}{2} \right )$ and not $ \sin \left ( \frac {B \plus{} C}{2} \right ) \equal{} \cos \left ( A \right )$

Oops  !! sorry for my stupid  mistakes  :oops:

But we can use in the last proof $ cos\frac {A}{2} > cosA > cos^2A$

and I think that the inequality is true for any acute triangle

[quote="Abdek"]Notice that :

$ a \plus{} b \minus{} c \equal{} a \plus{} b \plus{} c \minus{} 2c \equal{} abc \minus{} 2c \equal{} c(ab \minus{} 2)$

Therefore our inequality may be written as:

$ c(ab \minus{} 2).b(ac \minus{} 2).c(ab \minus{} 2) > \frac {4}{a \plus{} b \plus{} c}$
$ \Leftrightarrow (a \plus{} b \plus{} c)(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b) > 4$
$ \Leftrightarrow 16S^2 > 4$
$ \Leftrightarrow abc > 2R$
$ \Leftrightarrow a \plus{} b \plus{} c > 2R$

which is obvious  :)[/quote]
What are $ S$ and $ R\,?$ :maybe:

$ S$ is the area of the Triangle $ ABC$ and $ R$ is the circum-radius of this Triangle  :)

I made a mistake. The inequality should be
\[ (ab \minus{} 2)(bc \minus{} 2)(ca \minus{} 2)\ > \ \frac{16}{(a \plus{} b \plus{} c)^2}\]
Sorry. :(

$ (a \plus{} b \minus{} c)(b \plus{} a \minus{} c)(c \plus{} a \minus{} b)(a \plus{} b \plus{} c) > 16$

$ \sqrt {(a \plus{} b \minus{} c)(b \plus{} a \minus{} c)(c \plus{} a \minus{} b)(a \plus{} b \plus{} c)} \equal{} 4A(ABC)$

$ A(ABC) > 1$

$ a \equal{} 2RsinA,b \equal{} 2RsinB,c \equal{} 2RsinC$  $ \implies$  $ 2R(sinA \plus{} sinB \plus{} sinC) \equal{} 8R^{3}(sinAsinBsinC)$

$ \implies$  $ 4R^{2}(sinAsinBsinC) \equal{} sinA \plus{} sinB \plus{} sinC$

$ A(ABC) \equal{} 2R^{2}sinAsinBsinC \equal{} \frac {sinA \plus{} sinB \plus{} sinC}{2}$

if $ A(ABC) > 1$  $ \implies$  $ sinA \plus{} sinB \plus{} sinC > 2$  but i think it is not obvious

[quote="itiselizabeth"]I made a mistake. The inequality should be
\[ (ab \minus{} 2)(bc \minus{} 2)(ca \minus{} 2)\ > \ \frac {16}{(a \plus{} b \plus{} c)^2}\]
Sorry. :([/quote]
Once again: $ a\equal{}b\equal{}\sqrt{2},c\equal{}2\sqrt{2}$ is a counterexample. You need some other restriction, for instance that the triangle is acute. In that case we prove it the following way: It is equivalent to $ \sin A \plus{} \sin B \plus{} \sin C > 2$. $ x \to \sin x$ is concave on $ \left [ 0; \frac{\pi}{2} \right ]$ and since $ \left ( \frac{\pi}{2}, \frac{\pi}{2}, 0 \right ) \succ (A,B,C)$ we know by Karamata that $ \sin A \plus{} \sin B \plus{} \sin C > \sin \frac{\pi}{2} \plus{} \sin \frac{\pi}{2} \plus{} \sin 0 \equal{} 2$.

[quote="ValentineA"][quote="Mathias_DK"]Once again: $ a \equal{} b \equal{} \sqrt {2},c \equal{} 2\sqrt {2}$ is a counterexample.[/quote]
No, it is not. The lengths $ a,b,c$ have to form a triangle. $ \sqrt2,$ $ \sqrt2,$ $ 2\sqrt2$ are not the sides of a nondegenerate triangle.[/quote]

You don't understand . He means that ($ a,b,c$) are tending to ($ \sqrt{2},\sqrt{2},2\sqrt{2}$) . If they arbitraly close to the triplet ($ \sqrt{2},\sqrt{2},2\sqrt{2}$) , then both inequalities are wrong !

[quote="itiselizabeth"]I also forgot to specify that the triangle is an acute-angled triangle. Maybe this will eliminate Mathias’s scenario. :huh:[/quote]
[quote="Mathias_DK"]
Once again: $ a \equal{} b \equal{} \sqrt {2},c \equal{} 2\sqrt {2}$ is a counterexample. [b]You need some other restriction, for instance that the triangle is acute. [/b] In that case we prove it the following way: It is equivalent to $ \sin A \plus{} \sin B \plus{} \sin C > 2$. $ x \to \sin x$ is concave on $ \left [ 0; \frac {\pi}{2} \right ]$ and since $ \left ( \frac {\pi}{2}, \frac {\pi}{2}, 0 \right ) \succ (A,B,C)$ we know by Karamata that $ \sin A \plus{} \sin B \plus{} \sin C > \sin \frac {\pi}{2} \plus{} \sin \frac {\pi}{2} \plus{} \sin 0 \equal{} 2$.[/quote]

It certainly will :)


A sequence $(a_{n})$ of positive integers is given by $a_{n}=[n+\sqrt{n}+\frac{1}{2}]$. Find all of positive integers which belong to the sequence.

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=sequence&t=65732

the other way around and general case


integration
calculus
logarithms
calculus computations

How to show that $\frac{1}{n!}\int_{0}^{\infty}ln(\frac{x}{n}) x^{n}e^{-x}dx$ approaches to zero as n goes to ${\infty}$? It is possible to calculate integral directly, but thats pretty ugly.
Thanks  :).

I would do this:

Let $a_{n}$ be your sequence. Then,

$a_{n}=\frac{1}{n!}\int_{0}^{\infty}x^{n}\ln x e^{-x}dx-\frac{1}{n!}\ln n \Gamma(n+1)$

$=\frac{1}{n!}b_{n}-\ln n$, where

$b_{n}=\int_{0}^{\infty}x^{n}\ln x e^{-x}dx$.

Observe that:

$b_{n}=n!\left(-\gamma+1+\frac{1}{2}+\cdots+\frac{1}{n}\right)$.

It follows that

$a_{n}=1+\frac{1}{2}+\cdots+\frac{1}{n}-\ln n-\gamma\rightarrow 0$.


