For a gas, which goes from state $ p_1,V_1,T_1$ to state $ p_2,V_2,T_2$ (no change in the number of moles), how to express entropy increment? Based on formulas for isobaric and isochoric process, I think that the formula might be $ \Delta S\equal{}n\cdot C\cdot \ln{\frac{T_2}{T_1}}$ where $ n$ is the amount of the gas in moles, and $ C$ its molar heat capacity

How can I prove it (of course, for isothermal and adiabatic process it doesn't hold)?

Can I prove it at least for polytropic processes ($ pV^x\equal{}\text{const},x\in\mathbb{R}$)?

[hide="Hint"]
Entropy is a state function so however you go from one state to the other the change in entropy is the same.Consider a combination of isobaric and isochoric process to reach the final destination the change in entropy is just the sum[/hide]

You're missing the change in volume term. If you have a higher volume of gas at the same temperature, then you have higher entropy!

You seem to be working in the realm of classical thermodynamics, and [b]pardesi[/b]'s hint of $ S$ being a state function should be good enough for you to figure out the answer working in this realm. I'm going to give a stat-mech solution to this problem. I'll go slowly for readers who are not familiar.

The entropy of a state, in the stat-mech sense, is defined as $ \boxed{S \equal{} k \ln \Omega}$, where $ \Omega$ is the multiplicity of the state. Multiplicity is the number of possible microscopic-level setups that can give rise to that state (called microstates), and $ k$ is Boltzman's constant. I'll now employ two results:

[b]1.[/b] For an ideal gas, $ \Omega \equal{} F(N) V^N U^{\frac {f}{2}N}$, where $ V$ is volume, $ U$ is internal energy, $ f$ is number of degrees of freedom of each particle, $ N$ is number of particles, and $ F(N)$ is a particular function of $ N$. See 'Notes' at the end for details.

[b]2.[/b] $ U \equal{} \frac {f}{2} NkT$ for an ideal gas, as a result of the Equipartition Theorem.

Using the definition of entropy and the multiplicity formula:
$ \Delta S \equal{} k \ln \Omega_2 \minus{} k \ln \Omega_1 \equal{} k \ln \frac {F(N) V_2^N U_2^{\frac {f}{2}N}}{F(N) V_1^N U_1^{\frac {f}{2}N}} \equal{} Nk \ln \frac {V_2}{V_1} \plus{} \frac {f}{2}Nk \ln \frac {U_2}{U_1}$

[hide="Using these equalities"][b]1.[/b] $ R \equal{} A k$, where $ A$ is Avogadro's number, so $ Nk \equal{} nR$.

[b]2.[/b] $ C' \equal{} \frac {\partial U}{\partial T}$ is the definition of heat capacity $ C'$. From Equipartition Theorem, $ C' \equal{} \frac {f}{2} Nk$. Molar heat capacity is $ C \equal{} C'/n$, so $ nC \equal{} \frac {f}{2}Nk$

[b]3.[/b] Recall from Equipartition Theorem that for ideal gas, $ U \propto T$.[/hide], the above simplifies to $ \boxed{\Delta S \equal{} nR \ln \frac {V_2}{V_1} \plus{} nC \ln \frac {T_2}{T_1}}$

[hide="Notes"]This makes some intuitive sense:
[b]a.[/b] For each particle, the number of positions of possible positions is proportional to the volume. For N particles, the number of sets of positions is $ V^N$
[b]b.[/b] The total energy can be divided between $ fN$ degrees of freedom, somewhat explaining the exponent on $ U$. For a proper explanation, look up the relation. I know the book [i]Thermal Physics[/i] by Schroeder has a good explanation of this (it's a great book in general, too).[/hide]

Thanks both! But I still have some troubles...

The equality 2: Isn't the heat capacity defined as $ C' \equal{} \frac {\delta Q}{dT}$, where $ Q$ is the heat transferred to the gas (which, by the first law of thermodynamics is spent on internal energy increment and the work done by the gas)?

And if $ C \equal{} \frac {fR}{2}$, then as $ f \equal{} \frac {2}{\gamma \minus{} 1}$, we would get $ C \equal{} \frac {R}{\gamma \minus{} 1}$, which is in fact isochoric molar heat capacity (work is 0), and not for all processes. Maybe you forgot work done by the gas?
In fact, for polytropic processes, $ C \equal{} C_V \minus{} \frac {R}{n \minus{} 1}$ where $ n$ is the polytropic exponent.

I think that your formula is true if you replace $ C$ with $ C_V$, where $ C_V \equal{} \frac {fR}{2}$. Also, I managed to derive that modified formula with pardesi's hint.

And I tried my formula on polytropic processes, and some problems I was working on were correct with it. I'll post one:
[hide="Problem"]
Find the entropy increment of $ \nu \equal{} 2$ moles of an ideal gas whose adiabatic exponent is $ \gamma \equal{} 1.3$ if, as a result of a certain process, the gas volume increased $ a \equal{} 2$ times while the pressure dropped $ b \equal{} 3$ times.

[hide="Their solution"]
Based on pardesi's idea. They go through isochoric then isobaric process and get the entropy as $ \Delta S \equal{} \frac {\nu R}{\gamma \minus{} 1}(\gamma\ln a \minus{} \ln b) \equal{} \minus{} 10,94J/K$
[/hide]
[hide="My solution"]
As the entropy depends only on final and beginning state, the process can be written as $ pV^\frac {\ln b}{\ln a} \equal{} \text{const}$

$ C \equal{} C_V \minus{} \frac {R\ln a}{\ln b \minus{} \ln a}$

$ \Delta S \equal{} \nu R(\frac {1}{\gamma \minus{} 1} \minus{} \frac {\ln a}{\ln b \minus{} \ln a})\ln(\frac {a}{b}) \equal{} \minus{} 10,94J/K$
[/hide]
[/hide]

EDIT: Oh, I got it now. My formula can't be valid for all processes. Why? Because entropy is a state function, and if my formula is valid, then it's the same for all processes from point 1 to point 2, but the work on those processes isn't the same, so molar heat capacity also isn't, and then entropy increment wouldn't be the same every time.

But is my formula correct at least for polytropic processes?

Oh, oops, you're right, that's $ C_V$, my mistake, so my final result is flawed.  If you know the relation between pressure, volume and work done by the gas, then you can modify the formula appropriately without too much difficulty.

I hope at least that the concept was interesting :)

Yeah, it was pretty interesting (the statistical point of the view) :) Yeah, it's ok, because most of the processes that appear in problems at HS level are polytropic, so cool. Thanks!

Sidenote: there is one more question ( http://www.mathlinks.ro/viewtopic.php?t=319559 ), if you want to try it you can


USAMTS
inequalities
geometry
percent
number theory

Here is the prize system for this year's USAMTS.

Because of the unfortunate Round 2 debacle, there are potentially two ways that you can qualify for a prize.

You can earn a prize based on your total score from all 4 rounds as follows:

96-100: Gold
75-95: Silver
60-74: Bronze
40-59: Honorable Mention

If you originally submitted Round 2, you can also earn a prize based on your total score from Rounds 1, 3, and 4, as follows:

72-75: Gold
57-71: Silver
45-56: Bronze
30-44: Honorable Mention

You will earn the higher of the two prizes that you qualify for.

If you log onto http://www.usamts.org and go to your scores page, you should see your prize level and the date on which you can select your specific prize(s).  These dates are

Gold: Wednesday, April 22
Silver: Monday, April 27
Bronze: Thursday, April 30
Honorable Mention: Monday, May 4

The above dates are the [b]first[/b] date on which you can select prizes, but you don't have to do so on that exact date.  The deadline for selecting prizes is June 1.  Prizes will be mailed later in June.

Gold winners will get to select 3 prizes, silver and bronze winners will get to select 2 prizes, and honorable mentions will get to select 1 prize.  All winners will also receive a USAMTS T-shirt (you will be asked to specify your T-shirt size when you select your prize(s)).  All prizes are first come, first served, subject to availability.

We will be offering the following prizes:

- [i]Mathematica for Students[/i] software, generously donated by Wolfram Research.

- An assortment of books, including the following new prizes this year:
[i]The Millennium Problems[/i] by Keith Devlin
[i]A Mathematical Mosaic[/i] by Ravi Vakil
[i]Introduction to Number Theory and Inequalities[/i] by CJ Bradley
[i]A Mathematical Olympiad Primer[/i] by Geoff Smith
[i]Plane Euclidean Geometry[/i] by Gardiner and Bradley
...plus 8 other mathematics books to choose from.

Prizes will be mailed to the address in your profile on the USAMTS website, so log on and click on the "My Profile" tab to verify that your address is correct.  We will only mail prizes to US mailing addresses.  Also, you must have submitted a current year entry form in order to receive a prize; the "My Profile" page will indicate if you still need to do this.

[quote="mathemonster"]Does anyone have comments on "mathematical olypiad primer" and the book by CJ Bradley? I have Vakil's book already. silver yay!!![/quote]

Ehhhh... I don't think the book "Mathematical Olympiad Primer" by CJ Bradley is a prize...?

Wow a 44...argh so close to bronze. I shouldn't have messed up the signs on #5.

[quote="mathemonster"]Does anyone have comments on "mathematical olypiad primer" and the book by CJ Bradley? I have Vakil's book already. silver yay!!![/quote]
you mean geoff smith
i wonder if i can get a mathematica with bronze :rotfl:
i ptobably wont so ill end up getting two books

[quote="DPatrick"][quote="hurdler"]This doesn't really affect me, but should gold winners be extended to top 5 percent? I think that makes gold winners like 94-100 or something...[/quote]528 people submitted round 1.  There are 35 gold winners.  This is more than 5%.[/quote]

On the USAMTS website, it says that there were 393 scores for Round 1. Does that mean $ 528 \minus{} 393 \equal{} 135$ people scored $ 0$s?

Oh woops sorry I was thinking 10 percent apparently, when I computed :(

And I still somehow got a 78....

[quote="tenniskidperson3"]Is mathematica going to be v. 6 again, or the new v. 7?  Because I already have v. 6.

85 silver: Solidly silver, neither close to a bronze nor a gold. :|[/quote]
congrats haha

[quote="azndemon43"]Do you think silver medalists have a chance at mathematica?[/quote]

I was a silver medalist last year, and I had a chance. Also, there's only around 26 gold medalists, and a lot of them have been gold medalists before.

Is there any age restriction to the usamts

[quote="worthawholebean"]I got Mathematica last year as bronze. I had to wake up at 12 AM PDT on the day it opened, but it worked. There were no copies left at 6 AM when a friend of mine checked.[/quote]
youve improved alot in the last year haven't you?

So im guessing that the leaderboard wont be updated for people that did not turn in round 2?  It doesnt really matter for me since I didnt even get gold with updated scoring, but Im willing to bet there are some 70s that got gold but it looks like they only got bronze (like one could put an astreik)


Are there easier ways to this questions?
(I solved them by guess and check..)

18. By hitting a target, Juanita can score 16, 17, 23, 26, 29 or 40 points with each arrow. What is the fewest arrows she can shoot to score exactly 100 points?

26. In the addition problem shown, $ a$, $ b$, $ c$ represent three different digits. What is the three-digit sum?

[hide="#18"]We wish to use $ 40$ as much as possible in order to minimize the number of throws.  Unfortunately, we can not use it twice, as no other two numbers can add to $ 20$.  The smallest sum is $ 16 \plus{} 17 \equal{} 33$.  If we use $ 40$ once, we wish to find three numbers that add up to $ 60$.  However, notice that when any two of the others are added, the units digit is not $ 10$ minus any other one except for the case of $ 16,17,17$ and $ 29,26,29$.  The first three add up to $ 50$.  Two of these needs $ 6$ throws.  However, the case of $ 29,26,29$ only needs another $ 16$ to complete the sequence.
Thus, the fewest is $ 4$ throws.
[/hide]
[hide="#20"]It is easy to see that $ a < 5$.  Because $ c$ is not $ 0$, we can deduce that $ c \plus{} b > 10$ and $ c \equal{} 9$ in order for the tens digit to be $ b$.
Thus $ a \equal{} 4$, and $ b$=$ 5$.
The sum is then $ 459 \plus{} 495 \equal{} 954$[/hide]

[quote="CatalystOfNostalgia"]I haven't seen the AMC 8 but I sure hope that these are not from the 2007 AMC 8.[/quote]
don't worry, its not on the AMC 8 for this year.

[quote="CatalystOfNostalgia"]I haven't seen the AMC 8 but I sure hope that these are not from the 2007 AMC 8.[/quote]
I wouldn't post something from this year... That's against the AMC law..isn't it?


inequalities parameterization function calculus

can someone explain Jessen's in a more understandable way? (what's the sum parameters?)

if we have a function that is like a cup (the curve), then if we take some points on the graph, and find the corresponding function values, the average of those is greater or equal to the average of the x values of the points put through the function

[quote="Altheman"]if $f''(x)>0$, for $x\in [a,b]$, then given a set $\{x_i\}\subset [a,b]$, jensen gives us:

$\frac{\sum f(x_i)}{n} \ge f\left(\frac{\sum x_i}{n}\right)$

(you are forgetting the $n$)[/quote]

That's a specific case for it:

If $f$ is a convex function on the interval $I$, then for $x_i \in I$ and positive real weights $w_i$ satisfying $w_1+w_2+\cdots+w_n = 1$, we have

$w_1f(x_1)+w_2f(x_2)+\cdots+w_nf(x_n) \ge f(w_1x_1+w_2x_2+\cdots+w_nx_n)$.

ok, thanks. what part of calculus is this? I'm in AB right now..

Technically, the proof of Jensen's does not require calculus, and requires only a definition of convexity in terms of the position of a point on the graph relative to a secant line of two points to the left and right of it.

