Normal subgroup of automorphisms of a free group

Let $F_2=\langle X,Y\rangle$ be the free group of rank $2$ and consider $A,B,C\in Aut(F_2)$ given by: $$A(X,Y)\mapsto(YX^{-1}Y^{-1},Y^{-1})$$ $$B(X,Y)\mapsto(X^{-1},Y^{-1})$$ $$C(X,Y)\mapsto(X^{-1},XY^{-1}X^{-1})$$ The abelianisation of $F_2$ is isomorphic to $\mathbb{Z}^2$ whose automorphism group is $GL(2,\mathbb{Z})$. Thus we get an induced surjective map $m\colon Aut(F_2)\to GL(2,\mathbb{Z})$ which maps an element in $Aut(F_2)$ to the matrix associated to the corresponding element in $Aut(\mathbb{Z}^2)$. I would like to show that the subgroup $\langle A,B,C\rangle$ is normal in $Aut(F_2)$ by proving that $m(\langle A,B,C\rangle)$ is the centre of $GL(2,Z)$. So far I wasn't able to have any good idea... could you help me? Thanks, bye.

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Just calculate the matrices: m(A) = m(B) = m(C) = -1 –  Jack Schmidt Nov 22 '11 at 18:22
You may want to use \langle and \rangle instead of < and >; the spacing is correct for the former. –  Arturo Magidin Nov 22 '11 at 18:27
how did you calculate explicitly the matrices? $A$,$B$ and $C$ do not belong to $Aut(\mathbb{Z}^2)$ so I should compute their image in the quotient, first. Do I have to write how, say $A$, acts on the cosets $F_2'X$ and $F_2'Y$ in $F_2/F_2'$? P.S. I did use \langle and \rangle! :) –  fatoddsun Nov 22 '11 at 19:22
@fatoddsun: Since $[F_2,F_2]$ is fully invariant, any homomorphism $f\colon F_2\to F_2$ induces a homomorphism $f_{\rm ab}\colon \mathbb{Z}^2\to\mathbb{Z}^2$ by $f_{\rm ab}(\overline{X}) = \overline{f(X)}$ and $f_{\rm ab}(\overline{Y}) =\overline{f(Y)}$. So just "abelianize" the image. –  Arturo Magidin Nov 22 '11 at 20:24
@fatoddsun: Note, however, that having the image be normal is not sufficient; that will only show that the preimage under the abelianization map $\mathrm{Aut}(F_2)\to\mathrm{GL}(2,\mathbb{Z})$ is normal, and it's not immediately clear to me that this preimage is just $\langle A,B,C\rangle$. –  Arturo Magidin Nov 22 '11 at 20:57

Arturo Magidin's excellent answer explains how to show that $m(\langle A,B,C\rangle)=\{\pm 1\}$, and also why this does not in general imply that your subgroup is normal. However, I want to explain why in this case it is true that $\langle A,B,C\rangle$ is the full preimage of $\{\pm 1\}$, and thus is normal.

The key is a theorem of Nielsen (1) on what is now called $IA_n$: this is by definition the kernel of the natural surjection $Aut(F_n)\to GL_n\mathbb{Z}$, so it fits into a short exact sequence

$$1\to IA_n\to Aut(F_n)\to GL_n\mathbb{Z}\to 1$$

This group of automorphisms which are the Identity on the Abelianization has been well-studied, starting with Magnus (2), who proved that $IA_n$ is finitely generated by giving an explicit set of generators: type (A) which for given $i$ and $j$ conjugates $x_i$ by $x_j$ and fixes all other generators, and type(B) which for given $i$, $j$, and $k$ multiplies $x_i$ by $x_jx_kx_j^{-1}x_k^{-1}$ and fixes all other generators.

But anyway, we are interested here in $IA_2$, and Nielsen proves in (1) that $IA_2$ is just $Inn(F_2)$, the group of automorphisms obtained by conjugation. This is isomorphic to $F_2$ and generated by $conj_X\colon X\mapsto X, Y\mapsto XYX^{-1}$ and by $conj_Y\colon X\mapsto YXY^{-1}, Y\mapsto Y$.

We know that your group $\langle A,B,C\rangle$ subjects to $\{\pm 1\}$, so to show it is the full preimage of $\{\pm 1\}$ it suffices to show that $\langle A,B,C\rangle$ contains the full preimage of $\{1\}$, which is $IA_2$. But we can compute by hand that $AB=conj_Y$ and $CB=conj_X$. So by Nielsen's theorem your group contains $IA_2$, and thus fits into a short exact sequence

$$1\to IA_2\to \langle A,B,C\rangle\to \{\pm 1\}\to 1$$

In particular, your subgroup is normal in $Aut(F_2)$.

Finally, here is another method of proving that $\langle A,B,C\rangle$ is normal which does not require quoting any papers written in German. We know that the elementary Nielsen transformations provide a generating set for $Aut(F_n)$. This generating set becomes particularly simple in the case when $n=2$: we have three generators $R\colon X\mapsto X^{-1}, Y\mapsto Y$, $S\colon X\mapsto Y, Y\mapsto X$, and $T\colon X\mapsto XY,Y\mapsto Y$. To show that $\langle A,B,C\rangle$ is normal in $Aut(F_2)$ means that it is invariant under conjugation under any element of $Aut(F_2)$. But since $R,S,T$ generate $Aut(F_2)$, it is enough just to show that $\langle A,B,C\rangle$ is invariant under conjugation by $R$, by $S$, and by $T$.

To do this for $R$, for example, you just need to compute the conjugates $RAR^{-1}$, $RBR^{-1}$, and $RCR^{-1}$ and show that each can be written as a combination of $A$, $B$, and $C$. In this case this is quite easy: we have $RAR^{-1}=A$, $RBR^{-1}=B$, and $RCR^{-1}=C^{-1}$. Now all you have to do is do the same for $SAS^{-1}$, $SBS^{-1}$, $SCS^{-1}$, $TAT^{-1}$, $TBT^{-1}$, and $TCT^{-1}$, and you've proved that $\langle A,B,C\rangle$ is normal!

(1) Nielsen, Die Isomorphismen der allgemeinen unendlichen Gruppe mit zwei Erzeugenden, Math. Ann. 78 (1964), 385-397

(2) Magnus, Über n-dimensinale Gittertransformationen, Acta Math. 64 (1935), 353-367.

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Nice; thanks for filling the gap! –  Arturo Magidin Nov 22 '11 at 21:39

If $G$ is a group, $N$ is a normal subgroup of $G$, and $f\colon G\to G$ is a homomorphism with $f(N)\subseteq N$, then $f$ induces a homomorphism $\overline{f}\colon G/N\to G/N$ by $\overline{f}(gN) = f(g)N$. To see this, note that $\widehat{f}\colon G\to G/N$ given by $\widehat{f}(g) = gN$ is a homomorphism, and since $f(N)\subseteq N$, then $N$ is contained in the kernel of $\widehat{f}$, so $\widehat{f}$ factors through $G/N$.

Since $[F_2,F_2]$ is fully invariant, this argument will work for $A$, $B$, and $C$.

For $A$, $x$ is mapped to $yx^{-1}y^{-1}$, which gets mapped to $\overline{x}^{-1}$ in $F_2^{\rm ab}$; $y$ is mapped to $\overline{y}^{-1}$. So the induced map $A_{\rm ab}\colon F_2^{\rm ab}\to F_{2}^{\rm ab}$ sends $(a,b)$ to $(-a,-b)$.

This is the same as what $B_{\rm ab}$ does; again the same thing as $C_{\rm ab}$ does. So the image of the subgroup generated by $A$, $B$, and $C$ in $\mathrm{GL}(2,\mathbb{Z})$ is the subgroup generated by their images, which is just the subgroup generated by $-I$.

Note, however, that the fact that $m(\langle A,B,C\rangle) = \{I,-I\}$ does not show, by itself, that $\langle A,B,C\rangle$ is normal in $\mathrm{Aut}(F_2)$; it shows that the pre-image is normal. The preimage is the subgroup of $\mathrm{Aut}(F_2)$ generated by $A$, $B$, $C$, and all automorphisms that induce the identity on $F_2^{\rm ab}$. It is was not clear to me whether this is just $\langle A,B,C\rangle$. For this, see Tom Church's great answer.

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Thank you very much guys! –  fatoddsun Nov 23 '11 at 18:23


# Deriving Equations of a Parabola

1. May 29, 2008

### PFStudent

Hey,

1. The problem statement, all variables and given/known data
How do you derive the equations of the parabola from the general equation of a Conic Section?

2. Relevant equations
General Equation of a Conic Section,
$${{{A}{{x}^{2}}}+{{B}{x}{y}}+{{C}{{y}^{2}}}+{{D}{x}}+{{E}{y}}+{F}} = {0}$$

Where (for a parabola),
$${{\{}A,B,C,D,E,F{\}}}{{\,}{\,}{\,}}{\in}{{\,}{\,}{\,}}{\mathbb{R}}$$
$${{\{}A, C{\}}}{{\,}{\,}{\,}}{\neq}{{\,}{\,}{\,}}{0}$$
$${{{B}^{2}}-{{4}{A}{C}}} = {0}$$

3. The attempt at a solution
From the general equation of a Conic Section,
$${{{A}{{x}^{2}}}+{{B}{x}{y}}+{{C}{{y}^{2}}}+{{D}{x}}+{{E}{y}}+{F}} = {0}$$

How do I derive the following formulas for a parabola:
General Form for a Parabola,
$${{{{\left(}{{{H}{x}}+{{I}{y}}}{\right)}}^{2}}+{{J}{x}}+{{K}{y}}+{L}} = {0}$$
Where,
$${{{I}^{2}}-{{4}{H}{J}}} = {0}$$

Analytic Geometry Equations,
Vertical Axis of Symmetry
$${{{\left(}x-h{\right)}}^{2}} = {{{4}{p}}{{{\left(}y-k}{\right)}}}$$
$${y} = {{{a}{{x}^{2}}}+{{b}{x}}+{c}}$$
Horiztonal Axis of Symmetry
$${{{\left(}y-k{\right)}}^{2}} = {{{4}{p}}{{{\left(}x-h{\right)}}}}$$
$${x} = {{{d}{{y}^{2}}}+{{e}{y}}+{f}}$$

I read that every parabola is a combination of transformations of the parabola, $${{y}={{x}^{2}}}$$; but I'm not quite sure how that helps.

Thanks,

-PFStudent

Last edited: May 29, 2008
2. May 29, 2008

### Dick

I'm not precisely sure of what exactly you are supposed to do, but the only way that the quadratic terms Ax^2+Bxy+Cy^2 can be factored as (Hx+Iy)^2 is that B^2-4AC=0. Just multiply (Hx+Iy)^2 out and identify the terms. Does that help?

3. May 30, 2008

### HallsofIvy

First you need to find the "principal axes" (axis of symmetry and perpenpendicular to it through the vertex) and rotate the coordinate system so that new x' and y' axes are the principal axes.

There are two ways to do that, basically using the quadratic terms $Ax^2+ Bxy+ Cx^2$ (By the way, you don't have to put braces, { }, around every term. They are only necessary when you want an entire expression in a particular place. For example, [i t e x]e^{xy+ b}[/i t e x] gives $e^{xy+ b}$ while [i t e x]e^xy+ b[/i t e x] gives $e^xy+ b$.)

