calculus
integration
trigonometry
algebra
binomial theorem
calculus computations

I need help on getting something straightened out.

For the integral $ \int_0^\frac {1}{2}\cfrac{\sqrt {x}}{\sqrt {1 - x}}\: dx$, if I make the substitution $ \sqrt {x} = \sin{y}$, I get this:

$ x = \sin^2y$
$ dx = 2\sin{y}\cos{y}$

and for

$ \sqrt {0} = 0 = \sin{y}$ and $ \sqrt {\cfrac{1}{2}} = \cfrac{1}{\sqrt {2}} = \sin{y}$ to hold true,

$ y = 0$ and $ y = \cfrac{\pi}{4}$ respectively.

so:

\begin{align*} & \int_0^\frac {1}{2}\cfrac{\sqrt {x}}{\sqrt {1 - x}}\: dx \\
& = \int_0^\frac {\pi}{4}\cfrac{\sin{y}}{\sqrt {1 - \sin^2y}}\cdot{}2\sin{y}\cos{y}\: dy \\
& = 2\int_0^\frac {\pi}{4}\sin^2y\: dy \end{align*}

was the above procedure correct? In particular, did I change the bounds of integration in the correct step?

My grasp of this is still kinda shaky.. What is this called anyway? Replacing bounds? It'd be great if someone could tell me under what section of a calculus course I could find more information on this. Thanks! :)

Seems fine to me :) Usually the corresponding section in Calculus textbooks is called either "Change or variable" or "Finding integrals by substitution". Just look at the table of contents and see what looks like it.

we can't find any mistakes in your calculation, undefined 117. :)

How about this one?

Q. 1 $ \int_0^1 x\sqrt {1 \minus{} x}\ dx$.

Q.2 Let $ x \equal{} \cos ^3 t,\ y \equal{} \sin ^ 3 t\ \left(0\leq t\leq \frac {\pi}{2}\right)$.

$ \int_0^1 y\ dx$.

Thanks fedja - I found [url=http://en.wikipedia.org/wiki/Integration_by_substitution]this Wikipedia article[/url] :D

@kunny: hehe I spent 40 minutes filling up two pages trying to solve that first one by substituting $ x = \sin^2\theta$ :rotfl:

hmm idk is there a way to do it that way? I get $ \int_0^\frac {\pi}{2}2\sin^3\theta\cos^3\theta\: d\theta$, which looks a bit like Q. 2

okay so

[hide="Q. 1"]
$ \int_0^1x\sqrt {1 - x}\: dx$

$ u = 1 - x$, so $ x = 1 - u$ and $ du = - 1\: dx$.

\begin{align*} & - \int_1^0(1 - u)\sqrt {u}\: du \\
& = - \int_1^0(u^\frac {1}{2} - u^\frac {3}{2})\: du \\
& = - \int_1^0u^\frac {1}{2}\: du + \int_1^0u^\frac {3}{2}\: du \\
& = - \left[\cfrac{2}{3}u^\frac {3}{2}\right]_1^0 + \left[\cfrac{2}{5}u^\frac {5}{2}\right]_1^0 \\
& = \boxed{\cfrac{4}{15}} \end{align*}

w00t! :lol:
[/hide]
I'll need to think a little more about Q. 2 :|

Well, You could do the q.1 using that substitution ($ x = sin^2\theta$), but you won't get the integral you've written. You'll get $ \int_{0}^{\pi/2}2\sin^3\theta\cos^2\theta d\theta$. And now, you just need to transform it into the form suitable for integration. Of course, it is hard to do it (and quite unnecessary, as your first solution is elegant enough), but here's the form:

$ \frac {1}{8} (2\sin \theta + \sin 3\theta - \sin 5\theta)$

which gives you the same answer: $ \frac {4}{15}$.

mk I think I'll need a hint for Q. 2 :D
[hide="Q. 2"]
\begin{align*} \int_0^1y\: dx & = \int_0^\frac {\pi}{2}\sin^3t\cdot( - 3\cos^2t\sin{t})\: dt \\
& = - 3\int_0^\frac {\pi}{2}\sin^4t\cos^2t\: dt \\
& = - 3\int_0^\frac {\pi}{2}\sin^4t(1 - \sin^2t)\: dt \\
& = - 3\left(\int_0^\frac {\pi}{2}\sin^4t\: dt - \int_0^\frac {\pi}{2}\sin^6t\: dt\right) \end{align*}
err..
\begin{align*} & - 3\int_0^\frac {\pi}{2}(\sin^2t\cos{t})^2\: dt \\
& = - 3\int_0^\frac {\pi}{2}(\cos{t} - \cos^3t)^2\: dt \end{align*}
meh. :|
[/hide]

Well, let's stop here:

$ \minus{} 3\left(\int_0^\frac {\pi}{2}\sin^4t\: dt \minus{} \int_0^\frac {\pi}{2}\sin^6t\: dt\right)$

These two can be easily evaluated if you know:

$ \int\sin^n {ax}\;dx \equal{} \minus{}\frac{\sin^{n\minus{}1} ax\cos ax}{na} \plus{} \frac{n\minus{}1}{n}\int\sin^{n\minus{}2} ax\;dx$

But you can also use this nice result

[url]http://www.mathlinks.ro/viewtopic.php?t=214671[/url]

whoa.. that stuff looks crazy beastly, but thanks for reminding me about integration by parts.

hmm..

For $ \int\sin^3t( - 3\cos^2t\sin{t})\: dt$

$ u = \sin^3t$ and $ dv = - 3\cos^2t\sin{t}$
$ du = 3\sin^2t\cos{t}$ and $ v = \cos^3t$

so

\begin{align*} \int\sin^3t( - 3\cos^2t\sin{t})\: dt & = \sin^3t\cos^3t + 3\int\cos^4t\sin^2t\: dt \\
0 & = (\sin{t}\cos{t})^3 + 3\left(\int\cos^4t\sin^2t\: dt + \int\sin^4t\cos^2t\: dt\right) \\
0 & = (\sin{t}\cos{t})^3 + 3\int\sin^2t\cos^2t(\cos^2t + \sin^2t)\: dt \\
0 & = (\sin{t}\cos{t})^3 + 3\int\sin^2t\cos^2t\: dt \end{align*}

heh.. I'm probably going way off-course.

No need really for this kind of calculations, but I can't guarantee that you won't solve the problem this way :lol:  I thought a bit and found two more ways (well, the first one was the first I used, and the second one is more a showing off than a method):

1. $ \sin x\equal{}\frac{e^{ix}\minus{}e^{\minus{}ix}}{2i}$. Now, if you plug it in those two sine integrals, using binomial theorem reduces you to the easily integrable form shown in 2.

2. Theoretically, after using addition (multiple angle) trig formulae, it is possible to obtain:

$ \frac{1}{32}(2 \minus{} \cos2t \minus{} 2 \cos4t \plus{}\cos6t)$ which is the same you'll get in Method 1 using binomial theorem and proper grouping of elements.

I think I've seen that form $ \sin{x}\equal{}\cfrac{e^{ix}\minus{}e^{\minus{}ix}}{2i}$ before, but sorry I'm not familiar with it. What's it called? Hopefully I can look it up and familiarize myself :)

Well, I call it the complex form (or, even better-complex definition) of sine. $ x$ doesn't have to be real $ (x\in\mathbb{C})$.

ooh I found this page: [url]http://en.wikipedia.org/wiki/De_Moivre%27s_formula[/url]

that explains it in sorta easy terms..

I'll just need to eventually convince myself that $ e^{ix} = \cos{x} + i\sin{x}$

But yes! It works! :lol:
\[ - 3\left(\int_0^\frac {\pi}{2}\sin^4t\: dt - \int_0^\frac {\pi}{2}\sin^6t\: dt\right)
\]
[hide="First integrand"]
\begin{align*} \sin^4t & = \left(\cfrac{e^{ix} - e^{ - ix}}{2i}\right)^4 \\
& = \cfrac{1}{8}\cdot\cfrac{e^{4ix} + e^{ - 4ix}}{2} - \cfrac{4}{8}\cdot\cfrac{e^{2ix} + e^{ - 2ix}}{2} + \cfrac{6}{16} \\
& = \cfrac{1}{8}\cos{4x} - \cfrac{1}{2}\cos{2x} + \cfrac{3}{8} \end{align*}
[/hide]
[hide="Second integrand"]
I spent a bit more time on this one because I kept messing up pos/neg signs and the fractions. :P
\begin{align*} \sin^6t & = \left(\cfrac{e^{ix} - e^{ - ix}}{2i}\right)^6 \\
& = - \cfrac{1}{32}\cdot\cfrac{e^{6ix} + e^{ - 6ix}}{2} + \cfrac{6}{32}\cdot\cfrac{e^{4ix} + e^{ - 4ix}}{2} - \cfrac{15}{32}\cdot\cfrac{e^{2ix} + e^{ - 2ix}}{2} + \cfrac{20}{64} \\
& = - \cfrac{1}{32}\cos{6x} + \cfrac{3}{16}\cos{4x} - \cfrac{15}{32}\cos{2x} + \cfrac{5}{16} \end{align*}
[/hide]

So combining the two parts we see that the original expression equals:
\[ - 3\left[\int_0^\frac {\pi}{2}\left(\cfrac{1}{8}\cos{4x} - \cfrac{1}{2}\cos{2x} + \cfrac{3}{8}\right)\: dx - \int_0^\frac {\pi}{2}\left( - \cfrac{1}{32}\cos{6x} + \cfrac{3}{16}\cos{4x} - \cfrac{15}{32}\cos{2x} + \cfrac{5}{16}\right)\: dx\right]
\]
which, as hsiljak said, equals
\[ - 3\left[\int_0^\frac {\pi}{2}\cfrac{1}{32}(2 - \cos{2x} - 2\cos{4x} + \cos{6x})\: dx\right]
\]
Then simply:
\begin{align*} & - \cfrac{3}{32}\left(\int_0^\frac {\pi}{2}2\: dx - \int_0^\frac {\pi}{2}\cos{2x}\: dx - 2\int_0^\frac {\pi}{2}\cos{4x}\: dx + \int_0^\frac {\pi}{2}\cos{6x}\right) \\
& = - \cfrac{3}{32}\left(\left[\cfrac{}{}2x\right]_0^\frac {\pi}{2} - \left[\cfrac{1}{2}\sin{2x}\right]_0^\frac {\pi}{2} - \left[\cfrac{2}{4}\sin{2x}\right]_0^\frac {\pi}{2} + \left[\cfrac{1}{6}\sin{6x}\right]_0^\frac {\pi}{2}\right) \\
& = \boxed{ - \cfrac{3}{32}\pi} \end{align*}
yeah! :w00t:

Thanks for the help, hsiljak, and thanks for the problems kunny!

EDIT: oops, er I accidentally switched variables along the way, but you knew what I meant. :lol:


There are a few minor changes in the way the LaTeX code is rendered on the site.  For those of you unfamiliar with LaTeX, read the [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php]LaTeX guide[/url] provided by this site to learn more about typesetting mathematics.

First of all, all the formulas are rendered with the \displaystyle command from default. You need not add that command anymore. The \displaystyle command makes fractions, summations, etc., render more legibly, as shown below:

With \displaystyle: $ \sum_{k \equal{} 0}^n \frac 1{k}$

Without \displaystyle: $ \textstyle \sum_{k \equal{} 0}^n \frac 1{k}$.

To force your posts to be without \displaystyle you can use the \textstyle
 command (start your LaTeX code with \textstyle).

Second, we encourage all users to use the normal LaTeX syntax, i.e. the dollar signs, instead of the [tex ] and [/tex ] tags (without spaces there).

The [tex ] button will still be available and can still be used as before, but after rendering the text, the forum will automatically transform the [tex ] tags into equivalent dollar signs.

If you do not edit your post, you will "feel nothing". If you do, the [tex ] tags that you had used will be replaced by dollar signs.

PS I do not know if everyone knows this already, so I will mention it here again: on most browsers the latex code used to produce a block of math text (the whole latex code, including the dollar signs) can be viewed by holding the mouse a couple of seconds still above the picture.

[quote="RC-7th"]Is there a command for the thing that makes this $\textstyle \sum\limits_{k=0}^n \frac 1{k}$ different from this \[\textstyle \sum\limits_{k=0}^n \frac 1{k}\] ?[/quote] You probably didn't carefully read the text above. You are using the \limits command, which automatically acts as a \displaystyle. Without the \limits command (which is not necessary!), the code above would be
$ \textstyle \sum^n_{k=0} \frac 1k $.

The \limits thing is just a habit. I was wondering if there was a command equivalent of using double dollar signs [code]($$)[/code] on the ends.

next time, if you want to display dollars use [code ] (dollars) [/code ] (without spaces), and use all the time the function preview, to see if there are any errors in your messages ;)

as for your question, if you had carefully read the post above, you would have discovered that "we encourage all users to use the normal LaTeX syntax", that means any known LaTeX syntax, including double dollars as

\[ \frac 12 \cdot \frac 34 = \frac 38 \] does.

Please read carefully the general annoucements, as they usually contain all the answers you need. :)

Good luck "tExING" :D

The feature of some web browsers to show LaTeX code when scrolled over that v.v. remarked about on the original post does not work in Mozilla.

Sorry guys, I have just tested it on IE Explore, and Semaca Mobile browser for SE (where it works!).

It is curious why it does not work in Opera, or in Mozzila. I will try to see what is wrong ...  :maybe:

If anyone knows more about these browsers, you can check the source code yourselves and tell me what exactly is wrong :)  ;)

We might be able to add netscape to that list (or maybe my version is simply too old :? )

Ok, I've identified the problem. It simply seems that Opera, Mozzila and Netscape just do not accept the alt tag. Thus, there is only one solution. Tell me if it works ;)

Lemme try in Firefox:

\[
\frac{1}{2}! = \frac{1}{2}\Gamma{}\left(\frac{1}{2}\right) = \frac{\sqrt{\pi}}{2}
\]

if you type $$ and $$ at the ends of a sentence, does the code always come out as text in the center of a line and always occupy the whole line?
For example if you type "I have $$3$$ dollars", it will come out as

I have
                                                                                3
dollars.

