inequalities
geometry
3D geometry
inequalities proposed

I think the sign should be reversed :huh: , i.e.
$\sqrt[3]{(a+1)(b+1)(c+1)}+2\sqrt[3]{(a+3)(b+3)(c+3)}+\sqrt[3]{(a+5)(b+5)(c+5)}\leq 2\sqrt[3]{(a+2)(b+2)(c+2)}+2\sqrt[3]{(a+4)(b+4)(c+4)}$.
It is clear that if we can prove the inequality $\sqrt[3]{abc}+\sqrt[3]{xyz}\leq 2\sqrt[3]{(\frac{a+x}{2})(\frac{b+y}{2})(\frac{c+z}{2})}$, then the inequality follows directly.
By replacing all variables with their cubes, the above inequality is equivalent to
$(abc+xyz)^{3}\leq (a^{3}+x^{3})(b^{3}+y^{3})(c^{3}+z^{3})$
which is true by generalised Holder Inequality.


prove that for $ p$ where $ p$ is prime and $ p>5$
$ (p\minus{}1)!\plus{}1\equal{}p^k$ where $ k$ is a positive integer
has no solutions

[quote]classic?[/quote]
yeah i think it's well-known , solution :
[hide]See that for p>5 we got $ p\minus{}1>(p\minus{}1)/2>2$ , what means that $ (p\minus{}1)*2*(p\minus{}1)/2\equal{}(p\minus{}1)^2|(p\minus{}1)!$

$ (p\minus{}1)!\equal{}p^k\minus{}1$ -> $ (p\minus{}1)^2|p^k\minus{}1\equal{}(p\minus{}1)(p^{k\minus{}1}\plus{}p^{k\minus{}2}\plus{}...\plus{}1)$ , $ p\minus{}1|p^{k\minus{}1}\plus{}p^{k\minus{}2}\plus{}...\plus{}1$. Then just see that $ p^k\equiv 1 (mod p\minus{}1)$, -> $ p^{k\minus{}1}\plus{}p^{k\minus{}2}\plus{}...\plus{}1\equiv (k\minus{}1)\plus{}1\equal{}k (mod p\minus{}1)$.

So $ p\minus{}1|k$,$ p\minus{}1\equal{}<k$. Hence $ p^k>\equal{} p^{p\minus{}1}>(p\minus{}1)^{p\minus{}1}>(p\minus{}1)!$ - >$ p^k>(p\minus{}1)!\plus{}1$, contradiction :D :P :). [/hide].

[quote="tmy1018"]prove that for $ p$ where $ p$ is prime and $ p > 5$
$ (p \minus{} 1)! \plus{} 1 \equal{} p^k$ where $ k$ is a positive integer
has no solutions[/quote]

already posted


Consider a particle moving on X-axis in following way:
It starts at t=0 from origin and can make step of $ \plus{}1$ (probability=p) or $ \minus{}1$ (probability=q).Show that probability that after n steps the particle is at $ x\equal{}r$ is given by coefficient of $ x^r$ in the binomial expansion of $ {\left(px\plus{}\frac{q}{x}\right)}^n$

thanks

for evry step... me may move forward by one step with a probability p
and backward by one step with probability q

therefore, for one step,
we have
(px + q/x)
i.e. coefficient of x^1 (particle at 1) = probability p
and coefficient of x^(-1) (particle at -1) = probability q

but this is true for evry step,
so for n steps, raise it to the power of n.

i.e. (by induction)
suppose for n;
coefficient of x^r in (px + q/x)^n gives the probability of being at position x = r;

now, the position may increase by one with probability p, and may decrease by probability q
so, the probability of coming to the position x = r after n + 1 steps,
 = (probability of going ahead) * (probability of being at (r-1) at the end of n steps) + (probability of going backward) * (probability of being at (r+1) at the end of n steps)
 = p*[Coefficient of x^(r-1) in (px + q/x)^n] + [q*Coefficient of x^(r+1) in (px + q/x)^n]
 = coefficient of x^r in (px + q/x)^(n+1)


A quadrilateral ABCD with A + D > 180(degrees) is circumscribed about a
circle with centre I. A line through I meets AB in X and CD in Y .
Prove that if IX = IY, then AX · DY = BX · CY .

I have an idea, maybe it has something to do with ratios of areas of triangles formed within the quadrilateral. Then, using the given condition, ie IX=IY, maybe we can tell two triangle equal, and then add/subtract corresponding areas to get the required proof.

THanK YoU.

Let circle touch $ AB,DC$ at $ P,Q$. Since $ IP \equal{} IQ$ and $ IX \equal{} IY$,we have $ \angle IXA \equal{}\angle DYI$. Let$ \angle YXA \equal{} a,\angle XAD \equal{} 2b,\angle ADY \equal{} 2C,\angle DYX \equal{} a$.

Then $ a\plus{}2b\plus{}2c\plus{}a \equal{} 360, a\plus{}b\plus{}c \equal{} 180$. Since $ \angle IAX \equal{} b$, we have $ \angle AIX \equal{} c$. Similarly, $ \angle DIY \equal{} b$. Therefore, $ \triangle IAX$ and $ \triangle DIT$ are simlilarly.

So $ AX\cdot DY \equal{} IX\cdot IY$.Simlilarly,$ BX\cdot CY \equal{} IX\cdot IY$. so $ AX\cdot DY \equal{} BX\cdot CY$.

The same problem here: http://www.mathlinks.ro/Forum/viewtopic.php?t=149151

Please, please, please, do NOT ask the forum for help, chessfreak. It gives an undue advantage to the students who actually use the forum...

You only have to wait three more days, and then the last date will be over, and then you can post problems...

Till then, please stop asking for more solutions.


function
modular arithmetic
calculus
geometry
number theory

I don't really get what negative functions are..

Let's say that there's a function f(x)=10+x^2
Would -f(x)=-(10+x^2)  ?

Also, could someone clear up the meaning and use of the thing that looks like an equal sign, except with 3 lines instead of 2?

Yes you are correct.  -f(x) would mean -(10+x^2) in your case.

The equal sign with three lines is used in modular arithmetic.  It means congruent.  For example:
\[10\equiv 3 \pmod 7 \]

If you are not familiar with mods that means that when you divide 10 by 7 you get the same remainder when you divide 3 by 7.

Ok, what about this? This is from the book "Essential Calculus with Applications"

For example, the two functions
f(x) = 1/2 (x+1) - 1/2 (x-1)
and
g(x) = 1/2 (1+x) + 1/2 (1-x)
are both identically equal to the constant function 1. An equation like
1/2 (1+x) + 1/2 (1-x) :equiv: 1
involving the sign  :equiv: , is called an identity.

Nothing about modular arithmetic there...

"Congruent to" doesn't always have to be used with mods. It can even be used in geometry to say two shapes are congruent. All it means is that they are identical in whatever sense you are talking about (in mods, they are identical in that mod, in geometry, they are identical shapes, etc). In this case, the two expressions
1/2 (1+x) + 1/2 (1-x)
and
1

are congruent, meaning they are ALWAYS equal (for any x).

I learnt that is an identity sign. An equation such as 5x = 9y + 2 has an equals sign because the equation tells you information about the variables (such as values, or ratios). However something like 5x = 2x + 3x does not tell you any information, and is true for all values of x. When no information about x is given, it is an identity.

By no information, I mean if you were to treat these two equations as simultaneous equations, the bottom one wouldn't help at all in finding the value of y.

Could you also use the congruent sign like this-

1 :equiv:  3 - 2

when an equals sign could also be used?

You can understand by imagining an arithmetic sequence:

3, 5, 7, 9, 11 ..... Then 11=9=7=5=3 mod 2

Mod 2 means + OR - 2


Source: AHSME 1973 #3

The stronger Goldbach conjecture states that any even integer greater than 7 can be written as the sum of two different prime numbers.*  For such representations of the even number 126, the largest possible difference between the two primes is
$\text{(A)} \ 112 \qquad \text{(B)} \ 100 \qquad \text{(C)} \ 92 \qquad \text{(D)} \ 88 \qquad \text{(E)} \ 80$
[size=75]
*The regular Goldbach conjecture states that any even integer greater than 3 is expressible as a sum of two primes.  Neither this conjecture nor the stronger version has been settled.[/size]

since 126=113+13 (primes) we have 113-13=100 so the answer is $B$. :)

Another trick is to just go through all 5 answer choices to see which one is right (if you're really lazy, like me).  :D


group theory
abstract algebra
superior algebra
superior algebra unsolved

Suppose $ f: G \rightarrow H$ is a group homomorphism, $ N \triangleleft G$, and $ \text{ker}(f) \leq N$.

Prove that
$ G/N \cong f(G)/f(N)$.

I don't see how to do this. Any hints would be nice. Thank you.

use the fact that $ f(G)/f(N)\equal{}(G/\ker f)/(N/\ker f)\equal{}G/N$


If $b = 2000$, compute the value of the infinite sum:

$\displaystyle (\log_b 2)^0 (\log_b 5^{4^0}) + (\log_b 2)^1 (\log_b 5^{4^1}) + \ldots$

Good luck!

[hide]If we convert each logarithm of two to base two we get

$\frac {\log_2 2}{\log_2 2000} = \frac 1{(\log_2 2000)^{n-1}}$

Such that $n$ is the number of the current term.

We convert each logarithm of $5^{{4}^{n-1}}$ to base five where $n$ is the term

$\frac {4^{n-1}}{\log_5 2000}$

Multiplying the two together we have

$\frac 1{\log_5 2000} \cdot \frac {4^{n-1}}{(\log_2 2000)^{n-1}}$

Which is an infinite geometric sequence with a first term of $\frac 1{\log_5 2000}$ and a common ratio of $\frac {4}{\log_2 2000}$

Putting this in the formula for the sum of an infinite geometric sequence we then have

$\frac {\log_2 2000}{\log_5 2000 \cdot (\log_2 2000 - 4)} = \frac {\log_2 2000}{\frac {\log_2 2000}{\log_2 5} \cdot (\log_2 2000 - 4)} $
$=\frac {\log_2 2000 \cdot \log_2 5}{\log_2 2000 \cdot (\log_2 2000 - 4)}$
$=\frac {\log_2 5}{\log_2 125}$

So

$\frac {\log_2 2000}{\log_5 2000 \cdot (\log_2 2000 - 4)} = \frac 13$[/hide]


[quote="236factorial"]Given two sides of a triangle being x and y, what is the range of the other side of the triangle?[/quote]
Well since the two sides are called x and y, we will call the last one z.  Using the triangle inequality theorem, we get that x+y>z>x-y or x+y>z>y-x depending on which one of the sides is bigger.

[quote="ifai"]Let $z$ be the 3rd side. Then $|x-y|<z<x+y$.
I think it should be posted in the basic forum of middle school.[/quote]

Hey that's where I posted  :?  who moved it?


Euler
function
calculus
quadratics
calculus computations

I am writing an essay about numerical methods

I already know the euler approximation is $ y(y_0\plus{}h) \approx y(t_0)\plus{}hy'(t_0)$

but I am lost at analysing the error of euler method.

This is the quotation from wikipedia:

$ y(t_0 \plus{} h) \equal{} y(t_0) \plus{} h y'(t_0) \plus{} \frac{1}{2}h^2 y''(t_0) \plus{} O(h^3)$

I don't get the  $ \frac{1}{2}h^2 y''(t_0) \plus{} O(h^3)$ part

and I don't have a clue what $ O(x)$ function is and where $ \frac{1}{2}h^2y''(t_0)$ comes from.

I roughly know that error analysis begins from taylor expansion but I don't see any taylor expansion here.

Can you please explain that part means?

The $ O^{3}(x)$ just means "and higher order terms".  They're not important since they are dwarfed by those terms of magnitude $ h$ and $ h^{2}$.  To get $ \frac{1}{2}h^{2}y''(t_{0})$, just follow the formula, it's pretty explicit.

There's a great introduction to the O(x) notation at the top of the subforum Calculus Computations and Tutorials.

I'll give a brief (non-rigorous) explanation.

When we talk about the growth rate of functions for large values of input, we generally want to deal with only the largest growing term.

We denote the largest growing term (for sufficiently large inputs X) as O(that term)

For example, in a quadratic function the term that will dominate and grow the fastest as x gets large is the squared term.

If $ f(x) \equal{} x^2 \plus{} 5x \plus{} 1$, we say that $ f(x)$ is $ O(x^2)$.

Note that this Big-O notation (as it is commonly called) does not care whether there is a constant multiple on a term.

If $ g(x) \equal{} 3x^2 \plus{} 11x \minus{} 5$, we still say that $ g(x)$ is $ O(x^2)$.

So now, to your problem. You're right that this begins from Taylor Series expansion. What you're looking at on the right hand side are the first few terms of the Taylor Series expansion, truncated / abbreviated with the O(h^3) notation. It's basically saying: "There are more terms here, but we'll just point out the fastest growing one."

Hope that makes reasonable sense :P

Thank you both of you. now I understand  :)

I didn't think much about $ \frac{h^2}{2}f''(x)$ part eventhough I could

[quote="DiscreetFourierTransform"]We denote the largest growing term (for sufficiently large inputs X) as O(that term)[/quote]  Just a quick note: the fastest-growing term is what big-Oh notation is about, but "sufficiently large inputs" is wrong: big-Oh notation is context-dependent, and in the context of Taylor series we're actually looking at sufficiently small inputs (so, e.g., $ h^4 \in O(h^3)$ when $ h \to 0$, the situation we're in, but this is not true as $ h \to \infty$, when Taylor series often don't make sense, anyhow).

Aha! Great point JBL. That's totally right.


logarithms
limit
calculus
calculus computations

Prove (or disprove):
$\sum_{n=2}^{+\infty}\frac{\ln(n)}{n^{p}}\textrm{ converges if and only if }p>1$

I conjectured this since I knew that it was true for $p=1$ (diverges), and $p=2$ (converges).
For all $p\leq 1$, we may use a comparison test to show that it diverges.
For all $p\geq 2$, we may use a comparison test to show that it converges.

So really, I was wondering about the convergence of the series when $1<p<2$.

Thank you.

Suppose $p>1.$ Choose $s$ such that $1<s<p.$

Then $\frac{\ln n}{n^{p}}=\left(\frac{\ln n}{n^{p-s}}\right)\frac1{n^{s}}.$

But $\lim_{n\to\infty}\frac{\ln n}{n^{p-s}}=0,$ so in particular, $\frac{\ln n}{n^{p-s}}$ is bounded by some constant $M.$

Then $\sum_{n=1}^{\infty}\frac{\ln n}{n^{p}}\le\sum_{n=1}^{\infty}\frac{M}{n^{s}},$ which converges.