Rotating the axes by angle $\theta$, so that the new x'y'- axes are at angle $\theta$ to the old xy- axex, x and y are given, in terms of the new x', y' variables, by $x= x'cos(\theta)+ y'sin(\theta)$ and $y= -x'sin(\theta)+ y'cos(\theta). Replace x and y by those in $Ax^2+ Bxy+ Cx^2$ and choose$\theta[/itex] so that the coefficient of x'y' is 0. Use those formulas with the correct value of $\theta$ to replace x and y in the entire formula. If it really is a parabola, both the x'y' and y'2 terms should vanish.

The other way is to write it as a matrix problem:
$$Ax^2+ Bxy+ Cy^2= \left[\begin{array}{cc}x & y \end{array}\right]\left[\begin{array}{cc}A & \frac{B}{2} \\ \frac{B}{2} & C\end{array}\right]\left[\begin{array}{c}x \\ y\end{array}\right]$$

You can "diagonalize" that matrix (so writing the quadratic without the xy term) by finding the eigenvalues and eigenvectors of the coefficient matrix. The eigenvectors will point along the principal axes. Since this is a parabola, one of the eigenvalues should be 0.

4. May 31, 2008

### PFStudent

Hey,

Thanks for the replies: Dick and HallsofIvy.

Yea, to be honest I had not seen that general form for a parabola before until I saw it on wikipedia, maybe it is wrong? - here is the link, (Scroll down a bit to the section "General Parabola").

http://en.wikipedia.org/wiki/Parabola

Let me know what you think.

Hmm, ok that makes sense, I will try that. However, I thought since a parabola - in its' most general form - could be expressed as a conic section with the constraint that $${{B^2} = {4AC}}$$. Then shouldn't I be able to derive the more specific forms of a parabola from the general equation of a conic section with the constraint that $${{B^2} = {4AC}}$$?

Where the general equation of a Conic Section is,
$${{{A}{{x}^{2}}}+{{B}{x}{y}}+{{C}{{y}^{2}}}+{{D}{x}}+{{E}{y}}+{F}} = {0}$$

Also, yea I tend to use a bit too many braces ({ }), just a habit from first learning $LaTeX$ way back.

Thanks,

-PFStudent


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Updated on: 12 Jan 2019, 03:25
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How many nails did Rudy buy if he purchased them at a price of $0.25 per four nails, sold them at$0.22 per three nails, and made a profit of $2.60? A. 300 B. 240 C. 180 D. 160 E. 120 Originally posted by Nums99 on 12 Jan 2019, 01:42. Last edited by Bunuel on 12 Jan 2019, 03:25, edited 1 time in total. Renamed the topic, edited the question, moved to PS and added the OA. Senior Manager Joined: 15 Jan 2018 Posts: 371 Concentration: General Management, Finance GMAT 1: 720 Q50 V37 Re: How many nails did Rudy buy if he purchased them at a price of$0.25  [#permalink]

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12 Jan 2019, 02:12
1
Hi there,

Here's the solution.

Let the number of nails Rudy purchased be x.

Rudy purchased nails at a price of $0.25 per four nails. So, Cost Price of 4 nails =$0.25
Cost Price of 1 nail = = $0.25/4 Cost price of x nails = ($0.25/4)x

Also given that Rudy sold nails at $0.22 per three nails. So, Selling Price of 3 nails =$0.22
Selling Price of 1 nail = = $0.22/3 Selling price of x nails = ($0.22/3)x

Also, Rudy made a profit of $2.60. Profit = Selling Price - Cost Price$2.60 = ($0.22/3)x - ($0.25/4)x = x(0.88 - 0.75)/12
$2.60 = x(0.13)/12 x = 240 Hence, the the number of nails Rudy purchased is 240. Therefore, the Correct Answer is Option B. 240 _________________ New to GMAT Club or overwhelmed with so many resources? Follow the GMAT Club Study Plan! Not happy with your GMAT score? Retaking GMAT Strategies! Game of Timers - Join the Competition to Win Prizes Senior Manager Status: Manager Joined: 27 Oct 2018 Posts: 365 Location: Egypt Concentration: Strategy, International Business GPA: 3.67 WE: Pharmaceuticals (Health Care) Re: How many nails did Rudy buy if he purchased them at a price of$0.25  [#permalink]

### Show Tags

16 Jan 2019, 02:57
Nums99 wrote:
How many nails did Rudy buy if he purchased them at a price of $0.25 per four nails, sold them at$0.22 per three nails, and made a profit of $2.60? A. 300 B. 240 C. 180 D. 160 E. 120 Profit = Sales - Cost $$2.6 = \frac{0.22}{3}n - \frac{0.25}{4}n = \frac{0.88n}{12} - \frac{0.75n}{12}$$ $$\frac{13}{5} = \frac{0.13n}{12} = \frac{13n}{1200}$$ $$n = \frac{13*1200}{13*5} = 240$$ B _________________ RC Moderator Joined: 24 Aug 2016 Posts: 802 Location: Canada Concentration: Entrepreneurship, Operations GMAT 1: 630 Q48 V28 GMAT 2: 540 Q49 V16 Re: How many nails did Rudy buy if he purchased them at a price of$0.25  [#permalink]

### Show Tags

16 Jan 2019, 08:15
As we are dealing with two numbers 3 & 4 .... lets take the LCM of them i.e.,12 .
Cost price for 12 nails = $0.25*12/4=$0.75 & Selling price for 12 nails = $0.22*12/3=$0.88
Profit made per 12 nails = $0.88 -$0.75 = $0.13 So total '12 units' sold =$2.60/$0.13 =20 Total nails sold = 20*12 = 240 .... Thus Ans would be option B. _________________ Please let me know if I am going in wrong direction. Thanks in appreciation. Director Joined: 12 Feb 2015 Posts: 863 Re: How many nails did Rudy buy if he purchased them at a price of$0.25  [#permalink]

### Show Tags

25 Feb 2019, 09:26
Nums99 wrote:
How many nails did Rudy buy if he purchased them at a price of $0.25 per four nails, sold them at$0.22 per three nails, and made a profit of $2.60? A. 300 B. 240 C. 180 D. 160 E. 120 LCM approach is good in such questions:- Rudy purchased nails at a price of$0.25 per four nails, or $0.75 per 12 nails sold them at$0.22 per three nails, or $0.88 per 12 nails and made a profit of$2.60 (total) or $(0.88-0.75) =$ 0.13 per 12 nails

(2.60 * 12)/0.13 = 240 nails (Ans)
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How many nails did Rudy buy if he purchased them at a price of $0.25 [#permalink] ### Show Tags 28 Feb 2019, 11:09 Nums99 wrote: How many nails did Rudy buy if he purchased them at a price of$0.25 per four nails, sold them at $0.22 per three nails, and made a profit of$2.60?

A. 300
B. 240
C. 180
D. 160
E. 120

Cost Price of $$4$$ nails $$=  0.25$$

Cost Price of $$3$$ nails $$= \frac{0.25*3}{4} =  0.1875$$

Selling Price of $$3$$ nails $$= 0.22$$

Profit on $$3$$ nails $$= 0.22 - 0.1875 =  0.0325$$

Let $$"x"$$ be total number of nails to make profit of $$2.60$$.

$$\frac{0.0325}{3} = \frac{2.60}{x}$$

$$0.0325*x =3*2.60$$

$$x = \frac{3*2.60}{0.0325} = 240$$

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Re: How many nails did Rudy buy if he purchased them at a price of $0.25 [#permalink] ### Show Tags 02 Mar 2019, 10:08 Nums99 wrote: How many nails did Rudy buy if he purchased them at a price of$0.25 per four nails, sold them at $0.22 per three nails, and made a profit of$2.60?

A. 300
B. 240
C. 180
D. 160
E. 120

We can let n = the number of nails and create the equation:

(0.22/3)n - (0.25/4)n = 2.6

Multiplying by 12, we have:

0.22 x 4n - 0.25 x 3n = 31.2

0.88n - 0.75n = 31.2

0.13n = 31.2

n = 240

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Re: How many nails did Rudy buy if he purchased them at a price of $0.25 [#permalink] 02 Mar 2019, 10:08 Display posts from previous: Sort by # How many nails did Rudy buy if he purchased them at a price of$0.25

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# Aternating sum of an increasing sequence of positive integers

Suppose $$A = (a_n) = (a_1, a_2, a_3, . . .)$$ is an positive, increasing sequence of integers.

Define an $$A$$- expressible number $$c$$ if $$c$$ is the alternating sum of a finite subsequence of $$A.$$ To form such a sum, choose a finite subset of the sequence $$A,$$ list those numbers in increasing order (no repetitions allowed), and combine them with alternating plus and minus signs. We allow the trivial case of one-element subsequences, so that each an is $$A-$$expressible.

Definition. Sequence $$A = (a_n)$$ is an “alt-basis” if every positive integer is uniquely $$A-$$ expressible. That is, for every integer $$m > 0,$$ there is exactly one way to express $$m$$ as an alternating sum of a finite subsequence of $$A.$$

Examples. Sequence $$B = (2^{n−1}) = (1, 2, 4, 8, 16, . . .)$$ is not an alt-basis because some numbers are B-expressible in more than one way. For instance $$3 = −1 + 4 = 1 − 2 + 4.$$

Sequence $$C = (3^{n−1}) = (1, 3, 9, 27, 81, . . .)$$ is not an alt-basis because some numbers (like 4 and 5) are not C-expressible.

An example of an alt-basis is $$\{2^n-1\}=\{1,3,7,15,31,\ldots\}$$

Is there a fairly simple test to determine whether a given sequence is an alt basis?

I have attempted to solve this from a limited knowledge in sequences and have found out various kinds of sequences do not work but fail to see what it is that could make it work.

• Seems slightly (at least) related to this: encyclopediaofmath.org/index.php/Additive_basis But I think you know already about additive bases. – Masterphile Mar 25 at 0:37
• See math.stackexchange.com/questions/3579462. (Not an exact duplicate because it doesn't ask for a general test to determine whether a sequence is an alt-basis.) What is the source of this question? – joriki Mar 25 at 6:47
• Do you know of any alt-bases that are not of the form $\{b_1,b_2,\dots,b_k,a_{k+1},a_{k+2},\dots\}$ where $A=\{a_1,a_2,\dots\}=\{2^n-1\}$ and $\{b_1,\dots,b_k\}$ is some finite increasing set of integers? – Vepir Mar 25 at 21:54
• For example I believe that $\{5\cdot2^{n - 2} + (-1)^{n + 1} 2^{n - 2} - 1\}$ is an alt-basis. – Vepir Mar 25 at 22:50

I can’t answer the question, but I can at least give you a systematic large family of alt-bases.

If $$A$$ is a finite set of positive integers, let $$S(A)$$ be the set of $$A$$-expressible integers, and let $$S^+(A)$$ be the set of $$A$$-expressible positive integers. Then

$$S(A)=S^+(A)\cup\{-a:a\in S^+(A)\},$$

and if $$b>\max A$$, then

$$S^+\left(A\cup\{b\}\right)=S^+(A)\cup\{b-s:s\in S^+(A)\}\cup\{b\}.$$

Thus, if $$|A|=n$$, the maximum number of $$A$$-expressible positive integers is $$2^n-1$$, and $$\max S(A)=\max A$$.

Now suppose that $$A=\{a_n:n\in\Bbb Z^+\}$$, where $$a_n for each $$n\in\Bbb Z^+$$. For $$n\in\Bbb Z^+$$ let $$A_n=\{a_k\in A:1\le k\le n\}$$. Then each $$m\in S(A)$$ is uniquely $$A$$-expressible iff $$|S^+(A_n)|=2^n-1$$ for each $$n\in\Bbb Z^+$$. Moreover, $$S^+(A)=\Bbb Z^+$$ iff for each $$k\in S^+(A)$$ there is a minimal $$n(k)\in\Bbb Z^+$$ such that $$k\in S^+(A_{n(k)})$$. Note that either $$n(k)=1$$, or $$k\in S^+(A_{n(k)})\setminus S^+(A_{n(k)-1})=\{a_{n(k)}-s:s\in S^+(A_{n(k)-1})\}$$.

