Horizontal tangent line of a parametric curve

Suppose $x=t^2,y=t^3$ is a parametric curve. Here's a quote from my textbook:

The origin, which corresponds to $t=0$, is a singular point of the parametric curve, because $dx/dt=2t,dy/dt=3t^2$ are both zero when $t=0$.

So far so good.

But then they write:

However, the curve has a horizontal tangent line at the origin, because for all $t\neq 0$: $$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{3}{2}t$$ And thus: $$\lim_{t\to 0^+} \frac{dy}{dx}=\lim_{t\to 0^-} \frac{dy}{dx}=0$$

It looked a little odd for me. Nevertheless, I decided to use the same argument to show that the parametric curve $x=2\cos t - \cos (2t), y=2\sin t - \sin(2t)$ has a horizontal tangent line at $t=0$, that is at $(1,0)$.

However my professor said that this is wrong ("because the derivative is not zero" - indeed, $\frac{dy}{dx}\Big|_{t=0}$ is undefined - "$0/0$").

So who is right? Is the existence of the limit a sufficient condition for the (horizontal) tangent line to exist, as my textbook says, or not? I'm confused.

Thanks.

Short version of the question: can a parametric curve have a horizontal tangent line at a singular point?

I.e. is $\lim_{t\to 0} \frac{dy}{dx}=0$ a sufficient condition for a horizontal line to exist (at $t=0$)? (even if the derivative $\frac{dy}{dx}\Big |_{t=0}$ itself doesn't exist).

• Reading your comments to the answers below, I think you are unduly concerned about the distinction between $f(x_0)$ and $\lim_{x\to x_0}f(x)$. Where we have an isolated point $x_0$ for which $f(x_0)$ does not exist, but $\lim_{x\to x_0}f(x)$ does exist, it is common for people to assume without comment that we define $f(x_0)$ as the limit. – almagest May 3 '16 at 18:31
• what is your textbook ? – KonKan May 3 '16 at 18:44
• @KonKan - Calculus by Anton (5th ed.) – user239753 May 3 '16 at 18:53
• @almagest - unduly? This might be an obvious assumption for some, but not for me. The very fact that my professor dismissed it says that this is not trivial, and at least deserves a mention. – user239753 May 3 '16 at 19:02
• @user239753: i'm a bit confused: I'm searching in your book (it happens to have a copy of the 10th edition, available right in front of me) but i cannot find the quote you are mentioning. I am actually looking at Ch.10, par.10.1, p.692-705. The example you are mentioning is example 6, p.697. However, the authors simply mention that $t=0$ is a singular point. I cannot find the limit computation you are mentioning. Can you indicate the exact page ? – KonKan May 3 '16 at 19:21

Notice that both your curves are algebraic curves, of equations $x^3 - y^2 = 0$ and $(x^2+y^2-1)^2-4((x-1)^2+y^2)=0$ respectively.

We say that the line $ax + by + c = 0$ is tangent at the point $(x_0, y_0)$ to the curve given by $F(x,y) = 0$ if and only if $(x_0, y_0)$ is a multiple root of the system $\begin{cases} F(x,y) = 0 \\ ax + by + c = 0 \end{cases}$ (this is the definition that was originally used by algebraic geometers).

In your case, the points are $(0,0)$ in the first case and $(1, 0)$ in the second, and the line to be checked is $y=0$.

Plugging $y = 0$ into $x^3 - y^2 = 0$ gives $x^3 = 0$, which indeed has $x=0$ as a triple root, therefore $y=0$ is tangent to $x^3 - y^2 = 0$ at $(0,0)$.

Plugging $y = 0$ into $(x^2+y^2-1)^2-4((x-1)^2+y^2)=0$ gives $(x^2-1)^2 - 4(x-1)^2 = 0$, or equivalently $(x-1)^2 ((x+1)^2 - 4) = 0$, or again $(x-1)^3 (x+3) = 0$, which indeed has $x=1$ as a triple root (the root $-3$ being simple, meaning that at the point $(-3, 0)$ the line $y=0$ intersects the curve as a secant), therefore $y=0$ is tangent to $(x^2+y^2-1)^2-4((x-1)^2+y^2)=0$ at $(1,0)$.

Since the bounty giver is still unsure, let us follow Wikipedia and compute the intersection number in the first case. Wikipedia gives several methods, one of them is to consider the ideal generated by $y^2 - x^3$ and by $y$ in $\Bbb R[[x,y]]$ (the ring of formal series in $x$ and $y$) and compute the dimension of the quotient vector space $\Bbb R [[x,y]] / (x^3-y^2, y)$.

In the quotient space, both $y$ and $x^3 - y^2$ will become $0$, which implies that $x^3$ also becomes $0$. Therefore, the only powers of $x$ and $y$ that survive in the quotient set are $x^0 = y^0 = 1$, and $x$ and $x^2$ - a total of $3$ linearly independent powers, so $3$ will be the intersection number of $y=0$ and $x^3-y^2$ at $(0,0)$. Since everything $\ge 2$ means tangency, this shows that the line $y=0$ is tangent to $x^3-y^2$ at $(0,0)$.

The same thing could be done with what Wikipedia denotes by $I_{(0,0)}$, letting $P = y$ and $Q = x^3 - y^2$. Applying the properties that you see on that page (the numbers above equal signs are the Wikipedia properties that I apply),

$$I_{(0,0)} (y, x^3-y^2) \overset 6 = I_{(0,0)} (y, x^3) \overset 5 = I_{(0,0)}(x,y) + I_{(0,0)}(x,y) + I_{(0,0)}(x,y) \overset 4 = 1 + 1 + 1 = 3 .$$

The same computations could be done for the second example, it is just that they are more tedious. First, let us translate the curve such that the cusp moves from $(1,0)$ to $(0,0)$ (because, for simplicity, Wikipedia's formulae are given only for $(0,0)$). To do this we shall make the change of variable $x = u + 1$, which will lead to $(u^2 +2u +y^2)^2 -4(u^2+y^2)$. Next,

$$I_{(0,0)} ((u^2 +2u +y^2)^2 -4(u^2+y^2), y) \overset 6 = I_{(0,0)} (u^4 + 4u^3, y) = I_{(0,0)} (u^3 (u+4), y) \overset 5 = I_{(0,0)} (u^3, y) + \\ I_{(0,0)} (u+4, y) = I_{(0,0)} (u, y) + I_{(0,0)} (u, y) + I_{(0,0)} (u, y) + I_{(0,0)} (u+4, y) \overset {3, 4} = 1+1+1+0 = 3 .$$

Again, we obtain an intersection number $\ge 2$, which means tangency.

What others have correctly stated is that the two plane curves given in the problem are not differentiable submanifolds of $\Bbb R^2$ at the singular points under discussion, therefore they do not fit into the framework of diferential geometry, therefore we may not speak of their tangent spaces at the singular points as defined in differential geometry (a proof that an almost identical curve cannot be a smooth submanifold can be found here). Counterintuitively, though, they do have tangent lines! There is no contradiction, the thing is that we use two meanings for "being tangent" that are not synonymous: one comes from differential geometry ("tangent space"), and it doesn't apply here, the other from algebraic geometry ("intersection number"), and it does apply here. In the latter framework, we may speak about "a given line being tangent (or not) to a curve in a point" without speaking of "the tangent space at that curve in that point".

$y=0$ is tangent to those curves at the specified points. Those curves do not have nicely defined tangent spaces at those points.

• @user239753: Try these two: "Elementary Differential Geometry" by A.N. Pressley, and "Differential Geometry of Curves and Surfaces" by M.P. do Carmo. One last word: differential geometry in general is interested only in those spaces that are endowed with regular parametrizations (because one wants them to be invertible, with their inverses called "local charts"). Most theorems about parametrized curves assume the parametrization to be regular. – Alex M. Apr 28 '16 at 20:24
• @AlexM. Because this is all about subtile nuances: You wrote, the cardiod would be a manifold, because the defining function $F$ is smooth. Aren't you applying the regular value theorem (which is a close relative to the implicite function theorem) here, which means 0 needs to be a regular value for $F$? (Anyway: great answer!) – hase_olaf Apr 28 '16 at 21:31
• @AlexM. - thank you for your refined answer. A bit stupid question - what does "multiple root" mean? It has something to do with algebraic multiplicity? – user239753 May 3 '16 at 19:27
• @user239753: Yes, exactly. We say that $a$ is a root of order $k$ of $P$ (where $P$ is a polynomial) if and only if $(x-a)^k \mid P$ and $(x-a)^{k+1} \nmid P$. The analogue for algebraic curves is the concept of "intersection multiplicity". – Alex M. May 3 '16 at 19:41
• @AlexM: I have to confess .. I am the ignorant who downvoted your hard work ;) but give me a moment to explain. your answer (especially after your edit) seems valid in my eyes and you've obviously put hard work in it ! I give you credit for that. However, I feel it falls outside the spirit and the OP's question: it is a delicate question on a subtle point of parametric differentiation (from the calculus viewpoint) while you certainly view the situation from the algebraic/differential geometric eye. This is also why i rolled back your retagging. – KonKan May 5 '16 at 23:29

In order to find the tangent at some special point you should not try to compute the limit of $y'$ but the actual limit of secant directions at that point. Therefore in your first example you get $$m_+:=\lim_{t\to0+}{y(t)-y(0)\over x(t)-x(0)}=\lim_{t\to0+}{t^3\over t^2}=0\ ,$$ and similarly $m_-=0$. In your second example you have $${y(t)-y(0)\over x(t)-x(0)}={2\sin t-\sin(2t)-0\over 2\cos t-\cos(2t)-1}={2\sin t(1-\cos t)\over2\cos t(1-\cos t)}=\tan t\qquad(t\ne0)\ ,$$ and therefore $$m_+=m_-=\lim_{t\to0}{y(t)-y(0)\over x(t)-x(0)}=0\ .$$

You are referring to two different things here..

For the semi-cubical parabola curve $y=x^{3/2}$ there is horizontal tangent at the cusp y=0 or t=0.

For cardoid, depending on the cusp contact point chosen, slope of cusp tangent varies, look at all cusps of epicycloids:

EDIT 1:

Since it is passing through infinite curvature which is rate of change of tangent inclination, the slope is changing very fast and undefined.

It can be also seen at the point $(1,0)$ Changing the multiple angle

$$x=2\cos t - \cos (n t), y=2\sin t - \sin( n t)$$

has the effect of influencing curvature. It changes from (positive for circle) to negative for looped case, via the present central cardoid case of infinite curvature.

The direction of tangent is vertical, indeterminate and vertical respectively

• In both cases $\lim_{t\to 0} \frac{dy}{dx}=0$ yet $\frac{dy}{dx}\Big |_{t=0}$ is undefined. The question is whether in such a situation a horizontal tangent line exists. The textbook says it does, yet my professor claims that for a horizontal tangent line to exist the derivative must EXIST and be equal to zero, that is we must have $\frac{dy}{dx}\Big |_{t=0}=0$. – user239753 Apr 25 '16 at 6:53
• please read my comment again. I'm talking about the value of the derivative which does NOT exist. There is a difference between the limit (which indeed can be found with L'Hospital) and the actual value. – user239753 Apr 25 '16 at 7:28
• Note that you should not stop hunting for the L'Hospital limit if the first order evaluation fails. Do not conclude that it is undefined. At the cusp we are dealing with second orders and so limit of $f''(t)/g^{''}(t)$ should be next evaluated. – Narasimham Apr 25 '16 at 7:31
• unfortunately, you do not understand me. Here $dy/dx \neq 0$ even though $\lim_{t\to 0} \frac{dy}{dx}=0$. Do you know there is a difference between limit and value of the function? Take $$f(x)=\left\{\begin{matrix} 0 \text{ if } x\neq0\\ 1 \text{ if } x=0 \end{matrix}\right.$$. Here $$\lim_{x\to0}f(x)=0$$ yet $f(0)\neq 0$. The same goes here. $\lim_{t\to 0} \frac{dy}{dx}=0$ yet $\frac{dy}{dx} \Big |_{t=0}\neq 0$ ! Do you understand? – user239753 Apr 25 '16 at 7:37
• Ok, we resume after sometime,else moderator may indicate chat room. – Narasimham Apr 25 '16 at 8:34

the formula $$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$$ is valid for computing the derivative at points where $\frac{dx}{dt}\neq 0$. At singular points, i.e. points at which $$\frac{dy}{dt}=\frac{dx}{dt}=0$$ the above formula is not valid. However, this does not imply that the derivative does not exist; only that it is not computable by the above formula.

In such cases, a straightforward application of the definition of derivative (as already done by user Christian Blatter in his answer) can provide the answer: $$\frac{dy}{dx}\bigg|_{(0,0)}=\lim_{t\to 0^+}\frac{y(t)-y(0)}{x(t)-x(0)}=\lim_{t\to 0^-}\frac{y(t)-y(0)}{x(t)-x(0)}=\lim_{x\to0^\pm}\frac{t^3}{t^2}=0$$ which shows that the derivative exists at $(0,0)$ and so does the tangent line at $(0,0)$, which is no other than the horizontal axis itself.