More specifically:  http://mathworld.wolfram.com/ConvexFunction.html

In the interests of being even more general, it is not actually necessary that the weights add up to $1$:

$\frac{ \sum w_k f(x_k) }{\sum w_k} \ge f\left( \frac{ \sum w_k x_k}{ \sum w_k} \right)$

[quote="me@home"]if that isn't calculus, then what is it?
anyway, by weights what do you mean[/quote]

This is algebra.  Really hard algebra.  :D

The weights are the values of $w_1, w_2, ... w_n$.  Basically, instead of just taking the averages of $f(x_k)$ on the LHS and taking the averages of the $x_k$ on the RHS, you can take the "weighted average."

Average:  $\frac{ \sum x_k}{n}$
Weighted Average:  $\frac{ \sum w_k x_k}{\sum w_k}$

Thus, the Jensen's that I gave is really "Weighted Jensen's," and there's also "Weighted AM-GM," etc.

by really hard algebra, I can tell you mean "hard enough that the average calculus student would have only a vague idea of what you're talking about" :?

The algebra is surprisingly easy.
On a convex function(see picture), we have two points at $a$ and $b$ (where $a < b$) so for the convex function $f(x)$ we have $(a, f(a))$ and $(b, f(b))$. By the shape of the graph, we can intuitively tell that for any arbitrary value $v$ between $a$ and $b$, we see that $f(v)$ is less than where $x = v$ crosses the line between the other points. Say, this value $v$ is $\lambda$ of the way from $a$ to $b$. So, $v$ is just the value of $a$ plus $\lambda$ times the distance between $a$ and $b$, or $v = a + \lambda \cdot (b - a) = (1 - \lambda)a + (\lambda)b$. The line between the two points is $\frac{y - f(a)}{x - a} = \frac{f(b) - f(a)}{b - a}$. When $x = v$ we have:
$\frac{f(b) - f(a)}{b - a} = \frac{y - f(a)}{a + \lambda \cdot (b - a) - a} = \frac{y - f(a)}{\lambda \cdot (b - a)}$
$\Rightarrow \lambda(f(b) - f(a)) = y - f(a) \Rightarrow y = (1 - \lambda)f(a) + (\lambda)f(b)$
$\text{High Point} \boxed{\left((1 - \lambda)a + (\lambda)b, (1 - \lambda)f(a) + (\lambda)f(b)\right)}$
Now, the low point is just at $f(v)$ so:
$\text{Low Point} \boxed{\left((1 - \lambda)a + (\lambda)b, f((1 - \lambda)a + (\lambda)b)\right)}$
So, $\text{High Point} \geq \text{Low Point}$
$\Rightarrow \boxed{(1 - \lambda)f(a) + (\lambda)f(b) \geq f((1 - \lambda)a + (\lambda)b)}$

In other words, we have weights $w_1 = 1 - \lambda$, $w_2 = \lambda$, $a = x_1$, and $b = x_2$ such that $w_1 + w_2 = 1$ and $w_1 f(x_1) + w_2 f(x_2) \geq f(w_1 x_1 + w_2 x_2)$

With this nice little lemma, the rest follows by induction. So I'll leave that last bit to you me@home.

(P.S. the proof for concave functions is just as easy, it is just reversed since the secant line lies below the function)

Of course, the calculus is lurking; a convex function $f: \mathbb{R}^n\to\mathbb{R}$ is twice differentiable almost everywhere.


3_variables:

Let $ a,b,c > 0$, and $ a^{ \minus{} 2} \plus{} b^{ \minus{} 2} \plus{} c^{ \minus{} 2} \equal{} 1$,prove that:

$ a(b \plus{} c) \plus{} b(c \plus{} a) \plus{} c(a \plus{} b) \geq 2\sqrt{3}(a \plus{} b \plus{} c ).$

make equal condition?
BQ

Let $ a,b,c,d > 0$, and $ a^{ \minus{} 2} \plus{} b^{ \minus{} 2} \plus{} c^{ \minus{} 2} \plus{} d^{ \minus{} 2} \equal{} 1$,prove that:

$ a(b \plus{} c) \plus{} b(c \plus{} d) \plus{} c(d \plus{} a) \plus{} d(a \plus{} b) \geq 2\sqrt{4}(a \plus{} b \plus{} c \plus{} d).$

make equal condition?
BQ

[quote="xzlbq"]3_variables:
Let $ a,b,c > 0$, and $ a^{ \minus{} 2} \plus{} b^{ \minus{} 2} \plus{} c^{ \minus{} 2} \equal{} 1$,prove that:
$ a(b \plus{} c) \plus{} b(c \plus{} a) \plus{} c(a \plus{} b) \geq 2\sqrt {3}(a \plus{} b \plus{} c ).$
make equal condition?
BQ[/quote]
I  think i can prove it.

[quote="xzlbq"]3_variables:

Let $ a,b,c > 0$, and $ a^{ \minus{} 2} \plus{} b^{ \minus{} 2} \plus{} c^{ \minus{} 2} \equal{} 1$,prove that:

$ a(b \plus{} c) \plus{} b(c \plus{} a) \plus{} c(a \plus{} b) \geq 2\sqrt {3}(a \plus{} b \plus{} c ).$

make equal condition?
BQ[/quote]

$ \iff ab\plus{}bc\plus{}ca \ge \sqrt{3}(a\plus{}b\plus{}c).$

let $ x^2\equal{}a^{ \minus{} 2},y^2\equal{}b^{ \minus{} 2},z^2\equal{}c^{ \minus{} 2}$

$ \iff x^2\plus{}y^2\plus{}z^2 \equal{} 1, \frac{1}{xy} \plus{} \frac{1}{yz} \plus{} \frac{1}{zx} \ge \sqrt{3}\left(\frac{1}{x} \plus{} \frac{1}{y} \plus{} \frac{1}{z}\right)$

$ \iff \sqrt{\frac{x^2\plus{}y^2\plus{}z^2}{3}}\left(\frac{1}{xy} \plus{} \frac{1}{yz} \plus{} \frac{1}{zx}\right) \ge \frac{1}{x} \plus{} \frac{1}{y} \plus{} \frac{1}{z},$

because $ \sqrt{\frac{x^2\plus{}y^2\plus{}z^2}{3}} \ge \frac{x\plus{}y\plus{}z}{3}$, so we just need to prove that

$ \frac{x\plus{}y\plus{}z}{3}\left(\frac{1}{xy} \plus{} \frac{1}{yz} \plus{} \frac{1}{zx}\right) \ge \frac{1}{x} \plus{} \frac{1}{y} \plus{} \frac{1}{z}$

$ \iff (x\plus{}y\plus{}z)^2 \ge 3(xy\plus{}yz\plus{}zx),$

clear.

$ a,b,c,d,e > 0$, and $ a^{ \minus{} 2} \plus{} b^{ \minus{} 2} \plus{} c^{ \minus{} 2} \plus{} d^{ \minus{} 2} \plus{} e^{ \minus{} 2} \equal{} 1$,prove that:

$ a(b \plus{} c) \plus{} b(c \plus{} d) \plus{} c(d \plus{} e) \plus{} d(e \plus{} a) \plus{} e(a \plus{} b) \geq 2\sqrt{5}(a \plus{} b \plus{} c \plus{} d \plus{} e).$

make equal condition?
BQ

Let $ x_i > 0,i \equal{} 1,2,...,n,n\geq 3$, and $ \sum_{i \equal{} 1}^n{x_i^{ \minus{} 2}} \equal{} 1,x_{n \plus{} 1} \equal{} x_1,x_{n \plus{} 2} \equal{} x_2,$,prove that:

$ \sum_{i \equal{} 1}^n{x_i(x_{i \plus{} 1} \plus{} x_{i \plus{} 2})} \geq 2\sqrt {n}\sum_{i \equal{} 1}^n{x_i}.$

make equal condition?
BQ

Let $ x_i > 0,i \equal{} 1,2,...,n,n\geq 3$, and $ \sum_{i \equal{} 1}^n{x_i^{ \minus{} t}} \equal{} 1,x_{n \plus{} 1} \equal{} x_1,x_{n \plus{} 2} \equal{} x_2,t \geq 1$,prove that:

$ \sum_{i \equal{} 1}^n{x_i(x_{i \plus{} 1} \plus{} x_{i \plus{} 2})} \geq 2\sqrt [t]{n}\sum_{i \equal{} 1}^n{x_i}.$

make equal condition?
BQ

[quote="Zarif"]I`m sorry. I don`t know English. This inequality very easy by AM-GM.[/quote]

dear Zarif:
        solution in your Mother language，please.
BQ

[quote="xzlbq"][quote="xzlbq"]3_variables:

Let $ a,b,c > 0$, and $ a^{ \minus{} 2} \plus{} b^{ \minus{} 2} \plus{} c^{ \minus{} 2} \equal{} 1$,prove that:

$ a(b \plus{} c) \plus{} b(c \plus{} a) \plus{} c(a \plus{} b) \geq 2\sqrt {3}(a \plus{} b \plus{} c ).$

make equal condition?
BQ[/quote]

if hard,please prove that:

Let $ a,b,c > 0$, and $ a^{ \minus{} 3} \plus{} b^{ \minus{} 3} \plus{} c^{ \minus{} 3} \equal{} 1$,prove that:

$ a(b \plus{} c) \plus{} b(c \plus{} a) \plus{} c(a \plus{} b) \geq 2\sqrt [3] {3}(a \plus{} b \plus{} c ).$[/quote]

that's similar to my solution in 4#.

and you can also say that

if $ a,b,c > 0, t \ge 1$, and $ a^{ \minus{} t} \plus{} b^{ \minus{} t} \plus{} c^{ \minus{} t} \equal{} 1$, the following ineq holds:

$ a(b \plus{} c) \plus{} b(c \plus{} a) \plus{} c(a \plus{} b) \geq 2\sqrt[t] {3}(a \plus{} b \plus{} c ).$

[quote="kuing"][quote="xzlbq"][quote="xzlbq"]3_variables:

Let $ a,b,c > 0$, and $ a^{ \minus{} 2} \plus{} b^{ \minus{} 2} \plus{} c^{ \minus{} 2} \equal{} 1$,prove that:

$ a(b \plus{} c) \plus{} b(c \plus{} a) \plus{} c(a \plus{} b) \geq 2\sqrt {3}(a \plus{} b \plus{} c ).$

make equal condition?
BQ[/quote]

if hard,please prove that:

Let $ a,b,c > 0$, and $ a^{ \minus{} 3} \plus{} b^{ \minus{} 3} \plus{} c^{ \minus{} 3} \equal{} 1$,prove that:

$ a(b \plus{} c) \plus{} b(c \plus{} a) \plus{} c(a \plus{} b) \geq 2\sqrt [3] {3}(a \plus{} b \plus{} c ).$[/quote]

that's similar to my solution in 4#.

and you can also say that

if $ a,b,c > 0, t \ge 1$, and $ a^{ \minus{} t} \plus{} b^{ \minus{} t} \plus{} c^{ \minus{} t} \equal{} 1$, the following ineq holds:

$ a(b \plus{} c) \plus{} b(c \plus{} a) \plus{} c(a \plus{} b) \geq 2\sqrt [t] {3}(a \plus{} b \plus{} c ).$[/quote]

In this way could solve the n-variables?

I still did not prove that 4 variable  :huh:

[quote="spanferkel"]By the power mean ineq, it is sufficient to prove it for $ t \equal{} 1$.  :lol:[/quote]

yeah... :lol:

but how to prove the n-variables for $ t\equal{}1$?  :huh:


articles trigonometry Euler function logarithms ratio invariant

I was looking through some math articles and it sparked a thought. What is $ k^i$ or $ k^\pi$ where $ k \in \mathbb{R}$? What do those expressions [i]mean[/i]? Also, what happens if $ k$ is a complex number? What does it mean then?

Well, there's this interesting thing called [url=http://en.wikipedia.org/wiki/Euler%27s_formula]Euler's Formula[/url].

Well $ k$ can be represented as $ a\plus{}bi$ if $ k\in\mathbb{R}$ or $ k\in\mathbb{C}$ because $ \mathbb{R}\subset\mathbb{C}$. $ k\in\mathbb{R}$ iff $ b\equal{}0$.

Now, Euler's Formula states that any complex number $ \cos\theta\plus{}i\sin\theta$ ($ \text{cis }\theta$ for short) $ \equal{} e^{i\theta}$.

So if $ a\equal{}\cos\theta$ and $ b\equal{}\sin\theta$ (implying that $ \arccos a\equal{}\arcsin b$), then $ k\equal{}\text{cis }\theta$ so that $ k^i\equal{}e^{i^2\theta}\equal{}e^{\minus{}\theta}$.

Similarly, if $ k\equal{}\text{cis }\theta$, $ k^\pi\equal{}\text{cis}(\theta\pi)\equal{}e^{i\theta\pi}$.

Well, $ \pi$ is really just a number, namely 3.14159265... To raise $ k$ to the $ \pi th$ power can be simplified simply as
$ k^{3} \bullet k^{.14159265...}$. $ k^{.14159265...}$, of course, will just give back the number $ n$ such that $ n^\frac{1}{.14159265}$ will equal k. As for $ k^{i}$, and when $ k \in{C}$, I think gm_trtle does a good job in explaining it.

Thanks to all the people who posted. Now I have on more question: how did Euler derive such an equation?

I believe it comes from the Taylor series of $ e^x$, $ \sin x$, and $ \cos x$ for $ x\in\mathbb{C}$, and then comparing them.