How can you aviod this?

[quote="henryyangrui"]if you type \[ and \] at the ends of a sentence, does the code always come out as text in the center of a line and always occupy the whole line?
For example if you type "I have \[3\] dollars", it will come out as

I have
                                                                                3
dollars.

How can you aviod this?[/quote]What do you mean? This is how normal $\LaTeX$ works. If you want inline equations, use a single dolar sign at the beginning and the end of a formula: $\$ \frac 12 $\$ will reveal $ \frac 12 $ ;)


Hey..ive been bustin my brains with this problem.  It might be easy but im still having troubles

A equivalent triangle (A,B,C) with 10 cm as sides thus all angles are 60 degrees.
A half-circle passes trough point A and B passing trough the interior of the triangle, yet there is a small portion of the circle outside the triangle.  Find the area of that portion.

If you dont understand the question, just tell me. I'll try to do a drawing

BTW...
all i need is a guideline of how to do it.
I dont need the solution
thx in advance

hmm. i guess this is the picture, right??
[hide="hint"]
angle $AOM=120$ because $\triangle MOB$ is equalateral. with side length of $5$.
now find $S_{\widehat {AOM}+\triangle MOB}$ and subtract if from the area of the half circle[/hide]

sorry...i seem to have miss explainedthe question.
first of all, it is a half circle we are talking about...
second of all, i forgot to mention that the diameter of the circle is a side of the triangle so a portion of the circle would not be in the triangle....

i could do an image, but we're can i host it???

EDIT...
i am in grade 10 math, dont over complicate plz

here's a picture i drew...
find the area of the shaded part

BTW...each side is 10 cm

[hide="tostart you off.."]
i dont know if i made a correct assumption but anyways:
draw a line at top of circle to form a trapezoid at bottom of the equiliateral triangle (in other words, 3 smaller equilateral triangles of sidelength 5.
[/hide]
[hide="stilldontgetit?"]
The area of an equilateral triangle is
S^2*sqrt3/4 where s is sidelength, so the area of the three triangles is 75sqrt3/4
Since we know the diameter of the circle to be 10 we know that half the circle would be 25pi/2
the area o fthe two shaded things and thing at top of circle is 25pi/2-75sqrt3/4 <--
im assuming they are all the same so u multiply this value ------------------------------| by 2/3
im not gonna bother to actually calculate it but u get the idea :D[/hide]

Edit: sry i havent bothered to learn latex yet so i hope you can still read it  :P

I agree with sheepwarrior, the answer is \[\frac{2}{3}\left(\frac{25\pi}{2}-\frac{75\sqrt{3}}{4}\right).\] Here's a picture:


[b]Let ABCD be a convex quadrilateral inscribed in the circle[/b] $w$.[b] I note the [/b][b]intersections[/b]

$S\in AC\cap BD,\ P\in BB\cap DD,\ R\in AA\cap CC$. [b]Prove that:[/b]

$1.\ AB.CD=AD.BC\Longleftrightarrow R\in BD\Longleftrightarrow P\in AC$.[b] ($XX$ is the[/b] [b]tangent in the point[/b] $X\in w$[b]).[/b]

[b]2. The point S is the foot of a simedian-line at least in the one from the triangles ABC, ABD, ACD, BCD.[/b]

P = BB intersect DD ???  I think I am seeing deja vu... :(

could you explain?

Did you read the last sentence, al.M.V.? There, levi explains what he means by $XX$, in general. The post isn't that long, man, you could have read the whole thing.. :)


Prove or give a counterexample that a right angled triangle with integer sides can't have area as a perfect square.

[quote="usaha"]because one leg is odd the other is even.....and the odd cant be a perfect square without the even containing an odd multiple of it too[/quote]

[quote]$\cdots$ and the odd cant be a perfect square without the even containing an odd multiple of it too[/quote]
can you elborate more?

I welcome criticism of this proof as I am trying to become better at this sort of thing.

Proof:

[hide]
It can be shown that all primitive pythagorean triples (primitive in that there is no factor common to the lengths of all three sides) can be described for positive integers $a$ and $b$, where $a>b$, such that the two legs are $a^{2}-b^{2}$ and $2ab$ and the hypotnuse is $a^{2}+b^{2}$.

Thus the area of any such triangle is equal to half base times height or:
$A=\frac{1}{2}(a^{2}-b^{2})(2ab)=(a+b)(a-b)ab$

Consider non-primitive cases; all such cases can be described as a primitive case with each side multiplied by a common positive integer factor $k$. For such a non-primitive case, the area would be $A=k^{2}(a+b)(a-b)ab$ and for A to be a perfect square, the $k^{2}$ factors and the product $(a+b)(a-b)ab$ must still be a perfect square for the area to be a perfect square.

Thus, all we must do is show that the product $(a+b)(a-b)ab$ cannot be a perfect square to cover all primitive and non-primitive cases.

Thus the only cases to consider are:
1) $(a-b)=(a+b)ab$
2) $(a+b)=(a-b)ab$

Since a and b are both positive integers and a>b, case 1 is clearly rediculous.

Manipulating case 2 gives:
$a-b=\frac{1}{a}+\frac{1}{b}$
Since a and b are positive integers, this could only be possible if a-b=1. Thus:
$1=\frac{1}{b+1}+\frac{1}{b}$
And this clearly does not work for any positive integer b.

Thus the product $(a+b)(a-b)ab$ for relatively prime a and b cannot ever be a perfect square, and thus the area of a right triangle with integer sides cannot ever be a perfect square.[/hide]

[quote="usaha"]and the odd cant be a perfect square without the even containing an odd multiple of it too[/quote]

Not true.  Consider the triple $9, 40, 41$.

[quote="bthings_2000"]Thus, all we must do is show that the product $(a+b)(a-b)ab$ cannot be a perfect square to cover all primitive and non-primitive cases.

Thus the only cases to consider are:
1) $(a-b)=(a+b)ab$
2) $(a+b)=(a-b)ab$[/quote]

I have a problem with this narrowing down to two cases; what about the case where $a+b, a-b, a, b$ are all simultaneously square?

Hmm, good point t0rajir0u, and I've messed around with that for a while and not figured it out... input from anyone welcome.

Yea toujorou, you're right.

What i was trying to say is that (this is like a guess) if a leg is a perfect square, the other side must contain an odd power of an odd multiple (so 9, 40, 41 the 40 has only 1 multiple of 5), so when you do the area, it cant be a perfect square.  Also, for like generic cases like 5,12,13, im guessing that since one is even the other is odd, that when you multiply it up there will be an odd number power of an odd multiple, AI dont know what i just said.

Well, we can reduce it to the case of a primitive triangle (since dividing each side by $g$ divides area by $g^{2}$, so it remains a perfect square), and then one leg is twice a square and the other leg is a square. So we're looking at
\[4a^{4}+b^{4}=c^{2}\]
in positive integers.

Probably factoring the left hand side or using primitive Pythagorean triple formula could help here.

some treatment available here... external link.
[hide][url]http://mcraefamily.com/MathHelp/PythagTriangleAreaNotSquare.htm[/url][/hide]


function
inequalities
logarithms
induction
rearrangement inequality
algebra unsolved
algebra

It's not clear to me how you get the result from what you've written.

How about the following:

For $x < 1 < y$,
$f(x)f(y)=a^{2}+b^{2}+c^{2}+ab(x^{2}y+x y^{2})+ac(x^{2}+y^{2})+bc(x+y) \geq a^{2}+b^{2}+c^{2}+ab(x^{2}y^{2}+x y)+ac(x^{2}y^{2}+1)+bc(xy+1) = f(xy) f(1)$,

using the rearrangement inequality for each term.  So if the $x_{i}$ are not identically 1, we can convert a non-1 value to 1 while decreasing the product of the $f(x_{i})$.  Eventually all the $x_{i}$ are equal to 1, and the product of the $f(x_{i})$ is exactly 1.

I'm afraid I'm not following this either.

For example, if $y_{1}> 1$, then $\sigma(2) = 0$.

If $y_{k}>1$ we find little $x_{m}$ don't used before. If $y_{k}<1$ we find biggest $x_{m}$ don't used before. Define $m=\sigma (k+1)$.
It give $(y_{k+1}=y_{k}x_{\sigma (k+1)}, \ (y_{k}-1)(x_{\sigma (k+1)}-1)\le 0 \ \forall k .$

I just did this problem in a practice test (1990 All-Soviet Olympiad, I think?)

Say $e^{a_{i}}= x_{i}$. Let $g(x) = \log (a e^{2x}+b e^{x}+c)$. Then $g''(x) > 0$ for $x > 0$ (this is easy to confirm, unless I made an algebra error). Then we have $\sum a_{i}= 0$. We need to show $\sum g(a_{i}) = 0$. But Karimata gives $\sum g(a_{i}) \geq \sum g(0)$, and since $a+b+c = 1$, $g(0) = \log(1) = 0$, and we have the desired result.

Deedlit: Maybe I am missing something, but you appear to be using rearrangement backwards? Certainly $x+y \leq x y+1$ is not a true statement.

[quote="hieuchuoi@"]is there any solutions else? I dont like Rust's Solution  :wink:[/quote]

Was there a problem with mine?

[quote]Deedlit: Maybe I am missing something, but you appear to be using rearrangement backwards? Certainly x+y \leq x y+1 is not a true statement.[/quote]

Whoops, yes the inequality is the wrong way.  I'll fix it.

A nice solution  :D
Use this lemma:
$f(x).f(y) \geq f(\sqrt{xy})^{2}$
The rest is for you  :roll: !

sorry, i was wrong, and this is two new solutions
1) you can apply Holder ineq  :D
2) you can prove with n is odd, n is even and we have conclusion
if n is odd, we add $a_{n+1}=1$  :D

It is easy to prove that $f(x)f(y)\geq (f(\sqrt{xy}))^{2}$
Now suppose that $f(x_{1})...f(x_{n})\geq (f(\sqrt[n]{x_{1}..x_{n}}))^{n}$

is true for $n=2^{k}$ and prove it for $n=2^{k+1}$  :wink:

I can prove it forall n. But it is not trivial for $n\not =2^{k}$, for example for n=3.

Well, the easy way would be to take the logarithm and apply Jensen's inequality, which would make it essentially the same as 101101101's proof.

[quote="hieuchuoi@"]no, it's true for any positive integer $n$  :wink:[/quote]

Who said that ?  :wink:
After the proof for $n=2^{k}$  suppose $n$ is arbitary and $x_{1}..x_{n}=1$  . Let $k$ a positive integer such that $2^{k-1}<n\leq 2^{k}$ . Let us add ,if necessary, $x_{n+1}=x_{n+2}=...=x_{2^{k}}=1$. Since $f(x_{n+1})=....=f(x_{2^{k}})=1$ we may write
$f(x_{1})....f(n)=f(x_{1})...f(x_{2^{k}})\geq (f(\sqrt[2^{k}]{x_{1}...x_{2^{k}}}))^{2^{k}}=1$

 :wink:


number theory
prime numbers
number theory unsolved

I need a help with this problem (again on set N-[1]):

Let's divide all prime numbers into two infinite sets A,B so that each prime number is element of exactly one of sets A,B. Now let each composite number that is product of some primes only from A be element of A too and also let each composite number that is product of some primes only from B is element of B too. Let's denote the set of the remaining composite numbers (such that they aren't elements of A or B) as C. Prove that for each natural number n at least one of sets A,B or C contains n consecutive natural numbers.

Thanks for all your replies...

I think it is very simple.
Suppose $p_1$, ..., $p_n\in A$ and $q_1$, ..., $q_n\in B$. Then due to Chinese residue theorem we can find $m$ s.t. $p_iq_i\,|\,m+i$, $i=1..n$. So $C$ contains $n$ consecutive numbers.


A variable force F acts on a body which is free to move.The displacement of the body is proportional to $t^3$ where t is time.The power delivered by F to the body will be proprtional to what power of $t$?

[color=blue]I think the answer is $t^4$[/color]

0

$f = ct$
$a =\frac{ct}{m}$
$v = \frac{1}{m}\frac{ct^2}{2}$
$d = \frac{1}{m}\frac{ct^3}{3}$

$d$ is proportional to $t^3$

$p = \frac{1}{t}\int f(t)dx$
$p = \frac{ctd}{t}$
$p$ is proportional to $t^0$

$x = k t^3$

$v = 3 k t^2$

$a = 6 k t$

$F = 6 k m t$

$P = Fv = 18 k t ^3$

So the power is proportional to $t^3$.

[quote="oooooh ya"]isnt power work over time not force times velocity?[/quote]

yeah,my bad

$work = \int f dx$

$dist = vt$ if v is constant, not for a variable acceleration/force

[quote="phucnv87"]I think the answer is $t^4$[/quote]
post your work so we can see if it is

It seems we all seem to be getting different answers. I forgot to include $k^2$ and $m$ in my solution. Annie is on the right track until she says$P = W / t$. Its actually $P = \frac{dW}{dt} = \frac{d}{dt} \left( \frac{9}{2} k^2 m t^4 \right) = 18 k^2 m t^3$.

ooh I want to play too

$P = \frac{dW}{dt} = \frac{d}{dt} (F \cdot x) = F \frac{dx}{dt} + x \frac{dF}{dt} = ma \frac{dx}{dt} + x m\frac{da}{dt} = m \left( \frac{dx}{dt} \frac{d^2x}{dt^2} + x \frac{d^3x}{dt^3} \right)$. Let $x = kt^3$. Then the above is $m (3kt^2 \cdot 6kt + kt^3 \cdot 6k) = 24mk^2t^3$.

[b]MysticTerminator[/b], note that $\text dW = F \text dx$, not $\text dW = \text d(Fx)$  ;)

The answer is $P = 18mk^2t^3$...