P.S.1: Note that, when confined to each branch seperately: $$y(x_0)=y(x(0))=y(0)=0, \ \ x_0=x(0)=0 \\ y(x)=y(x(t))=y(t)=t^3, \ \ x=x(t)=t^2$$ So, on each one of the functions $y(x)=x^{3/2}$, $y(x)=−x^{3/2}$ i.e. on each one of the branches of the graph below, the following rates of change are identified: $$\frac{y\big(x(t)\big)−y(x_0)}{x(t)−x_0}=\frac{y(t)−y(0)}{x(t)−x(0)}$$ On the other hand, since $x=t^2\neq 0$ when $t\neq 0$ it should be clear that $$t\to 0\Leftrightarrow x\to 0$$ At this point, it remains to invoke a classical proposition on the limit of a composite function, claiming that:

Let the real functions $f,g$, for which $lim_{x\rightarrow x_0}g(x)=g_0\in \mathbb{R}$ and $lim_{g\rightarrow g_0}f(g)=l\in \mathbb{R}$.
If, furthermore $g(x)\neq g_0$ "close to $x_0$" (i.e. in some interval $(a,x_0))$, then $$\lim_{x\rightarrow x_0}f\big(g(x)\big)=\lim_{g\rightarrow g_0}f(g)=l$$

Seting $f(x):=\frac{y(x)−y(0)}{x−0}$ and $g(x):=x(t)$ and applying the proposition we get that: $$\lim_{t\to 0}\frac{y(t)−y(0)}{x(t)−x(0)}\equiv \lim_{t\to 0}\frac{y\big(x(t)\big)−y(0)}{x(t)−x(0)}=\lim_{x\to 0}\frac{y(x)−y(0)}{x−0}=\frac{dy}{dx}\bigg|_0$$ (where the far left term is computed using side limits: for the upper branch $t>0$ thus $t\to0\Leftrightarrow t\to0^+$ while for the lower branch $t<0$ thus $t\to0\Leftrightarrow t\to0^-$).

P.S.2: Regarding your final question: I do not believe that the existence of the limit is a sufficient condition, because generally, a differentiable function at $x_0$ need not have a continuous derivative at $x_0$: In other words, there may well be a situation at which: $\lim_{x\to x_0}f'(x)\neq f'(x_0)$.

However, this is not the case for the semicubical parabola, as we can see from its graph

which is probably what the author of your textbook had in his mind. He computes the limit of the derivative just as if we can imagine the tangent "slipping" accross either of the branches to "continuously" become horizontal. In my opinion, he implicitly uses the assumption that the derivative of the semicubical parabola is a continuous function; and he probably does that, based on the shape of the graph.

• Intuitively I understand. But in the definition of the derivative we use $\frac{y(x)-y(x_0)}{x-x_0}$ and not $\frac{y(t)-y(0)}{x(t)-x(0)}$. That is, we must find $y(x)$ first. Otherwise, can you prove rigorously that it works with $t$? – user239753 May 4 '16 at 7:11
• Please read carefully: There is nothing intuitive here. (Only the idea described in your textbook seems intuitive). When confined to each branch seperately: $$y(x_0)=y(x(0))=y(0), \ \ y(x)=y(x(t))=y(t) \\ x_0=x(0), \ \ x=x(t)$$ So, on each one of the functions $y=x^{3/2}$, $y=-x^{3/2}$ i.e. on each of the branches of the above graph, the rates of change you are writing are identified: $$\frac{y(x)-y(x_0)}{x-x_0}=\frac{y(t)-y(0)}{x(t)-x(0)}$$ On the other hand, since $x=t^2$ it should be clear that $$t\to0\Leftrightarrow x\to0$$ – KonKan May 4 '16 at 12:20
• i've added the above comment to the post – KonKan May 4 '16 at 12:51
• It should be $y(x(t))=(t^2)^3=t^6$, no? (I know that in this case it doesn't change the limit, but still). Thank you. P.S. - Also, I don't understand why somebody downvotes the answers here. – user239753 May 4 '16 at 14:28
• actually, for $t>0$: $$y(x(t))=\pm (x(t))^{3/2}=\pm (t^2)^{3/2}=\pm t^3$$ So, for $t\in (-\infty, +\infty)$: $$y(t)=t^3$$ – KonKan May 4 '16 at 19:14


Six dice, six different figures (model problem)

I am thinking about the same problem as posted here: Probability of All Distinct Faces When Six Dice Are Rolled.

If one rolls six dice, what is the probability of attaining six different figures. I do understand the solution but wondered why it is necessary to assume that the dice are distinguishable. If I assume they are not then I have exactly one desired result and

$$\binom{6 + 6 - 1}{6} = \binom{11}{6} = 462$$

possible results.

But I get a different probability in this case which leads me to the assumption that it is not working this way. But why exactly is that?

You have used a correct Stars and Bars argument to show that there are $462$ ways to distribute $6$ objects in $6$ boxes.

However, if we assume that the dice are fair, and do not influence each other, then these $462$ possibilities are not all equally likely.

Let us look at a much smaller example, two identical coins. There are $3$ ways to distribute these into $2$ boxes. However, in this case it is fairly clear that the probability of two heads is $\frac{1}{4}$ and not $\frac{1}{3}$. For whether the coins are distinguishable or not, it should make no difference to the probability if we toss the coins sequentially and not simultaneously. And sequential tossing tells us the probability is $\frac{1}{2}\cdot\frac{1}{2}$.

Roughly speaking, an extreme Stars and Bars case, such as all balls in the third box, has smaller probability than any specific more "even" distribution.

Remark: To see more informally that a probability model based on distinguishable dice is appropriate, assume that otherwise indistinguishable dice are made different by writing IDs on them with invisible ink. The writing should not affect the probability of events that do not involve the IDs, such as all six numbers being obtained.

There are $6!$ combinations of six different numbers, and there are $6^6$ combinations of six rolls. The probability is then: $$p = \frac{6!}{6^6}$$

• Yes, I know that. I was wondering why my way of thinking is wrong. Why can't I just record which numbers came up whithout recording from which dice it came? – Cyianor Jun 2 '14 at 14:03
• My guess would be you need to account for the fact all orders are the same, i.e. $123456$ is the same as $654321$ – Alex Jun 2 '14 at 14:10
• You mean I might not be able to apply Laplacian probability, i.e. p = good cases / total cases? – Cyianor Jun 2 '14 at 14:12
• Yes you are: here $6!$ is 'good cases' and $6^6$ are 'total cases' – Alex Jun 2 '14 at 14:13
• I meant in the case of my description above. If I say the dice are indistinguishable then the elements of my probability space have different probabilities and Laplacian probability is not applicable – Cyianor Jun 2 '14 at 14:17


# What is the least amount of straight cuts that should be made to get squares from a rectangle?

The problem is as follows:

Rachel has a rectangular piece of cloth which measures $$2$$ meters large and $$0.2$$ meters wide. She is to use a guillotin which can only makes cuts of $$60$$ cm of maximum in length. Another constrain in this guillotin is that it can only cut one layer of this cloth. Using this information, how many straight cuts minimum can Rachel make in order to get from such fabric $$10$$ square pieces of $$20\,cm$$ from each side?.

The alternatives given in my book are as follows:

$$\begin{array}{ll} 1.&\textrm{3 cuts}\\ 2.&\textrm{1 cut}\\ 3.&\textrm{2 cuts}\\ 4.&\textrm{4 cuts}\\ \end{array}$$

I'm not sure how to solve this question regarding cuts. But I believe that a way to reduce the number of cuts will require to fold the piece. But in this case this seems not to be possible because there is a condition in the problem which is forbidding to do such approach.

My initial approach was to make the necessary cuts to make some sort of grid which could I guess reduce the number of cuts but 10 is not a perfect square. Therefore what sort of logic should be here?.

I've went in circles for long on this question so I will really appreciate someone could give me a hand with this.

Therefore what should be the way to go here?. Can someone help me here please?. I'm confused.

• I think the question allows folding, but a guillotin cut cannot be through multiple layers. – peterwhy Mar 3 at 2:28
• I brute forced a 4 cut solution. If this is the answer, there must be a proof as to why this is the minimum required. – Andrew Chin Mar 3 at 2:33

To cut the strip of cloth, it only takes 9 cuts of 20 cm, and each cut of the guillotin cuts through 60 cm at most, so at least 3 cuts are necessary.

Number the cuts $$1$$ to $$9$$ sequentially, and the resultant squares $$0$$ to $$9$$.

Align cut $$1$$ under the guillotin, then fold two right angled triangles on the strip around cuts $$2$$ and $$3$$ so that the strip goes back under the guillotin, and align cut $$4$$ under the guillotin next to cut $$1$$.

(Fold twice like (B) in this random image found in Google: )

Do the same thing to align cut $$7$$ under the guillotin. With cut $$1, 4, 7$$ all aligned side by side, one guillotin cut of 60 cm would do all the 3 20-cm cuts.

Now squares 1-3, squares 4-6, squares 7-9 are on separated pieces of cloth, and they can be aligned freely to make the remaining 6 cloth-cuts in 2 guillotin cuts.

The optimal answer is 3 guillotin cuts.


# How can I find integers which satisfy $\frac{150+n}{15+n}=m$?

Here are some facts about myself:

1. In 2017, I was $$15$$ years old.
2. Canada, my country, was $$150$$ years old.

When will be the next time that my country's age will be a multiple of mine?

I've toned this down to a function. With $$n$$ being the number of years before this will happen and $$m$$ being any integer,

$$\frac{150+n}{15+n}=m$$

How would you find $$n$$?

• For extra credit, consider that Canada turned 150 on July 1, 2017. If your birthday is after that date, 14+n is also a valid denominator, as you will be both age 14 and age 15 during the 12 months that Canada is 150 years old. Likewise, if your 15th birthday is before that date, 16+n is a valid denominator, since you will be 15 when Canada turns 150, but turn 16 before Canada turns 151. – Nuclear Wang Dec 12 '18 at 4:51
• (1) Avoid using "interesting" in the title; (2) describe the problem in the title, not just your opinion and its topic. – Asaf Karagila Dec 12 '18 at 11:19

You want $$\frac{150+n}{15+n}=m$$, and clearing denominators gives us $$150+n=(15+n)m.$$ Subtracting $$15+n$$ from both sides give us $$135=(15+n)(m-1).$$ Now you are looking for the smallest $$n>1$$ for which such an $$m$$ exists, so the smallest $$n>1$$ for which $$15+n$$ divides $$135$$.

• Love it, thank you! Is 120 the only $n$ that exists for such a case? (Because $135\over270$ $=$ $1\over2$)? – Raymo111 Dec 11 '18 at 23:21
• What's wrong with $n=12$? The other solutions are $n=30$ and $n=120$. – Servaes Dec 11 '18 at 23:46
• Oh shoot, I mistakenly read your answer as $n+15=135$, my bad! Also, upvote if you think this question that I thought of was interesting, it's a real-life example of an age problem. – Raymo111 Dec 11 '18 at 23:48

First thing I would do is say that Canada is $$135$$ years older than you.

That gives you a simpler

$$\frac {135+n}{n} = k\\ \frac {135}{n} = k-1\\$$

It will happen every time your age is a factor of $$135.$$ It last happened when you were $$15.$$ It will next happen when you are $$27$$

• Do Canadians live past age 135? – richard1941 Dec 21 '18 at 16:51
• Almost never. But with advancements in medical technology, that may change. – Doug M Dec 21 '18 at 17:00

We want the smallest positive integer $$n$$ such that there is some (positive) integer $$k$$ such that $$\frac{150+n}{15+n}=k.$$ Note that $$k=1$$ can never work, so we can assume $$k-1\neq0.$$ Now we rearrange the above equation: multiplying both sides by $$15+n,$$ we get $$150+n=15k+nk;$$ now rearrange and factorize to get $$15(10-k)=(k-1)n;$$ and now divide both sides by $$k-1,$$ to get $$n=\frac{15(10-k)}{k-1}.$$ Since we want the smallest positive integer $$n,$$ we can just try values of $$k\in\{2,3,\ldots,9\},$$ starting from the largest and working our way down (because the function of $$k$$ on the right-hand side is decreasing in this range), until we arrive at an integer value of $$n.$$* When $$k=9,$$ $$8$$ or $$7$$ we get non-integer values of $$n;$$ when $$k=6$$ we find $$n=15\times4/5=12.$$

So last year, $$n$$ was $$0,$$ and the ratio of Canada's age to your age was $$k=150/15=10;$$ and $$11$$ years from now, $$n$$ will be $$12,$$ and the ratio of Canada's age to your age will be $$k=162/27=6.$$

*Incidentally, it is not obvious in advance that we will ever get an integer value of $$n;$$ if this were the case then the problem would simply have no answer. As it happens, the problem does have answers, namely $$(n,k)\in\{(0,10),(12,6),(30,4),(120,2)\}.$$

Alternatively:

$$\frac {150 + n}{15 + n} = \frac {150+ 10n}{15+n} +\frac {-9 n}{15+n}$$

$$=10 -\frac {9 n}{15+n}$$ (which is an integer for $$n=0$$ but when next?)

$$=10 - \frac {9n + 9*15}{15+n} + \frac {9*15}{15+n}=$$

$$=10 - 9 + \frac{3^3*5}{15+n}= 1 + \frac{3^3*5}{15+n}$$

which is an integer if $$15+n$$ is one of the factors of $$3^3*5$$.

