Given $ a, b, c \geq\ 0$.
Prove that:  $ (a^2\plus{}b^2\plus{}c^2)(a\plus{}b\plus{}c) \geq\ 3\sqrt[]{3abc(a^3\plus{}b^3\plus{}c^3)}$

It has posted on toanthpt.net  :)

From cheybychef's, $ (a^2\plus{}b^2\plus{}c^2)(a\plus{}b\plus{}c) \geq 3(a^3\plus{}b^3\plus{}c^3)$ and from $ A.M.\geq G.M.,$ $ a^3\plus{}b^3\plus{}c^3 \geq 3abc$
Combine the two inequalities....Done! :wink:

[quote="Sunkern_sunflora"]From cheybychef's, $ (a^2 \plus{} b^2 \plus{} c^2)(a \plus{} b \plus{} c) \geq 3(a^3 \plus{} b^3 \plus{} c^3)$ and from $ A.M.\geq G.M.,$ $ a^3 \plus{} b^3 \plus{} c^3 \geq 3abc$
Combine the two inequalities....Done! :wink:[/quote]
It is not true,you should check your "chebysev ineq"
In my ideal ,i let $ \sum ab \equal{} \frac {1 \minus{} q^2}{3},abc \equal{} r$
this ineq can be written $ 1 \plus{} 4q^2 \plus{} 4q^4\ge 243rq^2 \plus{} 729r^2$
$ RHS \minus{} LHS \equal{} f(r)$,$ f'(r) > 0$ then use $ r\le \frac{(1 \minus{} q)^2(1 \plus{} 2q)}{27}$ we will have the proof
I haven't finished it,sorry if it is wrong

Call $ X \equal{} ab^2\plus{}ac^2\plus{}ba^2\plus{}bc^2\plus{}ca^2\plus{}cb^2$. Then square both sides of the original inequality.

\[ (a^2 \plus{} b^2 \plus{} c^2)^2(a \plus{} b \plus{} c)^2 \geq 27abc(a^3 \plus{} b^3 \plus{} c^3)\]
\[ (a^3\plus{}b^3\plus{}c^3\plus{}X)^2\geq 27abc(a^3\plus{}b^3\plus{}c^3)\]
\[ (a^3\plus{}b^3\plus{}c^3)^2\plus{}(a^3\plus{}b^3\plus{}c^3)(2x\minus{}27abc) \plus{}X^2 \geq 0\]
\[ (a^3\plus{}b^3\plus{}c^3\plus{}\frac{2X\minus{}27abc}{2})^2\minus{}\left(\frac{2X\minus{}27abc}{2}\right)^2\plus{}X^2\geq0\]

Since $ (a^3\plus{}b^3\plus{}c^3\plus{}\frac{2X\minus{}27abc}{2})^2 \geq0$, the inequality is true if

\[ X \geq \left|\frac{2X\minus{}27abc}2\right|\]

This is always true since if $ \frac{2X\minus{}27abc}2$ is positive, since it results in $ 27abc \geq 0$. If $ \frac{2X\minus{}27abc}2$ is negative, then $ X\leq \frac{27abc}2$ implies:

\[ 2X\leq 27abc\]
\[ 4X\leq 2X\minus{}27abc\]
\[ 2X\leq \frac{2X\minus{}27abc}{2}\]
\[ \minus{}2X \leq \minus{}\frac{2X\minus{}27abc}{2}\]
\[ 2X \geq X \geq \left|\frac{2X\minus{}27abc}2\right|\]

This proves the inequality only using the fact that if $ a$ is real, $ a^2 \geq 0$.

[quote="Sunkern_sunflora"]From cheybychef's, $ (a^2 \plus{} b^2 \plus{} c^2)(a \plus{} b \plus{} c) \geq 3(a^3 \plus{} b^3 \plus{} c^3)$
Combine the two inequalities....Done! :wink:[/quote]
It is wrong. By Chebyshev ; we only have  $ (a^2 \plus{} b^2 \plus{} c^2)(a \plus{} b \plus{} c) \le 3(a^3 \plus{} b^3 \plus{} c^3)$ :wink:

I have just fonund that tranvanluan had a nice proof for it,post it,please

Yes, solution of tranvanluan is here. I haven't checked it. But if it is trues, it'll be a nice solution.
http://www.toanthpt.net/toanthpt/showthread.php?t=9244&page=10

[quote="arqady"][quote="nguoivn"]Given $ a, b, c \geq\ 0$.
Prove that:  $ (a^2 \plus{} b^2 \plus{} c^2)(a \plus{} b \plus{} c) \geq\ 3\sqrt []{3abc(a^3 \plus{} b^3 \plus{} c^3)}$

It has posted on toanthpt.net  :)[/quote]
SOS kills it. :wink:[/quote]
kill it,pls! :blush: I like SOS more than pqr

[quote="Math pro"][quote="arqady"][quote="nguoivn"]Given $ a, b, c \geq\ 0$.
Prove that:  $ (a^2 \plus{} b^2 \plus{} c^2)(a \plus{} b \plus{} c) \geq\ 3\sqrt []{3abc(a^3 \plus{} b^3 \plus{} c^3)}$

It has posted on toanthpt.net  :)[/quote]
SOS kills it. :wink:[/quote]
kill it,pls![/quote]
$ (a^2 \plus{} b^2 \plus{} c^2)(a \plus{} b \plus{} c) \geq\ 3\sqrt []{3abc(a^3 \plus{} b^3 \plus{} c^3)}\Leftrightarrow$ (after expanding )
$ \Leftrightarrow\sum_{sym}(a^6 \plus{} 4a^5b \plus{} 6a^4b^2 \plus{} 4a^3b^3 \plus{} 8a^3b^2c \plus{} 2a^2b^2c^2 \minus{} 25a^4bc)\geq0.$
But $ \sum_{sym}(a^6 \plus{} 4a^5b \plus{} 6a^4b^2 \plus{} 4a^3b^3 \plus{} 8a^3b^2c \plus{} 2a^2b^2c^2 \minus{} 25a^4bc) \equal{}$
$ \equal{} \sum_{cyc}(a^6 \minus{} a^5b \minus{} ab^5 \plus{} b^6) \plus{} 5\sum_{cyc}(a^5c \minus{} a^4bc \minus{} b^4ac \plus{} b^2c) \plus{}$
$ \plus{} 6\sum_{cyc}(c^4a^2 \minus{} 2c^4bc \plus{} c^4b^2) \plus{} 4\sum_{cyc}(c^3a^3 \minus{} c^3a^2b \minus{} c^3ab^2 \plus{} c^3b^3) \minus{}$
$ \minus{} 12abc\sum_{cyc}(a^4 \minus{} a^2b \minus{} ab^2 \plus{} b^3) \minus{} 4abc\sum_{cyc}(a^3 \minus{} abc) \equal{}$
$ \equal{} \sum_{cyc}(a \minus{} b)^2\left(6c^4 \plus{} 4(a \plus{} b)c^3 \minus{} 2abc^2 \plus{} (5a^3 \minus{} 9a^2b \minus{} 9ab^2 \plus{} 5b^3)c \plus{} a^4 \plus{} a^3b \plus{} a^2b^2 \plus{} ab^3 \plus{} b^4\right)\geq$
$ \geq\sum_{cyc}(a \minus{} b)^2\left(6c^4 \plus{} 4(a \plus{} b)c^3 \minus{} \frac {(a \plus{} b)^2}{2}c^2 \minus{} (a \plus{} b)^3c \plus{} \frac {5(a \plus{} b)^4}{16}\right).$
Let $ 2c\equal{}(a \plus{} b) t.$ Hence, it remains to prove that $ 6t^4 \plus{} 8t^3 \minus{} 2t^2 \minus{} 8t \plus{} 5\geq0,$ which is obvious.

[quote="nguoivn"]Given $ a, b, c \geq\ 0$. Prove that:

$ (a^2 \plus{} b^2 \plus{} c^2)(a \plus{} b \plus{} c) \geq\ 3\sqrt {3abc(a^3 \plus{} b^3 \plus{} c^3)}.$

It has posted on toanthpt.net  :)[/quote][quote="arqady"]SOS kills it. :wink:[/quote][quote="Math pro"]
kill it, pls! :blush:[/quote]
$ (a^2 \plus{} b^2 \plus{} c^2)^2(a \plus{} b \plus{} c)^2 \minus{} 27abc(a^3 \plus{} b^3 \plus{} c^3)$

$ \equal{} \sum{\frac {(b \minus{} c)^2}{6}\left[27a^4 \plus{} (b \minus{} c)^2\left(2a^2 \plus{} 16ab \plus{} 16ac \plus{} 3b^2 \plus{} 3c^2 \plus{} 8bc\right)\right]}\geq0.$

[quote="nguoivn"]I think the stronger seem still trues:
$ (a^2\plus{}b^2\plus{}c^2)^3 \geq\ \frac{9}{8}(a^3\plus{}b^3\plus{}c^3)(a\plus{}b)(b\plus{}c)(c\plus{}a)$[/quote]
You are right. It's true!  $uvw$ kills it. It enough to check only $b=c=1,$
which gives $(a-1)^2(4a^4-a^3-8a+14)\geq0,$ which is true.

[quote="nguoivn"]I think the stronger seem still trues:
$ (a^2\plus{}b^2\plus{}c^2)^3 \geq\ \frac{9}{8}(a^3\plus{}b^3\plus{}c^3)(a\plus{}b)(b\plus{}c)(c\plus{}a)$[/quote]

$ (a^2\plus{}b^2\plus{}c^2)^3 -\frac{9}{8}(a^3\plus{}b^3\plus{}c^3)(a\plus{}b)(b\plus{}c)(c\plus{}a)$

$=\sum_{cyc} (c^3b-8ac^2b-a^3b+4a^4+4b^4+9c^4-ab^3+c^3a)(a-b)^2\geq 0$

i proved the stronger inequality with an interesting proof.
 :D
$(a^2+b^2+c^2)(a+b+c)\ge (\sqrt{abc}+\sqrt{a^3+b^3+c^3+abc})^2$

[quote="red3"]
$(a^2+b^2+c^2)(a+b+c)\ge (\sqrt{abc}+\sqrt{a^3+b^3+c^3+abc})^2$[/quote]
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, our inequality is equivalent to $f(v^2)\geq0$, where
$f(v^2)=9u^2v^4+2uv^2w^3-12u^3w^3+w^6$.
We see that $f$ is increasing function.
thus, $f$ gets a minimal value, when $v^2$ gets a minimal value,
which happens, when two numbers from $\{a,b,c\}$ are equal.
Id est, it remains to prove our inequality for $b=c=1$, which gives $(a-1)^2\geq0$!  :)


geometric transformation
geometry
power of a point
radical axis
geometry proposed

Let $(O_{1},r_{1}),(O_{2},r_{2})$ be two circles and $A,B$ its touch points at common external tangent respectively $(r_{2}>r_{1})$. If $C,D$ are point of the radical axis at $O_{1}O_{2}$ and the insimilicenter of $(O_{1}),(O_{2})$ respectively, prove that $ABCD$ is cyclic.

If $F$ is the exsimilicenter of $(O_{1}), (O_{2}),$ now: $FA \cdot FB = FC \cdot FD.$ :wink:


In an obtuse-angled triangle ABC prove that

cot2A +cot2B +cot2C>0

Isnt' this true for acutes as well?

[edit: no, (-1)^3=-1...]

At first, Peter, for an acute triangle, the inequality is completely wrong. We actually have cot 2A + cot 2B + cot 2C < 0 for every acute triangle ABC and cot 2A + cot 2B + cot 2C > 0 for every obtuse triangle ABC.

[i]Proof.[/i] At first, we note:

[b]Lemma 1.[/b] For every triangle ABC, we have cot A + cot B + cot C > 0.

[i]Proof.[/i] If S is the area and a, b, c are the sides of triangle ABC, then $\displaystyle \cot A + \cot B + \cot C = \frac{a^2 + b^2 + c^2}{4S}$, what is obviously positive.

Now we are going to prove our two inequalities. Let A', B', C' are the feet of the altitudes of triangle ABC. Then, by Lemma 1, applied to triangle A'B'C', we have

  cot < C'A'B' + cot < A'B'C' + cot < B'C'A' > 0.

Now, if the triangle ABC is acute, then the angles of triangle A'B'C' are

  < C'A'B' = 180 - 2A;
  < A'B'C' = 180 - 2B;
  < B'C'A' = 180 - 2C,

so that

  cot 2A + cot 2B + cot 2C
  = - cot (180 - 2A) - cot (180 - 2B) - cot (180 - 2C)
  = - cot < C'A'B' - cot < A'B'C' - cot < B'C'A'
  = - (cot < C'A'B' + cot < A'B'C' + cot < B'C'A') < 0.

On the other hand, if the triangle ABC is obtuse, e. g. with an obtuse angle at A, then the angles of triangle A'B'C' are

  < C'A'B' = 2A - 180;
  < A'B'C' = 2B;
  < B'C'A' = 2C.

Hence,

  cot 2A + cot 2B + cot 2C = cot (2A - 180) + cot 2B + cot 2C
  = cot < C'A'B' + cot < A'B'C' + cot < B'C'A' > 0.

And this completes the proof.

You see, Peter, no dumb trigonometric calculations...

  Darij

Hello, I believe I have found another solution to this problem. Please tell me if I have some silly error (I did it pretty quickly this morning while procotoring a test and wrote it up this afternoon, so it might be bad):

Let C be the obtuse angle.
Let x = 2A, y = 2B <--> x+y < 180
cot2C = cot(2(180-A-B)) = cot(360-2A-2B) = cot(-2A-2B) = cot(-(x+y)) = -cot(x+y)
sinx, siny > 0, sin(x+y) > 0.
cot2A + cot2B + cot2C > 0 <--> cotx+coty cot(x+y) > 0
cotx +coty > cot(x+y) <--> cosx/sinx + cosy/siny >
(1-sinx/cosx*siny/cosy)/(sinx/cosx+siny/cosy) <-->
(cosxsiny+sinxcosy)/(sinxsiny) > (cosxcosy-sinxsiny)/(sinxcosy+cosxsiny) <-->
sin(x+y)/(sinxsiny) > cos(x+y)/sin(x+y) <-->
sin^2(x+y) > sinxsinycos(x+y)

Case 1: x+y >=90 --> cos(x+y) <=0. Thus, sinxsinycos(x+y) <=0, and sin^2(x+y) > 0. QED.