Let $ ABC$ acute triangle with $ AB>AC$. Let $ D$ on side $ BC$ such that $ AD$ perpendicular to $ BC$ and $ E,F$ on sides $ CA,AB$ respectively such that $ \angle ADE\equal{}\angle ADF$. Denote $ X$ is a point such that $ DX$ parallel to $ EF$. Let $ DX$ meet $ BE,CA$ at $ P,Q$ respectively
Prove that $ DP\equal{}DQ$

first we'll prove that $ BE,AD,CF$ are concurrent...
in order to prove this we'll prove the following lemma:

[b][u]lemma 1:[/u]let $ D$ be a point on side $ BC$ of $ \triangle ABC$ such that $ AD\perp BC$ and let $ P$ be an arbitary point on $ AD$ and let $ BP,CP$ intersect $ AC,AB$ at $ Y,X$ respectively,prove that $ \angle ADX\equal{}\angle ADY$.[/b]

[u]proof[/u]:[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=938415#938415[/url]

----------------------------------------

so according to the lemma we get that $ AD,BE,CF$ are concurrent,now let $ EF$ intersect $ BC$ at point $ K$...

now we know that point $ D$ beeing the midpoint of segment $ PQ$ is equivalent to $ E(F,P,D,Q)\equal{}\minus{}1$...

but $ EF,EP,ED,EQ$ intersect side $ BC$ at $ K,B,D,C$ so for proving $ E(F,P,D,Q)\equal{}\minus{}1$ its sufficient to show that $ (BC,DK)\equal{}\minus{}1$

[b][u]lemma 2:[/u]let $ D,E,F$ be three points on sides $ BC,CA,AB$ of triangle $ \triangle ABC$ such that $ AD,BE,CF$ are concurrent at point $ P$,let $ EF$ intersect $ BC,AP$ at $ K,S$ respectively prove that $ (BC,DK)\equal{}\minus{}1$[/b]

[u]proof[/u]:

we know that $ A(BC,DK)\equal{}A(FE,SK)\equal{}P(FE,SK)\equal{}P(CB,DK)$

$ \Rightarrow (BC,DK)\equal{}(CB,DK)\Rightarrow (BC,DK)\equal{}\minus{}1$

as wanted...


function
inequalities
inequalities unsolved

Here is something that I think it is absolutely difficult. My deepest respect for the one who will handle this question:
  Prove that if $ a_1,...,a_n$ are positive numbers with product 1 then
  $\sqrt {(n^2-1)a_1^2+1}+...+\sqrt{(n^2-1)a_n^2+1}\leq n(a_1+...+a_n)$ . Even the case n=3 was finally settled down by silversunboy, but the solution was extremely long and difficult, so... have fun, as Vess use to say.
 :D

the function $sqrt[{(n^2-1)x^2+1}]$ has just one point of inflection. Is it true that it is enough to prove the inequality just for the case where $n-1$ of the variables are on the convex part of the graph and just $1$ on the concave part?

Oh, sorry, too bad. I can also prove your inequalities, Vasc. I found them when creating this problem for n=3.

Can you please give me a hint to prove these inequalities, Harazi?

Thank you very much.

[quote="Vasc"]The inequality is not true for any n. I can prove that

$\displaystyle \sqrt {a_1^2+1}+...+\sqrt{a_n^2+1}\leq \sqrt{2}(a_1+...+a_n)$,

and

$\displaystyle \sqrt {2a_1^2+1}+...+\sqrt{2a_n^2+1}\leq \sqrt{3}(a_1+...+a_n)$,

and

$\displaystyle \sqrt {3a_1^2+1}+...+\sqrt{3a_n^2+1}\leq 2(a_1+...+a_n)$.[/quote]

It's very new for me.

Where is your General Problem,VASC?Is it:

$\displaystyle \sqrt {ka_1^2+1}+...+\sqrt{ka_n^2+1}\leq \sqrt{1+k}(a_1+...+a_n)$.

For $k\le n$?

Yes.
I proved it for $0<k\leq 3+2\sqrt{2}$. ;)

I had prove it for some constant $k$  :)

But I think the inequality still true for k is a function of $n$ ,as Harazi say.
And If that,the problem is more difficult!


risk is a very strategic game in my opinion but i have not found a way to play it online with other people and when i play on my pc against a cpu theres no strategy because the cpu sucks so if anyone likes risk and reads this post please post some strategies and websites to verse people online if any

um..
don't have any strategies, but here's a website
u have to make an account though
it free unless u want premium membership, which let's u play more games at one time and something...but u have to pay for that

[url]http://www.conquerclub.com/public.php?mode=home[/url]

I don't have any advice for it but it is a great game. Also, Settlers of Cataan is an awesome game a little like risk, but I think more interesting.

I don't have any advice or anything... but I like Risk. Like dynamo729 said, Settlers of Catan is kind of like Risk, but more interesting. If you haven't played it before I would recommend it.

hmmm, thanks for the site!  i used to play risk a lot with friends, but we haven't played in awhile, now I can play online!

What are your usernames on CC, I want to play you guys...
My username is chenhsi

my username is *surprise!* vallon22
what symbol thingy are you?

rank? private

do you have a game open?

we can play a quick game...

umm..
no i don't have an open game right now
i'm a cook, (though i can't cook for my life!)
oh well...

you've won [b]1[/b] out of twenty games???

i know, i suck
but in my defense, the people i played were pretty good
and i have absolutly no skills in risk

I suck a lot as well, but the first 6 or so games I won my team always won because of me, like I killed most of the opponents...

I got unhappy when this record was broken, even though we won ...

I love Risk, but sadly I suck at it.  :(


algebra
polynomial
superior algebra
superior algebra solved

Let $F$ be a field and $a$, $b$ are two algebraic of degree 3.
What are the possible values for $[F(a,b):F]$?

It is clear that two possible values are 3 and 9.
Can it be 6?

Yes it can, take for example $a$ and $b$ as distinct roots of the same polynomial, where this polynomial doesn't split by just adjuncting $a$ (e.g. $x^3-2$ over $\mathbb{Q}$)


calculus
integration
trigonometry
real analysis
function
calculus computations

Compuet the integral
\[\int_{-\infty}^{\infty}\frac{\sin{[5(t-u)]}\sin{(6u)}}{u(t-u)}du \]

with some sort of fourier transform, and PLEASE explain your method.
($t\in \mathbb{R}$)

Several facts to put together:

[Some details of these are sensitive to the question of where you put the $2\pi$ in defining the Fourier transform. I usually use $\widehat{f}(s)=\int_{-\infty}^{\infty}f(x)e^{-2\pi isx}\,dx.$]

1. Define convolution as $f*g(x)=\int_{-\infty}^{\infty}f(x-u)g(u)\,du.$

Then $\widehat{f*g}(s)=\widehat{f}(s)\widehat{f}(s).$

We also have that if $h(x)=f(x)g(x)$ then $\widehat{h}(s)=\widehat{f}*\widehat{g}(s).$

2. The Fourier transform of a "box" function is a "sinc" function.

If $\text{box}(x)=\begin{cases}1&\text{ if }|x|<\frac12\\0&\text{ otherwise}\end{cases}$ then

$\widehat{\text{box}}(s)=\frac{\sin \pi s}{\pi s}= \text{sinc}(s)$ with the understanding that $\text{sinc}$ is continuous at zero - in fact, an entire function.

Since these are real, even functions, it is also true that $\widehat{\text{sinc}}=\text{box}.$

3. For $a>0,$ if $g(x)=f(ax),$ then $\widehat{g}(s)=\frac1{a}\widehat{f}\left(\frac{x}{a}\right).$

---

So we have the convolution of two scaled sinc functions. The Fourier transform of this convolution is the product of appropriately scaled box functions. It should be easy enough to evaluate the product of box functions - we get a constant multiple of the one with the smaller box. Hence this convolution is itself a multiple of a sinc function.

Note, however, that we are treading on treacherous ground. $\text{sinc}\not\in L^{1}$ and the integral in this problem is not absolutely convergent. What I'm suggesting is essentially a formal or distributional computation; investigating convergence and under what terms is a different question.

