LaTeX
AMC
USA(J)MO
USAMO
AIME
geometry
AIME I

A sequence is defined as follows $a_1=a_2=a_3=1$, and, for all positive integers $n$, $a_{n+3}=a_{n+2}+a_{n+1}+a_n$. Given that $a_{28}=6090307$, $a_{29}=11201821$, and $a_{30}=20603361$, find the remainder when $\displaystyle \sum^{28}_{k=1} a_k$ is divided by 1000.

i messed around and got:

[hide] a_30 is equal to a_27+a_28 and i tried several other examples that fit the pattern. so i added a_28 to a _30 and divided by two.[/hide]

[quote="JSteinhardt"][hide]

S = a1 + a2 + a3 + ... + a27 + a28
3S = (a1 + a1 + a1) + (a2 + a2 + a2) + ...
     = (a1 + a1 + a2) + (a1 + a2 + a3) + (a2 + a3 + a4) + ... +
        (a26 + a27 + a28) + (a27 + a28 + a28)
     = (a1 + a2 + a3 + a4 + a5 + ... + a28) + a29 + a27 + a28 + a28
     = S + a28 + a30

3S = S + a28 + a30
2S = a28 + a30
S = (a28 + a30)/2

And I don't have the numbers in front of me, but that would IMO be the best way to do this problem.

[/hide][/quote]

Nice solution!

[quote="JSteinhardt"][quote="nat mc"][quote="JSteinhardt"][hide]

\begin{eqnarray*} {S} &=& {a_1}+{a_2}+{a_3}+\dots+{a_{27}}+{a_{28}} \\ {3S} &=& ({a_1}+{a_1}+{a_1})+({a_2}+{a_2}+{a_2})+\dots+({a_{28}}+{a_{28}}+{s_{28}}) \\ &=& ({a_1}+{a_1}+{a_2})+({a_1}+{a_2}+{a_3})+({a_2}+{a_3}+{a_4})+\dots+({a_{26} }+{a_{27}}+{a_{28}})+({a_{27} }+{a_{28}}+{a_{28}}) \\ &=& ({a_1}+{a_2}+{a_3}+\dots+{a_{27}}+{a_{28}})+{a_{29}}+{a_{27}}+{a_{28}}+{a_{28}} \\ &=& {S}+{a_{28}}+{a_{30}} \\ &\ & \\ {3S} &=& {S}+{a_{28}}+{a_{30}} \\ {2S} &=& {a_{28}}+{a_{30}} \\ {S} &=& {\frac{{a_{28}}+{a_{30}}}2} \end{eqnarray*}

And I don't have the numbers in front of me, but that would IMO be the best way to do this problem.

[/hide][/quote]
Just applied some $\text{\LaTeX}$. Good solution BTW[/quote]

Lol thx for making my format better =). Lucky for me recursive sequences happen to be one of about two things I'm actually good at mathematically (so I'll just have to hope both appear on the USAMO).[/quote]

Being good at only two things in math doesn't put you in the top 10 on TJ's math team.

And as a side note: adding makes me sad. Mostly because I can't do it correctly. At least for 28 numbers.

In the final step of this problem, when dividing the sum of a_28 and a_30 by 2, I saw that the last three digits were 668, and just quickly wrote 334 as my answer...  :blush:

[quote="me@home"]It should be noted that when JSteinhardt went from the 3rd equality to the fourth, he used the fact that $a_1=a_3$ to simplify it. For a little bit I thought that his anser was wrong before I noted that they were equal![/quote]

Oops, sorry about that, I'll make a note of that in my original solution.

@saturnlife: Yeah, someone from my school did that as well. I managed to [i]almost[/i] do something worse -- when I originally did the problem, I solved it and got (a28 + a30)/2, so I just took the first two big numbers I saw and added them together, then divided them by 2...luckily when I was checking my work I noticed that they gave us [i]three[/i] numbers (a28, a29, and a30).

[quote="me@home"]I'm just wondering, when you solved the problem during the actual test, did you use that fact ($a_1=a_3$) or did you not notice it and still get the right solution?  :P[/quote]

Well, I believe I actually had like $3S = S - 3 + 3 + a_{28} + a_{30}$. In other words I sort of just cut off the ends and added them back in to make sure I wasn't missing anything. Then to be double-sure (there was no way I was going to miss a problem with a solution this cool after finding it!) I checked the $S_n = \frac{a_n + a_{n+2}}{2}$ formula for a bunch of smaller cases to make sure it was actually true.

[quote="me@home"]yes, that's a good idea. I took the AIME 1 this year (it was my first year), but when the AIME 2 was posted I took a mock AIME at home with the problems. Anyway, when I solved this one I got the formula like you did and had everything the same but somehow got exactly 500 less than your answer. I guess I should have chekced for smaller cases.[/quote]

Did you just add the last three digits and then divide? If so you have the problem of 1000 not being a prime mod (so you have to consider the last [i]four[/i] digits).

[quote="nat mc"]I brute forced it with just the last 3 digits and got the right answer[/quote] Me too. :lol:

are you guys serious... without a calculator.. that's crazy (how long did it take)

That's worth is since you have 12 mins per problem if you do all.

I don't think I could concentrate that long...
anyways I was having a bad day and would give up on problems requiring just a bit of hard calculation.
I guess on a three hour test 15-20 minutes isn't such a waste, especially on a 10.

I didn't take that long while bruteforcing, but I didn't double check because I just made sure my a_28 and a_29 had the correct last 3 digits. When they matched up, I knew it had to be correct.

So it only took me around 5-10 minutes.

[quote="nat mc"]15-20 minutes with double and triple checking (yeah, way too long)[/quote] That is just 8 min above the average prob time.

I can offer another solution:

We'll denote $a_{n}$ as the $n$th term and $s_{n}$ as the sum of the first $n$ terms. I listed out a couple terms of both and realized the pattern: $s_{n}+s_{n+1}=a_{n+3}$. Proof: The base case is true- $s_{1}=1, s_{2}=2, a_{4}=3$. Assume that the formula is true for $n$.
$s_{n}+s_{n+1}=a_{n+3}$. Then \[s_{n}+a_{n+1}+s_{n+1}+a_{n+2}= a_{n+1}+a_{n+2}+a_{n+3}\\ s_{n+1}+s_{n+2}=a_{n+4}\]

[quote="beta"]Why did they make this so easily brute-forceable they can be like

a_1323023480237489459015702435892346095=3243(mod 10000), and it will be a lot less brute-forceable[/quote]
I think all AIME problems are brute force able, they just have a clever solution at the same time.

I also brute forced with just the last three digits. Except I started from the top as well as the bottom and met at like 20. When they didn't match up, I redid it from the bottom and easily found the problem.

[quote="nat mc"]15-20 minutes with double and triple checking (yeah, way too long)[/quote]

Hmm... not bad for an AIME problem for brute forcing...  :rotfl:

You can also "brute force" some of the geometry by using compass/protractor/ruler.

[quote="bpms"][quote="SplashD"][quote="Karth"]We're allowed to use compass, ruler, and protractor on geometry problems?:huh:  :mad: Dang... my AIME score could be so much higher on those 'find the angle problems' where the angle is an integer  :P[/quote]

you can get an estimate of the answer that way, but not an exact answer. :P[/quote]
You can construct it twice to double check :D. Seriously though, what if the answer isn't an integer and in says something like "multiply x by 100" :D.[/quote]

Well, then you're a bit dead, unless you know how to do the problem normally. If you know how to solve the problem (you shouldn't be doing constructions in the first place), go ahead and construct the problem once to get a rough estimate of the answer and then solve. If you have no idea, construct it like 10 times and see which number the angle comes closest too.  :lol:

I would have been dead on the AIME I #10 this year without the geometry tools. I had to draw out all the circles and play around with lines before I figured out that the line went through a certain point.

It's very easy to brute force taking modulo $125$ and $8$

This is a very easy problem to bash but it takes a little bit of time.

After bashing we get the answer $\boxed{834}$.

Let $x=\displaystyle \sum^{28}_{k=1} a_k$. We can find via repeated substitution that $x=\frac{x+a_{28}+a_{30}}{3}$. Solving for $x$ and taking $\pmod{1000}$ yields $\boxed{834}$.

[hide=Thoughts]A bit (way!?) too easy for p11. Solved this much faster than I would in an expected p11 (if i solve it at all).[/hide]

[hide=Post #39 by OlympusHero][url=aops.com/community/user/350767][b]OlympusHero[/b][/url] · 6 minutes ago [url=aops.com/community/p23947714](view)[/url][color=transparent]helo[/color]
p10: Let the sum be $p$. Multiplying $p$ by $3$ gives $3p = p + 2a_{28} + a_{27} + a_{29} = p+a_{28}+a_{30}$, so $p = \frac{a_{28} + a_{30}}{2} \equiv \boxed{334} \pmod {1000}$.

-----------
[color=#5b7083][aops]x[/aops] 39[color=transparent]hellloolo[/color] [aops]Y[/aops] 0 [color=transparent]hellloolo[/color] [/hide]

[tip=@OlympusHero][img]https://avatar.artofproblemsolving.com/avatar_350767.png?t=1638738494[/img]
AoPS User[/tip]

This is P11...

The numberings is me counting how many problems I have done today

[quote=OlympusHero]p10: Let the sum be $p$. Multiplying $p$ by $3$ gives $3p = p + 2a_{28} + a_{27} + a_{29} = p+a_{28}+a_{30}$, so $p = \frac{a_{28} + a_{30}}{2} \equiv \boxed{334} \pmod {1000}$.[/quote]

you sillied but i am honored

Fixed now.

[quote=asdf334]but i am honored[/quote]

:rotfl:


invariant
induction
combinatorics solved
combinatorics
Processes

On a blackboard we have the numbers $1, 2, 3, \ldots, 2001$. An operation is this: we erase $a$ and $b$ and we replace them with one number, $\frac{ab}{a+b+1}$. After 2000 operations, we are left with one number $k$. What is $k$?

Maybe an invariant works ? Just an idea, I didn't try it not yet but this seems to be a typical invariant problem to me.

Yeah, an invariant works. That's what's so annoying about these: you have to figure out the invariant. And by "figure out" I mean "guess" :D

Indeed, the invariant is not so obvious.
The ab in the numerator and the a + b + 1 suggest something like (a + 1)(b + 1) ...
But I don't have a solution not yet

Here's another one: we have the numbers 1, 1/2, .., 1/2001 and we replace a and b by (a^2+b^2-a^2*b^2)^1/2. The same questions.

let me get this straight there is a somewhat order in which you must do the operations to reach all numbers equal right?

it seems to me that if you manage to somehow prove that this CAN be done, you might easily derive the solution from your argument.

that's all for now

I'm REALLY sorry. I misunderstood. The operation is this: we erase a and b and we write ab/a+b+1 just ONCE, not TWICE, as I said in the first post. What we have to prove is that no matter what sequence of moves we do we end up with the same number. This is much easier now. An invariant works (I saw that in the soln at the end of the book where I found this prb), but we don't need invariants; it's enough to prove that the operation is associatve. This is really easy.

Sorry again for the confusing post.

thanks for sharing that ... I was getting quite nervous trying to prove that the problem is not correct :)

I think it is 1/2001

f(a,b) = ab/(1+a+b)

It is easy to verify that :

f(f(a,b),c) = f(f(a,c),b) = ...
f(f(a,b),f(c,d)) = f(f(a,c),f(b,d)) = ...

So we can do the operations in the order we want.

Now f(1/n,n+1) = 1/(n+1)

So by induction g(n) = f(f(f(f(f(...(1,2),3),4,......,n))))..))) = 1/n

Now g(2001) = 1/2001 is what we are looking for  :D

The invariant, by the way, is that each operation preserves \[ \prod_{x \in B} \left( \frac{1}{x} + 1 \right) \] where $B$ is the set of numbers on the blackboard.

Verification of this invariant, in painstaking detail:

\begin{align*} \left(\frac{1}{\left(\frac{ab}{a+b+1}\right)} + 1 \right) &= \frac{a+b+1}{ab} + 1 \\
&= \frac{ab+a+b+1}{ab} \\
&= \frac{(a+1)(b+1)}{ab} \\
&= \left(\frac{a+1}{a}\right)\left(\frac{b+1}{b}\right) \\
&= \left(\frac{1}{a} + 1\right)\left(\frac{1}{b} + 1\right) \end{align*}

Initially, our invariant is \[ \left(\frac{1}{1} + 1\right) \left(\frac{1}{2} + 1\right) \cdots \left(\frac{1}{2001} + 1\right) = \left(\frac{2}{1}\right) \left(\frac{3}{2}\right) \cdots \left(\frac{2002}{2001}\right) = 2002, \] so at the end the final value $k$ satisfies \[ \frac{1}{k} + 1 = 2002, \] and $k = \boxed{\frac{1}{2001}}$ is the answer.


Find the product of $ \left(1 \minus{} \frac{1}{3}\right)\left(1 \minus{} \frac{1}{4}\right)\left(1 \minus{} \frac{1}{5}\right)\cdot\ldots\cdot\left(1 \minus{} \frac{1}{76}\right)$ expressed as a common fraction.

hello, using that
$ \prod_{i\equal{}3}^{n}\left(1\minus{}\frac{1}{i}\right)\equal{}\frac{2}{n}$ we get as the result $ \frac{1}{38}$.
Sonnhard.

Or, we can write the products and see a pattern:

$ (\frac{2}{3})(\frac {3}{4})(\frac{4}{5})$ and so on. Note that the numerator and denominator cancel out between consecutive fractions. This leaves us with the last fraction:

$ \frac{2}{76}\equal{}\frac{1}{38}$

well for this problem we could first just list out the fraction answers

we get this

2/3*3/4*4/5

we can cancel and have

2/76=1/38

so our answer is 1/38 or $\boxed{1/38}$

 :nhl:  :nhl:   :nhl:    :nhl: '

may the force be with you


A coin tossed 2n times.
How many of the 2^2n outcomes have the number of heads and tails equalized for the first time after 2n tosses?

