function trigonometry algebra polynomial factorial

[b] To write in FORTRAN  or in C, C++,    a routine in order to compute , with at least $15$ exact decimals , the values  of
functions $e^x$ and  $\sin{x}$  for $x$  "[color=green] large [/color]    "
For instance , try to compute   $e^{1000}$ (why not ?)  and     $\sin{10^{40}}\; .$
What about  precision ?
[/b]

If you have no CAS in your computer I think the easiest way is to use some large precision arithmetic library. Then approximate your functions by Taylor polynomials of sufficient large degree so that error between the correct value and the approximation is enough small.

[quote="puuhikki"]... to use some large precision arithmetic library. .... approximate your functions by Taylor polynomials of sufficient large degree so that error  ....[/quote]
Taylor it's not good enough !

[quote="puuhikki"]What do you mean "is not good enough"? Is it too slow to calculate enough many terms?
 At least Taylor series of $\sin x$ and $e^x$ converges at every point in $\mathbb{R}$ as far as I remember.[/quote]
Right ( from mathematical point of view it's true ) .
Let $E(x)\approx e^x$ and $S(x)\approx \sin{x} .$  Suppose that you want to find $E(x)$  and  $S(x)$  such that $|E(x)-e^{x}|\le \varepsilon$  or  $\frac{|E(x)-e^x|}{e^x}|\le \varepsilon$
  [b] for $| x |$ large [/b] .   If, e.g. $y_0 \approx 10^{30} \; , \; x_0\approx 1000 \; \; ,\; \varepsilon={10}^{-10}.$ How do yo proceed to compute  $S(y_0)$ or $E(x_0)$
  [furnished by Taylor developments]  ?
 Another (interesting and perhaps of interest ) question is to find another class of functions , say ${\mathcal H}$ such that the if  , say
 $\left|T_{1178}(x_0) -e^{x_0\right| \le \varepsilon}$   where  $T_n(x)=\sum\limits_{k=0}^{n}\frac{x^k}{k!}$ then  in order to obtain the precision   $| E(x_0) - e^{x_0}|\le \varepsilon$ in the analytic expression of
$E(x)$   to appear only powers as , e.g $\{ x^0,x^1,....,x^7\} .$ Which is the "best" (in this sense)  class ${\mathcal H}$ ??
[size=75]
[b] It's class of rational functions ?  Perhaps developments furnished by continued fractions ? Or by Pade approximations ?[/b] [/size]
Some (concrete) questions: using a computer, e.g. a PC , try to find ,or  better to write a routine in order to find
approximative values for:
1)  $A: = e^{3000.17}$
2)   $B: = \sin{\left(.20008999999414444 \times 10 ^{50}\right)}$ .
[size=75] [color=blue] [b] If  your PC furnish  values $A , B$ how do you test the precision ?? [/b] [/color] [/size]

[quote="flip2004"]Suppose that you want to find $E(x)$  and  $S(x)$  such that $|E(x)-e^{x}|\le \varepsilon$  or  $\frac{|E(x)-e^x|}{e^x}|\le \varepsilon$[/quote]

You can calculate the number of terms in Lagrange's remainder. You can use this to show that remainder is smaller than $\epsilon/2$. Now you can approximate the polynomial such that error between the true value of polynomial in the given point and the approximation of polynomial in the given point is also less than $\epsilon/2$. Remember that every term causes numerical error so if you want to prove that total error in of the correct value and approximation is less than $\epsilon$ and your polynomial has $n$ terms, you have to prove that on average approximating every term in the polynomial the error should be less that $\epsilon/(n-1)$.

I'm not very experienced in computer programming or in numerical analysis so that I'm pretty sure you can do the exact error bound calculations and programming much faster than I can do on my own. I'm just so slow on my calculations and programming. But I'm sure you can find enough accurate appproximations if you prove always the maximum error you have. It is not hard but it is very boring. I made that a couple of times in my college.

I'm sure there might be some more sophisticated methods to prove the maximum error or approximate functions but I know just the very basics about numerical analysis. I should go to library and read some book first. I just answered because I know that Taylor polynomials gives enough accurate approximation if you take enough many terms.

It looks like there is some C++ library on http://www.apfloat.org/apfloat/ I have not tested that.

taylor's good enough.

$value = sin(x)$ or $value = exp(x)$ is nonsense because the two functions are produce irrationals.

you can do $| L - exp(x) | \leq \epsilon$ for epsilon is very small but you cant do it for all epilson, you would have an infinite number of digits to display (and store).

for a is very large $sin(a) = sin( mod(a,2\pi))$
(i think modulus only works for int, so you'd have to make your own function, (a/(2$\pi$) = n; truncate n; modulus = a - 2$\pi$n)

for exp(x) you can do
[code]
i = 0;
value = 0;
do     {
      term = x^i / fact(i);
      value += term;
      i++
}while(term > 0.000000001);
[/code]

you have a factorial in the bottom part of the fraction, and factorials grow like mad, so for even x very large the factorial will catch up in not too much time.  (and if you want to save time, $x^{i+1} = x*x^i$ and $(n+1)! = n! * (n + 1)$

also every term is less than the term before it, so the answer will be accurate to most likely 8 decimal places, certainly 6 or 7.

i forgot possibly the most important part

[quote="flip2004"]with at least $15$ exact decimals[/quote]

for the quotent of x^i / fact(i) to have 15 accurate decimal places, you need a lot of digits in value (if you have 10 digits greater than the units digit, the decimals will be off).

i started a program to calculate e a while back that i've picked back up to try to solve the thread, but its turning out to me much harder than i thought.

im making a class that contains an array, and each element in the array is a segment of digits of a single number.  i may post in a few days if i still cant figure out whats going on.


algorithm
geometry
geometric transformation
trigonometry
quadratics
conics
algebra

May be i have shifted from the topics where some mthematicians where cited.
Al khawarizmi has given algorithmic solUtion to second degree equation using general name for the unknown and for factors not as done before where you found only particular solution for particular problem.Before the algorithm was implicit and Alkhawarismi gave it explicitaly.

This probably belongs in the "Round Table" section, not the "Careers" section, unless you're talking about a career as a historian.

I'm teaching an "Early History of Mathematics" class this semester, and here's a comparison I've made for my class. Consider five civilizations in the world at around the year 1000 or 1100: western Christianity (Roman Catholic, with Latin as the language of scholarship), eastern Christianity (primarily the Byzantine Empire, with Greek as the language of scholarship), the Islamic world, India, and China.

In those five areas, who had a base-10 number system with effective arithmetical procedures? China, India, and the Muslim world (which had learned it from India). Not the Byzantines, and not the West, which was still stuck with the awkwardness of Roman numerals.

Where was geometry as in Euclid, Apollonius, Pappas, and so on studied? Fragments of Euclid existed in the West, and many of the documents could still be found in Byzantium, but the real activity and the scholarship that pushed this forward were in the Islamic world. That's where the real experts were. (Translations of some ancient texts were just starting to appear in the West, largely from Arabic sources with formerly Islamic parts of Spain as important centers for translation.)

Where was Ptolemy's astronomy studied? Well, the Muslims had combined Ptolemy with elements of Indian astronomy, Indian trigonometry, Indian numerals, and improved measurements to produce something more accurate than Ptolemy himself had ever managed. The Chinese and the Indians had their own reasonably accurate astronomy texts, as well.

How about solving equations? For quadratic equations, the notion of completing the square goes way back, to Babylonian times - but of course, al-Khwarizmi had that nice systematic exposition of it. For number-theoretic computations (linear equations, "Chinese remainder" problems), you'd be better off in India (after Brahmagupta) or China (especially after Qin Jiushao, although that's a little later - the 1200's). For cubic equations? The state of the art was solving cubics by intersecting conic sections, and Omar Khayyam gave the best exposition of that.

The first important Western mathematician was Fibonacci, and his importance is almost wholly derived from the fact that he was essentially an Arab mathematician writing for a European audience. He had been educated in North Africa, and his works are steeped in the traditions of al-Khwarizmi, among others. Fibonacci advocated the Indian base-10 number system, explained how to solve quadratic equations, and dabbled in cubic equations.

In short, if you looked at these five parts of the world in around 1100 and tried to take bets as to which one would be the home of nearly all of the important developments in mathematics 500 years later, it would have been very hard to place your bet on Western Europe. In particular, the Islamic countries had as its inheritance the best of both the Greek/Alexandrian accomplishments and those of India (as well as older Mesopotamian tendencies.) And the Islamic countries were generally richer, stronger, and had better technology than the West.

I don't really know why senouy is posting these threads. Celebrating the greatness of a culture may be one reason. But ask this: why is nearly every name he mentioned someone who worked before the year 1200? Why didn't the promise carry forward? How did the spirit of innovation die, and why was it left for Europeans to produce a Tartaglia, a Vieta, a Galileo, a Fermat, a Descartes?

One special note about al-Khwarizmi: while he did deal with all of the separate cases for the quadratic equation, his book, his [i]al-jabr[/i], was all verbal, rhetorical, not symbolic. And he did have to write down particular equations as stand-ins for the general case. The use of arbitrary parameters (as in $ ax^2 \plus{} bx \plus{} c \equal{} 0$) did not appear before the work of Vieta in the mid-1500's.

After reading smething in http://www.mathlinks.ro/viewtopic.php?t=125971
i tried to give an idea about arab and muslims mathmaticians , celbrating nothing other than right information .


Can somebody give me combination problems?

thanks in advance.

You mean combination as in nCr? (or $ \binom{n}{r}$ as you guys write it)

What if for part e the parents are crazy and name a girl Daniel?

how many ways can a diagonal intersect with another diagonal in an n-shaped polygon???

 :rotfl:  :rotfl:  :rotfl:

[hide]
for book problem
5!/2!

60

and @hunter

15?[/hide]

No,

[hide="hint"]
Notice no matter what you will always move 7 times.
[/hide]

dont u mean 7?

I did it again, got the same answer

[quote="Ihatepie"][quote="gauss1181"]A committee consists of 16 men and 12 women. How many ways can a committee of 7 be formed if...
a. all members must be women?
b. all members must be men?
c. there has to be at least one man?
d. there has to be at least one woman?
e. Daniel refuses to serve on the committee?
f. the committee must have 2 men and 5 women?

BIG, EVIL problem here! :roll:  :D[/quote]

[hide]
am I right?

I am assuming that each man would be considered a different person, and each woman would be considered a different person and that order counts.

a. 12!/5!

b. 16!/9!

c.  27!/21!

d. 27!/21!

e.27!/20!

f. 240 + 12!/7!

or am I way off with all?

(too lazy to do factorials out)

[/hide]

@hunter
can u only go north, and east?

because if not, there is infinite answers

also, by combinations, I meant like, there are 6 people, a,b,c,d,e and f. If they are sitting in a row, how many different ways can they be arranged?

1. a,b and c are best friends and have to sit next to each other
2. e and f want to sit next to each other also.[/quote]
No. You're WAY off on the combination formulas. Do you know the formula for combinations?

[quote="hunter34"]Yes I did mean 7.

[hide="Solution"]
For every point, you can choose 2 ways to go to the next point. So the answer is $ \binom{7}{2} \equal{} 21$.
[/hide][/quote]

when you are in one of the top four spots, or the left two spots (not counting destination) you can't go north, or east respectively. so that would take away 6 moves, making it my answer

@gauss

what I know is this

if i have to find total ways to arrange 5 people I do 5!

total ways to arrange say for example a,b,c,d,e and f

1. a,b, and c are best friends and want to sit together, I do 4!3!

say, same problem except

1. c and d don't want to sit next to each other

I do 6!-(5!2!)

or say

that there are 5 blocks that are h-a-p-p-y.

the total amount of combos are 5!/2!

also, if there is a word say, weird

the total amount of combos is 5!

finally, if say there are 5 different pairs of pants, 3 pairs of shirts and 4 pairs of

socks, the total outfits are 5*3*4

[quote="Ihatepie"][quote="ZhangPeijin"]How many ways can you place five identical objects on a shelf if the shelf only has room for 3 objects.
Yeah I know it's easy.
 :|[/quote]

just realized, if they are identical, 1 :D[/quote]
Dangit, I thought people would fall for my trick.
 :(

[quote="Ihatepie"][quote="hunter34"]Yes I did mean 7.

[hide="Solution"]
For every point, you can choose 2 ways to go to the next point. So the answer is $ \binom{7}{2} \equal{} 21$.
[/hide][/quote]

when you are in one of the top four spots, or the left two spots (not counting destination) you can't go north, or east respectively. so that would take away 6 moves, making it my answer
[/quote]

Or you could have taken the view of the [url=http://www.artofproblemsolving.com/Wiki/index.php/Combinatorial_identity#Hockey-Stick_Identity]Hockey Stick Identity[/url]
But yeah.

Ihatepie, you have to use the combinations formula to solve these, not your own silly imagination. The combination formula is: to arrange r items in groups with n possible items is C(n,r)=n!/r!(n-r)!

I know what you mean, but someone might say, is that $ \frac{n!}{r!}(n-r)!$ or $ \frac{n!}{(n-r)!r!}$?

It's the second one


I always confuse these four...well, maybe the first three, so here's a problem:
What is the mean, median, mode, and the range of set S if set S contains all possible solution to P/Q? (P and Q are each member pf the set {6, -1/2, -3, 1/3})

[hide]6/-1/2=-12

6/-3=-2

6/1/3=18

-1/2-3=1/6

-1/21/3=-3/2

-3/1/3=-9

Range=18- -12=38

Median=-1  3/4

Mode= None

Mean = -5/9

Rmr: mOde occurs the mOst (long O)

median is the middle number

mean is mean cuz it usually takes the longest

range is the difference of greatest and least[/hide]

Careful, you missed more than half the possibilities...there's a key word that's missing from my problem than the MATHCOUNTS problem before.


trigonometry
algebra
polynomial
function
domain
Functional Analysis
real analysis

Hi,

From basic functional analysis, we know that trig polynomials are dense in $ L^2([a,b])$, since we can show that the combined power of $ \sin(mx)$, $ \cos(mx)$, and $ 1$ form an orthonormal basis.

