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## Thursday, January 30, 2020
### Generic Callback Changes in CPLEX 12.10
CPLEX 12.10 is out, and there have been a few changes to the new(ish) generic callbacks. Rather than go into them in detail (and likely screw something up), I'll just point you to the slides for a presentation by Daniel Junglas of IBM at the 2019 INFORMS Annual Meeting.
I've written about half a dozen posts about generic callbacks since IBM introduced them (which you can find by typing "generic callback" in the search widget on the blog). A couple of things have been added recently, and I thought I would mention them. The generic callback approach uses a single callback function that can be called from a variety of contexts, including when CPLEX solves a node relaxation ("RELAXATION" context), when if finds a candidate solution ("CANDIDATE" context) and, now, when it is ready to split a node into children ("BRANCHING" context).
The branching context is one of the new features. It brings back most of the functionality of the branch callback in the legacy callback system. Unfortunately, it does not seem to have the ability to attach user information to the child nodes, which was a feature that was occasionally useful in the legacy system. You can get more or less equivalent functionality by creating a data store (array, map, whatever) in your global memory and storing the node information keyed by the unique index number of each child node. The catch is that you are now responsible for memory management (freeing up space when a node is pruned and the associated information is no longer needed), and for dealing with thread synchronization issues. | {
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Another new feature is that you can now inject a heuristic solution (if you have one) from all three of the contexts I mentioned above. CPLEX gives you a variety of options for how it will handle the injected solution: "NoCheck" (CPLEX will trust you that it is feasible); "CheckFeasible" (CPLEX will check feasibility and ignore the solution if it is not feasible); "Propagate" (Daniel's explanation: CPLEX will "propagate fixed variables and accept if feasible"); and "Solve" (CPLEX will solve a MIP problem with fixed variables and accept the result if feasible). I assume the latter two mean that you provide a partial solution, fixing some variables but not others. (Unfortunately I was unable to make it to Daniel's talk, so I'm speculating here.)
I'm not sure if those are the only new features, but they are the ones that are most relevant to me. I invite you to read through Daniel's slides to get a more complete picture, including both the reasons for switching from legacy callbacks to generic callbacks and some of the technical issues in using them.
## Tuesday, January 7, 2020
### Greedy Methods Can Be Exact | {
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## Tuesday, January 7, 2020
### Greedy Methods Can Be Exact
We generally sort optimization algorithms (as opposed to models) into two or three categories, based on how certain we are that solutions will be either optimal or at least "good". An answer by Michael Feldmeier to a question I posted on OR Stack Exchange neatly summarizes the categories:
• exact methods eventually cough up provably optimal solutions;
• approximate methods eventually cough up solutions with some (meaningful) guarantee regarding how far from optimal they might be; and
• heuristics provide no worst-case guarantees (but generally are either easy to implement, fast to execute or both).
I should explain my use of "meaningful" (which is not part of Michael's answer). A common way to estimate the "gap" between a solution and the optimum is to take $|z - \tilde{z}|/|z|$, where $z$ is the objective value of the solution produced by the algorithm and $\tilde{z}$ is some bound (lower bound in a minimization, upper bound in a maximization) of the optimal solution. Now suppose that we are minimizing a function known to be nonnegative. If we set $\tilde{s}=0$, we know that any method, no matter how stupid, will have a gap no worse than 100%. To me, that is not a meaningful guarantee. So I'll leave the definition of "meaningful" to the reader.
What brings all this to mind is a question posted on Mathematics Stack Exchange. The author of the question was trying to solve a nonlinear integer program. He approached it by applying a "greedy algorithm". Greedy algorithms are generally assumed to be heuristics, since it seldom is possible to provide useful guarantees on performance. In his case, though, the greedy algorithm is provably optimal, mainly due to the objective function being concave and separable. I'll state the problem and show a proof of optimality below (changing the original notation a bit). Brace yourself: the proof is a bit long-winded. | {
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You start with $N$ workers to be assigned to $M$ work stations. The output of workstation $m$, as a function of the number of workers $x$ assigned to it, is given by $$f_{m}(x)=a_{m}x+b_{m}-\frac{c_{m}}{x},$$
where $a_{m},b_{m},c_{m}$ are all positive constants. Since $f(0)=-\infty$, we can assume that each work station gets at least one worker (and, consequently, that $N>M$). Since $f_{m}'(x)=a_{m}+c_{m}/x^{2}>0$, each $f_{m}()$ is monotonically increasing. Thus, we can safely assume that all $N$ workers will be assigned somewhere. $f_{m}''(x)=-2c_{m}/x^{3}<0$, so $f_{m}()$ is strictly concave (which we will need later). We also note, for future reference, that the impact of adding one worker to a current staff of $x$ at station $m$ is $$\Delta f_{m}(x)=a_{m}+\frac{c_{m}}{x(x+1)}>0.$$
Similarly, the impact of removing one worker at station $m$ is $$\delta f_{m}(x)=-a_{m}-\frac{c_{m}}{x(x-1)}<0.$$We see that $\delta f_{m}(x)$ is an increasing function of $x$ (i.e., it gets less negative as $x$ gets bigger). We also note that $\Delta f_{m}(x)=-\delta f_{m}(x+1)$.
The IP model is easy to state. Let $x_{m}$ be the number of workers assigned to work station $m$. The model is
$$\max\sum_{m=1}^{M}f_{m}(x_{m})$$
subject to
$$\sum_{m=1}^{M}x_{m}\le N$$
with
$$x\in\mathbb{Z}_{+}^{M}.$$
The greedy algorithm starts with a single worker at each station ($x=(1,\dots,1)$) and, at each step, adds one worker to the workstation where that worker produces the greatest increase in objective value (breaking ties arbitrarily). It stops when all $N$ workers are assigned. To prove that it actually finds an optimal solution, I'll use proof by contradiction. | {
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Let $x^{(0)},x^{(1)},\dots,x^{(N-M)}$ be the sequence of solutions constructed by the greedy algorithm, with $x^{(0)}=(1,\dots,1)$, and let $x^{(k)}$ be the last solution in the sequence for which an optimal solution $x^{*}$ exists such that $x^{(k)}\le x^{*}$. The significance of the inequality is that if $x\le x^{*}$, it is possible to extend the partial solution $x$ to the optimal solution $x^{*}$ by adding unassigned workers to work stations. We know that $k$ is well defined because $x^{(0)}\le x^{*}$ for any optimal $x^{*}$. Since we are assuming that the greedy algorithm does not find an optimum, it must be that $k<N-M$.
Now identify the work station $j$ to which the greedy algorithm added a worker at step $k$, meaning that $x_{j}^{(k+1)}=x_{j}^{(k)}+1$ and $x_{i}^{(k+1)}=x_{i}^{(k)}$ for $i\neq j$. Since, by assumption, $x^{(k)}\le x^{*}$ but $x^{(k+1)}\not\le x^{*}$, it must be that $x_{j}^{(k)}=x_{j}^{*}$.
Next, since $x^{(k)}\le x^{*}$ and $x^{(k)}\neq x^{*}$ (else $x^{(k)}$ would be optimal), there is some work station $h\neq j$ such that $x_{h}^{(k)}<x_{h}^{*}$. Let $\tilde{x}$ be the solution obtained from $x^{(k)}$ by adding a worker to station $h$: $\tilde{x}_{h}=x_{h}^{(k)}+1$ and $\tilde{x}_{i}=x_{i}^{(k)}$ for $i\neq h$. Observe that $\tilde{x}\le x^{*}$. The greedy algorithm chose work station $j$ over work station $h$ at $x^{(k)}$, so it must be that
$$\Delta f_{j}(x_{j}^{(k)})\ge\Delta f_{h}(x_{h}^{(k)}). \quad (1)$$ | {
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Finally, let $\hat{x}$ be the result of starting from optimal solution $x^{*}$ and shifting one worker from station $h$ to station $j$. Since
$$x_{j}^{(k+1)}=x_{j}^{(k)}+1=x_{j}^{*}+1=\hat{x}_{j},$$ $$x_{h}^{(k+1)}=x_{h}^{(k)}<x_{h}^{*}\implies x_{h}^{(k+1)}\le\hat{x}_{h}$$
and
$$x_{i}^{(k+1)}=x_{i}^{(k)}\le x_{i}^{*}=\hat{x}_{i}\,\forall i\notin\{h,j\},$$
we have $x^{(k+1)}\le\hat{x}$. Under the assumption that $x^{(k)}$ was the last solution in the greedy sequence that could be extended to an optimal solution, it must be that $\hat{x}$ is not optimal. Thus the net change to the objective function at $x^{*}$ when shifting one worker from station $h$ to station $j$ must be negative, i.e.,
$$\Delta f_{j}(x_{j}^{*})+\delta f_{h}(x_{h}^{*})<0.\quad (2)$$
We showed previously that, under our assumptions, $x_{j}^{(k)}=x_{j}^{*}$, from which it follows that
$$\Delta f_{j}(x_{j}^{*})=\Delta f_{j}(x_{j}^{(k)}). \quad (3)$$
We also showed that $\delta f_{h}()$ is an increasing function. Since $\tilde{x}_{h}\le x_{h}^{*}$,
$$\delta f_{h}(x_{h}^{*})\ge\delta f_{h}(\tilde{x}_{h})=-\Delta f_{h}(x_{h}^{(k)}). \quad (4)$$
Combining (4) with (2), we have
$$\Delta f_{j}(x_{j}^{*})-\Delta f_{h}(x_{h}^{(k)})<0,$$
i.e.,
$$\Delta f_{j}(x_{j}^{*})<\Delta f_{h}(x_{h}^{(k)}). \quad (5)$$
Combining (3) with (5) yields
$$\Delta f_{j}(x_{j}^{(k)})<\Delta f_{h}(x_{h}^{(k)})$$ | {
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# Math Help - Determining the formula of this sequence?
1. ## Determining the formula of this sequence?
I was given the sequence
0, 1, 4, 11, 26, 57, 120, 247, 502, 1013, ...
and I'm looking for a method to determine an explicit formula for this.
I've tried doing difference columns, and I always end up with 2^n being the 2nd difference.
Anyone know a way to figure this out?
2. Originally Posted by paupsers
I was given the sequence
0, 1, 4, 11, 26, 57, 120, 247, 502, 1013, ...
and I'm looking for a method to determine an explicit formula for this.
I've tried doing difference columns, and I always end up with 2^n being the 2nd difference.
Anyone know a way to figure this out?
Good idea to look at differences. So you found:
$(u_{n+2}-u_{n+1})-(u_{n+1}-u_n)=2^n$,
assuming the first term is $u_1=0$.
If you sum the above equalities for $n$ ranging from 1 to $n$, you get a telescoping sum on the left, and a geometric sum on the right, which leads to:
$(u_{n+2}-u_{n+1})-(u_2-u_1)=2+2^2+\cdots +2^n = 2^{n+1}-2$,
hence
$u_{n+2}-u_{n+1}=2^{n+1}-1$.
Summing again this latter equality we have again a telescoping sum on the left and a geometric sum (minus constant terms) on the right, and we get:
$u_{n+2}-u_2=(2^2-1)+\cdots+(2^{n+1}-1)$,
hence (change of variable)
$u_n = (2^2-1)+\cdots +(2^{n-1}-1)+1$
and finally $u_n=2^n-n-1$. Same method works in many situations.
3. Originally Posted by paupsers
I was given the sequence
0, 1, 4, 11, 26, 57, 120, 247, 502, 1013, ...
and I'm looking for a method to determine an explicit formula for this.
I've tried doing difference columns, and I always end up with 2^n being the 2nd difference.
Anyone know a way to figure this out?
The Encyclopaedia of Integer Sequences gives: $2^n-n-1$
CB
4. Hello, paupsers!
Given the sequence: . $0, 1, 4, 11, 26, 57, 120, 247, 502, 1013, \hdots$ | {
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CB
4. Hello, paupsers!
Given the sequence: . $0, 1, 4, 11, 26, 57, 120, 247, 502, 1013, \hdots$
I'm looking for a method to determine an explicit formula for this.
I've tried doing difference columns, and I always end up with 2^n being the 2nd difference.
This is true . . . and should give you a big hint.
Consider the general form: . $f(n) = 2^n$
How does it compare with the given sequence?
$\begin{array}{c|cccccccccc}\hline
n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hline
\text{Sequence:} & 0 & 1 & 4 & 11 & 26 & 57 & 120 & 247 & 502 & 1013 \\ \hline
f(n) = 2^n & 2 & 4 & 8 & 16 & 32 & 64 & 128 & 256 & 512 & 1024 \\ \hline
\text{Error:} & +2 & +3 & +4 & +5 & +6 & +7 & +8 & +9 & +10 & +11\\ \hline
\end{array}$
If we use $f(n) = 2^n$, each term is too large by $(n+1)$
Therefore, the explicit formula is: . $F(n) \;=\;2^n - (n+1)$ | {
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# Can you use both sides of an equation to prove equality?
For example: | {
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$\color{red}{\text{Show that}}$$\color{red}{\frac{4\cos(2x)}{1+\cos(2x)}=4-2\sec^2(x)}$$ In high school my maths teacher told me To prove equality of an equation; you start on one side and manipulate it algebraically until it is equal to the other side. So starting from the LHS: $$\frac{4\cos(2x)}{1+\cos(2x)}=\frac{4(2\cos^2(x)-1)}{2\cos^2(x)}=\frac{2(2\cos^2(x)-1)}{\cos^2(x)}=\frac{4\cos^2(x)-2}{\cos^2(x)}=4-2\sec^2(x)$$$\large\fbox{}$At University, my Maths Analysis teacher tells me To prove a statement is true, you must not use what you are trying to prove. So using the same example as before: LHS = $$\frac{4\cos(2x)}{1+\cos(2x)}=\frac{4(2\cos^2(x)-1)}{2\cos^2(x)}=\frac{2(2\cos^2(x)-1)}{\cos^2(x)}=\frac{2\Big(2\cos^2(x)-\left[\sin^2(x)+\cos^2(x)\right]\Big)}{\cos^2(x)}=\frac{2(\cos^2(x)-\sin^2(x))}{\cos^2(x)}=\bbox[yellow]{2-2\tan^2(x)}$$ RHS =$$4-2\sec^2(x)=4-2(1+\tan^2(x))=\bbox[yellow]{2-2\tan^2(x)}$$ So I have shown that the two sides of the equality in$\color{red}{\rm{red}}$are equal to the same highlighted expression. But is this a sufficient proof? Since I used both sides of the equality (which is effectively; using what I was trying to prove) to show that $$\color{red}{\frac{4\cos(2x)}{1+\cos(2x)}=4-2\sec^2(x)}$$ One of the reasons why I am asking this question is because I have a bounty question which is suffering from the exact same issue that this post is about. ## EDIT: Comments and answers below seem to indicate that you can use both sides to prove equality. So does this mean that my high school maths teacher was wrong? $$\bbox[#AFF]{\text{Suppose we have an identity instead of an equality:}}$$ $$\bbox[#AFF]{\text{Is it plausible to manipulate both sides of an identity to prove the identity holds?}}$$ Thank you. • You did not use both side of the equality. You showed that each is equal to the same thing. That's ok. Using what you want to prove would be using the fact that both sides are equal, but you don't. – Captain Lama Apr 29 '16 at 10:59 • You | {
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be using the fact that both sides are equal, but you don't. – Captain Lama Apr 29 '16 at 10:59 • You can simplify both sides separately to get get to a common point – Archis Welankar Apr 29 '16 at 10:59 • Just another comment: proving that$a=b$is the same as proving that$a-b=0$. So there is no meaningful difference between "manipulating one side" and "manipulating both sides". – Nefertiti Apr 29 '16 at 11:10 • This has already been answered but I'd add that you are really using the fact that$=$sign is an equivalence relation and hence transitive. This might be an issue with more complicated proof whereby the relation is not transitive. – Karl Apr 29 '16 at 12:14 • You did not "use" either side. What does "use" mean here? It means making an assertion, stating a sentence. The sides of the Eq'n are not sentences. If you take either side's formula and show that it is equal to some other formula, without any unsupported or unwarranted assumptions, then you are logical. E.g.: The RHS is always equal to$4-4/(2 \cos^2 x)=4-4/(1+\cos 2 x)=4\cos 2 x/(1+\cos 2 x)$regardless of what is written on the LHS. – DanielWainfleet Apr 29 '16 at 14:07 ## 7 Answers There's no conflict between your high school teacher's advice To prove equality of an equation; you start on one side and manipulate it algebraically until it is equal to the other side. and your professor's To prove a statement is true, you must not use what you are trying to prove. As in Siddarth Venu's answer, if you prove$a = c$and$b = c$("working from both sides"), then$a = c = bby transitivity of equality. This conforms to both your teacher's and professor's advice. Both your high school teacher and university professor are steering you away from "two-column proofs" of the type: \begin{align*} -1 &= 1 &&\text{To be shown;} \\ (-1)^{2} &= (1)^{2} && \text{Square both sides;} \\ 1 &= 1 && \text{True statement. Therefore-1 = 1.} \end{align*} Here, you assume what you want to prove, deduce a true statement, and assert that | {
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= 1.} \end{align*} Here, you assume what you want to prove, deduce a true statement, and assert that the original assumption was true. This is bad logic for at least two glaring reasons: 1. If you assume-1 = 1$, there's no need to prove$-1 = 1$. 2. Logically, if$P$denotes the statement "$-1 = 1$" and$Q$denotes "$1 = 1$", the preceding argument shows "$P$implies$Q$and$Q$is true", which does not eliminate the possibility "$P$is false". What you can do logically is start ("provisionally", on scratch paper) with the statement$P$you're trying to prove and perform logically reversible operations on both sides until you reach a true statement$Q$. A proof can then be constructed by starting from$Q$and working backward until you reach$P$. Often times, the backward argument can be formulated as a sequence of equalities, conforming to your teacher's advice. (Note that in the initial phase of seeking a proof, you aren't bound by anything: You can make inspired guesses, additional assumptions, and the like. Only when you write up a final proof must you be careful to assume no more than is given, and to make logically-valid deductions.) • Suppose you start with$1=1$, take the square root of both sides except use$-1$on the left and$1$on the right, and come up with$-1=1$. It seems like there is more going on here than assuming what you want to prove. – Frank Hubeny Apr 29 '16 at 13:16 • @FrankHubeny To the best of my knowledge, you've just equated the two branches of a multifunction restricted to the real axis - which isn't valid. – QuantumFool Apr 29 '16 at 15:43 • @QuantumFool That's right. The problem is not with assuming what one has to prove but with invalid steps along the way. – Frank Hubeny Apr 29 '16 at 18:44 • @Frank: You're perfectly correct that assuming the conclusion is not the only fatal error in this argument; as noted, proving the converse is another. The (logically valid!) argument for "$-1 = 1$implies$1 = 1$" does not consist entirely of reversible steps, so (as | {
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valid!) argument for "$-1 = 1$implies$1 = 1$" does not consist entirely of reversible steps, so (as you note) does not prove that$-1 = 1$. I mentioned this example to illustrate what the OP has been cautioned against, and because a non-negligible fraction of American university students instinctively attempt to prove algebraic identities by starting with the desired conclusion and manipulating until they obtain a tautology. – Andrew D. Hwang Apr 29 '16 at 20:55 • @BLAZE: An "identity" (such as$\cos^{2} x + \sin^{2} x = 1$) is just an equation that holds for all values of one or more variables (possibly with a "small number of exceptions or restrictions"), so "yes", any proof technique that works for numerical equations also works for identities. – Andrew D. Hwang May 1 '16 at 0:04 It is enough.. Consider this example: To prove:$a=b$Proof: $$a=c$$ $$b=c$$ Since$a$and$b$are equal to the same thing,$a=b$. That is the exact technique you are using and it sure can be used. • I think we also have to consider the domain of each side and make this explicit up front. For example, let$a=(x-1)/(x-1)$and$b=1$. Then we algebraically manipulate this and get$a = 1$. If we do not consider the constraint that$x$is not equal to$1$in the domain of$a$, we get$a = b = 1$for all$x$. That is incorrect. – Frank Hubeny Apr 29 '16 at 15:12 To prove a statement is true, you must not use what you are trying to prove. The main problem is that you've misunderstood what the second teacher said (and possibly the meaning of the equals sign). What that second teacher is saying is do not take as prior facts what you're trying to prove. As one textbook puts it (Setek and Gallo, Fundamentals of Mathematics, 10th Edition, Sec. 3.8), "An argument, or proof, consists basically of two parts: the given statements, which are called premises, and the conclusion". Wikipedia says this (sourced from Cupillari, Antonella. The Nuts and Bolts of Proofs. Academic Press, 2001. Page 3.): "In mathematics and logic, a | {
