quadratics
number theory unsolved
number theory

I'm trying to find

i) all the odd primes p such that -3 is a quadratic residue mod p
ii) all the odd primes such that -3 is a quadratic nonreside mod p

n is a quadratic residue mod p if there exists x such that x squared is congruent to  n mod p.

source unknown

$\left( \frac{-3}{p}\right) = \left( \frac{3}{p}\right) \left( \frac{-1}{p}\right)$

Now,

$\left( \frac{3}{p}\right) = (-1)^{p-1}\left( \frac{p}{3}\right)$

Since $p$ is odd we have $(-1)^{p-1}= 1$.  Furthermore, $\left( \frac{p}{3}\right) = 1$ when $p = 3$ or $p \equiv 1 \bmod 6$.  Ignoring the first case,

$\left( \frac{3}{p}\right) = (-1)^{ \frac{p^{2}-1}{24}}$

And of course it is well-known that

$\left( \frac{-1}{p}\right) = (-1)^{ \frac{p-1}{2}}$

Hence for odd primes $p > 3$ we have

$\left( \frac{-3}{p}\right) = (-1)^{ \frac{(p-1)(p+13)}{24}}$

In other words, for odd primes $p > 3$:

- If $p \equiv 1, 11 \bmod 12$ then $-3$ is a QR $\bmod p$.
- If $p \equiv 5, 7 \bmod 12$ then $-3$ is a QNR $\bmod p$.

[quote="t0rajir0u"]
In other words, for odd primes $p > 3$:

- If $p \equiv 1, 11 \bmod 12$ then $-3$ is a QR $\bmod p$.
- If $p \equiv 5, 7 \bmod 12$ then $-3$ is a QNR $\bmod p$.[/quote]

Not really... Then -3 would be a quadratic nonresidue for p=7, while $2^{2}\equiv-3 (mod\ 7)$.

In fact:

\[\left(\frac{-3}{p}\right)=\left( \frac{3}{p}\right) \left( \frac{-1}{p}\right)=(-1)^{{{p-1}\over{2}}\cdot{{3-1}\over{2}}}\left({p\over 3}\right)\left(\frac{-1}{p}\right)=\]\[(-1)^{{p-1}\over{2}}\left({p\over 3}\right)(-1)^{{p-1}\over{2}}=\left({p\over 3}\right)\]
Therefore $-3$ is a quadratic residue for p>3 iff $p\equiv 1 (mod\ 3)$

[quote="Krzysztof Dorobisz"]Therefore $-3$ is a quadratic residue for p>3 iff $p\equiv 1 (mod\ 3)$[/quote]

Can't we simply state the better result $p \equiv 1 \mod 6$?

If you can explain me why this result is better...

I expressed myself badly. It's the same, of course.

I guess Keno20 may not know this fomula:
$(\frac{p}{q})(\frac{q}{p})=(-1)^{\frac{(p-1)(q-1)}{4}}$
so Keno20 will not understand the above answer...
by the way ,What is the Engilsh name of the above?I only know that it is proved by Gauss.

[quote="Hawk Tiger"]by the way ,What is the Engilsh name of the above?[/quote]

It's the Quadratic Reciprocity Law.


geometry
3D geometry
analytic geometry
search

Kyle will use four identical unit cubes to create a solid. Each cube must be glued to at least one other cube. Two cubes may only be glued together in such a way that a face of one cube exactly covers a face of the other cube. How many distinct solids could Kyle create? Two solids are considered to be the same if one solid can be repositioned to match the other solid.

[hide]
Answers says there're 8, but I can only find 7...
(The 5 2-d ones and then 2 others)[/hide]

It would help if you told us which solids you had found.  The five tetris pieces are obvious; then there are two distinct "twist" shapes (did you only count one? they can't be repositioned into each other) and a "corner" shape.

Edit:  In coordinate terms, the "3D" shapes I'm talking about are

- $ (0, 0, 0), (1, 0, 0), (1, 1, 0), (1, 1, 1)$
- $ (0, 0, 0), (1, 0, 0), (1, 1, 0), (1, 1, \minus{}1)$
- $ (0, 0, 0), (1, 0, 0), (0, 1, 0), (0, 0, 1)$

Oh right...
Arg I thought they could be repositioned into each other.

Ok, thanks.

The fact that they can't is actually fairly interesting; it's for the same reason that your left hand can't be repositioned into your right hand.  Chemists call such shapes [url=http://en.wikipedia.org/wiki/Enantiomer]enantiomers[/url] (a practical application of math!); there are "left-handed" and "right-handed" versions.


A shake from Jake's Shakes uses 2 tsp vanilla, 2 scoops ice cream, 4 oz or fresh bananas and 1/2 cup milk.  A banna (without the peel) weighs 6 oz on average.  How many whole bannas must jake order so that he can make 500 shakes?

[quote="A+MATH"][quote="risinglegend"]333.333~333.4[/quote]

Umm, you don't buy fractional bannanas so you round up to an integer.[/quote]

 :rotfl:  I can't believe he copied the answer wrong (he usuallydoes that), and he didn't round right anyway.  :D

Anyway, please stop putting 5 character posts  :roll:


Can we prove that :
If $ A$ is an open bounded set in a normed space $ E$ and there exist $ a \in A$ such that $ B(a, \frac{diam(A)}{2} ) \subset A$ then :
$ B(a, \frac{diam(A)}{2}) \equal{} A$

In an equivalent form: if $ A$ is open and properly contains the open unit ball, then $ \mathrm{diam}\,(A)>2$.
Yes, this is true: $ A$ contains an element $ x$ with $ \|x\|>1$, and the unit ball contains $ (\epsilon\minus{}1)\frac{x}{\|x\|}$. The diifference $ x\minus{}(\epsilon\minus{}1)\frac{x}{\|x\|}$ has norm $ \|x\|\plus{}1\minus{}\epsilon$.


My grand plan for this upcoming summer was to go to RSI, so of course I've asked for two teacher recommendations for that.  The problem is that I'm far from confident about my chances of getting in, so as a backup I thought it might be a good idea to apply to AwesomeMath.

Unfortunately, AwesomeMath [b]also[/b] requires recommendations, so I would probably end up having to ask the same teachers to write another recommendation for me, which I just think is too much for me to ask of them.

What I can do is wait until I hear back from RSI, which is March 31st, and if I don't make it in try to squeeze in an application to AwesomeMath before their regular postmark deadline (April 9th).  This is a pretty tight squeeze, and if get the letter from RSI late it might not be workable at all.

So what should I do?

[quote="t0rajir0u"]Unfortunately, AwesomeMath [b]also[/b] requires recommendations, so I would probably end up having to ask the same teachers to write another recommendation for me, which I just think is too much for me to ask of them.[/quote]

Unless the recommendation forms are [i]drastically[/i] different, it shouldn't be a problem. If they are, well, give them a nice present at the end of the year. Plus, get used to it, because college recommendations are just like that except several times worse. At least by now they've been done with college recommendations probably for a month or so.

They've already written it; presumably, they still have a copy on their hard drive.

The recommendation forms are different enough that I don't think I could recycle them (the AwesomeMath one asks a very specific question about three words that come to mind)...

Good teachers realize that writing recommendation letters is part of their job.  Don't be shy about asking for another letter.  Most teachers can pretty easily take their previous letter and do minor edits to make it appropriate for the new purpose.

However, as a courtesy to the teacher, you should provide them with self-addressed, stamped envelopes, and after the process is completed, you should write them a thank-you note.  :)

RSI is an [i]extremely[/i] competitive program to get into; there is [i]no one[/i] who should be "confident" about their chances of getting in.


I was looking through the book and came upon this problem, after I worked it out, I got a completely different answer than what is given, so I was wondering if my logic made sense.

Problem: In how many ways can we pick a group of 3 different numbers from the group 1,2,3,....,500 such that one number is the average of the other two? (the order in which we pick the numbers does not matter.)

given answer was
[hide]62250[/hide]

and I'm sure if this belongs in the Mathcount forum, if it doesn't, will somebody please move it for me?

The book's answer is correct.  What was your method?

Here is a simpler method than in the book (Problem 2.10).

[hide="Solution"]The smallest and largest numbers must be of the same parity (both even or both odd).  There are 250 even numbers from 1 to 500; thus there are $\binom{250}{2} = 31125$ even pairs.  There are 250 odd numbers; thus there are $\binom{250}{2} = 31125$ odd pairs.  Altogether there are $31125 + 31125 = 62250$ pairs.

Once we pick the smallest and largest numbers, the middle number is determined.  So the number of triples is 62250.[/hide]

Oo ok, thank you, I didn't really get it. I accidently did combination instead of permutation.

Thanks.


trigonometry
logarithms
integration
real analysis
real analysis unsolved

:D
$ \sum_{k\equal{}1}^{\infty}\frac{cos(ln(k))}{k^p}$

[color=red]where $ ln(k)$ means $ log_{e}k$[/color]

discuss the convergence of the above series :roll:

For $ p > 1,$ the series converges absolutely by comparison to $ \sum\frac1{k^p}.$

For $ p\le 0,$ the series diverges because its terms fail to go to zero.

That leaves the range $ 0 < p\le 1.$ We do not have absolute convergence, but do we have conditional convergence?

Assume $ 0 < p\le 1$ and write the series as

$ \sum_{k \equal{} 1}^{\infty}\frac {\sin(\ln(k \plus{} 1)) \minus{} \sin(\ln k)}{k^{p \minus{} 1}} \minus{} \sum_{k \equal{} 1}^{\infty}\frac1{k^{p \minus{} 1}}\left[\sin(\ln(k \plus{} 1)) \minus{} \sin(\ln k) \minus{} \frac {\cos(\ln k)}k\right]$

Let's start with that second sum. By one use of the mean value theorem, we have that

$ \sin(\ln(k \plus{} 1)) \minus{} \sin(\ln k) \equal{} \frac {\cos(\ln x)}{x}$ for some $ x\in(k,k \plus{} 1).$

By a second use of the mean value theorem, and noting that $ \frac {d}{dx}\,\frac {\cos(\ln x)}{x} \equal{} \frac { \minus{} \sin(\ln x) \minus{} \cos(\ln x)}{x^2},$ we then have that

$ \left|\frac {\cos(\ln x)}{x} \minus{} \frac {\cos(\ln k)}{k}\right|\le \frac2{k^2}.$

That means that the second series is bounded by $ \sum\frac {2}{k^{p \plus{} 1}}$ and converges absolutely.

We then turn to the first sum. Do a summation by parts, summing the telescoping terms $ (\sin(\ln (k \plus{} 1)) \minus{} \sin(\ln k))$ and differencing the $ \frac1{k^{p \minus{} 1}}.$

But that gives us

$ \sum_{k \equal{} 1}^n\frac {\sin(\ln(k \plus{} 1)) \minus{} \sin(\ln k)}{k^{p \minus{} 1}}$$ \equal{} \frac {\sin(\ln(n \plus{} 1))}{k^{p \minus{} 1}} \plus{} \sum_{k \equal{} 1}^{n \minus{} 1}\left(\frac1{k^{p \minus{} 1}} \minus{} \frac1{(k \plus{} 1)^{p \minus{} 1}}\right)\sin(\ln(k \plus{} 1))$

It's not yet clear what we're going to do with this.

Edit: on second thought, let's not do the summation by parts (which does hint at trouble). The terms $ \cos(\ln k)$ have the same sign - all positive or all negative - for a stretch of values ranging from $ ce^{an}$ to $ ce^{a(n \plus{} 1)}.$ On that stretch $ \sum\frac1{k^p}$ adds up to $ ce^{a(1 \minus{} p)n}.$ Note that that increases geometrically as $ n \to \infty.$ The partial sums of the original series are unbounded, and the series diverges.

Kent Merryfield, :lol:

In your reply:

you mean :  $ \sum_{k\equal{}ce^{an}}^{ce^{a(n\plus{}1)}}\frac{1}{k^{p}}\equal{}ce^{a(1\minus{}p)n}$

but i can't get this conclusion

could you explain it  carefully

$ \sum_{k = a + 1}^{b}\frac1{k^p}$ is comparable to $ \int_a^b\frac1{x^p}\,dx = \frac1{1 - p}(b^{1 - p} - a^{1 - p})$

If we then let ${ b = ce^{\alpha(n + 1}}$ and $ ce^{\alpha n},$ then

$ b^{1 - p} - a^{1 - p} = ce^{\alpha(n + 1)(1 - p)} - ce^{\alpha n(1 - p)}$

$ = ce^{\alpha n(1 - p)}(e^{\alpha(1 - p) - 1) = Ce^{\alpha n(1 - p)}.}$

Of course, I'm not saying this is an exact computation. I'll looking for rate of growth (or decay) and I have that much: the growth is exponential in $ n.$

I was assuming in all of this that $ 0 < p < 1.$

I realize now that the argument does not cover the $ p = 1$ case.

Edit: it takes care of the $ p=1$ case anyway. $ \sum_{k=ce^{an}}^{ce^{a(n+1)}\frac1k}$ is essentially a constant independent of $ n.$ The partial sums of the series will be bounded and will oscillate, but not converge.


For all you Ohio people who've done well in states, could anyone tell me what is the usual score needed to make nationals, or to make top 10? Also, what did you do to prepare?
Thanks

http://www.ohioengineer.com/programs/Kids_Programs.htm#MathCounts

All the info you need.


quadratics
number theory proposed
number theory

Prove that for any $ n$ there exist $ n$ consecutive positive integers, none of which can be written as the sum of two squares.

