vector
email
induction
inequalities
combinatorics proposed
combinatorics

8.8, 9.8, 11.8
a) 99 boxes contain apples and oranges. Prove that we can choose 50 boxes in such a way that they contain at least half of all apples and half of all oranges.
b) 100 boxes contain apples and oranges. Prove that we can choose 34 boxes in such a way that they contain at least a third of all apples and a third of all oranges.
c) 100 boxes contain apples, oranges and bananas. Prove that we can choose 51 boxes in such a way that they contain at least half of all apples, and half of all oranges and half of all bananas.
([i]I. Bogdanov, G. Chelnokov, E. Kulikov[/i])

Very nice problems.

For a):

Order the boxes according to the number of apples they contain ($b_{99}$ has the most apples and $b_1$ has the fewest). Then take: $b_{99}$, one of $b_{98}, b_{97}$, ..., one of $b_2, b_1$. It is clear that any such set contains at least half of all apples (in the worst case you have chosen $b_{99}, b_{97}, ..., b_1$ which has at least as many apples as $b_{98}, ..., b_2$). Now from each subset of 2 boxes (98-97, ..., 2-1) take the box with more oranges. Then the set also contains at least half of the oranges.

For b):

Order the boxes according to the number of apples they contain ($b_{100}$ has the most apples and $b_1$ has the fewest). Then take: $b_{100}$, one of $b_{99}, b_{98}, b_{97}$, ..., one of $b_3, b_2, b_1$. Now in the worst case we have chosen $b_{100}, b_{97}, ..., b_1$ which has at least as many apples as $b_{99}, b_{96}, ..., b_3$ and $b_{98}, b_{95}, ..., b_2$. So any such set contains at least a third of the apples. Now from each set of 3 boxes (99-98-97, ..., 3-2-1) take the box with the most oranges. This set satisfies the conditions.

well i have some ideas for this kind of problems but i cant finish them....

For every box $i$, identify the vector $v_i=(x,y,z)$  where $x$ is the number of apples on it, $y$ the number of oranges and $z$ the number of the other fruit on box $i$. the geometric interpretation is clear now....

OH,MY ROOMMATE AND I CAN ONLY SOLVE THE FIRST TWO PROBLEM,THE SOLUTION IS SIMILAR TO YOURS
I think the third problem is really hard for me.And I guess it must have a excelllent solution with a giant use of some lemma.Could anyone post the solution or email me:georgew_bush@163.com?THANKS! :D

Does anyone has the official solution of this beautiful problem ?I think some people from Russia sure. So please post it Thank you very much

[quote="Fedor Petrov"]c is very hard. There are many solutions, but none of our students managed to solve it. A hint: prove a lemma: if we have $2k$ boxes, we may partition them to two groups by $k$ boxes in such a way that weight of apples in both groups differ by at most the maximal weight of apples in a single box, and the same for oranges.[/quote]

How? I don't understand

[quote="indybar"][quote="BaBaK Ghalebi"][quote="amir2"]and what about c)  ?[/quote]
prove it with induction
 ;)[/quote]

Can you show?[/quote]
sure ASAP ;)

[quote="BaBaK Ghalebi"]
sure ASAP ;)[/quote]

hmm...
I don't think induction works here.  ;)

[quote="BaBaK Ghalebi"][quote="amir2"]and what about c)  ?[/quote]
prove it with induction
 ;)[/quote]
$2k-1$ box $\& k$
$n=k\to n=k+1$
if there were two boxes such that $a_1>a_2,b_1>b_2$ now by the assumption of the induction we ca n choose $k$ boxes from the $2k-1$ boxes left
now we can add the box that contains $a_1,b_1$
now if there werent such boxes then we have:
$a_1\geq a_2\geq ...\geq a{}_2{}_k_+_1$
$b_1\leq b_2\leq ...\leq b{}_2{}_k_+_1$
now we choose the boxes $1,3,5,...,2n+1$
$\Rightarrow$ $a_1+a_3+...+a{}_2{}_n_+_1\geq a_2+a_4+...+a{}_2{}_n$
 ;)
sorry if I didnt explain it clearly did you get it or should I explain it better and more carefully?? :D

I have no idea what you're trying to say.

What are the $a_i,b_i$ (I'm assuming they're the weights of apples, oranges resp. in the boxes) Then what about the bananas?

Please "explain it better and more carefully"  :D

c) part:

Lemma by [b]Fedor Petrov:[/b]

Any $ 2k$ pairs of positive reals $ (a_i,b_i)$ can be partitioned into $ 2$ groups of $ k$ pairs each, so that if $ x$ is a sum of $ a_i$ in the first group and $ y$ is a sum of $ a_i$ in the second group. Then $ |x - y|\leq\max_{i}a_i$. And the same inequality holds for the $ b_i$, i.e if $ m$ is a sum of $ b_i$ in the first group and $ n$ is a sum of $ b_i$ in the second group, then $ |m - n|\leq \max_i b_i$.

[b]Proof of the lemma[/b]:

The statement clearly holds for the case $ k = 1,2$. So assume that the statement holds for $ k = n$ and let's prove it for $ k = n + 1$.

WLOG assume that $ a_{2n + 2}\geq a_{2n + 1}\geq \max_{i = 1}^{2n}a_i$. Due to the induction assumptation used for the case $ 2n$ the following inequality must hold:

$ |x - y|\leq a_{2n + 1}$.

Consider two cases:

1.$ b_{2n + 2}\geq b_{2n + 1}$

WLOG suppose $ m\geq n$.

Then add the pair $ (a_{2n + 1},b_{2n + 1})$ to the first group, and $ (a_{2n + 2},b_{2n + 2})$ to the second, so

${ |x + a_{2n + 1} - y - a_{2n + 2}}|\leq |x - y| + |a_{2n + 1} - a_{2n + 2}|\leq a_{2n + 1} + a_{2n + 2} - a_{2n + 1} = a_{2n + 2}$

Now suppose that $ m + b_{2n + 1}\geq n + b_{2n + 2}$, then

$ |m + b_{2n + 1} - n - b_{2n + 2}| = m + b_{2n + 1} - n - b_{2n + 2}\leq\max_{i}^{2n + 2}b_i$,

if $ m + b_{2n + 1}\leq n + b_{2n + 2}$ then:

$ |m + b_{2n + 1} - n - b_{2n + 2}| = n + b_{2n + 2} - b_{2n + 1} - m\leq b_{2n + 2} - b_{2n + 1}\leq\max_{i}^{2n + 2}b_i$.

For the case $ b_{2n + 1} > b_{2n + 2}$ do the reverse addition, the rest is completely the same as in the considered case.

So the lemma is proved. $ \blacksquare$

The last step is pretty obvious, take two boxes, one with max number of apples weight and second with max number of oranges weights. So applying the lemma for $ 98$ ,for $ a_i$ take weights of apples and for $ b_i$ weights of oranges, and choose a group with bigger weight of bananas in it. And add both those boxes into this group.

I have a question: for part c, we were able to deal with apples, oranges, and bananas, but is there a way we can generalize this problem so that we can deal with, say, $n$ fruits instead of just $3$? Sorry if the question is vague (and for the massive bump), but I am super curious about this.

Here is what I believe is a different solution to part (c):

[hide=Solution]
First, select the box with the most apples and the box with the most bananas. (In the case these are the same box, select the box and another arbitrary box.) For the other $98$ boxes, let $a_{n+1}$ be the box with the $n$th most apples, and let $b_{n+1}$ be the box with the $n$th most bananas. We can turn any group of boxes among the $98$ boxes into a $98$-term sequence with the following rule: Write out $a_2,a_3,...,a_{99},$ and replace each $a_k$ with a $1$ if the box is selected, and a $-1$ if the box is not selected. Call this an apple sequence; proceed similarly with $b_k$ for a banana sequence.

Now, for any such sequence, define $S_k$ as the sum of the first $k$ terms. We will prove the following lemma:

If an apple/banana sequence of $n$ terms has $S_k$ nonnegative for all $k$ with $0 \leq k \leq n,$ then it must contain at least half of the apples/bananas.

Proof: We claim that in such a sequence, each $1$ can be paired up with a $-1$ later in the sequence. This implies that the sequence contains at least half of the fruits, since in each pair, the selected box contains more of the chosen fruit than the unselected box, meaning that in each pair, at least half of the fruits are selected, so overall at least half are selected.

To prove this, we show that by pairing each $1$ with the first unpaired $-1,$ we will have a desired sequence. Suppose there is some $1$ that ends up paired to an earlier $-1.$ Let it be the $k$th $1$ and the $n$th term of the sequence. Then it must also be the $k$th $-1.$ However, consider $S_{n-1}.$ This contains $k-1$ $1$s, but contains at least $k$ $-1$s, so it is negative, which is a contradiction. Therefore the lemma is proved.

Now consider all apple sequences that have the property such that for any positive integer $k$ from $1$ to $49,$ exactly one of $a_{2k},a_{2k+1}$ is a $1.$ We easily see that all of these sequences have $S_i=-1,0,1,$ since $S_{2i}=0$ for all positive integers $i$ and $S_{2i+1}$ can only be 1 away from $S_{2i}.$ Call these apple sequences good, and similarly define a good banana sequence in the same way.

We will now prove that there is some good apple sequence that can be paired with some good banana sequence. We use the following process to generate one: First, set the first unset $a_i$ to $1,$ and set the corresponding $b_j$ to $1.$ Next, set $b_{j+1}$ to $-1$ if $j$ is even, and set $b_{j-1}$ to $-1$ if $j$ is odd. Then set the corresponding $a_k$ to $-1,$ and similarly set $a_{k+1}$ or $a_{k-1}$ to $1$ when $k$ is even or odd, respectively. Repeat this until it fills in $a_{i+1},$ at which point we repeat the whole process for the next unset $a_i,$ and repeat until all terms are filled in.

Now we show that this process works. First, notice that the steps of going between corresponding terms in the different sequences and going between terms in pairs $a_{2k},a_{2k+1}$ and $b_{2k},b_{2k+1}$ are reversible. This means that following this process, we must eventually return to $a_i,$ and the preceding step must have been at $a_{i+1}.$ Also, notice that if some term was filled out at some point, then the other term in its pair must have also been filled out, either immediately before or immediately after, and note that each "loop" fills in the same number of terms in each sequence. Therefore, this algorithm generates a corresponding good apple sequence and good banana sequence, so a pair must exist.

Now consider the $99$-term sequence consisting of $a_1$ replaced by $1$ followed by a good apple sequence. Since for the good apple sequence, $S_k$ is always $-1,0,1,$ in our new sequence $S_k$ must always be $0,1,2,$ since the first $n+1$ terms of this sequence are $1$ followed by the first $n$ terms of the good apple sequence. Since $0,1,2$ are positive, this sequence contains at least half of the apples, and this means our $99$ term sequence contains $49+1=50$ boxes chosen that have at least half of the apples, since half of the terms in the $98$ term sequence are $1$s. Adding back the $b_1$ box, which is chosen, we get our $100$ boxes have a $51$ element subset with at least half of the apples. Similar logic shows that the same subset has at least half of the bananas, since the two good sequences are the same subset.

Now, consider the $98$-term apple and banana sequences that have a $1$ in a given place if and only if there are $-1$s in the same places in our already chosen good sequences. It is not hard to see that these are also good sequences, so $S_k$ also must be $-1,0,1.$ Now putting these sequences together with $a_1$ and $b_1$ gives us another $51$ element subset with at least half of the apples and bananas.

Now suppose neither of these two subsets contains at least half of the oranges. Consider the $49$-element subset that contains a box if and only if our first $51$-element subset doesn't contain it, and consider the corresponding $49$-element subset for our second $51$-element subset. Both of these must contain more than half of the oranges, but notice that they are defined such that their intersection is the null set. Therefore their union is just the $98$ boxes excluding $a_1,b_1.$ However, they contain more than the whole of the oranges in the $100$ boxes, which is a contradiction.

Therefore, there is some set of $51$ boxes containing at least half of each fruit.
[/hide]

Credit goes to one of my friends (not sure if he wants to be named):

Place the boxes in 3d space so that no four are coplanar. By the ham sandwich theorem, there is a plane which bisects the apples (treated as a single object) and does the same for the oranges and bananas. Choose the side of the plane with the least number of boxes. The worst case is when 3 boxes intersect the plane and there are 48 on one of the sides and 49 on the other. Thus you select at most 51 boxes. You have at least half of each type of fruit in this collection of boxes.


A half is a third of it. What is it?

[i]from book "The Moscow puzzles" by Boris A. Kordemsky[/i]
[u]Please hide Your solution . Write inside Hide ... /Hide[/u]

[hide="Solution"]

$ 1\frac{1}{2}$

[/hide]

[color=red][size=200][hide]1/2=1/3 X
this give us     X=3/2[/hide][/size][/color]

[hide][color=violet]1/2=1/3 x
x=3/2 or 1 1/2[/color][/hide]

Seriously guys, no need to post the exact same solution over and over again.


advanced fields
advanced fields unsolved

If A is not an empty finite set, show that Seq(A), the set of all finite sequences of elements of A, is countable.

huh?

if $ |A| \equal{} 1$, $ Seq(A)$ has only one element and is countable. but if $ |A| > 1$, clearly $ Seq(A)$ is uncountable (the power set of $ \mathbb{N}$ may be embedded into $ Seq(A)$).

