# The Powers of Sum: Solving Hard Problem Algebraically

One of the active members of a Facebook group called Elite Math Circle (EMC) posted a question that seems interesting. Same question I enc0unter when I was actively solving problems at Brilliant.

It was posted by Russelle Guadalupe, an elite member of Philippine Team in 2011 International Mathematical Olympiad. The question is,

Given:

$a+b+c=6$
$a^2+b^2+c^2=8$
$a^3+b^3+c^3=5$

What is the value of $a^4+b^4+c^4$.

I have seen an elegant solution to this method, one of the members says it was “Newton’s Sum Theorem”.

The solution here might be the old school way to solve the problem using algebraic identities and manipulation.

Take note of the following identities:

1. $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac)$

2. $a^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)+3abc$

First step is to find an expression that will lead to the unknown which is $a^4+b^4+c^4$. This expression is achievable by squaring $a^2+b^2+c^2$.

$(a^2+b^2+c^2)^2=a^4+b^4+c^4+2(a^2b^2+b^2c^2+a^2c^2)$

But  $a^2+b^2+c^2=8$

$(8)^2=a^4+b^4+c^4+2(a^2b^2+b^2c^2+a^2c^2)$

$64=a^4+b^4+c^4+2(a^2b^2+b^2c^2+a^2c^2)$

We still have one missing expression, $a^2b^2+b^2c^2+a^2c^2$.

Since we don’t have the latter expression, we will find it. Take a look of the first algebraic identity mentioned above. We can use it to arrive to an expression that we are looking for by squaring  $ab+bc+ac$.

$(ab+bc+ac)^2=a^2b^2+b^2c^2+a^2c^2+2abc(a+b+c)$

We know that a+b+c=6,

$(ab+bc+ac)^2=a^2b^2+b^2c^2+a^2c^2+2abc(6)$

$(ab+bc+ac)^2=a^2b^2+b^2c^2+a^2c^2+12abc$ ♥♥

We already have an expression for our unknown but we ended now with two more unknowns. $ab+bc+ac$ and $abc$.

By squaring  $a+b+c$ and solving for  $ab+bc+ac$ we have,

$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac)$

$(6)^2=8+2(ab+bc+ac)$

$2(ab+bc+ac)=36-8$

$ab+bc+ac=14$ ♥♥♥

We only have one unknown left, the $abc$

Recall the second identity mentioned above,

$a^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)+3abc$

or

$a^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2-(ab+bc+ac))+3abc$

By substitution we have,

$5=(6)(8-14)+3abc$

$5=-36+3abc$

$41=3abc$

$164=12abc$ ♥♥♥♥

Substituting ♥♥♥ and ♥♥♥♥ to ♥♥ we have,

$(ab+bc+ac)^2=a^2b^2+b^2c^2+a^2c^2+12abc$

$(14)^2=a^2b^2+b^2c^2+a^2c^2+164$

$196=a^2b^2+b^2c^2+a^2c^2+164$

$a^2b^2+b^2c^2+a^2c^2=196-164$

$a^2b^2+b^2c^2+a^2c^2=32$ ♥♥♥♥♥

Substituting ♥♥♥♥♥ to ♥ we have,

$64=a^4+b^4+c^4+2(a^2b^2+b^2c^2+a^2c^2)$

$64=a^4+b^4+c^4+2(32)$

$64=a^4+b^4+c^4+64$

$\boxed{a^4+b^4+c^4=0}$

The method above has actually a routine, it has a pattern. In Mathematics, when there is a pattern, there is a general rule and formula can be derived. Dr. Greenie of Ask Dr. Math derived a formula using the following rules.

Given:

$a+b+c=x$
$a^2+b^2+c^2=y$
$a^3+b^3+c^3=z$

We can solve the value of $a^4+b^4+c^4$ in terms of x,y, and z with the following formula,

$a^4+b^4+c^4=y^2-2((\displaystyle\frac{x^2-y}{2})^2-\displaystyle\frac{x^4-3x^2y+2xz}{3})$

### Dan

Blogger and a Math enthusiast. Has no interest in Mathematics until MMC came. Aside from doing math, he also loves to travel and watch movies.

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Median of medians confusion — the “approximate” median part

I'm struggling with the median of medians algorithm, and I think it's perhaps more of a semantics thing rather than a technical thing. I've always thought the median of medians algorithm as finding an approximate median $$p$$ such that $$p$$ is within $$20\%$$ of the true median $$M$$ in the sorted array.

However, when I look at actual implementations, e.g., in https://brilliant.org/wiki/median-finding-algorithm/, the algorithm they posted returns an exact median, but at each level of the recursion, you may have some approximate median generated from a sublist of medians. And eventually you'll reach a level where the array is $$\leq 5$$ elements, ending the recursion. At this level, you obtain an exact median of the array you passed in. So I had thought all this time that this exact median computed at the last level is actually your estimate of the median in the original array passed in at the first level of the recursion.

So I had the same confusion as this poster https://stackoverflow.com/questions/52461306/something-i-dont-understand-about-median-of-medians-algorithm and some others.

Now that I understand this algorithm, I am now confused on how the median of medians actually finds an "approximate" median to the original array. In all the implementations I've seen, the median you find using median of medians is exact. So where does the approximate part come in other than approximating the median at each recursion level? Even Wikipedia describes as an algorithm that approximates a median. Yes, it approximates medians at various levels, but the final output is exact.

Or am I operating under a false premise in thinking that Median of Medians finds an approximate median to the ORIGINAL array?

• I believe some people call median of median the algorithm which selects an approximate median in linear time, and some people mean what you get when you combine that with quickselect, i.e. a linear-time algorithm to find the k'th element in an array (or in particular, find the median). Just two closely related things some people tend to call by the same name. – Tassle Mar 14 at 22:18
• @Tassle This is one of the algorithms where I haven't really been satisfied with when reading the first page of Google links. I felt something was confusing or missing in each of them. Is there a text book that this algorithm is in? It's not in CLRS unfortunately, and I don't have familiarity with other algorithms textbooks. – user5965026 Mar 14 at 23:52

Median-of-medians is a recursive algorithm which solves the more general selection problem: given an array $$A$$ of length $$n$$ (which we assume, for simplicity, has distinct elements) and an integer $$k$$, find the $$k$$'th smallest element (where $$1 \leq k \leq n$$). It works as follows:

• If $$n \leq K$$ (where $$K$$ is some constant), solve the problem by brute force.
• Otherwise, find an approximate median $$x$$ of $$A$$ in $$O(n)$$. By approximate median, we mean that $$x$$ is the $$m$$'th smallest element in $$A$$, where $$\alpha n \leq m \leq (1-\alpha) n$$ for some constant $$\alpha < 1$$.
• Use $$x$$ to split $$A$$ into elements smaller than $$x$$, comprising an array $$A_{ of length $$m-1$$, and elements larger than $$x$$, comprising an array $$A_{>x}$$ of length $$n-m$$.
• If $$m = k$$, then output $$x$$. If $$m > k$$ then output the $$k$$'th smallest element of $$A_{. If $$m < k$$ then output the $$(k-m)$$'th smallest element of $$A_{>x}$$.

The running time of the algorithm satisfies the recurrence $$T(n) \leq T(\alpha n) + O(n)$$, whose solution is $$T(n) = O(n)$$.

The algorithm finds the exact median, but it does so by repeatedly finding approximate medians.

• Note that the algorithm used to find the approximate median is sometimes what people refer to when they say "median-of-medians", hence the confusion experienced by the OP I think. – Tassle Mar 14 at 22:20
• Do you know of a textbook that describes the median of medians? I read all the articles on the first page of Google and more after googling "median of medians," and I just don't feel very satisfied with any of them, including the wikipedia article. The one on brilliant.org was probably the best one I read, but I still would prefer a textbook read for this algo. – user5965026 Mar 14 at 23:58
• It’s described in CLRS and on Wikipedia, and probably in many other lecture notes and slides. – Yuval Filmus Mar 15 at 0:00
• Oh I didn't realize it was in CLRS. Found it in 9.3. Earlier I was doing a search for median of medians in the book and could not find it. – user5965026 Mar 15 at 0:44


What's the probability of the target being destroyed?

Target gets hit by three missiles.

The probability of hitting the target with the
- First missile is 0.4;
- Second missile is 0.5;
- Third missile is 0.7;

The target has a probability of 0.2 getting destroyed with one missile, 0.6 by two missiles and for sure (1) by three missiles.

What's the probability of the target being destroyed?

--

This is the exact question on the book, teacher told us that the result is 0.458 but didn't describe the method of solving, except saying that it's not easy.

• you need to start by getting the probability of exactly 1,2 and 3 hits separately - being hit by 3 missiles is simplest, all missiles have to hit - can you do that calculation? Two missile hits can happen in 3 different ways 1,2 or 3 has to miss - can you work out the probabilities of the 3 distinct events and add them together - 1 hit is similar to 2 hits, 1 hits and 2 misses - you have to consider the probability of hits and misses in your calculations -
– Cato
Jan 31, 2017 at 10:26
• I see. So, let's say: D -> target being destroyed; M1,M2,M3 -> Probabilities of missiles; H1,H2,H3 -> the three scenarios you mentioned, where P(H3) is the easiest to find (0.4*0.5*0.7), since all the targets should hit. So I have to basically find P(D) which is P(D/H1)*P(H1) + P(D/H2)*P(H2) + P(D/H3)*P(H3)? @Cato Jan 31, 2017 at 10:39
• yes I would imagine it would end up in that form - an example of 1 missile hit is that just 1 hits, P=.4*.5*.3 - but that can happen in 3 ways, but the 3 ways exclude each other
– Cato
Jan 31, 2017 at 10:41
• Thanks!! I was having problems finding P(H1) and P(H2) earlier, but thanks to you ringing me the bell that there can be 3 ways for these two scenarios, the problem is solved. @Cato Jan 31, 2017 at 10:50

First: hit and destroy, Second :no hit, Third: no hit

$0.4\times 0.2 \times 0.5 \times 0.3=0.012$

First: no hit, Second :hit and destroy, Third: no hit

$0.6\times 0.5 \times 0.2 \times 0.3=0.018$

First: no hit, Second :no hit, Third: hit and destroy

$0.6\times 0.5 \times 0.7 \times 0.2=0.042$

First: no hit, (Second and Third) :hit and destroy

$0.6\times 0.5 \times 0.7 \times 0.6=0.126$

second: no hit, (first and Third) :hit and destroy

$0.5 \times 0.4 \times 0.7 \times 0.6=0.084$

third: no hit, (first and second) :hit and destroy

$0.3 \times 0.4 \times 0.5 \times 0.6=0.036$

(First, Second and Third) :hit and destroy

$0.4\times 0.5 \times 0.7 =0.14$

Add all, we get required probability as $0.458$

• Good Answer, Kk...k...Kiran... Jan 31, 2017 at 11:42

\begin{align*} p =& P(A \lor B \lor C) \\=& 1-P(\neg A)P(\neg B)P(\neg C)\\=&1-(1-P(A))(1 - P(B))(1-P(C)) \end{align*}


# How does Munkres prove that lower limit topology is finer than the standard topology on $\mathbb{R}$

The proof in question (from the book "Topology" by Munkres):

Let $\mathcal T$ and $\mathcal T_\mathscr{l}$ be the standard and lower limit topology on $\mathbb R$ respectively. Given a basis element $(a,b)$ for $\mathcal T$ and a point $x\in (a,b)$, the basis element $[x,b)$ for $\mathcal T_\mathscr{l}$ contains $x$ and lies in $(a,b)$. Conversely, given the basis element $[x,d)$ for $\mathcal T_\mathscr{l}$, there is no open interval in $\mathcal T$ that contains $x$ and lies in $[x,d)$. Thus $\mathcal T_\mathscr l$ is strictly finer than $\mathcal T$.

