geometry induction calculus derivative function trigonometry integration

this problem can either be completely ridiculous and utterly trivial or quite interesting.

prove mathematically (don't give me the fingers solution) that 1+1=2. i heard that you can prove it using induction.

if this problem is ridiculous than i have just publicly embarrassed myself. :D

Most bogus proofs of this fact involve a sneaky division by zero or involves manipulating square roots of negative numbers as if they were real square roots.  There were several threads about this a couple of months ago, I think.
EDIT: Whoa! oops, I thought you were asking about 1+1=1 or 1=2 or something like that.  I just realized that 1+1=2 is not a bogus fact.

[quote="7h3.D3m0n.117"][quote="mihail911"]im not looking for a bogus proof. i was reading somewhere that as part of an advanced (beyond calc) math class, students were asked to prove 1+1=2.
i don't really find bogus math proofs that amusing.[/quote]

Well say mathematically you have one apple. And you now have another. How many do you have total? :D

Seriously, though, I always thought the operations in math were like postulates; kinda obvious and needs not any proof to its function.[/quote]
Maybe other things less obvious are defined as postulates, and operations in math are defined in terms of those other things...

also, you still have to prove that one apple + one apple = two apples mathematically with postulates if you use apples, and then divide by apple. It's simpler to just prove $ 1 \plus{} 1 \equal{} 2$, because then you won't have to prove that $ \frac {1 APPLE \plus{} 1 APPLE}{APPLE} \equal{} 1 \plus{} 1$ and $ \frac {2 APPLE}{APPLE} \equal{} 2$

I think http://mathforum.org/library/drmath/view/51551.html mentions 5 postulates to be used for that proof.

Saying obvious things are postulates is like saying that it's a postulate that right angles are congruent just because it's obvious, but actually that's a theorem, not a postulate, and uses other less obvious postulates to prove it.

[quote="ProtestanT"][quote="CircleSquared"]You need to read G&FF threads more  :wink:
[quote="CircleSquared"]
$ \sqrt { \minus{} 1} \equal{} \sqrt { \minus{} 1}$
$ \sqrt { \minus{} 1} \equal{} \sqrt {\frac {1}{ \minus{} 1}}$
$ i \equal{} \frac {1}{i}$
$ i^2 \equal{} 1$
$ \minus{} 1 \equal{} 1$
$ \frac { \minus{} 1}{4} \equal{} \frac {1}{4}$
$ \frac {1}{2} \equal{} 1$
$ 1 \plus{} 1 \equal{} \frac {1}{2} \plus{} \frac {1}{2} \equal{} 1$
[/quote][/quote]
That doesn't prove that $ 1 \plus{} 1 \equal{} 2$...

these sites have proofs:
http://mathforum.org/library/drmath/view/51551.html
http://www.mrtc.mdh.se/~icc/1+1=2.htm
http://www.mrtc.mdh.se/~icc/principia.gif[/quote]

The point is, I'm trying to disprove it  :D

Sort of like trying to prove the [url=http://en.wikipedia.org/wiki/Parallel_postulate]parallel postulate[/url], isn't it? That is, it's an axiom - a thing we just say is true and cannot prove or disprove. We could (correct me if I'm wrong) have built mathematics on the idea that $ 1 \plus{} 1 \equal{} 3$, and created a different but completely consistent system of rules and ideas that wouldn't be "wrong", [i]but[/i] just wouldn't have been as useful as $ 1 \plus{} 1 \equal{} 2$ (that is, this hypothetical math where $ 1 \plus{} 1 \equal{} 3$ wouldn't apply to real world situations quite as readily and would really be rather unintuitive).

If you like this sort of sideways question, you might like [url=http://en.wikipedia.org/wiki/Philosophy_of_mathematics]the philosophy of mathematics[/url] in general. Paul Erdos was an interesting guy (apparently God was hiding all the beautiful proofs from us in a book, namely, The Book).

[b]EDIT:[/b] That said, I've looked over some of the set theory based proofs, so, well, yes, then you can prove $ 1 \plus{} 1 \equal{} 2$ if you want to use set-theory. However, the problem is merely displaced - you [i]then[/i] need to prove the set theorems you used to prove it are true.

Set theory's a lot less accessible than "prove $ 1 \plus{} 1 \equal{} 2$" which is probably why people don't whip out their $ \mathbb{N}$'s, $ \mathbb{Q}$s, intersections and boolean algebra and ask you to prove something about sucessors. It doesn't have quite the same cool ring to it.

Point is, eventually, you will hit axioms. I hope. Being the set weenie I am, I just took $ 1 \plus{} 1 \equal{} 2$ as an axiom.

that would be an interesting topic if our math was defined around $ 1\plus{}1\equal{}3$ such that there are no contradictions...

Well, perhaps that wasn't the [i]best[/i] example. That'd get pretty hairy, since just about every operation ever would have to be redefined to accommodate this. But it could be made consistent - it would be total rubbish and really useless, of course, compared to real mathematics, but that doesn't matter because it's playing by a different set of rules.

Actually, a good example harkens back to the parallel postulate I mentioned earlier. You can have non-Euclidean geometry systems, where the parallel postulate is substituted with a different axiom: e.g. geometrical systems where the angles of a triangle don't add up to $ 180^{\circ}$. They work. They just don't necessarily have as much use as Euciledan geometry and can be a bit finicky.

Real mathematics is just a lot of rules that happen to work quite well in concert with reality: but it's just as 'made up', really, as $ 1 \plus{} 1 \equal{} 3$ math is.

This is probably why I don't think about the philosophy of mathematics very much, but if you can understand what I'm saying you should see it's not mad. Mathematics is a "useful fiction".

CircleSquared, does your proof really hold?

No, it doesn't.
I [i]think[/i] that $ \sqrt {\frac {a}{b}} \equal{} \frac {\sqrt {a}}{\sqrt {b}}$ only if a,b are positive.

Yes, $ a \geq 0$ and $ b > 0$. Same thing stops you from being able to prove $ 1 = - 1$:

\begin{align*}
i &= \sqrt { - 1}\\
i^2 &= \sqrt { - 1} \sqrt { - 1}\\
i^2 &= - 1
\end{align*}.

No big surprise there. But if we go back to the two square roots, we can combine them:

\begin{align*}
i^2 &= \sqrt { - 1 \cdot - 1}\\
i^2 &= \sqrt {1}
\end{align*}
Therefore, $ \boxed{- 1 = 1}$.

But this makes use of a rule that isn't defined for $ < 0$.

[quote="kh-"]Sort of like trying to prove the [url=http://en.wikipedia.org/wiki/Parallel_postulate]parallel postulate[/url], isn't it? That is, it's an axiom - a thing we just say is true and cannot prove or disprove.[/quote]

More or less.  I would call it a [b]definition[/b] of the symbol $ 2$.  ([i]What meaning can you ascribe to it otherwise?[/i])

[quote="ProtestanT"]that would be an interesting topic if our math was defined around $ 1 \plus{} 1 \equal{} 3$ such that there are no contradictions...[/quote]

Then $ 3$ would just be another name for $ 2$.  Alternately, you could be working $ \bmod 1$.

[b]To everyone:[/b]  This is a somewhat silly discussion, but the meta-issue raised here is of utmost importance in the discussion of many serious issues.  Things like this can't be properly discussed without a clear understanding of the [b]definitions[/b] of the words, symbols, and so forth involved.  Please don't try to discuss fascism or communism without an [b]agreed-upon[/b] definition of those two words.  Similarly, please don't try to discuss the statement $ 1 \plus{} 1 \equal{} 2$ without a definition of what $ 1$ is, what $ 2$ is, what $ \plus{}$ is, and what $ \equal{}$ is.

One definition:  Let $ 1$ be the cardinality of the smallest nonzero set.  Let $ 2$ be the cardinality of the next smallest nonzero set.  Define $ a \plus{} b$ to mean that we take a set with cardinality $ a$ and another set with cardinality $ b$ whose intersection is the empty set and find the cardinality of their union.  Then $ 1 \plus{} 1 \equal{} 2$ is the statement that the union of two disjoint sets with cardinality $ 1$ has cardinality $ 2$.

[quote="mihail911"]prove mathematically (don't give me the fingers solution) that 1+1=2. i heard that you can prove it using induction. [/quote]

This is absurd.  The proof that induction is valid is a much stronger result (depending on which set of axioms you're using).

[quote="0123456789"]kh, are you sure that works...I thought you could only do that with the square roots of two positive numbers....[/quote]

His point is that it doesn't work, which is correct.  Read his post more carefully.

Should be absolutely no chance, if you're relying on 'proper' logic and so on, as opposed to something silly. All proofs that $ 1 \equal{} 2$, $ 1 \plus{} 1 \neq 2$ etc all rely on an invalid step at some point, e.g. division by zero, using surd rules for negative integers, and so on. To be honest, I think it's pretty much a waste of time to attempt.

[quote="0123456789"]oh, ok...does anyone one know whether a valid proof might possbly exist that shows one plus one doesn't equal two...or is there absolutely no chance?[/quote]

There is absolutely no chance because it has already been proven (and/or defined) that $ 1 \plus{} 1 \equal{} 2$, and $ 1 \plus{} 1$ can't both equal and not equal $ 2$ at the same time, so any proofs that does prove the negation would have to have been based on logic that causes contradictions since $ 1 \plus{} 1\ne 2$ contradicts $ 1 \plus{} 1 \equal{} 2$.
[hide="why it contradicts"]
It's been proven that $ 1\plus{}1\equal{}2$, so it's a given.
If $ 1\plus{}1\ne 2$ is true as well,
then $ 1\plus{}1\ne 2 \implies 2\ne 2$ by substitution.
This contradicts the definition of $ \ne$, which means not equal to.
[/hide]

[quote="kh-"]To be honest, I think it's pretty much a waste of time to attempt.[/quote]

Actually, I find that considering invalid proofs is a very handy exercise.  The better you get at spotting the mistakes in proofs that lead to obviously incorrect conclusions, the stronger your grasp of the material.  For example, there is a silly little calculus proof that $ 1 \equal{} 2$, and it proceeds as follows:

$ \underbrace{x \plus{} x \plus{} ... \plus{} x}_{x \text{ times} } \equal{} x^2 \implies$
$ \frac{d}{dx} \underbrace{x \plus{} x \plus{} ... \plus{} x}_{x \text{ times} } \equal{} \frac{d}{dx} x^2 \implies$
$ \underbrace{1 \plus{} 1 \plus{} ... \plus{} 1}_{x \text{ times} } \equal{} 2x \implies$
$ x \equal{} 2x$ !

Spotting the mistake in this proof is a good test of how well you understand the principle behind differentiation.  :)

I dont understand the principle behind differentiation well.

That's a good point. You can learn something from them.

In the false proof, I don't see how one can differentiate both sides like that. Who said anything about their gradients/rates of change being equal?

$ 9x \equal{} 3$, for example, but it is not true that $ 9 \equal{} 0$.

I'm struggling to find the word for an operation that can't be applied to both sides and maintain equality, if there is one.

$ \underbrace{x \plus{} x \plus{} ... \plus{} x}_{x \text{ times} }$ is equal to $ x^2$...so that would be like differentiating $ x^2\equal{}x^2$. They do have the same rate of change, since they're equal...

I don't think that's the error, although I don't know the real one.

How would that work if $ x$ were a fraction? Would you have a fractional amount of 1's?

You'd think that that was true, wouldn't you? And I'd be inclined to agree: however, I looked at all the steps in turn. The first step is certainly true ($ x$ $ x$ times = $ x \cdot x$ = $ x^{2}$). The third step ($ 1$ $ x$ times = $ 1 \cdot x$ = $ x$) is true, it does equal $ x$ on the LHS. The last step's the obviously wrong conclusion $ x \equal{} 2x$ for $ x \neq 0$, so I figured the only place that could have a problem in it based on this flaky logic was differentiating both sides, something which I'm not intuitively comfortable with. Of course if the error now turns out to be somewhere else that'll say a whole lot about my intuition, but there you go.

[quote="kh-"]I don't see how one can differentiate both sides like that. Who said anything about their gradients/rates of change being equal?

$ 9x \equal{} 3$, for example, but it is not true that $ 9 \equal{} 0$.[/quote]

That's because it's not true that $ 9x \equal{} 3$ for every value of $ x$!  If two [b]functions[/b] are (identically) equal, then their derivatives will also be equal.  For example, it is true that

$ \cos 2x \equal{} 2 \cos^2 x \minus{} 1$

Taking the derivative of both sides, it is also true that

$ \minus{} 2 \sin 2x \equal{} \minus{} 4 \sin x \cos x$

Your confusion here is due more to the way mathematics is taught in school than anything else, and it is that a distinction is not made between [b]equations[/b] and [b]identities.[/b]  Similarly, a distinction is not made between [b]unknowns[/b] and [b]variables.[/b]

[quote="Icy"]How would that work if $ x$ were a fraction?  Would you have a fractional amount of 1s?[/quote]

;)

This is the real error.  Keep thinking along these lines.

[quote]Your confusion here is due more to the way mathematics is taught in school than anything else, and it is that a distinction is not made between equations and identities.[/quote]
^ Consider me guilty (again) of a somewhat rookie mistake. I am aware of the distinction between identities and equations, and while not too much importance is placed on it at school, it has certainly been pointed out to me more than once.

To the best of my knowledge, and I have always been taught, that 'unknowns' and 'variables' are terms used pretty much interchangeably to mean the same thing. I'm probably going to find this isn't the case now, of course, although a quick Google on this doesn't seem to turn up much.

It's my opinion (and I am not sure how common of a viewpoint this is) that such a distinction is necessary.

An [b]unknown[/b], to me, has a fixed but unknown (!) numerical value in its domain.  We establish equations to determine the value of unknowns.

A [b]variable[/b], on the other hand, is (again, to me) a generic input into a function, and assumes every value in its domain.  We establish identities that are valid for variables.

Nice "proof", t0r . . .
[hide]
The LHS is only defined when $ x$ is a positive integer, so it isn't continuous or differentiable
[/hide]

If $ x^2$ is defined over only integers is it correct to say it has a rate of change?
Is it also correct to say that rate is $ \frac{f(x\plus{}1)\minus{}f(x)}{x\plus{}1\minus{}x}\equal{}2x\plus{}1$?

[quote="Icy"]At the risk of sounding stupid, what is the difference between an identity and an equation? I'm pretty sure I know the difference, but I'm not 100% sure.[/quote]

An identity is "identically true," that is, if it contains a variable, it is true for all inputs, and if it does not contain a variable it is simply a statement that two computations produce the same numerical value.  For example, trigonometric identities are true regardless of the angle involved, and formulas for $ \pi$ are identities.

An equation is true for some fixed but unknown value of an unknown.  For example, $ 9x \equal{} 3$ is an equation saying that some quantity $ x$ exists such that $ 9$ times that quantity is $ 3$.