And the factors of $$3^3*5$$ are $$1, 3,9, 27, 5,15, 45, 135$$.

So this will occur when $$n = -14,-12, -10, -6,0, 12,30, 120$$

When you are $$1, 3, 5, 9, 15, 27, 45, 135$$ and canada is $$136, 138, 140, 144, 150, 162, 180, 270$$ and canada is exactly $$136,46, 28, 16, 10,6,4, 2$$ as old as you are.

(Enjoy your $$45$$ birthday when your country annexes my country after we collapse from the thirty year aftermath of the unrecoverable mistakes of the last two years.)

That's a fun problem. It's nice to see other people like to think about these things.

• That's nice, that's the way I usually think about these problems too; but you can shorten the process by writing $\dfrac {150+n}{15+n} = \dfrac {15+n}{15+n} + \dfrac{135}{15+n}$ – Ovi Dec 12 '18 at 2:34
• Hmmm.... I'm not sure why I did it the way I did. I think somehow I briefly thought the obviousness that 15 goes directly into 150 made me briefly think that they were the coeficients and I started doing it that way and just continued. Obviously it'd be shorter and direct to do the coefficients. – fleablood Dec 12 '18 at 15:40

Note that if $$k \mid a$$ and $$k \mid b$$, then $$k \mid a - b$$. In this particular instance, we have $$15 + n \mid 15 + n$$ and $$15 + n \mid 150 + n$$, so $$15 + n \mid 135$$. In other words, we are looking for $$n+15$$ to be the next factor of $$135$$ which is larger than $$15$$. Can you continue from here?

Servaes answer is excellent but let me add few cents for those who can't really make out something from dual variables and other factors (I know some people have problem with such things).

There is also a brute force solution, requiring a bit of thinking and a bit of simple calculation (the larger is the $$n$$ i $$n$$-times larger the more difficult it becomes though so I really suggest you try understanding the accepted solution anyway).

It's enough to check if there is an integer solution to equation $$\frac{150+n}{15+n}=9$$ If no, proceed with 8, 7 and so on. If you get to 2 and still have no anwser then the answer is never.

So a quick brute force check will like that: $$\frac{150+n}{15+n}=9$$ $$150+n=9\cdot(15+n)$$ $$150+n=135+9\cdot n$$ $$15=8\cdot n$$ $$n = \frac{15}{8}\notin\mathbb{Z}$$ So we continue with $$8$$ $$\frac{150+n}{15+n}=8$$ $$150+n=8\cdot(15+n)$$ $$150+n=120+8\cdot n$$ $$30=7\cdot n$$ $$n = \frac{30}{7}\notin\mathbb{Z}$$ You may continue from here. Note, a smart person using this method will notice some pattern that will make it even simpler to test (not requiring all the calculation, just a quick check that literally takes seconds) but I will not give a direct hint.

## Explanation

When you add the same number to both numerator and denominator of a fraction, the fraction decreases. In other words if you consider how the fraction of age of Canada and your age changes over time it will decrease.

On the other hand the fraction can never go to $$1$$ (or below) since the numerator will always be obviously greater than denominator (in our case by 135).

In other words you have a finite set of possible resultant fractions (namely $$\{2,3,4,5,6,7,8,9\}$$) that are potentially the sought "next case". From those you're looking for the largest one, so test them one by one, starting with $$9$$ and going down until you have an integer result or your options are gone.

## Remarks

1. Once again - this is a brute force solution and should not be treated as preferred if you can understand using two variables. But if you can't, it's still doable.
2. If you notice the relationship I mention above, I guess even testing situation where the initial fraction is 100 could be done.
3. You may try solving an interesting opposite question - when was the last time when such situation occurred (i.e. Canadas age was a multiply of your age). If you think it over well, you may use both Servaes and my approaches to that as well. Especially try rationalising that my approach will again have a finite and easily predictable number of tests to be performed.
4. A food for thoughts. Is there always a solution to the original question (with differently chosen ages)? Is there always a solution to the problem I state in remark 3?
5. Can you think of other related questions that will use similar approach (either Servaes's elegant one or mine brute force) to find a solution?

In general questions like this, where

$$\frac{x+n}{y+n}=m$$

$$m$$ will be an integer smaller than $$\frac{x}{y}$$. Since there are finitely many integers less than $$\frac{x}{y}$$, each value of $$m$$ can be checked one by one. Not all values will generally give an integer $$n$$.


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# When traveling at a constant speed of 32 miles per hour, a certain mot

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When traveling at a constant speed of 32 miles per hour, a certain mot  [#permalink]

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14 Jun 2016, 13:46
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When traveling at a constant speed of 32 miles per hour, a certain motorboat consumes 24 gallons of fuel per hour. What is the fuel consumption of this boat at this speed measured in miles traveled per gallon of fuel?

A) $$\frac{2}{3}$$

B) $$\frac{3}{4}$$

C) $$\frac{4}{5}$$

D) $$\frac{4}{3}$$

E) $$\frac{3}{2}$$

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Re: When traveling at a constant speed of 32 miles per hour, a certain mot  [#permalink]

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14 Jun 2016, 17:07
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AbdurRakib wrote:
When traveling at a constant speed of 32 miles per hour, a certain motorboat consumes 24 gallons of fuel per hour. What is the fuel consumption of this boat at this speed measured in miles traveled per gallon of fuel?

A) $$\frac{2}{3}$$
B) $$\frac{3}{4}$$
C) $$\frac{4}{5}$$
D) $$\frac{4}{3}$$
E) $$\frac{3}{2}$$

Dear AbdurRakib,
I'm happy to respond.

We have 32 mi/gal, and 24 gal/hr, and we want mile/gal. We need to divide (mi/hr) by (gal/hr) to get (mi/gal). Thus

fuel consumption = (32 mi/gal)/(24 gal/hr) = 32/24 mi/gal

In that fraction, cancel the common factor of 8.

fuel consumption = (32 mi/gal)/(24 gal/hr) = 32/24 mi/gal = 4/3 mi/gal.

Does this make sense?
Mike
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Re: When traveling at a constant speed of 32 miles per hour, a certain mot  [#permalink]

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15 Jun 2016, 00:46
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1
Distance travelled = 32 miles
Fel consumed = 24 gallons
So, 24 gallons lets motorboat travel 32 miles
So, miles traveled per gallon of fuel = 32/24 = 4/3
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Re: When traveling at a constant speed of 32 miles per hour, a certain mot  [#permalink]

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15 Jun 2016, 02:59
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In 1 hour the motorboat travels 32 miles
In travelling that 32 miles the motorboat uses 24 gallons of fuel in 1 hour

so basically we divide $$\frac{miles}{hr}$$ by $$\frac{gallons}{hr}$$ to get $$\frac{miles}{gallons}$$

$$\frac{32}{24}$$ =$$\frac{4}{3}$$ $$\frac{miles}{gallons}$$

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Re: When traveling at a constant speed of 32 miles per hour, a certain mot  [#permalink]

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13 Aug 2016, 11:05
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One way to solve this:--
The speed is 32 miles / hour. The fuel consumption is 24 gallons / hour. So in every hour two things are happening:-
The boat has covered 32 miles
The boat has consumed 24 gallons
What is the miles /gallon --> 32/24 => 4/3miles/gal. The choice is D
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Re: When traveling at a constant speed of 32 miles per hour, a certain mot  [#permalink]

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02 Dec 2016, 06:54
AbdurRakib wrote:
When traveling at a constant speed of 32 miles per hour, a certain motorboat consumes 24 gallons of fuel per hour. What is the fuel consumption of this boat at this speed measured in miles traveled per gallon of fuel?

A) $$\frac{2}{3}$$
B) $$\frac{3}{4}$$
C) $$\frac{4}{5}$$
D) $$\frac{4}{3}$$
E) $$\frac{3}{2}$$

We are given that a motorboat has a rate of 24 gallons per hour while traveling at a speed of 32 miles per hour. We must determine the boat’s fuel consumption measured in miles traveled per gallon of fuel.

Since the motorboat consumes 24 gallons when traveling 32 miles, the rate is 32/24 = 4/3.

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Re: When traveling at a constant speed of 32 miles per hour, a certain mot  [#permalink]

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23 Feb 2018, 12:17
mikemcgarry wrote:
AbdurRakib wrote:
When traveling at a constant speed of 32 miles per hour, a certain motorboat consumes 24 gallons of fuel per hour. What is the fuel consumption of this boat at this speed measured in miles traveled per gallon of fuel?

A) $$\frac{2}{3}$$
B) $$\frac{3}{4}$$
C) $$\frac{4}{5}$$
D) $$\frac{4}{3}$$
E) $$\frac{3}{2}$$

Dear AbdurRakib,
I'm happy to respond.

We have 32 mi/gal, and 24 gal/hr, and we want mile/gal. We need to divide (mi/hr) by (gal/hr) to get (mi/gal). Thus

fuel consumption = (32 mi/gal)/(24 gal/hr) = 32/24 mi/gal

In that fraction, cancel the common factor of 8.

fuel consumption = (32 mi/gal)/(24 gal/hr) = 32/24 mi/gal = 4/3 mi/gal.

Does this make sense?
Mike

hello there mikemcgarry, how are you ? you know what i dont understand, why isnt the answer B ? wouldnt it be logically correct to divide gallons by miles ?
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Re: When traveling at a constant speed of 32 miles per hour, a certain mot  [#permalink]

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03 Apr 2018, 10:01
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AbdurRakib wrote:
When traveling at a constant speed of 32 miles per hour, a certain motorboat consumes 24 gallons of fuel per hour. What is the fuel consumption of this boat at this speed measured in miles traveled per gallon of fuel?

A) $$\frac{2}{3}$$

B) $$\frac{3}{4}$$

C) $$\frac{4}{5}$$

D) $$\frac{4}{3}$$

E) $$\frac{3}{2}$$

Let's see what happens after ONE HOUR of traveling.

In ONE hour, the boat will travel 32 miles and will use 24 gallons of fuel
So, the fuel consumption rate is 32 miles per 24 gallons
Or we can write: fuel consumption rate = 32/24 miles/gallon
32/24 = 4/3
So, the fuel consumption rate = 4/3 miles per gallon

Cheers,
Brent
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Re: When traveling at a constant speed of 32 miles per hour, a certain mot  [#permalink]

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14 Apr 2018, 22:53
AbdurRakib wrote:
When traveling at a constant speed of 32 miles per hour, a certain motorboat consumes 24 gallons of fuel per hour. What is the fuel consumption of this boat at this speed measured in miles traveled per gallon of fuel?

A) $$\frac{2}{3}$$

B) $$\frac{3}{4}$$

C) $$\frac{4}{5}$$

D) $$\frac{4}{3}$$

E) $$\frac{3}{2}$$

It's (D)

$$\frac{Miles - traveled}{gallon-fuel}$$ = $$\frac{32}{24}$$ = $$\frac{4}{3}$$
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Re: When traveling at a constant speed of 32 miles per hour, a certain mot  [#permalink]

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25 Jan 2019, 18:10
mikemcgarry wrote:
AbdurRakib wrote:
When traveling at a constant speed of 32 miles per hour, a certain motorboat consumes 24 gallons of fuel per hour. What is the fuel consumption of this boat at this speed measured in miles traveled per gallon of fuel?

A) $$\frac{2}{3}$$
B) $$\frac{3}{4}$$
C) $$\frac{4}{5}$$
D) $$\frac{4}{3}$$
E) $$\frac{3}{2}$$

Dear AbdurRakib,
I'm happy to respond.

We have 32 mi/gal, and 24 gal/hr, and we want mile/gal. We need to divide (mi/hr) by (gal/hr) to get (mi/gal). Thus

fuel consumption = (32 mi/gal)/(24 gal/hr) = 32/24 mi/gal

In that fraction, cancel the common factor of 8.

fuel consumption = (32 mi/gal)/(24 gal/hr) = 32/24 mi/gal = 4/3 mi/gal.

Does this make sense?
Mike

Was checking through the solutions to see other solutions to this problem, isn't the units in red suppose to be in "hours", instead of gallons?
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Re: When traveling at a constant speed of 32 miles per hour, a certain mot  [#permalink]

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11 May 2019, 05:59
Speed of boat 32M/H and its consumption is 24 Gallon per hour.

Therefore boat consumes 24Gallon for every 32 Miles.
or for every Gallaon boat travels 32/24 miles
i.e 4/3 miles per gallon

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Re: When traveling at a constant speed of 32 miles per hour, a certain mot  [#permalink]

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11 May 2019, 22:55
AbdurRakib wrote:
When traveling at a constant speed of 32 miles per hour, a certain motorboat consumes 24 gallons of fuel per hour. What is the fuel consumption of this boat at this speed measured in miles traveled per gallon of fuel?