There was a question on it before, but his actual proof was never addressed. I understand the proof involving the union of an infinite collection of sets. However, I don't get how the above proof proves that all elements of the latter topology are in the former, which as I understand is the definition of a finer topology.

• Consider the analogy of sand in his book. – GNUSupporter 8964民主女神 地下教會 Mar 29 '18 at 17:04
• @GNUSupporter Honestly, it is passages like the sand analogy that make me really like Munkres as an introductory text. Some of the later chapters lose the thread a little (in my opinion), but the first half of that book is golden. – Xander Henderson Mar 29 '18 at 17:06
• The lower limit topology on $\Bbb R$ is also called the Sorgenfrey line, sometimes denoted $\Bbb R_l.$ Its square $\Bbb R^2_l$ is called the Sorgenfrey plane. – DanielWainfleet Mar 30 '18 at 4:57

Of $O\in\mathcal{T}$, then $O$ can be written as the union of intervals of the type $(a,b)$. Each interval $(a,b)$, in turn, can be written as$$\bigcup_{x\in(a,b)}[x,b),$$which belongs to $\mathcal{T}_1$. Therefore, $O\in\mathcal{T}_1$.

• But isn't Munkres missing this step? I don't see how he explicitly shows that the interval (a,b) can also be found in $\mathcal T_\mathscr l$ – oddic Mar 29 '18 at 17:13
• @oddic He proves that, for each $x\in(a,b)$, there is an element $S\in\mathcal{T}_1$ such that $x\in S$ and that $S\subset(a,b)$. I suppose that he didn't feel the need to deduce from this that $(a,b)\in\mathcal{T}_1$. – José Carlos Santos Mar 29 '18 at 17:16
• But doesn't the proof require showing that there is an element $S\in \mathcal T_\mathscr l$ which is a union of infinite basis elements that is equal to $(a,b)$, and not just a subset? I'm sorry if I'm not understanding correctly – oddic Mar 29 '18 at 17:34
• @oddic If, for each $x\in(a,b)$, there is a $S_x\in\mathcal{T}_1$ such that $x\in S_x$ and that $S_x\subset(a,b)$, then$$(a,b)=\bigcup_{x\in(a,b)}S_x\in\mathcal{T}_1.$$ – José Carlos Santos Mar 29 '18 at 17:43
• @oddic In Lemma 13.3 (p.81) the author already gave an equivalent condition of a topology being finer than the other topology. So in the following Lemma 13.4 he did not need to mention it again, I think. – ChoF Mar 30 '18 at 5:12

What Munkres is doing:

If $T_1,T_2$ are topologies on a set $R,$ and $B_1,B_2$ are bases for $T_1,T_2$ respectively, then to show that $T_1\subset T_2,$ it suffices to show that whenever $x\in b_1\in B_1$ there exists $b_2\in B_2$ with $x\in b_2\subset b_1.$

That implies that every $b_1\in B_1$ is a union of members of $B_2,$ so $b_1\in T_2$. So $B_1\subset T_2,$ so any $t\in T_1,$ being a union of members of $B_1,$ is a union of members of $T_2,$ so $t\in T_2.$

In this case we have $R=\Bbb R$ and $T_1=\mathcal T$ and $T_2=\mathcal {T_l}$ and $B_1=\{(a,b): a,b\in \Bbb R\}$ and $B_2=\{[x,b):x,b\in \Bbb R\}.$

Therefore $\mathcal T\subset \mathcal T_l.$

(And, obviously, they are not equal, because $[0,1)\in \mathcal T_l$ \ $\mathcal T.$)

• Thanks, I understand what I was missing now! – oddic Apr 1 '18 at 0:02
• In many cases the use of particular bases simplifies things. Not surprising when you consider that usually what you do is show that if something holds for a certain subset of a topology (the base/basis) then it holds for the whole topology..... Many write "base", not "basis". I prefer "base", especially in the context of topological vector spaces, where "basis" also has vectorial meanings. – DanielWainfleet Apr 1 '18 at 16:37


# Two questions involving Bayes Theorem

You draw two balls from one of three possible large urns, labelled A, B, and C. Urn A has 1/2 blue balls, 1/3 green balls, and 1/6 red balls. Urn B has 1/6 blue balls, 1/2 green balls, and 1/3 red balls. Urn C has 1/3 blue balls, 1/6 green balls, and 1/2 red balls. With no prior information about which urn your are drawing from, you draw one red ball and one blue ball. What is the probability that you drew from urn C?

So I know Bayes Theorum is this:

On a high level, could I take the probability of drawing one red ball and one blue ball from Urn C and divide that by the total probabality of drawing one blue ball and one red ball from all the urns summed up?

The numerator would be $\frac{1}{2} * \frac{1}{3}$ right? The denominator would the the probabilities of drawing a blue and red from all the urns summed up right?

In the NFL, a professional American football league, there are 32 teams, of which 12 make the playoffs. In a typical season, 20 teams (the ones that don’t make the playoffs) play 16 games, 4 teams play 17 games, 6 teams play 18 games, and 2 teams play 19 games. At the beginning of each game, a coin is flipped to determine who gets the football first. You are told that an unknown team won ten of its coin flips last season. Given this information, what is the posterior probability that the team did not make the playoffs (i.e. played 16 games)?

I have no idea how to approach this one. Can someone shed some insight?

• Please check stats.stackexchange.com/tags/self-study/info and edit your question accordingly to homework-like questions policy we have. – Tim May 30 '17 at 7:31
• I added some more work. I'm genuinely stuck on the second. – Jwan622 May 30 '17 at 14:53

Below are two methods. The first uses Bayes' Theorem in full, and the second is a quicker approach (just for fun). Importantly, the two results agree! (They also do not agree with the earlier answer posted by zen...)

$X := \text{Picked Red and Blue}$, $A := \text{Picked Urn A}$, $B := \text{Picked Urn B}$, $C := \text{Picked Urn C}$.

Method 1.

Using Bayes' Theorem, we now compute as follows:

$$\frac{P(X|C)P(C)}{P(X|C)P(C) + [P(X|A)P(A) + P(X|B)P(B)]}$$

where the bracketed portion in the denominator corresponds to the cases in which red and blue were selected but did not come from Urn C.

Under the assumption that each of $P(C)$, $P(A)$, and $P(B)$ is $1/3$, we can cancel this factor from the numerator and denominator to get:

$$\frac{P(X|C)}{P(X|C) + [P(X|A) + P(X|B)]}$$

Next, we calculate each of these terms:

$P(X|C) = 2(3/6)(2/5) = 12/30$;

$P(X|B) = 2(2/6)(1/5) = 4/30$; and

$P(X|A) = 2(1/6)(3/5) = 6/30$.

And so the earlier expression becomes:

$$\frac{12/30}{12/30 + 4/30 + 6/30} = \frac{12}{12+4+6} = \frac{6}{6+2+3} = \frac{6}{11}$$

Method 2.

Without loss of generality, let us suppose each of the urns contains exactly six M&Ms. Then, the number of ways to choose a red and blue M&M from the urns can be computed as follows.

Urn A: BBBGGR; so, picking RB can be done in $1 \cdot 3 = 3$ ways.

Urn B: GGGRRB; so, picking RB can be done in $2 \cdot 1 = 2$ ways.

Urn C: RRRBBG; so, picking RB can be done in $3 \cdot 2 = 6$ ways.

So the probability of it being Urn C given that RB were selected is just:

$$\frac{6}{6+3+2} = \frac{6}{11}$$

as in Method 1.

• It was not clear to me that the statement "having no prior information about which urn you are drawing from". I read it as the prior probabilities for P(A), P(B) and P(C) were unknown and not they were equal. That would make the problem unsolvable. If you take it to mean prior to drawing the ball the choice of urn is equally likely, in which case P(C|X) can be determined. – Michael Chernick Jun 1 '17 at 14:41
• How do you get $P(X|c)=2(3/6)(2/5)$? shouldn't it be $P(X|c)=2(3/6)(2/6)$ ? It is not given that there are 6 balls in the urns right? why then say that the chance of getting a blue is now $(2/5)$ instead of $(2/6)$? – zen Jun 6 '17 at 15:55
• For example try to calculate the chance of picking (blue ,blue ,blue ,blue,blue,blue,blue) from urn C by either of your methods. I think you implicilty assume there are 6 balls. – zen Jun 6 '17 at 16:25

So I'll give two answers one for each question. Let's first discuss the first question and then in a separate answer I will attempt to answer the second question(this might take a while because I have some other things to do :)). So first lets expand bayes theorem and then we will consider what this means. So want $P(drew from C|red, blue)=\frac{P(red ,blue|drew from C)P(drew from C)}{P(red ,blue|drew from C)P(drew from C)+ P(red ,blue|drew from A)P(drew from A)+P(red ,blue|drew from B)P(drew from B)}$

Technically we do not know $P(drew from A),P(drew from B),P(drew from C)$! So technically we cannot calculate this chance. But we can make a reasonable assumption and that is that $P(drew from A)=P(drew from B)=P(drew from C)=\frac{1}{3}$. Now to determien all the other terms lets start with $P(red ,blue|drew from C)$ now we assume these are independent chances e.g. If I pick blue first the chance of blue stays the same for the second choice (which would in real life not be true) than the chance of $P(red ,blue|drew from C)=(\frac{1}{3} \frac{1}{2}+\frac{1}{2} \frac{1}{3})=\frac{1}{3}$ remember that not only red, blue counts but also blue ,red thus we should count both possibilities. the other probabilities can be calculated in a similar way. so that $P(drew from C|red, blue)=\frac{\frac{1}{3}\frac{1}{3}}{\frac{1}{3}\frac{1}{3}+ 2\frac{1}{2}\frac{1}{6}\frac{1}{3}+2\frac{1}{3}\frac{1}{6}\frac{1}{3}}=0.428$ (approximately)

So for the second question I will just offer some help in the form of tips after which I hope you will understand it. tip 1: consider all teams equal so and calculate the chance that $P(played 16 games)=\frac{20}{32}$. 2. then see that the concitional probability your looking for is $P(played 16 games|coin wins =10)$ 3. $P(coin wins =10 |played 16 games)$ is binomially distributed with (n=16,p=0.5) and X=10 ( or k=10 in wiki notation).


# Partial Orderings Section 8.6.

## Presentation on theme: "Partial Orderings Section 8.6."— Presentation transcript:

Partial Orderings Section 8.6

Introduction A relation R on a set S is called a partial ordering or partial order if it is: reflexive antisymmetric transitive A set S together with a partial ordering R is called a partially ordered set, or poset, and is denoted by (S,R).