[quote="Fraenkel"]If $ x^2$ is defined over only integers is it correct to say it has a rate of change?
Is it also correct to say that rate is $ \frac {f(x \plus{} 1) \minus{} f(x)}{x \plus{} 1 \minus{} x} \equal{} 2x \plus{} 1$?[/quote]

This is not called a derivative, although it has a certain relation.  This is known as the [b]forward difference operator[/b], and the resulting function is referred to as the [b]first difference.[/b]

Thanks t0rajir0u for the help, and the post you originally gave should get uber awesome points for making me think alot.

It's cool that you can evaluate forward differences using the Nörlund-Rice integral.

http://en.wikipedia.org/wiki/N%C3%B6rlund-Rice_integral


How many of the first 50 natural numbers have an odd number of factors or exactly two factors?


For how many positive integer values of $k$ is $\sqrt{3k}$ an even integer less than 50?

[hide]Since we are are looking for all even values of $\sqrt{3k}$, we can start by solving for a possible positive value for k.  A positive even number can be written as $2n$, where $n$ is a positive integer.
$\sqrt{3k}=2n$
$3k=4n^2$
Solving this equation in positive integers using your favorite method gives us that $n=0\pmod3$.  But we need positive integers, so the minimum value of n is 3.  This means that the even values of $\sqrt{3k}=2n$ can be represented as $n=0\pmod6$, starting with 6.  We need to find out the number of solutions from 6 to 50:
$\displaystyle \left\lfloor\frac{50}{6}\right\rfloor=8$

Therefore, the number of positive even integers less than 50 that can be expressed as, $\sqrt{3k}$ is $\boxed{8}$[/hide]


Keiko tosses one penny and Ephraim tosses two pennies. The probability that Ephraim gets the same number of heads that Keiko gets is

$A.$ $1/4$
$B.$ $3/8$
$C.$ $1/2$
$D.$ $2/3$
$E.$ $3/4$

[hide]Either Keiko and Ephraim must both get no heads, or one head. The probability they get no heads is $\frac{1}{2}*\frac{1}{2}*\frac{1}{2}=\frac{1}{8}$. The probability they each get one is $\frac{1}{2}*\frac{1}{2}*1=\frac{1}{4}. \frac{1}{8}+\frac{1}{4}=\frac{3}{8}.$ Answer B.[/hide]

[hide]
Either Keiko gets no heads, or one head.
No heads: $\frac{1}{2}^{3}=\frac{1}{8}$
One head: $\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}$
$\frac{1}{8}+\frac{1}{4}=\frac{3}{8}$
$B$[/hide]

I thought that Iposted the answer here..

[hide="solution:"]
1/4+1/8=3/8, B[/hide]

Answers require solution. Please don't post an answer without some explanation of how you got it.

[hide] 1/2*1/2*1/2 gives the probability of all tails, and
the probability of heads is 1/2 (from the single coin) times 2/4 (from the double). thus, 1/4. adding the two :1/8+1/4 gives 3/8, our answer. [b]B[/b]. Btw, the 2/4 was gotten from the 2nd row of pascals triangle. 1-2-1...one head, one tail is represented in the 2, thats 2 possibilities out of 4.[/hide]


induction
combinatorics unsolved
combinatorics

Let $\mathcal{T}$ be the set of spanning trees of a connected graph $G$. Let $H$ be the graph with vertex set $T$ in
which $T_1, T_2 \in \mathcal T$ are joined by an edge if and only if $|E(T_1)\bigtriangleup E(T_2)| = 2$. Show that $H$ has diameter at most $n(G)-1$.

If $n(G)$ is what I think it is, namely the number of vertices of $G$, then I don't think it's true. If you have two spanning trees which share no edge, it's impossible to turn one of them into the other one through less than $n(G)-1$ transformations of the following type: you take an edge of a tree, delete it, and add another edge s.t. the new graph is still a spanning tree.

However, I think we can show that the diameter is $\le n(G)-1$, or, in other words, given any two spanning trees $T_1,T_2$, you can turn $T_1$ into $T_2$ through $\le n(G)-1$ transformations as described above. We can prove by induction on $|E(T_1)\bigtriangleup E(T_2)|$ that we can pick an edge in $E(T_1)\setminus E(T_2)$, delete it, and replace it with some edge in $E(T_2)\setminus E(T_1)$, and the conclusion that the distance between $T_1$ and $T_2$ is $\le n(G)-1=|E(T_1)|$ will follow.

In order to perform the induction step, simply eliminate any edge in $E(T_1)\setminus E(T_2)$. This disconnects the tree, and the new graph is a forest with two connected components, $C,C'$. There must be an edge in $E(T_2)$ which connects $C,C'$, and it must be different from the one we eliminated, and thus from any edge in $E(T_1)$, meaning that it belongs to $E(T_2)\setminus E(T_1)$. Add this edge to the forest to complete a spanning tree, and we're done.

Yes, it should be $n(G)-1$.

Now the second question:

Show that every edge of $H$ is contained in a Hamilton circuit of $H$.


Hey guys!
I'm having some tough time trying to figure this one out, so I was thinking maybe some of your brains might crack it.
I have the following situation:
[img]http://www.d-fox.eu/series1.jpg[/img]
where [b]i[/b] is a positive integer, [b]i[/b] and [b]r[/b] are [u]known numbers[/u], [b]p[/b] and [b]t[/b] are [u]unknown[/u].

What I'm after is [b]p[/b] and [b]t[/b]

Any thoughts on that? Any help is greatly appreciated!
Thanks,
b

It is vaguely similar to the [url=http://en.wikipedia.org/wiki/Subset_sum_problem]subset sum problem[/url], but is simpler because the set of subsets is restricted to subintervals. Unless the sequence $ a_n$ has some structure, I don't think you can do any better than compute partial sums and compare them. If a solution $ (p,t)$ exists, it will be found in time $ O(p\plus{}t)$.

Mlok, thanks for the reply. I'll look into that.

Cheers,
b


I was reading the solution to a problem and I had trouble understanding the last line of the solution, could someone please explain it to me?

Problem: Points A, B, C are given on a line in this order. Semicircles $ w$, $ w_1$, $ w_2$ are drawn on AC, AB, BC respectively as diameters on the same side of the line. A sequence of circles ($ k_n$) is constructed as follows: $ k_0$ is the circle determined by $ w_2$ and $ k_n$ is tangent to $ w$, $ w_1$, $ k_{n\minus{}1}$ for $ n \geq 1$. Prove that the distance from the center of $ k_n$ to AB is 2n times the radius of $ k_n$.

Solution: Invert with respect to a circle with center A and squared radius AB*AC. Points B and C exchange positions, $ w$ and $ w_1$ are transformed to the lines perpendicular to BC at C and B, and the sequence ($ k_n$) to the sequence of circles ($ k'_n$) inscribed in the region between the two lines. Obviously the distance from the center of $ k'_n$ to AB is 2n times its radius.

This is line I don't really get (particularly the meaning of homothetic) ...

Since circle $ k_n$ is homothetic to $ k'_n$ with respect to A, the statement immediately follows.

Help please?

http://mathworld.wolfram.com/Homothetic.html
Hope this helps


A muddy wheel translates on a horizontal surface with $ v_0$, rolling without slipping.
The mud over the wheel is continuously being released at all points on the wheel.
It is assumed that the mud causes no change in shape, mass or velocity of the wheel.
Find the value of the maximum displacement of a mud particle from its original position.

[hide="Maximum Displacement:"]If a particle is released from a point $ P$, it rises to some height and begins to fall down. What is the maximum displacement of this particle from the point $ P$ throughout its trajectory? The question wants the highest of this value amongst all the particles on  the wheel.

[hide="Note:"]It must be noted that the point $ P$ has moved along and its [b]instantaneous position should be taken into consideration, and not the original one.[/b]
[/hide]
[/hide]

Hey! No replies as yet! No takers for this one? Is it this tough???

Weird...

i) How does the wheel translates if it's rolling without slipping?!
ii) What means the instantaneous position? The final? then the displacement would be zero?!

[quote="Immanuel Bonfils"]
i) How does the wheel translates if it's rolling without slipping?!
[/quote]

The wheel is moving, rolling without slipping. Thats all I meant to say!

[quote="Immanuel Bonfils"]
ii) What means the instantaneous position? The final? then the displacement would be zero?

[/quote]

If a particle is shot out from some point $ P$ on the wheel. The point $ P$ moves some distance while the particle too does. At some time, the particle is farther from $ P$ than it was before or will be after throughout its journey.

didnt do it seriously
let
(3/2)*((3)^(1/2))=k

then my answer came out to be

kv^2/g

[quote="Attila the Hun"]is point P rolling with the wheel or stationary :D[/quote]

$ P$ is a point [b]on[/b] the wheel!

[quote="rohan2007"]
Let,

$ \frac{3}{2}(3^{\frac{1}{2}})\equal{}k$

Then my answer came out to be

$ \frac{kv^2}{g}$
[/quote]

Can you elucidate the way you proceeded a [b]little[/b] more in detail?

Actually that was just the answer

What i did was to generalise the displacement for a particle at an angle theta
as it breaks off tangentially , it gets a velocity along the plane v0 and the velocity due to the rotational motion .

the vector sum of both gives the actual direction
applying a bit of calculus i got the answer

[quote="rohan2007"]
What i did was to generalise the displacement for a particle at an angle theta
as it breaks off tangentially , it gets a velocity along the plane $ v_0$ and the velocity due to the rotational motion .

[/quote]

That is all understood, and I did mostly the same stuff.

${ \vec{v}_{net}} = \vec{v}_0 + (\vec \omega _0 \times \vec r)$

The relations between $ \theta$ of the radius of the particle with the horizontal gives the direction of $ \vec{v}_{net}$.

But that gets way too complicated. I'd be grateful if you post, if not entire, but parts of your work, especially the relations for the displacement (the ones you'd differentiate to get maxima/minima).

I hope you are aware that the point from which the particle was projected has travelled some distance not only along $ X$ but also along $ Y$.

I'm waiting for a solution my dear friends!


A triangle $ ABC$ has $ \angle BAC > \angle BCA$. A line $ AP$ is drawn so that $ \angle PAC \equal{} \angle BCA$ where $ P$ is inside the triangle. A point $ Q$ outside the triangle is constructed so that $ PQ$ is parallel to $ AB$, and $ BQ$ is parallel to $ AC$. $ R$ is the point on $ BC$ (separated from $ Q$ by the line $ AP$) such that $ \angle PRQ \equal{} \angle BCA$.

Prove that the circumcircle of $ ABC$ touches the circumcircle of $ PQR$.

[hide="Hidden by mod"][color=darkblue]   Let $ \omega_1, \omega_2$ be the circumcircles of triangles $ ABC$ and $ PQR$ respectively. Denote $ K in AP \cap \omega_2$. Thne we have $ \angle{QKP} \equal{} \angle{QRP} \equal{} \angle{BCA} \equal{} \angle{PAC}$, so $ QK \parallel AC$, but $ BQ$ also is parallel to $ AC$ so points $ B,Q,K$ are collinear. Then $ ABKC$ is an cyclic trapezoid, so it is isosceles and $ K$ lies on $ \omega_1$. So we have $ K \in \omega_1 \cap \omega_2$. We'll prove that this to circles are tangent at $ K$.
  Let $ d$ be a line tangent to $ \omega_1$ at $ K$. Then $ \angle{(BK,d)} \equal{} \angle{BCK} \equal{} \angle{BAK} \equal{} \angle{QPK} \equal{} \angle{QRK}$, so $ d$ is also tangent to $ \omega_2$ at $ K$.[/color][/hide]

Let $\{S\}\in AP\cap BQ$, then $BACS$ is an isosceles trapezoid, hence cyclic, and $\angle ASB=\angle ACB$, but we are given $\angle PRQ=\angle ACB$, so $BPRQ$ is cyclic as well; for the tangency, same as before.

Best regards,
sunken rock


In a box, there are red, blue, and green marbles.  There are twice as many blue marbles as green marbles and three times as many red marbles as blue marbles.  The probability of drawing 2 marbles of the same color is $\frac{4}{9}$.  What is the probability of drawing 3 marbles of the same color?

[hide="hint"]If you think through the problem, it's straight-forward algebra, combinations, and reading the problem.[/hide]

[hide] The answer is 5/21

I did this problem using guess and check :lol:

First, I checked the combination 1 green marble, 2 blue marbles and 6 red marbles.

The probability of drawing 2 marbles of the same color was 4/9 in that combination. (2/9*1/8+6/9*5/8)(I must be really lucky)

Then, I found the probability of getting 3 marbles of the same color which is 5/21. (6/9*5/8*4/7)

I also checked it with Alan at school  :no: [/hide]

Hope I did the hide thing correctly

[hide="um, peijin..."]Actually, that was your 4th guess
btw, you can visit my blog.
[url=http://www.artofproblemsolving.com/Forum/weblog.php?w=756]BLOG[/url]
[/hide]
Anyone want to come up with an algebra solution? (I never did it yet)

Ohh ya... I first did 6/9*5/9+2/9*1/9 for some weird reason and got 19/27 :|

Wait, you did, I thought you first checked if 2-4-12 worked?

:wallbash:  :wallbash_red:  [hide="why?????"] I solved it completely algebraically, and you had to make the ratio 1!. Guess and check would have been so much faster (im kinda slow at algebra, especially when im short on scratch paper). Anyways, here's the algebraic solution you requested (i haven't learned LateX, so it's messy):
Let G= # of green marbles:
Therefore, # of blue = 2G, #of red =6G

Probability of drawing 2 green will then be:
(G/9G) * (G-1)/(9G-1)

Prob. of drawing 2 blue=
(2G/9G)*(2G-1)/(9G-1)

Prob of drawing 2 red=
(6G/9G)*(6G-1)/(9G-1)

Sum all of them up and simplify, and you get
(41G-9)/(9(9G-1))=(4/9)
Cross multiply and solve for G, and you get G=1. Then find prob of selecting 3 greens in a row (0) + prob of selecting 3 blues in a row (also 0) + prob of selecting 3 reds in a row= (6/9)*(5/8)*(4/7)= [b]5/21[/b][/hide]

Apparently, everyone has the answer.

[hide="My Solution Then..."]Okay, so the probability is obviously
$\frac{\binom{x}{2}+\binom{2x}{2}+\binom{6x}{2}}{\binom{9x}{2}}$

If we write out the combinations and simplify the fraction, we have:
$\frac{41x-9}{81x-9}=\frac{4}{9}$

We then cross-multiply and simplify to get:
$45=45x$
So it's obvious that x=1.  Thus the marbles are 1-2-6. The desired answer is:

$\frac{\binom{6}{3}}{\binom{9}{3}}=\frac{20}{84}=\frac{5}{21}$[/hide]

:huh:
I sorta get it...

