ARML calculus algebra polynomial Vieta derivative

how to do this?.. this is from ARML 2003 Individual round #8

The graph of $f(x)=x^{4}+4x^{3}-16x^{2}+6x-5$ has a common tangent line at $x=p$ and $x=q$. Compute the product $pq$.

[hide="Wow, it took me awhile to figure out a non-calculus solution"] Ah, this is another one of those tricky ARML polynomial questions where we ignore the linear and constant terms.

Say the tangent line has equation $y = Ax+B$.  Then the polynomial $f(x)-Ax-B$ would have the $x$-axis as its common tangent, so then $p, q$ would both be double roots.

Ergo,

$x^{4}+4x^{3}-16x^{2}+(6-A) x-(5+B) = (x-p)(x-p)(x-q)(x-q)$

Vieta's sums give

$4 =-2p-2q \Leftrightarrow p+q =-2 \implies p^{2}+2pq+q^{2}= 4$
$-16 = p^{2}+q^{2}+4pq \implies-16 = 4+2pq \Leftrightarrow pq = \boxed{-10}$.

[b]Remark:[/b]  The statement to find the product $pq$ should scream Vieta's to a seasoned problem-solver.  The problem is to figure out what kind of polynomial would have roots $p, q$.   [/hide]

Wow, it took me too long to realize what it meant by a common tangent line..

the double roots solution is obvious once you see that though

t0rajir0u, nice solution.. very elegant.
but i still don't get why this is true

Say the tangent line has equation $y = Ax+B$. Then the polynomial $f(x)-Ax-B$ would have the $x$-axis as its common tangent, so then $p$, $q$ would both be double roots.

thanks. maybe it's because i am only a sophomore? lol

edit: oh now i get it.. $f(x)=Ax+B$ only has $2$ solutions.. which means $f(x)-Ax-B=0$ has $2$ solutions.... yeah.. now everything seems obvious that p and q are double roots

thanks again, t0rajir0u

[quote="Allan Z"]$f(x)=Ax+B$ only has $2$ solutions.. which means $f(x)-Ax-B=0$ has $2$ solutions.... [/quote]

Well, you need to rule out the case that $f(x)-Ax-B$ doesn't just have two other complex solutions.  The basic idea is that $f(x)-Ax-B$ has $p, q$ both as roots and as local extrema, which implies (either by checking signs in the neighborhood around $p, q$ or by calculus) that they are multiple roots.

The fact that the polynomial is 4th degree ensures that those are the only roots.

Is it possible to have a calculus-based solution (using derivatives, etc.)?

The calculus is [i]horrible[/i], you have to solve a very ugly system.  Basically, in order for there to be a common tangent at $p, q$, we require that

$f'(p) = f'(q) = \frac{f(p)-f(q)}{p-q}$

This is a system of two [i]cubic[/i] equations in two variables.  Don't bother.  The solution by Vieta's is much nicer.

[b]kunny:[/b]  Very nice!  It didn't occur to me to use the uniqueness of the common tangent to a given quartic, but of course when we complete the square...


inequalities
function
inequalities solved

Let x_i (i=1,..,n) (n \geq 2) be real positive numbers with sum s. Prove that for every real positive numbers a,b we have

(a+b/x_1)(a+b/x_2)...(a+b/x_n) \geq (a+nb/s)

Are you sure about the RHS? Is it not (a + nb/s) n ?
If I'm right the inequality follows directly from the convexity of he function f(x) = Ln(a+b/x) and from Jensen's inequality.

Pierre.


A ream of paper consists of 500 sheets.  If a standard box of 8$ \frac{1}{2}$'' by 11'' paper holds exactly ten reams with no extra space, and a sheet of paper is .004 inches thick, what is the box's volume in cubic inches?

This means there are 5000 sheets of paper. Each sheet is $ 0.04$ inches high, so the dimensions of the box are $ 8.5$ by $ 11$ by $ 5000 \times 0.004 \equal{} 20$. Multiplying these numbers yields $ \boxed{1870}$.

EDIT: Whoops, my mistake. Fixed.

[quote="isabella2296"]This means there are 5000 sheets of paper. Each sheet is $ 0.04$ inches high, so the dimensions of the box are $ 8.5$ by $ 11$ by $ 5000 \times 0.004 \equal{} 2$. Multiplying these numbers yields $ \boxed{187}$.[/quote]

$ 5000 \times .004 \equal{} 20$ not $ 2$, which means that your final answer needs to be multiplied by 10

the correct answer is 1870


inequalities trigonometry calculus derivative

If x and y are real and $ x^{2}\plus{}y^{2}\equal{}1$, compute the maximum value of $ (x\plus{}y)^{2}$.

Solution given: Since $ (x\plus{}y)^{2}\equal{}x^{2}\plus{}y^{2}\plus{}2xy\equal{}1\plus{}2xy$, $ \frac{x^{2}\plus{}y^{2}}{2}\geq \sqrt(x^{2}y^{2})\equal{}xy$ Thus, $ xy \leq 1/2$. So, $ 1\plus{}2(1/2)\equal{}2$

My solution: $ (x\plus{}y)^{2} \leq (x^{2}\plus{}y^{2})(1\plus{}1)$ (Cauchy's inequality)
                  $ (x\plus{}y)^{2} \leq (1)(1\plus{}1)$
                  $ (x\plus{}y)^{2} \leq 2$

I got the same answer, but im not sure if the process is right.

As far as I can see, yes, that solution does work.

[hide="On an unrelated note"]
This is my 500th post! I hope the majority my posts have been worthwhile and useful to the members of this forum.
[/hide]

Remember to state when equality holds.

You could also use trig to solve this:

Let $ x \equal{} \cos{\theta}$ and $ y \equal{} \sin{\theta}$. Then we have $ (x \plus{} y)^2 \equal{} (\cos{\theta} \plus{} \sin{\theta})^2 \equal{} \cos^2{\theta} \plus{} 2\sin{\theta}\cos{\theta} \plus{} \sin^2{\theta} \equal{} 1 \plus{} \sin{2\theta} \le 2$. Equality holds when $ \theta \equal{} \frac {\pi}{4}$ so when $ x \equal{} y \equal{} \frac {\sqrt {2}}{2}$. We see that we obtain the desired maximum value after we plug it into our expression.

[quote="TachyonPulse"]Remember to state when equality holds.[/quote]
Not exactly necessary... except you do have to show that your maximum is actually attainable for it to be a maximum. For example, I can say that $ 1\minus{}x^2 \le 2$, but 2 is definitely not the maximum of $ 1\minus{}x^2$.

Sorry for my random tangent.

You can assume that $ x,y$ are $ \ge 0$, because if they are not, the same values are got when we take $ \minus{}x$ and/or $ \minus{}y$

Absolutely brutal method:

Let $ f(x,y)\equal{}x^2\plus{}2xy\plus{}y^2\plus{}a(x^2\plus{}y^2\minus{}1)$

Using the method of Lagrange multipliers, we only need to check the boundaries (which are 0 and 1, because of the condition) and values for which partial derivatives of $ f$ are 0

Partial derivatives: for x - $ 2x\plus{}2y\plus{}2ax\equal{}0$, for y - $ 2x\plus{}2y\plus{}2ay\equal{}0$, for a - $ x^2\plus{}y^2\minus{}1\equal{}0$, which is the condition

This shows that the maximum is derived either at the boundaries or at the values for which the above equations hold. They hold for $ x\equal{}y$, so $ x\equal{}y\equal{}\frac{1}{\sqrt2}$. Now check the maximum, and it is when $ x\equal{}y$

Absolutely short method:

We need to find the maximum of $ 2xy$ (expand the square of the binom). But, $ 2xy\le x^2\plus{}y^2$ with equality for $ x\equal{}y$, so $ x\equal{}y$

[hide="Not so brutal Lagrange"]
Doing the stuffs we have $ 2(x\plus{}y)\equal{}2x\lambda\equal{}2y\lambda$ so $ x\equal{}y$
[/hide]


Time for some AIME. I've solved some of these

1. x, y, z are real numbers greater than 1 and w is a positive real number. If log_x w = 24, log_y w = 40 and log_xyz w = 12, find log_z w. (easy)

2. Find the product of the real roots of the equation x^2 + 18x + 30 = 2 :sqrt:(x^2 + 18x + 45). (pretty easy)

3.  w and z are complex numbers such that w^2 + z^2 = 7, w^3 + z^3 = 10. What is the largest possible real value of w + z?

4. What is the remainder on dividing 6^83 + 8^83 by 49?

5. 25 knights are seated at a round table and 3 are chosen at random. Find the probability that at least two of the chosen 3 are sitting next to each other.

[quote]5. 25 knights are seated at a round table and 3 are chosen at random. Find the probability that at least two of the chosen 3 are sitting next to each other.[/quote]

(25+25*21)/([size=75]25[/size]C[size=75]3[/size])=(25*22)/(25*23*4)=[b]11/46[/b].

I'm not sure if I'm correct though.

1.
1/logxyzw=

logwxyz= logwx +  logwy +  logwz = 1/12.

logwx= 1/logxw=1/24

logwy=1/logyw=1/40

1/24 + 1/40 + logwz = 1/12.
logwz=1/60

logzw=[b]60.[/b]

[quote="warpedklown"]
x  + 18x + 30 = u

u = 2(u+15)
u  = 4(u+15)
u  - 4u - 60 = 0
(u - 10)(u + 6) = 0
u = 10, -6

x  + 18x + 30 = 10
x  + 18x + 20 = 0
The product of the real roots is 20, since 18  - 4*20 > 0

x  + 18x + 30 = -6
x  + 18x + 36 = 0
The product of the real roots is 36, 18  - 4*36 > 0

Product of all the real roots = 20*36 = 720
[/quote]
I don't think that's correct...plug u=-6 back into your equation with u's, and you get -6 = 2(3)...

That's a very nice solution for three, thanks.

Here's another problem, if someone wants to work it...

Let x be the square of some integer k such that x has a units digit of one when written in bases 7, 8, and 9. Calculate the probability that the units digit of x is also one when x is written in decimal form. .

Yeah it's 20, whoops, I had done this problem before...

[quote="MagnusRex"]Let x be the square of some integer k such that x has a units digit of one when written in bases 7, 8, and 9. Calculate the probability that the units digit of x is also one when x is written in decimal form.[/quote]

k :^2: = x

x :equiv: 1 (mod 7)
k :equiv: 1, 6 (mod 7)

x :equiv: 1 (mod 8)
k :equiv: 1, 3, 5, 7 (mod 8)

x :equiv: 1 (mod 9)
k :equiv: 1, 8 (mod 9)

x :equiv: 1 (mod 504)

...I'll work on it later.  All you need to see is that x has to be a perfect square that will leave a remainder of one when divided by 504, and from there on it should be easier.

I do think there is an easier method than what I just tried, I call mine the "dumba** method."

Another method for 6^83+8^83 (mod 49) that is less ingenious but more natural

EulerPhi[49]=7*EulerPhi[7]=7*6=42.

So 6^83+8^83 \equiv 6^41+8^41. But 6x=1 (mod 49), has solution x=-8, so 6^41=-8. Similarly 8^41=-6. So 6^41+8^41 \equiv -8-6=-14=35 (mod 49).

it might take a while to type, but here's an outline.

From the first equation, you get sinx*cosx=1/10

Play around with the 2nd equation to get log10*(sin^2(x) + cos^2(x) + 2cosx*sinx) =log n.  Then substitute, and you get n=12.

Can you post the problem or something? (or were u talking about number 1 here or something..?)

hello wrote:I don't understand how you do problem 4 on the AIME 2003 1 with logs.
Master Confuted wrote:I can do that. Did it last night, actually.

The problem:

Given that log10(sin(x))+log10(cos(x))=-1 and that log10(sin(x)+cos(x))=(1/2)(log10(n) - 1), find n.

The rest is in a spoiler.

[hide]

First we apply some log rules to the first eqn.

log10(sin(x)*cos(x))=-1

sin(x)*cos(x)=10^(-1)=1/10

Then we do such things to the second eqn so it becomes awesome. (Note: this is not a technical term)

log10(sin(x)+cos(x))=(log10(:sqrt:(n)) - 1/2)

(sin(x)+cos(x))=:sqrt:(n)/:sqrt:(10)

Square it.

sin2(x)+cos2(x)+2sin(x)*cos(x)=n/10

Use identities and substitute.

1+2*(1/10)=n/10

1+2/10=n/10

12/10=n/10

n=12[/hide]

That wasn't that hard, it looked intimidating to start with though.

[hide]Find n such that log sin x + log cos x = -1, log (sin x + cos x) = (log n - 1)/2.

log cosx sinx = -1

cosx sinx = 1/10

2 log (sin x + cos x) = log n - 1

log (sin x + cos x) :^2: = log n - 1

log (sin :^2: x + 2 cos x sin x + cos :^2 x) = log n - 1

log (1 + 1/5) = log 6/5 = log n - 1

6/5 = n/10

n = 12[/hide]


calculus
integration
trigonometry
complex numbers
calculus computations

Integrate
(a) $e^{x}\cos x dx$
(b) $-e^{x}\cos x dx$

By parts twice: let $u=e^{x}\Rightarrow du=e^{x}dx$ and $dv=\cos(x)dx\Rightarrow v=sin(x)$, so $\int e^{x}\cos(x)dx=e^{x}\sin(x)-\int e^{x}\sin(x)dx$, now on the second integral let $w=e^{x}\Rightarrow dw=e^{x}dx$ and $dz=\sin(x)dx\Rightarrow z=-\cos(x)$, so now we have

$\int e^{x}\cos(x)dx=e^{x}\sin(x)-\left(-e^{x}\cos(x)-\int-e^{x}\cos(x)dx\right)$
so $\int e^{x}\cos(x)dx=e^{x}\sin(x)+e^{x}\cos(x)-\int e^{x}\cos(x)dx$. Solve for the integral to get $2\int e^{x}\cos(x)dx=e^{x}\sin(x)+e^{x}\cos(x)\Rightarrow \int e^{x}\cos(x)dx=\frac{1}{2}\left(e^{x}\sin(x)+e^{x}\cos(x)\right)+C$

Do the second one in a similar fashion.