Case 2: x+y < 90
sin^2(x+y) > sinxsinycos(x+y) <-->
(sinxcosy)^2+2sinxsinycosxcosy + (cosxsiny)^2 > sinxsinycosxcosy  (sinxsiny)^2 <-->
(sinxcosx)^2+(cosxsiny)^2+(sinxsiny)^2+sinxsinycosxcosy>0
sinx, siny, cosx, cosy > 0 because x,y <90, so
sinxcosx)^2+(cosxsiny)^2+(sinxsiny)^2+sinxsinycosxcosy>0 is true. QED

Thanks,
JTG


The formula for chromium chloride is CrCl[size=75]3[/size].  What would be the weight of the chlorine (Cl) in 1256 grams of chromium chloride (Cr, 52; Cl, 35)?

Problem from Saxon Algebra 2

[hide] The weight of one mole of CrCl3 is $ 52 \plus{} 35(3) \equal{} 157$g. We have $ 1256$g, so $ \frac {1256}{157} \equal{} 8$. Then, $ 8(3)(35) \equal{} \boxed{840}$.[/hide]


Problem : Let $ a,b,c >0 $ . Prove that :
 $ \frac{a}{\sqrt{a^2+8bc}} + \frac{b}{\sqrt{b^2+8ac}} + \frac{c}{\sqrt{c^2+8ba}} \geq 1 $

   I know a solution but I don't know how to find this solution .
  Please show me how to choose $ p $ such that the below ineq true for all $ a,b,c > 0 $ :
   $ \frac{a}{\sqrt{a^2+8bc}} \geq \frac{a^p}{a^p+b^p+c^p} $

FIrst of all, this is not an open problem!!
Anyway, you can do cross multiplication and expand it so that it remains
$a^2(b^{2p}+c^{2p}+a^pb^p+c^pa^p+b^pc^p+a^pb^p+c^pa^p+b^pc^p) \ge 8a^{2p}bc$
The simplest way for this is true is to apply amgm on the left hand side in which you get
$L.H.S. \ge 8a^2(a^{p/2}b^{3p/4}c^{3p/4}$
In the most ideal case, we have $p/2 + 2 = 2p$ and $3p/4 = 1$.
Luckily these two equations give a common solution $p=4/3$.

Another question : How can you find $ p $ such that :
 $ \sqrt{\frac{a^3}{a^3+(b+c)^3}} \geq \frac{a^p}{a^p+b^p+c^p}  \forall a,b,c >0 $

[quote="romano"]Dear Siuhochung ,
 I know p =2 and I can prove for this case , but I can't apply AMGM to find p . Can you show it to me ?[/quote]
this time i used amgm and power mean. I think there is no general method to find p.


[quote="manlio"]For $ a,b,c$ positive reals prove that

$ (\sum a^2) \sum \frac {1}{(a^2 \plus{} 2(b^2 \plus{} c^2))} \geq \sum \frac {b^2 \plus{} bc \plus{} c^2}{a^2 \plus{} (b \plus{} c)^2}$[/quote]
For along time, we haven't seen a solution for this one. Now, let me post my solution :)


I read the solutions for #24 and #25 of AMC-12B 2004, but I didn't understand the solutions:

24. In $\bigtriangleup ABC$, $AB=BC$, and $\overline{BD}$ is an altitude.  Point E is on the extension of $\overline{AC}$ such that BE=10.  The values of $tan \angle CBE$, $tan \angle DBE$, and $tan \angle ABE$ form a geometric progression, and the values of $cot \angle DBE$, $cot \angle CBE$, $cot \angle DBC$ form an arithmetic progression.  What is the area of $\bigtriangleup ABC$?

(See picture below.)
(A)16     (B)$\frac{50}{3}$     (C)$10\sqrt{3}$     (D)$8\sqrt{5}$     (E)18

25. Given that $2^{2004}$ is a 604-digit number whose first digit is 1, how many elements of the set $S=\{2^{0}, 2^{1}, 2^{2}, ..., 2^{2003}\}$ have a first digit of 4?

(A) 194     (B) 195      (C) 196     (D) 197     (E) 198

Without TeX, I prefer to leave #24 to someone who can present a solution in a halfway decent form.

For #25, observe the first several powers of 2.

1, 2, 4, 8 (1 digit)

16, 32, 64, (2 digits)
128, 256, 512 (3 digits)

1024, 2048, 4096, 8192 (4 digits)

16384, 32768, 65536 (5 digits)

Notice there are always 3 or 4 powers of 2 that are any digit length.

Let the number of different lengths for which there are only 3 powers of 2 that are exactly that many digits long be x.

Let the number of different lengths for which there are 4 powers of 2 that are exactly that many digits long be y.

x + y = 603, since the given 2^2004 is a 604-digit number whose first digit is 1.

Also, since there are 3 powers of 2 for each of the x digit lengths, and 4 powers of 2 for each of the y digit lengths, we have

3x + 4y = 2004 since there are 2004 powers of 2 being considered.

Solving this system gives y = 195.

Notice that only when there are 4 powers of 2 for a certain digit length will one of the powers have 4 as its leading digit. Therefore, by solving for y, we have found the answer of (B) 195.

25.
[hide="A"]$2^{2003}$ must have 603 digits since $2^{2004}$ has 604 digits, with its first being 1.
Calculate some of the smaller powers of 2, and notice their first digits:
1,2,4,8,
1,3,6,
1,2,5,
1,2,4,8,
...
I grouped them by the number of digits that power of 2 has. The first number of each group is always 1, since the highest an n-digit number doubled is an n+1 digit number with leading digit 1.
Also note that each group has 3 or 4 numbers. You can't have a group of only two numbers because in order to get an extra digit, you need to double at least a 5, and you can't get a 5 from doubling a 1 once.
Now, you know that given a specific number of digits, there can be exactly 3 or 4 powers of 2 with that many digits.
We also know that there are 2004 elements in S. If we subtract 3x603 from 2004, we get the number of groups with 4 digits. All of those groups have 4 as a member, so that is also the number of elements with 4 as the leading digit. Is the answer 195?[/hide]


[quote="math92"]Which of these is the fastest rate of speed?
A. 40 miles per hour
B. 15 yards per second
C. 1,000 inches per second
D. 200,000 feet per hour
E. .75 miles per minute[/quote]

A. $40 mph$
B. 45 feet per second = 2,700 feet per minute = 162,000 feet per hour $\approx 30.68\bar{18} mph$
C. 1K inches per second = 3,600,000 inches per hour $\approx 56.\bar{81} mph$
D. 200,000 feet per hour $\approx 37.\bar{87} mph$
E. .75 miles per minute $= 45 mph$

Did I screw up somewhere? I got C as the answer...

Yup, my bad! Dumb calculation mistake...


problem 1
A 40l sample of N2 gas containing SO2 gas as an impurity was bubbled through a 3% solution of H2O2. The SO2 was converted to H2SO4.
A 25ml portion of 0.01 mol/l NaOH was added to the solution, and the excess base was back titrated with 13.6 ml of 0.01 mol/l HCL.
Calculate the parts per million of SO2 if the density of SO2 is 2.85g/l

problem 2
4.32g of oxygen was required to completely burn a 2.16g sample of a mixture of methanol and ethanol.
1/ How many moles of ethanol are contained in the sample ?
2/ what is the percentage by weight of methanol in the sample ?

problem 3
A fatty acid was obtained by hydrolysis of some oil and fat. 15ml of 0.2mol/l aqueous solution of potassium hydroxide was required For the hydrolysis of 0.884g of the oil and fat.
1/ what is the molecular weight of the oil and fat
2/ what is the molecular weight of the fatty acid

Problem 1.

For the first part, start by writing the relevant redox half-reactions,

$SO_{2}+2H_{2}O \longrightarrow H_{2}SO_{4}+2e^{-}+2H^{+}$

$H_{2}O_{2}+2H^{+}+2e^{-}\longrightarrow 2H_{2}O$

and now add them to get the overall equation that describes the oxidation of sulphur dioxide into sulphuric acid:

$SO_{2}+H_{2}O_{2}\longrightarrow H_{2}SO_{4}$.

Next, when we add excess NaHO to the sulphuric acid solution, the following reaction occurs:

$H_{2}SO_{4}+2NaHO \longrightarrow Na_{2}SO_{4}+2H_{2}O$.

Finally, the titration of the excess NaHO with muriatic acid is described by the equation

$NaHO+HCl \longrightarrow NaCl+H_{2}O$.

Now, from the data of the titration, we know that the number of excess moles of NaHO is $n = 13.6 \times 10^{-3}L \cdot 0.010 mol/L = 13.6 \times 10^{-5}mol$, and so the number of moles of NaHO that reacted with sulphuric acid is $n(NaHO) = 25.0 \times 10^{-3}L \cdot 0.010 mol/L-13.6 \times 10^{-5}= 11.4 \times 10^{-5}mol$. From the stoicheometry of the reaction between H2SO4 and NaHO we know that 2 moles of NaHO react with 1 of acid, and so the $11.4 \times 10^{-5}mol$ of NaHO reacted with $5.7 \times 10^{-5}mol$ of H2SO4. As the SO2 react with H2O2 in 1:1, then the number of moles of sulphur dioxide present in the nitrogen gas sample is also $n(SO_{2}) = 5.7 \times 10^{-5}$, wich corresponds to a mass of $m(SO_{2}) = 5.7 \times 10^{-5}mol \cdot 64.06 g/mol = 3.65 \times 10^{-3}g$, and which corresponds to a volume of

$V(SO_{2}) = \frac{3.65 \times 10^{-3}g}{2.85 g/L}= 1.28 mL$.

Finally, the concentration of sulphur dioxide in the sample is given by

$[SO_{2}] = \frac{1.28 mL}{0.040 kL}= 32 ppm$.

[hide="Problem 2"]
Let's go ahead and write out our combustion rxn eqns:
$2CH_{3}OH+3O_{2}\rightarrow2CO_{2}+4H_{2}O$
$C_{2}H_{5}OH+3O_{2}\rightarrow2CO_{2}+3H_{2}O$

We have $(4.32gO_{2})(\frac{1molO_{2}}{32gO_{2}})=0.135molO_{2}$
Our rxns tell us that each mole of methanol requires 1.5 moles of oxygen, and each mole of ethanol requires 3 moles of oxygen.

Let $X$ be the number of moles of methanol in the mixture, and let $Y$ be the number of moles of ethanol in the mixture. We set up a system of eqns:

$XmolCH_{3}OH(\frac{3molO_{2}}{2molCH_{3}OH})+YmolC_{2}H_{5}OH(\frac{3molO_{2}}{1molC_{2}H_{5}OH})=0.135molO_{2}$

$XmolCH_{3}OH(\frac{32gCH_{3}OH}{1molCH_{3}OH})+YmolC_{2}H_{5}OH(\frac{46gC_{2}H_{5}OH}{1molC_{2}H_{5}OH})=2.16g(mixture)$

From this,

$\frac{3}{2}X+3Y=0.135$
$32X+46Y=2.16$

$32X+64Y=2.88$
$32X+46Y=2.16$

$18Y=0.72$
$Y=0.04$
$X=0.01$

$0.01molCH_{3}OH(\frac{32gCH_{3}OH}{1molCH_{3}OH})=0.32gCH_{3}OH$
$\frac{0.32gCH_{3}OH}{2.16g(mixture)}\approx0.148$

1) We have 0.04 moles of ethanol
2) The percent-by-mass of methanol is approximately 14.8%

[/hide]


Kim earned scores of 87, 83 and 88 on her first three mathematics examinations.
If Kim receives a score of 90 on the fourth exam, then by how much will her average increase?

The average of any series of data is the sum of the data divided by the number of terms. For the three original scores, the average is $ \frac{87\plus{}83\plus{}88}{3}\equal{}86$, while the average of the four scores when the $ 90$ is included is $ \frac{87\plus{}83\plus{}88\plus{}90}{4}\equal{}87$, so the average increases by $ 87\minus{}86\equal{}\boxed{1}$.


Can someone check my working for this question I made up?

[hide="question"]
Solve for $ \minus{} 360^{\circ} \leq x \leq 360^{\circ}$: $ 5 \tan (3x \plus{} 2) \equal{} 13$.
[/hide]

[hide="solution"]
For time's sake, let all units be in degrees.

Divide by 5 to yield $ \tan (3x \plus{} 2) \equal{} 13/5$.
A solution in quadrant 1 is $ \arctan 13/5 \equal{} 68.9625$.

Since $ \tan (3x \plus{} 2) > 0$, thus $ 3x \plus{} 2$ is in quadrant 1 or 3. Then we can find

$ 3x \plus{} 2 \equal{} 180 \plus{} 68.9625,68.9625, \minus{} 180 \plus{} 68.9625,540 \plus{} 68.9625,900 \plus{} 68.9625$, $ \minus{} 540 \plus{} 68.9625, \minus{} 900 \plus{} 68.9625$, since $ \minus{} 1078 < 3x \plus{} 2 < 1082$.

We can then proceed to find the values of $ x$.
[/hide]

Is there a faster way to solve trigonometric equations? This method is rather slow.

Also, I thought of solving equations like $ \cos (\sin x)\equal{}a$, where $ a$ is a given constant. How do we do these? The same way (except much more work)?

hello, the general solution is $ x\equal{}\minus{}\frac{2}{3}\plus{}\frac{1}{3}\arctan\left(\frac{13}{5}\right)\plus{}\frac{1}{3}k\pi$, $ k \in \mathbb{Z}$,
from here you can compute all possible solutions with $ \minus{}2\pi\le x \le2\pi$.
Sonnhard.

hello, from the equation $ 5\tan(3x\plus{}2)\equal{}13$ we have $ \tan(3x\plus{}2)\equal{}\frac{13}{5} \Leftrightarrow 3x\plus{}2\equal{}\arctan\left(\frac{13}{5}\right) \Leftrightarrow x\equal{}\minus{}\frac{2}{3}\plus{}\frac{1}{3}\arctan\left(\frac{13}{5}\right)$. We find the period by writing $ \tan(3x\plus{}2)\equal{}\tan\left(3\left(x\plus{}\frac{2}{3}\right)\right)$. Note that the period of $ \tan(x)$ is $ \pi$.
Sonnhard.