This is a use of the "simple" or "one-sided" comparison test. It's suggested by the knowledge the the logarithm grows slower than any power of $n,$ hence the logarithm can very nearly be ignored.

[quote="vishalarul"][quote="Kent Merryfield"]$\lim_{n\to\infty}\frac{\ln n}{n^{p-s}}=0$[/quote]

How would you prove that? If that could be proven, then your claim that:
[quote="Kent Merryfield"]the logarithm grows slower than any power of $n$, hence the logarithm can very nearly be ignored.[/quote]
would be justified.[/quote]
just use l'hopital:
\[\lim_{n\rightarrow \infty}\frac{\ln n}{n^{a}}=\lim_{n\rightarrow \infty}\frac{\frac{1}{n}}{an^{a-1}}\]
\[=\lim_{n\rightarrow \infty}\frac{1}{an^{a}}=0\]
for positive $a$.

or your series is aproximate equal to
$\frac{\ln n}{1-\alpha}n^{1-\alpha}$
when $\alpha<1$


I'm studying AoPS Volume 1 Ch.19 right now and don't think I can finish this book by Feb. 10th.
Should I study as many chapters as I can, or should I just study counting and sequence chapters and do lots of practice problems?  By the way, my current goal is getting at least 100 on AMC12.
Please advise.

P.S. Do they provide scratch papers on the test date?

I think you should briefly skim through the rest so if a question like that pops up, you'll atleast have an idea of what to do. After that, I would focus on my strengths so that I have complete mastery over the required/topics you're skilled at.

As to scratch paper, you take the AMC at your school, so yes, they will provide scratch paper unless they explicitly say otherwise.

In my opinion, Volume 2 is better than Volume 1 if you're studying for the AMC12.  I would make sure you are comfortable with as much stuff as you can get in.  If you feel comfortable with whatever you're on now, just move on and at least try to get familiar with everything.  Even if you can't master it, it's probably a good idea to get comfortable with a little of everything.

And scrap paper should be provided for you although you should probably ask just to make sure.

This thread is unlocked. Continue to provide help to him.

almost same question for me as OP:
im on chapter 18 of Vol. 2 & was wondering if i should do tons of aime/amc 12 problems to get ready and do the last few vol. 2 chapters after the amc 12; or try and bust out the last few chapters?
thx

Just as a note, if you're genuinely worried about doing well on the AMC, you should hold off on doing preparation for the few days before. Cramming may work for school assessments, but not for AMC contests.


[color=blue]Παραθέτω μια άσκηση θεωρίας αριθμών που δεν μπόρεσα να λύσω :blush: και θα παρακαλούσα όποιον μπορεί να βάλει τη λύση του :!: [/color]

Αποδείξτε ότι υπάρχουν άπειροι πρώτοι αριθμοί p που έχουν την εξής ιδιότητα: υπάρχει κάποιος φυσικός αριθμός n, που δεν διαρεί το $p-1$, τέτοιος ώστε το p να διαιρεί το $n!+1$

Βάζω την κύρια ιδέα .. :wink:
Πάρε έναν περιττό πρώτο  $q$ και επέλεξε $p \equiv 3 \ mod \ 4$ ώστε αυτός να είναι διαιρέτης του $(2q)!-1$.
Τότε $(p-2q-1)! \equiv (p-1)! \equiv-1 \ mod \ p$

άρα επιλέγουμε $n=p-2q-1$ . Επιπλέον υπάρχουν άπειροι τέτοιοι p kai n δεν διαιρεί το p-2q-1 (giati?  :wink: )

[quote="bilot4"][quote="silouan"]Τότε $(p-2q-1)! \equiv (p-1)!\ mod \ p$[/quote]

Σιλουανέ, μπορείς σε παρακαλώ να εξηγήσεις λίγο αυτή την ισοδυναμία :oops: Από πού τη συμπεραίνεις?  :huh:[/quote]
μοιαζει με αυτο  http://planetmath.org/encyclopedia/WilsonsTheoremResult.html
ουτε και εγω καταλαβαινω ,

Κώστα πολύ ωραίο αυτό το link! Πράγματι προκύπτει από αυτό το θεώρημα :wink:
Δηλαδή από ότι καταλαβαίνω έχουμε:
$(p-2q-1)!(2q+1-1)! \equiv-1 \ mod \ p$ κι αφού $(2q)!\equiv1 \ mod \ p$ συνεπάγεται ότι $(p-2q-1)! \equiv-1 \ mod \ p$

Βασικά δεν ήξερα ότι το Wilson's theorem είχε και προέκταση :oops:


Hi I'm given the following matrix, A:

1 2 1 0
-1 0 3 5
1 -2 1 1

I'm supposed to find it's row reduced echelon form, which I did, I'll call it R.

Now, I'm asked to find a 3x3 invertible matrix, P, such that R=PA

What's the procedure for this?

Essentially all you have to do it keep track of the row operations you use to reduce $ A$ to $ R$

$ A \equal{} \begin{bmatrix}1 & 2 & 1 & 0 \\
\minus{} 1 & 0 & 3 & 5 \\
1 & \minus{} 2 & 1 & 1\end{bmatrix}$

Left multiply by $ E_1 \equal{} \begin{bmatrix}1 & 0 & 0 \\
1 & 1 & 0 \\
\minus{} 1 & 0 & 1\end{bmatrix}.$

(Technically speaking, that's not an elementary matrix; it's the product of two elementary matrices. I combined them together to save space.)

$ E_1A \equal{} \begin{bmatrix}1 & 2 & 1 & 0 \\
0 & 2 & 4 & 5 \\
0 & \minus{} 4 & 0 & 1\end{bmatrix}$

Left multiply by $ E_2 \equal{} \begin{bmatrix}1 & 0 & 0 \\
0 & \frac12 & 0 \\
0 & 0 & 1\end{bmatrix}$

$ E_2E_1A \equal{} \begin{bmatrix}1 & 2 & 1 & 0 \\
0 & 1 & 2 & \frac52 \\
0 & \minus{} 4 & 0 & 1\end{bmatrix}$

Finally, left multiply by $ E_3 \equal{} \begin{bmatrix}1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 4 & 1\end{bmatrix}$

$ E_3E_2E_1A \equal{} \begin{bmatrix} 1 & 2 & 1 & 0 \\
0 & 1 & 2 & \frac52 \\
0 & 0 & 8 & 11\end{bmatrix} \equal{} R.$

That's in row echelon form; if you want to clean it out above the pivots, you have a few more things to left multiply by.

Now all you have to do is multiply out $ E_3E_2E_1.$ (And any more steps if you go on.) That's your $ P.$

Here's a better organization. Create the $ 3\times 7$ matrix $ [A\mid I].$ Row reduce it. The result is $ [R\mid P].$ That's really the same thing as I was doing above, but it's the right way to organize it.


find two constants  $a,b$ such that $1-x^k \ge a(1-x)x^b$and the equality occurs only and if only $x=\frac{1}{2}$
here $k>1,0<x<1$

and maybe it's hard to understand.
for example:
when $k=2$  then $a=3 \sqrt[3] {\frac{1}{4}},b=\frac{1}{3}$
I try to do it with series  expand,but I find it is too boring,and no result.
can somebody help me?
thanks in advance :)

as a starting point, $f(x)=1-x^k-a(1-x)x^b$ has a local minimum at $x=1/2$.
so $f(1/2)=0$, $f'(1/2)=0$ and $f''(1/2)\geq 0$.

ok.thanks.
and I have find that:
$b=\frac{2^k-k-1}{2^k-1}$
$a=\frac{1-(\frac 1 2)^k}{\frac 1 2 ({\frac 1 2})^b}$
can somebady help me prove when $0<x<1$
then $f(x)=1-x^k-a(1-x)x^b \ge 0$
can somebady help me?


For $x,y,z$ positive reals show that

$\frac{1}{x+y+z+1}-\frac{1}{(x+1)(y+1)(z+1)}\leq \frac{1}{8}$

Let $a=x+1,b=y+1,c=z+1$. Then the original inequality is equivalent to the following:
$\frac{1}{a+b+c-2}-\frac{1}{abc}\leq\frac{1}{8}$. ($a,b,c>1$)
Now let $a+b+c=u,abc=v$.
Then this is equivalent to $u \geq \frac{10v+16}{v+8}$.
We know that $u^{3}\geq27v$ by AM-GM.
So it is sufficient to prove that $27v \geq (\frac{10v+16}{v+8})^{3}$.
This is equivalent to $27v^{4}-352v^{3}+384v^{2}+6194v-4096\geq0$
$\left roundarrow (v-8)^{2}(27v^{2}+80v-64)\geq0$.
As $a,b,c>1$, $v>1$.
So $27v^{2}+80v-64>0$, and this completes the proof.
Equality holds when $a=b=c$ (by equality condition of AM-GM) and $v=8$, which means $x=y=z=1$.

Yess, it's right. My solution is the same.

Posted here: http://www.mathlinks.ro/Forum/viewtopic.php?t=81589 and http://www.mathlinks.ro/Forum/viewtopic.php?t=67113


function
calculus
integration
calculus computations

I think it's one of those functions you can't integrate.

my friend can integrate it using by parts.. integrate cos x^2.. any1 can help??

Well, using your friend's suggestion... I tried and the method failed in every possible way I could think of using integration by parts. In other words, your friend is wrong and made a mistake somewhere in his calculations.

I don't think the antiderivative is an elementary function. Isn't this related to Fresnel's Integrals? I know Fresnel's integrals do not have an elementary antiderivative.

PS: Can you prove that Fresnel's integrals are not elementary functions, :D ?

Try [url]http://integrals.wolfram.com[/url].  It says that the integral is related to the Fresnel integral.


function
number theory open
number theory

Start with any positive integer. Halve it if it is even; triple it and add 1 if it is odd. If you keep repeating this procedure, must you eventually reach the number 1?

This is the famous [b]Collatz conjecture [/b] that states that if we start with any positive integer, after a finite number of steps you will get always 1.
In other words the only possible cycle is 4 --> 2 --> 1 --> 4, no other cycle is possible! Arround '50-'60 many people in the american universities and research struggled to break it without success. In other form: if we start with some positive integer than there is an upper bound for all terms in the corresponding sequence. There are many results about... but still far away to prove it! You can find about it many things related on the web. It appers also in the form:
x/2 if x-even, (3x+1)/2 if x-odd
Note that some similar related functions may have more than one trivial cycle!


calculus
derivative
function
real analysis
real analysis unsolved

Define the function

f(x)=x^2 if x is rational , 0 if x is irrational.

Compute f'(0)

This was on my analysis midterm and I wasn't getting anywhere when trying to find the limit $ \frac{f(0) \minus{} f(x)}{0\minus{}x}$ as $ x\minus{}> 0$

When x is rational, $ f'(x)\equal{}2x$, so $ f(x)\equal{}0$.

For irrational, i was left to find the limit of $ \frac{f(x)}{x}$ as $ x\minus{}>0$

I concluded that limit does not exist.

I'm also thinking this function is not continuous, so not differentiable, and I just put too much work into it.

The function is discontinuous everywhere except zero, so its derivative doesn't exist at those places. Zero is the only place at which the derivative might possibly exist, and you have to try to calculate that limit.
What is it? $ \frac{f(x)\minus{}f(0)}{x\minus{}0}\equal{}\begin{cases}x&x\in\mathbb{Q}\\ 0&x\not\in\mathbb{Q}\end{cases}$
As $ x\to 0$, this goes to zero.

I understand that, but how would I go about calculating it?


[quote="bgbgbgbg"]Solve in R $ \sqrt{\frac{x\minus{}6}4}\plus{} \sqrt{\frac{x\minus{}7}3}\plus{} \sqrt{\frac{x\minus{}8}2}$ $ \equal{} \sqrt{\frac{x\minus{}4}6}\plus{} \sqrt{\frac{x\minus{}3}7}\plus{} \sqrt{\frac{x\minus{}2}8}$ [/quote]
Notice first that if a solution exists, it must be $ \ge 8$ in order all the terms are defined.

Let then $ f_{a,b}(x)\equal{}\sqrt{\frac{x\minus{}a}b}\minus{}\sqrt{\frac{x\minus{}b}a}$

$ f'_{a,b}(x)\equal{}\frac 1{2b\sqrt{\frac{x\minus{}a}b}}\minus{}\frac 1{2a\sqrt{\frac{x\minus{}b}a}}$ $ \equal{}\frac{\minus{}b\sqrt{\frac{x\minus{}a}b}\plus{}a\sqrt{\frac{x\minus{}b}a}}{2ab\sqrt{\frac{x\minus{}a}b}\sqrt{\frac{x\minus{}b}a}}$ $ \equal{}\frac{x(a\minus{}b)}{2ab\sqrt{\frac{x\minus{}a}b}\sqrt{\frac{x\minus{}b}a}(b\sqrt{\frac{x\minus{}a}b}\plus{}a\sqrt{\frac{x\minus{}b}a})}$

So $ f'_{a,b}(x)>0$ if $ a>b>0$ and $ x>0$

And so $ f_{6,4}(x)\plus{}f_{7,3}(x)\plus{}f_{8,2}(x)$ is strictly increasing for $ x>8$ and so has at most one zero in $ [8,\plus{}\infty)$

And since $ f_{6,4}(10)\plus{}f_{7,3}(10)\plus{}f_{8,2}(10)\equal{}0$, the required equation has a unique root $ \boxed{x\equal{}10}$

i have another solution
The equation is equivalent to
$ (x\minus{}10)(\frac{1}{24(\sqrt{\frac{x\minus{}6}{4}}\plus{}\sqrt{\frac{x\minus{}4}{6}})}\plus{}\frac{4}{21(\sqrt{\frac{x\minus{}7}{3}}\plus{}\sqrt{\frac{x\minus{}3}{7}})}\plus{}\frac{3}{8(\sqrt{\frac{x\minus{}8}{2}}\plus{}\sqrt{\frac{x\minus{}2}{8}})})\equal{}0$ so $ x\equal{}10$ is the only root

It's easy to see that $ x \equal{} 10$ is a solution.

