Why is tan 50 + tan 60 + tan 70 = tan 80?

hello, do you mean $ \tan(50^{\circ})\plus{}\tan(60^{\circ})\plus{}\tan(70^{\circ})\equal{}\tan(80^{\circ})$?
Sonnhard.

I think the problem is dual to proving the identity

$ \sqrt{3} \left(\frac{3\tan^2 80 \minus{}1}{\tan^2 80 \minus{}3}\right)\equal{}\tan80$

where the angle is in degrees (first note that the sum of tangents is equal to the product of tangents, and then express everything in terms of $ \tan80$), but am not sure how to prove it at this point...

There was a nice graphical proof of this which I cannot remember in a British teaching journal. (I have no memory for geometric proofs though :()

Given a triangle $ ABC$, prove that $ \tan A\plus{}\tan B\plus{}\tan C\equal{}\tan A\tan B\tan C$.

1986 Gakusyuin University entrance exam/Law

I have since seen a diagram (and been locked out for posting this separately, as it is the same problem).

ABCD is a ractangle (with ABCD going anti-clockwise from the bottom left corner). AD = BC = 1.
G and F are points on DC such that DG = tan 50, GF = tan 60 and FC = tan 70.  Prove that angle ACB = 80 degrees.

In questions like this the identity $ tan3x \equal{}tanx.(tan60\minus{}x).(tan60\plus{}x)$ is  usually helpful.

 Let x=20    then     $ tan60$ =$ tan20.tan40.tan80$         =$ \frac{1}{tan70}$.$ \frac{1}{tan50}$.$ tan80$

which gives $ tan80$ =$ tan50.tan60.tan70$ =$ tan50 \plus{}tan60 \plus{}tan70$(if A+B+C=180 then tanA +tanB +tanC=tanA.tanB.tanC)

There are similar useful facorizations for Sin3x and cos3x. See if you can derive them.

N.B. All angles above are in degrees.

Thanks. I was hoping to see a proof using the rectangle (which I misspelt).

Hi Michael.

For your factoring question, I get the following for sine:

sin 2x = 2. sin x. cos x
So sin 3x = sin (2x + x) = 2s.c^2 + (1-2s^2)s
=2s.(1-s^2) + s - 2s^2
=3s - 4s^3
=s(3 - 4s^2)
=s(3c^2 - s^2).

Sin (60 - x) . sin (60 + x) = 3(c^2 - s^2)/4
=(3-4s^2)/4.

So sin 3x = 4sin x. sin(60-x).sin(60+x).

I guess the factoring of cos 3x is much the same.

I make it cos 3x = 4. cos x. cos (60-x). cos (60 + x)

[quote="kunny"]Given a triangle $ ABC$, prove that $ \tan A \plus{} \tan B \plus{} \tan C \equal{} \tan A\tan B\tan C$.

1986 Gakusyuin University entrance exam/Law[/quote]
As $ A, B, C$are the angles of a triangle hence $ A \plus{} B \plus{} C \equal{} \pi$.
We know the formula:
$ 0 \equal{} \tan(A \plus{} B \plus{} C) \equal{} \frac {\tan A\plus{}\tan B\plus{}\tan C \minus{} \tan A\tan B\tan C}{1 \minus{} \sum\tan A\tan B}$
or, $ \tan A\tan B\tan C \minus{} \tan A\tan B\tan C \equal{} 0$
or, $ \tan A\tan B\tan C \equal{} \tan A\tan B\tan C$
QED.
Am I right, Kunny? :blush:
For those who do not know the formula of $ \tan(A \plus{} B \plus{} C)$, just write the formula of
$ \tan((A \plus{} B) \plus{} C)$....
{edited to fix a typo} :blush:  :(

[quote="Kouichi Nakagawa"]Hey, Sonnhard.
How can you misunderstand this trifling typo?
I feel your contribution idly.

\begin{eqnarray*} \tan 50^\circ + \tan 60^\circ + \tan 70^\circ & = & \tan (60^\circ - 10^\circ) + \tan 60^\circ + \tan (60^\circ + 10^\circ) \\
& = & \frac {\tan 60^\circ - \tan 10^\circ}{1 + \tan 60^\circ \tan 10^\circ} + \tan 60^\circ + \frac {\tan 60^\circ + \tan 10^\circ}{1 - \tan 60^\circ \tan 10^\circ} \\
& = & \frac {\sqrt {3} - \tan 10^\circ}{1 + \sqrt {3} \tan 10^\circ} + \sqrt {3} + \frac {\sqrt {3} + \tan 10^\circ}{1 - \sqrt {3} \tan 10^\circ} \\
& = & \frac {\sqrt {3}(\tan^2 10^\circ - 3)}{3\tan^2 10^\circ - 1} \end{eqnarray*}
And,
\begin{eqnarray*} \tan 80^\circ & = & \tan (60^\circ + 20^\circ) \\
& = & \frac {\tan 60^\circ + \tan 20^\circ}{1 - \tan 60^\circ \tan 20^\circ} \\
& = & \frac {\sqrt {3} + \tan 20^\circ}{1 - \sqrt {3} \tan 20^\circ} \\
& = & \frac {\sqrt {3} + \frac {2 \tan 10^\circ}{1 - \tan^2 10^\circ}}{1 - \sqrt {3} \frac {2 \tan 10^\circ}{1 - \tan^2 10^\circ}} \\
& = & \frac {\sqrt {3} \tan^2 10^\circ - 2 \tan 10^\circ - \sqrt {3}}{\tan^2 10^\circ + 2 \sqrt {3} \tan 10^\circ - 1} \end{eqnarray*}
And,
\begin{eqnarray*} \tan 80^\circ = \frac {1}{\tan 10^\circ} & \iff & \frac {\sqrt {3} \tan^2 10^\circ - 2 \tan 10^\circ - \sqrt {3}}{\tan^2 10^\circ + 2 \sqrt {3} \tan 10^\circ - 1} = \frac {1}{\tan 10^\circ} \\
& \iff & \frac {\sqrt {3} \tan^3 10^\circ - 3 \tan^2 10^\circ - 3 \sqrt {3} \tan 10^\circ + 1}{\tan 10^\circ (\tan^2 10^\circ + 2 \sqrt {3} \tan 10^\circ - )} = 0 \end{eqnarray*}

So,
\begin{eqnarray*} \tan 50^\circ + \tan 60^\circ + \tan 70^\circ - \tan 80^\circ & = & \frac {\sqrt {3}(\tan^2 10^\circ - 3)}{3\tan^2 10^\circ - 1} - \frac {1}{\tan 10^\circ} \\
& = & - \frac {\sqrt {3} \tan^3 10^\circ - 3 \tan^2 10^\circ - 3 \sqrt {3} \tan 10^\circ + 1}{ - 3 \tan^3 10^\circ + \tan 10^\circ} \\
& = & 0 \end{eqnarray*}

Thus, $ \boxed{\tan 50^\circ + \tan 60^\circ + \tan 70^\circ = \tan 80^\circ}$[/quote]

Better way : using kunny's hint and my proof for the hint/theorem, we see that
as $ 50^\circ+60^\circ+70^\circ=180^\circ$, the proof is trivial... :blush:

Any proofs just using the diagram of the three rectangles inside the large one?

[quote="Potla"][quote="kunny"]Given a triangle $ ABC$, prove that $ \tan A + \tan B + \tan C = \tan A\tan B\tan C$.

1986 Gakusyuin University entrance exam/Law[/quote]
As $ A, B, C$are the angles of a triangle hence $ A + B + C = \pi$.
We know the formula:
$ 0 = \tan(A + B + C) = \frac {\tan A\tan B\tan C - \tan A\tan B\tan C}{1 - \sum\tan A\tan B}$
or, $ \tan A\tan B\tan C - \tan A\tan B\tan C = 0$
or, $ \tan A\tan B\tan C = \tan A\tan B\tan C$
QED.
Am I right, Kunny? :blush:
For those who do not know the formula of $ \tan(A + B + C)$, just write the formula of
$ \tan((A + B) + C)$....[/quote]

I think,
\[ \tan (A+B+C) = \frac{\sum \tan A - \prod \tan A}{1 - \sum \tan A \tan B}\]

[b]Solution[/b]
\begin{eqnarray*}
\tan (B+C) = - \tan A
& \iff & \frac{\tan B + \tan C}{1 - \tan B \tan C} = - \tan A \\
& \iff & \tan B + \tan C = - \tan A + \tan A \tan B \tan C \\
& \iff & \sum \tan A = \prod \tan A
\end{eqnarray*}

[quote="Michael Niland"]In questions like this the identity $ tan3x \equal{} tanx.(tan60 \minus{} x).(tan60 \plus{} x)$ is  usually helpful.

 Let x=20    then     $ tan60$ =$ tan20.tan40.tan80$         =$ \frac {1}{tan70}$.$ \frac {1}{tan50}$.$ tan80$

which gives $ tan80$ =$ tan50.tan60.tan70$ =$ tan50 \plus{} tan60 \plus{} tan70$(if A+B+C=180 then tanA +tanB +tanC=tanA.tanB.tanC)

There are similar useful facorizations for Sin3x and cos3x. See if you can derive them.

N.B. All angles above are in degrees.[/quote]

I know $ \tan 3\theta \equal{} \frac{3\tan \theta \minus{} \tan^3 \theta}{1 \minus{} 3 \tan^2 \theta}$.
I can calculate $ \tan \theta \cdot \tan \left(\frac{\pi}{3} \minus{} \theta\right) \cdot \tan \left(\frac{\pi}{3} \plus{} \theta\right) \equal{} \frac{3\tan \theta \minus{} \tan^3 \theta}{1 \minus{} 3 \tan^2 \theta}$.
But, I never thought of identity.
Is this identity famous?

Thanks very good and very simple solution.
Excellent !!! I'm touched:)

[quote="Kouichi Nakagawa"][quote="Potla"][quote="kunny"]Given a triangle $ ABC$, prove that $ \tan A + \tan B + \tan C = \tan A\tan B\tan C$.

1986 Gakusyuin University entrance exam/Law[/quote]
As $ A, B, C$are the angles of a triangle hence $ A + B + C = \pi$.
We know the formula:
$ 0 = \tan(A + B + C) = \frac {\tan A+\tan B+\tan C - \tan A\tan B\tan C}{1 - \sum\tan A\tan B}$
or, $ \tan A\tan B\tan C - \tan A\tan B\tan C = 0$
or, $ \tan A\tan B\tan C = \tan A\tan B\tan C$
QED.
Am I right, Kunny? :blush:
For those who do not know the formula of $ \tan(A + B + C)$, just write the formula of
$ \tan((A + B) + C)$....[/quote]

I think,
\[ \tan (A + B + C) = \frac {\sum \tan A - \prod \tan A}{1 - \sum \tan A \tan B}
\]
[b]Solution[/b]
\begin{eqnarray*} \tan (B + C) = - \tan A & \iff & \frac {\tan B + \tan C}{1 - \tan B \tan C} = - \tan A \\
& \iff & \tan B + \tan C = - \tan A + \tan A \tan B \tan C \\
& \iff & \sum \tan A = \prod \tan A \end{eqnarray*}[/quote]

I made an ENORMOUS TYPO! :blush:  :(
I meant $ 0 = \tan(A + B + C) = \frac {\tan A+\tan B+\tan C - \tan A\tan B\tan C}{1 - \sum\tan A\tan B}$....

Sorry for the inconvenience.... :o

Let $ ABDC$ be a rhombus with $ \angle A = 20^\circ$, so $ \angle D = 20^\circ$ and $ \angle B = \angle C = 160^\circ$.  Extend $ AB$ and $ DC$ to $ E$ and $ F$, so that quadrilateral $ AEDF$ is a rectangle.  Assume that $ AF = 1$.  Then $ DF = \tan \angle DAF = \tan 80^\circ$.

[img]http://www.artofproblemsolving.com/Admin/latexrender/pictures/ae83e9d7bf0d1e2fbe93c2b29b3d8e54.png[/img]

Construct point $ P$ inside triangle $ BCD$ so triangle $ PBC$ is equilateral.  Since $ \angle DBC = 80^\circ$, $ \angle DBP = 20^\circ$.

Take point $ Q$ on $ BE$ such that $ BQ = BP = BC$.  Note that $ \angle QBC = 180^\circ - \angle ABC = 180^\circ - 80^\circ = 100^\circ$.  Therefore, $ \angle BQC = \angle BCQ = 40^\circ$.  Furthermore, $ \angle DBQ = \angle DBC - \angle DBC = 80^\circ - 60^\circ = 20^\circ$.  Then by SAS, triangles $ DBP$ and $ DBQ$ are congruent.  Therefore, $ \angle BDQ = \angle BDP = 10^\circ$.

Hence, $ \angle CDQ = \angle BDC + \angle BDQ = 20^\circ + 10^\circ = 30^\circ$, so $ \angle QDE = 60^\circ$.  Let $ R$ be the projection of $ C$ onto $ AE$.  Then $ QE = \tan \angle QDE = \tan 60^\circ$ and $ QR = \tan \angle CQR = \tan 50^\circ$.  Therefore, $ \tan 80^\circ = AE = AR + RQ + QE = CF + RQ + QE = \tan 70^\circ + \tan 50^\circ + \tan 60^\circ$.

Since $ BR = \tan 10^\circ$ and $ BE = \tan 70^\circ$, it also follows that $ \tan 70^\circ + \tan 10^\circ = EB + BR = EQ + QR = \tan 60^\circ + \tan 50^\circ$.

A classical geometer and his beautiful proofs. :)

In the rectangle introduced above (#8), E is directly below F (i.e. E is on AB such that EB = tan 70).

Angle DAG = 50 degrees; angle GEF = 60 degrees and angle CEF = 70 degrees.

Why in the diagram must angle ACB be 80 degrees?

ACB is isoceles, since ABDC is a rhombus.
Awesome proof, by the way.

