function
calculus
derivative
limit
algebra unsolved
algebra

Given is the function $f(x)=-x+\sqrt{(x+a)(x+b)}$, where $a$, $b$ are distinct given positive real numbers. Prove that for all real numbers $s\in (0,1)$ there exist only one positive real number $\alpha$ such that \[ f(\alpha)=\sqrt [s]{\frac{a^s+b^s}{2}} . \]

-f(x) is increase. ${f(0)=\sqrt{ab},f(\infty}=0.$ Therefore it is true if s<0, and it is wrong for s>0. For example a=1,b=4, f(x)<2, but ${\sqrt^s{\frac{a^s+b^s}{2}}}>\sqrt{ab}$, when s>0.

[quote="Rust"]-f(x) is increase. ${f(0)=\sqrt{ab},f(\infty})=0.$ Therefore it is true if s<0, and it is wrong for s>0. For example a=1,b=4, f(x)<2, but ${\sqrt^s{\frac{a^s+b^s}{2}}}>\sqrt{ab}$, when s>0.[/quote]I don't agree with you ,Rust.I think ${f(\infty})=(a+b)/2$but not ${f(\infty}=0$
for $f(x)=(a+b)/2-\frac{((a-b)^2)/4}{\sqrt{(x+a)(x+b)}}$
so this conclusion is right.

We claim that $f(x)$ is a strictly increasing function on the interval $[0,\infty)$. This can be shown by taking the derivative and doing a little algebra. We also know that $f(0) = \sqrt{ab}$.

Now we evaluate $\lim_{x \rightarrow \infty} (-x+\sqrt{(x+a)(x+b)}) = \lim_{x \rightarrow \infty} \left( \frac{(a+b)x+ab}{\sqrt{(x+a)(x+b)}+x}\right) = \frac{a+b}{2}$.

Since $\sqrt{ab} < \sqrt[s]{\frac{a^s+b^s}{2}} < \frac{a+b}{2}$ (by power mean, strict because $a,b$ distinct) and $f(x)$ is strictly increasing, there must be exactly one positive real $\alpha$ such that $f(\alpha) = \sqrt[s]{\frac{a^s+b^s}{2}}$, as desired.

Of course the more algebraic (and less calculus) way to do it is outlined in the above post, but this is what I came up with.

with a little calculate it can be obtained $\alpha=\frac{\sqrt[s]{(\frac{a^s+b^s}{2})^2}-ab}{(a+b)-2\sqrt[s]{\frac{a^s+b^s}{2}}}$
and obviously this fraction is positive and $\alpha$ is unique.


Prove that
\[ k^4+(k+1)^4+(k+2)^4=n^2+(n+1)^2 \]
has no integer solutions.

Take a look here my friend. I think this will help you
http://www.mathlinks.ro/Forum/viewtopic.php?t=56146

Another way by using mod

Taking mod 9 both side and after some analytic , you found that

LHS $\equiv 2,8 (\mod 9)$

RHS $\equiv 1,4,5,7 (\mod 9)$

so obviously no solution here


logarithms
calculus
integration
limit
geometric sequence
real analysis
real analysis solved

series for Ln(x) centered at 1
sum { (-1)^k (x-1)^k }  /  k!    n=1 to infinity

now this converges for 0<x<2 but it also converges for x = 2 (below series converges by alternating series test)
ln 2 = sum { (-1)^k+1 /  k }     for  n=1 to infinity  (if i plug in x=2 in above series)

the question is how can i show that it sums to ln 2?
that is proving either that x=2  is valid(which is not as x < 2)  or showing that
sum { (-1)^k+1 /  k }     for  n=1 to infinity  sums to ln 2.

The power series of $\ln(1+x)$ is
$x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots$
The power series converges to $\ln(1+x)$ on $(-1,1)$. Note that the series also converges at $x=1$ (Leibniz's Test), hence by Abel's Second Theorem, the power series is (left-hand side) continuous at $x=1$. Then it follows that
$1-\frac{1}{2}+\frac{1}{3}-\cdots=\ln 2$.

The geometric sequence

$\frac{1}{1 + x} = 1 - x + x^2 - x^3 + ...$

has quotient $-x$ and therefore, it has radius of convergence equal to 1. Inside the radius, we can integrate term by term and [b]the radius of convergence does not change[/b] in the process. On the radius of convergence, the series resulting from the integration may converge or diverge, but if it converges, the sum is still equal to $\int \frac{\text dx}{1 + x}$. For a convergence of an alternating sign series $\{a_n\}$, it is sufficient that $\lim_{n \rightarrow \infty} a_n = 0$ and $\lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} = A$, where $A$ is finite.

$\int \frac{\text dx}{1 + x} = \ln{(1 + x)} + C = \sum_{k=0}^{\infty}\frac{(-1)^k x^k}{k}$

The integration constant is determined for $x = 0$: $\ln 1 + C = 0$, $C = 0$. The sum is determined for $x = 1$:

$\ln 2 = \sum_{k=0}^{\infty}\frac{(-1)^k}{k}$

[quote="kunny"]Hint:

(1) Prove that $\frac{1}{1+x}=1-x+x^2-x^3+\cdots +(-1)^{n-1}x^{n-1}+(-1)^n\frac{x^n}{1+x}\ (x\neq -1)$

(2) Prove that $\int_0^1 \frac{x^n}{1+x}dx<\frac{1}{n+1}\ (n=1,2,\cdots)$[/quote]

yes this is what i was trying i have reached the last part
ln 2 + $(-1)^{n+1}\int_0^1 \frac{x^n}{1+x}dx$

for $\int_0^1 \frac{x^n}{1+x}dx<\frac{1}{n+1}$
i know it should it go to zero as n->$\infty$  i am guessing $\int_0^1{x^n}dx$ bounds the above integral but why ?( i dont have a concrete explanation x^n integral is bigger and goes to zero as n goes to $\infty$ so the above integral goes to zero as well??)
thanks,


[img]http://foto.terpe.lt/inkelti/20081118/apskritimai.JPG[/img]

Prove that AC=DB

AD=BC (it's semiperimeter of the triangle CDX where X is the intersection point of the inner tangents) hence result.

Let E,F be tangency points of inner tangents to T (assum E is the higher one). Lep P(ABX)=CD+DX+CX=CD+CX+DE+XE=CB+CF=2CB and analogously P(ABX)=2AD. Is it clear now?

Thank you very much, look's pretty easy  :)

I'm really bad at geometry  :oops:


Hello,
once again I would like to ask quite a serious question.
Rather than discussing policies of nations, or whether democracy is the best system, I would like to ask which democratic system has which flaws.  Judging from the huge amount of nations represented here, that should be quite interesting.  Let me explain :

In Belgium we elect a federal government (normally every four years).  Parliament seats are divided among the several sections of Belgium, and each section divides them according to the vote results in that section.
However the actual government (ministers and prime ministers) come from the coalition : a set of parties whose sum of votes is more than fifty percent.
We also have king but his purpose is to provide comic relief and sign laws, everyone is offended when the slighest thing like an opinion comes out of him and clever schemes are thought of when he refuses to sign something.

The main reason why I want to compare is that in Belgium we have the Vlaams Belang (Flemish Value/Importance).  The party is accused of racism, extreme right, and wants to split Belgium( it however is not the only party with that aim).  Couple of years back they were still Flemish Block, but that party got convicted for racism so they immediately formed a new party to escape that conviction.
The party becomes bigger and bigger, and last election it turned out be one of the biggest (30%).  However all other parties have an agreement, the 'cordon sanitaire' to keep them out of the 50% coalition, by never going in a government with them.
Many consider this undemocratic, to keep 30% voters out of politics, and to ignore the largest party.  On the other hand, 30% is still not 50%, so the majority did not vote on them.  Isn't this just the basic right of a group opposing something, to still be divided into factions among themselves?

In USA, parliament is, as I understand it, not chosen simultaneously with the president?
A thing that is different with the president there is that he in chosen by having the highest number of votes?  This creates a situation where two parties always will rule in my eyes, as voting for something like green party immediately means voting for the other end.

In France and Finland for instance, when no presidential candidate gets more than fifty percent, a second round is organised.  Last time in France, this caused many leftists to vote for right wing Chirac , because he was up against Le pen (far left).

So share your ideas.  Have I said something utterly incorrect or something you disagree with?  Please explain.

Let me just give you a primer on the American system: the American federal government has three elected portions, the Senate and the House of Representatives (which together make up Congress) and the Presidency.  (We don't call our legislature a Parliament.)  Each of the 435 Representatives is elected by the voters of an individual districts apportioned by population and serve 2 year terms.  Each of the 100 Senators is elected from a particular state (2 per state) and serve 6 year terms.  The president is elected through a confusing system: on election day, everyone goes and votes.  They aren't voting directly for the president, however: they are voting for Electors who have agreed to vote for one of the candidates for president.  In all but two small states, this happens on a "winner-takes-all" basis, so (for example) in New York, any candidate winning more than 50% of the vote gets all of New York's 31 electoral votes.  (Each state has a number of electoral votes equal to its number of Senators and Representatives combined; thus, roughly according to population but with a bonus tilt towards small states.)

So, there is your brief primer on American government.  There will be a quiz on this material next week ;-)

thanks for the explanation
Indeed parliament you don't use that word

I wouldn't really say I said anything wrong, but on the other hand I hadn't said that much about USA system.
But I am correct that all the ministres (in USA you call them secretaries?) are from one party, the one the president is from, and thus now republican?  Well in belgium that is so not true.

That was my point.  You can't go a bit more to the left than usual as it is synonymous with going much more to the right.  On the other hand you can imagine a system of ministers from three parties has its flaws as well  :(

Another interpretation of the American system: it is inefficient by design. The writers of the constitution wanted to restrict the government's power, and that was accomplished by splitting that power among several branches and bodies, which would rarely all be controlled by the same faction. When that wasn't enough, the Bill of Rights was added to list some things the government would never be allowed to do.

The other key feature of the American system: Nearly all legislative seats are based on districts, winner takes all. This makes it extremely difficult for a small party to win anything, so everything is split between the two major parties. A regional party could break through this system, but there are no regional parties in current politics.

I see.  So jmerry what exactly do you mean?  Could you elaborate a bit about the regional party (maybe i don't understand your words exactly)

And, although perhaps Belgian politics in particular aren't that  influential outside our tiny country, is it undemocratic what we do right now?  Personally I dont' think so.  I keep my opinion that one party should be surprised that the majority opposing, still has some more things to worry about constantly.

By "regional party" I mean a party that is very strong in one part of the country and has essentially no support elsewhere. A party of this type is small by national standards, but still has influence because it wins seats in its home region. The current American system has minor national parties instead that have small amounts of support everywhere, and never win anything.

Something that I thought about deeply as I used to work as an interpreter for official visitors to the United States is that political parties in the United States are much less ideological than in most other countries. For example, the Republican Party and the Democratic Party have EXCHANGED positions on many issues compared to their positions a century ago. One reason the two-party de-facto system in the United States is so stable is that whenever a third party discovers an appealing position on issues not supported by a major party, one major party or the other will change its position to gain support of the voters who have that position. There is a fairly strong third party, called the Independence Party, in my state, but it has very few positions on issues that cannot be adopted by either the Republican Party or Democratic-Farmer-Labor Party of my state. The biggest influence small parties have on United States politics is nudging the major parties into new ideological positions (at least until the next nudge comes along). That makes the whole overall national non-system very responsive to deeply felt popular concerns, with some time lag.

AFTER EDIT: The edit was only to correct a typing error that looked like a spelling error.

In 1995 (I think), President Clinton, (belonging to the Democratic Party) proposed a budget that the House of Representatives (which had a majority from the Republican Party) refused to pass. The impasse over the budget escalated to a temporary shutdown of many of the functions of the federal government. And yet Clinton continued as President for six more years, winning re-election in 1996. All the while the House of Representatives continued to have a Republican majority.

This sequence of events would be unimaginable in a parliamentary system. If a prime minister (a head of government) were ever to fail to get the parliament to pass such a basic measure as the yearly budget for the whole government, then his administration would collapse. It would have to collapse, and there would have to be new elections.

The system of separate election of legislatures and executives also exists at the state level in the U.S.A. Ask tokenadult if he can explain (at something less then book length) the political career of Jesse Ventura in a way someone from another country could understand. (I wouldn't even attempt such an explanation.)

The short answer about Jesse Ventura (a former professional wrestler who had been mayor of a suburban city) becoming a third-party governor of Minnesota for one term goes like this:

Ventura ran on a position saying that it was time to change and have an ordinary member of the public (although he is anything but ordinary) in office rather than a professional politician. Both of his opponents had been in politics for a long time. Many voters (I asked them) voted for him as a protest vote, "to stir things up," as they said. His previous experience as a mayor helped make him a credible candidate, and his campaign director put some GREAT ads on TV near the end of his campaign. His election victory was a stunning surprise. The other big surprise is that voter turnout was up for the election, and the Democratic-Farmer-Labor candidate got many fewer votes than expected. Ventura filled his state cabinet with mavericks from both major parties, and was moderately successful in attaining his state platform in a divided state legislature.

AFTER EDIT: Again, the edit was only to correct typing errors.

I had never heard of Jesse Ventura.  I found an interesting quote of him though :

[quote]"I believe patriotism comes from the heart. Patriotism is voluntary. It is a feeling of loyalty and allegiance that is the result of knowledge and belief. A patriot shows their patriotism through their actions, by their choice [such as voting, attending community meetings and speaking out when needed]. No law will make a citizen a patriot." [/quote]

I think that is true.  I , as a Belgian, find it a bit weird and totalitarian that kids needs to pledge allegiance and stuff.

Is he unique in his situation?

Slightly off topic : Tokenadult, you seem to be regarded by many as a person who knows a lot.  Are you professional historian/mathematician / professor or something, or several of them? ;)

[quote="fredbel6"]Tokenadult, . . . .  Are you professional historian/mathematician / professor or something, or several of them? ;)[/quote]

My undergraduate degree was in Chinese language, so my first professional training was in Sinology. I have lived in Chinese-speaking countries for six years of my life (1982-1985 and 1998-2001). I also have a law degree although I am not currently a lawyer in active practice. Anything I know about mathematics is through self-education, recreational reading I do to keep up with my son's interest in math. Like many Americans, I take advantage of lavish public library systems and a free press to keep up my knowledge of the world after graduating from school.


number theory
greatest common divisor
modular arithmetic

I don't know if they go in this forum, if not
Please move it to the correct place

1. Prove that if $gcd (a,m) = 1$, then the congruence $ax \equiv b\pmod m$ has the solution
$x \equiv a^{\phi(m) -1 } b \pmod m$

2. The definition of a primitive root can be extended to composite numbers. Say $w$ is a primitive root modulo $n$  if $\phi (n)$ is the smallest power of $w$ which is congruent to 1 modulo $n$.

a.) Find any primitive roots of 10.
b.) Show that 12 has no primitive roots

3. Prove that any multiple of a perfect number, other than the perfect number itself, is abundant.

fyi: Abundant numbers are those whose integer factors excluding the number itself, have a sum larger than the value of the number. For example take 20. The integer factors are 1, 2, 4, 5, 10 and 20 if you add all the factors (except 20) you get 22 which is larger than 20 therefore it is abundant.

Should I repost this in another place?

I feel that it may be a little difficult for Intermediate....

If anyone could help me out, I'd appreciate it

[hide="Solution to 1"]
\[
ax \equiv b \pmod m \iff a^{\phi (m)} x \equiv x \equiv a^{\phi m - 1} b \pmod m
\]
by Fermat's little theorem (Or is it Euler's theorem? I can't remember which...).
[/hide]
[hide="Solution to 3"]
Let $n = p_1^{e_1} p_2^{e_2} \cdots p_n^{e_n} \cdots$ be our perfect number where $p_1, p_2, \cdots$ are all the primes and $e_1, e_2,\cdots$ are infinitely many integer exponents.  Because the sum of $n$'s divisors is $n$, then we can say that
\[
n = \left(\frac{p_1^{e_1+1}-1}{p_1-1}\right) \cdots
\]
Suppose we are multiplying $n$ by a prime; WLOG let it be $p_1$.  So we get

\begin{eqnarray*}
np_1&=&p_1^{e_1 + 1} p_2^{e_2} p_3^{e_3} \cdots\\
 &=& \left(\frac{p_1^{e_1+2}-p_1}{p_1-1}\right) \cdots \\
&<& \left(\frac{p_1^{e_1+2}-1}{p_1-1}\right)\left(\frac{p_2^{e_2+1}-1}{p_2-1}\right) \cdots\\
&=& \mbox{the sum of the divisors of }np_1
\end{eqnarray*}
Since $np_1$ is less than the sum of its divisors, it is abundant.  This can easily be extended for general $nm$ with $m=p_1^{f_1} p_2^{f_2}\cdots$ with integers $f_i$ :P
[/hide]

Thanks for the help on those two

I really appreciate it

[quote="ankur87"]2. The definition of a primitive root can be extended to composite numbers. Say $w$ is a primitive root modulo $n$  if $\phi (n)$ is the smallest power of $w$ which is congruent to 1 modulo $n$.

a.) Find any primitive roots of 10.
b.) Show that 12 has no primitive roots
[/quote]
$\phi (10)=\phi (12)=4$
For a.) $w^4\equiv 1 \mod10$ and $w^3,w^2,w \neq 1 \mod10$. Only odds will work, so, $\pm 1, \pm 3$ work with $\pm 1$ being eliminated because $(\pm 1)^2 \equiv 1 \mod10$. So, 3 and 7 work.
For b.) $w^4\equiv 1 \mod12$ and $w^3,w^2,w \neq 1 \mod12$. Again only odds will work, so, $\pm 1, \pm 3, \pm 5$ may work. $\pm 1, \pm 5$ are eliminated because $(\pm 1)^2 \equiv 1 \mod12$ and $(\pm 5)^2 \equiv 1 \mod12$. $\pm 3$ is eliminated because $gcd(3,12)=3$ so any power of 3 will differ from 12 by a multiple of 3. So none can work.

In general:
$w^{\phi (n)} \equiv 1 \mod n$ tells us that $w^{\phi (n)}-nk = 1$. If  $gcd(w,n)=m$, $m \mid w^{\phi (n)} - nk$, therefore $w^{\phi (n)} \equiv 1 \mod n$ can only happen if $m=1$.

Thanks for the answer to that. I was going over it and kept on calculating things out...

Appreciate it.


linear algebra
matrix
trigonometry
inequalities
vector
linear algebra unsolved

[color=green] [b] Let $A$ be the real  matrix $A=\left(\begin{array}{cccc} 0&a&b&c\\ a&0&d&e\\ b&d&0&f\\ c&e&f&0\\ \end{array}\right)\; .$  where $a^2+b^2+c^2+d^2+e^2+f^2 = 6$.
Consider that $r_1 \le r_2 \le r_3 \le r_4$ are eigenvalues of $A$. It is known that  $r_4= 3.$
1) It's true that all eigenvalues are integer numbers ?
2) If $A$  is non-singular, which is the inverse matrix $A^{-1}$ ?
3) If  $f$ $: {\mathbb R} \to {\mathbb R}$ find the matrix   $f(A)$ (holonomic calculus). For instance consider cases  :
[/b] [/color]
\[ \begin{array}{c} f_1(x) = x^{n} \; \; , \; ( n\in {\mathbb N}^{*}) \; \; ,\;\; f_2 (x) =\sin{x} \\ f_3(x)=\arctan{x} \; \; ,\; \; f_4(x)=\left(1+x^2\right)^{-1}\; . \end{array}  \]

Here's the answer to part (1):

$\text{Tr}(A^2)=2(a^2+b^2+c^2+d^2+e^2+f^2)=12=r_1^2+r_2^2+r_3^2+r_4^2.$

So, with $r_4=3,$ we have

$r_1+r_2+r_3=-3$

$r_1^2+r_2^2+r_3^2=3$

Apply Cauchy's inequality to the vectors $(-1,-1,-1)$ and $(r_1,r_2,r_3).$ We get that

$(-r_1-r_2-r_3)^2\le3(r_1^2+r_2^2+r_3^2).$

$3^2\le3\cdot3$

Since this is an equality, we must satisfy the equality condition of Cauchy's inequality. Thus $r_1=r_2=r_3=-1$

So yes, all of the eigenvalues must be integers.

Since none of the eigenvalues are zero, then $A$ is invertible.

All that remains is to compute $A^{-1}.$

You posted this problem [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=15169]in 2004[/url], and it was solved then.

[quote="jmerry"]You posted this problem [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=15169]last year[/url], and it was solved then.[/quote] OOps ! Indeed !   I  make some non-important changes in the  original posted text.  Many thanks,Alex


If $ g(x)\equal{}$ |$ f(x)$| and $ p$ be the period of $ f$.Find the period of $ g$.

The answer isn't uniquely determined by the period of $ f$ alone.  If $ f$ has glide symmetry, as in the case of the sine or cosine, the period is cut in half, but it could also be unchanged or be cut even further (by $ n$ for any positive integer $ n$).

Also, $ g(x)$ may not have period because $ g(x)$ can be $ f(x)$ as well as $ \minus{}f(x)$ for every $ x$ .

Certainly it's possible for $ f$ to be periodic but for $ g$ to be the constant function; is that what you're saying?


calculus
derivative
search
number theory
relatively prime

1)Let  a,b element of  Z+
   a+b=12  then calculate  (1/a)+(4/b)  maximum value.(please don't use derivative)

2)Show that   i) p and q relatively prime  iff  p+q and p.q  relatively prime
                    ii)p and q relatively prime iff p-q and p.q relatively prime

Hi mathmaniack,

My attempt for (2) :

[hide]

(i) Let p and q are relatively prime. Now, assume d is a common prime dividing both p+q and pq. Since d is prime and it divides pq, so it either divides p or q. If it divides p (say), the as d | (p+q), so d | q too, and hence p and q aren't coprime.

Conversely, let p+q and pq be coprime. Assume d is a common divisor of p and q. Then

d| (p+q) and d|(pq), and hence p+q and pq aren't coprime.

(ii) Exactly similar to (i)

[/hide]

****************************************************

Sambit

To (1): I think you search for the minimum value because for a->0 you see there is no upper bound.

If my suggestion with the minimum value is true then here is the most elegant solution I found: 1/a+4/b=((a+b)/a+(4a+4b)/b)/12=(5+b/a+4a/b)/12. With AM-GM you have (5+b/a+4a/b)/12=>(5+2sqrt(4))/12=3/4. Equality holds for b/a=4a/b so for b=8,a=4.


vector
induction
advanced fields
advanced fields unsolved

Let $\mathbb{R}^n$ be the Euclidean space. Prove that for any vectors $a_1, a_2, ... ,a_{n+2}$ there exist $i\ne j \text{ such that } (a_i,a_j)\geq 0$.
To view all the problems visit [url]http://www.mathlinks.ro/Forum/topic-81099.html[/url]

Similar questions have been discussed before. For instance, the same holds for $2n+1$ instead of $n+2$ and $>$ instead of $\ge$.

To be specific, http://www.mathlinks.ro/Forum/viewtopic.php?t=39329 .

Provided this is false, I can take a set of such vectors {ai}.  Since (a1,aj)<0 any j>1, for all vectors v of the subspace spanned by {aj} j>1, which we denote by V1, we have (a1,v)<0.  But if dimV1=n, this would be impossible since a1 is contained in V1 itself.  So dimV1<=n-1.  But the number of non-zero vectors we have for V1 is n+1, so the induction concludes the problem.

I've proved for n=3:
Let be to the contrary that $\forall i\ne j \Rightarrow (a_i,a_j)< 0$.
If all the vectors are colleniar than there exist 2 vectors that will have same direction wich contradicts to our assumption.
So there exist 2 non-colleniar vectors a and b. Let the $\alpha$ be the plane such that vectors a and b belongs to it.
From our assumtion all the vectors are not ortogonal i.e $(a_i,a_j)\ne 0$.
So the projection to this plane of the rest 3 vectors is non zero vector and there exist a pair of vectors c and d such that they look to the same half-space.
Let $c=: pra_{\alpha} c + \gamma \text{ and } d=: pra_{\alpha} d + \delta$  such that   $(\gamma,a)=(\delta,a)=0$ and $(\gamma,\delta )>0$ since they are colleniar.
Then from assumtion we have that $0>(c,d)=(\gamma,\delta ) + (pra_{\alpha} c, pra_{\alpha} d )>(\gamma,\delta )$ so we have that there are 4 vectors in plane such that the angle between any 2 of them is >90 which s contradiction.

I am a beginner in a linear and euclidean space. I just wanted to ask whether my solution could be transformed to the general case?


function
calculus
integration
trigonometry
derivative
real analysis
real analysis unsolved

Hallo to everybody, I have some question that I can't an answer, if some of you can answer me I'll be very pleased of this
1) Let be f:[a,b]->R a derivable and bounded variation function on [a,b], is it true that f', the derivate of f is bounded on [a,b]?
2)Let be f:[a,b]->[c,d] an invertible and bounded variation function on [a,b], is it true that f^-1:[c,d]->[a,b] have to be a bounded variation function on [c,d]?
3)Let be f:[a,b]->[c,d] an invertible and Riemann integrable function on [a,b], is it true that f^-1:[c,d]->[a,b] have to be an integrable function on [c,d]?
Thanks to everybody bye

1. False.

Consider $ f(x) \equal{} x^{3/2}\sin\left(\frac1x\right)$ on $ [0,1].$ Note that $ f'(0) \equal{} 0$ and other than that, $ f(x) \equal{} \frac32x^{1/2}\sin\left(\frac1x\right) \plus{} x^{ \minus{} 1/2}\cos\left(\frac1x\right).$ This derivative is unbounded near zero, but since $ |f'(x)|\le Cx^{ \minus{} 1/2}$ is absolutely integrable, we do have that $ f$ is of bounded variation.

2. True for continuous functions, since an invertible continuous function has to be monotone, the inverse of a monotone function is monotone, and monotone functions on closed bounded intervals are of bounded variation.

As to whether you can cause any trouble with discontinuous invertible functions - I haven't worked through all of that just yet.

Yes, you can cause trouble with discontinuous functions. Let $ [a,b]\equal{}[\minus{}1,1]$ and $ [c,d]\equal{}[0,1]$. Define $ x_0\equal{}x_1\equal{}0, x_n\equal{}(\minus{}1)^n(1\minus{}\frac1n)$ for $ n\ge 2$ and $ y_n\equal{}\frac1n$ for $ n\ge 1$. Map $ (x_k,x_{k\plus{}2}]$ (or $ [x_{k\plus{}2},x_k)$; just make sure it's open on the $ x_k$ side and closed on the other) to $ [y_{k\plus{}2},y_{k\plus{}1})$ linearly for each $ k>0$. For the first intervals $ k\equal{}0$, do the same but omit the points $ x_0$ and $ y_1$; we've already accounted for the image of $ 0$. For the two remaining points, map $ \minus{}1$ to $ 0$ and $ 1$ to $ 1$.
The graph looks a bit like a kid's drawing of a Christmas tree.

This has bounded variation because it's increasing on $ (\minus{}1,0)$ and decreasing on $ (0,1)$. The total variation is $ 3$. The inverse doesn't have bounded variation, because it jumps back and forth between near $ \minus{}1$ and near $ 1$ infinitely often.


Can you find the next two terms in each of the following sequences?

THESE ARE PRETTY CHALLENGING

A)
2, 12, 1112, 3112, 132112, ..., ...

B) 1, 1, 1, 2, 3, 7, 23, ..., ...

C) 1, 1, 1, 2, 3, 5, 11, ..., ...

$ \text{HINT:}$ patterns B and C are very similar

I thought I had C (sum of two previous numbers), but then the 11 ended everytning.

C) 1, 1, 1, 2, 3, 5, 11, 26, 81

Each term is the sum of the previous term and the product of the two terms before it.

I just got B:

B) 1, 1, 1, 2, 3, 7, 23, 167, 3862

Each term is the sum of the product of the previous 2 terms and the product of the 2 terms before that.

For example:
                       a, b, c, d, e

e=ab +cd


linear algebra
matrix
algebra
polynomial

Let the value of the determinant be $ D$.  We make the following observations:
1) If $ a_1b_n \equal{} 0$ then $ A$ has an all-zero row and so $ D \equal{} 0$.
2) Rows 1 and 2 are almost multiples of each other.  If $ a_1b_2 \equal{} a_2b_1$ then actually they are linearly dependent.
3) Similarly, rows 2 and 3 are almost multiples of each other, and if $ a_2b_3 \equal{} a_3b_2$ then actually they are linearly dependent.
4) Et cetera.
Since $ D$ is a polynomial in the $ a_i$ and $ b_i$, we have that actually $ D$ must be divisible by $ a_1b_n$ and by $ a_ib_{i \plus{} 1} \minus{} a_{i \plus{} 1}b_i$ for $ i \equal{} 1, \ldots, n \minus{} 1$.  Thus, $ D \equal{} F \cdot a_1b_n \cdot \prod_{i \equal{} 1}^{n \minus{} 1} a_ib_{i \plus{} 1} \minus{} a_{i \plus{} 1}b_i$, where $ F$ is some other polynomial in the $ a_i$ and $ b_i$.  Comparing degrees on the left and right, we have that $ 2n \equal{} 2n \plus{} \deg F$ and thus $ F$ is just a constant.  Finally, the coefficient of $ \prod_{i \equal{} 1}^n a_ib_i$ in $ D$ is $ 1$, while (if I haven't made any mistake) the coefficient of $ \prod_{i \equal{} 1}^n a_ib_i$ in $ a_1b_n \cdot \prod_{i \equal{} 1}^{n \minus{} 1} a_ib_{i \plus{} 1} \minus{} a_{i \plus{} 1}b_i$ is $ (\minus{}1)^{n \minus{} 1}$.  It follows that $ F \equal{} (\minus{}1)^{n \minus{} 1}$ and so the requested determinant is
\[ (\minus{}1)^{n \minus{} 1} \cdot a_1b_n \cdot \prod_{i \equal{} 1}^{n \minus{} 1} a_ib_{i \plus{} 1} \minus{} a_{i \plus{} 1}b_i \equal{} a_1b_n \cdot \prod_{i \equal{} 1}^{n \minus{} 1} a_{i \plus{} 1}b_i \minus{} a_ib_{i \plus{} 1}.\]

(By the way, how did I actually figure this out?  Well, I guessed that it would be "nice" and so wrote out the cases $ n \equal{} 1, 2, 3$ by hand.  From here, it wasn't hard to guess a pattern, and then looking at the matrix it's pretty easy to see why the pattern holds.)


$ \sum_{k} {2n \plus{} 1 \choose 2p \plus{} 2k \plus{} 1} {p \plus{} k \choose k}$

The answer ([i]sans preuve[/i]) is $ 4^{n \minus{} p} \binom{2n \minus{} p}{p}$.


advanced fields
advanced fields unsolved

Prove that every Linear Continuum is normal.

this is a nice proof:

http://ocw.mit.edu/NR/rdonlyres/Mathematics/18-901Fall-2004/51A1BA10-27FB-4B44-ACD5-225EED9A15BD/0/notes_e.pdf


[hide="vishalarul"]
Line 2 should be:
$=4\cdot 2^{2x}...$, although this doesn't affect your answer as a similar mistake a few lines down cancels it out. Your main mistake is:

$4u^{2}-8-u=60$
It's $-u$. Just a typo really :P

Your answer works perfectly with the minus.
[/hide]


The general formula for the num. people in a group is $\text {total num of people } = \text { num in A } + \text { num in B } + \text { num in neither } - \text { num in both }$.

Would the formula for 3 groups be

$\text {total} = A + B + C + \text { neither } - (A,B) - (B,C) - (C,A) + (A,B,C)$

where $(x,y)$ means in both $x$ and $y$?

almost. You have to add (A, B, C) because if you look at someone in all three,

they get added three times at the beginning, then subracted 3 times because they are in each set of 2. This means they aren't being counted! so you muct add (A,B,C) to make it complete.

[quote="Hamster1800"]I just realised something.... This formula is the exact same thing that got moved to the getting started forum but instead of GCF, we're using groups with given intersections.

here's the general formula (but this could easily be moved)

let $\Omega()$ be the intersection of the groups in the input

with x groups:

[b]total people $=$ neither $+\sum_{i=1}^{x} (-1)^{i+1}($ sum of all $\Omega(\ldots)$ with $i$ groups in the input $)$ [/b]

sorry that I had to make up a name for the function...don't know how to do the intersection symbol.

Anyway, this means taht for 4 groups it would be:

[b]neither + A + B + C + D - (A,B) -(B,C)-(C,D)-(D,A)-(A,C)-(B,D)+(A,B,C)+(B,C,D)+(C,D,A)+(D,A,B)-(A,B,C,D)[/b][/quote]
Your sum there is kinda funny.  It just looks like there are more words than math symbols

yes it is more words than math symbols. Is that a problem? The (few) math symbols shorten it a lot. I never said (in this one) that is was mostly math symbols. Do sums have to have only math symbols? I don't think so!

or we could if you would prefer, define Q as what's in teh sum other than $(-1)^{i+1}$ so we have $\text{neither } + \sum_{i=1}^{x} (-1)^{i+1}(Q)$

Well, it works, and shortens up the notation.  But when your actually adding it up you'll have to expand it and sum it up like the old way, cause you can't just plug that into your calculator using the sum(seq(...)), which is very useful

well you can....  :P  :P

but you need programming with for loops. It would ask you how many groups, how many are in each group, how many are in each intersection of 2, how many....up to how many are in all groups and how many are in no groups.

I'm not going to write up a whole program but if you feel the need, go ahead.

[quote="nat mc"]That would be more difficult than actually doing the problem  :D[/quote]

yeah, but you write the program once, and you'll never have to do the problem again

Well lets say its more than n times more difficult than doing the problme, if you do it n times.  In other words, ITS REALLY HARD

No, it couldn't be that. That would mean that it gets harder the more problems you do :D instead of being [i]harder the more subsets there are[/i], which makes a lot more sense.

actually it doesn't seem that hard.... (30 element lists, anyone?)

basic layout:

you have a list:

{0,0,0,0,0,0,0}

This has 7 elements, so the groups are groups 1, 2, 3, 4, 5, 6, 7.

First, It asks you how many are in no groups, and adds that to 0.

Then it cycles through each time changing the last element until it is greater than 7 (the number of elements) Once that happens:

{0,0,0,0,0,0,8} -> {0,0,0,0,0,1,1}

continue like this until it becomes:

{7,7,7,7,7,7,7}

Then you'll have done it all. The total number of cycles needed is $n^n$ where $n$ is the number of groups. You figure out how to tell whether to add or subtract it. It's very easy but let's see you think about it.


geometry calculus function integration derivative algebra polynomial

I have a few questions about finding the area under very, very basic curves.   I have yet to take calculus but my teacher has taught me enough to solve most calculus questions that appear in competitions around here (using the helpful functions the TI calculators have).  Please correct any errors I might make while writing what I "know", also.

I know the power rule for finding integrals and derivatives of basic polynomials, so I can find the area under a curve of, say, $\int_{1}^{2}{x^2}$.  I find the integral of $x^2$ which is $\frac{1}{3}x^3$.  Then I plug in 2 for x, solve, and subtract from that the value when x is 1.  (again, correct me if that is wrong)

What I don't know how to do is find the area between curves.  Probably the most basic I can think of is the area between $f(x)$ and $g(x)$ where
$f(x) = x^2$
$g(x) = -x^2 + 8$
The intersection points will be $(-2, 4)$ and $(2,4)$, that doesn't require calculus to figure out.  I don't know how to find the area between them thouogh... $\int_{-2}^{2}{f(g(x))}$?  Perhaps $\int_{-2}^{2}{g(x)} - \int_{-2}^{2}{f(x)}$.  Actually I think that last one makes sense.  It would be the area under the whole thing, minus the area under the smaller one.  But does that always work?  Hm.

Yes, you're right. $\int_a^b g(x) dx-\int_a^b f(x) dx$
Just a technicality, remember to put $dx$ or whatever variable you are integrating with respect to at the end of the integral.  This is because each strip under the curve has a width $dx$, a height $f(x)$, and an area $dA$, so $dA=f(x)dx\Rightarrow A=\int_a^bf(x)dx$.
Here's an exercise, prove the power rule.

Okay, thanks.  So for any two functions $f(x), g(x)$ the area between them can be
$| \int_a^b{f(x)dx} - \int_a^b{g(x)dx} |$ where $a$ and $b$ are the x-values of the intersection points.  Well what if they don't intersect?  Then there are no values for $a$ and $b$.  I assume the area would be either $0$ or $\infty$

Prove the power rule?  I don't even know what the power rule IS, I just know that's what it's called when you manipulate coefficients and exponents to integrate/derivate polynomials.  I don't know what TO prove.  It's like saying "prove $log_2{8}$".  Okay, well I know that's equal to 3.  Maybe I should look up what the power rule actually is.  Then I'd have a better chance of understanding the chain rule...

In order to find the area between curves in a general sense, you should graph the curves and use the geometric principles that led you to $\int_a^bg(x)dx-\int_a^bf(x)dx$.

You said in your first post that you know the power rule for finding derivitives and integrals, if you know what a derivitive represents, you should be able to prove (basically derive) it.

Hint:
[hide]$f'(x)=\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$.[/hide]

For 2-dimentional calculus, the chain rule is $\frac{dy}{dx}=\frac{du}{dx} \cdot \frac{dy}{du}$, it basically allows you to use an intermediate function $u(x)$ in order to take the derivative.

Here's an example: $y=\sqrt{x^3+2}$
$u(x)=x^3+2$
$y=\sqrt{u}$.
$\frac{du}{dx}=3x^2$.
$\frac{dy}{du}=\frac{1}{2\sqrt{u}}=\frac{1}{2\sqrt{x^2+2}}$
$\frac{dy}{dx}=\frac{du}{dx} \cdot \frac{dy}{du}=\frac{3x^2}{2\sqrt{x^3+2}}$

When I said I knew the power rule, I meant I knew that the power rule is what you're using when finding the derivative of a simple polynomial.  I never saw that limit function you showed.  See, the academic team math tests our state makes goes into pretty deep calculus for the last, oh, 1/10 of the tests, and they brought my score from a great score to a decent one because the seniors had all taken calculus and I'm just a young'un who hasn't.  So our coach (also the calculus teacher) taught me how to use all these special functions the TI-83s have to solve these questions.  The premise is to plug in the equation, the calculator finds the derivative or integral based on the x you give it, then I plug in all the multiple choice answers there are with the x I used, and the one with the same answer is the right one.  I asked what dx and dy etc meant, and was told to just forget about them.

Since I don't know much about dy or dx, my understanding of the chain rule suffers.  Your example has made me understand it the most.  I see that, in your example, $\frac{du}{dx}$ is the derivative of $du$.  Those are terms I can understand.  But what are you doing with $\frac{dy}{du}$?  I see a correlation with the $y = \sqrt{u}$ that you put earlier, but I can't draw any conclusion from just that problem.

$\frac{dy}{du}$ is the derivative of $y$ with respect to $u$ as opposed to $x$.
The idea behind the limit I showed you is this, take two points on a function $a$ and $b$ let the cooridinates of $a$ be $(x,f(x))$, and let the coordinates of $b$ be $(x+h, f(x+h))$. Now, the slope of the line connecting these two points is referred to as the secant slope and is $\frac{f(x+h)-f(x)}{x-(x+h)}=\frac{f(x+h)-f(x)}{h}$.  So what if we wanted to find the slope of the line tangent to the curve at a certain point? (Also know as the instantaneous rate of change of the function), then $a$ and $b$ would have to become one point, meaning $h\to 0$, therefore $f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$, and this is what a derivative is (the slope of the tangent line at a given point).
For the power rule, plug in $f(x)=x^n$ and use the binomial expansion, you will see that all terms cancel except $nx^{n-1}$.

Here is another example that might help you with the chain rule.
$y=(3x^5+16)^{38}$ (I doubt that you would want to expand that).
$u=3x^5+16$.
$y=u^{38}$.
$\frac{du}{dx}=15x^4$
$\frac{dy}{du}=38u^{37}$
$\frac{dy}{dx}=\frac{du}{dx} \cdot \frac{dy}{du} = 15x^4(38u^{37})=15x^4(38(3x^5+16)^{37})=570x^4(3x^5+16)^{37}$

Here's a simple exercise to help you understand the concept of a derivative a little better.
Graph the function $f(x)=x^2$, with a ruler mark tangent lines at unit intervals of $x$ ($x=1,2,3,4,5,...$).
If you look at the slopes of all of these tangent lines, you will see that they are $2x$.

wow, thanks, the chain rule became a lot clearer with your first line.  The same with the purpose of the "d"s.  That last example was easier than the first, mainly because I don't know how to find the derivative of $\sqrt x$.  If I did, I might have made the connection that $\frac{dy}{du}$ is the derivative of dy with respect to u.  Now, what happens if there are two terms that I can take the derivative of together?  For example
$y=sin x cos x$
I know the derivative of sin x is cos x, and of cos x is -sin x (correct me there if I'm wrong), but how do I derivate when they're multiplied?
Now, for this particular equation, I could simplify to one term:
$y=\frac 1 2 (sin(2x))$
$y'=cos(2x)$
I've picked up on multiplying the 2 out from inside a trig function when derivating, although I'm not positive that is correct.
So, I solved that by simplifying and thus was a bad example, but how would I solve that if I didn't know I could simplify that?  Since the power rule is no use here, how could I use the chain rule (or some other rule I don't know)

Your answer is correct, but there is a product rule and a quotient rule for derivatives.
If $y=u(x)v(x)$, then $y'=u'(x)v(x)+v'(x)u(x)$.
If $y=\frac{u(x)}{v(x)}$, then $y'=\frac{u'(x)v(x)-v'(x)u(x)}{v(x)^2}$.
Here's your example:
$y=\sin(x)\cos(x)$ $u(x)=\sin(x)$ and $v(x)=\cos(x)$
$y'(x)=cos(x)^2-\sin(x)^2=\cos(2x)$.

For the derivative of $\sqrt{x}$, just use the power rule with $n=\frac{1}{2}$.

You know... I get it now.  thanks so much.  Do you know where I can find lists of derivatives/integrals of trig functions or logs etc?  That would be helpful.  Then I'll find some problems to do, or make some up and have the calculator check.
Again, thanks.


(this is a classic problem, modified to be easier...)

Three men and a monkey are shipwrecked on an island.  On the first day, they all gather mangoes from the mango trees on the island.  That evening, they put their mangoes in a pile and go to sleep.

During the night, the first man wakes up.  Since he is worried that someone will steal his share, he splits the mangoes up into three equal piles with one left over, which he gives to the monkey.  He hides one pile and puts the other two piles back together.

Later that night, the second man wakes up, splits the remaining mangoes into three equal piles with one left over, and gives the leftover mango to the monkey.  The second man hides his pile and puts the remainder back together.

Later that night, the third man wakes up and does the same thing; again, there is one mango left over which is given to the monkey.

The next morning, the men all wake up, split the remaining mangoes into three piles with one left over, and give the leftover mango to the monkey.

What is the minimum number of mangoes that were initially in the pile?

Isn't the original problem with five men and six iterations? It isn't really significantly harder since you'll still do exactly the same process.  (At least, that was how I solved it.) The answer suggests something very elegant to me, but I never figured out a neat combinatorial explanation.

[hide]Say we start off with a mangoes.  Next round we have 2(a-1)/3 mangoes.  Let's call this b.  After second time, we have 2(b-1)/3 mangoes, let's call this c.  After round 3 we get 2(c-1)/3, let's call this d, then after round four, each of the men will get (d-1)/3 mangoes, which we'll call e.  Substituting back in, we get c=(4(a-1)-6)/9.  Substituting that in, we get d=(8(a-1)-30)/27.  After our final round, we substitute again to get (8(a-1)-57)/81=e.  Now we just have a good ole fashioned Diophantine.  Solving for a we get (81e+65)/8.  We have to find the smallest integer value of e for which 8|(81e+65).  Just to overcomplicate things, let's use mods.  This statement is equivelent to 81e+65 :equiv: 0 (mod 8), which leads to e+1 :equiv: 0 (mod 8).  Clearly e=7.  Plugging that into the original equation, we get 79 mongoes.  Yay. [/hide]

ComplexZeta wrote:Does anyone know if there's a good reason for why it's 79=[hide]34-3+1[/hide]?

This would be true of the more complicated cases, too: that is, for five men and a monkey, there are [hide]5^4 - 5 + 1[/hide], for seven men and a monkey, there are [hide]7^4 - 7 + 1[/hide], etc.

This is explained in Gardner's Colossal Book of Mathematics.  From what I understand, that's because -2 is a special number when you have three men, -3 is a special number when you have four men, etc.

Suppose it were possible to have negative mangoes.  Then if we start with -2 mangoes, when the first man wakes up, there are enough to make 3 piles of -1, with one (positive) mango left over (which is given to the monkey).  The man takes one pile, leaving -2.  The second and third men repeat this, and the next morning, split -3 more mangoes.  So this would work with -2.  It should also work with any number of mangoes 3^4 more than -2, i.e. 79, since 3^4 more initial mangoes will not change the modular residues after 1, 2, or 3 "operations".  (For that matter, it would also work with 3^4 + 79 mangoes.)

We can extend this argument to other cases, too.  Just take the number of men, subtract one, tack on a negative sign, and add (number of men)^4.

ComplexZeta wrote:I don't think that's right. I think in general it's [hide]nn+1-n+1[/hide]. Maybe I'll work it out at some time if I don't have enough to do. (Too bad that never happens.)

You're right.  I was stuck in 3-mode.

[hide]

You can prove that nn+1-n+1 is the answer by induction.

Claim: If we start with nn+1-n+1 mangoes,

after k iterations of the process there will be nn+1-knk-n+1.

After n iterations there will be n(n-1)n-n+1, so each man gets (n-1)n-1 mangoes.

To show that this is minimal, say we start out with nn+1-n+1+M mangoes.  We see that nn+1 divides M.  So we are done.

[/hide]

--Dan

It occurred to me that this problem is a lot like AIMEB #6:

6. Three clever monkeys divide a pile of bananas. The first monkey takes some bananas from the pile, keeps three-fourths of them, and divides the rest equally between the other two. The second monkey takes some bananas from the pile, keeps one-fourth of them, and divides the rest equally between the other two. The third monkey takes the remaining bananas from the pile, keeps one-twelfth of them, and divides the rest equally between the other two. Given that each monkey receives a whole number of bananas whenever the bananas are divided, and the numbers of bananas the first, second, and third monkeys have at the end of the process are in the ratio 3:2:1, what is the least possible total for the number of bananas?


If the followings are given:

$ 0<x,y<90$ , $ 0<A\plus{}B<180$ and:

$ \frac{\sin(90\minus{}x)}{\sin(A)}\equal{}\frac{\sin(90\minus{}y)}{\sin(B)}\equal{}\frac{\sin(x\plus{}y)}{\sin(A\plus{}B)}$

Then can we conclude that $ 90\minus{}x\equal{}A$ ?

Let two triangle with angle $ 90 \minus{} x,90 \minus{} y,x \plus{} y$ with side $ a',b',c'$ and radical of circumcentre is $ R'$, and triangle with angle $ A,B,180 \minus{} A \minus{} B$ with side $ a,b,c$ radical of circumcentre is $ R$.
This condition mean:
$ \frac {a'.2R}{2R'.a} \equal{} \frac {b'.2R}{2R'.b} \equal{} \frac {c'.2R}{2R'.c}\Leftrightarrow \frac {a'}{a} \equal{} \frac {b'}{b} \equal{} \frac {c'}{c}$
Then two triangle are similar, from this $ 90 \minus{} x \equal{} A,90 \minus{} y \equal{} B$.


Under blog options, it says:
Shuffle Tag Cloud
Normall the tag cloud is displayed in alphabetical order. If you choose Yes, the order of the tags will be random.

Normally is missing a y.


algebra polynomial function ARML calculus integration geometry

1. Circle O has center (0,0) and radius 1, while circle Q has center (4,3) and radius 2. From point P secants are drawn to the circles, intersecting O at A and B, and Q at C and D. If P moves so that (PA)(PB) = (PC)(PD), then determine the intersection of the locus of P with the x axis.

2. Find all ordered triples (a,b,c) such that: a^2+8b^2+2c^2+1 :le: 4ab+4ac+2c

3. A polynomial function f(t) is defined by the following rule: f(t+1)-f(t)=2t and f(4)=6. Find a polynomial expression for f in terms of t.

3. [hide]The difference in terms is linear with respect to t, thus the polynomial should be quadratic.

Let f(t) = At2 + Bt + C

f(t+1) - f(t) = (2t+1)A + B = 2t

Because it holds for all t: 2A = 2 and A+B = 0. Then A = 1, B = -1. Given that f(4) = 6, we can find C = -6. Our polynomial is f(t) = t2 - t - 6.[/hide]

Quote:1. Circle O has center (0,0) and radius 1, while circle Q has center (4,3) and radius 2. From point P secants are drawn to the circles, intersecting O at A and B, and Q at C and D. If P moves so that (PA)(PB) = (PC)(PD), then determine the intersection of the locus of P with the x axis.

[hide]

I got x=11/4

First, using those power of a point theorems with a secant and a tangent, the tangent to the respective circle squared is equal to (PA)(PB) or (PC)(PD).

So um, P = (x,0), and the distance from P to the circle is y and then drawing in tangents and using a right triangle with Circle O,

1+y^2=x^2

and then with a triangle for Circle Q,

2^2+y^2 = (4-x)^2+3^2

so then -3= x^2-x^2+8x-16-9

22=8x

x=11/4

[/hide]

Mind disclosing where you got these problems from? Because if this was an ARML problem I dont think i would have come close to getting that in the allowed time.

Thanks for the solutions. This problem was on the team round of the MAML state meet a few years back. I'd say you could have spent 5-10 minutes on each depending on how good your team is :P

[quote="beta"]For #2, does a,b,c have to be integers?[/quote]

The problem doesn't state that they have to be integers, but when you solve, you do get integral solutions.

[quote="cats..."]tim, are you going to that tomorrow?[/quote]

Yeah, I'm going to the state meet. I'm on the Boston Latin team... are you going?

Boston Latin is going to New englands right?

Was anyone else there today?


cut cut cut cut cut cut cut cut cut cut
if you got lower than a 100 in a latin test(think 90% range)

i know!!! cut cut cut bleeeaaaat cut bleeatt
bllleeeeeddd
first under 100
cut cut cut cut cut cut cutcut

i am not really emo just feeling like being emo for those who are :ninja:

I am emo. cut cut cut cut cut cut cut. I'm probably the first emo goat you know! cut cut cut cut cut cut
p.s. This is not spam!!!!! :spam:  :spam:  :spam:

How do you get out of the Olympiad section?


It occured to me how hard it actually was to find old problems in the archives on MathLinks due to the poor choice of topic subjects. I mean, is it so hard to pick an appropriate topic which isn't that ambigous and which actually resembles something about the problem itself, not only; "difficult one", "really simple"; "try this one".

Thanks for letting me whine.

Good idea Mindspa, I will rename some threads in the Geometry forum.

  dg

yes I must admit that the subjects fields are not always original :D:D


calculus integration geometry derivative function rectangle limit

But how did they show that definite integration is the same as Riemann sums?

(The first time you learn about) definite integration (, it) is defined in terms of Riemann sums.

I think the question you actually want to ask is:  "why does indefinite integration give an antiderivative?"  In other words, what you seem to be looking for is a proof of the [url=http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus]fundamental theorem of calculus[/url].  Correct me if I'm wrong, though.

Yes, but I want a proof of the second part of the Fundamental Theorem of Calculus, not the first part. And how do u use Riemann sums to explain definite integration? Integration should be like anti-differentiation...why does it have any relation to area?

Basically, I don't see why integrals give you the area under a curve...what's the relation between the two?

Proofs are given in the article.  What you mean is "I don't see why antiderivatives give you the area under a curve."

If I can attempt to give an intuitive explanation:  differentiation takes a function $ f(x)$ to another function $ f'(x)$ that describes the rate of change of that function.  Integration is the process by which we find how much a function has changed by adding up its rates of change $ f'(x)$ over intervals that become smaller.  In terms of the graph of $ f'(x)$, this describes the sum of the areas of rectangles that approximate the area under the curve.

Perhaps something you're a bit more familiar with might help. Just for a moment, lets think physics: velocity is a rate, yes? One form of the particular rate is $ v\equal{}\frac{d}{t}$. So obviously $ vt\equal{}d$. Now lets extend this to a graph by allowing $ v\equal{}f(t)$. So the y-axis is velocity and the x-axis is time. Furthermore, for our purposes assume $ v\equal{}c$ where $ c$ is some constant greater than zero. Assume whatever this velocity is describing is traveling at 4 meters per second for three seconds.

Looks like the area under our velocity line from t=0 to t=3. In this case, it just so happened to be a rectangle. Very simple. But if you recall, a reimann sum is just a whole lot of rectangles right? The reason we developed limit summation and then antidifferentiation is because not all curves are so simple. It's easy to general the area of regions bounded by functions of degree two or less, but beyond that is troublesome.

Now to extend a wee bit further. $ f^{\prime}(x)$ is the rate of change in $ y\equal{}f(x)$. Exactly as $ v$ is to $ d$. Using notation similar to kinematics,
\[ y_b\equal{}y_a\plus{}y^{\prime}\Delta x\] Let $ \lim_{\Delta x \to 0} \Delta x \equal{}dx$, \[ y_b\equal{}y_a\plus{}y^{\prime}dx\] Now apply the fact that limit summation of these tiny dx values over an interval results in a finite value; Using the definition of a reimann sum we know this is a definite integral from a to b: \[ y_b\minus{}y_a\equal{}\int^b_a y^{\prime}dx\].

I think I kind of get it now.

This isn't exactly rigorous, but we know that $ \frac {dy}{dx} \equal{} f'(x)$, so $ dy \equal{} f'(x) \cdot dx$. The RHS is kind of like the formula for a rectangle, with a very very very small width, and height of $ f'(x)$. Therefore, if we add up all the $ dy$'s from $ a$ to $ b$, we get the sum of the area from $ a$ to $ b$. And the change in height from $ a$ to $ b$ is just simply $ f(b) \minus{} f(a)$.

Is that a way of looking at it?

Thanks for everyone's explanation. It was hard for me to visualize the concept, and most people's explanations were helpful, but failed to address why the Riemann sum can just be directly translated to a definite integral.

But igiul, your explanation really helped me to understand it. Thanks.

Yes sir that is. Keep in mind that though it may be explaining the change in $ y$ the whole reason the area gets involved is because if we graphed $ y$ in $ y$ we couldn't see a thing: that's one dimension. So we introduced a second axis along which the change in y is shown. Area happened.

I don't know whether you know this or not (and if you don't some people might disaprove (of me)). You won't be able to do much with it yet, but I figured it was pretty associated with the subject at hand.
[hide]Just for kicks, I'm gonna show you something interesting. You now know that the change if $ f$ is very much governed by the behaviour of $ f^{\prime}$... which is governed by the behaviour of $ f^{\prime \prime}$. And so on.

SO lets think differentials for a sec. As you just stated $ dy\equal{}f^{\prime}(c)\Delta x\Rightarrow f(x)\approx f(c)\plus{}f^{\prime}(c)\Delta x$. And as you probably already know, $ \Delta x\equal{}x\minus{}c$, so

$ f(x)\approx f(c)\plus{}f^{\prime}(c)(x\minus{}c)$ and $ f(x)\minus{}f(c)\approx \int^x_c f^{\prime}(c)dx$
($ f^{\prime}(c)$ is a constant.)
It's important you realize that this is an integral because we use this system to better approximate the first derivative using a differential of the second etc.

Rather, $ f(x)\equal{}\sum^{\infty}_{n\equal{}0} \int^n_{\delta} f^{[n]}(c)dx^n$

I know that was a big leap, but hold on. That represents the nth integral (you integrate the term repetitively n times) of the nth derivative evaluated at c. the little $ \delta$ represents the initial bounds of integration, b=x, c=a. But what is this doing? It's closing in on the error in the first approximation, $ f(x)\approx f(c)$. Another way of thinking of it is 'using one differential to approximate the one before it; then you integrate that (indefinitely) the necessary amount of times so that it affects the first approximation $ f(c)$'. If possible (I say this because of the first term where n=0), you're integrating $ n\minus{}1$ times indefinitely after one definite integral from c to x. It actually has a pattern so that we don't have to ugly up the general term of summation with an integrand.

$ f(x)\approx f(c)\plus{}\frac{f^{\prime}(c)(x\minus{}c)}{1}\plus{}\frac{f^{\prime \prime}(c)(x\minus{}c)^2}{1\cdot 2}\plus{}\frac{f^{\prime \prime \prime}(c)(x\minus{}c)^3}{1\cdot 2\cdot 3} \plus{}\ldots \plus{}\frac{f^{[n]}(c)(x\minus{}c)^n}{n!}$
And so the sum is
$ f(x)\equal{}\sum^{\infty}_{n\equal{}0} \frac{f^{[n]}(c)}{n!}(x\minus{}c)^n$
This is what's known as a taylor series centered at $ c$. They're very useful, but can also be quite problematic due to their covergent or divergent natures. You won't see them for another semester (If my assumption is correct. Sorry if I'm wrong) but they're great fun.

Try approximating the roots of $ x\minus{}\cos{x}\equal{}0$ with a second degree taylor series for cosine.[/hide]

[quote="igiul"]Perhaps something you're a bit more familiar with might help. Just for a moment, lets think physics: velocity is a rate, yes? One form of the particular rate is $ v \equal{} \frac {d}{t}$. So obviously $ vt \equal{} d$. Now lets extend this to a graph by allowing $ v \equal{} f(t)$. So the y-axis is velocity and the x-axis is time. Furthermore, for our purposes assume $ v \equal{} c$ where $ c$ is some constant greater than zero. Assume whatever this velocity is describing is traveling at 4 meters per second for three seconds.

Looks like the area under our velocity line from t=0 to t=3. In this case, it just so happened to be a rectangle. Very simple. But if you recall, a reimann sum is just a whole lot of rectangles right? The reason we developed limit summation and then antidifferentiation is because not all curves are so simple. It's easy to general the area of regions bounded by functions of degree two or less, but beyond that is troublesome.

Now to extend a wee bit further. $ f^{\prime}(x)$ is the rate of change in $ y \equal{} f(x)$. Exactly as $ v$ is to $ d$. Using notation similar to kinematics,
\[ y_b \equal{} y_a \plus{} y^{\prime}\Delta x
\]
Let $ \lim_{\Delta x \to 0} \Delta x \equal{} dx$,
\[ y_b \equal{} y_a \plus{} y^{\prime}dx
\]
Now apply the fact that limit summation of these tiny dx values over an interval results in a finite value; Using the definition of a reimann sum we know this is a definite integral from a to b:
\[ y_b \minus{} y_a \equal{} \int^b_a y^{\prime}dx
\]
.[/quote]

What do you mean the definition of a Riemann sum?
Sorry, but I don't understand how your explanation shows why Riemann sums is equivalent to anti-differentation, which leads to area. It seems that all you did in your algebra is definite what a definite integral is.

I basically want a way to visualize why the second part of the Fundamental Theorem of Calculus is true:

[quote]Let $ f$ be a continuous real-valued function defined on a closed interval $ [a, b]$. Let $ F$ be an antiderivative of $ f$, that is one of the infinitely many functions such that, for all $ x$ in $ [a, b]$,
\[ f(x) \equal{} F'(x).
\]
Then
\[ \int^b_a f(x)\,dx \equal{} F(b) \minus{} F(a).
\]
[/quote]

Try looking at the [url=http://en.wikipedia.org/wiki/Euler_method]Euler method[/url].  The intuitive idea is that to find the total change over an interval we add up the simultaneous changes over subdivisions of the interval.

Even though I'm not sure if this is what I'm looking for, could you explain to me what the $ f(t, y(t))$ means?

[url]http://en.wikipedia.org/wiki/Numerical_ordinary_differential_equations#The_Euler_method[/url]
[quote]Starting with the differential equation (1), we replace the derivative $ y'$ by the finite difference approximation
\[ y'(t) \approx \frac{y(t\plus{}h) \minus{} y(t)}{h},\]
which yields the following formula
\[ y(t\plus{}h) \approx y(t) \plus{} hf(t,y(t)).\][/quote]

The assumption of the Euler method is that we are trying to approximate the solution to a differential equation $ \frac{dy}{dx} \equal{} f(x, y)$.  In the special case that we are trying to understand a differentiable function $ \frac{dy}{dx} \equal{} f'(x)$, we merely have $ f(t, y(t)) \equal{} f'(t)$.

I still don't really see how this shows why Riemann sums imply anti-differentiation.

On Wolfram, [url]http://mathworld.wolfram.com/IndefiniteIntegral.html[/url], after displaying the definition of the definite integral, it says that you can use indefinite integrals to evaluate it. Then it says that while it is taught early on in elementary calculus, the result is actually a very deep result connecting the purely algebraic indefinite integral and the purely analytic (or geometric) definite integral.

How did they find that deep result?

How did they find out that definite integrals can be evaluated in terms of indefinite integrals?

Historically (to the best of my knowledge), the fundamental theorem was developed intuitively, driven by physical applications (which most of the examples here do a good job of conveying). As far as a rigorous proof goes, I'll post one if you'd like, but to really understand it takes a fair bit more work than what a typical single variable calculus course would give you, an intuitive notion of it should work fine for now.
I think I had the same issue as you when I first learned the Fundamental Theorem, and that's because I mistook the definition of a definite integral with the fundamental theorem itself. The definition of what it means for a function to be (in any way) integrable is in [b]no way[/b] connected to anti-derivatives, in fact, you can calculate some definite integrals without the fundamental theorem at all!
When you ask how anti-derivatives are connected with area, an even more fundamental question to ask first is, what is a rigorous definition of "area"? (I think that this is actually a much deeper question than I intend it to be, as I'm not versed in Measure Theory).
From what I know, I'd define area through the definite integral itself, so the anti-derivative's connection to area is explicitly expressed in the fundamental theorem.

But I don't understand why you can evaluate Riemann sums (definite integrals) using indefinite integrals. Like I know that $ \displaystyle\int^b_a f(x)\,dx$ is DEFINED as the area trapped between $ f(x)$ and the $ x$-axis, but I don't why you can evaluate this expression the same way as you would with a indefinite integral.

[hide="If you would like..."]
Let $ f$ be integrable on $ [a,b]$. Suppose that $ F$ is differentiable on $ [a,b]$ with $ F' \equal{} f$. Let $ \epsilon > 0$ be arbitrary. Choose $ \delta > 0$ so that for every partition $ X$ of $ [a,b]$ with the mesh of $ X$ less than $ \delta$, we have $ |\displaystyle\int^b_a f(x)\,dx \minus{} \displaystyle\sum_{i \equal{} 1}^{n}f(t_i)\Delta_i x| < \epsilon$ for any choice of sample points $ t_i \in [x_{i \minus{} 1},x_i]$. (Where $ \Delta_i x \equal{} (x_i \minus{} x_{i \minus{} 1})$, i.e. the inequality holds for every Riemann sum of $ f$ on $ X$).
Choose $ t_i \in [x_{i \minus{} 1},x_i]$ as in the Mean Value Theorem so that $ F'(t_i) \equal{} \frac {F(x_i) \minus{} F(x_{i \minus{} 1})}{x_i \minus{} x_{i \minus{} 1}}$, that is $ f(t_i)\Delta_i x \equal{} F(x_i) \minus{} F(x_{i \minus{} 1})$. Then $ |\displaystyle\int^b_a f(x)\,dx \minus{} \displaystyle\sum_{i \equal{} 1}^{n}f(t_i)\Delta_i x| < \epsilon$, and $ \displaystyle\sum_{i \equal{} 1}^{n}f(t_i)\Delta_i x \equal{} \displaystyle\sum_{i \equal{} 1}^{n}(F(x_i) \minus{} F(x_{i \minus{} 1}))$ which telescopes to $ F(b) \minus{} F(a)$.
So $ |\displaystyle\int^b_a f(x)\,dx \minus{} (F(b) \minus{} F(a))| < \epsilon$. Since $ \epsilon$ was arbitrary, $ |\displaystyle\int^b_a f(x)\,dx \minus{} (F(b) \minus{} F(a))| \equal{} 0$.

(not my own work, pulled more or less off my course notes)
[/hide]

That being said, I don't think this gives any better insight (in fact, probably in this case worst insight) than the intuitive arguments mentioned above.

Why is it so difficult to Google such results?

I kind of understand your proof, but where's the motivation for it? What's the motivation between the integral and Riemann sums?

[quote="123456789"]How do you prove that a definite integral yields the area under a curve? I never understood...[/quote]

A Riemann sum is defined as:

	 \[ \lim_{\Delta x\to 0}\sum_{i \equal{} 1}^{\infty} f(x_i)\Delta x
\]

In this sum $ \displaystyle{f(x_i)}$  is the height of the curve at $ \displaystyle{x_i}$ , an arbitrary point within a given interval that is $ \displaystyle{\Delta x}$  wide. Considering any particular rectangular strip, $ \displaystyle{f(x_i)\Delta x}$ , is the area of that strip by definition. Summing the areas of all such strips consequently yields the total area under the curve.

From the perspective of the definite integral, the antiderivative $ \displaystyle{F(a)}$  represents the area under the curve $ \displaystyle{f(x)}$  to the left of $ \displaystyle{a}$ ; likewise, $ \displaystyle{F(b)}$  represents the area under the curve $ \displaystyle{f(x)}$  to the left of $ \displaystyle{b}$ .  Hence, if $ \displaystyle{a\leq b}$  then $ \displaystyle{F(b) \minus{} F(a)}$  represents the area under the curve $ \displaystyle{f(x)}$  between $ \displaystyle{a}$  and $ \displaystyle{b}$ .

[quote="Dr. No"]From the perspective of the definite integral, the antiderivative $ \displaystyle{F(a)}$  represents the area under the curve $ \displaystyle{f(x)}$  to the left of $ \displaystyle{a}$[/quote]You write that the antiderivative represents the area, but why? I'm trying to get past just the definition.

The motivation behind the proof is that you're picking a sample point so that the area of a strip under $ f$ is equal to the change of $ F$ (the anti-derivative) over the width of the strip (which is why we employ the mean value theorem: an understanding of the mean value theorem is key to understanding our motivation). When you add these up all the strips in the interval it is equal to the change of the anti-derivative over the entire interval

[quote="123456789"]Thanks, I see the proof, but what was the original non-rigorous intuition that got the mathematicians to see the link?[/quote]

While Newton was developing the calculus he kept tabulated pairs of derivatives and their corresponding anti-derivatives in his notes. Leibniz, while developing the calculus independently of Newton, created the concept of what we know now as the Riemann sum, which he called [i]sum omnium[/i], the sum over all. It is from Leibniz that we got the notation $ \displaystyle{\int}$ to represent the integral (the integral symbol is a script "S" ). Leibniz was able to show that the Riemann sum of a given function is equal to its anti-derivative.

http://www.math.wpi.edu/IQP/BVCalcHist/calc1.html

In that website, it says that Wallis "noticed" the pattern (Power Rule). So there was no reasoning involved? Just got lucky with a pattern?

Am I asking dumb questions? I feel like none of the answers are what I'm looking for, or am I just not understanding?

Although I'm not sure this is necessarily the case, I think you'd be surprised how often noticing a pattern can lead someone on the right track, quoting from Zeitz: "It is a well-kept secret that much high level mathematical research is the result of low-tech plug and chug methods. The great Carl Gauss [...] painstakingly computed the number of integer solutions to $ x^2 \plus{} y^2 \le 90 000$."

[quote="123456789"]http://www.math.wpi.edu/IQP/BVCalcHist/calc1.html

In that website, it says that Wallis "noticed" the pattern (Power Rule). So there was no reasoning involved? Just got lucky with a pattern?[/quote]
Have you read the the entire thing?  There's rather a lot of reasoning involved in the approximation using Cavalieri's principle.

[quote="123456789"]Am I asking dumb questions? I feel like none of the answers are what I'm looking for, or am I just not understanding?[/quote]
I'm still not sure what you're looking for.  We've given proofs and we've given intuitions.  Does the concept still not make sense to you?  Let me go through the steps.

- Differentiation takes a function and returns its rate of change.
- Definite integration takes a function and returns the area under it on some interval.
- Indefinite integration takes a rate of change and returns the function it corresponds to.
- The area under a function that represents a rate of change is a sum of rates of change (Riemann sum), which is a definite integral.  On the other hand, because it calculates the difference between two values of a function (to which the rate of change corresponds), it is also a difference between two indefinite integrals.

It seems like the fourth step is what you're hung up on.  Again, do you need more in the way of proof or more in the way of intuition?  Do you understand why Riemann sums give definite integrals on one hand and indefinite integrals on the other?


[quote="ragnarok23"]Consider a function $f(x)$ that has zeroes $4$ and $9$. Given that Patrick randomly selects a number from the set $\{-10,-9,-8, ... 8, 9, 10\}$,  what is the probability that Patrick chooses a zero of $f(x^{2})$? Express your answer as a common fraction.[/quote]
[hide]Note that $f(x^{2})$ has the same zeros as $f(x)$. So there are 2 choices out of 21 numbers, giving you $\frac{2}{21}$[/hide]

[quote="bpms"]Note that $f(x^{2})$ has the same zeros as $f(x)$.[/quote]

Do think this over.

f(x^2) has zeros -2,2,-3,3

answer is 4/21

[hide]if $f(x)$ can be factored as $(x-4)(x-9)$ then $f(x^{2})$ can be factored as $(x^{2}-4)(x^{2}-9)=(x+2)(x-2)(x+3)(x-3)$ so the above answer is correct[/hide]

[quote="Farenhajt"][quote="bpms"]But the $f(x)$ can be factored as $(a-4)(a-9)$. So squaring this is $(a-4)(a-9)(a-4)(a-9)$. Unless I am misinterpreting something...[/quote]

You are - you think of $f^{2}(x)$, i.e. $f(x)\cdot f(x)$, instead of $f(x^{2})=f(x\cdot x)$[/quote]
You are right.
Let me point this out, $f^{2}(x)$ is actually $f(f(x))$, unlike the trig thing $\sin^{2}x=(\sin(x))^{2}$

The zeroes, for the $f(x^{2})$ function have been given to you.
There are four zeroes.

In the set of integers from $[-10,10]$
There are 21 elements.

Four of these are the zeroes.

Therefore, the probability is $P = \frac{4}{21}$

[quote="bpms"]
Let me point this out, $f^{2}(x)$ is actually $f(f(x))$, unlike the trig thing $\sin^{2}x=(\sin(x))^{2}$[/quote]

That depends on the context - if there's no confusion, $f^{2}(x)$ is used freely for $f(x)\cdot f(x)$, and if there CAN be a confusion, some use $f^{2}(x)$ for square and $f^{[2]}(x)$ for composition, or $f(x)^{2}$ for square and $f^{2}(x)$ for composition. But whatever notation you use, make sure it's clear what you mean.


search
combinatorics proposed
combinatorics

The top and bottom edges of a chessboard are identified together, as are the left and right edges, yielding a torus. Find the maximum number of knights which can be placed so that no two attack each other.

Posted:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=88102[/url]

(Found by doing a search on "knights" and "torus".)


trigonometry
inequalities
inequalities solved

Prove that
$\frac{cosA}{1-cosA}+\frac{cosA}{1-cosA}+\frac{cosA}{1-cosA}\geq 3$
Sorry if it has been posted before.

I think there must be some typo in it. Otherwise you could just take [tex]A=\frac{\pi}{2}[/tex]  :?

[quote="keira_khtn"]Use AM-GM for 3 real numbers and following well-known ineq:
[tex]cosAcosBcosC\geq\ (1-cosA)(1-cosB)(1-cosC)[/tex][/quote]

The sign is reserver!

It is enough to prove that:
$\sum\left(\frac{2\cos\alpha-1}{1-\cos\alpha}\right)\geq 0$
or
$\sum\frac{1}{\sin\alpha^2}\geq 12$
The last ineqaulity is true by Jensen


calculus
integration
pigeonhole principle
binomial coefficients
Pascal\u0027s Triangle
combinatorics proposed
combinatorics

[quote="Nguyen Van Linh"]How many ways are there to choose $ k$ numbers from $ n$ numbers $ 1,2,\cdots , n$ such that no two consecutive numbers is chosen?[/quote]
[color=blue]
Let the numbers be arranged in a line as $ \ 1,\ 2,\ 3,\cdots ,\ n$
Let there be $ \ x_1$ and $ \ x_{k \plus{} 1}$ numbers left respectively before the first and after the last selected number $ \ \implies x_1,\ x_{k \plus{} 1}\ge 0$ and $ \ y_{i \plus{} 1}$ numbers be left between $ \ i^{th}$ and $ \ (i \plus{} 1)^{th}$ selected numbers for  $ \ i \equal{} 1,\ 2,\ 3,\cdots,k \minus{} 1 \implies y_{i \plus{} 1} > 0\ \forall \ i \equal{} 1,\ 2,\ 3,\cdots,k \minus{} 1.$

$ \implies x_1 \plus{} y_2 \plus{} y_3 \plus{} \cdots \plus{} y_{k} \plus{} x_{k \plus{} 1} \equal{} n \minus{} k$

Let $ \ x_1 \equal{} y_1 \minus{} 1$ and $ \ x_{k \plus{} 1} \equal{} y_{k \plus{} 1} \minus{} 1\implies y_1,\ y_{k \plus{} 1} > 0$

$ \implies y_1 \plus{} y_2 \plus{} y_3 \plus{} \cdots \plus{} y_{k} \plus{} y_{k \plus{} 1} \equal{} n \minus{} k \plus{} 2$  ---(i)

Hence required number of ways = number of positive integral solutions of (i)$ \ \equal{} \binom {n \minus{} k \plus{} 1}{k}$
[/color]

Denote by $ F_{n.k}$ this number. It is obvious that $ F_{n,0} \equal{} 1$ (there is exactly one way to choose no element!), $ F_{n,1} \equal{} n$, and $ F_{n,k} \equal{} 0$ for $ k > (n \plus{} 1)/2$ (from pigeonhole principle). Now, looking at $ F_{n \plus{} 1,k}$, either the $ k$ elements are chosen from among the first $ n$ (therefore in $ F_{n,k}$ ways), or the element $ n \plus{} 1$ is chosen, so $ n$ is not, hence the remaining $ k \minus{} 1$ are chosen from among the first $ n \minus{} 1$ (in $ F_{n \minus{} 1,k \minus{} 1}$ ways). This leads to the recurrence relation $ F_{n \plus{} 1,k} \equal{} F_{n,k} \plus{} F_{n \minus{} 1,k \minus{} 1}$. But this is exactly the recurrence relation for some binomial coefficients; if one builds the array of its values, one will notice it is in fact Pascal's triangle, read diagonally. Therefore $ F_{n,k} \equal{} \genfrac {(} {)} {0pt} {} {n \minus{} k \plus{} 1} {k}$ (with the convention this value is zero for $ k>n\minus{}k\plus{}1$, i.e. $ k>(n\plus{}1)/2$).

[quote="mavropnevma"]Denote by $ F_{n.k}$ this number. It is obvious that $ F_{n,0} \equal{} 1$ (there is exactly one way to choose no element!), $ F_{n,1} \equal{} n$, and $ F_{n,k} \equal{} 0$ for $ k > (n \plus{} 1)/2$ (from pigeonhole principle). Now, looking at $ F_{n \plus{} 1,k}$, either the $ k$ elements are chosen from among the first $ n$ (therefore in $ F_{n,k}$ ways), or the element $ n \plus{} 1$ is chosen, so $ n$ is not, hence the remaining $ k \minus{} 1$ are chosen from among the first $ n \minus{} 1$ (in $ F_{n \minus{} 1,k \minus{} 1}$ ways). This leads to the recurrence relation $ F_{n \plus{} 1,k} \equal{} F_{n,k} \plus{} F_{n \minus{} 1,k \minus{} 1}$. But this is exactly the recurrence relation for some binomial coefficients; if one builds the array of its values, one will notice it is in fact Pascal's triangle, read diagonally. Therefore $ F_{n,k} \equal{} \genfrac {(} {)} {0pt} {} {n \minus{} k \plus{} 1} {k}$ (with the convention this value is zero for $ k > n \minus{} k \plus{} 1$, i.e. $ k > (n \plus{} 1)/2$).[/quote]

[color=blue]
Impressive solution!! good application of recurrence!

I propose an alternate using bijection principle:

Consider $ \ k$ identical dominoes and $ \ n \minus{} 2k \plus{} 1$ identical squares. arrange all of them in a line.

Number of ways = $ \frac {(n \minus{} 2k \plus{} 1 \plus{} k)!}{k!(n \minus{} 2k \plus{} 1)!} \equal{} \frac {(n \minus{} k \plus{} 1)!}{k!(n \minus{} 2k \plus{} 1)!} \equal{} \binom {n \minus{} k \plus{} 1}{k}$

Now after arranging all in a line, write $ \ 1,\ 2,\ 3,\cdots,\ n$ on them starting from left most position.
Note: Rightmost position will be blank as there are $ \ n\plus{}1$ positions and we are writing $ \ n$ numbers and on a domino we will write two numbers but if there is a domino at the end of the arrangement then its right place will be blank.
Finally pick left number of every domino and we are done with the problem.

$ \boxed{\boxed{\text{Hence desired answer} \equal{} \binom {n \minus{} k \plus{} 1}{k}}}$
[/color]

[quote="Nguyen Van Linh"]How many ways are there to choose $ k$ numbers from $ n$ numbers $ 1,2,\cdots , n$ such that no two consecutive numbers is chosen?[/quote]
Let the numbers be $ x_1,x_2,..,x_k$, wlog $ x_1 < x_2 < \dots < x_k$ and consider $ (y_1,y_2,..,y_k) \equal{} (x_1 \minus{} 1,x_2 \minus{} 2,...,x_k \minus{} k)$. Obviously $ 0 \le y_1 < y_2 < \dots < y_k \le n \minus{} k$. And there is a unique relationship between any $ y_1,..,y_k$ and $ x_1,..,x_k$. So we should find the number of ways to choose $ k$ different numbers from the set $ \{0,1,2,...,n \minus{} k\}$. This set has $ n \minus{} k \plus{} 1$ elements so the answer is $ \binom{n \minus{} k \plus{} 1}{k}$.

[quote="Mathias_DK"][quote="Nguyen Van Linh"]How many ways are there to choose $ k$ numbers from $ n$ numbers $ 1,2,\cdots , n$ such that no two consecutive numbers is chosen?[/quote]
Let the numbers be $ x_1,x_2,..,x_k$, wlog $ x_1 < x_2 < \dots < x_k$ and consider $ (y_1,y_2,..,y_k) \equal{} (x_1 \minus{} 1,x_2 \minus{} 2,...,x_k \minus{} k)$. Obviously $ 0 \le y_1 < y_2 < .. y_k \le n \minus{} k$. And there is a unique relationship between any $ y_1,..,y_k$ and $ x_1,..,x_k$. So we should find the number of ways to choose $ k$ different numbers from the set $ \{0,1,2,...,n \minus{} k\}$. This set has $ n \minus{} k \plus{} 1$ elements so the answer is $ \binom{n \minus{} k \plus{} 1}{k}$.[/quote]

[color=blue]
It is another application of bijection principle
Excellent!!

some typo error please fix it
[/color]

What if the numbers are written on a circle, so $ n$ and $ 1$ are considered consecutive?

[hide]Use the notations from my previous post. Denote by $ G_{n.k}$ this number. It is obvious that $ G_{n,0} \equal{} 1$ (there is exactly one way to choose no element!), $ G_{n,1} \equal{} n$, and $ G_{n,k} \equal{} 0$ for $ k > n/2$ (from pigeonhole principle). Now, for $ k>0$, looking at $ G_{n,k}$, either the $ k$ elements do not contain $ n$ (therefore $ F_{n\minus{}1,k}$ ways), or the element $ n$ is chosen, so $ 1$ and $ n\minus{}1$ are not, hence the remaining $ k\minus{}1$ are chosen from among the other elements $ 2,3,\ldots,n\minus{}2$ (in $ F_{n\minus{}3,k\minus{}1}$ ways). This leads to the relation $ G_{n,k} \equal{} F_{n\minus{}1,k} \plus{} F_{n\minus{}3,k\minus{}1} \equal{} \genfrac {(} {)} {0pt} {} {n\minus{}k} {k} \plus{} \genfrac {(} {)} {0pt} {} {n\minus{}k\minus{}1} {k\minus{}1} \equal{} \frac {n} {k} \genfrac {(} {)} {0pt} {} {n\minus{}k\minus{}1} {k\minus{}1}$.[/hide]


geometry
trapezoid
circumcircle
perpendicular bisector
geometry unsolved

Let ABCD  be a trapezoid such that AB IS  parallel to CD  and AB+CD=AD. Let P  be the point on AD  such that  AP=AB  and  PD=CD. Prove that:

a) angleBPC=90
b) Let Q be the midpoint of BC  and R the intersection of AD  with the circle through B, A and Q. Then B,P,R and  C  are concyclic.

a) is simply angle chasing.
angle BPC
=180 - angle BPA -angle CPD
=(360 - 2 angle BPA - 2 angle CPD)/2
=(360 - angle BPA - angle PBA - angle CPD - angle PCD)/2
=(angle BAP + angle CDP)/2
=180/2
90
Not sure about b) though.

The trapezoid angles $\angle DAB + \angle CDA = 180^o$ add up to a straight angle. The triangles $\triangle ABP, \triangle CDP$ are both isosceles with $AP = AB, DP = DC$. Hence,

$\angle APB + \angle DPC = 90^o - \frac{\angle DAB}{2} + 90^o - \frac{\angle CDA}
{2} = 180^2 - \frac{180^o}{2} = 90^o$

$\angle BPC = 180^o - (\angle APB + \angle DPC) = 90^o$

The perpendicular bisector of the segment $BP$ bisects the angle $\angle PAB \equiv \angle RAB \equiv \angle DAB$. Since it is perpendicular to the leg $BP$ of the right angle triangle $\triangle BPC$, it is parallel to the other leg $CP$. In addition, it cuts the leg $BP$ is half, hence it cuts the hypotenuse $BC$ also in half, i.e., at its midpoint $Q$. Since the angles $\angle BAQ = \angle RAQ = \frac{\angle DAB}{2}$ are equal, the chords $BQ = RQ$ of the circumcircle $(O)$ of the triangle $\triangle ABQ$ are also equal. Hence, the triangle $\triangle BRC$ has the median $RQ$ equal to half the opposite side $BC$, which implies that the angle $\angle BRC = 90^o$ is right. But then the circumcircle $(Q)$ of the right angle triangle $\triangle BPC$ is also the circumcircle of the right angle triangle $\triangle BRC$, i.e., the quadrilateral $BPRC$ is cyclic.

yetti that is the oficial solution :D  any others??? :lol:


geometry
parallelogram
geometry proposed

Dear everyone.

Please think of the following.

A circle with center O is internally tangent to two circles inside it, with centers O1 and O2, at points S and T respectively. Suppose the two circles inside intersect at points M, N with N closer to ST.
Show that S, N, T are collinear if and only if SO1/OO1 = OO2/TO2.

Thank you for reading. :)

$ O,O_1,S$ and $ O,O_2,T$ are collinear and triangles $ OST,O_1SN,O_2NT$ are isosceles whose  legs  are the radii of $ (O),(O_1),(O_2),$ respectively. If $ S,N,T$ are collinear, from $ \triangle O_1SN \sim \triangle O_2NT,$ due to $ \angle OST \equal{} \angle OTS$ and $ \angle O_1NS \equal{} \angle O_2NT,$ we obtain $ \frac {_{NO_1}}{^{TO_2}} \equal{} \frac {_{SO_1}}{^{NO_2}}.$ But it's easy to see that $ OO_1NO_2$ is a parallelogram $ \Longrightarrow$ $ NO_1 \equal{} OO_2 $ and $NO_2 \equal{} OO_1$ $ \Longrightarrow$ $ \frac {_{OO_2}}{^{TO_2}} \equal{} \frac {_{SO_1}}{^{OO_1}}.$

Assume that $ \frac {_{OO_2}}{^{TO_2}} \equal{} \frac {_{SO_1}}{^{OO_1}},$  which implies $ \frac {_{OO_2 \plus{} TO_2}}{^{TO_2}} \equal{} \frac {_{SO_1 \plus{} OO_1}}{^{OO_1}}$

$\Longrightarrow$ $ TO_2 \equal{} OO_1 \equal{} NO_2$ and similarly we'll have $ OO_2 \equal{} SO_1 \equal{} NO_1$ $ \Longrightarrow$ $ OO_1NO_2$ is a parallelogram.


Quite intreseting problem :

Here by Construction : i mean with just using $RULER$.

[color=blue]Problem[/color]A circle ,a point and a line ,are given in the plane.
construct a perpendicular line from the point to the line.

nice problem Arman today I solve it in the Physic class :D
Nice physic class :mad:


close, aznmonkey, there's one part you missed.

[quote="236factorial"]How many feet are in 3.5 bronessts, which is 4/5 of a third of a half of a mile?  :D[/quote]

[hide]No brainer.  :rotfl:

Anyways, 4/5*1/3*1/2 = 4/30 = 2/15

3 1/2*2/15 = 7/2*2/15 = 14/30 = 7/15 of a mile

5280/15 = 352 ft.

352*7= 2464 ft.[/hide]

[quote="236factorial"]How many feet are in 3.5 bronessts, which is 4/5 of a third of a half of a mile?  :D[/quote]
[hide]3.5* 4/5(5280*1/3*1/2)=2464 feet?[/hide]

correct, aznmonkey1992 just missed the half part  :P

[quote="anirudh"]State stuff clearer next time:

[quote="236Factorial and anirudh"]Is 4/5 of a third of a half of a mile 1 bronner or 3 broonners.  :bruce: [/quote][/quote]

now you can solve it since I fixed it  ;)

[quote="236factorial"]How many feet are in 3.5 bronessts, where a bronesst is 4/5 of a third of a half of a mile?  :D[/quote]

1bronner: 4/5*1/3*1/2*5280=2/15*5280=704

704*3.5=2464, unless my calculator made a mistake

your answer does not show up on my computer

[hide][quote="236factorial"]How many feet are in 3.5 bronessts, where a bronesst is 4/5 of a third of a half of a mile?  :D[/quote]

$\frac45\times\frac13\times\frac12=\frac{2}{15}$

$\frac{2}{15}\cdot3.5=                                                                                \frac{7}{15}$

$\frac{7}{15}\cdot5280=2464$

2464 feet[/hide]

[hide]2464 feet[/hide] did you make this up yourself?

[quote="jhollenbeck"] did you make this up yourself?[/quote]

Yep, as with most of the other problems I post  :)


trigonometry
limit
algebra unsolved
algebra

let sequence $ u_k\equal{}\frac{\cos1\minus{}\cos k.\cos (k\plus{}1)}{\cos k.\cos (k\plus{}1)}$.Defined $ a_k$,$ b_k$ are the number of postive number and negative number of sequence $ u_1,u_2,...,u_n$.
(i) Prove that $ a_k <b_k$
(ii) Find $ \lim \frac{a_n}{b_n}$

[quote="asish2000"]hint:
[hide]cos1 = cos(k+1-k) = cos(k+1)cosk - sin(k+1)sink

So, u(k) = -tank.tan(k+1)[/hide][/quote]
No, cos1=cos(k+1)cos(k)+sin(k+1)sink


function
real analysis
real analysis unsolved

[quote="Golden Rule"]Let E $ \subseteq\mathbb{R}$. A function f : E $ \rightarrow\mathbb{R}$ is said to be increasing on E if and only if $ x_1, x_2\in E$ and $ x_1 < x_2$ imply $ f(x_1)\le f(x_2)$. Suppose that f is increasing and bounded on an open, bounded, nonempty interval (a, b).
(a) Prove that f(a+) and f(b-) both exist and are finite.
(b) Prove that f is continuous on (a, b) if and only if f is uniformly continuous on (a, b).
(c) Show that (b) is false if f is unbounded. Find an increasing function g : (0, 1) $ \rightarrow\mathbb{R}$ that is continuous on (0, 1) but not uniformly continuous on (0, 1).[/quote]
(a) Try to prove that $ f(a\plus{}) \equal{} \inf_{(a,b)}f$ and $ f(b\minus{}) \equal{} \sup_{(a,b)}f$. The $ \inf$ and the $ \sup$ clearly exist.
(b) Using (a), extend $ f$ to a function that is defined on $ [a,b]$ and that is automatically continuous at the endpoints of the interval. Then remember that a continuous function defined on a closed and bounded interval is uniformly continuous.
(c) Try to find the easiest possible counterexample.

[quote="Golden Rule"](c) Show that (b) is false if f is unbounded. Find an increasing function g : (0, 1) $ \rightarrow\mathbb{R}$ that is continuous on (0, 1) but not uniformly continuous on (0, 1).[/quote]

In order to construct a counter-example, you must understand geometric meaning of uniformly continuity.


By what number is the number of cubic inches in the volume of a sphere multiplied when the number of inches in the radius is doubled?

The formula for volume of a sphere is: $ \frac{4}{3}\pi{r}^3$. If $ r$ is doubled, we have $ 2r$. Plug in $ 2r$ to get $ \frac{4}{3}\pi{2r}^3$

$ {2r}^3 \equal{} 8r$(the whole thing is being cubed, not just the $ r$).

$ \dfrac{\frac{4}{3}\pi{2r}^3}{\frac{4}{3}\pi{r}^3} \equal{} \boxed{8}$


rotation
geometry
geometric transformation
vector

here is 1 problem:-
Q)one disc is rotating with constant angular velocity 'w'.one body is just placed on it.all the surfaces in contact r frictionless.in the frame of ground body will n't have any motion. how do u interfere the problem in the frame of disc i.e, in this non inertial accelerating disc frame :?:
tell me the forces responsible for that motion :wink:

From the viewpoint of the observer rotating with the disc, the disc will be still and the body will rotate with angular velocity $-\omega$ (that is, in the direction opposite the direction of the disc rotation in the lab referential frame). From the disc frame, the body will have centripetal acceleration $\omega^{2}r$, where $r$ is the radius of the circle described by the body. This acceleration is due to the rotation of the body. A centrifugal force of intensity $m\omega^{2}r$ is acting on the body, but this force is antiparallel (has an opposite direction) to the centripetal acceleration and is not its source. (I won't bother writing minus signs from now on, but I'll say whether the force has a centrifugal or a centripetal direction.)

Actually, the force supplying this acceleration is the Coriolis force, which arises from the fact that not only is the disc system a non-inertial one, it is rotating rather than merely linearly accelerating. The Coriolis force is $F = 2m(\vec v \times \vec \omega)$, where $\vec v$ is the velocity of the body in the disc system. The intensity of this force is $2m\omega^{2}r$ as you can check, and it exactly cancels out the centrifugal force $m\omega^{2}r$ and supplies the centripetal acceleration $\omega^{2}r$.

I hope this made things a bit clearer -- if not, feel free to ask :) The existence of the Coriolis force can be deduced in several ways -- through conservation of angular momentum, or through lagrangian dynamics, for instance -- or, of course, through ordinary referential systems analysis.

[quote="Djole"] The intensity of this force is $2m\omega^{2}r$ as you can check, and it exactly cancels out the centrifugal force $m\omega^{2}r$ and supplies the centripetal acceleration $\omega^{2}r$.[/quote]

  hello!Did you forget that the coriolis force is in the direction perpendicular to the direction of radius vector :sleeping:
         How it can cancells with the force in radial direction :rotfl:

$\vec \omega$ is perpendicular to the plane of the disc. The body is rotating around the center of the disk, so $\vec v$ has a direction tangential to the radius vector. The cross product of these two has a radial direction.


search AMC USA(J)MO USAMO ARML Harvard college \/closed

Already posted this in the resources, but since it is quite relevant to suggestions, will post it here as well.  I would very much appreciate if the administators of this site would put some math problems in a certain file that can be easily accessed.

We'll be working on projects of which this will be a part in the coming years.  We won't, however, be able to include precise tests from other sources due to copyright law...  It'll be a while, but we expect to produce something like this eventually.

I know that several responses in the resources section mentioned something to this effect, but you should just search on google or something.  I know for a fact that there are lots of USAMO-level problems, for instance if you search for "________ Mathematical Olympiad" (inserting something like "Norwegian," "South African," "Flanders," "Russian," or any of 50 other country names, you will most likely be able to find problems.  Plus, there are plenty of websites that have compiled problems from various contests.  See Mu Alpha Theta's website ([url]http://www.mualphatheta.org[/url]), for example.  (Specifically, see [url]http://www.mualphatheta.org/Links/MAO_Links_Contests.htm[/url] for links to various national contests such as ARML, Math League, WMTS, Ole Miss POW, IMToT, etc.)

If you still can't find any problems, I can send you some.  I have found literally thousands of problems over the past few months this way, and I'm sure you will be able to also if you put your mind (or half...or a quarter... after all there are tons out there) to it. 8-)

Oh - and to Mr. Rusczyk... I am sure that you have a great number of problems (hopefully only the math type, heh), but if you're interested, I can send you a vast quantity of problems via e-mail or something.  That way if you do (excusing the potential legal complications of such an event) choose to post a "Large file of math problems" on this site, you would easily have the means to do so.

Another possibility is to create a "Links" page that would link to a bunch of sites with math problems, specifically a bunch of sites with, say, past IMO problems...or past Candian Mathematical Olympiad problems...or past Harvard-MIT Math Tournament Problems...or...you know what I mean.

If you're interested, I'd be more than willing to contribute the materials I have found to, as they say, spread the wealth around.

(Alas, the wealth is in math problems...not money.  Oh woe is me. :( )

mathfanatic - hook me up with that list of links if you have time.  Email them to rusczyk@artofproblemsolving.com - I'll create a links list on our site in the resources section for them (with proper accreditation, of course).

will do.  will include a list of everything I have.  :)

While you're at it... would you mind mailing me those links too? I don't mean to be a pain but I would appreciate it...

send it to idiotwithoutavillage@hotmail.com

Thx.
IW.OAV

I think many of us could benefit. Why not post them here?

They'll be up on the site in a few days.  In the resources section.

The list (or at least much of it) is up in the Resources section.  More coming in a bit.

Brilliant job on the resources section...this will certainly be very helpful to lots of AoPSers out there...

One thing I wanted to mention was PostScript.  A lot of those links have files in Adobe PostScript format (.ps).  The preferred viewer (which I use) is the Ghostscript-GSView Combo, available at [url]http://www.cs.wisc.edu/~ghost/doc/AFPL/get800.htm[/url]

Because there are a number of confusing links at this page, I would recommend placing a link to [url]ftp://mirror.cs.wisc.edu/pub/mirrors/ghost/AFPL/gs800/gs800w32.exe[/url] (for ghostscript) and [url]ftp://mirror.cs.wisc.edu/pub/mirrors/ghost/ghostgum/gsv44w32.exe[/url] (for GSView) - with a disclaimer, etc., of course. :wink:

Also, I think a lot of people on artofproblemsolving.com are interested in the AMC competitions; however, there are only a limited number of AMC-level problems available to them.  I've found that the COMC (http://www.cms.math.ca/Competitions/COMC/) and the VWO (Flanders Math Olympiad, at http://www.kulak.ac.be/vwo/uk/) are two competitions with AMC-level problems; they should be put in the Intermediate-Level section.  Thanks!

[quote]Also, I think a lot of people on artofproblemsolving.com are interested in the AMC competitions; however, there are only a limited number of AMC-level problems available to them. [/quote]

I definitely agree. I have found many many sites with countless olympiad-type proof problems, but almost none with short-answer problems. I guess part of the reason for this condition is that most people who are interested enough in math to make a problem site are pretty good at problem-solving, at the olympiad level.

Fierytycoon

Wow, he's fast (he changed it already!  See "Problems" on the Resources section)  Thanks, Mr. Rusczyk.

mathfanatic


let $x,y,z >0$  . find minimal value of expression :
$\frac{1}{x^2}+ \frac{1}{y^2}+ \frac{1}{z^2} + 8x^2 +108y^2 +16z^2$
          where $xy+yz+zx=1$

[quote="tranthanhnam"]let $x,y,z >0$  . prove that :
$\frac{1}{x^2}+ \frac{1}{y^2}+ \frac{1}{z^2} + 8x^2 +108y^2 +16z^2$
          where $xy+yz+zx=1$[/quote]

What exactly to prove ?  :huh:

Not [b]prove that [/b]but [b]find minimal value of expression [/b]


What are the limitations on value k when
x^4 + 4 x^3 + 4 x^2 + k
has 4 real roots

[hide]For f(x) = x^4 + 4x^3 + 4x^2 + k to have 4 real roots, two turning points must be either above or below the x-axis and one must be below or above the x-axis respectively.

Now, d(x^4 + 4x^3 + 4x^2 + k)/dx = 4x^3 + 12x^2 + 8x.  Turning points occur when f'(x) = 0.  4x^3 + 12x^32 + 8x = 0 ===> x(x+1)(x+2) = 0.  .'. x = 0, -1, -2.

Case 1:

f(-1) > 0, which means that f(-2) and f(0) < 0

The condition for this situation is k > -1 and k < 0, so -1 < k < 0.

Case 2:

f(-1) < 0, which means that f(-2) and f(0) > 0

The condition for this situation is k < -1 and k > 0, which is a paradox.  Therefore, the only condition for four real roots is -1 < k < 0.[/hide]

[hide="Solution"]This is equivalent to $ ((x\plus{}1)^2\minus{}1)^2\equal{}\minus{}k$, so $ k$ must be nonpositive. Let $ \minus{}k\equal{}m^2$; then each of the four real solutions must satisfy $ (x\plus{}1)^2\equal{}1\plus{}m$ or $ (x\plus{}1)^2\equal{}1\minus{}m$. But $ 1\plus{}k\equal{}1\minus{}m^2\equal{}(1\minus{}m)(1\plus{}m)\equal{}(x\plus{}1)^4\ge0$, so $ k\ge\minus{}1$ and our answer is $ \minus{}1\le k\le0$.[/hide]


7. The symbolism $ \lfloor x\rfloor$ denotes the largest integer not exceeding x. For example, $ \lfloor3\rfloor = 3$, and $ \lfloor\frac{9}{2}\rfloor = 4$. Compute $ \lfloor\sqrt{1}\rfloor + \lfloor\sqrt{2}\rfloor + \lfloor\sqrt{3}\rfloor +\ldots+\lfloor\sqrt{16}\rfloor$.

$ (A) 35 \qquad (B) 38 \qquad (C) 40 \qquad (D) 42 \qquad (E) 136$

This one is a lot easier if you know the trick
[hide]The floor function makes this problem nice. We know that everything between $ \sqrt{1}$ and $ \sqrt{3}$ is bigger than one but less than two, so it "floors" to 1. Similarly, the numbers between $ \sqrt{4}$ and $ \sqrt{8}$ simplify to 2. We get 1+1+1+2+2+2+2+2+3+3+3+3+3+3+3+4=3+10+21+4=[b]38, [size=150]B[/size][/b][/hide]


The bases of a trapezoid have lengths $10$ and $20$, and the lengths of its other two sides are $6$ and $8$. Find the area of the trapezoid.

k you can form a $6-8-10$ triangle clearly..if you draw it you'll see what i mean

then the height of the triangle therefore trapezoid is

$h=\frac{48}{10}=4.8$

So then the Area is

$A=\frac{(10+20)(4.8)}{2}=\boxed{72}$


I went to mu alpha theta competition and competed in algebra II division. i feel that even if i finished the entire Algebra II book, i can't do all of the problems. What do i need to learn to get a perfect score in MAO?

The purpose behind our online school here at AoPS is to help students reach a higher level of problem solving.  We specifically offer an Intermediate Algebra class that is great for students who are in or have completed Algebra II.

Also, have you seen the Art of Problem Solving books before?  I would suggest making sure that you know all of the topics in the first volume which covers most problem solving material tested on the AMC.  After that, the second volume will take you even farther.

Gee he's done all of those.. He's read the entire book and done research on things that are Pre-cal and so on.. He's asking how is it possible to find some hard Algebra II practice problems and things like that.. He always misses first place by a few points... that's why

If you're just missing first place by a couple points in MAO stuff, it's time to start working on harder problems.

Check out the Harvard-MIT materials, for example.

Don't worry about taking that last step in MAO from second to first.  Aim at learning more challenging material and that last step will take care of itself.

Yeah but he's GOTTA win at teams so I can leech off him so I can make my momma proud.

David, stop acting like an immature man....................... :D

Hmm...Rui, I feel your pain.

The same thing is happening to me right now, except I'm off by like 3 mistakes at every competition.

MAO has nothing to do w/ learning hard math. just do a million old tests, they always repeat questions anyway.
it works for me and gaku  :lol:

yeah MAO reuses material all the time.  also if you look at the old test you can see what kinds of things they ask at each point in the year.


Prove that

 \sum (m_a)(h_a) \leq s 2

where h_a,h_b,h_c and m_a,m_b,m_c  are heights and medians of a triangle of semiperimeter s.

it seems to me that you are missing something from the statement manlio, please edit your post to make it correct :)

Does anyone have a solution for this one? Dzeta gave me this problem and I couldn't solve it except for the case when the triangle is acute. For obtuse triangles I cannot prove it. And it's so nice. It seems that I'm unable to prove any inequality related to medians of a triangle. :(

Now solved in http://www.mathlinks.ro/Forum/viewtopic.php?t=21462 post #16. Argh, that was a piece of work...

  Darij