Well, the second one is just the same problem with a minus sign - no reason to do any work.

Two other methods:

Undetermined coefficients:

$\int e^{x}\cos x\,dx=Ae^{x}\cos x+Be^{x}\sin x+C$

Differentiate both sides, and equate like terms. Solve the resulting two linear equations for $A$ and $B.$

Complex numbers:

$\int e^{x}\cos x\,dx=\text{Re}\left(\int e^{(1+i)x}\,dx\right).$

OK. Now what about $e^{x}\cosh x$? I think by parts yield a contradiction...?

There's no contradiction- you're just mishandling signs.

You know, $\cosh x=\frac12(e^{x}+e^{-x})$; why not just distribute?

Why?

I think $\int e^{x}\cosh x dx = e^{x}\sinh x-\int e^{x}\sinh x dx$
and $\int e^{x}\sinh x dx = e^{x}\cosh x-\int e^{x}\cosh x dx$

Put that together and subtract $\int e^{x}\cosh x\,dx$ from both sides to get

$0=e^{x}\sinh x-e^{x}\cosh x+C$

But $\sinh x-\cosh x=\frac12(e^{x}-e^{-x}-e^{x}-e^{-x})=-e^{-x}$

so that now reads $0=-1+C,$ which is true but contains no useful information. There's no contradiction here, but there's also not a way to compute the integral.


Let $ A \equal{}$ $ \{$$ a^2 \plus{} a \plus{} 1,a^2 \plus{} a \plus{} 2,...,(a \plus{} n \plus{} 1)^2 \plus{} (a \plus{} n \plus{} 1) \plus{} 1$ $ \}$, where $ a,n$ are natural numbers.Prove that we can take $ x,y$ from $ A$ such that there exist the integers $ a,b,c$ with $ a \plus{} b \plus{} c \equal{} 0$ and $ 2xy \equal{} a^2 \plus{} b^2 \plus{} c^2$

Are there first and second variables "a" same?

you write $ A_{a,n}\equal{}[a^2\plus{}a\plus{}1,(a\plus{}n\plus{}1)^2\plus{}(a\plus{}n\plus{}1)\plus{}1]\cap\mathbb{N}$
if it's true for $ A_{a,0}$ then it's true for $ A_{a,n}$.


Prove that there is no integer $ a$ for which $ a^{2} \minus{} 3a \minus{} 19$ is divisible by $ 289$ . - This question carries 15 marks . This is the 2nd problem in the paper.

Again sorry for posting it again ..didnt notice keyree10 posting it anyway.
here is it http://www.mathlinks.ro/viewtopic.php?t=314837

A genius minded person's answer...
Assume 17 divides the thing. Subtract 17a and add 119 to get that 17 divides (a-10)whole square. Let a = 17k + 10
Substitution gives the expression = 289K + 51 for some K which is not divisible by 289. Thus proved


Let's take the real positif numbers $\ a,b,c$ whith the sum $\ 1$.Prove the inequality
$\sum_{cyc}\sqrt{\frac{c}{a(b-2)-c}}\geq\frac{4}{3}\cdot\frac{1}{a^2+b^2+c^2-abc}$.

This ineq is wrong.

in case $a=b=c=\frac{1}{3}$

the inner part of sqrt be negative.


Show that the number $x$ is rational if and only if three distinct terms that form a geometric progression can be chosen from the sequence
\[x, ~ x+1, ~ x+2,~ x+3,\ldots . \]

Hmm, this should work.  :)

[hide="Solution"]We first show that the statement holds if $x$ is rational. Let $x=m/n.$ Me must now find non-negative integers $a,$ $b,$ and $c$ such that
\[\frac{x+a}{x+b}=\frac{x+b}{x+c}\implies \frac{an+m}{bn+m}=\frac{bn+m}{cn+m}.\]
Let's start by testing the simplest case; $a=0.$ Next, let $b=km$ for some positive integer $k.$ We must now find a positive integer $c$ such that
\[\frac{bn+m}{cn+m}=\frac{an+m}{bn+m}=\frac{1}{nk+1}.\]
This is equivilant to showing
\[cn+m=m\left(\frac{cn}{m}+1\right)=m(nk+1)^{2}=m(n^{2}k^{2}+2nk+1).\]
Hence, $c=m(nk^{2}+2nk).$ and we are done.

Now we prove the only if statement. For the sake of contradticion, assume that
\[\frac{x+a}{x+b}=\frac{x+b}{x+c}\]
for some irrational $x$ and some nonnegative integers $a,b,c.$ It follows that
\[(x+b)^{2}=(x+a)(x+c)\implies 2bx+b^{2}=(a+c)x+ac.\]
Since $x$ is irrational, $2bx$ and $(a+c)x$ are also irrational, and we must have
\[2bx=(a+c)x~~\text{and}~~b^{2}=ac.\]
However, this implies that $a=c,$ a contradiction.
[/hide]

[hide]wlog a>b>c, for the geometric sequence, we have

$(x+b)^{2}=(x+a)(x+c)$, or $2bx+b^{2}=(a+c)x+ac$, $x=\frac{ac-b^{2}}{a+c-2b}$

so obviously $x$ is rational, for the other way,

we need a solution in distinct integers $a$, $b$, $c$, for $ac-b^{2}=m$ and $a+c-2b=n$...i don't know here...there should be a solution...[/hide]

[hide="If"]Let $x+a, x+b, x+c$ be the terms in the geometric sequence. Then we have $\frac{x+a}{x+b}=\frac{x+b}{x+c}$, which leads to $x^2+(a+c)x+ac=x^2+2bx+b^2 \rightarrow (a+c-2b)x=b^2-ac$ and $x=\frac{b^2-ac}{a+c-2b}$, which is rational since $a,b,c \in \mathbb{N}$[/hide]

[hide="Only if"]Let $x=\frac{m}{n}$, where $m,n \in \mathbb{N}$. Then it suffices to prove that there exist positive integers $a,b,c$ such that $\frac{m}{n}+a$, $\frac{m}n+b$, and $\frac{m}n+c$ form a geometric progression. Multiplying the desired sequence by $n$ gives $m+an$, $m+bn$, $m+cn$ as terms in the sequence, so it suffices to show that there exist positive integers $a,b,c$ such that $m+an,m+bn,m+cn$ form a geometric progression. Choosing $a=0,b=m,c=2m(n+2)$ accomplishes this;

$\frac{m+an}{m+bn}=\frac{m}{m+mn}=\frac{1}{1+m}=\frac{1+m}{1+2m+m^2}=\frac{m(1+m)}{m(1+2m+m^2)}=\frac{mn+m}{m+n(2m+mn)}=\frac{m+bn}{m+cn}$.

Therefore, $m+an,m+bn,m+cn$ form a geometric progression, and so do $\frac{m}{n}+a,\frac{m}{n}+b,\frac{m}{n}+c$ as desired [/hide]

[size=40]2500th post[/size]

$\Longleftarrow$: If the GP terms are $a,aq,aq^2$ (WLOG take it to be increasing), then $aq-a=m\land aq^2-aq=n\implies q={n\over m}\implies a={m\over q-1}\in\mathbb{Q}$, and as $a=x+K$ for some natural $K$, the claim follows.


function calculus derivative integration algebra domain

Prove that \[ \left( C_P \minus{} C_V \right) \left( \frac{\partial^2 T}{\partial p \partial V} \right) \plus{} \left( \frac{\partial C_p}{\partial p} \right)_V \left( \frac{\partial T}{\partial V} \right)_p \minus{} \left( \frac{\partial C_V}{\partial V} \right)_p \left( \frac{\partial T}{\partial p} \right)_V \equal{} 1 .\]

So, what would be the plan of attack of choice here? I'm asking only for the plan, since one would write a full solution iff his life depended on it.

BTW, this is under the heading "Thermodynamics", but it's hard to figure out why...

Are you assuming ideal gas conditions? Then $ C_p$ and $ C_V$ are independant of $ p$, $ V$ and $ T$.

That's the problem! I'm not! :noo: The most general definitions must be used, and that causes pain...

[quote="perfect_radio"]The most general definitions must be used[/quote]
Could you clarify that a bit?

$ C_V \equal{} \left( \frac{\delta Q}{dT} \right)_V$ and $ C_p \equal{} \left( \frac{\delta Q}{dT} \right)_p$. That's ugly :blush:

[quote="Shreyas_patankar"]Then $ C_P$ and $ C_V$ are independant of $ p$, $ V$ and $ T$.[/quote]

Wrong! Both heat capacities depend on T, because internal energy and enthalpy also depend on T.

[quote="Perfect_radio"]That's ugly[/quote]

Why ugly? Those are just definitions. Also, it would be better to consider instead

$ C_P \equal{} \left(\frac {\partial H}{\partial T}\right)_P$ and $ C_P \equal{} \left(\frac {\partial U}{\partial T}\right)_V$

The key to solve this problem is to start with Mayer's Relation,

$ C_P \minus{} C_V \equal{} T\left(\frac{\partial P}{\partial T}\right)_V \left(\frac{\partial V}{\partial T}\right)_P$

wow Carcul, that was a lot of help. :rotfl:
I mean when you see the difference of the heat capacities, the first thing that comes to your mind is the Robert-Mayer equation.... :wink:

we believe that you know it, but please could you show us something that is actually no trivial and useful here :?:

no offence there... :)

Yes, you are right, that was not a big help. This is a purely mathematical problem. As a side note, I think it would be a good idea to start by solving problems easier than this one.

Here's a much simpler exercise that may help you undertand how these kind of relations can be attacked:

Show that $ \left(\frac{\partial C_P}{\partial P}\right)_T \equal{} \minus{}T \left(\frac{\partial ^2 V}{\partial T^2}\right)_P$.

(the purpose of this is to see that, if you also find this exercise hard, then the posted one by Perfect_radio is harder)

yeahnigga, a nicer tone would help!

Anyway, I have a question about Robert Mayer's relation. Carcul says that it's \[ C_p \minus{} C_V \equal{} T\left(\frac{\partial P}{\partial T}\right)_V \left(\frac{\partial V}{\partial T}\right)_P ,\] whilst in my book it is proved that \[ C_p \minus{} C_V \equal{} \left( \left( \frac{\partial U}{\partial V} \right)_T \plus{} p \right) \left( \frac{\partial V}{\partial T} \right)_p .\]
Are the two equivalent? If yes, then how :?:

Yes, the two are equivalent. Just compute $ \left(\frac{\partial U}{\partial V}\right)_T$ starting from the fundamental relation,

$ dU \equal{} TdS \minus{} PdV$,

and then use the differential definition of the Helmholtz function,

$ dA \equal{} \minus{}SdT \minus{} PdV$.

Sorry for the rude tone there Carcul, hope no offence made... :)

Anyway, so here is something that look similar to me, maybe it can help:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=142614[/url]

Yes, the basic idea to solve these two problems (the one posted by Perfect_radio and the one posted by me) is on that post. The only difference is that the first gives more work. However, for Perfect_radio's problem you will also need to show Mayer's Relation.

So, are both problems already solved or are there still any doubts? Several related problems can be found [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=185862]here[/url].

[quote="perfect_radio"]Prove that
\[ \left( C_P - C_V \right) \left( \frac {\partial^2 T}{\partial p \partial V} \right) + \left( \frac {\partial C_p}{\partial p} \right)_V \left( \frac {\partial T}{\partial V} \right)_p - \left( \frac {\partial C_V}{\partial V} \right)_p \left( \frac {\partial T}{\partial p} \right)_V = 1 .
\]
So, what would be the plan of attack of choice here? I'm asking only for the plan, since one would write a full solution iff his life depended on it.

BTW, this is under the heading "Thermodynamics", but it's hard to figure out why...[/quote]

Starting with $ \Delta U = Q -p\Delta V$

We have  $ Q - \Delta U = p\Delta V$

We know that $ Q=C_p\Delta T\, ;\,\Delta U = C_V\Delta T$ .

It follows that $ \left(C_p - C_V\right)\Delta T = p\Delta V$ .

Let $ \Delta T = T - T_0\,,$  and let $ \Delta V = V - V_0$ .

Now,

	 \[ \frac{\partial^2}{\partial p \partial V}\left[\left(C_p - C_V\right)\Delta T\right]=\frac{\partial^2}{\partial p \partial V}\left(p\Delta V\right)
\]

On the RHS, $ \frac{\partial \Delta V}{\partial V} =\frac{\partial V}{\partial V} = 1\,\Rightarrow \text{ RHS } =1$ .

	 \[ \frac{\partial^2}{\partial p \partial V}\left[\left(C_p - C_V\right)\Delta T\right]= 1
\]

Now, it is merely a matter of differentiating the LHS.

	\[ \frac{\partial^2}{\partial p \partial V}\left[\left(C_p - C_V\right)\Delta T\right]=\left(C_p - C_V\right)\frac{\partial^2}{\partial p \partial V}\Delta T +\frac{\partial}{\partial V}\left(\frac{\partial C_p}{\partial p}\Delta T -\frac{\partial C_V}{\partial p}\Delta T\right)+\frac{\partial}{\partial p}\left(\frac{\partial C_p}{\partial V}\Delta T -\frac{\partial C_V}{\partial V}\Delta T\right)
\]

	\[ \frac{\partial^2}{\partial p \partial V}\Delta T = \frac{\partial^2 T}{\partial p \partial V}
\]

	\[ \left(\frac{\partial C_p}{\partial V}\right)_p=0
\]

	\[ \left(\frac{\partial C_V}{\partial p}\right)_V=0
\]

	\[ \therefore\frac{\partial^2}{\partial p \partial V}\left[\left(C_p - C_V\right)\Delta T\right]=\left(C_p - C_V\right)\frac{\partial^2 T}{\partial p \partial V} +\left(\frac{\partial C_p}{\partial p}\right)_V\left(\frac{\partial T}{\partial V}\right)_p-\left(\frac{\partial C_V}{\partial V}\right)_p\left(\frac{\partial T}{\partial p}\right)_V
\]

Finally,

	\[ \boxed{\left(C_p - C_V\right)\frac{\partial^2 T}{\partial p \partial V} +\left(\frac{\partial C_p}{\partial p}\right)_V\left(\frac{\partial T}{\partial V}\right)_p-\left(\frac{\partial C_V}{\partial V}\right)_p\left(\frac{\partial T}{\partial p}\right)_V=1}
\]

[quote="Carcul"]Dr. No, although you have reached the final expression, your reasoning is wrong from the start. First, why is $ \Delta U \equal{} Q \minus{} p \Delta V$? Second, why are $ Q \equal{} C_p \Delta T$ and $ \Delta U \equal{} C_v \Delta T$? And third, why are zero those partial derivatives of the heat capacities?[/quote]

What are you talking about?!?  :mad:

$ \Delta U \equal{} Q \minus{} p \Delta V$ is the first law of thermodynamics.

