calculus integration logarithms geometry trapezoid function derivative

find the sum by integration
\[\sqrt 1+\sqrt 2+\sqrt 3+............+\sqrt{100}\]

\[f(x) = x^{x^{x}}\]
\[\begin{array}{l}find \\ f''(x) \\ \end{array}\]

1st question: I don't understand.
2nd question : Put $g(x)=x^{x}$ then $\ln g(x)=x\ln{x}$ and $\frac{g'}{g}=\ln{x}+1$ . Similary , we can find $f'$ and $f''$.

1:
\[\sum_{x = 1}^{100}{\sqrt x }= 671.4629471 \]
can i found it by integration

2:
\[f'(x) = x^{x^{x}}\times \left({\left[{x^{x}\ln x\left({\ln (x)+1}\right)}\right]+\left[{\frac{{x^{x}}}{x}}\right]}\right) \]

You cannot really find the [b]exact[/b] value of the sum in question by integration, but you can find it approximately by replacing the sum with $\int_{0}^{100}\sqrt x\,dx=2000/3=666.6666\dots$. The precision is not too impressive though. It gets better if you use $\int_{0.5}^{100.5}$ instead: $0.5^{3/2}\approx 0.3$, $100.5^{3/2}\approx 1000+15\cdot 0.5=1007.5$, so you get about $(2/3)\cdot 1007.2\approx 2\cdot 335.7=671.4$ (here you are assumed to know that $2\sqrt 2$ is about $3$ and the linearization trick for finding the power $3/2$ for numbers close to perfect squares; if this is a problem coming from the standard calculus class, you should know both by the moment you cover integration). This is almost the best you can do without pen and paper.

Expanding on fedja's argument:
Since $\sqrt{x}$ is concave, $\sum_{n=1}^{100}\sqrt{n}>\int_{0.5}^{100.5}\sqrt{x}\,dx$ (the midpoint rule) and $\sum_{n=1}^{100}\sqrt{n}-\frac12(\sqrt{1}+\sqrt{100})< \int_{1}^{100}\sqrt{x}\,dx$ (the trapezoid rule).

If $S$ is the sum we care about, the first gives $S>671.4$ and the second gives $S-\frac{11}{2}<666$, so $S<671.5.$

$0.5^{3/2}$ is actually about $.35$, but since $671.4=\frac23\cdot 1007.1$, this isn't a problem; $671.4$ is still a strict lower bound.

Accuracy certified to within $\frac1{10}$ is pretty good for a sum of this size.

can we make rule or something for that
like
\[\sum_{n = a}^{b}{\sqrt n }\approx \int_{a-.5}^{b-.5}{\sqrt x }\cdot dx \]

Yes, except the upper limit is $b+0.5$. As jmerry pointed out, this is just the midpoint rule for integration. We have $\int_{k-0.5}^{k+0.5}\sqrt x\,dx\approx \sqrt k$ and all we need now is to add these approximate identities from $k=a$ to $k=b$. The concavity of $\sqrt x$ implies that the integral is actually less than the sum. If we want to know by how much less, we can use the second order Taylor formula $f(k+y)=f(k)+f'(k)y+\frac{1}{2}f''(k+\xi)y^{2}$ with $\xi$ between $0$ and $y$ for the function $f(x)=\sqrt x$. Since $f''(x)=-\frac{1}{4}x^{-3/2}$, we see that, for $|y|\le 0.5$, the error term in the $k$-th approximate identity is at most $\frac{1}{4}(k-1/2)^{-3/2}\int_{0}^{0.5}y^{2}\,dy=\frac 1{96}(k-1/2)^{-3/2}$. So, no matter how many terms we have in our sum, the error in the formula does not exceed $\frac 1{96}\sum_{k=1}^\infty (k-1/2)^{-3/2}$, which is less than $0.1$ (but greater than $0.01$), so jmerry's one tenth is not something special for the sum up to $100$. You can take the sum up to $1000000$ and still it will be the same one tenth. This is in striking contrast with what you get when using the right end rule $\sum_{a}^{b}\sqrt k\approx \int_{a-1}^{b}\sqrt x\,dx$. Here the error in the $k$-th term is governed by the first derivative, which is about $\frac 1{2\sqrt k}$. The corresponding series diverges, so the absolute error may now become large. The relative error is still small for large number of terms, though.

At last, I should say that there is nothing sacred about $\sqrt$ here: this technique works for many other functions as well. As a follow-up question, you may try to compute $\sum_{1000}^{1000000}\sqrt[3]k$ correctly up to the first decimal place using only pencil and a standard sheet of paper. If I replace $1000000$ by $1000000000000$ here, you'll be able to do it faster than your computer if you ask the latter to honestly add the numbers up. Isn't it nice? :)

To be exact, my $\frac1{10}$ is $\frac{\sqrt{2}}{6}-\frac16$. I get $\frac23N^{3}-\frac{N}{2}-\frac{\sqrt{2}}6<\sum_{n=1}^{N^{2}}\sqrt{n}<\frac23N^{3}-\frac{N}{2}-\frac16$ by the midpoint rule on the left and the trapezoid rule on the right.

The same kind of argument works for a wide range of functions; $\sqrt[3]{x}$ is certainly one of them.


inequalities
geometry
LaTeX
angle bisector
inequalities proposed

Prove that  $w_{a}w_{b}+w_{b}w_{c}+ w_{a}w_{c}\geq3A\sqrt{3}$, where $w_{a},w_{b},w_{c}$ denote the lengths of the interior angle bisectors of a triangle ABC and $A$ the area of the triangle.

The proof published in Crux was pretty extensive and i wonder whether someone in the forum could find something better.

Is  this  your problem??  [tex]w_aw_c+w_bw_a+w_cw_b\geq 3A\sqrt3 [/tex]
If  yes , then  just  use  the  fact  that  [tex] w_a\geq h_a\geq \sqrt{s(s-a)} [/tex] and use  AM-GM . Please learn  Latex  :)

Nickolas, your inequality should be reverted :
    $ m_{a}\geq\sqrt{p(p\minus{}a)}\geq w_{a}\geq h_{a}$ :P

$ w_{a}=\frac{2bc}{b+c} \sqrt{\frac{s(s-a)}{bc}} $ so $w_{a}\leq \sqrt{s(s-a)}$. Anyway, i do not believe that there is such a trivial solution to the inequality.

Oh, and I have written the inequality in Latex :)

I don't think it 's right for the case right issoceles triangle.

set $a=b=x$. Then $c=x\sqrt{2}$, $w_c=x\frac{\sqrt2}{2}$ and $A=\frac{x^2}{2}$. It follows easily by the interior angle bisector theorem that $w_b=w_a=x\sqrt{\frac{1}{(1+\sqrt{2})^2}+1}$. So in this case the ineq gives $\frac{1}{(1+\sqrt{2})^2}+1+\sqrt{2}\sqrt{\frac{1}{(1+\sqrt{2})^2}+1}\geq\frac{3\sqrt{3}}{2}$. $LHS=2,70$ and $RHS=2,59$, so i think that your counterexample is false!


[hide]
Note that since all the ai are either in (0,1) or (1,infinity), all of the logs are positive, and we can apply AM-GM to get that $ log_{a_1}a_2 \plus{} log_{a_2}a_3 \plus{} \ldots \plus{} log_{a_n}a_1 \ge n.$ We also have $ n(\log^2_{a_1}a_2\plus{}\log^2_{a_2}a_3\plus{}...\plus{}\log^2_{a_n}a_1) \geq (\log_{a_1}a_2\plus{}\log_{a_2}a_3\plus{}...\plus{}\log_{a_n}a_1)^2 \geq n(\log_{a_1}a_2\plus{}\log_{a_2}a_3\plus{}...\plus{}\log_{a_n}a_1),$ which implies the inequality we want to prove.[/hide]


inequalities
function
linear algebra
matrix
blogs

Let $ A,B,C$  are $ nxn$ Hermitian matrices and positive definite .$ \lambda_1,\lambda_2,\lambda_3$ are positive real numbers suth that  $ \lambda_1\plus{}\lambda_2\plus{}\lambda_3\equal{}1$
Prove that inequality :
       $ \det(\lambda_1A\plus{}\lambda_2B\plus{}\lambda_3C)\det(\lambda_1A^{\minus{}1}\plus{}\lambda_2B^{\minus{}1}\plus{}\lambda_3C^{\minus{}1}) \geq 1$
-- :P ---

Prove that : very nice !
1) $ Tr[(\lambda_{1}A \plus{} \lambda_{2}B \plus{} \lambda_{3}C)(\lambda_{1}A^{ \minus{} 1} \plus{} \lambda_{2}B^{ \minus{} 1} \plus{} \lambda_{3}C^{ \minus{} 1})] \geq n$
2) $ Tr(A(B \plus{} C)^{ \minus{} 1}A) \plus{} Tr(B(A \plus{} C)^{ \minus{} 1}B) \plus{} Tr(C(A \plus{} B)^{ \minus{} 1}C) \geq \frac {Tr(A \plus{} B \plus{} C)}{2}$
 General 1),2) Very nice ! :P

We use $ det(A\plus{}B)\geq((det(A))^{1/n}\plus{}(det(B))^{1/n})^n$ if $ A,B\geq{0}$.
Let $ E\equal{} \det(\lambda_{1}A\plus{}\lambda_{2}B\plus{}\lambda_{3}C)\det(\lambda_{1}A^{\minus{}1}\plus{}\lambda_{2}B^{\minus{}1}\plus{}\lambda_{3}C^{\minus{}1})$.
$ E\geq(\lambda_1(det(A))^{1/n}\plus{}\lambda_2(det(B))^{1/n}\plus{}\lambda_3(det(C))^{1/n})^n(\lambda_1(det(A))^{\minus{}1/n}\plus{}\lambda_2(det(B))^{\minus{}1/n}\plus{}\lambda_3(det(C))^{\minus{}1/n})^n$.
It remains to prove that $ F\equal{}(\lambda_1\alpha\plus{}\lambda_2\beta\plus{}\lambda_3\gamma)(\lambda_1/\alpha\plus{}\lambda_2/\beta\plus{}\lambda_3/\gamma)\geq{1}$ where $ \alpha,\beta,\gamma>0$.
We calculate the minimum of the function $ (\lambda_1,\lambda_2,\lambda_3)\rightarrow{F}$ defined on the compact subset  $ \lambda_1,\lambda_2,\lambda_3\geq{0},\lambda_1\plus{}\lambda_2\plus{}\lambda_3\equal{}1$. If we reason by recurrence then we may assume $ \lambda_1,\lambda_2,\lambda_3>0$ and we use the Lagrange method. We find that an extremum exists only if $ \alpha\equal{}\beta\equal{}\gamma$; in this case the function is $ 1$. Then other extrema are reached on the  edge and we conclude with the recurrence hypothesis.

Lemma 1) $ X,Y$ are $ n$x$ n$ Hermitian matrices and positive definite .Then $ SP_c(XY^{ \minus{} 1}) \subset \mathbb{(R^{ \plus{} })}$ Here $ SP_c(A)$ is spectrum(A)
 Lemma 2) $ A(B \plus{} C)^{ \minus{} 1}A \plus{} \frac {(B \plus{} C)}{4} \geq A$ ($ 1$),  $ B(A \plus{} C)^{ \minus{} 1}B \plus{} \frac {(A \plus{} C)}{4} \geq B$ ($ 2$),  $ C(B \plus{} A)^{ \minus{} 1}C \plus{} \frac {(B \plus{} A)}{4} \geq C$ ($ 3$),
Proof Lemma 1)
 We have $ Y^{ \minus{} 1}$ is $ n$x$ n$ Hermitian matrix and positive definite then the exist $ Z$   is $ n$x$ n$ Hermitian matrix and positive definite .Such that : $ Y^{ \minus{} 1} \equal{} Z^2$
We have  : $ SP_c(XY^{ \minus{} 1}) \equal{} SP_c((XZ).Z) \equal{} SP_c(Z(XZ)) \equal{} SP_c(ZXZ) \subset \mathbb{(R^{ \plus{} })}$ By $ ZXZ$ is Hermitian matrix and positive definite .
Proof Lemma 2) For ($ 1$),($ 2$),($ 3$) are similar . $ (B \plus{} C) \equal{} Udiag(\lambda_1,\lambda_2,...,\lambda_n)U^{*}$ , $ U$ is matrix Unita $ U^{*} \equal{} U^{ \minus{} 1}$,$ \lambda_k > 0 ,\forall k \in \{1,2,..,n\}$
 $ \forall x \in M_{n1}(\mathbb{C})$
Put $ y \equal{} U^{*}Ax \equal{} \begin{pmatrix}y_1 \\
y_2 \\
... \\
y_n\end{pmatrix},z \equal{} U^{*}x\begin{pmatrix}z_1 \\
z_2 \\
... \\
z_n\end{pmatrix}$
then we have :
 $ x^{*}(A(B \plus{} C)^{ \minus{} 1}A \plus{} \frac {(B \plus{} C)}{4})x \equal{} x^{*}AUdiag(\frac {1}{\lambda_1},\frac {1}{\lambda_2},..,\frac {1}{\lambda_n})U^{*}Ax \plus{} x^{*}Udiag(\frac {\lambda_1}{4},\frac {\lambda_2}{4},..,\frac {\lambda_n}{4})U^{*}x \equal{} y^{*}diag(\frac {1}{\lambda_1},\frac {1}{\lambda_2},..,\frac {1}{\lambda_n})y \plus{} z^{*}diag(\frac {\lambda_1}{4},\frac {\lambda_2}{4},..,\frac {\lambda_n}{4})z \equal{} \sum_{k \equal{} 1}^{n}{(\frac {|y_k|^2}{\lambda_k} \plus{} \frac {|\overline{z_k}|^2\lambda_k}{4})} \geq \sum_{k \equal{} 1}^{n}{|y_k\parallel{}\overline{z_k}|} \geq |\sum_{k \equal{} 1}^{n}{y_k\overline{z_k}}| \equal{} x^{*}Ax$
 Here $ z^{*}y \equal{} x^{*}Ax$,..The end .
Use lemma 1)
 We have : $ Tr[(\lambda_{1}A \plus{} \lambda_{2}B \plus{} \lambda_{3}C)(\lambda_{1}A^{ \minus{} 1} \plus{} \lambda_{2}B^{ \minus{} 1} \plus{} \lambda_{3}C^{ \minus{} 1})] \equal{} n(\lambda_1^2 \plus{} \lambda_2^2 \plus{} \lambda_3^2) \plus{} 2\lambda_1\lambda_2(Tr(AB^{ \minus{} 1}) \plus{} Tr(AB^{ \minus{} 1})^{ \minus{} 1} \plus{} \lambda_2\lambda_3(Tr(BC^{ \minus{} 1}) \plus{} Tr(BC^{ \minus{} 1})^{ \minus{} 1}) \plus{} \lambda_1\lambda_3(Tr(CA^{ \minus{} 1}) \plus{} Tr(CA^{ \minus{} 1})^{ \minus{} 1}) \geq n(\lambda_1^2 \plus{} \lambda_2^2 \plus{} \lambda_3^3 \plus{} 2\lambda_1\lambda_2 \plus{} 2\lambda_2\lambda_3 \plus{} \lambda_3\lambda_1) \equal{} n$
 Use lemma 2) We have
$ A(B \plus{} C)^{ \minus{} 1}A \plus{} B(A \plus{} C)^{ \minus{} 1}B \plus{} C(A \plus{} B)^{ \minus{} 1}C\geq\frac {(A \plus{} B \plus{} C)}{2}$
Implies $ Tr(A(B \plus{} C)^{ \minus{} 1}A) \plus{} Tr(B(A \plus{} C)^{ \minus{} 1}B) \plus{} Tr(C(A \plus{} B)^{ \minus{} 1}C)\geq\frac {Tr(A \plus{} B \plus{} C)}{2}$