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The Nuts and Bolts of Proofs. Academic Press, 2001. Page 3.): "In mathematics and logic, a direct proof is a way of showing the truth or falsehood of a given statement by a straightforward combination of established facts, usually axioms, existing lemmas and theorems, without making any further assumptions." Now consider a simple equation/identity like$6 = 2 \times 3$. The separate sides are expressions; if you just said "6" in English that's a sentence fragment, not an assertion of any fact. It can't be evaluated as either "true" or "false", because it has no assertive content. It cannot be used as a premise because it's not a proposition. What makes something a fully-formed statement in mathematical language is a relation, most commonly equals (but alternatively "is lesser than", "is greater than", etc., effectively the verbs of the language). Translating the equation$6 = 2 \times 3to English we get "6 is the same as 2 times 3", which is indeed a full sentence. This can be checked as being true or false; it makes an assertion. It can be used as a premise because it is a proposition of a particular fact. In conclusion, both your teachers are correct, and both of your proofs are correct (although most of us would prefer the more concise one). When one says "don't use what you're trying to prove" they're not talking about the appearance of any particular expression in an algebraic transformation; expressions are neither premises nor things that can be proven; they are sentence fragments. They're talking about an assertion of fact, which in math has to be a statement including a relational symbol (most commonly an equation). The fact that you didn't start by assuming that equality means that in both cases you've complied with your second teacher's warning. short answer: equality is symmetric, implication is not (both are however transitive) longer answer: • You are right: if you proof A = C and B = C for some terms/expressions/objects A,B,C then you are allowed to | {
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right: if you proof A = C and B = C for some terms/expressions/objects A,B,C then you are allowed to conclude that A = B (because "=" is transitive and symmetric) • Your teacher is right: if you prove that something true follows from A = B, i.e. A = B => true, you are not allowed to conclude the converse i.e. that A = B since "true is true" (because implication is not symmetric) I agree with @Siddharth Venu. If we have to prove a=b, At times, on proceeding from LHS you may end up in a stage(intermediate step) which you cannot solve any further and you continue to solve RHS to arrive at the same stage i.e, a=c and b=c so you can conclude that a=b these kind of equality problems often comes in trigonometry Many chapters like matrices and our normal algaebra always have a final step where you can directly establish a relation (a=b). There are two common errors that these methods are preventing: \begin{align} 1 &< 2 &&\text{True.} \\ 0 \cdot 1 &< 0 \cdot 2 &&\text{Um...} \\ 0 &< 0 &&\text{False.} \end{align} \begin{align} 1 &= 2 &&\text{False.} \\ 0 \cdot 1 &= 0 \cdot 2 &&\text{Um...} \\ 0 &= 0 &&\text{True.} \end{align} Much confusion ensues when the two "0$"s in the middle steps are obscured by being large, complicated expressions. For instance, is it evident that you are multiplying by zero when you multiply by$\sin(2x) - 2\sin(x)\cos(x)$? (This uses the double angle formula for sine to get an expression that is always zero.) The correct form of inference when you multiply or divide by something complicated is "$A = B$" becomes "$A/C = B/C$or$C = 0$". That is, you got what you expected or you have inadvertently done something crazy. It seems to take a lot of practice to remember that second clause. The example with$A$,$B$, and$C$can be altered somewhat without changing the need for the second clause. You can also multiply both sides by$C$. The relation need not be an equality. Note however, that the sense of an inequality may change if$C$is ever negative. | {
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not be an equality. Note however, that the sense of an inequality may change if$C$is ever negative. Unfortunately your high-school teacher (and some of the other answers) is wrong. It is false that proving something of the form "$A = B$" is always possible by starting from one side and algebraically manipulating it to get a sequence of equal expressions ending with the other side, not to say necessary. ## Not necessary Let us first deal with the false misconception that you can only go in one direction. Suppose you have proven the following, where$A,B,C$are any expressions:$A = C$.$B = C$. Then you can use the second sentence to substitute$C$for$B$in the first to obtain:$A = B$. This is logically valid because "$=$" means "is exactly the same as". ## Not always possible Let us now consider an example where it is simply impossible to manipulate from one side to the other to prove an equality!$\def\zz{\mathbb{Z}}\def\qq{\mathbb{Q}}\def\rr{\mathbb{R}}$Theorem: Take any$x \in \rr$such that$x^2 \le 0$. Then$x = 0$. Proof:$x = 0$or$x > 0$or$x < 0$. [by trichotomy] If$x > 0$:$x^2 = x \times x > 0$. [by positive multiplication] This contradicts$x^2 \le 0$. If$x < 0$:$0 < -x$. [by subtraction]$x^2 = (-x) \times (-x) > 0$. [by positive multiplication] This contradicts$x^2 \le 0$. Therefore$x = 0$. Here "positive multiplication" denotes "multiplying by a positive real", which preserves the inequality sign. Similarly subtraction preserves an inequality. The above theorem cannot be proven directly by algebraic manipulation from one side "$x$" to the other "$0$", simply because the only given condition is an inequality. Similarly the construction of$\sqrt{2}$in elementary real analysis, as shown below, does not permit a proof by algebraic manipulation. Theorem: Let$S = \{ r : r \in \qq_{\ge 0} \land r^2 \le 2 \}$. Then$S$is non-empty and has an upper bound in$\rr$. Let$x = \sup_\rr(S)$. Then$x^2 = 2$. Proof: [Exercise! Or see a good textbook like Spivak's.] ## General identities | {
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Then$x^2 = 2$. Proof: [Exercise! Or see a good textbook like Spivak's.] ## General identities To address the new sub-question in blue, it suffices to generalize the above theorem to the cube-root of arbitrary real numbers, namely:$\sup( \{ r : r \in \qq \land r^3 \le x \} )^3 = x$for any$x \in \rr$. This is an identity but cannot be proven by direct manipulation. You may not be satisfied with this counter-example, but in higher mathematics this kind of identity is in fact the usual kind that we are interested in, rather than identities that can be proven by algebraic manipulation (which are usually considered trivial). Here are some more examples:$\lceil \frac{m}{n} \rceil = \lfloor \frac{m-1}{n}+1 \rfloor$for any$m \in \zz$and$n \in \zz_{>0}$.$\sum_{k=0}^{n-1} 2^k = 2^n-1$for any$n \in \zz_{>0}$. Nevertheless, if you desire an arithmetic identity, the question becomes much more interesting and depends heavily on what you mean by "arithmetic" and what axioms you are allowed to use. Tarski's problem asked whether an arithmetic identity concerning positive integers can be proven using only the basic high-school identities about them, and Wilkie gave an explicit and simple identity that cannot be so proven, basically because any proof needs to use subtraction and hence negative integers. • It seems clear to me that the OP's teacher's advice, "To prove equality of an equation; you start on one side and manipulate it algebraically until it is equal to the other side.", was given in a particular context (avoiding "two-column proofs"), not asserted as the only way of establishing an equality. After re-reading every answer here, I can't find anyone claiming (explicitly or implicitly) that "proving something of the form "$A=B\$" is always possible by starting from one side and algebraically manipulating it to get a sequence of equal expressions ending with the other side [...]." – Andrew D. Hwang Apr 30 '16 at 13:31 | {
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• @AndrewD.Hwang: I know what you're saying, but let me explain why I make that comment. Imagine a student who is not good at mathematics whose teacher tells him/her exactly the words cited in the question. It's easy to see that the student will go away with the wrong impression as I specified. Indeed, standard English only allows it to be interpreted that way, because "To do X, you do Y." only means "In order to do X, you { have to / ought to / should / better } do Y." Furthermore, I have seen so many students with exactly that wrong conception. Any good answer ought to correct that. – user21820 Apr 30 '16 at 14:29
• @BLAZE: I didn't want to criticize the other answers too much, but actually Andrew's argument about avoiding two column proofs is in fact not so good because that is how we can justify a formal proof. If you learn natural deduction (especially Fitch-style, which can be extended to quantifiers like at math.stackexchange.com/a/1684204), you will understand what I mean. The issue in the 'proof' Andrew shows is not its two-column nature but because no axiom allows writing the first statement, and in some sense it doesn't address the actual issue of your high-school teacher's teaching. – user21820 May 1 '16 at 4:17
• @BLAZE: To be precise, it is in my opinion pointless to steer a student away from an incorrect proof form by saying things that aren't correct, and furthermore without even explaining what form is incorrect! Your university teacher at least said the right thing, though it is not well explained enough to dismantle the prior misconceptions of many students that they gain from high-school! – user21820 May 1 '16 at 4:20 | {
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# Is the probability of picking an Ace as the second card of a deck the same if the deck is shuffled after the first card?
Here's the situation:
You have a deck of 52 cards that is already shuffled. You pick the first card, and the first card is not an Ace. Is the probability of drawing an Ace as second card the same if
1. The second card is immediately drawn from the deck
2. The remaining deck is first shuffled without adding the first card back in, and then the second card is drawn.
For me, it is obvious that probability of drawing an Ace in the second case is 4/51, but I'm not entirely sure that the probability is the same in the first case.
Also, what would be the probability of the first case if we have drawn n cards and none of these are aces?
• yes, its the same. – supinf Sep 15 '15 at 11:54
• 4/(52-n). – Did Sep 15 '15 at 11:56
• As an aside, this is also the reason why it is pointless to doubly randomise a choice (e.g. tossing a coin to see who goes first for the real coin toss!) – Oscar Bravo Sep 15 '15 at 14:04
• Another aside... if you and a friend each draw a card, and then your friend reveals he didn't draw an ace... your odds of getting an ace are better if you replace your card with a new one from the 50 cards remaining. – kbelder Sep 15 '15 at 23:33
If it's not intuitively clear, you can check your calculation using contitional probability:
• $X$... the event "the second card is an ace"
• $Y$... the event "the first card is not an ace"
Then, $P(X\land Y) = \frac{48\cdot 4}{52\cdot 51}$ and $P(Y) = \frac{48}{52}$
Then $$P(X|Y) = \frac{P(X\land Y)}{P(Y)} = \frac{\frac{4\cdot 48}{52\cdot 51}}{\frac{48}{52}} = \frac{4}{51}$$ | {
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Given you know the first card of your shuffled deck, the remaining $51!$ ways of arranging the remaining $51$ cards are equally likely. For each of the $52$ possible values of the first card you have $51!$ equally likely arrangements of the remaining $51$ cards. This is consistent with there being $52\cdot 51!=52!$ arrangements of the whole deck to begin with.
You can conclude that, after drawing the first card, the remaining $51$ cards are equally likely to be in any shuffled order, and so shuffling them changes essentially nothing. This argument should lead to the conclusion that the probability is the same in either case.
Yes, the probability of drawing an Ace is the same if you shuffled the rest of the deck or didn't. You are correct that it's 4/51. If you drew n cards and none were Aces, then the probability of drawing an Ace next would be 4/(52-n).
In both cases, there are only 51 cards left, thus $Pr = \dfrac{4}{51}$
And if $n$ non-aces have been drawn, $Pr = \dfrac{4}{52-n}$ | {
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# Integrals over subset of measure space
Let $(X, \mathcal{M}, \mu)$ be a measure space. Suppose $E \in \mathcal{M}$ and $f \in L^+$ where $L^+$ is a space of measurable functions from $X$ to $[0, \infty]$. $\int_E f$ is defined by $\int_X f\chi_E$ where $\chi_E$ is a characteristic function of $E$. Now every time we want to use some property that is true for integrals over whole space $X$, we have to do manipulations with $\chi_E$. For example, let $f_n$ be a sequence in $L^+$ such that $f_n(x) \nearrow f(x)$ for all $x \in E$. Suppose we know that monotone convergence theorem is true for integrals over $X$ and we want to prove $$\int_E f = \lim_{n\to\infty} \int_E f_n.$$ Since $f_n\chi_E \nearrow f\chi_E$ we have $$\int_E f = \int_X f\chi_E = \lim_{n\to\infty} \int_X f_n\chi_E = \lim_{n\to\infty} \int_E f_n.$$ I know it's easy but is there a way to avoid such manipulations? I want to know that equality above and many other statements about integrals are true because $E$ is a measure space in its own right (seems much more natural).
More precisely, for $E \in \mathcal{M}$ define $$\mathcal{M}_E = \{ E \cap F \mid F \in \mathcal{M} \}.$$ It's easy to check that $\mathcal{M}_E$ is a $\sigma$-algebra. $(E, \mathcal{M}_E, \mu|_{\mathcal{M}_E})$ is a measure space and $\int_E f$ in space $(X, \mathcal{M}, \mu)$ is equal to $\int_E f|_E$ in space $(E, \mathcal{M}_E, \mu|_{\mathcal{M}_E})$ for every measurable $f$.
Is this a commonly accepted way of handling integrals over subset of measure space? If not, do I really have to use $\chi_E$ every time or there is some other way? | {
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• You are just using the measure $\mu|_E$, which leads to the same things. As long as you know this measure carries on the properties you want, you're fine. – Silvia Ghinassi Jul 1 '16 at 14:33
• @SilviaGhinassi intuitively it's very simple, but could you be more precise? – edubrovskiy Jul 1 '16 at 14:56
• I agree with you, and that's why I am afraid of getting into technicalities. I might easily say something wrong, so I'll leave it to someone else to provide you a good answer. – Silvia Ghinassi Jul 1 '16 at 15:01
There are basically two interesting types of functions
$$f:X \to [0,\infty] , \mu-\text{ measurable}$$
and
$$f : X \to [- \infty, \infty] \text{ which is integrable, i.e. } \int f(x) \mu(dx) < \infty .$$
These functions are interesting, because you can define for both the integral $\int_X f(x) \mu(dx)$, but maybe the integral is infinity in the first case.
Many theorems are true for either non-negative measurable functions (e.g. monotonic convergence theorem) or integrable functions (dominant convergence theorem).
Now if you have proven the monotonic convergece theorem for a general measurable space, then you don't need to prove it for integrals restricted on a set, simply because $(E, \mathcal{M}_E, \mu|_{\mathcal{M}_E})$ is again a measure space, and thus the monotonic convergence theorem is also true here. As well as all other properties and theorems that are known for integrals on a measurable space.
So in your case,if you have non-negative functions $f_n(x) \nearrow f(x)$ for all $x\in E$ you immediately get
$$\int f(x) \mu|_{\mathcal{M}_E}(dx) = \lim_{n\to \infty} \int f_n(x) \mu|_{\mathcal{M}_E}(dx)$$
However, to be sure that the theorem really holds for all functions restricted to $E$ you need to understand the relation between integrable functions of $\mu_{\mathcal{M}_E}$ and functions $f$ restricted from $X$ to $E$. The relation is as follows: | {
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Let $f:X \to [0,\infty]$ be $\mu$-measurable, or let $f:X \to [-\infty,\infty]$ be $\mu$ integrable. Define for some set $E \in \mathcal{M}$ the function $f':A \to [-\infty, \infty]$ with $f'(x) := f(x)$. Then, we have
$$\int f'(x) \mu_{\mathcal{M}_E}(dx) = \int_A f(x) \mu(dx)$$
• Could you give me a link to some book or paper where this approach is used? I've already checked relation between integrals on $X$ and on $E$. I just wanted to know that it's a commonly accepted approach. – edubrovskiy Jul 2 '16 at 7:06
• @edubrovskiy I can recommend the book **Measure and Integration Theory ** from Heinz Bauer. Its an old, solid and my personal favorite book. The relation between integrals on $X$ and integrals on $E$ is discussed at the end of §12. – Adam Jul 2 '16 at 8:53 | {
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How many triangles in picture
How many triangles in this picture:
-
I only count 40...and unless someone can explain me how this is a mathematics question I think I'm going to vote to close it as off-topic – DonAntonio Jun 26 '13 at 9:16
I do not think one needs mathematics to solve the puzzle. – Avitus Jun 26 '13 at 9:18
@DonAntonio: A systematic approach of the kind that’s second nature to most mathematicians works quite nicely and can even be generalized to larger diagrams of the same type. It’s certainly on-topic. – Brian M. Scott Jun 26 '13 at 9:20
That sounds sound, Brian. Not closing, then. Thanks. – DonAntonio Jun 26 '13 at 9:23
They can be counted quite easily by systematic brute force.
All of the triangles are isosceles right triangles; I’ll call the vertex opposite the hypotenuse the peak of the triangle. There are two kinds of triangles:
1. triangles whose hypotenuse lies along one side of the square;
2. triangles whose legs both lie along sides of the square and whose peaks are at the corners of the square.
The triangles of the second type are easy to count: each corner is the peak of $4$ triangles, so there are $4\cdot4=16$ such triangles.
The triangles of the first type are almost as easy to count. I’ll count those whose hypotenuses lie along the bottom edge of the square and then multiply that by $4$. Such a triangle must have a hypotenuse of length $1,2,3$, or $4$. There $4$ with hypotenuse of length $1$, $3$ with hypotenuse of length $2$, $2$ with hypotenuse of length $3$, and one with hypotenuse of length $4$, for a total of $10$ triangles whose hyponenuses lie along the base of the square. Multiply by $4$ to account for all $4$ sides, and you get $40$ triangles of the second type and $40+16=56$ triangles altogether.
Added: This approach generalizes quite nicely to larger squares: the corresponding diagram with a square of side $n$ will have $4n$ triangles of the first type and $$4\sum_{k=1}^nk=4\cdot\frac12n(n+1)=2n(n+1)$$ | {
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of the second type, for a total of $2n^2+6n=2n(n+3)$ triangles.
-
I know there's already an excellent answer from Brian M. Scott... Buuuuut...
Here's a visual supplement to the aforementioned systematic brute force.
-
Pretty! $\quad$ – Brian M. Scott Jun 27 '13 at 9:23 | {
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# Convergence of $\sum_{n=1}^{\infty}\left(\, \frac{1}{n} - \frac{1}{n + 2}\,\right)$
What criteria can I use to prove the convergence of $$\sum_{n=1}^{\infty}\left(\,{1 \over n} - {1 \over n + 2}\,\right)\ {\large ?}$$
My idea was to use ratio test: $$\displaystyle{1 \over n} - {1 \over n+2} = {2 \over n^{2} + 2n}$$ $$\displaystyle\frac{2}{\left(\, n + 1\,\right)^{2} + 2\left(\, n + 1\,\right)} \frac{n^{2} + 2n}{2} = \frac{n^{2} + 2n}{n^{2} + 4n + 3}$$
Of course $\displaystyle n^{2} + 2n \lt n^{2} + 4n + 3$ for all $\displaystyle n$ , but $$\displaystyle\lim \limits_{n \to \infty} \frac{n^{2} + 2n}{n^{2} + 4n + 3}=1$$ so I am not quite sure if I can apply ratio test.
You have $\frac{2}{n^2+n} \le \frac{2}{n^2}$. Thus you can use the direct comparision test to prove the convergence (if you already have proven in your course that $\sum_{n=1}^\infty \frac 1{n^2}$ and thus also $\sum_{n=1}^\infty \frac 2{n^2}$ converges).
Answer to your 2nd question: You never can apply the ratio test if the limit is 1 (as in this case).
I would write out the first few terms and see how they cancel.