A) $$\frac{2}{3}$$

B) $$\frac{3}{4}$$

C) $$\frac{4}{5}$$

D) $$\frac{4}{3}$$

E) $$\frac{3}{2}$$

In 24 gallons boat travel =36
In 1 " " " =36/24=4/3

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Re: When traveling at a constant speed of 32 miles per hour, a certain mot   [#permalink] 11 May 2019, 22:55
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Probability Theory

alakaboom1

New member
An automobile insurance company classifies drivers into 3 classes: class A, class B, and class C. The percentage of drivers in each class is: class A, 20%; class B, 65%; and class C, 15%. The probabilities that a driver in one of these classes will have an accident within one year are given by 0.01, 0.02, and 0.03 respectively. After purchasing an insurance policy, a driver has an accident within the first year. What is the probability that the driver is of class A?

I think that it's 1/6, because you'd do 0.01/0.06, since you already know that the driver has had the accident, but I'm not too sure. One other possibility that I'm considering would be 0.2 * 0.01, but I feel like this doesn't account for the fact that we are assuming that the driver is already in an accident...

galactus

Super Moderator
Staff member
You are looking for the probability the driver is from class A given they have an accident:

$$\displaystyle P(\text{class A}|\text{accident})$$

You can make a chart. Assume a number of drivers that is easy to work with. Say, 1000

Or, use Bayes Theorem. Do you know it?.

soroban

Elite Member
Hello, alakaboom1!

Here is a very primitive approach . . .

An automobile insurance company classifies drivers into 3 classes: class A, class B, and class C.
The percentage of drivers in each class is: class A, 20%; class B, 65%; and class C, 15%.
The probabilities that a driver in one of these classes will have an accident within one year are given by 0.01, 0.02, and 0.03 respectively.
After purchasing an insurance policy, a driver has an accident within the first year.
What is the probability that the driver is of class A?

Suppose there are 10,000 policyholders.

$$\displaystyle \text{20\% are class A drivers: }\:20\%\times 10,\!000 \,=\,2000$$
. . $$\displaystyle \text{1\% of them will have an accident: }\:1\% \times 2000 \,=\,20$$

$$\displaystyle \text{65\% are class B drivers: }\:65\% \times 10,\!000 \,=\,6500$$
. . $$\displaystyle \text{2\% of them will have an accident: }\:2\% \times 6500 \,=\,130$$

$$\displaystyle \text{15\% are class C drivers: }\:15\% \times 10,\!000 \,=\,1500$$
. . $$\displaystyle \text{3\% of them will have an accident: }\:3\% \times 1500 \,=\,45$$

$$\displaystyle \text{There is a total of: }\:20 + 130 + 45 \,=\,195\text{ accidents}$$
. . $$\displaystyle \text{of which 20 are class A drivers.}$$

$$\displaystyle \text{Therefore: }\(\text{class A}\,|\,\text{accident}) \;=\;\frac{20}{195} \;=\;\frac{4}{39}$$

galactus

Super Moderator
Staff member
That 'primitive
approach is a nice explanation.

Here is all I was getting at:

$$\displaystyle \frac{(.01)(.2)}{(.01)(.2)+(.02)(.65)+(.03)(.15)}=\frac{4}{39}\approx .1025$$

When confronted with a 'probability of something given something' problem, you can make a chart.

Assume 1000 policyholders.

$$\displaystyle \begin{array}{c|c|c|c|c}\text{}&A&B&C&\text{total}\\ \hline\text{has accidents}&2&13&4.5&19.5 \\ \hline\text{no accident}&198&637&145.5&980.5 \\ \hline\text{total}&200&650&150&1000\end{array}$$

Now, you can answer any problem they throw at you. In this case, we want

"Probability the driver is class A given they have an accident".

Go down the class A column and across the 'has accident' row. $$\displaystyle \frac{2}{19.5}=\frac{4}{39}$$

Say they asked for "probability the driver is from class C given they did not have an accident?"

Go down the C column and across the 'no accident' row and get $$\displaystyle \frac{145.5}{980.5}=\frac{291}{1961}\approx .148$$

and so on..............

alakaboom1

New member
Thanks a lot guys!

Out of curiosity, is this formula you guys mentioned known as Bayes Theorem? I've definitely used the formula before, but never knew it by name.


# Cosets of a group

One sentence from Amstrong's Group and Symmetry wrote the following to prove a group of order 6 is isomorphic to $\mathbf{Z}_6$:

The right cosets $\langle x\rangle$, $\langle x\rangle y$ give 6 elements $e$, $x$, $x^2$, $y$, $xy$, $x^2y$ which fill out $G$.

Where $x$ is of order $3$ and $y$ is of order 2. $x$, $y$ are both elements of $G$.

My confusion:

How can one guarantee that the right coset $\langle x\rangle$ and $\langle x\rangle y$ can fill out $G$? Because $\langle x\rangle$ and $\langle x\rangle y$ have no intersections? I am not sure about the properties of cosets. I also don't know if this is related to the fact that $|\langle x\rangle|=5$ and is precisely half the elements of $G$ and thus its left coset $y\langle x\rangle$ and right coset $\langle x\rangle y$ is exactly the same?

I don't know if you can understand my question. The question here actually arises from my vague understanding of "cosets". So everytime this term occurs I feel a steady uncertainty.

-
You may want to reread the question yourself and see where one or more letters can be added. – user21436 Mar 28 '12 at 1:36
@KannappanSampath: An artifact of using < and > in text; parser thought they were mark-up. – Arturo Magidin Mar 28 '12 at 2:10
This is specifically page 70 (and end of 69). – anon Mar 28 '12 at 2:12
Exactly. It is from page 70. Sorry for the disorganized order, I will check that next time before I submit. And thanks again to @ArturoMagidin for the edition. – Jinji Mar 28 '12 at 2:26
For more on cosets, see this answer. – Arturo Magidin Mar 28 '12 at 2:45

Let $H=\{e,x,x^2\}$ be the subgroup generated by $x$. The right cosets of $H$ are the sets of the form $Hz = \{hz\mid h\in G\}$ for fixed $z\in G$.

The right cosets of $H$ are known to form a partition of the set $G$; that is: any two cosets are either identical or disjoint, and their union is all of $G$.

Moreover, $Hz = Hw$ if and only if $zw^{-1}\in H$. This can be verified because $z\in Hz$ (obtained as $ez$), so there must exist some $h\in H$ such that $z = hw$. Hence $zw^{-1}=h\in H$, proving that if $Hz=Hw$, then $zw^{-1}\in H$. Conversely, if $zw^{-1}\in H$, then $z = (zw^{-1})w \in Hw$, so $Hz\cap Hw\neq\varnothing$; since right cosets are either disjoint or identical, and $Hz$ and $Hw$ are not disjoint, then they are identical.

Now, since $y$ is of order $2$, it is not in $H$ (every element of $H$ is either of order $3$ or of order $1$). That means that the cosets $He = H$ and $Hy=\{ey, xy, x^2y\}$ are distinct, hence they are disjoint. Since they are disjoint, $H$ and $Hy$, together, account for $6$ elements of $G$. Since $G$ has exactly $6$ elements by assumption, those are all the elements of $G$. That is: $$G = H\cup Hy = \{e,x,x^2\}\cup\{y,xy, x^2y\} = \{e,x,x^2,y,xy,x^2y\}.$$

The same is true for left cosets: left cosets of $H$ partition $G$, so any two left cosets are either disjoint or identical. Also, $zH = wH$ if and only if $z^{-1}w\in H$ (I'll leave the proof to you). That means that the two left cosets of $H$ are $eH=H$ and $yH$, since $yH\neq H$, so $yH\cap H=\varnothing$. Since $G$ has $6$ elements, $H$ has $3$, and $yH$ has another three, then $$G = H\cup yH = \{e,x,x^2\} \cup \{y, yx, yx^2\}.$$ Also, since $H\cap Hy=\varnothing$, then $Hy = G\setminus H$; similarly, $yH=G\setminus H$. So in fact, $Hy=yH$ as sets.

-
Thanks! Does the number of right cosets relate to the order of H? |G|/|H|=the number of right cosets? – Jinji Mar 28 '12 at 2:24
@Jinji: The number of cosets is, by definition, the index of $H$ in $G$, denoted $[G:H]$. Since the cosets all have the same size, namely $H$, and partition $G$, it follows that $|G|=|H|[G:H]$ (in the sense of cardinalities); in fact, if $K\lt H\lt G$, then $[G:K]=[G:H][H:K]$ in the sense of cardinalities. So in the special case in which $|G|$ is finite, it follows from $|G|=|H|[G:H]$ that $[G:H]=|G|/|H|$. – Arturo Magidin Mar 28 '12 at 2:37


# Proving that an integer is even if and only if it is not odd

There is this question, but the definition of "even" and "odd" that I am using uses integers instead of just natural numbers; i.e.,

• An integer $n$ is even iff there is some integer $k$ such that $n=2k$.
• An integer $n$ is odd iff there is some integer $k$ such that $n=2k+1$.

Here is what I have so far:

First we show that an integer $n$ is even or odd. We first use induction on the positive integers. For the base case, $1=2\cdot0+1$ so we are done. Now suppose inductively that $n$ is even or odd. If $n$ is even, then $n=2k$ for some $k$ so that $n+1=2k+1$ (odd). If $n$ is odd, then $n=2k+1$ for some $k$ so that $n+1=2(k+1)$ (even). This closes the induction, so every $n\in\mathbf{Z}^+$ is even or odd.

Now we show every $n\in\mathbf{Z}^-$ is even or odd. Let $n\in\mathbf{Z}^-$. Then $n=-k$ for some $k\in\mathbf{Z}$ (I think this follows immediately from most definitions of the integers.). Suppose $k$ is even. Then $k=2j$ for some $j$ so that $n=-k=-2j=2(-j)$ (even). Now suppose $k$ is odd. Then $k=2j+1$ for some $j$ so that $n=-k=-(2j+1)=-2j-1=-2j-1+1-1=-2j-2+1=2(-j-1)+1$ (odd).

For $0$, note that $0=2\cdot0$ (even).

Now we show that $n\in\mathbf{Z}$ cannot be both even and odd. Suppose for the sake of contradiction that $n\in\mathbf{Z}$ is both even and odd. Then there are integers $k,j$ such that $n=2k=2j+1$. This implies that $2(k-j)=1$ (like in the referenced question). So we must show that $1$ cannot be even in order to complete the proof.

This is where I am having trouble. I know that if I let $f:\mathbf{Z}\to\mathbf{Z};x\mapsto2x$ be a function, then $f$ is increasing so since $f(0)=0$ and $f(1)=2$ and $0<1<2$, there is no integer $m$ such that $f(m)=1$. But this seems complicated so I was wondering if there was an easier way to do this.

So my real question is: how can I show that $1$ is not even?

(This is not homework.)

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$f(m)=1 \implies 2m=1 \implies 2=\frac {1}{m}$ which is not a integer for any $m \in \mathbb Z$ – Theorem Jul 4 '12 at 20:57
@russell11, I think you meant a function $f:\mathbb{Z}^+ \to \mathbb{Z}^+$ (or whatever the appropriate notation is for 'nonnegative'). Otherwise you and Theorem converge on the right idea; suppose $1=2m$ for some integer $m$. $m>0$ comes immediately, as twice a non-positive integer is non-positive. $m<1$ comes from the function being increasing. – Eugene Shvarts Jul 4 '12 at 21:02

To show that $1$ is not even:

I assume you can prove or accept that $a \cdot 0 = 0$ for all $a \in \mathbb{Z}$, the product of a positive and negative number is negative, and $2a > a$ when $a >0$.

If $1$ is even, then there must exists $a < 1$ such that $2a = 1$. However, the only $a < 1$, which is an integer, is $0$ and clearly $2 \cdot 0 = 0$. So $1$ can not be be written as $2 \cdot a$ for any $a \in \mathbb{Z}$. $1$ is odd.

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Do you want to say, "So $1$ cannot be written as $2\cdot a$ for any $a\in\mathbb{Z}$" in the end? – Paul Jul 4 '12 at 21:13
@Paul Yes. Added the missing "not". – William Jul 4 '12 at 21:21

Hint $\$ Your induction step uses $\rm\:n\,$ even $\rm\,\Rightarrow\: n\!+\!1\:$ odd, and $\rm\:n\,$ odd $\rm\,\Rightarrow\:n\!+\!1\:$ even. The converses are both true, e.g. $\rm\,n\!+\!1\,$ odd $\,\Rightarrow$ $\rm\,n\!+\!1 = 2k\!+\!1\,$ $\Rightarrow$ $\rm\,n = 2k,\,$ since $\rm\,j\!+\!1 = k\!+\!1\,$ $\Rightarrow$ $\rm\,j = k\,$ by Peano axioms. Thus if $\rm\:n\!+\!1\,$ is both even and odd then so too is $\rm\,n.\,$ So, contrapositively, in your induction step you can lift "$\rm n\:$ is not both even and odd" from $\rm\,n\,$ up to $\rm\,n\!+\!1\,$ and hence prove by induction that every natural is even or odd, but not both (this is essentially my hint in the linked answer).