For $$n\in\Bbb Z^+$$ let

$$a_n=2^n-1=\underbrace{1\ldots 1}_n\text{ in binary},$$

and let $$A=\{a_n:n\in\Bbb Z^+\}$$. It’s not hard to see that

$$S^+(A_n)=\{1,\ldots,2^n-1\}$$

for each $$n\in\Bbb Z^+$$, so $$A$$ is, as you already observed, an alt-basis. For instance, working in binary, we see that

\begin{align*} 22&=10110_{\text{two}}\\ &=11111_{\text{two}}-1111_{\text{two}}+111_{\text{two}}-1_{\text{two}}\\ &=31-15+7-1\\ &=a_5-a_4+a_3-a_1. \end{align*}

Now let $$\ell,m\in\Bbb Z^+$$. For $$n=1,\ldots,\ell$$ let

$$\color{red}{a_n^{(\ell,m)}}=2^ma_n=\underbrace{1\ldots 1}_n\underbrace{0\ldots 0}_m\text{ in binary}.$$

For $$n=\ell+k$$, where $$k=1,\ldots,m$$, let

$$\color{blue}{a_n^{(\ell,m)}}=2^{m-k}a_n=\underbrace{1\ldots 1}_n\underbrace{0\ldots 0}_{m-k}\text{ in binary}.$$

Finally, for $$n>\ell+m$$ let $$a_n^{(\ell,m)}=a_n$$, and let $$A_{(\ell,m)}=\left\{a_n^{(\ell,m)}:n\in\Bbb Z^+\right\}$$; then $$A_{(\ell,m)}$$ is an alt-basis.

For example,

\begin{align*} A_{(4,2)}&=\{\color{red}{4},\color{red}{12},\color{red}{28},\color{red}{60},\color{blue}{62},\color{blue}{63},127,\ldots\}\\ &=\{\color{red}{100},\color{red}{1100},\color{red}{11100},\color{red}{111100},\color{blue}{111110},\color{blue}{111111},1111111,\ldots\}\text{ in binary}. \end{align*}

To verify this it suffices to show that $$S^+\left(\left\{a_n^{(\ell,m)}:1\le n\le \ell+m\right\}\right)=S^+(A_{\ell+m})$$. The argument is a bit messy to write out, but the idea is straightforward; I’ll illustrate it with $$A_{(4,2)}$$. First, it’s clear from the discussion of $$A$$ that

\begin{align*} S^+\left(\{4,12,28,60\}\right)&=S^+\left(4\{1,3,7,15\}\right)\\ &=4S^+\left(\{1,3,7,15\}\right)\\ &=4\{1,2,\ldots,15\}\\ &=\{4,8,12,\ldots,60\}\\ &=4S^+(A_4). \end{align*}

Then

\begin{align*} S^+(\{4,&12,28,60,62\})=\\ &4S^+(A_4)\cup\left\{|62-s|:s\in S^+(\{4,12,28,60\})\right\}\cup\{62\}=\\ &4S^+(A_4)\cup\left\{|62-s|:s\in\{4,8,12,\ldots,60\}\right\}\cup\{62\}=\\ &4S^+(A_4)\cup\{2,6,10,\ldots,58,62\}=\\ &\{2,4,6,8,\ldots,60,62\}=\\ &2S^+(A_5), \end{align*}

and a similar calculation shows that $$S^+(\{4,12,28,60,62,63\})=S^+(A_6)$$.

I did not collect the set of all alt-bases, but I did find some useful observations, including:

Alt-basis must contain an infinite number of terms of form $$a_k=2^{k}-1,k\in N\subseteq\mathbb N$$.

The converse does not hold. At the end, I give examples of alt-bases and not-alt-bases in this context.

Do correct me If I missed anything.

Let $$A=\{a_1,a_2,\dots\}$$ such that $$a_1\lt a_2 \lt \dots$$ are positive integers.

Definition. For $$A$$ to be an "alt-basis", we need to have both the "uniqueness" and "completeness". In other words, every number is expressible in exactly one way via alternating summation of subsets of $$A$$, which are summed in increasing order.

Definition. A finite (sub)sequence $$A|_n=\{a_1,\dots,a_n\}$$ is an "alt-prefix" if every integer in $$[1,2^{n}-1]$$ is uniquely expressible via alternating summation of subsets of $$A|_n$$ when summed in increasing order. The element $$a_n$$ is called an "anchor element".

Definition. "Anchor sequence" is a set $$\mathcal A(A):=\{a_{n_1},a_{n_2},\dots\}$$ of all "anchor elements" $$a_{n_1},a_{n_2},\dots$$

Notice that a set has $$2^n$$ subsets minus the empty set and that every subset can be rearranged in an increasing order. We want to assign a distinct value to each of those subsets via the alternating summation, to have an alt-basis. The alt-prefix is defined to cover exactly those $$2^n-1$$ subsets. It follows that:

Lemma. $$A$$ is an alt-basis $$\iff$$ $$A$$ is a union of alt-prefixes $$A=A|_{n_1}\cup A|_{n_2}\cup \dots$$

That is, $$A$$ is an alt-basis if and only if there exists a corresponding infinite anchor sequence $$\mathcal A(A)$$.

We add two more definitions to write all of this more easily:

Definition. Let $$s(\{b_1,\dots,b_n\})$$ be the result of the alternating summation of $$b_1\lt b_2\lt \dots\le b_n$$. Let $$s_+$$ and $$s_-$$ always start the alternating summation with $$+,-$$ respectively. Then $$s_+=-s_-$$. If $$n$$ is odd then $$s=s_+$$ and if $$n$$ is even then $$s=s_-$$. This guarantees $$s\gt 0$$ because the largest element $$b_n$$ will have a positive sign.

Definition. Define "$$n$$-th partial subset-sum set" of a positive increasing integer sequence $$A$$ as:

$$\mathcal S_n(A):=\{s(A_i):A_i\in\mathcal P(A|_n)\}$$

Where $$\mathcal P(A|_n)$$ is the set of all subsets of $$A|_n=\{a_1,a_2,\dots,a_n\}$$.

The set of all "anchor elements" $$\mathcal A(A)=\{a_{n_1},a_{n_2},\dots\}\subseteq A$$ satisfies $$S_{n_i}=[1,2^{n_i-1}-1]$$ for all $$n_i$$.

Corollary. $$A$$ is an alt-basis if and only if it "is covered by the anchor sequence": $$\max \mathcal A(A)\to \infty$$.

Notice that $$\max S_n = a_n$$. If $$a_n$$ is an anchor element, then $$\max S_n = 2^n-1$$. This gives:

Proposition. If $$a_n$$ is an anchor element, then $$a_n=2^n-1$$.

The converse does not hold. For example, in $$\{1,4,7\}$$ the $$a_3=7=2^3-1$$ but $$a_3$$ is not an anchor element, because $$S_3=\{1,3,4,6,7\}\ne[1,7]$$.

Example $$1$$. It is not hard to see that $$\mathcal A(\{2^n-1\})=\{2^n-1\}$$. This is because:

• $$S_1=\{(+1)\}$$ $$\implies$$ $$a_1$$ is an anchor element.

• $$S_2=\{(+1),(-1+3),(3)\}$$ $$\implies$$ $$a_2$$ is an anchor element.

• $$S_3=\{(+1),(-1+3),(3),(-3+7),(+1-3+7),(-1+7),(7)\}$$ $$\implies$$ $$a_3$$ is an anchor element.

• $$\dots$$ proceed via induction to show every $$a_n$$ is an anchor element.

Since $$\mathcal A(\{2^n-1\})$$ exists and covers the entire $$\{2^n-1\}$$, the $$\{2^n-1\}$$ is an alt-basis.

Example $$2$$. The $$\mathcal A(\{n\})=\{1\}$$ does not cover the entire $$\{n\}$$, hence $$\{n\}$$ is not an alt-basis.

It is not hard to see that $$\max S_n = n \lt 2^n-1\implies a_n$$ is not an anchor element, for every $$n\gt 1$$.

Example $$3.$$ We construct an alt-basis where every $$2$$nd element is an anchor element.

$$A=\begin{cases} 2^n-1, & n\text{ is even} \\ 2^n+2^{n-1}-1, & n\text{ is odd} \end{cases}$$

Use an inductive argument. Assume $$n=2k$$, $$a_{n}=2^{n}-1$$ is an anchor element, which means we have uniquely constructed all $$I_0=[1,2^n-1]=S_{n}$$ elements. Now we can subtract numbers in this interval from $$a_{n+1}$$ to see that:

• $$a_{n+1}=2^{n+1}+2^{n}-1$$ will cover $$I_1=[a_{n+1}-a_{n}, a_{n+1}]=[2^{n+1},2^{n+1}+2^{n}-1]$$

Here we see that $$I_0\cup I_1 \ne [1,2^{n+1}-1]$$ $$\implies$$ $$a_{n+1}$$ is not an anchor elemetn.

To see that $$a_{n+2}=2^{n+2}-1$$ is an anchor element, lets see what will we cover with it:

• $$a_{n+2}$$ combined with $$I_0$$ will cover $$I_2=[a_{n+2}-a_{n}, a_{n+2}]=[2^{n+1}+2^{n},2^{n+2}-1]$$

• $$a_{n+2}$$ combined with $$I_1$$ will cover $$I_3=[a_{n+2}-a_{n+1},a_{n+2}-2^{n+1}]=[2^n,2^{n+1}-1]$$

Now observe $$I=I_0\cup I_3\cup I_1\cup I_2$$ is equal to:

$$I=[1,2^n-1]\cup[2^n,2^{n+1}-1]\cup[2^{n+1},2^{n+1}+2^{n}-1]\cup[2^{n+1}+2^{n},2^{n+2}-1]=[1,2^{n+2}-1]$$

Implying $$a_{n+2}$$ covers $$I=[1,2^{n+2}-1]=S_{n+2}$$, $$\implies$$ $$a_{n+2}$$ is an anchor.

It is not hard to check base cases $$n=1,2$$, and we are done. We have:

$$\mathcal A\left(\left.\begin{cases} 2^n-1, & n\text{ is even} \\ 2^n+2^{n-1}-1, & n\text{ is odd} \end{cases}\right\}\right)=\{2^{2n}-1\}$$

So we have an alt-basis $$A$$.

Example $$4$$. It is not hard to show that:

$$A=\{2^k,2^k+1,2^k+3,2^k+7,\dots,2^k+2^{k}-1,2^{k+2}-1,2^{k+3}-1,\dots\}$$

Is an alt-basis for every $$k=0,1,2,\dots$$, whose anchors are all elements $$a_n,n\gt k$$.

Example $$5$$. The sequence of natural, triangular, tetrahedral,... numbers, or in general, any diagonal of the pascals triangle, is not an alt basis.

This is because for every fixed $$d$$, there exists $$n_0$$, such that for all $$n\ge n_0$$, we have $$\binom{n+d-1}{d}<2^n$$ implying that $$\max S_n\lt 2^n-1$$ for all $$n\ge n_0$$. This implies the sequence of anchors has at most $$n_0$$ elements, implying $$\max\mathcal A(A)\lt \infty$$, hence we do not have an alt-basis becuase of inevetable duplicates.

• I think => part of the initial lemme requires proof. – balcinus Mar 31 at 10:07


What is the sum of the $81$ products in the $9 \times 9$ multiplication grid?