However, I don't understand why this works for only sine or cosine series, alone, a. la. many introductory PDE textbooks.  It seems that a common practice is just to compute the so-called "Fourier Sine Series" of an $ L^2$ function, which magically converges.  Clearly, if $ \sin(mx)$, $ \cos(mx)$, and $ 1$ are all needed to form an orthonormal basis, then $ \sin(mx)$, by itself, cannot be a "basis," and similarly for the "cosine series."  Could someone enlighten me on this point?

Ah, I got it.  Thanks anyways.

Those variant series are working on different spaces. A linear combination of sines is an odd function, and a linear combination of cosines is an even function. This gives you a natural split; we can write any nice enough $ 2\pi$-periodic odd function $ f$ as $ f(x)\equal{}\sum_n a_n\sin(nx)$, and any nice enough $ 2\pi$-periodic even function $ g$ as $ g(x)\equal{}\sum_n b_n\cos(nx)$. Note that $ f(n\pi)\equal{}0$ and $ g'(n\pi)\equal{}0$ for all integer $ n$.

The PDE applications aren't working with the whole line as a domain, just an interval with some boundary conditions. On $ [0,\pi]$ with the boundary conditions $ f(0)\equal{}f(\pi)\equal{}0$, that sine series looks good; we can extend any such function to a $ 2\pi$-periodic odd function uniquely. Similarly, if the boundary conditions are $ f'(0)\equal{}f'(\pi)\equal{}0$, we can extend uniquely to a $ 2\pi$-periodic even function, and the cosine series works well.

But the periodicity of any $ L^2([0,\pi])$ function makes no difference in the $ L^2$ convergence of its Fourier series, right?  So I only need either Sine or Cosine when considering functions on $ [0,x]$.


Let $ a,b,c$ be nonnegative real numbers with sum $ 3$. Prove that
$ \frac{1}{5a^2 \plus{} ab \plus{} bc} \plus{} \frac {1}{5b^2 \plus{} bc \plus{} ca} \plus{} \frac {1}{5c^2 \plus{} ca \plus{} ab} \ge \frac {3}{7}.$
:)

[quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers with sum $ 3$. Prove that
$ \frac {1}{5a^2 \plus{} ab \plus{} bc} \plus{} \frac {1}{5b^2 \plus{} bc \plus{} ca} \plus{} \frac {1}{5c^2 \plus{} ca \plus{} ab} \ge \frac {3}{7}.$
:)[/quote]
$ \sum_{cyc}\frac {1}{5a^2 \plus{} ab \plus{} bc} \equal{} \sum_{cyc}\frac {(b \plus{} c)^2}{(b \plus{} c)^2(5a^2 \plus{} ab \plus{} bc)}\geq\frac {4(a \plus{} b \plus{} c)^2}{\sum(b \plus{} c)^2(5a^2 \plus{} ab \plus{} bc)}.$
But $ \frac {4(a \plus{} b \plus{} c)^2}{\sum(b \plus{} c)^2(5a^2 \plus{} ab \plus{} bc)}\geq\frac {3}{7}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(28a^4 \plus{} 85a^3b \plus{} 58a^3c \minus{} 156a^2b^2 \minus{} 15a^2bc)\geq0\Leftrightarrow$
$ \Leftrightarrow14\sum_{cyc}(2a^4 \plus{} 4a^3b \plus{} 4a^3c \minus{} 11a^2b^2 \plus{} a^2bc) \plus{} \sum_{cyc}(a^3b \plus{} a^3c \minus{} 2a^2b^2) \plus{}$
$ \plus{} \sum_{cyc}(28a^3b \plus{} a^3c \minus{} 29a^2bc)\geq0,$ which is obvious.
The following inequality is nicer.
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab\plus{}ac\plus{}bc\neq0.$ Prove that:

	\[ \frac {1}{a^2 \plus{} ab \plus{} bc} \plus{} \frac {1}{b^2 \plus{} bc \plus{} ca} \plus{} \frac {1}{c^2 \plus{} ca \plus{} ab} \ge \frac {9}{(a\plus{}b\plus{}c)^2}\]

[quote="Dimitris X"][quote="arqady"] following inequality is nicer.
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab \plus{} ac \plus{} bc\neq0.$ Prove that:
\[ \frac {1}{a^2 \plus{} ab \plus{} bc} \plus{} \frac {1}{b^2 \plus{} bc \plus{} ca} \plus{} \frac {1}{c^2 \plus{} ca \plus{} ab} \ge \frac {9}{(a \plus{} b \plus{} c)^2}
\]
[/quote]It is obvious from Adreescu's inequality :)[/quote]
What do you mean? Explain please. Thank you!

I think he refers to a variant of Cauchy-Schwartz due to Titu Andreescu.
I agree-
$ (\sum_{cyc} a^2\plus{}ab\plus{}bc)(\frac {1}{a^2 \plus{} ab \plus{} bc} \plus{} \frac {1}{b^2 \plus{} bc \plus{} ca} \plus{} \frac {1}{c^2 \plus{} ca \plus{} ab}) \ge (1\plus{}1\plus{}1)^2\equal{}9$
$ \implies$
$ \frac {1}{a^2 \plus{} ab \plus{} bc} \plus{} \frac {1}{b^2 \plus{} bc \plus{} ca} \plus{} \frac {1}{c^2 \plus{} ca \plus{} ab} \ge \frac {9}{(a \plus{} b \plus{} c)^2}$

Thank you, bambaman! This is also my proof. I did not know that here we use the Titu Andreescu's inequality.

Well, it is not really his inequality. It is exactly C-S:
$ \sum \frac {a_i^2}{b_i} \ge \frac {(\sum a_i)^2}{\sum b_i}$ (when $ b_i \ge 0, \sum b_i > 0$)
(you actually used it in your solution...)
Looking at the solution (of arqady's problem) again, we can interpret it as a use of AM-HM instead of Cauchy, which is more elementary.
All in all, can_hang2007's problem is more interesting, wondering whether there are more solutions.

[quote="bambaman"]I think he refers to a variant of Cauchy-Schwartz due to Titu Andreescu.
I agree-
$ (\sum_{cyc} a^2 \plus{} ab \plus{} bc)(\frac {1}{a^2 \plus{} ab \plus{} bc} \plus{} \frac {1}{b^2 \plus{} bc \plus{} ca} \plus{} \frac {1}{c^2 \plus{} ca \plus{} ab}) \ge (1 \plus{} 1 \plus{} 1)^2 \equal{} 9$
$ \implies$
$ \frac {1}{a^2 \plus{} ab \plus{} bc} \plus{} \frac {1}{b^2 \plus{} bc \plus{} ca} \plus{} \frac {1}{c^2 \plus{} ca \plus{} ab} \ge \frac {9}{(a \plus{} b \plus{} c)^2}$[/quote]
Exactly :)

Actually, we can prove by Cauchy-Schwarz that

$ \sum \frac 1{ka^2\plus{}ab\plus{}bc} \ge \frac 3{k\plus{}2}$

for any $ k \ge 0$. :wink:

[quote="Vasc"]Actually, we can prove by Cauchy-Schwarz that

$ \sum \frac 1{ka^2 \plus{} ab \plus{} bc} \ge \frac 3{k \plus{} 2}$

for any $ k \ge 0$. :wink:[/quote]
Yes.  :lol:

[quote="arqady"][quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers with sum $ 3$. Prove that
$ \frac {1}{5a^2 \plus{} ab \plus{} bc} \plus{} \frac {1}{5b^2 \plus{} bc \plus{} ca} \plus{} \frac {1}{5c^2 \plus{} ca \plus{} ab} \ge \frac {3}{7}.$
:)[/quote]
$ \sum_{cyc}\frac {1}{5a^2 \plus{} ab \plus{} bc} \equal{} \sum_{cyc}\frac {(b \plus{} c)^2}{(b \plus{} c)^2(5a^2 \plus{} ab \plus{} bc)}\geq\frac {4(a \plus{} b \plus{} c)^2}{\sum(b \plus{} c)^2(5a^2 \plus{} ab \plus{} bc)}.$
But $ \frac {4(a \plus{} b \plus{} c)^2}{\sum(b \plus{} c)^2(5a^2 \plus{} ab \plus{} bc)}\geq\frac {3}{7}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(28a^4 \plus{} 85a^3b \plus{} 58a^3c \minus{} 156a^2b^2 \minus{} 15a^2bc)\geq0\Leftrightarrow$
$ \Leftrightarrow14\sum_{cyc}(2a^4 \plus{} 4a^3b \plus{} 4a^3c \minus{} 11a^2b^2 \plus{} a^2bc) \plus{} \sum_{cyc}(a^3b \plus{} a^3c \minus{} 2a^2b^2) \plus{}$
$ \plus{} \sum_{cyc}(28a^3b \plus{} a^3c \minus{} 29a^2bc)\geq0,$ which is obvious.
The following inequality is nicer.
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab \plus{} ac \plus{} bc\neq0.$ Prove that:
\[ \frac {1}{a^2 \plus{} ab \plus{} bc} \plus{} \frac {1}{b^2 \plus{} bc \plus{} ca} \plus{} \frac {1}{c^2 \plus{} ca \plus{} ab} \ge \frac {9}{(a \plus{} b \plus{} c)^2}
\]
[/quote]

Why don't we use AM HM inequality? :roll:
$ \frac{\sum \frac{1}{a^2\plus{}ab\plus{}bc}}{3}\ge\frac{3}{\sum a^2\plus{}2\sum ab}$
$ \Longrightarrow \sum\frac{1}{a^2\plus{}ab\plus{}bc}\ge \frac{9}{(a\plus{}b\plus{}c)^2}$
 :lol:

[quote="Vasc"]Actually, we can prove by Cauchy-Schwarz that

$ \sum \frac 1{ka^2 \plus{} ab \plus{} bc} \ge \frac 3{k \plus{} 2}$

for any $ k \ge 0$. :wink:[/quote]
without any conditions??? :wink:   try, $ k\equal{}0$ , $ a\equal{}b\equal{}c\equal{}4$

[quote="peine"][quote="Vasc"]Actually, we can prove by Cauchy-Schwarz that

$ \sum \frac 1{ka^2 \plus{} ab \plus{} bc} \ge \frac 3{k \plus{} 2}$

for any $ k \ge 0$. :wink:[/quote]
without any conditions??? :wink:   try, $ k \equal{} 0$ , $ a \equal{} b \equal{} c \equal{} 4$[/quote]
We should have $ a\plus{}b\plus{}c\equal{}3$ :)

thank's a lot Mathias_DK . Now, I see it's true.


Let $ O$ be the center of a circle and $ A$ be a fixed interior point of the circle different from $ O.$ Determine all points $ P$ on the circumference of the circle such that the angle $ OPA$ is a maximum.

[quote="4everwise"]Let $O$ be the center of a circle and $A$ be a fixed interior point of the circle different from $O.$ Determine all points $P$ on the circumference of the circle such that the angle $OPA$ is a maximum.[/quote]
First, notice that $\angle OPA$ is less than $90^{o}$. That's because  $\angle OPA< \angle OAP$ because $OP>OA$, because $OP$ is radius, while $AO$ is less than the radius. So, $\angle OPA$ is less than $90^{o}$, so if $\angle OPA$ is maximized, $sin(OPA)$ is maximized.
By Sine Law in triangle $OAP$ we get:
$sin(OPA)=\frac{OA*sin(OAP)}{PO}$
Because $OA$ and $PO$ are fixed, $sin(OPA)$ is maximized if $sin(OAP)$. The maximimum value of $sin(OAP)$ is $1$, that's when $\angle OAP=90^{o}$, so $\angle OPA$ is maximized for all pts $P$ on circumcentre such that $\angle OAP=90^{o}$

[color=darkblue][b]Generalization.[/b] Given are one circle $w=C(O,R)$ and a two fixed points $A$ and $B$.
Ascertain the position of the point $P\in w$ for which the measure of the angle $\widehat{APB}$ is maximum (minimum).[/color]


[quote="lotrgreengrapes7926"][quote="Inspired By Nature"]Prove that, for all positive integer pairs $(a,b)$ where $b>2$, $2^{b}-1$ does not evenly divide $2^{a}+1$.[/quote] When $a=\frac{b}{2}$, they do divide evenly.  :maybe:[/quote]

Yes, the second divides the first in that case.  You need the first to divide the second.  For that to be possible, you automatically have to have $a \geq b$, so you don't need it as an extra condition.

[quote="lotrgreengrapes7926"][quote="Inspired By Nature"]Prove that, for all positive integer pairs $(a,b)$ where $b>2$, $2^{b}-1$ does not evenly divide $2^{a}+1$.[/quote] When $a=\frac{b}{2}$, they do divide evenly.  :maybe:[/quote]
$\frac{b}{2}$ may not be an integer.

[quote="t0rajir0u"][quote="Inspired By Nature"]$\frac{b}{2}$ may not be an integer.[/quote]

Irrelevant.  Consider the relative magnitudes of $2^{ \frac{b}{2}}+1$ and $2^{b}-1$.  ;)[/quote]
oh, ofcourse. :blush:


function
algebra
functional equation
algebra unsolved

Determine all functions :
$yf(2x)-xf(2y)=8xy(x^2-y^2)$

hi,

i think it isn't hard :

lets assume : $x=y$

$\Rightarrow 2xf(2x) = 0$

$\Rightarrow f(2x)=0 \Rightarrow  f(x) = 0$

 :D .

Think again. denote $g(x)=f(x)-x^3$ it sould seem easy from now on.

Yes,then $yg(2x)=xg(2y)$.When $x\neq0$,$\frac{g(x)}{x}$ is constant.
And $g(0)=0$
So $f(x)=kx+x^3(k\in R)$

the solution given by taking x = y is wrong because
the solution x = 0 was neglected. If it was considered then functional equation converts
as (taking x=0)
                           yf(0) = 0 now we have also to consider y = 0. so this mathod is wrong.

for the whole solution download the attachment


Let phi(n) denote the number of positive integers :le: n, coprime with n.
Prove that for infinitely many n, gcd(n,phi(n)) = 1.

let S be the set of all primes. As phi(p)=p-1, where p is a prime, and two consecutive numbers are always coprime, every member of S satisfies that condition. There are infinitely many members of S, so thus there are infinitely many numbers that satisfy that condition.