Thanks for your reply.
So how do I continue? please help.
Is this the way:
\[\mathcal{F}\left\{\frac{1}{30}\int_{-\infty}^{\infty}\frac{\sin{[5(t-u)]}\sin{(6u)}}{5\cdot6\cdot u(t-u)}du\right\}\]
is
\[\mathcal{F}\left\{\mbox{sinc}(5u)*\mbox{sinc}(6u)\right\}\]
and conv.Thm gives this as
\[\mathcal{F}\left\{\mbox{sinc}(5u)\right\}\cdot\mathcal{F}\left\{\mbox{sinc}(6u)\right\}=\frac{1}{30\cdot2\pi}\text{box}(\frac{\omega}{2\pi 5})\cdot\text{box}(\frac{\omega}{2\pi6}) \]
then we simplify the product
and take the inverse transform again to obtain the original integral $I$.
But I dont know how to continue from here.[color=red] Help please[/color].
The final answer is $\frac{\pi\sin{(5t)}}{t}$ for $t$ nonzero, for $t=0$ it is $5\pi$


hello, we have $ \frac{xy-y^2}{xy-x^2}=\frac{y(x-y)}{x(y-x)}=\frac{y(x-y)}{-x(x-y)}=-\frac{y}{x}$ for $ x %Error. "neqy" is a bad command.
$.
Sonnhard.

this is a flawed question, because i put y/-x, and it was considered wrong.

That expression IS wrong, as you put the negative sign in an inappropriate place.

To be more specific, the MathCounts form of answers says that

A "common fraction" is to be considered a fraction in the form $ \frac{\pm a}{b}$, where $ a$ and $ b$ are natural numbers and $ \text{GCF}(a,b)\equal{}1$.

Actually, it says $ \pm \frac{a}{b}$, but $ \frac{\pm a}{b}$ is ok as well.


Use energy conservation arguments to verify that a spherical sound (from a point source) wave must be written on the form.

$ \delta P(r,t) \equal{} \frac {\delta P_0}{r} cos(kr \minus{} wt)$

i  know of a pretty general proof using assumptions that in someway can be thought of as energy conservation...but what exacty is that u want

The problem is just like that.Could you please post that proof?

[quote="pardesi"]Actually we will use the very definition of spherical waves here...by definition they they can be defined only using the radius vector with respect to a certaing point that is the origin if the waves
so if $ \Psi(\vec{r}) \equal{} \Psi(r,\theta,\phi)$ represents the wavefunction then we can reduce that to $ \Psi(\vec{r}) \equal{} \Psi(r)$
hence we have $ \nabla^{2} \Psi(r) \equal{} \frac {\delta(r^{2} \frac {\delta \Psi}{\delta r})}{r^{2} \delta r} \equal{} \frac {\delta^{2}(r \Psi)}{r \delta r^{2}}$
then the differential wave-equation can be written as
$ \frac {\delta^{2}(r\Psi)}{r \delta r^{2}} \equal{} \frac {\delta^{2}(r \Psi)}{v^{2} \delta t^{2}}$
 which follows from the wave equation whcih in turn 'orginates' from the very fact that the waves move losslessly that is without losing energy
now this is nothing but a one diminsenional wave equation with wavefunction $ r \Psi$
so the solution is $ r \Psi(r,t) \equal{} f(r \minus{} vt)$
so $ \Psi(r,t) \equal{} \frac {f(r \minus{} vt)}{r}$
and if the wave is harmonic then we have $ f(r \minus{} vt) \equal{} A \cos(kr \minus{} \omega t)$
for some $ \omega$
this is a general derivation u can use the same thing for soundwaves[/quote]

$ \nabla^{2} \Psi(r) \equal{} \frac {\delta(r^{2} \frac {\delta \Psi}{\delta r})}{r^{2} \delta r} \equal{} \frac {\delta^{2}(r \Psi)}{r \delta r^{2}}$  -- Can you explain that part?

[quote="GravitoN"]

$ \nabla^{2} \Psi(r) \equal{} \frac {\delta(r^{2} \frac {\delta \Psi}{\delta r})}{r^{2} \delta r} \equal{} \frac {\delta^{2}(r \Psi)}{r \delta r^{2}}$  -- Can you explain that part?[/quote]
just compare the laplacians in spherical  form and that derived from x,y,z coordinates after transforming them to spherical coordinates...

Solution:

$ P\equal{}I*4\pi r^2$

$ I\minus{}\minus{}A^2$   (Intensity is proportional with square of Amplitude)

$ I\equal{}K*A^2$                K=constant

$ A\equal{}\frac{1}{r} \sqrt{\frac{P}{4\pi K}}\equal{}\frac{C}{r}$                 C is also constant.

$ \phi\equal{}A cos(kr\minus{}wt)\equal{}\frac{C}{r} cos(kr\minus{}wt)$


function
algebra
functional equation
algebra proposed

i have got a long solution to this functional equation (it's been long since i last tried olympiad stuff,so a bit rusted :) )
find all function from N to N  such that
f(m+f(n))=n+f(m+k) where m,n are natural numbers and k is a fixed natural numbers.
hope sum1 finds sumthing shorter.
best regards

Nice to have you back, Galois! :)

One question: does $\mathbb N$ contain zero? I'm asking because there are different conventions and notations in different countries..

Here is my solution:

\[f(m+f(n))=n+f(m+k)\quad (1)\]

For $m\to f(m)$ we get:

\[f(f(m)+f(n))=n+f(k+f(m))\overset{\text{by }(1)}=n+m+f(2k)\quad (2)\]

Suppose that $f(a)=f(b)$. By (2) we have:

\[a+b+f(2k)=f(f(a)+f(b))=f(f(a)+f(a))=a+a+f(2k)\]
hence
\[f(a)=f(b)\iff a=b\quad (3)\]

Now:

\[f(f(n)+f(n))=n+n+f(2k)=n+1+n-1+f(2k)=f(f(n+1)+f(n-1))\]

and by (3) we have

\[f(n)+f(n)=f(n+1)+f(n-1)\iff f(n+1)-f(n)=f(n)-f(n-1)\]

Now follows that $f$ is linear, $f(n)=pn+q$.

Putting this into (1) gives:

\[\begin{array}{rl}
f(m+f(n))=n+f(m+k)&\iff f(m+pn+q)=n+p(m+k)+q\iff\\
&\iff p(m+pn+q)+q=n+pm+pk+q\iff\\
&\iff p^2n+pq=n+pk\iff\\
&\iff (p,q)=(1,k)\end{array}\]

The only solution is

\[f(n)=n+k\]


function
calculus
derivative
algebra unsolved
algebra

let $ {x\in [\frac{ \pi}{12},\frac{5 \pi}{12}]}$ , then find the maximum and the mininum value of $ {f(x)\equal{}(sinx\plus{}2)(cosx\plus{}2)}$.

