# Thread: Analysis proof help needed

1. ## Analysis proof help needed

I've been driving myself crazy with the following proof:

Let E be an infinite set, and F be a finite subset of E. Let G = E\F (That is, E minus F). Prove that G is equivalent to E.

If I know that if E is countable, I can prove that G is countable, but what if E is uncountable?? Please help!

2. Originally Posted by stuckinanalysis
I've been driving myself crazy with the following proof:

Let E be an infinite set, and F be a finite subset of E. Let G = E\F (That is, E minus F). Prove that G is equivalent to E.

If I know that if E is countable, I can prove that G is countable, but what if E is uncountable?? Please help!
suppose E is uncountable but G is countable. we have $E=G \cup F.$ clearly G can't be finite. so to get a contradiction, you need to show that the union of an infinite countable set with a finite set is

infinite countable. to show this let $G=\{x_n: \ n \in \mathbb{N} \}, \ F=\{y_1, \cdots, y_m\}.$ define the map $f: G \cup F \longrightarrow \mathbb{N},$ by: $f(y_j)=j, \ f(x_n)=n+m.$ obviously $f$ is bijective and hence $G \cup F$ is countable.

3. You proved that G is uncountable, but does this prove G is equivalent to E? The set of real numbers and the power set of the real numbers are both uncountable, but they are not equivalent.

4. Originally Posted by stuckinanalysis
You proved that G is uncountable, but does this prove G is equivalent to E? The set of real numbers and the power set of the real numbers are both uncountable, but they are not equivalent.
you're right! i didn't see "equivalent"! ok, the problem is now much more interesting! so we have an infinite set $G$ and a finite set $F.$ we want to show that $G$ and $G \cup F$ are equivalent.

since we're dealing with cardinality here, we may assume that $G \cap F = \emptyset.$ now apply Zorn's lemma to find a set $G^*=\bigcup_{i \in I} G_i,$ with $G_i \subseteq G,$ which is maximal with respect to these properties:

1) every $G_i$ is countable,

2) $G_i \cap G_j = \emptyset$ for all $i, j \in I$ with $i \neq j.$

then maximality of $G^*$ implies that $G-G^*=H$ is finite. choose an $i_0 \in I$ and replace $G_{i_0}$ with $G_{i_0} \cup H.$ then: $G = \bigcup_{i \in I} G_i,$ and $G \cup F=(G_{i_0} \cup F) \cup \bigcup_{i \neq i_0}G_i.$ also it's clear that $G_i$ will still satisfy

the properties 1) and 2). thus, since $G_{i_0} \cup F$ and all $G_i$ are countable, they're equaivalent to $\mathbb{N},$ hence both $G$ and $G \cup F$ are equivalent to $I \times \mathbb{N}, \ \color{red}(*)$ and hence they must be equivalent. Q.E.D.

$\color{red}(*)$ the reason is that if $f_i: G_i \longrightarrow \mathbb{N}, \ i \in I,$ is a bijection, then you can define $f: \bigcup_{i \in I}G_i \longrightarrow I \times \mathbb{N}$ by: $f(g_i)=(i,f_i(g_i)).$ since $G_i$ are disjoint, $f$ is well-defined. bijectivity of $f$ is a trivial result

of bijectivity of each $f_i.$

5. Originally Posted by NonCommAlg
... now apply Zorn's lemma
But is that cheating?

There is a known result in ZFC which says if $X$ is an infinite set and $Y$ a subset with $|Y| < |X|$ then $|X - Y| = |X|$. This can be easily proven using axiom of choice and cardinality. However, it still relies on axiom of choice. When I saw this problem I wondered whether it is possible to solve it without any choice.

And if you are using choice then why use Zorn's lemma, simply put a well-ordering on $G$ and start mapping the least elements to the least elements. I think that might work and be a lot simpler.

6. Thanks. That gives me a lot to think about.

Here is a follow up/ similar question. Can it be done the same way?

Let X be an uncountable set, and let Y be a countable subset of X. How can I prove that X is equivalent to X U Y?

7. Originally Posted by stuckinanalysis
Let E be an infinite set, and F be a finite subset of E. Let G = E\F (That is, E minus F). Prove that G is equivalent to E.
I don't think you need the axiom of choice for this. Take a sequence $(x_n)$ of distinct points in E in which the elements of F constitute the first N elements of the sequence. Then define a bijection f from E to E\F by taking $f(x_n)=f(x_{N+n})$, and taking f to be the identity on $E\setminus\{x_n:n\in\mathbb{N}\}$.

8. Originally Posted by stuckinanalysis
Here is a follow up/ similar question. Can it be done the same way?

Let X be an uncountable set, and let Y be a countable subset of X. How can I prove that X is equivalent to X \ Y?
A similar method should work. Let Z be a countable set in X \ Y. Then Z and Z∪Y have the same cardinality, so let f be a bijection from Z to Z∪Y, and extend f to X \ Y by taking it to be the identity on X \ (Z∪Y).

9. Originally Posted by Opalg
Take a sequence $(x_n)$ of distinct points in E in which the elements of F constitute the first N elements of the sequence.
How?
We not supposed to be chosing.

I only see one way of doing this. Let $g$ be a choice function on $\mathcal{P}(E)$ i.e. $g(S) \in S$ for all $S\not = \emptyset$. Since $F$ is finite there is a bijection $f:n\to F$. Define a sequence, by recursion, $x: \mathbb{N} \to E$ by $x_k = f(k)$ for $k and $x_k = g( E - \{ x_i | i < n\})$. Then follow through with your argument.

10. Originally Posted by ThePerfectHacker
Originally Posted by Opalg
Take a sequence $(x_n)$ of distinct points in E in which the elements of F constitute the first N elements of the sequence.
How?
We not supposed to be choosing.
Just because you are making a choice, it doesn't necessarily follow that you require the axiom of choice. You only need to invoke that axiom if you want to make a simultaneous choice of an element from each of an infinite collection of sets. That is not the case here. The elements $x_n$ are chosen one at a time in an inductive fashion. You don't need the axiom of choice for that.

11. Originally Posted by Opalg
Just because you are making a choice, it doesn't necessarily follow that you require the axiom of choice. You only need to invoke that axiom if you want to make a simultaneous choice of an element from each of an infinite collection of sets. That is not the case here. The elements $x_n$ are chosen one at a time in an inductive fashion. You don't need the axiom of choice for that.
Are you sure? What you are doing is this:
$x_{n+1} = \text{ some element of }(E - \{ x_i | i < n\})$

The thing is that this function is not really defined.
That is why we need choice to define such a function.
---
If that (what I just did above) was really okay than we can apply the same argument to transfinite sequences:
As a consequence of which the axiom of choice would be a provable from the other axioms.

12. Originally Posted by ThePerfectHacker
Are you sure? What you are doing is this:
$x_{n+1} = \text{ some element of }(E - \{ x_i | i < n\})$

The thing is that this function is not really defined.
That is why we need choice to define such a function.
No, I'm just using the definition of "nonempty". If a set is nonempty then there exists an element in that set. I'm calling that element $x_{n+1}$.

Almost every proof in analysis starts by saying "Let ε>0." That involves choosing an element from the set of positive real numbers. But it is not true that every such proof requires the axiom of choice.

Originally Posted by ThePerfectHacker
If that (what I just did above) was really okay than we can apply the same argument to transfinite sequences:
As a consequence of which the axiom of choice would be a provable from the other axioms.
No, you could never get beyond a countable set that way, because you are only constructing one element at at time.


# Can the relation of $a^{p} \equiv a\pmod {2p}$ if (p prime and p>2) be added to Fermat's little theorem?

Fermat's little theorem

$a^{p} \equiv a\pmod p$ if p prime

I have noticed that the formula $a^{p} \equiv a\pmod {2p}$ (p prime and p>2) can be also written by using Fermat's little theorem

Proof: Let $a$ be any integer and $p>2$ be some prime number.

$a^{p} \equiv a\pmod p$

$a^{p}-a \equiv 0\pmod p$

$a(a^{p-1}-1) \equiv 0\pmod p$

$a(a^{p-1}-1) \equiv 0\pmod p$

$F(a)=(a^{p-1}-1)$ can be divided to $(a-1)$ because $F(1)=(1^{p-1}-1)=0$

$F(a)=(a^{p-1}-1)=(a-1).R(a)$ //R(a) is polinom that has degree p-2 and coefficients are Integer numbers.

$a (a-1) R(a) \equiv 0\pmod p$

$a (a-1)$ is always an even number so $a (a-1) R(a)$ will be always even number and also can be divided to $p$ because of Fermat's little theorem . Thus $a^{p} \equiv a\pmod {2p}$ if (p prime and p>2) .

I searched internet but I have not seen that relation in the internet.

Is it known formula?Please help if it is known relation(I would like to learn what the subject is )

Sorry if someone else asked the same question in here.

Thank you for answers and helps.

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If $a$ is any integer, and $n$ a positive integer, then $a^n \equiv a \mod 2$. Your equation is a combination of this fact and Fermat's little theorem. Both are known congruences, but I'm note sure there's a name for their combination. –  Joel Cohen Feb 4 '12 at 14:49

As Joel rightly observed :If $a$ is any integer, and $n$ a positive integer, then $a^n\equiv a \pmod2$

So : $a^p\equiv a \pmod2$ , and $a^p\equiv a \pmod p$

According to Chinese Remainder Theorem :

$a^p \equiv a \pmod { \operatorname{lcm} (2,p)}$

,and since $p$ is an odd prime it follows that : $\operatorname{lcm}(2,p) = 2p$ , therefore :

$a^p \equiv a \pmod {2p}$

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The answer, Maybe, can be perfect and if we add some additional steps to show how proof of Chinese Remainder Theorem. c,d,e are integers if $a^p\equiv a \pmod2$ then $a^p=a+2.c$ $a^p-a=2.c$ if $a^p\equiv a \pmod p$ then $a^p=a+p.d$ $a^p-a=p.d=2.c$ $d=2.e$ can be written because $2.c$ is even number Thus $a^p-a=p.d=2.c=2.p.e$ $a^p-a\equiv 0 \pmod{2p}$ $a^p\equiv a \pmod{2p}$ Thanks in advice –  Mathlover Feb 4 '12 at 19:32

You are correct I think, but the way you write it down makes it look more difficult than it is. Also you should be more specific on whether you are talking about one particular $a$ or "$\forall a$".

So let me show how I would write this down. First, let $a$ be any integer and $p>2$ be some prime number. Then

$$a^p \equiv a \pmod{2p} \iff 2p \mid a^p - a \iff 2\mid a^{p}-a \text{ and } p\mid a^p-a$$ The last equivalence holds because $2$ and $p$ are two different primes.