Thanks: I know that. You are referring to the diagram in #19 where nsato's impressive proof appears. I am referring to the diagram I described in #8 and #21. I would like to see a proof based on that diagram.

The diagram I was shown (and described in #8 and #21), as far as I remember, has AD marked as 1 unit long, includes the line segments AG, GE, EC, AC and EF and has the angle DAG marked as 50 degrees, GEF as 60 degrees and FEC as 70 degrees. I think it also has DG marked as tan 50, GF as tan 60 and FC as tan 70. The problem is to use this diagram to prove that angle ACB must be 80 degrees.

$ \boxed {\ \begin{array}{c}
\
\tan 50^{\circ} + \tan 60^{\circ} + \tan 70^{\circ} = \tan 80^{\circ}\\\
(50+60+70=180)\\\
\tan 50^{\circ} \cdot \tan 60^{\circ} \cdot \tan 70^{\circ} = \tan 80^{\circ}\\\
\tan 10^{\circ}=\tan 20^{\circ}\cdot\tan 30^{\circ}\cdot\tan 40^{\circ}\\\
\sqrt 3\cdot\tan 10=\tan 20\cdot \tan 40\\\
\sqrt 3\cdot\sin 10\cdot\cos 20\cdot\cos 40=\cos 10\cdot\sin 20\cdot\sin 40\\\
\sqrt 3\cdot\cos 40=4\cos^210\cdot\sin 20\\\
\sqrt 3\cdot\cos 40=2\cos 10\cdot(\sin 30+\sin 10)\\\
\sqrt 3\cdot\cos 40=\cos 10 +\sin 20\\\
\sqrt 3\cdot\cos 40=\sin 80 +\sin 20\\\
\sqrt 3\cdot\cos 40=2\sin 50\cdot\cos 30\\\
\cos 40=\sin 50\\\
\mathrm{O.K.}\\
\end{array}\ }$

Multumesc. Virgil. I like it but it doesn't help me solve the problem of proving that the angle is 80 degrees just from the diagram.

[color=darkblue]Just now I understood what you wish after I read all replies ![/color]

[quote] [color=darkred]Let $ ABCD$ be a rectangle for which $ AD \equal{} BC \equal{} 1$ . Suppose that exist the points $ \{E,F\}\subset (CD)$

so that $ E\in (DF)$ and $ DE \equal{} \tan 50^{\circ}$ , $ EF \equal{} \tan 60^{\circ}$ , $ FC \equal{} \tan 70^{\circ}$ . Then $ m(\angle ACB) \equal{} 80^{\circ}$ .[/color] [/quote]

[color=darkblue][b][u]Proof[/u].[/b] Denote $ X\in (AB)$ so that $ FX\perp AB$ . Therefore, $ m(\angle ACB) \equal{} 80^{\circ}$ $ \Longleftrightarrow$ $ XA \equal{} XC$ $ \Longleftrightarrow$ $ DE \plus{} EF \equal{} \frac {BC}{\sin \widehat {BXC}}$ $ \Longleftrightarrow$

$ \tan 50^{\circ} \plus{} \tan 60^{\circ} \equal{} \frac {1}{\sin 20^{\circ}}$ $ \Longleftrightarrow$ $ \sin 110^{\circ}\sin 20^{\circ} \equal{} \cos 50^{\circ}\cos 60^{\circ}$ $ \Longleftrightarrow$ $ 2\cos 20^{\circ}\sin 20^{\circ} \equal{} \cos 50^{\circ}$ $ \Longleftrightarrow$ $ \sin 40^{\circ} \equal{} \sin 40^{\circ}$ . [b][u]O.K.[/u][/b] [/color]


geometry
geometric transformation
superior algebra
superior algebra unsolved

If G is a set closed under operation * with the associative property, and if a*b=a*c implies b=c, and u*a=w*a implies u=w for a,b,c,u,w in G, then prove G is a group under *.

I'm trying to work my way through the text; already got stuck on the first day. Thanks!

We have to assume $G \neq \emptyset$. In other words, we have to show, that each non-empty finite monoid with injective right- and lefttranslations is a group. This follows from the more general (because injective maps from a finite set to itsself  are surjective) result:

[b]Every non-empty semigroup with surjective right- and lefttranslations is a group.[/b]

Proof. Pick an element $m$ of our monoid. Then there's an element $e$ with $me=m$, because lefttranslation with $m$ is surjective. And since the righttranslation with $m$ is surjective, each element $x$ has the form $ym$, hence $xe = yme=ym = x$. So, $e$ is an identity element.  For all elements $x$ there is a preimage of $e$ under the lefttranslation with $x$, which is an inverse element of $x$. Thus, we have a group. $\square$

By the way, of course, the converse is also true, and you cannot omit the assumption, that [i]both[/i] types of translations are surjective. This shows the example $a*b : = a$ or $a*b : = b$.


topology
geometry
3D geometry
sphere
integration
probability
invariant

$H$ a Hilbert space, $S$ the unit sphere of $H$, $T$ a normal operator.
Show that $D=\{(x;Tx), x \in S \}$ is convex and that the closure of $D$ is the convex hull of the spectrum of $T$.

For the first part, does the problem become easier when we only take into consideration normal operators? Because as far as I know, all known proofs for the corresponding assertion for arbitrary operators are computational, and may get messy. I can reduce it to $2\times 2$ matrices, but then the mess begins :).

For the second part, we can use the Spectral Theorem (is this how you solved it during the exam? :)):

According to the theorem, for every $x\in H$ there is a positive Borel measure $\mu_x$ on $\sigma=\sigma(T)$ such that $(Tx,x)=\int_\sigma\lambda\ \mbox d\mu_x(\lambda)\ (*)$. Moreover, the total weight $\mu_x(\sigma)$ is precisely $\|x\|^2$. In our case this is $1$, i.e. $\mu_x$ is a probability measure. $(*)$ now implies that for all $\|x\|=1,\ (Tx,x)$ is contained in the convex hull of $\sigma$, so the closure of $W$ (that set we're interested in) is a subset of the convex hull $\mbox{co}(\sigma)$ of $\sigma$.

Conversely, we know that $W$ is convex, so in order to prove that its closure contains $\mbox{co}(\sigma)$, it suffices to prove that it contains the boundary of $\sigma$. The boundary of the spectrum of an operator $A$ consists of points in the so-called "approximate point spectrum", i.e. complex numbers $\lambda$ such that we can find a sequence $(x_n)_n$ with $\|x_n\|=1$ and $(A-\lambda)x_n\to 0$. Let $\lambda$ be such a value for $T$, and let $(x_n)_n$ be a sequence  as above. Then $(Tx_n,x_n)-\lambda=((T-\lambda)x_n,x_n)\to 0$, so $\lambda$ does indeed belong to the closure of $W$.

no, I don't think the first part is easier for normal operator. Nonetheless our teacher gave us a proof without many nasty calculation. I will post it soon.

Here's a strengthening of the second part (I saw it in Halmos's "A Hilbert Space Problem Book"):

An operator $A$ on a Hilbert space $\mathcal H$ is called [i]subnormal[/i] if there is a Hilbert space $\mathcal K$ containing $\mathcal H$ and a normal operator $B$ on $\mathcal K$ such that $\mathcal H$ is invariant under $B$, and $A$ is the restriction of $B$ to $\mathcal H$. Clearly, a normal operator is subnormal.

Prove that the second part of the problem above holds for subnormal operators. That is: the closure of the numerical range $W(A)=\{(Ax,x)\ |\ \|x\|=1\}$ is equal to the convex hull of the spectrum of $A$ if $A$ is subnormal.


The average age of the three Wilson children is 7 years. If the two younger children are 4 years old and 7 years old, how many years old is the oldest child?

As I said in another post, you can notice that we must have $ 3\plus{}0\plus{}7\minus{}x\equal{}0$, giving $ x\equal{}\boxed{10}$.


suppose i carry out the reaction between a primary halide and water where the solvent itself is the water .say the temperature is bit higher than normal so we have sufficent dissociation of water molecules  so that we can study the product more easily.which path will the substitution primarily follow and why?

well what my actual aim is to know whether primary substrates can be made to undergo substituion via 1st order kinetics by changing the solvent that is by increasing the polarity to a large extent(say i carry this reaction out in trifluoro acetic acid) or no matter what the solvent is these will proceed via 2nd order kinetics primarily .
also the converse for tertiary substrates i.e whether they primarily follow 1st order kinetics or by simply changing  the solvent the we can change that to 2nd order kinetics (say i use DMSO)
 :?:

[quote="Pardesi"]suppose i carry out the reaction between a primary halide and water where the solvent itself is the water.[/quote]

But the organic halides have very low solubilities in water.

[quote="Pardesi"]say the temperature is bit higher than normal so we have sufficent dissociation of water molecules[/quote]

What do you mean? Although the reaction $2H_{2}O \rightleftharpoons H_{3}O^{+}+HO^{-}$ is endothermic, even for a temperature of 100ºC the $K_{w}$ is still only $51.3 \times 10^{-14}$, and the concentration of hydronium and hydroxide ions is only $7.2 \times 10^{-7}$.

[quote="Pardesi"]well what my actual aim is to know whether primary substrates can be made to undergo substituion via 1st order kinetics (...) we can change that to 2nd order kinetics (say i use DMSO)[/quote]

In my opinion, 1st order kinetics in a primary substracte would imply the formation of a primary carbocation, which is not a stable species no matter how polar is the solvent, because 2nd order kinetics (SN2 reaction) is to easy. For tertiary substractes, formation of a tertiary carbocation is to easy (don't forget that a tertiary carbon is too crowded for a SN2 reaction to take place at an appreciable rate).

for the first part i meant Solvolysis where the attacking reagent itself is the solvent .as far as i know these are special kinds of nucleophillic substitution reaction.

for the 2nd part i don't know what i wrote :wink:

for the third part that's what my question is that whether in the most polar of the
solvents like trifluoro actetic acids also primary substrates will exhibit $S_{N}2$
or will there be a change :?: (here in short will the product be a Racemic mixture or like most $S_{N}2$ reactions there will be complete inversion)

Also solvents like DMSO,DMF,HMPT are they unreactive towards alkyl halides or as such any other substrate which we normally use for substitution raections in general ?

Is non reactivity of the solvent with Substrate an important criteria for the selection of Solvent.


In trying to understand probability, and just have a tiny question..

I the probability for a outcome is, let say 51,27%.. and I want to know how many times I can run this action before i'm down to 1% chance that it will happen in a row - is it as simple as:

0,5127^X = 0.01 and then I solve X?

Thanks in advance!
/Johan

Well, certainly 1 way is to use logs, although it requires a calculator.
So,

0.5127^x = 0.01
log (0.5127^x) = log 0.01
x log 0.5127 = log 0.01
x = (log 0.01)/(log 0.5127)

All the logarithms are in base 10.

Thanks kab^2! I'm familiar with the log-solving, I just wanted a confirmation of the formula above. I soundend kind of "extreme" that after only 7 hits in a row, there is under 1% probabilty that it will it happen...

But now I know :-)

/Johan

I think you might have misinterpreted this result. if anything seems extreme to you you know there's a problem.
after 7 successes in a row, the probability of the 8th being successful is still 51.27%. The probability of 8 in a row bing successful is 1%. What I'm trying to say is, the 1% number is a probability describing the outcome of all 8 experiments, and teh fact that you may have gotten 7 in a row previously doesn't usually affect whether or not you get the 8th one. Does this make sense?

Yes, it make sense.. i know that that the 8th time isn't affected by the previous times.. what i wanted to know is how great the chance is to get 8 in a row.. :-)

Have a great day kab^2!


Let $ x$, $ y$, and $ z$ be integers satisfying the equation \[ \dfrac{2008}{41y^2}\equal{}\dfrac{2z}{2009}\plus{}\dfrac{2007}{2x^2}.\] Determine the greatest value that $ z$ can take.
[i]Budi Surodjo, Jogjakarta[/i]

we can easily see that $ 16 | y^2$ so we replace $ y$ by $ 4y$.
we have:
$ 4z\equal{}\frac{49 \times 251}{y^2}\minus{}\frac{2007 \times 2009}{x^2}$
since we are trying to find the greatest value of $ z$ we assume that $ z$ is positive and we know that it is an integer which leads to
$ y^2 | 251.49.x^2,x^2 | 2007.2009.y^2$
also none of $ 49$ and $ 251$ divides $ y$(note that these numbers are prime)
if not,then RHS will be at maximum $ 5,...$(it is attained when $ y\equal{}49$ and if $ y\equal{}251s$ then RHS will be lesser than $ 1$) and because LHS is divisible by $ 4$ so we must have $ z\equal{}1$ and
then we'll have
$ z\equal{}1, \frac{251}{49}\minus{}\frac{2007.2009}{x^2}\equal{}4 \implies \frac{55}{49}\equal{}\frac{2007.2009}{x^2}$
but we don't have $ 55 | 2007.2009.49$
so $ y$ divides $ x$ and then $ yt\equal{}x$
so we 'll have
$ 251.49.t^2\equal{}4zy^2t^2\plus{}2007.2009$
from these we can find all possible values of $ t$ since $ t^2 | 2007.2009$ or consequently $ t^2 | 9$(note that $ 2009\equal{}49.41$ and $ 2007\equal{}9.223$)  which implies $ t\equal{}1,3$.the rest is easy after finding $ y$ by putting $ t\equal{}1,3$.

We claim that the answer is $z = \boxed{789}.$ This is achieved by $x = 21, y = 4.$

We will now show that this is optimal. First of all, it's easy to see by $v_2$ considerations that $4|y.$ If $|y|>4$, then we have $\frac{2z}{2009} = \frac{2008}{41y^2} - \frac{2007}{2x^2} < \frac{2008}{41 \cdot 8^2} < \frac{2 \cdot 789}{2009}$, and so we're done in this case. Hence, assume WLOG that $y = 4$ from now on.