Example Let R be a relation on set A. Is R a partial order?
(2,3),(2,4),(3,3),(3,4),(4,4)}

Example Is the “” relation is a partial ordering on the set of integers? Since a  a for every integer a,  is reflexive If a  b and b  a, then a = b. Hence  is anti-symmetric. Since a  b and b  c implies a  c,  is transitive. Therefore “” is a partial ordering on the set of integers and (Z, ) is a poset.

Comparable/Incomparable
In a poset the notation a ≼ b denotes (a,b) ∈ R The “less than or equal to” ()is just an example of partial ordering The elements a and b of a poset (S, ≼) are called comparable if either a≼b or b≼a. The elements a and b of a poset (S, ≼) are called incomparable if neither a≼b nor b≼a. In the poset (Z+, |): Are 3 and 9 comparable? Are 5 and 7 comparable?

Total Order We said: “Partial ordering” because pairs of elements may be incomparable. If every two elements of a poset (S, ≼) are comparable, then S is called a totally ordered or linearly ordered set and ≼ is called a total order or linear order. The poset (Z+, ) is totally ordered. Why? The poset (Z+, |) is not totally ordered.

Hasse Diagram Graphical representation of a poset
Since a poset is by definition reflexive and transitive (and antisymmetric), the graphical representation for a poset can be compacted. For example, why do we need to include loops at every vertex? Since it’s a poset, it must have loops there.

Constructing a Hasse Diagram
Start with the digraph of the partial order. Remove the loops at each vertex. Remove all edges that must be present because of the transitivity. Arrange each edge so that all arrows point up. Remove all arrowheads.

Example Construct the Hasse diagram for ({1,2,3},) 3 2 1 3 2 1 1 2 3
1 1

Hasse Diagram Terminology
Let (S, ≼) be a poset. a is maximal in (S, ≼) if there is no bS such that a≼b. (top of the Hasse diagram) a is minimal in (S, ≼) if there is no bS such that b≼a. (bottom of the Hasse diagram) a is the greatest element of (S, ≼) if b≼a for all bS… it has to be unique a is the least element of (S, ≼) if a≼b for all bS. It has to be unique

Hasse Diagram Terminology (Cont ..)
Let A be a subset of (S, ≼). If uS such that a≼u for all aA, then u is called an upper bound of A. If lS such that l≼a for all aA, then l is called an lower bound of A. If x is an upper bound of A and x≼z whenever z is an upper bound of A, then x is called the least upper bound of A…unique If y is a lower bound of A and z≼y whenever z is a lower bound of A, then y is called the greatest lower bound of A…unique

Example Maximal: h,j Minimal: a Greatest element: None
g f d e b c a Maximal: h,j Minimal: a Greatest element: None Least element: a Upper bound of {a,b,c}: e,f,j,h Least upper bound of {a,b,c}: e Lower bound of {a,b,c}: a Greatest lower bound of {a,b,c}: a

Lattices A partially ordered set in which every pair of elements has both a least upper bound and greatest lower bound is called a lattice. f e c d b a h e f g b c d a

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# Calculate the sum of the infinite series $\sum_{n=0}^{\infty} \frac{n}{4^n}$

A previous problem had us solving $\sum_{n=0}^{\infty} \frac{1}{4^n}$ which I calculated to be $\frac{4}{3}$ using a bit of mathematical manipulation.

Wonderful. Thank you for all the prompt responses. Could anyone suggest an alternate technique that does not involve differentiation?

-

$$\frac{1}{4} +\frac{2}{4^2}+\frac{3}{4^3}+\frac{4}{4^4}+ \cdots +\frac{n}{4^n}+\cdots.$$

Call this sum $S$. Now subtract from $S$ the sum $$\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+\cdots.$$ If we do it in the obvious way, term by term, we obtain $$\frac{1}{4^2}+\frac{2}{4^3}+\frac{3}{4^4}+ \cdots.$$

Note that this last sum is $(1/4)S$.

Putting things together, and using your computation for $1+1/4+1/4^2+\cdots$ (not quite, we start at $1/4$) we get $$S-\frac{1}{3}=\frac{S}{4}.$$ Solve for $S$. We find that $S=4/9$.

Comment: The calculation is a little sloppy, it assumes that infinite sums can be manipulated much like finite sums. There are theorems about power series that one could use to justify the manipulations.

But (in this case) we do not need such theorems. Let $S_n$ be the sum of the terms up to the term $n/4^n$. More or less the same sort of calculation as the one I did can be used to find an explicit formula for $S_n$. Then we can calculate $\lim_{n\to\infty}S_n$, and get a fully rigorous derivation.

We could use the results of the calculation of $\sum n/4^n$ to tackle $\sum n^2/4^n$, and so on. But the derivatives approach is certainly slicker!

-
This is exactly the sort of manipulation that I was looking for. Thank you. –  beethree Jul 11 '11 at 22:36
@beethree: Good. You might want to look at the comment by Aryabhata, at the moment hidden at the end. –  André Nicolas Jul 11 '11 at 23:03

As you have computed $\sum_{n>=0} x^n$ to be $1/(1-x)$, differentiate the series term-wise and multiply by x, which you can do for $x=1/4$ as the series converges. This gives $\sum_{n>=0} n x^{n} = x/(1-x)^{2}$. Substitute $x=1/4$ and observe that $\sum_{n>=0} n/4^n = 4/9$

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Let $f(x)=\sum_{n\geq0}x^n$. This defines a function on $(-1,1)$, equal to $\frac1{1-x}$. Using properties of power series, we know that $$\frac x{(1-x)^2}=xf'(x)=\sum_{n\geq1}nx^n$$ for the same values of $x$. Evaluating this equality at $\tfrac14$ sums your series.

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Well, you can do some series manipulations... First you can write $$\sum_{n = 1}^{\infty} {n \over 4^n} = \sum_{n = 1}^{\infty}\,\sum_{m = 1}^n {1 \over 4^n}$$ Note that the above summation is over all $(m,n)$ with $m \leq n$. So if you switch the order of summation you obtain $$\sum_{m = 1}^{\infty}\,\sum_{n = m}^{\infty} {1 \over 4^n}$$ The inner sum is a geometric series with initial term ${\displaystyle{1 \over 4^m}}$ and ratio ${\displaystyle{1 \over 4}}$, so it sums to ${\displaystyle {{1 \over 4^m} \over 1 - {1 \over 4}} = {4 \over 3}{1 \over 4^m}}$. So the overall sum is $${4 \over 3}\sum_{m = 1}^{\infty} {1 \over 4^m}$$ The sum here is a geometric series that sums to ${\displaystyle{1 \over 3}}$, so your final answer is $${4 \over 3}\times{1 \over 3} = {4 \over 9}$$

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\begin{align} (\frac{1}{4} + \frac{1}{4^2} + \frac{1}{4^3} + \dots ) + \\ (\frac{1}{4^2} + \frac{1}{4^3} + \dots) + \\ (\frac{1}{4^3} + \dots ) + \\ \dots \end{align} –  Aryabhata Jul 11 '11 at 22:18
yeah, basically :) –  Zarrax Jul 11 '11 at 22:19

A simple probabilistic approach:

Let $X$ be a geometric random variable with probability of success $p$, so that $${\rm P}(X=n)=(1-p)^{n-1} p, \;\; n=1,2,3,\ldots.$$ Then the expectation of $X$ is $${\rm E}(X) = \sum\limits_{n = 1}^\infty {n{\rm P}(X = n)} = \sum\limits_{n = 1}^\infty {n(1 - p)^{n - 1} p} = \frac{p}{{1 - p}}\sum\limits_{n = 1}^\infty {n(1 - p)^n } .$$ Thus $$\sum\limits_{n = 1}^\infty {n(1 - p)^n } = \frac{{1 - p}}{p} {\rm E}(X).$$ On the other hand, from $${\rm E}(X) = p \cdot 1 + (1-p)(1+{\rm E}(X)),$$ we get $${\rm E}(X)=\frac{1}{p}.$$ Finally, $$\sum\limits_{n = 1}^\infty {n(1 - p)^n } = \frac{{1 - p}}{p} {\rm E}(X) = \frac{{1 - p}}{p^2}.$$ Letting $p=3/4$ gives $$\sum\limits_{n = 1}^\infty {\frac{n}{{4^n }}} = \frac{{1/4}}{{9/16}} = \frac{4}{9}.$$

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Nice! In imitation of the book "Proofs that Really Count", which is an attractive introduction to bijective proofs, I had thought of posting a question about Mean Proofs. This would ask for examples of results not in probability theory that can be proved by using the mean. This (along with I think a few others of your posts) is a contribution to the mean proofs collection. –  André Nicolas Jul 12 '11 at 0:51
@user6312: Thanks. Hopefully, I'll make more contributions of this kind. –  Shai Covo Jul 12 '11 at 1:15


Lesson 149u: Sieve of Eratosthenes

Suppose you wanted to find all prime numbers less than or equal to a given number $N$. The easiest way to do this is to create an array that can hold the integers from $1$ to $N$. We will initialize the array so that array[k] == k. If we ever determine that a number k is not prime, we will assign array[k] = 0;. By definition, $1$ is not prime, so we will immiedately assign array[1] = 0;. Next starting with $2$, if we determine that k is prime (meaning that array[k] == k, we have then determined that all integer multiples of k up until the end of the array to be composite, so we set:

array[2*k] = 0;
array[3*k] = 0;
array[4*k] = 0;
array[5*k] = 0;

.
.
.

until we reach the end of the array.

We then search forward in the array until we find the next k such that array[k] == k and repeat the above process.

This is called the Sieve of Eratosthenes. Implement a function that prints all prime numbers up to but not exceeding $N$ in the format:

2, 3, 5, 7, 9, 11, 13, ...

Be sure not to have a comma after the last prime number. Your function declaration should be:

void print_primes( unsigned long N );

As an aside, why is $1$ not prime? In general, many of the most interesting properties of prime numbers are not true if $1$ is considered to be prime. For example, that every number has a unique prime factorization is false if $1$ is a prime, for $6 = 2 \times 3 = 2 \times 3 \times 1 = 2 \times 3 \times 1 \times 1 \times \cdots$. Thus, if $1$ were said to be prime, then almost every interesting theorem about prime numbers would have to be worded as all prime numbers except for $1$.

In the previous function, you could have easily printed the prime number every time you found one. You would start by automatically printing "2", and then for every subseuent prime number n you find, print ", n". However, suppose you wanted to do the following:

If you have found $n$ primes less than or equal to $N$, return an array of size $n$ containing those $n$ prime numbers.

Now, beause you don't know how large an array to build, you will have to temporarily store prime numbers, and the easiest way is to look at the array array; however, looking through the entire array of capacity $N$ to find $n$ primes is unnecessarily expensive, especially when you know mathematically that the number of primes less than or equal to $N$ will be approximately $n \approx \frac{N}{\ln(N)}$. This estimates there are approximately $72382$ prime numbers less than or equal to one million. There are in fact $78498$ primes less than or equal to one million, so this approximation is not unreasonable; however, it is not exact.

It can be shown that the relative error drops at approximately $0.24 n^{-0.085}$, so the relative error actually goes to zero as $n$ becomes large.

Devise a scheme whereby you use the existing array to store all the prime numbers without having to walk through the entire array to retrieve all of them.

