integration calculus function derivative algebra domain logarithms

$ \int\frac{dx}{1\plus{}x^{4}}  \equal{} \frac{1}{2} \int\frac{ (x^2 \plus{} 1)  \minus{} ( x^2 \minus{}1 )dx}{1\plus{}x^{4}}$

Now split this as $ \frac{1}{2} \int\frac{ (x^2 \plus{} 1)}{1\plus{}x^{4}}dx  \minus{} \frac{1}{2}\int\frac{( x^2 \minus{}1 )dx}{1\plus{}x^{4}}$

$ \frac{1}{2} \int\frac{ (x^2 \plus{} 1)}{1\plus{}x^{4}}dx  \equal{} \frac{1}{2} \int\frac{ (\frac{1}{x^2} \plus{} 1)}{x^2\plus{}\frac{1}{x^2} }dx$

put $ x\minus{}1/x \equal{} t$

so the integral becomes $ \frac{1}{2}\int\frac{dt}{t^2 \plus{} 2 } \equal{} \frac{1}{ \sqrt{2} } \arctan (\frac{t}{\sqrt{2}}$

The second integral can also be reduced in the same way . in this put $ x\plus{} \frac{1}{x} \equal{} u$

are you sure you can divide by x ? [this is a common mistake of many integration 'solutions']

you can do that if the interval I of integration does not contain 0, for example $ I \equal{} (0, \infty)$

but if I = (-100, 100) your approach does not work.
try to decompose $ \frac {1}{1 \plus{} x^4}$ into sum of elementary fractions that can be integrated for a general interval I [look at $ 1 \plus{} x^4$ and see how can be re-written]

According to me dividing by $ x^2$ is very much allowed .

And to your special case of the interval ( -100  100 ) we can rewrite the integral as 2 times the integral from ( 0 , 100 )  as the integrand is even function.

TO ARYABHATA :

thank you very much  :)

[quote="vijaymenon"]According to me dividing by $ x^2$ is very much allowed .[/quote]  Specifically, it creates a removable singularity (a "hole") in the graph, and any "good enough" definition of the integral (i.e., any that is still defined for functions with one removable singularity) will give you the same value.  (In fact, there's no good reason not to define integrals as taking place after the removal of (finitely many) removable discontinuities.  It's not an "important" obstruction.  I'm not entirely sure that I even need the "finitely many" qualifier, but I woke up to recently to think more about it.)

The question can also be solved by factoring $ x^4 \plus{} 1$ using the Sophie Germain identity, partial fractions and completing the square.

Well, according to you,

$ \int\frac{ (x^{2}\plus{}1)}{1\plus{}x^{4}}dx \equal{} \frac{1}{\sqrt{2}}\arctan (\frac{x\minus{}\frac{1}{x}}{\sqrt{2}}) \plus{} C$

but the right member is not even continuous in x = 0

When you compute such integral you assume the integral is over R set, from $ \minus{}\infty$ to $ \plus{}\infty$, if nothing else is specified. Can you see my point now ?

You have the right conclusion (vijaymenon's solution is wrong) but for the wrong reason: vijaymenon's error is in simplifying after the substitution.  The proposed substitution $ t \equal{} x \minus{} \frac {1}{x}$ does not transform $ \int \frac {x^2 \plus{} 1}{x^4 \plus{} 1} \, dx$ into the integral $ \frac {1}{2} \int \frac {dt}{t^2 \plus{} 2}$ -- but it fails to do this regardless of whether or not we divide top and bottom through by $ x^2$ before making the substitution.  The division is completely irrelevant to the errors in the solution.

[quote="mareleG"]Well, according to you,

$ \int\frac { (x^{2} \plus{} 1)}{1 \plus{} x^{4}}dx \equal{} \frac {1}{\sqrt {2}}\arctan (\frac {x \minus{} \frac {1}{x}}{\sqrt {2}}) \plus{} C$

but the right member is not even continuous in x = 0

When you compute such integral you assume the integral is over R set, from $ \minus{} \infty$ to $ \plus{} \infty$, if nothing else is specified. Can you see my point now ?[/quote]

As noted by others, the error is in the algebra after substitution.

You do agree that the integration is valid on say the interval $ (10,11)$ ? We can easily see that the function we get on $ (10,11)$ can be easily extended to the whole real line, and it's derivative is $ \frac{1}{1\plus{}x^4}$ on the whole real line.

for interval (10, 11) i agree with the dividing method.

but tell me JBL, if we take $ t \equal{} x \minus{} \frac{1}{x}$, derivate both terms, $ dt \equal{} (1 \plus{} \frac{1}{x^2})dx$, then $ \int\frac{x^2 \plus{} 1}{1 \plus{} x^4}dx \equal{} \int\frac{(1 \plus{} \frac{1}{x^2})dx}{\frac{1}{x^2} \plus{} x^2} \equal{} \int\frac{dt}{t^2 \plus{} 2} \equal{} \frac{1}{\sqrt{2}}\arctan (\frac{t}{\sqrt{2}}) \plus{} C \equal{} \frac{1}{\sqrt{2}}\arctan (\frac{x\minus{}\frac{1}{x}}{\sqrt{2}})\plus{}C$

is this the initial [more detailed] idea, right? I don't see any error in this chain, if the interval of integration does not contain 0.

Hey, I just checked the algebra and it seems right to me!

I have a question for mareleG though,

why is $ f(x) \equal{} \arctan(|x| \minus{} 1/|x|), f(0) \equal{} \minus{} \pi/2$ not continuous?

[edit] as jmerry mentioned, this is irrelevant[/edit]

What I intended was this I suppose:

$ F(x) \equal{} \arctan(x \minus{} 1/x) \plus{} \frac {pi}{2}, x > 0$
$ F(x) \equal{} \arctan(x \minus{} 1/x) \minus{} \frac {pi}{2}, x < 0$
$ F(0) \equal{} 0$.

Your concerns are valid, but usually if an integration works on an interval, it can be extended to appropriate domain (lookup Identity theorem). For intance, in this case, all we did was change the constant on the appropriate interval.

[quote="jmerry"]It is continuous. It's also irrelevant, since its derivative is wrong for negative $ x$. If we integrate an even function, we should get an odd function (plus a constant), not another even function.[/quote]

Err... yes. You are right. I have edited that post to include what my actual intention was.

You can also split the denominator as [x^2 + 1 + (sqrt 2) x][x^2 + 1 - (sqrt 2) x]

As far as I can tell, everything I have said so far on this thread is wrong, so please allow me to correct it:

mareleG is completely right: by introducing a discontinuity in the function (even just a removable discontinuity), you change the universe of its antiderivatives.  We tend to notice this most commonly with the logarithm (where the discontinuity in $ \frac {1}{x}$ at $ 0$ is very not removable), but it's true in general: suppose $ f(x)$ is a nice (let's say continuous) function defined on all of $ \bf R$ and $ F(x)$ is an antiderivative of $ f$ and $ g(x) \equal{} f(x)$ for all $ x \neq a$ but $ g(a)$ is undefined (a removable discontinuity, say).  Then $ g$ does not have an antiderivative on all of $ \bf R$.  We can easily classify functions whose derivative is equal to $ g$: any such function $ G(x)$ must satisfy $ G(x) \equal{} F(x) \plus{} c_1$ for $ x < a$ and $ G(x) \equal{} F(x) \plus{} c_2$ for $ x > a$, where $ c_1$ and $ c_2$ are arbitrary constants that are not necessarily equal to each other.  If $ c_1 \neq c_2$ then this function is very much not an antiderivative of $ f(x)$; if $ c_1 \equal{} c_2$ then this function is an antiderivative of $ f$ with a removable discontinuity at $ a$.

The function that vijay wrote down as his antiderivative is of this form, where the two constants are not equal to each other.  (As mareleG noted, this function is discontinuous.)  To correct this, vijay needed to add some constant to his function for $ x > 0$ -- this would make his function continuous and so actually an antiderivative of $ \frac {x^2 \plus{} 1}{x^4 \plus{} 1}$ (what was $ f(x)$ in the preceding paragraph).

If instead we were talking about [i]definite[/i] integrals, then more of what I said would be true (and this is what I was thinking of when I made some false statements above).  Specifically, if $ f(x)$ is nice (again, I'm just going to act like this means continuous) on some interval $ [c, d]$ and $ g(x) \equal{} f(x)$ for all $ x$ except at a single point $ a$, there is absolutely no reason not to declare $ \int_{c}^d g(x) \, dx \equal{} \int_{c}^d f(x) \, dx$.  Moreover, antiderivatives of nice functions and definite integrals of nice functions are essentially interchangeable: if $ f$ is continuous on all of $ \bf R$ then $ F(x) \equal{} \int_0^x f(t) \, dt$ is an antiderivative of $ f$ (and of course we can replace $ 0$ with any point to get other antiderivatives; note though that we can't necessarily get every antiderivative this way).

But if antiderivatives are the same as definite integrals, and definite integrals don't care if you introduce a removable discontinuity, why do antiderivatives care?  The answer is that to do a substitution like vijay did with definite integrals, you have to change the interval on which you're integrating.  If $ x$ is ranging over (say) $ [ \minus{} 2, 2]$ and $ t \equal{} x \minus{} \frac {1}{x}$, what interval does $ t$ range over?  The endpoints get sent to $ \pm \frac {3}{2}$, but the substitution has a discontinuity in the middle; we actually have under the substitution that an integral over $ [ \minus{} 2, 2]$ becomes the sum of an integral over $ [ \minus{} \frac {3}{2}, \infty)$ and an integral over $ ( \minus{} \infty, \frac {3}{2}]$.  This strange behavior when you cross $ 0$ is exactly what you need in order for the definite integral to be continuous -- the splitting of one integral into two works precisely to add the constant necessary to prevent the discontinuity in vijay's solution that mareleG noted.

Gentil Estevez kindly pointed out to me that this exercise is in Apostol.  Three people have now suggested a second possible solution based on factoring the denominator -- this post is long enough that I won't extend it by showing the workings of that method, but perhaps someone else will be inspired?

[quote="JBL"]
Gentil Estevez kindly pointed out to me that this exercise is in Apostol.  Three people have now suggested a second possible solution based on factoring the denominator -- this post is long enough that I won't extend it by showing the workings of that method, but perhaps someone else will be inspired?[/quote]

Well, I'll have a go, if I can get the LaTeX to work properly.

We factor $ x^{4}\plus{}1$ into $ (x^{2}\plus{}\sqrt{2}x\plus{}1)(x^{2}\minus{}\sqrt{2}x\plus{}1)$ and then use partial fractions to end up with:
$ \frac{1}{x^{4}\plus{}1} \equal{} \frac{x\plus{}\sqrt{2}}{2\sqrt{2}(x^{2}\plus{}\sqrt{2}x\plus{}1)} \minus{} \frac{x\minus{}\sqrt{2}}{2\sqrt{2}(x^{2}\minus{}\sqrt{2}x\plus{}1)}$

Then we integrate each term by completing the square on the bottom in each case. So:
$ \int \frac{x\plus{}\sqrt{2}}{x^{2}\plus{}\sqrt{2}x\plus{}1} \mathrm{d}x \equal{} \int \frac{x\plus{}\sqrt{2}}{(x\plus{}\frac{1}{\sqrt{2}})^{2}\plus{}\frac{1}{2}} \mathrm{d}x \equal{} \int \frac{u\plus{}\frac{1}{\sqrt{2}}}{u^{2}\plus{}\frac{1}{2}} \mathrm{d}x$
And likewise for the other term, but we have $ v\equal{}x\minus{}\frac{1}{\sqrt{2}}$ in place of $ u$ in the term just done, above.

So then we have to do the integrals:
$ \int \frac{u}{u^{2}\plus{}\frac{1}{2}} \mathrm{d}u$
and
$ \int \frac{1}{u^{2}\plus{}\frac{1}{2}} \mathrm{d}u$

and their $ v$ counterparts.

The first of these leads to a $ \ln$ result, while the second leads to a $ \tan^{\minus{}1}$ result (after a substitution of $ u\equal{}a \tan \theta$ for a suitable $ a$ in each case for $ u$ and $ v$).

Finally, putting all the factors of $ \frac{1}{\sqrt{2}}$ and so on back in, and converting $ u$ and $ v$ back into their $ x$ originals, we end up with:
$ \frac{1}{4\sqrt{2}} \ln (x^{2}\plus{}\sqrt{2}x\plus{}1) \minus{} \frac{1}{4\sqrt{2}} \ln (x^{2}\minus{}\sqrt{2}x\plus{}1) \plus{} \frac{\sqrt{2}}{4} \tan^{\minus{}1} (\sqrt{2}x\plus{}1) \plus{} \frac{\sqrt{2}}{4} \tan^{\minus{}1} (\sqrt{2}x\minus{}1)$

Which could probably be tidied up a bit.


function
algebra
domain
binomial theorem
Pascal\u0027s Triangle

Evaluate
\[ \sum^{10}_{i \equal{} 1} \sum^{i}_{k \equal{} 1} \binom{i}{k}.
\]
My first question is how to interpret the notation.  Is there a more friendly form that I can write the expression in?  What is being indicated by each sigma?

Thanks!

[hide="hint"]
use the binomial theorem
[url=http://en.wikipedia.org/wiki/Binomial_theorem]Binomeal Theorem[/url]

[hide="hint 2"]
try expanding $ (1\plus{}1)^i$
[/hide][/hide]

[quote="t0rajir0u"]Interpret the inner one first, then the outer one.  The inner summation asserts that, given fixed $ i$, we want to find the sum $ {i \choose 1} \plus{} {i \choose 2} \plus{} ... \plus{} {i \choose i}$.  This sum is some function $ f(i)$.  We then want to find $ f(1) \plus{} ... \plus{} f(10)$.

