algebra polynomial calculus derivative trigonometry function search

[color=darkred]Let $ z\in\mathbb C$ for which exists $ n\in\mathbb N$ so that $ n\ge 2$ and $ 2^{n \minus{} 1}\cdot\left(z^n \plus{} 1\right) \equal{} (z \plus{} 1)^n$ . Prove that $ |z| \equal{} 1$ .[/color]

For $ n\geq 2$, the polynomial $ 2^{n\minus{}1}(x^n\plus{}1)\minus{}(x\plus{}1)^n$ has nonzero $ x^n$ coefficient, so it has $ n$ complex roots counting multiplicities. Now we can see (by taking derivatives, for example) that 1 is a double root of the polynomial, so it remains to find $ n\minus{}2$ other complex roots with absolute value 1. (If $ n\equal{}2$, we're already done.)

Such roots will have the form $ e^{\theta i}$ for $ \frac{\theta}{2\pi}\not\in \mathbb{Z}$. Now we have
\[ e^{\theta i}\plus{}1\equal{}e^\frac{\theta i}{2}(2\cos \frac{\theta}{2})\]
so after dividing out $ 2^n e^\frac{n\theta i}{2}$, it's equivalent to
\[ \cos \frac{n\theta}{2}\equal{}\cos^n \frac{\theta}{2}\]
We need to find $ n\minus{}2$ solutions in $ (0,2\pi)$. To show this, let $ f(\theta)\equal{}\cos \frac{n\theta}{2}\minus{}\cos^n \frac{\theta}{2}$. Then we have $ f\left( \frac{2\pi k}{n}\right)\equal{}(\minus{}1)^k\minus{}\cos^n \frac{\pi k}{n}$ for every integer $ k$. For the case $ 1\leq k\leq n\minus{}1$, $ f$ is strictly negative for $ k$ odd and strictly positive for $ k$ even. That's $ n\minus{}2$ sign changes, so the continuous $ f$ necessarily has $ n\minus{}2$ zeroes in $ \left[ \frac{2\pi}{n},\frac{2\pi (n\minus{}1)}{n}\right]$.

So if $ z$ were a complex number satisfying $ 2^{n\minus{}1}(z^n\plus{}1)\equal{}(z\plus{}1)^n$ for some $ n\geq 2$, then $ z$ is the root of a polynomial with all of its complex roots having absolute value 1. This forces $ |z|\equal{}1$, as desired.

[b][u]Please[/u], without continue functions or derivatives[/b], if it is possible !

I have an own such proof. See and the exercises from [url=http://www.artofproblemsolving.com/viewtopic.php?t=186913][color=red][b]here[/b][/color][/url] and from [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1438694044&t=42951][color=red][b]here[/b][/color][/url].

I"ll present here my proof [b][u]without Darboux's property or derivatives[/u].[/b]

[color=darkred]Let $ z\in\mathbb C$  for which exists $ n\in\mathbb N$ so that $ n\ge 2$ and $ 2^{n \minus{} 1}\left(z^n \plus{} 1\right) \equal{} (z \plus{} 1)^n$ . Prove that $ |z| \equal{} 1$ .[/color]

[b][u]Proof.[/u][/b] Observe that $ z\ne 0\ (z\equal{}0\implies n\equal{}1)$ . Suppose w.l.o.g. $ z\ne 1$ . Denote $ |z| \equal{} r$ . Thus, $ \overline z \equal{} \frac {r^2}{z}$ and

$ 2^{n \minus{} 1}\left(z^n \plus{} 1\right) \equal{} (z \plus{} 1)^n$ $ \Longleftrightarrow$ $ 2^{n \minus{} 1}\left(\overline z^n \plus{} 1\right) \equal{} (\overline z \plus{} 1)^n$ $ \Longleftrightarrow$ $ 2^{n \minus{} 1}\left(r^{2n} \plus{} z^n\right) \equal{} \left(r^2 \plus{} z\right)^n$ .

Suppose $ \mathrm {ad\ absurdum}$ that $ r\ne 1$ . Therefore, $ \left\{\begin{array}{c} 2^{n \minus{} 1}\left(z^n \plus{} 1\right) \equal{} (z \plus{} 1)^n \\
 \\
2^{n \minus{} 1}\left(r^{2n} \plus{} z^n\right) \equal{} \left(r^2 \plus{} z\right)^n\end{array}\ \right\|\ \boxed \minus{} \ \uparrow$ $ \implies$

$ 2^{n \minus{} 1}\left(r^{2n} \minus{} 1\right) \equal{} \left(r^2 \plus{} z\right)^n \minus{} (z \plus{} 1)^n$ $ \implies$ $ \frac {2^{n \minus{} 1}\left(r^{2n} \minus{} 1\right)}{r^2 \minus{} 1} \equal{} \left|\sum_{k \equal{} 1}^n\left(r^2 \plus{} z\right)^{n \minus{} k}(z \plus{} 1)^{k \minus{} 1}\right|\le$

$ \sum_{k \equal{} 1}^n\left(r^2 \plus{} r\right)^{n \minus{} k}(r \plus{} 1)^{k \minus{} 1} \equal{}$ $ (r \plus{} 1)^{n \minus{} 1}\cdot\sum_{k \equal{} 1}^nr^{n \minus{} k}$ $ \implies$  $ \frac {2^{n \minus{} 1}\left(r^{2n} \minus{} 1\right)}{r^2 \minus{} 1}\le (r \plus{} 1)^{n \minus{} 1}\cdot \frac {r^n \minus{} 1}{r \minus{} 1}$ $ \implies$

$ \boxed {\ 2^{n \minus{} 1}\left(r^n \plus{} 1\right)\le (r \plus{} 1)^{n}\ \ (*)\ }$ because $ \frac {r^n \minus{} 1}{r \minus{} 1} > 0$ . From the relation $ (*)$ and the well-known inequality

$ 2^{n \minus{} 1}\left(r^n \plus{} 1\right)\ge (r \plus{} 1)^{n}$ obtain $ 2^{n \minus{} 1}\left(r^n \plus{} 1\right) \equal{} (r \plus{} 1)^{n}$ , i.e. $ r \equal{} 1$ , what is $ \mathrm {absurd}$ . In concluzion,

$ 2^{n \minus{} 1}\left(z^n \plus{} 1\right) \equal{} (z \plus{} 1)^n$ $ \implies$ $ |z| \equal{} 1$ .

[b][u]Remark.[/u][/b] Here is an equivalent enunciation :

[color=darkred]Let $ \{u,v\}\subset \mathbb C$ for which exists $ n\in\mathbb N$ so that $ n\ge 2$ and $ 2^{n \minus{} 1}\left(u^n \plus{} v^n\right) \equal{} (u\plus{} v)^n$ . Prove that $ |u| \equal{} |v|$ .[/color]


geometry
perimeter
geometric transformation
reflection
geometry unsolved

Let $ P$ be an interior point of triangle $ ABC$, the cevian triangle of $ P$ is $ LMN$, determine $ P$ such that the perimeter of triangle $ LMN$ attains minimum.

It is well- known that this occurs for the orthocenter.
Sketch: We will actually show that the triangle with vertices as the feet of the altitudes gives the minimal perimeter of any triangle with vertices on the sides. Fix L on AB, let P and Q be the reflections of it over AC and BC, and X and Y the intersections of PQ with AC and AB. PX=LX and QY=LY then, so we see the perimeter of the inscribed triangle equals PQ, which is clearly minimized when M=X and N=Y. So we have to find the best choice for L. Since CP=CQ=CL, and angle PCQ is fixed, we find this occurs when CL is minimum, or L is the foot of the perpendicular from C to AB. We just have to show now that in this case, M and N also turn out to be the feet of the perpendiculars.

This was first given by Hungarian mathematician L Fejer in 1900, it is given in the book Geometric Problems on Maxima and Minima by Titu Andreescu, Oleg Mushkarov, and Luchezar Stoyanov

Exercise: Find all mistakes in this solution.

  darij

Is it because the triangle has to be acute  :P ? Or are there more mistakes?

[quote="dgreenb801"]Is it because the triangle has to be acute  :P ?[/quote]

Yes, this is the point where one starts wondering why the solution yields a wrong answer.
See http://www.cip.ifi.lmu.de/~grinberg/errata.html for details (scroll to "Page 169, §264, Proof of the last Theorem" in Johnson).

  darij


superior algebra
superior algebra solved

I am trying to find a noncommutative ring without a  unity...please help... thanks

noncommutative rings can easily be found by taking the ring of n dimensional square matrices over a ring $R$ , we denote this by $M_{n}(R)$

now for what you ask i suggest you take the ring of all 2 by 2 matrices, with entries even integers


The number 64 has the property that it is divisible by its units
digit. How many whole numbers between 10 and 50 have this property?

Why does 31 work though...it is not divisible by 3...

$ 1$ is the units digit for $ 31$, and the question stipulates divisibility by the units digit, which is obviously true for $ 31$.

If you think about it, you can tell that some numbers are always divisible by their units digit. Every integer is divisible by 1, so all integers ending in 1 will work. Every integer ending in 2 will work because all integers ending in 2 are even which means that they are divisible by 2. Every integer ending in 5 will work because all integers ending in 5 are divisible by 5.

Similarly, none of the integers ending in 0 will work because integers are not divisible by 0.

[hide=Alcumas Answer]Twelve numbers ending with 1, 2, or 5 have this property. They are 11, 12, 15, 21, 22, 25, 31, 32, 35, 41, 42, and 45. In addition, we have 33, 24, 44, 36, and 48, for a total of $\boxed{17}$. (Note that 20, 30, and 40 are not divisible by 0, since division by 0 is not defined.)[/hide]

[quote=BarbarianKing][hide=Alcumas Answer]Twelve numbers ending with 1, 2, or 5 have this property. They are 11, 12, 15, 21, 22, 25, 31, 32, 35, 41, 42, and 45. In addition, we have 33, 24, 44, 36, and 48, for a total of $\boxed{17}$. (Note that 20, 30, and 40 are[b] not divisible by 0, since division by 0 is not defined.)[/hide][/quote]

You spelled alucmas wrong. It is alc[b][b][b]u[/b][/b][/b]mus.

I think @BarbarianKing does it on purpose...it's been pointed out before -.-

[quote=rubik]let x be the tens digit, and y the unit digit.

Then, 10x+y is divisible by y.
$ \frac{10x\plus{}y}{y} \equal{} \frac{10x}{y} \plus{} y$ has to be an integer.
So $ \frac{10x}{y}$ is an integer.

We systematically check every integer between 10 and 50 for the property that 10x/y is an integer.
There aren't that many... the ones that work are:
11, 12, 15, 21, 22, 24, 25, 31, 32, 33, 35, 36, 41, 42, 44, 45, 48

Therefore, 17 integers satisfy this property.

EDIT: misread problem... corrected
This is a somewhat slow method... does anyone have there a better way to solve this problem?[/quote]

I did it this way too. But seems only slightly better than brute force. Amy other way to do it?

Is there a faster way to do this problem maybe with an equation because this problem was on the amc 8 and you only have 40 minutes to answer 25 questions on the amc 8
                    Thank you so much.

This way can be pretty fast, with practice.
[hide]We automatically see that numbers that end in 1,2, or 5 qualify, so that is 12 numbers so far. Then you can just check each of the other numbers by using the divisibility rules. You get 5 more possibilities, and 12+5=17. [/hide]

why isn't 24 in this? 24 is divisible by 4


We have $ \frac {n \plus{} 1}{2}a_n \plus{} \frac {a_{n \plus{} 1}}{n \plus{} 1} \equal{} \frac {n \plus{} 2}{2}a_{n \plus{} 1}$ then $ a_{n \plus{} 1} \equal{} \frac {(n \plus{} 1)^2}{n(n \plus{} 3)}a_n$.
From this we have $ a_n \equal{} \frac {3n}{(n \plus{} 1)(n \plus{} 2)}a_1$
Then $ \lim_{n\rightarrow\infty}(2008 \plus{} n)a_n \equal{} 1004$


[quote="SUPERMAN2"]Given a triangle ABC,knowing that there exists 3 points X,Y,Z lie on BC,CA ,AB respectively such that AX=BY=CZ and angle BAX=CBY=ACZ.Prove that ABC is equilateral.[/quote]
Help me with this problem,please.


function
combinatorics unsolved
combinatorics

Find the least natural number $ n$ ($ n\ge3$) with the following property: for any coloring in 2 colors of $ n$ distinct collinear points $ A_1,A_2,\ldots,A_n$ such that $ A_1A_2\equal{}A_2A_3\equal{}\cdots\equal{}A_{n\minus{}1}A_n$ there exist three points $ A_i,A_j,A_{2j\minus{}1}$ ($ 1\le i\le2j\minus{}i\le n$) which are colored in the same color.

Let $ f: A_i\rightarrow \{1,2\}$ be the function that associates every $ i$ th point $ A_i$ to one of the colors 1 or 2. Consider the triple $ (A_1,A_1,A_n)$ (which is actually a double $ (A_1,A_n)$), if $ f(A_1)\equal{}1$ wlog then $ f(A_2)\equal{}2$. Similarly, $ f(A_3)\equal{}1$ by considering the triple$ (A_2,A_2,A_3)$.

