function
vector
algebra
functional equation
algebra unsolved

Find all functions $ f: \mathbb{R^*} \to \mathbb{R^*}$ satisfying the functional relation
\[ f(f(x)\minus{}x)\equal{}2x, \quad \forall x\in \mathbb{R^*}.\]

hi!
First we consider the function g such that: $ g(x)\equal{}f(x)\minus{}x$ so the f.e become:

$ g(g(x))\plus{}g(x)\equal{}2x$ $ \forall x\in{\mathbb{R^{*}}}$

then we consider the sequence $ (x_{n})$ define by:$ x_{0}\equal{}x$ and $ g(x_{n})\equal{}x_{n\plus{}1}$ so if we solve this equation: $ x^{2}\plus{}x\minus{}2\equal{}0$ we find $ 1$ and $ \minus{}2$ ==>

 $ x_{n}\equal{}\alpha\plus{}(\minus{}2)^{n}\beta
 ;(\alpha,\beta)\in{\mathbb{R}}$ with $ x\equal{}\alpha\plus{}\beta$ ,then for $ n\equal{}1$ we find: $ g(x)\equal{}\alpha\minus{}
2\beta$ ==> $ f(x)\equal{}x\plus{}\alpha\minus{}2\beta$ $ (*)$.

We have: $ f(f(x)\minus{}x)\equal{}2x$ so by using $ (*)$ we find: $ \beta\equal{}0$ ==> $ f(x)\equal{}2x \forall x\in{\mathbb{R^{*}}}$

I fail to see why from the star (*) relation follows $ \beta \equal{} 0$ (other than it's convenient to be so).

we have $ f(f(x)\minus{}x)\equal{}2x$ and $ f(x)\equal{}x\plus{}\alpha\minus{}2\beta$ so:
$ f(x)\minus{}x\equal{}x\plus{}\alpha\minus{}2\beta\minus{}x\equal{}\alpha\minus{}2\beta$ and $ f(\alpha\minus{}2\beta)\equal{}2(\alpha\minus{}2\beta)$,we have $ x\equal{}\alpha\plus{}\beta$ too so we find : $ 2(\alpha\minus{}2\beta)\equal{}2(\alpha\plus{}\beta)$ ==> $ \beta\equal{}0$..

I still don't see why $ f(\alpha\minus{}2\beta)\equal{}2(\alpha\minus{}2\beta)$.

we have : $ f(x)\equal{}x\plus{}\alpha\minus{}2\beta$ so $ f(\alpha\minus{}2\beta)\equal{}\alpha\minus{}2\beta\plus{}\alpha\minus{}2\beta$ ==> $ f(\alpha\minus{}2\beta)\equal{}2(\alpha\minus{}2\beta)$

I don't think you can do that. Let's go back to where you started solving the recurrence relatiion: you started with $ x_0 \equal{} x$, etc. Then the coefficients $ \alpha$ and $ \beta$ depend on the value of the initial term, so in fact you have $ \alpha_x$ and $ \beta_x$.

Now, you got $ f(x) \equal{} x \plus{} \alpha_x \minus{} 2\beta_x$, and you're not allowed to replace $ x$ with any value you want, respectively $ \alpha_x \minus{} 2\beta_x$, because for this value you may have different coefficients $ \alpha$ and $ \beta$, in fact $ \alpha_{\alpha_x \minus{} 2\beta_x}$ and $ \beta_{\alpha_x \minus{} 2\beta_x}$.

My point is that coefficients $ \alpha$ and $ \beta$ are not "universal", over all $ x$.

yes,I undrestand your point, but I think that can correct my solution:
we have:

$ g(g(x)) \plus{} g(x) \equal{} 2x$

$ g(x) \equal{} x \minus{} 3\beta_{x}$

$ g(g(x)) \equal{} g(x) \minus{} 3\beta_{g(x)}$

then we conclude that: $ \minus{} 3(2\beta_{x} \minus{} \beta_{g(x)}) \equal{} 0$ ==> $ 2\beta_{x} \equal{} \beta_{g(x)}$, then by using this result we can find that: $ g(g(x)) \plus{} g(x) \equal{} 2x$<==>$ 2x \minus{} 12\beta_{x} \equal{} 2x$ ==> $ \beta_{x} \equal{} 0$ so $ g(x) \equal{} x$==>$ f(x) \equal{} 2x \forall x\in{\mathbb{R^{*}}}$

I hope that will be correct.

First ,$ f(x)\minus{}x \in R^* \forall x \in R^*$
$ g(x)\equal{}f(x)\minus{}x \Rightarrow g: R^* \rightarrow R^*  , g(g(x))\plus{}g(x)\equal{}2x$
LEt's consider the sequence :  $ u_0\equal{}x_0; u_{n\plus{}1}\equal{}g(x_n)$
$ \Rightarrow u_n>0 (*) \forall n \in N ;  u_(n\plus{}1)\plus{}u_n\equal{}2u_(n\minus{}1)$
$ \Rightarrow u_n\equal{}\alpha\plus{}(\minus{}2)^n .\beta$
* $ \beta <0 \Rightarrow u_{2n}<0$ when $ n$ is large enough (Contradition (*))
*$ \beta>0\Rightarrow u_{2n\plus{}1}<0$ when $ n$ is large enough (Contradition (*))
Hence :$ \beta\equal{}0$
$ \Rightarrow u_1\equal{}u_0\equal{}x_0$
$ \Rightarrow g(x_0)\equal{}x_0$
$ \Rightarrow f(x)\equal{}2x \forall x \in R^*$

[quote="Vincent Gilbert"]First ,$ f(x) - x \in R^* \forall x \in R^*$
$ g(x) = f(x) - x \Rightarrow g: R^* \rightarrow R^* , g(g(x)) + g(x) = 2x$
LEt's consider the sequence :  $ u_0 = x_0; u_{n + 1} = g(x_n)$
$ \Rightarrow u_n > 0 (*) \forall n \in N ; u_(n + 1) + u_n = 2u_(n - 1)$
$ \Rightarrow u_n = \alpha + ( - 2)^n .\beta$
* $ \beta < 0 \Rightarrow u_{2n} < 0$ when $ n$ is large enough (Contradition (*))
*$ \beta > 0\Rightarrow u_{2n + 1} < 0$ when $ n$ is large enough (Contradition (*))
Hence :$ \beta = 0$
$ \Rightarrow u_1 = u_0 = x_0$
$ \Rightarrow g(x_0) = x_0$
$ \Rightarrow f(x) = 2x \forall x \in R^*$[/quote]

What is the contradiction when ${ u_{2n}<0}$ or $ u_{2n+1}<0$? I think $ \mathbb{R}^*$ is the set of real numbers other than $ 0$, not the set of positive real numbers.

I think there should be some additional condition to this functional equation because of the following reasons:

1. It is clearly that we just need to solve an equivalent problem:
Find every function $ g: \mathbb{R}^*\to\mathbb{R}^*$ satisfies: $ g(g(x)) \plus{} g(x) \equal{} 2x,\forall x\in\mathbb{R}$.

2. By considering the sequence $ \{u_n\}_{n\equal{}0}^{\infty}$ defined by $ u_0 \equal{} x, u_{n\plus{}1} \equal{}g(u_n)$, it is known that $ u_n \equal{} \lambda_1 \plus{} \lambda_2 (\minus{}2)^n$, thus if $ \lambda_2\neq 0$, the sequence $ \{u_n\}_{n\equal{}0}^{\infty}$ takes both negative and positive values. Thus if we add the [i]continuity[/i] of $ g(x)$, we can deduce that $ \exists x: g(x)\equal{}0$, contradiction. Moreover, if $ g: \mathbb{R}^\plus{}\to\mathbb{R}^\plus{}$ or $ g: \mathbb{R}^\minus{} \to \mathbb{R}^\minus{}$, we can also find $ g$.

3. Still consider the sequence $ \{u_n\}$ above, we see that the definition of $ g$ is just in $ \{u_n\}$, therefore if we can point out a partion of $ \mathbb{R}^*$ into such disjoint sequences like $ \{u_n\}$, i.e. sequences satisfies $ u_{n\plus{}1} \plus{} u_n \minus{} 2u_{n\minus{}1} \equal{} 0$ then there is a function $ g$. This is just my sense that we can construct infinitely many functions $ g$ satisfies the problem in this way. :maybe:

[i]solution[/i] (Abhay Kumar Jha).

We begin with observation that $ f(x)\ge 0,\forall x\in R^*$.Define a new function $ g$ on $ R^*$ by setting $ g(x) \equal{} f(x) \minus{} x$ .Since $ f(x)\ge x$,we see that $ g(g(x))$ is meaningful.A simple computation gives $ g(g(x)) \equal{} f(g(x)) \minus{} g(x) \equal{} f(f(x) \minus{} x) \minus{} (f(x) \minus{} x) \equal{} 3x \minus{} f(x) \equal{} 2x \minus{} g(x).$

Thus we have to find $ g: R^* \to R^*$ satisfying

$ g(g(x)) \plus{} g(x) \equal{} 2x.$

Fix some $ a$ in $ R^*$ and define $ u_n \equal{} g^n(a)$ where $ g^n(x) \equal{} g(g^{n \minus{} 1}(x)).$Then $ < u_n >$ satysfies the recurrence relation $ u_{n \plus{} 2} \plus{} u_{n \plus{} 1} \plus{} \minus{} 2u_n \equal{} 0$ for all $ n \ge 0$,where $ u_o \equal{} a$.This diference equation for $ u_n$ has auxiliary equation $ x^2 \plus{} x \minus{} 2 \equal{} 0$ which has solutions $ x \equal{} 1$ and $ x \equal{} \minus{} 2$.Using the theory of difference equations,the general solution of the difference equation is given by

$ u_n \equal{} A(1)^n \minus{} B( \minus{} 2)^n$

for some constants $ A$ and $ B$.However, we see that $ g(x)\ge 0$ for all $ x\in R^*$,and hence $ u_n \equal{} g^n(a) \ge 0$.Since $ ( \minus{} 2)^n$ alters sign as $ n$ runs through natural numbers,we conclude that $ B \equal{} 0$.This forces $ u_n \equal{} A$ for all $ n$ and using initial condition,we can see that $ 2A \equal{} u_1 \plus{} u_2 \equal{} 2a$.We thus obtain $ A \equal{} a$ and hence $ u_n \equal{} a$ for all $ n$.But then $ g(a) \equal{} u_1 \equal{} a$ and we conclude that $ f(a) \equal{} 2a$.Since this is true for every $ a\in R^*$ we arrive at the solution $ f(x) \equal{} 2x$.

P.s.1
I cant understand the solution very well because i'm not familiar with sequences and thats why i post this problem on unsolved,hoping for an elementary solution....

P.s.2
The book uses the notation $ Ro$ and i guess he mean $ R^*$......

[quote="Dimitris X"][i]solution[/i] (Abhay Kumar Jha).

We begin with observation that $ f(x)\ge 0,\forall x\in R^*$.[/quote]
$ f(x) \equal{} \minus{}x$ is a solution too, so i don't get the point  :roll:

[quote="Dimitris X"]
We begin with observation that $ f(x)\ge 0,\forall x\in R^*$[/quote]

This is non-sense. If $ f(x) \geq 0,\forall x\in\mathbb{R}^*$ then we directly get the contradiction from the assertion $ f(f(x) \minus{} x) \equal{} 2x$ (just let $ x < 0$).

Maybe $ f: \mathbb{R_\plus{}} \to \mathbb{R_\plus{}}$^^

[quote="Mashimaru"][quote="FelixD"]Maybe $ f: \mathbb{R_ \plus{} } \to \mathbb{R_ \plus{} }$^^[/quote]

I think so, either. But could anyone prove that if $ f: \mathbb{R}^*\to \mathbb{R}^*$ then there are infinitely many functions $ f$ satisfies the problem? I have an idea in post #12 but I did not have time to work on it :([/quote]

Sure, take any Hamel basis of the $ \mathbb Q$-vectorspace $ \mathbb R$ and randomly choose either $ f(b_i)\equal{}2b_i$, either $ f(b_i)\equal{}\minus{}b_i$ for the elements of the basis and you find infinitely many solutions.

[quote="pco"][quote="Mashimaru"][quote="FelixD"]Maybe $ f: \mathbb{R_ \plus{} } \to \mathbb{R_ \plus{} }$^^[/quote]

I think so, either. But could anyone prove that if $ f: \mathbb{R}^*\to \mathbb{R}^*$ then there are infinitely many functions $ f$ satisfies the problem? I have an idea in post #12 but I did not have time to work on it :([/quote]

Sure, take any Hamel basis of the $ \mathbb Q$-vectorspace $ \mathbb R$ and randomly choose either $ f(b_i) \equal{} 2b_i$, either $ f(b_i) \equal{} \minus{} b_i$ for the elements of the basis and you find infinitely many solutions.[/quote]

Once you told me about the Hamel basis, Mr.[b]pco[/b]. But I still did not get how can we generate infinitely many functions satisfy the problem in this way. Could you make it more clearly, please? Thank you in advanced.

Just take one basis $ \{b_i\}$ of the $ \mathbb Q$-vector space $ \mathbb R$

Then define :
$ f_1(\sum q_ib_i)\equal{}\sum q_if(b_i)$ with $ f(b_1)\equal{}2b_1$ and $ f(b_i)\equal{}\minus{}b_i$ for all other elements of the basis. You got a solution.

$ f_2(\sum q_ib_i)\equal{}\sum q_if(b_i)$ with $ f(b_2)\equal{}2b_2$ and $ f(b_i)\equal{}\minus{}b_i$ for all other elements of the basis. You got another solution.

$ f_3(\sum q_ib_i)\equal{}\sum q_if(b_i)$ with $ f(b_3)\equal{}2b_3$ and $ f(b_i)\equal{}\minus{}b_i$ for all other elements of the basis. You got another solution.

And so on. Choosing randomly $ f(b_i)$ as either $ 2b_i$, either $ \minus{}b_i$, you build infinitely many different solutions to $ f(f(x)\minus{}x)\equal{}2x$


Four coins are picked out of a piggy bank that contains a collection of pennies, nickels, dimes, and quarters. Which of the following could [i]not[/i] be the total value of the four coins, in cents?

$ \textbf{(A)}\ 15 \qquad
\textbf{(B)}\ 25 \qquad
\textbf{(C)}\ 35 \qquad
\textbf{(D)}\ 45 \qquad
\textbf{(E)}\ 55$

We can never use pennies. 4 nickels is greater than 15, so A.

I think this is one of the only (and possibly first) problems that explicitly references the answer choices.

[b]

         You gotta BE KIDDING ME! Its A. Trust Me
[/b]

[quote="blackbelt14253"]you're kidding right?

this was #2 on the 10A, and it was #4 on the 12A? How does that work?[/quote]

Some of the problems for the AMC 10 are on the AMC 12 but not all of course, so the order is obviously not the same.

I think he was referring to the fact that it was labeled as "easier" on the 10A than on the 12A, because it was a lower number on the 10A.

(Yes, I'm now Riemann Hypothesis!)


geometry
function
ratio
geometry proposed

Find the largest $k>0$ with the property that for any convex polygon of area $S$ and any line $\ell$ in the plane, we can inscribe a triangle with area $\ge kS$ and a side parallel to $\ell$ in the polygon.