What is an easy way to solve this problem? I believe that the value in each box is the product of $x$ and $y$.

Suppose the 9 × 9 multiplication grid, shown here, were filled in completely. What would be the sum of the 81 products?

• I asked a question that used this result (and has a nice, clean answer!) several years ago: See MSE 226983 for more! – Benjamin Dickman Jun 12 '16 at 3:29
• (1+2+...+8+9)^2 – Evorlor Jun 12 '16 at 18:45

The sum of the products in the top row would just be $(1+2+3+4+5+6+7+8+9)=45$

Then the next row would be $(2+4+6+8+10+12+14+16+18) = 2\times45 = 90$

So the top two rows sum to $(1+2)\times 45 = 135$

Then it becomes obvious that the full sum of the products is the product of the sums, ie. $45\times45 = 2025$

The sum is essentially $$\sum_{a=1}^9 \sum_{b=1}^9 ab =\sum_{a=1}^9 a \sum_{b=1} ^9 b=\sum_{a=1}^9 a \frac{9\cdot 10}{2}=\left(\frac{9\cdot 10}{2}\right)^2$$

We want to find the following: $$\sum_{i=1}^9 \sum_{j=1}^9 ij$$ Factor out the $i$ from the first summation: $$\sum_{i=1}^9 i\left(\sum_{j=1}^9 j\right)$$ Note that $\sum_{j=1}^9 j=45$. $$\sum_{i=1}^9 i\cdot 45$$ Factor out the $45$: $$45\left(\sum_{i=1}^9 i\right)$$ Note that $\sum_{i=1}^9 i=45$. $$45\cdot 45=2025$$

Use the distributive principle. The sum of the entries in the $2$ column is $2$ times the sum of the numbers $1$ through $9$, so the sum of all the entries is the sum of the numbers $1$ through $9$ times (what?)

Hint $\quad \begin{eqnarray} &\color{#c00}{1+2+3}\\ +\ &\color{#0a0}{2+4+6}\\ +\ &\color{blue}{3+6+9}\end{eqnarray}$ $\quad =\quad \begin{eqnarray} &\color{#c00}1\,(1+2+3)\\ +\ &\color{#0a0}2\,(1+2+3)\\ +\ &\color{blue}3\,(1+2+3)\end{eqnarray}$ $\quad =\quad (\color{#c00}1 + \color{#0a0}2 + \color{blue} 3)(1+2+3)\ \ =\ \ 6\times 6$

Let's prove by induction that the sum of an ${n}\times{n}$ grid is $\frac{n^4+2n^3+n^2}{4}$:

First, show that this is true for $n=1$:

$\sum\limits_{x=1}^{1}\sum\limits_{y=1}^{1}xy=\frac{1^4+2\cdot1^3+1^2}{4}$

Second, assume that this is true for $n$:

$\sum\limits_{x=1}^{n}\sum\limits_{y=1}^{n}xy=\frac{n^4+2n^3+n^2}{4}$

Third, prove that this is true for $n+1$:

$\sum\limits_{x=1}^{n+1}\sum\limits_{y=1}^{n+1}xy=$

$\color\red{\sum\limits_{x=1}^{n}\sum\limits_{y=1}^{n}xy}+\left(\sum\limits_{x=1}^{n}x(n+1)\right)+\left(\sum\limits_{y=1}^{n}y(n+1)\right)+(n+1)(n+1)=$

$\color\red{\frac{n^4+2n^3+n^2}{4}}+\left(\sum\limits_{x=1}^{n}x(n+1)\right)+\left(\sum\limits_{y=1}^{n}y(n+1)\right)+(n+1)(n+1)=$

$\frac{n^4+2n^3+n^2}{4}+(n+1)\left(\sum\limits_{x=1}^{n}x\right)+(n+1)\left(\sum\limits_{y=1}^{n}y\right)+(n+1)(n+1)=$

$\frac{n^4+2n^3+n^2}{4}+(n+1)\left(\frac{n^2+n}{2}\right)+(n+1)\left(\frac{n^2+n}{2}\right)+(n+1)(n+1)=$

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

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

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

$\frac{n^4+6n^3+13n^2+12n+4}{4}=$

$\frac{(n+1)^4+2(n+1)^3+(n+1)^2}{4}$

Please note that the assumption is used only in the part marked red.

Therefore, the sum of a ${9}\times{9}$ grid is $\frac{9^4+2\cdot9^3+9^2}{4}=2025$.

I will focus on an intuition for finding the result. Once you have the formula, other answers are perfect in providing the means for proving it technically.

Let us start with a small size version. If you take the top $3\times 3$ table only with numbers $1$, $2$, $3$ (to save some electrons), the sum of products is:

$$s= 1\times 1 +1\times 2 + 1\times 3 + 2\times 1 +2\times 2 + 2\times 3 + 3\times 1 +3\times 2 + 3\times 3 \,.$$ You see that you can rewrite this sum of nine products as: $$s = (1+2+3)\times (1+2+3)\,,$$ using the distribution of multiplication over additions.

This form is more interesting, because you can see a pattern. The sum of the first integers, multiplied by itself.

It can be reassuring to verify it works: you get $36$, which you can check by hand ($6+12+18$). You can test it with the $2\times2$ or $4\times4$ matrix, to verify it is not a coincidence.

The hint is apparent in the squared shape of the table. The same works for bigger tables too. Each product of $i$ and $j$ in this order is contained once and only once in the product $(1,\ldots,i,\ldots,n)\times(1,\ldots,j,\ldots,n)$.

Now you have to find the sum of integers. If you forget the generic expression for the sum of the $n$ first integer, have this figure in mind:

You pack together two triangles which give you a $n\times (n+1)$ rectangle, of area $n(n+1)$, the double of the area of each triangle.

So finally the answer is $\left(n(n+1)/2\right)^2$. With $n=9$, you get $2025$.

Once you have the method, which is relatively simple, you can easily turn it into a more formal proof, via induction for instance, as given in other answers.

• The link to "I like visual proofs" is dead. Besides that, I think the name "visual proof" is not suitable here. Perception may be betraying, so visual remarks should be backed up by some other argument. This makes it not related to "visual"! – Jasper Jun 12 '16 at 16:23
• @Jasper I have modified the answer. The visual aspect was just a reminder for a formula given in other answers – Laurent Duval Jun 12 '16 at 16:49


# Absolute Value

## Definition

For a real number, the absolute value of $$x$$, denoted $$\lvert x \rvert$$, is defined as

$\left|x \right| = \begin{cases} x, & \text{if } x \geq 0 \\ -x, & \text{if } x < 0. \end{cases}$

## Technique

### Evaluate $\left| 3\left( 17-55 \right) \right|$.

\begin{aligned} \left| 3\left( 17-55 \right) \right| &= \left| 3\left( -38 \right) \right| \\ &= \left| -114 \right| \\ &= 114 _\square \end{aligned}

### If $x=\frac{5+\sqrt{11}i}{2}$ and $y=\frac{5 -\sqrt{11}i}{2}$, what is the value of $\left| x + 3y \right|^2$?

\begin{aligned} \left| x + 3y \right|^2 &= \left| \frac{5+\sqrt{11}i}{2} + 3\left( \frac{5 -\sqrt{11}i}{2} \right) \right|^2 \\ &= \left| \frac{5+\sqrt{11}i + 15 -3\sqrt{11}i}{2} \right|^2 \\ &= \left| \frac{20-2\sqrt{11}i}{2} \right|^2 \\ &= \left| 10 - \sqrt{11}i \right|^2 \\ &= \left( \sqrt{10^2 + \sqrt{11}^2} \right)^2 \\ &= \sqrt{111}^2 \\ &= 111 _\square \end{aligned}

### What is the absolute value of $2 - 10a$?

The answer depends on the value of $a$. From the definition above, we have

$\left|2-10a \right| = \begin{cases} 2-10a, & \text{if } 2-10a \geq 0 \\ -(2-10a), & \text{if } 2-10a < 0 \end{cases}$

Solving the inequality on the right, we obtain

\begin{aligned} 2-10a &\geq 0 \\ 2 & \geq 10a \\ \frac{1}{5} &\geq a. \end{aligned}

It follows that $2-10a \geq 0 \Longleftrightarrow a \leq \frac{1}{5}.$

Therefore,

$\left|2-10a \right| = \begin{cases} 2-10a, & \text{if } a \leq \frac{1}{5} \\ 10a - 2, & \text{if } a > \frac{1}{5} \end{cases} _\square$

Note by Arron Kau
7 years ago

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thank you very much :-)

- 6 years, 11 months ago

I like this format of note, I think I'll Start Share So !

Explanation More Exercises Solved !

- 7 years ago

I didn't quite understand the problem having $i$. At one stage the $i$ disappears. Can you please explain?

- 7 years ago

It is because the absolute value of a complex number ($a + bi$) is defined as $\sqrt{a^2 + b^2}$ , i.e., $|a + bi|$ = $\sqrt{a^2 + b^2}$. In the above question, $|10 - \sqrt11|$ = $\sqrt{10^2 + \sqrt{11}^2}$

- 7 years ago

Didn't know that, thanks!!

- 7 years ago

I am not sure but as far as my mind think I can say this is because when the sign of absolute value was removed.Then sign was changed to positive & i(iota) has nothing to do when there is is postive root.So,It would have been removed...

- 7 years ago

$i$ is defined as $\sqrt{-1}$ i.e. imaginary unit. So, those were complex numbers (real number + imaginary numbers).

Excellent...

- 7 years ago

In ending step ..note that to take modulus of imag eniry number is different than simple modulas

- 6 years, 5 months ago


# A randomized definition of the natural logarithm constant

The number $e$ is the base of the natural logarithm. It is an important constant in mathematics, which is approximately 2.718281828. This post discusses a charming little problem involving the the number $e$. This number can be represented in many ways. In a calculus course, the number $e$ may be defined as the upper limit of the following integral:

$\displaystyle \int_1^e \frac{1}{t} \ dt=1$

Another representation is that it is the sum $\displaystyle e=\sum_{n=1}^\infty \frac{1}{n!}$. Still another is that it is the limit $\displaystyle e=\lim_{n \rightarrow \infty} \biggl( 1+\frac{1}{n} \biggr)^n$. According to the authors of a brief article from the Mathematics Magazine [1], these textbook definitions do not give immediate insight about the number $e$ (article can be found here). As a result, students come away from the course without a solid understanding of the number $e$ and may have to resort to rote memorization on what the number $e$ is. Instead, the article gives six probability oriented ways in which the number $e$ can occur. These occurrences of $e$ are more interesting and, according to the authors of [1], can potentially increase students’ appreciation of the number $e$. In two of these six examples (Example 2 and Example 5), the number $e$ is defined by drawing random numbers from the interval $(0,1)$. This post discusses Example 2.

________________________________________________________________________

Random Experiment

Here’s the description of the random experiment that generates the number $e$. Randomly and successively select numbers from the interval $(0, 1)$. The experiment terminates when the sum of the random numbers exceeds 1. What is the average number of selections in this experiment? In other words, we are interested in the average length of the experiment. According to the article [1], the average length is the number $e$. The goal here is to give a proof of this result.

As illustration, the experiment is carried out 1 million times using random numbers that are generated in Excel using the Rand() function. The following table summarizes the results.