• of course; but to do this I have to prove absolute convergence first – Christian Dec 11 '14 at 18:42
• @Christian No, you do not. – Andrés E. Caicedo Dec 11 '14 at 18:45
• @Christian You can look at the partial sums and see how they cancel without worrying about absolute convergence. Your problem would be with $\sum \limits_{n=1}^{\infty} \frac1n - \sum \limits_{m=1}^{\infty}\frac{1}{m+2}$ – Henry Dec 11 '14 at 18:47
• It might be useful to mention Telescoping Series. – robjohn Dec 16 '14 at 13:04
• The reason that doing this works is that it allows you to calculate the partial sums directly, which you can use to show convergence of the series by definition (i.e. showing that the partial sums converge) – Joshua Mundinger Dec 16 '14 at 16:44
How about computing the partial sums? For any $N>2$ we have: | {
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How about computing the partial sums? For any $N>2$ we have:
$$\sum_{n=1}^{N}\left(\frac{1}{n}-\frac{1}{n+2}\right) = \frac{3}{2}-\frac{1}{N+1}-\frac{1}{N+2}$$ hence: $$\left|\frac{3}{2}-\sum_{n=1}^{N}\left(\frac{1}{n}-\frac{1}{n+2}\right)\right|\leq\frac{2}{N}$$ ensures convergence (towards $\frac{3}{2}$).
Here's another way to prove the convergence $$\sum \limits_{n=1}^{\infty} \left(\frac1n - \frac{1}{n+2}\right)= \sum \limits_{n=1}^{\infty} \frac{n+2-n}{n(n+2)}$$ $$= \sum \limits_{n=1}^{\infty} \frac{2}{n^2+2n}=2 \sum \limits_{n=1}^{\infty} \frac{1}{n^2+2n}$$ Also note that $$\left|\frac{1}{n^2+2n}\right|\leq \left|\frac{1}{n^2}\right|$$ And by the p-series test, we have $$\sum \limits_{n=1}^{\infty} \left|\frac{1}{n^2}\right| \Rightarrow \mbox{converges}$$ Which implies that $$\sum \limits_{n=1}^{\infty} \frac{1}{n^2} \Rightarrow \mbox{converges absolutely}$$ Therefore, by the direct comparison test $$2 \sum \limits_{n=1}^{\infty} \frac{1}{n^2+2n} \Rightarrow \mbox{converges absolutely}$$ Absolute convergence implies convergence. Also a convergent series multiplied by $2$ is still a convergent series.
• I hope the downvoter revisits this answer. I don't think it deserves a downvote, especially after the last edit. (+1) – robjohn Dec 16 '14 at 13:02
You can use the limit comparison test. Let $a_{n} = \frac{2}{n^{2}+2n}$ and consider $b_{n} = \frac{1}{n^{2}}$ then
$$\lim_{n\rightarrow\infty} \frac{a_{n}}{b_{n}}= \lim_{n\rightarrow \infty}\frac{2n^{2}}{n^{2}+2n} = 2$$
hence $\sum a_{n}$ and $\sum b_{n}$ converge or diverge together.. but $b_{n}$ is a convergent $p$-series.
That being said, my up vote is for Mark Bennet.
• with $\frac{1}{n^2}$ ? – Christian Dec 11 '14 at 18:42 | {
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# How to prove that a sequence diverges using the Cauchy definition?
It was a question of my integral calculus exam:
Write the Cauchy definition of $\lim_{n\rightarrow\infty} a_n = L$
So, I literally wrote:
" $\{a_n\}$ is a Cauchy sequence if given $\epsilon>0$ there exists $N \in \mathbb{N}$ such that $\forall\ m,n\in\mathbb{N}$ if $n,m\ge N$, then $|a_n - a_m|<\epsilon$, or with symbols:
$\forall\epsilon>0:\exists N\in\mathbb{N}:\forall\ m,n \in\mathbb{N}: m,n\ge N\implies |a_n - a_m|<\epsilon$.
And if $\{a_n\}$ is a Cauchy sequence, then $\lim_{n\rightarrow\infty} a_n = L$ ( i.e. $\{a_n\}$ converges) "
The following question was:
With this definition, find $\lim_{n\rightarrow\infty} (3n+2)$
And that is what I did:
I proposed that $\lim_{n\rightarrow\infty} (3n+2) = \infty$ i.e. $\{3n+2\}$ diverges. So I think that I have to prove that this limit is equal to infinity using the Cauchy definition.
But if a sequence $\{a_n\}$ diverges $\implies$ $\{a_n\}$ is not a Cauchy sequence.
And $\{a_n\}$ is not a Cauchy sequence if
$\exists\epsilon>0:\forall N\in\mathbb{N}:\exists\ m,n \in\mathbb{N}:$ $m,n\ge N$ and $|a_n - a_m|\ge\epsilon$ $(1)$
So, I must to prove $(1)$.
$(1)$ says that there exists $m,n\in\mathbb{N}$ such that ..., so $(1)$ has not to apply $\forall m,n\in\mathbb{N}$, and for this fact (I think) I can assign convinient values to $m,n$ in order to prove $(1)$:
If i choose $m=N+1,n=N$ then $|3(N+1)+2 -(3N+2)|=3>2$.
So exists $\epsilon=2>0$ such that $\forall N \in\mathbb{N}:$ there exists $m=N+1, n=N$ such that $N+1,N \ge N$ and $|3(N+1)+2 -(3N+2)|=3>\epsilon$.
Hence $\{3n+2\}$ is not a Cauchy sequence $\implies$ $\lim_{n\rightarrow\infty} (3n+2) = \infty$.
Is my proof correct? If it is not correct, could you help me proving that, please? | {
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Is my proof correct? If it is not correct, could you help me proving that, please?
• just reading quick, it looks perfectly good to me – Chessnerd321 Jun 28 '17 at 19:57
• I will note that, at the end, we cannot just say that $$\text{\{a_n\} not Cauchy\implies \lim_{n\to\infty}a_n=\infty}$$We can only say that $$\text{\{a_n\} not Cauchy\implies \lim_{n\to\infty}a_n\notin\Bbb R}$$The limit may be $\pm\infty$ or just non-existent. – Dave Jun 28 '17 at 20:00
• But if i i want to prove that the limit is $\infty$, what have i to do? – Sama Jun 28 '17 at 20:05
• @KarenSM You have to show that for each $M >0$, there is a natural number $N$ such that for all $n \geq N$, $a_n\geq M$. Quite trivial, you can take $N=[M]$, where $[M]$ is the “floor” of $M$, the unique integer satisfying $M-1< [M] \leq M$. – Li Chun Min Jun 28 '17 at 23:53
• @LiChunMin as far as I understand you, I cannot find the $\lim_{n\rightarrow\infty} (3n+2)$ with the Cauchy definition, right? – Sama Jun 29 '17 at 0:26
The definition of a Cauchy sequence is:
$\forall \epsilon > 0, \exists N \in \mathbb{N}: \forall m,n \in \mathbb{N}: m,n \ge N \implies |a_m-a_n| < \epsilon$, and the negation of this definition is:
$\exists \epsilon > 0, \forall N \in \mathbb{N}, \exists m,n \in \mathbb{N}: m,n \ge N, |a_m - a_n| \ge \epsilon$.
For your sequence $a_n = 3n+2, n \ge 1$, choose $\epsilon = 1$ (your choice is $3$), for any $N \in \mathbb{N}$, let $n =N, m = N+1$. Observe $m, n \ge N$, and $|a_m - a_n| = |a_{N+1} - a_{N}|= |3(N+1) - 2 - 3N+2|= 3 > 1= \epsilon$. Thus the sequence above is not Cauchy.
• But with $\epsilon=2$ is still fine, right? I'm nervous because I put this in my exam – Sama Jun 28 '17 at 20:22
• $\epsilon = 2$ works just fine. $2$ is better than $1$... – DeepSea Jun 28 '17 at 20:44 | {
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You did not write the def'n of $\lim_{n\to \infty}a_n=L.$ What you wrote in response to "Define what it means for $(a_n)_n$ to converge to $L$" was: " $(a_n)_n$ is a Cauchy sequence (which means ...) which converges to $L$."
The def'n of $\lim_{n\to \infty}a_n=L$ is $$\forall r>0\;\exists m\;\forall n>m\;(|L-a_n|<r).$$ I have omitted specifying that $m,n\in \mathbb N.$ It is conventional that in "$\lim_{n\to \infty}a_n$", the values of $n$ are restricted to members of $\mathbb N$ unless stated otherwise. | {
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# Which sets of sequence is countable and Uncountable.
Consider the sequences
$$\displaystyle X=\left\{(x_n): x_n \in \left\{0,1\right\},n \in \mathbb{N} \right\}$$ $$and$$ $$\displaystyle Y=\left\{(x_n)\in X:x_n=1 \;\;\text{for at most finitely many n} \right\}$$
I have to choose which is uncountable and which is countable.
Solution i tried- Here $$X$$ is set of sequence with enteries from $$\left\{0,1\right\}$$ thus it has number of elements $$2^{\aleph_0}$$ which is uncountable .
Now The set $$Y$$ it has all the sequences from the set $$X$$ but some of its elements of sequences is replaced by the only '$$1$$' so its Cardinality will be less then $$2^{\aleph_0}$$ ,but by $$\textbf{ continuum hypothesis}$$ there is no set having Cardinality in-between the $${\aleph_0}$$ and $$2^{\aleph_0}$$ so the set $$Y$$ will be countable
I write this proof but i don't even know this is correct or not but i am sure about set $$X$$ but not sure about $$Y$$ please help me with set $$Y$$
Thank you.
• You call both sets $X$, and in the 'second' $X$ you refer to $X$. But that does not make sense. I suppose that the sequences in your 'second' set, are supposed to be sequences of the first set, so they have at most finitly many n, and can only take 0 or 1? – Cornman Sep 28 '19 at 8:54
• The continuum hypothesis can not be proven, or disproven, so you should not use it. :) – Cornman Sep 28 '19 at 8:54
• my bad i will edit it – gaurav saini Sep 28 '19 at 8:56
• What is $x_n$ here? – orlp Sep 28 '19 at 9:01
• Representation of a sequence @orlp – gaurav saini Sep 28 '19 at 9:04
We know that the countable union of countable sets is countable.
Which means the countable union of countable unions of countable sets is a countable set.
And $$Y$$ is the countable union (indexed over how many $$1$$s the sequences have) of countable unions (indexed over how which positions the finite number of $$1$$s occupy) of countable sets.
Bear with me: | {
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Bear with me:
$$V_0=\{(0)=\{0,0,0,0,.....\}\}$$ is a set with one element.
$$V_1 =\{(x_n)$$ where one $$x_i=1$$ and all the rest are $$0\}$$ is countable as there is a one to one correspondence between the $$(x_n)$$ and the possible positions for $$x_i = 1$$.
$$W_{k,1} = \{(x_n): x_{j< k}=0; x_k = 1$$ and there is one $$x_{i>k}=1$$ but all the rest are $$0\}$$. $$\iota: W_{k,1}\leftrightarrow V_1$$ via for every $$(x_n)\in W_{k,1}$$, $$\iota((x_n)) = (w_n= x_{n-k})$$. That is if $$(x_n)$$ is the sequence where $$x_k=1$$ and $$x_{i>k} =1$$ and all else are $$0$$, the $$\iota((x_n))$$ is the sequence where $$x_{i-k}=1$$ and all else are $$0$$. This is clearly a bijection.
$$V_2=\{(x_n)$$ where two $$x_i=x_j=1$$ and all the rest are $$0\}$$. $$V_2=\cup_{i=1}^{\infty} W_1$$ so $$V_2$$ is countable as it is a countable union of countable sets.
Let $$V_m=\{(x_n)$$ where there are exactly $$m$$ $$1$$s and all the rest are $$0\}$$.
Let $$W_{k,m} = \{(x_n)$$ where $$x_{j< k}=0;$$ and there are $$m$$ $$x_{i>k}=1$$ and all the rest are $$0\}$$.
We can define the bijection $$\iota: W_{k,m}\leftrightarrow V_m$$ via $$(x_n) = (x_{n-k})$$.
And $$V_{m+1} = \cup_{i=1}^{\infty} W_{i,m}$$.
By induction if $$V_m$$ is countable then so is each $$W_{k,m}$$ and so $$V_{m+1}$$ as a union of countable sets.
Now your $$Y=\displaystyle Y=\left\{(x_n)\in X:x_n=1 \;\;\text{for at most finitely many n} \right\} = \cup_{i=o}^{\infty} V_i$$.
So $$Y$$ is a countable union of countable sets and thus countable.
$$X$$ is indeed uncountable and your proof is correct. Your proof for countability of $$Y$$ is incorrect:
Now the set $$Y$$ has all the sequences from the set $$X$$, but some of its elements of sequences are replaced by the only '$$1$$' so its cardinality will be less then $$2^{\aleph_0}$$ | {
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Even if you assume the continuums hypothesis (which is a strong assumption to make!), you don't know how many elements you really replaced or removed. How do you know that the cardinality became smaller? Did you construct an explicit injection for that failing to be surjective?
Still, $$Y$$ is countable. I'll explain how. Basically, the only information a sequence $$y \in Y$$ has is the finite, possibly disconnected, strip of ones it contains -- if any. You can see the zeroes as the default value with no information. Hence, we might posit the following wrong bijection:
$$Y \cong \{0,1\}^* \tag{wrong!}$$
E.g. we would do the following association:
• 1110000.... $$\mapsto$$ 111
• 1110100.... $$\mapsto$$ 11101
Certainly, this is injective. But is this surjective? No! We have to consider $$\{0,1\}^*$$ in such a way that padded zeroes at the right are disregarded. In other words, we have
$$Y \cong \{0,1\}^*/\sim$$
where $$\sim$$ is some equivalence. If you trust me that such a $$\sim$$ exists, then we actually don't need to work this out for the countability argument. Namely, since $$\{0,1\}^*$$ was already countable, so is every quotiened version of it.
Let $$A_n=\{1,\cdots,n\}$$. For each $$f \in \{0,1\}^{A_n}$$, define $$g_f:\Bbb N \to \{0,1\}$$ by $$g_f(x)=\begin{cases}f(x)&\text{if}\;x \in \{1,2,..,n\}\\0&\text{otherwise} \end{cases}$$ Then each $$g_f \in Y$$. Then $$Y$$ can be written as $$Y=\cup_{n=1}^\infty Y_n$$ where $$Y_n=\left\{g_f: f \in \{0,1\}^{A_n}\right\}$$ . Here each $$Y_n$$ is finite and hence $$Y$$ is countable!
• does $g_f:\Bbb N \to \{0,1\}$ should be $g_f:\Bbb N \to \{0,1\}^{A_n}$? – gaurav saini Oct 1 '19 at 9:57 | {
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# In how many different ways can a group of 8 people be divide
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Re: In how many different ways can a group of 8 people be divide [#permalink]
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22 Feb 2020, 05:18
Dear experts,
What if we had, FOR EXAMPLE, 3 teams of 1 individual in each and 1 team let's say consisting of 5 people. Will the answer be:
8C1*7C1*6C1*5C5/(4!/3!) ? Since we still have to unarrange , but have 3 repeating values.
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Re: In how many different ways can a group of 8 people be divide [#permalink]
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02 Mar 2020, 20:35
2
noboru wrote:
In how many different ways can a group of 8 people be divided into 4 teams of 2 people each?
A, 90
B. 105
C. 168
D. 420
E. 2520
Responding to a pm:
Quote:
What if we had, FOR EXAMPLE, 3 teams of 1 individual in each and 1 team let's say consisting of 5 people. Will the answer be:
8C1*7C1*6C1*5C5/(4!/3!) ? Since we still have to unarrange , but have 3 repeating values. I know from your post that if all four teams had different number of people we would not unarrange. But if some repeat?
Yes, consider various scenarios: | {
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Yes, consider various scenarios:
8 people and 2 teams (one of 3 people and the other of 5 people)
You just pick 3 of the 8 people for the 3 people team. Rest everyone will be in the 5 people team. No un-arranging required.
8 people and 2 teams (one of 4 people and the other of 4 people)
You pick 4 people for the first team (say A, B, C, D). The other 4 (E, F, G, H) belong to the other team of 4 people. So you calculate this as 8C4. Now within this 8C4 will lie the case in which you picked (E, F, G, H) for the first team and ( A, B, C, D) will belong to the other team. But note that this case is exactly the same as before. 2 teams one (A, B, C, D) and other (E, F, G, H). So you need to un-arrange here.
Similarly, take your case - 8 people - 3 teams of 1 individual in each and 1 team let's say consisting of 5 people
We select 5 people out of 8 for the 5 people team in 8C5 ways. Rest of the 3 people play individually and we need to create no teams so nothing to be done. Since we are not doing any selection for a team, we are not inadvertently arranging and hence un-arranging is not required.
Instead say we have 8 people - and we make 3 teams, one with 4 people and two teams with 2 people each
We select 4 people out of 8 for the 4 people team in 8C4 ways.
Then we select 2 people for the first two people team in 4C2 ways and the leftover 2 people are for the second two people team. But again, as discussed above, there will same cases counted twice here so we will need to un-arrange by dividing by 2.
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In how many different ways can a group of 8 people be divide [#permalink]
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15 Mar 2020, 00:16
Bunuel wrote:
manalq8 wrote:
In how many different ways can a group of 8 people be divided into 4 teams of 2 people each?
90
105
168
420
2520 | {
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90
105
168
420
2520
$$\frac{C^2_8*C^2_6*C^2_4*C^2_2}{4!}=105$$, we are dividing by 4! (factorial of the # of teams) as the order of the teams does not matter. If 8 people are - 1, 2, 3, 4, 5, 6, 7, 8, then (1,2)(3,4)(5,6)(7,8) would be the same 4 teams as (5,6)(7,8)(1,2)(3,4), as we don't have team #1, team #2...
For the first person we can pick a pair in 7 ways;
For the second one in 5 ways (as two are already chosen);
For the third one in 3 ways (as 4 people are already chosen);
For the fourth one there is only one left.
So we have 7*5*3*1=105
You can check similar problems:
http://gmatclub.com/forum/probability-8 ... %20equally
http://gmatclub.com/forum/probability-8 ... ide+groups
http://gmatclub.com/forum/combination-5 ... ml#p690842
http://gmatclub.com/forum/sub-committee ... ide+groups
There is also direct formula for this:
1. The number of ways in which $$mn$$ different items can be divided equally into $$m$$ groups, each containing $$n$$ objects and the order of the groups is not important is $$\frac{(mn)!}{(n!)^m*m!}$$.
2. The number of ways in which $$mn$$ different items can be divided equally into $$m$$ groups, each containing $$n$$ objects and the order of the groups is important is $$\frac{(mn)!}{(n!)^m}$$
Hope it helps.
we say we need to divide by 4! cus the order doesn't matter but when we use combinations arent we anyway implying that the order doesn't matter. I am not able to understand why we need to divide by 4! again when were using combinations and not permutations.
The second approach, I understand perfectly hence, I know I am missing something in the first one. Pls, help. This is really a confusing sub-topic for me under PnC
In how many different ways can a group of 8 people be divide [#permalink] 15 Mar 2020, 00:16
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# Help needed with different approach to a combinatorics question
Every day, the 15 students in Mr. Singh`s Advanced Chemistry class are randomly divided into 5 lab groups of 3 students each. What is the probability that three students - Linda, Martin, and Nancy - are in the same lab group today?
The answer to the question is to see it from the perspective of one of the students we are concerned with. From his/her point of view, 2 of the other 14 students are randomly chosen to be paired with her. This can be done in ${14}\choose{2}$ = 91 ways. Only one of those pairs are her "friends", so the answer is $\frac{1}{91}$.
This answer of course didn't come to me, I read it when I couldn't figure it out. Though curious if I could find the same answer with my approach, I tried to kind of work backwards, knowing the answer.
Here's what I tried to do: I instead thought of how many groups of three could be chosen from 15 students. I thought this could be done in ${15}\choose{3}$ = 455 ways. One of those contains the group of students we are interested in. But the answer is not $\frac{1}{455}$. How does 91 fit into this? Curious as I was, I divided 455 by 91, and lo and behold, I got 5. Did this 5 mean the 5 groups of three students? I wasn't sure.
If I divide 455 by 5, I get 91. But what is the meaning of this? I would rather think that, from the 455 available choices, we choose 5 things (in this case a thing is a group of 3 students). But this would mean ${455}\choose{5}$ = some huge number.
Let's say I have ABCDE, and I want to choose 2 of them. This would be ${5}\choose{2}$ = 10 combinations. If I would ask "how many groups of 5 can I make?", then the answer surely isn't $\frac{10}{5}$ = 2. I definitely can check with pen and paper and see that I can make quite more than 2 groups. So I know my thinking is faulty.