Take $ s$ such that $ s\minus{}i \equiv p_i \mod p_i^2$ for different primes $ p_i \equiv 3 \mod 4$.

Ok I know that such $ s$ exists but I tried  to prove that numbers $ s\minus{}1$ , $ s\minus{}2$ ... $ s\minus{}n$  satisfye  conditions from the problem and I failed

Could You give me some hint?

See the discussion [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=219017]here[/url].  Here it suffices to observe the following:  if $ s \minus{} i \equal{} a^2 \plus{} b^2$ is divisible by a prime $ p \equiv 3 \bmod 4$ with multiplicity $ 1$, then neither $ a$ nor $ b$ are divisible by $ p$, hence

$ a^2 \equiv \minus{}b^2 \bmod p \implies$
$ \left( \frac{a}{b} \right)^2 \equiv \minus{}1 \bmod p$

but this is a contradiction since $ \minus{}1$ is not a quadratic residue $ \bmod p$.


LaTeX
number theory
calculus
calculus computations

Dear Member,
                    To some who seemed 'Confused' of what i posted
couple of days ago related to "Number theory,"(Zero-Less Integer.),the question i really wanted to convey is this:

                 "Does 'Infinity,' 'Infinitesimal' implied the possibility
of existence of using(Theoretically,in practice.)a "Non-Zero Integer"(Less-Zero) from theoretical point of view in mathematical sciences?"

                    This question occured to me,intuitively while
reading a book on 'mathematical infinity.' I haven't mastered yet
the use of "LATEX TECHNIQUE" to express my ideas,mathematically.Hope,members will understand.

                                            Sincerly,
                                          Mikhail S.

I dont get this. Try to express yourself more clearly.

Some suggestions:

1) If this is meant as elaboration on an earlier post, you should have posted it as a reply in that thread.
2) Basic Latex is easy to learn, and there is a tutorial for it on this site.
3) You have to write mathematics to write mathematics.  If you have actual mathematical content to what you're saying, you should say it mathematically.
4) Ditto to what Ottem said.


probability
analytic geometry
geometry
geometric transformation
rotation
integration
combinatorial geometry

Select $ n$ points on a circle independently with uniform distribution. Let $ P_n$ be the probability that the center of the
circle is in the interior of the convex hull of these $ n$ points. Calculate the probabilities $ P_3$ and $ P_4$. [A. Renyi]

for $ P_3$:
one will work on the unit circle:
one can suppose that the first point $ Z_1$ has an argument equal to $ \alpha_0 = \pi$ (polar coordinate) and that the second point $ Z_2$ is in the upper half plane. (just do a rotation and a symmetry). Now that $ Z_1$ and $ Z_2$ are fixed, with $ 0 < \alpha_2 < \pi$, it is easy to see that the center of the circle $ O$ is inside the convex hull of $ Z_1Z_2Z_3$ iff the argument of $ Z_3$ satisfies: $ \alpha_2 - \pi < \alpha_3 < 0$. Hence, if $ Z_1$ and $ Z_2$ are known (as above), the (conditional) probability that the center is inside the convex hull of  $ Z_1Z_2Z_3$ is $ \frac {\pi - \alpha_1}{2 \pi}$. This leads to
\[ P_3 = \int_{\alpha_2 = 0}^{\pi} \frac {\pi - \alpha_1}{2 \pi} \frac {d \alpha_2}{\pi} = \frac {1}{4}
\]

Similar arguments (conditional probabilities) quickly lead to $ P_4$, but it gets nasty. Anyone has a neat expression of $ P_n$ ?

I think we can consider points distributed uniformly on the circumpherence (not inside!). And quite easily get something like $ \left(\frac{1}{2}\right)^{(n\minus{}1)}$. But i have note done this with pen and paper.

So here we are:
1. For each point inside circle we have corresponding point on the circumpherence. (the point with the same angle).
2. We will consider these points on the circumpherence.
3. If the convex hull contains the center then it's impossible to cover all $ n$ points with a hemicircle.
4. The probability that $ n$ points can be covered with a hemicircle is $ \frac {n}{2^{n \minus{} 1}}$. There are $ n$ possibility to choose initial point (the most clockwise). All other $ n \minus{} 1$ points should be in 180 degree counterclockwise interval.
5. We get $ 1 \minus{} \frac {n}{2^{n \minus{} 1}}$. For $ n \equal{} 3$ we get $ 1/4$.


This is for discussion of the great MMORPG Runescape only.

I sold my RS account for 20 dollars.

It was pretty fun to play when I was 12. Now I have moved onto to other games.

Um....what did your character even like have?

Level 83 member with 8 mil gp and a total level of around 1400.

The guy who bought it was only interested in the money. He took the money using BH and then dumped the account.

8mil gp was probably worth 40 dollars...

[quote="Meta_Knight"]i bet this thread won't get more than 2 pages :roll:[/quote]

1 1/2 max

but on the topic of RS, i sold my acc for 85 dollars....hehe stupid middle schoolers...no offense to any middle schooler here.

haha...before jagex made up that stupid rule where you couldn't give people huge amounts of money for free, this kid in my school who had no life kept buying RS money from me.  After the rule was made, though, I had to sell my account.

BH? I know I sound stupid, but what is BH?

I didn't know either, but

http://lmgtfy.com/?q=Runescape+BH

Runescape is great because its free to play and no d/l...I haven't played in a few years though, but plan on playing some this summer
__________________
[url=http://gamefriends.com/game_videos][color=red]Game Videos[/color][/url], [url=http://gamefriends.com/ask][color=blue]Ask a Gamer[/color][/url], [url=http://gamefriends.com/game_news][color=purple]Game News[/color][/url]

It gets boring when there isn't more equipment to upgrade to. I made a new account and started over.

:spider:

Hey coolts, is your avatar the album art from Disturbed's album "Indestructible"

Runescape is fun, at least to me. The only annoying thing is that it takes forever to make 8 mil for saradomin sword. I need another roughly 74k pure essence.


Here, all variables are be taken to be in the interval $ [0, 2\pi)$, and you are expected to give all solutions.

1. $ \cos^2{5x} \minus{} \cos^2{x} \equal{} (\minus{}\sin{4x})(\sin{6x})$
2. $ \frac{\cos{4x}}{\cos{x}} \equal{} 2\cos{3x} \minus{} 2$
3. $ 2\cos{2x} \plus{} 8\sin^2{x} \plus{} 2\sin^2{x} \equal{} \cos{4x}$
4. $ 9\sin^3{x} \minus{} 6\sin^2{x} \minus{} 3\sin{x} \plus{} 2 \equal{} 0$
5. $ (\sin{x} \plus{} \cos{x})^4  \minus{} 5\cos^2{x} \equal{} 1 \minus{} 4\cos^4{x}$
6.
$ \cos{x}\cos{y} \equal{} \frac{\cos{x}\cos{y} \plus{} \sin{x}\sin{y}}{2}$
$ \sin^2{x} \minus{} 5|\cos{y}| \plus{} 4 \equal{} 0$

If any of them are weird (e.g., have no solutions), tell me. I made up these problems myself (except for the first one) half a year ago and forgot how to solve some of them, though I do recall that when I made them all, they all had solutions. =D

[quote="Zhero"]
...
$ \sin^2{x} \minus{} 5|\cos{x}| \plus{} 4 \equal{} 0$
...[/quote]
$ \sin^2{x}\minus{}5|\cos{x} \plus{}4 \equal{} 0 \Leftrightarrow$
$ 1\minus{}\cos^2{x}\minus{}5|\cos{x}| \plus{}4 \equal{} 0$
if $ x\in [0,\pi/2)\cup (3\pi/2,2 \pi))$ then we have $ \minus{}\cos^2{x}\minus{}5\cos{x} \plus{} 5 \equal{} 0$, substituting $ w\equal{}\cos{x}$ it becomes simple.
Same way for if $ x\in [\pi/2,3\pi/2)$.

Uhh, the last two was a system of equations.

[quote="polysot"][quote="Zhero"]
...
$ \sin^2{x} \minus{} 5|\cos{x}| \plus{} 4 \equal{} 0$
...[/quote]
$ \sin^2{x} \minus{} 5|\cos{x}| \plus{} 4 \equal{} 0 \Leftrightarrow$
$ 1 \minus{} \cos^2{x} \minus{} 5|\cos{x}| \plus{} 4 \equal{} 0$
if $ x\in [0,\pi/2)\cup (3\pi/2,2 \pi))$ then we have $ \minus{} \cos^2{x} \minus{} 5\cos{x} \plus{} 5 \equal{} 0$, substituting $ w \equal{} \cos{x}$ it becomes simple.
Same way for if $ x\in [\pi/2,3\pi/2)$.[/quote]

2.$ \leftrightarrow \cos4x + 2\cosh = 2\cos3x\cosh = \cos4x + \cos2x$
$ \leftrightarrow 2\cosh = \cos2x$
$ \leftrightarrow 2\cosh = 2\cos^2x - 1$
substitute $ t= \cosh$  :)
3 . for the third i think there was some mistyping here , could you double-check it ? , thanks
4 notice that for number forth , it can be factorized as
$ (3\sinh - 2)(3\sin^2x - 1) = 0$:)
6. the first equation gives $ \sinh\tiny = \cosh\cosh$ which means $ \cosh = \tiny$
combine with the second equation gives
$ |\cos^2y| - 5|\cosh| + 4 = 0$  :)

[quote="HTA"]2.$ \leftrightarrow \cos 4x \plus{} 2\cos x \equal{} 2\cos 3x\cos x \equal{} \cos 4x \plus{} \cos 2x$
$ \leftrightarrow 2\cos x \equal{} \cos 2x$
$ \leftrightarrow 2\cos x \equal{} 2\cos^2x \minus{} 1$
substitute $ t \equal{} \cos x$  :)
3 . for the third i think there was some mistyping here , could you double-check it ? , thanks
4 notice that for number forth , it can be factorized as
$ (3\sin x \minus{} 2)(3\sin^2x \minus{} 1) \equal{} 0$:)
6. the first equation gives $ \sin x\sin y \equal{} \cos x\cos y$ which means $ \cos x \equal{} \sin y$
combine with the second equation gives
$ |\cos^2y| \minus{} 5|\cos y| \plus{} 4 \equal{} 0$  :)[/quote]
Generally, separating the x from \sin or \cos would be a good idea. :D


linear algebra
matrix
algebra
polynomial

Prove that det (I(n) +ABA*)=det(I(n)+A*BA) . :)

Use the multiplicativity of the determinant (that's enough) ($1=det(AA^{*})=det(A)det(A^{*})$), assuming that by $A^{*}$ you mean $A^{-1}$.

Let's assume he meant $A^*$ is the conjugate transpose and not the inverse.

A* is the adjoint of matrix A. sorry :blush:

Hi Ciprian! How are you?

You can prove the more general fact that:

\[ \det(I_n+AB)=\det(I_n+BA) \]

for any two matrices. From this follows your problem. I'll be here for more hints.

[quote="ciprian"]when A is inversable the problem is trivial but what about when it is not? :?[/quote]

For any matrix $A \in \mathcal M _ n ( \mathbb{C} )$ there is a sequence of invertible matrices $(X_n)_n$ which tend to $A$ (element by element). For a proof use the polynomial $\det (A-xI_n)$: there is $t \in \mathbb{R}$ s.t. $\det (A - x I_n) \neq 0$ for all $0 < |x| < t$ etc.

For your problem and xirti's, write the relation for invertible matrices $X_n$. Taking the limit of that we get the result.

I learned this method in class a few months ago. Baluna taught it.

BTW, how come you post problems at 3:29 in the afternoon?

Another method. (more elementary)

Prove the relation if $A$ invertible by using the fact that $AB$ and $BA$ have the same characteristic polynomial.(I don't take you the pleasure to prove it). :P

If $A$ is not invertible.
Consider $A(t)=A+tI_n$. Because $det (A-xI)$ is a polynomial we knew that $A(t)\neq 0$ for infinitely many values of $t$. Then we know that:
\[ \det(I_n+A(t)B)=\det(I_n+BA(t)) \]
for infinitely many values of $t$.
Since  $P(t)=\det(I_n+A(t)B)=det (I_n+AB+tB)$ is a polynomial in $t$ of degree at most n and $Q(t)=\det(I_n+BA(t))$ too, and they are equal for an infinity of values of t, we know that they coincide. So, $P(0)=Q(0)$. q.e.d.

Thank you cristi and perfect_radio for your solutions. :D Got the point :)  For perfect_radio: what grade are you now?

"I'm the dead and you're the death." Or was it the other way around?  :D


geometric series
real analysis
real analysis unsolved

I don't know if this is the right forum, I am not a native english speaker so not all the terms are familiar to me.

Anyway, what's been bugging me is this:

[code]X = 0.9999...     || * 10
10X = 9.9999...   || - X
9X = 9            || / 9
X = 1??
[/code]

What does $0.99999...$ mean? The normal interpretation of decimal notation would say it is
$0.999.... = \sum_{n=1}^\infty \frac{9}{10^{n}}$

This is a geometric series and so the sum is $\frac{9/10}{1-1/10}= 1$. :)

Here's another (more intuitive, perhaps) way to look at it: What's 1 - 0.99999....? It's just 0.000000...., right? Isn't that just 0? There will never be any other digit.