Consider the sequence $ \{a_\infty \}\equal{}\{3,5,9,\dots \}$ which obeys the recurrence relation $ a_{n\plus{}1}\equal{}2a_n \minus{}1$. We shall color the points so that
if $ f(A_i)\equal{}1$ then $ f(A_{i\plus{}1})\equal{}2$
or if $ f(A_i)\equal{}2$ then $ f(A_{i\plus{}1})\equal{}1$.

As we move from left of $ \mathbf{N}$ from 3 onwards we observe that either $ k$ th point belongs to a sequence else a new sequence starts from there (for example like $ A_4$) with the same recurrence relation (for example like $ 4,7,13,\dots$). Since every pair of form $ (j,2j\minus{}1)$ have different colours, we cant find a triple with all of them having same colour.

We are left to prove that no two sequences have a common element. Let $ \{a_\infty\}$ and $ \{b_\infty\}$ be two different sequences. Let $ a_k\equal{}b_r$ which implies $ a_1\equal{}b_{r\minus{}k\plus{}1}$ which means we are trying to start a sequence from a certain point when its already a part of sequence $ b$. Since this is not possible, we conclude that no two sequences have a common element.

Hence there does not exist a least natural number $ n$.

The coloring BWWBBWWB doesnt have any triple of points satisfying the condition.

If there are 9 points, there are 5 points in one color, let's say black. We can't color three consecutive points on the same color, and if there are no adjacent black points the arrangement is BWBWBWBWB which clearly satisfies the condition. So there are two adjacent points in black. Then, there are three cases  (WE DONT KNOW THE COLOR X):

- BBW --> BBWB --> BBWBXWW --> BBWBBWWXX --> B[b]B[/b]WB[b]B[/b]WW[b]B[/b]X  Contradiction.

- BBW --> BBWW --> BBWWB --> BBWWBXXWW--> BBW[b]WBXBW[/b]W If X = B or W there is BBB or WBWBW. Contradiction.

- WBBW --> WBBWXXB --> WBBWBXB --> WB[b]B[/b]W[b]B[/b]W[b]B[/b] Contradiction.

Then n=9


What is the smallest possible perimeter, in units, of a triangle whose side-length measures are consecutive integer values?

start trying some-
1,2,3, but that is degenerate
2,3,4 works

$ 2\plus{}3\plus{}4\equal{}9$


geometry
geometric transformation
rotation
reflection
AwesomeMath
summer program
pigeonhole principle

Find the six configurations that are unique under rotation and reflection of four distinct points on the plane such that among the $ \binom{4}{2} \equal{} 6$ line segments that may be drawn with the points as endpoints, there are only 2 distinct lengths. Prove that these are the only six configurations. (For example, four points on the vertices of a square would be one configuration, having only the lengths $ k$ and $ k\sqrt {2}$.)

EDIT: I know 6 configurations, but not how to prove that they are the only ones.

No responses?! :(  Is the problem worded poorly? Is the problem trivial? Is it too difficult?

(Problem from Zuming Feng, AwesomeMath 2008, don't know the original source though)

This seems like an ugly casework problem...

Take one vertex, it has 5 edges emanating from it. Since we have two lengths, we must have at least 3 edges of the same length by the pidgeonhole principle. From here, it should be pretty easy to finish: the endpoints of these edges make an isosceles triangle, etc.


[quote="21 7 15"]Prove that: $ \frac{n!}{r!(n\minus{}r)!}\plus{}\frac{n!}{(r\plus{}1)!(n\minus{}r\plus{}1)!}\equal{}\frac{(n\plus{}1)!}{r!(n\minus{}r\plus{}1)!}$[/quote]

[hide="solution"]$ \frac{n!}{r!(n\minus{}r)!}\equal{}\binom{n}{r}$

$ \frac{n!}{(r\plus{}1)!(n\minus{}r\plus{}1)!}\equal{}\binom{n}{r\plus{}1}$

$ \frac{(n\plus{}1)!}{r!(n\minus{}r\plus{}1)!}\equal{}\binom{n\plus{}1}{r}$

[[Pascal's Identity]][/hide]

In the original problem, shouldn't $ (r \plus{} 1)!$ be $ (r \minus{} 1)!$ instead?

hey plz explain how does (n/r)+(n/(r+1)=(n+1)/r

[hide="A combinatorial approach"]We are to count the number of teams of $ r$ members that can be formed from a group of $ n \plus{} 1$ people. Choose any one of the people and call him $ X$. The number of teams that already has $ X$ as a member is $ \binom{n}{r \minus{} 1}$, and the number of teams that does not have $ X$ as a member is $ \binom{n}{r}$. Together they sum up to the total number of teams $ \binom{n \plus{} 1}{r}$.[/hide]

The original identity should be written $ {n \choose r} \plus{} {n \choose r\minus{}1} \equal{} {n\plus{}1 \choose r}$.  The combinatorial proof is instructive, but the algebraic proof stays valid when $ n$ isn't a positive integer. $ n$ can actually be any complex number, where

$ {n \choose r} \equal{} \frac{n(n\minus{}1)...(n\minus{}(r\minus{}1))}{r!}$

and $ r$ is a non-negative integer.  This allows us to state the [i]generalized Binomial Theorem[/i], which is the expansion of $ (x \plus{} 1)^n$ for $ n$ not a non-negative integer (this expansion is infinite).


Solve these two equations to obtain f(x).

$ f(x) \plus{} f\left(\frac{1}{x}  \right ) \equal{} \frac{1}{a\plus{}b}\left [ x \plus{} \frac{1}{x}\minus{}10 \right ]$

$ f(x)\minus{} f\left ( \frac{1}{x} \right )\equal{} \frac{1}{a\minus{}b}\left [ \frac{1}{x}\minus{}x \right ]$

Answer : $ \frac{1}{a^2\minus{}b^2}\left \{ \frac{a}{x} \minus{} (b\times x) \right \}\minus{} \frac{5}{a\plus{}b}$

Thank you.

[quote="almonkez"][quote="ninni"]Solve these two equations to obtain f(x).

$ f(x) \plus{} f\left(\frac {1}{x} \right ) \equal{} \frac {1}{a \plus{} b}\left [ x \plus{} \frac {1}{x} \minus{} 10 \right ]$

$ f(x) \minus{} f\left ( \frac {1}{x} \right ) \equal{} \frac {1}{a \minus{} b}\left [ \frac {1}{x} \minus{} x \right ]$

Answer : $ \frac {1}{a^2 \minus{} b^2}\left \{\frac {a}{x} \minus{} (b\times x) \right \} \minus{} \frac {5}{a \plus{} b}$

Thank you.[/quote]

[url=http://www.servimg.com/image_preview.php?i=264&u=12899977][img]http://i89.servimg.com/u/f89/12/89/99/77/almonk13.jpg[/img][/url]
 :coolspeak:[/quote]

Thank you. I got that. I was solving it some different way, by that I could not reach to the solution. thanks. :-)


i got this calculator cable for TI-83 plus and i dont get how to use it

i downloaded the TI connect...now what

*sigh*...

what a n00b...

anyway, you then plug the calc in if you need to download something or another.

one end goes in the calculator, the other end goes in the USB port. then magic happens. or actually, use TI-Connect to send the downloads to the calculator.

well obviouslyi know how to plug my calculator in :D  :mad:

i figured it out but i didnt do it the way you did...

now on my TI-89 titanium, how do i activate programs?

[quote="elston"]Has anybody have had trouble syncing their TI-89 Titanium to their computer? I plug in the USB cable, and the computer recognizes it but the software doesn't, reinstalled the softare already, any hints?[/quote]

Are you using TI Connect?  Sometimes the computer takes a couple seconds to recognize that the calculator is plugged in.  If it doesn't start working I would download the most recent version of TI Connect (I think its freely available on the net) and install that.

I use TI Connect with the silver link cable on a TI-89 and I've never experienced any problems.


calculus LaTeX integration search \/closed

I want to give a soluiton for a calculus problem that will be a reference item

How are readers supposed to find the reference if they can't see it in the subject line??  For instance I just tried to post the latex code for$\int cos\theta d\theta$ in the subject line but it will not display

[quote]How are readers supposed to find the reference if they can't see it in the subject line?? [/quote]
There is a search button. Many people, including myself, consider it a bad form to use math symbols in the title of a math paper.

For reasons explained above (bad idea in a math paper to do that) and also for display reason that feature will not be implemented.


Who here is going to PAMO this year?
I just heard that I am going :wow:

congrats  :D

our team hasn't been selected  yet , but i think i'am going too :lol:

Sorry for the question, but where can I find problems of this olympiad???? Thanks...........

[quote="Melissa"]You can find them on this site's resources page :)
[url]http://www.mathlinks.ro/Forum/resources.php?c=1&cid=28[/url][/quote]

Thanks!!! Nice problems

I'm going!!!!!!! Like you didn't know that :D
can't wait!!


email
limit
trigonometry
LaTeX
calculus
derivative
calculus computations

I will paypal someone 20 dollars if they can work out these 13 problems and show the work step by step..... the questions can be found here...   http://www.southwestdental.net/cal.pdf .  I need these answeres by mid afternoon thursday 2-16-06 because my midterm is friday and I need to study them but very few people at my school are willing to help.  Please email me if you would like to do this or think you could possibly help.

skgallion@gmail.com
Steve

[quote="famtrecrew"]I will paypal someone 20 dollars [/quote].
Wrong place...
[quote="famtrecrew"]
I need these answeres by mid afternoon thursday 2-16-06
[/quote]
Wrong time...
[quote="famtrecrew"]
my midterm is friday and I need to study them but very few people at my school are willing to help.
[/quote]
That's a real pity. Maybe you should consider getting a private tutor. Here you can get help any time for free but you need to post problems one per time and let people solve them at their leisure. I will be not surprised if nobody gets interested in your offer but that's not because people are unwilling to help, just because you ask for too much and too soon at once...

As to the problems themselves, here are the  solutions for #1 and #13. Maybe somebody else will help you with other ones :)

#1. The limit of the sum is the sum of the limits (provided that both limits exist). Now $\lim_{x\to\infty}\frac{x^2+\sin x}{x^2-2x-3\cos x}=\lim_{x\to\infty}\frac{1+\frac{\sin x}{x^2}}{1-\frac 2x-3\frac{\cos x}{x^2}}$. Note that every expression in the numerator and in the denominator except $1$ tends to $0$, so the limit is $\frac 1 1=1$. The second limit $\lim_{x\to\infty} \frac{\sin 1/x}{1/x}=\lim_{y\to 0}\frac{\sin y}{y}=1$ The latter one of the few standard limits you should know by heart. So, the entire expression tends to $1+1=2$.

#13. Since $f(x)$ is continuous, it is enough to find two points where $f$ takes values of different sign. But the direct computation shows that $f(0)=-3<0$ while $f(2)=7>0$.

[quote="jmerry"]
The right way to ask for help on these:
[b]Work them first yourself[/b]. Then, when you get stuck on a few of them,...[/quote]
An excellent advice! I would sign under it except I'm afraid that we are dealing with the worst case scenario here, i.e., famtrecrew studied so well during the term that he got stuck on [b]all[/b] of them. Whether something can (or even should) be done to help a person in such a case, is a topic for the Round Table rather than for Computations and Tutorials. I'm inclined to give almost everybody a second chance, so, sometimes I would do it for my students. (Actually, I usually run a review session before each exam and they are allowed to ask questions about how to solve the practice problems. Of course, the assumption is that they have tried their best to solve them themselves, but I never try hard to verify this assumption). Anyway, as I said, the place and the time are wrong but if some people have free time and mercy, why not to give famtrecrew one time what he wants? If it helps him to study, that'll be great; if he is beyond any help by now, it wouldn't make the situation any worse. Not that I myself have time right away to write lengthy explanations for 11 problems (though, if he chooses just 2 more of them, I'll try to help).

[quote="jmerry"]Speaking as a TA:
I wouldn't do this for one of my own students; why would I do it for you?

The right way to ask for help on these:
[b]Work them first yourself[/b]. Then, when you get stuck on a few of them, ask for specific help on those problems. You'll find people are much more likely to answer a few specific questions than to do the whole list of practice problems for you.[/quote]

I did all of them myself...but with no answers to compare it to how do i know if they are right??? I would gladly post all the problems that I worked out but i dont know how to show the work in this forum the way you guys do....

You don't have to use $\LaTeX$ to post here- we can read plain text just fine, with appropriate parentheses, ^ for exponents, etc. Mathematical meaning is more important than looks.

ok so i have worked through the majority of the practice test and the problem that is holding me up the most is 4 and 9.  4 states: Find the second derivative d^2y/dx^2 , to the parametric curve defined by x = (2t+1)^(1/2) , y= 2t^2+2t when t=4.  When i work this problem out I am using the quotent rule and coming up with something like this=> (4t+2)/[(1/2)(2)^(-1/2) but past that (which i think is probably wrong anyway) i dont know what to do........

