I am trying to learn formal proofs of the properties of the limit.
Is there a proof using Epsilon-Delta?
(a_n and  b_n converge)

Subtract and add (lima_n)b_n and use triangle inequality.


real analysis
function
calculus
derivative
trigonometry
algebra
polynomial

Hi everyone, before my question, let me give you a definition:

[b]DEFINITION:[/b] Let $ f: R \to R$ be a real function.  We say that $ f$ is locally bijective, if for all $ x \in R$ there exists some $ U$ neighborhood of $ x$ such that $ f |_U$ is locally bijective, except for a set of measure zero (Lebesgue Measure).

so, my question is : is there a function of class C^1 which is NOT locally bijective?.

I know if the set of points where the derivative is zero has measure zero then this kind of function is locally bijective, so I could answer the question if I could construct a function which has a non countable set of points where the derivative is zero, but i cannot think in such an example.

If u could answer it if would be very helpful

Adriana

It's "bijective" with a j in English.
Also, the definition should state that the restriction of $ f$ to $ U\setminus X$ is 1-1 (injective) for some neighborhood $ U$ and some measure-zero set $ X$. "Bijective" implies surjection onto its codomain, which is definitely not what we want here.

I don't think you've asked the question you meant to- there are lots of locally bijective $ C^1$ functions, such as $ f(x)\equal{}x$.

A fact which might answer the question you meant to ask: a continuous function which is not strictly monotone is not locally bijective (look at a local maximum or minimum).

Sorry, I made a mistake.  The question should be : is there a function of class C^1 which is NOT locally bijective?.

$ \sin (x)$ for example, under my definition, is locally bijective, because the set which is NOT locally bijective has measure zero (namely $ S\equal{} \{\frac{ \pi}{2}\plus{}k\pi : k \in Z\}$).

Same as $ \cos (x)$, every polynomials, etc.  That's why i was thinking in some example which could have a non-countable set of points where the derivative equals zero.

Please let me know it if you don't understand something (specially the definition that i gave), because my english is not very good.

Thanks

Adriana

How about $ f(x)\equal{}0$ for all $ x$? :P

i thought in that example, in general if $ f(x)\equal{} c$ for all $ x$ it works, but this is kinda trivial, so i was trying in an example when $ f$ is not a constant.

Your definition has quantifiers in the wrong places- I didn't read that "except for a set of measure zero" as being about $ x$.

The definition you meant, as best I understand:
For almost every $ x$, there exists a neighborhood $ U$ of $ x$ such that $ f$ is injective on $ U$.

There are plenty of examples based on constants- any function which is constant on an interval isn't locally bijective.
A more interesting example:
Construct a Cantor-type set $ C$ as follows: Start with $ C_0\equal{}[0,1]$, and let $ S_0\equal{}\left(\frac38,\frac58\right)$, the open middle $ \frac14$ of the interval. Let $ C_1\equal{}C_0\setminus S_0$. For each $ n$, $ C_n$ is a disjoint union of closed intervals; let $ S_n$ be the union of the open middle $ \frac1{2^{n\plus{}2}}$ of each of these intervals, and then let $ C_{n\plus{}1}\equal{}C_n\setminus S_n$.
Define $ C\equal{}\cap C_n$. This $ C$ is a closed set with measure $ \left(1\minus{}\frac14\right)\left(1\minus{}\frac18\right)\cdots\left(1 \minus{}\frac1{2^{n\plus{}2}}\right)\cdots>\frac12$ and empty interior.

Now define a continuous function $ g$ as follows:
$ g(x)\equal{}0$ if $ x\in C$. Otherwise, there is some $ n$ and some subinterval $ (a,b)$ of $ S_n$ such that $ x\in (a,b)$. Let $ g(x)\equal{}(\minus{}1)^n\frac{(x\minus{}a)(b\minus{}x)}{b\minus{}a}$ in this case. We can show that $ g$ satisfies the Lipschitz condition $ |g(x)\minus{}g(y)|\le |x\minus{}y|$, so $ g$ is continuous. Note that for any $ x_0\in C$, $ g$ takes both positive and negative values in any neighborhood of $ x_0$.

Let $ f(x)\equal{}\int_0^x g(t)\,dt$. This $ f$ is $ C^1$, not constant on any interval, and not injective in any neighborhood of a point in $ C$.
If you want a function on all of $ \mathbb{R}$, just extend $ f$ to be periodic with period 1.

yeah, yesterday I saw that if the function was constant is some interval it cannot be locally bijective, but I was trying to think in some example not like that, and yours is perfect, thank you.


geometry
circumcircle
analytic geometry
ratio
vector

Let $ O$,$ I$ and $ H$ denote the circumcenter, incircle and orthocenter of $ ABC$. Prove

a) $ \vec{I}\equal{}\displaystyle\frac{a\vec{A}\plus{}b\vec{B}\plus{}c\vec{C}}{a\plus{}b\plus{}c}$

b) $ OH^2 \equal{} 9R^2\minus{}a^2\minus{}b^2\minus{}c^2$

The first is very easy if you use mass points/barycentrical coordinates, that is, set $ I \equal{} xA\plus{}yB\plus{}zC$ and figureout the ratios $ \frac{x}{y}$ etc. by considering the cevians passing through $ I$.

Second... you could just figure out their coordinates and compute everything the hard way. Not that it would be nice. ;)

How would one go about this problem using vectors? I'm kinda getting stuck when I tried.

Set $ X\equal{}\overline{BC}\cap\overline{AI}$, then we know that $ I \equal{} \lambda X \plus{} (1\minus{}\lambda) A$ for some $ \lambda\in\mathbb{R}$ but we also know that $ X \equal{} \frac{aA\plus{}bB}{a\plus{}b}$ so we get $ (a\plus{}b)I\equal{}\lambda aA\plus{}\lambda bB\plus{}(1\minus{}\lambda)(a\plus{}b)C$. Create similar equations for the other vertices and compare coefficients.

[quote="Tenoreoz"]Let $ O$,$ I$ and $ H$ denote the circumcenter, incircle and orthocenter of $ ABC$. Prove

a) $ \vec{I} \equal{} \displaystyle\frac {a\vec{A} \plus{} b\vec{B} \plus{} c\vec{C}}{a \plus{} b \plus{} c}$

b) $ OH^2 \equal{} 9R^2 \minus{} a^2 \minus{} b^2 \minus{} c^2$[/quote]
[hide="b"]
Yeah, graphing it out seems like a way to solve it (messy perhaps, though):
We call the origin (which will be the circumcenter) $ (0,0)\equal{}O$, $ A\equal{}(a,0)$, $ B\equal{}(b,c)$ and $ C\equal{}(d,e)$, where $ d^2\plus{}e^2\equal{}a^2\equal{}b^2\plus{}c^2$.  Hence, we have that $ H\equal{}(a\plus{}b\plus{}d,c\plus{}e)$.  Hence, $ 9R^2\minus{}AB^2\plus{}BC^2\plus{}AC^2\equal{}9a^2\minus{}[(a\minus{}b)^2\plus{}c^2\plus{}(b\minus{}d)^2\plus{}(c\minus{}e)^2\plus{}(a\minus{}d)^2\plus{}e^2]\equal{}9R^2\minus{}[a^2\plus{}b^2\minus{}2ab\plus{}c^2\plus{}b^2\minus{}2bd\plus{}d^2\plus{}c^2\plus{}e^2\minus{}2ce\plus{}a^2\plus{}d^2\minus{}2ad\plus{}e^2]\equal{}9R^2\minus{}[a^2\plus{}a^2\minus{}2ab\plus{}a^2\minus{}2bd\plus{}a^2\minus{}2ce\plus{}a^2\plus{}a^2\minus{}2ad]\equal{}3a^2\plus{}2ab\plus{}2bd\plus{}2ce\plus{}2ad\equal{}[a^2\plus{}b^2\plus{}d^2\plus{}2ab\plus{}2bd\plus{}2ad]\plus{}[c^2\plus{}e^2\plus{}2ce]\equal{}(a\plus{}b\plus{}d)^2\plus{}(c\plus{}e)^2\equal{}OH^2$
[/hide]

how do you get $ X \equal{} \displaystyle\frac {aA \plus{} bB}{a \plus{} b}$?

From what I see, $ X \equal{} \frac {BX}{BC}\cdot C \plus{} \frac {XC}{BC}\cdot B$.

EDIT: ah i see now  :)

[hide="2"]
Set the circumcenter to be the origin, then the orthocenter $ {\bf h}$ is given by $ {\bf a\plus{}b\plus{}c}$ (one way to verify this is to note
\[ {\bf (h\minus{}a)\cdot (b\minus{}c)\equal{}(b\plus{}c)\cdot (b\minus{}c)\equal{}|b|^2\minus{}|c|^2}\equal{}R^2\minus{}R^2\equal{}0\]
etc.) We then have
\[ OH^2\equal{}{\bf |h|^2\equal{}|a\plus{}b\plus{}c|^2\equal{}|a|^2\plus{}|b|^2\plus{}|c|^2\plus{}2(a\cdot b)\plus{}2(b\cdot c)\plus{}2(c\cdot a)}\]
\[ \equal{}{\bf 3|a|^2\plus{}3|b|^2\plus{}3|c|^2\minus{}(|b\minus{}c|^2\plus{}|c\minus{}a|^2\plus{}|a\minus{}b|^2)}\]
\[ \equal{}9R^2\minus{}(a^2\plus{}b^2\plus{}c^2)\]
[/hide]


The value you look for is $ \minus{} h$, and comes from having $ g(k) \equal{} 0$, i.e. $ f(k) \equal{} \minus{} h$.

However, this may very well be the maximum value of $ f$, for example if $ f$ were $ f(x) \equal{} \minus{} (x \plus{} 4)(x \plus{} 1)(x \minus{} 1)(x \minus{} 2)$
(it is a minimum if the leading coefficient of $ f$ is positive, and a maximum if the leading coefficient is negative).
Also, to be completely correct, $ k$ is the only root of $ g$ only in value (in fact is a double root, also said a root of multiplicity $ 2$).

Just sketch the possible graph for $ f$ and you'll understand.


trigonometry
inequalities
inequalities unsolved

Let $ a,b,c$ be real numbers. Prove the inequality $ \frac {\cos^2a}{\sin^2(b \minus{} c)} \plus{} \frac {\cos^2b}{\sin^2(c \minus{} a)} \plus{} \frac {\cos^2c}{\sin^2(a \minus{} b)}\geq 2$

I think it's post somewhere before, but i cannot find.


trust me on this: the easiest way is to count, but count systematically or else it will take a while.

i think the easiest way to do this is to count it out.

[quote="shinwoo"]how do you solve this kind of problem without actually counting?

How many combinations of pennies,nickels,and dimes, are there with a total value of 25cents ?[/quote]

$p+5n+10d=25$.

If $d=0$, then $p+5n=25$. 6 possible values of $n$.

If $d=1$, then $p+5n=15$. 4 possible values of $n$.

If $d=2$, then $p+5n=5$. 2 possible values of $n$.

$6+4+2=\fbox {12}$.

Complete list, where $p$ is the value of all the pennies, $n$ is the value of all the nickels, etc.

$(p,n,d)=$
$(25,0,0),(20,5,0),(15,10,0),(10,15,0),(5,20,0),(0,25,0)$
$(15,0,10),(10,5,10),(5,10,10),(0,15,10)$
$(5,0,20),(0,5,20)$.

See? You don't have to count them all! You just start with $d=0$, until $d$ is maximized. Then, for each $d$, you see how many values for $n$ there are. Then you add them all up.

[quote=chess64 wrote]If d=0, then p+5n=25. 6 possible values of n.

If d=1, then p+5n=15. 4 possible values of n.

If d=2, then p+5n=5. 2 possible values of n. [/quote]

that's same thing as counting isn't it?
how did you know right away there were 6 values, 4 values, and 2 values for each n;s?

[quote="shinwoo"][quote="chess64"]
If d=0, then p+5n=25. 6 possible values of n.

If d=1, then p+5n=15. 4 possible values of n.

If d=2, then p+5n=5. 2 possible values of n. [/quote]

that's same thing as counting isn't it?
how did you know right away there were 6 values, 4 values, and 2 values for each n;s?[/quote]

Because if $d=0$, then $n$ can be any number from 0 to 5. That's 6 numbers. Same thing with the others. It's not the same as counting  ;)

To find the number of solutions of

\[a_1+a_2+a_3+\cdots+a_m=n\]

is $_{(n+m-1)}C_{n}$.

This formula doesn't apply to this problem, but you can use generating functions...
[hide="Dont look... too hard for this forum..."]
We wish to find the number of solutions to

\[a+5b+25c=n\]

$C(z)=(1+z+z^2+\cdots+z^n+\cdots)(1+z^5+z^{10}+\cdots+z^{5n}+\cdots)(1+z^{25}+z^{50}+\cdots+z^{25n}+\cdots)$
$C(z)=\frac{1}{1-z} \cdot \frac{1}{1-z^5} \cdot \frac{1}{1-z^{25}}$
Then you would somehow express that as some sums, but I'm not sure how yet. I just started studying generated functions... :?
[/hide]

me neither  :lol:  there must be someone in here who does, perhaps you, the reader, can help us maybe...?  :P  :help:

I'm puzzled....  :huh: I really shouldn't have clicked. =/

why doesn't someone post this in the highschool or olympiad forums :).  im sure they'll be able to do this a+5b+25c=n problem quickly as ive heard its easy... just dont know how to do it myself  :(

[quote="Barnacle"]why doesn't someone post this in the highschool or olympiad forums :)[/quote]Wish granted :D

Okay I think I know how to do this now :D

The generating function is $(1+x+x^2+x^3+\cdots)(1+x^5+x^{10}+x^{15}+\cdots)(1+x^{10}+x^{20}+x^{30}+\cdots)$
$=\frac{1}{1-x} \cdot \frac{1}{1-x^5} \cdot \frac{1}{(1+x^{5})(1-x^{5})}$
$=\frac{1}{(1-x)(1-x^5)^2(1+x^{5})}$

We need to find the coefficient of $25.$ I can't figure out how to do that part though....
 :help:

$(1 + x + x^2 + x^3 + ... + x^{25})(1 + x^5 + x^{10} + x^{15} + x^{20} + x^{25})(1 + x^{10} + x^{20})$

Expand that and the coefficient of $x^{25}$ should be the number of combinations.  I've only read a little about generating functions so I'm not really sure how to simplify that expression.  chess64, could you explain how you changed it into those fractions?

[quote="dakyru"]chess64, could you explain how you changed it into those fractions?[/quote]

The sum of infinite series formula.