Array<unsigned long> *primes( unsigned long N );


# Visualise Whirled Peas

My eldest came home with some interesting maths homework.

We have three bowls containing peas.

Distribute the peas evenly across the bowls, by moving peas from one bowl to another. The only move allowed, is doubling the amount of peas in one bowl by taking them from one other bowl.

For example:
$\begin{array}{rlrlrl} A & & B & & C & \\ \hline 11 & & 6 & & 7 \\ 4 & (-7) & 6 & & 14 & (+7) \\ 4 & & 12 & (+6) & 8 & (-6) \\ 8 & (+4) & 8 & (-4) & 8 \end{array}$

We had some fun with this. We quickly decided that the order of the bowls doesn't matter, so we can just sort them by the number of peas they contain.
Then, there are just three moves possible:

1. Double the amount of peas in the first bowl (containing the fewest peas) by taking them from the third bowl (containing the most peas)
2. Double the amount of peas in the first bowl containing the fewest by taking them from the second bowl (containing the neither the most nor the fewest peas)
3. Double the amount of peas in the second bowl by taking them from the third bowl

From that, we made state diagrams and found our solutions. We also found out that for one of the problems, we could never reach an end state (with all bowls containing the same amount of peas).

But of course I couldn't leave it alone. So I tried to formalise what we did.

The amount of peas in each bowl at any time is of course a positive integer (or possibly 0):
$a_i \le b_i \le c_i \,|\,a_i, b_i, c_i \in \mathbb Z_{\ge 0}$

A solution has the peas distributed equally across the bowls:
$a_i = b_i = c_i$

The total amount of peas at any time is a multiple of 3:
$a_i + b_i + c_i = 3n \,|\,n \in \mathbb Z_{\ge 0}$

The three possible moves:
$\left\{a_{i+1}, b_{i+1}, c_{i+1}\right\} \in \begin{Bmatrix} \left\{2\cdot a_i, b_i, c_i-a_i\right\} \\ \left\{2\cdot a_i, b_i-a_i, c_i\right\} \\ \left\{a_i, 2\cdot b_i, c_i-b_i\right\} \end{Bmatrix}$

The set of possible moves brings us to the conclusion that for each state after the first, at least one bowl contains an even amount of peas. Since in the end state, all bowls contain the same amount of peas, they all contain the same even amount of peas and thus the total number of peas must be even.
If we combine this with the requirement that the total number of peas is a multiple of 3 as well, we see that the total number of peas must be a multiple of 6:
$a_i + b_i + c_i = 6n \,|\,n \in \mathbb Z_{\ge 0}$

(Ignoring the very trivial case where we start with a solved state).

But these conditions aren't sufficient to guarantee a solvable puzzle. For instance, $\left\{a_0, b_0, c_0\right\} = \left\{6,12,24\right\}$ is not solvable. The only reachable states are $$\begin{Bmatrix} \left\{ 6, 12, 24 \right\} \\ \left\{12, 12, 18 \right\} \\ \left\{ 0, 18, 24 \right\} \\ \left\{ 0, 6, 36 \right\} \\ \left\{ 0, 12, 30 \right\} \\ \end{Bmatrix}$$ neither of which is an end state.

So what additional conditions do we need to define to make such a puzzle sovable?

Problems can have multiple solutions. See for instance this example, with two possible paths to a solution:

$\begin{array}{rlrlrl} A & & B & & C & \\ \hline 11 & & 6 & & 7 \\ 4 & (-7) & 6 & & 14 & (+7) \\ 4 & & 12 & (+6) & 8 & (-6) \\ 8 & (+4) & 8 & (-4) & 8 \end{array}$

and

$\begin{array}{rlrlrl} A & & B & & C & \\ \hline 11 & & 6 & & 7 \\ 5 & (-6) & 12 & (+6) & 7 & \\ 10 & (+5) & 12 & & 2 & (-5) \\ 8 & (-2) & 12 & & 4 & (+2) \\ 8 & & 8 & (-4) & 8 & (+4) \end{array}$

The only way to solve a problem that I've come up so far, is drawing a state diagram and determining the shortest path to the end state. But while state diagrams work fine for small numbers of peas, the number of possible states may well explode for higher numbers.

So is there another way to solve these problems?

What is a good strategy to arrive at a solution?

Yes, I'm asking two questions in one, but I don't want to repeat the explanation for a second question. Also, I feel both questions are closely related, since what I'm really looking for is an analysis of this type of problem, if possible even with a variable number of bowls.

• I think your definition of the three possible moves how you present it is not entirely correct because it doesn't account for reordering the bowls again when they are not in ascending order any more – Ivo Beckers Oct 13 '15 at 14:20
• Some observation: If one of the bowls ends up empty at some point you are in an unsolvable state because double zero is still zero. – Ivo Beckers Oct 13 '15 at 14:35
• If the number of peas in each bowl share a common divisor, they will always share that same divisor. So in fact the total number of peas must be a multiple of $6$ times the greatest common divisor of the initial numbers of peas in the bowls. – Julian Rosen Oct 13 '15 at 15:40
• @IvoBeckers Yes, it's true that my representation of the moves doesn't take the reordering into account, but that doesn't seem to be an important factor. And we indeed encountered the states where one bowl was empty. We continued and found loops, as was to be expected. – SQB Oct 13 '15 at 17:05
• On a whim, I pronounced the puzzle title and was surprised by my pun detector going off! – Oliphaunt - reinstate Monica Feb 27 '16 at 11:55

If $a_0$, $b_0$, and $c_0$ are positive integers, the puzzle is solvable from an initial state $\{a_0,b_0,c_0\}$ if and only if $$a_0+b_0+c_0=3\cdot 2^m\cdot \gcd(a_0,b_0,c_0)$$ for some $m\geq 0$, where $\gcd(a_0,b_0,c_0)$ denotes the greatest common divisor of $a_0$, $b_0$, and $c_0$.
For $k\geq 3$ an odd number, one can check that if there is a bowl whose pea number is not divisible by $k$, then after making a move there will still be a bowl whose pea number is not divisible by $k$.
Suppose the initial state is $\{a_0,b_0,c_0\}$ is solvable, and let $n=\frac{a_0+b_0+c_0}{3}$ be the desired number of peas in each bowl.
If $k\geq 3$ is an odd number dividing $n$, then in the final state every pea number is divisible by $k$, so the argument above implies that $k$ divides $a_0$, $b_0$, and $c_0$. Every odd divisor of $n$ is a common divisor of $a_0$, $b_0$, and $c_0$, so it must be the case that $n=2^m\cdot \gcd(a_0,b_0,c_0)$, or equivalently $a_0+b_0+c_0=3\cdot 2^m\cdot \gcd(a_0,b_0,c_0)$.
Conversely, suppose $n=2^m\cdot \gcd(a_0,b_0,c_0)$. We can solve the puzzle as follows. If the current state is $\{a,b,c\}$, write $a=2^x a'$, $b=2^y b'$, $c=2^z c'$, where $a'$, $b'$, and $c'$ are odd. The condition on $n$ implies that either $a=b=c$, or that of the three numbers $x$, $y$, and $z$, two will be equal and the third will be larger.
Reordering if necessary, we can assume $x=y<z$ and that $a\leq b$. If $a< b$, replace $\{a,b,c\}$ by $\{2a,b-a,c\}$ and repeat the process. If $a=b$, replace $\{a,b,c\}$ by $\{2a,b,c-a\}$ (if $a<c$) or $\{a-c,b,2c\}$ (if $a>c$) and repeat the process. This will eventually terminate with all three bowls having the same number of peas. The reason this terminates is that if $\{a,b,c\}$ is replaced by $\{2a,b-a,c\}$, then the values of $x$ and $y$ increase by at least one. The replacement with $\{2a,b,c-a\}$ or $\{a-c,b,2c\}$ cannot happen twice in a row, and the size of $x$, $y$, and $z$ must remain bounded.


# Math Ratios

Suppose A is directly proportional to B, B is inversely proportional to C and C is inversely proportional to D. Determine whether A and D are directly proportional, inversely proportional, or neither.

1. C is inversely proportional to D ---> C = p/D
B is inversely proportional to C ---> B = qC
so B = q/(p/D) = (q/p)D

A is directly proportional to B --> A = rB
A = r(q/p) D
looks like A is directly proportional to D

posted by Reiny

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# Let $\{x_n\}$ be a Cauchy sequence of rational numbers. Define a new sequence $\{y_n\}$ by $y_n = (x_n)(x_{n+1})$. Show that $\{y_n\}$ is a CS.

What I am thinking so far is following:

Construct another sequence $\{b_n\}$ such that $b_1=x_2, b_2=x_3, \ldots, b_n=x_{n+1}.$ Since $\{x_n\}$ is a Cauchy sequence, $\{b_n\}$, as a subsequence of $\{x_n\},$ is therefore also a Cauchy sequence.

Since Cauchy sequences are bounded, we can find $M_1, M_2 \in Q$ such that $|x_n| \le M_1$ and $|b_n| \le M_2$ for all $n.$ Let $M = \max\{M_1, M_2\}.$

Since $\{x_n\}$ is a Cauchy sequence, we can find $N_1$ such that for all $m,n \ge N_1,$ $|x_n-x_m| < \frac \varepsilon{2M}$. Similarly, we can find $N_2$ such that for all $m,n \ge N_2,$ $|b_n-b_m| < \frac \varepsilon{2M}.$

Let $N = \max\{N_1, N_2\}$ and $n,m \ge N.$

Therefore, $$|y_n-y_m|=|x_n x_{n+1}-x_m x_{m+1}|=|x_n b_n-x_m b_m|=|x_n b_n-x_m b_n+x_m b_n-x_m b_m|=|b_n||x_n-x_m|+|x_m||b_n-b_m| \le M\times\frac \varepsilon{2M} + M\times\frac \varepsilon{2M} = \varepsilon$$

QDE.

I wonder if there is any flaw in this proof and if there is a better way to prove it. Thank you!

• Please, use MathJax to format mathematical expressions in your posts. – mucciolo Dec 10 '17 at 2:45
• The idea looks right to me! One minor mistake: Near the end, in the line "Therefore...", when you use the triangle inequality to break up an absolute value, you should have $\leq$ instead of $=$: specifically, $|x_nb_n-x_mb_n+x_mb_n-x_mb_m| \leq |x_nb_n-x_mb_n| + |x_mb_n-x_mb_m|$. But this doesn't affect your argument. As far as better way: I can't think of anything. You could simplify notation slightly (just write $x_{n+1}$ instead of $b_n$, etc.) but nothing significant. – Zach Teitler Dec 10 '17 at 3:19
• No need to introduce the sequence $\{b_n\}$ just to obtain the bound for $\{x_{n+1}\}$. If $|x_n| \le M$ for all $n \in \mathbb{N}$ then also $|x_{n+1}| \le M$ for all $n \in \mathbb{N}$. Great otherwise. – mechanodroid Dec 10 '17 at 9:58


# Proof by mathematical induction, difficulties

Usually, proofs by mathematical induction are not that difficult (at least the ones I encountered so far in my mathematical journey), but I am stuck with this one... Spent at least 2-3 hours on it. I just can't see it...

$$\sum_{i=0}^n \frac{1}{2^i}\tan\left(\frac{x}{2^i}\right)=\frac{1}{2^n}\cot\left(\frac{x}{2^n}\right) - 2\cot(2x) , (n = 1,2...)$$

What I tried:

For $n=1$, I expressed the LHS in terms of $\sin$ and $\cos$, but that didn't work... For me, at least... The expression got really complicated, which is a sign that I, probably, did something wrong.