[quote="Carcul"]Second, why are $ Q \equal{} C_p \Delta T$ and $ \Delta U \equal{} C_v \Delta T$?[/quote]

Because $ C_p\,\&\, C_V$ are defined such that, $ C_V \equal{}\frac{\partial U}{\partial T}\text{ and } C_p \equal{}\frac{\partial Q}{\partial T}$.

[quote="Carcul"] ...why are zero those partial derivatives of the heat capacities?[/quote]

Because at constant pressure $ C_p$ is constant -- that's what the subscript "p" means.  Likewise, at constant volume $ C_V$ is constant -- hence subscript "V."

Dr. No, the first law of thermodynamics is $ \Delta U \equal{} q \plus{} w$ (assuming there is no mass transfer). Only for a constant pressure process we can say that $ w \equal{} \minus{} P \Delta V$.

[quote="Dr. No"]Because $ C_p\,\&\, C_V$ are defined such that, $ C_V \equal{} \frac {\partial U}{\partial T}$ and $ C_p \equal{} \frac {\partial Q}{\partial T}$[/quote]

For $ C_V$ that definiton is right, but then only if $ U \equal{} U(T)$ we can write $ dU \equal{} C_V\,dT$, and then only if $ C_V \equal{} constant$ we can arrive at $ \Delta U \equal{} C_V \Delta T$. For $ C_p$ that definition is wrong, because as heat is not a state function you cannot differentiate it. The correct statement is $ C_P \equal{} \left(\frac {\delta Q}{dT}\right)_P$.

[quote="Dr. No"]Because at constant pressure $ C_p$ is constant -- that's what the subscript "p" means. Likewise, at constant volume $ C_V$ is constant -- hence subscript "V."[/quote]

Wrong again. The heat capacities are state functions, so $ C_P \equal{} C_P(P, V)$ and $ C_V \equal{} C_V(P,V)$. If one of the variables remains constant and the other changes, then also the heat capacity changes.

[quote="BanishedTraitor"]That is indeed wrong.

We may express $ C_p$ and $ C_v$ in the form $ \alpha + \beta T + \gamma T^{2}$. Cleary even if $ P$ or $ V$ are constant, we can change $ T$ to change the value of the heat capacities.[/quote]

Perhaps I should have been a bit more explicit. $ C_p$ is not a function of volume, at constant pressure. Likewise, $ C_V$ is not a function of pressure at constant volume.

Explicitly,

$ C_V\doteq (\frac{\delta Q}{dT})_V$

$ C_p\doteq (\frac{\delta Q}{dT})_p$

Consider, $ dU = \delta Q -pdV$

$ \Rightarrow \frac{\partial U}{\partial T} = \frac{\delta Q}{dT}-p\frac{dV}{dT}$

At constant volume, $ \frac{dV}{dT} = 0 \Rightarrow (\frac{\partial U}{\partial T})_V = (\frac{\delta Q}{dT})_V \equiv C_V$

Likewise consider, $ dH = dU +pdV +Vdp$

Substituting $ dU = \delta Q -pdV$ into the above, $ dH = \delta Q + Vdp$

$ \Rightarrow \frac{\partial H}{\partial T} =\frac{\delta Q}{dT} +V\frac{dp}{dT}$

At constant pressure, $ \frac{dp}{dT}=0 \Rightarrow (\frac{\partial H}{\partial T})_p = (\frac{\delta Q}{dT})_p \equiv C_p$

At this point it should be evident that $ \frac{\partial C_p}{\partial V} =\frac{\partial^2 H}{\partial V\partial T}|_p = \frac{dp}{dT}|_p \equiv 0$

Likewise, $ \frac{\partial C_V}{\partial p} = \frac{\partial^2 U}{\partial p\partial T}|_V = -\frac{dV}{dT}|_V \equiv 0$

[b]With this my original thesis is proved [u]unequivocably[/u]![/b]

Yes Kunny, but that's only for an ideal gas (and considering molar heat capacities).

[quote="kunny"]$ C_p \minus{} C_V \equal{} R$[/quote]

That is true, or rather, $ C_p \minus{} C_V \equal{} nR$, where $ n$ is the mass in moles, but the original problem statement is completely general; it does not concern ideal gases per se. To make that observation in the context of the given problem muddies the waters a bit, which I believe is why it was not done in the problem statement [i]a priori[/i].

Let's return to the original subject of this topic. Since no one has tried the hints I gave above, here is a solution that makes use of the second law:

[hide="Click here"]Start by consider the entropy S has a function of pressure P and volume V. Since S is a state function, we can write

$ dS \equal{} \left(\frac{\partial S}{\partial P}\right)_V dP \plus{} \left(\frac{\partial S}{\partial V}\right)_P dV$

Now using the chain rule we can write

$ dS \equal{} \left(\frac{\partial S}{\partial T}\right)_V \left(\frac{\partial T}{\partial P}\right)_V dP \plus{} \left(\frac{\partial S}{\partial T}\right)_P \left(\frac{\partial T}{\partial V}\right)_P dV$

$ \equal{}> dS \equal{} \frac{C_V}{T} \left(\frac{\partial T}{\partial P}\right)_V dP \plus{} \frac{C_P}{T} \left(\frac{\partial T}{\partial V}\right)_P dV \equal{}$

Now, as S is a state function then Schwartz's Theorem must hold, and so we must have

$ \left(\frac{\partial ^2 S}{\partial P \partial V}\right) \equal{} \left(\frac{\partial ^2 S}{\partial V \partial P}\right) \equal{}> \left[\frac{\partial}{\partial V}\left[\frac{C_V}{T} \left(\frac{\partial T}{\partial P}\right)_V\right]\right]_P \equal{} \left[\frac{\partial}{\partial P} \left[\frac{C_P}{T} \left(\frac{\partial T}{\partial V}\right)_P\right]\right]_V$

$ \equal{}> \left(\frac{\partial C_V}{\partial V}\right)_P \cdot \frac{1}{T} \left(\frac{\partial T}{\partial P}\right)_V \minus{} \frac{C_V}{T^2} \left(\frac{\partial T}{\partial V}\right)_P \left(\frac{\partial T}{\partial P}\right)_V \plus{} \frac{C_V}{T} \left(\frac{\partial ^2 T}{\partial V \partial P}\right)$

$ \equal{} \left(\frac{\partial C_P}{\partial P}\right)_V \cdot \frac{1}{T} \left(\frac{\partial T}{\partial V}\right)_P \minus{} \frac{C_P}{T^2} \left(\frac{\partial T}{\partial P}\right)_V \left(\frac{\partial T}{\partial V}\right)_P \plus{} \frac{C_P}{T} \left(\frac{\partial ^2 T}{\partial P \partial V}\right)$

Now multiply both sides by T and collect similar terms to get

$ \left(C_P \minus{} C_V\right) \left(\frac{\partial ^2 T}{\partial P \partial V}\right) \plus{} \left(\frac{\partial C_P}{\partial P}\right)_V \left(\frac{\partial T}{\partial V}\right)_P \minus{} \left(\frac{\partial C_V}{\partial V}\right)_P \left(\frac{\partial T}{\partial P}\right)_V \equal{}$

$ \equal{} \frac{C_P \minus{} C_V}{T} \left(\frac{\partial T}{\partial V}\right)_P \left(\frac{\partial T}{\partial P}\right)_V \equal{} \frac{C_P \minus{} C_V}{T \left(\frac{\partial V}{\partial T}\right)_P \left(\frac{\partial P}{\partial T}\right)_V}$

But according to Mayer's relation, $ C_P \minus{} C_V \equal{} T \left(\frac{\partial V}{\partial T}\right)_P \left(\frac{\partial P}{\partial T}\right)_V$, and so

$ \left(C_P \minus{} C_V\right) \left(\frac{\partial ^2 T}{\partial P \partial V}\right) \plus{} \left(\frac{\partial C_P}{\partial P}\right)_V \left(\frac{\partial T}{\partial V}\right)_P \minus{} \left(\frac{\partial C_V}{\partial V}\right)_P \left(\frac{\partial T}{\partial P}\right)_V \equal{} 1$

as we wanted.[/hide]

[quote="Carcul"]Let's return to the original subject of this topic. Since no one has tried the hints I gave above, here is a solution that makes use of the second law:
[/quote]

That works too  :yup:

[quote="Carcul"]Dr. No, the first law of thermodynamics is $ \Delta U \equal{} q \plus{} w$ (assuming there is no mass transfer). Only for a constant pressure process we can say that $ w \equal{} \minus{} P \Delta V$.[/quote]

I don't wish to keep beating a dead horse here, but I must clarify this bit.

It is simply not true that $ w \equal{} \minus{}P\Delta V$ only for constant pressure processes.

Allow me to retreat just a bit from thermodynamics to simple mechanics.

Work is defined as: $ W\doteq\int_{0}^{x}\vec{F}\cdot\, d\vec{s}$.

Therefore, $ \frac{dW}{dV} \equal{} \frac{d}{dV}\int_{0}^{x}\vec{F}\cdot\, d\vec{s}\equal{} F\left(x\right)\frac{\partial x}{\partial V}$

$ \frac{\partial x}{\partial V} \equal{} \frac{1}{\frac{\partial V}{\partial x}} \equal{} \frac{1}{A}$

$ \therefore \frac{dW}{dV} \equal{}\frac{F}{A}$

Pressure is defined as: $ p\doteq\frac{F}{A}$. This definition implies $ dW \equal{} pdV$ in general, which refutes the claim that this relation is valid only for constant pressure processes.

Now returning to thermodynamics, Carcul, you claim that the correct form of the first law of thermodynamics is $ \Delta U \equal{} q \plus{} w$.

If that is true then $ d\left(\Delta U\right) \equal{}\delta q \plus{} dw$

Let $ \Delta U \equal{} U\left(S,V\right)\minus{}U_0$ then $ d\left(\Delta U\right) \equal{} dU\left(S,V\right)\minus{}dU_0$.

$ dU_0 \equal{} 0$ and $ dU\left(S,V\right) \equal{} \frac{\partial U}{\partial S}dS \plus{}\frac{\partial U}{\partial V}dV$.

Therefore, we have $ \frac{\partial U}{\partial S}dS \plus{}\frac{\partial U}{\partial V}dV \equal{} \delta q \plus{} dw$.

Now we can expand the differentials on the right hand side as:

$ \delta q \equal{} \frac{\delta q}{\partial S} dS \plus{}\frac{\delta q}{\partial V}dV$

$ dw \equal{} \frac{\partial w}{\partial S} dS \plus{}\frac{\partial w}{\partial V}dV$

Ergo, $ \frac{\partial U}{\partial S}dS \plus{}\frac{\partial U}{\partial V}dV  \equal{} \left\{\frac{\delta q}{\partial S}\plus{}\frac{\partial w}{\partial S}\right\}dS \plus{}\left\{\frac{\delta q}{\partial V}\plus{}\frac{\partial w}{\partial V}\right\}dV$

$ \Rightarrow \frac{\partial U}{\partial S}\equal{} \left\{\frac{\delta q}{\partial S}\plus{}\frac{\partial w}{\partial S}\right\}$, and

$ \frac{\partial U}{\partial V} \equal{}\left\{\frac{\delta q}{\partial V}\plus{}\frac{\partial w}{\partial V}\right\}$.

$ \delta q \equal{} T \delta S\Rightarrow \frac{\delta q}{\partial S} \equal{} T\, ;\,\frac{\delta q}{\partial V} \equal{} T\frac{\partial S}{\partial V} \equal{} 0$.

Likewise, from the derivation made above, $ dw \equal{} pdV\Rightarrow\frac{\partial w}{\partial V} \equal{} p\,;\,\frac{\partial w}{\partial S} \equal{} p\frac{\partial V}{\partial S}\equal{}0$

Substituting these relations into those directly above them:

$ \Rightarrow \frac{\partial U}{\partial S}\equal{} T\text{ and }\frac{\partial U}{\partial V} \equal{} p$

Putting these back into the original (differential) statement of the 1st law:

$ dU \equal{} \frac{\partial U}{\partial S}dS \plus{}\frac{\partial U}{\partial V}dV \equal{} TdS \plus{} pdV$.

 :| [i]Oops, one of us has the wrong sign on the second term on the RHS.[/i]

I submit that the correct statement of the 1st law of thermodynamics is:

$ dU \equal{} \delta Q \minus{} dW \equal{} TdS \minus{} pdV$, where $ dW \equal{} pdV$ in [b]all[/b] cases.

Well isn't there anyone else who knows Thermodynamics? :roll:

Go ahead post your thoughts Carcul. I mean sooner or later I will check up with other sources and see which one's right.