Sorry, I forgot to mention one condition: prove that there are infinitely many [u]composite[/u] numbers n for which gcd(n, phi(n)) = 1.

Here's my solution.[hide]Let p be a prime congruent to 5 mod 6.  Let n = 3p.  Claim: n satisfies the condition.  First, phi(n) = phi(3p) = phi(3) phi(p) = 2(p-1).  Therefore,

gcd(n, phi(n)) = gcd(3p, 2(p-1)) :le: gcd(3, 2(p-1)) gcd(p, 2(p-1)) :le: gcd(3, 2) gcd(3, p-1) gcd(p, 2) gcd(p, p-1) = 1 * 1 * 1 * 1 = 1.

Because there are infinitely primes that are 5 mod 6, we are done.

[/hide]

Ok, that's nice.
Now I ask you to relate this problem to the following one:

Prove that infinitely many composite n satisfy:
If n|an -  1 for any integer a then n2|an - 1.

Of course it is a very difficult problem when you didn't see the problem about (phi(n), n) = 1 before. But now you have seen it, it should be trivial.


complex analysis
complex analysis unsolved

the shortest hyperbolic lenght that joins two points in the unit disc is called the
noneuclidean distance. Derive the following formula for the noneuclidean distance between z1 and z2:

1/2 log 1+|z1-z2|/|1-z1bar z2|  / 1-|z1-z2|/|1-z1bar z2|.

2) how many noneuclidean lenght in the upper half plane be defined?

thx anyway

[quote="ma_ghour"]2) how many noneuclidean lenght in the upper half plane be defined? [/quote]

It depends on what you mean by "non-Euclidean length", but:
Noneuclidean plane geometries (ie. on a 2-dimensional orientable surface with no holes)are classified, up to isometry, by one number: the curvature.


Prove that $\exists n \in N$ such that $2^{n}+3^{n}$ has exactly 23 prime divisors.

If you have anything to say about this problem, please do it there :
http://www.mathlinks.ro/Forum/viewtopic.php?t=17203


Prove that if $ a,b,c$ are the sides of the triangle,then the following inequality holds:

$ \sum a^6\plus{}abc(\sum a^3)\plus{}2(\sum a^3b^3)\leq\sum ab(a^4\plus{}b^4)\plus{}\sum a^2b^2(a^2\plus{}b^2)$.

[quote="arqady"][quote="Erken"]Prove that if $ a,b,c$ are the sides of the triangle,then the following inequality holds:

$ \sum a^6 \plus{} abc(\sum a^3) \plus{} 2(\sum a^3b^3)\leq\sum ab(a^4 \plus{} b^4) \plus{} \sum a^2b^2(a^2 \plus{} b^2)$.[/quote]
It's equivalent to $ \sum_{cyc}(a \minus{} b)^2(a^2 \plus{} c^2 \minus{} b^2)(b^2 \plus{} c^2 \minus{} a^2)\geq0,$
which not true for an isosceles blunt angle triangle . :wink:[/quote]
Oh,i'm sorry,i mean $ a,b,c$ are the sides of an acute angle triangle.Thank you for your solution.
P.S:
Pomnite ya vas spra6ival o tom kak raskaldivat' takogo roda neravenstva,i ne mogli bi vi  pokazat' na etom primere kak do etogo dogadatsya?
Spasibo.

For you, Erken!  :)

$ \sum_{cyc}(a^6 \plus{} a^4bc \plus{} 2a^3b^3 \minus{} a^5b \minus{} a^5c \minus{} a^4b^2 \minus{} a^4c^2)\leq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(2a^6 \plus{} 2a^4bc \plus{} 4a^3b^3 \minus{} 2a^5b \minus{} 2a^5c \minus{} 2a^4b^2 \minus{} 2a^4c^2)\leq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a^6 \minus{} a^5b \minus{} ab^5 \plus{} b^6) \minus{} \sum_{cyc}(a^5b \minus{} 2a^3b^3 \plus{} ab^5) \minus{}$
$ \minus{} \sum_{cyc}(a^4b^2 \minus{} 2a^3b^3 \plus{} a^2b^4) \minus{} \sum_{cyc}(c^4a^2 \minus{} 2c^4ab \plus{} c^4b^2)\leq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a \minus{} b)^2(a^4 \plus{} a^3b \plus{} a^2b^2 \plus{} ab^3 \plus{} b^4) \minus{} \sum_{cyc}(a \minus{} b)^2(a \plus{} b)^2ab \minus{}$
$ \minus{} \sum_{cyc}(a \minus{} b)^2a^2b^2 \minus{} \sum_{cyc}(a \minus{} b)^2c^4\leq0\Leftrightarrow\sum_{cyc}(a \minus{} b)^2(a^4 \minus{} 2a^2b^2 \plus{} b^4 \minus{} c^4)\leq0.$


Latex questions:

How to you make a table?

How you start a new paragraph?

How do you make different lines of equation?

For a new paragraph, you need to signal a new line and then indent.  To end the line, use either "\\" or "\newline".  Then, use "\indent" to indent it.

For tables, everything you need is here: http://en.wikibooks.org/wiki/LaTeX/Tables

And for lines of equations, look here: http://en.wikibooks.org/wiki/LaTeX/Mathematics

[quote="JRav"]For a new paragraph, you need to signal a new line and then indent.  To end the line, use either "\" or "\newline".  Then, use "\indent" to indent it.[/quote]The default is to always indent the first line of a paragraph so no need for \indent. A new paragraph is created by using a blank line in the editor - see [[LaTeX:Layout#Paragraphs]] for examples.


Solve the system of equation for all $a,b,c\in\mathbb{R^+}$ satisfying

$a+b+c=1$

$a\sqrt{bc}+b\sqrt{ac}+c\sqrt{ab}=1$

 :)

[quote="luimichael"]There are no solutions for the system of equations:
$1=(a\sqrt{bc}+b\sqrt{ca}+c\sqrt{ab})^2 \leq (a^2+b^2+c^2)(ca+ab+ca) =(1-2u)u$ if we denote $u=ab+bc+ca$
$=-2u^2+u=-2(u-\frac 1 4 )^2+\frac 1 8 \leq \frac 1 8$, which is a contradiction![/quote]

yes , you are right  :)

What if i change the condition $a+b+c=1$ to $ab+ac+bc=1$  :)

Sorry, I mis-interpreted the question....

Let me try:
Let $p=\sqrt{a}, q =\sqrt{b}, r = \sqrt{c}$.
The system becomes:
$p^2q^2+q^2r^2+r^2p^2 =1$ , and
$(p+q+r)pqr=1$.
However,
$(pq+qr+rp)^2=p^2q^2+q^2r^2+r^2p^2+2pqr(p+q+r)=1+2(1)=3 \implies pq+pr+rp=\sqrt{3}$.
Denote pqr = t, then $p+q+r=\frac 1 t$.
Now, p,q and r are the roots of the following cubic equation:
$X^3-{\frac 1 t}X^2+\sqrt{3}X-t=0$.---(*)

There should be a discriminant which check the nature of roots of a cubic, but I can't remember the expression....
Anyway, (*) will give the general solution to p,q and r, and hence a, b, c can be determined in this way.

[quote]
ab+bc+ca = 1
a\sqrt{bc} + b\sqrt{ac} + c\sqrt{ab} = 1
[/quote]

Set $x = bc$ etc.

we get $\sum x = 1 = \sum \sqrt{yz}$.

But $x+y \ge 2sqrt{xy}$ whence $1 = \sum x \ge \sum \sqrt{yz} = 1$ and x=y=z for equality

Hence ab=bc=ca whence it is clear a=b=c

hence $a = \sqrt{3}/3$


So a bunch of bad guys are going to get killed by an angry moograman.
The moograman will only let 1 person survive and the killing is as follows
Line them up all in a line and...
Kill the first one,
Spare the second one (for now)
Kill the next...
and spare the next
When you get to the end of the line the moograman starts over
kill the first and spare the next
until there is only 1 standing.
You are a bad guy and where would you stand?
For example:
There are 10 bad guys
round 1: x=die o=survives
xoxoxoxoxo
round 2: x=die o=survives
xxxoxxxxxx
so number 4 survives
[hide="can tell me why this works?"]
subtract the nearest power of 2 from the number of people. then multiply by 2.
I have found this due to a pattern. but this isn't a proof and I would like to see one.[/hide]
[size=150]EDIT THE MOOGRAMAN GOES KILL SPARE KILL SPARE. He does not start over back in the beginning.[/size]

[quote="mihail911"]no offense, but could you tune it down to about PG.  :D
im gonna have nightmares tonight. couldn't you have said the blissful moograman wanted to have a tea party with....
 :lol:[/quote]

well at least I didn't add the details and at least they weren't innocent people

[quote="moogra"]So a bunch of bad guys are going to get killed by an angry moograman.
The moograman will only let 1 person survive and the killing is as follows
Line them up all in a line and...
Kill the first one,
Spare the second one (for now)
Kill the next...
and spare the next
When you get to the end of the line the moograman starts over
kill the first and spare the next
until there is only 1 standing.
You are a bad guy and where would you stand?
For example:
There are 10 bad guys
round 1: x=die o=survives
xoxoxoxoxo
round 2: x=die o=survives
xxxoxxxxxx
so number 4 survives
[hide="can tell me why this works?"]
subtract the nearest power of 2 from the number of people. then multiply by 2.
I have found this due to a pattern. but this isn't a proof and I would like to see one.[/hide][/quote]
For 10 bad guys, shouldn't it be like this?
round 1: x=die o=survives
xoxoxoxoxo
round 2: x=die o=survives
xxxoxxxoxx
round 3: x=die o=survives
xxxxxxxoxx
So the last guy standing is #8

[quote="kh-"][quote]subtract the nearest power of 2 from the number of people. then multiply by 2. [/quote]
Don't think this always works. A counterexample is $ 9$ people:

$ 1, 2, 3, 4, 5, 6, 7, 8, 9$
$ 2, 4, 6, 8$
$ 4, 8$
$ 8$

And the nearest power of $ 2$ to $ 9$ is $ 8$, and $ (9 \minus{} 8) \times 2 \equal{} 2$. Which was killed off in the first iteration [whoops, did protestant just show this?]. What is obvious is that you need to be at an even place in the line, anyway, for every iteration.

[hide="Investigation"]This is a far cry from a proof, but after writing out some examples up to $ 16$ people there is a clear pattern emerging. It looks like this:

$ 1$ person $ \rightarrow 1$ survives
$ 2$ people $ \rightarrow 2$ survives
$ 3 \rightarrow 2$
$ 4 \rightarrow 4$
$ 5 \rightarrow 4$
$ 6 \rightarrow 4$
$ 7 \rightarrow 4$
$ 8 \rightarrow 8$
$ ...$
$ 15 \rightarrow 8$
$ 16 \rightarrow 16$

My conjecture, then, is that you take the number of people in the line and round down (floor?) to the nearest power of 2 to find the safe position. This is because the Moograman, in his infinite wisdom, systematically eliminates all the odd positions in each iteration. Your position being divisible by 2 is thus the criteria for successfully making it to the next iteration.

This means the highest power of two is always going to last the longest: it's 'most divisible by 2', if you like, as one's position is divided by 2 every iteration and the powers of two are fantastically good at being divided by 2, superseded in their divisibility by 2 only by the next power of 2.

Someone better than me can probably splice something of these thoughts into a rigourous proof. I can't think of a pretty formula to find the safe position, coincidentally.
[/hide][/quote]

Oh I meant, just keep skip a person everytime. So it would be
$ 1 2 3 4 5 6 7 8 9$
$ 2 4 6 8$
$ 2 6$
$ 2$

[quote="kh-"]So you've flipped it around now? I can't see what the Moograman is supposed to be doing now:

What you're saying:
$ 1, 2, 3, 4, 5, 6, 7, 8, 9$
$ 2, 4, 6, 8$ -- eliminated all [b]odd[/b] positions (kill, skip, kill, skip...)
$ 2, 6$ -- eliminated all [b]even[/b] positions (skip, kill, skip, kill)
$ 2$ -- eliminated last (skip, kill)

Another example you gave in the first post:
$ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10$
$ 2, 4, 6, 8, 10$ -- eliminated all [b]odd[/b] positions (kill, skip, kill, skip...)
$ 4, 8$ -- eliminated all [b]odd[/b] positions (skip, kill, skip, kill, skip)
$ 4$ -- eliminated last (skip, kill)

You've lost me as to the rule being used. I thought it was just knock out anyone in an odd numbered position, but now I don't see the consistency at all.[/quote]

no it's always been kill skip a person, kill, skip a person, kill, skip a person

OK, that's the premise I was working with. But there's a subtlety here I wish I had seen earlier so I didn't have to go through all this palaver: the Moograman is not 'resetting' after finishing an iteration on the line:

So he actually does this:
$ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10$
$ 2, 4, 6, 8, 10$
$ 4, 8$
$ 4$

Instead of:
$ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10$
$ 2, 4, 6, 8, 10$
$ 4, 8$
$ 8$

In which case nothing I've said in the investigation back there is very useful in this situation ha-ha-ha oh [i]God.[/i] :P

Well I listed a few and this is what I got
$ 1: no one lives$
$ 2: 2$
$ 3: 2$
$ 4: 4$
$ 5: 2$
$ 6: 4$
$ 7: 6$
$ 8: 8$
$ 9: 2$
$ 10: 4$
$ 11: 6$
$ 12: 8$
It is always even numbers living, because of the odds getting killed off first
I don't know why it always goes up by 2's though.


linear algebra
matrix
trigonometry
linear algebra unsolved

Assume that $ x_1,...,x_n$ are different integers greater than 1. Define $ a_{ij}=sin^2 (\frac{j\cdot\pi}{x_i})$. Prove that this matrix has non-zero determinant.


ratio
analytic geometry
vector
Euler
geometry

Is Ceva's Thm. the one that relates the ratios of sides cut by cevians (product equals 1) or is that Menalaus' thm. If what I am describing is in fact Ceva's thm., what is Menalaus' thm?

I was wondering how to prove that the altitudes of a triangle are concurrent with the thm. i was describing above. Any hints?

[quote="drunner2007"]Is Ceva's Thm. the one that relates the ratios of sides cut by cevians (product equals 1)[/quote]
That is true.
Menalaus's theorem is a little different:
http://mathworld.wolfram.com/MenelausTheorem.html

Ceva's theorem is the wrong way to go with the altitudes, as far as I know, as are coordinates, vectors, mass points, and Stewart's Theorem.  The Euler line is much more elegant.  (I was surprised when it wasn't in AoPS II.)  Mathworld completely ignores the elegant synthetic proof.

[hide="Hint"]The centroid is between the orthocenter and circumcenter.[/hide] By the way, cevians were named after Ceva, so that makes remembering the names easier.


function
floor function
real analysis
real analysis unsolved

How to prove that for every nonnegative $f\in L^{p}$, $1\leq p<\infty$ one can find simple functions $\phi_{i}: X\to [0,\infty]$ such that $\parallel f-\phi_{i}\parallel _{p}\to 0$ as $i\to \infty$?

Hints:
1) Show that $f$ can be approximated by a bounded function. Let
\[f^{M}(x) = \begin{cases}f(x), & \text{ if }f(x) \leq M;\\ M, & \text{ if }f(x) \geq M. \end{cases}\]
Show that $\|f(x)-f^{M}(x)\|_{p}\to 0$ as $M \to \infty$.
2) Assume now that $f(x)\in L_{p}[0,1]$ is bounded. Consider the function
\[\phi_{N}(x) = \frac{1}{N}\lfloor N \cdot f(x)\rfloor.\]
Prove that $\|f(x)-\phi_{N}(x)\|_{p}\leq 1/N$.

One more approximation step is needed if $X$ has infinite measure.


Hey let's start a new thread, this time on $ \pi$!! So basically we just list the digits of pi in order, and see how far we can get.
If you don't know the numbers, just google pi. At least that's how I figured out a few of them.

OK, I'll start off $ \Rightarrow$ [size=150][color=darkred]3[/color][/size]

1 [hide=" "]This is a secret code to all scientists out there. Scientists if you see this message then I've discovered a new element. This element is extremely radioactive, and perilous. DO NOT QUOTE THIS POST UNDER ANY CIRCUMSTANCES, but pm me and you will get a prize.[/hide]

5[hide=" "] There is a secret message hidden in my first post. DO NOT QUOTE THIS I REPEAT...DO NOT QUOTE THIS.[/hide]

2

The message is too small. Please make the message longer before submitting.

9... :D
 The message is too small. Please make the message longer before submitting.

$ 7$ :!:  :lol:

@k00lperson: lol. 100,000...believed as the world record of pi memorization (although it's controversial)

[size=200]2[/size]

Why not list the digits of $ e$ ?

[hide="digist"]
$ 2$
[/hide]

8

I thought about [i]e[/i] too, but I memorized more pi than e. Plus, the sequence of numbers in pi is more interesting than e. :wink:

:first:
The message is too big. Please make the message shorter before submitting.  :lol:

SIETE = spanish

SEPT = french

七 = chinese

Arabic Numeral: 1
Roman Numeral: I
English: one
Spanish: uno
Italian: uno
German: eins
French: un
Latin: unus
Hindi: एक
Chinese: 一
Korean: 일
Japanese: 一
 :first:  :first: (wow that looks like a bunch of language spam)

[quote="jampa"]I know i am repeating some numbers and adding alot more so here's what i got

3.141592653589793238462643 . . .

I think you made an error at position = 2112   :|


modular arithmetic
absolute value
number theory

1. For every triple $ (a,b,c)$ of nonzero real numbers, form the number $ \frac{a}{\|{a}}\plus{}\frac{b}{\|{b}}\plus{}\frac{c}{\|{c}}\plus{}\frac{abc}{\|{abc}}$.
What's the set of real numbers formed? (The denometers are absolute value)
2. Let a,b,c be positive integers. When is $ \sqrt{a\plus{}\frac{b}{c}}\equal{}a\sqrt{\frac{b}{c}}$?
3. How many three digit numbers contain 1 and 3 but none of the digits 0, 2, or 5?
4. If a, b, c are roots of $ 2x^3 \plus{}4x^2 \plus{}px\plus{}q\equal{}0$, find a+b+c.
5. What day of the week will it be $ 2005^{2005}$ days after a Monday?

[hide="4"]
Since the equation's roots are a, b, and c, and the coefficient of the $ x^{3}$ term is 2, the equation can be written as $ 2(x \minus{} a)(x \minus{} b)(x \minus{} c) \equal{} 0$
Expanding this gives $ 2x^3 \minus{} 2(a \plus{} b \plus{} c)x^2 \plus{} 2(ab \plus{} ac \plus{} bc)x \minus{} 2abc \equal{} 0$
Thus, $ 4 \equal{} \minus{} 2(a \plus{} b \plus{} c) \implies \boxed {a \plus{} b \plus{} c \equal{} \minus{} 2}$[/hide]

Stop posting MMPC problems.
[hide="1"]Each can be 1 or -1 depending if the numerator is positive or negative. $ {4,0, \minus{} 4}$.[/hide]

[hide="2"]Square, so $ a^2\frac{b}{c}\equal{}a\plus{}\frac{b}{c}$. Solving for c, it is $ \frac{b(a^2\minus{}1)}{a}$.[/hide]

[hide="5"] Fermat's Little Theorem says $ 2005^6\equiv1\pmod{7}$. So, $ 2005^{2005}\equiv3\pmod{7}$, so it is Thursday.[/hide]

[hide="3"]Casework when 1 and 3 are the first digit, 2nd digit, or 3rd digit. Of course, you overcount, so you have to use PIE. The final answer is 36.[/hide]


trigonometry
geometry
perimeter
algebra
difference of squares
special factorizations
AMC

A triangle has sides of length a, b, and c which satisfy the equation (a+b+c)(a+b-c) = 3ab.  Determine the measure of the angle opposite the side of length c.

(AHSME 1981, #21)

[hide]expand after noticing it is the difference of squares, move the c^2 onto the RHS and the 3ab onto the LHS.

you end with (1)a^2+b^2-ab=c^2, which looks strangely similar to the Law of Cosines.  the angle opposite c must satisfy 2cosx=1, since 2abCosx=ab (according to (1) and the law of cosines).

so cosx=1/2 and x=+/- 60.  angles are generally positive, so the angle is 60 degrees.[/hide]

[quote=darij grinberg]A slightly different solution:

[hide]Among the many advanced trigonometry formulas for triangles, I will use the following one: For every triangle ABC with sidelengths a, b, c and angles A, B, C, we have $\cos \frac{C}{2} = \sqrt{\frac{s \left(s-c\right) }{ab}}$, where $s=\frac{a+b+c}{2}$ is the semiperimeter (i. e. one half of the perimeter) of the triangle ABC. Using that $s=\frac{a+b+c}{2}$ and hence $s-c=\frac{a+b-c}{2}$, we can rewrite this formula as $\cos \frac{C}{2} = \sqrt{\frac{\frac{a+b+c}{2} \cdot \frac{a+b-c}{2} }{ab}} = \sqrt{\frac{\left(a+b+c\right) \left(a+b-c\right) }{4ab}}$. Now, given the condition (a + b + c) (a + b - c) = 3ab, we have $\frac{\left(a+b+c\right) \left(a+b-c\right) }{4ab} = \frac{3ab}{4ab} = \frac34$, so that $\cos \frac{C}{2} = \sqrt{\frac34} = \frac{\sqrt3}{2}$, and thus $\frac{C}{2} = 30^{\circ}$ (since $0^{\circ} < \frac{C}{2} < 90^{\circ}$), so that C = 60°.

Actually, I have posted the solution above not because it is shorter than the one using the cosine law, but because it is a good application of the formula

$\cos \frac{C}{2} = \sqrt{\frac{s \left(s-c\right) }{ab}}$

which holds for every triangle with sides a, b, c, angles A, B, C and semiperimeter s. Here are two similar important formulas:

$\sin \frac{C}{2} = \sqrt{\frac{\left(s-a\right) \left(s-b\right) }{ab}}$;

$\tan \frac{C}{2} = \sqrt{\frac{\left(s-a\right) \left(s-b\right) }{s \left(s-c\right) }}$.

And, of course, analogous formulas hold for $\frac{A}{2}$ and $\frac{B}{2}$. Proving these formulas is not difficult; just take the cosine law and apply the formulas $\sin \frac{\phi}{2} = \sqrt{\frac{1- \sin \phi}{2}}$ and $\cos \frac{\phi}{2} = \sqrt{\frac{1+ \sin \phi}{2}}$ (and the tan formula follows from the sin and cos formulas).[/hide]

  Darij[/quote]
Nice :)


Find all continuous functions $ f: \mathbb{R}\to\mathbb{R}$ such that ${ f(xy)=xf(y)+yf(x)}$, for all $ x,y\in\mathbb{R}$

we take for $ x\in \mathbb{R}$: $ h(x)=\frac{f(e^{x})}{e^{x}}$
$ \forall x,y\in \mathbb{R}: \ h(x+y)=\frac{e^xf(e^y)+e^yf(e^x)}{e^xe^y}=h(x)+h(y)$
then $ \exists a\in\mathbb{R}: \ h(x)=ax$
then $ \forall x\in\mathbb{R}^+: \ f(x)=h(ln(x))x=axln(x)$
and $ \forall x\in\mathbb{R}^-: \ f(x)=\frac{f(x^2)}{2x}=\frac{ax^2ln(x^2)}{2x}=axln(|x|)$
and $ f(0)=0f(0)+0f(0)=0$

$ f(x)=axln(|x|)\ if \ x\neq 0$ and $ f(0)=0$

If i change the condition of problem, it look strange:
Find all continuous functions $ f: (0,\infty)\to\(0,\infty)$ that satisfy $ f(x)f(y)=f(xy)+f\left(\frac{x}{y}\right)$

[hide="Hint"]Since it is symmetric in $ x$ and $ y$, $ f(x)f(y) \equal{} f(xy) \plus{} f\left( \frac {y}{x} \right) \implies f(x/y) \equal{} f(y/x)$ $ \implies f(x) \equal{} f(1/x)$

We can thus decompose $ f$ by defining another continuous function $ g$ such that $ f(x) \equal{} g(x)\plus{}g(1/x)$.

[/hide]

[hide="Another hint"]The solutions include $ f(x)\equal{}x^j \plus{} \frac{1}{x^j} \; \; \forall \; x \in \mathbb{R}$[/hide]


algebra
polynomial
calculus
calculus computations

The polynomial $ p(x)$ is of degree $ 9$ and $ p(x)\minus{}1$ is exactly divisible by $ (x\minus{}1)^{5}$.

(i) Find the value of $ p(1)$.

(ii) Show that $ p'(x)$ is exactly divisible by $ (x\minus{}1)^{4}$.

(iii) Given also that $ p(x) \plus{}1$ is exactly divisible by $ (x\plus{}1)^{5}$, find $ p(x)$.

Typing this as I go along..

$ p \minus{} 1$ being divisible by $ (x \minus{} 1)^5$ implies that $ p \equal{} q (x \minus{} 1)^5 \plus{} 1$, where $ q$ is of degree $ 4$.

We get $ p(1) \equal{} q(1) \cdot 0 \plus{} 1 \equal{} 1$.

$ p' \equal{} q' (x \minus{} 1)^5 \plus{} 5 q (x \minus{} 1)^4 \plus{} 0 \equal{} (x \minus{} 1)^4\left[q'(x \minus{} 1) \plus{} 5 q\right]$, thus $ p'$ is divisible by $ (x \minus{} 1)^4$.

$ p \plus{} 1 \equal{} q(x \minus{} 1)^5 \plus{} 2$, and should be of the form $ r (x \plus{} 1)^5$ where $ r$ is of degree $ 4$.

We thus get that we must have $ q (x \minus{} 1)^5 \plus{} 1 \equal{} r (x \plus{} 1)^5 \minus{} 1$

We must have $ r(1) 2^5 \minus{} 1 \equal{} p(1) \equal{} 1$, so $ r(1) \equal{} \frac1{32}$.

Furthermore, $ p'(1) \equal{} 0 \equal{} 5 r(1) 2^4 \plus{} r'(1) 2^5 \equal{} 80 r(1) \plus{} 32 r'(1)$. This then gives $ r'(1) \equal{} \minus{} \frac5{64}$.

Guess you can continue this or have some slicker way to get the result, but I have to go now.

Thanks, Tyl. I've done much the same as you; I haven't got any slicker way, nor have I found $ p(x)$ yet.

Part 3: $ p(x)\plus{}1$ is divisible by $ (x\plus{}1)^5$, so let $ p(x)\equal{}r(x\plus{}1)^5\minus{}1$. By similar reasoning from Part 2, $ p'(x)$ is also divisible by $ (x\plus{}1)^4$. Therefore, $ p'(x)\equal{}k(x\minus{}1)^4(x\plus{}1)^4$, where $ k$ is a constant. We can simplify and then integrate this:

$ p'(x)\equal{}k(x^2\minus{}1)^4$

$ p'(x)\equal{}k(x^8\minus{}4x^6\plus{}6x^4\minus{}4x^2\plus{}1)$

$ p(x)\equal{}k\left(\frac{x^9}{9}\minus{}\frac{4x^7}{7}\plus{}\frac{6x^5}{5}\minus{}\frac{4x^3}{3}\plus{}x\plus{}C\right)$

$ p(x)\equal{}k\left(\frac{35x^9\minus{}180x^7\plus{}378x^5\minus{}420x^3\plus{}315x\plus{}D}{315}\right)$

Now, if $ D\equal{}\minus{}128$, the numerator would be divisible by $ (x\minus{}1)^5$. Also, if $ D\equal{}128$, the numerator is divisible by $ (x\plus{}1)^5$. Therefore, $ D\equal{}0$ and $ \frac{k}{315}\equal{}\frac{1}{128}\Rightarrow k\equal{}\frac{315}{128}$ Our final answer would be:

$ p(x)\equal{}\frac{35x^9\minus{}180x^7\plus{}378x^5\minus{}420x^3\plus{}315x}{128}$.


function
complex analysis
real analysis
real analysis unsolved

No solution has been proposed for this problem ([b]U20[/b]) so far in the online publication.

Prove that one cannot find an entire function $f$ such that $f(f(x)) = e^{x}$ for all real number $x$ but can find an infinitely many times differentiable function with this property.


MATHCOUNTS
geometry
3D geometry
perimeter
calculus
derivative
ratio

I don't know if this was already posted, but here we go.

What is the Year in which you think that the Questions were the Hardest?  Also specify why it was hard and whether you are talking about School, Chapter, States, and Nationals.

Thanks

I hope I was specific.

[color=darkred][size=75]I added the 2003-2004 option into the poll.
-Neal[/size][/color]

[quote="mathgeniuse^ln(x)"]I was going to but I realized that there wouldn't be enough options to show the distribution.  But what can I add though?

Could I say 1999, 2000, 2001, 2002,..., 2006, and then say other?[/quote]

Yeah. You don't even have to go earlier than 1999 because the earlier competitions weren't at all hard.

1. 2004-2005
2. 1998-1999

2005 was generally hard and most people scored very low, 1999 Nats was pretty hard.

Heh, I just skipped 2003-04, I wll add that if I can.

I can't add 2003-04 to the poll.

Okay, due to my absent-mindedness I forgot 2003-2004.  Forgive me for my stupidity.  There should be 2003-2004 in there.  Sorry.

You can vote for "Other" if you believe that 2003-2004 is the hardest one.

I also agree.  But then again it is people's opinion on whether or not the test gets harder or easier throughout the years.  So I can't go against either of you people and I will remain fair.

But, I like 2005-2006, because I got 14th Place on it.

[quote="Ignite168"][quote="SplashD"]It's pretty obvious that overall, the problems get harder year by year.[/quote]

I disagree, 2006 was a lot easier compared to 2005 and 2000 was a lot easier than 1999.[/quote]

There have been some exeptions. I too agree 2005 was harder than 2006 but that's probably because of the 'pretty low' scores on the 2005 test.

[quote="mathgeniuse^ln(x)"]You shouldn't say "pitiful",  you are kind of hurting people's feelings.  Be a little more nicer.  I do agree that scores were pretty low though.  I got a 24 on that test.[/quote]

You should use correct grammer  :P

"Be a little more nicer."  should be "be a little nicer"

What about the 1994?  Personally, that is my favorite year.  I don't think it was as hard as last years, but it was definitely fairly hard.

Here is a sketch of the basic trend:

Really easy year (bad, in my opinion)
Really hard year
Moderate year (best)
Moderate year
easy year
really easy year
etc.

[quote="Ignite168"]2006 was like 12 times easier than 1999. I got a 16 combined on 1999 and our team got a 2. We did much better in 2006.[/quote]
1999 nats?

[quote="pascal_1623"]1993 was the hardest...
I think I did pretty good on the 2001.[/quote]

I think I practiced a 42 or 43 on 1993.

However, whenever you take a test during the actual nationals, it feels harder.  I bet when I take '06 again, it would not rank in my top 3.

[quote="bpms"][quote="mathgeniuse^ln(x)"]Yes that would be true.

I took the test again, and I got a 46, so it is pressure.

I wonder if we were being videotaped while we were taking the test.[/quote]
I actually tend to do better under pressure.[/quote]

one word: how?

I just focus better. THat's why I like the sprint

I like Sprint, only because they are easier than Target.

But yes, there is more pressure in Sprint, than Target.

I feel target is easier as you can use a calc and you have more time for each question

[quote="mathgeniuse^ln(x)"]Well, I believe that it is hard, only because there is more pressure, as well as the hardness of the problem.[/quote]
I agree. Target sucks. Sprint was my savior at states. My target score was crappy. Although my sprint score was below what i expected, My target score was awful

[quote="mathgeniuse^ln(x)"]But yes, there is more pressure in Sprint, than Target.[/quote]

[quote="mathgeniuse^ln(x)"]Well, I believe that it is hard, only because there is more pressure, as well as the hardness of the problem.[/quote]

you just contradicted yourself

[quote="SplashD"]yank the chain :yankchain:[/quote]
How come you can quote me? I have a url in my sig too.

Whatever...

Could we please get back on topic?

I like Sprint, and that is that.  I can't say why, I just do.


Suppose you know that there are two children in Mr. A's house. Now when Mr. B went to Mr. A's house, and greeted from outside of the door, there was the voice of a boy replying. What is the probability that this boy was the oldest boy of Mr. A?

If the other child is a girl, which is true half the time, then the child answering Mr. B is the oldest boy. Otherwise, he is only the oldest half the time. $ \frac{1}{2}*1 \plus{} \frac{1}{2}*\frac{1}{2} \equal{} \frac{3}{4}$.


I think it is:

$ \text{area of base}$ $ \times$ $ \text{height}$

If it was a regular tetrahedron, the volume would be $ s^3\frac {\sqrt2}{12}.$ I'll leave it to the OP to prove this(?).

Also AwesomeToad, the volume would be 1/3 of what you said, since it's a pyramid.


geometry
3D geometry
conics
parabola
rotation
geometric transformation
calculus

We have all seen the cliche related rates problem where water is draining into a cone and we must find the rate at which the water height is changing when the water is h units deep. Similar triangles.

Suppose, though, that instead of apex down, it is laying on its side. This is a tough nut. This is not homework, just a problem which was posed by a friend.

Here's something more definitive:

"A conical tank has radius of 10 feet and is 24 feet high. If water flows in at 20 ft^3/min, how fast is the depth of the water changing when the water is 6 feet deep?."

Normally, the cone would be apex down and easy to solve, but suppose its lying on its side.

Any ideas?.

You still just need to calculate the area of the cross-section, which is (for a given height) the area bounded by a certain parabola and a certain line.  So, first you have a geometry problem.

Thank you for your response. A parabola, that's what someone else said.

Truthfully, I fail to see a parabola anywhere.  :oops:

This is actually one of the standard [url=http://mathworld.wolfram.com/ConicSection.html]conic sections[/url].  You're intersecting a cone lying on its side with a horizontal plane.  Before we set the cone down on its side, it had some equation like $ z^2 \equal{} a^2(x^2 \plus{} y^2)$ and our plane had the equation $ z \equal{} ay \plus{} c$ for some $ a, c > 0$.  Then, we rotated everything around the $ x$-axis so that the plane becomes horizontal.  If you actually carry out this procedure, what you'll get some nice equation for a parabola, with $ c$ directly related to the height via the angle of rotation (which depends on $ a$).  Then you have to calcualte the bounds, etc.  Almost all of the work involves setting things up (the geometry problem) -- the actual integral you take at the end will be easy.

Duh, I knew that. I see what you mean. I should've seen that. Thank you.

May I trouble you some more?. I set the two equations equal:

$ (ay\plus{}c)^{2}\equal{}a^{2}(x^{2}\plus{}y^{2})$

I simplified this down to:

$ y\equal{}\frac{a}{2c}x^{2}\minus{}\frac{c}{2a}$

How do I go about finding a and c?.  Should I find my angle of rotation using the 'givens'?.

The portion of the cone that lies on the x-axis would have length $ \sqrt{10^{2}\plus{}24^{2}}\equal{}26$

The angle would be $ 2cos^{\minus{}1}(\frac{12}{13})\equal{}45.24$ degrees or 0.79 radians.

If I place the apex at the origin, the upper corner of the cone would have coordinates (18.31,18.46) and the other

corner lying on the x-axis has coordinates (26,0).  This gives a line equation of $ y\equal{}1.0082x$. This line

represents the side of the cone not lying on the x-axis. The other line representing the top of the cone has line

equation $ y\equal{} \minus{}2.4x\plus{}62.41$

Would these be relevant in solving the problem?.

May I ask how to go about finding a and c?.

I just thought this was a cool take on an old related rates problem.  I hope I am not being a pain.

Perhaps I do not know enough about conics. Getting the equation of the parabola made from the cross-sections of the cone is something I do not know how to do.


function
combinatorics open
combinatorics

For a given set P of n points, S proper superset P is a k-set if | S| = k and S =P intersects H for some open halfspace H. Even for points in two dimensions the problem remains open: The maximum number of k-sets as a function of n and k is known to be O(nk1/3) by a recent advance of Dey ,while the best lower bound is only slightly superlinear.
What is the maximum number of K-sets? :?:  :?:  :?: [/b]

So, this is an open problem in combinatorial geometry. Thus I move into the open problems section in combinatorics...

Pierre.


If (x+1)^2>5x-1 and <7x-3,find integral x.

[hide]First, $(x+1)^2>5x-1 \Rightarrow x^2-3x+2>0 \Rightarrow x>2 or x<1.$  Then, $(x+1)^2<7x-3 \Rightarrow x^2-5x+4<0 \Rightarrow x>1 and x<4$.  The only intersection is x=3.  [/hide]


analytic geometry
graphing lines
slope
vector
calculus
calculus computations

we know that if m1×m2=-1 then the two lines are perpendicular.how can we use this relation between x-axis and y-axis?

Is it enough to say that the coordinate axis are orthogonal, by definition?

thankyou very much.but can we prove it by this relation beween slopes

The slope of the x axis is 0, while of the y axis is infinity...

A better way is to correspond to each line in $ \mathbb{R}^2$ a vector with its direction and then to consider such vectors as elements of the Euclidean space $ \mathbb{R}^2$ with the usual inner product. On this space, two vectors $ u,v$ are orthogonal iff $ \langle u,v \rangle \equal{} 0$. With this understanding, u = (1,0) is a vector corresponding to the x axis (more precisely, it is a base for the subpace of $ \mathbb{R}^2$ generated by it), and v = (0,1) to the y axis, and $ \langle u,v \rangle \equal{} 0$, from which the conclusion follows.

Much in the same line as Carcul, the y-axis doesn't have a nice y = mx+b solution, because not all lines can be written in this form; a nicer way to consider lines is to write them in form ax + by = c; then the slope of a line is related to the vector $ (a,b)$.


We have this:

$ 10(\frac{4}{3})(\frac{6}{4})(\frac{8}{5})(\frac{10}{6})(\frac{12}{7})(\frac{14}{8})(\frac{16}{9})(\frac{18}{10})$

UGH!

Usually when you have a big fraction to simplify like this (or expression) you want to cancel things.  The 10 in the beginning (left hand side) and the 10 in the denominator of the rightmost term (18/10) can cancel.  Now you can rearrange the terms like this:

$ 4(\frac{6}{3})(\frac{8}{4})(\frac{10}{5})(\frac{12}{6})(\frac{14}{7})(\frac{16}{8})(\frac{18}{9})$

Look!  All those terms equal TWO.  You now have to simplify $ 4(2)(2)(2)(2)(2)(2)(2) \equal{} 4*2^7 \equal{} 2^9 \equal{} 512$

512 is your final answer.


[quote="ryan22"]$lim n \to 0 (5^n+1/5^n)$[/quote]

[hide]The limit is 2 I think[/hide]

The way it's written, it's $\infty$, as $1/5^n=\frac{1}{5^n}$

[quote="mathmanman"]Well, I guess he wanted to write (and so José would be right) :
[quote="ryan22"]$\lim_{n \to 0} \frac{5^n+1}{5^n}$[/quote]
Else, [b]Farenhajt[/b]'s answer is the good one.[/quote]

No, Farenhajt is lost in space... since in any case the limit is $2$. Apologize.

Wow, I got lost with you...
I really thought of it as 1/0...  :fool:

[quote="José"]Yes, I put 2 because I understood it was 5^N+(1/5)^N=1+1=2.[/quote]
Right, that's why you need put up a full solution for other math problems because if you understand it doesn't mean we (the audience) understand, albeit how easy the problem is.

[quote="Altheman"]i don't see how this is a limit, all you do is evaluate at n=0[/quote]
Indeed...
Well, let's say $n \to +\infty$, but that'd still be easy...


Let two positive integers $ a,b$ satisfying $ a^b \minus{} b^a \equal{} 1008$. Prove that $ a\equiv b (mod {1008})$

Only solution $ a\equal{}1009,b\equal{}1$, posted before.


inequalities
function
limit
inequalities proposed

Yes well known, our teacher gave us  some inequalities like this in
Terminale C 1985 in lyce Roland Garros, Runion Island

For LHS study the function
$f(x)=\frac{1}{3x+1} + x.ln(1+\frac{1}{x})$ for $x\geq 1$
strictly increasing $\lim_{+\infty}f = 1$

For RHS $g(x)=\frac{1}{2x+1} + x.ln(1+\frac{1}{x})$ for $x\geq 1$
strictly decreasing $\lim_{+\infty}g = 1$


[quote="jli"]i dont get how a line can be tangent to a point :huh:[/quote]
Tangent to [i]circle [/i]O. Sorry, I'll edit.

[quote="i_like_pie"][quote="Quevvy"]Congruent circles, centered at O and P, are externally tangent and a line through P is tangent to circle O at A. If AP=15 cm, how many centimeters are in the radius of circle P? Express your answer as a simplified radical.[/quote]
[hide]Draw triangle $AOP$. Let the radius of circle $P$ be $r$. The radius of circle $O$ is also $r$. We can now use the Pythagorean Theorem to find $r$.

$r^{2}+15^{2}=(2r)^{2}$

$r^{2}+225=4r^{2}$

$3r^{2}=225$

$r^{2}=75$

$r=\sqrt{75}=\boxed{5\sqrt{3}}$[/hide][/quote]
I totally forgot that the radius and the tangent made a right angle.

Also that triangle is a special triangle which would probably be faster.


Prove that $ \sqrt{5n\plus{}3} \notin \mathbb{Q}, \forall n \in \mathbb{N}.$


I don't know exactly what a finite field means.....and I want to know the solution of this problem.......our national olympiad is d-20 ...
problem...

If F is a finite field, prove that F[x] contains infinitely many irreducible polynimials...(in fact, it contains irreducibles of every positive degree, but this is harder to show..)

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

A finite field is a field with a finite number of elements; for example, $\mathbb{Z}_{p}$ for $p$ a prime.

The problem is fairly easy with a simple counting argument.  That is to say, count the [b]reducible[/b] polynomials of degree $k$ recursively, then count [b]every[/b] polynomial of degree $k$.

Well, the proof that there are some for each degree was already posted ( http://www.mathlinks.ro/Forum/viewtopic.php?t=30473 ), but note that there is a shorter proof for their infinity: just do exactly what you do if you proof that there are $\infty$ many primes via Euklid's way.

And better ;) :
[quote="t0rajir0u"]A finite field is a field with a finite number of elements; for example, $\mathbb{Z}/p \mathbb{Z}$ for $p$ a prime.[/quote]


algorithm
absolute value
combinatorics unsolved
combinatorics

Two players play the following game. At each turn the first player chooses a decimal digit, then the second player substitutes it for one of the stars in the subtraction  **** - **** . The first player tries to end up with the largest possible result, the second player tries to end up with the smallest possible result. Show that the first player can always play so that the result is at least 4000 and that the second player can always play so that the result is at most 4000.

Can't the first player guarantee much larger than 4000 by following the strategy of:

Choose 1 until the second player places it as the first digit in one of the two numbers, than choose 9 thereafter.

Maybe the absolute value shouldn't be there in the subtraction?

OK! I have repaired it! Subtraction haven't the absolute value!

The first player chooses $ 4$ until the difference of the numbers (stars will be considered as $ 0$-s) becomes negative.  Then he chooses $ 5$ until the difference is $ \ge0$, after $ 4$ again and so on.  This procedure continues until one of the first digits is taken. If the first number's first digit is taken, the first player chooses $ 0$-s until the end, if the second one's, he chooses $ 9$-s.  It's easy to see that the best number for both players with this algorithm is $ 4000$ and that player one is forced to apply this algorithm in order to obtain at least $ 4000$.

I don't know your solution! What is "difference", may be $ \neq$?
What is "negative"? May be ****-*4** or *4**-****?
If I lay number $ 4$ at *4**-****? What do you do to continue?

Sorry for my bad explanation. OK I will try again. The following algorithm stops if one of the first digits is taken: If the difference (for example **4*-***4, like I said before, we consider stars as $ 0$-s) is $ \ge0$ player one chooses $ 4$; if the difference is $ < 0$ (for example **5*-*4**) he chooses $ 5$. It's easy to see why. I've already written down clearly what happens if a first digit is taken. Clearly the problem can be generalized to $ n$ number of digits, $ n\ge2$, with the answer $ 4\cdot10^{n\minus{}1}$


calculus
integration
trigonometry
calculus computations

can you solve :D ( for $ 0<\alpha,\beta<1$ ) ,

$ \int_0^\infty\int_0^\infty\;x^{\alpha \minus{} 1}\;y^{\beta \minus{} 1}\;e^{ \minus{} (x \plus{} y)}\;\textbf dx\;\textbf dy$

use it to show $ \int_{ \minus{} \infty}^\infty\;e^{ \minus{} x^2}\;\textbf dx\; \equal{} \;\sqrt \pi$

I wonder why there was reply... because it really looks easy (or I'm wrong  :rotfl: )

$ \int_0^\infty\int_0^\infty\;x^{\alpha \minus{} 1}\;y^{\beta \minus{} 1}\;e^{ \minus{} (x \plus{} y)}\;\textbf dx\;\textbf dy \equal{}$
 $ \equal{} \int_0^\infty x^{\alpha \minus{} 1} e^{ \minus{} x} \, \textbf d x \cdot \int_0^\infty y^{\beta \minus{} 1} e^{ \minus{} y} \, \textbf d y \equal{} \Gamma ( \alpha ) \Gamma ( \beta )$

Now using $ \Gamma (a) \Gamma ( 1 \minus{} a) \equal{} \frac {\pi}{\sin a \pi}$ we have $ \Gamma \left( \frac {1}{2} \right) \equal{} \sqrt {\pi}$
and if in the integral
$ \int_0^{ \plus{} \infty} x^{\frac {1}{2} \minus{} 1}e^{ \minus{} x} \, \textbf d x \equal{} \sqrt {\pi}$
we substitute $ x \equal{} y^2$ we obtain
$ \int_0^\infty e^{ \minus{} y^2} \, \textbf d y \equal{} \frac {\sqrt {\pi}}{2}$
finally bacause $ \int_0^\infty e^{ \minus{} y^2} \, \textbf d y \equal{} \frac {1}{2} \int_{ \minus{} \infty}^\infty e^{ \minus{} y^2} \, \textbf d y$ it yields desired formula.


trigonometry
logarithms
geometry
3D geometry
tetrahedron
algebra
polynomial

Let's not post solutions until beta says it's OK. Maybe after he's posted the results.

1.  Since the three values that are squared must be $\geq 0$ (since they are squared) the only way the whole thing can equal 0 is if each of those things squared is 0.  So we have

\begin{eqnarray*}
a^2-1 &=& 0 \\
b^2-4 &=& 0\\
c^2-9 &=& 0
\end{eqnarray*}

So $a=\pm 1, b=\pm 2, and \ c=\pm 3$  So there are 2*2*2=8 ordered pairs.  D

2.  For this one note that \[1+3+5+\ldots+(2n-1)=n^2\] So we then have it equaling 21^2=441 C

3.  How many ordered pairs (x,y) satisfy x+4y=2004?

Note that y>0, and x>0 implies y<501.  Thus y=1,2,...500 yield all of the solutions and there are 500 pairs

4. A charity sells 140 benefit tickets for a total of 2001 dollars. Some tickets sell for full price (a whole dollar amount),
and the rest for half price. How much money (in dollars) is raised by the full-price tickets?

Set up an equation... let x = number of full price tickets, d = full price of a ticket.

We have,

xd + (140-x)d/2 = 2001
2xd + 140d - xd = 4002
xd + 140d = 4002
d(x+140) = 4002

Since x+140|4002 and 140 < x+140 < 280, we factor 4002=2*3*23*29, try some numbers, and conclude that x=2*3*29=174, so x=34, d=23, xd=782.

[quote="Rep123max"]23. Basically, we have to get $\cos5\theta$ in terms of $\cos\theta$ and $\sin\theta$.

By DeMoivere's, we have:
$\cos5\theta+i\sin\theta=(\cos\theta+i\sin\theta)^5$.

So basically, $\cos5\theta$ will be the real parts of $(\cos\theta+i\sin\theta)^5$.

When we expand it, the real parts are $\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta$.

We know that $\cos\theta=\frac{1}{4}$ and by the Pythagorean Identity, $\sin\theta=\sqrt{\frac{15}{16}}$.

Subbing this into what we got above, it turns out to be $\frac{61}{64}$.[/quote]

Because calculator use is permitted on the AMC, there is an easier way: evaluate Arccos(1/4), multiply by 5, and take the cosine of that :D.  Not very elegant, but probably quicker.

11.   $\cot{10^\circ}+\tan{5^\circ}=$

Express cot and tan in terms of sin and cos:
\[\begin{eqnarray*}
&=& \frac{\cos{10}}{\sin{10}}+\frac{\sin{5}}{\cos{5}}\\
&=& \frac{\cos{10} \cos{5}+\sin{10}\sin{5}}{\sin{10}\cos{5}}\\
&=& \frac{\cos{5}}{\sin{10}\cos{5}}\\
&=& \frac{1}{\sin{10}}\\
&=& \csc{10}\\
\end{eqnarray}\]
E is the answer.

14.  Using change of base we get:

\[\log_{100!}{2}+\log_{100!}{3}+\ldots+\log_{100!}{100}\]

which by the property of adding logs equals:

\[\log_{100!}{100!}=1\]

C

16.  I liked this one.

Square both the equations we are given to get:

\[\sin^2{x}-\sin^2{y}-2\sin{x}\sin{y}=\frac{1}{9}\]

and

\[\cos^2{x}-\cos^2{y}-2\cos{x}\cos{y}=\frac{1}{4}\]

Adding them and using the fact that $\sin^2{x}+\cos^2{x}=1$ and the cosine of the sum of angles formula we get(and moving a few terms):

\[\cos{x-y}=\frac{59}{72}\]
A

[quote="Rep123max"]23. Basically, we have to get $\cos5\theta$ in terms of $\cos\theta$ and $\sin\theta$.

By DeMoivere's, we have:
$\cos5\theta+i\sin\theta=(\cos\theta+i\sin\theta)^5$.

So basically, $\cos5\theta$ will be the real parts of $(\cos\theta+i\sin\theta)^5$.

When we expand it, the real parts are $\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta$.

We know that $\cos\theta=\frac{1}{4}$ and by the Pythagorean Identity, $\sin\theta=\sqrt{\frac{15}{16}}$.

Subbing this into what we got above, it turns out to be $\frac{61}{64}$.[/quote]

Why is $\cos5\theta+i\sin\theta=(\cos\theta+i\sin\theta)^5$.?

-interesting_move

#10:  Because $\frac{1}{\sqrt{a}+\sqrt{b}}=\frac{\sqrt{a}-\sqrt{b}}{a-b}$, the expression becomes $\sqrt9-\sqrt8+\sqrt8-\sqrt7+\sqrt7-\sqrt6+\sqrt6-\sqrt5+\sqrt5-\sqrt4+\sqrt4-\sqrt3=3-\sqrt3$, and we are done.

Shouldn't it be [tex]\cos 5\theta + i\sin 5\theta = (\cos\theta + i\sin\theta)^5[/tex]?

I fixed it.  I was having a lot of trouble getting the eqnarray to work.  I still have to get used to using that  :)

Number 7 - The Tricky One.

Realize that you form a 30 degree angle when you turn 150 degrees left, so you have a triangle with sides 3 and rt3, but that is it. The other side can be rt3, or it can be 2rt3 depending on whether or not the triangle is a 30-60-90 or a 30-120-30. So (E). Undetermined.

Number 9) Find the unit digit of the sum  [tex]1 + 2^5 + 3^5 + ... 2005^5[/tex].

Notice, that the continuation pattern for the units digit is at the most in groups of 4, so the 5th will always be the original digit. So, the last digit can be determined by 1+2+3+...2005 = 2005(2006)/2 = ... 5. The last digit is 5, or (C).

Aren't you the Mock AMC C creator? Why are you posting the solutions - just wondering.

For Number 13) To find the number of digits in 5^89, you take log_5^89. You know that log_2 = 0.3010, and you know that log_2 + log_5 = log_10 = 1, so log_5 = 0.699. log_5^89 = 89(log_5) = 89(0.699). This is if you don't have a scientific calculator on you. You get 62.something, and since it is 10^62.some, you have 63 digits. If you have a scientific calc, you can just plug in the numbers.

[quote="white_horse_king88"]Aren't you the Mock AMC C creator? Why are you posting the solutions - just wondering.
[/quote]

Why can't I post the solutions :maybe:

[quote="Rep123max"]23. Basically, we have to get $\cos5\theta$ in terms of $\cos\theta$ and $\sin\theta$.

By DeMoivere's, we have:
$\cos5\theta+i\sin5\theta=(\cos\theta+i\sin\theta)^5$.

So basically, $\cos5\theta$ will be the real parts of $(\cos\theta+i\sin\theta)^5$.

When we expand it, the real parts are $\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta$.

We know that $\cos\theta=\frac{1}{4}$ and by the Pythagorean Identity, $\sin\theta=\sqrt{\frac{15}{16}}$.

Subbing this into what we got above, it turns out to be $\frac{61}{64}$.[/quote]

Any short cut to expand and finding the real part?

I don't think you really need a shortcut to expanding it unless you intend to expand it the long way. You derive the coefficients for isin :theta: to the 0, 2, 4th powers, when the outcome is real. It's not that difficult - a little binomial to assist the coefficients part, and out comes the final polinomial by De Moivre's theorem.[/tex]

[quote="beta"]15)
Let the tetrahedron be ABCD, with BC and AD opposite, and let their midpoints be M and N, respectively. Note that $\Delta ABC  \Delta BCD$  are equilateral triangles. Therefore length of $AM=\frac{\sqrt3}{2} =BD$. Point M must lie on the perpendicular bisector of AD, so $MN \perp AD$  , ANM is a right triangle, $AN=\frac{1}{2} $ . By Pythagorean Theorem, $MN=\frac{\sqrt2}{2}$[/quote]

How do you know M must lie on the perpendicular bisector of AD?
It took 3 minutes for me to prove that.  Is it some sort of theorem, or did you also have to figure it out?

[quote="white_horse_king88"]It's an old theorem that you learn in most Euclidean Geometry classes. That if N is the midpoint of AD, and AM = MD, then M must lie on the perpendicular bisector of AD. Also, all points on the perpendicular bisector of AD are equidistant from A and D.  :)[/quote]

Oh, I get it.  It is the isoceles triangle thing.  The perpendicular bisector of the base contains the non-base corner.  Right?

[quote="white_horse_king88"]For Number 13)you know that log_2 + log_5 = log_10 = 1[/quote]

How do you figure this out?

Anyone wants to do my favorite ones (20, 21, 24, 25)?

You figure it out by loga_b + loga_c = loga_bc. You know by fact that log10_10 = 1.

[quote="Rep123max"]#24. I wish I had really attempted it during the actual contest. I just started to look at simpler cases or \[\sum_{k=0}^n(-1)^k{n\choose k}k\]

For n=1, it is -1. And then for everyone after that, it is 0, so the answer is C. I'll try to prove it later.[/quote]

What exactly does Sigma mean, in this case?  I've seen it being used, but never really understood it.

[b]#24[/b]. I also looked at simpler cases on the test, which is a very effective way to cheat, but here is the way to do it.

$\displaystyle\sum_{k=0}^n(-1)^k{n\choose k}k=\displaystyle\sum_{k=1}^n(-1)^k{n\choose k}k+\displaystyle(-1)^{0}{n\choose 0}\cdot 0=\displaystyle\sum_{k=1}^n(-1)^k\displaystyle\frac{n!}{(n-k)!(k)!}k$
$=\displaystyle\sum_{k=1}^n(-1)^k\displaystyle\frac{n!}{(n-k)!(k-1)!}=\displaystyle\sum_{k=1}^n(-1)^k\displaystyle\frac{n\cdot (n-1)!}{(n-k)!(k-1)!}$
$=\displaystyle\sum_{k=1}^n(-1)^k {n-1\choose k-1}\cdot n=\left(\displaystyle\sum_{k=0}^{n-1}(-1)^{k+1}{n-1\choose k}\right)\cdot n=-(1-1)^{n-1}\cdot n = 0$(C)

Edit: By binomial theorem, $(1-1)^{n-1}$ becomes $(1)^{n-1}(-1)^{0}\displaystyle{n-1\choose 0}+(1)^{n-2}(-1)^{1}\displaystyle{n-1\choose 1}+(1)^{n-3}(-1)^{2}\displaystyle{n-1\choose 2}+...+(1)^{1}(-1)^{n-2}\displaystyle{n-1\choose n-2}+(1)^{0}(-1)^{n-1}\displaystyle{n-1\choose n-1}=\displaystyle\sum_{k=0}^{n-1}(-1)^{k}{n-1\choose k}=(-1)(-1)\displaystyle\sum_{k=0}^{n-1}(-1)^{k}{n-1\choose k}=(-1)\left(\displaystyle\sum_{k=0}^{n-1}(-1)^{k+1}{n-1\choose k}\right)$

[b]#25[/b], consult my non-scale drawing. I solved using coordinates  :lol:. Always an ugly way to bash problems you are having trouble with. Anyways, $B = (0,0), C=(\sqrt{40},0), A=(\sqrt{10},\sqrt{40})$. To find D, note that BDC is a right triangle, so the altitude from D times BC times one half = BD times DC times one half, calculating areas in different ways, so we can conclude the y-coordinate of D $=\displaystyle\frac{12}{\sqrt{40}}=\displaystyle\frac{6}{\sqrt{10}}$. By pythagorean theorem, $(x-coord of D)^{2} + (\displaystyle\frac{6}{\sqrt{10}})^{2} = 2^{2} \Rightarrow x = \displaystyle\frac{2}{\sqrt{10}}$. So $D=(\displaystyle\frac{2}{\sqrt{10}},\displaystyle\frac{6}{\sqrt{10}})$ and the distance between A and D is $\sqrt{(\sqrt{10}-\displaystyle\frac{2}{\sqrt{10}})^{2}+(\sqrt{40}-\displaystyle\frac{6}{\sqrt{10}})^{2}}=\sqrt{26}$(B)

A good trig solution to #25:
Draw OM, where M is the midpoint of AC. Since $MA=\sqrt{10}$, Pythagorean Theorem we have $OM=\sqrt{40}, \tan OAC=2, \tan BAC=\frac{1}{3}$.
$\tan(OAC-BAC)=\frac{2-\frac{1}{3}}{1+2 \cdot \frac{1}{3}}=1=\tan OAB$.
So $\angle OAB=45^{\circ}, \cos OAB=\frac{\sqrt2}{2}$
By Law of Cosines,
$(\sqrt{50})^2+(6^2)-2(\sqrt{50})(6)\frac{\sqrt2}{2}=OB^2=26$.
$OB=\sqrt{26}$

[quote="Rep123max"]For the question about if there was a shortcut for expanding it, you won't have to write down any of the terms where $i\sin\theta$ is taken to an odd power, because that would have an $i$ in it. And if you were trying to say find $\sin5\theta$, you wouln't write down any of the terms where $i\sin\theta$ is taken to an even power.[/quote]

I got it, don't worry about it.  I was slightly confused at that moment.

Pork_Chop8,
I don't get the last step for the first problem you did; could you explain?

What number was the second problem?

[quote="beta"]Binomial Theorem,$\displaystyle \sum_{k=0}^{n} {n \choose k}x^ky^{n-k}=(x+y)^k$.

Sub in x=1, y=-1, we have $\displaystyle \sum_{k=0}^{n} (-1)^k(1)^{n-k}{n \choose k}=(1-1)^n=0$. Pork_Chop8's solution was the intended solution, I made #24 up.

The second problem is #25[/quote]

X= -1, y=1?

By the way, beta, how did you know MA was sqrt(10) for #25

[quote="Rep123max"]From the Pythagorean Theorem, he got $2^2+6^2=AC^2$ so $AC=\sqrt{40}=2\sqrt{10}$.

Since its the midpoint, it is half of that, or $\sqrt{10}$.[/quote]

Got it.  I think my brain cells died somehow :(

[quote="crayshot97"]20. n^2-3n-126=k^2, where k is an integer. First, note that if n=x is a solution, then n=3-x is also a solution. Therefore, when adding these we get 3, so the solution must be a positive multiple of 3. The only choice that satisfies this is 12(A). Now for an honest solution, we solve the quadratic n^2-3n-(126+k^2)=0. The discriminant must be a perfect square so we get 4k^2+513=q^2, where q is also an integer. Therefore (q-2k)(q+2k)=513=3^3*19. Therefore, we get 4 possible values for k. Each value of k gives 2 values for n which sum to 3. Therefore, the answer is 4*3=12.[/quote]

Good solution.  Especially the first one, I thought it was very interesting.  Now I have to ask you, how do you know if n=x is a solution, then n=3-x is a solution?  I know it works but how did you figure that out in the first place?


function
linear algebra
linear algebra unsolved

Right, this is probably very easy and well known but I enjoyed this funny problem, so those who don't know it yet can spend a few minutes solving this cute question:

Let $\mathbb{R}^{\mathbb{R}}$ be the space of all functions from $\mathbb{R}$ to $\mathbb{R}$. Let $\mathcal{E}$ and $\mathcal{O}$ be the subspaces of the even and odd functions respectively. Show that $\mathbb{R}^{\mathbb{R}} = \mathcal{E} \oplus \mathcal{O}$.

I always remember this using the cosh and sinh functions, for some reason.

First, we want to show $\mathcal{E}$ and $\mathcal{O}$ are disjoint, except for the zero function.  But this clearly holds, because if we have $f\in\mathcal{E}\cap\mathcal{O}$ with $f\not=0$, then we have $f(x)\not=0$ for some $x\not=0$.  But we have $f(-x)=-f(-x)\implies f(-x)=0$, and applying the same thing to $f(-x)$ gives $f(x)=0$.  So they are disjoint except for the zero function.

Next, given $f\in \mathbb{R}^{\mathbb{R}}$, take

$f_{\text{odd}}(x)=\frac{f(x)-f(-x)}{2}$

$f_{\text{even}}(x)=\frac{f(x)+f(-x)}{2}$

to construct the direct sum.


quadratics
number theory unsolved
number theory

Prove that : A square number incongruent to $ 3$ modulo $ 12p\plus{}5$ and $ 12p\minus{}5$ where $ p$ is a prime.

when is 3 a quadratic residue modulo a prime q? Think about the Quadratic Reciprocity law.


At a party, each man danced with exactly three women and each woman danced with exactly two men. Twelve men attended the party. How many women attended the party?

$ \textbf{(A)}\ 8\qquad
\textbf{(B)}\ 12\qquad
\textbf{(C)}\ 16\qquad
\textbf{(D)}\ 18\qquad
\textbf{(E)}\ 24$

[hide] Since each men danced with 3 women, there should be 3*12=36 women, but since each woman danced with 2 men, then you should divide this by 2 to get 18, which is our answer.  Thus, the answer is D.  [/hide]

[quote="#H34N1"]At a party, each man danced with exactly three women and each woman danced with exactly two men. Twelve men attended the party. How many women attended the party?
(A) 8 (B) 12 (C) 16 (D) 18 (E) 24[/quote]

[hide="soluiton"]There are a total of $3\cdot 12=36$ dances.  Each woman danced with 2 men so $\frac{36}{2}=\boxed{18}$[/hide]

I don't quite understand how this works. Couldn't each man also dance with 3 women if there were 6 women at the party, because $6C3$ is $> 12$, or even less than 6 women? This is why I am confused

Then the women wouldn't dance with exactly 2 men

But if there were 6 women, and 12 men, 12/6= 2

@above If there were 12 men and 6 women each man would only dance with 1 women and we need each man to dance with 3 women

wait I don't understand this either, the question doesn't say each man danced with 3 [b]different[/b] women. So therefore even if there were 100 men , 3 women would satisfy this condition.
As for each women dancing with 2 men, 3 women also satisfy this condition since there are more than 2 men at the party.

Therefore why couldn't the answer be 3 women??

Because each one of those women would dance with all 100 men. Each woman can only dance with 2 men, so we need to account for that.


calculus
integration
function
trigonometry
calculus computations

I believe the following integral cannot be evaluated either using elementary operations or with some other methods such as expanding into power series. What do you think?

$\int_\frac{pi}{2}^\frac{3pi}{4}{\sqrt{4+cos^{2}{x}dx}}$

Thanks and regards.

I heard this type of integral is closely related to the function called 'incomplete elliptic integral of the second kind'. This (non-elementary) function is defined as

$E ( \phi | m ) = \int_{0}^{\phi}\sqrt{1-m \sin^{2}\theta}\, d\theta$

where $m$ is called the parameter. Then your integral can be evaluated in terms of $E$:

$\sqrt{5}\left[E\left(\frac{3 \pi }{4}|\frac{1}{5}\right)-E\left(\frac{\pi }{2}|\frac{1}{5}\right) \right]$.

Thank you for the answer. I further studied on it as follows,
[url=http://imageshack.us][img]http://img520.imageshack.us/img520/8839/integral4tg6.gif[/img][/url]
Any comments?


trigonometry
real analysis
real analysis theorems

[quote="DanK"]I was doing some AP Calc homework the other day, and one of the questions had me find $\frac{d}{dx}\sin^{2}x$.  As I was doing the problem, I realized that $\frac{d}{dx}\sin^{2}x=2\sin x\cos x=\sin2x$.  Since $\frac{d}{dx}x^{2}=2x$, I was wondering if there was some kind of relationship involving this, like $\frac{d}{dx}\sin^{n}x=\sin^{n-1}nx$, or something like that.

I tried looking at it for $\sin^{3}x$, but it got really nasty, and I didn't prove anything...[/quote]
No such luck. The fact that $2\sin x \cos x = \sin 2x$ is a consequence of the angle addition formula for $\sin.$

$\sin\left(A+B\right) = \sin A \cos B+\cos A \sin B$

What happens when $A = B$ ?

Power series provide a glimmer of insight.

Near zero, $\sin x=x+$ (higher order terms).

Thus $\sin^{2}x=x^{2}+$ (higher order terms) so that

$\frac{d}{dx}\sin^{2}x=2x+$ (higher order terms)

and it's also true that $\sin2x=2x+$ (higher order terms).

In this particular case, it's even true that the higher order terms agree, but there's no reason to count on that in more general cases.

$\sin^{n}x=x^{2}+$ (higher order terms), so

$\frac{d}{dx}\sin^{n}x=nx^{n-1}+$ (higher order terms).

Your guess, $\sin^{n-1}(nx),$ looks like $n^{n-1}x^{n-1}$ near zero; that's the wrong coefficient.

Although the first coefficient could be adjusted by multiplication with a constant, a more serious problem remains: the period of $\sin nx$ is $2\pi/n$, while the period of $(\sin^{n}x)'$ is $2\pi$ for $n$ odd and $\pi$ for $n$ even.

You can easily prove the conjecture is false.

Conjecture:
$\frac{d \sin^{n}x}{dx}= \sin^{n-1}nx, \text{ for }n > 2$

$\Rightarrow n \sin^{n-1}x \cos x = \sin^{n-1}nx$

Let $x = \frac{\pi}{3}$ and let $n = 3.$

$\Rightarrow 3 \sin^{2}\frac{\pi}{3}\cos \frac{\pi}{3}= \sin^{2}3\cdot \frac{\pi}{3}$

$3\cdot \frac{1}{4}\cdot \frac{\sqrt{3}}{2}= \sin^{2}\pi$

$\frac{3\sqrt{3}}{8}= 0 %Error. "larrow" is a bad command.
%Error. "rarrow" is a bad command.
$

The conjecture leads to a contradiction; therefore the conjecture is false.

***

What you might be looking for are the following series:

$\cos nx =\sum_{k\text{ even}}^{n}i^{k}\binom{n}{k}\sin^{k}x \cos^{n-k}x.$

$\sin nx =\sum_{k\text{ odd}}^{n}i^{k-1}\binom{n}{k}\sin^{k}x \cos^{n-k}x.$

where $i\equiv \sqrt{-1}.$

These series are obtained from the binomial expansion of De Moivre's theorem.


What is the number of distinct prime factors of 96?

$ 96 \equal{} 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3$, which is 2 distinct prime factors.


vector
real analysis
real analysis unsolved

Give an example of closed subspaces such that their sum is not closed in any normed vector space(for example in $l_2$)

Huh? :? What am I missing here? The only open subspace of a normed space is the space itself. perhaps you meant we should find two closed subspaces whose sum is not closed?

Yes, of course i mean "not closed"-i've edited it.
Sorry


Suppose that $a$ and $b$ are distinct real numbers such that \[a-b, \; a^{2}-b^{2}, \; \cdots, \; a^{k}-b^{k}, \; \cdots\] are all integers. Show that $a$ and $b$ are integers.

First, we prove that a,b are rational numbers.
$a-b \in \mathbb{Q}$
$a+b = \frac{a^2-b^2}{a-b} \in \mathbb{Q}$
$a = \frac{(a+b)+(a-b)}2 \in \mathbb{Q}$
$b = \frac{(a+b)-(a-b)}2 \in \mathbb{Q}$

Then we can find three integers a',b',n such that:
$a = \frac{a'}n$
$b = \frac{b'}n$
$d \mid a' \land d \mid b' \land d \mid n \Rightarrow |d| = 1$
The hypotesis translates to $n^k \mid a'^k - b'^k \quad \forall k \in \mathbb{N}$.
Now, $n \mid a'-b'$ and we can suppose $(a',n) = 1$. Therefore $(b',n) = 1$.

Let d be the integer such that $n^d \mid \mid a'-b'$. First, we suppose d>1.
We have $n^{d+1} \mid a'^{d+1} - b'^{d+1}  = (a'-b')(a'^d + \ldots + b'^d)$. We get $n \mid (a'^d + \ldots + b'^d)$, but $\gcd(a'-b', a'^d + \ldots + b'^d) \mid n$, therefore $n \mid \mid (a'^d + \ldots + b'^d)$ and $n^{d+1} \mid \mid a'^{d+1} - b'^{d+1}$. (in this step we must remember that d>1).

Now we consider 2(d+1). We know that $n^{2d+2} \mid a'^{2d+2} - b'^{2d+2} = (a'^{d+1} + b'^{d+1})(a'^{d+1}-b'^{d+1})$. But $n^{d+1} \mid \mid a'^{d+1} - b'^{d+1}$, then $n^{d+1} \mid a'^{d+1} + b'^{d+1}$ and $n^{d+1} \mid (a'^{d+1} + b'^{d+1}) + (a'^{d+1} - b'^{d+1}) = 2a'^{d+1}$. But $(a',n) = 1$, then finally $n^{d+1} \mid 2$.
But therefore, n=1 and a,b are integers.

If $n \mid \mid a'-b'$, then $n^2 \mid (a'+b')(a'-b')$ and $n \mid a'+b'$ and $n \mid 2a'$, as before. Then n=2 and a',b' are odd integers such that a'-b' is a multiple of 2 but not of 4. Let us take d such that $2^d \mid \mid a'+b'$. Let us take a positive integer k such that $2^k - k > d$ (this obviously exists).
But then $2^{2^k} \mid a'^{2^k} - b'^{2^k} = (a'^{2^{k-1}} - b'^{2^{k-1}}) \cdots (a'^2+b'^2)(a'+b')(a'-b')$. We must find $2^k$ factors 2 in this product. But if x,y are odd integers, $2 \mid \mid x^2+y^2$. And $2 \mid \mid a'-b'$ by hypotesis, and $2^d \mid \mid a'+b'$ by hypotesis. Therefore we find exactly $d + k$ factors 2 in this product, which is too low.

I hope it is correct!

I hope it is correct!

Lets say we showed that a,b are rational (like edriv did). Let: $ a \equal{} \frac {x}{y}, b \equal{} \frac {s}{t}, (s,t) \equal{} (x,y) \equal{} 1$

$ a^n \minus{} b^n \in Z, n \in N \leftrightarrow (yt)^n|(xt)^n \minus{} (ys)^n$. If we show that $ yt|xt, ys$, we get: $ y|x$, $ t|s$, or: $ a,b$ are integers. It follows from the following lemma:

[u]Lemma:[/u] If $ a^n|b^n \minus{} c^n$ for all $ n \in N$, where $ a$,$ b \neq c$ are integers, then: $ a|b,c$.
[u]Proof:[/u] Let p be a prime divisor of a, $ g \equal{} gcd(b,c)$, ($ b' \equal{} b/g, c \equal{} c'/g$). We have, by "Lifting the Exponent" lemma:
$ ord_p(a^n) \le ord_p(b^n \minus{} c^n) \equal{}$
$ ord_p(g^n) \plus{} ord_p(b'^n \minus{} c'^n) \equal{}$
$ ord_p(g^n) \plus{} ord_p(b' \minus{} c') \plus{} ord_p(n) \equal{}$
$ ord_p(n(b \minus{} c)g^{n \minus{} 1})$
or (in the case p=2):
$ ord_p(a^n) \le ord_p(b^n \minus{} c^n) \equal{} ord_p(g^{n \minus{} 2}(b^2 \minus{} c^2)n/2)$.

We can put that together in the following way - $ ord_p(a^n) \le ord_p(g^{n \minus{} 1}(b^2 \minus{} c^2)n)$.

It implies that: $ a^n | g^{n}(b^2 \minus{} c^2)n$ for all n>1. Let $ gcd(a,g) \equal{} g'$. We have: $ (\frac {a}{g'})^n|(b^2 \minus{} c^2)n$, which implies that:
$ |(\frac {a}{g'})^n| \le |b^2 \minus{} c^2|n$ for all n>1. It can't happen for large n unless $ b^2 \minus{} c^2 \equal{} 0$ or $ a \equal{} \pm g'$.

First case - $ b^2 \equal{} c^2$. It means that $ b \equal{} \minus{} c$, or: $ a^{2n \minus{} 1}|2b^{2n \minus{} 1}$ for all n, which gives $ a|b$, and then also $ a|c$.
Second case - $ a \equal{} \pm g'$, or: $ a|b,c$.

Q.E.D.

[b]"Lifting the Exponent":[/b] If p is an odd prime, and a,b are integers, and: $ p|a \minus{} b, (a,b,p) \equal{} 1$, we have: $ ord_p(a^n \minus{} n^n) \equal{} ord_p(n(a \minus{} b))$
If p=2, $ p|a \minus{} b, (a,b,p) \equal{} 1$, we have: $ ord_p(a^n \minus{} n^n) \equal{} ord_p(n(a^2 \minus{} b^2)/2)$

I have dropped out some minor details, ask a question or point an error if you find one.

I assume it's [i]George T. Gilbert, Mark I. Krusemeyer, Loren C. Larson, The Wohascum County Problem Book, MAA[/i].

Here is a proof a little bit different from edriv. It is the same up to
$ n|a'\minus{}b'$, $ (n,a')\equal{}(n,b')\equal{}1$.

We can also assume $ n>0$ wlog.

Let $ a'\equal{}An\plus{}b'$, we know
$ n^k|a'^k\minus{}b'^k\equal{}\sum_{i\equal{}2}^k c_i(An)^ib'^{k\minus{}i}\plus{}kAnb'^{k\minus{}1}$, where $ c_i$ is binomial coefficients.
We can always find big $ k$ with $ (k,n)\equal{}1$.
Note the first term on RHS is multiple of $ (An)^2$. So we have $ n^2|kAnb'^{k\minus{}1}$, or $ n|A$.
Then the first term on RHS is multiple of $ (An)^2$, or $ n^4$. So we have $ n^4|kAnb'^{k\minus{}1}$, or $ n^3|A$.
Then the first term on RHS is multiple of $ (An)^2$, or $ n^8$. So we have $ n^8|kAnb'^{k\minus{}1}$, or $ n^7|A$.
....
We have $ n^m|A$ for any big integer $ m$, $ A$ being nonzero finite number implies $ n\equal{}1$. Hence $ a\equal{}a'$ and $ b\equal{}b'$

$ a = \frac {x}{z}, b = \frac {y}{z}, (x,y,z) = 1$
$p|z$
$p^n|x^n-y^n$ and $p|x-y$
by $LTE lemma$
                   $n \le v_p(x^n-y^n)=v_p(x-y)+v_p(n)$
                                         $n=p^k$
$p^k-k  \le v_p(x-y)$
contradiction


function
real analysis
real analysis unsolved

Please help me with this exercise...
esto sinartisi f:R :arrow: R me |f(x)-e^x|<=x^2.ean i f dio fores paragogisimi sto R kai i f'' sinechis sto 0 deixte oti:f''(x)<=3.
den echo tin paramikri idea pos bgeni... :(

[quote="koredim1989"]Please help me with this exercise...
function f:R :arrow: R and |f(x)-e^x|<=x^2.Both f' and f'' exist  in R and f'' continuous at 0.Prove that:f''(0)<=3.
I am not sure about the terms in english.With f'' continuous at 0 i mean that limf(x)=f(0)
               x :arrow: 0

Dear Koredim 1989, welcome.

I think probably you have been posted in a wrong Forum. But surely in a wrong area.

Kostas Vittas.

koredim1989 θες  να την γραψεις στο ελληνικο φορουμ ή στο φορουμ της αναλυσης ? αν θες θα στην γραψω σε λατεξ


[quote="warut_suk"]There are $n$ points on a circle. On every point, either $1$ or $-1$ is written. Then, between any two numbers, we write the product of them. Then we delete the original numbers. This process is repeated many times. Assume that for any arrangement of $1$ and $-1$ at the start, we get to a position where all numbers are $1$. Show that $n=2^{k}$.[/quote]

I'm not sure this is what you want the problem to say. Consider the following:

Take any collection of n points around a circle. Label exactly one point -1 and the rest of the points 1. Then no matter how you combine your products you will always have exactly one point labeled -1. This works for any number of n points thus the hypothesis is false.

Perhaps you mean if you assign the label 1 the same number of times you assign the label -1. Then this may be true.

cheers

-Q

No.

The points are on a circle. And when we write the new numbers, we delete the old numbers.

My apologies, I for some reason thought the new points were created one at a time.


geometry
geometric transformation
rotation
algebra
polynomial
factorial

pl explain how to solve problem of circular permutation of similar things . for ex suppose we have a four letters AABB .we have to arrange them in a circular order .how many ways are possible . i do like that
total ways of circular arrangements 3! but two pairs are identical therefore total ways are 3!/(2!2!) which is in fraction pl tell the fallacy.

I'm not sure how you would apply permutation to this one, but aren't there only 2? ABAB and AABB. All other cases are just those but rotated.

In your example, ayush_2008, there are 4! ways to rearrange it, if you simply look at each letter separately
Because you have 2 A's, there are half as many cases
and since there are also 2 B's, you have to divide by 2 again
so to calculate it, it would be 4!/(2x2) = 3! = 6

to calculate a more general case, you simply take the total number of permutations and then divide by how many repeated letters/numbers you have

i dont understand where you are getting 3!/(2!2!)

Ah, I know how to do it with permutations.
Linearly, there are $ 4!$ ways. However, you have to divide by $ 2!$ for the As and another $ 2!$ for the Bs. You also have to divide by $ (4\minus{}1)!$ to account for rotations.
So you get $ \frac{4!}{2!2!3!}\equal{}\boxed{2}$

oops, pascal12 is right, i forgot about rotation

The general method is via [url=http://qchu.wordpress.com/2009/06/13/gila-i-group-actions-and-equivalence-relations/]Polya theory[/url], specifically computations involving the cycle index polynomials of the cyclic groups.  The problem with just dividing out by the obvious symmetries is that certain orders are more symmetric than others.

pascal12, I don't understand why you're dividing by $ 3!$ to account for rotating four objects.  Your method shouldn't give the right answer for larger numbers of letters.

You're right. I just confused myself...  :huh:

Shouldn't you divide by $ 4$? I have no idea where the factorial on the $ 3!$ came from. This would give $ \frac {4!}{4\cdot2!2!} \equal{} \frac{3}{2}$ but this doesn't make sense.

You shouldn't be dividing by just one thing at all.  You're correct that there are $ \frac {4!}{2! 2!} \equal{} 6$ linear orderings where the two As and the two Bs are considered indistinguishable.  But these don't organize themselves into two groups of cyclic orderings of the same size.  The cyclic ordering AABB comes from [b]four[/b] linear orderings AABB, ABBA, BBAA, BAAB, whereas the cyclic ordering ABAB comes from [b]two[/b] linear orderings ABAB, BABA.  As I've already stated, the issue is that these two groups of linear orderings have different cyclic symmetries.

More precisely, in the language of group actions a typical combinatorial argument where you divide by factorials to avoid overcounting comes from a [b]free[/b] group action, and in a free group action every orbit has the same size as the group; in other words, in this situation you can divide by $ 4$.  Unfortunately, the group action here isn't free.

The correct generalization for $ k$ As and $ n \minus{} k$ Bs is given by [url=http://www.research.att.com/~njas/sequences/A047996]A047996[/url] and can be obtained using the methods in the post I linked to.

in circular permutation if there are n object then total no of arrangements are (n-1)!.so for four elements there are 3!ways .since two sets are similar therefore i divide by 2! 2!.

[quote="JBL"]Or, you could re-read the thread that precedes your post.  This would prevent you from
(1) repeating things others have already said, and
(2) being wrong.
:mad:   :roll:[/quote]
sir i am confused pl tell the simple formula with logic in circular permutation.

Your answer gives, as two (incorrect) answers have already given, $ \frac{3}{2}$.  I have already explained the correct method.

at last is there any general formula for circular permuation  when things are not distinct

That's exactly the question being addressed in this thread.  One can compute the number of circular permutations of an arbitrary multiset by Polya's theorem.


See if you can solve this!

What do the rich need, the poor have, is better than the best thing in the world, and if you eat it, you will die?

Wasn't this posted before?

[hide]Anyways, i think it's [b]nothing[/b][/hide]

[quote="SplashD"]how is nothing better than the best thing in the world?[/quote]

if it's better than the best thing in the world then nothing can be better than it. :)

Ohhh, I get it now.

That riddle is sorta dumb.

[hide="I think it is..."]A coffin? I heard something like this before.[/hide]

[quote="mathnerd314"]Another really old one:

The person who buys it doesn't use it; the person who uses it doesn't know he's using it.

What is it?[/quote]

[hide]A coffin[/hide]

[quote="SplashD"]Can't a person buy his own coffin if he knows he's going to die in 7 days?[/quote]

Yeah like what if the person watched The Ring. :)

Both the "coffin" and the "nothing" puzzles have already been posted in Aops. I really like the "nothing" puzzle. :)

What about those of us who believe in afterlives? The persons' spirit could possibly know that they are using the coffin to store the body.  :D