But $2\mid a^p-a$ for all $a$ and $p$: if $a$ is even, then clearly $a^p$ will be even as well; if $a$ is odd, then so is $a^p$ and hence $a^p-a$ will be even again. So this is indeed equivalent to

$$\dots \iff p\mid a^p-a \iff a^p \equiv a \pmod p.$$

-
Thanks for answer. –  Mathlover Feb 4 '12 at 14:46
No actually it's a continuation of the chain of equivalences up above. So it should be read as $$a^p \equiv a \pmod{2p} \iff \dots \iff a^p\equiv a \pmod{p}$$ –  Myself Feb 4 '12 at 14:51
This is the special case $\rm\ m = 2\:p,\ p\:$ odd, of the following generalization of Fermat's little theorem, quoted from Bill Dubuque's post
THEOREM $\$ For naturals $\rm\: k,m>1$
$\qquad\rm m\ |\ a^k-a\$ for all $\rm\:a\in\mathbb N\ \iff\ m\:$ is squarefree and prime $\rm\: p\:|\:m\: \Rightarrow\: p-1\ |\ k-1$


# Computing $\int_0^\pi \sin(x) \; dx$ using the definition.

A colleague of mine and I, in the course of teaching integral calculus for the umpteenth time, were wondering if we could expand the class of examples that our students are exposed to when computing Riemann integrals from the definition. Most such examples involve polynomial functions, and they work nicely because we have some well-known formulas, like

$\displaystyle\sum_{i=1}^n 1 = n$,

$\displaystyle\sum_{i=1}^n i = \frac{n(n+1)}{2}$,

and $\displaystyle\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$. (And certainly, there are others.)

But we wanted to expand the class of examples beyond polynomial functions, and started to think about trig functions. We came up with a computation for $\int_0^\pi \sin(x)\;dx$ using the definition of the Reimann integral, which I provide below.

Question: Is the computation below in the literature somewhere? If so, where?

Computation:

Using a Riemann sum with right endpoints, we have

$$\int_{0}^\pi \sin(x)\; dx = \lim_{n \to \infty}\sum_{i=1}^n \sin(x_i) \Delta x$$ where $\Delta x = \frac{\pi}{n}$ and $x_i = i\Delta x$.

The trick is to rewrite the Riemann sum $\sum_{i=1}^n \sin(x_i) \Delta x$ as a telescoping sum using the difference of cosines identity: $$\cos(b) - \cos(a) = 2 \sin\left(\frac{a+b}{2}\right)\sin\left(\frac{a-b}{2}\right)$$

Setting $a = \frac{2i+1}{2}\Delta x$ and $b = \frac{2i-1}{2}\Delta x$ yields

$$\cos\left(\frac{2i-1}{2}\Delta x\right) - \cos\left(\frac{2i+1}{2}\Delta x\right) = 2 \sin\left(i \Delta x\right) \sin\left(\frac{\Delta x}{2}\right)$$

Solving for $\sin(i \Delta x)$ gives

$$\sin(i \Delta x) = \frac{\cos\left(\frac{2i-1}{2}\Delta x\right) - \cos\left(\frac{2i+1}{2}\Delta x\right)}{2 \sin\left(\frac{\Delta x}{2}\right)}$$

The Riemann sum now is $$\sum_{i=1}^n \sin(i \Delta x) \Delta x = \frac{\frac{\Delta x}{2}}{\sin\left(\frac{\Delta x}{2}\right)} \sum_{i=1}^n \left[\cos\left(\frac{2i-1}{2}\Delta x\right) - \cos\left(\frac{2i+1}{2}\Delta x\right)\right]$$

The latter sum is telescoping, and so $$\sum_{i=1}^n \sin(i \Delta x) \Delta x = \frac{\frac{\Delta x}{2}}{\sin\left(\frac{\Delta x}{2}\right)}\left[ \cos\left(\frac{\Delta x}{2}\right) - \cos\left(\frac{2n+1}{2}\Delta x\right)\right]$$

Now recalling that $\displaystyle \Delta x = \frac{\pi}{n}$ and taking $n \to \infty$ we have

\begin{align} \int_0^\pi \sin(x)\;dx &= \lim_{n \to \infty} \frac{\frac{\Delta x}{2}}{\sin\left(\frac{\Delta x}{2}\right)}\left[ \cos\left(\frac{\Delta x}{2}\right) - \cos\left(\frac{2n+1}{2}\Delta x\right)\right]\\ & = 1 \cdot \left[ \cos(0)-\cos(\pi)\right]\\ & = 2 \end{align}

One nice thing about this computation is that you can replace $\pi$ with an arbitrary upper limit of integration $c$ and compute that $\int_0^c \sin(x) \;dx = 1 - \cos(c)$. From there, one can easily compute $$\int_a^b \sin(x) \; dx = \int_0^b \sin(x) \; dx - \int_0^a \sin(x) \; dx = -\cos(b) + \cos(a)$$ (as predicted by FTC).

Question: Is the computation above in the literature somewhere? If so, where?

• I don't know, but our teacher made us do this. $\int_a^b \sin(x)\,dx$ from reimann sum I mean. – Guy Apr 3 '14 at 18:43
• This answer contains telescoping ideas. – Ian Mateus Apr 3 '14 at 18:43
• I haven't seen it in a textbook or anywhere else, but I did it once myself in order to see how the limit $\sin x/x \rightarrow 1$ would play into the calculation. – Jason Zimba Apr 3 '14 at 18:49
• Not exactly "literature", but the sum $\sum \sin(n)$ is telescoped using a similar trick in this answer in order to apply the Dirichlet test to the series $\sum \frac{\sin(n)}{n}$. – Mike F Apr 4 '14 at 16:34

Below are some references I found in my "math library" at home this morning. Before posting just now, I quickly searched to see which were online and I've included links for those I found. The Boyer paper URL is a JSTOR item I don't have access to, but I've included the URL anyway because many here are probably at colleges/universities that have JSTOR access.

At the risk of pointing out the obvious, this method doesn't prove the Riemann integrability of ${\sin x},$ since equal length partitions are used along with endpoint evaluations. For instance, under this restriction the limit of the Riemann sums of the characteristic function of the rationals on a given compact interval exists. What this calculation does is give the value of the (Reimann) integral on a given compact interval under the assumption that the integral exists.

Direct quotes from original sources are indicated by italics. Within such a quote: (1) italics in the original is indicated by non-italics here; (2) brief additions and/or words of further explanation by me are indicated by non-italics between square braces "[stuff by me]"; (3) omissions in the original are indicated using "$[\dots]$".

[1] [Author not known], Sur la sinussoide [On the sinusoid], Question D'Examen, Nouvelles Annales de Mathématiques (1) 7 (1848), 436-437.

The (net) area above the $x$-axis and below the graph of $y = \sin{x},$ between $x = x_1$ and $x = x_2$ is found. The interval $[x_1, \, x_2]$ is divided into $n+1$ many abutting intervals each of length $h,$ so that $x_2 - x_1 = (n+1)h,$ and the identity $$h [\sin{x_1} + \sin{(x_1 + h)} + \sin{(x_2 + 2h)} + \cdots + \sin{(x_1 + nh)} \;\; = \;\; \frac{h \sin \left(x_1 + \frac{nh}{2}\right) \sin \frac{h}{2}(n+1)}{\sin{\frac{h}{2}}}$$ along with $\frac{nh}{2} = \frac{x_2 - x_1 - h}{2}$ is used to obtain the value $2\sin{\left(\frac{x_2 + x_1}{2}\right)} \sin{\left(\frac{x_2 - x_1}{2}\right)},$ which equals $\cos{x_1} - \cos{x_2}.$ Note: There is a typo at one point in the original, in which $\sin{(x_2 + h)}$ appears instead of $\sin{(x_2 + 2h)}.$ Also, the text only observes that the area's expression in terms of the sine function reduces to $1 - \cos{x_2}$ when $x_1 = 0.$

[2] William Elwood Byerly, Elements of the Integral Calculus, 2nd edition, Ginn and Company, 1892, xvi + 339 + 11 + 32 pages.

See Chapter VIII, Article 81, p. 76, Examples (2): By the aid of the trigonometrical formulas $$\cos{\theta} + \cos{2\theta} + \cos{3\theta} + \cdots + \cos{(n-1)\theta} = \frac{1}{2}\left[\sin{n\theta} \cot{\frac{\theta}{2}} - 1 - \cos{n\theta}\right],$$ [and] $$\sin{\theta} + \sin{2\theta} + \sin{3\theta} + \cdots + \sin{(n-1)\theta} = \frac{1}{2}\left[(1 - \cos{n\theta}) \cot{\frac{\theta}{2}} - \sin{n\theta}\right],$$ prove that $\;\;\int_{a}^{b}\cos{x}.dx = \sin{b} - \sin{a},\;$ and $\;\int_{a}^{b}\sin{x}.dx = \cos{a} - \cos{b}.$

[3] Carl Benjamin Boyer, History of the derivative and integral of the sine, Mathematics Teacher 40 #6 (October 1947), 267-275.

[4] Mark Bridger, A note on areas under a sine curve, The Pentagon 18 #1 (Fall 1958), 24-26.

At the time Mark Bridger was a student at the Bronx High School of Science (New York).

[5] Richard Courant and Fritz John, Introduction to Calculus and Analysis, Volume I, John Wiley and Sons (Interscience Publishers), 1965, xxiv + 661 pages.

See Chapter 2.2.e (Integration of $\sin x$ and $\cos x$), p. 135.

[6] Godfrey Harold Hardy, A Course of Pure Mathematics, 10th edition, Cambridge University Press, 1952, xii + 509 pages.

See Chapter VII, Article 164, p. 320, item 3. Calculate $\int_{a}^{b}x^2 \, dx,$ $\int_{a}^{b} \cos{mx} \, dx$ and $\int_{a}^{b} \sin{mx} \, dx$ by the method of Ex. 1. Note: Hardy's reference to "the method of Ex. 1" is simply the method of partitioning the interval $[a,b]$ into finitely many equal length subintervals, forming the relevant Riemann sum, and taking a limit. The needed trigonometric identities are not given as a hint, but I do notice that a closely related identity appears in Exercise 4 at the top of p. 323 for another purpose.

[7] Kenneth Sielke Miller and John Breffni Walsh, Elementary and Advanced Trigonometry, Harper and Brothers, 1962, xii + 350 pages.

See Chapter 10.3 (Solution of the Area Problem), pp. 215-217: (begins) We now turn to the second problem mentioned at the beginning of the chapter, namely, that of finding the area under one arch of a sine curve.

[8] Isidor Pavlovich Natanson, Summation of Infinitely Small Quantities, Topics in Mathematics, D. C. Heath and Company, 1963, viii + 59 pages.

See Chapter 6 (The Sinusoid), Articles 26-35, pp. 46-57. Specifically, see Article 29 (A trigonometric sum), Article 30 (A subsidiary inequality), Article 31 (The sine of an infinitely small angle), and Article 32 (The quadrature of the sinusoid) on pp. 46-51. The remaining articles in this chapter involve applications, mainly in calculating the volume of a rotating sinusoid and calculating effective (electrical) current.

[9] John Meigs Hubbell Olmsted, Intermediate Analysis. An Introduction to the Theory of Functions of One Real Variable, The Appleton-Century Mathematics Series, Appleton-Century-Crofts, 1956, xiv + 306 pages.

See Chapter 5, Article 503, p. 145, Exercise #19 & 20. The reader is asked to show that $\int_{a}^{b}\sin x \, dx = \cos{a} - \cos{b}$ and $\int_{a}^{b}\cos x \, dx = \sin{b} - \sin{a}$ by using equal length partitions of $[a,b].$ The relevant trig. identities are indirectly provided (by means of a hint, the reader is led to obtain these identities).

[10] Michael David Spivak, Calculus, 3rd edition, Publish or Perish, 1994, xiv + 670 pages.

See Chapter 15, p. 320, Problem 33.

[11] Isaac Todhunter, A Treatise on the Integral Calculus, 6th edition, MacMillan and Company, 1880, viii + 408 pages.