Prove that :
 $ Tr(\frac{A^m\plus{}B^m\plus{}C^m}{3}) \geq Tr(\frac{A\plus{}B\plus{}C}{3})^m$ ,$ \forall m \in \mathbb{N^{*}}$

1) I return to my post; $ F\geq{1}$ or $ log(\lambda_1\alpha\plus{}\lambda_2\beta\plus{}\lambda_3\gamma)\plus{}log(\lambda_1/\alpha\plus{}\lambda_2/\beta\plus{}\lambda_3/\gamma)\geq{0}$  can be proved easily as follows:
 $ \minus{}log$ is a convex function ; then  $ log(\lambda_1\alpha\plus{}\lambda_2\beta\plus{}\lambda_3\gamma)\geq\lambda_1log(\alpha)\plus{}\lambda_2log(\beta)\plus{}\lambda_3log(\gamma)$ and $ log(\lambda_1/\alpha\plus{}\lambda_2/\beta\plus{}\lambda_3/\gamma)\geq{\minus{}}\lambda_1log(\alpha)\minus{}\lambda_2log(\beta)\minus{}\lambda_3log(\gamma)$.
2) Hi QuyBac,
The trace function has nothing to do in your 2).
the result is : if $ A,B,C$ are non negative and if 2 matrices among $ A,B,C$ are definite then $ A(B\plus{}C)^{\minus{}1}A\plus{}B(A\plus{}C)^{\minus{}1}B\plus{}C(A\plus{}B)^{\minus{}1}C\geq\frac{(A\plus{}B\plus{}C)}{2}$

The first two results are implied by the following result:
if $ U$ is hermitian then let $ m(U)=inf(spectrum(U))$. We can prove easily that $ m(U+V)\geq{m(U)+m(V)}$.
Let $ S=(\lambda_{1}A+\lambda_{2}B+\lambda_{3}C)(\lambda_{1}A^{-1}+\lambda_{2}B^{-1}+\lambda_{3}C^{-1})$; then $ m(S)\geq{1}$.
Proof: $ S=({\lambda_1}^2+{\lambda_2}^2+{\lambda_3}^2)I+\lambda_1\lambda_2(AB^{-1}+BA^{-1})+\cdots$.
It remains to prove that $ m(AB^{-1}+(AB^{-1})^{-1})\geq{2}$. Let $ AB^{-1}=W$.
$ W,W^{-1}$ are simultaneously triangularizable (and diagonalizable !). Then $ spectrum(W+W^{-1})=(\sigma_i+1/\sigma_i)$ with $ \sigma_i>0$ and $ m(W+W^{-1})\geq{2}$.

[color=blue]Thanks loup blanc ![/color]
In math we have Inequality Nesbit for $ a,b,c > 0$
 $ \frac {a}{b \plus{} c} \plus{} \frac {b}{a \plus{} c} \plus{} \frac {c}{a \plus{} b} \geq \frac {3}{2}$
In Matrix Let $ A,B,C$ are $ n$x$ n$  Hermitian matrices and positive definite.I can prove that :
$ Tr( A^{\frac {1}{2}}(B \plus{} C)^{ \minus{} 1}A^{\frac {1}{2}} \plus{} B^{\frac {1}{2}}(A \plus{} C)^{ \minus{} 1}B^{\frac {1}{2}} \plus{} C^{\frac {1}{2}}(A \plus{} B)^{ \minus{} 1}C^{\frac {1}{2}})\geq\frac {3n}{2}$
But don't prove that :
  $ A^{\frac {1}{2}}(B \plus{} C)^{ \minus{} 1}A^{\frac {1}{2}} \plus{} B^{\frac {1}{2}}(A \plus{} C)^{ \minus{} 1}B^{\frac {1}{2}} \plus{} C^{\frac {1}{2}}(A \plus{} B)^{ \minus{} 1}C^{\frac {1}{2}}\geq\frac {3I_{n}}{2}$
I hope you interested it  :maybe:

Thanks loup blance !
 But $ Tr( A^{\frac{1}{2}}(B\plus{}C)^{\minus{}1}A^{\frac{1}{2}}\plus{}B^{\frac{1}{2}}(A\plus{}C)^{\minus{}1}B^{\frac{1}{2}}\plus{}C^{\frac{1}{2}}(A\plus{}B)^{\minus{}1}C^{\frac{1}{2}})\geq\frac{3n}{2}$ is true ! Do you have agree ?

We have :
 $ Tr( A^{\frac {1}{2}}(B \plus{} C)^{ \minus{} 1}A^{\frac {1}{2}} \plus{} B^{\frac {1}{2}}(A \plus{} C)^{ \minus{} 1}B^{\frac {1}{2}} \plus{} C^{\frac {1}{2}}(A \plus{} B)^{ \minus{} 1}C^{\frac {1}{2}}) \equal{} Tr(A(B \plus{} C)^{ \minus{} 1}) \plus{} Tr(B(A \plus{} C)^{ \minus{} 1}) \plus{} Tr(C(A \plus{} B)^{ \minus{} 1}) \equal{} \frac {1}{2}(Tr((A \plus{} B) \plus{} (B \plus{} C) \plus{} (C \plus{} A))((A \plus{} B)^{ \minus{} 1} \plus{} (B \plus{} C)^{ \minus{} 1} \plus{} (C \plus{} A)^{ \minus{} 1})) \minus{} 3Tr(I_n) \geq \frac {9n}{2} \minus{} 3n \equal{} \frac {3n}{2}$
You can read in blog of me .Thanks you are interested in  it !  :) I hope that you will check it ....

[quote]Prove that :
 $ Tr(\frac {A^m \plus{} B^m \plus{} C^m}{3}) \geq Tr(\frac {A \plus{} B \plus{} C}{3})^m$ ,$ \forall m \in \mathbb{N^{*}}$[/quote]
 Sorry It is correct with $ m\equal{}2$ and It is correct with $ \forall m \in \mathbb{N^{*}}$ If $ M\equal{}A\plus{}B\plus{}C$ such that $ MA\equal{}AM,MB\equal{}BM,MC\equal{}CM$
 With $ \forall m \geq 3$ I am not sure  :blush:


function
analytic geometry
graphing lines
slope
trigonometry
real analysis
real analysis unsolved

This problem is supposed to be like showing a kind of converse of the other one I posted is not true:

Show that a bounded continuous function on a bounded interval need not be uniformly continuous.

My solution idea:
Basically, i was going to construct a function with no limit at an endpoint.  I'm doing this problem from a book i'm following, and hardly any functions besides the exponential have been developed yet, so i want to try to stick to basic functions.  My only idea is to make a function $f$ that is composed of linear pieces with alternating slopes and increasing magnitudes over smaller and smaller intervals towards the endpoint.  Is there a much easier counter example? :P

You don't need to increase the magnitude of oscillation if you want the function to be bounded. Otherwise your idea is good. If you know trigonometric functions ($\sin$, $\cos$), use them to simplify the construction.

if the magnitude doesn't increase but the intervals for each linear piece get smaller towards the endpoint, woudn't it have a limit of 0 at the end point?

Maybe I wasn't clear:
I mean the magnitudes of the slopes are increasing, so that the function consistently reaches the same maximum on each of the linear intervals (which are shrinking)

OK, I misunderstood you. The magnitude of the slopes must grow indeed.


Let p be a prime number. Prove that there exists an integer a such that $p|(a^{2}-a+3)$ iff there exists an integer b such that $p|(b^{2}-b+25)$.


linear algebra
matrix
vector
calculus
integration
algorithm
number theory

What is the necessary and sufficient condition for two vectors in $Z^n$ in order to exist an integer matrix in $GL_n(Z)$ having the first two lines these vectors?

$v_1=(a_1,..,a_n)$ and $v_2=(b_1,..,b_n)$ these $2$ vectors.
If we just impose that $gcd(a_1,..,a_n)=gcd(b_1,..,b_n)=1$ and that $v_1$, $v_2$ are independant:
Consider the matrix $A$ with first line $v_1$, second line $v_2$ and whatever we want below for the moment.

We can find matrices $L_1$ and $R_1$ in $GL_n(Z)$ such that the first line of $L_1AR_1$ is $(1,0,..,0)$ because $gcd(a_1,..,a_n)=1$.
Because $v_1$, $v_2$ are independant it is sure that the second line of $L_1AR_1$ is different of $(a,0,..,0)$. From here it is clear that
we can find $L_2$ and $R_2$ in $GL_n(Z)$ such that the first $2$ lines of $L_2L_1AR_1R_2$ are $(1,0,..,0)$ and $(0,1,0,..,0)$.

Thus I think that $(L_1)^{-1}(L_2)^{-1}(R_2)^{-1}(R_1)^{-1}$ is a matrix of $GL_n(Z)$ with $v_1$ and $v_2$ as first $2$ lines.

alekk, I don't think these conditions suffice. Think about the vectors $(1,0,2),(1,2,0)$. Try to complete a $3\times 3$ matrix with integral entries having these two vectors as its first two rows and you'll see that the determinant is always even.

I can think of an obvious necessary and sufficient condition, but it's not pretty, and I'm sure this is not what the proposer had in mind :): $(a_1,\ldots,a_n)=1$, and if we perform some Euclidean algorithm-like operations on $(a_1,\ldots,a_n)$ in order to turn it into $(1,0,\ldots,0)$ (this sentence can be made precise :)), and perform the same operations on $(b_1,\ldots,b_n)$ getting $(b'_1,\ldots,b'_n)$, then $(b'_2,\ldots,b'_n)=1$.

I think the appropriate condition is that the two vectors are linearly independent mod $p$ for every prime $p$.

The possible determinants of an integer matrix with some fixed rows form an ideal; since this ideal clearly is not contained in $p\mathbb{Z}$ for any prime $p$, 1 is a possible determinant.

Here's another way to put it (which is a natural strengthening of the condition $(a_1,\ldots,a_n)=(b_1,\ldots,b_n)=1$):

For every $(p,q)=1$, we have $(pa_1+qb_1,\ldots,pa_n+qb_n)=1$.

thanx Grobber. Now, I understand where I made a mistake. Your condition with the euclidian algorithm is obviously true
and like you I'm sure it is not what the examinator was waiting for  :)

But I think what I wrote above is the correct condition, since that ugly thing with the Euclidean algorithm holds for such pairs of vectors.


Prove that for any $\delta$ greater than 1 and any positive number $\epsilon$, there is an $n$ such that $\left \vert \frac{\sigma (n)}{n} -\delta \right \vert < \epsilon$.

Proof is similar to the one of J8: http://www.mathlinks.ro/viewtopic.php?p=973615#973615
this time starting with the series
\[ \sum_{p\ \mbox{is prime}} \log(1 \plus{} \frac{1}{p}) \equal{} \infty.\]


A cube is created inside an octahedron such that each vertex of the cube is the center of an octahedron's face. Find the ratio of the volume between the octahedron and the cube.  ( I figured out how to do this problem but it took over 12 minutes so I was wondering if there was a more elegant way)

If you take two centres of adjacent octahedron faces, and drop perpendiculars to the common edge, I think it is clear that the side length of the cube is half the diagonal length of the octahedron. I imagine that this can then be used to get the answer slightly more quickly.


What is the 125th digit beyond the decimal point in the decimal representation of $ \frac47$?

$ \frac {4}{7} = 0.\overline{571428}$ repeating. Since $ 125 \equiv 5 \pmod 6$, our answer is $ \boxed{2}$. For a more in-depth explanation of modular arithmetic, see the [url=http://www.artofproblemsolving.com/Wiki/index.php/Modular_arithmetic]AoPS Wiki[/url] article.

For those who are familiar with modular arthimetic and want a little more practice, you could also find practice problems in Jason Batterson's book "Competition Math". (It's a great book.)
____________________
~pi man  :juggle:


geometry
geometric transformation
reflection
AMC
AIME
USA(J)MO
USAMO

This is in response to the recent wave of students who do really well on the AMC exams then complain about their performance.

Please stop and think for a minute before complaining about what are for most people excellent results.  Imagine you are a student who is just getting started and seeing someone complaining about a performance that is far, far better than what you are able to do.  How do you feel?  What do you think about the person who is complaining?

I'm guessing most of us don't like to see outright bragging, and we don't see much bragging on these boards.  However, most people read complaints about high scores as essentially equivalent to bragging (if not worse).

We'd like our community to be inclusive and attractive to beginning students.  The prevailing attitude we've seen recently of very strong students complaining about their excellent scores doesn't advance this goal (or, I think, reflect what the students complaining hope to reflect).

I understand these high scores may not be up to your standards, but complaining about doesn't help that!

I also realize that some of you may be tounge-in-cheek joking in your complaints, but it's not always obvious when that is the case.

Are we defining "high score" by the rest of the world's standards, or AoPS standards? This is not intended to be an insolent question; I know my scores are high by the rest of the world's standards but rather low by AoPS standards, and I don't want to mess up.

I mean here what you would call 'rest of the world' standards.

For your own happiness, of course, the most important standards are your own, which need not be anything near 'rest of world' standards.  I, for one, almost never meet my own standards.