[quote="CCMath1"]let $ {x\in [\frac {\pi}{12},\frac {5 \pi}{12}]}$ , then find the maximum and the mininum value of $ {f(x) \equal{} (sinx \plus{} 2)(cosx \plus{} 2)}$.[/quote]

$ f'(x)\equal{}(\cos(x)\minus{}\sin(x))(\sin(x)\plus{}\cos(x)\plus{}2)$ and its sign is sign of $ \cos(x)\minus{}\sin(x)$

So $ f(x)$ is increasing on $ [\frac {\pi}{12},\frac {3 \pi}{12}]$, maximum at $ \frac{\pi}4$ and decreasing on $ [\frac {3\pi}{12},\frac {5 \pi}{12}]$

So max value is $ f(\frac{\pi}4)\equal{}\frac92\plus{}2\sqrt 2$

And min value is $ \min(f(\frac{\pi}{12}),f(\frac{5\pi}{12})$ $ \equal{}\sqrt 6\plus{}\frac{17}4$

[quote="CCMath1"]Nice ！
But can we solve it without using derivative function？ :)[/quote]

We can say that $ f(x)\equal{}f(\frac{\pi}2\minus{}x)$ and so we only have to study $ f(x)$ over $ [\frac{\pi}{12},\frac{\pi}{4}]$

And $ f(x)$ may be written : $ f(x)\equal{}4\plus{}\frac 12\sin(2x)\plus{}2\sqrt 2\cos(x\minus{}\frac{\pi}4)$ and so is the sum of two increasing functions (over $ [\frac{\pi}{12},\frac{\pi}{4}]$)

And so min value is $ f(\frac{\pi}{12})$ and max value is $ f(\frac{\pi}{4})$

Hence the results

WONDERFUL！
Thanks a lot pco ， I have learned a lot


define $dig_{b}(x)=dig_{b}(a_{0}+a_{1}*b+...a_{n}*b^{n})=a_{0}+a_{1}+...a_{n}$
For any positive integer y, denote its divisors as $d_{1}, d_{2}, ... d_{p}$ (these include 1 and y)

Define $F_{b}(y)=dig_{b}(d_{1})+dig_{b}(d_{2})+...dig_{b}(d_{p})$

Show that for all b, the sequence $F_{b}(y), F_{b}(F_{b}(y))...$ converges.

This is number theory, not calculus. Please try to post threads in the right forum.

A counterexample: If $b=7$, the sequence starting at 2 cycles: $2\to 3\to 4\to 10\to 2$. There are plenty of others.


function
trigonometry
algebra
domain
analytic geometry
calculus
derivative

How would you find the turning points of $ f(x)\equal{}\sin(1\plus{}\sin x)$?

I'm guessing the relative extremes of the function where the domain would be the period.

[url=http://www.teacherschoice.com.au/Maths_Library/Calculus/stationary_points.htm]Turning Point[/url] - I don't know if this is the most mathematically "updated" term for the concept, but it is the one my textbook (circa 1942) uses.

Specifically, the problem asks for the $ x$ coordinate of the turning points P, Q, and R which are defined as the first 3 turning points of the function in the positive $ x$ axis.

After graphing it, [url=http://img406.imageshack.us/img406/3654/picture1yci.png]this picture[/url] shows the first desired turning point.

I found the $ x$ coordinates by finding what value of $ x$ makes $ 1 \plus{} \sin x \equal{} \frac {\pi}{2}$, because the first turning point occurs when $ \sin (1 \plus{} \sin (x)) \equal{} 1$, but this probably isn't the best way (especially because it requires looking at the graph first).

EDIT: In addition, my method gives me the $ x$ coordinate as $ x\equal{}0.6074$, whereas the graph shows $ x\equal{}0.6137$ - I'm not sure if we can discount the difference to be just rounding errors.

It can be attributed to the resolution of the graph, and to the fact that the derivative is changing continuously, and your eyes might not be able to judge perfectly what looks like "flat".

Your x-value is right, although it should actually round up to 0.6075.

I don't understand what you find problematic about your approach. It might help if you actually post your solution.


geometry
trapezoid
parallelogram
symmetry
geometric transformation
reflection
trigonometry

I have just found in my archive a nice problem posted by Mathmaniack and lost in the hacker attack. I will post a solution shortly.

Let ABCD be an isosceles trapezoid with AB = CD and AD || BC. Let E be a point on the extension of the segment BC beyound C. Let the segments AE and CD intersect at a point F. Assume that AB = EF and $\measuredangle BAE=2\cdot\measuredangle DAF$. Prove that $AD\cdot BC=CE^2$.

  Darij

And here is my solution:

Let the parallel to the line CD through the point E meet the line AD at a point T. Then, ET || CD. On the other hand, DT || CE (this is just another way to say AD || BC). Hence, the quadrilateral CDTE is a parallelogram, so that DT = CE and CD = ET. From CD = ET, AB = CD and AB = EF, it follows that ET = EF; hence, the triangle FET is isosceles, so that its base angle < ETF equals $\measuredangle ETF=\frac{180^{\circ}-\measuredangle FET}{2}$.

Since the quadrilateral ABCD is an isosceles trapezoid, < ADC = < DAB. Since ET || CD, we have < ATE = < ADC. Thus, < ATE = < DAB. In other words, < ATE = < TAB. Together with AT || BE (this is equivalent to AD || BC), this yields that the quadrilateral ABET is an isosceles trapezoid. The perpendicular bisector of its base AT must be the symmetry axis of this trapezoid.

Now, denote < DAF = x; since, $\measuredangle BAE=2\cdot\measuredangle DAF$, we then have < BAE = 2x. Consequently, < DAB = < DAF + < BAE = x + 2x = 3x. Thus, < ATE = < DAB = 3x. Hence, by the sum of angles in triangle AET, we get < AET = 180° - < EAT - < ATE = 180° - < DAF - < ATE = 180° - x - 3x = 180° - 4x; in other words, < FET = 180° - 4x, and thus 180° - < FET = 4x. Hence, $\measuredangle ETF=\frac{180^{\circ}-\measuredangle FET}{2}=\frac{4x}{2}=2x$. Hence, < FTA = < ATE - < ETF = 3x - 2x = x = < DAF = < FAT. Thus, the triangle AFT is isosceles, so that AF = TF, and the point F lies on the perpendicular bisector of the segment AT. But this perpendicular bisector, as we know, is the symmetry axis of the isosceles trapezoid ABET. Thus, the reflection with respect to this perpendicular bisector maps the point A to the point T and the point E to the point B; also, it must map the point F to itself (since F lies on this perpendicular bisector). Now, as the points A, E and F are collinear, their images under this reflection, namely the points T, B and F, must also be collinear, and thus, since AD || BC, Thales yields $\frac{AD}{DT}=\frac{CE}{BC}$. Since DT = CE, this becomes $\frac{AD}{CE}=\frac{CE}{BC}$, and thus $AD\cdot BC=CE^2$.

  Darij

[b]Very nice problem ![/b] The my solution isn't attractively, but is fast.
[hide="My solution"]
I note $AD=a,\ BC=b,\ AB=CD=EF=c,\ m(\widehat {DAF})=x$.
From the theorem of sinus we obtain the following relations:
$CE=c\cdot \frac{\sin 4x}{\sin 3x},\ FC=c\cdot \frac{\sin x}{\sin 3x},\ FD=a\cdot \frac{\sin x}{\sin 4x}$. But $a=b+2c\cos 3x\ \ (1)$ and
$CD=CF+FD\Longleftrightarrow c=c\cdot \frac{\sin x}{\sin 3x}+a\cdot \frac{\sin x}{\sin 4x}$$\Longleftrightarrow a\sin 3x=2c\cos 2x\sin 4x\ \ (2)$.
From the relations (1) and (2) we obtain: $b=c\cdot \frac{\sin 2x}{\sin 3x},\ a=2c\cdot \frac{\cos 2x\sin 4x}{\sin 3x}$.
Therefore, $CE^2=AD\cdot BC\Longleftrightarrow c^2\cdot \frac{\sin ^2 4x}{\sin ^2 3x}=ab\Longleftrightarrow$
$c^2\cdot \frac{\sin^2 4x}{\sin^2 3x}=$ $2c^2\cdot \frac{\cos 2x\sin 4x}{\sin 3x}\cdot \frac{\sin 2x}{\sin 3x}$, what is true.
[/hide]

[i]Thanks, Darij[/i], for this interesting problem !