Well ,guys dont laught at this ok:
But If somehow,We may be able to consider the circle A $\infty-$gon then ,we can found a bound for $k$.
Cause the area of an inscribed triangle (with a side parallel to a given line) inside a circle with area $S$ has a bound .
[hide="bound"]
$K \le Max(\frac{2Cos^2(\frac{\alpha}{2})Sin(\alpha)}{\pi}),\ when  0< \alpha <180$[/hide]
I am sure this is useless but if you want to know [hide="why"]
[color=blue]Problem.Let the $C$ be a circle with area $S$ .Find the maxim area of an inscribed tringle ,with a side parallel to a given line .[/color]

Well,Let $ABC$ be such a trinagle ,clearly $ABC$ must be isosceles ,$(AB=AC)$, let $\angle BAC=\alpha$ ,
So we can calculte the $S[ABC]$ by a function by $\alpha $
Which is  $S[ABC]=2R^2Cos^2(\frac{\alpha}{2})Sin(\alpha)$[hide="proof"]$S[ABC]=\frac12 AB*AC*Sin(\alpha),
but by the law o the sinus in trinagle ABC,\frac{AB=AC=2RSin(90-frac{\alpha}{2})=2RCos(\frac{\alpha}{2})}$
,So the $S{ABC]=2R^2Cos^2(\frac{\alpha}{2})Sin(\alpha)}$[/hide]
So the ration between the area of ABC and the circle will be a bound for $K$ which is
$\frac{2R^2Cos^2(\frac{\alpha}{2})Sin(\alpha)}{R^2 \pi}=\frac{2Cos^2(\frac{\alpha}{2})Sin(\alpha)}{\pi}$

So a pound for $K$ is :
$K \le Max(\frac{2Cos^2(\frac{\alpha}{2})Sin(\alpha)}{\pi}),\ when  0< \alpha <180$[/hide]

Sorry for the revive of this topic, but I've also wondered about this problem.  A theorem of Sas in 1939 guarantees an inscribed triangle such that the constant in question is equal to 3sqrt(3)/(4pi) but I believe that with the extra parallel side condition the constant should be 3/8.  If I find a proof I'll post it, but meanwhile, does anyone have any ideas?

Is this in any way related to IMO shortlist 2004 Geo 6?


inequalities
trigonometry
inequalities unsolved

these problems  are  not easy
i will take a

[quote="yzymat"]9.
$ 0 < x,y,z < \frac {\pi}{2},$ $ \sin^2x \plus{} \sin^2y \plus{} \sin^2z \equal{} 1,$ then
$ \frac {1}{\cos^2x \plus{} \cos^2y} \plus{} \frac {1}{\cos^2x \plus{} \cos^2z} \plus{} \frac {1}{\cos^2y \plus{} \cos^2z}\leq$
$ \leq\frac {1}{4}\left(\frac {1}{\sin x\sin y} \plus{} \frac {1}{\sin x\sin z} \plus{} \frac {1}{\sin y\sin z}\right)$[/quote]
$ \sum_{cyc}\frac {1}{4\sin x\sin y}\geq\sum_{cyc}\frac {1}{2(\sin^2x \plus{} \sin^2y)}.$
Let $ \sin^2y \plus{} \sin^2z \equal{} a,$ $ \sin^2x \plus{} \sin^2z \equal{} b$ and $ \sin^2x \plus{} \sin^2y \equal{} c.$
Hence, it remains to prove that $ \sum_{cyc}\frac {1}{2a}\geq\sum_{cyc}\frac {1}{a \plus{} b},$ which is true because by Cauchy-Schwartz we obtain:
$ \frac {1}{a} \plus{} \frac {1}{b}\geq\frac {(1 \plus{} 1)^2}{a \plus{} b} \equal{} \frac {4}{a \plus{} b}.$

[quote="yzymat"]10.
$ a_i > 0,$ $ i \equal{} 1,2,3,4,5.$ Determine the minimal value of
\[ \frac {a_1}{a_2 \plus{} 3a_3 \plus{} 5a_4 \plus{} 7a_5} \plus{} \frac {a_2}{a_3 \plus{} 3a_4 \plus{} 5a_5 \plus{} 7a_1} \plus{} ... \plus{} \frac {a_5}{a_1 \plus{} 3a_2 \plus{} 5a_3 \plus{} 7a_4}
\]
[/quote]
If $ a_1 \equal{} a_2 \equal{} ... \equal{} a_5$ then the expression equal to $ \frac {5}{16}.$
We'll prove that $ \frac {5}{16}$ is a minimal value of the expression.
Let $ a_{i \plus{} 5} \equal{} a_i$ for all $ i\in\{1,2,3,4\}.$
Then by Cauchy-Schwartz we obtain:
$ \sum_{i \equal{} 1}^5\frac {a_i}{a_{i \plus{} 1} \plus{} 3a_{i \plus{} 2} \plus{} 5a_{i \plus{} 3} \plus{} 7a_{i \plus{} 4}} \equal{} \sum_{i \equal{} 1}^5\frac {a_i^2}{a_ia_{i \plus{} 1} \plus{} 3a_ia_{i \plus{} 2} \plus{} 5a_ia_{i \plus{} 3} \plus{} 7a_ia_{i \plus{} 4}}\geq$
$ \geq\frac {\left(\sum a_i\right)^2}{\sum(a_ia_{i \plus{} 1} \plus{} 3a_ia_{i \plus{} 2} \plus{} 5a_ia_{i \plus{} 3} \plus{} 7a_ia_{i \plus{} 4})} \equal{} \frac {\left(\sum a_i\right)^2}{\sum(8a_ia_{i \plus{} 1} \plus{} 8a_ia_{i \plus{} 2})}.$
Id est, it remains to prove that $ \frac {\left(\sum a_i\right)^2}{\sum(8a_ia_{i \plus{} 1} \plus{} 8a_ia_{i \plus{} 2})}\geq\frac {5}{16},$
which is equivalent to $ \sum_{1\leq i < j\leq5}(a_i \minus{} a_j)^2\geq0.$

[quote="yzymat"]11.
$ a, b, c > 0.$  Determine the minimal value of $ \frac {a \plus{} 3c}{a \plus{} 2b \plus{} c} \plus{} \frac {4b}{a \plus{} b \plus{} 2c} \minus{} \frac {8c}{a \plus{} b \plus{} 3c}.$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15779

[quote="yzymat"]12.
Let $ a,$ $ b,$ $ c$ are positive integers, and $ \frac{a}{b\plus{}c}\equal{}\frac{b}{a\plus{}c}\minus{}\frac{c}{a\plus{}b},$ then $ \frac{b}{a\plus{}c}\geq\frac{\sqrt{17}\minus{}1}{4}.$[/quote]
I think it's true for all positive $ a,$ $ b$ and $ c.$ :wink:


Find all positive integral solution of the equation
2(x+y+z) +9=3xyz

[hide]2(x+y+z)=3xyz-9 so 2(x+y+z)=3(xyz-3) or 2/3=(xyz-3)/(x+y+z).  This means that x+y+z>xyz-3.  Case 1: 2 variables are equal to 1.  In this case, the non-1 variable might as well be x.  x-3/x+2=2/3.  Or 3x-9=2x+4.  The x is 13.  Case 2: 1 variable equals 1.  In this case, let the 1 variable be z.  We have xy-3/x+y+1=2/3.  Now, cross multiply and simplify to get 3xy-2x=2y+11 or x(3y-2)=2y+11 or x=2y+11/3y-2.  If y is even, x is not an integer because the top is always odd, but the bottom is even.  This also prevents x from being even.  If x is 1, we have Case 1.  However, if y is too large, 2y+11/3y-2 is less than 1.  Since Case 1 dictates if x is 1, y=13, anything larger than 13 results in a fraction less than 1 and if y=1, x=13.  If y=3, x=17/7 y=5, x=21/13 y=7, x=25/19 y=9, x=29/25 and y=11, x=33/31.  So there's no possiblity with just one 1.  Case 3: no ones.  We again must find the threshold for 2/3=(xyz-3)/(x+y+z).  Since there's no ones, we might as well try 2s.  2^3-3/2(3)=5/6.  Since the product xyz will be larger than x+y+z when the numbers are made larger, the fraction grows larger.  (Just test 2,2,3 and you get 9/7.)  Therefore, the only soutions are (13,1,1) (1,13,1) (1,1,13). [/hide] Oh, another question, can (xyz-3)/(x+y+z) generate every rational number when x,y,z are integers or is there some that can't be generated?

dengmi, seriously, space out your solutions, and use latex, you have been at this site long enough already

but one way to look at this would be in $\mod 3$, then find possibilities for $x,y,z$ in $\mod 3$
$x+y+z \equiv 0 \mod 3$, so we can have the mods be

$0,0,0$
$1,1,1$
$2,2,2$
$0,1,2$

then do more case work, and repeat looking in $\mod 3$, not elegant, but it would work...

also looking in mod 2, we see that they have to be odd


hello, the result is zero, use
$ \log_5(\tan(\pi/9))\equal{}\log_5(\sin(\pi/9))\minus{}\log_5(\cos(\pi/9))$
Sonnhard.

Actually, unless I'm making a mistake somewhere, it doesn't matter what $ x$ is.  The answer should still always be $ 0$.

We can simplify the expression using the addition and subtraction rules for logarithms (this is what Dr Sonnhard Graubner did to arrive at the answer):

$ \log_5 \frac{\sin x}{\cos x}\minus{}\log_5 \sin x\equal{}\log_5 \frac{\sin x}{\sin x \cos x}\equal{}\log_5 \frac{1}{\cos x}$.

Adding in the $ \log_5 \cos x$ term, we have:

$ \log_5 \frac{1}{\cos x}\plus{}\log_5 \cos{x}\equal{}\log_5 1\equal{}\boxed{0}$.

[quote="Stokes93"]This doesn't seem like a hard problem, so I'm probably missing something pretty obvious, but I can't figure out a way to solve the following question:

Given $ x \equal{} 20^\circ$, what is the value of $ log_5(tan x) \minus{} log_5(sin x) \plus{} log_5(cos x)$?[/quote]

$ log_5(tan x) \minus{} log_5(sin x) \plus{} log_5(cos x)$

$ log_5(\frac{sin x}{cos x}) \minus{} log_5(sin x) \plus{} log_5(cos x)$

Using the fact that $ log_x(yz) \equal{} log_x(y) \plus{} log_x(z)$, we get;

$ log_5((sin x) \times (\frac{1}{cos x})) \minus{} log_5(sin x) \plus{} log_5(cos x)$

$ log_5(sin x) \plus{} log_5(\frac{1}{cos x}) \minus{} log_5(sin x) \plus{} log_5(cos x)$

$ log_5(\frac{1}{cos x} \plus{} log(cos x)$

Now we can use the same formula but backwards.

$ log_5(\frac{cos x}{cos x})$

$ log_5(1)$

$ 0$


group theory
abstract algebra
analytic geometry
vector
superior algebra
superior algebra unsolved

Hi,
I wonder how many subgroups of index 2 does a direct product Z60*Z45*Z12*Z36 have? I found three but there seems to be more.
Thanks

edit:typo. order 2->index 2

An element in a direct product has order dividing $ k$ if and only if each of its coordinates has order dividing $ k$. There are thus $ 2\cdot 1\cdot 2\cdot 2\equal{}8$ elements of order dividing 2. One is the identity, and the other seven have order 2. Each such element combines with the identity to form a 2-element subgroup.

oops,  :oops: sorry guys, for some reason i typed ORDER but what i meant was INDEX 2

In an arbitrary $ \mathbb{Z}$-module $ M$, index-2 subgroups correspond exactly to codimension-1 subspaces of the vector space $ M/2M$. Once again, $ M/2M\equal{}(\mathbb{Z}/2\mathbb{Z})\times\{0\}\times(\mathbb{Z}/2\mathbb{Z})\times(\mathbb{Z}/2\mathbb{Z})$, and there are exactly 7 such subspaces.
It's not a coincidence that the answer is the same.

Thank you jmerry. but since i've only covered up to free groups in dummit, i'm not reli sure about the terms you used. what confuses me about this question is the following. Any subset of the product is again a product of the same form, i.e A1*A2*A3*A4. Suppose now this subset is a subgroup indeed. Does it imply that A1 is a subgroup of Z60 etc? it seems plausible to me but from what you have said, an index 2 subgroup could be indecomposable.

The language I was using is quite basic, but it's not from group theory at all. You should know vector spaces, and modules are essentially the same thing with a ring in place of a field. Abelian groups always have a natural structure as a module over the integers, defining multiplication by integers with repeated addition. This is very frequently useful.
Next, we quotient by everything divisible by 2; if $ x \equal{} 2y$ for some other $ y$, $ x$ must be in any (normal) subgroup of index 2. Everything interesting happens in that quotient, which is much smaller (8 elements) and easier to understand.

[quote="andy_tok"]Any subset of the product is again a product of the same form, i.e A1*A2*A3*A4.[/quote] This is false, except in the sense that it's non-uniquely isomorphic to something of that form. For example, $ \{(a,b): a \plus{} b\text{ is even}\}$ is an index-2 subgroup of $ \mathbb{Z}\times\mathbb{Z}$, and it's not generated by its elements on the axes.


Check out http://www.mathlinks.ro/Forum/weblogs.php :)


I'm working with aligning equations, and here is what I have:

\begin{align*}
\displaystyle \frac {2}{35} &= \displaystyle \frac {x}{595}\\
(2)(595) &= (35)(x)\\
1190 &= 35x\\
x &= 34
\end{align*}

Is there any way to have more space between each individual line?  They look too crammed-in to me.

     -- Pawel

hi

I couldn't compile with the align* environnement, in which package is this?
anyway, I've put them in eqnarray and I used

\doublespacing

in the package "setspace" before the equations and it was better.
for more space, maybe you can use \vspace{3ex} or any size but i don t think this is the right solution.

ledurt

There are a few ways to insert space that I can think of, but they always look artificial and end up worse than you started with. I'll keep searching for a natural way to insert space into equations like this.

Ledurt, the align environment is one of the many equation-type environments provided by the amsmath package. These alternate environments are more flexible than eqnarray, and as a result i have completely eliminated eqnarray from my vocabulary. I also always include the amsthm and amssymb packages in my documents. Documentation can be found [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuidePack.php#ams]here[/url].

One more comment: in displayed equation environments such as align, the equation is put in displaystyle by default; you don't have to manually type it in.

Try \renewcommand{\arraystretch}{2}, where the "2" is a multiplicative factor (so that, for example, this command will give you twice as much space between rows).

Put this command before the \begin{align*} statement.

That works very well!  Thanks!

     -- Pawel


Does anyone know any good textbooks for algebra II?

Do you want to buy one?  I suggest the Heath book, it is EXTREMELY good.  It was the chosen textbook for the Duke TIP classes lat year.

One of the most challenging textbooks, which my school uses and pretty much no public schools use, is supposed to be the houghton mifflin text; "Algebra II and Trignometry". Another good but much easier text is Algebra II by Larson, which my old school used. I'm not much of a textbook expert (nor do I use them much) but I think these are supposed to be some of the better text.