Any help would ne appreciated.
Thanks!

So if you have $ 2n $ tosses, $ n $ are heads and $ n $ are tails. Thus it is just $ \binom{2n}{n} $ ways to arrange $ n $ of either among $ 2n $ tosses.


calculus
integration
trigonometry
logarithms
calculus computations

Find the integral

$ \int_{0}^{\pi/2}\frac{\cos^4x}{\sqrt2\minus{}\sin2x}\cdot{dx}$

Observe

[hide]$ \cos^4 x \equal{} \cos^2 x (1 \minus{} \sin^2 x) \equal{} \cos^2 x \minus{} \frac{1}{4}\sin^2 2x \equal{} \frac{1}{4}\left(2 \minus{} 2 \plus{} 4 \cos^2 x \minus{} \sin^2 2x\right)$

$ \equal{} \frac{1}{4}(2 \plus{} 2(2\cos^2 x \minus{} 1) \minus{} \sin^2 2x) \equal{} \frac{1}{4}(2 \plus{} 2 \cos 2x \minus{} \sin^2 2x)$

$ \equal{} \frac{1}{4}((\sqrt{2} \minus{} \sin 2x)(\sqrt{2} \plus{} \sin 2x) \plus{} 2 \cos 2x)$.

Therefore,

$ \frac{\cos^4 x}{\sqrt{2} \minus{} \sin 2x} \equal{} \frac{1}{4}\left(\sqrt{2} \plus{} \sin 2x \plus{} \frac{2 \cos 2x}{\sqrt{2} \minus{} \sin 2x}\right)$.

Now it is a trivial matter to integrate:

$ \int_{x\equal{}0}^{\pi/2} \frac{\cos^4 x}{\sqrt{2} \minus{} \sin 2x} \, dx \equal{} \frac{1}{4}\left[x\sqrt{2} \minus{} \frac{\cos 2x}{2} \minus{} \log(\sqrt{2} \minus{} \sin 2x) \right]_{x\equal{}0}^{\pi/2}$

$ \equal{} \frac{1}{4}\left(\frac{\pi}{\sqrt{2}} \plus{} \frac{1}{2} \minus{} \log \sqrt{2} \plus{} \frac{1}{2} \plus{} \log \sqrt{2} \right) \equal{} \frac{2\plus{}\pi\sqrt{2}}{8}$.[/hide]


geometry
calculus
integration
calculus computations

I have a problem that asks me to find the area under (x/((x^2)-(2x)-3)) from x=0 to x=5.

I used partial fraction decomposition.  If I take the integrals of the parts, one part will integrate to the log of a negative number.

Is the area infinite?

$\int_{0}^{5}\frac{x}{x^{2}-2x-3}\,dx= \int_{0}^{5}\frac{x}{(x-3)(x+1)}\,dx$

The demoninator is zero when $x=3.$ This makes this an improper integral, and it is improper for two reasons: the left side of 3 and the right side of 3. We must break it into pieces.

$\int_{0}^{3}\frac{x}{(x-3)(x+1)}\,dx+\int_{3}^{5}\frac{x}{(x-3)(x+1)}\,dx$

Now look at the pieces. The first integral, from 0 to 3, is the integral of a negative function. Compute it, and we find that it diverges to $-\infty.$

The second integral, from 3 to 5, is the integral of a positive function. It also diverges, to $+\infty.$

The conventions surrounding improper integrals are that when an integral is improper in more than one region, each region must separately yield a convergent integral in order for us to accept the whole as convergent. Therefore, we say that this is a divergent integral. With both a $-\infty$ and a $+\infty,$ we don't call it infinite - we don't call it anything at all, other than divergent.

Side note: the formula $\int\frac1x\,dx=\ln|x|+C$ is really two different formulas that have been combined in one line for convenience. One formula is valid for $x>0,$ and the other is valid for $x<0.$ We shouldn't be using both formulas for the same integral. If that's what we appear to be doing, it's usually becuase we're ignoring that we really have an improper integral and aren't yet dealing with it properly.


function
number theory
least common multiple
induction
arithmetic sequence
algebra unsolved
algebra

A function $f : N \longrightarrow N$ satisfies $f(n) = f(n + f(n))$ for all $n \in N$.
(a) Prove that if the range of $f$ is finite, then $f$ is periodic.
(b) Give an example of a non-periodic function $f$ with this property.

Welcome every body assign to solve this difficult problem

a) It is obviosly. Let $A_i=\{n|f(n)=a_i\},i=1,2,...k,a_i\in N$. It is easy to show that $A_i$ are arithmetic progression, and $T=lcm (a_1,a_2,...a_k)$ is period.
b) Let $f(n)=2^{1+ord_2(n)}.$

[quote="Rust"]a) It is obviosly. Let $A_i=\{n|f(n)=a_i\},i=1,2,...k,a_i\in N$. It is easy to show that $A_i$ are arithmetic progression, and $T=\text{lcm} (a_1,a_2,...a_k)$ is period.
b) Let $f(n)=2^{1+ord_2(n)}.$[/quote]
Dear Rust!  How to show that $A_i$ are arithmetic progression? Please write it clearly for me! I'm sory for my stupid!

Let me show you how it works. From our equation we can easily prove by induction that for any natural $k$ : \[ f(n+k \cdot f(n))=f(n) \] . Now,as the set of possible values is finite, there exists a number $T$ which is a product of all different values of function $f$. We 'll show that this number is a period of our function . Let's take any natural number $m$. We can write $T$ in a form $T=l \cdot f(m)$ for some natural $l$. We have then: \[ f(m+T)=f(m+l \cdot f(m))=f(m) \] Which means that our function is periodic.

Now i understand the solution. Thank you very much!

How can we prove this using induction. And where do we use that the range of f is finite?

a) It is obviosly. Let $T=lcm(f(n)),f(n+T)=f(n+lf(n))=f(n)$. I don't undestand what is needed to prove.

we have that f(n)=f(n+f(n)), not f(n)=f(n+l.f(n))

$f(n)=f(n+f(n))=f(n+f(n)+f(n+fn))=f(n+f(n)+f(n))$
are you clear now?


Oh, strange. But It can you the same method as in my problem, Furthermore, this way can be used for almost problem 4-variables. :D

I gave a similar solution, with AC-Method, in the book
Algebraic Inequalities - Old and New Methods.
The editor, Mircea Lascu - Gill Publishing House, will be present with this book at IMO-2006, in Slovenia.


calculus
integration
logarithms
function
calculus computations

No, the second does not make sense even as a principal value- it is simply divergent at 1, and the integral is $\infty$. Did you stop to think how a positive function could have a negative integral?

$ \int_{0}^{1}\frac{dx}{(\pi^{2}+\ln^{2}x)(1+x^{2})}=\int_{-\infty}^{0}\frac{e^{\pi t}\pi dt}{(\pi^{2}+\pi^{2}t^{2}) (1+e^{2\pi t})}=\frac{1}{4\pi}\int_{-\infty}^\infty \frac{\text{sech}\, \pi t}{1+t^{2}}\, dt$

Define $ f(z)=\frac{\text{sech}\, \pi z}{z}$ then $ \oint f dz=\int_{-R-i}^{R-i}+\int_{R-i}^{R+i}+\int_{R+i}^{-R+i}+\int_{-R+i}^{-R-i}\to I+0+I+0=2I$ for $ R\to \infty$

$ I=\int_\Bbb{R}\frac{\text{sech}(\pi t-i\pi)}{t-i}dt=-\int_\Bbb{R}\frac{\text{sech}\, \pi t}{t-i}dt =\int_\Bbb{R}\frac{\text{sech}\, \pi t}{t+i}dt$

$ \Rightarrow 2I=\int_\Bbb{R}\text{sech}\, \pi t \, \left(\frac{1}{t+i}-\frac{1}{t-i}\right) dt =-2i \int_\Bbb{R}\frac{\text{sech}\, \pi t}{t^{2}+1}dt$

$ 2I=2\pi i \left( \underbrace{\text{Res}\left(f,z=-\frac{i}2\right)}_{=-\frac{2}{\pi}}+\underbrace{\text{Res}\left(f,z=0\right)}_{=1}+\underbrace{\text{Res}\left(f,z=\frac{i}2\right)}_{-\frac{2}{\pi}}\right)$

$ \frac{1}{4\pi}\int_\Bbb{R}\frac{\text{sech}\, \pi t}{t^{2}+1}dt =\frac{1}{4\pi}\frac{2I}{-2i}=\frac{2\pi i \left(1-\frac{4}{\pi}\right)}{-2\pi i \cdot 4}=\frac{1}{\pi}-\frac14$


A state with ten million cares plans to issue license plates which consist of any four letters followed bt an n-digit number.  IF the state wants to have enough distinct license plates for all of the cars, what is the minimum possible value for n?

[hide="Answer"]With four letters we can have $26^4$ combinations. Since we need 10 million combinations, we can divide 10 million by $26^4=456976$. \[\frac{10000000}{456976}=22\] With one number, we can get 10. With two numbers, we can get $10^2$ combinations of numbers. So, we need two numbers after the four letters.[/hide]


[quote="Thaakisfox"]solve the following system of equations. $x$ , $y$ are a function of $t$.
$\ddot x-2\dot y+2x =0$
$3\dot x+\ddot y-8y =0$

If $x(0)=0$, $\dot x (0) =1$ , $y(0)=1$ , $\dot y (0)=-1$[/quote]
Do this by the method of Laplace Transforms.
Let $L(x) = F(s)$ and $L(y) = G(s)$

Then,
$[-s+s^{2}F(s)]-2[-1+sG(s)]+2F(s) = 0$
$3[-sF(s)]+[-1+s+s^{2}G(s)]-8G(s) = 0$

$F(s)[s^{2}+2]+G(s)[-2s] = s-3$
$F(s)[-3s]+G(s)[s^{2}-8] =-s+1$

Now use Cramer's Rule and all the boring work.


limit
function
inequalities proposed
inequalities

Imagine that you have to write expresion using n (positive integer) chars. Only chars you can use are 0,1,2,3,4,5,6,7,8,9,+,-,*,/,(,)
What is the maximal value you can get ?
I think the answer is 999...9 (n nines).

But you can add also "!" (factorial). After that I think that maximum values are:
n=1: 9
n=2: 9!
n=3: 9!!
n=4: 9!!!
...

Or you can add just "^" (without factorial). Then I think maximum values are:
n=1: 9
n=2: 99
n=3: 9^9
n=4: 9^99
n=5: 9^9^9
n=6: 9^9^99
...

My question is what will happen if you add both "!" and "^" ?
What will be maximal values?
And also are my thoughts for first three cases all right?
Has anyone got a nice proof for them?
What about introducing new chars?

I'm pretty sure for $n$ characters it's just 9!...! with n-1 factorials. Obviously, there's no point to having more than one consecutive 9 since we can replace any consecutive 9's beyond the first with, say, !'s.  Now, we prove the Lemma

k^9!...! < k!!!...!

for any k at least 9 where both sides have the same number of characters. It is obviously true for k^9 vs k!!. If we use Stirling's approximation, known to obey

$\lim_{n\rightarrow \infty} \frac{n!}{\left(\frac{n}{e}\right)^n\sqrt{2\pi n}} = 1$

Then we can write

$k!!!...! \sim \left(\frac{k!!\dots !}{e}\right)^{k!!\dots !}\cdot\sqrt{2\pi k!!\dots !} >> \left(\frac{k!!\dots !}{e}\right)^{k!!\dots !} >> k^{k!!\dots !} >> k^{9!\dots !}$

I don't know the bounds for Stirling's approximation, but if it's true that $n! > \epsilon \cdot \left(\frac{n}{e}\right)^{n}\sqrt{2\pi n}$ for any reasonable $\epsilon < 1$ (All I need is like .001; does someone know such an $\epsilon$?), then, since all instances of taking an "exponent-9"  have been shown to be non-optimal, the entire expression collapses into 9 plus a bunch of factorials.

Thanx a lot Mildorf.
2Manlio: If you like (or have nothing to do :)) you can add some new chars (for example "%" and make some function for it) and delete some of them. I don't know whether it leads to some "amazing" results. I have it from my head, so it can be really unuseful.  :D


function
integration
calculus
calculus computations

I need to integrate x^n-1 *exp(-x/2) from the bounds 0...infinity  for x >0 n >1

my book gets  2^n *Gamma Function(n).

I get 2*Gamma Function (n)

 int(x^n-1 *exp(-x/2))  =  2*int((x/2)^n-1  *exp(-x/2)  = 2*Gamma FUnction

Can someone show me what I'm doing wrong?  thanks.

$ \int_0^{\infty}x^{n\minus{}1}e^{\minus{}x/2}\,dx$
Substitute $ t\equal{}x/2$:
$ \int_0^{\infty}(2t)^{n\minus{}1}e^{\minus{}t}\cdot 2\,dt$
$ \equal{}2^n\int_0^{\infty}t^{n\minus{}1}e^{\minus{}t}\,dt$.

It looks like you didn't actually do the substitution.

When you pull out your 2n,  shouldn't you have pulled out 2^(n-1)

... and one more. There are twos in more than one place.

Adding anything here is not necessary because your question has already been answer, but anyway:
you have with $ t \equal{} x/2$, we have $ 2dt \equal{} dx$ and $ x^{n\minus{}1} \equal{} 2^{n\minus{}1}t^{n\minus{}1}$ and so we get
$ e^{\minus{}x/2}*x^{n\minus{}1}dx \equal{} e^{\minus{}t}*2^{n\minus{}1}t^{n\minus{}1}*2dt \equal{} e^{\minus{}t}*2^{n}t^{n\minus{}1}dt$.
So the book's is the correct answer.


Putnam AMC USA(J)MO USAMO calculus derivative USAMTS

I am currently a junior in high school who is preparing for the USAMO and other Olympiad level mathematics. I have unfortunately realized that I started "real math" too late (only freshman year) and want to do well on the Putnam exam, so I'm deciding to start early. I have a few questions:

What is the best way to start preparation for the Putnam examination (books, etc.)?