I suck at algebra  :|


hello, for your first sum mathematica says $ \sqrt{\frac{\pi}{2}}\text{Fresnel}C[x^2]$ and the second one $ \sqrt{\frac{\pi}{2}}\text{Fresnel}S[x^2]$ and the searched limit is given by $ \sqrt{\frac{\pi}{2}}$
Sonnhard.


function
LaTeX
calculus
calculus computations

This is mainly an exercise in using the chain rule.  In the case of three functions, it says that $ (f_{1}\circ f_{2}\circ f_{3})'\equal{}f_{1}'\circ f_{2}\circ f_{3}\cdot f_{2}'\circ f_{3}\cdot f_{3}'$.

I've worked through the problem twice and gotten 2 different answers.

Can you confirm one of them:

[175t^2 + 130t - 18] / [5(5t^2 + 4t + 1)^2]

or

[2(-35t-13)(175t^2 + 130t + 23)] / [25(5t^2 + 4t + 1)^3]

Thanks

Instead, why don't you show your workings (choose the method you think is more plausible) and we can figure it out from there.

Please learn a bit of [[LaTeX]] (see also http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690 )


geometry
trigonometry
perpendicular bisector
trig identities
Law of Sines
angle bisector
area of a triangle

In triangle $ ABC$, side $ AC$ and the perpendicular bisector of $ BC$ meet in point $ D$, and $ BD$ bisects $ \angle ABC$.  If $ AD \equal{} 9$ and $ DC \equal{} 7$, what is the area of triangle $ ABD$?

$ \textbf{(A)}\ 14 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 28 \qquad \textbf{(D)}\ 14\sqrt5 \qquad \textbf{(E)}\ 28\sqrt5$

[hide="My Solution"]Let $ DX$ be the perpendicular bisector to $ BC$ where $ X \in BC.$ Let $ \angle DBX = \theta.$ Since $ \angle DXB = \angle DXC = \pi /2$ and $ BX = XC,$ triangles $ \Delta DXB$ and $ \Delta DXC$ are congruent, which implies that $ \angle DAB = 180-3\theta, \angle ADB = 2\theta, DB = 7.$ Using the Law of Sines, we see that

\[ \frac{7}{\sin{180-3\theta}}= \frac{9}{\sin \theta}\iff \sin \theta = \frac{\sqrt{5}}{3}\iff [ADB] = 14\sqrt{5}. \][/hide]

another way would be use angle bisector formula and find DB off it in terms of AB and BC, then use the fact that DB is 7 and find AB and BC with it (since you write 7 interms of AB or BC), then you can use herons to find the area of ABD

We have $ \angle ABD = \angle DBC = \angle ACB$

$ \underline{\underline{\triangle ABD, \triangle ACB}}:$

$ \begin{array}{ll}\angle ABD = \angle ACB \\ \angle BAD = \angle CAB\ \text{ (common})\end{array}\}\Rightarrow \triangle ABD\sim \triangle ACB \Rightarrow$

$ \frac{AB}{AD}= \frac{AC}{AB}\Rightarrow AB^{2}= AC\cdot AD = 16\cdot 9 \Rightarrow AB = 4\cdot 3 = 12$

Now from heron in the triangle $ ABD$ which is $ 12-7-9$ we find the answer

$ \sqrt{14(14-12)(14-7)(14-9)}=14\sqrt 5$

[quote="erdogankerem123"][quote="Karth"]
$ \frac{7}{\sin{180\minus{}3\theta}}\equal{}\frac{9}{\sin\theta}\iff\sin\theta \equal{}\frac{\sqrt{5}}{3}\iff [ADB] \equal{} 14\sqrt{5}.$
[/quote]
How did you solve for $ \sin\theta$?[/quote]

Hint: try and express $ \sin 3\theta$ in terms of $ \sin\theta$.

[quote="erdogankerem123"][quote="Karth"]
$ \frac{7}{\sin{180-3\theta}}=\frac{9}{\sin\theta}\iff\sin\theta =\frac{\sqrt{5}}{3}\iff [ADB] = 14\sqrt{5}.$
[/quote]
How did you solve for $ \sin\theta$?[/quote]

[hide="Look here"]$ \frac{7}{\sin(\pi-3\theta)}=\frac{9}{\sin\theta}$

$ \frac{7}{\sin 3\theta}=\frac{9}{\sin\theta}$

$ 9\sin 3\theta = 7\sin\theta$

Let's find a triple angle formula:
\begin{eqnarray*}\sin 3\theta &=&\sin (\theta+2\theta)\\ &=&\sin\theta\cos 2\theta+\sin 2\theta\cos\theta\\ &=&\sin\theta(1-2\sin^{2}\theta)+2\sin\theta\cos^{2}\theta\\ &=&\sin\theta-2\sin^{3}\theta+2\sin\theta (1-\sin^{2}\theta)\\ &=& 3\sin\theta-4\sin^{3}\theta\end{eqnarray*}

Hence,

$ 27\sin\theta-36\sin^{3}\theta = 7\sin\theta$

$ \sin\theta (9\sin^{2}\theta-5)=0$

So either $ \sin\theta = 0$ or $ \sin\theta =\pm\frac{\sqrt{5}}{3}$.

But the $ \sin\theta$ can't be $ 0$, or else the expression would be undefined. It can't be negative, since we are looking at angles in a triangle. Hence, $ \sin\theta =\frac{\sqrt{5}}{3}$.[/hide]

I started to use Stewart's, then I was like nah, that's too much work, then I couldn't find an easier way, then I thought an AMC 12 #23 is pretty difficult, I might as well  :D.

We find BD=7 since DK is a perp bisector. By Angle Bisector, $ \frac{AB}{9}\equal{}\frac{BC}{7}$. By Stewart's, $ 7AB^2\plus{}9BC^2\equal{}49 \cdot 16\plus{}63 \cdot 16$

After subbing in and crunching the numbers, AB=12. Then use Heron's and the answer is $ 14\sqrt{5}$.

Sorry for such an old revive - but my method will (hopefully) justify it.

For those of you wonder what stewart's theorem, heron's formula, or even the triple angle formula is (and wondering if you have to use such advanced theorems - although this is an AMC 12 #23), you could solve this a more elementary, although slightly longer way.

[hide="my elementary but long solution"]
I will outline my solution. Note that triangles (let E be the intersection of the perpendicular from D to BC and BC) BDE and BDE are similar (by SAS). Thus angle DCB congruent to angle DBC. By a version of the exterior angle theorem, angle ADB is equal to the sum of angle DBC and angle DCB.

Now introduce the angle bisector of angle ADB, letting the angle bisector intersect AB at point M. Now triangles ADM is similar to triangle ACB. Thus we know that $ \frac{MD}{2EC} = \frac{9}{16}$.

Now we let the intersection of the perpendicular from M to DB to be point Z. Thus triangles MZD is similar to triangle DEC. Because of similarity and the equation we derived in the paragraph before, we get that EC = $ \frac{14}{3}$. By Pythagorean Theorem on triangle DEC, we get that DE = $ \frac{7 \cdot \sqrt{5}}{3}$.

Now we introduce the perpendicular line from point A to BC, letting it intersect at a point Y. Now triangles AYC and DEC are similar. Thus we get that $ \frac{7}{16} = \frac{\frac{7 \cdot \sqrt{5}}{3}}{AY}$ From this we get that AY = $ \frac{16 \cdot \sqrt{5}}{3}$.

By the area of a triangle formula, namely A = $ \frac{1}{2} \cdot b \cdot h$, (using BC = $ \frac{28}{3}$), we get the area of the triangle to be $ \frac{8 \cdot \sqrt{5} \cdot 28}{9}$.

Now finally introduce the height from B to AC, letting it intersect at a point X. Now area of triangle ABC is also $ \frac{1}{2} \cdot 16 \cdot BX$. So we obtain the equation $ \frac{8 \cdot \sqrt{5} \cdot 28}{9} = \frac{1}{2} \cdot (9 BC + 7 BC)$ from which we find that BX = $ \frac{28 \sqrt{5}}{9}$.

So...

the area of triangle ABD is $ \frac{1}{2} \cdot 9 \cdot \frac{28 \sqrt{5}}{9}$, which is really $ \longrightarrow (D) \fbox {14 \sqrt{5}}$.
[/hide]

[quote="Karth"]Let $ DX$ be the perpendicular bisector to $ BC$ where $ X \in BC.$ Let $ \angle DBX = \theta.$ Since $ \angle DXB = \angle DXC = \pi /2$ and $ BX = XC,$ triangles $ \Delta DXB$ and $ \Delta DXC$ are congruent, which implies that $ \angle DAB = 180-3\theta, \angle ADB = 2\theta, DB = 7.$ Using the Law of Sines, we see that

\[ \frac{7}{\sin{180-3\theta}}= \frac{9}{\sin \theta}\iff \sin \theta = \frac{\sqrt{5}}{3}\iff [ADB] = 14\sqrt{5}. \][/quote]

So..now what? Sine area formula?... Sorry a bit new to trig.

^ Last implication?  If so, \[
	[ADB] = \frac{1}{2} AD \cdot BD \cdot \sin2\theta
	= AD \cdot BD \cdot \sin\theta \cos\theta \]
Now plug in the givens.  (You get $\cos\theta$ from $\sin\theta^2 + \cos\theta^2 = 1$.)

I solved a different way that is synthetic,

[asy]
real s1=(7/3)*sqrt(5);
real s2=22/3;
real x=14/3;
pair C=(0,0),B=(-2*x,0),A=(-x-s1,s1+s2),D=extension(B,bisectorpoint(A,B,C),A,C),P=foot(D,B,C),Q=foot(D,A,B);
D(MP("A",A,N)--MP("B",B,W)--MP("C",C,E)--cycle);
D(MP("D",D,E)--MP("P",P,S));
D(D--MP("Q",Q,W));
D(rightanglemark(D,Q,A,30));D(rightanglemark(D,P,C,30));D(B--D);
MA(P,B,D,1);
MA(D,B,Q,1.2);
MA(D,C,P,1);
label("$7$",(D+C)/2,E);
label("$7$",(D+B)/2,N,p=fontsize(6pt));
label("$9$",(A+D)/2,E);
label("$x$",(B+P)/2,S);
label("$x$",(C+P)/2,S);
label("$x$",(B+Q)/2,W);
label("$\sqrt{32+x^2}$",(A+Q)/2,W,p=fontsize(8pt));
label("$\sqrt{49-x^2}$",(D+Q)/2,N,p=fontsize(8pt));
[/asy]

Let $P$ be the foot of the perpendicular bisector. Let $Q$ be the foot of the altitude from $D$ to $AB$. let $BP=PC=x$. By SAS congruency, $\triangle PBD\cong\triangle PCD$ so $\angle C$ is congruent to the angles of the bisector. So $\triangle DBC$ is isosceles. Thus, we have $BD=7$ as well. We have by AAS congruency, $\triangle PDB\cong\triangle QDB$ so $BQ=x$. Then $QD=\sqrt{49-x^2}$ and $AQ=\sqrt{32+x^2}$ all from Pythagorean Theorem.

Now by the Angle Bisector Theorem, we have \[\frac{BC}{DC}=\frac{AB}{AD}\implies\frac{2x}{7}=\frac{x+\sqrt{32+x^2}{9}.\] Solving for $x$ yields \begin{align*}18x=7x+7\sqrt{32+x^2}&\implies 11x=7\sqrt{32+x^2}\\&\implies 121x^2=49(32+x^2)\\&\implies72x^2=49\times32\\&\implies x^2=\frac{7^2\cdot2^5}{2^3\cdot3^2}\\&\implies x=\frac{14}{3}.\end{align*} Note that \[ [ABD]=\frac{(AB)(QD)}{2}=\frac{\sqrt{49-x^2}(x+\sqrt{32+x^2})}{2}.\] We have, \begin{align*}\sqrt{49-x^2}=\sqrt{49-196/9}&=\frac{7\sqrt{5}}{3}\\ \sqrt{32+x^2}=\sqrt{32+196/9}&=\frac{22}{3}.\end{align*} So \[ [ABD]=\frac{(14/3+22/3)(7/3)(\sqrt{5})}{2}=\frac{12(7/3)\sqrt{5}}{2}=14\sqrt{5}\] as desired.

I have solved in a different way.

It is obvious that BD=DC.
So, BD=7.
Let ,angle ACB=angle DBC=angle ABD= a.
So,angle ADB=2a.
Now from INMO 1992, Q.1.
You get AB^2=9*(9+7)=144
So,BD=12.
And by Heron's formula
Area of ADB=14*square root of 5.

You can apply Stewart's theorem to triangle ABC where by the angle bisector theorem, AB=9x, BC=7x, AD=9, DC=7, and BD=7. You could then apply Heron's with 9x, 9, and 7.

No way, this was fun?!? Am I becoming a ... GEO MAIN *shudders*

[b]Solution:[/b] Let the perpendicular bisector of $\overline{BC}$ intersect it at $X$. Notice that triangles $CDX$ and $BDX$ are congruent, so $BD = 7$. Now we can use stewart's theorem along with angle bisector and get that $BC = \frac{28}{3}$. Notice that, by Pythagorean theorem, we can now find $XD$ to be $\frac{7\sqrt{5}}{3}$. Now we know that the area of triangle $CBD$ is $\frac{98\sqrt{5}}{9}$. Notice that the area of $ABD$ is just $\frac{9}{7}$ times this, we get our answer of D.


algebra
polynomial
number theory theorems
number theory

This came to my mind last weak ,Do we have any thoerem in this nature ?
Let $f(x)=\sum_{i=0}^{n}a_{i}x^{i}, f \in \mathcal{Z}[x]$. let $a_{0}=1$.
If for any $m \in \mathcal{Z},f(m)$ is not prime then there exist polynomails like $p,q \in \mathcal{Z}[x] ,deg[p]*deg[q] \neq 0$
that \[f(x)=p(x)q(x)\]

Hello,

I think it's false : for example, we have : \[\forall n\in Z, n^{2}\equiv 1\,(mod.3)\] so \[\forall n\in Z, 8n^{2}+1\equiv\,0(mod.3)\] Moreover, for all $n\in Z$, $8n^{2}+1\neq 3$. Consequently, $8n^{2}+1$ is never a prime. But, $f(X)=8X^{2}+1$ is an irreducible polynomial.

Olivier.

That example doesn't work: $73=8\cdot 3^{2}+1$ is prime.

I'd use something like $n^{2}+n+16$. It's always even, and it's never 2.

Lets state it in this way :

For Every irreducible polynomail in $\mathcal{Z}[x]$ with the last(not leading) coeffiecent equal to $1$ ,there exist a natural $m$ that $f(m)$ is prime .

$x^{2}+x+2$ is a counterexample ;)

But this problem is after some correction simply open and probably the most often posten open problem here (I would give a link, but here works as good as nothing...).