If $P$ and $Q$ are monic polynomials such that $P(P(x))=Q(Q(x))$, prove that $P=Q$.


function
real analysis
real analysis unsolved

I found a nice identity involving digamma function: $\psi_0 (1-z)-\psi_0 (z)=\pi \cot \pi z$. Can someone give hint how to prove that?

The problem was easier than I expected. All you need to do is to write $\psi (z)=\frac{\Gamma^\prime (z)}{\Gamma (z)}$ and the identity follows easily from the fact $\Gamma (1-z)\Gamma (z)=\frac{\pi}{\text{sin }\pi z}$.


geometry
calculus
limit
analytic geometry
geometry solved

Given three points $M_0$, $M_1$ and $M_2$ in the plane. Construct a sequence of points $M_i$ such that for every i, the point $M_i$ is the centroid of the triangle $M_{i-3}M_{i-2}M_{i-1}$. Prove that this sequence converges, and its limit is the point $M=\frac{M_0+2M_1+3M_2}{6}$.

Michel Garitte says he knows some solutions of this problem but searches for a purely geometric solution. I, personally, have nothing but a very ugly solution (and even this one thanks to my father). Could somebody help?

  Darij

Well, we have $m_n=\alpha+\beta z^n+\gamma \bar z^n,\ \forall n\ge 0$, where $z=\frac{-1}3+i\frac{\sqrt 2}3$, and $\bar z$ is the conjugate of $z$ (this is because $1,z,\bar z$ are the roots of the characteristic eqn of the recurrence). Since $|z|=|\bar z|<1$, we get $\lim_{n\to\infty}m_n=\alpha$, and we can compute the coefficients $\alpha, \beta,\gamma$ wrt $m_0,m_1,m_2$ (a system of three eqns), and I bet we will find $\alpha=\frac{m_0+2m_1+3m_2}6$.

Is this the solution you have, Darij?

I wouldn't dare attack it synthetically, though.. :)

[quote="grobber"]Is this the solution you have, Darij?[/quote]

Quite the same (my solution actually had some superfluous parts but it is actually equivalent to yours). Maybe I underestimated the difficulty of the problem - it seemed so obvious...

  Darij

You are so complicated guys! ;)

It is easy to see that $\frac{M_i+2M_{i+1}+3M_{i+2}}{6}$ is independent of $i$, moreover $\frac{5}{6}\cdot P_{M_iM{i+1}M_{i+2}}>P_{M_{i+3}M{i+4}M_{i+5}}$.
It follows that sequence of embeded triangles $M_iM{i+1}M_{i+2}$ converges to unique point $M=\frac{M_0+2M_{1}+3M_{2}}{6}$.

Oh my god! :) Thanks, Myth, that's really cool!

[quote="Myth"]It is easy to see that $\frac{M_i+2M_{i+1}+3M_{i+2}}{6}$ is independent of $i$, moreover $\frac{5}{6}\cdot P_{M_iM{i+1}M_{i+2}}>P_{M_{i+3}M{i+4}M_{i+5}}$.
It follows that sequence of embeded triangles $M_iM{i+1}M_{i+2}$ converges to unique point $M=\frac{M_0+2M_{1}+3M_{2}}{6}$.[/quote]

Can you please explain me this proof more precisely? What is $P_{XYZ}$ ? Is it the longest side of triangle XYZ? And why does it follow that the sequence has a limit?

Thanks!
  Darij

$P_{XYZ}$ is a girth of $XYZ$, i.e.  $P_{XYZ}=XY+YZ+ZX$.
Any sequence $K_i$ of embeded compacts has nonempty limit $K$, $K=\cap K_i$. Since diameter of triangles tends to 0 we conclude that intersection of all compacts can't have more than one point. This is a basic result of analysis.

I though that my proof is very simple and clear. :?

I didn't say it wasn't simple and clear. But I just didn't know the analysis results you used. Honestly speaking, I don't know them now either, so I don't really understand your proof completely. But thanks anyway, once again I saw how a real calculus proof looks like  :lol:

Together with my parents I generalized the problem today. Here is the generalization:

[i]Let $N_1$, $N_2$, ..., $N_n$ be n arbitrary points in $\mathbb{R}^m$. Construct a sequence $N_i$ such that for every i, the point $N_i$ is defined as the centroid of the points $N_{i-n}$, $N_{i-n+1}$, ..., $N_{i-1}$. Then this sequence converges, and its limit is $N=\frac{1\cdot N_{1}+2\cdot N_{2}+...+n\cdot N_{n}}{1+2+...+n}$.[/i]

The proof is similar to the one Grobber gave for the original problem, but somewhat harder. Probably you can find an advanced calculus proof again  :)

  Darij

Ok!
It is clear we just need to consider x-coordinates of initial points. Let $x_1$, ... be x-coordinates, $y_k=\min_{i=k,...,k+n-1} x_i$ and $Y=\max_{i=k,...,k+n-1} x_i$. Point $\hat{x}=\frac{x_1+2x_2+\dots+nx_n}{1+2+...+n}$ is independent of our transformation, $y_k\leq x\leq Y_k$, $y_k$ is nondecreasing sequence and $Y_k$ is nonincreasing sequence.
Thus we have sequence of embedded segments $[y_k,Y_k]$ $\Rightarrow$ $y_k\to y$ and $Y_k\to Y$. Suppose $y<Y$ $\Rightarrow$ $\exists k$ s.t. $y-y_k<(Y-y)/4n$ and $Y_k-Y<(Y-y)/4n$, but using this assumption we easily obtain that after $n$ transormation of $n$-tuple ($x_k$,$x_{k+1}$,...,$x_{k+n-1}$) all points will be inside $(y,Y)$ -- contradiction.
So $Y=y$ $\Rightarrow$ $x=y=Y$, and it is mean that all points converge to $x$.


What is $ 1^2\plus{}(\frac{1}{2})^2\plus{}(\frac{1}{4})^2\plus{}(\frac{1}{8})^2\plus{}(\frac{1}{8})^2\plus{}(\frac{1}{16})^2\plus{}...$?

*...=The numbers keep going on forever until infinity*

Hint- Make it equal $ x$:)

Thanks!!

-LHM

[quote="LHM"]What is $ 1^2 \plus{} (\frac {1}{2})^2 \plus{} (\frac {1}{4})^2 \plus{} (\frac {1}{8})^2 \plus{} (\frac {1}{8})^2 \plus{} (\frac {1}{16})^2 \plus{} ...$?

*...=The numbers keep going on forever until infinity*

Hint- Make it equal $ x$:)

Thanks!!

-LHM[/quote]

[hide]
This sequence contains the squares of $ 2^{\minus{}n}$, where $ n$ is a whole number.  $ (2^{\minus{}n})^{2}\equal{}4^{\minus{}n}$, so the series is $ 1\plus{}\frac{1}{4}\plus{}\left(\frac{1}{4}\right)^{2}\plus{}\left(\frac{1}{4}\right)^{3}\plus{}...$
By letting this be x, we get
$ x\equal{}1\plus{}\frac{1}{4}\plus{}...$
$ 4x\equal{}4\plus{}1\plus{}\frac{1}{4}\plus{}...$
Subtracting the first from the second yields
$ 3x\equal{}4$
so $ x\equal{}\boxed{\frac{4}{3}}$.
[/hide]

[quote="LHM"]What is $ 1^2 \plus{} (\frac {1}{2})^2 \plus{} (\frac {1}{4})^2 \plus{} (\frac {1}{8})^2 \plus{} (\frac {1}{8})^2 \plus{} (\frac {1}{16})^2 \plus{} ...$?

*...=The numbers keep going on forever until infinity*

Hint- Make it equal $ x$:)

Thanks!!

-LHM[/quote]

[quote="alanchou"]Subtracting the first from the second yields
$ 3x \equal{} 4$
so $ x \equal{} \boxed{\frac {4}{3}}$.
[/quote]
I get all the parts except subtracting the first from the second and getting $ 3x \equal{} 4$.
Can you explain it?

[quote="xpmath"]In the second equation, he multiplied the first by 4.
Subtracting  the first equation from the second causes all the terms beside the 4 to cancel off, and then he can divide by 3.[/quote]

Oh! I didn't think those were separate equations.
I get it now.
Thanks

could somebody give me the formula for sum of infinite sequences??

i keep forgetting,
and is there a formula for sum of finite sequences??

don't just memorize the formula. understand why its true. ill give u the formula for the sum of of finite series with a common ratio and you see if you can derive the one for an infinite series. it follows from the one im going to give you
[hide="formula"]formula for sum of the first n terms of a finite sequence with common ratio r and first term a is
$ \dfrac{a\minus{}ar^n}{1\minus{}r}$[/hide]

[quote="kimmystar94"]okay...
i'll look it up on google and look at how it's derived from, if i can't find it, i'll try doing so myself...[/quote]

It helps more if you do that the other way around, so you understand it more than if you just found it in a book, web page, etc.

[quote="LHM"]What is $ 1^2 \plus{} (\frac {1}{2})^2 \plus{} (\frac {1}{4})^2 \plus{} (\frac {1}{8})^2 \plus{} (\frac {1}{8})^2 \plus{} (\frac {1}{16})^2 \plus{} ...$?

*...=The numbers keep going on forever until infinity*

Hint- Make it equal $ x$:)

Thanks!!

-LHM[/quote]
wait, you mean "What is $ 1^2 \plus{} (\frac {1}{2})^2 \plus{} (\frac {1}{4})^2 \plus{} (\frac {1}{8})^2 \plus{} (\frac {1}{16})^2 \plus{} (\frac {1}{64})^2 \plus{} ...$?, right?

[quote="minicon"]He probably copied and pasted one to many times.  Anyway, he was looking for this.  You multiplied by 4 to get 64, where it should be 32.

What is $ 1^2 \plus{} \left(\frac {1}{2}\right)^2 \plus{} \left(\frac {1}{4}\right)^2 \plus{} \left(\frac {1}{8}\right)^2 \plus{} \left(\frac {1}{16}\right)^2 \plus{} \left(\frac {1}{32}\right)^2 \plus{} ...$?[/quote]
oh yeah, sorry that's what I meant I meant.
lol, i such a hypocrite...
I didn't really get how to do it otherwise though


geometry
geometric transformation
linear algebra
set theory
linear algebra unsolved

Ok, I wonder (and I hope it is correct) that if we have a system of $n-1$ equations and $n$ variables, and also each  equation contains at least $2$ variables in it, then this system has a solution $(x_1,\ldots ,x_n)$ with $\prod_{i=1}^n x_i\neq 0$.
This problem came from the following combinatorial problem :
We have $n-1$ subsets from a set with $n$ elements, and each subset has at least $2$ elements. Now, we must prove that we can color the elements of the set in $2$ colors, so that each of the sets contains both colors.
I think that it is obvious how we cand reduce the combinatorial one to the linear algebra one (btw, is there a non-algebric solution?)

You can have a nontrivial solution, but you don't get one with nonzero entries. Consider the homogeneous system $x+y=0,x-y=0$ on variables $x,y,z.$

You oversimplified in translation, and the problem needs the condition that the subsets cover the set as well.

You don't need all of the elements nonzero- just enough that none of the subsets is identically zero.

Where did you find the problem, DusT? I don't even think it's true, not even with the additional necessary condition pointed out by jmerry: the $n-1$ subsets cover the $n$-element set.

Think about the following example:

We have the $5$ elements $a,b,c,d,e$, and the $4$ subsets $\{a,b\},\{b,c\},\{c,a\},\{d,e\}$. There's no way you can color the three elements $a,b,c$ so that the first three sets in the list are all bicolor.

[quote="grobber"]Where did you find the problem, DusT? I don't even think it's true, not even with the additional necessary condition pointed out by jmerry: the $n-1$ subsets cover the $n$-element set.[/quote]
First of all, it is very funny of you to ask grobber where did I find the problem, considering this http://www.mathliks.ro/Forum/viewtopic.php?t=40741#
:?  :rotfl:
[quote="grobber"]
Think about the following example:

We have the $5$ elements $a,b,c,d,e$, and the $4$ subsets $\{a,b\},\{b,c\},\{c,a\},\{d,e\}$. There's no way you can color the three elements $a,b,c$ so that the first three sets in the list are all bicolor.[/quote]
So I think this example actually a bit disproves that your assumption in that post is a bit wrong. Anyway, I think the combinatorial problem is wrong, maybe we need some more assumptions.
About the linear algebra one, what should be the additional conditions we should impose? Maybe we could translate them in set theory :?

Unfortunately, I cannot access that thread, because my network restrictions do not allow me to access MathLinks. Could you please post the Artofproblemsolving version of the thread? Thank you.

[quote="grobber"]I haven't thought about the problem you've actually posted (the one with the multisets), but I found the original one (I mean the one with sets instead of multisets) quite nice in itself, so I'm posting a solution for it.

Actually, a more general problem: what's the minimal number of questions necessary to guess any binary number with $n$ digits?

I think it's $n$ (which is obviously achievable by asking the number placed in each position separately).

Suppose the contrary, i.e. assume we can do it in $n-1$ questions. Then we may as well assume that no question is concerned with only one position, since then the problem would be reduced to numbers with $n-1$ digits, and we can use induction.

It's very easy to prove by induction that given any $n-1$ subsets $A_1,\ldots,A_{n-1}$, each one with at least $2$ elements, of $\{1,2,\ldots, n\}$, we can color $\{1,2,\ldots, n\}$ with $2$ colors s.t. each of the $A_i$'s is also bicolored.