But where? What does the 5 in $\frac{455}{5}$ = 91 mean? And what should my line of thought be when starting from ${15}\choose{3}$ to get to the answer? | {
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There are $\binom{15}3=455$ possible $3$-person groups, so if we were picking just one group of $3$ at random, the probability would indeed be $\frac1{455}$ that we picked this specific group. However, we’re not just picking a single group: we’re dividing the $15$ students into five groups of $3$, so we actually have $5$ chances to get the desired group, not just one. And $\frac5{455}$ is, as you noticed, $\frac1{91}$.
You might reasonably worry that picking a random partition of the students into $5$ groups of $3$ students isn’t really the same as picking $5$ groups of $3$ at random with replacement from the collection of all $455$ $3$-person groups. We can do a more elaborate calculation to show that it really does work out right. There are
$$\binom{15}3\binom{12}3\binom93\binom63\binom33$$
ways to pick a first group of $3$, then pick a second group of $3$ from the $12$ people who remain, and so on until we have all $5$ groups. However, this counts each of the $5!$ permutations of the $5$ groups separately, so the number of ways to partition the students into $5$ groups of $3$ is really only
$$\frac1{5!}\binom{15}3\binom{12}3\binom93\binom63\binom33\;.$$
How many of these partitions have the desired trio as one of its parts? The same reasoning shows that if we set Linda, Martin, and Nancy aside and form $4$ groups of $3$ from the remaining $12$ students, we can do this in
$$\frac1{4!}\binom{12}3\binom93\binom63\binom33$$
different ways. The probability that a randomly chosen partition has our trio as one of its groups is therefore
$$\frac{\frac1{4!}\binom{12}3\binom93\binom63\binom33}{\frac1{5!}\binom{15}3\binom{12}3\binom93\binom63\binom33}=\frac{5!}{4!\binom{15}3}=\frac5{\binom{15}3}=\frac5{455}=\frac1{91}\;.$$ | {
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• "5 chances to get the desired group" - I try to think about this, but I'm having a hard time. I keep connecting the 5 groups with the 455 things, which I probably shouldn't do. As an aside, how long did it take you to be fluent in this, i.e., have the insight quick and the ability to see it from multiple angles? – Garth Marenghi Oct 20 '16 at 19:37
• @Garth: When you divide the students into $5$ groups of $3$, you really are picking $5$ groups, not just one, and one of those $5$ could be the one that we want. I really can’t answer that last question: it’s been over $50$ years since I first encountered these ideas. I don’t think that I ever had any trouble with them, but I know from having taught the subject that my experience is not typical. – Brian M. Scott Oct 20 '16 at 19:46 | {
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# How many ways are there to divide elements into equal unlabelled sets?
How many ways are there to divide N elements into K sets where each set has exactly N/K elements.
For the example of 6 elements into 3 sets each with 2 elements. I started by selecting the elements that would go in the first set (6 choose 2) and then those that would go into the second as (4 choose 2) and then the 2 remaining elements into the third set. This gives, (6 choose 2) * (4 choose 2). In general
(N choose N/K) * (N-(N/K) choose N/K) * (N-(2*N/K) choose N/K) * ... * 1
• Any thoughts on the problem? You aren't a new user here. You should know that questions on this site require at least some input on your efforts towards the problem. Otherwise you are destined for downvotes or being placed on hold – jameselmore Mar 12 '15 at 21:10
• Added. I'm concerned that I'm assuming some kind of ordering. @jameselmore – nickponline Mar 12 '15 at 21:16
Your approach, as noted by Ross Millikan's answer, is effective. Another way to approach such a problem would be to consider "interpreting" a permutation as such a partition - like, if we wrote the elements in the order: $$eabcfd$$ we might just group them into pairs as $\{\{e,a\},\{b,c\},\{f,d\}\}$ - where we just "fill" the expression $\{\{\_,\_\},\{\_,\_\},\{\_,\_\}\}$ by drawing from the order in which we wrote the elements. | {
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However, $eabcfd$ and $aebcfd$ represent the same partition, as the pairs are unordered - and $bceaf\!\,\!d$ also represents the same partition, as the order of the pairs does not matter. Proceeding thusly, we can see that we can reorder within each of the $K$ sets in $(N/K)!$ ways without affecting the partition, and we can reorder the order in which the sets appear in the partition in $K!$ ways - and that, these are the only transformations which do not affect the partition. Thus, dividing the number of permutations of the elements by the number of permutations representing any given partition yields that there are $$\frac{N!}{(N/K)!^k K!}$$ such partitions.
(This can also be found by expanding ${a \choose b}=\frac{a!}{b!(a-b)!}$ and looking at cancellations in your expression)
• Thanks! Is there a way to generalize this such that: if I make D draws with replacement from N distinct items. how many ways are there to get C copies of any K distinct items. I tried to generalize with your analogy of "filling" in the bracket. But for D > C*K it's unclear how to no double count groups. – nickponline Mar 20 '15 at 16:26
Hint: You are close. As the sets are unlabeled, choosing $\{a,b\},\{c,d\},\{e,f\}$ is the same as choosing $\{e,f\},\{c,d\},\{a,b\}$, but you have counted them both. | {
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• Ah, so I should divide by the number of ways of ordering the sets = K! – nickponline Mar 12 '15 at 21:20
• Awesome :), is there a more compact formula that avoids the product? – nickponline Mar 12 '15 at 21:24
• If you look at the factorial expression for the binomial coefficients, there is a lot of cancellation. Let $M=\frac NK$ Your first two terms are $\frac {N!}{(N-M)!M!}\cdot \frac {(N-M)!}{(N-2M)!M!}$ and you can see the cancellation. You wind up with $\frac {N!}{(M!)^KK!}$ There is a nice combinatorial interpretation. There are $N!$ ways to put the elements in order. Reordering elements within a group can be done in $M!$ ways and you have $K$ of those. The last $K!$ is the ordering of the groups. – Ross Millikan Mar 12 '15 at 21:54
In response to your comment on Ross's answer, there is a more compact form: $$\text{number of partitions }=P_{n,k}=\frac{n!}{(n/k)!^{k}k!}$$ To prove this, we instead prove $$n!=P_{n,k}\cdot (n/k)!^k\cdot k!$$ To prove that, ask how many ways are there permute $n$ elements? There are certainly $n!$ such ways. But another to form a permutations is this: first, divide the $n$ elements into $k$ equal sets $(P_{n,k}$ ways to do this). Then, permute each element within each of the sets (they all have size $n/k$, and there are $k$ of them, so there are $(n/k)!^k$ ways. Finally, put the $k$ sets themselves into some order $(k!$ ways).
You can check that $P_{6,3}=\frac{6!}{2!^3\cdot3!}=15$, and that indeed there are $5$ choices for what the set containing $1$ is, and then $\binom{4}2=3$ ways to partition the remaining 4 elements into sets. | {
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# Representing a linear transformation with respect to a new basis
I have a question about the following exercise:
Let $$f:\mathbb{R}^3\to\mathbb{R}^3$$ be the linear transformation such that $$f(x,y,z)=(x-y+2z,-2x+y,x+z).$$ Represent the transformation with respect to the basis $$\{(1,1,0),(0,1,-1),(1,1,1)\}$$.
What I have done:
I found the images of the basis vectors given:
\begin{align}f(1,1,0) &= (0,-1,1)\\ \\ f(0,1,-1) &= (-3,1,-1)\\ \\ f(1,1,1) &= (2,-1,2) \end{align}
and then I found how these vectors can be written as a linear combination of the basis vectors given:
\begin{align}(0,-1,1)&=\fbox{0}\cdot (1,1,0)+\fbox{(-1)}\cdot (0,1,-1)+\fbox{0}\cdot (1,1,1) \\ \\ (-3,1,-1)&=\fbox{-6}\cdot (1,1,0)+\fbox{4}\cdot (0,1,-1)\;\;\,\,+\fbox{3}\cdot (1,1,1) \\ \\ (2,-1,2)&=\fbox{3}\cdot (1,1,0)+\fbox{(-3)}\cdot (0,1,-1)+\fbox{-1}\cdot (1,1,1)\end{align}
thus the matrix which represents $$f$$ with respect to this basis should be $$\begin{bmatrix}0 & -6 & 3\\ -1 & 4 & -3\\ 0 & 3 & - 1\end{bmatrix}.$$
Is this correct? Or should I write the matrix with respect to the canonical basis of $$\mathbb{R}^3$$ and then make the change of basis $$C^{-1}AC$$?
• How did you calculate the coefficients in the boxes? Some of them, as currently written, are incorrect. Jul 10 at 12:33
• Both methods are correct, and should give you the same answer – if you do them right. Jul 10 at 13:05
• @shoteyes they should be correct now, thank you for pointing that out. For each vector $v$, I set up a system to find out which linear combination of basis vectors is equal to it, ie I found out which $c_1,c_2,c_3$ are such that $c_1 v_1+c_2v_2+c_3v_3=v$ and solved it using Gaussian elimination. Jul 10 at 15:48
• Any thoughts on the comments, or on Dan's answer? Jul 12 at 13:47
• @Gerry Myerson they have all been very useful, I have upvoted the comments and accepted Dan's answer, thank you very much Jul 12 at 13:57 | {
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This is the same. Maybe the square diagram in the sequel shows in the "simplest" way why.
First i have to say something about the used convention for vectors. Because this is the "canonical impediment" when dealing with base change.
We work with column vectors, and matrices act on them by left multiplication. The linear map of left multiplication with a matrix $$A$$ will be denoted below (abusively) also by $$A$$. So $$x$$ goes via $$A$$ to $$A\cdot x=Ax$$, displayed as $$x\overset{A}\longrightarrow Ax\ .$$ "Most of the world" uses column vectors. (Some authors write notes or books (e.g. in Word), and find it handy to use row vectors, so they can be simpler displayed in the book rows. In this case linear maps induced by matrices use the multiplication from the right with such matrices. As long as we need in computations only linear combinations the convention is not so important, but it becomes when we use linear maps induced by matrices.)
We will work in the "category" of (finite dimensional) vector spaces (over $$\Bbb R$$) with a fixed bases. The space $$V:=\Bbb R^3$$ comes with the canonical base $$\mathcal E=(e_1,e_2,e_3)$$, where $$e_1,e_2,e_3$$ are the columns of the matrix $$E$$ below, $$E= \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}\ .$$ We write this object as $$(V,\mathcal E)$$. By abuse, we may want to write $$(V,E)$$ instead.
We start with two objects in this category. For our purposes let them have the same underlying vector space $$V=W=\Bbb R^3$$, first object is $$(V,\mathcal B=(b_1,b_2,b_3))$$, and the second object is $$(W,\mathcal C=(c_1,c_2,c_3))$$. | {
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A linear map $$g:V\to W$$ is defined "abstractly", and has no need for chosen bases. But in practice, $$g$$ is usually given bases-specific in the following way. Let $$v$$ be a vector in $$V$$. We write it w.r.t. $$\mathcal B$$ as $$v=x_1b_1+x_2b_2+x_3b_3$$, and write this data as a column vector: $$v = \begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}_{\mathcal B} :=x_1b_1+x_2b_2+x_3b_3 \ .$$ Then we consider a matrix $$M=M_{\mathcal B, \mathcal C}$$ and build the matrix multiplication vector: $$\begin{bmatrix}y_1\\y_2\\y_3\end{bmatrix} = M \begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}\ .$$ Then we consider the vector $$w\in W$$ which written in base $$\mathcal C$$ has the $$y$$-components, so $$w = \begin{bmatrix}y_1\\y_2\\y_3\end{bmatrix}_{\mathcal C} :=y_1c_1+y_2c_2+y_3c_3 \ ,$$ and the map $$g$$ is mapping linearly $$v$$ to $$w$$.
This concludes the section related to conventions and notations.
Let $$\mathcal C$$ be the base from the OP, the base with vectors which are columns of $$C= \begin{bmatrix} 1 & 0 & 1\\ 1 & 1 & 1\\ 0 & -1 & 1 \end{bmatrix}\ .$$ Let $$A$$ be the matrix for the given linear map $$f$$ w.r.t. the canonical base $$\mathcal E$$. $$A=\begin{bmatrix}1&-1&2\\-2&1&0\\1&0&1\end{bmatrix}\ .$$ Consider now the diagram: $$\require{AMScd}$$ $$\begin{CD} (V,E) @>A>f> (V,E) \\ @A C A {\operatorname{id}}A @A {\operatorname{id}}A C A\\ (V,C) @>f>{C^{-1}AC}> (V,C) \\ \end{CD}$$
Indeed, $$C$$ is the matrix of the identity seen as a map $$(V,\mathcal C)\to(V,\mathcal E)$$. For instance, $$c_1=\begin{bmatrix}1\\0\\0\end{bmatrix}_{\mathcal C} \qquad\text{ goes to }\qquad c_1 =\begin{bmatrix}1\\1\\0\end{bmatrix}_{\mathcal E} =C\begin{bmatrix}1\\0\\0\end{bmatrix}_{\mathcal E}\ .$$
It remains to compute explicitly the matrix $$C^{-1}AC$$. Computer, of course: | {
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It remains to compute explicitly the matrix $$C^{-1}AC$$. Computer, of course:
sage: A = matrix(3, 3, [1, -1, 2, -2, 1, 0, 1, 0, 1])
sage: C = matrix(3, 3, [1, 0, 1, 1, 1, 1, 0, -1, 1])
sage: A
[ 1 -1 2]
[-2 1 0]
[ 1 0 1]
sage: C
[ 1 0 1]
[ 1 1 1]
[ 0 -1 1]
sage: C.inverse() * A * C
[ 0 -6 3]
[-1 4 -3]
[ 0 3 -1]
And we check the result both ways:
$$\bf(1)$$
$$\require{AMScd}$$ $$\begin{CD} c_j=\begin{bmatrix}1\\1\\0\end{bmatrix}_{\mathcal E}\ ,\ \begin{bmatrix}0\\1\\-1\end{bmatrix}_{\mathcal E}\ ,\ \begin{bmatrix}1\\1\\1\end{bmatrix}_{\mathcal E} @>A>f> fc_j=\begin{bmatrix}0\\-1\\1\end{bmatrix}_{\mathcal E}\ ,\ \begin{bmatrix}-3\\1\\-1\end{bmatrix}_{\mathcal E}\ ,\ \begin{bmatrix}2\\-1\\2\end{bmatrix}_{\mathcal E} \\ @A C A {\operatorname{id}}A @A {\operatorname{id}}A C A\\ c_j=\begin{bmatrix}1\\0\\0\end{bmatrix}_{\mathcal C}\ ,\ \begin{bmatrix}0\\1\\0\end{bmatrix}_{\mathcal C}\ ,\ \begin{bmatrix}0\\0\\1\end{bmatrix}_{\mathcal C} @>f>{C^{-1}AC}> fc_j=\begin{bmatrix}0\\-1\\0\end{bmatrix}_{\mathcal C}\ ,\ \begin{bmatrix}-6\\4\\3\end{bmatrix}_{\mathcal C}\ ,\ \begin{bmatrix}3\\-3\\-1\end{bmatrix}_{\mathcal C} \end{CD}$$
Or simpler, using block matrices, and ignoring the knowledge of the bases:
$$\require{AMScd}$$ $$\begin{CD} C @>A>f> AC \\ @A C A {\operatorname{id}}A @A {\operatorname{id}}A C A\\ E @>f>{C^{-1}AC}> C^{-1}AC \end{CD}$$
$$\bf(2)$$ In the spirit of the OP, using copy+pasted+corrected row vector computations:
\begin{aligned} (0,-1,1) &=\boxed{0}\cdot (1,1,0)+\boxed{(-1)}\cdot (0,1,-1)+\boxed{0}\cdot (1,1,1) \ , \\ \\ (-3,1,-1) &=\boxed{-6}\cdot (1,1,0)+\boxed{4}\cdot (0,1,-1)+\boxed{3}\cdot (1,1,1) \ , \\ \\ (2,-1,2) &=\boxed{3}\cdot (1,1,0)+\boxed{(-3)}\cdot (0,1,-1)+\boxed{-1}\cdot (1,1,1) \ . \end{aligned} | {
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# Finding orthonormal basis from orthogonal basis
The following is example C.5 from Appendix C (Linear Spaces Review) of Introduction to Laplace Transforms and Fourier Series, Second Edition, by Phil Dyke:
Example C.5 Show that $\{\sin(x),\cos(x)\}$ is an orthonormal basis for the inner product space $V=\{a\sin(x)+b\cos(x); a,b\in\mathbb R, 0\le x\le\pi\}$ using as inner product $$\langle f,g \rangle = \int_0^1 fg dx, \qquad f,g\in V$$ and determine an orthonormal basis.
Solution $V$ is two dimensional and the set $\{\sin(x),\cos(x)\}$ is obviously a basis. We merely need to check orthogonality. First of all, \begin{align}\langle\sin(x),\cos(x)\rangle=\int_0^\pi\sin(x)\cos(x)\,dx&=\frac{1}{2}\int_0^\pi\sin(2x)\,dx\\ &=\left[-\frac{1}{4}\cos(2x)\right]_0^\pi \\ &=0.\end{align} Hence orthogonality is established. Also, $$\langle\sin(x),\sin(x)\rangle=\int_0^\pi\sin^2(x)\,dx=\frac{\pi}{2}$$ and $$\langle\cos(x),\cos(x)\rangle=\int_0^\pi\cos^2(x)\,dx=\frac{\pi}{2}.$$ Therefore $$\left\{\sqrt{\dfrac{2}{\pi}}\sin(x),\sqrt{\dfrac{2}{\pi}}\cos(x)\right\}$$ is an orthonormal basis.
I understand that, for orthonormality, we require that $\| \mathbf{a} \| = 1$. However, I'm unsure of how the orthonormal basis was found at the bottom of the proof?
I would appreciate it if people could please take the time to clarify this. | {
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I would appreciate it if people could please take the time to clarify this.
• $\lvert \lvert \sin(x) \rvert \rvert = \sqrt{\langle \sin(x), \sin(x) \rangle} = \sqrt{\frac{\pi}{2}}$. So in order to make an orthonormal basis from this orthogonal basis, we devide by the length, – Tim Dikland Jul 3 '18 at 10:09
• @TimDikland You mean $||\sin(x) ||$? – The Pointer Jul 3 '18 at 10:11
• Exactly, editted. – Tim Dikland Jul 3 '18 at 10:11
• Whenever you require an orthogonal basis to be ortonormal, just divide each vecotr by its norm. It remains an orthogonal basis (because of the properties of the inner product), but the norm of each vector is 1. – LuxGiammi Jul 3 '18 at 10:15
• Is that a typo in your example where you are using the integral over $[0,1]$ to define the inner product? If not, that changes things. – Disintegrating By Parts Jul 5 '18 at 3:07
As mentioned in the comments to the main post, $\lVert \sin(x) \rVert = \sqrt{\langle \sin(x), \sin(x) \rangle} = \sqrt{\frac{\pi}{2}}$. We then divide the orthogonal vectors by their norms in order convert them into orthonormal vectors. This gets us the orthonormal basis mentioned in the textbook excerpt. | {
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# Computing the product $(\frac{d}{dx}+x)^n(-\frac{d}{dx}+x)^n$
I want to compute the product $$(\frac{d}{dx}+x)^n(-\frac{d}{dx}+x)^n,$$ for a natural number $$n$$. For $$n$$ equal to 0 or 1, the computation is very simple obviously, but for such a low number as 2 the brute force calculation begins to be rather cumbersome and I cannot see any pattern emerging. I tried to find some connection with the Rodrigues' formula for the Hermite polynomials but I could not.
These operators come up in the algebraic approach to the quantum harmonic oscillator.