-
Thanks, this is really neat. I didn't understand your hint in the other question, but now I do. – russell11 Jul 5 '12 at 1:18


Consider sorting $$n$$ numbers stored in array $$A$$ by first finding the smallest element of $$A$$ and exchanging it with the element in $$A[1]$$. Then find the second smallest element of $$A$$, and exchange it with $$A[2]$$. Continue in this manner for the first $$n-1$$ elements of $$A$$. Write pseudocode for this algorithm, which is known as selection sort. What loop invariant does this algorithm maintain? Why does it need to run for only the first $$n-1$$ elements, rather than for all $$n$$ elements? Give the best-case and worst-case running times of selection sort in $$\Theta$$-notation.

#### Pseudocode

$$\textsc {Selection-Sort }(A)$$\begin{aligned}1& \quad \textbf {for }i=1\textbf { to }A.length-1 \\2& \quad \qquad minIndex=i \\3& \quad \qquad \textbf {for }j=i+1\textbf { to }A.length \\4& \quad \qquad \qquad \textbf {if }A[j]<A[minIndex]\text { and }j\not=minIndex \\5& \quad \qquad \qquad \qquad minIndex=j \\6& \quad \qquad \text {swap }A[i]\text { with }A[minIndex] \\\end{aligned}

A python implementation of the above pseudocode is shared at end of the page, you can verify the workings for yourself.

#### Loop Invariant

At the start of the each iteration of the outer for loop of lines 1-6, the subarray $$A[1..i - 1]$$ consists of $$i - 1$$ smallest elements of $$A$$, sorted in increasing order.

#### Why only first n - 1 elements

The algorithm needs to run for only the first $$n - 1$$ elements, rather than for all $$n$$ elements because the last iteration will compare $$A[n]$$ with the minimum element in $$A[1 .. n - 1]$$ in line 4 and swap them if necessary. So, there is no need to continue the algorithm for all the way to the last element.

#### Running Times

For both the best-case (sorted array) and worst-case (reverse sorted array), the algorithm will anyway take one element at a time and compare it with all the other elements. In other words, each of the $$n$$ elements will be compared with rest of the $$n - 1$$ elements. So, the running times for both scenario will be $$\Theta(n^2)$$.

The above reasoning should be sufficient to understand or convey why the runtime would be $$\Theta(n^2)$$. However, for the sake of completeness, an exhaustive mathematical proof is given below.

#### Runtime Analysis

Let’s assume the inner for loop in line 3-5 is executed for $$t_j$$ times for $$j = 2, 3, \ldots, n$$, where $$n = A.length$$. Now note that, line 5 will be executed less than $$t_j - 1$$ times in the average case, but it’ll still be of the order of $$n$$.

For the sake of simplicity let’s assume the worst case, i.e. a reverse sorted array, when it’ll be executed exactly $$t_j - 1$$ times. Note, this assumption is only for that particular line, which is not going to change our overall analysis, it will only make our calculation easier.

We can now calculate the cost and times for individual lines of the pseudocode as follows …

Line Cost Times
1 $$c_1$$ $$n$$
2 $$c_2$$ $$n - 1$$
3 $$c_3$$ $$\sum_{j = 2}^n t_j$$
4 $$c_4$$ $$\sum_{j = 2}^n (t_j - 1)$$
5 $$c_5$$ $$\sum_{j = 2}^n (t_j - 1)$$
6 $$c_6$$ $$n - 1$$

Now, for any arbitrary value of $$j$$, the inner for loop (line 3-5) compares the previously computed minimum value with all elements in the subarray $$A[j..n]$$. So the inner for loop executes $$n - j + 1$$ times, i.e. $$t_j = (n - j + 1)$$ for $$j = 2, 3, \ldots, n$$. So, $$t_2 = n - 1, t_3 = n - 2, \ldots t_n = 1$$. We can calculate the summations for line 3-5 as follows …

\begin{aligned} \sum_{j = 2}^n t_j & = (n - 1) + (n - 2) + \cdots + 1 \\ & = \frac {n(n - 1)} 2 \\ \\ \sum_{j = 2}^n (t_j - 1) & = \sum_{j = 2}^n t_j - \sum_{j = 2}^n 1 \\ & = \frac {n(n - 1)} 2 - (n - 1) \\ & = \frac {(n - 3)(n - 2)} 2 \end{aligned}

Therefore, we can calculate the running time as follows..

$T(n) = c_1(n - 1) + (c_2 + c_6)n + c_3 \frac {n(n - 1)} 2 + (c_4 + c_5) \frac {(n - 1)(n - 2)} 2$

For best-case scenario, as the array is already sorted, line #5 will not be executed ever. So, $$c_5 = 0$$. Even with that (and without that for worst-case) the expression of $$T(n)$$ will be reduced to the form $$an^2 + bn + c$$, i.e. the algorithm will run at $$\Theta(n^2)$$ time.

#### Python Code

# Selection Sort def SelectionSort(A): for i in range(len(A)): minIndex = i for j in range(i + 1, len(A)): if A[j] < A[minIndex] and j != minIndex: minIndex = j A[i], A[minIndex] = A[minIndex], A[i] # Test import random num_failed = 0 total_tests = 100 for i in range(total_tests): length = random.randint(2, 50) lst = [random.randint(0, 100) for _ in range(length)] SelectionSort(lst) # Check if list has been sorted for i in range(len(lst) - 1): if lst[i] > lst[i + 1]: num_failed += 1 print(f"Test #{i:<2}: List is not sorted") break if num_failed > 0: break if num_failed > 0: print(f"\nFailed") else: print(f"Passed {total_tests}/{total_tests} tests")


# Math Help - [SOLVED] Vectors Question...

1. ## [SOLVED] Vectors Question...

Question:
In the diagram OABCDEFG is a cube which the lenght of each edge is $2$ units. Unit vectors i, j, k are parallel to $\vec{OA}$, $\vec{OC}$, $\vec{OD}$ respectively. The mid-points of $AB$ and $FG$ are $M$ and $N$ respectively.

(i) Express each of the vectors $\vec{ON}$ and $\vec{MG}$ in terms of i, j and k. [3 Marks]

(ii) Find the angle between the directions of $\vec{ON}$ and $\vec{MG}$, are correct to the nearest $0.1^o$.
[4 Marks]

Attempt:

(i) $\vec{ON} = i + 2j + 2k$
$\vec{MG} = -2i + j + 2k$

are they right? and will I be rewarded 3 marks for showing no steps? becuase I don't there are any steps needed!

(ii) Need Help!

2. Your answers to (i) seem right. As for how many marks you'll get: I don't know, I'm not an examiner. I can't see what working you COULD show, really. I know I wouldn't. That's just my opinion though.

(ii)

In maths, there's a fun thing called the scalar product (the "dot product" they call it at school).

It's defined like this:

a DOT b = |a| * |b| * cos(angle)

where a and b are vectors, |a| means the magnitude of a, and angle is the angle between the two of them.

So you can rearrange that to get

cos(angle) = [ a DOT b ] / [ |a| * |b| ]

So now all you need to know is what a DOT b is:

I'll show that with an example:

(1, 2, 3) DOT (4, 5, 6) = 1 x 4 + 2 x 5 + 3 x 6 = 4 + 10 + 18 = 32

All you do is multiply the two "firsts" (x-components) together, then the two seconds, then the thirds and add them all up.

Work that out, work out the two magnitudes, substitute it all in and do an inverse cosine and you're away.

Enjoy.

3. Originally Posted by Fedex
Your answers to (i) seem right. As for how many marks you'll get: I don't know, I'm not an examiner. I can't see what working you COULD show, really. I know I wouldn't. That's just my opinion though.

(ii)

In maths, there's a fun thing called the scalar product (the "dot product" they call it at school).

It's defined like this:

a DOT b = |a| * |b| * cos(angle)

where a and b are vectors, |a| means the magnitude of a, and angle is the angle between the two of them.

So you can rearrange that to get

cos(angle) = [ a DOT b ] / [ |a| * |b| ]

So now all you need to know is what a DOT b is:

I'll show that with an example:

(1, 2, 3) DOT (4, 5, 6) = 1 x 4 + 2 x 5 + 3 x 6 = 4 + 10 + 18 = 32

All you do is multiply the two "firsts" (x-components) together, then the two seconds, then the thirds and add them all up.

Work that out, work out the two magnitudes, substitute it all in and do an inverse cosine and you're away.

Enjoy.
Thanks for the explanation and the example

$a.b = \left|a\right|\times \left|b\right| \times \cos\theta$

$\theta = \cos^{-1} \left( \frac{a.b}{\left|a\right| \times \left|b\right|} \right)$

$a.b = \left(\begin{array}{c}1\\2\\2\end{array}\right) \times \left(\begin{array}{c}-2\\1\\2\end{array}\right) = (1 \times -2) + (2 \times 1) + (2 \times 2) = 4$

$\left|a\right|\times \left|b\right| = \left(\begin{array}{c}1\\2\\2\end{array}\right) \times \left(\begin{array}{c}2\\1\\2\end{array}\right) = (1 \times 2) + (2 \times 1) + (2 \times 2) = 8$

$\theta = \cos^{-1} = \frac{4}{8}$

$\theta = 60^o$

Are my steps correct? and is the answer correct?

4. Okay, a few mistakes there.

You wrote your two column vectors like this:

You SHOULD write them with a dot in between them. When you put a multiply sign in between them it means something else (in fact, a cross represents the "cross" or "vector" product of them - that's something else entirely!)

So write it with a dot, yeah?

Also, when you worked out the magnitude of the vectors, you seem to have gone wrong.

The MAGNITUDE of a vector (its "length") is just a number, not a vector.

For for the vector a = (1,2,3)

|a| = SQUARE ROOT OF [ 1^2 + 2^2 + 3^2 ]

Are you happy with that? It comes straight from a 3D version of Pythagoras' Theorem.

5. Originally Posted by Fedex
Okay, a few mistakes there.

You wrote your two column vectors like this:

You SHOULD write them with a dot in between them. When you put a multiply sign in between them it means something else (in fact, a cross represents the "cross" or "vector" product of them - that's something else entirely!)

So write it with a dot, yeah?

Also, when you worked out the magnitude of the vectors, you seem to have gone wrong.

The MAGNITUDE of a vector (its "length") is just a number, not a vector.

For for the vector a = (1,2,3)

|a| = SQUARE ROOT OF [ 1^2 + 2^2 + 3^2 ]

Are you happy with that? It comes straight from a 3D version of Pythagoras' Theorem.
Thanks again

$a.b = \left|a\right|.\left|b\right|.\cos\theta$

$\theta = \cos^{-1} \left( \frac{a.b}{\left|a\right|.\left|b\right|} \right)$

$a.b = \left(\begin{array}{c}1\\2\\2\end{array}\right). \left(\begin{array}{c}-2\\1\\2\end{array}\right) = (1 \times -2) + (2 \times 1) + (2 \times 2) = 4$

$\left|a\right| = \sqrt{(1^2 + 2^2 + 2^2} = 3$
$\left|b\right| = \sqrt{(-2^2 + 1^2 + 2^2)} = 1$

$\left|a\right|.\left|b\right| = 3\times1 = 3$

$\theta = \cos^{-1} \frac{4}{3}$ (Invalid)

Where did I go wrong?

6. What's (-2)^2 again?

Check out that part of your working for the magnitude of b.

I think that's your problem :-)

7. Originally Posted by Fedex
What's (-2)^2 again?

Check out that part of your working for the magnitude of b.

I think that's your problem :-)
$a.b = \left|a\right|.\left|b\right|.\cos\theta$

$\theta = \cos^{-1} \left( \frac{a.b}{\left|a\right|.\left|b\right|} \right)$

$a.b = \left(\begin{array}{c}1\\2\\2\end{array}\right). \left(\begin{array}{c}-2\\1\\2\end{array}\right) = (1 \times -2) + (2 \times 1) + (2 \times 2) = 4$

$\left|a\right| = \sqrt{(1^2 + 2^2 + 2^2} = 3$
$\left|b\right| = \sqrt{(-2^2 + 1^2 + 2^2)} = 3$

$\left|a\right|.\left|b\right| = 3\times3 = 9$

$\theta = \cos^{-1} \frac{4}{9} = 63.6^o$


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# Tutor profile: Yannic V.

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Yannic V.
Ph.D. in mathematics
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## Questions

### Subject:Discrete Math

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Question:

Find a formula for the sum $\sum_{k=0}^n k \binom{n}{k},$ by means of the binomial theorem. What can be deduced about the average number of elements in a random subset of $\{1,2, \hdots, n\}$?

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Yannic V.