[quote="furious"][quote="dakyru"]$(1 + x + x^2 + x^3 + ... + x^{25})(1 + x^5 + x^{10} + x^{15} + x^{20} + x^{25})(1 + x^{10} + x^{20})$[/quote]
supposedly, it's easier to solve if you don't limit your generated function and let it go on the infinity so you can create infinite sums.
but like chess64, i don't know how to find the $x^{25}$ term either[/quote]

Well think about it.  Which is easier to work with:  $\frac{x^{30}-1}{x^5-1}$ or $\frac{1}{x^5-1}$?  The second one obviously is and that is why it is preferable to let your generating function go on infinitely rather than truncating it [i]when possible[/i] (sometimes it isn't logical to let it go on infinitely due to restrictions in the problem).

Although you can model the problem with gf's I'm pretty sure that there is no easy/elementary way to find the desired coefficient.  I would agree with Hamster1800 on this one.

yeah DPatrick's formula doesn't relaly owrk in this case... brute expansion and serching for pattern is prolly best


geometry
incenter
circumcircle
projective geometry
geometry proposed

The triangle $ABC$ has incenter $I$. A circle lies inside the circumcircle and touches it. It also touches the sides $CA$, $BC$ at $D$, $E$ respectively. Show that $I$ is the midpoint of $DE$.

[color=darkblue][size=134] :!: [b]An usual lemma.[/b][/size] The circle $C_{i}$ is interior tangent in the point $T$ to the circle $C_{e}$. For any point $M\in C_{i}$ denote the intersections $X,Y$
between the tangent in $M$ to the circle $C_{i}$ with the circle $C_{e}$. Then the ray $[TM$ is the bisector of the angle $\widehat{XTY}\ .$

[b]Proof of the proposed problem.[/b] Denote the interior circle $C_{i}$ which is tangent in the point $T$ to the circumcircle of the given triangle  and is tangent to the sidelines $CA$, $CB$ in the points $D$, $E$ respectively. Using the above lemma results that the intersections $M\in BI\cap TD$ and $N\in AI\cap TE$ belong to the circumcircle of the triangle $ABC$. Apply the Pascal's theorem to the hexagon $BCANTM$ : the intersections $I\in MB\cap AN$, $D\in CA\cap TM$, $E\in NT\cap BC$ are collineary, i.e. the point $I$ belongs to the line $DE$ and $CD=CE$ $\Longrightarrow$ $ID=IE\ .$[/color]


Hi everyone,

I need to determine the characteristics of this PDE to classify it and then put it in canonical form to show that its solution is of the form:

u(x,t) = x f(x-t/k) + g(x-t/k)

uxx + 2k uxt + k^2 utt = 0

I tried to solve it. I think it's parabolic. But my problem is that I don't know what transformation to use to put it to canonical form. I appreciate any help.

Thanks

[quote="V.V."][quote="AngelaS"]Hi everyone,

I need to determine the characteristics of this PDE to classify it and then put it in canonical form to show that its solution is of the form:

u(x,t) = x f(x-t/k) + g(x-t/k)

uxx + 2k uxt + k^2 utt = 0

I tried to solve it. I think it's parabolic. But my problem is that I don't know what transformation to use to put it to canonical form. I appreciate any help.[/quote]

$1\cdot dt^{2}-2kdxdt+k^{2}dx^{2}=0$ - equation for characteristics.
Hence, $(dt-kdx)^{2}=0$ and $t=kx$ is characteristic.

We have $D=a_{11}a_{22}-a_{12}^{2}=0$. Thus it's parabolic equation.

Let $\alpha=x-t/k$, $\beta=x$. ($\beta$ is arbitrary function, non dependent by $\alpha$).

We have $u_{\beta\beta}=0$.
Then $u=A(\alpha)\beta+B(\alpha)$, where $A$ and $B$ - arbitrary functions.[/quote]

Thanks a lot V.V. But here you know that the U should be in the form of
u(x,t) = x f(x-t/k) + g(x-t/k) so you choose alfa as x-t/k and beta as x. But what if we didn't know this? How can we find the form of the transformation?
I also don't understand why  $u_{\beta\beta}=0$ and how you conclude that  $u=A(\alpha)\beta+B(\alpha)$ ?
Thanks again

[quote="AngelaS"]
But here you know that the U should be in the form of
u(x,t) = x f(x-t/k) + g(x-t/k) so you choose alfa as x-t/k and beta as x. But what if we didn't know this? How can we find the form of the transformation?
I also don't understand why  $u_{\beta\beta}=0$ and how you conclude that  $u=A(\alpha)\beta+B(\alpha)$ ?
Thanks again[/quote]

We solved char. equation. $\alpha$ is characteristic.
From there $\alpha=x-t/k$.

$u_{\beta\beta}=0$
Then $u_\beta=A(\alpha)$.
Then $u=A(\alpha)\beta+B(\alpha)$.


function
geometry
rectangle
real analysis
integration
real analysis unsolved

Suppose $ F$ is a measure on a closed rectangle, $ B \subset \mathbb{R}^N$.  Is it necessarily the case that for a sequence of measures $ F_n$ (on the Lebesgue-measurable sets of $ B$) such that $ DF_n \rightarrow DF$ in the $ \sup$ norm (where $ DG$ denotes the distribution function $ DG(x) \equal{} G(\left\{p \in B\; |\; p_i \leq x_i \forall 1 \leq i \leq n \right\})$), $ \int_B f dF_n \rightarrow \int_B f dF$ for all $ C^{\infty}$ functions, $ f$?

What if $ F$ is absolutely continuous or $ C^{\infty}$?

Assuming that $ F$ is finite and all measures are positive, the answer is "yes". Just note that continuous functions can be uniformly approximated by finite linear combinations of characteristic functions of half-open rectangles.


There are many circles on a plane. Each is divided into four parts by two mutually perpendicular diameters. Each part is painted either red, yellow or blue. No matter how the circles are rotated in the plane, they are different from one another. At most how many circles are painted with all three colors?

[hide="semi brute force"]
Split into three categories based on which color is doubled:
Circle sections in clockwise order:
[two reds]
RRYB
RYRB
RRBY
[two blues]
BBYR
BYBR
BBRY
[two yellows]
YYBR
YBYR
YYRB

Totals to nine.

[/hide]

It is slightly faster to just consider what arrangements of colors there are, then count the actual red, yellow, blue afterward. I haven't looked at the details of your solution yet, so I don't know how correct it is, but it seems ok.


function
calculus
derivative
real analysis
real analysis theorems

[b]Show that[/b]
\[ f(xy, z - 2x) = 0
\]
[b]satisfies, under certain conditions, the equation[/b]
\[ \#x{\frac {\partial \#z}{\partial \#x}} - \#y{\frac {\partial \#z}{\partial \#y}} = 2x
\]
[b]What are these conditions?[/b]

First of all I rewrite the equations to make clear what is a function and what is a variable (place,point, etc.).

Let $ u(x,y,z): \equal{}xy$, $ v(x,y,z): \equal{}z\minus{}2x$.
Furthermore $ f: \mathbf{R}^2 \to \mathbf{R}$.

Let $ w: \equal{}(u,v)$, $ w: \mathbf{R}^3\to\mathbf{R}^2$.

Let $ F: \equal{}f\circ w$ be a composite function.

If I understand well, then $ \frac{\partial z}{\partial x}\equal{}D_1 F$, where $ D_1$ is the usual differential operator.

If the above is correct, then you can calculate the desired partial derivatives by the chain rule, or by computer algebra system.
So what you need is the existence of partial derivatives of $ f$.


I;m interested in a diffrent proof than mine...

http://www.mathlinks.ro/viewtopic.php?p=1041567#1041567

It's not necessary to post the same problem three times  :mad:

Domi

i've posted in 3 places because nowone posted a proof and I need another one diffrent than mine  :wink:


Here is a simmilar problems.. perhaps it will help! :D http://www.mathlinks.ro/phpBB/viewtopic.php?t=287


geometry
3D geometry
ceiling function
inequalities

I don't get this problem at all but it looks to me like the answer would be $\lceil x \rceil$ where x is found through this equation $2^{x}=509$

[hide]The number of different cones, if he uses x scoop cones is $2^{x}$, since each of the x scoops can be either vanilla or chocolate. We need at least 509 different cones. The smallest x such that $2^{x}\ge 509$ is $9$.[/hide]

[quote="wee"]My older brother told me the answer the problem below was 9 but he won't tell me how he got it. I worked backwards and I still could not figure it out. Any help will be great. Thank you

There are 509 students.  Each student desires a
different ice cream cone, but the same number of
scoops.  Vanilla and chocolate are the only two ice
cream flavors on sale.  Ike knows he can make four
different cones, if he uses double scoop cones. How
many scoops of ice cream is needed to make a different
cone with an equal number of scoops for each student?[/quote]

[hide="solution"]The answer would be $2^{x}\ge 509=\boxed{9}$ because there are 2 kinds of scoops, and the x amount of scoups can be selected from the flavors of vanilla or chocolate. There are 509 students, so you need 509 cones. I hope this helps.  :) [/hide]

anirudh: How do you have an inequality and an equality in the same expression?

I think (and I might be wrong) that he means that the answer to the inequality is nine.  :wink:

[quote="wee"]one question, how did you get that logarithm. I understand the 2 but how did you come up with 2^x?[/quote]
Let's make a sequence of $x$ terms, using the letters C (chocolate) and V (vanilla). For example, if $x=3$, then the sequence "CVV" corresponds to chocolate on the top, vanilla in the middle, and vanilla on the bottm. For each of the $x$ terms in the sequence, we have two choices: C or V. Since we need to make this choice $x$ times to construct a sequence (and each sequence made in this way is different) the number of possible sequences is $2 \cdot 2 \cdot 2 \cdot \dots \cdot 2$ (with $x$ 2's), or $2^{x}$.


Use the Intermediate Value Theorem to show that each polynomial function has a zero in the given interval.

NOTE: What is the [b]Intermediate Value Theorem [/b]all about anyway?  Can someone explain it to me in simple terms?

(1) f(x) = 3x^3 - 10x + 9; [-3, -2]

(2) f(x) = x^5 - 3x^4 - 2x^3 + 6x^2 + x + 2; [1.7, 1.8]

The Intermediate Value Theorem basically states that if a continuous function $f(x)$ has the property that $f(a)$ and $f(b)$ have different signs (one is positive, one is negative) then $f(c) = 0$ for some $c$ between $a, b$.

(Edit:  The full statement is that for $c \in (a, b)$ a continuous function $f(x)$ must take on every value $f(a) < f(c) < f(b)$.)

That is because continuous functions can't "cheat."  You need to be able to draw a line from some point below the $x$-axis to some point above the $x$-axis, and if you don't lift your pencil then you have to pass through the $x$-axis.  That's the basic idea.

[i]Why is your teacher not telling you these things?[/i]

Can you at least solve the first one?

Give me a guide to use.

[b]One.[/b]  $f(-3) =-42$ but $f(-2) = 5$.  The sign is different, so somewhere in between there must be a zero.

Why did you select -3 and -2?  Why those two particular negative numbers?

You also said:

"The sign is different, so somewhere in between there must be a zero."

This statement is [b]NOT [/b]clear.  What does it mean that there is a zero in between -3 and -2?

Do you mean a [color=red]HOLE[/color] in the graph of the function?

[quote="Interval"]Why did you select -3 and -2?  Why those two particular negative numbers?[/quote]

They're... given in the problem.

[quote="Interval"]What does it mean that there is a zero in between -3 and -2?[/quote]

A zero is a root.  In other words, if $c$ is a zero of the function $f(x)$ then $f(c) = 0$.

I mean that there exists some number $c$ such that $-3 < c <-2$ and such that $f(c) = 0$.  That is the content of the Intermediate Value Theorem.

[i]Again, why is your teacher not doing his/her job and teaching you?[/i]

I thank you.

To answer your question about [color=red]MY TEACHER[/color], here is the answer:  [color=green]THERE IS NO TEACHER.[/color]

I am 40 years old preparing (on my own) for an upcoming math state exam to get my license for full-time teaching.  Okay?

Ah.

I apologize for any rudeness on my part that that particular misconception may have caused.  :oops:


Let $ ABC$ be an acute triangle, $ AD$, $ BE$, $ CF$ its altitudes, with $ D$ in $ BC$, $ E$ in $ CA$, and $ F$ in $ AB$. Let $ M$ be the midpoint of $ BC$. Circumcircle of $ AEF$ meets $ AM$ at $ A$ and at $ X$. $ AM$ meets $ CF$ at $ Y$. $ AD$ meets $ BX$ at $ Z$. Prove that $ YZ$ is parallel to $ BC$.

Denote by $ T$ the intersection of $ BX$ and $ CY$. Since the segments $ EF$ and $ BC$ are antiparallel and since the $ A$-symmedian of triangle $ AEF$ is the locus of the midpoints of the segments intercepted by the antiparallels to $ EF$ with $ AB$ and $ AC$, it follows that $ AM$ is the $ A$-symmedian of $ AEF$. In this case, the quadrilateral $ AEXF$ is harmonic, and thus
$ (YZ \cap BC, B, M, C)\equal{}Y(Z,B,X,Y)\equal{}(Z,B,X, T)$
$ \equal{}H(Z,B,X, T)\equal{}H(A,E,X,F)\equal{}(A,E,X,F)\equal{}\minus{}1$. We conclude that $ YZ \cap BC \equal{} \infty$, i.e. $ YZ \| BC$.

Denote H , O  is the orthocentre and circumcenter of △ABC , respectively
P is midpoint of AH , Q is the center of nine-point circle
MP is the diameter of nine-point circle
MF^2=MX*MA=MB^2  => △BMX∽△AMB  => ∠BXM=∠ABC => YXHZ are concyclic  =>
YZ⊥AD => YZ∥BC


function
quadratics
algebra
rational function
complex analysis
complex analysis unsolved

let $ c_{n}$ be the nth fibonacci no. show that the $ c_{n}$ are taylor coefficents of a rational function and determine a closed expression for $ c_{n}$

the closed expression is well known....so i thought the fn defined by $ f(z)=\sum_{n=0}^{\infty}\((c_{n}.z^{n})$ will have $ c_{n}$'s as its taylor coeff in the disk of convergence. by the way can someone help me with the radius of convergence of that series? i computed and got $ \frac{2}{1+\sqrt5}$... i doubt if thats correct..the qn asks for a rational fn.. i'm not sure what to do. i'll be glad to have some hints.thanks for any help

That's the right radius of convergence. The denominator is a quadratic polynomial.

thanks for your help jmerry. how did you guess that the denominator was a quadratic? after posting the problem here i worked on it again.. and did this

let $ S\equal{}c_{0}\plus{}c_{1}z\plus{}c_{2}z^{2}\plus{}...$

 then i did $ \minus{}zS$ and -$ z^{2}S$ and added them and got the series as a rational function. but i dont know how to justify......