On problem 9 it states.  A drop of mist is a perfect sphere.  The drop picks up moisture at a rate proportional to its surface area.  Show that under these circumstances the drop's radius increases at a constant rate.  I am completly confused on how to even approach this problem.  If someone could post a hint on where to start I would appreciate it.. it does not have to be a full answer just a starting point would help.

[quote="famtrecrew"]
I did all of them myself...but with no answers to compare it to how do i know if they are right??? [/quote]
If all you need is answers, that is easy (but why didn't you say so in the first place? :?)
Here goes:

2) $f'(x)=\frac{x-3}{(x-1)x}+\frac{x+2}{(x-1)x}-\frac{(x+2)(x-3)}{(x-1)^2 x}-\frac{(x+2)(x-3)}{(x-1)x^2}$. Linearization at $2$: $y=\frac 92(x-2)-2$ (of course, your answers may be written in a different form, but they should be the same up to algebraic manipulations)

3) a) No
    b) The limit is $0$

4) $54$

5) $2x+\cos((x^2+1)^2)\cdot 2(x^2+1)\cdot 2x$ (the expression in the problem is written in an ambiguous way. I understood it as $\sin((x^2+1)^2)$. If  you interpreted it as $(\sin(x^2+1))^2$, the answer becomes $2x+2\sin(x^2+1)\cos(x^2+1)\cdot 2x$.

6)  $\frac{16}3$ feet per second.

7) $-36$

8) a) $x=1, y=1$ or $x=-1, y=1$.
    b)  There must be some misprint here: there are no real $x$ such that $x^2+1+1=1$

10) The tangent line is $y-2=-\frac 14(x-1)$; the normal is $y-2=4(x-1)$

11) About $1\%$

12) $2$

I hope I haven't made a stupid mistake in my computations. If some of your answers disagrees with mine, we can go over the corresponding problem in detail. This leaves you with 2 unanswered questions: the third part of 3 and 9. Before answering them, I'd like to see how you did them.

4: $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{4t+2}{2\cdot\frac12(2t+1)^{-\frac12}}= 2(2t+1)^{\frac32}$
Now $\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dt} \left(\frac{dy}{dx}\right)\cdot\frac1{\frac{dx}{dt}}=4\cdot\frac32\cdot(2t+1)^{\frac12}\cdot \frac1{(2t+1)^{-\frac12}}=6(2t+1)$

9: Write $V=\frac{4\pi}3r^3$. Differentiate with respect to $t$. Write $A=4\pi r^2$. Notice anything?


Find all pairs of integers $ (m,n)$ such that:

$ |(m^2\plus{}2000m\plus{}999999)\minus{}(3n^3\plus{}9n^2\plus{}27n)|\equal{}1.$

Let $ x\equal{}m\plus{}1000$, then $ f(x,n)\equal{}x^2\minus{}1\minus{}3n(n^2\plus{}3n\plus{}9), |f(x,n)|\equal{}1$. If $ 3\not x$, then $ 3|f(x,n)$ - contradition with $ f(x,n)\equal{}\pm 1$.
If $ 3|x$, then $ 3|n, 9|x$ , and $ x\equal{}9t,n\equal{}3y$ give $ t^2\equal{}y(y^2\plus{}y\plus{}1)$.
If $ t\equal{}0$ we get solution $ m\equal{}\minus{}1000,n\equal{}0$,else $ y\equal{}k^2, k^4\plus{}k^2\plus{}1\equal{}l^2$ have not solution, because
$ k^2<l<k^2\plus{}1$.


this is not a very new problem, but I can't seem to solve.  can someone help?

There are $100$ towns in a country and some of them are connected by airlines.  It is known that one can reach every town from any other (perhaps with several intermediate steps).  Prove that you can fly around the country and visit all the towns making no more than
a) $198$ flights;
b) $196$ flights.

Thanks.

We have a connected graph of $n$ vertices and want to touch them all with at most $2n-4$ steps. Obviously, if it is possible for a graph $G$, it's possible for any graph obtained from $G$ by adding edges, so we can remove edges from $G$ till we get a tree (if we don't have a tree, find a cycle and take away an edge; it remains connected).

So it's enough to prove it for trees. But now it's obvious: since a tree has $n-1$ edges, the sum of the degrees is $2n-2$, so there's a vertex with degree at most $1$, call it $X$ and let its neighbor be $Y$. Remove $X$ with its edge and take the path that covers the graph with $n-1$ vertices in $2n-6$ steps; to this path, add edges $YX, XY$ where $Y$ appears in the sequence of steps, and we're done.

why does it follow that from having a degree sum of $2n-2$ that there is a vertex with degree at most $1$? (I know that is true for trees in general but you seemed to derive it from the fact that the sum of degrees was $2n-2$). how do we know that the $n-1$ vertices has a path of $2n-6$?  sorry for being a noob :).
thanks.

It's ok. Well, if the sum of the degrees is $2n-2$, since there are $n$ vertices and each has degree at least $1$ (there are no isolated vertices), there has to be a vertex of degree $1$, as otherwise the sum would be at least $2 \times n>2n-2$. The existence of a path of length $\leq 2n-6$ for the graph with $n-1$ vertices is just inductive hypothesis (note that $2n-6=2(n-1)-4$).

cool thanks. btw, do you know what $n$ it is first true for?

For $n=2$, we need a flight, but $2n-4=0$. For $n=3$, we can do with $2$ flights, and $2n-4=6-4=2$, so $3$ is the smallest $n$.


geometry
3D geometry
analytic geometry
symmetry
algebra
polynomial
combinatorics open

I thought I posted this before, but I couldn't find it, so I apologize if this a double-post.

Consider the unit cube placed at the origin so that each vertex of the cube can be uniquely specified by coordiantes $ (x,y,z)$ where each $ x,y,z$ are either $ 0$ or $ 1$.

The problem: how many ways are there to traverse from the vertex $ (0,0,0)$ to $ (1,1,1)$ without passing through the same vertex twice.

In this case of the 3-dimensional cube, the solution is easy. By symmetry, we may assume our first step is:

1). $ (0,0,0)$ to $ (1,0,0)$

and then multiply our result by 3. Similarly, we may assume our second step is:

2). $ (1,0,0)$ to $ (1,1,0)$

and then multiply our result by 2. Given this, it is easy to count by simply looking at a picture that there are 3 ways to travel from $ (1,1,0)$ to $ (1,1,1)$ without going through either $ (0,0,0)$, $ (1,0,0)$, or $ (1,1,0)$. Thus, the answer is $ 3\cdot 2\cdot 3\equal{}18$.

Now, the real problem: how many ways are there to do this for the 4-dimensional hypercube? That is, how many ways are there to travel from $ (0,0,0,0)$ to $ (1,1,1,1)$ without passing through a vertex twice?

I don't think anybody really knows how to solve self-avoiding walk problems in polynomial time.  I tentatively suggest that the problem for $ d$ dimensions is exponential in $ d$.  See, for example, [url=http://cat.inist.fr/?aModele=afficheN&cpsidt=1020105]this paper[/url].


calculus
derivative
invariant
geometry
geometric transformation
rotation
algebra

Can anything general be said about the solutions of a functional equation of the form:
$ f^{n}(x)+a_{n-1}f^{n-1}(x)+\dots+a_{2}f^{2}(x)+a_{1}f(x)+a_{0}x = 0$ (f - real-valued, $ f^{k}(x) = f(f(f(\dots(x))\dots)))$ without additional assumptions?

If the "characteristic" equation $ \lambda^{n}+a_{n-1}\lambda^{n-1}+\dots+\alpha_{0}=0$ has a real root $ \lambda_{0}$, then $ f(x)=\lambda_{0}x$ solves the functional equation. This condition is also necessary for the existence of a solution that fixes $ 0$ and has a derivative there.

Let's consider an example. If $ f(f(x))+x=0$, then the graph of $ f$ must be invariant under clockwise rotation by $ 90$ degrees -- indeed, along with each point $ (x,y)$ the graph contains $ (y,-x)$. Rotating it twice, we find that the graph is centrally symmetric, which means that $ f$ is an odd function. But then $ f\circ f$ is even*, a contradiction.

Does anyone have an example where the "characteristic" equation has no real roots, yet the functional equation has a solution?

(*) Okay, that is wrong. I still think that $ f(f(x))+x=0$ has no solutions.


Of course not. Take for example $ f(x)\equal{}\minus{} \sqrt {x(1 \minus{} x)}$

I edited my post ,FelixD ,and sorry for my translating  mistake.

[quote="enndb0x"]Prove that if function $ f: [0,1]\rightarrow \mathbb{R}$ satisfy conditions

$ i) f(0) \equal{} f(1) \equal{} 0$

$ ii) f\left(\frac {a \plus{} b}{2} \right) \leq f(a) \plus{} f(b) ,\forall a,b \in [0,1]$

then equation $ f(x) \equal{} 0$ have infinitely many roots in $ [0,1]$ .Is there any such function different from $ 0$ function[/quote]

$ \forall_{a\in [0,1]}b \equal{} a\Longrightarrow 0\leq f(a)$ , $ \ \forall_{a\in [0,1]}b \equal{} 0 \Longrightarrow 0\leq f(\frac a2) \leq f(a)$ , $ \ \forall_{a\in [0,1]}b \equal{} 1\Longrightarrow 0\leq f(\frac {a\plus{}1}2) \leq f(a)$
Now it's easy to see that forall $ n$ (natural number) we have  $ f(\frac 1{2^n}) \equal{} 0$
Exemple  of a such function : $ f(x) \equal{} 0$ if $ (x\in [0,1]\cap Q)$ (rational), $ \ f(x) \equal{} 1$ if $ x\in ([0,1] \minus{} Q)$ (irational)
 :cool:


Suppose $\triangle{ABC}$ is a triangle with 3 acute angles A, B, and C. Then the point $(\cos{B}-\sin{A}), (\sin{B}-\cos{A})$ lies in which quadrants?

[hide="Answer"]I kind of know how to prove it rigorously, but can't figure out how to express it. Testing gives $\boxed{\text{II}}$ as the answer.[/hide]


function
inequalities
complex analysis
superior algebra
superior algebra unsolved

if $||xy||\leq C||yx||$ for every pair of elements $x,y\in B$,where $B$ is banach algebra ,with some constant $C$ ,then $B$ is commutative algebra.

Hint: this has practically nothing to do with superior algebra, it's analysis.  ;)

isn't this standard ? e.g.,

Let $ x,y$ be arbitrary elements of the banach algebra and pick any linear functional $ \phi$ of $ B$. Now define an entire function via $ f(z)\equal{}\phi\left(e^{zx}ye^{\minus{}zx}\right)$ and observe the following inequality:

$ \left|f(z)\right|\leq \parallel \phi\parallel \,\left|\left|e^{zx}ye^{\minus{}zx}\right|\right|\leq C\,\parallel \phi\parallel \,\parallel ye^{\minus{}zx}e^{zx}\parallel \equal{}C\,\parallel \phi\parallel \,\parallel y\parallel $

As a bounded entire function $ f$ must be a constant, i.e. $ 0\equal{}f'(z)\equal{}\phi\left(e^{zx}(xy\minus{}yx)e^{\minus{}zx}\right)$ and setting $ z\equal{}0$ we see that the commutator $ xy\minus{}yx$ must vanish on each linear functional, which is only possible for $ xy\equal{}yx$.


Can anyone prove in plain english why the angle inscribed in a semicircle is a right angle?

Join the angle with the centre of the semi-circle. Two isoceles triangles appear. Calulcate the angles.

You can also complete the circle and use the fact that it cuts off a 180 degree arc, and since the point is on the circle, the angle is 180/2 = 90 degrees.

Since the inscribed angle is one half the arc it intercepts, 1/2*180 = 90.


prove that :
$ \sqrt{\frac{a}{b\plus{}c}}\plus{}\sqrt{\frac{b}{c\plus{}d}}\plus{}\sqrt{\frac{c}{d\plus{}a}}\plus{}\sqrt{\frac{d}{a\plus{}b}}>2$

Presumably the domain is meant to be positive reals (so that the left-hand side is always meaningful).

[quote="The Zuton Force"]The strict condition is not true -- consider $ (a,b,c,d) \equal{} (1,0,1,0)$[/quote]
sorry
$ a,b,c,d \in R^\plus{}$

[quote="mathson"]prove that :
$ \sqrt {\frac {a}{b \plus{} c}} \plus{} \sqrt {\frac {b}{c \plus{} d}} \plus{} \sqrt {\frac {c}{d \plus{} a}} \plus{} \sqrt {\frac {d}{a \plus{} b}} > 2$[/quote]
we have $ a \plus{} b \plus{} c \plus{} d \geq 2\sqrt {a(b \plus{} c \plus{} d)}$ leads to $ \frac {1}{\sqrt{a(b \plus{} c \plus{} d)}} \geq \frac {2}{a \plus{} b \plus{} c \plus{} d}$ which means $ \sqrt {\frac {a}{b \plus{} c \plus{} d}} \geq \frac {2a}{a \plus{} b \plus{} c \plus{} d}$
setting up 3 other similar one we have the following inequality
$ \sqrt {\frac {a}{b \plus{} c \plus{} d}} \plus{} \sqrt {\frac {b}{c \plus{} d \plus{} a}} \plus{} \sqrt {\frac {c}{d \plus{} a \plus{} b}} \plus{} \sqrt {\frac {d}{a \plus{} b \plus{} c}} > 2$
however , notice that $ \sqrt {\frac {a}{b \plus{} c }} > \sqrt {\frac {a}{b \plus{} c \plus{} d}}$ ...

finally we have desired result :)

[quote="polskimisiek"][quote="HTA"] leads to $ \frac {1}{a(b \plus{} c \plus{} d)} \geq \frac {2}{a \plus{} b \plus{} c \plus{} d}$ [/quote]

I think you missed something :wink:[/quote]
thanks polskimisiek , it was a typo , i edited it .