If $ a > 5$ then $ \frac {x \minus{} a}{10 \minus{} a} > \frac {x \minus{} 10 \plus{} a}{a} \iff x > 10$ and $ \frac {x \minus{} a}{10 \minus{} a} < \frac {x \minus{} 10 \plus{} a}{a} \iff x < 10$. (And $ a\equal{}6,7,8$ corresponds to the three squareroots on the left and the right)
So $ x > 10 \Rightarrow LHS > RHS$ and $ x < 10 \Rightarrow LHS < RHS$. So $ x \equal{} 10$ is the only solution.


The number of diagonals in a regular polygon is equal to the number of sides. What is the number of degrees in the sum of all interior angles of the polygon?

Number of diagonals: ((n-3)(n)/2)
Number of sides: n
To equate:
((n-3)(n)/2)=n
n^2/2-3n/2=n
n^2/2=5n/2
n^2=5n
n=5

So a pentagon follows this description.

The sum of all the interior angles of the polygon is (n-2)180=[b]540[/b].

Number of diagonals in $n$-gon $=\frac{n(n-3)}{2}$

$\frac{n(n-3)}{2}=n$
$n(n-3)=2n$
$n^2=5n$
$n=0, 5$

$n$ can't be $0$ so $n=5$ making the polygon a pentagon.

The sum of the interior angles of a pentagon is $\boxed{540}$ degrees.


Fie a,b,c,d nr. reale distincte oricare 2 si n>2 nr. natural.
Demonstrati ca ecuatia: $X^n  = \left( {\begin{array}{*{20}c}
   a & 0 & 0  \\
   0 & b & 0  \\
   d & 0 & c  \\
\end{array}} \right)
$, $X \in M_3(R)$ nu are solutii.

Could you translate this one in english ?

Prove that the equation above has no solutions.
De fapt nu e chiar atât de grea pe cat pare, ba chiar e greşită, în condiţiile date de tine aici. Concret pentru $a=1$, $b=2^n$, $c=3^n$, $d=0$ avem soluţia: $X= \left ( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \\ \end{array} \right) $ ;)


Let a,b are positive integer numbers

such that a^3 + b^3 = 2

Prove that:

                a^9 + b^9 + 5(a^12 + b^12) >= 8+ 2(a^15 + b^15).

You may homogenize the inequality:

$$\frac{(a^9 + b^9)(a^3 + b^3)^2}{4} + \frac{5(a^{12} + b^{12})(a^3 + b^3)}{2} \ge \frac{(a^3 + b^3)^5}{4} + 2(a^{15} + b^{15})$$

Now bunching and use the Muirhead ideas!  :P


em... :? what about $(x,y,z)=(k,2k,k)$?
I think you have missed out something...

:blush:  :oops:  :blush:  I wanted to make problem harder but it seems that it's obvius problem :lol:

If u can find one soln, then you can find infinitely many etc...

Bomb

How? how can we make from one sol infinty answrs.

[quote=Mathx]prove that$x^3+y^3=9$ have infinity answers in $Q$[/quote]
:| Conjunction
$\boxed{(x,y)=(k,\sqrt[3]{9-k^3})}$

[quote=SS20]We have to find rational solutions[/quote]

Obviously from conjuction. I can figure in with functions.

[quote=targo___]Can anyone say what is conjunction? :(
is it about finding solution of parametric form? :maybe:[/quote]

It says that it can takes any value...

[quote=ZetaX]Just to solve this problem to bring it to the solved section ;) :
find a point on the corresponding elliptic curve and begin adding points to get infinetely many.[/quote]

Which elliptic curve do we use? Aren't elliptic curves with one square one cube, not two cubes?

[quote=StealthEllipticCurve]
Which elliptic curve do we use? Aren't elliptic curves with one square one cube, not two cubes?[/quote]
Yes, but you can substitute $z=x+y$, so that the equation is an elliptic curve in $z$ and $x$.

can someone pls give a detailed s
olution of this bcz i am weak in NT

[quote=Mathx]prove that$x^3+y^3=9$ have infinity answers in $Q$[/quote]

We use the formula given by Binet to prove this and for constructing solutions for this. Firstly, note that $(2, 1)$ satisfies. This will be $(x_0, y_0)$.

Now, for all $n \geq 0$, we let $(t_n, u_n) = \left ( \frac{x_n (x_n^3 + 2y_n^3)}{x_n^3 - y_n^3}, \frac{y_n (2x_n^3 + y_n^3)}{x_n^3 - y_n^3} \right )$. Then. we have $(x_{n+1}, y_{n+1}) = \left (\frac{t_n (t_n^3 - 2u_n^3)}{t_n^3 + u_n^3}, \frac{u_n (2t_n^3 - u_n^3)}{t_n^3 + u_n^3} \right )$.

This gives solutions $(2, 1), \left ( -\frac{36520}{90391}, \frac{188479}{90391} \right ), \ldots$.

[quote=mathwizard9]what is binet formula but that is related to fibonacci numbers right[/quote]

Binet has done much work in many different fields including elliptic curves. He proposed the above formula I mentioned as a construction for rational solutions.


AMC
AIME
modular arithmetic
algebra
difference of squares
special factorizations

How many of the integers between 1 and 1000 can be expressed as the difference of two perfect squares?

I don't know if it will help, but my sig contains an equation for finding $ y^{2}\minus{}x^{2}$. It is:
$ y^{2}\minus{}x^{2}\equal{} x\plus{}2\left(\sum_{z \equal{} x\plus{}1}^{y\minus{}1}(z)\right)\plus{}y$

[quote="Temperal"]It's inclusive.[/quote]

In general, however, between implies exclusive, because 1000 is not between 1 and 1000.

[hide="Solution"][b]Lemma[/b] Every positive integer except those of the form $ 4k\plus{}2$ can be expressed as a difference of two squares for every non-negative integer $ k$.

[b]Proof[/b]

$ 4k \equal{} (k\plus{}1)^{2}\minus{}(k\minus{}1)^{2}$.
$ 4k\plus{}1 \equal{} (2k\plus{}1)^{2}\minus{}(2k)^{2}$.
$ 4k\plus{}3 \equal{} (2k\plus{}2)^{2}\minus{}(2k\plus{}1)^{2}$.

Now to prove that $ 4k\plus{}2\neq x^{2}\minus{}y^{2}$, consider the equation $ \mod 4$. All squares are either congruent to $ 0$ or $ 1\mod 4$. So $ x^{2}\minus{}y^{2}\equiv 0, 1,$ or $ \minus{}1\pmod 4\not\equiv 2\pmod 4$. This completes the proof.$ \square$

Now the number of integers between $ 1$ and $ 1000$ that cannot be written as $ 4k\plus{}2$ is: $ 1000\minus{}\frac{998\minus{}2}{4}\plus{}1 \equal{}\boxed{750}$.
[/hide]

[quote="Icy"][quote="xpmath"][hide="If inclusive"]So, with the  two numbers a and b, we can write the difference of squares as $ (a\plus{}b)(a\minus{}b)$. Let's break the problem up into cases where we get evens, and where we get odds. For the odds, let the number be $ 2x\plus{}1$. We can let $ a \equal{}\frac{2x\plus{}2}{2}\equal{} x\plus{}1$ and let the $ b \equal{} x$, making $ (a\plus{}b)(a\minus{}b) \equal{} (2x\plus{}1)(1) \equal{} 2x\plus{}1$ as we wanted. That takes care of the odds.
For evens, both a and b must either be odd or even, other wise $ a\plus{}b$, and $ a\minus{}b$ are odd, and we end up with an odd number. So if they're both odd or even, we get the product of two even numbers, which is divisible by 4, taking care 250 numbers. In all, we can get 500+250=750 numbers.[/hide]
[hide="exclusive"] Well, 1 is an odd number, and 1000 is divisible by 4, so we get 748 instead.[/hide][/quote]

When finding the value of $ a$ in the odd case, where does the $ \frac{2x\plus{}2}{2}$ come from? Also, I'm not sure where the 500 is from. I get that 250 is 1000 divided by 4, but why do we divide 1000 by 2?[/quote]
Well, dince any odd integer can be expressed as $ 2x\plus{}1$, which can be written as $ x\plus{}(x\plus{}1)$, I decided, in the difference of squares to make $ (a\minus{}b)$ equal to 1, which happens when $ a\equal{}x\plus{}1$, and $ b\equal{}x$


algebra
polynomial
inequalities
algebra proposed

Let $ P$ be a polynomial with positive coefficients. Prove: If the inequality $ P\left( \frac {1}{x}\right) \geq \frac {1}{P\left( x\right) }$ is valid for $ x \equal{} 1$, then it is valid for all $ x > 0$, too.

Isn't it just Cauchy or am I doing a stupid mistake ?

If $P = \sum a_i \times x^i$

Hypothesis is $(\sum a_i)^2 \geq 1$

By Cauchy then $(\sum a_i \times x^i) \times (\sum a_i \times 1/x^i) \geq (\sum a_i)^2 \geq 1$

Don't worry, it's correct, the problem is very simple indeed.

quite right ,its just that simple ,but its a little funny decoration of cauchy ? Isnt it? :)

[quote="lomos_lupin"]but its a little funny decoration of cauchy ? Isnt it? :)[/quote]

Sure !

The problem is a bit intimidating at first, espcially if you think that Titu Andreescu is one more romanian genius like those found in this forum  ;)

This is quite pointless now - but nevertheless it was posted before:

http://www.mathlinks.ro/Forum/viewtopic.php?t=5443


Order the numbers $1,2,\cdots,n$ in a sequence $a_1,a_2,\cdots ,a_n$ such that
$|a_1-a_2|+|a_2-a_3|+|a_3-a_4|+\cdots |a_{n-1}-a_n|+|a_n-a_1|$ attains minimum value.

Claim: The minimum should be 2(n-1)= 2n -2.
Proof: The expression is equal to the total distance travelled when we go through every point in the set {1,2,3,...,n}. Suppose you start at j , then you have to hit n once, and then 1 and back to j.
The total length of this loop is at least 2n-2. Similary, if you start from j and then hit 1 first, you still have to hit n, then back to j. The total distance travelled is at least 2n-2.
The following sequence shows how to attain the minimum:
1=>2=>3=>...=>n=>1.

Please point out my mistake if my guess is incorrect.  :)


[quote="mathstud"]the sum of the reciprocals of the first four consecutive positive integers is greater than two  What is the least number of consecutive positive integers necessary to make the sum of the reciprocals greater than three?[/quote]
[hide]
1/1+1/2+1/3+1/4
1/1=1
1/2+1/3+1/4 > 1
1/5+1/6+1/7 > 1/2
1/8+1/9+1/10+1/11 is almost 1/2
the other numbers are over their respective 1's and 1/2's enough to make up for the difference.
so it takes 11

[/hide]


prove
$ f_{2n\plus{}2}\equal{}\sum_{i\ge 0}\sum_{j\ge 0}\binom{n\minus{}i}{j}\binom{n\minus{}j}{i}$

Recall that $ f_{n \plus{} 1} \equal{} \sum_{i \ge 0} {n \minus{} i \choose i}$.  This is because $ f_{n \plus{} 1}$ is the number of ways to tile a $ 1 \times n$ grid with $ 1 \times 1$ and $ 1 \times 2$ tiles, so given such a tiling with $ n \minus{} i$ total tiles we choose $ i$ of them to be $ 1 \times 2$ tiles.  Alternately, we can choose $ n \minus{} 2i$ of them to be $ 1 \times 1$ tiles.

$ f_{2n \plus{} 2}$ is therefore the number of ways to tile a $ 1 \times 2n \plus{} 1$ grid.  There are $ 2n \plus{} 1 \minus{} s$ total tiles and $ s$ different $ 1 \times 2$ tiles, hence $ 2n \plus{} 1 \minus{} 2s$ different $ 1 \times 1$ tiles.  Distinguish the center $ 1 \times 1$ tile and suppose that there are $ n \minus{} i$ tiles to the left and $ n \minus{} j$ tiles to the right for $ i \plus{} j \equal{} s$.  By construction there are $ n \minus{} i \minus{} j$ different $ 1 \times 1$ tiles on both sides, so there are $ {n \minus{} i \choose n \minus{} i \minus{} j}$ ways of tiling the left and $ {n \minus{} j \choose n \minus{} i \minus{} j}$ ways of tiling the right.  This gives

$ f_{2n \plus{} 2} \equal{} \sum_{i \ge 0} \sum_{j \ge 0} {n \minus{} i \choose n \minus{} i \minus{} j} {n \minus{} j \choose n \minus{} i \minus{} j}$

as desired.

I have to remember this method, very tricky and useful

thanks very much :)


Let the sequence $(a_{n})$ with $a_{1}= 1 , a_{2}= 3$ and $a_{n}a_{n+2}= a^{2}_{n+1}+8, \forall n \in N^{*}$.
a) Find the general term $a_{n}$ of this sequence as a function of $n$,
b) Prove that the number $A_{n}= a_{n}+(-1)^{n}$ is a perfect square for all $n \in N^{*}$.

[u]Babis[/u]

[hide]
a1 = 1 = 0^2 + 1,
a2 = 3 = 2^2 - 1,
a3 = 17 = 4^2 + 1,
a4 = 99 = 10^2 - 1
a5 = 577 = 24^2 + 1
and the sequence b1, b2, b3, ... = 0, 2, 4, 10, 24, ...is suggestive since each term b(n) = 2b(n-1) + b(n - 2).   *
The characteristic equation for * is  x^2 - 2x - 1 = 0 giving
x = 1 + sqrt(2) or y = 1 - sqrt(2).
b(n) therefore takes the form cx^n + dy^n for constants c and d.  Substituting for b1 and b2 gives c = 1/sqrt(2) = -d.
Hence b(n) = [(1 + sqrt(2))^n - (1-sqrt(2))^n]/sqrt(2).      **
Then a(n) = [b(n-1)]^2 + (-1)^(n+1).      ***
Substituting ** into a(n) and simplifying leads to:
a(n) = [(1 + sqrt(2))^(2n-2) + (1 - sqrt(2))^(2n-2)]/2
This formula is easily proved by induction.

The second part follows immediately from *** since
A(n) = a(n) + (-1)^n
       = [b(n-1)]^2 + (-1)^(n+1) + (-1)^n
       = [b(n-1)]^2
       = perfect square since each of the b(n) is a term in the sequence
          generated by * and therefore clearly an integer.[/hide]

[quote="Rust"]$a_{n+2}=6a_{n+1}-a_{n}$ [/quote]

 Dear Rust,
  please tell me : How did you find the above interesting relation? By suspection or did you make any algaibraic trick ?