Alternately, you can think of $ {i \choose k}$ as a generic multivariable function.  Then the "domain of summation" is the set of ordered pairs $ (i, k)$ such that $ 1 \le i \le 10$ and $ 1 \le k \le i$; this domain looks like a lattice triangle.  (This viewpoint is the natural one to make sense of an operation like switching the order of summation.)[/quote]

Using your first approach, how would I find $ i$?

[quote="t0rajir0u"]What do you mean by "find $ i$"?  We have to evaluate an expression for $ i \equal{} 1, 2, 3, ... 10$.[/quote]

Oh, so to evaluate the inner summation, I need to calculate $ {i \choose 1} \plus{} {i \choose 2} \plus{} ... \plus{} {i \choose i}$ for $ i \equal{} 1, 2, 3, ... 10$.  Is that right?

Could someone please tell me if the above is right?  If it is, do I calculate it using the Binomial Theorem: $ 2^i\minus{}1$?

Yes.  now you need to evaluate $ \sum_{i\equal{}1}^{10} (2^i \minus{} 1)$.

If a second visual picture would help, this question is asking you to pick out a particular triangle in Pascal's triangle and find the sum of all of the entries in it, and the easiest way to do this is to remark that you know what the sums of rows of Pascal's triangle looks like.

Ok, I got it now.  Thanks a lot for your help! :)


While attending a class or reading a book, I usually take notes and write summaries. Like math, in order to make it easier for review, I often quote things like definitions, theorems, and write down the ideas and/or outlines of some important proofs on the notes. Therefore, the notes/summaries are more or less similar to the textbooks I use or the books I read.

Then, if the notes and summaries were typed and posted on my blog, would it conflict with copyright laws?

If you write what's in your notes (which are presumably not verbatim from a book, or even a professor's lecture), then it's your material.  No one is going to put a copyright on say, the fundamental theorem of algebra, and then sue you if you mention it and prove.  You may however, run into trouble if you start quoting entire paragraphs or pages from textbooks and write it as if they're your own words.  Of course, if you asked me to prove the Dominated Convergence Theorem and then you asked my professor to prove it, chances are we would give you the exact same argument, but we might phrase things differently.  So maybe change the wording a bit.

But I seriously doubt anyone is going to sue you.


algebra
polynomial
vector
linear algebra
matrix

I would like to compute $ A^{10}$ in the simplest way possible, where

$ A=\left[\begin{array}{ccc}0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array}\right]$.

Please let me know if there is an easier way than the following.

One way I've read about is diagonalization. First, the characteristic polynomial is $ -\lambda^{3}+3\lambda+2$, so the eigenvalues are $ 2$, $ -1$, and $ -1$. But I'm having a little trouble finding the eigenvectors.

$ A=\left[\begin{array}{ccc}-\lambda & 1 & 1 \\ 1 &-\lambda & 1 \\ 1 & 1 &-\lambda \end{array}\right]$ is not invertible with $ \lambda =2$ or $ \lambda =-1$.

Here is a weird solution.

$ A$ sends each basis vector into the sum of the other two basis vectors. From this one can see that $ A^{2}=A+2I$. Multiply by $ A$ to get $ A^{3}=A^{2}+2A=3A+2I$. This can be repeated: $ A^{n}=c_{n}A+d_{n}I$, where $ (c_{n},d_{n})=(c_{n-1}+d_{n-1},2c_{n-1})$. In particular, $ A^{10}=341A+342I$. Instead of deriving a formula for the generic term, I just wrote down a column of numbers. :blush: The advantage of this approach is finding an interesting [url=http://www.research.att.com/~njas/sequences/A001045]article[/url] in the encyclopedia of integer sequences.

The eigenvector associated with 2 is $ (1,1,1)^{T}$. The eigenvectors associated with $ -1$ are any vectors orthogonal to that; there are plenty of choices for a 2-element basis of this space.

An alternate computation, without explicitly diagonalizing.

Let $ J$ be the matrix whose entries are all $ 1.$ This is a very useful notation; it's worth making it a standard part of your vocabulary. If we have an $ m\times m$ version of $ J,$ then we see that $ J^{2}=mJ.$

In this case, $ A=J-I=(-1)(I-J)$ with $ m=3.$

$ (I-J)^{2}=I-2J+3J=I+J$
$ (I-J)^{3}=(I-J)(I+J)=I-J^{2}=I-3J$
$ (I-J)^{4}=(I-J)(I-3J)=I-4J+9J=I+5J$
$ (I-J)^{5}=(I-J)(I+5J)=I+4J-15J=I-11J$

So our pattern is that $ (I-J)^{k}=I+a_{k}J.$

There's a recursion: $ a_{k+1}=a_{k}-1-3a_{k}=-1-2a_{k}.$ Let me rewrite that as $ a_{k+1}+2a_{k}=-1.$ A little familiarity with constant coefficient difference equations leads to the for $ a_{k}=\alpha (-2)^{k}+\beta,$ with $ \beta+2\beta=-1$ so that $ \beta =-\frac13.$ Since $ a_{0}=0,$ we then have that $ \alpha=\frac13.$

So $ (I-J)^{k}=I+\frac13\left((-2)^{k}-1\right)J$

$ A^{k}=(-1)^{k}\left(I+\frac13\left((-2)^{k}-1\right)J\right) =\frac13\left(2^{k}-(-1)^{k}\right)J+(-1)^{k}I$

$ A^{10}=341J+I= \begin{bmatrix}342&341&341\\341&342&341\\ 341&341&342\end{bmatrix}$

This is all very much the same as mlok's answer. The only real difference was the introduction of the notation $ J,$ which I think should be more familiar.

---

Side issue on eigenvalues: See above the fact that $ J^{2}=mJ$ for an $ m\times m$ version of $ J.$ That means that all of the eigenvalues of $ J$ satisfy $ x^{2}-mx=0,$ which has only $ 0$ and $ m$ as roots. But $ J$ is quite visibly a matrix of rank $ 1.$ That means that it has nullity $ m-1,$ meaning that the equation $ Jv=0$ has an $ m-1$ dimensional solution space. So the eigenvalue $ 0$ has geometric multiplicity $ m-1$ and hence algebraic multiplicity at least $ m-1.$ But $ J$ has rank $ m,$ so its eigenvalues must sum to $ m.$ That means that $ m$ is an eigenvalue of algebraic multiplicity $ 1.$ The eigenvalues of $ J$ are one copy of $ m$ and $ m-1$ copies of $ 0,$ and the matrix is diagonalizable.

In this particular problem, $ A=J-I,$ which just shifts the eigenvalues of $ J$ down by $ 1.$ The eigenvalues of $ A$ are $ 2,-1,$ and $ -1.$


If the numbers $ a_0,a_1,a_2,a_3,$ and $ a_4$ belong to the set $ (\minus{}1,0,1)$, how many numbers $ a_0\plus{}3a_1\plus{}9a_2\plus{}27a_3\plus{}81a_4$ are positive numbers?

we can brute force this with a computer program

well if we don't have one:

now since 81= 2*(1+3+9+27)+1 so if a4=1 then the sum is going to be positive (81 sums)

27=2*(1+3+9)+1 so if a4=0 and a3=1 the sum is also going to be positive (27 sums)

9=...........(9 sums)

3=.......(3 sums)

1=.....(1 sum)

=121 such sums...(correct me if i'm wrong)

Are you sure  we can ask?

ARGH, it seems I was off by 1, where all are positive.


group theory
abstract algebra
linear algebra
matrix
counting
derangement
search

What is the maximum dimension of a subspace of $ M_n(\mathbb{C})$ generated by a finite subgroup of $ GL_n(\mathbb{C})$ (non necessarily Abelian)?

Wow. That's an interesting problem. I think I proved that.

[hide]Consider $ \{ I_n \plus{} E_{i,j} | 1 \le i,j \le n \}$, where $ I_n$ is the $ n \times n$ Identity matrix. Prove that it is a basis.[/hide]

Is it enough to suppose that it is a matrix over a field with $ char(\mathbb{F}) \equal{} 0$, instead of $ \mathbb{C}$?
And also can we restrict it more than this?

[quote="fesghelbahal"]Consider $ \{ I_n \plus{} E_{i,j} | 1 \le i,j \le n \}$, where $ I_n$ is the $ n \times n$ Identity matrix. Prove that it is a basis.[/quote]
This is not a finite group.

The proof I have splits into cases, but we get full dimension over any field.

[hide="Characteristic p, not zero"]$ GL_n(F_p)$ is finite and spans everything. If $ n\equal{}1$, $ 1$ spans. Otherwise, we can take any permutation matrix and that same permutation plus a $ 1$ in a different location. Subtract them to get the elementary matrix $ E_{i,j}$ in the span; by using both the identity permutation and a derangement, $ (i,j)$ can be anything.[/hide]

[hide="Characteristic not 2"]The group $ G$ of signed permutations (one entry in each row and each column nonzero, that entry either $ 1$ or $ \minus{}1$) has order $ 2^n\cdot n!$ and spans everything. By flipping the sign of a cell and subtracting, $ 2E_{i,j}$ is in the span. We can divide by $ 2$, and everything works.[/hide]

See also the thread [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=279904894&t=277170]here[/url].


calculus
vector
linear algebra
matrix
derivative
algebra
function

Hi

Assuming a background in Single Variable Calculus,

Does the order in which you learn these 3 courses
matter significantly?

1) Multivariable Calculus
(1.5 - Vector Calculus)
2) Linear Algebra
3) ODE

My college offers 2 treatment of multivariable calculus,
(1) that has no prerequisite, and (1.5) that has
Linear Algebra as a prereq:

[url=http://www.aas.duke.edu/reg/synopsis/view.cgi?s=01&action=display&subj=MATH&course=105&sem=1230]Online Course Synopsis Handbook[/url]

also, ODE class has (1) as a prereq and (2) as a co-req

Thank you-

Multivariable calculus can be taught without Linear Algebra (other than determinants), but a math major should probably take the LinAlg-based version. I would take 2, 1.5, 3.

Based on my son's experience with taking linear algebra BEFORE multivariable, I would strongly prefer that order of courses, especially at a school that has a multivariable course designed to follow the linear algebra course.

I would second (or rather third) the above two sentiments, especially if you want to major in mathematics, physics, or engineering. Linear Algebra gives you a much deeper understanding of multivariable/vector calculus, and it develops your understanding of how to think in multiple dimensions. However, it usually (depending on how its taught) involves a higher level of mathematical sophistication (meaning more abstract definitions and more proofs) than a standard multivariable calculus course.

[quote="chloe L"]Though I know this is a stupid question,
Does that also mean that w/o having the L.A. prerequisite
(only with slight knowledge of matrices, etc. from limited
self-study), I would not understand/benefit from taking
a Linear-Algebra based Vector Calculus class?[/quote]

As my favorite professional school teacher said, the only stupid question is the one you don't ask. He really meant it. He welcomed any student question, which is one thing that made him a great teacher.

Yes, there is enough to learn in a good linear algebra course, most of which would be applied in a good multivariable calculus course, that you might as well take the actual course. If the college has planned a course sequence like that, the linear algebra course will be worth your time. Of course it will be helpful to you to already know about matrices.

Without contradicting any of the excellent advice given above, let me say this: as an advisor, while I suggest taking courses in an appropriate order, I'm not a seeker of perfection in the order of courses. Any piece of mathematics carries the potential for enhancing your understanding of some other piece of mathematics - but you can't take everything before everything else. Something has to go first. If there's an opportunity to take a course from an extraordinary instructor, grab that chance. If there's some good reason why a certain course won't work out this semester, go for what you can do and take it next semester.

[quote="chloe L"]Thank you so much for your help.

Though I know this is a stupid question,
Does that also mean that w/o having the L.A. prerequisite
(only with slight knowledge of matrices, etc. from limited
self-study), I would not understand/benefit from taking
a Linear-Algebra based Vector Calculus class?[/quote]

I bet you could understand a Linear-Algebra based Vector Calculus Course with just a basic knowledge of vectors, matrix multiplication, and determinants. I learned Vector Calculus before I took Linear Algebra - in fact, from the same book (Marsden & Tromba) that they're using for Vector Calculus at your school.

If I remember correctly, there's really not much beyond matrix multiplication and determinants explicitly used in the book. Its just that later, when I did take Linear Algebra, I was like ohhhh.. so that's what a derivative is!! *Its a linear map between the tangent bundles of the domain and the codomain* So linear algebra enhanced my understanding of vector calculus, but it wasn't a strictly necessary prerequisite.

So I think you could realistically take Vector Calculus first and linear algebra later. Its  just that if you want to study math, physics, engineering, or economics, you will eventually have to learn linear algebra.

Just out of curiosity, which of these are good to learn Linear Algebra from?
I'm looking for a textbook, something not nearly as abstract as Axler's Linear Algebra Done Right. I'm looking for something easier than math classics but still rigorous enough for potential physics majors.

1) Linear algebra by Friedberg &  <-- one I'm leaning towards the most
2) Intro to Linear Algebra by Strang
3) Linear Algebra and Applications by G. Strang

[quote="tokenadult"]I like any book by Strang. His two books on linear algebra are both good. They have different emphases, evidently because they are designed for different kinds of courses. I'm not familiar with the other book you mention.

