# 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: . $0, 1, 4, 11, 26, 57, 120, 247, 502, 1013, \hdots$

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: . $F(n) \;=\;2^n - (n+1)$


# Can you use both sides of an equation to prove equality?

For example:

$\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 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$a = c = bby transitivity of equality. This conforms to both your teacher's and professor's advice. Both your high school teacher and university professor are steering you away from "two-column proofs" of the type: \begin{align*} -1 &= 1 &&\text{To be shown;} \\ (-1)^{2} &= (1)^{2} && \text{Square both sides;} \\ 1 &= 1 && \text{True statement. Therefore-1 = 1.} \end{align*} Here, you assume what you want to prove, deduce a true statement, and assert that the original assumption was true. This is bad logic for at least two glaring reasons: 1. If you assume-1 = 1$, 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 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 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 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. 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 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
• @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


# 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 $$P(X|Y) = \frac{P(X\land Y)}{P(Y)} = \frac{\frac{4\cdot 48}{52\cdot 51}}{\frac{48}{52}} = \frac{4}{51}$$

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, $Pr = \dfrac{4}{52-n}$


# 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?

• 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:

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


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 $$4\sum_{k=1}^nk=4\cdot\frac12n(n+1)=2n(n+1)$$

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


# 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:

$$\sum_{n=1}^{N}\left(\frac{1}{n}-\frac{1}{n+2}\right) = \frac{3}{2}-\frac{1}{N+1}-\frac{1}{N+2}$$ 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


# 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?

• 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

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.


# 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:

$$V_0=\{(0)=\{0,0,0,0,.....\}\}$$ 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 $$2^{\aleph_0}$$

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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02 Mar 2020, 20:35
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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:

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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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

$$\frac{C^2_8*C^2_6*C^2_4*C^2_2}{4!}=105$$, 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?

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

$$\frac{\frac1{4!}\binom{12}3\binom93\binom63\binom33}{\frac1{5!}\binom{15}3\binom{12}3\binom93\binom63\binom33}=\frac{5!}{4!\binom{15}3}=\frac5{\binom{15}3}=\frac5{455}=\frac1{91}\;.$$

• "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


# 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.

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.

• 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.


# 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

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 $$(W,\mathcal C=(c_1,c_2,c_3))$$.

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:

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}


# 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.

• $\lvert \lvert \sin(x) \rvert \rvert = \sqrt{\langle \sin(x), \sin(x) \rangle} = \sqrt{\frac{\pi}{2}}$. 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.


# 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 $$\frac{d}{dx}\circ x=1+x\frac{d}{dx}$$

• 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 $$n\in\mathbb N_0$$ $$(a+b)^n(a-b)^n=\prod_{j=1}^n (a^2-b^2+(2j-1))$$

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

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 $$\frac{\partial^{8}}{\partial x^{8}} - 4 x^{2} \frac{\partial^{6}}{\partial x^{6}} - 16 \frac{\partial^{6}}{\partial x^{6}} - 24 x \frac{\partial^{5}}{\partial x^{5}} + 6 x^{4} \frac{\partial^{4}}{\partial x^{4}} + 48 x^{2} \frac{\partial^{4}}{\partial x^{4}} + 42 \frac{\partial^{4}}{\partial x^{4}} + 48 x^{3} \frac{\partial^{3}}{\partial x^{3}} + 192 x \frac{\partial^{3}}{\partial x^{3}} - 4 x^{6} \frac{\partial^{2}}{\partial x^{2}} - 48 x^{4} \frac{\partial^{2}}{\partial x^{2}} - 36 x^{2} \frac{\partial^{2}}{\partial x^{2}} + 48 \frac{\partial^{2}}{\partial x^{2}} - 24 x^{5} \frac{\partial}{\partial x} - 192 x^{3} \frac{\partial}{\partial x} - 216 x \frac{\partial}{\partial x} + x^{8} + 16 x^{6} + 42 x^{4} - 48 x^{2} - 39$$.

For $$n = 5$$, the result is $$-\frac{\partial^{10}}{\partial x^{10}} + 5 x^{2} \frac{\partial^{8}}{\partial x^{8}} + 25 \frac{\partial^{8}}{\partial x^{8}} + 40 x \frac{\partial^{7}}{\partial x^{7}} - 10 x^{4} \frac{\partial^{6}}{\partial x^{6}} - 100 x^{2} \frac{\partial^{6}}{\partial x^{6}} - 130 \frac{\partial^{6}}{\partial x^{6}} - 120 x^{3} \frac{\partial^{5}}{\partial x^{5}} - 600 x \frac{\partial^{5}}{\partial x^{5}} + 10 x^{6} \frac{\partial^{4}}{\partial x^{4}} + 150 x^{4} \frac{\partial^{4}}{\partial x^{4}} + 150 x^{2} \frac{\partial^{4}}{\partial x^{4}} - 150 \frac{\partial^{4}}{\partial x^{4}} + 120 x^{5} \frac{\partial^{3}}{\partial x^{3}} + 1200 x^{3} \frac{\partial^{3}}{\partial x^{3}} + 1800 x \frac{\partial^{3}}{\partial x^{3}} - 5 x^{8} \frac{\partial^{2}}{\partial x^{2}} - 100 x^{6} \frac{\partial^{2}}{\partial x^{2}} - 150 x^{4} \frac{\partial^{2}}{\partial x^{2}} + 1500 x^{2} \frac{\partial^{2}}{\partial x^{2}} + 975 \frac{\partial^{2}}{\partial x^{2}} - 40 x^{7} \frac{\partial}{\partial x} - 600 x^{5} \frac{\partial}{\partial x} - 1800 x^{3} \frac{\partial}{\partial x} - 600 x \frac{\partial}{\partial x} + x^{10} + 25 x^{8} + 130 x^{6} - 150 x^{4} - 975 x^{2} - 255$$.

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"

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 $$\left( xy\right) ^{n}=\sum\limits_{k=0} ^{n} \genfrac{\{}{\}}{0pt}{0}{n+1}{k+1} y^{k}x^{k}$$.

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}\\ & =\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. $$\blacksquare$$

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 $$\left( yx\right) ^{n}=\sum\limits_{k=0} ^{n} \genfrac{\{}{\}}{0pt}{0}{n}{k} y^{k}x^{k}$$.

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 Proposition 2. $$\blacksquare$$

• 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*}

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


# 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.


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.

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 $5\%$.

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


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.

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

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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

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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