Explicit Example
To avoid any misunderstanding, I am going to show explicitly the computation for the case $$n=1$$: $$(\frac{d}{dx}+x)(-\frac{d}{dx}+x)=-\frac{d^2}{dx^2}+1+x\frac{d}{dx}-x\frac{d}{dx}+x^2=-\frac{d^2}{dx^2}+x^2+1.$$
One can think of a function $$f$$ the operators are acting on. For example, $$(\frac{d}{dx}\circ x) f= (\frac{d}{dx}x)f+x\frac{d}{dx}f=(1+\frac{d}{dx})f,$$ then $$\frac{d}{dx}\circ x=1+x\frac{d}{dx}$$ | {
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• Since this is physics related, can I ask what the stand-alone differential expression $\mathrm{d}/\mathrm{d}x$ means for your context ? – Rebellos Feb 15 at 19:47
• That simplification doesn't work though as operators don't commute. It simplifies to $(-\frac{d^2}{dx^2}-x\frac{d}{dx}+\frac{d}{dx}x+x^2)^n$. – Nick Guerrero Feb 15 at 20:00
• @Rebellos, $d/dx$ is simply the derivative operator. In this context, it is supposed that the operators are acting on some function. For example, the commutator between $d/dx$ and $x$ is computed as $[d/dx,x]f=d/dx(xf)-x(d/dx f)=f+d/dx f-d/dx f=f$, then $[d/dx,x]=1$. – jobe Feb 15 at 20:15
• @NickGuerrero Inasmuch as these operators ($\mathscr{L}^+$ and $\mathscr{L}^-$) do not commute, how would you show, as you asserted, that $$\left(\mathscr{L}^+\right)^n \left(\mathscr{L}^-\right)^n=(\mathscr{L}^+\mathscr{L}^-)^n?$$ – Mark Viola Feb 15 at 20:32
• @jobe See THIS, which provides a discussion for $(A+B)^n$ where $[A,B]\ne 0$ (i.e., the operators do not commute). – Mark Viola Feb 15 at 20:47
For convenience let us rewrite $$x,\partial_x$$ with $$a,b$$ so $$[a,b]=x\partial_x-\partial_x x=-1$$ (in the operator sense on the Schwartz space).
Lemma. For all $$n\in\mathbb N$$ $$[(a+b)^n,a-b]=2n(a+b)^{n-1}$$
Proof. As darij pointed out, one has $$[a+b,a-b]=2$$ (i.e. the case $$n=1$$). The trick then is $$\begin{split} [(a+b)^{n+1},a-b]&=(a+b)[(a+b)^n,a-b]+[a+b,a-b](a+b)^n\\ &=(a+b)2n(a+b)^{n-1}+2(a+b)^{n}=2(n+1)(a+b)^{(n+1)-1} \end{split}$$ which concludes the proof via induction. $$\square$$
Proposition. For all $$n\in\mathbb N_0$$ $$(a+b)^n(a-b)^n=\prod_{j=1}^n (a^2-b^2+(2j-1))$$ | {
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Proposition. For all $$n\in\mathbb N_0$$ $$(a+b)^n(a-b)^n=\prod_{j=1}^n (a^2-b^2+(2j-1))$$
Proof. ($$n=0$$ is obvious). Note that $$(a-b)(a+b)=a^2-b^2-1$$. Using the previous lemma $$\begin{split} (a+b)^{n+1}(a-b)^{n+1}&=[(a+b)^{n+1},a-b](a-b)^n+(a-b)(a+b)(a+b)^n(a-b)^n\\ &=2(n+1)(a+b)^n(a-b)^n +(a^2-b^2-1)(a+b)^n(a-b)^n\\ &=\big( a^2-b^2+2n+1)(a+b)^n(a-b)^n=\prod_{j=1}^{n+1} (a^2-b^2+(2j-1)) \end{split}$$ which again concludes the proof via induction. $$\square$$
This result reproduces the cases (aside from $$n=0$$, obvious)
• $$(a+b)(a-b)=a^2-b^2+1$$
• Making use of $$[a^2,b^2]=-4ab-2$$ (similar techniques) one gets $$\begin{split} (a+b)^2(a-b)^2=(a^2-b^2+1)(a^2-b^2+3)&=a^4-a^2b^2-b^2a^2+4a^2-4b^2+b^4+3\\ &=a^4-2a^2b^2-4ab+4a^2-4b^2+b^4+1 \end{split}$$
etc... I feel like this formula is the best thing one can hope for in terms of structure.
Edit: Thanks darij for the +200 rep!
• Great job! Pretty sure that a factorization into degree-$2$ factors is an answer to the OP. – darij grinberg Feb 15 at 21:26
• Thank you! I like how we both independently considered a commutator of the form $[x^m,y]$ (my Lemma / your eq. (1)) to approach the problem... – Frederik vom Ende Feb 15 at 21:30
• Thank you for your answer! I am very happy my question has received so amazing answers, but I am having a hard time choosing which one to accept. Your answer seems to be the more explicit that is possible without becoming too complicated. I would like to have the derivative on the right side of each term, but probably that would made the result very complicated as the obtained by @Sangchul Lee. – jobe Feb 17 at 1:12
• Glad to hear so! Of course one may use further tricks to shift all the derivatives ($b$'s) to the right such as $[x^2,\partial_x^2]=-4x\partial_x-2$ etc. although I'm not sure how simple or applicable that'd end up being. Anyways, just accept whatever answer helps (or will likely end up helping) you the most! – Frederik vom Ende Feb 17 at 7:35 | {
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Just some Sage-generated data to play around with:
For $$n = 0$$, the result is $$1$$.
For $$n = 1$$, the result is $$-\frac{\partial^{2}}{\partial x^{2}} + x^{2} + 1$$.
For $$n = 2$$, the result is $$\frac{\partial^{4}}{\partial x^{4}} - 2 x^{2} \frac{\partial^{2}}{\partial x^{2}} - 4 \frac{\partial^{2}}{\partial x^{2}} - 4 x \frac{\partial}{\partial x} + x^{4} + 4 x^{2} + 1$$.
For $$n = 3$$, the result is $$-\frac{\partial^{6}}{\partial x^{6}} + 3 x^{2} \frac{\partial^{4}}{\partial x^{4}} + 9 \frac{\partial^{4}}{\partial x^{4}} + 12 x \frac{\partial^{3}}{\partial x^{3}} - 3 x^{4} \frac{\partial^{2}}{\partial x^{2}} - 18 x^{2} \frac{\partial^{2}}{\partial x^{2}} - 9 \frac{\partial^{2}}{\partial x^{2}} - 12 x^{3} \frac{\partial}{\partial x} - 36 x \frac{\partial}{\partial x} + x^{6} + 9 x^{4} + 9 x^{2} - 3$$.
For $$n = 4$$, the result is $$\frac{\partial^{8}}{\partial x^{8}} - 4 x^{2} \frac{\partial^{6}}{\partial x^{6}} - 16 \frac{\partial^{6}}{\partial x^{6}} - 24 x \frac{\partial^{5}}{\partial x^{5}} + 6 x^{4} \frac{\partial^{4}}{\partial x^{4}} + 48 x^{2} \frac{\partial^{4}}{\partial x^{4}} + 42 \frac{\partial^{4}}{\partial x^{4}} + 48 x^{3} \frac{\partial^{3}}{\partial x^{3}} + 192 x \frac{\partial^{3}}{\partial x^{3}} - 4 x^{6} \frac{\partial^{2}}{\partial x^{2}} - 48 x^{4} \frac{\partial^{2}}{\partial x^{2}} - 36 x^{2} \frac{\partial^{2}}{\partial x^{2}} + 48 \frac{\partial^{2}}{\partial x^{2}} - 24 x^{5} \frac{\partial}{\partial x} - 192 x^{3} \frac{\partial}{\partial x} - 216 x \frac{\partial}{\partial x} + x^{8} + 16 x^{6} + 42 x^{4} - 48 x^{2} - 39$$. | {
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For $$n = 5$$, the result is $$-\frac{\partial^{10}}{\partial x^{10}} + 5 x^{2} \frac{\partial^{8}}{\partial x^{8}} + 25 \frac{\partial^{8}}{\partial x^{8}} + 40 x \frac{\partial^{7}}{\partial x^{7}} - 10 x^{4} \frac{\partial^{6}}{\partial x^{6}} - 100 x^{2} \frac{\partial^{6}}{\partial x^{6}} - 130 \frac{\partial^{6}}{\partial x^{6}} - 120 x^{3} \frac{\partial^{5}}{\partial x^{5}} - 600 x \frac{\partial^{5}}{\partial x^{5}} + 10 x^{6} \frac{\partial^{4}}{\partial x^{4}} + 150 x^{4} \frac{\partial^{4}}{\partial x^{4}} + 150 x^{2} \frac{\partial^{4}}{\partial x^{4}} - 150 \frac{\partial^{4}}{\partial x^{4}} + 120 x^{5} \frac{\partial^{3}}{\partial x^{3}} + 1200 x^{3} \frac{\partial^{3}}{\partial x^{3}} + 1800 x \frac{\partial^{3}}{\partial x^{3}} - 5 x^{8} \frac{\partial^{2}}{\partial x^{2}} - 100 x^{6} \frac{\partial^{2}}{\partial x^{2}} - 150 x^{4} \frac{\partial^{2}}{\partial x^{2}} + 1500 x^{2} \frac{\partial^{2}}{\partial x^{2}} + 975 \frac{\partial^{2}}{\partial x^{2}} - 40 x^{7} \frac{\partial}{\partial x} - 600 x^{5} \frac{\partial}{\partial x} - 1800 x^{3} \frac{\partial}{\partial x} - 600 x \frac{\partial}{\partial x} + x^{10} + 25 x^{8} + 130 x^{6} - 150 x^{4} - 975 x^{2} - 255$$. | {
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For $$n = 6$$, the result is $$\frac{\partial^{12}}{\partial x^{12}} - 6 x^{2} \frac{\partial^{10}}{\partial x^{10}} - 36 \frac{\partial^{10}}{\partial x^{10}} - 60 x \frac{\partial^{9}}{\partial x^{9}} + 15 x^{4} \frac{\partial^{8}}{\partial x^{8}} + 180 x^{2} \frac{\partial^{8}}{\partial x^{8}} + 315 \frac{\partial^{8}}{\partial x^{8}} + 240 x^{3} \frac{\partial^{7}}{\partial x^{7}} + 1440 x \frac{\partial^{7}}{\partial x^{7}} - 20 x^{6} \frac{\partial^{6}}{\partial x^{6}} - 360 x^{4} \frac{\partial^{6}}{\partial x^{6}} - 540 x^{2} \frac{\partial^{6}}{\partial x^{6}} + 120 \frac{\partial^{6}}{\partial x^{6}} - 360 x^{5} \frac{\partial^{5}}{\partial x^{5}} - 4320 x^{3} \frac{\partial^{5}}{\partial x^{5}} - 8280 x \frac{\partial^{5}}{\partial x^{5}} + 15 x^{8} \frac{\partial^{4}}{\partial x^{4}} + 360 x^{6} \frac{\partial^{4}}{\partial x^{4}} + 450 x^{4} \frac{\partial^{4}}{\partial x^{4}} - 9000 x^{2} \frac{\partial^{4}}{\partial x^{4}} - 6525 \frac{\partial^{4}}{\partial x^{4}} + 240 x^{7} \frac{\partial^{3}}{\partial x^{3}} + 4320 x^{5} \frac{\partial^{3}}{\partial x^{3}} + 15600 x^{3} \frac{\partial^{3}}{\partial x^{3}} + 7200 x \frac{\partial^{3}}{\partial x^{3}} - 6 x^{10} \frac{\partial^{2}}{\partial x^{2}} - 180 x^{8} \frac{\partial^{2}}{\partial x^{2}} - 540 x^{6} \frac{\partial^{2}}{\partial x^{2}} + 9000 x^{4} \frac{\partial^{2}}{\partial x^{2}} + 31050 x^{2} \frac{\partial^{2}}{\partial x^{2}} + 9180 \frac{\partial^{2}}{\partial x^{2}} - 60 x^{9} \frac{\partial}{\partial x} - 1440 x^{7} \frac{\partial}{\partial x} - 8280 x^{5} \frac{\partial}{\partial x} - 7200 x^{3} \frac{\partial}{\partial x} + 8100 x \frac{\partial}{\partial x} + x^{12} + 36 x^{10} + 315 x^{8} - 120 x^{6} - 6525 x^{4} - 9180 x^{2} - 855$$.
Sage code:
A.<x> = DifferentialWeylAlgebra(QQ)
x, dx = A.gens()
def r(n):
return (dx + x) ** n * (-dx + x) ** n
for i in range(7):
print "For $$n = " + str(i) + "$$, the result is $$" + latex(r(i)) + "$$.\r\n" | {
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for i in range(7):
print "For $$n = " + str(i) + "$$, the result is $$" + latex(r(i)) + "$$.\r\n"
Note that it is easily seen that $$\left[\dfrac{\partial}{\partial x} + x, - \dfrac{\partial}{\partial x} + x\right] = 2$$. Thus, the operators $$\dfrac{\partial}{\partial x} + x$$ and $$- \dfrac{\partial}{\partial x} + x$$ themselves generate an isomorphic copy of the Weyl algebra, except for a scalar factor of $$2$$.
In view of the relation $$\left[\dfrac{\partial}{\partial x} + x, - \dfrac{\partial}{\partial x} + x\right] = 2$$, perhaps the following copypasta from some of my old homework will come useful.
Let $$\mathbb{N} = \left\{0,1,2,\ldots\right\}$$.
Now we need an easy fact from quantum algebra:
Proposition 1. Let $$A$$ be a ring (not necessarily commutative). Let $$x\in A$$ and $$y\in A$$ be such that $$xy-yx=1$$. Then, \begin{align} \left( xy\right) ^{n}=\sum\limits_{k=0}^{n} \genfrac{\{}{\}}{0pt}{0}{n+1}{k+1} y^{k}x^{k}, \end{align} where the curly braces denote Stirling numbers of the second kind.
Proof of Proposition 1. First of all, it is easy to see that $$$$x^{m}y=mx^{m-1}+yx^{m}\ \ \ \ \ \ \ \ \ \ \text{for every }m\in\mathbb{N} \label{darij1.pf.xy-yx.1} \tag{1}$$$$ (this allows $$m=0$$ if $$0x^{0-1}$$ is interpreted as $$0$$). Indeed, the proof of \eqref{darij1.pf.xy-yx.1} proceeds by induction over $$m$$ and is straightforward enough to be left to the reader.
We will now prove Proposition 1 by induction over $$k$$. The induction base is obvious, so we step to the induction step:
Let $$n>0$$. Assuming that $$\left( xy\right) ^{n-1}=\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n}{k+1} y^{k}x^{k}$$, we need to show that $$\left( xy\right) ^{n}=\sum\limits_{k=0} ^{n} \genfrac{\{}{\}}{0pt}{0}{n+1}{k+1} y^{k}x^{k}$$. | {
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We have \begin{align*} \left( xy\right) ^{n} & =\left( xy\right) ^{n-1}xy=\left( \sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n}{k+1} y^{k}x^{k}\right) xy\ \ \ \ \ \ \ \ \ \ \left( \text{since }\left( xy\right) ^{n-1}=\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n}{k+1} y^{k}x^{k}\right) \\ & =\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n}{k+1} y^{k}\underbrace{x^{k}x}_{=x^{k+1}}y=\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n}{k+1} y^{k}\underbrace{x^{k+1}y}_{\substack{=\left( k+1\right) x^{k} +yx^{k+1}\\\text{(by \eqref{darij1.pf.xy-yx.1}, applied to }m=k+1\text{)}}}\\ & =\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n}{k+1} y^{k}\left( \left( k+1\right) x^{k}+yx^{k+1}\right) \\ & =\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n}{k+1} \left( k+1\right) y^{k}x^{k}+\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n}{k+1} \underbrace{y^{k}y}_{=y^{k+1}}x^{k+1}\\ & =\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n}{k+1} \left( k+1\right) y^{k}x^{k}+\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n}{k+1} y^{k+1}x^{k+1}\\ & =\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n}{k+1} \left( k+1\right) y^{k}x^{k}+\sum\limits_{k=1}^{n} \genfrac{\{}{\}}{0pt}{0}{n}{k} y^{k}x^{k}\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{here, we substituted }k-1\text{ for }k\text{ in the second sum}\right) \\ & =\sum\limits_{k=0}^{n} \genfrac{\{}{\}}{0pt}{0}{n}{k+1} \left( k+1\right) y^{k}x^{k}+\sum\limits_{k=0}^{n} \genfrac{\{}{\}}{0pt}{0}{n}{k} y^{k}x^{k}\\ & \ \ \ \ \ \ \ \ \ \ \left( \begin{array}[c]{c} \text{here, we extended both sums by zero terms, using the fact}\\ \text{that } \genfrac{\{}{\}}{0pt}{0}{n}{n+1} = \genfrac{\{}{\}}{0pt}{0}{n}{0} =0\text{ whenever }n>0 \end{array} \right) \\ & =\sum\limits_{k=0}^{n}\underbrace{\left( \genfrac{\{}{\}}{0pt}{0}{n}{k+1} \left( k+1\right) + \genfrac{\{}{\}}{0pt}{0}{n}{k} \right) }_{\substack{= \genfrac{\{}{\}}{0pt}{0}{n+1}{k+1} \\\text{(by the recursion formula for Stirling numbers}\\\text{of the second kind)}}}y^{k}x^{k}\\ | {
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\\\text{(by the recursion formula for Stirling numbers}\\\text{of the second kind)}}}y^{k}x^{k}\\ & =\sum\limits_{k=0}^{n} \genfrac{\{}{\}}{0pt}{0}{n+1}{k+1} y^{k}x^{k}. \end{align*} This completes the induction step, and thus the inductive proof of Proposition 1. $$\blacksquare$$ | {
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Proposition 2. Let $$A$$ be a ring (not necessarily commutative). Let $$x\in A$$ and $$y\in A$$ be such that $$xy-yx=1$$. Then, \begin{align} \left( yx\right) ^{n}=\sum\limits_{k=0}^{n} \genfrac{\{}{\}}{0pt}{0}{n}{k} y^{k}x^{k}, \end{align} where the curly braces denote Stirling numbers of the second kind.
Proof of Proposition 2. Just as in the proof of Proposition 1, we show that \eqref{darij1.pf.xy-yx.1} holds.