We have that:

$$\frac{2z}{2009} = \frac{251}{82} - \frac{2007}{2x^2}.$$

Observe that we must have $3|x$ by $v_3$ considerations, and actually that $v_3(x) = 1.$ We also have that $v_7(x) \le 1$. Finally, we know that for any prime $p \neq 3, 7$, $p \nmid x$, as otherwise we'd have that $v_p(RHS) \le -2$, contradiction.

Hence, $x \le 21$, and so the RHS is at most $\frac{251}{82} - \frac{2007}{2 \cdot 21^2} = \frac{2 \cdot 789}{2009},$ as desired.

$\square$


Is given an isoscele trapezium. The largest base is $ a$, if the acute angle is $ \alpha$ then find its perimeter .

I  found $ 2a(1\plus{}tan^{2}(\frac{\alpha}{2}))$

You are right... the problem being who's right??


geometry
3D geometry
tetrahedron
MATHCOUNTS
geometric transformation
rotation

Are you sure it's $\boxed{10}$. I got $120$... $P(5,4)$? Unless our perception of distinct is different. I'm considering a tetrahedron to be distinct relative to its sides not to the viewer's image.

:roll: I think the surfaces of the tetrahedron are the same so We have P(5,4)=120

Math is my life :!:

Wait...
Did you count rotation? Maybe it's $C(5,4)$.

[quote="cat810930"]Colors: [black, white, red, blue, green]
If your are to color the four sides of a tetrahedron(a 4-sided 3-D object, each side is a triangle) with four different colorschosen from the five colors above, how many distinct tetrahedrons can you make?

When I draw it out I got 10 as the answer, but I would like to know how to get the answer without all the drawings. I believe one of the MATHCOUNTS posters has a similar problem with a cube, so I would assume this to be middle school level.[/quote]

Well if 10 is the answer, then since there are 5 ways to choose the 4 colors, maybe each combination of colors has 2 ways to arrange it on a tetrahedron.

You have to be clear, in knowing exactly how you "count" .  Suppose  you have a tetrahedron  called A. (and you put it down with, say the red face down)
Now you take an identical tetrahedron,  but put you down with different color face down.

These two would look different,  (If they are, say  photographed)  but if you [b]allow  rotation  [/b] they can be called identical.

In this case  each tetrahedron is equivalent to  12  (You can select any face to be called down (4 possibilities) and then have one of the 3 face to face you) .

So answer is $(5 \cdot 4\cdot 3\cdot 2 ) / 12 = 10$
Hope this helps.


geometry
incenter
inequalities
geometry proposed

Let $ABC$ a triangle.Let $AD$ altitude of the trinagle,$D\in BC$.Let $I$ the incenter of the triangle $ABC$.
 Prove that:
                                           $r\leq DI \leq R-r$ ;)

A interesting inequality!. If $M\in BC\cap (I,r)$ then $DI\le R-r\Longleftrightarrow DM\le OI.$
Thus, the your inequality becomes very strongly.
I hope to be true. I will be come again afterwards.

[b]Remark.[/b] $DM\le HI,\ HI\le 2\cdot OI.$ (I have a nice solution for the second). Can say something.

Levi,this inequality is true for any triangle, at least I think so. :?
Maybe, by a counterexample......if you have one. :)


Digit $ b$ is the units digit in the six-digit number $ 213,20b$, which is divisible by 11. What is the value of $ b$?

Using the divisibility rule for 11,

2-1+3-2+0-b=2+b=11

b=9

213209 is not divisible by 11
do it on a calc.....

anyway, using the divisiblity rule for 11, we get

$ 2\minus{}1\plus{}3\minus{}2\plus{}0\minus{}b\equal{}2\minus{}b$

therefore, $ b\equal{}2$

[quote]do it on a calc..... [/quote]

I never like to hear that :P  "If I drop this apple, it will fall to the ground... check any physics textbook!"

To be more to the point, we should simply say that mewto55555 accidentally dropped the sign next to the b (hey, it happens to everyone once in a while).

One should also note when doing this it's not always $ \equal{}11$, rather it has to be a multiple of 11.

If your last step is $ C\minus{}b$, because $ 0\leq b \leq 9$, we have that $ C\minus{}9\leq C\minus{}b\leq C$.  In this case C=2, so $ \minus{}7\leq 2\minus{}b\leq 2$.  We need to pick the only multiple of 11 between -7 and 2... that is, 0!  So 2-b=0 gets us b=2.

Another example, using the same test on $ 19191b$, your last step will be $ \minus{}15\minus{}b$.

So $ \minus{}24\leq \minus{}15\minus{}b\leq \minus{}15$

The only multiple of 11 between -24 and -15 is -22.

Thus $ \minus{}15\minus{}b\equal{}\minus{}22\implies b\equal{}7$.

[quote="RightAnnihilator"][quote]do it on a calc..... [/quote]

I never like to hear that :P  "If I drop this apple, it will fall to the ground... check any physics textbook!"

To be more to the point, we should simply say that mewto55555 accidentally dropped the sign next to the b (hey, it happens to everyone once in a while).

One should also note when doing this it's not always $ \equal{} 11$, rather it has to be a multiple of 11.

If your last step is $ C \minus{} b$, because $ 0\leq b \leq 9$, we have that $ C \minus{} 9\leq C \minus{} b\leq C$.  In this case C=2, so $ \minus{} 7\leq 2 \minus{} b\leq 2$.  We need to pick the only multiple of 11 between -7 and 2... that is, 0!  So 2-b=0 gets us b=2.

Another example, using the same test on $ 19191b$, your last step will be $ \minus{} 15 \minus{} b$.

So $ \minus{} 24\leq \minus{} 15 \minus{} b\leq \minus{} 15$

The only multiple of 11 between -24 and -15 is -22.

Thus $ \minus{} 15 \minus{} b \equal{} \minus{} 22\implies b \equal{} 7$.[/quote]

Its much faster to use the divisibility rule for 11, but how did you get b=7?

@dragon

the answer to his example is 7, the answer to this problem is 2, if that's what your asking

BTW, when i said do it on a calc, i meant check mewto's answer on a calc...it doesn't work
i gave my solution after

Oh. I didn't even read what he was doing. I thought he was doing the problem. Sorry about that!

I'm a bit old fashioned... I think using your calculator to perform division of a number with only 6 digits is like driving your Hummer to your mailbox :D

But that probably comes from being bitter toward my students poor math skills...

Anyhow yeah, before you ask a question about a post it's a good idea to read it!

Yea. I will (hopefully). This is getting a little off topic. Please lock this thread izzy.


I have this trouble. When I get someone posted on this on my topic, I receive a reply. Then when someone replies to it again, I don't receive mail for that topic again.

And today, I had this weird problem. I received a mail saying I have a new private message but I didn't get one.

Somebody fix this. I need to get this fixed because people will send me the answers for my contest this Friday and I need it to be fixed.

Thanks.

The email for the new private message was almost certainly someone sending you a PM then deleting it before you retrieved it.  We'll look into the other error.


group theory
abstract algebra
superior algebra
superior algebra unsolved

I need hint to show this problem using Sylow technique :

if $ G$ is a p-group , and $ H \triangleleft G$ ,$ H\ne\{e\}$ then $ H\cap Z(G)\ne \{e\}$.

I don't see how Sylow techniques will help you. Sylow's Theorem deals with the maximal $ p$-group in a given group, and our whole group is already a $ p$-group.

I will give you a hint though. Have you seen the proof that a $ p$-group has a non-trivial center? Try to modify that proof ;)

[hide="More Hint"]Think conjugacy classes.[/hide]

Thank you calc ;
I found this solution for what u said dude  :D

http://planetmath.org/encyclopedia/ProofOfANontrivialNormalSubgroupOfAFinitePGroupGAndTheCenterOfGHaveNontrivialIntersection.html

But depending on the following lemma :

If $ G$ is a $ p \minus{} group$ , then $ |X|\equiv |X_o| \ mod \ p$, where $ X$ is a $ G \minus{} set$,

Where $ X_o \equal{} \{x\in X: g.x \equal{} x , \forall g\in G\}$

by putting $ H \equal{} X$ and using conjugation action on elements , I got the result trivially . :wink:

Yep. Those two solutions are actually equivalent. The class equation is simply a way of breaking up the group into orbits under conjugation. Similarly, the proof of that lemma involves breaking up $ X$ into orbits.


There are 47 students in a classroom with seats arranged in 6 rows $ \times$ 8 columns, and the seat in the $ i$-th row and $ j$-th column is denoted by $ (i,j).$ Now, an adjustment is made for students’ seats in the new school term. For a student with the original seat $ (i,j),$ if his/her new seat is $ (m,n),$ we say that the student is moved by $ [a, b] \equal{} [i \minus{} m, j \minus{} n]$ and define the position value of the student as $ a\plus{}b.$ Let $ S$ denote the sum of the position values of all the students. Determine the difference between the greatest and smallest possible values of $ S.$

[quote="Andy Loo"][quote="feliz"]Isn't it $ |a| \plus{} |b|$?[/quote]
No! It's $a+b$.[/quote]

So a student moved from $(1,5)$ to $(5,1)$ has the position value of $0$?

[quote=vsathiam]Idk if this can constitute as a proof, but the problem's summation of position values is essentially equivalent to the the difference between initial total state value and final total state value. (Here I am defining state value as the sum of the coordinates of a space.)

Our maximum total position value is acquired for any configuration with an initial empty space at (1,1) and final empty space at (6,8). This gives a value of 12.

The minimum occurs when the above two states are flipped, and gives -12.

So the difference is 24.

*2 things: can someone please tell me if the above solution is legit? Here's an example to help illustrate my solution:

Define position value similarly for initial sequences (1,2,3...n) and $(p_1,p_2,p_3...p_n)$ where second set is a permutation of the former one. Sum of both sequences are fixed, the total position value should always be 0.[/quote]
Specifically $S=\sum_{\text{filled old}}i+\sum_{\text{filled old}}j-\sum_{\text{filled new}}i-\sum_{\text{filled new}}j=i_{\text{empty new}}+j_{\text{empty new}}-i_{\text{empty old}}-j_{\text{empty old}}$
Then the max is clearly $14-2$, and the min is $2-14$ giving $\boxed{24}$


Hi,

I have one suggestion for a new feature, a feature I have seen implemented on a site with similar forum software. That is a button choice in the private message areas to save all the private messages in a specified format (I would prefer plain text, but other choices are possible) to the user's personal computer. The site where this feature exists, and where I have used this feature, is

http://talk.collegeconfidential.com

and I suspect that there is some fairly standard implementation of this feature somewhere in the PhPBB community.

Being able to save ALL private messages, outgoing and incoming, without using lots of your server space would be very helpful to me in my task as a forum moderator. I have the lawyer's habit of always wanting to save an authenticated copy of everthing I write  :)  .  Could you see how much bother there is in implementing this feature and maybe move it up on your implementation queue? Thanks for your help with this and for all the fine features of this site.

It's been added to the queue now. But now it´s really not the time. Sent me a private message / email in a couple of weeks and we'll see what we can do about it then.


MATHCOUNTS
probability
geometry
3D geometry
counting
distinguishability

I thought about starting a thread called "MathCounts Tip of the Week" where, every week, a tip for taking MathCounts  tests or doing MathCounts-like problems is discussed. I would put a link to where someone could find more information about the tip, like "the Shoelace Theorem" and maybe a problem, and then it would be discussed.

I would probably need helpers for this. If you are interested in helping, you can post here or PM me.

Would you want to participate in the discussions? No signups are needed, but I want to see if there will be inactivity or not. Please answer by posting or voting in the poll.

@Mods: Is this forum a good place for this?

i like it. i wanna help. I have a few tricks/concepts. I'll pm u more information.

Here is some more information on how it will work:

I am hoping that others and I learn from more experienced people. For example, Person A could be good at Counting and Person B could be good at Similar Triangles. Person A teaches Person B good techniques for counting and Person B teaches Person A about how to use and identify Similar Triangles.

Instead of the above example, there would be many people involved and the subject will change weekly. Then, if someone still doesn't understand it, they could ask about it in another topic.

Good idea. I think a lot of people could benefit, as long as it doesn't sink down the threads list. Is it topics and concepts, or also strategy, stupid mistakes, speed, pressure, etc.?

[hide="Helpers (2)"]
HiDN428
pythag011
[/hide]

The tips will be about topics and concepts as well as strategy, stupid mistakes, speed, pressure, etc. The strategy, stupid mistakes, speed, pressure, etc tips will probably be posted a few at a time, because one isn't enough for one week's discussion.

Also, you could make requests and suggestions.

I am hoping to start it next week or sometime soon. I might post one every weekend, so it would start one week from now.

Keep the comments, suggestions, ... coming.

I also NEED HELPERS.

Update:

[hide="Helpers"]
HiDN428
pythag011
Ihatepie
[/hide]

[hide="Helpers (4)"]
HiDN428
pythag011
Ihatepie
nikeballa96
[/hide]

Thanks for helping.

Does anybody have any suggestions? Anything other than "I like it"?

I don't like it. :P

/me helps

Don't put it in a blog. Nobody will notice it then.

I was planning to put it in the Middle School Contests forum.
@1=2: Is that a joke or real? I am assuming it is a joke from the smilie.

[hide="Helpers (5)"]
HiDN428
pythag011
Ihatepie
nikeballa96
1=2
[/hide]

hey... I might want to be some sort of helper

I will put you for Week 2.

Helpers:
[hide="Week 1 (10)"]
HiDN428
pythag011
Ihatepie
nikeballa96
1=2
mathgenius1
Mewto55555
batteredbutnotdefeated
james4l
RoFlLoLcOpT
[/hide]
[hide="Week 2 (3)"]
FlyAgaric
Thunder365
Sabien5
[/hide]

Helpers:

[hide="Week 2 (5)"]
FlyAgaric
Thunder365
Sabien5
pytheagle
BOGTRO
[/hide]

I wont post week 1 helpers again until I post the tip.