See Chapter IV, Article 38, pp. 52-53: (begins) Suppose we wish to find the integral of $\sin x$ between limits $a$ and $b$ immediately from the definition. By Art. 4 we have to find the limit $[\ldots]$ Note: The copy I found online is dated 1863, but what I've quoted seems to be the same calculation and it's in the same location in the book.

[12] Edwin Bidwell Wilson, Advanced Calculus, Ginn and Company, 1911, x + 566 pages.

See Chapter I, p. 30, Exercise 9. With the aid of the trigonometric formulas $[\ldots]$ show $(\alpha) \; \int_{a}^{b} \cos{x}\,dx = \sin{b} - \sin{a},\;\;$ [and] $\;\;(\beta) \; \int_{a}^{b} \sin{x}\,dx = \cos{a} - \cos{b}.$

(ADDED 3 DAYS LATER) I thought it would be useful to archive in my answer some references for evaluating definite integrals of other functions by Reimann sums. However, because I want to get this finished during my lunch hour and I don't want to substantially increase the length of this already long answer, I will use briefer citations.

Harding/Barnett, Solution to Calculus Problem #369, Amer. Math. Monthly 22 #6 (June 1915), 208-210.

Two solutions for the evaluation of $\int_{a}^{b} \ln{x}\,dx$. The second solution makes use of Fermat's method of geometric subdivision of the interval $[a,b].$

Barnett, Solution to Calculus Problem #366, Amer. Math. Monthly 22 #5 (May 1915), 168-169.

One solution for the evaluation of $\int_{a}^{b}{\sin}^{-1}x\,dx.$

Eli Maor, On the direct integration $\int_{a}^{b}x^{n}\,dx,$ Int. J. Math. Educ. Sci. Technol. 5 (1974), 199-200.

The method involves working with the sum of the left endpoint sums and the right endpoint sums. The method might be similar to what Otto Dunkel does in Note on the quadrature of the parabola, Amer. Math. Monthly 27 #3 (March 1920), 116-117.

Finally, the following are similar to the kind of things found in Natanson's Summation of Infinitely Small Quantities:

W. R. Longley, Some limit proofs in solid geometry, Amer. Math. Monthly 31 #4 (April 1924), 196-202.

Joseph B. Reynolds, Some applications of algebra to theorems in solid geometry, Mathematics Teacher 18 #1 (January 1925), 1-9.

Jos. B. Reynolds, Finding plane areas by algebra, Mathematics Teacher 21 #4 (April 1928), 197-203.

• This is a great list of references! Thanks! – wckronholm Apr 4 '14 at 14:39

I suspect Tom Apostol's calculus textbook has this (i.e., finds $\int_0^\pi\sin x\,dx$ by using limits of Riemann sums without the fundamental theorem).

I question the place of Riemann sums in the curriculum. Rigorous definitions are unsuitable for a calculus-for-liberal-education course unless the students are unusual. Such a course should acquaint students with the reason why calculus is important in the course of human events over recent and coming centuries, and with the fact that it overcame difficulties like how to define $\text{rate} = \dfrac{\text{distance}}{\text{time}}$ when the distance and the time are both $0$, and how to to find $$\sum\text{force}\times\text{distance}$$ when there are infinitely many infinitely small distances, each with its own value of "force". The conventional calculus course is a watered-down version of a course for students who come in with a prior desire or a pre-identified need to understand calculus, rather than for students who are there in order to pay a price in homework for a grade to impress employers, and whom one should be trying to seduce into another course of action.

If you expect students to marvel at the fact that it's possible to find this integral by using limits of Riemann sums, I expect the way they will think of it and remember it is "We did some technical stuff and turned it in and got graded", and if you have them plod through finding the integral by antidifferentiating and substituting endpoint values, then the way they will think of it and remember it is "We did some technical stuff and turned it in and got graded", and they won't know the difference. (I'm hoping for a malpractice suit against every university that knowingly encourages unqualified students to take calculus, to the point where those are 99.9% of the ones who show up. They bring in tuition money.)

Another occasion for Riemann sums is numerical integration, but that doesn't seem to be your purpose.

• Apostol's book gives the formulas for $\int_a^b \sin(x)\;dx$ and $\int_a^b \cos(x)\;dx$ followed by the statement "we shall not prove these formulas at this stage because they will be derived by an easier method in Chapter 2, with the help of differential calculus." – wckronholm Apr 3 '14 at 18:55
• @wckronholm : OK. Apostol's book may still be a place to look if anyone wants similar material. Also, maybe Otto Toeplitz' "genetic approach" calculus book. – Michael Hardy Apr 3 '14 at 18:58
• $99.9\%$ may be unduly pessimistic. Perhaps I will be accused of being a rosy-eyed optimist, but $95\%$ seems more reasonable. – André Nicolas Apr 3 '14 at 19:15
• @AndréNicolas : Depends on which institution. There are colleges that think of themselves as selective, at which the students are hard-working and intelligent, but have attitudes about math courses: They exist in order to provide an opportunity to show that they can follow instruction and work hard and get a grade, and they've always gotten "A"s in math, and they demand their right to get a course consisting of algorithms for them to follow to get an "A". – Michael Hardy Apr 3 '14 at 19:36
• @MichaelHardy I actually had my class run through this computation in small groups in class today. While it's true that many of them were longing for the Fundamental Theorem of Calculus they "learned" in high school and could give them the answer in about 5 seconds, there were more than a few who really were appreciating all of the pieces that went into the computation. In general, I think students benefit from exposure to this sort of activity because of the potential to broaden their reasoning skills. – wckronholm Apr 3 '14 at 20:15

Alternatively, you can use $e^{i \theta} = \cos(\theta) + i \sin(\theta)$ and the partial sum formula for Geometric series. Using left endpoint approximation instead, the sum we want to evaluate is $$\sum_{k=0}^{n-1} \sin(k \, \Delta x)\,\Delta x$$ where $\Delta x = \frac{\pi}{n}$. Note that $\sum_{k=0}^{n-1} \sin(k \,\Delta x)$ is the imaginary part of $\sum_{k=0}^{n-1} e^{i k \,\Delta x}$, that is $\sum_{k=0}^{n-1} z^k$ where $z = e^{i \,\Delta x}$. This last sum has the closed form $\frac{1-z^n}{1-z} = \frac{2}{1-z}$ using $z^n = e^{i \pi} = -1$. We get $$\sum_{k=0}^{n-1} \sin(k \,\Delta x)\,\Delta x =\mathrm{Im}\left( \frac{2}{1-e^{i\, \Delta x}} \right) \,\Delta x = \mathrm{Im}\left( \frac{2\,\Delta x}{1-e^{i \,\Delta x}} \right).$$ Finally, letting $n \to \infty$ so that $\Delta x \to 0$ we get $$\lim_{\Delta x \to 0} \frac{2\,\Delta x}{1-e^{i \,\Delta x}} = -2\left( \lim_{\Delta x \to 0 } \frac{e^{i \,\Delta x} - 1}{\Delta x} \right)^{-1} = -2\left( \frac{d}{dt} e^{it} \big|_{t=0} \right)^{-1} = -2i^{-1} = 2i$$ so that $$\lim_{n \to \infty} \sum_{k=0}^{n-1} \sin(k \,\Delta x)\,\Delta x = \operatorname{Im}(2i) = 2.$$

• Thanks. We were aware of this method already, and wanted to find an approach which did not rely on complex numbers. – wckronholm Apr 4 '14 at 16:05
• @wckronholm: That makes sense, just thought I'd add it. – Mike F Apr 4 '14 at 16:28
• @wckronholm: It has probably occurred to you already, but if you don't want to use complex numbers, you can still do integrals like $\int_0^1 e^x \ dx$ using geometric series. – Mike F Apr 4 '14 at 16:29
• Thanks, that has indeed occurred to me, and I will be showing this to my students in about 20 minutes! – wckronholm Apr 4 '14 at 16:39


# Binomial formula modulo a prime.

Let $p$ be a prime. I see that modulo $p$, the binomial formula reduces to $$(x+y)^p\equiv x^p+y^p \pmod p$$ because $p \choose k$ is a multiple of $k$ whenever $k=1..p-1$, but don't we have by Fermat's little theorem that $a^p\equiv a \pmod p$ for any integer $a$ and so this is true for $a=x+y$ hence the binomial formula reduces to $(x+y)^p\equiv x+y \pmod p$. Thanks for your help!

• Yes, but you also get $(x+y)^p\equiv x+y \mod p$ straight from FLT without binomial theorem at all. – user281392 Oct 30 '15 at 21:24
• Both are true. $(x+y)^p$ is congruent to $x^p + y^p$ modulo $p$. Now by Fermat's little theorem, $x^p$ is congruent to $x$ and $y^p$ is congruent to $y$ modulo $p$, and so $x^p + y^p$ is congruent to $x + y$ modulo $p$. – Dylan Oct 30 '15 at 21:24
• @user281392. Very nice observation. +1 – Shailesh Oct 31 '15 at 2:56

Yes, everything that you have written is true. It is the case that $$(x+y)^p \equiv x^p+y^p \mod p$$ using the Binomial Theorem, and it is also true that $$(x+y)^p \equiv x+y \mod p$$ by Fermat's Little Theorem.

There is not contradiction here because by Fermat's Little Theorem, we have that $$x^p \equiv x \mod p \quad\text{ and }\quad y^p \equiv y \mod p$$ and so $$x^p+y^p\equiv x+y \mod p$$ as we expect.

Your observation that $$(x+y)^p \equiv x^p+y^p \mod p$$ by the Binomial Theorem is sometimes used to prove Fermat's Little Theorem by induction. The (complete) proof is as follows:

First we show, as noted by you, that $p \mid \binom{p}{k}$ for $1 \leq k < p$. We can prove this either by considering the factors of $p$ in the numerator and denominator of $\binom{p}{k}$, or by noticing that $$\binom{p}{k} = \frac{p}{k}\binom{p-1}{k-1}$$ Thus $$k \mid p\binom{p-1}{k-1}$$ and since $k$ and $p$ are relatively prime, we get that $$k \mid \binom{p-1}{k-1}$$ showing that $$\frac{1}{k}\binom{p-1}{k-1}$$ is an integer, and so $$\binom{p}{k} = p \cdot \frac{1}{k}\binom{p-1}{k-1}$$ is an integer divisible by $p$.

As you noted in your question, $$(x+y)^p = \sum_{k=0}^p \binom{p}{k} x^k y^{p-k} = x^p + y^p + \sum_{k=1}^{p-1} \binom{p}{k} x^k y^{p-k}$$ Since every term in the final sum is divisble by $p$, we get that $$(x+y)^p \equiv x^p + y^p \mod p$$ and this holds for all integers $x$ and $y$.

We can now prove by induction that $$x^p \equiv x \mod p$$ for all integers $x$. The base case of $x=0$ is straightforward since $0^p=0$. Now suppose that $x^p \equiv x \mod p$ for some $x$. Then we have that $$(x+1)^p \equiv x^p + 1^p \equiv x+1$$ using our observation above, and the inductive hypothesis.

Thus the claim holds for $x+1$ as well, and hence for all natural numbers by induction. For the negative integers, we can note that if $x$ is a positive integer then $$(-x)^p = (-1)^p x^p \equiv -x \mod p$$ Here, we use that $x^p \equiv x \mod p$, and also that $(-1)^p \equiv -1 \mod p$, since if $p$ is odd then $(-1)^p = -1$, and if $p=2$ then $(-1)^p = 1 \equiv -1 \mod 2$.