Let $x,y \in \mathbb{R}$. Remember that the binomial theorem sates that $(x+y)^n = \sum_{k=0}^n \binom{n}{k}x^{k}y^{n-k},$ for all $n \in \mathbb{N}$. In particular, for $y=1$ we have $(x+1)^n=\sum_{k=0}^n \binom{n}{k}x^k.$ Considering $x$ as a variable, we differentiate both sides of the above equation to obtain $n(x+1)^{n-1}= \sum_{k=0}^n k\binom{n}{k}x^{k-1}.$ It remains to plug in $x=1$: $\sum_{k=0}^n k \binom{n}{k} = n2^{n-1}.$ The average number of elements in a random subset of $\{1,2, \hdots, n\}$ is thus $\frac{\sum_{k=0}^n k \binom{n}{k}}{\sum_{k=0}^n \binom{n}{k}} = \frac{n 2^{n-1}}{2^n}=\frac{n}{2}.$ This is not surprising, since the number of subsets of size $k$ is exactly the same as the number of subsets of size $n-k$, for any $k$.

### Subject:Calculus

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Question:

Solve the integral $\displaystyle \int 3 \sec(2x-1)\tan(2x-1) \,dx$.

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Yannic V.

As the arguments of both functions in the integral are not single variables, we can start to consider a change of variable in order to just have one variable in the secant and tangent function. This can be made using the change of variable $$\label{cambio1} \left\lbrace\begin{array}{l} y = 2x-1 \\ dy = 2 \,dx \end{array}\right.$$ Notice that the derivative $dy$ is easy to obtain from the expression of the integral: $dx$ can be expressed in terms of $dy$ as $$\label{eq1} dx = \frac{dy}{2}.$$ Substituting the change of variable \eqref{cambio1} in the integral, we obtain \begin{align} \int 3 \sec(2x-1)\tan(2x-1)\, dx &= \int 3 \sec(y)\tan(y) \,\frac{dy}{2} \tag{\footnotesize by \eqref{cambio1} and \eqref{eq1}}\\ &= \frac{3}{2} \int \sec(y)\tan(y) \,dy. \label{da1} \end{align} We have now reduced our problem to a more simple integral. We want to integrate the expression $\sec(y)\tan(y)$, which is the derivative of\footnote{This comes from the following calculation: $(\sec(y))'= \left(\frac{1}{\cos(y)} \right)' = \frac{(0\cos(y))-1(-\sin(y))}{\cos(y)^2}= \frac{\sin(y)}{\cos(y)^2} = \frac{1}{\cos(y)}\frac{\sin(y)}{\cos(y)}=\sec(y)\tan(y).$ } the function $\sec(y)$. It is then appropriate to consider the following change of variable: $$\label{cambio2} \left\lbrace\begin{array}{l} u = \sec(y)\\ du = \sec(y)\tan(y) \,dy \end{array}\right.$$ We can now replace \eqref{cambio2} into the expression \eqref{da1}. Remark that \eqref{da1} just contain the derivative part of\footnote{This is not a problem. When doing a change of variable, the most important part is that the expression corresponding to the derivative of the change of basis appears in the integral (up to a scalar). } \eqref{cambio2}. Thus, we have: \begin{align} \int 3 \sec(2x-1)\tan(2x-1)\, dx &= \frac{3}{2} \int \sec(y)\tan(y) \,dy. \tag{\footnotesize{by \eqref{da1}}}\\ &= \frac{3}{2} \int du \tag{\footnotesize{by \eqref{cambio2}}}\\ &= \frac{3}{2}\, u + C, \label{da2} \end{align} where $C$ is a constant. Finally, we just have to substitute $u$ by its value in terms of $y$ (given by \eqref{cambio2}), and replace $y$ by its value in terms of $x$ (given by \eqref{cambio1}): \begin{align} \int 3 \sec(2x-1)\tan(2x-1)\, dx &= \frac{3}{2}\, u + C \tag{\footnotesize{by \eqref{da2}}}\\ &= \frac{3}{2}\sec(y) + C \tag{\footnotesize{substitute $u=\sec(y)$}}\\ &= \frac{3}{2}\sec(2x-1) + C. \tag{\footnotesize{substitute $y=2x-1$}} \end{align}

### Subject:Algebra

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Question:

If $G$ is a finite group, show that the number of elements in $G$ of order greater than $2$ must be even. In particular, prove that any group of even order must contain an element of order 2.

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Yannic V.

We will solve the question by a counting argument as follows. Let $X$ be the set of all elements of G with order greater than two. If $X = \emptyset$ then $|X|=0$ and we’re done. Suppose now that $X \neq \emptyset$. Since the orders of $g$ and $g^{-1}$ coincides for any $g \in G$, we have that $g \in X \Longleftrightarrow g^{-1} \in X.$ Moreover, if $g \in X$ then $g^2\neq e_G$. From here, $g \neq g^{-1}$. It follows that $X$ can be written as the disjoint union of two element sets of the form $\{g, g^{-1}\}$ {x,x−1}, and hence that $|X|$ is even. Suppose that $|G|$ is even. Since $e_G$ has order $1$, $e_G \notin X$. It follows that $G \setminus X \neq \emptyset$. So $0 < |G \setminus X|=|G|-|X|$. Since $|G|$ and $|X|$ are both even, it follows that $|G\setminus X|$ is a nonzero even integer, i.e. is at least 2. Thus, there is an $z \in G \setminus X$ such that $z \neq e_G$ . Since $X$ consists of all elements in $G$ of order greater than $2$, it must be the case that $|z|=2$.

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# How to integrate $\int \frac{dx}{(1+x^2)^2}$?

I need to integrate this to finish an old STEP problem I'm doing, but I'm stuck here, at the very end:

$$\int_0^\infty \frac{dx}{(1+x^2)^2}$$

The result should be $\pi\over 4$ . I don't know how to approach this. *Somehow, this question doesn't seem to've been posted here ever (at least I couldn't find it).

Also, Wolfram tells me:

$$\int \frac{dx}{(1+x^2)^2} = \frac{1}{2}\left(\frac{x}{x^2+1}+\tan^{-1}x\right)+c$$

but I don't see how one can derive this without knowing the result beforehand.

Somehow, this question doesn't seem to've been posted here ever (at least I couldn't find it).

EDIT: If you're interested, the problem in question is: STEP II - problem 4 (year 2014).

• use $\frac 1{(1+x^2)^2 }=\frac 1{1+x^2} - \frac{x^2}{(1+x^2)^2}$ and do an integration by parts $\int \frac{x^2}{(1+x^2)^2} \, dx$ – abel Jun 2 '16 at 18:04
• I tried $x = \sinh(z)$ since $1+x^2 = \cosh^2(x)$ but I got stuck at $$\int \frac{1}{\cosh^3(z)} \, {\rm d}z$$ – ja72 Jun 2 '16 at 18:12

Use $x = \tan t$. Then $(1 + x^2)^2 = \sec^4 t$, and $\dfrac{dx}{dt} = \sec^2 t$, so the integral becomes \begin{align*} \int_0^{\pi/2} \cos^2 t \,\mathrm dt = \dfrac 1 2 \times \dfrac {\pi} 2 = \dfrac{\pi}{4}. \end{align*}

• Nice. Thanks. It's been a while and I forgot about trig substitions :( – I want to make games Jun 2 '16 at 18:08

For any $\alpha>0$, let: $$I(\alpha)= \int_{0}^{+\infty}\frac{dx}{\alpha^2+x^2} = \frac{\pi}{2\alpha}.$$ By differentiating both sides with respect to $\alpha$ we get: $$\int_{0}^{+\infty}\frac{2\alpha}{(\alpha^2+x^2)^2}\,dx = \frac{\pi}{2\alpha^2}$$ and by evaluating at $\alpha=1$: $$\int_{0}^{+\infty}\frac{dx}{(x^2+1)^2} = \color{red}{\frac{\pi}{4}}.$$

• How do you get these expressions? – ja72 Jun 2 '16 at 18:07
• @ja72: What's obscure? I think my answer is pretty straightforward to follow. The technique is known as differentiation under the integral sign, or "Feynman's trick". – Jack D'Aurizio Jun 2 '16 at 18:08
• I'm sure that's common knowledge, but how (a very general idea would do) do you prove that differentiating an integral is the same as differentiating the thing that's integrated? – I want to make games Jun 2 '16 at 18:09
• @M.Vinay: thanks, I was just writing the same thing :) – Jack D'Aurizio Jun 2 '16 at 18:12
• @ja72: I assure you that my memorization skills are close to being awful (that is the main reason beyond my appraise for modern technology, that allow us to store tons of useful data in portable devices), but when it comes to mathematical practice, many things become natural. Practice makes perfect. – Jack D'Aurizio Jun 2 '16 at 18:20

Use $u=tan x$ $$(1+x^2)^2=(1+tan^2u)^2=(sec^2u)^2=sec^4u$$ $$dx=sec^2u du$$ The integral becomes $$\int{\frac{1}{sec^2u}du=\int{cos^2udu}}$$

• functions need to be upright characters. Use \sin instead of sin. Also the differential needs to be upright. Use {\rm d} instead of d. – ja72 Jun 2 '16 at 18:14

Add and subtract $x^2$ from the numerator, you get $\arctan$ on one hand and an $\int \frac{x^2}{(1 + x^2)^2}dx$, which you can do by parts ($u = x$ and $dv = \frac{x}{(1+x^2)^2}dx$)

$$\int_0^\infty\frac{1}{\left(1+x^2\right)^2}\space\text{d}x=\lim_{n\to\infty}\int_0^n\frac{1}{\left(1+x^2\right)^2}\space\text{d}x=$$

Substitute $x=\tan(u)$ and $\text{d}x=\sec^2(u)\space\text{d}u$.

So $\left(1+x^2\right)^2=\left(1+\tan^2(u)\right)^2=\sec^4(u)$ and $u=\arctan(x)$.

This gives a new lower bound $u=\arctan(0)=0$ and upper bound $u=\arctan(n)$:

$$\lim_{n\to\infty}\int_0^{\arctan(n)}\cos^2(u)\space\text{d}u=$$

Use:

$$\cos^2(x)=\frac{1+\cos(2u)}{2}$$

$$\frac{1}{2}\lim_{n\to\infty}\left[\int_0^{\arctan(n)}1\space\text{d}u+\int_0^{\arctan(n)}\cos(2u)\space\text{d}u\right]=$$

Substitute $s=2u$ and $\text{d}s=2\space\text{d}u$.

This gives a new lower bound $s=2\cdot0=0$ and upper bound $s=2\arctan(n)$:

$$\frac{1}{2}\lim_{n\to\infty}\left[\left[u\right]_0^{\arctan(n)}+\frac{1}{2}\int_0^{2\arctan(n)}\cos(s)\space\text{d}s\right]=$$ $$\frac{1}{2}\lim_{n\to\infty}\left[\left[u\right]_0^{\arctan(n)}+\frac{1}{2}\left[\sin(s)\right]_0^{2\arctan(n)}\right]=$$ $$\frac{1}{2}\lim_{n\to\infty}\left[\left(\arctan(n)-0\right)+\frac{\sin(2\arctan(n))-\sin(0)}{2}\right]=$$ $$\frac{1}{2}\lim_{n\to\infty}\left[\arctan(n)+\frac{\sin(2\arctan(n))}{2}\right]=\frac{1}{2}\cdot\frac{\pi}{2}=\frac{\pi}{4}$$

let $I_n=\int_{0}^{\infty}\frac{1}{(1+x^2)^n}dx$ then $$I_{n+1}=\frac{2n-1}{2n}I_n$$ we know $I_1=\int_{0}^{\infty}\frac{1}{1+x^2}dx=\frac{\pi}{2}$, so $$I_2=\frac{2(1)-1}{2(1)}I_1=\frac{\pi}{4}$$

• (+1) I see in you a future star of MSE. – Jack D'Aurizio Jun 2 '16 at 20:54

Here's a pretty solution. Notice that $$I=\int_{0}^{\infty}\frac{1}{(1+x^2)^2}dx=\frac{1}{2}\int_{-\infty}^{\infty}\frac{1}{(1+x^2)^2}dx=\frac{1}{2}\int_{\partial\Omega}\frac{1}{(1-iz)^2(1+iz)^2}dz$$ where $\Omega$ is the top part of the complex plane. Since $f(z)$ decays faster than $\frac{1}{z^2}$, $$I=\pi i \lim_{z \to i}\frac{d}{dz}\frac{(z-i)^2}{(1-iz)^2(1+iz)^2}=\frac{\pi}{4}$$.


# Estimating error when calculating $\pi^2$ with $8 + \frac{8}{3^2} + \frac{8}{5^2} + \frac{8}{7^2} + \cdots$

While answering a CodeReview question (Approximating constant $\pi^2$ to within error), I noticed that when calculating the sum $$8 + \dfrac{8}{3^2} + \dfrac{8}{5^2} + \dfrac{8}{7^2} + \dfrac{8}{9^2} + \cdots$$

and stopping at the smallest term larger than $\varepsilon$, the difference to $\pi^2$ seems very close to $\sqrt{2\varepsilon}$.