[color=red]I've done something as you wished  :wink: [/color]

[quote="ZetaX"]$x^{2}+x+\boxed{2}$ is a counterexample.
I've done something as you wished
[/quote]
Great effort ha ?
No its not ,Read the posts above.
As for the problem, yes  i found out that it is open ,seems Johann Dirichlet first intended it.


calculus
geometric series
geometric sequence

Evaluate: $ \displaystyle\sum_{n\equal{}1}^{\infty} \frac{n^2\plus{}n\plus{}1}{5^n}$

[hide="Answer"]$ \frac {x}{1 \minus{} x} \equal{} \sum_{n \equal{} 1}^{\infty}x^n$ (geometric series with r = x). Now differentiate both terms,

$ \frac {1}{(1 \minus{} x)^2} \equal{} \sum_{n \equal{} 1}^{\infty}nx^{n \minus{} 1}$,

and multiply both terms by x,

$ \frac {x}{(1 \minus{} x)^2} \equal{} \sum_{n \equal{} 1}^{\infty}nx^{n}$.

Now differentiate again and multiply by x:

$ \frac {1 \plus{} x}{(1 \minus{} x)^3} \equal{} \sum_{n \equal{} 1}^{\infty}n^2x^{n \minus{} 1} \equal{} > \frac {x(1 \plus{} x)}{(1 \minus{} x)^3} \equal{} \sum_{n \equal{} 1}^{\infty}n^2x^{n}$.

Hence: $ \sum_{n \equal{} 1}^{\infty}x^n \plus{} \sum_{n \equal{} 1}^{\infty}nx^{n} \plus{} \sum_{n \equal{} 1}^{\infty}n^2x^{n} \equal{} \sum_{n \equal{} 1}^{\infty}(n^2 \plus{} n \plus{} 1)x^n \equal{}$

$ \equal{} \frac {x(1 \plus{} x)}{(1 \minus{} x)^3} \plus{} \frac {x}{(1 \minus{} x)^2} \plus{} \frac {x}{1 \minus{} x}$

$ \equal{} > S_x \equal{} \sum_{n \equal{} 1}^{\infty}(n^2 \plus{} n \plus{} 1)x^n \equal{} \frac {x^3 \minus{} 2x^2 \plus{} 3x}{(1 \minus{} x)^3}$.

Now make x = 1/5 and we get $ S_{1/5} \equal{} \sum_{n\equal{}1}^{\infty}\frac{n^2\plus{}n\plus{}1}{5^n} \equal{} \frac {33}{32}$.[/hide]

There should be a non-calculus solution... I think... I hope  :(

This is a standard technique for sums of the form:

$ \sum_{n \equal{} 1}^{\infty} \frac {P(n)}{k^n}$

[hide="The Procedure"]
Let your sum be $ S$.

Pull out a factor of $ k^{\minus{}1}$ to get: $ \sum_{n \equal{} 1}^{\infty} \frac {P(n)}{k^n} \equal{} \frac{1}{k} \sum_{n \equal{} 1}^{\infty} \frac {P(n)}{k^{n\minus{}1}}$

Slide your index - replace $ n$ with $ n\plus{}1$ and reduce the bounds by 1. Why does this work? You are subtracting one from each value while you are looking at the bounds, but adding it back when you evaluate that expression at each value. Harmless.

$ \frac{1}{k} \sum_{n \equal{} 1}^{\infty} \frac {P(n)}{k^{n\minus{}1}} \equal{} \frac{1}{k} \sum_{n \equal{} 0}^{\infty} \frac {P(n\plus{}1)}{k^n}$

The sum from $ 0$ to $ \infty$ is equal to the expression evaluated at $ 0$ plus the sum from $ 1$ to $ \infty$:

$ \frac{1}{k} \sum_{n \equal{} 0}^{\infty} \frac {P(n\plus{}1)}{k^n} \equal{} \frac{P(1)}{k} \plus{} \frac{1}{k} \sum_{n \equal{} 1}^{\infty} \frac {P(n\plus{}1)}{k^n}$

Now, $ P(n\plus{}1) \equal{} P(n) \plus{} (P(n\plus{}1)\minus{}P(n))$, so:
$ \frac{P(1)}{k} \plus{} \frac{1}{k} \sum_{n \equal{} 1}^{\infty} \frac {P(n\plus{}1)}{k^n} \equal{} \frac{P(1)}{k} \plus{} \frac{1}{k} \sum_{n \equal{} 1}^{\infty} \frac {P(n)}{k^n}\plus{} \frac{1}{k} \sum_{n \equal{} 1}^{\infty} \frac {P(n\plus{}1)\minus{}P(n)}{k^n}$

The first term is a constant, the second term is a multiple of the original sum, and the third term is a sum with a lower degree polynomial. Keep doing this until you reach a geometric sequence, and then evaluate the whole expression.

[/hide]

[hide="Application"]We can employ a power-reducing index shift:

$ S \equal{} \sum_{n \equal{} 1}^{\infty} \frac {n^2 \plus{} n \plus{} 1}{5^n} \equal{} \frac {1}{5} \sum_{n \equal{} 1}^{\infty} \frac {n^2 \plus{} n \plus{} 1}{5^{n \minus{} 1}} \equal{} \frac {1}{5} \sum_{n \equal{} 0}^{\infty} \frac {n^2 \plus{} 3n \plus{} 3}{5^n} \equal{} \frac {3}{5} \plus{} \frac {1}{5} \sum_{n \equal{} 1}^{\infty} \frac {n^2 \plus{} 3n \plus{} 3}{5^n}$

$ \equal{} \frac {3}{5} \plus{} \frac {1}{5} \sum_{n \equal{} 1}^{\infty} \frac {n^2 \plus{} n \plus{} 1}{5^n} \plus{} \frac {1}{5} \sum_{n \equal{} 1}^{\infty} \frac {2n \plus{} 2}{5^n}$ $ \equal{} \frac {3}{5} \plus{} \frac {1}{5} S \plus{} \frac {2}{5} \sum_{n \equal{} 1}^{\infty} \frac {n \plus{} 1}{5^n}$

$ \iff \frac {4}{5} S \equal{} \frac {3}{5} \plus{} \frac {2}{5} \sum_{n \equal{} 1}^{\infty} \frac {n \plus{} 1}{5^n}$ $ \iff S \equal{} \frac {3}{4} \plus{} \frac {1}{2} \sum_{n \equal{} 1}^{\infty} \frac {n \plus{} 1}{5^n}$

We can similarly say that:

$ R \equal{} \sum_{n \equal{} 1}^{\infty} \frac {n \plus{} 1}{5^n} \equal{} \frac {1}{5} \sum_{n \equal{} 1}^{\infty} \frac {n \plus{} 1}{5^{n \minus{} 1}} \equal{} \frac {1}{5} \sum_{n \equal{} 0}^{\infty} \frac {n \plus{} 2}{5^n} \equal{} \frac {1}{5} \sum_{n \equal{} 0}^{\infty} \frac {n \plus{} 2}{5^n} \equal{} \frac {2}{5} \plus{} \frac {1}{5} \sum_{n \equal{} 1}^{\infty} \frac {n \plus{} 2}{5^n}$

$ \equal{} \frac {2}{5} \plus{} \frac {1}{5} \sum_{n \equal{} 1}^{\infty} \frac {n \plus{} 1}{5^n} \plus{} \frac {1}{5} \sum_{n \equal{} 1}^{\infty} \frac {1}{5^n} \equal{} \frac {2}{5} \plus{} \frac {1}{5} R \plus{} \frac {1}{5} \left( \frac {\frac {1}{5}}{1 \minus{} \frac {1}{5}} \right)$

$ \iff \frac {4}{5} R \equal{} \frac {2}{5} \plus{} \frac {1}{20} \implies R \equal{} \frac {9}{16}$

Now we can substitute back in:

$ S \equal{} \frac {3}{4} \plus{} \frac {1}{2} R \equal{} \frac {33}{32}$[/hide]

How about $ S \equal{} \sum_{n\equal{}0}^{\plus{} \infty} \frac{n^2 \plus{} 3n \plus{} 2}{5^n}$?


[quote="bpms"][quote="espark52"]Rocky is standing on a sidewalk made ofcement squares. He will toss a coin 4 times. For each head, he'll move one square forward. For each tail, he'll move one square backward. What is the probablility that after 4 tosses he'll be back in the square he started? express as a common fraction.[/quote]
[hide]$\frac{3}{8}$ ?[/hide] If I'm right I'll post a solution[/quote]

yea, ur right

You are really finding the distinct arrangements of llrr, divided by the total possibilities which is 6/16=3/8

[quote="espark52"]Rocky is standing on a sidewalk made ofcement squares. He will toss a coin 4 times. For each head, he'll move one square forward. For each tail, he'll move one square backward. What is the probablility that after 4 tosses he'll be back in the square he started? express as a common fraction.[/quote]
[hide]

1 way to get 0 heads
4 ways to get 1 head
6 ways to get 2 heads ******
4 ways to get 3 heads
1 way to get 4 heads

so..16 ways, and there's 6 of what u want...so the fraction is 3/8[/hide]
-jorian

edit:im just showing the solution (i know how to do it in a couple of seconds)

[quote="jhredsox"][quote="espark52"]Rocky is standing on a sidewalk made ofcement squares. He will toss a coin 4 times. For each head, he'll move one square forward. For each tail, he'll move one square backward. What is the probablility that after 4 tosses he'll be back in the square he started? express as a common fraction.[/quote]

[quote]
1 way to get 0 heads
4 ways to get 1 head
6 ways to get 2 heads ******
4 ways to get 3 heads
1 way to get 4 heads

so..16 ways, and there's 6 of what u want...so the fraction is 3/8
[hide]
-jorian[/hide][/quote][/quote]
Why are you quoting yourself?
Anyways, Here is a faster way to find the total number of ways. Notice that for each flip there are two possibilities, a head or a tail. So there are $2^{4}$ possibilities.

And 1, 4, 6, 4, 1. What do you notice about those numbers. And what do those kinds of numbers sum to?

Well, the numbers are integers...

And they sum up to an even number...

[hide]$\binom{4}{n}$
$2^{k}$[/hide]

In general $ \sum^{n}_{k=0}\binom{n}{k}=2^{n}$

[hide="solution"]For Rocky to be on the square he started on, he has to get two heads and two tails. There are 6 combinations:

hhtt 1/16
htht 1/16
htth 1/16
thht 1/16
thth 1/16
tthh 1/16

So we add: 6/16=3/8[/hide]


geometry
trapezoid
rectangle
number theory
relatively prime

Let $ABC$ be a triangle with sides 3, 4, and 5, and $DEFG$ be a 6-by-7 rectangle. A segment is drawn to divide triangle $ABC$ into a triangle $U_1$ and a trapezoid $V_1$ and another segment is drawn to divide rectangle $DEFG$ into a triangle $U_2$ and a trapezoid $V_2$ such that $U_1$ is similar to $U_2$ and $V_1$ is similar to $V_2$. The minimum value of the area of $U_1$ can be written in the form $m/n$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

[hide]
i think the only possible way is to draw a line from a corner of the rectangle the opposite side with length 6, and to draw a line parallel to the smaller leg of the right triangle, with length 3, so that it is perpendicular to the other leg.

$GF = DE = 6\\ GD = FE = 7\\ AB=5, BC=4, AC=3\\$

then,
$JC=\frac{7}{2}, BJ=\frac{1}{2}\\ AI=\frac{35}{8}, IB=\frac{5}{8}\\ IJ=\frac{3}{8}, HF=\frac{3}{4}\\ GH=\frac{21}{4}, GD=\frac{28}{4}\\ DGH=\frac{1}{2}(\frac{21}{4})(\frac{28}{4})=\frac{147}{8}\\ m+n=155$
[/hide]

Hm..I used the same idea, and got a triangle with legs [hide]$\frac{1}{2}$ and $\frac{3}{8}$, thus creating area $\frac{3}{32}$ and a final answer $35$[/hide]

Sorry to bump up such an old topic, but I need to know, which one of these answers is correct?

umm.. according to Kalva K81o7 is correct:
[url]http://www.kalva.demon.co.uk/aime/soln/answ04a.html[/url]

haha

EDIT: oh ok.. what i did wrong was i didn't consider another case. although the problem explicitly said "minimum value"

http://www.unl.edu/amc/e-exams/e7-aime/archiveaime.shtml
Has the official answers for everything in like the past ten years.


geometry
area of a triangle
Heron\u0027s formula

#25. What is the area of the region bounded by the three lines with equations $2x+y=8$, $2x-5y=20$ and $x+y=10$?

I need help with this one.

[quote="Klebian"]One way to do it:
Find the intersection of each of the lines using a system of equations.
Now you know the 3 end points of the triangle.
Then, find the distance between each fo the points, and use Heron's formula.[/quote]

[hide]True
But the distances turn out to be radicals, and it wouldn't be a good idea to use Heron's Formula since you can't use calculators on the Sprint Round.

The best way probably is to graph the three lines.  Then find the intersection points of the three lines which turns out to be A(5,-2), B(10,0), and C(-2,12).  Call a new point D(4,0), the point which 2x+y=8.  Now you can draw an altitude from C to BD, which has a length of 12, and you can draw an altitude from A to BD which has a length of 2.  So the area of ABD is 6*2/2=6, and the area of CBD is 12*6/2=36.  So the area of ABC is 42

$42$[/hide]


Μιας και απ'ότι ξέρω σήμερα-αύριο φεύγει η αποστολή μας για το Βιετνάμ για τη Διεθνή Ολυμπιάδα Μαθηματικών, θα ήθελα να ευχηθώ καλή επιτυχία σε όλα τα παιδιά και καλά μετάλλια!! Ειδικά θα ήθελα να ευχηθώ στον Σιλουανό που συμμετέχει πρώτη φορά σε διεθνή διοργάνωση! Είμαι σίγουρος όμως ότι μπορεί (και το αξίζει) να πάει πάρα πολύ καλά!

Ας σχολιάσουμε επίσης εδώ, τα θέματα του διαγωνισμού, τις παρατηρήσεις μας και τις (ενδεχόμενες) λύσεις μας!
[size=150][color=red]
[b]ΚΑΛΗ ΕΠΙΤΥΧΙΑ ΠΑΙΔΙΑ!!!![/color][/size][/b]

Αλέξανδρος

Παιδιά , εύχομαι σε όλους σας

[b][size=150] [color=red]  ΚΑΛΗ ΕΠΙΤΥΧΙΑ !!![/color][/size][/b]

 και να αποδώσετε σύμφωνα με τις δυνάμεις σας , που σίγουρα αξίζουν μετάλλια!.

Μπάμπης

Παιδια,

πραγματικα σας ευχομαι καλή επιτυχία! Είμαι σίγουρος ότι θα πάρουμε αρκετά μετάλλια!!

Υπάρχουν κάτι φήμες ότι έφεραν ειδικούς trainers από Αγγλία, καταγωγή από Ζάκυνθο και έφτασαν στα γραφεία της ΕΜΕ μέσω Πάτρας....Αληθεύουν?

Τέσπα, κοιτάξτε να πάρετε ό,τι περισσότερο μπορείτε από τον Αρετάκη-αν και είμαι σίγουρος ότι δε λέει ποτέ τα μυστικά...