EDIT: I think Carcul is right. I just consulted a text book which states that $ w \equal{} \minus{} P. \Delta V$ only if we are considering $ P \minus{} V$ work and that the pressure is constant.

[quote="Carcul"]If anyone is interested, I can make some remarks about the incorrect statements that have just been posted.

Also for those interested, I can provide a simple reasoning that shows the statements $ \left(\frac {\partial C_P}{\partial V}\right)_P \equal{} 0$ and $ \left(\frac {\partial C_V}{\partial P}\right)_V \equal{} 0$ to be wrong.[/quote]

Give it your best shot, Big Guy.

I don't want to get into a flame war with you, but given the nature of your challenges I find it nearly unavoidable. It has been said that you are a teacher, as am I; therefore, might I suggest that we each bear in mind the learning of those who read what we write? rather than trying cast aspersions upon each other.

[quote="Dr. No"]might I suggest that we each bear in mind the learning of those who read what we write[/quote]

That's precisely why I was trying (without success) to call your atention to what you write. Saying incorrect statements to students should not be the behaviour of a teacher (specially when giving an university course).

And I am not trying to annoy you, I don't know why you say that. Having someone that is ready to make critics about us is good and should be seen as a good oportunity to learn, not to fight.

About your post above:

1) The first law of thermodynamics is a result that was originally made on the basis of experimental observations regarding [i]finite[/i] processes. Here's a statement of the first law that is not perhaps very known: [i]"When a system changes from an initial state 1 to a final state 2, the sum of the work W with the heat Q, which the system exchanges with its surroundings, is determined solely by states 1 and 2: it does not depend on the intermediate process. This means that there is a certain quantity Z, such that its difference between states 1 and 2 is given by $ Z_2 \minus{} Z_1 \equal{} W \plus{} Q$. This quantity is called 'internal energy' and is simbolized by 'U', and so the above is written as $ U_2 \minus{} U_1 \equal{} \Delta U \equal{} W \plus{} Q$".[/i]
Then, when we apply the law to an infinitesimal process we can then write $ dU \equal{} \delta q \plus{} \delta W$.

2) The work W and the heat Q are not state functions, so partial derivatives of W or Q have no meaning. Part of your error is precisely there.

3) You say that "correct statement of the 1st law of thermodynamics is: $ dU \equal{} \delta Q \plus{} dW \equal{} TdS \minus{} pdV$", but what is the sense of giving such an equation for the first law if it contains the function entropy S, which only appears as a consequence of the second law?

4) You also say that "$ dW \equal{} pdV$ in [b]all[/b] cases". What is true is that $ \delta W \equal{} \minus{} PdV$ for every reversible process, but for an irreversible change of state that does not hold. Also, what you said in the other post was $ W \equal{} \minus{} P \Delta V$ for all cases, which is not the same as $ \delta W \equal{} \minus{} PdV$ (see the difference between an infinitesimal process and finite one). Therefore, only if the pressure remains constant and if the process is reversible then we can write $ W \equal{} \minus{} P \Delta V$. Otherwise, the form of the state equation $ P \equal{} f(T,V)$ must be known in order for integration to be possible.

Regarding the heat capacities:

$ \left(\frac{\partial C_V}{\partial P}\right)_V \equal{} \left(\frac{\partial C_V}{\partial T}\right)_V \left(\frac{\partial T}{\partial P}\right)_V \equal{} \left(\frac{\partial C_V}{\partial T}\right)_V \cdot \frac{\kappa_T}{\alpha}$

Now, for a process at constant volume its reasonable to accept that $ C_V$ changes with temperature. One can also show that the compressibility factor is always positive, and, as far as I know, $ \alpha$ cannot be infinite. The conclusion follows. For $ C_P$, a similar reasoning can be made.

[quote="Dr. No"]might I suggest that we each bear in mind the learning of those who read what we write[/quote]

[quote="Carcul"]That's precisely why I was trying (without success) to call your atention to what you write. Saying incorrect statements to students should not be the behaviour of a teacher (specially when giving an university course).

And I am not trying to annoy you, I don't know why you say that. Having someone that is ready to make critics about us is good and should be seen as a good oportunity to learn, not to fight.[/quote]

As I said before, I don’t mind criticism when that criticism is legitimate. The whole point of doing mathematics is to discover the truth and acknowledge it. Further it is not a matter of ego. When I make mistakes, and I have, I admit them and try to learn something from them. Nonetheless, your criticisms of my arguments ever since we first encountered one another have lacked substance. That might be my own fault. Perhaps I was not communicating clearly. If so, forgive me, please. But in the overall, unless I say something that is blatantly false, I have no need of being upbraided by you.  I am not one of your students, and I am not some 14 year old prodigy who comes here for help with his homework.  I am an engineer, and a part-time teacher of physics and physical science at the local community college. My specialty is physics, not math; as a result, what I post is often written with a practical eye.  I don’t get bent out of shape if someone writes $ \Delta$ when they really mean $ \delta$ or God forbid, $ d$, and it doesn’t bother me if someone differentiates an integral function with respect to the same variable over which it was integrated. These kinds of distinctions do not indicate one’s intelligence, they indicate that one is more interested in being a nuisance than a pedagogue.

[quote="Carcul"]About your post above:

1) The first law of thermodynamics is a result that was originally made on the basis of experimental observations regarding [i]finite[/i] processes. Here's a statement of the first law that is not perhaps very known: [i]"When a system changes from an initial state 1 to a final state 2, the sum of the work W with the heat Q, which the system exchanges with its surroundings, is determined solely by states 1 and 2: it does not depend on the intermediate process. This means that there is a certain quantity Z, such that its difference between states 1 and 2 is given by $ Z_2 - Z_1 = W + Q$. This quantity is called 'internal energy' and is simbolized by 'U', and so the above is written as $ U_2 - U_1 = \Delta U = W + Q$".[/i]
Then, when we apply the law to an infinitesimal process we can then write $ dU = \delta q + \delta W$.[/quote]

I have no disagreement with this statement, except maybe with the sign of the $ W$ term. I prefer to write: $ dU = \delta Q -\delta W$ I have some texts that express it as you have, and some that express it as I have. It is a question of style really, a personal choice, but for me it a lot more clear to say “The change in internal energy is equal to the heat added to the system minus the work done by the system on its surroundings” than it is to say, “The change in internal energy is equal to the heat added to the system plus the work done on the system by its surroundings.” How does one necessarily distinguish between what is work and what is heat if one puts it in this way?

[quote="Carcul"]2) The work W and the heat Q are not state functions, so partial derivatives of W or Q have no meaning. Part of your error is precisely there.[/quote]

That is untrue. It violates no mathematical principle to write the partial derivative of any function, whether said function is a thermodynamic state variable or not. If it is the case that the function is independent of the variable with respect to which it was differentiated then the partial derivative is zero. (In the circumstance where a particular partial is not defined at some point, then one may always restrict the domain to exclude the singularity.)

[quote="Carcul"]3) You say that "correct statement of the 1st law of thermodynamics is: $ dU = \delta Q + dW = TdS - pdV$", but what is the sense of giving such an equation for the first law if it contains the function entropy S, which only appears as a consequence of the second law?[/quote]

Entropy is an extrinsic variable. The second law of thermodynamics explains the behavior of entropy; but, not having appealed to the second law does not mean that entropy does not exist. Further, that the first and second laws are interdependent does not render either the first or second law invalid in the absence of an expression for the other.

[quote="Carcul"]4) You also say that "$ dW = pdV$ in [b]all[/b] cases". What is true is that $ \delta W = - PdV$ for every reversible process, but for an irreversible change of state that does not hold.[/quote]

First, the sign for $ dW$, or $ \delta W$ if you prefer, as I’ve already pointed out is a matter of personal choice. Second, irreversibility does not affect the nature of work. Work is defined as: $ W =\int \vec{F}\cdot\, d\vec{s}$. Therefore, $ \frac{dW}{dV} = \frac{\vec{F}}{A}\cdot\hat{n}$.

You are free to call $ \frac{\vec{F}\cdot\hat{n}}{A}$ either $ p$ or $ -p$ as you prefer, but it is a pressure nonetheless.  The use of $ \delta$ as in $ \delta W$ comes from the concept of virtual work, not irreversibility, which is why I prefer to write $ dW$ rather than $ \delta W$. The $ \delta$ in $ \delta Q$ implies that the differential heat is inexact, and thus in the case of an irreversible process, $ \delta Q < TdS$

[quote="Carcul"]Also, what you said in the other post was $ W = - P \Delta V$ for all cases...[/quote]

I don’t recall having put it exactly that way, but if you say I did then I probably did.

[quote="Carcul"]... which is not the same as $ \delta W = - PdV$ (see the difference between an infinitesimal process and finite one). Therefore, only if the pressure remains constant and if the process is reversible then we can write $ W = - P \Delta V$. Otherwise, the form of the state equation $ P = f(T,V)$ must be known in order for integration to be possible.[/quote]

For the record, $ \delta W = -p\delta V$ - virtual work, again. The expression $ W= -P\Delta V$ is indeed strictly valid only for isobaric (constant pressure) processes. But having said this, consider the context in which I said it. I believe it was in one the various discussions of how to explain thermodynamics to a child, and if that is the case, I, for one, am not going to confuse the issue by discussing the subtleties of differential calculus with a youngster who ostensibly has only recently learned how to multiply three-digit numbers together.  Furthermore, nothing I’ve said in any thread here should be construed as a denial of the fact that $ P=f(T,V)$

[quote="Carcul"]Regarding the heat capacities:

$ (\frac {\partial C_V}{\partial P})_V = (\frac {\partial C_V}{\partial T})_V (\frac {\partial T}{\partial P})_V = (\frac {\partial C_V}{\partial T})_V \cdot \frac {\kappa_T}{\alpha}$

Now, for a process at constant volume its reasonable to accept that $ C_V$ changes with temperature. One can also show that the compressibility factor is always positive, and, as far as I know, $ \alpha$ cannot be infinite. The conclusion follows. For $ C_P$, a similar reasoning can be made.[/quote]

Let us consider the definition:

$ C_V\doteq(\frac{\delta Q}{dT})_V = (T\frac{dS}{dT})_V$

 $ \Rightarrow \frac{\partial C_V}{\partial p} =  \frac{\partial T}{\partial p}\frac{dS}{dT}|_V$

$ T\doteq\frac{\partial U}{\partial S}\Rightarrow \frac{\partial^2 U}{\partial p\partial S}=\frac{\partial T}{\partial p}$

$ \frac{\partial U}{\partial p} = -V$

$ \frac{\partial^2 U}{\partial S\partial p} =-\frac{\partial V}{\partial S}$

$ \frac{\partial T}{\partial p}= \frac{\partial^2 U}{\partial p\partial S}=\frac{\partial^2 U}{\partial S\partial p}=-\frac{\partial V}{\partial S}$

$ \Rightarrow \frac{\partial C_V}{\partial p} = -\frac{\partial V}{\partial S}\frac{dS}{dT}|_V = 0$

It is obvious to me that you have a great love for this subject. But that does not imply that everyone must see it precisely as you do, nor does it imply that anyone with a different insight than yours is necessarily wrong.

[quote="Dr. No"]as I’ve already pointed out is a matter of personal choice.[/quote]

You are right: the sign convention depends if we take the point of view of the system or of the surroundings.

[quote="Dr. No"]Second, irreversibility does not affect the nature of work.[/quote]

No, it doesn't. But it affects its magnitude and possibility of calculating it with thermodynamics. What I as trying to say was this: $ \delta W = - PdV$ only for a reversible process, where P is the pressure [i]of the system[/i]. For an irreversible process we cannot equate the external pressure with the pressure of the system.

[quote="Dr. No"]The use of $ \delta$ as in $ \delta W$ comes from the concept of virtual work, not irreversibility[/quote]

The idea that I have is the following: $ \delta$ is used in two different contexts, 1) in identifying inexact differentials, and 2) when discussing virtual processes. However, virtual processes are commonly discussed in the context of equilibrium and stability, which was not what I was doing.

[quote="Dr. No"]$ \Rightarrow \frac {\partial C_V}{\partial p} =  \frac {\partial T}{\partial p}\frac {dS}{dT}|_V$[/quote]

Shouldn't it be

$ (\frac {\partial C_V}{\partial P})_V = \frac {\partial}{\partial P} [T (\frac {\partial S}{\partial T})_V]_V = (\frac {\partial T}{\partial P})_V (\frac {\partial S}{\partial T})_V + T \frac {\partial ^2 S}{\partial P \partial T}$

$ = (\frac {\partial S}{\partial P})_V + T \frac {\partial ^2 S}{\partial P \partial T}$

?

Actually I thought $ \delta W \equal{} \minus{}P. dV$ the $ P$ stands for external pressure  :maybe: .

Yes, that's right, but only for a reversible process, in which there is mechanical equilibrium and so equality of pressures along the process. What we are mainly interested in thermodynamics is to be able to compute quantities through the use of properties of the system, not of the surroundings.