Yes, good point. If $x$ and $y$ are integer, we can simply write $(x+y)^p = x + y \mod p$.

However, the formula $(x+y)^p \equiv x^p + y^p \mod p$ is also interesting because it applies not only to elements of $\mathbb{Z}/p\mathbb{Z}$, but also to any commutative algebra over $\mathbb{Z}/p\mathbb{Z}$ (for example, you may apply it to polynomials with coefficients in $\mathbb{Z}/p\mathbb{Z}$, or for field extensions of $\mathbb{Z}/p\mathbb{Z}$).

You have to show that $p \mid \binom{p}{k}$ when $1 < k < p$. We have: $k!(p-k)! \mid p!=p(p-1)!$ Using unique prime factorization we can write: $k!(p-k)! = a_1^{n_1}\cdot a_2^{n_2}\cdots a_m^{n_m}$ with the $a_i$'s are distinct primes smaller than $p$, and none of the factors divide $p$ since $a_i\nmid p$. Thus $a_i^{n_i} \mid (p-1)!$,and since these factors are pairwise coprime, we have $k!(p-k)! \mid (p-1)!$, this shows $p \mid \binom{p}{k}$, and the conclusion follows from this fact.


# Distribution over the product of three, or n, independent Beta random variables

I would like to calculate the PDF for the product of three independent Beta random variables. Specifically, I would like to find the distribution of the product of the following: $X_1\sim Beta(1,3/2)$, $X_2\sim Beta(3/2,1)$ and $X_3\sim Beta(2,1/2)$.

I can manage to get Mathematica to output a distribution for the case of the the product $X_1 X_2$ by doing the following:

PDF[TransformedDistribution[
u v w, {u \[Distributed] BetaDistribution[1, 3/2],

This takes a few minutes to evaluate. However, when I move to $X_1 X_2 X_3$ Mathematica seems to be eternally stuck on the calculation:

PDF[TransformedDistribution[
u v w, {u \[Distributed] BetaDistribution[1, 3/2],

I could approximate this by a large number of draws from the distributions of $X_1$, $X_2$ and $X_3$; multiplying each set together. However, I would like the analytic form.

Does anyone have any idea how I can do this? The reason I state $n$ in the question is because I would like eventually to generalise this calculation to the product of more $Beta$ random variables.

Best,

Ben

• Might have better luck getting an answer on current SOTA for this over at Math.StackExchange.Com, then writing appropriate MMA code. I don't expect you'll find a compact/low computational complexity solution. – ciao Aug 11 '15 at 22:09
• It has been a while since I've done something like this but if I recall correctly you may be able to do this quicker using the moment generation function of the Beta distributions, performing the product on these, and then matching that form to the form of a know distribution's moment generating function. – Edmund Aug 11 '15 at 22:16
• Here's a useful reference - extends earlier work that required integer parameters to (relatively) arbitrary parameters... – ciao Aug 11 '15 at 22:23
• @Edmund good luck with that. – wolfies Aug 12 '15 at 8:06

Here is an answer that does not use a 3rd party package and works for an arbitrary amount of Beta distributions.

You can make use of a closed form for the product of n Beta distributions from the Handbook of Beta Distribution and Its Applications, Products and Linear Combinations, I. Products, B. Exact Distributions as found on page 57.

This expresses a product of n Beta distributions as the product of a constant $K$ (a function of the Beta distributions parameters in $\Gamma$ functions) and a Meijer-G function of the Beta distribution parameters.

ClearAll[pdfProductBeta];
pdfProductBeta[
α_ /; VectorQ[α, NumericQ],
β_ /; VectorQ[β, NumericQ],
x_Symbol] /; Dimensions[α] == Dimensions[β] :=
Module[{k},
k = Times @@ (Gamma[Plus @@ #]/Gamma[#[[1]]] & /@ (Transpose@{α, β}));
k MeijerG[{{}, α + β - 1}, {α - 1, {}}, x]
]

For the three Beta distributions in the question

a = {1, 3/2, 2}; b = {3/2, 1, 1/2};
pdfProductBeta[a, b, x]
(* 27/32 π MeijerG[{{}, {3/2, 3/2, 3/2}}, {{0, 1/2, 1}, {}}, x] *)

This result matches to @wolfies answer above that makes use of the mathStatica 3rd party package.

The two Beta case plots quickly so we can compare this easily to the TransfromedDistribution PDF from the built in Mathematica functions.

tpdf = PDF[
TransformedDistribution[
u v, {u \[Distributed] BetaDistribution[1, 3/2],

a = {1, 3/2}; b = {3/2, 1};
mpdf = pdfProductBeta[a, b, x];

GraphicsRow[
Plot[#, {x, 0, 1}, PlotRangePadding -> None] & /@ {tpdf, mpdf}]

For the three Beta case we need to limit the upper bound in the plot as it takes very long to calculate there.

a = {1, 3/2, 2}; b = {3/2, 1, 1/2};
mpdf3 = pdfProductBeta[a, b, x];
(* evaluate pdf at a point *)
mpdf3 /. x -> 0.5
(* 0.479319 *)
Plot[mpdf3, {x, 0, 0.9}, Exclusions -> {0}, AxesOrigin -> {0, 0},
PlotRangePadding -> None, PlotRange -> Full]

With the pdfProductBeta function you can construct the pdf for the product of an arbitrary number of Beta distributions without the need of a third party package.

Hope this helps.

• Very nice indeed to have a general solution from the Handbook of Beta Distribution -- I would go further and say that your answer not only constructs the pdf of the product of Beta random variables without the need of an add-on package, ... it does so without needing Mathematica either. – wolfies Aug 12 '15 at 14:03
• @wolfies Well ... I wouldn't want to try constructing, evaluating, or <cringe> plotting it without Mma. :-) It quickly gets process heavy for small $n$. – Edmund Aug 12 '15 at 14:09
• Note that this is no longer needed in version 11 and above as TransformedDistribution directly uses the above method for a product of BetaDistributions. – Edmund Apr 8 '17 at 13:18

Let random variable $X_i \sim Beta(a_i,b_i)$, with pdf $f_i(x_i)$. The OP is interested in 3 specific parameter combinations:

The pdf of $Y = X_2 X_3$, say $g(y)$, is:

where I am using the TransformProduct function from the mathStatica package for Mathematica, and where domain[g] = {y,0,1}.

The pdf of $Z = X_1 X_2 X_3 = Y* X_1$, say $h(z)$, is then:

All done.

Quick Monte Carlo check

It is always a good idea to check symbolic solutions with Monte Carlo methods.

The following plot compares:

• the exact symbolic solution pdf obtained $h(z)$ [ red dashed curve ], to

• an empirical Monte Carlo simulation of the pdf [ blue squiggly curve ]

All looks good.

• That Meijer's function had to show up indicates that things are not too simple. Tho, those parameters look familiar… what happens if you apply FunctionExpand[]? – J. M. will be back soon Aug 12 '15 at 9:15
• Applying FunctionExpand returns the same MeijerG function. I might add that it appears tractable for most of the domain ... but trying to plot it for $x$ close to 1 becomes very slow (which is also why that is 'left out' in the diagram above). – wolfies Aug 12 '15 at 11:44
• Ah, that part I can explain; the contour integration being done under the hood converges rather slowly when the argument is near $1$. A pity that there does not seem to be a simpler form (and I was so sure that there was)… – J. M. will be back soon Aug 12 '15 at 11:46
• @J. M. What is the reason the contour integration converges so slowly near 1? I am trying to generate points of the pdf near there, and was wondering if you had any ideas? – ben18785 Aug 12 '15 at 14:19
• @ben, it's an inherent limitation of the contour integral method for the $G$-function. (The situation is similar to the one for generalized hypergeometric functions; arguments near $1$ are often very difficult cases, numerically speaking.) That's why I was hoping there might be expressions in terms of simpler special functions. I do not have Mathematica at hand to investigate further, unfortunately. – J. M. will be back soon Aug 12 '15 at 15:04

The problem can indeed be solved explicitly for the product of n = 3 Beta-distributed variables and the explicit parameters of the OP.

In part 1 I show only the results, and turn later, in part 2, to the details of calculation in Mathematica, part 3 is discussion.

Part 1 Results

The PDF of the Beta distribution is given by

f[x_, a_, b_] = Simplify[PDF[BetaDistribution[a, b], x], 0 < x < 1]

(*
((1 - x)^(-1 + b) x^(-1 + a))/Beta[a, b]
*)

Let the random variables and their rescpective distributions be X1 ~ Beta(1,3/2) , X2 ~ Beta(3/2,1) and X3 ~ Beta(2,1/2), and let

f2n(x) = dist( X1*X2 )
f3n(x) = dist( X1*X2*X3 )

be the requested distributions.

Then we find

f2n[x_] = 9/2 (Sqrt[1 - x] - Sqrt[x] ArcTan[Sqrt[-1 + 1/x]]);

f3n[x_] = 27/4 (EllipticE[1 - x] - x EllipticK[1 - x]) -
27/16 \[Pi] MeijerG[{{}, {3/2, 3/2, 3/2}}, {{1/2, 1, 1}, {}}, x];

The functions are normalized

Integrate[f2n[x], {x, 0, 1}]

(* 1 *)

Integrate[f3n[x], {x, 0, 1}]

(* 1 *)

A plot of the two functions is shown here

Plot[{f2n[x], f3n[x]}, {x, 0.0001, 0.9}, PlotRange -> {{0, 1}, {0, 4}},
PlotLabel ->
"Distributions of the product of\ntwo (yellow) and three (blue)\nbeta \
distributed random variables", AxesLabel -> {"x", "f(x)"}]
(* 150812_Plot_Prod_Beta_dist.jpg *)

Since with the Meijer function Mathematica requires very long calculation times close to x = 1 I have left this region out.

Observations

1) I believe that the case of general n should be tackled using the Mellin transformation, as is natural for products (as is Fourier for sums). Our result for n = 3, the MeijerG function, already exhibits this pattern.

Part 2: Derivation

The distributions were calculated here using the general formula

  fn(x) = Integrate( Prod( du f(u,p)) DiracDelta(x-Prod(u)) )

Details will be given later.

Part 3: Discussion

Let me rather start a brief discussion suggested by a comment of Guess who it is.

2.1) The result of wolfie (which I saw only after having finished my calculations) is

f3wolfie[x_] :=
27/32 \[Pi] MeijerG[{{}, {3/2, 3/2, 3/2}}, {{0, 1/2, 1}, {}}, x]

which is not identical at first sight with my result

f3wolfgang[x_] :=
27/4 (EllipticE[1 - x] - x EllipticK[1 - x]) -
27/16 \[Pi] MeijerG[{{}, {3/2, 3/2, 3/2}}, {{1/2, 1, 1}, {}}, x]

But the functions indeed coincide as can be seen in a plot, or some clever FullSimplify?

For f3n[x], the last step in my calculation was this integral

Integrate[27/8 (
Sqrt[-((w x)/(-1 + w))] (Sqrt[-1 + w/x] - ArcCos[Sqrt[x/w]]))/w, {w, x, 1},
Assumptions - 0 < x < 1]

and the two terms in f3wolfgang correspond to the two terms in the this integrand.