It wasn't a guess- I've seen it before. Also, it's a standard example of a standard generating function method- converting a recursion into an equation involving the function.


function
probability
integration
calculus
calculus computations

1. Suppose that X1 and X2 are independent r.v, each having density function f(x)=2x, 0<x<1. 0, otherwise
Find the cdf of the random variable Z=X1+X2. Use the result to obtain the density function of

Hmm... exactly the material I'm doing in my class this semester (Chapter 6 of the Wackerly, Mendenhall, Sheaffer textbook.)

The joint density is $f(x_{1},x_{2})=4x_{1}x_{2}$ on the square $[0,1]\times[0,1].$

Now we need to start drawing pictures with that square. We're trying to compute $P(Z\le z)=P(X_{1}+X_{2}\le z)$ There are four cases. Draw the diagonal line $x_{1}+y_{1}=z;$ we're trying to find the probability of being below and to the left of this line.

If $z\le 0,$ then this probability is zero. If $z\ge 2,$ then this probability is one. Then there are two cases that need specific computation.

If $0\le z\le 1,$ then

$P(X_{1}+X_{2}\le z)=\int_{0}^{z}\int_{0}^{z-x_{1}}4x_{1}x_{2}\,dx_{2}\,dx_{1}$

$=\int_{0}^{z}2x_{1}(z-x_{1})^{2}\,dx_{1}= \int_{0}^{z}2z^{2}x_{1}-4zx_{1}^{2}+2x_{1}^{3}\,dx_{1}$

$=z^{4}-\frac43z^{4}+\frac12z^{4}=\frac16z^{4}.$

Now suppose $1\le z\le 2.$ We'll calculate the complement:

$P(X_{1}+X_{2}>z)= \int_{z-1}^{1}\int_{z-x_{1}}^{1}4x_{1}x_{2}\,dx_{2}dx_{1}$

$=\int_{z-1}^{1}2x_{1}\left(1-(z-x_{1})^{2}\right)\,dx_{1}= \int_{z-1}^{1}2x_{1}-2x_{1}z^{2}+4x_{1}^{2}z-2x_{1}^{3}\,dx_{1}$

$\left(1-z^{2}+\frac43z-\frac12\right)-\left((z-1)^{2}-(z-1)^{4}+\frac43(z-1)^{4}-\frac12(z-1)^{4}\right)$

(more details later)

By some details I'd rather not write out (there's some messy calculus going on here), I get that

$P(Z\le z)=\begin{cases}0&\text{for }z\le 0\\ \frac16z^{4}&\text{for }0\le z\le 1\\ 1-\frac83z+2z^{2}-\frac16z^{4}&\text{for }1\le z\le 2\\0&\text{for }z\ge 2\end{cases}$

Differentiating, we get a density function

${f(z)=\begin{cases}\frac23z^{3}&\text{for }0\le z\le 1}\\-\frac83+4z-\frac23x^{3}&\text{for }1\le z\le 2\\ 0&\text{otherwise}\end{cases}$


[quote="cezar lupu"]If $a,b,c$ are positive real numbers such that $a+b+c\geq abc.$ Then the following inequality holds:

                                              $a+b+c\geq 3sqrt3$[/quote]

Take a = 1 and b and c close enough to 0 for a counterexample.

  Darij

And reversed ineq. is not true either, take $a=\frac{4}{3},b=2,c=2$

[quote="cezar lupu"]If $a,b,c$ are positive real numbers such that $a+b+c\geq abc.$ Then the following inequality holds:

                                              $a+b+c\geq 3sqrt3$[/quote]

It should be If $a,b,c$ are positive real numbers such that $a+b+c\leq abc.$ Then the following inequality holds:

                                              $a+b+c\geq 3\sqrt3$. But it's trivial

Namdung,yes you're right.The inequality is trivial. The interesting case is when $a+b+c=abc$ but this is also trivial.


In the "Complicated Equations" part of the LaTeX tutorial, you contrast the following bits of LaTeX code:

[code]\documentclass[11pt]{article}
\begin{document}
Since each $x_i$ can range from 1 to 12, the generating function for each is $x + x^2 + x^3 + \cdots + x^{12} = x(1+x+x^2+\cdots+x^{11}) =
\frac{x(1-x^{12})}{(1-x)}$.
\end{document}[/code]

and

[code]\documentclass[11pt]{article}
\begin{document}
Since each $x_i$ can range from 1 to 12, the generating function for each is $x + x^2 + x^3 + \cdots + x^{12} = x(1+x+x^2+\cdots+x^{11}) = \frac{x(1-x^{12})}{(1-x)}$.
\end{document}[/code]

I realize that in the latter one, there is supposed to be a carriage return between "the generating function for each is" and "$x + x^2 + ...".  However, it isn't apparent on the page in question, probably because we reach the end of a line anyway at the word 'is'.

Interesting.  Looks fine on my page.  Hmm...  Have to think about how to prevent the line break there.

Thanks for pointing this out; keep 'em coming.

"comiple" in the Maxwell's Equations example paragraph should be "compile"
http://www.artofproblemsolving.com/LaTeX/AoPS_L_BasicExam.php


calculus
integration
inequalities
triangle inequality

if the length of 2 sides of an obtuse triangle with integral sides are 4 and 6, find the sum of all the possible values of a third side

Call the third side $x$.  By triangle inequality, we must have

$2<x<10$ and since it is obtuse we must have

$x^{2}<20$ or $x^{2}>52$ from here we see that the only integral solutions are $3, 4, 8$ and $9$, so the sum is $24$

[quote="Fanatic"]I basically did this problem the same way jhcreinhardt did it, but I thought that since the numbers were so small, I might as well use a logical guess and check approach, checking if they met the criteria, rather than laying out the inequalities and seeing which numbers met the criteria.[/quote]

While this may be quicker in some cases and perhaps in this one, the purpose of posting solutions is to demonstrate the proper way to solve the probelms, giving readers of the post a better general knowledge of the problem and its background.

Yes, but since you already posted the solution, there's no point for me to post the exact same thing. The only reason I even bothered to post my "solution" was because I was responding to imaretard and I didn't want my post to be completely off topic.

Umm, back to the topic. In an obtuse triangle $a^{2}+b^{2}<c^{2}$ where a,b, and c are the sides and a and b are the short ones. No one stated this fact although most think it is obvious.[/hide]


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $H$ be a maximal subgroup of a group $G$ and $H \triangleleft G$ . Prove that $G/H$ is a cyclic group of prime order .

 :)

$G/H$ cannot contain any nontrivial subgroups otherwise by corresopndence theorem it contradicts the maximality of H. The result follows.

[quote="Soarer"] corresopndence theorem [/quote]

???  :?

For a normal subgroup $H$ of $G$, subgroups of $G/H$ and subgroups of $G$ containing $H$ has a one-to-one correspondence.


Consider the following two complexes: $ [Fe(ox)_{3}]^{3-}$ and $ [Ru(ox)_{3}]^{3-}$.

a) What are their geometries?
b) Write their names (don't forget to write the Shock number).
c) What is the coordination number of each?
d) Write their most probable electronic configurations.

a) Both of them must be octahedral
b) (I think it is Stock number notation explaining the oxidation state)
c) Obviously 6 for both
d) I think Fe pairing doesnt take place and it has d5 and sp3d2 whereas in Ru pairing takes place and we get d2sp3

[quote="Carcul"]The Shock number is the oxidation state of a metal written in front of it, between brackets, and in Roman numerals. Their names?[/quote]

As Chemrock said its called stock number here  :D


quadratics
algebra
quadratic formula
algebra unsolved

Solve the equation: \[ 8x^2\plus{}10x\plus{}3\equal{}\frac{1}{4x^2\plus{}7x\plus{}3}\]

hello, $ x_1\equal{}\minus{}\frac{3}{4}\minus{}\frac{1}{8}\sqrt{2\plus{}2\sqrt{33}}$,
$ x_2\equal{}\minus{}\frac{3}{4}\plus{}\frac{1}{8}\sqrt{2\plus{}2\sqrt{33}}$
Sonnhard.

Hi, thank you for your help. But I need a concrete proof. Thanks a lot.

[hide]
$ (8x^2\plus{}10x\plus{}3)(4x^2\plus{}7x\plus{}3)\equal{}1$

$ 24(x^2\plus{}\frac{10}{8}x\plus{}\frac{3}{8})(x^2\plus{}\frac{7}{4}x\plus{}\frac{3}{4})\equal{}1$

$ ((x\plus{}\frac{5}{8})^2\minus{}\frac{1}{64})((x\plus{}\frac{7}{8})^2\minus{}\frac{1}{64})\equal{}\frac{1}{24}$

Let $ y\equal{}x\plus{}6/8$

$ ((y\minus{}\frac{1}{8})^2\minus{}\frac{1}{64})((y\plus{}\frac{1}{8})^2\minus{}\frac{1}{64})\equal{}\frac{1}{24}$

$ (y^2\minus{}\frac{1}{4}y)(y^2\plus{}\frac{1}{4}y)\equal{}\frac{1}{24}$

$ y^4\minus{}\frac{1}{16}y^2\minus{}\frac{1}{24}\equal{}0$

$ 48y^4\minus{}3y^2\minus{}2\equal{}0$

I'm not one to bash, but you get eq. in y^2 with quadratic equation, which solves for 2 solutions of y, which gives 4 solns of x( not necessarily distinct.[/hide]

Wow, I feel stupid. And tired. The latter is why I did that. Sry.

Solve the equation :)
$ \sqrt [3]{24x^2 \plus{} 2x \plus{} 1993} \equal{} x^3 \minus{} 69x^2 \plus{} 674x \minus{} 5383$
It makes I am remembered ...  :)

if you factorise it in the beginning you have

(4x + 3)(2x + 1) = 1/(4x + 3)(x + 1)

multiplying you have

(4x + 3)(2x + 1)(4x + 3)(x + 1)= 1

which is

(4x + 3)(2x + 1)(4x + 3)(x + 1) - 1=0

then you could have

y = (4x + 3)

and plug that back in the original equation

which could make it easier

[quote="April"]Solve the equation:
\[ 8x^2 + 10x + 3 = \frac {1}{4x^2 + 7x + 3}
\]
[/quote]
Here goes a complete solution :)

Setting, $ y=4x^2+5x$ we get,
\[ 8x^2+10x+3=\frac{1}{4x^2+7x+3}
\\
\Leftrightarrow 2y+3=\frac{1}{y+2x+3}
\\
\Leftrightarrow 2y^2+(4x+9)y+(6x+8)=0
\]
Using quadratic formula for, $ y$ we get,
\[ y=\frac{-(4x+9) \pm \sqrt{(4x+9)^2-8(6x+8)}}{4} \]

Putting the value of $ y$ and then after some calculation we get,
\[ 16x^2+24x+9=\pm \sqrt{16x^2+24x+17} \]
Here, $ 16x^2+24x+9=(4x+3)^2>0$
So, \begin{eqnarray}16x^2+24x+9=\sqrt{16x^2+24x+17}\end{eqnarray}
(Here I  assumed that we want the solution for $ x \in \mathbb{R}$, if it is not the case we can do in the same way)
Setting $ (4x+3)^2=z$ in $ (1)$we get,
$ z=\sqrt{z+8}$
Solving this we get,
$ z=\frac{1+\sqrt{33}}{2}$
And putting the value of, $ z$ we get,
$ x=\frac{-3 \pm \sqrt{\frac{1+\sqrt{33}}{2}}}{4}$

:o  :(  :maybe:
I can't say anything but Wow!!! just have a look here and put in the box [code]8x^2 + 10x + 3 - 1/(4x^2 + 7x + 3)=0 [/code][url]http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=equations&s2=solve&s3=basic[/url]
It solved the equation within a sec....huh...so why did i work for solving it??
 :ninja:


geometry
incenter
circumcircle
geometry solved

Maybe your problem is wrong,I think A' should be the circumcenter of the triangle IBC,etc.

[quote="marko avila"]Let triangle ABC  have incenter I . Let A'  the incenter of triangle IBC,  B' of IAC and C' of  IAB  prove that ABC  and A'B'C' have the same circumcenter[/quote]

Indeed, this is wrong; however, if you actually wanted to assume A', B', C' to be the circumcenters of triangles IBC, IAC and IAB (and not incenters), then it is trivial: According to http://www.mathlinks.ro/Forum/viewtopic.php?t=6095 post #2, the midpoint of the arc BC on the circumcircle of triangle ABC is the circumcenter of triangle BIC. In other words, the circumcenter A' of triangle BIC is the midpoint of the arc BC on the circumcircle of triangle ABC. Hence, the point A' lies on the circumcircle of triangle ABC. Similarly, the points B' and C' lie on this circumcircle as well, and consequently, the triangles ABC and A'B'C' have the same circumcircle, and thus the same circumcenter.

  darij

no wonder i couldnt solve it!!!!! very sorry darij and thanks!!! :blush:


I looking to construct (in Geo. Sketchpad) a Steiner Chain with 4 circles chained around the inner circle.  Also, the inner circle and the outer circle are NOT concentric.

I know inversions are involved, but I was hoping to get a little more detailed help on this one.

Any thoughts?

Invert wrt a circle centered at a limit point and you get concentric circles.  From here it's easy to construct the inverses of the desired circles, and then invert back.

Check out the "Steiner Circles" page in the example sketch that comes with Sketchpad, located in

Sketchpad/Samples/Sketches/Geometry/Example Theorems.gsp

This may have more than what you need, but you can try "Show All Hidden" and use Object Properties to explore how the construction is put together.

Best,

Nick Jackiw
Sketchpad Projects
KCP Technologies

Nick,

Thanks for the tip.  I found that file in Sketchpad.  That should help me get through my problem.


Prove that $ arctan(1)\plus{}arctan(2)\plus{}arctan(3)\equal{}180$ degrees given only:

$ arctan(1)$ is the angle opposite 1 in a right triangle with catheti $ 1$ and $ 1$

$ arctan(2)$ is the angle opposite 2 in a right triangle with catheti $ 1$ and $ 2$

$ arctan(3)$ is the angle opposite 3 in a right triangle with catheti $ 1$ and $ 3$

Using No Trigonometry.

Is this what you're looking for?