No, the power set cannot be embedded. In fact, we have an embedding of $ Seq(A)$ into $ \mathbb N$ if we have $ A \equal{} \{0,1,...,n \minus{} 1\}$ and map the sequence $ a_0,...,a_m$ to $ a_0 \plus{} a_1 \cdot n \plus{} ... \plus{} a_m \cdot n^m$. All other cases reduce to this.

sorry, I did not read *finite* sequences ...


geometry
rectangle
trigonometry
incenter
angle bisector

[quote="fishythefish"]For the second one, yes, quite obviously. Just look at a square.[/quote]

I require a proof.

Perhaps you should clarify that second question -- are you asking whether all quadrilaterals have an incircle, or if there is any single one that has one?  I suspect you are asking the former, while fishy is answering the latter.

While all triangles have inscribed circles, and so do regular polygons, but most others don't.

I can disprove that every quadrilateral has an inscribed circle with an example.

A rectangle with width 1, and length 2. Because the inscribed circle must be tangent to all of the sides, the greatest length the diameter could be is 1, otherwise it would intersect the length twice on each side. But then the circle cannot meet the other two sides. Case closed.

(I answered both questions now. :P )

[quote="ThetaPi"]Prove that the angle bisectors of the interior angles of any triangles are concurrent at a single point.

Does any quadrilateral have an inscribed circle?[/quote]

number 2 is trivial by trig cevas

[quote="JavaMan"]For 1 you could use the [url=http://en.wikipedia.org/wiki/Angle_bisector_theorem]Angle bisector theorem[/url] together with [url=http://en.wikipedia.org/wiki/Ceva%27s_theorem]Ceva's Theorem[/url].  :)[/quote]

Prove Ceva's Theorem. I didn't know there existed such a theorem.

EDIT: Oh. Wikipedia has a proof. Never mind then.

Ceva's Theorem should be viewed as a significant generalization of this result.  The fact that the angle bisectors intersect at a unique point follows directly from the uniqueness of the incircle, so what you're asking for is a proof that the incircle is unique.

[hide="Proof"] The feet of the perpendiculars from an incenter of a triangle divides up its side lengths into quantities $ x, y, z$ such that the side lengths are $ x \plus{} y, y \plus{} z, z \plus{} x$.  The three values $ x, y, z$ are uniquely determined from the side lengths because

$ y \equal{} \frac {(x \plus{} y) \plus{} (y \plus{} z) \minus{} (z \minus{} x)}{2}$

and cyclic permutations.  Therefore, given a triangle, the feet of the perpendiculars from any of its incenters is uniquely determined.  The intersection of the perpendiculars themselves then uniquely determines the incenter. [/hide]
Perhaps now you'd like a proof that the incircle [i]exists[/i]?

Proofs of common results are easily available online; that's what Google is for.

Edit:  The above argument generalizes to quadrilaterals with an inscribed circle (I forget what the term is).  Specifically, there exist positive reals such that the side lengths are $ x \plus{} y, y \plus{} z, z \plus{} w, w \plus{} x$.  Unlike the case of triangles, this system is not solvable in general because $ (x \plus{} y) \plus{} (z \plus{} w) \equal{} (y \plus{} z) \plus{} (w \plus{} x)$; in other words, opposite sides in a quadrilateral with an inscribed circle have the same sum.  This characterizes such quadrilaterals.


ratio
absolute value
calculus
calculus computations

[quote="keta"]Can anyone explain the following problem?

Determine the radius of convergence and the interval of convergence of the following series:

(Sigma from n=0 to infinity.) {nx n}/{(4  n )(n 2 +1)}.

I don't understand which convergent test to use for this problem.

Thx.[/quote]

If I understood well your problem, you mean $f(x)=\sum_{n=0}^{\infty}\frac{nx^n}{4^n(n^2+1)}$. In this case, I believe the regular n-th root test, or ratio test should work. Either way, If I am not mistaken, the radius of convergency is 4, and the interval is centred at 0, if the problem is the one I wrote.

Best,

The two tests djimenez mentioned - the root test and the ratio test - are nothing more than somewhat automated comparisons to geometric series. A power series always has an $x^n$ (or $(x-x_0)^n$) so it always has something geometric - and that means the root and ratio tests are always in order, with the choice between them a matter of convenience.

But the root or ratio test, while it should tell you the radius of convergence, is helpless at telling you whether the series converges on the boundary of the interval of convergence. In this case, the radius of convergence is 4. If $x=\pm4$, then the absolute value of the terms is $\frac{n}{n^2+1}$, which is too large for absolute convergence. The series diverges when $x=4$ (terms all positive) but converges by the alternating series test when $x=-4.$ The interval of convergence is $[-4,4).$


with a number like -67, are the digits 6 and 7, -6 and -7, or -6 and 7?

6 and 7

in most competitions, digits r defined to be a 1 digit positive integers, i think :wink:

I would think -60 and -7... Place value ftw.

-67 = -(60+7)
-67 = -60 + (-7)

Unless your question is different from what it seems to be to me.


Let $a,b,c>0$ such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2$.Prove that:

                     $\frac{bc}{a(b+c)}+\frac{ca}{b(c+a)}+\frac{ab}{c(a+b)}\geq\frac{ab+bc+ca}{a+b+c}$ ;)

After homogenizing, we need only show that

$2\sum \frac {bc}{a(b+c)}\cdot (abc(a+b+c))\geq (ab+bc+ca)^2$

Write the LHS as

$\left(\sum\frac {2bc}{a(b+c)}\right)\left(\sum \frac {abc(b+c)}{2}\right)$

By Cauchy, this is $\geq $

$\left(\sum \sqrt{\frac {2bc\cdot abc\cdot (b+c)}{2a(b+c)}}\right)^2=\left(\sum bc\right)^2$

as desired.

Note that
\[
ab+bc+ca=abc\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=2abc
\]
Upon divide the bothsides of the inequality by $abc$ we obtain an equivalent inequality
\[
\frac{1}{a^2(b+c)}+\frac{1}{b^2(c+a)}+\frac{1}{c^2(a+b)}\geq \frac{2}{a+b+c}
\]

Now
\[
LHS=\frac{1/a^2}{b+c}+\frac{1/b^2}{a+c}+\frac{1/c^2}{a+b}\geq \frac{\left(1/a+1/b+1/c\right)^2}{2(a+b+c)}=\frac{2}{a+b+c}
\]
and we are done.


Using Cauchy inequlity directly , by the way, n should > 1.

$(n-1)(S_2-a^2_k) \geq (S_1-a_k)^2$

So $\frac{S_2-a^2_k}{S_1-a_k} \geq \frac{S_1-a_k}{n-1}$

Adding n such inequalies, the result follows.

thanks zhaoli for rewrite this exercuce with latex  :bye:


A refrigerator is offered at sale at $ \$ 250.00$ less successive discounts of $ 20\%$ and $ 15\%$. The sale price of the refrigerator is:
$ \textbf{(A)}\ 35\% \text{ less than } \$250.00 \qquad\textbf{(B)}\ 65\% \text{ of } \$250.00 \qquad\textbf{(C)}\ 77\% \text{ of } \$250.00 \qquad\textbf{(D)}\  68\% \text{ of } \$250.00 \qquad\textbf{(E)}\ \text{none of these}$

A discount of $ 20\%$ represents bringing the price down to $ \frac{4}{5}$ of the original value. Likewise, a discount of $ 15\%$ lowers the price to $ \frac{17}{20}$ of the original value. Multiplying the two together, we get $ \frac{4}{5}(\frac{17}{20})$, making $ \frac{17}{25}$, or $ 68\%$ of the original price. Thus, the answer is $ \fbox{D}$.

[hide="Similar..."]
Let $ D_1$ and $ D_2$ be the successive discounts, and $ D$ be the final discount.  Then we have $ D\equal{}D_1\plus{}D_2\minus{}D_1D_2 \implies D\equal{}20\plus{}15\minus{}3\equal{}32$.  Thus, the sale price, $ S$, equals $ 250\minus{}(.32)(250)\equal{}(.68)(250)$, which is $ 68 \%$ of $ 250$, or $ \boxed{\textbf{(D)}}$.
[/hide]


Pascal\u0027s Triangle
algebra
binomial theorem

Find the sum of the coefficients in the expansion of $(4x+2y)^{3}$.

[hide]$(4x+2y)^{3}=64x^{3}+96x^{2}y+48xy^{2}+8y^{3}$

$64+96+48+8=\boxed{216}$

I'm not sure, but can the sum of the coefficients of $(ax+by)^{z}$ be expressed as $(a+b)^{z}?$[/hide]

to help solve future problems like the one u posed u remembe rthe following definition:

coefficiant- $(ax+by)$ $a$ and $b$ are the coefficiants and $x$ and $y$ are the variables

[hide="Hint"]How to reduce the monomial $cx^{m}y^{n}$ to the coefficient $c$? Put $x=1, y=1$[/hide]

[hide="Answer"]Sum of the coefficients of $(4x+2y)^{3}$ is $(4+2)^{3}=216$[/hide]

Yep!  :)
My math teacher tends to give problems like this a lot, and finding simple shortcuts makes math homework go that much faster. All it is is a matter of informal proof, just for your own use.

I have to stuff that in my memory somehow. Do you know the proof? It must be pretty cool too...

It's not rigorous. I just proved it for a few simple cases, stuff we get in math class. I don't know the rigorous proof. I'm assuming it's fairly straightforward.

[b][u]Another example :

For $(2x+y)^{n}\equiv A_{0}x^{n}+A_{1}x^{n-1}y+\ldots+A_{n-1}xy^{n-1}+A_{n}y^{n}$ ascertain the sums : $\{\begin{array}{c}1.\blacktriangleright\ A_{0}-A_{1}+A_{2}-\ldots\ .\\\\ 2.\blacktriangleright\ A_{0}+A_{2}+A_{4}+\ldots\ .\\\\ 3.\blacktriangleright\ A_{1}+A_{3}+A_{5}+\ldots\ .\\\\ \ldots\ldots\mathrm{\ a.s.o.\ }\ldots\ldots\ .\end{array}$

Whenever you hear sum of coefficients think f(1)  :D

[hide="quick solution"]By Pascal's Triangle:

[code]   1
  11
 121
1331[/code]

the expansion for $\left(x+y\right)^{3}=x^{3}+3x^{2}y+3xy^{2}+y^{3}$. We plug in 4x and 2y:

$\left(4x+2y\right)^{3}=64x^{3}+96x^{2}y+48xy^{2}+8y^{3}$

We sum those up:

$64+96+48+8=160+56=\boxed{216}$[/hide]

Very clever, Farenhajt.

Yup, just imagine expanding something like $(231x-384y+154z)^{457}$ in order to find the sum of the coefficients.

Ooooo I know this , plug in x=1, y=1, and z=1, and we get 1^457=1

[hide]Let x=y=1 and the expansion will leave the sum of the coefficients.  $(4x1+2x1)^{3}=6^{3}=216$.[/hide]

[quote="i_like_pie"]
I'm not sure, but can the sum of the coefficients of $(ax+by)^{z}$ be expressed as $(a+b)^{z}?$[/quote]

By Newton's Binomial Theorem, we have $(a+b)^{n}=\sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^{k}$.
Applying this to our case, we get $(ax+by)^{n}=\sum_{k=0}^{n}\binom{n}{k}(ax)^{n-k}(by)^{k}$, whiche can we rewritten as $(a+b)^{n}=\sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^{k}x^{n-k}y^{k}$.
Now it's evident that the sum of the coefficients we're looking at is $\sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^{k}$, but that's $(a+b)^{n}$! This demonstrates that it suffices to assume $x,y=1$ in the original binomial to calculate the sum of the coefficients of the resultant polynomial...
Hope I've been clear!  :wink:

The Binomial Theorem is unnecessary, since the result generalizes to any polynomial.

Let $P(x) = a_{0}+a_{1}x+...+a_{n}x^{n}$ be any polynomial.  Then the sum of its coefficients is

$a_{0}+a_{1}+...+a_{n}$

Which is, of course, $P(1)$. This result generalizes to an arbitrary number of variables.


function
real analysis
real analysis solved

I have no idea whether this is true or not. Can someone give me some references?

Let A and B be 2 sets. If we can define injections from A to B and from B to A, then can we define a bijection from A to B?

A related question:

If we can define a bijection from A to a subset of A, A', then can we define a bijection from A to any subset of A which contains B?

The first question is called Canthor's Theorem, and yes it is true :).

Is there a solution I could understand? Maybe it's easier than I think? Or not?

Never mind, I found about it on the Internet. It's caleed th Cantor-Bernstein-Schroeder Theorem, and there was a short proof there:

f:A->B and g:B->A are injections.

C0=A\g[B] and Cn+1=g[f[Cn]] for all n>=0. C=the reunion of all Cn with n from 0 to 00. It's easy to show that h:A->B, h(x)=f(x) if x\in C and g-1(x) if x not \in C is a bijection.

Sorry for the repeated posts, especially since they answer my own question :D. The second question is also true. Let B be a subset of A such tht B contains A'. The restriction of f (the bijection from A to A') to B is an injectije function :B->A'. The identical function is an injective function :A'->B. By using question 1, we get that A' and B have the same cardinality, and since the same can be said about A' and A, it means that A and B have the same cardinality.

Don't worry Grobber, it is never worthless to talk about Cantor-Bernstein' theorem. It is useful one.