$\left[\begin{array}{rrrrrr} \text{Length of} & \text{ } & \text{ } & \text{ } & \text{Relative} \\ \text{Experiment} & \text{ } & \text{Frequency} & \text{ } & \text{Frequency} \\ \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \\ 2 & \text{ } & 500777 & \text{ } & 0.500777 \\ 3 & \text{ } & 332736 & \text{ } & 0.332736 \\ 4 & \text{ } & 124875 & \text{ } & 0.124875 \\ 5 & \text{ } & 33465 & \text{ } & 0.033465 \\ 6 & \text{ } & 6827 & \text{ } & 0.006827 \\ 7 & \text{ } & 1130 & \text{ } & 0.001130 \\ 8 & \text{ } & 172 & \text{ } & 0.000172 \\ 9 & \text{ } & 14 & \text{ } & 0.000014 \\ 10 & \text{ } & 4 & \text{ } & 0.000004 \\ \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \\ \text{Total} & \text{ } & 1000000 & \text{ } & \text{ } \end{array}\right]$

The average length of experiment in these 1 million experiments is 2.717001. Even though the rate of convergence to the number $e$ is fairly slow, the simulated data demonstrates that on average it takes approximately $e$ number of simulations of numbers in the unit interval to get a sum that exceeds 1.

________________________________________________________________________

A proof

Let $U_1,U_2,U_3,\cdots$ be a sequence of independent and identically distributed random variables such that the common distribution is a uniform distribution on the unit interval $(0,1)$. Let $N$ be defined as follows:

$\displaystyle N=\text{min}\left\{n: U_1+U_2+\cdots+U_n>1 \right\} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$

The objective is to determine $E(N)$. On the surface, it seems that we need to describe the distribution of the independent sum $X_n=U_1+U_2+\cdots+U_n$ for all possible $n$. Doing this may be possible but the result would be messy. It turns out that we do not need to do so. We need to evaluate the probability $P(X_n \le 1)$ for all $n$. We show that

$\displaystyle F_n(x)=P(X_n \le x)=\frac{x^n}{n!} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$

for $0 \le x \le 1$ and for $n=1,2,3,\cdots$. This is accomplished by an induction proof. This is true for $n=1$ since $X_1=U_1$ is a uniform distribution. Suppose (2) holds for the integer $n-1$. This means that $\displaystyle F_{n-1}(x)=P(X_{n-1} \le x)=\frac{x^{n-1}}{(n-1)!}$ for $0 \le x \le 1$. Note that $X_n=X_{n-1}+U_n$, which is an independent sum. Let’s write out the convolution formula for this independent sum:

\displaystyle \begin{aligned} F_n(x)&=\int_{0}^x F_{n-1}(x-y) \cdot f_{U_n}(y) \ dy \\&=\int_{0}^x \frac{(x-y)^{n-1}}{(n-1)!} \cdot 1 \ dy \\&=\frac{1}{(n-1)!} \int_0^x (x-y)^{n-1} \ dy \\&=\frac{x^n}{n!} \end{aligned}

The above derivation completes the proof of the claim. We now come back to the problem of evaluating the mean of the random variable $N$ defined in (1). First, note that $N>n$ if and only if $X_n=U_1+U_2+\cdots+U_n \le 1$. So we have the probability statement $\displaystyle P(N>n)=F_n(1)=\frac{1}{n!}$.

As a result, the following is the probability function of the random variable $N$.

\displaystyle \begin{aligned} P(N=n)&=P(N>n-1)-P(N>n) \\&=F_{n-1}(1)-F_n(1) \\&=\frac{1}{(n-1)!}-\frac{1}{n!} \\&=\frac{n-1}{n!} \end{aligned}

Now evaluate the mean.

\displaystyle \begin{aligned} E(N)&=\sum \limits_{n=2}^\infty \frac{n(n-1)}{n!} \\&=\sum \limits_{n=2}^\infty \frac{1}{(n-2)!} \\&=\sum \limits_{m=0}^\infty \frac{1}{m!} \\&=e \end{aligned}

With the above derivation, the proof that $e=$ 2.718281828… is the average number of random numbers to select in order to obtain a sum that exceeds 1 is completed.

________________________________________________________________________

Reference

1. Shultz H. S., Leonard B., Unexpected Occurrences of the Number e,Mathematics Magazine, October 1989, Volume 62, Number 4, pp. 269–271.

________________________________________________________________________
$\copyright \ \text{2015 by Dan Ma}$


# Can the product of infinitely many elements from $\mathbb Q$ be irrational?

I know there are infinite sums of rational values, which are irrational (for example the Basel Problem). But I was wondering, whether the product of infinitely many rational numbers can be irrational. Thank you for your answers.

• Do you know Wallis's product for $\pi$? Mar 28, 2018 at 13:23
• $e=\lim_{n\to \infty}{(1+1/n)^n}$ Mar 28, 2018 at 13:29
• @Vasya Correct limit of rationals, but not an infinite product. Mar 28, 2018 at 13:29
• @DzamoNorton Not a limit of partial products. The number of factors increases each time, but the factors change: $(1 + 1/2)(1+1/2)$, $(1+1/3)(1 + 1/3)(1+ 1/3)$ and so on. Mar 28, 2018 at 19:51
• @toliveira: An "infinite product" is not multiplication. It's a limit. Fundamentally, multiplication has two operands; you can inductively extend that to any finite number of operands. But "any finite number" does not include "infinitely many". Mar 29, 2018 at 15:10

Yes, it can.

Consider any sequence $(a_n)$ of non-zero rational numbers which converges to an irrational number. Then define the sequence $b_n$ by $b_1 = a_1$ and $$b_n = \frac{a_n}{a_{n-1}}$$ for $n > 1$.

We then have that $$b_1 b_2 \cdots b_n = a_1 \frac{a_2}{a_1} \frac{a_3}{a_2} \cdots \frac{a_n}{a_{n-1}} = a_n.$$

We thus see that every term of $(b_n)$ is rational, and that the product of the terms of $(b_n)$ is the same as the limit of $a_n$, which is irrational.

• Great answer. Very simple. Mar 29, 2018 at 9:57
• But it does look like a "cheat" since everything is divided out by itself. So, is there an infinite product which does not contain a collection of $\frac{a_n}{a_n}$ ? both your answer and mohammad Riazi's have this "cheat", while Kumar's does not. Mar 29, 2018 at 18:56
• @Carl This isn't a cheat. In fact you can rewrite any infinite product in this way. Mar 29, 2018 at 19:23
• +1 But I think it needs at least a short explanation (or link to one, maybe the question is already out here too?) that such a sequence "of non-zero rational numbers which converges to an irrational number" exists. It is easier too see than the initial question (I think) but I guess some of those who will fins the initial question interesting may not see this as obvious Mar 29, 2018 at 20:08

Yes, every irrational number is an infinite product of rationals.

We can write an infinite sum of rationals as an infinite product of rationals.

\begin{align} a&=a,\\ a+b&=a\times\frac {a+b}{a}\\ a+b+c &= a \times \frac {a+b}{a}\times\frac {a+b+c}{a+b}\\.\\.\\.\\.\end{align}

For example, $$\sqrt 2 =1.414213....=1+.4+.01+.004+.....=$$

$$1\times \frac {1.4}{1}\times \frac {1.41}{1.4}\times\frac {1.414}{1.41}\times .....$$

• Your answer is chosen for a review audit of mine, which I upvoted (+1) because it is simple, generalized and easy to understand. Mar 29, 2018 at 5:10
• @TrầnThúcMinhTrí Thanks for your attention. Mar 29, 2018 at 11:19
• I took the liberty to align equalities at '=' sign. Feel free to rollback if you don't like it. Mar 30, 2018 at 7:29
• Thanks, I like it Mar 30, 2018 at 11:12

Yes!

$\cfrac{\pi}{2} = \cfrac{2}1 \cfrac 23 \cfrac 43 \cfrac 45 \cfrac 65 \cfrac 67 \cdots$

• This looks cool, but what is the geometric picture to go along with it? I an almost see polygons in there... Mar 28, 2018 at 16:40
• @MikeWise I din't understand which polygon you are talking about. Mar 28, 2018 at 16:42
• General term is $\cfrac{2n}{(2n – 1)}\cfrac{2n}{(2n + 1)}$ Mar 28, 2018 at 16:43
• @MikeWise: I was curious about that as well. Here's a short note on how when you draw rectangles of those areas, they add up to a quarter circle in the limit: math.chalmers.se/~wastlund/monthly.pdf Mar 28, 2018 at 17:39
• Cool, thanks for that, will print it out and read it on my flight in the morning. :) Mar 28, 2018 at 17:47

Too big to be a comment: it should be noted that the order is more crucial in infinite products than in infinite sums, which is strikingly seen on the example cited many times already:

\begin{align*}\cfrac{\pi}{2}&=\cfrac{2}1 \cfrac 23 \cdot \cfrac 43 \cfrac 45 \cdot\cfrac 65 \cfrac 67\cdot \ldots\\ &= \cfrac{2^2}{2^2-1}\cdot \cfrac{4^2}{4^2-1}\cdot \cfrac{6^2}{6^2-1}\ldots\\ \end{align*} is an infinite product with partials starting at $\frac43$ and increasing towards $\frac\pi 2$ (every factor is greater than $1$), whereas the seemingly identical

\begin{align*}0&=\cfrac{2}3 \cfrac 23 \cdot \cfrac 45\cfrac 45\cdot\cfrac 67 \cfrac 67 \cdot\ldots\end{align*}

starts below $1$ and decreases, towards $0$. All that happened was a shift of denominators one step to the left.

• Products are isomorphic to sum of logs, so any order effect that shows up in one can be generated in the other. By allowing the denominators to be shifted independently of the numerators, you are treating the fraction (a/b) as ab^-1, so when you take the log, it's log(a)-log(b), which is an alternating series, and of course order is more important in alternating series than monotonic ones. It's the alternation generated by treating the numerator and denominator separately, not the product, that makes order important. Mar 28, 2018 at 15:00
• Allowing the denominators to be shifted independently of the numerators can also be interpreted as replacing each term a/b by two terms, a and 1/b. Thus when you take logs, you get two terms, log a and - log b. Mar 28, 2018 at 22:10
• These explanations are of course correct, but the point is that it is not obvious to realise that it can matter, easy to make a mistake here. Mar 29, 2018 at 8:32
• I think the claim is not that one shouldn't point out that order matters in an infinite product, but rather that one shouldn't claim that it matters more in infinite products than in infinite sums, since (as @Acccumulation points out) for products of positive real numbers and sums of real numbers it is literally the same phenomenon. Mar 29, 2018 at 22:38
• @LSpice when I said that it matters more, what I meant is that one should beware of the intuition that a permutation of terms acting only locally does not affect the final result (you need a violent perturbation of terms to give $\sum{(-1)^n\over n}$ a different limit). Of course here what is happening is that the terms are not only reordered, but broken to pieces and the pieces reordered. *This natural way to think of a rational factor as being made of two pieces does not happen with series. * Mar 29, 2018 at 23:44

Consider the Riemann-Zeta Function: $$\sum_{n=1}^{\infty}\frac{1}{n^{s}}=\prod_{p\text{ prime}}\frac{1}{1-p^{-s}}.$$ For $s=2$, the infinite sum on the left is $\pi^{2}/6$, which is irrational. Thus, $\pi^{2}/6$ is an infinite product of rationals.

There is a simple way to obtain any irrational number as an infinite product:

• take any sequence $s_n$ of rational numbers converging to the targeted irrational one (say the approximations of $\pi$ to $n$ decimals);

• form the product of the numbers $f_n:=\dfrac{s_{n+1}}{s_n}$, with $f_0=1$.

$$\pi=\prod_{n=0}^\infty f_n=\frac{31}{10}\cdot\frac{314}{310}\cdot\frac{3141}{3140}\cdot\frac{31415}{31410}\cdot\frac{314159}{314150}\cdots$$

• Ooops, I just noticed that my answer is a duplicate.
– user65203
Mar 30, 2018 at 15:29


# A gravitation problem with two masses

1. Apr 11, 2014

### toothpaste666

1. The problem statement, all variables and given/known data

A uniform sphere has mass M and radius r. A spherical cavity (no mass) of radius r/2 is then carved within this sphere (the cavity's surface passes through the sphere's center and just touches the sphere's outer surface). The centers of the original sphere and the cavity lie on a straight line, which defines the x axis.