[hide][img]http://img401.imageshack.us/img401/997/proofkt6.gif[/img][/hide]


limit
logarithms
calculus
calculus computations

$ 1\blacktriangleright\ \ \lim_{n\to\infty} n^{a_n} \equal{} 1\ ,$ where $ a_n \equal{} \frac {1}{\sqrt [n] {n!}}\cdot\left(\frac 11 \plus{} \frac 12 \plus{} \ldots \plus{} \frac 1n\right)\ .$

$ 2\blacktriangleright \ \ \lim_{n\to\infty}n^{b_n} \equal{} e\ ,$ where $ b_n \equal{} \frac {\ln n!}{n\cdot\left(\frac 11 \plus{} \frac 12 \plus{} \ldots \plus{} \frac 1n\right)^2}\ .$

because for big $ n$ we have $ a_{n}$~$ \frac {e\ln n}{n}\minus{}>0$ after is simpl.
and in seccond we have
$ b_{n}$~$ \frac {\ln \frac {n}{e}}{(\ln n )^{2}}$
therefore we have
$ n^{b_{n}}$~$ e^{\frac {(\ln n )(\ln \frac {n}{e})}{(\ln n)^{2}}} \minus{} > e$ obviusly


Given a regular 2010 polygon. Let $ A,\ B,\ C$ be the distinct vetices of the polygon. Find the number of triangle such that all internal  angles of the triangle are some integer times $ 1^\circ$. Note that any pairs of rearrangements of $ A,\ B,\ C$ are considered as same pair.

[hide]For a regular polygon with $ n$ sides and distinct vertices $ A,\ B,\ C, \ \angle ABC \equal{} \frac {180k}{n},$ for some positive integer $ k \le n \minus{} 2.$ Inserting $ n \equal{} 2010$ gives
$ \frac {180k}{2010} \equal{} \frac {6k}{67}$. Therefore, for $ \angle ABC$ to have an integral value, $ 67 | k$, meaning there are $ \lfloor\frac {2008}{67}\rfloor \equal{} 29$ possible values for $ k$, namely the set $ \{67, 2(67), 3(67), ... , 29(67)\}$.
Since we are looking for distinct triples of angles of a triangle, this problem is the same as finding all distinct triples of positive integers below $ 30$ that add up to $ 30$, allowing for repetition of numbers. One brute-forcing session later, I came up with the answer $ \boxed{75}$.[/hide]


Angle BAC and the exterior angle CBD of triangle ABC are each bisected. If the angle formed by these bisectors when they meet is 50 degreees, find the number of degrees in angle C of triangle ABC.

Billy

there must be some rule I don't know or else I wouldn't be stumped like this.

I posted a message saying the question was worded wrong, when really i kept reading BAC as ABC.  maybe I'm becoming dyslexic :(

I am sure.

Note that it is not isosceles.

I got it from the All-Time Greatest book

Should I post the answer?

Billy

[hide]After doing the problem with the CORRECT triangle, I arrived at the same answer lol. Set the angles created by the angle bisector of CAB to both y and the angles formed by the angle bisector of CBD both to x. This makes angle CBA = 180 - 2x. This also makes angle C = 50 + x - y and angle A = 2y. Angle C + A + B = 180 so 50 + x - y + 2y + 180 - 2x = 180. This simplifies to x - y = 50 and x = 50 + y. By substitution, angle C is now 50 + 50 + y - y = 100.[/hide]

[quote="cj"][hide=" my simple way"] We call 1/2 angle BAC = m, 1/2 angle CBD = y, angle C = x. Then we have y=m+1/2 x and m+x=y+50. Solving this systerm equation, we get right way x=100.[/hide][/quote]

nice

Yes, after several 2-variable equations, I finally figured out that the measure of angle C was something like 100 degrees.

wow. That was idoitic
(commenting on his own post)

If you think you were idiotic, then I don't know how idiotic I am. At first, I said that the measure of that angle was 280 degrees because I assumed that a straight angle measured 360 degrees!!!!!!!!!!!!!!!!!


trigonometry
function
calculus
calculus computations

Calculate the Fourier series of a particular sine function over an appropriate $ 2\pi$ inteveral.  Maybe that's not regular Calculus though.  Probably another way you're looking for.


search
number theory unsolved
number theory

Let \[ f(x) \equal{} x^3  \plus{} 17\] . Prove that for every integer n >= 2 there exists a
natural number x for which f(x) is divisible by 3^n but not by 3^n+1


find the definite integral of 1/x from x=1 to x=2 using riemann sum method

It's rather difficult to do with a Riemann sum because you end up with $ \sum_{k\equal{}1}^{n}\frac{n}{n\plus{}1}$.

But, indeed, the correct answer is $ \int_1^2\frac{1}{x}\ dx\equal{}\ln2\minus{}\ln1\equal{}\ln2\minus{}0\equal{}\boxed{\ln2}$.

Also, wrong forum. Look here: http://www.artofproblemsolving.com/Forum/index.php?f=297


Five years ago, a computer that worked half as fast as a current
computer cost twice as much money. The current computer's
speed to price ratio is what percent of the older computer's
speed to price ratio?

Half the money: $ \frac{1}{2}$.  Double the speed: $ 2$.  Our ratio is thus $ \frac{2}{\frac{1}{2}}\equal{}4$, or $ \boxed{400 \%}$.


complex analysis
arithmetic sequence
number theory solved
number theory

Here's one:

Prove that there are infinitely many primes ending in a 9. I haven't given it MUCH thought, but I don't have any ideas.

I just came with this idea! Lets make the problem even more interesting! :D

1)Are there infinitely many primes ending in 1?
1)Are there infinitely many primes ending in 3?
1)Are there infinitely many primes ending in 7?

cheers! :D :D

A well-known and difficult theorem of Dirichlet states that for every coprime integers a,b, there arithmetic progression {a*n+b / n in } contains infinitely many primes.

Using it with a = 10 and b in {1,3,7,9}, the answers to the above questions is 'yes', in each case.

Pierre

I have found only non-elementary proofs of Dirichlet's theorem concerning the primes...does anybody know of an elementary proof?..

It depends on what you mean by 'elementary'...Usually, it means 'which only uses real numbers' (and does not use compex number and holomorphic functions).
In that case, an elementary proof of Dirichlet's theorem has been found by Selberg (he also found an 'elementary' proof of the prime number theorem), but it is long and complicated. Thus, it is absolutely not 'elementary' in the sense of 'simple'...

Pierre.

I didn't expect it to be simple. Thank you.

I found this in Hojoo Lee's NT problem file (you can find the link in the "Math Links" section). It said there: " solve blablabla (the problem) without Dirichlet's theorem".

Ok guys, I think i've got a soln:

It's obviously equivalent to the existence of infinitely many primes of the form 5k-1.

I used the law of square reciprocity (I DON'T know a proof of this one, but I bet you can find one on the Net):

(p/5)*(5/p) = (-1)^[(p-1)/2]*[(5-1)/2] = (-1)^(p-1) = 1, so p is a square residue mod 5 iff 5 is a square residue mod p. We know that p is a square residue mod 5 iff p=+-1 (5) (1 or -1 (mod 5)), so, from these 2,  p|x^2-5 for some x iff p = +-1 (5), so for a natural x, all primes which divide x^2-5 are +-1 (5). Let's assume there are finitely many primes of the form 5k-1, and let's call them p1, p2, p3,.., pn. We can take x s.t. p1*..*pn|x. This means that all primes which divide x^2-5 have form 5k+1. From the Chinese Remainder Thm we can choose x s.t. x=0(p1)=0(p2)=..=0(pn)=2(5), meaning that x^2-5 is -1(5). This is false because all primes which divide x^2-5 are supposed to be 1(5). We have a contradiction, so there are infinitely many primes of the form 5k-1 Q.E.D.


inequalities
AMC
USA(J)MO
USAMO
function
pigeonhole principle
inequalities unsolved

Prove that the inequality
$\left(a^{2}+2\right)\left(b^{2}+2\right)\left(c^{2}+2\right) \geq 3\left(a+b+c\right)^{2}$
holds for all positive reals $a$, $b$, $c$.

This is an extension of the APMO 2004 inequality, which states that

$\left(a^{2}+2\right)\left(b^{2}+2\right)\left(c^{2}+2\right) \geq 9\left(ab+bc+ca\right)$ .

For this APMO inequality, see
http://www.mathlinks.ro/Forum/viewtopic.php?t=21835
http://www.mathlinks.ro/Forum/viewtopic.php?t=19666 (generalization)
http://www.mathlinks.ro/Forum/viewtopic.php?t=4830 ;
note that Harazi's post in the third of these topics also proves the extension!

  darij

[quote="conceive"][quote="songoku2006"]We have a stronger inequalities:
$\(a^2+2)(b^2+2)(c^2+2)\geq\ 3(a^2+b^2+c^2)$ :)[/quote]

I think your generalization is dry. The following one is better:

$(a^2+2)(b^2+2)(c^2+2)\ge 9(a^2+b^2+c^2)$[/quote]

I dont think that is correct .

Try $a=b=\sqrt{0.1}$ and $c=\sqrt{2}$ .... LHS=17.64 , RHS=19.8

[quote="shobber"]Prove that the inequality
$ \left(a^{2} \plus{} 2\right)\left(b^{2} \plus{} 2\right)\left(c^{2} \plus{} 2\right) \geq 3\left(a \plus{} b \plus{} c\right)^{2}$
holds for all positive reals $ a$, $ b$, $ c$.[/quote]
My solution:
Put $ t \equal{} \frac {b \plus{} c}{2}$
By AM-GM we have:
$ (b^2 \plus{} 2)(c^2 \plus{} 2) \geq 3 \plus{} 2bc \plus{} 2(b^2 \plus{} c^2) \geq 3 \plus{} 6t^2$ and $ t^2 \geq bc$
We need prove that:
$ (a^2 \plus{} 2)(3 \plus{} 6t^2) \geq 9(2at \plus{} t^2)$
<=>$ (a \minus{} t)^2 \plus{} 2(at \minus{} 1)^2 \geq 0$
This inequality is true
Equality occurs when  $ a \equal{} b \equal{} c \equal{} 1$

This is my soln...

there are two beetween $ a,b,c$ (WLOG $ a$ and $ b$) such that $ (a^2,1,1)$ and $ (b^2,1,1)$ are equally sorted (beetween $ 3$ numbers there are always two of them on the same side of 1). So by the Chebychev inequality we get:

\[ (a^2\plus{}1\plus{}1)(1\plus{}1\plus{}b^2) \geq 3(a^2\plus{}1\plus{}b^2)\]

Now by CS we have:

\[ (a^2\plus{}1\plus{}b^2)(1\plus{}c^2\plus{}1) \geq (a\plus{}b\plus{}c)^2\]

And recolletting all together we'll get: $ (a^2\plus{}2)(b^2\plus{}2)(c^2\plus{}2) \geq 3(a\plus{}b\plus{}c)^2$

[quote="Simo_the_Wolf"]This is my soln...

there are two beetween $ a,b,c$ (WLOG $ a$ and $ b$) such that $ (a^2,1,1)$ and $ (b^2,1,1)$ are equally sorted (beetween $ 3$ numbers there are always two of them on the same side of 1). So by the Chebychev inequality we get:
\[ (a^2 \plus{} 1 \plus{} 1)(1 \plus{} 1 \plus{} b^2) \geq 3(a^2 \plus{} 1 \plus{} b^2)
\]
Now by CS we have:
\[ (a^2 \plus{} 1 \plus{} b^2)(1 \plus{} c^2 \plus{} 1) \geq (a \plus{} b \plus{} c)^2
\]
And recolletting all together we'll get: $ (a^2 \plus{} 2)(b^2 \plus{} 2)(c^2 \plus{} 2) \geq 3(a \plus{} b \plus{} c)^2$[/quote]
 I don't think so . If a,b,c <1 , the problem is right  . We can't  use  Chebychev . I think the problem is hard

Factoring the RHS, we have a bunch of stuff plus $ 4(a^2 \plus{} b^2 \plus{} c^2)$ which is obviously greater than $ 3(a^2 \plus{} b^2 \plus{} c^2)$. By the inequality having to do with convex functions (the name of which I can't remember), $ 3(a^2 \plus{} b^2 \plus{} c^2)\geq 3(a \plus{} b \plus{} c)^2$. Q.E.D.

Also, this can be generalized using the same procedure to $ (x_1^2 \plus{} 2)(x_2^2 \plus{} 2)\cdots (x_n^2 \plus{} 2) \geq n (x_1 \plus{} x_2 \plus{} \cdots \plus{} x_n)^2$ for $ x_i\in \mathbb{R}^ \plus{}$.

p.s. I hope I didn't make any mistakes. >_<

[quote="shobber"]Prove that the inequality
$ \left(a^{2} \plus{} 2\right)\left(b^{2} \plus{} 2\right)\left(c^{2} \plus{} 2\right) \geq 3\left(a \plus{} b \plus{} c\right)^{2}$
holds for all positive reals $ a$, $ b$, $ c$.[/quote]

Here is my proof : ( I hope it is different from the previous ones )

WLOG,assume that $ (b^2 \minus{} 1)(c^2 \minus{} 1) \geq 0.$Using Caushy-Schwarz inequality,we have:

$ (a^2 \plus{} 2)(b^2 \plus{} c^2 \plus{} 1) \equal{} (a^2 \plus{} 1 \plus{} 1)(1 \plus{} b^2 \plus{} c^2) \geq (a \plus{} b \plus{} c)^2$

Hence,It suffices to prove that:

$ (b^2 \plus{} 2)(c^2 \plus{} 2) \geq 3(b^2 \plus{} c^2 \plus{} 1)$

Which is equivalent to $ : (b^2 \minus{} 1)(c^2 \minus{} 1) \geq 0$

Done !

[hide="Solution"]First of all note that
$(a^2+2)(b^2+2)(c^2+2)-3(a+b+c)^2=$
$2(ab-1)^2+3(bc-1)^2+3(ca-1)^2+(a-b)^2+c^2(a^2-1)(b^2-1)$

But by the pigeonhole principle we have that either there are two numbers that are at least $1$ or there are two numbers that are at most $1$, so WLOG $(a^2-1)(b^2-1)\geq 0$, which proves the inequality.

Equality holds iff $ab=bc=ca=1\Rightarrow a=b=c=1$[/hide]


integration
calculus
probability
function
counting
distinguishability
absolute value

[quote="Kent Merryfield"]Let $g(z)=\frac1{\sqrt{2\pi}}\int_{z}^{\infty}e^{-x^{2}/2}\,dx$

1. Show that, for $z>0,\ g(z)<\frac1{\sqrt{2\pi}}e^{-z^{2}/2}\cdot \frac1z.$

2. Sharpen that estimate to

$\frac1{\sqrt{2\pi}}e^{-z^{2}/2}\cdot \frac1z-\ ?\ <g(z)< \frac1{\sqrt{2\pi}}e^{-z^{2}/2}\cdot \frac1z,$

where the question mark stands for some useful expression that goes to zero as $z\to\infty$ faster than the other term in this expression.