[quote="Carcul"]

[quote="Dr. No"]$ \Rightarrow \frac {\partial C_V}{\partial p} =  \frac {\partial T}{\partial p}\frac {dS}{dT}|_V$[/quote]

Shouldn't it be

$ (\frac {\partial C_V}{\partial P})_V = \frac {\partial}{\partial P} [T (\frac {\partial S}{\partial T})_V]_V = (\frac {\partial T}{\partial P})_V (\frac {\partial S}{\partial T})_V + T \frac {\partial ^2 S}{\partial P \partial T}$

$ = (\frac {\partial S}{\partial P})_V + T \frac {\partial ^2 S}{\partial P \partial T}$

?[/quote]

Indeed you can represent it that way, but notice

	\[ \frac{\partial C_V}{\partial p} = \frac{\partial}{\partial p}(T\frac{dS}{dT})_V =(\frac{\partial S}{\partial p})_V + T\frac{\partial^2 S}{\partial p\partial T}|_V
\]

From Maxwell’s relations:

       \[ (\frac{\partial T}{\partial V})_S = -(\frac{\partial p}{\partial S})_V =\frac{\partial^2 U}{\partial p \partial V}
\]

That indicates

	\[ \frac{\partial C_V}{\partial p} = -(\frac{\partial V}{\partial T})_S - T\frac{\partial^2 V}{\partial T^2}|_S
\]

Therefore,

	 \[ (\frac{\partial C_V}{\partial p})_V = -(\frac{\partial V}{\partial T})_{S,V} - T\frac{\partial^2 V}{\partial T^2}|_{S,V} = 0
\]

As no one (including Dr. No) as noted it, here is a comment on the above. In the first line there is a slight but important error: in the partial derivative $ \left(\frac{\partial S}{\partial T}\right)_V$, the entropy is being considered as a function of T and V; therefore you cannot differentiate directly with respect to the pressure, but you must use the chain rule.

Dr No -you have quoted that Cp-Cv=n*R........

Are't you contradicting Meyer's relationship which says that Cp-Cv=R.[/quote]

It's meaningless...

There are different types of heat capacities, like molar heat capacity etc.

Dr. No used molar heat capacity, for which meyers relation is that...

Valeriummaximum, the Mayer's relation for ideal gases is given by $ C_P \minus{} C_V \equal{} nR$. However, some authors symbolize the molar heat capacities by $ C_P$ and $ C_V$ also, and in this case $ C_P \minus{} C_V \equal{} R$.


Find x if 10^(3x-8)=2^(5-x)

I don't know where to get started on this.

[hide="hint"]
Try taking the log of both sides

[hide="bigger hint"]
isolate x after you take the log

[hide="answer"]

take the log
$ 3x\minus{}8\equal{}(5\minus{}x) log2$
isolate x
$ 3x\plus{}xlog2\equal{}5log2\plus{}8$

$ x\equal{}\frac{5log2\plus{}8}{3\plus{}log2}$
[/hide]
[/hide]
[/hide]

Or you can simplify 10 to 2*5 and work from there, assuming you know the laws of exponents.

-rts2007

[hide="Alternative solution"]$ 10^{3x\minus{}8}\equal{}2^{5\minus{}x}$

$ 2^{3x\minus{}8\minus{}5\plus{}x}\cdot 5^{3x\minus{}8}\equal{}1$

$ 2^{4x\minus{}13}\equal{}5^{8\minus{}3x}$

$ 4x\ln 2\minus{}13\ln 2\equal{}8\ln 5\minus{}3x\ln 5$

$ x(4\ln 2\plus{}3\ln 5)\equal{}8\ln 5\plus{}13\ln 2$

$ \boxed{x\equal{}\frac{8\ln 5\plus{}13\ln 2}{4\ln 2\plus{}3\ln 5}}$[/hide]

Ihatepie, you probably do know how to do it but was distracted by the fact that you have $ 2$ to some power equals $ 5$ to some power.  The way to solve these equations are just by taking logs of both sides, and to solve for $ x$.


inequalities
function
inequalities unsolved

a,b,c are reals.prove that:
a^2+b^2+c^2-4(ab+bc+ac)+3(abc)^2/3>=0

The "4" should be replaced by a "2". In other words, your inequality should be

$a^{2}+b^{2}+c^{2}-2\left(ab+bc+ac\right)+3\sqrt[3]{\left(abc\right)^{2}}\geq 0$.

Here is my proof:

We call $x=a^{2}$, $y=b^{2}$, $z=c^{2}$. These numbers x, y, z are clearly nonnegative.

Let X = ln x, Y = ln y and Z = ln z. Then, $x=e^{X}$, $y=e^{Y}$ and $z=e^{Z}$. By the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=430]Popoviciu inequality[/url], applied to the convex function $f\left(t\right) = e^{t}$, we have

$e^{X}+e^{Y}+e^{Z}+3e^{\frac{X+Y+Z}{3}}\geq 2e^{\frac{X+Y}{2}}+2e^{\frac{Y+Z}{2}}+2e^{\frac{X+Z}{2}}$.

Using the power laws ($e^{u+v}=e^{u}e^{v}$ and $e^{uv}=\left(e^{u}\right)^{v}$), this transforms into

$e^{X}+e^{Y}+e^{Z}+3\sqrt[3]{e^{X}e^{Y}e^{Z}}\geq 2\sqrt{e^{X}e^{Y}}+2\sqrt{e^{Y}e^{Z}}+2\sqrt{e^{X}e^{Z}}$.

Equivalently,

$x+y+z+3\sqrt[3]{xyz}\geq 2\sqrt{xy}+2\sqrt{yz}+2\sqrt{xz}$.

Since $x=a^{2}$, $y=b^{2}$, $z=c^{2}$, this takes the form

$a^{2}+b^{2}+c^{2}+3\sqrt[3]{\left(abc\right)^{2}}\geq 2\left|ab\right|+2\left|bc\right|+\left|ac\right|$.

Since for any real number t, we have $\left|t\right| \geq t$, this entails

$a^{2}+b^{2}+c^{2}+3\sqrt[3]{\left(abc\right)^{2}}\geq 2ab+2bc+2ac=2\left(ab+bc+ac\right)$.

And thus,

$a^{2}+b^{2}+c^{2}-2\left(ab+bc+ac\right)+3\sqrt[3]{\left(abc\right)^{2}}\geq 0$.

$\blacksquare$

[b]EDIT:[/b] See also http://www.mathlinks.ro/Forum/viewtopic.php?t=19076 for a proof of that inequality in a dehomogenized version.

  Darij

[quote="Amir.S"] $ a,b,c$ are reals, prove that:

$ a^2 \plus{} b^2 \plus{} c^2 \plus{} 3(abc)^{\frac {2}{3}}\geq 2(bc \plus{} ca \plus{} ab).$[/quote]$ a^2 \plus{} b^2 \plus{} c^2 \minus{} 2(bc \plus{} ca \plus{} ab) \plus{} 3(abc)^{\frac {2}{3}}$

$ \equiv x^6 \plus{} y^6 \plus{} z^6 \minus{} 2\left(y^3z^3 \plus{} z^3x^3 \plus{} x^3y^3\right) \plus{} 3x^2y^2z^2$

$ \equal{} \prod{(y \minus{} z)^2} \plus{} \sum{x^2\left(x^2 \minus{} y^2 \minus{} z^2 \plus{} yz\right)^2}\geq0.$

See also : http://www.mathlinks.ro/viewtopic.php?t=50745


geometry
function
integration
calculus
calculus computations

hi im having touble with this problem.
Sketch the region enclosed by x+y^2=42 and x+y=0. Decide whether to integrate with respect to x or y. Then find the area of the region.

well switched the x and y and got the equations y=(-x^2)+42 and y=-x. then i got the intersection points x=7 and x=-6. then i did the intergration of the equations from -6 to 7 and got the equations ((-x^3)/3)+42x+((x^2)/2) computed at -6 and 7, i got 496.1666667 for my answer, is this correct?

Sketch the region enclosed by y=e^(3x), y=e^(4x), and x=1. Decide whether to integrate with respect to x or y. Then find the area of the region.
can anyone help me with this question too.

thanks for your help.

[quote="mr.yellow"]
Sketch the region enclosed by x+y^2=42 and x+y=0. Decide whether to integrate with respect to x or y. Then find the area of the region.[/quote]

Hello,mr.yellow.

No.You are wrong.

You can integrate with respect to $y$.

The intersection points of two graphs are $(y,x)=(-6,6),(7,-7)$.Since $-y^2+42\geqq -y$ at $-6\leqq y\leqq 7$.

Therefore the desired area is $\int_{-6}^7 \{-y^2+42-(-y)\}dy=-\int_{-6}^7 (y+6)(y-7)dy=\frac{1}{6}\{7-(-6)\}^3=\frac{2397}{6}=399.5$

ok i think i got the second one.
but can anyone help me with the second one.

thanks

[quote="mr.yellow"]hi im having touble with this problem.
Sketch the region enclosed by x+y^2=42 and x+y=0. Decide whether to integrate with respect to x or y. Then find the area of the region.[/quote]

mr. yellow,

1. Write $x=f(y)$ for each curve.

2. Write $g(y) =f_1(y) - f_2(y)$

3. The area of interest is then $\text{Area}=\int_{y_1}^{y_2} g(y) dy$


inequalities trigonometry Support quadratics function geometry calculus

Hi, all mathlinkers,
long time ago I posted the following problem:

[b]It is given triangle ABC with points M, N, P on its sides (internally). U, V, W are intersection points of AN and CM, BP and AN, CM and BP. Triangle UVW is equaliteral. If AM=BN=CP is it true that ABC is equaliteral?[/b]

but no one had solved it yet. It is the reason to double post it.
It was my original post:
http://www.mathlinks.ro/Forum/viewtopic.php?t=171622
it is not usefull at all but I just mention it.

This problem is that my math teacher told us.
It is Construction Approximately Equal triangle.

The detials I don't remember.I will think about it.

:) I know it is difficult. If it wasn't I would not say it is difficult.

I "feel" that the triangle is equaliteral.
Approaches to prove I have in mind are.
1) To use some lemma that you may invent and a triangle sides inequalities
(may be called vittasko's approach)
2) Trigonometry and inequalities
(as my previous problem with equaliteral triangle - http://www.mathlinks.ro/viewtopic.php?t=205626)
3) Other approach - maybe inversion or circumscribed circle and/or additional constructions...

I think 2) will be the best way to start ... but I may be wrong.

actually this problem is not as tough as it seems
yes the triangle is equilateral
this problem was in the tst2008 of [size=200] INDIA [/size]
i found 2 solutions good
i'm posting one of them
set appropriate distances and then
apply menelaus to triangle UVW and transversals APB, BMC and CNA
some algebraic manipulations gives us the solution
 :)
i'm quite lazy
 :roll:

I cannot understand you. Please be more detailed. May you scan or attach the solutions you have?

the scanner does not allow me to scan my brain
here comes the next post

I'll be happy to see well written solution to this problem. And I'll also be happy to know problem, invented by me is used in some competition.

AM=BN=CP=l
side of triangle UVW =r
i have assumed common convention
AV=a
WB=b
UC=c
 to seg ADB we get
(l-c)(a)(b)=(l-r-c)(a+r)(b+r)
i.e let abc=k
substitute c=k/(ab)
applying menelaus to other three we get symtric equations
so assume a>b
we get a contradiction
so a=b=c
you are done

Thank you for the solution. I'll check it very carefully. What material other solution uses?
What are the other problems from INDIA TST 2008. May you post them?

i told you im lazy!!!!! :)
one more thing great problem

I'm sorry. I cannot understand your solution.

1. You talk about ADB - we don't have defined a point D nowhere?!
2. Another question. How you express with your notation BM, CN, AP segments?

I'm lazy too, but it is not good. It is similar to say I should not invent new problems and post them in mathlinks.ro and people cannot use them - it is stupid. Peoples like mathlinks.ro are intended for people to support each other and to share their knowledge. I'm happy I had the ability to enter there I knew lot of new friends and made popular lots of bulgarian math problems. Please, think about it.

i dont think any of our professers (paper setters) are on mathlinks so i think it might be a double discovery
 :)

I think they should be there  - it is an interesting place.

i meant APB for ADB
a better representation
so you get the expression above now
let (a+r)(b+r)(c+r)=m
so set (a+r)(b+r)=m/(c+r)
and ab=k/c
so you get a quadratic in c
this quadratic is same as the one you get in a and b using the others as the transversal
so a=b=c

now idont think it should be a problem

You wrote:
AV=a
WB=b
UC=c

and you use a, b, c for the sides of the triangle ABC as I understand from here the confusion comes.

There is no way to apply Menelaus here without using - AP, BM, CN - how you express them?

To eliminate the confusion, please write how you get one of the three equations you use - detailed. Then write clear all three equations and how you use them and then it will be OK. I'm sorry I cannot understand you  :( .

[quote="gauravpatil"]
the inequalities can easily be verified[/quote]
I hate to say it, but gauravpatil's general attitude towards posting solutions is hardly acceptable. Almost nobody would mind occasional "obviously" and "easy to see" from a person who is almost always correct but sometimes a bit sloppy in his/her explanations but when somebody guilty of posting 2 (or even 3) wrong solutions in just one thread responds a direct request for clarification of a particular spot in his last post with "it is easy", in my humble opinion he would better not respond at all. If it is easy, show us how. Well, I know I sound like Myth today :roll: but really, gauravpatil, if this behavior continues, nobody will bother to read your posts. There are certain unwritten rules of posting on this forum and one of them is "Respond to direct questions with as clear answers as you can".

OK, enough said. So, again, why are the ratios less than $ 1/2$?

OK forget it let the three solutions be wrong
the first was a misunderstanding
the second one is BADLY wrong(my appologies) :oops:
 my appologies for the third wrong solution(this solution is correct for me altleast so im not posting the rest of the solutions) :)
also iwill give an idea
try to see angles and show that the sum of angles of $ \Delta ABC >  \pi$

EDITED sincere thanks to fejda :oops:

[quote="gauravpatil"]
also i will give an idea
try to see angles and show that the sum of angles of $ \Delta ABC > 2 \pi$[/quote]
Well, if you want to continue teasing people instead of posting your arguments, I am not interested in reading your posts any more (borislav_mirchev was surprisingly patient with you, but don't expect other people to be like him). I know at least one solution to the problem (whoever is interested in it may look at my recent post in the Proposed and Own Problems section) and I just wanted to see if you had a different valid approach from that of mine and that of Akashnil to at least one of the cases. I have neither time, nor desire to try to follow the hints that are next to impossible to understand. If you want to just indicate the idea instead of posting the full argument, it is OK, but then your post should be far more specific than this criptic remark of yours with an obvious misprint in it (I bet that regardless of what you might mean, $ 2\pi$ should be $ \pi$) and it should really help people, not just irritate them. So bye-bye, see if other people will tolerate your posting style much longer :P.

dosent matter to me whether they tolerate or not
(it dosent matter to me to not post)
i am really sorry if i hurt you
it wasnt meant to be a remark :surrender:
im once again sorry i had no meaning to hurt you  :)
please accept my appologies if i could not satisfy you (this was the meaning of my last post :( )

gauravpatil - I think you are an intelligent man - but fedja is partially right - it is not enough only to know and to have good ideas... it is valuable how you express your ideas ... I think it will be important for your future life.