I am currently enrolled in Calculus BC and thus have no knowledge of mathematics above or equal to Calculus apart from derivatives and basic integration. In other words, I haven't taken linear algebra and differential equations are effectively.

What is the best way to learn these materials? I will be taking a course dealing with linear algebra next year as a senior, but is there any way I can excel in the Putnam exam?

Since you can only do Putnam four times, unless you're planning on blasting through undergrad at a high rate, there's no reason to start doing Putnam while you're still in high school.

That said, I'd say the best way to prepare is to work problems. Of course, this alone won't do you any good, but as you're going through old tests, you will see what kind of material they like to include. You will see what sort of methods they like you to employ. You can find out what your weaknesses are.

There are tons of threads about this in the forum already, but a good book is Larsen's "Problem Solving Through Problems."

It can't hurt you to study linear algebra and some more interesting techniques of analysis (like differential equations). A good book for lin alg is Hoffman and Kunze. Again, there are threads where people more knowledgeable than myself discuss the merits of college-level textbooks.

You're not taking the Putnam next year. Wait until you get to college. For now, deal with general problem-solving, notably things that require proof. Do USAMTS; if there are other mail-in proof contests ([i]e.g.[/i], Pomona/Wisconsin) look at them as well.

Yes, there are calculus-based problems on the Putnam and a good grasp of that subject helps - but it's not a calculus exam. The Putnam usually isn't too highly technical (although problems are borrowed unanounced from advanced topics on the assumption that they can be solved without necessarily knowing the topic they come from.) But for now it's more important that you deal with difficult problems than it is whether those problems are college level or high school level. There are books, of course - Zeitz, Engel, Larsen. But pull from whatever sources serve to stimulate you.


calculus
integration
geometry
conics
parabola
calculus computations

Determin integers $ m,\ n\ (m>n>0)$ for which the area of the region bounded by the curve $ y\equal{}x^2\minus{}x$ and the lines $ y\equal{}mx,\ y\equal{}nx$ is $ \frac{37}{6}$.

$ m \equal{} 3 ,   n \equal{}2$

On drawing the graph we find that the area is $ \frac{ m (m\plus{}1)^{2} }{2} \minus{} \frac{n(n\plus{}1)^2}{2} \minus{} \int_{1\plus{}n}^{1\plus{}m} x^{2} \minus{} x   dx$

On evaluating this we get $ (1\plus{}m)^3 \minus{} (1 \plus{} n)^3 \equal{} 37$

On comparing this we get $ m\equal{} 3 , n\equal{}2$

Thank you

That's correct answer. :)

Here is my solution:

Let $ S_{m}$ be the area of the region bounded by the parabola $ y \equal{} x^2 \minus{} x$ and $ y \equal{} mx$, we have $ S_{m} \equal{} \minus{} \int_0^{m \plus{} 1} x\{x \minus{} (m \plus{} 1)\}\ dx \equal{} \frac {1}{6}(m \plus{} 1)^3$ by the reduction formula $ \boxed{\int_{\alpha}^{\beta} (x \minus{} \alpha)(x \minus{} \beta)\ dx \equal{} \minus{} \frac {1}{6}(\beta \minus{} \alpha)^3\ (\alpha < \beta)}$.

Thus since $ m > n > 0$ the area given in the problem is $ S_{m} \minus{} S_{n} \equal{} \frac {37}{6}\Longleftrightarrow \frac {(m \plus{} 1)^3}{6} \minus{} \frac {(n \plus{} 1)^3}{6} \equal{} \frac {37}{6}$

$ \Longleftrightarrow (m \plus{} 1)^3 \minus{} (n \plus{} 1)^3 \equal{} 37$. Let $ a \equal{} m \plus{} 1,\ b \equal{} n \plus{} 1$ we are to determine the pair of positive integer $ (a,\ b)\ (a > b > 0)$ such that

$ (a \minus{} b)(a^2 \plus{} ab \plus{} b^2) \equal{} 37$.

$ 37$ is prime, so we have only one pair $ a \minus{} b \equal{} 1,\ a^2 \plus{} ab \plus{} b^2 \equal{} 37$ because $ a^2 \plus{} ab \plus{} b^2\geq 7$.

Therefore we obtain $ (a,\ b) \equal{} (4,\ 3)$, yielding $ (m,\ n) \equal{} (3,\ 2)$.


Find all $ n$ such $ n^4\minus{}4n^3\plus{}22n^2\minus{}36n\plus{}18$ is a perfect square. I want, if possible, a solution using congruence. I know it can be done using inequality.

Congruence can only help if there are no solutions.  If there are any solutions (and there are; $ n \equal{} 1$), then congruence won't give you any information.

Are there any restrictions on what kind of numbers $ n$ are allowed, such as $ n$ must be real, positive, integer, natural, etc.?

[hide]Anyway, It might be helpful to substitute $ n \equal{} x \plus{} 1$ to get:
$ n^{4} \minus{} 4n^{3} \plus{} 22n^{2} \minus{} 36n \plus{} 18 \equal{} (x \plus{} 1)^{4} \minus{} 4(x \plus{} 1)^{3} \plus{} 22(x \plus{} 1)^{2} \minus{} 36(x \plus{} 1) \plus{} 18$ $ \equal{} x^4 \plus{} 16x^2 \plus{} 1$.

If $ n$ must be an integer, then so must $ x$, and the obvious solution is to note that:
$ x^4 \plus{} 2x^2 \plus{} 1 \le x^4 \plus{} 16x^2 \plus{} 1 < x^4 \plus{} 16x^2 \plus{} 64$

$ (x^2 \plus{} 1)^2 \le x^4 \plus{} 16x^2 \plus{} 1 < (x^2 \plus{} 8)^2$

And so, we only have $ 7$ cases to test: $ x^4 \plus{} 16x^2 \plus{} 1 \equal{} (x^2 \plus{} j)^2$ for $ j \equal{} 1,2,3,4,5,6,7$.
We get the solutions $ (x,j) \equal{} (0,1); ( \pm 2,5)$

So, the values of $ n$ are, $ n \equal{} \minus{} 1,1,3$

As for a congruence solution, this may be hard as there are finitely many solutions, but a congruence will narrow down the possible values of $ n$ to either none or infinitely many. [/hide]


geometry
incenter
trigonometry
geometry open

[u][b](Sorry to my bad English)[/b][/u]

This is my own problem:
Let ABC be a triangle and its incenter at I
 Incircle touched BC, CA, AB at D,E,F.
DD', EE', FF' are concur at I, $\frac{D'I}{DI}=\frac{E'I}{EI}=\frac{F'I}{FI}=k$.
Prove that AD', BE', CF' are concur.

It is easy. I used trig ceva.
Let AD', BE', CF' are concur at J.
When k varied, find the locus of the J.

I think it is hyperbola. (GSP)
Furthermore, it is a Feuerbach Hyperbola.
I, H, Gergonne point, Nagel point, Mitten point, Schiffler point, X90, X177...
are on this locus.
Can you prove it? please..

Yes, it is the Feuerbach hyperbola. See the PS in my post at http://www.mathlinks.ro/Forum/viewtopic.php?t=30114 .

  Darij

Thank you,
but it is maybe too hard for me.
Can you give a hint?


search
algebra
polynomial
trigonometry
number theory
prime factorization
algebra unsolved

[quote="tuandokim"]Find $ x \in R$ so that:
a)$ sin(x) \in Q$
b)$ cos(x) \in Q$[/quote]

You just have to search for any $ p,q$ rationals such that $ p^2\plus{}q^2\equal{}1$

Since the general solution of this equation is $ (\frac{a^2\minus{}b^2}{a^2\plus{}b^2})^2\plus{}(\frac{2ab}{a^2\plus{}b^2})^2\equal{}1$, $ a,b\in\mathbb{Z}$, the values you are looking for are :

$ \arcsin(\frac{a^2\minus{}b^2}{a^2\plus{}b^2})$ $ \forall a,b\in\mathbb{Z}$, a and b not together zero

[quote="JK-"]it's obvious..... :rotfl:[/quote]

Maybe it's obvious for you, but not for the original poster. Better to give a solution or a hint than this kind of messge, dont you think ?

[quote="tuandokim"]thanks pco
can you find f(x) that sin(x) is the root of f(x)
for ex:   sin(20) and sin(25)
$ f(x) \equal{} x^3 \plus{} ax^2 \plus{} bx \plus{} c$
find a,b,c so that sin 20 is the root of f(x)[/quote]

I suppose you want $ f(x)$ to be a polynomial with all coefficents in $ \mathbb{Q}$. Else, just write f(x)=x - sin(20)

Then, the answer is no in general case. Since no polynomial with coefficients in $ \mathbb{Q}$ has transcendental roots and since $ \sin(x)$ may have transcendental values :

For example, no polynomial with coefficients in $ \mathbb{Q}$ can have as root $ \sin(\arcsin( \pi \minus{} 3))$

In the specific case of $ \sin(20)$, and supposing you mean $ \sin(20$ degrees), so $ \sin\frac {\pi}{9}$, we have :

$ \sin(3x) \equal{} 3\sin(x) \minus{} 4(\sin(x))^3$ and so : $ \sin\frac {\pi}{9}$ is root of $ \sin\frac {\pi}{3} \equal{} 3x \minus{} 4x^3$

And so $ \sin\frac {\pi}{9}$ is root of $ 3x \minus{} 4x^3 \equal{} \frac {\sqrt 3}{2}$

And so $ \sin\frac {\pi}{9}$ is root of $ 4(3x \minus{} 4x^3)^2 \equal{} 3$

And so $ \sin\frac {\pi}{9}$ is root of $ 64x^6 \minus{} 96x^4 \plus{} 36x^2 \minus{} 3 \equal{} 0$

In fact, it's rather easy to show that, for any $ p\in\mathbb{Q}$ it always exists polynomial with coefficients in $ \mathbb{Z}$ whose $ \sin(p\pi)$ is a root.

[quote="JK-"]it's obvious..... :rotfl:[/quote]

That's not necessrarily obvious. :wink:

[quote="tuandokim"]Find $ x \in R$ so that:
a)$ sin(x) \in Q$
b)$ cos(x) \in Q$[/quote]
i think what he did mean is find $ x \in Q$ so that $ sin(x) \in Q$

oh,I made a big mistake.
find $ x\in Q$ that $ sin(x) \in Q$

Only $ x \equal{} 0$.  What you mean is $ \sin \pi x \in \mathbb{Q}$, of course :)

If you want both $ \cos \pi x$ and $ \sin \pi x$ to be rational, you want a root of unity that is also in $ \mathbb{Q}[i]$.  It is a standard result that $ 1, \minus{}1, i, \minus{}i$ are the only possibilities; one way to see that no others exist is prime factorization in $ \mathbb{Z}[i]$, and another is a standard result on cyclotomic polynomials which implies that $ 2 \cos \pi x \in \mathbb{Q}$ if and only if $ x \equal{} 0, \frac{1}{3}, \frac{1}{2}, \frac{2}{3}, 1$, etc.

If you only want $ \sin \pi x$ to be rational, the same result above implies that $ x \equal{} 0, \frac{1}{6}, \frac{1}{2}, \frac{5}{6}, 1$, etc.


vector
combinatorics proposed
combinatorics

let $A$ and $B$ be families of the subsets of an n-element set with the property that $|a\cap b|$ is odd for all $a\in A$ and $b\in B|$. prove that then $|A||B|\leq 2^{n-1}$.

Translating the problem into vector language we have two sets of vectors $A,B \subset \{0,1\}^n$ with $uv=1$ whenever $u \in A, v \in B$.
Now let $b_1,\cdots,n_m$ be a basis of $B$. Then we have $|B|\leq 2^{m-1}$ because obviously the sets $B, B+b_1$ are disjoint subsets of the vectorial space spanned by $b_1,\cdots,b_m$.
Now we are left to prove that $|A|\leq 2^{n-m}$. But since $b_i$ are linearly independent there exist vectors $b_i'$ with $b_ib_j'=1$ iff $i=j$.
We see that $b_1',\cdots,b_m'$ are linearly independent because if $\sum e_i b_i'=0$ then multiplying by $b_i$ we get $e_i=0$ for any $i$.
Hence $H$ the subspace spanned by $b_1',\cdots,b_m'$ satisfies $|H|=2^m$.
Now we see that the sets $A+x, x \in H$ are all disjoint subsets of $\{0,1\}^n$. Thus we get $|A|\leq 2^{n-m}$
QED


topology
real analysis
real analysis unsolved

Prove that an open (closed) subset of a metric space $E$ is connected if and only if it is not the disjoint union of two nonempty open (closed) subsets of $E$.

THis problem is extremely difficult. Can anyone help me out.

Mucho gracias.

What is the definition of "connected"? Please quote the exact words your textbook uses to define that concept.

Then look carefully at that definition, and consider what the contrapositive would say.

Text: "A metric space $E$ is connected if the only subsets of $E$ which are both open and close are $E$ and $\emptyset$. A subset $S$ of a metric space is a connected subset if the subspace $S$ is connected.

If a metric space $E$ can be written $E = A \cup B$, where $A$ and $B$ are disjoint, nonempty, open subsets of $E$, then $E$ is not connected.

Well, here's my stab at the problem:
If part: assume that an open subset $S$ of a metric space $E$ is not the disjoint union of two nonempty open subsets of $E$. Call these two subsets $A$ and $B$. That means that $S$ is not equivalent to $A \cup B$. Our job is to prove that $S$ is connected. $S$ can be expressed as the union of a collection of mutually nondisjoint subsets. B/c $S$ is open, this collection of mutually nondisjoint subsets must be a covering for $S$, whether finite or not. I'm stuck here:

There's a theorem in my book that says: Let $\{S_{i}\}_{i \in I}$ be a collection of connected subsets of a metric space E. Suppose there exists $i_{0}\in I$ such that for each $i\in I$ the sets $S_{i}$ and $S{}_{i}{}_{0}$ have a nonempty intersection. Then $\cup_{i\in I}S_{i}$ is connected.

