According to this topology article, we could replace $b$ in $a + b\Bbb{Z}$ with $b^{k}$ for any fixed $k \gt 0$ and we should still have a topology.
For instance unions of subsets of the form $a + b^k \Bbb{Z}$. I verified the $\cap$ property for bases.
You could go further and even do unions o...
I'm working on a proof that $s_n\rightarrow\infty$ then $(s_n)^2\rightarrow\infty$ (introductory analysis), so I need to show that for all numbers $M$ there is $N$ s.t. $(s_n)^2\geq M$, whenever $n\geq N$
in taking the hypothesis on similar problems involving proofs for converging series, it generally involved constructing a manipulation of the hypothesis condition that could arrive at what is needed, but it was not quite clear to me how it can be done in a diverging case such as this
this is homework so I'd rather just have a nudge in the right direction, feel free to ask me a question
@BalarkaSen, I showed that it was compact, but not separable. Then I proved that all compact metric spaces are separable. Thus, if it was metrizable, then it would necessarily be separable, which is a contradiction.
well, a language. you can't do much topology over algebraic curves. the real useful analogy goes back to Grothendieck which i know nothing of. so i am not the right person to ask.
We have the equation $$2u_{xx}-u_{tt}+u_{xt}=f(x, t)$$
This is equal to $$\left (\frac{2\partial^2}{\partial{x^2}}-\frac{\partial ^2}{\partial{t^2}}+\frac{\partial ^2}{\partial{x}\partial{t}}\right )u=f$$
To find the characteristics do we solve the homogeneous equation $$\frac{2\partial^2}{\p...
Is it right that the integral curve of a zero point of a vector field is always constant? in other words zeros of vector fields are fixed points of flow?
You have already done that third part, notice that if $a = n \in \mathbb{N}$ then $a_{i} = 0$ for all $i \ge n+1$. So you have a polynomial of degree $n$ as a solution.
For the last part, just some ideas - try differentiating the Leguerre polynomial and substituting it into the equation. Or, co...
the way we define $+\infty$ is "something that is greater than all the other real numbers" and $-\infty$ being "something that is smaller than all the other real numbers". you can do this very rigorously by constructing a set by adjoining two elements with the set of all reals and appropriately defining a partial/total order.
that $\lim_{x \to 0^+} 1/x = +\infty$ doesn't mean $1/0 = +\infty$.
off-topic : i've usually done these problems by using partial differentiation. there's a result that tells you that total derivative of a function $df(x, y)$ is always $\frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy$. now you can differentiate $x^y$ using this very easily and subbing in $x = y$ gives you the desired. it's a cool trick.
It isn't. Only the calc sequence, linear algebra, one semester of abstract algebra, and one of (real analysis / complex analysis / differential geometry) are required @MaryStar.
Beyond that, I can choose whatever courses interest me.
the choice of L would seem dependent on the forward statement being at least convergent, would a converse here simply be referring to a sequence like 1,-1,1,-1,1,... able to meet the hypothesis of the converse? (despite having no obvious choice for "L"?)
there it is ^
I already proved the forward case relatively trivially with a variation of the triangle inequality
I remember seeing a question that asked for different visualizations of mathmathical principles. I'm pretty sure I saw it here on Math.SE. It demonstrated things like Sin/Cos in intuitive ways.
where divergence to infinity is that for every number $M$ there is a $N\in\mathbb{Z}$ such that $s_n\geq M$ for $n\geq N$
I've done most of these practice problems with fanciful manipulation of relevant inequalities, but have spent a while without noticing anything fruitful here, or that whatever I think of has a glaring issue
@GBeau I believe that it is true. $\lim_{n \to +\infty} a_n=L$ means that $\forall \epsilon>0, \exists n_0 \in \mathbb{N}$ such that $\forall n \geq n_0: |a_n-L|< \epsilon \Rightarrow L-\epsilon< a_n<L+\epsilon$. Since $- \epsilon-L<L-\epsilon<a_n< \epsilon+L$ we deduce that $|a_n|< \epsilon+L \leq \epsilon+|L|$. Thus: $||a_n|-|L|| \leq |\epsilon+|L|-|L||= \epsilon$. So $|a_n| \to |L|$.
@GBeau I think that we could prove it as follows. $s_n\rightarrow\infty\implies (s_n)^2\rightarrow\infty$
We know that $s_n \to +\infty$. That means that $\forall M>0, \exists n_0 \in \mathbb{N} $ such that $\forall n \geq n_0: s_n>M$
$s_n^2-k^2=(s_n-k)(s_n+k)$ where $k$ any positive integer.
From the above definition, $\exists n_0$ such that $\forall n \geq n_0$: $s_n>k \Rightarrow s_n-k>0$.
Also since $s_n \to +\infty$ we have that $s_n >0 \text{ for some } n \geq n_1$ and so it holds that $s_n+k>0 \Rightarrow s_n>-k,k>0$ for $n \geq n_1$.