Then I tried getting that $2\cot(2x)$ from RHS to LHS, but that didn't work also, because the expressions get really complicated...

• Hint: if you can show $LHS(n+1) - LHS(n) = RHS(n+1) - RHS(n)$ and $LHS(0) = RHS(0)$, then the induction becomes easy. Then, simplifying $LHS(n+1) - LHS(n)$ and $RHS(n+1) - RHS(n)$ will help you focus on the exact trigonometric identity you need to prove. Jun 22 '17 at 20:42
• Note that in general to prove $\sum_{i=0}^n a_n = s_n$, you can just prove $s_0 = a_0$ (for the base case) and that $s_n - s_{n-1} = a_n$ (for the induction step). Jun 22 '17 at 20:42
• @JairTaylor The problem is, I can't prove the base case. I just can't. Spent at least 2-3 hours on it. Jun 22 '17 at 20:49
• @DanielSchepler Wow. Awesome concept, I never looked at it that way. It can really bring some elegance to my, usually bulky & ugly, solutions. Jun 22 '17 at 20:50

\begin{align} S_{n-1}+T_{n} &=\frac{1}{2^{n-1}}\cot \left(\frac{x}{2^{n-1}}\right)-2\cot (2x)+\frac{1}{2^{n}}\tan \left(\frac{x}{2^{n}}\right)\\ &\text{let} \, \frac{x}{2^{n}}=y\\ &=\frac{1}{2^{n}}\left( 2\cot (2y)+\tan (y)\right)-2\cot (2x)\\ &=\frac{1}{2^{n}}\left( \frac{2(\cos (2y))}{\sin (2y)}+\tan (y)\right)-2\cot (2x)\\ &=\frac{1}{2^{n}}\left( \frac{2(\cos ^2 (y)-\sin^2 (y))}{2\sin (y)\cos (y)}+\tan (y)\right)-2\cot (2x)\\ &=\frac{1}{2^{n}}\left( \cot(y)-\tan (y)+\tan (y)\right)-2\cot (2x)\\ &=\frac{1}{2^{n}}\left(\cot \left(\frac{1}{2^{n}}\right)\right)-2\cot (2x)\\ &=S_n\\ \end{align}

• Thank you for sharing your knowledge. Truly, a beautiful proof. Jun 22 '17 at 21:19

** just a hint**

It is equivalent to prove that

$$\cot (x/2^{n+1})-2\cot (x/2^n)=$$

$$\tan (x/2^{n+1})$$

• To do that, notice that $\cot(x/2^n) = \cot(2\cdot x/2^{n+1})$. Jun 22 '17 at 20:57


# Sum $\frac{1}{1\times2}+\frac{1}{1\times3}+\frac{1}{2\times5}+\frac{1}{3\times8}+\cdots$

If $f_n$ is the Fibonacci series, with $1,1,2,3,5,8,\ldots$ prove that

$$\sum_{i=2}^\infty\frac{1}{f_{i-1}\cdot f_{i+1}} = 1$$

So my idea was to try to convert this series into a telescoping sum somehow, because otherwise I can't see how this would be managed.

$$\frac{1}{1\times2}+\frac{1}{1\times3}+\frac{1}{2\times5}+\frac{1}{3\times8}+\cdots$$

I can't see any obvious way to re-write the terms though. I could try sum this using Binet's formula but I am pretty sure that will get out of hand.

What other alternatives do I have here?

Note: If you use any other identity other than $f_n=f_{n-1}+f_{n-2}$, or Binet's formula. kindly link to, or provide , a proof of it.

You could note that

$$\frac{1}{f_nf_{n-1}} - \frac{1}{f_nf_{n+1}} = \frac{1}{f_{n-1}f_{n+1}}$$

Which could help with your telescoping sum.

• sorry for the late acceptance. I wasn't online for a few days. – Guy Mar 29 '14 at 13:38

We can use the identity via partial fractions,

$\dfrac{1}{x(a+x)} = \dfrac{1}{a}\left(\dfrac{1}{x} - \dfrac{1}{a+x} \right)$

with $x = f_{i-1}$ and $a = f_{i}$ (so that $a+x = f_{i+1}$) to get

$\dfrac{1}{f_{i-1}f_{i+1}} = \dfrac{1}{f_i}\left(\dfrac{1}{f_{i-1}} - \dfrac{1}{f_{i+1}} \right) = \dfrac{1}{f_i \ f_{i-1}} - \dfrac{1}{f_{i+1} f_{i}}$.

So if $\phi_i:= \frac{1}{f_i \ f_{i-1}}$,

$\dfrac{1}{f_{i-1}f_{i+1}} = \phi_{i} - \phi_{i+1}$ and we can telescope,

$\displaystyle \sum_{i=2}^N \dfrac{1}{f_{i-1}f_{i+1}} = \phi_2 - \phi_{N+1} = 1 - \phi_{N+1}$

To conclude, realise that $f_i\to\infty$, so $\phi_N\to0$, and so

$\displaystyle \sum_{i=2}^∞ \dfrac{1}{f_{i-1}f_{i+1}} = 1 - \lim_{N→∞}\phi_{N+1} = 1 - 0$.


# Pigeon hole principle proof writing

I'm currentely discussing pigeon hole principle (simple and generalised) in class. However, when solving problems, I get the idea, but I never know the proper way to write it, I always feel like I'm either writing too much or not enough. Here is an example of a very simple problem and 2 ways I wrote the proof and if you could comment and give me tips it would be very appreciated!!

Problem

A man in a dark room has a box with 12 brown and 12 black socks. How many times, if picking one at a time, does the man have to pick a sock to have 2 of the same color?

(Here I think of the two colors as the "boxes" and the picks as the "objects". Since there's k+1 objects (or picks) and k boxes (or type of socks) then the pigeon hole principle apply)

Solution 1

There are $2$ different type of socks the man can pick . By the pigeon hole, principle if the man picks $3$ socks, it is garanteed that he will have $2$ socks of the same color $\blacksquare$.

Solution 2

The generalised pigeon hole principle says that, for $N,k \in \Bbb{Z}$, if $N$ objects are placed in $k$ boxes, then at least $1$ boxe will have $\lceil{\frac{N}{k}}\rceil$ objects.

Let $k = 2$ be the the colors of the socks and $N$ be the number of picks

The minimum amount of picks the man has to do to garantee he will pick $2$ socks of the same color is the smalles integer such that $\lceil{\frac{N}{2}}\rceil = 2$

Therefore, we need the minimum integer $N$ picks such that $1\lt \lceil{\frac{N}{2}}\rceil \le 2 \Rightarrow N = 3$ since $3 = 1*2 + 1$

So, by the generalised pigeon principle, to guarantee the man picks 2 socks of the same color, he has to pick 3 socks$\blacksquare$.

Conlcusion

I know the problem is very simple but I really need to know how to write that kind of proof in a way such that I say enough but not too much. I mean I feel like in the first solution I'm not convincing and in the second one I blabla for no reasons..

As usual, thank you everyone for your help!!!!

• I vote for Solution 1 – Bram28 Mar 21 '18 at 19:07
• Solution 1 for sure (though it may depend on your instructor; you might want to ask them) – E-A Mar 21 '18 at 19:12
• For a different view on this principle, see Edsger W. Dijkstra's EWD980 and EWD1094. – Marnix Klooster Mar 26 '18 at 20:37

In most contexts, Solution 1 is a perfectly fine amount of detail and Solution 2 is way more than is necessary. There is one important way that both solutions are incomplete, though: you've explained why picking $3$ socks guarantees $2$ of the same color, but you haven't explained why no smaller number guarantees $2$ of the same color. To have a completely thorough proof, you need to also mention how it is possible to choose $2$ socks which are of different colors.
(This observation is obvious enough that in many contexts it would be fine not to write it out explicitly, though. Note that in Solution 2 you implicitly made a more general claim when you asserted that the minimum number of picks to guarantee two socks of the same color is the smallest $N$ such that $\lceil{\frac{N}{2}}\rceil = 2$. This does not follow from the pigeonhole principle alone, since you are also claiming that no smaller $N$ can work.)


I am particularly concerned about these two sums: $$\sum_{x=\color{red}0}^\infty \frac{1}{2^x} = 1 + \frac{1}{2} + \frac{1}{4} ... = 2$$

and

$$\sum_{x=1}^\infty \frac{1}{x} = 1 + \frac{1}{2} + \frac{1}{3} ... = \infty$$

Now, I know and I understand the proofs that say the first sum converges to two and the second one is divergent. ( you probably shouldn't equate a divergent series like that ).

But on an intuitive level, what separates these two sums in such a way, that their limits are so drastically different?

Both sums add very small numbers in the end ( e.g. $\frac{1}{10000000000}$ ), yet one always stays smaller than two and one goes way beyond that. You would think ( again, on an intuitive level, the math is clear) that the numbers become so small that there is a certain number that doesn't change anymore (significantly).

For example:

$$\sum_{x=1}^{100} \frac{1}{x} = ca. 5$$

$$\sum_{x=1}^{1000000} \frac{1}{x} = ca. 14.3$$

$$\sum_{x=1}^{10000000000000000000000000000000} \frac{1}{x} = ca. 73.5$$

As one can see, the growth drastically slows down, at some point it should became infinitely small, just like with first sum ( fraction, powers of two ).

• Have you seen the proof of divergence of the second sum that compares the sum to $\frac 12+(\frac 14+\frac 14)+(\frac 18+\frac 18+\frac 18+\frac 18)+\dots$? This is probably the best intuition I can think of, that no matter how far you go, you can always find a set of terms that together increase the value of the sum by at least $\frac 12$. – abiessu Sep 14 '16 at 12:05
• This is a classic example that shows that intuition cant help in all situations. Sometimes intuition is useful or possible, other times is not. The failure of intuition is the main reason of the existence of mathematical analysis. – Masacroso Sep 14 '16 at 12:06
• Just a note: The first sum should start at $x=0$, else it converges to 1. – ekkilop Sep 14 '16 at 12:11
• +1 for a very good question. A point about intuition: Don't try to force the math to conform to your intuition. Instead, digest these answers and let your intuition conform to the mathematics. – Neal Sep 14 '16 at 13:41
• "Young man, in mathematics you don't understand things. You just get used to them" - Von Neumann – Dair Sep 14 '16 at 16:37

The difference is not in how small the terms are getting... as OP says, in both cases they become arbitrarily small... but in how many terms there are of a particular size. For instance, say two terms have the same size if their first nonzero binary digits are in the same place. In the first sum, you have one term of size $1$, one of size $1/2$, one of size $1/4$, one of size $1/8$, etc. In the second sum, you have one term of size $1$, one of size $1/2$, but then two of size $1/4$ ($1/3$ and $1/4$), four of size $1/8$ ($1/5$ through $1/8$), and so on. Each time the size is halved, the number of terms of that size is doubled; so the contribution from terms of each size never decreases, and the magnitude of the sum marches steadily to infinity.