1. a balloon has a volume of 5.00 liters of 22 degree celsius and 1.00 atmosphere of pressure.  What will be the balloons's volume in liters if the pressure is increased to 3.00 atmospheres at 22 degree celsius?

a. 1.67    b. 0.600  c. 1.30  d. 0.800  i chose a.

#2.  a 3.2 liter sample of gas is at 40 degree celsius and 1.0 atmosphere of pressure. If the temperature decreases to 20 degree celsius and the pressure decreases to 0.60 atmospheres, what is the new volume in liters.

a. 5.7   b. 5.0    c. 1.8   d. 0.20   i chose b.

#3  A gas sample has a volume of 500ml at 20 degree celsius.  What celsius temperature would the gas have a volume half as large.

a 10  b. 147    c.  -147      d.   -127  i chose d, but wasn't sure how to work problem

All your answers are correct. In problem #3, use Charles' Law:

$\frac{V_{1}}{T_{1}}= \frac{V_{2}}{T_{2}}$.

One can generalize these situtations with the ideal gas law:

$PV = nRT$

It follows from this that:

$\frac{P_{1}V_{1}}{n_{1}T_{1}}= \frac{P_{2}V_{2}}{n_{2}T_{2}}$


But even if technology steps in and all that, there might be a huge debate on the ethics and all that stuff of using technology to alter life by itself..... {just look at the debate on human cloning...l..}......So I kind of feel two decades is too short....that stuff might quite continue for at least about 50-60 years from now.

[quote="Dog of Justice"][url=http://www.foreignpolicy.com/story/cms.php?story_id=3376]http://www.foreignpolicy.com/story/cms.php?story_id=3376[/url]

My take: this stands to be a dominant historical force over the next few decades.  (Perhaps longer, but I'm optimistic that advanced technology can eventually step in and radically change the rules of the game.)[/quote]

From the article

[quote]
The greatly expanded childless segment of contemporary society, whose members are drawn disproportionately from the feminist and countercultural movements of the 1960s and 70s, will leave no genetic legacy. Nor will their emotional or psychological influence on the next generation compare with that of their parents.[/quote]

Ironically these are the same people who have been calling their opponents `dinosaurs'.

[quote]In addition to the greater fertility of conservative segments of society, the rollback of the welfare state forced by population aging and decline will give these elements an additional survival advantage, and therefore spur even higher fertility.[/quote]

Welfare leads to more conservatives. Put that in your pipe and smoke it, hippies! (that was a joke)

General comments:

There seems to be no good argument in the article for why `patriarchy', and not some wider cause, is connected to high birthrates. He seems to be assuming implicitly that only women who are controlled by men reproduce in any significant proportion, which does not seem obvious

anyway, to help fight the problem of educated people not reproducing enough, I intend to become a sperm donor. Hopefully my progeny will be as strong and intelligent and agressive as I am, but more importantly, if they are as fruitful as me, well, there just might be hope.....

as for the return of patriarchy-- you make it sound like that's a [i]bad[/i] thing.  :lol:

Anyway, a more serious and related topic is demographic conquest-- certain groups of people are being encouraged by their leaders to reproduce in greater numbers, and this not only may threaten our western `liberal`values, but our conservative ones as well.

If one assumes that a downward sloping demand curve applies to children as well as almost everything else ever analyzed, one can ask why the US tax law provides economic incentive to have children, unless one has income beyond a certain level.

The only logical assumptions that can be behind this are:

A) Politicians have so little backbone and class warfare is so strong that they refuse to do the right thing at political cost today,

B) Those drafting the tax code don't understand basic economics and how the tax code influences behavior,

C) Politicians believe that the economic system in the US is random and there are no inheritable characteristics that are preferentially accumulated in higher income families, or

D) Politicians want to have the leadership of this country taken over by immigrants.

Please let me know if you can find another logic alternative that I've overlooked.

This is dismal.  The government has few tools to address this issue and they are using one to make it worse rather than reduce the significance of the problem with respect to the higher income "subclass" of their country.


Let [i]p[/i] be an odd prime and a be any integer which is not congruent to 0 modulo [i]p[/i]. Prove that the congruence x^2 = -a^2 mod p has solutions if and only if p=1 mod 4

I don't have any idea how to tackle this problem...

Thank you TUAN!!!

can i have some more detailed explanation..?

$x^{2}\equiv-a^{2}\pmod{p}\Longleftrightarrow (xa^{-1})^{2}\equiv-1\pmod{p}.$ ($a^{-1}$ exist because $\gcd (a,p)=1$).

It is the Legendre Symbol, not the usual division.  :wink:

[quote="bthings_2000"][quote="N.T.TUAN"]$x^{2}\equiv-a^{2}\pmod{p}\Longleftrightarrow (\frac{-1}{p})=1\Longleftrightarrow p\equiv 1\pmod{4}.$[/quote]

I do not understand what you mean by $(\frac{-1}{p})=1$ because that implies p=-1... or am I missing notation? Please explain.[/quote]

That fraction-like thingy is actually a Legendre symbol:
\[\left( \frac{a}{p}\right) =\left\{ \begin{array}{rl}0,&p|a\\ 1,&p\not| a\text{ and }x^{2}\equiv a\pmod{p}\text{ has solution}\\-1,&\text{otherwise}\\ \end{array}\right.\]
(I've actually seen the Legendre symbol and a fraction appear in the same line in a proof...  :rotfl: )

The key in the problem is that $x^{2}\equiv-1\pmod{p}$ has solution for odd prime $p$ if and only if $p\equiv 1\pmod{4}$. For the "only if" part, raise both sides of $x^{2}\equiv-1$ to the $\frac{p-1}{2}$ power. For the "if" part, you can construct a square root using primitive roots (if $g$ a primitive root modulo $p$, then $g^\frac{p-1}{4}$ is a square root of -1) or Wilson's Theorem (when $p\equiv 1\pmod{4}$, we have $\left( \frac{p-1}{2}\right) !$ (that's a fraction here) as a square root of -1).


Hello!

  Let ABCD be a convex quadrilateral. Let P be the intersection point of AB and CD and let I_1 and I_2 be the incenters in the triangles ABC and BCD respectively.Let E be the intersection point of I_1I_2 and AB and F the intersection point of I_1I_2 and CD.
Prove that PE=PF if and only if ABCD is cyclic.

Can you help me with this problem, please?

This is from CWMO 2004.
Try to prove $I_1I_2CB$ is cyclic.

Or look at http://www.mathlinks.ro/Forum/viewtopic.php?t=58690 (at least for the direction $PE=PF$ $\Rightarrow$ the quadrilateral ABCD is cyclic; the other direction is analogous).

  Darij


limit
calculus
integration
calculus computations

Evaluate the limit:

$ \lim_{n \to \infty} \left( \sum_{k \equal{} 0}^{n}{\frac {1}{\binom{n}{k}} } \right) \equal{} ?$

Where $ \binom{n}{k}\equal{} \frac{n!}{k!(n\minus{}k)!}$.

http://mathworld.wolfram.com/BinomialSums.html

(31) and on.

Thanks :).

Somebody knows a proof without using integrals?

[hide=" Another approach... (click here)"]Let $ a_n \equal{} \sum_{k \equal{} 0}^n {\frac {1}{\binom{n}{0} }}$.

It is clear that $ a_n \geq 2$.

We have that $ a_n \equal{} 1 \plus{} \frac {1}{\binom{n}{1}} \plus{} \frac {1}{\binom{n}{2}} \plus{} \dots \plus{} \frac {1}{\binom{n}{n \minus{} 2}} \plus{} \frac {1}{\binom{n}{n \minus{} 1}} \plus{} 1$.

 But then $ a_n < 2 \plus{} \frac {2}{n} \plus{} \frac {n \minus{} 3}{\binom{n}{2}} \equal{} 2 \plus{} \frac {2}{n} \plus{} \frac {(n \minus{} 3) \cdot 2! \cdot (n \minus{} 2)!}{n!} \equal{} 2 \plus{} \frac {2}{n} \plus{} \frac {2(n \minus{} 3)}{n(n \minus{} 1)}$.

So,
\[ 2 < \lim_{n \to \infty}{a_n} < \lim_{n \to \infty} \left(2 \plus{} \frac {2}{n} \plus{} \frac {2(n \minus{} 3)}{n(n \minus{} 1)} \right) \equal{} 2.
\]
Then $ \lim_{n \to \infty}{a_n} \equal{} 2$[/hide]


Let W =Sqrt(3) +i , be a fixed number in the complex plane.

Z is a variable complex number such that  I (Z-W) I <IWI  (where I...I denotes the complex absolute value or modulus)

Find a value of A such that  A< argZ < A+(Pi)

you can use |x| for modulus/absolute value, on querty it should be shift+\. Or in latex you can do $|x|$ or $\left| x \right|$ as well.


induction function algebra polynomial inequalities trigonometry

The functions $ T_n (x)$, $ n \equal{} 0, 1, 2, \ldots$, satisfy the recurrence relation
\[ T_{n\plus{}1} (x) \minus{} 2x T_n(x) \plus{} T_{n\minus{}1} (x) \equal{} 0.
\]
Show by induction that
\[ (T_n(x))^2 \minus{} T_{n\minus{}1}(x) T_{n\plus{}1}(x) \equal{} f(x),
\] where $ f(x) \equal{} (T_1(x))^2 \minus{} T_0(x) T_2(x)$.

Induction?  The useful idea here is to think of $ x$ as a constant and apply standard techniques from the theory of linear recurrences.  The really useful idea here is a certain identity I'll motivate by pointing you [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=175257]here[/url] or alternately you can see why this problem isn't really any harder than proving the corresponding identity for Fibonacci numbers by reading some of the relevant comments [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=217361]here[/url].

If you're wondering, these functions have a name and a purpose:  after giving them certain initial conditions, they are called the [url=http://en.wikipedia.org/wiki/Chebyshev_polynomials]Chebyshev polynomials[/url] of the first or second kind.  (Fun things happen to the recursion when you set $ x \equal{} \cos \theta$ for $ \theta$ a rational multiple of $ \pi$, which is a way to motivate some of the results you'll find.)


pigeonhole principle
greatest common divisor
number theory
relatively prime

In each square of an $ 8$ by $ 8$ board, Michael writes a positive integer not exceeding $ 10$ such that each two numbers that appear in adjacent or diagonally adjacent squares are relatively prime. Prove that some number appears at least $ 11$ times.

[hide="solution"]Divide the 8x8 grid into 16 2x2 squares. If two numbers that are both even appear in the same 2x2 square, they are not relatively prime but they are adjacent. Similiarly, only one number divisible by 3 can appear in each 2x2 square. Therefore, even numbers can only appear 16 times. Numbers divisible by 3 can also only appear in total 16 times, so the other numbers appear 64-16-16=32 times. The other numbers are 1, 5, and 7. There are 3 of these numbers. Using pigeonhole, one number must appear 11 times[/hide]

I'm surprised that the bound is so sharp.

One thing that would be appropriate is to show that you can actually do such a tiling. It isn't totally obvious that such a numbering exists.

grey = 8
white=2
yellow =3
red=9
green=1
blue=5
purple=7

I just realized that it is possible to put $ 1$ in every square so I guess it is trivial.

[quote="pythag011"][hide="solution"]The other numbers are 1, 5, and 7. There are 3 of these numbers. Using pigeonhole, one number must appear 11 times[/hide][/quote]

But isn't every number NOT relatively prime to 1? I thought it'd be impossible to put a 1 on there.

Two numbers are relatively prime if their gcf is 1, so 1 is relatively prime with all integers.

Then why does $ \phi(p) \equal{} p\minus{}1$ for all primes p?

$ 1, 2, 3, ... p\minus{}1$ are relatively prime to $ p$.  What is the issue?

Oops nevermind...1 being relatively prime to everything would let it be on the board I guess.


Find a point(s)  M inside triangle ABC so that the  3
  incircles  of  AMB,BMC,CMA  have a common tangent line.

   Pestich.

It follows from continuity reasons. Take $M$ near to $A$ and then move it toward to $BC$


algebra
polynomial
linear algebra
matrix
vector
complex numbers
linear algebra theorems

It exists I know. It's a theorem. The book I use is Sheldon-Axler he uses the factorization of a polynomial linear transform to prove this. I was wondering how I can see this more intuitively... can anyone suggest a way to do this?

You mean complex linear transformation.  Axler's proof is considered, at least by Axler, to be more intuitive than the usual proof, which is based on the fact that the characteristic polynomial has a complex root.  You should probably develop your intuition about linear transformations.

o.O this is really interesting stuff. Could you elaborate on complex linear transformations? Sorry I just had a course in introductory linear algebra and am now fascinated by the topic.