@FlyAgaric: You are already a Week 2 helper, and it is okay if you don't do a lot.

This will start tomorrow. I will post the rules, including a helper's jobs, and then the first tip in a different post.

Helpers: You can also suggest topics for the next week.

The Tip of the Week is pretty much dead. Should I still continue it? If you want it continued, then please contribute to the discussion. Also, sign up as a helper if you are actually going to do something. Vote by posting here saying whether I should continue it or not.

Yes. I would like it to continue but i don't have the time to help. Sorry PowerofPi.

I think that it should continue because it has really taught me stuff that i didn't know. And a suggestion for the next tip is Balls and Urns

Okay, tomorrow I will post week 3 (counting arrangements) and balls and urns for week 4. If inactivity continues, I will pause it again. PLEASE POST!

Here's a balls and urns problem: how many ways can two identical cubes be given to three different people?

@powerofpi, I'm sorry I didn't post earlier, I kinda forgot :blush:

But.. you never explained balls and urns

Balls and Urns is a term used when distributing objects into groups. For example, say we have 4 balls and we need to put it into 2 urns. How many ways can we do this?

Let * denote a ball. Thus, our balls look like this:

****.

We want to cut this into 2 groups. Let | be the cut. Now we have

|****.

We need to arrange these (Note that **|** denotes 2,2, *|*** denotes 1,3, etc.). There are 5 objects, and 1 divider. Thus there are $ \binom{5}{1}\equal{}5$ ways to arrange them. Checking, we get the solutions (0,4), (1,3), (2,2), (3,1), (4,0) using algebra, and there are in fact 5 solutions.

This works only if

1. The balls are indistinguishable
2. The urns are distinguishable.

This can be applied to harder problems. For example, if I have 8 apples and 3 oranges, and I want to give them to 4 national mathcounts participants, but 1 of them doesn't eat apples and a different one doesn't eat oranges, we can now take care of this easily. We focus on the apples and oranges seperately. There are 3 participants eating apples, so we have 8 balls, and 2 dividers (make sure you understand why there are only 2 dividers).

There are 10 total objects, and 2 dividers, so there are $ \binom{10}{2}\equal{}45$ ways to give out the apples.

We also have 3 oranges to give to 3 kids. Using the same method as before, there are 5 objects and 2 dividers, making $ \binom{5}{2}\equal{}10$ ways to give out the oranges.

Our final answer is just $ 45 \cdot 10\equal{}450$. Balls and urns made an imtimidating looking problem managable.


yeah...my mom always gets mad when i twirl my pen and almost every smart math person i know does it.  do you?

Yep... :D

I do the one called "Sonic Normal" all the time, except I start between my ring finger and pinky:
http://www.pentrix.com/videos/videolist.php

yep sonic and thumbaround but i havent learned the reverse of either yet =( it became a big fad at our school for a period of time.

well i use pencils instead of pens but i'm kind of fidgety with those

mainly i just twirl my eraser though.

I don't twirl them so much as I toss them up in the air constantly. People are always getting pissed b/c of it.

I lose my pencils... so I can't really learn how to twirl them.

How DO you twirl them anyway?

Eh, I can only do the thumb around, and only the forward direction at that.

I attempted to teach myself to thumbaround, and I could do it about once every three tries, but my history teacher's gonna teach the class on the last day of school, so hopefully I'll be able to do it by then.

It's more of a skill acquired with practice rather than someone just showing you...it takes roughly a week's worth of practice(assuming you only do it during class =P) to get it close to a 100% of the time.

I can't twirl my pencil; I just put one between my index and my thumb and then move my fingers up and down, creating an hourglass.

Yup. In between index and middle. I use a PhD Ultra pencil and since it's fatter than regular #2 pencils its harder to twirl (but for some reason I can't twirl #2 pencils that well...). My friend does it with his drum sticks (well all percussionists to it) and we started trying to do it (and dropping them) at the MD state band adjudication  :D

and darnit nearly all those vids are for right handed people...grr I'm left handed!

lol

i do that thing where you flick it around your thumb

except i do it twice instead of once

looks shinier

lol i tried once
...
ya it didnt turn out too great... x_x
but almost every asian @ my school does it...

Other times, I just put my pencil between index and middle, lead point on the palm side of my hand, and shake my hand so that it hits it. Great for annoying people!!! :P

I just can't do it. It looks really cool and I want to do it but I fail every time. :blush:

I twirl it with my 2nd,3rd, and 4th fingers. Nowadays, I do it without noticing usually while taking a test or working on something or thinking.

I twirl it around my fingers and thumb..nothing to fancy...i usualyl do it to kill time in class..i picked it up from some asian ppl in like grade 10..lol..

I really need to learn this valuable skill! But I don't have any guide! HELP :surrender:

I also need to juggle four balls at once.

I've never heard of a single person who twirls their pen and frankly, it looks pointless.

i can only do the sonic. there's this one kid at my school who's really good

Well I'm about the only kid at my school that twirls, and I can only do the forward thumbaround, but I can do it with almost all that I've tried (except TV remotes...grr)...I haven't tried drumsticks though...too bad I play the icky viola

[quote="iostream.h"]i can do sonics and thumbarounds forwards and backwards.  i'm getting pretty good at fingerpassing up and down very quickly.

i actually picked up this skill in debate, not math/physics contests.  debaters are wayyyy more intense about pen twirling than mathletes.[/quote]

Apparently, thats what every single debater does, I swear. At our school i think every debater knows how to pen spin


algebra
polynomial
group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $F$ be a field of characteristic not $2$, let $K$ be a Galois extension with $[K: F]=4$. Prove that if $Gal(K/F)=\mathbb Z_{2}\times \mathbb Z_{2}$, then $K=F(\sqrt a,\sqrt b)$ for some $a,b \in F$.

$\exists \alpha\in F$ s.t. $K=F(\alpha)$; let $P$ be the $\alpha$'s minimal polynomial of degree 4.
Necessarily $Gal(K/F)$ is the permutation subgroup $V=\{id,(12)(34),(13)(24),(14)(23)\}$ because the other groups isomorphic to $\mathbb{Z}_{2}.\mathbb{Z}_{2}$ don't act transitively on the $P$-roots. $V$ admits 3 subgroups $H_{1}=\{id,(12)(34)\},H_{2}=\{id,(13)(24)\},H_{3}=\{id,(14)(23)\}$. With the Galois correspondance theorem we obtain 3 distinct $K$-subfields $F(H_{i})$ s.t. $[F(H_{i}),F]=2$. $F(H_{i})=F(\sqrt{a_{i}})$ where $a_{i}\in{F}$ isn't a square. Thus $K=F(\sqrt{a_{1}},\sqrt{a_{2}})$.

Thanks loup blanc! What do you mean by "the other groups isomorphic to $\mathbb Z_{2}\times \mathbb Z_{2}$ don't act transitively on the $P$ roots"?

Hi captain; I wasn't clear. Let $S_{4}$ be the permutations of the $P$-roots: $\{1,2,3,4\}$.
Among the $S_{4}$-subgroups of order 4 only $V$ acts transitively on the $P$-roots ($\forall{i,j},\exists\sigma\in{V}$ s.t. $\sigma(i)=j)$. For example $U=\{id,(12),(34),(12)(34)\}$ is a $S_{4}$-subgroup of order 4 but $U(1)=2$ and $U$ can't be the Galois's group of $P$.

A precision: in my last post I speak about $S_{4}$-subgroups of order 4 which are isomorphic to $\mathbb{Z}_{2}.\mathbb{Z}_{2}$.


[hide]The general term of each term of this sum is:

$\frac{1}{(n+1)\sqrt{n}+n\sqrt{n+1}}=\frac{(n+1)\sqrt{n}-n\sqrt{n+1}}{(n+1)n}=\frac{\sqrt{n}}{n}-\frac{\sqrt{n+1}}{n+1}$ for $1\le n \le 99$ a natural number.

with this we know that two consecutive sum term cancel each other and the remainders are the first and last term,in consequence the sum is:

$\frac{\sqrt{1}}{1}-\frac{\sqrt{100}}{100}=\frac{100-\sqrt{100}}{100}=\frac{9}{10}$[/hide]

Maple says it's 0.8999999995.. computers with their floating point arithematic..  :P


linear algebra
matrix
topology
linear algebra unsolved

if $A \in M_{2n}$,and such that,$A\left( \begin{matrix}0 & I_{n}\\-I_{n}& 0 \end{matrix}\right)A^{T}= \left( \begin{matrix}0 & I_{n}\\-I_{n}& 0 \end{matrix}\right)$,Prove:det(A)=1

switching two columns changes the sign of the determinant

switch the first column with the n+1 th
the second with the n+2th

and the n-th with the last

now you have

$\left( \begin{matrix}I_{n}& 0\\0&-I_{n}\end{matrix}\right)$

the determinant of this is $(-1)^{n}$

this gives the desired result

Sorry,I am not good in Linear algebra.
Then I can't understand your proof.
Do you make the columns changes both side?
And what will the left side happen?
Can you(Or somebody)explain to me?Thank you.

This is a very famous and classical property of simplectic matrices, but the proof is definitely more involved than the previous one.

What do you mean?What about fredbel6's proof?

fredbel6 you did not prove that det(A)=1, you just swicht column of the
matrix RHS. You did not look what happen on the LHS wich is a product
of three matrices.

Here some hint

[hide]

First case
A=
(X  Y)
(Z  T)  with T inversible use it in the statement you will get relations

on X,Y,Z,T

try to write A as a product of two triangular matrices by blocks wich involves X,Y,Z,T

then you 'll get det(A)=1 [/hide]

---------------
As harazi said this symplectic matrix wich is used in physics
for instance quantic mecanic.
For advance see sympletic manifold

[quote="zhaobin"][quote="Soarer"]I think fredbel6 means you can switch the rows in that way, so that the det required is $(-1)^{n}$ times of the new matrix. But for a matrix of that form we can directly compute the determinant by $det(I_{n})det(-I_{n}) = (-1)^{n}$, so..[/quote]
Thank you.
But do you mean after switching,the LHS's determinant is $(-1)^{n}det A$
How can I see it?[/quote]

Oh, now I see I interpreted the problem wrongly.
I thought you mean A equals that matrix and you made a typo in $A^{T}$.

Here the proof when  T is invertible

[hide]

A=
(X Y)
(Z T)  is symplectic with T invertible

K=
(0 I)
(-I 0)

the conditions $A K A^{t}=K$  where $A^{t}$ is the transpose of $A$

gives

$X.Y^{t}=Y.X^{t}$
$Z.T^{t}=T.Z^{t}$
$X.T^{t}-Y.Z^{t}=I_{n}$

on the other hand you can write $A$ as a product of two trinagular matrices by blocks

A=
(I  Q) ((X-QZ)      0)
(0  I) (  Z            T)    with $Q=Y.T^{-1}$

take the determinant and use the three relations aboves

$det(A)=det(X-QZ)det(T) = det(X.T^{t}-Y.Z^{t})=det(I_{n})=1$

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

Now you have to examine the case T is not invertible.

[/hide]

Zhaobin was it exam problem ?

I think fredbel means that we have by multiplicity of determinant that $det A \cdot 1\cdot det A^{T}= 1$, saying that $\det A = \pm 1$. From the comments I assume it's highly nontrivial to exclude the case $-1$ though.

@ moubi: T means transposition, right? how can that not be invertible?  :oops:

i changed notation peter vdd

I note transpose $^{t}$

$T$ is right down corner in the matrix A by decomposition by block
$T$ is a  matrix

[quote="Peter VDD"]

@ moubi: T means transposition, right? how can that not be invertible?  :oops:[/quote]
If T is not be invertible,we have Y is invertible(This is because $det(A)\neq 0$
Thus:
we can write:
A=
(Q I) (Z T)
(I 0) (0 (Y-QT))
Here I define $Q=XZ^{-1}$

To Moubinool:
It is a problem I see in the Chinese forum. :)
Can you tell me why:
$det(X-QZ)det(T)=det(XT^{t}-YZ^{t})$
Thank you :)

$det(X-QZ)det(T)=det(X-YT^{-1}Z)det(T^{t})=det(XT^{t}-YT^{-1}ZT^{t})$ but $ZT^{t}$ is symetric =$TZ^{t}$

you get $det(XT^{t}-YZ^{t})=det(I_{n})=1$

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

what it the link for chinese forum seems maths problems from there are interesting

:blush:
 And the link is:
[url]http://www.bossh.net/forums[/url]

 :)

Just found this thread when searching for an old matrix problem.

The quickest solution to this one uses the formula $\operatorname{pf}\left(UVU^T\right) = \det U \cdot \operatorname{pf}V$ for any $2n\times 2n$-matrix $U$ and any skew-symmetric $2n\times 2n$-matrix $V$, where $\operatorname{pf}$ means the Pfaffian. (See Lemma 1 in [url=http://mathoverflow.net/questions/47923/relationship-between-determinants/47975#47975]my answer to MathOverflow question #47923[/url] for a quick and dirty proof of this.) This can be applied to $A=U$ and $V=\left( \begin{matrix}0 & I_{n}\\-I_{n}& 0 \end{matrix}\right)$, and it remains to recall that $\operatorname{pf} \left( \begin{matrix}0 & I_{n}\\-I_{n}& 0 \end{matrix}\right) = \pm 1 \neq 0$.


calculus
trigonometry
integration
function
derivative
calculus computations

Hi, I didnt know where else to put this, so here it is...I think this is the 1969 AP Calculus AB #1 for the FRQ section, but I can't solve it whatsoever...and the solutions arent online:

A particle moves along the y-axis with velocity given by $v(t) = t\sin(t^{2})$ for $t \ge 0$.

a.) In which direction (up or down) is the particle moving at time t = 1.5? Why?

b.) Find the acceleration of the particle at time t = 1.5. Is the velocity of the particle increasing at t = 1.5? Why or why not?

c.) Given that y(t) is the position of the particle at time t and that y(0) = 3, find y(2).

d.) Find the total distance traveled by the particle from t = 0 to t = 2

THANKS A LOT FOR ALL THE HELP NEEDED IN ADVANCE

Kunny, thanks for your help!