With $$n = \left\lfloor\frac{\sqrt{\frac{8}{\varepsilon}} - 1}{2} \right\rfloor,$$

it looks like:

$$\sum_{i=0}^n \frac{8}{(2i+1)^2} \approx \pi^2 - \sqrt{2\varepsilon}.$$

For example, with $\varepsilon = 10^{-10}$:

$n = 141421$

$\sum = 9.86959025 \dots$

$\pi^2 - \sum \approx 1.4142170*10^{-5}$

$(\pi^2 - \sum)^2 \approx 2.0000099*10^{-10} \approx 2\varepsilon$

I don't think it's a coincidence, but I don't know where to begin to link

$$\sum_{i=n+1}^\infty \frac{8}{(2i+1)^2}$$ to $\sqrt{2\varepsilon}$.

Wolfram Alpha expresses this sum in terms of a polygamma function but I don't know anything about it.

Is the approximation correct? Is there any simple way to prove it?

• The usual method of estimating the error in such series is the Euler-Maclaurin formula. – Lord Shark the Unknown Mar 8 '18 at 15:58
• I have changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance. – GNUSupporter 8964民主女神 地下教會 Mar 15 '18 at 16:35

One has $${2\over (2k+1)^2}<{1\over 2k}-{1\over 2k+2}\ ,$$ and therefore (teleskoping sum!) $$0<\sum_{k=n}^\infty{2\over(2k+1)^2}<{1\over 2n}\ ,$$ or $$0<\sum_{k=n}^\infty{8\over(2k+1)^2}<{2\over n}\ .$$ Maybe this will bring you over the top.

• Thanks a lot. If I understand it correctly, we still need another inequality to prove that the sum isn't too far from $\frac{2}{n}$, right? – Eric Duminil Mar 8 '18 at 21:04
• Applying the same logic with ${1\over 2k+1}-{1\over 2k+3} < {2\over (2k+1)^2}$, we obtain $\frac{2}{n+1} < \sum < \frac{2}{n}$, right? That would be a pretty good approximation. – Eric Duminil Mar 9 '18 at 8:12

Recall from calculus the integral test:

$$\displaystyle\sum a(n)\text{ converges if and only if }\int a(x)\text{ converges}$$ where the sequence $a(n)=a_n>0$ and $a_n$ is decreasing, and the function $a(x)$ is continuous.

From the proof of this statement, we can deduce the inequality: if $\sum_{n=1}^\infty a_n=L$, then $$\left|L-\sum_{i=0}^n a_n\right|\leq\int_{n}^\infty a(x)\,dx,$$(see here for help with providing visual intuition as to why the inequality is true) where the notation $a(x)$ is used to represent $a_n$ with the $n$ replaced by $x$. Thus, performing the necessary calculations, we find

\begin{align*} \int_{n}^\infty\frac{8}{(2x+1)^2}\,dx&=\frac{4}{2n+1}\\ &\leq\frac{4}{\sqrt{\frac{8}\varepsilon}-1}\\ &=\frac{1}{\sqrt{\frac{1}{2\varepsilon}}-\frac14}\\ &=\frac{\sqrt{2\varepsilon}}{1-\frac{\sqrt{2\varepsilon}}{4}}\\ &\approx \sqrt{2\varepsilon},\end{align*} where I've taken $$n=\left\lfloor\frac{\sqrt{\frac{8}{\varepsilon}}-1}{2}\right\rfloor.$$

• This looks good, thanks. I'll try to do it again on a piece of paper and come back to comment/accept. – Eric Duminil Mar 8 '18 at 17:07
• Thanks, @EricDuminil. There was one place I had something like $\frac{1}{\sqrt{\frac{1}{2\varepsilon}}+1}$ where I had to use some algebra to simplify (hence I didn't want to type it, at least, not yet haha). – Clayton Mar 8 '18 at 17:15
• First, the integral test has hypotheses that you need to mention. And, although the estimate you give for the error is correct (under the same hypotheses!), I don't see how we can deduce that error estimate from the integral test itself. How does that go? – David C. Ullrich Mar 8 '18 at 17:20
• @DavidC.Ullrich: I suppose it is more from the proof of the integral test than the integral test itself, but the idea is that the sequence/function is positive and decreasing, so we can set it up so the integral is an overestimate (if you prefer to think in terms of Riemann sums, it is like using right endpoints for the estimate). – Clayton Mar 8 '18 at 17:33
• @DavidC.Ullrich: You're right about the test having hypotheses that I ought to mention. I knew I wouldn't have much time when I submitted the answer. At any rate, I'm at fault for not being explicit enough. I have about an hour now, so I'll provide more explanations. If you have any other suggestions, please let me know. – Clayton Mar 8 '18 at 17:36


# Polynomial uniqueness proof

• Nov 6th 2012, 10:30 AM
Polynomial uniqueness proof
This problem is from my Gelfand's Algebra book.

Problem 164. Prove that a polynomial of degree not exceeding 2 is defined uniquely by three of its values.

This means that if $P(x)$ and $Q(x)$ are polynomials of degree not exceeding 2 and $P(x_1)=Q(x_1),P(x_2)=Q(x_2),P(x_3)=Q(x_3)$ for three different numbers $x_1,x_2,\text{ and } x_3,$ then the polynomials $P(x)$ and $Q(x)$ are equal.

I'm not very good at proofs, so I have questions. If I show that if $P(x_1)=Q(x_1),P(x_2)=Q(x_2),P(x_3)=Q(x_3)$ is true and $P(x)$ and $Q(x)$ are polynomials of degree not exceeding 2, then $P(x)$ and $Q(x)$ are equal, will it prove that a polynomial of degree not exceeding 2 is defined uniquely by three of its values?

Is this the way to go?

Let $V(x)=P(x)-Q(x)$, After that $V(x_1)=V(x_2)=V(x_3)=0$, because $P(x_1)=Q(x_1),P(x_2)=Q(x_2),P(x_3)=Q(x_3)$, so $x_1,x_2,x_3$ are roots of $V(x)$ which can't have more than two roots. Here I'm confused, I shouldn't have assumed that $P(x_1)=Q(x_1),P(x_2)=Q(x_2),P(x_3)=Q(x_3)$ was true, right?
• Nov 6th 2012, 11:11 AM
Deveno
Re: Polynomial uniqueness proof
actually, your proof is fine. a polynomial of degree n, cannot have more than n roots.

so you have proved that V(x) is NOT of degree 0,1, or 2. what's left?
• Nov 6th 2012, 11:17 AM
emakarov
Re: Polynomial uniqueness proof
I don't know what comes before this exercise in the book, but relying on the fact that a polynomial of degree n cannot have more than n roots seems to easy. It is possible that the real problem is to prove that fact.
• Nov 6th 2012, 11:18 AM
Re: Polynomial uniqueness proof
Quote:

Originally Posted by Deveno
actually, your proof is fine. a polynomial of degree n, cannot have more than n roots.

so you have proved that V(x) is NOT of degree 0,1, or 2. what's left?

Yes, but does that prove that polynomial of degree not exceeding 2 is defined uniquely by three of its values?

Quote:

Originally Posted by emakarov
I don't know what comes before this exercise in the book, but relying on the fact that a polynomial of degree n cannot have more than n roots seems to easy. It is possible that the real problem is to prove that fact.

Yes, I think that's right, I'm assuming too much and I don't really understand how I can prove that polynomial of degree not exceeding 2 is defined uniquely by three of its values.
• Nov 6th 2012, 11:40 AM
Deveno
Re: Polynomial uniqueness proof
Quote:

Originally Posted by emakarov
I don't know what comes before this exercise in the book, but relying on the fact that a polynomial of degree n cannot have more than n roots seems to easy. It is possible that the real problem is to prove that fact.

well this is only true in a field, of course, but presumably we are talking about polynomials in R[x].

and one can argue by degree:

proof(by induction): if p(x) in R[x] is of degree n > 0 then p has at most n real roots.

(the reason for requiring n > 0 will be explained later).

base case: n = 1.

in this case p(x) = ax + b, which has the sole root x = -b/a (we may assume a ≠ 0, or else p is not of degree 1). since exactly one root is a stronger condition that at most one root, the theorem holds in this case.

(strong) induction hypothesis: suppose whenever 0 < k < n, deg(p) = k implies p has at most k roots.

let deg(p) = n.

it could be that p has no roots at all. since 0 < n, the theorem holds in this case.

otherwise, let a be a root of p.

writing p(x) = q(x)(x - a) + r(x), where either r(x) is identically 0, or deg(r) < deg(x-a) = 1, we see that r is a constant polynomial, or the 0-polynomial.

so p(x) = q(x)(x - a) + r, for some real number r. thus:

0 = p(a) = q(a)(a - a) + r = q(a)0 + r = r.

hence p(x) = q(x)(x - a).

since deg(p) = deg(q) + deg(x-a), we have:

n = deg(q) + 1, that is:

deg(q) = n-1 < n, so q has at most n-1 roots.

lemma: if p(x) = f(x)g(x) for real polynomials p,f, and g, then if a is a root of p, either a is a root of f, or a is a root of g.

proof: 0 = p(a) = f(a)g(a), so either f(a) = 0, or g(a) = 0.

main proof continued:

let b be any root of p(x) = q(x)(x - a).

then either b is a root of q(x) (of which we have at most n - 1), or b is a root of x - a (which has the single root a).

thus we have at most n - 1 + 1 = n roots of p, QED.

*********

you have proved that if P = Q at 3 points (and deg(P), deg(Q) < 3), P = Q at every point.

the only thing remaining to show is that given any 3 points, there is a polynomial of degree < 3 that goes through those 3 points. can you do this?

**********

OOPS! i almost forgot my explanation for why we take n > 0:

a polynomial of degree 0 is a constant polynomial:

p(x) = c.

unless c = 0, p has no roots at all (which is ok according to our theorem).

but if p(x) = 0, for all x, we have infinitely many roots (very bad).

on the other hand:

p(x) = 0 = 0 + 0x = 0 + 0x + 0x2, etc.

is sort of a SPECIAL polynomial, it really doesn't "have" a degree. it is this special case we seek to avoid by stating deg(p) > 0. depending on which book you read you get:

deg(0) = undefined, or sometimes:

deg(0) = -∞ (this is to make the formula deg(fg) = deg(f) + deg(g) always work out).
• Nov 6th 2012, 12:10 PM
Re: Polynomial uniqueness proof
Quote:

Originally Posted by Deveno

you have proved that if P = Q at 3 points (and deg(P), deg(Q) < 3), P = Q at every point.

the only thing remaining to show is that given any 3 points, there is a polynomial of degree < 3 that goes through those 3 points. can you do this?

I will try to prove it now, thank you for help!
• Nov 6th 2012, 12:25 PM
HallsofIvy
Re: Polynomial uniqueness proof
Because you are specifically asked about "polynomials of degree not greater than 2", I don't think you need to be very "sophisticated"! You can use the fact that any such polynomial can be written in the form $f(x)= a_1x^2+ b_1x+ c_1$. So suppose $f(x_1)= y_1$, $f(x_2)= y_2$, and $f(x_3)= y_3$ and that $g(x)= a_2x^2+ b_2x+ c$ takes on those same values. That gives three equations you can solve to show that $a_1= a_2$, $b_1= b_2$, and $c_1= c_2$.
• Nov 6th 2012, 01:22 PM
Deveno
Re: Polynomial uniqueness proof
Quote:

Originally Posted by HallsofIvy
Because you are specifically asked about "polynomials of degree not greater than 2", I don't think you need to be very "sophisticated"! You can use the fact that any such polynomial can be written in the form $f(x)= a_1x^2+ b_1x+ c_1$. So suppose $f(x_1)= y_1$, $f(x_2)= y_2$, and $f(x_3)= y_3$ and that $g(x)= a_2x^2+ b_2x+ c$ takes on those same values. That gives three equations you can solve to show that $a_1= a_2$, $b_1= b_2$, and $c_1= c_2$.

the nice thing about this suggestion is that it also suggests a way to prove there is a polynomial which goes through our 3 given points.

(think: system of linear equations)
• Nov 8th 2012, 08:21 AM


# Does the uniform continuity of $f: X \rightarrow \mathbb{R}$ imply $f: A \rightarrow \mathbb{R}$ is also uniformly continuous, when $A \subset X$?

I've been preparing for the prelim in August, and was working on a problem involving uniform continuity and restriction of functions. I absentmindedly assumed the above by considering the contrapositive: if $f: A \rightarrow \mathbb{R}$ isn't uniformly continuous, that implies $\exists \ \epsilon$ such that no $\delta$ satisfies $d(x,y) < \delta \implies d(f(x),f(y)) < \epsilon, \,\,\ \forall x,y \in A$, and this failure of $\epsilon$'s existence shouldn't change when I "add more points" by considering $f: X \rightarrow \mathbb{R}.$

However, if this is true, we obtained a lot of results I consider to be strangely powerful. For example, if a function is continuous on $\mathbb{R}$, it is uniformly continuous on any bounded interval I, as it's uniformly continuous on $\overline{I}$ which is compact by Heine-Borel. Hence, if $f$ is a real-valued function continuous on a subset $A$ of $R$, it's uniformly continuous on any bounded subset $X$ of $A$.