Interesting think with problems like this is they are tricky ... people underestimate them and make errors (I also wrote 3-4 wrong solutions) ... these problem are looking similar but they require to be solved different approaches and lots of ingenuity... it is the reason I think they are valuable and I post them.

fedja - I think you are experienced man and you gave many valuable remarks I omited. If gauravpatil is a schoolboy don't expect so many things from him. There are probably 10 years difference in age between you and him. I'm interested to see a solution to this problem if it is possible after some time, please, post your solution. It also will be interesting to see if it is correct - it is human to err.
If it is correct I'll be very happy. And I'll learn something new.

I think there is no sence to have conflicts we are here to spend our time and to see something beautiful and interesting to collaborate not to make confrontation ... so enjoy mathlinks.

[quote="borislav_mirchev"]If gauravpatil is a schoolboy don't expect so many things from him. There are probably 10 years difference in age between you and him.[/quote]
Yes. I've been wandering about the attitude of gaurav as well. He is only in his eleventh standards (as I know. Sorry if I'm wrong). I think he has more intelligence than experience. That's the problem.

[hide="To Gaurav"]I'm actually junior to you, reading in my tenth standards. I think I had the same reason for not qualifying in the INMO, as you had for not qualifying in the TST. We lack maturity. And I'm trying to improve that. You are an asset for our country, qualifying the INMO so early. Being junior, I should not advice you. But you are a genius in my opinion. And for the sake of the country, you should listen to others. I had the misconception that in Olympiads finding the main idea behind a problem and the way towards a solution is enough. And I believed for some time that I had solved all six problems in the INMO :P , until the result came out :( . One is required to give complete fool-proof step-by-step solutions such that everyone can understand. In mathlinks I'm trying to practice that. The less people misunderstands a part of a solution, the better. Enough is said. You are much better in maths and you have the full right not to listen to me. I hope you will some day bring a gold for the country. That's all[/hide]

[quote="borislav_mirchev"]
If gauravpatil is a schoolboy don't expect so many things from him. There are probably 10 years difference in age between you and him.[/quote]
Ah, I'm afraid it is even more than that :(. Well, schoolboy or no schoolboy, one has to learn the rules some day and the earlier, the better :) What I expect is very simple and reasonable: if you have something to say, say it as clearly as you can. I'll take all my words back if there will be a single person who finds gauravpatil's last hint understandable and useful :roll:. There is just no point in making posts of that type, which essentially amount to "I can solve the problem, but I'm not telling you how; see if you can guess it! :P".
[quote]
 I'm interested to see a solution to this problem if it is possible after some time, please, post your solution. It also will be interesting to see if it is correct - it is human to err.
[/quote]
See [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=214127[/url]
[quote]
I think there is no sence to have conflicts
[/quote]
No sense, indeed. That's exactly why there are some unwritten rules here :) Anyway, I might be a little bit harsh, but I hope my point is clear ;).

You've done lots of work! Thank you. I'll study it very carefully. It seems beautiful and promising. You guess what were my intention to post for next problem.

sorry thats not my attitude
i just am to slow at typing and dont like it at all
again SOrry fejda and all here
i know about this problem of mine (almost everyone tells me this)
but i think it may take time (atleast now i have started posting semi-solutions)
before i hardly posted (the same reason)
THANKS for the friendly advices and comments AKSHANIL and borislav_mirchev
they will surely help me improve
I'M improving (and trying further :wink: )

Some ideas:

Notation: Let $ ABC$ be a triangle, $ M$, $ N$, $ P$ on $ BC$, $ CA$, $ AB$. Then $ U\in BN\cap CP$, etc.

Let $ S$ denote the center of spiral similarity that maps $ BN$ to $ CP$. Since $ BN\equal{}CP$, this spiral similarity is a rotation. The angle between $ BN$ and $ CP$ is $ 120$ so $ \angle BSC\equal{}\angle NSP\equal{}120$.

Now this uses two important parts of the diagram, but we have not used $ AM\equal{}BN$, and the other angles of the equilateral triangle.

There are a few other things that we can deduce, like $ U\in BN$, so the image of $ U$, call it $ V'$ lies on $ CP$. I am not sure if these considerations help.

Perhaps we can define other centers of spiral similarity for $ AM$ and $ BN$, for example.

The composition of a homothety and rotation with the same center, as far as I understand. You can move any 2 distinct points to any other 2 distinct points in this manner.

Fedja, I rode your solution it is rather interesting - you introduce absolutely new terms and generalizations - where you know these terms from?
I think if some of these problems is given on a math competition lots of the competitors will have no chances.

Those are just very basic analysis techniques. Any decent analysis course covers them. I don't know what exactly your educational background is but one normally learns such stuff somewhere between the last year at school and the second year in college (depending on the country, school, and other things).

As to the competitions, you just should choose a suitable one. There is no point in giving anything like that on the first round of the district school olympiad but on the Mathlinks contest, these might be some of the easiest problems. I don't think they are suitable for competitions that last just a few hours (it took me about 6 hours to find those solutions; I'm not as quick nowdays, of course, as I used to be but still...) but as a problem for a competition that lasts a couple of weeks, each of them would make sense. I remember much harder problems given at mathematical fights in Leningrad when I was a 10th grader (the last school year in Russia).

Anyway, I'm glad to hear that you liked my approach. If you have any questions about it, feel free to ask. :)

I know some techniques in analysis ... math is only my hobby I graduated university long time ago. Now I'm working in IT area. I mean it is interesting to see a clever application of analysis in problems that looks like "pure geometric".

About your approach - I'll try to understand every row written by you and it require me to be in shape. At the moment I'm under pressure - I'm starting a new job and I have not slep from few nights - it is the reason for me to not be very productive. I'll try till Monday to solve the problem - proposed by you and to invent some other problems as well - next week to publish my friend's solution or different solution invented by me.


geometry
perimeter
calculus
derivative
function
calculus computations

A piece of wire 30 m long is cut into two pieces. One piece is bent into a square and the other is bent into a circle. (Give your answers correct to two decimal places.)

(a) How much wire should be used for the circle in order to maximize the total area?

(b) How much wire should be used for the circle in order to minimize the total area?

how do i solve this?

Let $ x$ be the length of the perimeter of the circle, we have the radius is $ \frac {x}{2\pi}$ and a side length of the square is $ \frac {30 \minus{} x}{4}\ (0 < x < 30).$

If you set the derivative of the appropriate function equal to zero, you'll get one critical point in the interior of the interval that kunny just mentioned. But don't forget that you must also consider the endpoints: all circle/no square and all square/no circle.

can i see the full solution, cause i still dont get this

Next step: with $ x$ meaning what kunny defined it as (the circumference of the circle) then

$ f(x)\equal{}$ (area of circle) + (area of square)

$ f(x)\equal{}??$

(You supply the formula.)

General principle for this board: when you show us something we recognize as a fairly routine exercise (commonly assigned as homework) from a fairly typical Calculus I class, we're not going to just write out the solution for you. We need some positive participation on your part.


Prove that a rectangle with a certain perimeter has a maximum area when it is a square.

a+b=c where a and b are side lengths.

a=(c-b)

So the area is (c-b)(b)=-b^2+bc=-(b^2-bc) which is maximized when b=c/2

Thus a=c-b=c-c/2=c/2 so a=b so it is a square.

[hide]
Let the dimensions of the rectangle be $ l,w$ such that $ 2l \plus{} 2w \equal{} P$.
By AM-GM, $ P \equal{} 2l \plus{} 2w\geq 4\sqrt {lw}$. From this, we find that $ A \equal{} lw\leq\frac {P^2}{16}$, so the maximum value of $ A$ occurs when $ l \equal{} w$.
[/hide]

let x,y be 2 sides of the rectangle, so x+y=p/2 with p is its perimeter.
Using Cauchy ineq, we have xy=<(x+y)^2/4=p^2/16
Equality occurs when x=y=p/4

$ (x\plus{}n)(x\minus{}n) \equal{}x^2\minus{}n^2 \le x^2$ Equality holds when n=0.

[quote="dgreenb801"]$ (x \plus{} n)(x \minus{} n) \equal{} x^2 \minus{} n^2 \le x^2$ Equality holds when n=0.[/quote]
Too smart!I'm starting feeling that I'm a dumb...


Euler
geometry
circumcircle
geometric transformation
power of a point
radical axis

$D,E,F$ are the feet of the A-,B- and C- altitudes and $H$ is the orthocenter of $\triangle ABC.$ Circles $\omega_a, \omega_b, \omega_c$ with diameters $AU,BV,CW$ have common point $P$ and go through $D,E,F,$ respectively. Since $\overline{HA} \cdot \overline{HD}=\overline{HB} \cdot \overline{HE}=\overline{HC} \cdot \overline{HF},$ it follows that $H$ has equal power WRT $\omega_a, \omega_b, \omega_c$ $\Longrightarrow$ $\omega_a, \omega_b, \omega_c$ are coaxal. Therefore, $\omega_a, \omega_b, \omega_c$ meet again at a point $Q$ lying on $HP$ and their centers $A_0,B_0,C_0,$ lying on the sidelines $B'C',C'A',A'B'$ of the medial triangle $\triangle A'B'C',$ are collinear. Hence, the intersections $U \equiv AA_0 \cap BC,$  $V \equiv BB_0 \cap CA$ and $W \equiv CC_0 \cap AB$ are collinear $(*).$ Let $O$ be the circumcenter of $\triangle ABC$ and assume that $P \in OH.$ Denote $L$  the midpoint of $OA$ and $T$ the perpendicular foot from $U$ to $OA,$ i.e. the second intersection of $\omega_a$ with $OA.$  Then

$p(O,(U))=UO^2-UA^2=TO^2-TA^2=(TO-TA)(TO+TA)=2R \cdot LT$

$p(O,\omega_a)=OP \cdot OQ=OT \cdot R=R \left ( \frac{R}{2}- LT \right)$

$\Longrightarrow \ OP \cdot OQ= \frac{_1}{^2}[ R^2-p(O,(U))]$

Similarly, we'll get the expressions

$OP \cdot OQ= \frac{_1}{^2}[ R^2-p(O,(V))] \ , \ OP \cdot OQ= \frac{_1}{^2}[ R^2-p(O,(W))]$

We deduce that $p(O,(U))=p(O,(V))=p(O,(W))$ $\Longrightarrow$ $O$ has equal power WRT $(U),(V),(W).$ Together with $(*),$ we conclude that $(U),(V),(W)$ are coaxal and their common radical axis goes through the circumcenter of $\triangle ABC.$ Conversely, if $(U),(V),(W)$ are coaxal, let $O'$ be the intersection of $HP$ with the radical axis of $(U),(V),(W).$ We obtain by analogous reasoning that

$O'P \cdot O'Q= \frac{_1}{^2}[ O'A^2-p(O',(U))]$

$O'P \cdot O'Q= \frac{_1}{^2}[ O'B^2-p(O',(V))] $

$O'P \cdot O'Q= \frac{_1}{^2}[ O'C^2-p(O',(W))]$

Since $p(O',(U))=p(O',(V))=p(O',(W))$ $\Longrightarrow$ $O'A=O'B=O'C$ $\Longrightarrow$ $O \equiv O',$ i.e. $P \in OH.$

Hi!
This problem is also true:
We denote $M_a$ be the midpoint of $BC, (U)$ is the circle with center $U$ and radius $UM_a$. Similar for $(V)$ and $(W)$.
If $(U), (V), (W)$ are coaxal then $P$ lies on Euler line of triangle $ABC$ and if $P$ lies on Euler line of triangle $ABC, (U), (V), (W)$ are coaxal or don't have a common point.