But when discussing issues in a public forum like this, please understand that others don't understand (or agree with) the yardstick you measure yourself by!

Really great post by Mr. Ruscyk here. I've been thinking about trying to say something along the same lines but didn't know how to do it appropriately.

Basically, try to put yourself in others' shoes. I remember when I was in 8th grade and I took the AIME and I was talking to this person who got a 13, but missed problems 1 and 9. He couldn't stop complaining about his disappointment. And there I was thinking to myself about how I would have killed to get a 10 in any way shape or form.

Here at the forums, it's great when people post their scores and talk about how they're happy or maybe discuss how they're unhappy about their scores. But I think it's really got to be disheartening to some people to hear something like (loosely paraphrased) "I got a 7 on Mildorf's Mock AIME but I missed 3 to such STUPID MISTAKES oh my goodness I hate myself". How does that make the person with a 4 feel?

As for the comment "by AOPS standards", I understand what you mean, but let me digress and say that let's not get carried away here. This is by far the best online math community in the history of the world :). However, I don't see Mr. Feng posting here with great frequency (not as of yet at least). I don't see Reid, Tiankai, etc. There's a ton of elite students and teachers posting here, but let's keep in mind that not EVERY SINGLE BRILLIANT PERSON posts at AOPS. Just like how every one of us can continue to improve, this community can still get better. As this community grows (as it undoubtedly will, I personally have let just about every mathematically oriented person I know know about AOPS), perhaps one day we'll have the whole IMO team as frequent posters. Until then, spread the word about AoPS, and don't forget that even as a math community, we can always do better.

[quote="darktreb"]
Here at the forums, it's great when people post their scores and talk about how they're happy or maybe discuss how they're unhappy about their scores. But I think it's really got to be disheartening to some people to hear something like (loosely paraphrased) "I got a 7 on Mildorf's Mock AIME but I missed 3 to such STUPID MISTAKES oh my goodness I hate myself". How does that make the person with a 4 feel?
[/quote]

I believe that was me :blush: . Anyway I sincerely apologize for my attitudes toward my scores on practices and the real AIME in the last couple of weeks. I didn't mean to hurt anyone's feeling, it was just my own disappointments when I felt I could've done better. I will try to be more considerate in future.

i hope people don't get to offended by a little "bragging". i'm a little ashamed of my amc-10 score and by reading all of the posts, it makes me embarassed and silly, but it also makes me want to do better and strive to that high score and continuously raise the bar. i hope people who are offended see such high scores as a goal rather than something to discourage them.

Yeah no knocks against anybody in particular. I'm sure I've been guilty of it on several occassions too. The point that is important to be made is that it's all about improvement, both as an individual problem solver, and for the community as a whole.

As an old Baby Boomer who liked math as a kid, but who never heard about how to get into math contests until well after graduating from high school, I am happy for the young people here who have the chance to try out the contests at a young age. Yeah, not everyone can be the national champion, but I've tried a few of those contest problems from time to time (not often enough) to improve my coaching, and they are HARD, period. Don't be hard on yourselves when you are already feeding yourself hard problems.

Yes, AoPS is definitely raising the standard of math competition in the United States and around the world, and I think that is wonderful. The most valuable thing that AoPS does is give us all a chance to have some perspective: it's not the end of the world to get such-and-such a score on a math contest. When the math contests are fun, and when you enjoy challenging YOURSELF to a higher level, they can be very good for you. But ultimately you will go on to other challenges in life like how to find time to enjoy your spouse's company while still making a living, or how to get your kids to do their homework, or how to help your favorite candidate for public office win an election. You gain the most from competitive activities if they teach you how not to be crushed when you experience a momentary setback.

Good luck in your next round of competition.

"...but it also makes me want to do better and strive to that high score and continuously raise the bar. i hope people who are offended see such high scores as a goal rather than something to discourage them."

The thing to remember is that not everyone has your perspective, and that's okay; we'd like to include them in the mathematical community as well.  Maybe later, after they've been working on this stuff for longer, they'll see high scores as goals rather than intimidations.  But we want them to keep working on mathematics long enough to get there!

I used to have a similar attitude (though about a different aspect of the mathematical community than contests, 'cause I've never been a contest person) as you express above.  My feeling was basically that because I was comfortable with the way the mathematical community operates, then there was no problem for anyone else either.  But then someone said to me, "But smbelcas, not everyone should have to be like _you_ in order to become mathematicians."  That really struck me---I know I have a very strong personality, for example, and this often enables me to ask questions where others might be shy---and convinced me almost immediately that there are times when the mathematical community is less than welcoming to some people.  And... it's hard to know who those people are, because they're usually quiet about it.  It's a continuing struggle for me (over many many years!) to be aware of ways in which I might be accidentally making younger or less experienced math people get discouraged.  But it's a worthy struggle, and so I keep doing it---I want more people to become and remain excited about mathematics.  (Don't we all?)

I'd just like to say that I apologize for all such posts that I've made in the past IF I've made them since I really don't remember anything I've posted.  I totally understand your argument Mr. Rusczyk, and I would have been quite upset if I saw students complaing about their 10+ scores on the AIME when I was in 5th grade and scored a 1 on the AIME.  Whether I have posted such uncompassionate things in the past, which I'm pretty sure I haven't, or not, I will make sure not to post anything without thinking in the future.   :)

[quote="pascal012"]Hear, hear. Don't make people like me feel even more stupid.  :)[/quote]

exactly!

Though these contests are fun and challenging, and high scorers are to be commended for their math ability, low scorers should not automatically rule out a career in math.  Many contests emphasize speed and breadth.  For successful mathematicians,  who typically work on a single problem for weeks, months, years at a time, patience and depth are more important criteria.  Moreover the ability to create new problems is as important as the ability to solve them.

If you love math but can't seem to get the highest scores on these contests, don't give up.  Passion, diligence, stubbornness, and a good math education are the necessary traits for a productive career in math.

Well said wild octagon. I seem to fall in that category. Althought I got a pitful 4 on the AIME this year, I will work for 10+ next year.... it just needs a little desire, and hard work.

"All you need is ganas."

Ok, It might be degrading to hear about someone complaining about a high score, but would not that just be motivation to be at a level where you score as high? I think if you are really interested in math, and want to solve problems, hearing a few people complain about 'lofty' scores and 'ridiculous easy AIMEs' should not discourage you.

Bah, I guess I just hate feeling stupid.

I know someone who considered getting a 144 on the AMC 12, pwnage...but by the test.

=/...

[quote="eryaman"][quote="JesusFreak197"]probability, what year were you in Nationals?[/quote]He's a freshman now, so that would be 2004 Nationals.[/quote]

And 2003 Nationals as a 7th grader.

[quote="darktreb"]Continuing on this train of thought, I've always had this personal theory that if you were looking for the most humble people in the world, you should go to those who are aware of the fact (loosely used word here) that they are the second most talented/smart/able/whatever person in any specific field/area/whatever. For this person must have spent innumerable hours dedicating themself to their craft and honing their skills. This person must have put in effort unimaginable to the common person. And yet, this person knows that he/she, for all that effort, dedication, torture, is still not the number one most talented/smart/able/whatever person in that field/area/whatever. This person understands what he/she has been through, and that even with all that, there still exists someone better. Now that must truly be the most incredibly humbling feeling in the world. We can only imagine.[/quote]

That's a wonderful thought, and I would totally agree with your theory. I think that situation also induces 100% genuine humility, which is great as well. darktreb, I'm definitely going to be thinking more about that idea.

One belief of mine is that complaints of really high scores may come from the wrong mindset that one has of test-taking itself. I don't mean in terms of ability, I mean in terms of how one approaches the test itself and how positively one thinks. I got a 5 on the AIME this year, not even close to qualifying for USAMO, but I was still very proud of myself, as it was my first time taking the AIME, and it was a cool experience to work with math for 3 hours straight.

Really, a job well done is a job well done. I have seen people kill themselves over scores from careless errors, and watched them wonder what their scores could've been, but I think even [u]that[/u] is a part of the learning experience. If you can look at silly mistakes as a part of learning, then I believe that will also help in the long run. I understand that some might disagree because they, for example, added two numbers wrong. Lesson learned there: Add more carefully!
I know my thoughts are a bit jumbled, but all in all: Anything can be viewed as a learning experience, and that is what makes learning truly beautiful.

I am always disappointed to see that I have scored 1 place, 1 point, whatever below some huge milemarker. For instance, I recently realized, to my great surprise, that I have only won two competitions (regional mathcounts)! However, I have plenty of second place trophies! I don't care though. I would really like to make the IMO team not because I want to prove myself or want to win some scholarship, but because I want to take the IMO under real circumstances. The reward of taking a math test is seeing new and interesting problems and being able to take another math test! I have found, and it must be for this reason, that I am less apprehensive about national contests than state contests (though I'm not so about regional since our region isn't particularly competitive :) ).

Along with this thinking I was very disappointed to make an 8 on the AIME this year, not because it was a bad score (in fact I tied for second among middle schoolers on the AIME2), but just because it was one point below qualifying for the USAMO, not because it was a bad score. I think that under this sort of thinking, it is healthy and inspiring to be disappointed about your score, but if you post about it, say "I got an 8, which was disappointing considering I wanted to make the USAMO" instead of "Oh no! I made an 8! I can't believe my stupidity!" In fact, I believe the titles you give yourself carry over to your listeners.

Finally, darktreb and ALMN568, if I were this person, I would not be humiliated or humbled at all! Being the best and being among the best are just as good. In fact, I would prefer to be categorized as among the best. I would have people to talk to who understood me, and if I wanted fame and fortune, I would still have it in plenty! It must be a boring life indeed to have nobody to talk to about your achievements, no incentive to get better, no improvements to even make!

Yes, I know exactly how it feels to be one point under. For example, this year I got an 8 on the AIME, and got 14th place in mathcounts, tying for 12th. So yes, when I say I am upset I got 14th I am upset that I was so close, not that I did badly.

[quote]Finally, darktreb and ALMN568, if I were this person, I would not be humiliated or humbled at all! Being the best and being among the best are just as good. In fact, I would prefer to be categorized as among the best. I would have people to talk to who understood me, and if I wanted fame and fortune, I would still have it in plenty! It must be a boring life indeed to have nobody to talk to about your achievements, no incentive to get better, no improvements to even make![/quote]

Get there and we'll see.

And there are always improvements to be made, even if you are the best.

Because you can always add one to the highest number. . . .

In most any endeavor, we can always say there is someone better.  We could say Leonhard Euler was the best mathematician because he produced so much.  Or, we could say Riemann or Gauss or Ramanujan for different reasons.  They were all very brilliant.  And, each contributed in a different way.  I don't know if we can very well compare their aptitudes either, because they are mostly unprecedented.

This is not only true for math, but for sports, music, art, ... even rubiks cubing.  Yes, I know, Macky seems to dominate, but one could still argue.  There are others that can do different things with Rubiks cubes than Macky can.

Thus, I do not believe this second place person exists.  If you were to be second place, you would probably be classified as the best anyway.

However, there can be a best in a certain competition, at a certain time, and that is what gets annoying. But then, what would the point of competition be?

[quote]Thus, I do not believe this second place person exists. If you were to be second place, you would probably be classified as the best anyway.[/quote]

I think you guys are taking the statement too literally.

I know I'm late, but that was a great post! I took the AIME for a first time this year as a Junior in high school and I got a 5. I was feeling ok about it, then I saw an "ignorant" freshman "fresh out of my Mathcounts days" complaining about his 6. Reeaaalllyyy annoying.

[quote="sapphyre571"]Well I was furiated at my 6 this year as an 8th grader for three reasons: 3 problems that I missed by three or less (I think It was Problems 4, 11, and 15). I mean I still had tied for second among Texas eighth graders on AIME I, but a three fewer stupid mistakes and that means USAMO in eighth grade, which I would [i]not[/i] be dissapointed in...
 :oops_sign:[/quote]

*Slaps forehead* 6 as an 8th grader is still very good, so don't complain.

And to answer your question, because they're different contests  :P

WHY IN THE WORLD DO YOU GUYS WANT TO REVIVE THE OLD THREAD!

Okay, I'm not angry but seriously, the important rule for this forum is that you do not revive an old thread unless it is very very important such as you want to ask question to solution (hence, do not post if you're posting alternate solution). And clearly, posting your thoughts for this type of thing is totally ridiculous and not allowed.

LOCKED


geometry
rectangle
quadratics
special factorizations

Joe is trying to make a rectangular garden.  He plans to use the wall of his house as one side (pretend that the wall can be any measure it just has to correspond with its opposite side of the garden).  He wants to have the biggest garden possible but only has 60 meters of fencing to use.

a)  Find the dimensions of the biggest garden possible.
b)  Prove that this is the biggest garden possible.
c)  Generalize this proof.

[hide="Hint"]
We want to maximize $ ab$ while $ 2a\plus{}b\equal{}60$.
[/hide]

[hide="Solution"]$ 2a \plus{} b \equal{} 60$
$ 2a \plus{} b\geq 2\sqrt {2ab}$
$ 60\geq 2\sqrt {2ab}$
$ 450\geq ab$

a. So the maximum $ ab$ is $ 450$.
b. I just proved it with AM-GM.
c. If $ 2a \plus{} b \equal{} n$, then we'd have the max $ ab$ as $ \frac {n^2}{8}$.[/hide]

Remember that you have to show that it's attainable!
[hide="Other Part of the Solution"]
According to the equality condition for AM-GM, we have $ 2a \equal{} b$. Plugging that into the original equation, we have

$ 2b \equal{} 60 \implies b \equal{} 30$

Subbing $ b$ back in to get $ a$, we have

$ 2a \plus{} 30 \equal{} 60 \implies a \equal{} 15$

$ ab \equal{} (15)(3) \equal{} 450$. Thus, the area of $ 450$ units squared is attainable.
[/hide]

Or, you can use quadratic optimization:
$ 2a\plus{}b\equal{}60 \longmapsto b\equal{}60\minus{}2a$.  We are trying to find the largest possible value of $ ab\equal{}60a\minus{}2a^{2}$.  Let the area of the pen be $ A$.  Therefore, we have that:
$ A\equal{}60a\minus{}2a^{2}\equal{}\equal{}>\frac{\minus{}A}{2}\equal{}a^{2}\minus{}30a$.  Completing the square by adding $ (\frac{30}{2})^{2}$ to both sides and then simplifying, we have that $ A\equal{}\minus{}2(a\minus{}15)^{2}\plus{}2(15^{2})$.
The greatest possible value of this area is when $ a\equal{}0$, because when it's not the area gets continually smaller because of the $ \minus{}2$.  Therefore, the answer is $ 2 \bullet 15^{2}\equal{}\boxed {450}$.


geometry
3D geometry
sphere
ARML
Asymptote
Pythagorean Theorem

Let T = TNYWR.  The difference in surface areas of two concentric spheres is T.  A circular disk is tangent to the inner sphere and its circumference is on the other sphere.  Compute the [u]diameter[/u] of the disk.