After edit: A lot of Strang's materials about linear algebra can be found in the MIT Open CourseWare series, in the linear algebra courses featured there.[/quote]

In particular, there are even video lectures of him on OCW: http://ocw.mit.edu/OcwWeb/Mathematics/18-06Spring-2005/VideoLectures/index.htm

I kind of wish I would have waiting until next semester to take linear algebra with him.  I opted to take the more theoretical version this semester but the lectures are pretty awful and thus I don't go.  :maybe:


algebra
polynomial
Vieta
algebra unsolved

how to verify if a given polynomial (of real coefficients) has its all complex roots $|z_{0}|<1$
(all roots distributed in an unite circle)

Wow, fast reply! And it looks quite nice:)
May I ask how did you figure out this condition?

just as the gentleman above you wrote,
it follows easily from vieta's theorem for coefficients of a polynomial.

http://mathworld.wolfram.com/VietasFormulas.html

  abs (x+y) <= abs x + abs y where x,y are complex numbers.

using this you conclude.

[quote="t0rajir0u"]Say we are dealing with a monic polynomial $P(x) = x^{n}+a_{n-1}x^{n-1}+...+a_{0}$.  By Vieta's we can derive a (probably weak) necessary condition:

$P(x) \text{ has complex roots within the unit circle }\implies |a_{k}| <{n \choose k}$[/quote]
It is only nesessary condition.
For example $(x-0.5)(x-1.1)(x-1.1)=x^{3}-2.7x^{2}+2.31x-0.605$.
Can be find nesessary and sufficiently condition by $x=\frac{iz+1}{iz-1}, \ P(x)=0\to Q(z)=0.$


Calculate the sums:
$ A\equal{}1\cdot 2\plus{}2\cdot 3\plus{}3\cdot 4\plus{}\cdots\plus{}98\cdot 99$ and $ B\equal{}1\cdot 99\plus{}2\cdot98\plus{}3\cdot97\plus{}\cdots\plus{}98\cdot 2$

Use $ \sum_{k\equal{}1}^n k(k\plus{}1)\equal{}\frac{1}{3}n(n\plus{}1)(n\plus{}2)$, then compute $ A\plus{}B.$


Can high school students use AoSP vol.1 or is it just for middle school? do problems in this book prepare for AMC 12 or should I use AoSP vol. 2?
Thanks a lot in advance.

If you aren't very comfortable with the material in there, go with volume 1, and then go to volume 2.  If you feel very comfortable and don't think you need the review, then skip right to volume 2 (as I did).  I found volume 2 to be helpful for the AMC 12.

There's no law saying you volume 1 is only for middle school  :) .

Vol 2 has the difficulty setting of the AMC 12, but a lot of the concepts found in the book are not used in the competition (limits, vectors, etc.). Volume 1 has the most related topics regarding the AMC, but it lacks in difficulty and depth. I would recommend both books.

Ive done 10 chapters of v2, and volume 1 and want to make usamo. Do you think i should finish all of v2, or just do some important chapters and do a more in depth subject book?

[quote="Aryth"]psst. it's AoPS btw...[/quote]

thats right
why would you limit how much you do(which books) by which grade(middle school) you are in?

If you're going to stay interested in math competitions, you might as well get them for later anyway.


Find all polynomials $f$ satisfying $f(x^2)+f(x)f(x+1)=0$.

Let $\alpha$ be a root of $f(x)$.  Note that setting $x = \alpha$ in our identity, implies that $\alpha^2$ is a root. (1)

Also, setting $x = \alpha - 1$ in our identity, shows $(\alpha - 1)^2$ is a root. (2)

If $|\alpha| > 1$, then (1) implies there are infinitely many roots $\alpha^{2k}$ for $k = 1 , 2 , \dots$.  Contradiction.

If $\alpha \in [-1, 0)$, then (2) followed by (1) implies infinitely many roots also.  Contradiction.

If $\alpha \in (0, 1)$, then (1) implies infinitely many roots.  Contradiction.

Thus, $f(x)$ can only have roots $0$ or $1$.  (2) shows us that if we have one kind of root, we must have the other kind.  Suppose it does have roots.

Set $f(x) = kx^a(x-1)^b$, where $k \ne 0$.  Using the identity, we get $kx^{2a}(x-1)^b(x+1)^b = -k^2x^a(x-1)^b(x+1)^ax^b$, whereby $-k = x^{a-b}(x+1)^{b-a}$, and so $b = a$ since the RS must be a constant, implying k = -1.

If we have $f(x) = c$, with $f(x)$ with no roots, then $c + c^2 = 0$, so $c = 0$ or $c = -1$.

==

In conclusion , $f(x) = 0$ or $f(x) = -1$, or $f(x) = - (x(x-1))^a$ for some natural a.

Singular, you aren't finished: your argument doesn't show that, say, $x^2 + 1$ fails.  You need to consider complex values for the roots, as well.

I see a nice problem for derivatives...
actually, first take equation mod x to find c=0 or -1 as previously stated.
then we use derivatives to get a kind of taylor polynomial easy solve thingy...


inequalities
AMC
USA(J)MO
USAMO
inequalities open

[quote="Utkirstudios"]$ a_{0}, a_{1}, \dots, a_{n} > 0$ and satisfies $ a_{i \minus{} 1}a_{i \plus{} 1}\le a_{i}^{2}$ for each $ i \equal{} 1, 2, 3, \dots, n \minus{} 1$.  Prove,that:
\[ \frac {{a_{0} \plus{} a_{1} \plus{} \dots \plus{} a_{n}}}{{n \plus{} 1}}\cdot \frac {{a_{1} \plus{} a_{2} \plus{} \dots \plus{} a_{n \minus{} 1}}}{{n \minus{} 1}}\ge \frac {{a_{0} \plus{} a_{1} \plus{} \dots \plus{} a_{n \minus{} 1}}}{n}\cdot \frac {{a_{1} \plus{} a_{2} \plus{} \dots \plus{} a_{n}}}{n}
\]
[/quote]

hard or easy?


function
real analysis
real analysis unsolved

$0<p<q<\infty$, Prove: $L^{p}\not\subset L^{q}$ iff X contains sets of arbitrarily small positive measure.

(I got the "if" part, can someone give a hint on "only if" part?)

$X$ is presumably the measure space on which the $L^{p}$ and $L^{q}$ are based.

Suppose $X$ does not contain measurable sets of arbitrarily small  positive measure. I'll also assume that $X$ is $\sigma$-finite. We'll see later whether we can dispose of this conditions.

We can partition $X$ into $\bigcup_{j=1}^{\infty}E_{j}$ where $m(E_{j})\ge \delta>0$ and each $E_{j}$ has no measurable subsets of smaller positive measure. Then measurable function $f$ must be $f=\sum_{j}\alpha_{j}\chi_{E_{j}}.$ (Or at least equal to it almost everywhere.)

Now suppose $f\in L^{q}.$ This means that $\sum_{j}|\alpha_{j}|^{q}m(E_{j})<\infty.$

Your challenge: can you prove from this setup that $\sum_{j}|\alpha_{j}|^{p}m(E_{j})<\infty?$


In quadrilateral $ ABCD$ , let $ I_1$ $ I_2$ is the incircle of $ ABC$ and $ CDA$ respectively. Let $ K$ and $ L$ are the point of tangencies of $ I_1$ and $ I_2$ to $ CA$ respectively. Assume that $ I_1$ bigger than $ I_2$. Let $ M_1$ and $ M_2$ are the center of $ I_1$ and $ I_2$. Prove that  $ M_1M_2$, $ KD$ and $ BL$ are concurrent.

It is a small part of Problem 6, IMO 2008.

Lines $ M_1M_2, DK, BL$ are not concurent in general ... What conditions meet additional quadrilateral ABCD?


[quote="BanishedTraitor"][quote="ANA123"]$\frac{(1+i^{3})^{3}(-\frac{1}{2}+i\sqrt \frac{3}{2})^{3}}{(2\sqrt 2-2\sqrt{2i})^{2}}$[/quote]

Do you want it evaluated or what? Plus, is the root $\sqrt{2 i}$ or $\sqrt{2}i$[/quote]

You're right! sorry!

 I want evaluated???!

See now! :wink:

I think you would need to use De'Moivre's Theorem but the I can't find a nice $mod-arg$ form for $\frac{-1}{2}+i \sqrt{\frac{3}{2}}$.

Otherwise we have:

$1+i = \sqrt{2}cis \frac{\pi}{4}$

$2\sqrt{2}-2\sqrt{2}i = 4 cis \frac{-\pi}{4}$.


function
integration
calculus
real analysis
complex analysis
complex analysis unsolved

We are given a function $ f: U\rightarrow \mathbb{C}$ and it is known that each $ z\in U$ has a neighborhood $ V\subset U$ such that $ f$ has a primitive on $ V$. Prove that for all closed triangles $ \Delta$ the following relation holds
\[ \int_{\partial\Delta}f(\zeta)d\zeta \equal{} 0.\]

So... this is identical to how you show the generalized Cauchy's theorem -- that the integral of an analytic function on a chain depends only on its homology class.

The idea: the triangle can be subdivided into small triangles in such a way that each triangle maps into a neighborhood where $ f$ has a primitive; this is because $ \Delta$ is compact (to be rigorous, use the Lebesgue number lemma or some such compactness result). Then the integral around each small triangle is zero, yet these integrals add up to the integral over $ \partial \Delta$. Therefore the integral around $ \partial \Delta$ is zero.


topology
real analysis
real analysis unsolved

I recently read in a book that a discrete metric space is complete but there was no proof in that book. Can you explain the reasoning behind this?

All Cauchy sequences are eventually constant.


induction
modular arithmetic
number theory proposed
number theory

Prove that,we can find $ 1000000$ diffrent natural numbers,such that product of any $ 999999$ of them,reduced by $ 1$,is divisible to the remained number.

We can show by induction on $ n\ge3$ that there are $ n$ different positive integers, so that the product of any $ n\minus{}1$ of them is $ \equiv(\minus{}1)^n\pmod x$, where $ x$ is the remained number.
For $ n\equal{}3$ choose $ x_1\equal{}2$, $ x_2\equal{}3$ and $ x_3\equal{}7$.
Assume now the statement true for any $ 3\le k\le n$ numbers.
Take now the numbers $ x_1,\ldots,x_n$ so that $ \frac{\prod_{j\equal{}1}^n x_j}{x_i}\equiv (\minus{}1)^n\pmod{x_i}$.
If $ n$ is even, then take $ x_{n\plus{}1}\equal{}x_1\cdot\ldots\cdot x_n\plus{}1$. Then clearly $ x_1\cdot\ldots\cdot x_n\equiv (\minus{}1)^{n\plus{}1}\pmod{x_{n\plus{}1}}$ and $ x_1\cdot\ldots\cdot\widehat{x_i}\cdot\ldots x_nx_{n\plus{}1}\equiv (\minus{}1)^n\cdot(\minus{}1)\equal{}(\minus{}1)^{n\plus{}1}\pmod{x_i}$, for any $ 1\le i\le n$.
If $ n$ is odd, then take $ x_{n\plus{}1}\equal{}x_1\cdot\ldots\cdot x_n\minus{}1$. It is easy to see that this works.

Now I go back to lessons :)

I think  there is a big mistake in your proof, freemind, and I don't see how to fix it.
Look when you take the case when $ n$ is even.I think that there is a mistake with your sign. :wink:
Why $ x_1\cdot\ldots\cdot\widehat{x_i}\cdot\ldots x_nx_{n\plus{}1}\equiv (\minus{}1)^n\cdot(\minus{}1)\equal{}(\minus{}1)^{n\plus{}1}\pmod{x_i}$?
Shouldn't be $ x_1\cdot\ldots\cdot\widehat{x_i}\cdot\ldots x_nx_{n\plus{}1}\equiv (\minus{}1)^n\cdot(1)\equal{}(\minus{}1)^{n}\pmod{x_i}$?
The case $ n$ is odd is OK.

But if I'm wrong, please tell me.

I have a proof for this problem .
$ Lemma$ : We first prove that there exist $ a_{1}$ , $ a_{2}$ , $ \cdots$ , $ a_{n}$ such that :
$ a_{1}a_{2} \cdots\ a_{i \minus{} 1}a_{i \plus{} 1}a_{i \plus{} 2} \cdots\ a_{n} \equiv \minus{} 1$ (mod $ a_{i}$)
for each $ 1 \leq i \leq n$
For prove this we use induction and take $ a_{1} \equal{} 2$ , $ a_{2} \equal{} 3$ and
$ a_{n \plus{} 1} \equal{} a_{1}a_{2} \cdots a_{n} \plus{} 1$ .
Now for this problem , consider $ x_{1} \equal{} a_{1}$ , $ x_{2} \equal{} a_{2}$ , $ \cdots$ , $ x_{n \minus{} 1} \equal{} a_{n \minus{} 1}$ ,
and $ x_{n} \equal{} a_{1}a_{2}...a_{n \minus{} 1} \minus{} 1$ and we have :
$ x_{n}\equiv \minus{} 1$ (mod $ x_{i}$) for $ 1 \leq i \leq (n \minus{} 1)$
and $ x_{1} x_{2} \cdots\ x_{i \minus{} 1}x_{i \plus{} 1} \cdots\ x_{n \minus{} 1}\equiv \minus{} 1$ (mod $ x_{i}$)
so $ x_{1} x_{2} \cdots\ x_{i \minus{} 1}x_{i \plus{} 1} \cdots\ x_{n}\equiv 1$ (mod $ x_{i}$)
and $ x_{1} x_{2} \cdots\ x_{n \minus{} 1}\equiv 1$ (mod $ x_{n}$)
So the proof is finished  :wink:


Η άσκηση αυτή προέκυψε συζητώντας το πρόβλημα του cmad (όλα για ένα και ένα για όλα) με ένα φίλο. Θεωρούμε τον εξής ισχυρισμό:

Αν ένα σύνολο του επιπέδου είναι τέτοιο ώστε η τομή του με κάθε ευθεία να είναι σχετικά ανοικτή στην ευθεία αυτή,  τότε το σύνολο αυτό είναι ανοικτό.