Thank you again - Babis

Ok, the problem is solved.
The sequence is $1,3,17,99,577,3363,19601,114243,...$

and the formula $a_{n}= \frac{(1+\sqrt 2)^{2n-2}+(1-\sqrt 2)^{2n-2}}{2}$ is nice!

I wonder if we can find another way of solution using the following thoughts.
While I was trying to solve the problem, I found an interesting property:

Let $x=3+\sqrt 8$

$x^{0}=1+0\cdot\sqrt 8$
$x^{1}=3+1\cdot\sqrt 8$
$x^{2}=17+6\cdot\sqrt 8$
$x^{3}=99+35\cdot\sqrt 8$
$x^{4}=577+204\cdot\sqrt 8$
$x^{5}=3363+1189\cdot\sqrt 8$
$x^{6}=19601+6930\cdot\sqrt 8$
$x^{7}=114243+40391\cdot\sqrt 8$

So, $\boxed{x^{n-1}= a_{n}+b_{n}\sqrt 8}$

where $\{a_{n}\}$ happens to be our sequence
and $\{b_{n}\}$ is another sequence with
$b_{1}=0$
$b_{2}=1$
$b_{n+2}=6b_{n+1}-b_{n}$

[hide="The recursive formula"]
$(3+\sqrt 8)(a+b\sqrt 8) = (3a+8b)+(a+3b)\sqrt 8$

So we get a recursive formula for $a_{n},b_{n}:$
$a_{n}=3a_{n-1}+8b_{n-1}$
$b_{n}=a_{n-1}+3b_{n-1}$, or simply

$\left(\begin{array}{c}a_{n}\\ b_{n}\\ \end{array}\right)= \left(\begin{array}{cc}3 & 8 \\ 1 & 3 \\ \end{array}\right) \left(\begin{array}{c}a_{n-1}\\ b_{n-1}\\ \end{array}\right)\Rightarrow $

$\left(\begin{array}{c}a_{n}\\ b_{n}\\ \end{array}\right)= \left(\begin{array}{cc}3 & 8 \\ 1 & 3 \\ \end{array}\right)^{n}\left(\begin{array}{c}1 \\ 0 \\ \end{array}\right)$[/hide]

[color=darkred]I wonder if someone can explain the relation between the sequence $\{a_{n}\}$ and the powers of $x$.
Can we use this property of $x$ to solve the problem?[/color]


What is the least positive integer value of $ p$ such that $ 7p$ divided by 11 has a remainder of 1?


$f(x)$ is defined on $[a,b]$. If for any $\epsilon > 0$ there exists $\delta > 0$, it holds that
$\left|\sum_{i=1}^n f(y_i)-f(x_i) \right| < \epsilon$
for any non-overlapping intervals $[x_i,y_i]\subset [a,b] (i=1,2,\dots,n)$ which satisfy $\sum_{i=1}^n (y_i-x_i) < \delta$
Prove that $f(x)$ is absolutely continuous on $[a,b]$

[quote="grobber"]What's the role of $\delta$ here? Maybe the non-overlapping intervals have the length of their union $<\delta$?[/quote]
Yes, you are right.

We will find $\delta$ for $\epsilon/2$, then we partion sum $S=\sum_{i=1}^n |f(y_i)-f(x_i)|$ into two sums: $S_1$ for positive $f(y_i)-f(x_i)$ and $S_2$ for negative $f(y_i)-f(x_i)$, then $S=S_1+S_2$ and $S_1,S_2\leq \epsilon/2$.

[quote="grobber"]What's the role of $\delta$ here? Maybe the non-overlapping intervals have the length of their union $<\delta$?[/quote]
Yes... I forget to say it.  :(
There has been so many times that I missed a condition....  :(


For $x=y$ it's easy to show that there are no solutions.

For $x \neq y$ we can write the equation as $f(x)=f(y)$ where $f(x)= x^{4}-8x^{3}-8x$ .

I was wondering if this could help somehow :maybe:

Nice solution,I think We can deduce that the equation has only finit solutions from this.

Let $u=x+y$ and $v=xy$ .Then our equation is equivalent to
$u^{3}-2uv=8u^{2}-8v+8$ . We can find that u=2t and now it is equivalent to
$2t^{3}-tv=8t^{2}-2v+2$ .Solving for $v$ we can find that
$v=2t^{2}-4t-8-\frac{18}{t-2}$
So $(t-2)|18$  so (x,y)=(8,2)  and (y,x)=(2,8)
are the only solutions

The initial equation equivalent to $(x+y)(x^{2}+y^{2}) = 8(x^{2}+xy+y^{2}+1) > 0$ $\cdots (1)$
From $(1)$ we have
i) $x+y > 0 \Rightarrow x+y \geq 1$
ii) $x+y$ is even
iii) $x+y = 8+\frac{8(xy+1)}{x^{2}+y^{2}}\leq 12+\frac{4}{xy}\leq 16$
iv) $x,y \neq 0$
v) Let $x+y = A, xy =B$, we have $B = \frac{A^{2}}{2}-2A-8-\frac{36}{A-4}$
Hence, $2 \leq A \leq 16$ and $A-4|36$, and the possible values of $A$ are $2,6,8,10,16$
a) $A=2 \Rightarrow B=8$
b) $A=6 \Rightarrow B=-20$
c) $A=8 \Rightarrow B=-1$
d) $A=10 \Rightarrow B=16$
e) $A=16 \Rightarrow B=84$
Only d) gives integer solutions, which are $(x,y) = [(2,8),(8,2)]$

Maybe this isn't a nice solution, but I just want to make sure that this is legal....  :roll:

@ RDeepMath : isnt this silouan solution ???

@ Silouan : Nice solution  :)

Oops... I'm sorry. I just didn't look whether there was any similar solution with mine...  :oops:


When will the button allowing you to convert a list of problems in the problems section to a .pdf be fixed? It would greatly help me if it was fixed. I don't want to print off the entire border and navigation info and waste ink.

It's on the list of things to fix.

For now, try cut-and-paste over to a word processor before printing.


The sequence $ (a_n)$ satisfies $ a_1 \equal{} 1$ and $ \displaystyle 5^{(a_{n\plus{}1}\minus{}a_n)} \minus{} 1 \equal{} \frac{1}{n\plus{}\frac{2}{3}}$ for $ n \geq 1$. Let $ k$ be the least integer greater than $ 1$ for which $ a_k$ is an integer. Find $ k$.

[hide="i missed this one of the actual thing :("]
We can simplify this expression to get $ a_{n\plus{}1}\equal{}a_n\plus{}\log_5 \frac{3n\plus{}5} {3n\plus{}2}$. It is possible to show by induction that $ a_n\equal{}1\plus{}\log_5 \frac{3n\plus{}2} {5}\equal{}\log_5 (3n\plus{}2)$. We want 3k+2 to be a power of 5. It cannot be $ 5^1$ since k>1. $ 3k\plus{}2\equal{}5^2$ does not yield an integer k. $ 3k\plus{}2\equal{}5^3$ gives us $ k\equal{}\boxed{041}$.[/hide]

[hide]Telescoping multiplying, we have
$ 5^{a_{n\plus{}1}\minus{}a_n}\cdot 5^{a_{n}\minus{}a_{n\minus{}1}}...\cdot 5^{a_{2}\minus{}a_{1}} \cdot 5^{a_1} \equal{} \frac{3n\plus{}5}{3n\plus{}2}\cdot \frac{3n\plus{}2}{3n\minus{}1}\cdot ...\cdot \frac{8}{5} \cdot 5^{1}$
which simplifies to $ 5^{a_{n\plus{}1}}\equal{}3n\plus{}5$ or $ 5^{a_n}\equal{}3n\plus{}2$. It's easy to verify that $ a_n\equal{}3$ is the smallest possible value which yields $ n\equal{}41$. (Almost made an off-by-one error!)
[/hide]

[hide="Invariant"]

Let $ b_n \equal{} \frac{5^{a_n}}{n\plus{}\frac{2}{3}}$

The given equation simplifies to

$ b_{n\plus{}1} \equal{} b_n$

Therefore,

$ b_n \equal{} b_1 \equal{} 3$ for all $ n$.

So we have

$ 5^{a_n} \equal{} 3n \plus{} 2$,

The smallest value of $ n$ for which $ a_n$ is an integer would be

$ \frac{125 \minus{} 2}{3} \equal{} 41$

[/hide]

[hide="Mashing"]

For uncreative people like me...
Seriously, it probably was very easy to notice that $ a_1 \equal{} \log_5{5}, a_2 \equal{} \log_5{8}, a_3 \equal{} \log_5{11}, a_n \equal{} \log_5{3n \plus{} 2}$, so we have to find the smallest positive integer n such that $ 3n \plus{} 2$ is a perfect power of 5. Some more mashing brings up 125, which is $ 3\cdot 41 \plus{} 2$, so the answer is $ \boxed{41}$

The way I did it (mostly the same):
I spent a while jiggling around with the algebra and making sure I dont get my superscripts and subscripts mixed up xD. Anyways, I calculated $ a_2$, saw a logarithm, and gave up trying to work with $ a_n$, and tried to mash it with $ 5^{a_n}$. It was considerably easier

We know that $ 5^{(a_{n \plus{} 1} \minus{} a_n)} \minus{} 1 \equal{} \frac {1}{n \plus{} \frac {2}{3}}$. A little reordering brings $ \frac {5^{a_{n \plus{} 1}}}{5^{a_n}} \equal{} \frac {3n \plus{} 5}{3n \plus{} 2}$. And then I brute forced, and I got something like $ \text{multiplying n products of} : \ 5\left(\frac {8}{5}\right)\left(\frac {11}{8}\right)\left(\frac {14}{11}\right)\left(\frac {17}{14}\right)\left(\frac {20}{17}\right)\cdots \equal{} a_n$. And since $ a_n$ must be a power of 5 that is 2 more than a multiple of three, except for 5 itself, 125 must equal the product, so $ 3n \plus{} 2 \equal{} 120$, so $ n \equal{} \boxed{41}$.
[/hide]

i put 40.............off by one error..how stupid
made mistake when finding expression for a_k (substituted k instead of k-1), so i instead found the expression for a_(k+1), so when i found k = 40, i just put down 40, thinking that my expression was a_k

Plugging in $n = 1$ to our equation and simplifying gives $5^{a_2}=8$. Plugging in $n = 2$ to our equation and simplifying gives $5^{a_3} = 11$. From this we can conjecture that $5^{a_k}=3k+2$, which can be shown by induction. The smallest number equivalent to $2 \pmod 3$ that is a multiple of $5$ is just $125$, which gives an answer of $\frac{125-2}{3} = \boxed{41}$.


Let $ x,y,z,t>0$ satisfy $ x\plus{}y\plus{}z\plus{}t<1$.Prove that:
$ \frac{1}{(1\plus{}x)(1\plus{}y)(1\plus{}z)(1\plus{}t)}\plus{}\frac{1}{1\minus{}x\minus{}y\minus{}z\minus{}t} > 2$.

[quote="SUPERMAN2"]Let $ x,y,z,t > 0$ satisfy $ x \plus{} y \plus{} z \plus{} t < 1$.Prove that:
$ \frac {1}{(1 \plus{} x)(1 \plus{} y)(1 \plus{} z)(1 \plus{} t)} \plus{} \frac {1}{1 \minus{} x \minus{} y \minus{} z \minus{} t} > 2$.[/quote]

Let : $ x\plus{}y\plus{}z\plus{}t \equal{} p$

$ LHS \ge \frac{1}{(1\plus{}\frac{p}{4})^4}\plus{}\frac{1}{1\minus{}p}\equal{} f(p)$ , $ 0 <p <1$

$ f(p)\minus{}$ increase $ (0,1)$ , so $ f(p) > f(0)\equal{}2$

[quote="SUPERMAN2"]Let $ x,y,z,t > 0$ satisfy $ x \plus{} y \plus{} z \plus{} t < 1$.Prove that:
$ \frac {1}{(1 \plus{} x)(1 \plus{} y)(1 \plus{} z)(1 \plus{} t)} \plus{} \frac {1}{1 \minus{} x \minus{} y \minus{} z \minus{} t} > 2$.[/quote]
Let $ x\plus{}y\plus{}z\plus{}t\equal{}a$,then applying AM-GM's inequality we have:

$ \frac {1}{(1 \plus{} x)(1 \plus{} y)(1 \plus{} z)(1 \plus{} t)} \geq  \frac{256}{(4\plus{}a)^4} \geq \frac{32}{(4\plus{}a)^2}\minus{}1 \geq \frac{16}{4\plus{}a}\minus{}3$

Hence,It suffices to prove that :

$ \frac{16}{4\plus{}a} \plus{} \frac{1}{1\minus{}a} \geq 5$ which is equivalent to $ a^2 \geq 0$

Done !

Equality doesn't occur !

A simple solution
x, y, z and t are all real numbers in the interval (0 , 1)

Therefore $ (1\minus{}x)(1\minus{}y)(1\minus{}z)(1\minus{}t) >1\minus{}x\minus{}y\minus{}z\minus{}t$ and $ (1\minus{}x^2)(1\minus{}y^2)(1\minus{}z^2)(1\minus{}t^2) <1$

So , $ \frac{1}{(1\plus{}x)(1\plus{}y)(1\plus{}z)(1\plus{}t)}$ +  $ \frac{1}{1\minus{}x\minus{}y\minus{}z\minus{}t}$  > $ \frac{1}{(1\plus{}x)(1\plus{}y)(1\plus{}z)(1\plus{}t)}$ +$ \frac{1}{(1\minus{}x)(1\minus{}y)(1\minus{}z)(1\minus{}t)}$ > $ \frac{2}{\sqrt{(1\minus{}x^2)(1\minus{}y^2)(1\minus{}z^2)(1\minus{}t^2)}}$ >2 by the AM-GM inequality


quadratics
number theory unsolved
number theory

Let p be a prime had the form 4k+1.Prove that$p=a^2+b^2$ with:
$a=1/2\sum(\frac{i(i^2-s)}{p})$    i=0 to p-1.
$b=1/2\sum(\frac{i(i^2-r)}{p})$    i=0 to p-1.
With r is a quadratic residue of $Zp$
With s is a non-quadratic residue of $Zp$

It's not hard to show that $p|a^2+b^2$, but that's not quite enough.