3. What more can you say?

[This is well-known, of course. The question is mostly directed to those who are learning calculus.][/quote]
Well, I am not a beginner when it comes to basic calculus;still, I would like to try this problem which is new to me.

(1)
$g(z)=\frac1{\sqrt{2\pi}}\int_{z}^{\infty}e^{-x^{2}/2}\,dx =-\frac1{\sqrt{2\pi}}\int_{z}^{\infty}\frac{1}{x}\,d(e^{-x^{2}/2})$

$=-\frac1{\sqrt{2\pi}}\Big[ \frac{1}{x}e^{{-x^{2}/2}}\Big|_{z}^{\infty}+\int_{z}^{\infty}\frac{1}{x^{2}}e^{-x^{2}/2}\,dx \Big]$ ... (Integration by parts)

$= \frac1{\sqrt{2\pi}}\frac{1}{z}e^{{-z^{2}/2}}-\frac1{\sqrt{2\pi}}\int_{z}^{\infty}\frac{1}{x^{2}}e^{-x^{2}/2}\,dx$ ... $(*)$

$\Rightarrow g(z) < \frac1{\sqrt{2\pi}}\frac{1}{z}e^{{-z^{2}/2}}$

(2) $(*)$ further gives us, after employing "integration by part" once more,

$g(z) = \frac1{\sqrt{2\pi}}\frac{1}{z}e^{{-z^{2}/2}}+\frac1{\sqrt{2\pi}}\Big[-\frac{1}{z^{3}}e^{-z^{2}/2}+3\int_{z}^{\infty}\frac{1}{x^{4}}e^{-x^{2}/2}\,dx \Big]$

$\Rightarrow \frac1{\sqrt{2\pi}}\frac{1}{z}e^{{-z^{2}/2}}-\boxed{\frac1{\sqrt{2\pi}}\frac{1}{z^{3}}e^{-z^{2}/2}}< g(z)$ ... $(**)$

(3) I am not sure what the question means.

(3) was an invitation to continue the integration by parts, and then to decide what it means.

If we continue the integration by parts, I think that we get that \[g(z) = \frac{e^{-z^{2}/2}}{\sqrt{2 \pi}}\cdot \sum_{n=0}^\infty (-1)^{n}\frac{(2n-1)!!}{z^{2n+1}}.\] I was taken by the wave :surf: It looked like it would be true (at first sight). Guess I should write things down next time  :blush:

Look again at that series- does it even converge?

Since no one else has reacted to jmerry's post in the last week, I'll step back in.

Indeed that series diverges for all $z.$ However any fixed partial sum is in error by less than the next term.

The $N+1$st partial sum is, for large enough $z,$ a better estimate for the integral than the $N$th partial sum. But that phrase, "large enough $z,$" is not uniform. That is, for a larger $N$ we have to go to a larger $z$ for the claim to be true.

This is typical behavior for asymptotic series.

We have two series available for estimating the cumulative normal probability.

The first is the Taylor series centered at zero. As a series of real numbers, this converges for all $z.$ The convergence is not uniform - for larger values of values of $z,$ you need more terms to reach the same error tolerance. But I chose that phrase carefully - as a series of real numbers, in which arithmetic can be done exactly. How does it behave as a series of floating-point numbers, in which arithmetic is subject to roundoff error? The answer is that for large enough $z,$ the series is unusable. We're attmepting to write a very small number as an alternating sum of a large number of large numbers - and the roundoff error swallows the computation. This makes an interesting programming exercise: working in a system that lets you draw graphs, write a program that adds up this power series in a conditional loop, exiting the loop when the term gets below some reasonable tolerance. Do this for a (reasonably incremented) range of values of $z,$ and then plot the resulting graph. For small values of $z$ you get a nice smooth curve, but beyond a certain point (probably somewhere between $z=3$ and $z=6$), the graph breaks up into random noise.

The second series is the asymptotic series, as represented in perfect_radio's post. This series diverges for all $z.$ What we can do is write a program that add's up terms while the terms are growing smaller, then exits the loop before adding a term that is bigger (in absolute value) than the preceding term. The larger $z$ is, the more terms we get to add. On the other hand, for large $z,$ the cumulative-from-the-right probability, what we have called $g(z),$ is a very small number. All a reasonable graph would show is a function indistinguishable from zero for large enough $z.$ The asymptotic series is quite useless for small $z,$ but for $z$ bigger than 2 or 3, it starts becoming very useful.

Rather than relying forever on tables photocopied from books, I decided a few years ago to just build my own normal probability table. I used the Taylor series for small $z$ and the asymptotic series for large $z.$ (The practical boundary between small and large was about $z=3.$)

One final warning about the normal probability for large values of $z:$ very few, if any, things in nature are exactly normally distributed. The only reason that many things in nature appear to be normally distributed is that they are subject to many influences, and the Central Limit Theorem describes how these influences aggregate. But the normal approximation promised by the Central Limit Theorem becomes less accurate as an approximation for large values of $z.$


Παιδιά θέλω να αποδείξω ότι :

[b](ab')+(bc')+(ca') = (a+b+c)(a'+b'+c')[/b]

προφανώς το + υποδιλώνει το λογικό ΟR και ο πολλ/μος το λογικό AND .

Αρχίζω από το δεύτερο μέρος με άλγεβρα bool για να καταλήξω στο πρώτο αλλά μάταια δεν βγαίνει με τπτ , αν έχετε κάποιες υποδείξεις θα είμουν ευγνώμον. Ευχαριστώ

[quote="Durandal"]Δείξε το αναπαριστώντας τις προτάσεις $ a,b,c$ με σύνολα (οπότε η πρόσθεση αντιστοιχεί στην ένωση και ο πολ/σμός στην τομή)... κάνοντας ένα διάγραμμα Venn δε θα πρέπει να δυσκολευτείς πολύ...

Cheerio,

Durandal 1707[/quote]

Βασικά με άλγεβρα boole θέλει να το δείξουμε, αλλά πίστεψε με, έχω δοκιμάσει τα πάντα, και δεν μου βγαίνει , υποψιάζομαι ότι υπάρχει λάθος στην άσκηση... αν ξέρει κάποιος, κάτι ή έχει κάποια ιδέα, ας απαντήσει.

Ευχαριστώ

proton.. otan leei kati o durandal exei dikio!

deuteron...se periptosi pou o durandal exei adiko, efarmozetai to proto! oxi plaka kano..to deutero einai oti tha sou elega na dokimaseis karnaugh maps-pisteuo oti prepei na to kseftilizei sta grhgora..

xpd, η άσκηση σωστή είναι... αν το ζωγραφίσεις πραγματικά φαίνεται αμέσως.

Τώρα, αν θέλεις να το δείξεις με βάση τους κανόνες της άλγεβρας Boole... το δοκίμασα και βαρέθηκα μετά από δυο τρεις γραμμές πράξεις αλλά με υπομονή νομίζω βγαίνει. Αν καταλήγεις σε λάθος, θα σου συμβούλευα να μη χρησιμοποιείς τον συμβολισμό $ \plus{} ,\cdot$ αλλά τον $ \vee,\wedge$ γιατί παραείμαστε συνηθισμένοι να χρησιμοποιούμε  το $ \plus{} ,\cdot$ με τον "κανονικό" τρόπο και είναι πολύ πιθανό να σου βγαίνει λάθος κάπου από εκεί...

Επίσης +1 για τα Karnaugh maps (xpd, μπορεί να τα ξέρεις και ως Veitch diagrams)

@Nick: ο durandal έχει δίκιο όταν:
1. Διαβάσει σωστά την άσκηση.
2. Καταλάβει σωστά την άσκηση.
3. Λύσει σωστά την άσκηση.
4. Γράψει σωστά αυτό που θέλει να πει.
5. Δεν ξεχάσει αυτό που θέλει να πει στα μισά του δρόμου.
6, 7$ \cdots$

Corollary: αν είχα μία δεκάρα για κάθε βλακεία που έχω πει... ;)

Cheerio,

Durandal 1707

Αν αναπτύξεις το δεύτερο μέλος και αλλάξεις τη σειρά βγαίνει:
(ab’ + a’b) + (ac’+a’c) +(bc’+b’c)
Εδώ αξίζει να προσέξεις ότι κάθε παρένθεση είναι η αποκλειστική διάζευξη των όρων που συμμετέχουν πχ η πρώτη είναι αποκλειστική διάζευξη των a και b (a XOR b). Από ότι φαίνεται η παράσταση που δίνεις είναι κάτι σαν γενίκευση της αποκλειστικής διάζευξης για 3 όρους, όπου το βασικό σκεπτικό είναι το εξής: «Είναι αληθής αν δεν είναι όλοι ίδιοι μεταξύ τους δηλαδή αν υπάρχει ζευγάρι διαφορετικών». Πρόσεξε ότι διαισθητικά αυτό λένε και τα 2 μέλη.

Από ότι φαίνεται το πρώτο μέλος είναι η πιο συμπαγείς γραφή της και το δεύτερο κάτι σαν πιο αναλυτικό.

Η βασική ιδέα για μια αμιγώς αλγεβρική απόδειξη είναι να πας και τα 2 μέλη σε μια ακόμα πιο αναλυτική μορφή που περιέχει σε κάθε όρο και τα 3 (a, b, c) ή τις αρνήσεις τους ελπίζοντας ότι θα αποκαλυφθεί η κρυμμένη ισότητα. Για να το πετύχεις αυτό πολλαπλασίασε τον κάθε όρο με τη διάζευξη του όρου που λείπει με τον αντίθετό του. Δηλαδή το ab’ θα το πολλαπλασιάσεις με το (c+c’) που κάνει 1. Ομοίως τον ca’ θα τον πολλαπλασιάσεις με (b’+b). Νομίζω κατάλαβες τι θέλω να πω. Αυτό θα το κάνεις για όλους και στο τέλος θα έχεις και στα 2 μέλη «αθροίσματα» (κανονικά διαζεύξεις) γινομένων (κανονικά συζεύξεων) των a, b, c ή των αρνήσεών τους.
Εκεί θα εφαρμόσεις στο δεύτερο μέλος ότι το x+x=x και θα διώξεις αυτά που επαναλαμβάνονται. Έτσι θα μείνουν και τα 2 μέλη ίσα. Στο χαρτί μου βγήκε. Αν δε μου ξέφυγε κάτι αυτός πρέπει να είναι ένας τρόπος.


if $ x\in R$(positive) then can $ x^2\plus{}\frac{1}{4x}$ be $ \sqrt{3}\minus{}1$ and why?

Well if it's true, then $ 4x^3\minus{}4x(\sqrt{3}\minus{}1)\plus{}1\equal{}0$ and solve the cubic equation.
Look at the dicriminant, if it's positive or 0, then it can, if it's negative then it can't be.

anyone else?
maybe proving $ x^2\plus{}\frac{1}{4x} > \sqrt{3}\minus{}1$

[hide="Hint"]Use AM-GM on $ x^2$, $ \frac{1}{8x}$, and $ \frac{1}{8x}$.[/hide]

[quote="nsato"][hide="Hint"]Use AM-GM on $ x^2$, $ \frac {1}{8x}$, and $ \frac {1}{8x}$.[/hide][/quote]
Yes thats it  x>=3/4  :lol:

Nice one!
Sorry, in my method, if discriminant<0 then it has 1 real root


It seems that the following inequality is also true.
Let $a$, $b$ and $c$ are non-negative numbers. Prove that:
\[\sum_{cyc}\sqrt[3]{a^3+\frac{b^3+c^3}{2}}\geq\sum_{cyc}\sqrt[3]{a^3+b^2c}\]

I used the Vasc's inequality: $(a^2+b^2+c^2)^2\geq3(a^3b+b^3c+c^3a)$.


geometry
circumcircle
search
cyclic quadrilateral
geometry proposed

[color=darkred]A circle $ \Omega_1$ is tangent outwardly to the circle $ \Omega_2$ of bigger radius. Line $ t_1$ is tangent at points $ A$ and $ D$ to the circles $ \Omega_1$ and $ \Omega_2$ respectively. Line $ t_2$, parallel to $ t_1$, is tangent to the circle $ \Omega_1$ and cuts $ \Omega_2$ at points $ E$ and $ F$. Point $ C$ belongs to the circle $ \Omega_2$ such that $ D$ and $ C$ are separated by the line $ EF$. Denote $ B$ the intersection of $ EF$ and $ CD$. Prove that circumcircle of $ ABC$ is tangent to the line $ AD$.[/color]

DE=DF=2R*2r=DJ*DK=AD
AD^2=DE^2=DB*DC
we are done

An other solution . Let X be the tangent point of the line to circle $ \omega_2$ .Because $ I_2A,I_2X$are perpendicular to AD so $ I_2,A,X$ are collinear .Let A' be the intersect point of IA and ${ \omega_2}$ ,then $ I_1,B,A'$ are collinear .We have (all angles take mod pi )
$ \angle{BXI}=\angle{XAI}=\angle{DA'I}=\angle{DCI}=\angle{BCI}$ ,therefore $ CBIX$ is a cyclic quadrilateral . It gives $ DC.DB=DI.DX=DA^2$ so DA is the tangent line of $ (ABC)$

TTsphn, nice..
That was also mine solution at the TST...

Let $ F$ the foot of the perpendicular from $ D$ to the line $ BC$.
Let $ R$ and $ r$ the big radius and the small radius, respectively.
For prove that $ AD$ is tangent to the circumcircle of $ ABC$, it's enough to prove that $ AD^{2}\equal{}DC.DB$ .
Consider $ <CDA\equal{}x$.By the sine's law, we have that: $ CD\equal{}2R.sinx$.
Since $ t_{1}$ is parallel to $ t_{2}$ and $ DE$ is parallel to the diameter of $ \Omega_{1}$, we have that: $ DE\equal{}2r$.
Then, in the right triangle $ BDE$, $ BD\equal{}\frac{2r}{sinx}$.From this, we get that $ BD.CD \equal{} 4Rr$.
But it's a well-known fact (only application of Pythagoras) that $ AD^{2}\equal{}4Rr$.
Hence, $ AD^{2}$=$ BD.CD$, as desired.
It concludes the proof.