So.. Do you mean because the value of 0.9999... is infinitely close to 1, it IS 1? I'm just more confused now  :blush:

In the real numbers, to put it shortly, yes. You have to identify those two numbers with each other because otherwise their difference should be a nonzero real. But the difference between them is clearly smaller than any positive real so it must be zero...


[b][size=100][color=DarkBlue]We draw a circle $ (O),$ taken as cord the given segment $ AB$ and let $ M$ be, the midpoint of the arc $ AB.$ We define a fixed point $ P$ on $ (O),$ between $ A,$ $ M$ and let $ C$ be, a mobile point on the circle $ (O).$ Determine the point $ C,$ in order to be $ MA\cdot MB\cdot MC \equal{} PA\cdot PB\cdot PC$[/color][/size][/b].

Kostas Vittas.

PS. This not difficult problem is inspired from the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=190749]Extremum problem[/url], of [b][size=100]Virgil Nicula[/size][/b].

$\frac {PC}{MC}= \frac {MA \cdot MB}{PA \cdot PB} \equal{} k$ $\Longrightarrow$ the wanted points $ C$ satisfy $ \frac {PC}{MC} \equal{} k.$

$\Longrightarrow$ $C$ lies on the Apollonius circle relative to $P,M$ with ratio $k.$ Note that we do not need $ M$ to be the midpoint of the arc $ AB,$ unless you mean something else.

Thank you very much dear [b][size=100]Luis[/size][/b], for your interest.

I have not in mind a different solution of this easy problem and as you said, it is not necessary to be the point $ M,$ as the midpoint of the arc $ AB.$

$ \bullet$ This problem has been arisen, in trying to solve ( unfortunately without any success ) the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=190749]Extremum problem[/url], of [b][size=100]Virgil Nicula[/size][/b].

I conjectured the wanted maximum product there, at the point $ N,$ as the midpoint of the arc $ PM$ ( please see the schema t=190749 there ).

So, if we can determine the point $ P,$ such that $ PA\cdot PB\cdot PC \equal{} MA\cdot MB\cdot MC,$ when the triangle $ ABC$ is given ( consider the point $ B$ in the schema t=190749, as  the point $ C$ in the schema t=191359 ) then, we can solve the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=190749]Extremum problem[/url].

Best regards, Kostas Vittas.


geometry
3D geometry
tetrahedron
geometry proposed

A tetrahedron $OABC$ satisfies the following conditions.

[a] $OA\perp{BC},OB\perp{AC},OC\perp{AB}$
[b] The area of four face of the tetrahedron are all equal.

Prove that the tetrahedron is a regular tetrahedron.

Let $E$ be the feet of altitude from $B$ to $OC$. We have $OC\perp BE$ and $OC\perp AB$. It follows $OC\perp ABE$, so $AE\perp OC$. Since $S_{OAC}=S_{OBC}$ and side $OC$ is common for triangles $OAC$ and $OBC$, we conclude $AE=BE$.
Let $F$ be the midpoint of $AB$. Since $AE=BE$, it yields $EF\perp AB$, so plane $OCF\perp AB$. It means $OF$ is altitude and median from $O$ to $AB$, hence $OA=OB$.
Analogously we will prove equality of all sides.

Oh ,fantastic!. It's the very "Seeing is believing",Thank you, Myth!

Incidentaly, my solution is using vector. Because I'm not good at Geometry.

kunny


Let p >= 5 be a prime number and let r be the number of ways of placing
p identical checkers on a p × p checkerboard so that not all checkers are
in the same row (but they may be in the same column). Show that r is
divisible by p^5.


does anyone know where i can find a physics forum similar to mathlinks.ro? thank you.

try this website http://www.advancedphysics.org/ I know someone who uses it

wow, thanks  :lol:  you're the only one that answered.


Bueno ak subo un archivo de desigualdades publicadas en la revista $Crux Mathematicorum$.

Posteen las soluciones a las desigualdades que ya hayan resuelto o que resuelvan.

Si $a,b,c>0$ y $a<b+c$.

Demostrar que: $\frac{a}{1+a}< \frac{b}{1+b}+\frac{c}{1+c}$

Expandiendo obtenemos que la desigualdad es equivalente a: $(b+c-a)+bc+abc > 0$

Esto es obviamente verdadero ya que estos tres términos son positivos.


[quote="moin1998"][quote="BOGTRO"][quote="GameBot"]Bocephus has a bag full of nickels and dimes. If there are 3 times as many dimes as nickels, and he has $ \$36.05$ in his bag, how many nickels does he have?[/quote]

There are n nickels and 3n dimes.

$ 35n\equal{}3605 \implies n\equal{}103$.[/quote]

Why 35?[/quote]

This is because each nickel is worth 5 cents and there are $n$ nickels, and each dime is worth 10 cents and there are $3n$ dimes.  So $5 \cdot n + 10 \cdot 3n = 5n + 30n = 35n$.

[quote="moin1998"][quote="BOGTRO"][quote="GameBot"]Bocephus has a bag full of nickels and dimes. If there are 3 times as many dimes as nickels, and he has $ \$36.05$ in his bag, how many nickels does he have?[/quote]

There are n nickels and 3n dimes.

$ 35n\equal{}3605 \implies n\equal{}103$.[/quote]

Why 35?[/quote]

Each nickel is worth $5$ cents, and each dime is worth $10$ cents. Thus the number of cents he has is terms of $n$ is equal to $5n+3(10n)=35n$.


x, y, z - positive reals
$ x\plus{}y\plus{}z\equal{}2$
$ A\plus{}B\plus{}C\equal{}\pi$
Prove that $ x^2.\sin^2{A}\plus{}y^2.\sin^2{B}\plus{}z^2.\sin^2{C}\leq 1$.

I forgot to write that $ A\in \left(0;\frac{\pi}{2}\right), \ B\in \left(0;\frac{\pi}{2}\right), \ C\in \left(0;\frac{\pi}{2}\right)$.

Noone?

My teacher also couldn't solve it  :oops:

It doesn't seem to be right.
Pick
$ x\equal{}2, y\equal{}0, z\equal{}0$
$ A\equal{}\frac{\pi}{2}, B\equal{}\frac{\pi}{4}, C\equal{}\frac{\pi}{4}$

Then we have: $ 4 \le 1$, contradiction.

Of course you wanted $ x$, $ y$ and $ z$ be positive but making small changes
to the variables here won't change LHS much.


Prove that, for all real $x_{i}>-1$, where all $x_{i}$ have the same sign, the following inequality holds:

\[(1+x_{1})(1+x_{2})...(1+x_{n}) \geq 1+x_{1}+x_{2}+...+x_{n}\]

Note that the case where all $x_{i}$'s are equal gives us the Bernouilli Inequality.

Why is it a generalisation? This IS bernouilli, no? (and it's been posted several times in Ineqs, theorems & formulae)

Bernoulli's inequality is generally seen $(1+x)^{\alpha}\ge 1+\alpha x$.

But it ain't helping us much for things like $a,b,c>0, a^{2}+b^{2}+c^{2}=1 \Rightarrow (2-3ab)(2-3bc)(2-3ca) \geq 1$. ;)

Usually I've seen both together called bernouiili, but this version is far stronger for olympiad purposes than $(1+x)^{\alpha}\ge 1+\alpha x$.

Yes.  I was probably unclear.  I meant to say that he wrote that it is a generalizing relative to the normal version mentioned above.  Indeed, this version is far stronger. :)

It's not entirely generalised, since $\alpha$ can be [i]real[/i] in your case, not just integer. That is also why I am not happy with the name 'generalised', they're both independent ineqs, but in olympiad context I have seen the sum-product version usually referred to as just 'Bernouilli'.

Isn't the initial inequality trivial by induction?

\begin{eqnarray*}(1+x_{1})\dots(1+x_{n})(1+x_{n+1}) &\geqslant& (1+x_{1}+\dots+x_{n})(1+x_{n+1})\quad[1+x_{n+1}>0]\\ &=& 1+x_{1}+\dots+x_{n}+x_{n+1}+x_{n+1}(x_{1}+\dots+x_{n})\\ &\geqslant& 1+x_{1}+\dots+x_{n}+x_{n+1}\end{eqnarray*}

as $x_{n+1}$ and $x_{1}+\dots+x_{n}$ have the same sign.

Why do $x_{n+1}$ and $x_{1}+\cdots+x_{n}$ have to have the same sign? Your steps wouldn't hold if, say, we had $x_{1}=-\frac{1}{2}$, $x_{2}=-\frac{1}{2}$, and $x_{3}= 2$.

[quote="JBL"][quote="paladin8"]Why do $x_{n+1}$ and $x_{1}+\cdots+x_{n}$ have to have the same sign?[/quote]
[quote="Jan"]Prove that, for all real $x_{i}>-1$, [b]where all $x_{i}$ have the same sign[/b] ...[/quote][/quote]

Ha. I knew I couldn't read.

[quote="Peter VDD"]The inequality is even false at all if the signs can be mixed. :)[/quote]

In fact, it's false in the particular example paladin8 gave (which, I suppose, should make us feel even better that the proof doesn't cover that case :wink: ).


trigonometry
trig identities
Law of Cosines
geometry unsolved
geometry

Is there an infinite set of points in the plane such that no three points are collinear, and the distance between any two points is rational?

Yeah, $A_i(\cos2\alpha_i,\sin2\alpha_i)$,where $\sin\alpha_i=\frac{i^2-1}{i^2+1},\cos\alpha_i=\frac{2i}{i^2+1}$.($i\in\mathbb{N}$)
It's obvious that $A_i$ is on the unit circle,so no three is collinear.
$A_iA_j=2\sin|\alpha_i-\alpha_j|\in\mathbb{Q}$(consider the angle $A_iOA_j$)

[quote=dgreenb801] Can anyone show there are infinitely many rationals $ (x,y)$ that satisfy this?[/quote]

Looks like Eisenstein integers?

[quote=dgreenb801]I'm not really sure if this is the same as the previous proof.
We will use induction. Consider a circle radius $ \frac {\sqrt {3}}{2}$, and an equilateral triangle inscribed in it with vertices A,B,C, side length 1. Clearly no three points on the circle are collinear.
These three points obviously work. Now suppose a point D on minor arc AB is a rational distance from A and B. By Ptolemy, it is a rational distance from C. Now if we choose another point F on minor arc AB a rational distance from A and B, also by Ptolemy FC is rational but then we have cyclic quad FDBC, so by Ptolemy FD is rational. It remains to show there are infinitely many points on arc AB a rational distance from A and B.
But if N is a point on minor arc AB a distance of x and y from A and B, respectively, then $ \angle ANB \equal{} 120$, so by the law of cosines, $ x^2 \plus{} y^2 \plus{} xy \equal{} 1$. Can anyone show there are infinitely many rationals $ (x,y)$ that satisfy this?[/quote]

About this solution, I am not sure if such a point on minor arc $AB$ whose distance from both $A,B$ is rational will always exist. You can guarantee that the point has rational distance from at least one of $A,B$, but not both (since in that equation, if $x$ is rational, it might still happen that $y$ is irrational).

@above, read the next post ;)

[quote=dgreenb801]($ \frac {2n \plus{} 1}{n^2 \plus{} n \plus{} 1}$, $ \frac {n^2 \minus{} 1}{n^2 \plus{} n \plus{} 1}$) works for the above proof.[/quote]

[quote=Wolstenholme]These answers don't seem to have any motivation so let me provide motivation for shobber's proof.  Let $ S = \{(1, \frac{2s}{s^2 - 1}) : s \in \mathbb{Q}\} $.  Consider the inversion centered about the origin $ O $, with radius $ r \in \mathbb{Q} $.  From now on the inverse of a point $ X $ will be denoted by $ X' $.  It suffices to show that for all $ P, Q \in S $ that $ P'Q' \in \mathbb{Q} $ because then we will have found an infinite set of points, no three collinear (they all lie on a circle), with pairwise rational distances.

Now note that $ P'Q' = \frac{r^2 \cdot PQ}{OP \cdot OQ} $.  It is clear that for all $ A, B \in S \cup {(0, 0)} $ we have that $ AB \in \mathbb{Q} $ and this implies that $ P'Q' \in \mathbb{Q} $ as desired.[/quote]

Why do $P'$ and $Q'$ lie on the  unit circle?


analytic geometry
combinatorics proposed
combinatorics

I don't know how to classify this problem , it's from Iran 1999. Anyone knows any other solution than the one from the book with 1999-2000 contests ? That one is kind of artificial ..

Suppose that n(r) denotes the number of points with integer coordinates on a circle of radius r>1. Prove that :

n(r) < 6 (\pi r^2)1/3

problem 289 in http://www.cms.math.ca/Competitions/MOCP/2004/sol_jan.html has a solution


[quote="Nameless"][quote="barra"]Let $a,b,c$ be positive real numbers such that $a+b+c=3$. Show that
\[ 21(a^3+b^3+c^3)\geq 36abc+(ab+bc+ca)^3  \][/quote]
I think this ineq is very easy.
We have $12(a^3+b^3+c^3)\ge 36abc$ then we have to prove
$9(a^3+b^3+c^3)\ge (ab+bc+ca)^3\Leftrightarrow (a^3+b^3+c^3)(a+b+c)^3\ge 3(ab+bc+ca)^3$
But $(a+b+c)^2\ge 3(ab+bc+ca)$
       $(a^3+b^3+c^3)(a+b+c)\ge(a^2+b^2+c^2)^2\ge(ab+bc+ca)^2$[/quote]

Sorry to disappoint you  :blush: . Originally the inequality was proving
\[ 3(a^3+b^3+c^3)\geq 3abc+2(a^3c+b^3c+abc^2)  \]
whenever $a+b+c=3$. I just rearrange to make it nicer. I hope this one is not as easy as the former (though I don't know if it is really more difficult).