Conclusions such as this seem too strong! Is there a flaw in my reasoning, and if so, where is it?

-
No, you're perfectly right. If $f: X \to \mathbb{R}$ is uniformly continuous then the restriction of $f$ to any subset $A \subset X$ is also uniformly continuous. This can be seen by writing the condition of uniform continuity on $X$ as $\forall\, \varepsilon \gt 0\;\exists\;\delta \gt 0: \forall\, x,y \in X : \ldots$. If you replace $X$ by $A$ and restrict $f$ to $A$ then the last condition becomes weaker as it then reads $\forall\,x,y \in A$. The "too strong" conclusions are correct. –  t.b. Jul 4 '11 at 6:45
Thank you! I went back and thought about why I considered the conclusions "too strong", and I realized I was subconsciously only thinking about Lipschitz continuous functions. (Also, the statements gave off a sense of "if the conditions were this nice, they'd mention it in class" to me, which made me suspicious.) –  JakeR Jul 4 '11 at 6:55
I hope you're aware that Lipschitz is much stronger than uniform continuity. One standard example for showing this is $x \mapsto \sqrt{x}$ on $[0,\infty)$ which is uniformly continuous but not Lipschitz (because the slopes become arbitrarily steep near the origin). –  t.b. Jul 4 '11 at 6:57
Sadly, I can't claim that I wasn't conscious of that. I'm still not sure why I was implicitly assuming my functions were Lipschitz. –  JakeR Jul 4 '11 at 7:13
There is a rather difficult theorem due to Rademacher indicating a further strong distinction between Lipschitz and uniform continuity: It asserts that a Lipschitz continuous function on $\mathbb{R}^{n}$ is differentiable "almost everywhere" (a precise technical term that should make sense intuitively), while you probably came across examples of continuous functions on $$0,1$$ that aren't differentiable anywhere. –  t.b. Jul 4 '11 at 7:22

It is true, and your conclusion that every continuous $f:\mathbb{R}\to\mathbb{R}$ is uniformly continuous on bounded subsets of $\mathbb{R}$ is also correct.
One could go further in saying precisely why it is true, which might help to convince you. Suppose $f:X\to \mathbb{R}$ is uniformly continuous and $A\subset X$. Given $\varepsilon>0$, by uniform continuity of $f$ there exists $\delta>0$ such that for all $x,y\in X$, $d(x,y)<\delta$ implies $d(f(x),f(y))<\varepsilon$. Now this same $\delta$ works for the restriction $f\vert_A$, because if you have $x,y\in A$ with $d(x,y)<\delta$, then $x$ and $y$ are also in $X$, so $d(f(x),f(y))<\varepsilon$.
@JakeR: Yes that's it exactly. The precise statement is: Let $A \subset X$ be a subset of a metric space and let $Y$ be a complete metric space. If $f: A \to Y$ is uniformly continuous then there exists a unique extension of $f$ to a continuous function on $\overline{A}$. If $\overline{A}$ is compact then the condition is also necessary by your observations in your question. The proof proceeds exactly in the way you describe but of course you need to check continuity of the extension. –  t.b. Jul 4 '11 at 7:09
Sorry, I had deleted the comment accidentally. We were talking about the condition above for continuous extensions, and I attempted a sketch of a proof: Let $x \in \overline{A}$ but not $A$. Consider sequences in $A$ converging to $x$: these sequences are Cauchy, and since $f$ is uniformly continuous, evaluating each of these sequences termwise by $f$ are also Cauchy. They converge uniquely (interweave two, and we have convergent subsequences in a Hausdorff space) to a point $y$. Associating $f(x) = y$, and doing this for all limit points, gives a continuous extension to $\overline{A}$. –  JakeR Jul 4 '11 at 7:18
In regards to your "too strong" conclusions there's a slightly more general statement one can make. Let $f: (X,d_1) \rightarrow (Y,d_2)$ be a continuous function and $A \subset X$ compact. Consider $g:=f|_A$ then $g$ is continuous (with A given the metric induced from $X$). Then we have that $g$ is a continuous function whose domain is compact, so $g$ is uniformly continuous. That is $f$ is uniformly continuous on $A$.


# >The value of $\mathop{\sum\sum}_{0\leq i<j\leq n}\binom{n}{i}\cdot \binom{n}{j}$

Find the value of $$\mathop{\sum\sum}_{0\leq i<j\leq n}\binom{n}{i}\cdot \binom{n}{j}$$

I get the result: $$\frac{1}{2}\left(2^{2n}-\binom{2n}{n}\right)$$ via a numeric argument.

My question is: Can we solve it using a combinational argument?

My Numeric Argument: $$\left(\sum^{n}_{r=0}\binom{n}{i}\right)^2=\sum^{n}_{r=0}\binom{n}{i}^2+2\mathop{\sum\sum}_{0\leq i<j\leq n}\binom{n}{i}\cdot \binom{n}{j}$$

So here $$\displaystyle \sum^{n}_{r=0}\binom{n}{i} = \binom{n}{0}+\binom{n}{1}+.....+\binom{n}{n} = 2^n$$

and $$\displaystyle \sum^{n}_{r=0}\binom{n}{i}^2=\binom{n}{0}^2+\binom{n}{1}^2+.....+\binom{n}{n}^2 = \binom{2n}{n}$$

above we have calculate Using $$(1+x)^n = \binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+.....+\binom{n}{n}x^n$$

and $$(x+1)^n = \binom{n}{0}x^n+\binom{n}{1}x^{n-1}+\binom{n}{2}x^{n-2}+.....+\binom{n}{n}x^0$$

Now calcualting Coefficient of $x^n$ in $$(1+x)^n\cdot (x+1)^n = (1+x)^{2n} = \binom{2n}{n}$$

So we get $$\mathop{\sum\sum}_{0\leq i<j\leq n}\binom{n}{i}\cdot \binom{n}{j} = \frac{1}{2}\left[2^{2n} - \binom{2n}{n}\right]$$

Thanks

• Really sloppy to write $(1+x)^{2n}=\binom{2n}{n}$. – Thomas Andrews Feb 27 '16 at 17:35
• I don't agree on how you obtain $\binom{2n}{n}$. What did you do with the mixed terms? EDIT: Now, I see what you did. You only look at the coefficients for $x^n$. Nice. – Friedrich Philipp Feb 27 '16 at 17:36
• Mixed terms? @FriedrichPhilipp – Thomas Andrews Feb 27 '16 at 17:36
• Why don't you ask up front for a combinatorial proof, if that is your main question, rather than forcing answerers to read a non-combinatorial proof? – Thomas Andrews Feb 27 '16 at 17:39
• @Thomas Andrews: Yes. In $(1+x)^n(1+x)^n$. But I edited my comment. – Friedrich Philipp Feb 27 '16 at 17:39

Consider two sets, $A$ and $B$ each with $n$ elements. All elements are considered distinct.

$\displaystyle \sum_{0 \leq i < j \leq n} \binom{n}{i} \binom{n}{j}$ can be interpreted as the number of ways to pick a non-empty subset of $A \cup B$ with the requirement that the number of elements from $A$ who are picked is strictly smaller than the number of elements from $B$ who are picked.

$2^{2n}$ counts the total number of ways to pick a subset of any size from $A \cup B$. The number of cases where the same number of elements are picked from $A$ and $B$ (including the empty set) is obtained from the sum $\displaystyle \sum_{i=0}^n \binom{n}{i}^2$.

By symmetry, half of the $\displaystyle 2^{2n} - \sum_{i=0}^n \binom{n}{i}^2$ cases have more elements from $A$ compared to $B$.

The identity $\displaystyle \sum_{i=0}^n \binom{n}{i}^2 = \binom{2n}{n}$ matches the result with yours.

I do not know of a combinatorial argument for this last identity though. Does anyone have any?

• The last can be rewritten $\sum \binom{n}{i}\binom{n}{n-i}$ and the terms represent the number of ways to choose $i$ elements from $A$ and $n-i$ elements from $B$, which, when summed, is the number of ways to choose $n$ elements from $A\cup B$. – Thomas Andrews Feb 27 '16 at 17:44
• How many ways to choose $n$ people from $n$ boys and $n$ girls? For every decision about which $i$ boys we will pick, there are $\binom{n}{i}$ ways to decide which girls we will not pick. – André Nicolas Feb 27 '16 at 17:48

A bijective correspondence can be established between this issue and the following one:

[Dealing with the LHS of the equation :]

Let $$S$$ be a set with Card(S)=n.

Consider all (ordered) pairs of subsets $$(A,B)$$ such that

$$A \subsetneqq B \subset S. \ \ (1)$$

[Dealing with the RHS of the equation :]

Consider all subsets of a set $$T$$ with $$2n$$ elements, then exclude a certain number of them (to be precised later), $$T$$ being defined as :

$$T:=S \cup I \ \ \ \ \text{with} \ \ \ \ \ I:=\{1,2,\cdots n\}.$$

Let $$C$$ be any subset of $$T$$. We are going to establish (in the "good cases") a correspondence between $$C$$ and an ordered pair $$(A,B)$$ verifying (1).

Let us define first a certain fixed ordering of the elements of $$S$$ :

$$a_1 < a_2 < \cdots < a_n. \ \ (2)$$

Let $$B:=T \cap S$$ and $$J:=T \cap I$$. Three cases occur :

• If $$Card(J), $$J$$ is the set of indices "selecting" the elements of $$B$$ that belong to $$A$$ in the ordered set $$S$$.

• If $$Card(J)>Card(B)$$, we switch the rôles of indices and elements. This accounts for the half part of the formula: indeed this second operation will give the same sets $$(A,B)$$.

• If $$Card(J)=Card(B)$$, which happens in $$2n \choose n$$ cases, such cases cannot be placed in correspondence with a case considered in (1), thus have to be discarded.

I know this could be written in a more rigorous way, but I believe the main explanations are there.


# How to determine the number of combinations in a conditional group?

I have a probability/combinatorics question that is giving me trouble specifically I don't understand why there is a need for the denominator $6$ "the different orderings of $3$ people" was the explanation but I'm not sure why that is necessary. I've put the question below and my thought process as well.

Question:

If a committee of $3$ people is to be selected from among $5$ married couples so that the committee does not include two people who are married to each other, how many such committees are possible?

Thought Process:

1. First person of the $3$ selected can be any one among the $10$ folks $$\frac{10!}{(1!)(9!)}$$

2. Second Person of the $3$ selected must not be married to the first person so lowers availability to $8$ folks to choose from $$\frac{8!}{((1!)(7!)}$$

3. Third Person of the $3$ selected must not be married to either the first or second person - lowers availability to $6$ folks to choose from $$\frac{6!}{(1!)(5!)}$$

4. Number of possible ways a committee can be formed: $10 \times 8 \times 6 = 480$

***So I'm able to get to the $480$ but the answer explanation says that I need to divide this $480$ by $6$ and I don't know why because it sounds illogical to do so. Any help would be amazing.

• Please read this tutorial on how to typeset mathematics on this site. – N. F. Taussig Apr 24 '18 at 10:26
• Consider a related but easier problem: how many ways can you choose a committee of $5$ men from the $5$ married couples? Let's do it your way: choose any man, $5$ choices; choose another man, $4$ choices, and so on. So there are $5\times4\times3\times2\times1=520$ ways to choose a committee of $5$ men. Does that seem like the logical answer? Or should we divide by $5!$ and get a final answer of $1$? – bof Apr 24 '18 at 10:44

you can do this way too

trio committee with no couples togeather= (total no. of committees - number of committees with couples togeather)

$$=(^{10} C_{3}-^5 C_{1}.8)=80 ways$$

foot note:

$^5 C_{1}:-$ for choosing $1$ pair or couple(i.e, $2$ peoples) and after we've chosen $2$ peoples we are left with $8$ choices for $3rd$ person to form committee that is why we multiply them to get

$^5 C_{1}.8$

• Thank you so much!! This different approach was interesting and let me think of it differently 🙏😃 – Omar Apr 24 '18 at 12:06

Suppose the three people who are selected are Anne, Barbara, and Charles.

There are six orders in which those same three people could be selected:

Anne, Barbara, Charles

Anne, Charles, Barbara

Barbara, Anne, Charles

Barbara, Charles, Anne

Charles, Anne, Barbara

Charles, Barbara, Anne

However, all six choices constitute the same committee. Therefore, you need to divide your answer by the $3! = 6$ orders in which you could obtain the same three people, which yields the answer $$\frac{10 \cdot 8 \cdot 6}{3!} = \frac{10 \cdot 8 \cdot 6}{6} = 10 \cdot 8 = 80$$

Alternate Approach: Choose three of the five couples from which to choose the committee members in $\binom{5}{3}$ ways, then choose one of the two members from each selected couple, which can be done in $2^3$ ways. Thus, the number of possible selections is $$\binom{5}{3} \cdot 2^3 = 10 \cdot 8 = 80$$

• Thank you so much mate this was super super helpful 🙏😃 – Omar Apr 24 '18 at 12:05