We just use the well-known facts:
$1:\ \left(\frac ap \right)\equiv a^{\frac{p-1}2}$;
$2:\ \sum_{i=1}^{p-1}i^k$ is $p-1$ if $p-1|k$ and $0$ otherwise (I mean $0$ in $\mathbb Z_p$);
$p|r^{\frac{p-1}2}+s^{\frac{p-1}2}$ (actually, this follows from $1$).

Now we can expand $(i^3-ri)^{\frac{p-1}2}$ and $(i^3-si)^{\frac{p-1}2}$ to get $p|a^2+b^2$.

We need to show $a,b<\sqrt p$, and I have no idea how to do that..


function
real analysis
real analysis unsolved

Let $ \mu$ be a finitely additive set function on the algebra $ \mathcal{A}$, then $ \mu(E\cup F) \equal{} \mu(E)\plus{}\mu(F)\minus{}\mu(E\cap F)$. Can you write a corresponding formula for $ \mu(E\cup F\cup G)$? For arbitrary finite unions?


Find, with proof, the minimum number of equilateral triangles of side 1 which are needed to cover a square of side 1 entirely.

[hide](Assuming that the triangles must be inside the square) Firstly, the square has 4 sides, at 90 degree angles. We must have triangles with sides coinciding with sides of the square to ensure that each side is covered. This leads to a minimum of 4 triangles, as a square has 4 sides. If we place 4 equilateral triangles onto the edges of the square, we see that indeed, they cover the whole square (can be made rigorous by...analytical geometry is what I'm thinking)
[/hide]


I have seen this problem many times before, but I am kind of a beginner to these types of problems:
Factor $ 6^7\plus{}6^2\plus{}1$. Thanks a lot!

hello, we get $ 6^7\plus{}6^2\plus{}1\equal{}17\cdot 43\cdot 383$
Sonnhard.

Dear Dr. Sonhard, could you please tell me how you attacked the problem? From where did you get the idea of using $ 6^2\plus{}6\plus{}1$ as part of the factorization? I can understand most of your work, but can you please tell me how you got your ideas?

Once again, we see that $ (x^{2} \plus{} x \plus{} 1)$ is a factor and, dividing, we get

$ x^{7} \plus{} x^{2} \plus{} 1 \equal{} (x^{2} \plus{} x \plus{} 1)(x^5 \minus{} x^4 \plus{} x^2 \minus{} x\plus{}1)$

Substituting $ x\equal{}6$, we have $ 6^{7} \plus{} 6^{2} \plus{} 1 \equal{} (43)(6511) \equal{} (43)(17)(383)$ (and $ 383$ is prime.)

Yeah, thanks, but I already know about the substitution part. I was just asking whether there is any method to understand $ (6^2\plus{}6\plus{}1)|(6^7\plus{}6^2\plus{}1)$ so quickly

Consider the polynomial $ P(x) \equal{} x^7 \plus{} x^2 \plus{} 1$, and $ \omega_k \equal{} \cos\frac {2k\pi}{3} \plus{} i\sin\frac {2k\pi}{3}$, for $ k \equal{}1,2$, a primitive root of order $ 3$ of the unit, so $ \omega_k^3 \equal{} 1$ and $ \omega_k^2 \plus{} \omega_k \plus{} 1 \equal{} 0$. Then $ P(\omega_k) \equal{} \omega_k^7 \plus{} \omega_k^2 \plus{} 1 \equal{} \omega_k \plus{} \omega_k^2 \plus{} 1 \equal{} 0$. Therefore $ P(x)$ is divisible by $ x^2 \plus{} x \plus{} 1$. All that remains is make $ x\equal{}6$.
Of course this works for many more other situations.

because x^2+x+1 has all the thrid roots of unity w^2+w+1=0 w^7=w(e^2pi(i)/3) so w^7+w^2+1 has x^2+x+1 as a factor


Do we have to use induction to solve this problem?

Just to note it, the squares suggest cauchy to me, but I'm horrible at inequalities.

No takers?  :(

[hide="An idea"] If we let $r = \sqrt{M}$ and assume WLOG that $a_{1}\le a_{2}\le ... \le a_{n}$, then

$a_{1+k}\ge a_{1}+kr$ [/hide]


Given $ \log_2{3} \equal{}a$ and $ \log_3{7} \equal{}b$, write $ \log_{14}{56}$ in terms of a and b.

Using $ \log$ properties, we have $ \log_23\cdot\log_37 \equal{} \log_27 \equal{} ab$. Adding $ \log_22$ to both sides, we have $ \log_2{14} \equal{} ab \plus{} 1\implies\log_{14}2 \equal{} \frac1{ab \plus{} 1}$.

Since $ \log_{14}{56} \equal{} \log_{14}{14\cdot4} \equal{} 1 \plus{} \log_{14}4 \equal{} 1 \plus{} 2\log_{14}2$, and
$ \log_{14}2 \equal{} \frac1{ab \plus{} 1}$, we have $ 1 \plus{} \frac2{ab \plus{} 1} \equal{} \boxed{\frac {ab \plus{} 3}{ab \plus{} 1}}$ as our answer.

EDIT: Fixed mistakes.

Speaking of logs how do you do this problem?

$ \log_3{75} \cdot \log_5{45} \minus{} (\log_{3}{25} \plus{} \log_{5}{9})$

Uhh I dont understand how you got this:

$ \log_3{25} \equal{} 2\log_3{5} \log_5{9} \equal{} 2\log_{5}{3}$

Oh, i get it==>
By saying this:
log3(25) = 2log3(5) log5(9) = 2log5(3)
he really meant
log3(25) = 2log3(5) and log5(9) = 2log5(3)

Even with this, when I evaluate it I keep getting 4.

(I'm assuming that when there is no space between the logs you are multiplying)

I edited my post; what I originally thought he meant isn't correct.

i wasnt multiplying logs

what i meant was

log3(25) = 2log3(5)
and
log5(9) = 2log5(3)


limit
algebra
polynomial
function
rational function
complex analysis

1) limit at infinity of h(z) = |z|/ z
2) limit at infinity of h(z) = Arg z

are both of the limit don't exist???
since for
1) let z=t , t positive would give 1 and t negative would give -1

2) as z-> infinity, z still could be at any angle????
    or answer for this limit is just Arg z?? since Arg Z is a constant...

help...plz...I'm confused...

If there is no mistake in the formulation of the problem, then you are perfectly right: both limits do not exist  :) .

good..thanks...
so what is the general method for doing limits of complex number as z-> infinity?????
how should I think infinity in complex plane in general???

[quote="SHWW"]
so what is the general method for doing limits of complex number as z-> infinity?????[/quote]
Just use the definition: $\lim_{z\to\infty} h(z)=L$ iff for every $\varepsilon>0$ there exists $R>0$ such that $|h(z)-L|<\varepsilon$ whenever $|z|>R$. Another good trick is to note that $\lim_{z\to\infty} h(z)=\lim_{w\to 0}h\left(\textstyle \frac 1w \right)$.

[quote="SHWW"]how should I think infinity in complex plane in general???
[/quote]
One very powerful mental image is that of the "Riemann Sphere." Think of $\mathbb{C}\cup\{\infty\}$ as a sphere. Sets of the form $\{z: |z|>R\}\cup\{\infty\}$ are neigborhoods of the point at infinity.

Thinking this way, if $P(z)$ is a nonconstant polynomial, then $\lim_{z\to\infty}P(z)=\infty.$ We could also say that any non-constant polynomial has a pole at infinity.

But  $\lim_{z\to\infty}e^z$ does not exist. The exponential has an essential singularity at infinity.

$\lim_{z\to\infty}\frac{3z^2+2z-4}{z^2-1}=3.$ We would say that this particular rational function has a removable singularity at infinity.


can anyone tell me from where can i get used n cheap copies of AOPS textbooks..
AOPS VOL 1
AOPS VOL 2
AOPS INTRODUCTION TO COUNTING AND PROBABILITY

thanks

Ebay probably.

This belongs in a different forum though.

aops vol 1 for 5 bucks, 2 dollars for solutions, and you pay for shipping.

in brand new condition.

PM me to talk out how to get it done.


I'm taking pre-calculus now (8th grade) so I'll be taking Calculus next year. In the article The Calculus Trap, http://www.artofproblemsolving.com/Resources/AoPS_R_A_Calculus.php, RR. says that we shouldn't rush through the standard curriculum.

But even though I'm in a math club, I still have to take Calculus at school, don't I? They don't have any alternatives at [i]school[/i]. So what should I do?

i think the point of that article is that you shouldn't be breezing thru courses just to think you are ahead. you maybe missing out things that you should have learned. im taking AP Chem and AP Calc I next year as a sophomore (did i spell it right?), and im going to be the one of the two sophomore students taking AP Calc I. and im the only sophomore taking AP Chem. and yes now i feel a bit regret because it looks like im going miss out a lot of things in trigonometry. i saw previously a couple of posts in here on some trigonometry problems and i couldn't solve them! it's going to be a whole lot busier once school starts in september. i didn't even make it to AIME when i was in Myanmar (aka Burma) as a freshman. i know i can do it but when it comes to competitions, im usually slow and never gets to even see what the last few questions were... take some time to meditate while taking calculus if you will, that will help you ease what that article is going to talk about. take courses at a steady pace, don't "rush"... :)

With the way that the curriculum is set up, you basically have to take calculus before you can progress far in math (for example, I don't know if many colleges will let you take a class in abstract algebra if they see you don't know calculus). I think Richard's point in that article is that you shouldn't let all your math energy go to standard classes. So if you're learning calculus but also working on interesting problems in, say, combinatorics, then you're fine.

yea, i agree with ComplexZeta because a lot of times when i am doing some math problems, i recall things related to the problem that i learned [i]outside[/i] of standard classes, for example, like warut_suk's problem that i solved last two weeks ago, that problem was hard and it is in no way i can find a method of solving it if i used my algebra and even trigonometry class notes to do it... i hope you get what ComplexZeta and i am saying here  ;)

I agree with the above statements. If you want to follow the article's logic, after you take calculus, you don't have to jump into college math classes in sophomore year right away. You can go study math that is not cirricular, and take such courses when you have developed your "problem-solving" skill further. That is my interpretation of the article's messages.


Can anyone tell me from definition is 0.99999999 (to the infinity) equal to 1?

Is 0.99999999999999(to infinity) a real number?  can it be expressed in fraction?

The decimal expansion "0.99999999999999999 . . ." (followed by an unlimited number of zeros) is taken to be the same real number as the number written "1." Similarly, "1.0999999999999 . . . " (followed by an unlimited number of zeros) is taken to be the same real number as "1.1," and so on for an unlimited number of other examples. This is one of the most frequently asked questions on the sci.math newsgroup on Usenet.

1 and 1.1 are real numbers, and are also, for that matter, rational numbers (fractions).

JP, I've already asked you to continue the discussion in [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=19502]this thread[/url]. Please do so, we don't need 5 topics on the same issue :)


Prove that: If $f(x)$ is period function then $f'(x)$ is period function, too


Suppose there are x socks in a drawer; some of them white some of them black. It is the case that when two socks are drawn without replacement, there is a probability of exactly 1/2 that either both are black or both are white. If x is at
most 2006, what is the minimum value x can take?

Wait..
If X is maxed out at 2006, then the largest value it can take is 2006..

Can you rephrase the question?

sorry, I meant minimum, but I changed it now.


Does anybody know a way that I can get better at writing summations with those weird greek letters, like $ \sum$ and $ \prod$?

I have the most trouble knowing how to write a given series/sequence in this form.

If you have AoPS Vol. 1, you can go through all the practice problems in the summation chapter, and solve the examples along with the book (write everything down as the book explains a problem.  Other than that, just remember: "Practice makes better."

Do you mean actually converting from written out form to sigma/pi form?

Then, once you get the hang of it, practice will gradually steady your hand. Its very straightforward actually. Find the common parts, the lower bound, and the upper bound.

[quote="mihail911"]thanks for the advice guys. and archimedes, i do have aops vol 1, that's where i realized i was having a lot of trouble with it, and now iv run out of problems to try... :(
anywayz, buzzer11, what do u mean about finding the lower and upper bound...i do mean writing a sequence like 1,2,3, 4..in sigma/pi form.

thanks again![/quote]

You don't write sequences in sigma/pi form. You write series in that form. There's a big difference.

A sequence is something like
$ a_{1},a_{2},a_{3},...$
where each $ a_{i}$ is a real number. Sequences can be denoted as $ (a_{n})$. Finding the general term of a sequence is commonly referred to as finding the pattern. For example, the sequence
$ 2, 2, 8, 18, 50, 128, 328$
has the pattern "twice the square of the Fibonacci numbers", and the general term is $ 2\cdot F_{n-1}^{2}$
A series is something like
\[ \sum_{i=1}^{+\infty}\frac{1}{i^{2}}\].

When buzzer talked about finding an upper and lower bound, he/she meant that if you have something like this:
$ \frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\ldots+\frac{9}{10}$,

look at the first term and the last term, and try to compare them in some meaningful way. We see that each term is of the form $ \frac{i}{i+1}$, for $ 1\le i\le 9$. We have just determined our bounds; they are $ 1$ and $ 9$. Now, we can easily write in sigma form:
\[ \sum_{i=1}^{9}\frac{i}{i+1}\].

Here are some practice problems:

1) Write in sigma form:
$ \frac{1}{2}+\frac{3}{4}+\frac{5}{8}+\frac{7}{16}+\ldots+\frac{19}{1024}$
Bonus: Without a calculator, calculate that sum.

2) Write in pi form:
$ \frac{1}{24}\cdot \frac{1}{35}\cdot \frac{1}{48}\cdots \frac{1}{624}$.
[hide="Hint"]Add one to the denominators, and you should see something special...[/hide]

[quote="vishalarul"]
When buzzer talked about finding an upper and lower bound, he/she meant
[/quote]

He. :)

[quote="vishalarul"]
\[ \sum_{i=1}^{9}\frac{i}{i+1}\]
[/quote]

Note that this is exactly the same as:

\[ \sum_{i=2}^{10}\frac{i-1}{i}\]

[quote="buzzer11"][quote="vishalarul"]
\[ \sum_{i=1}^{9}\frac{i}{i+1}\]
[/quote]

Note that this is exactly the same as:
\[ \sum_{i=2}^{10}\frac{i-1}{i}\]
[/quote]

This is a very useful technique. In general,
\[ \sum_{i=k}^{n}f(i) \]

is exactly the same as:

\[ \sum_{i=k+r}^{n+r}f(i-r) \]

This is called [b]re-indexing[/b].