Here's a proof:
[hide="Simplest proof I know of"]
Let $ f(x)\equal{}e^{\minus{}ix}(\cos x\plus{}i\sin x)$
$ f'(x)\equal{}\minus{}ie^{\minus{}ix}(\cos x\plus{}i\sin x)\plus{}e^{\minus{}ix}(\minus{}\sin x\plus{}i\cos x)$
$ \equal{}e^{\minus{}ix}(\minus{}i\cos x\plus{}\sin x)\plus{}e^{\minus{}ix}(\minus{}\sin x\plus{}i\cos x)$
$ \equal{}e^{\minus{}ix}(\minus{}i\cos x\plus{}\sin x\minus{}\sin x\plus{}i\cos x)$
$ \equal{}0$

If $ f'(x)\equal{}0$ then $ f(x)\equal{}c$ for some constant $ c$. ($ f(x)$ is a constant function).
Thus, $ f(x)\equal{}f(0)$
$ f(0)\equal{}e^{0}(\cos 0\plus{}i\sin 0)\equal{}1$
Therefore, $ c\equal{}1$ and so $ f(x)\equal{}1$.

This implies that $ e^{\minus{}ix}(\cos x\plus{}i\sin x)\equal{}1$
Multiplying by $ e^{ix}$ we get: $ e^{ix\minus{}ix}(\cos x\plus{}i\sin x)\equal{}e^{ix}$
$ \implies \cos x\plus{}i\sin x\equal{}e^{ix}$, $ \mathbb{QED}$.
[/hide]

The previous posts have discussed how we perform exponentiation computationally over the complex numbers, although they are neither correct nor complete. Consult [url=http://en.wikipedia.org/wiki/Exponentiation]"Exponentiation" (Wikipedia)[/url] for an overview of various computational techniques and identities and descriptions of when various identities are valid.

I will give you a different angle of exponentiation -- I will develop it from simplistic notions of what it is and try to answer your question: "what does it [i]mean[/i] to raise a number to a real or complex power?" After all, the notion of "repeated multiplication" won't help you evaluate $ e^{i \pi}$ or such expressions!

[b]Outline:[/b]
1. What does it [i]mean[/i] to raise a number to an irrational power?
2. [i]The[/i] exponential function
3. Extension of exp(x) and log(x) into $ \mathbb{C}$
4. A Concrete Example
5. Conclusion

Please bear with the length. It is faster to read through than it seems!

[hide="CLICK TO READ"]

[b]1. What does it [i]mean[/i] to raise a number to an irrational power?[/b]
- You know what, say, $ 5^3$ means... it is $ 5 \cdot 5 \cdot 5$.
- You know what $ 5^{\frac {3}{4}}$ means; it is the number $ r$ such that $ r^4 \equal{} 5^3 \equal{} 5 \cdot 5 \cdot 5$.

[color=darkblue][u]Note:[/u] In fact, there are [i]four[/i] values of $ r$ that satisfy this; we usually agree to assign the [i]principle[/i] value or a particular real value, which I will call the [i]positive real[/i] value to such expressions. For example, we might assume $ 1^{\frac {1}{2}} \equal{} 1$ and not $ \minus{} 1$. In different cases, it may be most useful to say $ 1^{\frac {1}{4}} \equal{} 1$ or $ \equal{} i$. Until we introduce the role of complex numbers in this post, let's agree to use the positive real value..[/color]

To extend the naive definition of exponentiation (repeated multiplication/root taking) to something that can be applied to even the real numbers, you have to think of real numbers as limits of [url=http://en.wikipedia.org/wiki/Cauchy_sequence]Cauchy sequences[/url].

It turns out that every real number can be expressed as a [url=http://en.wikipedia.org/wiki/Limit_of_a_sequence]limit[/url] of a Cauchy sequence of rational numbers -- ie, the value it approaches.

For example, the number $ \pi$ can be expressed as the limit of the sequence $ 3, \; 3.1, \; 3.14, \ldots$
We can make more interesting Cauchy sequences, too -- for example, $ \frac {3}{2} , \; \frac {7}{5}, \; \frac {17}{12}, \; \frac {41}{29} \ldots$ is a sequence that converges on $ \sqrt {2}$

- Let us agree for now that as a number $ a$ approaches another number $ b$, $ k^a$ should approach $ k^b$

Suppose the sequence of rational numbers $ a_1, \; a_2, \; a_3, \; \ldots$ converges on a limit $ A$. We then can define the value $ k^A$ as the limit of the sequence $ k^{a_1}, \; k^{a_2}, \; k^{a_3} ,\; \ldots$, which is a computable real number. What's more, since each $ a_i$ is a rational, each of the terms of this sequence can be computed by repeated multiplication and taking integer roots.

[b]2. [i]The[/i] exponential function:[/b]

I will now step aside to define a totally new and seemingly unrelated function, which I will call $ \text{exp} \; x$:
\[ \boxed{\text{exp} \; x \equal{} \sum_{n \equal{} 0}^{\infty} \frac {x^n}{n!}}
\]
It is not difficult to show that the series definition above converges for all real $ x$. It is also monotone increasing, which means it has an inverse function. We name the inverse the "natural logarithm," denoted $ \log x$

[color=darkblue][u]Note 1:[/u]  There are other suitable definitions, as well.
[u]Note 2:[/u] We can derive other equivalent expressions for $ \log x$, allowing us to compute it without using the series definition of $ \text{exp}$.
[u]Note 3:[/u] Some textbooks insist on using $ \ln x$ for the natural logarithm, reserving $ \log x$ for the [i]base-10[/i] logarithm. Let us agree for the rest of this post that $ \log x$ refers to the natural logarithm and not the base-10 logarithm.[/color]

The interesting thing about $ \text{exp} \; x$ is that the ratio $ \frac {\text{exp} \; (x \plus{} 1)}{\text{exp} \; x}$ is [i]invariant[/i] of $ x$. This means that there exist real numbers $ C$ and $ e$ such that:
\[ \text{exp} \; x \equal{} C e^x
\]
Substituting $ x \equal{} 0$ into the series for $ \text{exp} \; x$ and into $ C e^x$ gives $ C \equal{} 1$, therefore [i][u]there exists a real number[/u][/i] $ e$ such that:
\[ \boxed{\text{exp} \; x \equal{} e^x }
\]
This serves as a bridge between the world of exponential functions and the world of infinite series:
\[ a^x \equal{} (e^{\log a})^x \equal{} e^{\log a \cdot x} \equal{} \text{exp} \; \left( \log a \cdot x \right)
\]
[b]3. Extension of exp(x) and log(x) into \mathbb{C}:[/b]

It turns out that (and the proof of this is non-trivial) the series definition for $ \text{exp} \; x$ converges and is continuous for complex $ x$, also. This is great news -- it means we can use our "bridge" to evaluate exponential functions through this infinite series! How easy!

Not quite. Our "bridge" relied on the new $ \log$ function, which was introduced as the inverse of $ \text{exp}$. However, $ \text{exp}$ is no longer a one-to-one function -- for example, $ \text{exp} \; x \plus{} 2 \pi i \equal{} \text{exp} \; x$. Thus, the complex natural logarithm is [i]multivalued[/i].

We can still define exponentiation over the complex numbers as follows:
\[ \boxed{ a^b \equal{} \text{exp} \; \left( \log a \cdot b\right) \qquad \forall a, b \in \mathbb{C}}
\]
Unfortunately, since we relied on a multivalued function, the result is usually multivalued. The exception is when $ b$ is an integer (1+2i is [i]not[/i] an example of an integer) so that adding $ 2 \pi i$ to the value of $ \log a$ adds a multiple of $ 2 \pi i$ to the overall exponent, and this is undone anyway by the $ \text{exp}$ function. In the next section, I will illustrate the multivalued nature of complex exponentiation.

[b]4. A Concrete Example:[/b] Find $ i^i$.
We use our definition of complex exponentiation: $ i^i \equal{} e^{\log i \cdot i}$
Note that $ e^{\frac {\pi i}{2}} \equal{} i$, so one value is generated by $ \log i \equal{} \frac {\pi i}{2}$. Thus, we can generate all values of $ \log i$ with the expression $ \frac {\pi i}{2} \plus{} 2n \pi i$. Plugging this back into our expression:
\[ i^i \equal{} e^{\left( \frac {\pi i}{2} \plus{} 2n \pi i \right) i } \equal{} e^{ \minus{} \frac {\pi}{2} \minus{} 2n \pi}
\]
Thus, $ n \equal{} 0 \implies i^i \equal{} e^{ \minus{} \frac {\pi}{2}} \approx 0.20788$ is one possibility.
We obtain a vastly different result by chosing $ n \equal{} \minus{} 1$; namely, $ e^{\frac {3}{2} \pi} \approx 111.318$

[b]5. Conclusion:[/b]

I hope you've come to appreciate the motivation behind exponentiation in the real numbers and complex numbers, and their various quirks. Make sure to read the Wikipedia article linked at the beginning of this post for precise techniques on simplifying terms with exponents.

The main points:
1. We can obtain multiple values when we raise a real number to a rational power. We usually have to agree (explicitly or implicitly) from beforehand on how to evaluate these expressions.
2. Exponentiation in the real numbers comes from taking limits of sequences with rational exponents which we know how to handle from the naive definition of exponentiation.
3. The exponential function $ \text{exp} \; x$ and it's inverse function, $ \log x$, can be used to express exponents via infinite series
4. $ \text{exp} \; x$ and $ \log x$ can be extended into the complex numbers [i]because of[/i] the series definition, which allow us to extend exponentiation into the complex numbers as well by [i]defining all exponentiation[/i] in terms of $ \text{exp}$ and $ \log$
4. However, $ \text{exp}$ is no longer one-to-one when we are in the complex numbers, and thus $ \log$ becomes multivalued. This means exponentiation in the complex numbers is usually multivalued. We usually have to agree on how to evaluate these ahead of time, based on the context (or else, out of sheer necessity).[/hide]

/////////////////


calculus
derivative
real analysis
real analysis unsolved

How can I prove the equation ?
$ if \ F(x,y)\equal{}f(y\plus{}ax)\plus{}g(y\minus{}ax) \ ,then \  \frac{\partial^2 F}{\partial x^2}\equal{}a^{2}\frac{\partial^2 F}{\partial y^2}$
I always get the result as the below ,

$ \frac{\partial^2 F}{\partial x^2} \equal{}a^2\frac{\partial^2 F}{\partial f^2}\minus{}2a^2\frac{\partial^2 F}{\partial f \partial g} \plus{}a^2\frac{\partial^2 F}{\partial g^2}$

$ \frac{\partial^2 F}{\partial y^2} \equal{}\frac{\partial^2 F}{\partial f^2}\plus{}2\frac{\partial^2 F}{\partial f \partial g} \plus{}\frac{\partial^2 F}{\partial g^2}$

or else  $ \frac{\partial^2 F}{\partial x^2} \neq a^{2}\frac{\partial^2 F}{\partial y^2}$ ?

Why don't you just find the derivatives of $ F$? You have:

$ \frac{\partial^2 F}{\partial x^2}(x,y) \equal{} a^2 \frac{d^2 f}{d x^2}(y\plus{}a x) \plus{} a^2  \frac{d^2 g}{d x^2}(y\minus{}a x)$

$ \frac{\partial^2 F}{\partial y^2} (x,y)\equal{} \frac{d^2 f}{d y^2}(y\plus{}a x)\plus{} \frac{d^2 g}{d y^2}(y\minus{}a x)$

The equality you need is quite obvious from this..


Find all integers $n\geq 1$ such that \[ \frac{1}{n(n+3)} \] is a decimal number. :)

It is equavalent $n=2^k,n+3=5^m, \ or \ n=5^m,n+3=2^k$, because (n,n+3)=1.
In first case k=1,m=1 (if k>1 $5^m=3(mod \ 4)$ give contradition).
In second case solution is m=3,k=7.

It is simple. $2^k|(5^m+3) \Longrightarrow m=3+l*2^{k-2},k>7,l\ge 1$


What is the most general form of the rational function which has modulus one on the unit circle?

$ f(z)g(1/z)$, where $ f$ and $ g$ are finite Blaschke products.


An alarm clock runs 4 minutes slow every hour. It was set right $ 3\frac{1}{2}$ hours ago. Now another clock, which is correct, shows noon. In how many minutes, to the neatest minute, will the alarm clock show noon ?

[size=75]
from book "The Moscow puzzles" by Boris A. Kordemsky

Please hide Your solution . Write inside Hide ... /Hide
[/size]

[hide=" Solution"]
The alarm clock is now $ 3\frac{1}{2}\times \equal{} 14$ minutes slow. It loses another minute in $ 15$ minutes. So it wlll show noon in $ 15$ minutes time.
[/hide]


Find the flaw in the scene below! Post your guesses in HIDE TAGS.

For those who ask "describe what you were doing": Apparently I was doing my homework, and I stumbled across something.

(No it was not me posting at 11:32 PM. That is normal for me.)
(No it is not the red scribbles. That is my name, and I rather not give it out.)

[hide=" "]Look at the top. Look at the sides. Try the bottom. Do not tell anyone of this hint.[/hide]

[hide="Maybe..."]Word says you're on "Page 2 of 2" when you're really on the first (and only) page of your document?[/hide]

No, he's on Page 2, if you look on the right.

[hide="A guess"]It says"Microsoft Word non-commercial use". Is that whats wrong? [/hide]

[hide="maybeNOT"]How many words are in your name? Did you make red scribbles not just for anonymity or to trick us with the # of words? Because you should have 2 words in the name (usually), but looking at the word count, it's supposed to have 3.[/hide]

[hide="Q12"]The words, if you zoom in enough, seem to have a faint coloring/highlight around them?[/hide]

[hide=" "]"trai" has wrong color underline[/hide]

[quote="PhireKaLk6781"][hide="What was hidden"]"trai" has wrong color underline[/hide][/quote]

I was about to type trait.

cf was half right. ernie was right. Yes, at first glance, it looks like I'm on the first page. But if you look at the right, then you will see I'm not on the first page. Good job.

remy: nope, it was for anonymity.
Chompy: that is a problem with prt-scrning it, and saving. I can do nothing about it.