A linear transform (from a space into itself) is a type of operator. Every linear transform can be represented in matrix form in straightforward manner. Thus, when we refer to "linear transform" we are really referring to a matrix being treated as a transform (i.e., we care about its "transformative" properties, like eigenvalues). The adjective "complex" means (specifically) that the base field of this matrix (that is, the base field of the vector space we are transforming) is the complex numbers.

However, if I understand correctly which theorem is being spoken about, the reliance on the field of complex numbers is superficial, as any algebraically closed field will do.


geometry
geometric transformation
reflection
rotation

Does anyone have any information about the contested problem last year like what number it was, why it was contested, and what other answers were accepted?

If I remember correctly it was what is the smallest two digit number $x$ for which the last two digits of $x^3$ is $x$. The common answer was $24$. However, Texas pointed out that negative numbers weren't considered. So $24$ was NOT accepted. ONLY $-99$ was. Most answer keys, such as the $1$-year competition set, have this changed.

The one posted above was one although it wasn't truly contested. As soon as everyone saw the answer i'm sure they didn't argue.

Anyway, the one that WAS contested was this:

Some random person randomly wanted to find out how many ways the first two turns of tic-tac-toe could be played. She randomly decided that reflections and rotations wouldn't count, but switching X and O would. How many ways could the first two turns be played?

The answer key had the wrong answer on it, but they were grading it with the right answer. We sent in a complaint even though we got it wrong both ways -_-.

PS. If it wasn't uberly obvious, I changed the wording a bit.

Sorry to bump the thread, but does anyone know what the correct answer was supposed to be?


MATHCOUNTS email LaTeX probability AMC AIME geometry

I have NO idea if this will work, but if anyone from the JLS MathCounts Club followed directions and signed up for AoPS, this topic is for discussing the week 1 class and homework problems, or other club-related material. Also feel free to start your own topic if you think it's appropriate. Um. Also tell me how I can be a better teacher. I have the strangest suspicion that you'll have LOTS of ideas there. Um. Yeah.

[quote="solafidefarms"]Eh, so you've taken up teaching? :)[/quote]
Yeah...I'm supposed to be JLS's assistant coach. I spent today trying to explain permutations and combinations. I'm not sure if I made sense to anyone who didn't know them already, but I tried.

[quote="Bluestalk"]um...SURE. Anyway. I'm following directions for once, but I'm supposed to be doing other stuff.  :maybe:

SO. Now what? I still don't really get this site. And the title of this forum's going to expire VERY SOON. Any bright ideas, bookaholic?  :?:

I think you and I are going to be the only ones for awhile, unless "blondeazn" is going to actually register.[/quote]
If she doesn't, make her. I'll probably remind her by email, too. Make the other MathCounts people you know do it too.

What specifically don't you get about this site?

[quote="Edz504"]...just a question...what happens if we don't do our homework?[/quote]
I can't force you to, but in the future I'm going to assume that you know everything I've already talked about, so unless the problems aren't challenging at all, you should probably make sure you can do them so you don't get lost in the future. If the homework is too hard I can help you.

1. who r Edz504 and junggi? i MIGHT think i know who u are, but from past experience....NO.

2. WHAT IS LATEX, HOW DO U USE IT, and WHAT GOOD IS IT?
(gotta do social studies, so....i can't really read the site...)

3. where are the practice problems, and how do you use them?

[quote="junggi"]oh lol bookaholic you made a hw discussion topic? :lol:
omg I wanna be a coach too
edit:I'm coming next week[/quote]
Gaaaah. Why did I ever tell you two? Mean bullies. Okay, you can come if you stay very silent and do not bother me with questions like "If we have four choices for each letter in the word, why aren't there 4+4+4+4 instead of 4*4*4*4 different words?" That and generally don't laugh at me. Please.

[quote]1. who r Edz504 and junggi? i MIGHT think i know who u are, but from past experience....NO. [/quote]
Junggi was the single non-Chinese person on the school team last year. I don't know Edz504.

[quote]2. WHAT IS LATEX, HOW DO U USE IT, and WHAT GOOD IS IT?
(gotta do social studies, so....i can't really read the site...)[/quote]
LaTeX is a program that lets you type math expressions and symbols that you can't do that well with just the regular keyboard. To use it, you first have to go to the LaTeX page on AoPS and download everything it tells you to. Then go through the tutorials. You don't need to know it that well for it to be useful.

[quote]3. where are the practice problems, and how do you use them?[/quote]
There are links with all kinds of practice problems in the Resources section. The other forums on the message board also have a lot of problems. Also, have you borrowed one of Ms. Beddoes' binders with old tests yet? If not you should.

[quote="bookaholic"]
[quote]1. who r Edz504 and junggi? i MIGHT think i know who u are, but from past experience....NO. [/quote]
Junggi was the single non-Chinese person on the school team last year. I don't know Edz504.
[/quote]

haha...I feel special

POP QUIZ
FILL IN THE BLANK
Edz504=____
Bluestalk=_____

Im guessing Bluestalk is susan or linda?
edit:
my sister's(kimberly) username on aops is juyoung or something. I don't think she ever posted

I'm going to use this topic to criticize your teaching. :D  You need to articulate yourself better.  You were basicly just going through the calculations as if everyone knew what you were doing.  If I really had that question (why do you multiply the ways?  why not add?), your explanation really wouldn't have cut it.  You could have "illustrated" the method by drawing a tree diagram or something of the sort.

Just a few thoughts on what I saw.

Have fun torturing people with Mathcounts. :)

[quote]...just a question...what happens if we don't do our homework?[/quote]

You die a painful death.  It's your decision to not do your homework, but you will have to live (or die) with the consequences.

edit: added crapcrap pun

[quote="ccy"][quote]...just a question...what happens if we don't do our homework?[/quote]

You die a painful death.  It's your decision to not do your homework, but you will have to live with the consequences.[/quote]
 :rotfl:

btw, if you want to do more practice problems, try the mathcounts forum. I havent been there much but it should be good for practice.
BUT! ignore every post by anirudh. He is the best spammer ive seen in my lifetime(except that guy who sent me spam mail with a picture of spam in it)

Lol  Susan, you practically [i]live[/i] on AoPS now. :P

SHUT UP. It's not like I WANT TO.

PLUS since AIM and YIM are blocked, pming is the closest thing I have. JEEZ.

3.1415926535 is 10 digits (decimal) remmber that

yeah no one posted for 3 hours so i umpec it

this is getting TOO SPAMMY OMG.
moogra, PLEASE stop spamming this forum AND the other forums. PLEASE.

ok to make this a non spam post:
try this one

A license plate in a certain state consists of 4 digits, not necessarily distinct, and 2 letters, also not necessarily distinct. These six characters may appear in any order, except that the two letters must appear next to each other. How many distinct license plates are possible(AMC 10 A 2006)?

[quote="i_like_pie"][quote="junggi"][quote="Bluestalk"]I did that and totally guessed. Hold on, I'm going to see if I can do it.

Um.....

How are we supposed to do the part where the letters are next to eachother?

((shows you just how sad I am))[/quote]
[hide="hint"]make the letters into a block. ie consider the 2 adjacent letters as one thingy.
or you can just write down the cases since theres not much[/hide]
ps. GD my wording sucks[/quote]
NOT MUCH? :wallbash:  :stretcher:

[hide="Wild Guess"]$6600000$[/hide][/quote]
uh no. and leave it in factored form its much nicer :)
and uh by "not much" i was referring to where the letters can be(eg ####AA, ###AA#,##AA## BLABLABLA.)

[hide]There are 10000 ways for the digits to be arranged $\left(10^{4}\right).$

The letters can be arranged in 3380 ways $\left(26^{2}\cdot5\right).$

Multiplying that, we get $33800000$ total combinations.[/hide]

[hide]10^4*26^5 yeah. ok i read i like pie's post but i get it now. I was lazy and didn't do it :( [/hide]

oh adn 400th post! :)


Find an expression for the radius x of a circle externally tangent to 3 tangent circles of radius a, b, c.
Equivalent statement: In a triangle with sides a+b, b+c, and a+c, find x such that segments of length a+x, b+x, c+x drawn from the vertices of the triangle are concurrent (with bounds on x of course, such that this is possible).

I had a wierd idea for solving last year's USAMO number 2, but I couldn't get this...

Oh snap thanks, I didn't know that's what Soddy Circles were...
Hmmm okay knowing that makes number 2 on last year's usamo really easy then...[/hide]

Many people used Descartes' theorem (which Soddy's formula is based on) on the USAMO last year. A lot of them did not get full credit because it is a little tricky to reduce the problem to the case where three circles of the covering are tangent.


trigonometry
geometry
rhombus
similar triangles
geometry proposed

A circle is tangent to the sides of a square AB, BC, CD, DA at K, L, M, N respectively. Let U be on (DM) and V on (DN) s.t. BU||KV. Prove that UV is tangent to the circle (the circle inscribed in the square). It's kind of easy, but I posted it anyway  :D

It is enough to prove NV+UM=VU.
If DU=x then
UM=1/2 - x;
UC=1-x ==> AV=1/(2-2x) (since BCU and AKV are similar triangles)
NV=1/(2-2x) - 1/2
VD=1-1/(2-2x)

Thus
UV=\sqrt(VD2+DU2)=1/(2-2x)-x=NV+UM.

would be to prove that, if $ O$ is the center of the square, then $ \angle VOU \equal{} 45^\circ$.
For this issue we note $ AB\equal{}2a$, $ UM\equal{}x$, $ VD\equal{}y$ and $ \frac{x\plus{}a}{2a} \equal{} m$ ( 1 ). From (1) we easily get $ \frac{x}{a}\equal{}2m\minus{}1$ ( 2 ) and $ \frac{a\minus{}y}{a}\equal{}\frac{1}{m}\minus{}1$ ( 3 ), but (2) and (3) represent $ tan \angle MOU$ and $ tan \angle VON$ respectively and, compiling we get $ tan (\angle MOU \plus{}\angle VON) \equal{} 1$, i.e. $ \angle VON \plus{} \angle MOU \equal{} 45^\circ$, q.e.d.

Best regards,
sunken rock

[quote="grobber"] [color=darkred]A circle $ w = C(O)$ is tangent to the sides $ [AB]$ , $ [BC]$ , $ [CD]$ , $ [DA]$ of the square $ ABCD$ at $ K$ , $ L$ , $ M$ , $ N$

respectively. Consider  $ U\in (DM)$ and $ V\in (DN)$ so that  $ BU\parallel KV$ . Prove that $ UV$ is tangent to $ w$ .[/color] [/quote]

[color=darkblue][b][u]Proof[/u].[/b] Suppose w.l.o.g. $ AK = 1$ . Denote $ T\in UV$ so that $ OT\perp UV$ and $ DV = x$ , $ DU = y$ . Therefore,

$ BU\parallel KV\Longleftrightarrow$ $ \triangle AKV\sim\triangle CUB\Longleftrightarrow$ $ \frac {AK}{AV} = \frac {CU}{CB}\Longleftrightarrow$ $ \frac {1}{2 - x} = \frac {2 - y}{2}\Longleftrightarrow$ $ \boxed {2 + xy = 2(x + y)}\ \ (*)$ .

Since $ [ONDM] = [ONV] + [OUV] + [UVD] + [OMU]$ obtain that $ UV$ is tangent to the circle $ w$ $ \Longleftrightarrow$ $ OT = 1$ $ \Longleftrightarrow$

$ 2 = (1 - x) + \sqrt {x^2 + y^2} + xy + (1 - y)$ $ \Longleftrightarrow$ $ x + y - xy = \sqrt {x^2 + y^2}$ $ \Longleftrightarrow$ $ \boxed {2 + xy = 2(x + y)}$ .

[b]Remark.[/b] Denote $ m(\angle NOV) = \alpha$ , $ m(\angle MOU) = \beta$ , $ m(\angle UOV) = \phi$ $ \implies$ $ \alpha + \beta = 90^{\circ} - \phi$ . Therefore, $ \tan\alpha = 1 - x$ ,

$ \tan\beta = 1 - y$ and $ \tan\phi = \frac {1 - \tan\alpha\tan\beta}{\tan\alpha + \tan\beta} =$ $ \frac {1 - (1 - x)(1 - y)}{(1 - x) + (1 - y)}\stackrel{(*)}{\ = \ }$ $ \frac {(x + y) + 2 - 2(x + y)}{2 - (x + y)} = 1\implies$ $ \phi = 45^{\circ}$ .

[u]Otherwise[/u], the point $ O$ is the $ D$-exincentre of $ \triangle UDV$ . Thus, $ m(\widehat {UOV}) = 90^{\circ} - \frac 12\cdot m(\widehat {UDV}) = 45^{\circ}$ .

We can extend this problem to a rhombus ! I"ll come back soon. [/color]


The following identity holds for every "x"  real number:
$(a+x)^{2}+(b+x)^{2}=(c+x)^{2}+(d+x)^{2}$

Prove that:
a)$a^{2}+b^{2}=c^{2}+d^{2}$
b)$a+b=c+d$
c)$a^{n}+b^{n}=c^{n}+d^{n}$

A,b,c,d are real numbers and n is natural.