Dear Mathlinkers,
two ideas for a more synthetic proof...
1. the circle (1) centered at D and passing through A is orthogonal to the first circle (omega 1).
2. the circle passing through A,B, C is orthogonal to omega 1.
Sincerely
Jean-Louis

That's an old problem  :(  for e.g. see  http://www.mathlinks.ro/viewtopic.php?search_id=803892854&t=53277


let A be a $ m\times n$ matrix, rank(A)=1. prove that there exists 2 matrices $ B_{m\times 1}$ and $ C_{1\times n}$ such that A=BC

If the (column) rank of A is 1 then each column must be a scalar multiple of some vector (or $ m\times1$ matrix) B. Take C to be the $ 1\times n$ matrix consisting of those scalar multiples.


i asked my physics teacher to register physics olympiad about a month ago...
but, now he is saying that he forgot to register............................
wtff

I am just wondering that is there any way to take the test? not next year though....................

You should probably contact the AAPT, which is the organization that runs it.


calculus
integration
trigonometry
logarithms
calculus computations

Ascertain the relation $ \mathrm {R}(a,b,c) \equal{} 0$ so that  $ \int_{ \minus{} \infty}^{\infty}\frac {x^2 \plus{} a}{x^4 \plus{} bx^2 \plus{} c^2}\ \mathrm{dx} \equal{} \pi$ , where $ \{a,b,c\}\subset (0,\infty )$ .

after expressing it as

$ I\equal{}4\left[\int_0^{\infty}\frac {(\sqrt 2\, x)^2 \plus{}b}{\left((\sqrt 2\, x)^2 \plus{}b\right)^2\minus{} \left(\sqrt{b^2\minus{}4c^2}\right)^2}\;\textbf dx\plus{}\frac{(2a\minus{}b)}{\sqrt{b^2\minus{}4c^2}}\int_0^{\infty}\frac {\sqrt{b^2\minus{}4c^2}}{\left((\sqrt 2\,x)^2 \plus{}b\right)^2\minus{} \left(\sqrt{b^2\minus{}4c^2}\right)^2}\;\textbf dx\right]$

and then making the usual substitutions, you will eventually get

$ I\equal{}\sqrt{\frac{2}{b^2\minus{}4c^2}}\left[ \frac{ \sqrt{b^2\minus{}4c^2} \plus{}(2a\minus{}b)}{\sqrt{b\minus{}\sqrt{b^2\minus{}4c^2}}}\;\tan^{\minus{}1}\left( \frac{\sqrt 2\,x}{\sqrt{b\minus{}\sqrt{b^2\minus{}4c^2}}}\right)\plus{}\right.$

[color=white]..................................................[/color] $ \left. \frac{ \sqrt{b^2\minus{}4c^2} \minus{}(2a\minus{}b)}{\sqrt{b\plus{}\sqrt{b^2\minus{}4c^2}}}\;\tan^{\minus{}1}\left( \frac{\sqrt 2\,x}{\sqrt{b\plus{}\sqrt{b^2\minus{}4c^2}}}\right)
\right]\quad\bigg|_0^{\infty}$

$ \Rightarrow \pi \equal{} \sqrt{\frac{2}{b^2\minus{}4c^2}}\;\;\frac{\pi}{2} \;\; \left[ \frac{ \sqrt{b^2\minus{}4c^2} \plus{}(2a\minus{}b)}{\sqrt{b\minus{}\sqrt{b^2\minus{}4c^2}}}\plus{} \frac{ \sqrt{b^2\minus{}4c^2} \minus{}(2a\minus{}b)}{\sqrt{b\plus{}\sqrt{b^2\minus{}4c^2}}}\right]$

$ \Rightarrow   1 \equal{} \frac{1}{\sqrt 2} \;\; \frac{1}{\sqrt{b^2\minus{}4c^2}}\;\;\frac{1}{2c} \left[ \sqrt{b\plus{}\sqrt{b^2\minus{}4c^2}}\cdot\left(\sqrt{b^2\minus{}4c^2} \plus{}(2a\minus{}b)\right)\plus{}\sqrt{b\minus{}\sqrt{b^2\minus{}4c^2}}\cdot\left(\sqrt{b^2\minus{}4c^2} \minus{}(2a\minus{}b)\right)\right]$

i don't want to do this question anymore.

[quote="misan"] [color=darkblue]... I don't want to do this question anymore. [/color][/quote]
[color=darkblue]I am sorry, [b]Misan[/b] ![/color]

[quote="Virgil Nicula"] [color=darkred]Ascertain the relation $ \mathrm {R}(a,b,c) \equal{} 0$ so that  $ K\equiv \int_{ \minus{} \infty}^{\infty}\frac {x^2 \plus{} a}{x^4 \plus{} bx^2 \plus{} c^2}\ \mathrm{dx} \equal{} \pi$ , where $ \{a,b,c\}\subset (0,\infty )$ .[/color] [/quote]

[color=darkblue][b][u]Proof.[/u][/b]

$ \blacktriangleright\ \boxed {\ I\equiv\int_0^{\infty}\frac {x^2 \plus{} c}{x^4 \plus{} bx^2 \plus{} c^2}\ \mathrm{dx}\ } \equal{}$ $ \int_0^{\infty}\frac {1 \plus{} \frac {c}{x^2}}{x^2 \plus{} \frac {c^2}{x^2} \plus{} b}\ \mathrm {dx} \equal{}$ $ \int_0^{\infty}\frac {\left(x \minus{} \frac cx\right)'}{\left(x \minus{} \frac cx\right)^2 \plus{} 2c \plus{} b}\ \mathrm {dx}$ .

Thus, $ I \equal{} \left\|\frac {1}{\sqrt {2c \plus{} b}}\arctan\frac {x^2 \minus{} c}{x\sqrt {2c \plus{} b}}\right\|_0^{\infty} \equal{}$ $ \frac {\pi}{\sqrt {b \plus{} 2c}}$ $ \implies$ $ \boxed {\ I \equal{} \frac {\pi}{\sqrt {b \plus{} 2c}}\ }$ .

$ \blacktriangleright\ \boxed {\ J\equiv\int_0^{\infty}\frac {x^2 \minus{} c}{x^4 \plus{} bx^2 \plus{} c^2}\ \mathrm{dx} \ } \equal{}$ $ \int_0^{\infty}\frac {1 \minus{} \frac {c}{x^2}}{x^2 \plus{} \frac {c^2}{x^2} \plus{} b}\ \mathrm {dx} \equal{}$ $ \int_0^{\infty}\frac {\left(x \plus{} \frac cx\right)'}{\left(x \plus{} \frac cx\right)^2 \minus{} 2c \plus{} b}\ \mathrm {dx}$ .

If $ b \equal{} 2c$ , then $ J \equal{} \minus{} \left\|\frac {x}{x^2 \plus{} c}\right\|_0^{\infty} \equal{} 0$ . If $ b > 2c$ , then $ J \equal{} \left\|\frac {1}{\sqrt {b \minus{} 2c}}\arctan\frac {x^2 \plus{} c}{x\sqrt {b \minus{} 2c}}\right\|_0^{\infty} \equal{} 0$ .

If $ b < 2c$ , then $ J \equal{} \left\|\frac {1}{2\sqrt {2c \minus{} b}}\ln\frac {x^2 \minus{} x\sqrt {2c \minus{} b} \plus{} c}{x^2 \plus{} x\sqrt {2c \minus{} b} \plus{} c}\right\|_0^{\infty} \equal{} 0$ . Therefore, for any $ b > 0$ , $ c > 0$ we have $ \boxed {\ J \equal{} 0\ }$ .

$ \blacktriangleright\ \boxed {\ U\equiv\int_0^{\infty}\frac {x^2}{x^4 \plus{} bx^2 \plus{} c^2}\ \mathrm{dx}\ } \equal{} \frac 12\cdot (I \plus{} J) \equal{} \frac 12\cdot I$ and $ \boxed {\ V\equiv\int_0^{\infty}\frac {1}{x^4 \plus{} bx^2 \plus{} c^2}\ \mathrm{dx}\ } \equal{} \frac {1}{2c}\cdot (I \minus{} J) \equal{} \frac {1}{2c}\cdot I$ .

Therefore, $ K\equal{}2\cdot \int_{0}^{\infty}\frac {x^2 \plus{} a}{x^4 \plus{} bx^2 \plus{} c^2}\ \mathrm{dx} \equal{} 2(U \plus{} aV) \equal{}$ $ \left(1 \plus{} \frac ac\right)\cdot I$ $ \implies$ $ \boxed {\boxed {\boxed {\ K \equal{} \left(1 \plus{} \frac ac\right)\cdot \frac {\pi}{\sqrt {b \plus{} 2c}}\ }}}$ .

In conclusion, $ \boxed {\ K \equal{} \pi\ \Longleftrightarrow\ \left(1 \plus{} \frac ac\right)^2 \equal{} b \plus{} 2c\ }$ .

[b][u]Remark.[/u][/b] Show easily that [b]if[/b] $ b > 2c$, then my relation $ \boxed {\ \left(1 \plus{} \frac ac\right)^2 \equal{} b \plus{} 2c\ }$ is equivalently with "[b]Misan[/b]'s relation, i.e.

$ \boxed {\ \frac {1}{\sqrt 2} \;\; \frac {1}{\sqrt {b^2 \minus{} 4c^2}}\;\;\frac {1}{2c}\left[ \sqrt {b \plus{} \sqrt {b^2 \minus{} 4c^2}}\cdot\left(\sqrt {b^2 \minus{} 4c^2} \plus{} (2a \minus{} b)\right) \plus{} \sqrt {b \minus{} \sqrt {b^2 \minus{} 4c^2}}\cdot\left(\sqrt {b^2 \minus{} 4c^2} \minus{} (2a \minus{} b)\right)\right] \equal{} 1\ }$ .[/color]


[color=darkblue]Let $ x,y,z,t \in R$ and $ \begin {cases} x \plus{} y \plus{} z \plus{} t \equal{} 3 \\
x^2 \plus{} y^2 \plus{} z^2 \plus{} t^2 \equal{} 9 \\
x^3 \plus{} y^3 \plus{} z^3 \plus{} t^3 \equal{} 27 \\
x^4 \plus{} y^4 \plus{} z^4 \plus{} t^4 \equal{} 81 \end {cases}$

Find $ S \equal{} x^{2008} \plus{} y^{2009} \plus{} z^{2010} \plus{} t^{2011}$[/color]

[color=darkblue]Yes. You are true, but can you post full solution?[/color]

[quote="rofler"]$ (x^2 \plus{} y^2 \plus{} z^2 \plus{} t^2)^2 \equal{} x^4 \plus{} y^4 \plus{} z^4 \plus{} t^4$

So after expanding, we see that in every pair of numbers, at least one must be 0. Therefore, 3 of the numbers must be 0. Then by the first equation, the 4th number is 3. These yield the 4 values of s: $ 3^{2008}$, $ 3^{2009}$, $ 3^{2010}$, and $ 3^{2011}$. Q.E.D.[/quote]
$ (x^2 \plus{} y^2 \plus{} z^2 \plus{} t^2)^2 \equal{} x^4 \plus{} y^4 \plus{} z^4 \plus{} t^4$ [color=darkred]is OK.[/color]


[color=darkred] [u][b]Problem.[/b][/u] The incircle with center $I$ of triangle $ABC$ touches it's sides $BC,\,CA,\,AB$ in $D,\,E,\,F$ respectively.

Prove that: $DI,\,EF$ and median $AM$ of triangle $ABC$ are concurrent.[/color]

[color=indigo][b]Remark.[/b] From this problem, we have two new and nice results about concurrent lines:[/color]

[color=darkred][u][b]Result 1.[/b][/u] The incircle with center $I$ of triangle $ABC$ touches it's sides $BC,\,CA,\,AB$ in $D,\,E,\,F$ respectively. Let $G$ be the centroid of triangle $ABC.$
Define: $\begin{cases}M=AG\cap EF\\ N=BG\cap FD\\ P=CG\cap DE\end{cases}$ then we have: $DM,\,EN,\,FP$ are concurrent.

[u][b]Result 2.[/b][/u] The incircle with center $I$ of triangle $ABC$ touches it's sides $BC,\,CA,\,AB$ in $D,\,E,\,F$ respectively.
Define: $\begin{cases}R=DI\cap EF\\ S=EI\cap FD\\ P=FI\cap DE\end{cases}$ then we have $AR,\,BS,\,CT$ are concurrent. [/color]

Let $ (O)$ be circumcircle of the $ \triangle ABC.$ Circle $ (O')$ through $ I$ and tangent to $ (O)$ at $ A$ cuts $ AB, AC$ at $ B', C'.$ $ A$ is external similarity center of the circumcircles $ (O'), (O)$ $ \Longrightarrow$ $ \triangle AB'C' \sim \triangle ABC$ are centrally similar with center $ A$ $ \Longrightarrow$ $ B'C' \parallel BC$ and $ AM$ cuts $ B'C'$ at its midpoint $ M'.$ $ AI$ bisects $ \angle C'AB'$ $ \Longrightarrow$ $ I$ is midpoint of the arc $ B'C'$ of $ (O')$ opposite to $ A.$ $ EF$ is Simson line of $ \triangle AB'C'$ with the pole $ I$ $ \Longrightarrow$ $ M' \equiv AM \cap B'C' \cap DI$ is on $ EF.$

[color=indigo][size=150]Thank you, [b]yetti[/b] for your solution! And now, I'll present my proof.[/size]  :P [/color]
[quote="April"][color=darkred] [u][b]Problem.[/b][/u] The incircle with center $I$ of triangle $ABC$ touches it's sides $BC,\,CA,\,AB$ in $D,\,E,\,F$ respectively.

Prove that: $DI,\,EF$ and median $AM$ of triangle $ABC$ are concurrent.[/color][/quote]
 [color=indigo] [b]Proof.[/b] Let $K=DI\cap EF$
Let a line passed through $K$ and parallel to $BC$ meets $AB,\,AC$ at $R,\,S,$ respectively.
We have the quadrilaterals $IKRF$ and $IKES$ are cyclic, then:
$\widehat{IRK}=\widehat{IFK}=\widehat{IEK}=\widehat{ISK}$
Thus $\triangle IRS$ is isosceles $\implies KR=KS$
So $AK$ passed throught midpoint $M$ of $BC$

Hence: $DI,\,EF,\,AM$ are concurrent at $K$[/color]

[quote="April"][color=darkred] [u][b]Problem.[/b][/u] The incircle with center $I$ of triangle $ABC$ touches it's sides $BC,\,CA,\,AB$ in $D,\,E,\,F$ respectively.