This means that for any set of $n-1$ questions, each one concerned with at least $2$ positions, we can partition $\{1,2,\ldots,n\}$ in two classes $A,B$ (forming the bicoloration which we spoke about in the paragraph above) s.t. each one of the $n-1$ sets cuts both $A$ and $B$. If the second player receives the answer $\{0,1\}$ to each one of these $n-1$ questions, he cannot say whether the number has $0$ on each position in the set $A$ and $1$ everywhere else, or vice-versa.
I hope I did not misunderstand the problem.[/quote]

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=40741

I hope it works....and that I understood correctly what you said.......

Yes, that proof is wrong. Sorry about that. It's not the first time and I'm sure it won't be the last time I post a false proof :(.. I'll delete it.

Edit:

Actually, after some thought, I believe this should be a sufficient condition for bicoloration (and it has a certain elegance, if it is correct :)):

If we have $n$ finite sets s.t. for any $0<p\le n$ of them, their union has at least $p+1$ elements, we can color the elements of the union of all the sets in two colors s.t. all the sets are bicolor.

I think I can prove it, but, as usual, I'm not sure it's correct. I'll leave it here (even though it belongs in the "Combinatorics" section) to see what the others (DusT? :)) think.


For any $z_1$ and $z_2$ , prove that $|\dfrac{z_1 - z_2 }{1 - \overline{z_1}z_2} | < 1$ if $|z_i | < 1$

I don't think its correct.  For example, take .99 and -.99.


calculus
integration
function
real analysis
real analysis unsolved

Suppose that $a,b\in\mathbb{R}^+$ which $a+b<1$ and $f:[0,+\infty) \rightarrow [0,+\infty) $ be the increasing function s.t. $\forall x\geq 0 ,\int _0^x f(t)dt=\int _0^{ax} f(t)dt+\int _0^{bx} f(t)dt$. Prove that $\forall x\geq 0 , f(x)=0$

Because $f$ is positive and increasing, it's easy to see that

$\forall x,u>0 : uf(x)  \leq \int _x^{x+u} f(t)dt=\int _{ax}^{a(x+u)} f(t)dt+\int _{bx}^{b(x+u)} f(t)dt $ $\leq auf(a(x+u)) + buf(b(x+u))$

so $f(x)  \leq af(a(x+u)) + bf(b(x+u))$ .

I take $u = \frac{1-a-b}{a+b}x \Longrightarrow  (a(x+u) \leq x)\& ( b(x+u) \leq x) $ $\Longrightarrow (f(a(x+u)) \leq f(x))\& ( f(b(x+u)) \leq f(x))$ , so $f(x) \leq (a+b)f(x)$

But $0 < a+b < 1$ and $ 0 \leq f(x)$, so we have to  have $\forall x>0, f(x)=0$ and $0 \leq f(0) \leq f(1)=0 \Longrightarrow f(0)=0$
 :cool:  q.e.d.= c.q.f.d. :cool:

[quote="enescu"]If you know the solution, then this is not the place to post the problem. Try "Algebra Proposed and Own Problems" instead.
If you don't know the solution (but, still, you find it "not hard but cool...") I'll post it.[/quote]
 :yup:     :lol:


group theory
abstract algebra
superior algebra
superior algebra unsolved

G is a group such that the order of G is o(G) = m * p^a, where p is prime, a,m are integers such that a>=0, m>0, and p does not divide m. A is a subgroup of G of order n * p^b, where b is an integer such that 0 <= b < a, and n is a positive integer such that p does not divide n. Of course, n divides m. Let B be the only p-Sylow subgroup of G, so o(B) = p^a. Does o(AB) = n * p^a? If yes, prove it to be so.

Any help is appreciated.

$ B$ is the unique $ p$-Sylow subgroup of $ AB$, i.e. for some divisor $ k$ of $ m$ we get (note, that $ B\leq AB\leq G$):

$ kp^b\equal{}|AB|\equal{}|A|\underbrace{\frac{|B|}{|A\cap B|}}_{\equal{}\,p^t\;for\;some\;t}\equal{}np^{a\plus{}t}\Longrightarrow k\equal{}n$

(alternatively, one can show that $ A\cap B$ is the unique $ p$-Sylow subgroup of $ A$ of order $ p^a$)

[quote="hjbrasch"]$ B$ is the unique $ p$-Sylow subgroup of $ AB$, i.e. for some divisor $ k$ of $ m$ we get (note, that $ B\leq AB\leq G$):

$ kp^b \equal{} |AB| \equal{} |A|\underbrace{\frac {|B|}{|A\cap B|}}_{ \equal{} \,p^t\;for\;some\;t} \equal{} np^{a \plus{} t}\Longrightarrow k \equal{} n$

(alternatively, one can show that $ A\cap B$ is the unique $ p$-Sylow subgroup of $ A$ of order $ p^a$)[/quote]

Wow, that looks very nice and simple! I'm not sure I follow all of it though... I guess I'll start with my first question, since the answer to that might explain the rest: Why is B the unique p-Sylow subgroup of AB, and how would you get k * p^b = |AB| from that? Especially since |B| = p^a and a > b, it doesn't seem like the latter could be true..

Would you mind showing it the alternative way you mentioned too? Different approaches to the proof might help me understand it more intuitively.

The answers to your questions:

[i]Why is $ B$ the unique $ p$-Sylow subgroup of $ AB$ ?[/i]

Because $ B$ is the unique $ p$-Sylow subgroup of $ G\geq AB\geq B$.

[i]and how would you get $ kp^b \equal{} |AB|$ from that? Especially since $ |B| \equal{} p^a$ and $ a > b$, it doesn't seem like the latter could be true.. [/i]

Your $ a>b$ argument is true, because I made a typo. The correct line is $ kp^a\equal{}|AB|\equal{}|A|\underbrace{\frac{|B|}{|A\cap B|}}_{\equal{}p^t\;for\;some\;t}\equal{}np^{b\plus{}t}\Longrightarrow k\equal{}n$. Now, for your "how"-question: $ B\leq AB\Longrightarrow p^a\equal{}|B|$ divides $ |AB|\Longrightarrow |AB|\equal{}kp^a$ with some $ p\not| k$. Finally, $ AB/B\leq G/B$ gives $ k|m$ ($ B$ is the unique, i.e. normal $ p$-Sylow subgroup)

[i]Would you mind showing it the alternative way you mentioned too?[/i]

Let $ S$ be a $ p$-Sylow subgroup of $ A$ (of order $ p^b$). Using Sylow's theorem and $ B$'s normality we can find $ g\in G$ with $ S^g\leq B\Longrightarrow S\leq B^{g^{\minus{}1}}\equal{}B\Longrightarrow S\subseteq A\cap B\Longrightarrow S\equal{}A\cap B$, the last step since $ A\cap B$ is a $ p$-subgroup of $ A$.

Therefore, $ |AB|\equal{}\frac{|A\parallel{}B|}{|A\cap B|}\equal{}\frac{|A\parallel{}B|}{|S|}\equal{}\frac{np^bp^a}{p^b}\equal{}np^a$

Ok, using your approach, I came up with the following:

A and B are subgroups of AB (since the identity element is in both A and B), so we have both:

|AB| >= n*p^b, and n*p^b divides |AB|
|AB| >= p^a, and p^a divides |AB|

Then AB a subgroup of G gives:

|AB| <= m*p^a, and |AB| divides m*p^a

These three together give:

|AB| = k*n*p^a for some (positive) integer k.

Now I get something similar to what you had:

k*n*p^a = |AB| = |A||B|/|A intersect B|

I'm still not sure how to deal with |A intersect B| yet though...

EDIT: Sorry, didn't see your response. Reading now...

Awesome! I think I got it! Thank you so much for your help, the feeling of understanding is great!

Sorry for the triple post, but I just realized there's still something lacking. (Or at least something I'm missing!) This all assumes that AB is a subgroup of G. Can we show that it is? Otherwise, your line (for instance) that, "B is the unique p-Sylow subgroup of G [which has as a subgroup] AB [which has as a subgroup] B", is false, isn't it?

[quote="Jan"]This follows from the following lemma:

If $ N$ is a normal subgroup of $ G$ and $ H$ an arbitrary subgroup of $ G$, then $ NH \equal{} HN$ and this is a subgroup of $ G$.

Here, the unique Sylow $ p$-subgroup is clearly a normal divisor.[/quote]

Ah, great, thanks!


[quote="April"]Let $ D$, $ E$, $ F$ be the tangency points of the incircle $ \omega (I)$ of triangle $ ABC$ with the sides $ BC$, $ CA$, $ AB$, respectively. Let $ M$, $ N$, $ P$ be the middle points of the segments $ IA$, $ IB$, $ IC$. Prove that $ DM$, $ EN$, $ FP$ are concurrent.[/quote]

Lema:kariya theorem
http://www.mathlinks.ro/viewtopic.php?search_id=376406392&t=30114

Let $ D^{'},E^{'},F^{'}$ be the tangency points of the incircle of triangle $ MNP$. Then $ D^{'},E^{'},F^{'}$ is midpoint of $ ID,IE,IF$. Therefore, by kariya theorem in triangle MNP, $ DM,EN,FP$ is concurrent


function
analytic geometry
graphing lines
slope
calculus
derivative
Gauss

Okay, this is going to turn into something long.  Mostly, I just want to know if what I'm writing down has been thought of before, and, as it likely has, by whom (and when).  It's just stuff I've done in my spare time.

The idea is to generalize the concept of "parallel" to include continuous, differentiable curves other than lines.  (Actually, I need my function to be at least C:^2: for this to work).  A nice example of curves with properties of parallelism are concentric circles -- they never meet, any line perpendicular to one is perpendicular to the other, and the distance between them along these perpendiculars is constant.  So we'll take that as our definition of parallel: any line perpendicular to one is perpendicular to the other, and the distance between them along these perpendiculars is constant.

So, we start with a function y = g(x), and we want to find a function y = f(x) such that f is "parallel" to g under the definition given.  In particular, this means that for any point x1, there is a corresponding point x2 such that g'(x1) = f'(x2) and (x1, g(x1)) and (x2, f(x2)) are two points on a line perpendicular to the tangents of the two curves, so it has slope -1/g'(x1).  (Note that this condition is very simple when we have a curve of zero slope, since the corresponding x-value of the parallel curve will be exactly the same as the x-value of our curve.  Thus, we can only consider those points where g'(x) is non-zero and we should end up with at worst removable discontinuities.)  Also, we want the point x2 to move in a nice fashion as x1 moves in a nice fashion.  So, I'll just impose a bit more structure on my system and let x2 = :phi:(x1), where :phi: is also a C1 function.

Then, plugging into the equations for the line and for equal derivatives using :phi:, we get that f'(:phi:(x)) = g'(x) and that
(f(:phi:(x)) - g(x))*g'(x) = -(:phi:(x) - x), or by rearranging,
f(:phi:(x)) = g(x)  + (x - :phi:(x))/g'(x).  This is an equation among nice differentiable functions, and it holds for all x.  So we can differentiate both sides to get that:
f'(:phi:(x))*:phi:'(x) = g'(x) + ((1-:phi:'(x))g'(x) - (x - :phi:(x))g''(x))/g'(x)2
g'(x)*:phi:'(x) = g'(x) + ((1-:phi:'(x))g'(x) - (x - :phi:(x))g''(x))/g'(x)2
g'(x)3*:phi:'(x) = g'(x)3 + ((1-:phi:'(x))g'(x) - (x - :phi:(x))g''(x))

Okay, so keep messing around.  You get a differential equation in x which has a complete solution
:phi:(x) = x + c*g'(x)/:sqrt:(g'(x)2 + 1)

Then, theoretically, we can find the inverse of this function and substitute it back in to find f.  The problem is that it's almost never invertible, except for when we have that g is linear or circular (which we knew the answer to in advance, anyhow.)  And if we choose g' to make it invertible, then it almost always is not integrable to get the original value of g(x).

So, seperately you can do the parametric case (particularly simple for functions defined in polar, actually) which opens up the opportunity for the curves Gauss mentioned, such as self-intersecting curves or basically any nice curve in the plane.  I haven't finished this yet, though.

But really what I want to know is if any of you have heard of anything like this before, if you think it's either interesting or significant (I'm mostly just hoping for interesting), and if you have any suggestions for it.

Is the perpendicular condition really necessary? We don't consider that to be very important in non-Euclidean geometry. I think an interesting idea would be to find C :inf:  (maybe C1, but then they won't be analytic in C) functions f and g such that f-g is always nonzero over the complex numbers. Clearly f=g+k works for some complex k, but that's not very interesting. A more interesting example is f(z)=g(z)+e^z. Then again, that's just reducing the problem to looking for analytic functions that don't have complex zeros.

No, but those types of functions just don't intersect.  If we restrict it to real numbers, we can get much cooler effects than non-intersection.

The first question is to define your new notion of parallel. Under your new definition is parallelism transitive? - Meaning, if curve C1 is parallel to C2, and C2 is parallel to C3, is C3 necessarily parallel to C1?

Yes, it is transitive.  Because perpendicularity is transitive.

Be careful. Are you sure? Your definition of parallel is not just about perpendicularity. It's about curves that come from the loci of another curve.

And what about self-intersecting curves? Are you allowing those?

I'm just dealing with functions right now, so it leaves out a bunch of possibilities -- for example, the curve y = x2 doesn't have any parallel functions on its interior, because they would need to cross themselves (and not be functions) down around x = 0.

I am fairly certain that my parallel is going to be transitive -- I'll try and write this up tonight. Have to go do a paper now

Edit:  It's all pretty much done now.  See if you still think there are problems, Gauss.  And anyone else, feel free to comment.

I'm not going to go step by step through your calculations, but I think there are some errors. Just to bring up a few points - I don't think you ever used the fact that you want the distance between curve g and f to remain constant. I'm not sure that the "perpendicular" structure you imposed necessarily implies that it will be (though it may).

I'm also not convinced about this removable discontinuities thing by looking only at the points where g' is non-zero and trying to glue all the pieces together.