Βάλτε του και καμμια δύσκολη Γεωμετρία, γιατί μου φαίνεται ότι μόνο με το tennis ασχολείται τώρα τελευταία...(προσπαθεί απεγνωσμένα να μου πάρει ποντο)

Τεσπα, για τον Αρετάκη θα μπορούσα να σας πω πολλά...Λέω το τελευταίο και κλείνω: Είμαι σίγουρος οτι εχει γίνει προπαγάνδα απο τον Αρετάκη και από έναν άλλο γνωστό trainer της ΕΜΕ οσον αφορά στις Ασιάτισες γυναίκες...Απλούστατα μην τους ακούτε!!!

Αυτά για να ξεμουδιάσουμε λιγο,

αντε ΚΑΛΗ ΕΠΙΤΥΧΙΑ

υ.γ. ρίξτε καμια σπόντα

και κάτι άλλο: είναι μία από τις χρονιές που μπορούμε να ελπίζουμε για χρυσο ή χρυσά.Βάλτε τα δυνατά σας!

ΚΑΛΗ ΕΠΙΤΥΧΙΑ ΠΑΙΔΙΑ.

Βάζω ένα θεματάκι για εξάσκηση αν έχετε τώρα πια χρόνο.

Δώδεκα μουσικοί Μ1, Μ2 κτλ. συγκεντρώνονται σε ένα φεστιβάλ μουσικής δωματίου διαρκείας μίας βδομάδας. Κάθε μέρα, υπάρχει μία προγραμματισμένη συναυλία και κάποιοι από τους μουσικούς παίζουν ενώ οι υπόλοιποι μουσικοί ακούν σαν ακροατήριο. Για i=1,2,...12 ορίζουμε Τi τον αριθμό των συναυλιών κατά τις οποίες ο Μi παίζει και ορίζουμε επίσης Τ=Τ1+Τ2+....+Τ12. Προσδιορίστε τι μικρότερη δυνατή τιμή του Τ τέτοια ώστε να είναι δυνατό για κάθε μουσικό να ακούσει, σαν μέλος του κοινού, όλους τους άλλους μουσικούς.

Αύριο η λύση αν δεν υπάρξει ανταπόκριση.

ΚΑΛΗ ΕΠΙΤΥΧΙΑ ΚΙ ΑΠΟ ΜΕΝΑ ΠΑΙΔΙΑ ΜΕ ΤΗΝ ΝΙΚΗ

Εύχομαι καλό ταξίδι και καλή επιτυχία στα παιδιά που θα πάνε στο Βιετνάμ και να γυρίσουν πίσω με μετάλλια.

[quote="Nick Rapanos"] ...Αληθεύουν?
Τέσπα, κοιτάξτε να πάρετε ό,τι περισσότερο μπορείτε από τον Αρετάκη-αν και είμαι σίγουρος ότι δε λέει ποτέ τα μυστικά...
 Είμαι σίγουρος οτι εχει γίνει προπαγάνδα απο τον Αρετάκη και από έναν άλλο γνωστό trainer της ΕΜΕ οσον αφορά στις Ασιάτισες γυναίκες...Απλούστατα μην τους ακούτε!!!
Αυτά για να ξεμουδιάσουμε λιγο,
[/quote]

Πάντως μυστικά λέει έτσι και ίσως πιο πολλά από ... :D  .
Για την προπαγάνδα για πες γιατί δεν ξέρουμε τίποτα .
Τι παίζει και ποιος είναι ο άλλος γνωστός trainer . Ή μάλλον με τι ασχολείται ? :wink:

[quote="Nick Rapanos"]Silouane yponoeis oti o aretakis sas leei pio polla mystika apo emena??

ton dorodokisate me kamia giaponeza???

ps. an sou po me ti asxoleitai o trainer, einai san na sou leo to onoma tou!!!ki ego den to ksera oti exei ki autos ekthmhsh stis giaponezes, alla mou to pan prosfata!

kata t' alla einai fysiologikos anthropos![/quote]
ΟΙ ΘΕΩΡΙΕΣ ΠΑΓΚΟΣΜΙΑΣ ΣΥΝΟΜΟΣΙΑΣ ΤΟΥ ΝΙΚΟΥ ΡΑΠΑΝΟΥ  :rotfl:  :rotfl:
ΠΟΥ ΣΑΙ ΡΕ ΜΕΓΑΛΕ ΜΑΥΡΑ ΜΑΤΙΑ ΝΑ ΣΕ ΔΟΥΜΕ ΠΙΑ
ΧΑΙΡΕΤΩ

[quote="Nick Rapanos"]Silouane yponoeis oti o aretakis sas leei pio polla mystika apo emena??[/quote]

Όχι ρε ...  :)  . Ότι σου έχω ζητήσει μου το έχεις δώσει . Αν σκεφτείς εγώ σου χρωστάω αυτή με τις 10 μεταβλητές  :D  . (μπορώ να πω με περιφάνεια ότι είμαι ο μοναδικός που έχεις πει μυστικά  :wink: )(αρμονικά πχ έμαθα από εσένα)

Πάντως χθες αποκαλύφθηκε ο trainer .... :P  . Αλλά δεν έχει προτίμηση στις γιαπωνέζες αλλά στις βιετναμέζες

Αύριο φεύγουμε ... See you

Παιδιά, σας εύχομαι κι εγώ ολόψυχα καλή επιτυχία στην ΙΜΟ !!!

ΟΕΟ, ΕΟΕ, περιμένουμε χρυσό !!!

@ Δημήτρη ( bilot4 ) και Σιλουανέ ( Silouan )

[hide]Ε, κοιτάξτε να βρείτε και καμιά Βιετναμέζα να της μάθετε το θεώρημα... "Greek Lover"

Και ένα ανέκδοτο:

Πάει ένας τυπάς και λέει σε μια τουρίστρια:

"Hello, do you have weather for a coffe ?"

Απαντάει αυτή:

"Are you serious?"

Απαντάει αυτός¨:

"No, I am Greek"[/hide]

οραίο ανέκδοτο του μοστελ
παιδιά για άλλη μια φορά ΚΑΛΟ ΤΑΞΙΔΙ ΚΙ ΚΑΛΗ ΕΠΙΤΥΧΙΑ


Ahhh... the ARML.

since we won the Division B last year, we are competing in division A. Personally I think I should be in team A again (although I did practically nothing last yr), and it should be a lot of fun, knowing that I can actually help the team in solving Power Round and Individual Round.

I hope some of you good math people from Washington can come again this year and hopefully hoist the division A Title!!! (Although that is practically impossible, because of Thomas Jefferson.... not my school, but the fancier one down in Virginia.  :( )

how do you get on the Washington ARML team?

this might be stupid, but what is ARML???
(i'm in WA :blush: )

American Regional Math League. goto arml.com for more info. Washington team for this year has been chosen already, so sorry for the guys who were looking to be in our team but couldnt. Maybe next year  ;)

i dont care about competing this year. how did you get selected for it or find out how to sign up for it?

maybe i could do it next year...

it was suppposed to be organized by my coach, Mr. Norris, but he was too busy with Mu Alpha Theta, so Garfield High took over. Usually USAMO-ers are auto-invites, and high scores on the AIME helps. Some of the invitees could not go, so the organizer made waiting list and invited accordingly.

[quote="jeffrey"]oh, so they just get a list of high aime scores from washington and invite the people on the list?[/quote]
pretty much. and through local and state competitions we know whos good and such.

Who were all of the WA students who went ARML? I know Brian Basham and Rachel Hinman - was Edward Chang also there?

Interesting - a lot of the students at ARML know about SIMUW - we definitely need to advertise more about SIMUW at other competitions. Like the Washington State Math Championships. Of course that would mean a more competitive SIMUW and umm..yeah..

brian said edward wasn't on there.
(i might be wrong though.)

Here is the roster that went to ARML from Washington:

Adam, Nathan, Jeffrey, Young, Justin C., Justin (the other one), Rachel, Brian, Pavan, Aaron, Seungsoo, Seungjun, Qiaochu, Derek, Joel.

There were 3 seniors on the team, so we should be better next yr, although i will miss them regardless. We had people that I thought were good that did not respond to going this year. This one person said that the email was trashed to junk mail in his account, so he did not receive information until we came back.

i'm surprised i didn't see Zach (from TJ) on there.
and there are many people that i dont even know. hopefully this year i will :)

Zach didnt want to go, and to be honest, none of the other TJers besides me are interested in harder competitions.

just wondering, how strong is TJ's 11/12 team for next year.(i mean 2005-2006)
i know Bellevue has a very strong team this year.

Not much.

EDIT: forgot i am a Junior now. That adds some relief to the team.

Bellevue will have a pretty good team I think. And pretty much everyone will be Calc BC next year; we will dominate that class test again :D. Hopefully, to finish off HS for me the year after we'll have a strong team, too. After that we may be a little lacking :P.


Find the solution of the linear systems

 X+2Y-5Z=1
2X-2Y+ Z=2
4X-2Y+3Z=3

For some reason i cant figure this out, the problem before this had the same left side but different right. I solved that one and used those steps to solve the right side of this and for some reason its not working out. I even start from scratch and solved this on its own and still cann get it. please help me, thanks

[quote="worthawholebean"]This should go in HSB or Classroom Math.

[hide="solution"]
Start by subtracting the second from the third:
\begin{eqnarray*}2x+2z&=&1 \\ x+z&=&\frac{1}{2}\\ x&=&\frac{1}{2}-z \\ 2y-6z&=&\frac{1}{2}\\-2y+3z&=&1 \\-3z&=&\frac{3}{2}\\ z&=&-\frac{1}{2}\\ x&=&\frac{3}{2}\\ 3-2y-\frac{1}{2}&=& 2 \\ y&=&\frac{1}{4}\end{eqnarray*}
[/hide][/quote]
Solution is wrong. Mistake is found at the line $-2y+3z=1$


function
analytic geometry
rotation
vector
trigonometry
exterior angle
trig identities

1 NEW EDIT: Added question 3.

Hello,

I'm going through my textbook, and there are a few things I don't quite understand, so I was wondering if people would be willing to help me out. This is not homework related, I merely want to understand as much as possible in the textbook. I will continue adding questions as I see necessary, although I will probably do so on a new post in the same thread, so as to bump my thread back into a noticeable location. I will also edit and mark satisfactorily answered questions so that there is less redundancy. Thank you!

$ I$. If there is a body attached to a spring, the force exerted on it at a displacement $ x$ from equilibrium may be written:

$ F(x) \equal{} \minus{} (k_1x \plus{} k_2x^2 \plus{} k_3x^3 \plus{} \cdots)$

My question is:
I've always learned Hooke's Law to simply be $ F(x) \equal{} \minus{} kx$. My way of rationalizing this new form is that the one I've been familiar with for so long is the same as the above stated, with the exception of the fact that the displacement $ x$ must be very small, and therefore cancels out the rest of the terms. However, how do we know definitively that the above power series is the correct, true equation for force? Can someone link me to or provide me with a derivation? I've never encountered it before.

$ II$. The solution to the 2nd order ODE

$ m\frac {d^2x}{dt^2} \equal{} \minus{} kx$

is $ x \equal{} Asin(\omega t \plus{} \phi_0)$, where $ \omega \equal{} \sqrt {\frac {k}{m}}$

My question is:
Why did people originally choose to substitute $ \omega$ for $ \sqrt {\frac {k}{m}}$ in the first place? Did it simplify the math? Did they know that the angular frequency would be equal to the spring constant divided by mass, or did they do it for convenience and find that this was so later? Also, can someone either provide me with or link me to a page showing the derivation of the equation that gives position as a function of time? I have not yet learned to solve 2nd order ODE's, but I would like to see this derivation anyway.

$ III$. I'm having some troubles understanding the mathematics behind this:

Suppose we have two SHM's described by the following equations:

$ x_1 \equal{} A_1cos(\omega t \plus{} \alpha_1)$
$ x_2 \equal{} A_2cos(\omega t \plus{} \alpha_2)$

The two can be summed and expressed as:

$ x \equal{} Acos(\omega t \plus{} \alpha)$

The book then proceeds to prove this geometrically by graphing $ x_1$ and $ x_2$ on the same coordinate axes, with lengths $ A_1$ and $ A_2$, rotated from the x-axis in a positive direction by $ \omega t \plus{} \alpha_1$ and $ \omega t \plus{} \alpha_2$ respectively.

The two vectors are then summed, resulting in a triangle with sides $ A_1$, $ A_2$, and $ A$.

This is what makes no sense to me: The magnitude of $ A$ can be found in terms of $ A_1$ and $ A_2$ with the law of cosines. In general, this is:

$ c^2 \equal{} a^2 \plus{} b^2 \minus{} 2abcos(\theta)$, where $ \theta$ is the angle between the two sides of length $ a$ and $ b$.

The textbook, however, uses the following equation:

$ A^2 \equal{} A_1^2 \plus{} A_2^2 \plus{} 2A_1A_2cos(\alpha_2 \minus{} \alpha_1)$

I don't see why the term with cosine isn't negative. I don't understand why the angle used is $ \alpha_2 \minus{} \alpha_2$, which is the exterior angle to the angle formed by the two legs. Is there some property of the law of cosines I'm forgetting?

[quote="Icy"]2. The solution to the 2nd order ODE

$ m\frac {d^2x}{dt^2} \equal{} \minus{} kx$

is $ x \equal{} Asin(\omega t \plus{} \phi_0)$, where $ \omega \equal{} \sqrt {\frac {k}{m}}$

My question is:
Why did people originally choose to substitute $ \omega$ for $ \sqrt {\frac {k}{m}}$ in the first place? ......
I have not yet learned to solve 2nd order ODE's, but I would like to see this derivation anyway.[/quote]
An explanation without differential equations:
Consider a point $ (P)$ moving in a circle of radius $ A$ around the origin $ (O)$ in the $ xy$ plane.
For simplicity (to simplify my picture), I am going to change the phase shift of your equation and put my point at $ (A,0)$ initially $ ($at $ t\equal{}0)$. (I just start my timer at a more convenient point, but that does not change the reasoning).
The angle swept by the ray $ OP$ is:
$ \theta \equal{} \omega t$
The coordinates of point $ P$ are:
$ x\equal{}cos(\omega t)$
$ y\equal{}sin(\omega t)$
The velocity vector for point $ P$ has a magnitude $ A \omega$ and is perpendicular to $ OP$.
The acceleration vector has a magnitude $ A \omega^2$ and is directed towards $ O$ (centripetal acceleration).
A bit of trigonometry allows you to find the $ x$ components of the velocity and acceleration:
$ v_x \equal{} \minus{}A \omega sin(\omega t)$
$ a_x \equal{} \minus{}A \omega^2 cos(\omega t)$
The "shadow" of point $ P$ on the $ x$ axis is moving in one dimension, with position, velocity and acceleration given by
$ x\equal{}cos(\omega t)$
$ v \equal{} \minus{}A \omega sin(\omega t)$
$ a \equal{} \minus{}A \omega^2 cos(\omega t)$
The motion is "simple harmonic motion," and we can rearrange to get:
$ a \equal{} \minus{} \omega^2 [Acos(\omega t)] \equal{} \minus{} \omega^2 x$
The acceleration is a linear function of the position and opposes the displacement away from the origin. You have a "linear restoring force," like the one generated by the spring.
The spring supposedly caused an acceleration given by:
$ \frac {F}{m} \equal{} a \equal{} \minus{} \frac {k}{m} x$
Compare that to the equation $ a \equal{} \minus{} \omega^2 x$ above and you will find that
$ \frac {k}{m} \equal{} \omega^2$

III.
In the law of cosines, when the cos enters with his sign changed (as in your formula) it's refers to the internal opposite angle. In the vectors (or phasors), the angle is the external, or the supplement of the former.