[quote="livetolove212"] We denote $M_a$ be the midpoint of $BC, (U)$ is the circle with center $U$ and radius $UM_a.$ Similar for $(V)$ and $(W).$ If $(U), (V), (W)$ are coaxal then $P$ lies on Euler line of triangle $ABC$ and if $P$ lies on Euler line of triangle $ABC,$ $ (U), (V), (W)$ are coaxal or don't have a common point.[/quote]
Dear Linh, nice result, but  coaxal circles don't necessarily have to be secant, thus we don't have to distinguish that (U),(V),(W) are either coaxal or non-intersecting circles since they're always coaxal by [url=http://mathworld.wolfram.com/CoaxalCircles.html]definition[/url]. For convenience, rename the circles with centers $U,V,W$ that pass through $M_a,M_b,M_c$ as $\omega_1,\omega_2,\omega_3$ in order to use the same notation of my previous post. Assume $P \in OH.$ Let $N$ be the 9-point center of $\triangle ABC$ and $R$ the midpoint of $ON.$ By Stewart theorem for the cevian $UR$ in $\triangle UHO$ we have

$UR^2=\frac{_3}{^4} UO^2+\frac{_1}{^4} UH^2-\frac{_3}{^{16}}OH^2$

$p(R,\omega_1)=UM_a^2-\frac{_3}{^4}UO^2-\frac{_1}{^4} UH^2+\frac{_3}{^{16}}OH^2$

$p(R,\omega_1)=\frac{_1}{^4}( OU^2-UH^2-AH^2)+\frac{_3}{^{16}}OH^2$

$p(R,\omega_1)=\frac{_1}{^4}(UA^2-p(O,(U))-UH^2-AH^2)+\frac{_3}{^{16}}OH^2$

$p(R,\omega_1)=\frac{_1}{^4}[2HA \cdot HD -p(O,(U))]+\frac{_3}{^{16}}OH^2 \ \ (1)$

By cyclic exchange of elements we get the expressions:

$p(R,\omega_2)=\frac{_1}{^4}[2HB \cdot HE -p(O,(V))]+\frac{_3}{^{16}}OH^2 \ \ (2)$

$p(R,\omega_3)=\frac{_1}{^4}[2HC \cdot HF -p(O,(W))]+\frac{_3}{^{16}}OH^2 \ \ (3)$

We already know that $(U),(V),(W)$ are coaxal and their radical axis goes through $O,$ thus  powers of $O$ WRT $(U),(V),(W)$ are equal. Since $HA \cdot HD= HB \cdot HE= HC \cdot HF,$ then from $(1),(2),(3)$ it follows that powers of $R$ WRT $\omega_1,\omega_2,\omega_3$  are equal $\Longrightarrow$ $\omega_1,\omega_2,\omega_3$ are coaxal and their common radical axis goes through the 9-point center of $\triangle M_aM_bM_c.$

[quote="yetti"]Let $ P$ be a point in the plane of a (non-equilateral) $ \triangle ABC.$ Perpendiculars to $ AP, BP, CP$ at $ P$ cut the lines $ BC, CA, AB$ at $ U, V, W,$ respectively. ($ UVW$ is orthotransversal of $ P$ WRT $ \triangle ABC.$) Let $ (U), (V), (W)$ be circles centered at $ U, V, W$ with radii $ UA, VB, WC.$ Show that $ (U), (V), (W)$ are coaxal $ \Longleftrightarrow$ $ P$ is on Euler line of the $ \triangle ABC.$[/quote]
My proof was:
Let $A', B', C'$ be midpoints of $BC, CA, AB.$ Let $(U'), (V'), (W')$ be circles on diameters $AU, BV, CW,$ intersecting at $P.$ They have equal powers to the orthocenter $H$ of $\triangle ABC$ $\Longrightarrow$ they are coaxal with common radical axis $PH$ $\Longrightarrow$ their centers $U' \in B'C', V' \in C'A', W' \in A'B'$ are collinear $\Longrightarrow$ $\frac{\overline{UB}}{\overline{UC}} \cdot \frac{\overline{VC}}{\overline{VA}} \cdot \frac{\overline{WA}}{\overline{WB}} = \frac{\overline{U'C'}}{\overline{U'B'}} \cdot \frac{\overline{V'A'}}{\overline{V'C'}} \cdot \frac{\overline{W'B'}}{\overline{W'A'}} = 1$ $\Longrightarrow$ $U, V, W$ are collinear. Circles $(U), (V), (W)$ are internally tangent to circles $(U'), (V'), (W')$ at $A, B, C.$

Let $P \in OH.$ Let $AO, BO, CO$ cut $(U), (V), (W)$ at $X, Y, Z$ and $(U'), (V'), (W')$ at $X', Y', Z'$  $\Longrightarrow$ $\overline{AX} = 2\overline{AX'},$ $\overline{BY} = 2\overline{BY'},$ $\overline{CZ} = 2\overline{CZ'}.$ $O \in PH$ $\Longrightarrow$
$\overline{OA} \cdot \overline{OX'} = \overline{OB} \cdot \overline{OY'} = \overline{OC} \cdot \overline{OZ'}$  $\Longrightarrow$
$\overline{OX'} = \overline{OY'} =\overline{OZ'}$ $\Longrightarrow$ $\overline{AX'} = \overline{BY'} =\overline{CZ'}$ $\Longrightarrow$
$\overline{AX} = \overline{BY} =\overline{CZ}$  $\Longrightarrow$  $\overline{OX} = \overline{OY} =\overline{OZ}$  $\Longrightarrow$
$\overline{OA} \cdot \overline{OX} =  \overline{OB} \cdot \overline{OY} = \overline{OC} \cdot \overline{OZ}.$
Combined with  collinearity of their centers $\Longrightarrow$  circles $(U), (V), (W)$ are coaxal.

Conversely, let $(U), (V), (W)$ be coaxal and assume $P \not\in OH.$ Let perpendicular bisectors $OA', OB', OC'$ of $BC, CA, AB$ cut the radical axis $PH$ of $(U'), (V'), (W')$ at $R, S, T.$ Let $BR, CR$ cut $(V),(W)$ at $Y, Z$ and $(V'), (W')$ at $Y', Z'.$ $R \in PH$ $\Longrightarrow$
$\overline{RB} \cdot \overline{RY'} = \overline{RC} \cdot \overline{RZ'}$  $\Longrightarrow$ $\overline{RY'} =\overline{RZ'}$ $\Longrightarrow$
$\overline{BY'} =\overline{CZ'}$ $\Longrightarrow$ $\overline{BY} =\overline{CZ}$  $\Longrightarrow$
$\overline{RY} =\overline{RZ}$  $\Longrightarrow$ $\overline{RB} \cdot \overline{RY} = \overline{RC} \cdot \overline{RZ}$ $\Longrightarrow$
$R$ is on radical axis of $(V), (W)$ and similarly, $S, T$ are on radical axes of the circle pairs $(W), (U)$ and $(U), (V).$ Since $(U), (V), (W)$ are coaxal, $RST \equiv PH$ is their common radical axis.
Let $A^*, B^*, C^*$ be reflections of $A, B, C$ in $BC, CA, AB.$ Powers of $H$ to $(U), (V), (W)$ are
$\overline{HA} \cdot \overline{HA^*} = \overline{HB} \cdot \overline{HB^*} = \overline{HC} \cdot \overline{HC^*}$  $\Longrightarrow$
$HA^2 = HB^2 = HC^2$ $\Longrightarrow$ $\triangle ABC$ is equilateral, which contradicts the problem condition.


geometry perimeter circumcircle trigonometry

I have a question from my geometry teacher (im in 10th
grade), and he told us to prove that a triangle would be equal if
the perimeters of three triangles' made by the intersection of 3
altitudes of the triangle (H) and two of the vertices (AB, AC, BC)
are equal. But, I'm getting stuck. Please help.

I know if I find out if the altitudes bisect the angle or the side,
I would be able to prove that it is equilateral, I have been trying
for three days, but I have yet not got how to make the "first move"
on it so I can continue.

R is the circumradius, H is the orthocenter.

$a = BC = 2R \sin A,\ b = CA = 2R \sin B,\ c = AB = 2R \sin C$

If the triangle is right or acute,

$AH = 2R \cos A,\ BH = 2R\cos B,\ CH = 2R \cos C$

If the triangle is obtuse, WLOG $C > 90^\circ$,  $\cos C < 0$ and

$CH = -2R \cos C$

In any case, the angles $A, B$ are acute.

$p(\triangle BCH) = p(\triangle CAH)$

$BC + BH + CH = CA + CH + AH$

$BC + BH = CA + AH$

$2R \sin A + 2R \cos B = 2R \sin B + 2R \cos A$

$\sin A + \cos B = \sin B + \cos A$

$\cos A - \sin A= \cos B - \sin B$

$\cos A \cos 45^\circ - \sin A \sin 45^\circ = \cos B \cos 45^\circ - \sin B \sin 45^\circ$

$\cos (A + 45^\circ) = \cos (B + 45^\circ)$

$A = B$ for acute angles, because cosine is decreasing in $(45^\circ, 135^\circ)$.

If the triangle is acute, similarly, $B = C, C = A$, so the triangle is equailateral. If the triangle is obtuse with $C > 90^\circ$, then

$p(\triangle BCH) = p(\triangle ABH)$

$BC + BH + CH = AB + AH + BH$

$BC + CH = AB + AH$

$2R \sin A - 2R \cos C = 2R \sin C + 2R \cos A$

$\sin A - \cos C = \sin C + \cos A$

$\sin A - \cos A = \sin C + \cos C$

$\sin A \cos 45^\circ - \cos A \sin 45^\circ = \sin C \cos 45^\circ + \cos C \sin 45^\circ$

$\sin (A - 45^\circ) = \sin (C + 45^\circ) = \sin(135^\circ - C)$

$A = 180^\circ - C$ for acute angle A and obtuse angle C, because sine is increasing in $(-45^\circ, +45^\circ)$. Hence, $A = B = 0,\ C = 180^\circ$, so that an obtuse triangle satisfying the given conditions does not exist.


In triangle $ABC$ AM is the median, BL is the bisector of angle B and CH is perpendicular to AB. Prove that if AM,BL and CH intersect in one point then $c^{2}(c-a)=(b^{2}-a^{2})(a+c)$ where AB=c,BC=a and CA=b

let the intersect point be N. BM=MC=a/2.  MB/BA=MN/NA=(a/2)/c now, by menelaos theroem: (CM/CB).(BH/HA).(AN/NM)=1. so, (BH/HA)=a/c.
by ratio properties, BH/a=HA/c=(BH+HA)/(a+c).therefore, BH=ac/(a+c),HA=(c^2)/(a+c). since CHB and CHA are right triangles, CB^2-BH^2=AC^2-HA^2. this is equvalent to our relation.


i read some previous posts, and if you write a program on mathematica or matlab or something, and you prove that your program will give the right answer and you include the program in your proof, and the rest of the proof is perfect, is it still possible to get a 5/5 on that problem? and if you write the program, but you are unable to prove that the program will yield the right answer, then about how many points will you lose. i think of this because last year (2007-08 USAMTS round 1) on a problem, i had the right answer, and a perfect solution, but i wrote a program on matlab and at the time i didn't know that i was required to prove that it yields the right answer, and i got a 3/5, is that about right? again this has nothing to do with any of the 2008-09 round 1 problems, just a general question.... :)

This is not a question we can answer in general, as the amount of deduction will vary from problem to problem, depending on how critical the program is to the solution to the problem.

Keep in mind, the USAMTS is intended to be a math contest, not a programming contest.  Although exploration with a computer may help with you a particular problem, the problems are all designed to be solvable without the use of a computer.

Here is my advice:

Programming is, in general, more work than its worth.  I used to program to do counting problems (with javascript or java), and got 5's on all of them.  But I spent way more time programming then I spent once I started using combinations, permutations, sums and the like to count.  Casework also works very nicely.

(Also, I believe programs are graded harsher for partial credit.)

On very rare occasions, programs can also be very helpful.  When I solved 5/4/19, I had to use a program to find the answer (approximately) and then set what I knew the answer to be equivalent to and set them equal, worked some algebra to get both sides equal, and then worked backwards to find what substitutions I needed to prove the answer using only algebra and no program.  See for yourself.  I got a commendation for it :)


[quote="mahbub"]can any one help me to solve twofive and hidden password problem at section 5.5?[/quote]

http://train.usaco.org is I think expressly set up for this purpose.

However, I can help you in this case I think.

Hidden Password is a very clever DP. Also you can write an extremely efficient recursively branching data structure that just flat-out sorts stuff and [i]may[/i] run in time (it took me like 15 tries to make it efficient enough). From the Analysis:

"For each i, let V[i] be the length of the longest substring starting at i which is the lexicographically least substring in SS of that length. For example, in "abcabab", the lexicographically least 2-letter substring is "ab", the least 3-letter substring is "aba", so V[0]=2, as the suffix starting at 0 goes "abc...". "

That should get you started I think.

Twofive I have not myself solved yet, but it seems to quite obviously be some clever pre-computation. Of course, that hasn't got me anywhere yet, so who knows?

well... i also thought that.. but say if the case is:

aaaa... 1,00,000 times that is worst case,

now how can the algo be efficient? bcoz every point is a starting point.

am i missing some point?

ok at last i have done it. not that much hard after u walk in right path. i obtained a  bit diff way. the idea of repetation of substring


What is the least prime number which is the sum of two distinct positive perfect squares?

hello, i think it is $ 5$, because $ 5\equal{}1^2\plus{}2^2$
Sonnhard.


function
geometry
3D geometry
LaTeX
number theory
prime factorization
algebra

There are three classes A, B and C.
Class A has 2 more students than class B.
Class B has 1 more student than class C.
The product of the numbers of students in the three classes is 99360.
How many students are there in class A?

[hide="Hint"]The equations are
\[ a\equal{}b\plus{}2 \\ b\equal{}c\plus{}1 \\ abc\equal{}99360\][/hide]

[hide]We have a system of equations:
\[ b \equal{} a \minus{} 2 \\
c \equal{} b \minus{} 1 \\
abc \equal{} 99360
\]
Writing $ b$ and $ c$ in terms of $ a$, we have:
\[ b \equal{} a \minus{} 2 \\
c \equal{} a \minus{} 3 \\
abc \equal{} 99360
\]
Substituting in, we have:
\[ a(a \minus{} 2)(a \minus{} 3) \equal{} 99360
\]
We can expand and solve the cubic, but that would take too much work. Instead, we see that $ a$, $ a \minus{} 2$, and $ a \minus{} 3$ are all relatively close to each other, and we find that $ \sqrt [3]{99360} \approx 46.31665578$. We also know that the prime factorization of $ 99360$ is $ 2^5\cdot 3^3\cdot 5\cdot 23$.

With some intelligent educated guessing, we find that $ a \equal{} 2^4\cdot 3 \equal{} \boxed{48}$.

Or, you can use your TI-89 like me, or graph the function and have the calculator approximate the zeroes.[/hide]

took me a while but I found it out 48times46times45is 99360 and class A has the most people therefore 48 is your anwer!

[quote="123456789"][hide]We can expand and solve the cubic, but that would take too much work. Instead, we see that $ a$, $ a \minus{} 2$, and $ a \minus{} 3$ are all relatively close to each other, and we find that $ \sqrt [3]{99360} \approx 46.31665578$. We also know that the prime factorization of $ 99360$ is $ 2^5\cdot 3^3\cdot 5\cdot 23$.