[quote="barra"]Sorry to disappoint you  :blush: . Originally the inequality was proving
\[ 3(a^3+b^3+c^3)\geq 3abc+2(a^3c+b^3c+abc^2)  \]
whenever $a+b+c=3$. I just rearrange to make it nicer. I hope this one is not as easy as the former (though I don't know if it is really more difficult).[/quote]

   I just prove your inequal in case $a \geq b \geq c$ but I will try another cases as well :D
$(a+b+c)(a^3+b^3+c^3) \geq (a+b+c)abc+2(a^3c+b^3c+abc^2)$
$(a-b)[(a^2+2ab+2b^2)(a-c)+(b-c)(3b-c)(b+c)]+ (b-c)^2(c+2b)^2 \geq 0$


I am so happy that i find this forum because I can ask questions here.

I have two questions from the book introduction to mathematical statistics fifth version.

I hope somebody can help to answer my questions.

Questions are in the attachment. I don't know how to edit formula here.

Your intuition for the first question is wrong. It would be like saying "the limit of the sequence $ \frac {1}{2}$, $ \frac {1}{3}$, $ \frac {1}{4}$, $ \frac {1}{5}$, $ \ldots$ cannot equal zero because the elements of the sequence are always greater than zero" -- which I hope you see is a false argument.

The correct way to think about this question would be to ask yourself, "can I name a value, x, that is in every single one of the sets $ A_k$, no matter how big $ k$ is?" If not, then the intersection is the empty set.

Thank you very much, Gauss202. Your answer is really helpful. I think I need to read the definition of limit again to really understand the  meaning.

Also I hope somebody would like to answer my second question.

[quote="deepnorth"]Also I hope somebody would like to answer my second question.[/quote]

google the words 'simplex' and 'volume'.

Thank you very much blahblahblah, i find the clue to understand the problem. I have to re-read my calculus book. Thanks.


function
real analysis
real analysis unsolved

Prove that a continuous function which takes on no value more than twice must take on some value exactely once.

heres a tentative approach, i didnt check all details.

so assume the contrary, ie it doesnt take any value exactly once- which means it takes every value exactly twice. you consider a closed interval$[a,b]$. f must assume a maximum value and a minimum value somwhere on this interval(  infact twice by the hypothesis). so some extrema, either a maximum or a minimum must be attained in the interior, lets say a maximum at the point c. now draw the line $y=f(c)-\epsilon$ for sufficiently small $\epsilon$. since f is continuous it has the intermediate value property, so it assumes this value atleast 3 times which is a contradiction. ( draw a picture, it will be clear) . i dont think i've explained properly. take a look, see if you understand, or else i will try to explain in a better way

[quote="kouboy"]since f is continuous it has the intermediate value property, so it assumes this value atleast 3 times which is a contradiction. ( draw a picture, it will be clear) .[/quote]

Sorry i didn´t understand that final part.
I startes to solve that problem exactely like you, but i didn´t do the conclusion.

----------------------
thank you to help me:)

The value $f(c)-\epsilon$ is taken at least 3 times: 2 times for the interior extremal point and at least once (once because the other extremal point may be one of the endpoints) for the other extremal point (extremal in the same sense) :)

Ow my god, i think i´m in trumble  :rotfl:  because Í can´t understand why for the interior extremal there are two values that are mapped to $f(c)-e$

I though by this way,if  $f(a) < f(c)-e <f(c)$ so, the intermediate value property says that there are ONE value d such that $f(d)=f(c)-e$.Where are the second one????
And if $f(k)$ is the other maximum so for $b$ such that $f(b)<f(c)-e<f(k)$ there are ONE value p such that $f(p)=f(c)-e$.
And that don´t lies in a contradiction

You take the intervals $[c-\epsilon_{1},c+\epsilon_{1}]$ for the interior extremal point and $I_{2}=[d-\epsilon_{2},d]$ or $I_{2}=[d,d+\epsilon_{2}]$ for the other extremal point. I say 'or' because d can be one of the endpoints, so I choose the interval $I_{2}$ convenably. Now I look at the intersections of the following images: $f([c-\epsilon_{1},c]$, $f([c,c+\epsilon_{1}])$and $f(I_{2})$. For sufficiently small $\epsilon_{1}$ and $\epsilon_{2}$ you find that a point different from c appears in each intersection...contradiction :) Make an arbitrary graphic :) Maybe somone could choose the epsilons more rigurously.


[quote="Interval"]A circular courtyard has an area of 10 – 2x^2. There are two rectangular flower beds in the courtyard. Which of the following expressions represents the green lawn area.[/quote]
Where's the following expressions?

So sorry but here is the complete question:

A circular courtyard has an area of 10 – 2x^2. There are two rectangular flower beds in the courtyard. Which of the following expressions represents the green lawn area.

(1)...(10 – 2x^2) – (2x^2 + 3x^2)= 10 + 3x^2
(2)....(10 – 2x^2) – 2x^2 – 3x^2 = 10 – 7x^2
(3)....(10 – 2x^2) – 2x^2 – 3x^2 = 10 – 3x^2

Can you be more specific?  I don't really understand the question...

I'm sorry b-flat but the question is what it is.

I typed it exactly as it written in the math study book.

[hide]
Take the area of the circle and subtract the area of both rectangles, which are 2x^2 and 3x^2.

So the answer is 10 - 7x^2 which is 2).[/hide]

[quote="Interval"]Okay here it is again:

A circular courtyard has an area of 10 – 2x^2. There are two rectangular flower beds in the courtyard. Which of the following expressions represents the green lawn area.

(1)...(10 – 2x^2) – (2x^2 + 3x^2)= 10 + 3x^2
(2)....(10 – 2x^2) – 2x^2 – 3x^2 = 10 – 7x^2
(3)....(10 – 2x^2) – 2x^2 – 3x^2 = 10 – 3x^2

NOTE:

There is a picture of a cicular garden and inside the garden there are two rectangles.

One rectangle has the following dimensions: Length = 2x; Width = x.

The other rectangle has: Length = 3x and Width = x.[/quote]
The lengths of the two rectangles really helped. I think thats what b-flat wanted. Anyway its just
[hide]The whole area minus the areas of both rectangles. So it is 2.[/hide]

[hide]
Take the two rectangles.  One rectangle has length 3x and width x.  Therefore, the area of it is $3x^{2}$.  Take the other rectangle.  This one has length 2x and width x.  Therefore, its area is $2x^{2}$.  Hence, the total area of the two rectangles is $2x^{2}+3x^{2}=5x^{2}$.  Now, since the total area of the circle is $10-2x^{2}$, we have that the area that is part of the circle, but not part of the rectangles is $10-2x^{2}-5x^{2}=10-7x^{2}$.  Thus, the answer is 2.  [/hide]

You guys (and b-flat) are very smart people.

You are all amazing in math.

Thanks!


let ABC  given acute angled triangle ,P any point inside it,show,
 ((AP+BP+CP)^2)>= (root3)(AP*BC+BP*CA+CP*AB)


It's supposed to be a comma.  By "the two conditions are independent," I assume you mean that you are asking two completely different questions.  (i) could be restated, "Prove that for for every graph $ G$, either $ G$ or its complement contains a Hamiltonian cycle."  It is also clearly false: choose a single vertex, color all edges leading to that vertex red and color all other edges blue.

[quote="Nbach"]Each edge and each diagonal of a $ n$-gon is coloured by red or blue.Prove that vertices of polygon can be labbel  as
$ A_{1},A_{2},...,A_{n}$ such that
i) $ A_{1}A_{2},A_{2}A_{3},...,A_{n-1}A_{n}mA_{n}A_{1}$ are  same colour
ii) There exist $ 1<k<n$ such that $ A_{1}A_{2},A_{2}A_{3},...,A_{k-1}A_{k}$ are blue and $ A_{k}A_{k+1},A_{k+1}A_{k+2},...A_{n-1}A_{n},A_{n}A_{1}$ are red
( Note that two above conditions are independent)[/quote]

So you want either i) or ii) to hold? This is http://www.mathlinks.ro/Forum/viewtopic.php?t=111249 .

  darij


inequalities
search
inequalities proposed

Let a,b,c,x,y,z be positive real numbers such that: \[(a+b+c)(x+y+z)=(a^{2}+b^{2}+c^{2})(x^{2}+y^{2}+z^{2})=4.\] Prove that \[\displaystyle abcxyz < \frac{1}{36}.\]

by Vasile C$\^{\i}}$rtoaje and Mircea Lascu

Has anyone tried to solve this FANTASTIC inequality?

Thank you very much. :maybe:

The solution is published in the last book published by GIL Publishing House, "Old and New Inequalities". The authors of the book are: Titu Andreescu, Vasile Cirtoaje, Gabriel Dospinescu and Mircea Lascu.

I am very interested to purchase this  book.

I think that every people keen on inequalities should have it on his writing-desk as the authors

[quote]Titu Andreescu, Vasile Cartoaje, Gabriel Dospinescu and Mircea Lascuare [/quote]

have created very BEAUTIFUL and  MAGNIFICENT inequalities for all friends of the forum and very GENIAL and INGENIOUS methods of solutions.

I am sure  this book will become an USEFUL and ABSOLUTELY NECESSARY tool of work for lovers of inequalities like me as it will help us to get new BRILLIANT and ORIGINAL ideas for solving inequalities.

I would like to know whom I must  contact to purchase the book :

   [quote]<< OLD  and NEW INEQUALITIES >>   published by GIL Publishing House[/quote]

Thank you very much to all authors

I don't know if you're laughing or you're trying to get a free copy of it :D or just exaggerating but anyway that's not the right way to make a book popular ;)

[quote="manlio"]very BEAUTIFUL and MAGNIFICENT [...] very GENIAL and INGENIOUS [...] USEFUL and ABSOLUTELY NECESSARY [...] BRILLIANT and ORIGINAL [...] so PRECIOUS and ADMIRABLE[/quote]

Well, I don't think it's about laughing, the book is really very good. But you see I wrote just "very good", and not FANTASTIC, GREAT, INGENIOUS, BRILLIANT, ABSOLUTELY NECESSARY and INCREDIBLE !!!!!!! For me, "very good" is already a strongly positive characterization of something. With all respect, Manlio, I really don't see any reason to overemphasize things in a way that they lose their signification and begin sounding ridicolous. If you call a very good, well-written and strongly helpful book GREAT INGENIOUS BRILLIANT INCREDIBLE without even having seen it, then I wonder what you would say if you see a really INCREDIBLE work - let's say a three-pages proof of the Riemann conjecture, just imagine! Manlio, I have done a search for the newest messages posted by you:

[quote="manlio"]Your solution is INCREDIBLE and IMPOSSIBLE TO FIND so I would like to know if it exists a more natural solution[/quote]

[quote="manlio"]Has anyone tried to solve this FANTASTIC inequality?[/quote]

[quote="manlio"]You can find the ASTONISHING, SURPRISING and FANTASTIC solution of Harazi in the topic : "After a famous CRUX inequality" in Solved Problems Section[/quote]

Actually I find the solutions you spoke about ingenious and very beautiful. I think all of us understand that these are words people usually not throw around on every purpose, and they are already considered as strong expressions of admirance. Then why WRITING IN CAPITALS? In the online community, excessive writing in capitals is considered to be impolite; in fact, posting in forums is often compared with talking, and writing in capitals corresponds to shouting. Even if I really admire someone's solution, I would not shout at him!

I don't deny you have many interesting posts, you are one of the most active forum members in posting problems, but I think the praises and compliments you utter (mostly in the "inequalities" forum) are much too exaggerated, they cause an inflation of words like "wonderful", "beautiful", "great" without actually giving any pleasure to the ones you are praising (at least not after they have seen how carelessly you handle these words), and they just disturb people trying to solve problems...

Sorry for the critics, nothing against your many contributions to ML and inequalities. I just wanted to put some things at the right place.

BTW, on your question whom you should consult to get the Old and New Inequalities book: I have it from Mircea Lascu ("Gil" on MathLinks). He visits Germany from time to time, so it's possible to buy his books in Germany, but unfortunately I don't know how often he is in Italy...

PS. I think I should also say something about some habits of other MathLinkers, namely closing sentences with rows of exclamation signs !!!!!!! and putting three smilies after each problem :D  :?  :)

  Darij

[quote="sqing"]Let  $ a,b,c,x,y,z$ be nonnegative real numbers such that
 $a+b+c+x+y+z=1$ and $abc+xyz =\frac{1}{36}$.
Prove that \[abz+bcx+cay\le\frac{1}{108}.\][/quote]
When does the equality hold? :maybe:

$a=b=\frac{1}{3},c=\frac{1}{4},x=y=0,z=\frac{1}{12}.$

[url=http://www.wmtc-math.net/cn/questions2011.html]It is the probleme 0f  WMTC  2012[/url]


calculus
real analysis
real analysis unsolved

If you posted something here and can't find it, then your topic may have been moved.

If you are taking AP Calculus (AB or BC) or a college "regular" calculus course, or are self-studying calculus, and you posted something that could have been a classroom topic (or quiz, or homework exercise) from that course, then it has been moved to the "Calculus Computations and Tutorials" forum. Look for it there.