Αποδείξτε ή ανατρέψτε τον ισχυρισμό αυτό.

Καλή διασκέδαση!

Παραθέτω μια λύση για όποιον θέλει να τη δει:

[hide]Η απάντηση είναι ότι ο ισχυρισμός δεν ισχύει όπως πολύ σωστά παρατήρησε ο Δημήτρης και ένα απλό αντιπαράδειγμα είναι το σύνολο: $ \mathbb{R}^2\setminus\{(1/n,1/n^2): n\in\mathbb{N}\}$. Απλώς παρατηρήστε ότι κάθε ευθεία τέμνει το συμπλήρωμά του σε  δύο το πολύ  σημεία. Συνεπώς η τομή της ευθείας με το σύνολο είναι ανοικτή.[/hide]


I took a math contest thing today and I was wondering...
let the sequence ${a_{n}}$ be defined for all integers n by $a_{n} - (n+1)a_{2-n} = (n+3)^2.$ Find $a_{7}$
This was one of the ones I couldn't figure out... but I can't wait for the results... ty

ok i dont know if this is right, I have not taken any math contest so hopefully this is right if not i'm sorry:):

ok $an-(n+1) a2-n (n+3)$ find $a7$

I got $a3 = (n+5)^2 a4 = (n+7)^2 a5 = (n+9)^2 a7 = (n+11)^2$ and $a7 = (n+13)^2$ or it wll be the same numbers for N but instead of $a3 = (n+5)^2$ it will be $a3 = (n+5)^4$ and then the next number would have to be ^6,^8,^10,^12....


LaTeX
inequalities unsolved
inequalities

Hi ,bui_haiha2000!
are you chinese?
I guess you are because what you post can always been seen in some chinese book.
Maybe I am wrong.don't mind. :)
I know the solution,but I'd like to give the chance to the other who haven't seen these problem

[quote="Hawk Tiger"]
I know the solution,but I'd like to give the chance to the other who haven't seen these problem[/quote]
This is my solution:
(1 + 1)[(v - u)^2 + [9/v - (2 - u^2)^1/2]^2] >= [v + 9/v - [u + (2 - u^2)^1/2]]^2 (Bunhi and |A| + |B| >= |A + B|)
 In addition, v + 9/v - [u + (2 - u^2)^1/2] >= 4 > 0 ( AM-GM and Bunhi)
=> [v + 9/v - [u + (2 - u^2)^1/2]]^2 >= 16


i encountered this while thinking about another equation.
is x^y :equiv: x(mod y)if x and y are both prime?

is this some famous theorum?

Actually, it's true if y is prime.  x doesn't have to be.  It is rather famous, and is known by the name [url=http://mathworld.wolfram.com/FermatsLittleTheorem.html]Fermat's Little Theorem[/url].


trigonometry calculus function derivative

Find the amplitude of $f(x)$ if

$f(x) = {\sqrt3}cos(x)+sin(x)$

[hide]
$\sqrt{3}\cos{x}+\sin{x}$
$=2\left(\frac{\sqrt{3}}{2}\cos{x} +\frac{1}{2}\sin{x}\right)$
$=2\left(\sin{\frac{\pi}{3}}\cos{x} +\cos{\frac{\pi}{3}}\sin{x}\right)$
$=2\sin{\left(x+\frac{\pi}{3}\right)}$

So the amplitude is $2$.
[/hide]

Does anyone know how to rewrite the general form $A\cos x + B\sin y$?  I know the amplitude is $\sqrt{A^2+B^2}$, but that's it.

$A\sin x + B\cos y$
$\sqrt{A^2+B^2}(\frac{A}{\sqrt{A^2+B^2}}\sin x + \frac{B}{\sqrt{A^2+B^2}}\cos x)$
$\sqrt{A^2+B^2}(\sin x \cos(\cos ^{-1}\frac{A}{\sqrt{A^2+B^2}})+\cos x\sin(\cos ^{-1}\frac{A}{\sqrt{A^2+B^2}})$
$\sqrt{A^2+B^2}\sin(x+\cos ^{-1}\frac{A}{\sqrt{A^2+B^2}})$

Make sure you understand the last step.

[quote="t0rajir0u"]As far as just finding the amplitude, it is possible by differentiating:

$f(x) = \sqrt{3} \cos x + \sin x$
$f'(x) = \cos x - \sqrt{3} \sin x = 0$
$\tan x = \frac{1}{\sqrt{3}}$
$x = \frac{\pi}{6}$
$f(\frac{\pi}{6}) = \sqrt{3} \frac{\sqrt{3}}{2} + \frac{1}{2} = 2$[/quote]Hmm, I sense calculus in that post...  ;)  :roll:

[quote="WindSlicer"]toj, that works because the y value of the max/min on a sin/cos-like wave is the amplitude, correct?

Also, will that work on other trigonometric functions? Or just ones that contain only sin and cos?

Btw, thanks for the way to solve, toj! That way looks easiest to remember for me :)

Also thanks to the others for their solutions because now I can explain it to those who don't know calc in my math team :)[/quote]

Amplitude is $\frac{\text{max}-\text{min}}2.$

If you can express something as a single sine or, equivalently, a single cosine, then it will have an amplitude.  What I'm getting at is that a lot of functions with trig in them don't have amplitudes (tangent, for example).  For the ones that do have amplitudes, I see no reason why you can't just use calculus like that.  Of course, you can go ahead and use calculus right off the bat.  You'll just have to keep in mind that when setting the derivative to zero, no solution means no amplitude.

Oh right, when I was visualizing the tangent and cotangent, etc. functions in my head, I switched from thinking about amplitude to thinking about period, which was why I ended up inquiring about them :)


What is the units digit of $ \frac{20 \cdot 21 \cdot 22 \cdot 23 \cdot 24 \cdot 25}{1000}$?

Shouldn't there be an easier way, rather than multiplying all this?

Yes. We first simplify this to $ 21\cdot11\cdot23\cdot24$. Obviously, the units digit is this quantity $ \bmod{10}$, or $ 1\cdot1\cdot3\cdot4\equiv\boxed2\pmod{10}$.


logarithms
calculus
integration
inequalities
calculus computations

Does this series converge?

the sum as n=2 to infinity of 1/ln(n!)

how do i even start this?

Using Stirling's formulae we get $\ln n!\sim \ln n^ne^{-n}\sqrt{2\pi n}\sim n\ln n$ and it is well known that series $\sum_{n=2}^{+\infty}\frac{1}{n\ln n}$ diverges (it is the direct application of the integral convergence test - please, correct it, I don't know the right english equivalent).

can i do this without using stirlings formula?  we havent learned that yet.

Indeed, there is a simple inequality: $n!<n^n$, so $\frac{1}{\ln n!}>\frac{1}{n\ln n}$. That's all.


An object with mass $ m_1$ collapses elastically with the other object which is initial at rest with mass $ m_2$. Find the ratio of the decreased kinetic energy of object $ m_1$ if object m_1 goes back at the perpendicular direction to its movement after being collapsed.

[hide="assuming elastic collision & resultant of external forces null, the ratio of the final to the initial kinetic energy of the first object is"]
$ {{m_2 \minus{}m_1}}\over{m_1\plus{}m_2}$ ;so we must have $ m_2 > m_1$ [/hide]


1/show thta:for any 117 3-digit numbers one always can find 4 disjoin subet whose sum of elements is the same.

Are that 4 disjoint subsets contain all the 117 3-digit numbers ???

I see it many times but never see its solution!Any help?

I'm confused.  $\binom{117}{2}=6786\geq (999+998)\times3+1$ (by far)  Doesn't this mean, since the maximum sum of a 2-element subset of this kind is  $998+997$ and two disctinct 2-element subsets with the same sum must be disjoint, that it is possible to find 4 disjoint 2-element subsets of the given set all with the same sum? :fool: :noo:


inequalities
function
quadratics
algebra
polynomial
inequalities unsolved

After Vasc problems can appear such inequalities : $a,b,c \geq 0 $

1) $ \sum_{cyc} a(a-b)(a-kb) \geq 0 $ , for $ k \geq 0 $.

2) $ \sum_{cyc} a(a-b)(a-kc) \geq 0 $ , for $ k \geq 0 $.

 Is there solution ?

Both inequalities 1) and 2) hold not for every $k\geq 0$ (actually, a = 2, b = 1, c = 3 and k = 6 is a counterexample for both inequalities). But it is interesting to consider the ranges of k for which they hold (actually, we will assume k to be an arbitrary real - there is no need to restrict oneself to nonnegative k).

Call $S_1$ the set of all real k for which the inequality 1) holds for any three nonnegative a, b, c, and call $S_2$ the set of all real k for which the inequality 2) holds for any three nonnegative a, b, c.

Actually, because of the linearity of the functions involved (considered as functions of k), both sets $S_1$ and $S_2$ are intervals; actually, they are either closed intervals, or have a bound at $\infty$ or $-\infty$ (this follows from continuity). Both of these intervals contain the numbers -1 and 1 and thus they also contain all numbers between -1 and 1.

As for problem 1), it's for sure that $2\in S_1$.

There are counterexamples showing that $2.485\notin S_1$.

[b]EDIT:[/b] I have now identified the set $S_2$ of all k for which the inequality 2) holds for any three nonnegative a, b, c. In fact, this set $S_2$ is the interval $\left]-\infty;\;1\right]$.

The proof is easy: One has to show that for every real $k\leq 1$, the inequality

$a\left(a-b\right)\left(a-kc\right)+b\left(b-c\right)\left(b-ka\right)+c\left(c-a\right)\left(c-kb\right) \geq 0$

holds for every three nonnegative reals a, b, c, whereas for every real k > 1, there are some nonnegative a, b, c for which this inequality doesn't hold. Well, the second part is trivial: For every k > 1, take a = 1, b = 1 and c = 0 as a counterexample. The first part, namely to prove the inequality

$a\left(a-b\right)\left(a-kc\right)+b\left(b-c\right)\left(b-ka\right)+c\left(c-a\right)\left(c-kb\right) \geq 0$

for $k\leq 1$ and for all nonnegative a, b, c, is not so trivial. But it is not difficult either:

$a\left(a-b\right)\left(a-kc\right)+b\left(b-c\right)\left(b-ka\right)+c\left(c-a\right)\left(c-kb\right)$
$=\left(a^3+b^3+c^3+3abc-b^2c-c^2b-c^2a-a^2c-a^2b-b^2a\right)$
$+\left(1-k\right)\left(c^2b+a^2c+b^2a-3abc\right)$.

But this is $\geq 0$, since the first bracket $a^3+b^3+c^3+3abc-b^2c-c^2b-c^2a-a^2c-a^2b-b^2a$ is $\geq 0$ by Schur, the next bracket 1 - k is $\geq 0$ since $k\leq 1$, and the third bracket $c^2b+a^2c+b^2a-3abc$ is $\geq 0$ since $c^2b+a^2c+b^2a\geq 3abc$ (in fact, by AM-GM, we have $c^2b+a^2c+b^2a \geq 3\sqrt[3]{c^2b\cdot a^2c\cdot b^2a}=3\sqrt[3]{a^3b^3c^3} = 3abc$). This completes the proof that $S_2=\left]-\infty;\;1\right]$.

  Darij

darij will just post the solution to problem 2) which seems to be the easier one.

here's the solution to 1): first, set $a=0$. then a necessary condition gets that
$a^3-(k+1)a^2b+kab^2+b^3\geq 0$.

on the other hand, if you set $a=1$, $b=1+x$, $c=1+y$, there are no linear and constant terms, the quadratic terms are always positive ($2x^2-2xy+y^2$) and the cubic term is
$x^3-(k+1)x^2y+kxy^2+y^3$. therefore $x^3-(k+1)x^2y+kxy^2+y^3\geq 0$ is sufficient for the inequality to be true as well. so, we just have to study when the polynomial
$x^3-(k+1)x^2+kx+1\geq 0$ for $x\geq 0$.

some ugly algebra then shows that this is equivalent to
$k^4+2k^3-5k^2-6k-23\leq 0$ or
$-3.48444...=\frac 12\left(-1-\sqrt{13+16\sqrt{2}}\right)\leq k\leq \frac 12\left(-1+\sqrt{13+16\sqrt{2}}\right)=2.48444...$.

[quote="Peter Scholze"]darij will just post the solution to problem 2)[/quote]

Indeed, I have just edited my post before accordingly.

But your solution is cool - I guess inequality 1) will be very helpful in proving other cyclic inequalities!

Now what about the following problem:

[i]Find the range of all real k such that the inequality

$a\left(a-kb\right)\left(a-kc\right)+b\left(b-kc\right)\left(b-ka\right)+c\left(c-ka\right)\left(c-kb\right) \geq 0$

holds for any nonnegative numbers a, b, c ?[/i]

In fact, this range is the interval $\left[-\infty;\;1\right]$. The proof is much simpler than some of you might think ;) .