With some intelligent educated guessing, we find that $ a \equal{} 2^4\cdot 3 \equal{} \boxed{48}$.[/hide][/quote]

Can you teach or explain that intelligent educated guessing O_O

I've seen many people do this, is there a trick or something to know this?  :(
because I absolutely have no idea.

Okay first lets make it simpler...

C=n
B=n+1
A=n+3

okay.. Now we're going somewhere...

So now heres our equation: n(n+1)=$ N^2$+N

Okay. then we multiply through by A. (N+3)$ N^2$+N=$ N^3$+3$ N^2$+$ N^2$+3N

Woo. Thats done. Now lets simplify...

So that would be:  $ N^3$+4$ N^2$+3N.

Okay so after that we need to factor that  :roll:

so first we need the first terms

$ (N^2,    )(N    )$

Then we need to do the outer and inner. These need to add up to $ 4N^2$

So we get:
$ (N^2,N)(N,3)$

And the las term as we wanted is 3N.

So when we factored we got:

$ (N^2,N)(N,3)$

After that i have no clue how to solve the problem sorry  :(

HEy if someone could post how to finish my work that would be great because I have absolutely no clue lolz. :rotfl:

That would be great.

[hide]Class C = X
Class B = X + 1
Class A = X + 3

We see that class C, B, and A are all very close to each other.  So we find the cube root of 99360 is 46.31665.  Then with some guess and check, we see that Class C has 45, B has 46, and class A has 48.[/hide]

lol why are you replying to this?  He posted it in february...so it was probably homework.

i got a complete solution for ur question but as am a new user i dont know how to express cubes and squares , can u tell how to type or express powers here

Well, usually people understand the caret symbol "^" as "to the power of," but the most common way to express these things is to use LaTeX. LaTeX can be used by simply embedding a formula between dollar signs like:

[code]$formula here$[/code]

For example, x^3 would become $ x^3$.

The LaTeX guide is always handy:

http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:About

let X be a natural number
let the number of students in   classA = X+3
                                          classB = X+1
                                          classC = X
given that, X(X+1)(X+3) = 99360
we can understand that,
               X^3 < X(X+1)(X+3) < (X+3)^3
               X < cuberoot of {X(X+1)(X+3)} < (X+3)

let us compare (X+1)^3 and X(X+1)(X+3)
               X^3 + 3X^2 + 3X +1 and X^3 + 4X^2 +3X
         (X^3 + 3X^2 + 3X) + 1 and (X^3 + 3X^2 + 3X) + X^2
      as X > 0, X^2 > 1 , hence
        {(X^3 + 3X^2 + 3X) + 1} <{(X^3 + 3X^2 + 3X) + X^2}
       therefore, (X+1)^3 < X(X+1)(X+3)
                     (X+1) < cuberoot  of X(X+1)(X+3) --- eq.1

let us compare (X+2)^3 and X(X+1)(X+3)
      X^3 + 6X^2 + 12X + 8 and X^3 + 4X^2 + 3X
 (X^3 +4X^2 + 3X) + (2X^2 + 9X + 8) and (X^3 + 4X^2 + 3X)
from above as (2X^2 + 9X + 8) is a positive quantity , then
(X^3 +4X^2 + 3X) + (2X^2 + 9X + 8) > (X^3 + 4X^2 + 3X)
therefore, (X+2)^3 > X(X+1)(X+3)
                  (X+2) > cuberoot of X(X+1)(X+3) ----eq.2

from eq.1 and eq.2 ,
                  (X+1) < cuberoot  of X(X+1)(X+3) < (X+2)
 as given that X(X+1)(X+3) = 99360
                  (X+1) < cuberoot of 99360 < (X+2)
                  (X+1) < 46.316 < (X+2) ---- eq.3
     as X is a natural number X+1 , X+2 are also consecutive natural numbers
then from eq.3 we know that
     (X+1) = 46
           X = 45

there fore number of students in class A = (X+3) = 48
                                             class B = (X+1) = 46
                                             class C = X =45

Santosh, I would strongly advise you to click on the link in my previous post. It is WAY easier to express your math when you can use latex.

Ex.

Instead of having to say "cuberoot ---> 3" or "cuberoot of 3" or something like that, you can simply say:

$ \sqrt[3]{3}$


if p is a prime of the form 4k+1, then prove that the sum of the quadratic residues of p in the range$ [1,p)$
is $ p(p\minus{}1)/4$

Hint 1:
[hide]
prove that if r is a residue then p-r is a residue
[/hide]

Hint 2:
[hide]
prove that there are $ (p\minus{}1)/2$ quadratic residues
[/hide]

My hints are based on quadratic reciprocity
I would love to learn other new methods

You don't need quadratic reciprocity to prove the first lemma; all you need is that $ \minus{}1$ is a quadratic residue $ \bmod p$, and all you need for that is Euler's lemma and the observation that clearly $ (\minus{}1)^{ \frac{p\minus{}1}{2} } \equiv 1 \bmod p$.

Yeah sry i posted quadratic reciprocity by mistake. :D

My first hint can be proved by elementary number theory
If r is a quadratic residue mod p then we have for some $ x$, $ x^2$ is congruent to r mod p
But we observe that $ (p\minus{}x)^2$ is congruent to $ \minus{}x^2$ mod p which is congruent to -r mod p
Which implies p-r is a quadratic residue too :D

That's not true; $ (p \minus{} x)^2 \equal{} p^2 \minus{} 2px \plus{} x^2 \equiv x^2 \bmod p$.  All you've shown is that $ x$ and $ \minus{} x$ are both square roots of $ r$, which says nothing about $ \minus{} r$.

What's not elementary about Euler's lemma?

[quote="t0rajir0u"]That's not true; $ (p - x)^2 = p^2 - 2px = x^2 \equiv x^2 \bmod p$.  All you've shown is that $ x$ and $ - x$ are both square roots of $ r$, which says nothing about $ - r$.

What's not elementary about Euler's lemma?[/quote]

well i was wrong as usual. :D ur proof is excellent.
to prove -r is a residue
what i meant was $ (-r/p)$ is 1
But $ (-r/p)$=$ (r/p)$*$ (-1/p)$
But we know $ (-1/p)$ is $ (-1^{p-1/2})$
but since p is 4k+1  $ {(-1^{p-1/2})}$ is 1
which implies -r is a quadratic residue as well.

[quote="befuddlers"]But we know $ ( \minus{} 1/p)$ is $ ( \minus{} 1^{p \minus{} 1/2})$[/quote]
This is exactly the content of Euler's lemma; $ \left( \frac{a}{p} \right) \equal{} a^{ \frac{p\minus{}1}{2} }$.  And one does not need the Legendre symbol to show that the product of two quadratic residues is a quadratic residue.  (I am only being as elementary as possible to emphasize the point that quadratic reciprocity is not necessary here.)


[quote="martini"]OK, I am trying to assist with some 5th grade Fraction work...I am stuck on a word problem subtracting fractions...anyone able to help out?[/quote]Well, the problem is ... we can't help you if the math problem is missing in your post!  :lol:  Please post the problem so we can help you because there is nothing we can help if there is no problem to be tackled.

just make the denominators the same using equivalent fractions, then subtract the numerators, and simplify and thats all there is to it.

EXAMPLE:

$\frac{70}{100}$ - $\frac{50}{100}$

Since the denominators r the same, you can subract the numerators and eventually get $\frac{20}{100}$ or $\frac{1}{5}$ in simplified terms.

uhh... martini.. would you kindly show us the problem? We would then gladly help you with it.


function
superior algebra
superior algebra unsolved

what is sqrt(14) factored into irreducibles in ring of integers Z[sqrt(14)] of Q[sqrt(14)]?

sqrt(14)--mean square root of 14.

It helps to use the norm function $ N(a\plus{}b\sqrt{14})\equal{}a^2\minus{}14b^2$. The norm is multiplicative, takes integer values, $ N(\alpha )$ is 0 if and only if $ \alpha$ is 0, and $ N(\alpha)$ is $ \pm 1$ if and only if $ \alpha$ is a unit.

In this case, $ N(\sqrt{14})\equal{}\minus{}14$ so any nontrivial factorization of $ \sqrt{14}$ would factor into two irreducibles, one with norm $ \pm 2$ and one with norm $ \mp 7$.

Here it's not hard to notice that $ 4\pm \sqrt{14}$ have norm 2, so hopefully one of them divides $ \sqrt{14}$:
\[ \frac{\sqrt{14}}{4\pm \sqrt{14}}\equal{}\frac{\sqrt{14}(4\mp \sqrt{14})}{2}\equal{}2\sqrt{14}\mp 7\]
It turns out they both do! It doesn't matter which one we choose: a factorization is
\[ \sqrt{14}\equal{}\boxed{(4\plus{}\sqrt{14})(\minus{}7\plus{}2\sqrt{14})}\]


logarithms
absolute value
complex numbers

Absolute Value of "ln(x)" = 6

and

e^x-70e^-x=3
i would know how to solve this one except there is a -x in the power for -70e^-x.

Multiply through by $e^{x}$, then make the substitution $u=e^{x}$, now you have a quadratic.

For the first one, you have $|\ln x| = 6$, so we have two eqautions $\ln(x) = 6$ or $\ln(x) =-6$, so $x=e^{6}$ or $x=e^{-6}$.

[quote="Jimmy"]Multiply through by $e^{x}$, then make the substitution $u=e^{x}$, now you have a quadratic.[/quote]
[hide] $e^{x}-70e^{-x}=3$
$e^{2x}-70=3e^{x}$
$e^{2x}-3e^{x}-70=0$
Let $u=e^{x}$
$u^{2}-3u-70=0$
$(u-10)(u+7)=0$
$u=10$ or $u=-7$
$e^{x}=10$ or $e^{x}=-7$ (no solution)
$x=\boxed{\ln10}$

EDIT. Actually, $e^{\pi i}=-1$ so $-7=e^{\ln7}e^{\pi i}=e^{\ln7+\pi i}\Longrightarrow x=\boxed{\ln7+\pi i}$[/hide]

[quote="lotrgreengrapes7926"]
EDIT. Actually, $e^{\pi i}=-1$ so $-7=e^{\ln7}e^{\pi i}=e^{\ln7+\pi i}\Longrightarrow x=\boxed{\ln7+\pi i}$[/quote]

Not quite. $-1=e^{i(\pi+2k\pi)}\implies\ln(-7)=\ln 7+i(2k+1)\pi, k\in\mathbb{Z}$. Logarithms of negative reals and of complex numbers don't have uniquely defined values.


function
modular arithmetic
number theory
totient function

1.) What is the last three digits of $ 2^{10000}$?
2.) What is the last 4 digits of $ {9^9}^9$?

[hide="The first one"]
I will use a generalization of Fermat's Little Theorem

$ a^{\phi(n)} \equiv 1 \mod n$ where phi represents Euler's totient function (coprimes leq)

Letting $ a\equal{}2$ and $ n\equal{}1000$

$ 2^{\phi(1000)} \equiv 1 \mod 1000$

Since $ \phi(1000) \equal{} \phi(5^3)\phi(2^3) \equal{} (5^3\minus{}5^2)(2^3\minus{}2^2) \equal{} 400$
So
$ 2^{400} \equiv 1 \mod 1000$
Raising both sides to the 25th

$ 2^{10000} \equiv 1 \mod 1000$
so the last three digits are $ \boxed{001}$
[/hide]

How can 2 raised to any power end in 1?

[hide]The last digits of 2 raised to a power follow the pattern: 2,4,8,6,2,4,8,6.  Since 10000 is 0 (mod 4), the last digit of 2^10000 is 6.

That's as far as I can get.[/hide]

The last digit of 2) has to be 9.
[hide="Hmmm"]
9 raised to powers.
Last 2 digits: 09 81 29 61 49 41 69 21 89 01 repeat
Note palindromic-ness of tens-place digits...
I believe the last two digits should be 89.
[/hide]
[hide="Number one"]
2 to powers last 2 digits: 02 04 08 16 32 64 28 56 12 24 48 96 92 84 68 36 72 44 88 76 52 04 ...
adding up to a loop of 20 last-digit pairs.
so the last digits are I think 76?
[/hide]

Strange how it seems so easy.

(EDIT: I meant at first.)

Number 1
@chickendude: 2 and 1000 are not coprime!
[hide]
Let $ a \equal{} 2^{10000}$, then obviously $ a\equiv 0\pmod 8$. By Euler's Theorem, $ 2^{\phi(125)}\equiv 1 \pmod {125}$, which simplifies to $ 2^100\equiv 1\pmod {125}$. Hence $ a\equiv 1 \pmod {125}$.
Let $ a \equal{} 8x$, and $ 8x \plus{} 1 \equal{} 125y$, so
$ x \equal{} 15y \plus{} \frac {5y \plus{} 1}{8}$.
Now let $ 5y \plus{} 1 \equal{} 8z$, we can quickly find $ (3,2)$ is a solution (else you can continue this process if you're so unlucky...). Plugging back in them, we find $ a\equiv 376 \pmod {1000}$.

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solve an equation
$ \sqrt{1\minus{}x^2}\plus{}2\sqrt[3]{1\plus{}x^2}\equal{}2$

[quote="devil_maths"]solve an equation
$ \sqrt {1 \minus{} x^2} \plus{} 2\sqrt [3]{1 \plus{} x^2} \equal{} 2$[/quote]

Would you try to square and then cube the equation?

(Basically raising to 6th power. But I'm only saying that to get rid of square roots.)

[quote="devil_maths"]solve an equation
$ \sqrt {1 \minus{} x^2} \plus{} 2\sqrt [3]{1 \plus{} x^2} \equal{} 2$[/quote]

There are no real roots.

http://www.quickmath.com

Equations:  Solve

From the $ \sqrt{1\minus{}x^2}$, you can tell that if there is a real root, $ x \in [\minus{}1,1]$.