If your post did not involve calculus or analysis at all, then it has been moved somewhere else. These often wind up in one of the high school forums. One way to locate it: Look up your own profile, and click on the "show all posts" option. You should be able to recognize and locate your topic that way.

I am just curious, but do you move solved problems as well so that if a problem is solved it is moved from the "Unsolved and proposed problems" section to "Calculus tutorials".


calculus
integration
algebra
polynomial
Vieta
complex numbers
system of equations

i can't do this [color=red][b]thing[/b][/color] !  :w00t:

$\int_{-\infty}^{\infty}\;\left( \frac{x^{2}}{1+4x+3x^{2}-4x^{3}-2x^{4}+2x^{5}+x^{6}}\right) \;dx\;=\;\boxed{\pi}$

can [color=blue][b]you[/b][/color]  ?  :rotfl:

The problem can be done in 3 steps:

1) Factor the polynomial $Q(x)$ in the denominator as $P(x)\overline{P(x)}$ where $P(x)=x^{3}+(1+i)x^{2}-(2-i)x-1$. (Can you see how to guess it?)

2) Prove that all the roots of $P(x)$ lie in the lower half-plane $\text{Im\,}z<0$ (Can you do that?)

3) Use the residue theorem and Vieta formulae to compute the integral (Do you have enough patience for this step?)

If I haven't mistaken anywhere, the answer is $\pi$, indeed :).

[quote="Fedja"]Can you see how to guess it?[/quote]

Question 1: How can we see that?

Question 2: Can't that polynomial be factorized in others polynomials with real coefficients?

Thanks.

The idea was to use the residue theorem to say that the integral in question is "just" $2\pi i$ times the sum $\sum_{k=1}^{3}\frac{\xi_{k}^{2}}{Q'(\xi_{k})}$ where $\xi_{1,2,3}$ are the roots of $Q$ with positive imaginary part (so the full list of roots consists of these three roots and their conjugates). Now, it is clear that $Q(x)=P(x)\overline{P(x)}$ where $P(x)=(x-\xi_{1})(x-\xi_{2})(x-\xi_{3})$ and that the sum to find is symmetric in $\xi_{1},\xi_{2},\xi_{3}$. So, our only hope is that the coefficients of $P(x)$ are something nice, and "nice" in this context would mean Gaussian integers. Now, assuming that $P(x)=x^{3}+ax^{2}+bx+c$ and writing down the system of equations for the coefficients, we get
\[\begin{cases}a+\bar a=2 \\ |a|^{2}+b+\bar b=-2 \\ a\bar b+\bar a b+c+\bar c=-4 \\ |b|^{2}+a\bar c+\bar a c=3 \\ b\bar c+\bar b c=4 \\ |c|^{2}=1 \end{cases}\]
Now, if the coefficients are Gaussian integers, $c$ may be $\pm 1$ or $\pm i$. $c=1$ doesn't work because then the first and the fourth equation together yield $|b|=1$, so we have no chance for the fifth equation to hold. $c=-1$ allows to conclude that $b+\bar b=-4$, $|b|^{2}=5$, $|a|^{2}=2$, after which we have a finite number of options and one of them works. $c=\pm i$  result in unsolvable systems as well (I leave the details to you). So, we get the factorization I mentioned in my previous post (up to notation). Now we have to pray that our factors, indeed, separate the roots the way we want, which brings us to the second step.


If ABCD is a quadrilateral which is circumscribed about a circle and AB=4, BC=5, and CD=6, then AD=

A.5
B.6
C.7
D.8
E.9

[hide]By the theorem that states that two tangents to a circle drawn from the same point are congruent, we can make the equations:

$ x\plus{}y\equal{}4$ (1)
$ y\plus{}z\equal{}5$ (2)
$ z\plus{}a\equal{}6$ (3)
$ a\plus{}x\equal{}?$ (4)

With the tangents drawn from A = $ x$
With the tangents drawn from B = $ y$
With the tangents drawn from C = $ z$
With the tangents drawn from D = $ a$

Now, its a simple system of equations:

(3)-(2)= $ a\minus{}y\equal{}1$ (5)

(5)+(1)= $ a\plus{}x\equal{}5$

So our answer is $ \boxed{A}$ or 5.

[/hide]


Construct a function $ f: \mathbb{R}\rightarrow\mathbb{R}^{\plus{}}$ such that every rational number $ q$ is a local minimum of $ f$.

You can just take the function that is identically $ 0$. You probably wanted to ask about a function having a [b]strict[/b] local minimum at every rational point. ;)

[quote="Jumbler"]Construct a function $ f: \mathbb{R}\rightarrow\mathbb{R}^{ \plus{} }$ such that every rational number $ q$ is a local minimum of $ f$.[/quote]

Hint: Let $ q_1, q_2, ...$ be the rationals and set $ f(q_n) \equal{} n$. That has the desired properties on the rationals, not the reals, but it's something to think about.

Just set $ f(q_n)\equal{}2\minus{}\frac{1}{n}$ and $ f(x) \equal{} 2$ for $ x \notin \mathbb Q$.

[quote="calc rulz"]Just set $ f(q_n) \equal{} 2 \minus{} \frac {1}{n}$ and $ f(x) \equal{} 2$ for $ x \notin \mathbb Q$.[/quote]

Sure, but I was trying to leave something for the OP to do.


Hey this came up on my math test. What is $ \sqrt{ \minus{} 8}*\sqrt{ \minus{} 40}$

[quote="hongsquare"]Hey this came up on my math test. What is $ \sqrt { \minus{} 8}*\sqrt { \minus{} 40}$[/quote]

$ \sqrt { \minus{} 8}*\sqrt { \minus{} 40}$

$ \equal{} i \sqrt{8} \times i \sqrt{40}$

$ \equal{} 2i \sqrt{2} \times 2i \sqrt{10}$

$ \equal{} (2i)^2 \times \sqrt{2 \times 10}$

$ \equal{} 4 \times \minus{}1 \times 2 \sqrt{5}$

$ \equal{} \minus{} 8 \sqrt{5}$

What if you do

$ \sqrt{\minus{}8} * \sqrt{\minus{}40} \equal{} \sqrt{8 * 40} \equal{} 8\sqrt{5}$ ?

Would it mean that the answer is $ \pm 8 \sqrt{5}$ ?

iirc, $ \sqrt{a}*\sqrt{b}$ ≠ $ \sqrt{ab}$ when $ a$ and $ b$ are negative.

I think I also remember that "rule," but can't figure out why it is true. Why is that true?

[quote="Ihatepie"]iirc, $ \sqrt {a}*\sqrt {b}$ ≠ $ \sqrt {ab}$ when $ a$ and $ b$ are negative.[/quote]
Really? But I thought the only reason we could say that $ \sqrt{\minus{}4} \equal{} 2i$ is because $ \sqrt{\minus{}4} \equal{} \sqrt{\minus{}1} \cdot \sqrt{4} \equal{} 2i$. So clearly, we need that identity to hold for at least $ a,b \equal{} \minus{}1$, otherwise we wouldn't be able to perform operations as shown.

[quote="TachyonPulse"][quote="Ihatepie"]iirc, $ \sqrt {a}*\sqrt {b}$ ≠ $ \sqrt {ab}$ when $ a$ [b]and[/b] $ b$ are negative.[/quote]
Really? [/quote]Yes, really, notice the conjunction '[b]and[/b]'.

wasn't that a problem when they first "adopted" not-real numbers?

they surpassed it by making a rule that as soon as you have sqrts with negative numbers they have to be converted in terms of the imaginary unit


Hi
This maybe a simple sum. :maybe:
A Man may take a step forward,backward left or right with equal probability.
Find the probability that after nine steps he will be just one step away from his initial position.

Here's a starter:  The set of all locations that he may end up at is a square with side length 18 steps.  So there's only 324 possible locations he can be.

[hide="Solution"]The man must take an even number of steps in one direction and an odd number of steps in the other direction.  WLOG assume he takes an even number of forward/backward steps.  If he takes $ 2n$ forward/backward steps, then he must take $ n$ forward steps and $ n$ backward steps.  The probability of this is $ \frac{\binom{2n}{n}}{2^{2n}}$.  The man takes $ 9\minus{}2n$ left/right steps, so he must either take $ 4\minus{}n$ steps to the left or $ 5\minus{}n$ steps to the left.  The probability of this is $ \frac{\binom{9\minus{}2n}{4\minus{}n}\plus{}\binom{9\minus{}2n}{5\minus{}n}}{2^{9\minus{}2n}}\equal{}\frac{\binom{10\minus{}2n}{5\minus{}n}}{2^{9\minus{}2n}}$.  Summing over all possible $ n$, our answer is

$ \sum_{n\equal{}0}^{4}\frac{\binom{2n}{n}}{2^{2n}}\cdot\frac{\binom{10\minus{}2n}{5\minus{}n}}{2^{9\minus{}2n}}\equal{}\sum_{n\equal{}0}^{4}\frac{\binom{2n}{n}\binom{10\minus{}2n}{5\minus{}n}}{2^{2n}}$

which can be computed pretty easily.  :) [/hide]

But the ans is wrong
ans given was
$ \frac{3969}{16384}$

matt276's problem involves a failure to consider the positions in which ups and downs appear relative to lefts and rights.

Here's how I would go about it: [hide]such paths are in obvious bijection with paths of length 10 returning to the original location.  The total number of such paths is $ \sum_{i \equal{} 0}^5 \binom{10}{2i}\binom{2i}{i}\binom{10 \minus{} 2i}{5 \minus{} i}$.  (The first factor in this product is what matt276's solution didn't take into account.)  Then the answer is just this number divided by $ 4^9$.[/hide]

Thanks JBL, I see my mistake.  Your solution looks right.  :)


Well......I think there might be pictures in here...I just don't know!

Hey guys, here's a look at my school :) . Jorian

what does this mean

ur prob wondering how i came to put those numbers in... :P
-jorian


geometry
3D geometry
algebra
linear equation

Substitution, then algebra. x_x.

$\sqrt{x}=y$ so  $\sqrt{y+2}-\sqrt{y-2} = \frac{2}{3}.$

Upon squaring both sides, we get $2y - 2\sqrt{y^2-4} = \frac{4}{9}.$

Now rearrange and divide by 2 to get $y-\frac{2}{9} = \sqrt{y^2-4}.$

Square both sides, and we yield $y^2 - \frac{4}{9}y + \frac{4}{81} = y^2-4.$

Now one can see that the $y^2$ terms cancel out, so you have a linear equation where you can easily find that $y = \frac{82}{9}.$

Therefore, $x = y^2 = \frac{6724}{81}.$  Tah dah.

3√(√x +2)- 3√(√x-2) =2

  I mean √x +2 and √x-2    is  2 seperate not inside of the square root.

[quote="pkothari13"]Substitution,......
so  $\sqrt{y+2}-\sqrt{y-2} = \frac{2}{3}.$
[/quote]
One trick which makes it a little simple is to remember $(a+b) = \frac{a^2-b^2}{a-b}$
$(y+2) - (y-2) = 4$
and  $\sqrt{y+2}-\sqrt{y-2} = \frac{2}{3}$
so   $\sqrt{y+2}+\sqrt{y-2} = \frac{4 \cdot 3}{2}= 6$
Adding and dividing by 2,  we get $\sqrt{y+2}= \frac{10} 3$  and you get the value of $y=\frac{100} 9 -2$, perhaps , with a little less  work :)
(Hope this is useful)

PS - This trick is useful  in many such cases ...

$(\sqrt{x}+2)^\frac{1}{3}-(\sqrt{x}-2)^\frac{1}{3}=2$

Correct me if I'm wrong but is this the problem you want solved xcidav?


logarithms
integration
real analysis
real analysis unsolved

Mathematica says that
\[ \sum_{n\in N}\frac {1}{2^n n^2} \equal{} \frac {\pi^2 \minus{} 6(\log 2)^2}{12}
\]
which facts should be used to prove it ?

This has already been solved on this forum. In short:

1) Set $ f(x) \equal{} \sum_{n \equal{} 1}^{ \plus{} \infty}\frac {x^n}{n^2}$. We want f(1/2), and we know f(1).

2) Then $ f'(x) \equal{} \sum_{n \equal{} 1}^{ \plus{} \infty}\frac {x^{n \minus{} 1}}{n} \equal{} \minus{} \frac {\ln(1 \minus{} x)}{x}$.

3) Also $ f(1 \minus{} x) \equal{} \sum_{n \equal{} 1}^{ \plus{} \infty}\frac {(1 \minus{} x)^n}{n^2}$, and so $ f'(1 \minus{} x) \equal{} \frac {\ln x}{1 \minus{} x}$.

4) Hence: $ f'(x) \plus{} f'(1 \minus{} x) \equal{} \minus{} \frac {\ln(1 \minus{} x)}{x} \plus{} \frac {\ln x}{1 \minus{} x}$.

5) No integrate both sides between 1 and t and in the final make t = 1/2.


geometry
incenter
circumcircle
geometric transformation
geometry proposed

3 congruent circles inside a triangle such that:
1. Each circle touches two sides of the triangle.
2. The circles all pass through a common point $ O$.

Prove that $ O$, the incenter, and the circumcenter of the triangle are  collinear.

Their common point $ O$ is actually the exsimilicenter of the circumcircle and incircle. This problem appears as Problem 2137, No. 3, Volume 23, Crux Mathematicorum.


Let A be a finite set of positive integers.
Prove that there exists a finite set B which satisfies :
(1) all elements of B are positive integers ;
(2) A is a subset of B ;
(3) :Sigma:x is in Bx2 = :pi:x is in Bx
where the :pi: is the product sign.

Well, isn't this interesting enough ?  :?

It looks pretty interesting to me, but I don't know how to solve it :D  :D This is one of the problems from a round of the contest on Mathlinks. I don't know if the solution can be found there, but I want to try to solve it  before I look.

I solved it in http://www.mathlinks.ro/Forum/viewtopic.php?t=24756 .
I think official solution is the same.

You had the same first solution as I did Myth.  But I just came up with a second, easier one.  Add two's to the set until P>S.  Then, just add ones.  Each 1 will add 1 to S, and leave the product unchanged.

Is that legit?

Last I checked, the sets $\{2\}$, $\{2,2\}$, $\{2,2,2,2\}$ and so on are just the same set

Depends on the context.  I wouldn't think that's true here.  Myth's soln. also fails then, because he added 2's as well.

No, a set always either contains an item (once) or not at all.  It cannot contain an item multiple times.  (If the problem wanted that concept, it would have said multiset.)

You're right that Myth originally used multiple copies of 2, but later in the thread that flaw was corrected.


Let $ p>2007$ be a prime number and $ n$ is an integer exceeding $ 2006p$.Prove that $ C^{2006p}_{n}\minus{}C^{2006}{k}$ is a mutiple of $ p$,where $ k\equal{}[\frac{n}{k}]$ is the largest integer qhich is not larger than $ \frac{n}{p}$

This is the same problem which has been locked on two other threads because it is (I think) an active contest problem.

Yes , it is from M&Y . :)
[quote="toan hoc"]Let $ p > 2007$ be a prime number and $ n$ is an integer exceeding $ 2006p$.Prove that $ C^{2006p}_{n}\minus{}C^{2006}{k}$ is a mutiple of $ p$,where $ k \equal{} [\frac{n}{k}]$ is the largest integer qhich is not larger than $ \frac{n}{p}$[/quote]
toan hoc,
Please go to the link http://www.mathlinks.ro/Forum/viewtopic.php?p=931847#931847 .

Who's toan hoc? Where're the solution for this problem?


algebra
polynomial
Vieta
algebra unsolved

Prove that the product of the two real roots of $x^{4}+x^{3}-1 = 0$ is a root of $x^{6}+x^{4}+x^{3}-x^{2}-1 = 0$

[hide="hint"]
If $a,b,c,d$ are the roots of $x^{4}+x^{3}-1 = 0$, show that $ab, ac, ad, bc, bd, cd$ are the roots of $x^{6}+x^{4}+x^{3}-x^{2}-1 = 0$. Use Vieta's formulas.[/hide]


I know this one is beyond the scope of the forum but at least adorn this forum :D
Let a bilinear operation $f(v,w)=v\times w \in \mathbb{R}^{n}$ be defined in $\mathbb{R}^{n}$,$n\geq 3$,satisfying the following properties:$v\time w$ is perpendicular to $v$ and $w$ and $|v\time w|^{2}=|v|^{2}|w|^{2}-(v.w)^{2}$.then $n=3,7$ :)

Indeed, this is not easy at all... it uses a deep result of Hurwitz. If I manage to find a nicer solution (although this is probably impossible), I will try to post it. Otherwise, I will present the steps to be made.


I have seen this problem in "kiran kedlaya's notes from the MOP"

How should the numbers 1,2,...,n be arranged in a circle to make the sum of the products of pairs of adjacent numbers as large as possible? As small as possible?

[hide]I think I got it, even if it's anything but a clean proof.

Everything comes from :

if a<b and c<d then ac+bd > ad+bc (rearrangement or just (a-b)(d-c)>0).

So to have maximum sum we have to put the big ones with the big ones and the small ones with the small ones.

Hence (for n=8) 7 5 3 1 2 4 6 8 gives the maximum value (look the zig-zag distribution).

To have minimum sum, it's the reverse.

Hence 4 6 2 8 1 7 3 5 gives the minimum value.[/hide]


Evaluate the sum:

$ \sum_{k \equal{} 2}^n \frac {k^3 \minus{} 3k^2 \minus{} 3k \plus{} 2}{(k \plus{} 1)!} \equal{} ?$.

hello, prove by induction that
$ \sum_{k \equal{} 2}^n \frac {k^3 \minus{} 3k^2 \minus{} 3k \plus{} 2}{(k \plus{} 1)!} \equal{} \frac{1}{2}\frac{6\plus{}2n\minus{}2n^2\minus{}3(n\plus{}1)!}{(n\plus{}1)!}$.
Sonnhard.

Thank you .
I would like to know also how do you came up with that result.

You know maybe another way, a tricky calcul of the sum?

[hide="Hint!"]Write $ \frac {k^3 \minus{} 3k^2 \minus{} 3k \plus{} 2}{(k \plus{} 1)!}\equal{} \frac{A}{(k\minus{}2)!}\plus{} \frac{B}{(k\minus{}1)!}\plus{} \frac{C}{k!}\plus{} \frac{D}{(k\plus{}1)!}$. It is easy now to find A,B,C,D and to deal with the sum.[/hide]

Telescoping Method! :lol:

$ \sum_{k\equal{}2}^n \frac{k^3\minus{}3k^2\minus{}3k\plus{}2}{(k\plus{}1)!}$

$ \equal{}\sum_{k\equal{}2}^n \left(\frac{k^2\minus{}3k\minus{}1}{k!}\minus{}\frac{(k\plus{}1)^2\minus{}3(k\plus{}1)\minus{}1}{(k\plus{}1)!}\right)$


If a, b, c are real numbers, f(x) = x^4 + ax^2 + bx + c has -3, 1+i as zeros, find a, b, c.

Because the polynomial has real coefficients, you can conclude that the conjugate of $ 1\plus{}i$ is also a root of the polynomial. You have three roots now; Vieta's tells you the sum of all of them is $ 0$, so that's how you get the last one. From there you can determine the coefficients.

[quote="Domenic182"][quote="MellowMelon"]Because the polynomial has real coefficients, you can conclude that the conjugate of $ 1 \plus{} i$ is also a root of the polynomial. You have three roots now; Vieta's tells you the sum of all of them is $ 0$, so that's how you get the last one. From there you can determine the coefficients.[/quote]I might be doing this totally wrong, but I think you're telling me to do
(-3) + (1 + i) + X = 0
and solving that comes to
X = 2 - i

Now if I'm correct on this (which I'm not feeling good about), what's the next step?[/quote]
You are wrong :P. The polynomial has four roots. We know $ \minus{}3$ and $ 1 \plus{} i$. What he says is that when a polynomial with real coefficients have a complex root, then the conjugate root is also included. So it also has the root $ 1 \minus{} i$. Then to find the last you use vietas formulas :)

[quote="Domenic182"][quote="Mathias_DK"]
You are wrong :P. The polynomial has four roots. We know $ \minus{} 3$ and $ 1 \plus{} i$. What he says is that when a polynomial with real coefficients have a complex root, then the conjugate root is also included. So it also has the root $ 1 \minus{} i$. Then to find the last you use vietas formulas :)[/quote]
Ah, thank you Mathias. So the roots are $ \minus{} 3$, $ 1 \plus{} i$ and $ 1 \minus{} i$.
Now, I looked online for Vieta's formula, but all the sources made it look quite confusing. Can you show me how you would set the roots up in the formula?[/quote]
When you have a polynomial like $ x^2 \plus{} ax \plus{} b$ you can factorise it. If you know the roots, $ r_1$, $ r_2$, you say:
$ x^2 \plus{} ax \plus{} b \equal{} (x\minus{}r_1)(x\minus{}r_2) \equal{} x^2 \minus{} (r_1\plus{}r_2)x \plus{} r_1r_2$. Thus $ a \equal{} \minus{}(r_1\plus{}r_2)$ and $ b \equal{} r_1r_2$.
Third degree gives:
$ x^3 \plus{} ax^2 \plus{} bx \plus{} c \equal{} (x\minus{}r_1)(x\minus{}r_2)(x\minus{}r_3) \equal{} x^3 \minus{} (r_1 \plus{} r_2 \plus{} r_3)x^2 \plus{} (r_1r_2 \plus{} r_2r_3 \plus{} r_3r_1)x \minus{} r_1r_2r_3$ so
$ a \equal{} \minus{}(r_1 \plus{} r_2 \plus{} r_3)$ $ b \equal{} r_1r_2 \plus{} r_2r_3 \plus{} r_3r_1$ $ c \equal{} \minus{}r_1r_2r_3$.

I guess you catched the system by now? :)


calculus
derivative
function
geometry
geometric transformation

[b]First, how do you study?[/b]

I usually:
1. Take notes on any material given or any lectures
2. Quiz myself on the material
3. Ask any questions I have about the material
4. Repeat Step 2

[b]How do you take notes on lectures or chapters in textbooks?[/b]

I usually:

-Fold my sheet of paper in half and write vocabulary words or ideas on the left side and the definitions or explanations on the right side.

[b]This takes a while and takes up a lot of paper, but I can easily quiz myself. Does anyone have a better method?
[/b]

For quizzing myself, I look online and in my textbook for practice questions. I use my notes to quiz myself and ask my mom or siblings to quiz me on the material.

For studying foreign languages, I use a bunch of flash cards to remember those funky Chinese characters. I trace the characters with tracing paper and try the practice problems.

For preparing for presentations in school, I must practice in front of people. I never know how I'll do unless I'm nervous and under pressure beforehand.

I need to get more efficient. What methods do you use to learn things in school or to learn things in general?

Hello aufha,

here is a site with a strong emphasis on note taking methods, especially mind mapping:
[url]http://heuristixx.wordpress.com/[/url]

For online mind mapping try
[url]http://www.mindomo.com[/url]

Finally, here is a mindomo map on math problem solving:
[url]http://www.mindomo.com/view?m=3ff8623d06851945926c6d8cb7ba122d[/url]

Regards,
Thomas

Are you comprehending what you are writing down? Rephrasing the definitions/ideas as you write them down?

If you do that you will have a much easier time retaining the information, as it will shows that you understand the material, and rephrasing into shorthand often makes the job easier. In addition, the process of writing something down (NOT as quick as possible) in your own words helps memory retention immensely.

If you can't remember something, write it down a few times.  That will help you retain it.  It worked when I was trying to find out how to solve a rubik's cube. :)

I usually study at the last minute, or I don't study at all. I don't advise it, even though I still get A's. It makes me feel like I don't know the material, even though it isn't that hard yet.

I take a lot of notes in class, so I can refer back to the teacher's [almost] exact words and I know the proper equations to solving algebra, etc. Sometimes, I bring my math textbook home and re-read the lessons, do the practices, review again, etc.

But I'm talking about math here...

For History, Social Studies, whatever you might call it, I just read up on notes.

For English or L/A, I have no clue how to study. Read up on notes, maybe.

I'm not sure I have a better method, it's basically the same as yours...

1. Read up on material
2. Do a few exercises
3. Re-read notes, see if the text didn't hold a specific example
4. See if I had trouble on anything
5. Quiz myself on that specific thing
6. Re-read notes and text for good luck! =D

[quote="D3m0n Shad0w"][...]I don't study at all. I don't advise it, even though I still get A's.[...][/quote]

Agreed.

[quote="D3m0n Shad0w"][...]It makes me feel like I don't know the material, even though it isn't that hard yet.
[...][/quote]

Disagreed. I know the content very well from class already, though, since I've never studied in my life, maybe studying gives you deep profound revelations about the material...? I doubt it.

Yeah, I never study for anything much school related. During class, where we're forced to listen, I can remember everything.

I don't study, unless it's history class. In which case I just summarize into semi-sentences and write arrows that signify cause and effect, i.e. monkeys $ \Rightarrow$ ice cream.

There are some classes in which you must fully understand a concept (science and biology) and others in which you need sheer memorization (english and history). For science, I read the section in my textbook a few times, and my teacher will explain a little bit deeper so everything becomes clear.

[quote="7h3.D3m0n.117"]sheer memorization (english and history).[/quote]

I disagree that sheer memorization is required for english. English is about critical thinking and writing... you can't write a good essay by just memorizing methods... you need to be able to apply writing techniques. Also, you have to understand any books you read, etc.

Well, maybe not the writing/literature part. But the grammar parts require memorization (predicates, for example).

Everything is science. Math is the science of numbers, english is the science of language (well, our language, I mean), reading is the science of literature, and history is the science of... uh... history,.  :ninja: I don't see why history is all memorization, some you need to understand it.

[quote="Temperal"]I don't see why history is all memorization, some you need to understand it.[/quote]

What happened in the Third Servile War? You spent the majority of history class memorizing historical facts than understanding why or how.

math and science are also memorization, then.

How do you find the derivative on any function?

What's the power rule?

What does being continuous mean?

What is the formula for specific heat?

What is the derivative of work?

What's the SI unit for power?

All memorization.

Those, perhaps, are school kind of questions. Problems that you'd find on any "real" math (such as any competition math) aren't just asking for answers like that. You actually need to think for a creative way to find the answer.

But "real" history isn't about memorizing a list of dates and names, either.

Then what is "real" history? I've never taken a history course that required understanding of concepts.

That's a tough one. But (right now) all I have to know is when Columbus founded the Americas (1492!) and like what century the colonies were started. We didn't really have to memorize everything, just one important date.

Does that count? I'm having a feeling I'll have to memorize a lot more in the future..or just this school-year.

Well, culture is a concept you must understand. And the motivation behind certain acts, foreign policies, etc.

Let me give you an example from a lecture I sat in on last year.  The professor was Laurel Thatcher Ulrich.  You may have heard a poem (written by Longfellow) about a ride taken by one [url=http://poetry.eserver.org/paul-revere.html]Paul Revere[/url].  It's known that this poem was based on an actual historical event and also known that the poem is far from historically accurate, so there's a question left open of what actually happened.  As an example, Paul Revere says in the point, "One if by land, and two if by sea;  And I on the opposite shore will be ...."  So, one might ask, where did Paul Revere actually begin his ride?  A second question one might ask is, where did he get his horse?  (Records exist indicating that Revere, a city boy, owned neither horse nor stables.)  You can build a sequence of puzzles like this, and you can also begin to answer them based on documents and records from the time.  (For example, Professor Ulrich made the case that Revere probably began his ride in Charlestown.)  This is just one tiny example from one single lecture in an undergraduate class for non-majors, but I think it represents a clearly different approach to knowledge than rote memorization.  Of course, to study any field in depth requires a great deal of knowledge, and part of the purpose of a high school education is to provide students with sufficient background that they may have a real choice about which of many possible avenues of study to pursue.

To elaborate on JBL's point, and also sort of to explain the approach of high school teaching:

As you go back in history, less and less is known about what was actually going on. We often have written historical accounts, but these have gone through many recopyings, translations, etc. (and were often written by very biased sources, too, or under oppressive regimes). And sometimes we don't even have written historical accounts, or written accounts don't describe the things that we're actually interested in.

For example, what if we want to study economic history? 2000 years ago, nobody thought to write an analysis of the economy of the Roman empire. What can we say about the economy of imperial Rome during its early expansion? Historians can piece together various kinds of evidence. There are some papyri from Egypt which record specific transactions. These can be used to gauge relative levels of business activity over time. The quantity and purity of precious metal coinage at a given time is a useful economic indicator (finds of Roman denarii in India provide evidence of the extent of Roman economic influence in the East).

But in order to make this kind of analysis, you have to have a lot of background knowledge already. That's why you memorize a lot of facts in high school. The idea is that when you graduate from high school, you should be prepared to start doing more interesting academic work. I'm not going to comment on whether this is the best teaching strategy, but I do believe that some amount of memorization is necessary in any decent education. Don't treat it as an abstract list of facts that you just memorize, dump out on an exam, and forget. Those things you memorize are a foundation for you to understand your language (and other languages), the society you live in, the governments and cultures around the world, how the things around us came to be as they are, etc.

Maybe you think something like "well, what if I just want to be a scientist/mathematician and don't care about what's going on outside of my fields?" But that's very socially irresponsible and closed-minded. Look at Grothendieck (very politically active, and one of the greatest mathematicians of the 20th century). Or how about Linus Pauling, who won Nobel prizes in both chemistry and peace? This is your world. Don't lock yourself in a tiny corner of it.

I understand the importance and values of other subjects such as history, I just disagree with the way it's being taught.

Back on the subject of studying:

1.) Never cram everything the night before the test. Do a review each night.

2.) Take a short break after 20-30 minutes of studying. Studies have shown that a teenager loses interest and his attention span sags after prolonged studying.

3.) Don't compartmentalize. Especially in science. Be able to pull back something from the beginning of the year and describe what it is, and how it relates to the subject you are doing now.


Let $ n>2$ be an odd positive integer. Prove that there are no positive integers $ x,y,z$ such that $ x\plus{}y$ is prime and \[ x^n \plus{} y^n \equal{} z^n\]

[quote="NogarD"]Apply Fermat's last theorem, and we have the answer. Trivially :P
No really, i just had to say it ;)[/quote]
well done, such an elegant proof

 :)

[quote="t0rajir0u"]Tsk tsk.

[hide="Hint"] If $ n$ is odd, $ x \plus{} y | x^n \plus{} y^n$. [/hide][/quote]

[hide="thanks for the hint"]
assume there are such positive integers for which this equation is true
we can use t0ra's hint, say LHS is divisible by x+y
now RHS must be divisible by x+y
because x+y is prime, that means z is divisible by x+y
now RHS is $ a^n(x\plus{}y)^n$ where a is a positive integer
by whatever its called inequality $ (x\plus{}y)^n>x^n\plus{}y^n$ because they're positive integers
so that means RHS is greater than LHS
this is no longer an equation
contradiction
so there are no such x, y, z

[/hide]

I agree that (x + y)^n > x^n + y^n is always true; however, I would like to see a proof.
Anyone?


Minor base $ BC$ of trapezoid $ ABCD$ is equal to side $ AB$, and diagonal $ AC$ is equal to base $ AD$. The line passing through $ B$ and parallel to $ AC$ intersects line $ DC$ in point $ M$. Prove that $ AM$ is the bisector of angle $ BAC$.

[i]Author: A.Blinkov, Y.Blinkov[/i]

Because BC//AD and BW//AC we can get :△BMC∽△ACD；We can get ：BC=BM=BA；SO∠BAM=（∠BMA=∠MAC）[This is because ：BM//AC]；so ∠BAM=∠MAC...Q-E-D..


let $ AA_2,BB_2,CC_2$ are bisectors of ABC triangle.consider $ A_3,B_3,C_3$ are symmetries of $ A_1,B_1,C_1$ toward $ AA_2,BB_2,CC_2$ respectively.
1)prove $ AA_3,BB_3,CC_3$ are crossover
2)if ($ AA_3,BB_3)\equal{}M$ and AB=BE=AD prove that : $ IM \bot DE$

since $ AA_1,BB_1,CC_1$ are concurrent so are $ AA_3,BB_3,CC_3$.also note that the points $ A_3,B_3,C_3$ are on $ (I)$.
for the second part we must prove that $ I,M,O$ are collinear ($ O$ is the circumcenter of $ \Delta ABC$ and it's known that $ IO \perp DE$.)
also u can prove that $ \Delta A_3B_3C_3$ and $ \Delta ABC$ are homogeneous and since $ I$ and $ O$ are the circumcenters of these two triangles so they are collinear with the point of intersection of $ (AA_3,BB_3,CC_3)\equal{}M$.so we are done .

Dear Mathlinkers,
for the first part and for more precision you can see
http://perso.orange.fr/jl.ayme   vol. 3  Cinq theoremes de Nagel p. 5-9
and also Le théorème de Steinbart
For the second part: nice proof from Shoki.
Sincerely
Jean-Louis


calculus
derivative
function
logarithms
calculus computations

Hey all

I have a question on my homework that says "A function is harmonic on a set $ S$ if $ f_{xx}$ and $ f_{yy}$ are continuous and $ f_{xx}+f_{yy}= 0$ throughout $ S$. What is the largest set on which $ f(x,y) = \ln{(x^{2}+y^{2})}$ is harmonic?"

I don't really understand how to do it.

I did my calculations and found that $ f_{xx}= \frac{-2(x^{2}-y^{2})}{(x^{2}+y^{2})^{2}}f_{yy}= \frac{-2(y^{2}-x^{2})}{(x^{2}+y^{2})^{2}}$

If I add those two together, I get that $ f_{xx}+f_{yy}= 0$ always, but the $ f(x,y)$ is not always continuous since it is not defined for $ f(0,0)$

If someone knows what this question is asking, that would be a great help, I have made some progress but I do not know what to do...

Thanks,
Ankur

The function is not even defined at $ (0,0)$, so it does not have derivatives there. Therefore, $ (0,0)$ is not in the set $ S$ that you are looking for. But all other points are, as your calculations show.


trigonometry
geometry
Triangle
perpendicular bisector
IMO Shortlist

Let $\alpha, \beta$ and $\gamma$ denote the angles of the triangle $ABC$. The perpendicular bisector of $AB$ intersects $BC$ at the point $X$, the perpendicular bisector of $AC$ intersects it at $Y$. Prove that $\tan(\beta) \cdot \tan(\gamma) = 3$ implies $BC= XY$ (or in other words: Prove that a sufficient condition for $BC = XY$ is $\tan(\beta) \cdot \tan(\gamma) = 3$). Show that this condition is not necessary, and give a necessary and sufficient condition for $BC = XY$.

The other case is $\tan (\beta)\cdot \tan(\gamma)=-1$. It's easy to show that if $\angle B,\ \angle C<\frac {\pi}2$ then the necessary and sufficient condition is that which was already stated ($\tan (\beta)\cdot \tan(\gamma)=3$), and if one of them is obtuse, then $\tan (\beta)\cdot \tan(\gamma)=-1$. I guess the necessary and sufficient condition is $\tan (\beta)\cdot \tan(\gamma)$ being in the set $\{-1,\ 3\}$.


inequalities
logarithms
inequalities proposed

[color=blue]
 Let $f: I\to (0,\infty)$ be a concave function. Then the following inequality holds:

\[ f^{3}(\frac{x+y+z}{3})\cdot f(x)f(y)f(z)\leq f^{2}(\frac{x+y}{2})f^{2}(\frac{y+z}{2})f^{2}(\frac{z+x}{2}).  \], for any $x, y, z\in I$. ;)

 [/color]

[quote="Ezbakhe Yassin"][quote="Soarer"]It's just popoviciu applied to $g(x) = \ln f(x)$ isn't it.[/quote]

What does [b]popoviciu[/b] say? :?:[/quote]
[u][b]Popoviciu's Inequality[/b][/u]
If $f(x)$ is convex on $[a,b]$, then for $x,y,x\in [a,b]$, we have
$f(x)+f(y)+f(z)+3f(\frac{x+y+z}{3})\geq 2[f(\frac{x+y}{2})+f(\frac{y+z}{2})+f(\frac{z+x}{2})]$.
Equality holds when $x=y=z$.

[quote="Ezbakhe Yassin"][quote="leepakhin"]If $f(x)$ is convex on $[a,b]$, then for $x,y,x\in [a,b]$, we have
$f(x)+f(y)+f(z)+f(\frac{x+y+z}{3})\geq 2[f(\frac{x+y}{2})+f(\frac{y+z}{2})+f(\frac{z+x}{2})]$.
Equality holds when $x=y=z$.[/quote]
Are you sure it is like that?  :huh:

I have found that it is this one:

$f(x)+f(y)+f(z)+3f(\frac{x+y+z}{3})\geq 2[f(\frac{x+y}{2})+f(\frac{y+z}{2})+f(\frac{z+x}{2})]$[/quote]
Sorry.


Ok, I took this problem from USSR Olympiad Problem Book. I don't have any problems with its solution. But, I don't know what's wrong with my solution either.

Problem: Given 13 gears, each weighing an integral # of grams. It is known that [b]any[/b] 12 of them can be placed on a pan balance, [b]6 on each[/b], such that the scale will be in equilibrium. Prove that all gears must be of equal weight.

Solution: Let us divide the 13 gears into 2 groups of 6. Let's say that the two groups have weight x. If I remove a gear of weight A from Group 1, the total weight of Group 1 would be x-A. If I replace the 13th gear which has weight B, then the total weight of group 1 would be x-A+B. Because it's given that we can choose any 12 such that group 1 and group 2 would be in equilibrium, then x-A+B=x => A=B. Doing this process repeatedly, we would see that all of them have the same weight.

why did the book still consider parity and etc.? I'd appreciate if you explain what's faulty with my solution in a detailed way. thanks!

You've just proved that because $ 1 \plus{} 4 \equal{} 2 \plus{} 3$ and $ 1 \plus{} 5 \equal{} 2 \plus{} 4$, we must have $ 5 \equal{} 3$ ;)

Here are some places this problem has appeared before:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=224724
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=142892
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=48475
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=224567
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=224623

I found this by searching variously for subsets of the words "football weight basketball height parity."


Why is 0/0 indeterminate?
My friend offered me the following reason:
Say $\frac{0}{0}$ is some number $z$, then
$\frac{0}{0} = z \Rightarrow 0 = 0 \cdot z$
Thus $z$ could be any number and so it is impossible to determine the value of z. Hence 0/0 is indeterminate.

But here's my worry. To go from $\frac{0}{0} = z$ to $0=0 \cdot z$, you are actually multiplying both sides by 0! Thus,
$0 \cdot \frac{0}{0} = 0 \cdot z$ but you can't simplify the left hand side can you?!

An indeterminate number is one with indeterminable value.

So if $\frac{0}{0}$ is not indeterminate, please tell me its value.

1) It could be 0 since the numerator is 0.

2) It could be 1 since the numerator and denominator are equal.

3) It could be undefined since its denominator is 0.

So which one is it?

That's why we call it indeterminate.

yeah
here are some that i do not know why they are inderminate

$1^{\infty}$, $0^\infty$

So is there a more precise definition of what it means for something to be indeterminate?

There is no such thing as an 'indeterminate number'.  There are indeterminate limits, of the form $0/0$, $1^{\infty}$, $\infty-\infty$, etc.

I've been taught that it can be $0$, $1$, or infinity, but I interpreted it as your friend said.

Think of it this way:
Why is $\displaystyle\frac{n}{0}$ undefined, when $n$ is any non-zero number?  It is because by the definition of division, the quotient is some number $x$ for which $0*x = n$.  Since that does not exist, $\displaystyle\frac{n}{0}$ is undefined.

Take that logic and apply it back to the indeterminate form.  By the definition of division, $\displaystyle\frac{0}{0}$ is a number $x$ for which $x*0 = 0$.  however, $x$ can take on any value so the indeterminate form "needs more work" (when dealing with limits, at least).

To blahblahblah:
But what about the time when you are just doing some simple algebra and you come across $\frac{0}{0}$??

To seesaw:
Did you read my original post? What is your opinion on the argument I gave for the invalidity of this type of analysis?

For your explanation of $\frac{n}{0}$ you are actually multiplying both sides of ypur equation by $0$. Thus, on the left side you are simplifying $\frac{0}{0}$ to $1$ (same mistake that I pointed out above). This kind of argument dioesn't prove (informally or formally) anything!

[quote="swapnillium"]To blahblahblah:
But what about the time when you are just doing some simple algebra and you come across $\frac{0}{0}$??[/quote]

Then you are doing the algebra wrong.

To blahblahblah:
Do you mean to say that there is no way you can get $\frac{0}{0}$ when your are just doing algebra?

$\frac{0}{0}$ simply don't exists! Our dear mathematics, first of all, says: - Do not divide by zero!
In your conversation, you multiplied both sides of the equation by zero. This is very very wrong! Every equality becames true when both sides are mutiplied by zero.

And a lot of people forgets that infinity IS NOT a real number. We can't work infinity like a real number. Look at this...

$k = 1 + 2 + 4 + 8 + 16 + \dotsc$
$k = 1 + 2(1 + 2 + 4 + 8 + \dotsc)$
$k = 1 + 2k$
$k = -1$ ????

It happens because the first equality says that $k = \infty$...

Definitions aren't allegory!

$\displaystyle\frac{0}{0}$ does not exist because $\frac{n}{0}$ does not exist.  The function $f(x) = 0 \cdot x$ is not one-to-one, so it does not have an inverse.

For to know what is $ \frac ab$ , first you must know what is $ \frac 1b$ and in this case $ \frac ab\equal{}a\cdot\frac 1b$ !

For example, $ \frac {1}{\plus{}0}\equal{}\plus{}\infty$ , $ \frac {1}{\minus{}0}\equal{}\minus{}\infty$ . What mean these, later ... at the calculus.

There is no difference between $ \plus{}0$ and $ \minus{}0$. Even in calculus this is wrong.
It's essential not to mistake real numbers with limits.

There is some way to speak of infinitesimal numbers (surreal/hyperreal analysis), but even there you can not divide through $ 0$.

i think the $ \infty$ is something useless and wrong.
but my brain says $ \frac{0}{0}$ does exist. just because we dont know what it is, doesnt mean it doesnt exist.

[quote="binomial_4eva"]my brain says $ \frac {0}{0}$ does exist. just because we dont know what it is, doesnt mean it doesnt exist.[/quote]

Your brain has nothing to do with it.  Mathematicians spent decades wrestling with the foundations of arithmetic, and when they were done, they set up [url=http://en.wikipedia.org/wiki/Axiom]axioms[/url] on which they built up modern mathematics in a rigorous way, and in standard arithmetic, division by zero is [b]not allowed.[/b]

There is no question of "we don't know what it is."  The issue is that the quantity $ \frac{0}{0}$ cannot be assigned a consistent value in a meaningful way.  Axioms are set up to make systems [i]powerful[/i] and [i]general.[/i]  The inclusion of a value $ \frac{0}{0}$ (which you cannot do anything to) is not powerful, nor is it general, and every statement in arithmetic would have to be rewritten to include that exception.  This is not the way we do mathematics.

you know what a number is , but if u want me to say what a number is , you may look at a dictionary.
$ pony \neq number$

Most dictionarys surely do not have a mathematical definition of "number". And in my definition, a pony can be a number.

That discussion is beyond the point. The point is that claiming that a number exists completely based on intuition is both absurd and wrong.

this is easily one of the five worst threads i've ever had the misfortune of reading on AoPS and one of only a handful i can ever remember unsubscribing from.

Temperal: That's completely true! (and same for nonexistence of some numbers like "pony"  :wink: )
It's just that binomial_4eva wants to base everything on his intuition and feelings, and he is often far away from "reality"...
I agree that intuition is very important even in mathematics, but you only get that intuition by already knowing what you are talking about!
So please, dear binomial_4eva, stop talking like this.

Come on people... binomial_4eva is a troll. Stop feeding the troll.

He just seems very distrustful of anyone's mathematical conclusions except his.  The problem being that mathematics is one of the hardest subjects to learn if you don't trust what other people tell you about it...  :|

[quote="ZetaX"]No! Trolls are rare creatures! Don't kill them!
I always wanted one as a pet  :D

But in fact, I'm not completely sure about him really being a troll.[/quote]
With enough LSD, anything is possible :)

This thread has gotten off topic though, and I think that if the people visiting don't agree that $ \frac{x}{0}$ for all x is undefined, then I don't think anything we say will convince them.

I don't get it...  :maybe: what's a troll?

[i](and why hasn't this thread been locked yet)[/i]

See http://en.wikipedia.org/wiki/Troll_%28Internet%29 .
But this thread may still get funny like the other one  :wink:

I am not a troll, i really dont get it,
[quote]He just seems very distrustful of anyone's mathematical conclusions except his[/quote]
i dont distrust others mathematical conclusions. i just dont get it . i just say my opinion, nothing else.
i am really confused.how can a pony be a number
pony means a small type of horse, doesnt it ?
how can a small type of horse mean number?

it can be if we say how many numbers of pony , but pony isnt a number.

A number is an element of a ring. Any we can make any set into a ring, e.g. the set of ponys  :wink:

The problem is that you argue by your feelings, while you ignore comments by others. Please ask if you don't understand something and tell us what you don't understand.

Calling someone a troll typically means that you believe they are intentionally disruptive, rude and difficult.  binomial4eva can be difficult, but unintentionally so.

I feel there should be some joke here about ponies and multiplication, but I don't know what it is.

Incidentally, I don't think ponies can be numbers because I don't think the set of all ponies is well-defined.

We only need a set containing one single pony  :wink:

But well, I'm waiting for binomial_4eva's questions...


1. You have already invested  450 in a stock with an annual return of 9.00%. How much of an additional  900 should be invested at 10.50% and how much at 6% so that the total return on the entire  1350 is 8.67%?
 ........?????....... at 10.50%
 ........?????...... at 6.00%

--------- ------------------
2. If you borrow 450 from a credit union at 13% annual interest and 400 from a bank at 15% annual interest, what is the effective annual interest rate (that is, what single rate of interest on 850 would result in the same total amount of interest)?

.....??????.... %


function
algebra unsolved
algebra
Functional Equations

Let $f: \mathbb N \rightarrow \mathbb N$ be a function satisfying :
i ) $\forall n \in \mathbb N : f(n+f(n))=f(n)$
ii ) $\exists n_0 \in \mathbb N : f(n_0)=1$
Show that $f(n)=1$ for every $n \in \mathbb N$

Trivial.
since f(n0) =1 we have that f(n0 +i)=1 for all i>=1.
Now f(n0 -1) >=1 then f(n0-1) = f(f(n0-1) + n0-1) =1 since the exp. inside >= n0.
similarly f(n0-2)=1 ,....
-A


[quote="Beat"]Solve in integers $8x^{3}+8x^{2}y+8xy^{2}+8y^{3}=15(x^{2}+y^{2}+xy+1)$[/quote]

Multiplying by $x-y$ each side, we have $8(x^{4}-y{4})=15(x-y)+15(x^{3}-y^{3})$, or $8x^{4}-15x^{3}-15x=8y^{4}-15y^{3}-15y$

So we have to find integers $m$ and $n$ such that $f(m)=f(n)$ with $f(x)=8x^{4}-15x^{3}-15x$

It is easy to see that $f(x)$ has 2 roots $0$ and $2<r<3$ and that $f(x)\geq 0$ and f monotonous increasing for $x\geq r$ .

So, if $m \geq 3$, $f(m)=f(x)$ $\Rightarrow $ $x<0$.

Hence the solutions, if they exist, must be in $[0,r]\subset [0,3)$. So, $m$ and $n$ may only be 0, 1 or 2.
Simple check gives $f(1)=f(2)=-22$

So the only solutions are $(1,2)$ and $(2,1)$

--
Patrick


Why are you posting number theory questions in the Biology forum?

Well, to answer the question, $7|2^{n}-1 \iff 2^{n}\equiv 1\mod{7}\iff 3|n$ for $n \ge 0$.


function
number theory
relatively prime
totient function
modular arithmetic

$ 7^{7^{1000}}$

We need to find $ 7^{1000}$ mod phi(10)=4 which is 1

Units digit is therefore $ 7^1 \equal{} 7$

What is [b]mod phi(10)=4[/b]  I mean  phi(10)=4 what is it  ?

Well, phi(n)= The number of integers in [1,n] that are relatively prime to n.

A very well known result is that if $ gcd(a,n) \equal{} 1$ Then $ a^{phi(n)}$ is congruent to 1 mod n

Therefore, in this problem, we have $ 7^{4}$ is congruent to 1 mod 10.

Now, since $ 7^{1000}$ is congruent to $ ( \minus{} 1)^{1000}$ which is congruent to 1 mod 4, we know that $ 7^{1000}$ is equal to 4k+1 for some non-negative integer k.

$ 7^{7^{1000}} \equal{} 7^{4k \plus{} 1} \equal{} (7^4)^k*7$ which is congruent to $ 1^k*7 \equal{} 7$ mod 10. So the units digit is 7.

hmmm...
...I would do this (personally):
7^1=7
7^2=49
7^3=343
7^4=2401
7^5=16807
7^6=117649
7^7=823543
7^8=5764801

By now, I would have come to the conlusion that the pattern for units digits goes: 7,9,3,1,7,9,3,1,...etc

Therefore I would conclude that 7^71000 has a units digit of 1 as it is a multiple of 4 and every 4th power seems to have a units digit of 1.

(This is probably not the best way to answer the problem but for this particular problem -at least- it works)

No, he wants $ 7^{7^{1000}}$ and my way is the standard way.

uh i did in this way:

its $ 7^{7^{1000}}$

and $ 7^2$ is congruent to -1(mod10)

so $ 7^{1000}$ is congruent to $ (\minus{}1)^{500}\equal{}1$ (mod10)

therefore, the answer is $ 7^1\equal{}7$.

You can't take the exponent mod 10, you have to take it mod phi(10). It just so happened to work out to the same thing.

Firstly, we need to understand congruences. [url=http://en.wikipedia.org/wiki/Modular_arithmetic]Wikipedia[/url] has a pretty good introduction to it.

[b]Pre-problem 1.1.[/b] Is it true that $ 6\equiv5 \mod 1$?

[hide="sol"]
Yes. Also, we can have $ 6\equiv6 \mod 1$, $ 6\equiv0 \mod 1$ and so on. We can also have $ 6\equiv \minus{}1 \mod 7$.
[/hide]

Like rofler did it, we use [url=http://mathworld.wolfram.com/EulersTotientTheorem.html]Euler's totient theorem[/url] (also see [url=http://www.artofproblemsolving.com/Wiki/index.php/Euler%27s_Totient_Theorem]this[/url]), which makes use of [url=http://www.artofproblemsolving.com/Wiki/index.php/Euler%27s_totient_function]Euler's totient function[/url].

Knowing all these theorems, try solving this..

[hide="solution"]
Then $ 7^{7^{1000}}\equiv 7^1\equiv 7 \mod 10$.

Note that $ \phi(100) \equal{} 40$. The $ \phi(n)$ function is multiplicative. (A very difficult proof is provided in AoPS volume 2)

[/hide]

By the way, the topic should be Number Theory. This isn't really arithmetic. Even [i]Mathematica[/i] couldn't compute $ 7^{7^{1000}}$.

Oh wait, we use modular arithmetic. Is modular arithmetic arithmetic?

Some follow-up problems:

[list]Find the units digit of $ 3^{1001}7^{1002}{13}^{1003}$. (1983 AHSME)[/list]
[list]What is the tens and units digit of $ 7^{2008}\plus{}4^{2009}\plus{}8^{2001}$?[/list]
[list]A number which when divided by 10 leaves a remainder of 9, when divided by 9 leaves a remainder of 8, by 8 leaves a remainder of 7, etc., down to where, when divided by 2, leaves a remainder of 1. Find this number. (1951 AHSME)[/list][/b]


[color=darkblue]Find $ x \in R$ of this equation: $ \boxed{5^{1\plus{}x}\plus{}5^{1\minus{}x}\plus{}25^{x}\plus{}25^{\minus{}x}\equal{}x}$[/color]

[quote="thanhnam2902"][color=darkblue]Find $ x \in R$ of this equation: $ \boxed{5^{1 \plus{} x} \plus{} 5^{1 \minus{} x} \plus{} 25^{x} \plus{} 25^{ \minus{} x} \equal{} x}$[/color][/quote]

. Case 1 : $ x < 0  \rightarrow LHS >0> RHS$

. Case 2 : $ x \ge\ 0$ . Let$ f(x) \equal{} 5^{1 \plus{} x} \minus{} x$

$ \rightarrow f '(x)\equal{} 5^{1 \plus{} x}.ln5 \minus{} 1 > 0 \rightarrow f(x) \ge\ f(0) \equal{} 5  > 0$

$ \rightarrow LHS > RHS$

So Equality doesn't hold.

It's nice. Thank [color=darkblue]onlylove_math [/color]very much.