We will now prove Proposition 2 by induction over $$k$$. The induction base is obvious, so we step to the induction step:
Let $$n>0$$. Assuming that $$\left( yx\right) ^{n-1}=\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n-1}{k} y^{k}x^{k}$$, we need to show that $$\left( yx\right) ^{n}=\sum\limits_{k=0} ^{n} \genfrac{\{}{\}}{0pt}{0}{n}{k} y^{k}x^{k}$$. | {
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We have \begin{align*} \left( yx\right) ^{n} & =\left( yx\right) ^{n-1}yx=\left( \sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n-1}{k} y^{k}x^{k}\right) yx\ \ \ \ \ \ \ \ \ \ \left( \text{since }\left( yx\right) ^{n-1}=\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n-1}{k} y^{k}x^{k}\right) \\ & =\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n-1}{k} y^{k}\underbrace{x^{k}y}_{\substack{=kx^{k-1}+yx^{k}\\\text{(by \eqref{darij1.pf.xy-yx.1}, applied to }m=k\text{)}}}x\\ & =\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n-1}{k} y^{k}\left( kx^{k-1}+yx^{k}\right) x\\ & =\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n-1}{k} ky^{k}\underbrace{x^{k-1}x}_{=x^{k}}+\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n-1}{k} k\underbrace{y^{k}y}_{=y^{k+1}}\underbrace{x^{k}x}_{=x^{k+1}}\\ & =\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n-1}{k} ky^{k}x^{k}+\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n-1}{k} ky^{k+1}x^{k+1}\\ & =\sum\limits_{k=0}^{n-1} \genfrac{\{}{\}}{0pt}{0}{n-1}{k} ky^{k}x^{k}+\sum\limits_{k=1}^{n} \genfrac{\{}{\}}{0pt}{0}{n-1}{k-1} ky^{k}x^{k}\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{here, we substituted }k-1\text{ for }k\text{ in the second sum}\right) \\ & =\sum\limits_{k=0}^{n} \genfrac{\{}{\}}{0pt}{0}{n-1}{k} ky^{k}x^{k}+\sum\limits_{k=0}^{n} \genfrac{\{}{\}}{0pt}{0}{n-1}{k-1} ky^{k}x^{k}\\ & \ \ \ \ \ \ \ \ \ \ \left( \begin{array}[c]{c} \text{here, we extended both sums by zero terms, using the fact}\\ \text{that } \genfrac{\{}{\}}{0pt}{0}{n-1}{n} = \genfrac{\{}{\}}{0pt}{0}{n-1}{-1} =0\text{ whenever }n>0 \end{array} \right) \\ & =\sum\limits_{k=0}^{n}\underbrace{\left( \genfrac{\{}{\}}{0pt}{0}{n-1}{k} k+ \genfrac{\{}{\}}{0pt}{0}{n-1}{k-1} \right) }_{\substack{= \genfrac{\{}{\}}{0pt}{0}{n}{k} \\\text{(by the recursion formula for Stirling numbers}\\\text{of the second kind)}}}y^{k}x^{k}\\ & =\sum\limits_{k=0}^{n} \genfrac{\{}{\}}{0pt}{0}{n}{k} y^{k}x^{k}. \end{align*} This completes the induction step, and thus the inductive proof of | {
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y^{k}x^{k}. \end{align*} This completes the induction step, and thus the inductive proof of Proposition 2. $$\blacksquare$$ | {
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• Thank you very much for this code! I really needed something like this! The only point is that I am desperately looking for a closed formula for arbitrary $n$. Do you have any thoughts about that? – jobe Feb 15 at 20:36
• @jobe: My output doesn't give me much hope for a closed formula -- but of course, the Weyl algebra has several bases in which these elements could be expanded, and maybe one of them looks better. – darij grinberg Feb 15 at 20:39
• Thank you for the reference , but I could not open it. – jobe Feb 15 at 20:45
Let $$X, Y$$ be operators which are central, meaning that $$[X,Y] = XY-YX = c$$ for some scalar $$c$$. Then
1. As a form of binomial theorem, we have
$$(X+Y)^n = \sum_{\substack{a,b,m \geq 0 \\ a+b+2m=n}} \frac{n!}{a!b!m!2^m} c^m Y^b X^a = \sum_{a=0}^{n} \binom{n}{a} P_{n-a}(c,Y)X^a,$$
where $$P_n$$ is defined by the following sum
$$P_n(c, x) = \sum_{m=0}^{\lfloor n/2\rfloor} \frac{n!}{(n-2m)!m!2^m} c^m x^{n-2m} = \left( c\frac{d}{dx} + x \right)^n \mathbf{1}.$$
In particular, if $$c = -1$$ then $$P_n(-1, x) = \operatorname{He}_n(x)$$, where $$\operatorname{He}_n$$ is the probabilists' Hermite polynomial. Similarly, if $$c = 1$$, then $$P_n(1, x) = i^{-n} \operatorname{He}_n(ix)$$.
2. Under the same condition, for any polynomials $$f, g$$ we have
$$f(X)g(Y) = \sum_{m \geq 0} \frac{c^m}{m!} g^{(m)}(Y)f^{(m)}(X).$$
In our case, $$[\frac{d}{dx}, x] = 1$$, and so, we can use both formulas to give a complicated, but still explicit expression for the product of $$(\frac{d}{dx}+x)^n$$ and $$(-\frac{d}{dx}+x)^n$$. Combining altogether,
\begin{align*} &\left( \frac{d}{dx} + x \right)^n \left( -\frac{d}{dx} + x \right)^n \\ &= \sum_{p \geq 0} \frac{1}{p!} \Bigg( \sum_{\substack{a_i, b_i, m_i \geq 0 \\ a_i+b_i+2m_i = n-p}} \frac{(-1)^{a_2+m_2} (n!)^2}{a_1!a_2!b_1!b_2!m_1!m_2!2^{m_1+m_2}} x^{b_1+b_2} \left( \frac{d}{dx} \right)^{a_1+a_2} \Bigg). \end{align*} | {
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The following is a sample Mathematica code, comparing this formula with the actual answer for the case $$n = 3$$.
Coef[n_, c_, l_] := n!/(l[[1]]! l[[2]]! l[[3]]! 2^l[[3]]) c^l[[3]];
T[n_] := FrobeniusSolve[{1, 1, 2}, n];
(* Compute (x+d/dx) (x-d/dx)^n f(x) *)
n = 3;
Nest[Expand[x # + D[#, x]] &, Nest[Expand[x # - D[#, x]] &, f[x], n], n]
Sum[1/p! Sum[ Sum[ Coef[n, 1, l1] Coef[n, -1, l2]
(-1)^l2[[2]] x^(l1[[1]] + l2[[1]]) D[f[x],
{x, l1[[2]] + l2[[2]]}], {l1, T[n - p]}], {l2, T[n - p]}], {p, 0, n}] // Expand
Clear[n];
• Impressive! Your answer gives an explicit expression for the product as I have asked for, but it is indeed complicated! I am still trying to figure out how using it effectively in the problems I have in mind, but that is my fault. Thank you! – jobe Feb 17 at 1:19
• @jobe, Glad it helped! This formula may be still far from being qualified as generally useful, but a form of combinatoric interpretation of the coefficients in the expansion of $(X+Y)^n$ is available, which may possible be useful for some applications: $$\frac{n!}{(n-2m)!m!2^m}=[\text{# of m-matchings in \{1,\cdots,n\}}].$$ So $P_n(c,x)$ computes the sum of weights over all possible matchings on $\{1,\cdots,n\}$, where each matching is weighted by the factor $c^m x^{n-2m}$, i.e. each matched pair receives the weight $c$ and each unmatched vertice receives the weight $x$. – Sangchul Lee Feb 17 at 1:32 | {
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# Pendulum dilemma
## Homework Statement
I am suppose to find a linear graph for the equation T=2pie(√L/√g)
## The Attempt at a Solution
The best linear graph I could think of was L/T^2.
Am I doing it right? Thanks.
Yes.
T$^{2}$ on y-axis and L on x-axis.
Yes.
T$^{2}$ on y-axis and L on x-axis.
Thank you very much. This has been like a huge thorn in my finger all day!
I thought the slope was always y/x, so why isn't the slope of the graph T^2/L?
Slope is NOT y/x but slope = $\Delta$y/($\Delta$x)
or more exactly slope = dy/(dx).
T = 2∏√(L/g)
therefore T$^{2}$ = 4π$^{2}$L/g
so d(T$^{2}$)/dL = 4π$^{2}$/g = slope
T = 2∏√(L/g)
therefore T$^{2}$ = 4π$^{2}$L/g
so d(T$^{2}$)/dL = 4π$^{2}$/g = slope
oh that makes much more sense. thank you. | {
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what is the argument for not using the average of an average?
I want to disprove someone's calculation of percentage of cash sales for a year by taking summing percentage of cash sales by month and dividing by 12. I sense the correct way is to take total cash sales for the year and divide by total sales for year to arrive at percentage of sales but need correct reasoning why it is wrong to use an average of an average.
-
What is 'percentage of cash sales'? How is it calculated? – Michael Biro Jan 4 at 21:54
You need a counter example. E.g.
Mon Sales Cash Percent
Jan 100 70 70%
Feb 10 1 10%
Mar 10 1 10%
Apr 10 1 10%
May 10 1 10%
Jun 10 1 10%
Jul 10 1 10%
Aug 10 1 10%
Sep 10 1 10%
Oct 10 1 10%
Nov 10 1 10%
Dec 100 70 70%
Total sales are $300$, cash sales are $150$ so the overall percentage is $50\%$. But the average of the monthly percentages is $20\%$.
-
Thank you very much. I will be back to this site I'm sure. – denise Jan 5 at 14:42
You are correct, here's a mathematical example:
Let $s_1, s_2, s_3 \cdots s_{12}$ be the sales of the months.
Let $c_1, c_2, c_3 \cdots c_{12}$ be the cash sales of the months.
$$\frac{1}{12} \sum_{k=1}^{12} \frac{c_k}{s_k}$$
Is the same as: $$\frac{\sum_{k=1}^{12} c_k}{\sum_{k=1}^{12} s_k}$$
You can show it is false by plugging in some numbers.
-
Thanks for taking the time to respond. Very helpful. – denise Jan 5 at 14:41
The average of the averages leads to the right average only when the samples have the same size.
If each month has exactly the same number of sales, then taking the average of the averages would be right. But if the number of sales per month is different, then the average of averages leads in general to the wrong answer. | {
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Simple situation: First month 100 sales of 1 each. Average 1. Second month 1 sale of 100. Average 100 .
The average of the averages is 50.50, but there were 101 sales and only 200 income....
-
Right, otherwise you need a weighted average. A simple unweighted arithmetic mean would be wrong. – Fixed Point Jan 5 at 1:27
Thank you very much. Language very helpful. – denise Jan 5 at 14:40
Suppose in January I sell \$100 worth of product, and all of it is sold for cash. Every other month I sell \$1 worth of product, and none of it is sold for cash. I have then sold \$111 dollars worth of product, of which \$100 was sold for cash, so my percentage of cash sales is $\frac{100}{111}\approx90\%$. Calculated the other way, we would get that for $1$ month $100\%$ of my sales were cash, while for the other $11$, $0\%$ of my sales were cash, giving $\frac{100\%}{12}\approx 8.3\%$. This is clearly wrong.
-
Thanks so much for responding. Examples are so helpful! – denise Jan 5 at 14:41
It is clear to you that taking cash sales for a year, and dividing by total sales for the year, will give you the correct proportion of cash sales for that year.
The problem with taking monthly proportions and averaging them (average of averages) is that that procedure could produce a quite different number.
Let's take a simple numerical example. January to October: total sales each month $\$1000$, cash sales each month$\$500$. November and December (Christmas season, people are buying a lot, mostly on credit): total sales each month $\$10000$, cash sales each month$\$500$.
So total sales for the year, 30000, cash sales 6000. Thus the yearly proportion of cash sales is $\dfrac{6000}{30000}=20\%$. This is the correct percentage.
Now let's compute the monthly averages. For January through October, they are $50\%$. For each of November and December, they are $5\%$. | {
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To find the average of the monthly proportions, as a percent, we take $\frac{1}{12}(50+50+50+50+50+50+50+50+50+50 +5+5)$. This is approximately $42.5\%$, which is wildly different from the true average of $20\%$.
For many businesses, sales exhibit a strong seasonality. If the pattern of cash sales versus total sales also exhibits seasonality, averaging monthly averages may give answers that are quite far from the truth.
-
Exactly what I needed! Thanks very much. – denise Jan 5 at 14:39 | {
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Definition 4.21.1. Let $I$ be a set and let $\leq$ be a binary relation on $I$.
1. We say $\leq$ is a preorder if it is transitive (if $i \leq j$ and $j \leq k$ then $i \leq k$) and reflexive ($i \leq i$ for all $i \in I$).
2. A preordered set is a set endowed with a preorder.
3. A directed set is a preordered set $(I, \leq )$ such that $I$ is not empty and such that $\forall i, j \in I$, there exists $k \in I$ with $i \leq k, j \leq k$.
4. We say $\leq$ is a partial order if it is a preorder which is antisymmetric (if $i \leq j$ and $j \leq i$, then $i = j$).
5. A partially ordered set is a set endowed with a partial order.
6. A directed partially ordered set is a directed set whose ordering is a partial order.
## Comments (4)
Comment #2102 by Keenan Kidwell on
I just noticed that in the definition of a partially ordered set, the condition of antisymmetry is omitted. Isn't it more standard to call a transitive, reflexive relation a preorder? I'm guessing the antisymmetry condition doesn't play a role in considerations involving e.g. colimits over a directed set, so this is technically moot, but I'm curious if the decision to use a (as far as I can tell) not-quite-standard definition was intentional.
Comment #2129 by on
OK, somebody else has mentioned this previously... and at the time there was a proof somewhere using the weaker notion... but I cannot find it now. The motivation for the weaker notion is that it is exactly enough conditions to turn $I$ into a category, so you can define (co)limits. I think that for almost all statements in the Stacks project it does not make a difference which definition you use. For example the proof of Lemma 4.21.5 seems to produce a partially ordered set in the stronger sense. | {
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OK, so maybe we should change this (if you are reading this and agree please leave a comment). Then a directed set will not be a partially ordered set in general. So lot's of lemmas should get multiple statements some with directed partially ordered sets and some with just directed sets... This is lot's of work, so I will only do this if more people chime in.
Comment #2138 by Keenan Kidwell on
I was actually very happy when I saw this definition as it made me realize that the formalism of colimits and limits over directed sets worked just fine without the antisymmetry condition. I guess what it really showed me (which is something actually more standard) is that "directed sets" need not be "partially ordered" in the stronger sense. Since you don't actually seem to need the stronger sense, and it would require a non-trivial amount of work if you were to switch, I think you should leave it as it is.
Comment #2146 by on
OK, I decided to revert to the standard definition of partially ordered sets and now a directed set is a preordered set with upper bounds for finite subsets. Going through all the corrections I found that it absolutely does not matter at all! It very slightly shortens the Stacks project text because it allows us to say "directed set" instead of "directed partially ordered set" in many lemmas, propositions, etc. But I am not sure if the use of the terminology "directed set" is completely standard. I found places where people suggest one should say "directed proset" where "proset" is an abbreviation for "preordered set" but that is just plain ugly. Oh well.
Big set of changes can be found here.
There are also:
• 2 comment(s) on Section 4.21: Limits and colimits over preordered sets
## Post a comment
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# What is x% more than y? (1) x% of y is 9 (2) y% more than x is 24
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What is x% more than y? (1) x% of y is 9 (2) y% more than x is 24 [#permalink]
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05 Mar 2018, 04:04
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What is x% more than y?
(1) x% of y is 9
(2) y% more than x is 24
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Joined: 04 Jan 2018
Posts: 39
Re: What is x% more than y? (1) x% of y is 9 (2) y% more than x is 24 [#permalink]
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05 Mar 2018, 09:32
2
ans is C | {
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### Show Tags
05 Mar 2018, 09:32
2
ans is C
1) xy/100=9..........................(1)
2) x+xy/100=24....................(2)
subst 1 in 2
we get 'x'
subst x in 1 or 2.
we get y
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What is x% more than y? (1) x% of y is 9 (2) y% more than x is 24 [#permalink]
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19 Mar 2018, 12:35
SOLUTION
We need to find the value which is $$x$$% more than $$y$$.
Thus, we need to find the value of $$y$$+$$\frac{x}{100}$$ * $$y$$.
Statement-1: “$$x$$% of $$y$$ is $$9$$”.
• $$\frac{x}{100}$$ * $$y$$=$$9$$……………………(1)
Since we don’t know the value of $$x$$ and $$y$$, statement 1 alone is not sufficient to answer the question.
Statement-2: “$$y$$% more than $$x$$ is $$24$$”.
• $$x$$+ $$\frac{y}{100}$$*$$x$$=$$24$$…………………….(2)
Since we don’t know the value of $$x$$ and $$y$$, statement 2 alone is not sufficient to answer the question.
Combining both the statements:
• $$x$$+ $$\frac{x*y}{100}$$ = $$24$$
• $$x+9=24$$
• $$x=15$$
Putting the value of $$x$$ in equation $$1$$
• $$\frac{x*y}{100}$$=$$9$$
• $$15$$*$$\frac{y}{100}$$=$$9$$
• $$y=60$$
We now have the value of $$x$$ and $$y$$ both. Hence, we can find the asked value.
Statement (1) and (2) together are sufficient.
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# Euclidean algorithm
I found something interesting about solving this question:
If p is the number of pairs of values (x, y) that can make the equation 84x + 140y = 28 true, what is true about p?
I noticed while solving that I could simplify the equation:
84x - 140y = 28
42x – 70y = 14
21x – 35y = 7
3x – 5y = 1
Noting that 3x = 5y + 1 means that 3x must end in 5 + 1 = 6 or 0 + 1 = 1 to allow y to be divisible by 5.
This means 3x should end in either 1 or 6 to be equal to 5y + 1
Finding the following values for x and matching values for y:
3x = 6, 21, 36, 51, 66, 81, 96, 111, 126, 141, 156
5y = 5, 20, 35, 50, 65, 80, 95, 110, 125, 140, 155
Then finding the series of x and y:
X = 2, 7, 12, 17, 22, 27, 32, 37, 42, 47, 52
Y = 1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31
X seems to be increasing by 5, and Y by 3.
Which means that you can find pairs of x and y by the formulas:
x = 2+5a and y = 1+3a
Leading to a sequence that will satisfy 84x + 140y = 28.
Would this also be more generally true? I also thought it might be an interesting additional question to the quiz.
On another note, this text editor does not seem to like enters in paragraphs. Any way I can fix this?
Note by Sytse Durkstra
1 year ago
MarkdownAppears as
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MathAppears as
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## Comments
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Yes, you just applied Euclidean algorithm.
To enter $$\LaTeX$$, type like
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but remove the spaces.
- 1 year ago
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# Arc length question
##### Member
The length of the minor arc of a circle is 10cm, while the area of the sector AOB is 150cm2.
a) Form two equations involving r and θ, where θ is measured in radians.
b) Solve these equations simultaneously to find r and θ.
Help to solve? Cant understand the question very well.
I think the arc length formula was
length=$$\displaystyle \frac{n}{360}\cdot2\pi(r)$$
$$\displaystyle \therefore10=\frac{n}{360}\cdot2\pi(r)$$
The question states, Form two equations involving r and θ, where θ is measured in radians.
So would we have to arrange the formula to find the r and $$\displaystyle \theta$$
#### MarkFL
Staff member
The formulas you want are:
Circular arc-length:
$$\displaystyle s=r\theta\tag{1}$$
Area of circular sector
$$\displaystyle A=\frac{1}{2}r^2\theta\tag{2}$$
Now, if we solve (1) for $\theta$, we find:
$$\displaystyle \theta=\frac{s}{r}$$
Now, substituting this into (2), we obtain:
$$\displaystyle A=\frac{1}{2}r^2\left(\frac{s}{r}\right)=\frac{rs}{2}\implies r=\frac{2A}{s}\implies\theta=\frac{s^2}{2A}$$
Now, just plug in the given values for $A$ and $s$.
##### Member
The formulas you want are:
Circular arc-length:
$$\displaystyle s=r\theta\tag{1}$$
Area of circular sector
$$\displaystyle A=\frac{1}{2}r^2\theta\tag{2}$$
Now, if we solve (1) for $\theta$, we find:
$$\displaystyle \theta=\frac{s}{r}$$
Now, substituting this into (2), we obtain:
$$\displaystyle A=\frac{1}{2}r^2\left(\frac{s}{r}\right)=\frac{rs}{2}\implies r=\frac{2A}{s}\implies\theta=\frac{s^2}{2A}$$
Now, just plug in the given values for $A$ and $s$.
Im finding trouble understanding the question.
"Form two equations involving r and $$\displaystyle \theta$$" Basically means transpose the formula or make the equation equal to r and $$\displaystyle \theta$$?
Since the arc length = $$\displaystyle s=r\theta$$ (I thought arc length was length=$$\displaystyle \frac{n}{360}\cdot2\pi(r)$$?
r=$$\displaystyle s\theta$$? | {
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r=$$\displaystyle s\theta$$?
For b) we just sub in the values? Where s is the length?
#### MarkFL
Staff member
You are given values for the arc-length $s$ and the area $A$, and so you want to use the formulas relating $r$ and $\theta$ to these given values (which I gave as (1) and (2)) to be able to express both $r$ and $\theta$ as functions of $s$ and $A$ (which I did using some algebra), so that you can use these given values to determine $r$ and $\theta$.
Once you have $r$ and $\theta$ as functions of $s$ and $A$, it is simply a matter of using the given values to evaluate $r$ and $\theta$.
#### SuperSonic4
##### Well-known member
MHB Math Helper
Im finding trouble understanding the question.
"Form two equations involving r and $$\displaystyle \theta$$" Basically means transpose the formula or make the equation equal to r and $$\displaystyle \theta$$?
It's asking to make two equations for the length of an arc and the area of a sector. As long as your equations have $$\displaystyle r$$ and $$\displaystyle /theta$$ in them then you're fine for this step.