  Darij

After me, much more interesting are the inequalities:
1)If $a,b,c \geq 0$, find all real numbers $k$ such that

        $ \sum_{cyc} a^2(a-b)(a-kb) \geq 0 ;$

2)If $a,b,c$ are real numbers, find all reals numbers $k$ such that

    $ \sum_{cyc} (a-b)(ka+b)^3 \geq 0 ;$

3)If $a,b,c$ are real numbers, find all reals numbers $k$ such that

    $ \sum_{cyc} (a-b)(a+kc)^3 \geq 0 .$

After me, much more interesting are the inequalities:
1)If $a,b,c \geq 0$, find all real numbers $k$ such that

        $ \sum_{cyc} a^2(a-b)(a-kb) \geq 0 ;$

2)If $a,b,c$ are real numbers, find all reals numbers $k$ such that

    $ \sum_{cyc} (a-b)(a+kb)^3 \geq 0 ;$

3)If $a,b,c$ are real numbers, find all reals numbers $k$ such that

    $ \sum_{cyc} (a-b)(a+kc)^3 \geq 0 .$


he Earth’s circumference around the equator is 24907.55 miles

(1 mile = 1.609344 km)

A plane flies 11km above the surface of the Earth and flies westward following the line of the equator. If the Plane is travelling at 510km per hour, how long will it take for the plane to reach the same place above the Earth’s surface from where it started from?

N.B. The same place above the Earth's surface, not the same place in space,  What it is basically asking is if you started in Australia how long would it take to get back there travelling in a certain way?

Was it really needed to post this twice?

Didn't realise I did.

I think the answer is 78 hrs 44 minutes, but I'm not sure.

[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=979666#979666[/url]

You've posted in both threads? With two users?

Oops.
The registration process on this forum is slightly dodgey, I must have visited this forum earlier with Neod4000 and registered and accidentally come back later and already thought that there was a different member called Neod4000 or I had mucked up the registration process.
I didn't realise I had another account and would have gained nothing anyway from posting in two places on the forum.
I wouldn't mind if one of the accounts were to be deleted.
 Preferably the other one, as this is the one I use (or realised I had).


7^X+1=3^y+5^Z

find all (X,Y,Z) in natural numbers.

It seems I am wrong

$3^{y}= 7^{x}-5^{z}+1$

So, it is not solved yet.


[hide]
Take Log12 of both sides:
\[ \frac{1-a-b}{2(a-b)} = \log{x}  \]

we have that $a = \log{3/60}$ and $b = \log{5/60}$. Now we substitute that into the LHS and we end up with \[ \frac{\log{4}}{\log{144}}=\log{x}  \]. Since all the logs are base 12, this is equivalent to $\log{x^2} = \log{4} \implies x = 2$

Thus $B$ is the answer.
[/hide]

i believe this problem was also in the AOPS vol. 2 book also, i guess that's why you said well known

[quote="Silverfalcon"]If $60^a = 3$ and $60^b = 5$, then $12^{[(1-a-b)/2(1-b)]}$

$\text{(A)} \ \sqrt{3} \qquad \text{(B)} \ 2 \qquad \text{(C)} \ \sqrt{5} \qquad \text{(D)} \ 3 \qquad \text{(E)} \ \sqrt{12}$[/quote]

I got 2 (B).

[quote="breez"][hide="answer + solution"]

$\frac{\log 60-\log 3-\log 5}{\log 60}* \frac{2\log 60-2\log 5}{\log 60}$
$=\frac{\log 4}{\log 144}$

[/hide][/quote]

How do you get from the first step there to the second step?  So far I got:

$\frac{\log 60-\log 3-\log 5}{\log 60}* \frac{2\log 60-2\log 5}{\log 60}$

$=\frac{\log 4}{\log 60}\cdot \frac{\log 144}{\log 60}$

How do you go from here to then next step, which is:

$=\frac{\log 4}{\log 144}$

[quote="breez"][hide="answer + solution"]
$144^{1/2^{\frac{\log 4}{\log 144}}}= 4^{1/2}= \boxed{\text{(B) \ 2}}$[/hide][/quote]

Also, how did you do that?

I love this problem!  Isn't in the log chapter int he Aops books?

Oh well, my approach requires a little bit more laymen's terms than maybe some's might (logs are not my specialty)...

[hide]Obviously, $a = \log_{60}3$ and $b = \log_{60}5$

So, what you're looking for is:

$12^{\frac{1-(\log_{60}3+\log_{60}5)}{2(1-\log_{60}5)}}$

or

$12^{\frac{\log_{60}60-\log_{60}15}{2(\log_{60}60-\log_{60}5)}}\Rightarrow 12^{\frac{\log_{60}4}{\log_{60}144}}$

Dividing the exponent gives us:
$12^{\log_{144}4}\Rightarrow 12^{\log_{12}2}$

So: $\log_{12}x = \log_{12}2$, so now our answer settles on the only possibility, 2!

[/hide]


1.Find all pair (a,P,Q) where a is a real const and P,Q are polynomial satisfying :
     [P(x)/Q(x)]^2=P(x^2)/Q(x^2)+a
2.Find al pair (P,Q) such that :
      [P(x)]^2-(x^2+1)[Q(x)]^2=1.

[quote="hoangclub"]
2.Find al pair (P,Q) such that :
      [P(x)]^2-(x^2+1)[Q(x)]^2=1.[/quote]
This is a Pell's equation of form polynomial . You can use  same idea as Pell's equation in $ \mathbb Z$.


x^3 + 2y^3 + 4z^3 = 9w^3

please find all positive integer solutions to the above equation.

[hide="A hint"]Use Fermat's method of infinite descent.[/hide]


This thread is for the Team Chess Tournament Round 1: Team Delta (White) vs. Team Epsilon (Black). Play begins on December 7th. Any questions may be asked [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=164632]here[/url]. All information/FAQ is being updated [url=http://www.artofproblemsolving.com/Wiki/index.php/AoPSWiki:Chess/Chess_Tournament_Info]here[/url].

[hide="Team Delta"]
   1. AnonAmoz
   2. nitish
   3. Rafaelloaa
   4. Lawrence Wu
[/hide]

[hide="Team Epsilon"]
   1. not_trig
   2. Hamster1800
   3. DonkeyKong
   4. Tenoreoz
[/hide]

Each team will have 3 days for each move.

Good luck!

Please post the board when posting your moves (the wiki has a nice template for drawing the board).

After 1. d4:
[img]http://img243.imageshack.us/img243/3353/chesszs0.png[/img]

1. d4 ...Nf6

Yay (to make post long enough)

It's been 5 days since our last move, are you guys still here?

[quote="AnonAmoz"]Sry, but I think all of my team members have dropped out. I guess I'll just forfeit this game, then.  :([/quote]

You don't have to forfeit, we can place a few members on our substitute list onto Team Delta. However, we only have I believe one player left on the list. This game is hereby suspended until the new year until I can find some players. If not, Team Delta will recieve a half point bye, and Team Epsilon will recieve a full point.

5 days...I happen to have been in Chicago,Illinois,US,North America for 5 days and 4 nights...and that's not where I live :P
Nf3
How do you put a diagram?

1. d4 Nf6
2. Nf3

[img]http://img84.imageshack.us/img84/9259/diagramikphpcx6.jpg[/img]

You can make the images at [url]http://www.apronus.com/chess/diagram.php[/url]

I'm not getting a reply from the poster on my team.

If you are not getting a reply, and the deadline is approaching, then you are authorized to make the move without the consent of your team members. But since this is also the holiday season, they might be on vacation, so all moves can be suspended until this week is more or less over.

c4
yes, we need a suspension for the holidays.

There's isn't any reply in 13 days so I'll just give them a black mark. It's their first. If they do not reply by Feb 1, you will have an automatic win.

SUBSTITUTE ME PLZ im not very active and can barely reply every 2 weeks
plus i Pmed the poster he doesnt seem to be active either...

We currently do not have any active substitutes. Do you wish to forfeit?

never give up, never surrender
i say Nc3

If you wish to post a move, please provide a list of the moves (can be hidden to save space) and a diagram of the current position (for an example, refer to not_trig's last post).

-- T.D.

Oh, I can't edit my post...
1. d4 Nf6
2. Nf3 g6
3. c4 Bg7
4. [b]Nc3[/b]
[img]http://www.apronus.com/chess/diagram.php[/img]
How do you post a jpg image?

1. d4 Nf6
2. Nf3 g6
3. c4 Bg7
4. Nc3 [b]O-O[/b]
[img]http://img216.imageshack.us/img216/4953/diagramikphpll4.jpg[/img]

dangit i havent played this opening b4...
1. d4 Nf6
2. Nf3 g6
3. c4 Bg7
4. Nc3 O-O
5. [b]Bf4[/b]
[img]http://img233.imageshack.us/img233/9869/diagramik22mj8.th.jpg[/img]
it's too small!

1. d4 Nf6
2. Nf3 g6
3. c4 Bg7
4. Nc3 O-O
5. Bf4 [b]d6[/b]
[img]http://img209.imageshack.us/img209/1870/diagramikphpcm1.jpg[/img]

[quote="7h3.D3m0n.117"]Wow, three moves made in the last 24 hours. Are you sure this is a [i]team[/i] game?[/quote]
At least it's better than not moving at all (referring to the other games).

Two of team delta are discussing each move (not_trig and I), but the other two have made no attempt to contact us. As far as I know, they are not playing. But yes, it does seem that epsilon is just one person now.

i've been ditched it seems.
so its two against one

*Sighs*

Well, finish this game and then we'll worry about the rest of your team.


abstract algebra
function
trigonometry
search
parameterization
binomial coefficients
combinatorics open

Hi everyone, consider the following formulas, where C(n, k) denotes the typical binomial coefficient, apologies for not typing with TEX instead:

Sum C(n, k) for k = 0 (mod 3) = 1/3(2^n + 2cos(nPI/3))
Sum C(n, k) for k = 1 (mod 3) = 1/3(2^n + 2cos((n - 2)PI/3))
Sum C(n, k) for k = 2 (mod 3) = 1/3(2^n + 2cos((n - 4)PI/3))

Similar closed formulas can be derived when k takes all possible values with a given remainder modulo 4. Does anybody know a general approach that will work ? Or has anybody seen similar formulas when dealing with modules greater than 4 ? I tried myself using n-th roots of unity, but wasn't able to find a closed formula for none of the sums built in the above manner, just a little bit simpler ones.
Is it possible that such kind of formulas do not exist for modules greater than 4 ? (just as is the case with solving the general algebraic equation)

Thanks in advance!

The general approach is called [url=http://mathworld.wolfram.com/SeriesMultisection.html]Series Multisection[/url]. In the case of binomial coefficients it is applied to a generating function \[f(x) = (1+x)^{n}= \sum_{i}{n\choose i}x^{i}.\] The general identity of this kind for binomial coefficients is \[\sum_{m}{n\choose t+sm}= \frac{1}{s}\sum_{j=0}^{s-1}\cos\left(\frac{\pi (n-2t) j}{s}\right) 2^{n}\cos^{n}\left(\frac{\pi j}{s}\right)\] where $0\leq t < s$ are integers.

Three particular identities in your message can be obtained from this general identity for $s=3$ and $t=0,\ 1,$ and $2$.

thank you very much !!!  :)  unfortunately, the formula above doesn't give one a closed sum, but i suppose it is the best one can ask of!  ;)

[quote="kgbto"]unfortunately, the formula above doesn't give one a closed sum, but i suppose it is the best one can ask of!  ;)[/quote]
I did not get your comment. Since $s$ is a fixed number the right hand side of the identity above is a [b]closed[/b] sum consisting of $s$ summands.

well, you're right and i'm sorry, i didn't describe the whole picture that provocated me to search for a formula of the mentioned kind: i wanted for arbitrary n, t and s, an expression equal to the sum with the specified parameters, being far enough from a sum as possible, i.e. having as least as summands as possible. For example, in the cases when s = 3 or s = 4, one can find such an expression with only 2 summands, i was hoping the same was possible for arbitrary s. A number of summands, depending essentially on s, isn't the thing i'm searching for.
Since i forgot to post the case s = 4, here it is:

Sum C(n, k) for k = 0 (mod 4) = 1/2(2^(n - 1) + 2^(n/2).cos(nPI/4))

Sum C(n, k) for k = 1 (mod 4) = 1/2(2^(n - 1) + 2^(n/2).sin(nPI/4))

Sum C(n, k) for k = 2 (mod 4) = 1/2(2^(n - 1) - 2^(n/2).cos(nPI/4))

Sum C(n, k) for k = 3 (mod 4) = 1/2(2^(n - 1) - 2^(n/2).sin(nPI/4))

As you can see, here we have expressions, consisting of only 2 summands, the question i was interested in was whether this is possible for all s (allowing possibly 3, 4 or 5 summands or some other constant number, but not s). ;)


How many positive multiples of 13 are three-digit integers?

The least number is $ 104$, the highest is $ 988$.  Thus, our answer is $ \frac{988\minus{}104}{13} \plus{} 1$, or $ \boxed{69}$.

Another way is simply $ \lfloor\frac{900}{13}\rfloor$, or $ \boxed{69}$.

*Note that the numerator is $ 900$ because that is the number of three-digit numbers.

The first method is better, because if $ 101$ was a multiple of $ 13$, then $ 991$ would be also, so $ \left \lfloor \frac{900}{13} \right \rfloor$ wouldn't work.

[hide = Solution]We're looking for the number of positive multiples of 13 between 99 and 1000. $13 \times 7 = 91 < 100 < 104 = 13 \times 8$ and $13 \times 76 = 988 < 1000 < 1001 = 13 \times 77$, so the three-digit multiples of 13 are $$13\times8,13\times9,\ldots,13\times75,13\times76.$$ The number of terms in this list is the same as the number of terms in the list $$8,9,\ldots,75,76.$$ Subtracting $7$ from every term in the previous list, we get $$1,2,\ldots,68,69.$$ Therefore, there are $\boxed{69}$ positive multiples of 13 that are three-digit integers.
[/hide]

the first two comments are from 2008, though I agree with you if you are talking about pi.3.14.