I'm having trouble figuring out what this diagram looks like.  I know that the concentric spheres look like:

[img]http://chartersart.com/Beale/images/3DSPHRS.GIF[/img]

But what is the "disk" and where does it go?

Thanks in advance!

[size=75]EDIT: Sorry about this over-sized image.  I couldn't find any better ones.[/size]

Consider a projection of the figure onto the plane; then it's just two concentric circles. Then the analogous figure to the disk would be a tangent line to the smaller circle which is a chord of the larger circle. In three dimensions, it's just a circular cross-section of the larger sphere which touches the smaller sphere at a point.

(That also told you how to solve the problem.)

[hide="my solution"]
Let $ R$ be the radius of the outer sphere, $ r$ the radius of the inner sphere, and $ x$ the radius of the disk.

Then, by pythagorean theorem, we get:

$ r^2 \plus{}x^2 \equal{} R^2$

So, $ x^2 \equal{} R^2 \minus{} r^2$ and $ x \equal{} \sqrt{R^2 \minus{} r^2}$.

The diameter of the disk therefore is $ 2\sqrt{R^2 \minus{} r^2}$.

The formula for surface area of a sphere is $ 4\pi r^2$. Therefore, $ T \equal{} 4\pi R^2 \minus{} 4\pi r^2 \equal{} 4\pi (R^2 \minus{} r^2)$. Solving this for $ R^2 \minus{} r^2$, we get $ R^2 \minus{} r^2 \equal{} \frac{T}{4\pi}$.

Substituting that expression into the expression for the diameter of the disk, we get:

$ 2\sqrt{\frac{T}{4\pi}} \equal{} \sqrt{\frac{T}{\pi}}$.[/hide]

[b]Interpretation:[/b] Take an arbitrary point on the inner sphere. Draw the tangent plane at that point. Take the circular section of this plane that lies inside the larger sphere.

This is obviously from an ARML relay round, or something similar.

[hide="Relay-Appropriate Solution"]The problem statement (correctly) implies that only the difference in surface areas is relevant, and the actual radii are not.

Therefore consider the case in which the smaller sphere is degenerate, and the larger sphere has surface area T.

The circular disk we want is equatorial cross-section of the larger sphere.

The computation is now trivial.

Pass your answer, and then check it more rigorously (see previous post).[/hide]

[b]Note to the OP[/b] -- [i]next time, please do one of the following:
1. Make a smaller image
2. Add the image as an attachment instead of using remote hosting
3. Draw the diagram in Asymptote
This way, the page width does not get extended, and readers won't have to scroll horizontally to read subsequent posts. Thanks![/i]


Θα παραθέσω μια θεωρία που λύνει τα χέρια σε πολλές ασκήσεις άλγεβρας: αυτή της ομογενοποίησης.

Η θεωρία λέει ότι, αν πολλαπλασιάζοντας με τυχαίο μη μηδενικό αριθμό λ όλες τις μεταβλητές μιας ισότητας ή ανισότητας, αυτή παραμένη αναλοίωτη,τότε μπορεί να επιλεγεί συγκεκριμένη τιμή για οποιαδήποτε παράσταση που είναι ομογενές πολυώνυμο των μεταβλητών. (Η τιμή αυτή πρέπει να είναι θετική αν το λ είναι μόνο θετικός και αρνητική όταν είναι μόνο αρνητικός.)

Θα δώσω ένα παράδειγμα:
H ανισότητα $ \frac{a}{b}\plus{}\frac{a}{b} >\equal{} 2$ για θετικούς, μπορεί να αποδειχτεί και έτσι.
Έστω $ k>0$, θέτουμε όπου a to ka και όπου b το kb. Τότε το πρώτο μέλλος παραμένει το ίδιο(τα k απλοποιούνται). Άρα η ανισότητα είναι ομογενής, άρα μπορούμε να θέσουμε (θεωρήσουμε σαν δεδομένο) ότι a+b=1 ή ab=1, ή ακόμα και (a^2+ab+b^2 = 2007). Η ανισότητα που προκύπτει με αντικατάσταση τώρα έχει λιγότερους αγνώστους και δουλεύεται ποιο εύκολα.

Η θεωρία αυτή είναι πολύ χρήσιμη σε ιδιαίτεα δύσκολα προβλήματα.

Πολύ ωραία !

 ΑΥτό το διάβασες κάπου ή το ανακάλυψες στην προσπάθεια να λύσεις κάποιο πρόβλημα ;

 Μπάμπης

Βασικά, το είχα αποδείξει σε μια άσκηση και το έψαξα επειδή μου έκανε εντύπωση. Υπάρχει στο βιβλίο των ανισοτήτων των εκδόσεων Σαββάλας.

Σε περίπτωση που έχουμε $ f(a, b ,c ) >= g(a, b, c) (Ι), f(ka, kb, kc) = f(a, b, c), g(ka, kb, kc) = g(a, b, c)$ για κάθε $ k$, μπορούμε να θέσουμε:
$ d = ia$ και
$ e = ib$ και
$ f = ic$ και
$ d + e + f = 1$,
που, αφού είναι σύστημα 4 εξισώσεων με τέσερις αγνώστους (oi περιπτώσεις ίσες με μηδέν βγαίνουν), υπάρχουν πάντα τα $ d, e, f, i$
άρα η $ (I)$ μετασχηματίζεται σε:
$ f(d, e, f ) >= g(d ,e ,f )$ με $ d + e + f = 1$

Με τον ίδιο τρόπο αποδεικνύεται οποιαδήποτε συνθήκη θεωρηθεί ως δεδομένη.

Αυτό που ξέχασα να αναφέρω στη θεωρία και το βλέπω τώρα είναι ότι μπορούμε να θέσουμε αντί αυτών, μια συγκεκριμένη τιμή σε μία μεταβλητή. (Στο αρχικό παράδειγμα θα μπορούσαμε να θέσουμε b=1.)

Όλα αυτά ισχύουν μόνο στην περίπτωση που η άσκηση δεν μας έχει δώσει και άλλο δεδομένο που να είναι ισότητα (όπως $ a \plus{} b \plus{} c \equal{} 0$. Αν είναι ανισότητα, όπως $ a \plus{} b \plus{} c > \equal{} 0$, εφαρμόζεται, αρκεί να μην έρχεται σε αντίθεση με αυτή. Για παράδειγμα αν $ a \plus{} b \plus{} c > \equal{} 0$ και είναι ομογενής η ανισότητα, τότε μπορούμε να θέσουμε $ a \plus{} b \plus{} c \equal{} S > 0$). Σε αυτήν την περίπτωση το σύστημα είναι 5 εξισώσεων με 4 αγνώστους που δεν έχει πάντα λύση.


function
integration
probability and stats

Let $ X_1^0, X_2^0,...,X_n^0$ denote the i.i.d failure times of $ n$ items that are censored on the right by i.i.d. random variables $ U_1,U_2,...,U_n$, which are independent of the $ X_i^0$. Denote the common distribution function of $ X_i^0$ by $ F^0$ and the distribution of the $ U_i$ by $ H$. Assume that $ F^0$ is absolutely continuous with density $ f^0$ and $ H$ is continuous.
The observed data are $ X_i, i \equal{} 1,2,...n$ where $ X_i \equal{} {\rm min}\{X_i^0, U_i\}$. The distribution of the $ X_i$'s, $ F(x)$, satisfies $ 1\minus{}F(x) \equal{} (1\minus{}F^0(x))(1\minus{}H(x)).$
I believe that one can transform the censored sample $ X_i, i\equal{}1,\dots,n$ to the complete exponential sample $ Y_1,\dots, Y_n$ by applying the transformation $ \Lambda^0(x) \equal{} \int_0^x \lambda^0(u)\,du$ where $ \lambda^0(x) \equal{} f^0(x)/[1 \minus{} F^0(x)]$ (the hazard rate function).
[b]The proof is:[/b] for any $ i \equal{} 1, \dots,n$ and for $ y\geq 0$
$ P\{Y_i\leq y\} \equal{} P\{\Lambda^0(X_i) \leq y\} \\
\phantom{P\{Y_i\leq y\}} \equal{} P\{ \minus{} \log(1 \minus{} F^0(X_i))\leq y\} \\
\phantom{P\{Y_i\leq y\}} \equal{} P\{\log(1 \minus{} F^0(X_i))\geq \minus{} y\} \\
\phantom{P\{Y_i\leq y\}} \equal{} P\{1 \minus{} F^0(X_i) \geq e^{ \minus{} y}\} \\
\phantom{P\{Y_i\leq y\}} \equal{} P\{F^0(X_i) \leq 1 \minus{} e^{ \minus{} y}\} \\
\phantom{P\{Y_i\leq y\}} \equal{} 1 \minus{} e^{ \minus{} y},$
where in the second step above the well known relation $ \Lambda^0(X_i) \equal{} \minus{} \log(1 \minus{} F^0(X_i))$ is used. Also, in the last step the fact that $ Z \equal{} F^0(X)$ has uniform $ U[0,1]$ distribution is used. Of course, for $ y < 0, F^0(y) \equal{} 0$ and therefore $ \Lambda^0(y) \equal{} 0$.

I believe that this proof is right but I'm not sure whether I am missing something or not, so naturally the question is if anybody sees anything wrong (or not).

Thanks in advance

Erm... In your computation censoring has completely disappeared. Isn't that suspicious?

Sorry for the confusing notation. I edited the message and I hope that things are now clearer.
Thanks for pointing this out fedja.

OK, let me be more specific.
[quote]
$ P\{F^0(X_i) \leq 1 \minus{} e^{ \minus{} y}\} \equal{} 1 \minus{} e^{ \minus{} y}$
[/quote]
Where does this equality come from (it is true with $ X$ replaced by $ X^0$ for an obvious reason but I do not see why it should hold after censoring).

What I meant is that neither your transform, nor the outcome of it involves any information about censoring. If you replace $ F^0$ by $ F$ in your transform definition and your computation, everything will become correct.

Ok, I can see the mistake now. In fact $ Z\equal{}F^0(X)$ is not uniform when $ X$ corresponds to censoring (i.e. when $ U_i \leq X_i$).  In other words such a transformation would make exponential only part of the sample, the part that corresponds to $ F^0$.


At breakfast, lunch, and dinner, Joe randomly chooses with equal probabilities either an apple, an orange, or a banana to eat. On a given day, what is the probability that Joe will eat at least two different kinds of fruit?

The probability that Joe eats only one kind of fruit is $ \frac{3}{3}\cdot\frac{1}{3}\cdot\frac{1}{3}\equal{}\frac{1}{9}$. So the answer is $ 1\minus{}\frac{1}{9}\equal{}\frac{8}{9}$.

[quote=ianrod]I think there is an easier way to solve this problem. What if we do complementary counting? For an example, we can say that Joe eats only one kind of fruit the whole time. So there are 3 ways he can do that. Then, we figure total possibilities, being $3^3$ choices. So then we have: $27 - 3$ ways that he can eat two or more fruit, so then our answer being $24/27$ = $8/9$.[/quote]

uh @2above did complementary counting

Wouldn't this be 7/9 as there are $6!$ ways to order the fruits?


Find all pairs $ (x,y)$ of real numbers such that: $ y^2\minus{}[x]^2\equal{}19.99$ and $ x^2\plus{}[y]^2\equal{}1999.$

Let $ n\equal{}[y]$. If $ |n|\le 31$, then $ x^2\ge 1036\to |x|>32$  give contradition with $ y^2\equal{}[x]^2\plus{}19,99$.
If $ |n|\ge 33$, then $ x^2\le 910$ give contradition with $ y^2\equal{}[x]^2\plus{}19.99$,
Therefore $ n\equal{}\pm 32,x\equal{}\pm \sqrt{1075}$. If $ x<0$, then $ y^2\equal{}1089\plus{}19,99$  contradition,
If $ x>0$, then $ x\equal{}\sqrt{1075}, y\equal{}\sqrt{1043,99}$.

[quote="Rust"]Let $ n \equal{} [y]$. If $ |n|\le 31$, then $ x^2\ge 1036\to |x| > 32$  give contradition with $ y^2 \equal{} [x]^2 \plus{} 19,99$.
If $ |n|\ge 33$, then $ x^2\le 910$ give contradition with $ y^2 \equal{} [x]^2 \plus{} 19.99$,
Therefore $ n \equal{} \pm 32,x \equal{} \pm \sqrt {1075}$. If $ x < 0$, then $ y^2 \equal{} 1089 \plus{} 19,99$  contradition,
If $ x > 0$, then $ x \equal{} \sqrt {1075}, y \equal{} \sqrt {1043,99}$.[/quote]

Hemmm , i've not checked the demo, just the result (BTW, you should do it more often, Rust) and the result appears to be wrong :

$ x^2\plus{}[y]^2\equal{}(\sqrt {1075})^2\plus{}[\sqrt {1043,99}]^2\equal{}1075\plus{}32^2\equal{}2099\ne 1999$

I am sorry for my mistake. Because $ [y]\equal{}\pm 32$ we get $ x^2\equal{}1999\minus{}32^2\equal{}975 \not \equal{}1075$.
If $ x>0$, then $ y^2\equal{}961\plus{}19,99, y<0\to x\equal{}\sqrt{975}, y\equal{}\minus{}\sqrt{980,99}$.
If $ x<0$, then $ y>0$ and $ x\equal{}\minus{}\sqrt{975},y\equal{}\sqrt{1043,99}.$


John received a cake with diameter one foot for his birthday. He wants to cut it into as many pieces as possible using three straight cuts. What is the greatest number of pieces that can be made?

[hide]can he stack the cake?
like he cuts it in 2 peices, can he put the two peices on top of each other before cutting again?
and so on?
If so I think it's 8[/hide]

[hide]The answer's 8. First he cuts the cake horizontally, then he makes two veritcal cuts.[/hide]

Really? Can you show how 8 pieces can be made? (And the cake cannot be stacked, the cake remains on the board and cannot be moved. When was the last time you successfully picked up a half-cake?!)