Pierre.

Well, then, here's just another observation (:D):

(1)<=>(2) ((1) and (2) are the 2 questions I asked in my first post in this topic). We saw that (1)=>(2). To show that (2)=>(1):

Let f:A->B and g:B->A be injections. Then A is in bijection with g(f(A)) and B is in bijection with g(B). It's clear that g(B) contains g(f(A)) because B contains f(A). From (2) we get that A is in bijection with g(B) because, as I said, g(B) contains g(f(A)), so both A and B are in bijection with g(B), so they're in bijection with each other.


function
combinatorics proposed
combinatorics

assume that k,n are two positive  integer $k\leq n$count the number of permutation $\{\ 1,\dots ,n\}\ $ st for any $1\leq i,j\leq k$and any positive integer m we have $f^m(i)\neq j$ ($f^m$ meas iterarte function,)

I suppose you mean for all $i\ne j$, since there's always some $m$ s.t. $f^m(i)=i$, ($i$ is always part of a cycle) right?


Let $ a,b>1$ . Solve the equation: $ (a^3\plus{}b)^x\minus{}a^{3x}\equal{}3a^xb(a^x\plus{}b)\plus{}b^3$ .

[hide="Solution"]Note that $ (a^x \plus{} b)^3 \equal{} a^{3x} \plus{} 3a^xb(a^x \plus{} b) \plus{} b^3$

So we can rewrite it as:

$ (a^3 \plus{} b)^x \equal{} (a^x \plus{} b)^3$

So we have $ x \equal{} 3$ for all $ a,b$[/hide]


If $ a, b,c$ are positive real numbers, then

\[ 3(a^2b^2\plus{}b^2c^2\plus{}c^2a^2)(a^2\plus{}b^2\plus{}c^2)\ge (a^2\plus{}ab\plus{}b^2)(b^2\plus{}bc\plus{}c^2)(c^2\plus{}ca\plus{}a^2)\]

Do you want us to prove that?

(btw, it isn't high school basics level.)

I`m sorry, I didn`t know where to post it. I thought posting it here will be ok. Anyway, a moderator can move this post.

[hide="Brute Force"]
Multiply out, every term is of 6th degree, so AM-GM solves it (and maybe with a combination with Schur) [/hide]


inequalities
algebra
polynomial
inequalities proposed

Prove that with$a,b,c>0$ and$p\ge3+\sqrt{7}$we have
$\frac{1}{pa^2+bc}+\frac{1}{pb^2+ca}+\frac{1}{pc^2+ab}\ge \frac{9}{(p+1)(ab+bc+ca)}$
 :roll:
I think my solution is right.

[quote="hungkhtn"]Hi. It's not difficult. We only need to check if $a=b$. Use some computing. But maybe you shouldn't post it in new topic. Try to seach it in this forum.[/quote]

it can be solved by sum of squares again?
Just wonder, are there any general methods that can be used to write an inequality into sum of squares?

[quote="hungkhtn"]Hi. It's not difficult. We only need to check if $a=b$. Use some computing. But maybe you shouldn't post it in new topic. Try to seach it in this forum.[/quote]
I think it is not easy.  :lol: I wait somebody post a solution.

Oh,It is a interesting problem.If you said that,can you display your proof?

I haven't a nice solution.
I proved it by expanding way, which is laborious enough.

[quote="hungkhtn"]Oh. A big mistake when I saw this problem. I didn't remember and I thought that:

$\frac{1}{a^2+kbc}+\frac{1}{b^2+kac}+\frac{1}{c^2+kab}\ge \frac{9}{(k+1)(ab+bc+ca)}$

And... I don't know why $k\ge 3+\sqrt{3}$? Afterthat, I see that $k\le {\frac{3}-\sqrt{7}}{2}$. But two this one is the same. I use the method sum of square.[/quote]
I can display the most important in my solution :
Suppose that$a \le b \le c$we have$x=\frac{a}{b},y=\frac{b}{c}$and define$z=\frac{c}{a}$if you want,change our ineq.($z=\frac{1}{xy}$)
$t\in(0,1]: 1+x+xy=1+t+t^2$fixed $t$and$x=x(y)=\frac{t(1+t)}{1+y}$
Check it for me! :roll:

Indeed, the ineq $(a-b)^2(b-c)^2(c-a)^2 \ge 0$ is the key. :lol:

mumath's solution is long and it doesn't solve the following ineq:
find all p>0;
$\sum \frac{1}{\sqrt{pa^2+bc}} \ge \frac{3\sqrt{3}}{\sqrt{(p+1)(ab+bc+ca)}}$

[quote="nthd"]I can display the most important in my solution :
Suppose that$a \le b \le c$we have$x=\frac{a}{b},y=\frac{b}{c}$and define$z=\frac{c}{a}$if you want,change our ineq.($z=\frac{1}{xy}$)
$t\in(0,1]: 1+x+xy=1+t+t^2$fixed $t$and$x=x(y)=\frac{t(1+t)}{1+y}$[/quote]
I don't understand the end of your solution. Can you give us some additional details ?

[quote="Vasc"]Let $ a,b,c > 0$ and $ p\ge 3 \plus{} \sqrt {7}.$ Prove that

$ \frac {p \plus{} 1}{pa^2 \plus{} bc} \plus{} \frac {p \plus{} 1}{pb^2 \plus{} ca} \plus{} \frac {p \plus{} 1}{pc^2 \plus{} ab}\ge \frac {9}{bc \plus{} ca \plus{} ab}.$

I proved it by expanding way, which is laborious enough.[/quote]$ \sum{\frac {p \plus{} 1}{pa^2 \plus{} bc}} \minus{} \frac {9}{bc \plus{} ca \plus{} ab}\equiv\frac {F(a,b,c,p)}{(pa^2 \plus{} bc)(pb^2 \plus{} ca)(pc^2 \plus{} ab)(bc \plus{} ca \plus{} ab)}.$

$ F(a,b,c,p) \equal{} F(a,a \plus{} s,a \plus{} t,k \plus{} 5) \equal{} (k \plus{} 3)(k \plus{} 6)(5k \plus{} 24)\left(s^2 \plus{} t^2 \minus{} st\right)x^4$

$ \plus{} (s \plus{} t)\left[2\left(2k^3 \plus{} 32k^2 \plus{} 162k \plus{} 261\right)(s \minus{} t)^2 \plus{} \left(8k^3 \plus{} 106k^2 \plus{} 450k \plus{} 603\right)st\right]x^3$

$ \plus{} \left\{(s \minus{} t)^2\left[(k \plus{} 5)(4k \plus{} 15)\left(s^2 \plus{} t^2\right) \plus{} 3\left(2k^3 \plus{} 32k^2 \plus{} 162k \plus{} 261\right)st\right] \plus{} \left(18k^3 \plus{} 226k^2 \plus{} 899k \plus{} 1101\right)s^2t^2\right\}x^2$

$ \plus{} st(s \plus{} t)\left[(k \plus{} 5)(4k \plus{} 15)\left(s^2 \plus{} t^2\right) \plus{} \left(4k^3 \plus{} 38k^2 \plus{} 91k \plus{} 6\right)st\right]x$

$ \plus{} (k \plus{} 5)s^2t^2\left[(k \plus{} 6)(s \minus{} t)^2 \plus{} \left(k \plus{} 2 \minus{} \sqrt {7}\right)\left(k \plus{} 2 \plus{} \sqrt {7}\right)st\right]\geq0,$

which is clearly true for $ a \equal{} \min\{a,b,c\}$ and $ k\geq \sqrt {7} \minus{} 2 \equal{} 0.64575\cdots.$
[quote="nthd"]Find all $ p$ such that

$ \sqrt {\frac {p \plus{} 1}{pa^2 \plus{} bc}} \plus{} \sqrt {\frac {p \plus{} 1}{pb^2 \plus{} ca}} \plus{} \sqrt {\frac {p \plus{} 1}{pc^2 \plus{} ab}} \ge 3\sqrt\frac { 3}{bc \plus{} ca \plus{} ab}.$[/quote]It holds for positive $ a,b,c$ if and only if $ p \equal{} 0\vee p\geq 8.4971\cdots$, which is a root of the following irreducible polynomial

$ p^4 \minus{} 60p^3 \plus{} 460p^2 \minus{} 192p \plus{} 16.$


calculate the work done to locate a 150kg satellite into a geostationary orbit.

I'm guessing i need to use ep=- GMmG/r
but it would be possitive right? oR negative

the radius will be 35800 km? or 42,245 cause the 35800 is the altitude for the satellite?????
m and M is mass of earth and the satellite
so is this how to work out the right answer

6.67*10^-11 (mass of eath)(150)
-----------------------------------------
35800000 or 42,245 000
Is this the correct way.

PLease help

Also is this the right forumla to figure the orbital radius for a geostationary satellite for any planet?
r =  Gm    ^1/3
   ---
    4 (pie)^2

Beware, work is the CHANGE of energy. And yes, there is a minus in the formula. Then,

$W = (-\frac{GMm}{r}) - (-\frac{GMm}{R_0})$

where $R_0$ is the starting radius (probably radius of the Earth) and $r$ is the terminating radius (i.e. the radius of the geostationary orbit), gives a positive value.

Keep in mind also that a satelite in orbit has a velocity, and hence has kinetic energy.

Yes, the potential energy looks right. Don't forget the kinetic energy though.

So i will have 2 answers, one for Potential energy and one for knitic energy.
The forumlua for kenitic energy is .5 mv^2
right. How would i calculate the velocity. Thanks

You can use a fairly well-known formula

$a=\frac{v^2}{r}$

where $a$ must be equal to the gravitational acceleration (in the proper height).


linear algebra
matrix
linear algebra unsolved

Consider $ A,B,C \in M_n(\mathbb{C})$ such that $ A \cdot B \cdot C\equal{}I_n$. Prove that if $ I_n\plus{}A\plus{}AB$, $ I_n\plus{}B\plus{}BC$, $ I_n\plus{}C\plus{}CA$ are reversible, then the sum of their reversed matrix is $ I_n$.


Find a positive integer that is $ 1.5$ more than its reciprocal.

Setting the number as x, we get the equation $ x \equal{} \frac1x \plus{} \frac32$
$ 2x^2 \minus{} 3x \minus{} 2 \equal{} 0$
$ x \equal{} \minus{}\frac12$ or $ 2$
We want a positive integer:
$ \boxed2$


Show that for any $i=1,2,3$, there exist infinity many positive integer $n$, such that among $n$, $n+2$ and $n+28$, there are exactly $i$ terms that can be expressed as the sum of the cubes of three positive integers.

Our "Maths discussion group" seems solve the case of $i=1,3$:
First,a lemma,$a^{3}+b^{3}+c^{3}$ can't be $4,5(mod 9)$
Brief proof:$x^{3}=0,1,-1 (mod9)$ and try all case of $a,b,c$ in mod 9 and result is out.
For $i=1$,take $n=(9a+1)^{3}+(9b+1)^{3}+(9c+1)^{3}=3 (mod9)$ then
$n+2=5(mod9)$ and $n+28=4(mod9)$,both can't be sum of 3 cubes.
For $i=3$, take $n=k^{3}+(6k)^{3}+(8k)^{3}$, then $n+2=(9k)^{3}+1+1$ and $n+28=(9k)^{3}+27+1$
Also, one of our "group member" think that if we can prove there are infinite $d$ such that $d^{3}=a^{3}+b^{3}+c^{3}$ have no integral solution, then the case of $i=2$ is also solved

ychjae, i=2 is solved.
Observed that 5^3+5^3 = 6^3 + 2^3 + 26
Therefore, we can take
n+28 = (3k-1)^3 +5^3+5^3
n+2 = (3k-1)^3 + 6^3 + 2^3
n =(3k-1)^3 + 6^3 + 2^3 -2=(3k-1)^3 + 222 which can't be the sum of three cubes by the lemma.

By the way, i think Rust's solution is wrong.

For $i = 1$ take $n = 1^3+1^3+(3k+1)^3$ for any $k \in \mathbb{N}.$ Then, using the fact that any sum of three cubes is congruent to $\pm 3, \pm 2, \pm 1, 0$ (mod $9$), we know that $n+2, n+28$ are not sums of three cubes.

For $i = 2$ take $n = (3k+1)^3 + 10^3 + 18^3$ for any $k \in \mathbb{N}.$ In this way, $n+28 = (3k+1)^3 + 1^3+19^3$ and $n+2 \equiv 4$ (mod $9$), so that $n, n+28$ are sums of three cubes whereas $n+2$ is not from the earlier fact.

For $i = 3$ take $n = 216k^3.$ In this way, $n = (3k)^3 + (4k)^3 + (5k)^3, n+2 = (6k)^3+1^3+1^3, n+28 = (6k)^3+1^3+3^3.$
$\square$


Does the product of all terms in the infinite geometric sequence with starting term 9 and ratio 2/3 converge?

Hmm...

$ \lim_{k\to\infty}9\cdot\left(\frac{2}{3}\right)^k\equal{}0$, so I'd say the terms eventually become infinitesimally small, and the product simply becomes 0.

oh woops i was kinda of thinking of something else. Can we show that \[ \frac 32 \times \frac 54 \times \frac 76 \times frac 98 \times \cdots \times \frac{2n \plus{}1}{2n}\] converges? thanks

No.  $ \prod_{i\equal{}1}^{n} \left( 1 \plus{} \frac{1}{2i} \right) > \sum_{i\equal{}1}^{n} \frac{1}{2i}$, which diverges.

