The iron content of hemoglobin can be determined by converting the iron to Fe2+ and titrating using MnO4-. A 5.0 g sample of hemoglobin requires 21.6 mL of 0.00310 M potassium permanganate. What is the percent of iron in the hemoglobin sample?

Start by writing the balanced equation of this redox titration.


ok, we will be starting a new round of boys vs. girls: The Revival. lets play to 5, ok? if you dont know the rules, click here: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=295442[/url]

and, since there are more boys than girls, i'll start on 20. since the girls have isabella.

20

[b]15[/b]

lol i'm a girl too, and meewhee, and westiepaw, and angel... XD


inequalities
function
number theory unsolved
number theory

Find all positive integers x,y such that: $ x^y\minus{}y^x\equal{}1$

its a special case of catalan's conjecture.
only solution isx=3,y=2.
and there 'does' exist elementary solution.

This is a particular case Mihailescu's theorem which states that the only solution in positive integers of the equation $ x^{p}\minus{}y^{q}\equal{}1$ is
$ (x,y,p,q)\equal{}(3,2,2,3)$.

if we don't consider trivial cases(like $ x\equal{}1,2,3,4$) then because of the well-know inequality we must have $ y>x$.
also we can prove that $ y\equal{}x\plus{}1$
for this ,we have just to prove that
$ x^y\minus{}y^x < x^{y\plus{}1}\minus{}(y\plus{}1)^x$
it means that when $ x$ is an integer $ >3$ then the function $ f(n)\equal{}x^n\minus{}n^x$
increases strictly in $ [x\plus{}1,\plus{} \infty)$ which implies that $ f(n)$ reaches its minimum in $ x\plus{}1$ and so $ y\equal{}x\plus{}1$.
to prove the inequality mentioned we supose that
$ x^y\minus{}y^x \ge x^{y\plus{}1}\minus{}(y\plus{}1)^x \leftrightarrow (y\plus{}1)^x\minus{}y^x \ge x^y(x\minus{}1)$
if we divide both sides by $ x^y$ then we get

$ \frac{(y\plus{}1)^x}{x^y} > \frac{(y\plus{}1)^x\minus{}y^x}{x^y} \ge x\minus{}1$ (**)

but we have

$ \frac{(y\plus{}1)^x}{x^y}$

$ \equal{} \frac{y^x\plus{}y^{x\minus{}1}(x)\plus{}....\plus{}yx\plus{}1^x}{x^y} <$

$ < \frac{y^x\plus{}y^x\plus{}....\plus{}y^x\plus{}y^2.\binom{x}{2}\plus{}yx\plus{}1^x}{x^y}$

where the expression $ y^x$ appears $ x\minus{}2$ times in the above sum(note that we obviously have $ \binom{x}{i} < y^i$)
so we get

$ \frac{(y\plus{}1)^x}{x^y}<$

$ < \frac{y^x\plus{}y^x\plus{}....\plus{}y^x\plus{}y^2.\binom{x}{2}\plus{}yx\plus{}1^x}{x^y} <$

$ <(x\minus{}2)\plus{}\frac{y^2.\binom{x}{2}\plus{}yx\plus{}1^x}{x^y} <$

$ < (x\minus{}2)\plus{}1\equal{}x\minus{}1$

which is a clear contradiction with (**).
so we must have $ y\equal{}x\plus{}1$ and now we must solve
$ x^{x\plus{}1}\minus{}(x\plus{}1)^{x}\equal{}1$
now divide both parts by $ x^x$ and use the fact that for all $ n$ we have $ (1\plus{}\frac{1}{n})^n<3$.
done!


probability
number theory
least common multiple

Two distinct fractions are chosen randomly from the set $ \left\{\frac 13, \frac 15, \frac{7}{10}, \frac 56\right\}$. What is the probability that their sum is less than 1? Express your answer as a common fraction.

1/3

You just count all the possibilities; they're only 6. However, I'm still to lazy to list them out.

multiply each fraction by the lcm of the denominators: 3,5,10,6 which is 30 and get the new set containing
{10,6,21,25} Since everything is multiplied by 30, then the new question becomes What is the probability that the sum is less than 30? This is much easier to see because now the numbers are in integer form. since the total amount of pairs is 4 choose 2= 6 and there are two pairs of numbers that have a sum less than 30: 10+6<30 and 6+21<30, the answer is 2/6=1/3

[hide="Solution"]Listing all the sums, we get:

$\frac13+\frac15 = \frac{8}{15}$

$\frac13+\frac{7}{10} = \frac{31}{30}$

$\frac13+\frac56 = \frac{7}{6}$

$\frac15+\frac{7}{10} = \frac{9}{10}$

$\frac15+\frac56 = \frac{31}{30}$

$\frac{7}{10}+\frac56 = \frac{23}{15}$

Only 2 of the sums: $\frac{8}{15}$ and $\frac{9}{10}$ are less than 1. There are 4 choose 2 = 6 total sums.

So the probability is $\frac{2}{6} = \boxed{\frac13}$.[/hide]


Sharks, drowning, and that duck from Aflac. lol

[quote="tcjy"]What's your worst fear? Mine:

Spam
Rick Astley
Pokeplushies
living dead carcasses which rise fromn the dead and hide in your closet
high voltage towers

POKEPLUSHIES

MUD KIPZ

and  :P
Yes, all are true.[/quote]

all of them are found in my blog except for
living dead carcasses which rise fromn the dead and hide in your closet
high voltage towers
i'm scared of high voltage towers too...

[quote]
Art of problem solving shutting down randomly....... FOREVER  :(  :(  :( [/quote]

Me too.
and spiders *shivers*

I'm not that scared of spiders :D
\
But I'm scared of :spam:

[quote="louiethechewy123"]Art of problem solving shutting down randomly....... FOREVER  :(  :(  :([/quote]

Meh, I still have DATS.us to go to. The Internet is not AoPS.

Um, my worst fear... is the people on XKCD forums, for reasons not explained.

In order:

Spiders, big or small but especially the large ones
Cockroaches *thinks of that awful time in that museum, in the bug section...*
Other bugs
AoPS shutting down
Failing math forever

AoPS shutting down
getting 5th place in states
Not making states
Not making the school team

SPIDERS. SPIDERS. SPIDERS.  :spider:
God I hate them.

Other than spiders:
Fanfiction.net shutting down forever. (I LIVE on that site!)

Gmail shutting down forever. (I live there, too.)

YouTube shutting down forever. (I also live there.)

Word Processor crashing. (Writing is my life, and all my stories are on the computer.)

Disappearance of the Internet. (omigod that'd be hell...)

youtube shutting down (i can't live without it)

phone companies shutting down (i also can't live without it)

I'm (deathly) afraid of (in this order)....

Spiders.  Not the little ones.  Just the medium-huge ones.

Any gross bug.  I don't like bugs...

Failing ANYTHING. >_<

(just 'cause I need a spammy one...)
My calculator dying when I really need it.

 :D

Mine is getting a B or lower in math, astronomy.

My worst fiction fears: Shades, Galbatorix, Shruikan, Kull, Urgals in general

My worst hypothetical fears: The notion of losing my Eragon books
                                        The notion of AoPS shutting down forever
                                        The notion of not being able to memorize more than 18 digits of pi

My worst real fears: Anything that is NOT tame that bites, spins webs, stings, or annoys.

[quote="pianogirl"]My worst fiction fears: Shades, Galbatorix, Shruikan, Kull, Urgals in general [/quote]

lolz its moved from LOTR to Inheritance series eh?

and didnt u read brisingr? if u havent i wont give it away but yea

im surprised no ones said heights yet. im epicly acrophobic...

BUGS!
I HATE BUGS!
They scare me soo...
anything creepy and crawly scare the heck out of me.

 :wacko:  :what?:

Oh and another fear that I haven't mentioned before;

SMILEY SPAM.

THAT SERIOUSLY HAS TO STOP KILLING SHOUTBOXES.

My greatest fear is that I might not find anything scary enough to actually frighten me....ever in my life :(  :rotfl:

How about death? Are you scared of that?

Sadly enough , though it might be hard to believe, i am not even scared of death.

Oh I actually figured out what I am scared of :(  :o

$ TIME$ :|

[quote="Xaenir"]Oh I actually figured out what I am scared of :(  :o

$ TIME$  :|[/quote]

Omigosh me too!  :) Seriously though, time is terrifying..... imagine being trapped in it or something.... [shudders] I sincerely hope humans never learn how to travel through time. The consequences aren't as simple as sci-fi movies make it.


conics
parabola
geometry
3D geometry
parameterization

Find all R's such that the circle given by y² + x² = R² is tangent to the parabola y = x² + 2.

http://www.artofproblemsolving.com/Forum/feed.php?w=509

EDIT: Prove your claim.  :P

My approach is way too easy...

[hide]Both are centered at the origin.  The parabola has a vertex at (0,2).  Perhaps R=2 is the only solution since any other circle at the origin will intersect the parabola twice and not at all.[/hide]

[hide="Proof"]$ x\equal{}0$, or otherwise we'd also have an intersection at $ \minus{}x$. So $ R\equal{}2$.[/hide]

[b]worthawholebean[/b], [color=blue]any point of tangency is also a point of intersection[/color], so we [i]want[/i] to have an intersection. If the circle were instead $ y \equal{} x^2 \color{red}{ \minus{}} 2$, we would have $ R \in \left \{ 2, \sqrt {\frac {7}{4}} \right \}$ (correct me if I'm wrong). For the second value, imagine dropping a ball into an upside-down cone -- it's tangent to the sides.

Here is a different solution that involves utilizing the sign of the x-value of the point of intersection.

[hide="Solution"]Let us consider the circle to be $ c(x) \equal{} \sqrt {R^2 \minus{} x^2}$ with $ R$ a parameter, since the bottom half will obviously not intersect the parabola.
$ \blacktriangleright$ For $ x > 0$, $ c(x)$ is decreasing with $ x$, but the parabola is increasing with $ x$, so we cannot have tangency.
$ \blacktriangleright$ For $ x < 0$, $ c(x)$ is increasing with $ x$ and the parabola is decreasing with $ x$, and again, we cannot have a point of tangency.

This leaves us to check the case of $ x \equal{} 0$, and find the unique value $ \boxed{R \equal{} 2}$[/hide]


Since \\ \\ auto-indents and it's annoying to fix it, I just use

[tex ][ /tex]

[tex ][ /tex]

In math mode, \; makes a space. I think that's a slightly smaller space than usual, but I don't know what the normal space is. Specifically, \; is useful for integrals since

[tex]\displaystyle \int f(x)\; dx[/tex]

looks much better than

[tex]\displaystyle\int f(x) dx[/tex].

\, makes a space, \; makes a space, \quad makes a space, \qquad makes a space, and I think \ makes a tiny tiny space, but I'm not sure.

[tex]
\displaystyle
\backslash ,

\displaystyle\int_a^b f(x) \, dx

\backslash ;

\displaystyle\int_a^b f(x) \; dx

quad

\displaystyle\int_a^b f(x) \quad dx

qquad

\displaystyle\int_a^b f(x) \qquad dx
[/tex]

Here are just a few examples of various horizontal spaces.

[tex]a \  b[/tex]
[tex]a \, b[/tex]
[tex]a \: b[/tex]
[tex]a \; b[/tex]
[tex]a \! b[/tex]
[tex]a \quad b[/tex]
[tex]a \qquad b[/tex]
[tex]a \hspace{1cm} b[/tex]
[tex]a \hspace{1in} b[/tex]

[code]a \  b
a \, b
a \: b
a \; b
a \! b
a \quad b
a \qquad b
a \hspace{1cm} b
a \hspace{1in} b[/code]

\quad = 1 space according to current type size
\qquad = 2 quads
\, = small space =3/18 of a quad
\: = medium space = 4/18 of a quad
\; = large space = 5/18 of a quad
\! = negative space = -3/18 of a quad
\hspace can be any size both positive and negative and can use as units
cm, mm, in, pt, bp (=1/72 in), pc (12 pt), dd (1238/1157 pt), cc (12 dd), em (width of M), ex (height of x)

[quote="zabelman"]Is there a way to skip a single line of text? I somewhat cheated by using a line with 'invisible' text by using the \phantom command, but there HAS to be a way to say "\skipline" or something. Any suggestions?[/quote]

The TeX commands \smallskip, \medskip, and \bigskip create stretchy vertical spaces of sizes 3 points, 6 points, and 12 points, respectively. They work in LaTeX, too. For larger spaces, use the \vspace{} command, such as
\vspace{20pt plus 1pt} or \vspace{\baselineskip}.

Erin Schram

[quote="hatsup123"]can't you use \paragraph ????[/quote]

\paragraph is a section-defining command that adds a heading to a paragraph and an entry to the table of contents. Perhaps you meant \par, the command to end the current paragraph?

\text{Testing }\begin{minipage}[t]{1in}One\par Two\par\par Three\end{minipage}

[tex]\text{Testing }\begin{minipage}[t]{1in}One\par Two\par\par Three\end{minipage}[/tex]

As you can see, two \par commands in a row do not create a blank line.

Erin Schram


hello, defining $ f(x)=2^y+y^2-17$ we have $ f^'(y)=2^y\ln(2)+2y=0 \Leftrightarrow y_{Min}=-\frac{\text{LambertW}\left(\frac{1}{2}\ln(2)^2\right)}{\ln(2)}$. Further we get $ \lim_{y \to \pm\infty}f(y)=+\infty$, hence we have to intersection points with the $ x$ axses, $ y_1\approx -4,116106516$ or $ y_2=3$.
Sonnhard.


http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=1995

What's the question asking for?  I don't understand why the phrase "at least" is used...

That just means the area of their union, which is the area that is covered by at least one of the five squares.  (It may be covered by two or more, but it still counts only once in the union.)

Ahh thanks.  It's so clear now.

Is the problem a simple matter of calculating the total area and then subtracting the intersections of the areas once?


calculus
function
derivative
analytic geometry
graphing lines
slope
integration

So i was expirementing with my Ti-89 today and i came across the calc functions of "[i]d[/i] Differentiate" and A freaky wavy line that says "integrate " next to it.  I also see "limit" and "ln(".  Can someone please explain theese concepts to me or give me a link to somewhere where i can learn about them?  Also i have another question: What is so important about the number e?  What is e anyways?

[quote="daermon"]this needs to be moved :maybe:
no calc in HS basics[/quote]
sorry, i didnt know where to put it.

Well, this is in the wrong forum, but I'll answer your question anyway since it will probably be moved to the correct forum.

Differentiation is finding the instantaneous rate of change, or the slope, of any function.  If you have a linear equation
$y=mx+b$, then the slope is $m$, right?  What about functions that aren't linear?  Do they have slopes too?  The answer is yes, and differentiation is basically trying to find those.  There are many more advanced applications to economics, the sciences, optimization, etc.  But that is the basic idea.

Integration is trying to find the area under a specific region of a graph, or the area between the graph and the x-axis.  It gets complicated when the function is negative, but that's it at a high level.  You can Wikipedia it if you want.

The ln, or natural logarithm, is using $e$ as the base.  It just tells you what $e$ needs to be raised to to get your number.  $e$ is an important number that ranges from graphs to finance and interest.  The graph of $e^{x}$ increases more quickly than any polynomial, and the slope of the graph is the same as the graph.  There are many definitions, but one is the limit of x as x approaches infinity of $(1+1/x)^{x}$

[quote="bpms"]The other common definition of e is
$e=\sum^{\infty}_{n=0}\frac{1}{n!}$[/quote]

Related subject: Well, we know that your definition depends on the Taylor series of $e^{x}$, which relies on the fact that $\frac{d}{dx}e^{x}=e^{x}$. Can you prove that fact given the definition $e=\lim_{n\rightarrow\infty}(1+\frac{1}{n})^{n}$? In other words, can you prove that your definition and Aegor's compound-interest definition are one and the same?

This should probably be moved to Calculus now.

[quote="bpms"]The other common definition of e is
$e=\sum^{\infty}_{n=0}\frac{1}{n!}$[/quote]

Or as the unique $a>1$ such that $\int_{1}^{a}{\frac1x}dx = 1$ ...


geometry
inequalities solved
inequalities

If $ a$, $ b$, $ c$ are nonnegative reals, then prove that:
$ \left(a^2\plus{}b^2\plus{}c^2\right)\left(\minus{}a\plus{}b\plus{}c\right)\left(a\minus{}b\plus{}c\right)\left(a\plus{}b\minus{}c\right)\leq abc\left(ab\plus{}bc\plus{}ca\right)$.

It is obvious that we can assume that a,b,c are sides of a triangle. Let S its area and R its circumradius. Then (b+c-a)(c+a-b)(a+b-c)=16S 2 /(a+b+c)=(abc) 2 /R 2 (a+b+c). Using that R 2 \geq (  \sum a 2 )/9 and that (a+b+c)(ab+bc+ca) \geq 9abc, the conclusion easily follows.

[quote="harazi"] Using that R 2 \geq (  \sum a 2 )/9 [/quote]

I had the rest of the proof, but I didn't know this one. How's this named? Pretty interesting thing if this exists :D

It follows from Leibnitz relation OG 2 =R 2 -1/9* \sum a 2


function
combinatorial geometry
advanced fields
advanced fields unsolved

Let $E$ be a Banach space.
A function $f: E \to E$ is called compact if the image of any bounded set is relatively compact. Show that a continuous and compact function $f: E \to E$ that satisfies $|f(x)|_{E}\leq a|x|_{E}+b$, where $0 \leq a < 1$, has a fixed point.

For large enough $r,\ f$ is a self-map of the closed ball $B_{r}$ of $E$ of radius $r$ centered at the origin.

[url=http://planetmath.org/encyclopedia/SchauderFixedPointTheorem.html]Schauder's Fixed Point theorem[/url] states that a continuous self-map $\varphi$ of a compact convex subset $B$ in a normed space $X$ has a fixed point. Apply this to $X=E$ with $B$ being the closed convex hull of $f(B_{r})$, and $g$ being the restriction of $f$. The conclusion follows.


probability
combinatorics unsolved
combinatorics

hello
a lecturer gives 10 question in an exam, students have to answer 5
so the answer is there are 10C5 combinations yes.....

A student has prepared 7 of these. what are her chances of answering all the q's on the exam? is this 7C5/10C5, i doubt it as the probability of this is very low

(ii) at least 4 questions on the exam
is this 7C4*3C1/10C5 + 7C5/10C5

(iii) a student wants to have a probability of .75 of being able to answer all the questions on the paper. how many out of 10 must she prepare?
is this .75/x= 10C5

Thanks

You're answers to the first two are correct (as far as I can see) - but the answer to the third one is that if she wants a 0.75 chance of being able to answer all five questions, she'll need to prepare them all.

Her chances of being able to answer all five questions having prepared $n$ of them are ${n\choose 5}\left/{10\choose 5}\right.$, which increases with respect to $n$ - but ${9\choose 5}\left/{10\choose 5}\right.=\frac{1}{2}<0.75$, so even nine questions prepared is not enough to get the required ratio.

If you meant that she wants to have enough questions prepared that her expected score would be at least 75%, then the answer becomes eight questions, which may have been what you had in mind.


[quote="bigknick"]The incircle with radius 1 of triangle ABC touches that midsegment of the triangle whichis parallel to side AC. Find the length of AC if the cos of angle ABC=.8[/quote]

(AM=MB , BN=NC ) MN midsegment  parallel to side AC  -----> MN+AC=AM+NC (AB=c , BC=a , AC=b )  <---> $ \frac{b}{2}\plus{}b\equal{}\frac{a}{2}\plus{}\frac{c}{2}$ --->  $ 3b\equal{}a\plus{}c$
($ \angle ABC\equal{}\beta$) $ sin\beta\equal{}\sqrt{1\minus{}cos^{2}\beta}\equal{}\sqrt{1\minus{}0.8^{2}}\equal{}0.6$  --->

$ r\equal{}\frac{S}{P}\equal{}\frac{\frac{1}{2}acsin\beta}{\frac{a\plus{}b\plus{}c}{2}}\equal{}\frac{0.6ac}{4b}\equal{}1$

$ ac\equal{}\frac{4b}{0.6}\equal{}\frac{20b}{3}$

$ b^{2}\equal{}a^{2}\plus{}c^{2}\minus{}2ac cos\beta\equal{}(a\plus{}c)^{2}\minus{}2ac\minus{}2ac*0.8\equal{}(a\plus{}c)^{2}\minus{}3.6ac\equal{}(3b)^{2}\minus{}3.6*\frac{20b}{3}\equal{}9b^{2}\minus{}24b$ ---> $ 24b\equal{}10b^{2}$ ----> $ b\equal{}2.4\equal{}AC$


group theory
abstract algebra
superior algebra
superior algebra solved

Show that a group $ G$ is not the union of two proper subgroups. Can a group be the union of three proper subgroups? [page 105]

First one is very well known: If $ U_{1}\cup U_{2}$ is a subgroup of G iff $ U_{1}\leq U_{2}$ or $ U_{2}\leq U_{1}$.

For the second question consider $ G\equal{}V_{4}\leq S_{4}$. In this group we have:
$ V_{4}\equal{}\{id,(12)(34)\}\cup\{id,(13)(24)\}\cup\{id,(14)(23)\}$


calculus
integration
Diophantine equation
Diophantine Equations

Let $a, b, c, d, e, f$ be integers such that $b^{2}-4ac>0$ is not a perfect square and $4acf+bde-ae^{2}-cd^{2}-fb^{2}\neq 0$. Let \[f(x, y)=ax^{2}+bxy+cy^{2}+dx+ey+f\] Suppose that $f(x, y)=0$ has an integral solution. Show that $f(x, y)=0$ has infinitely many integral solutions.

We has a result in Pell's equation:
[b]1.[/b]Consider the equation:
$ ax^2 \minus{} by^2 \equal{} n$ where $ a,b\in N$,$ n$ is diffent from 0 and $ ab$ is not a square .
If the equation has a positive solution then it has infinite solution.
Consequence:
If $ c$ is different from 0,$ a$ is a solution and isn't a perfect square,$ b^2 \minus{} 4ac$ different from 0 then the equation $ ax^2 \plus{} bx \plus{} c \equal{} y^2$ if has a positive solution then it has infinite solution .(1)
Now consider the $ f(x,y)$
$ f(x,y) \equal{} ax^2 \plus{} x(by \plus{} d) \plus{} cy^2 \plus{} ey \plus{} d$
$ \Delta \equal{} (by \plus{} d)^2 \minus{} 4a(c^2 \plus{} ey \plus{} f) \equal{} (b^2 \minus{} 4ac)y^2 \plus{} 2y(bd \minus{} 2ac) \plus{} d^2 \minus{} 4af$
We prove that exist infinite $ y$ to $ \Delta \equal{} m^2$
Enough to prove it satisfy condition (1)
a)Because the equation has a solution then exist an integer $ y$ to $ \Delta$ is a perfect square
b)$ b^2 \minus{} 4ac > 0$ and it is not an perfect square.(From the condition of this problem).
c)$ \Delta_{1}' \equal{} \minus{} 4a(acf \plus{} bde \minus{} ae^2 \minus{} cd^2 \minus{} fb^2)$ is diferent from 0.
So it has many integer solution.

What do you mean by '$ a$ is solution'? And how do you transform to a pell equation on that same line?

On the $ \Delta$: how do you know $ (b^2\minus{}4ac)$ isn't a perfect square?

On $ \Delta_1$, can it be you have a typo there? Else I don't understand what you want to say...


1. Find Daryl's age, in years, given that he is two-thirds as old as he will be eight years before he is twice as old as he is now.

2. Given $ 7a^2\plus{}3a^2b\plus{}14ab\minus{}2ab^2\equal{}61$ and $ 7b^2\minus{}3a^2b\plus{}2ab^2\equal{}2$, find the value of $ |a\plus{}b|.$

[hide]1. 16 years
2. 3 [/hide]

2. Add them up: You get 7(a+b)^2 = 63, so a+b = +-3 and |a+b| = 3.

[hide="2."]
Given $ 7a^2 + 3a^2b + 14ab - 2ab^2 = 61$ and $ 7b^2 - 3a^2b + 2ab^2 = 2$, find the value of $ |a + b|.$

First off, this problem is a lot easier than it looks. Set it up like this:

  $ 7a^2 + 3a^2b + 14ab - 2ab^2 = 61$
[u]+$ 7b^2 - 3a^2b + 2ab^2 = 2$[/u]
$ 7a^2 + 7b^2 + 14ab = 63$

(Notice how the $ + 3a^2b, - 3a^2b, + 2ab^2, and - 2ab^2$ all drop out. Nice, huh? :wink:)

Now what you're left with isn't so bad. Divide by 7.

${{{ frac\{{7a^2}} {7} + frac\{7b^2}{7} + frac\{14ab} {7} = \frac {63}{7}$

Now you have: $ a^2 + b^2 + 2ab = 9$.

Once you rearrange it, it might look familiar. Now it looks like this:

$ a^2 + 2ab + b^2 = 9$

Now you factor, giving you:

$ (a + b)^2 = 9$

Obviously, (a+b) is now [u]+[/u] 3. Since it's the absolute value, it's a $ + 3$.

This --[$ (a + b)^2 = a^2 + 2ab + b^2$] is known as the sum of squares (I think), and is a typical pattern. It's a good idea to memorize it.

[/hide]

I'll fix the LaTeX later. My parents are yelling at me to get to sleep.


Let $ x,y,z>0$ and $ xyz\equal{}1$. Prove that: $ \frac{1}{1\plus{}x^3\plus{}y^3}\plus{}\frac{1}{1\plus{}y^3\plus{}z^3}\plus{}\frac{1}{1\plus{}z^3\plus{}x^3} \le 1$

Maybe this substitution could work:

Let $ a = x^3, b = y^3, c = z^3$

Therefore, $ abc = (xyz)^3 = 1$

The inequality then becomes:
$ \frac {1}{1 + a + b} + \frac {1}{1 + b + c} + \frac {1}{1 + a + c}\g\le1$

Hi Carbon57. Can you prove this ineq:
$ \frac {1}{1 + a + b} + \frac {1}{1 + b + c} + \frac {1}{1 + a + c}\g\le1$ with $ abc=1$.

Maybe...

$ \frac {1}{1 \plus{} a \plus{} b} \plus{} \frac {1}{1 \plus{} b \plus{} c} \plus{} \frac {1}{1 \plus{} a \plus{} c}\le\frac {\sqrt [3]{c} \plus{} \sqrt [3]{a} \plus{} \sqrt [3]{b}}{3}$, with the given restrictions, with equality when $ a \equal{} b \equal{} c \equal{} 1$

Since

$ \frac {\sqrt [3]{c} \plus{} \sqrt [3]{a} \plus{} \sqrt [3]{b}}{3}\ge1$, with equality when $ a \equal{} b \equal{} c \equal{} 1$, the max of $ \frac {1}{1 \plus{} a \plus{} b} \plus{} \frac {1}{1 \plus{} b \plus{} c} \plus{} \frac {1}{1 \plus{} a \plus{} c}$ is 1, and thus,

$ \frac {1}{1 \plus{} a \plus{} b} \plus{} \frac {1}{1 \plus{} b \plus{} c} \plus{} \frac {1}{1 \plus{} a \plus{} c}\le1$

I think you are wrong. Because since $ \frac {\sqrt [3]{c} \plus{} \sqrt [3]{a} \plus{} \sqrt [3]{b}}{3}\ge1$,  why you $ \Longrightarrow$ the max of $ \frac {1}{1 \plus{} a \plus{} b} \plus{} \frac {1}{1 \plus{} b \plus{} c} \plus{} \frac {1}{1 \plus{} a \plus{} c}$ is $ 1$ ?

Yeah..you're right. That's not necessarily true...I wasn't really sure about that method in the first place

[quote="thanhnam2902"]Let $ x,y,z > 0$ and $ xyz = 1$. Prove that: $ \frac {1}{1 + x^3 + y^3} + \frac {1}{1 + y^3 + z^3} + \frac {1}{1 + z^3 + x^3} \le 1$[/quote]

[hide]Observe that $ x^{3}+y^{3}=(x+y)(x^{2}-xy+y^{2}) \ge (x+y)xy$, by a Trivial Inequality Application.

So $ \frac{1}{1+x^{3}+y^{3}} \le \frac{1}{xyz+(x+y)xy}$

Observe then, that $ \frac{1}{1+x^{3}+y^{3}+\frac{1}{1+y^{3}+z^{3}}+\frac{1}{1+z^{3}+x^{3}} \le \frac{1}{xyz+(x+y)xy}+\frac{1}{xyz+(y+z)yz}+\frac{1}{xyz+(z+x)zx}$

So if we add all these up, we get that the LHS is less than or equal to $ \frac{z+x+y}{(x+y+z)xyz}=\frac{1}{xyz}=1$.[/hide]

[quote="mathgeniuse^ln(x)"][quote="thanhnam2902"]Let $ x,y,z > 0$ and $ xyz \equal{} 1$. Prove that: $ \frac {1}{1 \plus{} x^3 \plus{} y^3} \plus{} \frac {1}{1 \plus{} y^3 \plus{} z^3} \plus{} \frac {1}{1 \plus{} z^3 \plus{} x^3} \le 1$[/quote]

[hide]Observe that $ x^{3} \plus{} y^{3} \equal{} (x \plus{} y)(x^{2} \minus{} xy \plus{} y^{2}) \ge (x \plus{} y)xy$, by a Trivial Inequality Application.

So $ \frac {1}{1 \plus{} x^{3} \plus{} y^{3}} \le \frac {1}{xyz \plus{} (x \plus{} y)xy}$

Observe then, that $ \frac {1}{1 \plus{} x^{3} \plus{} y^{3}} \plus{} \frac {1}{1 \plus{} y^{3} \plus{} z^{3}} \plus{} \frac {1}{1 \plus{} z^{3} \plus{} x^{3}} \le \frac {1}{xyz \plus{} (x \plus{} y)xy} \plus{} \frac {1}{xyz \plus{} (y \plus{} z)yz} \plus{} \frac {1}{xyz \plus{} (z \plus{} x)zx}$

So if we add all these up, we get that the LHS is less than or equal to $ \frac {z \plus{} x \plus{} y}{(x \plus{} y \plus{} z)xyz} \equal{} \frac {1}{xyz} \equal{} 1$.[/hide][/quote]


[quote="ch1n353ch3s54a1l"]If A, B, C, and D represent distinct single nonzero base ten digits (0-9) for which AAAA+BBB+CC+D = 1995, then A x B x C x D = ?[/quote]

[hide]
A has to equal 1, because AAAA<1995.
Now, we have BBB+CC+D=884.
B can't equal 8 (888>884).  Let's try 7.
Now, CC+D=107.
It's clear that C has to equal 9.  Now, D equals 8.
1$\cdot$7$\cdot$9$\cdot$8=$504$.
[/hide]

[hide]The easiest digit to find is A, which equals one (if it were anything higher, then it would be larger than 1995. For B, we need to find a number where AAAA+BBB will be large enough so 1995-(AAAA+BBB) will be less than 108, which is the largest possible sum of CC+D. We find that B=7 works. Consequently, C=9 and D=8. 1*7*9*8=[b]504[/b].[/hide]

[hide]
First, $A$ has to be one as $AAAA<1995$.
So now we have
$1111+BBB+CC+D=1995$ or
$BBB+CC+D=884.$
$B$ is clearly 8 or 7.
But it would be 8 only if the tens part had no carrying, but if it is 8 then one of $C,D$ has to be 0.
So $B$ is 7, and $777+CC+D=884$
$CC+D=107$
From here we see that $D+C=7$ or 17, but if it equaled 7, no combinations work, so it has to be 17
$8,9$
Therefore C and D are $9,8$.
The product is thus:
$1\times7\times9\times8=504$.[/hide]


Assume that S = {x in R | x in (0,1)}. Why the limit points of S are all points in [0,1]? More specifically, why {0} and {1} are included?

Just use the definition of a limit point. There are two standard definitions used in most textbooks: one using neighborhoods and one using sequences.
If you prefer the one with neighborhoods, take any neighborhood of $0$, say $(-\delta,\delta)$ with $\delta>0$. Then this neighborhood contains a point in the interval $(0,1)$ (for example, the point $\frac12\min\{\delta,1\}$ is there). Thus $0$ is a limit point of the interval $(0,1)$.
If you prefer the one with sequences, just note that there exists a sequence of points of the interval $(0,1)$ converging to $0$ (for example, the sequence $\frac 12,\frac13,\frac14,\dots$).
That's why $0$ is included. The explanation of why $1$ is included is similar.


calculus
integration
trigonometry
logarithms
calculus computations

Evaluate $ \int_0^{\frac{\pi}{6}} \frac{\sqrt{1\plus{}\sin x}}{\cos x}\ dx$.

[quote="kunny"]Evaluate $ \int_0^{\frac {\pi}{6}} \frac {\sqrt {1 + \sin x}}{\cos x}\ dx$.[/quote]

note that $ \sqrt {1 + \sin x}=\cos\frac{x}{2}+\sin\frac{x}{2},\cos x=\cos^2\frac{x}{2}-\sin^2\frac{x}{2}$,the given integral can be translated into

$ \int_0^{\frac {\pi}{6}} \frac{1}{\cos\frac{x}{2}-\sin \frac{x}{2}} dx=-\frac{1}{\sqrt{2}}\int_{0}^{\frac {\pi}{6}}\frac{1}{\sin(\frac{x}{2}-\frac{\pi}{4})}dx$

let $ t=\frac{x}{2}-\frac{\pi}{4}$,integral=$ -\frac{1}{\sqrt{2}}\int_{-\frac{\pi}{4}}^{-\frac{\pi}{6}}\frac{1}{\sin t} dt=
\frac{1}{\sqrt{2}}(-\frac{1}{2}\ln(2)+\ln(2-\sqrt{2})-\ln(2-\sqrt{3}))$

That's correct. My answer is $ \sqrt{2}\ln (2\sqrt{2}\plus{}\sqrt{6}\minus{}\sqrt{3}\minus{}\sqrt{2})$.

Both Kunnys and Deng Qunyi`s answers are wrong :/ They do not even evaluate to the same number
(kunnys, gives 1.07 and Deng gives 0.30811)
Not saying that the calculations are wrong, just the answrs.

The correct answer value for this problem equals

\[ I \, = \, \sqrt{2} \left[ \ln \left( 2 + \sqrt{3} \right) - \coth^{-1}\left( \sqrt{2} \right) \right] \, \approx \, 0.6161\]

Solution of Kunny:
$ \sqrt{2}\ln (2\sqrt{2}+\sqrt{6}-\sqrt{3}-2) $
Solution of Deng:
let $ t=\frac{x}{2}-\frac{\pi}{4} $, then $dx = 2 dt$; $2$ is neglected in the result.


geometry
incenter
analytic geometry
vector

My teacher gave us an extra credit assignment over the weekend,
We have to find an algebraic method to determine the incenter of a triangle [b]given the coordinates of the vertices[/b].
And i'm flat out stuck..i've tried so hard to find a method, but I can't D:
Does anyone know a solution?
Thanks in advanced for the help ~

Vectors can probably be used to find this.

Also, maybe using the inscribed circle in a triangle drawn from the vertices will help.

The coordinates of the incenter are a weighted average of the coordinates of the vertices:

http://en.wikipedia.org/wiki/Incenter#Coordinates_of_the_incenter

Basically, you need to know the side lengths of the triangle as well as the coordinates.  However, given the coordinates, you can use distance formula to find the length of the sides.


quadratics
number theory
relatively prime
Diophantine equation

Let $ n\in N^*$. A permutation $ (a_1,a_2,...,a_n)$ of the numbers $ (1,2,...,n)$ is called [i]quadratic [/i] iff at least one of the numbers $ a_1,a_1\plus{}a_2,...,a_1\plus{}a_2\plus{}a\plus{}...\plus{}a_n$ is a perfect square. Find the greatest natural number $ n\leq 2003$, such that every permutation of $ (1,2,...,n)$ is quadratic.

[hide="Solution"]If $ 1 \plus{} 2 \plus{} \cdots \plus{} n$ is a square, then every permutation is quadratic.

Assume that $ 1 \plus{} 2 \plus{} \cdots \plus{} n$ is not a square.  We try to construct a non-quadratic permutation.  We start with the integers equivalent to 2 modulo 4.  Note that $ 2 \plus{} 6 \plus{} 10 \plus{} \cdots \plus{} (4k \minus{} 2) \equal{} 2(1 \plus{} 3 \plus{} 5 \plus{} \cdots \plus{} (2k \minus{} 1)) \equal{} 2k^{2}$, which is not a square.  Therefore if we start with the 2 mod 4 integers in increasing order, none of our sums will be squares.  We want an odd number of these 2 mod 4 at the beginning, so if necessary leave out our largest 2 mod 4 integer.

Now we just throw in the rest of the integers in order, adding $ 1, 3, 4, 5, 7, 8, 9, 11, 12,\ldots$ to the end of our permutation.  The sum up to this point is equivalent to 2 modulo 4, and the remaining sums modulo 4 are $ 3, 2, 2, 3, 2, 2, \ldots$, none of which are squares.  At the end of the sequence we can reinsert the largest 2 mod 4 integer if we left it out earlier, since we are assuming $ 1 \plus{} 2 \plus{} \cdots \plus{} n$ is not a square.  This gives us a non-quadratic permutation.

Now the problem reduces to finding the largest $ n\le 2003$ such that $ 1 \plus{} 2 \plus{} \cdots \plus{} n$ is a perfect square.  We want $ n(n \plus{} 1) \equal{} 2k^{2}$.  Since $ n$ and $ n \plus{} 1$ are relatively prime, either $ n \equal{} a^{2}$ and $ n \plus{} 1 \equal{} 2b^{2}$ or $ n \equal{} 2b^{2}$ and $ n \plus{} 1 \equal{} a^{2}$.  The first case gives the Pell equation $ a^{2} \minus{} 2b^{2} \equal{} \minus{} 1$, and the second case gives another Pell equation, $ a^{2} \minus{} 2b^{2} \equal{} 1$.  The initial solution to the first equation is $ a \equal{} b \equal{} 1$, which is represented by $ 1 \plus{} \sqrt {2}$, so all solutions to either equation are given by $ (1 \plus{} \sqrt {2})^{n}$.  Expanding gives

$ 1 \plus{} \sqrt {2},3 \plus{} 2\sqrt {2},7 \plus{} 5\sqrt {2},17 \plus{} 12\sqrt {2},41 \plus{} 29\sqrt {2},99 \plus{} 70\sqrt {2}$

The last term here produces $ n$ that is too large.  Since $ 41 \plus{} 29\sqrt {2}$ is an odd numbered term here, it corresponds to the first case above, so the corresponding $ n$ is $ a^{2} \equal{} 41^{2} \equal{} 1681$.  Since $ 17 \plus{} 12\sqrt {2}$ is an even numbered term, it corresponds to the second case, so the corresponding $ n$ is $ 2b^{2} \equal{} 288$, which is less than 1681.  Therefore, 1681 is our desired answer.

Hope I did that right...[/hide]


inequalities
logarithms
function
calculus
calculus computations

Let $a_{n} \ge 0 \quad \forall n = 1, 2, 3 ... $

If $\sum_{n=1}^{\infty} a_{n}$ converge show that $\sum_{n=1}^{\infty} \frac{\sqrt{a_{n}}}{n}$ converges.

Help :)

Cauchy-Schwarz:
$\left(\sum \frac{\sqrt{a_n}}n\right)^2 \le \sum a_n\sum \frac1{n^2}<\infty$.
In general, if $\sum a_n^2$ and $\sum b_n^2$ converge, $\sum a_nb_n$ converges.

But now I see that Cauchy-Schwarz is close to the same thing just with different comparison method.

Edit: more detail than here will follow in later posts here.

[quote="adidasty"]Could you also go about it by squaring both series and then using direct comparison?

$\sum a_{n}^2 \geq \sum \frac {a_{n}}{n^{2}} \quad \forall n = 1, 2, 3 \ldots$

So then since the former series is always less, it converges by direct comparison?

[/quote]

Why is this true? The first series doesn't necessarily dominate the second series.

nope. But from a certain number n onwards it is limited by that.

(since you know the first one converges, so a can't keep ascending.)

No, it's only true if $a_n\geq \frac {1}{n^2}$

Lemma: If $\sum a_n$ converges, and $\sum b_n$ diverges, then $\sum\frac{a_n}{n^2\cdot b_n}$ converges.
Proof of lemma: trivial.  [the common epsilon/(>>0) stuff]
[yeah b>0 but that is fulfilled from a certain n onwards by divergence]

Proof:

 --- IS $\sum \dfrac{\sqrt{a_n}}{n}$ convergent? --- [ this is $\dfrac{a_n}{n^2\cdot \left(\frac{\sqrt{a_n}}{n}\right)}$ ]

==> YES
Proven. :D

==> NO
By the Lemma, and knowing $\sum a_n$ converges and $\sum \frac{\sqrt{a_n}}{n}$ diverges, we know that the requested $\sum \frac{\sqrt{a_n}}{n}$ cannot diverge. Divergent and cannot diverge => Contradiction.

[quote="Peter VDD"]Lemma: If $\sum a_n$ converges, and $\sum b_n$ diverges, then $\sum\frac{a_n}{n\cdot b_n}$ converges.[/quote]
Counterexample: $a_n=\frac1{n\cdot(\ln n)^{\frac32}}, b_n=\frac1{n\cdot(\ln n)^{\frac34}}$
Then $\frac{a_n}{n\cdot b_n}=\frac1{n\cdot(\ln n)^{\frac34}}$, which diverges.

[quote="Peter VDD"]That said, I think you can claim much more powerful convergences here. I think you can even claim $\sum \sqrt{\frac{a_n}{n}}$ is convergent! [/quote]
No way. Counterexample: $a_n=\frac1{n\cdot(\ln n)^2}$

This was a simple problem, which really only has one solution. That solution has many names, but it's the only way to do it, and it always comes with the same estimate for the sum.

I knew I had a typo somewhere, on my paper I had twice the same function.

Sorry for the bogus lemma :D I think this one should do better. The previous one was only for the row to converge, this one also makes the sum of the row converge.

[quote="Peter VDD"]Lol... you probably mean smaller than that, you are copying adidasty's mistake ;)

I'm not claiming his inequality is true either, just that it is limited. I removed my previous proof and will try to post a clearer / more detailed one here.[/quote]

No, I'm saying that his assertion only holds under those conditions.  Which is an absurdity, of course; $\zeta(s)<\infty$ for all $s>1$, not just $s\geq 2$.  Of course, this says nothing about series that behave differently than the zeta function but still converge, which is another glaring flaw.  I agree that there is essentially only one way to do this problem; you can probably apply different inequalities

e.g $a_n+\frac {1}{n^2}\geq 2\frac{\sqrt{a_n}}{n}$.  Now, since these are absolutely convergent series, we can rearrange them, and so on, and so on, so the result follows by AM-GM

Yes, that's the way for smart people of course. :D


function
algebra
domain
calculus
integration
topology
Functional Analysis

Let $ A\subset \mathbb C$ be a closed and countable set. Prove that if the analytic function $ f: \mathbb C\backslash A\longrightarrow \mathbb C$ is bounded, then $ f$ is constant.

Since any isolated singularity must be removable, we can assume without loss of generality that $ A$ is perfect.  But perfect sets are empty or uncountable.

I knew you could remove all the isolated ones, but it was driving me crazy what to do if there was an accumulation point. This is an excellent observation.

Note there is a much harder way to do this using Runge's Theorem.

I was asked if some derived set (set of accumulation points) of a countable set must be perfect, and indeed, it seems that my solution is woefully inadequate.

However, it should be enough to show that if $ S_0 \equal{} S$ is some closed set, and $ S_{n \plus{} 1}$ is the derived set of $ S_n$ for all $ n$, then $ T \equal{} \cap S_n$ is perfect.

EDIT: Removed, because what I wrote made absolutely no sense...

I think your solution is still inconsistent. You say that $ W_y$ cover $ K$ which is not true because they cover only $ K\setminus \{x\}$.

You're quite right... so we're still lacking a solution here.  If my topological lemma is false, I'm quite curious about a counterexample...

From something lorincz sent me, and other stuff I've seen online, it appears that we really need to talk about ordinals to make this proof work.  I'm not comfortable doing this with complete rigor, but here's the idea:

The lemma I gave above is false.  It is possible that $ \cap S_n$ is not perfect, in which case we need to repeat the process.  We can of course repeat the process infinitely many times if we want, and so forth.  Ordinals provide the right measure of this repetition, and also give us a clue of how to keep expanding the domain of our function in a consistent manner, while eliminating the removable poles.

It's a theorem that we only need a countable ordinal to end up with a perfect set (implying that every closed set is the union of a countable set and a perfect set) but we may not even need this fact--if we just look at a large enough ordinal then it is clear our chain stabilizes at some point, giving us a perfect set.

This is the last I'm going to say on this topic--somebody who understands this stuff a little better should weigh in.

[quote="1234567a"]@ ehsan2004 :
Could you please post your solution using Zorn lemma here?
thanks[/quote]

I'll do so but these days I haven't got enough time. Have you seen the proof of Hahn-Banach theorem? the main idea was based on the order of domains and showing that the maximal element is $ \mathbb{C}$.

Formalizing what has been said:

Suppose that $ f$ is bounded by $ M$. Consider the collection $ \mathcal{S}$ of pairs $ (X,g)$ of open sets $ X$ such that $ \mathbb{C} \setminus A \subset X \subset \mathbb{C}$ and analytic extensions $ g$ of $ f$ to $ X$ with $ |g| \leq M$. Order this collection so that $ (X_1, g_1) \leq (X_2, g_2)$ iff $ X_1 \subset X_2$ and $ g_1(z) \equal{} g_2(z)$ for $ z \in X_1$.

If $ \{(X_i,g_i)\}$ is a chain in $ \mathcal{S}$, then clearly $ (\cup_{i} X_i, g)$, where $ g(z) \equal{} g_i(z)$ if $ z \in X_i$, is an upper bound of the chain. So by Zorn's lemma there is a maximal element $ (Y,h) \in \mathcal{S}$.

$ \mathbb{C} \setminus Y$ is closed and countable, and so is by Baire's theorem either empty or has an isolated point. If it is empty we are done, because we have extended $ f$ boundedly to the whole plane, and so it is constant. If there is an isolated point we could extend $ h$ to this point and there would be a contradiction because $ (Y, h)$ would not be maximal.

Instead of just considering sets I chose to consider pairs of sets and functions, because I am not sure how things work out with connectedness - I worry that there could perhaps be distinct analytic extensions. Probably there is not, but I took the safe route.

There is a completely different route: cover the exceptional set with open disks with a very small sum of radii and use the Cauchy integral formula ;)


integration
real analysis
real analysis unsolved

$\int ln(x)dx$ (1)

$u = ln(x)$, $du = \frac{1}{x}$ (2)
$dv = dx$, $v = x$ (3)

$\int ln(x)dx = xln(x) - x + C$ (4)

$\int ln(a + bx) dx$ (5)
$\frac{1}{b}\int ln(a + bx) bdx$ (6)

$let: u = a + bx$ (7)
$du = bdx$ (8)

$\frac{1}{b}\int ln(u)du = uln(u) - u + C$

therefore...
$\frac{1}{b}\int ln(a + bx) dx = \frac{1}{b}[(a + bx)ln(a + bx) - a - bx] + C$
which is false

i've redone the math with a + bx instead of u ([hide="here"]
$\int ln(a + bx) dx$ (5.0)

$u = ln(a +bx)$, $du = \frac{1}{a + bx}$ (5.1)
$dv = dx$, $v = x$ (5.2)

$\frac{1}{b}\int ln(a + bx) dx = xln(a + bx) - \int \frac{x}{a + bx}$(5.3)

[/hide]). and figured out why it was false, but im wondering what i'm doing wrong with substitution, is there a rule with substitution i'm breaking?

Where did that extra $\frac1b$ come from in the last statement?
Every numbered statement is correct. The last two unnumbered statements are not. You can't just move constants around ignoring their meaning.

It is easy to verify that $\frac1b((a+bx)\ln(a+bx)-a-bx)$ is an antiderivative for $\ln(a+bx)$


[quote="M.A"][quote="onlylove_math"]Put a;b;c is 3 bases  of an  triangle . Find min :

 $ \frac {a^2}{\sqrt { a^3b \plus{} b^3c}} \plus{} \frac {b^2}{\sqrt { b^3c \plus{} c^3a}} \plus{} \frac {c^2}{\sqrt { c^3a \plus{} a^3b}}$[/quote]
Only cauchy :wink:[/quote]
Really?? :o I don't think so . Can U post your proof ?

[quote="M.A"][quote="onlylove_math"]Put a;b;c is 3 bases  of an  triangle . Find min :

 $ \frac {a^2}{\sqrt { a^3b \plus{} b^3c}} \plus{} \frac {b^2}{\sqrt { b^3c \plus{} c^3a}} \plus{} \frac {c^2}{\sqrt { c^3a \plus{} a^3b}}$[/quote]
Only cauchy :wink:[/quote]
my proof is used Holder ineq and an ineq which has form shur ineq,it is not hard,but your proof was amazed to me,can you post it here


I don't understand something about phi and pi.  Supposedly phi and pi are related by an equation that looks like this: $\pi=x\phi^2$ for some value x, where x is a rational number, supposedly the reciprical of a perfect square.  Check this out http://members.ispwest.com/r-logan/index.html for that assertion.  Other places also cite it as true.

Now, again, phi can be found to be $(1+\sqrt(5))/2$ or $(-1+\sqrt(5))/2$, I'm not sure which, but regardless, it is one of the solutions to the equation $\phi^2=\phi+1$.  This means that it isn't transcendental, but still irrational.  Pi, on the other hand, is well known to be transcendental.  However, that would mean, by the formula cited in the first paragraph, that pi isn't transcendental?!  Can anyone find my mistake?

um...that definately can't be true,

if pi were equal to x*phi^2 for x rational, then that would make pi algebraic over the integers, but its been proven that pi is transcendental.

as a good rule of thumb, if a paper puts quotes from the bible at the end of the proof, then the paper is probably not correct :-)

I find the first paragraph of the proof horribly laughable. The author first makes $i$ the exponent of the exponent (instead of the exponent of the entire expression), so now $a^{bc} = a^{\left(b^c\right)}$. He also distributed an exponent over addition, so $(a + b)^c = a^c + b^c$. Oh yeah, this WORKS.

This article is akin in style to the Chinese high school teacher's "proof" of the Goldbach Conjecture in the Fun and Games section.


Alia's digital clock read 7:15 a.m. when she left for school. When she returned home 7 hours and 15 minutes later, the clock read 5:55 a.m. because the power had gone off during the day. If her clock automatically reset to 12:00 a.m. when power was restored at what time that morning did the power return?

The clock reset 5 hours and 55 minutes before she got home, which is 80 minutes after she left at 7:15. So the clock reset at 8:35.


number theory
modular arithmetic
algebra
binomial theorem
AMC

Well, Fermat's Little Theorem is the only way that will guarantee you that the answer will be nice.  Otherwise, there's a quite longish process to go through.
(equals signs mean "equivalent to mod 41"
One way:
4^1 = 4
4^2 = 16
4^4 = 16^2 = 256 = 10
4^8 = 10^2 = 18
4^16 = 18^2 = 324 = -4
4^32 = (-4)^2 = 16
Wow.  That was convenient.  Now we know
4^64 = 10
4^128 = 18
4^256 = -4
4^512 = 16
4^1024 = 10

4^2001 = 4^(1024 + 512 + 256 + 128 + 64 + 16 + 1)
= 10*16*(-4)*18*10*(-4)*4
= 10^2 * (-4)^2 * 18 * 16 * 4
= 18 * 16 * 18 * 16 * 4
= 18^2 * 16^2 * 4
= (-4) * 10 * 4
= (-4) * 40 = (-4) * (-1) = 4.

Or, you could examine 4^1, 4^2, 4^3, 4^4, . . . in hopes of a pattern.  That method sucks, though.  Fermat Little Theorem is really the way to go.

[quote="cats..."]This was from 2001 MAML Olympiad.

Find the remainder when 45^2001 is divided by 41.

I realize this is the same as finding the remainder when 4^2001 is divided by 41, but don't know where to go from there.  The solution says to use Fermat's little theorem, but can someone give me an alternative way?

Thanks.[/quote]
Can you show that finding the remainder when 45^2001 is divided by 41 is the same thing as findingthe remainder when 4^2001 is divided by 41?

And where did you find the solution? I only have the answers...

Where were you getting your questions from?  Here's where i got them: http://www.maml.org/wwwroot/sampleproblems/complete_sets/completesets.htm

Well, if you don't know anything about modular arithmetic, here's a neat way.

We can use the binomial theorem and write 45^2001 as (41+4)^2001.  This is equal to 41^2001+2001*41^2000*4^1+....2001*41^1*4^2000 + 4^2001.  Every term except the 4^2001 will have a factor of 41, so they will all leave a remainder of 0 when divided by 41.  Therefore, all we're concerned with is  the remainder when 4^2001 is divided by 41.

...and if you [i]do[/i] know about modular arithmetic, then you know that

if a :equiv: b mod c, then a^n :equiv: b^n mod c.

45 :equiv: 4 mod 41
so
45^2001 :equiv: 4^2001 mod 41

Actually, the modular arithmetic theorem a :equiv: b (mod m)  implies
a^n :equiv: b^n (mod m) for n a positive integer is proved using the argument that cats gave.


quadratics conics parabola function algebra polynomial rotation

How would you convert a quadratic equation in standard form ($ x \equal{} a(y \minus{} h)^2 \plus{} k$) to basic form ($ y \equal{} ax^2 \plus{} bx \plus{} c$)?

That wouldn't be a quadratic equation - it would be like a parabola reflected over $ y\equal{}x$, like a double-square-root function.

And also, I believe $ ax^2 \plus{}bx \plus{}c \equal{} y$ is standard form...?

[quote="qwertythecucumber"]I'm uncertain about what Yongyi said (although he's usually right). However, I'm pretty sure $ x \equal{} a(y \minus{} h)^2 \plus{} k$ is standard form.
If what Yongyi said is right (which is probably the case) then I would just like to know how to solve for $ y$ anyway.[/quote]

This is not a quadratic equation because when expanding:

$ x \equal{} a(y^2 \minus{} 2yh \plus{} h^2) \plus{} k$
$ x \equal{} ay^2 \minus{} 2ayh \plus{} ah^2 \plus{} k$

$ x$ here is a function of $ y$, which means when expressing $ y$ as a function of $ x$, we need to [i]solve[/i] for $ y$ as follows:

$ x \equal{} a(y \minus{} h)^2 \plus{} k$
$ x \minus{} k \equal{} a(y \minus{} h)^2$
$ \frac {x \minus{} k}a \equal{} (y \minus{} h)^2$
$ \sqrt {\frac {x \minus{} k}a} \equal{} y \minus{} h$
$ y \equal{} \sqrt {\frac {x \minus{} k}a} \plus{} h$

Note that this is not a quadratic equation at all.

However, I assume you mean $ y \equal{} a(x \minus{} h)^2 \plus{} k$, which expands to

$ y \equal{} a(x^2 \minus{} 2xh \plus{} h^2) \plus{} k$
$ y \equal{} ax^2 \minus{} 2axh \plus{} h^2 \plus{} k$
$ y \equal{} ax^2 \minus{} (2ah)x \plus{} (h^2 \plus{} k)$

What you did is to express $ x$ as a polynomial in terms of $ y$, while the standard quadratic equation dictates that $ y$ has to be in terms of $ x$.

That equation is (almost) in vertex form for a parabola (interpreted as a conic section) opening to the left or the right. You can't convert it to standard form because standard form can also be a function, where this isn't, and also because standard form has an "x squared" where this one has a "y squared".

Standard Form: $ y\equal{}ax^2\plus{}bx\plus{}c$

Vertex Form: $ y\equal{}a(x\minus{}h)^2\plus{}k$

If the $ y$ value is squared and $ x$ is not, then the graph would look like a sideways parabola.

And if you had these two equations:

$ y\equal{}a(x\minus{}h)^2\plus{}k$

$ y\equal{}a(x^2\minus{}2hx\plus{}h^2)\plus{}k$

$ y\equal{}ax^2\minus{}2hx\plus{}h^2\plus{}k$

So essentially, when converting vertex form to standard form,

$ a\equal{}a$

$ b\equal{}\minus{}2h$

and $ c\equal{}h^2\plus{}k$.

[quote="Yongyi781"]That wouldn't be a quadratic equation - it would be like a parabola reflected over $ y \equal{} x$, like a double-square-root function.[/quote]

To simplify, the y term has a higher power than the x term.

This would not be a quadratic , yes yongi is right.
If treat this as
y= ax^2 + C then it represents a parabola.

Though the graph of this would be similar to a quadratic equation , on expanding it'll become clear its not.

I believe that you mean ax^2+bx+c. And yes, I think we all understand that it is a parabola rotated 90 degrees, due to the x and y values being interchanged.

For more information on this subject, I would study up on conic sections.

[quote="fishythefish"]
For more information on this subject, I would study up on conic sections.[/quote]

Without linear algebra and orthogonally diagonalizable matrices, there's not much else you can do with parabolas unless their axes are parallel to the $ x\minus{}$ and $ y\minus{}$ axes, since your equation will be of the form $ Ax^2\plus{}Bxy\plus{}Cy^2\plus{}D\equal{}0$, and that $ xy$ prevents you from doing any analysis of the parabola since it is not a function.  :(

[quote="isabella2296"]And also, I believe $ ax^2 \plus{} bx \plus{} c \equal{} y$ is standard form...?[/quote]That ($ ax^2 \plus{} bx \plus{} c \equal{} y$) is [b]general form[/b].

Standard form is $ y \equal{} n(x \minus{} h)^2 \plus{} k$ (It's also known as vertex form, since the vertex is $ (h,k)$).

*This information is from an Algebra 2 textbook

[quote="fishythefish"]jvezuela, in the equation you gave, either a or c would have to be 0. Otherwise it's not a parabola, it's one of the other conics.[/quote]

Are you sure?

Below is the graph of $ 5x^2\plus{}10xy\plus{}5y^2\minus{}x\minus{}15y\plus{}3\equal{}0$. Note that I omitted the $ Dx$ and $ Ey$ terms in the previous post, as they only serve to translate the axes. However, it is [i]not[/i] true that either of $ A$ or $ C$ in $ Ax^2\plus{}Bxy\plus{}Cy^2\plus{}Dx\plus{}Ey\plus{}F\equal{}0$ be $ 0$. It is the discriminant $ B^2\minus{}4AC$ that has to be $ 0$ in order for the conic to be a parabola.


limit
logarithms
algebra proposed
algebra

For positive integers $ x$, let $ f(x)\equal{}\sum_{i\equal{}x}^{\infty} \frac{1}{i}$. Find $ \lim_{x\rightarrow \infty} f(x)$

$ f(x)$ is defined nowhere, so I'm not sure what the point of this question is.  The question you really want to be asking is about $ f_k(x) \equal{} \sum_{i\equal{}x}^{kx} \frac{1}{i}$.

[hide="Answer"] $ \ln k$.  Why? [/hide]


Solve this cryptarithm, where each of the capital letters in bold represents a different base ten digit from 1 to 9 and, [b]R[/b]  > [b]U[/b] > [b]X[/b].

([b]P[/b] + [b]Q[/b])*[b]R[/b] + ([b]S[/b] + [b]T[/b])*[b]U[/b] + ([b]V[/b] + [b]W[/b])*[b]X[/b] = 70

The solution of the above problem which conforms to all the rules as stated is given below:
(4 + 5) * 3 + (7 + 6) * 2 + (8 + 9) * 1 = 70
where
P=4, Q= 5,[b] R =3 [/b], S=7, T=6, [b] U =2 [/b], V=8, W=9 and [b] X=1 [/b].
Clearly R>U>X.
where '*' denotes multiplication


calculus
integration
geometry
3D geometry
calculus computations

$ \int 3(ice)^2 \;{\rm d}ice \equal{} ?$

 :D  :D

$ (ice)^{3}$.  I assume there's supposed to be some hidden joke but it's not funny.

it isn't $ (ice)^3$

it is [hide]

$ \int 3(ice)^2 \; {\rm d}ice \\
\equal{} (ice)^3 \plus{} C \\
\equal{} ice cube \plus{} sea \\
\equal{} iceberg$

[/hide]

That was a really stupid joke! :D
[hide="Ha! Ha! Ha!"]
This deserves [hide="an"]$ The$ $ Poorest$ $ Joke$ $ of$ $ The$ $ Year$[/hide] award!!!
[/hide]


calculus
derivative
quadratics
calculus computations

can range of the expression (t^2+4t+9)/(t^2-4t+9) (where   t=tanx)be calculated with the help of AM and GM property.ANSWER IS [1/5,5]
sorry

You left out a /. I'm pretty sure you mean $ \frac {t^2 \plus{} 4t \plus{} 9}{t^2 \minus{} 4t \plus{} 9}.$

And let's stipulate that some pretty ordinary calculus - use the quotient rule to compute the derivative, set the numerator of the resulting fraction equal to zero - works without any great difficulty. You're asking for an alternative solution.

Here's one alternative solution:

We seek a positive number $ A$ such that $ t^2\plus{}4t\plus{}9\le A(t^2\minus{}4t\plus{}9).$

Multiply this out: $ 0\le (A\minus{}1)t^2\minus{}(4A\plus{}4)t\plus{}9(A\minus{}1).$

Consideration of the behavior of the quotient as $ t\to\pm\infty$ leads to the knowledge that $ A>1.$ So we can safely divide by $ A\minus{}1$ to get

$ 0\le t^2\minus{}4\left(\frac{A\plus{}1}{A\minus{}1}\right)t\plus{}9.$

In order for that to be true for all $ t,$ we must have $ 4\left(\frac{A\plus{}1}{A\minus{}1}\right)\le 6.$ (Since $ t^2\minus{}6t\plus{}9$ is a perfect square.)

This rearranges to $ 5\le A.$ So we can say that the quotient is $ \le 5.$ At this point, replacing $ t$ by $ \minus{}t$ in the quotient leads to the other side of the inequality.

Let $ f(t) \equal{} \frac{t^2\plus{}4t\plus{}9}{t^2\minus{}4t\plus{}9}$

We note that $ f(\minus{}t) \equal{} \frac{1}{f(t)}$

Now, f(0) = 1

We take up the case when t>0. Let $ t \equal{} y^2,y>0$

Then $ f(t) \equal{} g(y) \equal{} \frac{y^4 \plus{} 4y^2\plus{}9}{y^4\minus{}4y^2\plus{}9} \equal{} \frac{y^2 \plus{} \frac{9}{y^2} \plus{} 4}{y^2\plus{} \frac{9}{y^2}\minus{}4}$

So if we let $ z \equal{} y \plus{} \frac{3}{y}$ then the expression becomes $ \frac{ \left(y \plus{} \frac{3}{y} \right)^2 \minus{} 2}{\left(y \plus{} \frac{3}{y} \right)^2 \minus{} 10} \equal{} 1 \plus{} \frac{8}{\left(y \plus{} \frac{3}{y} \right)^2 \minus{} 10}$

But from AM-GM, $ y \plus{} \frac{3}{y} \ge 2 \sqrt 3$ and so $ \left(y \plus{} \frac{3}{y} \right)^2 \ge 12 \Rightarrow 1 \plus{} \frac{8}{\left(y \plus{} \frac{3}{y} \right)^2 \minus{} 10} \le 5$

Hence $ 1<f(t) \le 5$ when t>0.

$ f(\minus{}t) \equal{} \frac{1}{f(t)} \Rightarrow \frac{1}{5} \le f(t) < 1$ for t<0.

Thus the range of f(t) is $ \left[\frac{1}{5}, 5 \right]$

$ y\equal{}\frac{t^{2}\plus{}4t\plus{}9}{t^{2}\minus{}4t\plus{}9}$

Rearranging, we have

$ (y\minus{}1)t^{2}\minus{}4(y\plus{}1)t\plus{}9(y\minus{}1)\equal{}0$

For real values of $ t$,

$ 16(y\plus{}1)^{2}\minus{}36(y\minus{}1)(y\minus{}1)\geqslant 0\vspace{1pt}$

Hence $ \frac{1}{5}\leq y\leq 5$

$ y\equal{}\frac{t^2\plus{}4t\plus{}9}{t^2\minus{}4t\plus{}9}\equal{}1\plus{}\frac{8}{t\plus{}\frac{9}{t}\minus{}4}$ for $ t\neq 0$ then you can use $ |t|\plus{}\frac{9}{|t|}\geq 6$. When $ t\equal{}0$, $ y\equal{}1$.

[quote="IndoChina"]$ y \equal{} \frac {t^{2} \plus{} 4t \plus{} 9}{t^{2} \minus{} 4t \plus{} 9}$

Rearranging, we have

$ (y \minus{} 1)t^{2} \minus{} 4(y \plus{} 1)t \plus{} 9(y \minus{} 1) \equal{} 0$

For real values of $ t$,

$ 16(y \plus{} 1)^{2} \minus{} 36(y \minus{} 1)(y \minus{} 1)\geqslant 0\vspace{1pt}$

Hence $ \frac {1}{5}\leq y\leq 5$[/quote]

Why can we use $ D\geq 0$?

Because D>= 0 will guarantee real zeroes of the quadratic equation.

I do know that, but why $ (y\minus{}1)t^2\minus{}4(y\plus{}1)t\plus{}9(y\minus{}1)\equal{}0$ is a quadratic equation?

[quote="kunny"]I do know that, but why ... is a quadratic equation?[/quote]
Considered as an equation in $ t.$

This is actually the same as my argument up in post #2 - can you understand my explanation in that post?

Yes. Your solution is good, but Japanese high school students didn't notice it.

We should argue whether the coefficient of $ t^2$ , $ y \minus{} 1$ is 0 or not.


Given a triangle $ OPQ$ such that $ PQ\equal{}3, P,\ Q$ are on the boarder of the origin $ O$ and the region $ D\equal{}\{(x,\ y)|0\leq y\leq \sqrt{3}x\}$ on the $ x\minus{}y$ plane. Find the maximum value of the inradius of $ \triangle{OPQ}.$


There is a set of 1000 switches, each of which has four positions, called $ A, B, C,$ and $ D.$  When the position of any switch changes, it is only from $ A$ to $ B,$ from $ B$ to $ C,$ from $ C$ to $ D,$ or from $ D$ to $ A.$  Initially each switch is in position $ A.$  The switches are labeled with the 1000 different integers $ 2^x3^y5^z,$ where $ x, y,$ and $ z$ take on the values $ 0, 1, \ldots, 9.$  At step $ i$ of a 1000-step process, the $ i$th switch is advanced one step, and so are all the other switches whose labels divide the label on the $ i$th switch.  After step 1000 has been completed, how many switches will be in position $ A$?

Well I knew that it had to move a multiple of 4 times but I cant quite figure out how many numbers divide each one without brute forcing everyone of them (like for a number $ 2^x3^y5^z$ I could find out if $ (10 \minus{} x)(10 \minus{} y)(10 \minus{} z) \equal{} 4n$ since for it to divide a number, the number's power for the $ 2$ has to be greater or equal to the $ x$ or less than or equal to the $ 9$ and that is $ 10 \minus{} x$ numbers, symmetry for $ y$ and $ z$ and they are between $ 0$ and $ 9$), but I didnt know how to find which $ x,y,z$ worked so I brute forced and found no answer.

Is there an easier way?  Thanks

BTW this was a number 7, I thought this would be like a number 10+

You're right in saying that $ 4|(10\minus{}x)(10\minus{}y)(10\minus{}z)$ if and only if it will end up in position A. This is true for all cases except (odd,odd,odd), (2 mod 4, odd, odd) and symmetric. (odd,odd,odd) is the case for 125 of the switches, and the others apply to 75 switches each (2, 6, 10 are the 2 mod 4 possibilities), for a total of 350 switches that are not in position A, so 650 are in position A.

You can also see [url=http://www.kalva.demon.co.uk/aime/soln/answ99.html]Kalva[/url] for another solution. That solution goes the other direction, counting all the switches that are in position A.

ah thanks, i forgot about complementary counting

i was stuck when i was trying to do PIE too many times.


$f(x)=\sqrt{x^{2}-2x+6}$ had unique fixed poind x=3. Because from equation we get $x,y,z>\sqrt 5$ and $3>f(x)>x$ if $\sqrt 5 <x<3$. If $x>3$ then $3<f(x)<x$. Therefore solution is only $x=y=z=3$.


function
algorithm
articles
algebra
system of equations

[code]So I am given three non linear equations x1(x2-2)=0,
                                                                x2(x3-3)=0,
                                                                x3(x1-1)=0
and i need to solve using newtons method using 100 randomly selected initial guesses for x1^(0), x2^(0) and x3^(0)[/code]

I don't know anything about matlab - try asking in the math forums. Maybe it has some function which will do exactly what you need, and maybe it doesn't. Maybe it is easy to implement the Newton's algorithm in matlab and maybe it's next to impossible (although I doubt this).

So I will assume we're doing this in C. First of all, what is Newton's method? Here's a nice article that quickly explains it: http://www.numericalmathematics.com/solution_of_equations.htm . Your question sounds like a school assignment - if it is, I would also recommend reading your lecture notes in case you're supposed to do it differently.

I will rename your variables $ x_1,x_2,x_3$ to $ x,y,z$ if you don't mind. From now on I will follow the article I gave you a link to. In our case:

$ F(x,y,z)=x(y-2)$
$ G(x,y,z)=y(z-3)$
$ H(x,y,z)=z(x-1)$

$ \begin{array}{lll}
F_x=y-2 & F_y=x & F_z=0 &
G_x=0 & G_y=z-3 & G_z=y &
H_x=z & H_y=0 & H_z=x-1 &
\end{array}$

So to get from $ x^{(k)},y^{(k)},z^{(k)}$ to $ x^{(k+1)},y^{(k+1)},z^{(k+1)}$ we need to solve the following system of equations:

$ \left(\begin{array}{ccc}
y^{(k)}-2 & x^{(k)} & 0 &
0 & z^{(k)}-3 & y^{(k)} &
z^{(k)} & 0 & x^{(k)}-1 &
\end{array}\right)
\left(\begin{array}{c}
\Delta x &
\Delta y &
\Delta z &
\end{array}\right)
=
\left(\begin{array}{c}
x^{(k)}(y^{(k)}-2) &
y^{(k)}(z^{(k)}-3) &
z^{(k)}(x^{(k)}-1) &
\end{array}\right)$

for variables $ \Delta x,\Delta y,\Delta z$ and then

$ x^{(k+1)}=x^{(k)}+\Delta x$,
$ y^{(k+1)}=y^{(k)}+\Delta y$,
$ z^{(k+1)}=z^{(k)}+\Delta z$.

OK, that was a pretty long post. The easiest way to solve the system of three linear equations if you have nothing but a C compiler is in my opinion with determinants (Cramer's rule). The code of the complete algorithm would look something like this:

[code]
// 100 initial guesses
for (int t = 0; t < 100; t ++) {
	double x, y, z;

	// Should probably choose a different range for the initial values
	x = rand() / RAND_MAX;
	y = rand() / RAND_MAX;
	z = rand() / RAND_MAX;

	// Let's do 250 iterations
	for (int i = 0; i < 250; i ++) {
		double det, detx, dety, detz, dx, dy, dz;

		// Calculate the determinants as required in Cramer's rule
		det = ...
		detx = ...
		dety = ...
		detz = ...

		// We can't continue if the determinant is zero
		if (abs(det) < 0.0001) {
			break;
		}

		dx = detx / det;
		dy = dety / det;
		dz = detz / det;

		x += dx;
		y += dy;
		z += dz;
	}

	// What we now have in x, y, z *might* be a solution
	printf("Possible solution: %f %f %f\n", x, y, z);
}
[/code]

If you would like a book to learn about all sorts of numerical methods, I highly recommend [url=http://www.amazon.com/Numerical-Recipes-C-Scientific-Computing/dp/0521431085/ref=artofproblemsolv-20?ie=UTF8&s=books&qid=1201217174&sr=8-2]Numerical Recipes in C[/url]. While expensive, it's the most wide-ranging numerical computation book I've ever seen, and in general one of my four favorite computer science books.


quadratics
modular arithmetic
number theory unsolved
number theory

Given:
1. q,p odd primes.
2. 2 a non-(quadratic)residue of q
3. $ \frac{q^{(p\minus{}1)}\minus{}1}{p}$ a perfect square

To prove: q,p must = 3,5.

This is exercise 10.14.10 in Uspensky and Heaslett's Elementary Number Theory.  Chapter 10 introduces Legendre and Jacobi symbols and proves the quadratic reciprocity law.

Request help.

at first we note that $ \left(\frac {p}{q}\right) \equal{} \minus{} 1$ and also $ q \equiv \pm3 \pmod {8}$.
we can rewrite the fraction in this form: $ \frac {(q^{\frac {p \minus{} 1}{2}} \minus{} 1).(q^{\frac {p \minus{} 1}{2}} \plus{} 1)}{p} \equal{} x^2$.
but we obviously have $ \gcd(q^{\frac {p \minus{} 1}{2}} \minus{} 1,q^{\frac {p \minus{} 1}{2}} \plus{} 1) \equal{} 2$,which implies that we have 2 cases:
1) $ \left(\frac {q}{p}\right) \equal{} \minus{} 1 \implies q^{\frac {p \minus{} 1}{2}} \minus{} 1 \equal{} 2z^2,q^{\frac {p \minus{} 1}{2}} \plus{} 1 \equal{} 2pl^2 ,2zl \equal{} x$.
so we get
$ 2pl^2 \minus{} 2z^2 \equal{} 2 \implies \left(\frac { \minus{} 1}{p}\right) \equal{} 1 \implies p \equiv 1 \pmod {4}$
and we have
$ \left(\frac { \minus{} 2}{q}\right) \equal{} 1,\left(\frac {2}{q}\right) \equal{} \minus{} 1 \implies \left(\frac { \minus{} 1}{q}\right) \equal{} \minus{} 1 \implies q \equiv \minus{} 1 \pmod {4} \implies q \equiv \plus{} 3 \pmod{8}$
but we have $ (q^{\frac {p \minus{} 1}{4}})^2 \minus{} 2z^2 \equal{} 1 \implies (q^{\frac {p \minus{} 1}{4}} \minus{} 1)(q^{\frac {p \minus{} 1}{4}} \plus{} 1) \equal{} 2z^2$.
Let's suppose $ p \equiv 1 \pmod {8}$,u'll get $ q^{\frac {p \minus{} 1}{4}} \plus{} 1 \equal{} 2s^2$.which is a contradiction with
$ \left(\frac {2}{q}\right) \equal{} \minus{} 1$.so we must have $ p \equiv \minus{} 3 \pmod {8}$ and so
$ q^{\frac {p \minus{} 1}{4}} \minus{} 1 \equal{} 2f^2,q^{\frac {p \minus{} 1}{4}} \plus{} 1 \equal{} m^2,fm \equal{} z$.
so we have $ q^{\frac {p \minus{} 1}{4}} \equal{} (m \minus{} 1)(m \plus{} 1)$.but this equation can easily be solved.
u'll have $ p \equal{} 5,q \equal{} 3$.
2) $ \left(\frac {q}{p}\right) \equal{} 1 \implies q^{\frac {p \minus{} 1}{2}} \minus{} 1 \equal{} 2pz^2,q^{\frac {p \minus{} 1}{2}} \plus{} 1 \equal{} 2l^2 ,2zl \equal{} x$.
but in this case $ \left(\frac {2}{q}\right) \equal{} 1$ which is a contradiction.
the problem was DELLLLLLICIOUS,THX! :D

Intriguing response, thanks.  Questions:

(A)  How do you know that $ \left(\frac {p}{q}\right) \equal{} \minus{} 1$?

(B)  Given $ \left(\frac {q}{p}\right) \equal{} \minus{} 1,$ why is it impossible that $ q^{\frac {p \minus{} 1}{2}} \minus{} 1 \equal{} 2pz^2, q^{\frac {p \minus{} 1}{2}} \plus{} 1 \equal{} 2l^2$?

(C)  Given $ \left(\frac {q}{p}\right) \equal{} 1,$ why is it impossible that $ q^{\frac {p \minus{} 1}{2}} \minus{} 1 \equal{} 2z^2, q^{\frac {p \minus{} 1}{2}} \plus{} 1 \equal{} 2pl^2$?

(D)  In case 1, immediately after proving that $ p \equiv 1 \pmod {4},$ why must $ \left(\frac { \minus{} 2}{q}\right) \equal{} 1$?

(E)  In case 1, immediately after proving that $ (q^{\frac{p\minus{}1}{4}}\minus{}1)(q^{\frac{p\minus{}1}{4}}\plus{}1) \equal{} 2z^{2},$  why does the supposition that $ p \equiv 1 \pmod {8},$ imply that $ q^{\frac {p \minus{} 1}{4}} \plus{} 1$ is of form $ 2s^2.$

(F)  Similar to question (E),  why does the supposition that $ p \equiv \minus{}3 \pmod {8},$ imply that $ q^{\frac {p \minus{} 1}{4}} \minus{} 1$ is of form $ 2f^2.$

Thanks again for Shoki's response.  Although I can't duplicate all of his
reasoning, I am able to resolve all my questions and thus validate his proof.

(A) Suppose $ \left(\frac {p}{q}\right) \equal{} 1.$ Then $ \exists a \ni p \equiv a^2 \pmod{q}.$  Therefore, since $ q^{p\minus{}1} \minus{} 1 \equal{} px^2, \left(\frac {\minus{}1}{q}\right) \equal{} 1,$ and obviously, $ \left(\frac {1}{q}\right) \equal{} 1.$
Therefore, regardless of whether $ 2pl^2 \equal{} q^{\frac{p\minus{}1}{2}} \plus{} 1$ or $ 2ps^2 \equal{} q^{\frac{p\minus{}1}{2}} \minus{} 1, \left(\frac {2p}{q}\right) \equal{} 1 \implies \left(\frac {2}{q}\right) \equal{} 1$ : contradiction.

I resolved (B) and (C) using a slightly different approach than Shoki.  The contradiction generated in case 2 does not depend on the presumption that $ \left(\frac {q}{p}\right) \equal{} 1,$ so this contradiction simply establishes that it is impossible for $ q^{\frac {p \minus{} 1}{2}} \plus{} 1 \equal{} 2l^2.$

Therefore, $ q^{\frac {p \minus{} 1}{2}} \minus{} 1 \equal{} 2z^2, q^{\frac {p \minus{} 1}{2}} \plus{} 1 \equal{} 2pl^2.$ Thus, the first part of case 1 is established, except for the presumption that $ \left(\frac {q}{p}\right) \equal{} \minus{} 1.$  Under this scenario, the implication that $ p \equiv 1 \pmod{4}$ is still valid, which implies that  $ \left(\frac {q}{p}\right) \equal{} \left(\frac {p}{q}\right) \equal{} \minus{} 1.$

(D) $ 1 \minus{} q^{\frac {p \minus{} 1}{2}} \equal{} \minus{}2z^2 \implies \left(\frac {\minus{}2}{q}\right) \equal{} 1.$

I also resolved (E) and (F) by using a slightly different approach than Shoki. I totally ignored the issue of whether $ p \equiv 1 \pmod {8}$ or  $ p \equiv \minus{}3 \pmod {8}$ and noticed that the contradiction generated by the assumption that $ q^{\frac{p\minus{}1}{4}} \plus{} 1 \equal{} 2f^2$ still holds.

Therefore, $ q^{\frac{p\minus{}1}{4}} \plus{} 1 \equal{} m^2$ is established, and the remainder of Shoki's argument does not depend on the advance presumption that $ p \equiv \minus{}3 \pmod {8}.$  That is, since $ (m\minus{}1)$ and $ (m\plus{}1)$ both divide $ q,$  $ m\plus{}1$ must  $ \equal{} q \equal{} 3$ which implies that $ {\frac{p\minus{}1}{4}} \equal{} 1.$

[quote="zugzwang"]Intriguing response, thanks.  Questions:

(A)  How do you know that $ \left(\frac {p}{q}\right) \equal{} \minus{} 1$?

(B)  Given $ \left(\frac {q}{p}\right) \equal{} \minus{} 1,$ why is it impossible that $ q^{\frac {p \minus{} 1}{2}} \minus{} 1 \equal{} 2pz^2, q^{\frac {p \minus{} 1}{2}} \plus{} 1 \equal{} 2l^2$?

(C)  Given $ \left(\frac {q}{p}\right) \equal{} 1,$ why is it impossible that $ q^{\frac {p \minus{} 1}{2}} \minus{} 1 \equal{} 2z^2, q^{\frac {p \minus{} 1}{2}} \plus{} 1 \equal{} 2pl^2$?

(D)  In case 1, immediately after proving that $ p \equiv 1 \pmod {4},$ why must $ \left(\frac { \minus{} 2}{q}\right) \equal{} 1$?

(E)  In case 1, immediately after proving that $ (q^{\frac {p \minus{} 1}{4}} \minus{} 1)(q^{\frac {p \minus{} 1}{4}} \plus{} 1) \equal{} 2z^{2},$  why does the supposition that $ p \equiv 1 \pmod {8},$ imply that $ q^{\frac {p \minus{} 1}{4}} \plus{} 1$ is of form $ 2s^2.$

(F)  Similar to question (E),  why does the supposition that $ p \equiv \minus{} 3 \pmod {8},$ imply that $ q^{\frac {p \minus{} 1}{4}} \minus{} 1$ is of form $ 2f^2.$[/quote]

i said $ \left(\frac {p}{q}\right) \equal{} \minus{} 1$ but i didn't prove!(though it was totally useless to prove it and i even don't use this in my solution!!!)
sorry, but i didn't totally understand what u said in your 3rd post.
my answer to (A) is obvious,
it was useless to say it and ....forget it... :D
(B)if  $ \left(\frac {q}{p}\right) \equal{} \minus{} 1$ then by the well-known theorem we must have $ q^{\frac {p \minus{} 1}{2}} \equiv \minus{} 1 \pmod {p}$.so that's why it's impossible  that $ q^{\frac {p \minus{} 1}{2}} \minus{} 1 \equal{} 2pz^2$!.
my answer to (C) is same as (B).
(D)note that in this case i assumed
$ \left(\frac {q}{p}\right) \equal{} \minus{} 1$ which implies $ \implies q^{\frac {p \minus{} 1}{2}} \minus{} 1 \equal{} 2z^{2},q^{\frac {p \minus{} 1}{2}} \plus{} 1 \equal{} 2pl^{2},2zl \equal{} x$
but we have from the first equality  $ \plus{} 1 \equiv \minus{} 2z^2 \pmod {q}$ and so $ \left(\frac { \minus{} 2}{q}\right) \equal{} 1$! :)
(E)if $ p\equiv 1\pmod{8}$ then $ \frac {p \minus{} 1}{4}$ is even and so $ 2 \parallel{} q^{\frac {p \minus{} 1}{4}} \plus{} 1$ but we have
$ \gcd(q^{\frac {p \minus{} 1}{4}} \plus{} 1,q^{\frac {p \minus{} 1}{4}} \minus{} 1) \equal{} 2 \implies \gcd(\frac {q^{\frac {p \minus{} 1}{4}} \plus{} 1}{2},q^{\frac {p \minus{} 1}{4}} \minus{} 1) \equal{} 1$.
and also we have
$ (\frac {q^{\frac {p \minus{} 1}{4}} \plus{} 1}{2}).(q^{\frac {p \minus{} 1}{4}} \minus{} 1) \equal{} z^2$ so we get
$ \implies \frac {q^{\frac {p \minus{} 1}{4}} \plus{} 1}{2} \equal{} s^2$.
my answer to (F) is the same as (E),use  $ \gcd$ and what we got before:
$ \left(\frac { \minus{} 2}{q}\right) \equal{} 1,\left(\frac {2}{q}\right) \equal{} \minus{} 1\implies\left(\frac { \minus{} 1}{q}\right) \equal{} \minus{} 1\implies q\equiv \minus{} 1\pmod{4}\implies q\equiv \plus{} 3\pmod{8}$
and $ \frac {p \minus{} 1}{4}$ is odd because of $ p \equiv \minus{} 3 \pmod {8}$ so $ 4 | q^{\frac {p \minus{} 1}{4}} \plus{} 1 \implies 2 \parallel{} q^{\frac {p \minus{} 1}{4}} \minus{} 1$ . :)
hope u got what u wanted....

Yes, this cleared up my questions...  Thanks.

Peripherally, I had trouble originally attacking this problem and I could not have originated your conjuring.  I initially resolved the similar problem linked below by googling{perfect square}, after googing{quadratic residue} failed.

Link: http://www.mathlinks.ro/viewtopic.php?t=304139

I'm curious what other keywords I might google to see solutions of similar problems.  Googling{Legendre} or {Jacobi} doesn't work.


Can anybody post the solution? Thank you!

[quote="arqady"]For collection  :D
Let $ a,$ $ b$ and $ c$ are positive numbers. Prove that
$ \sum_{cyc}\frac {a \plus{} b}{\sqrt {c^{2} \plus{} 3ab}}\geq3.$[/quote]

How to deal with?

[quote=shfdfzhjj][quote="shfdfzhjj"][quote="Wonder"]$ a,b,c \in R^{ \plus{} }$, prove that:
$ \sum{\sqrt {\frac {a^{2} \plus{} ab \plus{} b^{2}}{c^{2} \plus{} ab}}}\geq \frac {3\sqrt {6}}{2}$[/quote]

And,How about this,i try it again,but failed :([/quote]

no one?[/quote]

Let$$q(a,b,c)=\frac{a^2+ab+b^2}{c^2+ab},d(a,b,c)=\frac{\sqrt{6}}{2}$$we can prove
        $$\sum_{cyc}{q(a,b,c)}\geq{\sum_{cyc}{d(a,b,c)}}$$
$$\sum_{cyc}{q(a,b,c)q(b,c,a)}\geq{\sum_{cyc}{d(a,b,c)d(b,c,a)}}$$
$$\prod_{cyc}{q(a,b,c)}\geq{\prod_{cyc}{d(a,b,c)}}$$
Threfore,we have
      $$\sum_{cyc}{\sqrt{q(a,b,c)}}\geq{\sum_{cyc}{\sqrt{d(a,b,c)}}}$$

[quote=arqady]For collection  :D
Let $ a,$ $ b$ and $ c$ are positive numbers. Prove that
$ \sum_{cyc}\frac{a\plus{}b}{\sqrt{c^{2}\plus{}3ab}}\geq3.$[/quote]

Let $ a,$ $ b$ and $ c$ are positive numbers. Prove that
$$ \sum_{cyc}\frac{a\plus{}b}{\sqrt{c^{2}\plus{}3ca}}\geq \sum_{cyc}\frac{a\plus{}b}{\sqrt{c^{2}\plus{}3ab}}\geq3$$

[quote=luofangxiang][quote=arqady]For collection  :D
Let $ a,$ $ b$ and $ c$ are positive numbers. Prove that
$ \sum_{cyc}\frac{a\plus{}b}{\sqrt{c^{2}\plus{}3ab}}\geq3.$[/quote]

[img]http://s10.sinaimg.cn/middle/006ptkjAzy743Sqx3GV09&690[/img][/quote]

the first is also wrong


[quote="chien than"]Given $a;b;c>2$ such that:
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1.$
Prove that:
$(x-2)(y-2)(z-2) \leq 1$[/quote]
The inequality $(x-2)(y-2)(z-2) \leq 1$ is true
for all positive $x,$ $y$ and $z$ such that $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1.$ :wink:

[quote="arqady"]

The inequality $(x-2)(y-2)(z-2) \leq 1$ is true
for all positive $x,$ $y$ and $z$ such that $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1.$ :wink:[/quote]

I denote $\frac{1}{x}=a$ ,$\frac{1}{y}=b$ and $\frac{1}{z}=c$ $\rightarrow$ $a+b+c=1$ and the inequalitie becomes
$(a+b-c)(b+c-a)(a+c-b)\leq abc$ $\Leftrightarrow$ $\sum a(a-b)(a-c)\geq 0$


group theory
abstract algebra
superior algebra
superior algebra unsolved

If G is finite group and non-solvable, then whether or not G has a non-solvable  minimal normal subgrop?

I might be a fool, but isn't the answer obvious?
 By assumptions the set of normal non solvable subgroups is non empty and finite.

@Fadalbala:
G is non-solvable, but the assumptions don't say anything about the solvability of the minimal normal subgroups.

There are groups that have only non-solvable minimal normal subgroups, e.g. $ S_{n}$ with $ n\geq 5$.
But there are also groups that aren't solvable and have solvable normal subgroups: Let $ G=SL_{n}(\mathbb{F})$ with $ n>2$ or $ |\mathbb{F}|>3$. Then the factorgroup $ PSL_{n}(\mathbb{F})$ is a simple, nonabelian group and  therefore non-solvable. Hence G is non-solvable too.
But in general the center of G is nontrivial, so there are abelian minimal normal subgroups. (in fact: in those cases is [i]every[/i] minimal normal subgroup a subgroup of the center)

I see: I misunderstood the question, he did not talk about non solvable normal subgroups that are minimal for these two properties, for that is trivial,

 sorry  rulin :blush:


advanced fields
advanced fields theorems

Hello ! I need somebody who can help me by telling me a book where I can find the proof of $ \pi_1(M)\equal{}Z$, where $ M$ is the Moebius strip.

FS

the moebius strip deformation retracts to $ S^1$.

Why do you need a book for this¿ Take a look at any book on fundamental groups (e.g. Hatcher) to see the general proof for deformation retracts (and with some luck you may find that special example). But what's the problem in showing the rest (= that it really deformation retracts) itself¿ There is no book only about this  :wink:

how can we show it doesn't deformation retract to its boundary?

Let $ i: S^1 \to M$ be the boundaries embedding and $ r: M \to S^1$ be the retraction, so we have $ r \circ i\equal{} \mathrm{id}_{S^1}$. Then the induced map $ i_*: \mathbb Z \cong \pi_1(S^1) \to \pi_1(M) \cong \mathbb Z$ is multiplication by $ 2$ (check this be retracting $ M$ to the middle $ S^1$ and compare generators). But then $ r_* \circ i_*$ can't be the identity on $ \mathbb Z$, contradiction.


$\text{Find all } x \text{ that satisfy } \sqrt{x + 2\sqrt{x-1}} + \sqrt{x+1-2\sqrt{2x-2}} = 1 + \sqrt{2}$

The answer is:
[hide]
$1 \le x \le 3$
[/hide]

[hide="Very brief solution summary"]I only got 3, but then I realized that in my method of solving, I inadvertedly left out values of x less than 3. What I did was factor the radicands into perfect squares so the outermost radicals would disappear. The rest is easy algebra and casework.[/hide]

Yes, factoring and solving the answer is 3.  However, the answer the poster gave is different, and plugging in 2, part of the given answer, the equation holds true.  If I saw this on a test, I'd certainly answer 3 and go on.
How would I know without plugging in 2 arbitrarily that 3 is not the only answer?

because say $\sqrt{x^2}=y$.    Is $x=y$ the only possibility?

So you've gotta try some casework.  Also keep in mind the domain of the square root function.

[quote="djimenez"][quote="pkothari13"]$\text{Find all } x \text{ that satisfy } \sqrt{x + 2\sqrt{x-1}} + \sqrt{x+1-2\sqrt{2x-2}} = 1 + \sqrt{2}$
[/quote]

Something that can help is the following observation

[hide="Observation"]
$x + 2\sqrt{x-1}=(\sqrt{x-1}+1)^2$
$x+1-2\sqrt{2x-2} = (\sqrt{x-1}-\sqrt2)^2$
[/hide][/quote]

Well, one possibility is the following:
[hide="Click"]
$x + 2\sqrt{x-1}=(\sqrt{x-1}+1)^2$; the square root is $\sqrt{x-1}+1$
$x+1-2\sqrt{2x-2} = (\sqrt{x-1}-\sqrt2)^2$ The square root can be $-\sqrt{x-1} + \sqrt{2}$

Of course, the latter can be simplified to two things, but by taking the negative square root, you will just get $1 + \sqrt{2} = 1 + \sqrt{2}$. Therefore, I suppose you will have to take into consideration domain restrictions, and solve from there, as djshowdown2 said.[/hide]

My first post wasn't very clear. I just wanted to reiterate what everyone is saying.
The thing to realize here is that $\sqrt{x^2}=|x|$, not necessarily $x$.
So you must consider two cases: $x\ge 0$ and $x<0$.


calculus
integration
trigonometry
function
calculus computations

then i get
-1
--------------------
40 + 40*u*cos(40x)

then how do i get the anti D of this?

When you do a substitution, you must do it completely. If there's still anything dependent on the original variable, you haven't finished.

ok i get that one, but if the question were

x^28
--------   dx
1+x^58

what would u be?

In general, you have

$ c\displaystyle\int \dfrac{f'(x)}{1 \plus{} (f(x))^2)} \, dx \equal{} c \tan^{\minus{}1} [f(x)] \plus{} D$

Can you see the link between $ x^{58}$ and $ x^{28}$?

what about

216x^2
---------
root(1-6x)

what would u = ?

or which trig function can i relate this to?

Hi. I would substitute $ u\equal{} 1\minus{}6x$, so that $ du \equal{} \minus{}6dx$ and the integral becomes

$ \minus{}36 \int \frac{(1\minus{}u)^{2}}{36 \sqrt{u}}\ du$.

We can then expand the denominator to get $ u^{\minus{}\frac{1}{2}} \minus{} 2u^{\frac{1}{2}} \plus{} u^{\frac{3}{2}}$ and so on.

can u explain all your steps, cause i am lost

Hey,
you can substitute $ u\equal{}x^{29}.$

hello, we get $ \frac{1}{40(1\plus{}\cos(40x))}\plus{}C$ as the result.
Sonnhard.

[hide="Solution"]
$ I \equal{} \int \frac {sin(40x)}{1 \plus{} (cos(40x))^2}dx$

Put $ cos(40x) \equal{} t$

$ \minus{} 40sin(40x)dx \equal{} dt$

$ I \equal{} \int \frac { \minus{} 1}{40}\frac {dt}{1 \plus{} t^2}$

$ I \equal{} \frac { \minus{} 1}{40}tan^{ \minus{} 1}{t} \plus{} C$

$ I \equal{} \frac { \minus{} 1}{40}tan^{ \minus{} 1}{cos(40x)} \plus{} C$[/hide]


limit
real analysis
real analysis solved

I will let Moubi post a solution for this one.  :D  In France, this is a classical puzzle for oral X.  ;)

[quote="harazi"] In France, this is a classical puzzle for oral X.  ;)[/quote]
I don't know what this is mean?

They use to give this problem in oral examination quite often.

you mean it is easy :)
but I really don't know how to solve. ;)

I never said it is easy. I will post solution soon, if you want.  ;)

By the way, look about some other problems with source Oral X or ENS and you will see how "easy" they are. An for this problem, consider $x_n=a_1^2+...+a_n^2$. Obviously, $s_n$ goes to infinity (if not $a_n$ tends to 0 and the hypothesis is not verified), so $a_n$ goes to 0. Thus, $ s_{n+1}/s_n$ tends to 1, thus they are equivalent. Next, from the statement of the problem and the above observation we easily see that we must have $ s_{n+1}^3-s_n^3$ tends to $3$. Thus,by Cesao we have $ s_n^3/n$ tends to 3 and since $a_n$ is equivalent to $1/s_n$, we conclude.

In my opinion, there is a big chance to not solve it on oral exam without exterior help.

Yes, indeed, and what is even more difficult is that at X you don't get many indications. But the truth is that comparing to other problems, this is almost nothing.

$harazi$,your solution is very nice.thank you. :)
and I think this problem can be writen as general form. :)


Consider a triangle $ ABC$ and the points $ D,E,Z$ on the sides $ AB,BC,CA$ respectively such that

$ \Vec{AD}\equal{}\frac{1}{3}\vec{AB},\vec{BE}\equal{}\frac{1}{3}\vec{BC},\vec{CZ}\equal{}\frac{1}{3}\vec{CA}$.

If $ AE$ meets $ CD$ in $ M$, $ AE$ meets $ BZ$ in $ N$ and $ BZ$ meets $ CD$ in $ L$ prove that

$ (areaLMN)\equal{}\frac{1}{7}(areaABC)$

P.S I am interesting for a solution with vectors.

Hmm.. I guess areal co-ordinates are related to vectors in that they can be defined in terms of vectors, so see here for a solution using areals. It's the third post.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=300344


conics
hyperbola
analytic geometry
Gauss

Consider the conic $ \frac{13(x\minus{}11)^2}{47}\minus{}\frac{7(y\minus{}91)^2}{61}\equal{}1$.
Find the exact coordinates of the foci.

I rewrote it as $ \frac{(x\minus{}11)^2}{\frac{47}{13}}\minus{}\frac{(y\minus{}91)^2}{\frac{61}{7}}\equal{}1$

so $ a^2\equal{}\frac{61}{7}$
and $ b^2\equal{}\frac{47}{13}$
with hyperbolas, $ a^2\plus{}b^2\equal{}c^2$ so $ c^2\equal{}\frac{1122}{91}$
Where do I go from here?

[hide="Solution"]

This problem deals with shifting of origin...

$ \frac{(x\minus{}11)^2}{\frac{47}{13}} \minus{} \frac{(y\minus{}91)^2}{\frac{61}{7}} \equal{} 1$

If origin of the coordinate system is shifted to[b] (h, k)[/b] from[b] (0, 0),[/b]

you can write eqn of hyperbola as

$ \frac{(x\minus{}h)^2}{a^2} \minus{} \frac{(y\minus{}k)^2}{b^2} \equal{} 1$

So, here

$ h \equal{} 11 \\ k \equal{} 91 \\ a^2 \equal{} \frac{47}{13} \\ b^2 \equal{} \frac{61}{7}$

Take

$ X \equal{} x \minus{} h \ and \ Y \equal{} y \minus{} k  \\
\\
\\
\Rightarrow \ \frac{X^2}{a^2} \minus{} \frac{Y^2}{b^2} \equal{} 1$

Coordinates of focii are given by

$ ( X, Y ) \equiv ( \pm ae, 0 ) \ where \ e \equal{} \sqrt{1\plus{}\frac{b^2}{a^2}}$

or

$ (x,y) \equiv ( \pm ae \plus{}h, k)$  [i]... (1)[/i]

$ ae \equal{} \sqrt{a^2\plus{}b^2} \equal{} \sqrt{\frac{1122}{91}}$

[b]Focii[/b] ([i]using (1)[/i])  =

$ ( 11 \pm \sqrt{\frac{1122}{91}} , 91 )$

[/hide]

I take offense to the term "ugly numbers". All numbers should be loved equally! Just because some numbers are a little bigger than others, or look a little different, or are a little irrational - that shouldn't make a difference. All numbers are beautiful!

[b]True ....
I agree with gauss [/b]   :yup:


The measure of the area of the square is 81-90x+25x^2 where x is a positive integer. What is the smallest possible perimeter measure for the square?

Yes, we a function f(x) = (5x - 9)^2

That means the length of the side is (5x - 9).
The perimeter (P) is 4(5x - 9) = 20x - 36.
Since P > 0, then 20x - 36 > 0 ==> x > 36/20 (=1.8).
Since x is a positive integer, x must be 2 and you can solve for the perimeter.

I have another problem concerning perimeter and area.

A farmer has 50 m of fencing and wishes to build a rectangular yard by adding to an existing 10 m section of fence. What is the maximum area that can be enclosed?

[quote="Jean-Louis"]A farmer has 50 m of fencing and wishes to build a rectangular yard by adding to an existing 10 m section of fence. What is the maximum area that can be enclosed?[/quote]
[hide]The largest area is achieved when the rectangle is a square. With 60 m of fence, the square will have sides of length 15 m. The maximum area is therefore 225 m².[/hide]


inequalities
logarithms
function
inequalities unsolved

Set $a>b>0$ , Prove : $a^{a}b^{b}>(ab)^\frac{a+b}{2}$

But don't do it by prove $\frac{a^{a}b^{b}}{(ab)^\frac{a+b}{2}}>1$ , See how many other methods you can find

[quote="kg2100"]Set $a>b>0$ , Prove :

But don't do it by prove $\frac{a^{a}b^{b}}{(ab)^\frac{a+b}{2}}>1$ [/quote]
$a^{a}b^{b}>(ab)^\frac{a+b}{2}\Leftrightarrow\left(\frac{a}{b}\right)^{\frac{a-b}{2}}>1.$ :D
[quote="kg2100"]
See how many other methods you can find[/quote]
Method number 1:
$f(x)=\ln x$ is concave function and $g(x)=x\ln x$ is convex function. :wink:

I think this question can be generalize as follow :

Set $a_{1}\geqslant a_{2}\geqslant a_{3}\geqslant...\geqslant a_{n}>0$

Is there anyone could prove that $\prod_{i=1}^{n}a_{i}^{a_{i}}\geqslant(\prod_{i=1}^{n}a_{i})^{\frac{1}{n}\sum_{i=1}^{n}a_{i}}$

Don't need to set $a_{1}\geqslant a_{2}\geqslant\cdots\geqslant a_{n}$ and the conclusion is still right.

Actually $f(x)=x\ln x$ is convex and with Jensen we have
$\frac{f(a_{1})+f(a_{2})+\cdots+f(a_{n})}{n}\geqslant f\left(\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}\right)\\ \Longleftrightarrow\ln\prod_{i=1}^{n}a_{i}^{a_{i}}\geqslant\ln\left(\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}\right)^{a_{1}+a_{2}+\cdots+a_{n}}$
And notice that with AM-GM we have
$\left(\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}\right)^{a_{1}+a_{2}+\cdots+a_{n}}\geqslant(a_{1}a_{2}\cdots a_{n})^{\frac{1}{n}(a_{1}+a_{2}+\cdots+a_{n})}$
Then it's done :wink:

[quote="kg2100"]Set $ a,b > 0$, prove that $ \sqrt [a \plus{} b]{a^{a}b^{b}} \geq \sqrt {ab}.$[/quote]I think this inequality can be strengthened  as follows :

$ \sqrt [a \plus{} b]{a^{a}b^{b}} \geq \sqrt {\frac {a^2 \plus{} b^2}{2}}.$


if you drop a ball of mass m and it hits the floor with a velocity v and rebounds with a velocity -v how is momentum conserved...isn't there a change in momentum of the ball-floor system (given the floor has no motion before, during, and after the impact) of -2mv between the moment right before the impact and the moment right after the impact?

I think you have to look at the system as a whole, so the entire Earth is actually moving.

I don't know what you mean...could you explain?

Because the Earth is pulling on the ball, and for every force there is an equal and opposite force (Newton's law), then the ball is also pulling back on the Earth. It doesn't move the Earth much, since it's incredibly massive, but that force is there. Therefore, you actually have the ball and the Earth moving towards each other, and after the collision, they are both moving away from each other.

You could think about it this way:
The ball has mass m with velocity v before hitting the ground and mass m with velocity (-v) after hitting the ground.  The ground has mass M and 0 velocity before being hit by the ball and velocity V after being hit by the ball.

mv + M(0) = m(-v) + MV
(2mv/M) = V
Since M >> m, 2mv/M = approximately 0.


In this diagram a scheme is indicated for associating all the points of segment $ \overline{AB}$ with those of segment $ \overline{A'B'}$, and reciprocally. To described this association scheme analytically, let $ x$ be the distance from a point $ P$ on $ \overline{AB}$ to $ D$ and let $ y$ be the distance from the associated point $ P'$ of $ \overline{A'B'}$ to $ D'$. Then for any pair of associated points, if $ x \equal{} a,\, x \plus{} y$ equals:
[asy]defaultpen(linewidth(.8pt));
unitsize(.8cm);

pair D= (0,9);
pair E = origin;
pair A = (3,9);
pair P = (3.6,9);
pair B = (4,9);
pair F = (1,0);
pair G = (2.6,0);
pair H = (5,0);

dot((0,0));dot((1,0));dot((2,0));dot((3,0));dot((4,0));dot((5,0));
dot((0,9));dot((1,9));dot((2,9));dot((3,9));dot((4,9));dot((5,9));
draw((D+(0,0.5))--(0,-0.5));
draw(A--H);
draw(P--G);
draw(B--F);
draw(F--H);
draw(A--B);

label("$D$",D,NW);
label("$D'$",E,NW);
label("0",(0,0),SE);
label("1",(1,0),SE);
label("2",(2,0),SE);
label("3",(3,0),SE);
label("4",(4,0),SE);
label("5",(5,0),SE);
label("0",(0,9),SE);
label("1",(1,9),SE);
label("2",(2,9),SE);
label("3",(3,9),SW);
label("4",(4,9),SE);
label("5",(5,9),SE);
label("$B'$",F,NW);
label("$P'$",G,S);
label("$A'$",H,NE);
label("$A$",A,NW);
label("$P$",P,N);
label("$B$",B,NE);[/asy]
$ \textbf{(A)}\ 13a\qquad \textbf{(B)}\ 17a \minus{} 51\qquad \textbf{(C)}\ 17 \minus{} 3a\qquad \textbf{(D)}\ \frac {17 \minus{} 3a}{4}\qquad \textbf{(E)}\ 12a \minus{} 34$

well, you could let x=3 or 4 and then just brute force it since it implies that all points between 3 and 4 work  :P  :P  :P

Just notice that the triangles are similar, with ratio $ 1: 4$ and you get

$ y\equal{}1\plus{}4(4\minus{}a)$

$ \therefore x\plus{}y \equal{} a \plus{} 1 \plus{} 4(4\minus{}a) \equal{} 17\minus{}3a \longrightarrow \boxed{C}$


I like Obama's idealistic "the system is screwed up" message. However, before I actually cast a vote I would have to look more closely at what Hilary and Obama actually want to do policy-wise. I have a feeling that Hilary is better at policy stuff in general just due to experience, and also I suspect that I'd agree more with her fiscal plans than I would with Obama's. I don't know, though -- maybe Obama would fight harder for individual freedoms and other social good stuff, which is also of value to me...

There's only two candidates as of now, and out of those two, I would choose Obama. For me, I don't think Hillary could lead our country properly economically or foreign policy wise, much less deal with all of the domestic issues. She seems like she relies too much on Bill, and seems like she would make the wrong decision in terms of stress in a difficult situation. Obama is much calmer, and smarter, and better. :)

I actually do not really care who wins between Hillary and Obama.
I, personally, thought that Hillary did way better than Obama in their debate.
Either one is great.
And same as Xevarion- I like how he thinks that the system or whatever is screwed up.

I like what Hillary said in the debate: " It took one clinton to clean up the mess of the first Bush. It's going to take another Clinton to clean up the mess of the second Bush" something like that. Lawl.

[quote="hunter34"]Actually Bush did a good job.[/quote]
At what?

[quote="hunter34"]Remember: Hillary as a president $ \implies$ no military.[/quote]
What's wrong with that?

[quote="hunter34"]Remember: Hillary as a president $ \implies$ no military.[/quote]

Remember from when?  Your absurd non sequiturs $ \neq$ reality.

[quote]Actually Bush did a good job.

Remember: Hillary as a president  no military.[/quote]
................?
this is a democrat forum plus not even republicans like him, and
military is dumb.

[quote="7h3.D3m0n.117"][quote="korammath"]military is dumb.[/quote]

I think a military is quite important. Or would you prefer terrorists to reside next to your home because we don't have the Marines to push them out?[/quote]

hahahahahahahahaha

really come on

e: if you're fakeposting well done sir

[quote="hunter34"]Actually Bush did a good job.

Remember: Hillary as a president $ \implies$ no military.[/quote]Die.  Now.

[quote="SBS1"]WE NEED TO REMEMBER THAT WE DON'T LOG IN TO EXPRESS POLITICAL OPINIONS.[/quote]

I AM GLAD THAT YOU, SBS1, REGISTERED FOR THESE FORUMS YESTERDAY TO TELL US EXACTLY WHAT WAS ACCEPTABLE TO POST ABOUT.  THANK YOU FOR YOUR GUIDANCE GOOD SIR.

Are either of you moderators? just wondering.

i go for either Obama or Mcain (dunno how to spell that.) i know its really weird since one is dem and other is repub, but i just dont like hillary. personally, i think she is relying too much on bill clinton. obama is ok but a bit inexperienced. i'm not sure what he can do about the iraq war. Mcain may be republican, but he has good ideals, like helping the environment. i just hope he doesnt follow in bush's steps.

[quote="dreambold22"]I just hope he doesn't follow in Bush's footsteps.[/quote][Edited slightly]
If elected, McCain [b]will[/b] continue Bush's policies. He agrees with Bush on almost everything, especially the war. Have you heard that "hundred years" bit? There will be no end to the mess in Iraq under McCain, since he refuses to leave without an impossible victory.

Early on, I liked both Dem candidates and leaned toward Clinton. As the election went on, I found out more about Obama, and now prefer him.

Obama, universal health care ftw. :)

not ftw: Obama being pro-choice :(

I THINK OBAMA IS THE ONE WHO DESERVES TO BE THE PRESIDENT.

[quote="Brut3Forc3"]Obama got enough delegates from South Dakota to bring his total to at least 2119, i.e., he will be the Democratic nominee.[/quote]

According to the news...... You are correct!!! Yippee!!!!!

CELEBRATION!!!!  :jump:  :trampoline:  :coolspeak:

yay obama won the democrats votes!!!!!!! now against mccain!!!!!! not that i think mccain stands a chance. :lol:


probability
geometry
geometric transformation

[quote="usaha"][hide]93[/hide][/quote]
How would you get that?

Here is a straightforward translation from the written problem to two equations.  From here, you might attempt to solve with algebra or with a guess and check table.

[hide]
$\frac{G-3}{B+G-3}=\frac{2}{5}$

$\frac{B+7}{B+G+7}=\frac{5}{8}$
[/hide]

[quote="usaha"]i did the problem wrong.. i forgot to add 7 and subtract 3 from the denominators.. do what rcv did and solve the system of equations.[/quote]
Me too...

So then you would get $G=15$ and $B=\boxed{18}$?

[quote="Quevvy"]A bag contains only blue and green marbles. If three green marbles are removed from the bag, the probability of drawing a green marble from the remaining marbles would be $\frac{2}{5}$. If, instead, seven more blue marbles are added to the bag, the probability of obtaining a blue marble would be $\frac{5}{8}$. What was the number of blue marbles in the bag before any changes were made?[/quote]
[hide]
Let x be green
y be blue
3x=2y+9
5x=3x+49
102 blue marbles
[/hide]


modular arithmetic
number theory proposed
number theory

hi,

i'll be happy if anyone help me to solve this problem :

Find all integers x such that 7 divides $x^{2}+x+1$.

thanks in advance.

thank you very much Rust .but can you explain more ,please?

Well, what you want is $1+x+x^{2}= 0 \pmod 7$.
It is equivalent to $\frac{1-x^{3}}{1-x}= 0 \pmod 7$.
So it yields : $x^{3}= 1 \pmod 7$, with $x \neq 1$.
Hence the two solutions by Rust.

thank you very much mathmanman for you clarifications  :) .


All of David's telephone numbers have the form $ 555\minus{}abc\minus{}defg$, where $ a$, $ b$, $ c$, $ d$, $ e$, $ f$, and $ g$ are distinct digits and in increasing order, and none is either $ 0$ or $ 1$. How many different telephone numbers can David have?

$ \textbf{(A)}\ 1\qquad \textbf{(B)}\ 2\qquad \textbf{(C)}\ 7\qquad \textbf{(D)}\ 8\qquad \textbf{(E)}\ 9$

OK, so in the integers $ 2$-$ 9$, we must exclude one.  This is simply $ \binom{8}{1}$, or $ \boxed{8}$.


geometry
3D geometry
group theory
abstract algebra
ratio
number theory unsolved
number theory

Let m be an integer $ m>1$, let n be the sum of the elements of a subgroup {$ 1,2,...m$}. Find all pairs $ (m,n)$ in order to have $ (m^{4}\plus{}mn)/(m^{2}n\plus{}1)$ as an integer.

I noticed that $ m$ must be odd. I think it might help.

$ \frac{m(m^{3}\plus{}n)}{m^{2}n\plus{}1}\in\mathbb{Z}$.  But $ (m, m^{2}n\plus{}1) \equal{} 1$.  Therefore, $ \frac{m^{3}\plus{}n}{m^{2}n\plus{}1}\in\mathbb{Z}$
Clearly both the numerator and denominator here are positive.  Hence, $ m^{3}\plus{}n\ge m^{2}n\plus{}1\implies m^{3}\minus{}1\ge n(m^{2}\minus{}1)\implies n\le m\plus{}\frac{1}{m\plus{}1}\implies n\le m$.

Finally, notice that the numerator is congruence to $ n$ modulo $ m$, while the denominator is $ 1$ modulo $ m$.  This means that their ratio is congruent to $ n$ modulo $ m$.  So we have, for some nonnegative integer $ k$,

$ \frac{m^{3}\plus{}n}{m^{2}n\plus{}1}\equal{} km\plus{}n$
$ m^{3}\plus{}n \equal{} (m^{2}n\plus{}1)(km\plus{}n)$
Clearly, if $ k\ge 1$, then this is impossible.  Therefore, we have $ k \equal{} 0$.  So
$ m^{3}\plus{}n \equal{} m^{2}n^{2}\plus{}n$
$ m \equal{} n^{2}$.

So we must have that $ m$ is $ n^{2}$.

EDIT: And $ n\equal{}0$ is always a solution as well.  So the set of solutions $ (m, n)$ is

$ \{(n^{2}, n)|n\in\mathbb{N}\}\cup\{(m, n)|n\equal{}0, m\in\mathbb{N}\}$

It's a nice solution! But I didn't understand why you set $ n\equal{}0$ as a solution, because $ n$ is the sum of a subgroup, thus it cannot be zero.

[quote="danilorj"]It's a nice solution! But I didn't understand why you set $ n \equal{} 0$ as a solution, because $ n$ is the sum of a subgroup, thus it cannot be zero.[/quote]

Thanks.  It's a nice problem.  $ n \equal{} 0$ is the sum of an empty subset though...so if an empty subset is not allowed, then $ n \equal{} 0$ is not a solution.

By the way, be careful what terms you use.  Here, it didn't matter, because the meaning was understood.  But the word "group" doesn't mean the same thing as the word "set" in mathematics.


symmetry
geometry
geometric transformation
reflection

Consider the polar graph $ r\equal{}f(\theta)$...

a. What kind of symmetry is observed if $ r$ is replaced by $ \minus{}r$?

b. What kind of symmetry is observed if $ \theta$ is replaced by $ \minus{}\theta$?

a) reflection through the origin

b) reflection over the x-axis