If I could somehow show that $S$ can be expressed this way, I can show it is connected.

I'm really sleepy now. I'll post the only if part tomorrow.

here's my take on the "only if" part:

Assume an open subset $S$ of a metric space $E$ is connected. That means it cannot be expressed as the union of two open, nonempty subsets of $E$... wait I'm confused, when we talk about connectness, how do we give a definition for the connectedness of a subspace of a metric space? FOr example, $S$ is connected, so $S$ can't be wriiten as the disjoint union of two open, nonempty, subsets of $S$ or subsets of $E$?

I am totally confused and any guidance is vastly appreciated.

THank you all so much!

The definition of connectedness of a subspace:

Let $S$ be a subset of $E$. Give $S$ the relative topology (the same metric as E). If $U$ is an open set in $E$, we can show that $U\cap S$ is open in $S$; this would be the definition in a more general topological space.

$E$ is connected if it cannot be dissected into two nonempty relatively open sets. The set $\{0,1\}\subset \mathbb{R}$ is not connected, even though it has no subsets which are open in $\mathbb{R}$.


AMC
AIME
modular arithmetic
floor function
Vieta
probability
blogs

[quote="SnowEverywhere"]Hello. I am a newbie to the AMC contests this year and I made the AIME on the AMC 10. I want to do somewhat well and do not feel very confident in my ability to find practice problems which will accurately test the skills I need to get around a 6 or higher (what I want to get). I thought that maybe to prepare for the AIME if enough people respond, we could start a AIME practice problem marathon. Even though its an old problem, here's one to start the marathon off.

[b]PROBLEM 1[/b]
For how many values of k is 12^12 the least common multiple of the positive integers 6^6, 8^8 and k?[/quote]

$ 12^{12}\equal{}2^{24} \cdot 3^{12}$. Since $ 6^6\equal{}2^6 \cdot 3^6$, and $ 8^8\equal{}2^24$, k cannot have a power of 2 larger then 24, for obvious reasons.

Also, k must have a exponent of 12 in its 3 factorization. Since k cannot contain any other primes, There are 25 possible values of k, as its power of 2 can be 0 to 24.

I'm not sure about that solution, so if it's wrong tell me.

2. Ina  parlor game the "magician" asks one of the participants to think of a three-digit number (abc), which represents the three-digit number with digits a,b,c. The "magician" then asks this person to make the numbers (acb), (bac), (bca), (cab), and (cba), and to add them, revealing the sum N. When told the value of N, the "magician" can identify the origianl number, (abc). Determine (abc) when N=3194.

BONUS: Prove that the magician can find (abc) for any N.

[quote="BOGTRO"][quote="SnowEverywhere"]Hello. I am a newbie to the AMC contests this year and I made the AIME on the AMC 10. I want to do somewhat well and do not feel very confident in my ability to find practice problems which will accurately test the skills I need to get around a 6 or higher (what I want to get). I thought that maybe to prepare for the AIME if enough people respond, we could start a AIME practice problem marathon. Even though its an old problem, here's one to start the marathon off.

[b]PROBLEM 1[/b]
For how many values of k is 12^12 the least common multiple of the positive integers 6^6, 8^8 and k?[/quote]

$ 12^{12} \equal{} 2^{24} \cdot 3^{12}$. Since $ 6^6 \equal{} 2^6 \cdot 3^6$, and $ 8^8 \equal{} 2^24$, k cannot have a power of 2 larger then 24, for obvious reasons.

Also, k must have a exponent of 12 in its 3 factorization. Since k cannot contain any other primes, There are 25 possible values of k, as its power of 2 can be 0 to 24.

I'm not sure about that solution, so if it's wrong tell me.

2. Ina  parlor game the "magician" asks one of the participants to think of a three-digit number (abc), which represents the three-digit number with digits a,b,c. The "magician" then asks this person to make the numbers (acb), (bac), (bca), (cab), and (cba), and to add them, revealing the sum N. When told the value of N, the "magician" can identify the origianl number, (abc). Determine (abc) when N=3194.

BONUS: Prove that the magician can find (abc) for any N.[/quote]

Clearly, N=122a+212b+221c=222a+222b+222c-100a-10b-c.

Since abc<1000, and the possible values for 222a+222b+222c are 3330 (abc=136), 3552 (abc=358), 3774 (abc=580), and 3996 (abc=802)

now you can just check to find that the answer is 358.

By the above method, there are at most 4 possible values.

Another way is for $ N\equal{}30459834095843$ or any other sufficiently large number.

[quote="AIME15"]Another way is for $ N \equal{} 30459834095843$ or any other sufficiently large number.[/quote]

??? How would the magician recieve THAT number? The maximum is $ 999 \cdot 6\equal{}5994$.

For your bonus, you wrote for "any N".

I guess complex numbers etc. also work.

Yes. N denotes the sum of acb, bca, etc. Thus, N cannot be the number you mentioned (some like 20 digit number)

dude keep the difficulty level down to like 1-8 type problems

snoweverywhere was not asking for that.

[quote="AIME15"]/end spam

Find the largest possible value of $ k$ for which $ 3^{11}$ is expressible as the sum of $ k$ consecutive positive intgers.

1987 AIME Problem 11[/quote]

No coincidence I just did that AIME.

I'll leave it for someone else solving for the first time.

@blackbelt: Then what was he asking for? This problem would probably show up around #4-6 on AIME's now anyway.

[hide="Solution"]
A necessary condition for a number $ N$ being expressible as the sum of $ k$ positive integers is that $ N\geq k(k\plus{}1)/2$, and:

If $ 2|k$, then $ N \equiv (k/2) \pmod k$.
If $ 2 \not| k$, then $ N \equiv 0 \pmod k$.

First, the upper bound for $ k$ is the largest positive integer such that $ k(k\plus{}1)/2<3^{11}$. Thus $ k(k\plus{}1)<2\cdot 3^{11}\equal{}2\cdot 177147\equal{}354294 \implies k\leq 594$.

The only odd $ k$ that will work are powers of 3, and the largest such number is 243.
The only even $ k$ that work are twice powers of 3, and the largest such number is 486.

Thus the answer is $ \boxed{486}$.
[/hide]

[hide]Let the first number be $ a$, we have
$ (a\plus{}a\plus{}k\minus{}1)k/2 \equal{} 3^{11}$
$ (b\plus{}k)k\equal{}2\cdot 3^{11}$
where $ b\equal{}2a\minus{}1$ is an odd positive integer.
Note $ k^2 \leq (b\plus{}k)k\equal{}2\cdot 3^{11} < 3^{12}$, so $ k < 3^6$. Since $ k|2\cdot 3^{11}$, maximum $ k\equal{}2\cdot 3^5\equal{}486$. This gives $ b\plus{}k\equal{}3^{6}$, so $ b\equal{}3^{6}\minus{}2\cdot 3^{5}\equal{}3^{5}$ which satisfies our condition. Therefore, the answer is $ 486$.
[/hide]

New problem:

An integer $ k$ is added to each of the numbers 36,300,596. The square roots of these numbers then create an arithmetic sequence. Find k.

Okay let the squareroots be a, a+b, and a+2b.

Then,

2ab+b^2=264
2ab+3b^2=296

Therefore, b=4, and a=31

31^2=961, and thus k=961-32=929 EDIT: i meant 961-36-925 X.X

[hide]I got 925 and my calculator agrees with me. So we know that $ \sqrt {36 \plus{} k} \plus{} d \equal{} \sqrt {300 \plus{} k}$ and $ \sqrt {300 \plus{} k} \plus{} d \equal{} \sqrt {596 \plus{} k}$. Thus $ 2 \sqrt {300 \plus{} k} \minus{} \sqrt {36 \plus{} k} \equal{} \sqrt {596 \plus{} k}$. Squaring and doing arithmetic yields a k of 925. I'm curious, what problem number would this be?[/hide]

Problem: Let $ P$ be the product of the non-real roots of  $ x^4 \minus{} 4x^3 \plus{}6x^2 \minus{}4x \equal{} 2005$. Find $ \lfloor P \rfloor$

Crazypianist, I got the same answer as you doing it by hand.

EDIT:
[hide="Answer to Problem"]
You recognize that by adding $ 1$ to both sides, you can factor the LHS.

$ x^4 \minus{} 4x^3 \plus{} 6x^2 \minus{} 4x \equal{} 2005\\
(x \minus{} 1)^4 \equal{} 2006\\
x \equal{} 1 \pm \sqrt[4]{2006}$
Because there are complex $ 4$th roots of $ 2006$, the other two roots of the equation must be non-real. Thus, by Vieta's we have:

$ r_1r_2r_3r_4 \equal{} \minus{}2005\\
r_1r_2(1 \minus{} \sqrt{2006}) \equal{} \minus{}2005\\
r_1r_2 \equal{} 1 \plus{} \sqrt{2006}$

We know that $ 44 < \sqrt{2006} < 45$ by quickly checking squares so the answer is $ \boxed{045}$.
[/hide]

oops... its 961-36=925.. thanks for catching my mistake crazypianist!

Adding one and factoring and taking the fourth root:

(x-1)=fourthrt2006,fourthrt2006i,-fourthrt2006, or -fourthrt(2006)i

clearly, for it to be nonreal, x=fourthrt2006i+1 or -fourthrt2006i+1.

Multiplying gives 1+sqrt(2006), which rounds down to 44+1=45

[quote="crazypianist1116"][hide]I got 925 and my calculator agrees with me. So we know that $ \sqrt {36 \plus{} k} \plus{} d \equal{} \sqrt {300 \plus{} k}$ and $ \sqrt {300 \plus{} k} \plus{} d \equal{} \sqrt {596 \plus{} k}$. Thus $ 2 \sqrt {300 \plus{} k} \minus{} \sqrt {36 \plus{} k} \equal{} \sqrt {596 \plus{} k}$. Squaring and doing arithmetic yields a k of 925. I'm curious, what problem number would this be?[/hide]

Problem: Let $ P$ be the product of the non-real roots of  $ x^4 \minus{} 4x^3 \plus{} 6x^2 \minus{} 4x \equal{} 2005$. Find $ \lfloor P \rfloor$[/quote]

It was #7, but it would be earlier by now in all likelyhood.

Problem:

$ (x\minus{}1)^4\equal{}2006$
$ (x\minus{}1)^2\equal{}\sqrt{2006}, \minus{}\sqrt{2006}$
$ x\minus{}1\equal{}\sqrt[4]{2006}, \minus{}\sqrt[4]{2006}, \sqrt{\minus{}\sqrt{2006}}, \minus{}\sqrt{\minus{}\sqrt{2006}}$.

$ (1\plus{}\sqrt{\minus{}\sqrt{2006}})(1\minus{}\sqrt{\minus{}\sqrt{2006}})$
$ 1\minus{}(\sqrt{\minus{}\sqrt{2006}}^2)\equal{}1\plus{}\sqrt{2006}\equal{}P$
$ \lfloor P \rfloor \equal{} 45$.

New Problem: In a sequence of coin tosses, one can keep a record of the number of instances when a tail is imemdiately followed by a head, a head is immediately followed by a head, etc. We denote these by TH,HH, etc. For example, in the sequence HHTTHHHHTHHTTTT of 15 coin tosses, we observe that there are 5 HH, 3 HT, 2 TH, and 4 TT subsequences. How many different sequences of 15 coin tosses will contain exactly 2 HH, 3 HT, four TH, and five TT subsequences.

Note: This is a #13.

[hide]The given means we have 5 H's in 1-14 tosses and 6 H's in 2-15 tosses, so the last one is H. Similarly, the first one is T. We see the sequence is of form:
T...HT...HT...HT...H
We have 2 H's and 5 T's left to make 2 HH's and 5 TT's. To place H's, consider T's as dividers, which means we have $ \binom{5}{3}\equal{}10$ ways. Similarly, to place T's, consider H's as dividers, giving $ \binom{8}{3}\equal{}56$ ways. Thus, there are totally $ 560$ sequences.
[/hide]

[quote="BOGTRO"]@blackbelt: Then what was he asking for? This problem would probably show up around #4-6 on AIME's now anyway.[/quote]

meh, i just saw #11 and said dude snow is looking for a 6 not a 10

Find the least positive integer $ k$ for which the equation $ \lfloor\frac{2002}{n}\rfloor\equal{}k$ has no integer solutions for n.
2002#8

Same as no solution to $ k \leq \frac{2002}{n} < k\plus{}1 \equiv \frac{2002}{k\plus{}1} \leq n < \frac{2002}{k}$.

$ \frac{2002}{k} \minus{} \frac{2002}{k\plus{}1} < 1$ which means $ k$ is greater than $ 44$. Then just test $ k$ until u get $ 49$ to work.

Problem:

A solitarire game is played as follows. Six distinct pairs of matched tiles are placed in a bag. The player randomly draws tiles one at a time from the bag and retains them, except that matching tiles are put aside as soon as they appear in the player's hand. The game ends if the player ever holds three tiles, no two of which match; otherwise the drawing continues until the bag is empty. The probability that the bag will be emptied is $ p/q$ where $ p$ and $ q$ are relatively prime positive integers. Find $ p\plus{}q$.

[quote="BOGTRO"][quote="MellowMelon"][quote="BOGTRO"]BONUS: Prove that the magician can find (abc) for any N.[/quote]
$ N \equal{} 3000$?[/quote]

$ 222(a \plus{} b \plus{} c) \equal{} 3000 \plus{} abc$
$ 3000 \equiv 780 \equiv 114 \pmod 222$

$ abc \equal{} 114,336,558,780$

$ 222 \cdot 6 \neq 3114$
$ 222 \cdot 12 \equal{} 2220 \plus{} 444 \equal{} 2664 \neq 3336$
$ 222 \cdot 18 \equal{} 4440 \minus{} 444 \equal{} 3996 \neq 3558$
$ 222 \cdot 15 \equal{} 2220 \plus{} 1110 \equal{} 3330 \neq 3780$.