I was wondering why the formula looked different ...  ;)


limit
number theory proposed
number theory

Find all $a,b,c \in R$ such that $[na]\cdot [n^{2}b]=[n^{3}c]$ for all $n \in \mathbb N$.

My solution isn't short so i want the other solution :)  :)

[i]Edited by Myth[/i]

Please, don't call topics like "nice but easy"!

Read http://www.mathlinks.ro/viewtopic.php?p=140098

We have $ab=\lim_{n\to\infty} \frac{[na]\cdot [n^2b]}{n^3}=\lim_{n\to\infty}\frac{[n^3c]}{n^3}=c$, so $c=ab$.

Consider the case $a,b>0$. Let $a=a_0,a_1a_2a_3...$ and $b=b_0,b_1b_2b_3...$ (binary representation).
Choose large $n$ s.t. $2^na>10$ and $2^{2n}b>10$. Then $[2^na]\cdot [2^{2n}b]=\overline{a_0a_1...a_n}\cdot \overline{b_0b_1...b_{2n}}$ and $2^{3n}ab>\overline{a_0a_1...a_n}\cdot \overline{b_0b_1...b_{2n}}+10\cdot 0,a_{n+1}a_{n+2}...+10\cdot 0,b_{2n+1}b_{2n+2}...$. It follows that $10\cdot 0,a_{n+1}a_{n+2}...<1$ and $10\cdot 0,b_{2n+1}a_{2n+2}...<1$, i.e. $a_{n+1}=0$ and $b_{2n+1}=b_{2n+2}=0$, so $a_n=b_n=0$ for all large $n$. And we can choose arbitrary base instead of 2. Therefore $a$ and $b$ are integer.

I applied the same analysis for cases $(a>0,b<0)$, $(a<0,b>0)$ and $(a<0,b<0)$. However, to write it here is to long.


Exista vreun sir real marginit $\left( x_{n}\right)_{n \geq 1}$ astfel incat $\left| \left( x_{i}-x_{j}\right) \left( i-j \right) \right| \geq 1, \, \forall i \neq j$?

Asta e problema 6 de la olimpiada sovietica 1978.
Un exemplu de astfel de şir este $x_{n}=4\{n\sqrt{2}\}.$


if a+b+c=2 then prove that:

3(a^3+b^3+c^3)+10(ab+bc+ca)>=16

after homogenisation it becomes $a^3+b^3+c^3+3abc\geq ab(a+b)+bc(b+c)+ac(a+c)$.
You may suppose $a\geq b\geq c$.
If $c\geq 0$, it is true by Schur.
If $c<0$; put x=a+b. Then the conclusion follows rapidly from x>2


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ G$ be a finite group of order $ n.$  Suppose $ G$ has a simple subgroup of order $ >n/3$ and also has a proper normal subgroup of order $ 3.$  Then $ n\equal{}6.$

Suppose that $ H\leq G$ is simple, $ N\lhd G$, $ |N|\equal{}3$, $ |H| > \frac{n}{3}$.

If $ H \equal{} G$, then $ G$ is simple so $ N\equal{}G$ but then $ N$ is not a proper subgroup, contradiction.

So $ |H|\equal{}\frac{n}{2}$.  Let $ N \equal{} \langle x \rangle$, then the cosets $ H$, $ xH$, $ x^2 H$ cannot be distinct, so we have $ x\in H$.  Hence $ N \lhd H$, so $ N\equal{}H$ and $ 3 \equal{} \frac{n}{2}$.


I'm having trouble with this problem. Can someone help me, not necessarily giving me the answer but hints would be nice.  :)  thanks!
[b]
For each of n=84 and n=88, find the smallest integer multiple of n whose base 10 representation consists entirely of 6's and 7's.[/b]

[hide]
Well, you should prime factorize them, and then look at each factor by itself.
For example, 88 is 11*8, so you should see how you can arrange the last 3 digits to make it divisible by 8, then see which of these arrangements you can add a digit or two to to make it divisible by 11 and have it be the smallest.

And, 84 is 7 * 4 * 3, so you look at arrangements of the last 2 digits, and you see it must be 76. So, obviously there are have to be 3 7s. (to be divisible by 3), so see how you can add 2 more 7's and as few 6s as possible to the left of the 76.
[/hide]

wow! that makes a lot of sense, i never would have thought of that...thanks a lot klebian!!! :)


algebra
polynomial
linear algebra
linear algebra unsolved

For each $n$, denote
\[ P_n(X)=nX^{n-1}+(n-1)X^{n-2}+...+2X+1.  \]
Find all $n$ such that the polynomial $P_n(X)$ is irreducible over $Q$.

I can prove that it's prime for $n$ prime... Valentin Vornicu has a similar problem, for $n=2004$.

http://www.mathlinks.ro/Forum/viewtopic.php?t=75053&postorder=asc

[quote="spix"]As I assume everyone is well aware $P_n(X)=(X+X^2+...+X^n)'$ so I made a few computations and got that $P_n(X)=\frac{X^{n+1}-2X^n+1}{(X-1)^2}$ (assuming I made the computations correct)... now what?!... does this help?[/quote]
Maybe, but I thinks that sense is difficult to finish.

[quote="harazi"]Guys, don't kid yourself, the problem is still open as far as I know.[/quote]
I want to discuss this problem, and if any has solutions for some n, please write them here.
for n<20000, it has been done by computer aid.

for $n$ prime it's not hard... Use the lemma posted here
http://www.mathlinks.ro/Forum/viewtopic.php?p=440913#p440913


calculus
integration
calculus computations

A particel is moved along the x-axis by a force that measures 10/(x+1)^2 pounds at a point x feet from the origin. Find the work done in moving the particle from the origin to a distance of 9 feet.

can you give me the example and answer, here's my work

i antidifferentiated 10/(1+x)^2 into U^-1 or 10U^-1 if you want to leave the 10 their, then i put in 1+x into the u,

1/2 integral x=0, x=9, 10(1+9)^-1= 1  1 times 1/2 equals 1/2

then i plugged in 0

and i got 5, then i subtracted 1/2 - 5=and this equals -9/2
and the answer is 9 ft-lb

so what did i do wrong, and can u show me how u got the answer

[quote="afcweswarrior"]i antidifferentiated 10/(1+x)^2 into U^-1 or 10U^-1[/quote] Which should have been $-10U^{-1}$ or $-10(1+x)^{-1}$.

[quote="afcweswarrior"]1/2 integral[/quote] Why did you put $1/2$ here?

when i substituted (
1+x)

oh man, that's what i did wrong i substituted 1+x^2
i was suppost to let u=1+x


Este problema esta sencillo, pero bueno

Halle todas las soluciones enteras al sistema de ecuaciones
$a+b+c=1$
$a^3+b^3+c^2=1$

Bonito problema. Tengo dos soluciones, pero prefiero no publicarlas todavia. Ahora, doy una "pista" bastante evidente que puede ayudar:

[hide="Pista"]
1. Sirve aplicar productos especiales, reacomodando los terminos y dividiendo en casos. No son tantos.
[/hide]

Ok, coloco mi solucion

[hide=""Solucion""] Primero, si c=1, obtenemos que
$a+b=0\Rightarrow a^3+b^3=(a+b)(a^2-ab+b^2=0=1-c^2$
Entonces todas las triplas (a,b,c)=(x,-x, 1) cumplen. Ahora si $c\not=1$ podemos escribir la segunda ecuacion como como
$(a+b)(a^2-ab+b^2)=(1-c)(1+c)$
Como $c-1\not=0$ podemos dividir a ambos lados por $c-1=a+b$ y nos queda
$a^2-ab+b^2=1+c$
Ademas, sabemos que
$a^2+2ab+b^2=(1-c)^2$
Restabdi ka segunda menos la primera, obtenemos que
$3ab=c(c-3)$
Entonces vemos que $3|c$, de donde $c=3k$ y queda que
$ab=3k(k-1)$
$a+b=1-3k$
Entonces se tiene que $a,b$ son las raices de la ecuacion
$X^2-(1-3k)X+3k(k-1)=0$
Donde $\Delta=-3k^2+6k+1$. Queremos que $\Delta>0$, y con simples calculos obtenemos que $0\leq k\leq 2$. Por ultimo vemos que en cada uno de estos casos, $\Delta$ es un cuadrado perfecto, entonces resolvemos y queda
Si $k=0$, entonces $\{a,b\}=\{0,1\}$ y obtenemos las triplas $(a,b,c)=(0,1,0);(1,0,0)$. Si $k=1$, $\{a,b\}=\{-2,0\}$ de donde se tienen las triplas $(a,b,c)=(-2,0,3);(0,-2,3)$. Por ultimo, cuando $k=2$, obtenemos $\{a,b\}=\{-3,-2\}$, dandonos las triplas $(a,b,c)=(-3,-2,6);(-2,-3,6)$. Con esto se tienen todas las triplas y se sigue lo pedido.[/hide]


let a,b,c are the sides of a triangle.prove that:
\[ \sum_{a,b,c}(\frac{\sqrt{b}\plus{}\sqrt{c}\minus{}\sqrt{c}}{\sqrt{b\plus{}c\minus{}a}})\leq3\]

[quote="stvs_f"]let a,b,c are the sides of a triangle.prove that:
\[ \sum_{a,b,c}(\frac {\sqrt {b} \plus{} \sqrt {c} \minus{} \sqrt {c}}{\sqrt {b \plus{} c \minus{} a}})\leq3
\]
[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=130812


Hi,

May be proved by noting that   \[{{n+1} \choose {i}} = \frac{n+1}{i} {{n} \choose {i-1}}\]

(or by integrating binomial sum)


For each of following systems of lin.equations Ax = B, factor A as LU, solve LC = B for c using forward substitution, solve UX = C be back substitution.

1)  2x + y = -3
     6x + 4y = -4

2) x + 2y + 0z = 3
    2x + 6y -z =7
    5x - 4y + 10z =4

Thank you

Can someone show me how to forward and backward substitute?

So do you mean elimination and back substitution? I am just learning so even if that is not so, may I try to solve?

FOR #1 I think:

1) 2x + y = -3
6x + 4y = -4

the matrix is(sorry i don't have latex)

2 1 -3
6 4 -4

so we mult. by -3 and add to row 2(note that 2 & 4 are pivots..):

2 1 -3
0 1  5

and thus we have

2x+y=-3
y=5
so
x=-4

If you are still on this site please tell me if this is what you were looking for and I can show you #2.

First you need to perform the LU factorization.  For small systems with nice numbers you can usually do this in your head:

Factorization:
$A = LU \implies \begin{bmatrix}2 & 1 \\ 6 & 4 \end{bmatrix}= \begin{bmatrix}1 & 0 \\ 3 & 1 \end{bmatrix}\begin{bmatrix}2 & 1 \\ 0 & 1 \end{bmatrix}$

Forward substitution goes from the top row to the bottom row.  Each row you solve gives you just enough values to solve for the next row:
$L\widehat{c}= B \implies \begin{bmatrix}1 & 0 \\ 3 & 1 \end{bmatrix}\begin{bmatrix}c_{x}\\ c_{y}\end{bmatrix}= \begin{bmatrix}-3 \\-4 \end{bmatrix}\implies \begin{bmatrix}c_{x}\\ c_{y}\end{bmatrix}= \begin{bmatrix}-3 \\ 5 \end{bmatrix}$

Backward substitution is like forward substitution, except you start from the bottom row and work your way up:
$U\widehat{x}= C \implies \begin{bmatrix}2 & 1 \\ 0 & 1 \end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}-3 \\ 5 \end{bmatrix}\implies \begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}-4 \\ 5 \end{bmatrix}$

You solve the second problem the same way you solve the first:
$A = LU \implies \begin{bmatrix}1 & 2 & 0 \\ 2 & 6 &-1 \\ 5 &-4 & 10 \end{bmatrix}= \begin{bmatrix}1 & 0 & 0 \\ 2 & 1 & 0 \\ 5 &-7 & 1 \end{bmatrix}\begin{bmatrix}1 & 2 & 0 \\ 0 & 2 &-1 \\ 0 & 0 & 3 \end{bmatrix}$

I've leave the rest up to you.

I like the way you do it! I'm going to remember it...


Determine all natural numbers $ n$ and $ k\geq 1$ such that $ k$ divides each of the numbers

$ \binom{n}{1},\binom{n}{2},...,\binom{n}{n\minus{}1}$

Let $ p$ is prime divisor of k. Then $ p|\binom{n}{1}\equal{}n$. Let $ p^m|n, p^{m\plus{}1}\not |n$.
If $ n\not \equal{}p^m$ we consider $ p\not |\binom{n}{p^m}$. Therefore $ n\equal{}p^m$. It give , that k had only one prime divisor p.
If $ k\equal{}p^l,l>1$ we consider $ p^l\not |\binom{p^m}{p^{m\minus{}1}}$. It give $ k\equal{}p,n\equal{}p^m$.


The left expression has 3 variables a, b, c but in the right expression has 3 variables x, y, z.
Can you write exactly the problem?