$ \left( \frac 32 \times \frac 54 \times \frac 76 \times \frac 98 \times \cdots \times \frac{2n \plus{}1}{2n} \right)^2$
$ > \left( \frac 32 \times \frac 54 \times \frac 76 \times \frac 98 \times \cdots \times \frac{2n \plus{}1}{2n} \right) \left( \frac 43 \times \frac 65 \times \frac 87 \times \cdots \times \frac{2n \plus{}2}{2n\plus{}1} \right)$
$ \equal{} \frac 32 \times \frac 43 \times \frac 54 \times \cdots \times \frac{2n\plus{}2}{2n\plus{}1}$
$ \equal{} n\plus{}1$

Thus the original sequence of products is greater than the sequence $ \sqrt{n\plus{}1}$, which diverges.

[quote="t0rajir0u"]No.  $ \prod_{i \equal{} 1}^{n} \left( 1 \plus{} \frac {1}{2i} \right) > \sum_{i \equal{} 1}^{n} \frac {1}{2i}$, which diverges.[/quote]
Hmm, how do we prove this inequality? I initially tried to rewrite the lhs as e to the sum of $ \ln(1\plus{}1/2i)$ but it didn't help much...

If you think about how it (the LHS) expands,you can see that the RHS is the LHS's terms where you pair off each 1/2i with infinitely other 1's

I proved it with induction, just to be silly.

[hide="Blah"][b]Base cases[/b]:
$ n\equal{}1$: $ 1\plus{}1/2>1/2$.
$ n\equal{}2$: $ (1\plus{}1/2)(1\plus{}1/4)>\frac{1}{2}\plus{}\frac{1}{4}$.
$ n\equal{}3$: $ (1\plus{}1/2)(1\plus{}1/4)(1\plus{}1/8)>\frac{1}{2}\plus{}\frac{1}{4}\plus{}\frac{1}{6}$.

[b]Inductive step[/b]: Assume it holds for $ n\equal{}m$ for some $ m>3$. That is, $ \prod_{i\equal{}1}^{m}\left(1\plus{}\frac{1}{2i}\right)>\sum_{i\equal{}1}^{m}\frac{1}{2i}$.

Multiply both sides by $ 1\plus{}\frac{1}{2(m\plus{}1)}$ to get:

$ \prod_{i\equal{}1}^{m\plus{}1}\left(1\plus{}\frac{1}{2i}\right)>\left(1\plus{}\frac{1}{2(m\plus{}1)}\right)\sum_{i\equal{}1}^{m}\frac{1}{2i}$.

Therefore it suffices to prove that $ \left(1\plus{}\frac{1}{2(m\plus{}1)}\right)\sum_{i\equal{}1}^{m}\frac{1}{2i}\geq \sum_{i\equal{}1}^{m\plus{}1}\frac{1}{2i}$.

Subtract $ \sum_{i\equal{}1}^{m}\frac{1}{2i}$ from both sides and factor:

$ \frac{1}{2(m\plus{}1)}\sum_{i\equal{}1}^{m}\frac{1}{2i}\geq \frac{1}{2(m\plus{}1)}$.

Divide:

$ \sum_{i\equal{}1}^{m}\frac{1}{2i}\geq 1,$

which is true for any $ m>3$.

Therefore, $ \prod_{i\equal{}1}^{m\plus{}1}\left(1\plus{}\frac{1}{2i}\right)>\left(1\plus{}\frac{1}{2(m\plus{}1)}\right)\sum_{i\equal{}1}^{m}\frac{1}{2i}\geq \sum_{i\equal{}1}^{m\plus{}1}\frac{1}{2i}$, which completes the induction.[/hide]


$\triangle ABC$ is an isosceles triangle with $AB=AC$. If the angle bisector of $\angle B$ hits $AC$ at $D$ and $BD+AD=BC$, find $\angle DBC$.

trig solution[hide]
Let $\angle DBC=x$ rewrite $BD+AD=BC$ as $1+\frac{AD}{BD}=\frac{BC}{BD}$ using sine theoreme $1+\frac{sinx}{sin4x}=\frac{\sin3x}{\sin2x}$ this becomes $sin 4x +sinx = 2\sin3x\cos2x$ or $\sin5x=sin4x$ so $x=20$ [/hide]

That's correct, but can you explain your solution a bit more thoroughly? Thanks :)

Any other solution?

Hope this is more clear now :D
[hide]Let  $\angle DBC=x$ then $\angle DCB=2x$ and $\angle BAC=180-4x$ from  $BD+AD=BC$ we obtain
$1+\frac{AD}{BD}=\frac{BC}{BD}$. Using sine theoreme in $\Delta ADB$ we have $\frac{AD}{BD}=\frac{sinx}{sin(180-x)}=\frac{sinx}{sin4x}$ and from sine theoreme in $\Delta BDC$ we get $\frac{BC}{BD}=\frac{sin(180-3x)}{sin2x}=\frac{sin3x}{sin2x}$
The rest is just simple trigonometry. Hope i made it more clear this time[/hide]


Is it possbile to re-write the following expression:

$\frac{1}{a \cdot (a+z)}$

  I know this is a very general or rather vague question but to say anything more would immediately reveal the solution as obvious, not to mention rob the reader of the enjoyment of solving the problem without aid.

  O' and $\frac{1}{a^2 +az}$ is not a valid answer, as it is only simplifying the expression, not re-writing it.

you could rightit as $\frac{\frac1a-\frac1{a+z}}z$ which is $\frac1{az}-\frac1{az+z^2}$

is that what you're looking for?

[quote="cj"]I think he is looking for this form:

 $\frac1{z} (\frac1{a}-\frac1{a+z})$

(very simple partial fraction decomposition)[/quote]
look at the previous post (exactly what I did)

Another possibility is $\frac{1}{a}\cdot \frac{1}{a+z}$, but that seems a little too obvious. I would guess that your answers are correct, although there could be other ways.

Yes, nat mc's solution was what I was initally thinking of, but as for the existence of other solutions I am completely clueless.


topology
advanced fields
advanced fields unsolved

Define a topology T,on the set of real numbers,R,as follows:T contains the empty set,and a nonempty subset O of R is in T if and only if,the complement of O,is a finite set.Which of the following statements is/are true?
[size=150]1.Every infinite subset of R is closed
2.Every subset of R is compact.
3.R is Hausdorff[/size]

1. Definitely not. $ I\equal{}(\minus{}\infty, 0)\cup (0,\infty)$ is infinite and open since the complement is $ \{0\}$ a finite set. In fact, the only closed sets are finite ones.

2. The space itself is compact, so this should be true.

3. It is not Hausdorff.

For the second question: let $ X$ be a subset of $ \mathbb{R}$ and $ \{U_{\alpha}\}_{\alpha\in A}$ an open cover. If $ \alpha$ is any index, then $ U_{\alpha}^{c}$ is a finite set. Since it is a finite set, $ U_{\alpha}^{c}\cap X$ is finite, and it is clear that there are finitely many sets in the open cover, $ U_{\alpha_{1}},\ldots,U_{\alpha_{n}}$ which cover this set (for each point in $ U_{\alpha}^{c}\cap X$, just choose an element of the open cover containing that point). But then $ \{U_{\alpha},U_{\alpha_{1}},\ldots,U_{\alpha_{n}}\}$ is a finite subcover of $ X$. In other words, $ X$ is compact.


Hey...Ok so i just read like all 11 pages of the Mole game,  I would have applied but i didnt see it fast enough, so im sure that there are others like me who have seen the post and wish that they could be in a game.  I would be willing to start a mole game 2 if people would be interested.  Please pm me or post below if you would be interested in participating!!!

@perfect--im not trying to copy but i want to offer the game to players that might have been rejected the first time around--thanks

There will be another Mole game after the first one is done.

I'm currently working on tasks for the second one...I've got some good ones for next time.


Consider the pattern

1,9,36,100,.....

Find the 15th number in the sequence.

[hide] The sequence adds the next cubic number starting from 2^3.

Therefore, the 15th term would be 1+2^3+3^3+4^3...+15^3.

Now lets just add them up with the nifty built in calculator on my computer and get a total of 14400.

Hope i'm right :lol:
[/hide]

[hide]Yes you are, but there's a slicker solution that involves triangular #s (you don't have to think about a calculator)(I asked this question at lunch today)[/hide]

Yeah, you got the right answer, but here's a much simpler method.
[hide="simpler solution"]
It's just the square of all the triangular numbers. $1^{2}. (1+2)^{2}=9. (1+2+3)^{2}=36.$ ETC
So $(1+2+3...15)^{2}=(\frac{(15)(16)}{2})^{2}=(120)^{2}=14400.$[/hide]

Alan we didn't have the same lunches today  :|

[hide]If you want to talk about lunch periods, click on the link.

(Sorry, link non-existent, just go to my blog and post in the triangle theorems thing)[/hide]


I am having major difficulties with this topic and I do need to understand it for the ap test.  I was wondering if someone could help me with the set up for the following problems:

1.  A raft 6 m x 4 m in area is floating on a river.  When a loaded car pulls the raft, it sinks 3.0 cm lower in the water.  What is the weight of the car?

2. In an aorta with a cross-sectional area of 16 cm squared the blood (density = 1030 kg/m^3) flows with a speed of 30 cm/s.  What is the flow rate, in kilograms per second, of blood in the aorta?  The aorta branches to form a large number of fine capillaries with a combined cross-sectional area of 2 x 10^3 cm squared.  What is sthe speed of blood flow in the capillaries?

3.  Hydraulic stamping machines exert tremendous forces on a sheet of metal to form it into the desired shape.  Suppose the input force is 900 N on a piston that has a diameter of 1.8 cm.  The output force is exerted on a piston that has a diameter of 36 cm.  How large a force does the press exert on the sheet being formed?

Thank you so much for all the help!

By Archimedes law of buoyancy, the weight of the car $F_G$ is equal to the weight of the water pushed out, which is equal to the volume of this water $\triangle V = S \triangle h$ (S is the raft area) times its density $\rho_w$ times the gravitational acceleration $g$: $F_G = \triangle V \rho_w g$.

The flow rate of blood (volume per time) is equal to $\frac{\triangle V}{\triangle t} = S_a v_a = S_c v_c$, where $S_a, S_c$ are cross-sections of the aorta and capillaries and $v_a, v_c$ the corresponding velocities, because as a good approximation, blood is considered incompressible.

Hydrostatic pressure in hydraulic liquid is the same in every direction. Neglecting the difference (if any) of piston heights above the floor, $p = \frac{F_1}{S_1} = \frac{F_2}{S_2}$.


Prove that; for $ a,b,c>0$ we have:
$ \frac{a^2}{b}\plus{}\frac{b^2}{c}\plus{}\frac{c^2}{a}\geq (a\plus{}b\plus{}c)\plus{}\frac{4(a\minus{}b)^2}{a\plus{}b\plus{}c}$
I have assumed $ a\plus{}b\plus{}c\equal{}1$ and then proceeded, but it did not help much (apparently).

[hide="source"]Inequalities Through Problems; Hoojoo Lee[/hide]

hello, rewriting the given inrquality t0
$ \frac{(a\minus{}b)^2}{b}\plus{}\frac{(b\minus{}c)^2}{c}\plus{}\frac{(c\minus{}a)^2}{a}\geq\frac{4(a\minus{}b)^2}{a\plus{}b\plus{}c}$
this is equivalent to
$ (a\plus{}b\plus{}c)\left(\frac{(a\minus{}b)^2}{b}\plus{}\frac{(b\minus{}c)^2}{c}\plus{}\frac{(c\minus{}a)^2}{a}\right)\geq4(a\minus{}b)^2$
Now using Cauchy and we can prove
$ (a\plus{}b\plus{}c)\left(\frac{(a\minus{}b)^2}{b}\plus{}\frac{(b\minus{}c)^2}{c}\plus{}\frac{(c\minus{}a)^2}{a}\right)\geq4\left(\max(a,b,c)\minus{}\min(a,b,c)\right)^2$
Sonnhard.

[quote="Potla"]Prove that; for $ a,b,c > 0$ we have:
$ \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a}\geq (a \plus{} b \plus{} c) \plus{} \frac {4(a \minus{} b)^2}{a \plus{} b \plus{} c}$
I have assumed $ a \plus{} b \plus{} c \equal{} 1$ and then proceeded, but it did not help much (apparently).

[hide="source"]Inequalities Through Problems; Hoojoo Lee[/hide][/quote]
$ <\equal{}>\frac{(a\minus{}b)^2}{b}\plus{}\frac{(b\minus{}c)^2}{c}\plus{}\frac{(c\minus{}a)^2}{a} \ge \frac{4(a\minus{}b)^2}{a\plus{}b\plus{}c}$
It's true by cauchy-swarchz :)


Could anyone explain modular numbers in semi-plain English?

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=76228 . Locked in favor of that topic.


hehehe...

A tip:  Don't use so many words.  It gets hard to read.

I think it's fine... it does become a problem if there are lots of frames though. With only three frames, I don't think people would get turned off so quickly.

This is a comic creating system that offers little variation in the
body positions and/or facial expressions of some (if not all) of the
characters.  In a different comic creating system, I could relatively
easily have a character rolling his eyes in one particular panel with
no dialogue at all to communicate something.  Also, more of the
background would be seen in that panel as a bonus.