This problem was given at a Romanian olympiad in 1993.

Isn't this very easy for a Romanian test?  :maybe:
Unless I am missing something, here's a sketch of my solution:

[hide]It's equivalent to \[a^{2}+b^{2}+2x(a+b) = c^{2}+d^{2}+2x(c+d)\]
For a) just set $x=0$
For b) we set $x=\frac{1}{2}$ and use a)

Then c)
$(a+b)^{2}-(a^{2}+b^{2}) = (c+d)^{2}-(c^{2}+d^{2})$ so $ab = cd$

Now we can use induction on $n$
If $a^{i}+b^{i}= c^{i}+d^{i}$ for $1\leq i\leq n$, then \[(a^{n}+b^{n})(a+b)=(c^{n}+d^{n})(c+d)\]
\[\iff a^{n+1}+b^{n+1}+ab(a^{n-1}+b^{n-1})=c^{n+1}+d^{n+1}+cd(c^{n-1}+d^{n-1})\]
So finally also $a^{n+1}+b^{n+1}= c^{n+1}+d^{n+1}$
[/hide]

It was a 7th grade district round  :P .

Grrrrr, i was unsure if we can set x a certain number...   :blush:

[quote="alex__ib"]i was unsure if we can set x a certain number...   :blush:[/quote]

Certainly if an identity holds for every real $x$ then it must hold for any specific real $x$   :P


Let $ a,b,c$ be nonnegative real numbers, prove that
\[ \left( \frac{2a}{c}\minus{}1\right)(a\minus{}b)^2 \plus{} \left( \frac{2c}{b}\minus{}1\right)(c\minus{}a)^2  \plus{}\left( \frac{2b}{a}\minus{}1\right)(b\minus{}c)^2  \ge 0.\]
:)

[quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers, prove that
\[ \left( \frac {2a}{c} \minus{} 1\right)(a \minus{} b)^2 \plus{} \left( \frac {2c}{b} \minus{} 1\right)(c \minus{} a)^2 \plus{} \left( \frac {2b}{a} \minus{} 1\right)(b \minus{} c)^2 \ge 0.
\]
:)[/quote]
For more special, $ 2$ is also the smallest constant so that the inequality
\[ \left( \frac {ka}{c} \minus{} 1\right)(a \minus{} b)^2 \plus{} \left( \frac {kc}{b} \minus{} 1\right)(c \minus{} a)^2 \plus{} \left( \frac {kb}{a} \minus{} 1\right)(b \minus{} c)^2 \ge 0
\]
holds. :)


summer program
Mathcamp
AMC
AIME
USA(J)MO
USAMO
MathPath

Have you guys received any letters from AMC or the contributors of AMC? Because you did a good job on the contest?
I received one invitation from Mathcamp last week.
I hope I can receive a letter from AMC, because there is one in the Teacher's Manual.

When we filled in the answer form for AMC and AIME for "Do you want some univ or orgnizations to collect your info from the test?", does it only mean those unimportant orgnizations? Or will some well-known univ or scholarship orgs may send us some letters? Or they may have already sent some, but I didn't receive? What have you guys recently received?

Any PAST(excluding this year) USAMO qualifier or AIME High-scorer, what's your opinion and experience? (This is my first year to make it.) Are we going to receive something that is really valuable?
Thanks.

I know Mathpath & Mathcamp both send invites to top scorers on various math competitions, but I haven't heard about any letters from AMC's sponsors.

I got  a "Dear USAMTS top-scorer," letter from Mathcamp last week.

I got one for Mathcamp saying "Dear USAMTS top scorer"

[quote="chess64"][quote="nat mc"]Yeah I received a Canada/USa Mathcamp pamhplet and the letter said "Dear AMC top-scorer"[/quote]

Me too.[/quote]Yep, same here.

Doesn't look like I'll quallify for USAMO if they have too many 7's then, because I haven't gotten one.

That question is completely irrelevant to this topic!  :lol:

[quote="solafidefarms"]Would the AMC people contact the USAMO qualifiers individually by email before they released the USAMO qualifier list?[/quote]

I'm pretty sure the answer is no.

[quote="calc rulz"][quote="pieterminate"]Didn't every one who qualified for the aime get a letter from mathcamp?[/quote]

Hmm, I qualified for AIME, but only got a letter saying "Dear top USAMTS scorer," not one saying "Dear top AMC scorer."[/quote]

Maybe they don't like wasting stamps and only send 1 per person?

Well they don't send an invite to every AIME qualifier, probably just the top couple hundred for both AMC's.

I just received a letter from National Honor Roll, but it doesn't say anything about AMC. Did you guys also receive one? What is it? Ad? or Real things?

National Honor Roll is the biggest scam... They get you to fill out a detailed form, even down to spots for you to list Grandparents and Grandparent's Addresses. Then they send ads to said grandparents to buy the book their child is in. Same goes for parents. And did I mention that they use you in the book, but you have to buy the book at $30 to see it?!

:D

[quote="solafidefarms"]National Honor Roll is the biggest scam... They get you to fill out a detailed form, even down to spots for you to list Grandparents and Grandparent's Addresses. Then they send ads to said grandparents to buy the book their child is in. Same goes for parents. And did I mention that they use you in the book, but you have to buy the book at 30 to see it?!

:D[/quote]

So it's truly an ad?

I also got one from Awesome Math.  I don't think there are any restrictions on who can apply.  My AMC score was 130, by the way.

[quote="Go Beyond"][quote="kimby_102"]I got one from Mathcamp and the other from Awesome math program :D[/quote]

Did you receive that recently?
What is Awesome math?
And who can qualify that? Very Top AMC scorers? above 140?[/quote]

Awsome Math is an excellent math camp as well (looking at faculty).
The letter is just an invitation to apply, and many probably received the letter(since even i got 1).

[quote="kimby_102"]I got one from Mathcamp and the other from Awesome math program :D[/quote]

Me too.

An August snowball fight sounds cool since in GA, we don't even have December snowball fights.

BTW: A dodecahedron is not a hair accessory (No offence).

$\$$3000 dollars is a little out of my price range, so If I don't qualify for MOSP, I will have to go with my parents to some place in Asia for no apparent reason.

Got one from Mathcamp and an email from the ARML people... :)

I just received a letter from [i]Who's Who American High School Students[/i], due to AMC or AIME or USAMO qualifer, I am not sure.

When we filled in the answer form for AMC and AIME for "Do you want some univ or orgnizations to collect your info from the test?", does it only mean those unimportant orgnizations? Or will some well-known univ or scholarship orgs may send us some letters? Or they may have already sent some, but I didn't receive? What have you guys recently received?

Any PAST(excluding this year) USAMO qualifier or AIME High-scorer, what's your opinion and experience? (This is my first year to make it.) Are we going to receive something that is really valuable?
Thanks. :)

I'm thinking that you got Who's Who from PSAT, which is where you'll get college letters, and acedemic stuff like NAtional Honor Roll.

But stuff you get from getting good score s on the AMC/AIME is usually math camp type stuff, which is usually worth it.

But yea, so far, I've only gotten a letter from MathCamp.

[quote="nirvanatear"]so what did you get on usamo, Go Beyond?
i wonder if mathpath or mathcamp would invite me[/quote]

nothing important yet,
so far, I have got mathcamp invitation, national honor roll, who'who American High school Students, AMC congratulation letter with student number, college magazine.

I just received one invitation from HCSSiM.

I got a 126 on AMC 12 A, and 137 on AMC 12 B, and 9 on AIME.... I haven't gotten any letters... boo!

 :rotfl:

Ah well, its not like I was really planning on attending any camps this summer, I've got classes to worry about.

BTW, you don't need an invitation to go anywhere.

I got ones from Mathpath, Mathcamp, and AwesomeMath. I was already aware of all of the programs, but I probably wouldn't have applied to Mathcamp without the pamphlet. (Was going to apply to Mathpath only.) Having a hard copy of the admissions problems around while I was trying to procrastinate on homework motivated me to try the problems :lol:

[quote="pieterminate"]Joining it, I got one from mathcamp, one from Awsome math, and one from HCSSIM.  And my amc scores were terrible, so... :roll:[/quote]

Terrible = ?

[quote="calc rulz"]
Terrible = ?[/quote]

Barely enough.  :(

[quote="probability1.01"][quote="MysticTerminator"]hm, my mom said there was some letter from Titu (I guess that would be AwesomeMath?)[/quote]

Whoa maybe he's going all like "be instructor here at AwesomeMath... we caged all turkeys already."[/quote]

oh. yea that would make more sense . . . I was wondering why he'd want me to go. Now maybe I should try to find that letter? hm

Just received one invitation from Boston U's PROMYS math program.
However, I have been accepted by mathcamp, so I will not consider much to go there.


[quote="kunny"]For positive integer $ n$, let $ f(n)$ be the number of the L.C.M. of positive integers $ a,\ b$ with $ a\leq b$. Find $ f(600)$.[/quote]

I don't think I understand the question. Does $ f(n)$ return the number of distinct values of $ LCM(a,b)$ where $ 0<a\leq b\leq n$?


Compute: $ \left(\dfrac{4}{5}\right)^{\minus{}2}$ Express your answer as a common fraction.


Here is a problem from my friend, it seems fun. I'm going to think about it with u. Ok? ;)

Suppose A is a set of one hundred distinct natural numbers with the property that given a, b, c in A (distinct or not), there is a triangle with no angle greater than a right angle, and with sides of lengths a, b, c. What's the smallest possible value for the largest element of A?

[i]Mod Edit - Please be careful not to post problems from current competitions. Even if this was unintentional, check your problem sources.[/i]

- [i]mod edit - problem censored since it is on Mathcamp quiz[/i]

I think this thread should be stopped this problem is on the 2005 Mathcamp quiz.

- [i]Thanks - thread locked. - tetrahedr0n.[/i]


[quote="isabella2296"]The Brazilian stamps cost $ 12 \times 6 \equal{} 72$ cents, the French stamps cost $ 12 \times 6 \equal{} 72$ cents, the stamps from Peru cost $ 6 \times 4 \equal{} 24$ cents, and the Spanish stamps cost $ 13 \times 5 \equal{} 65$ cents.

The sum of those numbers is 233, divided by 4 is $ 58.2$ cents, to the nearest tenth.[/quote]

The error is in the last line, where izzy divided the cost in cents ($233$) by $4$. Instead, she should have divided it by $43$, which is the total number of stamps, since the question asked for average price per stamp. This will give you the correct answer of $5.4$

Missed this one because I did not read "...to the nearest tenth." :(


modular arithmetic
ratio
number theory proposed
number theory

Find all positive integers $A$ which can be represented in the form: \[ A = \left ( m - \dfrac 1n \right) \left( n - \dfrac 1p \right) \left( p - \dfrac 1m \right) \]
where $m\geq n\geq p \geq 1$ are integer numbers.

[i]Ioan Bogdan[/i]

Solution
[hide]$\left ( m - \frac 1n \right) \left( n - \frac 1p \right) \left( p - \frac 1m \right) = \frac{(mn-1)(np-1)(pm-1)}{mnp}$
$\implies$
$p \lvert mn-1 \implies mn \equiv 1 \pmod p$
$m \lvert np-1 \implies np \equiv 1 \pmod m$
$n \lvert pm-1 \implies pm \equiv 1 \pmod n$

Let
$p = \frac{mn-1}{k}$
$k<m$ to preserve the ordering of $m,n,p$

$n(mn-1) \equiv k \pmod m$
$m(mn-1) \equiv k \pmod n$

$n+k \equiv 0 \pmod m$
$m+k \equiv 0 \pmod n$

$m+k < 2n \implies m+k=n$
Therefore $n < 2m$
Therefore $n+k < 3k \implies n+k = 2m \implies m+k+k=2m \implies m=2k$

$m\frac{mn-1}{k} \equiv 1 \pmod n \implies 2(mn-1) \equiv 1 \pmod n \implies -2 \equiv 1 \pmod n \implies n=3$ or $n=1$

$m \le n \implies m=2$ or $m=1$

This leaves us with three ordered pairs $(m,n)$
$(1,1), (1,3), (2,3)$

$(m,n)=(1,1) \implies A = 0$
$(m,n)=(1,3) \implies p = 1$, contradicting the ordering.
$(m,n)=(2,3) \implies p = 5$, and $A=21$

Therefore, the two possible values of $A$ are $0$ and $21$[/hide]

Alternate solution:
[hide]
The product equals $ mnp\minus{}m\minus{}n\minus{}p\plus{}\frac{mn\plus{}np\plus{}pm\minus{}1}{mnp}$, or $ mnp\leq mn\plus{}np\plus{}pm\minus{}1$.  If $ p\geq3$, it must hold $ 3m\plus{}3n\leq 2mn\leq 3m\plus{}3n\minus{}1$, absurd, yielding $ p\equal{}1$ or $ p\equal{}2$.