8 can not be made. By ladyknights way you get 6, not 8.

7 can be made. Make a vertical and horizontal cut, then a diagonal cut that passes through the first 2 cuts at diff points.

No... assuming the cake is tall enough to cut horizontally, you do that and then make two perpendicular cuts across diameters vertically. That gives eight pieces.

Oh come on, cutting it through the middle???
Is that even possible?
I mean, how the heck [i]do[/i] you cut a cake through the middle middle???
It won't even freaking reach.

By my way you get 8, and the problem doesn't say how long the knife is.

oops..................Treething never [i]did[/i] say how tall the cake was, nor where it could be cut... :oops:

[quote="Treething"]ok, it's two feet tall.[/quote]]

Two feet tall?  :o

[quote="Klebian"]can it be cut through the middle?
(Like in ladyknights drawing)[/quote]

Maybe he didn't specify for a reason.  :lol: so basically it's 8 if yes, 6 if no..

Ok... Who would want to eat the bottom (messed up) part of the cake? ;)

I think it would be 7...

[quote="ch1n353ch3s54a1l"]6 if no..[/quote]

7 if no.  Two cuts make 4, and then the third but makes 7 total.

Let's just agree on this:

3-d cake (which this problem is): [hide] 8 [/hide]

2-d cake (which this problem isn't): [hide] 7 [/hide]

Cutting the cake through the middle:
Pros:
1 extra piece of cake
Cons:
4 pieces of cake (on the bottom) that nobody wants to eat
A big mess

Which would you want? :)

con:
necessity of a freaking huge knife.

[quote="Klebian"]necessity of a freaking huge knife.[/quote]

I think it's actually called a sword.

In the Math world, nobody cares about what each piece tastes like.
Let's assume that every piece has the same features, but different  shape and size

And also, no, you wouldn't need a word for cutting. Maybe, 5 inch blade is all. Cut a radius, circle around the center. Tada!

Even better, although not as aesthetically or tastingly pleasing:
Cut from top, diagonal to other side, bottom.
Do that a couple of times, many pieces  :winner_first:

Aryth, check your pm box :)

Also, this topic has gone from math to uhh.. cake-ometry?

[hide="Maximum number of pieces with n cuts in two dimensions"]
If there are $k_n$ pieces after the $n^{th}$ cut, then the $(n+1)^{th}$ cut can pass through $n$ other cuts to form, maximally, $n+1$ new pieces.  There are $k_0 = 1$ pieces after the $0^{th}$ cut.  Hence the maximum number of pieces is $\frac{ n(n+1) }{2} + 1 = {n \choose 2} + {n \choose 1} + {n \choose 0}$.  This evaluates to $\boxed{7}$ when $n = 3$.
[/hide]

[hide="Maximum number of pieces with n cuts in three dimensions"]
The argument is too annoyingly long to give here, but by extension the formula is ${n \choose 3} + {n \choose 2} + {n \choose 1} + {n \choose 0}$ where $m < n, {m \choose n} = 0$.  This evaluates to $\boxed{8}$ when $n = 3$.
[/hide]

[hide="Maximum number of pieces with n cuts in m dimensions"]
By extension:  $\sum_{k=0}^{m} {n \choose k}$, where, again, $m < n, {m \choose n} = 0$.
[/hide]

Wow, t0ra, that's really usefull -- thx.

[quote="robinhe"]$m$ dimensions?

What does the $\sum$ thing mean?[/quote]

Sure - for example, you could consider the number of pieces in which you could slice a 4-dimensional cake, which is $m = 4$.  :)

$\sum$ is the sigma, or summation, symbol.  It basically works like this:  the part on the bottom, usually something like $k = 0$, sets the 'dummy variable,' which changes from $k = 0$ to $k = 1, 2, 3, ...$ all the way up to the part on the top, which is the upper bound.  For example, $\sum_{k=0}^{3} k^2 = 0^2 + 1^2 + 2^2 + 3^2$ and $\sum_{i=13}^{16} (i^3) = 13^3 + 14^3 + 15^3 + 16^3$.

Hence $\sum_{k=0}^{m} {n \choose 0} = {n \choose 0} + {n \choose 1} + {n \choose 2} + {n \choose 3} + ... + {n \choose m-2} + {n \choose m-1} + {n \choose m}$

Notice another interesting property of this expression, which is that if $m = n$ the expression is equivalent to $2^n$, which makes sense - if you had a new dimension for every new cut you made, all you have to do is cut all the pieces you have in half along the new dimension.  :)

(Oh, also, a partial explanation for the ${n \choose k}$ expression:  It's kind of hard to explain, but let's try in two dimensions.  In two dimensions, first, put all the lines on a coordinate grid and look at the pieces that have a "lowest point," that is, if the cake wasn't there, they would still have a lower bound.  (The pieces that are bounded by the lower part of the cake don't have a "lowest point.")  Every point that is made by an intersection of the cuts is the "lowest point" of exactly one piece, and because two lines determine a point, there are ${n \choose 2}$ points, so ${n \choose 2}$ pieces with lowest points.  How many pieces don't have lowest points?  Consider all the pieces without lowest points that are bounded to the left (we are assuming none of the cuts are parallel to the $x$-axis).  Every cut that is made is the left-bound of one piece without a lowest point, so there are ${n \choose 1}$ of those, and then there is one piece that doesn't have a left-bound, which is the "leftmost piece," or ${n \choose 0}$, so the total number of pieces is ${n \choose 2} + {n \choose 1} + {n \choose 0}$ as stated.  The argument is extensible to higher dimensions.)

[quote="t0rajir0u"][quote="robinhe"]$m$ dimensions?

What does the $\sum$ thing mean?[/quote]

Sure - for example, you could consider the number of pieces in which you could slice a 4-dimensional cake, which is $m = 4$.  :)[/quote]

I just asked about $m$ if $m>3$.  Imagine a cake in a 5 dimension world.  :D

But thanks for your explanation about $\sum$, even though that doesn't belong here.

I know, you're wondering why I asked the question in the first place.

[hide]I got 7.  :) [/hide]

That was for 2 dimensions


Rigorously find the sum of the factors of n. :evilgrin:

[hide="This is hardly complete, much less rigorous"]

$\sigma (n)$, the sum of factors of $n$, is a [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=classic&t=117784]multiplicative function[/url].
(i.e., if $\gcd(n,m)=1$, $\sigma(mn)=\sigma(n)\sigma(m)$)
Hence if there are no repeated prime factors, and $p_{1}, p_{2}...,p_{k}$ are the prime factors of $n$, then
$\sigma (n)= \sigma(p_{1}) \sigma(p_{2})...\sigma(p_{k})=(p_{1}+1)(p_{2}+1)....(p_{k}+1)$

If there are repeated prime factors, then things get a bit more tricky. I'd be surprised if there was any simple formula in the general case. It wouldn't be too hard to do manually for each specific case though.[/hide]

Aren't repeated prime factors about the same thing?

$\sigma(n)=\frac{(p_{1}^{a_{1}+1}-1)(p_{2}^{a_{2}+1}-1)\cdots(p_{n}^{a_{n}+1}-1)}{(p_{1}-1)(p_{2}-1)\cdots(p_{n}-1)}$

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=670902[/url]

I'm not sure if that's what you were meaning.

here's what I meant:

If

$x=b_{1}^{e_{1}}\times b_{2}^{e_{2}}\times \cdots \times b_{n}^{e_{n}}$

Then prove that the sum of the factors of n is equal to

$\left(b_{1}^{e_{1}}+b_{1}^{e_{1}-1}+\cdots+b_{1}+1\right)\left(b_{2}^{e_{2}}+b_{2}^{e_{2}-1}+\cdots+b_{2}+1\right)\cdots\left(b_{n}^{e_{n}}+b_{n}^{e_{n}-1}+\cdots+b_{n}+1\right)$

for all b_n, e_n, and x.

Well, I'm not really good at proofs, but let me see.

In my previous equation $(b_{n}^{e_{n}}+b_{n}^{e_{n}-1}+\cdots+b_{n}+1)=\frac{b_{n}^{e_{n}+1}}{b_{n}-1}$, because just multiple the LHS by $b_{n}-1$ and you will see that it telescopes to the right hand side.

Now basically, when you are multiplying those out, you are multiplying each single factor prime by every combination with every other single prime factor, hence you have obtained all of the total factors possible, and thus the equation sums them.

Probably doesn't sound very rigourous though...

I feel so stupid now. :fool: That was one of the shortest proofs that I've ever seen!

[hide="A completion of tims proof"] Let $\sigma(n)$ denote the sum of the divisors of $n$.

[b]Lemma:[/b] $gcd(a, b) = 1 \implies \sigma(mn) = \sigma(m) \sigma(n)$

Proof:  Let the factors of $m$ be $a_{1}, ... a_{k}$ and let the factors of $n$ be $b_{1}, ... b_{l}$.  Then every factor of $mn$ must be of the form $a_{i}b_{j}$.  Summing,

$\sigma(mn) = \sum_{i \le k, j \le l}a_{i}b_{j}= \left( \sum_{i=1}^{k}a_{i}\right) \left( \sum_{j=1}^{l}b_{j}\right) = \sigma(m) \sigma(n)$

As desired.

[b]Lemma:[/b]  Let $p$ be a prime.  Then $\sigma(p^{k}) = 1+p+p^{2}+...+p^{k}$

Proof:  Obviously those are the only divisors of $p^{k}$.

Combining the two lemmas provides the common formulation of $\sigma(n)$. [/hide]


I solved a problem and noticed the rating for difficulty bar at the bottom.  How is that intended to work?  The poor, fair, etc system that shows when you mouse over it doesn't seem to tell me how I should rate it.

In other words, what is a "norm" for rating difficulty?  Is a poor supposed to be something like "What is 3+5?"  What is meant by "perfect" for difficulty?

Thanks for any help!

My System:

Very Good-Perfect: A Problem Which I Did Not Get (depending on how it was solved)
Fair-Good: A Problem Which Was Somewhat Challenging (but I know how to do it)
Very Poor-Poor: Something like $ \binom{10}{2}$.

Yes but if they gave us a problem like finding all solutions to 10th degree polynomials, then isnt that so hard that it becomes poor?

basically, ernie is saying that he gives hard problems a high rating, medium problems and medium rating, and very easy problems a low rating

considering your example, mewto, i would say that would probably contridict this system
maybe leave a comment on it in the forums or something?

anyway, i doubt alcumus would have those types of problems

Considering the fact that Alcumus is currently revolved around Intro to C & P...

The rating system makes much more sense now.


I think there may have the condtion that $a,b,c,d >0$
my solution
let reference frame in the plane.
let $O=(0,0),A=(a,b),B=(a+c,b+d)$
with $S=\frac 1 2 |ac-bd|$
let $C$ be the perpendicular  from A to OB
and the inequality is equal to $OB+2AD \ge OA+AB$
by:$OD+ AD \ge OA$,$BD+AD \ge AB$
the inequlity is proved(and you should show that $\angle AOB <\frac{\pi}{2}$ and $\angle ABO <\frac{\pi}{2}$

or you can prove with the following way directly:
$\sqrt{\frac{(a^2+b^2)((a+c)^2+(b+d)^2)-(ad-bc)^2}{(a+c)^2+(b+d)^2}}+\frac{|ad-bc|}{\sqrt{(a+c)^2 + (b+d)^2}} \ge \sqrt{a^2+b^2}$
and
$\sqrt{\frac{(c^2+d^2)((a+c)^2+(b+d)^2)-(ad-bc)^2}{(a+c)^2+(b+d)^2}}+\frac{|ad-bc|}{\sqrt{(a+c)^2 + (b+d)^2}} \ge \sqrt{c^2+d^2}$
and
$\sqrt{\frac{(a^2+b^2)((a+c)^2+(b+d)^2)-(ad-bc)^2}{(a+c)^2+(b+d)^2}}+\sqrt{\frac{(c^2+d^2)((a+c)^2+(b+d)^2)-(ad-bc)^2}{(a+c)^2+(b+d)^2}}=\sqrt{(a+c)^2 + (b+d)^2}$
 :)

[quote="zhaobin"]I think there may have the condtion that $a,b,c,d >0$
[/quote]
i'm sorry ( nice solution )


Express $ -1+\sqrt{3}i$ in mod-arg form. Use De Moivre's theorem to show that
$ (-1+\sqrt{3})^{n}$ + $ (-1-1\sqrt{3}i)^{n}$ = $ 2^{n+1}(cos{\frac{2n\pi}{3})}$, n is a positve integer

the books soln.
[hide][url=http://img245.imageshack.us/my.php?image=95538261kk4.png][img]http://img245.imageshack.us/img245/6615/95538261kk4.th.png[/img][/url][/hide]

How did they obtain this step

$ 2Re(z^{n})=2^{n+1}cos(\frac{2n\pi}{3})$

Before answering your question, please edit your text, bos1234.

Let $ z=-1+\sqrt{3}i=2\left(\cos \frac{2\pi}{3}+i\sin \frac{2\pi}{3}\right)$, by de Moivre's thorem,

$ z^{n}=2^{n}\left(\cos \frac{2n\pi}{3}+i\sin \frac{2n\pi}{3}\right)$.

$ \therefore (-1+\sqrt{3}i)^{n}+(-1-\sqrt{3}i)^{n}=z^{n}+({\overline{z}}^{n})=z^{n}+\overline{(z^{n})}$

$ =2Re(z^{n})=2\cdot 2^{n}\cos \frac{2n\pi}{3}=2^{n+1}\cos \frac{2n\pi}{3}$.

kk thanks                                         .


You have the misfortune to own an unreliable clock. This one gains exactly 15 minutes every hour. It is now showing 2:00pm and you know that is was correct at midnight, when you set it. The clock stopped 7 hours ago, what is the correct time now?

Assume the correct time now is = t
Correct time 7 hours before will be t - 7

Currently the clock shows time as 02:00 PM = 14:00 hours (milletry time)

The clock gains 15 min in 1 hour
So if correct 60 min is elapsed the then clock shows as 75 min elapsed
So if 60 min is elapsed in clock then true time elapsed = (60/75)x60 = 48 min.

So the clock gains 60-48 = 12 min every faulty clock hour.

So after 14:00 hours clock gains 14x12 min = 168 min = 2 hours 48 min.

