vector
topology
induction
algebra
polynomial
function
linear algebra

In linear algebra, we have a theorem:

[b] Every vector space has a basis. [/b]

Consider the following vector space  over $ \mathbb{R}$ (the space of all real sequences) :
$ V\equal{}\{ (a_1, a_2, ... ) : a_i \in \mathbb{R} \}$
in which addtion and scalar multiplication are performed element wise.

I try my best but could not find a sound basis for that vector space. Could someone give me a construction of it ?

[hide] It seems that the basis is uncountable.  :) [/hide]

Hmm, what is wrong with the [url=http://en.wikipedia.org/wiki/Schauder_basis]Schauder basis[/url] $ \bigcup_{k \in \mathbb{N}} \{x_k \equal{} (0,0,..,1,...) : x_k \text{ is 1 in the k'th position and 0 everywhere else} \}$?

This is clearly talking in the Hamel sense- there is no topology involved. If we were looking at some $ \ell^p$, that would be a reasonable answer.

Proving the existence of the basis requires the Axiom of Choice. There is no construction, and there can't be a construction.
Anything that relies on the Axiom of Choice is inherently nonconstructive.

Can I ask another question. Is the set $ \bigcup_{k \in \mathbb{N}} \{x_k \equal{} (0,0,..,1,...) : x_k \text{ is 1 in the k'th position and 0 everywhere else} \}$ a basis for the original question?

No, because the linear span of this set consists only of vectors that have finitely many non-zero elements. As jmerry pointed out, there's no topology given, so it doesn't make any sense to add infinitely many vectors. In the ordinary sense, a linear combination is just a linear combination of finitely many elements.

It is indeed true that every vectors space has a basis even in this restricted sense of "basis". In our case, there exists a set $ S$ of sequences of real numbers such that every sequence of real numbers can be written uniquely as a linear combination of[i] finitely many[/i] members of $ S$. So $ S$ is necessarily uncountable and cannot be explicitly described, due to the inherent use of AC, as jmerry mentioned.

I went away and read up on this stuff again and it makes perfect sense now. Thanks everyone for the info.

Is a basis of $ \mathbb{R}^\mathbb{N}$ countable or not ?

Uncountable. There is (up to vector space isomorphism) only one infinite dimensional space with a countable basis, and it's been mentioned on this thread as the space of "finitely supported" sequences. (I'm referring to the first sentence in randomgraph's post.) You could also think of this (in another isomorphic version) as the the vector space of all polynomials.

Here's a challenge: can someone produce an uncountable independent set in the space of all sequences?

[quote="Kent Merryfield"]the space of "finitely supported" sequences. (I'm referring to the first sentence in randomgraph's post.) You could also think of this (in another isomorphic version) as the the vector space of all polynomials.[/quote]
I like to think of it as the space of sequences that converge to $ 0$ in the discrete metric.  In any case, this is far from constructive, but transfinite induction might work:  start with a set of sequences, and at the next step pick a sequence that isn't in the linear span of the sequences you currently have.  My concern is that this process is guaranteed to continue as long as you only have a countable number of sequences, but I'm not sure exactly when transfinite induction will halt here.

Edit:  D'oh.  Pick the sequences $ \{ (1, \alpha, \alpha^2, ... ), \alpha \in \mathbb{R} \}$.

Edit 2:  This is an interesting example; the linear span of the sequences I gave above is the set of sequences that obey a homogeneous linear recursion given by an arbitrary real polynomial with distinct real roots, which is both clearly uncountable and clearly a very "small" portion of the entire space.

[quote="Kent Merryfield"]Here's a challenge: can someone produce an uncountable independent set in the space of all sequences?[/quote]

We will find such set for each space $ \mathbb{F}^{\mathbb{N}}$ over $ \mathbb{F}$, where $ \mathbb{F}$ is an arbitrary field.

Let $ X$ be a set of cardinality $ \aleph_{0}$, and let $ A, B\subset X$, $ \#A, \#B \equal{} \aleph_{0}$. We call $ A, B$ almost disjoint if $ A\cap B$ is finite.

Now take $ X: \equal{} \{(a_{1}, \ldots, a_{n}) \ : a_{i}\in \{0,1\}, \ n\in \mathbb{N}_{1}\}$.
Clearly $ \#X \equal{} \aleph_{0}$ and function:
$ f: X \ni (a_{1}, \ldots, a_{n})\mapsto 2^{n} \plus{} \sum_{i \equal{} 1}^{n}a_{i}2^{i \minus{} 1}\in \mathbb{N}_{2} \equal{} \{2, 3, 4, \ldots\}$
is bijection.

We will give a construction of a family of $ 2^{\aleph_{0}}$ pairwise almost disjoint subsets of $ X$.
Let $ \chi: \mathbb{N}_{1}\to \{0, 1\}$.
We define $ A_{\chi}: \equal{} \{x \equal{} (a_{1}, \ldots, a_{n})\in X \ : \ \forall i \equal{} 1, 2, \ldots, n : \ \chi(i) \equal{} a_{i}\}$
It is easy to see that for each $ \chi$ set $ A_{\chi}$ is countable, and if $ \chi_{1}\neq \chi_{2}$ then $ A_{\chi_{1}}\cap A_{\chi_{2}}$ is finite.
Therefore $ \mathcal{A} : \equal{} \{A_{\chi} : \ \chi: \mathbb{N}_{1}\to \{0,1\} \}$ is family satisfying our conditions.

Now come back to the space $ \mathbb{F}^{\mathbb{N}}$, and let $ \mathcal{I} : \equal{} \{f(A) \ : A\in \mathcal{A}\}$. Then $ \mathcal{I}$ is a family of $ 2^{\aleph_{0}}$ pairwise almost disjoint subsets of $ \mathbb{N}$.
For $ I\in \mathcal{I}$ consider $ f_{I}\in \mathbb{F}^{\mathbb{N}},\ f_{I}(n) : \equal{} \begin{cases}0, \ n\in \mathbb{N}\setminus I \\
1, \ n\in I\end{cases}$
The set $ \{f_{I}\ : \ I\in \mathcal{I}\}$ has cardinality $ \#\mathcal{I} \equal{} 2^{\aleph_{0}}$, and it is linear independent set, since for all pairwise distinct $ I_{1}, \ldots, I_{k} \in \mathcal{I}$ the set $ I_{1}\cap (I_{2}\cup \ldots\cup I_{k})$ is finite.


geometry
geometric transformation
reflection
incenter
geometry unsolved

consider ABC is an right triangle that <C=90.O is the center of scribed circle of ABC traingle.
AB is tagent to scribed circle at M.consider a circle that it 's center is M and pass O intersect
the bisectors of <A and <B at K and L respectively.prove that K and L are respectively centers of scribed circles
of ACD and BCD, that CD is altitude of ABC triangle.

Solution:

Let $ M'$ br the reflection of $ M$ on the bisector of $ \angle B$.
Clearly $ M'$ is also the point where the incircle of $ ABC$ touch $ BC$.
Note that $ M'LMO$ is a rhombus. Hence $ M'L\parallel{}OM\parallel{}CD$
Since $ CO$ is the bisector of $ \angle C \equal{} 90^0\implies \angle M'CO \equal{} 45^0\implies M'O \equal{} M'C$
Hence $ M'C \equal{} M'O \equal{} M'L\implies M'CL$ is isosceles.
$ \angle DCM' \equal{} \angle LM'B \equal{} 2\angle LCM'$
We already had $ BL$ is the bisector of $ \angle DBC$. Now we have $ CL$ the bisector of $ \angle BCD$.
Thus $ L$ is the incenter of $ BCD$
Similarly it can be proved that $ K$ is the incenter of $ ACD$
$ QED$


Find all pairs of $i,j$ that for every $n$ there is a number like $m$ that just consist $i,j$ and $n|m$. ;)

I think it is'nt hard :huh:
but if you want hint say

What does $m$ contains $i,j$ mean?  $i| m$ and $j| m$?

Also can anyone tell me how to do the "divides" sign in Latex?

[quote="eryaman"]What does $m$ contains $i,j$ mean?  $i| m$ and $j| m$?[/quote]
No, I mean that  every digit is of $m$ is $i$ or $j$.

[quote="eryaman"]Also can anyone tell me how to do the "divides" sign in Latex?[/quote]
It is  $\frac{abc}{xyz} \Rightarrow$ \frac{abc}{xyz}

i think he meant to say "divisible", not "divides". I am sure he knows how to use fractions.

code for the Divisibility sign $a|b$ is

[code]a|b[/code]

you can find that bar by shifting on the key to the left of backspace.[/code]

Ok Let's solve the problem,I give you 3 day, if anyone can solve ;)

[quote="amirhtlusa"]i'm not understanding the problem,
cant just $i,j$ be the prime divisors of $n$? ;)[/quote]
OK.I will explain it again:
Find all pair of $(i,j)$ that for every $n$ there is $m$ such that $n|m$ and $m$ consist only $i,j$ (I mean that every digit of $m$ is $i$ or $j$).
i.e if $i=1$ & $j=0$ it is obviously correct. ;)

Ok
I'm going to send solution:
first I prove that for $(0,1),(0,2),...,(0,9)$ it is correct .
Consider $1111...1$($n$ time)
              $1111...1$($n-1$ time)
              .           .
              .         .
              .       .
              $111$
              $11$
              $1$
there are n number so two of them are equivalent  mod p. so there is number like $ m =1111...1000...0$ that $n|m$. and we can prove for $(0,2),...,(0,9)$ in this way.
now I want to prove that there is no other pair. if we consider 10 if $10|m$ then m should have $0$ at first digit. ;)

[quote="eryaman"]Also can anyone tell me how to do the "divides" sign in Latex?[/quote]This thread's problem doesn't make much sense to me but I can answer that question.  :D

a \mid b $\rightarrow{a}\mid{b}$

and

a \nmid b $\rightarrow{a}\nmid{b}$.


geometry
circumcircle
inradius
trigonometry
geometry unsolved

(B.Frenkin) The product of two sides in a triangle is equal to $ 8Rr$, where $ R$ and $ r$ are the circumradius and the inradius of the triangle. Prove that the angle between these sides is less than $ 60^{\circ}$.

Unexpectedly easy,isn't it?
Suppose that these sides are $ AB,AC$ and for convenience denote them $ c,b$ respectively.Let $ \alpha$ be the angle between these two sides.Our goal is to prove that $ \alpha < 60^{\circ}$.Assume the contrary,i.e $ \alpha\geq 60^{\circ}$,therefore,$ \cos{\alpha}\leq\frac {1}{2}$
By the well-known formula of area of $ \triangle ABC$,it follows that:
\[ \frac {abc}{4R} \equal{} a\cdot 2r \equal{} pr \equal{} \frac {a \plus{} b \plus{} c}{2}\cdot r
\]
It means that $ b \plus{} c \equal{} 3a$.Squaring the last relation gives us:
$ b^2 \plus{} 2bc \plus{} c^2 \equal{} 9a^2 \equal{} 9(b^2 \plus{} c^2 \minus{} 2bc\cos{\alpha})$,or

$ 8(b^2 \plus{} c^2) \equal{} bc(2 \plus{} 18\cos{\alpha})\leq 11bc$,what is clearly impossible,as it contradicts to am-gm inequality...
P.S:
It seems to me that I missed something,because if my solution would be correct we can even prove that $ \alpha\leq 45^{\circ}$
But I can't figure out where I did the mistake :maybe:

everytihng is ok

This is the first problem of second day

also we have $ \alpha \leq \arcsin(\frac {3\sqrt3}{8})$

in fact from the fact that $ b\plus{}c\equal{}3a$ it follows that $ a\equal{}s\minus{}a<\min\{b,c\}$, which implies that $ \angle A<60^{\circ}$..

Here is my solution:

$ bc\equal{}8Rr\Rightarrow \frac{b}{2R}\frac{c}{2R}\equal{}2\frac{r}{R}$

$ \Rightarrow \sin B\sin C\equal{}8\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$

$ \Rightarrow \cos \frac{B}{2}\cos \frac{C}{2}\equal{}2\sin\frac{A}{2}$

$ \Rightarrow\frac{1}{2}\ge \sin\frac{A}{2}$

$ \Rightarrow 60\ge A$


calculus
videos
vector
calculus computations

Hello,

I would like to know if any of you knows where can i find calculus books with problems and their solutions, any dvds or videos that can tutor how to use integerals.
any idea will be appreciate it, thanks

I recomend Schaum's Outline of Advanced Calculus by Murry Spiegel. Also, you can find a bunch of free online math text books at [url=http://mathbooks.library.cornell.edu:8085/Dienst/UIMATH/1.0/GenHome]the Cornell Univeristy Library[/url].

Are you looking for a full text?  Marsden and Tromba offer a fairly complete treatment and you can the solutions manual separately.

ok I will look into these you guys gave me, is there any ISBN number or book names?

J. Marsden and A. Tromba, Vector Calculus, W. H. Freeman and Company, New York, 2003 (Fifth Edition).

Vector Calculus, 3rd Edition
Susan Jane Colley, Prentice Hall ISBN-10: 0131858742 ISBN-13: 9780131858749


[color=White]-----,[/color]VW
[color=White]-...[COLOR=Black]*[/color]-,[/COLOR]GT
_________
[color=White]-----[/color]IW
[color=White]---[/color]VW
_________
[color=White]---[/color]TVW

Uh.. that looks really messed up.  :lol:

Haha yeah, especially if you highlight :)

weird.... and what does $\text{cryptarithmic}$ mean?