Prove that: $DI,\,EF$ and median $AM$ of triangle $ABC$ are concurrent.[/color][/quote]

Another clone of [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=124774#124774]http://www.mathlinks.ro/Forum/viewtopic.php?t=5830 post #4 Theorem 1[/url] :P

  darij


$ x$ is a positive number. Find the minimum of $ x\plus{}\frac{1}{x}$.

Extension: $ x$ is a positive number. Find the minimum of $ x^{2}\plus{}\frac{2}{x}$.

AM-GM IS CHEATING (for the first one at least)

[hide="first one"]Minimum is 2. If x is bigger than 1, then x becomes too big, and when x is less than one, 1/x becomes too big. Well maybe im kinda using the concept of AM-GM... though i didnt use it directly.[/hide]

[hide="2nd one"]Since $ x^{2}$ is a function that has a value increasing constantly as x increases, and $ \frac{2}{x}$ has a positive value for all positive x, and because where $ x^{3}$ is greater than 2, $ x^{2}$ is greater than $ \frac{2}{x}$, and when x=2, the function has a greater value then when x=1. Basically I mean that the function is always increasing, so the smallest value is when x=1, giving a result of 1. (If you included decimals buzzer11's reasoning works.) [/hide]

Those aren't very rigorous...

[hide="1"]$ (x\minus{}\frac{1}{x})^{2}\ge0\Rightarrow x^{2}\plus{}\frac{1}{x^{2}}\ge2$, $ (x\plus{}\frac{1}{x})^{2}\equal{}x^{2}\plus{}\frac{1}{x^{2}}\plus{}2\ge4\Rightarrow x\plus{}\frac{1}{x}\ge2$, so the minimum is 2. We see that we can indeed obtain the minimum, by setting $ x\equal{}1$.[/hide]

[hide="2"]$ x^{2}\plus{}\frac{2}{x}\equal{}x^{2}\plus{}\frac{1}{x}\plus{}\frac{1}{x}\ge3\sqrt[3]{x^{2}\cdot\frac{1}{x}\cdot\frac{1}{x}}\equal{}3$, and we can get this by setting $ x\equal{}1$.[/hide]


geometry
circumcircle
geometric transformation
reflection
trapezoid
homothety
power of a point

Suppose that $ P$ is a point on the plane of $ \Delta ABC$. Let $ (O)$ be the circumcircle of $ \Delta ABC$ then denote by $ A_2$ the second intersection of $ PA$ with $ (O)$. Also denote by $ A_1$ the second intersection of the altitude from $ A$ to $ BC$ with $ (O)$. Denote by $ A'\equal{}A_1A_2 \cap BC$. Define similarly $ B',C'$. Prove that $ A'B'C'$ lie on a line which is perpendicular to $ HP'$. Where $ H$ is the orthocenter of $ \Delta ABC$ and $ P'$ the isogonal conjugate of $ P$ wrt triangle $ ABC$.

Reflections $A_0,B_0,C_0$ of $A_2,B_2,C_2$ across $BC,CA,AB$ and orthocenter $H$ of $\triangle ABC$ are concyclic on the P-Hagge circle, whose center $V$ is the midpoint of the segment connecting $H$ with the anticomplement of the isogonal conjugate $U$ of $P$ WRT $\triangle ABC.$ For a proof you may see the general lemma in the problem [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=320075]Brocard and Hagge circle[/url]. Since $HA_0$ and $A_1A_2$ are symmetric about $BC,$ then $A' \equiv HA_0 \cap A_1A_2.$  Furthermore, from the isosceles trapezoid $HA_1A_2A_0$ with legs $HA_0=A_1A_2,$  it follows that ${A'H} \cdot {A'A_0}={A'A_1} \cdot {A'A_2}$ $\Longrightarrow$ $A'$ has equal power with respect to P-Hagge circle $(V)$ and circumcircle $(O)$ of $\triangle ABC.$ Likewise, $B',C'$ have equal power with respect to $(V)$ and $(O)$ $\Longrightarrow$ points $A',B',C'$ are collinear on the radical axis of $(V),(O).$ Thus, $OV \perp A'B'C'$ and it remains to show that $OV \parallel HP'.$

Let $D \equiv UP' \cap OV.$ Since $H,O$ and $P',U$ are homologous under the homothety with center $G$  (centroid of ABC) and coefficient $\frac{_{\overline{GO}}}{^{\overline{GH}}}=\frac{_{\overline{GP'}}}{^{\overline{GU}}}=-\frac{_1}{^2},$ then by Menelaus' theorem for $\triangle UHG \cup \overline{ODV},$ we get

$\frac{\overline{UD}}{\overline{DG}} = \frac{\overline{OH}}{\overline{OG}} \cdot \frac{\overline{VU}}{\overline{HV}}=3 \ \Longrightarrow$ $\overline{UD}=3\overline{DG}=\overline{DP'}$ $\Longrightarrow$ $D$ is midpoint of $UP'$

$OV$ is the U-midline of $\triangle UHP'$ $\Longrightarrow$ $HP' \parallel OV,$  but $OV \perp A'B'C',$ therefore $HP' \perp A'B'C'$ and the proof is completed.


Two identical objects of mass m are placed at either end of a spring of spring constant k and the whole system is placed on a horizontal frictionless surface. At what angular frequency ω does the system oscillate?

It would be $ {\omega = 2 \pi \sqrt {\frac {\mu}{k}}}$

where $ \mu = \frac {m \times m}{m + m} = \frac {m}{2}$ is the [url=http://en.wikipedia.org/wiki/Reduced_mass]reduced mass[/url]

[i]EDIT: OH!!  Sorry my mistake
I should've written -   $ {\omega = \sqrt {\frac {k}{\mu}}}$[/i]

I think you meant that equals $ T$, and then $ \omega$ is $ \frac {2\pi}{T}$.

I don't understand the solution, nor the the answer.
The answer choices are only
a) $ \sqrt{\frac{k}{m}}$
b) $ \sqrt{\frac{2k}{m}}$
c) $ \sqrt{\frac{k}{2m}}$
d) $ 2\sqrt{\frac{k}{m}}$
e) $ \frac{\sqrt{\frac{k}{m}}}{2}$

[hide="Symmetry"]Consider the frame of the c.o.m. of the spring-mass system. Note that the middle of the spring remains stationary. Thus, you can consider the movement of the left half of the spring and left mass independent of that of the right half of the spring and the right mass (why?). Thus, this problem is identical to the case of one mass and one fixed end, except with half the spring length, and thus double the spring constant.[/hide]

[b]Warning:[/b] There is some degree of subtlety to these kinds of arguments -- for any problem [i]make sure[/i] you think about whether or not this kind of argument makes physical sense before you pursue such a solution.

That said, you can come up with some very nice solutions and insights using symmetry.

so the total energy of the system stays constant:

1/2 *(2m) * (v^2) + 1/2 * k * x^2 = C

v = dx/dt so:

m *(dx/dt)^2 + 1/2* k*x^2 = C

divide through by m:

(dx/dt)^2 + k/(2m) * x^2 = C.

Now knowing the pattern of sin and cos, we guess that x = A* sin(wt+phi).

this means that w^2 = k/2m

w = sqrt(k/2m)

If we change the problem so that one side's mass is $ m_1$ and the other is $ m_2$, I don't understand the reasoning behind how to solve it.

Again, look from the c.o.m. frame, which is inertial. The c.o.m. isn't moving, so if you were to clamp the spring at the c.o.m. (splitting it into two spring with one fixed end), it wouldn't make a difference. Both of the two resulting spring-mass systems would oscillate at the same frequency. Can you find this frequency?

This approach gives you the same result as the [url=http://en.wikipedia.org/wiki/Reduced_mass]reduced mass[/url] that [b]savy[/b] mentioned up above.


Can someone explain to me in detail, how to solve a system of congruence using the Chinese Remainder Theorem??

Let's say, we want to solve: $x\equiv a\pmod m$ and $x\equiv b\pmod n$ where $\gcd(m,n)=1$.

Find a $y$ such that $y\equiv 1\pmod m$ and $y\equiv 0\pmod n$

and find a $z$ such that $z\equiv 0\pmod m$ and $z\equiv 1\pmod n$

Then $x=ay+bz + kmn$ is a solution for any integer $k$.

[quote]
Let's just deal with one system, how about: $2\pmod 3$ and $4\pmod 7$. ...
[/quote]

In the above example, $y=7$, $z=15$, then $x=74+21k$

This method can be extended  to more than two congruences.

THANX so much!! I keep messing everything up.. :P


I have a question on computer science in U of T

 I know that in UW I can go co-op for good,

 but in UofT I heard the # of students is 50-60

 which sounds pretty good.

 Since I'm also thinking about computer science,

 I want to make sure which university is better.

 (I almost forgot, but I decided to take M.D. at least even if I go into computer science)

 So, could you tell me which university is better in terms of rank, US university preference, ... etc?

i think UW's computer science is much better than UT's...
i went to UW during the Open House, i think its math faculty is fantastic...

I balieve that UW's CompSci is indeed better than that of U of T's.

If you have any more questions regarding this, feel free to ask us at compsci.ca (A forum dedicated to Canadian Computer Scientists  :lol: )


conics
parabola
geometry
perimeter
analytic geometry

What is the right triangle with the smallest perimeter that can be inscribed in the parabola y=x^2? Proof would be nice, by the way! Give the coordinates of the vertices. I guess that's the best way to do it.

Should the vertices have integer coordinates ?

I don't think thats a necessary condition.. though it may turn out to be true..

As TripleM said, it is not necessary, but it may or may not turn out to be true.

People don't like this problem? Should I move it to advanced?

I think so.  It didn't seem intermediate to me.

Hmm. Managed to solve this with quite a bit of algebra - letting the x coordinates of the vertices be a, b, c, with right angle at a then we get (a+b)(a+c) = -1 and want to minimise the perimeter which I'm not gonna write here - sum of three square roots. If you just consider the first 2 and simplify them, the (a+b)(a+c)=-1 helps simplify things quite nicely. I also used a substitution x=a+b, y=b+c, z=c+a which helped things quite a bit.. with xz=-1.

Anyway, in the end I think it comes down to 2+rt2.. not too much of a surprise.
Coordinates (1,1), (0,0), (-1,1)

Yup, that's right. I should mention that this comes from a contest where there are four problems you have to do, but for each problem there are two choices. The time limit is 80 minutes, which is rather harsh for problems like this, but they allow you to use calculators. In general the problems aren't quite as hard as this one. It was quite convenient that the one time I took this contest (only one per school can take it each year, and it's part of a contest with many different categories, so this is only one of seven tests you can take), I recognized two of the problems. One was ``well-known,'' and the other was a problem I had asked myself about two years earlier and had solved then. Therefore I had only two questions to do and the writeups and the tiebreaks.


Well, this problem is not actually mind,but I thought it was a good problem to post it here.(Resource= Mathcounts Forum)

The longest side of an isosceles right triangle has length $ \sqrt [3]{2}$ . Another side has length $ \ 2^{a}$. Find the value $ \ a$ expressed as a common fraction.

I won't post the solution until someone posts a right answer..sorry.. :)

P:S Everyone can try EXCEPT Martin, because he will probably crack it less than 30 sec. :lol:

i figured it out halfway
and im stuck

 :idea:
i can just ask the same problem in another forum and get the answer.

also
how do u make the radical signs and stuff?[/hide]

[quote="spiders_hate_chestnut"]a= -1/6

is it right?

memo: i did not post it to another forum and just pasted the answer they gave me.[/quote]
Yes! You are right!
Good Job!
If anyone wants to know the solution, go to search and type [u]Tiger3[/u] and look at the post called [u]Three Questions[/u]. I posted my solution there..

This forum is for everyone in Oklahoma, it's just that Jenks trying to use it in their training.


linear algebra
matrix
vector
complex numbers
linear algebra unsolved

$ A$ is an n x n symetric matrix, $ E$ is an n x n orthogonal matrix such that $ A'\equal{}EAE^{\minus{}1}$ is diagonal

how to show:

1. $ tr(A)\equal{}tr(EAE^{\minus{}1})$

2. $ \sum_{i,j\equal{}1}^n(a_{i,j})^2\equal{}\sum_{i,j\equal{}1}^n(a'_{i,j})^2$

#1 is true in great generality, without any of the restrictions of this problem. [color=green][Which is what kunny just said.][/color]

The key theorem is that whenever $ AB$ and $ BA$ are both defined (note that we don't even need $ A$ or $ B$ to be square), then $ \text{tr}(AB) \equal{} \text{tr}(BA).$

The proof can be mechanical - just write out the definition of the product of matrices and look at the definition of trace. This way, the proof comes down to interchanging the order of summation  in a double (finite) sum.

As for the second:

(Note that we seem by default to be working over the real numbers. Essentially the same result should be true over the complex numbers, but we would need some notational changes.)

If $ E$ is orthogonal and $ x$ is a column vector, then $ \sum_i[(Ex)_i]^2 \equal{} \sum_ix_i^2.$ That's a vital property of orthogonal matrices - they preserve the length of the vectors they act on.

If we then consider $ A$ to be made up of column vectors stacked side-by-side, we see that

$ \sum_i[(EA)_{i,j}]^2 \equal{} \sum_i(a_{i,j})^2$ for each $ j.$ Add that up over all $ j$ to get

$ \sum_j\sum_i[(EA)_{i,j}]^2 \equal{} \sum_j\sum_i(a_{i,j})^2$

Now consider rows instead of columns and do exactly the same thing for multiplying on the right by $ E^{ \minus{} 1} \equal{} E^T$ and we have our result:

$ \sum_j\sum_i[(EAE^T)_{i,j}]^2 \equal{} \sum_j\sum_i(a_{i,j})^2$

Incidentally, we made no use of $ A'$ being diagonal; the result is true for any similarity by orthogonal matrices. It's just that if $ A'$ is diagonal, we can write

$ \sum_i\sum_j(a_{i,j}')^2 \equal{} \sum_{j \equal{} 1}^n(a_{j,j}')^2$ is the sum of the squares of the eigenvalues.

[quote="kalazealot"]$ A$ is an n x n symetric matrix, $ E$ is an n x n orthogonal matrix such that $ A' \equal{} EAE^{ \minus{} 1}$ is diagonal

how to show:

2. $ \sum_{i,j \equal{} 1}^n(a_{i,j})^2 \equal{} \sum_{i,j \equal{} 1}^n(a'_{i,j})^2$[/quote]
That's the same as showing that $ tr(A^2) \equal{} tr((A')^2)$.


It looks like it's just exponential or logarithmic functions to me, you're just taking a small section of them...

Maybe it's a family of functions such that each equation gives a slightly different graph.