"I commit to participate actively in Health for at least three months, especially during the private beta, and to ask or answer at least ten questions."
@Silent If $ab = ac = bc = abc$, then it means that all of them will have to be zero. Supposing that all of them are non-zero, we have $b = c$ from $ab = ac$ and $a = b $ from $ac = bc$. So we must have all of them equal to each other and this means that they're not distinct.
Just taught my class Gauss-Jordan elimination. They get confused easily, so I am trying to force myself to follow the algorithm as prescribed to a T, even if it results in fractions in inbetween steps. /sigh
"Is it a zero? No? Good. Is it a one? Yes? Good. Then what do I do next?..."
"Is it a zero? Yes? Bad. So what do I do next?..."
I'd like to think they all are getting it though. Even the football players had good questions.
The hardest part will be interpreting word problems, as usual.
Unfortunately not @Mary I have not looked into Partial Differential Equations yet, and am rather rusty even with ODE's. I can often help students looking for guidance on setting up a problem in ODE, but the method of actually solving I have to look up with them.
I focus mostly on combinatorics and graph theory, and have been taking what analysis and algebra courses seem to be necessary to switch into the PhD program here.
Yes, though just for a freshman undergrad course at the moment. I am currently in the Master's program here, and teach what are called 'recitations' where we briefly review the material that the professor lectured about and then run through several examples.
As for my own studies, the problem we are working on in combinatorics currently is that of Leveled Posets which have Symmetric Chain Partitions (SCP's), Hamiltonian Cycles (HC's), Hamiltonian Cycle Symmetric Chain Partitions (HCSCP's), and what are known as strong HCSCP's.
@JMoravitz So is your master related to combinatorics and graphs? I am taking this semester the subject Algorithms and Complexity and there we see graphs and Dynamic Programming.. Could I ask you something about it?
So, "why is polynomial time good and exponential time bad?" Well., given an input of $n$ pieces of information, we want to run some sort of algorithm on it to get a particular output to tell us something about our data. When it runs in polynomial time, for sizable $n$ it will finish the algorithm in a "reasonable" amount of time (e.g. trying to find the maximum element in a set of $n$ elements requires $n-1$ operations)
Comparing this to something that is nonpolynomial time: given a sizable $n$ it might finish the algorithm in an absurdly large amount of time (perhaps requiring more time than the universe is expected to have been in existence).
For example, computing what the graph ramsey number $Ram(K_7,K_7)$ is.
(What is the smallest $n$ for which a complete graph on $n$ vertices for which in any 2-coloring of the edges, there exists a monochromatic $K_7$)
If we were to take a guess, say we guess $k=32$, we need to look at every 2coloring of a graph on 32 vertices, (which there are $2^{32}$ such graphs), and determine if there is a monochromatic $K_7$ in that graph.
@evinda as for the fourth one, yes, heapsort runs in $O(n\log n)$ time, however despite the fact that that is not exactly a "polynomial"
it is better than some polynomial, therefore we call it "polynomial time" anyways.
Something that runs in $O(f(n))$ time is "NP (non-polynomial time)" iff for every polynomial $p(n)$ you have some $N$ for which every $n\geq N$ implies that $f(n)> p(n)$
in particular, $n\log n \leq n^2$ since $\log n\leq n$
so $O(n\log n)$ is not NP, implying that it is polynomial time
The last one is rather interesting, "Each problem that belongs to NP is solved in exponential time" is definitely false, but I would give a very different reason than you.
@evinda there are times even worse than exponential time. Take for example the question of calculating Van der Werden numbers, or other results in Ramsey Theory. Some of these numbers that appear in practice in calculating an upper bound for the $n^{th}$ number are ACKERMANIC.
You have polynomial time (uses a polynomial of the form $n^k$), you have exponential time (of the form $a^n$), you have towerian (of the form $a^{b^{\ddots^n}}$, you have Wowzer functions ( yea.. not going to bother writing this one), and then a class even bigger than that, Ackermanic
In particular, a towerian function, $t(n)$ is such that for every exponential function $e(n)$ you have some $N$ such that $n\geq N$ implies $t(n)>e(n)$
similarly for comparing wowzer and ackermanic functions too
@evinda you're being rather quiet. Is what I'm saying making sense?
Also, at the rest of the peanut gallery, $\ddots$ does diagonal topleft to down right, could someone remind me what the code is for bottom left to topright dots?
@JMoravitz For the first point: You said " When it runs in polynomial time, for sizable $n$ it will finish the algorithm in a "reasonable" amount of time". With it are you referring to the polynomial time? Also could you explain me further what a monochromatic $K_7$ is?
by a "reasonable" amount of time, yes, we consider "polynomial in $n$" to be "reasonable". Its not a very precise statement unfortunately, because even polynomial time algorithms can take an absurdly long time to run if the constants or powers are large enough
As for what I mean by a monochromatic $K_7$, a $K_n$ is the complete graph on $n$ vertices. (i.e. you have $n$ dots and a line between each and every one of the dots)
By monochromatic, I mean that every edge is the same color.