[quote="cornoth"]
[color=White]-----,[/color]VW
[color=White]-...[/color][color=Black]*[/color][color=white]-,[/color]GT
_________
[color=White]-----[/color]IW
[color=White]---[/color]VW
_________
[color=White]---[/color]TVW[/quote]

Caution: Answer below-
[hide][color=White]----,,[/color]25
[color=White]...[/color][color=Black]*[/color][color=white]-,[/color]13
_________
[color=White]-----[/color]75
[color=White]---,[/color]25
_________
[color=White]---,[/color]325

I'll explain if nobody else wants to[/hide]

Oh, now I see. :P I didn't get the original problem though...

I see what cornoth did, he put white dashed lines so he could put spaces to make the problem look nice. I thought it was subtraction, but the solution format was multiplication. djshowdown2, thanks for clarifying.


function
calculus
integration
real analysis
real analysis unsolved

Consider the function:
$ f(x) \equal{} \begin{cases} 1 & \text{if } x \in \textbf{Q} \\
0 & \text{if } x \in \textbf{R \minus{} Q} \end{cases}$

I know this function is a classic function in Intro to Analysis classes.

Consider:
  $ \displaystyle\int^{10}_0 f(x)\,dx \equal{} a$?
 Does $ a \equal{} 10$ ?

Is this function even integrable (I'm not exactly sure about all the finer points about integration since I wasn't able to fit in the second semester Intro to Analysis class in my schedule)?

What if the function was:
$ f(x) \equal{} \begin{cases} 2 & \text{if } x \in \textbf{Q} \\
1 & \text{if } x \in \textbf{R \minus{} Q} \end{cases}$What about$ \displaystyle\int^{10}_0 f(x)\,dx \equal{} b$

Does $ b \equal{} 30?$

 That's seems weird to me.

Any help would be most appreciated![/b]

actually $ a\equal{}0$ and $ b\equal{}10$ :roll:

What integral are you currently studying? If it is the Riemann integral (which I assume from the context of your post), then both functions are simply not integrable.

Take the first case, one on the rationals and zero on the irrationals. Take any partition of $ [0,10]$ whatsoever. Then every interval of that partition contains both rational and irrational points. We find that for any partition $ P,$ the upper sum is $ U(f,P)\equal{}10$ and the lower sum is $ L(f,P)\equal{}0.$ The infinum of the upper sums does not equal the supremum of the lower sums, so the integral does not exist.

-----

Albanian Eagle answered the question as if you were studying the Lebesgue integral. In a Lebesgue setting, your original function is equal to zero "almost everywhere" and your second function is equal to $ 1$ almost everywhere. That is, for the purposes of integration, we cannot tell any practical difference between your first function and the zero function.


Let $ a,b,c$ be possitive real numbers. Prove that: $ \large\sum_{cyc}^{a,b,c}\sqrt [3]{\frac{a^{2}\plus{}bc}{b^{2}\plus{}c^{2}}}\geq 3$

[quote="bibobeo"]Well, it's supposed to be $ \geq 3$[/quote]
Try $ a\equal{}b\equal{}1$ and $ c\rightarrow0^{\plus{}}.$ :wink:

My proof for the given Ineq:

Applying Holder's Ineq to the LHS, we have:
$ \large\sum_{cyc}^{a,b,c}\frac{\sqrt [3]{(a^{2}\plus{}bc)(b^{2}\plus{}c^{2})(c^{2}\plus{}b^{2})}}{b^{2}\plus{}c^{2}}\geq\sum_{cyc}^{a,b,c}\frac{(abc)^{2/3}\plus{}bc}{b^{2}\plus{}c^{2}}$
So all we need to do is prove: $ \large\sum_{cyc}^{a,b,c}\frac{(abc)^{\frac{2}{3}}\plus{}bc}{b^{2}\plus{}c^{2}}\geq\frac{3}{2}$.
WLOG, assume that $ abc \equal{} 1$, it is equalavent to: $ \large\sum_{cyc}^{a,b,c}\frac{1\plus{}bc}{b^{2}\plus{}c^{2}}\geq\frac{3}{2}$, which can be obtained from Iran1996!

Are there any other proof for it??

[quote="Mashimaru"]
So all we need to do is prove: $ \large\sum_{cyc}^{a,b,c}\frac{(abc)^{\frac{2}{3}}\plus{}bc}{b^{2}\plus{}c^{2}}\geq\frac{3}{2}$.
[/quote]
But it's wrong for $ a\equal{}b\equal{}1$ and $ c\rightarrow0^{\plus{}}.$ :(

[quote="Mashimaru"]Let $ a,b,c$ be possitive real numbers. Prove that: $ \large\sum_{cyc}^{a,b,c}\sqrt [3]{\frac{a^{2}\plus{}bc}{b^{2}\plus{}c^{2}}}\geq\frac{3}{2}$[/quote]
Where did you find that problem?Are you sure that this inequality is true;
I know one inequality quite similar with yours,maybe it can help :) :
$ \large\sum_{cyc}^{a,b,c}\sqrt [3]{\frac{a^{2}\plus{}bc}{b^{2}\plus{}c^{2}}}\geq\frac{3(a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}ab\plus{}bc\plus{}ca)\sqrt [3]{abc}}{\sum a(b^{2}\plus{}c^{2})}$ :wink:


evaluate the function

.                                              ........
                                         .......
                              ^sqrt(2)
                  ^sqrt(2)
        ^sqrt(2)
sqrt(2)

I'm not sure how else to do it other than to compute some values using $ \sqrt{2}\approx 1.414$
It seems to converge to $ 2$.

I did some research and there doesn't seem to be a way to solve this without using the Lambert W-function.

$ \sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\hdots}}}=x$

$ \sqrt{2}^x=x$
$ 2^\frac{x}{2}=x$
$ x=2$.

Note that this does not imply that $ 2^\sqrt{2}=2 \implies \sqrt{2}=1 \implies 1=2$, since then there would be a "last $ \sqrt{2}$" at the end of the "power chain", which there isn't since this is infinite.

Is there a good way to solve something of the form $ n^x\equal{}x$ for $ x$?

[quote="grn_trtle"]Is there a good way to solve something of the form $ n^x \equal{} x$ for $ x$?[/quote]

Well, not really. You kinda have to get lucky. Which problem writers will definately make so on a contest.

$ x\equal{}\log_3x$ I guess.

Yeah, that's what I thought.

Though if you introduce the Lambert W function, you have
$ x\equal{}\minus{}\frac{W(\minus{}\ln n)}{\ln n}$

But that's not appropriate for high school competitions, I think.


Given equilateral triangle $ABC$ , $M,N$ are two arbitrary point on plane, $A_{m},A_{n}$ is projection of $M,N$ on line $BC$ similarly $B_{m},B_{n},C_{m},C_{n}$ prove that $\sqrt{A_{m}A_{n}},\sqrt{B_{m}B_{n}},\sqrt{C_{m}C_{n}}$ are three side of a triangle, and calculate its area follow $MN$ and $ABC$ if you can!

Are you saying the area of the triangle is a function of MN?  It doesn't seem to be.

Maybe... but we can see the area of that triangle must depend on $MN$ surely? :maybe:

[quote="PDatK40SP"]It's depend on $\overrightarrow{MN}$, I think so. We have if $\overrightarrow{M_{1}N_{1}}=k\overrightarrow{M_{2}N_{2}}$, then $A_{m_{1}}A_{n_{1}}=kA_{m_{2}}A_{n_{2}}$ ... So we can suppose that $M \equiv A$ and $N \in BC$ ... I can't post the solution because now I'm busy, sorry ... But I think now it's easier.[/quote]
what's problem you said easy, prove three sides or calculate area ? Follow me calculating area is so hard than you thought!


W denotes a complex fifith root of 1.
Find the value of

(W^2 -1).(W^2 -W).(W^2 -W^3).(W^2 -W^4)

$f(x) = \prod_{i=1}^4 (x - \omega^i) = x^4 + x^3 + x^2 + x + 1$ $\Rightarrow \frac{df}{dx} = \prod_{i=1}^4 (x - \omega^i) \cdot \left( \sum_{i=1}^4 \frac{1}{x-\omega^i} \right) = 4x^3 + 3x^2 + 2x + 1$. Plugging in $\omega^2$ yields $(\omega^2 - 1)(\omega^2 - \omega)(\omega^2 - \omega^3)(\omega^2 - \omega^4) = 4\omega + 3\omega^4 + 2\omega^2 + 1$, I think.

[quote="mli"]Have you seen how to prove the value for cos(36)?[/quote]
Letting $x=36^\circ$, we have: $\cos 2x=\cos 3x$
And then use double angle, triple angle, etc., right?

[quote="mli"]Have you seen how to prove the value for cos(36)?  That should pretty much give the answer.[/quote]
I've seen the proof for 18 so I guess you could just use the double angle formula to get 36.


LaTeX ratio search trigonometry function calculus Gmaas

How well do you know LaTeX? Do you use it on the forum?

Hmm valentyn I have my doubts about polling this: people can't change their vote. People who come here without latex knowlegde (most of them) will say no but will learn it here. But then if they know more about it, there vote here stays fixed to no.

Just my 50 cents...

i barely kno how to use it but, Mathew told me!

I know what it is, but I don't know how to use it. (I'm planning to learn very soon, though.)

Know what it is.. refuse to .. well too lazy to learn right now...

How can you use Latex on the forum? I mean, when I upload a file to the forum (jeje, it sounds like if I had done that thousands of times) usually is a pdf genereated with latex, but inside the forum itself, how and where can you use it?

Best,

David, you can use $\LaTeX$ anywhere on the forum. Most commands are allowed. Just a few ones are forbidden due to security issues. Just browse the forum. If you see longer formulae, especially in the [url=http://www.mathlinks.ro/Forum/index.php?f=217]Olympiad[/url] and in the [url=http://www.mathlinks.ro/Forum/index.php?f=218]College section[/url] you see lots of $\LaTeX$, you can click on it and you will see the $\LaTeX$ source code displayed.  :)

I like it too, but for some reason I didn't know that you would use it in the forum, now that I know, this is great. Does anyone knows if you can use pictex in here??

Best,

[quote="djimenez"]Does anyone knows if you can use pictex in here??[/quote]You can try out stuff in the Test Forum. :)

Hello,

Just in case, I introduced this figure in the test forum (hidden, and without the double dollar sign)

[hide]
\begin{array}{c}
\beginpicture
    \setplotarea x from 0 to 200, y from 0 to 100

    \circulararc 360 degrees from  50 0 center at  50 50
    \ellipticalarc axes ratio 3:1 180 degrees from 0 50  center at 50 50
    \setdots <3pt>
    \ellipticalarc axes ratio 3:1 180 degrees from 100 50  center at 50 50
    \setsolid

    \arrow <5pt> [.5,1] from    110 50 to  130 50
    \plot 140 30 140 70 /

    \circulararc 360 degrees from  140 0 center at  140 15
    \ellipticalarc axes ratio 3:1 180 degrees from 125 15  center at 140 15
    \setdots <3pt>
    \ellipticalarc axes ratio 3:1 180 degrees from 155 15  center at 140 15
    \setsolid

    \circulararc 360 degrees from  140 70 center at  140 85
    \ellipticalarc axes ratio 3:1 180 degrees from 125 85  center at 140 85
    \setdots <3pt>
    \ellipticalarc axes ratio 3:1 180 degrees from 155 85  center at 140 85
    \setsolid

    \arrow <5pt> [.5,1] from    155  7.5 to 170 12
    \arrow <5pt> [.5,1] from    155 22.5 to 170 18
    \arrow <5pt> [.5,1] from    155 77.5 to 170 82
    \arrow <5pt> [.5,1] from    155 92.5 to 170 88
    \arrow <5pt> [.5,1] from    145   50 to 160 50

    \put {$e_1^2$}    at  180  15
    \put {$f(e_1^2)$} at  190  85
    \put {$S^1$}      at  170  50

\endpicture\\
\mbox{ Figure \#2}\\
\end{array}
[/hide]

And it says

[Unparseable or potentially dangerous latex formula. Error 5 : 1032x375]

AKA, PiCTeX does not work in the forum.

Best regards,

i love $\text{\LaTeX}$
\begin{eqnarray}
\sum_{k=1}^{n} & = & 1^2+2^2+\cdots+n^2\nonumber\\
               & = & \frac{n(n+1)(2n+1)}{6}
\end{eqnarray}
i have no idea what that means

I like $\LaTeX$ too, it's a really beautiful tool, it makes work seem so, "mathematically" presentable - provided you have at least basic knowledge.  As you get better at it, your work goes faster.  Also then your work looks professional, which needless to say, gives your lecturers a sense that you take your work seriously.  Mind you, I just use it because I detest typing mathematics in Word.  Took me 40 minutes to do 1 summation in Word.  Then I used $\LaTeX$, and took me 30 seconds, after that we got married and had kids. (just kidding  :rotfl: )

-Ixeus

Latex makes everything easier to read.