MarkFL has given you the equations you need (his TeX is also better than mine! )
$$\displaystyle \displaystyle s=r\theta\tag{1}$$
$$\displaystyle \displaystyle A=\frac{1}{2}r^2\theta\tag{2}$$
As you see both of these equations have $$\displaystyle r$$ and $$\displaystyle \theta$$ in them so Part A of the question has been solved already
Since the arc length = $$\displaystyle s=r\theta$$ (I thought arc length was length=$$\displaystyle \frac{n}{360}\cdot2\pi(r)$$?
r=$$\displaystyle s\theta$$?
$$\displaystyle \frac{n}{360}\cdot 2\pi r$$ is the equation for arc length where $$\displaystyle n$$ is in degrees. Since the question asks for radians you want $$\displaystyle s = r\theta$$. See the spoiler below for why they are equivalent
By definition a full circle has 360 degrees and $$2\pi$$ radians so you can say $$360^{\circ} = 2\pi$$ (we omit the c symbol for radians) | {
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Your formula for degrees is $$s = \frac{\theta}{360} \cdot 2\pi r$$ but since we know that $$360^{\circ} = 2\pi$$ it can be subbed in
$$s = \frac{\theta}{2\pi} \cdot 2\pi r = r \theta$$ - the radian expression
For b) we just sub in the values? Where s is the length?
Sub in the values given (10cm and 150cm²) for s and A respectively so you have two equations and two unknowns ($$\displaystyle r$$ and $$\displaystyle \theta$$).
You can now solve simultaneously using either the substitution method MarkFL showed in post 2. He found $$\displaystyle \theta$$ in terms of $$\displaystyle r$$ which means he was able to substitute $$\displaystyle \frac{s}{r}$$ wherever $$\displaystyle \theta$$ appeared.
He did in terms of $$\displaystyle s$$ in post 2 but if you find it easier you may use $$\displaystyle s=10$$ for your example
##### Member
It's asking to make two equations for the length of an arc and the area of a sector. As long as your equations have $$\displaystyle r$$ and $$\displaystyle /theta$$ in them then you're fine for this step.
MarkFL has given you the equations you need (his TeX is also better than mine! )
$$\displaystyle \displaystyle s=r\theta\tag{1}$$
$$\displaystyle \displaystyle A=\frac{1}{2}r^2\theta\tag{2}$$
As you see both of these equations have $$\displaystyle r$$ and $$\displaystyle \theta$$ in them so Part A of the question has been solved already
$$\displaystyle \frac{n}{360}\cdot 2\pi r$$ is the equation for arc length where $$\displaystyle n$$ is in degrees. Since the question asks for radians you want $$\displaystyle s = r\theta$$. See the spoiler below for why they are equivalent
By definition a full circle has 360 degrees and $$2\pi$$ radians so you can say $$360^{\circ} = 2\pi$$ (we omit the c symbol for radians)
Your formula for degrees is $$s = \frac{\theta}{360} \cdot 2\pi r$$ but since we know that $$360^{\circ} = 2\pi$$ it can be subbed in
$$s = \frac{\theta}{2\pi} \cdot 2\pi r = r \theta$$ - the radian expression | {
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$$s = \frac{\theta}{2\pi} \cdot 2\pi r = r \theta$$ - the radian expression
Sub in the values given (10cm and 150cm²) for s and A respectively so you have two equations and two unknowns ($$\displaystyle r$$ and $$\displaystyle \theta$$).
You can now solve simultaneously using either the substitution method MarkFL showed in post 2. He found $$\displaystyle \theta$$ in terms of $$\displaystyle r$$ which means he was able to substitute $$\displaystyle \frac{s}{r}$$ wherever $$\displaystyle \theta$$ appeared.
He did in terms of $$\displaystyle s$$ in post 2 but if you find it easier you may use $$\displaystyle s=10$$ for your example
Thanks this really did help
##### Member
The formulas you want are:
Circular arc-length:
$$\displaystyle s=r\theta\tag{1}$$
Area of circular sector
$$\displaystyle A=\frac{1}{2}r^2\theta\tag{2}$$
Now, if we solve (1) for $\theta$, we find:
$$\displaystyle \theta=\frac{s}{r}$$
Now, substituting this into (2), we obtain:
$$\displaystyle A=\frac{1}{2}r^2\left(\frac{s}{r}\right)=\frac{rs}{2}\implies r=\frac{2A}{s}\implies\theta=\frac{s^2}{2A}$$
Now, just plug in the given values for $A$ and $s$.
Beginning to understand this a lot easier.
Just a question on the last part.
How did you rearrange to get $$\displaystyle \theta$$?
It went from, $$\displaystyle A=\frac{1}{2}r^2\left(\frac{s}{r}\right)$$ which I understand to $$\displaystyle =\frac{rs}{2}$$?
#### MarkFL
Staff member
Beginning to understand this a lot easier.
Just a question on the last part.
How did you rearrange to get $$\displaystyle \theta$$?
$$\displaystyle s=r\theta$$
Divide through by $r$
$$\displaystyle \frac{s}{r}=\frac{\cancel{r}\theta}{\cancel{r}}$$
$$\displaystyle \theta=\frac{s}{r}$$
It went from, $$\displaystyle A=\frac{1}{2}r^2\left(\frac{s}{r}\right)$$ which I understand to $$\displaystyle =\frac{rs}{2}$$?
$$\displaystyle A=\frac{1}{2}r^2\left(\frac{s}{r}\right)=\frac{\cancel{r}\cdot r\cdot s}{2\cancel{r}}=\frac{rs}{2}$$ | {
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# Optimal route consisting of rowing then walking
Problem
You're in a boat on point A in the water, and you need to get to point B on land. Your rowing speed is 3km/h, and your walking speed 5km/h.
See figure:
Find the route that takes the least amount of time.
My idea
I started by marking an arbitrary route:
From here, I figure the total time is going to be $$T = \frac{R}{3\mathrm{km/h}} + \frac{W}{5\mathrm{km/h}}$$
Since this is a function of two variables, I'm stuck.
A general idea is to express $W$ in terms of $R$ to make it single-variable, and then apply the usual optimization tactics (with derivatives), but I'm having a hard time finding such an expression.
Any help appreciated!
EDIT - Alternative solution?
Since the distance from A to the right angle (RA) is traveled 3/5 times as fast as the distance between RA and B, could I just scale the former up?
That way, I get A-RA being a distance of $6\cdot\frac53 = 10\mathrm{km}$, which makes the hypotenuse $\sqrt{181}$ the shortest distance between A and B. And since we scaled it up, we can consider it traversable with walking speed rather than rowing speed!
Thoughts? | {
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Thoughts?
• but .. where is the river / land border ? – G Cab Oct 5 '17 at 21:56
• The horizontal leg of the triangle would simultaneously be the border and the walking route. It sounds strange, but the idea is that the distance between the shoreline and the walking path is "negligible", thus they can be seen as one-and-the-same. – Alec Oct 6 '17 at 8:27
• Thanks, now it is clear. Another question, do you know (accept) trigonometric approach? – G Cab Oct 6 '17 at 21:12
• @GCab - Yes. I've reached an answer of 3.4hours with that approach. However, $\frac{\sqrt{181}}{5}$ doesn't give the same answer, and I can't figure out why that wouldn't work, and if there's some adjustment that needs to be made. Or if it needs to be discarded completely. – Alec Oct 7 '17 at 11:01
• I think the answer is here, so wikipedia gets the bonus: en.wikipedia.org/wiki/Fermat%27s_principle – Ethan Bolker Oct 8 '17 at 14:23
• a) the solution
The formula has already been indicated by wgrenard and AdamBL $$T = {1 \over 3}\sqrt {36 + \left( {9 - W} \right)^{\,2} } + {1 \over 5}W$$
differentiating that $${{dT} \over {dW}} = {{5W + 3\sqrt {36 + \left( {9 - W} \right)^{\,2} } - 45} \over {15\,\sqrt {36 + \left( {9 - W} \right)^{\,2} } }}$$ and equating to $0$ gives \eqalign{ & {{dT} \over {dW}} = 0\quad \Rightarrow \quad 3\sqrt {36 + \left( {9 - W} \right)^{\,2} } = 45 - 5W\quad \Rightarrow \cr & \Rightarrow \quad 9\left( {36 + \left( {9 - W} \right)^{\,2} } \right) = 25\left( {9 - W} \right)^{\,2} \quad \Rightarrow \cr & \Rightarrow \quad 9 \cdot 36 = 16\left( {9 - W} \right)^{\,2} \quad \Rightarrow \quad W = 9 - \sqrt {{{9 \cdot 36} \over {16}}} = 9 - {9 \over 2} = {9 \over 2} \cr} which is a minimum, because the function is convex as already indicated.
Thus $$\left\{ \matrix{ W_m = 9/2 \hfill \cr T_m = 17/5 \hfill \cr R_m = 15/2 \hfill \cr} \right.$$
• b) Scaling | {
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• b) Scaling
Your idea of scaling according to speed is quite entangling.
That means (if I understood properly) that you are transforming the triangle from space to time units.
But, by introducing different scaling factors for the two coordinates, you undermine the Euclidean norm, which does not " transfer" between the two systems (if assumed valid in one, shall be modified in the other).
Consider for example the transformation sketched below.
From the mathematical point of view it is a linear scale transformation $$\left( {\matrix{ {y_1 } \cr {y_2 } \cr } } \right) = \left( {\matrix{ {1/v_1 } & 0 \cr 0 & {1/v_2 } \cr } } \right) \left( {\matrix{ {x_1 } \cr {x_2 } \cr } } \right)$$
Now, with constant $v_1, \,v_2$, any path in $x$ will transform in the corresponding path in $y$ (going through corresponding points).
If the path is a curve parametrized through a common parameter $\lambda$, not influencing the $v_k$'s, then, at any given value of $\lambda$ the point on the $x$ plane will transform into the corresponding point in $y$ plane $$\left( {\matrix{ {y_{1}(\lambda) } \cr {y_{2}(\lambda) } \cr } } \right) = \left( {\matrix{ {1/v_1 } & 0 \cr 0 & {1/v_2 } \cr } } \right) \left( {\matrix{ {x{_1}(\lambda) } \cr {x_{2}(\lambda) } \cr } } \right)$$ and the minimal path in one plane will be the corresponding minimal path in the other.
But we shall also have that $$\frac{d}{{d\lambda }}\left( {\matrix{ {y_{1}(\lambda) } \cr {y_{2}(\lambda) } \cr } } \right) = \left( {\matrix{ {1/v_1 } & 0 \cr 0 & {1/v_2 } \cr } } \right) \frac{d}{{d\lambda }}\left( {\matrix{ {x{_1}(\lambda) } \cr {x_{2}(\lambda) } \cr } } \right)$$ that is that the "velocities" compose vectorially.
Therefore if $\lambda$ is the time, you shall go from $A$ to $C$ with a composition of a vertical rowing speed and a horizontal walking speed (a "$\infty$-thlon"), which takes the same time as rowing $AH$ and walking $HC$. | {
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When, instead, you just row on $AC$, then you shall change the above matrix - for that segment only - according to the $\angle AC$, and of course you loose the correspondence minimal to minimal as based on the Euclidean norm (straight line $A'B'$).
• The question has recieved a nice handful of good answers. I'm still somewhat perplexed as to the alternative solution, which I was hoping to focus on with the bounty, but since you're the only one who even mentioned it, the bounty goes here. I would be interested to learn more about the entangling nature of that strategy though. Do you have any links, or simply a concept I should look into? – Alec Oct 12 '17 at 7:23
• Thanks for appreciation. I will try and add more elaboration to clear the "scaling" matter. – G Cab Oct 12 '17 at 8:48
• @Alec: I expanded on the 2nd part: wish I succeeded and make it more clear and .. convincing. – G Cab Oct 12 '17 at 13:31
The portion of the line on the top left is $9-w$. So by the Pythagorean theorem $R^2 = 36 + (9-w)^2$. I think this is the relationship you are looking for.
You do not need to worry about the hypotenuse in the larger right triangle.
Consider the right triangle that has R as the hypotenuse. Left side of this triangle will have length $6$ and the top side will have length $(9-W)$. By the Pythagorean theorem, $R^{2} = 36+(9-W)^2$.
Solving for R and inserting in your original equation yields $T= \frac{1}{3} \sqrt{36+(9-W)^{2}} + \frac{1}{5} W$. This function is strictly convex, so you can simply minimize it by solving $\frac{d\ T}{d\ W}=0$ (the 'usual optimization tactics').
Regarding your alternative solution, I am unsure what you are arguing. The shortest distance between the points A and B will always be the line connecting them. As you have shown in your drawing, $|AB|=3\sqrt{13} \ne\sqrt{181}$. | {
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• No, I mean if we've scaled the left-most side to 10km (by multiplying 5/3), the hypotenuse will be $\sqrt{10^2 + 9^2} = \sqrt{181}$. – Alec Oct 4 '17 at 7:39
• Yes, that is correct. If I understand you correctly, though, you are arguing that it will always be optimal to row the entire way, since the hypotenuse will always be the shortest distance from A to B. That's obviously not the case. – AdB Oct 5 '17 at 15:37
• Nono, I'm not arguing that it's faster to row always. I'm saying that if we scale the rowing distance up by 5/3, then we can consider that distance to be traversable by walking speed instead of rowing speed. – Alec Oct 5 '17 at 17:09
Snell's Law is usually applied to optics, but it is based on the quickest path through two media in which the speed of light differs. Snell's Law says that $$n_1\sin(i_1)=n_2\sin(i_2)\tag1$$ where $i_k$ is the angle of incidence to the boundary of the path and $n_k$ is inversely proportional to the speed in the particular medium ($n_kv_k=c$).
We can adapt this to the current situation by noting that $(1)$ is equivalent to $$\frac{\sin(i_1)}{v_1}=\frac{\sin(i_2)}{v_2}\tag2$$ If we travel at all along the shore, $\sin(i_2)=1$ (the angle of incidence is $90^\circ$). Since $v_1=3$ and $v_2=5$, we must have $\sin(i_1)=\frac35$, which implies that $\tan(i_1)=\frac34$.
If $\tan(i_1)=\frac34$, and the width of the river is $6$ km, then the downriver distance must be $\frac34\cdot6=4.5$ km.
This path takes $\frac{7.5}3+\frac{4.5}5=3.4$ hours. | {
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This path takes $\frac{7.5}3+\frac{4.5}5=3.4$ hours.
• Wait, so we must find an angle such that we only travel 4.5km by water? I must be interpreting that wrong, because the shortest possible water-distance must be 6km, right? – Alec Oct 8 '17 at 20:50
• No, the downriver distance is $4.5$ km, the river is still $6$ km wide, so that makes that water distance $\sqrt{4.5^2+6^2}=7.5$ km. – robjohn Oct 8 '17 at 22:55
• Ah, gotcha! Yeah, that was a nice take on this problem. I never would have connected this to Snell's Law on my own. – Alec Oct 9 '17 at 7:47 | {
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# Magnetic Dollars
Imgur
Suppose that I have two urns, with one magnetic dollar in each urn. I begin randomly throwing more magnetic dollars in the general vicinity of the urns, and the coins fall in the urns by a simple rule:
Let's say that Urn A has $$x$$ coins, and Urn B has $$y$$ coins. The probability that the next coin falls into Urn A is $$\dfrac{x}{x+y}$$, and the probability that the next coin falls into Urn B is $$\dfrac{y}{x+y}$$.
You keep throwing magnetic dollars until there is a total of $$1,000,000$$ magnetic dollars in total.
What would you bet would be the price on average, of the urn with the smaller amount of coins?
$$\ 10$$, perhaps? Maybe $$\ 100$$ or $$\ 500$$? Post your bet or write it down on a piece of paper before looking at the next section.
The less-priced urn, on average, is actually worth a grand total of a quarter of a million dollars! Don't worry if you guessed wrong; many professional mathematicians also guessed much lower than this. In fact, when a group of mathematicians were asked this question and were asked to bet, most people only bet $$\ 10$$ and only one person bet over $$\ 100$$.
But why does the lower-priced urn price so high? You may want to try the problem out yourself before I go over a very nice and elegant solution. See if you can find it!
Tried it out yet? In the case that you have, let's see how this problem can be so elegantly solved, as I claimed.
Suppose that you have a deck of cards; one red, and $$999,998$$ white. Currently, you just have a red card. Now every turn, you place a white card in any available slot. For example, in the first move, you have $$2$$ available slots: one above the red card, and one below. In the second move, you have $$3$$ available slots, and so on. | {
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But wait! Let's say that the empty slots above the red card are the magnetic dollars in Urn A, and the empty slots below the red card are the magnetic dollars in Urn B. Notice that if you had $$x$$ empty slots above the red card and $$y$$ empty slots below the red card, then the probability that the next card will be above the red card is $$\dfrac{x}{x+y}$$, and the probability that the next card will be below the red card is $$\dfrac{y}{x+y}$$! We've found a one-to-one correspondence between the original magnetic dollar problem and this new card problem!
Finally, we know that in a random placements of white cards in this fashion will result in a uniform distribution of where the red card is, every single final position is of equal probability. This means that in the original problem, the probability of $$42$$ magnetic dollars being in Urn A is the same as the probability of $$314,159$$ magnetic dollars in Urn A. Therefore, the average price of the lower priced urn is clearly $$250,000$$. $$\Box$$
Note by Daniel Liu
4 years, 8 months ago
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The result is certain not intuitive, nor obvious. | {
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Sort by:
The result is certain not intuitive, nor obvious.
You can look at this problem John's Red And Blue Balls, which is based off the same idea.
Staff - 4 years, 8 months ago
I know some of you have been waiting for me to post for a long time. Sorry for the inactivity, I had a bunch of homework and stuff.
Well, here is my next #CosinesGroup posts. The problem in my opinion is a very cool problem because of the awesome elegant solution.
This problem has also been called "Polya's Urns", but I looked that up and "Polya's Urns" actually covers a lot of random complicated stuff, so I kept it at magnetic dollars.
Hope you enjoy! Feedback is appreciated.
- 4 years, 8 months ago
This is the solution in Peter Winkler ' Mathematical Puzzles. An excellent book.
- 2 years, 2 months ago
How do you do this?
- 4 years, 7 months ago
I'm curious to know why only one mathematician guessed over $100. I personally guessed around$150,000 becuase while the probability might be pretty low at some point, in A MILLION tries, it would be bound to happen quite a bit.
- 4 years, 8 months ago
Well, their reasoning would probably be then assuming a sort of "reverse" normal distribution curve. The average of all the games would probably be pretty low.
They were asked to bet immediately, depriving them of the chance to think it over. Their first instinct told them that the more coins that get in a urn, the higher probability of getting more coins in, giving a feedback loop. This means that the smaller urn probably won't really have any coins at all.
- 4 years, 8 months ago | {
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# Asymptotic behavior of $x_{n+1}=x_n+\frac{1}{x_n}, \space\space\space x_0=1$
I have defined a sequence $x_n$ as follows: $$x_{n+1}=x_n+\frac{1}{x_n}, \space\space\space x_0=1$$ After convincing myself that there is no nice closed-form formula for $x_n$, I decided to try and find an asymptotic formula for $x_n$ (unfortunately, I have very little experience with asymptotic formulae). I noticed that the recurrence is equivalent to $$\Delta x_{n}=\frac{1}{x_n}$$ and so a solution can be approximated by solving the corresponding differential equation for $y(t)$: $$y'=\frac{1}{y}$$ This differential equation (with initial value $y(0)=1$) yielded $$y=\sqrt{2t+1}$$ which led me to believe that $$x_n\approx \sqrt{2t+1}$$ When I plotted $x_n$ and $y(n)$ side by side, it did indeed appear that they were very close to each other. However, this isn't enough for me... I would like to prove that $y(n)$ is a good approximation for $x_n$, either by proving that $$\lim_{n\to\infty}(x_n-\sqrt{2n+1})=0$$ ...or, even better, by finding a zero-approaching function $f$ satisfying $$x_n=\sqrt{2n+1}+O(f(x))$$ This is where I got stuck. How do can I prove (or disprove?) the statement in the first equality, and find $f$ satisfying the second?