If $ p\equal{}2$, then $ 2mn$ must divide $ mn\plus{}2m\plus{}2n\minus{}1$, and if their ratio is at least $ 2$, then $ 2mn\plus{}2m\plus{}2n\leq4mn\leq mn\plus{}2m\plus{}2n\minus{}1$ absurd, hence $ mn\equal{}2m\plus{}2n\minus{}1$, and $ m,n$ must clearly be odd.  If $ n\geq5$, then $ 2m\plus{}2n\plus{}m\leq mn\equal{}2m\plus{}2n\minus{}1$, absurd, and $ n\equal{}3$, yielding $ m\equal{}5$.  We then find the product equal to $ \frac{14}{3}\frac{5}{2}\frac{9}{5}\equal{}21$.

If $ p\equal{}1$, then $ mn$ must divide $ mn\plus{}m\plus{}n\minus{}1$, or $ mn$ must divide $ m\plus{}n\minus{}1$.  If $ n\geq2$, then $ m\plus{}n\leq mn\leq m\plus{}n\minus{}1$, absurd, yielding $ n\equal{}1$, or one of the factors is $ 0$ and the product is $ 0$.

The only values are $ 0$ and $ 21$.

[/hide]


Show that the Cleavance centre is identical with Spieker centre.

dear pleurestique,I know that url but I can not see any proof there...

Here's a nice argument:

The intersection of the cleavers must be the centroid of the border of $ABC$, so in order to prove the desired result, we have to show that the Spieker center is the centroid of the border of $ABC$. If $D,E,F$ are the midpoints of $BC,CA,AB$ respectively, then the centroid of the border of $ABC$ is the centroid of the system formed by the three points $D,E,F$ if we give $D,E,F$ the masses $EF,FD,DE$ respectively. It's easy to see that this means that the point must be the incenter of $DEF$, and this is the Spieker center.

You can find some properties of cleavers and of the Spieker center of a triangle in the very first pages of Honsberger's "Episodes in Nineteenth and Twentiethth Century Euclidean Geometry". These first pages are avaliable online at http://www.amazon.com/exec/obidos/tg/detail/-/0883856395?v=glance ("sample pages").

  Darij


function
calculus
inequalities
inequalities proposed

[quote="nntien"]Let $ x \in R$, find minimum of following function:

$ f(x) \equal{} \sqrt {\frac {1}{2}x^2 \plus{} 2} \plus{} \sqrt {\frac {1}{2}x^2 \minus{} \frac {4}{5}x \plus{} \frac {8}{5}} \plus{} \sqrt {\frac {1}{2}x^2 \minus{} 4x \plus{} 10} \plus{} \sqrt {\frac {1}{2}x^2 \minus{} \frac {16}{5}x \plus{} \frac {32}{5}}$

Don't use calculus!
:)[/quote]
$ 2f(x)\equal{} \sqrt{(x\minus{}2)^{2}\plus{}(x\plus{}2)^{2}}\plus{}\sqrt{(x\plus{}\frac{4}{5})^{2}\plus{}(x\minus{}\frac{12}{5})^{2}}\plus{}\sqrt{(x\minus{}6)^{2}\plus{}(x\minus{}2)^{2}}\plus{}\sqrt{(x\minus{}\frac{24}{5})^{2}\plus{}(x\minus{}\frac{8}{5})^{2}}$
Because $ (\minus{}2,2),(2,2),(6,2)$ are colinear, and $ (\minus{}\frac{4}{5},\frac{12}{5}),(2,2),(\frac{24}{5},\frac{8}{5})$ are colinear, and $ 2f(x)$ is the distance between the point $ (x,x)$ and $ (\minus{}2,2),(6,2),(\minus{}\frac{4}{5},\frac{12}{5}),(\frac{24}{5},\frac{8}{5})$, $ 2f(2)$ is the minimum value of $ 2f(x)$. Therefore, the minimum of $ f(x)$ is $ f(2)$.

[quote="msceok"][quote="nntien"]Let $ x \in R$, find minimum of following function:

$ f(x) \equal{} \sqrt {\frac {1}{2}x^2 \plus{} 2} \plus{} \sqrt {\frac {1}{2}x^2 \minus{} \frac {4}{5}x \plus{} \frac {8}{5}} \plus{} \sqrt {\frac {1}{2}x^2 \minus{} 4x \plus{} 10} \plus{} \sqrt {\frac {1}{2}x^2 \minus{} \frac {16}{5}x \plus{} \frac {32}{5}}$

Don't use calculus!
:)[/quote]
$ 2f(x) \equal{} \sqrt {(x \minus{} 2)^{2} \plus{} (x \plus{} 2)^{2}} \plus{} \sqrt {(x \plus{} \frac {4}{5})^{2} \plus{} (x \minus{} \frac {12}{5})^{2}} \plus{} \sqrt {(x \minus{} 6)^{2} \plus{} (x \minus{} 2)^{2}} \plus{} \sqrt {(x \minus{} \frac {24}{5})^{2} \plus{} (x \minus{} \frac {8}{5})^{2}}$
Because $ ( \minus{} 2,2),(2,2),(6,2)$ are colinear, and $ ( \minus{} \frac {4}{5},\frac {12}{5}),(2,2),(\frac {24}{5},\frac {8}{5})$ are colinear, and $ 2f(x)$ is the distance between the point $ (x,x)$ and $ ( \minus{} 2,2),(6,2),( \minus{} \frac {4}{5},\frac {12}{5}),(\frac {24}{5},\frac {8}{5})$, $ 2f(2)$ is the minimum value of $ 2f(x)$. Therefore, the minimum of $ f(x)$ is $ f(2)$.[/quote]

That's right! I use Minkowski's inequality! Thanks!


[quote="matt276eagles"]I sent my entry form in about a month ago via snail mail, but according to my profile on the USAMTS site I haven't sent in an entry form yet.  Should I send in another one?[/quote]

Not yet.  We haven't filed the forms we've received yet.  We'll do that in September some time.


hello, let $ a\equal{}1234$ then we have $ \frac{a}{(a\plus{}1)^2\minus{}a(a\plus{}2)}\equal{}\frac{a}{a^2\plus{}2a\plus{}1\minus{}a^2\minus{}2a}\equal{}a$
Sonnhard.


Show that there are infinite numbers of rational numbers in the interval $ (0,1)$

[hide]
Consider rational numbers in the form $ \frac {1}{n} \longrightarrow n \in \mathbb{Z} \neq 1$
$ \Rightarrow 0 < \frac {1}{n} < 1$
and there are infinitely many counting numbers $ > 1$
[/hide]

Hello
Here is one more proof::
Draw a segment $ AB$ where A correspond to $ (0,0)$ and B correspond to $ (1,0)$.Now draw one more segment $ A'B'$ parallel and above it.Taking $ A'B'$ as diameter draw a circle .If u draw a line through the mid-point of segment $ AB$ it will also pass through the mid-point of segment $ A'B'$ too.
If u draw a line perpendicular to the segment $ AB$ passing in b/w of midpoint and $ B$.Let this line cut the circumference of semicircle at M.Now Join M to center of semicircle.U can clearly observe that the  line will not cross the segment $ AB$.
  Now take a perpendicular line which tends to pass through $ B$ will touch the circumference at a point $ M'$ which tends to touch the point $ B'$.Now if we join M' and center it will meet x-axis at infinity.
Thus every infinite number outside this segment corresponds to a number inside this segment.
Thank u.

:| I dont understand this at all. Could you label your diagram please. Thanks

hello
Here is my second attempt.
First  choose a line segment on x-axis $ AB$ where A is $ (0,0)$ and B is $ (1,0)$.
Now above this line segment draw one more segment $ A'B'$ parallel to AB such that A' comes just above A and B' just above B.
Now taking $ A'B'$ as diameter draw a semi-circle with centre $ O$.
Draw a line perpendicular to $ AB$ passing through its midpoint$ (0.5,0)$.U can easily observe that it will also pass through $ O$ and would bisect the circumference of semi-circle.
Now the main part::
1.Draw II perpendicular line passing $ AB$ through a point in b/w midpoint$ (0.5,0)$ and$ B$$ (1,0)$. Let this line meet the semicircle at point $ M$ .Now if u join $ O$ and $ M$ the line $ OM$ will meet x-axis outside the segment.
2.Draw III perpendicular line passing through a point which is tending to $ B$ and will touch the semicircle at a point $ M'$ which is tending to $ B'$.Now if u join the point $ O$ and $ M'$,the line $ OM'$ is just parallel to $ AB$ and $ A'B'$.Therefore the line $ OM'$ will meet the x-axis at infinity.

[color=green]Conclusion[/color]
We now there are infinite numbers outside the segment $ AB$ and each of these infinite numbers correspond to a number inside the segment.Thus we have infinite numbers in this segment.
Hope this time my explanation is somehow better.
Thank u.
Note::Click on the picture to get a magnified picture.


counting
derangement
combinatorics unsolved
combinatorics

Count the number of 2-regular graphs on n vertices.

Let this number be f(n)

Labelled vertices, I assume.  If we allow self-loops and multiedges, the answer is $ n!$.  If we allow multiedges but not self-loops, the answer is $ D_n$, the number of derangements of an $ n$-element set (permutations with no fixed points).  If we allow neither multiedges nor self-loops, the answer is the number of permutations of an $ n$-element set with all cycles of length 3 or greater, counted by [url=http://www.research.att.com/~njas/sequences/A038205]A038205[/url] in the OEIS.  There's not a nice closed form, although there is a relatively simple recursion and a very nice exponential generating function.


Find all functions $ f: \mathbb{N}\rightarrow \mathbb{Z}$ such that

$ f(x\plus{}|f(y)|)\equal{}x\plus{}f(y)$, for all natural numbers $ x,y$.

[hide="A rather mean problem if you ask me"]Suppose there exists a positive integer $ m$ such that $ f(m) \equal{} 0$. Then $ y \equal{} m$ gives $ f(x) \equal{} x$. This is a valid solution and it also contradicts the assumption.

Suppose there exist positive integers $ m,n$ such that $ f(m) \equal{} \minus{}n$. Then $ x \equal{} n, y \equal{} m$ gives $ f(m\plus{}n) \equal{} 0$. Again, by the previous paragraph, contradiction.

Therefore, $ f$ only takes on positive integer values. The equation is now reduced to $ f(x\plus{}f(y)) \equal{} x \plus{} f(y)$ with $ f : \mathbb{N} \rightarrow \mathbb{N}$.

Let $ a$ be the minimum of $ \{ f(n) | n \in \mathbb{N} \}$. $ m$ exists by the well-ordering principle of $ \mathbb{N}$. Let $ b$ be a value for which $ f(b) \equal{} a$. By putting in $ y \equal{} b$, we know that $ f(x\plus{}a) \equal{} x\plus{}a$. That is, $ f(x) \equal{} x$ for all $ x > a$

Now for all $ x \leq a$, we can define $ f(x)$ to be anything greater than or equal to $ a$ without contradiction. Because $ a$ is the minimum value $ f$ takes, $ x \plus{} f(y) > a$, always. Since we have that $ f$ is fixed for all values greater than $ a$, the original equation will hold.

So we have a ridiculous amount of answers that I'm not going to list.[/hide]


In a triangle $ ABC$, the angle bisector at $ B$ intersects $ AC$ at $ D$ and the circumcircle again at $ E$. The circumcircle of the triangle $ DAE$ meets the segment $ AB$ again at $ F$. Prove that the triangles $ DBC$ and $ DBF$ are congruent.

We have $ \angle DBF \equal{} \angle DBC$ because $ DB$ is the bisector of $ \angle FBC$.
We also have $ \angle DFB \equal{} \angle DEA \equal{} \angle BCD$ (because $ DEFA$ and $ BCEA$ are cyclic quadrilaterals).

Hence, triangles $ DBF$ and $ DBC$ are similar. But, since they share the common side $ BD$, they are congruent.

easy,$\angle DFA+\angle DEA=180$,so $\angle DFB=\angle C$($\angle AEB=\angle C$),$BD$ is the angle bisector,hence done.


WHERE IS EVERYBODY?!

We should start posting things here soon. Math Problems, contest results, you know, stuph like that!

Or is everybody just [b]antisocial[/b]?

:huh: What contests have taken place so far?

How can we Upstate New Yorkers possibly hope to have such a vibrant, lively forum as NYC?

-looks- Oh, wait, their forum is dead too  :|

Contests that have taken place so far: All AMCs, PUMaC, HMMT, MC Chapter. NYSML, AIME, and MC State are coming up soon.

Proposal on how to keep this forum [b]alive[/b]: First of all, we need to check this forum a lot more regularly. Second, we need to post stuff so that this forum doesn't die. Third, we need to point Upstate New York AoPS users to this forum, so that they can use it.

Here's a list of AoPSers that I know are in Upstate NY:

me
you
xgames43
pianogirl
SmitSchu
Hexahedron
BlueMouse
sailingdog
dilberto
seashell1082

These people should know about this forum, and should visit it once every [b]white[/b] moon.

OK!
HERE'S A PROBLEM:
Give the first 90 numbers of the Fibonacci Sequence
oh and about the signature
12.9+ refers to college level work, meaining that it would be recommended for you to take AP courses in what ever subject that's in or all subjects depending on the exam you recieved the score is

${\text{Find } 1+1. \text { Express your answer in terms of } \pi.}$

the answer is pi-(pi -2).

