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# Boxing Pythagoras

## More on 0.999…=1

In my last post, I discussed a particular video which I found to be more than a bit misleading. The discussion centered around a simple, but extremely counterintuitive notion of mathematics: the fact that the number 0.999…, or zero-point-nine-repeating, is equal to 1.

Well, as I mentioned, the very counterintuitive nature of the result led at least one of my readers to question its validity. As such, I thought I would lay out one proof of this concept, in order to make it easier for those who do not accept the result to pinpoint exactly where they disagree. I’ll break my proof down into numbered steps, to ease in that venture.

## (1) Definition of 0.999…

By the symbol, 0.999…, I mean to say an infinite decimal expansion in which all digits to the right of the decimal place are 9’s. Mathematically, we can express this as:

$0.999... \equiv \sum_{n=1}^{\infty} 9\times (\frac{1}{10})^{n}$

## (2) Partial Sums

Those of you who remember your Calculus might immediately recognize this Summation as a textbook example of a convergent geometric series. However, for those who do not, let’s work through the steps to determine the limit of this expression.

Provided the series converges, we say that the value of the summation is equal to that series’ limit.

(a) $\sum_{n=1}^{\infty} 9\times (\frac{1}{10})^{n}=\lim_{k\rightarrow \infty} \sum_{n=1}^{k} 9\times (\frac{1}{10})^{n}$

Similarly, if convergent, the limit of the series is equal to the limit of the partial sums of the series. In general, the nth partial sum of our series can be seen to be:

(b) $s_n=1-(\frac{1}{10})^n$

So, as long as the series converges, we can see that:

(c) $\sum_{n=1}^{\infty} 9\times (\frac{1}{10})^{n}=\lim_{k\rightarrow \infty} s_k$

## (3) Convergence

If the partial sums form a convergent sequence, then the whole series converges. A convergent sequence is one which has an existent, finite limit.

(a) $\lim_{k\rightarrow \infty} s_k=\lim_{k\rightarrow \infty} [1-(\frac{1}{10})^k]$

(b) $\lim_{k\rightarrow \infty} s_k=(\lim_{k\rightarrow \infty} 1) -[\lim_{k\rightarrow \infty} (\frac{1}{10})^k]$

(c) $\lim_{k\rightarrow \infty} s_k=1 -[\lim_{k\rightarrow \infty} (\frac{1}{10})^k]$

## (4) Limit

So, now, if $\lim_{k\rightarrow \infty} (\frac{1}{10})^k$ exists and is finite, then it follows that $\lim_{k\rightarrow \infty} s_k$ exists and is finite. And, if that is the case, then it is evident that our summation series from (1) converges.

We say that the limit of some function, $\lim_{k\rightarrow \infty} f(k)$, exists and is finite if, as k is made arbitrarily large, then $f(k)$ becomes arbitrarily close to some Real number, L.

So, our question now becomes, as is made any arbitrarily large value, does $(\frac{1}{10})^k$ become arbitrarily close to any single Real number? It’s fairly obvious that, the larger k we utilize, the closer $(\frac{1}{10})^k$ gets to 0. We can, in fact, make $(\frac{1}{10})^k$ as close to 0 as we want, simply by choosing a large enough value of k.

In general, it will always be true that:

(a) $0<(\frac{1}{10})^{k+1}<(\frac{1}{10})^k$

…for any Real number, k. Additionally, for any Real number r such that $r>0$, we can choose a value of k which would make it true that:

(b) $0<(\frac{1}{10})^k

And for all k, it is true that:

(c) $(\frac{1}{10})^k>0$

Therefore:

(d) $\lim_{k\rightarrow \infty} (\frac{1}{10})^k=0$

Since 0 is a finite, Real number, it is clear that the limit exists.

## (5) Evaluating the Summation Series

Now, we have everything we need to evaluate our initial summation.

(a) $0.999... \equiv \sum_{n=1}^{\infty} 9\times (\frac{1}{10})^{n}$

(b) $\sum_{n=1}^{\infty} 9\times (\frac{1}{10})^{n}=\lim_{k\rightarrow \infty} \sum_{n=1}^{k} 9\times (\frac{1}{10})^{n}$

(c) $\lim_{k\rightarrow \infty} \sum_{n=1}^{k} 9\times (\frac{1}{10})^{n}=\lim_{k\rightarrow \infty} [1-(\frac{1}{10})^k]$

(d) $\lim_{k\rightarrow \infty} [1-(\frac{1}{10})^k]=1-\lim_{k\rightarrow \infty} (\frac{1}{10})^k=1-0=1$

(e) $\sum_{n=1}^{\infty} 9\times (\frac{1}{10})^{n}=\lim_{k\rightarrow \infty} [1-(\frac{1}{10})^k]=1$

(f) $0.999... = 1$

## 18 thoughts on “More on 0.999…=1”

1. Just for fun I googled “definition of real number”. Pity for the high school students who go on to study math at college. This is what I got from the first page and a half:

Real numbers are numbers that can be found on the number line.
The Real Numbers did not have a name before Imaginary Numbers were thought of. They got called “Real” because they were not Imaginary. That is the actual answer!
A ‘real number’ is any element of the set R, which is the union of the set of a rational numbers and the set of irrational numbers.
A real number is a rational number or the limit of a sequence of rational numbers.

Seriously, this is worth a look:

I would be interested in a proof that every real number, as defined by Dedekind, has a constructable decimal expansion. The simple proof of uncountability depends on this.

Finally, and this is fundamentally important in math, what is written in symbols is only a representation of a number, and so one can probably live with the fact that 1.000000(rec) and 0.99999(rec) are two representaions of the multiplicative identity of the real number system.

• Jolly interesting stuff.
What would it mean for a decimal expansion to be constructible in this context? My immediate reaction would be to say that uncountability implies that most real numbers wouldn’t have a decimal expansion that is ‘constructible’, but it depends what you mean

• Following Dedekind’s definition of “The Cut”, it requires that every rational number can be assigned to one only of the two sets “above the cut” and “below the cut”, and this is all very well in theory, but only possible in practice if there is a finite description of the “Cut”, as in ” It is the square root of 2″, and hence a finite description of the decision algorithm. So in this case, and of course in many actual cases, the algorithm can allow the decision to be made: Take a rational number, square it, if more than 2 the number is in the “above” set. Now we are left with the problem of the countability of the set of all algorithms, and here I stop !!!!!

2. I can’t let go of this !
The difficulty with the simple proof is the subtraction, as it is only formally possible, so I thought of adding:
0.99999(rec) + 0.100000 = 1.099999(rec) a finite doable operation
0.99999(rec) + 0.010000 = 1.009999(rec) a finite doable operation
and each time the sum is nearer to 1
With a loose version of epsilon/delta I can say “Give me a number bigger than 1 and I can find a rational number which makes the above sum less than your number”
Or, in simpler terms, anything you add to 0.9999(rec) gives you a number bigger than 1
Hence 0.99999(rec) and 1.00000(rec) are representations of the unit 1
And also, if 3+4 = 7 then 0.99999(rec) = 1.00000(rec) = 1, with the commonly accepted meaning of the equals sign.

3. I looked into how real numbers were defined because I wanted to understand how a concept as intangible as ‘infinity’ could be used in a supposedly logic-based subject like mathematics.

Attempts at defining real numbers include declaring them as equivalence classes of Cauchy sequences of rational numbers, Dedekind cuts, or certain infinite “decimal representations”. These definitions all require the acceptance of ‘infinitely many’ iterations or occurrences.

But when mathematicians say “for n = 1 to infinity” or “infinitely many” what number type are they referring to (do they even know or think about it – I doubt it)?

It can’t be natural numbers or integers because these cannot “go to infinity”. By definition they can only take finite values. It can’t even be ‘reals’ because these do not allow infinity as a value.

So presumably they must be talking about something like the Hyperreal numbers or the Surreal numbers? Note that these are extensions to the reals, which they both take as already being well-defined.

So all definitions of ‘reals’ are based on the assumption that ‘reals’ are already well-defined. In short, real numbers are not well-defined at all.

This is a glaring problem in ‘Step(1)’ of the main article above, where the notion of “for n = 1 to infinity” has been used as if this is obviously an acceptable thing to use.

This comes as no surprise because mainstream mathematics cannot define ‘infinity’. It merely assumes it exists and treats any weird consequences as simply discovering the strange nature of infinity rather than indicating a mistake.

As for the simple proof of ‘10x – x’, it is scandalous that any mathematician should claim this is in any way valid (but they all still do). The flaw in the so-called proof is clearly shown in this video from 06:00 to 08:45 (minutes): https://www.youtube.com/watch?v=–HdatJwbQY .

In order to make the subtraction work, you have to ‘borrow’ an extra term from ‘infinity’ for one of the two endless series. If you accept infinity, then there are ‘infinitely many’ other ways you could ‘borrow’ different amounts of terms ‘from infinity’ and the subtraction would not cancel out completely (thus there are ‘infinitely many’ ways 0.999… does not =1).

4. Another big problem with the main article above is in step (2) where it is stated: “Provided the series converges, we say that the value of the summation is equal to that series’ limit”.

You can define the notion of a ‘limit’ and you can say the limit has a value of 1 and that this limit is associated with the series 0.999…, but you cannot simply assert that the value of the summation is equal to the limit. This is to simply assume what you set out to prove.

By the way, you can define 0.999… without using ‘infinitely many’; you can say it is the series where the sum to the n-th term is 1 minus (1 / (10 to the power n)).

This approach can be taken for all series, whether ‘converging’ or not. The ‘fixed part’ will equate to what you call the ‘limit’ for series that are said to ‘converge’. Here we have a property with a clear meaning that applies to any type of series and which does not require us to assert the series must equate to it.