Thank you both for constructive criticism (- & +).


Suppose the integers $1,2,\ldots 10$ are split into two disjoint collections $a_1,a_2, \ldots a_5$ and $b_1 , \ldots b_5$ such that $a_1 <a _2 < a_3 <a_4 <a _5 , b_1 < b_2 < b_3 < b_4 < b_5$
(i) Show that the larger number in any pair $\{ a_j, b_j \}$ , $1 \leq j \leq 5$ is at least $6$.
(ii) Show that $\sum_{i=1} ^{5} | a_i - b_i|$ = 25 for every such partition.

Are you sure the conditions are right? Why can't the a's be 1,3,5,7,9 and b's be 2,4,6,8,10? Then the larger of {a1,b1} is 2 and the sum is 5.
Please clarify. Thanks.

As Bictor717 pointed out 2.5 years ago, something's not right about the conditions of the problem. Here is the 'renewed' problem that perhaps would make some sense:

Suppose the integers $ 1,2,\dots,10$ are split into two disjoint collections $ a_1,a_2,\dots,a_5$ and $ b_1,\dots,b_5$ (without the inequalities). Prove that the following two statements are equivalent:
(i) The larger number in any pair $ \{a_j,b_j\}$, $ 1\leq j\leq5$ is at least $ 6$.
(ii) $ \sum_{i\equal{}1}^5|a_i\minus{}b_i|\equal{}25$.

It's easy to prove (i) implies (ii) - if (i) is true then each you can change the sum to simply the 5 greater numbers (which are the numbers 6-10) minus the 5 lesser numbers (which are the numbers 1-5)

not sure about the other way around, but i guess you could work something out involving 25 being the max sum

The problem is correct except for the fact that a_i and b_i are oppositely sorted.

It has been solved in at least two other places: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=209254

The Given Question above is wrong. Some please correct the problem in the [b]Contest Section[/b].

The [b]Corrected Question : [/b]
Suppose the integers $1,2,\ldots 10$ are split into two disjoint collections $a_1,a_2, \ldots a_5$ and $b_1 , \ldots b_5$ such that $a_1 <a _2 < a_3 <a_4 <a _5 , b_1 > b_2 > b_3 > b_4 > b_5$
(i) Show that the larger number in any pair $\{ a_j, b_j \}$ , $1 \leq j \leq 5$ is at least $6$.
(ii) Show that $\sum_{i=1} ^{5} | a_i - b_i|$ = 25 for every such partition.

Considering the correct problem:\\

i) FTSOC assume $\max(a_{j} , b_{j})<6$\\

So $\frac{1}{2} \cdot \left( a_{j}+b_{j}+|a_{j}-b_{j}|\right)<6$ $\forall$ $j \in \{1,2,3,4,5\}$\\

$\iff  \frac{1}{2} \cdot \left( \sum _{j=1}^{5} (a_{j}+b_{j})+\sum_{j=1}^{5}(|a_{j}-b_{j}|)\right)<\sum_{j=1}^{5} 6$\\

$\iff \frac{55}{2} +\frac{1}{2}\cdot \left(\sum_{j=1}^{5} |a_{j}-b_{j}|\right)<30 \iff \sum_{j=1}^{5} |a_{j}-b_{j}|<5$ but also since $b_{1}>b_{2}>\cdots >b_{5}$ and $a_{1}< a_{2} \cdots <a_{5}$\\

We have $|a_{j}-b_{j}| \geqslant 1$ $\forall$ $j \in \{1,2,3,4,5\} \implies \sum_{j=1}^{5} |a_{j}-b_{j}| \geqslant 5$ which is a contradiction. Hence $\max(a_{j}, b_{j}) \geqslant 6$  $\square$\\

ii)  $\sum_{j=1}^5\left|a_j-b_j\right|=\Omega$
consider $\frac{\Omega+\sum_{j=1}^{5} \left(a_j+b_j\right)}{2}=\frac{\Omega+55}{2}=\sum_{j=1}^{5} \max (a_{j}, b_{j})$,
clearly since a $a_5>a_4>a_3>a_2>a_1$ and
$$
b_1>b_2>b_3>b_4>b_5
$$
$$
\max \left\{a_1, b_1\right\} < \max \left\{a_2, b_2\right\}<\ldots
$$
$$
<\max \left\{a_5, b_5\right\}
$$
or vice versa
so from (i) we have
$(6+7+8+9+10)$ as only possible value
which is 40
So $\quad \Omega+55=80$
$\Omega=25$ , Hence we always have $\sum_{j=1}^{j=5} |a_{j}-b_{j}|=25$ $\blacksquare$


function
limit
real analysis
real analysis solved

Hi all.

Doing a little self-study on limits, continuity, etc.

I have the following function:

$f(x)=\{\begin{array}{cc}x,&\mbox{ if } x \mbox{ is rational}\\0, &\mbox{ if } x \mbox{ is irrational}\end{array}$

How would one prove that this function is continuous at $x=0$?

Looking at a plot of this function, I would say that it's discontinuous everywhere but 0.

Would it be showing that the limit exists at $x=0$?

Thanks for the help.

Bailey

Hi.

Show that the limit exists at $0$ and this limit is $f(0)=0$.

This one makes more sense if you go ahead and quote the formal definition.

Informally, $f$ is continuous at $0$ iff you can make $f(x)$ arbitrarily close to $f(0)$ by making $x$ arbitrarily close to $0.$ The formalization of that mostly involves specifying what we mean by "arbitrarily close":

$f$ is continuous at $0$ iff $\forall\,\epsilon>0\,\exists\,\delta>0$ such that if $|x-0|<\delta$ then $|f(x)-f(0)|<\epsilon.$

Can you see how to make that work for your example? The core of a proof of continuity (or of a limit) will be a description of how you pick $\delta$ once you know $\epsilon.$ In this particular case, you can simply set $\delta=\epsilon.$

While we're at it, can we also prove the other thing you claimed? Namely, that for $a\ne 0,\,f$ is discontinuous at $a.$ Restating the definition,

$f$ is discontinuous at $a$ iff $\exists\,\epsilon>0$ such that $\forall\,\delta>0,\,\exists \,x$ with $|x-a|<\delta$ such that $|f(x)-f(a)|\ge\epsilon.$

Do you see how to apply that to your example?

You can use the definition of limit as well.

take ε>0. For |x| <ε  ,then ,  you have

 |f(x) - f(0)| = |f(x)| = ( |x| if x is rational or 0 , if x is irrational)<ε.
This means that limf(x) = f(0) as x tends to 0.

  Babis

thanks all!  I'm starting to see the light.  Let me put pencil to paper to see if I understand it clearly.

Best wishes.

Bailey

okay, let me get this straight...I want to prove: $\lim\limits_{x\rightarrow 0}f(x)=f(0)$, where $f(0)=0$.

So, I need to show that:

$\forall \ \epsilon > 0 \ \exists\ \delta > 0$   such that

$0 < |x-0| < \delta$   implies   $\ |f(x)-f(0)| < \epsilon$

Proof:

Letting $\delta=\epsilon$     (which I see how to get)

$0 < |x-0| < \delta$ $\Longrightarrow$ $0 < |x-0| < \epsilon$

$\Longrightarrow$ $|x-0| < \epsilon$

$\Longrightarrow$ $|f(x)-f(0)| < \epsilon$ $\square$

Is it that simple or am I still missing something?

Bailey

It is that simple and you are correct.

Now, why is this function discontinuous everywhere else?

Try with the negation of the definition. Let [b]a[/b] different of 0.
 <<  there exist ε>0 , such that for all δ>0  there exists  x with  /x/<δ  and

    $/f(x) - f(a)/ \geq  \epsilon$                   >>

 For example , if you take [b]a[/b]  as rational and  ε = (/f(a)/) /2  , then very - very near to [b]a[/b] you can find irrational x. For this x you will have  /f(x) - f(a)/ >= ε.

The same argument works also if a is irrational( very near to the irrational a you can find x rational( Q is dense to IR or something like that)).

 [u]Babis[/u]

  I was in Uni before 1980 , so I have forgotten some things(the most !!!) , so tis is hint which probably will help you.

Thanks for the help everyone...and the extra stuff to think about.

Bailey


There was a steel beam 50 feet long. The entire construction crew (total weight: 6000 pounds) was standing at the center of the beam. Assume the modulus of elasticity is 29,000,000 pounds per square inch, the moment of inertia is 850 inches^4, the beam has simple pin connections at either end, and all loads other than the weight of the crew are disregarded. :D

There doesn't appear to be a question asked.  Either way, I'm sending this off to Other Problem Solving Topics, where it belongs.


calculus
derivative
quadratics
algebra
polynomial
Putnam
quadratic formula

For which real numbers c is there a straight line that
intersects the curve x^4 + 9x^3 + cx^2 + 9x + 4
in four distinct points?

$ y\equal{}x^4\plus{}9 x^3\plus{}c x^2\plus{}9 x\plus{}4$

Making $ x\to x\minus{}\frac{9}{4}$

$ y\equal{}x^4\plus{}\left(c\minus{}\frac{243}{8}\right) x^2\plus{}\left(\frac{801}{8}\minus{}\frac{9 c}{2}\right) x\plus{}\frac{81
   c}{16}\minus{}\frac{23843}{256}$

If we could elliminate the $ x^1$ variable we could make $ x^2 \equal{} y$

$ \frac{801}{8}\minus{}\frac{9 c}{2}\equal{}0\Rightarrow c\equal{}\frac{89}{4}$

and we would have a straigth line. so, with this c, we would have solved the problem.

could possible be more values for $ c$, i didnt solved that

Claim: such a line is possible if and only if the quartic curve has two distinct inflection points. Since the second derivative is $ 12x^2 \plus{} 54x \plus{} 2c,$ by the quadratic formula, this is possible precisely when $ c < \frac {243}{8}.$

(Yes, $ \frac {89}{4} < \frac {243}{8}.$ But it's not unique. However, if we look at lordykelvin's work, we see that he had the necessary computations for the full answer, had he only realized what he had.)

As for proving the claim:

Suppose there is a line $ L(x)$ such that $ f(x) \minus{} L(x) \equal{} 0$ has four distinct solutions. Then, by Rolle's Theorem, $ f'(x) \minus{} L'(x) \equal{} 0$ has three distinct solutions and $ f''(x) \minus{} L''(x) \equal{} f''(x) \equal{} 0$ has two distinct solutions. But since $ f''(x)$ is a quadratic polynomial, if it has two distinct roots, then it changes sign at those roots and hence each such root is an inflection point.

Conversely, suppose that the curve does have two distinct inflection points. As a quartic with positive leading coefficient, it must be concave downward between these two points and concave upward to the outside of them. Draw the chord line $ y \equal{} L(x)$ connecting these two points of inflection. By the concavity just explained, we must have $ f(x) > L(x)$ between the two inflection points and $ f(x) < L(x)$ on some interval just to the outside of those points. But the growth of $ x^4$ must mean that there will be two more crossings further to the outside.

One line solution:

Since $ x \equal{} 0$ is not a root, we can rewrite the given equation as $ \minus{} x^2 \minus{} 9x \minus{} \frac {9}{x} \minus{} \frac {4}{x^2} \equal{} c$.

Therefore we are to find the condition for which $ y \equal{} \minus{} x^2 \minus{} 9x \minus{} \frac {9}{x} \minus{} \frac {4}{x^2}$ has 4 distinct intersection points with $ y \equal{} c$.

Let $ f(x) \equal{}\minus{} x^2 \minus{} 9x \minus{} \frac {9}{x} \minus{} \frac {4}{x^2}$, we have $ f'(x) \equal{} \minus{} \frac {1}{x^3}(2x^4 \plus{} 9x^3 \minus{} 9x \minus{} 8)$, but I can't find the roots of $ f'(x) \equal{} 0$ by factorization. :(

kunny, it looks like you are trying to answer a different question than the one asked. You seem to be trying to find conditions for $ f(x)\equal{}0$ to have four roots, which is very different than conditions for which there exist $ a$ and $ b$ such that $ f(x)\minus{}ax\minus{}b\equal{}0$ has four roots.

Oh, I see. Thank you, Kent Merryfield.

kunny

The problem is Putnam 1994 B2. As such you can assume that arguments at least as good as (probably better than) what I provided in the last two paragraphs of my long post above would be required.


[b]Problem 4.[/b] Let $k$ be the circumcircle of $\triangle ABC$, and $D$ the point on the arc $\overarc{AB},$ which do not pass through $C$. $I_A$ and $I_B$ are the centers of incircles of $\triangle ADC$ and $\triangle BDC$, respectively. Proove that the circumcircle of $\triangle I_AI_BC$ touches $k$ iff \[ \frac{AD}{BD}=\frac{AC+CD}{BC+CD}.  \]

[i] Stoyan Atanasov[/i]

Let $CI_1$ and $CI_2$ meet the circumcircle of $\triangle{ABC}$ at $X$, respectively $Y$.
Then $\frac{AC+CD}{AD}=\frac{CX}{XD}$ and $\frac{BC+CD}{BD}=\frac{CY}{YD}$ (1).
The tangency relation between the two circles (from the statement) is equivalent to the fact that quadrilateral $CXDY$ is harmonical (or $I_1I_2||XY$).
However, that's the case when the fractions in (1) are equal and we are done.