So correct time when the clock stopped is = 14:00 - 2 hours 48 min = 11 hours 12 min

So t - 7 = 11 hours 12 min
So t = 18 hours 12 min = 06:12 PM CST

Thanks,
http://www.teacherone.com


I wondering about the technique to solve question of this type:

w+x+y+z=20

A) # of solutions for (w,x,y,z) if w,x,y,z>=0

B) # of solutions for (w,x,y,z) if w,x,y,z>0

I think it had something to do with dividers but it's been a looong time since I was saw one of these.

EDIT: never mind, it just came back to me again.

A) Ordering a string of 20 1's and 3 0's. 23!/(20!3!).

B) Ordering a string of 18 1's and 3 0's. 21!/(18!3!).

[quote="t0rajir0u"]Are you sure about the second one?  It reduces to the first if you consider $(w-1)+(x-1)+(y-1)+(z-1) = 16$.[/quote]

Oops, I don't know what I was thinking. Thanks for pointing that out.


Several segments are given on a straight line, and it is known that any two of them intersect. Prove that some point on the line belongs to all the segments.

[hide]
Lets assume that there is a segment (A) which appears first, (if there are many pick a random one). It must overlap the 'last' segment (B) (that is the one which disappears last). Consider any random segment, let us assume it doesn't overlap the point where A and B overlap. It must overlap with A though, and also B. But since it can't have a break, it must overlap their overlap[/hide]


Express in terms of $ \pi$ the number of square units in the area of a semicircle with diameter 24.

If the diameter is $ 24$, the radius is $ 12$.

Let's find the area of the circle first, then find half that for the semicircle.
The area of the circle is $ \pi r^2$, which here is $ 144\pi$. Half of that, the area of the semicircle, is $ \boxed{72\pi}$.

By the way, the review says $ 144\pi$; I reported the error.


number theory
least common multiple
symmetry
AMC
AIME

Let $[r,s]$ denote the least common multiple of positive integers $r$ and $s$.  Find the number of ordered triples $(a,b,c)$ of positive integers for which $[a,b] = 1000$, $[b,c] = 2000$, and $[c,a] = 2000$

[hide]First, rewrite $a$, $b$, and $c$ as $a=2^A5^R$, $b=2^B5^S$, and $c=2^C5^T$. Because the prime factorizations of $1000$ and $2000$ both contain $5^3$, at least two of $R$, $S$ and $T$ must be $3$. There are 10 ways that this can occur: $(R,S,T)$ can be permutations of $(3,3,0)$ (3 permutations), $(3,3,1)$ (3 permutations), $(3,3,2)$ (3 permutations), and $(3,3,3)$ (1 permutation). Similarly, there so we need only worry about arranging $A$ and $B$. $(A,B)$ can be permutations of $(0,3)$ (2 permutations), $(1,3)$ (2 permutations), $(2,3)$ (2 permutations), and $(3,3)$ (1 permutation). In total, our answer is $7\cdot 10=70$.
[/hide]

[quote="cdymdcool"][quote="4everwise"][hide]First, rewrite $ a$, $ b$, and $ c$ as $ a \equal{} 2^A5^R$, $ b \equal{} 2^B5^S$, and $ c \equal{} 2^C5^T$. Because the prime factorizations of $ 1000$ and $ 2000$ both contain $ 5^3$, at least two of $ R$, $ S$ and $ T$ must be $ 3$. There are 10 ways that this can occur: $ (R,S,T)$ can be permutations of $ (3,3,0)$ (3 permutations), $ (3,3,1)$ (3 permutations), $ (3,3,2)$ (3 permutations), and $ (3,3,3)$ (1 permutation). Similarly, there so we need only worry about arranging $ A$ and $ B$. $ (A,B)$ can be permutations of $ (0,3)$ (2 permutations), $ (1,3)$ (2 permutations), $ (2,3)$ (2 permutations), and $ (3,3)$ (1 permutation). In total, our answer is $ 7\cdot 10 \equal{} 70$.
[/hide][/quote]

But 2000 contains 2^4, in this solution are we only considering 2^3?[/quote]
I think he skipped saying this, but, adopting the same notation as the solution above, we have $ \max(A,B)\equal{}3, \max(B,C)\equal{}\max(C,A)\equal{}4$. Clearly, $ A,B\leq 3$, so we must have $ C\equal{}4$. Then the rest of the solution follows.

[quote="Brut3Forc3"][quote="cdymdcool"][quote="4everwise"][hide]First, rewrite $ a$, $ b$, and $ c$ as $ a \equal{} 2^A5^R$, $ b \equal{} 2^B5^S$, and $ c \equal{} 2^C5^T$. Because the prime factorizations of $ 1000$ and $ 2000$ both contain $ 5^3$, at least two of $ R$, $ S$ and $ T$ must be $ 3$. There are 10 ways that this can occur: $ (R,S,T)$ can be permutations of $ (3,3,0)$ (3 permutations), $ (3,3,1)$ (3 permutations), $ (3,3,2)$ (3 permutations), and $ (3,3,3)$ (1 permutation). Similarly, there so we need only worry about arranging $ A$ and $ B$. $ (A,B)$ can be permutations of $ (0,3)$ (2 permutations), $ (1,3)$ (2 permutations), $ (2,3)$ (2 permutations), and $ (3,3)$ (1 permutation). In total, our answer is $ 7\cdot 10 \equal{} 70$.
[/hide][/quote]

But 2000 contains 2^4, in this solution are we only considering 2^3?[/quote]
I think he skipped saying this, but, adopting the same notation as the solution above, we have $ \max(A,B) \equal{} 3, \max(B,C) \equal{} \max(C,A) \equal{} 4$. Clearly, $ A,B\leq 3$, so we must have $ C \equal{} 4$. Then the rest of the solution follows.[/quote]

But do we have to consider the case that A and B have different values?

[quote="ThinkFlow"]Since $ [b,c] \equal{} 2000$, and $ [c,a] \equal{} 2000$ isn't it true that a and b are the same? :maybe:[/quote]
not necessarily, for all cases? I'm not sure though :maybe:
it's kinda like if you have lcm (8, 10)=40 and lcm (8, 20)=40?
meh :maybe:

A more PIE based solution:

[hide="Solution"]
Note that $c$ has a power of $2^4$ in it, if it does not then $a$ and $b$ must but that means $\text{lcm}[a,b]$ has a power of $2^4$ in it, absurd.

Now $a=2^{e_1}*5^{e_2}, b=2^{e_3}*5^{e_4}, c=2^4*5^{e_5}$.

We first off find the choices for $e_1, e_3$.  $e_3=3$ gives us $4$ choices for $e_1$ and $e_1=3$ gives us $3$ new choices for $e_3$ (or by PIE it's $4+4-1$), hence we have $7$ choices for $e_1/e_3$.

Now we have to consider powers of $5$.  Among $e_2, e_4$ at least one must be $3$. Same with among $e_4$ and $e_6$ and $e_2$ and $e_6$.

We use PIE here by counted when $e_2=3$, $e_4=3$ or $e_6=3$.

$P(e_2\cup e_4\cup e_6)=P(e_2)+P(e_4)+P(e_6)-P(e_2\cap e_4)-P(e_2\cap e_6)-P(e_4\cap e_6)+P(e_2\cap e_4\cap e_6)$.  $P(e_2)$ implies $e_2=3$ giving us $7$ choices for $e_4/e_6$ (same logic as before).  By symmetry, $P(e_2)=P(e_4)=P(e_6)$.  Now, $P(e_2\cap e_4)$ will have satisfied all the equations above, so we have $4$ choices for $e_6$ and so this is $P(e_2\cap e_4)$ which equals other two element intersections by symmetry.  Lastly, $P(e_2\cap e_4\cap e_6)=1$ so hence we get $3*7-3*4+1=10$.

Our final answer is $10*7=\boxed{70}$. [/hide]

The notation MAX(a,b) = 3 refers to the fact that the highest power is three correct?

Hmmmm......this is an AIME question.

Here's the link: http://www.artofproblemsolving.com/Wiki/index.php/1987_AIME_Problems/Problem_7

Hmm here's a bit of a twist on the solution that hasn't been posted yet

[hide="solution"]

We use the fact that $lcm(a,b,c) = lcm(lcm(a,b),c)$ (try proving it yourself!). Thus we have that $lcm(a,b,c) = lcm(1000,c) = 2000$.

We have that $1000 = 2^3*5^3$ and $2000 = 2^4*5^3$. Thus $c = 2^45^C$ where $C \le 3$.

There are actually only 2 cases we have to consider from here:

[b]Case 1[/b] C < 3

Let $a = 2^{A_1}5^{A_2}$ and $c = 2^45^C$. We know that $A_1 \le 3$ (since $gcd(a,b) = 1000 = 2^3*5^3$). Thus LCM(a,c) = $2^{max(A_1,4)}5^{maxA_2,C} = 2^4*5^3$. Thus $A_2 = 3$. We use this same logic on $b$.

Now we know that $a = 2^{A_1}5^3$ and $b=2^{B_1}5^3$. We see that $lcm(a,b) = 2^{max(A_1,B_1)}5^3 = 2^35^3$. Thus $\text{max}(A_1,B_1) = 3$. If $A_1 = 3$, then there are 4 possible values for $B_1$ (which would be $0-3$) and vice versa for if $B=3$. But we overcount $A=B=3$. Thus there are $4+4-1 = 7$ for every value $C < 3$ which would be $3$ values. That makes $7*3 = 21$ pairs in total.

[b]case 2[/b] C=3

Now $a = 2^{A_1}5^{A_2}$ where $A_2$ is not necessarily less than 3. The same is said about $b = 2^{B_1}5^{B_2}$. We have that $\text{lcm}(a,b) = 2^{max(A_1,B_1)}*5^{max(A_2,B_2)} = 2^3*5^3$

From before we have that there are $7*7 = 49$ total ways this is possible.

That means the total number of triples is $21+49 = \boxed{70}$

[/hide]


Pat wants to buy four donuts from an ample supply of three types of
donuts: glazed, chocolate, and powdered. How many different
selections are possible?

Let the donuts be $ *'s$. We can line the donuts up in a row:
$ ****$. Now we need to place $ 2$ "dividers" to distinguish between different donuts. There are $ 5$ spaces to place them (on the outsides too), the answer is $ \binom{5}{2}\equal{}\boxed{10}$ ways for Pat to buy the donuts.

I got 18.  (wrong)

Let's call the different donuts a, b, and c.

The different cases are:

aaaa          aaab          aabb          aabc
3               6              6               3.

Total = 18.  (?!?!)

I claim that you're both wrong. Rofl... the choices for the dividers don't need to be in distinguishable spots.  bwu, aabb has 3 possible ways not 6. AABB is the same as BBAA. The answer is 15.

[hide="In other words"]
the answer is
$\dbinom{6}{2}=15$[/hide]


ugh summations.  :) I will attempt this and edit this post later.

Someone try this cause it's very nice :P

I had a supershort proof...
Hint : logarithmic mean :)


Solve equation  $ 2x\plus{}\sin(\sin x)\equal{}3\sin x$

_________________________________________
Azerbaijan Land of Fire  :ninja:

Well,a graphing calculator check shows that $ x\equal{}0$ is the only solution, but I'm wondering if somebody can find a more mathematical solution i.e not one with graphing calculators :) .


geometry
parallelogram
analytic geometry
geometric transformation
rotation
vector
reflection

Given Arbitrary triangle ABC, construct squares externally on AB and BC, with centers O and P respectively, Let M be the midpoint of AC. Find all possible values of angle OMP.

I'm interested in seeing the wide array of solutions that will come through on this...

So far I'm putting my money on [hide]the angle is always 90 degrees,[/hide] but I don't have a proof yet.  Everything I've tried so far is too nasty.

[hide]Construct midpoints E and F of AB and BC respectively. Join OE, PF, ME and MF. We have:

OE :perp: AB and OE = 1/2 AB;

PF :perp: BC and PF = 1/2 BC;

ME // BC and ME = 1/2 BC; and

MF // AB and MF = 1/2 AB.

<AEM = <MFC = <ABC by parallel lines. Thus <OEM = <MFP = 90o + <ABC.

BEMF is a parallelogram, therefore <EMF = <ABC. Since OE = MF, EM = FP, <OEM = <FMP, we have OEM :cong: MFP, which implies that <EOM = <FMP and <EMO = <FPM.

Notice that <EOM + <EMO + <AEM + 90o = 180o, which implies that <FMP + <EMO + <EMF = <OMP = 90o. QED.

[/hide]

Okay, haven't read Osiris's yet, so if this overlaps, my apologies.  (Just solved during lunch.)

[hide]Set your triangle in the coordinate plane.  But M at (0, 0), A at (a, 0), C at (-a, 0), and B at (x, y).  Then the midpoint of AB is at ((a + x)/2, y/2) and the midpoint of BC is at ((x - a)/2, y/2).  Since we know the points O and P are determined by perpendiculars to the sides of length half the side, we can simply add the appropriate amounts to the x and y coordinates of both of these points to get the centers of the squares.  That gives us centers located at ((x - 1 - y)/2, (x + 1 + y)/2) and ((x + 1 + y)/2, (y - x + 1)/2), and we can clearly see from this that the lines between each of those points and the origin, M, are perpendicular.  Yipee.[/hide]

Ah, after having read Osiris's proof: that's really nice.  It's what I was trying to get but wasn't able to do

My solution uses rotations and only a few lines, but its nice to see there is a simple non-rotational solution.

Vector solution (which isn't easy in my opinion): (use the same diagram)

[hide]MO = ME + EO

MP = MF + FP

MO . MP = ME . MF + ME . FP + MF . EO + EO . FP

ME :perp: FP, MF :perp: EO, so ME . FP = MF . EO = 0

If you extend OE and PF to intersect at X, you see that the quadrilateral EBFX is cyclic; thus <OXP = 180o - <ABC

|EO| = |MF|, |FP| = |ME|

So MO . MP = ME . MF + EO . FP

= |ME||MF| cos ABC + |MF||ME| cos (180o - <ABC)

= |ME||MF| cos ABC - |MF||ME| cos ABC

= 0.

Thus MO :perp: MP.

[/hide]

Rotational Proof:

[hide]

Consider the three rotations r1, r2, and r3, centered about O, M and P respectively, such that r1 is a rotation of 90 degrees clockwise, r2 is a rotation of 180 degrees, and r3 is a rotation of 90 degrees clockwise.

The sum of the rotations, r4(X) = r3(r2(r1(X))), has measure 90 + 180 + 90 = 360 degrees, and is therefore a translation in the plane.  However:

r1(B) = A.

r2(A) = C.

r3(C) = B.

therefore, r4(B) = B and the translation is the null vector; in other words, the rotation is the identity.