With what gravitational force will the hollowed-out sphere attract a point mass m which lies on the x axis a distance d from the sphere's center? [Hint: Subtract the effect of the "small" sphere (the cavity) from that of the larger entire sphere.]

2. Relevant equations

$F = G\frac{m_1m_2}{r^2}$

3. The attempt at a solution

Let mass $M_f$ = the mass of the sphere without the cavity, $M_c$ = the mass of the cavity, and $R_f$ = the distance between the two masses.

$M_f = M - M_c$
$R_f = d + R_2$
$F = G\frac{(M_f) m}{(R_f)^2}$

is this right?

2. Apr 11, 2014

### SammyS

Staff Emeritus
Did you intend for MC to be negative?

I interpret the Hint to be to calculate the gravitational force that would be exerted on the point mass if the sphere were not hollow, but had the same density. From that, subtract the gravitational force that would be exerted on the point mass by an isolated uniform sphere the size of the cavity, also having the same density.

That would give MF = M + MC .

When you say, "Rf = the distance between the two masses", what two masses are you referring to ?

3. Apr 12, 2014

### toothpaste666

Here is the picture. The two masses are the small particle mass on the right and the sphere on the left with the cavity

#### Attached Files:

• ###### physicsprob.png
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4. Apr 12, 2014

### haruspex

The hint does not say subtract the masses, it says subtract the gravitational effects.
What force would be exerted by the complete sphere?
Suppose we take the "complement" of the cavitated sphere, i.e. throw away the larger sphere and instead have just the smaller sphere (at the position of the cavity). What force would that exert?

5. Apr 13, 2014

### toothpaste666

so if M is the mass of the sphere before the cavity was taken out and Mc is the mass of the cavity then we have:

$G\frac{Mm}{d^2} - G\frac{M_c m}{(d-r_2)^2}$

6. Apr 13, 2014

### SammyS

Staff Emeritus
Yes, if M is the mass of the sphere before the cavity is taken out. That is as the problem is stated. (So, I believe I was incorrect in my previous post.)

You can determine Mc in terms of M.

7. Apr 13, 2014

### toothpaste666

If the mass is evenly distributed wouldn't subtracting the volume be the same as subtracting the mass?

8. Apr 13, 2014

### SammyS

Staff Emeritus
What are you subtracting from what?

What is the mass of a sphere the size of the cavity? How is that related to the mass of the large sphere before the cavity is formed in the sphere?

9. Apr 13, 2014

### toothpaste666

Subtracting the volume of the cavity from the volume of the original sphere i mean. Wouldn't the mass be the same as the volume if its evenly distributed?

10. Apr 13, 2014

### SammyS

Staff Emeritus
Mass is be proportional to volume in this case.

What fraction of the initial solid sphere's mass is removed to make the cavity ?

11. Apr 13, 2014

### toothpaste666

To find the volumes let Vc = Volume of cavity:

$r_2 = \frac{r}{2}$

$V_c = \frac{4pi(\frac{r}{2})^3}{3}$

$V_c = \frac{\frac{4pi(r^3)}{8}}{3}$

$V_c = \frac{\frac{pi(r^3)}{2}}{3}$

$V_c = \frac{pi(r^3)}{6}$

since $V_M =\frac{4pi(r^3)}{3}$

we have

$V_M = 8V_c$

does this get me closer to expressing Mc in terms of M?

12. Apr 14, 2014

### haruspex

That's it.

13. Apr 14, 2014

### toothpaste666

but how do i relate that back to mass?

14. Apr 14, 2014

### haruspex

Through density, which is the same for both.

15. Apr 14, 2014

### toothpaste666

$D = \frac{M}{V_M} = \frac{3M}{4(pi)r^3}$

$M_c = D V_c = D \frac{(pi)r^3}{6} = (\frac{3M}{4(pi)r^3})(\frac{(pi)r^3}{6}) = \frac{M}{8}$

which gives

$G\frac{Mm}{d^2} - G\frac{Mm}{8(d-r_2)^2}$

how do we know the density is equal?

16. Apr 14, 2014

### haruspex

See the OP:

17. Apr 14, 2014

Thank you.


# Limits and substitution

As in @RobertZ's answer to this question, we often perform substitutions when evaluating limits. For instance, if you're asked to show that $$L = \lim_{t \to 0} \frac{\sin t^3}{t^3} = 1,$$ it's pretty common to say "Let $x = t^3$; then as $t \to 0$, we have $x \to 0$, so $$L = \lim_{x \to 0} \frac{\sin x}{x}$$ which we know is $1$, and we're done."

What's going on here in general is an application of the following "Theorem"

Theorem 1: If the function $g$ satisfies [fill in missing properties] and $$\lim_{t \to a} g(t) = b,$$ then $$\lim_{t \to a} f(g(t)) = \lim_{x \to b} f(x),$$ i.e., one limit exists if and only if the other does, and if they both exist, they're equal.

In the example above, $f(x) = \frac{\sin x}{x}$ and $g(t) = t^3$ and $a = b = 0$.

There's an alternative form, in which we're asked to show that $$L = \lim_{t \to 0} \frac{\sin \sqrt[3]{t}}{\sqrt[3]{t}} = 1,$$ it's pretty common to say "Let $t = x^3$; then as $t \to 0$, we have $x \to 0$, so $$L = \lim_{x \to 0} \frac{\sin x}{x}$$ which we know is $1$, and we're done."

In this case, the implicit theorem is very similar to the other, but with the role of $g$ reversed (i.e., we're substituting $t = x^3$ instead of $x = t^3$, so the natural form of the theorem puts $g$ on the other side):

Theorem 2: If the function $g$ satisfies [fill in missing properties] and $$\lim_{x \to b} g(x) = a,$$ then $$\lim_{t \to a} f(t) = \lim_{x \to b} f(g(x)).$$

In the second example above, we have $a = b = 0$, $f(t) = \frac{\sin \sqrt[3]{t}}{\sqrt[3]{t}}$, and $g(x) = x^3$.

The two theorems are obviously the same: if you swap $a$ and $b$, $x$ and $t$, and reverse the equality in the last line, they're identical. But each represents a different approach to working with limits, so I've stated both.

In the second form, it's clearly important that $g$ be surjective near $a$ (i.e., for every small enough interval $I = (b-\epsilon, b + \epsilon)$, there's an interval $I' = (a-\delta, a + \delta)$ such that $I- \{b\} \subset g(I' - \{a\})$. (Hat-tip to MathematicsStudent1122 for the observation that I need to delete $a$ and $b$ from those intervals). Otherwise you could use things substitutions like $s = t^2$, which would turn a two-sided limit into a one-sided one (or vice versa), in which case one limit might exists and the other might not.

Addendum to clarify why this might matter, for @MathematicsStudent1122:

Consider $$f(x) = \begin{cases} 1 & x \ge 0 \\ 0 & x < 0 \end{cases}.$$

and look at $L = \lim_{x \to 0} f(x^2)$. It's clear that $L$ exists and is $1$. But if we substitute $t = x^2$, then we get $L = \lim_{t \to 0} f(t)$, which does not exist; hence this "substitution" is not valid: I've turned what amounts to a 1-sided limit (which exists) into a two-sided limit (which does not exist). The domains of $f$ and $g$ are both all of $\Bbb R$.

My question is this:

What is a reasonable set of missing properties for each of these theorems? (I can work out the exact properties easily enough by running through the definitions, but they don't seem to be very helpful/checkable.)

One answer might be "$g$ is locally a bijection", but that rules out things like $y = x + x\sin \frac{1}{x}$ near $x = 0$, so it seems too limited. (It also rules out things like $x \mapsto x + \sin x$ for limits as $x \to \infty$, which is a pity.)

I recognize that this is not a strictly mathematical question. But my goal is to come up with a "calculus student's theorem", one that says "if you're trying to work out a limit, which may or may not exist, then it's OK to do substitutions of this sort along the way," and which will cover the vast majority of the problems that they might encounter in a standard calculus book, or even in Spivak's book.

This question gives two theorems, but both have assumptions about the existence of limits. This one comes a little closer, but still isn't entirely satisfactory.

I'd love any nice-enough condition to be broadly useful. In particular, I think it's completely reasonable to require, for instance, that the "substitution function" $g$ be continuous, and perhaps even differentiable (although I doubt that's of much use).

• – Jack D'Aurizio Aug 20 '17 at 15:03
• Great point, @JackD'Aurizio. That may be the first place where this particular stunt bothered me. :) – John Hughes Aug 20 '17 at 15:04
• So that's part of the quest for the Holy Grail of all lazy students, the Grand Unified Formula Of All Textbook Exercises? ;-) – Professor Vector Aug 20 '17 at 15:06
• Substitutions like $t=x\sin \frac{1}{x}$ simply don't always work. If we let $g(x) = x\sin \frac{1}{x}$ and $f(x) = 1$ for $x=0$ and $f(x) = 0$ for $x \neq 0$, then we can note that $$\lim_{x \to 0} f(g(x))$$ does not exist, but $$\lim_{y \to 0} f(y)$$ does exist. – MathematicsStudent1122 Aug 20 '17 at 15:56
• @ParamanandSingh: Well...I've mused about it on miscellaneous occasions for about 45 years, but I thank you for pointing out that I've bene wasting my time. :) Your answer in that linked case is the easy one --- the one that only goes one direction. I want to know that if I make a substitution and the origiinal limit does NOT exist, then the new limit is also guaranteed to not exist. – John Hughes Aug 20 '17 at 18:07

I'm afraid this isn't quite what you're after, but it seems to be a decent starting point. These conditions, while restrictive, appear to be necessary for the following proof.

Assume $\lim_{x\to a}g(x)=b$, where the function $g$ also satisfies: $$\text{there exists a neighborhood U of a such that b\not\in g(U\setminus\{a\})} \tag{1}$$ and for each $(y_n)$ converging to $b$ with $y_n\ne b$, $$\text{there exists (x_n) such that g(x_n)=y_n for large n and x_n\to a}. \tag{2}$$

Then we show that $$\lim_{x\to a}f(g(a)) = \lim_{y\to b}f(y).$$

Proof:

Let $L_1:=\lim_{x\to a}f(g(a))$ and $L_2:=\lim_{y\to b}f(y)$, either of which may or may not exist. We will use the sequential criterion.

First suppose $L_2$ exists and let $(x_n)$ be a sequence such that $x_n\to a$ and $x_n\ne a$ for all $n$. Since $\lim_{x\to a}g(x)=b$, we have $g(x_n)\to b$. Due to $(1)$, we have $g(x_n)\ne b$ for $n$ large enough. Then, because $L_2$ exists, we have $f(g(x_n))\to L_2$. This proves by the sequential criterion that $L_1$ exists and $L_1=L_2$.

Now assume $L_1$ exists and let $(y_n)$ be a sequence such that $y_n\to b$ and $y_n\ne b$ for all $n$. By $(2)$ there exists a sequence $(x_n)$ such that $g(x_n)=y_n$ for large $n$ and $x_n\to a$. Since $y_n\ne b$ for all $n$, we have $x_n\ne a$ for large $n$. Then $$\lim_{n\to\infty}f(y_n)=\lim_{n\to\infty}f(g(x_n))=L_1,$$ and so the sequential criterion implies $L_2$ exists and $L_2=L_1$.