So no solutions (?)

I must have made a mistake in my solution. If anyone can point it out that would be nice.

EDIT: Just realized. This means that the magician will never recieve the number 3000, since there are no such abc.[/quote]

You started perfectly but abc equals -3000 (mod 222) and not +3000(mod 222). But it's a fact that 3000 has no solution. Try your method with 2950, it will work.

[quote="GANITEE"][/quote][quote="BOGTRO"][quote="MellowMelon"][quote="BOGTRO"]BONUS: Prove that the magician can find (abc) for any N.[/quote]
$ N = 3000$?[/quote]

$ 222(a + b + c) = 3000 + abc$
$ 3000 \equiv 780 \equiv 114 \pmod 222$

$ abc = 114,336,558,780$

$ 222 \cdot 6 \neq 3114$
$ 222 \cdot 12 = 2220 + 444 = 2664 \neq 3336$
$ 222 \cdot 18 = 4440 - 444 = 3996 \neq 3558$
$ 222 \cdot 15 = 2220 + 1110 = 3330 \neq 3780$.

So no solutions (?)

I must have made a mistake in my solution. If anyone can point it out that would be nice.

EDIT: Just realized. This means that the magician will never recieve the number 3000, since there are no such abc.[/quote][quote="GANITEE"]

You started perfectly but abc equals -3000 (mod 222) and not +3000(mod 222). But it's a fact that 3000 has no solution. Try your method with 2950, it will work.[/quote]

ahh... thank you!

However, even with this error, 3000 won't have a solution. Try my bonus again, slightly reworded:

Define N=acb+bca+bac+cba+cab, where each number is three digit with the first number in the hundreds place, and so on (so 354=3 hundred and fifty four, not 3*5*4=60). For any N that yields a possible value for abc, show that this abc is unique.

Well, let me try!
N + abc = 222 ( a + b + c )
If ABC is another solution,
N + ABC = 222 ( A + B + C )
The two solutions may differ by 222, 444, 666 or 888. Accordingly their sums must differ by 1, 2, 3 or 4 respectively.
But it can be checked that addition of 222 to abc leads to addition of 6, -3 or -12 to a + b + c.
Similarly for 444 it's 12, 3 or -6, for 666 it's 18, 9 or 0, for 888 it's 24, 15 or 6 where actually addition of 1, 2, 3 and 4 is needed.
So ABC can't be another solution.
Hence if solution exists, it's unique.
( I think it's convincing, but there may be more elegant solution.)

[quote="soltixnj"]New Problem:

Find the first three digits after the decimal point of n if n^2 = 2009.[/quote]
821

post a solution  :roll:
anyway guess a random number (i'd guess 45)
average 2009/45 and 45
then average the result with 2009/result
and repeat.

long division makes me sad.

Make it easier by turning it into 7sqrt(41).

The answer is already posted: 821.

New problem (stolen from my blog):

Square $ ABCD$ has coordinates $ A(0,2)$ and $ C(6,4)$. Square $ EFGH$ has coordinates $ E(4, 4)$ and $ G(14,2)$. If $ ABCD$ and $ EFGH$ are configured such that the figure $ BDFH$ is a quadrilateral, then find the area of that quadrilateral.

Correct. New problem:

If $ 2^{2^{2^{2^{2}}}}$ can be expressed in the form $ a\cdot 10^n$ where $ 1\leq a < 10$ and $ n$ is an integer, find the remainder when $ n$ is divided by 1000 given that $ \log_{10} 2\approx 0.30103$.

Hmm.. On that $ \sqrt {41}$ problem I was positive there was a better way than root-approximation algorithms (I was trying generalized binomial theorem) but I guess not...

Anyway, [hide="Solution to that log 2 problem"]
$ 65536\cdot\log 2 \equal{} 19728.30208 \equal{} \log a \plus{} n\implies 10^{19728.30208 \minus{} n} \equal{} a$
We must have $ 19728.30208 > n > 19727.30208$ to satisfy the bounds for $ a$, so $ n \equal{} 19728$, and our answer is $ \boxed{728}$.
[/hide]

New problem:

Given that $ \sum_{k \equal{} 1}^{35}\sin 5k \equal{} \tan\frac {m}{n}$, where angles are measured in degrees, and $ m$ and $ n$ are relatively prime positive integers that satisfy $ m/n < 90$, find $ m \plus{} n$. (AIME 1999 #11).


$ a_{1} \equal{}1$, $ a_{n\plus{}1} \equal{}\frac{a_{n} }{\sqrt{1\plus{}4a_{n}^{2} } }$ for $ n\ge 1$. What is the least $ k$ such that $ a_{k} <10^{\minus{}2}$ ?

$\textbf{(A)}\ 2501  \qquad\textbf{(B)}\ 251  \qquad\textbf{(C)}\ 2499  \qquad\textbf{(D)}\ 249  \qquad\textbf{(E)}\ \text{None}$

[hide]$ a_{n+1}^2={a_n^2\over 1+4a_n^2}\implies {1\over a_{n+1}^2}={1\over a_n^2}+4$

Hence $ {1\over a_n^2}$ is an arithmetic sequence, so $ {1\over a_n^2}={1\over a_1^2}+4(n-1)\implies {1\over a_n^2}=4n-3$

Therefore $ a_n={1\over\sqrt{4n-3}}$

Now

\begin{eqnarray*}{1\over\sqrt{4n-3}}<0.01\iff\sqrt{4n-3}>100\iff 4n-3>10000\iff n-{3\over 4}>2500\iff n>2500+{3\over 4}\end{eqnarray*}

Therefore $ n\geqslant 2501$.[/hide]


calculus Asymptote integration geometry limit logarithms function

This is probably a stupid question, but I haven't done calculus in a year and a half and recently ran into something that made me think - while, if I remember correctly, it's not valid to integrate through a vertical asymptote or other discontinuity, is there some level of abstraction at which something like $ \int_{\minus{}n}^{n}\frac{1}{x}dx$ for some real n is defined and 0? Thinking of the integral as taking the area, it seems clear that equal and opposite halves to either side of the y-axis should add to zero even if both sides are not individually integrable.

To intgrate over a discontinuity, you split the interval at the discontinuity and integrate as an improper integral. But I don't think that the example you give is valid, since the antiderivitive of $ \frac{1}{x}$ is not defined for negative x.

Oh, that's right that ln x is undefined for x < 0, of course, oops (although it doesn't change the intuitive feeling that the area from -n to n should be 0 due to symmetry). How about $ x^{\minus{}3}$? You can just take that integral from -n to n and it works, although the proper definition may be to split it at 0? And if that's the case, [i]is[/i] that an improper integral, since at no point is a limit being explicitly taken? (Sorry for the sloppy calculus)

There is an object which would be zero, but it's not "some level of abstraction" of the integral. Rather, it's a separate, specifically defined object called the principal value.

By definition, $ \text{p.v.}\int_{\minus{}n}^n\frac1x\,dx\equal{}\lim_{\delta\to0^\plus{}}\left(\int_{\minus{}n}^{\minus{}\delta}\frac1x\,dx\plus{}\int_{\delta}^n\frac1x\,dx\right)\equal{}0$

But the principal value isn't the integral - not the Lebegue integral (which doesn't exist for this example), and not the improper Riemann integral (which also doesn't exist for this example). And there are a number of theorems about integrals which aren't true or make no sense for principal values; you have to be very careful when using it, and you really have to treat it as something other than an integral.

The way I phrase it in a Calculus II class: this integral is improper for two reasons - the right hand side of zero and the left hand side of zero. If you're asking whether the improper integral exists, then you must split up the original integral into as many pieces as it takes for each piece to only be improper for one reason. We can still accept such a thing as having an improper integral provided each piece converges separately. For instance:

$ \int_{\minus{}1}^1\frac1{\sqrt[3]{x}}\,dx\equal{}0,$ as an improper integral.

But by those standards $ \int_{\minus{}n}^n\frac1x\,dx$ does not exist as an improper integral because when we split it into the required two pieces, both pieces diverge (and just one diverging would be enough to kill it.)

A principal value is an attempt to balance two reasons for which an integral is improper by imposing convergence "at the same rate". And it's delicate. For instance,

$ \lim_{\delta\to0^\plus{}}\left(\int_{\minus{}n}^{\minus{}2\delta}\frac1x\,dx\plus{}\int_{\delta}^n\frac1x\,dx\right)\equal{}\ln 2$ and $ \lim_{\delta\to0^\plus{}}\left(\int_{\minus{}n}^{\minus{}\delta^2}\frac1x\,dx\plus{}\int_{\delta}^n\frac1x\,dx\right)\equal{}\infty.$

[quote="KBriggs"]... since the antiderivitive of $ \frac {1}{x}$ is not defined for negative x.[/quote]
Say what? Of course $ \frac1x$ has an antiderivative for negative $ x.$ It's a continuous function, and all continuous functions have antiderivatives. The specific formula is:

for $ x<0,\ \int\frac1x\,dx\equal{}\ln(\minus{}x)\plus{}C.$

So there's one formula valid for $ x>0$ and another formula valid for $ x<0.$ Of course, it doesn't have an antiderivative at zero, and there's no "global" antiderivative valid on the whole line.

It is conventional to bundle the two formulas together as

$ \int\frac1x\,dx\equal{}\ln|x|\plus{}C.$

But that's still two formulas, just combined to make them easier to memorize.

Thanks, Kent! Principal value is exactly what I was thinking of from what you said; is that a calculus topic, or real analysis outside of calculus? Wikipedia has a more general entry under complex analysis.

And why do you say that either side of $ \int_{ \minus{} 1}^{1}\frac {1}{\sqrt [3]{x}}\,dx \equal{} 0,$ converges, while those of the other functions do not? It seems that the asymptotes of both $ \frac {1}{\sqrt [3]{x}}$ and $ \frac {1}{x}$ go to infinity on either side of the y-axis, just at different rates, no?

And about not thinking about $ \int\frac {1}x\,dx \equal{} \ln|x| \plus{} C.$, that was just silly.  :oops:

The presence of that vertical asymptote (or other kind of singularity) merely means that the integral is improper; it doesn't tell you about convergence. Some improper integrals converge and some diverge.

$ \int_0^1\frac1{\sqrt[3]{x}}\,dx\equal{}\left.\frac32x^{\frac23}\right|_{0^\plus{}}^1\equal{}\frac32$ converges.

Similarly, $ \int_{\minus{}1}^0\frac1{\sqrt[3]{x}}\,dx\equal{}\minus{}\frac32,$ so $ \int_{\minus{}1}^1\frac1{\sqrt[3]{x}}\,dx\equal{}\minus{}\frac32\plus{}\frac32\equal{}0$ converges.

But $ \int_0^1\frac1x\,dx\equal{}\ln x\bigg{|}_{0^\plus{}}^1\equal{}0\minus{}(\minus{}\infty)\equal{}\infty$ diverges.

Similarly, $ \int_{\minus{}1}^0\frac1x\,dx\equal{}\minus{}\infty,$ so $ \int_{\minus{}1}^1\frac1x\,dx$ diverges, because we can make no sense of $ \infty\minus{}\infty.$

Another pair of examples:

$ \int_0^{\infty}\frac{x}{(1\plus{}x^2)^2}\,dx$ converges by limit comparison to $ \int_1^{\infty}\frac1{x^3}\,dx,$ and the same can be said for $ \int_{\minus{}\infty}^{0}\frac{x}{(1\plus{}x^2)^2}\,dx.$ Therefore we can say that $ \int_{\minus{}\infty}^{\infty}\frac{x}{(1\plus{}x^2)^2}\,dx$ converges, and since it is the integral of an odd function over a symmetric interval, it equals zero.

But $ \int_0{\infty}\frac{x}{1\plus{}x^2}\,dx$ diverges to $ \plus{}\infty$ by limit comparison to $ \int_1^{\infty}\frac1x\,dx.$ Similalry, $ \int_{\minus{}\infty}^{0}\frac{x}{1\plus{}x^2}\,dx\equal{}\minus{}\infty.$ Hence all we can say about $ \int_{\minus{}\infty}^{\infty}\frac{x}{1\plus{}x^2}\,dx$ is that the integral diverges. There is another form of principal value under which we could say that it is zero - but once again, that is a principal value, not a convergent integral.


This is kind of a blunt topic, but something to keep the board moving along.  :)

My pick, Saturn.  Those rings are spectacular.

:huh: It will not let me edit the poll.
Of course that should be another option though.

Valentin, can you edit the poll please?

I personally like the inner planets because it is where we live.

I also like the outer stars, because if you stare at them long enough, you get this sense how small you are compared to the universe, and that you should know more there then here.

I like the outer planets the best. Something about looking at a planet so different from our own and yet so relatively close is very cool. Plus, we're always discovering new things about them, like another moon or a new massive storm. There's just so much we still don't know, and it's exciting and interesting to think of all the possibilities in our own "backyard," so to speak.

My favorite planet, besides Earth, is Kepler 186f. It is interesting because it basically doesn't have a day and night cycle, it takes so long to rotate, that days last for eons. It also orbits a red dwarf. My favorite star (Although theoretical) is a Quasi Star, it is theorized to be the largest star in the known universe, it might have been responsible for the creation of the "Supermassive" black holes at the center of most galaxies. Also, who doesn't love neutron stars? My computer background is that of two neutron stars colliding from Melodysheeps awesome video: Timelapse of the future"

My favorite cosmic object is Saturn. Absolutely majestic.

[quote=A_Crabby_Crab]there is no such thing as a dwarf planet. only planets[/quote]

Pluto? Ceres?

Anything that is big enongh to can make its mass into a ball on its own is a planet. So yes pluto and ceres are planets. Trappist-1 is my favorite  :)

Saturn and Neptune, apart from Earth, are my favorites! I would choose "Outer Planets."