$x=m(\widehat {EAD}),\ G\in (BE),\ \widehat {BAG}\equiv \widehat {EAG}\Longrightarrow$$GA=GE\Longrightarrow \triangle ABG\equiv \triangle EFG \Longrightarrow$
$GB=GF,\ m(\widehat {BGF})=$ $180^{\circ}-2x,\ m(\widehat {GBF})=m(\widehat {GFB})=x\Longrightarrow FB=FE=AB$
a.s.o.


Pascal\u0027s Triangle
algebra
binomial theorem

I know you can use pascal's triangle to solve this, but you wouldn't want to generate down to the 12th row in a countdown round.  Is there any other trick?

If you know the formula for the binomial theorem, you can easily just skip to the term you need.

By the binomial theorem, the coefficient of a^6*b^6 when you expand (a+b)^12 is:

$C(12,6)=924$

Well, that's just the solution.  If you want to know more about the binomial theorem and why it works, you can go here: [url]http://www.artofproblemsolving.com/Wiki/index.php/Binomial_theorem[/url]


[quote="alex2008"]Let be $ a,b,c > 0$ . $ (\forall)n\in \mathbb{N}$ let be :

$ S_n \equal{} \frac {a^nb^n}{a^{2n \plus{} 3} \plus{} b^{2n \plus{} 3} \plus{} a^nb^n} \plus{} \frac {b^nc^n}{b^{2n \plus{} 3} \plus{} c^{2n \plus{} 3} \plus{} b^nc^n} \plus{} \frac {c^na^n}{c^{2n \plus{} 3} \plus{} a^{2n \plus{} 3} \plus{} c^na^n}$

i)Show that $ S_n\le S_{n \minus{} 1}\ ,\ (\forall)n\ge 1$.

ii)If $ abc \equal{} 1$ prove that $ S_n \le 1\ ,\ (\forall) n \ge 1$.[/quote]
Nice inequality
i) Since $ (a^{2n\plus{}2}\minus{}b^{2n\plus{}2})(a\minus{}b) \ge 0$ ,we have: $ a^{2n\plus{}3}\plus{}b^{2n\plus{}3} \plus{} a^nb^n \ge ab(a^{2n\plus{}1}\plus{}b^{2n\plus{}1}\plus{}a^{n\minus{}1}b^{n\minus{}1})$
So $ \frac {a^nb^n}{a^{2n\plus{}3}\plus{}b^{2n\plus{}3}\plus{}a^nb^n} \le \frac{a^{n\minus{}1}{b^n\minus{}1}}{a^{2n\plus{}1}\plus{}b^{2n\plus{}1}\plus{}a^{n\minus{}1}b^{n\minus{}1}}$
Similar,we have Q.E.D
ii) We have :$ S_1\equal{}\sum \ \frac{ab}{a^5\plus{}b^5\plus{}ab}$
But $ a^2\plus{}b^2 \ge 2ab$,so $ a^5\plus{}b^5 \ge a^2b^2(a\plus{}b)$
Then $ \frac{ab}{a^5\plus{}b^5\plus{}ab} \le \frac{ab}{ab\plus{}a^2b^2(a\plus{}b)} \le \frac{1}{1\plus{}ab(a\plus{}b)}\equal{}\frac{c}{a\plus{}b\plus{}c}$
Similar,we have Q.E.D :)


geometry
3D geometry
quadratics
modular arithmetic

Show that if an integer is simultaneously a square and a cube, then it must be either of the form $ 7k$ or $ 7k\plus{}1$.

[hide="Solution"]
One should note that the quadratic residues in $ \pmod{7}$ are $ \{0,1,2,4\}$ while the cubic residues are $ \{0,1,6\}$ and the intersection of both sets is $ \{0,1\}$ which implies that if an integer is simultaneously a square and a cube the it is of the form $ 7k$ or $ 7k\plus{}1$  :ninja:
[/hide]


Dear friends,

As you see, our site (I like to use this word "our site") is growths from day to day. And we thank our administrator Valentin Vornicu, our moderators Orl, Langragia, Amfulger, Iandrei, our activists Manlio, Harazi, Grobber, Arne, Pbornsztein, Moubinool, Myth ... for their contributions.

But now, in our site there are so many unsolved problems left. So, I would like to invite you to "kill" them in Xmax holidays. Anyone support me?

Namdung

Namdung I cannot agree with you more :) . "Our site" is a very good term. I also want to thank all of you who have made this possible ...

Happy solving everyone :D

I'm agree with Pierre. So, our dear moderators, pls, move the solved problems to their appointed places.

Namdung

When I read the line "our activists Manlio, Harazi, Grobber, Arne, Pbornsztein, Moubinool, Myth", I had a weird thought in mind: namely that if you asked most MathLinks members to name 10 users on this site, chances would be that 8 of the 10 would occur again and again. :)

Or maybe you will only have around 10 answers? :D

Pierre.

That's true...maybe only the 12 people Namdung named (and Namdung himself) would be answering ;)

[quote="pbornsztein"]As far as I see, many problems have been solved but still are in the 'to be solve'-sections, waiting for better solutions.
There are many topics to check, so first thing to do is to list all the really unsolved problems...

Pierre.[/quote]
============
We can see that almost solved problems allready moved to its places. But now, somewhere there are some inadequate-stated problems (with counterexample). So please remove them for an easier reading.

Namdung

Up to now I moved all the problems from the misc, number theory and algebra section. The other sections will be done in due course unless the other moderators are quickly. Please write me a message and say which problems you refer to when stating: "But now, somewhere there are some inadequate-stated problems (with counterexample). So please remove them for an easier reading." !

[quote="Namdung"]But now, in our site there are so many unsolved problems left. So, I would like to invite you to "kill" them in Xmax holidays. Anyone support me?
Namdung[/quote]

The Xmax holiday is over and gone. Unfortunately I couldn't support Namdung's initiative. I'm from the city of Constanta and I study in Bucharest. When I'n not in Bucharest, where I have a LAN connection to the internet, I use a dial-up connection and then I don't feel very much like posting or moderating.

About the solved problems in the unsolved section: Many of then have long solutions, that are hard to follow and therefore it will take some time before they are moved. If you really want certain problems to be moved, just ask, and we (the moderators) will check them and move if correct. Please just understand the inconveniences. There was a problem in the NT section. It was solved by anegut. The solution had two posts one page each. How could I have checked that?

1) Christmas is over...the Santa hat at the top can be removed :)

2) Does anybody think that 29 categories is a bit too much?

xmas hat gone :(
what do you mean by  29 categories?

Well, parts of the forum, i.e. Algebra, Combinatorics, Geometry etc.

Vali I think it looks quite cool this way with the Xmas hat only when you rollover :)


This is the answer:

[hide]40 mins.[/hide] The guys were slooooow!