One can show that $Ram(K_3,K_3) = 6$. Worded in another way, at a party with six people, where each guest at the party either is friends with or is not friends with each other partyguest (and "is/isnot friends with" is reflexive), you can gaurantee that there is a group of 3 people who either are all simultaneously friends, or are all simultaneously not friends
Similarly, one can show that to be able to gaurantee a clique of four friends (every two people in the clique are friends) or a co-clique of four non-friends (every two people in the co-clique are not friends) (i.e. a square with diagonals), it will require 18 people.
relating this idea of friends and nonfriends to the definition I gave before, just color friendships with blue edges and nonfriendships with red edges
a clique of $n$ people will be a blue $K_n$ and a coclique of $n$ people will be a red $K_n$
The next number in the sequence however, $Ram(K_5,K_5)$ is still unknown. It is somewhere between 43 and 49. We do know that every ramsey number exists and is finite however, and have upper bounds.
Erdos was quoted as saying, if an alien race came and threatened the Earth with extinction unless we could produce $Ram(K_5,K_5)$, if we pooled together all of our resources, we might be able to calculate it exactly within a year, however if they ask us for $Ram(K_6,K_6)$ our only choice would be to launch a preemptive attack.
@JMoravitz So if we want to compute for example $Ram(K_3,K_3)$ can we consider that there two groups of three people each of the group and we check in each group if the persons are friends are not? Or have I understood it wrong?
To prove $Ram(K_3,K_3)$, it suffices to show that $n=5$ is not large enough (i.e. there exists a 2coloring of $K_5$ where there is neither a red triangle nor a blue triangle), and that $n=6$ is always sufficient.
For a proof: Consider $K_5$ colored with the outer edges blue and the inner edges red. I.e. a pentagram with outside blue and the star in the middle red
You can convince yourself that there is no monochromatic triangle
To prove 6 is sufficient, look at a specific vertex. Either that vertex has five edges leaving it. Either more than half will be blue or more than half will be red. Suppose it was blue
Look at the special vertex, and three of it's neighbors along blue edges. If there is at least one blue edge between two of its neighbors that forms a blue triangle and we are done. Else, there are only red edges between its three neighbors, that is a red triangle
The proof for $Ram(K_4,K_4)$ is much longer I'm afraid, so don't ask for it.
@JMoravitz I am confused now... :/ I haven't understood how we compute Ram(K_3, K_3). Do we draw 3 edges at one side and 3 other edges at the opposite side? And if so, what do we check?
Furthermore, you can construct a 5-regular graph on 6 vertices ($K_6$), so it follows that you can construct a 5-regular graph on any even number of vertices with $n\geq 6$
Suppose you have a 5-regular graph on $n\geq 6$ vertices. Look at two special vertices, $x$ and $y$ and to of each of their neighbors $x_l, x_r, y_l, y_r$ such that all six of these are different vertices (we can always find such a collection of six vertices. why?)
Try to construct your graph on $n+2$ vertices in the following way. Add two new vertices, $u$ and $v$. Delete the edges between $x,x_l$ and $x,x_r$ and $y,y_l$ adn $y,y_r$.
So, the degree of $x$ and $y$ each dropped by two, whereas the degree of each of $x_l,x_r,y_l,y_r$ each dropped by one
I wouldn't think so., it doesn't seem to be strong enough to imply anything
So, add an edge between $u$ and each of $x,x_r,y,y_r$, add an edge between $v$ and $x,x_l,y,y_l$, and add an edge between $u$ and $v$.
In doing so, the degrees of $x$ and $y$ each increased back up two, to return to being of degree 5, the degrees of each of $x_l,x_r,y_l,y_r$ each increased back up 1, to return to being of degree 5,
According to this topology article, we could replace $b$ in $a + b\Bbb{Z}$ with $b^{k}$ for any fixed $k \gt 0$ and we should still have a topology.
For instance unions of subsets of the form $a + b^k \Bbb{Z}$. I verified the $\cap$ property for bases.
You could go further and even do unions o...
We have the equation $$2u_{xx}-u_{tt}+u_{xt}=f(x, t)$$
This is equal to $$\left (\frac{2\partial^2}{\partial{x^2}}-\frac{\partial ^2}{\partial{t^2}}+\frac{\partial ^2}{\partial{x}\partial{t}}\right )u=f$$
To find the characteristics do we solve the homogeneous equation $$\frac{2\partial^2}{\p...
You have already done that third part, notice that if $a = n \in \mathbb{N}$ then $a_{i} = 0$ for all $i \ge n+1$. So you have a polynomial of degree $n$ as a solution.
For the last part, just some ideas - try differentiating the Leguerre polynomial and substituting it into the equation. Or, co...