If you don't get it, write out all the terms and see that they coincide. This is useful when you have something like:

\[ \sum_{i=1}^{n}\frac{1}{i+1}+\sum_{i=5}^{n+4}\frac{i-4}{i-3}\]

Just re-index the second summation:

\begin{eqnarray*}\sum_{i=1}^{n}\frac{1}{i+1}+\sum_{i=1}^{n}\frac{i}{i+1}&=&\sum_{i=1}^{n}\left[\frac{1}{i+1}+\frac{i}{i+1}\right]\\ &=&\sum_{i=1}^{n}1\\ &=& n \end{eqnarray*}

And, that trivializes the problem! Of course, you could just start writing out the terms, but after some practice, this is faster and more formal.

i have one more question:
u know how there's always a number below and above the summation/product in pi/sigma form. usually goes something like $ \sum_{i=5}^{10}$
well, can those numbers begin as fractions are must they begin and end as integers. also, does the top number always have to be an increase over the bottom number or can the summation be all numbers from $ i=5$ to $ i=-7$ or something like that?
the book didn't explain that.

They must be integers. otherwise, you wont know the interval to go up in.

also, if you want to write "from 5 to -7", you might as well just write "from -7 to 5", because thats the same thing...

oh...i get it...thanks buzzer...(man, im such a noob)


Here's a cute problem:

Consider all the $r$ element subsets of the set $1,2,3,...,n$. Take the least element from each of these subsets. Prove that the average of each of these least elements is $\frac{n+1}{r+1}$

Very interesting and easy. Cmon, show us some takers by the intermediate community pls  :D

You just have to prove that:
$\sum_{k=1}^r = {k*{n \choose r-k}} * \frac{1}{{n \choose r}} = \frac{n+1}{r+1}$


function
inequalities proposed
inequalities

[quote="arqady"][quote="SUPERMAN2"]Let $ a_i$ be the positive real numbers($ i \equal{} 1,2,...,10$) such that $ a_1 \plus{} a_2 \plus{} a_3 \plus{} ... \plus{} a_{10} \equal{} 10$.
Prove that
$ a_1^2 \plus{} a_2^2 \plus{} ... \plus{} a_{10}^2 \leq \frac {1}{a_1^2} \plus{} \frac {1}{a_2^2} \plus{} ... \plus{} \frac {1}{a_{10}^2}$[/quote]
Use Vasc's LCF-theorem. :wink:[/quote]
What is Vasc's LCF-theorem?Is it similar to mixing variables?

[quote="KDS"]
What is Vasc's LCF-theorem?Is it similar to mixing variables?[/quote]
It's the following theorem.
Let $ f$ defined on an interval $ I$ and $ f$ is a concave function for $ u\leq s,$ $ s\in I.$
If $ \sum_{i\equal{}1}^{n}f(x_i)\leq nf\left(\frac{1}{n}\sum_{i\equal{}1}^nx_1\right)$ for all $ x_i\in I$ such that $ \frac{1}{n}\sum_{i\equal{}1}^nx_1\equal{}s$ and $ x_1\equal{}x_2\equal{}...\equal{}x_{n\minus{}1}\leq s,$
then $ \sum_{i\equal{}1}^{n}f(x_i)\leq nf\left(\frac{1}{n}\sum_{i\equal{}1}^nx_1\right)$ holds for all $ x_i\in I$ such that $ \frac{1}{n}\sum_{i\equal{}1}^nx_1\equal{}s.$


$ (\forall)n\in \mathbb{N}$ let be $ p(n)$ the biggest perfect square such that $ p(n)\le n$ . Find all the real numbers $ a$ such that $ p\left(\frac{a\plus{}1}{2}\right)\equal{}\frac{a\plus{}2}{3}$

[hide]Obviously $ p(n) \equal{} \lfloor\sqrt {n}\rfloor^2$, so the equation becomes

$ \left\lfloor\sqrt {a \plus{} 1\over 2}\right\rfloor^2 \equal{} {a \plus{} 2\over 3}$

Put $ \sqrt {a \plus{} 1\over 2} \equal{} n \plus{} \alpha$ where $ n$ is the integer part and $ \alpha$ the fractional part.

Then $ a \equal{} 2(n^2 \plus{} 2n\alpha \plus{} \alpha^2) \minus{} 1$, hence the equation becomes

$ 3n^2 \equal{} 2n^2 \plus{} 4n\alpha \plus{} 2\alpha^2 \plus{} 1\iff 2\alpha^2 \plus{} 4n\alpha \minus{} n^2 \plus{} 1 \equal{} 0$

The non-negative solution is $ \alpha \equal{} {\sqrt {6n^2 \minus{} 2} \minus{} 2n\over 2}$

We must have $ 0\leqslant\alpha < 1$, which yields $ 1\leqslant n^2 < 4n \plus{} 3$. The right-hand part yields $ n\in(2 \minus{} \sqrt {7},2 \plus{} \sqrt {7})$ hence $ 0\leqslant n\leqslant 4$, but $ n \equal{} 0$ doesn't satisfy the left-hand part.

$ \sqrt {a \plus{} 1\over 2} \equal{} n \plus{} \alpha \equal{} {\sqrt {6n^2 \minus{} 2}\over 2}\iff a \equal{} 3n^2 \minus{} 2$, hence $ a\in\{1, 10, 25, 46\}$

NOTE: If the arguments of $ p(\cdot)$ can ONLY be natural numbers, then $ a\in\{1,25\}$[/hide]

[quote="Farenhajt"][hide]Obviously $ p(n) \equal{} \lfloor\sqrt {n}\rfloor^2$, so the equation becomes

$ \left\lfloor\sqrt {a \plus{} 1\over 2}\right\rfloor^2 \equal{} {a \plus{} 2\over 3}$

Put $ \sqrt {a \plus{} 1\over 2} \equal{} n \plus{} \alpha$ where $ n$ is the integer part and $ \alpha$ the fractional part.

Then $ a \equal{} 2(n^2 \plus{} 2n\alpha \plus{} \alpha^2) \minus{} 1$, hence the equation becomes

$ 3n^2 \equal{} 2n^2 \plus{} 4n\alpha \plus{} 2\alpha^2 \plus{} 1\iff 2\alpha^2 \plus{} 4n\alpha \minus{} n^2 \plus{} 1 \equal{} 0$

The non-negative solution is $ \alpha \equal{} {\sqrt {6n^2 \minus{} 2} \minus{} 2\over 2}$

We must have $ 0\leqslant\alpha < 1$, which yields $ 1\leqslant n^2 < 3$, hence $ n \equal{} 1$. That in turn gives $ \alpha \equal{} 0$, so $ \sqrt {a \plus{} 1\over 2} \equal{} 1\iff a \equal{} 1$ is the only solution.[/hide][/quote]

Please verify again your solution . You missed something .

My solution :

Obviously $ a$ is odd . We have also that :

$ \frac {a \plus{} 2}{3} \equal{} k^2\ ,\ k\in \mathbb{Z}$ .

$ \Rightarrow a \equal{} 3k^2 \minus{} 2$($ k$ is odd) .

From $ k^2\le \frac {a \plus{} 1}{2} \equal{} \frac {3k^2 \minus{} 1}{2}\le k^2 \plus{} 2k \plus{} 1\Rightarrow k(k \minus{} 4) < 3\Rightarrow k\in \{1,3\}\Rightarrow a\in \{1,25\}$

Edit : Sorry , i posted in the same time as you edited , now you're solution is good

[quote="alex2008"]Edit : Sorry , i posted in the same time as you edited , now you're solution is good[/quote]

Yup, an "$ n$" omitted led to a disaster :)


Find integers $x,y,z$ with  $12x+15y+20z=1$.

fixed $z$ and see http://mathworld.wolfram.com/DiophantineEquation.html


From the given bar graph of monthly rainfall for Dry Gulch, find the number of inches in the mean rainfall per month.

[asy]size (10cm,10cm);
pair A,B,C,D,E,two,four,six,eight,ten,twelve,fourteen,sixteen,eighteen,twenty,jan1,jan2,jan3,jan4,feb1,feb2,feb3,
feb4,mar1,mar2,mar3,mar4,apr1,apr2,apr3,apr4,may1,may2,may3,may4,jun1,jun2,jun3,jun4,jul1,jul2,jul3,jul4,
aug1,aug2,aug3,aug4,sep1,sep2,sep3,sep4,oct1,oct2,oct3,oct4,nov1,nov2,nov3,nov4,dec1,dec2,dec3,dec4;
A=(0,0);
B=(36,0);
C=(0,20);
D=(0,20.5);
E=(37,0);

two=(0,2);
four=(0,4);
six=(0,6);
eight=(0,8);
ten=(0,10);
twelve=(0,12);
fourteen=(0,14);
sixteen=(0,16);
eighteen=(0,18);
twenty=(0,20);

jan1=(1,0);
jan2=(1,2);
jan3=(3,2);
jan4=(3,0);

draw (jan1--jan2--jan3--jan4);

label ("$J$", midpoint(jan1--jan4),S);

label ("$2$", midpoint(jan2--jan3),N);

feb1=(4,0);
feb2=(4,3);
feb3=(6,3);
feb4=(6,0);

draw (feb1--feb2--feb3--feb4);

label ("$F$",midpoint(feb1--feb4),S);

label ("$3$", midpoint(feb2--feb3),N);

mar1=(7,0);
mar2=(7,6);
mar3=(9,6);
mar4=(9,0);

draw (mar1--mar2--mar3--mar4);

label("$M$",midpoint(mar1--mar4),S);

label ("$6$", midpoint(mar2--mar3),N);

apr1=(10,0);
apr2=(10,8);
apr3=(12,8);
apr4=(12,0);

draw (apr1--apr2--apr3--apr4);

label("$A$",midpoint(apr1--apr4),S);

label ("$8$", midpoint(apr2--apr3),N);

may1=(13,0);
may2=(13,7);
may3=(15,7);
may4=(15,0);

draw (may1--may2--may3--may4);

label ("$M$",midpoint(may1--may4),S);

label ("$7$", midpoint(may2--may3),N);

jun1=(16,0);
jun2=(16,4);
jun3=(18,4);
jun4=(18,0);

draw (jun1--jun2--jun3--jun4);

label("$J$",midpoint(jun1--jun4),S);

label ("$4$", midpoint(jun2--jun3),N);

jul1=(19,0);
jul2=(19,2);
jul3=(21,2);
jul4=(21,0);

draw (jul1--jul2--jul3--jul4);

label ("$J$",midpoint(jul1--jul4),S);

label ("$2$", midpoint(jul2--jul3),N);

aug1=(22,0);
aug2=(22,1);
aug3=(24,1);
aug4=(24,0);

draw (aug1--aug2--aug3--aug4);

label ("$A$", midpoint(aug1--aug4),S);

label ("$1$", midpoint(aug2--aug3),N);

sep1=(25,0);
sep2=(25,3);
sep3=(27,3);
sep4=(27,0);

draw (sep1--sep2--sep3--sep4);

label ("$S$",midpoint(sep1--sep4),S);

label ("$3$", midpoint(sep2--sep3),N);

oct1=(28,0);
oct2=(28,5);
oct3=(30,5);
oct4=(30,0);

draw (oct1--oct2--oct3--oct4);

label ("$O$",midpoint(oct1--oct4),S);

label ("$5$", midpoint(oct2--oct3),N);

nov1=(31,0);
nov2=(31,3);
nov3=(33,3);
nov4=(33,0);

draw (nov1--nov2--nov3--nov4);

label ("$N$",midpoint(nov1--nov4),S);

label ("$3$", midpoint(nov2--nov3),N);

dec1=(34,0);
dec2=(34,2);
dec3=(36,2);
dec4=(36,0);

draw (dec1--dec2--dec3--dec4);

label("$D$",midpoint(dec1--dec4),S);

label ("$2$", midpoint(dec2--dec3),N);

draw (A--B);
draw (A--C);
draw (C--D);
draw (B--E);

add(pathticks(A--C,1,0.1,7,25));
add(pathticks(A--C,1,0.2,7,25));
add(pathticks(A--C,1,0.3,7,25));
add(pathticks(A--C,1,0.4,7,25));
add(pathticks(A--C,1,0.5,7,25));
add(pathticks(A--C,1,0.6,7,25));
add(pathticks(A--C,1,0.7,7,25));
add(pathticks(A--C,1,0.8,7,25));
add(pathticks(A--C,1,0.9,7,25));
add(pathticks(A--C,1,1,7,25));

label ("$2$", two,W);
label ("$4$", four,W);
label ("$6$", six,W);
label ("$8$", eight,W);
label ("$10$", ten,W);
label ("$12$", twelve,W);
label ("$14$", fourteen,W);
label ("$16$", sixteen,W);
label ("$18$", eighteen,W);
label ("$20$", twenty,W);

label(rotate(90)*"\textbf{Rainfall (inches)}",shift(-5,0)*(A--D));[/asy]

But the answer said that it was 4. I said 23/6 and it marked me incorrect.


geometry
geometric transformation
reflection

i have applied to u of toronto, sfu, waterloo in the program of actuarial science.  My average is around 92%, is it likely for me to get accepted in all three?  Which one is the best out of the three? What actually is actuarial science all about? I applied simply because I am good with numbers.  Any advices will be greatly appreciated

o dam danny applied for Waterloo...guess im not going to be top of the class. o did u get ur acceptance letters yet?

I applied to AS in Waterloo too and probably don't know much more on this than you guys do. But here is what I know:

Actuarial Science deals with the calculating of risks and profits in financial situations; an actuary's job is generally determining what to do in order to maximize the profits while maintaining a manageable level of risk, such as calculating insurance rates, determining which project to invest money in etc. A large part of the study will be based on statistics, math modeling and probabilities.

Hopefully this helps!

@hhung: In my opinion, Actuarial Science is not that competitive a major and a 92% average should be fine for all these schools.

[quote="soledad42"]
Actuarial Science deals with the calculating of risks and profits in financial situations; an actuary's job is generally determining what to do in order to maximize the profits while maintaining a manageable level of risk, such as calculating insurance rates, determining which project to invest money in etc. A large part of the study will be based on statistics, math modeling and probabilities.

Hopefully this helps!
[/quote]

Thanks skel, that helps so much!...NOT. Obviously I am asking for more specific information.

If you need more specific information, why don't you check out the Waterloo website?

This private message was mysteriously sent to me today:
[quote="Soledad42"]

[quote="Mathgod"]

Ya, I have also applied to Actuarial Science at Waterloo, but I am clueless of what it is all about. Can anyone offer some insight please?