• But when mathematicians say “for n = 1 to infinity” or “infinitely many” what number type are they referring to (do they even know or think about it – I doubt it)?

I can absolutely assure you that mathematicians have put an incredible amount of thought into this subject, including a great many volumes of erudite work and quite a bit of heated discussion. And they have done so for no less than 400 years.

When a mathematician says “for n=1 to infinity” or “as n goes to infinity,” they are not proposing that there is any time when the integer n will equal infinity– that would be entirely incoherent. Rather, these phrases act as a shorthand for “let us explore the trend in the change in value of a particular function as n is made arbitrarily larger with no upper bound.”

This comes as no surprise because mainstream mathematics cannot define ‘infinity’.

This is simply not true. I’ve given one definition of “infinity” above, when discussing infinity in the evolution of a function. When discussing “infinity” as a property of numbers, there is a very simple definition: for all real numbers r, an infinite number N is one such that 0<|r|<|N|.

As for the simple proof of ‘10x – x’, it is scandalous that any mathematician should claim this is in any way valid (but they all still do)…

In order to make the subtraction work, you have to ‘borrow’ an extra term from ‘infinity’ for one of the two endless series

Once again, Mr. Peny’s video is entirely incorrect. I find it exceedingly curious that you acknowledge the nigh-unanimous agreement of professional mathematicians on this subject, but you would instead trust the flawed analysis of a YouTube video produced by someone who is not a mathematician.

There is no “borrowing” of a term from infinity, any more than you “borrow” a zero from infinity when you say that 1*10=10.

You can define the notion of a ‘limit’ and you can say the limit has a value of 1 and that this limit is associated with the series 0.999…, but you cannot simply assert that the value of the summation is equal to the limit. This is to simply assume what you set out to prove.

It is not to simply assume what we set out to prove. Rather, it is to define the notion of equality for functions. What does it mean to say that f(x) is equal to g(x)? We are defining this notion in a consistent manner and showing that, based upon this definition, 0.999… is equal to 1.

By the way, you can define 0.999… without using ‘infinitely many’; you can say it is the series where the sum to the n-th term is 1 minus (1 / (10 to the power n)).

You could define it as such, but then you would not be talking about the same thing which mathematicians are talking about when they refer to 0.999…; mathematicians mean precisely what I wrote in (1), when they use the symbol 0.999… in a discussion. Your alternate definition, here, does not define a single number, but rather a whole set of numbers– one which, ironically, contains an infinite number of terms.

5. You said: “they are not proposing that there is any time when the integer n will equal infinity”.

And: “let us explore the trend in the change in value of a particular function as n is made arbitrarily larger with no upper bound”

But if n is an integer, the ‘trend’ is immaterial because we know that n cannot reach ‘infinity’ and so we know the value of the function (the sum to the n-th term of the series) cannot equal the ‘limit’ for any n. Thus your starting definition in (1) is false.

You said: “there is a very simple definition: for all real numbers r, an infinite number N is one such that 0<|r|<|N|.”

I could just as easily declare a Ponderer number, P is one such that 0<|r|<|N|<|P|
And because I am not showing you how I construct a P number, and because I claim that any counter-intuitive answers that result from using P are just down to the nature of P, you will be unable to prove that P is not a valid number. I will have denied you the ability to prove it is invalid through my vague definition.

You said: “Once again, Mr. Peny’s video is entirely incorrect.”

But once again, you fail to identify the flaw in his argument.

You said: “We are defining this notion in a consistent manner and showing that, based upon this definition, 0.999… is equal to 1”

If you think of 1 as being the series 1 +(zero times n), then the sum to the n-th term will always be 1. This allows the two series to be compared. The series for 1 is clearly different to the series where the sum to the n-th term is 1 minus (1 / (10 to the power n)).

You have chosen to compare two functions by asserting they must both equate to a single numeric value. You cannot prove the sum will somehow equate to the limit after ‘infinitely many’ iterations (which cannot be realised, in n is an integer). So you simply assert what you want to prove, that the sum equals the series limit.

You said: “You could define it as such, but then you would not be talking about the same thing which mathematicians are talking about when they refer to 0.999…”

But then why have mathematicians chosen to use a definition that makes no sense (for the reasons I have just explained) rather than this one that does?

• But if n is an integer, the ‘trend’ is immaterial because we know that n cannot reach ‘infinity’

I don’t see why you think this is relevant. I very clearly defined what I meant by “limit,” in the main article, and that definition does not require that n ever equal infinity. As I said, I completely agree that n cannot ever equal infinity, and that this would not even be a cogent statement. That’s completely irrelevant to what it means for a particular number to be the limit of a particular function.

I could just as easily declare a Ponderer number, P is one such that 0<|r|<|N|<|P|

You absolutely could proffer such a definition. It would then be incumbent upon you to show that such numbers are useful and can be discussed by some consistent mathematics– for example, the way that Abraham Robinson expounded upon the Hyperreal numbers back in the 1960’s.

But once again, you fail to identify the flaw in his argument.

The flaw comes around 8:05 in the video which you keep posting, in which Mr. Peny decides to simply omit the ellipsis denoting that the sequence is an infinite expansion, and then pretends that he’s discussing the same number which he had been before.

You cannot prove the sum will somehow equate to the limit after ‘infinitely many’ iterations (which cannot be realised, in n is an integer).

Nothing in my argument, in the original article, ever required that we perform infinitely many iterations of any iterative function, so I am not sure why you think this is at all relevant.

But then why have mathematicians chosen to use a definition that makes no sense (for the reasons I have just explained) rather than this one that does?

They haven’t. They’ve chosen to utilize a definition which is perfectly sensible and which is consistent with all the mathematics developed over the last several millennia of human scholarship.

• You said: “I completely agree that n cannot ever equal infinity, and that this would not even be a cogent statement. That’s completely irrelevant to what it means for a particular number to be the limit of a particular function”

Then which of these statements is wrong:

(1) After n terms have been added, we cannot add ‘infinitely many’ more terms because however far we extend n, it can only contain a finite value
(2) Any finite value for n will result in a sum that is less than 1
(3) As n cannot reach ‘infinity’, the sum cannot reach the value of 1
(4) Asserting that the sum must equal the limit based on the ‘trend’ directly contradicts (3)

And which of these statements is wrong:

(1) The sequence of positive non-zero terms (in 0.999…) is endless, meaning there is no last term
(2) As there is no ‘last term’, any attempt to calculate the sum as a single fixed value must fail

Or does this logic make perfect sense?
(1) Despite the notion of ‘endlessness’ indicating we cannot have a completed infinity, we will assume you can
(2) Despite not being able to prove the sum can have a fixed value, we will assume it can
(3) We will examine the trend as n increases and we will assert that the sum must equal the limit, and we will call this proof

You said: “Mr. Peny decides to simply omit the ellipsis denoting that the sequence is an infinite expansion, and then pretends that he’s discussing the same number”

Let me explain the flaw in another way to add clarity…

You start of with x = 0.999…, here we have ‘infinitely many’ terms
When we multiply by 10 to get 10x, we still have ‘infinitely many’ terms but each term has been multiplied by 10.
The illusion here is that we have created an extra term because the decimal representation is 9.999…, but we have not. This is easier to visualise if you think in terms of a series of individual terms as shown in the video. You have not created an extra term by multiplying by 10!
The 9.999… series needs to have 1 more term than the 0.999… series in order for the endless parts of 10x and x to match exactly (and thus cancel out) when performing the subtraction.
Now if we are allowed to borrow extra terms from infinity, then there are ‘infinitely many’ ways that different numbers of terms could be borrowed and where 0.999… would appear to not equate to 1 (using this same so-called proof).

• Then which of these statements is wrong:

(1) After n terms have been added, we cannot add ‘infinitely many’ more terms because however far we extend n, it can only contain a finite value
(2) Any finite value for n will result in a sum that is less than 1
(3) As n cannot reach ‘infinity’, the sum cannot reach the value of 1
(4) Asserting that the sum must equal the limit based on the ‘trend’ directly contradicts (3)

We’ll start with (1), which is wrong. The fact that the nth partial sum is finite does not imply that we cannot add infinitely many more terms. After that, (2) is correct. However, (3) is incorrect; once again, the definition of a limit is not dependent upon n reaching infinity. Since (3) is wrong, it then follows that (4) is also wrong.

And which of these statements is wrong:

(1) The sequence of positive non-zero terms (in 0.999…) is endless, meaning there is no last term
(2) As there is no ‘last term’, any attempt to calculate the sum as a single fixed value must fail

Here, statement (2) is wrong. The fact that there is no final term does not imply that “any attempt to calculate the sum as a single fixed value must fail.” I defined a method for calculating a single, fixed value for this series which does not fail, which is rigorous, and which is consistent with the rest of mathematics.

Or does this logic make perfect sense?
(1) Despite the notion of ‘endlessness’ indicating we cannot have a completed infinity, we will assume you can
(2) Despite not being able to prove the sum can have a fixed value, we will assume it can
(3) We will examine the trend as n increases and we will assert that the sum must equal the limit, and we will call this proof

Here, I see no reason to accept the premise of (1), that the concept of endlessness implies we cannot have a completed infinity. I disagree with (2), and have shown that we can prove the sum has a single, fixed, and meaningful value which is consistent and useful in the rest of mathematics. In (3) we are not asserting that the sum must equal the limit, we are defining what we mean when we say two functions are equal. Once again, this definition yields results which are entirely consistent with the rest of mathematics.

Let me explain the flaw in another way to add clarity…

You start of with x = 0.999…, here we have ‘infinitely many’ terms
When we multiply by 10 to get 10x, we still have ‘infinitely many’ terms but each term has been multiplied by 10.

I agree so far.