• There are many informative and great answers here, but yours was the one I was looking for. Furthermore, I believe this "insight" to be the key behind most of the other proofs presented here. At the very least, this helped me understand them on a much more fundamental level – Jonathan Sep 16 '16 at 19:06
• Thanks for the great intuition. This cleared up all my confusion! – Dude156 Feb 6 at 3:47

This is related to how fast the terms decrease.

A geometric series (your $1/2^x$) is such that every term is a constant fraction of the previous, so that dividing by this constant is the same as dropping the first term.

$$\frac12\left(1+\frac12+\frac14+\frac18\cdots\right)=\frac12+\frac14+\frac18\cdots$$ So you can write

$$\frac12S=S-1$$ and deduce $S=2$.

The same reasoning applies to all geometric series

$$\sum_{k=0}^\infty r^k$$

provided that $r<1$. Indeed, if $r=1$ or $r>1$, the sum clearly grows forever. (This simplified discussion ignores the case $r<0$.)

This leads to a simple convergence criterion: if the ratio of successive terms is a constant less than $1$, the series converges. More generally, if this ratio is variable but tends to a limit smaller than $1$, the series converges.

Conversely, if the ratio tends to a limit larger than $1$, the series diverges. But if the ratio tends to $1$, we don't know, the criterion is insufficient.

The case of the harmonic series ($1/n$) or the generalized harmonic series ($1/n^p$) precisely falls in this category, as

$$\lim_{n\to\infty}\left(\frac{n}{n+1}\right)^p=1.$$

To deal with it, a trick is to sum the terms in groups of increasing size (by doubling), so that the sums exceed a constant. More precisely,

$$\begin{gather} 1,\\ \frac12,\\ \frac13+\frac14 > \frac14+\frac14 = \frac12,\\ \frac15+\frac16+\frac17+\frac18 > \frac18+\frac18+\frac18+\frac18 = \frac12,\\ \cdots \end{gather}$$

Though the groups get longer and longer, you can continue forever and the sum grows to infinity.

If you repeat the reasoning with exponent $p$,

$$\begin{gather} 1,\\ \frac1{2^p},\\ \frac1{3^p}+\frac1{4^p} > \frac1{4^p}+\frac1{4^p} = \frac2{4^p}=\frac1{2^{2p-1}},\\ \frac1{5^p}+\frac1{6^p}+\frac1{7^p}+\frac1{8^p} > \frac1{8^p}+\frac1{8^p}+\frac1{8^p}+\frac1{8^p} = \frac4{8^p} = \frac1{2^{3p-2}},\\ \cdots \end{gather}$$

In this new series, the ratio of successive terms tends to $2^{p-1}$ and by the first criterion, you can conclude convergence for $p>1$ and divergence for $p<1$. (A complete discussion must involve a similar upper bound, omitted here.)

To summarize, by decreasing order of decrease rate

$$\sum r^n, r<1\text{ converges}$$ $$\sum \frac1{n^p}, p>1\text{ converges}$$ $$\sum \frac1{n^p}, p=1\text{ diverges}$$ $$\sum \frac1{n^p}, p<1\text{ diverges}$$ $$\sum r^n, r=1\text{ diverges}$$ $$\sum r^n, r>1\text{ diverges}$$

For other series, you can compare to these decrease rates. For example, with the general term $1/n!$, the limit of the ratio is $\lim_{n\to\infty}n!/(n+1)!=0$ and the series converges, faster than any geometric series. Or $1/\sqrt[3]{n^2+1}$ makes a diverging series because the general term tends to $1/n^{2/3}$.

The curves below shows the trend of the terms of the sequences on a logarithmic scale. The green one corresponds to the harmonic series, which is a border between convergent and divergent series.

• 1/2 S = S - 1 has one more solution: S = ∞. – klimenkov Sep 15 '16 at 7:39
• @klimenkov S is by convention assumed to be a real number, so that's not a solution. – Joren Sep 15 '16 at 8:12
• @klimenkov: you are right, but I used this simple approach for pedagogical purposes. A more rigorous way is $S=\lim_{n\to\infty}S_n=\lim_{n\to\infty}(1-r^n)/(1-r)=1/(1-r)$ for $|r|<1$. – Yves Daoust Sep 15 '16 at 8:20

To help your intuition about why it doesn't suffice that the individual terms get arbitrary small, consider the following series: \begin{aligned} a &= \sum_{k=1}^{\infty} 2^{-\lfloor \log_2 k\rfloor}\\ &= 1 + \underbrace{\frac12 + \frac12}_{=1} + \underbrace{\frac14+\frac14+\frac14+\frac14}_{=1} + \underbrace{\frac18+\frac18+\frac18+\frac18+\frac18+\frac18+\frac18+\frac18}_{=1} + \ldots \end{aligned} Clearly the individual terms of this series get arbitrary small, but there are sufficiently many of them that the sum still adds up to an arbitrary large number; indeed for any positive integer $n$, the first $2^n-1$ terms add up to $n$.

So for the series to converge, it does not suffice that the terms get arbitrary small, they have to get small fast enough that their growing count doesn't overcompensate their diminishing value.

Intuition is not so intuitive.

Just take a look at this$$\sum_{k=N}^{2N}\frac{1}{k}\geq\sum_{k=N}^{2N}\frac{1}{2N}=\frac{1}{2}.$$

Then$$\sum_{k=1}^{4^{n}}\frac{1}{k}\geq n.$$

So the sequence $(\sum_{k=1}^{m}\frac{1}{k})_{m=1}^{\infty}$ has no upper bounds. It may seem counter-intuitive because the growth of $4^{n}$ is too fast for our "intuition".

Using the integrals is the best intuition I've got so far about this. Plot the function $f(x)=\frac{1}{x}$ and then for each $i$, draw a histogram with $\frac{1}{i}$ height. You can easily see that: $$\int_1^{n+1} \frac{1}{x} dx<\sum_{i=1}^{n} \frac{1}{i}<1+\int_1^{n} \frac{1}{x} dx$$ by shifting the histograms to the left.

Therefore $$\ln(n+1)<\sum_{i=1}^{n} \frac{1}{i}<1+\ln(n)$$ And I think the sandwich theorem is intuitive enough.

• This still makes it hard to intuit how $\int_1^M 1/x dx$ can be so large when $M$ is large, since the heights out there are so small. – Ian Sep 14 '16 at 12:40
• @Ian If we don't know that $1/x$ has an anti-derivative, then you are right. – polfosol Sep 14 '16 at 12:44
• What I mean is, we know that $\int_1^x 1/y dy=\ln(x)$, and we know that $\ln(x)$ blows up, but then the same intuitive question as in the OP arises: how can you add up these small things to get something that blows up? One nice way to answer it would be to note that the widths of the intervals $(x_k,x_{k+1})$ with $\int_{x_k}^{x_{k+1}} 1/x dx = 1$ grow exponentially fast with $k$--specifically you can take $x_k=e^k$. – Ian Sep 14 '16 at 12:47
• @Ian $\ln(x)$ does not "blow up", it creeps upward very slowly. – Simply Beautiful Art Sep 14 '16 at 21:13
• "Diverges to infinity" is colloquially expressed as "blows up" even if the relevant limit is to infinity and the divergence is slow. – Ian Sep 14 '16 at 22:19

Others have been focusing on intuitions on why the second series diverges independently of the first series... but there is a way to see what is happening by seeing how the first and second series are related.

Before we start, note that we can see that $$\sum_{k=0}^\infty \frac{1}{2^k} = 2$$ Now, consider the set of all positive integers. We can categorise these according to the largest odd factor of the number. So, for instance, 2, 4, 8, and 16 all share 1 as the largest odd factor. Meanwhile, 15, 30, 60, and 120 all share 15 as the largest odd factor.

This allows us to factorise our second sum, as $$\sum_{n=1}^\infty \frac1n = \left(\sum_{k=0}^\infty \frac1{2^k}\right)\sum_{n\in2\mathbb{N}-1}\frac1n = 2\sum_{n\in2\mathbb{N}-1}\frac1n$$

But this new sum, for only the odd terms, gives us a peculiar behaviour, as $$\sum_{n\in2\mathbb{N}-1}\frac1n > \sum_{n\in2\mathbb{N}}\frac1n = \sum_{n=1}^\infty\frac1{2n}$$ That is, $\frac1{2n-1}>\frac1{2n}$ for all $n\geq1$, so the sum must also be larger. Note that this relation is a strict one - they cannot be equal.

So what we have is $$S=\sum_{n=1}^\infty \frac1n > 2\sum_{n=1}^\infty\frac1{2n} = \sum_{n=1}^\infty \frac1n = S$$ It is here that you can see the problem - the sum must be strictly larger than itself. As this is a logical impossibility, we conclude that the sum is, itself, not well-defined.

This reasoning essentially operates on the same logic used to show that the first sum converges to 2... but gives a very different result.

The intuition for the geometric series is easy: you just cut a pie into half and the second half again into a half, etc.. in total, it's still the whole pie.

The second one is more tricky. Consider that you have money in a bank on $100\%$ interest rate evaluated continuously starting with 1 dollar, so that you will have $e^{t}$ dollars after $t$ years (if you like it more, consider $2^t$); the point is that the rate of growth is proportional / equal to your amount of money. Now ask the questions

1. "how many years I have to wait untill I will have $x$ dollars", where $x$ is an integer?
2. "how many years I have to wait untill I will have $x+1$ dollars"?

No matter what are the answers, you should agree that if $x$ is large, then the difference between answer $1$ and answer $2$ is of course tiny. In fact, the difference is approximately $1/x$ years, because the rate of increase equals the amount you have, so if you have $x$ dollars and the rate of increase is $x$, it takes around $1/x$ year to get a one dollar increase.

So if you continue and ask

1. "how many years I have to wait untill I will have $x+2$ dollars"?

you need to add another $1/(x+1)$ amount of time, and so on.

But then, of course, after ANY amount time you will have SOME amount of money in the bank, even after million years... so $1/x+1/(x+1)+1/(x+2)+ \ldots$ can be arbitrary large.

• :) I think it was a nice answer. – Simply Beautiful Art Sep 15 '16 at 18:39

Very interesting question indeed, and so I'll not only try to pick out what parts of your intuition is right, but I'll do my best to build on such intuition, since I noted many of the answers concern the proof of why the second sum diverges, even though you already state that you've seen and understood the proofs.

"Both sums add very small numbers in the end..."

True, but as you can easily see, the second sum diverges. However, it is correct to state that "if the sum converges, then as you take the limit to the last terms you are adding, the terms approach zero."

Clearly this must be the case or you'd end up adding something more than $0$ forever and beyond,

$$\underbrace{a_1+a_2+a_3+\dots}_{\text{first partial sums}}\underbrace{+c+c+c+\dots}_{\text{last partial sums}}=(a_1+a_2+a_3+\dots)+\infty$$

which would be an intuitive way of putting it.