[quote="Mike33"]1.

Sam emptied his collection of nickels and dimes on his bed. If his 46 coins were worth 3.10, how many coins of each type did he have?

2.

Jonathan and Tim started running toward each other. Jonathan at 16 yards per minute and Tim at 18 yards per minute. If they were 1,360 yards apart when they started running. how long it take them to collide?

3.

Three times the mystery number was increased by 100, and the result was 12 less then the number. Find the mystery number.[/quote]

[hide]1. 30 nickles and 16 dimes because if he had just nickels he would have 2.30 so he needs 80 cents more which means 16 less nickels.

2. 1360/(16+18)= 40 mins

3.  3x+100=x-12

     2x=-112

       x=-56[/hide]

[hide]1. 30 nickles and 16 dimes
2. 40 minutes
3. -56=mystery number[/hide]

[hide="1"]Write these two equations, where n is the number of nickels and d is the number of dimes:
n+d=46,.05n+.1d=3.10. Solve for n and d and you get n=30, d=16.
[/hide]
[hide="2"]The trick to this is seeing that when both boys are running together, they cover 34 yards per minute. So, 1360/34=40 minutes.They run slower than I do. :D [/hide]
[hide="3"]Write an equation like this, with x being the mystery number:3x+100=x-12.Solve for x and you get x=-56.[/hide]


I want to get a book on inequalities to help prepare for the Putnam exam. Its common for me to spend about half an hour getting frustrated on a practice problem and then find that the only way to do the problem is with an inequality that I have never heard of. Does anyone have any reccomendations? Here are some I found on amazon:

http://www.amazon.com/Introduction-I...62368&sr= 1-2
http://www.amazon.com/Cauchy-Schwarz...0962111&sr=8-2
http://www.amazon.com/Inequalities-C...0962431&sr=1-1

Classical inequalities are a much smaller part of the Putnam than they are of most high school olympiads.

I enjoyed the book [i]Cauchy-Schwarz Master Class[/i] by Steele.  Well-written.


limit
logarithms
calculus
integration
calculus computations

Write it as $-\lim_{n\to\infty}\frac{1}{n}\sum_{t=1}^{n}\ln\frac{t}{n}$ and think of Riemann sums.

am I right?
$\sum$~$1-\frac{\ln{2\pi}}{2n}-\frac{\ln{n}}{2n}$ :wink:
when $n\to oo$

Use the Fundamental Theorem of Calculus you must.
\[-\lim_{n\to\infty}\frac{1}{n}\sum_{t=1}^{n}\ln\frac{t}{n}=-\int_{0}^{1}\ln x\,dx=-(x\ln x-x)\bigg|_{0}^{1}=1\] This is an improper integral, but it converges.

[quote="§outh§tar"]oops. missed a (-) sign[/quote]
where?  :maybe:

I think that §outh§tar mistook $\left(\frac{n}{t}\right)$ for $\binom{n}{t}$.


[quote="chien than"]Prove that:$2005^{2}+2^{2005}$ is relatively prime to 2005[/quote]
[hide]Ok, just show that it isn't divisible by 5 or 401.
$2^{2005}\neq 0 \mod 5$
$2^{2005}\neq 0 \mod 401$[/hide]
So clearly the number relatively prime to 2005.

[hide]Anything added to $2005^{2}$ will make it relatively prime to 2005 if whatever added to it is not divisible by any factors of 2005.

Because

$2^{n}\mod (401) \neq0$

and

$2^{n}\mod (5) \neq0$

$2005^{2}+2^{2005}$ is relatively prime to 2005[/hide]


If a six sided dice is rolled six times, what is the probability of two or more 5's?

[hide]12281/46656[/hide]

[quote="ZzZzZzZzZzZz"]Why did you ask a question if you already found the answer?[/quote]
B/c I wanna know how they got the answer. I found hte answer in an answer key, not a solutions manual. hte answer is on HERE b/c people who post solutions can check to see if their answer is right.


calculus logarithms function limit integration derivative geometry

1. Discuss whether [tex]\displaystyle{ \sum^{\infty}_{n=3} \frac {1}{n \ln n (\ln \ln n)}}[/tex] converges or diverges.

2. Do the same for [tex]\displaystyle{ \sum^{\infty}_{n=3} \frac {1}{n \ln n (\ln \ln n)^2}}[/tex].

Comment in spoiler (don't read since it contains something about the answers):

[hide]That first sequence -- and the potential to extend it infinitely -- intrigued me a lot, since you can theoretically construct sequences in which the bottoms are functions getting consistently bigger, all bigger than just the harmonic sequence, and whose sums still diverge.  I've always thought that was really cool -- although the fact that you have to lose the e^e^e^... at the beginning is kind of a downside.[/hide]

[hide]If the integral converges then so does the summation. The integral of the first one is not to bad, as one can do u-substitution with ln(ln(x)), whose derivative is 1/(xln(x)). So, the integral of that is just 1/u, or ln(ln(ln(x))), which clearly diverges. For the second, perform the same u-substituion, yielding the integral of 1/(u^2), which is grows as 1/u. So, the second converges while the first does not.[/hide][/hide]

Hi, Bucephalus, I like your screenname. You can get invisible text by highlighting text in your reply and choosing the Spoiler button above your message composition window. That hides solutions away from participants who want to think up the solutions themselves.

Thanks for sharing.

[quote="Bucephalus"]If the integral converges then so does the summation.[/quote]

Is this necessarily true?

-- James

It's only true if the function is eventually nonincreasing (and positive).

For sums which steadily increase or decrease, you can simply compare the integral to one particular set of rectangles to see -- just make the rectangles either all above or all below the function.  Doesn't work so well for functions with lots of ups and downs, though.

These are the Bertrand's series, it's well explained at http://www.sosmath.com/calculus/improper/series/series.html

OK, for the first one.

Let u = lnln(n)

Then du = 1/ln(n)*1/n

so the sum can be written as du/u

The improper integral of this equals

lim  lnlnln(n)  evaluated from 3 to a
a-> :inf:

This obviously equals infinity. Because this can be used as the lower bound for the series, the series approaches infinity.

For the second

This series can be bounded by the improper integral

du(u)^(-2) where u again equals lnlnn

The integral of this is -1/u evaluated from 3 to a, as a approaches infinity

Therefore this one converges because we get -1/a (which equals zero) +(1/3)

You could also simply see, for both that when the functions are written in terms of u, they are both just simply p series. One is (1/u) the other is (1/u^2)

Therefore the first diverges and the second converges.

I didn't check out that site belanos, is that what it says?