For part d, I have a question. The answer that you got $\frac{1}{2}(1+\cos{4})$ doesn't seem to match with the integral:

$\int_{0}^{2}|t\sin{t^{2}}|dt$. If you integrate this integral, using any method (calculator or by hand), you obtain: 1.17318, and when you integrate:

$\int_{0}^{\sqrt{\pi}}t\sin t^{2}dt-\int_{\sqrt{\pi}}^{2}t\sin t^{2}dt$, you get .17318

Is there a mistake?

Furthermore, for part b.) is the acceleration of the particle increasing or decreasing? I assume it is decreasing because the acceleration is negative right?

[quote="prime_lover"]
for part b.) is the acceleration of the particle increasing or decreasing? I assume it is decreasing because the acceleration is negative right?[/quote]

Note that the question asks whether the [i]velocity[/i] is increasing or decreasing, not the acceleration.  To tell whether the velocity (or any function for that matter) is increasing or decreasing at a point, evaluate its derivative at that point (It just so happens that acceleration is the derivative of velocity.).  If the acceleration is positive, the velocity is increasing; if the acceleration is negative, the velocity is decreasing; and if the acceleration is zero, the velocity is not changing.

[quote="Zakary"]I'm curious... How did you get that question?  From a teacher, adviser, etc.? online? a book?  Could I somehow manage to get questions from the same source?[/quote]

AP Teachers have access to all of the Free Response Questions, and 8 years of the Multiple Choice Questions going back to 1969. Some of the much older questions now appear in various AP study guides.


Find the standard matrix for the orthogonal projection onto the line with parametric equations
$ x\equal{}2t,y\equal{}\minus{}t,z\equal{}4t;t\in\mathbb{R}$.

Since the basis $ A\equal{}(2,\minus{}1,4)$, I tried using the formula for the standard matrix $ A(A^TA)^{\minus{}1}A^T$ but it gives me 1.  :huh:

In what you wrote, $ A$ should be a column vector, $ A\equal{}\begin{bmatrix}2\\\minus{}1\\4\end{bmatrix}.$

Then $ A^TA$ is $ 1\times 1$ and equals $ 21$ so that $ (A^TA)^{\minus{}1}\equal{}\frac{1}{21}.$

That makes

$ P\equal{}A(A^TA)^{\minus{}1}A^T\equal{}\frac1{21}\begin{bmatrix}2\\\minus{}1\\4\end{bmatrix}\begin{bmatrix}2&\minus{}1&4\end{bmatrix}\equal{}\frac1{21}\begin{bmatrix}4&\minus{}2&8\\\minus{}2&1&\minus{}4\\8&\minus{}4&16\end{bmatrix}.$

You can check for yourself that $ P^2\equal{}P.$

Oops, I missed up in multiplying out the last matrix... but thanks anyways.


conics
trigonometry
ellipse
parabola
hyperbola
geometry
geometric transformation

I am going to try answering this but it might take me a few mins to type out.

EDIT: Here goes.

Okay you have to be familiar with the polar forms of conic equations (most are only familiar with the rectangular form).  These equations are in the form:

[tex]r=\frac{ep}{1{\pm}e\sin\theta}[/tex]

or

[tex]r=\frac{ep}{1{\pm}e\cos\theta}[/tex]

Now each of these equations, when you change the cos and sin, or the  :pm: , these change the way the graph is oriented, how wide it is, or w/e I don't remember play around w/polar graphs on the TI to figure it out.

However, its the value of e, meaning eccentricity, that gives the TYPE of conic it is:

0<e<1  gives you a ellipse
e=1 gives you a parabola
e>1 gives a hyperbola

(note that a circle is just r=a constant)

So your goal is to get [tex]r=\frac{\cos\theta-\sin\theta}{2\cos^2\theta}[/tex] into that form.

Be back, Ill finish how to solve after dinner (If I can figure it out!)

EDIT:

Okay I am back but I can't figure it out.  I thought you might be able to do it using trigonometric identities and such.  Maybe this is the rotation of axis ones.  I am unsure.

[quote]I haven't tried this problem yet but couldn't you just convert to rectangular?[/quote]

Well, you would have to get the equation in standard form that I mentioned to convert to rectangular.  Getting it into standard form is the problem.

[quote]Err... 2cos^2 x = 1 + cos 2x..[/quote]

That could be helpful, could you explain a little more?

If you substituted it in it would look [i]more[/i] like the standard form.

I didn't have a solution  :oops:

But another thing you might use is

cos x - sin x = sqrt(1 - sin 2x)

Don't take me too seriously though.  I don't really know what I'm doing on this one.

Try converting everything to x's and y's. Remember that x = r cos :theta: and y = r sin :theta:.

(spoiler)[hide]You should come out with y = x - 2x2. So it's a parabola with eccentricity 1.[/hide]

[quote="Rep123max"]I think I might have it, what through me off (sadly) was [tex]\tan\theta[/tex].

Find the value of the eccentricity of the conic described by the polar equation:
[tex]\displaystyle r=\frac{\cos\theta-\sin\theta}{2\cos^2\theta}[/tex].

We cross multiply and get [tex]2r\cos^2\theta=\cos\theta-\sin\theta[/tex].

Then we divide by [tex]\cos\theta[/tex] and get [tex]\displaystyle 2r\cos\theta=1-\frac{\sin\theta}{\cos\theta}[/tex].

We simplify to get [tex]\displaystyle 2x=1-\frac{y}{x}[/tex] and after cross multiplying we get [tex]2x^2-x=-y[/tex] so [tex]y=x-2x^2[/tex] so it is a parabola which has eccentricity of 1.
[/quote]

Technically, I think you're supposed to multiply sin/cos by r/r, but it really doesn't matter.

[quote]
Stupid [tex]\tan\theta[/tex].[/quote]

lol


calculus
integration
inequalities
function
triangle inequality
absolute value
advanced fields

f(z) is analytic on and within the unit circle. Also, |f(z)| < 1 for |z| <= 1 and f(0) = 0. Show that f(z) < |z| for |z|<= 1.

I know this can be done by the max mod principle, but I'm following a way which doesn't seem to use it.

After showing f(z)/z is analytic, I am to express [f(z_0)/z_0]^n by the Cauchy integral formula, so that maybe later I would come with |f(z_0)/z_0|^n <= 1. I still can't get any expression via the Cauchy integral formula that does this. Maybe someone has an idea to do this?

Any help would be greatly appreciated! Thanks..

I'm not quite sure why you're trying to avoid the maximum modulus principle, since that principle is so naturally suited to the proof of this theorem. I need to guess what you mean by "[f(z_0)/z_0]^n".

Maybe this is what you're looking for: Suppose that $|h(\zeta)|<1$ for $|\zeta|= 1$ and note that for $\zeta$ on the boundary circle, $\frac1{|\zeta-z|}\le\frac1{1-|z|}$. Put this into the Cauchy Integral Formula for $h$ to get that $|h(z)|<\frac1{|2\pi i|}\int_{|\zeta|=1}\frac{|h(\zeta)||d\zeta|}{|\zeta-z|} < \frac1{1-|z|}$. Now let $h(z) = (g(z))^n$ where $g(z)=\frac{f(z)}z$. Since this satisfies the boundary estimate no matter what $n$ is, we get that $|g(z)|^n<\frac1{1-|z|}$   $\forall n$. This estimate can only be true independent of $n$ if $|g(z)|\le 1$. That's our proof. It also seems to be an extra-hard way to prove the maximum modulus principle in the first place.

wow! that was great kent. thanks a lot! :lol:

just some minor questions..

how did you get the inequality $ |{z_1 - z_2}|\ge\,||z_1| - z_2| $, and did you mean $ |h(z)| \le \frac{1}{|2 \pi i|} \displaystyle\int_{|\zeta|=1}\frac{|h(\zeta)||d\zeta|}{|\zeta - z|} $

i cant use the max mod becaue it wasnt proved in the book.. its a math physics text, unfortunately.  :(

again, thanks a lot!

I've just realized a mistake in what I wrote: that shouldn't be $ \frac1{|\zeta-z|}\le\frac1{|1-z|} $; instead it should be $ \frac1{|\zeta-z|}\le\frac1{1-|z|} $. I'll go back and edit this to fix it.

As to where that inequality involving $|h(z)|$ and the integral: that simply is the Cauchy Integral Formula, followed by applying the theorem (think of it as an analogue of the triangle inequality) that the absolute value of an integral is less than or equal to the integral of the absolute value. (Have you seen the proof of that for complex valued functions?)

I still don't particularly like this proof, becuase I would prefer to first prove, then use, the Maximum Modulus Principle. I see Max Mod as a more elementary and more accessible theorem than Schwarz, so why not prove it first?

thanks for clearing that up. :)  we just did it via max mod in class this morning, and yup, it was more straightforward and much easier.

thanks for helping out! it was nice knowing two proofs of the same theorem.


trigonometry
inequalities
inequalities unsolved

[quote="sylow_theory"][quote="phuong"]Let  $ABC$ be a triangle. Prove that: $\sin\frac{A}{2}+\sin\frac{B}{2}+\sin\frac{C}{2}\geq\frac{4}{3}\left(1+\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\right)$[/quote]

The inequality,
$\sin \frac{A}{2}+\sin \frac{B}{2}+\sin \frac{C}{2}\geq \frac{3}{2}$
(It was mentioned here before several times).

Thus, if you can show that,

$\frac{3}{2}\geq \frac{4}{3}\left(1+\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\right)$
Iff,
$\frac{1}{8}\geq \sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$
Then it works.

Of show that (iff),
$1\geq (1-\cos A)(1-\cos B)(1-\cos C)$

It seems to be true. Cuz,
For an equaliteral triangle this is equality.[/quote]

it should be $\sin \frac{A}{2}+\sin \frac{B}{2}+\sin \frac{C}{2}\leq  \frac{3}{2}$ direct jensen result

The sign is reversed. :wink:

[quote="andy_tok"]

it should be $\sin \frac{A}{2}+\sin \frac{B}{2}+\sin \frac{C}{2}\leq \frac{3}{2}$ direct jensen result

[/quote]
Right

[b]A similar problem[/b]:
$\quad\cos\frac{A}{2}+\cos\frac{B}{2}+\cos\frac{C}{2}\ge\frac{4}{\sqrt{3}}\left(1+\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\right)$

But now no ideas to kill it :|

[color=purple]No idea ? I don't think so.Now i don't have time to post all solution when i free i'll post it [/color]

May be the following rezult relating with above ine:
$2(\sin A\sin B+\sin B\sin C+\sin C\sin A)\ge \cos A+\cos B+\cos C+\frac{8}{\sqrt{3}}\sin A\sin B\sin C$

This ineq is sharp.
it's equivalent to$2\sum\sin^{2}{\frac{A}{2}}+3\sum\sin{\frac{A}{2}}\geq 6$
can't get nowhere after that :blush: but i'll try

[quote="hien"]
\[2(\sin A\sin B+\sin B\sin C+\sin C\sin A)\ge \cos A+\cos B+\cos C+\frac{8}{\sqrt{3}}\sin A\sin B\sin C\quad (*) \]
[/quote]
[hide="Hint."]
\[(*)\Longleftrightarrow \frac{4}{\sqrt 3}S\le\frac{1}{2}(ab+bc+ca)-R(R+r) \]
And we have to prove that:
\[\Longleftrightarrow \frac{4}{\sqrt 3}pr\le 2r^{2}+5Rr \] :wink:
[/hide]

When A,B approach 0 degree, the inequality fails.
So ABC shouldn't be a blunt triangle?

[quote="April"][quote="hien"]
\[2(\sin A\sin B+\sin B\sin C+\sin C\sin A)\ge \cos A+\cos B+\cos C+\frac{8}{\sqrt{3}}\sin A\sin B\sin C\quad (*) \]
[/quote]
[hide="Hint."]
\[(*)\Longleftrightarrow \frac{4}{\sqrt 3}S\le\frac{1}{2}(ab+bc+ca)-R(R+r) \]
And we have to prove that:
\[\Longleftrightarrow \frac{4}{\sqrt 3}pr\le 2r{2}+5Rr \]
:wink:
[/hide][/quote]

$\Longleftrightarrow \frac{4}{\sqrt 3}pr\le 2r^{2}+5Rr$
$\Longleftrightarrow 3(5R+2r)^{2}\ge 16p^{2}$
$\Longleftrightarrow 3(5R+2r)\ge 4(a^{2}+b^{2}+c^{2}+2ab+2bc+2ca$
$\Longleftrightarrow 3(5R+2r)^{2}\ge 4(2(a^{2}+b^{2}+c^{2})+(2ab+2bc+2ca-a^{2}-b^{2}-c^{2}))$ (1)
Because $2(ab+bc+ca)-a^{2}-b^{2}-c^{2}=4r(r+4R)$
Therefor $(1) \Longleftrightarrow 3(5R+2r)^{2}\ge 4(2(a^{2}+b^{2}+c^{2})+4r(r+4R))$ (2)
From the known inequality $a^{2}+b^{2}+c^{2}\le 9R^{2}$ then
$RHS (2)\le 4(18R^{2}+4r(r+4R))=3(5R+2r)^{2}-(3R^{2}-4Rr-4r^{2})=3(5R+2r)^{2}-(R-2r)(3R+2r)$
Hence $R\ge 2r$ deducing $RHS (2) \le 3(5R+2r)^{2}$. Done!


probability
search
probability and stats

We choose $ n$ points on priemter of a circle , find the probability that the center of circle lie inside of $ n$-gon.


So any of you guys writing the US physics team exam on Tuesday?