I.
Power series ( Taylor or McLaurin series) can describe any "good behaved" function, since wee choose the appropriate coefficients ($ k_n$).

[quote="Icy"]1 NEW EDIT: Added question 3.

Hello,

I'm going through my textbook, and there are a few things I don't quite understand, so I was wondering if people would be willing to help me out. This is not homework related, I merely want to understand as much as possible in the textbook. I will continue adding questions as I see necessary, although I will probably do so on a new post in the same thread, so as to bump my thread back into a noticeable location. I will also edit and mark satisfactorily answered questions so that there is less redundancy. Thank you!

$ I$. If there is a body attached to a spring, the force exerted on it at a displacement $ x$ from equilibrium may be written:

$ F(x) \equal{} \minus{} (k_1x \plus{} k_2x^2 \plus{} k_3x^3 \plus{} \cdots)$

My question is:
I've always learned Hooke's Law to simply be $ F(x) \equal{} \minus{} kx$. My way of rationalizing this new form is that the one I've been familiar with for so long is the same as the above stated, with the exception of the fact that the displacement $ x$ must be very small, and therefore cancels out the rest of the terms. However, how do we know definitively that the above power series is the correct, true equation for force? Can someone link me to or provide me with a derivation? I've never encountered it before.[/quote]
[hide="I"]
This question assumes that the potential energy function of the spring is some generalized power series: $ V(x) \equal{} \sum_{n \equal{} 0}^{\infty}a_n x^n$. (This might well be the case if the spring is non-linear.)

Then the spring force is given by: $ \vec{F} \equal{} \minus{} \nabla V(x) \implies F \equal{} \minus{} \sum_{n \equal{} 1}^{\infty} na_nx^{n \minus{} 1}$.

If we let $ n \equal{} m \plus{} 1$ then $ F \equal{} \minus{} \sum_{m \equal{} 0}^{\infty}(m \plus{} 1)a_{m \plus{} 1}x^m$.

Since the coefficients $ a_{m \plus{} 1}$ are yet undetermined we may replace them by letting $ (m \plus{} 1)a_{m \plus{} 1} \equal{} k_{m}$ then $ F \equal{} \minus{} \sum_{m \equal{} 0}^{\infty}k_{m}x^m$.

Further we observe that if $ k_{0}\neq 0$ then there is a constant force due to the mere presence of the spring; that is an unphysical solution, therefore $ k_0 \equal{} 0$ and $ F \equal{} \minus{} \sum_{m \equal{} 1}^{\infty}k_{m}x^m$.[/hide]

[quote="Icy"]$ II$. The solution to the 2nd order ODE

$ m\frac {d^2x}{dt^2} \equal{} \minus{} kx$

is $ x \equal{} Asin(\omega t \plus{} \phi_0)$, where $ \omega \equal{} \sqrt {\frac {k}{m}}$

My question is:
Why did people originally choose to substitute $ \omega$ for $ \sqrt {\frac {k}{m}}$ in the first place? Did it simplify the math? Did they know that the angular frequency would be equal to the spring constant divided by mass, or did they do it for convenience and find that this was so later? Also, can someone either provide me with or link me to a page showing the derivation of the equation that gives position as a function of time? I have not yet learned to solve 2nd order ODE's, but I would like to see this derivation anyway.[/quote]
[hide="II"]
Consider the function $ x(t) \equal{} A\cos(\omega t \minus{} \phi)$.

Then

$ \frac {d^2x}{dt^2} \equal{} \minus{} \omega^2 A\cos(\omega t \minus{} \phi) \implies \frac {d^2x}{dt^2} \equal{} \minus{} \omega^2 x \\
\implies \frac {d^2x}{dt^2} \plus{} \omega^2 x \equal{} 0$

Now consider the ODE that you wrote: $ m\frac {d^2x}{dt^2} \equal{} \minus{} kx$,

Adding $ k x$ to both sides you get: $ m\frac {d^2x}{dt^2} \plus{} kx \equal{} 0$.

Now, divide through by $ m$ to obtain $ \frac {d^2x}{dt^2} \plus{} \frac {k}{m}x \equal{} 0$.

Due to the uniqueness theorem for differential equations if two differential equations have the same form their solutions are equivalent. Thus by inspection you have $ \omega^2 \equal{} \frac {k}{m}$[/hide]

[quote="Icy"]$ III$. I'm having some troubles understanding the mathematics behind this:

Suppose we have two SHM's described by the following equations:

$ x_1 \equal{} A_1cos(\omega t \plus{} \alpha_1)$
$ x_2 \equal{} A_2cos(\omega t \plus{} \alpha_2)$

The two can be summed and expressed as:

$ x \equal{} Acos(\omega t \plus{} \alpha)$

The book then proceeds to prove this geometrically by graphing $ x_1$ and $ x_2$ on the same coordinate axes, with lengths $ A_1$ and $ A_2$, rotated from the x-axis in a positive direction by $ \omega t \plus{} \alpha_1$ and $ \omega t \plus{} \alpha_2$ respectively.

The two vectors are then summed, resulting in a triangle with sides $ A_1$, $ A_2$, and $ A$.

This is what makes no sense to me: The magnitude of $ A$ can be found in terms of $ A_1$ and $ A_2$ with the law of cosines. In general, this is:

$ c^2 \equal{} a^2 \plus{} b^2 \minus{} 2abcos(\theta)$, where $ \theta$ is the angle between the two sides of length $ a$ and $ b$.

The textbook, however, uses the following equation:

$ A^2 \equal{} A_1^2 \plus{} A_2^2 \plus{} 2A_1A_2cos(\alpha_2 \minus{} \alpha_1)$

I don't see why the term with cosine isn't negative. I don't understand why the angle used is $ \alpha_2 \minus{} \alpha_2$, which is the exterior angle to the angle formed by the two legs. Is there some property of the law of cosines I'm forgetting?[/quote]
[hide="III"]
No, you are not forgetting anything. As has already been explained, when you have a sum of vectors then $ (\vec{x}_1 \plus{} \vec{x}_2)\cdot (\vec{x}_1 \plus{} \vec{x}_2) \equal{} x_{1}^{2} \plus{} 2\vec{x}_1\cdot\vec{x}_2 \plus{} x_{2}^2 \equal{} x_{1}^{2} \plus{} 2x_1x_2\cos\phi_{ext} \plus{} x_{2}^2$.

But $ \phi_{ext}$ is the supplement of the angle between the sides that one typically learns in high school trigonometry. Consequently if you substitute this fact, $ \phi_{ext} \equal{} \pi \minus{} \theta$, into the above equation the result in the law of cosines in standard form.

You may also derive the amplitude of the resultant wave by using complex numbers.

Consider $ Ae^{i(\omega t \plus{} \alpha)} \equal{} A_1e^{i(\omega t \plus{} \alpha_1)} \plus{} A_2e^{i(\omega t \plus{} \alpha_2)}$.

Then

$ |A|^2 \equal{} (A_1e^{i(\omega t \plus{} \alpha_1)} \plus{} A_2e^{i(\omega t \plus{} \alpha_2)})(A_{1}^{\ast}e^{ \minus{} i(\omega t \plus{} \alpha_1)} \plus{} A_{2}^{\ast}e^{ \minus{} i(\omega t \plus{} \alpha_2)}) \\
\equal{} |A_1|^2 \plus{} 2\Re{\{A_1A_{2}^{\ast}\}}\cos(\alpha_1 \minus{} \alpha_2) \plus{} |A_2|^2$

Since, $ A,\,A_1,\, \text{and } A_2$ are each real this final expression reduces to
\[ \boxed{A^2 \equal{} A_{1}^{2} \plus{} 2A_1A_2\cos(\alpha_1 \minus{} \alpha_2) \plus{} A_{2}^{2}}
\]
[/hide]


inequalities
trigonometry
analytic geometry
symmetry
geometry
3D geometry
sphere

[quote="silouan"] $(a^2+b^2+c^2)(a+b+c)^2\geq 2(a^4+b^4+c^4+6abc)$

which is true by AM-GM[/quote]
please check this one...

[quote="Ezbakhe Yassin"][quote="silouan"]By Cauchy in engel form

$L.H.S\geq\frac{(a+b+c)^2}{a^4+b^4+c^4+6abc}$.

So it sufices to prove that $(a^2+b^2+c^2)(a+b+c)^2\geq 2(a^4+b^4+c^4+6abc)$

which is true by AM-GM[/quote]

1. What is [b]Cauchy in engel form[/b]?

2. How did you get this: $L.H.S\geq\frac{(a+b+c)^2}{a^4+b^4+c^4+6abc}$ ?

THANKS![/quote]
this is because $(a^4+b^4+c^4+6abc)L.H.S=(a^4+2abc+b^4+2abc+c^4+2abc)(\frac{a^2}{a^4+2abc}+\frac{b^2}{b^4+2abc}+\frac{c^2}{c^4+2abc}) \ge (a+b+c)^2$

[quote="Kummer"]Sorry but really i don't know at which value of a,b,c the equality occured? Can anyone find it?[/quote]
when $b=c=k$ $a=\sqrt{1-2k^2}$
$k \to 0$

Well, my approach is proving this equalivent inequality:
$\frac{1}{2x+yz}+\frac{1}{2y+xz}+\frac{1}{2z+xy} \geq 2$
where $xy+yz+zx=1$  :)

[quote="Anh Cuong"]Well, my approach is proving this equalivent inequality:
$\frac{1}{x+2yz}+\frac{1}{y+2xz}+\frac{1}{z+2xy} \geq \frac{5}{2}$
where $xy+yz+zx=1$  :)[/quote]

Hi
Have you solved it ,too??

If so, present your solution!  :)

[quote="Anh Cuong"]Prove this inequality for $a,b,c \geq 0$ satisfying $a^2+b^2+c^2=1$
$\frac{a}{a^3+2bc}+\frac{b}{b^3+2ca}+\frac{c}{c^3+2ab} \geq 2$  :)[/quote]

I suck at inequalities, but I feel like trying.  Okay.  So.

By Jensen's, $\sum_{cyc} \frac{a^2}{a^4 + 2abc} \ge \frac{1}{a^6 + b^6 + c^6 + 2abc}$.  Now it remains to prove $a^6 + b^6 + c^6 + 2abc \le \frac{1}{2}$ which is... err... false.  Clearly not the way to go.

[quote="arqady"][quote="Anh Cuong"]Well, my approach is proving this equalivent inequality:
$\frac{1}{x+2yz}+\frac{1}{y+2xz}+\frac{1}{z+2xy} \geq \frac{5}{2}$
where $xy+yz+zx=1$  :)[/quote]
But is wrong: $x=y=z=\frac{1}{\sqrt3}.$  ;)
Id est, are you mean that no approach? :lol:[/quote]
Oh, sorry, I have edited it.  :blush:

[quote="Anh Cuong"]Prove this inequality for $a,b,c \geq 0$ satisfying $a^2+b^2+c^2=1$
$\frac{a}{a^3+2bc}+\frac{b}{b^3+2ca}+\frac{c}{c^3+2ab} \geq 2$  :)[/quote]

I have an approach, but am unable to complete it.

Spherical coordinates.  $\alpha$ is the x-y angle, and $\beta$ is the y-z angle.  We have $a = \cos \beta \cos \alpha$ (x-coordinate), $b = \cos \beta \sin \alpha$ (y-coordinate), $c = \sin \beta$ and the inequality becomes, after some cancelling,

$\frac{\cos \alpha}{\cos^2 \beta \cos^3 \alpha + 2 \sin \beta \sin \alpha} + \frac{\sin \alpha}{\cos^2 \beta \sin^3 \alpha + 2 \sin \beta \cos \alpha} + \frac{\sin \beta}{\sin^3 \beta + 2 \sin \alpha \cos \alpha \cos^2 \beta} \ge 2$

With equality $\alpha = \frac{\pi}{4}, \beta \to \frac{\pi}{2}$.  However, this destroys the symmetry, and it's very ugly.  But does anyone feel like attacking this new problem?  It seems like classical inequalities won't work on the old one...

(I was considering expressing the LHS as a sum of lines that could be proven by a geometric argument to be larger than the diameter of the unit sphere, but this argument doesn't hold for $a^2 + b^2 + c^2 = r^2$ so...)

So it is still unsolved, try more, oke  :) .

[quote="Anh Cuong"]Prove this inequality for $a,b,c \geq 0$ satisfying $a^2+b^2+c^2=1$
$\frac{a}{a^3+2bc}+\frac{b}{b^3+2ca}+\frac{c}{c^3+2ab} \geq 2$  :)[/quote]
This is wrong.
Try $a=0.9,$ $b=c=\sqrt{0.19}$ ;)

[quote="kar"]...arqady, it should be $b=c=\sqrt{\frac{0.19}{2}}$ if $a=0.9$.[/quote]
You are right, Karina. Thank you! :)

[quote="Anh Cuong"]Prove this inequality for $a,b,c \geq 0$ satisfying $a^2+b^2+c^2=1$
$\frac{a}{a^3+2bc}+\frac{b}{b^3+2ca}+\frac{c}{c^3+2ab} \geq 2$  :)[/quote]
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, our inequality is equivalent to $f(w^3)\leq0$, where $f$ is a convex function.
Hence, by [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=55&t=278791]uvw[/url] it's enough to check two cases only:
1. $w^3=0$;
Let $c=0$.
Hence, we need to prove that $\frac{1}{a^2}+\frac{1}{b^2}\geq2$, where $a^2+b^2=1$, which is obvious.
2. $b=c$.
Hence, $a^2+2b^2=1$ and we need to prove that $\frac{a}{a^3+2b^2}+\frac{2b}{b^3+2ab}\geq2$,
which is $\frac{a}{a^3+1-a^2}+\frac{4}{1-a^2+4a}\geq2$, which is equivalent to
$(a-1)(2a^4-8a^3+a^2+5a-2)\geq0$, which is true because
$2a^4-8a^3+a^2+5a-2\leq0$ is true for all $0\leq a\leq1$.
The last inequality is an interesting enough. :wink:


Find the positive difference between the sum of the first 100 positive multiples of 5 and the sum of the first 100 positive multiples of 6.
Generalize (Find the positive difference between the sum of the first x positive multiples of a and the sum of the first x positive multiples of b.)