Then finally, you introduced this :phi: function in a kind of ad hoc way. There's no indication that such a :phi: exists, or that it is even unique. And I don't think your final equation for :phi: is correct. Try applying it to the case where g(x) = :sqrt:(1-x2) - a half-circle. Your :phi:(x) should equal x*r, where r is the radius of your new "parallel" half-circle; but it doesn't.

To answer your basic question though - Yes, this sort of thing has been considered before. It's basically the geometry of loci (if I am using that term correctly, I may not be). But it is the study of how new curves are created from the set of points a set distance away from other curves. It was a major area of study to the ancient greeks, even before calculus.

I still find this very interesting though, and I think it's definitely worth still studying and considering. It's also something that you will brush on briefly in Differential Geometry when you study the Parallel Transport of a vector along a curve (via the connection). [See any diff. geo. book on parallel transport]. That's essentially what you're doing here, only you're just keeping track of the endpoints of these vectors as you parallel transport them to create new curves.

Unless I typed something up wrong, those calculations will give the correct value when you start with the function g = :sqrt:(1 + x2).  I did it out by had earlier and it worked fine.  And I just checked it again on mathematica.

Sorry, I forgot to add the bit about being equidistant.  If you look at the form of the :phi: function, it will impose a constant-distance criteria.  This actually surprised me a lot when I was doing it, but it does work very nicely.  The horizontal piece has length :phi:(x) - x = c*g'(x)/:sqrt:(g'(x)2 + 1), and the line has slope -1/g'(x), so the vertical portion must be -c/g'(x), which gives the distance between the two curves along the perpendicular of c, a constant.

As for transitivity: what this amounts to is "sliding" the function along the lines perpendicular to it.  This sliding action is pretty clearly reversible -- that is, if one function is parallel to another, the second function is parallel to the first.  We can thus "slide" from the third function back to the second and from the second to the first, so that transitivity works.  I know this isn't too clear, but when you work out what the perpendicular portion of the parallel thing works out to, it should make sense.

I know I pulled :phi: out of nowhere.  It won't necessarily exist, but if things are going to be parallel in any nice sort of way, it should be.  (That is, if we have one function of x parallel to another function of x, the corresponding points must be connected by a nice-looking function.  Otherwise there would be holes in the second function somewhere, or parallelness wouldn't be symmetric.  If I really wanted, I could make it a condition of two functions being parallel, since it's true in the specific cases (lines and circles) that I'm trying to generalize.

Okay, not knowing anything about differential geometry, that last bit you said does sound quite a lot like my procedure.  Would I be able to just "pick up" a book on differential geometry, or will I need to do a lot of other work first?  And do you have any recommendations for a good differential geometry book?

Okay, you're right. I agree with you on all of your points. And I agree that your :phi:(x) also works for the half-circle (I made a mistake when I calculated it). Though I think you could do it a lot better using the parametric equations of a curve. Then you don't have to deal with all this funny business about g being single valued, or flat in certain places, etc.

Differential geometry shouldn't be too hard to dive into from the point that you're at now. The only prereqs are linear algebra and a solid basis in calculus. The bibles of differential geometry are [u]A Comprehensive Introduction to Differential Geometry, Vols. 1 & 2[/u] by Michael Spivak. Part of what makes them so great is that they are written like lecture notes. They are slow moving, and basically walk you through diff. geometry with the motivations for things and telling some of the history of it's development. It's not a quick way to learn diff. geometry, but it's thorough and accessible. These are the books that I used to learn diff. geometry basically on my own.

Another good book is [u]Elementary Differential Geometry[/u] by Barett ONeill. It gets more quickly to the point, without as much of the "talky"-motivation and history of Spivak.

Any book on differential geometry will talk about parallel transport though. It's one of the basic topics in diff. geometry.

Oh, and I almost forgot. A really good book on differential geometry is Frank Morgan's [u]Riemannian Geometry: A Beginner's Guide[/u]. It's probably the best really introductory book that I've seen on diff. geometry. I'd say this is the best place to start.

Okay, great, thanks.  I'll see if the library has them.

I'm using Pressley's [i]Elementary Differential Geometry[/i] for a class. It's nice in that it's easy to read, and it's not too expensive as far as math books go.

The parametric solution to this problem turns out to be pretty nice and simple - with very few requirements on your original curve. You should see if you can solve it this way.

Okay -- I started out that method a couple of days ago but I haven't carried it through.  I'll see what happens.


probability geometry logarithms function trigonometry quadratics calculus

at my schools math club, I do pretty well compared to others but some people just know so much. Then I come on here and realize that the people who I thought were super smart, arent even compared to some of you in here and that im really dumb in comparison  :o

Where/how did some of you learn so much? Im in grade 11 uni math and I currently have a mark of 104 and the class is super boring and we dont do anything that would increase my knowledge of anything when it comes to contests..

So basicly.. in order to do good on any of these contests you have to take learning math into your own hands I'm guessing and im wondering where to start doing this myself..

I was thinking of ordering one of the CNML contest books (5 years worth of contests + solutions), but aside from that is there anything else I can do?

Thanks guys

I suggest ordering some books like Art Of Problem Solving: vol.1 and 2. Follow that up with Art and Craft Of Problem Solving. Look through the following list,
[url]http://www.artofproblemsolving.com/Books/AoPS_B_Item.php[/url]
They have some very good titles available. :)
Aops vol.1 is just basic stuff to get a solid base. Aops vol.2 and acops get more advanced. Keep in mind its not all about knowing a lot of theorems and formulas, its about knowing how to solve problems efficiently.

those books are all pretty expensive, do you know of any web links to information like combinatorics, number theory, sequences??

I just started getting into contests and there is so much that I don't know  :oops:

The AoPS books as far as textbooks go are relatively inexpensive. There are a lot of e-books and websites about certain topics scattered around the forum, you're sure to find them with a search. Here's one that somebody posted that I found just the other day as an example:

[url]http://www.cs.cmu.edu/~adamchik/21-127/lectures/recursion_4_print.pdf[/url]

about how thick and in depth are they?

I mean, i'm a 'noob' when it comes to topics like number theory, sequences, combinations and probability and if I am to invest in such books Id like to know that the solutions are fully explained rather than solutions like "this = this so this = this ... this = this" which often times leave me still stumped :wink: I'm in grade 11, but I assume I should get volume1 as I am certain there is stuff in there that I am clueless about, and also tricks and theorems that probably arent in volume2 because its assumed that the reader would already know them?

I dont ever plan on making it to cmo or anything like that, but I like problem solving and it gets me thinking which I really like. I wish I'd joined the math club and started participating much earlier.

I plan to hammer the books over the summer and to show up next year and surprise the heck out of everyone at my school  :roll:.. its a good goal I think anyway.

[quote="Glen Mann"]
I plan to hammer the books over the summer and to show up next year and surprise the heck out of everyone at my school  :roll:.. its a good goal I think anyway.[/quote]

lol nice plan

those books are great, and they provide very good strategies and analysis
even the solutions

it's a great idea to get them trust me

they're not too hard for the beginner

Volume 1 will definitely have some stuff you've probably never seen before if you're doing grade 11 in Ontario (a thorough introduction to euclidean geometry, permutations/combinations, modular arithmetic, methods of proof, logarithms(?), complex numbers (they took that out of the curriculum this year right?), probably some other stuff I can't remember right now). But it will also be lacking stuff you have done (trigonometric functions, trigonometric identities, etc.)

My advice to you is to try problems from every chapter, as even if you've been doing, for example, linear equations for a while, some of the harder questions in the problem sets will definitely provide you with some insights.

yeah theyve really dumbed down the math curriculum in Ontario. We're doing trigonometry now in my math class and I havnt learned a single new thing, we just had a quiz on the Sine and Cosine laws  :rotfl:.. I understand I am "advanced" compared to people in my class but, did we really have to "learn" things over again which we should already have known from last year? maybe some review.. but not a whole unit.

the teacher said we are done for trig now.. which means no identities, nothing about radians.. We spent a week on sine and cosine function translations.. even though we already spent a year on that for quadratic functions earlier this year (theyre the same rules!)..

The things we're doing next I think are sequences and series, and then 'business' math.. basicly stuff about interest etc, and then she mentioned something about exponential equations but thats another yawn.

Luckily my teacher taught us about complex numbers this year even though she wasnt supposed to :lol:.. but yes no permutations/combinations, no modular arithmetic, no proofs, no logarithms..

They have lowered the math standard quite a bit in Ontario over the years. Its because many people are failing math, and this is there 'quick fix' to get the marks going. :roll:

which really bugs me.. because if youre taknig university level courses, you should be planning on going to university, thus actually attempting to learn the stuff in highschool.

If kids cant cut the university level courses in highschool, how do they plan on cutting it in university? lowering standards does nothing but make dumb kids look smarter, thus making it even harder on them when they get into university because they looked better than they really are.

my opinion, there's no need to make courses easier, there will always be a proportion of students who doesn't understand, and there will be a proportion who excel. frankly if u belong in the bottom half, u don't deserve to go into fields that involve mathematics (i.e. engineering, math, cs, science) nor do i trust that they will be competent professionals after graduation from universities. surely, i will prefer not my depend my life on the things they design/develop.

like one of my profs said, no matter how hard he makes the exam or how shitty his explanations r, there will be ppl who understands him and do well on exams. no matter how easy he makes the exam and how well his explanations r, there will always be ppl who do poorly.

dumbing down the courses is not the best idea to help students adjust, but it is the easiest and probably the least costy one.

And now they are merging algebra and discrete together to make the course even easier, even though it already is quite basic...
If you want to train for olympiads in Canada you will get no help from your school teachers. :wink:

Alex, you changed your avatar. lol, still Lotr. :)
I agree with your philosophy J-I-M-B-O, making the courses easier doesn't do much to help the students. We don't even have a geometry course offered at our school anymore. :(

Listen to how the mucked up the grade12 courses in toronto..

There used to be as most of you may know

Advanced functions & intro calc
Geometry & discrete mathematics

now theyve changed the courses into

Advanced functions
Calculus & Vectors

both grade12 courses, except you need advanced functions as a prerequisite to calculus & vectors..

So at my school, which isnt semestered, this kinda messes thigns up a wee bit

I can only imagine what has been thrown out of the courses  :|

Instead of having to fulfill the prerequisite you can take both courses at the same time. Actually, other than the fact that some things will probably get taken out of the courses, I don't mind the way it got changed at all. For one, not nearly enough kids were taking Discrete and they opted for Data Management instead (which, IMO, is the most useless math course ever designed). Now that the only way to take calculus is to take 2 grade 12 maths, more kids will study useful things and less people will be taking Data  :D. Also, a lot of kids were severely under-prepared for calculus; (Ontario is one of few places where calculus is the fourth highschool course taken in math rather than the 5th) for example, I have tutored kids in calculus that have forgotten the quadratic formula, or the equation of a circle, or even [u]basic[/u] geometry. Hopefully, when the final curriculum comes out, stuff will have just been shifted around rather than fully omitted (but knowing Ontario... stuff will have been omitted.)

[quote="Glen Mann"]so what do you do rem? I've noticed that you really seem to know your stuff around here..

Ive also noticed that my mathclub at school is a joke.. the teacher running it never teaches us anything.. he'll occasionally give us an old CNML contest to practice with, but we never learn anything new :mad:.. and then he wonders how other schools get pefect scores[/quote]
Practicing with old CNML contests... :rotfl:  :rotfl:  :rotfl: lol this is ridiculous.

Well, practicing old contests helps, but not cnml... like past olympiads. Then I have several Russian books with a lot of good olympiad problems. I have learned enough theorems and formulas ( i think) then it's just training your skills for "handling the problem" and seeing through what it can be solved. :)

lol.. past olympiads eh

sadly.. there are many a thing on the CNML's that I do not understand.. ive been learning a lot lately though and i've been practicing old comc's and Ive been doing fairly well on them actually.. I didnt even know COMC existed because my mathclub teacher never mentioned it at all. My school is such a joke.

perhaps by the end of the summer I'll have moved on to practicing olympiads :oops:

and for the record I am definatly getting those AoPS books.. not any time soon since I cant afford it but they will be in my library soon enough!

what grade are you in and what school do you go to?

btw another useful book for preparing for Waterloo contests is PROBLEMS PROBLEMS PROBLEMS.

[quote]btw another useful book for preparing for Waterloo contests is PROBLEMS PROBLEMS PROBLEMS.[/quote]

You posted this about 3 months ago so I decided to steal (borrow?) a couple (unused) volumes from my school (vol 3 and 5) and they're DEFINITELY worth working through for CEMC contests. If you work through the entirety of volumes 3 and 5 (excluding the calculus sections and some problems on matrices if you wish) you'll be prepared to take any of the CEMC contests. The moral of my story is, take rem's advice, they're worth getting.

I've looked at volume 8 and in my opinion although it's supposed to cover the same topics it's just not the same (probably because it lacks old Descartes and CIMC level problems from back in the 80s, which are the best problems in 3 and 5.)

I didn't order them online, so I can't help you with this. Sorry.

did you just go to UW and purchase them there?

I saw them getting sold at Saint David's and i just bought them...

LOL my math teacher can solve some olympiad problems
and Glen u can go to the 'getting started' subforum to see some questions(or intermediate if u think that's too easy)  :)

good luck for those of u taking Apmo and Aime

It blows how curricular math material in highschool is so "un-thought-provoking."  It's just the usual "plug in this, plug in that, get answer."  I hope highschools lower the difficulty of the material so much that universities completely take over the education of math, or even science at that lol.  But then again, not all university classes are that informative, either. What a dilemma  :P

[quote="Glen Mann"]Well, I've found I can answer a number of questions in the beginner forum, and the odd one in the intermediate forum.

Sometimes in things like geometry I know how to find answers using the sin cos tan ratios and their laws when there are other ways to get exact answers and not just decimal approximations.