2.2) Behaviour close to x = 1

As I haven't found a quick answer in the literature I tackled the poblem "experimentally":

Numerically my solution can be respresented as

f3nn[x_]:=NIntegrate[
27/8 ( Sqrt[-((
w x)/(-1 + w))] (Sqrt[-1 + w/x] - ArcCos[Sqrt[x/w]]))/w, {w, x,
1}]

Plotting this close to 1 with a negative power of (1-x) attached

Plot[1/(1 - x)^k f3nn[x], {x, 0.5, 1.1}]

and playing with k close to 2 shows that f3nn[x] ~= const (1-x)^2. I suspect even that k = 2 is exact because only for this value the function exhibits a sharp shoulder.

Afterwards I found (after some lengthy study) the exact behaviour of the constant to be 27 Pi/16, i.e. the distribution function has the series Expansion

f3n[x] = 27/64 \[Pi] (1 - x)^2 + O[1 - x]^3

Also, the MeijerG-function has the series expansion

MeijerG[{{}, {3/2, 3/2, 3/2}}, {{1/2, 1, 1}, {}}, x] == (1 - x) -
1/8 (1 - x)^2 + O[1 - x]^3

Remark: the developemnts here show that there is a simple integral representation of the MeijerG function, much simpler than the complex integral version. We have

MeijerG[{{}, {3/2, 3/2, 3/2}}, {{1/2, 1, 1}, {}}, x] = 2/Pi Integrate[
27/8 ( Sqrt[-((w x)/(-1 + w))] (Sqrt[-1 + w/x]
- ArcCos[Sqrt[x/w]]))/w, {w, x, 1}]

The difficulties of Mathematica calculating the numerical values close to 1 have been overcome by this representation and the series expansion.

• "using the Mellin transformation" - with the Meijer result, this is what you're implicitly doing anyway, since the $G$-function is effectively an inverse Mellin transform… – J. M. will be back soon Aug 12 '15 at 12:29
• I have been busy doing the calculations and the editing so I did't see the resuls of wolfie. Sorry for that. – Dr. Wolfgang Hintze Aug 12 '15 at 12:30
• @J. M.: that's what I was - cautiously - saying ;-) – Dr. Wolfgang Hintze Aug 12 '15 at 12:34
• The reason why I had mentioned that the Meijer result looked familiar in the other answer is that it resembled some of the Mellin-Barnes representations for elliptic integrals that I remember; your result, which at least involves the complete elliptic integrals, seems to be tantalizing. – J. M. will be back soon Aug 12 '15 at 12:41
• Some nice reference to MeijerG is ams.org/notices/201307/rnoti-p866.pdf pointing out specifically the closure under convolutions. – Dr. Wolfgang Hintze Aug 12 '15 at 13:20


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Three men and eight machines can finish a job in half the time taken

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31 Mar 2020, 02:27
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Three men and eight machines can finish a job in half the time taken by three machines and eight men to finish the same job. If two machines can finish the job in 13 days, then how many men can finish the job in 13 days?

A. 10
B. 11
C. 12
D. 13
E. 14

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Re: Three men and eight machines can finish a job in half the time taken  [#permalink]

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31 Mar 2020, 02:49
rate of one machine ; 13*2 ; 26 days
and rate of man ; x
3/x+8/26 = 2*(8/x+3/26)
solve for x
we get x= 169
so in 13 days ; 169/13 ; 13 men would be required
IMO D

Bunuel wrote:
Three men and eight machines can finish a job in half the time taken by three machines and eight men to finish the same job. If two machines can finish the job in 13 days, then how many men can finish the job in 13 days?

A. 10
B. 11
C. 12
D. 13
E. 14

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31 Mar 2020, 02:50
1
time taken by 3Men + 8 Machines = half * time taken by 3 machines + 8 men
In other words time taken by 3 men + 8 machines = time taken by 6 machines + 16 men
Two machines can finish the work in 13 days means that one machine can finish the work in 26 days.

let the time take for a man to finish the work be x, then

3/x + 8/26 = 6/26 + 16/x
2/26 = 13/x
x = 13*13
Hence 13 men will take 13*13/13 = 13 days.

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Re: Three men and eight machines can finish a job in half the time taken  [#permalink]

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31 Mar 2020, 03:01
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Bunuel wrote:
Three men and eight machines can finish a job in half the time taken by three machines and eight men to finish the same job. If two machines can finish the job in 13 days, then how many men can finish the job in 13 days?

A. 10
B. 11
C. 12
D. 13
E. 14

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31 Mar 2020, 03:33
1
Bunuel wrote:
Three men and eight machines can finish a job in half the time taken by three machines and eight men to finish the same job. If two machines can finish the job in 13 days, then how many men can finish the job in 13 days?

A. 10
B. 11
C. 12
D. 13
E. 14

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Solution:

Let A and B be the per day work of one man and one machine respectively
• Work done by 3 men and 8 machines in one day = 3A + 8B
• Work done by 8 men and 3 machines in one day = 8A + 3B
o Time taken to complete the work by 3 men and 8 machines is half time by 8 men and 3 machines
o Work done by 3 men and 8 machines in one day is twice of the work done by 8 men and 3 machines in one day
o $$3A + 8B = 2(8A + 3B)$$
o $$3A +8B = 16A + 6B$$
o $$8B-6B = 16A – 3A$$
o $$2B = 13A$$
• Work done by 2 machines in one day is equal to the work done by 13 men in one day.
• 2 machines can complete the job in 13 days
o 13 men can complete the job in 13 days
Hence, the correct answer is Option D.
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Re: Three men and eight machines can finish a job in half the time taken  [#permalink]

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31 Mar 2020, 04:17
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Since, TEAM-1 (3 men and 8 machines) can finish a job in half the time taken by TEAM-2 (3 machines and 8 men) to finish the same job, we can express the TWO TEAM'S CAPACITY through the following equation:
3 MEN+ 8 MACHINES= 2 (3 MACHINES+ 8 MEN)
=> 3 MEN+8 MACHINES= 6 MACHINES+16 MEN
=>8 MACHINES- 6 MACHINES= 16 MEN- 3 MEN
=> 2 MACHINES=13 MEN
Therefore, we can infer that THE CAPACITY OF TWO MACHINES IS EQUIVALENT TO THAT OF 13 MEN.
Hence, it requires 13 men to complete what can be done by 2 machines in the same amount of time.

Bunuel, please convey me whether my approach is efficient.
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Re: Three men and eight machines can finish a job in half the time taken  [#permalink]

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31 Mar 2020, 09:20
1
Bunuel wrote:
Three men and eight machines can finish a job in half the time taken by three machines and eight men to finish the same job. If two machines can finish the job in 13 days, then how many men can finish the job in 13 days?

A. 10
B. 11
C. 12
D. 13
E. 14

two machines can finish the job in 13 days
--> 1 machine can complete a job in 26 days
Let 1 man complete a job in '$$m$$' days

Three men and eight machines can finish a job in half the time taken by three machines and eight men to finish the same job
--> $$\frac{3}{m} + \frac{8}{26} = 2(\frac{3}{26} + \frac{8}{m})$$
--> $$\frac{8}{26} - \frac{6}{26} = \frac{16}{m} - \frac{3}{m}$$
--> $$\frac{1}{13} = \frac{13}{m}$$
--> $$m = 169$$ days

Number of men required to finish the job in 13 days = $$\frac{169}{13} = 13$$ men

Option D
Re: Three men and eight machines can finish a job in half the time taken   [#permalink] 31 Mar 2020, 09:20


# Definition of continuity not clear

According to my book continuity is (informally) defined as follows:

Function $f$ is continuous at $c$ iif it is both right continuous and left continuous at $c$.

Now according to the book the function $f(x) = \sqrt{4 - x^2}$ is continuous at every point of its domain, $[-2, 2]$. I understand this holds for $(-2, 2)$,but what about the endpoints?

If I look at the definition and consider the left and right endpoints, then how could it be left continuous at its left endpoint or right continuous at its right endpoint? Since there's no function beyond $-2$, then how can there be a limit from the left?

• When we say that $f$ is continuous on a closed interval $[a,b]$, we mean that it is both left and right continuous at each point of $(a,b)$, right continuous at $a$, and left continuous at $b$. – Brian M. Scott Oct 20 '16 at 16:59
• If you make an answer of it, I can upvote it. – Apeiron Oct 20 '16 at 17:15
• Similar to question math.stackexchange.com/questions/1969405/… – LutzL Oct 20 '16 at 20:27

It’s a matter of convention. When we say that a function $f$ is continuous on a closed interval $[a,b]$, we mean that it is both left and right continuous at each point of the open interval $(a,b)$, right continuous at $a$, and left continuous at $b$. In other words, it has every possible one-sided continuity at each point of the closed interval.

Added: The real culprit here is the assertion that $f$ is continuous at a point $c$ if and only if it is both left and right continuous at $c$. This is fine if the domain of $f$ is the whole real line or an open interval, but it’s not quite right for arbitrary domains. Suppose that $f$ has domain $D$, and $c\in D$. Then $f$ is continuous at $c$ if it satisfies the following two conditions:

• if $c$ is a limit point of $\{x\in D:x<c\}$, then $f$ is left continuous at $c$, and
• if $c$ is a limit point of $\{x\in D:x>c\}$, then $f$ is right continuous at $c$.

When $D$ is a closed interval $[a,b]$, any $c\in(a,b)$ is a limit point of both $$\{x\in[a,b]:x<c\}=[a,c)$$ and $$\{x\in[a,b]:x>c\}=(c,b]\;.$$ When $c=a$, however, the first of these sets is empty, so $a$ isn’t a limit point of it, and we don’t require $f$ to be left continuous at $a$. When $c=b$, the second set is empty, and we don’t require $f$ to be right continuous at $c$.

Many standard first-year calculus texts are a bit sloppy about such details.

• It's a language abuse to say that $f$ is continuous at $[a,b]$. – hamam_Abdallah Oct 20 '16 at 17:26
• @Abdallah: No, it isn’t; it’s a convention that agrees with the use of the word continuous in more general topological settings. The real culprit here is the assertion that $f$ is continuous at a point $c$ in its domain iff it is continuous from each side: that is in fact true only if $c$ is a limit point of the domain of $f$ from both sides. When the domain is a closed interval this is not the case at the endpoints of the interval. – Brian M. Scott Oct 20 '16 at 17:32
• Thanks, @BrianM.Scott, this was really helpful. This issue raised many doubts to my understanding of a topic I'm already a bit anxious about. I will update the definition accordingly in my concepts notes. – Apeiron Oct 21 '16 at 7:27
• @Apeiron: You’re welcome; glad it helped. – Brian M. Scott Oct 21 '16 at 13:51