• You're right --- that's not quite what I'm after. For one thing, it's not a condition that most calc students could test. :( But I agree that it's a decent starting point -- thanks. – John Hughes Aug 21 '17 at 19:58


Prove that if $A$ is normal, then eigenvectors corresponding to distinct eigenvalues are necessarily orthogonal (alternative proof)

The problem statement is as follows:

Prove that for a normal matrix $A$, eigenvectors corresponding to different eigenvalues are necessarily orthogonal.

I can certainly prove that this is the case, using the spectral theorem. The gist of my proof is presented below.

If possible, I would like to find a simpler proof. I was hoping that there might be some sort of manipulation along these lines, noting that $$\langle Av_1,A v_2\rangle = \langle v_1,A^*Av_2\rangle = \langle v_1,AA^*v_2\rangle = \langle A^* v_1,A^* v_2 \rangle$$

Any ideas here would be appreciated.

My proof:

Let $\{v_{\lambda,i}\}$ be an orthonormal basis of eigenvectors (as guaranteed by the spectral theorem) such that $$A v_{\lambda,i} = \lambda v_{\lambda,i}$$ Let $v_1,\lambda_1$ and $v_2,\lambda_2$ be eigenpairs with $\lambda_1 \neq \lambda_2$. We may write $v_1 = \sum_{i,\lambda}a_{i,\lambda}v_{i,\lambda} .$ We then have $$0 = Av_1 - \lambda_1 v_1 = \sum_{i,\lambda}(\lambda - \lambda_1)a_{i,\lambda}v_{i,\lambda}$$ So that $a_{i,\lambda} = 0$ when $\lambda \neq \lambda_1$. Similarly, we may write $v_2 = \sum_{i,\lambda}b_{i,\lambda}v_{i,\lambda}$, and note that $b_{i,\lambda} = 0$ when $\lambda \neq \lambda_2$. From there, we have $$\langle v_1,v_2 \rangle = \sum_{i,\lambda}a_{i,\lambda}b_{i,\lambda}$$ the above must be zero since for each pair $i,\lambda$, either $a_{i,\lambda}=0$ or $b_{i,\lambda} = 0$.

• "Let $v_{\lambda,i}$ be an orthonormal basis of eigenvectors..." I'd guess that this might be something very close to what would need to be proved. – Algebraic Pavel May 3 '14 at 2:13
• @PavelJiranek I was worried that I had used circular logic at some point. However, the existence of such a basis (i.e. the spectral theorem) comes directly from the Schur triangularization theorem, which says nothing about normal matrices in particular. – Omnomnomnom May 3 '14 at 19:11

Assume $\;\lambda\neq \mu\;$ and

$$\begin{cases}Av=\lambda v\;\,\implies\; A^*v=\overline \lambda v\\{}\\Aw=\mu w\implies A^*w=\overline\mu w\end{cases}$$

From this we get:

$$\begin{cases}\langle v,Aw\rangle=\langle v,\mu w\rangle=\overline\mu\langle v,w\rangle\\{}\\ \langle v,Aw\rangle=\langle A^*v,w\rangle=\langle\overline\lambda v,w\rangle=\overline\lambda\langle v,w\rangle \end{cases}$$

and since $\;\overline\mu\neq\overline\lambda\;$ , we get $\;\langle v,w\rangle =0\;$

Question: Where did we use normality in the above?

• How de we know that $v$ is also an eigenvector for $A^*$? – Berci May 2 '14 at 22:55
• @Berci, read and think of the question at the end of the answer. – DonAntonio May 2 '14 at 22:56
• From the other two answers this follows, of course, using that $A$ is normal. But can you show it directly? – Berci May 2 '14 at 22:57
• Why directly? Use the definition and work it out. I usually don't give full answers and, after all, the OP is trying to get a more or less simpler proof to the claim than the one using the spectral theorem... – DonAntonio May 2 '14 at 23:00
• I like it! Showing that $Av=\lambda v\;\,\implies\; A^*v=\overline \lambda v$ seems to be the step in the spirit of what I was looking for. Thanks. – Omnomnomnom May 2 '14 at 23:14

Specializing your identity to $v_1=v_2=v$, we get $\|Av\|=\|A^*v\|$. Hence $\ker A=\ker A^*$. Recalling that $\ker A^* = (\operatorname{ran} A)^\perp$ for general $A$, we conclude that the kernel and range of a normal matrix are mutually orthogonal.

It remains to apply the above conclusion to $A-\lambda I$ where $\lambda$ is an eigenvalue of $A$.

• Neat! Although DonAntonio's proof is more along the lines I was thinking of, this is a nice perspective. – Omnomnomnom May 2 '14 at 23:16

I try to give another simple proof to $$T^*v=\bar{\lambda}v ~\text{ if }~ Tv=\lambda v$$ where $T$ is a normal operator on a Hilbert space $H$.

Suppose $V=\ker(T-\lambda I)$. Since $T^*$ communicate with $T$, $$T^*V\subset V.$$ Because $$\langle v,T^*v\rangle =\langle Tv,v\rangle =\langle \lambda v,v\rangle=\langle v,\bar{\lambda}v\rangle ~~\forall v \in V,$$ $\langle u,T^*v\rangle =\langle u, \bar{\lambda}v\rangle ~\forall u,v\in V$ by polarisation identity, and thus $T^*v=\bar{\lambda}v.$

REMARK: Let $\sigma:V\times V\to W$ be a sesquilinear form, where $V$ and $W$ are linear vector spaces over $\mathbb{C}$. The follwing formula is called Polarisation Identity : $$\sigma(u,v)=\sum_{k=0}^3 i^k\sigma(u+i^k v, u+i^kv).$$

• I don't really see how this answer addresses the question being asked. Also, the question being asked was answered 3 years ago. – Omnomnomnom Jun 12 '17 at 15:01
• @Omnomnomnom This is why $Av=\lambda v\Rightarrow A^*v=\bar{\lambda}v$ using no spectral theorem which is the point to the question I think. – C.Ding Jun 13 '17 at 2:37
• That is not the point to the question, actually. The point is to show that eigenvectors corresponding to different eigenvalues are necessarily orthogonal. – Omnomnomnom Jun 13 '17 at 2:58
• Oh, sorry to disturb you. As I give some additional remarks to the right answer you have accepted. – C.Ding Jun 13 '17 at 3:04

A normal matrix is unitarily similar to diagonal matrix.

$$A = UDU^{-1}$$ where $U$ is Unitary matrix.

Eigen decompositions tells that $U$ is a matrix composed of columns which are eigenvectors of $A$. And matrix $D$ is Diagonal matrix with eigenvalues on diagonal.

Property: Columns of Unitary matrix are orthogonal.

So, columns of $U$ (which are eigenvectors of $A$) are orthogonal.


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# Within a rectangular courtyard of length 60 feet, a graveled path,

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Within a rectangular courtyard of length 60 feet, a graveled path,  [#permalink]

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24 Sep 2018, 10:25
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Within a rectangular courtyard of length 60 feet, a graveled path, 3 feet wide, is laid down along all the four sides. The cost of graveling the path is Rs 2 per sqft. If the path had been twice as wide, the gravek would have cost Rs 984 more. The width of the courtyard is :

A 24
B 30
C 40
D 45
E 54
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Within a rectangular courtyard of length 60 feet, a graveled path,  [#permalink]

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24 Sep 2018, 11:12
Being x the width of the rectangle, we have the following equation:

$$2[2(60*3)+2*3(x-6)] + 984 = 2[2(60*6)+2*6(x-12)]$$

where $$2(60*3)$$ is the area of the path next to the longer side (60 ft long and 3 ft wide), and $$2*3(x-6)$$ the area of the path next to the shorter side, avoiding to count the area from the corner twice. Then we need to repeat the same idea with the bigger path. Both areas are multiplied by 2 because we are measuring the cost, which is Rs 2 per sq ft. We have to consider that the path with double size is Rs 984 more expensive, so we add this into our equation. Then, we can solve the equation:

$$2[360+6x-36]+984=2(720+12x-144)$$
$$360+6x-36+492=720+12x-144$$
$$324+6x+492=576+12x$$
$$816-576=6x$$
$$240=6x$$
$$x=40$$

Thus, C is the correct answer.
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Re: Within a rectangular courtyard of length 60 feet, a graveled path,  [#permalink]

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24 Sep 2018, 11:33
sultanatehere wrote:
Within a rectangular courtyard of length 60 feet, a graveled path, 3 feet wide, is laid down along all the four sides. The cost of graveling the path is Rs 2 per sqft. If the path had been twice as wide, the gravek would have cost Rs 984 more. The width of the courtyard is :

A 24
B 30
C 40
D 45
E 54

A = Area of outer rectangle = 60 x Width
Ax = Area of inner rectangle with width of 3 = 54 x (Width-6)
Ay = Area of inner rectangle with twice the width, i.e 6 = 48 x (Width - 12)

Area of gravel with normal width = A-Ax
Area of gravel with twice width = A-Ay

Given, cost of gravelling twice width - cost of gravelling normal width = 984.
Given, cost of gravelling = Rs2/sq ft.

2(A-Ay) - 2(A-Ax) = 984.
A-Ay-A+Ax=492.
Ax-Ay=492.

On substituting Ax and Ay from above and solving the above equation, we get width=40.

Hence C.

Cheers!
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Re: Within a rectangular courtyard of length 60 feet, a graveled path,  [#permalink]

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24 Sep 2018, 13:48
Tried to solve using an Ans choices and got lucky with the 1st option
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Re: Within a rectangular courtyard of length 60 feet, a graveled path,  [#permalink]

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27 Sep 2018, 17:20
sultanatehere wrote:
Within a rectangular courtyard of length 60 feet, a graveled path, 3 feet wide, is laid down along all the four sides. The cost of graveling the path is Rs 2 per sqft. If the path had been twice as wide, the gravek would have cost Rs 984 more. The width of the courtyard is :

A 24
B 30
C 40
D 45
E 54

We can let w = the width of the courtyard. Therefore, the courtyard, including the graveled path, has an area of 60w sq ft. Excluding the path, the area of the courtyard would be (60 - 2(3))(w - 2(3)) = 54(w - 6) = 54w - 324 sq ft. So the path alone has an area of 60w - (54w - 324) = 6w + 324 sq ft. Since the cost of graveling the path is Rs 2 per sq ft, the cost of graveling the path is Rs 12w + 648.

If the path had been twice as wide, then the courtyard, including the path, would still have an area of 60w sq ft. However, excluding the path, the area of the courtyard would be (60 - 2(6))(w - 2(6)) = 48(w - 12) = 48w - 576 sq ft. So the path alone would have an area of 60w - (48w - 576) = 12w + 576 sq ft, and the cost of the path would be Rs 24w + 1152. We are told that this would cost Rs 984 more, so we can set up the following equation to solve for w:

12w + 648 + 984 = 24w + 1152

12w + 1632 = 24w + 1152

480 = 12w

40 = w

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Re: Within a rectangular courtyard of length 60 feet, a graveled path, &nbs [#permalink] 27 Sep 2018, 17:20
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# 5 drill machines can drill 15 meters in 5 hours.

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5 drill machines can drill 15 meters in 5 hours.  [#permalink]

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06 Nov 2017, 07:17
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5 drill machines can drill 15 meters in 5 hours. Assuming the speed of each machine to be the same, how many machines would be needed to drill 60 meters in 2 hours?

A)8
B)16
C)23
D)50
E)60
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Re: 5 drill machines can drill 15 meters in 5 hours.  [#permalink]

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06 Nov 2017, 07:27
Mac 5 meters 15 hours 5
Mac x meters 60 hours 2
Do it with proportion.
X/5= 60/15 * 5/2
X=50

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Re: 5 drill machines can drill 15 meters in 5 hours.  [#permalink]

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06 Nov 2017, 07:32
ManishKM1 wrote:
5 drill machines can drill 15 meters in 5 hours. Assuming the speed of each machine to be the same, how many machines would be needed to drill 60 meters in 2 hours?

A)8
B)16
C)23
D)50
E)60

to drill 15 mtr in 5 hrs, 5 machines are required..
so to drill 15*4 mtr in 5 hrs, 5*4 machines are required..
to drill 60 mtr in 1 hrs, 20*5 machines are required..
finally to drill 60 mtr in 2 hrs, 100/2 or 50 machines are required..

D
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5 drill machines can drill 15 meters in 5 hours.  [#permalink]

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06 Nov 2017, 09:17
1
ManishKM1 wrote:
5 drill machines can drill 15 meters in 5 hours. Assuming the speed of each machine to be the same, how many machines would be needed to drill 60 meters in 2 hours?

A)8
B)16
C)23
D)50
E)60

Find a machine's individual rate from the first information, and use it to solve the question.

W = (# of machines) * (indiv.) rate * time
W = 15, # of machines = 5, rate = ??, time = 5

15 = 5 * rate * 5
15 = 25 * rate

Individual rate = $$\frac{15}{25}=\frac{3}{5}$$ in meters/hour

At that rate, how many machines are needed to drill 60 meters in 2 hours?

W = (# of machines) * (indiv.) rate * time
W = 60, # of machines = ??, rate = $$\frac{3}{5}$$, time = 2

60 = (# of machines) * $$\frac{3}{5}$$ * 2
60 = (# of machines) * $$\frac{6}{5}$$

# of machines = $$\frac{60}{(\frac{6}{5})}= 60 * \frac{5}{6}=$$ 50

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Re: 5 drill machines can drill 15 meters in 5 hours.  [#permalink]

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12 Nov 2017, 08:47
1
1
ManishKM1 wrote:
5 drill machines can drill 15 meters in 5 hours. Assuming the speed of each machine to be the same, how many machines would be needed to drill 60 meters in 2 hours?

A)8
B)16
C)23
D)50
E)60

The rate for 5 drill machines is 15/5 = 3 meters per hour. We need to determine how many drill machines we need for a rate of 60/2 = 30 meters per hour. Since 30 is 10 times 3, we would need 5 x 10 = 50 drill machines.

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Re: 5 drill machines can drill 15 meters in 5 hours. &nbs [#permalink] 12 Nov 2017, 08:47
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# What is the average (arithmetic mean) of eleven consecutive

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What is the average (arithmetic mean) of eleven consecutive [#permalink]

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23 Apr 2010, 18:05
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What is the average (arithmetic mean) of eleven consecutive integers?

(1) The average of the first nine integers is 7
(2) The average of the last nine integers is 9
[Reveal] Spoiler: OA
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Re: Average ( arithmetic mean ) of eleven consecutive integers? [#permalink]

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23 Apr 2010, 18:11
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Some how i got E

1. (63+ X10 + X11) / 11 = ?? ..sitill missing two numbers? so insuff??
2. (X1+X2 +81)/ 11 = ?? Still missing two numbers so Insuff???

and combining 1 and 2 still missing some info ? .. Sorry friends, i need help here.
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Re: Average ( arithmetic mean ) of eleven consecutive integers? [#permalink]

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24 Apr 2010, 06:10
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zz0vlb wrote:
What is the average ( arithmetic mean ) of eleven consecutive integers?

1.The avg of first nine integers is 7
2. The avg of the last nine integers is 9
[Reveal] Spoiler:
D

friends, please explain this to me.

Consecutive integers represent evenly spaced set. For every evenly spaced set mean=median, in our case $$mean=median=x_6$$.

(1) $$x_1+x_2+...+x_9=63$$ --> there can be only one set of 9 consecutive integers to total 63. Sufficient.

If you want to calculate: $$(x_6-5)+(x_6-4)+(x_6-3)+(x_6-2)+(x_6-1)+x_6+(x_6+1)+(x_6+2)+(x_6+3)=63$$ --> $$x_6=8$$.

OR: Mean(=median of first 9 terms=5th term)*# of terms=63 --> $$x_5*9=63$$ --> $$x_5=7$$ --> $$x_6=7+1=8$$

(2) $$x_3+x_4+...+x_{11}=81$$ --> there can be only one set of 9 consecutive integers to total 81. Sufficient.

If you want to calculate: $$(x_6-3)+(x_6-2)+(x_6-1)+x_6+(x_6+1)+(x_6+2)+(x_6+3)+(x_6+4)+(x_6+5)=81$$ --> $$x_6=8$$.

OR: Mean(=median of last 9 terms=7th term)*# of terms=81 --> $$x_7*9=81$$ --> $$x_7=9$$ --> $$x_6=9-1=8$$

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Re: Average ( arithmetic mean ) of eleven consecutive integers? [#permalink]

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24 Apr 2010, 07:11
Thanks for clarifying this Bunuel. +1Kudos to you.
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Re: What is the average (arithmetic mean ) of eleven consecutive [#permalink]

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12 Aug 2013, 04:00
Bumping for review and further discussion.
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Re: What is the average (arithmetic mean ) of eleven consecutive [#permalink]

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12 Aug 2013, 04:39
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zz0vlb wrote:
What is the average (arithmetic mean ) of eleven consecutive integers?

(1) The avg of first nine integers is 7
(2) The avg of the last nine integers is 9

Here is a neat little trick for such kind of problems:

Will be better illustrated using a numerical example: take the set {2,3,4,5,6}. Here the common difference (d)=1. The initial average = 4. Now, the averge of the set, after removing the last integer of the set(i.e. 6)will be reduced by exactly $$\frac{d}{2} units \to$$ The new Average = $$4-\frac{1}{2} = 3.5$$

Again, for the new set of {2,3,4,5} the average is 3.5 . Now, if the last integer is removed, the new average will again be = 3.5-0.5 = 3.

Similarly, for the same set {2,3,4,5,6}, if we remove the first integer from the given set, the average increases by 0.5 and so on and so forth.

Back to the problem:

From F.S 1, we know that the average of the first 9 integers is 7. Thus, the average with the original 11 integers must have been 7+0.5+0.5 = 8. Sufficient.

From F.S 2, we know that the average of the last 9 integers is 9, thus the average with the initial 11 integers must have been 9-0.5-0.5 = 8. Sufficient.

D.
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Re: What is the average (arithmetic mean ) of eleven consecutive [#permalink]

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12 Aug 2013, 23:15
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zz0vlb wrote:
What is the average (arithmetic mean ) of eleven consecutive integers?

(1) The avg of first nine integers is 7
(2) The avg of the last nine integers is 9

As a general rule whenever there is a AP the average of the series is always the median of the series. Here it is a AP with difference 1

1. First 9 integers average is 7 . So the median that is the 5th digit is 7. Hence we can easily find the series and the average of the 11 consecutive digit series. Sufficient
2. Average of last 9 integers is 9 hence we know that for this subset of 9 integers the 5th integer would be 9 and we can find the series on the basis of this and the average. Sufficient

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Re: What is the average (arithmetic mean) of eleven consecutive [#permalink]

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28 Aug 2014, 09:43
Hello from the GMAT Club BumpBot!

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Re: What is the average (arithmetic mean) of eleven consecutive [#permalink]

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09 Mar 2015, 19:15
I considered following approach

if the smallest number in set is x , then sum of 11 consecutive numbers = 11x+(1+2+...10)=11x+55--->A
if largest number in set is x ,then sum of 11 consecutive numbers=11x-(1+2+10)=11x-55

Now as per statement 1 , average of first 9 numbers is 7 i.e sum =63
sum of 11 numbers =63+x+9+x+10----->B

Equating A& B
11X+55=63+X+9+10 ,which can be solved to get x=3

statement I is sufficient
similar approach for Statement II
11X-55=8+2X-19 ,can be solved to get X=13

statement 2 is sufficient

OA=D
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Re: What is the average (arithmetic mean) of eleven consecutive [#permalink]

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09 Mar 2015, 19:26
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Hi All,

When you look at this question, if you find yourself unsure of where to "start", it might help to break down everything that you know into small pieces:

1st: We're told that we have 11 consecutive integers. That means the 11 numbers are whole numbers that are in a row. If we can figure out ANY of the numbers AND it's place "in line", then we can figure out ALL of the other numbers and answer the question that's asked (the average of all 11 = ?)

2nd: Fact 1 tells us that the average of the FIRST 9 integers is 7. For just a moment, ignore the fact that there are 9 consecutive integers and let's just focus on the average = 7.

What would have to happen for a group of consecutive integers to have an average of 7?

Here are some examples:

7

6, 7, 8

5, 6, 7, 8, 9

Notice how there are the SAME number of terms below 7 as above 7. THAT'S a pattern.

With 9 total terms, that means there has to be 4 above and 4 below:

3, 4, 5, 6,.......7.......8, 9, 10, 11

Now we have enough information to figure out the other 2 terms (12 and 13) and answer the question. So Fact 1 is SUFFICIENT

With this same approach, we can deal with Fact 2.

The key to tackling most GMAT questions is to be comfortable breaking the prompt into logical pieces. Don't try to do every step at once and don't try to do work in your head. Think about what the information means, take the proper notes and be prepared to "play around" with a question if you're immediately certain about how to handle it.

[Reveal] Spoiler:
D

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Re: What is the average (arithmetic mean) of eleven consecutive [#permalink]

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10 Mar 2016, 03:12
Hello from the GMAT Club BumpBot!

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Re: What is the average (arithmetic mean) of eleven consecutive [#permalink]

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29 Jun 2016, 03:10
wow such complex explanations for such a simple problem?

given :

11 consec integers

let them be x,x+1,x+2,...,x+10

Q: what is their mean?

mean is (11x+55)/11 = x+5.
Q becomes what is x+5

1) mean first 9 is 7.

so (9x+36)/9 = x+4 = 7 , so x+5 =8 ,--> sufficient A or D

2) mean of last 9 is 9.

so (9x+54)/9 = x+6= ---> x+5=8, sufficient . so D

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Re: What is the average (arithmetic mean) of eleven consecutive [#permalink]

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15 Jul 2016, 00:10
zz0vlb wrote:
What is the average (arithmetic mean) of eleven consecutive integers?

(1) The average of the first nine integers is 7
(2) The average of the last nine integers is 9

For odd number of consecutive integer the mean and median both is the middle value. Use this property to solve th question
(1) The average of the first nine integers is 7
7 will be the middle value; there will be 4 consecutive integers to the left and also to the right of 7
we will have {3,4,5,6,7,8,9,10,11}
now we can add last two consecutive integer after 11, they will be 12,13
our new set will become = {3,4,5,6,7,8,9,10,11,12,13}
again since the number of total elements in the set is odd, Mean will simply be the middle value = 8
SUFFICIENT

(2) The average of the last nine integers is 9
Again number of element in the set are odd, 9 will be the middle value; 4 consecutive integers will lie to its left and right
Middle value will be
{5,6,7,8,9,10,11,12,13}
Add 3,4 at the start of the set
new set = {3,4,5,6,7,8,9,10,11,12,13}
Mean will be 8
Sufficient

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Re: What is the average (arithmetic mean) of eleven consecutive   [#permalink] 15 Jul 2016, 00:10
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