I've been practicing mathcounts questions (one link in resources links to a site where it generates mathcounts quizzes) and a lot of topics come up that leave me absolutely clueless sometimes especially the 'team' questions. This makes me feel really.. and I mean really stupid considering the fact that I am in grade 11 and these tests are aimed towards middle schoolers.

geometry and algebra are my strongest areas I think.. the rest. well it flat out stinks. I guess I am doing alright for myself, I mean I'm sure i'd be just as good as most of the mathcounts kids if id been introduced to higher math at that age.

just gotta keep on truckin..[/quote]

Canada's math education really fall far behind compare to USA, there is no way you can get contest help from ur math teachers

[quote="princevegetam"]that's true

canadian math education is just terrible

that's why usually canadian IMO kids are immigrants from other countries
they've had better training before coming to canada[/quote]

True.

Indeed, the same can be said with most of the other science olympiads ...

[quote="SM4RT"]just wondering if you order those books online how long does it take for you to receive them.[/quote]

kk i found out it takes atleast 10 days if you are in Canada and 4-6 weeks if you are outside of Canada

if you live in the area you get the books in like 6 days  :D..

go read this: [An introduction to the Theory of Numbers] BY G. H. HARDY AND E. M. WRIGHT

and get lost, u will improve a lot for sure

[quote="GlenthemanN"]Thanks for the book recommendation :D

I got one titled finite mathematics from my math teacher and its helped me a lot with series', permutations/combinations and probability. If I knew what I know now 2 months ago I would have done A LOT better on the Fermat.

I stink in number theory alot though :o

This is the correct one right?
http://www.chapters.indigo.ca/books/item/books-978019853171/0198531710/An-Introduction-to-the-Theory-of-Numbers?ref=Books%3a+Search+Top+Sellers[/quote]

LOL, it is FREE online actually

http://modular.fas.harvard.edu/scans/papers/hardy/Hardy-Wright-Theory_of_Numbers.pdf

have fun!

oh wow thanks scarfy! :D

but I think when I view it there are some symbols missing or something sometimes :huh:

its at my library so im gonna try to get it

[quote="princevegetam"]that's true

canadian math education is just terrible

that's why usually canadian IMO kids are immigrants from other countries
they've had better training before coming to canada[/quote]

yeah, i agree with you but if you guys all think ontario's math sucks, just wait til you come to bc. everyone gets perfect on the math 12 provincial it becomes a game of who can make less mistakes.


If $ a\plus{}b\plus{}c\equal{}x\plus{}y\plus{}z\equal{}0$ prove that

$ 4(ax\plus{}by\plus{}cz)^3\minus{}3(ax\plus{}by\plus{}cz)(a^2\plus{}b^2\plus{}c^2)(x^2\plus{}y^2\plus{}z^2)\minus{}2(a\minus{}b)(b\minus{}c)(c\minus{}a)(x\minus{}y)(y\minus{}z)(z\minus{}x)\equal{}54abcxyz$

:)

[quote="manlio"]If $ a \plus{} b \plus{} c \equal{} x \plus{} y \plus{} z \equal{} 0$ prove that

$ 4(ax \plus{} by \plus{} cz)^3 \minus{} 3(ax \plus{} by \plus{} cz)(a^2 \plus{} b^2 \plus{} c^2)(x^2 \plus{} y^2 \plus{} z^2) \minus{} 2(a \minus{} b)(b \minus{} c)(c \minus{} a)(x \minus{} y)(y \minus{} z)(z \minus{} x) \equal{} 54abcxyz$

:)[/quote]

I proved it showing that
$ P(x,y,z,a,b,c)\equal{}4(ax \plus{} by \plus{} cz)^3 \minus{} 3(ax \plus{} by \plus{} cz)(a^2 \plus{} b^2 \plus{} c^2)(x^2 \plus{} y^2 \plus{} z^2) \minus{} 54abcxyz$

is null for $ x\equal{}y$ or $ y\equal{}z$ or $ z\equal{}x$ or $ a\equal{}b$ or $ b\equal{}c$ or $ c\equal{}a$ and so

$ P(x,y,z,a,b,c)\equal{}k(x\minus{}y)(y\minus{}z)(z\minus{}x)(a\minus{}b)(b\minus{}c)(c\minus{}a)$

and to find $ k$ substitute $ (a,b,c)\equal{}(z,x,y)\equal{}(1,2,\minus{}3)$ and find $ k\equal{}2$.


LaTeX
inequalities unsolved
inequalities

I don't know about anyone else but I find attached .doc files to be really sketchy. Can't you just post the problem on the board directly?

Can you tell me the way to post directly? I don't know so I must do that

Type in the problem. In this case:

[quote="duc 95"]
Over all $ a,b,c > 0$ satisfying $ ab \plus{} bc \plus{} ca \equal{} 1$, find the minimum value of $ A \equal{} \alpha a^2 \plus{} \beta b^2 \plus{} \gamma c^2$, where
$ \alpha, \beta,$ and $ \gamma$ are constants.
[/quote]

[quote="duc 95"]can you tell me details?[/quote]

Learn $ \text{\LaTeX}$....
Just click ''quote'' on others post and see the code....
It is really easy....


Can you relate the eigenvalues of $ P$ to that of $ M \equal{} PP^{*}$ where $ P$ is a symmetric matrix.
[hide="Comments"]
I am not sure the relation if any is simple but a hunch is(which seems too strong with the given conditions) if $ \mu$ belongs to spectra of P,then $ |\mu|^{2}$ belongs to spectra of M.[/hide]

Is your question:
if $ P\in\mathcal{M}_n(\mathbb{C})$ is symmetic (not hermitian !) then what can we say about the eigenvalues of $ PP^*$ ?
$ PP^*$ is a non negative hermitian matrix.The singular values of $ P$ are the non negative real $ \sigma_i(P)$ (the square roots of the eigenvalues of $ PP^*$) where $ \sigma_i(P)\geq\sigma_{i\plus{}1}(P)$. Let $ spectrum(P)\equal{}(\lambda_i)$ where $ |\lambda_i|\geq|\lambda_{i\plus{}1}|$.
These inequalities stand: for all $ k\leq{n}$,   $ \Pi_{i\leq{k}}|\lambda_i|\leq\sqrt{\Pi_{i\leq{k}}\sigma_i(P^2)}\leq\Pi_{i\leq{k}}\sigma_i(P)$.
Moreover if $ p>0$, $ \sum_{i\leq{k}}|\lambda_i|^p\leq\sum_{i\leq{k}}\sigma_i(P)^p$.
$ P$ is a normal matrix iff for all $ i$, $ |\lambda_i|\equal{}\sigma_i(P)$ iff $ \sum_{i\leq{n}}|\lambda_i|^2\equal{}\sum_{i\leq{n}}\sigma_i(P)^2$.
     Here $ P$ is a symmetric matrix but the answer is definitively NO (except if $ P$ is normal and symmetric, that is $ P$ is real symmetric) because any complex matrix is similar to a symmetric matrix.
If $ P\equal{}\begin{pmatrix}1&2\plus{}i\\2\plus{}i&0\end{pmatrix}$ then $ |\lambda_i|\approx{2.73},1.83$ and $ \sigma_i(P)\approx{2}.79,1.79$.

yes that was the question i was talking about.thanks :)

My argument "because any complex matrix is similar to a symmetric matrix" doesn't stand because if $ P\equal{}QSQ^{\minus{}1}$ where $ S$ is symmetric  and $ Q$ is also symmetric (we can do that) then $ P^*\not\equal{}QS^*Q^{\minus{}1}$.
Yet the instance chosen in my preceding post and the result about normal matrices show that the result about the $ \lambda_i,\sigma_i$ is false even if $ P$ is symmetric.


[color=white]Could you increase the maximum avatar file size?  When I make my avatar images I somehow always end up with a picture that's 6.42 KB or something and I can't upload it until I crop off some stuff.  10 KB would be great.[/color]

[color=white]OH BOY!  Thanks! :D [/color]


function
calculus
derivative
integration
calculus computations

[b]1. The problem statement, all variables and given/known data[/b]
Question 5(10 marks)
Assume that a function f(x) is continuous on the interval [0,1].Express the following derivatives as formulae in terms of
$ x,f(x),f'(x) and \int_{a}^{x}f(t)dt$

$ a)\frac {d}{dx}\int_{0}^{x}tf(t)dt$          $ b)\frac {d}{dx}\int_{0}^{x}xf(t)dt$

       $ c)\frac {d}{dx}\int_{0}^{x}xf(x)dt$

[b]3. The attempt at a solution[/b]

I got 6 marks from 10.Can you help me to find my mistakes?:frown:

a)$ xf(x)$

b)$ \frac {d}{dx}\int_{0}^{x}xf(t)dt \equal{} \int_{0}^{x}f(t)dt \plus{} x\frac {d}{dx}\int_{0}^{x}f (t)dt \equal{} \int_{0}^{x}f(t)dt \plus{} xf(x)$

c)$ \frac {d}{dx}\int_{0}^{x}xf(x)dt \equal{} \int_{0}^{x}f(x)dt \plus{} f'(x)x\int_{0}^{x}dt \plus{} xf (x)\frac {d}{dx}\int_{0}^{x}dt \equal{} f(x)x \plus{} f'(x)x^2 \plus{} xf(x) \equal{} 2xf(x) \plus{} x^2f'(x)$

[quote="azatkgz"]Assume that a function f(x) is continuous on the interval [0,1][/quote]

this cannot deduce the existence of $ f'(x)$  :oops: .


inequalities
calculus
derivative
inequalities unsolved

Prove that there is the following packing up inequality ([u]without the derivatives[/u]):
$n\in N^*\Longrightarrow\underline {\overline {\left| \ \left(\frac{n}{n+1}\right)^{n+1}<\sqrt [n+1] {(n+1)!}-\sqrt [n] {n!}\ \right| }}<\left(\frac{n}{n+1}\right)^n$[hide="Remark."][u]Remark.[/u] It is wellknown that the [b]Lalescu's sequence[/b]
$L_n\equiv \sqrt [n+1] {(n+1)!}-\sqrt [n] {n!},\ n\in N^*$ respects the inequality $L_n<\left(\frac{n}{n+1}\right)^n.$
See http://www.mathlinks.ro/Forum/viewtopic.php?t=67063 [/hide]

This bilateral inequality is one of the best inequalities from diferential calculus. The left inequality was examined only on the computer with the affirmative answer until $n=50$ !.


[quote="GameBot"]A $ \text{\textit{perfect number}}$ is a counting number that is equal to the sum
of its proper factors. What is the sum of the two smallest perfect
numbers?[/quote]

The trick to find perfect numbers is take a power of 2 greater than 2 (call it $ x$), subtract 1 from it, and check if the result is a prime number. [i][u][b]If it is[/b][/u][/i], multiply this result $ (x\minus{}1)$ by $ \frac{x}2$. The result is a perfect number.

The two smallest perfect numbers are 6 $ (4\minus{}1)(4/2)$ and 28 $ (8\minus{}1)(8/2)$, so 6+28=$ \boxed{34}$

[quote="l Bob l"][quote="GameBot"]A $ \text{\textit{perfect number}}$ is a counting number that is equal to the sum
of its proper factors. What is the sum of the two smallest perfect
numbers?[/quote]

The trick to find perfect numbers is take a power of 2 greater than 2 (call it $ x$), subtract 1 from it, and check if the result is a prime number. [i][u][b]If it is[/b][/u][/i], multiply this result $ (x \minus{} 1)$ by $ \frac {x}2$. The result is a perfect number.

The two smallest perfect numbers are 6 $ (4 \minus{} 1)(4/2)$ and 28 $ (8 \minus{} 1)(8/2)$, so 6+28=$ \boxed{34}$[/quote]

Im kinda confused. Why do you have to multiply the result $ (x \minus{} 1)$ by $ \frac {x}2$. :maybe:

I suppose to explain this first requires a bit of background knowledge.

A prime that is of the form $ 2^p\minus{}1$, where $ p$ is a prime, is known as a Mersenne prime.  We could just discard the requirement that $ p$ is a prime, because the expression would obviously be composite or $ 1$.

It was proven that all numbers of the form $ 2^{p\minus{}1}(2^p\minus{}1)$, where $ 2^p\minus{}1$ is a Mersenne prime, are perfect numbers.  You can find the proof in multiple places here on AoPS.

Just for reference, the proof I wrote some while back:

Say we have a prime number $ p$ in the form $ 2^k \minus{} 1$. Let $ a \equal{} 2^{k \minus{} 1}(2^k \minus{} 1)$. We want to show that the sum of $ a$'s factors is $ 2a$.

The sum of the factors of is $ p$ is $ p \plus{} 1 \equal{} 2^k$. We then have:

$ \sigma(a) \equal{} \sigma(2^k \minus{} 1) \cdot \sigma(p) \equal{} (2^k \minus{} 1)2^k \equal{} 2a$. QED.

[quote=isabella2296]Just for reference, the proof I wrote some while back:

Say we have a prime number $ p$ in the form $ 2^k \minus{} 1$. Let $ a \equal{} 2^{k \minus{} 1}(2^k \minus{} 1)$. We want to show that the sum of $ a$'s factors is $ 2a$.

The sum of the factors of is $ p$ is $ p \plus{} 1 \equal{} 2^k$. We then have:

$ \sigma(a) \equal{} \sigma(2^{k \minus{} 1}) \cdot \sigma(p) \equal{} (2^k \minus{} 1)2^k \equal{} 2a$. QED.[/quote]

Did you mean that because I was reading over this and I got really confused haha (just in the very last line). Thank you so much for this though!


What are your opninons on a mathematical essay / investigation ( not really since its already been dissected) on Poker probabilty of Texas hold'em?

Would it prove to be of Calculus BC level? / higher?

Thoughts/suggestions for changes and/or math essay topics will be highly appreciated! thxs

It would not require any techniques more complicated than simple combinatorial reasoning.  However, the calculations are likely to be fairly tedious, and there are lots and lots of them to make.  The difficulty is in the amount of work to do, not any advanced mathematics required to do it.  In fact, this is a much better programming problem than it is a math problem.

I agree with JBL.  I would suggest doing something with abstract math.  Possibly ring theory or something of that sort?  But that's beyond B.C. Calc... beyond high school math.

You could also do physics, if you enjoy physics that is.  All of Newton's laws are calculus.  That would be Multi-Variable Calculus level.

From personal experience, one good problem for someone just past BC (or AB) Calculus is to prove with integrals several of the moment of inertia formulas for objects, as seen [url=http://en.wikipedia.org/wiki/List_of_moments_of_inertia]here[/url]. Some of them need multiple integrals though.

I personally did this during the physics AP test rather than memorize the formulas.


calculus
integration
logarithms
trigonometry
real analysis
real analysis unsolved

First integral

[url= https://mathworld.wolfram.com/HarmonicNumber.html] used (22)[/url]
https://mathworld.wolfram.com/HarmonicNumber.html

We start by observing the following:
\begin{align}
\int_{-\pi/2}^{\pi/2} \left(e^{ix}+e^{-ix}\right)^s e^{iwx}\,dx&= \frac{1}{i} \oint \limits_{\gamma} \left(z+z^{-1}\right)^s z^{(w-1)}\,dz\\
&=  \frac{1}{i} \oint \limits_{\gamma} \left(z+z^{-1}\right)^s z^{(w-1)}\,dz\\
&=  \frac{1}{i} \oint\limits_{\gamma} \left(1+z^{2}\right)^s z^{(w-s-1)}\,dz
\end{align}
where $\gamma$ a semicircle with in the upper positive halfplane with $\displaystyle{\substack{|z| = 1\\ -\pi/2 <\arg(z) < \pi/2}}$ recesses at the points $z = \pm i$. We complete the semicircle with a straight section that joins the points $z=-i$ and $z=i$ apart from the recesses $z=-i$,$0$, i. Then:
\begin{align}\oint\limits_{\gamma} \left(1+z^{2}\right)^s z^{(w-s-1)}\,dz &= e^{i\frac{\pi}{2}(w-s)}\int_{0}^{1} (1-x^2)^sx^{w-s-1}\,dx+e^{i\frac{\pi}{2}(w-s)}\int_0^{-1} (1-x^2)^sx^{w-s-1}\,dx\\
&= e^{i\frac{\pi}{2}(w-s)}\int_{0}^{1} (1-x^2)^sx^{w-s-1}\,dx-e^{-i\frac{\pi}{2}(w-s)}\int_0^{1} (1-x^2)^sx^{w-s-1}\,dx\\
&= 2i\sin \left(\frac{\pi (w-s)}{2}\right)\int_0^1 x^{w-s-1}(1-x^2)^s\,dx\\
&= i\sin \left(\frac{\pi (w-s)}{2}\right)B\left(\frac{w-s}{2},s+1\right)\end{align}
Consequently
$$\displaystyle{\int_{-\pi/2}^{\pi/2} \left(e^{ix}+e^{-ix}\right)^s e^{iwx}\,dx = \sin \left(\frac{\pi (w-s)}{2}\right)B\left(\frac{w-s}{2},s+1\right)}$$
Producing as to w we have
\begin{align}&\int_{-\pi/2}^{\pi/2} x\left(e^{ix}+e^{-ix}\right)^s e^{iwx}\,dx\\
&=-i\frac{d}{dw} \sin \left(\frac{\pi (w-s)}{2}\right)B\left(\frac{w-s}{2},s+1\right) \\&= -iB\left(\frac{w-s}{2},s+1\right)\left(\frac{1}{2}\sin \left(\frac{\pi (w-s)}{2}\right)\left(\psi\left(\frac{w-s}{2}\right) - \psi\left(\frac{w+s}{2}+1\right)\right) + \frac{\pi}{2}\cos \left(\frac{\pi (w-s)}{2}\right)\right)\end{align}
So for the original integral we have:
\begin{align}I&=\int_{0}^{\pi/2}\frac{x^2}{x^2+\log^2 (2\cos x)}\,dx \\&= \frac{1}{2}\int_{-\pi/2}^{\pi/2}\frac{x^2}{x^2+\log^2 (2\cos x)}\,dx\\
&= \frac{1}{2}\mathfrak{I}\int_{-\pi/2}^{\pi/2}\frac{x}{\log (2\cos x) - ix}\,dx\\
&= \frac{1}{2}\mathfrak{I}\int_{-\pi/2}^{\pi/2}\frac{x}{\log \left(1+e^{-2ix}\right)}\,dx\\
&= \frac{1}{2}\mathfrak{I}\int_{-\pi/2}^{\pi/2}\int_0^1 \frac{x(1+e^{-2ix})^s}{(1+e^{-2ix}) - 1}\,ds\,dx\\
&= \frac{1}{2}\mathfrak{I}\int_0^1\int_{-\pi/2}^{\pi/2} x(e^{ix}+e^{-ix})^se^{i(2-s)x}\,dx\,ds
\end{align}
Setting $w=2-s$ and using formulas $\displaystyle{B(1-s,1+s) = \frac{\pi s}{\sin \pi s}}$ and $\displaystyle{\psi(1-s) - \psi (s) = \pi\cot \pi s}$ the result is simplified in
\begin{align}I &= -\frac{\pi}{4}\int_0^1 s(\psi (1-s) - \psi (2) - \pi \cot \pi s)\,ds \\
&= -\frac{\pi}{4}\int_0^1 s(\psi (s) - \psi (2))\,ds\\
&= -\frac{\pi}{4}\left(-\frac{1}{2}+\frac{\gamma}{2} - \int_0^1 \log \Gamma(s)\,ds\right)\\
&= \frac{\pi}{8}\left(1-\gamma + \log (2\pi)\right)
\end{align}


A robot moving forward at a constant speed takes 2.5 hours
to travel 1 kilometer. Moving forward at this same constant
speed, it takes the robot 90 seconds to travel the length of a
particular hallway. How many meters long is the hallway?

so moving at a constant speed, the robot goes (1*1000)-> 1000 meters
in (2.5*60*60)-> 9000 seconds.

How long in meters a robot goes in 90 sec is...
90/9000 = x/1000

so this means...

1/100 = x/1000

cross multiply...

1000 = 100x

$ \text{x} \equal{} \mathfrak{10}$

so the robot goes 10 meters in 90 seconds.[/hide]


We'll say that a sequence $f_n : X \to \mathbb{R}$or $\mathbb{C}$ tends to $f(x)$ quasi-uniformly if $X$ we can represent as a countable union of sets on which $f_n$ converges to $f$ uniformly. Prove that there exists sequences $f_n : [0,1] \to \mathbb{R}$, such that $f_n \to f \equiv 0$ pointwise but not quasi-uniformly.

To each point $x$ associate the non-decreasing sequence of positive integers $(S^x_n)_{n\ge 1}$, where $S^x_n$ is the smallest $m_0$ such that $m\ge m_0\Rightarrow f_m(x)\in\left(-\frac 1n,\frac 1n\right)$. Conversely, for every collection of $c$ (continuum cardinality) non-decreasing sequences of positive integers, we can find a sequence of functions defined on $I=[0,1]$ so that the correspondence above holds.

Since there are $c$ non-decreasing sequences of positive integers, we can construct a sequence of functions $f_n$ such that the corresponding sequences $(S^x_n)_n$ as $x$ varies over $I$ are all possible non-decreasing sequences. Now suppose $I$ can be written as a union $A_1\cup A_2\cup\ldots$ such that $f_n\to 0$ on each $A_i$ uniformly (assume also that $A_1\subset A_2\subset\ldots$). We now use a diagonal argument to get a contradiction: uniform convergence on a set $X$ translates to the fact that for every $n$, the set $(S^x_n)_{x\in X}$ be bounded; there is an $x_0\in I$ such that for all $n\ge 1,\ S^{x_0}_n>\max_{x\in A_n}S^x_n$, and this $x_0$ won't belong to any of the $A_n$.


Okay this is probably completely stupid and doesn't even deserve a place in the Round Table.

Are

-Emil Fischer (Nobel, Chemistry)
-Hans Fischer (Nobel, Chemistry)
-Ernst Otto Fischer (Nobel, Chemistry)
-Edmond H. Fischer (Nobel, Medicine and Physiology)

Related at all? Or is $ \text{Fischer}$ just a really common last name?

According to a stranger with no credibility, Emil Fischer was some kind of an advisor of Hans Fischer. It may be more accurate to say that Hans was Emil's postdoc.

We kid in our family about not being related to Nobel Prize winners. I am not, as far as I know, related to R. Bruce Merrifield. And my wife, whose name was Laue, is not, as far as she knows, related to Max von Laue.

There are a million Browns, Smiths, and some other really common name out there. Fischer is probably one of them.

Their biographies can be found at nobelprizes.org . Besides the above-mentioned connection between Hans Fischer and Emil Fischer, there is another trivia bit: the third son of Emil Fischer, Hermann Otto Laurenz Fischer, was Professor of Biochemistry at UC-Berkeley. Which is almost as good as winning a Nobel Prize.  :)

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

fischer looks pretty common


geometry
circumcircle
trigonometry
inequalities
search
function
geometry unsolved

In a triangle $ABC$ with side $a,b,c$ and the radius of the circumcirlce$R$ , radius of the incircle $r$,and $s=\frac{a+b+c}{2}$
we have:$\cos A\cos B\cos C=\frac{s^{2}-(2R+r)^{2}}{4R^{2}}$
and $s^{2}\leq 4Rr+4R^{2}+3r^{2}$
I think these  two are famous.
And I hope somebody can list some equality and  inequality about the sides of the triangle and the the radius of the circumcirlce and incircle and all that stuff.

actually:
$16Rr-5r^{2}\le s^{2}\le 4R^{2}+4Rr+3r^{2}$
is better :wink:

Sorry,I can't prove this one .Could you give me some hint?

Use the search function with key word: Gerretsen.


Hi,

I'm doing this research and because my computer has been attacked by virus recently, I can't use Google or any other search engines to continue my research. :( My report is due very soon and I got all the information except ONE!

The one thing that I need help is when was this disease called genital wart first identified..

I used http://www.health.yahoo.com and it gave me all those effects and other stuffs but I can't find it when it was identified.

If anyone can tell this, I'll be SUPER appreciated!  :( I really need to know that information because that's only one thing I couldn't find.

P.S. I did post on the science forum but I have not received any answers and hopefully, because of close reactions of math and science, I can get the answer from here.

Thank you a lot!

this site says 1907 but it might be talking about warts in general
http://www-micro.msb.le.ac.uk/3035/Papillomaviruses.html

this site says 1907 too.. and its talking about genital.. so its probably 1907
http://www.emedicine.com/med/byname/condyloma-acuminatum.htm

And to solve your computer problems, you might want to use Ad-Aware. It can be downloaded at http://www.lavasoft.com

[quote="joml88"]A virus that wont let you use search engines but yet you can still get on this site and other sites?[/quote]

Yeah, this is WEIRD. The thing is that I can get on the internet with specific site I typed (so I type http://www.artofproblemsolving.com, I can get to it.) but I can't use search engine.

Also, I have problem with my start page because I have virus called Start.Page.Trojan which changes the home page.

I'm in safe mode now and this really bothers me because I can only do very basic functions.

temperily, u could try using proxy servers to access sites.. maybe that'll help

Either that or getting a virus scan and ad-ware scan.  For ad-ware I'd reccomend [url=http://ww.lavasoftusa.com/]Ad-Aware[/url].  If you need a virus scan, I know that McAfee and Norton have free-after-rebate thing.  I could also give you a virus scan to use.

I hope your project doesnt include pictures... :?

For virus I suggest Norton. It works quite well for me.

Lavasoft's Adaware

Norton Security

And Spybot Search and Destroy keep my computer healthy

If your computer is really messed up, look into how to use Hijack This, but be careful you don't wipe out what you need with that program

CWShredder is also good, as well as ZoneAlarm for a firewall and AVG free for AV, as well, as Avast, BUT DON'T USE AVAST TOGETHER WITH NORTON OR AVG!!!!!


calculus
integration
limit
function
trigonometry
calculus computations

Hi, I was wondering if anyone knows how to evaluate this integral.

Question:  $ \displaystyle\int_0^{\infty}\frac{x^2}{(a^2\plus{}x^2)(b^2\plus{}x^2)}dx$

There are many methods. For a first-year calculus method, we want to use partial fractions.

But note that you see $ x^2$ with no $ x.$ My suggestion for cutting down on the work:

Substitute $ u\equal{}x^2.$ (This is an algebra substitution, not a calculus substitution.) Find $ A$ and $ B$ so that

$ \frac{u}{(a^2\plus{}u)(b^2\plus{}u)}\equal{}\frac{A}{a^2\plus{}u}\plus{}\frac{B}{b^2\plus{}u}.$

Then reverse the substitution so you have $ \frac{A}{a^2\plus{}x^2}\plus{}\frac{B}{b^2\plus{}x^2}.$

Then integrate (there will be arctangents).

hello, write your integrand in the form
${{ \frac{1}{a^2-b^2}(\frac{1}{1+\frac{x^2}{a^2}}}-\frac{1}{1+\frac{x^2}{b^2}}})$.
Sonnhard.

red_dog: Your method contains the tacit assumption that $ a\ne b.$

What would happen if $ a\equal{}b?$ In particular, would your final answer still be valid in that case?

[quote="Kent Merryfield"]red_dog: Your method contains the tacit assumption that $ a\ne b.$

What would happen if $ a \equal{} b?$ In particular, would your final answer still be valid in that case?[/quote]

If $ a\equal{}b$ then $ I_n\equal{}\int_0^n\frac{x^2}{(a^2\plus{}x^2)^2}dx\equal{}\minus{}\frac{1}{2}\int_0^nx\left(\frac{1}{a^2\plus{}x^2}\right)'dx\equal{}$

$ \equal{}\left. \minus{}\frac{x}{2(a^2\plus{}x^2)}\right|_0^n\plus{}\frac{1}{2}\int_0^n\frac{dx}{a^2\plus{}x^2}\equal{}\minus{}\frac{n}{2(a^2\plus{}n^2)}\plus{}\frac{1}{2a}\arctan\frac{n}{a}$

$ \lim_{n\to\infty}I_n\equal{}\frac{\pi}{4a}$

Nicely done and thank you. Of course, I knew the answer to "would the answer still be valid in that case?" since it's clear that the integral is a continuous function of $ a$ and $ b$ as long as you stay away from either one being zero.

Thank you everyone for the nice solutions.  Here is another method:

Consider $ f(t) \equal{} e^{ \minus{} at}$ and $ g(t) \equal{} e^{ \minus{} bt}$, then the Fourier sine transforms of $ f(t)$ and $ g(t)$ are given by $ \displaystyle \int_0^{\infty}f(t)\sin (tx) dt$.
Thus, $ \displaystyle\mathcal{F}_S\{f(t)\} \equal{} \frac {x}{a^2 \plus{} x^2}$ and $ \displaystyle\mathcal{F}_S\{g(t)\} \equal{} \frac {x}{b^2 \plus{} x^2}$
By Parseval's identity, $ \displaystyle\frac {2}{\pi}\int_0^{\infty}\left(\frac {x}{a^2 \plus{} x^2}\right)\left(\frac {x}{b^2 \plus{} x^2}\right)dx \equal{} \int_0^\infty e^{ \minus{} (a \plus{} b)t}dt$
Therefore, $ \displaystyle\frac {2}{\pi}I \equal{} \frac {1}{a \plus{} b}\Rightarrow \boxed{I \equal{} \frac {\pi}{2(a \plus{} b)}}$


Δινεται η καμπυλη $ C$ με εξισωση  $ e^k(e^{\sqrt{x}}\plus{} e^{\sqrt{y}}\minus{}1)\plus{}(1\minus{}e^k)e^{\sqrt{x}\plus{}\sqrt{y}}\equal{}0$.
N..δ.ο. για καθε τιμη του θετικου πραγματικου $ k$ στην γραφικη παρασταση της $ C$ ανηκουν πεπερασμενα σημεια $ M(x,y)$ οπου $ x,y$ ειναι θετικοι ακεραιοι

Could you translate the problem into English?

Thanks in advance.

kunny

Let  $ C$  a curve with equation  $ e^k(e^{\sqrt {x}} \plus{} e^{\sqrt {y}} \minus{} 1) \plus{} (1 \minus{} e^k)e^{\sqrt {x} \plus{} \sqrt {y}} \equal{} 0$.
Prove that for each value of the real number $ k$ there exist only finite points $ M(x,y)$ with positive integer coordinates on $ C$

Thank you ,silouan.

$ e^k(e^{\sqrt {x}} \plus{} e^{\sqrt {y}} \minus{} 1) \plus{} (1 \minus{} e^k)e^{\sqrt {x} \plus{} \sqrt {y}} \equal{} 0\Longleftrightarrow e^k\equal{}\frac{e^{\sqrt{x}}\cdot e^{\sqrt{y}}}{(e^{\sqrt{x}}\minus{}1)(e^{\sqrt{y}}\minus{}1)}$

@kunny: Let me to give a hand to show him why :D

So we can consider the case $ k > 0$ (because $ k\leq 0$ is trivial true) and let $ c \equal{} e^{k/2} > 1$. The function $ f(t) \equal{} \frac {e^t}{e^t \minus{} 1}$ then by a quick computation, we see that $ f$ is increasing. Let us assume that $ x\geq y$ then $ c^2\leq f(\sqrt x)^2$ so $ c\leq f(\sqrt x)$. This implies $ e^{\sqrt x}\leq \frac {c}{c \minus{} 1}$. This shows that $ x$ and so $ y$ is, is finitely.

Sorry, I gave just a hint. My solution is almost same as yours. :)

[quote="kunny"] My solution is almost same as yours. :)[/quote]

Really ? Then your solution is also incorrect . Obsiously the function $ f$ which was defined above is decreasing and not increasing . Try again  :wink:

Μια λυση
Εστω οτι υπαρχουν απειρα σημεια $ M(x_i,y_i)$  στην $ C$ τοτε η εξισωση ισοδυναμα μετατρεπεται στην $ f(x)f(y) = e^k$ (1) οπου $ f(x) = \frac {e^\sqrt {x}}{{e^\sqrt {x}} - 1}$
Εστω $ f(c_i) = max{(f(x_i),f(y_i)})$ .Ευκολα βρισκουμε οτι η $ f$ εχει συνολο τιμων το $ (1, +\infty)$ οποτε η (1) μετατρεπεται στην $ f^2(c_i)\geq{f(a)}$ (2). Αφου υπαρχουν απειρες λυσεις $ c_i$ υπαρχει ακολουθια λυσεων $ {c{}_n}{}_k$  με
$ \lim{{c{}_n}{}_k} = +\infty$.  H (2) τώρα για $ c_i$ to $ c{}_n{}_k$  και οριο του $ n\rightarrow +\infty$ δινει
$ 1\geq{f(a)} > 1$ που ειναι ατοπο


[quote="mdk"]Let u be an irrational real number. Let S be the set of all real numbers of the form a+bu, where a and b are integers. Show that S is dense in the real numbers.[/quote]

[hide]
{a + bu | a, b in Z, u in R / Q} = {u | u in R / Q}?
[/hide]

@mdk: this problem is obviously not intermediate level (at least without defining what being dense means).

@tactics: obviously not as setting $ b=0$ shows (you get all of $ \mathbb R$). And this has nothing to do with the problem as $ u$ is fixed.

[quote="ZetaX"]@tactics: obviously not as setting $ b=0$ shows (you get all of $ \mathbb R$). And this has nothing to do with the problem as $ u$ is fixed.[/quote]

Oh my, 0 wins again.

It is enough to show that if $ \alpha$ is an irrational number, then the sequence $ \{n\alpha\}$ is dense in $ [0,1]$.  I believe this is known as Kronecker's theorem, but in any event, the proof is fairly simple; $ \{n\alpha\}$ is an infinite bounded sequence of distinct terms, so it contains a convergent subsequence, and by taking differences (modulo a bit of fussing), we can find a subsequence tending to $ 0$, and build up from that to get the result.


How can  I prove that:

Let  $ F(x): R^m\rightarrow R^n$ be a function such that  at $ \mathbf{rank}\Big[\frac {\displaystyle \partial F}{\displaystyle \partial x}(a)\Big]  \equal{} k$, then for some open neighborhood of $ a$, $ \mathbf{rank}\Big[\frac {\displaystyle \partial F}{\displaystyle \partial x}(x)\Big]\ge k$

Thanks in advance

Ajoye

If the rank is at least $ k$, then there is a $ k\times k$ submatrix with nonzero determinant. Since the determinant is a continuous function, it is also nonzero in some neighborhood of the point.

There was a little typo. I modified it as follows

Let  $ F(x): R^m\rightarrow R^n$ be a function such that  at $ \mathbf{rank}\Big[\frac {\displaystyle \partial F}{\displaystyle \partial x}(a)\Big] \equal{} k$, then for some open neighborhood of $ a$, $ \mathbf{rank}\Big[\frac {\displaystyle \partial F}{\displaystyle \partial x}(x)\Big]\ge k$

My question is why can't $ \mathbf{rank}\Big[\frac {\displaystyle \partial F}{\displaystyle \partial x}(x)\Big]\le k$ ? The rank can increase but why cant it decrease in the open set? I am kinda having hard time to see this simple thing.  :(

Ajoye[/quote]

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=167019]here[/url]. If $ A$ is a matrix of rank $ k$, there are invertible matrices $ R,C$ such that $ RAC\equal{}\begin{bmatrix}I_k&0\\0&0\end{bmatrix}$; these $ R,C$ represent row and column operations.
In that standard form, it's easy to see why we can't make the rank smaller with small perturbations.

Your statement should include that $ F$ is (at least) $ C^1$. Weird things can happen if we don't assume continuity of the derivative.


calculus
integration
trigonometry
limit
search
real analysis
calculus computations

I can't find limes of this sequence
\[\int_{0}^{\pi}\frac{\sin x}{(\cos(nx))^{2}+1}dx\]
can you help ?

A general theorem: If $g$ is continuous and periodic with period $P$, and $f$ is reasonably nice, $\lim_{n\to\infty}\int_{a}^{b}f(x)g(nx)\,dx=\int_{a}^{b}f(x)\,dx\cdot\frac1P\int_{0}^{P}g(t)\,dt$- the integral of $f$ multiplied by the average value of $g$.

I know it's been discussed before, but a search of the forums for Riemann-Lebesgue turns up way too many false positives.

The integral of $\sin$ from zero to $\pi$ is 2; that leaves the calculation of the average value $\frac1{\pi}\int_{0}^{\pi}\frac1{\cos^{2}t+1}\,dt$.

This integral can be rewritten as
$\frac1{\pi}\int_{-\frac{\pi}2}^{\frac{\pi}2}\frac{\sec^{2}t}{1+\sec^{2}t}\,dt= \frac1{\pi}\int_{-\frac{\pi}2}^{\frac{\pi}2}\frac1{2+\tan^{2}t}\cdot \sec^{2}t\,dt=\frac1{\pi}\int_{-\infty}^{\infty}\frac1{2+u^{2}}\,du=\frac1{\sqrt{2}}$.

The limit is $\sqrt{2}$.


hey,
could smbdy pleees giv me a link to some nice ebook on polynomials?
thnks in advance

check out http://web.mit.edu/rwbarton/Public/mop/polynomials.pdf

or http://mathcircle.berkeley.edu/poly.pdf

   I don't know if there aere books on the web but these handouts (particularly the second one by Gabriel Carroll) are quite comprehensive.


I have found this inequality on an old algebra's book and also on a web site.
I cannot prove it.

Let a,b,c the sides of a triangle. Prove that

(a+b-c)^a(b+c-a)^b(c+a-b)^c  \leq (a^a)(b^b)(c^c)

It is easy, instead, to show that

(b+c-a)^a(c+a-b)^b(a+b-c)^c  \leq (a^a)(b^b)(c^c)

[quote="manlio"]I have found this inequality on an old algebra's book and also on a web site.
I cannot prove it.

Let a,b,c the sides of a triangle. Prove that

(a+b-c)^a(b+c-a)^b(c+a-b)^c  \leq (a^a)(b^b)(c^c) (1)

It is easy, instead, to show that

(b+c-a)^a(c+a-b)^b(a+b-c)^c  \leq (a^a)(b^b)(c^c)[/quote]

=======
Inequality is not changed if we change (a,b,c) -> (ka, kb, kb)
So, we can assume that a, b, c < 1
Now (1) is equivalent to

(1+(b-c)/a)^2.(1+(c-a)/b)^b.(1+(a-b)/c)^c \leg 1  (2)

By Bernoulli inequality (1+(b-c)/a)^a <= 1+b-c
Similarly  (1+(c-a)/b)^b <= 1 + c-a, (1+(a-b)/c)^c <= 1+a-b.

Multiplying these inq's, we get  LHS of (2) <= (1+b-c)(1+c-a)(1+a-b)<=1

(The last one is AM-GM inequality)

Have joy.

Namdung

[quote="manlio"] Let $a,b,c$ be the sides of a triangle. Prove that $(a+b-c)^{a}(b+c-a)^{b}(c+a-b)^{c}\leq a^{a}b^{b}c^{c}$.[/quote]
Using the weighted arithmetic mean - geometric mean inequality,

$\sqrt[a+b+c]{\left(\frac{a+b-c}{a}\right)^{a}\left(\frac{b+c-a}{b}\right)^{b}\left(\frac{c+a-b}{c}\right)^{c}\right}$

$\leq\frac{1}{a+b+c}\left(a\frac{a+b-c}{a}+b\frac{b+c-a}{b}+c\frac{c+a-b}{c}\right)=1$.
[hide="Conjecture."]The following inequality
$(a+b-c)^{a}(b+c-a)^{b}(c+a-b)^{c}\leq \left(a-\frac{4}{5}b+\frac{4}{5}c\right)^{a}\left(b-\frac{4}{5}c+\frac{4}{5}a\right)^{b}\left(c-\frac{4}{5}a+\frac{4}{5}b\right)^{c}$
holds for the sides $a,b,c$ of a triangle.[/hide]


Solve in positive real numbers the following equation:

                 \[ x^{-y}+y^{-x}=4-x-y.  \]

x=y=1  :!: . But i can't solve. Everybody, did you have any solution?

The equation has infinitely many solutions. I don't think they can be expressed in a nice way. Try this one, in exchange:
Solve in positive real numbers the equation
\[ x^y+y^x=1+xy. \]

[quote="enescu"]
Solve in positive real numbers the equation
\[ x^y+y^x=1+xy.  \][/quote]
Use http://www.artofproblemsolving.com/Forum/viewtopic.php?t=42452


What is the least positive integer such that it is a multiple of 16
and the product of its digits is a multiple of 12?

I guess we start from the smallest multiple of 16.

[u]The number[/u] [u]The product of digits[/u]
0 (does 0 count? If yes, this is the answer) 0
16 ------------------------ 6
32 ------------------------ 6
48 ------------------------ 32
64 ------------------------ 24

Therefore, the smallest number would either be 0 or 64

its 64, just test diffffferent multiples

i dont really understand the question  :blush:

The directions said positive integer, so it cant be 0.

[quote="isabella2296"]Basically, we're looking for the smallest, positive integer that fits these two requirements:

1. A multiple of 16

2. The product of the digits is a multiple of 12

So we start by checking the smallest multiples of 16.

16 --> product of digits = 6, not a multiple of 12
32 --> product of digits = 6, not a multiple of 12
48 --> product of digits = 32, not a multiple of 12
[color=red]64 --> product of digits = 24, which is a multiple of 12[/color]

So $ \boxed{64}$ is our answer.[/quote]

Erm...hypocritical???? LOOK WHAT YOU SAID EARLIER!!!
[quote="isabella2296"]This was already answered. Actually [b]read[/b] the above posts please. Don't spam like this.

And you copied my avatar![/quote]

grrrrr
at least im not a hypocrite

[quote="herefishyfishy1"]I think that Isabella provided an explanation because myyellowducky said
"i dont really understand the question"[/quote]

Yes. Sorry if it was spam, but I thought it would be okay because she didn't understand, so I tried to rephrase it.