So, lets look at r4(O).

r1(O) = O.

r2(M) = O', where O' is the reflection of O over M.

Therefore, r3(O') must be O to preserve the identity of the rotation r4.

This means OPO' is a right isosceles triangle with right angle at P. M is the midpoint of OO', so PM is the median to OO', which is perpendicular to OO' since OP = PO'. Therefore, PM :perp: OM.

QED.

[/hide]

r2(B) = C should be r3(B) = C?

Also, my diagram has to be flipped horizontally for your proof to work; all rotations need to be CCW with my diagram.

Just a heads-up. Nice proof nonetheless.

Ok I fixed my solution; I named them in the wrong orientation so things got funny.

Diagram of problem (and aid for solution).

Geometer's Sketchpad is a beautiful thing.


geometry
trigonometry
vector
analytic geometry
trig identities
Law of Cosines

Given a quadrilateral $ABCD$ such that $AB^2 + CD^2 = AD^2 + BC^2$, prove that $AC \perp BD$.

We call $P$ the intersection of $AC$ and $BD$.
Now by the law of cosines, we get:
$- 2APBP\cos(BPA) -2CPDP\cos(BPA)= 2 APDP \cos(BPA) + 2 BPCP \cos(BPA)$

So \[ 2 \cos(BPA).(APDP + CPDP + APBP + BPCP) = 0 \]
\[ \iff \cos(BPA) = 0 \Rightarrow \angle BPA = \frac{\pi}{2} \]

That's one way to do it. There are plenty of other ways: vectors, coordinates, proof by contradiction, ... or, the nicest way: you can assume, without loss of generality, that $\angle{APB} \geq 90^{\circ}$ so that $AB^2 \geq AP^2 + BP^2$, $CD^2 \geq CP^2 + DP^2$, and then, since $\angle{BPC} \leq 90^{\circ}$, $BC^2 \leq BP^2 + CP^2$ and $AD^2 \leq AP^2 + DP^2$. Hence \[ AB^2 + CD^2 \geq AP^2 + BP^2 + CP^2 + DP^2 \geq AD^2 + BC^2, \] and since equality holds, we must have $\angle{APB} = 90^{\circ}$, which is what we want :)


function
topology
calculus
calculus computations

I need a function such that $ d(f(x),f(y))<d(x,y)$ in a complete metric space $ M$.  In other words, show that both conditions of the contraction principle are necessary by finding counter examples.

Would this work?
$ M\equal{}[1,\infty)$, $ f(x)\equal{}\frac{1}{2}x$

$ f$ is not a function from $ M$ to itself.  If you want to show that completeness is necessary, you want $ M \equal{} (0, \infty)$.  $ f(x) \equal{} \frac {x}{2}$ has no fixed point in $ M$.

Edit:  Oh, sorry, I misread your question.  The above $ M, f$ still don't work; the function you chose is Lipschitz with constant $ \frac {1}{2} < 1$.  The point is that you want to find a function that satisfies $ d(f(x), f(y)) < d(x, y)$ but that does [i]not[/i] satisfy $ d(f(x), f(y)) < (1 \minus{} \epsilon) d(x, y)$ for any $ \epsilon > 0$.  One such function is $ f(x) \equal{} \sqrt {x^2 \plus{} 1}$, which satisfies the desired property on $ \mathbb{R}$ but has no fixed points.

I spoke to my teacher about this problem and function only two days ago. Is there a metric in which that function is a contraction?

In a contraction, the iteration converges to the fixed point. Here, the iteration diverges to $ \infty$. If you want a metric in which this function is a contraction, the completion of the space under that metric should include $ \infty$. Can you construct it now?

I'll have to think about what including infinity means.


abstract algebra
Ring Theory
superior algebra
superior algebra unsolved

Give an example of a finitely generated but not finitely presented module such that the following canonical homomorphism
  $ S{}^-{}^1 Hom(M,N) --> Hom_S{}^-{}^1(S{}^-{}^1 M,S{}^-{}^1 N)$
is 1-1 but onto.

I am trying with the ring C[0,1] and the maximal ideal ={f: f(.5)=0}
but no success

$ R: \equal{}C[0,1]$

Then $ \mathbb R$ is a $ R$-module certainly generated by one element, but it is not of finite presentation.

Proof: Let's assume that there is a finite presentation of $ \mathbb R$, there is a maximal ideal $ \mathfrak{m}$ in $ R$ which is finitely generated. Now every maximal element in $ R$ comes from a unique element $ x_0\in[0,1]$, i.e. $ \mathfrak{m}\equal{}\{f\in R|f(x_0)\equal{}0\}$. If $ f\in\mathfrak{m}$ is non-negative, then $ f$ is a square in $ \mathfrak{m}$ by taking the square root of $ f$, thus $ f\in\mathfrak{m}^2$ in this case. For general $ f$ decompose into the positive and negative part ($ f\equal{}f^\plus{}\minus{}f^\minus{}$) and get \[ \mathfrak{m}\equal{}\mathfrak{m}^2\] Assuming that $ \mathfrak{m}$ be finitely generated yields that the localization $ \mathfrak{m}_\mathfrak{m}$ vanishes, by Nakayma. That means that if $ f\in R$ vanishes in $ x_0$, then it vanishes in an open neighborhood of $ x_0$ - which is absurd.

Altogether this shows that $ \mathfrak{m}$ is not finitely generated and $ \mathbb R$ is not of finite presentation over $ R$.

To get an example you want show that every module is zero: $ Hom_R(\mathfrak{m}^2,\mathbb R)\equal{}0$ and $ Hom_{S^{\minus{}1}R}(S^{\minus{}1}\mathfrak{m}^2,\mathbb R)\equal{}0$ for $ S\equal{}R\setminus \mathfrak{m}$. Note that $ S^{\minus{}1}\mathbb R\equal{}\mathbb R$. [Do use the squares here. ;)]

how do you define the module structure? f r := f(0) r?

Correct, it is induced by the short exact sequence \[ 0\to\mathfrak{m}\to R \to \mathbb{R}\to 0\,.\]


inequalities
function
inequalities unsolved

\[ \text{Let}\; x,y,z\; \text{be positive real numbers such that} \; x+y+z=1,\; \text{prove that} \]
\[ \;\;\; \frac1{\sqrt{1+xy}}+\frac1{\sqrt{1+yz}}+\frac1{\sqrt{1+zx}}\geq\frac{9}{\sqrt{10}},\\ \;\text{and}\;\; \, \frac1{\sqrt{1-xy}}+\frac1{\sqrt{1-yz}}+\frac1{\sqrt{1-zx}}\geq\frac{9}{\sqrt{8}}.  \]

The first one:
$\frac1{\sqrt{1+xy}}+\frac1{\sqrt{1+yz}}+\frac1{\sqrt{1+zx}}\ge\frac{9}{\sqrt{9+3(\sum yz}}\ge\frac{9}{\sqrt{10}}$

[quote="pvthuan"]\[ \text{Let}\; x,y,z\; \text{be positive real numbers such that} \; x+y+z=1,\; \text{prove that}  \]
\[ \;\;\; \frac1{\sqrt{1+xy}}+\frac1{\sqrt{1+yz}}+\frac1{\sqrt{1+zx}}\geq\frac{9}{\sqrt{10}}  \][/quote]

We will use this inequality: $(x+y+z)^2\geq 3(xy+yz+zx)$

The function $f(x)=\frac1{\sqrt{1+x}}$ is convex, so Jensen's yields that:

$\frac1{\sqrt{1+xy}}+\frac1{\sqrt{1+yz}}+\frac1{\sqrt{1+zx}}\geq \frac3{\sqrt{1+\frac{xy+yz+zx}{3}}}=\frac9{\sqrt{9+3(xy+yz+zx)}}$$\geq \frac9{\sqrt{9+(x+y+z)^2}}=\frac9{\sqrt{10}}$

[quote="Ezbakhe Yassin"]
$\frac9{\sqrt{9-3(xy+yz+zx)}}\geq \frac9{\sqrt{9-(x+y+z)^2}}$[/quote]
sign reversed

[quote="pvthuan"]These inequalites are not that easy. I like them very much. Has anyone got a solution?[/quote]
In fact I don't think the second one is true. What if you take x -> 1, y=z->0?

The first one was solved by Cauchy and Minkowski 's ineqs and the second one is wrong .

[quote="pvthuan"]These inequalites are not that easy. I like them very much. Has anyone got a solution?[/quote]

For the first one I posted a solution.

The second one is not true. Take $x=y=\frac1{16}$


Points P(-3,14), Q(5,10), R(3,0), and S(-1,-4) are given. The midpoints of quadrilateral PQRS are connected to form a new quadrilateral. Find the area of this new quadrilateral.

Without calculator.. :)

[quote="nat mc"]We see the midpoints are (1,12),(4,5),(1,-2),(-2,5)
We could use shoelace, but it nicely lines up so that it is the area of two triangles with easily findable height and base.
They both have base 6 and height 7 so the area is 42[/quote]

You can look at that way but it is also rhombus so we can apply $A = d_1 d_2 \frac{1}{2}$.  :)


real analysis
calculus
integration
function
real analysis theorems

which is lebesgues theorem about Riemann integrals?
it is required for the romanian county math olympiad, as i saw on http://www.edu.ro

if a function f:I->R is continuous on all points of I, except some points found in a set D included in I. If D is neglijable Lebesgue then f is integrale Rieman.

One can use this to prove for example that if f(x) is integrable then f(x^2) is also integrable.

oh and another thing. You say that a set L is neglijable Lebesgue if for each e>0 there is a set of intervals {In} such that \sum l(In)<e  and L < Un >1  In, where l(I) is the length of the interval I.

Lebesgue's Theorem:

A bounded function f on [a, b] is Riemann integrable if and only if the set of points where f is not continuous has measure zero.

[quote="Valentin Vornicu"]If D is neglijable Lebesgue then f is integrale Rieman.
[/quote]
neglijable -> negligible


vector
linear algebra
matrix
quadratics
probability and stats

$ (X,Y,Z)\sim N_3(m,\Sigma)$ where
$ \Sigma \equal{} \left( \begin{array}{ccc} 4 & 1 & 2 \\
1 & 1 & 1 \\
2 & 1 & 4 \end{array} \right), m \equal{} \left( \begin{array}{c} 1 \\
1 \\
0 \end{array} \right)$
Find $ Var((X \minus{} Y)Z)$ :)

if $ v$ is a gaussian vector with covariance matrix $ \Sigma$ then how can you compute the covariance matrix of the gaussian vector $ Av$, where $ A$ is any fixed matrix. From here, you should not be to far from the solution ..

[quote="alekk"]if $ v$ is a gaussian vector with covariance matrix $ \Sigma$ then how can you compute the covariance matrix of the gaussian vector $ Av$, where $ A$ is any fixed matrix. [/quote]
To answer just this: The covariance of $ Av$ is $ E(Avv^TA^T)\minus{}E(Av)E(v^TA^T)$

$ \equal{}A[E(vv^T)\minus{}E(v)E(v^T)]A^T\equal{}A\Sigma A^T.$

In this particular case, $ U\equal{}X\minus{}Y$ and $ V\equal{}Z$ means that $ \begin{bmatrix}U\\V\end{bmatrix}\equal{} \begin{bmatrix}1&\minus{}1&0\\ 0&0&1\end{bmatrix} \begin{bmatrix}X\\ Y\\ Z\end{bmatrix}.$

Then $ \begin{bmatrix}1&\minus{}1&0\\ 0&0&1\end{bmatrix} \begin{bmatrix} 4&1&2\\ 1&1&1\\ 2&1&4\end{bmatrix} \begin{bmatrix} 1&0\\ \minus{}1&0\\ 0&1\end{bmatrix}\equal{} \begin{bmatrix}3&1\\ 1&4\end{bmatrix}.$

So $ X\minus{}Y$ and $ Z$ are both normal with mean zero (that was a separate computation). The variance of $ X\minus{}Y$ is $ 3,$ the variance of $ Z$ is $ 4,$ and the covariance of the two is $ 1.$ The expectation $ E((X\minus{}Y)Z)$ is just the covariance, namely $ 1.$ However, the variance of this is not just a computation of a quadratic form, and will require a little more work.

So, how to calculate $ E[((X\minus{}Y)Z)^2]$? Or any other idea to find $ Var((X\minus{}Y)Z)$ :)


Jack bought an item for $ \$$300. It had been discounted by 20$ \%$.  How many dollars was the discount?

We have $ \frac{4}{5}x\equal{}300$. Thus $ x\equal{}300*\frac{5}{4}\equal{}375$. That is a discount of $ 375\minus{}300\equal{}\boxed{75}$.


Hello,

I remember that the first time I solved this problem(almost ten years ago), I was very naive, and did a very complicated proof of it, even when afterwards I found a very easy solution. Nevertheless, I think that I learned more from doing the complicated solution, than from the easyone. Maybe some of you can find it ilustrative too.

Compute the following sum,

\[
\sum_{k=1}^{n}\left\lfloor\frac{n+2^{k-1}}{2^k}\right\rfloor
\]

I hope you'll enjoy it.

Best regards,

$\displaystyle \sum_{k=1}^{n}\frac{n+2^{k-1}}{2^k}$

$=\displaystyle \sum_{k=1}^{n} \frac{n}{2^k}+\displaystyle \sum_{k=1}^{n} \frac{1}{2}$

$=\displaystyle \sum_{k=1}^{n} n\displaystyle(\frac{1}{2^k})+\frac{n}{2}$

$=2n+\frac{n}{2}$

$=\frac{5n}{2}$

$\lfloor x \rfloor$ is the greatest integer function (you round down $x$).

Hello jli,

[quote="jli"]
\[\displaystyle \sum_{k=1}^{n}\frac{n+2^{k-1}}{2^k}=\displaystyle \sum_{k=1}^{n} \frac{n}{2^k}+\displaystyle \sum_{k=1}^{n} \frac{1}{2}=\displaystyle \sum_{k=1}^{n} n\displaystyle(\frac{1}{2^k})+\frac{n}{2}=2n+\frac{n}{2}=\frac{5n}{2}\]
[/quote]

Two things, firs than all, as someone already said, $\lfloor x\rfloor$ means floor or integer part. You round (always below) $x$. On the other side, note that

\[\sum_{k=1}^{n} n\displaystyle(\frac{1}{2^k})\neq2n\]

It is not an infinite series, and any way, the sumatory starts in $k=1$ not $k=0$.