How many degrees are in a hexagon?

[quote="anonymous0"]Does anyone who lives in upstate NY exist anymore?[/quote]
Of course not.

Nope, all people of Upstate New York mysteriously vanished. Survivors suddenly became antisocial.

Problem: Find the circumference of a circle with diameter $1$. Give your answer rounded to the first 1337 decimal places.

Hmmm....1=2 doesn't post here anymore either...We should actually post something other than in this INACTIVITY! Bah! thread...

[quote="1=2"]Here's a list of AoPSers that I know are in Upstate NY:

me
you
xgames43
pianogirl
SmitSchu
Hexahedron
BlueMouse
sailingdog
dilberto
seashell1082

[/quote]
You didn't know eavqg.whbeg?

[quote="Draco"]1=2 doesn't post here anymore either[/quote]
I'm not interested in posting in a forum where less than 5 people are interested in posting.

1. 1=2
2. willwang123
3. anonymous0
4. draco
5. hrithikguy

Hrithikguy lives in Upstate NY?

Updated list of Upstate New Yorkers

Draco
1=2
willwang123
Anonymous0
Sammymax
Hanes
jasonion
Tim14
Trigfan
Tareyza
Jeffshen

...not very many. And half of them don't even post.

Also, 5 unique people have posted in this forum this year.

hrthikguy is from Tennessee... anyway,

Draco
1=2
willwang123
Anonymous0
Sammymax
Hanes
jasonion
Tim14
Trigfan
Tareyza
Jeffshen
Pattycakechichi

Does trigfan count as being from upstate New York?

OK. You three need something productive to do instead of spamming this thread. I created a MathCounts Marathon thread in this forum for you guys to have fun with.

Hmmm, I don't think the MC marathon worked... Now that competition season is about to start, hopefully we'll have something that's not spam to post about.


inequalities
function
triangle inequality
inequalities proposed

Let $a,b,c$ are real numbers.Prove the following inequality.

\[\frac{|a|}{1+|a|}+\frac{|b|}{1+|b|}\geq \frac{|a+b|}{1+|a+b|}\]

Sorry for that I can't understand.Please explain in detail. :?

kunny

Thank you for your reply,Soarer.Now I can understand your solution.

Here is my solution.

Since $y=\frac{x}{1+x}=1-\frac{1}{1+x}$ is an increasing function for $x\geq 0$, use the triangle inequality,

$|a+b|\leq |a|+|b|$, so we have $\frac{|a+b|}{1+|a+b|}\leq \frac{|a|+|b|}{1+|a|+|b|}=\frac{|a|}{1+|a|+|b|}+\frac{|b|}{1+|a|+|b|}\leq \frac{|a|}{1+|a|}+\frac{|b|}{1+|b|}$.

kunny

Here's another possibly similar solution.

Since $x/(1+x)$ is increasing for $x\geq 0$, we may assume that $a,b$ have the same sign, and consequently are both $\geq 0$.  But then

$\frac {x}{1+x}+\frac {y}{1+y}\geq \frac {x+y}{1+x+y}=\frac {|x+y|}{1+|x+y|}$


conics
hyperbola
rotation
geometry
geometric transformation
reflection
Asymptote

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14110[/img]

 :(

[size=150]\[ \left\{ {_{xy \equal{} a^2 }^{x^2 \plus{} y^2 \equal{} d^2 } } \right.
\][/size]

$ s\equal{}x\plus{}y\equal{}\sqrt{d^2\plus{}2a^2}$ that can be costructed by pytagora, so x, y are the solution of the equation $ x^2 \minus{} sx \plus{} a^2$ so they are $ \frac{s \pm \sqrt{s^2\minus{}4a^2}}{2}$ that can be costructed by pytagora.

i want the [b]s[/b] construction  and  [b]x [/b], [b]y[/b]  construction

 :(

$ x^2 \plus{} y^2 \equal{} d^2$ is a circle centered at the origin $ O$. $ xy \equal{} a^2$ is a rectangular hyperbola centered at the origin, x-, y-axes are asymptotes. Rotate the hyperbola by $ \minus{} 45^\circ.$ The circle does not change, the hyperbola equation becomes $ x^2 \minus{} y^2 \equal{} 2a^2.$ Adding the equations, $ 2x^2 \equal{} d^2 \plus{} 2d^2.$ Construct intersection(s) of the circle with the hyperbola. ($ E, F$ are foci. You do not have to draw the hyperbola, just the asymptotes.) $ OABC$ is square with the side $ OA \equal{} a,$ diagonal $ OB \equal{} a \sqrt 2.$ $ OL \perp OB$ and $ OL \equal{} d.$ $ BL^2 \equal{} d^2 \plus{} 2a^2.$ $ K$ is the midpoint of $ BL.$ Perpendicular bisector of $ BL$ cuts the circle $ (K)$ with diameter $ BL$ at $ M, N.$ The distance $ x \equal{} \sqrt {\frac {d^2 \plus{} 2a^2}{2}} \equal{} \frac {BL}{\sqrt {2}} \equal{} LM$ has to be transfered along $ OB.$ For example, reflect $ MN$ in the line $ KCA$ into $ UV$ (also a diameter of $ (K)$). Circle with radius $ OU \equal{} OV$ cuts the ray $ OB$ at $ W$ and perpendicular to $ OB$ at $ W$ cuts the circle $ (O)$ at $ Q$ one intersection of the circcle $ (O)$ with the hyperbola. $ P, R$ are feet of perpendiculars from $ Q$ to the hyperbola asymptotes $ OA, OC.$ Rectangle $ OPQR$ has area $ a^2 \equal{} OA \cdot OC \equal{} OP \cdot OR$ and diagonal $ d \equal{} OQ.$

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14740[/img]

 :thumbup:


It's tough to explain how to do this, but just visualize the building in your head and you see the answer is D.


Gauss
function
set theory
advanced fields
advanced fields theorems

Can someone explain me how can we use the concept of infinity with mathematical rigor, because i still cant understand!
Thanks!

I want to hear an explanation from Valentin!

Infinity is a very subtle thing. I'll give you some ideas and see if this can help out.

At an intuitive sense, infinity is like a 'thing' very very VERY large or somewhere far far far... away. Infinity is NOT an exact value. Infinity is rather a mathematical device sort of like denoting something that never ends, like a sequence, like the natural numbers! There are different kinds of infinities but we are not getting into that.

At a more formal level (and I believe someone in here can do better at this than I do), infinity is as follows: suppose we have a real number system R. We make an [i]extended real number system[/i] R* as the real number system R with the following 2 'extra' conditions:

1)For every x \in R, denote - \infty to be the greatest lower bound of x.
2)For every x \in R, denote + \infty to be the least upper bound of x.

I defined it under the S-bound properties because those are most commonly used definitions in analysis. If any of the other readers see anything wrong here, let me know. I didn't check this  :)

Sorry that I'm not Valentin, (is he on now?) but if you insist on getting an explanation from him, it's worth the wait. He explains better than I do  :)

Well, this is a very controversed subject.It put into difficulty even famous mathematicians.There are two kinds of infinity:actual and potential.For example Gauss and Kronecker never accepted the actual infinity ie there exists actual infinite sets.
Cantor dedicated  a big part of his life for rigorously fundamenting the concept of infinity.When studying thi subject he encountered many "paradoxes".He found out that   there exists different types of infinity .How many?An infinity!The complicated thing with the infinity is that man's intuition doesnt work very well at this chapter.It is more easy to imagine finite sets than infinite sets,so when working with infinite sets everything wich may seem intuitive clear must be tested by rigorous proof.
Formally, with the basic set theory axioms we can't prove that there exists actual infinite sets, so we must introduce a new axiom:the axiom of infinity! This fundaments the infinite sets.(for more details one can consult som set theory books-.The infinite set with the" fewest" elements is the set of natural numbers.Then we can extend the number system-first we construct integer and rationals and then irrationals(this extending to irrationals can be made in many ways-using Cauchy sequences,Dedekind cuts,etc).
Although, the problems with infinity don't stop here.We need other axioms such the axiom of choice.If we hadn't the axiom of choice we couldn't prove, for example the equivalence between the classic definition of the limit of a function and the definition with sequences(Heine).
Another important concepts wich deal with infinity are cardinal numbers wich make possible comparing the infinite sets using bijections.

So infinity can be mathematically rigorously founded.
This is sufficient for the mathemathical use of infinity.If we dont stop here we may get into philosophy.I may give an example:in his last years of life Cantor gave up mathematics and began teaching phylosophy.

I heard that Cantor went beyond limit and was ... well crazy? ... by constant attacks from Kronecker. Poor fella I can tell you that.

Cantor couldn't publish his work before his death, because Kronecker had influence at most mathematical magazines of the time and stopped his articles from being published.So he never knew that his work will influence so much the mathematics of XX century.

That's sort of sad, isn't it? I had heard about the conflict between the two, but I had no idea it was so serious..

shame on Kronecker.
if he wasn't agree with Cantor, why did he stop his articles ?wasn't there a way
more kind to "solve" this problem ?
this is not a good way to be famous on mathematics !

One problem with explaining infinity formally is that the concept of "infinity" is used in a lot of completely different contexts mathematically. But each individual instance can actually be defined quite concretely (sort of).

For example, in calculus/analysis, when you talk about a limit of a function or sequence as the variable "goes to infinity", you're not talking about anything outside of the ordinary real or natural numbers. The standard $n_0$-delta definition of sequence convergence ($\forall \epsilon > 0, \exists n_0, \forall n>n_0, |x_n - L| < \epsilon$) is simply a statement about every positive real number. Sorry if you've never seen that sort of thing (it probably looks like complete gibberish if so), but the point is that $\infty$ is just a shorthand way of saying "arbitrarily large". In some way it's all a matter of notation.

Infinity has a completely different sort of feel in set theory. It's still definable (probably by saying that an infinite set is one that can't be put into one-to-one coorespondence with a set of the form {0,1,2,...,n} for some natural number n (or the empty set)). Of course, there are several different sizes of infinity that sets can be. It gets quite complicated, to say the least. I guess the real question comes down to "how can set theory be logically definable with such craziness as infinite sets?". As for that, I have nothing to say other than that axiomatic set theory is quite rigorous and you can still make some kind of living in philosophy.

I hope some small piece of these ill-concieved ramblings helps you to better comprehend the wonders of infinity.  :) Good luck.


A base-$ 10$ three-digit number $ n$ is selected at random. Which of the following is closest to the probability that the base-$ 9$ representation and the base-$ 11$ representation of $ n$ are both three-digit numerals?

$ \textbf{(A)}\ 0.3 \qquad
\textbf{(B)}\ 0.4 \qquad
\textbf{(C)}\ 0.5 \qquad
\textbf{(D)}\ 0.6 \qquad
\textbf{(E)}\ 0.7$

There are a total of 900 three digit base ten numbers. The largest base nine three digit number is 888, which is equivalent to 728. The smallest three digit base eleven number is 100, which is equivalent to 121. The number of integers between 121 and 728 inclusive is 608. So the answer is 608/900 which is equivalent to .675555555... This is closest to E.

[quote="Wickedestjr"]This is equivalent to .66222222... This is closest to D.[/quote]

It's closer to E)0.7  :wink:

[hide]

Clearly, the largest three digit base 11 number exceeds the largest three digit decimal number, as well as the smallest three digit base 9 number being less than the smallest three digit decimal number, so we only need to count the numbers between $ 100_{11}$ and $ 888_9$.

$ 100_{11}\equal{}11^2\equal{}121$

$ 888_9\equal{}(9^2)\plus{}8(9)\plus{}8(1)\equal{}8(91)\equal{}728$

Hence, there are $ 728\minus{}121\plus{}1\equal{}608$ numbers. $ 608/900\approx .68\approx0.7\Longrightarrow\mathrm{ (E) \ }$

[/hide]

[quote="sunehra"]He's right.

Your solution was good until the part where you said the smallest three digit base eleven number is 111. It's actually 100. Converting that to base 10, we have 121. So the number are be between 121 and 728, inclusive. So the probability is $ \frac {608}{900} \approx \frac {6}{9} \equal{} \frac {2}{3} \equal{} 0.\overline{6}$. Thus the closest is $ 0.7 (E)$.[/quote]

Oh oops. I can't believe I made two mistakes in one problem.  :blush:


Do anyone know how to use TI-89 to graph slope field?
such as y'=-x-2 and initial condition point(-1,1).

Look it up in the manual.  It should already be built in as a program so the manual should have it.

The attachment is an example of graph the solution to the logistic 1st-order differential equation y'=.001y*(100-y).
In Y=Editor, the manual ask us put in .001*y1*(100-y1).
but I don't know how to get slope field if y'=10x+2y which has both x and y.
I have try just put 10*x+2*y in the calculator, it tells me undefined variable.

There said "undefined variable". Do anyone know if I can define the variable in calculator?

[quote="JRav"]Does it have to be on a calculator?  These are rather simple slope fields to do by hand.[/quote]
Yes, it have to be done on the calculator because I want to know how to graph it by both calculator and hand and that is the reason for me to buy TI-89 calculator.