Thanks skel, that helps so much!...NOT. Obviously I am asking for more specific information.  [/quote]

In the first passage, your use of the word "clueless" and the inquiry for insight have effectively created a tone of ignorance and curiosity. However, much less obviously, you have also subtly created suspense with this post, leaving the reader a lot of room for different interpretations as seen from the second passage. The most prominent technique in the second passage is irony. The verbal irony used in the first sentence of the second passage accurately reflects a tone of sarcasm mixed with disappointment; interestingly enough, this passage also fully displays situational irony, where a kindhearted effort directed at helping the author was sarcastically rejected and ridiculed. Following the use of irony, you immediately destroy this atmosphere with an all-capitalized word "NOT", which reinforces the disappointment. The last sentence in this passage serves to point out the real meaning of the first sentence, conclude that the interpretation of the reader is "obviously" incorrect and unveil the previously unclear desired answers for the inquiry. Interestingly enough, when we juxtapose this sentence with the previous words such as "clueless", we have discovered that as unlikely as it may sound, the author have acquired a significant amount of knowledge towards the subject of Actuarial science in approximately 23 hours and 44 minutes. Through the uses of irony and the shift of tone from one passage to the next, you use this misinterpretation to exhibit the theme that confusion may lead to unexpected, and sometimes undesired, results.

P.S.: Just finished a commentary in English yesterday. Can't stop this mode.

[/quote]

Sorry Skely, I can't help it. xD. You can scold me on Sunday. :)

[quote="rem"]92% should definitely be enough, especially at UofT, which admits almost everyone but then there's a gigantic dropout rate in 1st yrs (especially in lifesci :rotfl: )
I think when it comes to Actuarial Science, UW is the best, then UofT, then SFU.
Yeah, actuarial science is basically using math and stats and I guess some programming to calculate risks. Pretty good job I think.
use google to find more details about actuarial science :D[/quote]
when u say there is a gigantic dropout rate, do u mean they got cut, or they just don't feel like taking it anymore, sorry for my ignorant :blush:

Well some get cut, others just don't want to continue with a low 50 in the course...
However, if you work hard and keep up with the work, I am sure this will not be a problem for you.

If you guys are all planning on going into ActSci at Waterloo but don't actually know what it is, maybe Mathematical Finance would be something interesting to look into as well. It's a joint program between the departments of pure math and actuarial science and is a much more challenging program.

Math Finance is like taking the fun pure math courses, and not having to take the boring Actsci ones (you only have to do the more interesting/useful ones =P)  On another note, you don't write the Actsci exams.

I am told that the SFU Actsci program is really good, is harder than the Waterloo ones, but they also have a better hiring-upon-graduation rate.  (Don't quote me on this, this is from a friend who graduated from SFU a couple years ago).

Soledad's info on actuary's was really good actually, what do you mean by more specific information?

Just don't get stuck doing taxes for your first waterloo co-op job.  That kinda sucks.

[quote="eddyliu"]and no one answered my question...either of u got ur acceptances yet?[/quote]

Yes, I got accepted.

[quote="eddyliu"]wait wat was ur average?[/quote]

Excluding English, around 96-97.

[quote="eddyliu"]wait wat was ur average?[/quote]
I think Danny's average would not matter for acceptance  :D

[quote="eddyliu"]yo danny did u apply for US schools?[/quote]
I think this is a forum, if you want to discuss this kind of questions, it might be better to do it over PM/facebook/email :wink:
Thanks

waterloo's act sci is best in north america, its not even close. sfu can't be near it, as west coast's entry level actuarial positions are very very small. toronto and waterlerloo are the insurance capitals of canada.

I can answer some questions indirectly about the program, I was going into it but got swayed by the beauty of pure math. Waterloo's act sci program is very large, you will have lots of peers (and competition).


hey ppl help me.....
i m literally new 2 internet n stuff...
how did u rite d question in the understanable way..pls help me
n tell me n e thing u found difficult wen u joined..
thanks

you don't look new to the internet,already using the language(ppl,etc ;) )

obviously, if you look at his other posts, he can write perfectly well...this should be locked

( i wish i had the awesome mod powers to do that...  :lol: )

This forum is for problems, so if you have any questions on phrasing problems, please ask on a different forum.

Wait do you mean [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php]$\underline{\text{\LaTeX}}$[/url]?


real analysis
integration
calculus
derivative
algebra
polynomial
function

$D_{a}(t)=\frac{sin2 \pi at}{\pi t}\in L^{2}(R)$ is the fourier transform of $\chi_{[-a,a]}$.

Prove:
(1).$\int_{-a}^{a}f(\xi)e^{-2\pi i \xi t}d\xi=\int_{R}\hat{f}(t)D_{a}(x-t)dt$ for all $f \in L^{p}(R)$, $1\leq p\leq 2$
(2).is it ture for $p>2$?

For (1):

so far the only rigorous proof I found is for $p=2$ only, it is Folland "real analysis" Lemma 8.34.

For $1\leq p<2$ how do we even know that RHS is finite a.e. ?

For(2): maybe difficult.

To start with, there's no doubt that this is formally true - that it's the right formula. (Oops - not quite the right formula - that should be an $x$ in the exponent on the left hand side, not a $t.$) For instance, suppose that $f$ belongs the the Schwartz class (overkill, since you will only need a few derivatives and some polynomial rate of decay). Then you shouldn't have any trouble proving this true. And the Schwartz class is dense in $L^{p}$ for $1\le p<\infty.$

The case of $p=1$ is problematic for the reasons you've indicated. I will be able to show that, as an improper integral, $\int_{\mathbb{R}}\hat{f}(t)D_{a}(x-t)\,dt$ is Cesaro summable or Abel summable to the left hand side. But, as you mentioned, we're unable to conclude that it exists as a Lebesgue integral.

Now, as for the case $p>2,$ here's a serious question: what does it even mean?

For $f\in L^{1}(\mathbb{R}),$ we can define the Fourier transform $\hat{f}$ directly. The integral defining it converges.

For $f\in L^{1}\cap L^{2},$ we have $\hat{f}$ defined as before, and we can prove the Plancherel Theorem: $\|\hat{f}\|_{2}=\|f\|_{2}.$ Given that, and given that $L^{1}\cap L^{2}$ is dense in $L^{2},$ we can extend the definition of the Fourier transform to all of $L^{2}.$ What we're doing is extending the definition of a uniformly continuous function to the completion of its original domain.

For $1<p<2,$ we have the Hausdorff-Young inequality: $\|\hat{f}\|_{q}\le\|f\|_{p}$ for $\frac1p+\frac1q=1.$ Again, this allows us to extend the definition of the operator from $L^{1}\cap L^{p}$ to $L^{p}$ by continuity, and the meaning of $\hat{f}$ is clear.

But on $L^{p}$ for $p>2?$ Sure, we can define the Fourier transform on $L^{1}\cap L^{p},$ and sure, we can define, as a tempered distribution, the Fourier transform of a tempered distribution. But we have no Hausdorff-Young inequality - no norm inequality that allows us to continuously or uniquely extend the Fourier transform operator to all of $L^{p}.$

----

The original question would make more sense with the roles reversed:

Show that $\int_{-a}^{a}\hat{g}(\xi)e^{-2\pi i\xi x}\,d\xi=\int_{\mathbb{R}}g(t)D_{a}(x-t)\,dt$ for all $g\in L^{p}(\mathbb{R}),\ 1\le p\le 2.$

[quote="Kent Merryfield"]
The original question would make more sense with the roles reversed:

Show that $\int_{-a}^{a}\hat{g}(\xi)e^{-2\pi i\xi x}\,d\xi=\int_{\mathbb{R}}g(t)D_{a}(x-t)\,dt$ for all $g\in L^{p}(\mathbb{R}),\ 1\le p\le 2.$[/quote]
Yes, indeed. In this form, Let $1/p+1/2=1/r+1$, then since $1\leq p\leq 2$, $r>0$. So RHS is weak $L^{r}$ by Folland 8.9(b), thus finite a.e. But will this form imply the form given in problem?

[quote="Kent Merryfield"]
For instance, suppose that f belongs the the Schwartz class (overkill, since you will only need a few derivatives and some polynomial rate of decay). Then you shouldn't have any trouble proving this true. And the Schwartz class is dense in L^{p} for 1\le p<\infty. [/quote]
using $f_{n}\in S$ and $f_{n}\to f$, $\hat{f_{n}}\to \hat{f}$ works fine for Plencherel, (uniforma continuity), however maynot work for pointwise (a.e) equality, since can't use LDCT, what is the dominated function for LHS and RHS? Do we have to choose $|f_{n}|<|f|$?


I'm trying to find a solution... here's first step:
[b]Easy ineq[/b] \[a+b+c\geq 3\sqrt[3]{abc}\] So if $\sqrt[6]{\frac{a+b}{a+c}}+\sqrt[6]{\frac{b+c}{b+a}}+\sqrt[6]{\frac{c+a}{c+b}}\le q$ 1, we are done.
So assume it is greater, now show stuff...

Nobody has posted here? I'm really curious about this one (btw my previous post is nothing... please ignore it...)

I guess I can't think of anything good because i'm a beginner so i will try to expand everything but it won't be fun...

[quote="ductrung"]For any positive real numbers $a,b$ and $c$, prove that \[\sqrt[6]{\frac{a+b}{a+c}}+\sqrt[6]{\frac{b+c}{b+a}}+\sqrt[6]{\frac{c+a}{c+b}}\le \frac{a+b+c}{\sqrt[3]{abc}}\] [/quote]
By Holder's: \[LHS^{3}\leq 3\sum_{cyc}\sqrt{a+b}\sum_{cyc}{1 \over \sqrt{a+b}}=9+3\sum_{sym}\sqrt{a+b \over a+c}.\] We are now to prove that: \[{(a+b+c)^{3}\over abc}\ge 9+3\sum_{sym}\sqrt{a+b \over a+c}.\] And we also have: \[{2(a+b+c)^{3}\over 9abc}\geq \sum_{sym}\sqrt{a+b \over a+c}\\ \iff{(a+b+c)\sum_{cyc}(b-c)^{2}+6\sum_{cyc}a(b-c)^{2}\over 9abc}\ge \sum_{sym}{(b-c)^{2}\over \sqrt{a+b}\sqrt{a+c}\left(\sqrt{a+b}+\sqrt{a+c}\right)^{2}}.\] But: \[{(7a+b+c)\sqrt{a+b}\sqrt{a+c}\left(\sqrt{a+b}+\sqrt{a+c}\right)^{2}}\ge 9abc.\] Proof complete.

\[LHS^{3}\leq 3\sum_{cyc}\sqrt{a+b}\sum_{cyc}{1 \over \sqrt{a+b}}=9+3\sum_{sym}\sqrt{a+b \over a+c}.\] I've been researching Holders and I'm pretty sure that's not it... but close (not sure where the constant three comes from...)

[quote="me@home"] \[LHS^{3}\leq 3\sum_{cyc}\sqrt{a+b}\sum_{cyc}{1 \over \sqrt{a+b}}=9+3\sum_{sym}\sqrt{a+b \over a+c}.\] I've been researching Holders and I'm pretty sure that's not it... [/quote]
$\sqrt[3]{3\sum_{cyc}\sqrt{a+b}\sum_{cyc}{1 \over \sqrt{a+b}}}=$
$(1^{3}+1^{3}+1^{3})^\frac{1}{3}\left(\left(\sqrt[6]{a+b}\right)^{3}+\left(\sqrt[6]{b+c}\right)^{3}+\left(\sqrt[6]{c+a}\right)^{3}\right)^\frac{1}{3}\left(\left(\frac{1}{\sqrt[6]{a+c}}\right)^{3}+\left(\frac{1}{\sqrt[6]{b+a}}\right)^{3}+\left(\frac{1}{\sqrt[6]{c+b}}\right)^{3}\right)^\frac{1}{3}\geq$
$\geq\sum_{cyc}1\cdot\sqrt[6]{a+b}\cdot\frac{1}{\sqrt[6]{a+c}}=LHS.$ :wink:

Wow, very nice!! :o
Thank you so much, except I am still a little confused: can't you always insert more '1' terms and introduce new constants?

That would change the powers, wouldn't it?

[quote="Lovasz"] \[{2(a+b+c)^{2}\over 9abc}\geq \sum_{sym}\sqrt{a+b \over a+c}\] .[/quote]
Should it be \[{6(a+b+c)^{3}\over 9abc}\geq \sum_{sym}\sqrt{a+b \over a+c}\] . :?:

[quote="Lovasz"][quote="zhaobin"][quote="Lovasz"] \[{2(a+b+c)^{3}\over 9abc}\geq \sum_{sym}\sqrt{a+b \over a+c}\] .[/quote]
Should it be \[{6(a+b+c)^{3}\over 9abc}\geq \sum_{sym}\sqrt{a+b \over a+c}\] . :?:[/quote]
The RHS contains 6 terms![/quote]
Oh,I am wrong,but you are wrong，either. :lol:
Then it should be: \[{2(a+b+c)^{3}\over 9abc}\geq \sum_{sym}\sqrt{a+b \over a+c}\]

[quote="zhaobin"][/quote][quote="Lovasz"][quote="zhaobin"][quote="Lovasz"] \[{2(a+b+c)^{3}\over 9abc}\geq \sum_{sym}\sqrt{a+b \over a+c}\] .[/quote]
Should it be \[{6(a+b+c)^{3}\over 9abc}\geq \sum_{sym}\sqrt{a+b \over a+c}\] . :?:[/quote]
The RHS contains 6 terms![/quote][quote="zhaobin"]
Oh,I am wrong,but you are wrong，either. :lol:
Then it should be: \[{2(a+b+c)^{3}\over 9abc}\geq \sum_{sym}\sqrt{a+b \over a+c}\] [/quote]
Sorry but I edited it before your last post! Thank you!

[quote="ductrung"]Nice proof, Lovasz.

Actually, the following stronger result holds \[\sqrt[3]{\frac{a+b}{a+c}}+\sqrt[3]{\frac{b+c}{b+a}}+\sqrt[3]{\frac{c+a}{c+b}}\le\frac{a+b+c}{\sqrt[3]{abc}}\] [/quote]
Let's Holder it agian! \[\left(\sqrt[3]{\frac{a+b}{a+c}}+\sqrt[3]{\frac{b+c}{b+a}}+\sqrt[3]{\frac{c+a}{c+b}}\right)^{3}\\ \le 6(a+b+c)\left({1 \over a+b}+{1 \over b+c}+{1 \over c+a}\right) \\ \le{3(a+b+c)\left({1 \over a}+{1 \over b}+{1 \over c}\right).}\] The rest is: \[{(a+b+c)^{3}\over abc}\ge{3(a+b+c)}\left({1 \over a}+{1 \over b}+{1 \over c}\right) \\ \iff (a+b+c)^{2}\ge 3(ab+bc+ca).\] Proof complete.