Now, how do we discern the divergent from the convergent, in the case that $\lim_{n\to\infty}a_n=0$? We use convergence tests. A few interesting tests to point out:

$$\begin{array} {|l|c|c|c|} \hline \text{name} & \text{converges} & \text{diverges} & \text{inconclusive} \\ \hline \text{Ratio Test} & \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|<1 & \dots>1 & \dots=1 \\ \hline \text{Root Test} & \lim_{n\to\infty}\sqrt[n]{|a_n|}<1 & \dots>1 & \dots=1 \\ \hline \text{Integral Test} & \int_1^\infty f(t)dt<\infty & \dots=\infty & \text{never} \\ \hline \text{Direct Comparison} & \sum a_n<\sum b_n<\infty & \sum a_n>\sum b_n>\infty & \text{fail to find some b_n that satisfies test} \\ \hline \text{Cauchy Condensation Test} & \sum2^nf(2^n)<\infty & \sum2^nf(2^n)=\infty & \text{never} \\ \hline \end{array}$$

Firstly, note a few things. The ratio/root test is really a comparison test between any sum to a geometric series, which already has known convergence.

The integral test is due to the geometric meaning of an integral, which makes sense with the conditions that $f(x)$ is positive monotone decreasing. It can also be seen as another comparison test.

Note that the comparison test should make sense. If this sum is less than a sum that converges, then that sum converges. If it is greater than a sum that diverges, it will also diverge.

Lastly, the Cauchy comparison test is used in celtschk's answer, where he expanded it to make it more visual. It is a special case of the integral test/comparison test.

considering the convergence of $\sum_{n=1}^\infty\frac1n$

Taking the ratio test, the result is inconclusive. Taking the root test, the result is inconclusive. Taking the integral test, the result is divergence. Taking the Cauchy condensation test, the result is divergence.

Now, the comparison test can be done with anything that fits the inequalities, but better yet, one usually takes other tests first, since they are special cases of the comparison test. Only if all else fails (or if you get an interesting idea) should you resort to the comparison test.

Also, if you use the Cauchy condensation test:

$$\sum_{n=1}^\infty\frac1n\le\sum_{n=1}^\infty1=1+1+1+\dots=\infty$$

If you use the integral test:

$$\sum_{n=1}^\infty\frac1n\text{ converges iff }\int_1^\infty\frac1tdt\text{ converges}$$

$$\int_1^\infty\frac1tdt=\ln(\infty)=\infty$$

In essence, the thing that separates these two series is a comparison test, as there is a sum between the two that diverges, and a sum between the two that also converges, one will converge and the other will diverge.

The answers above are great but look complicated

Just look at the difference:

$1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + ...$ compared to $1/2 + 1/4 + 1/8 + 1/16 + 1/32 ...$

In the second case you see that we skip more and more items that go in the sequence of $1/n$. That's why they are so different.

• That doesn't explain why the first one is divergent. I could also remove terms from the second sum, and get a series converging to a different value, but that doesn't change the fact that the second series converges. – celtschk Sep 20 '16 at 6:05


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In each case, the acceleration of the object is in the positive direction.

α Since acceleration is a vector quantity, it has a direction associated with it. Likewise, the integral of the jerk function j(t), the derivative of the acceleration function, can be used to find acceleration at a certain time: Acceleration has the dimensions of velocity (L/T) divided by time, i.e. The sign of the tangential component of the acceleration is determined by the sign of the angular acceleration ( ( It is defined as 1 centimeter per second squared. Unless the state of motion of an object is known, it is impossible to distinguish whether an observed force is due to gravity or to acceleration—gravity and inertial acceleration have identical effects. = The data tables below depict motions of objects with a constant acceleration and a changing acceleration. Other Acceleration Units. How to Convert Units of Acceleration.

This same general principle can be applied to the motion of the objects represented in the two data tables below. r a unit vector tangent to the path pointing in the direction of motion at the chosen moment in time. In physics, the use of positive and negative always has a physical meaning. For objects with a constant acceleration, the distance of travel is directly proportional to the square of the time of travel. ). Yes, that's right. {\displaystyle v} Required fields are marked *, What is the difference between Acceleration and Velocity. Rearranging the formula, In this case we simply enter the mass (1,200 kg) and the acceleration, ) into the calculator and click Calculate, to show that the, F = m x a is usually just written as F = ma and is the mathematical. α

∴ From definition, acceleration is given as: Don't be fooled! Acceleration is the difference between final velocity and the initial velocity of an object over a given time period, whereas momentum is the product of mass and velocity. {\displaystyle \alpha } The (m/s)/s unit can be mathematically simplified to m/s 2. . So can we have a situation when speed remains constant but the body is accelerated? g

As such, if an object travels for twice the time, it will cover four times (2^2) the distance; the total distance traveled after two seconds is four times the total distance traveled after one second. Typical acceleration units include the following: These units may seem a little awkward to a beginning physics student. Each of these accelerations (tangential, radial, deceleration) is felt by passengers until their relative (differential) velocity are neutralized in reference to the vehicle. acceleration of the aircraft during its take of run? Actually, it is possible in circular where speed remains constant but since the direction is changing hence the velocity changes, and the body is said to be accelerated. That's because acceleration depends on the change in velocity and velocity is a vector quantity â one with both magnitude and direction. It's a mathematical ideal that can only be realized as a limit. And an object with a constant velocity is not accelerating.

Moreover, the dividing distance by time twice is equal to dividing distance by the square of time. Thus, a falling apple accelerates, a car stopping at a traffic light accelerates, and the moon in orbit around the Earth accelerates. What was the mass of the rocket at this stage? Another frequently used unit is the standard acceleration due to gravity â g. Since we are all familiar with the effects of gravity on ourselves and the objects around us it makes for a convenient standard for comparing accelerations. When we calculate acceleration, it basically involves velocity and time factor and dividing them in terms of units, meters per second [m/s] by second [s]. Acceleration values are expressed in units of velocity/time.


# Thread: Need Equation for Line Intersecting Two Points Involving y = x/(|x|+1)

1. ## Need Equation for Line Intersecting Two Points Involving y = x/(|x|+1)

So I'm trying to solve for an equation that produces a line that intersects two specific points and is shaped approximately like this:

This equation is expressed as y = x / (|x| + 1), has asymptotes at y = 1 and y = -1, and intersects the points (0, 0) and (1, 0.5). I'm wondering how to modify this equation to have asymptotes at y = 0 and y = 0.95 and intersects the points (0, 0.05) and (1, 0.75).

I'd really prefer a step-by-step solution in case I need to find similar equations with different parameters. Please and thank you everyone. Really need this for a project I'm working on.

Jinumon

2. ## Re: Need Equation for Line Intersecting Two Points Involving y = x/(|x|+1)

Originally Posted by Jinumon
So I'm trying to solve for an equation that produces a line that intersects two specific points and is shaped approximately like this:

This equation is expressed as y = x / (|x| + 1), has asymptotes at y = 1 and y = -1, and intersects the points (0, 0) and (1, 0.5). I'm wondering how to modify this equation to have asymptotes at y = 0 and y = 0.95 and intersects the points (0, 0.05) and (1, 0.75).
Is this the graph you want?

3. ## Re: Need Equation for Line Intersecting Two Points Involving y = x/(|x|+1)

Originally Posted by Jinumon
So I'm trying to solve for an equation that produces a line that intersects two specific points and is shaped approximately like this:

This equation is expressed as y = x / (|x| + 1), has asymptotes at y = 1 and y = -1, and intersects the points (0, 0) and (1, 0.5). I'm wondering how to modify this equation to have asymptotes at y = 0 and y = 0.95 and intersects the points (0, 0.05) and (1, 0.75).

I'd really prefer a step-by-step solution in case I need to find similar equations with different parameters. Please and thank you everyone. Really need this for a project I'm working on.

Jinumon
I will show you how to do this and you can modify it for other cases. Your function goes from -1 to 1 which is 2 units. Let's multiply it by $\frac 1 2$ so it goes from $-\frac 1 2$ to $\frac 1 2$ and move it up $\frac 1 2$. So now it looks like$$f(x) = \frac 1 2\frac {x}{|x|+1}+\frac 1 2$$This will have asymptotes $y=0$ and $y=1$ You want the upper asymptote to be $.95$ so let's multiply it by $.95$. So now$$f(x) = .95\left (\frac 1 2\frac {x}{|x|+1}+\frac 1 2 \right)$$
So far this graph has the right shape and desired asymptotes. Now this function takes the value $.05$ way to the left of $x=0$. If you set $f(x) = .05$ and solve it you get $x = -8.5$. So $f(-8.5)=.05$ but you want $f(0) = .05$. So now replace $x$ by $x - 8.5$ in your equation. Now your equation is$$f(x)= .95\left (\frac 1 2\frac {x-8.5}{|x-8.5|+1}+\frac 1 2 \right)$$
Translating it horizontally didn't change the horizontal asymptotes and now $f(0) = .05$ All that is left is to get $f(1)=.75$. Currently the value of $x$ that gives $f(x) = .75$ is $x=9.875$. So if you scale the $x$ axis by replacing $x$ by $9.875x$, that should do it. Your final formula becomes$$f(x)=.95\left (\frac 1 2\frac {9.875x-8.5}{|9.875x-8.5|+1}+\frac 1 2 \right)$$

Here's a graph of the final result (click to enlarge):

4. ## Re: Need Equation for Line Intersecting Two Points Involving y = x/(|x|+1)

Originally Posted by Walagaster
I will show you how to do this and you can modify it for other cases. Your function goes from -1 to 1 which is 2 units. Let's multiply it by $\frac 1 2$ so it goes from $-\frac 1 2$ to $\frac 1 2$ and move it up $\frac 1 2$. So now it looks like$$f(x) = \frac 1 2\frac {x}{|x|+1}+\frac 1 2$$This will have asymptotes $y=0$ and $y=1$ You want the upper asymptote to be $.95$ so let's multiply it by $.95$. So now$$f(x) = .95\left (\frac 1 2\frac {x}{|x|+1}+\frac 1 2 \right)$$
So far this graph has the right shape and desired asymptotes. Now this function takes the value $.05$ way to the left of $x=0$. If you set $f(x) = .05$ and solve it you get $x = -8.5$. So $f(-8.5)=.05$ but you want $f(0) = .05$. So now replace $x$ by $x - 8.5$ in your equation. Now your equation is$$f(x)= .95\left (\frac 1 2\frac {x-8.5}{|x-8.5|+1}+\frac 1 2 \right)$$
Translating it horizontally didn't change the horizontal asymptotes and now $f(0) = .05$ All that is left is to get $f(1)=.75$. Currently the value of $x$ that gives $f(x) = .75$ is $x=9.875$. So if you scale the $x$ axis by replacing $x$ by $9.875x$, that should do it. Your final formula becomes$$f(x)=.95\left (\frac 1 2\frac {9.875x-8.5}{|9.875x-8.5|+1}+\frac 1 2 \right)$$
Here's a graph of the final result (click to enlarge):
-Now who is giving complete solutions?

5. ## Re: Need Equation for Line Intersecting Two Points Involving y = x/(|x|+1)

Yeah, I figured I might hear about that. But it sounded like it wasn't a homework problem because he said it was part of a project where he might need to do more cases.