I know I just finished it a few hours ago.

What are everyone's thoughts on it? It was hard for me, mostly because I didn't study enough.

Are we supposed to talk about it yet? [I don't remember; I did it the last two years, but not this year.]

According to [url=http://www.aapt.org/olympiad2009/]this[/url] there is a contest window, (January 19 to February 2).

Wheeler didn't register in time, so I am trying to find somewhere else to take it.

You are not supposed to talk about the contest until February 20, I think.

Yes, since this is more or less the equivalent of an AMC for Physics, try to act like it's an AMC (since there is a long contest window, it's especially important not to discuss).


Study the convergence of the following sequence: (prove if it is convergent or not).
$ x_{n\plus{}1}\equal{}\sqrt{3\plus{}2x_n},$ where $ x_0\equal{}\sqrt{3}$.

$ x_{n\plus{}1}\minus{}3\equal{}\sqrt{3\plus{}2x_n}\minus{}3\equal{}\frac{2(x_n\minus{}3)}{\sqrt{3\plus{}2x_n}\plus{}3}$.

$ \therefore |x_{n\plus{}1}\minus{}3|\equal{}\frac{2}{\sqrt{3\plus{}2x_n}\plus{}3}|x_n\minus{}3|\leq \frac{2}{3}|x_n\minus{}3|$.

$ \therefore 0\leq |x_n\minus{}3|\leq \left(\frac{2}{3}\right)^{n}|x_0\minus{}3|$, yielding $ \lim_{n\to\infty} x_n\equal{}3$.

We clearly have $ x_n\equal{}\sqrt{3\plus{}2\sqrt{3\plus{}2\sqrt{3\plus{}\dots}}}$. Hence for large $ n$ we get $ x_n\equal{}\sqrt{3\plus{}2x_n}$ or $ x_n^2\minus{}2x_n\equal{}3$, which implies $ x_n\equal{}3$, since $ x_n>0$. Thus the sequence converges to $ 3$.

Mhmm kunny I don't understand what you're doing. It seems like you transformed the question into an epsilon-delta question and are using the definition of a limit. But what I don't understand is how you chose $ 3$ at the beginning, since you need to know that to prove the limit. :S

If the limit $ L$ exists, you can find it by solving $ L \equal{} \sqrt{3 \plus{} 2L}$ and using the answer which satisfies the initial condition (in this case one of the answers wouldn't work because the sequence is monotonic and starts after it). If you assume the limit exists in this way it gives you a good idea of what to work with to show boundedness.

[quote="nayel"]We clearly have $ x_n \equal{} \sqrt {3 \plus{} 2\sqrt {3 \plus{} 2\sqrt {3 \plus{} \dots}}}$.[/quote]
This doesn't mean anything.  The limit of the sequence is the definition of this expression, and you haven't proven that that limit exists.

Let $ f(x) \equal{} \sqrt{3 \plus{} 2x}$.  Then

$ \frac{f(x) \minus{} f(y)}{x \minus{} y} \equal{} \frac{\sqrt{3 \plus{} 2x} \minus{} \sqrt{3 \plus{} 2y}}{x \minus{} y} \equal{} \frac{2x \minus{} 2y}{(x \minus{} y)(\sqrt{3 \plus{} 2x} \plus{} \sqrt{3 \plus{} 2y})} \le \frac{1}{ \sqrt{3} }$

so $ f$ is a [url=http://en.wikipedia.org/wiki/Contraction_mapping]contraction mapping[/url] and has a unique fixed point, which can only be $ 3$.

Isn't that a little strong? Wouldn't the monotone convergence theorem suffice?

As the problem is written, but once you know that $ f$ is a contraction mapping you know that the limit doesn't depend on the initial value (as long as it's positive), which can be greater than $ 3$.  I suppose in that case $ x_n$ is monotone decreasing.  Either way, kunny's argument is the most elementary if we are going for clarity.

[quote="Fermat's Little Turtle"]Mhmm kunny I don't understand what you're doing. It seems like you transformed the question into an epsilon-delta question[/quote]

As I have frequently said so far, epsilon-delta doesn't appear to Japanese high school class, so I wrote the solution for them.

[quote="Fermat's Little Turtle"]
But what I don't understand is how you chose $ 3$ at the beginning, since you need to know that to prove the limit. :S[/quote]

Needless to say, as  max_tm showed, I suppose the existence of $ \lim_{n\to\infty} x_n$, or $ \lim_{n\to\infty} x_n\equal{} \lim_{n\to\infty} x_{n \plus{} 1} \equal{} \alpha \in{\mathbb{R}}$.

Alternative solution:

For $ f(x) \equal{} \sqrt {3 \plus{} 2x}$ let $ x_{n \plus{} 1} \equal{} f(x_n)$, since $ f(3) \equal{} 3$, by Mean Value Theorem, for the constant $ c$ such that $ x_n < c < 3$ or $ 3 < c < x_n$, we have

$ |x_{n \plus{} 1} \minus{} 3| \equal{} |f(x_n) \minus{} f(3)| \equal{} \left|\frac {f(x_n) \minus{} f(3)}{x_n} \minus{} 3\right\parallel{}x_n \minus{} 3|$

$ \equal{} |f'(c)\parallel{}x_n \minus{} 3| < \frac {1}{\sqrt {3}}|x_n \minus{} 3|$, yielding $ \lim_{n\to\infty} x_n \equal{} 3$.

[quote="t0rajir0u"][quote="nayel"]We clearly have $ x_n \equal{} \sqrt {3 \plus{} 2\sqrt {3 \plus{} 2\sqrt {3 \plus{} \dots}}}$.[/quote]
This doesn't mean anything.  The limit of the sequence is the definition of this expression, and you haven't proven that that limit exists.[/quote]

Sorry I meant to have $ n$ square roots in that expression which I should have stated more clearly.

[hide="non-calculus solution"]After a few iterations, we see that the total is always less than three. Let's try to prove it:

$ \sqrt{3\plus{}2x}\equal{}\sqrt{9}\equal{}3$ when $ x\equal{}3$, so when $ x<3$, then $ \sqrt{3\plus{}2x}\equal{}\sqrt{y}<\sqrt{9}\equal{}3$, or $ \sqrt{3\plus{}2x}<3$, where $ y<9$. Thus since $ x_0$ is less than three, $ x_n$ is less than three for all $ n$.

Now notice that $ \sqrt{3\plus{}2x_n}>x_n$. This is because if this is true, then $ 0>x_n^2\minus{}2x_n\minus{}3$, which has roots at -1 and 3. The region between -1 and 3 is negative, and thus $ x_n^2\minus{}2x_n\minus{}3$ is negative. Thus $ \sqrt{3\plus{}2x_n}>x_n$. Therefore this sequence is forever increasing, and is never equal to or greater than 3, so $ x_n$ becomes arbitrarily close, but doesn't touch, to 3.[/hide]

The problem with not using "calculus" is that you can't complete your solution.  You've shown that the sequence is monotonically increasing and bounded above by $ 3$.  To deduce from here that the sequence converges requires "calculus" (by which I really mean some basic lemmas in analysis or an argument like kunny's), and to deduce that the sequence actually converges to $ 3$ (and not some number less than $ 3$) requires some more "calculus."


Solve the system of equations \[ \{\begin{array}{lll} x^3-8x&=&y^3+2y\\ x^3-3&=&3(y^2+1)\end{array}\]

[hide]From the second equation,

$ x^3\minus{}6\equal{}3y^2$
$ y^2\equal{}\frac{x^3}{3}\minus{}2$

Substituting into the first equation,

$ x^3\minus{}8x\equal{}y((\frac{x^3}{3}\minus{}2)\plus{}2)$
$ y\equal{}3\minus{}\frac{24}{x^2}$
$ y^2\equal{}9\minus{}\frac{144}{x^2}\plus{}\frac{576}{x^4}$

Transitively,

$ \frac{x^3}{3}\minus{}2\equal{}9\minus{}\frac{144}{x^2}\plus{}\frac{576}{x^4}$

Obviously $ x$ does not equal $ 0$, so

$ x^7\minus{}33x^4\plus{}432x^2\minus{}1728\equal{}0$

From which point it seems a calculator is necessary...[/hide]

Is there a nicer way/does anyone know a non-calculator approach from here?

I can't find one. Although, it doesn't seem [i]that[/i] obvious that $ x \neq 0$, in fact, that actually gives the clean solutions $ (x,y) \equal{} (0, \pm i \sqrt{2} )$


factorial
function
algebra
polynomial
binomial coefficients
combinatorics proposed
combinatorics

Hi! :D I spent some time trying to solve this problem, and the only idea I came up with was replacing the rising factorials (you know, that symbol $(x)_{m}$), with binomial coefficients by means of the formula \[\binom{n}{k}=\frac{n^\underline{k}}{k!}\] Where $n^{\underline{k}}=n(n-1)\ldots(n-k+1)$ stands for the "falling factorial" (this is the counterpart of the "rising factorial"; as far I know you may think it's another kind of Pochhammer's symbol). And then I noticed it could be nice to use some generating functions, but I didn't succeed :roll: Could you tell me if I went astray? :maybe: Because I don't know about another approach we could use when facing problems of this kind :) Thanks!

Note first that for fixed $n$, both sides of the equation are polynomial in $x$ of degree at most $n$.  Thus, it suffices to show that the given equation is true for $n+1$ values of $x$.

First, for $x = 0$ we have on the left $\frac{n!}{n!}= 1$ while on the right we have 0 in every term except $k = 0$, for which we have $\frac{1}{n}\cdot\frac{1}{0!}\cdot n\cdot 1^{n}= 1$, so that's good.

Second, for $x =-1$ we have on the left side 0 because the last term in the product is $nx+n$.  On the right side, $(1+x)^{n-k}$ gives us 0 in every term except $k = n$, but then $n-k$ gives us 0, so the right side is also 0.

Now, let $x =-\frac{m}{n}$ for an integer $m$, $1 \leq m \leq n-1$.  The left side has as a term in its product $nx+m$, so the left side is 0.  Now we work with the sum $S = \sum_{k=0}^{n}\frac{(nx)_{k}}{k!}(n-k)(1+x)^{n-k}$.  Note that if $k \geq m+1$, the factor of $(nx)_{k}$ is 0.  So in fact we need terms only up to $k = m$.  Making this change and several substitutions,
\[S = \sum_{k=0}^{m}\frac{(-m)(-m+1) \cdots (-m+(k-1))}{k!}(n-k)\left(\frac{n-m}{n}\right)^{n-k}= \]

\[= \frac{(n-m)^{n}}{n^{n}}\sum_{k = 0}^{m}(-1)^{k}\binom{m}{k}(n-k) \left(\frac{n}{n-m}\right)^{k}\]

\[=\frac{(n-m)^{n}}{n^{n}}\cdot\left(n \sum_{k=0}^{m}\binom{m}{k}\left(\frac{-n}{n-m}\right)^{k}-m \sum_{k = 1}^{m}\binom{m-1}{k-1}\left(\frac{-n}{n-m}\right)^{k}\right) \]

\[= \frac{(n-m)^{n}}{n^{n}}\cdot\left(n \left(1+\frac{-n}{n-m}\right)^{m}-m\left(\frac{-n}{n-m}\right)\left(1+\frac{-n}{n-m}\right)^{m-1}\right) = 0 \]
and so we're done.

Alternate solutions would be more than welcome.  My initial thought was to consider both sides as generating functions in $x$ and solve for their coefficients, but that didn't seem very illuminating.  It also seems like there should be some combinatorial proof for positive integers $x$, since then the Pochhamer symbol is a binomial coefficient.


How do I do arithmetic with numbers that have different bases?

Example:
What is (444 base 5)+(333 base 4)-(222 base 3)?

it happens that your examples are pretty easy but in a normal case, you should convert to a single base (preferrably base 10, you might find it easier)

$ 444(5)\equal{}4\cdot 5^2\plus{}4\cdot 5\plus{}4\cdot 1(10)\equal{}124(10)$
$ 333(4)\equal{}3\cdot 4^2\plus{}3\cdot 4\plus{}3\cdot 1(10)\equal{}63(10)$
$ 222(3)\equal{}2\cdot 3^2\plus{}2\cdot 3\plus{}2\cdot 1(10)\equal{}26(10)$
Number $ x$ in base $ y$ I write as $ x(y)$.
Stewenmeow is right, you should convert to a single base, and the best is $ 10$.
How do you do it:

$ a_ka_{k\minus{}1}...a_1a_0(b)\equal{}a_k\cdot b^k\plus{}a_{k\minus{}1}\cdot b^{k\minus{}1}\plus{}...\plus{}a_1\cdot b\plus{}a_0(10)$

Bugi, if you want to denote what base a number is in, you can do this:

For a number $ a$ in base $ x$, you write $ a_x$.

So in this example, the numbers are $ 444_5, 333_4, 222_3$.

P.S. This is my 500th post!!! :lol:


geometry
geometric transformation
rotation
trigonometry

A spring with constant k=6 N/m is connected to a wall and to a spring with k=18 N/m. The latter is connected to a 4 kg mass. On its other side, this mass is connected to a wall by a k=1.5 N/m spring. What is $ \omega$ for this mass? Use this to find the maximum acceleration of the mass, given that the mass is released from rest with displacement A=0.04 m.

I believe $ \omega$ is called the [i]angular frequency[/i] with units rad/s (to distinguish from the frequency $ f$, units Hz).

Oh, okay.  But I'm not really sure why there would be any angular frequency, because $ \omega$ is a quantity typically used in rotations, not oscillations.  It  measures, as you say, how many radians an object rotates in some time.  So I'm  not really sure where it comes into play.

I'm not too clear on how to do this problem, but

http://en.wikipedia.org/wiki/Angular_frequency

there is something about $ \omega ^2 \equal{}\frac{k}{m}$...if that helps.