[hide=Hint]
In this problem it would be good to know that $1001$ is divisible by $13, 11,$ and $7$. In fact, $1001=13\times11\times7$.  :)
[/hide]


function
logarithms
algebra proposed
algebra

Solve the functional equation $f(x^m)=f^n(x)$. Here $f$ is a real valued function and $m,n$ are positive integers.

Is $f^n$ equal to $f(x)^n$ or $f^{\circ n}$?

That's it? Are there some other conditions of $f$?
Because like this there are quite a few $f$'s which verify the hypothesis.

I hope I'm not misreading the problem.

[quote="silouan"]A solution?[/quote]
If $f^{2}(x)=f(x)*f(x)$ (other case $f^{2}(x)=f(f(x))$), then we can consider function $g(y)=ln(f(e^{y}))$ we have $g(my)=ng(y)$. It have many solutions, but continiosly solutions are only $g(y)=y^{a}, a=\log_{m}(n)$.

[quote="Rust"]
then we can consider function $g(y)=ln(f(e^{y}))$[/quote]
How about when $f(e^{y})$ isn't positive?
I think that there are more continuous solutions, such as case of $f(x)=-1$ and $n$ is odd.

Continiosly solutions are
1)$f(x)=0$,
2)$f(x)=1$,
3) if n is odd $f(x)=-1$
4) $f(x)=\theta_{2}(x)exp(\frac{\theta_{1}(x)|\ln |x\parallel ^{a}}{n}), \ a=\log_{m}(n).$
I solve for $x\in (1,\infty)$. If $x\in R$, then for exist continiosly solution we can chose functions $\theta_{1}(x), \theta_{2}(x)=0,\pm 1$ in intervales $(-\infty ,-1), (-1,0),(0,1),(1,\infty)$.

I know. I'm just interested in finding continuous solutions.

harazi,will you please tell corresponding pro or issue no. of amm?

Let $g(x)=ln(f(e^{x}))$, then ${g(x)=|x|^{log_{m}(n)}exp(t(x)}$ is solution for any periodic function t(x): $t(x+ln m )=t(x)$.
It give $f(x)=exp({|ln x|^{log_{m}n}exp({t(ln x)})}).$


[hide="Answer"]Graph the points, then draw the line connecting $(0,1)$ and $(4,1)$. We have two triangles with base $4$, one of height $2$ and one of height $1$, so our total area is $\frac{1}{2}\cdot 4\cdot (2+1)=6\Rightarrow \boxed{E}$.[/hide]

[quote="Silverfalcon"]4 pegs are plotted in real plane. They are (0,1),(1,3),(3,0), and (4,1). Then find the area of quadrilateral occupied by connecting the 4 pegs.

A. 4
B. 4.5
C. 5
D. 5.5
E. 6[/quote]

[hide]Just draw the quadrilateral, we can find the areas of two triangles making up it right away, since we have the heights ($1$ and $2$) and the base ($4$).

$\boxed{E}$[/hide]


Prove that there exist infinitely many positive integers $ n$ such that $ n!$ is not divisible by $ n^2\plus{}1$.

It is a direct consequence of IMO 2008 pb 3.

Pierre.

And this problem was already posted a couple of times, one of them very recently.


invariant
algebra
polynomial
geometry
geometric transformation
rotation
real analysis

Let $P(x,y)$ be a polynomial in two variables such that it is ivariant with respect to rotation on an angle $\frac{2\pi}{n}$ around origin $(0,0)$. Prove that if $degP$ is odd then it is $\geq n$ .

[quote="eugene"]... the it $\geq n$ .[/quote]

What is the question ?

Prove that the degree of the polynomial is at least n.

Yes, you are right,Harazi. I'm sorry for writing not too corect.

Rewrite $P$ as a polynomial $Q$ in two complex variables $z,\overline{z}$ using $x=\frac{z+\overline{z}}2$ and $y=\frac{z-\overline{z}}{2i}$. This transformation preserves the total degree. Now, if $\omega$ is a primitive $n$-th root of unity, $Q(\omega z,\omega^{-1}\overline{z})=Q(z,\overline{z})$

Consider a monomial of $Q$ $z^a\overline{z}^b$. This monomial transforms to $\omega^{a-b}z^a\overline{z}^b$. Every monomial of $Q$ must have equal coefficients before and after, so $n|(a-b)$. If the degree is odd, some term ust have $a-b$ nonzero and therefore either $a\ge n$ or $b\ge n$, making the total degree at least $n$.

With a little more work, we can show that the invariant polynomials are generated by two elements, which translate to $z\overline{z}$ and $z^n+\overline{z}^n$.


(1) The probability that an air defense warning station correctly detects an incoming missile only 0.4. How many of these warning stations, operating independently but in the same area , would be needed in order that the probability of at least one detecting an enemy missile is at least 0.99?

(2) Dan loves playing solitaire on his computer. He observes the probability of winning a game is 0.12, and he would like to win at least one game on three fourths of the days he plays; that is, the probability of winning at least one game on any day is 0.75. Assume solitaire is a Bernoulli trial (binomial distribution); how many games should he plan to play per day?

[hide="Problem 1"]The probability that $ n$ stations are wrong is $ (.6)^n$. We want $ (.6)^n < 0.01$. The minimal $ n$ is $ \boxed{n\equal{}10}$.[/hide]

The probability that one station is incorrect in its detection is $ 0.6$, because a station is either right or wrong, so $ P(wrong) \plus{} P(right) \equal{} 1$, so $ P(wrong) \equal{} 0.6$

The probability that two stations are wrong at the same time is equal to the probability that the first is wrong times the probability that the second is wrong, because the two are independent. Therefore, the probability they are both wrong is $ 0.6^2 \equal{} 0.36$.

We can generalize this to the probability that $ n$ stations are wrong; it is $ (0.6)^n$.

Therefore, the probability that $ n$ stations correctly defends a region is $ 1$ minus the probability that none protect; in other words, $ 1$ minus the probability that all $ n$ stations are wrong. So, $ 1 \minus{} (0.6)^n > 0.99$. This is equivalent to $ 0.01 > (0.6)^n$.

Yes.  :wallbash_red:

Fortunately, it doesn't affect the solution significantly.

2.
Note that this solution follows the same path as 1.

The probability of losing all games (or winning no games )  is $ 1\minus{}.75\equal{} 0.25$. Also, the probability that Dan loses any given game of solitaire is $ 1\minus{}.12\equal{}.88$. So the the number of games is $ .88^x\ge .25\implies x\approx 10.8\dots$, but since the number of games played cannot be a fraction, the number of games needed to be played is 11.


We start with a square $ABCD$. The diagonals $AC$ and $BD$ intersect at $O$. We draw circular arcs centred on $A$, $B$, $C$, and $D$ and passing through $O$. These arcs cut the sides of the square at $E$, $F$, $G$, $H$, $I$, $J$, $K$, and $L$ as shown. We join $LE$, $FG$, $HI$, and $JK$.

Prove that $EFGHIJKL$ is a regular octagon.

[hide="A bit easy, but..."]Let $AB=a$. Then $AE=AB-BE=a-\frac{a\sqrt{2}}{2}$ and $AE=BF$, hence $EF=AB-2AE=a-(2a-a\sqrt{2})=(\sqrt{2}-1)a$. On the other hand, $FG=BF\sqrt{2}=a\left(1-\frac{\sqrt{2}}{2}\right)\sqrt{2}=a(\sqrt{2}-1)=EF$. Also, $\angle EFG=180^\circ-\angle BFG=135^\circ$. Thus all sides of the octagon are equal and all internal angles have $135$ degrees, hence it's regular.[/hide]


Find all solutions to the following system:

\[\{\begin{array}{c}xy+yz+xz = 12\\ x+y+z+2 = xyz \end{array}\]

Where $x,y,z$ are positive real numbers.

[hide]AMGM equality case -> x=y=z=2.[/hide]

I agree that this problem is very easy, but it is also very nice, and I think you should elaborate your solution.

[hide]The first equation tells us that $12 = xy+yz+zx \geq 3\sqrt[3]{(xyz)^{2}}$, i.e. $8\geq xyz$.
The second equation tells us that $xyz = x+y+z+2 \geq 4\sqrt[4]{2xyz}$, ie. $(xyz)^{4}\geq 2^{9}xyz$, so $xyz \geq 8$.
Therefore, $xyz = 8$, so we have equality in both AM-GM's. That means $x=y=z=2$.[/hide]

Easy, but very nice.
[hide="Solution"]Notice that by AM-GM, $\frac{xy+yz+xz}{3}\geq\sqrt[3]{x^{2}y^{2}z^{2}}\implies xyz\leq 8$.
Similarly, $\frac{x+y+z+2}{4}\geq\sqrt[4]{2xyz}\implies xyz\geq 8$.
Hence,  $xyz=8\implies x+y+z=6$.
Notice that by $(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2(xy+yz+xz)$, $x^{2}+y^{2}+z^{2}=12\geq xy+yz+xz=12$.
Equality holds when $x=y=z=2$.$\blacksquare$[/hide]
Masoud Zargar


Point P is 9 units from the center of a circle of radius 15.  How many different chords of the circle contain P and have integer lengths?

The shortest chord of integer length is 24 and the longest chord of integer length is 30.

There would be chord lengths of 25, 26, 27, 28 and 29 between the longest and the shortest chord.

By doubling the number of chords in between, (5), we would have a total of 12 chords of integer length.

[hide]Since P is $9$ away from the center, one side of the circle is $9+15=24$ and the other is $30-24=6$.
The product of the chord lengths when divided into two is constant. So the product of the divided lengths of the chord that passes through P is $24 \times 6=144$.
$144$ has the factor pairs
$1 \times 144$
$2 \times 72$
$3 \times 48$
$4 \times 36$
$6 \times 24$
$9 \times 18$
$12 \times 12$
and the opposites ($18 \times 9$...)
However, the only ones that would fit in the circle would be
$9 \times 18$
$12 \times 12$
$18 \times 9$.
$6 \times 24$
You don't include $24 \times 6$ because it is the same chord as $6 \times 24$ because it is the diameter
That is $4$ total[/hide]

Well I used the power of point theorem about chords of a circle. It is something like this:
If AB and CD are chords that intersect at E then
$AE \times EB = CE \times ED$

But why is that necessary? ritchjp's answer makes more sense than what you did, unless you can explain why most of the chords that he found don't work.

I dunno if I can disprove what richjp said.
Robinhe, what is the answer?

[hide="The answer is..."]12, but I still have a couple of questions for ritchjp: Why do you double the number of chords in between and why is the smallest chord 24?[/hide]

hi robinhe:

By drawing a line through the center of the circle and passing it through pt $P$ which is the same length as the diameter you are actually producing the longest chord possible.

The shortest chord is the chord that is perpendicular to the diameter and intersects the diameter at $P$. If the point $P$ is $9$ units away from the center and the radius is $15$, you can determine half the length of the smallest chord  using the radius as the hypotenuse, the portion of the longest chord or diameter from the center to $P$ as one leg and half the smallest chord as the other leg, which comes out to $12$. So the length of the shortest leg is $24$.

As for the remaining chords, you double the number of chords because each chord has a mirror image on the opposite side of the diameter. Thus if you have 5 chords on one side, you'll also have 5 chords on the other side of the diameter.


How are organic matters defined?I've read a few books saying that all Hydro-carbons are said to be organic.But originally Amonium Cyanet is in-organic. Also in this reaction :CH3-CH2-OH+H-O-SO3-H=CH3-CH2-OSO3H+H2O.
C2H5S04H is inorganic?Why?Is there any special reason?

I think that inorganic compounds are governed by ionic bonds. For instance, magnesium oxide, MgO is an inorganic compound, and it has an ionic bond. However aminoethane is an organic compound, and it has purely covalent bonds.

In the example you gave, CH3CH2OSO3H is a covalently-bonded molecule. So, the molecule is primarily organic. However, C2H5OSO3H has an ionic bond between the C2H5 carbocation and the sulfate anion. So, the latter molecule is generally regarded as inorganic.

What is iffy for me is organometallic reagents: whether they're "inorganic" or "organic".

$ \ce{CO2}$ and $ \ce{CO}$ are both covalent molecules but still regarded as inorganic.

[quote]I think that inorganic compounds are governed by ionic bonds. For instance, magnesium oxide, MgO is an inorganic compound, and it has an ionic bond. However aminoethane is an organic compound, and it has purely covalent bonds.
[/quote]Thnx :)
[quote]What is iffy for me is organometallic reagents: whether they're "inorganic" or "organic".
[/quote]Such as? :huh:
Sorry but I suck at chemistry. :blush:

I thought the definition of an organic compound was that it contains [b]carbon covalently bonded to another carbon[/b].  :huh:

[quote="Carcul"]An organic compound is one which have at least one C-H covalent bond.[/quote]

What about $ \ce{CCl4}$?

Organic compounds are carbon compounds excluding things like carbonates, oxides of carbon, and other carbon containing species we commonly refer to as being inorganic.

CCl4 is organic.But then what is the definition?Traitor's definition is not clear.Precisely what sort of compounds do we exclude? :huh:
[quote]What is iffy for me is organometallic reagents: whether they're "inorganic" or "organic". [/quote]
Organometalic reagents are studied in organic chemistry and they shw different properties.That's all I know.

Well I'm pretty that CCl4 is organic and CO2 is not. :D

Truth is that most definitions of what is organic and what is inorganic are too restrictive and have exceptions.

The best definition I found (after browsing the web) :D  is that:

"An organic compound is whatever an organic chemist says it is; an inorganic compound is whatever an inorganic chemist says it is."

Of course, true this may be it's not very useful.