onto, to have an inverse, since if it is not surjective, the function's inverse's domain will have some elements left out which are not mapped to any element in the range of the function's inverse. Not all functions have inverse functions. This means, for instance, that no parabola (quadratic function) will have an inverse that is also a function. So y = m * x + b, where m and b are constants, is a linear equation. The graph of this function contains all ordered pairs of the form (x,2). To have an inverse, a function must be injective i.e one-one. do all kinds of functions have inverse function? In fact, the domain and range need not even be subsets of the reals. No. The function f is defined as f(x) = x^2 -2x -1, x is a real number. We did all of our work correctly and we do in fact have the inverse. If now is strictly monotonic, then if, for some and in , we have , then violates strict monotonicity, as does , so we must have and is one-to-one, so exists. Please teach me how to do so using the example below! There is one final topic that we need to address quickly before we leave this section. It is not true that a function can only intersect its inverse on the line y=x, and your example of f(x) = -x^3 demonstrates that. Other types of series and also infinite products may be used when convenient. but y = a * x^2 where a is a constant, is not linear. Two functions f and g are inverse functions if for every coordinate pair in f, (a, b), there exists a corresponding coordinate pair in the inverse function, g, (b, a).In other words, the coordinate pairs of the inverse functions have the input and output interchanged. Question 64635: Explain why an even function f does not have an inverse f-1 (f exponeant -1) Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website! The horizontal line test can determine if a function is one-to-one. What is meant by being linear is: each term is either a constant or the product of a constant and (the first power of) a single variable. In this section it helps to think of f as transforming a 3 into a … This is clearly not a function (for one thing, if you graph it, it fails the vertical line test), but it is most certainly a relation. The graph of inverse functions are reflections over the line y = x. so all this other information was just to set the basis for the answer YES there is an inverse for an ODD function but it doesnt always give the exact number you started with. let y=f(x). Inverse Functions. A function must be a one-to-one function, meaning that each y-value has a unique x-value paired to it. There is an interesting relationship between the graph of a function and its inverse. Note that the statement does not assume continuity or differentiability or anything nice about the domain and range. Consider the function f(x) = 2x + 1. Answer to (a) For a function to have an inverse, it must be _____. viviennelopez26 is waiting for your help. if i then took the inverse sine of -1/2 i would still get -30-30 doesnt = 210 but gives the same answer when put in the sin function Add your … If the function is linear, then yes, it should have an inverse that is also a function. This means that each x-value must be matched to one and only one y-value. how do you solve for the inverse of a one-to-one function? It should be bijective (injective+surjective). Problem 86E from Chapter 3.6: Functions that meet this criteria are called one-to one functions. Imagine finding the inverse of a function … Definition of Inverse Function. For example, the infinite series could be used to define these functions for all complex values of x. Only one-to-one functions have inverses, as the inverse of a many-to-one function would be one-to-many, which isn't a function. A function may be defined by means of a power series. Suppose is an increasing function on its domain.Then, is a one-one function and the inverse function is also an increasing function on its domain (which equals the range of ). Problem 33 Easy Difficulty. Statement. Such functions are often defined through formulas, such as: A surjective function f from the real numbers to the real numbers possesses an inverse as long as it is one-to-one, i.e. Restrictions on the Domains of the Trig Functions A function must be one-to-one for it to have an inverse. Logarithmic Investigations 49 – The Inverse Function No Calculator DO ALL functions have Sin(210) = -1/2. As we are sure you know, the trig functions are not one-to-one and in fact they are periodic (i.e. Explain.. Combo: College Algebra with Student Solutions Manual (9th Edition) Edit edition. Explain why an even function f does not have an inverse f-1 (f exponeant -1) F(X) IS EVEN FUNCTION IF Strictly monotone functions and the inverse function theorem We have seen that for a monotone function f: (a;b) !R, the left and right hand limits y 0 = lim x!x 0 f(x) and y+ 0 = lim x!x+ 0 f(x) both exist for all x 0 2(a;b).. Not all functions have inverses. Explain your reasoning. There is one final topic that we need to address quickly before we leave this section. An inverse function is a function that will “undo” anything that the original function does. The inverse relation is then defined as the set consisting of all ordered pairs of the form (2,x). We know how to evaluate f at 3, f(3) = 2*3 + 1 = 7. For example, we all have a way of tying our shoes, and how we tie our shoes could be called a function. Inverse of a Function: Inverse of a function f(x) is denoted by {eq}f^{-1}(x) {/eq}.. Such functions are called invertible functions, and we use the notation $$f^{−1}(x)$$. I know that a function does not have an inverse if it is not a one-to-one function, but I don't know how to prove a function is not one-to-one. For example, the function f(x) = 2x has the inverse function f −1 (x) = x/2. x^2 is a many-to-one function because two values of x give the same value e.g. Before defining the inverse of a function we need to have the right mental image of function. We did all of our work correctly and we do in fact have the inverse. There is an interesting relationship between the graph of a function and the graph of its inverse. Question: Do all functions have inverses? So a monotonic function must be strictly monotonic to have an inverse. This is what they were trying to explain with their sets of points. Let us start with an example: Here we have the function f(x) = 2x+3, written as a flow diagram: The Inverse Function goes the other way: So the inverse of: 2x+3 is: (y-3)/2 . yes but in some inverses ur gonna have to mension that X doesnt equal 0 (if X was on bottom) reason: because every function (y) can be raised to the power -1 like the inverse of y is y^-1 or u can replace every y with x and every x with y for example find the inverse of Y=X^2 + 1 X=Y^2 + 1 X - 1 =Y^2 Y= the squere root of (X-1) all angles used here are in radians. Not every element of a complete residue system modulo m has a modular multiplicative inverse, for instance, zero never does. So a monotonic function has an inverse iff it is strictly monotonic. Suppose that for x = a, y=b, and also that for x=c, y=b. For instance, supposing your function is made up of these points: { (1, 0), (–3, 5), (0, 4) }. For a function to have an inverse, the function must be one-to-one. Thank you! if you do this . their values repeat themselves periodically). Define and Graph an Inverse. View 49C - PowerPoint - The Inverse Function.pdf from MATH MISC at Atlantic County Institute of Technology. \begin{array}{|l|c|c|c|c|c|c|} \hline x & -3 & -2 & -1 & 0 & 2 & 3 \\ \hline f(x) & 10 & 6 & 4 & 1 & -3 & -10 \\ \h… An inverse function goes the other way! Yeah, got the idea. While it is not possible to find an inverse of most polynomial functions, some basic polynomials do have inverses. Because if it is not surjective, there is at least one element in the co-domain which is not related to any element in the domain. No. Really clear math lessons (pre-algebra, algebra, precalculus), cool math games, online graphing calculators, geometry art, fractals, polyhedra, parents and teachers areas too. Answer to Does a constant function have an inverse? Hello! Suppose we want to find the inverse of a function … both 3 and -3 map to 9 Hope this helps. Thank you. as long as the graph of y = f(x) has, for each possible y value only one corresponding x value, and thus passes the horizontal line test.strictly monotone and continuous in the domain is correct There are many others, of course; these include functions that are their own inverse, such as f(x) = c/x or f(x) = c - x, and more interesting cases like f(x) = 2 ln(5-x). Warning: $$f^{−1}(x)$$ is not the same as the reciprocal of the function $$f(x)$$. Inverting Tabular Functions. The inverse of a function has all the same points as the original function, except that the x's and y's have been reversed. Does the function have an inverse function? Does the function have an inverse function? This implies any discontinuity of fis a jump and there are at most a countable number. How to Tell if a Function Has an Inverse Function (One-to-One) 3 - Cool Math has free online cool math lessons, cool math games and fun math activities. 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# Sequence Word Problem

#### 0313phd

##### New member
Problem: The number of bricks in the bottom row of a brick wall is 49. The next row up from the bottom contains 47 bricks, and each subsequent row contains 2 fewer bricks than the row immediately below it. The number of bricks in the top row is 3. If the wall is one brick thick, what is the total number of bricks in the wall?

I am not getting the correct answer. Is this an arithmetic sequence? Sn= n/2 (a1 +an)
so that since there are 46 steps between the bottom and top steps, 46/2(3 + 49). However, the correct answer is 624. Is this a geometric sequence? Any help will be greatly appreciated. 0313

#### soroban

##### Elite Member
Hello,0313phd!

The number of bricks in the bottom row of a brick wall is 49.
The next row up from the bottom contains 47 bricks,
and each subsequent row contains 2 fewer bricks than the row immediately below it.
The number of bricks in the top row is 3.

If the wall is one brick thick, what is the total number of bricks in the wall?

I am not getting the correct answer.
Is this an arithmetic sequence? . Yes!
. . . $$\displaystyle S_n\:=\: \tfrac{n}{2}\left(a_1 + a_n\right)$$

So that since there are 46 steps between the bottom and top steps. . No, there are 24 rows.

Counting from top, we have:

. . $$\displaystyle \begin{array}{|c||c|c|c|c|c|c|} \hline \text{Row} & 1 & 2 & 3 & 4 & \hdots & 24 \\ \hline \text{Bricks} & 3 & 5 & 7 & 9 & \hdots & 49 \\ \hline\end{array}$$

$$\displaystyle \text{So we have: }\:S_{24} \;=\;\tfrac{24}{2}(3 + 49) \:=\12)(52) \:=\:624$$

#### TchrWill

##### Full Member
0313phd said:
Problem: The number of bricks in the bottom row of a brick wall is 49. The next row up from the bottom contains 47 bricks, and each subsequent row contains 2 fewer bricks than the row immediately below it. The number of bricks in the top row is 3. If the wall is one brick thick, what is the total number of bricks in the wall?

I am not getting the correct answer. Is this an arithmetic sequence? Sn= n/2 (a1 +an)
so that since there are 46 steps between the bottom and top steps, 46/2(3 + 49). However, the correct answer is 624. Is this a geometric sequence? Any help will be greatly appreciated. 0313
An alternate thought:

Add a brick atop those below and you have rows of 49, 47, 45, 43,...........7, 5, 3, 1.

The sum of the first "n" odd numbers is n^2.

Therefore, the sum of the 25 rows of bricks is 25^2 or 625.

Remove the single brick added for conveniance, and the total number of bricks is 624.

J

#### JeffM

##### Guest
0313phd said:
Problem: The number of bricks in the bottom row of a brick wall is 49. The next row up from the bottom contains 47 bricks, and each subsequent row contains 2 fewer bricks than the row immediately below it. The number of bricks in the top row is 3. If the wall is one brick thick, what is the total number of bricks in the wall?

I am not getting the correct answer. Is this an arithmetic sequence? Sn= n/2 (a1 +an)
so that since there are 46 steps between the bottom and top steps, 46/2(3 + 49). However, the correct answer is 624. Is this a geometric sequence? Any help will be greatly appreciated. 0313
1 + 3 + 5 + ... + 49 = S[sub:241ucnbg]49[/sub:241ucnbg]
3 + 5 + ... + 49 = T[sub:241ucnbg]49[/sub:241ucnbg].
So, T[sub:241ucnbg]n[/sub:241ucnbg] = S[sub:241ucnbg]n[/sub:241ucnbg] - 1.

All clear so far?

There is a formula for the sum of the first n odd numbers, but maybe you do not know it. So derive it. (It is tricky to decide what to memorize. My tendency is to remember concepts and techniques and as few formulae as possible, but that is me. I was (sort of) trained as an historian so you cannot rely on me for math advice UNLESS I PROVE IT.)

Let's do a few examples. (This is not the only way to solve such problems, but, WHEN IT WORKS, it is the simplest.)

1 = 1.
1 + 3 = 4.
1 + 3 + 5 = 9.
1 + 3 + 5 + 7 = 16.

Do you see a pattern?

Not looking for a fancy proof of the pattern, but let's confirm it through an informal mathematical induction.

What is the standard formula for an odd number?

Assume [(2 * 1) - 1] + ... (2k - 1) = k[sup:241ucnbg]2[/sup:241ucnbg].
Then [(2 * 1) - 1] + ... [(2k) - 1] + [2(k + 1) - 1] = k[sup:241ucnbg]2[/sup:241ucnbg] + 2k + 2 - 1 = k[sup:241ucnbg]2[/sup:241ucnbg] + 2k + 1 = (k + 1)[sup:241ucnbg]2[/sup:241ucnbg].