[quote="ComplexZeta"]Here is a problem that showed up on a mailing list to which I subscribe:

Show that if [tex]a_1 + a_2 + \cdots[/tex]  is an absolutely convergent series, then there exists another absolutely convergent series [tex]b_1 + b_2 + \cdots[/tex] such that

[tex]\displaystyle\lim_{n\to\infty} \Bigg(\frac{a_n}{b_n}\Bigg)=0.[/tex][/quote]

I think this works :

bn = an / R(n-1)^a with Rn =  :Sigma: k>=n ak and 0 < a < 1


If $ 2^5 \times 8^3 \times 16^2 \equal{} 4^m$, what is the value of $ m$?


let $ a;b;c\geq 0$ ; prove that:
$ \sum a^3\plus{}3abc\geq\sum \frac{ab(a^2\plus{}b^2)}{a\plus{}b}$

[quote="rofler"]$ a(a \minus{} b)(a \minus{} c) \plus{} b(b \minus{} a)(b \minus{} c) \plus{} c(c \minus{} a)(c \minus{} b) \ge 0$

$ a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc \ge a^2b \plus{} a^2c \plus{} b^2a \plus{} b^2c \plus{} c^2a \plus{} c^2b$

Now, it suffices to show that $ a^2b \plus{} b^2a\ge \frac {ab(a^2 \plus{} b^2)}{a \plus{} b}$

$ a^3b \plus{} b^3a \plus{} 2a^2b^2\ge a^3b \plus{} b^3a$

$ 2a^2b^2\ge0$[/quote]

ok very goog

[quote="nguoivn"]I think the stronger is also trues:
$ a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc \geq\ \frac {2ab(a^2 \plus{} b^2)}{a \plus{} b} \plus{} \frac {2bc(b^2 \plus{} c^2)}{b \plus{} c} \plus{} \frac {2ca(c^2 \plus{} a^2)}{c \plus{} a}$
It's stronger than Schur  :wink:
And the stronger still trues:
$ a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc \geq\ \frac {2ab(a^3 \plus{} b^3)}{a^2 \plus{} b^2} \plus{} \frac {2bc(b^3 \plus{} c^3)}{b^2 \plus{} c^2} \plus{} \frac {2ca(c^3 \plus{} a^3)}{c^2 \plus{} a^2}$[/quote]

I think, too. But this ineq is false
a=1;b=c=2


Show that if the two roots of $ax^{2}+2bx+c=0$ are the same, then a, b, and c form a geometric sequence.

[hide]Same root $\Longleftrightarrow \Delta=0 \Longleftrightarrow (2b)^{2}-4ac=0 \Longleftrightarrow b^{2}=ac$.[/hide]

[hide]Use the Discriminant, it should be pretty obvious.[/hide]

[hide] Take the discriminant.

$D = 4b^{2}-4ac = 0$

$4b^{2}= 4ac$

Dividing both sides by 4, we get

$b^{2}= ac$

Thus, $b$ is the geometric mean of $a$ and $c$.
This implies that $a, b, c$ form a geometric sequence, as desired. QED[/hide]

We also have to say:
$b^{2}=ac$ does not represent a geometric sequence when $b=0$ and either $a$ or $c$ is $0$.

If $a,b=0$ and $c\ne0$, we have $c=0$. Obviously not true.

If $b,c=0$, $a\ne0$, we have $ax^{2}=0\Longrightarrow x^{2}=0 \Longrightarrow x=0$. It has a double root, and $a,0,0$ is a geometric sequence.


there are 101 natural numbers

from 1 to 101

there are two groups A and B, such that group A has m natural numbers and group B has the remaining.

When 40 is removed from one of the groups and put into the other.

the average of both groups is raised by 0.5

find the value of m.

Suppose we call the sum in A as $S_A$ while sum in B as $S_B$ . So we have $S_A+S_B=(101)(102)/2=5151$

WLOG , assume 40 is originally in A . So the average before 40 is moved to B is

$\frac{S_A}{m}$

while the average in B is $\frac{S_B}{101-m}$

After 40 is moved to B from A , the average in A becomes

$\frac{S_A-40}{m-1}$

while average in B becomes

$\frac{S_B+40}{102-m}$

Since we know the new average increase by 0.5 , hence

$\frac{S_A-40}{m-1}-\frac{S_A}{m}=\frac{1}{2}$

$S_A-40m=\frac{m(m-1)}{2}$ ...............(i)

While for B

$\frac{S_B+40}{102-m}-\frac{S_B}{101-m}=\frac{1}{2}$

$4040-40m-S_B=\frac{(102-m)(101-m)}{2}$ ........(ii)

(i) - (ii) ; $S_A+S_B-4040=\frac{m(m-1)}{2}-\frac{(102-m)(101-m)}{2}$ ..........(iii)

Recall that $S_A+S_B=5151$

Hence , substitute into (iii) and you will get your $m$  :)  .The rest is easy from here .

yep

[hide]

basically i did sort of the same thing..

we have four equations and four variables and voila.. you can solve for each equation.

firstly, let the number of elements in group A be n
therefore, the number of elements in B must be 101 - n.

and the sum of 1 + 2 + 3.. 101 = 5151

so let the sum of all elements of A be t
therefore, the sum of B must be 5151 - t.

let the average of group A be x and average of group B be y

therefore, t/n = x --> [b]equation 1[/b]
and
(5151 - t)/(101 - n) = y --> [b]equation 2[/b]

so these are the first two equations.

now we are told that when 40 is taken from one group and brought to the other; the average is raised by 0.5 for both groups.

so we have two more equations to fiddle with.

wlog let the 40 be taken from group A to be put into group B.

(t - 40)/(n - 1) = x + 0.5 --> [b]equation 3[/b]

(5191 - t)/(102 - n) = y + 0.5 --> [b]equation 4[/b]

so we have our four equations...

now if you combine equation 1 and 3,

and then you combine equations 2 and 4,

you should end up with a result of x - y = 50.5

now if you let y = x - 50.5 ; we are down to three variables.

substitue all y values to become x - 50.5 in equations 2 and 4

which should yield x = 70.5

so y = 20

and now you are down to two variables, very simple from here on in..

[/hide][/hide]


Rectangle ISPE is drawn with IS>3PS, IS=20, EI=3. There are two circles centered at P and S such that the two circles are tangent to each other. A semicircle with center at N, which is the midpoint of EI, and radius 2, is drawn so that it is tangent to the rectangle and does not cross EP or IS. Another circle circumscribes the whole figure. What is the radius of this circle?

If I understood you correctly, then it is impossible to draw a semicircle of radius 2 and center N that is tangent to the rectangle. The dimensions are 3 by 20, right?


real analysis
calculus
integration
real analysis unsolved

Let $ C \subset [0,1]$ be the Cantor ternary set. Define $ f(x): \Re\rightarrow\Re,$
$ f(x)\equal{}x$ is $ x \notin C,$   $ 0$ if $ x\in C.$

Prove that $ f$ is Lebesague measurable.

ABM, I'd prefer to keep measure theory and Lebesgue integration, even straightforward classroom/textbook material, out of the "Computations and Tutorials" section. That's why I moved this.

What about this more general question: $ f$ is measurable. $ g\equal{}f$ almost everywhere. Then $ g$ is measurable.


Nine cards are numbered 1 through 9. What is the probability of selecting a card with a number greater than four or an even number? Express your answer as a common fraction.

It could show 2456789, 7 possibilities, ior 7/9


This is an exercise from AoPS2.

Suppose only one percent of the population has the disease mathphoberia. We design a test which diagnoses the disease successfully in afflicted patients 90% of the time. Unfortunately, 20% of the time it reports that the person has mathphoberia, when in fact they don't. The test reports that I have mathphoberia. What is the probability that the test is correct?

I got a rather ugly answer. Would like to see if anyone else gets the same thing.

[hide]

.01*.9 = P(you actually have the disease and the test says you do)

.99*.2=p(you don't have the disease, but the test says you do)

[.01*.9]/[.01*.9+.99*.2]  :approx: .043478  which is 1/23[/hide]

Ah! okay good. I caught my error.