From 2pi/3(4/5+4x) to $\frac{2\pi}{3(\frac{4}{5} + 4x)}$

latex does look much nicer, but i can only type short expressions, not long equations :blush:

What makes latex hard for the normal user, it's that it has no graphical interface as Words and other text formatting software, and latex need to read a lot of tutorial and documentations...

Latex is cool but I failed to learn it, I'm still learning.

I got my latex bu learning what other people said and following the helps

LATEX is cool when other people type it. I can't type LATEX. No more room in my memory bank.

$I'm just learning Latex(I normaly copy and past my answer!!)$

I like $\LaTeX$, so I quickly learned it and I still like it. I think I know it well enough to do problems, etc.

[quote=Nerd_of_the_Ages]$ I b_{a}r^{e}l_{y}k_{n}o^{w}a^{n}y L^{A}T_{E}X, b_{u}t I'^{m}t_{o}o l^{a}z_{y}t^{o}l_{e}a^{r}n m_{o}r^{e}.$[/quote]

ur gonna give me a heart attack

[quote=jackdillon]$I'm just learning Latex(I normally copy and past my answer!!)$[/quote]

you can use the \text{} command for $\LaTeX$ to render your spaces.

$$I'm just learning Latex(I normally copy and past my answer!!)$$
$$\text{vs.}$$
$$\text{I'm just learning Latex(I normally copy and past my answer!!)}$$


Euler geometry circumcircle trigonometry

[i]I do not know the solution so excuse me if it is too easy[/i] :blush:

Let $ ABC$ be a triangle whose euler line is parallel to line BC. Prove that $ tan{B}\plus{}tan{C}\equal{}2tan{A}$

[u]Solution.[/u] Let $ A',F,O$ be the midpoint of $ BC$, foot of $ A$'s altitude, and the circumcenter, respectively.  Since the euler line is parallel to $ BC$, $ OA'\equal{}\frac{AF}{3}$.  Since $ AF\equal{}b\sin C\equal{}2R\sin B\sin C$ (sine law) and $ OA'\equal{}R\cos A\equal{}R(\sin B\sin C\minus{}\cos B\cos C)$,
\[ 3\equal{}\frac{AF}{OA'}\equal{}\tan B\tan C.\]
Using the identity (easy to prove) $ \sum\tan A\equal{}\prod\tan A$, we have $ 2\tan A\equal{}\tan B\plus{}\tan C$. $ \Box$


integration
inequalities
calculus
function
real analysis
real analysis unsolved

[quote="Xevarion"]The RHS is just simple integration (as I'm sure you already realized).

The LHS is an application of Fubini's Theorem. The easiest way to see what they did is to graph the region over which you integrate in the first inequality. If you want to change the order of integration, what are the new bounds? Once that's done, you can integrate in $x$ with impunity because $f$ is a function of $t$. And so we have their second inequality.[/quote]

What is Fubini's Theorem? :blush:  :?

[quote="mikeynot"]In their solution, they have:

$\int^1_0 \int^1_x f(t) \,dt \,dx\geq \int^1_0 (1-x^2)/2\,dx$ or

$\int^1_0 t*f(t)\,dt \geq 1/3$

I don't understand how they got the second inequality from the first inequality. Can someone explain, especially the LHS?
Thanks.[/quote]

$\$\int^1_0 \int^1_x f(t) \,dt \,dx=$\int^1_0 x' \int^1_x f(t) \,dt \,dx=x \int^1_x f(t) \,dt \,dx\mid^1_0+$\int^1_0 (-\int^1_x f(t) \,dt)' \,dx=int^1_0 t*f(t)\,dt$

[quote="mikeynot"]In their solution, they have:

$\int^1_0 \int^1_x f(t) \,dt \,dx\geq \int^1_0 (1-x^2)/2\,dx$ or

$\int^1_0 t*f(t)\,dt \geq 1/3$

I don't understand how they got the second inequality from the first inequality. Can someone explain, especially the LHS?
Thanks.[/quote]

$\int^1_0 \int^1_x f(t) \,dt \,dx=\int^1_0 x' \int^1_x f(t) \,dt \,dx=x \int^1_x f(t) \,dt \,dx\mid^1_0+\int^1_0 (-\int^1_x f(t) \,dt)' \,dx$=$\int^1_0 t \cdot f(t)\,dt$, $x \int^1_x f(t) \,dt \,dx\mid^1_0=0$ ;)


What are the complex numbers?
Simple but I want to know it.  :lol:

What is the Harmonic mean?
Give me a brief introduction to it please.

Thanks in advance  :)

The complex numbers are numbers of the form $ a \plus{} bi$, where $ a$ and $ b$ are real numbers and $ i$ is the imaginary unit, $ \sqrt{ \minus{} 1}$.

The harmonic mean of two numbers $ a$ and $ b$ is defined as $ \frac {1}{\frac {1}{2}(\frac {1}{a} \plus{} \frac {1}{b})}$.  (In words, "the reciprocal of the average of the reciprocals")

This can be generalized to $ \frac {1}{\frac {1}{n}(\frac {1}{a_1} \plus{} \frac {1}{a_2} \plus{} \cdots \plus{} \frac {1}{a_n})}$.

The harmonic mean can be used in problems dealing with average rate.  Here as an example:

If I drive from my home city to a city $ 200$ miles away at a rate of $ 60$ miles per hour, and then drive back to my home city at a rate of $ 40$ miles per hour, what is my average speed?

[hide="hint"]
It's not $ 50$.
[/hide]

[hide="solution"]
On the drive down, I am going $ 60$ miles per hour, so it takes me $ \frac {200}{60}$ hours to drive down.

On the drive back, I am going $ 40$ miles per hour, so it takes me $ \frac {200}{40}$ hours to drive back.

Adding the two fractions, we find that it takes me $ \frac {25}{3}$ hours to drive to and from the city.  The trip was a total of $ 400$ miles, so $ \frac {400}{\frac {25}{3}} \equal{} \boxed{48}$ miles per hour.

Notice that $ \frac {1}{\frac {1}{2}(\frac {1}{60} \plus{} \frac {1}{40})} \equal{} \boxed{48}$.
[/hide]

Harmonic mean of the numbers $ a_1,a_2,a_3,...,a_n$ is equal to $ \frac {n}{\frac {1}{a_1} \plus{} \frac {1}{a_2} \plus{} \frac {1}{a_3} \plus{} ... \plus{} \frac {1}{a_n}}$ which is the same as Discrete math wrote.

Thank you very much guys.
Thanks for the example "Discrete_Math".   :)


[quote="dima ukraine"]Prove for all $ a,b,c > \frac {1}{\sqrt {6}}$ satisfying $ a^2 + b^2 + c^2 = 1$
${ \sum_{cyc}\frac {1 + a^2}{\sqrt {2a^2 + 3ab - c^2}}}\ge2(a + b + c)$[/quote]
See here: http://www.mathlinks.ro/viewtopic.php?t=155354 :)


geometry trigonometry perimeter AMC USA(J)MO USAMO circumcircle

Sorry my presentation on Finite Geometry last weekend was not very coherent. Hopefully these sites will explain it more clearly: http://en.wikipedia.org/wiki/Finite_geometry and http://home.wlu.edu/~mcraea/Finite_Geometry/TableOfContents/table.html It really is a cool subject. It is at the intersection (no pun intended) of foundational geometry and combinatorics. There are also some interesting open and investigative type problems in Finite Geometry that could make good research projects.

Also, let's use this thread to post some proofs of the important theorems of geometry from your packet. Let's start with proofs for the formulas for areas of triangles. (The more creative the proof the better!)

$ A \equal{} \frac {1}{2}bh$
$ A \equal{} \frac {1}{2}ab\sin \theta$
$ A \equal{} rs$, where r is the radius of the inscribed circle and s is the semi-perimeter (i.e. half of the perimeter)
$ A \equal{} \sqrt {s(s \minus{} a)(s \minus{} b)(s \minus{} c)}$ (this is called Heron's formula)
$ A \equal{} \frac {abc}{4R}$, where R is the radius of the circumscribed circle

The above formulas show up so often that you should be familiar with all of them. The following formula is not as essential on regional and AMC level contests, but if you are preparing for the USAMO you should be able to prove it and be familiar with it.

$ A \equal{} (s \minus{} a)r_a$, where $ r_a$ is the radius of the excircle opposite vertex A.

How do you actually prove the 3rd-5th formulas? I notice you can prove the 1st two with trig...

[hide="Third Formula"]
Assuming an inscribed circle exists, draw lines from the center of the inscribed circle and the vertices of the polygon.  This splits the polygon into however many triangles, each with a drawn height as a radius of the inscribed circle.  Taking the first formula as proven, we can now add the triangle areas together, as is it were one triangle with the height as $ r$ and a base equal to the perimeter of the original polygon, $ p$.

Thus, the area is $ \frac{1}{2}pr$=$ rs$.[/hide]

Hello
[hide="fifth"]
consider a circle with center O and radius R circumcircle to triangle $ ABC$.Join the center O to B and C.Now join the mid points with midpoints of BC and CA.
U can easily prove $ BD \equal{} BOsinBOD$
                          $ \frac {a}{2} \equal{} RsinA$
       we know $ sinA \equal{} \frac {2}{bc}\sqrt {s(s \minus{} a)(s \minus{} b)(s \minus{} c)} \equal{} \frac {2Area}{bc}$
  using both   $ R \equal{} \frac {abc}{4Area}$

[/hide]
[hide="sixth"]
Well consider a triangle ABC with an excircle opposite vertex A touching BC at D,extended AB at F and extended AC at E, with center O and radius $ r_a$.
Area of triangle(A)=area of OCA +area of OAB -area of OBC
                        $ A \equal{} r_a[\frac{b \plus{} c \minus{} a}{2}] \equal{} r_a(s \minus{} a)$[/hide]
Thank u.

For the fourth, just look google "Heron's Formula."  Wikipedia has a proof using the law of cosines and the second area formula posted here.


Starting at O, red posts are planted every 1 metre apart, blue posts are planted every $ \phi$, the golden ratio, metres apart and yellow posts are planted $ \phi ^{2}$ metres apart. When is the first time that a blue and a yellow post are between two red posts, that is how far from O?

[quote="AndrewTom"]Starting at O, red posts are planted every 1 metre apart, blue posts are planted every $ \phi$, the golden ratio, metres apart and yellow posts are planted $ \phi ^{2}$ metres apart [in one straight line]. When is the first time that a blue and a yellow post are between two red posts, that is how far from O?[/quote]

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Am I right in suspecting that this never happens: that if $ n\phi$ is between $ a$ and $ a\plus{}1$, where $ a$ is a positive integer, then $ m\phi^{2} \equal{} m(\phi \plus{}1)$ can't be?

Thanks for the reply confirming my suspicions; not yet:

I've thought of assuming that $ a < n\phi < a \plus{} 1$ and $ a < m\phi^{2} < a \plus{} 1$ and using $ \phi^{2} \equal{} \phi \plus{} 1$,

$ \frac {a}{n} < \phi < \frac {a \plus{} 1}{n}$,

$ a < m\phi \plus{} m < a \plus{} 1$

$ a \minus{} m < m\phi < a \plus{} 1 \minus{} m$

$ \frac {a}{m} \minus{} 1 < \phi < \frac {a \plus{} 1}{m} \minus{} 1$, and using this to get a contradiction.

Or, perhaps, considering multiples of $ \frac {1 \plus{} \sqrt {5}}{2}$ and of $ \frac {3 \plus{} \sqrt {5}}{2}$.

I've had a number of ideas but haven't got very far yet.

Thanks MellowMellon, I divided by the wrong numbers : I should have divided by $ \phi$ and $ \phi^{2}$ and I didn't realise that $ \phi^{2} \equal{} \phi \plus{} 1$ means $ 1\equal{} \frac {1}{\phi} \plus{} \frac {1}{\phi^{2}}$. I do now!  Thanks again.


geometry
rectangle
perimeter
complex numbers
combinatorics proposed
combinatorics

I found many topics here about tiling a rectangle, but couldn't find this one:

Prove that rectangle $a\cdot b$ could be tiled with rectangles $1\cdot d$ if and only if $d|a$ or $d|b$.

[quote="diophantus"]Simple.  If d|a and d|b is not true, then there will be a portion of a 1*d retangle passing beyond the perimeter of the a*b retangle.  Which in effect means it does not tile.[/quote]

I have to say, that's basically repeating the question.

Bojan Basic: You probably meant that $a, b$ are integers. Then the problem is a consequence of [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=18613]this[/url].

Sorry, I was picturing the rectangles in evenly aligned rows rather than turned and shifted.

A long time ago, when I tried to solve this, I did the following:

Start from the upper left corner, and put a $e^{\frac {2\pi i}d}$ (actually, we can put $1$ or anything else which is non-zero) in it. Then, whenever you go to the left or down, put the number you were standing on multiplid by $e^{\frac {2\pi i}d}$ on the new square. If the rectangle can be tiled with $1\times d$ tiles, then the sum of the complex numbers in it is $0$. However, if none of $a,b$ are divisible by $d$, then the sum is non-zero (I think :)).