Now can you solve your problem?

THANKS, 0313

3 + 49 = 52
5 + 47 = 52
....


# Inequality from Chapter 5 of the book *How to Think Like a Mathematician*

This is from the book How to think like a Mathematician,

How can I prove the inequality $$\sqrt[\large 7]{7!} < \sqrt[\large 8]{8!}$$

without complicated calculus? I tried and finally obtained just $$\frac 17 \cdot \ln(7!) < \frac 18 \cdot \ln(8!)$$

• Are we allowed complex roots and negative roots to prove the contrary ;-) Feb 28, 2015 at 19:22
• I don't think that it is in the proposition Mar 2, 2015 at 11:24
• 7! is the product of seven positive numbers, all of which are smaller than 8. $8^7$ is the product of 7 eights. So the inequality follows. Dec 7, 2021 at 22:52
• Be sure to write your argument starting from $7! < 8^7$ and ending with what you want to prove, not backwards the way you discovered where to start. Dec 7, 2021 at 22:53
• If you have a dataset of $n$ real numbers, and you add an $n+1$st number greater than the average of the $n$ numbers, the new set has a greater average. (Hint: logarithms.) Dec 7, 2021 at 22:53

Your inequality is equivalent to $$(7!)^8 < (8!)^7$$ divide it by $(7!)^7$, and get $$7! < 8^7$$ and this is clear, since $$1 \cdots 7 < 8 \cdots 8$$

• In the spirit of the question title, I'd note that the first line of this answer is key. Maths does of course love cleverness, but also thrives on knowing to do really simply things such as taking powers of both sides here to remove those ugly roots. Feb 27, 2015 at 3:02
• You are right, great ! I would have of to think there. Feb 28, 2015 at 17:39
• Elegant! really clever! May 27, 2015 at 16:19

Think of

$${\ln(7!)\over7}={\ln(1)+\cdots+\ln(7)\over7}$$

as the average of seven numbers and

$${\ln(8!)\over8}={\ln(1)+\cdots+\ln(8)\over8}$$

as the average when an eighth number is added. Since the new number is larger than the previous seven, the average must also be larger. (E.g., if you get a better score on your final than on any of your midterms, your grade should go up, not down.)

• You are right, but it is clear to me only when i compute ln(8) Feb 28, 2015 at 18:08
• @Gwydyon, the logarithm is an increasing function, so $\ln(8)$ is larger than its predecessors. Feb 28, 2015 at 20:00
• Yes I know that. Mar 2, 2015 at 11:21
• @Gwydyon, I guess I don't understand your previous comment then. Mar 2, 2015 at 15:22
• Because x/8 is lower than y/7, if x=y, that is not the case but ln(7!) and ln((7+1)!) are near close, for x>=1 Mar 3, 2015 at 16:41

Note that $$\sqrt[7]{7!} < \sqrt[8]{8!} \iff\\ (7!)^8 < (8!)^7 \iff\\ 7! < \frac{(8!)^7}{(7!)^7} \iff\\ 7! < 8^7$$ You should find that the proof of this last line is fairly straightforward.

• You are right but someone has writen the same statements Feb 28, 2015 at 18:11

$8\ln (7!) < 7\ln (8!) \Rightarrow \ln (7!) < 7\ln 8 \iff \ln 1 + \ln 2 +\cdots \ln 7 < 7\ln 8$ which is clear.

• great! you achieved my second statement Feb 28, 2015 at 18:53

You have already turned the comparison of two geometric means into the comparison of two arithmetic means. So consider a more general comparison: show that appending a larger number always raises the geometric mean of a list of positive numbers by showing the effect on the arithmetic mean. Suppose the $x_i$ are real and $x_{n+1}$ is strictly largest. \begin{equation*} \begin{split} (1/(n+1)) \sum_{i=1}^{n+1} x_i &= (1/(n+1)) (x_{n+1} + \sum_{i=1}^{n} x_i) \\ &=(1/(n+1) (n x_{n+1}/n + n \sum_{i=1}^{n} x_i / n) \\ &> (1/(n+1) (\sum_{i=1}^{n} x_i/n + n \sum_{i=1}^{n} x_i / n) \\ &= (1/(n+1) ((n+1) \sum_{i=1}^{n} x_i / n) \\ &= \sum_{i=1}^{n} x_i / n \end{split} \end{equation*}

Note that we really only needed $x_{n+1}$ to be larger than the previous mean.

The solution occurs just by doing simple calculations,
Lets start, $$\sqrt[7]{7!}<\sqrt[8]{8!}$$
iff $$(\sqrt[7]{7!})^{7\cdot8}<(\sqrt[8]{8!})^{7\cdot8}$$
iff $$(7!)^{8}<(8!)^7$$
iff $$(7!)^8<(7!\cdot8)^7$$
iff $$(7!)^8<8^7\cdot(7!)^7$$
iff $$(7!)<8^7$$
iff $$1\cdot2\cdots6\cdot7<8\cdot8\cdot8\cdot8\cdot8\cdot8\cdot8$$
wich is obviously true since $$1<8,2<8,\ldots,7<8$$

Consider this

$$x=\ln 8!-\frac87\ln7!=\ln8-\frac17\ln7!=\frac17\left(7\ln8-\ln7!\right)=\frac17\ln\frac{8^7}{7!}>0$$ Hence, since the exponential is a monotonically increasing function: $$e^{x/8}>1\implies (8!)^{1/8}>(7!)^{1/7}.$$


Main Content

# graphplot

Plot Markov chain directed graph

## Description

example

graphplot(mc) creates a plot of the directed graph (digraph) of the discrete-time Markov chain mc. Nodes correspond to the states of mc. Directed edges correspond to nonzero transition probabilities in the transition matrix mc.P.

example

graphplot(mc,Name,Value) uses additional options specified by one or more name-value arguments. Options include highlighting transition probabilities, communicating classes, and specifying class properties of recurrence/transience and period. Also, you can plot the condensed digraph instead, with communicating classes as supernodes.

graphplot(ax,___) plots on the axes specified by ax instead of the current axes (gca) using any of the input argument combinations in the previous syntaxes. The option ax can precede any of the input argument combinations in the previous syntaxes.

example

h = graphplot(___) returns the handle to the digraph plot. Use h to modify properties of the plot after you create it.

## Examples

collapse all

Consider this theoretical, right-stochastic transition matrix of a stochastic process.

$P=\left[\begin{array}{ccccccc}0& 0& 1/2& 1/4& 1/4& 0& 0\\ 0& 0& 1/3& 0& 2/3& 0& 0\\ 0& 0& 0& 0& 0& 1/3& 2/3\\ 0& 0& 0& 0& 0& 1/2& 1/2\\ 0& 0& 0& 0& 0& 3/4& 1/4\\ 1/2& 1/2& 0& 0& 0& 0& 0\\ 1/4& 3/4& 0& 0& 0& 0& 0\end{array}\right].$

Create the Markov chain that is characterized by the transition matrix P.

P = [ 0   0  1/2 1/4 1/4  0   0 ;
0   0  1/3  0  2/3  0   0 ;
0   0   0   0   0  1/3 2/3;
0   0   0   0   0  1/2 1/2;
0   0   0   0   0  3/4 1/4;
1/2 1/2  0   0   0   0   0 ;
1/4 3/4  0   0   0   0   0 ];
mc = dtmc(P);

Plot a directed graph of the Markov chain.

figure;
graphplot(mc);

Consider this theoretical, right-stochastic transition matrix of a stochastic process.

$P=\left[\begin{array}{cccc}0.5& 0.5& 0& 0\\ 0.5& 0& 0.5& 0\\ 0& 0& 0& 1\\ 0& 0& 1& 0\end{array}\right].$

Create the Markov chain that is characterized by the transition matrix P. Name the states Regime 1 through Regime 4.

P = [0.5 0.5 0 0; 0.5 0 0.5 0; 0 0 0 1; 0 0 1 0];
mc = dtmc(P,'StateNames',["Regime 1" "Regime 2" "Regime 3" "Regime 4"]);

Plot a directed graph of the Markov chain. Identify the communicating classes in the digraph and color the edges according to the probability of transition.

figure;
graphplot(mc,'ColorNodes',true,'ColorEdges',true)

States 3 and 4 compose a communicating class with period 2. States 1 and 2 are transient.

Create a "dumbbell" Markov chain containing 10 states in each "weight" and three states in the "bar."

• Specify random transition probabilities between states within each weight.

• If the Markov chain reaches the state in a weight that is closest to the bar, then specify a high probability of transitioning to the bar.

• Specify uniform transitions between states in the bar.

rng(1); % For reproducibility
w = 10;                              % Dumbbell weights
DBar = [0 1 0; 1 0 1; 0 1 0];        % Dumbbell bar
DB = blkdiag(rand(w),DBar,rand(w));  % Transition matrix

% Connect dumbbell weights and bar
DB(w,w+1) = 1;
DB(w+1,w) = 1;
DB(w+3,w+4) = 1;
DB(w+4,w+3) = 1;

db = dtmc(DB);

Plot a directed graph of the Markov chain. Return the plot handle.

figure;
h = graphplot(db);

Observe that the state labels are difficult to read. Remove the labels entirely.

h.NodeLabel = {};

## Input Arguments

collapse all

Discrete-time Markov chain with NumStates states and transition matrix P, specified as a dtmc object. P must be fully specified (no NaN entries).

Axes on which to plot, specified as an Axes object.

By default, graphplot plots to the current axes (gca).

### Name-Value Arguments

Specify optional pairs of arguments as Name1=Value1,...,NameN=ValueN, where Name is the argument name and Value is the corresponding value. Name-value arguments must appear after other arguments, but the order of the pairs does not matter.

Before R2021a, use commas to separate each name and value, and enclose Name in quotes.

Example: 'ColorEdges',true,'ColorNodes',true colors the edges to indicate transition probabilities and colors the nodes based on their communicating class.

Flag for labeling nodes using state names, specified as the comma-separated pair consisting of 'LabelNodes' and a value in this table.

ValueDescription
trueLabel nodes using the names in mc.StateNames.
falseLabel nodes using state numbers.

Example: 'LabelNodes',false

Data Types: logical

Flag for coloring nodes based on communicating class, specified as the comma-separated pair consisting of 'ColorNodes' and a value in this table.

ValueDescription
trueNodes in the same communicating class have the same color. Solid markers represent nodes in recurrent classes, and hollow markers represent nodes in transient classes. The legend contains the periodicity of recurrent classes.
falseAll nodes have the same color.

Example: 'ColorNodes',true

Data Types: logical

Flag for labeling edges with the transition probabilities in the transition matrix mc.P, specified as the comma-separated pair consisting of 'LabelEdges' and a value in this table.

ValueDescription
trueLabel edges with transition probabilities rounded to two decimal places.
falseDo not label edges.

Example: 'LabelEdges',true

Data Types: logical

Flag for coloring edges to indicate transition probabilities, specified as the comma-separated pair consisting of 'ColorEdges' and a value in this table.

ValueDescription
trueColor edges to indicate transition probabilities. Include a color bar, which summarizes the color coding.
falseUse the same color for all edges.

Example: 'ColorEdges',true

Data Types: logical

Flag for condensing the graph, with each communicating class represented by a single supernode, specified as the comma-separated pair consisting of 'Condense' and a value in this table.