NOTE: There may be a closed-form that I wasn't able to find. In fact, the similar sequence $$y_{n+1}=\frac{y_n}{2}-\frac{1}{2y_n}$$ has closed form formula $$y_n=\frac{1}{\tan(2^n\arctan(y_0^{-1}))}$$ ...but even if you do find a closed form, I would still like to know how to prove the above statements, since the techniques would be useful to know for future problems. | {
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• I think $\int_n^{n+1} f(x) dx = \frac 1 {\int_0^n f(x)}$ is more accurate. But its probably nastier to solve. – mathreadler Nov 29 '17 at 0:15
• See OEIS sequence A073833 and related sequences. – Robert Israel Nov 29 '17 at 0:21
• Probably the first step, based on this guess, would be to write a recurrence for $a_n := x_n^2$: $a_{n+1} = a_n + 2 + \frac{1}{a_n}$. So, to start a bootstrapping argument, it would be easy to prove $a_n \ge 2n - 1$ for each $n$... And then $a_{n+1} \le a_n + 2 + \frac{1}{2n-1}$ so also $a_n \le 2n + \frac{1}{2} \log(n) - C$, etc. ... – Daniel Schepler Nov 29 '17 at 0:28
• Here is a class of arguments that could work. You can try to prove inductively both an upper and a lower bound simultaneously, say $\sqrt{2n+a} \le x_n \le \sqrt{2n+b}$ for some $a, b$, because when bounding $x_{n+1}$ in terms of $x_n$ you need both a lower and an upper bound on $x_n$ to establish either an upper bound or a lower bound on $x_n + \frac{1}{x_n}$. – Qiaochu Yuan Nov 29 '17 at 0:55
• Possible duplicate of Closed form for the sequence defined by $a_0=1$ and $a_{n+1} = a_n + a_n^{-1}$ – Aryabhata Nov 29 '17 at 3:10 | {
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You can show it's a quite nice bound by doing the following. First let $$f(x)=x+\frac{1}{x}$$ and then define $b_n=\sqrt{2n+1}$, $\epsilon_n=x_n-b_n$, and $\xi_n=f(b_n)-b_{n+1}$. First notice that $$f(x+\epsilon)-f(x)\le \epsilon$$ when $x,\epsilon \ge 0$ (we will later show that $\epsilon_n\ge 0$, for now assume this). We then have that \begin{aligned} \epsilon_{n+1} &= x_{n+1}-b_{n+1} \\ &= f(x_n)-f(b_n)+\xi_n \\ &= f(b_n+\epsilon_n)-f(b_n)+\xi_n \\ &\le \epsilon_n+\xi_n \end{aligned} Combined with $\epsilon_0=0$, this tells us that $$0\le \epsilon_{n}\le \sum_{i=0}^{\infty}\xi_i\approx 0.56$$ The above being a convergent series (according to Wolfram by comparison). This of course gives $$\sqrt{2n+1}\le x_n\le \sqrt{2n+1}+0.56$$ for all $n$, making it a very good approximation.
To show that $\epsilon_n\ge 0$ we first see that $\epsilon_0=0$. Then assume $\epsilon_n\ge 0$, by the above we have $\epsilon_{n+1}=f(b_n+\epsilon_n)-f(b_n)+\xi_n$. It is easy to see that $f$ is increasing on $x\ge 1$, and that $b_n\ge 1$, thus since $\epsilon_n\ge 0$ we see that $f(b_n+\epsilon_n)-f(b_n)\ge 0$ finally yielding $\epsilon_{n+1}\ge \xi_n\ge 0$, where $\xi_n\ge 0$ is easy to verify since we have an explicit formula for $\xi_n$. This completes the induction.
The bound numerically looks like it can be improved and it might also be good to very find the series that Wolfram is comparing ours to. And this still doesn't resolve whether $\lim_{n\to \infty}(x_n-b_n)=0$. | {
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# Thread: Help in combinatorics exercise
1. ## Help in combinatorics exercise
Hey everyone I'm stuck in the exercise, I will be happy to work out a solution
Find how many integers - n, have the attribute:
n is divisible by 7 and is not divisible by any natural number - k that sustains:
Thanks friends.
2. ## Re: Help in combinatorics exercise
You are looking for numbers that have 7 as a factor, but do not have 2,3, or 5, as a factor.
Let $A_k = \{n \in \mathbb{N}|1 \le n \le 7770\text{ and }k\text{ divides }n\}$
You are looking for:
\begin{align*}\left|A_7 \setminus (A_2 \cap A_3 \cap A_5)\right| = & |A_7| - |A_{7\cdot 2}| - |A_{7\cdot 3}| - |A_{7\cdot 5}| \\ & + |A_{7\cdot 2\cdot 3}| + |A_{7\cdot 2\cdot 5}| + |A_{7\cdot 3\cdot 5}| \\ & - |A_{7\cdot 2\cdot 3\cdot 5}|\end{align*}
This is because if $p,q$ are distinct primes, then $A_p\cap A_q = A_{pq}$, and also from the Inclusion/Exclusion principle. Additionally, for any prime $p$, you have:
$|A_p| = \left\lfloor \dfrac{7770}{p} \right\rfloor$
I hope you get the idea. So, the answer is:
$\dfrac{7770}{7}-\left(\dfrac{7770}{7\cdot 2} + \dfrac{7770}{7\cdot 3} + \dfrac{7770}{7\cdot 5}\right) + \left(\dfrac{7770}{7\cdot 2\cdot 3} + \dfrac{7770}{7\cdot 2\cdot 5} + \dfrac{7770}{7\cdot 3\cdot 5} \right) - \dfrac{7770}{7\cdot 2\cdot 3\cdot 5} = 296$
Edit: To verify, you can use Excel.
In cell A1, type the formula: =MOD(ROW(),2)>0
(this will be TRUE if the row number is NOT divisible by 2)
In cell B1, type: =MOD(ROW(),3)>0
C1: =MOD(ROW(),4)>0
D1: =MOD(ROW(),5)>0
E1: =MOD(ROW(),6)>0
F1: =MOD(ROW(),7)=0
(notice this one you want it to be equal to zero because that means the row number is divisible by 7)
G1: =MOD(ROW(),8)>0
H1: =MOD(ROW(),9)>0
I1: =MOD(ROW(),10)>0
J1: =AND(A1:I1)
Then select cells A1:J1 and copy the formulas down to row 7770. Then, in any open cell, add the formula: =COUNTIF(J1:J7770,TRUE) and it will tell you how many rows satisfied the conditions you set for divisibility. | {
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3. ## Re: Help in combinatorics exercise
Thanks a friend for helping me very much
4. ## Re: Help in combinatorics exercise
Edit: To verify, you can use Excel.
In cell A1, type the formula: =MOD(ROW(),2)>0
(this will be TRUE if the row number is NOT divisible by 2)
In cell B1, type: =MOD(ROW(),3)>0
C1: =MOD(ROW(),4)>0
D1: =MOD(ROW(),5)>0
E1: =MOD(ROW(),6)>0
F1: =MOD(ROW(),7)=0
(notice this one you want it to be equal to zero because that means the row number is divisible by 7)
G1: =MOD(ROW(),8)>0
H1: =MOD(ROW(),9)>0
I1: =MOD(ROW(),10)>0
J1: =AND(A1:I1)
Then select cells A1:J1 and copy the formulas down to row 7770. Then, in any open cell, add the formula: =COUNTIF(J1:J7770,TRUE) and it will tell you how many rows satisfied the conditions you set for divisibility.[/QUOTE]
This site shows me something interesting that I did not know, can you please repeat the last part?
5. ## Re: Help in combinatorics exercise
Originally Posted by yossa
Edit: To verify, you can use Excel.
In cell A1, type the formula: =MOD(ROW(),2)>0
(this will be TRUE if the row number is NOT divisible by 2)
In cell B1, type: =MOD(ROW(),3)>0
C1: =MOD(ROW(),4)>0
D1: =MOD(ROW(),5)>0
E1: =MOD(ROW(),6)>0
F1: =MOD(ROW(),7)=0
(notice this one you want it to be equal to zero because that means the row number is divisible by 7)
G1: =MOD(ROW(),8)>0
H1: =MOD(ROW(),9)>0
I1: =MOD(ROW(),10)>0
J1: =AND(A1:I1)
Then select cells A1:J1 and copy the formulas down to row 7770. Then, in any open cell, add the formula: =COUNTIF(J1:J7770,TRUE) and it will tell you how many rows satisfied the conditions you set for divisibility.
This site shows me something interesting that I did not know, can you please repeat the last part?
I'm not sure what you are asking. If you do not have Microsoft Excel, then any OpenOffice spreadsheet will work.
6. ## Re: Help in combinatorics exercise | {
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6. ## Re: Help in combinatorics exercise
Originally Posted by SlipEternal
I'm not sure what you are asking. If you do not have Microsoft Excel, then any OpenOffice spreadsheet will work.
I'll explain myself, I have Microsoft Excel, I was able to place everything in a simple phase a - "Then select cells A1:J1 and copy the formulas down to row 7770. Then, in any open cell, add the formula: =COUNTIF(J1:J7770,TRUE) and it will tell you how many rows satisfied the conditions you set for divisibility." not so figured out
7. ## Re: Help in combinatorics exercise
Select cells A1:J1. Right click and choose Copy (or CTRL+C on the keyboard). On the keyboard, press CTRL+G (this will bring up a dialog that says GoTo in the title. In the Reference line, put A7770. Now, while holding the Shift key (do not let go), do the following:
1. Press and release the End key
2. Press and release the Up arrow key
3. Press and release the right arrow key nine times (so you will have every cell in the range A1:J7770 highlighted).
Release the Shift key, right click, and press Paste (or CTRL+V on the keyboard).
8. ## Re: Help in combinatorics exercise
Originally Posted by SlipEternal
Select cells A1:J1. Right click and choose Copy (or CTRL+C on the keyboard). On the keyboard, press CTRL+G (this will bring up a dialog that says GoTo in the title. In the Reference line, put A7770. Now, while holding the Shift key (do not let go), do the following:
1. Press and release the End key
2. Press and release the Up arrow key
3. Press and release the right arrow key nine times (so you will have every cell in the range A1:J7770 highlighted).
Release the Shift key, right click, and press Paste (or CTRL+V on the keyboard).
i get it thanks | {
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• Predator Prey Model Simulation Matlab | {
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• Design, simulation and analysis in Simulink. Aggregate models consider a population as a collective group, and capture the change in the size of a population over time. The link to this assignment on github is here. ) Wilensky, U. GIBSON1*, DAVID L. I have a program called Predator Prey that's in the collection of programs that comes with NCM, Numerical Computing with MATLAB. This dataset includes the source computer code and supporting data files for the predator-prey simulation model (parameterized for summer flounder, Paralichthys dentatus) developed to investigate bottom-up effects defined to be temporal pulses in prey abundance on predator growth, production, and fisheries management. What are synonyms for Predator and prey?. This model is common, e. The predator-prey population-change dynamics are modeled using linear and nonlinear time series models. As a mathematical consequence of the herd behavior, they considered competition models and predator-prey systems. This video will show you the basics and give you an idea of what working in MATLAB looks like. Lotka-Volterra predator prey model. It was developed independently by Alfred Lotka and Vito Volterra in. While creating a model for combined predator and prey strategy would inform an estimation of overall fitness throughout an animal’s lifetime, an overarching model of this sort would be extremely complex and is beyond the scope of our paper. In the model system, the predators thrive when there are plentiful prey but, ultimately, outstrip their food supply and decline. However, the organisms in Holland’s simulation are very simple and do not involve any behavioral model. Describing the dynamics of such models occasionally requires some techniques of model analysis. Simulation of CTMC model I Use CTMC model to simulate predator-prey dynamics I Initial conditions are X(0) = 50 preys and Y(0) = 100 predators 0 5 10 15 20 25 30 0 50 100 150 200 250 300 350 400 Time Population Size X (Prey) Y (Predator) I Prey | {
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5 10 15 20 25 30 0 50 100 150 200 250 300 350 400 Time Population Size X (Prey) Y (Predator) I Prey reproduction rate c 1 = 1 reactions/second I Rate of predator consumption of prey c 2 = 0:005. Student Challenge: Set up a level of hunting that keeps the populations of predators and prey at healthy levels. Predator and Prey Interactions Level A: Up and Down in the Wild: Predator and Prey. Both a detailed large eddy simulation of the dynamics and microphysics of a precipitating shallow boundary layer cloud system and a simpler model built upon basic physical principles, reproduce predator-prey behavior with rain acting as the predator and cloud as the prey. In particular, it describes how to structure a model and how to define species - that are the key components of GAMA models. They will, however, also be modified during this exercise. Related Data and Programs: FD_PREDATOR_PREY, a MATLAB program which solves a pair of predator prey ODE's using a finite difference approximation. The model is derived and the behavior of its solutions is discussed. Here is how Volterra got to these equations: The number of predatory shes immediately after WWI was much larger than. The traditional mathematical model describing the predator-prey interactions consists of the following system of differential equations. C=Death rate of predators. Click on the link below to download a zipped archive of the original iThink (. Models of interacting populations. Yes, it is agent-based model. We now replace the difference equation model used there with a more sophisticated differential equation model. This example implements best practices with MATLAB and Robotics System Toolbox. I have a Predator-Prey Model: dR/dt = λR - aRF dF/dt = -μF + bRF Where λ and μ are growth rates of rabbits (R) and foxes (F) respectively, treated in isolation. Read "A Fractional Predator-Prey Model and its Solution, International Journal of Nonlinear Sciences and Numerical Simulation" on DeepDyve, the largest online | {
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Journal of Nonlinear Sciences and Numerical Simulation" on DeepDyve, the largest online rental service for scholarly research with thousands of academic publications available at your fingertips. PY - 2009/1. In this research article, we considered an ecological prey predator fishery model system with a generalized case where both the patches are accessible to both prey and predator. Let's use the example function lotka that uses $\alpha = 0. For more information about iThink or to download a free trial version, visit www. Open a diary file in Matlab in order to save your work. These are ordinary differential equations that are straightforward to solve. An if-then. One such pair of systems is the population of Foxes and Rabbits. The total prey population is divided into two subdivisions, namely susceptible prey population and infected prey population. We show that food availability and predators' densities influence patterns of prey distribution. Use model blocks to import, initialize, and simulate models from the MATLAB ® environment into a Simulink model. Lotka was born in Lemberg, Austria-Hungary, but his parents immigrated to the US. The WATOR simulation was one of the first of these. October 30, 2017 Post source code In this post, I'll explore using R to analyze dynamical systems. , model the interactions of two industrial sectors. EE 5323 Homework 2. Learn to use MATLAB and Simulink for Simulation and other science and engineering computations. Usage of Boids for a prey-predator simulation. For a systematic approach to some of this we turn to Dynamical systems theory. This is to be able to compare with the behaviour of a corresponding stochastic and dynamic model. 2Mathematics Department , Faculty of Science , Al-Azhar University, Assiut 71511, Egypt. GIBSON1*, DAVID L. Usage of Boids for a prey-predator simulation- + Dailymotion. As the population of the prey increases then the predator population will increase. Models of interacting populations. The model is | {
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then the predator population will increase. Models of interacting populations. The model is used to study the ecological dynamics of the lion-buffalo-Uganda Kob prey-predator system of Queen Elizabeth National Park, Western Uganda. This will help us use the lotka model with different values of alpha and beta. Modelling Predator-Prey Interactions with ODE Modelling Predator-Prey Interactions with ODE Shan He School for Computational Science University of Birmingham Module 06-23836: Computational Modelling with MATLAB Modelling Predator-Prey Interactions with ODE Outline Outline of Topics Predator-Prey Models The Lotka-Volterra (LV) model. Simulate Identified Model in Simulink. Simulate Identified Model in Simulink. Or copy & paste this link into an email or IM:. fd1d_predator_prey_test. Determine the equilibrium points and their nature for the system. Aggregate models consider a population as a collective group, and capture the change in the size of a population over time. Originally, the HANDY Model is derived from a predator-prey model as indicated in [1]. ODEs are frequently used in biology to model population dynamics. 1007/s11859-015-1054-4. Save the first part of this model. Introduction This chapter, originally intended for inclusion in [4], focuses on mod-eling issues by way of an example of a predator-prey model where the. Yes, it is agent-based model. Vector with the named parameters of the model: k1. DYNAMICS OF A MODEL THREE SPECIES PREDATOR-PREY SYSTEM WITH CHOICE by Douglas Magomo August 2007 Studies of predator-prey systems vary from simple Lotka-Volterra type to nonlinear systems involving the Holling Type II or Holling Type III functional response functions. Matlab code for the examples discussed below is in this compressed folder. If no predators, prey population grows at natural rate: for some constant a > 0,. Since we are considering two species, the model will involve two equations, one which describes how the prey population changes and the | {
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the model will involve two equations, one which describes how the prey population changes and the second which describes how the predator population changes. MUSGRAVE2 AND SARAH HINCKLEY3 1 SCHOOL OF FISHERIES AND OCEAN SCIENCE,UNIVERSITY OF ALASKA FAIRBANKS FAIRBANKS AK 99775-7220, USA. A STAGE-STRUCTURED PREDATOR-PREY MODEL HAL SMITH 1. My book that's available on the MathWorks website. The x_t denote the number of snow hares (prey) and y_t be the. But these functions also arise in the other sciences. In recent years, many authors have explored the dynamic relationship between predators and their preys. Some predator-prey models use terms similar to those appearing in the Jacob-Monod model to describe the rate at which predators consume prey. Finally, you will see a demonstration of the concepts above through an autonomous object tracking example. There is a simulation speed slider on the bottom of the model page (I think it has. What is the carrying capacity for moose in the simulation model of Isle Royale, prior to any changes in the weather?. Predator Prey Models in Real Life. Di erential Equations (Aggregate) Models with MATLAB and Octave A Predator-Prey Example Di erential equations in biology are most commonly associated with aggregate models. Participants are assigned a role in the food chain, participate in the simulation, collect and analyze results, and assess factors affecting their survival. Back to Eduweb Portfolio. In the special case in which both predator and prey are from the same species, predation is called cannibalism. Part 1: create the model. We define a prey (mouse) and predator (cat) model. However, the organisms in Holland’s simulation are very simple and do not involve any behavioral model. Lotka was born in Lemberg, Austria-Hungary, but his parents immigrated to the US. Many of our resources are part of collections that are created by our various research projects. The Predator-Prey Model Simulation. The prey–predator algorithm was used | {
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various research projects. The Predator-Prey Model Simulation. The prey–predator algorithm was used to evaluate the best performance of the heat sinks. This model reflects the point in time where the predator species has evolved completely and no longer competes for the initial food source. Software Programming And Modelling For Scientific Researchers. Fall 2017 Math 636 Predator-Prey Models 1. Using the Lotka-Volterra predator prey model as a simple case-study, I use the R packages deSolve to solve a system of differential equations and FME to perform a sensitivity analysis. In the Lotka Volterra predator-prey model, the changes in the predator population y and the prey population x are described by the following equations: Δxt=xt+1−xt=axt−bxtyt Δyt=yt+1−yt=cxtyt−dyt. The prey population increases when there are no predators, and the predator population decreases when there are no prey. We will assume that the predators are Greater Californian Killer Foxes and the prey are Lesser Fluffy Rabbits. Predator-Prey Models in Excel. This paper describes the GA model using a new selection method inspired by predator-prey interactions. I The main hypothesis: The prey-predator interaction is the only factor. What are synonyms for Predator and prey?. in pursuit of her chosen prey. Predator-Prey Cycles. Predator-Prey Simulation Lab. Then, the model was further developed to include density dependent prey growth and a functional response of the form developed by C. ’ Note that this model can be considered as a simple predator-prey model. For the eBay data, we have >>polyfit(year,income,1) ans = 100. Simulink is also practical. 2Modeling Predator-Prey Dynamics and Climate Influence 2. PloS one, 7, e28924 [paper3, 1 predator - 2 prey model] Unfortunately, there are very few technical documents available on how to implement point process IBMs and corresponding moment equations. Antonyms for Predator and prey. A STAGE-STRUCTURED PREDATOR-PREY MODEL HAL SMITH 1. PREDATOR PREY MODELS IN | {
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