Yes, Lovasz  :lol:  it is essentially the same as my proof, which concludes with a slightly stronger form: \[\sqrt{ \frac{(a+b+c)^{3}}{3abc}}\ge\sqrt{\frac{a+b}{a+c}}+\sqrt{\frac{b+c}{b+a}}+\sqrt{\frac{c+a}{c+b}}\] First, I have 2 different proofs to the original 6-root version (both different from yours) and then found the same proof as yours for the 3-root version. I suspect that the 2-root version also holds \[\frac{a+b+c}{\sqrt[3]{abc}}\ge\sqrt{\frac{a+b}{a+c}}+\sqrt{\frac{b+c}{b+a}}+\sqrt{\frac{c+a}{c+b}}\]

ductrung, I don't know what's happening exactly  :o , but it is very extraordinary when I realized you, darij grinberg and I had created some inequalities with many similarities. Before starting the new school year, I also found the following inequalities, for all positive reals: \begin{eqnarray*}{(x+y+z)^{3}\over 18xyz}&\geq&{x \over y+z}+{y \over z+x}+{z \over x+y}, \\{x+y+z \over \sqrt[3]{xyz}}&\geq& \sqrt{2x \over y+z}+\sqrt{2y \over z+x}+\sqrt{2z \over x+y}. \end{eqnarray*} The second one asserts your 2-root inequality!  :)

[quote="Lovasz"] \begin{eqnarray*}{(x+y+z)^{3}\over 18xyz}&\geq&{x \over y+z}+{y \over z+x}+{z \over x+y}, \\{x+y+z \over \sqrt[3]{xyz}}&\geq& \sqrt{2x \over y+z}+\sqrt{2y \over z+x}+\sqrt{2z \over x+y}. \end{eqnarray*} The second one asserts your 2-root inequality!  :)[/quote]

Lovasz, could you explain why the 2nd above implies the 2-root inequality I posted?

If so, that would imply all the version I posted above and we only need a proof for that one.

PS: The first can be proved this way \[\frac{(x+y+z)^{3}}{18xyz}\ge \frac{1}{2}+\frac{x^{2}+y^{2}+z^{2}}{xy+yz+zx}\ge\sum_{cycl}\frac{x}{y+z}\]

ductrung, I thought it is Rearrangement! But finally I found it wrong.
However, I got a solutions for the second one last night. But I believe there's at least 2 or 3 proofs!

[quote]I suspect that the 2-root version also holds \[\frac{a+b+c}{\sqrt[3]{abc}}\ge\sqrt{\frac{a+b}{a+c}}+\sqrt{\frac{b+c}{b+a}}+\sqrt{\frac{c+a}{c+b}}\] [/quote]

Yes, it's true, too :)

And yes, Lovasz, I found 2 different proofs for the 2nd inequality of yours.

[quote="Lovasz"]Would you mind posting the [u]harder[/u] proof (not trivial solution by AM-GM) for both two 2nd-root inequalities, ductrung?

I wonder if ours are different.[/quote]
We use the following lemma \[\frac{3(a+b+c)}{2(ab+bc+ca)}\ge\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\] which is easy to prove. \[\left( \frac{a+b+c}{\sqrt[3]{abc}}\right)^{2}\ge \frac{3(a+b+c)^{2}}{ab+bc+ca}\ge 2(a+b+c)\sum_{cycl}\frac{1}{a+c}= \sum_{cycl}(a+b)\sum_{cycl}\frac{1}{a+c}\]  \[\ge \left(\sum_{cycl}\sqrt{\frac{a+b}{a+c}}\right)^{2}\] The first inequality is due to the AM-GM and the last Cauchy-Schwarz.

Exactly the same idea solves your 2-root problem; and :) \[\frac{a+b+c}{\sqrt[3]{abc}}\ge \sqrt{ \frac{2a}{a+b}}+\sqrt{ \frac{2b}{b+c}}+\sqrt{ \frac{2c}{c+a}}\] I found the 6-root version first and then proved the 3-root and then get stuck for a while with the 2-root version. If I found the 2-root first, I'd not have posted the other versions.

[quote="ductrung"]We use the following lemma \[\frac{3(a+b+c)}{2(ab+bc+ca)}\ge\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\] [/quote]
Mine is different (a bit only)!
[quote="ductrung"] \[\frac{a+b+c}{\sqrt[3]{abc}}\ge \sqrt{ \frac{2a}{a+b}}+\sqrt{ \frac{2b}{b+c}}+\sqrt{ \frac{2c}{c+a}}\] [/quote]
Hmm... It's rearrangement since my 2-nd root. Anyhow, we have: \[\sum_{cyc}\sqrt{2a \over a+b}\leq 3,\] Moreover, \[{a+b+c \over \sqrt[3]{abc}}\ge 3 \ !\] Finally, our solutions aren't same at all!

Could anyone show me how to cope with this inequality $ {(a\plus{}b\plus{}c)^{3}\over abc}\ge 9\plus{}3\sum_{sym}\sqrt{a\plus{}b \over a\plus{}c}$

because I don't know how to invent presented proof.

[quote="matex (L)(L)(L)"]Could anyone show me how to cope with this inequality $ {(a \plus{} b \plus{} c)^{3}\over abc}\ge 9 \plus{} 3\sum_{sym}\sqrt {a \plus{} b \over a \plus{} c}$

because I don't know how to invent presented proof.[/quote]

Using AM-GM Inequality we have
\[ {(a \plus{} b \plus{} c)^{3}\over 3abc}\ge 9
\]
It suffices to prove that
\[ {2(a \plus{} b \plus{} c)^{3}\over 9abc}\ge \sum_{sym}\sqrt {a \plus{} b \over a \plus{} c}
\]
By Cauchy-Schwarz Inequality we have
\[ \left(\sum_{sym}\sqrt {a \plus{} b \over a \plus{} c}\right)^2\le 4(a \plus{} b \plus{} c)\left(\sum_{sym}\frac1{a \plus{} b}\right)
\]
Easy to prove that
\[ \sum_{sym}\frac1{a \plus{} b} \equal{} 2\sum_{cyc}\frac1{a \plus{} b}\le\frac {8(a \plus{} b \plus{} c)^2}{3(a \plus{} b)(b \plus{} c)(c \plus{} a)}\le\frac {(a \plus{} b \plus{} c)^2}{3abc}
\]
It suffices to prove that
\[ \left({2(a \plus{} b \plus{} c)^{3}\over 9abc}\right)^2\ge \frac {4(a \plus{} b \plus{} c)^3}{3abc}
\]

\[ \Leftrightarrow (a \plus{} b \plus{} c)^3\ge27abc
\]
Which true by AM-GM Inequality.  So we have done.:)

thank You very very much

There are really great mathematics on this forum.

Here is my result:

If $ 2b\plus{}2c \geq a \geq b,c >0$, then

$ \sqrt[4]{\frac{a\plus{}b}{b\plus{}c}} \plus{} \sqrt[4]{\frac{b\plus{}c}{c\plus{}a}}\plus{}\sqrt[4]{\frac{c\plus{}a}{a\plus{}b}} \leq \frac{26(a^2\plus{}b^2\plus{}c^2)\plus{}22(ab\plus{}bc\plus{}ca)}{7(a^2\plus{}b^2\plus{}c^2)\plus{}9(ab\plus{}bc\plus{}ca)}$.

i think it is very hard.

[quote="ductrung"]For any positive real numbers $ a,b$ and $ c$, prove that
\[ \sqrt [6]{\frac {a \plus{} b}{a \plus{} c}} \plus{} \sqrt [6]{\frac {b \plus{} c}{b \plus{} a}} \plus{} \sqrt [6]{\frac {c \plus{} a}{c \plus{} b}}\le \frac {a \plus{} b \plus{} c}{\sqrt [3]{abc}}
\]
[/quote]
For any positive real numbers $ a,b$ and $ c$, $ k\ge 2$.We also have

\[ \sqrt [k]{\frac {a \plus{} b}{a \plus{} c}} \plus{} \sqrt [k]{\frac {b \plus{} c}{b \plus{} a}} \plus{} \sqrt [k]{\frac {c \plus{} a}{c \plus{} b}}\le \frac {a \plus{} b \plus{} c}{\sqrt [3]{abc}}
\]

[quote="nguoivn"][quote="Nak"]Here is my result:

If $ 2b \plus{} 2c \geq a \geq b,c > 0$, then

$ \sqrt [4]{\frac {a \plus{} b}{b \plus{} c}} \plus{} \sqrt [4]{\frac {b \plus{} c}{c \plus{} a}} \plus{} \sqrt [4]{\frac {c \plus{} a}{a \plus{} b}} \leq \frac {26(a^2 \plus{} b^2 \plus{} c^2) \plus{} 22(ab \plus{} bc \plus{} ca)}{7(a^2 \plus{} b^2 \plus{} c^2) \plus{} 9(ab \plus{} bc \plus{} ca)}$.

i think it is very hard.[/quote]

Dear my friend, sorry but indeed I think it's ugly and very weak. By Am-Gm, we are easily to prove the stronger and nicer:
With $ a, b, c \geq\ 0$, we have: $ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \plus{} 1 \geq\ \frac {2}{3} (\frac {a \plus{} b}{b \plus{} c} \plus{} \frac {b \plus{} c}{c \plus{} a} \plus{} \frac {c \plus{} a}{a \plus{} b})$
 :maybe:[/quote]
please, explain why it is stronger.


geometry
circumcircle
search
geometry proposed

A regular pentagon $MNPQR$ is inscribed in a circumference, and $F$ is a point on arc $BC$. Prove that
\[\overline{FM}+\overline{FQ}=\overline{FN}+\overline{FP}+\overline{FR}\]

This is a well known old problem which is also true in the configuration of a regular [b][size=100]n-gon,[/size][/b] where [b][size=100]n[/size][/b] is any odd number.

So if $A_{1}A_{2}A_{3}...A_{2n+1},$ is a regular polygon, where [b][size=100]n[/size][/b] any integer and $M,$ an arbitrary point on its circumcircle, then the follows equality is true: $MA_{1}+MA_{3}+MA_{5}+...+MA_{2n+1}= MA_{2}+MA_{4}+MA_{6}+...+MA_{2n}.$

But I am not good in references, I don't remember the proof and I am so lazy to search about it, in ny archives.

Kostas Vittas.

[quote="vittasko"]This is a well known old problem which is also true in the configuration of a regular [b][size=100]n-gon,[/size][/b] where [b][size=100]n[/size][/b] is any odd number.

So if $A_{1}A_{2}A_{3}...A_{2n+1},$ is a regular [b][size=100]n-gon,[/size][/b] where [b][size=100]n[/size][/b] any integer and $M,$ an arbitrary point on its circumcircle, then the follows equality is true: $MA_{1}+MA_{3}+MA_{5}+...+MA_{2n+1}= MA_{2}+MA_{4}+MA_{6}+...+MA_{2n}.$

But I am not good in references, I don't remember the proof and I am so lazy to search about it, in ny archives.

Kostas Vittas.[/quote]

one correction: is one (2n+1)-gon!  :D

to prove this, see that, by Ptolomeu, $MA_{i}+MA_{i+2}= 2MA_{i+1}cos(\frac{360}{2n+1})$, for all i's (from here is easy to deduce the result)

Thank you [b][size=100]e.lopes[/size][/b] for the remark. I was inattentive in ''copy-paste''.

kostas Vittas.


induction
combinatorics proposed
combinatorics

let $ P_1,P_2,...,P_n$ are n point in plane and  M=Maxl$ P_iP_j$l  such that repeated in K times.
prove that :\[ n \geq K \geq 1\]

P. Erd�s:  On sets of distances of $ n$ points,  Amer. Math. Monthly  53 (1946), 248--250 MR7,471c; Zentralblatt 60,348.

http://www.renyi.hu/~p_erdos/1946-03.pdf

I hope somebody gives a non inductive proof.


calculus
integration
calculus computations

Hello everyone!

I am having lots of trouble with a problem I can't figure out and was hoping someone could help. The problem is located in this website...

[url]http://www.kent.k12.wa.us/pcpow/questions/calc/current/index.html[/url]

Thank you very much!

our high school a.p. calculus teacher gave us this same problem to do for extra credit. i'm having a really hard time with this one! does anybody know how to do it??? i would really appreciate the help!!! :help:

That's a standard problem about integrals and derivaives. Let me just say that the distance of the speed racer from the origin at the moment $t$ is $\int_0^t v(s)\,ds$ and the acceleration at the moment $t$ is $v'(t)$. What makes the problem unpleasant is that, in order to find the answers, you'll need to solve various equations like $e^t-t-1=100$, etc., which cannot be solved analytically, so there will be a lot of numerical computations (fortunately, we have calculators nowdays to do them) and the answers are going to be some ugly decimals.

Thank You Very Much!
Hey, and sorry about the post in that other forum!


absolute value
complex numbers
superior algebra
superior algebra solved

Do you know any examples of algebraically closed fields other than $\mathbb{C}$?

you mean, an algebraically closed field $K$ not defined as an algebraic closure of some other field?
of course, this construction leads to as many algebraically closed fields as you need...

the only "constructive" algebraically closed fields i know, more than $\mathbb{C}$, are $\bar{\mathbb{F}}_{p}= \bigcup_{n\ge1}\mathbb{F}_{p^{n}}$...

The field $\mathbb{C}_{p}$, the completion of $\overline{\mathbb{Q}_{p}}$, the algebraic closure of the $p$-adic field $\mathbb{Q}_{p}$, is also algebraically closed (well,  when $p$ is the "prime" (=valuation) given by the standard absolute value on $\mathbb{Q}$, this yields $\mathbb{C}$).
The same holds for the spherical complete field $\Omega_{p}$.

Is it me, or have none of you guys mentioned ... the algebraic closure of $\mathbb{Q}$ ?  :o

That is a subfield op $\mathbb{C}$, consisting of all algebraic complex numbers

[quote="ZetaX"]
The same holds for the spherical complete field $\Omega_{p}$.[/quote]What's that?