[quote="NguyenDungTN"][quote="evarist"][color=purple] Let a,b,c are three positive real numbers satisfy $ a \plus{} b \plus{} c \equal{} 1$. Prove that :
    $ (1 \plus{} a)\sqrt {\frac {1 \minus{} a}{a}} \plus{} (1 \plus{} b)\sqrt {\frac {1 \minus{} b}{b}} \plus{} (1 \plus{} c)\sqrt {\frac {1 \plus{} c}{c}}\ge\frac {3\sqrt {3}}{4}.\frac {(1 \plus{} x)(1 \plus{} y)(1 \plus{} z)}{\sqrt {(1 \minus{} x)(1 \minus{} y)(1 \minus{} z)}}$
  It's of Walther Janous welcome hungkhtn, arqady, can_hang2007, Vacs, hope you can help me to prove this problem  :) [/color][/quote]

I think correct problem is:
$ (1 \plus{} a)\sqrt {\frac {1 \minus{} a}{a}} \plus{} (1 \plus{} b)\sqrt {\frac {1 \minus{} b}{b}} \plus{} (1 \plus{} c)\sqrt {\frac {1 \plus{} c}{c}}\ge\frac {3\sqrt {3}}{4}.\frac {(1 \plus{} a)(1 \plus{} b)(1 \plus{} c)}{\sqrt {(1 \minus{} a)(1 \minus{} b)(1 \minus{} c)}}$[/quote]

It's   $ (1 \plus{} a)\sqrt {\frac {1 \minus{} a}{a}} \plus{} (1 \plus{} b)\sqrt {\frac {1 \minus{} b}{b}} \plus{} (1 \plus{} c)\sqrt {\frac {1 \plus{} c}{c}}\ge\frac {3\sqrt {3}}{4}.\frac {(1 \plus{} a)(1 \plus{} b)(1 \plus{} c)}{\sqrt {(1 \minus{} a)(1 \minus{} b)(1 \minus{} c)}}$

or   $ (1 \plus{} a)\sqrt {\frac {1 \minus{} a}{a}} \plus{} (1 \plus{} b)\sqrt {\frac {1 \minus{} b}{b}} \plus{} (1 \plus{} c)\sqrt {\frac {1 \minus{} c}{c}}\ge\frac {3\sqrt {3}}{4}.\frac {(1 \plus{} a)(1 \plus{} b)(1 \plus{} c)}{\sqrt {(1 \minus{} a)(1 \minus{} b)(1 \minus{} c)}}$ ?

I think the second is problem which evarist wants to post  :wink:

[color=purple] thank you Dung and doduclam the corriction problem must be as you posted. Sorry ervery one  :) [/color]


Let $ a,b,c,d$ be reals such that $ a^2\plus{}b^2\plus{}c^2\plus{}d^2\equal{}1$.

Prove that

$ \frac{1}{1\minus{}ab}\plus{}\frac{1}{1\minus{}ac}\plus{}\frac{1}{1\minus{}ad}\plus{}\frac{1}{1\minus{}bc}\plus{}\frac{1}{1\minus{}bd}\plus{}\frac{1}{1\minus{}cd}\leq 8$

[u]Solution[/u]
\[ S \equal{} \frac {1}{1 \minus{} ab} \plus{} \frac {1}{1 \minus{} bc} \plus{} \frac {1}{1 \minus{} cd} \plus{} \frac {1}{1 \minus{} ca} \plus{} \frac {1}{1 \minus{} bd} \plus{} \frac {1}{1 \minus{} da}
\]

\[ \therefore S \minus{} 6 \equal{} \frac {ab}{1 \minus{} ab} \plus{} \frac {bc}{1 \minus{} bc} \plus{} \frac {cd}{1 \minus{} cd} \plus{} \frac {ca}{1 \minus{} ca} \plus{} \frac {bd}{1 \minus{} bd} \plus{} \frac {da}{1 \minus{} da}
\]

\[ \equal{} \sum_{cyc}\frac {2ab}{2 \minus{} 2ab}
\]

\[ \equal{} \sum_{cyc}\frac {2ab}{2a^2 \plus{} 2b^2 \plus{} 2c^2 \plus{} 2d^2 \minus{} 2ab}
\]

\[ \leq \sum_{cyc}\frac {2ab}{a^2 \plus{} b^2 \plus{} 2c^2 \plus{} 2d^2}
\]

\[ \equal{} \sum_{cyc}\frac {2ab}{(a^2 \plus{} c^2 \plus{} d^2) \plus{} (b^2 \plus{} c^2 \plus{} d^2)}
\]

\[ \leq \frac {1}{2}\sum_{cyc}(\frac {a^2}{a^2 \plus{} c^2 \plus{} d^2} \plus{} \frac {b^2}{b^2 \plus{} c^2 \plus{} d^2}) \equal{} 2
\]
QED.


There is a secret message in this post. First one to decode it gets absolutely nothing.

[color=white]zngu92, gung jnf ernyyl rnfl.[/color]

The secret message is
zngu92, gung jnf ernyyl rnfl.

Yay, I get absolutely nothing!  :lol:  Hey, wait a sec.

Haha, it is halfway almost very entertaining. A secret encoded crytopuzzle resides elegantly throughout my e-missive, so solve :P. A great encryption truly, only occasionally. Doesn't really always turn just while done...

:D

alan and ra, you guys are wrong
your things can be deciphered more  :)

[quote="ra5249"]to mathnerd314 or watever:
zngu92, gung jnf ernyyl rnfl.[/quote]

Math92, that was really easy...

Nobody's solved mine yet :P

[hide="Solution"]
h i i h a v e a s e c r e t m e s s a g e t o o d r a t j w d
The conclusion follow immediately.  :P
[/hide]

[quote="bookaholic"]"Hi, I have a secret message too. Dratjwd." What's with the last part?[/quote]

Hi, I have a secret message too. Drat. JWD

I used to have this one for a signature, though as far as I know, no one has decrypted it. It's easy though if you just see the little trick... two words and you'll find it trivial.

**N**0**1**7**P**Y**R**C**N**3**Y**7**5**4**N**5**1**H**7**3**3**5**U**N**4**C**

[quote="Peter VDD"]**N**0**1**7**P**Y**R**C**N**3**Y**7**5**4**N**5**1**H**7**3**3**5**U**N**4**C**[/quote]
[hide="My response to that, which leads to an obvious answer"]Yes, I can see that nasty encryption.[/hide]

Huh, I used to wonder about that, and then I looked at again today, and got it without even looking at hint...

Okay, there's a secret message hidden in this post too.  Though I kinda stole it from somebody else.

EDIT: Wait, I forgot to put the actual message in.[hide=" :blush: "][color=white][size=9]This is the message.[/size][/color][/hide]

[color=white]This is not the message.  Keep looking.[/color]

OK, to miyomiyo, it is "This is the message".
And to PeterVDD, I still can't figure out ur code...  :blush:

Aargh, I finally get it!

(For the record: I didn't see the post by sapphyre571)

[quote="math92"]There is a secret message in this post.  First one to guess the secret message gets a prize.

 :D
[hide=" :D "] The secret message is: The elephant ate my balloon. [/hide]
 :D[/quote]

Lol, math92. I get it...It took me a whole day, but I get it... :rotfl:

In case someone wants one: [hide=" A hint "]Try to reply with a quote...[/hide]

[hide=" The answer "] The elephant ate my balloon. ra5249 wasn't kidding. :wink: [/hide]

[quote="math92"]Uh...good job guys on decoding the post! You guys should all be cryptologists for the CIA.[/quote]

:roll:

Fgiweoru bjerm qwivk iowp jmvegf.

[hide="decode this"]
public void paint (Graphics g)
	{
		g.setColor(Color.YELLOW);
		Polygon ears=new Polygon();
		ears.addPoint(33,7);
		ears.addPoint(77,33);
		ears.addPoint(118,32);
		ears.addPoint(174,26);
		ears.addPoint(133,47);
		ears.addPoint(63,45);
		g.fillPolygon(ears);

		g.setColor(Color.BLACK);
		Polygon blackEars=new Polygon();
		blackEars.addPoint(43,14);
		blackEars.addPoint(48,28);
		blackEars.addPoint(34,8);
		g.fillPolygon(blackEars);

		Polygon blackEars2=new Polygon();
		blackEars2.addPoint(161,28);
		blackEars2.addPoint(174,26);
		blackEars2.addPoint(155,37);
		g.fillPolygon(blackEars2);

		//face
		g.setColor(Color.YELLOW);
		g.fillOval(119,41,20,49);
		g.fillOval(59,40,20,49);
		g.fillRect(67,40,61,49);

		//eyes
		g.setColor(Color.BLACK);
		g.fillOval(71,56,12,12);
		g.fillOval(117,55,12,12);
		g.setColor(Color.WHITE);
		g.fillOval(73,57,6,6);
		g.fillOval(119,56,6,6);

		//cheeks
		g.setColor(Color.RED);
		g.fillOval(63,70,12,12);
		g.fillOval(125,70,12,12);

		g.setColor(Color.BLACK);
		Polygon nose=new Polygon();
		nose.addPoint(100,73);
		nose.addPoint(97,70);
		nose.addPoint(103,70);
		g.fillPolygon(nose);

		//mouth
		g.drawArc(82,62,20,20,250,70);
		g.drawArc(98,62,20,20,220,70);

		//body
		g.setColor(Color.YELLOW);
		g.fillOval(58,87,32,68);
		g.fillOval(111,87,32,68);
		g.fillRect(74,87,53,68);

		//arms
		g.setColor(Color.BLACK);
		g.drawLine(75,102,87,129);
		g.drawLine(87,129,83,128);
		g.drawLine(83,129,83,131);
		g.drawLine(83,131,80,129);
		g.drawLine(80,129,80,134);
		g.drawLine(80,134,64,122);
		g.drawLine(121,106,114,127);
		g.drawLine(114,127,118,128);
		g.drawLine(118,127,119,130);
		g.drawLine(119,130,121,129);
		g.drawLine(121,129,123,131);
		g.drawLine(123,131,137,113);

		//tail
		Color brown=new Color(102,0,0);
		g.setColor(Color.YELLOW);
		Polygon tail=new Polygon();
		tail.addPoint(143,120);
		tail.addPoint(148,115);
		tail.addPoint(143,107);
		tail.addPoint(154,95);
		tail.addPoint(142,82);
		tail.addPoint(186,36);
		tail.addPoint(191,68);
		tail.addPoint(167,88);
		tail.addPoint(173,96);
		tail.addPoint(155,109);
		tail.addPoint(159,117);
		tail.addPoint(149,125);
		tail.addPoint(150,130);
		tail.addPoint(143,134);
		g.fillPolygon(tail);

		g.setColor(brown);
		Polygon brownTail=new Polygon();
		brownTail.addPoint(143,120);
		brownTail.addPoint(148,115);
		brownTail.addPoint(156,112);
		brownTail.addPoint(159,117);
		brownTail.addPoint(149,125);
		brownTail.addPoint(150,130);
		brownTail.addPoint(143,134);
		g.fillPolygon(brownTail);

		//feet
		g.setColor(Color.YELLOW);
		Polygon leftFoot=new Polygon();
		leftFoot.addPoint(77,155);
		leftFoot.addPoint(72,160);
		leftFoot.addPoint(76,160);
		leftFoot.addPoint(77,160);
		leftFoot.addPoint(78,161);
		leftFoot.addPoint(81,163);
		leftFoot.addPoint(87,155);
		g.fillPolygon(leftFoot);

		Polygon rightFoot=new Polygon();
		rightFoot.addPoint(127,155);
		rightFoot.addPoint(135,159);
		rightFoot.addPoint(131,159);
		rightFoot.addPoint(132,160);
		rightFoot.addPoint(129,160);
		rightFoot.addPoint(128,162);
		rightFoot.addPoint(118,155);
		g.fillPolygon(rightFoot);
[/hide]

[quote="bubka"][quote="KrazyNerd"]ra5249 wasn't kidding.[/quote]Who claimed he was, huh? :huh:

As for peter's quote, I had tried leetspeek for some time now, even before seeing the hints, though why I had always started at the front will now remain a mystery. :)[/quote]
i don't get it
what is the message?

[quote="KrazyNerd"]Oh, no...I don't have the program to run that, but I have a bad feeling...

It's not a pikachu, is it?[/quote]you win

@iamagenius: read backwards (obviously).

I haven't got mathnerd's yet. It is not a Caesar shift, it may be a transposition.

[quote="bubka"]I haven't got mathnerd's yet. It is not a Caesar shift, it may be a transposition.[/quote]

It's unbreakable.

[hide]I just banged on the keyboard randomly and made it look convincingly like a code. Sorry about that.[/hide]

[quote="laughinghead505"]I have one! No one will find this one!! :D  :D[/quote]Because there is none...

[quote="laughinghead505"]I have one! No one will find this one!! :D  :D[/quote]Hmm, by trial and error, it's not by clicking the sixth box....

[quote="iamagenius"]C 4 N U 5 3 3 7 H 1 5 N 4 5 7 Y 3 N C R Y P 7 1 0 N

can you decipher ......... cryption...?

i don't really get it?[/quote]C4N = can
U = you
533 = see
7H15 = this
N457Y = nasty
3NCRYP710N = encryption

--> Can you see this nasty encryption?

[quote="sapphyre571"]@ra5429

think 1337speak[/quote]
lol
i just figured it out  :blush:


superior algebra
superior algebra solved

Hello

How can you prove mathematically that a complete bipatite graph
can only have a even number of edges?

I can explain this by words or intuition but Im having trouble formulating
this is methematical forms.

Thanks!
X

The complete bipartite graph $B_{1,1}$ has only one edge...

[quote="eXKoR"]great I've found the answer:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15325[/quote]

answer to what?

To the problem given there, what he wrote is just nonsense.


geometry
rectangle
calculus
integration
real analysis
invariant
geometric transformation

I have no idea where to put this, so I will put this here:

If a finite collection of points has an area of 0, and sometimes even an infinite collection of points (i.e. a line) has an area of 0, what justifies that a different infinite collection of points (like a rectangle) to have an area that is not 0?

Definition of area.  If a rectangle has area=0, then what is area?

Infinity is not always the same.  It's sort of weird but it's true.  Consider this example: The set of rational numbers, $ \mathbb{Q}$ is obvioulsy infinte, as is the set of irrational numbers.  However, the set of irrational numbers is much, much larger.  Check this link out.  Interesting stuff.

http://en.wikipedia.org/wiki/Countable_sets

Good question.

Notions such as length, area, and volume (known as [u]measure[/u]) are not the same as the notion of [u]cardinality[/u]. They're both notions of "amount", but measure typically tells you more about the relative stucture/geometry of a set than it's cardinality.