[quote="JRav"]because $ \omega$ is a quantity typically used in rotations, not oscillations[/quote]

What do rotations and oscillations have in common? Angular frequency is not only used in rotations and oscillations, but also in AC circuits (where the circuit quantities V and I are not constant but oscillate).

Ok so you can find the effective spring constant of the two springs on the left by taking half the harmonic mean.

So the differential equation is $ m\cdot \frac{d^2x}{dx^2}\equal{}\minus{}6x$.

Solve for $ x(t)$ and somehow get $ \omega$.

Good, you found the equation of motion
\[ \frac {d^2x}{dt^2} \plus{} \frac {6}{4}x \equal{} 0
\]
from the theory of ordinary differential equations, the homogenous solution to this equation (just a fancy term for the right hand side equalling 0) is found by substituting in $ x(t) \equal{} e^{rt}$ into the equation then solving for $ r$.  So we get
\[ \left(r^2 \plus{} \frac {3}{2}\right)e^{rt} \equal{} 0
\]
and since $ e^{rt}$ can never be zero, we find $ r^2 \plus{} \frac {3}{2} \equal{} 0$.  Thus $ r \equal{} \pm i\frac {\sqrt {6}}{2}$ and substituting back into $ x \equal{} e^{rt}$ we get $ x \equal{} e^{\pm i\frac {\sqrt {6}}{2}t}$ so using Euler's fomula
\[ e^{i\theta} \equal{} \cos{\theta} \plus{} i\sin{\theta}
\]
we find $ x \equal{} A\cos\left(\frac {\sqrt {6}}{2}t\right) \plus{} iB\sin\left(\frac {\sqrt {6}}{2}t\right)$.  Ok, so the reason we put $ A$ and $ B$ is because any arbitrary constant that we multiply by the cosine and since terms in the answer will not matter when we plug the solution back into the original equation.  Now, in ODE theory, if a solution has an imaginary coefficient, we just don't consider the $ i$ (just sort of accept this for now then verify it later).  Thus the solution is
\[ x(t) \equal{} A\cos\left(\frac {\sqrt {6}}{2}t\right) \plus{} B\sin\left(\frac {\sqrt {6}}{2}t\right)
\]
where $ A,B$ are constants.

Just a comment, if you have an ordinary differential equation of the form
\[ \frac {d^2x}{dt^2} \plus{} \omega^2x \equal{} 0
\]
(for $ \omega^2 > 0$) then the solution is $ x(t) \equal{} A\cos{\omega t} \plus{} B\sin{\omega t}$.  (Verify this!)


Solve system  $ \sqrt{x}\plus{}y\equal{}3, x\plus{}\sqrt{y}\equal{}4$

This' not nice.

A changes of variables defined by $ u\equal{}\sqrt x$ and $ v\equal{}\sqrt y$ would derive into a polinomial which miraculously has a rational root. After that, the polinomial it's reduced into another one which its roots are very nasty.


calculus
integration
trigonometry
function
algebra
rational function
calculus computations

Evaluate $ \int_{0}^{\pi }\frac{2\cos nt\plus{}\cos (n\minus{}1)t}{5\plus{}4\cos t}dt$, given that  $ n$ is a positive integer.

May I ask where you found this problem?  I can't think of anything and my computer algebra system says that the result cannot be expressed in terms of elementary functions.

Note the definite integral.

Worth asking: what is the Fourier series representation of $ \frac1{5\plus{}4\cos t}$ (on an interval of length $ 2\pi$)? Worth asking because this is a linear combination of two Fourier coefficients.

Your (Stank) CAS is assuming that $ n$ represents an arbitrary real number. As an integer, this does come out elementary; if nothing else, the substitution $ s \equal{} \tan \frac {t}{2}$ transforms it into the integral of a rational function.

Now, that rational function is very complicated. I'd rather try something else: complex analysis.
First this is an even function, so the value $ I(n)$ we're trying to find is the same as $ \frac12\int_{ \minus{} \pi}^{\pi}\frac {2\cos nt \plus{} \cos (n \minus{} 1)t}{5 \plus{} 4\cos t}\,dt$
Set $ z \equal{} e^{it}, dz \equal{} ie^{it}\,dt$; we get
$ I(n) \equal{} \frac12\int_{|z| \equal{} 1}\frac {z^n \plus{} \overline{z}^n \plus{} \frac12(z^{n \minus{} 1} \plus{} \overline{z}^{n \minus{} 1})}{5 \plus{} 2z \plus{} 2\overline{z}}\,\frac{dz}{iz}$
$ I(n) \equal{} \frac1{4i}\int_{|z| \equal{} 1}\frac {2z^n \plus{} z^{n \minus{} 1} \plus{} z^{1 \minus{} n} \plus{} 2z^{ \minus{} n}}{z(2z \plus{} 5 \plus{} 2z^{ \minus{} 1})}\,dz$
$ I(n) \equal{} \frac1{4i}\int_{|z| \equal{} 1}\frac {2z^n \plus{} z^{n \minus{} 1} \plus{} z^{1 \minus{} n} \plus{} 2z^{ \minus{} n}}{(2z \plus{} 1)(z \plus{} 2)}\,dz$

We want to evaluate this integral by the Residue theorem. There are two poles inside the circle, at $ \minus{} \frac12$ and at zero. The former is easy to deal with; it's a simple pole, and the residue is $ \frac12\frac {2z^n \plus{} z^{n \minus{} 1} \plus{} z^{1 \minus{} n} \plus{} 2z^{ \minus{} n}}{z \plus{} 2}$, evaluated at $ z \equal{} \minus{} \frac12$ for $ (\minus{}1)^n\cdot 2^{n\minus{}1}$. The latter is a pole of order $ n$, and we have to go through with the division (or something equivalent) to find the residue.
First, we simplify a bit with partial fractions. $ \frac3{(2z \plus{} 1)(z \plus{} 2)} \equal{} \frac {2}{2z \plus{} 1} \minus{} \frac {1}{z \plus{} 2}$, so $ \frac {z^{1 \minus{} n} \plus{} 2z^{ \minus{} n}}{(2z \plus{} 1)(z \plus{} 2)} \equal{} \frac13\left( \minus{} \frac {z^{ \minus{} n}(z \plus{} 2)}{z \plus{} 2} \plus{} \frac {2(z^{1 \minus{} n} \plus{} 2z^{ \minus{} n})}{2z \plus{} 1}\right)$
The $ z^{ \minus{} n}$ term contributes nothing to the residue unless $ z \equal{} 1$. That leaves us with $ \frac23\cdot\frac {z^{1 \minus{} n} \plus{} 2z^{ \minus{} n}}{1 \plus{} 2z} \equal{} \frac23\cdot z^{ \minus{} n}\cdot (z \plus{} 2)(1 \minus{} 2z \plus{} 4z^2 \minus{} 8z^3 \plus{} \cdots \plus{} ( \minus{} 2)^kz^k \plus{} \cdots)$
The $ z^{ \minus{} 1}$ terms of that Laurent series are $ \frac23\cdot z^{ \minus{} n}\cdot z\cdot ( \minus{} 2)^{n \minus{} 2}z^{n \minus{} 2} \equal{} ( \minus{} 1)^{n \minus{} 2}\cdot\frac {2^{n \minus{} 1}}{3}\cdot z^{ \minus{} 1}$ and $ \frac23\cdot z^{ \minus{} n}\cdot 2\cdot ( \minus{} 2)^{n \minus{} 1}z^{n \minus{} 1} \equal{} ( \minus{} 1)^{n \minus{} 1}\cdot\frac {2^{n \plus{} 1}}{3}\cdot z^{ \minus{} 1}$. Add them together for a residue of $ ( \minus{} 1)^{n \minus{} 1}\cdot 2^{n \minus{} 1}$, as long as $ n > 1$. When $ n \equal{} 1$, a simpler calculation gives a residue of $ 1$, fitting this same formula.

Finally, we put the pieces together: $ I \equal{} \frac1{4i}\cdot 2\pi i\cdot \left((\minus{}1)\cdot 2^{n\minus{}1} \plus{} ( \minus{} 1)^{n \minus{} 1}\cdot 2^{n\minus{}1}\right) \equal{} 0$.

Now, is there a clever way to do this with less work?

[Edit-fixed an error in the residue at $ \frac12$]


I know that users belong into 4 classes: Regular, Bots, Moderators, and Administrators. What exactly are the bots?

The Search Engine is your [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1205226179&t=99267]friend.[/url]


Let $y$ denote the sum $1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{1000000}}$. What is $\lfloor y \rfloor$ ?

Calculus isn't the only way to do this, you can also use
[hide="something"]
bounds which can be summed easily
[/hide]

[quote="yagaron"]but it is too superior for a high school post[/quote]
why so i am too a high-school student an i have understund well ur solution. :huh:

[quote="yagaron"]this isnt a intermediate problem!!!

I can proof that
${\left\lfloor \sum_{n=1}^{10^{6}}\frac{1}{\sqrt{n}}\right\rfloor =\left\lfloor \int_{1}^{10^{6}}\frac{1}{\sqrt{n}}\, dn\right\rfloor =1998}$

but it is too superior for a high school post[/quote]

Can u post the proof?


Hi guys...this year I'll go to MIT where is RSI 2006! If anyone else go there or anyone is interested please write here ;)

[i]Edited by orl to prevent that you send me to many love messages.[/i]  ;)

MIT means Massachusett's Institute of Technology and RSI is Research Sciece Institute . RSI is a program of Center for Excellence in Education in the USA. You can see more info here http://cee.org/rsi/index.shtml

[quote="joml88"]You have a slight typo in your first post that you might want to fix ;)[/quote]

Indeed.

[quote="keitoki"]I'm a freshman now, but RSI is definitely something I'd like to try. If I'm elibible.

Do you know if permanent residents can apply?[/quote]
Err.... why wouldn't they be able to? (Though I might be misinterpreting your question)
I'm also going to try to get into RSI in a few years.

Well, it says in the brochure that you have to be a citizen.

But on the application form it asks "immigration status if not citizen." And it's a bit confusing.    :?:

I know a US-resident non-citizen that went two years ago.

[quote="bilarev"]paladin8 Congratulations for your participation ;)  Maybe you are going ot make reserch in math or???
 :P[/quote]

Thanks. I'm planning on doing research in Math and Computer Science.

[quote="bilarev"][quote="paladin8"]
I'm planning on doing research in Math and Computer Science.
[/quote]
You are going to do research in the both... :o Is it permissible???In what field in math?[/quote]

Yup, of course it's permissible. They don't tell you to write both your primary AND secondary fields for nothing :P... I'm planning on researching something in the area of Number Theory/Cryptography, but this could very well change after I get there. We'll see, though!

[quote="agolsme"]Wait, do you know(or have an idea) who your mentor will be? I mean, how is it that you get both your primary and secondary fields? The form I got only listed my secondary field. Did you specify something extra in the research sheet ?[/quote]

No, I don't know yet. I was contacted however (for a completely different reason; there was a mistake in my info) and the person who contacted me was talking about several mentor options concerning both fields. A friend of mine who went in 2003 had a project in both math and computer science.

I'll be there, doing molecular (cell) bio research with a professor at BU.


limit
real analysis
real analysis theorems

Stolz Cesaro:

Let (x_n) and (y_n) be two sequences with the properties:
a)y_n>0 for any n natural;
b) y_n<y_(n+1) for any n
c) (y_n) is not bounded superior;
d) there exists a=lim(n-->oo) (x_(n+1)-x_n)/(y_(n+1)-y_n)
then there exists lim(n-->oo) x_n/y_n and even more
lim(n-->oo) x_n/y_n=a
-------------------------------------------------------------------------
Useful theorem in many sequences problems

No proof?
There's proof in nearly every math analysis books in China.

First consider (x_n-x_{n-1})/(y_n-y_{n-1})  \lim L, L is definite.
for any given  \varepsilon > 0, there exists N such that for all n>N it holds

|(x_n-x_{n-1})/(y_n-y_{n-1})-L| <  \varepsilon/2

or

L-\varepsilon/2 < (x_n-x_{n-1})/(y_n-y_{n-1})-L < L+\varepsilon/2

it means that for all n>N, all the fractions

(x_{N+1}-x_N)/(y_{N+1}-y_N),
(x_{N+2}-x_{N+1})/(y_{N+2}-y_{N+1}),
...
(x_{n-1}-x_{n-2})/(y_{n-1}-y_{n-2}),
(x_n-x_{n-1})/(y_n-y_{n-1})

are in the range. Since y_n is increasing, and the fractions above have positive denominators, so the range includes

(x_n-x_N)/(y_n-y_N)

because its denominator is the sum of all the denominators of the fractions above and so is the numerator.

hence for n > N it holds that

|(x_n-x_N)/(y_n-y_N) - L| < L+\varepsilon/2

from

x_n / y_n - L = (x_n - L y_n)/y_n + (1-y_N/y_n)((x_n-x_N)/(y_n-y_N) - L)

we conclude that

|x_n / y_n - L| \leq  |(x_n - L y_n)/y_n| + |(x_n-x_N)/(y_n-y_N) - L|
 = A + B

As we know, B is less than \varepsilon/2 when n>N. Since y_n \lim +\infty, so when n>N'>N, A<\varepsilon/2 too.

this gives |x_n / y_n - L| < \varepsilon for all n>N'.

Now suppose (x_n-x_{n-1})/(y_n-y_{n-1})  \lim +\infty.
Just as the argument above,
we can get x_n-x_{n-1} > y_n - y_{n-1} for n large enough.
y_n\lim +\infty tells us that x_n\lim +\infty and x_n is increasing.

thus apply the theorem we have proved to y_n/x_n

lim y_n/x_n = lim (y_n-y_{n-1})/(x_n-x_{n-1}) = 0

finally we get lim x_n/y_n = +\infty

A similar argument works for (x_n-x_{n-1})/(y_n-y_{n-1})  \lim -\infty.