[quote=jerrywang55]I like Kepler 22b it is the closest inhabitable planet next to the earth.   :jump:[/quote]

actually isn't proxima centauri b the closest exoplanet that is capable of supporting life?

[quote=dbnl]There should also be a choice that says Exoplanets. I love exoplanets, especially Kepler 452b and Kepler 1649c.[/quote]

i wuv exoplanets 2!


conics
parabola
geometry
geometric transformation
reflection
geometry solved

A point is given on axis of a parabola. Find the shortest distance from it
    to parabola.

    Maj. Pestich

This might be totally off, but I think it's something like this:

If the point is closer to the vertex than the reflection of the vertex in the focus, then the smallest distance is the one from the point to the vertex. In the remaining situations, the distance is equal to the distance from the focus to the projection of $P$ on the directrix, where $P$ is a point on the parabola whose distance to the focus (and to the directrix) is equal to the distance between our point and the focus (sorry, no computations :)).

Does it stand a chance of being ok? :)

I got the same as grobber (I think):

points: V=vertex, F=focus, X=point on axis, G=reflection of V through F

G is the center of the osculating circle to the parabola at V. So if X is between V and G, then then the circle with center X and radius XV is contained inside this osculating circle and is tangent to the parabola at V, so V is the closest point to X.

Assume X is not between V and G. Let P be a point on the parabola so that d(X,F)=d(P,F), and let Q be the projection of P onto the directrix. Then PQ=PF (by properties of parabolas) and PF=FX, and since XF is parallel to PQ, PQFX is a parallelogram. Also, from the congruent isosceles triangles, we have equal angles galore. So by chasing angles, <XPF=<XPT where T is a point on line PQ with P between T and Q. So by the reflexive property of parabolas (namely that FP and TP form equal angles with the tangent at P), XP is orthogonal to the tangent at P. Since also XP=XF<XV, P is the closest point.


function
induction
algebra proposed
algebra

find all function $ f: Q^\plus{} \minus{}> Q^\plus{}$such that $ f(x)\equal{}f(x\plus{}1)\equal{}\frac {1}{x}f( \frac {1}{x})$ for all $ x \in Q^\plus{}$

We will prove by induction that for each positive rational $ x \equal{} \frac p q$ where $ \gcd( p,q) \equal{} 1$, we have $ f(x) \equal{} qf(1)$.

Clearly, $ f$ is $ 1$-periodic.
It follows that, for each integer $ n>0$ we have $ f(n)\equal{}f(1)$.

Now, let $ q>1$ be an integer. Assume that for each positive integers $ p,k$ such that $ \gcd(p,k)\equal{}1$ and $ k \leq q\minus{}1$, we have $ f(\frac p q) \equal{} qf(1)$.
Then, let $ p$ be any positive integer such that $ \gcd(p,q)\equal{}1$.
Let $ p\equal{}aq\plus{}r$ be the euclidean division of $ p$ by $ q$. Note that $ 1 \leq r \leq q\minus{}1$ since $ q$ cannot divide $ p$.
Clearly, $ f(\frac p q) \equal{} f( \frac r q)$ and that $ \gcd (r,q)\equal{} 1$.
But $ f(\frac r q) \equal{} \frac q r f(\frac q r ) \equal{} \frac q r \times rf(1)$ from the induction hypothesis.
Thus $ f(\frac p q) \equal{} qf(1)$, and we are done.

Pierre.


What is the probability that in the state countdown round, I will win, assuming that:
There are 10 people (including me);
I have a 1/10 chance of getting placed in any spot at the start;
I have a 1/2 chance of winning any given round.

Bonus:
assume I am better than 5 of the people. No 2 people are tied in skill. I still have the same chance of being placed in any given spot. A lower-skilled person will win with a probability of 1/4. What is the probability that I will win now?

Are you talking about a bracket countdown or the type of countdown that goes 10th vs 9th, then (previous winner) vs 8th, etc.

10th vs 9th, winner vs 8th, (state or regional)

What?????
You mean $ \approx\frac {1}{10}$ right?
and besides, think simpler.
Any given person has a 1/10 chance of winning because everyone has an equal chance of winning.

By "win", do you mean winning just one game?

no, and I found my mistake in the simpler technique.
I mean win the tournament.

Funny idea/problem. I think the terms of the equation are correct, but you didn't add correctly.


How can u get good at math??? What are u supposed to do to get top placements at tournaments and all? Read lots of books? what kind of books? what do u do?

The only way to get good is by doing more and more math! Things eventually fall in place.

First of all, you won't be anything by reading math books....

the main stuff to do is practice.. Practice, Practice , Practice...

When I was in korea, I had many math cocaches.. They all stunk... They are no better than those smart high schoolers in AoPS....

So, after awhile, I selftaught math....( with some of my dad's help..He lived with me back then )

All I did was practice school math..I did math at least 1 hour a day..that constantcy helps alot...

so, I've been to national tournament...I've been selected in school to go to math tournaments..
( I did okay..I usually went up to finals )

but

I had to move to USA because of my dad's business,

so, I haven't actually did math for 2 years, and I'm doing over the whole math concepts again...

The most important thing is practicing constantly...

there is no such thing as math secrets...

well, first off, you start by doing ALL of your math team homework, and when you get to math count, there's TONS of homework from that

and there's tons of problems online

like this site

and you can go to the official math count page, think it's mathcount.org

and then there's some probability problems from mu alpha theta (is that how you spell it?), even tho it's high school math  team... but there's froms from the probabilty section that is easy enough for you, I did some last year.

ask math teachers, friends stuf flike that if they have math problems.  DO searches online too for problems.  And btw its mathcount[b]s[/b].org

they have previous competitions and stuff if you seek that sort of help.  Basically you can self teach yourself the basics, and develop your own sorts of shortcuts.  Everyone has some.  And nothing else practice.  And PAY ATTENTION in math class... it goes far ways... or so I heard.[/b]

[quote]Oooo so you practice lots and lots! What do u practice? where do u get the probs to practice?[/quote]

I used to buy bunch of basic math books...
When you keep solving similar math problems, you would eventually learn how to deal with them....

[quote]The AoPs books are an excellent source of problems. If you work through by yourself every single problem in the books, your math skills will improve. It may seem like you're not making any progress, but every problem that makes your think will improve your math skills.[/quote]

I would like to recomand AoPS BASICS!!!!
It would be alittle difficult, but I'm pretty sure you could handle it..
( the only problem is, you need more practice problems-not only AoPS Basics..you need much more practices than that...the book is more like meant to understand concepts of math )

Math is like anything else -- you learn it by practice.  Some people will need more practice, some less, but you learn by doing.  The more time you spend doing math, the better you will get and hopefully the more you will enjoy.  Some people prefer solving problems, others prefer acquiring knowledge -- either way, through working in a class or on your own, you'll get good at math the more you do it.

[quote]They must've been pretty smart then 0.0[/quote]

I'm telling you, those people are just some kind of monsteres who take up money and give me no knowledge....

bet they are much stupider than you, Treething...

[quote="kueh"]may i suggest: problem solving strategies by Arthur Engel! If you can do all the problems in there, you are well on your way (=[/quote]

is that a book ?

No, it's a movie :P .

I'm pretty sure it's a book, and probably a very good one, even though I've never heard of it.

[quote="kueh"]may i suggest: problem solving strategies by Arthur Engel! If you can do all the problems in there, you are well on your way (=[/quote]

Note however, that this is an awesome [b]Olympiad[/b] book, not getting started

[quote="kueh"]may i suggest: problem solving strategies by Arthur Engel! If you can do all the problems in there, you are well on your way (=[/quote]

This is *not* a good book for a beginner.  It's for considerably experienced students.  I have yet to meet the middle schooler who is really ready for this book.

[quote="Treething"][quote]When I was in korea, I had many math cocaches.. They all stunk... They are no better than those smart high schoolers in AoPS.... [/quote]
They must've been pretty smart then 0.0[/quote]

I think you are underestimating the capabilities of the high school students on the board. There are some really intelligent people out there.

[quote]I think you are underestimating the capabilities of the high school students on the board. There are some really intelligent people out there.[/quote]

that's exactly what I'm talking about...

most highschoolers in AoPS are all from different countries and they are like really smart people who wins in all the great international competitions...

Again...

my coaches stunk...

take summer classes and hang out with smart people

Practice more bee style questions
 and do the AOPS books


Do there exist a set $A\subset [0,1]$ such that  (i) $A$ is a finite union of segments of total length 1/2 (ii) the symmetric difference of $A$ and $B:=A/2\cup(A/2+1/2)$ is a union of segment of the total length less than 0.0001 ?

[b]proposed by Konstantin Kokhas [/b]

[quote="Myth"]Was the problem solved on 239 MO?[/quote]

Hi Myth,

have a look [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=20575]here.[/url]  Fedor Petrov gives the answer: The problem was solved by one student.  :)


Baseball is not scored in "points".  And what about the possibility that the NL wins (although I want the AL to win).

fine, "runs". WTH, its not like it matters.

That game was beast! I thought NL actually had a shot at it this year...

I was at the game and it was pretty crazy.  Merciless booing of Papelbum and the run he allowed was just icing.

are you going? ( well it may be a little early to decide...but?)

probably not...

yay! WA has posted 10 states!!!

It's in St. Louis, but it might now be at [b]BUsch[/]b Stadium, seeing as AB no longer exists! :furious:

T-T at first I thought AL was Alabama......

[asy]size(500);void kn(int n) {pair[] points;for(int i=0; i<n; ++i) {points[i]=dir(360.0/n*i);}for(int i=0; i<n; ++i) {for(int j=i+1; j<n; ++j) {draw(points[i]--points[j]);}}}kn(45);[/asy]

sorry spasm!

[asy]size(600);void kn(int n) {pair[] points;for(int i=0; i<n; ++i) {points[i]=dir(360.0/n*i);}for(int i=0; i<n; ++i) {for(int j=i+1; j<n; ++j) {draw(points[i]--points[j]);}}}kn(45);[/asy]

[quote="not_trig"]dude that is a nice picture[/quote]

Yes, us Missourians are beastly at copy-paste!

For a second I didn't doubt your programming skills. :P


algorithm
number theory
greatest common divisor
induction
absolute value

u know that equation of finding the GCD(a,b)
A=Bn1+r1
B=r1(n2)+r2
r1=r2(n3)+r3
...

until the remainder r equals zero, and the remainder before tat is the GCD..

how do u prove tat or explain tat?

the method is called the euclidean method

heres an explaination.. but i dont really understand it.. the blue parts

The proof of this algorithm is not difficult. Suppose a and b are the numbers whose gcd has to be determined. And suppose the remainder of the division of a by b is c. Therefore a = qb + c where q is the quotient of the division. [color=blue]Now any common divisor of a and b also divides c (since c can be written as c = a − qb); [/color]similarly, any common divisor of b and c will also divide a. [color=blue]Thus the greatest common divisor of a and b is the same as the greatest comon divisor of b and c.[/color] Therefore it is enough if we continue the process with the numbers b and c. Since c is smaller in absolute value than b, we will reach c = 0 after finitely many steps.

heres some more.. someone explain to me!

Let a > b. The proof is by induction on Eulen(a, b). If Eulen(a, b) = 1, i.e., if b|a, then a = bu for an integer u. Hence, a + (1 - u)b = b = gcd(a, b). We can take s = 1 and t = 1 - u.

Assume the Corollary has been established for all pairs of numbers for which Eulen is less than n. Let Eulen(a, b) = n. Apply one step of the algorithm: a = bu + r. Eulen(b, r) = n - 1. By the inductive assumption, there exist x and y such that bx + ry = gcd(b,r) = gcd(a,b). Express r as r = a - bu. Hence, ry = ay - buy; bx + (ay - buy) = gcd(a, b). Finally, b(x - uy) + ay = gcd(a, b) and we can take s = x - uy and t = y.


MC 08 State Sprint #29:

Given that $ x^2 \minus{} 5x \plus{} 8 \equal{} 1$, what is the positive value of $ x^4 \minus{} 10x^3 \plus{} 25x^2 \minus{} 9$?

I got that $ x \equal{} \frac{5\pm{i}\sqrt{3}}{2}$, which is the correct value, but I have no idea how to equate the second polynomial with a shortcut. Please help!

[hide="Hint"]Square $ x^2 \minus{} 5x$.[/hide]

What is $ (x^2\minus{}5x)^2$? Can you use the first equation to substitute for the $ x$ terms in the second expression?

Thank you! I tried that before, but I got $ x^4 \minus{} 5x^3 \plus{} 25x^2$ due to a negligent FOIL:-)


$ \sqrt[3]{2} \plus{} \sqrt[3]{3} \plus{} \sqrt[3]{5}$ is Irrational ? WHY ?
THX

The sum of algebraic integers are algebraic integers

[quote="mewto55555"]Think about it. When you add irrational numbers you (almost) always get another irrational.[/quote]

Thus, mewto55555, any statement with "almost" without any qualifier does not qualify to be used in a mathematical proof such as this.

SimonM, rational numbers are also algebraic numbers (of degree 1).

Algebraic numbers aren't algebraic integers

This is pretty difficult,I think.

Show that $ \sqrt[3]{2}\plus{}\sqrt[3]{3}\plus{}\sqrt[3]{6}$ is irrational.

This is easier.

If you divide any of these cube roots by another, the result is irrational, so no like terms can be combined. Since no like terms can be combined, no irrationals can cancel out, so the expression is irrational.

[quote="mathelete"]no like terms can be combined. Since no like terms can be combined, no irrationals can cancel out[/quote]
This is not at all sufficient proof.  You don't know that the only way a sum of irrational numbers can be rational is if certain terms cancel out.  This turns out to be true, but it is [b]much[/b] more difficult to prove than you think.

One standard way to do this is to compute the degree of the field extension $ \mathbb{Q}[ \sqrt [3]{2} \plus{} \sqrt [3]{3} \plus{} \sqrt [3]{5} ] : \mathbb{Q}$ and show that it is greater than $ 1$.  It should, in fact, be $ 27$.  There are well-known techniques for doing this; they are beyond the level of this forum.