This problem, as well as my other one "The Root of the Problem", should probably be moved to the Intermediate Forum. I think students there could probably benefit from them more.

I am sort of confused about conditional probability.

If P(A | B) = P(A ^ B) (as defined in AoPS 2) and P(A ^ B) = P(A)P(B) (as defined in AoPS 2). Why can't you just cancel out the P(B)s, and get P(A)?

I assume you meant P(A | B) = P(A ^ B)/P(B).

The reason you can't just substitute in like you said is because P(A^B) = P(A)P(B) [b]only if A and B are independent[/b]. And of course, if A and B are independent, P(A|B) [b]is[/b] equal to P(A), since whether B happens or not has no effect on A.[/b]


superior algebra
superior algebra unsolved

Let A be a ring with $ 0 \not \equal{} 1$ and K be the set
$ K\equal{}${$ x\in A | x^2\minus{}x\plus{}1\equal{}0$}.
1) If $ x\in K$ then $ x$ is invertible and $ x^{\minus{}1}\equal{}x^5$
2) If $ K\equal{}A$ \ {0,1}  then A  is isomorphic with $ Z_{3}$ or $ 1\plus{}1\equal{}0$ .

1) is simple...$ x^6\minus{}1\equal{} (x^2\minus{}x\plus{}1)( x^2\plus{}x\plus{}1)(x\minus{}1)(x\plus{}1)\equal{}0$ . A solution for 2) ?

If $ 1 \neq \minus{}1$ then $ x^2 \minus{} x \plus{} 1 \neq x^2 \plus{} x \plus{} 1$, so $ A$ has no elements of order $ 3$ and therefore has no elements of order $ 6$.  It follows that $ K$ contains only elements of order $ 2$, but

$ x^2 \equal{} 1 \implies x^2 \minus{} x \plus{} 1 \equal{} 0 \Leftrightarrow x \equal{} 2$

so $ A \equal{} \{ 0, 1, 2 \}$ and is a division ring, hence isomorphic to $ \mathbb{Z}/3\mathbb{Z}$.  Otherwise, $ 1 \equal{} \minus{}1$.


geometry
trapezoid
3D geometry
similar triangles

In trapezoid $ ABCD$, $ AB$ is parallel to $ CD$, $ P$ is on $ BC$ and $ Q$ is on $ DA$ such that $ PQ$ is parallel to $ AB$ and $ PQ$ divides the trapezoid into two smaller trapezoids of equal area.  If $ AB \equal{} a$ and $ CD \equal{} b$, compute the length $ PQ$.

In the frustrum of a circular cone, one base has area $ A$ and the other base has area $ B$.  What is the area of the slice of the frustrum that is parallel to the two bases and equidistant from them?

[hide="First Part"]Let the desired quantity be $ x$. Let the two non-parallel sides of the trapezoid intersect at $ E$. Let the distance from $ E$ to $ AB$, $ PQ$, and $ CD$ be $ h_{1}$, $ h_{2}$, and $ h_{3}$. Using the formula for the area of the trapezoids,

$ (a \plus{} x)(h_{2} \minus{} h_{1}) \equal{} (b \plus{} x)(h_{3} \minus{} h_{2})$

Dividing by $ h_{2}$ and using the facts that $ h_{1}/h_{2} \equal{} a/x$, $ h_{3}/h_{2} \equal{} b/x$ (from similar triangles),

$ (a \plus{} x)(1 \minus{} a/x) \equal{} (b \plus{} x)(1 \minus{} b/x)$

Multiplying through, we get $ x^{2} \equal{} \frac {a^{2} \plus{} b^{2}}{2}$. Square-rooting both sides gets us the desired.[/hide]

[hide="Second part"]Take the cross section that goes through the vertex of the cone (let that be $ R$) and contains both diameters of the bases. Let the distance from $ R$ to the base of area $ A$ be $ h_{1}$, $ R$ to base area $ B$ be $ h_{3}$, and $ R$ to desired base be $ h_{2}$. We have $ h_{3} \minus{} h_{2} \equal{} h_{2} \minus{} h_{1}$. Dividing through by $ h_{2}$, $ h_{1}/h_{2} \plus{} h_{3}/h_{2} \equal{} 2$.

Let the desired quantity be $ X$. Let the radii of the bases be $ a$, $ b$, and $ x$ (corresponding to area of $ A$, $ B$, $ X$). By similar triangles, $ h_{1}/h_{2} \equal{} a/x$, $ h_{3}/h_{2} \equal{} b/x$. Hence, $ x \equal{} (a \plus{} b)/2$, and since $ \pi a^{2} \equal{} A$ (and similarly for $ b$, $ x$), we can now solve for $ X$:

$ X \equal{} \left(\frac {\sqrt {A/\pi} \plus{} \sqrt {B/\pi}}{2}\right)^{2} \pi \equal{} \left(\frac{\sqrt{A} \plus{} \sqrt{B}}{2}\right)^{2}$[/hide]

EDIT: Thanks, JBL

Is this merely coincidence, or is there some sort of connection between power means and these geometry problems?

Yes, although you can multiply the $ \pi$s away in the second expression.


Let X1, X2... be iid with E|X1|< infinity.
Let Mn = max (X1, X2....., Xn). Prove that (Mn)/n converges to 0 almost surely.

I've shown that (Xn)/n converges to 0 almostly surely. Does it implies that (Mn)/n converges to 0 a.s. also?? Why?

If $ M_n$ is eventually a constant sequence, then, since $ M_n < \infty$ a.s. for all $ n$, the result follows immediately.  Otherwise we can find an infinite, distinct subsequence $ X_{n_i}$ such that $ M_{n_i} \equal{} X_{n_i} < X_{n_j} \equal{} M_{n_j}$ for $ i<j$.

On one hand, $ X_{n_i}/n_i \to 0$ a.s. as a subsequence of a convergent sequence.

On the other hand, given $ \varepsilon>0$, pick $ N$ so large that $ j\geq N$ implies $ |X_{n_j}/n_j| < \varepsilon$.  Then if $ k \geq n_N$, and $ j$ is the largest index for which $ k \geq n_j$, then we have $ |M_k/k| \leq |X_{n_j}/n_j| < \varepsilon$, which means that $ M_n/n\to 0$ a.s.

This problem was previously discussed [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=170224]here[/url]


In the picture shown bellow, the capacitor is initially discharged. In a given instant, the circuit is closed at both switches.

(a) Find the currents in the circuit in the initial instant, when the capacitor is still discharged.

(b) After some time equilibrium is attained, and there are only stationary currents in the circuit. At this moment, the capacitor is charged at is maximum. Find, under these conditions, the currents in the circuit.

(c) Compute the potential difference between the plates of the capacitor. What is the negative plate?

(d) Evaluate the energy stored in the capacitor.

[b](a) Find the currents in the circuit in the initial instant, when the capacitor is still discharged.[/b]

At the initial instance, the capacitor acts as a perfect (pure) conductor.

[b](b) After some time equilibrium is attained, and there are only stationary currents in the circuit. At this moment, the capacitor is charged at is maximum. Find, under these conditions, the currents in the circuit. [/b]

There is no current through the capacitor now, hence the whole line (with $ 3\Omega,\;6V,\;2\mu F$) is useless.

Now you can solve the others too!