ValueDescription
trueNodes are supernodes containing communicating classes. Node labels list the component states of each supernode. An edge from supernode i to supernode j indicates a nonzero probability of transition from some state in supernode i to some state in supernode j. Transition probabilities between supernodes are not well defined, and graphplot disables edge information.
falseNodes are states in mc.

Example: 'Condense',true

Data Types: logical

## Output Arguments

collapse all

Handle to the graph plot, returned as a graphics object. h is a unique identifier, which you can use to query or modify properties of the plot.

## Tips

• To produce the directed graph as a MATLAB® digraph object and use additional functions of that object, enter:

G = digraph(mc.P)

• For readability, the 'LabelNodes' name-value pair argument allows you to turn off lengthy node labels and replace them with node numbers. To remove node labels completely, set h.NodeLabel = {};.

• To compute node information on communicating classes and their properties, use classify.

• To extract a communicating class in the graph, use subchain.

• The condensed graph is useful for:

• Identifying transient classes (supernodes with positive outdegree)

• Identifying recurrent classes (supernodes with zero outdegree)

• Visualizing the overall structure of unichains (chains with a single recurrent class and any transient classes that transition into it)

## References

[1] Gallager, R.G. Stochastic Processes: Theory for Applications. Cambridge, UK: Cambridge University Press, 2013.

[2] Horn, R., and C. R. Johnson. Matrix Analysis. Cambridge, UK: Cambridge University Press, 1985.

[3] Jarvis, J. P., and D. R. Shier. "Graph-Theoretic Analysis of Finite Markov Chains." In Applied Mathematical Modeling: A Multidisciplinary Approach. Boca Raton: CRC Press, 2000.

## Version History

Introduced in R2017b


# Find the smallest positive integer divisible by 63 such that the sum of its digits is also divisible by 63.

TASK: Find the smallest positive integer divisible by 63 such that the sum of its digits is also divisible by 63.

MY WORK: Let the number be $$A=\overline{x_n x_{n-1} x_{n-2} \cdots x_1 x_0}$$. Since $$63|(x_n+x_{n-1}+\cdots+x_1+x_0)$$, we have that $$x_n+x_{n-1}+\cdots+x_0\ge63\cdots(*)$$ and since $$x_0,x_1,\cdots,x_n$$ are n+1 digits, we have that $$x_n+x_{n-1}+\cdots+x_0\le9+9+\cdots+9=9(n+1)$$ which then means that $$9(n+1)\ge63\Leftrightarrow n+1\ge7$$ i.e that the number $$A$$ has at least seven digits. If it has $$7$$ digits, all of them would have to be $$9$$ to satisfy the inequality $$(*)$$ which would mean that $$A=9999999$$. But then the condition $$63|A$$ would not be satisfied. So the number A doesn't have seven digits - it has at least eight digits.

However, I do not know where to go from here.

• So check numbers like 18999999 and 19899999 and 19989999 and so on. – Gerry Myerson Nov 20 '18 at 8:15
• You get divisibility by $9$ for free so you only need be concerned about checking for $7$ – Mark Bennet Nov 20 '18 at 8:44

Assume the the number is $$1$$ with 6 $$9$$s and one $$8$$.

Now $$19999999\equiv 5\pmod 7$$

If we subtract $$10^k$$ we will get aus a number with a $$1$$ , 6$$9$$ and one $$8$$ anda different equivalence. so we need to find the $$10^k\equiv 5\mod 7$$.

$$10\equiv 3$$

$$100\equiv 30\equiv 2$$

$$1000\equiv 20\equiv 6$$

$$10,000\equiv 60\equiv 4$$

$$100,000 \equiv 40\equiv 5$$

So $$19,999,999-100,000=19,899,999\equiv 0\pmod 7$$.

And that's that. It's digits add to $$63$$ so it's divisible by $$9$$ and it's divisible by $$7$$. And beginning with $$1$$ and the only such divisible by $$7$$ it's the smallest such number.

====

I thought I made it clear why this is the smallest.

No element with $$7$$ digits or fewer exist as the OP figured out. For a group with $$8$$ digits the smallest would start with a $$1$$. If you have an $$8$$ digit number beginning with $$1$$ and whose digits add to $$63$$ the remaining digits must be six $$9$$s and one $$8$$. Such a number can be written as $$19,999,999 - 10^k$$ where $$0\le k \le 7$$. For such a number to be divisible by $$63$$ we must have $$10^k \equiv 5 \pmod 7$$. The ONLY such $$k$$ is $$k = 5$$ and $$10^k =100,000$$ and the number is $$19,899,999$$. So this is the only such number divisible by $$63$$ whose digits add to $$63$$ in the smallest possible category of types of numbers that can have such numbers. So this is the smallest such number.

• Is that the smallest such number? [no criticism of your answer intended... but OP originally asked for smallest.] (+1 on answer.) – coffeemath Nov 20 '18 at 9:01
• Sorry @coffeemath , I had a typo. – Akash Roy Nov 20 '18 at 10:18
• Yes, it's the smallest. And I explained why. There can't be one with 7 digits, so the smallest has eight or more. This has 8 so it is in the smallest group. the smallest digit it can begin with is 1 so this is the smallest of the smallest. If number starts with 1 and has 8 digits and adds to $63$ then is must have 6 nines and one 8. So smallest number would be $19,999,999 - 10^k$ where $10^k\equiv 5 \mod 7$. The only such option for $0\le k \le 7$ is $k=5$. So this is the only solution with 8 digits begining with $1$. And no smaller number is a solution. – fleablood Nov 20 '18 at 17:05

(This is essentially the same solution as @fleablood 's; but doubts were raised whether it is actually the smallest.)

Such a number has at least $$8$$ digits. Since the prescribed digit sum is $$63$$ we have to deduct exactly $$9$$ units from writing eight nines. Trying with $$x_1=1$$ as first digit, and all other nines, we have given away $$8$$ units, one more to go. Divisibility by $$9$$ is taken care of automatically. Now $$19\,999\,999=5$$ mod $$7$$; therefore we need to find a $$k\in[0..6]$$ with $$10^k=5$$ mod $$7$$, or we are bust with $$x_1=1$$. Fortunately $$10^5=5$$ mod $$7$$. It follows that $$19\,899\,999$$ is the smallest number with the required properties.

• @coffeemath: Sorry for the typo. Thank you for reporting it. – Christian Blatter Nov 20 '18 at 10:15

The smallest number whose digits all sum to a multiple of $$63$$ is $$9{,}999{,}999$$. The next smallest is $$18{,}999{,}999$$, then $$19{,}899{,}999$$, then $$19,989{,}999$$, and so on. All of these are clearly divisible by $$9$$, so it suffices to check for divisibility by $$7$$. As it happens, the first two are not, but $$19{,}899{,}999/7=2{,}842{,}857$$ (and, just to doublecheck, $$19{,}899{,}999/63=315{,}873$$).

Remark: It's not a priori obvious that any of the numbers described here will turn out to be divisible by $$7$$. You could say we just got lucky. Or you could do a modular argument to show that luck had nothing to do with it. One thing is obvious: the smallest number sought for is certainly no greater than $$777{,}777{,}777$$.

• The modular argument is not hard. The numbers are all of the form $19,999,999 - 10^k$ so we need $19,999,999 - 10^k \equiv 0 \mod 7$ or $10^k \equiv 5 \mod 7$. As $10$ and $7$ is relativley prime and $7$ is prime the $10^k; 0\le k < 7$ are distinct modulo $7$ and $10^5$ is the only one that works. – fleablood Nov 20 '18 at 17:28
• @fleablood, are you saying, more generally, that if $10$ and $p$ are relatively prime (with $p$ a prime), then $10^k$ for $0\le k\lt p$ are distinct modulo $p$? It's true that $10$ is a primitive root mod $7$, but not because it's relatively prime to $7$. – Barry Cipra Nov 20 '18 at 22:46
• Yeah, I guess I worded it incorrectly. $10\equiv 3$ is a primitive root is what I meant. – fleablood Nov 20 '18 at 23:08

I also verify that the Number 19,899,999 is the smallest number that is divisible by 63 and Also its sum is divisible by 63. Here a program I wrote in Python 3. To find this out by brute forcing :

for i in range(9999990, 19900000, 63):
sum_is = sum(int(d) for d in str(i))
if sum_is%63==0:
print(i)

Prints :
19899999

Also this time I started from 9999990 because this number the last number divisible by 63 that has sum less than 63.

This can also be used to do the same with any number.

Just change the range and 63 as you want.

Hope this Helps!

[EDIT SUMMARY] For more Optimization and better Understanding. Added direct summing instead of sum_digits as request/suggested by @Paul Evans. Also added jump of 63 as Suggested by @Kyle Kanos.

• Instead of the mysterious sum_digits(i) you could write python code: sum(int(d) for d in str(i)) – Paul Evans Nov 20 '18 at 16:01
• You can also save time by starting at 9999990 and incrementing by 63, eliminating the need to test i%63==0. – Kyle Kanos Nov 20 '18 at 19:47
• Yes! That's more optimized. Thanx for advice. – FightWithCode Nov 21 '18 at 13:20


321 views

For an undirected graph $G=(V, E)$, the line graph $G'=(V', E')$ is obtained by replacing each edge in $E$ by a vertex, and adding an edge between two vertices in $V'$ if the corresponding edges in $G$ are incident on the same vertex. Which of the following is TRUE of line graphs?

1. the line graph for a complete graph is complete
2. the line graph for a connected graph is connected
3. the line graph for a bipartite graph is bipartite
4. the maximum degree of any vertex in the line graph is at most the maximum degree in the original graph
5. each vertex in the line graph has degree one or two
retagged | 321 views
+2
Option B). is true.
+2
yes, Agreed. (y)
0
why not option C?
0
you can check with K(2,2). The line graph obtained is not biparitite.

option B is the right answer

We can solve this question by eliminating options

Option (A) False

Let us take a complete graph of  $4$ vertices:

In a line graph, no. of edges is

$\sum_{i=0}^{n}$  ${}^{d_{i}}C_{{2}},$     $d_{i}$is degree of each vertex

=$\frac{3\times 2}{2}+\frac{3\times 2}{2}+\frac{3\times 2}{2}+\frac{3\times 2}{2} = 3\times 4=12$

No. of vertices in line graph = No. of edges in original graph
No. of vertices in line graph $= 6$
So, no. of edges to make complete graph with $6$ vertices $= \frac{6\times 5}{2} = 3\times 5=15$
But for given line graph from complete graph of $4$ vertices we have only $12$ edges.

Option (B)   True

1. Smallest line graph for original graph one edge

which is also connected graph

If a graph is connected with more then one edge, it will never be disconnected

Option (C)    False

This cannot be 2-colorable and hence is not bipartite.

Option (D)    False

Because line graph degree of vertex depends on the attribute

e.g., $\left [ A,B \right ]$ as point in line graph, then degree of this vertex depends on degree of $A$ and degree of $B$ in the original graph.

I'm drawing degree for a point $[AB]$ in line graph.

So, this is wrong.

Option (E)    False (wrong as proved in above option (D))

edited

The line graph of a connected graph is connected. If G is connected, it contains a path connecting any two of its edges, which translates into a path in L(G) containing any two of the vertices of L(G). Therefore, option B is correct.

We can also do this question using elimination of options.

edited
0
but your L(G2): e1---e2 it is bipertite right?e1 may be in one partition and e2 on another...it is bipertite i guess.
0
Hi,

I have corrected the example, and updated the file.
0
great example you have updated...proving line of tree is not a tree too :p