Wow.  I was right while rejecting someone.  LOL.

Lookie at the one called "Multi!". Very little FTW information is required. If you need a hint, look at the second picture (about isabella2296). Note how isabella's rating tends to range from 1600s to 1800s.

[hide="Possibly?"][i]Is[/i] there a "Guests" username?

Also, why are there two "dragon96"'s? Is that allowed?[/hide]

For guess 1: If you go to FTW right now, you will see Guests | <number>

For guess 2: Two dragon96's = one is me, one is a multi.

How is that possible?
I thought that you couldn't repeat usernames...

[quote="dragon96"]Hint: So in the first picture, you see two dragon96's. This is part of the flaw. How does this differ from the case of isabella2296?[/quote]

The fact that the isabella2296 in the picture is only a multi? And that if the real user were on there, their names would be grouped together...

[hide="a thought on the izzy 2 pic"]Note how Jongao's rating is in the 1700's.
First, now he's called Jongy
Second, now he's 2000+
We can conclude it's an old picture
Perhaps that's why?[/hide]

It looks like this one was way too hard. Quote this post for my answer.

[hide=" "]Answer: If both isabella2296's were in teh lobby, you would see one listed under the other. In FTW, the reproduceable letter is "l" and "I". In dragon96, neither a capital i nor a lowercase l appears in the username, so there is no such multi.[/hide][/hide]

i see my name.
i see my rating.
i see my rating totally awesome :)
i see that dragon doesn't realize many things about makig multi accounts.

Correction: You used to see you awesome rating, but it disappeared.

And what exactly was flawed about this "multiple account"? Since I have a user of dragon96, no one can change his/her rating to dragon96. Even if I changed usernames.

Nice try. I don't think I'll fall for the disconnected d.

[hide="duh blue"]you also spelled "clragon" wrong :| [/hide]

[quote][hide="guesses"][list]clargon96
no posts
not rated
new member[/list][/hide][/quote]

But in the FTW list, it doesn't look like a "cl" to me... it's exactly the same width as a d.

if you look closely, you see that there is a space
click on the image and you'll see.

You weren't supposed to guess on that. :P


agar $a,b,c$ addadi mosbate koochektar az yek bashand va$a+c\le 1$ bashand oon vaght:
$\sqrt{a^2+b^2}+\sqrt{c^2+1}+\sqrt{(1-b)^2+(a+c)^2}\geq (1+\sqrt{5})(a+bc)$

belakhare be man nagofti ke in soalaye khafan ro az koja miareee? :D

[quote="ashegh"]fekr konam to jabr e safa bashe[/quote]

 :noo:  :harhar:

[quote="mehdi"]mitoonin ba bordar halleshoon konin(fasle haftom e safaa idasho tozih dade).dar zemn akharaye bakhshe namosavihaye safa dar in ide bishtar bahs shode.
agar ham ahle raveshe  "bi rahmaane" hastin(hama ro be tavaan beresoonin va saade konin) shayad hal she
rasti man ye chizayi too ketaabe "nabarabariha,parviz shahriyari" didam ke kheili kheili keyf kardam.oonja chand ta ideye halle namosaviha ro ba moshtagh va makhsoosan antegral ovorde ke nesbatan toope vali hichi safa nemishe.[/quote]

age betooni hallesh koni khoshhal mishim.

[quote="mehdi"]mitoonin ba bordar halleshoon konin(fasle haftom e safaa idasho tozih dade).dar zemn akharaye bakhshe namosavihaye safa dar in ide bishtar bahs shode.
agar ham ahle raveshe  "bi rahmaane" hastin(hama ro be tavaan beresoonin va saade konin) shayad hal she
rasti man ye chizayi too ketaabe "nabarabariha,parviz shahriyari" didam ke kheili kheili keyf kardam.oonja chand ta ideye halle namosaviha ro ba moshtagh va makhsoosan antegral ovorde ke nesbatan toope vali hichi safa nemishe.[/quote]
ajab ideye jalebi. :idea:

man mokhlese hameye shoma hastam.badjoore badjoor.faghat ye rize alan saram shoolooghe ke ghol midam age beresam be hamin zoodiha hallesho bezaram.
faghat ino begam ke faaseleye noghteye $(a,b)$ az mabdae mokhtasaat baraabare $\sqrt{a^2+b^2}$ hast.
hala age ino dar halle masale lahaaz konid kheili az masale haye radicalie in tippy kheili raahat hal mishan(az namosavie mosalas va minkowsky ghaafel nashin ke bazi vaghta kheili be daade aadam miresan).
rastesh nemitoonam ba in zaboone alkanam begam ke cheghadr mokhlese hamatoon hastam :)
bedrood

salam ideye jalebi daadi agha mehdi;faghatnamosaaviye minkowsky dige chiye

vali be nazare man ba hichkoodoom az inha hal nemishe.
I am sure.


Here is a problem I got on a pre-calc test i took a couple of days ago.  My teacher and i have different answers for it, but I am sure I'm correct.  Can yall give me yalls solutions, you can see mine here.
[b]QUESTION[/b]
The midpoints of a unit square (square with a side length of 1) are joined to form a new square.  This procedure is repeated to form another, then another, and so on.  Find the sum of the area of all the squares formed this way.
[url=http://schoolhelpnetwork.com/midpoints.gif]Picture of this Square[/url]
(it came with that picture)

I got one.  My teacher got two.
Its really a simple problem. but i wanna get a consensus.

Maybe your teacher counted the original unit square and you didn't? I'm not sure, I haven't worked on the problem yet, but getting twice as big of an answer is slightly odd.

EDIT: Each square has half the area of the previous - if you count the original, you get $1+\frac{1}{2}+\frac{1}{4}+... = 2$. If you don't, you get $\frac{1}{2}+\frac{1}{4}+... = 1$.

ya, if you count the original unit square, your teacher is correct, if you dont, you are.

Solution
[hide] Well, you can see that the area of the original unit square is $1$. Also, you know that the areas of the squares will form a geometric progression. Then to find the area of the next square, you take midpoint of one of the sides (length of $\frac{1}{2}$) and multiply by $\sqrt{2}$ to get the length of the side of the next square, which is: $\frac{\sqrt{2}}{2}$, then you square it to get the area of the square, which is $\frac{1}{2}$, then you can conclude that the common ratio between successive areas is $\frac{1}{2}$. Then, using the formula $\frac{a}{1-r}$ where $a$ is the first term and $r$ is the common ratio, one can conclude that the sum of the areas is equal to $\frac{1}{\frac{1}{2}}$ or $2$.[/hide]

edit - Paladin8 beat me to it  :blush:

I had this exact same question on a test not too long ago!

If you are absolutely certain that you are right, I don't see how getting a consensus would help you get points back or whatever. If you feel the question was ambiguous, then there might be a chance that you can convince your teacher that your way is also correct. Otherwise, explain your reasoning and maybe your teacher would see that you know what you're doing and give you some credit.
Approach the teacher or drop the issue. The last thing you want to do is let this hang over you and ruin the rest of your math class.

of course, you HAVE to count the first square.
the area is :$\sum_{n=0}^{\infty}(\frac{1}{2})^n=2$
your teacher is correct. :)

When I first looked at the problem, I immediately arrived at 2, but upon reading it more closely, it says find the sum of the areas of all squares [i]formed this way[/i] right after describing the midpoint thing.  So if you count the ones formed using midpoints it is 1.  It's certainly a point to argue, just explain how you interpretted it, and you will probably get credit.

When I first looked at the problem, i read it carefully, and then realized that the original square was NOT being counted because it says "all squares formed this way".  Clearly, the first square was not formed this way, according to the information given.  I agree with djshowdown2: you're correct.  Have you told your teacher how you solved it as yet?  If your teacher sees that you did it the correct way, but just interpreted it differently, you will probably get credit.  Clearly, you knew how to solve the problem, and you did by your interpretation: that's what counts i guess...

Well, if it's the sum of all the squares formed this way, then the original square doesn't count, because it was not formed that way. Therefore, if your teacher counted the original square(which I'm assuming is how they got 1 more than you, I haven't actually don't the problem), then you'd be right.

It's ambiguous whether to count the first square or not.  It's a poorly worded question.

I'd say that either one of you are right depending on interpretation.

your all pretty much correct depending on the interpretation i suppose.  But my teacher did NOT give me any sorts of credit.  When i tried to talk to her about it she was like "No Carlos, its one" and then looked away and ignored me... I firmly believe because the question is worded so poorly, that i shoudl receive full credit on the answer since i did answer the question.  Thnx for all your replies.

She just turned away and ignored you?  :huh:  Some teacher...
Maybe you need a nother way to present it to her.  Say something like "I realize that I misunderstood the problem, but I want to improve myself and not do this again in future.  Can you tell me what I can do in future to correct this problem?"  Then it will seem as though you're asking for help, rather than arguing.  You still may not get the point, but at least the misunderstanding can be avoided in future...

I really don't see that it is poorly worded but I suppose I could understand the misinterpretation. Still, it says all the squares, and the original is one of the squares. First of all, if it's just a simple question that will bring your grade up by .01%, just leave it. If it's worth a lot, be persistent and chance are your teacher will understand... or be forced to ;) .

It doesnt say [quote]all the squares[/quote] it says [quote]the area of all the squares formed this way[/quote].  The original unit square was NOT formed by connecting the midpoints of a previous square.  and it would bring my grade up 1 point which would bring me from 79 t0 80 :(.  There is not point in arguing, shes not gonna change it :(


trigonometry
geometry
trig identities
Law of Sines

Find the values of $ x,y,$ and $ z$ knowing that $ xy \plus{} xz \plus{} yz \equal{} 1$

$ 3\left(x \plus{} \frac {1}{x}\right) \equal{} 4\left(y \plus{} \frac {1}{y}\right) \equal{} 5\left(z \plus{} \frac {1}{z}\right)$

[hide="Triquality??"]Yes, triquality, 3 simultaneous equalities.[/hide]

Well, it's two equalities implying that three things are equal.

(Edit:  This wasn't intended to be a response to the question, just a comment on your choice of terminology.)

[hide]Hint:
Let $ x\equal{}tan\theta$, $ y\equal{}tan\alpha$, $ z\equal{}tan\beta$(these angles are from 0 to 90 degrees). What can u infer from these substitution?[/hide]

It does simplify to

$ \frac{3}{\sin \alpha \cos \alpha}\equal{}\frac{4}{\sin \beta \cos \beta}\equal{}\frac{5}{\sin \gamma \cos \gamma}$

But what do I do with the $ \tan \alpha \tan \beta\plus{}\tan \beta \tan \gamma\plus{}\tan \alpha \tan \beta\equal{}1$ ?

xy+yz+xz=1 and if x=tan(a/2) so it is a triangle.And it is a 3-4-5 triangle you can find x,y,z values from there.

Hi, How do you mean :$ 3\minus{}4\minus{}5$ triangle.Are you saying that $ x: y: z \equal{} 3: 4: 5$ .Thats not correct.

To clear up mehmetcantu's suggestion, note that $ \sin \alpha \cos \alpha \equal{} \frac12 \sin 2\alpha$.

$ \frac6{\sin 2\alpha} \equal{} \frac8{\sin 2\beta} \equal{} \frac {10}{\sin 2\gamma}$

Which is the law of sines for a 6-8-10 triangle with $ 2\alpha$, $ 2\beta$, $ 2\gamma$ angles (or a 3-4-5 triangle, if you prefer).

(As for why using the second equation, $ 2\alpha$, $ 2\beta$, and $ 2\gamma$ are angles of a triangle if and only if

$ \tan \alpha \tan \beta \plus{} \tan \beta \tan \gamma \plus{} \tan \alpha \tan \gamma \equal{} 1$

Looks familiar?)

with this try
tanA*tanB*tanC = tanA + tanB + tanC


function
vector
calculus
derivative
geometry
3D geometry
sphere

Let $n$ be a positive integer .Let $A$ be proper closed subset of $\mathbb{R}^n$ ,and let $\delta_{A}(x)$ denote the euclidean distance between $A$ and $x$.prove that $\delta _{A}(x)$ is diffrentiable at $b$ not belonging to $A$ iff there is one $a$ such that $\delta_{A}(b)=|a-b|$.

I'm afraid it didn't come out very pretty :oops: , so I'll just sketch it.

First of all, notice that for all $b$ there always is at least one $a$ s.t. $\delta(b)=\|b-a\|$. This is because by restricting our attention to the intersection of $A$ with some closed ball centered at $b$ which cuts $A$ in at least a point, we may assume that $A$ is actually compact, so there is at least one point $a$ in it where $\|b-a\|$, as a function of $a$, reaches its minimum.

In order to prove that if there are at least two points $a_1,a_2\in A$ for which $\|b-a_i\|=\delta(b)$, then $\delta$  is not differentiable at $b$, it suffices to prove the case when we're working in $\mathbb R^2$, by looking only at the plane generated by $b,a_i$.

The situation is this: we have a circle in the plane centered at $b$, and two points $a_1,a_2$ on this circle. We want to prove that $\delta$, which maps a point in the plane to its distance from a closed set which contains $a_1,a_2$ and no points inside the circle, is not differentiable at $b$. If we suppose the contrary, let $T$ be $f'(b):\mathbb R^2\to\mathbb R$. By making $x\to b$ on the segments $[ba_1],[ba_2]$, with $ba_1=ba_2$, we can easily see that we must actuall have $Ta_1=Ta_2$, so $T$ is actually a function of the projection of a vector on the perpendicular bisector $d$ of $a_1a_2$. If for some $x$, this projection is $x_d$ (measured in some direction on $d$ which we consider positive), then we have $Tx=kx_d$ for some constant $k$.

By making $x\to b$ on the segment $[ba_1]$, we will find a value of $k$ which does not agree with what we get by making $x\to b$ along the prependicular bisector of $a_1a_2$ (I did say I would just sketch it, didn’t I? :)).

For the converse, we can prove that, in fact, if $a\in A$ is the unique point for which $\delta(b)=\|b-a\|$, then the the derivative of $\delta$ at $b$ equals the derivative at $b$ of the function $\varphi:\ x\mapsto \|x-a\|$. In order to get this, we need to prove that $\frac{|\varphi(b+h)-\delta(b+h)|}{\|h\|}\to 0\ (*)$ as $h\to 0$.

Given $h$, let $b_h$ be a point for which $\|b+h-b_h\|=\delta(b+h)$, and let $b_h'$ be the intersection of the segment connecting $b+h$ and $b_h$ with the sphere $(b,\|b-a\|)$. It's easy to see, from the uniqueness of $a$, that $b_h,b_h'\to a$ as $h\to 0$.

$\delta(b+h)$ lies between $\|b+h-a\|=\varphi(b+h)$ and $\|b+h-b_h'\|$, so in order to get $(*)$ it suffices to prove that $\frac{|\|b+h-a\|-\|b+h-b_h'\||}{\|h\|}\to 0\ (**)$ as $h\to 0$.

By applying Ptolemy's inequality for the four points $a,b,b+h,b_h'$, we get $\|b+h-b_h'\|\cdot \|b-a\|\le\|b-b_h'\|\cdot\|b+h-a\|+\|h\|\cdot\|a-b_h'\|$. Since $\|b-b_h'\|=\|b-a\|$, we find $\|b-a\|(\|b+h-b_h'\|-\|b+h-a\|)\le \|h\|\cdot\|a-b_h'\|\ (1)$. In the same way we find $\|b-a\|(\|b+h-a\|-\|b+h-b_h'\|)\le \|h\|\cdot\|a-b_h'\|\ (2)$, and $(1),(2)\Rightarrow (**)$, since $\|a-b_h'\|\to 0$ as $h\to 0$.

I do hope it's correct, but I doubt anyone will ever read it :).


[hide="solution"]
There are 1000 minutes in a day, so the hour (little) hand moves 360/1000 degrees per minute. There are 100 minutes in an hour and the minute (big) hand goes around each hour so it moves 360/100 degrees per minute. The measure of the angle between the hands after x minutes is: x(360/100 - 360/1000) mod 360
plugging in for x = 675 gives us: [b]27 degrees[/b][/hide]

[hide="part 2"] Actual time:

because there are 1440 minutes in a day, each "minute" is 1440/1000 of a real minute. so the actual time past in minutes is 1440x/1000. plugging in for 675 gives us 972 which is 4:12 pm.[/hide]

You've answered the first question correctly, how about the second one?

Yeah, I got the same answer for part 1
Solution to Part 2:[hide] 11:15 am[/hide]

how did you get that? 675 minutes out of 1000 is definately past 12:00 noon....

[quote="G-UNIT"]One clock maker designed an analog clock on which one day has 10 hours, and one hour has 100 minutes. On this clock the short hand goes around once a day, and the long hand goes around in an "hour", which means that it goes around 10 times a day. What is the angle between the two hands of this clock when it shows 6 hours 75 minutes counting from midnight? What time is it really then?[/quote]

well, for 1.. i just have to say as a side note, theres .. i think 1440 minutes in  aday, so the minutes would have to be adjusted for such a clock to work anyway.. but i got

11:15.. cuz its 675 minutes after midnight - midnight is meaning real time so 12:00. 675/60 is 11.25.. soo.. adding 11.25 REAL hours to 12:00 midnight, gets you to 11 and .25(1/4) of an hour is 15 minutes

My answer is after 675 minutes past 12 midnight real time it is 11:15 (i agree with 4everwise)

EDIT: also, 675 minutes.. its real minutes, as i said before, the clock isnt running on a real 1440 minute day, and u cant convert it or he shouldnt use the term 'minute' for that time

You're saying that each minute is a real minute. But he said that each day has 10 hours on this clock. Since there are 100 minutes in a hour, that means there are $\frac{1440}{100} = 14.4 \text{ hours}$ but there are only 10 hours. Thus the "minutes" can not be real minutes

Right, so this clock wouldnt work, but im assuming he means 'minute' as a real 'minute', or else he should state otherwsie. The clock, with 14.4 real hour days, its a bad clock.. ull wake up at 8 AM that clock, and itll really be midnight, but the question doesnt have to do with whether the clock is a good clock..

[quote="math92"]same here  :ninja:[/quote]

how come whenever you use emoticons, they never work?

If a minute is a real minute then why isn't an hour a real hour?

Your solution kinda falls apart when asked that.

I think neither the minute nor the hour are real with the new clock

I think that the clocks work like this. Midnights always coincide, and the minutes and hours are adjusted to the correct proportions on the new clock

if the hour hand moves around ten times a day, that mens that the minutes are different

Yeah, exactly.  A day is always a day, but they are split up differently according to different clocks

or maybe a minute is a minute and the weirdo clock stays at midnight for 7 60-minute hours and 20 minutes

I'm pretty sure that its a day is a day, because then you wouldn't be able to compare time between the clocks

you could if the weirdop clock stopped and stayed at midnight for 440 minutes

I'm pretty sure that G-UNIT means that there are 1,000 NewMinutes in a day instead of our ordinary 1440.

yeah, just what I said, so the answers are right

It took me a while to figure out why it wasn't 11:15, but now I see the 4:12.

I don't know how you would think it was 1115 in the first place


How to remove duplicate numbers from an unsorted list in O(N) time?
N is the number of elements in the list.

This is also an interview question.

You could use a hash table if the numbers are large, and check whether each number has been used or not... On average, it would take about O(N) time.


Two moving points $A,\ B$ are on the concentric circles centered the origin $(0,\ 0)$ with radii $a,\ b$ respectively on $xy$ plane.
Let $C=(1,\ 0).$

Assumed that the area of $\triangle{ABC}$ takes the maximum value when $A$ is $A_{0}=\left(a\cos \frac{3\pi}{4},\ a\sin \frac{3\pi}{4}\right),$ and $B$ is $B_{0}=\left(b\cos \frac{4\pi}{3},\ b\sin \frac{4\pi}{3}\right).$

(1) Find $a,\ b.$
(2) Find the radius $R$ of the circumcircle of $\triangle{A_{0}B_{0}C}.$

[hide]If $A$ is fixed the triangle with maximum area (when moving point $B$) is obtained when $CO \perp AC$ ($O$ is the origin). With the same reasonig we conclude that the triangle with max area (when moving both points) must have $O$ for an orthocenter.
From $AB \perp CO$ we get $a \cos 45^{o}= b \cos 60^{o}$ or $b=a \sqrt2$.
From line $CA \perp BO$ (writing the equation of $CA$) we obtain $a=\frac{\sqrt 2(\sqrt3+1)}2$. So $b=\sqrt3+1$.
$\angle AOB=105^{o}=180^{o}-\angle ACB$ so $\angle ACB=75^{o}$. From the formula $CO=2R \cos \angle ACB$ we get $R=\frac{1}{2 \cos75^{o}}$[/hide]


If $ p$ is prime number of form $ 4k\plus{}3$ that does not divide integers $ x$ and $ y$ prove that $ x^{2}\plus{}y^{2}$ is not divisible by $ p$.

[hide="Equivalently..."]$ x^2\not\equiv \minus{}1 \pmod{p}$

[hide="Hint"]Use Wilson's theorem and group the numbers in a good way.[/hide][/hide]

Can anyone give a complete proof to this question?

If $ p\mid x^2\plus{}y^2$ then $ x^2\equiv \minus{}y^2(mod\ p)$ but $ n^2\equiv 0,1(mod\ 4)$ so $ x\equiv y\equiv 0(mod\ p)$ but $ p$ does not divide $ x$ or $ y$ contradiction  :o

[quote="pelao_malo"]If $ p\mid x^2 \plus{} y^2$ then $ x^2\equiv \minus{} y^2(mod\ p)$ but $ n^2\equiv 0,1(mod\ 4)$ so $ x\equiv y\equiv 0(mod\ p)$ but $ p$ does not divide $ x$ or $ y$ contradiction  :o[/quote]

I don't understand how you get from $ x^{2}\equiv \minus{} y^{2}\pmod{p}$ and $ n^{2}\equiv 0,1\pmod{4}$ to $ x\equiv y\equiv 0\pmod{p}$.  :huh:

[hide="My Solution"]We have $ p|x^{2} \plus{} y^{2}$, so $ x^{2}\equiv \minus{} y^{2}\pmod{p}$.  Now we raise both sides of this equivalence to the $ \frac {p \minus{} 1}{2}$ power, which is odd since $ p\equiv 3\pmod{4}$.  Therefore $ x^{p \minus{} 1}\equiv \minus{} y^{p \minus{} 1}\pmod{p}$.  But by FLT, since neither $ x$ nor $ y$ are divisible by $ p$, this becomes $ 1\equiv \minus{} 1\pmod{p}$.  This is only true for $ p \equal{} 2$, a contradiction.[/hide]

[hide="Using Wilsons theorem"]As above, we have $ p|x^2\plus{}y^2\implies (xy^{\minus{}1})^2\equiv \minus{}1 \pmod p$. Now, build sets of the form $ x,\minus{}x,x^{\minus{}1},\minus{}x^{\minus{}1}$ where $ x^{\minus{}1}\in\{1,\dots,p\minus{}1\}$ with $ xx^{\minus{}1}\equiv 1\pmod p$. Since $ p>2$ we can't have $ x\equiv\minus{}x$; furthermore, the only solutions to $ x\equiv x^{\minus{}1}$ are $ 1$ and $ p\minus{}1$.

By Wilson's theorem, we know $ 2\cdot3\cdots(p\minus{}2)\equiv 1\pmod p$. The product contains $ 4k$ numbers; for all those numbers $ x$ with $ x\not\equiv \minus{}x^{\minus{}1}\pmod p$ we can remove the numbers $ a,\minus{}a,a^{\minus{}1},\minus{}a^{\minus{}1}$ since these are pairwise different and their product is one. Now if there is a solution to  $ x \equiv x^{\minus{}1}\pmod p$ we have on the left side the numbers $ x$ and $ \minus{}x^{\minus{}1}$ with product -1 and on the right side the number 1, a contradiction. [/hide]


It wasn't even that ... it was
"weapons of mass destruction-related program activities."

"Look, an Iraqi government scientist sneezed!  This is clear evidence of a biological weapons-related program activity!"  ;)


The Princeton University Math Club would like to invite you to the third annual Princeton University Mathematics Competition (PUMaC), which will take place on Saturday, January 31, 2009.

PUMaC is a team-oriented math contest for high school students from across the nation. While this competition is designed primarily for high school students, we welcome younger students to participate as well.

Registration for PUMaC will take place online and should be submitted by Friday, November 14, 2008. There is an initial registration limit of two teams per school. However, a waitlist will be made for schools who would like to register more teams. More information about registration will be available in the coming weeks.

The regular registration fee is $ \$$10 per student. We require all payment by Friday, December 12, 2008 and accept checks only. Checks should be made out to Princeton University Undergraduate Math Club. Please note that if registration or payment is late, we may be unable to assure your team’s participation.

For more information about PUMaC, please visit our website, https://cgi.math.princeton.edu/mathclub/index.php/Princeton_Math_Competition_2008. If you have any questions, please feel free to contact us at pumac@math.princeton.edu.

All the best, and we hope to see you on January $ 31^{st}$ , 2009!

[url=http://webscript.princeton.edu/~mathclub/pumacos/index.php]Registration[/url] is now open! Let us know if you have any problems.

I don't know about the rest of the teams, but I know that team Quagga is dying to know the full results/see the solutions. Do you know at what point in time they will be available?

Thanks,
-diophantient


Prove that \[\frac{n^{2}}{3}+\frac{n}{2}+\frac{1}{6}\geq \sqrt[n]{(n!)^{2}}\forall n\in\mathbb{N}.\]


Find the smallest positive integer $x$ such that when divided by $10$, it leaves a remainder of $9$, when divided by $9$, it leaves a remainder of $8$, when divided by $7$, it leaves a remainder of $6$, when divided by $5$, it leaves a remainder of $4$........, all the way till when divided by $2$, it leaves a remainder of $1$.

[quote="mna851"]-1?[/quote]

Nope....by the way, it's positive...use mods

[quote="Iversonfan2005"][quote="chess64"]The smallest number that has 10, 9, 8, $\cdots$, and 2 as factors, plus 1:

$2 \cdot 3 \cdot 2 \cdot 5 \cdot 7 \cdot 2 \cdot 3 +1 = 2521$[/quote]

Why plus 1?  ;)[/quote]

B/c it leaves a remainder of 1.

No, it should be minus $1$, here's why:

$x=9(mod 10)$
$x=8(mod 9)$
$x=7(mod 8)$
.......
$x=1(mod 2)$

By adding $1$ to each side, the equation still holds its congruency, thus:

$x+1=10(mod 10)=0(mod 10)$
$x+1=0(mod 9)$
........
$x+1=0(mod 2)$

Thus you have to find the smallest number divisible by integers $2$ through $10$, so you need $2,3,2,5,7,2,3$ which is equal to $2520$ therefore $x+1=2520$ so $x=2519$.

I know this problem! It was in one of [b]The Art of Problem Solving[\b] books.

We can think of this as several congruences
$x\equiv 9(mod10)$
$x\equiv 8(mod9)$
$x\equiv 7(mod8)$
$x\equiv 6(mod7)$
$x\equiv 5(mod6)$
$x\equiv 4(mod5)$
$x\equiv 3(mod4)$
$x\equiv 2(mod3)$
$x\equiv 1(mod2)$
~~~~~~~~~~~~~~~~~~~~~~~~~~~
Therefore, all we need is x+1 to be divisible by all numbers from 2 to 10.

$x+1=\frac{10!}{2*6*4*3*10}$
$x+1=2520$
$x=2519$

Therefore, the largest integer is 2519.

Ah, you got to it before I finished typing.  :(

[quote="Iversonfan2005"]No, it should be minus $1$, here's why:...[/quote]

Oh, I misread the problem. It looked like something I saw before, but it was actually different. Each leaves a remainder of $-1$, not $1$, so $2520+-1=2519$.  :blush:

[quote="Jan"][quote]$x+1=\frac{10!}{2*6*4*3*10}$[/quote]
I don't see how you get to this equation, could you please explain how you got it from the fact that the number is divisible by $10,9,8...$ ?[/quote]

well... $10!=10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1$
so $10!$ is divisible by every number from $1$ till ten $10$

however, this is not the smallest number
prime factor the whole thing and you get $2^8\cdot 3^4\cdot 5^2\cdot 7$
but it only needs to be divisible by $2^3\cdot 3^2\cdot 5\cdot 7$, so you can cancel the rest out...

or in other words: since $10!$ is divisble by $8$, it is divisble by $4$ and $2$ (--> divide by $4\cdot 2$)
since $10!$ is divisble by $9$, it is divisible by $3$ (--> divide by $3$)
since $10!$ is divisble by $9\cdot 8$, it is divisble by $6$ (--> divide by $6$)
since $10!$ is divisible by $8\cdot 5$, it is divisible by $10$ (--> divide by $10$)

Easier way:

You need a 2, a 3, a 2 (4 has two 2s), a 5, nothing for 6 (we already have 2,3), 7, a 2 (one more 2 for 8), a 3 (one more 3 for 9), and nothing for 10 (we already have 2,5)

So $2 \cdot 3 \cdot 2 \cdot 5 \cdot 7 \cdot 2 \cdot 3 =2520$.


calculus
integration
trigonometry
calculus computations

Put $ t \equal{} x \minus{} 1003$ than
\[ \int_0^{2006} {x(x \minus{} 1)(x \minus{} 2)...(x \minus{} 2006)dx} \equal{} \\
\equal{} \int_{ \minus{} 1003}^{1003} (t \plus{} 1003) \cdot (t \plus{} 1002) \cdot \ldots \cdot t \cdot \ldots \cdot (t \minus{} 1002) \cdot (t \minus{} 1003) \; dt \equal{} 0
\]
:) because integrand is an odd function.

else  this problem:Prove that
$ \int\limits_{ \minus{} \pi }^\pi  {\cos 2x\cos 3x\cos 4x...\cos 2005xdx}  > 0$


I see a problem in the JMO. I may be misunderstanding the scoring of the test and/or the qualifications for red mosp, and if I am, correct me.

If someone is mainly concerned with making it into red mosp, then he or she would approach the JMO by first trying to solve problems J3 and J6 because these are the problems that mainly determine entrance into red mosp. On the other hand, if someone just wants to get as high a score as possible, he or she will solve the problems in order, 1-3 and 4-6. Thus, someone who could solve problems 1, 2, 4, and 5 but who is concerned about mosp may work on J3 and J6 and never get to 1, 2, 4, and 5, and maybe not even solve J3 or J6, and thus get a lower score than someone who approached the test by going in order and who they could have possibly done better than.

Essentially, there are two very distinct things someone may want to get out of this test: getting into red mosp or getting a high score. Each of these things requires a very different approach. This test has two measures of achievement, but because everyone is still graded in the same pool, distinctions between competitors may not actually indicate their comparative ability or achievement on the exam.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=280201]Please keep discussion in one place.[/url]


$ [x^{2n}](1 \plus{} x)^{2n} (1 \minus{} x)^{2n} \equal{} [x^{2n}](1 \minus{} x^2)^{2n}$.  Presumably, this means there's a nice combinatorial solution.

Again nobody notices the generalization:
http://www.mathlinks.ro/viewtopic.php?t=262752

  darij

Suggestion for a combinatorial proof (which should also work for the generalization):

[hide]
The sum in question counts (up to sign) the number of pairs of subsets such that $ |S| \equal{} |T| \equal{} k$.    Let $ i$ be the first number on which the sets agree (i.e. either $ i$ is in both sets or in neither), and either move $ i$ into or out of both sets.  This creates a pairing that changes the sign of $ ( \minus{} 1)^k$, so the sum cancels everywhere this map is defined.

The only place it isn't defined is when $ S$ and $ T$ are compliments, so there's $ ( \minus{} 1)^n \binom{2n}{n}$ left over (that many choices for $ S$, at which point $ T$ is forced by the assumption they're compliments).
[/hide]

Very nice.  I had the same idea but wasn't sure if it was actually a correct involution.


Find the largest positive $ n$ such as $ n^{2}\plus{}2009n$ is perfect square .

My answer : [hide] After analyzeing the possibles values of $ n$ which are $ n\equal{}{1,7,49,41,287,2009}$ I conlude that  there is no $ n$ such as $ n^{2}\plus{}2009n$ be a perfect square. [/hide]

Take n=16 for the first counterexample to your claim.

Let $ n^{2}\plus{}2009n\equal{}r^{2}$ we have $ n\plus{}2009\equal{}\frac{r^{2}}{n}$ so we have that $ n|r^{2}$ hence $ n|r$ .

Let $ r\equal{}pn$ we have $ n\plus{}2009\equal{}p^{2}n$ and $ \frac{n\plus{}2009}{n}\equal{}p^{2}$ hence $ n|2009$.

[quote="Elio (n)"]so we have that $ n|r^{2}$ hence $ n|r$. [/quote]
This is not valid. For example, $ 27|81$ but $ 27\not{|}9$.

Can anyone  give a rational solution for the problem .

[hide]Let $ n^2 \plus{} 2009n \equal{} r^2$.  Then $ 4n^2 \plus{} 8036n \plus{} 2009^2 \equal{} 4r^2 \plus{} 2009^2$ and $ (2n \plus{} 2009)^2 \minus{} r^2 \equal{} 2009^2$. We can factor the LHS to give $ (2n \plus{} 2009 \plus{} r)(2n \plus{} 2009 \minus{} r)$.  Now 2009^2 has factors $ 7^k41^j$ where $ 0\leq k\leq4$ and $ 0\leq j\leq2$.  Every pair of these gives a solution by plugging the factors into the expression, to give the ordered pairs $ (n, r)\equal{}$
$ (1008016, 1009020) \\
(143143, 144144) \\
(19600, 20580) \\
(2023, 2856) \\
(16, 180) \\
(23616, 24600) \\
(2583, 3444) \\
(0, 0)$

These are all solutions to the problem.[/hide]


Prove that for positive reals $ x,y,z$ the following inequality holds -
$ (x^2\minus{}yz)^2 \ge \frac{27}{8}xy(xy\minus{}z^2)(xz\minus{}y^2)$

hello, i have simplifyied to $ 8(x^2\minus{}yz)^2\plus{}27xy\left(y(y\minus{}x)\plus{}x(y\minus{}z)\right)\left(x(y\minus{}z)\plus{}z(x\minus{}z)\right)\geq0$, this is true for
$ y \geq x \geq z$ or $ y \le x\le z$, for the other cases i have no idea.
Sonnhard.

continued here - http://www.mathlinks.ro/Forum/viewtopic.php?t=299899&start=160

hello, ok, thank you for the link.
Sonnhard.

With the condition -$ xz\ge y^{2},xy\ge z^{2}$this obviously implies the LHS is also positive.
make this substitution $ a\equal{}\frac{x}{y}$,$ b\equal{}\frac{y}{z}$,$ c\equal{}\frac{z}{x}$
the inequality gets transformed into -
$ 8(a\minus{}c)^{2}\ge27(a\minus{}b)(b\minus{}c)$ or
$ 2((a\minus{}c)^{2})^{\frac{1}{3}}\ge 3 ((a\minus{}b)(b\minus{}c))^\frac{1}{3}$we have $ a\ge b\ge c$
multiply both sides by $ (a\minus{}c)^\frac{1}{3}$, and we get
$ 2(a\minus{}c)\ge 3((a\minus{}b)(b\minus{}c)(a\minus{}c))^\frac{1}{3}$but this is just AM-GM if we write the LHS as $ 2(a\minus{}c) \equal{} a\minus{}b\plus{}b\minus{}c\plus{}a\minus{}c$ so we are done and the inequality is true with equality holding  :) when $ a\equal{}b\equal{}c$ :)


A truck driver earns $ 20$ cents per mile. She drove 1000 miles in 20 hours. What is her hourly wage in dollars?

The driver earns .2*1000=200 dollars for the total ride.
200/20 hours=10 dollars an hour.


Determine all surjective functions f: N -> N such that m|n <=> f(m)|f(n).

Let f be such a function.

- Since 1|n for each n, we deduce that f(1)|f(n) for each n. Since f is surjective, it follows that f(1) divides 1, therefore f(1) = 1.

- Let m,n such that f(u) = f(n). Then f(m)|f(n) and f(n)|f(m) so that m|n and n|m, thus m = n.
It follows that f is injective.
Since f is surjective, we deduce that f is bijective.
Let g be the inverse function of f. Thus m|n iff g(m)|g(n), for all m,n.

- Let p be a prime. If n divides f(p) then g(n) divides p and then g(n) = 1 or p, that leads to n = 1 or f(p). Thus f(p) is a prime.

- Let p be a prime and k be a positive integer. If q is a prime which divides f(p k ), then g(q) is a prime dividing p k . It follows that g(q) = p and q = f(p). Thus f(p k ) = p t  for some t.
But each of the numbers 1,p,..,p k is a divisor of p k , so that f(p k ) has at least k+1 distinct divisors, and then k  \leq  t. With a similar reasoning on g, we deduce that t  \leq  k, and then k = t. That is :
f(p k ) = ((f(p)) k  for each prime p and all integer k  \geq  0.

- Let a,b be two coprime integers. Then, f(u) and f(v) has to be coprime.
Moreover, we have f(u)|f(uv) and f(v)|f(uv) thus f(u)f(v)|f(uv). The same reasoning with g leads to f(uv)|f(u)f(v) so that f(uv) = f(u)f(v). Thus :
f is multiplicative.

Let P be the set of the prime numbers. From above, the restriction of f to P is a bijection from P onto P, and if n = (p_1) a_1 (p_2) a_2 ...(p_k) a_k  is the prime decomposition of n, then f(n) = (f(p_1)) a_1 f((p_2)) a_2 ...f((p_k)) a_k  is the prime decomposition of f(n).

It is easy to see that conversely, if h is any bijection from P onto P and f is defined by f(n) = (h(p_1)) a_1 h((p_2)) a_2 ...h((p_k)) a_k  where n = (p_1) a_1 (p_2) a_2 ...(p_k) a_k  is the prime decomposition of n, then f is a solution of the problem.

Pierre.


Gagnes finally blew a save. He had converted 84 straight saves before that. Does anyone think any closer will ever come close to that record?

I haven't been paying much attention to baseball lately but all records fall at some time or another.

"RT-2PM2 Topol-M Silo-based intercontinental ballistic missile hits streak"

/BRO WHY ARE YOU GUYS BUMPING OLD THHREADS


Could someone give me some pointers on solving clock problems such as the ones on MC?
Thanks.

Sure:  Keep these things in mind:

With the hour and minute hand remember there are 360 degrees in a clock.  So at 9:00 the hour hand is at 9 and the minute hand is at 0.  Since there are 30 degrees per hour, the measure of the small angle is 90 degrees, and the large angle is 270 degrees.  Of course that was an easy problem.  Also, remember to express things as fractions when necessary.  Here's a page of MC with an example of a clock problem from a Target Round:

thanks very much.
Will someone please lock this, I accidentally posted it twice.
 :blush:  :blush:


is there an easier way to find the sum of all of the prime #s? I NEED TO KNOW THIS B4 competition....thanx :lol:

You mean those problems that are like, "Find the sum of the prime numbers between $50$ and $70$"?

One way would be to just memorize two-digit primes.  This isn't a bad idea because they show up on competitions a lot.

The particular problem I gave has the really nice property that it's symmetric around $60$.  The thing about $60$ is that $2, 3, 5$ all divide it, whicih means if $60-x$ is a prime, then $60+x$ is [b]probably[/b] a prime (unless $7$ divides it), and vice versa.  So, the primes in that range are

$53, 59, 61, 67$

Or

$60-7, 60-1, 60+1, 60+7$

Which add up to $240$.  :)

This won't work for all problems of this type, and generally you just have to find them.

[b]Avoiding numbers that look prime that aren't:[/b]

A lot of beginners trip on weird multiples of $3$, such as $57$.  Just remember that if $3$ divides the sum of the digits, then it divides the number.

Also, you may want to memorize weird multiples of $7$, such as $91, 119$.

[quote="t0rajir0u"]
A lot of beginners trip on weird multiples of $3$, such as $57$.  Just remember that if $3$ divides the sum of the digits, then it divides the number.

Also, you may want to memorize weird multiples of $7$, such as $91, 119$.[/quote]

Is there a site where it displays the rules for dividing by numbers from $2-9$? I can't always remember some of them  :blush:

2------- last number is divisible by 2

3-------the sum of the digits is divisible by 3

4-------the last two didgits are divisible by 4

5-------last digit must be 5 or 0

6------multiple of 3 and 2

7-------take the last digit of a number and multiply it by 2.  then subtract that value from the rest of the number and see if the rsult is diviible by 7.   FOR XAMPLE:  given a number ABC.  just check if the integer AB-2C is a multiple of 7.  repeat this process until it is easy to tell if a number is divisible by 7.

8-----last 3 digits is a multiple of 8

9-------sum of digits is multiple of 9

 :D  :D i finished

well, i think you are talking about the divisibilty of 7.

just learn from examples.     pick a random number:  347

double the last digit: 7X2=14.  subtract 14 from 34, which is 20.

obviously,  20 is not a multiple of 7, so 347 is not a multiple of 7.

EX. 42[b]36[/b] is divisible by 4 because the last two digits 36 is divisible by 4.

[quote="7h3.D3m0n.117"]umm when you mean the last [blank] digits do you mean that number with [blank] digits or ...?[/quote]
For 8, any number where the last 3 digits (formed into a number) are divisible by 8, then the number is divisible by 8.

Example: for the number xxxx...x392, it is divisible by 8 because 392/8=49

[quote="kyyuanmathcount"]It's often not worth remembering any divisibility rules besides $2^{n}$, $3$, $9$, and $11$.  You should just know how to divide very quickly in your head.[/quote]

This is good advice.  The way I check divisibility for larger numbers is just fast mental arithmetic.

For example, is the number $18578$ divisible by $7$?  Rather than use the rule for $7$, I write

$18578 = 14000+4200+350+28$

All of which are clearly divisible by $7$.  :)  (In my head, I would have subtracted $14000$ away, then $4200$, and so forth.)

[quote="tridecalogism_lover"]maybe to make myself a little bit more clear: i meant like the problems where they ask the sum of all the prime numbers.....[/quote]

they won't ask questions like that because the answer is obvious

for example, if I ask you to find the sum of the numbers of all positive integers?

simply put the answer as

$\infty$ because prime numbers go on forever

ok. thank you everyone....i'll see if it works! :)

[quote="t0rajir0u"]Well, here's a fun question:  What is the sum of all the [b]even[/b] prime numbers?[/quote]
2 - it's the only one!


Find all complexe polynomial $ P$ wich satisfy $ P(\mathbb{C}) \subset \mathbb{R}$.
good luck.

If deg(P)>0, then the equation P(x)=i has a solution in C (Fundamental theorem of algebra). Hence, deg(P)=0, which implies that P is a real constant.
 :maybe:


Are  there anybody who know a web link that is including waterloo Bernoulli Trials competition questions? Thanks.

Some not-so-recent ones are [url=http://www.stats.uwaterloo.ca/~cgsmall/problem.archive.html]available[/url]. Seems like a pretty strange thing to be looking for, though.


${f=\left(\frac{1}{y}+\frac{1}{z}+\frac{1}{x}\right)^{n}}$
${x,y,z,n)>0}$
${n\in \mathbb{R}}$
${x+y+z=\theta }$

Proove that

${f\geq \left(\frac{9}{\theta }\right)^{n}}$

$\theta$ is a constant

By AM-GM we have $(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\geq 9$, and we have done :D


This is cool and read carefully or you might not understand- http://www.google.com/googlegulp/

Check out the gmail website http://www.gmail.com too!  :D

My email storage is raising by the second! =D

[quote="h_s_potter2002"]Is the storage really a joke though?[/quote]

The storage is no joke; 2gigs.  Google Gulp is a joke :D

[quote="probability1.01"]Hey, any of you have Gulp Caps?  If not, let's plan to raid Google for one :D.[/quote]

People were selling Gulp Caps on Ebay yesterday.  I'm not sure if they are still doing that, though...

GoogleGulp is excelent, I will order a sixpack of each flavor  :D  :D  :D .

Nevertheless, it is amaizing the amount of space they are giving rightnow  :?  :huh:  :what?:  :alien: .

[quote="djimenez"]Nevertheless, it is amaizing the amount of space they are giving rightnow  :?  :huh:  :what?:  :alien: .[/quote]

It looks like the counter is slowing down.

From April 1st to April 2nd, the space increased 1 gig.
From April 2nd to April 3rd, space increased maybe 4 megabytes.


inequalities
Cauchy Inequality
inequalities proposed

[quote="zhaobin"]
Try,$a=1,b=-1,c=0$[/quote]
You are right, zhaobin. Thank you.
But for all non-negative $a,$ $b$ and $c$ follow inequality holds:
$a^{6}+b^{6}+c^{6}-3a^{2}b^{2}c^{2}\geq \frac{9}{2}(a-b)^{2}(b-c)^{2}(c-a)^{2}.$
I am sorry.

That's ok. anyhow this inequality is very simple, while there are many equality conditions like $(a,b,c)=(k,k,k),(k,k,-k),(k,0,-k)$ for all real number $k$, and all the cyclic form of these.

That's what I mean  :lol: and all of the equality conditions appears from the condition of cauchy inequality, namely $(b^{2}-c^{2})b=(c^{2}-a^{2})a$ and so on. nice job!

But for all non-negative $a,$ $b$ and $c$ follow inequality holds:
$a^{6}+b^{6}+c^{6}-3a^{2}b^{2}c^{2}\geq \frac{9}{2}(a-b)^{2}(b-c)^{2}(c-a)^{2}.$
I am sorry.[/quote]

The following inequality is also true
[img]http://s9.sinaimg.cn/middle/0018zOAxgy710o1ISjSd8&690[/img]

See also here:
http://www.artofproblemsolving.com/community/c6h621716


Of 100 students surveyed, 80 were taking a math class and 75 were taking a science class. If
70 students were taking both math and science, how many were taking neither?

$ 75\minus{}70\equal{}5$ students took science only.

$ 80\minus{}70\equal{}10$ students took math only.

$ 70\plus{}10\plus{}5\equal{}85$ students took math or science.

$ 100\minus{}85\equal{}\boxed{15}$ students took neither.

[quote] students took science only.

 students took math only.

 students took math or science.

 students took neither.
[/quote]


A bowling ball is a solid ball with a spherical surface and diameter 30 cm. To custom fit a bowling ball for each bowler, three holes are drilled in the ball. Bowler Kris has holes drilled that are 8 cm deep and have diameters of 2 cm, 2 cm, and 3 cm. Assuming the three holes are right circular cylinders, find the number of cubic centimeters in the volume of the fitted bowling ball. Express your answer in terms of $ \pi$.

The volume of the whole bowling ball is $ \frac43 \pi (15)^3$, or $ 4500\pi$.  The volume of the smaller holes is $ \pi (1)^2 \times 8$, or $ 8\pi$.  The larger hole's volume is $ \pi (1.5)^2 \times 8$, or $ 18\pi$.

Thus, the total volume is $ 4500\pi \minus{} 2(8\pi) \minus{} 18\pi$, or $ \boxed{4466\pi}$.


In a large crate of oranges, 100[i]p[/i]% are bad. A random sample of 10 oranges produces 2 bad ones. On the assumption that this is the most likely number of bad oranges to find in the sample, what can you say about the possible values of [i]p[/i]?

[hide]Let $ P(X\equal{}k)$ be the probability of getting k bad oranges out of a group of 10.

In order for 2 to be the most likely choice, you want $ P(X\equal{}2) > P(X\equal{}1)$ and $ P(X\equal{}2) > P(X\equal{}3)$ [think about what a histogram of binomial probability density looks like].

Since $ P(X\equal{}k) \equal{} \binom{10}{k}p^k(1\minus{}p)^{10\minus{}k}$, we want:

$ \binom{10}{2}p^2(1\minus{}p)^8 > \binom{10}{1}p(1\minus{}p)^9$
$ \Rightarrow 45p^2(1\minus{}p)^8 > 10p(1\minus{}p)^9$
$ \Rightarrow p > \frac{2}{11}$

and

$ \binom{10}{2}p^2(1\minus{}p)^8 > \binom{10}{3}p^3(1\minus{}p)^7$
$ \Rightarrow 45p^2(1\minus{}p)^8 > 120p^3(1\minus{}p)^7$
$ \Rightarrow p < \frac{3}{11}$

Therefore, $ \boxed{\frac{2}{11} < p < \frac{3}{11}}$.[/hide]

gauss202 where did you learn that stuff?

I'm a high school teacher. But ideally, one would learn about binomial probability distributions in either an AP/Introductory Statistics class, or possibly in an Algebra II, Precalculus, or Data Analysis class.


Given triangle $ ABC$ and $ (I)$ is incenter assume to exist  a circle is through $ B,C$ and touch $ (I)$ at $ A'$, similarly we get $ B',C'$. Prove that $ AA' ,BB',CC'$ are concurrent.


inequalities
geometry
circumcircle
geometry proposed

Given $\triangle ABC$ with orthocenter $H$ circumcenter $O$ and circumradius $R$ prove that
$OH\le HA+HB+HC-3R<3R$

I think that the first $<$ should be turned into $\leq$.

yes you are right, it should be "$\le$" thanks :)

I don't think this is true since $\sum HA-3R = R(2 \sum cosA-3) \leq 0$

ok you are right treegoner, I have a stupid mistake while I created it sorry, but when $OH^{2}=9R^{2}-(a^{2}+b^{2}+c^{2})<9R^{2}\rightarrow OH<3R$ is the ineq that too weak then what is the way to do stronger ?

I think the way to make it stronger is $HA+HB+HC \geq OH$  :P

ok that is a good idea treegoner because
$\vec{HA}+\vec{HB}+\vec{HC}=2\vec{HO}\rightarrow HA+HB+HC>2HO$ we also have $HA+HB+HB<6R$

Right. In fact, I forgot the constant $2$. But anyway, since you have $HA+HB+HC=2(R+r)$, your right inequality is far too weak. :)


calculus
integration
logarithms
trigonometry
real analysis
real analysis unsolved

show, for $ m,n\;>\;0$, :D

$ \textbf{(i)}\;\;\int_{0}^{\infty}\;\;\ln x\,\cdot\,\sin\, (mx)\,\cdot\, e^{\minus{}nx}\;\;\textbf dx\;\;\equal{}\;\;\boxed{\frac{1}{m^{2}\plus{}n^{2}}\cdot\left(n\cdot\tan^{\minus{}1}\left(\frac{m}{n}\right)\minus{}m\cdot\gamma\minus{}\frac{m}{2}\cdot\ln (m^{2}\plus{}n^{2})\right)}$

$ \textbf{(ii)}\;\;\int_{0}^{\infty}\;\;\ln x\,\cdot\,\cos\, (mx)\,\cdot\, e^{\minus{}nx}\;\;\textbf dx\;\;\equal{}\;\;\boxed{\minus{}\,\frac{1}{m^{2}\plus{}n^{2}}\cdot\left(m\cdot\tan^{\minus{}1}\left(\frac{m}{n}\right)\plus{}n\cdot\gamma\plus{}\frac{n}{2}\cdot\ln (m^{2}\plus{}n^{2})\right)}$

Differentiate $ \int_{0}^{\infty}x^{z\minus{}1}\, e^{\minus{}Rx}\, dx \equal{}\frac{\Gamma(z)}{R^{z}}$ with respect to $ z$

$ \int_{0}^{\infty}x^{z\minus{}1}\ln x\; e^{\minus{}Rx}\, dx \equal{}\frac{R^{z}\,\Gamma'(z)\minus{}\Gamma(z)\, R^{z}\,\ln R}{R^{2z}}$ and set $ z \equal{} 1$

$ \int_{0}^{\infty}\ln x\; e^{\minus{}Rx}\, dx \equal{}\frac{\minus{}\gamma\minus{}\ln R}{R}$ . After that set $ R \equal{} n\plus{}im$

$ \int_{0}^{\infty}\ln x\; e^{\minus{}imx}\, e^{\minus{}nx}\, dx \equal{}\frac{\minus{}\gamma\minus{}\frac{1}{2}\ln(n^{2}\plus{}m^{2})\minus{}i\,\arctan\frac{m}{n}}{n\plus{}im}$

$ \equal{}\frac{\minus{}1}{n^{2}\plus{}m^{2}}\left[ (n\minus{}im)\left(\gamma\plus{}\frac{1}{2}\ln(n^{2}\plus{}m^{2})\plus{}i\,\arctan\frac{m}{n}\right)\right]$

$ \equal{}\frac{\minus{}1}{n^{2}\plus{}m^{2}}\left(n\gamma\plus{}\frac{n}{2}\ln(n^{2}\plus{}m^{2})\plus{}m\,\arctan\frac{m}{n}\right)\plus{}\frac{i}{n^{2}\plus{}m^{2}}\left(m\gamma\plus{}\frac{m}{2}\ln(n^{2}\plus{}m^{2})\minus{}n\,\arctan\frac{m}{n}\right)$


Let a,b,c≥0,prove that:
$ \sum_{cyc}a\sqrt{a^2\plus{}2bc}$≥$ \sqrt{3}(ab\plus{}bc\plus{}ca)$

[quote="SUPERMAN2"]Let a,b,c≥0,prove that:
$ \sum_{cyc}a\sqrt {a^2 + 2bc}$≥$ \sqrt {3}(ab + bc + ca)$[/quote]
 square two position we have:
$ \sum{a^4} + 2\sum{ab\sqrt {(a^2 + 2bc)(b^2 + 2ca)} \geq 3\sum{a^2b^2} + 4abc\sum{a}}$
 to apply cauchuy-shawrs inequality:
$ 2\sum{ab\sqrt {(a^2 + 2bc)(b^2 + 2ac)}} \geq 2\sum{a^2b^2} + 4abc\sum{\sqrt {ab}}$
 we must prove that:$ \sum{a^4} + 4abc\sum{\sqrt {ab}} \geq 2\sum{a^2b^2} + 4abc\sum{a} < = > \sum{(a^2 - b^2)^2} - 4abc\sum{(\sqrt {a} - \sqrt {b})^2} \geq 0$
   $ \sum{(a^2 - b^2)^2} = \sum{(\sqrt {a} - \sqrt {b})^2(\sqrt {a} + \sqrt {b})^2(a + b)^2} \geq \sum{4(\sqrt {a} - \sqrt {b})^2ab(a + b)}$
 prove that:$ \sum{(\sqrt {a} - \sqrt {b})^2(4ab(a + b) - 4abc)} \geq 0$
            <=>$ \sum{(\sqrt {a} - \sqrt {b})^2ab(a + b - c)} \geq 0$
   let $ Sa = bc(c + b - a)$
       $ Sb = ac(a + c - b)$
       $ Sc = ab(a + b - c)$
 to suppose $ a \geq b \geq c$.we have $ Sc \geq 0$,$ Sb \geq 0$
  apply S.O.S we must prove that:$ Sb + Sa \geq 0 < = > ac(a + c - b) + bc(b + c - a) \geq a^2c + b^2c - 2abc \geq 0$
 equality is a=b=c


Let $ a$, $ b$, and $ c$ be nonzero, real numbers such that $ a \plus{} \frac{1}{b} \equal{} 5, b \plus{} \frac{1}{c} \equal{} 12, c \plus{} \frac{1}{a} \equal{} 13$. Find $ abc \plus{} \frac{1}{abc}$.

I played around for a while and ended up with $ abc \equal{} \frac{30}{a \plus{} b \plus{} c} \minus{} 1$ and $ (5 \minus{} a) \equal{} \frac{1}{b}, (12 \minus{} b) \equal{} \frac{1}{c}, (13 \minus{} c) \equal{} \frac{1}{a}$. Hints only, please, on where to go from here.

Sigh, I got it now (fail.) Thanks to Fifi. :P

Yeah lol..I was just about to ask that..
Whats fifi?

hello, we have $ \left(a\plus{}\frac{1}{b}\right)\left(b\plus{}\frac{1}{c}\right)\left(c\plus{}\frac{1}{a}\right)\equal{}abc\plus{}\frac{1}{abc}\plus{}a\plus{}b\plus{}c\plus{}\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}$, hence we get
$ 5\cdot12\cdot13\equal{}abc\plus{}\frac{1}{abc}\plus{}30$ and from here we get the result $ 750$.
Sonnhard.


calculus
integration
function
geometry
derivative
complex numbers
complex analysis

I am curious: in single variable calculus, can we use complex numbers in derivatives/integrals and simply treat i as a constant? I tried doing this with the integral of cos(x) by writing cos(x) = (e^ix+e^-ix)/2 and integrating that, and got (e^ix-e^-ix)/2i = sin(x) as my result! However, I am wondering whether this can always be done, and, if so, whether there is any justification for it.

Like you, i haven't made it past single variable calc yet, but I think you're right about the complex numbers. I tried a few examples and it seemed to work fine. Oh, and the term "radius of convergence" now has meaning! (an actual circle rather than just an interval on the number line)

Yes, you can use complex numbers in integrals and derivatives. It would be awfully fishy if you couldn't.

[quote="ComplexZeta"]It would be awfully fishy if you couldn't.[/quote]

Alot of things in math are fishy... until you actually understand them. Im not at that understanding phase quite yet.

Yeah..it wouldn't seem very fishy to me. Is there a simple way to justify the use of complex numbers like this?

The reason complex numbers work in the same places real numbers work is that they basically have all the same important topological properties as far as integrating along the real line is concerned.  There can be some problems in doing it, though.  For example, you probably don't want to involved complex numbers when you're integrating real functions of real numbers, such as :sqrt:(x) from -1 to 1, for example :)

But the point is that I can! It doesn't matter which path I take when evaluate [tex]\int_{-1}^1 \sqrt{x}\; dx[/tex]. If I want, I can go around an arc of a circle. Or I could go around a square. Or I could take some ridiculous path. But the only thing that matters is that I start at -1 and end at 1. That's the wonderful thing about complex numbers (or actually about analytic functions).

But they aren't talking about line integrals in the complex plane -- they're talking about real integrals with [i]i[/i] involved in them, which is different.  And you couldn't just take [i]any[/i] path -- the path through 0 is problematic, isn't it?  (Since :sqrt: isn't analytic there?)

Hm. I'm not sure about that. (I should be ashamed.) Where do we make the branch cut of [tex]\sqrt{x}[/tex] anyway?

Anyway, when you're trying to do [tex]\int_{-1}^1 \sqrt{x}\; dx[/tex], I can guarantee that you'll be using complex numbers.

Mathematically, you are able to integrate in the complex numbers over any interval... [tex]\displaystyle{\int_{-1}^{1} \sqrt{x} \; dx = \frac{2}{3} + \frac{2i}{3}}[/tex].... however I do not see a practical use for this result... the point is that it can be done.

Being able to integrate in the complex plane is really cool. Complex analysis, which studies this sort of thing, is filled with incredible results such as these:

If f is analytic within a region C and on the boundary of C in the complex plane and if a is a point inside C, then

[tex]\displaystyle{\oint_C f(z)\; dz=0}[/tex], and
[tex]\displaystyle{f(a)=\frac{1}{2\pi i}\oint_C \frac{f(z)}{z-a}\; dz}[/tex].

And for everyone who doesn't know:
Analytic means differentiable in the complex plane.

Actually, analytic means differentiable.  Functions don't have to be very nice to integrate, but they do have to be rather nice to differentiate.

Not quite. A function can be integrable even if it's not analytic ([tex]\bar{z}[/tex] is an example; its integral over a simple closed curve is twice the area enclosed by the curve). What analytic means is that it can be approximated by power series. So you might think that practically everything is analytic, but consider this function:

[tex]\displaystyle{f(x)=\begin{cases} e^{-\frac{1}{x^2}} & x\not= 0, \\ 0 & x=0.\end{cases}}[/tex]

Check that all the derivatives of f at 0 are 0. Thus at 0, f isn't approximated well by power series since power series make it look like the 0 function, which it isn't.

In terms of being able to check whether a function is analytic, here are the simple criteria, known as the Cauchy-Riemann equations: Let f(z)=f(x+iy)=u(x,y)+iv(x,y), where u and v are real-valued real functions. Then f is analytic if

[tex]\begin{eqnarray*} \frac{\partial u}{\partial x} & = & \frac{\partial v}{\partial y} \\ \frac{\partial u}{\partial y} & = & \frac{\partial v}{\partial x}.\end{eqnarray*}[/tex]

Some people use the Cauchy-Riemann equations as the definition of analyticity, whereas some people use other definitions such as the conformality of the derivative.

Although Magnara was right in the basic point that all analytic functions are integrable.  (It just doesn't go the other way.)

[quote="Magnara"][quote="ComplexZeta"]Mathematics is very rarely fishy. Complex numbers almost always work in place of real numbers unless there's a very good reason why they shouldn't. Take a look at Needham's book [i]Visual Complex Analysis[/i], which is practically an encyclopedia in the form of a textbook on complex analysis.[/quote]

I saw VCA at my local Borders.  I'm pretty sure it was the Dover edition, which means the cheaper version, but sometimes w/o diagrams and stuff.  Is it still such a great encyclopedia?[/quote]

Oops. I didn't see this before. VCA is really cool because it has so much stuff that most other complex analysis books don't have. But VCA without the diagrams would be terrible. It really is a visual sort of thing. I'm reading up on chapter 6 in VCA (the chapter on non-euclidean geometry) right now since I'm going to be giving a lecture on pseudospheres in my differential geometry class. It's not easy reading, but it's definitely understandable (more so than Coxeter's [i]Non-Euclidean Geometry[/i], in my opinion).

On a very tangential note, does anyone know where I could purchase a (wooden?) pseudosphere? I know of a website that sells Klein bottles, but I couldn't find out that sells pseudospheres.


geometric transformation
geometry proposed
geometry

Let $ABC$ be triangle with its Apollonius outer circle of excircles $(O_1)$. If $P,Q$ and $U,V$ are the touch points of the excircles $(I_c), (I_b)$ with $(O_1)$ and $BC$ respectively, prove that:$<UPH=<MQV$, where $H$ is the foot of $AH$ altitude and $M$ is midpoint of $UV$.

[quote]Apollonius outer circle of excircles [/quote]
Juancarlos, can you tell me what is this?

See:
http://mathworld.wolfram.com/ApolloniusCircle.html

The circle which touches all three excircles of a triangle and encompasses them
is often known as the Apollonius circle too.

Let $ D \equiv I_bI_c \cap BC$ be the exsimilicenter of $ (I_b) \sim (I_c).$ $ P$ and $ Q$ are the exsimilicenters of $ (I_c) \sim (O_1)$ and $ (I_b) \sim (O_1).$ Then by Monge-D'Alembert theorem $ PQ$ passes through $ D.$ Since the division $ (I_c,I_b,A,D)$ is harmonic, so is its orthogonal projection on $ BC,$ namely $ (U,V,H,D).$  Hence, $ DU \cdot DV \equal{} DH \cdot DM$ $ \Longrightarrow$ $ H,M$ are inverse points under the inversion with center $ D$ taking $ (I_b),(I_c)$ into each other as $ P,Q$ and $ U,V$ are also pairs of inverse points. Therefore since  $ D \equiv PQ \cap HM,$ we deduce that $ \angle UPH$ and $ \angle MQV$ are equal and oppositely directed.


Let $ x,y,z>0$ such that $ 3(x\plus{}y\plus{}z)\equal{}xy\plus{}yz\plus{}zx$ . Prove that :

i) $ 3(x\plus{}y\plus{}z)\ge xyz$

ii) $ xy\plus{}yz\plus{}zx\ge 27$

iii) $ x^2\plus{}y^2\plus{}z^2\ge 3\sqrt[5]{9x^2y^2z^2}$

[hide=" one"] By Cauchy

$ \left(3(x \plus{} y \plus{} z)\right)^2 \equal{} (xy \plus{} yz \plus{} zx)(yz \plus{} zx \plus{} xy) \ge \left(\sqrt {xyz}(x \plus{} y \plus{} z)\right)^2 \Leftrightarrow 9 \ge xyz$

Using that result
$ 3(x \plus{} y \plus{} z) \equal{} \sqrt {3(x \plus{} y \plus{} z)(yz \plus{} zx \plus{} xy)} \ge 3\sqrt {xyz} \ge xyz$[/hide]

[hide="two"] $ 9 \equal{} 3\left(\frac {xy \plus{} yz \plus{} zx}{x \plus{} y \plus{} z}\right) \le \frac {(x \plus{} y \plus{} z)^2}{x \plus{} y \plus{} z} \equal{} x \plus{} y \plus{} z$

$ \therefore xy \plus{} yz \plus{} zx \equal{} 3(x \plus{} y \plus{} z) \ge 27$[/hide]

solution of third
5(x²+y²+z²)≥(x²+y²+z²)+(xy+yz+zx)+3(xy+yz+zx)≥x²+y²+z²+27+81 (by second ineq) ≥5{(27×81x²y²z²)^(1/5)}=15{(9x²y²z²)^(1/5)} (by AM-GM)
∴$ x^{2} \plus{} y^{2} \plus{} z^{2}\ge 3\sqrt [5]{9x^{2}y^{2}z^{2}}$
I think ≥ has to change for ＞. (27=81 never correct)
I'm sorry about my bad English and bad numerical formula.


Find all function $ f: R\to R$ satisfy: f continuous at 0 and $ f(x)\equal{}f\left (  \frac{x^2}{x^2\plus{}1}\right )$

[quote="Narutovn"]Find all function $ f: R\to R$ satisfy: f continuous at 0 and $ f(x) \equal{} f\left ( \frac {x^2}{x^2 \plus{} 1}\right )$[/quote]

Let $ g(x)\equal{}\frac{x^2}{1\plus{}x^2}$. The equation is $ f(x)\equal{}f(g(x))$ and so $ f(x)\equal{}f(g^{(n)}(x))$ ($ n^{th}$ composition).

It's immediate to show that $ \lim_{n\to \plus{}\infty}g^{(n)}(x)\equal{}0$ and, since $ f(x)$ is continuous at $ 0$, $ f(x)\equal{}f(0)$ $ \forall x$

Hence the solution : $ f(x)\equal{}a$ $ \forall x$


Two related questions:

Let the set of all triples $ (m, n, k) \in \mathbb{N}^3$ such that $ \frac{1}{m} + \frac{1}{n} = \frac{1}{k}$ be denoted by $ S$.

(1) For how many $ k$ does this happen? Specifically, if $ K_N = \#\{k : k < N \text{ and } \exists m, n \text{ st } (m, n, k) \in S\}$, then what is the asymptotic behavior of $ K_N$ as $ N \to \infty$?

(2) For how many triples does this happen (i.e. does it eventually happen a lot of times for the same $ k$)? Specifically, if ${ A_N = \#\{(m, n, k) \in S : \text{ and } m, n < N}\}$, then what is the asymptotic behavior of $ A_N$ as $ N \to \infty$?

All solutions are $ m \equal{} (a \plus{} b)ac,n \equal{} (a \plus{} b)bc,k \equal{} abc$. We can see only $ (a,b) \equal{} 1$.
1)$ \frac{1}{2k}\plus{}\frac{1}{2k}\equal{}\frac 1k.$ (for any k).
2)Number of terms $ A_N$ is $ CNln(N)\plus{}O(N).$