Best,


An equiangular octagon ABCDEFGH has sides of length 2, 2 :sqrt: 2, 4, 4 :sqrt: 2, 6, 7, 7, 8. Given that AB = 8 , find the length of EF .

Here's the idea: the points $(A,B)$, $(C,D)$, $(E,F)$ and $(G,H)$ are, respectively, on the four sides of a rectangle $XYZT$ (assume $A\in XY$). Moreover, the triangles $BYC,\ CZE,\ FTG,\ HXA$ are isosceles. By using the fact that $XY=ZT$ and $XT=YZ$ we get, respectively, the equations $a+\frac {b+h}{\sqrt 2}=e+\frac {d+f}{\sqrt 2}$ and $c+\frac {b+d}{\sqrt 2}=g+\frac {f+h}{\sqrt 2}$, where $a=AB, b=BC\ldots$.

From the first equation we can observe that if (a) $e$ is irrational, then we will have $(d+f)-(b+h)=8\sqrt 2+r$ where $r$ is rational, and that's impossible, so we must have (b): $e$ is rational (integer, actually). The second thing we observe is that $(d+f)-(b+h)$ must be $\sqrt 2$ multiplied by a natural number $a-e$. This number can be $1,2,4,6$. It certainly can't be $6$ because we don't have that many $\sqrt 2$'s and it can't be $1$ because we only have even multiples of $\sqrt 2$, so it can only be $2$ or $4$. It's easy to dismiss the case $a-e=4$ (exercice :)), so $a-e=2\Rightarrow e=6$.

Grobber,i didn't understand the last part of your solution.From "if [b][i]e[/i][/b] is irrational...till the end.
Could you please explain it to me with details(if this doesn't bother you).Thanks.

Ok man, I'll try.

In the first paragraph I wrote two equalities. Just focus on the first one, the one involving $a$ and $e$. Now assume $e\in \{2\sqrt 2,\ 4\sqrt 2\}$. We have $(a-e)\sqrt 2=(b+h)-(d+f)=8\sqrt 2-e\sqrt 2$. $e\sqrt 2$ is natural, so the coefficient of $\sqrt 2$ in the difference $(b+h)-(d+f)$ must be $8$, but that's simple impossible because we have at most $4\sqrt 2+2\sqrt 2=6\sqrt 2$ which can be formed with the given numbers, which means that $e$ is natural, so it's one of $7,6,4,2$.

(1) Assume it's $7$. In this case $(b+h)-(d+f)=\sqrt 2$ and that's also impossible. From $4\sqrt 2$ and $2\sqrt 2$ we simply can't get $\sqrt 2$ by subtractions and additions/

(2) Assume $e=2$. We must have $(b+h)-(d+f)=6\sqrt 2$. This can only happen if $\{b,h\}=\{4\sqrt 2,2\sqrt 2\}$, but in this case we also have to subtract the natural numbers $d,f$, getting $6\sqrt 2+t$, where $t$ is a non-zero integer, so we have a contradiction again.

(3) Assume $e=4$. We get $(b+h)-(d+f)=4\sqrt 2$. This can only be done if one of $b,h$ is $4\sqrt 2$ and the other one is the sum $d+f$. Assume $b=4\sqrt 2$. We thus have $h=d+f$. This can only happen if $h=6,\ \{d,f\}=\{2,4\}$, but $e=4$, so $4$ must be used two times, which isn't allowed because only one side of the octagon is $4$: another contradiction.

We're only left with the case $e=6$. Note that's at this point the problem is solved, but if you want to be at peace with yourself you can quite easily construct the octagon by using both the equalities I've written.

I hope it's clearer this time.


Estw $ \pi(x)$ o ari8mos twn prwtwn pou den xepernaei to x kai $ n$ akeraios $ n \geq 2$.
a)ndo $ \pi(n! \plus{} 2n) \plus{} \pi(n) \leq \pi(n!\plus{}n) \plus{} \pi(2n)$
b)Exoume $ \frac{\pi(n)}{n} > \frac{\pi(n\minus{}1)}{n\minus{}1}$. ndo o $ n$ einai prwtos

[quote="basted"]Estw $ \pi(x)$ o ari8mos twn prwtwn pou den xepernaei to x kai $ n$ akeraios $ n \geq 2$.
a)ndo $ \pi(n! \plus{} 2n) \plus{} \pi(n) \leq \pi(n! \plus{} n) \plus{} \pi(2n)$
b)Exoume $ \frac {\pi(n)}{n} > \frac {\pi(n \minus{} 1)}{n \minus{} 1}$. ndo o $ n$ einai prwtos[/quote]

Gia to a exume isodinama $ \pi(n! \plus{} 2n)\minus{} \pi(n! \plus{} n) <\equal{} \pi(2n)\minus{} \pi(n)$
opote arki ndo
"An o $ n!\plus{}n\plus{}k$ ine protos (me $ 0<\equal{}k<\equal{}n)$ ,tote kai o $ n\plus{}k$ ine protos" kai 8a exume teliosi..
Omos an o $ n!\plus{}n\plus{}k$ ine protos auto simeni oti ine diaforos midenos mod2 ,diaforos midenos mod 3 .... diaforos midenos mod n kai epidi  to n! ine 0 mod2 mod3, ..mod n ,pernume
oti to $ n\plus{}k$ ine diaforo tu midenos mod2 ,mod3, ...mod n ,pragma pu simeni oti o $ n\plus{}k$ ine protos gt diaforetika 8a ixe ena proto diereti $ q<\equal{}\sqrt{n\plus{}k}<\equal{}\sqrt{2n}<\equal{}n$
 pu 8a itan 0 mod q ,atopo...

Gia to b , an $ n$ sin8etos , $ \pi(n)\equal{}\pi(n\minus{}1)$ opote $ n\minus{}1>n$ ,atopo. Ara $ n$ protos...


geometry
trigonometry
circumcircle
symmetry
trig identities
Law of Sines

Ok, here's the deal, find the measure of angle x. You can only use ELEMENTARY methods of solving the problem, so you can't use law of sines or cosines or etc.

Good LUCK!!!!

[img]http://thinkzone.wlonk.com/MathFun/Triangle1.gif[/img]

btw, I solved this problem in less than 5 minutes. :D:D:D

This picture was from http://thinkzone.wlonk.com/MathFun/Triangle.htm

[quote="AndrewTom"]I first drew a line through $ D$ parallel to $ AB$ and then had some isosceles triangles etc. It took me a LOT longer than $ 5$ minutes. I got $ 20$ degrees.

Ihatepie's post arrived while I was typing - after a day of silence.[/quote]
Can you post yhe whole solution?

$ F$ is circumcenter of $ \triangle BDC$.
$ G$ is center of $ \triangle ABC$.

[hide="Answer"]$ x\equal{}20^\circ$[/hide]

[quote="Yavor Tsvetanov 52"]Well,i want to ask you how you proov that the points F,G and D lie on one line?
And how the point G lie on the circumcircle of the triangle CAD?
Thanks![/quote]

$ \boxed{F \ \text{on} \ AC}$
$ \angle FCD \equal{} \angle FDC \equal{} 20^\circ$
$ \angle FCB \equal{} \angle FBC \equal{} 60^\circ$
So $ \angle FBD \equal{} \angle FDB \equal{} 10^\circ$
Thus $ F$ on $ AC$.

$ \boxed{GC \ \text{is line of symmetry} \ \triangle FBC}$
$ \angle GAB \equal{} \angle GAC \equal{} 20^\circ$
$ \angle GBC \equal{} \angle GBA \equal{} 40^\circ$
$ \angle GCA \equal{} \angle GCB \equal{} 30^\circ$
And $ \angle FCB \equal{} \angle FBC \equal{} 60^\circ$
So $ \triangle FBC$ is equilateral triangle.
Thus $ GC$ is line of symmetry $ \triangle FBC$.

$ \boxed{D, F, G \ \text{are collinear}}$
$ \angle CBG \equal{} \angle CFG \equal{} 40^\circ$
$ \angle CFG \equal{} \angle CDF \plus{} \angle FDC \equal{} 40^\circ$
Thus $ D$, $ F$, $ G$ are collinear.

$ \boxed{A, G, C, D \ \text{are concyclic}}$
$ \angle GAC \equal{} \angle GDC \equal{} 20^\circ$
Thus $ A$, $ G$, $ C$, $ D$ are concyclic.

$ \boxed{\angle ADB \equal{} 20^\circ}$
$ \angle ADG \equal{} \angle ACG \equal{} 30^\circ$
Thus $ \angle ADB \equal{} \angle ADG \minus{} (\angle BDC \minus{} \angle GDC) \equal{} 30^\circ \minus{} (30^\circ \minus{} 20^\circ) \equal{} 20^\circ$.

EDIT:$ BD \to GC$. I'm typo :(

[quote="Yavor Tsvetanov 52"][quote="Kouichi Nakagawa"]
$ \boxed{BD \ \text{is line of symmetry} \ \triangle FBC}$
$ \angle GAB \equal{} \angle GAC \equal{} 20^\circ$
$ \angle GBC \equal{} \angle GBA \equal{} 40^\circ$
$ \angle GCA \equal{} \angle GCB \equal{} 30^\circ$
And $ \angle FCB \equal{} \angle FBC \equal{} 60^\circ$
So $ \triangle FBC$ is equilateral triangle.
Thus $ BD$ is line of symmetry $ \triangle FBC$.
[/quote]
What is line of symmetry? :huh:
And i think that this part is unuseful?
That's because we know that tr. FBC is an equilateral and CG is the bisector of the angle BCA.Thus we have angle CFG=40.
Correct me if i'm wrong.[/quote]

Sorry.
I'm typo.
This proof is $ GC$.
I think that you understand it if you have you watch contents of the proof.

I'm sorry that I do a confusing mistake.


Let $p=\overline{a_na_{n-1}...a_1a_0}$ be a prime number $p\ge 5$. Prove that the polynomial $f(x)=\sum_{k=0}^na_kx^k$ is irreductible in $\mathbb Z[X]$.

See topic http://www.mathlinks.ro/Forum/viewtopic.php?t=20661
I think, we can easily generalize attached solution.


Probably an easier way to do this but.
[hide]50 - 1
450 - x
Cross multiply and find the time.
450/50 = x
9 = x

50 - 1
475 - y
Cross multiply again.
9.5 = y

9.5 - 9 = .5 hours.
It will take a half hour less.[/hide]

hehe

[hide]

450 miles = 9 hours
475 miles = 9.5 hours

9.5-9= .5 hours...or 30 minutes

[/hide]

[hide]$rt = d, \frac{rt}{r}= \frac{d}{r}, t = \frac{d}{r}= \frac{450}{50}= 9+(-\frac{475}{50}) = 9+(-9.5) =-0.5,$[/hide]

[size=75][color=green]Mod edit: Changed latex (use dollar signs instead of [code]) and added hide tags[/color][/size]

[hide="solution"]We an take the difference of the distances, and get 25.  [b]30[/b]  minutes.[/hide]

[quote="anirudh"]A car is averaging 50 miles per hour. If the car maintains this speed, how many minutes less would a 450-mile trip take than a 475-mile trip?[/quote]

[hide] 30 mins [/hide]

what is really weird is were working on something similar to this for my science class.


Can any of u suggest any good Geometry book for the Olympiads?
Thanks
-Chang

There's of course "Geometry Revisited" by Coxeter and Greitzer;
you can also try Ross Honsbergers work :)
And there are online geometry notes by Kiran Kedlaya and Paul Yiu.


Given $ a,b,c>0$ such that $ a\plus{}b\plus{}c\equal{}abc$ , show that :

$ \frac{a^2\plus{}b^2}{(a\plus{}b)^4}\plus{}\frac{b^2\plus{}c^2}{(b\plus{}c)^4}\plus{}\frac{c^2\plus{}a^2}{(c\plus{}a)^4}\le \frac{1}{8}$

It follows from $ \frac {8(a^2 \plus{} b^2)}{(a \plus{} b)^4}\le \frac {1}{ab}$ since summing up 3 of these yields the result.

$ 8(a^3b \plus{} b^3a)\le a^4 \plus{} b^4 \plus{} 4(b^3a \plus{} a^3b) \plus{} 6a^2b^2$

$ 0\le a^4 \plus{} b^4 \minus{} 4(a^3b \plus{} b^3a) \plus{} 6a^2b^2$

$ 0 \le (a \minus{} b)^4$

Q.E.D.


Let $ (O)$ be a circle contact $ AB$ at $ M$ such that $ MA \equal{} MB$.Draw secant line $ ACD$ and $ BEF$ to $ (O)$
The lines $ FC$ and $ DE$ cut $ AB$ at $ K;L$,respectively.Prove that $ KA \equal{} LB$

It is just the inverse of the problem you can find here:

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=288035728&t=181191


THis is another part of a question I was stuck on, while I can probably do it I am a little bit tired and so have decided to put it here so I can learn of a proper method to actually tackle such problems (since what I usually do is just mess around with the numbers until I can get to a solution).

the problem is: [quote]T=(25/w)-(25/(w-800)), make w the subject of the equatio[/quote]
I am sort of annoyed that to solve such a problem up to this date I have been simply playing around with the numbers trying to solve it and would very much appreciate it if someone could give me step by step instructions on how I should tackle such a problem.

thank you all :)

We want to solve for $w$ right? I don't think $T$ is an injective function of $w$, so you won't get $w$ as a single-valued function of $T$.

$T=\frac{25}{w}-\frac{25}{w-800}\Longrightarrow{T}w(w-800)=-20000\Longrightarrow{w}^2-800w+\frac{20000}{T}=0.$ Then $w=\frac{800\pm\sqrt{640000-\frac{80000}{T}}}{2}=\frac{800\pm200\sqrt{16-\frac{2}{T}}}{2}=400\pm100\sqrt{16-\frac{2}{T}}.$ I don't have a calculator handy but you could try graphing these to make sure they look like inverses.

[quote="behemoth571"]Basically just get $ w $ out of the denominator and set it up into something useful like AntonioMainenti's quadratic.[/quote]
yeah I see what he's done, the problem I have with these is always trying too hard to solve it when the question itself is relatively easy.


inequalities
trigonometry
inequalities proposed

Let $x,y,z\in\mathbb{R}$ such that: $E(\alpha)= cos \alpha x+cos \alpha y+cos \alpha z$,$\alpha\in\mathbb{R}$.If $E(1)=E(3)=0$ then$E(2)\leq 0$
  ;)

What if we take $\cos x = 1$, $\cos y = -1$, $\cos z = 0$?