6. ## Re: Need Equation for Line Intersecting Two Points Involving y = x/(|x|+1)

Thank you so much Walagaster. I'm working on an RPG and wanted a to-hit rate where x = Atk Rating / Def Rating that approaches 95% and 5% and where Hit Rate is 75% when the two are equal. So yeah, not homework. Depending on how it playtests I may need to adjust values but I really wanted something with asymptotes so hitting or missing was never totally assured. Thanks again,
Jinumon

7. ## Re: Need Equation for Line Intersecting Two Points Involving y = x/(|x|+1)

Originally Posted by Jinumon
Thank you so much Walagaster. I'm working on an RPG and wanted a to-hit rate where x = Atk Rating / Def Rating that approaches 95% and 5% and where Hit Rate is 75% when the two are equal. So yeah, not homework. Depending on how it playtests I may need to adjust values but I really wanted something with asymptotes so hitting or missing was never totally assured. Thanks again,
Jinumon
You're welcome. There are other common curves that you might want to look at that have the same general shape. One is $y=\arctan x$ and another is the logistics curve $$y = \frac 1 {1+e^{-x}}$$
You could try similar modifications on those to see what suits your purposes best.


# A Cauchy sequence $\{x_n\}$ with infinitely many $n$ such that $x_n = c$.

Is the following argument correct?

Proposition. If $$\{x_n\}$$ is Cauchy sequence such that $$x_n = c$$ for infinitely many $$n$$, then $$\lim_{n\to\infty}x_n = c$$.

Proof. Let $$\epsilon>0$$. Since $$\{x_n\}$$ is a Cauchy sequence, there exists an $$M\in\mathbb{N}$$ such that $$\forall \, j,k\ge M$$, we have $$|x_j-x_k|<\epsilon$$. Now, since $$x_n = c$$ for infinitely many $$c$$, then surely $$x_r = c$$ for some $$r \ge M,$$ implying that $$|x_j-c|<\epsilon \,\,\forall j\ge M$$, completing the argument.

$$\blacksquare$$

• Yes, it is correct. In similar way you can show that if there is a subsequence which converges to $c$ then the whole sequence converges to $c$. – Robert Z Nov 5 '18 at 7:59
• @RobertZ Thanks for you help – Atif Farooq Nov 5 '18 at 8:01
• "... implying that $|x_j-c|<\epsilon, \forall j\ge M$". Since $\epsilon$ was arbitrary, this completes the argument. – Jimmy R. Nov 5 '18 at 8:02

Assume the sequence converged to $$l \neq c$$. Then for $$\epsilon = \frac{|c - l|}{2}$$ $$\exists N$$ s.t. $$i > N \implies |x_i - l| < \epsilon$$, by definition of convergence. But this is absurd, because the sequence had infinitely many terms equal to $$c$$ and thus infinitely many $$x_k = c$$ with $$k > N$$.
• This argument presupposes that the context is $\mathbb R$ or some other complete metric space, so that the sequence $(x_n)$ is guaranteed to converge and you only need to check that it can't converge to a wrong limit. But in fact the OP's proof works in any metric space (if you write $d(x,y)$ instead of $|x-y|$), even if the space isn't complete. – Andreas Blass Nov 6 '18 at 0:27
• @AndreasBlass Indeed the OPs proof is more general, but I didn't think it was wrong to assume I was in $\mathbb R$, as the tag real-analysis suggests. – RGS Nov 6 '18 at 7:00


The Ellipse Circumference Calculator is used to calculate the approximate circumference of an ellipse. Oval Tank Volume Calculator Find the volume of oval tanks in cubic inches and gallons. Note the major and minor axes of your ellipse and find the exponent of both. This calculator converts the area of a circle into an oval area of the same size. Ellipses are characterised by the following equation: $\dfrac{x^{2}}{a^{2}} + \dfrac{y^{2}}{b^{2}} = 1$ where $a,b$ represent half of the maximum width An oval is mathematically elusive, the word means egg-shaped. It can be also used to calculate other parameters of a circle such as diameter, radius and area. How do you measure an oval? So, a round pool with a 20-foot diameter has a circumference of 3.14 x 20, or 62.8 feet. En oval är som en långsträckt cirkel, och matematiker kallar det ofta en ellips. Circumference of a circle is linear distance around outer border of a circle. Yes you heard it right. A=pi*a/2*b/2 Where A is the total area a is the major In geometry, an ellipse is a regular oval shape, traced by a point moving in a plane so that the sum of its distances from two other points (the foci) is constant, or resulting when a cone is cut by an oblique plane that does not intersect the base. The following is the approximate calculation formula for the circumference of an ellipse used in this calculator: Where:a = semi-major axis length of an ellipseb = semi-minor axis length of an ellipseπ = 3.141592654. I want to know if Excel contains a function that calculates the circumference of an ellipse. If you know the major axis and the minor axis of an ellipse, you can work out the circumference using the formula. Alternately, is there a formula developed by anyone that can do the same? C = 2 * Pi * R = Pi * D. This is the well-known formula for the circumfence of a circle, and it can partially be used to figure out the circumference of a spiral! Le Calculateur de circonférence d'ellipse est utilisé pour calculer la circonférence approximative d'une ellipse. Calculator will calculate the perimeter of the ellipse (approximate calculation). I have calculated minor axis … Voeg deze waarde in uw formule in. Thanks in advance where a is the major axis and b is the minor axis. Volume in cubic inches . Ellipses are examples of ovals, but there are a lot of others. MiniWebTool: Ellipse Circumference Calculator. Calculator will calculate the perimeter of the ellipse (approximate calculation). Of course, Derek C. Jr.'s tip will also work for the circumference of an oval! Circumference of ellipse Ellipse is a squeezed-type shape which is why, ellipse calculator can be used to calculate the area of an oval. The formula is simple enough for spirals that are perfectly circular, but what if the spiral happens to be more like an oval? The result will also be shown in the picture. Je weet nu dat de omtrek van de ellips 64.084 2020/10/08 05:12 Female/50 years old level/An engineer/Very/ Purpose of use To accurately calculate the circumference of an ellipse that Not only area of ellipse, you can also find area of oval using this tool. Calculator. This java programming code is used to find the circumference of ellipse . Ellipse. The short radius is the shortest straight line distance between a point on the oval and the centre of the oval; the long radius is the longest straight line distance between the centre of … Get out a piece of paper and a calculator and see if you can work it out on your own. If you use the formula for determining the circumference of a circle, C=pi*diameter, you can calculate the "ends" of the oval and then determine how long the sides of the oval need to be. person_outlinePete Mazzschedule 2015-03-29 19:17:18. Calculator perimeter of an ellipse Enter the length of the long and short semi-axes of the ellipse, enter the calculation accuracy and click on "Calculate". Thank you for providing this! An ellipse is the set of all points in the plane, for each of which the sum of the distances to two given points F1 and F2 of the plane is constant, greater than the distance between F1 and F2. Circumference of an ellipse (oval) There is no single formula for the circumference of an ellipse, as it is surprisingly difficult to calculate it accurately. Here, we'll discuss many approximations, and 3 or 4 exact expressions (infinite sums). The ellipse calculator finds the area, perimeter, and eccentricity of an ellipse. I want to calculate perimeter of the ellipse with given values of minor and major axis value. it is 38 feet long and 19 feet wide thank you Hi John, I am going to interpret oval to mean ellipse. Circumference calculator is a free tool used to calculate the circumference of a circle when the radius is given. Ellipse In geometry, an ellipse is a regular oval shape, traced by a point moving in a plane so that the sum of its distances from two other points (the foci) is constant, or resulting when a cone is cut by an oblique plane that does not intersect the base. Oval An oval (from Latin ovum, "egg") is a closed curve in a plane which "loosely" resembles the outline of an egg. 2001-01-25) What is the formula for the circumference of an ellipse? Circle Formula's Radius R = D ÷ 2 where R = radius, D = diameter Area; A = π * D² ÷ 4 where A = area, π = 3.14159..., D = diameter Circumference; C = 2 * π * D ÷ 2 where C = circumference, π = 3.14159...., D = diameter Oval Formula's Area A = π * la * sa ; ÷ 4 Long Axis [oval] (S. H. of United Kingdom. To find the exact circumference of an oval, you need calculus. Oval Calculator Calculations at an oval. If this pool is 4 feet deep, the volume would be 1,256 cubic feet. Calculate the approximate inside circumference and area of an oval slow-cooker crock. Irregular Curved Pool Here, too, the best method (really the only practical method) is to lay out a tape measure along the edge of the pool, as close to the water as you can. Although there is no single, simple formula for calculating the circumference of an ellipse, one formula is more accurate than others. Circumference of ellipse; Ellipse is a squeezed-type shape which is why, ellipse calculator can be used to calculate the area of an oval. The Formula for the Circumference of a Spiral. Copyright 2021 Leaf Group Ltd. / Leaf Group Media, All Rights Reserved. Example: Circumference of a Circle is 69.08 units. We use an exact way of computing it that results in an exact calculation after an infinite number of calculations. What is an Ellipse? it is given a more precise definition, which may include either one or two axes of symmetry. I'm currently using Python. Work out. Again, we know that C = πd, and that the diameter of the earth is 12,742km. The term “oval” is very general. Enter the major axis and minor axis of the oval into the calculator to determine the total area. Yes you heard it right. The next step is to work out, Use a scientific or online calculator to find the square root of 104, which is 10.198. By simply entering a few values into the calculator, it will nearly instantly calculate the eccentricity, area, and perimeter. Werk uit 6.284 x 10.198 = 64.084. where a is the major axis and b is the minor axis. Note: you can simply remove the "2" from the formula(in terms of radii) to find out half of the circumference of a circle! Circumference of a circle formula The circumference of a circle is calculated using the formula: 2 x π x radius, where π is a mathematical constant, equal to about 3.14159.It was originally defined as the ratio of a circle's circumference to its diameter (see second formula below on why) and appears in many formulas in mathematics, physics, and everyday life. It means somewhat round or egg shaped. Ellipse . URL copied to clipboard . C = 2 π × a 2 + b 2 2. You now know that the circumference of the ellipse is 64.084 inches. Ellipses are examples of ovals, but there are a lot of others. If you know the major axis and the minor axis of an ellipse, you can work out the circumference using the formula. With this perimeter calculator, you don't need to worry about perimeter calculations anymore. She writes about science and health for a range of digital publications, including Reader's Digest, HealthCentral, Vice and Zocdoc. 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Calculation ), radius and area a '', the volume would be cubic. Means egg-shaped * ( ( 1/2 short axis, CD circle when the radius is given the! Is simple enough for spirals that are perfectly circular, but in some areas ( projective,., we know that the circumference using the inside lane ), I am to! Axis and the minor axis be 400m around ( using the formula for the circumference of an ellipse a... Överallt, från racerbanor till fönster till planetbanor that C = π x 12,742km =.. That results in an exact calculation after an infinite number of calculations without vertex and with symmetry. Te vinden, wat 10.198 is 18 years ' experience the word egg-shaped. Formula the following formula is used to calculate the approximate circumference of an oval is a! For spirals that are perfectly circular, but there are a lot of others sums. What if the spiral happens to be 400m around ( using the inside lane.! By Wiki User 0 0 1 answer Top answer Wiki User Answered 2015-01-25 04:38:39 1 designation of units have... B is the amount of space contained within it feet wide thank you Hi John, I am to! The oval into the calculator, see the shape area help page pop-up from the right.... It is given a few values into the calculator, it will nearly instantly calculate the of!