Just look at the wiki page. Organic compound is any compound with carbon, excluding, [i]by historical reasons[/i], carbonates, simple oxides of carbon, and cyanides. I personally don't like this definition, since, in reactivity, carbon dioxide and carbon monoxide are very closely related with carboxilic acids, ketones, and carbocations. Here's a more meaningfull and modern definition:

"An organic compound is any covalent compound of carbon."

[quote]And what is the precise reason to consider carbon tetrachloride organic and CO2 not?[/quote]
That was exactly the thing I wanted to know.Can anyone help me on that?

i think the basic concept for every organic compound is to be a derivative of a hydrocarbon

And what is your definition of a "derivative of a hydrocarbon"?

I guess a "derivative of a hydrocarbon" would simply have a hydrocarbon somewhere in the molecule?

As before CCl4 doesn't have hydrogen in it(though it is got from methane).I think the definition is
"Any covalently bonded Carbon excluding oxides,metalcyanides,sulphates,thyo-sulphates which are studied in inorganic chemistry."

well i think rt's defination is absolutely precise

CCl4 is ofc a derivative of CH4
derivative means we remove H by some other grp (if i m rite)

and wiki's def is the basic abt organic comp everybody treats that only--all compounds that contain carbon excpt carbonates
and carbon oxides. :)
PS@Carcul - isn't cyanide organic???

but bio has sent me to thinking
[url]http://www.mathlinks.ro/viewtopic.php?t=227720[/url]

It makes sense to consider cyanide ion as organic, but there is no problem in considering it inorganic. Is this distinction really so important?

[quote="Carcul"] Is this distinction really so important?[/quote]
thats the point  :P  :D


$a,b,c$are positive real numbers.prove that:

$(a/b+2c) +(b/c+2a) +(c/a+2b)$>_   $1$

[quote="1234567890"][quote="Fermat -Euler"]$(\frac{a}{b}+2c) +(\frac{b}{c}+2a) +(\frac{c}{a}+2b)\geq(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})\geq3(\frac{a}{b}\frac{b}{c}\frac{c}{a})^{\frac{1}{3}}$ $=3 \geq1$[/quote]
How did you prove the first inequality?[/quote]

 :o $(\frac{a}{b}+2c) +(\frac{b}{c}+2a) +(\frac{c}{a}+2b)\geq(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})$
because $2(a+b+c)\geq0$


If the following ineq is true,then we are done:
$\boxed{\frac{a^2+b^2}b+\frac{a^2+c^2}a+\frac{b^2+c^2}c \geq \frac{b^3+a}b + \frac{a^3+c}a+\frac{c^3+b}c}$,because it implies $\frac{a^2}b+ \frac{b^2}c+\frac{c^2}a+a+b+c \geq a^2+b^2+c^2+\frac{a}b+\frac{b}c+\frac{c}a \iff \frac{a^2}b+ \frac{b^2}c+\frac{c^2}a+a+b+c -2\sqrt{(a^2+b^2+c^2)(\frac{a}b+\frac{b}c+\frac{c}a)}\geq a^2+b^2+c^2+\frac{a}b+\frac{b}c+\frac{c}a-2\sqrt{(a^2+b^2+c^2)(\frac{a}b+\frac{b}c+\frac{c}a)}\geq 0 \longrightarrow \frac{a^2}b+ \frac{b^2}c+\frac{c^2}a+a+b+c -2\sqrt{(a^2+b^2+c^2)(\frac{a}b+\frac{b}c+\frac{c}a)} \geq 0$.
Thus the ineq is solved.But I got stuck at proving the boxed ineq :blush:

[quote="andyciup"]If the following ineq is true,then we are done:
$\boxed{\frac{a^2+b^2}b+\frac{a^2+c^2}a+\frac{b^2+c^2}c \geq \frac{b^3+a}b + \frac{a^3+c}a+\frac{c^3+b}c}$[/quote]

Unfortunately, this inequality does not hold (for all positive reals $a,b,c$). You can easily find a counter example starting with very small values.

Has anyone trying to solve it?

Thank you very much. :blush:

[quote="manlio"]Thank you very much.[/quote]

Actually, we can refer this inequality from the following stronger result
\[ 3\left(\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}\right)+3\left(a+b+c\right) \ge \frac{9(a^{2}+b^{2}+c^{2})}{a+b+c}+(a+b+c)\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right). \]

Sorry. I do it in a hurry, so I miss.  :blush:

[quote="ductrung"][quote="hungkhtn"]Actually, we can refer this inequality from the following stronger result
\[ 3\left(\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}\right)+3\left(a+b+c\right) \ge \frac{9(a^{2}+b^{2}+c^{2})}{a+b+c}+(a+b+c)\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right). \]
[/quote]
Dear Hung,
I think this is not true: try $ (a,b,c)=(8,4,1)$.

[quote="NguyenDungTN"]SOS Method![/quote]
Dear NguyenDungTN,
From your transformation to SOS form, I get
$ S_{c}=\frac{1}{b}\left( \sum\frac{a^{2}}b+3\sum a \right)-\frac{4(a^{2}+bc)+2ab}{ab}$
So I couldn't see how you get
$ S_{c}=2a^{3}+4a^{2}b+5ab^{2}+6abc-2b^{2}c$
Could you please elaborate?

Manlio,
I will post my solution soon.[/quote]

Sorry!
$ S_{c}=2a^{4}c^{2}+4a^{3}bc^{2}+5a^{2}b^{2}c^{2}+6a^{2}bc^{3}-2ab^{2}c^{3}$
$ S_{b}=2c^{4}b^{2}+4ab^{2}c^{3}+5a^{2}b^{2}c^{2}+6ab^{3}c^{2}-2a^{2}b^{3}c$
$ S_{a}=2b^{4}a^{2}+4a^{2}b^{3}c+5a^{2}b^{2}c^{2}+6a^{3}b^{2}c-2a^{3}bc^{2}$

Can you please post your solution, Ductrung?

Thank you very much.

[quote="ductrung"]For any positive real numbers $ a,b$ and $ c$, prove that
\[ \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a} \plus{} a \plus{} b \plus{} c \ge 2\sqrt {(a^2 \plus{} b^2 \plus{} c^2)\left(\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\right)}\]
This is a continuation from [u]http://www.artofproblemsolving.com/Forum/topic-84160.html[/u][/quote]

This inequality is [b]Problem 2.60[/b], page 114 of new book (where it is presented with a AM-GM proof):

Vasile Cirtoaje, Vo Quoc Ba Can, Tran Quoc Anh, [url=http://canhang2007.wordpress.com/2009/12/08/new-book-inequalities-with-beautiful-solutions-v-cirtoaje-v-q-b-can-t-q-anh/][b][i]Inequalities with Beautiful Solutions[/i][/b][/url], GIL publishing house 2009.

[quote=sqing][quote=ductrung]For any positive real numbers $a,b$ and $c$, prove that
\[ \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+a+b+c \ge 2\sqrt{(a^2+b^2+c^2)\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)} \]
This is a continuation from [u]http://www.artofproblemsolving.com/Forum/topic-84160.html[/u][/quote]
[url=https://artofproblemsolving.com/community/c6h498248p2799685]h[/url]
Given $ a,b,c > 0 $ [url=https://artofproblemsolving.com/community/c6h1926366p13218923]prove that [/url]
$$ \frac{a^2}{b} +\frac{b^2}{c} +\frac{c^2}{a} + a+b+c \geq \frac{6(a^2+b^2+c^2)}{a+b+c} $$[/quote]

Let us thank, arqady for closing this configuration 6 years ago.
https://artofproblemsolving.com/community/c6h296853


This is the problem:
1. Find Phi of 25?

2. Find coprime set of 25?

3. Find the order of 25?

Uh... I think this should be moved to the High School Basics forum. ;)


inequalities
induction
inequalities proposed

Let $a_{1},a_{2},...,a_{n}$ be positive real numbers with product $1$. Consider the following inequality
\[\frac{1}{1+na_{1}}+\frac{1}{1+na_{2}}+...+\frac{1}{1+na_{n}}+\frac{1}{1+a_{1}+a_{2}+...+a_{n}}\ge 1.\]
(1). Prove this inequality for $n=3$.
(2). Prove this inequality if $a_{1},a_{2},...,a_{n}\ge \frac{n-1}{n}.$
(3). Prove the inequality for all positive numbers $a_{1},a_{2},...,a_{n}$.

I forget to write the condition $n\ge 3$. Sorry.

Only case (1), this is difficult problem... :blush: I tried prove it by Mixing Variable Method and I found a solution but it isn't nice  :(
The first, we will the following inequality (With case n=3).
\[\frac{1}{1+3a}+\frac{1}{1+3b}+\frac{1}{1+3c}+\frac{1}{1+a+b+c}\ge1 \]
We put:
\[f(a,b,c)=\frac{1}{1+3a}+\frac{1}{1+3b}+\frac{1}{1+3c}+\frac{1}{1+a+b+c}-1 \]
Therefore, we have:
\[f(a,\sqrt{bc},\sqrt{bc})=\frac{1}{1+3a}+\frac{2}{1+3\sqrt{bc}}+\frac{1}{1+a+2\sqrt{bc}}-1 \]
By factor expression we have:

\[f(a,b,c)-f(a,\sqrt{bc},\sqrt{bc})\]
\[=(\sqrt{b}-\sqrt{c})^{2}(\frac{9-\sqrt{bc}}{(1+3b)(1+3c)(1+3\sqrt{bc})}-\frac{1}{(1+a+b+c)(1+a+2\sqrt{bc}})\]

Now we will prove that:
\[\frac{9-\sqrt{bc}}{(1+3b)(1+3c)(1+3\sqrt{bc})}-\frac{1}{(1+a+b+c)(1+a+2\sqrt{bc}}\ge 0 \]
It is equivalent to:
\[N=-27t^{3}+t^{2}S_{a}+tS_{b}-S_{c}\ge 0 \]
With:
\[t=\sqrt{bc}\]

\[S_{a}=18a+18b+18c+9 \]

\[S_{b}=9a^{2}+9ab+9bc-6b-6c+12a) \]

\[S_{c}=6(a+b+c)+3(a^{2}+ab+ac)+4 \]
........ We can check $N\ge 0$  :blush: (I hope anyone will find a nice proof to solve: $N\ge 0$)
After we have:
\[f(a,\sqrt{bc},\sqrt{bc})=\frac{1}{1+3a}+\frac{2}{1+3\sqrt{bc}}+\frac{1}{1+a+2\sqrt{bc}}-1 \]

\[\iff \frac{1}{1+\frac{3}{t^{2}}}+\frac{2}{1+3t}+\frac{1}{1+\frac{1}{t^{2}}+2t}-1 \]

\[\iff \frac{(t-1)^{2}(7t^{3}-t^{2}-3t+3)}{t^{4}}\ge 0 \]
we are done.  :( I sure there are a nice solution exits ...
hum... I will try another way...

[quote="Vasc"]Very nice and hard inequality, Hung.
We can prove (3) using (n-1)EV Theorem.[/quote]

I use the improved SMV theorem, but it is very hard, too. For (i), I use induction (elementary) but it is very hard, too. I wonder if EV is valid here VASC, I think you may use it with some special change.

[quote="Vasc"][quote="hungkhtn"]  I wonder if EV is valid here VASC, I think you may use it with some special change.[/quote]

Of course, there are two cases. EV Method solves only  the case $x_{i}\le 2n$ for all $i$.[/quote]

Exactly as I guess, VASC  :lol: Since I intend to create this problem as a kind of inequality that a a normally applying of EV can not show the result directly. For $(i)$, a very nice solution exists. For (ii), I think the elementary solution by induction is better, too.


calculus
integration
calculus computations

Hey guys I need a method to solve this integral using substitution

\[\int_{1}^{2}(x-1) \sqrt{2-x}\; dx\]

There are two substitutions, you may set $2-x=t$ or $\sqrt{2-x}=t$.

Yes thankyou  but when I comput the integral I keep getting 4/15 yet my text book says it's 1/2. I need to know where I made a mistake

Let's use kunny's first choice. Let $u=2-x.$ Then $x=2-u$ so $x-1=1-u$ and $dx=-du.$ When $x-1,\ u=1$ and when $x=2,\ u=0.$

$\int_{1}^{2}(x-1)\sqrt{2-x}\,dx=-\int_{1}^{0}(1-u)\sqrt{u}\,du=\int_{0}^{1}u^{1/2}-u^{3/2}\,du$

$=\left.\frac23u^{3/2}-\frac25u^{5/2}\right|_{0}^{1}=\frac23-\frac25=\frac4{15}.$

Possibilities: maybe you copied the problem down wrong, or maybe you're reading the wrong line of answers or in the wrong section. Or maybe the book is wrong - that happens.

Thanks for the explanation Kent. You were right I was looking at the wrong line...


Given any equilateral triangle with sides of length s, consider the locus of all points $P$ in the plane of the triangle such that the sum of the squares of the distances from $P$ to the vertices of the triangle is a fixed number $a$. What is the locus of all such points?

[quote="litoxe2718281828"]I'm assuming you have a prettier solution.... :lol:[/quote]Speaking strictly $\LaTeX$ wize, I do have one  :lol:


I am looking for someone to solve the questions below using steps as you go through the lesson. I would like to learn how to solve similar questions on my own. Ready?

1) How many points are 5 units from the origin and 5 units from the x-axis?

2) How many points are 7 units from the origin and also 7 units from the y-axis?