A simpler version of the proof:  suppose $ \sqrt {2} \plus{} \sqrt {3}$ were rational.  Then

$ (\sqrt {2} \plus{} \sqrt {3})^3 \equal{} 11 \sqrt {2} \plus{} 9 \sqrt {3}$

would be rational.  But since $ 9 \sqrt {2} \plus{} 9 \sqrt {3}$ would be rational, we would have $ 2 \sqrt {2}$ rational; contradiction.  Hence $ \sqrt {2} \plus{} \sqrt {3}$ is irrational.  In other words, $ \mathbb{Q}[\sqrt {2} \plus{} \sqrt {3}] \equal{} \mathbb{Q}[ \sqrt {2}][ \sqrt {3} ]$.

But you can imagine how computationally intensive it would be to carry this proof out here!


At which angle does a projectile acheive its maximum range? WHY?

well let us solve this together . :)
do u know the formulae for range of a projectile if yes do write it

Solution

Lets say the projectile is fired at speed $ u$ at an angle $ \theta$ to the horizontal.

Lets consider the vertical component of motion only for now, the upwards direction is considered positive:

The vertical component of initial velocity is given by $ u_{y}\equal{} u\sin{\theta}$, so the vertical component of final velocity (just before the projectile lands is) $ v_{y}\equal{}\minus{}u\sin{\theta}$. This follows from conservation of energy. The gravitational acceleration is $ \minus{}g$. let the time taken for the flight be $ t$.

From the fact that constant acceleration is the change in velocity per unit time:

$ \minus{}g \equal{}\frac{v_{y}\minus{}u_{y}}{t}$

$ \Rightarrow t \equal{}\frac{v_{y}\minus{}u_{y}}{\minus{}g}\equal{}\frac{\minus{}u\sin{\theta}\minus{}u\sin{\theta}}{\minus{}g}\equal{}\frac{2u\sin{\theta}}{g}$

Now the projectile will continue at the same velocity in the horizontal direction, as no forces act in this direction (besides air resistance which we consider negligible). Let the horizontal range be $ s_{x}$. The horizontal component of velocity is given by $ u_{x}\equal{} v\cos{\theta}$.

$ \Rightarrow s_{x}\equal{} u_{x}t \equal{} u\cos{\theta}\frac{2u\sin{\theta}}{g}\equal{}\frac{u^{2}(2\sin{\theta}\cos{\theta})}{g}\equal{}\frac{u^{2}\sin{2\theta}}{g}$

Hence $ s_{x}$ is maximum when $ \sin{2\theta}$ is a maximum of $ 1$.

This occurs when $ 2\theta \equal{}\frac{\pi}{2}\Rightarrow\theta \equal{}\frac{\pi}{4}\equal{} 45^{o}$


I'm currently a rising Junior who will be taking Calculus BC. I am looking for a good text to supplement my course. Anyone have any recommendations? I am looking for a text that will present calculus in an eloquent manner and provide me with thorough understanding of the subject.

Around 8 topics down the page is...

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=84782


complex analysis
algebra
polynomial
function
domain
ratio
complex numbers

Suppose that $N>0$, $a_{0},a_{1},...,a_{N}$ are pairwise different complex numbers and $r_{0},r_{1},...,r_{N}$ are rational numbers that add up to $1$. Let $K$ be a connected compact set containing all $a_{i}$. Prove that there exists a holomorphic determination of $(z-a_{0})^{r_{0}}...(z-a_{N})^{r_{N}}$ in the complement of $K$.

Quite simple. The conclusion holds if (your question) and [i]only if[/i] $\sum r_{i}\in \mathbb{Z}$. This is equivalent to: let $f \in \mathbb{C}[z]$ be a polynomial and $K$ a compact connected set containing its roots. Then, $f(z)$ has a holomorphic $m$-th power root in $\mathbb{C}-K$ if (your question) [i]and only if[/i] $m \mid \deg{f}$. To show the [i]if[/i] part, take any boundary point $b$ of $K$, and join it with the infinitely distant point $\infty$ by a simple curve $\ell$ disjoint from $K$. Let $a_{1},\ldots,a_{n}$ be the zeros of $f$, taken with multiplicities. The domain $\mathbb{C}-K-\ell$ is simply connected, and hence each $\sqrt[m]{z-a_{i}}$ has a holomorphic determination there. There are exactly $m$ such, any two differing by constant multiplication by an $m$-th root of unity. Take any fixed such determination for each $i$. This defines a holomorphic determination of $h(z) : = \sqrt[m]{(z-a_{1})\cdots (z-a_{n})}= \sqrt[m]{f(z)}$ on $\mathbb{C}-K-\ell$. For each $z_{0}\in \ell$, the ratio of the limit of this $h(z)$ as $z$ approaches $z_{0}$ clockwise to the limit of $h(z)$ as $z$ approaches $z_{0}$ anticlockwise is $1 / (e^{2\pi i/m})^{n}= e^{\frac{-2\pi in}{m}}= 1$, which gives the analytic continuation of $h(z)$ on $\mathbb{C}-K$.

Yes, people try to solve human problems from time to time, too...


Find all positive integer  $n$  such that   $n|(2n-1)$

Don't you think it's a little easy for Pre-Olympiad?   :huh:

[hide="Solution"]$(n, 2n-1) = (n, n-1) = 1$; $n=1$ is the only solution. [/hide]

[hide]$\frac{1}{x}$ is only an integer when $x=\boxed{1}$[/hide]


This problem appears in our IMO training problem set 1.
The problems of the problem set should be solved without external help, and posts of this type are not allowed to post on mathlinks.
green, DO NOT post any problems of this kind in mathlinks again!


LaTeX
number theory unsolved
number theory

Fiven n is positive integer. Find an integer x such  that $x^n(x+1)^{n+1}+(x+2)^{n+2} =0$ :(

[Admin Edit: Made Latex compliant].

I don't understand Thanhliem,x is positive number,isn't it???

thank you ! I think x is a integer n is a positive integer. :)

Trivial : x+1 must divide 1 , so   (x+1 = 1 ==> x=0)  or (x+1 = -1 ==>x=-2)
Manifestly x=0 and x=-2 are not solutions, so no integer solution .
 :D


Let f(x) in R[x], deg f = n satisfy that:
    f(0),f(n+1) in Z and f(i)= $\frac{i}{i+1}$      i=1,2...n
Prove that n is the odd number


A good basketballplayer should jump 120 cm . What is the initial  valocity he will use to begin the moving?

$ \frac {1}{2}gt^2 \equal{} h\Longleftrightarrow t \equal{} \sqrt {\frac {2h}{g}}\therefore v_0 \equal{} gt \equal{} \sqrt {2gh}.$

or by Conservative Law of Energy, $ \frac {1}{2}mv_0^2 \equal{} mgh\Longleftrightarrow v_0 \equal{}\sqrt {2gh}.$

A better question: what is then the force that the basketballplayer must exert on the ground in order to achive that height?

can we reply based on the impulse monmentum theorem....
the momentum that the player imparts to the ground=F*t..........
using that we can determine the force imparted by the player....

As a side note, there are lots lots of great players. But a 120 cm jump? Not even Jordan can jump that high.


[size=150][b]Problem [/b][/size]
	Consider two identical homogeneous balls, A and B, with the same initial temperatures. One of them is at rest on a horizontal plane, while the second one hangs on a thread (Fig. 6). The same quantities of heat have been supplied to both balls. Are the final temperatures of the balls the same or not? Justify your answer. (All kinds of heat losses are negligible.)

[img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=258[/img]

I dont think the temperatures will be same cuz while heatin the ball on the table,its centre of mass rises,hence its potential energy increases while in the other case,the potential energy decreases,so some heat is being used up for the ball on the table to increase its P.E,hence it will be colder than the other ball..... :rotfl:  :rotfl: is this right ??:rotfl:  :rotfl:

When I first heard this problem from [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=6828]kubus[/url], I had really no ideas. I don't think I could even solve it now if I didn't know the answer already but nevertheless, I like it a lot. Isn't it amazing how can a ball sacrifice some temperature to move itself a little higher?

Theoretically, it's true.

But I don't think you could observe this kind of phenomenon. The effect is miserably small and any kind of loss / disruption of the system are usually much higher. And, I don't know how would you measure the temperature so precisely. Maybe through the black-body radiation? (No usual conventional thermometer could measure that small change in T without actually changing it, I think...)

[quote="Rushil"]Yup, you're right. this is a question from the first IPhO( so it's quite simple). Did you read the answer somewhere??
Anyways, I did find this question to be really interesting and ingenious!!![/quote]

Now that i,a kid gets it,it becomes simple,eh????This is such an old question which i've already posted over here if u noticed.....Now u think people younger to u cant get good conceptual questions like these?????

[quote="Rushil"]Don't get offended kiddo!! Sorry I didn't notice you had posted this earlier. DOnt cry!!!![/quote]
I dont mean to start a fight with ya Rushil but who r u tellin not to cry!!!WHo said "DId u read the answer somewhere?".That shows u didnt get it and were cryin because I GOT IT!!!!

Please calm down! Just a bit of fun - I believe this is not the right place. So Sorry! I suggest you shouldn't post anything else here lest it be spamming!


analytic geometry
geometry
geometric transformation
reflection
graphing lines
slope
function

1) $a^2 - a = (a-2)^2 - (a-2)$
$a^2 - a = a^2 - 4a + 4 - a + 2$
$4a = 6$
$a = 3/2$

2)  This will only change the sign of the function the new line will remain the same distance away from the x-axis and will "move" at the same slope, it will just be positive where it was negative and vice versa.  The new slope is 4/5.

3) The graph would be a parabola opening upwards (a U shaped graph).  There is not much else you can tell about it without b and c being defined.  You could determine the coordinates of the vertex in terms of b and c though.

Oh oh, this is posted twice in getting started, mod please delete this or something...

[quote="Silentfyre"]Last few questions:

Let [x] be defined as [x]= x square - x for all values of x. If [a] = [a-2]. what is the value of a?

In the xy-coordinate plane, line m is the reflection of line L about the x-axis. If the slope of line m is -4/5, what is the slope of line L?

I was also wondering how the graph of f would look like if the function f is defined by
f(x) = x square + bx + c, where b and c are constants.

Thanks.[/quote]

[hide="Question 1"]
${a^2}-a={(a-2)^2}-(a-2)$
${a^2}-a={{{a^2}-4a+4}-{a+2}}$
${-a}={-4a}+4-a+2$
$4a=6$
$a=3/2$[/hide]

[hide="Question 2"]
The slope is $\frac{4}{5}$
Drawing a picture will help
[/hide]


calculus
integration
trigonometry
function
logarithms
calculus computations

Let the definite integral $ I_n\equal{}\int_0^{\frac{\pi}{4}} \frac{dx}{(\cos x)^n}\ (n\equal{}0,\ \pm 1,\ \pm 2,\ \cdots )$.

(1) Find $ I_0,\ I_{\minus{}1},\ I_2$.

(2) Find $ I_1$.

(3) Express $ I_{n\plus{}2}$ in terms of $ I_n$.

(4) Find $ I_{\minus{}3},\ I_{\minus{}2},\ I_3$.

(5) Evaluate the definite integrals $ \int_0^1 \sqrt{x^2\plus{}1}\ dx,\ \int_0^1 \frac{dx}{(x^2\plus{}1)^2}\ dx$ in using the avobe results.

You are not allowed to use the formula of integral for $ \sqrt{x^2\plus{}1}$ directively here.

hello kunny, $ I_0 \equal{} \frac {\pi}{4}$, $ I_{ \minus{} 1} \equal{}\frac{1}{\sqrt{2}}$ and $ I_2\equal{}1$.
Sonnhard.

Guten Tag, Dr Sonnhard Graubner.

You are wrong for $ I_0$.

Now the answer is correct.

hello kunny,yes i thought that was $ I_n\equal{}\int_{0}^{\frac{\pi}{2}}\frac{1}{\cos(x)^n}\,dx$, i have corrected it,sorry.
Sonnhard.

No problem, everyone sometimes make mistakes. :lol:

I'll solve this in other fashion. Of course if you don't mind.