[quote=Sailor]
quadrilateral $CXDY$ is harmonical (or $I_1I_2||XY$).[/quote]

Sorry for gravedigging this post, but why is this equivalent?

[quote=Duarti]but why is this equivalent?[/quote]

$XI_1=DX$


This seems to be pretty easy (I guess that's why it's problem 1 :)).

We can definitely find 3 of our 8 numbers which are in the same progression, and we consider the cases:

[b]case 1: the 3 numbers form a progression with common difference 1[/b]

It's clear that we're done in this case.

[b]Case 2: the 3 numbers form a progression with common difference 2[/b]

If the progression is formed with even numbers, we're through. If it's formed with odd numbers, we may assume it's 1,3,5 (it's the same if it's 3,5,7). The rest of the numbers are 2,4,6,7,8. The numbers 2,4,6,8 can't belong to the first progression, because that only has odd numbers, so they must belong to the other 2 progressions. If 4,6 or 6,8 belong to the same progression we're done, otherwise it means that 4,8 belong to the same progression, which thus contains all multiples of 4, including 1980, and that's it.

[b] Case 3: the 3 numbers form a progression with common difference 3[/b]

The 3 numbers can either be 2,5,8 or 1,4,7. Let's assume they're 1,4,7 (the other case is the same). The numbers 2,3,5,6,8 can't belong to the first progression, so they must be distributed in the other 2 progressions, so there exists one progression which contains 3 of these 5 numbers. These 3 numbers can only be 2,5,8, and the other progression must contain 3 and 6, so it contains all multiples of 3, including 1980.

1. warm-up
The warden rewards a prisoner a chance to escape.
There are two boxes, 10 white balls, and 10 black balls.
Later, he will randomly pick a box, and randomly pick a ball in the box. How should he arrange the balls into the boxes to maximize his chance of getting out?
You need a white ball to get out.
2. problem (If you heard, tell me and don't write anything)
The warden rewards 100 prisoners. Every day, the warden picks a random prisoner. He is led to a room only with a lightbulb and switch. He will do one of four things:
1. Turn the light on if it was off
2. Turn the light off if it was on
3. Do nothing
4. Declare "All 100 prisoners have been taken to this room".

The lightbulb starts off.

When the prisoner leaves, the lightbulb is not touched. If he declares, and is right, everybody is freed. If he is wrong, all would be killed.

The prisoners only get to talk with each other once during the first day before the choosing. After, no communication is allowed.

How should the prisoners avoid being killed? This doesn't mean thet they do nothing. THEY WANT OUT!!!!!!

Answers in spoiler
Not anymore since no one is replying; the answer will be posted later.

[quote="RC-7th"]What do you mean by "if he declares"? Declare what?[/quote]

Declare choice #4

In the first problem, does he get out by picking a black or a white ball?

In the second, to avoid being killed, they should simply turn the light on and off, since it is effectively doing nothing.

[quote="towbomb"]In the second, to avoid being killed, they should simply turn the light on and off, since it is effectively doing nothing.[/quote]
Somebody has to declare or they'd do it every day until they rot in prison, and what they really want is to go free, even at risk of execution...at least that's what I assume pakagawa meant, since the way he/she worded it the problem's completely pointless...

towbomb wrote:In the first problem, does he get out by picking a black or a white ball?

White, but if it were black, it would be [hide]one black in a box and everything else in the other box.[/hide]

It's a white ball to get out.

NO!!!!!!

The prisoners want OUT!
Not to play with the light bulb!!?!?!!!1

Oh yeah, and here's what I think for the first one:

[hide]Put a white ball in one box and the nineteen others in the second box. That gives you 50% for picking the first box times 100% of picking the white one = 50%  For the second box it's 9 white and 10 blacks = 9/19 chance times 50% of picking the second box = 23.7%  50% + 23.7% =  73.7% or about 3/4 chance to get out.[/hide]

Most other ways it's around 50% and I'm pretty sure this is right.

wereallgonnadie wrote:I think I got something for the second one:
[hide]If the prizoners are picked randomly, some of them can be picked twice or more. Thus, those that are picked for their first time should change the state of the lightbulb. Those that get picked again should do nothing.
Also, since there are 100 prizoners they can only declare after 100 days because that is the least amount of days it can take. Then it depends whether the light was on or off to begin with? [/hide]

a guy who does not get picked does not know who is picked or how many times the lightbulb was changed. But, you said an important thing: The 100 days is a limit. Think about this.

Dear administrator:

Can you move this to the Intermediate level? I'm not getting good answers... I was going to post it there anyway.

hmmmm, I'm thinking, give me a chance, I'll come up with something later, I have to go.

Mabye, if the prisoners switched the bulb if it is their first time and start doing so only after 100 days?

I'm getting pissed.

Say theres prisoner A. He goes in on the 100th day, turns the light on. he goes in again, and again. Then, on the 103th day, prisoner B goes in. He does not know if it's AAAB or ABCD. (it's on so it can't be ABAC, etc...)

So does prisoner A switch the light each time he goes in?

And why start on the 100th day? Why not the 1st?

I think i have an idea for the second one... not the best way... but:

[hide]

Assign one prisoner the job of turning off the lightbulb, and all 99 others, the job of turning on the lightbulb. Each time (since they are chosen at random, they are bound to be chosen more than once) any of the 99 goes in and finds the lightbulb off AND if he never turned on the lightbulb before, he turns on the lightbulb. Hence, whenever that one prisoner assigned the job to turn off the lightbulb goes into the room and sees the lightbulb turned on, he then turns it off, and counts how many times he had turned off the lightbulb, thus, when he counts up to 99 (eventually...), it is safe for him to declare "all 100 prisoners have been taken to this room".

[/hide]

That was my first idea too...

But then I'm haunted by the counterargument that this game can easily stretch for an infinite time...

The actual "answer" takes only half the time. Yours will take about 40 years, and mine only 15. (it might be inaccurate)...

[hide]Hint: What could the prisoners to to save time before the 100th day?[/hide]

I'm bored, so here are the answers in spoiler.

[hide]put all the black balls and 9 white balls in one box and 1 white ball only in the other box.

2. have the light off. After the same guy goes in the second time, he turns the light on. He is the counter. He counts the numbers of days other people came. After the 100th day, if a guy goes in and the light is on, he keeps it off. Then, if it's his first time in, he turns it back on. If it's still off, he wins by declaring.

Then,  after the control goes again, he turns the light off. When a new guy goes in, he turns the light on. if a new guy goes in and the light is on, he is still a new guy. Then, the control turns it off and counts another person. Continue this until all 100 people are counted.[/hide]


Math: 780
CR:  560
WS: 690 (10 Essay)

Could it get any worse :(

Has anyone taken the Princeton Review or Kaplan classroom SAT course? and if so, does it really help, I know it's a 200 point improvement, but I really don't think i can boost my CR score  :( I made a 140 point improvement (all on WS)  from the November SAT by just doing Practice tests my self so is the Course worth it?

Well, I got a 2160 in March 2005 (650/800/710 reading/math/writing).  I took Princeton Review during the summer/fall and got 2290 on the January test (800/800/690).  The writing pisses me off though, lol.

It's hard to say if the improvement is solely the course.  Personally, I don't think the course did anything from me; most of my improvement on critical reading just came from doing practice tests on my own.  Hope that helps.

[quote="dakyru"]Well, I got a 2160 in March 2005 (650/800/710 reading/math/writing).  I took Princeton Review during the summer/fall and got 2290 on the January test (800/800/690).  The writing pisses me off though, lol.[/quote]

lol, so you must've gotten your money back for the course right since it "guarantees" a 200 point improvement.

Haha, really?  I'll go ask them about it.

I'm sure there's a disclaimer for initial scores of 2000+ :)

Nice perfect scores though, dakryu

Did anybody get a message that their scores are not available today and check again on Feb.18th. Here is the quote:

Please check back after Saturday, February 18 for January SAT scores

Although most scores for the January SAT are online now, a small number are not yet available. Learn why your scores may not be available and check back after Saturday, February 18.

Their standard explanation does not really explain anything.
Is this something to worry about?

I don't think the score delay announcement (which my son got) is anything to worry about. Most likely it means that there was a split in scoring the essay, so that the essay is being referred to a third scorer to break the split.

i took the test in october...800 math 800 cr 780 writing.

i think that the most important thing in CR is to build up a strong vocabulary, - thats what i did. and for writing, most of it isnt too hard so long as you defend what you write with examples, logic, etc. bc all they ask for is an opinion essay, so you really cant be wrong...and dont be afraid to make up stuff lol, they dont take off for that. unless its like. blatantly obvious and illogical - then they might.

...and math. i guess most of us here are already 700+ scorers in SAT math.  :lol:

[quote="juicybooty911"]math level 2    800 :D[/quote]
oh yeah, i heard that for math level 2 that its around 88th percentile to get an 800...is that true?

most people who bother to take level 2 are good math students and with the curve, 800 is common

[quote="MysticTerminator"]2390

writing :mad:[/quote]

How did you bother to miss one on the writing section?

Impressive scores though.

[quote="dkushnir"]My son (7th grade) got 700 in Math (before 13th Birthday).  Anybody in here is in Johns Hopkins SET program?[/quote]

Yes, several AoPSers are in [url=http://www.jhu.edu/gifted/set/index.html]SET[/url]. There is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=6882]a little discussion of SET[/url] on the AoPS Other U.S. Contests & Programs Forum.

i'm in SET and I've never seen the AOPS thread on it. Thanks tokenadult for pointing that out

[quote="kheg_k"]i  never got my jan SAT scores back. :mad:[/quote]

check online, the mailed form should be back in 2 or 3 more days

[quote="kheg_k"][quote="dkushnir"]My son (7th grade) got 700 in Math (before 13th Birthday).  Anybody in here is in Johns Hopkins SET program?[/quote]

i got 730. and i m pretty young myself[/quote]

Yeah well guess what... I got 840 before my -6th birthday... so obviously I'm way smarter than all of you. OK let's not brag, especially about the SAT..

I agree. Bragging about standardized tests is pretty useless. It is a great accomplishment to get into SET (I'm a member too :D) but it's really not... that big. Get onto USA IMO team and come back :) (not that I have, of course!)

[quote="aznphatso"]I agree. Bragging about standardized tests is pretty useless.[/quote]

I agree with this principle too.

Nonetheless, I think the statement appearing earlier in the thread was by way of giving background for asking a question about [url=http://www.jhu.edu/gifted/set/index.html]SET[/url], and that's a good thing, in my opinion, because not all AoPSers know about SET.

Thank you talkenadult. Exactly.
And if this program is good for something, it is worth to know.
So far (when it came to something serious) it looks like they run some kind of counceling services for a fee.


Are you sure that's correct?  There are squares in the denominator of the first term but not in the later ones.

Also, do you know how to use html?  It makes reading things much easier.  For example, if you wanted to write exponents, you can do

n

except without the spaces inside the <>.  The you can get things like xn.  Also, if you write "sub" instead of sup, you get subscripts.  x37.  Very nice, no?  Easy to read.

Oh, and putting spaces on either side of + signs is helpful, too.

On an aside, wouldn't it make more sense if "sup" was "sub," and vice-versa?  Since that's the direction the letters point in, after all.  Or maybe not.  I can see it argued both ways.

I do want to know if there was an error in the question, though.

sorry the correct statement:

Find out xi (xi natural \{0}) i=1,n so that:
[x12/(x2 + x3)] + [x22/(x3 + x4)] + .. + [xn2/(x1 + x2)] = [(n-1)/(x1 + x2 + .. + xn)];
[x] denotes the integer part.

[edited by mod to make it look nice :)]

Thanks for the formatting, mod.

[hide]

Just a question on the way to a solution:  the right side is 0 necessarily, right?  I mean, we have n natural numbers, whose sum is definitely bigger than (n - 1).   So then every term on the left should be zero, right?  And from there I think I can show that (1, 1, ..., 1) will be the only solution without a whole lot of trouble.  Am I missing something?

[/hide]

Oh, we also have to have n :ge: 3, yes?


If $\frac{1}{x} + \frac{1}{y} = 3$ and $\frac{1}{x} - \frac{1}{y} = -7$ what is the value of $x + y$.  Express your answer as a common fraction.

This was in a target round 1993. I know this:

[hide="solution"]
let's denote $a= \frac{1}{x}$, $b=\frac{1}{y}$
then
$a+b =3$
$a-b = -7$

adding two together, we have $2a = -4, a = -2$
$b = 5$.

So $x= \frac{1}{a} = -\frac{1}{2}$, $y=\frac{1}{b} = \frac{1}{5}$

x+y = $-\frac{3}{10}$
[/hide]

[hide="click here if you wish to do so..."]
-3/10[/hide]

[hide]I did it a strange way...
y+x=3xy
y-x=-7xy
2y=-4xy
2=-4x
-1/2=x
then I solved for y to get -3/10 as the final answer.[/hide]


function
number theory unsolved
number theory

Let $n$ be a positive integer, $n\geq 2$. Let $M=\{0,1,2,\ldots n-1\}$. For an integer nonzero number $a$ we define the function $f_{a}: M\longrightarrow M$, such that $f_{a}(x)$ is the remainder when dividing $ax$ at $n$. Find a necessary and sufficient condition such that $f_{a}$ is bijective. And if $f_{a}$ is bijective and $n$ is a prime number, prove that $a^{n(n-1)}-1$ is divisible by $n^{2}$.