[quote="wolfram"]Do know of any function that could capture the following behavor?:

[img]http://www.angelfire.com/planet/wolfram/graf6.JPG[/img]

I mean depending on the parameters, the graphs of the function could look like line the lines in the picture (except of the red middle line that passes X-axle). The green and purple graphs above the middle line should in reality not go to X=0 like in the picture but bend in an S-shape close to X=0 and come closer and closer to X-the larger the X is.

Thank you for any help[/quote]

$f(x) = (1-cx)^k$ for well-chosen reals $k$ ?

Thank you t0rajir0u the function you wrote is usefull, but Im looking for something else.
The Im looking for is should be a sigmoid that passes or starts at y=1 x=0 un never is under y=0. Do you know any good sigmoid functions?

[quote="wolfram"]A sigmoid function is shaped like an S or Z.[/quote]

[url=http://mathworld.wolfram.com/SigmoidFunction.html]Sigmoid Function[/url]


What is the probability that in 10 flips of a coin, there is at least one series of 3 consecutive heads?

Can you generalize this to n flips?

There was a very similar topic posted [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=own+zargar+masoud&t=100676]here[/url]

However, in that topic no-one found a closed form for the general answer, in terms of $n$.


logarithms
geometry
geometric transformation
dilation
quadratics
algebra

HCl is a strong acid, meaning it dissociates 100% in solution.

So wouldn't the pH just be $-\log\left(6.5\cdot 10^{-8}\right)=7.19$?

[quote="Zakary"]No.  When the acid is so dilute, one has to make adjustments -- there must be equal total concentrations of corresponding cations and anions.  An acid solution obviously cannot have a basic pH![/quote]
Yeah...I was like what the heck the pH is greater than 7???  :blush:

EDIT: Ok, I'm going to refrain from making any posts in this forum until I've read a chemistry textbook from front to back.

That's right Zakary, so the answer is...

Calculate the pH of a $1.0\times 10^{-7}M$ solution of NaOH in water.

So...I guess this is using the same concept? How is it solved?

[quote="me@home"]So NaOH completely dissociates (of course) so $pOH=-\log [OH^{-}]=7$
Using $K_{w}=10^{-14}$ yields $pH=7$

I know you were posting this to illustrate a point but I fail to see why anything should be different than the usual problem. Does it come from approximating the quadratic for smal numbers?[/quote]
The pH is not 7.

This was a challenge problem in Zumdahl, and the answer is given as 7.20 at the back of the book. Unfortunately I don't see how they got that.

[quote="Carcul"]Compute the pH of a solution of HCl with concentration $[HCl] = 6.5\times10^{-8}M$, at 25ºC.[/quote]

Is it around $pH= 6.78$?

Very close BanishedTraitor, the correct answer is pH = 6.86. The [H3O+] in solution is given by

$[H_{3}O^{+}] = \frac{\sqrt{C^{2}+4K_{w}}+C}{2}$

where C is the concentration of the solution (in this case $6.5 \times 10^{-8}$) and Kw the ionic product of water.

[quote="Rnwang2"]This was a challenge problem in Zumdahl, and the answer is given as 7.20 at the back of the book. Unfortunately I don't see how they got that.[/quote]

The reasoning is the same and we get that

$[H_{3}O^{+}] = \frac{2K_{w}}{\sqrt{C^{2}+4K_{w}}+C}$

where C is concentration of the NaHO. Substitute C by $1 \times 10^{-7}$ and Kw by $1 \times 10^{-14}$ and you will get the correct answer (pH = 7.20).

How can we arrive at these expressions?
Begin by ask yourself what are the concentration of H3O+ and HO- in water before the preparation of HCl solution.

[hide="Answer"]$[H_{3}O^{+}] = [HO^{-}] = 1 \times 10^{-7}$ (at 25ºC)[/hide]

Now ask what are the concentrations of those ions in the instant [i]immediatly[/i] after the preparation of the HCl solution of concentration C.

[hide="Answer"]$[H_{3}O^{+}] = 1 \times 10^{-7}+C$, $[HO^{-}] = 1 \times 10^{-7}$[/hide]

What effect in the ionization equilibrium of water, $2H_{2}O(l) \rightleftharpoons H_{3}O^{+}(aq)+HO^{-}(aq)$, will have the H3O+ from the HCl? So, what will be the equilibrium concentrations of the ions?

[hide="Answer"]$[H_{3}O^{+}] = 1 \times 10^{-7}+C-x, [HO^{-}] = 1 \times 10^{-7}-x$.

To simplify the calculations make $1 \times 10^{-7}-x = t$, and then use the Kw expression:

$[H_{3}O^{+}][HO^{-}] = K_{w}$.

Then solve the resultant quadratic equation for t, and then $[H_{3}O^{+}] = C+t$.[/hide]

A similar reasoning can then be made for the NaHO solution.

Okay, so it does come from approximating the quadratic :wink:
now in your post, you mean $[Cl]$ everywhere instead of $C$, right?

oops you wrote that at the top I guess

Wow thanks very much for the explanation Carcul, I was going to ask something similar  :lol:


geometry trapezoid rectangle trigonometry ratio

the formula for area of a trapezoid is 1/2(b1+b2)*h, where b1 is base 1, b2 is base 2, and h is height.  it will be [hide]4.5 (from the two bases)[/hide] times height, which i can't remember how to get from the angles right now.  :(  do you have the height?

[quote="mr. math"]how do you do the problem:

find the area of the trapezoid ABCD if angle A= 30 degrees, angle d= thirty degrees, side BC= 4, and AD= 5

please include explanation

 :ninja:  :rotfl:  :starwars:[/quote]
[hide]
Since angle A=angle D then the triangles on the sides are congruent making the side of the triangle on the side with length 5 $.5$. Then the side opposite the angle A is $\frac{.5\sqrt{3}}{3}=\frac{\sqrt{3}}{6}$. The area of the two triangles is $\frac{\sqrt{3}}{6}\cdot.5=\frac{\sqrt{3}}{12}$. The area of the rectangle in the middle is $4\cdot\frac{\sqrt{3}}{6}=\frac{2\sqrt{3}}{3}$ so the total area is $\frac{2\sqrt{3}}{3}+\frac{\sqrt{3}}{12}=\boxed{\frac{3\sqrt{3}}{4}}$[/hide]

how do you get the other side?  do you have to use trig?

when it is a 30 degree angle no.
if you have a 30-60-90 triangle, the ratio of the sides is
side opposite 30 degree angle : side opposite 60 degree angle : side opposite 90 degree angle =
x : x*sqrt3 : 2x
whenever you see 30 or 60 degree angles you should have warning signs flash around in your head.

[hide]
You see that the base angles are $30^{\circ}$ each.
You can make 30-60-90 triangles
The part of the base that is the side of the triangles is the length of the lower base minus the upper base. That is 5-4=1
The total base of the triangles is 1, and since there are two, the base of each is $\frac 12$.
The height of the triangles is thus $\frac{\sqrt{3}}{2}$
Using the formula $A=\frac 12 h(b_1 + b_2)$
We know that the height is $\frac{\sqrt{3}}{2}$
The bases are 5 and 4
Applying the formula,
$A=\frac 12 \frac{\sqrt{3}}{2}(5+4)$
Simplifying,
$A=\frac{9\sqrt{3}}{4}$[/hide]

[hide]You can make a 30-60-90 triangle using a leg, the height, and the length 1/2 (since AD-BC/2=1/2). The sides are in a ratio of 1:rt3:2 each side opposite 30:60:90 respectively. As 1/2 is in the ratio to the height rt3:1, we divide 1/2 by rt3, rationalize, and get rt3/6. Put the height into the formula for area of a trapezoid(B1+B2*h/2) and you get 3rt3/4.[/hide]

[quote="ritchjp"]hi easyas3.14159...

I think you were on the right track when you said that the ratio of $\frac{1}{2}$ to the height of the trapezoid is $\sqrt{3} \ : \ 1$. But $\frac{\frac{1}{2}}{\sqrt{3}} = \frac{1}{2 \sqrt{3}}$.

So the height of the trapezoid is $\frac{1}{2 \sqrt{3}}$.

The area of the trapezoid is $\frac{1}{2}(5 + 4)(\frac{1}{2 \sqrt{3}})$ or $\frac{9}{4 \sqrt{3}}$ which is approximately $1.3$.

By golly, it is the same as $\frac{3 \sqrt{3}}{4}$[/quote]
I just rationalized the height.

[quote="mr. math"]i get it that the base is 1/2, but how do you find the height from that(of the triangles), and by the way how do you put a picture under your username :yoda:  :ddr:[/quote]

Find the height from the length of the base? Since the base is opposite to the $60^o$ angle and the ratio of a 30-60-90 degree triangle is $\frac{1}{2},\frac{\sqrt{3}}{2},1$, you divide the length of the base (opposite the 60 degree angle) by $\sqrt{3}$...because...the ratio says to...if that's what you mean. So $\frac{\frac{1}{2}}{\sqrt{3}}=\frac{1}{2\sqrt{3}}=\frac{\sqrt{3}}{6}$.
For the picture thingy, go to profile and at the bottom there's an avatar control panel and you can load a picture from your computer or from a URL.

[quote="biffanddoc"]I don't see how my answer is wrong.
If the base of the triangles are each $\frac 12$, then isn't the height of each $\frac{\sqrt{3}}{2}$[/quote]

No.... since the height is opposite the [b]30[/b] degree angle, it is $\frac{\frac{1}{2}}{\sqrt3}$.


:blush: I'm sorry that I'm poor in english(i'm a senior school student).Who can introduce some programmes for translation?

[quote="wuhao_jshj"][quote="shobber"]The best tanslation program is your brain.[/quote]
I DO NOT MEAN THAT.I MEAN IF I CAN FIND A PROGRAM LIKE A DICTIONARY.[/quote]
There are lots of online dictionary. Simply google for it.


Gauss
function
algorithm
search
number theory
prime numbers

Have anyone heard of the book called "Prime numbers :Most mysterious figures in mathematics" by David Wells and Recreation in the theory of Numbers by Albert Beiler.
If so can you please pour your reviews and comments regarding it .

Here's a link to some customer reviews for both books.

Prime Numbers: [url]http://www.amazon.com/Prime-Numbers-Most-Mysterious-Figures/product-reviews/0471462349/ref=dp_top_cm_cr_acr_txt?ie=UTF8&showViewpoints=1[/url]

Recreation in the Theory of Numbers: [url]http://www.amazon.com/Recreations-Theory-Numbers-Albert-Beiler/product-reviews/0486210960/ref=dp_top_cm_cr_acr_txt?ie=UTF8&showViewpoints=1[/url]

published
June 10th 2005 by Wiley

binding
Hardcover, 288 pages

isbn
0471462349    (isbn13: 9780471462347)

description
A fascinating journey into the mind-bending world of prime numbers

Cicadas of the genus Magicicada appear once every 7, 13, or 17 years. Is it just a coincidence that these are all prime numbers? How do twin primes differ from cousin primes, and what on earth (or in the mind of a mathematician) could be sexy about prime numbers? What did Albert Wilansky find so fascinating about his brother-in-law's phone number?

Mathematicians have been asking questions about prime numbers for more than twenty-five centuries, and every answer seems to generate a new rash of questions. In Prime Numbers: The Most Mysterious Figures in Math, you'll meet the world's most gifted mathematicians, from Pythagoras and Euclid to Fermat, Gauss, and Erd¿o¿s, and you'll discover a host of unique insights and inventive conjectures that have both enlarged our understanding and deepened the mystique of prime numbers. This comprehensive, A-to-Z guide covers everything you ever wanted to know—and much more that you never suspected—about prime numbers, including:

The unproven Riemann hypothesis and the power of the zeta function
The "Primes is in P" algorithm
The sieve of Eratosthenes of Cyrene
Fermat and Fibonacci numbers
The Great Internet Mersenne Prime Search
And much, much more

Source http://www.goodreads.com/book/show/145689.Prime_Numbers_The_Most_Mysterious_Figures_in_Math

(You can also find reviews by people and the cheapest book new/used, if you have trouble finding it just let me know)


The first four are from Junior TST and the other from IMO TST of that year.

The first problem can be done via Cauchy, I think.

Socrates (my favorite philosopher by the way :D ) can you post the problems in separate topics so we can discuss them separately? (post solutions and etc.) :) Thanks.

on a closer look, these are kinda trivial....for example, i think try to find all n for which 111...111 (n digits) is a primes is much more interesting... :D


Let $ c$ be an arbitrary integer number,consider the following sequence:
$ a_{n} \equal{} \frac {a_{n \minus{} 1}}{2}$ if $ a_{n \minus{} 1}$ is even and $ a_n \equal{} 3a_{n \minus{} 1} \plus{} 1$ otherwise.($ a_0 \equal{} c$)
Determine $ f(c) \equal{} min\{i\in\mathbb{N}|a_i \equal{} 1\}$?(at least show the way to determine this value)

[quote="randomgraph"]...and once you find a way to do that, write up your proof, submit it to Inventiones Mathematicae, and wait for the offers of tenured positions to flow in :)

The [url=http://en.wikipedia.org/wiki/Collatz_conjecture]Collatz Conjecture[/url] is the statement that $ f(c)$ is well defined, that is, that $ a_i$ actually is 1 for some $ i$, for any starting value $ c$. Referring to this conjecture, Paul Erdős once said: "Mathematics is not yet ready for such problems."

Of course, for any given $ c$, there is a way to determine the value: iterate the operation until you get 1. This is strictly not an algorithm, just a semi-algorithm, since a-priori it's not clear that it'll ever stop. There are many partial results on this conjecture, some of which are mentioned in the Wikipedia article.[/quote]
well...and how long should I wait? :) I was going to write down the proof right here,but possibility of being left behind does not let me to do it...kidding  :)
I've been perplexed by how this problem is famous...furthermore Paul Erdős quotation doesn't leave me any chance to succeed in solving of this problem...
big thanks for a such nice links(wikipedia rules...somtimes  :) )


AMC
AIME
USA(J)MO
USAMO
function
pigeonhole principle
\/closed

I was wondering when you think the summer/fall classes will be posted?

I was wondering what level class would be most beneficial for me if I was getting about 4-7 on AIMEs and 125+ on AMCs, with a goal of getting into the double digits for AIME.

Yes, Intermediate material includes most AIME-level problems, hard AMC problems, and easy USAMO problems.

[quote="math88fun"]Maybe Intermediate Counting would be a good choice.[/quote]

Perhaps so.  If you're interested in learning how to use recursion to solve problems like this year's AIME #11, as well as inclusion-exclusion, generating functions, 1-1 correspondence, pigeonhole principle, and other topics.  If that sounds like a fun set of topics and you don't feel like you've mastered them already, that's a perfect class for you.