See [url=http://en.wikipedia.org/wiki/Measure_Theory]Measure Theory[/url]. (Note: this is an abstract presentation of the idea of measure - something you wouldn't normally learn about until well into College.)

Actually, I was reading this in Apostol (the single variable calculus volume) where he defines area (for defining integration) firstly if a set is "measurable." Perhaps my question would be better satisfied if someone also explained what the important of this is.

Here's the importance: You're reading about integration, and integrals (at their basic level) are based on limits of sums of areas. Well... what is the definition of area? This is a complicated question and a very important question.

The full blown answer to the question begins with the definition of a measurable set with respect to a measure (usually the Lebesgue measure). It turns out that there are some sets which are not (Lebesgue) measurable - basically meaning that they don't have a well-defined area in any usual sense.

If you think that area is an obvious concept that doesn't need definition, ask yourself the question, what exactly does it mean for the area of a circle to be $ \pi r^2$ [i]square units[/i]?
What is the area of the set $ S \equal{} \{(x,y) | x$ has no 2's in its decimal representation and $ y$ has at least one 2 in it's decimal representation$ \} \cap [0,1]X[0,1]$?

I know area is not an obvious concept and hence needs axioms to describe it, which is why I did not have an obvious answer to my own question. :) When I said importance, I meant the importance of measure and being measurable in this definition of area that he is describing, not the importance of defining area.

After reading through those Wikipedia entries, it seems that measure is a $ \sigma$-algebra, and hence its unions must countable. Thus the union of infinite points with area zero is not zero. Am I right? Thus area must be defined in a different manner...

Actually, someone can explain it to me anyway...

If you try to find a measure (volume) with "expected" properties (like: a rectangle/cuboid/etc has the canonical volume, it's invariant under rotation and translation, the volume of a disjoint union of the sum of the individual measures) for all sets of $ n$-dimensional space, then you get problems. The probably best known example is the Banach-Tarski-Pardoxon ( http://en.wikipedia.org/wiki/Banach-Tarski ).
The solution is simple: just don't try to measure everything.
One still want's to measure "most" sets, e.g. all open or closed sets. Then one can have the mentioned properties, finally leading to the Borel- and the Lebesgue-measure (or in more general manner to general measure theory). See http://en.wikipedia.org/wiki/Measure_(mathematics) .
Then this has lots of good properties, and one can build integration theory solely on this (without much more analysis). This kind of integration has much more properties than "standard" integration.

I will say that right now you shouldn't worry about measurable sets and all that. That will come later when you study Real Analysis. Just be aware for now that there are some crazy subsets of $ \mathbb{R}$ and $ \mathbb{R}^2$ that don't have well-defined "length" or "area".

Since you're studying integration, here are some not-too-theoretical, but fun, integration questions:

1. What is the average distance of all the points in a circle from the center?
2. What is the average distance of all the points in a sphere from the center?
3. Prove that the gravity inside a hollow sphere of mass M equals 0 (physics knowledge required).

Question 3 is true by the shell theorem and can be shown for electrostatics as well.  Another interesting piece on area is the Dirac Delta function.  This is a function such that it is infinitesimally thin, yet has an area of 1.

http://en.wikipedia.org/wiki/Dirac_delta_function

It isn't a function at all (and thus has no "area"). It's more a set of functions, the real limit function has area $ 0$ (or better: should not be considered).

[i]
1. What is the average distance of all the points in a circle from the center?
2. What is the average distance of all the points in a sphere from the center? [/i]

[hide]I got 2/3 the radius for the first question and 3/4 the radius for the second question. What I did was look at all the concentric circles or spheres inside the given circle or sphere, multiply each concentric circle or sphere by its radius, add up all these values, and then divide by the area for #1 and the volume for #2.[/hide]

The dirac delta is only treated as a function for convenience in physics, it fits in better as an elementary example of a generalised function (or as they are called, a distribution).

Area is notoriously difficult to define in elementary terms (the best we usually get away with is by defining axiomatically that a unit area exists and then describing the notion of area by a set of its properties), and better left till you're exposed to a rigorous theory of measures.

What is the definition of infinite volume/area? It means that at least one of its dimensions must stretch indefinitely (in informal terms). Thus its surface area/perimeter is always...

[quote="mathking123"]i know this may be somewhat off-topic, but you may be interested in looking up the Koch star. Similar to Gabriel's trumpet, it has the property of finite area with infinite perimeter.

Gabriels trumpet is a solid of revolution generated by rotating the graph of y=1/x about the x axis, and considering it from x=1-infinity. it has finite volume,(of pi if i remember correctly?), while it has infinite surface area.

Unlike these two objects, Can any object have infinite volume(area), while still having finite surface area(perimeter)?[/quote]

How about (and I'm speaking loosely here) rotating the Dirac Delta 'function' about the x-axis? The resulting surface area would be infinite (basically a disc of infinite radius) but the volume would be finite I think.

But I'm having trouble thinking of a standard function to fit this description.


How can you create a second blog?
I believe you could...

I am fairly certain that each user can only create one blog. You may, however, change the name of your original blog and then delete all the entries and shouts, so it would be a new blog, sort of.

[quote="isabella2296"]I'm confused - Nerd_of_the_Ages apparently has two avatars.[/quote]

This was an issue that was dealt with before your time.

But this blog seems to have 'orginated' relatively recently...

[quote="#H34N1"]Its a clone[/quote]

Thanks, I was trying not to give them any clever ideas.  :dry:


When two lines are drawn in a plane, they can create four non-overlapping regions.  If five straight lines are drawn in the plane, what is the largest number of non-overlapping regions they can create?

The formula for maximum intersections of $ n$ lines in a plane is $ 1$ more than the $ n$th triangular number.

This formula can be written mathematically as $ \frac {n(n \plus{} 1)}{2} \plus{} 1$.

So we plug in for $ n \equal{} 5$.

$ \frac {5(5 \plus{} 1)}{2} \plus{} 1$

$ \frac {30}{2} \plus{} 1$

$ 15 \plus{} 1$

The answer is $ \boxed{16}$ intersections.

[quote="Textangle"]The formula for maximum intersections of $ n$ lines in a plane is $ 1$ more than the $ n$th triangular number.
...
The answer is $ \boxed{16}$ intersections.[/quote]

Err... this question isn't asking anything about intersections of lines, it's talking about regions of the plane.

I'm also curious where you found this formula... please let me know, thanks!

The most regions I've been able to cook up is 15 :/  EDIT: and then like 30 seconds later, 16

i'm guessing text was solving for regions, but put intersections by mistake (because that's the right answer)

Wouldn't the maximum intersection of lines be the same as the number of regions?

You revived!  :mad:

Anyway, no it isn't. The $n$th triangular number is the maximum number of intersections created by $n$ lines, however, the number of regions created is the $n$th triangular number plus $1$ (as said by Textangle).

Edit: Facepalm.

[quote="$LaTeX$"]You revived!  :mad:  [/quote]

what do you mean?

Usually, you don't want to post in a thread that hasn't been posted on in three years. Or four months, for that matter...


A huge-room has $ 1000$ doors, each with a lock provided and numbered $ 1$ to $ 1000$. A mad-man enters and at the first round, he unlocks all doors. At the second round, he locks only doors that have numbers the multiples of $ 2$. At the third round, he operates locks of doors which have a multiple of three as their number-he unlocks locked doors and locks doors which are unlocked. And this goes on for long. At his $ 1000^{th}$ round, how many doors are locked?

it does not really belong this sub-forum, does it ?

Every door changes it's state as many times, as many divisors it has. So, the unlocked doors have odd number of divisors, so they are perfect squares.

i'm confused...what happens in round 1000?


Is it bad to take, say, SAT II Physics after only taking physics honors? is it necessary to learn teh AP material? Thanks.

i have no idea what [i]your[/i] school's honors physics class's curriculum is, but the SAT II tests are made for testing high school learning, not freshman college learning, which is what the AP tests (are supposed to) do. So no matter what physics class you took, as long as you took a high school course, you should be fine with some reviewing on the SAT IIs.

If I were you I would try to get a SAT II Physics review book, either from the library or purchase one, and see how much you know of it.  I agree with khan-it depends on the quality of your schools honors program.

I cant speak for Physics, but I took first year chemistry at my high school, got an A for the final grade, and I took the SAT IIs in June and got a 760.

don't take it unless you are going to study very HARD by yourself, very hard


[quote="Drunken_Math"][hide]
$f(x)=(x-1)^2+3(x-1)+5$[/hide]
[/quote]

[hide="clarification"]
He just substituted $(x-1)$ for $(x+1)$ to get $f((x-1)+1)=f(x)$
[/hide]


articles
geometry
geometric transformation
reflection

The cover edition of the latest edition of business week is found [url=http://www.businessweek.com/magazine/content/06_04/b3968001.htm]here[/url].

It's an interesting article talking about more opportunities for math in the world, particularly in business and analyzing data.

Any comments?

I notice the small numbers and the fact that we are finally having sufficient computer power and information available to make this a potential future area of great interest.  No market driven company like Ford or P&G can avoid the potential to be on the ground floor for the very small cost it takes to form such a group.

It may well be like those modeling the planet's climate -- the model is at least several orders of magnitude too simple to have meaningful results.  I'm hopefull, but expect this to be a dead end for at least a while.  Some areas, limited in scope and time, will probably work, but there will be large mistakes (model doesn't reflect real market).  For example with the test markets the army of MBAs unleached on businesses -- Polaroid almost lost the company when its test market for SX-70 failed to note that some folks who did not live in the test markets bought cameras anyway and the factory was built ten times too large.

It'll be great to look back 30 years from now and see how mislead we were in discussing this potential.  I'm sure we will be wrong -- I just don't know which way.


floor function
number theory unsolved
number theory

The sum of $ k$ different even and $ l$ different odd natural numbers equals
1997. Find the maximum possible value of $ 3k + 4l$ .

The sum of the odd numbers is at least $k^2$ (the sum of the first $k$ odd numbers), and the sum of the even numbers is at least $l^2+l$ (the sum of the first $l$ even numbers), so $3l+4k\le3l+4\lfloor\sqrt{1997-l^2-l}\rfloor\le221$ for $0\le l\le44$ ($l<45$ since $45^2+45=2060>1997$).

This bound is achievable, taking all the even numbers less than $56$, all the odd numbers less than $67$, and the number $85$ (for instance).

[quote="Diarmuid"]  $3l+4k\le3l+4\lfloor\sqrt{1997-l^2-l}\rfloor\le221$ for $0\le l\le44$ ($l<45$ since $45^2+45=2060>1997$).

This bound is achievable, taking all the even numbers less than $56$, all the odd numbers less than $67$, and the number $85$ (for instance).[/quote]
So the maximum value is 436?And an another how do you find the number 221?

$436$? No, the maximum value is $221$.

I found it by making a table of the relevant values, and looking (I did it in Excel, but it's small enough that you could make it by hand in an exam).  I don't know of any smarter way of doing it.

[quote="Diarmuid"]
This bound is achievable, taking all the even numbers less than $56$, all the odd numbers less than $67$, and the number $85$ (for instance).[/quote]
How do you find this?

Oops - that should be $54$, not $56$.  The set chosen is the first $27$ even numbers, the first $34$ odd numbers, and $85$ as the remainder necessary to make up the sum.

The bound I found was that the maximum value of $3k+4l$ was $221$.  This can be achieved with $k=27$, $l=35$ (or with $k=23$, $l=38$, but we need an odd $l$ to have an odd sum).  So we take the first $27$ even numbers, and the first $35$ odd numbers, and add them up - but this is only $1981$, so we need another $16$ -  I got this by adding $16$ to one of the numbers (in particular, $69$).


ratio
geometry
trigonometry
area of a triangle

Triangle ABC is equilateral  Extend $ AC$ through $ C$ to a point $ D$ such that $ AC \equal{} DC$.  Let $ E$ be the midpoint of $ BC$.  Draw the segment from $ D$ to $ E$.  Extend $ DE$ to hit side $ AB$ at $ F$. Find the ratio of $ BF$ to $ FA$.

[hide]
First, just set a side of $ ABC$ to be 2 (for the sake of ease) and $ BF$ to be $ x$.
Now, we will find the ratio of the area of $ AFD$ to $ AED$ in two different ways.

First, the altitude from F to AD has length $ \frac{(2\minus{}x)\sqrt{3}}{2}$ since it is the longer leg of a 30-60-90 triangle.
The altitude from E to AD is $ \frac{\sqrt{3}}{2}$ since it is the shorter leg of a 30-60-90 triangle.
Now, since $ AFD$ and $ AED$ have the same base of 4, the ratio of the areas is just $ 2\minus{}x$.

Next, a different way to find this ratio:
$ [AFE] \equal{} \frac{2\minus{}x}{4}[AED]$ so $ \frac{[AFD]}{[AED]} \equal{} \frac{6\minus{}x}{4}$.

Equating the 2 ratios and solving, we have $ x \equal{} \frac{2}{3}$ so $ \frac{BF}{FA} \equal{} \frac{1}{2}$.[/hide]

Pretty complicated for a MC problem...

[quote="alkjash"]

Next, a different way to find this ratio:
$ [AFE] \equal{} \frac {2 \minus{} x}{4}[AED]$ so $ \frac {[AFD]}{[AED]} \equal{} \frac {6 \minus{} x}{4}$.

Pretty complicated for a MC problem...[/quote]

I think you should probably explain this.  Or I will =)... So $ \angle EAD\equal{}\angle EAB$.  Since the area of a triangle can be computed by $ \frac{1}{2}ab\sin\alpha$, the ratio of the two areas is simply the ratio of the two corresponding sides.  Since $ AE$ is a common side, we simply need $ \frac{AB}{AD}\equal{}\frac{2\minus{}x}{4}$.

Nice solution by the way.  I made the problem up xD. Use menelaus if you know it.  Gives you the answer in one step.

Ahh... right looked something like that..
Haven't gotten around to memorizing Menelaus.  :blush:
Doesn't have as nice a pattern as Ceva, you know?