Strange problem...
Evident, that $f_a$ bijective iff $gcd(a,n)=1$(if $gcd(a,n)=1$, then $ai\not=aj$ for all $0\leq i<j\leq n-1$, if $gcd(a,n)=d>1$, then $a*\frac{n}{d}\rightarrow0$ and $a*0\rightarrow0$)
If $n$ is prime and $f_a$ bijective, then $a$ isn't divisible by $n$ :arrow: by Little Fermat's theorem $a^{n-1}\equiv1(mod n)$ and $a^{n(n-1)}-1=(a^{n-1}-1)(a^{(n-1)^2}+a^{(n-2)(n-1)}+...+a^{n-1}+1)$ is divisible by $n^2$.


Compute the smallest number d such that fewer than half of the positive integers with d-digits have all distinct digits.

Deleted. - Overlaps...

[hide]Wouldn't that be $1$...[/hide]

The second one I believe was posted in the practice series.

For the second one can't you just examine the equation wihtout sub scripts mod 5

The second one is part of a set of problems that I posted that will last 2 weeks on my contest...sry  :blush:

The question is kind of hard to understand.

By common sense, though, it has to be 5. I'm not exactly sure how to explain it, but if you think about it, that has to be right.

It is 4
The case should be
5*10^d>9Pd
So d=4


Is it possible to draw a diagonal on each small square on the surface of a Rubik's cube so that the lines form a continuous non-intersecting path through all the squares?

I think that it's not possible, but why?

Do they need to join back together at the end?

OK, never mind. I've figured it out myself.
One can only ever get to half the corners of each square on the cube. Let these corners be vertices, and the lines that connect them, be the edges in a graph.
Now each vertex has 4 edges, and is of even degree, except the ones at the corners of the cube, which each have 3 edges.
For an Eularian path to exist, all vertices need to be of even degree, with the exception that the starting and ending vertices may be of odd degree. However, there are 4 vertices of odd degree (the ones at the corners of the cube; remember we can only over get to half the corners of each square), so an Eularian path cannot exist.


A Fat Man's Prayer:

Lord, My soul is ripped with riot
incited by my wicked diet.
"We Are What We Eat," said a wise old man!
and, Lord, if that's true, I'm a garbage can.
I want to rise on Judgment Day, that's plain!
But at my present weight, I'll need a crane.
So grant me strength, that I may not fall
into the clutches of cholesterol.

May my flesh with carrot-curls be sated,
that my soul may be polyunsaturated
And show me the light, that I may bear witness
to the President's Council on Physical Fitness.

And at oleomargarine I'll never mutter,
for the road to Hell is spread with butter.
And cream is cursed; and cake is awful;
and Satan is hiding in every waffle.

Mephistopheles lurks in provolone;
the Devil is in each slice of baloney,
Beelzebub is a chocolate drop,
and Lucifer is a lollipop.

Give me this day my daily slice
but, cut it thin and toast it twice.
I beg upon my dimpled knees,
deliver me from jujubees.

And when my days of trial are done,
and my war with malted milk is won,
Let me stand with the Saints in Heaven
In a shining robe--size 37.

I can do it Lord, If You'll show to me,
the virtues of lettuce and celery.
If You'll teach me the evil of mayonnaise,
of pasta a la Milannaise
potatoes a la Lyonnaise
and crisp-fried chicken from the South.

Lord, if you love me, shut my mouth.

Guys i'm not going to post jokes if you aren't. We need laughter.

I DON'T CARE IF YOU DON'T KNOW ANY JOKES JUST GOOGLE THEM AND POST!

[youtube]MIaORknS1Dk[/youtube]


ARML
AMC
AIME
USA(J)MO
USAMO
geometry
analytic geometry

[quote="not_trig"]
ARML is not that hard, but it's hard to make it, especially in NC.
:wink:[/quote]

Are there enough interested and capable math students in NC to justify additional ARML teams?

Some other states/schools send more teams - but most of these places are a bit closer to competition sites and probably spend less in travel expenses.  Still, funding an additional ARML team or two might be possible.

If we did have additional ARML team(s), should the members come from one school, perhaps NC School of Science and Math, or should we create a NC C and D Teams in much the same way the B team has been created in the past.

Another alternative would be to regionalize B, C, and D teams within the state.  This might permit additional practice sessions and encourage the formation and use of math circles in communities where such circles are weak or non-existent.

Additional thoughts?

-Greg

additional teams should be created the same way

NCSSM already gets a huge advantage with its walk-on policy in the state math contest

although an East ARML team would be pretty awesome since we would still be pretty owningish

We're good with two. Arnav: if you make an East ARML team and split NC into ECH and everything else, we would completely forfeit any advantages. That's basicly dividing the team in half.

It really seems kind of excessive to have 45 people going from one state. First of all, we would probably need two buses. But also, you would have to assume that everybody on the B team is at least as good as anybody on the C team. I really don't think that works out.

Besides, are those extra 31-45 really interested in ARML (it seems like some of the B team isn't)? I would probably be more open if a student who would have made such a team suggested this.

I'll just briefly say that if we had extra teams, I like the idea of a state A team, but regional B/C/D teams like Greg said. That would make these teams more likely to do well.

[quote="jb05"]We're good with two. Arnav: if you make an East ARML team and split NC into ECH and everything else, we would completely forfeit any advantages. That's basicly dividing the team in half.[/quote]

yeah I know but it would be funny

NCSSM students get to go to the state math contest without qualifying through regional contests?

eh I don't think it's that hard to make ARML like we do take 2 teams after all

[quote="gt59"]The question might be analyzed using prior NC contest scores and some knowlege of the difficulty of the ARML related to the other contests.

Approximately 60 NC people scored 5 or above on the AIME last year while just 30 made an NC ARML team.  A few in the top 20 at the State Contest did not take the AIME or did not score 5 or above.  There are a significant number of highly capable math students, even USAMO qualifiers, who don't qualify for the State Math Contest at regional competitions.

I'm not sure if this is a deep enough pool to justify another ARML bus or not.  There may be many qualified folks that can't go.

While getting a 5 on the AIME may not seem like much of a feat to some in talent-rich communities, among the ~80,000 high school students in the Winston-Salem, Greensboro, and High Point area (Triad) there were just a couple of people who managed a 5 on the AIME (I don't think either one made one of the ARML teams.)  There are wide swaths of the state that have no realistic chance of sending a student to ARML in a typical year.

- Greg[/quote]

yea we need to open up the regionals a bit and send more people to state...it's sort of stupid when you have MOPpers not making it past regionals

[quote="NeverOddOrEven"]who decides whether or not we will have more ARML teams?[/quote]

I think just about anyone can sponsor a team as long as they are willing to pay all expenses for the students (not a trivial matter with the travel from NC.)  As far as I know, this is the only significant limitation on the number of teams.  If the Florida site becomes a reality, it should provide additional room at the Penn State site - but I don't think that site facilities have limited the number of teams in the past.

Some elite schools and larger cities send 3 or more teams.  On the other hand, there are probably several states unrepresented.

If anyone wanted to create another team, it would be a good idea to coordinate it with the sponsors and coaches of the existing ARML teams.  The NC Teachers of Mathematics organization may provide the bulk of the present funding for the NC ARML teams.

- Greg

[quote="gt59"]While getting a 5 on the AIME may not seem like much of a feat to some in talent-rich communities, among the ~80,000 high school students in the Winston-Salem, Greensboro, and High Point area (Triad) there were just a couple of people who managed a 5 on the AIME (I don't think either one made one of the ARML teams.)[/quote]

Admittedly, a 5 is good percentage-wise, but ARML is a national contest. There were 2363 people who made a 5+ on the AIME last year, and those are not necessarily the top 2363 in the country. If all of these people attended ARML in addition to the ones who already do, we would need a lot more sites and a lot more room.

I think we do a good job of making our ARML teams. If we do make more teams, I don't think that this is the best way to do it.

I don't think we need more teams.

Making ARML should be hard; the harder, the better for the team. Rigorous selection yields better team members.

If you really want to have a C team, form it after the pool for the A and B teams has been chosen.  Choose the best people left after that.

Do people think that just because the A team won last year, we need to have more teams?

dude having more teams is obviously not going to affect the teams already in place

[quote="jb05"]
Admittedly, a 5 is good percentage-wise, but ARML is a national contest. There were 2363 people who made a 5+ on the AIME last year, and those are not necessarily the top 2363 in the country. If all of these people attended ARML in addition to the ones who already do, we would need a lot more sites and a lot more room.

I think we do a good job of making our ARML teams. If we do make more teams, I don't think that this is the best way to do it.[/quote]

Last year there were close to 1700 participants at ARML's three sites - with room to spare.  If ARML expands to four sites next year, then there would be room for at least 2400 students - roughly one seat for everyone who made 5 or better on the 2006 AIME.

I didn't intend to suggest that the AIME score determine team membership or that everyone with a 5 on AIME should be invited to ARML.  Sorry about the confusion.

If the rest of the country sends an [u]average[/u] of one person to ARML for every score of "5" or better on the AIME, then we might consider doing the same.  Sixty NC students scored 5's or better on AIME last year.  Talent-wise, I think we could support 4 teams and that the worst of the four would still perform respectably at ARML.

Funding might be another matter. :wink:

-- Greg

I think that right now, we have a good balance.  Remember that everyone who is going to ARML from NC is better at math than the average high-schooler.  As we approach that norm, though, by adding more teams, it will get harder and harder to distinguish between who should go and who shouldn't.  Also, do we want to send people to ARML who might not do as well and thus would not want to participate again.


Hi,

I'm using makeindex in my documents but the index do not appear in my table of content even when I tried the following code :

\printindex
\addcontentsline{toc}{chapter}{Index}

does anybody knows solution to my problem ? thanks

Lilian

The best way is to use the [url=http://www.tex.ac.uk/tex-archive/macros/latex/contrib/tocbibind/]tocbibind package[/url], see the [url=http://www.tex.ac.uk/tex-archive/macros/latex/contrib/tocbibind/tocbibind.pdf]documentation[/url]. There's a full explanation at the [url=http://www.tex.ac.uk/cgi-bin/texfaq2html?label=tocbibind]TeX FAQ[/url].


Can anyone tell me the results of chinese team?

I only know that some chinese contestant has got 41 points (he got a 6 in problem 1 !!). The Chinese team is the first with 220 points.

Do you have the detail score of each contestant?

The format of the name is : "Family name, Given name"

Huang, Zhiyi (黄志毅)  6 7 7 7 7 7
Zhu, Qingsan (朱庆三)  6 7 4 7 7 7
Li, Xianying (李先颖)  7 7 2 7 7 7
Lin, Yuncheng (林运城)  6 7 1 7 7 7
Peng, Minyu (彭闽昱)  6 6 4 7 5 7
Yang, Shiwu (杨诗武) 6 7 3 7 4 7

[quote="liyi"]The format of the name is : "Family name, Given name"

Huang, Zhiyi (黄志毅)  6 7 7 7 7 7 very funny how it is possible to solve all and have a mistake at easiest question :maybe:

That's not a real mistake, that's a bagatelle. The problem was simple, hence the jury decided to bitch around on this one. I'd got 6 points for it, too. See http://www.mathlinks.ro/Forum/viewtopic.php?p=115015 .

  Darij

try to guess what am I going to write....
Yes U guessed....ME TOO....
Unfortunately I had a small mistake as many of participicants

Oh sorry what are your scores mine are not so impressive. I had to have more but I failed this IMO unfortunately and that was my last in my life. Now I am at Kazakh-British Technical University studying Petrolleum Engineering.
My  scores at IMO 6 5 0 0 1 0 heh very bad isn't that?

[quote="Rafal"]I'm amused by the results achieved by the Chinese team each year , how do
they manage to score such high scores ?! Is this a matter of excellent training methods :) or a matter of genius ... or maybe both ... However I noticed that
if a country's population is enourmous - the chances for a super contestant are
statistically much bigger :) . However Romania, Bulgaria and Vietnam are a good exception :) .[/quote]

I don't think Vietnam is an exception. Do you know what's the population of Vietnam?


Profits of 26,000 dollars are to be divided among three partners in the ratio of 3: 4: 6. Compare with the smallest share, the largest on is

1. 1 1/6 times as great
2. 1 1/3  times as great
3. 1 1/2  times as great
4. 2 times as great

NOTE: I want to know the steps needed to solve this question.

Is it this way:

3x + 4x + 6x = 26,000 and solve for x?

Then I compare 3x with 6x after finding the value of x, right?

I don't quite understand your problem.. could you try rewording it ?

Well, well Sharkman, correct at last. The assumption is that in a ratio of $ 3: 4: 6$, just as any other ratio, they are all proportional to one another about a constant. In this case, the constant you chose is $ x$.

But that's not the [i]fastest[/i] way to solve. assume you find $ x$, then make the rational fraction, but guess what,  $ x$ cancels, as it is on the bottom and top.
[hide="leaving behind"]$ \frac{6}{3}= 2$ times as large.[/hide]

igiul is right, there's absolutely no need to solve for $ x$.  Just notice that the ratio of smallest to largest is 3:6 =  1:2, therefore the largest share is twice the smallest.


Thought this was a cool geometry problem.