Yup, now try to generalize. It should be a breeze since you found the shortcut.

P.S. It's not really my problem since I just tweaked a problem in the [url=http://mathcounts.org/GoFigure/html/default.htm]Go Figure! Math Challenge[/url]...


geometry
incenter
symmetry
trigonometry
geometry unsolved

Let $I$ be the incenter of a triangle $ABC$ with $\angle BAC = 60^{\cdot}$. A line through $I$ parallel to $AC$ intersects $AB$ at $F$. Let $P$ be the point on the side $BC$ such that $3BP = BC$. Prove that $\angle BFP = \frac{1}{2}\angle ABC$.

I solved it...use a ugly method.I think that you won't like it.

Let $G$ be the intersection of $AC$ with line parallel to $AB$ passing through $I$. Let us draw lines $k$ and $l$ intersecting side $BC$ such that $\angle (k,AB)=\angle CBA /2$ and $\angle (l,AC)=\angle BCA /2$. Obviously, $k\parallel l$. Another obvious thing: $\triangle AFG$ and $\triangle IFG$ are equilateral. Therefore, the height of $\triangle IFG$ from $F$ is perpendicular to $AB$ and so it is the symmetry axis of lines $l$ and $BI$. So if $H$ is the intersection of $l$ and $AB$, we have $BF=FH$. Let $P'$ and $Q'$ be the intersections of $k$ with $BC$, and $l$ with $BC$ respectively. We have thus proved that $BP'=P'Q'$. The same way we prove that $CQ'=Q'P'$, so $3BP'=BC$, hence $P'=P$.

[color=darkred][b][u]Lemma.[/u][/b] Let $ABC$ be a triangle. Then $B=2A\Longleftrightarrow$ $b^{2}=a(a+c)\ .$[/color]

[color=darkblue][b]Proof.[/b] $B=2A$ $\Longleftrightarrow$ $\tan A=\tan \frac{B}{2}$ $\Longleftrightarrow$ $\frac{\sin A}{\cos A}=\frac{\sin B}{1+\cos B}$ $\Longleftrightarrow$ $a\cdot(1+\frac{a^{2}+c^{2}-b^{2}}{2ac})=$ $b\cdot\frac{b^{2}+c^{2}-a^{2}}{2bc}$ $\Longleftrightarrow$ $a^{2}+c^{2}+2ac-b^{2}=b^{2}+c^{2}-a^{2}$ $\Longleftrightarrow$ $b^{2}=a(a+c)\ .$

[b]Proof of the proposed problem.[/b]

Prove easily that $BP=\frac{a}{3}$ , $BF=\frac{c(a+c)}{a+b+c}$ and $A=60\Longleftrightarrow a^{2}=b^{2}+c^{2}-bc\ .$

Therefore, using the above lemma, in the triangle $BFP$ we "ll have :

$m(\widehat{BFP})=\frac{B}{2}$ $\Longleftrightarrow$ $FP^{2}=BP\cdot(BP+BF)$ $\Longleftrightarrow$

$BF^{2}+BP^{2}-2\cdot BF\cdot BP\cdot\cos B=BP^{2}+BP\cdot BF$ $\Longleftrightarrow$

$\frac{BF}{BP}=1+2\cdot\cos B$ $\Longleftrightarrow$ $\frac{3c(a+c)}{a(a+b+c)}=1+\frac{a^{2}+c^{2}-b^{2}}{ac}$ $\Longleftrightarrow$

$\{\begin{array}{c}3c^{2}(a+c)=(a+b+c)(a^{2}+c^{2}-b^{2}+ac)\\\\ a^{2}=b^{2}+c^{2}-bc\end{array}\|$ $\Longleftrightarrow$

$3c^{2}(a+c)=(a+b+c)(2c^{2}-bc+ac)$ $\Longleftrightarrow$

$3c(a+c)=[a+(b+c)][a+(2c-b)]$ $\Longleftrightarrow$

$\{\begin{array}{c}3c(a+c)=a^{2}+3ca+(b+c)(2c-b)\\\\ a^{2}=b^{2}+c^{2}-bc\end{array}$ $\Longleftrightarrow$

$3c^{2}=(b^{2}+c^{2}-bc)+2c^{2}+bc-b^{2}\ ,$ what is truly.[/color]

[quote="Virgil Nicula"][color=darkred][b][u]Lemma.[/u][/b] Let $ABC$ be a triangle. Then $B=2A\Longleftrightarrow$ $b^{2}=a(a+c)\ .$[/color]
[/quote]
[b] Another proof.[/b] Take the point $D$ lie in the external of the ray $[BA$ so that: $BD=a.$
We have: $\triangle{ACD}\sim\triangle{DCB}\Longrightarrow\frac{AC}{DC}=\frac{CD}{CB}\Longrightarrow b^{2}=a(a+c)$


trigonometry
complex numbers
linear algebra

We can multiply by the Vandermond{\epsilon_k} and see the result


[hide]
You look for integers that multiply to get 3 and add up to -4.

$ -3$ and $ -1$ work because $ -3 \cdot-1 =3$ and $ -3+-1=-4$.

Thus the polynomial factors into $ (x-3)(x-1)$ (expand it out and see for yourself!).[/hide]

indeed! I might as well suggest a different approach? (even though it's not really appropriate for this situation)

>> Equality of Polynomials

$ f(x) = x^{2}-4x+3$
$ f(1) = (1)^{2}-4(1)+3 = 0$
therefore, $ (x-1)$ is a factor of f
so,
$ (x-1)(mx+c) = x^{2}-4x+3$
expanding gives,
$ mx^{2}+cx-mx-c = x^{2}-4x+3$
simplfying gives,
$ mx^{2}+(c-m)x-c = x^{2}-4x+3$
for this equality to be true the polynomials must be equivalent to each other
$ m = 1$ ,
$ c-m =-4$
$ c-1 =-4$
therefore,
$ c =-3$
hence,
$ (x-1)(x-3)$ is the factorised version of f.

same as buzzer11 but from different way that i  found my self.
$ x^{2}-4x+3=0$
what number multiply by what other number gives 3.
$ xy=3$
what number plus what other number gives -4
$ x+y=-4$
so we have a system of equation to solve
$ xy=3$
$ x+y=-4$
by substitution,
$ y=\frac{3}{x}$
$ x+\frac{3}{x}=\frac{x}{1}+\frac{3}{x}=\frac{x^{2}+3}{x}=-4$
$ \therefore$  $ x^{2}+4x+3=0$ now if we solve this equation it gives the $ -3$ and $ -1$
which is the $ (x-1)(x-3)$
just changing the sign in $ a_{n-1}$ from + to - or from - to +
this applies when factorising any monic quadratic polynomial ,

[quote="binomial_4eva"]same as buzzer11 but from different way that i  found my self.
$ x^{2}-4x+3=0$
what number multiply by what other number gives 3.
$ xy=3$
what number plus what other number gives -4
$ x+y=-4$
so we have a system of equation to solve
$ xy=3$
$ \boxed{x+y=-4}$
by substitution,
$ y=\frac{3}{x}$
$ x+\frac{3}{x}=\frac{x}{1}+\frac{3}{x}=\frac{x^{2}+3}{x}=-4$
$ \therefore$  $ x^{2}+4x+3=0$ now if we solve this equation it gives the $ -3$ and $ -1$
which is the $ (x-1)(x-3)$
just changing the sign in $ a_{n-1}$ from + to - or from - to +
this applies when factorising any monic quadratic polynomial ,[/quote]

$ x+y=-4$

is incorrect. It should be
$ x+y=4$

You're just going around in circles. You went from
$ x^{2}-4x+3 = 0$
to
$ x^{2}-4x+3 = 0$ (after the correction)

Why not graph it and estimate the intercepts and then plug in?
buzzer11's first post was enough, I don't know why people are still posting in this topic.

[quote="Funakoshi"]I can't seem to factor the equation x^2 - 4x +3=0. Can anyone help me?[/quote]

[hide]just find out which 2 numbers are common in the equation: $ x^{2}$ $ -4x$ $ +3$ = $ 0$ In this case, it is:

$ (x-3) (x-1)$

Find out which 2 numbers add up to the middle term, and which too multiply to -3.[/hide]

[quote="binomial_4eva"]vishalarul i am not wrong. you have to add to numbers to get -4 as buzzer11 said[/quote]

Actually, your answer is correct, even though your solution is incorrect.
[b]The sum of the roots of the equation is [/b]$ 4$.
[b]The product of the roots of the equation is [/b]$ 3$.

You went from
$ x^{2}-4x+3 = 0$

to

$ x^{2}+4x+3 = 0$

I believe buzzer11 was looking for numbers such that $ r,s$
$ (x+r)(x+s) = x^{2}-4x+3$

while, you looked for numbers such that
$ (x-r)(x-s) = x^{2}-4x+3$

In other words, buzzer looked for the negatives of roots of the equation, while you looked for the roots of the equation.

Besides, your solution just went in circles. When you claimed that the solutions of $ x^{3}+4x+3 = 0$ were $ -3$ and $ -1$, you just jumped through the intent of the problem.

You complicated the problem even more.

[quote="7h3.D3m0n.117"]By substituting 1 for x, aren't you just basically guess-and-checking your way through the factoring method?[/quote]

Yeah lol  :D    :oops:

Applying a little meta-problem solving, if you are relatively sure that factoring is going to yield integer roots, then one can systematically check the factors of the $ c/a$, including negative factors, for two which add up to $ b/a$.  In this case, $ c/a = 3$ and $ b/a =-4$, so the possible factorizations of $ 3$ including negative factors are $ 1, 3$ and $ -1,-3$, from which it is clear that $ -1+-3 =-4$.


calculus
integration
trigonometry
calculus computations

This problem is proposed in an exam but it seems weird and intriguing.

Evaluate the following

$ \int_{\minus{}1}^{1} \frac{x^2\plus{}1}{(x^4\plus{}x^2\plus{}1)(e^x\plus{}1)}$

Let $ t\equal{}x\minus{}\frac{1}{x}$
At x = -1, t = 0, at x = 1, t = 0
So I = 0...

[quote="chuong01"]Let $ t \equal{} x \minus{} \frac {1}{x}$
At x = -1, t = 0, at x = 1, t = 0
So I = 0...[/quote]

What is the value you would assign to t when $ x \equal{} 0$?

I think when $ x \equal{} 0$ then $ t \equal{} \minus{}\infty$.

From picture74, we have:
$ 2I\equal{}\int_{\minus{}1}^{1}\frac{x^{2}\plus{}1}{x^{4}\plus{}x^{2}\plus{}1}dx \implies I \equal{} \int_{0}^{1} \frac{x^{2}\plus{}1}{x^{4}\plus{}x^{2}\plus{}1}dx \equal{} \int_{0}^{1} \frac{\left(1 \plus{} \frac{1}{x^2} \right)}{(x \minus{} \frac{1}{x})^{2} \plus{} 3} dx$

Setting $ t \equal{} x \minus{} \frac{1}{x}$ gives:

$ I \equal{} \minus{} \int_{0}^{\infty} \frac{1}{t^2 \plus{} 3} dt \equal{} \minus{} \frac{1}{\sqrt{3}}\left[\tan^{\minus{}1} t \right]_{0}^{\infty} \equal{} \frac{\pi}{2\sqrt{3}}$

Numerical estimation shows that

$ \int_{\minus{}1}^1\frac{x^2\plus{}1}{x^4\plus{}x^2\plus{}1}\,dx\approxeq 1.813799364$

That's $ \frac{\pi}{\sqrt{3}},$ not $ \frac{\pi}{2\sqrt{3}}$

Recheck your lower limit of integration, and take a look at vijaymenon's most recent post.

You've also got a completely extraneous negative sign. The integral is positive, and $ t\equal{}x\minus{}\frac1x$ is an increasing mapping, not a decreasing mapping.

My calculation shows that
$ \int_{0}^{1}\frac{x^{2}\plus{}1}{x^{4}\plus{}x^{2}\plus{}1}\,dx\ \equal{} \frac{\pi}{2\sqrt{3}}$ and so multiplying by 2 gives the correct answer so you are right and I am right too.

@akech: in your calculation, you didn't write $ e^x\plus{}1$, any explanation please?

[quote="pr1nc3 bhp"]@akech: in your calculation, you didn't write $ e^x \plus{} 1$, any explanation please?[/quote]

That cancels out. See the first few lines of picture74's thread.

[quote="great math"]This problem is proposed in an exam but it seems weird and intriguing.

Evaluate the following

$ \int_{ \minus{} 1}^{1} \frac {x^2 \plus{} 1}{(x^4 \plus{} x^2 \plus{} 1)(e^x \plus{} 1)}$[/quote]

Let $ f(x) \equal{} \int_{ \minus{} x}^{x} \frac {g(t)}{e^{t} \plus{} 1} dt$ where $ g(x) \equal{} \frac {x^2 \plus{} 1}{x^4 \plus{} x^2 \plus{} 1}$, By Fundamental Theorem of Calculus, since $ g( \minus{} x) \equal{} g(x)$,

$ f'(x) \equal{} \frac {g(x)}{e^{x} \plus{} 1} \plus{} \frac {g( \minus{} x)}{e^{ \minus{} x} \plus{} 1}$

$ \equal{} g(x)\left(\frac {1}{e^{x} \plus{} 1} \plus{} \frac {1}{e^{ \minus{} x} \plus{} 1}\right)$

$ \equal{} g(x)\left(\frac {1}{e^{x} \plus{} 1} \plus{} \frac {e^{x}}{e^{x} \plus{} 1}\right)$

$ \equal{} g(x)$

Let $ f(x) \equal{} \int g(x)\ dx\equiv G(x) \plus{} C$, $ f(0) \equal{} 0\Longleftrightarrow G(0) \plus{} C \equal{} 0\Longleftrightarrow C \equal{} \minus{} G(0)$, yielding $ f(x) \equal{} G(x) \minus{} G(0)$.