The illusion here is that we have created an extra term because the decimal representation is 9.999…, but we have not. This is easier to visualise if you think in terms of a series of individual terms as shown in the video. You have not created an extra term by multiplying by 10!

I’m confused– who has ever asserted that we have created an “extra term?” Assuming x is the series which I defined in (1) of the article, I completely agree that 10x has the same number of terms as x. This seems entirely irrelevant.

The 9.999… series needs to have 1 more term than the 0.999… series in order for the endless parts of 10x and x to match exactly (and thus cancel out) when performing the subtraction.

No, it doesn’t. Again, both series have exactly the same number of terms. Both 9.999… and 0.999… have a 9 for the digit in the tenths place. Both have a 9 for the digit in the hundredths place. Both have a 9 in the thousandths place. And, in general, both have a 9 in the 10^(-k) place. Since 9 minus 9 always equals zero, the 10^(-k) places from the two numbers will always cancel with one another. This doesn’t require that 9.999… have any more terms in its series than does 0.999…

• If we really give this some wider thinking we will see that 1.0000(rec)… is not the same thing as 1

• You said: “The fact that the nth partial sum is finite does not imply that we cannot add infinitely many more terms”

You agree that n is an integer and cannot become infinite, and you have defined your summation using n, and yet you still claim you can add an ‘infinite amount’ of terms. Can you really not see a problem?

Then you said; “(3) is incorrect; once again, the definition of a limit is not dependent upon n reaching infinity”

But point (3) does not mention a limit. It says “As n cannot reach ‘infinity’, the sum cannot reach the value of 1”. It says nothing about the definition of a limit being dependant on reaching infinity. You are protesting against something you have asserted it to be saying, not what it actually says.

You said: “I defined a method for calculating a single, fixed value for this series”

In my first two replies I pointed out why I believe you did not.

You said: “In (3) we are not asserting that the sum must equal the limit, we are defining what we mean when we say two functions are equal.”

Surely the most sensible way to compare a number with an endless series is to consider the number as an endless series [e.g. where the n-th sum is 1 plus (zero times n)] and then compare the n-th sums).
Rather than elevating the simpler object to the same structure as the more complicated object for comparison, you are deciding to extract a property from the more complicated object (the limit) that happens to be of the same type as the simpler object, and you are using that as the basis for your comparison.
I am not disputing that you believe it is better to consider the series as being equal to some fixed value, or that you have defined the way to do this is to assert that the sum equals the series limit.
You can argue that this is the accepted or ‘defined’ way to compare functions, but just saying something is defined in a certain way does not make it the best way of doing something.
I am not disputing that the limit is 1.
Therefore by asserting/defining that the sum equals the series limit, you have asserted/defined what you set out to prove.

You said: “I’m confused– who has ever asserted that we have created an “extra term?””
And later you said: “in general, both have a 9 in the 10^(-k) place”

But here you have created an extra term without even realising it.
If you start off with a series where the n-th term is n/(10 to the power n) then when you multiply the series by 10 you just get a series where the n-th term is 10n/(10 to the power n).
In the so-called proof, you have to compare the first (n+1) terms from the 10x series with the first n terms from the x series in order for the trailing parts to cancel each other out.

The same flawed logic can be used with diverging series. For example,

Series1 = 1+2+4+8+…
Series2 = 2 x Series1

Series1 = Series 2 – Series1
=
2+4+8+…
-1-2-4-8-…
= -1

Therefore 1+2+4+8+… = -1

And by subtracting -1 from both sides: 2+4+8+… = -2

But, if we are allowing this logic where we can line-up whatever terms we want to cancel out the trailing parts (which it appears we are), then
1+2+4+… = 1+(1+1)+(1+1+1+1)+…=1+1+1+…
And by the same logic
2+4+8+… = 1+1+1+…

Therefore
-1 = 1+1+1+… = -2

• Correction: I meant to say “by subtracting 1 from both sides” not -1 obviously!

6. I start with the reasonable assertion that 0.333(rec)… is the result of dividing 1 by 3, and 1 divided by 3 is one third, a well defined number.
So let us divide 2 by 2, weel it’s easier to see what is going on if we divide 20 by 2 using a valid but non standard variation of the “standard algorithm”.
Here goes:
Write 20 as 20.00000(rec)…
Then start the algorithm: – 2 goes into 20 9 times with remainder 2
Bring down the first decimal zero and get remainder 2.0
2 goes into 2.0 9 times, with remainder 0.2
…and so on…
At each step we get another 9 in the decimal expansion (quotient as a decimal), and so the process is unending, and gives 9.9999(rec)…

But we “know” that 20 divided by 2 is 10, so 9.9999(rec)… and 10 are representations of the same number.
One could start with 2 divided by 2 but this requires a bit more suspension of belief.

Clearly the problem lies with the general repeating decimal expansion process, as for 1/7 etcetera. 0.9999(rec) is not “special”.

7. You said: “I start with the reasonable assertion that 0.333(rec)… is the result of dividing 1 by 3, and 1 divided by 3 is one third”

Hold on a minute, this is far from reasonable.

After n digits have been processed by the decimal expansion of 1/3, you have 0.333…3 (i.e. a string of n ‘3’ digits) plus (1/3)(1/(10 to the power n)).
You need the last term for equality.
Unless you know how to change from a finite number of iterations into an infinite number of iterations, you can only ever process a finite number of digits.

So you are doing two very unreasonable things here.
First you are assuming that ‘infinitely many’ can somehow be achieved.
Secondly you are asserting that when this mysterious value is reached, the last expression (that I pointed out above) will somehow disappear.
You cannot prove any of these claims.

With your 20 divided by 2 example, after processing n digits you need to add an expression that takes account of your remainder in order to achieve equality.

So after two digits, 9.9 you would have to add 0.1. However far you go you will not achieve ‘infinitely many’ nines and have no remainder, at least not in any mathematically rigorous sense.

• It would seem that you do not accept the notion of an infinite decimal expansion at all, and so the discussion of the equality or not between 0.999(rec) and 1 is somewhat pointless.
No-one is arguing that 0.9999(as many as you like + the correct bit left over is not equal to 1.

Here’s a little verse I posted in August 2014

Infinity, a place beyond.

That most strange place, infinity,
Is somewhere I don’t want to be.
I’d rather stay with Brouwer
In his ivory tower.

8. I would gladly accept the notion of an infinite decimal expansion if it could be show how a value expressed in totally finite terms (like 1/3, or the square root of 2, or pi = limit of the circumference of a circle with diameter 1 etc.) can be transformed into a decimal containing ‘infinitely many’ digits.

I don’t need to see all the digits, I just want to know how the algorithm can end and achieve equality with the starting value. I fail to understand how anyone can accept the notion without this proof.

The great thing about trying to prove 0.999… equals 1 is that if it could be done, then it would clearly show how a finite object can transform into an infinite object with the same value. But sadly, this has never been proved; people just assume it to be true.

By saying you’d rather stay with Brouwer I assume you have some sympathy with his Intuitionism, in which all infinity is considered to be potential infinity (= a mathematical procedure in which there is an unending series of steps).

But you can still have endlessness without the use of expressions like ‘potential infinity’. Regardless of how precisely such terms are mathematically defined, they will always carry mysterious connotations which does not help matters. I believe the philosopher David Hume wanted all notions of infinity removed from mathematics. The words used in mathematics should help to provide clarity and assist understanding, not create confusion and hint at mysticism.


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On Simplicity of Reduced Group $C^\ast$-algebras

Botao Long  (ECNU)

13:00 pm to 14:00 pm, November 14th, 2016   Science Building A1510

Abstract:

A group is said to be $C^\ast$-simple if its reduced group $C^\ast$-algebra is simple. In this talk I will introduce some recent progress on $C^\ast$-simplicity and its relative properties (unique trace property, trivial amenable radical, etc.).

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# How to create a labeled incremental reference in MS Word?

I want to do cross-references manually in microsoft word 2010 (similarly to latex \label - \ref or \cite - \bibliography). I found that the Field function does almost excactly what I want (the syntax is a bit weird). If I type the following to the document (wave brackets are field marks produced by ctrl+f9):

{set dischargeRate {seq Figure}}Figure {ref dischargeRate}: Discharge rate vs. hole diameter. Figure is from Reference [{ref authorA}].

The results are shown in Figure {ref dischargeRate} and published previously in [{ref authorB}] and [{ref authorA}].

References:

{set authorA {seq cites}}[{ref authorA}] author, title, journal, year

{set authorB {seq cites}}[{ref authorB}] author, title, journal, year

the above produces:

Figure 1: Discharge rate vs. hole diameter. Figure is from Reference [1].

The results are shown in Figure 1 and published previously in [2] and [1].

References:

[1] author, title, journal, year

[2] author, title, journal, year

Is there a way to define increment and reference with one command instead of those three commands: set, seq and ref? Or how do I create a macro that does this for me. I am looking something like {setOrRef sequencename labelname} that shows a number i+1 that can be later referenced by {setOrRef sequencename labelname}.

Also there should be a check that labels are not redefined. For example: If the label does not exist, the sequence (Figure or cites) is incremented by one and that number is assigned to the label. If the label exists the existing number for the label is used.

I consider this as a programming question as it so close to macros and automating Ms Word.

-

If you want to use field codes and nothing else, I don't think there is a practical way to avoid using SEQ, SET and REF.

There are a number of difficulties detecting the existence/non-existence of bookmarks. I think you can sidestep them using the following set of nested fields, but I cannot say I have tried this "for real". Personally, I would try to avoid this kind of complexity in field coding.

All you need to do is insert the bookmark name you want once in the QUOTE field, i.e. by substituting the "bookmark" name you want instead of "bm". Here, I use "s" as the name that provides sequential reference numbers. It isn't actually a bookmark name, by the way, but a SEQ name.

As usual, all the {} have to be the special field code braces that you can insert using ctrl-F9 on Windows Word. You will still need to be careful about your naming of references, e.g. don't use "AuthorA" and "AuthorA1". You will need to avoid using any other SEQ names such as AuthorA2, and avoid using SEQ names s and s1 elsewhere.

{QUOTE {SET b bm}{SEQ "{b}{SEQ {b}}" \r{SEQ "s{={SEQ {b}}-3 \#"1'';''"}}" \h}{SEQ "{b}1" \c}}

Since it's not obvious how this works, I'll step through. Suppose you name your reference "AuthorA". Then you would insert

{QUOTE {SET b AuthorA}{SEQ "{b}{SEQ {b}}" \r{SEQ "s{={SEQ {b}}-3 \#"1'';''"}} \h}{SEQ "{b}1" \c}}

Bookmark "b" is set to "AuthorA"

{SEQ "{b}{SEQ {b}}"}

expands to

{SEQ "AuthorA{SEQ AuthorA}"}

In the first such set of fields for AuthorA , this expands to

{SEQ "AuthorA1" }

In subsequent sets, it expands to

{SEQ "AuthorA2" }

and so on.

The final field (the one actually inserted by the QUOTE)

{SEQ "{b}1" \c}

expands to

{SEQ "AuthorA1" \c}

i.e. it is always the value returned by the first set of fields in the document for AuthorA.

The value for AuthorA1 is set by

\r{SEQ "s{={SEQ {b}}-3 \#"1'';''"}}

{SEQ {b}}

expands to

{SEQ AuthorA}

which will actually be 2 in the very first set of fields for AuthorA (this is one of Word's SEQ field evaluation quirks). So

{={SEQ AuthorA}-3}

will be -1 and the numeric switch

\#"1'';''"

will return ''. So AuthorA1 is set to

\r{SEQ "s"}

i.e. the next value of the sequence s.

Subsequently,

{={SEQ AuthorA}-3}

is greater than 0, the numeric switch returns '1' and so AuthorA2 will be set to

\r{SEQ "s1"}

But we don't care what AuthorA2 is set to. The purpose of this bit of numeric formatting is to ensure that { SEQ s } only increments the first time time we insert one of these sets of fields for a particular "reference name".

For inserting such stuff, you could create an autotext/building block that would insert the block. All you would need to do is change "bm". A piece of VBA to prompt for a bookmark name and insert the same block would not be hard, but I leave you to look around for that.

As always, even using the general technique suggested, there may be a way to simplify the fields. Personally, I think I would go for a design that let me insert some kind of placeholder and required me to run some "pre-publishing VBA" to set them up correctly.

-


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# Help with calculating the product of cosets of a polynomial ring.

Description of Ring and Ideal

First I'm sorry this is my first post and I don't how to entirely format math equations yet, but I'm trying to figure it out.

But I am having a difficult time understanding this specific case of a quotient ring. I know that generally speaking the product of elements of a quotient ring are

$${(a + I)(b + I) = (ab + I)}$$

where I is an ideal in the ring.

So for my specific question in the quotient ring R[x]/((x^2)-4) the problem states that each coset has the unique form $${a + bx + (x^2-4)}$$ given that a and b are real numbers. The problem then asks to express the product of two of these cosets

$${[a + bx + (x^2-4)] * [c + dx + (x^2-4)]}$$

in the same form, that is, to find the product in the form $${e + fx + (x^2-4)}$$ and write out what e and f are in terms of a,b,c, and d.

What I have so far is this:

I'm essentially taking $${(a + I)(b + I) = (ab + I)}$$ to be

$${(a + bx + (x^2-4))(c + dx + (x^2-4))}$$

and then

$${(a + bx)(c + dx) = ac + x(ad + cb) + bdx^2}$$

But from my understanding of how the quotient ring works is that since the principal ideal is $${(x^2-4)}$$ then that leads to an additional condition that we have $${x^2-4 = 0}$$ which implies $${x^2 = 4}$$. So then my equation above becomes

$${ac + x(ad + cb) + bdx^2 = ac + x(ad + cb) + 4bd}$$

and so in the form the question is asking me it would then be that for the product of the two cosets in terms of a,b,c, and d we have that

$${e + fx + (x^2-4)}$$ $${e = 4bd + ac}$$ $${f = ad + cb}$$

Is this the correct way of going about multiplying two elements of this quotient ring? Or am I entirely missing the process here?

• What do you doubt about your work? – Bill Dubuque Dec 7 '20 at 9:24
• Looks good to me. – Gerry Myerson Dec 7 '20 at 9:37
• I was both unsure of only taking the first two expressions and leaving out the (x^2-4) and having the condition of x^2 = 4. I have been having an incredibly difficult time understand abstract algebra and wrapping my head around the concepts. – Z Ham Dec 7 '20 at 19:08

You can't do $$x^2=4$$, only “sort of”. What you do know is that $$x^2-4\in(x^2-4)=I$$
and therefore $$x^2+I=4+I$$.
So when you multiply $$(a+bx+I)(c+dx+I)=ac+(ad+bc)x+bdx^2+I$$ you can use $$bdx^2+I=4bd+I$$ and you end up with $$(a+bx+I)(c+dx+I)=(ac+4bd)+(ad+bc)x+I$$


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# The 16th International Workshop on Tau Lepton Physics (TAU2021) (Virtual Edition)

September 27, 2021 to October 1, 2021
Indiana University
America/Indiana/Indianapolis timezone

## Semi- and di-tauonic B decays and LFUV at future $e^+e^-$ colliders

Oct 1, 2021, 10:50 AM
2h
Virtual (Indiana University)

### Virtual

#### Indiana University

Poster contribution Tau2021 Abstracts

### Speaker

Lingfeng Li (Brown U.)

### Description

$b\to s\tau^+\tau^-$ measurements are highly motivated for addressing lepton-flavor-universality (LFU)-violating puzzles such as $R_{K^{(\ast)}}$ anomalies. The anomalies of $R_{D^{(*)}}$ and $R_{J/\psi}$ further strengthen their necessity and importance, given that the LFU-violating hints from both involve the third-generation leptons directly. $Z$ factories at the future $e^-e^+$ colliders stand at a great position to conduct such measurements because of their relatively high production rates and reconstruction efficiencies for $B$ mesons at the $Z$ pole. To fully explore this potential, we pursue a dedicated sensitivity study in four $b\to s\tau^+\tau^-$ benchmark channels, namely $B^0\to K^{\ast 0} \tau^+ \tau^-$, $B_s\to\phi \tau^+ \tau^-$, $B^+ \to K^+ \tau^+ \tau^-$ and $B_s \to \tau^+ \tau^-$, at the future $Z$ factories. We develop a fully tracker-based scheme for reconstructing the signal $B$ mesons and introduce a semi-quantitative method for estimating their major backgrounds. The simulations indicate that branching ratios of the first three channels can be measured with a precision $\sim \mathcal O(10^{-7} - 10^{-6})$ and that of $B_s \to \tau^+ \tau^-$ with a precision $\sim \mathcal O(10^{-5})$ at Tera-$Z$. The impacts of luminosity and tracker resolution on the expected sensitivities are explored. The interpretations of these results in effective field theory are also presented.

### Primary author

Lingfeng Li (Brown U.)


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Multiple linear regression with restricted ranges in R

I'm trying to do a simple least-squares fit on some data. The formula is simply:

$Y = C_0X_1 + C_0X_2 + C_0X_3 + C_0X_4 + C_0X_5$

I have 24 rows of $Y$'s and $X$'s, and I'm trying to fit the $C$'s.

The problem, however, is that I need to constrain the ranges of the $C$'s to be non-zero, regardless of the quality of the fit. I've read the docs on lm in R, but I cannot find the option to constrain the variable ranges. Any ideas?

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What is the reason to constrain the ranges? If it is just a technical question you should maybe ask on stackoverflow, since this is more of a stats theoretical site. If the reason for constraint is a inheret to your model, I'd advise you to consider a (semi)-logarithmic model depending on what you are doing. Cutting out data point entirely tends to be a really bad idea for inference. – IMA Jan 21 at 8:07
I'm trying to fit some biological data, and need to constrain the coefficients such that they are not nonsensical... – user19869 Jan 21 at 8:17
Oh I completely misunderstood you then. Sorry. May I ask: Is the linear model necessarily appropriate? What results do you get when estimating right now? – IMA Jan 21 at 8:21
no problem....linear model's a start...right now I get negative values for one of the coefficients, which I can't have. – user19869 Jan 21 at 8:26
I fixed up the formatting, but do you really mean to have $C_0$ five times? This would be most unusual; I think you want $C_1$, $C_2$ etc. Also, you didn't include an intercept; was that deliberate? – Peter Flom Jan 21 at 11:48
show 1 more comment

I had the same problem and found two solutions.

1) First option is that you need to obtain the likelihood by hand and then optimize in R by using the optim command. When you use this command, you need to specify starting values for the parameters to be estimated. As far as I know, R uses 0 for all the coefficients. You need to try different initial values. Of course, you will highly unlikely get the positives coefficients if you determine these initial values by hand. My suggestion is to generate random numbers using rnorm (or some other distribution) for the initials values and then use repeat until break command. You need to break the repetition when all of the coefficients are positive. Then, you need to save these numbers so that you no longer need to run the repeat until break command. For more info, see http://www.r-bloggers.com/logistic-regression/

Another solution would be to use a Bayesian approach. If you specify uniform priors whose lower limits are 0 for your coefficients, you will get positive coefficients. You can conduct this analysis in WinBugs in minutes. Here is an example:

model{
# model’s likelihood
for (i in 1:n){
time[i] ~ dnorm( mu[i], tau ) # stochastic componenent
mu[i] <- beta1 * cases[i] + beta2 * distance[i]
}
# prior distributions
tau ~ dgamma( 0.01, 0.01 )
beta1 ~ dunif( 0, 1000)
beta2 ~ dunif( 0, 1000)
# definition of sigma
s2<-1/tau
s <-sqrt(s2)
}

-

The issue could be not including an intercept, as mentioned above. You can add another $X_i$ and set it to 1 throughout if you need it in variable form.
Not including an intercept can make your results invalid if the mean of $Y$ truly does deviate from 0.

If that is not the case then I strongly suspect misspecification to be the issue here. There are two caveats that are of special interest, but I don't know on what level your knowledge is so I will keep this brief initially since it is pretty basic stuff.

First: Missing variable bias. Any variable with a strong influence (or in this case, correlation) on the dependent variable which is also correlated with the exogenous variable in question can not be left out - otherwise your estimator $C_i$ will be invalid.
This could explain an unexpectedly negative $C_i$ if there is another variable which you left out of the model.
In this case one would suspect a variable that influences $Y$ negatively, yet is correlated with your $C_i$. Its influence will be attributed to $C_i$, maybe pulling it just into negative range.

Second: The underlying effect is not linear.
I would like you to plot the exogenous variables against the endogenous for a first hint. Then please go ahead and run a RESET-Test. The R function is this here
Please don't do this by hand.

This tests your model against several more general Taylor approximations. If the relationship is not linear, this could be a good indicator. The test will generally "fail" your model if those other models do a better job at explaining the variation of $Y$ for whatever reason.

Of course in the end any estimation problem which makes your estimates biased could be the problem. In fact the RESET test may trigger because of other things than just the linearity of the model - but I suspect those two.

Edit: Because there was another approach posted above let me reiterate that you do NOT want to use some quick fix to force your coefficients into beeing positive. All this does is make your inference nonsense.
Using this model, especially using lm, there is a REASON why that coefficient is estimated as negative. If that can not be in reality, then it is almost certain that the whole model is errornous. Switching to a different method or trying to fix the "error" until you reach an acceptable value is very dangerous - statistical inference absolutely mandates that if you get bogus results then you have to take a step back, not sideways.

Also consider that you literally have 19 or 18 degrees of freedom on this regression. To use OLS you HAVE to have the complete set of small-samples assumptions in check, otherwise it comes as no surprise your regression goes astray.

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 Some negative coeff. estimates? Maybe some of the X's are highly correlated and a multicollinearity problem shows up. Try PCA or exclude some of the highly correlated variables - if any. – Pantera Jan 21 at 22:53


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# Jacobian row of zeros. Why?

I have a Simulink model that has a standard Simulink block that determines the Jacobian for a given Rigid Body Tree model. The Rigid Body Tree model is a built in model of the Franka Emika Panda robot and is loaded into the workspace using:  panda_matlab = loadrobot("frankaEmikaPanda").

The input to the block consists of the robots configuration (config), which is determined by the joint angles $$q$$. In this case all the joint angles are $$0$$. This means that the robot is in an upright position and joints 3 and 5 depend linearly on joint 1 Which means that columns 1, 3 and 5 of the Jacobian should be identical. This is indeed the case, However the Jacobian also has one row of zeros. I do not understand where this row of zeros come from.

I also used my own, simplified Panda model with fewer links. In the same upright position with, the same linearly dependent links. The jacobian matrix of this model is shown below. This also has a row of zeros in the Jacobian.

What is causing the jacobian to have this row of zeros?


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# Evaluation and Benchmarking¶

The evaluation of Community Discovery algorithms is not an easy task. cdlib implements two families of evaluation strategies:

• Internal evaluation through fitness scores;
• External evaluation through partitions comparison.

Moreover, cdlib integrates both standard synthetic network benchmarks and real networks with annotated ground truths, thus allowing for testing identified communities against ground-truths.

Finally, cdlib also provides a way to rank clustering results generated by a set of algorithms over a given input graph.

Note

The following lists are aligned to CD evaluation methods available in the GitHub main branch of cdlib.

## Internal Evaluation: Fitness scores¶

Fitness functions allows to summarize the characteristics of a computed set of communities. cdlib implements the following quality scores:

 avg_distance(graph, communities, **kwargs) Average distance. avg_embeddedness(graph, communities, **kwargs) Average embeddedness of nodes within the community. average_internal_degree(graph, community, …) The average internal degree of the community set. avg_transitivity(graph, communities, **kwargs) Average transitivity. conductance(graph, community, summary) Fraction of total edge volume that points outside the community. cut_ratio(graph, community, summary) Fraction of existing edges (out of all possible edges) leaving the community. edges_inside(graph, community, summary) Number of edges internal to the community. expansion(graph, community, summary) Number of edges per community node that point outside the cluster. fraction_over_median_degree(graph, …) Fraction of community nodes of having internal degree higher than the median degree value. hub_dominance(graph, communities, **kwargs) Hub dominance. internal_edge_density(graph, community, summary) The internal density of the community set. normalized_cut(graph, community, summary) Normalized variant of the Cut-Ratio max_odf(graph, community, summary) Maximum fraction of edges of a node of a community that point outside the community itself. avg_odf(graph, community, summary) Average fraction of edges of a node of a community that point outside the community itself. flake_odf(graph, community, summary) Fraction of nodes in S that have fewer edges pointing inside than to the outside of the community. scaled_density(graph, communities, **kwargs) Scaled density. significance(graph, communities, **kwargs) Significance estimates how likely a partition of dense communities appear in a random graph. size(graph, communities, **kwargs) Size is the number of nodes in the community surprise(graph, communities, **kwargs) Surprise is statistical approach proposes a quality metric assuming that edges between vertices emerge randomly according to a hyper-geometric distribution. triangle_participation_ratio(graph, …) Fraction of community nodes that belong to a triad. purity(communities) Purity is the product of the frequencies of the most frequent labels carried by the nodes within the communities

Among the fitness function a well-defined family of measures is the Modularity-based one:

 erdos_renyi_modularity(graph, communities, …) Erdos-Renyi modularity is a variation of the Newman-Girvan one. link_modularity(graph, communities, **kwargs) Quality function designed for directed graphs with overlapping communities. modularity_density(graph, communities, lmbd, …) The modularity density is one of several propositions that envisioned to palliate the resolution limit issue of modularity based measures. modularity_overlap(graph, communities, weight) Determines the Overlapping Modularity of a partition C on a graph G. newman_girvan_modularity(graph, communities, …) Difference the fraction of intra community edges of a partition with the expected number of such edges if distributed according to a null model. z_modularity(graph, communities, **kwargs) Z-modularity is another variant of the standard modularity proposed to avoid the resolution limit.

Some measures will return an instance of FitnessResult that takes together min/max/mean/std values of the computed index.

 FitnessResult(min, max, score, std)

## External Evaluation: Partition Comparisons¶

It is often useful to compare different graph partition to assess their resemblance. cdlib implements the following partition comparisons scores:

 adjusted_mutual_information(first_partition, …) Adjusted Mutual Information between two clusterings. adjusted_rand_index(first_partition, …) Rand index adjusted for chance. f1(first_partition, second_partition) Compute the average F1 score of the optimal algorithms matches among the partitions in input. nf1(first_partition, second_partition) Compute the Normalized F1 score of the optimal algorithms matches among the partitions in input. normalized_mutual_information(…) Normalized Mutual Information between two clusterings. omega(first_partition, second_partition) Index of resemblance for overlapping, complete coverage, network clusterings. overlapping_normalized_mutual_information_LFK(…) Overlapping Normalized Mutual Information between two clusterings. overlapping_normalized_mutual_information_MGH(…) Overlapping Normalized Mutual Information between two clusterings. variation_of_information(first_partition, …) Variation of Information among two nodes partitions.

Some measures will return an instance of MatchingResult that takes together mean and standard deviation values of the computed index.

 MatchingResult(score, std)

## Synthetic Benchmarks¶

External evaluation scores can be fruitfully used to compare alternative clusterings of the same network, but also to asses to what extent an identified node clustering matches a known ground truth partition.

To facilitate such standard evaluation task, cdlib exposes a set of standard synthetic network generators providing topological community ground truth annotations.

In particular, cdlib make available benchmarks for:

• static community discovery;
• dynamic community discovery;
• feature-rich (i.e., node-attributed) community discovery.

All details can be found in the dedicated page.

## Networks With Annotated Communities¶

Although evaluating a topological partition against an annotated “semantic” one is not among the safest path to follow [Peel17], cdlib natively integrates well-known medium-size network datasets with ground-truth communities.

Due to the non-negligible sizes of such datasets, we designed a simple API to gather them from a dedicated remote repository transparently.

All details on remote datasets can be found on the dedicated page.

## Ranking Algorithms¶

Once a set of alternative clusterings have been extracted from a given network, is there a way to select the best one given a set of target fitness functions?

cdlib exposes a few standard techniques to address such an issue: all details can be found in the dedicated documentation page.

 [Peel17] Peel, Leto, Daniel B. Larremore, and Aaron Clauset. “The ground truth about metadata and community detection in networks.” Science advances 3.5 (2017): e1602548.


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# Discovering $2D$ to $3D$ projection

I have a two dimensional plane with $U, V$ coordinates from $0,0$ to $1.0,1.0$.

The $U$ coordinate is projected to $\mathbb{R}^3$ and primarily contributes to an $X$ coordinate. However, the rate of change along $U$ is not uniform to the changes in $X$. The same is true for the $V$ coordinate in the plane and it's relation to the $Y$ coordinate in my $3D$ space.

More concretely I have a $2D$ plane of $UV$ coordinates and I am calculating the way light refracts through an object. In the simplest case the refraction indices of my two mediums would be equal, and all the refraction vectors for each $UV$ coordinate would simply be $0,0,-1$, or straight lines through the plane. Essentially what you see when you look out a thing glass window.

In a more complicated case, like with a sphere of some other material the refraction vectors differ. I can calculate my data points but I would like to derive a function of the following form:

$$f(u,v) = \overrightarrow{V_r} = (x,y,z)$$

Assuming I have a non-uniform set of values for $x, y$, and $z$ in relation to the $(u,v)$ coordinates how can I go about discovering the function to describe the $x,y,z$ values.

I hope this is clear, let me know if other details are necessary.

Edit: Let me try a different likely more meaningful explanation.

I have a $2$ dimensional surface. Each point on the surface corresponds to a $3$ dimensional vector. The surface is square and has length and width equal to $1$; though this point might actually be irrelevant. Let's start with a simple case, in which the surface points mapped to the $3D$ vectors are symmetric along both the $X$ and $Y$ axes. I don't have the function that maps my $2D$ surface point to the corresponding $3D$ vectors, that is what I want to find, but in a simple case the mapping is 1:1. I can calculate, through a series of long, complicated and resource heavy calculations the $3D$ vector for any point in my $2D$ surface. I have done that for $1024*1024\approx1,000,000$ points. So I have a million $3D$ vectors. How can I go about deriving an approximation function that will give me the $3D$ vector given some $2D$ point in my surface.

Image I have the surface (not the one shown in the image), I want to derive the function of the surface in terms of the horizontal and axial axes. As opposed to the image, the Horizontal and axial values may not equal the "$X$,$Y$" coordinates of my surface. That is I am not only looking for "height", that we'll call the $Z$ axis.

-
Hi, I'm not sure what you mean by coordinates of a plane, are they the coordinates of the generating vectors, or all points on the plane? Also your objective is not clear, do you want to derive a refraction law? I think it would be helpful to rewrite the question, adding some definitions and clear mathematical statements. –  Artur Gower Sep 10 '13 at 13:38


text:
The Remainder Term and Error Estimation

If the Taylor series for a function is truncated at the term, what is the difference between and the value given by the Taylor polynomial? That is, what is the error involved in using the Taylor polynomial to approximate the function?

Suppose you expand f around c, and that f is -times continuously differentiable on an open interval containing c. Taylor's Theorem with the Remainder Term says that if x is another point in this interval, then

where z is a number in the open interval between x and c.

is the degree Taylor polynomial at c. The other term on the right is called the Lagrange remainder term:

The appearance of z, a point between x and c, and the fact that it's being plugged into a derivative suggest that there is a connection between this result and the Mean Value Theorem. In fact, for the result says

where z is between x and c. This is the Mean Value Theorem.

On the one hand, this reflects the fact that Taylor's theorem is proved using a generalization of the Mean Value Theorem. On the other hand, this shows that you can regard a Taylor expansion as an extension of the Mean Value Theorem.

There is also an expression for the error which involves an integral. I won't discuss it here.

Example. Compute the Remainder Term for .

For the third remainder term, I need the fourth derivative:

The Remainder Term is

where z is a number between x and 1.

Example. Compute the Remainder Term for .

Since I want the Remainder Term, I need to find an expression for the derivative. I'll compute derivative until I see a pattern:

Notice that it's easier to see the pattern if you don't multiply out the power of 4.

Thus,

The Remainder Term is

where z is a number between x and 3.

There are several things you might do with the Remainder Term:

1. Estimate the error in using to estimate on a given interval . (The interval and the degree n are fixed; you want to find the error.)

2. Find the smallest value of n for which approximates to within a given error ("tolerance") on a given interval . (The interval and the error are fixed; you want to find the degree.)

3. Find the largest interval on which approximates to within a given error ("tolerance"). (The degree and the error are fixed; you want to find the interval.)

Example. The Maclaurin series for is

What is the largest error which might result from using the first three terms of the series to approximate , if ?

The remainder term is

where . I want to estimate the maximum size of . I take absolute values, because I don't care whether the error is positive or negative, only how large it is.

, and you can check by taking derivatives that . Thus, . So

Since I want the largest possible error, I want to see how large the terms and could be.

Remember that z is between 0 and x, and . So

First, means that

How large can be, given that ? As z goes from 0 to 1, decreases, so it is largest if . This means that

In general, to estimate the z-term you'd have to find the absolute max on the interval for z. If you know that the z-term is either increasing or decreasing, you can check its value at the interval endpoints, and take the largest.

Using the estimates for and , I have

The error is no greater than .

I can check this by plotting the difference between the degree Taylor polynomial and .

From the picture, it looks as though the maximum error is around 0.15 (in absolute value). The estimated error was pretty conservative.

Example. (a) Compute for , and express using and the remainder term.

Since I want , I need the fourth derivative:

Thus,

Now

Therefore,

where z is between 0 and x.

(b) Use to approximate the largest error that occurs in using to approximate for .

I have

I'll estimate the z and x-terms one at a time.

Since , I have

Since and z is between 0 and x, it follows that . On this interval, decreases, so it attains its largest value at . Therefore,

Thus,

The error is no greater than .

Example. Find the smallest value of n for which the degree Taylor series for at approximates on the interval with an error no greater than .

Notice that

So

First, I'll estimate how large the z and x-terms can be. Since and since is an increasing function, I have

Since and is an increasing function, I have

Thus,

Therefore, I want the smallest n for which

I can't solve this inequality algebraically, so I'll have to use trial-and-error:

The smallest value of n is .

You can also use the Remainder Term to estimate the error in using a Taylor polynomial to approximate an integral.

Example. Calvin wants to impress Phoebe Small by using the MacLaurin series for to approximate to within 0.0001. How many terms of the series should he use?

The Maclaurin series for is

(Substitute in the standard series for .) I want to know how many terms of the series to use to approximate the integral.

Since , , , and in general, . Therefore,

In the integral, x goes from 0 to 0.5, and z is a number between 0 (the expansion point) and x. Therefore, I know that z is a number between 0 and 0.5. Taking the worst possible case, the largest could be is . Replace with e to obtain

Insert this into the integral (remembering to multiply by x):

I want the smallest value of n for which this ugly mess is less than 0.0001. The easiest way to do this is by trial: Plug in successive values of n until you discover that is the smallest value that works.


text:
## tmagloire1 one year ago A box is to be constructed from a sheet of cardboard that is 20 cm by 50 cm by cutting out squares of length x by x from each corner and bending up the sides. What is the maximum volume this box could have? (Round your answer to two decimal places. Do not include units, for example, 10.22 cm would be 10.22.)

1. SolomonZelman

calc I prob optimization, correct?

2. SolomonZelman

|dw:1437076558138:dw|

3. tmagloire1

Yep optimization

4. tmagloire1

I got 2029.95 but that isn't correct

5. SolomonZelman

|dw:1437076685341:dw|

6. tmagloire1

Yep that's how i set it up

7. SolomonZelman

do you know why I subtracted 2xfrom 20 and 50?

8. SolomonZelman

So you do know why we subtract 2x from 20 and 50, right/

9. tmagloire1

Yes i understand

10. SolomonZelman

So, now, we will create the function for volume. Find critical numbers Find the maximum of it (with limits we will later provide/make)

11. SolomonZelman

The volume would be height•lenght•width

12. tmagloire1

I got 4.468

13. SolomonZelman

and in this case v(x)=2x•(20-2x)•(50-2x)

14. SolomonZelman

where V stands for volume

15. tmagloire1

Why would it be 2x instead of just x

16. SolomonZelman

u need v(x)

17. SolomonZelman

Why would it be 2x instead of just x, because you bend a cut-out of length x from both sides, don't you?

18. SolomonZelman

|dw:1437076945620:dw|

19. tmagloire1

ohh okay

20. SolomonZelman

so the side 20 and side 50 is decreased by 2x. Got that, good:)

21. tmagloire1

So now take the derivative of the function we got?

22. SolomonZelman

now we will do a regular process once we got the function V(x)=2x•(20-2x)•(50-2x)

23. SolomonZelman

I would perhaps simplify it a little instead of doing a product twice, but you can find the derivative as you wish:)

24. SolomonZelman

25. SolomonZelman

V(x)=x•(20-2x)•(50-2x) ^ ^ just x, not 2x.

26. SolomonZelman

but 20-2x and 50-2x are correct measures

27. tmagloire1

Okay so take the derivative of that then?

28. SolomonZelman

V(x)=x•(20-2x)•(50-2x) V(x)=x•(1000-40x-100x+4x²) V(x)=x•(1000-140x+4x²) I would do derivative, product now

29. SolomonZelman

V$$\rm ~'$$(x) = 1•(1000-140x+4x²)+x•(0-140+8x)

30. tmagloire1

So it will be 12x^2-280x-1000?

31. SolomonZelman

I would use wolfram, actually, because I am the laziest person on this planet.

32. tmagloire1

Hahaha i just used Wolfram for it because i didn't feel like taking the derivative myself xD

33. SolomonZelman

V'(x)=4(3x²-70x+250) by wolfram.... I did it like this because wolfram can give a handy version faster

34. SolomonZelman

set V'(x)=0 to find your critical number

35. SolomonZelman

numbers*

36. SolomonZelman

remember tho, that 0>x>10 or else your volume is going to be =0 or negative.

37. tmagloire1

I got 11.66 :/

38. SolomonZelman

nope

39. SolomonZelman

that wouldn't make sense, because you get a negative volume if you plug it in our volume function.

40. SolomonZelman

As again 0<x<10

41. SolomonZelman

find the roots of 4(3x²-70x+250)=0

42. tmagloire1

26.48 and -3.146 which still doesn't make sense right?

43. SolomonZelman

And I will tell you will get the ±, and only take the minu one, because the one with the + is greater than 10

44. SolomonZelman

what crtical number(s) did you get?

45. tmagloire1

I am getting -60 now :// sos

46. SolomonZelman

|dw:1437077863587:dw|

47. SolomonZelman

take the -, because + will get x>10

48. SolomonZelman

35-5√19 ≈ 4.4

49. tmagloire1

Oh I must have been doing some crazy math error when taking the crit #

50. tmagloire1

so take 4.4 and use it in the equation at the beginning?

51. SolomonZelman

yes,

52. tmagloire1

i got -2000 ._.

53. SolomonZelman

you did something wrong again-:(

54. tmagloire1

2030.3*

55. SolomonZelman

Yup, that is right

56. SolomonZelman

2030.336

57. tmagloire1

Oh okay! Thank you so much of all the help. I think I understand it now!!

58. SolomonZelman

Yes.... overview. 1. We created a function V(x)=x • (20-2x) • (50-2x) 2. Constraint x>0, because otherwise volume is negative or 0 (based on the first, x, component of V(x)). x<10, because otherwise (based on the 20-2x part), the volume would be 0 or negative. 3. We differentiated the function to get 2 critical numbers (that is two solutions of V'(x)=0) NOTE: Critical numbers (usually) are also values for which v'(x) is undefined, if defined for initial function v(x) - (but no such values in this case). Also, critical numbers usually can be closed interval boundaries, but our interval is open (we didn't include 0 or 10), so that is not our case. So in our case only solutions to v(x)=0 are critical numbers. 4. We excluded the (35+5√19)/3 solution based on our constraint (because it is greater than 10), and rather used the only solution remaining, i.e. (35-5√19)/3. 5. We got the answr

59. SolomonZelman

yw


text:
## Intermediate Algebra for College Students (7th Edition)

$-\frac{4}{x}$
$\frac{20x^3}{-5x^4}$ Apply the fraction rule: $\frac{a}{-b} = -\frac{a}{b}$ Hence, $\frac{20x^3}{-5x^4}$ = $-\frac{20x^3}{5x^4}$ $=-\frac{4x^3}{x^4}$ Recall that $\frac{a^m}{a^n} = \frac{1}{a^{n-m}}$ Thus, $-\frac{4x^3}{x^4}= -\frac{4}{x^{4-3}}$ $=-\frac{4}{x}$


text:
# What are the reactants in the reaction 2NO + 2CO -> N_2 + 2CO_2?

$\text{2NO"+"2CO}$$\rightarrow$${\text{N"_2"+2CO}}_{2}$


text:
## Sender text not always copied when replying/fwd - Win client

I did a fresh install of GW8.01H1 on WinXPP SP3

Occasionally when replying to an open email, the sender text is not included. (I know there is an option not include this text but I never choose that option.)

To overcome this, I have to close the reply and then try replying (sometime Reply All instead of just Reply) again several times before the sender text is included.

The problem also sometimes happens when forwarding messages.

In short I have no idea why it stops working and why it starts working again.

Has anyone come across this/fixed it?

TIA, Tomi


text:
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Partially Ordered Set
Get Partially Ordered Set essential facts below. View Videos or join the Partially Ordered Set discussion. Add Partially Ordered Set to your PopFlock.com topic list for future reference or share this resource on social media.
Partially Ordered Set
The Hasse diagram of the set of all subsets of a three-element set {x, y, z}, ordered by inclusion. Distinct sets on the same horizontal level are incomparable with each other. Some other pairs, such as {x} and {y,z}, are also incomparable.

In mathematics, especially order theory, a partially ordered set (also poset) formalizes and generalizes the intuitive concept of an ordering, sequencing, or arrangement of the elements of a set. A poset consists of a set together with a binary relation indicating that, for certain pairs of elements in the set, one of the elements precedes the other in the ordering. The relation itself is called a "partial order." The word partial in the names "partial order" and "partially ordered set" is used as an indication that not every pair of elements needs to be comparable. That is, there may be pairs of elements for which neither element precedes the other in the poset. Partial orders thus generalize total orders, in which every pair is comparable.

Formally, a partial order is any binary relation that is reflexive (each element is comparable to itself), antisymmetric (no two different elements precede each other), and transitive (the start of a chain of precedence relations must precede the end of the chain).

One familiar example of a partially ordered set is a collection of people ordered by genealogical descendancy. Some pairs of people bear the descendant-ancestor relationship, but other pairs of people are incomparable, with neither being a descendant of the other.

A poset can be visualized through its Hasse diagram, which depicts the ordering relation.[1]

## Formal definition

A (non-strict) partial order[2] is a binary relation set P satisfying particular axioms which are discussed below. When a b, we say that a is related to b. (This does not imply that b is also related to a, because the relation need not be symmetric.)

The axioms for a non-strict partial order state that the relation reflexive, antisymmetric, and transitive. That is, for all a, b, and c in P, it must satisfy:

1. a a (reflexivity: every element is related to itself).
2. if a b and b a, then a = b (antisymmetry: two distinct elements cannot be related in both directions).
3. if a b and b c, then a c (transitivity: if a first element is related to a second element, and, in turn, that element is related to a third element, then the first element is related to the third element).

In other words, a partial order is an antisymmetric preorder.

A set with a partial order is called a partially ordered set (also called a poset). The term ordered set is sometimes also used, as long as it is clear from the context that no other kind of order is meant. In particular, totally ordered sets can also be referred to as "ordered sets", especially in areas where these structures are more common than posets.

For a, b, elements of a partially ordered set P, if a b or b a, then a and b are comparable. Otherwise they are incomparable. In the figure on top-right, e.g. {x} and {x,y,z} are comparable, while {x} and {y} are not. A partial order under which every pair of elements is comparable is called a total order or linear order; a totally ordered set is also called a chain (e.g., the natural numbers with their standard order). A subset of a poset in which no two distinct elements are comparable is called an antichain (e.g. the set of singletons {{x}, {y}, {z}} in the top-right figure). An element a is said to be strictly less than an element b, if a b and a?b. An element a is said to be covered by another element b, written a?b (or a<:b), if a is strictly less than b and no third element c fits between them; formally: if both ab and a?b are true, and acb is false for each c with a?c?b. A more concise definition will be given below using the strict order corresponding to "

## Examples

Standard examples of posets arising in mathematics include:

## Extrema

 Nonnegative integers, ordered by divisibility
 The figure above with the greatest and least elements removed. In this reduced poset, the top row of elements are all maximal elements, and the bottom row are all minimal elements, but there is no greatest and no least element. The set {x, y} is an upper bound for the collection of elements {{x}, {y}}.

There are several notions of "greatest" and "least" element in a poset P, notably:

• Greatest element and least element: An element g in P is a greatest element if for every element a in P, a g. An element m in P is a least element if for every element a in P, a >= m. A poset can only have one greatest or least element.
• Maximal elements and minimal elements: An element g in P is a maximal element if there is no element a in P such that a > g. Similarly, an element m in P is a minimal element if there is no element a in P such that a < m. If a poset has a greatest element, it must be the unique maximal element, but otherwise there can be more than one maximal element, and similarly for least elements and minimal elements.
• Upper and lower bounds: For a subset A of P, an element x in P is an upper bound of A if a x, for each element a in A. In particular, x need not be in A to be an upper bound of A. Similarly, an element x in P is a lower bound of A if a >= x, for each element a in A. A greatest element of P is an upper bound of P itself, and a least element is a lower bound of P.

For example, consider the positive integers, ordered by divisibility: 1 is a least element, as it divides all other elements; on the other hand this poset does not have a greatest element (although if one would include 0 in the poset, which is a multiple of any integer, that would be a greatest element; see figure). This partially ordered set does not even have any maximal elements, since any g divides for instance 2g, which is distinct from it, so g is not maximal. If the number 1 is excluded, while keeping divisibility as ordering on the elements greater than 1, then the resulting poset does not have a least element, but any prime number is a minimal element for it. In this poset, 60 is an upper bound (though not a least upper bound) of the subset {2,3,5,10}, which does not have any lower bound (since 1 is not in the poset); on the other hand 2 is a lower bound of the subset of powers of 2, which does not have any upper bound.

## Orders on the Cartesian product of partially ordered sets

 Reflexive closure of strict direct product order on N×N. Elements covered by (3,3) and covering (3,3) are highlighted in green and red, respectively.
 Product order on N×N
 Lexicographic order on N×N

In order of increasing strength, i.e., decreasing sets of pairs, three of the possible partial orders on the Cartesian product of two partially ordered sets are (see figures):

All three can similarly be defined for the Cartesian product of more than two sets.

Applied to ordered vector spaces over the same field, the result is in each case also an ordered vector space.

## Sums of partially ordered sets

Hasse diagram of a series-parallel partial order, formed as the ordinal sum of three smaller partial orders.

Another way to combine two posets is the ordinal sum[4] (or linear sum[5]), Z = X ? Y, defined on the union of the underlying sets X and Y by the order a Z b if and only if:

• a, b ? X with a X b, or
• a, b ? Y with a Y b, or
• a ? X and b ? Y.

If two posets are well-ordered, then so is their ordinal sum.[6] The ordinal sum operation is one of two operations used to form series-parallel partial orders, and in this context is called series composition. The other operation used to form these orders, the disjoint union of two partially ordered sets (with no order relation between elements of one set and elements of the other set) is called in this context parallel composition.

## Strict and non-strict partial orders

In some contexts, the partial order defined above is called a non-strict (or reflexive) partial order. In these contexts, a strict (or irreflexive) partial order "<" is a binary relation that is irreflexive, transitive and asymmetric, i.e. which satisfies for all a, b, and c in P:

• not a < a (irreflexivity),
• if a < b and b < c then a < c (transitivity), and
• if a < b then not b < a (asymmetry; implied by irreflexivity and transitivity[7]).

Strict and non-strict partial orders are closely related. A non-strict partial order may be converted to a strict partial order by removing all relationships of the form aa. Conversely, a strict partial order may be converted to a non-strict partial order by adjoining all relationships of that form. Thus, if "irreflexive kernel given by:

a < b if a b and a ? b

Conversely, if "<" is a strict partial order, then the corresponding non-strict partial order "reflexive closure given by:

a b if a < b or a = b.

This is the reason for using the notation "

Using the strict order "<", the relation "a is covered by b" can be equivalently rephrased as "a<b, but not a<c<b for any c". Strict partial orders are useful because they correspond more directly to directed acyclic graphs (dags): every strict partial order is a dag, and the transitive closure of a dag is both a strict partial order and also a dag itself.

## Inverse and order dual

The inverse (or converse) of a partial order relation converse of =, it is the relation that satisfies x >= y if and only if y x. The inverse of a partial order relation is reflexive, transitive, and antisymmetric, and hence itself a partial order relation. The order dual of a partially ordered set is the same set with the partial order relation replaced by its inverse. The irreflexive relation > is to >= as < is to

Any one of the four relations =, and > on a given set uniquely determines the other three.

In general two elements x and y of a partial order may stand in any of four mutually exclusive relationships to each other: either x < y, or x = y, or x > y, or x and y are incomparable (none of the other three). A totally ordered set is one that rules out this fourth possibility: all pairs of elements are comparable and we then say that trichotomy holds. The natural numbers, the integers, the rationals, and the reals are all totally ordered by their algebraic (signed) magnitude whereas the complex numbers are not. This is not to say that the complex numbers cannot be totally ordered; we could for example order them lexicographically via x+iy < u+iv if and only if x < u or (x = u and y < v), but this is not ordering by magnitude in any reasonable sense as it makes 1 greater than 100i. Ordering them by absolute magnitude yields a preorder in which all pairs are comparable, but this is not a partial order since 1 and i have the same absolute magnitude but are not equal, violating antisymmetry.

## Mappings between partially ordered sets

 Order isomorphism between the divisors of 120 (partially ordered by divisibility) and the divisor-closed subsets of {2,3,4,5,8} (partially ordered by set inclusion)
 Order-preserving, but not order-reflecting (since f(u)f(v), but not uv) map.

Given two partially ordered sets (S,T,f: S -> T is called order-preserving, or monotone, or isotone, if for all x and y in S, xy implies f(x) f(y). If (U,f: S -> T and g: T -> U are order-preserving, their composition (g?f): S -> U is order-preserving, too. A function f: S -> T is called order-reflecting if for all x and y in S, f(x) f(y) implies xy. If f is both order-preserving and order-reflecting, then it is called an order-embedding of (S,T,f is necessarily injective, since f(x) = f(y) implies x y and y x. If an order-embedding between two posets S and T exists, one says that S can be embedded into T. If an order-embedding f: S -> T is bijective, it is called an order isomorphism, and the partial orders (S,T,isomorphic. Isomorphic orders have structurally similar Hasse diagrams (cf. right picture). It can be shown that if order-preserving maps f: S -> T and g: T -> S exist such that g?f and f?g yields the identity function on S and T, respectively, then S and T are order-isomorphic. [8]

For example, a mapping f: N -> P(N) from the set of natural numbers (ordered by divisibility) to the power set of natural numbers (ordered by set inclusion) can be defined by taking each number to the set of its prime divisors. It is order-preserving: if x divides y, then each prime divisor of x is also a prime divisor of y. However, it is neither injective (since it maps both 12 and 6 to {2,3}) nor order-reflecting (since 12 doesn't divide 6). Taking instead each number to the set of its prime power divisors defines a map g: N -> P(N) that is order-preserving, order-reflecting, and hence an order-embedding. It is not an order-isomorphism (since it e.g. doesn't map any number to the set {4}), but it can be made one by restricting its codomain to g(N). The right picture shows a subset of N and its isomorphic image under g. The construction of such an order-isomorphism into a power set can be generalized to a wide class of partial orders, called distributive lattices, see "Birkhoff's representation theorem".

## Number of partial orders

Sequence A001035 in OEIS gives the number of partial orders on a set of n labeled elements:

Number of n-element binary relations of different types
Elements Any Transitive Reflexive Preorder Partial order Total preorder Total order Equivalence relation
0 1 1 1 1 1 1 1 1
1 2 2 1 1 1 1 1 1
2 16 13 4 4 3 3 2 2
3 512 171 64 29 19 13 6 5
4 355 219 75 24 15
n 2n2 2n2-n ?n
k=0

k! S(n, k)
n! ?n
k=0

S(n, k)
OEIS A002416 A006905 A053763 A000798 A001035 A000670 A000142 A000110

The number of strict partial orders is the same as that of partial orders.

If the count is made only up to isomorphism, the sequence 1, 1, 2, 5, 16, 63, 318, ... (sequence in the OEIS) is obtained.

## Linear extension

A partial order * on a set X is an extension of another partial order X provided that for all elements x and y of X, whenever x y, it is also the case that x * y. A linear extension is an extension that is also a linear (i.e., total) order. Every partial order can be extended to a total order (order-extension principle).[9]

In computer science, algorithms for finding linear extensions of partial orders (represented as the reachability orders of directed acyclic graphs) are called topological sorting.

## In category theory

Every poset (and every preordered set) may be considered as a category where, for objects x and y, there is at most one morphism from x to y. More explicitly, let hom(x, y) = {(x, y)} if x y (and otherwise the empty set) and (y, z)?(x, y) = (x, z). Such categories are sometimes called posetal.

Posets are equivalent to one another if and only if they are isomorphic. In a poset, the smallest element, if it exists, is an initial object, and the largest element, if it exists, is a terminal object. Also, every preordered set is equivalent to a poset. Finally, every subcategory of a poset is isomorphism-closed.

## Partial orders in topological spaces

If P is a partially ordered set that has also been given the structure of a topological space, then it is customary to assume that {(a, b) : ab} is a closed subset of the topological product space ${\displaystyle P\times P}$. Under this assumption partial order relations are well behaved at limits in the sense that if ${\displaystyle a_{i}\to a}$, ${\displaystyle b_{i}\to b}$ and ai bi for all i, then a b.[10]

## Intervals

An interval in a poset P is a subset I of P with the property that, for any x and y in I and any z in P, if x z y, then z is also in I. (This definition generalizes the interval definition for real numbers.)

For a b, the closed interval [a, b] is the set of elements x satisfying a x b (i.e. a x and x b). It contains at least the elements a and b.

Using the corresponding strict relation "<", the open interval (a, b) is the set of elements x satisfying a < x < b (i.e. a < x and x < b). An open interval may be empty even if a < b. For example, the open interval (1, 2) on the integers is empty since there are no integers I such that 1 < I < 2.

The half-open intervals [a, b) and (a, b] are defined similarly.

Sometimes the definitions are extended to allow a > b, in which case the interval is empty.

An interval I is bounded if there exist elements a and b of P such that I ? [a, b]. Every interval that can be represented in interval notation is obviously bounded, but the converse is not true. For example, let P = (0, 1) ? (1, 2) ? (2, 3) as a subposet of the real numbers. The subset (1, 2) is a bounded interval, but it has no infimum or supremum in P, so it cannot be written in interval notation using elements of P.

A poset is called locally finite if every bounded interval is finite. For example, the integers are locally finite under their natural ordering. The lexicographical order on the cartesian product N×N is not locally finite, since (1, 2) . Using the interval notation, the property "a is covered by b" can be rephrased equivalently as [a, b] = {a, b}.

This concept of an interval in a partial order should not be confused with the particular class of partial orders known as the interval orders.

## Notes

1. ^ Merrifield, Richard E.; Simmons, Howard E. (1989). Topological Methods in Chemistry. New York: John Wiley & Sons. pp. 28. ISBN 0-471-83817-9. Retrieved 2012. A partially ordered set is conveniently represented by a Hasse diagram...
2. ^ Simovici, Dan A. & Djeraba, Chabane (2008). "Partially Ordered Sets". Mathematical Tools for Data Mining: Set Theory, Partial Orders, Combinatorics. Springer. ISBN 9781848002012.
3. ^
4. ^ Neggers, J.; Kim, Hee Sik (1998), "4.2 Product Order and Lexicographic Order", Basic Posets, World Scientific, pp. 62-63, ISBN 9789810235895
5. ^ Davey, B. A.; Priestley, H. A. (2002). Introduction to Lattices and Order (Second ed.). New York: Cambridge University Press. pp. 17-18. ISBN 0-521-78451-4 – via Google Books.
6. ^ P. R. Halmos (1974). Naive Set Theory. Springer. p. 82. ISBN 978-1-4757-1645-0.
7. ^ Fla?ka, V.; Je?ek, J.; Kepka, T.; Kortelainen, J. (2007). Transitive Closures of Binary Relations I. Prague: School of Mathematics - Physics Charles University. p. 1. Lemma 1.1 (iv). Note that this source refers to asymmetric relations as "strictly antisymmetric".
8. ^ Davey, B. A.; Priestley, H. A. (2002). "Maps between ordered sets". Introduction to Lattices and Order (2nd ed.). New York: Cambridge University Press. pp. 23-24. ISBN 0-521-78451-4. MR 1902334..
9. ^ Jech, Thomas (2008) [1973]. The Axiom of Choice. Dover Publications. ISBN 978-0-486-46624-8.
10. ^ Ward, L. E. Jr (1954). "Partially Ordered Topological Spaces". Proceedings of the American Mathematical Society. 5 (1): 144-161. doi:10.1090/S0002-9939-1954-0063016-5. hdl:10338.dmlcz/101379.