You may want to try a College Algebra book (not really college algebra -- it is really just algebra 2).  There are quite a few out there and they are pretty much the same (the only author I can think of is Sullivan).

The college algebra books might be good, but I would not recommend them for self study.


Cities of a certain country are connected by air routes (each connecting two cities) served by $ 2k \plus{} 1$ companies, where the first company serves one route, the second serves two, etc. According to a law in this country, a company cannot serve more than one route from the same city. Some day, the companies agreed to redistribute the routes between themselves so that each company serves the same number of routes. Show that this can be done without violating the law.


Sean $k$ un entero positivo y $P$ un polinomio con coeficientes enteros. Pruebe que existe un entero positivo $n$ tal que $P(1)+P(2)+....+P(n)$ es divisible por $k$.

Como tenemos que:
$a\equiv b (\bmod k)$ implica que:
$P(a) \equiv P(b) (\bmod k)$ tenemos que:

$P(1)+P(2)+...+P(k) \equiv P(tk+1)+P(tk+2)+...+P((t+1)k) (\bmod k)$ esto ultimo quiere decir que:

$P(1)+P(2)+...+P(k^{2}) \equiv k(P(1)+P(2)+...P(k)) \equiv 0 (\bmod k)$

QED

cuenca

podrias  poner los demas  problemas  ???  :D


abstract algebra
modular arithmetic
induction
group theory
number theory
relatively prime
superior algebra

1.
find all  numbers $ n$ st.:

if $ G$ is a group and $ |G|\equal{}n$, then $ G$ is a cyclic group.

2.
find all  numbers $ n$ st.:

if $ G$ is an abelian group and $ |G|\equal{}n$, then $ G$ is a cyclic group.

2 holds iff $ n$ is squarefree: it's a consequence of the Fundamental Theorem of Finite Abelian Groups. (Every finite abelian group is a direct product of cyclic groups of prime-power order.)

1 is much more difficult: the $ n$ for which it holds are squarefree $ n$ such that for all primes $ p,q$ dividing $ n$, $ p\not\equiv 1\pmod{q}$.

[hide="noncyclic groups for the other values of n"]
If $ n\equal{}p^2m$, the group $ \mathbb{Z}_{pm}\times \mathbb{Z}_p$ is not cyclic.

If $ n\equal{}pqm$ where $ p,q$ primes and $ p\equiv 1\pmod{q}$, then there is a nonabelian group $ H$ of order $ pq$, so that $ H\times \mathbb{Z}_m$ is not cyclic. (To construct $ H$, there exists $ r$ such that $ r^q\equiv 1\pmod{p}$ but $ r\not\equiv 1\pmod{p}$, then $ H$ can be the semidirect product $ \mathbb{Z}_p\rtimes \mathbb{Z}_q$ under the homomorphism mapping $ y$ to multiplication by $ r^y$. In other words, take ordered pairs $ (x,y)$ where $ x\in \mathbb{Z}_p$ and $ y\in \mathbb{Z}_q$ and define $ \oplus$ by
\[ (x_1,y_1)\oplus (x_2,y_2)\equal{}(x_1\plus{}r^{y_1}x_2,y_1\plus{}y_2)\]
[/hide]

Of course, there's still the proof that all groups of the nice orders are cyclic. I think this can be done with induction on the number of prime factors of $ n$: try to show that if $ G$ is a group of a composite nice order, then $ G$ is not simple, using the fact that all proper subgroups are cyclic. Then if $ \left< h\right>$ is a proper nontrivial normal subgroup of $ G$...
[hide="rest of proof"]
$ G/\left< h\right>$ is cyclic. Let $ |h|\equal{}m$ and $ n\equal{}mr$, so $ G/\left< h\right>$ has an element $ k^*\left< h\right>$ of order $ r$. Raising $ k^*$ to an appropriate power gives an element $ k$ of order $ r$.

Conjugation by $ k$ is an automorphism of $ \left< h\right>$. The order of the automorphism divides both $ r$ and $ \varphi (m)$ (the number of automorphisms of $ \left <h\right>$. But $ r$ and $ \varphi (m)$ are relatively prime, so the conjugation has order 1. This means $ h,k$ commute, so $ hk$ has order $ mr\equal{}n$, making $ G\equal{}\left< hk\right>$.
[/hide]


limit
logarithms
calculus
calculus computations

$ \lim_{x\to\infty}\left(\sqrt [3]{x^{3}\plus{}x^{2}\plus{}x\plus{}1}\minus{}\frac{\sqrt{x^{2}\plus{}x\plus{}1}\cdot\ln(e^{x}\plus{}x)}{x}\right)$

But beacause $ \lim_{x\to\infty}\frac{\ln(e^{x}\plus{}x)}{x}\equal{}\lim_{x\to\infty}\frac{x\plus{}\ln\left( 1\plus{}\frac{x}{e^{x}}\right)}{x}\equal{} 1$
it remains only to evaluate
$ \lim_{x\to\infty}\left(\sqrt [3]{x^{3}\plus{}x^{2}\plus{}x\plus{}1}\minus{}\sqrt{x^{2}\plus{}x\plus{}1}\right)$

Let $ a \equal{} x^{3}\plus{}x^{2}\plus{}x\plus{}1$ and $ b \equal{} x^{2}\plus{}x\plus{}1$
we have $ \sqrt [3]{a}\minus{}\sqrt{b}\equal{}\frac{\sqrt [3]{a^{2}}\minus{}b}{\sqrt [3]{a}\plus{}\sqrt{b}}\equal{}\frac{a^{2}\minus{}b^{3}}{(\sqrt [3]{a}\plus{}\sqrt{b})(\sqrt [3]{a^{4}}\plus{}\sqrt [3]{a^{2}}b\plus{}b^{2})}$

But $ a^{2}\minus{}b^{3}\equal{} x^{6}\plus{}2x^{5}\plus{}o(x^{5})\minus{}x^{6}\minus{}3x^{5}\minus{}o(x^{5}) \equal{}\minus{}x^{5}\plus{}o(x^{5})$

and  $ (\sqrt [3]{a}\plus{}\sqrt{b})(\sqrt [3]{a^{4}}\plus{}\sqrt [3]{a^{2}}b\plus{}b^{2}\equal{} 6x^{5}\plus{}o(x^{5})$

Consequently $ \lim_{x\to\infty}\left(\sqrt [3]{x^{3}\plus{}x^{2}\plus{}x\plus{}1}\minus{}\sqrt{x^{2}\plus{}x\plus{}1}\right) \equal{}\minus{}\frac{1}{6}$

[quote="Extremal"]your answer is right.
but this is not correct
[quote="Svejk"]

But $ a^{2}\minus{}b^{3}\equal{} x^{6}\plus{}2x^{5}\plus{}o(x^{5})\minus{}x^{6}\minus{}3x^{5}\minus{}o(x^{5}) \equal{}\minus{}x^{5}\plus{}o(x^{5})$

and  $ (\sqrt [3]{a}\plus{}\sqrt{b})(\sqrt [3]{a^{4}}\plus{}\sqrt [3]{a^{2}}b\plus{}b^{2}\equal{} 6x^{5}\plus{}o(x^{5})$
[/quote]
aslo you can use taylor and it will be more easy[/quote]

Maybe I should have set $ E_{1}(x) \equal{} x^{3}\plus{}x^{2}\plus{}....$ and $ E_{2}(x) \equal{} x^{2}\plus{}x\plus{}1$ instead of $ a$ and $ b$  :blush:  it's more correctly.Is this the issue ? If not please point out my mistake.

[quote="Svejk"]$ \lim_{x\to\infty}\left(\sqrt [3]{x^{3}\plus{}x^{2}\plus{}x\plus{}1}\minus{}\frac{\sqrt{x^{2}\plus{}x\plus{}1}\cdot\ln(e^{x}\plus{}x)}{x}\right)$

But beacause $ \lim_{x\to\infty}\frac{\ln(e^{x}\plus{}x)}{x}\equal{}\lim_{x\to\infty}\frac{x\plus{}\ln\left( 1\plus{}\frac{x}{e^{x}}\right)}{x}\equal{} 1$
it remains only to evaluate
$ \lim_{x\to\infty}\left(\sqrt [3]{x^{3}\plus{}x^{2}\plus{}x\plus{}1}\minus{}\sqrt{x^{2}\plus{}x\plus{}1}\right)$[/quote]
This reasoning is not correct.  Just because $ \lim h(x) \equal{} 1$, it does not follow that  $ \lim [f(x)\minus{}g(x) h(x)]$ is equal to $ \lim [f(x)\minus{}g(x)]$.

To Ravi B:
in this case we can do it
To Svejk: now you are right,but we use   the next expression :
$ a^{2}\minus{}b^{3}\equal{} x^{6}\plus{}2x^{5}\plus{}P(x)$ where $ deg P(x)\leq 4$
remember
by definition we can wrote $ f(x)\equal{}o(x)$ if $ \lim_{x\to 0}\frac{f(x)}{x}\to 0$ (not $ x\to\plus{}oo$)

[quote="Extremal"]

remember
by definition we can wrote $ f(x) \equal{} o(x)$ if $ \lim_{x\to 0}\frac{f(x)}{x}\to 0$ (not $ x\to\plus{}oo$)[/quote]

I don't agree with you.

we may have: $ f(x)\equal{}o(g(x))$ if $ \lim_{x\to\infty}\frac{f(x)}{g(x)}\equal{}0$ as well.It's not necessary $ x\to 0$

[quote="Svejk"][quote="Extremal"]

remember
by definition we can wrote $ f(x) \equal{} o(x)$ if $ \lim_{x\to 0}\frac{f(x)}{x}\to 0$ (not $ x\to\plus{}oo$)[/quote]

I don't agree with you.

we may have: $ f(x) \equal{} o(g(x))$ if $ \lim_{x\to\infty}\frac{f(x)}{g(x)}\equal{} 0$ as well.It's not necessary $ x\to 0$[/quote]
you are right

[quote="the_wonder"]Ascertain $ \lim_{x\to\infty }f(x)$, where ${ f(x) =\sqrt [3]{x^{3}+x^{2}+x+1}-\sqrt{x^{2}+x+1}}\cdot\frac{\ln\ (e^{x}+x)}{x}$.[/quote]
[color=darkblue][b]Lemma[/b] ([u]memorize this remarkable limit[/u] !). $ \boxed{\ g(x) =\sqrt [n]{x^{n}+px^{n-1}+\ldots}-x\implies\lim_{x\to\infty }g(x) =\frac{p}{n}\ }$. Indeed,

$ g(x) =\frac{\left(\frac{x^{n}+px^{n-1}+\ldots}{x^{n}}\right)^{\frac{1}{n}}-1}{\frac{x^{n}+px^{n-1}+\ldots}{x^{n}}-1}\cdot\frac{px^{n-1}+\ldots}{x^{n-1}}\implies\lim_{x\to\infty}g(x) =\frac{p}{n}$ because $ \lim_{t\to 1}\frac{t^{\alpha}-1}{t-1}=\alpha$, where $ \alpha\in\mathcal R^{*}$. In the our case, $ \alpha =\frac{1}{n}$.

[b]Proof.[/b] Observe that $ f(x) =$ $ \left(\sqrt [3]{x^{3}+x^{3}+x+1}-x\right)-\left(\sqrt{x^{2}+x+1}-x\right)+\frac{\sqrt{x^{2}+x+1}}{x}\ \cdot\ \left[\ x-\ln\ \left(x+e^{x}\right)\ \right]$. In conclusion,

using the upper lemma, $ \boxed{\ \lim_{x\to\infty }f(x) =-\frac{1}{6}\ }$ because $ h(x) = x-\ln\ (x+e^{x}) =-\ln\ \left(1+\frac{x}{e^{x}}\right)$ and $ \lim_{x\to\infty}\frac{x}{e^{x}}= 0\implies$ $ \lim_{x\to\infty }h(x) = 0$.[/color]


Given are $\triangle ABC,M$ inside it, $d_{a},d_{b},d_{c}$ is distances from $M$ to line $BC,CA,AB$ prove that
$sin\frac{A}{2}MA+sin\frac{B}{2}MB+sin\frac{C}{2}MC\ge d_{a}+d_{b}+d_{c}$
 :roll:

This problem is quite obvious followed immediately from $sin \frac{A}{2}\geq \frac{cos \measuredangle{BAM}+cos \measuredangle{CAM}}{2}$

yes you are right treegoner it's too easy, It has a corollary:
$sin\frac{A}{2}MA^{2}+sin\frac{B}{2}MB^{2}+sin\frac{C}{2}MC^{2}\ge 2(d_{a}d_{b}+d_{b}d_{c}+d_{c}d_{a})$ :)
a similar problem
$cos\frac{A}{2}MA+cos\frac{B}{2}MB+cos\frac{C}{2}MC\ge p$
 :)

doesn't the first one follow easily since by rearrangement $d_{a}^{2}+...\geq d_{a}d_{b}...$
then you just multiply through in the identity treegoner posted?

yes it is quite easy too:
$\frac{3}{2}(\sum sin\frac{A}{2}MA^{2})\ge (\sum sin\frac{A}{2})(\sum sin\frac{A}{2}MA^{2})\ge (\sum sin\frac{A}{2}MA)^{2}\ge (d_{a}+d_{b}+d_{c})^{2}\ge 3(d_{a}d_{b}+d_{b}d_{c}+d_{c}d_{a})$
similar
$\sum cos\frac{A}{2}MA^{2}\ge \frac{2}{\sqrt{3}}(ab+bc+ca)$
they are too weak but maybe they are quite nice! :)


Below is a $4\times4$ grid. We wish to fill in the grid such that each row, each column, and each $2\times2$ square outlined by the double lines contains the digits 1 through 4. The first row has already been filled in. Find, with proof, the number of ways we can complete the rest of the grid.

\[ \begin{tabular}{||c|c||c|c||}\hline\hline 1 & 2 & 3 & 4\\ \hline &&&\\ \hline\hline &&&\\ \hline &&&\\ \hline\hline \end{tabular}  \]

This is problem #1. ;)

Don't have time to write up a solution now...


Bonjour,

j'ai appris le français pendant huit ans, mais il faut constater que je ne comprends pas encore ces "liaisons".  Ca veut dire : prononcer un consonant à la fin du mot parce que le mot suivant commence avec une voyelle..
J'ai lu cet article et ces règles, mais.... :(

S'il vous plaît, dites-moi s'il faut faire une liaison :

1je veu[b]x [/b]un chien
2.mai[b]s [/b]il vient
3.mai[b]s[/b] enfin
4.Il étai[b]t[/b] amoureux.
5.J'étai[b]s[/b] amoureux.
6.deu[b]x [/b]arbres
7.deu[b]x[/b] hommes
8.troi[b]s[/b] arbres
9.troi[b]s [/b]hommes
10.di[b]x [/b]arbres
11.di[b]x [/b]hommes
12.nous acheton[b]s [/b]une voiture
13.nous achete[b]z [/b]une voiture
14. je veu[b]x[/b] écouter.

S'il y a une règle magnifique qui explique tout cela, ce serait bienvenue  :( .