Anyone have problem watching the video "Counting is as Easy as 24, 28, 32"?  It keeps stopping on me - Every few seconds the video l stops and is replaced by the advertisement for the AoPS "Counting/Probability" book.

I don't believe it is a connection issue as I just finished watching the other video and had no problems.   Seems to be just this video.

appreciate help.  thanks.

Please try it again after you have restarted your browser and let us know if the problem continues.

The difficulty is probably the buffer. press play, then pause. wait till the green bar is full and then watch it.


logarithms algebra function domain trigonometry

[quote="ifai"]Solve the equation $(\log x)^{\log x}=x$.[/quote]

(log x)^(log x)=x

Log X=K if and only if 10^K=X

K^K=10^K  if and only if K=K√(10^K) if and only if  K=10

10^K=X

10^10=X

2^10*5^10=X=10000000000

[quote="ifai"]Solve the equation $(\log x)^{\log x}=x$.[/quote]

$\log x=y$, $x=10^y$

$y^y=10^y \implies y=10 \implies x=10^{10}$, i don't know about the complex roots, it depends on your domain for the log function, also complex answers might not work anyway

$(\log x)^{\log x}=x\Rightarrow \log x(\log (\log x))=\log x\Rightarrow \log (\log x)=1$
$\log x=10\Rightarrow x=10^{10}$

By the definition of logs, $\log_{b}a=x$ if $b^{x}=a$.  Then,

$\log_{\log x}x=\log x\Rightarrow \log x=10\Rightarrow x=10^{10}$, where we note that the bases must be the same because the arguments are the same.

[quote="ifai"]Solve the equation $(\log x)^{\log x}=x$.[/quote]

[hide="Solution... and... complex roots?"]
$u = \log x$
$u^u = 10^u$

So obvious solution when $u = 10 \implies x = 10^{10}$.  Now, for the complex solutions...

$z = re^{ix}$
$z^z = r^{z} e^{izx} = e^{r \ln r( \cos x + i \sin x) } e^{-x \sin x + ix \cos x} = r^{r \cos x} e^{-x \sin x} e^{i (r \ln r \sin x + x \cos x} = ... \left( \cos (r \ln r \sin x + x \cos x) + i \sin (r \ln r \sin x + x \cos x) \right)$
$10^z = e^{z \ln 10} = e^{r \ln 10 \cos x + ir \ln 10 \sin x} = 10^{r \cos x} \left( \cos (r \ln 10 \sin x) + i \sin (r \ln 10 \sin x) \right)$

Equating real and imaginary parts, $r^{r \cos x} e^{-x \sin x} \cos (r \ln r \sin x + x \cos x) = 10^{r \cos x} \cos (r \ln 10 \sin x)$
$r^{r \cos x} e^{-x \sin x} \sin (r \ln r \sin x + x \cos x) = 10^{r \cos x} \sin (r \ln 10 \sin x)$

This hurts my brain.   :(
[/hide]


superior algebra
superior algebra unsolved

The following is the multiplication table of some group. Fill in the gaps according to the group laws.

\begin{tabular}{|c|c|c|c|c|c|c|}
  \hline
     \ast & e & v & w & x & y & z\\
  \hline
     e & e & v & w & x & y & z\\
  \hline
     v & v & w & e & y & - & -\\
  \hline
     w & w & - & - & - & - & -\\
  \hline
     x & x & z & - & - & - & v\\
  \hline
     y & y & - & - & - & - & -\\
  \hline
     z & z & - & - & - & - & -\\
  \hline
\end{tabular}

[quote="HilbertThm90"]It is $ S_3$, is it possible to identify which is what element?[/quote]

There is not a unique way to identify these as the elements of $ S_3$, but there is way to match them to elements of $ S_3$


A baseball hit at an angle of $ x$ to the horizontal with initial velocity $ v$ has horizontal range, $ R$, given by $ R \equal{} \frac {v}{g}sin(2x)$.

Here $ g$ is the acceleration due to gravity. Sketch R as a function of $ x$ for $ 0\le x\le \frac{\pi}{2}$. What angle $ x$ gives the maximum range? What is the maximum range $ R$?

I really don't feel like sketching right now (sorry), but:

[hide="Calculus"]
$ \frac{dR}{dx}\equal{}2\cdot\frac{v}{g}\cdot\cos{2x}\equal{}0\implies\cos{2x}\equal{}0\implies 2x\equal{}\frac{\pi}{2}\implies \boxed{x\equal{}\frac{\pi}{4}}$.

$ R\equal{}\frac{v}{g}\cdot\sin\frac{\pi}{2}\equal{}\boxed{\frac{v}{g}}$.
[/hide]

[hide="Non-calculus"]
We know that $ \minus{}1\leq\sin{x}\leq 1$, so the maximum $ R$ will occur when $ \sin{2x}\equal{}1\implies \boxed{R\equal{}\frac{v}{g}}$.
If $ \sin{2x}\equal{}1$, $ 2x\equal{}\frac{\pi}{2}\implies \boxed{x\equal{}\frac{\pi}{4}}$.
[/hide]


Is there anyone that knows the formula to find the number of rectangles in a 2x8 units grid just like the one below. Thank you in advance for those who shares.

   . . . . . . . . .
   . . . . . . . . .
   . . . . . . . . .

There are three possible horizontal lines and nine possible vertical lines, and a rectangle is formed by picking two of each, so the answer is $ {3 \choose 2} \cdot {9 \choose 2} \equal{} 108$.
In general, the formula is $ {m\plus{}1 \choose 2} \cdot {n\plus{}1 \choose 2}$

btw, he said formula. And, this should not go in intermediate.


This question is from the AMC.

"Let $ P(x) \equal{} (x\minus{}1)(x\minus{}2)(x\minus{}3)$. For how many polynomials $ Q(x)$ does there exist a polynomial $ R(x)$ of degree 3 such that $ P(Z(x)) \equal{} P(x) * R(x)$?

[hide="Correct Answer, not Solution"]
22
[/hide]

My question is [b]can anyone explain to me where I made a mistake in solving the question (because I got a different answer)?[/b]

\[ P(Q(x)) \equal{} (Q(x) \minus{} 1)(Q(x) \minus{} 2)(Q(x) \minus{} 3) \equal{} (x\minus{}1)(x\minus{}2)(x\minus{}3)R(x)\]
\[ Q(x)Q(x)Q(x) \minus{} 6Q(x)Q(x) \plus{} 11Q(x) \minus{} 6 \equal{} R(x)x^3 \minus{} 6R(x)x^2 \plus{} 11R(x)x \minus{} 6R(x)\]

Because $ Q(x)Q(x)Q(x) \equal{} R(x)x^3$, we get tat Q(x) is a quadratic. So let $ Q(x) \equal{} ax^2 \plus{} bx \plus{} c$

Now we group the terms (by considering their degrees)...
$ [Q(x)Q(x)Q(x) \minus{} R(x)x^3] \minus{} [6Q(x)Q(x)\minus{}  6x^2(r(x))] \plus{} [11Q(x) \minus{} R(x)11x] \minus{} [6 \minus{} R(x)6] \equal{} 0$

From which we get that (from the last pairing)...
\[ 6 \equal{} R(x)6\]
\[ R(x) \equal{} 1\]

But we are given that R(x) is of degree 3.
Therefore, there are no polynomials Q(x) that gives the required R(x).

But the answer is not 0, though...

Note: Upon retrospect, I noticed that I didn't really use the fact that $ Q(x)$ is a quadratic. Maybe that has something to do with the real solution...

I thank you in advance.

How did you group the terms?

If you grouped them according to their degrees, it should have been
$ [Q(x)Q(x)Q(x) \minus{} R(x)x^3] \plus{} [6x^2\cdot R(x)] \minus{} [6Q(x)Q(x) \plus{} R(x)\cdot 11x] \plus{} [6\cdot R(x)] \plus{} [11Q(x)] \minus{} [6] \equal{} 0$.

Plus, from your grouping, just because
$ [Q(x)Q(x)Q(x)\minus{}R(x)x^{3}]\minus{}[6Q(x)Q(x)\minus{}6x^{2}(r(x))]\plus{}[11Q(x)\minus{}R(x)11x]\minus{}[6\minus{}R(x)6] \equal{} 0$
does not apply that $ 6\minus{}6\cdot R(x)\equal{}0$.

$ [Q(x)Q(x)Q(x)\minus{}R(x)x^{3}]\minus{}[6Q(x)Q(x)\minus{}6x^{2}(r(x))]\plus{}[11Q(x)\minus{}R(x)11x]\minus{}[6\minus{}R(x)6] \equal{} 0$

That's not grouping the terms by their degrees. If $ Q(x) \equal{} x^2 \plus{} 1$, you'll have constant terms and squared terms and 4th power terms in that first expansion. There's no concluding each bracketed expression is identically 0.

Thanks!

I guessed that I merely assumed that the pairings would work - I never really considered the degrees.

Thanks again!

[quote="Arrange your tan"]"does not apply that $ 6 \minus{} 6\cdot R(x) \equal{} 0$."

You meant it does not imply $ 6 \minus{} 6\cdot R(x) \equal{} 0$.[/quote]
Blah yes.


Find all positive integer nubers $n$ such that the following equation:
$x^3+y^3+z^3=nx^2y^2z^2$
has positve integer solutions

[quote="pavel kozlov"]Also from Russian Summer test selection team 2005.
It is very easy problem for China.[/quote]
Hi [b]Pavel[/b] ;) .
I want ask you :? ,can you post same of Russian TST? :?


Ashley's old dresser has nine drawers. Six of the drawers each have
a volume of 2 ft$ ^3$, and three drawers each have a volume of 1 ft$ ^3$. Her new dresser has the same total volume in five equal-sized drawers.
What is the volume in cubic feet of each drawer of her new dresser?

Her old drawer had a volume of $ 6(2) \plus{} 3(1) \equal{} 15$ because there were 6 drawers of volume 2 and 3 drawers of volume 1.

Since this volume of $ 15$ is divided into $ 5$ drawers, the volume of each of her new drawers is $ \frac{15}{5} \equal{} \boxed{3}$


prove that the centroid,circumcentre and the orthocentre of a triangle are collinear.
(using Euclidean geometry)

thank you...

How is it possible??? :huh:

It is possible only in case of equilateral triangle i suppose.

no, i think for equilateral triangle they coincide.
but for any other triangle they are collinear...

please help someone...


Find the formula for the sum:

$\sum_{n=1}^{m}{n2^{n-1}}$

If you know Calculus, for a generalization let $S(x)=\sum_{n=1}^{m}nx^{n-1}$, find $\int S(x)\ dx$, and then differentiate.

[b]Follow-up:[/b]  Can a counting proof of the formula be found?

Where can I find more problems like these? I need a lot of practice.


[quote="tenniskidperson3"]Sorry if this has been asked before, but I searched and couldn't find anything on it.  What is the World Primary Mathematics Competition?  I read on the Mathcounts Website that Bobby Shen got a perfect score at it last year, and I want to learn more about it.  Google was unhelpful (only 10 links, mostly talking about competitors and only briefly mentioning the contest itself).[/quote]
"primary" - probably for middle school. Are u seriously still in MS?
edit: it is for 1-6th graders, and has a website. Clearly u did not search too hard.

No, I'm in 10th grade.  Would you be so kind as to direct me to the website?

http://www.mathscompetition.com/

first hit on google when I searched "world primary mathematics competition"  :wink:  :P

OK, I usually type it in with quotes to eliminate stuff like

"World Primary Economics world"
The primary idea in economics is competition, as the
mathematics  behind it clearly shows...

and other things of that nature where the words are in there but not arranged how I want them to be.  Thanks for your help, though.  :blush:


actually this is Schur,I think.But I also think that the identity is useful:

   $x^3+y^3+z^3- 3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$ what do you think? :?

A constant term in right hand sind is making problem to the way to schur. :?

Well so where that identity takes us? I am not sure. Can you show me?

[quote="mahbub"]4. let $x,y,z \geq 0$ with $x+y+z=1$. Prove that
$x^3+y^3+z^3+6xyz \geq \frac{1}{4}$[/quote]

this inequality isn't the best possible, because \[ x^3 + y^3 + z^3 + \dfrac{15}{4}xyz \geq \frac14 \] also holds for any nonnegative $x,y,z$ s.t. $x+y+z=1$.

proof:

homogenize the ineq and obtain:

$x^3+ y^3 + z^3 + \frac{15}{4} xyz \geq \frac{(x+y+z)^3}{4}$ with no conditions now.

$\Leftrightarrow 4x^3+ 4y^3 + 4z^3 + 15xyz \geq (x+y+z)^3$

$\Leftrightarrow 4 \sum x^3 + 15xyz \geq \sum x^3 + 3 \sum x^2y + 3 \sum xy^2 + 6xyz$

$\Leftrightarrow 3 \sum x^3 + 9xyz \geq 3 \sum x^2y + 3 \sum xy^2$

$\Leftrightarrow \sum x^3 + 3xyz \geq \sum x^2y + \sum xy^2$

$\Leftrightarrow \sum x(x-y)(x-z) \geq 0$, which is schur, true.

eq holds iff $x=y=z=\dfrac13$ or $x=0$ and $y=z=\frac12$ or the kind.

I found this solution:
we know
$x^3+y^3+z^3=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)+3xyz=(x+y+z)^2-3(xy+yz+zx) +3xyz =1-3(xy+yz+zx)+3xyz$,
 Then we most prove this new inequality in this form:
$\frac 14\geq (xy+yz+zx)-3xyz\Longrightarrow xy+yz+zx-3xyz=xy(1-z)+yz(1-x)+zx(1-y)=xy(x+y)+yz(y+z)+zx(z+x)=x^2(y+z)+y^2(z+x)+z^2(x+y)=x^2(1-x)+y^2(1-y)+z^2(1-z)\leq \frac {x+y+z}{4}=\frac 14$.
 In the end inequality we use $x(1-x)\leq\frac 14$. :lol: