## Is 0/0 undefined or indeterminate?

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### Is 0/0 undefined or indeterminate?

I'm confused, my book says that it's undefined, but I've been reading on the internet that it's indeterminate.
krum
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### Re: Is 0/0 undefined or indeterminate?

krum » August 16th, 2015, 7:37 pm wrote:I'm confused, my book says that it's undefined, but I've been reading on the internet that it's indeterminate.

The way I learned it in school:

a/0 is 'undefined' when a≠0,
a/0 is 'indeterminate' when a=0.

Thus, unless nomenclature has changed since I was in school, the answer to your question is 'indeterminate'.

Darby
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### Re: Is 0/0 undefined or indeterminate?

Calculus books will refer to "indeterminate forms", because they play a special role in certain infinite series.

$0^{0}$ is one of them. As is

$\frac{0}{0} ,$

$\infty ^{\infty },$

$\frac{\infty }{\infty },$ and

$\infty ^{0}$

However, 1/0 is not an indeterminate form, and for that reason it can't be used in certain theorems. If those "undefined" a/0 things appear in certain limits of infinite series, math students can justifiably say the the answer is "infinity", and be technically correct. Depending on things like the textbook or the professor, the right answer may also be saying that the series (or limit) "diverges."

hyksos
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### Re: Is 0/0 undefined or indeterminate?

krum,

What's $\frac{x}{0}$? Okay, let's ask that question mathematically by:
1. Call the definition $k$.
2. Write that $k=\frac{x}{0}$.
3. Solve for $x$:
• $k=\frac{x}{0}$;
• $0{\times}k=0{\times}{\frac{x}{0}}$
• $0{\times}k=x$
• $x=0$.
4. Consider $x=0$.
• No definition $k$ would lead to a contradiction.
• So all possible definitions $k$ may be consistent.
• So we cannot determine a particular definition that $k$ must be.
• Call this inability to determine "indeterminate".
5. Consider $x{\neq}0$.
• Any definition $k$ would lead to a contradiction.
• So there is no possible definition $k$.
• Call this inability to define "undefined".

Therefore $\frac{x}{0}$ is:
1. indeterminate when $x=0$.
2. undefined when $x{\neq}0$.

Historically some mathematicians have failed to rigorously consider branching cases when they had fractions. For example, if their proof contained $\frac{x}{y}$ at some intermediate step but was later removed (e.g. by multiplication by $y$), then it may be hard to tell, but the proof may be flawed through reliance on $0{\times}{\frac{x}{0}}=x$. This hidden division-by-zero fallacy was the first major math fallacy that I was warned about.
Natural ChemE
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### Re: Is 0/0 undefined or indeterminate?

krum » Sun Aug 16, 2015 6:37 pm wrote:I'm confused, my book says that it's undefined, but I've been reading on the internet that it's indeterminate.

I hope this isn't off topic, but undefined suggests it can be, but just hasn't yet been defined. Would not undefinable be more accurate, in the context of the OP?

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### Re: Is 0/0 undefined or indeterminate?

Watson » August 17th, 2015, 12:58 pm wrote:
krum » Sun Aug 16, 2015 6:37 pm wrote:I'm confused, my book says that it's undefined, but I've been reading on the internet that it's indeterminate.

I hope this isn't off topic, but undefined suggests it can be, but just hasn't yet been defined. Would not undefinable be more accurate, in the context of the OP?

No. Classical Mathematics is actually quite clear and unambiguous on this point. It's been well covered in the previous posts.

Darby
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### Re: Is 0/0 undefined or indeterminate?

To put it more philosophically, x/x means a ratio of things - litterboxes to cats, doorknobs to doors, etc. Zero isn't anything, it isn't a real quantity, it is simply an absence, viz. nothing. Seen clearly, it has no place in a ratio. When you have no cats or litterboxes, then their ratio is indeterminate because there are no things to relate to each other. When you have ten cats and no litterboxes, then you have a ratio that is undefined - 10/0. You can't define a ratio because you only have one sort of quantity. And a yard full of cat poop. Does this sound right?

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### Re: Is 0/0 undefined or indeterminate?

So it is undefinable. I'm not sure how this is somewhat the same as indeterminable.

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### Re: Is 0/0 undefined or indeterminate?

Thay are not the same ... they are different terms each with their own specific definitions.

The inelegant but tried and true school methodology of learning may come in handy herr ... if understanding is not instantly forthcoming, rely on memorization until it does. Worked for me when I needed it.

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### Re: Is 0/0 undefined or indeterminate?

Natural ChemE » Mon Aug 17, 2015 1:26 am wrote:krum,

What's $\frac{x}{0}$? Okay, let's ask that question mathematically by:
1. Call the definition $k$.
2. Write that $k=\frac{x}{0}$.
3. Solve for $x$:
• $k=\frac{x}{0}$;
• $0{\times}k=0{\times}{\frac{x}{0}}$
• $0{\times}k=x$
• $x=0$.
4. Consider $x=0$.
• No definition $k$ would lead to a contradiction.
• So all possible definitions $k$ may be consistent.
• So we cannot determine a particular definition that $k$ must be.
• Call this inability to determine "indeterminate".
5. Consider $x{\neq}0$.
• Any definition $k$ would lead to a contradiction.
• So there is no possible definition $k$.
• Call this inability to define "undefined".

Therefore $\frac{x}{0}$ is:
1. indeterminate when $x=0$.
2. undefined when $x{\neq}0$.

Historically some mathematicians have failed to rigorously consider branching cases when they had fractions. For example, if their proof contained $\frac{x}{y}$ at some intermediate step but was later removed (e.g. by multiplication by $y$), then it may be hard to tell, but the proof may be flawed through reliance on $0{\times}{\frac{x}{0}}=x$. This hidden division-by-zero fallacy was the first major math fallacy that I was warned about.

Outstanding! Perhaps the clearest explanation of this that I have seen.
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### Re: Is 0/0 undefined or indeterminate?

Watson » August 17th, 2015, 4:34 pm wrote:So it is undefinable. I'm not sure how this is somewhat the same as indeterminable.

Yup, "undefinable" would be a good way to think about it.

Mathematicians say "undefined" because they mean to say that $\frac{x}{0}$ is undefined within elementary algebra. However, they see $\frac{x}{0}$ as definable since they're not constricted to elementary algebra.

If we were to take a more abstract algebra approach, we could just state
${\text{Watson}}{\left(x\right)}{\equiv}{\frac{x}{0}}$.
And I know what you're thinking: how's that help, right? Well, simple: the ${\text{Watson}}{\left(x\right)}$ function doesn't return a normal number, but rather some new mathematical entity. This new mathematical entity obeys
$0{\times}{\text{Watson}}{\left(x\right)}=x$,
so it doesn't suffer from the automatic inconsistency that all elementary entities do in the post that I made above.

What other properties does this new mathematical entity have? I dunno, I'm too lazy to make them up right now. And, heck, it's named after you, so you make them up. You're allowed to make up whatever you want so long as you rigorously avoid contradicting yourself.

To sum this up, yeah, you're right that "undefinable" would make more sense to most folks since we usually think about math in the context of elementary algebra. However, since mathematicians don't see things that way, to them it's moreso "undefined" than "undefinable".

-----

Separately, I forgot to address your question about how undefined and indeterminate are somewhat the same. Well, in a way, they're actually exact opposites:
1. undefined: No elementary description fits.
2. indeterminate: The set of elementary descriptions which fit includes non-equivalent members.
However, these two terms are similar in that they refer to cases in which there isn't exactly one known set of equivalent elementary representations.

By which I mean:
1. Most elementary entities are defined by only one set of equivalent representations. Examples of representations for $k=2$:
• $k=0+2$
• $k=0+1+1$
• $k=3-1+1-3+2.5-0.5$
2. Undefined entities have no elementary representations. Examples of representations for $k={\frac{2}{0}}$:
• [None. $\frac{2}{0}$ can't even represent itself because it leads to a contradiction; you're literally not even allowed to write it! It's meaningless gibberish.].
3. Indeterminate entities have more than one possible set of equivalent representations. Examples of possible sets of equivalent representations for $k=\frac{0}{0}$:
• Set of representations in which $k=0$, including:
• $k=0$
• $k=0+0$
• $k=1-1$
• $k=0*0$
• $k=0.5-\frac{1}{2}$
• Set of representations in which $k=1$, including:
• $k=1$
• $k=1+0$
• $k=2-1$
• $k=1*1$
• $k=1.5-\frac{1}{2}$
• Set of representations in which $k=2$, including:
• $k=2$
• $k=1+1$
• $k=2-0$
• $k=1*2$
• $k=1.5+\frac{1}{2}$
• [Etc., since $k$ can be just about anything in this case.]

To sum this up, "undefined" and "indeterminate" are similar in that they both indicate that there's not exactly one set of equivalent elementary representations. However, "undefined" indicates zero sets while "indeterminate" indicates more than one set.
Natural ChemE
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### Re: Is 0/0 undefined or indeterminate?

Exhaustively well said, but I think it might be easier for many to consider that undefined & indeterminate (or undefinable and indeterminable, as Watson put it) are somewhat similar only in a purely linguistic sense, whereas they are quite dissimilar in a purely mathematical sense ... kinda like 'alien' and 'illegal alien' are superficially similar linquistically, and yet are completely very different idiomatically.

Mathematics and linquistics share a lot of common ground, and indeed the former was birthed out of the latter, but it is contextually important to remember their differences and that they are not interchangeable - mathematics is more rigorously symbol oriented, whereas linguistics is generally more meaning oriented.

Math is a language unto itself.

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### Re: Is 0/0 undefined or indeterminate?

Watson » August 17th, 2015, 5:58 pm wrote:I hope this isn't off topic, but undefined suggests it can be, but just hasn't yet been defined. Would not undefinable be more accurate, in the context of the OP?

Here you have been shown that no mathematician would ever use a term like "undefinable", because if something is not defined within a domain of mathematics then a good mathematician would rapidly create a new domain where the thing IS defined (such domain would probably start from that definition itself).

For example they would use your (algebraically undefined) ratio 0/0 to represent a discontinuity in the function x/|x|: for x>0 the function has constant value 1, and its limit for x--> 0 (from the right) also is 1; for x<0 the function has value -1 and that is also its limit for x-->0 (from the left). Since at zero the function instantaneously jumps from -1 to +1, its value cannot be determined there. Still, it is defined, as you can plot and integrate the function with no problems.

[Forget this second part - it only is an attempt at pretending I am also a "good mathematician", simply because I share that attitude: if you find something that does not fit into your model, then claim that that's only a part of your model, you do have a better, more complex one!]

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### Re: Is 0/0 undefined or indeterminate?

if something is not defined within a domain of mathematics then a good mathematician would rapidly create a new domain where the thing IS defined (such domain would probably start from that definition itself).

Yeah you can do that, but it depends on the professor teaching it, and on the context where the expression appears during the course. If you have a series where the numerator approaches a finite value, say 2 , and the denominator is trending to zero
$\frac{2}{0}$
you can justifiably define a "new domain" where the real answer to the problem is $\infty$

Confusion arrises in introductory courses where the students are asked to instead provide the answer as "Does not exist".

The differentiation between "undefined" and "indeterminate" goes beyond pathological issues presented by Natural ChemE. In l'Hopital's Rule, the expression is required to be indeterminate. Otherwise, in expressions that are undefined, the theorem does not hold. (more importantly) you cannot use the technique to solve the limit.

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### Re: Is 0/0 undefined or indeterminate?

hyksos,

That's a common misconception, but L'Hôpital's rule isn't meaningfully related to this topic.

L'Hôpital's rule is an alternative way to evaluate the limit of fractions. While it's usually more work than simpler methods, sometimes it's actually easier or/and provides a better result.

The commonly discussed case for L'Hôpital's rule is when just substituting in the approached value would result in either $\frac{0}{0}$ or $\frac{{\infty}}{{\infty}}$. The problem with both of these results is that they're not very descriptive, i.e. they're indeterminate. Math books often recommend considering L'Hôpital's rule when this happens since it might provide a determinate result.
Natural ChemE
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### Re: Is 0/0 undefined or indeterminate?

Natural ChemE.

L'Hôpital's rule is not just for $\frac{0}{0}$ and$\frac{ \infty }{ \infty }$ as you have claimed. The rule will work on any situation that is an indeterminate form. You have given two examples . There are many more, such as $0^0$ and several more others.

However, $\frac{2}{0}$ is NOT indeterminate! That's where the danger comes in. L'Hôpital's rule cannot be applied to situations that are "formally infinity", or in limits and series that genuinely diverge.

The main thing I wanted to communicate here is : yes, I agree with you that students are told to plug-and-chug L'Hôpital's rule, as you have described. However, dig in the literature a little deeper. L'Hôpital's rule is buttressed by two historical proofs and their corresponding theorems. This differentiation between indeterminate and undefined is very real in mathematics, and it has real consequences in theorems.

hyksos
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### Re: Is 0/0 undefined or indeterminate?

hksos,
Natural ChemE » August 31st, 2015, 7:55 pm wrote:The commonly discussed case for L'Hôpital's rule is when just substituting in the approached value would result in either $\frac{0}{0}$ or $\frac{{\infty}}{{\infty}}$.
hksos » September 12th, 2015, 4:17 pm wrote:L'Hôpital's rule is not just for $\frac{0}{0}$ and$\frac{ \infty }{ \infty }$ as you have claimed.
How the heck did you get from my statement to your representation of it?

There's a lot wrong with what you've said. Unless you can demonstrate some relevance, I'll just split 'em off to avoid polluting a good thread.
Natural ChemE
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### Re: Is 0/0 undefined or indeterminate?

[quote="[url=http://www.sciencechatforum.com/viewtopic.php?p=286235#p286235]Natural ChemE » August 17th, 2015, 1:26 am[/url]"]krum,

What's $$\frac{x}{0}$$? Okay, let's ask that question mathematically by:[list=1]
[*]Call the definition $$k$$.
[*]Write that $$k=\frac{x}{0}$$.
[*]Solve for $$x$$:
[list]
[*]$$k=\frac{x}{0}$$;
[*]$$0{\times}k=0{\times}{\frac{x}{0}}$$
[*]$$0{\times}k=x$$
[*]$$x=0$$.[/list]
[*]Consider $$x=0$$.
[list]
[*]No definition $$k$$ would lead to a contradiction.
[*]So all possible definitions $$k$$ may be consistent.
[*]So we cannot determine a particular definition that $$k$$ must be.
[*]Call this inability to determine "indeterminate".[/list]
[*]Consider $$x{\neq}0$$.
[list]
[*]Any definition $$k$$ would lead to a contradiction.
[*]So there is no possible definition $$k$$.
[*]Call this inability to define "undefined".[/list][/list]

Therefore $$\frac{x}{0}$$ is:[list=i]
[*]indeterminate when $$x=0$$.
[*]undefined when $$x{\neq}0$$.[/list]

Historically some mathematicians have failed to rigorously consider branching cases when they had fractions. For example, if their proof contained $$\frac{x}{y}$$ at some intermediate step but was later removed (e.g. by multiplication by $$y$$), then it may be hard to tell, but the proof may be flawed through reliance on $$0{\times}{\frac{x}{0}}=x$$. This [url=https://en.wikipedia.org/wiki/Division_by_zero#Fallacies]hidden division-by-zero fallacy[/url] was the first major math fallacy that I was warned about.[/quote]

The bottom line is that 0/0 is not equal to any number and a/0 where a is not 0 is not any number.
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### Re: Is 0/0 undefined or indeterminate?

It is indeterminate.
williamjohn
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### Re: Is 0/0 undefined or indeterminate?

n/0 does not exist, for all n.

1. x/y defined as (the z such that y*z = x).
2. 0*z = 0, for all z.

n/0 = (the z: 0*z = n).

If ~(n=0) then there is no z such that 0*z = n.

e.g. (1/0) = (the z such that 0*z = 1).
But 2 asserts that all values of 0*z are equal to 0,
i.e. there is no value of z for which 0*z=1,

If there is no value of z such that 0*z=1 then
(the z such that 0*z = 1) does not exist.

(the z: 0*z = n) is not unique for all n > 0.

If n=0 then there is more than one value of z such that 0*z=0.
0*z=0 is true for all z.

i.e. there is no unique value of z such that 0*z = 0.

That is to say, n/0 does not exist for all values of n,
even though n/0 is defined.
Owen
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# How to pick $10$ people from $13$ such that at least $1$ is a woman

Thirteen people on a softball team show up for a game. Of the $13$ people who show up, $3$ are women. How many ways are there to choose $10$ players to take the field if at least one of these players must be a woman?

The given answer is calculated by summing the combination of $1$ woman + $9$ men, $2$ women + $8$ men, and $3$ women + $7$ men.

My question is, why can't we set this up as the sum $\binom{3}{1} + \binom{12}{9}$ - picking one of the three women first, then picking $9$ from the remaining $12$ men and women combined? The only requirement is that we have at least one woman, which is satisfied by $\binom{3}{1}$, and that leaves a pool of $12$ from which to pick the remaining $9$. The answer this way is close to the answer given, but it's $62$ short. I get that it's the "wrong" answer but I'm wondering why my thinking was wrong in setting it up this way. Thanks.

• You must multiply $C(3,1)$ with $C(12,9)$ because for every choice of a woman you have $C(12,9)$ choices for the other people. But this does not work either because you count arrangements multiple times. – Peter Jun 12 '16 at 22:15

Firstly you multiply, not add, if you are thinking of $\dbinom31$ and $\dbinom{12}{9}$

Secondly, this approach will over count. Suppose you chose Alicia , and then you chose $9$ from the remaining $12$, you would also have combos where Britney was first chosen, and Alicia was chosen from the $12$ group.

Thirdly, your book approach is correct, but unnecessarily tedious.

The best way is to compute
[All possible combos] - [All male combos ] $\;=\dbinom{13}{10} - \dbinom{10}{10}$

• I get what you are saying, but the constraint is "at least one", so wouldn't having a combination where Alicia is picked first, followed by a combination where Britney is picked and then Alicia is picked, be perfectly valid? Because there is still "at least one" in each combination. Unless I misunderstood you. – Dave Jun 12 '16 at 22:51
• $A$ followed by $B$, and $B$ followed by $A$ are valid, but you are counting them twice (apart from other overcounts). Problems of the "at least one" type are best tackled using the complement. You can verify for yourself that the book method, and the above method give the same answer. – true blue anil Jun 13 '16 at 4:44

Among $13$ people, you can choose $10$ players in $$\binom {13}{10}$$ ways.

Now, suppose there is no woman in the squad, then you can choose that team in $\binom{10}{10}=1$ way.

So, number of ways of choosing a team with at least $1$ woman is $$\binom{13}3-1=286-1=285$$ ways.

When you are counting several countings(call it nested counting), you should apply the multiplication principle. Here counting ways of choosing $1$ woman and $9$ men, you are in some kind of this nested counting, where, for any $9$ men among the $10$, the woman can occur, so, have to multiply this.

And upon deriving each choices by multiplying, you add those terms. Because, in one counting, there is $1$ woman , in some other count, there are $2$, and in other, there are $3$, women.

So, all are separate cases, so add them, like $$\binom31 \binom{10}9+\binom 32\binom {10}8+\binom33\binom {10}7=285$$

• This is a better solution than the one on the supplied answer sheet. On the other hand, it doesn't explain why OP's method fails. – David K Jun 12 '16 at 22:18


# Convex hull of idempotent matrices

What is the convex hull of the set of $n\times n$ (potentially asymmetric) idempotent matrices?

Consider the set $S:=\{A\in\mathbb{R}^{n\times n}:A^2=A\}$, the set of idempotent matrices. Is there a simple description of the convex hull of $S$?

By convex hull, I mean: $C:=\{\sum_i a_i M_i:\sum_ia_i=1,a_i\geq0,M_i\in S\}$

• At the very least, we know that every matrix in the hull has a trace between $0$ and $n$. For all I know, we might get every matrix with trace between $0$ and $n$. – Omnomnomnom Oct 9 '15 at 18:34
• A condition on trace seems quite loose, however. Is there a smaller convex set containing the idempotent matrices? – Justin Solomon Oct 9 '15 at 18:42
• I really don't know. By the way, it looks like your question is about to be closed because it "needs more context". – Omnomnomnom Oct 9 '15 at 18:46
• I think the problem that people may have is that it looks like you didn't put enough work into answering the question yourself. – Omnomnomnom Oct 9 '15 at 18:52
• Sure, but the people of the site and the powers that be tend to expect some kind of partial work, though not necessarily a partial result. I happen to think your question is fine as is. – Omnomnomnom Oct 9 '15 at 18:55

First I'll do the $2 \times 2$ case. Consider a $2 \times 2$ matrix of trace $1$:
$$A = \pmatrix{a & b\cr c & 1-a\cr}$$ For any real $r$, every point $(b,c)$ in the plane is a convex combination of two points on the hyperbola $x y = -r$ (the two axes in the case $r=0$). Thus $A$ is a convex combination of two idempotent matrices of the form $$\pmatrix{ a & x\cr y & 1-a\cr}$$ where $xy = a-a^2$. The only idempotent matrix of trace $0$ is $0$, and the only idempotent matrix of trace $2$ is $I$. Any matrix of trace strictly between $0$ and $1$ is a convex combination of $0$ and a matrix of trace $1$, while any matrix of trace strictly between $1$ and $2$ is a convex combination of $I$ and a matrix of trace $1$. Thus the convex hull in the $2 \times 2$ case consists of $0$, $I$ and all $A$ with $0 < \text{tr}(A) < 2$.

This leads to a solution in the $n \times n$ case for any $n > 2$. We have idempotent matrices of trace $1$ with the form $$\pmatrix{a & x & 0 & \ldots & 0\cr y & 1-a & 0 & \ldots & 0\cr 0 & 0 & 0 & \ldots & 0\cr \ldots & \ldots & \ldots & \ldots & \ldots\cr 0 & 0 & 0 & \ldots & 0\cr}$$ where $xy = a - a^2$, and convex combinations of these give all $n \times n$ matrices of trace $1$ that are $0$ outside the top left $2 \times 2$ submatrix. By permuting indices, we get all $n \times n$ matrices of trace $1$ that are $0$ outside some principal $2 \times 2$ submatrix. Convex combinations of these give us all $n \times n$ matrices of trace $1$. Now any $n \times n$ matrix of trace strictly between $0$ and $1$ is a convex combination of $0$ and a matrix of trace $1$, while any matrix of trace strictly between $1$ and $n$ is a convex combination of $I$ and a matrix of trace $1$. Thus we find that the convex hull consists of $0$, $I$ and all $A$ with $0 < \text{tr}(A) < n$.

• so do we know that there won't be any matrices with trace $2$ besides $I$? – Omnomnomnom Oct 9 '15 at 19:14
• Also, what about matrices with trace $0$? – Omnomnomnom Oct 9 '15 at 19:17
• An idempotent matrix of trace $k$ is a projection on a subspace of dimension $k$. The only subspace of dimension $2$ is the whole space, and a projection on that is the identity. The only subspace of dimension $0$ is $\{0\}$, and a projection on that is $0$. – Robert Israel Oct 9 '15 at 19:25
• I see, there is a unique idempotent element with trace $2$ (or $0$), and so the convex hull must also have a unique such element. – Omnomnomnom Oct 9 '15 at 19:32
• Wow! Thanks to both of you for your help -- it's much appreciated. I'm surprised the convex hull is so big. – Justin Solomon Oct 9 '15 at 19:50

The result can be generalized by induction. Let $C_n$ denote the convex hull over $n \times n$ matrices.

First of all, note that if $A \in C_{n-1}$ and $x \in \Bbb R^n$, then the elements $$\pmatrix{A&0\\0&0_{1\times 1}},E_x := \pmatrix{I_{n-1}&x\\0&0}$$ are all in $C_n$.

Next, note that $C_n$ is closed under similarity (anything similar to an element of $C_n$ is in $C_n$).

Now, take any upper triangular $A \in C_{n-1}$. That is, $$A = \pmatrix{\lambda_1 & * \cdots \\ &\ddots\\ && \lambda_{n-1}}$$ with $\sum\lambda_i \in (0,n-1)$. Note that any convex combination of $A$ and $I$ is in $C_n$. Also, any convex combination of a matrix $C_n$ with $E_x$ is in $C_n$.

Conclude that $C_n$ contains every upper triangular matrix with trace (strictly) between $0$ and $n$ and a non-negative bottom-right entry. Because every matrix is upper-triangularizable, conclude that $C_n$ contains every matrix with trace strictly between $0$ and $n$.


A very interesting question: intersection point of $x^y=y^x$

I've been investigating the Cartesian graph of $$x^y=y^x$$. Obviously, part of the graph is comprised of the line $$y=x$$ but there is also a curve that is symmetrical about the line $$y=x$$. (We can prove this symmetry by noting that the function $$x^y=y^x$$ is self-inverse; all self-inverse functions are symmetrical about the line $$y=x$$.)

An image of the graph is shown below:

I decided to find the intersection point and arrived at an intriguing result: the intersection point between the two curves is at $$(e,e)$$.

The following is my method: If the gradient of the line $$y=x$$ is $$1$$, the gradient of the curve at the intersection point must be $$-1$$ as it's a normal to the line (as it's symmetrical about the line). This means that at that point $$\frac{dy}{dx}=-1$$. Now to find $$\frac{dy}{dx}$$.

We have $$x^y=y^x$$. I then used a very powerful tehnique for differentiating these sorts of functions. We know that eg $$x^y=e^{\ln{x^y}}=e^{y\ln{x}}$$ and $$\frac{d}{dx}e^{f(x)}=f'(x)e^{f(x)}$$ Applying it to our function $$x^y=y^x$$ and using implicit differentiation and the product rule gives us: $$(\frac{dy}{dx}\times \ln x +\frac{y}{x})x^y=(\ln y +\frac{dy}{dx}\times \frac{x}{y})y^x$$ So $$\frac{dy}{dx}x^y\ln x +yx^{y-1}=\frac{dy}{dx}xy^{x-1}+y^x \ln y$$ Extensively rearranging gives: $$\frac{dy}{dx}=\frac{y^x \ln y -yx^{y-1}}{x^y \ln x -xy^{x-1}}$$ Let $$\frac{dy}{dx}=-1$$: $$y^x \ln y -yx^{y-1}=xy^{x-1}-x^y\ln x$$ But we know $$x=y$$ since we are at the intersection point with the line $$y=x$$: $$x^x \ln x -x^x=x^x-x^x\ln x$$ So $$2x^x \ln x -2x^x=0$$ $$2x^x(\ln x -1)=0$$ This means either $$2x^x=0$$ or $$\ln x -1=0$$ but we know $$x^x$$ is always greater than $$0$$ so $$\ln x =1$$, leaving us with: $$x=y=e$$ So I have my answer, but is there any other method of getting it? I have heard there is. Any help will be very welcome.

• The solution to $y^x=x^y$ can be expressed in terms of Lambert's W function as $$y=-xW\left(-\log(x)/x\right)/\log(x)$$Lambert's W function has been studied extensively. [Here](en.wikipedia.org/wiki/Lambert_W_function) is a link to a Wikipedia article that might be of interest. Note in particular that $W(-1/e)=-1$. Jun 18 '20 at 14:42
• Does this answer your question? Where does the curve $x^y = y^x$ intersect itself? Jun 18 '20 at 15:13
• Another possible duplicate: Given that $x^y=y^x$, what could $x$ and $y$ be?. Jun 18 '20 at 15:14
• Does this answer your question? Given that $x^y=y^x$, what could $x$ and $y$ be? Jun 18 '20 at 16:11
• Only the question titled 'Where does the curve $x^y=y^x$ intersect itself?' is helpful, but it still isn't very detailed when explaining the limits.Thanks for the suggestions though! Jun 18 '20 at 18:15

A Simple Approach

The simplest approach that I have found is to look at the intersections of $$y=tx\qquad\text{and}\qquad x^y=y^x\tag1$$ That is, \begin{align} x^{tx}&=(tx)^x\tag{2a}\\[3pt] x^t&=tx\tag{2b}\\[3pt] x^{t-1}&=t\tag{2c}\\ x&=t^{\frac1{t-1}}\tag{2d}\\ y&=t^{\frac t{t-1}}\tag{2e} \end{align} Explanation:
$$\text{(2a)}$$: $$x^y=y^x$$
$$\text{(2b)}$$: raise to the $$\frac1x$$ power
$$\text{(2c)}$$: divide by $$x$$
$$\text{(2d)}$$: raise to the $$\frac1{t-1}$$ power
$$\text{(2e)}$$: $$y=tx$$

Now, if we wish to find where $$x=y$$, let $$t\to1$$. That is, \begin{align} x &=\lim_{t\to1}t^{\frac1{t-1}}\tag{3a}\\ &=\lim_{n\to\infty}\left(1+\frac1n\right)^n\tag{3b}\\[6pt] &=e\tag{3c} \end{align} Explanation:
$$\text{(3a)}$$: $$x=y$$ when $$t=1$$
$$\text{(3b)}$$: $$t=1+\frac1n$$
$$\text{(3c)}$$: evaluate the limit

Further Musings

We can also compute \begin{align} y &=\lim_{t\to1}t^{\frac t{t-1}}\tag{4a}\\ &=\lim_{n\to\infty}\left(1+\frac1n\right)^{n+1}\tag{4b}\\[6pt] &=e\tag{4c} \end{align} Explanation:
$$\text{(4a)}$$: $$x=y$$ when $$t=1$$
$$\text{(4b)}$$: $$t=1+\frac1n$$
$$\text{(4c)}$$: evaluate the limit

Using the results from this answer, we see that $$\left(1+\frac1n\right)^n$$ is increasing and $$\left(1+\frac1n\right)^{n+1}$$ is decreasing, which means that $$\left(1+\frac1n\right)^n\le e\le\left(1+\frac1n\right)^{n+1}$$.

Thus, as $$t\to1^+$$, $$(4)$$ and $$(5)$$ show that $$x\to e^-$$ and $$y\to e^+$$.

Furthermore, if we substitute $$t\mapsto1/t$$, we get \begin{align} t^{\frac1{t-1}} &\mapsto(1/t)^{\frac1{1/t-1}}\tag{5a}\\ &=t^{\frac t{t-1}}\tag{5b} \end{align} and \begin{align} t^{\frac t{t-1}} &\mapsto(1/t)^{\frac{1/t}{1/t-1}}\tag{6a}\\ &=t^{\frac1{t-1}}\tag{6b} \end{align} That is, substituting $$t\mapsto1/t$$ swaps $$x$$ and $$y$$.

This means that, as $$t\to1^-$$, $$x\to e^+$$ and $$y\to e^-$$.

Relation to the Graph

Here is where these points sit on the graph as $$t\to1^+$$:

• Incredibly clear, thank you!!! Jun 23 '20 at 14:33
• @robjohn Your answers always have fantastic graphics and explanations. Nicely done, +1 from me. Jul 7 '20 at 5:35

Not sure to understand well the question but if it provides any help, I think there is something missing in your analysis.

Your equation can be seen as $$g(x,y) = x^{y} = y^{x} = f(x,y)$$. You are looking for intersecting points or surfaces (x,y) in $$\mathbb{R}^{3}$$, since both $$f(\cdot,\cdot)$$ and $$g(\cdot,\cdot)$$ are 2D scalar fields. Otherwise, if you are thinking the expression $$x^{y} = y^{x}$$ as the equation of a curve in $$\mathbb{R}$$ (somehow you figure out to express $$y = f(x)$$ or $$x = f(x)$$, it doesn't make sense to talk about an intercept (or interception set) since you must provide another curve or surface.

The same applies with the fields $$g$$ and $$f$$. You must provide another equation since you have two unknowns.

The answer you got $$(x,y) = (e,e)$$ is the trivial answer, and works for any real $$k \in \mathbb{R}$$: $$(x,y) = (k,k)$$.

I don't know if this is what you are looking for, but I would go like this.

First rewrite the equation : $$x^y=y^x\Leftrightarrow y\ln(x)=x\ln(y)\Leftrightarrow \frac{\ln(x)}{x}=\frac{\ln(y)}{y}$$ Now consider the function $$f(x)=\frac{\ln(x)}{x}$$ defined on $$(0,+\infty)$$ and whose derivative is $$f'(x)=\frac{1-\ln(x)}{x^2}$$. This shows that $$f$$ induces two bijections $$f_1:(0,e]\to (-\infty,\frac{1}{e}]$$ and $$f_2:[e,+\infty)\to[\frac{1}{e},0)$$.

Now the locus $$f(x)=f(y)$$ can be divided in two curves, namely $$y=f_1^{-1}(f(x))$$ and $$y=f_2^{-1}(f(x))$$. (Note however that these two curves are not the obvious ones). And we are looking their intersection. Since the ranges of $$f_1^{-1}$$ and $$f_2^{-1}$$ are respectively $$(0,e]$$ and $$[e,+\infty)$$, the only possible value of $$y$$ for the intersection is $$y=e$$ which happen when $$x=e$$.

• Actually, this method shows that for a function $f$ which has a single change of variation (hence a unique local extrema $x_0$), then $f(x)=f(y)$ has two branches whose intersection is exactly $(x_0,x_0)$. Jun 18 '20 at 15:46

This is not exactly an answer, but part of your reasoning was a bit hand-wavy and I want to make it fully rigorous.

Your reasoning about the slopes of the two parts of the plot is intuitively correct, but needs proof. So it is clear that the slope of the first part of the plot is $$1$$ since this encompasses the solutions $$x^x=x^x$$, i.e, the identity map which has slope $$1$$ by definition. However, showing that the other branch of the plot has slope $$-1$$ is a bit more difficult. We need to show that

1: The relation $$y^x-x^y=0$$ is its own inverse

2: If a relation is its own inverse, its slope at a point $$(x,y)$$ is $$-1$$ if $$x=y$$, with the exception of the identity map.

The first point is rather easy and I'll leave it for you to do yourself. The second part, however, requires a bit of care. Any two variable relation can be summarized by the general equation $$f(x,y)=0$$ Well, not quite any, but I hope you get the idea. For example, the equation of a circle: $$x^2+y^2-r^2=0$$ We want to use information about $$f$$ to deduce $$\frac{\mathrm{d}y}{\mathrm{d}x}$$. Using a bit of knowledge from multivariable calculus, we know that the total derivative of a two-variable function $$f$$ in terms of an arbitrary variable $$u$$ is $$\frac{\mathrm{d} f}{\mathrm{d} u}=\frac{\partial f}{\partial x}\frac{\mathrm{d}x}{\mathrm{d}u}+\frac{\partial f}{\partial y}\frac{\mathrm{d}y}{\mathrm{d}u}$$ In the special case that $$u=x$$, $$\frac{\mathrm{d} f}{\mathrm{d} x}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\frac{\mathrm{d}y}{\mathrm{d}x}$$ And similarly $$\frac{\mathrm{d} f}{\mathrm{d} y}=\frac{\partial f}{\partial x}\frac{\mathrm{d}x}{\mathrm{d}y}+\frac{\partial f}{\partial y}$$ If the relation is its own inverse, then $$f(x,y)=f(y,x)=0$$ Thus $$\frac{\mathrm{d} f}{\mathrm{d} x}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d} f}{\mathrm{d} y}=\frac{\partial f}{\partial x}\frac{\mathrm{d}x}{\mathrm{d}y}+\frac{\partial f}{\partial y}$$ So $$\frac{\partial f}{\partial x}\left(1-\frac{\mathrm{d}x}{\mathrm{d}y}\right)=\frac{\partial f}{\partial y}\left(1-\frac{\mathrm{d}y}{\mathrm{d}x}\right)$$ But again, since $$f(x,y)=f(y,x)=0$$, $$\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}$$ if $$x=y$$, thus $$1-\frac{\mathrm{d}x}{\mathrm{d}y}=1-\frac{\mathrm{d}y}{\mathrm{d}x}$$ $$\frac{1}{\frac{\mathrm{d}y}{\mathrm{d}x}}=\frac{\mathrm{d}y}{\mathrm{d}x}$$ Therefore $$\frac{\mathrm{d}y}{\mathrm{d}x}=\pm 1$$. Excluding the only self-inverse relation with slope $$1$$, the identity map, we conclude $$\frac{\mathrm{d}y}{\mathrm{d}x}\big|_{(x,x)}=-1.$$

Following this, the rest of your reasoning is correct. Others have stated that this problem can be tackled using the product log, so I can't really think of another way of going about this problem. Good question though!

• Thanks for the detailed answer; I realized I didn't have ironclad proof for the second point, but that explanation is very helpful and welcome! Thanks a s well for the final compliment :) Jun 18 '20 at 18:05

We are looking for solutions to $$\exp(y\log x) = \exp(x\log y) \, .$$ Taking logs of both sides and rearranging, this equation becomes $$\frac{x}{\log x}=\frac{y}{\log y} \, .$$ Consider the graph of $$f(z)=z/\log z$$:

Its unusual shape can be explained by a few observations:

• For $$0, $$\log z$$ is negative, so $$f(z)$$ is negative.
• As $$z\to1$$, $$\log z \to 0$$, and so the graph has an asymptote at $$z=1$$.
• For $$z>1$$, $$\log z$$ is positive, so $$f(z)$$ is positive.
• $$f(z)$$ is large when $$\log z$$ is small, or when $$z$$ is very large.

It is the final observation that gives us a hint as to why $$f(z)$$ has a local minimum. We can study the behaviour of $$f(z)$$ more closely by considering $$f'(z) = \frac{\log z - 1}{\log^2z} \, .$$ For $$1, $$f'(z)$$ is negative; at $$z=e$$, $$f'(z)$$ is zero; and for $$z>e$$, $$f'(z)$$ is positive.

Now that we have proven that $$f(z)$$ has a local minimum at $$(e,e)$$, we can begin to examine your question more closely. Let $$(x,\log x)$$ and $$(y,\log y)$$ be arbitrary points that lies on the graph.

• Suppose $$0. Then, because $$f(z)$$ is one-to-one on the interval $$(0,1)$$, the only time when $$f(x)=f(y)$$ is when $$x=y$$.
• If $$x>1$$, then things are more complicated. Clearly, if $$x=y$$, then we still have $$f(x)=f(y)$$. However, if $$1, then there is a unique number $$y\in(e,\infty)$$ such that $$f(x)=f(y)$$. This is obviously still the case if the roles of $$x$$ and $$y$$ are interchanged.
• If $$x$$ is slightly smaller than $$e$$, then $$y$$ is either equal to $$x$$, or is only slightly larger than $$e$$. This means that the two sections of the graph get very close to each other as $$x$$ tends to $$e$$.
• In the limit, when $$x=e$$, the two parts of the graph coincide.

The equation is equivalent to

$$\frac{\log x}x=\frac{\log y}y.$$

The function

$$\frac{\log x}x$$

has a maximum at $$x=e$$ and for $$x>0$$

$$\frac{\log x}x=\frac{\log y}y$$ has two solutions in $$x$$.

Using Lambert's function,

$$-\frac1x\log\frac1x=\frac{\log y}y$$ and

$$e^{\log\frac1x}\log\frac1x=-\frac{\log y}y$$

and

$$\log\frac1x=W\left(-\frac{\log y}y\right).$$


# A tangent to the ellipse meets the $x$ and $y$ axes . If $O$ is the origin, find the minimum area of triangle $AOB$.

The question is that : A tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ meets the $x$ and $y$ axes respectively at $A$ and $B$. If $O$ is the origin, find the minimum area of triangle $AOB$.

What I have attempted:

Because the equation of an ellipse is $$\frac{xx_1}{a^2}+\frac{yy_1}{b^2}=1$$

We have that the $x$ intercept is $x=\frac{a^2}{x_1}$ and $y$ intercept is $y=\frac{b^2}{y_1}$

Hence the area of a triangle $(AOB)$ is given by $A=\frac{a^2b^2}{2x_1y_1}$ now I am stuck, how should I proceed?

• What is the derivative $y'$? Aug 4, 2016 at 3:41
• Notice affine transformation preserves tangency and ratio of areas, you can use an affine transform to map the ellipse to the unit circle. The smallest triangle $AOB$ will be mapped to a right angled isoceles triangle of side $\sqrt{2}$ and area $1$. This means the smallest triangle $AOB$ has area $ab$. Aug 4, 2016 at 4:00

$A = \frac {a^2 b^2}{2xy}$ constrained by $\frac {x^2}{a^2} + \frac {y^2}{b^2} = 1$

$\frac{dA}{dx} = \frac {-2a^2b^2 (y + x y')}{(2xy)^2} = 0\\ y + x y' = 0\\ y' = -\frac yx$

Differentiating the constraint.

$\frac x{a^2} + \frac {y y'}{b^2} = 0\\ y' = -\frac {b^2x}{a^2y}\\ \frac {b^2x}{a^2y} = \frac yx\\ b^2 x^2 = a^2 y^2$

Again from the constraint:

$b^2 x^2 + a^2 y^2 = a^2 b^2\\ 2a^2 y^2 = a^2 b^2\\ y^2 = \frac {b^2}{2}\\ y = \frac b{\sqrt {2}}\\ 2xy = ab\\ A = \frac {a^2b^2}{2xy} = ab$

alternate

$x = a \cos t\\ y = b \sin t\\ A = \frac {ab}{2} \csc t \sec t\\ \frac{dA}{dt} = \frac {ab}{2} \sec t \csc t ( -\cot t + \tan t) = 0\\ \cot t = \tan t\\ t = \frac {\pi}{4}\\ A = ab$

That is much nicer.

With the change of coordinates $x=au, y=bv$ the problem boils down to the same problem where the original ellipse is replaced by a unit circle. In such a case it is trivial that the minimum area of $AOB$ is $1$. Since the determinant of the Jacobian of the linear map $x=au, y=bv$ is $ab$, the answer to the original question is $\color{red}{ab}$, plain and simple.


unit:PlanckTime

Type
Description

In physics, the Planck time, denoted by $$t_P$$, is the unit of time in the system of natural units known as Planck units. It is the time required for light to travel, in a vacuum, a distance of 1 Planck length. The unit is named after Max Planck, who was the first to propose it. $$\\ t_P \equiv \sqrt{\frac{\hbar G}{c^5}} \approx 5.39106(32) \times 10^{-44} s$$ where, $$c$$ is the speed of light in a vacuum, $$\hbar$$ is the reduced Planck's constant (defined as $$\hbar = \frac{h}{2 \pi}$$ and $$G$$ is the gravitational constant. The two digits between parentheses denote the standard error of the estimated value.

Properties
0.0000000000000000000000000000000000000000000539124
$$t_P$$
Annotations
In physics, the Planck time, denoted by $$t_P$$, is the unit of time in the system of natural units known as Planck units. It is the time required for light to travel, in a vacuum, a distance of 1 Planck length. The unit is named after Max Planck, who was the first to propose it. $$\\ t_P \equiv \sqrt{\frac{\hbar G}{c^5}} \approx 5.39106(32) \times 10^{-44} s$$ where, $$c$$ is the speed of light in a vacuum, $$\hbar$$ is the reduced Planck's constant (defined as $$\hbar = \frac{h}{2 \pi}$$ and $$G$$ is the gravitational constant. The two digits between parentheses denote the standard error of the estimated value.
Planck Time(en)

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# Generalized eigenvector for product of commuting matrices

Suppose $$A,B$$ are commuting invertible matrices with a common generalized eigenvector $$v$$ with eigenvalues $$a,b$$ respectively. That is, suppose there exist positive integers $$K,L$$ such that $$(A-aI)^K v= 0$$ and $$(B-bI)^Lv=0$$. Is it true that there exists a positive integer $$M$$ such that $$(AB-abI)^Mv = 0$$?

A bit of context: I came across this while thinking about the analogous version of this for eigenvalues (when $$K=1$$ and $$L=1$$). This is easy to show. In fact, the above statement is also easy to show if only one of $$K=1$$ or $$L=1$$ is true (that is, $$v$$ is an eigenvector for one matrix and a generalized eigenvector for the other).

Proof: WLOG suppose $$L=1$$. Then we have \begin{align} (AB-abI)^Kv &= \sum_{j=0}^K {K \choose j} A^{K-j}B^{K-j}a^jb^jv\\ &= \sum_{j=0}^K {K \choose j} A^{K-j}a^jb^Kv\\ &= b^K(A-aI)^Kv\\ &= 0 \end{align}

A natural question: is it true if $$v$$ is a generalized eigenvector for both matrices?

• Welcome! Note that askers are expected to provide context for their questions, as is explained here. For example, it would be helpful if you could edit your questions to address some of the following. What motivated this problem, or where did you encounter it? What are your thoughts on the problem? What have you tried so far? – Ben Grossmann Jun 22 at 16:20
• A potentially helpful way to reframe the problem: let $P = A - aI$ and $Q = B - bI$. We could equivalently ask: if $PQ = QP$ and $K,L$ are such that $P^Kv = Q^Lv = 0$, then does there necessarily exist an $M$ such that $$([P + aI][Q + bI] - ab I)^Mv = 0?$$ – Ben Grossmann Jun 22 at 16:25
• The answer is yes: the statement is true. Once you provide some context, I can leave a proof as an answer. – Ben Grossmann Jun 22 at 16:46
• Just added some context. Thanks for pointing the rules out btw, I'm a new poster here. – Underbounded Jun 22 at 16:47
• I figured as much, thanks for obliging. – Ben Grossmann Jun 22 at 16:48

Here is a more abstract perspective. Let $$V$$ be the smallest subspace of our vector space which contains $$v$$ and is closed under the actions of $$A$$ and $$B$$. Let $$R$$ be the subring of the endomorphism ring of $$V$$ generated by the scalar matrices together with $$A$$ and $$B$$. By assumption, this $$R$$ is commutative, and $$(A-a)^K$$ and $$(B-b)^L$$ are equal to $$0$$ in this ring (since they annihilate $$v$$ and $$v$$ generates $$V$$ under the action of $$R$$). The question then becomes simply:

Let $$R$$ be a commutative ring and $$A,B,a,b\in R$$. Suppose $$A-a$$ and $$B-b$$ are nilpotent. Then must $$AB-ab$$ be nilpotent?

But now the answer is very easy using the fact that the set $$N$$ of nilpotent elements of $$R$$ is an ideal. By assumption, $$A=a$$ and $$B=b$$ in the quotient ring $$R/N$$. Thus $$AB=ab$$ in that quotient ring. That is, $$AB-ab\in N$$.

• This is a really neat approach! – Underbounded Jun 22 at 21:23

Yes. We can proceed as follows.

Begin by reframing the question: if $$PQ = QP$$ and $$K,L$$ are such that $$P^Kv = Q^Lv = 0$$ and if $$a,b$$ are arbitrary scalars, then does there necessarily exist an $$M$$ such that $$([P + aI][Q + bI] - ab I)^Mv = 0?$$ Note that $$[P + aI][Q + bI] - ab I = PQ + bP + aQ.$$ Now, take $$m = \max\{K,L\}$$ and $$M = 2m-1$$; note that $$P^mv = Q^mv = 0$$. Apply the multinomial theorem: $$(PQ + bP + aQ)^Mv = \sum_{k_1 + k_2 + k_3 = M} \binom{M}{k_1,k_2,k_3} a^{k_3}b^{k_2}P^{k_1+k_2}Q^{k_1+k_3}v \\= \sum_{k_1 + k_2 + k_3 = M} \binom{M}{k_1,k_2,k_3} a^{k_3}b^{k_2} Q^{k_1+k_3}P^{k_1+k_2}v.$$ Now, for all triples of non-negative integers $$k_1,k_2,k_3$$ with $$k_1 + k_2 + k_3 = 2m-1$$, we see that $$k_1 + k_2$$ and $$k_1 + k_3$$ are non-negative integers with sum greater than or equal to $$2m-1$$. It follows that either $$k_1 + k_2 \geq m$$ or $$k_1 + k_3 \geq m$$. In either case, we find that $$P^{k_1+k_2}[Q^{k_1 + k_3}v] = Q^{k_1+k_3}[P^{k_1+k_2}v] = 0.$$ Thus, the above expansion of $$(PQ + bP + aQ)^Mv$$ must be equal to zero.

• Binomial theorem suffices if you write $AB-abI$ as $(A-aI)B+a(B-bI)$. This also lowers the value of $m$ to $K+L-1$. – user1551 Jun 22 at 17:45
• @user1551 That's neat! Actually, I realize now that $M = K +L - 1$ suffices for me as well; we only need $k_1 + k_2 \geq K$ or $k_1 + k_3 \geq L$ – Ben Grossmann Jun 22 at 17:57


# What do the lines in a ContourPlot Mean?

If you plot a two variable function like $x^2+y^2$, you get a series of circles with lines and colored areas. My question is, what do the lines and the colored areas represent?

• Try this experiment: ContourPlot[Cos[x] + Cos[y], {x, 0, 4 Pi}, {y, 0, 4 Pi}], followed by Plot3D[Cos[x] + Cos[y], {x, 0, 4 Pi}, {y, 0, 4 Pi}] and compare the two. Now try ContourPlot[x^2 + y^2, {x, -4, 4 }, {y, -4, 4 }] and Plot3D[x^2 + y^2, {x, -4, 4 }, {y, -4, 4 }]. Got it? ContourPlot by default generates colorized grayscale output, in which larger values are shown lighter. You may also like to read: en.wikipedia.org/wiki/Contour_line and itl.nist.gov/div898/handbook/eda/section3/contour.htm – Moo Apr 23 '16 at 12:50
• @Moo, OK, so now I know why there are areas of different colors but what is the meaning if the lines bounding each colored area? Do they represent something? – MrDi Apr 23 '16 at 12:56
• It's explained well here: en.wikipedia.org/wiki/Contour_line The value of the function doesn't change as you move along a contour. – Szabolcs Apr 23 '16 at 13:10
• @MrDi: There needs to be a way to select peaks and valleys and to bound them somehow, so they use some algorithm that picks those areas. I think you can find some details on the Wiki link I provided. it sounds like you get it, think of it as trying to collapse a 3D image onto a 2D surface. These diagrams can be extremely useful. The lines are just separating areas where something is changing based on some algorithm. – Moo Apr 23 '16 at 13:13
• @Moo, OK, thanks for help. – MrDi Apr 23 '16 at 13:14

It might be easier for you to see the correspondence between a contour plot and a surface plot if you plot them side-by-side like this (code adapted from here):

peaks[x_, y_] := 3 (1 - x)^2 Exp[-x^2 - (y + 1)^2] - 10 (x/5 - x^3 - y^5) Exp[-x^2 - y^2] -
Exp[-(x + 1)^2 - y^2]/3

With[{c = Subdivide[-5, 5, 10]},
GraphicsRow[{ContourPlot[peaks[x, y], {x, -3, 3}, {y, -3, 3},
ColorFunction -> "DarkTerrain",
Contours -> c, PlotRange -> All],
Plot3D[peaks[x, y], {x, -3, 3}, {y, -3, 3}, BoundaryStyle -> None,
Boxed -> False, ColorFunction -> "DarkTerrain",
MeshFunctions -> {#3 &}, Mesh -> {c}, PlotRange -> All,
ViewPoint -> {0, -2.7, 3.}]}]]

where I took the opportunity to use a function with more interesting contours. (Unfortunately, I don't know of any way to refine the meshlines produced by MeshFunctions; if you know a way, let me know in the comments!)

As noted in the comments above, the contours in ContourPlot[] are so-called isoclines; that is, lines of constant altitude ($z$ value). The interpretation of the colors depends on what colormap you're using; in this case, I used the "DarkTerrain" color gradient, which goes like this:

where the bluish color ("water") corresponds to the valleys, the whitish color ("snowcap") corresponds to the peaks, and the earthy colors corresponds to intermediate values. If you are familiar with map reading, running along isoclines corresponds to staying at a constant (flat) altitude, and crossing the isoclines corresponds to either ascent or descent.

A biographical segue: I was a Boy Scout once upon a time, and map reading was de rigueur for Scouts who wanted to have camping experience.


# Non-diagonal n-th roots of the identity matrix

Q. Are there $$n$$-th root analogs of this non-diagonal cube-root of the $$3 \times 3$$ identity matrix?

\begin{align*} \left( \begin{array}{ccc} 0 & 0 & -i \\ i & 0 & 0 \\ 0 & 1 & 0 \\ \end{array} \right)^3 = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) \end{align*}

I am looking for $$A^n=I$$, where $$I$$ is any dimension $$\le n$$.

(A naive question: I am not an expert in this area.)

• I'm not sure what features you want of your "analogue" here. Are you just looking for a non-diagonal $A$ such that $A^n = I$? Does it have to be $3 \times 3$? Or maybe $n \times n$? Does it have to match some feature of how the non-zero terms are laid out (e.g. one non-zero term in each row/column)? It would be good to get some more information about what you'd like to see in an answer. Jul 1 at 20:26
• The problem I see is that a matrix can have n different nth roots of a matrix. How do you decide which you want? Jul 1 at 20:38
• @GeorgeIvey "The problem I see is that a matrix can have n different nth roots of a matrix." More, usually, and occasionally fewer. Jul 1 at 20:41
• The companion matrix of the $n$th cyclotomic polynomial has order $\phi(n)$, which is less than $n$. Does that work for you?
– lhf
Jul 1 at 20:48
• Hi, Joseph. The permutation matrix of a cyclic permutation always works. Apparently you can alter the final pair of 1's into $i$ and $-i$ and still get $A^n = I$ Jul 1 at 21:19

We want to solve $$A^n = I$$ where $$A$$ and $$I$$ are $$m \times m$$

Let $$A = P \Lambda P^{-1}$$

where $$\Lambda = \text{diag}(\omega_1, \omega_2, \dots , \omega_m)$$

where the $$\omega_k$$'s are $$n$$-roots of unity. That is, $$\omega_k = \exp\big(\dfrac{i 2 \pi j}{n} \big)$$ ,$$k = 1,2,\dots,m$$ and $$j \in \{ 0, 1, 2, ...., n-1 \}$$

and $$P$$ is any invertible $$m \times m$$ matrix.

Then $$A^n = \big(P \Lambda P^{-1}\big)\big(P \Lambda P^{-1}\big)\dots \big(P \Lambda P^{-1}\big) = P^{-1} \Lambda^n P = P^{-1} I P = I$$

• Yours is the most general example, even in dimension $n$. Can you prove it?
– Ruy
Jul 2 at 0:26
• Thanks. Please check my updated solution for the proof. Jul 2 at 1:13

You can take a rotation matrix that rotates $$\phi=2\pi/n$$ around some axis, for example in 3 dimensions:

$$R_\phi=\begin{pmatrix} \cos \phi & \sin\phi & 0 \\ -\sin\phi & \cos\phi &0 \\ 0 & 0 & 1\\ \end{pmatrix}$$

This won't work for $$n=2$$ though, because $$R_\phi$$ is diagonal in that case like $$R_\phi=\operatorname{diag}(-1,-1,1)$$.

Then $$R_\phi^n=I$$ and $$R_\phi^k\neq I$$ for any $$0. You can build new matrices using any invertible matrix $$A$$ and conjugate like

$$R_{\phi,A}:= AR_\phi A^{-1} \tag 1$$

then obviously:

$$R_{\phi,A}^n = (AR_\phi A^{-1})^n = A R_\phi^nA^{-1} = I$$

(I am not sure whether this could fix the case $$n=2$$ and produce some valid (non-diagonal) results.)

• You can use Wolfram Alpha to check that {{1,1,0},{0,0,1},{1,0,0}}^-1 * {{-1,0,0},{0,-1,0},{0,0,1}} * {{1,1,0},{0,0,1},{1,0,0}} is not diagonal. Of course, for n=2 you can also use a reflection instead of a rotation, I.e. exchange two basis vectors. Jul 2 at 14:39

If $$A^n = I$$, i.e., $$A^n - I = 0$$, then the minimal polynomial $$m_A$$ of $$A$$ divides $$p(x) := x^n - 1$$. (For $$n > 1$$) one nondiagonal solution is the companion matrix of $$p$$ itself, namely, $$C_p := \pmatrix{\cdot & 1 \\ I_{n - 1} & \cdot}$$ (indeed, $$p = m_{C_p}$$).

Notice that $$C_p$$ is the permutation matrix for the permutation (in fact $$n$$-cycle) $$\pmatrix{1 & 2 & \cdots & n - 1 & n}$$ of $$n$$ elements. More generally, the permutation matrix of any permutation of $$n$$ elements of order dividing $$n$$ (and not the identity permutation) yields another solution.

There are uncountably many solutions for any $$n > 1$$: Given any solution $$A$$ and any invertible matrix $$S$$, $$SAS^{-1}$$ is another solution provided it is not diagonal (since diagonality is a closed condition that by hypothesis is not always satisfied).

The most general solution of the equation $$A^n=1$$ among $$m\times m$$ complex matrices is $$A=SDS^{-1},$$ where $$S$$ is an invertible matrix and $$D$$ is a diagonal matrix whose diagonal entries are $$n^{th}$$ roots of unity.

The reason is that the minimal polynomial of such a matrix $$A$$ divides the polynomial $$x^n-1$$, and hence has distinct roots, which in turn implies that $$A$$ is diagonalizable.


# Find a cubic polynomial.

If $f(x)$ is a polynomial of degree three with leading coefficient $1$ such that $f(1)=1$, $f(2)=4$, $f(3)=9$, then $f(4)=?,\ f(6/5)=(6/5)^3?$

I attempt:

I managed to solve this by assuming polynomial to be of the form $f(x)=x^3+ax^2+bx+c$, then getting the value of $a,b,c$, back substituting in the equation and so on....

But we can see:

$f(x)=q_1(x-1)+1\\f(x)=q_2(x-2)+4\\f(x)=q_3(x-3)+9$

Also when we put $x=1$ we get $f(1)=1^2$, when $x=2$ then $f(2)=2^2$, when $x=3$ then $f(3)=3^2$ but $f(4)\neq4^2$ (from answer).

Can this information be used to reproduce $f(x)$ directly without using the step I described in very first line of my solution?

Consider $g(x) = f(x)-x^2$. Then $g(1) = g(2) = g(3) = 0$ and $g$ is also a cubic polynomial and has leading coefficient 1. Thus $g(x) = (x-1)(x-2)(x-3)$ and hence $f(x) = (x-1)(x-2)(x-3)+x^2$. It now follows that $f(4) = 22$. Other values can be calculated.

• should $f(4)=22$ Jul 11 '17 at 3:58
• The OP wrote "with leading coefficient $1$". Jul 11 '17 at 3:59
• I assumed that $f(6/5)=(6/5)^3$ to be proved from the given information that $f$ is monic.
– user348749
Jul 11 '17 at 4:00

Classic long method:

Let $f(x) = x^3 + b x^2 + c x + d$ with $f(1) = 1$, $f(2) = 4$, $f(3) = 9$, which leads to \begin{align} f(1) &= 1 = 1 + b + c + d \hspace{10mm} \to d = -b - c \\ f(2) &= 4 = 8 + 4 b + 2 c + d = 8 + 3b + c \hspace{10mm} \to c = -4 - 3b, \, d = 4 + 2b \\ f(3) &= 9 = 27 + 9b + 3c + d = 19 + 2b \end{align} from which $b = -5$, $c = 11$, and $d = -6$ and $$f(x) = x^3 - 5 \, x^2 + 11 \, x -6.$$

With $f(x)$ then \begin{align} f(4) &= 64 - 80 + 55 -6 = 22 \\ f\left(\frac{6}{5}\right) &= \left(\frac{6}{5}\right)^{3} - \frac{36 - 66 + 30}{5} = \left(\frac{6}{5}\right)^{3}. \end{align}

It may also be noticed that $f(x)$ can be seen in the form $$f(x) = \left(x - \frac{5}{3}\right)^{3} + \frac{8}{3} \, \left( x - \frac{5}{3}\right) + \frac{83}{27}.$$ From this it is easy to see that \begin{align} f\left(\frac{5}{3}\right) &= 3 + \frac{2}{27} \\ f\left(\frac{5}{6}\right) &= \frac{59}{216}. \end{align}

• Looking at your last form of $f(x)$ if I put $5/3$ in then I would not get the same answer as you have for $f(5/3)$ just after - the former gives $3+2/27$... I've not chased the maths to see which is right, just happened to notice that discrepancy (or I've done something stupid which is also possible). Jul 11 '17 at 8:47

If

• $$f(x) = x^3 + ax^2 + bx + c$$

• $$f(1) = 1$$

• $$f(2) = 4$$

• $$f(3) = 9$$

then

\begin{align} f(x+1) - f(x) &= (3x^2+3x+1) + a(2x+1) + b \\ \hline 4-1 &= f(2)-f(1) \\ 3 &= 7 + 3a + b \\ \hline 9-4 &= f(3) - f(2) \\ 5 &= 19 + 5a + b \\ \hline 3a + b &= -4 \\ 5a + b &= -14 \\ \hline a &= -5 \\ b &= 11 \\ c &= -6 \end{align}

S0 $$f(x) = x^3 - 5x^2 + 11x - 6$$.

You could make a difference table

$$\begin{array}{c} 1 && 4 && 9 \\ & 3 && 5 \\ && 2 \end{array}$$

Using $$f(x) = x^3 + ax^2 + bx + c$$, this corresponds to

$$\begin{array}{c} 1+a+b+c && 8 + 4a + 2b + c && 27 + 9a + 3b + c \\ & 7+3a+b && 19 + 5a + b \\ && 12+2a \end{array}$$

Then, comparing entries...

\begin{align} 12+2a=2 &\implies a = -5 \\ 7+3a+b = 3 &\implies b=11 \\ 1+a+b+c = 1 &\implies c = -6 \end{align}


# Analyzing $\biggl\lfloor{\frac{x}{5}}\bigg\rfloor=\bigg\lfloor{\frac{x}{7}}\bigg\rfloor$

How many non negative integral values of $x$ satisfy the equation :$$\biggl\lfloor{\dfrac{x}{5}}\bigg\rfloor=\bigg\lfloor{\dfrac{x}{7}}\bigg\rfloor$$.

My try:

Writing few numbers and putting in the required equation.

$\underbrace{0,1,2,3,4}_{\implies 0=0},\underbrace{5,6}_{\displaystyle\times},\underbrace{7,8,9}_{\implies 1=1},\underbrace{10,11,12,13}_{\displaystyle\times} ,\underbrace{14}_{\implies 2=2},\underbrace{15,16,17,18,19,20}_{\text{LHS gives 3 but RHS gives 2}},\underbrace{21,22,23,24}_{\text{LHS gives 4 but RHS gives 3}}$.

The point which I wish to make here is that after $14$ we'll get $5's$ multiple before than $7's$, hence $\forall x \geq 15 \implies \biggl\lfloor{\dfrac{x}{5}}\bigg\rfloor>\bigg\lfloor{\dfrac{x}{7}}\bigg\rfloor$

Hence the only solution are: $\{0,1,2,3,4,7,8,9,14\}$. Does these make sense?

I need to know other solutions that don't incorporate such analyzing, means using some properties and framing it to solve above.

By the division rule,

$$x=5q+r=7q+s$$ with $0\le r<5,0\le s<7$.

From this we draw $$2q=r-s\in[-6,4],$$

thus the only possibilities are with $q=0,1,2$.

• $q=0$: $r=s$, possible for $x\in[0,4]\cap[0,6]=[0,4]$;

• $q=1$: $5+r=7+s$, possible for $x\in[5,9]\cap[7,13]=[7,9]$;

• $q=2$: $10+r=14+s$, possible for $x\in[10,14]\cap[14,20]=[14]$.

• why intersection? – mnulb Jan 25 '17 at 16:52
• Because the two bracketings must hold. What else would you do ? – Yves Daoust Jan 25 '17 at 16:53
• did you came up with those intersections by hit and trials or something different, and if something new, please don't mind adding there. – mnulb Jan 25 '17 at 17:08
• @Ayushakj: the intervals are drawn from the equations and inequations. – Yves Daoust Jan 25 '17 at 17:11

Rewrite the given condition as $$5n\le x<5n+5,\quad 7n\le x<7n+7$$ where $n$ is an integer. Since $x\ge0$, $n\ge0$ as well, so the above is equivalent to $7n\le x<5n+5$. The only nonnegative integers $n$ satisfying $7n<5n+5$ are $0$, $1$, and $2$. And so just a little cleanup is needed.

For a solution the LHS and RHS are a positive integer, let's call this $c$: $$c = \left \lfloor \frac{x}{5} \right \rfloor = \left \lfloor \frac{x}{7} \right \rfloor$$

Then, there are integers $0 \leq a < 5, 0 \leq b < 7$, s.t.: $$x = 5c + a = 7c + b$$ Thus: $$a - b = 2c$$

With the restrictions to $a,b$, this is enough to find all solutions.

Analysis:

Let us take the general form where $p>q$ :

$$\left\lfloor{\dfrac{x}{q}}\right\rfloor=\left\lfloor{\dfrac{x}{p}}\right\rfloor$$

Solution for above equation exists when $\dfrac{x}{q}-\dfrac{x}{p}<1\Rightarrow x<\dfrac{p\cdot q}{p-q}$

So there is no need to check for integers greater than$\left\lfloor\dfrac{p\cdot q}{p-q}\right\rfloor$

Also $\left\lfloor{\dfrac{x}{q}}\right\rfloor=\left\lfloor{\dfrac{x}{p}}\right\rfloor=k$ implies \begin{align} &x\in [k\cdot p,(k+1)p) \cap\ [k\cdot q,(k+1)\cdot q)\\ \Rightarrow\ \ &x\in [k\cdot p,(k+1)\cdot q) \end{align}

The integers which satisfy the equation are in $$[0p,1q)\ \cup\ [1p,2q)\ \cup\ [2p,3q) \cup\ ...\ \cup\ [(n-1)p,nq)$$ where $nq<\left\lfloor\dfrac{p\cdot q}{p-q}\right\rfloor$ .

• Well done but you are missing the case $k=0$, giving the interval $[0,q)$. – Yves Daoust Jan 26 '17 at 7:41


# Let $a$ and $b$ commute. If $m$ and $n$ are relatively prime, then ord($ab$) = $mn$. [duplicate]

This is exercise $$10.E.4$$ from Pinter:

Let $$a$$ and $$b$$ be elements of a group $$G$$.

Let ord($$a$$) = $$m$$ and ord($$b$$) = $$n$$.

Prove:

Let $$a$$ and $$b$$ commute.

If $$m$$ and $$n$$ are relatively prime, then ord($$ab$$) = $$mn$$.

(HINT: Use $$10.E.2$$.)

Here is $$10.E.2$$, which Pinter suggests that we use:

If $$m$$ and $$n$$ are relatively prime, then no power of $$a$$ can be equal to any power of $$b$$ (except for $$e$$).

I'm also going to use $$10.E.1$$:

If $$a$$ and $$b$$ commute, then ord($$ab$$) is a divisor of lcm($$m$$,$$n$$).

As well as $$B.T6.i$$ (Theorem $$6.i$$ from Appendix $$B$$. A fact from basic number theory.):

If gcd($$m$$,$$n$$) = 1 then lcm($$m$$,$$n$$) = $$mn$$

Let's begin.

We are given that $$m$$ and $$n$$ are relatively prime which means that:

$$\text{gcd}(m,n) = 1$$

By $$B.T6.i$$:

$$\text{lcm}(m,n) = mn \tag{1}$$

By $$10.E.1$$:

$$\text{ord}(ab)\ |\ \text{lcm}(m,n)$$

Substituting $$(1)$$:

$$\text{ord}(ab)\ |\ mn$$

Which means that there is an integer $$x$$ such that:

$$\text{ord}(ab) x = mn$$

This is so close! For the theorem to be true, we'd have to show that $$x = 1$$.

However, the original exercise statement says to use $$10.E.2$$.

• Is that helpful in showing that $$x = 1$$?
• Or is there some other completely different approach whereby $$10.E.2$$ is used?

The proof uses the following fact:

If $$a|c$$ and $$b|c$$ then $$lcm(a,b)|c$$.

For this exercise, I wanted to only use theorems that had been presented in the book up to that point. And, I didn't seem to recall seeing this theorem. (If anyone spots this in Pinter, please comment below with the location.)

However, I did notice that the following similar fact is in Appendix B (REVIEW OF THE INTEGERS) as exercise B.9:

If $$a|c$$ and $$b|c$$ and $$gcd(a,b) = 1$$ then $$ab|c$$.

So yeah, it looks like the approach shown below is definitely a valid way to go if you want to stick to what's presented in the book.

• – Angina Seng Jul 26 '19 at 4:35
• – Dietrich Burde Jul 26 '19 at 11:36
• The 2nd $\Rightarrow$ in my answer in the dupe is essentially a proof of 10.E.2, so replace that by an invocation of 10.E.2 to obtain the hinted proof. This is a well-known proof and likely occurs here many times. – Bill Dubuque Jul 26 '19 at 14:47

### $$\boxed{\textit{You can show mn\ \big|\ \text{ord}(ab) using 10.E.2:}}$$

Since $$a$$ and $$b$$ commute we can distribute common powers of $$a$$ and $$b$$: $$e = (ab)^{\text{ord}(ab)} = a^{\text{ord}(ab)}b^{\text{ord}(ab)}$$ This implies $$a^{\text{ord}(ab)} = b^{-\text{ord}(ab)}$$ or, in other words, some power of $$a$$ equals some power of $$b$$. But $$10.E.2$$ says that since $$m = \text{ord}(a)$$ and $$n = \text{ord}(b)$$ are relatively prime, no power of $$a$$ can equal any power of $$b$$ unless those powers of $$a$$ and $$b$$ both evaluate to $$e$$.

So we must have $$a^{\text{ord}(ab)} = e = a^m$$ and $$b^{-\text{ord}(ab)} = e = b^n$$.

But then, $$m\ \big|\ \text{ord}(ab)$$. This is because $$m$$ is the order of $$a$$ i.e. the least positive integer power of $$a$$ that evaluates to $$e$$; so if any other positive integer power of $$a$$ evaluates to $$e$$ (like $$\text{ord}(ab)$$ here), then $$m$$ must divide that integer. For a similar reason, $$n\ \big|\ \text{ord}(ab)$$.

Therefore since both $$m$$ and $$n$$ divide $$\text{ord}(ab)$$, their least common multiple $$\text{lcm}(m,n)$$ must divide $$\text{ord}(ab)$$. But in our case $$\text{lcm}(m,n) = mn$$ so we finally have $$mn\ \big|\ \text{ord}(ab)$$.

Along with the result you already established $$\text{ord}(ab)\ \big|\ mn$$, this suggests $$\text{ord}(ab) = mn$$.

P.S. If you don't believe the bolded statements above, use Euclidean Division. E.g. to see $$m\ \big|\ \text{ord}(ab)$$, use Euclidean Division on the integer pair $$\big(m, \text{ord}(ab)\big)$$ to get $$\text{ord}(ab) = km + r$$ for some $$k, r \in \Bbb Z$$ with $$0 \leq r < m$$. Do you see why $$r$$ must be $$0$$?

• Excellent and very clear presentation! I learned a few things from your proof. Thanks very much for explaining each part. I've updated my answer with some comments regarding your approach. – dharmatech Jul 26 '19 at 20:05


Complex analysis: residue integration question

Tags:
1. Nov 9, 2014

Wheelwalker

I'm asked to evaluate the following integral: $\int_{c} \frac{30z^2-23z+5}{(2z-1)^2(3z-1)}dz$ where c is the unit circle. This function has a simple pole at $z=\frac{1}{3}$ and a second order pole at $z=\frac{1}{2}$, both of which are within my region of integration. I then went about computing the residues using the following formula (which I, mistakenly perhaps, thought was general and applied to any mth-order pole): $Res f(z) = \frac{1}{(1-1)!}\lim_{z \to 1/2}{\frac{d}{dz}\{[2z-1]^2 \frac{30z^2-23z+5}{(2z-1)^2(3z-1)}}\}$ for which I got 2 (this was just the residue for the first pole). The residue I got for the other singularity was 6. They are supposed to be 1/2 and 2, respectively. What am I doing wrong, exactly? Does this residue formula not apply to this case? If so, why not? Thanks in advance.

Last edited: Nov 9, 2014
2. Nov 9, 2014

lurflurf

The formula you should have used is $$Res f(z) = \frac{1}{(2-1)!}\lim_{z \to 1/2}{\frac{d}{dz}\{[z-\dfrac{1}{2}]^2 \frac{30z^2-23z+5}{(2z-1)^2(3z-1)}}\}$$
That is why you got 2=4(1/2) instead of 1/2 and 6=3*2 instead of 2

3. Nov 9, 2014

Wheelwalker

Oh, wow that makes a lot of sense. I got in the habit of multiplying $f(z_0)$ by whichever term was creating the singularity in the denominator because a lot of the first problems we did would have something simple like $(z-1)$ there. So then, $(z-z_0)$ would equal exactly what was there. Anyway, thank you so much!

4. Nov 9, 2014

Wheelwalker

Ah, and yes that factorial term definitely should have been $(2-1)!$, although of course it doesn't matter in this case.

5. Nov 9, 2014

Wheelwalker

One more question for you actually. So, is this the most general residue formula? This should apply to nearly any case?

6. Nov 10, 2014

mathwonk

I am not very good at computations, but I am getting residues opposite to what you seem to say, i.e. res = 1/2 at z=1/2, and res = 2 at z= 1/3.

My reasoning is as follows: the residue at a, is the coefficient of 1/(z-a) in the Laurent expansion in powers of (z-a). So at z= 1/3 we should have 1/(z-1/3) . g(z), where g(z) = (30z^2 -23z + 5)/3(2z-1)^2. Then when we expand g in powers of (z-1/3), the constant term will be the residue, i.e. the coefficient of 1/(z-1/3) after we multiply by the 1/(z-1/3) out front. so this is just g(1/3) = (30/9 - 23/3 + 5)/3(1/3)^2 = 2.

Then for the other one, at z= 1/2, the function can be written as 1/(z-1/2)^2 h(z), where h(z) = (30z^2-23z+5)/4(3z-1). then the residue after multiplying by the factor 1/(z-1/2)^2, will be the linear coefficient of the expansion of h(z), i.e. should be the derivative of h(z). By the quotient rule, this is (1/4)(7/2 - 3)/(1/4) = (1/8)/(1/4) = 1/2.

Am I missing something? My point in doing it this way is to emphasize not memorizing formulas. but going back to the meaning of the concept. But of course one wants to get it right, and I am not super confident after apparently getting a different answer. Or maybe I just misunderstood what was said.

7. Nov 10, 2014

Wheelwalker

Those are the correct residues. I made a mistake in finding them the first time.

8. Nov 10, 2014

mathwonk

Thank you. My recommended point of view, as I said, is to write the function as a product of form 1/(z-a)^m . g(z), where g is holomorphic at a.

Then we want the coefficient of (z-a)^(m-1) in the Taylor expansion of g. this is of course just g^(m-1)(a) /(m-1)!, i.e. the (m-1) st derivative of g at a, divided by (m-1)!

I suggest this because your error was apparently in recalling the residue formula incorrectly. I hope this method makes that less likely, since all you have to recall this way is that you want the coefficient of 1/(z-a). In the product 1/(z-a)^m . g(z), that will be the coefficient of (z-a)^(m-1) in the expansion of g.


# Absolute value inequality with variable on both sides

I am trying to solve the following inequality:

$$|3-5x| \le x$$

I am not familiar with inequalities including one absolute value with variables on both sides. I tried to solve it as follows:

$$-x \le 3-5x \le x$$

Then I solved for each side separately,as follows:

$$3-5x \le x$$

$$x \ge (1/2)$$

$$-x \le 3-5x$$

$$x \le \frac34$$

I know my solution is incorrect and that it actually lies between $\frac12$ and $\frac34$, but I wanted to know what is wrong with my method and what is the appropriate approach to solving such inequalities.

Thanks,

This is how I learned it back in Algebra I.

You can split it into two equations. Since the sign is $\le$, it will be that Equation 1 and Equation 2 are true.

Here are the steps:

$$|3-5x| \le x$$

$$3-5x \le x \;\;\;\;\;| \;\;-3+5x\le x$$

$$6x \ge 3 \;\;\;\;\;\;\;| \;\;-4x \ge -3$$

$$x \ge \frac 12 \;\;\;| \;\;\;\;x \le \frac 34$$

$$\frac 12 \le x \le \frac 34$$

• This should be sufficient, but for further elaboration just reply Sep 8 '16 at 2:38
• This approach is very specific towards linear inequalities (but definitely works, I think)--it doesn't work (I believe) for higher order ones. Sep 8 '16 at 3:53
• Yes, I know. I was trying to give a simple easy-to-understand solution. Sep 8 '16 at 4:02
• Your "$|$" symbol is for AND or OR? In either case you have a logical issue as you start of with an OR and end with and AND. Sep 8 '16 at 4:15
• @Maccavity And. And please point out the issue. Sep 8 '16 at 4:16

Divide $-x \leq 3-5x \leq x$ by $x$ and go from there. Although I think what you have is correct, the solution IS the interval $\dfrac{1}{2} \leq x \leq \dfrac{3}{4}$.

You need to break up the absolute value into its intervals:

$$|x| = \begin{cases} x & x > 0 \\ -x & x < 0 \\ 0 & x = 0 \end{cases}$$

Therefore for $|3-5x|$ you need to find the interval when it's less than zero and when it's greater than zero:

$$|3-5x| = \begin{cases} 3 - 5x & 3 - 5x > 0 \rightarrow 3 > 5x \rightarrow x < \frac{3}{5}\\ 5x - 3 & 3 - 5x < 0 \rightarrow 3 <5x \rightarrow x > \frac{3}{5} \\ 0 & 3 - 5x = 0 \rightarrow 3 = 5x \rightarrow x = \frac{3}{5} \end{cases}$$

Now you solve the inequality in each case:

1. $x < \frac{3}{5} \rightarrow |3 - 5x| = 3 - 5x$ $$3 - 5x \leq x \\ 3 \leq 6x \\ x \geq \frac{1}{2}$$
2. $x > \frac{3}{5} \rightarrow |3 - 5x| = 5x - 3$ $$5x - 3 \leq x \\ 4x \leq 3 \\ x \leq \frac{3}{4}$$
3. $x = \frac{3}{5} \rightarrow |3 - 5x| = 0$ $$0 \leq x \\ x \geq 0$$

Now you need to analyze each case:

1. $x < \frac{3}{5} \wedge x \geq \frac{1}{2}$

It is true that $\frac{3}{5} = \frac{6}{10} \geq \frac{5}{10}$. Therefore this particular interval is true for $\frac{1}{2} \leq x < \frac{3}{5}$.

1. $x > \frac{3}{5} \wedge x \leq \frac{3}{4}$

Since $\frac{3}{5} = \frac{12}{20}$ and $\frac{3}{4} = \frac{15}{20}$, this is true for $\frac{3}{5} < x \leq \frac{3}{4}$.

1. $x = \frac{3}{5} \wedge x \geq 0$

$\frac{3}{5} > 0$ therefore this is trivially satisfied--thus $x = \frac{3}{5}$ is allowed.

When we combine these results, we find that $\frac{1}{2} \leq x \leq \frac{3}{4}$.

PS Edit:

It's probably easier to use:

$$|x| = \begin{cases} x & x \geq 0 \\ -x & x \leq 0 \\ \end{cases}$$

In that case you get the two intervals:

$$\frac{1}{2} \leq x \leq \frac{3}{5}$$

and

$$\frac{3}{5} \leq x \leq \frac{3}{4}$$

Which clearly combines to give: $\frac{1}{2} \leq x \leq \frac{3}{4}$.

Your solution was correct except you needed to use the word AND correctly. The inequality $-x \le 3-5x \le x$ is equivalent to the inequalities

\begin{align} -x \le 3-5x \quad &\text{AND} \quad 3-5x \le x \\ 4x \le 3 \quad &\text{AND} \quad 3 \le 6x \\ x \le \frac 34 \quad &\text{AND} \quad \frac 12 \le x \\ \frac 12 \le x \quad &\text{AND} \quad x \le \frac 34 \\ x &\in\left[\frac 12, \frac 34 \right] \end{align}


# Let $T^{m-1}v\neq 0$ but $T^mv=0$. Show that $v,Tv,\ldots,T^{m-1}v$ are linearly independent

Let $T\in\mathcal L(V)$ and $T^{m-1}v\neq 0$ but $T^mv=0$ for some positive integer $m$ and some $v\in V$. Show that $v,Tv,\ldots,T^{m-1}v$ are linearly independent.

I had written a proof but Im not sure if it is correct. And in the case it would be correct I dont know how to write it better and clearly. So I have two questions:

1. It is the proof below correct?

2. If so, how I can write it better using the same ideas?

### The attempted proof:

1) If $T^{m-1} v$ would be linearly dependent of $T^{m-2} v$ then exists some $\lambda\neq 0$ such that

$$T^{m-1}v=\lambda T^{m-2}v\implies T^mv=\lambda T(T^{m-2}v)=\lambda T^{m-1}v=0\implies \lambda=0$$

Then $T^{m-2}v$ is linearly independent of $T^{m-1}v$.

2) Now observe that

$$\lambda_1v+\lambda_2Tv+\lambda_3T^2v=0\implies T^{m-2}(\lambda_1v+\lambda_2Tv+\lambda_3T^2v)=\lambda_1T^{m-2}v+\lambda_2T^{m-1}v=0$$

so $\lambda_1,\lambda_2=0$ as we had shown previously, so the original equation reduces to

$$\lambda_3T^2v=0\implies \lambda_3=0$$

thus $v,Tv,T^2v$ are linearly independent.

3) Repeating recursively the analysis in 2) for longer lists of vectors of the form $v,Tv,\ldots,T^kv$ for $k< m$ we can show that the list $v,Tv,\ldots,T^{m-1}v$ is linearly independent.

• This is a related post – Bijesh K.S Jul 11 '17 at 19:15
• The if and only if in the first line looks problematic as $T$ isn't invertible. – DMath Jul 11 '17 at 19:19
• @DMath thank you... I overlook it completely. – Masacroso Jul 11 '17 at 19:20
• I'm afraid that the last part is the bulk of the proof and you just cut it off by saying “repeating the analysis”. Note that for $\{v,Tv,T^2v,T^3v\}$ you need to go down to $T^{m-3}$, which is not as easy as the initial step for $\{T^{m-2}v,T^{m-1}v\}$. – egreg Jul 11 '17 at 20:33

Hint

$$a_0v+a_1Tv+\ldots+a_{m-1}T^{m-1}v=0 \quad(1)$$

multiply by $T^{m-1}$, then

$$a_0T^{m-1}v=0\to a_0=0$$

so, from $(1)$,

$$a_1Tv+\ldots+a_{m-1}T^{m-1}v=0$$

multiply by $T^{m-2}$, then

$$a_1T^{m-1}v=0\to a_1=0$$

can you finish?

• ok but, it is my proof correct? – Masacroso Jul 11 '17 at 19:17
• @Masacroso: it seems correct to me. You just need induction to finish it. – Arnaldo Jul 11 '17 at 19:19

See below for a simple proof.

Suppose that they are not linear independent; that is for some choice of $\alpha_{k}$,

$T^{m-1}v = \sum_{k=0}^{m-2}\alpha_{k}T^{k}v$.

Then $0 =T^{m}v = T\cdot T^{m-1}v = \sum_{k=0}^{m-2}\alpha_{k}T^{k+1}v$,

so $Tv,...,T^{m-1}v$ are linearly dependent.

But that means that $T^{m-1}v = \sum_{k=1}^{m-2}\beta_{k}T^{k}v$.

Inductively repeat the argument to eventually conclude that for some $\gamma \neq 0$, $T^{m-1}v = \gamma T^{m-2} v$, but then we would have to say that $0 = T T^{m-1} v = \gamma T^{m-1} v \neq 0$, a contradiction.

More simply put: if you had a linear combination $c_0 v + c_1 T v + \ldots c_{m-1} T^{m-1} v = 0$ with $c_j$ not all zero, let $c_i$ be the first nonzero coefficient. Then $$T^i v = - \sum_{j=i+1}^{m-1} (c_j/c_i) T^j v$$ Applying $T^{m-i-1}$ to both sides, $$T^{m-1} v = - \sum_{j=i+1}^{m-1} (c_j/c_i) T^{m+j-i-1} v = 0$$ But all terms on the right are $0$, so $T^{m-1} v = 0$, contradiction!


# Thread: a couple of trigonometric integrals

1. ## a couple of trigonometric integrals

$\displaystyle \int_{0}^{\pi} x \sin (\cos^{2}x)\cos(\sin^{2}x) \ dx$

$\displaystyle \int_{0}^{\pi} x \Big(\sin^{2}(\sin x) + \cos^{2} (\cos x) \Big)\ dx$

My initial thought was to introduce a parameter and differentiate, but that approach didn't get me anywhere.

2. Originally Posted by Random Variable
$\displaystyle \int_{0}^{\pi} x \sin (\cos^{2}x)\cos(\sin^{2}x) \ dx$

$\displaystyle \int_{0}^{\pi} x \Big(\sin^{2}(\sin x) + \cos^{2} (\cos x) \Big)\ dx$
Here's how you do the first one:

Spoiler:
Let

$\displaystyle \displaystyle I=\int_0^\pi x\sin\left(\cos^2(x)\right)\cos^2(\sin(x))\text{ }dx$

Then, letting $\displaystyle z=\pi-x$ we find that

\displaystyle \displaystyle \begin{aligned}I &=\int_0^\pi (\pi-z)\sin\left(\cos^2(z)\right)\cos\left(\sin^2(z)\ri ght)\text{ }dz\\ &=\pi\int_0^\pi\sin\left(\cos^2(z)\right)\cos\left (\sin^2(z)\right)\text{ }dz-I\end{aligned}

Thus,

$\displaystyle \displaystyle \frac{2}{\pi}I=\int_0^\pi \sin\left(\cos^2(z)\right)\sin\left(\cos^2(z)\righ t)\text{ }dz$

Recall though that

$\displaystyle \displaystyle \sin(\phi)\cos(\theta)=\frac{\sin(\phi+\theta)+\si n(\phi-\theta)}{2}$

And thus

\displaystyle \displaystyle \begin{aligned}\frac{2}{\pi} I &= \int_0^\pi \sin\left(\cos^2(z)\right)\cos^2\left(\sin(z)\righ t)\text{ }dz\\ &= \frac{1}{2}\int_0^\pi\left(\sin\left(\cos^2(z)+\si n^2(z)\right)+\sin\left(\cos^2(z)-\sin^2(z)\right)\right)\text{ }dz\\ &=\frac{1}{2}\pi\sin(1)+\frac{1}{2}\int_0^\pi \sin(\cos(2z))\text{ }dz\end{aligned}

Note though that

$\displaystyle \displaystyle \int_0^\pi\sin(\cos(2z))\text{ }dz=\int_0^{\frac{\pi}{4}}\sin(\cos(2z))\text{ }dz+\int_{\frac{3\pi}{4}}^\pi\sin(\cos(2z))\text{ }dz+\int_{\frac{3\pi}{4}}^{\pi}\sin(\cos(2z))\text { }dz$

Note though that by letting $\displaystyle \displaystyle u=z-\frac{3\pi}{4}$ one gets that

$\displaystyle \displaystyle \int_{\frac{3\pi}{4}}^{\pi}\sin(\cos(2z))=\int_0^{ \frac{\pi}{4}}\sin(\cos(2u))\text{ }du$

And thus

$\displaystyle \displaystyle \int_0^\pi\sin(\cos(2z))\text{ }dz=2\int_0^{\frac{\pi}{4}}\sin(\cos(2z))\text{ }dz+\int_{\frac{\pi}{4}}^{\frac{3\pi}}{4}\sin(\cos (2z))\text{ }dz$

But, letting $\displaystyle w=z-\frac{\pi}{2}$ in this second integral one arrives at

$\displaystyle \displaystyle \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}\sin(\cos(2z) )=-\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}\sin(\cos(2w))\text{ }dw$

and since the integrand is even this is equal to

$\displaystyle \displaystyle -2\int_0^{\frac{\pi}{4}}\sin(\cos(2w))\text{ }dw$

Therefore,

\displaystyle \displaystyle \begin{aligned}\int_0^\pi\sin(\cos(2z))\text{ }dz &= 2\int_0^{\frac{\pi}{4}}\sin(\cos(2z))+\int_{\frac{ \pi}{4}}^{\frac{3\pi}{4}}\sin(\cos(2z))\text{ }dz\\ &= 2\int_0^{\frac{\pi}{4}}\sin(\cos(2z))\text{ }dz-2\int_0^{\frac{\pi}{4}}\sin(\cos(2z))\text{ }dz\\ &=0\end{aligned}

Thus, finally we arrive at

\displaystyle \displaystyle \begin{aligned}\frac{2}{\pi}I &=\frac{\pi}{2}\sin(1)+\frac{1}{2}\int_0^\pi\sin(\ cos(2z))\text{ }dz\\ &=\frac{\pi}{2}\sin(1)\end{aligned}

and thus $\displaystyle \displaystyle I=\frac{\pi^2 \sin(1)}{4}$.

3. The ODE you had was solvable.

4. Originally Posted by Random Variable
The ODE you had was solvable.
See my new method. It's more elementary.

5. Originally Posted by Random Variable
$\displaystyle \int_{0}^{\pi} x \sin (\cos^{2}x)\cos(\sin^{2}x) \ dx$

$\displaystyle \int_{0}^{\pi} x \Big(\sin^{2}(\sin x) + \cos^{2} (\cos x) \Big)\ dx$

My initial thought was to introduce a parameter and differentiate, but that approach didn't get me anywhere.
The second one is interesting. I have a solution, but it's a tad long. Do you have any ideas as of yet?

6. to prove that $\displaystyle \displaystyle\int_0^\pi\sin(\cos 2x)\,dx=0,$ split it up into two integrals with $\displaystyle 0\le x\le\dfrac\pi2$ and $\displaystyle \dfrac\pi2\le x\le\pi$ then apply substitution $\displaystyle t=x-\dfrac\pi2$ on the second piece and you'll see is equal to minus the first piece, hence, the result.

7. Here's my solution to the second

Spoiler:

First use the double angle formula so

$\displaystyle \displaystyle \int_0^{\pi} x\left( \dfrac{1-\cos \left(2 \sin x\right)}{2} + \dfrac{1 + \cos \left(2 \cos x\right)}{2} \right)\, dx$

$\displaystyle \displaystyle = \int_0^{\pi} x\,dx + \frac{1}{2}\int_0^{\pi} \cos \left(2 \cos x\right) - \cos \left(2 \sin x\right)\, dx$

We will show the second integral is zero by considering it in two parts. (**)

First, that $\displaystyle \displaystyle \int_0^{\pi/2} \cos \left(2 \cos x\right) - \cos \left(2 \sin x\right)\,dx = \int_{\pi/2}^{\pi} \cos \left(2 \cos x\right) - \cos \left(2 \sin x\right)dx$

Let $\displaystyle x = \pi -u$ in $\displaystyle \displaystyle \int_{\pi/2}^{\pi} \cos \left(2 \cos x\right) - \cos \left(2 \sin x\right)dx$

Next that $\displaystyle \displaystyle \int_0^{\pi/4} \cos \left(2 \cos x\right) - \cos \left(2 \sin x\right)\,dx = -\int_{\pi/4}^{\pi/2} \cos \left(2 \cos x\right) - \cos \left(2 \sin x\right)dx$

Let $\displaystyle x = \pi/2-u$ in$\displaystyle \displaystyle \int_{\pi/4}^{\pi/2} \cos \left(2 \cos x\right) - \cos \left(2 \sin x\right)\, dx$

Thus, we are left with $\displaystyle \displaystyle \int_0^{\pi} x\,dx = \dfrac{\pi^2}{2}$

8. here's mine to the second:

put $\displaystyle t=\pi-x$ on the integral to see that is equal to $\displaystyle \displaystyle\frac\pi2\int_0^\pi\left( {{\sin }^{2}}(\sin x)+{{\cos }^{2}}(\cos x) \right)\,dx.$

i claim the latter integral equals $\displaystyle \pi,$ hence the original integral equals $\displaystyle \dfrac{\pi^2}2,$ so in order to see that write the integral as

$\displaystyle \displaystyle\int_{0}^{\frac{\pi }{2}}{\left( {{\sin }^{2}}(\sin x)+{{\cos }^{2}}(\cos x) \right)\,dx}+\int_{\frac{\pi }{2}}^{\pi }{\left( {{\sin }^{2}}(\sin x)+{{\cos }^{2}}(\cos x) \right)\,dx},$

and put $\displaystyle t=x-\dfrac\pi2$ on the second one, then add them up and you'll see that the claimed integral achieve the aforesaid value, as required.

another solution for the first one:

we consider $\displaystyle \displaystyle\int_{0}^{\pi }{\sin \left( {{\sin }^{2}}x \right)\cos \left( {{\cos }^{2}}x \right)\,dx},$ (1) then write it as $\displaystyle \displaystyle\int_{0}^{\frac{\pi }{2}}{\sin \left( {{\sin }^{2}}x \right)\cos \left( {{\cos }^{2}}x \right)\,dx}+\int_{\frac{\pi }{2}}^{\pi }{\sin \left( {{\sin }^{2}}x \right)\cos \left( {{\cos }^{2}}x \right)\,dx}$ and put $\displaystyle t=x-\dfrac\pi2$ on the second one then we get $\displaystyle \displaystyle\int_{0}^{\frac{\pi }{2}}{\sin \left( {{\sin }^{2}}x \right)\cos \left( {{\cos }^{2}}x \right)\,dx}+\int_{0}^{\frac{\pi }{2}}{\sin \left( {{\cos }^{2}}x \right)\cos \left( {{\sin }^{2}}x \right)\,dx},$ so when adding those we note that the integrand is actually $\displaystyle \sin(\sin^2x+\cos^2x)$ so (1) equals $\displaystyle \dfrac\pi2\sin1.$

on the original integral was easy to show that it's equal to $\displaystyle \displaystyle\frac\pi2\int_{0}^{\pi }{\sin \left( {{\sin }^{2}}x \right)\cos \left( {{\cos }^{2}}x \right)\,dx},$ thus by (1) we conclude.


Combinations problem: Choosing ways to select $8$ questions out of $12$.

In an exam, there are $12$ questions in total. He has to attempt $8$ questions in all. There are two parts: Part A and Part B of the question paper containing $5$ and $7$ questions respectively. How many ways are there to attempt the exam, such that you attempt eight questions and from each part you attempt at least three questions?
Please don't answer with the three case method. What I want is the error in my method?
My method: For three questions from each part, we do: $${5 \choose 3}*{7\choose3}$$. But 2 questions are lett, so for that, we multiply the expression by $${6\choose2}$$. Because 6 questions are left to choose from. But this does not give the answer, why?

As you are aware, the actual number of ways to select eight questions from the examination given the restrictions that at least three questions must be selected from each part is $$\binom{5}{3}\binom{7}{5} + \binom{5}{4}\binom{7}{4} + \binom{5}{5}\binom{7}{3}$$ The alternative method you proposed of selecting three questions from part A, three questions from part B, and two additional questions counts the same selection more than once.

If we choose three questions from part A and five questions from part B, the second method counts the same selection of questions ten times since there are $\binom{5}{3}$ ways to choose three of the five selected questions from part B.

To make this concrete, suppose the selected questions are $A_1, A_2, A_3, B_1, B_2, B_3, B_4, B_5$. We can make this particular selection once. However, the second method counts this particular selection ten times. \begin{align*} & \color{red}{A_1}, \color{red}{A_2}, \color{red}{A_3}, \color{blue}{B_1}, \color{blue}{B_2}, \color{blue}{B_3}, B_4, B_5\\ & \color{red}{A_1}, \color{red}{A_2}, \color{red}{A_3}, \color{blue}{B_1}, \color{blue}{B_2}, B_3, \color{blue}{B_4}, B_5\\ & \color{red}{A_1}, \color{red}{A_2}, \color{red}{A_3}, \color{blue}{B_1}, \color{blue}{B_2}, B_3, B_4, \color{blue}{B_5}\\ & \color{red}{A_1}, \color{red}{A_2}, \color{red}{A_3}, \color{blue}{B_1}, B_2, \color{blue}{B_3}, \color{blue}{B_4}, B_5\\ & \color{red}{A_1}, \color{red}{A_2}, \color{red}{A_3}, \color{blue}{B_1}, B_2, \color{blue}{B_3}, B_4, \color{blue}{B_5}\\ & \color{red}{A_1}, \color{red}{A_2}, \color{red}{A_3}, \color{blue}{B_1}, B_2, B_3, \color{blue}{B_4}, \color{blue}{B_5}\\ & \color{red}{A_1}, \color{red}{A_2}, \color{red}{A_3}, B_1, \color{blue}{B_2}, \color{blue}{B_3}, \color{blue}{B_4}, B_5\\ & \color{red}{A_1}, \color{red}{A_2}, \color{red}{A_3}, B_1, \color{blue}{B_2}, \color{blue}{B_3}, B_4, \color{blue}{B_5}\\ & \color{red}{A_1}, \color{red}{A_2}, \color{red}{A_3}, B_1, \color{blue}{B_2}, B_3, \color{blue}{B_4}, \color{blue}{B_5}\\ & \color{red}{A_1}, \color{red}{A_2}, \color{red}{A_3}, B_1, B_2, \color{blue}{B_3}, \color{blue}{B_4}, \color{blue}{B_5}\\ \end{align*}

By similar reasoning, if we choose five questions from part A and three questions from part B, the second method counts the same selection of questions ten times since there are $\binom{5}{3}$ ways to choose three of the five selected questions from part A.

If we choose four questions from part A and four questions from part B, the second method counts the same selection sixteen times since we count the same set of four questions from part A four times for each of the $\binom{4}{3}$ ways we choose three of those four selected questions and the same set of four questions from part B four times for each of the $\binom{4}{3}$ ways we choose three of those four selected questions.

Note that $$10\binom{5}{3}\binom{7}{5} + 16\binom{5}{4}\binom{7}{4} + 10\binom{5}{5}\binom{7}{3} = \binom{5}{3}\binom{7}{3}\binom{6}{2}$$

• I've edited my answer to make what I'm saying more concrete. – N. F. Taussig Oct 8 '15 at 11:43
• Okay, we see that this is happening. But why is this happening? Can we add something so that my method works? – Aditya Agarwal Oct 8 '15 at 11:45
• As you can see, the method you attempted to use does not work since it counts the same set of selected questions every time you choose $3$ of the selected questions. To avoid overcounting, we use cases. I do not see a way to make the method you attempted work. – N. F. Taussig Oct 8 '15 at 11:55

You are counting certain combinations of questions more than once. This is not so easy to see for your example, thus, let's simplify it to a trivial example: you have to choose at least 5 questions out of 5 of part A (i.e. all of A), and at least 1 out of 7 of part B, and, as before, you have to choose 8 questions in total.

We can trivially see that we indeed have to choose 3 questions out of B to satisfy both conditions, since we have already chosen all of A. Thus, the correct answer would be $\left(\begin{matrix}5\\5\end{matrix}\right)\left(\begin{matrix}7\\3\end{matrix}\right)=1\cdot 35=35$.

Your way of solving this question would however say that there are $\left(\begin{matrix}5\\5\end{matrix}\right)\left(\begin{matrix}7\\1\end{matrix}\right)\left(\begin{matrix}6\\2\end{matrix}\right)=1\cdot 7\cdot 15=105$ possibilities of choosing.

Edit: to clarify the problem in this simplified example: assume you choose questions 1,2 and 3 from B. In your calculations there are three possibilities to do so: you choose question 1 because you have to choose at least one from B (Condition 1), and you choose questions 2 and 3 because you have to choose in total 8 (Condition 2). Or you choose question 2 to satisfy Condition 1, and questions 1 and 3 for Condition 2. Finally, you can choose question 3 to satisfy Condition 1, and questions 1 and 2 for Condition 2. In other words, you count each combination 3 times, and, indeed $35\cdot3=105$.


# Solution of a given linear diophantine equation.

I was solving the following linear diophantine equation :

$56x + 72y = 40$ in integers.

My attempt: I got that 8 is the gcd of 56 and 72 and $8|40$ and hence a solution exists and I can write:

$8 = 56 - 16 *3$

$\implies$ $8= 56 - (72 -56*1)*3$

$\implies$ $8= 4*56 - 3*72$.

So my answer is $x = 4$ and $y = -3$. But in book its showing $x = 20$ and $y = -15$. Where I went wrong? Kindly help.

• Remember that $20=5 \times 4$ and $-15=5 \times 3$. Both answers are correct
– user284001
Nov 15, 2016 at 16:49
• @Bacon, So why did they multiply by 5? Nov 15, 2016 at 16:50
• At the outset you noted that $8|40$ and this is...
– user284001
Nov 15, 2016 at 16:50

$56x+72y=40$
$7x+9y=5$
Select the term with the smaller of the two coëfficients and isolate it on the LHS.
$7x=5-9y$
$\text{[1] }x=\frac{5-9y}7=1-y+\frac{-2-2y}7$
$\text{New variable }a=\frac{-2-2y}7$
$-2-2y=7a$
$-2-7a=2y$
$\text{[2] }y=-\frac{2+7a}2=-1-4a+\frac{a}2$
$\text{New variable }b=\frac{a}2$
$a=2b$
Substitute $a\text{ into [2].}$
$y=-\frac{2+7(2b)}2=-\frac{2+14b}2=-1-7b$
Substitute $y\text{ into [1].}$
$x=\frac{5-9(-1-7b)}7=\frac{5+9+63b}7=\frac{14+63b}7=2+9b$
$\text{So, }(x,y)=(2+9b,-1-7b).$
There is an infinitude of integer answers; but your textbook, for some reason, favors $b=2.$

write the equation as $$56x\equiv 40\mod 72$$ or $$x\equiv \frac{40}{76}=\frac{5}{7}\equiv \frac{77}{7}\equiv 11 \mod 72$$ thus $$x=11+72K$$ with $$k \in \mathbb{Z}$$

You can solve $56x+72y=40$ using convergent.

Note: If $\alpha,\beta$ are integral solutions to $ax+by=c$ for $x,y$ respectively, the other solutions take the form$$x=\alpha-bt\\ y=\beta+at\tag{1}$$

Here, $a=56,b=72$ so the $x$ values model the form $\alpha-72t$ and $\beta+56t$ for $y$.

The nearest converging fraction to $\dfrac {9}{7}$ is $\dfrac 43$ and hence, we have$$7\cdot 4-3\cdot 9=1$$

For which we can multiply both sides by $40$ to get$$40\cdot 7\cdot 4-3\cdot 9\cdot 40=40\implies 56\cdot 20+72(-15)=40\tag2$$ Thus, we have $\alpha=20,\beta=-15$. And by $(1)$, we get$$x=20-72t\\y=-15+56t$$

Note that we also get the solutions that the book says along the way!


# The probability that it is a woman, or a man who is not bringing a dessert?

John invites 12 friends to a dinner party, half of which are men. Exactly one man and one woman are bringing desserts. If one person from this group is selected at random, what is the probability that it is a woman, or a man who is not bringing a dessert?

I know that the two events are mutually exclusive; so that, the answer is simply $6/12 + 5/12 = 11/12$. However, when I looked at the answer in the book that I study from, I found the same result but an approach that's really weird for me. Here is it literally:

$$P(woman) = 6/12 = 1/2$$ $$P(not\ bringing\ a\ dessert\ ) = 10/12 = 5/6$$ $$P(woman\ and\ not\ bringing\ a\ dessert) = 1/2\ *\ 5/6=5/12$$ $$P(woman\ or\ a\ man\ not\ bringing\ a\ dessert) = 1/2 + 5/6 - 5/12 = 11/12$$

May anyone explain the reasoning behind that approach? And is it technically right?

Your approach is certainly right. The book's approach is right as well. It is just that they should have said $P(\text{woman OR man not bringing dessert}) = P(\text{woman OR not bringing dessert})$ (this works because there are just two genders). Then you don't have mutually exclusive events, so you have to use inclusion-exclusion. $$P(\text{woman OR not bringing dessert}) = P(\text{woman}) + P(\text{not bringing dessert}) - P(\text{woman} \cap \text{not bringing dessert}).$$

The term $P(\text{woman} \cap \text{not bringing dessert}) = 5/12$, thus they have $1/2 + 10/12 - 5/12 = 11/12$.

What you are doing is you are breaking up your space into mutually exclusive events. You are saying it is equivalent to calculating $P(\text{woman} \sqcup \left(\text{not woman AND no dessert}\right))$. That way you have $1/2 + 5/12$.

There's another way you can solve this problem. This probability is equal to $1 - P(\text{man bringing dessert}) = 1 - 1/12 = 11/12$.

NOTE: Edited after a misinterpretation pointed out by @Abdu Magdy

• But P(man not bringing a dessert) should have equalled $5/12$ not $10/12$, Consequently, P(woman ∩ man not bringing dessert) should have equalled $5/12 \ *\ 1/2 = 5/24$ – Abdu Magdy Oct 21 '17 at 18:52
• I interpret $P(\text{man not bringing a dessert})$ as $P(\text{not bringing a dessert} | \text{man}) = \frac{P(\text{not bringing dessert} \cap \text{man})}{P(\text{man})} = \frac{\frac{5}{12}}{\frac{1}{2}} = \frac{5}{6}$, but YMMV. In my experience with GRE problems, sometimes the language is indeed ambiguous. – Abhiram Natarajan Oct 21 '17 at 19:02
• I'm sorry but I really need to understand... Based on the definition of mutually inclusive events; if we only are to choose one person, May we choose a woman and a man not bringing a dessert at the same time? – Abdu Magdy Oct 21 '17 at 19:23
• >> May we choose a man ... same time? You certainly cannot choose that way. It's more like when you have "not bringing dessert", it can be either man or woman. Let's try this way. They should have said $P(\text{woman OR not bringing dessert})$. That way, you don't have mutually exclusive events. So by inclusion exclusion, you have $1/2 + 10/12 - 5/12$. I said this wrong in my answer. I shall edit it right away. – Abhiram Natarajan Oct 21 '17 at 20:41
• Sorry about the confusion. – Abhiram Natarajan Oct 21 '17 at 21:00


Negating “Zach blocks e-mails and texts from Jennifer”

I am reviewing some basic propositional logic. The question that I have come across that has given some confusion is Zach blocks e-mails and texts from Jennifer where I am asked to find the negation of this proposition. Here is what they provided as an answer in the book: Zach does not block e-mails from Jennifer, or he does not block texts from Jennifer.

Why couldn't the answer simply be It is not true that Zach blocks e-mails and texts from Jennifer? Why did they have to introduce the disjunction or?

-
Of course your answer is fine, indeed clearer. They presumably want to test whether you know how to negate a conjunction and obtain a disjunction. But if this is a test question, surely you should not be expected to be a mind reader. –  André Nicolas Dec 11 '12 at 2:00

What you gave is one possible answer, the book gives another which is logically equivalent, although slightly cleaner statement since it is further reduced. They are the same because of one of De-Morgan's laws; $\neg(P \land Q) \equiv (\neg P) \lor (\neg Q)$

-

Zach blocks emails and texts from Jennifer means Zach blocks emails from Jennifer and Zach blocks texts from Jennifer. Negating this gives Zach does not block emails from Jennifer or Zach does not block texts from Jennifer.

It is true that the negation is also It is not true that Zach blocks emails and texts from Jennifer but the point of the exercise is to make one rewrite it in a useful way.

Note that in general, (not (A and B)) is logically equivalent to ((not A) or (not B)).

-
Or does "Zach blocks emails and texts from Jennifer" mean "Zach blocks emails and Zach blocks texts from Jennifer"? Don't you love the ambiguities of natural language? (Of course, the translation you give is almost certainly the intended meaning.) –  Code-Guru Dec 11 '12 at 4:11
• $P(x, y)$: x blocks emails from y
• $Q(x, y)$: x blocks texts from y
• $z$: Zach
• $j$: Jennifer

$P(z, j)$: Zach blocks emails from Jennifer;
$Q(z, j)$: Zach blocks texts from Jennifer.

You essentially negated: $P(z, j) \land Q(z, j)$.

So did the text.

The text applied "distribution of negation over conjunction" (one of DeMorgan's Laws):

• "It is not the case that $[(P(z, j)$ and $Q(z, j)]$" $$\iff \lnot[P(z, j) \land Q(z, j)]$$ $$\iff \lnot P(z, j) \lor \lnot Q(z, j)\tag{DeMorgan's law}$$
• which is to say: not $P(z, j)$, or, not $Q(z, j)$.
You are correct with your statement that the sentence translates to $\;\;\lnot[P(z,j) \land Q(z, j)]$;
It's logically equivalent to the text's answer: $\;\lnot P(z, j) \lor \lnot Q(z, j)$.