Ummm...
Not necessarily...
At least the way I saw it proved...


function
inequalities
algebra unsolved
algebra

Let $f$ be a function from $R$ to $R$ sastisfying $\exists f''(x)\ \ \ \forall x\in R$
Assume that : $|f(x)|\le a$
$|f'(x)|\le b$
$|f''(x)|\le c$
Where $a,b,c$ are constant
Prove the inequality : $b\le \sqrt{2ac}$


prove that in a triangle we have
sinA*sinB+sinB*sinC+sinC*sinA<=cos(A/2)*cos(B/2)+cos(B/2)*cos(C/2)+cos(C/2)*cos(A/2);
Do you have a strongert ineq.?

$2\sum{sinAsinB}=\sum(cosC+cos(A-B))$
$2\sum{cos(A/2)cos(B/2)}=\sum(sin(C/2)+cos\frac{A-B}{2})$
But $\sum cosC \le \sum sin(C/2)$
$\sum cos(A-B) \le \sum cos\frac{A-B}{2}$

Am I wrong?


if $ A_n$ is not = 0 and $ {[A_n]}^{\infty}_1$ Converges $ \lim_{n \rightarrow \infty} \frac{A_{n\plus{}1}}{A_n} \equal{} L \Rightarrow |L| \leq 1$
how can i prove it


function
trigonometry
limit
integration
calculus
calculus computations

Let $f,g\;: \mathbb R\rightarrow\mathbb R$ with $g(x)=\sqrt{\{x\}-\{x\}^2}\qquad$   and    $\qquad f(x)=\arccos(\cos x)$; $\{x\}$ is the usual fractional part of x.
I) Prove that the two functions are integrable in every conpact set of $\mathbb R$
II) Compute
\[ \lim_{x\to \infty}\frac{\int_0^xf(t)\,dt}{\int_0^xg(t)\,dt}  \]

You can easily compute the upper integral for $x=k\pi$, and the lower one for $x$ positive integer. Since the numerator goes to infinity as $x \rightarrow + \infty$, you can compute the quotient and you'll find $4$.


Define a function f from the interval I = [0, 2π) of R to R2 by setting f(t) = (cos t, sin t).
Show that the inverse function  from f(I) to I is not continuous.

[quote="buddhu"]f : I -> R^2 has image f(I) = S^1 which is a compact subset of R^2; but the inverse map from f(I) to R maps S^1 to [0, 2 \pi ) which is not compact  in R. therefore as the inverse image of a compact set is not compact, f inverse is not continuous.  :maybe:[/quote]

This works in this case since we are showing that the [i]inverse[/i] map is not continuous, but I just want to note the following: the continuous image of a compact set is compact (this is the result we used to show $ f^{\minus{}1}$ is not continuous), but the preimage of a compact set under a continuous map is not necessarily compact. If the preimage of every compact set under a continuous map is always compact, then the map is called a proper map.

In other words, this exercise asks to show that $ f$ is not a proper map. I think buddhu understood this already, but I just wanted to make sure diophantine interprets the solution correctly.


function
floor function
trigonometry
ceiling function
logarithms
calculus
integration

Alright, so I'm looking for a way to rewrite the floor function as $x-$ (some composition with trigonometric functions)

One idea I have is using Arcsin(sin(x)) or something of that nature somehow. Any help in constructing an equivilent to [x] would be appreciated  :)

What's the problem with the current definition?

Nothing, I just want to see if floor can be written using trigonometric functions.

Well you're not going to be able to write it with continuous functions or any composition of such, because it's discontinuous at infinitely many points.

Alright, so I've made a bit of progress, but now I'm stuck:

So far I have:

$\lfloor x\rfloor=x-\sqrt{\left(\frac{\sin^{-1}(\cos (\pi x))}{\pi}-\frac{\sin^{-1}(\sin (\pi x))}{2\sqrt{(\sin^{-1}(\sin (\pi x)))^{2}}}\right)^{2}}$

Here's the thing: it works for all non-integers, but whenever I put in an integer, it encounters an indeterminate from the $\sin (\pi x)$.

The only reason that whole thing is there is to take care of the fact that the $\frac12$ is either positive or negative depending on the value of $\sin (\pi x)$...

[quote="lotrgreengrapes7926"]I was wondering if it was possible to define the ceiling function in terms of $\lfloor x\rfloor$ and $[x]$, since I don't think there's a ceiling function on my graphing calculator.[/quote]
$\lceil x\rceil =-\lfloor-x \rfloor$
I like to imagine it as a composition of reflections.

I'm not sure I understand exactly what you want. Can you tell me what's wrong with simply $\lfloor x \rfloor=x-\frac{sin^{-1}(sin(2\pi x))}{2\pi}$

Because its not, since Arcsin(sin(x)) still increases and decreases.

So your suggestion would resemble the attached:

$\frac{\tan \pi x/2 }{|\tan \pi x/2 |}= (-1)^{\lfloor x \rfloor}$, and then you just have to take a logarithm base $-1$ ...  :P

[quote]...more "accurately" defined functions. [/quote]

There is no sense in which $\sin$, $\sin^{-1}$ or any other trigonometric function is "more accurately defined" then the plain old floor function.  (There is still, of course, the interest in doing something for the sake of doing it.)

I guess its more of that, just trying to do it for the sake of doing it.

See, the thing about your tan thing is, tangent function becomes zero at some integral values, so you'd get an indeterminate (which is what's wrong with
$\lfloor x\rfloor=x-\sqrt{\left(\frac{\sin^{-1}(\cos (\pi x))}{\pi}-\frac{\sin^{-1}(\sin (\pi x))}{2\sqrt{(\sin^{-1}(\sin (\pi x)))^{2}}}\right)^{2}}$
as well...

Do you just want a [i]finite[/i] function?  Because $x-\lfloor x \rfloor$ has a very simple Fourier series.

[quote="t0rajir0u"][quote="rnwang2"][quote="lotrgreengrapes7926"]I was wondering if it was possible to define the ceiling function in terms of $\lfloor x\rfloor$ and $[x]$, since I don't think there's a ceiling function on my graphing calculator.[/quote]
$\lceil x\rceil=\lfloor x\rfloor+1-\left\lfloor 1-x+\lfloor x\rfloor\right\rfloor$.[/quote]

Bictor's answer

[quote="Bictor717"] $\lceil x \rceil =-\lfloor-x \rfloor$ [/quote]

seems a little nicer to me.   :P[/quote]
 :rotfl:


geometry
circumcircle
incenter
Euler
power of a point
radical axis
geometry unsolved

Let P be the intersection of the diagonals AC and BD of the convex quadrilateral ABCD for which AB=CA=BD.
Let O and I be the circumcenter and incenter of the triangle ABP.
Prove that if $ O \neq I$, then $ OI \perp CD$.
 :help: please~

If $ a \equal{} BP, b \equal{} PA, p \equal{} AB$ are side lengths of the $ \triangle ABP,$ $ \overline{CP} \equal{} p \minus{} b, \overline{DP} \equal{} p \minus{} a.$ External bisectors of the angles $ \angle PAB, \angle ABP$ cut $ BP, AP$ at $ X, Y$ and $ \frac {\overline{XP}}{\overline{XB}} \equal{} \frac {b}{p},$ $ \frac {\overline{YP}}{\overline{YA}} \equal{} \frac {a}{p}$ $ \Longrightarrow$ $ \overline{XP} \equal{} \frac {ab}{p \minus{} b},$ $ \overline{YP} \equal{} \frac {ab}{p \minus{} a}$ $ \Longrightarrow$ $ \frac {\overline{DP}}{\overline{CP}} \equal{} \frac {p \minus{} a}{p \minus{} b} \equal{} \frac {\overline{XP}}{\overline{YP}}$ $ \Longrightarrow$ $ YX \parallel CD.$

If $ I_a, I_b, I_p$ are excenters of the $ \triangle ABP,$ $ A, B, P$ are the altitude feet of the excentral $ \triangle I_aI_bI_p,$ $ I$ is its orthocenter, $ (O)$ its 9-point circle, $ OI$ its Euler line. The circumcenter $ O'$ of the $ \triangle I_aI_bI_p$ is on $ OI.$ $ API_aI_p$ and $ BPI_bI_p$ are cyclic because of the right angles $ \angle I_pAI_a, \angle I_pPI_a$ and $ \angle I_pBI_b, \angle I_pPI_b$ $ \Longrightarrow$ $ \overline {YP} \cdot \overline {YA} \equal{} \overline {YI_a} \cdot \overline {YI_p}$ and $ \overline {XP} \cdot \overline {XB} \equal{} \overline {XI_b} \cdot \overline {XI_p}$ $ \Longrightarrow$ $ YX$ is the radical axis of $ (O), (O')$ perpendicular to the center line $ O'O$ identical with $ OI$ $ \Longrightarrow$ $ CD \perp OI.$

Thanks a lot ~ I am happy for your solution.
:coolspeak:


This is the thread that mock Mathcounts that I'll make would be posted.

I'm not quite confident whether I'll be able to hold any timed contest (since I have other stuff to do), here is what I'll do.

I'll post the questions on PDF.

After a day or two, I'll post the solution packet on another post (PDF also).

I'll probably include all sprint, target, team, countdown (I'm thinking of around 40 questions? Is that sound okay? :) ), and if I have good idea, maybe a master round (I have All-Time Greatest Mathcounts Problems so I might be able to know what they require in there) and like always, those solutions are what I think, not what you should have. :lol: There are always more than one way to solve the problems.

Before the answer packet is posted, you can discuss the questions or whatever you prefer to do.

If you're okay or happy with this idea, please post here.

Thanks and Good luck on your National MC.

Silverfalcon

Should we follow regular time rules (40 min sprint 24 target 20 team)?

It depends on you.

If you want to be real, do it that way.

If you just want to see how it goes, do whatever you prefer.

I think this would be great! Old national problems are so hard to access  . . . I need more problems to practice on  . . ..

Around what difficulty will this test be?

I thought he was in 6th grade--I could be wrong.

[quote="Max300"][quote="DarkKnight"]Hmm, this sounds like fun. :D I need practice so I can even GO to mathcounts next year.[/quote]

Aren't you in high school?[/quote]

lol, akash, i think you have the wrong andrew...

[quote="DarkKnight"][quote="Max300"][quote="DarkKnight"]Hmm, this sounds like fun. :D I need practice so I can even GO to mathcounts next year.[/quote]

Aren't you in high school?[/quote]

Uhh? What? I'm in 6th grade. How'd you get the idea I was in high school?[/quote]

there was a guy who got 5th in state last year named Andrew Wang (at least in texas)

[quote="tarquin"]Yeah Andrew qualified for the USAMO.[/quote]

Yeah. I was looking at the USAMO qualifiers and that guy had the same name as me. :P I'm only in 6th grade.

i guess it's an easy mix up, but last year, andrew and mark were the two to beat all year in mathcounts, they both went to sartartia and always got like 1 and 2 with a 46 and 45, now mark moved to fort settlement and has dennis on his team instead...lol, hard to imagine mark zhang getting a non-perfect in mathcounts


Posts related to Inorganic and Analytical Chemistry.


The surface of a $ 4'' \times 5'' \times 6''$ block is painted yellow. If the block is separated into one-inch cubes, what is the number of cubes with exactly one yellow face?

Uhh.... isnt this one just (2*3+3*4+2*4)*2=52????

The answer on FTW was 26, and i don't see what i did wrong.

I don't see what you did wrong either.  Maybe the answer-typer forgot to multiply by 2?


Determine the maximum value that is obtained by multiplying together a set of positive integers which are all different and whose sum is 70.

[hide=" I think it helps.."]AM - GM gives us for the set of positive integers $ \{a_1, a_2, ...,a_n\}$

$ \frac{a_1 \plus{} a_2 \plus{} ... \plus{} a_n}{n} \geq \sqrt[n]{a_1a_2...a_n}$

$ \frac{70}{n} \geq \sqrt[n] {a_1a_2...a_n}$

$ \left( \frac{70}{n} \right) ^n \geq a_1a_2...a_n$

Maximum occurs when there is equality. So now the problem is just how to optimize $ \left( \frac{70}{n} \right) ^n$.[/hide]

[hide="comment"]Nice methodology. However in terms of the given conditions the set of positive integers must be all different, and thus a(1), a(2),...., a(n) cannot be equal.[/hide]

Solution:
[hide]Mark the numbers we choose for the set like this:
_, 2, 3, 4, 5, _, _, _, 9, 10, 11, 12, _, 14

Then if we ever have a blank after a number (like the 5) and then a different blank before a bigger number (like the 9), we can move those two numbers closer together (in this example changing them to 6 and 8). This will increase the product by AM/GM or Jensen's or whatever inequality you want to use.

We can perform this operation whenever there are two different blanks somewhere inside our set of numbers.

The set with maximal product, therefore, can have at most one blank inside it, so it must be a sequence of consecutive numbers with one of the numbers missing.

Also our set must have at least one of 2 and 3; otherwise we could just put 2 into the set and decrease the lowest odd number by 2, which increases the product. So, the string of consecutive numbers starts with either a 2 or a 3.

From there getting the solution is easy:
2, 3, 4, 5, 6, _, 8, 9, 10, 11, 12[/hide]

The AM-GM approach would still work if done in cases.  If we want $ k$ numbers that multiply to a maximal value, we can make them as close as possible to $ \frac{70}{k}$ while still being distinct.  I'll have to check the cases on that though...