[quote="kunny"]
(5) Evaluate the definite integral $ \int_0^1 \frac {dx}{(x^2 + 1)^2}\ dx$.
[/quote]
Since the integrand is a continuous function on $ [0,1]$ then is integrable. The thing is that I'll put $ x\to\frac1x$ and that will transfer our convergent integral into an improper one, so $ \int_{0}^{1}{\frac {dx}{{{\left( {{x}^{2}} + 1 \right)}^{2}}}} = \int_{1}^{\infty }{\frac {{{x}^{2}}}{{{\left( {{x}^{2}} + 1 \right)}^{2}}}\,dx}.$

Of course this last integral does converge since for all $ x\ge1,$ it's $ \frac {{{x}^{2}}}{{{\left( {{x}^{2}} + 1 \right)}^{2}}}\le \frac {{{x}^{2}}}{{{x}^{4}}} = \frac {1}{{{x}^{2}}}$ then we conclude the convergence by direct comparison. Finally:

\begin{eqnarray*} \int_{1}^{\infty }{\frac {{{x}^{2}}}{{{\left( {{x}^{2}} + 1 \right)}^{2}}}\,dx} & = & \int_{1}^{\infty }{\left( - \frac {1}{2\left( {{x}^{2}} + 1 \right)} \right)'x\,dx} \\
& = & \left. - \frac {x}{2\left( {{x}^{2}} + 1 \right)} \right|_{1}^{\infty } + \frac {1}{2}\int_{1}^{\infty }{\frac {dx}{{{x}^{2}} + 1}} \\
& = & \frac {1}{4} + \frac {\pi }{8} = \frac {2 + \pi }{8}. \end{eqnarray*}

-----------

I just realized that $ \frac{1}{{{\left( {{x}^{2}}+1 \right)}^{2}}}=\frac{1}{{{x}^{2}}+1}-\frac{{{x}^{2}}}{{{\left( {{x}^{2}}+1 \right)}^{2}}},$ which leads the same solution.

hello, for $ I_1$ i have got $ I_1\equal{}\ln\left(\frac{2\plus{}\sqrt{2}}{\sqrt{2}}\right)$.
Sonnhard.

hello
[hide="3"]
$ I_{n} \equal{} \int^\frac{\pi}{4}_{0} \frac {dx}{cos^nx}$
$ I_{n} \equal{} \int^\frac{\pi}{4}_{0} sec^2x\frac {dx}{cos^{n \minus{} 2}}$
$ I_{n} \equal{} [sec^{n \minus{} 2}xtanx]^\frac{\pi}{4}_{0} \minus{} \int^\frac{\pi}{4}_{0} (n \minus{} 2)sec^{n \minus{} 3}xsecxtan^2xdx$
$ I_{n} \equal{} [sec^{n \minus{} 2}xtanx]^\frac{\pi}{4}_{0} \minus{} (n \minus{} 2)I_{n} \plus{} (n \minus{} 2)I_{n \minus{} 2}$
$ I_{n} \equal{} \frac {[sec^{n \minus{} 2}xtanx]^\frac{\pi}{4}_{0}}{n \minus{} 1} \plus{} \frac {n \minus{} 2}{n \minus{} 1}I_{n \minus{} 2}$
$ I_{n \plus{} 2} \equal{} \frac {[sec^{n}xtanx]^\frac{\pi}{4}_{0}}{n \plus{} 1} \plus{} \frac {n}{n \plus{} 1}I_{n}$
$ I_{n\plus{}2}\equal{}\frac{(\sqrt2)^{n}}{n\plus{}1}\plus{}\frac {n}{n \plus{} 1}I_{n}$
[/hide]
thank u

putting n=1 in the above relation:
$ I_{3}\equal{}\frac{\sqrt2}{2}\plus{}\frac{I_{1}}{2}\equal{}\frac{\sqrt2}{2}\plus{}\frac{log(\sqrt2\plus{}1)}{2}$
thank u

From Q1 to Q4 has been solved. :lol:

Now We have only one question of Q5.

Well, it can be shown by a quick substitution that $ x \equal{} \tan \theta \implies \int_0^1\sqrt {1 \plus{} x^2}\,dx \equal{} \int_0^\frac {\pi}{4}\sec^3\theta\,d\theta \equal{} \int_0^\frac {\pi}{4}\frac {d\theta}{\cos^3\theta} \equal{} I_3$

The same substitution also shows that $ \int_0^1\frac {dx}{(x^2 \plus{} 1)^2} \equal{} \int_0^1\cos^2\theta\,d\theta \equal{} I_{ \minus{} 2}$

That's right. Japanese　High School students don't study the integral of $ \sqrt {1 \plus{} x^2}$ as formula and in same fashion so is $ \int \frac {1}{1 \plus{} x^2}\ dx$.

That's why we applicants of Japanese university entrance examination solve these integrals with some hints. Say, [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=252810&search_id=935839245]such as this[/url].


If a,b,c are positive real numbers such that $ ab\plus{}bc\plus{}cd\plus{}da\equal{}4$, then
$ (a\plus{}b)(b\plus{}c)(c\plus{}d)(d\plus{}a) \ge abcd(a\plus{}b\plus{}c\plus{}d)^2$
I can't solve it now, hep me!

[quote="N.M.Dung"]If a,b,c are positive real numbers such that $ ab \plus{} bc \plus{} cd \plus{} da \equal{} 4$, then
$ (a \plus{} b)(b \plus{} c)(c \plus{} d)(d \plus{} a) \ge abcd(a \plus{} b \plus{} c \plus{} d)^2$
I can't solve it now, hep me![/quote]
It follows from $ (a \plus{} b)(b \plus{} c)(c \plus{} d)(d \plus{} a)\geq(abc\plus{}abd\plus{}acd\plus{}bcd)(a\plus{}b\plus{}c\plus{}d)$
and $ (abc\plus{}abd\plus{}acd\plus{}bcd)(ab\plus{}bc\plus{}cd\plus{}da)\geq4abcd(a\plus{}b\plus{}c\plus{}d).$

[quote="arqady"][quote="N.M.Dung"]If a,b,c are positive real numbers such that $ ab \plus{} bc \plus{} cd \plus{} da \equal{} 4$, then
$ (a \plus{} b)(b \plus{} c)(c \plus{} d)(d \plus{} a) \ge abcd(a \plus{} b \plus{} c \plus{} d)^2$
I can't solve it now, hep me![/quote]
It follows from $ (a \plus{} b)(b \plus{} c)(c \plus{} d)(d \plus{} a)\geq(abc \plus{} abd \plus{} acd \plus{} bcd)(a \plus{} b \plus{} c \plus{} d)$
and $ (abc \plus{} abd \plus{} acd \plus{} bcd)(ab \plus{} bc \plus{} cd \plus{} da)\geq4abcd(a \plus{} b \plus{} c \plus{} d).$[/quote]
Thanks, argady!


integration
logarithms
quadratics
geometry
perimeter
Pythagorean Theorem

After how many terms of the summation $ 4\sum^{\infty}_{n \equal{} 0}\frac {( \minus{} 1)^{n}}{2n \plus{} 1}$ will the sum be within .01 of pi?

[quote="fibointepi"]After how many terms of the summation $ 4\sum^{}_{n \equal{} 0}\frac {( \minus{} 1)^{n}}{2n \plus{} 1}$ will the sum be within .01 of pi?[/quote]

Oh man... Alternating Series Test... so like when $ \frac{4}{2n\plus{}1}<.01$, which is when $ 2n\plus{}1>400 \implies n\ge 200$ so 200. (of course this is because we know the whole thing sums to $ \pi$)

[quote="not_trig"]so like when $ \frac {4}{2n \plus{} 1} < .01$, which is when $ 2n \plus{} 1 > 400 \implies n\ge 200$ so [size=150][b]200[/b][/size]. [/quote]

? so that? also this is a pretty high bound... it may converge faster than that

[quote="fibointepi"]and the answer is...[/quote]

$ 100$

(maybe $ 99$ works too)

It doesn't converge much faster than that- a factor of $ 2$, but that's all. The error bound from the alternating series is the next term, and the actual error in this case is approximately half the next term.

If you want faster convergence, use a point that's not on the boundary of the region of convergence. The series $ \sum_{k\equal{}0}^{\infty} \frac1{\sqrt{3}}\frac{(\minus{}1)^k}{(2k\plus{}1)3^k}$ for $ \arctan\left(\frac1{\sqrt{3}}\right)$ converges to $ \frac{\pi}6$ much faster, for example.

Or use some trigonometric identities. A classic is the following. Since it uses numbers closer to zero than what jmerry just said, it converges even faster:

$ \frac{\pi}4\equal{}4\arctan\left(\frac15\right)\minus{}\arctan\left(\frac1{239}\right).$

Hello guys, 100 members of $ \sum^{\infty}_{n \equal{} 0}\frac {4( \minus{} 1)^{n}}{2n \plus{} 1}$ to obtain $ \pi$ within .01?!..

Try another serie $ \sum^{\infty}_{n \equal{} 0}\sqrt {n \plus{} 2}$   :lol:

[quote="digger"]Hello guys, 100 members of $ \sum^{\infty}_{n \equal{} 0}\frac {4( \minus{} 1)^{n}}{2n \plus{} 1}$ to obtain $ \pi$ within .01?!..
[/quote]

Yeah. That sequence is a [i]extremely[/i] slow approximation of pi.

Ramanujan's identity for $ \frac{1}{\pi}$ adds what, [i]eight[/i] digits to the actual answer with every term?  That's the quickest way I know (although you do have to compute the inverse of the sum once you stop).

And now there's talk of finding more general forms of Ramanujan's identity that converge even quicker.  Wow...

[quote="jmerry"]That diverges to $ \infty$ rapidly. Why say anything about it?[/quote]
Right, it diverges evidently, but can you prove that 100 members of
$ \sum^{\infty}_{n \equal{} 0}\frac {4( \minus{} 1)^{n}}{2n \plus{} 1}$ approximates $ \pi$ better than 2 members of $ \sum^{\infty}_{n \equal{} 0}\sqrt {n \plus{} 2}$?   :lol:
When to say without smile there is great amount of other series which are much more fit for approximation - this one is one of the worst.

It may be slow, but at least it converges to the right number. Convergence is the standard requirement for any numerical method. Summing a divergent series to estimate a particular number is pointless.*

As for faster convergence- see my first post on this thread. I was even limiting myself to something that could use the same proof.

*Yes, I know about asymptotic series like [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=670944]this[/url]. Your example does not belong to this category.

[quote="jmerry"] Your example does not belong to this category.[/quote]
Right, it doesn't. It belonges to category named joke:
$ \sqrt2 \approx 1.414$
$ \sqrt3 \approx 1.732$
==============
$ \sqrt2 \plus{} \sqrt3 \approx 3.146 \approx \pi$ within .004.  :lol:

[quote="Teki-Teki"]And now there's talk of finding more general forms of Ramanujan's identity that converge even quicker.  Wow...[/quote]

as far as I recall, these correspond to [url=http://en.wikipedia.org/wiki/Class_number_problem]imaginary quadratic fields with class number one[/url], and there are only a finite number of those.  Ramanujan's corresponds to $ \mathbb{Q}[\sqrt { \minus{} 67}]$; there is one more, $ \mathbb{Q}[\sqrt { \minus{} 163}]$, and the [url=http://en.wikipedia.org/wiki/Chudnovsky_brothers]Chudnovsky's[/url] found the corresponding formula and used it to set some $ \pi$ calculation records in the 1980s.

Here's something for which the core idea is thousands of years old, but I have one modern twist.

Let $ S_n$ be the side length of a regular $ n$-gon inscribed in a unit circle. Then $ P_n\equal{}nS_n$ is the perimeter of the polygon, and $ P_n/2$ will be an approximation of $ \pi$ that gets closer as $ n$ increases.

We need a starting point; either $ S_4\equal{}\sqrt{2}$ or $ S_6\equal{}1.$ Now, what happens if we double the number of sides? As a geometry exercise requiring only that you know the Pythagorean theorem, we can derive the following recursion:
\[ S_{2n}\equal{}\sqrt{2\minus{}2\sqrt{1\minus{}\left(\frac{S_n}{2}\right)^2}}\]
That will give us a sequence of numbers that converges to $ \pi.$ We'll get a column of numbers $ a_n\equal{}\frac{P_n}{2}$ where $ n$ doubles with each row of our table.

The modern twist is an idea from numerical analysis called Richardson extrapolation. This allows us to accelerate the convergence of a sequence provided we know what exponents appear in its asymptotics. In this particular case, we can show that we have only even powers:
\[ \frac{P_n}{2}\equal{}\pi\plus{}\alpha n^{\minus{}2}\plus{}\beta n^{\minus{}4}\plus{}\gamma n^{\minus{}6}
\plus{}\cdots\]
That's all we need to know to use Richardson extrapolation - the exponents. We don't need to know the coefficients $ \alpha,\beta,\gamma,\dots.$

We form a second column $ b_{2n}$ by letting $ b_{2n}\equal{}\frac{4a_{2n}\minus{}a_n}{3}.$ And then we form a third column by letting $ c_{2n}\equal{}\frac{16b_{2n}\minus{}b_n}{15}.$ If we go to a fourth column, we'd have $ 256$ and $ 255$ instead of $ 16$ and $ 15$; the next column would have $ 1024$ and $ 1023.$

Here's a very small version of this table:

$ \begin{array}{ccccc}n&S_n&P_n/2&&\\ 6&1.00000000&3.00000000&&\\ 12&0.51763809&3.10582854&3.15110472&\\ 24&0.26105238&3.13262861&3.14156197&3.14159245\end{array}$

In doing that, I computed four square roots (two for each row of the table after the first), two squarings, and 20 or 30 additions and/or multiplications and divisions by rather small integers. It's not a huge amount of calculation and I already have $ \pi$ accurate to an error of about $ 2\times 10^{\minus{}7}.$ Push the table to one more row (a 48-gon) and one more column, and the error goes down to about $ 10^{\minus{}11}.$


Let $a,$ $b$ and $c$ be distinct positive real numbers. Prove the following inequality:
$\frac{a^{2}b+a^{2}c+b^{2}a+b^{2}c+c^{2}a+c^{2}b-6abc}{ a^{2}+b^{2}+c^{2}-ab-bc-ac}\geq \frac{16abc}{ (a+b+c)^{2}}$

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From the two identities
\[2(a^{2}+b^{2}+c^{2}-ab-bc-ca)=(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\]
\[a^{2}b+a^{2}c+b^{2}a+b^{2}c+c^{2}a+c^{2}b-6abc=c(a-b)^{2}+a(b-c)^{2}+b(c-a)^{2}\]
the inequality is equivalent to
\[\sum_\text{cyc}[c(a+b+c)^{2}-8abc](a-b)^{2}\geq 0\]
WLOG let $a\geq b\geq c$ and use $(a-c)^{2}=(a-b)^{2}+(b-c)^{2}+2(a-b)(b-c)$ so that the left hand side of the above becomes
\[[(b+c)(a+b+c)^{2}-16abc](a-b)^{2}+[(a+b)(a+b+c)^{2}-16abc](b-c)^{2}\]
\[+2b[(a+b+c)^{2}-8ac](a-b)(b-c)\]
The first two bracketed terms are both nonnegative, since for all positive reals $x,y,z$,
\[(x+y)(x+y+z)^{2}\geq 4z(x+y)^{2}\geq 16xyz\]
by AM-GM. For the third term, we have
\[(a+b+c)^{2}\geq 4a(b+c)\geq 8ac\]
by AM-GM and $b\geq c$. So everything is nonnegative.
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Yes, scorpius119!
Iurie Boreico and Ivan Borsenco have formulated this problem for natural numbers. :D