387. In the diagram, the semicircles centered at P and Q are tangent to each other and to the large semicricle, and their radii are 6 and 4 respectively. Line LM is tangent to semicircles P and Q. Find LM. (Mandelbrot #3)

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=721578#p721578

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=464943#p464943

lol this question is uses the same idea as two different AIME problems (4 on 1995 AIME, and 9 on 2006 AIME II)..

I wonder if this question or the 1995 one came first....

oh well, I would just extend LM until its on the same line as PQ, then three similar triangles...(after drawing in the center of the biggest circle, call it O and connecting that to LM)...

Does anyone know the actual answer to this question?

Wait is this this years Mandelbrot round 3

wait this post was created in 2007 wowowowow :|

It should be mentioned that several people have not yet taken this year's Mandelbrot...

Also, it's probably from 2007.


ok so i just installed miktex and whatnot, and finished installing texnic center...  i've started the configuration wizard for texnic center, and its asking me to type in the complete path of the directory where the 'executables' are located...   now i might be a math guy, but i'm not really too much of a computer guy.  i'm a bit confused...  sum1 help please!

It's the directory that latex.exe (and a zillion other exe files) is located in - probably C:\something_or_other\miktex\bin

If you installed it in C:\texmf,

this is what:  C:\texmf\miktex\bin

Then, if you installed it in C:\Program Files\MiKTeX 2.5\, you can locate it in

C:\Program Files\MiKTeX 2.5\miktex\bin.

Its in %Application Directory%/Miktex/Bin/LaTeX.exe

This was a problem with the directions, and it took me a few minutes to learn how to get around it.


Segment $AB$ is both a diameter of a circle of radius 1 and a side of an equilateral triangle $ABC$. The circle also intersects $AC$ and $BD$ at points $D$ and $E$, respectively. The length of $AE$ is

$\displaystyle \text{(A)} \ \frac{3}{2} \qquad \text{(B)} \ \frac{5}{3} \qquad \text{(C)} \ \frac{\sqrt 3}{2} \qquad \text{(D)} \ \sqrt{3} \qquad \text{(E)} \ \frac{2 + \sqrt 3}{2}$

[hide="Solution"]
As AB is the diameter of the circle, AEB must be 90 degrees. Therefore we have a right triangle with angle 60 degrees (opposite to AE) and hypotenuse of length 2.

Therefore,
sin 60 = AE / 2
AE = 2 sqrt[3] / 2 = sqrt[3]

So the answers D sqrt[3][/hide]

[hide="Answer"]Since $AB$ is a diameter of the circle, $\triangle ABE$ is a right triangle, and it is also a 30-60-90 triangle because $\angle ABE=60^\circ$ by the given. Therefore, $AE=\frac{AB\sqrt{3}}{2}=\sqrt{3}\Rightarrow \boxed{D}$.[/hide]


algebra
polynomial
trigonometry
real analysis
real analysis unsolved

let  P[0,1] be set of polynomials resticted to [0,1].show f(x)=x(1-x) isnt interior of P[0,1] in C[0,1],whats interior of P[0,1] in C[0,1]? :D

Isn't the interior empty? $p(x)+\epsilon \sin x$ is in $C[0,1]\setminus P[0,1]$ for any $\epsilon\ne 0$.


喂!哈锣![电话]喂喂  每个人;各个人,

With all the respect possible, the point of this forum is for you to discuss in [b]Chinese[/b] not in English, so that Chinese students that are not aces at English may have  a change of integrating.

All the English discussions should be taken in the main forums.

谢谢你。

我?知?很多汉语  :blush: 但我知?一些 :P
 ... 因此请 使用汉语语言在这个地方 ;)

sorry I made an English thread on this forum
but the replies are in Chinese,  :D

it seems that he translated directly into Chinese.

[quote="Valentin Vornicu"]我丿知鿓很多汉语  :blush: 但我知鿓一些 :P
 ... 因此请 使用汉语语言在这个地方 ;)[/quote]
oh my God
Valentin懂得多少种语言？Amazing!!!
好厉害呀！
罗马尼亚学中文么？
just wondering!!!! :P

liyi说过了...他直接把英文或是什么的翻译成中文.
看他的语序都不对.

Yeah
I just realized that.
Sorry for pointing out your... Valentin!(It's brilliant of you still!)
That's normaly what I do to Korean... :D

[quote="liyi"][quote="Valentin Vornicu"]它是好 ;)[/quote]
从这个看出显然那是机器翻译的  :D
没有中国人这么说话的[/quote]
它是好=that's great?

[quote="Valentin Vornicu"]喂!哈锣![电话]喂喂  每个人;各个人...
[/quote]
Call 翻译成了 电话  笑死人！！

“每个人;各个人...”是everyone和everybody吧。
不过，Valentin有这份心我们还是很感谢的……

[quote="dingdongdog"]:D Oh,it may be the opppsite of "Chinglish".[/quote]Well I tried  :blush:  :P

[quote="Valentin Vornicu"]我丿知鿓很多汉语  :blush: 但我知鿓一些 :P
 ... 因此请 使用汉语语言在这个地方 ;)[/quote]
...so please use Chinese in this place :D  :D  :D

valentine u r so funny...
well we all can understand ur broken chinese and...
thank you for using chinese here...
晕...
不小心又写成英文了-_-
那些中文挺好玩的 ...

这也能让在美国长大的中国人多练习读中文。我的中文水水平级差。 :blush:

[quote="4everwise"]这也能让在美国长大的中国人多练习读中文。我的中文水水平级差。 :blush:[/quote]
能打字说明还有救。。。。。。 :lol:

haha~很少見有羅馬尼亞人打中文寫的  :lol:
不過第1步已經不錯了 :)


function
ratio
calculus
calculus computations

I'm studying Taylor series right now and I got confused at a somewhat subtle point, namely the values of the variable for which the expansion converges. Particularly about the binomial expansion (when the exponent is not an integer)
\[ (1 \plus{} x)^\alpha \equal{} 1 \plus{} \alpha x \plus{} \cdots \plus{} \binom{\alpha}{n}x^n \plus{} \cdots\]
In all my books the following sentence is attached to the formula: "This series converges for all all $ |x| < 1$", condition that apparently does not depend on $ \alpha$ (at least I used to think so). However when I was reading the solution to a random exercise on finding the Taylor expansion near $ x \equal{} 0$ of the function
\[ f(x) \equal{} \frac {1}{\sqrt {1 \plus{} x}},\]
the values of $ x$ for which the series converge were $ ( \minus{} 1/2,1)$, and not $ ( \minus{} 1,1)$ as I expected from the condition for convergence of the binomial expansion of $ f(x) \equal{} (1 \plus{} x)^{ \minus{} 1/2}.$ The way my teacher computed the values was by finding the remainder term of the Taylor expansion and then analyzing when it would tend to zero, and it makes perfect sense for me. However I used to think that the convergence of the binomial series was already established for all $ |x| < 1$. So can you please explain me the flaw in my reasoning? Or maybe it means that the convergence of the binomial series also depends on the particular value of $ \alpha$ (in this case -1/2)?? :maybe:

Thanks for your help.

[quote="Jutaro"] The way my teacher computed the values was by finding the remainder term of the Taylor expansion and then analyzing when it would tend to zero[/quote]  Without having thought too hard: this is a [i]sufficient[/i] condition for convergence, but it is not [i]necessary[/i] -- the remainder term puts an upper bound on the error, but it's not necessarily sharp.  So, you can have convergence of the Taylor series even if the error term does not go to 0.

If you're actually interested in computing the radius of convergence, use the ratio test (or the root test.  In principle this only works with "nicely behaved" Taylor series (you need the limit to converge), but in practice it works with just about everything you might be interested it.

Also, a power series centered at 0 must have interval of convergence symmetric around 0, except possibly at the endpoints.

[quote="JBL"]Without having thought too hard: this is a [i]sufficient[/i] condition for convergence, but it is not [i]necessary[/i] -- the remainder term puts an upper bound on the error, but it's not necessarily sharp.  So, you can have convergence of the Taylor series even if the error term does not go to 0.
[/quote]

Now I come to think about it, the statement of the exercise was "prove that the series converges in this interval", but no claim was made that it would not outside it. It all makes sense now, thank you. :)


calculus integration trigonometry function algebra polynomial domain

Let $f(x)=e^{\sqrt[4]{x}}\sin \sqrt[4]{x}$. Prove that

$\int_{0}^{\infty}f(x)x^{n}dx=0$, for all natural numbers $n$.

I think that 4 can be replaced by any natural number of the form $4m$, where $m$ is a fixed natural number.

With the positive exponent on your exponential, you have a divergent integral. How do you intend to fix that?

Ohhhh..., I am very sorry for the mistake , thank you very much Kent, it should be a negative exponent.

Let $f(x)=e^{-\sqrt[4]{x}}\sin(\sqrt[4]{x})$.

Prove that

$\int_{0}^{\infty}f(x)x^{n}dx=0$ for all natural numbers $n\geq 0$.

Define $f$ on $\mathbb{R}$ by letting $f(x)$ be your formula for $x\ge 0$ and $f(x)=0$ for $x<0.$ Take the Fourier transform $\widehat{f}(\xi).$ Your claim, if true, is equivalent to saying that $\left(\frac{d}{d\xi}\right)^{n}\widehat{f}(0)=0$ for all $n\ge 0.$ This in turn would make $\widehat{f}$ a function that is $C^{\infty}$ but not analytic.

Nothing, yet - just another way of looking at it. I haven't yet worked through whether or not it's true.

Here is my solution to this problem:

$\int_{0}^{\infty}f(x)x^{n}dx=4\int_{0}^{\infty}e^{-t}(\sin t )t^{4n+3}dt$ which can be obtained via the substitution $x=t^{4}$.

Let $A=\int_{0}^{\infty}e^{-t}(\sin t ) t^{4n+3}dt$

let $B=\int_{0}^{\infty}e^{-t}(\cos t ) t^{4n+3}dt$
then we get that

$B-iA=\int_{0}^{\infty}e^{-(1+i)t}t^{4n+3}dt=\frac{\Gamma(4n+4)}{(1+i)^{4n+4}}\in R$ via the substitution $t(1+i)=y$

Thus, $A=0$.

This problem is a counterexample to the following problem:

Let $f$ be a continuous function defined on a finite interval of the form $[a,b]$. Then if $\int_{a}^{b}f(x)x^{n}dx=0$ for all natural numbers $n$ then  $f$ is the constant function 0.

The question is whether the same statement holds for functions defined on intervals of infinite form. The problem above shows that this nice property does not hold for the case when the interval is not of finite form.

If you asked me to prove that result on $[a,b],$ my proof would start with the polynomials being dense in some reasonable function space ($L^{2},$ perhaps) on $[a,b].$ That line of proof is not available on $[a,\infty).$

I referred to the fact that your function $f$ has a Fourier transform which has a zero of infinite order at the origin. A function on $[a,b]$ would have a Fourier series, which is to say a Fourier transform whose domain is $\mathbb{Z}.$ The concept of a "zero of infinite order" makes no sense for a sequence.


$ a,b \in \mathbb{R}$, $ a < 0,b > 0$, satisfying that:

$ a\sin{\theta} \plus{} b\cos{\theta}\geq 0$
$ a\cos{\theta} \minus{} b\sin{\theta}\geq 0$

Find the Maximum of $ \sin{\theta}$ :oops:

[quote="Rust"]$ tg \theta = \frac {b}{a}$[/quote]

I'm sorry that the answer is not correct... :blush:

That was a hint, not an answer.  Let $ \phi$ be an angle such that $ \tan \phi \equal{} \frac{b}{a}$ and rewrite the system in terms of $ \phi$.


For $ x,y,z>0$,$ x+y+z=1$
Find min $ S= \frac{x}{x+yz}+\frac{2y}{y+zx}+\frac{3z}{z+xy}$

[quote="darkmaster"]For$x,y,z \geq 0$ ,$ x+y+z=1$
Find min $ S= \frac{x}{x+yz}+\frac{2y}{y+zx}+\frac{3}{z+xy}$[/quote]                                                                                $ 6-S=\frac{yz}{(x+y)(x+z)}+\frac{2xz}{(x+y)(y+z)}+\frac{3xy}{(z+y)(x+z)}$                                                                                And easy to find that : $6-S \le 3$                                                          The equality holds iff $z=0$ IF $x,y,z \geq 0$                                          S haven't the minimum

[quote="darkmaster"]Are you sure that $ 6-S \leq 3$?[/quote]
Yes!
$6-S \le 3 <->3xy(x+y)+yz(y+z)+2xz(x+z) \ge 3(xy(x+y)+yz(y+z)+xz(x+z) +2xyz)$

Let $ x,y,z>0 ,x+y+z=1$ and $ xy+yz+zx\neq 0.$ Prove that$$2\leq \frac{x}{x+yz}+\frac{2y}{y+zx}+\frac{z}{z+xy}\leq\frac{25}{8} $$
$$3\leq \frac{x}{x+yz}+\frac{2y}{y+zx}+\frac{2z}{z+xy}\leq 4 $$
$$3\leq \frac{x}{x+yz}+\frac{2y}{y+zx}+\frac{3z}{z+xy}\leq 5$$
$$\frac{x}{x+yz}+\frac{y}{y+2zx}+\frac{ z}{z+2xy}  \leq 2$$
$$\frac{x}{x+yz}+\frac{2y}{y+2zx}+\frac{ z}{z+2xy}  \leq 3$$