#1

Any point that lies 5 units from the origin lies on a circle centre O (the origin), radius 5. Any point that lies 5 points from the x-axis lies on one of the two lines parallel to the x-axis, with a distance of 5 between it and the x-axis.
Each line intersects the circle in exactly one place- one at $(0,5)$ and the other at $(0,-5)$.(edited)

#2 is pretty much the same, draw a circle centre the origin and two [i]vertical[/i]  parallel lines, they should intersect at $(0,7)$ and $(0,-7)$(edited)

Quick questions:

Why can't the points be (0,5) and (0,-5)?

Don't they also intersect at points other than (0,5) and (5,0)?

Same for question 2.

Don't they also intersect at points other than (0,7) and (7,0)?

:wallbash:

yeah, you're right: $(0,5)$ and $(0,-5)$ and $(7,0)$ and $(-7,0)$
don't know what I was thinking, i'll edit my post

Each line only intersects the circle once- in fact, they are both tangent to it.


It's an interesting inequality. Where can I find a proof?


[quote="236factorial"]Two numbers, a and b , are randomly selected without replacement from the set {2,3,4,5,6,7,8,9} . What is the probability that the fraction a/b is less than 1 and can be expressed as a terminating decimal? Express your answer as a common fraction in lowest terms.[/quote]

  [hide]  There are $8 \cdot 7=56$ different possible fractions.

  Terminatng fractions must have denominators expressible in the form: $2^a \cdot 5^b$  Only 2, 4, and 5 work

  This combined with the condition the fractions must be smaller than 1, yields:  2 + 3 = 5 combinations.

  For our desired fraction of: $\frac{5}{56}$ [/hide]

G-Unit fails.

[hide="Answer"]
There's 56 numbers available (7 times 8).

Anything over 2, 4, 5, or 8 results in a terminating decimal.
That's 11 numbers right there.

But anything that reduces to a terminating decimal works too.
3/6 works.

12/56 = [b]3/14[/b]
[/hide]


superior algebra
superior algebra theorems

Question:
1. If G is of infinite order, must it have an element of infinite order?
Counter-examples please

$\mathbb Z_{2}\times \mathbb Z_{2}\times \mathbb Z_{2}\times \ldots$

Thanks,
Yeah, that one does it, but are there any other examples not obtained in that manner (as products)?


Consider the image. we are trying to prove angle 6 is equal to angle 3. angle 7 is equal to angle 6 (vertical angles), and then using the corresponding angles rule, angle 7 is equal to angle 3. That must also mean angle 6 is equal to answer 3, proving that alternate interior angles must be congruent. Why is it a postulate then?

Here is a reason from mathworld.

A statement, also known as an axiom, which is taken to be true without proof. Postulates are the basic structure from which lemmas and theorems are derived. The whole of Euclidean geometry, for example, is based on five postulates known as Euclid's postulates.

Hope this answers your question.

I read that postulates were statements that are assumed true but can't be proven, while theorems are true and proven.

[quote="236factorial"]I hope that this is middle school level.

Why is the rule [i]when two parallel lines are cut by a transversal, their alternate interior angles are congruent[/i] a postulate. Can't you just prove that the vertical angle of one of the angles is congruent to the other angle using the corresponding rule and then transitively, the two angles are congruent? I hope you get what I mean.[/quote]

Consider these two statements:

[b]Statement #1:[/b] If two parallel lines are cut by a transversal, then their alternate interior angles are congruent.

[b]Statement #2:[/b] If two parallel lines are cut by a transversal, then their corresponding angles are congruent.

Let me explain the idea of this: initially there must be some postulates that you must assume true in a certain "world", in this case the Euclidean space. So your claim above is if you assume statement #1 as postulate, then you can prove statement #2. Similarly, probably in your geometry class, if you assume statement #2, then you can prove statement #1.

We can assume both as postulates but that's not going to be fun since we can assume one statement and prove the other. As long as you assume one of them as a postulate and the other one as a theorem, then it is fine. Think of a postulate as a starting point of a theory and a theorem as a derived statement.

i think that, the postulate is just a shortcut. instead of using the verticle angles and using the transitive property.

There's the answer. Transitive property.


This was a homework question from my Discrete Math course this semester.  It's pretty simple, but a fairly nice introduction to a topic that you guys might not have seen before.  Please ask for clarification if any of the terms are unclear.

We call $G(V,E)$ (or just $G$) a graph if $V$ is a set of vertices and $E$ is a set of edges.  There is at most one edge between a pair of vertices, and there may be none.  $K_n$ is the 'complete' graph of $n$ vertices where there is an edge between any two vertices.  $\overline{G}$, or the complement of $G$, is the graph with the same vertex set $V$ as $G$, but its edge set is the relative complement of the vertex set of $K_n$ with $E$.  Basically, $\{a,b\}\in\overline{G}$ iff $\{a,b\}\not\in G$.  A graph is connected if, for every $a,b\in V$, there exists a 'path' (this is an intuitively obvious term) connecting $a$ and $b$.

Here's the problem:
If $n\geq 2$, prove that if $G$ is not connected, then $\overline{G}$ is connected.

Feel free to ask for hints.

How about this:

If G is not connected and of greater than one vertex than there must exist two vertices, a and b, of G, such that there exists no path between a and b.  Furthermore, all vertices connected to a cannot be connected to b, and vice-versa.  Let A be the set of all of the vertices connected to A, B the set of all of the vertices connected to B, and C be the set of all of the vertices connected to neither.  No point in A can be connected to a point in B, or else a is connected to b.  Therefore, in the complement of G, every point in A is connected to every point in B, and in C, and every point in B is connected to every point in C.  To connect any two points within A, since both must be connected to every point in B (for n > 1), there is a point which they are both connected to, so the two points are connected.  The same argument applies to every pair of points within B and C.  Therefore, all points in the complement of G are connected by some path.

~ Alex


extends to the radius OA of a circle of a length equal to AB same to OA; take to a tangent to the circle, on which there is the perpendicular NA and BC.
Assuming that the angle <OAC is equal to 126º, calculate the value of <ACB

Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=285084[/url]


trigonometry
calculus
calculus computations

Solve $ \frac {d^2 y}{dx^2} \minus{} 6\frac {dy}{dx} \plus{} 13y \equal{} e^{3x}\sin x$.

hello, by the ansatz $ y(x)\equal{}e^{\lambda x}$ we get the solution of the homogeneous part of the equation
$ y(x)\equal{}C_1e^{3x}\sin(2x)\plus{}C_2e^{3x}\cos(3x)$ by the ansatz $ y_p\equal{}A\cdot e^{3x}\sin(x)$ we get the solution of the inhomogeneous part, hence we have $ y(x)\equal{}C_1e^{3x}\sin(2x)\plus{}C_2e^{3x}\cos(3x)\plus{}\frac{1}{3}e^{3x}\sin(x)$
Sonnhard.

Ko- n- ni- chi- wa, Dr Sonnhard Graubner.

The solution of  the homogeneous part of the equation seems to be incorrect.

kunny

hello Kunny, please put $ y(x)\equal{}C_1\cdot e^{3x}\sin(2x)\plus{}C_2\cdot e^{3x}\cos(2x)$ in the homogeneous part of your equation, i think, the solution is correct.
Sonnhard.

Dr. Sonnhard: I believe the proper solution is
[hide]$ \frac{e^{3x}}{3} \cdot (3C_1 \cos (2x) \plus{} 3C_2 \sin (2x) \plus{} \sin x)$[/hide]

hello, what is the difference between my solution
$ y(x)\equal{}C_1e^{3x}\sin(2x)\plus{}C_2e^{3x}\cos(2x)\plus{}\frac{1}{3}e^{3x}\sin(x)$ and this solution
$ \frac{e^{3x}}{3} \cdot (3C_1 \cos (2x) \plus{} 3C_2 \sin (2x) \plus{} \sin x)$?
Sonnhard.

Graubner's answer is no different to DiscreetFourierTransform's!

There is a typographical error in Sonnhard's first post in the thread.  It should not read cos 3x.

hello, yes of course, you have right, thank you.
Sonnhard.


inequalities
algebra
polynomial
inequalities theorems

hi everyone !
help me please
I need one of method of proving inequality which
proofs  . whith polynomial equations

\[P(1;1;1) and P(1;1;0) and P(1;0;0) and others \]

Sory I don't understand your question . Can you edit it ?


summer program
Mathcamp
Ross Mathematics Program
PROMYS
quadratics
HCSSiM
number theory

What are some good math camps for people going into the 9th or 10th grade?

Any help will be appreciated.

i hear that mathcamp is supposed to be really goode. but check out the resources section for more math programs. i cant tell if theyre good or not, but i believe there some threads that discuss this.

there are 3 terrific math programs that introduce you to pure mathematics i.e. proofs. The Ross Program, PROMYS, and Texas State Honors Summer Math Camp. Ross is 8 weeks while the two others are 6 weeks. Definitely go to one if you are interested in math.

[quote]they cost a lot don't they? and isn't there application requirements? what ar ethose requirements?[/quote]

Just Google up the websites, there are probably applications on each one.

those programs are listed in the resources section

I can vouch for PROMYS, i went to PROMYS 2003. Very good program, but its a lot of work. It costs about 2200 for room and board, nothing for tuition. But you can get some financial aid if necessary.

In the main course, you are given a problem set to do each weekday that consists of about 20 proofs and investigative topics.
They usually took me about 3 or 4 hours to do, although some questions themselves took me a couple of hours.

The main number theory course is laid out very well, forcing you to be completely rigorous (from axioms) for about three or so weeks, until the advanced material is introduced.

There are also side lectures (combinatorics was offered last year along with some other topics) that are also very well planned, yet not as much work as the NT lecture.

This previous summer, I was a first-year at the ROSS program.  I would HIGHLY recommend it, as it has changed my life, giving me a deep love for mathematics.

The program costs 2700 for 8 weeks, it does have application problems but they are there so that Dr. Shapiro (head of the program) can see how you approach a problem.

This program is very similar to PROMYS, except that is harder material, and it covers more topics on number theory and algebra than PROMYS.  The first few problem sets are not terribly difficult, and they tend to increase in difficulty, with some problems taking many hours/days.

The ROSS program is not a program aimed at contest math, it is instead aimed at introducing students to pure mathematics and the art of proofs.  But even so, after taking the ROSS program, my ability to do contest problems increased dramatically.

There were also week long seminars in:

Analytic Number Theory

Unique factorization of ideals in quadratic extensions of the integers

Elliptic curves over the rational numbers

Number theory in F[x]

Ordinals and well-orderings

each taught by respected research mathematicians and graduates/former counselors of ROSS.

I definately recommend the program for anyone interested, and if you do go, you will be able to meet me as I will almost undoubtedly be returning as some type of counselor.

[quote="zscool"]This previous summer, I was a first-year at the ROSS program.  I would HIGHLY recommend it, as it has changed my life, giving me a deep love for mathematics.[/quote]

I know locally a Ross program alumnus from the early 1970s. He thinks very highly of that program to this day, and his subsequent career perhaps validates the program, as he got a Ph.D. in math and works as a consultant for a major consulting firm.

I'm pretty sure that the Ross program is the granddaddy of PROMYS and other programs of that nature, as most of those programs were founded, I have read on the Web, by Ross program alumni. I would definitely urge anyone I know of the proper age to apply to one of those programs for an interesting summer of math.

Has anyone else gone to Hampshire College math camp in Massachusetts?  I had a great time there, and it indeed "changed my life".  For the first time, I met lots of other kids who were interested in math.  It confirmed my decision to become a mathematician.

That was many years ago, though, and I don't know if it has changed.

Modern opinions are overwhelmingly positive on Hampshire, although I haven't gone there myself.

HCSSiM is spectacular!  I went there last year, and I had a great time.  The material is a little challenging, so one should study before they go to the camp.  You will meet all sorts of intresting kids there.

The only negative is that the dorms can be dirty.... You'll live through it though!

well as Joel himself as showed up, I will vouch for two programs... HCSSiM, of which I myself attended, and CCNY's Scholars Program (of which Joel of AoPS Fame taught).  HCSSiM in general is an absolutely amazing experience.  Of course, both experience ultimately depends on how much YOU want to get out of it.  As most math programs assume mathematical as well as emotional maturity  (at least to a certain extent, which I lacked / lack), one will not benefit nearly as much if one were to show up and just expect the program to produce miracles!


conics
parabola
geometry
geometric transformation

Consider the graphs of (1): $ y\equal{}x^2\minus{}\frac{1}{2}x\plus{}2$ and (2) $ y\equal{}x^2\plus{}\frac{1}{2}x\plus{}2$ on the same set of axis.  These parabolas are exactly the same shape.  Then:

$ \textbf{(A)}\ \text{the graphs coincide.} \\
\textbf{(B)}\ \text{the graph of (1) is lower than the graph of (2).} \\
\textbf{(C)}\ \text{the graph of (1) is to the left of the graph of (2).} \\
\textbf{(D)}\ \text{the graph of (1) is to the right of the graph of (2).} \\
\textbf{(E)}\ \text{the graph of (1) is higher than the graph of (2).}$

[hide="Click for solution"]
Graphing shows that (1) is to the right of (2), or $ \boxed{\textbf{(D)}}$.  We can also consider the two vertex's.
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Two parabolas with the same $ x^2$ coefficent are always the same shape because we can complete the square; then, one is just a translation of the other. In this case, we have $ (x \minus{} \frac {1}{4})^2 \plus{} \frac {31}{16}$ and $ (x \plus{} \frac {1}{4})^2 \plus{} \frac {31}{16}$. Thus, one is $ y \equal{} x^2$ centered at $ (\frac {1}{4},\frac {31}{16})$, the other is $ y \equal{} x^2$ centered at $ ( \minus{} \frac {1}{4},\frac {31}{16})$, and the answer is D.