We get the same effect if we color squares into $d$ colors by diagonals. Then if $d\nmid a$ and $d\nmid b$ then numbers of squares of distinct colors will not be the same.


inequalities
trigonometry
inequalities unsolved

Let $ ABC$ is triangle with $ \angle A<90^0, \angle B<90^0, \angle C<90^0$. Prove that $ \boxed{tan^nA\plus{}tan^nB\plus{}tan^nC \ge 3\sqrt{3^n}, n \in N}$

There are some similars problems here. use Jensen and it's donne (Actually $ n$ can be a Real number greater than 1)

$ \tan ^n(a) \plus{} \tan ^n(b) \plus{} \tan ^n(c)\geq 3 \tan ^n\left(\frac {1}{3} (a \plus{} b \plus{} c)\right) \equal{} 3 \tan ^n\left(\frac {\pi }{3}\right) \equal{} 3^{\frac {n}{2} \plus{} 1}$

I think we use two following lemma:
1/  $ tanA\plus{}tanB\plus{}tanC \geq 3\sqrt{3}$  ( Note that we have $ tanA \plus{} tanB\plus{} tanC\equal{} tanA.tanB.tanC$)

2/  For a,b,c be positive real numbers, we have $ \frac{ a^n\plus{}b^n\plus{}c^n}{3} \geq \frac{(a\plus{}b\plus{}c)^n}{3^n}$ ( n be natural number)

[quote="ElChapin"]Is $ tan^n(x)$ convex for any real greater than one?

 :)  :?:[/quote]

$ \frac {\partial ^2\tan ^n(x)}{\partial x^2} \equal{} n \left((n \minus{} 1) \csc ^2(x) \plus{} 2\right) \sec ^2(x) \tan ^n(x)$

$ \frac {n \minus{} 1}{\sin ^2(x)} \plus{} 2 > 0$

Especial solution if $ n\geq 1$


inequalities
function
algebra unsolved
algebra

Find $ p \in R$ that $ \exists \ x \in R$ satisfy:
$ \sqrt{2p\plus{}1\minus{}x^2} \plus{}\sqrt{3x\plus{}p\plus{}4} \equal{} \sqrt{x^2\plus{}9x\plus{}3p\plus{}9}$

[quote="duytungct"]Find $ p \in R$ that $ \exists \ x \in R$ satisfy:
$ \sqrt {2p \plus{} 1 \minus{} x^2} \plus{} \sqrt {3x \plus{} p \plus{} 4} \equal{} \sqrt {x^2 \plus{} 9x \plus{} 3p \plus{} 9}$[/quote]

$ p\equal{}x\equal{}0$

No no! I said it is strange! And it has many roots
At least, let's try $ p\equal{}x^2\plus{}x$ with $ x\geq \minus{}1$  :mad:

[quote="duytungct"]Find $ p \in R$ that $ \exists \ x \in R$ satisfy:
$ \sqrt {2p \plus{} 1 \minus{} x^2} \plus{} \sqrt {3x \plus{} p \plus{} 4} \equal{} \sqrt {x^2 \plus{} 9x \plus{} 3p \plus{} 9}$[/quote]

I'm sorry but you said "Find $ p \in R$ [such] that $ \exists \ x \in R$ satisfy ..." .

I found one p (p=0) such that $ \exists \ x$ ($ x\equal{}0$) satisfy ...

So I answered your question (except if I misunderstood).

Now may be you want something else (all p for example) but you should say it more clearly.

[quote="pco"][quote="duytungct"]Find $ p \in R$ that $ \exists \ x \in R$ satisfy:
$ \sqrt {2p \plus{} 1 \minus{} x^2} \plus{} \sqrt {3x \plus{} p \plus{} 4} \equal{} \sqrt {x^2 \plus{} 9x \plus{} 3p \plus{} 9}$[/quote]

I'm sorry but you said "Find $ p \in R$ [such] that $ \exists \ x \in R$ satisfy ..." .

I found one p (p=0) such that $ \exists \ x$ ($ x \equal{} 0$) satisfy ...

So I answered your question (except if I misunderstood).

Now may be you want something else (all p for example) but you should say it more clearly.[/quote]

Oh! I see, sorry pco.
It means I must change something, right?  :D
[quote]Find [b]all [/b] $ p \in R$ that $ \exists \ x \in R$ satisfy:
$ \sqrt {2p \plus{} 1 \minus{} x^2} \plus{} \sqrt {3x \plus{} p \plus{} 4} \equal{} \sqrt {x^2 \plus{} 9x \plus{} 3p \plus{} 9}$[/quote]

[quote="duytungct"]Find [b]all [/b]$ p \in R$ that $ \exists \ x \in R$ satisfy:
$ \sqrt {2p \plus{} 1 \minus{} x^2} \plus{} \sqrt {3x \plus{} p \plus{} 4} \equal{} \sqrt {x^2 \plus{} 9x \plus{} 3p \plus{} 9}$[/quote]

I don't have the full answer to this question but at least I can show that :

$ 1)$ No such $ x$ exists for $ p<\minus{}\frac{1}{2}$
Obvious since then $ 2p \plus{} 1 \minus{} x^2<0$ and then the first square root is meaningless.

$ 2)$ Such $ x$ exists for any $ p\geq 0$ :
Let $ f(x)\equal{}\sqrt {2p \plus{} 1 \minus{} x^2} \plus{} \sqrt {3x \plus{} p \plus{} 4}$ and $ g(x)\equal{}\sqrt {x^2 \plus{} 9x \plus{} 3p \plus{} 9}$

$ 2.1)$ $ \forall x\in [0,\sqrt{2p\plus{}1}]$, all square roots are defined :
We have $ x^2\geq 2p\plus{}1$ and so $ 2p \plus{} 1 \minus{} x^2\geq 0$ and so the first square root is defined.
We have $ 3x \plus{} p \plus{} 4\geq 4>0$ and so the second square root is defined.
We have $ x^2 \plus{} 9x \plus{} 3p \plus{} 9\geq 9>0$ and so the thitd square root is defined.

$ 2.2)$ $ f(\sqrt{2p\plus{}1}) \leq g(\sqrt{2p\plus{}1})$
Since $ x\equal{}\sqrt{2p\plus{}1}$ is equivalent to $ x^2\equal{}2p\plus{}1$ and $ x\geq 0$, the requisted inequality is equivalent to :

$ \sqrt {0} \plus{} \sqrt {3x \plus{} p \plus{} 4}\leq \sqrt {9x \plus{} 5p \plus{} 10}$  and $ x\geq 0$

$ 3x \plus{} p \plus{} 4\leq 9x \plus{} 5p \plus{} 10$  and $ x\geq 0$

$ 0\leq 3x \plus{} 2p \plus{} 3$  and $ x\geq 0$

Which is true since $ x\geq 0$ and $ p\geq 0$

$ 2.2)$ $ f(0) \geq g(0)$. This is equivalent to :

$ \sqrt {2p \plus{} 1} \plus{} \sqrt {p \plus{} 4}\geq \sqrt {3p \plus{} 9}$

$ 3p\plus{}5\plus{}2\sqrt {(2p \plus{} 1)(p\plus{}4)} \geq 3p \plus{} 9$

$ \sqrt {2p^2\plus{}9p\plus{}4} \geq 2$

$ 2p^2\plus{}9p\plus{}4 \geq 4$

$ 2p^2\plus{}9p\geq 0$, which is always true since $ p\geq 0$

$ 2.3)$ Requested $ x$ exists for any $ p\geq 0$ :
$ f(x)$ and $ g(x)$ are continuous functions defined on $ [0,\sqrt{2p\plus{}1}]$
$ f(0) \geq g(0)$ and $ f(\sqrt{2p\plus{}1})\leq g(\sqrt{2p\plus{}1})$

And so $ \exists x\in[0,\sqrt{2p\plus{}1}]$ such that $ f(x)\equal{}g(x)$

$ 3)$ It remains the case $ p\in[\minus{}\frac{1}{2},0)$
To be continued ... .

[quote] $ x^2\geq 2p \plus{} 1$ [/quote] $ \to$ [quote] $ x^2\leq 2p \plus{} 1$ [/quote]  :)

I only found that
1) $ x^2\plus{}9x\plus{}3p\plus{}9 \geq (2p\plus{}1\minus{}x^2)\plus{}(3x\plus{}p\plus{}4)$ then $ (x^2\plus{}3x\plus{}2)\geq 0 \to x \leq \minus{}2$ or $ x\geq\minus{}1$
2)If $ p \geq \minus{} \frac {1}{4}$ then $ \exists \ x_0\geq \minus{}1  : p \equal{} x_0^2 \plus{} x_0$ and $ x_0$ is always a root of the first equation
So there  is only one case left  $ \minus{} \frac {1}{2} \leq p < \minus{} \frac {1}{4}$..... :|

He he, I can finish the last case now!
Assume with $ p < \minus{} \frac {1}{4}$ still $ \exists \ x$ satisfy
Then $ \sqrt {(2p \plus{} 1 \minus{} x^2)(3x \plus{} p \plus{} 4)} \equal{} (x^2 \plus{} 3x \plus{} 2)$
$ \to (x^2 \plus{} 3x \plus{} 2) < \sqrt {(\frac {1}{2} \minus{} x^2)(3x \plus{} \frac {15}{4})}$
$ \to (x^2 \plus{} 3x \plus{} 2) < \sqrt {(\frac {3}{2} \minus{} 3x^2)(x \plus{} \frac {5}{4})} \leq \frac {1}{2}(\frac {3}{2} \minus{} 3x^2 \plus{} x \plus{} \frac {5}{4})$
$ \to 5x^2 \plus{} 5x \plus{} \frac {5}{4} < 0$
$ \to 5(x \plus{} \frac {1}{2})^2 < 0$ Wrong!  :)

Uhm, but I still think it isn't easy like that.What is the part of the number $ \minus{} \frac {1}{4}$?

[quote="duytungct"]Find [b]all [/b]$ p \in R$ that $ \exists \ x \in R$ satisfy:
$ \sqrt {2p \plus{} 1 \minus{} x^2} \plus{} \sqrt {3x \plus{} p \plus{} 4} \equal{} \sqrt {x^2 \plus{} 9x \plus{} 3p \plus{} 9}$[/quote]

Without, first, any check about validity of square roots we can raise to power 2 to get necessary conditions :

$ 2p\plus{}1\minus{}x^2\plus{}3x\plus{}p\plus{}4\plus{}2\sqrt {(2p \plus{} 1 \minus{} x^2)(3x \plus{} p \plus{} 4)} \equal{} x^2 \plus{} 9x \plus{} 3p \plus{} 9$

$ \sqrt {(2p \plus{} 1 \minus{} x^2)(3x \plus{} p \plus{} 4)} \equal{} x^2 \plus{} 3x  \plus{} 2\equal{}(x\plus{}1)(x\plus{}2)$

Once again raise to power 2 :

$ (2p \plus{} 1 \minus{} x^2)(3x \plus{} p \plus{} 4) \equal{} (x\plus{}1)^2(x\plus{}2)^2$
$ 2p^2 \minus{}p(x^2\minus{}6x\minus{}9) \minus{}x(x\plus{}1)(x^2\plus{}8x\plus{}9)\equal{}0$

$ 2(p\minus{}x(x\plus{}1))(p\plus{}\frac{x^2\plus{}8x\plus{}9}{2})\equal{}0$

And so only two possibilities : $ \boxed{p\equal{}x(x\plus{}1)}$ or $ \boxed{p\equal{}\minus{}\frac{x^2\plus{}8x\plus{}9}{2}}$

A quick check shows that the first is always possible (all data whose square root is taken are non-negative).
The same quick check shows that the second case is always impossible ($ 2p \plus{} 1 \minus{} x^2<0$ or $ 3x \plus{} p \plus{} 4<0$)

So the only possibility is $ p\equal{}x(x\plus{}1)$, and so $ p\geq \minus{}\frac{1}{4}$


calculus
derivative
function
analytic geometry
graphing lines
slope
logarithms

1. The derivative of y=x^2 is 2x, so the tangent line to x^2 at x=2 is y=2x.

How can I get the derivative of y=2^x by using a calculator?  I am not saying to get a graph, I want a derivative function of y+2^x.

To your first question, $ y' \equal{} 2x$ is an equation that gives the slope of the tangent line at any point, not the equation of the tangent line.  The slope at $ x \equal{} 2$ is in fact $ 4$, and the tangent line to $ x^2$ at $ x \equal{} 2$ actually has the equation $ y \equal{} 4x \minus{} 4$.

To your second question, you need to know that $ \frac{d}{dx} e^x \equal{} e^x$.  Then

$ \frac{d}{dx} 2^x \equal{} \frac{d}{dx} e^{x \ln 2} \equal{} \ln 2 \cdot e^{x \ln 2} \equal{} 2^x \ln 2$

You want to find the derivative of $ y \equal{} 2^{x}$
$ y \equal{} 2^{x} \equal{} e^{x\ln2}$

$ y' \equal{} e^{u}*u' \equal{} e^{x\ln2}*ln2 \equal{} ln2*2^{x}$ (by chain rule).

 On my calculator, here is how you enter this: d(2^x,x)

Edited: t0rajir0u was way faster


algebra
polynomial
linear algebra
linear algebra unsolved

Let $ x, y, z, X, Y, Z$ be real numbers (they are different, not equal). Prove that there exist only one polynominal $ w$ with grade not higher than $ 2$ such as $ w(x)\equal{}X$, $ w(y)\equal{}Y$, $ w(z)\equal{}Z$.

Yes, it's easy.
If $ p(x)$ and $ q(x)$ - polynomials with this property, then $ p(x) \minus{} q(x)$ has three roots, but $ deg[p(x) \minus{} q(x)] \leq 2$. So, $ p(x) \minus{} q(x) \equal{} 0$ for all $ x$.

And this polynom exists:

$ p(u) \equal{} \frac {(u \minus{} y)(u \minus{} z)}{(x \minus{} y)(x \minus{} z)} X \plus{} \frac {(u \minus{} x)(u \minus{} z)}{(y \minus{} x)(y \minus{} z)}Y \plus{}\frac {(u \minus{} x)(u \minus{} y)}{(z \minus{} x)(z \minus{} y)}Z$.

Sorry for my absence.

You say that $ p(x)\minus{}q(x)$ has three roots and also $ deg[p(x)\minus{}q(x)]\leqslant 2$. Hmmm... maybe I am stupid (o, yes I am ;]) but how the hell $ p(x)\minus{}q(x)$ has three roots ??

And how did you find the formula of $ p(u)$??

(To be sure - I am not university studet, I am just junior high school student, but I' d like to learn some new things.)
I mean, I just cannot understand your (too) short solution; could you explain that please??
Thank you

[quote="Invariant"]You say that $ p(x) \minus{} q(x)$ has three roots and also $ deg[p(x) \minus{} q(x)]\leqslant 2$. Hmmm... maybe I am stupid (o, yes I am ;]) but how the hell $ p(x) \minus{} q(x)$ has three roots ??[/quote]
If $ p(x)$ and $ q(x)$ both have the desired property then $ p(x) \equal{} q(x) \equal{} X, p(y) \equal{} q(y) \equal{} Y, p(z) \equal{} q(z) \equal{} Z$, which means that
$ p(x) \minus{} q(x) \equal{} p(y) \minus{} q(y) \equal{} p(z) \minus{} p(z) \equal{} 0$.

So $ x, y, z$ are the three roots of $ p(x) \minus{} q(x)$.  The actual expression for the unique polynomial is given by [url=http://mathworld.wolfram.com/LagrangeInterpolatingPolynomial.html]Lagrange interpolation[/url].

Yes, t0rajir0u had answered for your question. I can only add, that you can see here, too:
http://en.wikipedia.org/wiki/Lagrange_polynomial


number theory
relatively prime
number theory proposed

Let $S(n)$ be sum of digits of positive integer $n$ and $d(n)$ number of digits of $n$.

Show that for each $0\leq\alpha<\beta\leq 9$ where $\alpha$ and $\beta$ are real numbers and each positive integer $n$ there exist positive integer $k$ for which:

$\alpha<\frac{S(kn)}{d(kn)}<\beta$

So for each triple $\alpha,\beta,n$ find such k.

Is that meant to be for all n? Or something else?

Yeah, I think now it's ok, I've edited it.

If my solution is not completely wrong, then I think that for given n, the set of those fractions is dense in $[0,9]$. Indeed, consider $ w$ between 0 and 9 and write $ w=9e$. It is not hard to show that we can assume that $ n$ is relatively prime with 10. Then define $  x=p\cdot \phi(n)$, where p will be chosen later. Let also $ g=[ep\phi(n)]$.Then take $k_p=\frac{(10^g-1)(1+10^x+10^{2x}+...+10^{(pn-1)x})}{n}$. Then $s(k_p\cdot n)=9gpn$. Moreover, $ d(k_p\cdot n)$ is equivalent to $ (pn-1)p\cdot\phi(n)$. Thus for very large p the sequence $\frac{s(k_p n)}{d(k_p n)}$ converges to w. At least, it seems to me now.

Yeah Harazi, it's 0 to 9...
I've edited it.
It's hard to me to say whether your solution is right or not, because I don't see in it, but I'll try to understand :)
I'll also post my solution later...

Idea of my solution:

Let $n=a*10^l$ where $a$ is not divisible by 10. We know that $10^{\phi(a)}-1=999\ldots 9$ ($\phi(a)$ nines) is divisible by $a$. From this we know that there exist a multiple of $n$ in form $999\ldots 9000\ldots 0$. If we multiply it with number $1000\ldots 1$ with $\phi(n)-1$ zeros we will have number with same number of zeros and twice more nines.

So, one more time in other words: We can construct multiple of $n$ which have constant number of zeros and as many nines at the beginning as we want. Now to fulfil $\alpha<\frac{S(kn)}{d(kn)}<\beta$ just make this number having enough nines and then start adding some zeros (multiplying by 10) to fit between $\alpha$ and $\beta$. This is just explanation how to construct this multiple of $n$. It can be written formaly, but It would be hard to read  :D


number theory
least common multiple
greatest common divisor
modular arithmetic

[quote]Show that given any 3 multiples of 2, exactly one is a multiple of 3[/quote]
Maybe you mean three [i]consecutive[/i] multiples of two?

I believe this question should be worded something like "Given 3 consecutive mulitples of 2" because a counterexample would be 6,12,18.....

err...is it?

[hide]
3 = 2 + 1
3n = 2n + n
3n = even # + n
The nth multiple of 3 is even if n is even and odd if n is odd.
The LCM of 2 & 3 is 6, so every 6 intervals the multiples of 2 and 3 coincide.
There is 2 consecutive multiples of 3 in an interval of 6. One has to be even and the next odd (because the nth multiple of 3 is even if n is even and odd if n is odd.)
For every 2 consecutive multiples of 3, there is 3 consecutive multiples of 2.
So for every 3 consecutive multiples of 2, there is one multiple that is divisible by 3 (the multiple that coincides)
Sorry if this is confusingly written...
:?

[/hide]

[quote="eryaman"]Show that given any 3 consecutive multiples of 2, exactly one is a multiple of 3.

[color=blue][size=75][mod edit][/size][/color][/quote]

[hide="Answer"]

We want to show that exactly one of the following terms is divisble by 3.

$2(n), 2(n+1), 2(n+2)$

  Thus it is either,

$n$ or $n+1$ or $n+2$

  Now we know that among any three consecutive integers these exists one and only one that is divisible by 3.  Thus among any three consectuive multiple of 2, there exisst one and only one which is divisble by 3. [/hide]

[hide]
well i'll test a few
2, 4, 6
6 is a multiple of 3
8, 10, 12
12 is a multiple of 3
So the gcf of 2 and 3 is 6
6/2=3 so every three multiples of 2 are multiples of 3.
i'm trying to get out of the classroom math and basic forums so don't laugh at me if my proofs are stupid.[/hide]

Sorry, I did mean consecutive multiples of 2.  Woops.  :blush:

[hide]
numbers:
2k, 2k+2, 2k+4

in mod 3, 2k is 2, 1, or 0.

multiples of 2 in mod 3 repeat this pattern:
2, 1, 0, 2, 1, 0, ...

thus, we have 3 consecutive multiples of 2, which can be represented in mod 3 as:

2, 1, 0 leading to the last term to be =0 mod3
1, 0, 2 leading to the second term to be = 0 mod 3
0, 2, 1 leading to the first term to be =0 mod3

AWD.
[/hide]

[hide]Let the numbers be k, k+2, k+4. If k is a multiple of three, the case is automatically true.

IF k-1 is a multiple of 3, k=1 (mod 3), k+2=0(mod3), the case is true also.

If k+1 is a multiple of 3, k+2=1 (mod 3), and k+4= 0 (mod 3). Also works.

[/hide]

Here's a question, similar to the one given -

Show that given any 3 consecutive multiples of x, exactly one is a multiple of 3.

If this is too hard for the Getting Started, then a mod can move it up to Intermediate.

[hide]Let $x$ be an integer.  Then all multiples of $x$ are in the form $kx$ for some integer $k$.  Then the three consecutive multiples of $x$ are $kx$, $(k+1)x$, and $(k+2)x$.  Consider three cases:

$k\equiv 0\pmod{3}$:
Then $kx\equiv 0\cdot x\equiv 0\pmod{3}$, so $kx$ is a multiple of $3$.

$k\equiv 1\pmod{3}$:
Then $k+2\equiv 0\pmod{3}$, so $(k+2)x\equiv 0\pmod{3}$, so $(k+2)x$ is a multiple of $3$.

$k\equiv 2\pmod{3}$.
Then $k+1\equiv 0\pmod{3}$, so $(k+1)x\equiv 0\pmod{3}$, so $(k+1)x$ is a multiple of $3$.[/hide]


Let $ a_1 \geq a_2 \geq \ldots \geq a_n > 0$. Prove the inequality
$ a_1a_2\ldots a_{n \minus{} 1} \plus{} (2a_2 \minus{} a_1)(2a_3 \minus{} a_2) \ldots (2a_n \minus{} a_{n \minus{} 1}) \geq 2a_2a_3 \ldots a_n.$

Let $ b_i\equal{}\frac{a_{i\plus{}1}}{a_i}$.  Notice that $ 1\ge b_i\ge0$  Divide both sides by $ \prod_1^{n\minus{}1}a_i$ to get

$ 1\plus{}\prod(2b_i\minus{}1)\ge2\prod b_i$

$ \Leftrightarrow\prod(2b_i\minus{}1)\ge2\prod b_i\minus{}1$

Now we can induct on $ n$.  The base case is easy enough.  By the inductive hypothesis we have

$ \prod_1^{n}(2b_i\minus{}1)\ge(2\prod_1^{n\minus{}1} b_i\minus{}1)(2b_n\minus{}1)$

Having the right hand side greater than or equal to $ 2\prod_1^n b_i\minus{}1$ is equivalent to

$ \prod_1^nb_i\plus{}1\ge b_n\plus{}\prod_1^{n\minus{}1}b_i$

This is simple rearrangement with $ 1\ge\prod_1^{n\minus{}1}b_i$ and $ 1\ge b_n$.


vector
linear algebra
linear algebra unsolved

Prove that if K : $\mathbb{R}^{p}\rightarrow \mathbb{R}^{q}$ and L : $\mathbb{R}^{q}\rightarrow \mathbb{R}^{r}$ are linear transformations,
then
$\parallel L$ o $K \parallel \leq \parallel L\parallel \parallel K\parallel$

Since there are several possible norms for linear operators (matrices), this is several possible questions. Do you mean to specify a norm? Or is the problem to show that this is true for the operator norm induced by any valid norm on $\mathbb{R}^{n}?$

A linear transformation L from a normed vector space X to a normed vector space Y is said to be bounded if the set

$\{\frac{\parallel L(x) \parallel }{\parallel x \parallel}: x \in X, x \not= 0 \}$
is bounded above. In this case, the least upper bound of this set is called the operator norm of L and is denoted $\parallel L \parallel$.

So I'll take that as the operator norm induced by any valid norm on $\mathbb{R}^{n},$ of which there are several well-known and useful ones.

There's actually almost nothing to this proof.

We have that for all $y,\ \|L(y)\|\le\|L\|\cdot\|y\|$ and that for all $x,\ \|K(x)\|\le\|L\|\cdot\|x\|.$

Hence, $\|L\circ K(x)\|\le\|L\|\cdot\|K(x)\|\le \|L\|\cdot\|K\|\cdot\|x\|.$

Now, for all nonzero $x,$ divide through by $\|x\|,$ then take the supremum over $x,$ and you have the result.

Followup: The condition number of an invertible operator $L$ is defined as $\|L\|\cdot\|L^{-1}\|.$ What is the smallest possible value of a condition number?


geometry
circumcircle
incenter
trigonometry
geometry proposed

Bisectors of triangle ABC of an angles A and C intersect with BC and AB at points A1 and C1 respectively. Lines AA1 and CC1 intersect circumcircle of triangle ABC at points A2 and C2 respectively. K is intersection point of C1A2 and A1C2. I is incenter of ABC. Prove that the line KI divides AC into two equal parts.

[color=darkblue]We have

$ \dfrac{C_2C_1}{C_1I} \equal{} \dfrac{AC_2}{AI}\cdot \dfrac{\sin{\angle C_2AB}}{\sin \angle C_1AI} \equal{} \dfrac{2R\sin \frac {C}{2}}{AI}\cdot \dfrac{\sin\frac {C}{2}}{\sin \frac {A}{2}} \equal{} \dfrac{2R\sin^2 \frac {C}{2}}{r}$

Similiarly

$ \dfrac{A_2A_1}{A_1I} \equal{} \dfrac{2R\sin^2 \frac {A}{2}}{r}$

Let $ D \in KI \cap A_2C_2$. By Ceva in $ \triangle C_2A_2I$ we have

$ \dfrac{C_2D}{A_2D} \equal{} \dfrac{C_2C_1}{C_1I}\cdot \dfrac{IA_1}{A_1A_2} \equal{} \dfrac{\sin^2\frac {C}{2}}{\sin^2\frac {A}{2}}$

Then

$ \dfrac{\sin{\angle C_2ID}}{\sin\angle A_2ID} \equal{} \dfrac{C_2D}{A_2D}\cdot \dfrac{IA_2}{IC_2} \equal{} \dfrac{\sin^2\frac {C}{2}}{\sin^2\frac {A}{2}} \cdot \dfrac{\sin \frac {A}{2} }{\sin \frac {C}{2} } \equal{} \dfrac{\sin\frac {C}{2}}{\sin\frac {A}{2}}$

If we denote $ M \in KI \cap AC$. Then

$ \dfrac{AM}{MC} \equal{} \dfrac{AI}{IC}\cdot \dfrac{\sin \angle AIM}{\sin \angle CIM} \equal{} \dfrac{AI}{IC}\cdot \dfrac{\sin\angle A_2ID}{\sin{\angle C_2ID}} \equal{} \dfrac{AI}{IC}\cdot \dfrac{\sin\frac {A}{2} } {\sin\frac {C}{2} } \equal{} \dfrac{r}{r} \equal{} 1$

And we are done...[/color]

[quote="Virgil Nicula"] [color=darkred][b][u]An easy extension[/u].[/b] Let $ P$ be an interior point of $ \triangle ABC$ and denote the points $ D\in (BC)$ , $ E\in (CA)$ , $ F\in (AB)$ so

that $ P\in AD\cap BE\cap CF$ . Define the second intersections $ M$ , $ N$ of the lines $ BE$ , $ CF$ respectively with the circumcircle

$ w$ of $ \triangle ABC$ and $ K\in MF\cap NE$ , $ L\in KP\cap BC$ . Prove that $ AA\cap MN\cap EF\ne\emptyset$ and $ \frac {LB}{LC} \equal{} \left(\frac {DB}{DC}\cdot\frac bc\right)^2$ .

[b][u]Remark[/u].[/b] Denote $ T\ \in\ (AP\ \cap\ w$ and $ S\ \in\ TT\ \cap\ BC$ . Then the division $ (\ B\ ,\ C\ ;\ S\ ,\ L\ )$ is harmonically ![/color] [/quote]

[quote="Virgil Nicula"][color=darkred][b][u]Lemma I[/u].[/b] Let $ ABC$ be a triangle. Consider the points  $ D\in (BC)$ , $ E\in (CA)$ ,

$ F\in (AB)$ and $ X\in AD\cap EF$ . Then $ \boxed {\ \frac {XF}{XE} \equal{} \frac {DB}{DC}\cdot \frac {AF}{AE}\cdot \frac {AC}{AB}\ }$ .[/color]
[hide="Proof."]
[color=darkblue]Denote $ \left\|\begin{array}{c} m(\widehat {DAB}) \equal{} x \\
 \\
m(\widehat {DAC}) \equal{} y\end{array}\right\|$ . Then $ \left\|\begin{array}{c} \frac {XF}{XE} \equal{} \frac {AF}{AE}\cdot\frac {\sin x}{\sin y} \\
 \\
\frac {DB}{DC} \equal{} \frac {AB}{AC}\cdot\frac {\sin x}{\sin y}\end{array}\right\|\ : \downarrow\ \implies$ $ \frac {XF}{XE} \equal{} \frac {DB}{DC}\cdot \frac {AF}{AE}\cdot \frac {AC}{AB}$ .[/color] [/hide]

[color=darkred][b][u]Lemma II[/u].[/b] Letr $ ABC$ be a triangle and $ D\in (BC)$ , $ E\in (CA)$ , $ F\in (AB)$ be three points

so that $ AD\cap BE\cap CF\ne\emptyset$ . Denote $ X\in AD\cap EF$ . Then $ \boxed {\ \frac {XF}{XE} \equal{} \frac {BF}{BA}\cdot\frac {CA}{CE}\ }$ .[/color]
[hide="Proof."] [color=darkblue]Apply the upper lemma $ \frac {XF}{XE} \equal{} \frac {DB}{DC}\cdot\frac {AC}{AB}\cdot \frac {AF}{AE}$ and the [u][b]Ceva[/b]'s theorem[/u] : $ \frac {DB}{DC}\cdot\frac {EC}{EA}\cdot\frac {FA}{FB} \equal{} 1$ .

Thus, $ \frac {XF}{XE} \equal{} \left(\frac {EA}{EC}\cdot\frac {FB}{FA}\right)\cdot\frac {AF}{AE}\cdot\frac {AC}{AB}$ , i.e. $ \frac {XF}{XE} \equal{} \frac {BF}{BA}\cdot\frac {CA}{CE}$ .[/color][/hide]

[color=darkred][b][u]Lemma III[/u].[/b] Let $ ABCD$ be a convex quadrilateral for which denote $ I\in AC\cap BC$ .

Consider $ X\in (BC)$ , $ Y\in (AD)$ . Then $ \boxed {\ I\in XY\ \Longleftrightarrow\ \frac {IA}{IC}\cdot\frac {IB}{ID} \equal{} \frac {XB}{XC}\cdot\frac {YA}{YD}\ }$ .[/color]
[hide="Proof."]
[color=darkblue]Denote $ \left\|\begin{array}{ccc} m(\widehat{XIB}) \equal{} \alpha & ; & m(\widehat {XIC}) \equal{} \beta \\
 \\
m(\widehat {YIA}) \equal{} \gamma & ; & m(\widehat {YID}) \equal{} \delta\end{array}\right\|$ , where $ \alpha \plus{} \beta \equal{} \gamma \plus{} \delta$ . Using the well-known relations $ \left\|\begin{array}{c} \frac {XB}{XC} \equal{} \frac {IB}{IC}\cdot\frac {\sin \alpha}{\sin\beta} \\
 \\
\frac {YA}{YD} \equal{} \frac {IA}{ID}\cdot\frac {\sin \gamma}{\sin\delta}\end{array}\right\|$

obtain equivalencies : $ \frac {IA}{IC}\cdot\frac {IB}{ID} \equal{} \frac {XB}{XC}\cdot\frac {YA}{YD}$ $ \Longleftrightarrow$ $ \frac {IA}{IC}\cdot\frac {IB}{ID} \equal{} \frac {IB}{IC}\cdot\frac {\sin \alpha}{\sin\beta}\cdot\frac {IA}{ID}\cdot\frac {\sin \gamma}{\sin\delta}$ $ \Longleftrightarrow$  $ \sin \alpha\cdot\sin \gamma \equal{} \sin\beta\cdot\sin \delta$ $ \Longleftrightarrow$

$ \cos (\alpha \minus{} \gamma ) \minus{} \cos (\alpha \plus{} \gamma ) \equal{} \cos (\beta \minus{} \delta ) \minus{} \cos (\beta \plus{} \delta )$ $ \Longleftrightarrow$ $ \left\|\begin{array}{c} \alpha \plus{} \gamma \equal{} \beta \plus{} \delta \\
 \\
(\ \alpha \plus{} \beta \equal{} \gamma \plus{} \delta\ )\end{array}\right\|$ $ \Longleftrightarrow$ $ \alpha \equal{} \delta\ \wedge\ \beta \equal{} \gamma$ $ \Longleftrightarrow$ $ I\in XY$ . [/color][/hide][/quote]

Notice A2C2 is perp bisector of BI hence B2C2 cuts BB at F and FI=FB so FI//BC. Pascal for (BAA2C2CB) so FC1A1 collinear so -1=I(FEC1A1)=I(FDAC) but IF//BC so D is midpoint of BC so q.e.d


You're right. Intersections can't be imaginary.

Um..this is just two points so no intersections....and I think your fridnd meant, (0, 2i) and (0,-2i), which are imaginary solutions to the two systems..

[quote="7h3.D3m0n.117"]On my math test today here was the bonus question:

Find the intersection of the equations-

$(x+2)^{2}+y^{2}=0$
$(x-2)^{2}+y^{2}=0$

My friend said it was $(0,i \sqrt{3})$ or something. I said there was no intersection. Second of all how can and intersection be imaginary?[/quote]
[hide]
$(x+2)^{2}=(x-2)^{2}$
$(x+2)^{2}-(x-2)^{2}=0$
$4(2x)=0$ (conjugate binomial)
$x=0$
$4+y^{2}=0$
$y=2i$[/hide]
Did the teacher specify the real plane or no?

y can equal -2i too..

but just out of curiousity, is this from the circles section of algebra II?

Whenever $x,y$ are real $(x+2)^{2}+y^{2}=0$
$(x-2)^{2}+y^{2}=0$
$\implies y=0, x+2=x-2=0$ which is impossible so all solutions are imaginary.
Complex solutions don't make sense in the context because that would require a four dimensional graphing space, which isn't expected from an algebra class in school.
Of course there's no real problem with the equations but I'm pretty sure that the intended answer was no solution.

intersection means real...unless your teacher said otherwise...

I think that it was a trick question, because it was a bonus...

[quote="Art of Owna"]but just out of curiousity, is this from the circles section of algebra II?[/quote]

yes it is from the conics section of Alg II. my friend shut up for 15 minutes straight (hes a talker) when he was wrong.

thanks =) got a 95 on the test because of time pressure and stupid mistakes  :blush: ... ($\frac{18}{20}+$ bonus)


Find function $f: R \to R$ satisfy:

Problem 1: $f(x+y)f(xy)= x^{2}f(y)+y^{2}f(x)$;

Problem 2: $f(x)f(y)f(z)= \frac{f(x)}{(x-y)(x-z)}+\frac{f(y)}{(y-x)(y-z)}\frac{f(z)}{(z-x)(z-y)}$
With $x \neq y \neq z \neq x$

[quote="Denbox"]Problem 2: $f(x)f(y)f(z)= \frac{f(x)}{(x-y)(x-z)}+\frac{f(y)}{(y-x)(y-z)}\frac{f(z)}{(z-x)(z-y)}$
With $x \neq y \neq z \neq x$[/quote]
Maybe you meant
$f(x)f(y)f(z)= \frac{f(x)}{(x-y)(x-z)}+\frac{f(y)}{(y-x)(y-z)}\boxed{+}\frac{f(z)}{(z-x)(z-y)}$

For problem 1 i could guess two possibilities (but there may be others)

[hide]
$f(x) = x, f(x) = 0$
[/hide]


meh. stupid me i should read the questions better next time...

[quote="ml2"]Mystic, your ruling is over, mine has began!!![/quote]

right well if there's another contest (ahem...as in tomorrow) and i read the questions better (stupid me) we'll see.

edit- Never mind, I see where I made the mistake.

ahem. <i> YOUR </i> contest?

Actually its the AMC's contest :)

And you used the same scoring!!!!

eh? I would like to remind you that these problems are completely original (up to a point ;))

Hmm...most of these problems I did jack from the AHSME, yes, but I changed them a bit to make them [b]mine[/b] :-D.  One or two I didn't take from there.  And I meant, my contest wasn't easy, but it's quite tricky.


Look [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=233460]here.[/url]

I tried to post this message. A perfectly good balanced-dollar-sign formula got interpreted as unparseable and with unbalanced dollar signs, and the "hide" bb-code got scrambled around. I had to add a completely extraneous dollar sign at the end to get it to post at all.

Can anyone see what is wrong with it?

Added in edit: it seems to be the "hide" code that is messing things up. Or is it the \approxeq command inside the hide code?

Bug confirmed. It has nothing to do with \approxeq
Everything works just fine except when using the [code][hide=Hint for 1.][/code] expression, that gets mis-parsed somewhere. I will take a look.

I'm on it Kent. Don't worry about it anymore, I'll track this bug down and fix it ;)

It had nothing to do with latex. The hide tags in general were messed up (now they are fixed :D )


Let $ a,b,c$ be positive reals such that $ a^2\plus{}b^2\plus{}c^2\equal{}3$. Find the greatest possible constant $ k$ such that

$ ab^2\plus{}bc^2\plus{}ca^2\geq k(ab\plus{}bc\plus{}ca)^2$

[quote="indybar"]Let $ a,b,c$ be positive reals such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$. Find the greatest possible constant $ k$ such that

$ ab^2 \plus{} bc^2 \plus{} ca^2\geq k(ab \plus{} bc \plus{} ca)^2$[/quote]
Let $ a\equal{}b\equal{}c\equal{}1$, then we have $ k \le \frac{1}{3}$, we will show that it's the value we need to find, i.e,
\[ ab^2\plus{}bc^2\plus{}ca^2 \ge \frac{1}{3}(ab\plus{}bc\plus{}ca)^2\]
The Cauchy Schwarz Inequality gives us
\[ (ab^2\plus{}bc^2\plus{}ca^2)(a\plus{}b\plus{}c) \ge (\sqrt{ab^2\cdot a}\plus{}\sqrt{bc^2\cdot b}\plus{}\sqrt{ca^2\cdot c})^2 \equal{}(ab\plus{}bc\plus{}ca)^2\]
Hence
\[ ab^2\plus{}bc^2\plus{}ca^2 \ge \frac{(ab\plus{}bc\plus{}ca)^2}{a\plus{}b\plus{}c}\]
It suffices to prove that
\[ 3 \ge a\plus{}b\plus{}c\]
But this inequality is true because
\[ a\plus{}b\plus{}c \le \frac{a^2\plus{}1}{2}\plus{}\frac{b^2\plus{}1}{2}\plus{}\frac{c^2\plus{}1}{2}\equal{}\frac{a^2\plus{}b^2\plus{}c^2\plus{}3}{2}\equal{}3\]
We have done. :)


I am not sure if all of you got the rules straight ... so please adress any specific misunderstandings that you might have here ... I'll try to answer as quick as I can ....

I wish there were chat rooms on AoPS so mathletes could communicate easier.

Have I missed something? Why have you posted in a nearly 4-year old topic (one of the oldest around, I believe   :maybe: ) in order to make an irrelevant comment?

Is there an inscription here? Where are the problems posted or how do I get to participate in this contest?

dear mathlinkers

I would like to know the rules to this contest


vector
linear algebra
matrix
advanced fields
advanced fields unsolved

Let $||x||=\sqrt{x^{H}x}$ that is $l_2 - norm$
Let $M \in C^{nxn}$ be a hermitian positive definite, and consider the vector norm:
$||x||_{M}=\sqrt{x^{H}Mx}$.
Find an expression for the matrix norm $||A||_M$ induced by this vector norm, show also that
$||A||_M \le \sqrt{K(M)}||A||$, where $||K||=||M||||M^{-1}||$

I can't see beyond this..
$||A||_M= \sup_{x\neq0}\frac{||Ax||_M}{||x||_M}=\sup_{x\neq0}\frac{\sqrt{(Ax)^{H}M(Ax)}}{\sqrt{x^{H}Mx}}$

I don't know what to do with the M..

[quote="jmerry"]I'm no ready to post an answer, but matrix questions like your two topics here should be asked in the Superior Algebra section.[/quote]

I guess ineed it should be posted elsewhere.. but in the superior algebra they couldn't help me so far.. And it's more numerical analysis than Algebra.. I think..

I found this:
Since M is h.p.d there exists a matrix U upper triangular, non-singular, unique, such that
$M = U^{H}U$
so we can say $||A||_{M} = ||A||_{U^{H}U} = \sqrt{A^{H}U^{H}UA} = ||UA|| $

All I can get from here is that $||A||_M \leq \kappa(U) ||A|| $
But I still can't find a way to link this with $\kappa(M)$
and I also know that since M is normal I have
$K(M) = \frac{\abs{\lambda_{max}(M)}}{\abs{\lambda_{min}(M)}}$

edit:
Ok, I've found the solution at last:
The singular values of U are the square roots of the eigenvalues of M since M is normal.
Then I get $\kappa(M) = \kappa(U)^2$


Let $ A \subset [0,1]$ be a measurable set of positive measure. Prove that there exists $ x, y \in A$ such that $ x \neq y$ and $ x \minus{}y$ is rational.

Just consider the sets $ A\plus{}q$, $ q\in \mathbb Q\cap[0,1]$. Two of them must intersect (otherwise $ [0,2]$ would have infinite measure), hence the conclusion.

Is the following the reason as to why [0,2] must have an infinite measure?
If for each $ i\in \{1,2,...,n\}$, $ \bigcap_{i \equal{} 1}^{n} A \plus{} r_{i} \equal{} \emptyset$.
Then one has,
$ \bigcup_{i \equal{} 1}^{n} A \plus{} r_{i} \subset [0,2]$ implying
$ m[0,2] \geq m(\bigcup_{i \equal{} 1}^{n} A \plus{} r_{i}) \equal{} m(\sum_{i \equal{} 1}^{n}A \plus{} r_{i}) \equal{} \sum_{i \equal{} 1}^{n}m(A \plus{} r_{i}) \equal{} \sum_{i \equal{} 1}^{n}m(A) \equal{} m(A)n$.
Since m(A) is positive, the measure of [0,2] is infinite.

[quote="akech"]
$ m[0,2] \geq m(\bigcup_{i \equal{} 1}^{n} A \plus{} r_{i}) \equal{} m(\sum_{i \equal{} 1}^{n}A \plus{} r_{i}) \equal{} \sum_{i \equal{} 1}^{n}m(A \plus{} r_{i}) \equal{} \sum_{i \equal{} 1}^{n}m(A) \equal{} m(A)n$.[/quote]

What does $ m(\sum_{i \equal{} 1}^{n}A \plus{} r_{i})$ even mean? Certainly, $ n$ sets having empty intersection does imply pairwise empty intersections!

However, the contrapositive of fedja's assertion would be that for all pairs of rationals the corresponding sets do not intersect. So, yea, I guess.


geometry college MIT Stanford Princeton Harvard

I wanted to know which one among IISER and NISER/CBS is a better institute..Both are almost same in terms of curriculum...Then why has government opened two different institutes??

as per I know
they are exactly the same and having the same purpose except the funding organisations
unlike the IISERs, which will be governed by India's Ministry of Human Resources Development, NISER/CBS operates under the umbrella of the Indian Department of Atomic Energy (DAE).

that's it
they were even planned under the same scheme

I think if they were set up under the same scheme and for the same purpose then the labs/facilities might be equivalent
though actually I have no idea :blush:

as far as par is concerned
last years posts indicate that he is still in IIT-M
though I'm not sure again :rotfl:

[quote="savy2020"]Thanks

Which would be better in terms of facilities (laboratories etc)?

[hide="btw"] what's par (sir) doing now?[/hide][/quote]
As abhitayal said i am  [b]still[/b] at IIT M.
ok,now to the question which one is better.To start with i would like to know what you people are looking for/want to do in life which because that is relevant to teh answer.Assuming you want to pursue research in pure Sciences(Math,Physics,Chemistry,Biology or any mixture of these) the IISER's and the NISER's seem to be good in terms of their goal.
Next year the first batch from IISER will pass out.So what they end up doing will definitely set a benchmark for the batches to come and will locally tell you the sucess of such programmes.As [b]abhitayal[/b] mentioned they differ in their funding agencies and afaik the NISER students will be considered for employment directly after theit masters into the DAE centres across India so that's a perk for students studying there.
But apart from that at IISER one has to pursue two year of Common classes before specializing while at NISER(like all other institutes in the country) it's just one year.
IISER Mohali has a pretty good physics dept.,I have been to NISER Bhubaneswar which is growing fast though a common 'problem' in all these new institutions is the lack of permanent faculty members which i guess will get solved in a year or so.

Sir,
Can you tell us what you think of the 4 yr B.S proposed by IISC?
As it is an established institute compared to IISER's i would prefer it.
I want to do research in Chem(or phy if i fail)

Hi

As I work closely with CBS and with NEST examination, let me share few things.

1. All 8 institutes IISERs at Pune, Kolkata, Thiruvanantapuram, Mohali, Bhopal, NISER at Bhubaneswar, CBS at Mumbai and IIST at Thiruvanatapuram are pretty new. Till first batch gets their jobs / further admissions, it would be too hard to compare them.

2.most of them are working from makeshift campuses but some have given more attention to their makeshift infrastructure (IISER-K, NISER) to make sure initial batches are also comfortable.

3. Almost all of them are on recruiting spree to get best faculties but most faculties in most places are new hence untested. We can't comment on their quality either way. IISER-K picked up quiet a few experienced physics teachers from Jadhavpur hence they are bit ahead of others. CBS relies on getting visiting faculty members from TIFR, BARC, HBCSE, IIT(B) as well as Mumbai colleges hence they are comfortable too.

4. Academic structure of these places is varied. CBS and NISER have one common year. The advanced courses in these institutes are tailored to DAE needs (e.g. "reactor physics", "radiation biology" etc.). Similarly IIST has courses tailored for ISRO needs. IISERs have two common years. IISER-P has no concept of "major" and people will simply get degrees "master of science" without specifying subject. Other IISERs do have subjectwise streams.

5. IIST requires students sign a bond to work for ISRO for 3 years after completion. ISRO will offer jobs to IIST graduates with CGPA of 7 and above. Similarly, BARC will offer jobs to CBS and NISER graduates with CGPA of 7.5 and above. No bond at CBS / NISER or any IISERs.

Visva-Bharati, santiniketan has a similar 5 year programme with admissions through NEST. IISc programme is due to start in 2011, so nothing can be said about it yet.

Yes thank you sir.
well looks like IISER(K) is quite good. Not bad as it is close to where i stay.
However is it possible to study basic science properly at IIT?

diptarka u cn do msc shemistry from iit b ...

Yup I know but i am worried about standard and research focus. I have not heard much about iit and basic science(specially chem) going hand in hand.

Sir,
Thanks a lot for your information.
Could you tell how these 8 institutions compare with Chennai Mathematical Institute (CMI) and ISI?

i have a question (pretty silly it may seem) which is
hoe to get in these colleges pl tell elaborately

@eldiablo
For NISER/CBS/Visva-Bharati, santiniketan - NEST http://www.nestexam.in
For IISER - either of 3 ways mentioned here - http://iiser-admissions.in/
For IIST - ISAT http://www.iist.ac.in/IIST/ISAT2010/ (I think deadline is over)

[quote="wmap"]4. Academic structure of these places is varied. CBS and NISER have one common year. The advanced courses in these institutes are tailored to DAE needs (e.g. "reactor physics", "radiation biology" etc.). Similarly IIST has courses tailored for ISRO needs. IISERs have two common years. IISER-P has no concept of "major" and people will simply get degrees "master of science" without specifying subject. Other IISERs do have subjectwise streams.
[/quote]
This makes me think ......

thnx savy always helping
check ur pm i have some questions

@pardesi
are you kv shudhodhan from dav cspur.
just curious :P

[quote="LOTRfanbbsr"]
are you kv shudhodhan from dav cspur.
just curious :P[/quote]
Yes  he is :D

[hide="Proof"]
[url=http://www.mathlinks.ro/viewtopic.php?p=1017760#1017760]See this[/url]
[url=https://docs.google.com/viewer?url=http://www.shaastra.org/2009/site/sites/default/files/Online%2520Math%2520Results.pdf]OR this[/url]  :P [hide="in case"] you don't believe the first one[/hide]

[/hide]

[hide="btw"]
Our mod seems to author [url=http://arxiv.org/abs/0910.4504]this pretty famous :P  paper[/url]
[/hide]

[quote="savy2020"]Thanks sir :)
Yes I want to pursue research in pure science(esp Physics).

And what is the difference between courses offered at IISER/NISER/CBS and the [url=http://www.physics.iitm.ac.in/index_files/courses.html]Integrated M.S (Physics) Course[/url] offered at IIT M.  Also how do these differ from B.Tech EP at IITM?

[hide="P.S."] What was the reason behind your preferring EP at IITM against IISER/NISER as you've mentioned [url=http://www.mathlinks.ro/viewtopic.php?t=189356]here[/url] that that's your preference.

[hide="Also"]Is it possible to transfer to foreign universities after joining Integrated M.S.? I think with B.Tech EP, it may be possible but abt M.S I've no idea[/hide]

[/hide][/quote]
Sir?

how does cmi stand as compared to iit,bits,isi,niser/iiser. :?:

Sorry for the delay :)
Ok first a disclaimer my opinions will probably hold true only for Maths and Physics,none of my immediate acquaintances are pursuin a biology/Chemistry course hence i cannot really tell you much.
First about I.I.T Madras.
We offer two Undergraduate degrees in the Physics department:
1.B.Tech Engineering Physics in association with the Electrical Department I.I.T Madras
2.M.S.(5 years) in Physics

Both the degrees have a good amount of physics by default(that is without the electives which form a major part of your study).One can choose a lot of electives offered by the physics department in all areas in physics,the present range of electives is huge and one can pretty much learn all phyiscs he /she requires for grad school.The Theoretical Physics faculty by far one of the best in the country(along with EP IIT Bombay).We have people working on AdS/CFT(String theory),Dynamical systems,Quantum Chaos,Condensed Matter (Field Theories),Complex Fluids,Quantum Information and courses are offered in all these fields except String theory instead a course on General Relativity and Cosmology is offfered which is a rare thing at an undergraduate level. The M.S. course was a dream of Dr.Balakrishanan,professor Emeritus  I.I.T Madras one of the renowned physicst in the world which is quite different and 'modern' as compared to the one existing at I.I.T Kanpur.The present batch of the students in the M.S course(the first batch) is brilliant.
We also have a physics group called [url=http://boltzmann.wikidot.com/]Boltzmann[/url] which meets every Saturday and discuss physics.Every semester one topic is chosen and students/professors give lecture s on the various aspects of the chosen topic.We have done Ising Models,Quantum Information theory,Random Walks,QED in the past,this sem we are doing General Relativity.You get to meet like minded people and that was one thing why I.I.T Kanpur became what it is now that is there was great group cohesion.

Now lets focus on some other places:
IISER:As i told before Mohali seems to have a good physics faculty,i have been to IISER kolkatta they also seem to have a good faculty especially the ones from the nearby Saha  and Bose institutes.In any case Kolkattta is great place for learning pure sciences.

NISER:I have been to the IOP,Bhubaneswar many a times(the place where NISER's transit campus is located).IOP is good place but the areas of research are not very diverse (atleast as of now) may be after the NISER starts they will diversify it,but there are some really good people.

CMI:This is a wonderful place for Maths and CS.Very flexible curicullum,a brilliant director and constant interaction with IMSc makes this place great.You can actually choose courses offered at IMSc.The physics department is upcoming.They have had really good batches in BSc physics hence i guess with time the department will only get better.

ISI:
ISI Bangalore:Very good place for pure math.Very focussed research oriented place.Good professors and vicinity to IISc are the plus points.They only offer B.Mat and not B.Stat
ISI Kolkatta:No idea but supposed to be a great place for Statistics.

@LOTR:yes i am Shuddhodan do i know you?

@pardesi
i am in dav cspur class 12.
you are a big icon at school. :lol:

[quote="Lakshya"]Well, I wanted to ask few more things, sir and other people. I want to do research work in Physics. My doubts are all asked by myself keeping only Physics in mind.
What is probably a good path? Direct away with B.Sc. physics or rather first B.Tech. and then M.Sc.? Although the first one is preferable but the second one was successfully followed by the likes of R. Shankar, Madhur Tulsiani etc.
Is the education in India self sufficient to carry out research work in basic and fundamental physics? Or probably the likes of Berkely, MIT, Caltech, Stanford, Princeton, Cambridge, Harvard are more suitable? Also I would thank if someone would provide any particualr info about these particular universities?
Do these above mentioned universities provide any scholarships at bachelor or masters level? I know I can get this on the university sites but the info given there is not clear and they often talk about scholarships to the native students and ignore international applicants.
I have also heard that international olympiad participants also get scholarship offers from these universities, any body has any info about that?[/quote]
Anybody??


$x,y,z,t \geq 0\\
\\
\frac{xy+xt+zy+zt}{x+y+z+t} \geq \frac{xy}{x+y}+\frac{zt}{z+t}$

Hmm, this looks like the Vietnam 62 inequality,

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=1216


Let $ A$ and $ B$ be distinct n x n matrices with real entries. If $ A^{3}=B^{3}$ and $ A^{2}B=B^{2}A$, can $ A^{2}+B^{2}$ be invertible?

$ (A^{2}+B^{2})A=A^{3}+B^{2}A.$

$ (A^{2}+B^{2})B=A^{2}B+B^{3}=A^{3}+B^{2}A.$

Hence, $ (A^{2}+B^{2})A=(A^{2}+B^{2})B.$

If $ A^{2}+B^{2}$ were invertible, we could left-multiply the last equation by its inverse and deduce that $ A=B.$ Since we were given that $ A\ne B,$ we must then have that $ A^{2}+B^{2}$ is not invertible.


Three numbers are chosen at random between 1 and 0. What is the probability that the difference between the greatest and the least is $\frac{1}{3}$?


The solution for AIME II 2007 #1 was confusing for me, specifically the case with two zeroes.

[quote]A mathematical organization is producing a set of commemorative license plates. Each plate contains a sequence of five characters chosen from the four letters in AIME and the four digits in 2007. No character may appear in a sequence more times than it appears among the four letters in AIME or the four digits in 2007. A set of plates in which each possible sequence appears exactly once contains N license plates. Find \frac{N}{10}.
[edit]
Solution

There are 7 different characters that can be picked, with 0 being the only number that can be repeated twice.

    * If 0 appears 0 or 1 times amongst the sequence, there are \frac{7!}{(7-5)!} = 2520 sequences possible.
    [b]* If 0 appears twice in the sequence, there are {5\choose2} = 10 places to place the 0s. There are \frac{6!}{(6-3)!} = 120 ways to place the remaining three characters. Totally, that gives us 10 \cdot 120 = 1200. [/b]

Thus, \displaystyle N = 2520 + 1200 = 3720, and \frac{N}{10} = 372. [/quote]

The first case, I did fine.

Second case, I couldn't really get the solution...I tried the following and got it right.

First, I simply separated the zeroes from the collection of characters, which can only be done in one way (obviously).  "Set these zeroes aside;" just don't do anything with them for now.

Now, only six characters are available to be chosen from, and this is simply a permutation.  Now we have to select three more characters.  Therefore, $6*5*4.$

Next, we can place one of the zeroes in four locations.  Finally, we can place the last zero in five locations.  These events are all independent; the three characters I chose don't affect where I place the first zero, and  where I place my first zero doesn't affect where I place the second.

All of the possibilities in this case have two zeroes in them; we have used permutations, so we have counted each possibility twice.  Divide by $2.$

$6*5*4*4*5/2 = 1200$

Is this logically sound, and is this way of solving it equally as effective in most cases?

I understand that the combination method used in the AoPS Wiki solution takes care of the overcounting; however, it says that because two of the numbers we are choosing out of the five are zeroes, there are 10 places to put the 0s.  However, what if there were 100000 characters to choose from?  The combination portion of the product is still 5 choose 2, so there will still be only 10 ways to place the zeroes...but there would clearly be a lot more, wouldn't there?

I understand the use of the combination...but I don't see how there would be 10 ways to "place the zeroes."  What does that even mean?

[quote="mathzilla"]I don't see how there would be 10 ways to "place the zeroes."  What does that even mean?[/quote]

We are attempting to produce a sequence of five characters by imagining that we have five blanks, like so:

_ _ _ _ _

We are placing the characters $A, I, M, E, 2, 0, 0, 7$ in these five blanks.  Now, the solution suggests that in the second place we first place both $0$s.  There are ${5 \choose 2}$ ways to do this; for example,

0 _ _ 0 _

Among the three remaining blanks we then place three more characters, which can be done in $6P3$ ways.

Your solution is valid, but this solution is faster.

The quesion says you want a sequence of [b]5[/b] and even if there were 1000000000 characters to choose from, if there are only 2 zeros out of that many characters, and 5 places to put them its still 5C2.


find all positive integer solutions to $3^x+4^y=5^z$.

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=19256]here[/url], [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=42173]here[/url] and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=72587]here[/url] for this and related questions.


Prove that under Lebesgue measure on $ R^2$
every circle has measure zero.

If Circle $ X$ with center O has measure $ x$,
then $ \sum_{n \equal{} 1}^{\infty}\mu(\frac {1}{\sqrt {n}}X) \equal{} \sum_{n \equal{} 1}^{\infty}\frac {x}{n} < \infty$, because $ X$ is bounded.
Then $ x \equal{} 0$.

Hi PeterP

as im a beginner in measure theory can you give a description of the sets
of the form $ \frac{1}{\sqrt{n}}X$

im not sure on that notation either, but what hes saying in the proof is that you can squeeze countably infinitely many circles inside of your original circle X.

if they are inside of circle X, they must be a smaller circle, hence if circle X has measure x, then they must be of the form $ \alpha x$ for some $ \alpha \in [0,1)$

defining $ X_n$ as the circle with measure $ \frac{x}{n}$ with the same center as X, then we can fit

$ Y\equal{}\cup_{n\equal{}1}^\infty X_n$ inside of the shaded original circle X

since a shaded circle X has finite measure in R^2, we know that the measure of Y must be smaller than it

hence the sum he mentioned must converge, which it only does at x=0.

[quote="inertia"]
as im a beginner in measure theory can you give a description of the sets
of the form $ \frac {1}{\sqrt {n}}X$[/quote]

For a set $ X$, and $ u>0$
$ uX: \equal{} \{ux : x \in X \}$
It is easy to show that $ \mu (uX) \equal{} u\mu (X)$.


group theory
abstract algebra
superior algebra
superior algebra solved

We have the groups (Q,+) and (Z,+) and form the factor group Q/Z.Find all the subgroups of Q/Z wich have n elements.

An example is simple-we take G the group of the classes of the folllowing elements:0,1/n,2/n,...,n-1/n.Is this the only one?

Let a1/b1+Z, a2/b2+Z...an/bn+Z be the elements of an n-element subgroup, G, of Q/Z.

Let m=lcm(b_1...b_n). Then G< <1/m+Z> <Q/Z.
<1/m+Z> is isomorphic to Zm=Z/mZ, and G is a subgoup of it.

Lets look at the subgoups of Z/mZ. It is known that the subgroups of G'/H are in one-to-one correspondence to the subroups of G' containing H for any commutative group G' and H-one of its subgroups.

So the subgoups of Z/mZ are {dZ/mZ | d|m}. So all the subgoups of <(1/m)+Z> are cyclic. So G is cyclic. If G has n elements, this means G[u]~[/u]Zn.

Let a/b+Z be a generator of G, gcd(a,b)=1. Then a/b has order n, so na/b is in Z, so b|na, so b|n. If b<n, then b*a/b is in Z, so a/b has order smaller or equal to b<n (a contradiction). So b=n. Because a and b=n are coprime, there exists k in Z such that ka=1(mod n), so 1/n+Z is in G, so G=<(1/n)+Z>. So xxxxtt's example is the only one.


function
integration
search
geometry
real analysis
real analysis unsolved

I have a question about the following problem:
Let $f: [0,1]\to \mathbb{R}$ a continuous functin such that $\int_{0}^{1}f(x)g(x)dx=\int_{0}^{1}f(x)dx\cdot\int_{0}^{1}g(x)dx$ for any $g: [0,1]\to \mathbb{R}$ continuous but not derivable. Prove that $f$ is constant.
Well I have a solution but my first idea was this question (maybe stupid :blush: ):
Can we find a sequence of functions $g_{n}$ continuous but not derivable that converge uniformly to $f$? And if so can we take the limit and conclude that $\int_{0}^{1}f(x)^{2}dx={\int_{0}^{1}f(x)dx}^{2}$ i.e. $f(x)=c$?

PS: I think I asked for this in another post...I am not sure that I received an answer: Could someone reccomend me some good online text about uniform convergence/uniform continuity? I would be really grateful :)

PPS: [u]my question (about my argument on this problem) is still unanswered[/u]

[quote="ciprian"]

PS: I think I asked for this in another post...I am not sure that I received an answer: Could someone reccomend me some good online text about uniform convergence/uniform continuity? I would be really grateful :)[/quote]
Search, I have
[url=http://www.lookforbook.com/cgi-bin/search.cgi?lang=en&st=real+analysis&network=1]Real Analysis[/url] :D

@N.T.TUAN: Thx a lot :) I also searched the forum but I have not founf it :(  And of those books could you recommend one?
What about the question from this topic?
And another question? What program dio I need to open those files? I do not hnow the extension  :maybe:

I will choose W.Rudin's books :)

[quote="ciprian"] What program dio I need to open those files? I do not hnow the extension  :maybe:[/quote]
With pdf file you need adobe , with djvu file you need DjVuWebBrowserPlugin :)

Well unfortunatly those books need special authorization...And I can't open the others because I do not have a progarm for .djvu extension. But I think I asked the question about sources for uniform convergence in the wrong area :blush:  I think going back to the topic would be the best thing to do :)  If someone would give a link  in  PM regardinfg my PS (with a common extension -pdf/doc)  would be also great :)
Again sry for such many questions :blush:
So:
[quote="ciprian"]I have a question about the following problem:
Let $f: [0,1]\to \mathbb{R}$ a continuous functin such that $\int_{0}^{1}f(x)g(x)dx=\int_{0}^{1}f(x)dx\cdot\int_{0}^{1}g(x)dx$ for any $g: [0,1]\to \mathbb{R}$ continuous but not derivable. Prove that $f$ is constant.
Well I have a solution but my first idea was this question (maybe stupid :blush: ):
Can we find a sequence of functions $g_{n}$ continuous but not derivable that converge uniformly to $f$? And if so can we take the limit and conclude that $\int_{0}^{1}f(x)^{2}dx={\int_{0}^{1}f(x)dx}^{2}$ i.e. $f(x)=c$?

[/quote]

Nobody? Please tell me if my first argument is silly or if it has some relevance ...

Ciprian,

Here are some observations:

1) if your function $f$ is only continuous then take $g=f$ and you are done.

2) If $f$ is continuous and differentiable then definitely this argument does not work and for this case your approach with the sequence of functions would work.

I think that the answer to your question is yes. Namely it is possible to construct a sequence of continuous and nondifferentiable functions which converge uniformly to $f$. Here is the explanation:

In Royden's book Real analysis on page 50 the following problem can be found.

47. A continuous function $\psi$ on $[a,b]$ is called a polygonal function ( or piecewise linear) if there is a subdivision $a=x_{0}<x_{1}\cdots<x_{n}=b$ such that $\psi$ is linear on each interval $[x_{i}, x_{i+1}]$. Let $f$ be an arbitrary continuous function on $[a,b]$ and $\epsilon$ a positive number. Show that there is a polygonal function $\psi$ on $[a,b]$ with $|f(x)-\psi(x)|<\epsilon$ for all $x\in [a,b]$.

Now if $\epsilon =\frac{1}{n}$ you have your sequence $\psi_{n}$ of functions with the desired property.

Is the following true?

Given any real continuous function defined on a compact interval, there is a sequence $\left\{ f_{n}\right\}_{n \geq 1}$ of continuous nowhere differentiable functions such that $f_{n}\to f$.


calculus
integration
function
geometry
trapezoid
inequalities
real analysis

$\frac{1}{17}\le \frac{1}{1+x^{4}}\le\frac1{x^{4}}$ for $1\le x\le 2.$

The interesting challenge isn't so much finding some estimates as guessing exactly where those particular estimates come from.

The function is decreasing and concave up. Hence any left-hand Riemann sum will be an overestimate and any right-hand Riemann sum will be an underestimate (the $\frac1{17}$ is a right-hand Riemann sum.) Any trapezoidal rule will be an overestimate and any midpoint Riemann sum will be an underestimate. Using the latter and the trivial partition, we get

$\frac{16}{97}\le\int_{1}^{2}\frac{1}{1+x^{4}}\,dx\le\frac{9}{34}$

[color=green][Arithmetic? Bah.][/color]

That should be $\frac{16}{97}$ and $\frac{9}{34}$ in the last line, assuming single-interval trapezoid and midpoint estimates.

The left inequality is done because we obtain:(x.^4)+1>=17
thus the right side is proven for using ::(x.^4)+1>=2*(x.^2).
then : integ(2*(x.^-2))  from:x=1 to x=2 ,is equal to:0.25<7/24
your suprimum must be corrected.
See here:
Stewart.Calculus for single variables (5e)Thomson-2003 -Page 373.