$ \therefore \int_{ \minus{} 1}^1 \frac {t^2 \plus{} 1}{(t^4 \plus{} t^2 \plus{} 1)(e^{t} \plus{} 1)}\ dt \equal{} f(1) \equal{} G(1) \minus{} G(0) \equal{} [G(x)]_0^1 \equal{} \int _0^1 g(x)\ dx$

$ \equal{} \int_0^1 \frac {x^2 \plus{} 1}{x^4 \plus{} x^2 \plus{} 1}\ dx$

$ \equal{} \frac {1}{2}\left(\int_0^1 \frac {1}{x^2 \minus{} x \plus{} 1} \plus{} \int_0^1 \frac {1}{x^2 \plus{} x \plus{} 1}\right)\ dx$

$ \equal{} \frac {1}{2}\left(\int_0^1 \frac {1}{\left(x \minus{} \frac {1}{2}\right)^2 \plus{} \frac {3}{4}} \plus{} \frac {1}{\left(x \plus{} \frac {1}{2}\right)^2 \plus{} \frac {3}{4}}\right)\ dx$

$ \equal{} \frac {1}{2}\cdot \frac {2}{\sqrt {3}}\left[\tan ^ { \minus{} 1} \frac {2}{\sqrt {3}}\left(x \minus{} \frac {1}{2}\right) \plus{} \tan ^ { \minus{} 1} \frac {2}{\sqrt {3}}\left(x \plus{} \frac {1}{2}\right)\right]_0^1$

$ \equal{} \frac {1}{\sqrt {3}}\left(\tan ^{ \minus{} 1} \frac {1}{\sqrt {3}} \plus{} \tan ^{ \minus{} 1} \sqrt {3}\right)$

$ \equal{} \frac {1}{\sqrt {3}}\left(\frac {\pi}{6} \plus{} \frac {\pi}{3}\right) \equal{} \boxed{\frac {\pi}{2\sqrt {3}}}$

[color=blue]Solution for those who don't study arc tangent, say, Japanese High Scool Students[/color]

$ \int_0^1 \frac {x^2 \plus{} 1}{x^4 \plus{} x^2 \plus{} 1}\ dx$

$ \equal{} \frac {1}{2}\left(\int_0^1 \frac {1}{x^2 \minus{} x \plus{} 1} \plus{} \int_0^1 \frac {1}{x^2 \plus{} x \plus{} 1}\right)\ dx$

$ \equal{} \frac {1}{2}\left(\int_0^1 \frac {1}{\left(x \minus{} \frac {1}{2}\right)^2 \plus{} \frac {3}{4}} \plus{} \frac {1}{\left(x \plus{} \frac {1}{2}\right)^2 \plus{} \frac {3}{4}}\right)\ dx$

$ \equal{} \frac {1}{2}\left(\int_{ \minus{} \frac {1}{2}}^{\frac {1}{2}} \frac {1}{x^2 \plus{} \frac {3}{4}} \plus{} \int_{\frac {1}{2}}^{\frac {3}{2}} \frac {1}{x^2 \plus{} \frac {3}{4}}\right)$

$ \equal{} \frac {1}{2}\int_{ \minus{} \frac {1}{2}}^{\frac {3}{2}} \frac {1}{x^2 \plus{} \frac {3}{4}}\ dx$ Let $ x \equal{} \frac {\sqrt {3}}{2}\tan \theta$

$ \equal{} \frac {1}{2}\int_{ \minus{} \frac {\pi}{6}}^{\frac {\pi}{3}} \frac {1}{\frac {3}{4}(1 \plus{} \tan ^ 2 \theta)}\cdot \frac {\sqrt {3}}{2}\cdot \frac {1}{\cos ^ 2 \theta}\ d\theta$

$ \equal{} \frac {1}{\sqrt {3}}\left(\frac {\pi}{3} \plus{} \frac {\pi}{6}\right) \equal{} \boxed{\frac {\pi}{2\sqrt {3}}}$


In a cyclic quadrilateral $ ABCD$ the sides $ AB$ and $ AD$ are equal, $ CD>AB\plus{}BC$. Prove that $ \angle ABC>120^\circ$.

Let $AB=AD=x, BC=y$ and $CD=x+z$ with $y < z$.

Let $\theta = \angle ADC$ so $\angle ABC = 180^o-\theta$. It's easy to see that because $CD > BC$ then $\angle ABC > \angle ADC$ thus $0 < \theta < 90^o$

Now $AC^2 = x^2 + y^2 + 2xy \cos \theta = x^2 + (x+z)^2 - 2x(x+z)\cos \theta$, simplify it to get:
$\cos \theta = \frac{(x+z)^2-y^2}{2x(x+y+z)} = \frac{x+z-y}{x}$

But since $\theta$ is an acute angle, $\theta < 60^o \iff \cos \theta > 1/2$
So we need to show
$x+z-y > x$ which is true because $z > y$

[quote="CoBa_c_Kacka"]Firstly, it's clear that $ AC$ is the bisector of $ \angle BCD,$ because $ AB\equal{}AD.$ Let $ R$ be an internal point on $ CD$ such that $ BC\equal{}CR.$ Hence we have that $ \Delta ABC\cong \Delta ARC.$ From here $ AR\equal{}AB\equal{}AD$ and $ \angle ABC\cong \angle ARC.$ So we have to prove that $ \angle ARD<60^\circ.$ But from the condition $ CD>AB\plus{}BC$ we have that $ DR>AR\equal{}AD$ which means that $ \Delta ARD$ is isosceles with base bigger than his leg. Hence $ \angle ARD<60^\circ.$[/quote]

Very nice solution. That was a brilliant introduction of R.


inequalities
function
limit
real analysis
real analysis unsolved

[color=green][b]  Let $-\infty<a<b<\infty$ and $f: I \to {\mathbb R}$ where $I: =(a,b)$ or $I: =[a,b] .$  If
\[ f\left(\frac{x+y}{2}\right) <\frac{f(x)+f(y)}{2} \; \; \; ,\; \; \; \forall x,y\in I ,  \]
the function $f$ is continuous on $I$  ? [/b] [/color]

The problem was proposed last year at RMO 11-th grade. You can easily prove that the function $g(x)=\frac{f(x)-f(a)}{x-a}$ is monotone. It will have lateral limits in a. Denote them by $\ell_1$ and $\ell_2$. We will have that $\lim_{x\to a,x<a}f(x)-f(a)=\lim_{x\to a,x<a}[g(x)-g(a)](x-a)=\ell_1\cdot 0=0$ the same argument holds for $\lim{x\to a, x>a}$. So, the function is continuous in $a$. Since $a$ is chosen arbitrarly, $f$ is continuous on $\mathbb{R}$.

[quote="xirti"] ... the function is continuous .... [/quote]
Thanks for interest but please take  care !   However a question : do you appreciate that the situation is the same as in the case of functions $f: I\to {\mathbb R}$  which are convex in the sense of Jensen , i.e.
\[ f\left(\frac{x+y}{2}\right) \le \frac{f(x)+f(y)}{2} \; \; \; ,\; \; \; \forall x,y\in I \; ?  \]
[color=red] [b] Unfortunatelly do you have not convinced me ! [/b] [/color]

The short answer- the function doesn't have to be continuous. Try adding something to an example you already know.

f(x)=0 on (a,b] and f(a)=1 is counterexample for the closed interval. But I think that the function must be continuous for the open interval.

No, it doesn't have to be continuous.

Let $g: (0,1)\to\mathbb R$ be the restriction of a non-continuous additive function to $(0,1)$, and take $f(x)=g(x)+x^2$.

I'm confused :? my argument works for a function defined on the entire real line. Why it didn't work for an interval? is it a problem with the strict convexity?


A solid, uniform cylinder with mass M and radius R is initially at rest. It is gently placed onto a conveyor belt at constant velocity v. The cylinder's axis is perpendicular to the belt's velocity. The coefficients of static and kinetic friction are nonzero constants. (Ignore other energy losses such as rolling resistance).
(a). What is the final speed of the cylinder's center of mass?
(b). If a cylinder with twice the radius and four timest the mass had been used in place of the original cylinder, what would have been its final speed?
(c). If a ring with the same mass and radius had been used in place of the cylinder, what would have been its final speed? Assume ring has negligible thickness.

[quote="integration"]You can solve this problem using conservation of angular momentum.  Set the reference frame of the conveyor belt as "stationary".  Then, the cylinder is initially moving as a speed $ v$ relative to the belt.

Its initial angular momentum is:
\[ MvR \equal{} L_i\]
The only net force acting on the cylinder is friction.  If you compute the angular momentum relative to a point on the surface of the belt, then friction provides no net torque: Angular momentum about that center is conserved.

We also set the "no sliding" condition:
\[ v_f \equal{} R \omega_f\]
Hence we get:
\[ L_i \equal{} L_f\]

\[ MvR \equal{} Mv_f R \plus{} \frac {I {v_f}}{R}\]
Which leads to:
\[ v_f \equal{} \frac {MvR}{MR \plus{} \frac {I}{R}}\]
relative to the conveyor belt

Or relative to the initially dropping point:
\[ V_f \equal{} v \minus{} v_f \equal{} v(1 \minus{} \frac {MR}{MR \plus{} \frac {I}{R}})\]
Now for a: plug in $ \frac12 MR^2 \equal{} I$.

For b: Plug in $ I \equal{} \frac 12 (4M)(2R)^2$ and compare

For c: Plug in $ I \equal{} MR^2$[/quote]

Thanks integration for your reply! I had a question however. In your line when you say
\[ MvR \equal{} Mv_fR \plus{} \frac{I{v_f}}{R}\]

I understand that you are using conservation of angular momentum, but how is the right side of the equation $ L_f$? It's basically the same as saying $ MvR \equal{} Mv_f R \plus{} I\omega_f$. Isn't that redundant?

Thanks for the help!

My solution:

Same reference frame stuff so the problem is equivalent to a cylinder placed onto a stationary conveyor belt with a speed v to the right. There is a torque due to friction acting on it. Since $ F_{fric} \equal{} C, \tau_{fric} \equal{} C$. Then $ Ft \equal{} \delta p \equal{} mv \minus{} mv_f$ where $ F$ is the magnitude of force of friction (when its rolling with slipping its to the left). Also $ FRt \equal{} \delta L \equal{} I\omega_f$ since $ I\omega_i \equal{} 0$. Then we get by algebra $ Ft \equal{} mv \minus{} mv_f \equal{} \dfrac{I\omega_i}{R}$ and from there you do the same as the above solution

[quote="integration"]Yep, that's correct. I just converted $ \omega_f \equal{} \frac {v_f}{R}$[/quote]

right, but what i'm asking is why the $ L_f$ is equal to the sum of the two terms you presented? what does each one signify?

[quote="mihail911"][quote="integration"]Yep, that's correct. I just converted $ \omega_f = \frac {v_f}{R}$[/quote]

right, but what i'm asking is why the $ L_f$ is equal to the sum of the two terms you presented? what does each one signify?[/quote]

I had that up until a friend told me it was wrong and made me delete it.

From the definition of $ L=r\times p$, we can reduce this to $ L=I\omega + R\times MV$ where R and V are the position vectors and velocity of the center of mass respectively. The $ R\times MV$ term reduces to $ MvL$ if the wheel has a radius of L. The R represents position vector from the origin whereas the L represents the radius (from evaluating the magnitude of the cross product).

Alternatively, we can also get this from claiming that $ L=I\omega$ about the rim and by the parallel axis theorem, we get $ I_0=mL^2/2$ and there is the additional $ mL^2\omega=mL^2v/L=mvL$ term which we get also.

The $ Mv_f R$ term that integration refers to is dealing with the translational velocity whereas the $ I\omega$ term refers to the rotational aspect.

[quote="integration"]Oh, sorry, I didn't understand what you were asking.

I am computing angular momentum relative to a center fixed onto the platform.  The angular momentum of a rigid body about a certain point, is the angular momentum of its center of mass about that point + the angular momentum of the body about its center of mass (theorem, which can be prove fairly easily with some vector calc).

THus:
\[ L \equal{} MvR \plus{} I_{CM} \omega\]
, where $ MvR$ is the angular momentum of center of mass about a point fixed on the platform (Remember $ \vec L \equal{} \vec r \times \vec p$)[/quote]

Now it makes sense. I was not aware of that theorem therefore would not have thought of that. Thanks for the help!


Sorry,its better is:

[quote="dduclam"]Let $ a,b,c$ be [b]positive[/b] real numbers. Prove that
$ \frac {a^2}{bc} \plus{} \frac {b^2}{ca} \plus{} \frac {c^2}{ab}\ge \sqrt [9]{\frac {a^9 \plus{} b^9}{2c^9}} \plus{} \sqrt [9]{\frac {b^9 \plus{} c^9}{2a^9}} \plus{} \sqrt [9]{\frac {c^9 \plus{} a^9}{2b^9}}$[/quote]

[quote="Albanian Eagle"]I haven't worked out all the computations of SOS, but I think it can be solved by Mildorf's lemma
http://www.mathlinks.ro/viewtopic.php?p=395004&search_id=115266224#395004[/quote]

I don't think so. My solution only using AM-GM,very brief !

[quote="isomorphism"]Is $ (x \plus{} y)^9 \geq 2^8 (x^3 \plus{} y^3)x^3y^3$??
It looks like true... but I dont know how to prove it?

P.S: I reduced your inequality to this form[/quote]
It's here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=176919
( the mine or the scorpius119's proof. )

[quote="dduclam"]Let $ a,b,c$ be non-negative real numbers. Prove that
$ \frac {a^2}{bc} \plus{} \frac {b^2}{ca} \plus{} \frac {c^2}{ab}\ge \sqrt [9]{\frac {a^9 \plus{} b^9}{2c^9}} \plus{} \sqrt [9]{\frac {b^9 \plus{} c^9}{2a^9}} \plus{} \sqrt [9]{\frac {c^9 \plus{} a^9}{2b^9}}$[/quote]

Yes we can prove this using only AM-GM, it is homogenus so assume abc=1
So we have to prove $ a^3 \plus{} b^3 \plus{} c^3 \geq \sum_{cyc}\sqrt[9]{\frac{a^{18}b^9 \plus{} a^9b^{18}}2}$
[hide]It suffices to prove $ a^3 \plus{} b^3 \geq 2 \sqrt[9]{\frac{a^{18}b^9 \plus{} a^9b^{18}}2}$ then we can add  it for all cyclic terms to get required inequality. It is equivalent to:
$ (a^3 \plus{} b^3)^9 \geq 256(a^{18}b^9 \plus{} a^9b^{18})$
$ \leftrightarrow (x \plus{} y)^9 \geq 256(x^3 \plus{} y^3)x^3y^3$
$ \leftrightarrow (x^2 \plus{} y^2 \plus{}2xy)^4 \geq 256(x^2 \plus{} y^2 \minus{} xy)x^3y^3$
which is true (by "scorpius119 substitution" :wink: )
"If we let $ X\equal{}xy$ and $ Y\equal{}x^2 \plus{} y^2 \minus{} xy$, then AM-GM gives $ (3X\plus{}Y)^4\geq 256X^3Y$"[/hide]

Hope my solution is right and nice  :wink:

[quote="dduclam"]Let $ a,b,c$ be non-negative real numbers. Prove that
$ \frac {a^2}{bc} \plus{} \frac {b^2}{ca} \plus{} \frac {c^2}{ab}\ge \sqrt [9]{\frac {a^9 \plus{} b^9}{2c^9}} \plus{} \sqrt [9]{\frac {b^9 \plus{} c^9}{2a^9}} \plus{} \sqrt [9]{\frac {c^9 \plus{} a^9}{2b^9}}$[/quote]
Can we change 9 to any other positive intergers greater than 2 ? :wink: