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# Bijections, Prufer Codes and Cayley’s Formula

I’m currently at the training camp in Cambridge for this year’s UK IMO squad. This afternoon I gave a talk to some of the less experienced students about combinatorics. My aim was to cover as many useful tricks for calculating the sizes of combinatorial sets as I could in an hour and a half. We started by discussing binomial coefficients, which pleasingly turned out to be revision for the majority. But my next goal was to demonstrate that we are much more interested in the fact that we can calculate these if we want than in the actual expression for their values.

Put another way, my argument was that the interpretation of $\binom{n}{m}$ as the number of ways to choose m objects from a collection of n, or the number of up-and-right paths from (0,0) to (m,n) is more useful than the fact that $\binom{n}{m}=\frac{n!}{m!(n-m)!}$. The opening gambit was to prove the fundamental result underlying the famous construction of Pascal’s triangle that

$\binom{n+1}{m+1}=\binom{n}{m}+\binom{n}{m+1}.$

This is not a hard result to prove by manipulating factorials, but it is a very easy result to prove in the path-counting setting, for example.

So it turned out that the goal of my session, as further supported by some unsubtly motivated problems from the collection, was to convince the students to use bijections as much as possible. That is, if you have to count something awkward, show that counting the awkward thing is equivalent to counting something more manageable, then count that instead. For many simpler questions, this equivalence is often drawn implicitly using words (“each of the n objects can be in any subset of the collection of bags so we multiply…” etc), but it is always worth having in mind the formal bijective approach. Apart from anything else, asking the question “is this bijection so obvious I don’t need to prove it” is often a good starting-point for assessing whether the argument is in fact correct!

Anyway, I really wanted to show my favouriite bijection argument, but there wasn’t time, and I didn’t want to spoil other lecturers’ thunder by defining a graph and a tree and so forth. The exploration process encoding of trees is a strong contender, but today I want to define quickly the Prufer coding for trees, and use it to prove a famous result I’ve been using a lot recently, Cayley’s formula for the number of spanning trees on the complete graph with n vertices, $n^{n-2}$.

We are going to count rooted trees instead. Since we can choose any vertex to be the root, there are $n^{n-1}$ rooted trees on n vertices. The description of the Prufer code is relatively simple. Take a rooted tree with vertices labelled by [n]. A leaf is a vertex with degree 1, other than the root. Find the leaf with the largest label. Write down the label of the single vertex to which this leaf is connected, then delete the leaf. Now repeat the procedure, writing down the label of the vertex connected to the leaf now with the largest label, until there are only two vertices remaining, when you delete the non-root vertex, and write down the label of the root. We get a string of (n-1) labels. We want to show that this mapping is a bijection from the set of rooted trees with vertices labelled by [n] to $[n]^{n-1}$.

Let’s record informally how we would recover a tree from the Prufer code. First, observe that the label of any vertex which is not a leaf must appear in the code. Why? Well, the root label appears right at the end, if not earlier, and every vertex must be deleted. But a vertex cannot be deleted until it has degree one, so the neighbours further from the root (or ancestors) of the vertex must be removed first, and so by construction the label appears. So know what the root is, and what the leaves are straight away.

In fact we can say slightly more than this. The number of times the root label appears is the degree of the root, while the number of times any other label appears is the degree of the corresponding vertex minus one. Call this sequence the Prufer degrees.

So we construct the tree backwards from the leaves towards the root. We add edges one at a time, with the k-th edge joining the vertex with the k-th label to some other vertex. For k=1, this other vertex is the leaf with maximum label. In general, let $G_k$ be the graph formed after the addition of k-1 edges, so $G_1$ is empty, and $G_n$ is the full tree. Define $T_k$ to be the set of vertices such that their degree in $G_k$ is exactly one less than their Prufer degree. Note that $T_1$ is therefore the set of leaves suggested by the Prufer code. So we form $G_{k+1}$ by adding an edge between the vertex with label appearing at position k+1 in the Prufer sequence and the vertex of $T_k$ with maximum label.

Proving that this is indeed the inverse is a bit fiddly, more because of notation than any actual mathematics. You probably want to show injectivity by an extremal argument, taking the closest vertex to the root that is different in two trees with the same Prufer code. I hope it isn’t a complete cop out to swerve around presenting this in full technical detail, as I feel I’ve achieved by main goal of explaining why bijection arguments can reduce a counting problem that was genuinely challenging to an exercise in choosing sensible notation for proving a fairly natural bijection.

# Mixing Times 6 – Aldous-Broder Algorithm and Cover Times

In several previous posts, I’ve discussed the Uniform Spanning Tree. The definition is straightforward: we choose uniformly at random from the set of trees which span a fixed underlying graph. But for a dense underlying graph, there are a very large number spanning trees. Cayley’s formula says that the complete graph K_n has $n^{n-2}$ spanning trees, so to select from this list is impractical.

We seek a better algorithm. In a post about a year ago, I presented the result that the path between two fixed points x and y in the UST is distributed as the path generated by Loop-Erased Random Walk, for which we start at x and delete cycles as they appear. An initial problem might be that this only gives us a single path, which might be enough in some contexts, but in general we will want to specify the whole tree. Wilson’s Algorithm is an unsurprising but useful extension to this equivalence which does just that. You start by constructing the LERW between two vertices, then you add the LERW which connects some other vertex to the path you already have. Then you take a further vertex not currently explored and start LERW there, continuing until you hit the tree that you already have. Iterate this process, which must terminate after at most n steps when there are no vertices which to start from. The tree thus obtained is the UST. The tricky part is proving that the method for selecting which unused vertices to start from has no effect on the distribution of paths between two fixed points.

I want to consider a different algorithm, discovered roughly simultaneously by Aldous and Broder. Start a random walk on the underlying graph at some particular vertex. Every time we traverse an edge which takes us to a vertex we haven’t yet explored, add this edge to the tree. For now I don’t want to give a proof that this algorithm works, but rather to talk about how fast it works, because it ties in nicely with something from the Mixing Times book we’ve been reading recently. It is clear that the algorithm terminates at the first time the random walk has visited every vertex. This is a stopping time, called the cover time of the Markov chain. If we are working with an underlying complete, then we notice that this is annoying, because it means that the cover time will increase like n.log n. That is, it will take an increasingly long time to gather the final few vertices into the tree. Perhaps some combination of Aldous-Broder initially then Wilson’s method for the final o(n) vertices might be preferable?

I want to discuss how to treat this cover time. Often we have information about the hitting times of states from other states $\mathbb{E}_x T_y$. A relationship between S, the hitting time, defined to be the maximum of the previous display over x and y, and the expected cover time would be useful, especially for a highly symmetric graph like the complete graph where the expected hitting times are all the same.

Matthews’ Method relates these two for an irreducible finite Markov chain on n states. It says:

$t_{cov}\leq t_{hit}\left(1+\frac12+\ldots+\frac 1 n\right).$

We first remark that this agrees with what we should get for the random walk on the complete graph. There, the hitting time of x from y is a geometric random variable with success probability 1/n, hence expectation is n. The cover time is the standard coupon collector problem, giving expectation n log n, and the sum of reciprocals factor is asymptotically a good approximation.

The intuition is that if we continue until we hit state 1, then reset and continue until we hit state 2, and so on, by the time we hit state n after (n-1) iterations, this is a very poor overestimate of the cover time, because we are actually likely to have hit most states many times. What we want to do really is say that after we’ve hit state 1, we continue until we hit state 2, unless we’ve already done so, in which case we choose a different state to aim for, one which we haven’t already visited. But this becomes complicated because we then need to know the precise conditional probabilities of visiting any site on the way between two other states, which will depend rather strongly on the exact structure of the chain.

Peres et al give a coupling proof in Chapter 11 of their book which I think can be made a bit shorter, at least informally. The key step is that we still consider hitting the sites in order, only now in a random order.

That is, we choose a permutation $\sigma\in S_n$ uniformly at random, and we let $T_k$ be the first time that states $\sigma(1),\ldots,\sigma(k)$ have all been visited. This is a random time that is measurable in the product space, and for each $\sigma$ it is a stopping time.

The key observation is that $\mathbb{P}(T_{k+1}=T_k)=1-\frac{1}{k+1}$. This holds conditional on any path of the Markov chain because the requirement for the event is that $\sigma(k+1)$ is visited after $\{\sigma(1),\ldots,\sigma(k)\}$. The statement therefore holds as stated as well as just pathwise. Then, by the SMP, conditional on $\{T_{k+1}>T_k\}$, we have

$T_{k+1}-T_k \leq_{st} t_{hit}.$

Note that by the definition of $t_{hit}$, this bound on the hitting time $T_{k+1}$ is unaffected by concerns about where the chain actually is at $T_k$ (since it is not necessarily at $\sigma(k)$).

So, removing the conditioning, we have:

$\mathbb{E}\Big[T_{k+1}-T_k\Big]\leq\frac{1}{k+1}t_{hit},$

and so the telescoping sum gives us Matthews’ result.

One example is the cover time of random walk on the n x n torus, which turns out to be

$O(n^2(\log n)^2).$

If anyone remembers that Microsoft screensaver from many years ago which started with a black screen and a snake leaving a trail of white pixels as it negotiated the screen, this will be familiar. The last few black bits take a frustratingly long while to disappear. Obviously that isn’t quite a random walk, but it perhaps diminishes the surprise that it should take this long to find the cover time.

There are a couple of interesting things I wanted to say about electrical networks for Markov chains and analytic methods for mixing times, but the moment may have passed, so this is probably the last post about Mixing Times. Plans are in motion for a similar reading group next term, possible on Random Matrices.

# Minimum Spanning Trees

In my last post, I discussed the Uniform Spanning Tree. To summarise very briefly, given a connected graph on n vertices, a tree is a subgraph, that is a subset of the edges, which is connected, but which contains no cycles. It turns out this requires the tree to have n-1 edges.

We are interested in natural mechanisms for generating randomly chosen spanning trees of a given graph. One way we can always do this is to choose uniformly at random from the set of possible trees. This UST is in some sense canonical, but it is worth knowing about some other measures on trees that might be of interest.

A family of natural problems in operations research concerns an arbitrary complex network, with some weight or cost associated to each connection. The question is how to perform some operation on the network so as to minimise the resulting cost. Perhaps the most famous such problem is that of the Travelling Salesman. The story is that a salesman needs to visit n locations and wants to do the trip as efficiently as possible. This might be thought of as some sort of financial or time cost, but proably the easiest way to set it up is to imagine he is trying to minimise the distance he has to travel. It is not hard to see why this problem might genuinely arise in plenty of real-world situations, where a organisation or agent is trying to be as efficient as possible.

It might be the case that it is not possible to travel between every pair of locations, but we needn’t assume that for now. So if he knows the distance between any pair of cities, he wants to know which of the possible routes gives the shortest overall distance. The problem is that there are n! routes, and this grows roughly like n^n, which is faster than exponential, so for as few as 20 cities it has turned into a comparison which is too large to compute.

There are various algorithms which reduce the number of routes that must be checked, and some approximation methods. But if you want the exact answer, it is not currently possible to calculate this in polynomial time.

Minimal Spanning Trees and Uniqueness

For the travelling salesman, we were looking for the minimal cost spanning path. In the case of the complete graph, this is the same as the minimal cost non-repeating path of length n-1. Such paths are a subset of the set of spanning trees on the underlying graph. So what if we look instead for the minimal cost spanning tree? This exists as after all, there are only finitely many spanning trees.

So far, this has been deterministic, but we were looking for a random spanning tree. We can achieve this by choosing the weights at random. Anything other than assigning the weights as an IID sequence seems likely to be complicated, but there isn’t a canonical choice of the distribution of the weights. Our first question will be whether the distribution of the weights affects the distribution of the induced MST. In fact it will turn out that so long as the distribution is continuous, it has no effect on the distribution of the MST. The continuous condition might seem odd, but it is present only to ensure that the weights almost certainly end up generating a unique MST.

It turns out that there is a straightforward greedy algorithm to find the MST once the weights are known. We will examine some consequences of this algorithm in the random setting. First we check uniqueness. The condition required for uniqueness is that the weights be distinct. Note that this is slightly weaker than the statement that all of sums of (n-1)-tuples be distinct, which immediately implies a unique MST.

We now prove this condition. Suppose we have distinct weights, and an associated MST. If the underlying graph is a tree, then the result is clear. Otherwise, add some extra edge e, with weight w(e). By the definition of a tree, this generates exctly one cycle. Consider the other edges, say $e_1,\ldots,e_k$ in this cycle. If any of $w(e_i)>w(e)$ then we can replace e_i with e to get a spanning tree with smaller weight, a contradiction of the claim that we started with an MST. So by distinctness of weights, we conclude that $w(e)>w(e_i)$ for all i.

Conversely, suppose we remove some edge e which IS in the MST. We end up with exactly two connected components. Consider all the edges in the underlying graph between the two components, and suppose that one of these f satisfies w(f)<w(e). Then if we add in edge f, which is by construction not in the original MST, we end up with a smaller total weight than we started with, a further contradiction.

We can summarise this in a neat form. Given an edge e between x and y, consider the set of all edges in the underlying graph with weight LESS THAN w(e). Then if x and y are in different components, the edge e must be in the MST. Since we have an explicit description of which edges are present, it follows that the MST is unique. The problem is that working out the component structure of the graph with higher weights removed is computationally rather intensive. We want a slightly faster algorithm.

Kruskal’s Algorithm

Several rather similar algorithms were developed roughly simultaneously. Prim’s algorithm is a slight generalisation of what we will discuss. Anyway, for now we consider Kruskal’s algorithm which has the advantage that it can be described without really needing to draw a diagram.

We start by ordering the weights. Without loss of generality, we might as well relabel the edges so that

$w(e_1)< w(e_2)<\ldots< w(e_{|E|}).$

Now, by the condition derived in the argument for uniqueness, we must have e_1 and e_2 in any MST. Now consider e_3. Unless doing so would create a cycle, add e_3. Then, unless doing so would create a cycle, add e_4. Continue. It is clear that the result of this procedure is acyclic. To check it is actually a spanning tree, we show that it is also connected. Suppose not, and two of the components are A and B. Let e be the edge between A and B with minimal weight. According to the algorithm, we should have included e in our MST because at no point would adding it possibly have created a cycle. So we have proved that this greedy algorithm does indeed give the (unique) MST.

A useful consequence of this is that we know the two edges with overall minimum weight are definitely in the MST. In the search for a random measure on spanning trees, what is most important is that we didn’t use the actual values of the weights in this construction, only the order. In other words, we might as well have assumed the weights were a random permutation from $S_{|E|}$. This now answers our original question about how the random weight MST depends in distribution on the underlying edge weight distribution. So long as with probability one the weights are distinct (which holds if the distribution is continuous), then the distribution of the resulting spanning tree is constant.

It’s not too hard to show this isn’t the same as UST: n=4 suffices as a counterexample. But the difference in asymptotic behaviour of properties such as the diameter is of interest, and will be explored in the next post.

# Uniform Spanning Trees

For applications to random graphs, the local binomial structure and independence means that the Galton-Watson branching process is a useful structure to consider embedding in the graph. In several previous posts, I have shown how we can set up the so-called exploration process which visits the sites in a component as if the component were actually a tree. The typical degree is O(1), and so in particular small components will be trees with high probability in the limit. In the giant component for a supercritical graph, this is not the case, but it doesn’t matter, as we ignore vertices we have already explored in our exploration process. We can consider the excess edges separately by ‘sprinkling’ them back in once we have the tree-like backbone of all the components. Again, independence is crucial here.

I am now thinking about a new model. We take an Erdos-Renyi process as before, with edges arriving at some fixed rate, but whenever a cycle appears, we immediately delete all the edges that make up the cycle. Thus at all times the system consists of a collection (or forest) of trees on the n vertices. So initially this process will look exactly like the normal E-R process, but as soon as the components start getting large, we start getting excess edges which destroy the cycles and make everything small again. The question to ask is: if we run the process for long enough, roughly how large are all the components? It seems unlikely that the splitting mechanism is so weak that we will get true giant components forming, ie O(n) sizes, so we might guess that, in common with some other split-merge models of this type, we end up with components of size $n^{2/3}$, as in the critical window for the E-R process.

In any case, the scaling limit process is likely to have components whose sizes grow with n, so we will have a class of trees larger than those we have considered previously, which have typically been O(1). So it’s worth thinking about some ways to generate random trees on a fixed number of vertices.

Conditioned Galton-Watson

Our favourite method of creating trees is inductive. We take a root and connect the root to a number of offspring given by a fixed distribution, and each of these some offspring given by an independent sample from the same distribution and so on. The natural formulation gives no control over the size of the tree. This is a random variable whose distribution depends on the offspring distribution, and which in some circumstances be computed explicitly, for example when the offspring distribution is geometric. In other cases, it is easier to make recourse to generating functions or to a random walk analogue as described in the exploration process discussion.

Of course, there is nothing to stop us conditioning on the total size of the population. This is equivalent to conditioning on the hitting time of -1 for the corresponding random walk, and Donsker’s theorem gives several consequences of a convergence relation towards a rescaled Brownian excursion. Note that there is no a priori labelling for the resulting tree. This will have to be supplied later, with breadth-first and depth-first the most natural choices, which might cause annoyance if you actually want to use it. In particular, it is not obvious, and probably not true unless you are careful, that the distribution is invariant under permuting the labels (having initially assumed 1 is the root etc) which is not ideal if you are embedding into the complete graph.

However, we would like to have some more direct constructions of random trees on n vertices. We now consider perhaps the two best known such methods. These are of particular interest as they are applicable to finding random spanning trees embedded in any graph, rather than just the complete graph.

Uniform Spanning Tree

Given a connected graph, consider the set of all subgraphs which are trees and span the vertex set of the original graph. An element of this set is called a spanning tree. A uniform spanning tree is chosen uniformly at random from the set of spanning trees on the complex graph on n vertices. A famous result of Arthur Cayley says that the number of such spanning trees is $n^{n-2}$. There are various neat proofs, many of which consider a mild generalisation which gives us a more natural framework for using induction. This might be a suitable subject for a subsequent post.

While there is no objective answer to the question of what is the right model for random trees on n vertices, this is what you get from the Erdos-Renyi process. Formally, conditional on the sizes of the (tree) components, the structures of the tree components are given by UST.

To see why this is the case, observe that when we condition that a component has m vertices and is a tree, we are demanding that it be connected and have m-1 edges. Since the probability of a particular configuration appearing in G(n,p) is a function only of the number of edges in the configuration, it follows that the probability of each spanning tree on the m vertices in question is equal.

Interesting things happen when you do this dynamically. That is, if we have two USTs of sizes m and n at some time t, and condition that the next edge to be added in the process joins them, then the resulting component is not a UST on m+n vertices. To see why, consider the probability of a ‘star’, that is a tree with a single distinguished vertex to which every other vertex is joined. Then the probability that the UST on m vertices is a star is $\frac{m}{m^{m-2}}=m^{-(m-3)}$. By contrast, it is not possible to obtain a star on m+n vertices by joining a tree on m vertices and a tree on n vertices with an additional edge.

However, I think the UST property is preserved by the cycle deletion mechanism mentioned at the very start of this post. My working has been very much of the back of the envelope variety, but I am fairly convinced that once you have taken a UST and conditioned on the sizes of the smaller trees which result from cycle deletion. My argument is that you might as well fix the cycle to be deleted, then condition on how many vertices are in each of the trees coming off this cycle. Now the choice of each of these trees is clearly uniform among spanning trees on the correct number of vertices.

However, it is my current belief that the combination of these two mechanisms does not give UST-like trees even after conditioning on the sizes at fixed time.

# Random Walks and Spanning Trees

Introduction

In this post, I’m going to talk about probability distributions on graphs. In particular, consider paths on a graph, and how to assign a distribution to these in a natural way. One way is to consider a standard random walk, treating the vertices as states of a discrete Markov chain. But, in many situations, this isn’t really enough. We might want paths, that is, walks which are self-avoiding. But even in regular lattices like $\mathbb{Z}^d$, it is hard to say a great deal about the set of self-avoiding walks (SAWs) of length n. We would prefer a distribution which has a natural product form, or which at least we can sample from without large combinatorial calculations.

A spanning tree is a connected graph without cycles. The set of edges is maximal, in the sense that adding any further edge creates a cycle, and also minimal, as removing one will disconnect the graph. Counting the number of spanning trees on labelled vertices is harder than one might suspect, and is possibly worth a post by itself. However, in general the uniform distribution on spanning trees is a useful object to consider. Any spanning tree contains a unique path between two vertices of the graph, and so a Uniform Spanning Tree (UST) induces a distribution on paths.

An alternative construction is to take a random walk and remove the cycles. This is not well-defined unless you specify a canonical order in which to remove them, and the obvious option is to remove the cycles in the order in which they appear. That is, every time you end up at a vertex which you have already visited, you remove all the edges traversed since you were last at v. This gives a Loop-Erased Random Walk (LERW), and from this another measure on paths between two vertices (subject to connectivity conditions which guarantee that the LERW almost surely hits the target vertex eventually).

At an informal level, the difference between these measures is significant. Inducing a measure down from a natural but potentially unwieldy uniform distribution is theoretically natural, but hard to work with. On the other hand a computer could sample from LERW, just by performing a random walk and storing the history suitably, which immediately makes it useful. So, the following theorem is elegant in its own right, but in particular as a bridge between these two frameworks.

Theorem: The measures on paths from to in a finite graph G induced by UST and generated by LERW are the same.

Proof: Some books give proofs which involve coupling a LERW construction with a sequence of STs directly, driven by an underlying random walk on the graph. The difficulty in this approach is that to prove that the uniform distribution on spanning trees is stationary often requires the random walk to be in equilibrium, a criterion which causes difficulties when it come to starting at x later in the proof. Essentially, the difficulty in many LERW constructions is that the edges being removed are incident to the wrong vertex in the random walk, and so RW equilibrium is required to show that the spanning tree transitions are reversible. Instead, we proceed by an argument motivated by the fact that LERWs appear ‘backwards’ in UST constructions. Continue reading


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# Need help on this 2

$\large \sqrt[3]{\frac{(a+b)(b+c)(c+a)}{abc}}\geq\frac{4}{3}\bigg(\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{c^2}{c^2+ab}\bigg)$ If $$a,b$$ and $$c$$ are positive reals, prove the inequality above.

I've proven $\sqrt[3]{\frac{(a+b)(b+c)(c+a)}{abc}}\geq 2$ So now I have to prove $\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{c^2}{c^2+ab}\leq\frac{3}{2}$ I've tried to prove it but then realised that what I was trying to prove is wrong, can somebody help me?

Note by Gurīdo Cuong
1 year, 12 months ago

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Oops, I believe you've made an error. This inequality is not trivial, because, you showed,
$$\dfrac{a^2}{a^2+bc}+\dfrac{b^2}{b^2+ac}+\dfrac{c^2}{c^2+ab} \ge \dfrac{(a+b+c)^2}{a^2+b^2+c^2+ab+bc+ca} \\$$ and $$\dfrac{3}{2} \ge \dfrac{(a+b+c)^2}{a^2+b^2+c^2+ab+bc+ca}$$
But this does not imply $$\dfrac{3}{2} \ge \dfrac{a^2}{a^2+bc}+\dfrac{b^2}{b^2+ac}+\dfrac{c^2}{c^2+ab}$$ which is not true for all (a, b, c).

- 1 year, 11 months ago

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Yes you are right. Do you know how to proceed with this problem?

- 1 year, 11 months ago

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I'm working on it, with some normalisation technique.

- 1 year, 11 months ago

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I proved it up to the lhs is greater than or equal to 427abc/( (a+b+c)^3+27abc)

- 1 year, 12 months ago

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can you show me how?

- 1 year, 12 months ago

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If you are on Slack, I can send you there.

- 1 year, 12 months ago

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yes I am, can you send it to me?

- 1 year, 12 months ago

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I am busy with my finals so I would send you it later make sure you check it

- 1 year, 12 months ago

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alright, good luck dude

- 1 year, 12 months ago

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Thanks, but please check it out later, but it may contain flaws and feel free to spot the mistakes

- 1 year, 12 months ago

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How did you proved the first part

- 1 year, 12 months ago

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First can be proved easily using AM-GM inequality.

- 1 year, 12 months ago

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Yes just saw that

- 1 year, 12 months ago

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# ergodic definition

• Of or associated with particular methods that, provided plenty of time, at some point come back to previously skilled condition.
• Of or regarding a process in which every sequence or sample of sufficient dimensions are equally representative of this entire.
• good recurrent aperiodic state of stochastic systems; tending in probability to a limiting type which in addition to the preliminary problems
• Of or associated with particular systems that, given enough time, at some point return to previously skilled state.
• Of or concerning an activity where every sequence or sample of enough size is equally representative associated with entire.
• positive recurrent aperiodic condition of stochastic systems; tending in probability to a limiting form that's independent of the initial problems
• Of or regarding certain systems that, provided plenty of time, at some point come back to previously experienced condition.
• Of or associated with a process in which every series or test of adequate dimensions are similarly representative associated with entire.
• good recurrent aperiodic condition of stochastic systems; tending in probability to a limiting form that's independent of the preliminary conditions
• Of or concerning specific systems that, offered enough time, will eventually come back to previously experienced condition.
• Of or relating to a procedure by which every series or sample of enough dimensions are equally representative of this whole.
• positive recurrent aperiodic state of stochastic methods; tending in likelihood to a limiting type that's independent of the preliminary circumstances

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# Which of the following statements best describes what we can infer from the table about the... 1 answer below »

Which of the following statements best describes what we can infer from the table about the relative strength of the $US compared with currencies for countries in which BMY does business: WebAssign Changes in the exchange rates of the$US vis-à-vis other world currencies resulted in a reduction in BMY sales for 2007 vis-à-vis 2006
Foreign exchange fluctuations increased BMY sales in 2007. This is consistent with a strengthening of the $US. Foreign exchange fluctuations increased BMY sales in 2007. This is consistent with a weakening of the$US.
Information in the table does not provide insight into the relative strength of the $US vis-à-vis other world currencies. We must look elsewhere in financial statements for that information. Document Preview: WebAssign Page 1 of 9 Marie-Cecile Heinze WebAssign MBA 501, section 1, Fall 2011 Module 5 (Homework) Instructor: Diann Ferrell Current Score :– / 100 Due : Sunday, September 25 2011 11:59 PM EDT 1. –/10 points EasManAcct2 5.17.rand. Interpreting Foreign Currency Translation Disclosure Bristol-Myers Squibb (BMY)† reports the following table in its 10-K report relating to the year-over-year change in sales. Analysis of % Change Total Change Volume Price Foreign Exchange 2007 vs. 2006 11% 8% 1% 2% 2006 vs. 2005 (7)% (8)% 1% — (a) Did sales increase or decrease in 2006 and 2007? By what percentage? What amount of this change was attributable to fluctuations in the value of foreign currencies vis-à-vis the$US? Sales Percent due to fluctuations Year Percent Change in the value of foreign currencies Increase Decrease 2006   % % 2007   % % (b) Which of the following statements best describes what we can infer from the table about the relative strength of the $US compared with currencies for countries in which BMY does business: Changes in the exchange rates of the$US vis-à-vis other world currencies resulted in a reduction in BMY sales for 2007 vis-à-vis 2006  Foreign exchange fluctuations increased BMY sales in 2007. This is consistent with a strengthening of the $US.  Foreign exchange fluctuations increased BMY sales in 2007. This is consistent with a weakening of the$US.  Information in the table does not provide insight into the relative strength of the $US vis-à-vis other world currencies. We must look  elsewhere in financial statements for that information. http://www.webassign.net/web/Student/Assignment-Responses/last?dep=3343942 17/09/2011WebAssign Page 2 of 9 2. –/20 points EasManAcct2 5.25.rand. Constructing and Assessing Income Statements Using Percentage-of-Completion On March 15, 2007, Frankel Construction contracted to build a shopping center at a contract price of$145 million. The...

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Sciencemadness Discussion Board » Fundamentals » Reagents and Apparatus Acquisition » Things That Frustrate As A DIY Chemist Select A Forum Fundamentals   » Chemistry in General   » Organic Chemistry   » Reagents and Apparatus Acquisition   » Beginnings   » Responsible Practices   » Miscellaneous   » The Wiki Special topics   » Technochemistry   » Energetic Materials   » Biochemistry   » Radiochemistry   » Computational Models and Techniques   » Prepublication Non-chemistry   » Forum Matters   » Legal and Societal Issues   » Test Forum

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Author: Subject: Things That Frustrate As A DIY Chemist
xfusion44
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 Quote: Originally posted by byko3y What frustrates me is that I can acutally make explosives, while it's hard for me to obtain acetone and toluene (and also phosphorus chlorides, elemental phosphorus and permanganate). But I can freely purchase iodine, iodides and hypophosphites. God only knows why some particular chemicals are banned in some countries.

Yeah, this is also well known to me. In USA you can buy drain cleaners with H2SO4 anywhere, but not here in the EU, but there is probably something that is easier to get here than in the US. You just can't get all of the chemicals in one country.

The Volatile Chemist
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Nitrates and nitrites are a pain to get here. I've never seen a good source in the mid- US.

unfrozen
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Posts: 19
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Location: Wyoming, US
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Mood: Waiting for an indicator

 Quote: Originally posted by The Volatile Chemist Nitrates and nitrites are a pain to get here. I've never seen a good source in the mid- US.

Is this anywhere useful? Prices are better in larger sizes.

Sodium Nitrite, $7.50 / 2lb http://www.dudadiesel.com/choose_item.php?id=2sniteF Sodium Nitrate,$3.75 / 1 lb
Potassium Nitrate, 3.95 / 1 lb
szuko03
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 Quote: Originally posted by The Volatile Chemist Nitrates and nitrites are a pain to get here. I've never seen a good source in the mid- US.

I can get potassium nitrate at a home depot like store called Lowes, sorry I am never sure what store names are national and which arent but you can get stump remover that is pure potassium nitrate. I got mine for 7 dollars a pound, which seems just a little higher then ordering it online. You may want to look at the stump removers in your area just be careful because stupid me bought one that wasnt nitrate eons ago.

Chemistry is a natural drive, not an interest.
Texium

Posts: 3522
Registered: 11-1-2014
Location: Salt Lake City
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Mood: Triturated

Quote: Originally posted by szuko03
 Quote: Originally posted by The Volatile Chemist Nitrates and nitrites are a pain to get here. I've never seen a good source in the mid- US.

I can get potassium nitrate at a home depot like store called Lowes, sorry I am never sure what store names are national and which arent but you can get stump remover that is pure potassium nitrate. I got mine for 7 dollars a pound, which seems just a little higher then ordering it online. You may want to look at the stump removers in your area just be careful because stupid me bought one that wasnt nitrate eons ago.
Well luckily for you that other stump remover that you bought is likely sodium metabisulfite, which is very useful as a reducing agent for cleaning up nasty oxidizing stuff like dichromate, and also used for making the peculiar Chevreul's salt. I was fortunate enough to find both that and the potassium nitrate version side by side on the shelf at Lowe's, so I happily bought both. Although, the KNO3 one is really quite overpriced and not super pure, so when I run out of it (probably soon) I'll buy some from Duda Diesel instead. Their products are of quite good quality, affordable, and they have great service too.

Come check out the Official Sciencemadness Wiki
They're not really active right now, but here's my YouTube channel and my blog.
szuko03
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^that is true but by eons ago I mean I wasnt even into that side of chemistry and was just trying to surprise my friends. And its purity is good enough for sugar rockets which was what I was interested in at the time. I am also lucky in the sense that nothing gets thrown away so that sodium metabisulfite, which it is I remember the name, is still in the garage.

To be honest it is sometimes surprising what you can just buy, its like how can they "regulate" something and then sell it on the shelf in a public place for cash.

[Edited on 1-6-2015 by szuko03]

Chemistry is a natural drive, not an interest.
SimpleChemist-238
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Mood: Chlorine Trifloride Flame Thrower

Man, I can't stand the black helicopters always flying by. But when at the end of a hard day I can always go to this forum and laugh at the humor and terrible chemistry jokes. It is the strange people on this forum who dream of destroying the Earth that I call my closest friends. Also it looks like our emails are safe from the NSA for a few days well Congress argues about the Freedom Act!

We are chemists , we bring light to the darkness. Knowledge to ignorant, excitement to the depressed and unknowing. we bring crops to broken fields and water to the desert. Where there is fear we bring curiosity.

The Volatile Chemist
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Quote: Originally posted by zts16
Quote: Originally posted by szuko03
 Quote: Originally posted by The Volatile Chemist Nitrates and nitrites are a pain to get here. I've never seen a good source in the mid- US.

I can get potassium nitrate at a home depot like store called Lowes, sorry I am never sure what store names are national and which arent but you can get stump remover that is pure potassium nitrate. I got mine for 7 dollars a pound, which seems just a little higher then ordering it online. You may want to look at the stump removers in your area just be careful because stupid me bought one that wasnt nitrate eons ago.
Well luckily for you that other stump remover that you bought is likely sodium metabisulfite, which is very useful as a reducing agent for cleaning up nasty oxidizing stuff like dichromate, and also used for making the peculiar Chevreul's salt. I was fortunate enough to find both that and the potassium nitrate version side by side on the shelf at Lowe's, so I happily bought both. Although, the KNO3 one is really quite overpriced and not super pure, so when I run out of it (probably soon) I'll buy some from Duda Diesel instead. Their products are of quite good quality, affordable, and they have great service too.

I have a Lowe's nearby. What were the brand names? I'd looked on lowe's site, but they don't have anything tagged with 'nitrate'.

 Quote: Originally posted by SimpleChemist-238 Man, I can't stand the black helicopters always flying by. But when at the end of a hard day I can always go to this forum and laugh at the humor and terrible chemistry jokes. It is the strange people on this forum who dream of destroying the Earth that I call my closest friends. Also it looks like our emails are safe from the NSA for a few days well Congress argues about the Freedom Act!

What was the result of this, did they decide yet? I first heard about this on Wikileak's twitter page. It seemed like it was a long time ago.

Texium

Posts: 3522
Registered: 11-1-2014
Location: Salt Lake City
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Mood: Triturated

Quote: Originally posted by The Volatile Chemist
Quote: Originally posted by zts16
Quote: Originally posted by szuko03
 Quote: Originally posted by The Volatile Chemist Nitrates and nitrites are a pain to get here. I've never seen a good source in the mid- US.

I can get potassium nitrate at a home depot like store called Lowes, sorry I am never sure what store names are national and which arent but you can get stump remover that is pure potassium nitrate. I got mine for 7 dollars a pound, which seems just a little higher then ordering it online. You may want to look at the stump removers in your area just be careful because stupid me bought one that wasnt nitrate eons ago.
Well luckily for you that other stump remover that you bought is likely sodium metabisulfite, which is very useful as a reducing agent for cleaning up nasty oxidizing stuff like dichromate, and also used for making the peculiar Chevreul's salt. I was fortunate enough to find both that and the potassium nitrate version side by side on the shelf at Lowe's, so I happily bought both. Although, the KNO3 one is really quite overpriced and not super pure, so when I run out of it (probably soon) I'll buy some from Duda Diesel instead. Their products are of quite good quality, affordable, and they have great service too.

I have a Lowe's nearby. What were the brand names? I'd looked on lowe's site, but they don't have anything tagged with 'nitrate'.
The potassium nitrate one is Spectracide brand, and comes in a black bottle with a pointy lid that works as a sort of pour spout. The metabisulfite one is Bonide brand and comes in a white, squarish bottle with a purple lid.

Come check out the Official Sciencemadness Wiki
They're not really active right now, but here's my YouTube channel and my blog.
aga
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The Most frustrating thing is NOT knowing when to run away screaming instead of putting on the gas mask and trying to take control.

It's a fine line methinks.

[Edited on 4-6-2015 by aga]

binbin
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 Quote: Originally posted by Pickardjr I dont like the phobia of today associated with home chemistry. If someone was to see all my equipment set up, then I'd be in for a lot of explaining . Im just preparing solvents/reagents for other experiments. Average people really don't understand, and assume its drug related because they believe what the T.V tells them. My other peeve is not having enough time to complete procedures that are time sensitive, thus having to postpone. I could go on but that the top two for me.

Most people don't have hobbies. They suspect if a grown man/woman is performing chemistry, that is it for economic gain on the side. Breaking Bad has enforced this stereotype. Do I blame them? Not at all! I think they are just looking out for the community. It's still infuriating though.

And here lies the one reason I have a tough time making male friends in my community... nobody has interesting hobbies past "lets get drunk and watch the game." Chemistry? Astronomy? Metal working? Tinkering? Computers/Tech? There aren't many nerds or interesting people in my city of 250,000.

However I think the most frustrating/stressful things as a home chemist are:

1. Control of fumes with proper equipment. Fume hoods are just too damn expensive.

2. Fear of chemical fires, and/or fear of proper storage of your chemicals.

[Edited on 5-6-2015 by binbin]
Loptr
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Mood: Grateful

Quote: Originally posted by binbin
 Quote: Originally posted by Pickardjr I dont like the phobia of today associated with home chemistry. If someone was to see all my equipment set up, then I'd be in for a lot of explaining . Im just preparing solvents/reagents for other experiments. Average people really don't understand, and assume its drug related because they believe what the T.V tells them. My other peeve is not having enough time to complete procedures that are time sensitive, thus having to postpone. I could go on but that the top two for me.

Most people don't have hobbies. They suspect if a grown man/woman is performing chemistry, that is it for economic gain on the side. Breaking Bad has enforced this stereotype. Do I blame them? Not at all! I think they are just looking out for the community. It's still infuriating though.

And here lies the one reason I have a tough time making male friends in my community... nobody has interesting hobbies past "lets get drunk and watch the game." Chemistry? Astronomy? Metal working? Tinkering? Computers/Tech? There aren't many nerds or interesting people in my city of 250,000.

However I think the most frustrating/stressful things as a home chemist are:

1. Control of fumes with proper equipment. Fume hoods are just too damn expensive.

2. Fear of chemical fires, and/or fear of proper storage of your chemicals.

[Edited on 5-6-2015 by binbin]

Actually, I would say that there has been a resurgence in various hobbies due to the maker movement. When I was in Charlottesville (UVA) there was a huge group that would get together in the hacker/open spaces to do just this; not to mention the un-conferences.

Just look at all the young people interested in hobbies that are know sharing it with the world in maker magazines, youtube, instructables, etc.

We are many.

[Edited on 5-6-2015 by Loptr]
Magpie
lab constructor

Posts: 5939
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Mood: Chemistry: the subtle science.

Recently I have made several electro-mechanical devices to support my chemistry hobby. I talk about these in a tangential way with my friends at morning coffee. Of course they always ask: and what is the purpose of this device?

I give a circumspect answer and they drop it. One electrical engineer was quite interested in my stepper motor stirrer. He wanted me to bring it to Starbucks, plug it in to the wall outlet, and give a demonstration. I replied. "No way. I'm not bringing a black box with wires coming out of it! Someone would call the bomb squad." His answer: "And well they should." This is a sad state of affairs.

The single most important condition for a successful synthesis is good mixing - Nicodem
The Volatile Chemist
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Have there ever been SM organized chem hobby conferences/events?

aga
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Nope.

We're all Odd and live all over the planet, and don't really do 'people events' as well as we can type stuff into computers or stuff reagents into beakers before blasting them with heat.

A bit like You i guess.

P.S. an Event would cost a huge amount of money, simply down to the air-fares, hotels etc

P.P.S. the more Capable of the SM membership would be a bit paranoid as well, and would think it a FBI/DEA/CIA event designed to flush them out.

P.P.P.S. Cool ! That means that i can come to the party !

Sulaiman
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the main thing that frustrates me as a D.I.Y. Chemist is my lack of any real depth of knowledge

and I've not yet discovered the joy of cleaning glassware
or the skill of not using so much at once.
byko3y
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Mood: dooM

I'm so pissed because of lack of well developed/known and easy methods of preparation of common reagents.
As you may know, some time ago I was forced to develop anthranilic acid synthesis all over again (because I can't just leave my 5% yield as is and pretend I've succeded), until s.c.wack found a 100 years old article with absolutely the same procedure which had been forgotten, and instead tonns of non-reproducible procedures were spread all over the internets.
Same thing happenned to elemental phosphorus thread - it is filled from the begginning to the end with procedures, which are mostly untested and the rest suits for industrial production, but not for a home chemist. In thirteen (13) years nobody managed to find an answer, it's like a huge bin that collects all the useless ideas to avoid flooding the forum with dozens of threads. I hope soon I will prove that you can actually make phosphorus by means of kitchenware and basic OTC chemicals.
Aluminium isopropoxide, acetonitrile, pyridine, acetic anhydride, sulfuric acid, sulfuryl chloride, thionyl chloride, dimethyl carbonate - those are usefull and potentially easy to make reagents, but scantily explored. I was going to put oxalyl chloride into the list, but there were few members on the forum who actually performed extensive research on the subject and some experiments, however, there's still a lot to explore in the oxalyl chloride synthesis.
I want to remind you about a famous aluminium reduction of nitroalkenes, which was unknown for so many years, while in fact it was well known to chemists hundred years ago.
I'm starting to think there are some agents on forums who intentionally discuss useless procedures to hide the truth, which lies in the fact nobody knows shit and the usable method is yet to be discovered or rediscovered.
Some people blame Lerner for being a dick, but let's look at facts - he's one of few members who actually tested the procedures. Half of them are not usable at home, but who cares?
The Volatile Chemist
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That's why I continually push for some sort of organizational measure. The SM Wiki is a good idea, but it doesn't have the expansive qualities to organize large amounts of information and procedures. I'm looking for some good model to use for such a measure, but I haven't found one that works.

byko3y
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That's pretty much what rhodium was trying to do and sucked hard, because he did not perform most of the procedures he gather into his site. I'm talking about these articles:
http://bitnest.ca/Rhodium/chemistry/nitro2amine.mg-n2h4-hcoo...
http://bitnest.ca/Rhodium/chemistry/oxime2amine.mg-af.html
http://bitnest.ca/Rhodium/chemistry/debenzylation.zn-af.html
Some1 could gather his private site the verified information about procedures his collegues and him already performed. Like google docs.
But the other problem that bothers me - not much of people seem to care about researching basic reagents, and those who actually do, seem to have too few knowledge and/or equipment to perform the correct analysis, like it was in the TEMPO case, where I still got no confirmation that the resulting compound is actually a TEMPO and not substituted compound or some mix.
I remember my attempt to post a message to professional chemists about a need for developing procedures for basic reagents preparation, and I was laughed at, because YOU CAN'T MAKE MONEY ON IT. Why anyone would bother himself with chemistry if he can't make money, right, Jessie?
And thanks god today we have enough access to science journals, because otherwise we would be like blind kittens having no idea where to go and what is this world. But still a lot of knowledge is obscured, e.g. few people seem to know about dithionite and thiourea peroxide, while those reagent are so much OTC, and ther's no way amateur chemist can make NaBH4 ro LAH.
Once again in case you haven't understood: you can't take dithionite or thiourea peroxide away from people, just like you can't ban hydrogen peroxide or sulfuric acid, because those reagent are used a lot in different industries. But you can ban borohydrodes easily, because the only branch of industry, that uses it, is drug synthesis, and the companies making drugs are forced to obtain licenses. So, basically, if you want to be an advanced chemist and you think that status quo is okay, then you are forced to attain some colledge or make your business and obtain a license for precursors (because in most cases an individual person is either not eligible to have this license or nobody would sell suspicious chemicals to an individual).
I'm worried about bans on chemical equipment, but I don't think you can ban the equipment made from steel, because otherwise a lot of people will become potentially suspicious and the whole idea loses sense.
AFAIK, the restriction on equipment is implemented only in australia-NZ and some US states, probably because a lots of people actually have glassware, including schools, thus there's no sense in controling sales of this equipment, because no government have resources to do that. Even in australia/US i believe those are violated a lot, taking into account their formulation is vague.
Zephyr
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Location: Seattle, WA
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Byko3y:

1. Are you saying I faked my yield?!?! Just because you didn't get the same result once doesn't mean everyone else is wrong.
2. Many people did succeed in the phosphorus thread, mostly using OTC materials.
3. You yourself discuss some pretty useless stuff...
4. Decent amounts of research has been done into all of the reagents you've listed, you just need to search deeper.
5. Are you saying scientific journals make you dumber????

I'm done... I can't read this any longer....

Sciencemadness Patches for sale! U2U me if you are interested.
byko3y
International Hazard

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Mood: dooM

Pinkhippo11, describe your first position fully, because I have no idea what you are talking about there.
2. I know a lot of people succeeded, but their devices are far from OTC and/or the yields are miserable. The amount of phosphorus that can be further used for more than one experiment starts somewhere at 20g - I doubt anyone in the phosphorus thread made that much.
4. Yes and no. Yes, the researches have probably been done, no, there's no people on the forum who can share the results of the research;
5. No.
Amos
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@byko3y : Why are you on this forum? What do you do to contribute, and what gives you the platform to continually belittle members that do contribute?

byko3y
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If I succeed in my upcoming phosphorus experiments, then I would have a huge platform to detract those contributions.
But that will happen not earlier than my recovery from bromine intoxication, which makes me so irritable.
What do I do to contribute to what? Did you mean phosphorus thread? Speaking about it, I found no convenient procedure in the thread, so I was forced to spend a lot, A LOT of time searching for other ways, while the discussion in the thread goes through the same methods over and over again. And I can barely remember all the things which are written there, because there's really a lot of text, and I'm not a phosphorus manufacture specialist.
I can disclose the way I'm gonna prepare the phosphorus - I'm gonna use something phosphine-like, but I have not made decision about details. I don't say it has to be gaseous phosphine, countrary, it should be dissolved in a solvent. I don't see much reason to talk about it, because most likely nobody would try to develop the method.

[Edited on 11-6-2015 by byko3y]
j_sum1

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Mood: Taking opportunities to breathe.

I intend to do a foray into extracting P. I will use a variation of Rogeryermaw's method using sodium hexametaphosphate, SiO2, Al with a NaCl flux. The set-up is not that hard although my furnace design is going to be different from his. It all looks rather doable. he has posted in the P thread (scattered over several pages) He also shows his procedure in a video on his youtube channel.
If that helps at all byko3y.
Amos
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 Quote: Originally posted by byko3y If I succeed in my upcoming phosphorus experiments, then I would have a huge platform to detract those contributions. But that will happen not earlier than my recovery from bromine intoxication, which makes me so irritable. What do I do to contribute to what? Did you mean phosphorus thread? Speaking about it, I found no convenient procedure in the thread, so I was forced to spend a lot, A LOT of time searching for other ways, while the discussion in the thread goes through the same methods over and over again. And I can barely remember all the things which are written there, because there's really a lot of text, and I'm not a phosphorus manufacture specialist. I can disclose the way I'm gonna prepare the phosphorus - I'm gonna use something phosphine-like, but I have not made decision about details. I don't say it has to be gaseous phosphine, countrary, it should be dissolved in a solvent. I don't see much reason to talk about it, because most likely nobody would try to develop the method. [Edited on 11-6-2015 by byko3y]

So let me get this straight: You're complaining that people on the forum have had to use too-specialized procedures that take too much work (waaah) to make phosphorus with decent yields. And since they're not OTC enough for you, you're going to show everyone how OTC and simple you can be by generating and making use of phosphine gas. As opposed to simply heating a pretty simply welded apparatus using the correct heating element and power supply, keeping everything in a nearly closed system. I think it'd be great to try out your new method for producing phosphorus, and I have the highest hopes for you. But if you're complaining that existing methods aren't accessible, and that everyone's posting untested procedures(false) much like the one you're discussing now, then this is a new level of hypocrisy. Conduct your research, share anything interesting you find, and make the site a better place yourself.

Pages:  1  2    4

 Sciencemadness Discussion Board » Fundamentals » Reagents and Apparatus Acquisition » Things That Frustrate As A DIY Chemist Select A Forum Fundamentals   » Chemistry in General   » Organic Chemistry   » Reagents and Apparatus Acquisition   » Beginnings   » Responsible Practices   » Miscellaneous   » The Wiki Special topics   » Technochemistry   » Energetic Materials   » Biochemistry   » Radiochemistry   » Computational Models and Techniques   » Prepublication Non-chemistry   » Forum Matters   » Legal and Societal Issues   » Test Forum


text:
# Question a82c7

Jun 17, 2015

New volume: $\text{3.2 L}$

#### Explanation:

STP conditions imply a pressure of 100 kPa and a temperature of 273.15 K. Under these conditions, * 1mole* of any ideal gas occupies exctly 22.7 L - this is known as the molar volume of a gas at STP.

So, in order to determine the volume of the sample at STP, all you really need to know is how many moles of gas you're dealing with.

To do that, use the ideal gas law equation for the initial conditions of the sample.

$P V = n R T \implies n = \frac{P V}{R T}$

n_(N_2) = (6.2cancel("atm") * 525 * 10^(-3)cancel("L"))/(0.082(cancel("atm") * cancel("L"))/("mol" * cancel("K")) * (273.15 + 12)cancel("K")) = "0.140 moles" ${N}_{2}$

This means that the new volume of the sample will be

0.14cancel("moles") * "22.7 L"/(1cancel("mole")) = "3.18 L"#

Rounded to two sig figs, the answer will be

${V}_{{N}_{2}} = \textcolor{g r e e n}{\text{3.2 L}}$

SIDE NOTE Many online sources and textbooks still list the old STP conditions. more precisely 1 atm and 273.15 K, which makes the molar volume of a gas equal to 22.4 L.

If you are supposed to use the old value of the molar volume of a gas at STP, simple replace 22.7 with 22.4 in the final calculation.


text:
# A bolt from the blue

The following game was played in the ICC 5-minute pool against an opponent whose FIDE rating is 2071. I don’t normally annotate or show blitz games that I’ve played for obvious quality reasons, but I uncorked an extremely snazzy tactic in this game that I can’t help but share. (To my opponent’s credit, he was very gracious and also congratulated me on the move.) A steady diet of exercises based on ‘invisible’ moves seems to have finally paid off!

*Spoiler alert* The key positions is after White plays 18.g3, so you can try to work it out for yourself if you like! Of course, knowing ahead of time the critical moment makes any tactic easier to spot, which is part of why finding moves like this can be so difficult!

[Event "5-minute pool"]
[Site "Internet Chess Club"]
[Date "2013.05.14"]
[Round "?"]
[White "Pavel"]
[Black "Aaron"]
[Result "0-1"]
[WhiteELO "2071"]
[BlackELO "2098"]

1.e4 e5 2.Nf3 Nc6 3.Bb5 Nf6 4.O-O Bc5
( 4...Nxe4 5.d4 Nd6 6.Bxc6 dxc6 7.dxe5 Nf5 8.Qxd8+ Kxd8 {is the
classical Berlin.} )
5.c3
5...O-O
( 5...d6 $2 6.d4 Bb6 7.d5 a6 8.Ba4 {and White wins material.}) 6.d4 Bb6 7.Re1 7...d6 8.Bxc6$6
{Since White does not win a pawn on e5 as a result of this trade due
to tactical reasons, this move is premature.}
8...bxc6 9.dxe5 Ng4
{The point. With the double attack on e5 and f2 Black retains material
equality.}
10.Be3
( 10.Rf1 $2 Ba6 ) 10...Bxe3 11.fxe3 Nxe5 12.Nxe5 dxe5 {The position is approximately even, assuming White can find a good square for his knight.} 13.Na3 {This maneuver is a bit slow.} 13...Qh4 {Pressuring e4 and eying the kingside.} 14.Qf3 f5$5
( 14...Be6 {was safer, but Black plays for more.} )
15.exf5 Bxf5 16.Qxc6
{White accepts the pawn. But now Black gains time to coordinate an attack on
the king.}
16...Be4
( 16...Bd3 $2 17.Qd5+ ) 17.Qe6+ ( 17.Qc4+ Kh8 18.Qe2 {was a safer alternative.} ) 17...Kh8 18.g3$2
( 18.Rf1 {and White is alive and kicking. The point of 18.g3 is that
18...Qh3 is impossible. However...
} )
18...Rf2 \$1
{Once the initial shock wears off, this move is actually pretty simple
to calculate. The myriad checkmating threats force White to capture
either the rook or the queen (apart from trivial spite moves like
Qc8+/Qe8+/Qg8+/Qh6). But both captures lead to forced mate!}
( 18...Rf2 19.gxh4
( 19.Kxf2 Qxh2+ 20.Kf1 Qg2# )
19...Rg2+ 20.Kf1
( 20.Kh1 Rg3# )
20...Rf8+ 21.Qf7 Rxf7# )
0-1

# Barkley and Doppler

Recently we’ve learned a bit about physical modeling of PDEs in my pattern formation class. Here is a video depicting Barkley turbulence, which is a type of reaction-diffusion system.

On an unrelated note, here are a few of my results from implementing a simulation of the Doppler effect. In our first example, a source is traveling at about Mach $$\frac{1}{10}$$ horizontally from right to left across a distance of $$200$$ meters. When it is directly in front of us, the distance between us and the source is $$10$$ meters.

In the second example, the source travels twice as fast, twice as far, and we are twice as close. The shift is correspondingly more dramatic.

One of the more counterintuitive aspects of the Doppler effect is the situation in which the source is coming directly at us. We never really experience this in real life (unless we are actually getting run over by the source). In this situation, the source does not in fact slide continuously from a higher frequency to a lower frequency but rather it does a discrete jump. (This is because the Doppler effect depends on the projections of the velocity vectors of the source and the receiver onto the line connecting the two, so unless the source and receiver are traveling directly toward each other, these projections vary continuously. However, when the source and receiver travel directly toward each other, the projections are constant until they meet, at which point they discretely jump to a new value.) In the particular example I have implemented, the source is traveling at Mach $$\frac{1}{4}$$. (Warning: this example might be somewhat uncomfortable to listen to!)

Notice that we heard what sounded like a major sixth! This is actually a pretty straightforward consequence of the mathematics behind the Doppler shift. The emitted frequency, $$f$$, is distorted by a frequency factor of $$\frac{c}{c + v_s}$$, where $$c$$ is the speed of sound in the relevant medium and $$v_s$$ is the velocity of the source approaching you, with the convention that it is negative as it approaches you and positive as it recedes. Hence, the frequency ratio between the approaching sound and the receding sound is given by $$\frac{c + |v_s|}{c – |v_s|}$$. In particular, at Mach $$\frac{1}{4}$$, this ratio is $$\frac{\frac{5}{4}}{\frac{3}{4}} = \frac{5}{3}$$, which is a just major sixth. Other musical intervals are, of course, also easily obtainable — Mach $$\frac{1}{3}$$ yields an octave, for instance. In general, the reader can verify with some basic algebra that in order to get the frequency ratio $$P:Q$$, the source must travel at Mach $$\frac{P – Q}{P + Q}$$.

You might have noticed the expression $$c – v_s$$ in the denominator above and wondered about the situation in which we are traveling Mach $$1$$ or faster, but that’s another story for a future post.


text:
# DSolve cannot solve for certain branches of the solution

Could somebody please tell me, why I am always getting an error message when trying to run the following code:

DSolve[y'[x] == c*(d - y[x]) - b*(d - y[x])*y[x] && y[0] == y0, y, x]

The error message:

DSolve::bvfail: For some branches of the general solution, unable to solve the
conditions.

-

I can offer a small workaround. Your problem is equvalent to

sol=FullSimplify[DSolve[{y'[x] == A0 + A1 y[x] + A2 y[x]^2, y[0] == y0}, y[x], x]]

By expanding and comparing with your variables:

y'[x] == c d - (c + b d) y[x] + b y[x]^2
y'[x] == A0 + A1 y[x] + A2 y[x]^2

We can get your formulation by the substitution:

PowerExpand[FullSimplify[sol /. {A0 -> c d, A1 -> -(c + b d) , A2 -> b}]]

You can check now by direct substitution that this is indeed solution to your differential equation.

==== Edit: answering "why does not work?" question ===

I can try to guess the trouble of your formulation - I think it is in your choice of parameters. As Sjoerd C. de Vries in his answer noticed a general solution leads to

DSolve[y'[x] == c*(d - y[x]) - b*(d - y[x])*y[x], y, x]

Now Solve cannot "solve" your initial value problem:

Using Reduce you can arrive to a complex conditions set for the solution:

Which looks glorious ;-) but not simple. With a bit different formulation above (via A0, A1, A2) you do not run into this problem - Solve can handle easily your initial condition. This is rather a rare case - you were lucky to hit exactly problematic choice of parameters. This was some quick thinking - it's subject to verification.

-
Thank you very much for your great idea! I will use this workaround, but do you by any chance have any idea why I get the error message? I did forget to mention that if I take a piece of paper and a pencil, I get the solution of the equation within 3 minutes. –  Rieux May 6 '12 at 21:43
Thank you very much for the quick reply –  Rieux May 7 '12 at 0:37

The equation doesn't have a solution for arbitrary values of y0. Let's see what happens if we leave away the boundary condition:

sol = DSolve[{y'[x] == c*(d - y[x]) - b*(d - y[x])*y[x]}, y, x][[1, 1]]

(* ==>
y ->  Function[{x}, (d E^(c x + c C[1]) +
c E^(b d x + b d C[1]))/(E^(c x + c C[1]) +
b E^(b d x + b d C[1]))
]
*)

y[0] /. sol

(* ==> (d E^(c C[1]) + c E^(b d C[1]))/(E^(c C[1]) + b E^(b d C[1])) *)

so only for these specific values of y0 do you get a solution. Let's try it for C[1]=1:

DSolve[{y'[x] == c*(d - y[x]) - b*(d - y[x])*y[x],
y[0] == (d E^c  + c E^(b d ))/(E^c  + b E^(b d ))}, y,
x] // FullSimplify

Solve::ifun :  "Inverse functions are being used by Solve,
so some solutions may not be found; use Reduce for
complete solution information.>>"

(*
==> {{y ->
Function[{x}, (
d E^(c x + (c (-b d + Log[E^c]))/(c - b d)) +
c E^(b d x + (b d (-b d + Log[E^c]))/(c - b d)))/(
E^(c x + (c (-b d + Log[E^c]))/(c - b d)) +
b E^(b d x + (b d (-b d + Log[E^c]))/(c - b d)))]}}
*)

So, indeed you get a solution now.

-
Hi dear Sjoerd! Thank you very much for the explanation, it would have never occured to me to spare the boundary condition. However, looking at it a bit closer came the next question, which was that the solution was a continuous function of C[1], so I could not see why Mathematica still had problems with finding the right one. Then I realized, that unfortunately it was a bounded function of C[1]. Now I understand what caused the problem, but still I would ask sg else, namely I cannot understand what Mathematica needs Inverse Functions even with correct parameters for. –  Rieux May 6 '12 at 23:58
Because however I am not entirely sure I still suppose that the solution is an injective function of C[1] for any given x value, so after finding a suitable C[1] there is not supposed to be any lost solution. Or does this notification of inverse functions refer to sg else? –  Rieux May 7 '12 at 6:59
@Rieux You might want to read this piece of MMA help text about branch cuts –  Sjoerd C. de Vries May 7 '12 at 10:46
Thank you for the link! –  Rieux May 7 '12 at 12:51