[quote="cyberspace"]If we want $ k$ numbers that multiply to a maximal value, we can make them as close as possible to $ \frac {70}{k}$ while still being distinct.[/quote]
In that case, the obvious thing to do is to assume that the numbers are consecutive. Suppose the numbers are $ n,\,n \plus{} 1,\,n \plus{} 2,\,\ldots,\,n \plus{} k \minus{} 1.$ The sum of these numbers is

[list]$ \frac {k(n \plus{} n \plus{} k \minus{} 1)}2\ \equal{} \ 70$[/list]
giving $ k(2n \plus{} k \minus{} 1) \equal{} 140.$ So $ k$ is a divisor of 140. Combining this with the given conditions, we find that the following are the only possibilities:

[list]$ k \equal{} 4,\ n \equal{} 16 \\
[2mm] k \equal{} 5,\ n \equal{} 12 \\
[2mm] k \equal{} 7,\ n \equal{} 7$[/list]
Hence the maximum product for a selection of consecutive integers is

[list]$ 7\times8\times9\times10\times11\times12\times13 \equal{} 8\,648\,640.$[/list]

So, are there non-consecutive integers whose product is higher than this (or can we prove that there aren’t)? ;)


Where x,y,z are positive reals, show that $x^{2}(x-y)(x-2y)+y^{2}(y-z)(y-2z)+z^{2}(z-x)(z-2x)\geq 0$

Can this be generalised Schur-style to $x^{r}(x-y)(x-2y)+y^{r}(y-z)(y-2z)+z^{r}(z-x)(z-2x)\geq 0$?

Sorry if either of these questions is obvious, I have only just started learning inequalities properly.

the first ineq is a very strong ineq proposed by vasc.

Yeah sorry, I should have said the source of the first is Vasc (via the Hojoo Lee sheet). The second is my own hypothesis based on the first.

Anyone got a proof?

http://www.mathlinks.ro/Forum/viewtopic.php?p=251877#p251877
There are three proofs of the first ineq there, one by Namdung, one by Vasc, oen by Stefan V. (posted by darij)

For the record, the generalisation is false. :(


1.  Find positive integers k, n such that kn! = (((3!)!)!/3! and n is as large as possible.

I got this off Kalva, but I think this question is bugged....

Answer and my solution in spoiler:

[hide]

I don't think Kalva had this problem correct.

(((3!)!)!)/3!=720!/3!... So the largest n would be 720 while k would be 6.... I dunno where I went wrong.

[/hide]

[quote="Ragingg"]1.  Find positive integers k, n such that kn! = (((3!)!)!/3! and n is as large as possible.
[/quote]

k*n!=720!/6
k*n!=120*719!

So:
k=120
n=719

Oh... Hmm I read the problem wrong. Woops! Wow... That was a stupid mistake.  :oops:

[quote="captcha000"]The question is indeed bugged. The original question (I hope AMC doesn't get mad at me for copyright issues) was
1. Given that ((3!)!)!/(3!)=k*n!, where k and n are positive integers and n is as large as possible, find k+n.[/quote]

I don't think the question was bugged; he just misinterpreted it for a second.

Yeah theres nothing wrong with the question.. thats exactly the same thing that you posted captcha :)


[quote="happyme"]What is the largest possible product of a set of positive integers whose sum is 20?[/quote]

Edit: hmm...
Moldytape pointed that out...
I should think about it more ...

[hide]Actually, 2^7x3^2=1152 is bigger and 2x7+3x2=20[/hide]

[hide]3s optimize, so put in as many 3s as possible (note that 2 2s is preferable to a 3 and a 1).[/hide]

bpms and DarkKnight are both right, while you can get 2^10 and 4^5, they both equal 1024.  So 3^6*2=1458. :D

[quote="happyme"]What is the largest possible product of a set of positive integers whose sum is 20?[/quote]

Can you please be a bit more specific on 'set'?


How many integer solutions does $ xy \plus{} 9(x \plus{} y) \equal{} 2006$ have?

[quote="rofler"]Rewrite as $ (x \plus{} 9)(y \plus{} 9) \equal{} 1925 \equal{} 7*11*25$.

$ x \plus{} 9$ has to be one of the positive or negative factors of this number. Since there are 8 positive factors, there are 8 negative factors, making 16 solutions.[/quote] You have to add $ 81$ not subtract it. So,$ (x \plus{} 9)(y \plus{} 9) \equal{} 2087$
And $ 2087$ is a prime. Which implies that $ 2$ positive and $ 2$ negative solutions and $ 4$ solutions in total.
Moreover,I think that $ 1925 \equal{} 7.11.5^2$ has $ 12$ positive divisors. BTW nice try. :)


Find the smallest integer between 100 and 999 which is equal to the sum of the factorials of its digits.

[hide]I used guess and check and I got 145.[/hide]

If only I can simplify this: $100a+10b+c=a!+b!+c!$....., which is kind of hard...

P.S. Robinhe, how did you just guess and check?

Well, I knew 5! was 120, so I thought that the number might be close to 120. So, I tried 125, 135, and I came to 145, which is my answer.  I just got lucky.

I think what Robinhe did is what I did. You want the lowest three digit one, and to get into the hundreds you need a 5, which would best go in the ones place to make it smallest. Since 5! is 120, and you need two more digits before the five, you try 1_5 and it works once you get to 145. Does this make sense? It's not the most elegant solution. Any better ideas anyone?
[b]EDIT[/b] Beaten to it yet again.

Okay I understand, but still there should be a way to simplify it enough to deduce an answer....

If you start with 5 as your highest digit, you can make a table of factorial values to help you find an answer.

5!+1!+2!=123
5!+1!+3!=127
...

THere's only like 10 possibilities.


real analysis
function
integration
limit
advanced fields
advanced fields unsolved

I am just learning about distributions on my own for an interesting problem of how generally I may apply some idea. Anyway, my personal story is not of much importance to this question.

In all that follows I am thinking of tempered distributions in particular.

I have just read that we cannot in general define pointwise product or convolution for distributions. In fact we can define it when one of the distributions has compact support. For example, the Dirac distribution is the identity operator for convolution, so why shouldn't that hold true when we convolve it with a distribution instead of a function? From this it follows we can only define pointwise product for distributions if the Fourier transform of one of them has compact support.

I am told that Schwartz proved this ages ago. But I don't really see why it is so. What problem do we encounter? Thanks...

[quote="Xevarion"]What problem do we encounter? Thanks...[/quote]
I give you two versions of the same problem.

1. Consider the Dirac delta, $\delta.$ What is the pointwise product of $\delta$ with itself?

If $f$ is a smooth function, then the pointwise product $f\cdot \delta$ is simply $f(0)\delta.$ But since $\delta(0)$ makes no sense, we clearly can't do that.

2. Let $1$ be the function on $\mathbb{R}$ whose constant value is $1.$ How can you possibly define $1*1?$

You can define convolution when one of the distributions is compactly supported - and that makes $\delta$ the identity element for convolution even with distributions. Actually, with tempered distributions, you can go a little weaker than compactly supported, if you can make some sense out of what "vanishes rapidly at infinity" should mean for a distribution.

You can define pointwise multiplications when one of the distributions is repesented by a (tempered) smooth function - and that makes $1$ the identity element for pointwise multiplication.

Thanks, Professor Merryfield.

Okay, so I guess the idea is that if I apply a distribution to a test function, it is no longer necessarily a test function. So then if we try to apply the second distribution, it doesn't make sense.

Ok, but how about this? Suppose we make a sequence of functions (nice functions, maybe $L^{2}$) that converges to the distribution. Then we can do stuff like pointwise products or convolutions with impunity. So what happens when we look at the limit of the pointwise product, inner product-ed with a test function? If the pointwise product makes sense for the distribution, will this converge to what we want, but when it doesn't, there is no limit?

Sure, let's do that. The only kind of limits that make sense are weak limits - that is, limits in the sense of distributions.

Let $\varphi$ be a test function such that $\int\varphi=1$ and $\varphi(0)=1.$ We should be able to manufacture something like that. $\varphi*\varphi=\psi$ is also a test function with $\int\psi=1.$

Now let $f_{n}(x)=n\varphi(nx)$ and $g_{n}(x)=\varphi\left(\frac{x}{n}\right).$ Take the limits: $f_{n}\to\delta$ and $g_{n}\to1.$

The pointwise product $f_{n}\cdot f_{n}$ is the function $n^{2}\varphi^{2}(nx).$ If $\eta$ is any test function such that $\eta(0)\ne0,$ then

$\lim\int n^{2}\varphi^{2}(nx)\eta(x)\,dx=\lim n\int\varphi^{2}(u) \eta(u/n)\,du$ does not exist.

Thus collapses our effort to use such limits to find $\delta\cdot\delta.$

We can compute that $g_{n}*g_{n}(x)=n\psi(x/n).$ That is not going to have a weak limit, which collapses our effort to use such limits to find $1*1.$

That's the easy part. The question you're really asking is this: had those limits existed, would they be well-defined? Let me think about that.


function
trigonometry
algebra unsolved
algebra

Find the maximum value of the function:

$ f(x)\equal{}\frac{x}{x^2\plus{}9}\plus{}\frac{1}{x^2\minus{}6x\plus{}21}\plus{}\cos {2 \pi x}, x>0.$

$ f(x) \equal{} \frac {x}{x^2 \plus{} 9} \plus{} \frac {1}{x^2 \minus{} 6x \plus{} 21} \plus{} \cos {2 \pi x}\ (x > 0).$

$ \leq \frac {x}{(x \minus{} 3)^2 \plus{} 6x} \plus{} \frac {1}{(x \minus{} 3)^2 \plus{} 12} \plus{} \cos 2\pi x\leq \boxed{\frac {5}{4}}$ when $ x \equal{} 3.$

You can also use AM-GM.


hi everyone, i just joined to your forum  :)


Find the point P in triangle ABC,to let the triangle PBC 、PAC and PAB have the same radius.
 :(  My English is so poor!!!

I think that P is not constructible with ruler-compass since it must satisfy

$ \frac {[\triangle APC]}{PC \plus{} PA \plus{} AC} \equal{} \frac {[\triangle BPC]}{PB \plus{} PC \plus{} BC} \ , \   \frac {[\triangle APB]}{PA \plus{} PB \plus{} AB} \equal{} \frac {[\triangle BPC]}{PB \plus{} PC \plus{} BC}$

Both loci are  not constructible by R & C.

I got this problem from a math magazine....I guess the point maybe the internal point...
Thank you all the same...


Inegalitatea Medilor

\[(\sum{x_i}\frac{1}{n})^n\geq{\prod{x_i}}\]

Inegalitatea Cauchy

\[\sum{x^2_i}\sum{y^2_i}\geq(\sum{x_iy_i})^2\]

Inegalitatea Cebasev

\[n\sum{x_iy_i}\geq\sum{x_i}\sum{y_i} \\ unde \ x = \ ( \ x_1 \ , \ x_2 \ , \ ..... , \ x_n ) \ \ si \ \  y = \ (  \ y_1  \ , \ y_2 \ , \ ... \ y_n ) \ sunt \\ n-uple \ la \ fel \ ordonate \ \\ daca \ \ \ \ n-uplele \ sunt \ invers \ ordonate \ atunci \ avem : \\n\sum{x_iy_i}\leq\sum{x_i}\sum{y_i}\]

Inegalitatea Minkovsky

\[ \sqrt{\sum( x_i \ + \ y_i \ )^2}\leq\sqrt\ \sum(x_i)^2} +\sqrt\ \sum(y_i)^2 \]

Inegalitatea Schur

\[ a^r(a-b)(a-c)+b^r(b-a)(b-c)+c^r(c-a)(c-b)\ge 0 \]

Inegalitatea Holder

\[ \sum(p_ix_iy_i)\leq(\sum(p_ix_i^\alpha)^\frac{1}{\alpha})(\sum(p_iy_i^\beta)^\frac{1}{\beta})\\\ unde \frac{1}{\alpha}+\frac{1}{\beta}=1\]

Inegalitatea Jensen

\[ f( \sum(p_ix_i))\leq\ \sum(p_if(x_i)) \ \ unde \ f \ este \ convexa \ pe \ intervalul \ I \\ \sum(p_i)=1 , p_1,p_2,....,p_n\geq0 \ \ x_1,x_2,....,x_n\ \in \ I \]

Daca tot vrei sa faci o lista, vezi si [url=http://www.animath.fr/cours/inegalites.pdf]materialul asta.[/url]

[quote="rockymarcianocosmin99"]Inegalitatea Medilor

\[ (\sum{x_i}\frac{1}{n})^n\geq{\prod{x_i} }\]

Inegalitatea Cauchy

\[ \sum{x^2_i}\sum{y^2_i}\geq(\sum{x_iy_i})^2 \]

Inegalitatea Cebasev

\[ n\sum{x_iy_i}\geq\sum{x_i}\sum{y_i} \\ unde \ x = \ ( \ x_1 \ , \ x_2 \ , \ ..... , \ x_n ) \ \ si \ \ y = \ ( \ y_1 \ , \ y_2 \ , \ ... \ y_n ) \ sunt \\ n-uple \ la \ fel \ ordonate \ \\ daca \ \ \ \ n-uplele \ sunt \ invers \ ordonate \ atunci \ avem : \\n\sum{x_iy_i}\leq\sum{x_i}\sum{y_i} \]

Inegalitatea Minkovsky

\[ \sqrt{\sum( x_i \ + \ y_i \ )^2}\leq\sqrt\ \sum(x_i)^2} +\sqrt\ \sum(y_i)^2  \]

Inegalitatea Schur

\[ a^r(a-b)(a-c)+b^r(b-a)(b-c)+c^r(c-a)(c-b)\ge 0  \]

Inegalitatea Holder

\[ \sum(p_ix_iy_i)\leq(\sum(p_ix_i^\alpha)^\frac{1}{\alpha})(\sum(p_iy_i^\beta)^\frac{1}{\beta})\\\ unde \frac{1}{\alpha}+\frac{1}{\beta}=1 \]

Inegalitatea Jensen

\[ f( \sum(p_ix_i))\leq\ \sum(p_if(x_i)) \ \ unde \ f \ este \ convexa \ pe \ intervalul \ I \\ \sum(p_i)=1 , p_1,p_2,....,p_n\geq0 \ \ x_1,x_2,....,x_n\ \in \ I  \][/quote]

     Ai uitat Inegalitatea lui Bergstrom! :mad: Este des utilizata!


The war has to be brought speedily to a successful conclusion and the destruction of Japanese cities by means of atomic bombs may very well be an effective method of warfare. We feel, however, that such an attack on Japan could not be justified in the present circumstances. We believe that the United States ought not to resort to the use of atomic bombs in the present phase of the war, at least not unless the terms which will be imposed upon Japan after the war are publicly announced and subsequently Japan is given an opportunity to surrender.
If such public announcement gave assurance to the Japanese that they could look forward to a life devoted to peaceful pursuits in their homeland and if Japan still refused to surrender, our nation would then be faced with a situation which might require a re-examination of her position with respect to the use of atomic bombs in the war.
Atomic bombs are primarily a means for the ruthless annihilation of cities. Once they were introduced as an instrument of war it would be difficult to resist for long the temptation of putting them to such use.  Both sides possess certain weapon capabilities (for example, poison gas), which, though not expressly outlawed, are understandably forbidden.  Every nation is held responsible for resisting the use of these tempting, yet inhumane, weapons.  If one nation was to commit a wartime atrocity, then all other nations would be justified in the
There are several alternates to the atomic bomb, which would be just as effective or perhaps more effective.
The last few years show a marked tendency toward increasing ruthlessness. At present our Air Forces, striking at the Japanese cities, are using the same methods of warfare which were condemned by American public opinion only a few years ago when applied by the Germans to the cities of England. Our use of atomic bombs in this war would carry the world a long way further on this path of ruthlessness.
Atomic power will provide the nations with new means of destruction. The atomic bombs at our disposal represent only the first step in this direction and there is almost no limit to the destructive power which will become available in the course of this development. Thus a nation which sets the precedent of using these newly liberated forces of nature for purposes of destruction may have to bear the responsibility of opening the door to an era of devastation on an unimaginable scale.

What do you guys think?

I think that you should read the book of Revelation in the bible which will tell you what will happen in the (possibly near) future.

Do not think that anybody actually supports the use of atomic bombs or similar WMD. But the way things are moving, a nuclear war may become inevitable - it may even come in our lifetime!
So enjoy life and do your good deeds while you have time. If anything can persuade the beasts in the US Administration to stay away from the inhumane deeds, it won't be our innocent opinions. It will be more advisable to concentreate on games and maths!

If a nuclear war breakes out, I hope you all are saved through Jesus!  Otherwise it would be a dark day in your life.  Eventually someone is going to be completely defeated through normal warfare, and in desperation destroy the world with bombs, bialogical diseases, and the like.  I think it was the right descision for us to bomb Japan, because if we had not done so, they would have continued fighting until every last one of them had been killed off (Examples at the many islands in the Pacific Ocean, the Japanese totally fought until they were all gone)  That is my opinion on the matter.

filletwho, I have an interest in history too, but I really hope a moderator moves this to other topics->round table, this is more the place to chill and talk about games, and ask random questions

US officials sometimes argued that if the war had continued without the use of the bombs, more japanse would have died

I can't estimate whether or not this is true.

Point is, the bomb was developed because of fear the germans were developing one too.  Later inquiries after the liberation of western europe apparently pointed out the germans were never that close.  On the other hand, some historians claim that germany actually tested a small nuclear device during the war.
Anyway, after huge spending in the manhattan project, I think they needed to show that it had paid off actually.  Also it was a good way of demonstrating superiority towards the russians.

One question, maybe a bit morbid, don't you think the russians actually saved thousands, perhaps millions of lives, by developing nuclear weapons too.  During the Korean war, general McArthur proposed bombing several chinese cities (he was later replaced because he appeared out of control).  The plan was rejected, partially because of fear of intervention of the Soviet Union, at that time already a nuclear power.

Come to think of it, it must have been pretty scary for the soviet union to see the usa becoming the ultimate superpower, as the only nations with such weapons, even though it was only for a couple of years.

One thing bothers me, the USA is the first to point fingers at nations developing nuclear weapons, and at the same time it is the only nation to have ever bombed another country. And as despicable as you may find nuclear weapons, it is a fact that nations like Pakistan or North Korea are less likely to be attacked, because they DO have nukes.  (On the other hand, the conflict at the end of sixties between Soviet Union and China shows that two nuclear nations clashing does not immediately imply a nuclear war)

I am interested in your thoughts about this too.


i send you the second problem.
i wish you'll help me
Here the problem,
Solve equation:
$ \sqrt{1\plus{}\sqrt{1\minus{}x^2}}\equal{}x(1\plus{}2\sqrt{1\minus{}x})$

Many thanks to you!
Best, Luan.

[hide]
Well $ x\equal{}0$ $ x\equal{}1$ are solutions.
I believe there are two others.[/hide]

Fraenkel, $ x\equal{}0$ is an extraneous root.

[quote="10000th User"]Fraenkel, $ x \equal{} 0$ is an extraneous root.[/quote]

lol my bad.
i was seeing the wrong equation when i said that.

Thanks to interest the problem!
But i saw that, this equation has a root $ x\equal{}1$. i can't find any other roots.can you see again and help me solve it.
Best, Luan.

do you mind doing the algebra for your solution in latex for us?

Wait... did you just "guess and check" or actually do algebra to get $ x\equal{}1$ ?

Because I saw that $ 1$ works, but it's too tedious to do it the long way... :D

lol i just looked at the equation and said x=1 is a solution.

I think there is another real root. Having trouble finding it.


[quote="bilarev"]Is it true that if $a_{i}\in[-1;1]$ and $\sum_{i=1}^{2n}a_{i}^{3}=0$ then $\sum_{i=1}^{2n}a_{i}\leq 0?$[/quote]

Isn't this absurd? We can replace all $a_{i}$ by $-a_{i}$, the first condition remains true but the second changes its sign (and it's easy to see that equality can't always hold)?

  darij

Yeah, I saw it after I posted the problem. It was stupid.
Thanks Darij :roll:

Let $a_{i}$=sin $x_{i}$
With $\sum_{i=1}^{n}a_{i}^{3}$=0
We can prove |$\sum_{i=1}^{n}a_{i}$| $\leq$ $\frac{1}{3}$

It's simple,use sin3x=3sinx-4sin$x^{3}$ only

you should replace $\frac{1}{3}$ with $\frac{n}{3}$ :wink:


$x-y=(a-b)(a+b-c)$, not $(a-b)(a+b+c)$.
So your solution is wrong.  :(

Sorry, that's a big mistake.  :(  I will try it again.

I think you are wrong.
When $a=0.08, b=1, c=1$ -> $LHS-RHS=0.000605067 >0$.
Please check your calculation.
If $a=b$ -> $LHS-RHS = (3-2\sqrt{2})(a-c)^{2}c^{2}((1+\sqrt{2})c-a)^{2}>0$.
I checked my solution, But I can't find any mistakes.

I'm not sure how it could be wrong, I'll double check the file but I'm pretty sure there is no mistake...

I think your program is not precise.
I used Mathematica to calculate it.
What program did you use?

[quote="me@home"]In case no one's figured out yet, its false.
[img]8631[/img]
and that would have been an amazing proof, by the way :P[/quote]

if $a=0.08$ $b=1$ $c=1$ then $x=1.0064$ and not $x=1.01$

and moreover $LHS = 9.977726743....$ and $RHS = 9.977121676....$

so $LHS-RHS=6.0506742*10{}^{-}{}^{4}$

Sorry for posting an irrelevant subject.
I am planning to buy computer software for math.
Is Mathematica better or Maple better?
They seem to be very expensive, so I'm thinking about it.


trigonometry
geometry
geometric transformation
reflection
power of a point
Pythagorean Theorem

$AB$ is a diameter of circle $O$. $AB$ is extended to $C$ so that $BC=2$. ($A$, $B$ and $C$ are collinear in this order.) $D$ is a point on circle $O$ so that $CD$ is tangent to the circle. If $AB=6$, find the length of $BD$.

[hide="Hint 1"] Power-of-a-Point Theorem  :) [hide="Then, a second hint..."] [hide="Only if you really cannot figure out what to do next  :P "] Stewart's Theorem.  :wink:  [/hide] [/hide] [/hide]

Yeah, Stewart's Theorem works, but you don't even need to use it. There's an easier way :wink:

[quote="cincodemayo5590"]Well you could be lazy and use [hide="a piece of pie"] Law of Cosines[/hide] also.[/quote]

lol sam...you cannot use that if your solution sucks...random words are reserved for arnav/alex zhai who totally pwn

i guess the fastest i see is using pythagorean theorem to get $AD^{2}+BD^{2}=36$ and $\triangle ADC\sim\triangle DBC$ to get $DC=4$ and $\frac{AD}{DB}=2$...then solve from there

i suppose since i feel like showing off random geometry...here are two more solutions...

reflect $B$ and $C$ over $AD$ to $B'$ and $C'$. Let $C'D$ intersect $AC$ at $E$. By menelaus' theorem, $\frac{AE}{EB}*\frac{BD}{DB'}*\frac{B'C'}{C'A}=1$, $\frac{AE}{EB}=4$...and $AE+EB=6$, so $AE=\frac{24}{5}$.

we also have $\angle BDC=\angle B'DC'=\angle EDB=\angle DAB$, so $\triangle DAB\sim\triangle EDB$, so $DB^{2}=EB*BA=6*6*4*\frac{1}{5}$ so $DB=\frac{12\sqrt{5}}{5}$

--

or drop a perpendicular from D to AC, say D'...then AD'BC are in a harmonic range (consider other tangent from C, etc) $\frac{AD'}{D'B}*\frac{BC}{CA}=1$, $\frac{AD'+D'B}{D'B}=5$, and $AD'+D'B=6$, so $AD'=\frac{24}{5}$...using similarity, $\triangle ADB\sim\triangle DD'B$, so $DB^{2}=D'B*AB$...same answer...

Um i dont know where i am going wrong but here

[hide]you know that $OD=3$ and that $CD=4$. Now drop an altitude to $OC$ which has length $\frac{12}{5}$. Call the foot of this altitude $N$. Now you know that $NC=5-\frac{9}{5}=\frac{16}{5}$ Now you know that since $BC=2$ that $NB=\frac{16}{5}-2=\frac{6}{5}$ now you have a right triangle $DNB$ with legs $\frac{6}{5}$ and $\frac{12}{5}$ so then $DB=\frac{6\sqrt5}{5}$[/hide]

[hide="I am just going to use Stewart for this..."]$c=3$

$b=4$

$m=3$

$n=2$

$cnc+bmb=dad+man$

$18+48 = d^{2}5+30$

$d^{2}= \frac{36}{5}$

$d=\frac{6\sqrt{5}}{5}$
[/hide]
WHAT??? I'm getting the same answer as sonny.


If x is greater than 0, how large  must x be so that sqrt(x^2+x)-x shall differ from 1/2 by less than 0.02?

Firstly, we need to show that $ \lim_{x \to+\infty}({\sqrt{x^{2}+x}-x}) = \frac{1}{2}$.

Let ${ y = \sqrt{x^{2}+x}-x}$

$ \Longrightarrow y+x ={\sqrt{x^{2}+x}}$
$ \Longrightarrow y^{2}+2xy+x^{2}= x^{2}+x$
$ \Longrightarrow y^{2}+2xy = x$
$ \Longrightarrow x(1-2y) = y^{2}$
$ \Longrightarrow x = \frac{y^{2}}{1-2y}$

As $ y \to \frac{1}{2}$ from below, $ x \to+\infty \Longleftrightarrow$ as $ x \to+\infty$,  $ y \to \frac{1}{2}$ from below.

Having shown that $ \lim_{x \to+\infty}(\sqrt{x^{2}+x}-x) = \frac{1}{2}$, we now require the solution to $ {\sqrt{x^{2}+x}-x}> 0.5-0.02 = 0.48$

$ \Longrightarrow x > \frac{0.48^{2}}{1-2{\cdot}0.48}$
$ \Longrightarrow x > 5.76$

[quote="sludgethrower"]Firstly, we need to show that $ \lim_{x \to+\infty}({\sqrt{x^{2}+x}-x}) = \frac{1}{2}$.[/quote]

This isn't actually necessary to address the stated problem, although it is the result that the problem suggests.

[hide="One line"] $ \sqrt{x^{2}+x}-x = \frac{x}{ \sqrt{x^{2}+x}+x}= \frac{1}{ \sqrt{1+\frac{1}{x}}+1}$ [/hide]

[quote="davidyko"]We solve the equation $ \sqrt{x^{2}+x}-x=.48$, as it is apparent that $ \sqrt{x^{2}+x}$ is always greater than $ x=\sqrt{x^{2}}$.

$ \sqrt{x^{2}+x}-x=.48$
$ x^{2}+x=(x+.48)^{2}$
$ .04x=.2304$
$ x=5.76$

So $ x$ must be greater than $ 5.76$ for the expression above to differ from $ \frac{1}{2}$ by less than $ 0.02$.[/quote]

The problem with this is that you assumed $ f(x) = \sqrt{x^{2}+x}-x$ was an increasing function; without this knowledge, you also need to solve $ \sqrt{x^{2}+x}-x = 0.52$, and even then you're assuming that $ f(x)$ is monotonic.

[quote="t0rajir0u"][hide="One line"] $ \sqrt{x^{2}+x}-x = \frac{x}{ \sqrt{x^{2}+x}+x}= \frac{1}{ \sqrt{1+\frac{1}{x}}+1}$ [/hide][/quote]

Dammit, it's solutions like these that make me want to learn more.

And I suppose you could calc bash it into submission, as you can easily get the fact that it's monotonically increasing by taking the first derivative.

Fortunately, the fact that it's monotonic follows pretty easily from the rewrite I gave, which is perfectly elementary.   :)

And if you're wondering about the motivation, I've seen many problems use that conjugation concept; for example, to telescope a summation.

"Take the conjugate" is a title of one of the chapters of [i]Math Olympiad Treasures[/i] by Titu Andreescu. Consequently, it is a technique that you should be able to recognize quickly.