Consonant = consonne. :)

Liaison : 3, 6, 7, 8, 9, 10, 11
Pas de liaison : 1, 2, 4, 5, 12, 13, 14. (enfin je pense ^^)

Tiens, ceci te conviendra peut-être : http://www.etudes-litteraires.com/regles-de-liaison.php

[quote="mathmanman"]Consonant = consonne. :)
[/quote]
Woops :D  "une consonne" alors? (je n'ai aucun "feeling" pour le gendre de ces substantifs)

Merci, je vais regarder ce site et peut-être revenir pour demandres quelques questions supplémentaires.
Je voudrais utiliser ce site [url]http://demo.acapela-group.com/[/url]
C'est un site très intéressant (ils ont même créé deux formes de néerlandais : la forme Belgique et celle aux Pays-Bas  :) ) mais regardez ce qu'on écoute quand on essaie "peut-être" :huh: .

C'est un peu étonnant qu'il n'y ait pas de réactions des Français ici :
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=109824[/url]
J'ai même aidé quelqu'un à écrire une traduction... :|  mais je ne suis pas sûr du tout si c'est correct..

Yep, une consonne. :)

Ok.

Hmm, je ne vois pas le problème avec "peut-être"..

Quant au dernier lien, personnellement, je ne lis jamais les posts dans "Games & Fun Factory", donc..

Chez moi, ce "bot" fait quelque chose totalement bizarre avec "peut-être" :maybe:

Mais ce n'est pas encore totalement évident pour moi... :|

15. nous avon[b]s[/b] un chat
16. ils veulen[b]t [/b]aider
17. nous somme[b]s [/b]amoureux
18. nous avon[b]s[/b] aimé
19. de petite[b]s[/b] organisations
20. ils on[b]t [/b]aimé
21. je veux vou[b]s[/b] aider
22. je ne veux plu[b]s[/b] écouter
23. je veu[b]x[/b] encore gagner
24. il ne veut jamai[b]s[/b] écouter
25. commen[b]t [/b]allez-vous.
26. commen[b]t[/b] aller à Paris?
27. je sui[b]s[/b] amoureux
28. je sui[b]s[/b] au huitième étage.

Mais je ne pense pas que je comprends tout ces termes :

"le déterminant" de "mes amis", c'est "mes", n'est-ce pas, mais un déterminant, ça peut être quoi? (c'est toujours un pronom? :ninja: )

une forme verbale composée, c'est quoi exactement? J'ai cherché sur Wikipedia, mais ce n'était pas très clair

Il n'y a pas une "liaison officielle" pour "t"? [url]http://www.etudes-litteraires.com/regles-de-liaison.php[/url] ne l'explique ( cette question a quelque chose à faire avec numéro 20) Pour exemple, dans "ont-ils" le "t" c'est comme un "t" normal?

Merci beaucoup,
fredbel6

"Nous avon[b]s[/b] un chat" -ici on fait la liaison ! :roll:
Pour 18 et 20 aussi !

Je crois que les "liaisons" en français sont assez subtiles, moi-même au bout de 15 ans je ne les maîtrise pas ! :mad:

[quote="mstoenescu"]"Nous avon[b]s[/b] un chat" -ici on fait la liaison ! :roll:[/quote]
 :maybe:  Merci, mais c'est le français général?

C'est déjà si difficile pour moi. :(

[quote="mstoenescu"]Je viens de regarder dans [b]Dictionnnaire de l'orthographe[/b] d'André Jouette, paru en 1991 chez France Loisirs (ISBN:2-7242-5263-2).
La page 373 est dédiée aux "liaisons du français" ! Il y a des liaisons obligatoires, d'autres facultatives et des interdits de liaison !
C'est un super livre, tu pourrais le trouver chez ebay ou priceminister ou ...
Je l'utilise très souvent ! :roll:[/quote]
Ca, ce serait un peu trop pour le moment (373 pages!) mais merci! :wink:

Si vous savez le réponse aux questions que j'ai posées dans ma deuxième messages, vous pouvez l'essayer aussi bien-sûr!

Mais... votre nom (mstoenescu).. vous n'êtes pas français?  Vous venez de la Romanie?

[quote="fredbel6"]
Ca, ce serait un peu trop pour le moment (373 pages!)

Mais... votre nom (mstoenescu).. vous n'êtes pas français?  Vous venez de la Romanie?[/quote]

Il y a [b]que[/b] la page 373 oû ils parlent des règles de liaison !
On dit "Vous venez [b]de[/b] Roumanie"

Oui, je suis roumain et français ! :roll:

J'ai ajouté quelques questions (22,23,24,25,26,27,28).  Vous voulez jeter un coup d'oeil? :(
 :o
[quote="mstoenescu"][quote="mstoenescu"][b]Dictionnnaire de l'orthographe[/b] d'André Jouette, paru en 1991 chez France Loisirs (ISBN:2-7242-5263-2).
[/quote]

Je viens de voir que [b]priceminister[/b] l'offre à...[b]2.99[/b] euros ! :roll:
et il fait presque [b]800[/b] pages ![/quote]
Uhm, je vais le considérer, merci. :)

J'ai découvert que mon livre d'école et ce site ne disent pas la même chose. :(

[quote]On dit "Vous venez de Roumanie" [/quote]
 :oops_sign:  :| J'avais oublié qu'on n'écrit pas d'articles pour les pays "féminins".

[quote="fredbel6"]J'ai ajouté quelques questions (22,23,24,25,26,27,28).  Vous voulez jeter un coup d'oeil ?[/quote]

Je ferais partout la liaison mais pas pour 24. Le motif: cela sonne [b]mal[/b]
à mon oreille (francisée depuis 15 ans :wink: )

Puis tu as dit : "Je vais le considérer". Plus correctement serait peut-être "je vais le [b]prendre en considération[/b]"

C'est un peu difficile quand on parle de "partout mais..." parce que j'ai écrit plusieurs listes.

Donc qu'est-ce que vous (=mstoenescu) feriez avec  :| :
[quote]15. nous avon[b]s[/b] un chat
16. ils veulen[b]t [/b]aider
17. nous somme[b]s [/b]amoureux
18. nous avon[b]s[/b] aimé
19. de petite[b]s[/b] organisations
20. ils on[b]t [/b]aimé
21. je veux vou[b]s[/b] aider
22. je ne veux plu[b]s[/b] écouter
23. je veu[b]x[/b] encore gagner
24. il ne veut jamai[b]s[/b] écouter
25. commen[b]t [/b]allez-vous.
26. commen[b]t[/b] aller à Paris?
27. je sui[b]s[/b] amoureux
28. je sui[b]s[/b] au huitième étage.[/quote]

je pense que ce serait beaucoup plus clair si vous feriez :
78.  blabla avon[b]s[/b] aime-> oui
ou
78.  blabla avon[b]s[/b] aime-> non

Merci! (c'est vraiment difficile, j'ai cherché partout mais tous les sites disent autre chose :!: )

[quote="fredbel6"]15. nous avon[b]s[/b] un chat
16. ils veulen[b]t [/b]aider
17. nous somme[b]s [/b]amoureux
18. nous avon[b]s[/b] aimé
19. de petite[b]s[/b] organisations
20. ils on[b]t [/b]aimé
21. je veux vou[b]s[/b] aider
22. je ne veux plu[b]s[/b] écouter
23. je veu[b]x[/b] encore gagner
24. il ne veut jamai[b]s[/b] écouter
25. commen[b]t [/b]allez-vous.
26. commen[b]t[/b] aller à Paris?
27. je sui[b]s[/b] amoureux
28. je sui[b]s[/b] au huitième étage.[/quote]

Il y a quelques jours, j'ai fait une faute de liaison au téléphone avec ma grand-mère et elle m'a directement corrigé. Donc sois prévenu: je ne suis pas du tout sûr de mes réponses.  :maybe:

Oui: 17, 18, 19, 21, 22, 25, 26, 27
Non: 15, 16, 20 (mais il faut une liaison chez "ils ont"), 23, 24

Je crois que 28 peut être dans "oui" et "non".

Hier, il y avait "BBC Learning" sur la télévision.  Cette nuit, on pouvait apprendre un peu de français ("Savoir dire").
Je me rappelle qu'on faisait la liaison tout le temps pour "c'es[b]t [/b]une baguette" :o

Si vous me permettez, quelques nouveaux numéros  :| (ce n'est pas du tout facile!):

29. J'étai[b]s[/b] au huitième étage.
30. J'étai[b]s[/b] en France.
31. J'étai[b]s[/b] amoureux.
32. les chanson[b]s[/b] en français
33. c'es[b]t [/b]au huitième étage
34. C'es[b]t[/b] en France.
35. C'es[b]t[/b] en Allemagne.

Je ferais la liaison uniquement pour 33,34 et 35...

Les autres [b]sonnent mal[/b] à l'oreille. :mad:

[quote="Kurt Gödel"]Je suis d'accord avec mstoenescu, mais je ferais peut-être quand même la liaison au numéro 31.[/quote]
Oui Kurt Gödel, il y a ici une émission "Lez amours" donc, on invite les gens à prononcer le [b]z[/b] de liaison dans " je sui[b]s[/b] amoureux"
Il y aussi le refrain: "quand je vois tes yeux, je suis amoureux" et le "[b]z[/b]" s'entend ! :blush:
Donc, d'accord aussi pour 31.

Quand j'étais petit, j'allais chez ma grand-mère pendant les vacances et à midi, elle regardait toujours "Lez amours"  :roll:

Vous êtes sûrs de ce [i]18. nous avon[b]s [/b]aimé[/i] ?
 :maybe:

J'ai quelques nouvelles questions :surrender::

36. e[b]n[/b] Allemagne
37. ils n'on[b]t [/b]aucun moyen
38. il fau[b]t[/b] apprendre une autre langue
39. aucu[b]n[/b] appareil

Pourquoi est-ce que tu as des doutes, Frédéric?

Pour la nouvelle série:
Liaison: 36, 39
Pas de liaison: 38

Je ne suis pas sûr pour 37. Moi, je ferais la liaison, mais bon...  :)

mstoenescu, je vous remercie mais (pour moi  :| ) c'est très difficile quand on dit "les autres" parce que j'ai écrit plusieurs listes.

J'ai des doutes parce qu'il semble qu'il n'y ait pas de règles!

On ne peut quand même pas mémoriser la liaison pour toutes les combinations possibles? :mad:

Je sais que ce n'est pas comme les mathématiques (beacoup d'exceptions..) mais ... n'y at-il pas de règles élémentaires  :o

J'ai pris ma grammaire pratique de l'école sécondaire :
[quote]
On fait la liaison pour :

1. l'article et le substantif (le[b]s[/b] enfants)
2. l'adjectif et le substantif (des bon[b]s [/b]enfants, di[b]x [/b]arbres, se[b]s[/b] enfants)
3. le pronom personnel et le verbe( je le[b]s[/b] ai vus,nou[b]s[/b] allons,..)
4. la préposition( sauf selon et vers) et le mot suivant
5. "certains" adverbes et l'adjectif (plu[b]s[/b] aimable) (c'est quoi ça, "certains" :huh: ?)[/quote]

On ne discute jamais les cas comme "nous avon[b]s [/b]aimé" et "quan[b]d [/b]ont-ils fait ça"
Je m'énerve un peu, pourquoi nos livres de l'école ne sont-ils pas clairs, si on n'avait que ce livre, on penserait que c'est un liste complet? :mad:

à kurt gödel : ça m'arrive encore de regarder "lez amours" :blush:
à fredbel6 : je vais rendre visite, dans les prochains jours, à un super ami qui est [b]prof de français[/b] à la retraite. Je vais lui poser la question des liaisons et je vais te répondre plus "savamment"... :wink: (il est français de "pure souche" !)

Si vous me permettez  :(  :

40. Je veux vou[b]s[/b] aider.
41. les Etat[b]s[/b]-Unis
42. Il avai[b]t[/b] attendu.
43. Ils avaien[b]t[/b] attendu.
44. Nous avon[b]s[/b] attendu.
45. Pourquoi ces garçon[b]s [/b]ont-ils attendu?

J'apprécie toutes les réponses de tout le monde :) .  Vous êtes tous très gentils.

mstoenescu, vous connaissez quelqu'un qui est prof de français  :!:  Oui, cela pourrait être très intéressant, mais il ne faut pas que vous vous fatiguiez pour moi. :wink:

Bonjour fredbel6 !

Tu vois, je ferrais la liaison pour 40,41 et 44 mais pas pour 42,43 et 45.
Tout simplement car [b]cela ne sonne bien à mon oreille ![/b]. Et mon ami
m'a complété en disant que c'est l'usage et la "sonorité" qui déterminent s'il
y a liaison ou pas ! Mais cela parmi les liaisons facultatives ! :)

Merci, mais je ne pense pas que je comprends ce que ça veut dire. :|  :!:

J'ai encore quelques questions si vous me permettez  :o  :blush:
46. Voous vou[b]s[/b] en avez bu?
47. Vou[b]s[/b] en êtes au courant?
48. Ils son[b]t[/b] en France.
49. Ils von[b]t [/b]à Paris.
50. Nous somme[b]s[/b] à Paris.

C'est vraiment difficile, ces liaisons.  Le problème, c'est que pendant les huit années à l'école, j'ai LU en français.  Mon professeur parlait en français en classe, mais quand j'étudiais à la maison, il n'y avait aucun moyen pour reconnaître les liaisons, et quand le professeur nous disait de parler en français, c'était seulement pour nous évaluer, il écrivait notre score sans jamais nous dire ce qu'on devait amérioler. :ninja:  :?:  :mad:

Merci!

[quote="fredbel6"]Merci, mais je ne pense pas que je comprends ce que ça veut dire[/quote]

Plus simple serait : " Je ne pense pas comprendre ce que cela veut dire".

En ce qui concerne tes dernières (et non pas "ultimes" !) questions, je ferais volontiers la liaison pour 46 à 49 mais pour 50 parfois OUI parfois NON !
Je m'explique:Si je veux me donner "plus d'importance" je dirais:
Nous somme[b]z [/b] à Paris. (c'est un peu ronflant ,d'accord ?)
Sinon, sur un ton neutre, je dirais "Nous sommes à Paris" sans faire de liaison ...J'insiste et je signe : La langue française n'est pas un language (comme les maths !), elle est [b]malléable[/b] et...surprenante !
Tu ne pourrais pas écouter une émisssion à la radio (France culture, ou France inter ?). Ils sont des sites internet, tu peux "podcaster" et faire attention aux liaisons ! Je pense que le fait de faire des liaisons [b]facultatives[/b] tient aussi de la liberté française  :wink:

Hmmmm, je vous remercie.  Oui, je sais qu'une langue n'est pas comme les mathématiques, mais... on ne peut pas nier qu'il y a beaucoup de choses qui sont totalement incorrectes.  Comme "il[b]s[/b] travaillent"

Français sera toujours ma troisième langue, pour le moment, je voudrais éviter les liaisons incorrectes  , et d'oublier les liaisons obligées  :( .

Je connais cet article sur Wikipedia, j'ai déjà lu la version anglaise et la version française, mais ce n'est pas si facile, d'abord ils énumèrent les liaisons obligés et facultatives....mais puis on écrit qu'on peut le faire quand ce n'est pas interdit. :huh:

Donc

"Je ne suis pa[b]s[/b] amoureux."
"Je ne suis pa[b]s[/b] en France."
"Je ne veux pa[b]s[/b] attendre."
C'est tout facultatif? :?:
 :!:

Le français est vraiment difficile.  Il n'y a aucun lien entre les lettres et la prononciation.
"un oeuf, des oeufs" est un exemple parfait  :o

En anglais c'est la même chose, comparez "bare" et "bear", ou "wear"/"where"/"were"

Ce n'est pas le ças en néerlandais, ma langue maternelle.  Quand on écrit une consonne, il faut la prononcer, c'est très simple. :wink:

Oui, je reçois quatre chaînes francophones: RTBF I et II (deux chaîne publiques de l'autre communauté belge (les francophones)), France 2 et RTL-TVI (une chaîne commerciale de la communaute française en Belgique)

Je l'écoute souvent, mais ils parlent si vite :mad: .  Il y avait un interview  avec l'ambassadeur d'Afghanistan, et le journaliste lui forçait de parler en français et il posait des questions très longues et il parlait très vite, c'était quand même évident que l'ambassadeur avait des difficultés.  Mais parce que l'ambassadeur parlait très calmement, très lentement (il devait se concentrer)... je pouvais le comprendre beaucoup plus facilement.  C'est bizarre mais c'est vrai, pour moi, comprendre les non-francophones en français est plus facile. :huh:  :(

[quote="mstoenescu"][quote="fredbel6"]

Ce n'est pas le ças en néerlandais, ma langue maternelle.  Quand on écrit une consonne, il faut la prononcer, c'est très simple.[/quote]

En roumain aussi ! :)[/quote]
Hmm, peut-être je devrais apprendre roumain!  :D  Si j'ai déjà appris le latin, serait-ce difficile? :ninja:

51. Pourquoi ces garçons ont-il[b]s[/b] attendu?

Je dirais : pas de liaison? Est-ce correct? :maybe:
Merci!

Salut fredblel6 !

Le roumain possède
-toutes les declinaisons du latin,
-des changements [b]à l'intérieur du mot[/b] "comme en allemand" (exemple: garçon = baiat (il doit y avoir un "chapeau" sur le premier "a" qui le fait lire comme le deuxième "e" du mot allemand "lern[b]e[/b]n"), puis garçons= baieti (on rajoute un "i" et on change le "a" en "e" !!)
-le fameux (?) son "i" avec un chapeau qui se prononce comme un "i" mais...en émettant le son du fond de la bouche et en soulevant les narines (explication personnelle pour les étrangers !) qui rébutent le plus !?
-un cocktail de mots latins, slaves, grecs, turcs, hongrois et...français

A part ça, pour le 51 je ne ferais pas de liaison (cela ne me sonne pas bien !) :maybe:


The sequence $ (a_n)_{n \equal{} 1}^{\infty}$ is such that$ a_1>0$ and $ a_{n\plus{}1}\equal{}\frac{a_n}{1\plus{}a_n^2}$ for $ n\geq 1$. Prove that

(a)$ a_n\leq \frac{1}{\sqrt{2n}}$
(b) there exists $ n$ such that $ a_n > \frac {7}{10\sqrt {n}}$

Let $ b_n\equal{}\frac{1}{a_n^2}$, then $ b_{n\plus{}1}\equal{}b_n\plus{}2\plus{}\frac{1}{b_n}, \ b_1\equal{}\frac{1}{a_1^2}>0$.
Therefore $ b_n\equal{}2n\plus{}\frac 12 lnn\plus{}c\plus{}O(\frac{ln(n)}{n}$.

Well, Rust, if you want to post, please post a full solution that everybody can understand, not just do it halfway like this...  :wink:

Let $ b_n\equal{}2n\plus{}c_n,c_1>\minus{}2$. Then $ c_{n\plus{}1}\equal{}c_n\plus{}\frac{1}{2n\plus{}c_n}>c_n$. Let $ d_n\equal{}c_n\minus{}\frac 12 ln(n)$. Then
$ d_{n\plus{}1}\equal{}d_n\plus{}\frac{1}{2n\plus{}0.5ln(n)\plus{}d_n}\minus{}0.5ln(1\plus{}\frac 1n )$.  Because $ d_{n\plus{}1}\minus{}d_n\equal{}O(\frac{ln(n)}{n^2})$, we get exist $ d_{\infty}$, suth that $ d_n\equal{}d_{\infty }\plus{}O(\frac{ln(n)}{n}$.
It give $ b_n\equal{}2n\plus{}\frac 12 ln(n)\plus{}d_{\infty}\plus{}O(\frac{ln(n)}{n})$ and
$ a_n\equal{}\frac{1}{\sqrt{2n\plus{}\frac 12 ln(n)\plus{}d_{\infty}\plus{}O(\frac{ln(n)}{n})}}$.

And by the way, there must be a more elementary solution  :D

Actually the problem statement is a little bit wrong. Part (a) should be $ a_n\leq \frac{1}{\sqrt{2n}}$ for $ n\geq 2$.  :)


How do you label a line segment? Like [i]AB[/i] with the line over it?

How do you make a greater than or less than with the equals part?

[code]$\le$[/code]
$ \le$

[code]$\ge$[/code]
$ \ge$


induction
graph theory
combinatorics unsolved
combinatorics

$G$ is a connected graph and $\delta$ is the leat degree in $G$,prove that if $G$ has more than $2 \delta$ vertices,then $G$ contains a path with length at least $2\delta$

I am not sure if I understand the problem correctly. What do you mean by "path"?
Is it allowed to visit edges more than once?
(Then I could just go round cycles to obtain paths of any length.)
Or is it not allowed? Then consider the complete graph with 3 vertices. It has $\delta = 2$, but
does not contain a path of length $4$, as there are only $3$ edges...

In graph theory, a path in a graph is a sequence of vertices such that from each of its vertices there is an edge to the next vertex in the sequence.(wikipedia)

sorrrrrrrrrrrrrrrrrrrrrrrrrrrrry,it was a typo...
I ment that $G$ contains more than $2 \delta$ vertices
I edited the problem,sorry

Hmmm...
You may use induction on the number $n$ of vertices
The first step is to prove that if $G$ is not $2$-regular, the result is quite easy (consider a vertex $A$ whose removal disconnect the graph, and use the induction hypothesis on each connected component. Then, concatenate two paths with two edges with endpoint $A$).

Then, we are leaving to prove the result in the case where $G$ is $2$-connected. this is now a classical exercise in graph theory to prove that such a graph has an hamiltonian circuit or a circuit with length at least $2\delta$.

Pierre.

how could I miss that...
I almost forgot about the $2-connectivity$ of $G$...
thanx anyway


Is the equation a^n*b+a*b^n able to be factored?

Can it be divided by something? Besides a and b

It can be factored by [hide]ab[/hide], if that counts, but you can go farther if you go farther if you use complex numbers. It also depends on n.

factor out ab. then if n-1 is odd then it can be further factored to ab(a+b)(a^n-2-a^n-3*b+a^n-4*b^2...+b^n-2)

if n-1 is even do the same thing except replace every a by a^2 and b by b^2 and adjust the pattern in the last factor accordingly...

[quote="jeffrey"]
if n-1 is even do the same thing except replace every a by a^2 and b by b^2 and adjust the pattern in the last factor accordingly...[/quote]

Can you be more specific? This is what I need.

Thanks!

sorry, i wasn't specific enough.

if n-1 is an integer power of two, theres not much you can do. but if its got any odd factor you can make a substitution. call the odd factor x and the even factor y. the you can call it (a^y)^x + (b^y)^x.

for example

a^6+b^6= (a^2)^3+ (b^2)^3= (a^2+b^2)(a^4-a^2b^2+b^4)


with a counting argument show,

\[ \binom{2n\plus{}2}{n\plus{}1}\equal{}\binom{2n}{n\plus{}1}\plus{}\binom{2n}{n\minus{}1}\plus{}2\binom{2n}{n}\]

I'll use $ (a,b) \equal{} \binom{a}{b}$ so that I don't have to LaTeX...

[hide="Argument"][b]Situation:[/b] How many ways are there to arrange 2n+2 kids, indistinguishable beyond gender, in a line, if there are n+1 girls and n+1 boys?

[b]Method 1:[/b] There are (2n+2, n+1) ways to pick spots for the n+1 girls. Boys fill in the remaining spots.

[b]Method 2:[/b] Suppose we consider the leftmost 2 kids and the rightmost 2n kids separately.

[i][b]Case 1:[/b] [/i]The left 2 kids are GG. There is 1 arrangement. There are n-1 girls (and n+1 boys) in the right 2n kids, so there are (2n,n-1) ways of arranging them.

[i][b]Case 2:[/b] [/i]The left 2 kids are BB. There is 1 arrangement. There are n+1 girls in the right 2n kids, so there are (2n,n+1) ways of arranging them.

[i][b]Case 3:[/b][/i] The left 2 kids are GB. There are 2 arrangements. There are n boys and n girls in the right 2n kids, so there (2n,n) ways of arranging them. This gives a total of 2*(2n,n) arrangements in this case.

[b]Conclusion:[/b] The sum of the 3 terms of method 2 correspond to the 3 terms in the RHS, and value from method 1 corresponds to the LHS. Since they both count the same thing, they must yeild the same value.[/hide]


limit
trigonometry
floor function
function
calculus
calculus computations

Can anyone help me with this one?
Prove that :
${{\lim_{n\to\infty}sin(n)}}$ doesn't exist

We have $f(n)=\sin{n}$, so just take 2 sequences that tends to same value($\infty$) like: $a_{n}=\frac{\pi}{2}+2n\pi$ and $b_{n}=2n\pi$ and $\lim_{n\to\infty}{\sin(a_{n})}\neq\lim_{n\to\infty}{\sin(b_{n})}$ thus $\lim_{n\to\infty}{\sin(n)}$ does not exist. (for $n\to-\infty$ it's the same thing)

That assumed $n$ was a real variable. What if $n$ is an integer variable?

For all positive integer $m$ we have
\[ \frac{\pi}{2} + 2 \pi m \leq 2 + 2 \pi m \leq 3 + \lfloor 2 \pi m \rfloor \leq 3 + 2 \pi m \leq \pi + 2 \pi m  \]
Recall that $\sin x$ is decreasing on $\left[ \frac{\pi}{2} + 2 \pi m ; \pi + 2 \pi m \right]$ for all integers $m$. Hence
\[ 0.91 > \sin ( 2 ) \geq \sin \left( 3 + \lfloor 2 \pi m \rfloor \right) \geq \sin ( 3 ) > 0.14  \]

Similarly for all positive integer $m$,
\[ 2 \pi m \geq \lfloor 2 \pi m \rfloor \geq 2 \pi m - 1  \]
$\sin x$ is increasing on $\left[ 2 \pi m - 1 ; 2 \pi m ] \right.$ for all integers $m$. So
\[ \left. \left. 0 = \sin ( 2 \pi m \right) \geq \sin \left( \lfloor 2 \pi m \rfloor \right) \geq \sin ( 2 \pi m - 1 \right) = - \sin ( 1 ) > - 0.85  \]

Concluding, for $m$ a positive integer,
\begin{eqnarray*} 0 \geq \sin ( \lfloor 2 \pi m \rfloor ) > - 0.85 & \mbox{and} & 0.91 > \sin \left( 3 + \lfloor 2 \pi m \rfloor \right) > 0.14 \end{eqnarray*}
with $\lfloor 2 \pi m \rfloor$ and $3 + \lfloor 2 \pi m \rfloor$ integers.Therefore the limit
\[ \lim_{n \rightarrow \infty} \sin n  \]
cannot exist.

I don't know if this statement is true or not.
If $lim_{x\rightarrow \infty} (sin(x))$ then $lim_{n\to \infty} (sin(n))$ doesn't exist.If this one is correct then the solution of Slizzel is corect.
     About solution of {x}, I don't understand :blush:

  I try to prove $lim_{n \to \infty}sin(n)$ by this way.
  Suppose that $lim_{n \to \infty}(sin(n)=L)$
   -Since $-1 \leq sin(n) \leq 1$ so $-1 \leq L \leq 1$
   -Acording to definition of limit. For any $\epsilon > 0$ there exist the number N such that if $n > N$ then
    $| sinn -L | < \epsilon$ or $L-\epsilon <sin n < L+\epsilon$ for all of  $\epsilon > 0$. But if we draw the graph of this function $f(x)=sin(x)$ we see that there isn't posibility to trap all of the value of $sinn$ in the region $L-\epsilon$ and $L+\epsilon$
   Is it correct?

[quote="Gibbenergy"]
If $lim_{x\rightarrow \infty} (sin(x))$ [color=red][does not exist][/color] then $lim_{n\to \infty} (sin(n))$ doesn't exist.
[/quote]
No it's not true. (This is why jmerry remark's holds).
Consider for example, the function defined by $f(x) = \mbox{[} x \mbox{ is integer]}$ (written in Iverson's notation)
[i]C'est-a-dire[/i], if $x$ is an integer, then $f(x)=1$, and if $x$ is not an integer then $f(x)=0$.
Then of course $\lim_{x\rightarrow \infty} f(x)$ cannot exist, but $\lim_{n\to \infty} f(n)$ does exist and is equal to $1$.
Homever the converse of your statement is true. For any function $f$, if $\lim_{n\to \infty} f(n)$ does not exist, then $\lim_{x\rightarrow \infty} f(x)$ does not exist.

Also, tell me where my argument gets obscur. (I just show that we cannot trap $sin$ in a "small region", since it "lies infinitely often in two distinct region")

[quote="Gibbenergy"]But if we draw the graph of this function $f(x)=\sin(x)$we see that there isn't posibility to trap all of the value of $\sin n$ [that should be $\sin x$] in the region $L-\epsilon$ and $L+\epsilon$ [/quote]
That may be a vague indication of where one might start looking to find a proof, but that is not itself a proof.

But beyond that, even if we do successfully argue that $\lim_{x\to\infty}\sin x$ does not exist, that doesn't have to tell us anything about a particular subsequence. After all, $\lim_{n\to\infty}\sin(n\pi)$ does exist.

{x}'s proof is correct. Here's a slightly more streamlined version.

There exists a sequence of integers $n_j$ such that $\left|2j\pi+\frac{\pi}2-n_j\right|\le\frac12.$

There exists a sequence of integers $m_k$ such that $\left|2k\pi+\frac{3\pi}2-m_k\right|\le\frac12.$

This is just finding the closest approximating integer.

Then $\sin n_j\ge\sin\left(\frac{\pi}2-\frac12\right)$ and $\sin m_k\le-\sin\left(\frac{\pi}2-\frac12\right).$

Hence $|\sin n_j-\sin m_k|\ge a$ for a fixed positive number $a$ and the sequence $\sin n$ is not a Cauchy sequence and thus cannot converge.

[quote="Kent Merryfield"]streamlined[/quote]
What does it (EDIT: this word) mean  :oops: ? I guess it's something meliorative  but (EDIT : I can guess) nothing beyond that...  :roll:
EDIT: I was just asking about the meaning of the word "streamlined", since it was unknown to me. Due to my bad formulation, I probably got misunderstood. Anyway I looked it up in a dictionary, so the problem is fixed :)

I was just trying to make it shorter and maybe a tiny bit easier to read. Apart from insignificant details, it's the same as your proof. And I was working on my reply to Gibbenergy and didn't notice that you had also replied; we both said essentially the same things to him.

Thank you all of you for replying me.
Now, I understand the solution of {x}
   Basicly, {x} tried to show that there has infinity number $a_n$ lies between 2 distince regions. So there will exist one $\epsilon$  that can't trap all of the value of $sin(a_n)$ in it. Is it your argument {x}.
   In Kent Merryfield solution, you try to prove that the distance between 2 sequences (that your already construct) is $a$ (which in this particular case $a=2sin (\frac{\pi}{2}-\frac12)$ ) and let to the fact that there will exist one $\epsilon$ can't satisfy it. Right?
   And also, in Kent Merryfield solution, you mention about Cauchy sequences. Please let me know what is it?
Thank you a lot.

A Cauchy seqence, also known as sequence that "converges within itself," is a sequence $(a_n)$ with this property:

For all $\epsilon>0,$ there exists $N$ such that for all $n,m>N$ (note: $n$ and $m$ need not be close to each other), $|a_n-a_m|<\epsilon.$

A short, straightforward proof shows that a convergent sequence must be Cauchy. (The rough paraphrase of the proof: $a_n$ and $a_m$ are both close to the limit, so they're close to each other.)

And that's what I used, in its contrapositive form: a sequence that is not Cauchy cannot converge.

In the real numbers, every Cauchy sequence converges. This is deep, and is one way to say that the real numbers are complete. It is logically equivalent to other forms of completeness, such as the least upper bound axiom.

[hide="Here is a classical proof."][color=blue]Let the sequence $\boxed {a_n=\sin n\ ,\ n\in N^*}\ .$ From the property $\(\forall\right)n\in N^*\ ,$ $\left|\sin n\right|\le 1$ (the sequence $a_n$ is bounded) results $a_n\rightarrow a\in R$ (convergent) or the sequence $a_n$ hasn't limit . Suppose that the sequence $a_n$ has limit, i.e. $\lim_{n\to\infty}\sin n=a\in R\ .$ In this case we have :
$1.\blacktriangleright a_{n+2}-a_n=\sin (n+2)-\sin n=2\sin 1\cdot \cos (n+1)$ $\Longrightarrow$ $\boxed {\cos n\rightarrow 0}\ .$
$2.\blacktriangleright a_{2n}=\sin 2n=2\sin n\cos n\rightarrow 0$ $\Longrightarrow$ $a_{2n}\rightarrow 0$ $\Longrightarrow$ $a_n=\boxed {\sin n\rightarrow 0}\ .$
But $1=\sin^2 n+\cos^2 n\rightarrow 0\ ,$ what isn't truly. Therefore, we  come to the conclusion that the sequence $a_n$ hasn't limit .[/color][/hide]

EDIT: [quote="Gibbenergy"] Basicly, {x} tried to show that there has infinity number a_n lies between 2 distince regions. So there will exist one $\epsilon$ that can't trap all of the value of $\text{sin}(a_n)$ in it. Is it your argument {x} ? [/quote]
Exactly, this was my purpose. I'm sorry for having not made it explicit immediately.


geometry
perimeter
inequalities
triangle inequality

How many non-congruent triangles with perimeter 7 have integer side lengths?

We must have $ a\plus{}b\plus{}c\equal{}7$ and $ a\plus{}b>c$ if $ a\geq b\geq c$. Listing out is really easy; we have $ (3,3,1)$ and $ (3,2,2)$ as the only solutions, and our answer is $ \boxed2$.

[quote="Python54"]I don't understand why the longest side can't be greater than 3. Could someone please explain?

Thank you in advance.[/quote]
[hide]If one of the sides is greater than $3$, it would not satisfy the triangle inequality, which states that the sum of any $2$ sides must be greater than the third side.  Thus, if a side of the triangle measures $4$, then the sum of the remaining two lengths is $7-4=3$, which is less than $4$ and does not satisfy the triangle inequality.[/hide]

[font=TIMES NEW ROMAN][hide=2 simple steps]
[b]Step 1 - Gather[/b]
The distinct sets of three numbers that add up to $7$ are $(1,1,5), (1,2,4), (1,3,3)$ and $(2,2,3).$
[b]Step 2 - Filter[/b]
The first two sets can't be triangles, so there are only $\boxed{2}$ possible triangles. $$\definecolor{color}{rgb}{0.7, 0.9, 0.3} \scriptsize{\colorbox{color}{\color{black}{A pifinity solution}}}$$

[quote=Littlelachy]OK I've had a problem with anything relating to triangles.

Most of the time it never states whether I should count degenerate triangles so should I or should I not when doing these types of problems.[/quote]

no. but there actually aren't degenerate triangles here because the perimeter is odd.


Today I tried proving this with induction:

[tex]\displaystyle{F_n=\frac{1}{\sqrt{5}}[(\phi)^n-(-\frac{1}{\phi})^n]}[/tex]

but I had some trouble with how to go about it.  Could someone offer me some hints as to if it is doable for me as someone who has very little experience with induction(as I just learned how to do it)?

p.s.  I wasn't going to use the  :phi: symbols but when using latex I couldn't get the 1+sqrt5/2 stuff to work.  I had it all fine until the last part.  Don't know.  Maybe someone could help me with that also since I am not familiar with latex.

Hint:

[hide] If you add F_(n-1) + F_(n-2), you get F_n.  Also, keep in mind that (1+sqrt5)/2 and (1-sqrt5)/2 are the roots to the equation x^2-x-1=0.

[/hide]

Establishing that it is true for n = 1 [i]and[/i] n = 2 shouldn't be too hard. After that, assume that it's true for n = k-1 and n = k. We need to show that [tex]F_{k-1} + F_k = F_{k+1}[/tex].

The key is to note that [tex]\displaystyle{1 + \phi = \phi^2}[/tex] and [tex]\displaystyle{1 + (- \frac {1}{\phi}) = (- \frac {1}{\phi})^2}[/tex], because both are roots of [tex]x^2 - x -1 = 0[/tex].

[tex]\displaystyle{F_{k-1} + F_k = [\phi^{k-1} - (- \frac {1}{\phi})^{k-1}] + [\phi^k - (- \frac {1}{\phi})^k]}[/tex]
[tex]\displaystyle{= [\phi^{k-1} + \phi^k] - [(- \frac {1}{\phi})^{k-1} + (- \frac {1}{\phi})^k]}[/tex]
[tex]\displaystyle{= \phi^{k+1} - (- \frac {1}{\phi})^{k+1} = F_{k+1}}[/tex]

[quote="gauss202"]Establishing that it is true for n = 1 shouldn't be too hard. After that, assume that it is true for n = k-1 and n = k. We need to show that Fk-1 + Fk = Fk+1.
[/quote]

Thats not a complete proof. You need to establish two base cases if you are to assume two cases in your inductive hypothesis.

[quote="Nukular"]
Thats not a complete proof. You need to establish two base cases if you are to assume two cases in your inductive hypothesis.[/quote]

Good point.

[quote="JBL"]Also, you left out your :rt5:'s, but since they're in every term, it doesn't matter too much.  :)[/quote]

Another good point.

Yes.  In fact, the sequence that begins 2, 1, ... is known as the Lucas sequences (the Lucas numbers).

Here's a good problem: prove that other than 1, 2 and 3, there is no integer which is both a Lucas number and a Fibonacci number.

Here's another good problem: find a closed form for the nth Lucas number.

Using induction we can show that F_n+F_(n-2)=L_n. We can show by induction that
F_n+F_(n-2), which is F_(n-1)+2F_(n-2),  can never be a Fiboncci number.

[quote="Magnara"]Would it be possible to make an arbitrary formula that will give you the nth "Fibbonacci" number for a series that doesn't start with 1,1?  For example, the one that starts with 2,5 is
2,5,7,12,19,31,...[/quote]

The way in general to solve f(n+2) = af(n+1) + bf(n) is to get the roots of t^2 = at + b. When a=b=1, this is where the phi comes in. Then the final formula is constant * first root ^ n + constant * second root ^ n (slightly different when the roots are equal but I'm too tired to write that out). So basically if you change the first numbers, the formula is going to look very similar to the Fibonacci formula.


function
complex analysis
complex analysis theorems

I have a function where I'm suppose to find Rn when n goes to infinity. Rn is the residu where z = 1.

$ f \equal{}\frac {e^z} {(z\minus{}2)(z\minus{}1)^n}$

I try to Talyor expand the function and look at the sums like this:

$ e^z \equal{} e^{1\plus{}(z\minus{}1)}  \equal{} e \displaystyle\sum_{k\equal{}0}^\infty \frac {(z\minus{}1)^k}{k!}     ;k \equal{} 0, 1, 2 ...$

$ \frac {1}{z\minus{}2} \equal{} \frac {1}{(1\minus{}(z\minus{}1))}  \equal{} \minus{} \displaystyle\sum_{k\equal{}0}^\infty (z\minus{}1)^k$

These two with the factor $ \frac {1}{ (z\minus{}1)^n}$ this gives me:

$ \minus{}e  \displaystyle\sum_{k\equal{}0}^\infty (\frac {(z\minus{}1)^{2k}}{k!} \frac {1}{(z\minus{}1)^k}) \equal{} \minus{}e  \displaystyle\sum_{k\equal{}0}^\infty (\frac {(z\minus{}1)^{2}}{k!} \frac {1}{1})$

-e is hunky dory, but I can't make anything out of the sum-part.

Question: How do I get further. Or if theres something wrong, where do I go wrong?

Edit, just found out about the [code] $  - sign[/code] and it's function in this forum.

[quote="yberfuzz"]I have a function where I'm suppose to find Rn when n goes to infinity. Rn is the residu where z = 1.
[/quote]

Would you Please let me know what is the meaning of residu


Let $ a,b,c>0$. Prove or disprove that:

a)$ \frac{a^3\plus{}b^3\plus{}c^3}{3abc}\plus{}\frac{8abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}\ge 2$

b)$ \frac{a\plus{}b}{\sqrt{ab}}\plus{}\frac{b\plus{}c}{\sqrt{bc}}\plus{}\frac{c\plus{}a}{\sqrt{ca}}\plus{}\frac{(a\plus{}b\plus{}c)^2}{a^2\plus{}b^2\plus{}c^2}\ge 9$

[quote="Pain rinnegan"]Let $ a,b,c > 0$. Prove or disprove that:

a)$ \frac {a^3 \plus{} b^3 \plus{} c^3}{3abc} \plus{} \frac {8abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}\ge 2$

b)$ \frac {a \plus{} b}{\sqrt {ab}} \plus{} \frac {b \plus{} c}{\sqrt {bc}} \plus{} \frac {c \plus{} a}{\sqrt {ca}} \plus{} \frac {(a \plus{} b \plus{} c)^2}{a^2 \plus{} b^2 \plus{} c^2}\ge 9$[/quote]
a) It is wellknown that $ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \plus{} \frac {8abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \ge 2$. And by Muirhead it is obvious that $ \frac {a^3 \plus{} b^3 \plus{} c^3}{3abc} \ge \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}$

Edit: An easier solution:
By AM-GM: $ \frac{a^3\plus{}b^3\plus{}c^3}{3abc} \plus{} \frac{8abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)} \ge 2\sqrt{\frac{8(a^3\plus{}b^3\plus{}c^3)}{3(a\plus{}b)(b\plus{}c)(c\plus{}a)}} \ge 2 \iff$
$ 8(a^3\plus{}b^3\plus{}c^3) \ge 3(a\plus{}b)(b\plus{}c)(c\plus{}a)$ which is obvious by Muirhead.

for a) we have 2 more different solutions. take a look here

http://www.mathlinks.ro/viewtopic.php?t=237479

for b) we have from Cauchy:

$ \sum\frac {a \plus{} b}{\sqrt {ab}} \plus{} \frac {\left(a \plus{} b \plus{} c\right)^{2}}{a^{2} \plus{} b^{2} \plus{} c^{2}} \equal{} \sum\frac {\left(a \plus{} b\right)^{2}}{(a \plus{} b)\sqrt {ab}} \plus{} \frac {\left(a \plus{} b \plus{} c\right)^{2}}{a^{2} \plus{} b^{2} \plus{} c^{2}}\geq \frac {9\left(a \plus{} b \plus{} c\right)^{2}}{\sum(a \plus{} b)\sqrt {ab} \plus{} \sum a^2}$

So, it remains to prove that $ \frac {9\left(a \plus{} b \plus{} c\right)^{2}}{\sum(a \plus{} b)\sqrt {ab} \plus{} \sum a^2}\geq 9\Longleftrightarrow ...\Longleftrightarrow 2\sum ab\geq \sum (a \plus{} b)\sqrt {ab}$.

But, from AM-GM we get $ 2\sum ab \geq2\sum a\sqrt{bc}\geq \sum (a\plus{}b)\sqrt{ab}$ which is true by rearrangement. Am i correct :?:  :|

[quote="Γιώργος"]for a) we have 2 more different solutions. take a look here

http://www.mathlinks.ro/viewtopic.php?t=237479

for b) we have from Cauchy:

$ \sum\frac {a \plus{} b}{\sqrt {ab}} \plus{} \frac {\left(a \plus{} b \plus{} c\right)^{2}}{a^{2} \plus{} b^{2} \plus{} c^{2}} \equal{} \sum\frac {\left(a \plus{} b\right)^{2}}{(a \plus{} b)\sqrt {ab}} \plus{} \frac {\left(a \plus{} b \plus{} c\right)^{2}}{a^{2} \plus{} b^{2} \plus{} c^{2}}\geq \frac {9\left(a \plus{} b \plus{} c\right)^{2}}{\sum(a \plus{} b)\sqrt {ab} \plus{} \sum a^2}$

So, it remains to prove that $ \frac {9\left(a \plus{} b \plus{} c\right)^{2}}{\sum(a \plus{} b)\sqrt {ab} \plus{} \sum a^2}\geq 9\Longleftrightarrow ...\Longleftrightarrow 2\sum ab\geq \sum (a \plus{} b)\sqrt {ab}$.

But, from AM-GM we get $ 2\sum ab \geq2\sum a\sqrt {bc}\geq \sum (a \plus{} b)\sqrt {ab}$ which is true by rearrangement. Am i correct :?:  :|[/quote]
$ \sum 2ab \ge (a\plus{}b)\sqrt{ab}$ is wrong. In fact $ 2ab \minus{} (a\plus{}b)\sqrt{ab} \equal{} \minus{}\sqrt{ab}(\sqrt{a}\minus{}\sqrt{b})^2 \le 0$

[quote="Mathias_DK"][quote="Γιώργος"]for a) we have 2 more different solutions. take a look here

http://www.mathlinks.ro/viewtopic.php?t=237479

for b) we have from Cauchy:

$ \sum\frac {a \plus{} b}{\sqrt {ab}} \plus{} \frac {\left(a \plus{} b \plus{} c\right)^{2}}{a^{2} \plus{} b^{2} \plus{} c^{2}} \equal{} \sum\frac {\left(a \plus{} b\right)^{2}}{(a \plus{} b)\sqrt {ab}} \plus{} \frac {\left(a \plus{} b \plus{} c\right)^{2}}{a^{2} \plus{} b^{2} \plus{} c^{2}}\geq \frac {9\left(a \plus{} b \plus{} c\right)^{2}}{\sum(a \plus{} b)\sqrt {ab} \plus{} \sum a^2}$

So, it remains to prove that $ \frac {9\left(a \plus{} b \plus{} c\right)^{2}}{\sum(a \plus{} b)\sqrt {ab} \plus{} \sum a^2}\geq 9\Longleftrightarrow ...\Longleftrightarrow 2\sum ab\geq \sum (a \plus{} b)\sqrt {ab}$.

But, from AM-GM we get $ 2\sum ab \geq2\sum a\sqrt {bc}\geq \sum (a \plus{} b)\sqrt {ab}$ which is true by rearrangement. Am i correct :?:  :|[/quote]
$ \sum 2ab \ge (a \plus{} b)\sqrt {ab}$ is wrong. In fact $ 2ab \minus{} (a \plus{} b)\sqrt {ab} \equal{} \minus{} \sqrt {ab}(\sqrt {a} \minus{} \sqrt {b})^2 \le 0$[/quote]

sorry then for the incorect solution :(

The second seems true and hard. I tried to solve it by Muirhead, but i failed.

The second inequality is wrong. Take for example $ a \equal{} 5$, $ b \equal{} 6$, $ c \equal{} 7$. Then you get $ LHS\minus{}RHS\equal{}\minus{}0.01191565 \dots$.


We have a square. Inside of it we take a point and form that point pararels at swill and at diagonals. So we must demonstrate that the black and the white area is equal. Please excuse my english:)

Complete the triangles like in figure.

We add the same quantity to the dark part and to the light part: it's easy to show that the external-up-right and external-down-left triangles are congruent; the same for the other tro triangle I added.

Now, it'easy to see that we can form four pairs of equal dark and light triangles.  :)

[quote="edriv"]
We add the same quantity to the dark part and to the light part: it's easy to show that the external-up-right and external-down-left triangles are congruent [/quote]

Can you prove it? I couldn't... :help:


$ I$ is the incenter of triangle $ ABC$,$ A\equal{}60$ . the line that is parallel to $ AC$ and pass through $ I$ intersect $ AB$ at $ F$. $ P$ is on $ BC$ such that $ BC\equal{}3BP$. prove that $ \hat{BFP}$ $ \equal{}$ $ \frac{1}{2}$$ \hat{ABC}$

Pls see:
http://www.mathlinks.ro/viewtopic.php?t=300986
http://www.mathlinks.ro/viewtopic.php?t=316752

Best regards,
sunken rock


[hide="Hint"]Let y = x sin x and apply AM-GM.[/hide]

Can't get it even with a really obvious hint?? :P

[hide="Solution"]
Let $y=x\sin{x}$, as 1234567890 suggested.

We can rewrite the expression as $\frac{9y^2+4}{y}=9y+\frac{4}{y}$.

Since $x>0$ and $\sin{x}>0$ ($\because 0< x<\pi$), we have $y>0$.
So we can apply AM-GM:
$9y+\frac{4}{y}\ge 2\sqrt{9y\cdot\frac{4}{y}}=12$

The equality holds when $9y=\frac{4}{y}\Longleftrightarrow y^2=\frac{4}{9} \Longleftrightarrow y=\frac{2}{3}$

Therefore, the minimum value is $\boxed{12}$ (when $x\sin{x}=\frac{2}{3}$).

[/hide]

EDIT- Sorry for stupid mistakes... :oops:  Edited now.

frt, that is almost correct except for 2 things
you forgot to divide by 2 for the am and
you have to show that the minium exists (what if there was a stronger inequality than that?)

[hide="Solution"]
Let $y=x\sin{x}$.

We can rewrite the expression as $\frac{9y^2+4}{y}=9y+\frac{4}{y}$.

Since $x>0$ and $\sin{x}>0$ because $0< x<\pi$, we have $y>0$.
So we can apply AM-GM:
$9y+\frac{4}{y}\ge 2\sqrt{9y\cdot\frac{4}{y}}=12$

The equality holds when $9y=\frac{4}{y}\Longleftrightarrow y^2=\frac49\Longleftrightarrow y=\frac23$

Therefore, the minimum value is $12$ (when $x\sin{x}=\frac23$). [/hide]

Well , I must admit AM-GM is faster but anyway I use more elementary way  :P
[hide]
Let $x\sin x=t$ and set $\frac{9t^2+4}{t}=y$  . Simplifying we get

$9t^2-yt+4=0$ . Since $y$ is real , the discriminant should be $\geq 0$ , hence

$D=y^2-144\geq 0$ $\implies y\geq 12$  ( since $t>0$ and $y>0$) . Hence the minimum is when $y=12$ which is when $t=6$ .
 :)
[/hide]

[quote="Zellex"]Sorry for reviving a dead thread...but I was just reading this solution and something is confusing me.

[quote="4everwise"][hide="Solution"]
The equality holds when [b]$9y=\frac{4}{y}\Longleftrightarrow y^2=36\Longleftrightarrow y=6$[/b]

Therefore, the minimum value is $12$ (when $x\sin{x}=6$). [/hide][/quote]

How does $9y=\frac4y \rightarrow y^2=36$? I thought $9y=\frac4y\rightarrow y^2=\frac49 \rightarrow y=\frac23$. Also, if $x\sin{x}=6$, wouldn't the answer be something like $\frac{164}{3}$? Or is there something that I'm completely missing?

*EDIT* When $y=\frac23$, we get $\frac{(9)(\frac49)+4}{\frac23}\rightarrow\frac{4+4}{\frac23}\rightarrow\frac{24}{2}\rightarrow\boxed{12}$[/quote]Yeah, that was an old mistake somebody caught in my soluition document.

I see..I just find it odd that everyone got $x\sin x=6$.

It seems that the solution in 4everwise's solution document was mine (post #4)... Sorry for the careless mistake :oops:  :ewpu:

(I edited my solution, just to make sure...)

[quote="pkerichang"]just curious...Why do they have such a trivial problem with Calculus, I know it's not required, but this gives some group of people advantages on the test...[/quote]Well, the 1983 test was just messed up. :P

Actually, I think it's because fewer high school students took calculus courses back then.

[quote="pkerichang"]just curious...Why do they have such a trivial problem with Calculus, I know it's not required, but this gives some group of people advantages on the test...[/quote]

No problem in AMC requires the use of calculus. So, you can use calculus but that's not the "intended" method that problem creaters had.

this problem is obvious and straight forward with calculus...i think it is a bad problem for that reason

I know this has been an old thread for a while now, but I was just reading over the solution.
I'm sorry to ask, but what is AM-GM?

AM-GM says the average (arithmetic mean) of a bunch of numbers is greater than or equal to the geometric mean. They are only equal when all the numbers are the same. In this case, with only two numbers, $ \frac {a \plus{} b}{2} \ge \sqrt {ab}$, or $ a \plus{} b \ge 2\sqrt {ab}$.

Oh, ic. Thanks, now the problem makes sense :]

[quote="chess64"]Find the minimum value of
\[\frac{9x^2 \sin^2 x + 4}{x \sin x}\]
for $0 < x < \pi$.[/quote]

[hide="Solution"]
For $0 < x < \pi$, notice that $x,\sin x > 0$. Therefore,
\[\frac{9x^2 \sin^2 x + 4}{x \sin x} = 9x \sin x + \dfrac{4}{x \sin x}  \ge 2\sqrt{36} = 12 \]
from AM-GM. Equality is achieved when $9x \sin x = \dfrac{4}{x \sin x}$, or $x \sin x = \dfrac{2}{3}$, which is clearly possible because $x \sin x = 0$ when $x=0$, and $x\sin x = \dfrac{\pi}{2}$ when $x = \dfrac{\pi}{2}$, so $x \sin x$, due to its continuity, must hit every value is achieved.
[/hide]

Also, people are saying that this is a bad problem because it is trivial with calculus, and gives people who know calculus an unfair advantage. I disagree, because this problem is also trivial with AM-GM :P

Last ditch effort and just throw all the calc you now at it or AM-GM :).

How did you even find this? This is reallly old...

[hide="Solution Without AM-GM"]The expression $9x^2\sin^2{x}+4$ is really close to the square of a binomial. So we write it as $(3x\sin x-2)^2+12x\sin x$, and split the fraction into $$\frac{(3x\sin x-2)^2}{x\sin x}+12.$$ The minimum value of $(3x\sin x-2)^2$ is $0$, which occurs when $x\sin x=\frac{2}{3}$, and this is achievable (this is why we use the binomial $3x\sin x-2$ instead of $3x\sin x+2$). Hence, the minimum value of the entire expression is $\boxed{012}$.[/hide]

it is too easy it follows by am gm inequality as sine function is positive in the above question

$\frac{9x^2\sin^2x+4}{x\sin x}=9x\sin x+\frac4{x\sin x}\ge12$
by AM-GM. Equality is at $x\sin x=\frac23$. Let $f(x)=x\sin x-\frac23$. Since $f$ is continuous, we use IVT and get a root in the interval $(0.5,1)$, thus equality happens.

Storage. An easy problem.

[hide=Solution]Let $y = x \sin x$. Then we have $\frac{9y^2+4}{y}$, which is $9y+\frac{4}{y}$. We want to minimize this. Applying AM-GM gives $9y+\frac{4}{y} \geq 2\sqrt{9y \cdot \frac{4}{y}}=12$. Equality holds when $x \sin x = \frac{2}{3}$, which is obviously possible. Hence, the answer is $\boxed{12}$.[/hide]

Make the substitution $y=x\sin x$. AM-GM gives $$9y+\frac{4}{y} \geq \boxed{12},$$ which can clearly be achieved by checking the equality case. $\blacksquare$

[quote=jayevvms123][hide="Calculus Solution"]

If we substitute $y$ for $xsinx$, then the function becomes $9y + \frac{4}{y}$. If we take $\frac{d}{dx}[9y + \frac{4}{y}]$, we get $9 - \frac{4}{y^2}$.
From here we can set this equal to zero as such: $$9 - \frac{4}{y^2} = 0$$ Adding $ \frac{4}{y^2}$ to both sides yields us $$9 = \frac{4}{y^2}$$ We can then multiply both sides by $y^2$ to get $$9y^2 = 4$$ Next dividing both sides by 9 gets us $$y^2 = \frac{4}{9}$$ Square rooting both sides gives us $$y = \frac{2}{3}$$ Finally, we plug this $\frac{2}{3}$ back into our original expression, which was $9y + \frac{4}{y}$.
This would look something like this: $$9(\frac{2}{3}) + \frac{4}{\frac{2}{3}} = 12$$ This gives us our final answer of $\boxed{12}$.

P.S - This was my first proof ever so please give me suggestions on how to make my solutions/proofs better :)

It equals to $9x\sin x+\frac{4}{x\sin x}$ , and we can use the AM-GM to that

[quote=smo]Sorry to revive an old thread, but I had a conceptual question. On these kinds of problems, what guarantee is there that some kind of inequality will yield the [b][i]absolute[/i][/b] minimum. What I mean is that in some cases we may have some value $HM < x < GM$ as the minimum, can't we? So the only sure-fire way is to use calculus. Of course I'm probably mistaken, so if someone would help me with my confusion, I'd be really grateful.[/quote]

Am-gm eguality case gives the best

AM-GM: $\[\frac{9y + \frac{4}{y}}{2} \ge \sqrt{9y \cdot \frac{4}{y}} = 6,\]$
$9y + \frac{4}{y} \ge \boxed{12}$
The minimum value is achievable; AM-GM holds when all terms are equal. $y = x \sin x = \tfrac{2}{3}.$ must have a solution for the interval $0 < x < \pi$

Let $y=x\sin(x)$, hence, by $AM-GM$, we have $\frac{9y + \frac{4}{y}}{2} \ge \sqrt{9y \cdot \frac{4}{y}} = 6$, so we multiply by $2$ to get $\boxed{12}.$

[hide = AM-GM Solution]
Let $y=x\sin{x}$. We can rewrite the expression as $\frac{9y^2+4}{y}=9y+\frac{4}{y}$.

Since $x>0$, and $\sin{x}>0$ because $0< x<\pi$, we have $y>0$. So we can apply AM-GM:

\[9y+\frac{4}{y}\ge 2\sqrt{9y\cdot\frac{4}{y}}=12\]
The equality holds when $9y=\frac{4}{y}\Longleftrightarrow y^2=\frac49\Longleftrightarrow y=\frac23$.

Therefore, the minimum value is $\boxed{012}$.
[/hide]

Let $k=x\sin x$. The given expression becomes $\frac{9k^2+4}{k} \ge \frac{12k}{k} = \boxed{12}$.

[hide=9]We start by simplifying the expression into $9x\sin x +\frac{4}{x\sin x}$. Since $x\sin x$ is always positive when $0 \leq x \leq \pi$, we can use AM-GM on this expression to get that $\frac{9x\sin x +\frac{4}{x\sin x}}{2} \geq \sqrt{9x\sin x \cdot \frac{4}{x\sin x}}=6$. Therefore, the minimum value of our sum is $12$ and it is achieved when $x\sin x =\frac{2}{3}$, which does occur within our bounds for $x$.

Key idea: Split the fraction up to get something we can use AM-GM on. Something that tips off that $x\sin x$ is not important is that its hard to find the maximum value of, which tells us that we probably don't have to maximize it, we just have to make sure that it can be a certain value that minimizes the expresssion.[/hide]


Let $k,m,n$ be natural numbers such that $m+k+1$ is a prime greater than $n+1$. Let $c_s=s(s+1)$. Prove tha the product \[ (c_{m+1}-c_k)(c_{m+2}-c_k)\cdots(c_{m+n}-c_k)  \] is divisible by $c_1c_2 \cdots c_n$.


abstract algebra
group theory
superior algebra
superior algebra unsolved

We say that an abelian group G has the property (P) if for any abelian group H, any subgroup $H'\le H$ and any morphism $f':H'\to G$, there exists a morphism $f:H\to G$ so that $f_{|H'}=f'$.
1) Prove that the group $(\mathbb Q^*,\cdot)$ does not have the property (P).
2) Prove that the group $(\mathbb Q,+)$ has the property (P).

1) is rather easy: we consider a cyclic group $H:=\{e,a,a^2,a^3\}$ and $H^{'}=\{e,a^2\}$ and the morphism $f': H^{'}\rightarrow \Bbb Q^{*}, f'(a^2)=-1$ we have no morphism $f$ with the required property since for each $y:=f(a)$, $-1=f(a^2)=f(a)^2=y^2$ must hold which is impossible in $\Bbb Q$.

I will think about 2)

Misha

Just an observatiopn: it can be shown that property $P$ is equivalent to the divisibility of the group. Since $(\mathbb Q,+)$ is clearly divisible, that's that. :)

what does divisibility of the group mean? :?

A group is divisible if for any element $a$ and any natueal $n$ there is an element $b$ s.t. $nb=a$.

P.S. In my previous post, I was talking about abelian groups.


Source: MMPC 1995, Round 1 (#33)

The three stooges (Larry, Moe, and Curly Joe) make fruity purchases.  Larry buys one banana, two oranges, and two apples for $2.00.  Moe buys three bananas, one orange, and one apple for $1.50.  Which of the following purchases by Curly Joe would enable you to compute the cost of an apple?

(A) One banana for $.20
(B) One orange and one apple for $0.90
(C) Four bananas, three oranges, three apples for $3.50
(D) One banana, one orange, two apples for $1.60
(E) Two bananas, three oranges, three apples for $3.10

This is a neat problem, and tests understanding rather than computation.  Fun!  [hide]Because the coefficients for orange and apple in the two equations are linearly related (those in the first equation are k (in this case two) times those in the second), it is as though the two equations are in terms of two variables:  bananas and "orange and apple"s.  This means that the price of a banana or the price of an "orange and apple"  can be computed.  So, A and B are definitely not answers... this is information we could have solved for, had we wanted to.  Option C is four of one thing we could find the price of and three of another thing we could find the price of, so this isn't new information, either.  Option E is similar, so option D is our answer.[/hide]


i was in class today, and we were given a problem where we had to find the number of factors of a large number. My friend pulled out this formula, but she had no idea why it works. It's been bothering me and it would be great if any of you could explain to me.

The formula says this:

To find the number of factors of a positive integer A, first find the prime factorization of the number. Take the exponents of each prime, and add one to each. Multiply them to get the number of factors.

So basically, if the positive integer A has the prime factorization $a^{x}*b^{y}*c^{z}$, then it has $\left({x+1}\right)\left({y+1}\right)\left({z+1}\right)$ number of factors.

We can demonstrate that this is true with smaller numbers. For example, 36 has 9 factors $\left\{{1,2,3,4,6,9,12,18,36}\right\}$. Its prime factorization is $2^{2}*3^{2}$. Using the formula, we get $\left({2+1}\right)\left({2+1}\right) = 9$.

Can anyone explain to me why this works? Thanks.

[hide]Think about it like this:

If $A$ has prime factorization $A=a^{x}\cdot b^{y}\cdot c^{z}$, then it's factors will be of the form:

$A=a^{x_{1}}\cdot b^{y_{1}}\cdot c^{z_{1}}$

where

$0\leq x_{1}\leq x$
$0\leq y_{1}\leq y$
$0\leq z_{1}\leq z$

It is easy to check that the first contributes $x+1$ factors, the second contributes $y+1$ factors and the third contributes $z+1$ factors.

So the answer is $(x+1)(y+1)(z+1)$[/hide]

That's true if a number has 3 prime factors... what about if it has more or less factors?

Let the prime factorization of A = $p_{1}^{a_{1}}\cdot p_{2}^{a_{2}}\cdot p_{3}^{a_{3}}\cdots$

The simplest way I can think of doing this is you can use $p_{1}$ $a_{1}+1$ times (use it 0 times, 1, 2, 3....).  Same for all of the other ones.

Hope this helps!

if you have a large number
like say $240$
It's prime factorization is
$2^{4}* 3 * 5$

all factors of $240$ can be factored into some combination of those prime factors.

For example:
$60$ is a factor of $240$ and
$60 = 2^{2}* 3 * 5$

each factor of $240$ will have certain prime "contributors" from $240$. $60$ has $2$,$2$,$3$, and $5$

now, to find out how many factors $240$ has, we look at how many ways you can select "contributors"
the $2$ can contribute anywhere from 0 to 4 times (5 ways)
the $3$ can contribute either 0 or 1 time (2 ways)
the $5$ can contribute either 0 or 1 time (2 ways)

the number of ways you can select contributors is $5 * 2 * 2 = 20$ factors

when generalized, each prime factor: $p^{n}$ can contribute in n+1 different ways, which is why you add one to each exponent and take the product

hope this helps

[hide]If you have $a^{x}*b^{y}*c^{z}$

the factors will be just changing the exponents to anything lower than or equal to $x$, $y$, or $z$

So the number of ways you can change x is from all the positive integers from 1 up to x, but then you have to include 0. So its $x+1$

Same for the others, then by the fundamental theorum of counting (or w/e you don't need to know the name) you just multiply them together.[/hide]

[hide]

we consider  that   the prime factorization of the number is [img]http://www.mathlinks.ro/Forum/latexrender/pictures/7/7/2/7724425f3d46d63e69417f37140955189d62bfe7.gif[/img]
and we consider E={a,a,a....a,b,b,b,b....,b,c,c....c}
a apear x time
b apear y time
c apear z time

then the number of factors is the cardinal of P(E)
2^(x+y+z)
[/hide]

[quote="goldbach"]2^(x+y+z)[/quote]

No. We're not counting subsets.

plz can you mention the error in my proof

[quote="minsoens"][quote="goldbach"]we consider  that   the prime factorization of the number is [img]http://www.mathlinks.ro/Forum/latexrender/pictures/7/7/2/7724425f3d46d63e69417f37140955189d62bfe7.gif[/img]
and we consider E={a,a,a....a,b,b,b,b....,b,c,c....c}
a apear x time
b apear y time
c apear z time

then the number of factors is the cardinal of P(E)
2^(x+y+z)[/quote]i think that would work when $a\ne a\ne a\ne \cdots \ne a$, $b\ne b\ne b\ne \cdots \ne b$, $c\ne c\ne c\ne \cdots \ne c$..  :wink:[/quote]

A set, by definition, can have no repetitions.

(That rhymes.)

Try looking at the other proofs in this thread. :)

[quote="mathnerd314"]

A set, by definition, can have no repetitions.

(That rhymes.)

Try looking at the other proofs in this thread. :)[/quote]

thanx for the information


function
geometry
3D geometry
sphere
complex analysis
real analysis
real analysis unsolved

Let $ \ell_{1},\ell_{2}$ be two rays in the plane, both starting from the origin. Call the pair $ (\ell_{1},\ell_{2})$ "good" if there is a non-zero harmonic function $ u$ on the plane which vanishes on $ \ell_{1}\cup\ell_{2}$.

Find the good pairs of rays in the plane.

This is more of complex analysis. Let $ f$ be the entire function such that $ \mathrm{Re}\,f\equal{}u$ and $ f(0)\equal{}0$. Expand $ f(z)\equal{}c_{n}z^{n}\plus{}\dots$ with $ c_{n}\ne 0$. We have $ \mathrm{Re}(c_{n}z^{n})\equal{}0$ for $ z\in\ell_{1}\cup\ell_{2}$. Therefore, the angle between $ \ell_{1}$ and $ \ell_{2}$ is a multiple of $ \pi/n$.
This necessary condition is also sufficient, because we can take $ u(z)\equal{}\mathrm{Re}(c_{n}z^{n})$.

grobber is this the problem of the miklos schweitzer competition that you were referring to?

Let $ e_1, ..., e_n$ be semilines on the plane starting from a common point.
Prove, that if there is no $ u\not \equiv 0$ harmonic function on the whole plane
that vanishes on the set $ e_1 \cup ... \cup e_n$, then there exists a pair i, j of
indices such that no $ u \not \equal{} 0$ harmonic function on the whole plane exists
that vanishes on $ e_i \cup e_j$ .

A natural question: given $ n\ge 3$, what is the maximal number $ M_n$ such that any $ M_n$ half-lines starting from the origin lie in the zero set of a nonzero harmonic function on $ \mathbb R^n$? We know that $ M_2 \equal{} 1$.

Since it's past 2 am here, I'll just take a wild guess: $ M_n \equal{} 2n \minus{} 3$ for all $ n$. :maybe:

Equivalently, $ M_n$ is the maximal integer such that for any $ M_n$ points on the sphere $ S^{n\minus{}1}$ there is a nonzero [url=http://en.wikipedia.org/wiki/Spherical_harmonics]spherical harmonic[/url] that vanishes at all these points. I guess the answer is known already to people who work with spherical harmonics.

i thought it is better to continue the discussion here.

[quote]We have $ \mathrm{Re}(c_{n}z^{n})\equal{}0 for z\in\ell_{1}\cup\ell_{2}$[/quote]

i am not sure i understand this properly. it should be Re ($ c_n z^n\plus{}...$)=0 ... or is it because near the origin, all the other terms are very small in comparison with the first one.

[quote]near the origin, all the other terms are very small in comparison with the first one.[/quote]
Exactly.

Another explanation: the map $ f(z)\equal{}c_nz^n\plus{}\dots$ can be written as $ g(z)^n$, where $ g'(0)\ne 0$. Just factor out $ z^n$ from $ f(z)$ and take the $ n$th root of the rest, which does not vanish near $ 0$. In a neighborhood of $ 0$, the map $ g$ is a holomorphic bijection between the zero set of $ u$ and the set $ \mathrm{Re}(\zeta^n)\equal{}0$ which is easy to understand.


Two runners began running in the same direction on a quarter-mile track. The first runner began five feet ahead of the second runner. The first runner's pace is 6 minutes per mile, while the second runner's pace is 8 minutes per mile. Both runners continue at their respective paces for 119 minutes. How many times will the first runner pass the second runner?

The first runner will be able to complete a lap in $ \frac32$ minutes, and the second one will be able to in $ 2$. So The first runner will finish $ 78$ laps, and the second will finish $ 59$. $ 78\minus{}59\equal{}19$


geometry
cyclic quadrilateral
geometry unsolved

Let $ABCD$ be a quadrilateral with $AD=BC$ and $\widehat{A}+\widehat{B}=120^{0}$. Let us draw equilateral $ACP,DCQ,DBR$ away from $AB$ . Prove that the points $P,Q,R$ are collinear.

The condition $AD = BC,$ is not necessary.

We denote as $E,$ the intersection point of the sidelines $AD,$ $BC,$ of $ABCD.$ So, we have that $\angle CED = 60^{o}.$

From cyclic quadrilaterals $ACEP,$ $CDEQ,$ $BDER,$ we have that

$\angle QEC+\angle CED+\angle AEP = 60^{o}+60^{o}+60^{o}= 180^{o}$ $\Longrightarrow$ $Q,$ $E,$ $P,$ are collinear $(1).$

$\angle QEC = \angle QDC = 60^{o}= \angle RDB = \angle REB$ $\Longrightarrow$ $R,$ $Q,$ $E,$ are collinear $(2).$

From $(1),$ $(2)$ we conclude that the points $P,$ $Q,$ $R$ are collinear and the proof is completed.

Kostas Vittas.

The contition $AD = BC,$ is necessary for the Line's problem.

So, if $K,$ $L,$ $M,$ are the midpoints of the segments $AC,$ $CD,$ $BD$ respectively, we have that $\angle KLM = \angle AEB = 60^{o}$ $($ because of $KL\parallel AE$ and $LM\parallel BE$ $)$ and also $KL = \frac{AD}{2}= \frac{BC}{2}= LM.$ Hence the triangle $\bigtriangleup KLM$ is equilateral and the proof is completed.

Kostas Vittas.

The triangle $BAQ$ is echilateral as well. Any nice proofs?

[quote="Line"]that's an equilic quadrilateral!

I've got another one:

let $ABCD$ a equilic quadrilateral with $\measuredangle{\text{ }}A+\measuredangle{\text{ }}B = 120^\circ$, and $\overline{AD}= \overline{BC}$. Let $O$, $P$ and $Q$ be midpoints of $\overline{AC}$, $\overline{BD}$ and $\overline{CD}$ respectively. Prove that $\triangle OPQ$ is equilateral.[/quote]

The condition $AD = BC,$ also is necessary for the andyciup’s problem.

From cyclic Quadrilateral $CDEQ$ $\Longrightarrow$ $\angle ECQ = \angle EDQ$ $,(1)$

From $(1)$ $\Longrightarrow$ $\angle ADQ = \angle BCQ$ and so, the triangles $\bigtriangleup ADQ,$ $\bigtriangleup BCQ$ are congruent, because they have also $AD = BC$ and $DQ = CQ.$

Hence , we have that $AQ = BQ$ $,(2)$ and $\angle EAQ = \angle EBQ$  $,(3)$

From $(3)$ $\Longrightarrow$ the quadrilateral $ABQE$ is cyclic and hence, we have that $\angle AQB = \angle AEB = 60^{o}.$

That is the triangle $\bigtriangleup AQB,$ is equilateral and the proof is completed.

Kostas Vittas.


Posted before http://www.mathlinks.ro/Forum/viewtopic.php?p=183465#183465 .


for which integers $n$ does the equation $\frac{3xy-1}{x+y}=n$ have a solution in integers $x$ and $y$.

If $3 \nmid n$, it clearly has no solution.
If $n \equiv 1 \mod 3$, take $x=-\frac{n^{2}-n+3}{3}, y= \frac{n-1}{3}$.
If $n \equiv-1 \mod 3$, take $x=\frac{n^{2}+n+3}{3}, y= \frac{n+1}{3}$.

It give $x=\frac{yn+1}{3y-n}$. If $3\not |n$ exist y, suth than $3y-n=1$ or $3y-n=-1$, it give integer solution. If $3|n$ no solution.


The triangle ABC has AB = 1995, BC = 1993, CA = 1994. CX is an altitude. Find the distance between the points at which the incircles of ACX and BCX touch CX.

If both inradius can be found... but how...

[hide]This is really inefficient, but given enough time it'll work:

Suppose BX = x, AX = y, and CX = h. Let the inradius of CBX be r1, and the inradius of CAX be r2. We have:

x2 + h2 = 19932

y2 + h2 = 19942

x + y = 1995

From these equations you can figure out what x, y and h are.

Recall that the inradius is given by r = k / s, where k is the area of the triangle and s the semiperimeter (semiperimeter = (a + b + c)/2). (This is easy to prove.) Then,

r1 = (xh / 2) / ((1993 + x + h) / 2) = xh / (1993 + x + h)

r2 = (yh / 2) / ((1994 + y + h) / 2) = yh / (1994 + y + h)

Sub in x, y and h, and you should have the inradii of the triangles; find their difference to get the final answer.[/hide]

yes that was it...
i forgot how to find inradius when all three sides are given...

[quote="xirtamax"]i forgot how to find inradius when all three sides are given...[/quote]

You can use Heron's forumula to find the area:
s=(a+b+c)/2
A=sqrt(s(s-a)(s-b)(s-c))

Then the inradius formula:
A=rs

Or:

Construct the incentre of the triangle, and join it to the three vertices. This divides the triangle into 3 smaller triangles. They have areas ar / 2, br / 2 and cr / 2 respectively, therefore k = (a + b + c)r / 2 = sr. Therefore r = k / s.

There is no need to actually calculate $x$ and $y$. $r_x=\frac{x+h-AC}{2}, r_y=\frac{y+h-BC}{2}$, the required segment being equal to the difference of the 2 radii.
We have $x+y=AB, x^2-y^2=AC^2-BC^2$, hence $x-y=\frac{3987}{1995}$, consequently the required segment is $\frac{332}{665}$.

Best regards,
sunken rock