# Three Times Larger: Idiom or Error?

Having just written about issues of wording with regard to percentages, we should look at another wording issue that touches on percentages and several other matters of wording. What does “three times larger” mean? How about “300% more”? We’ll focus on one discussion that involved several of us, and referred back to other answers we’ve given.

## Percent increase vs. factor

The question, from 2006, started with the idea of a percent increase:

Percent Increase and "Increase by a Factor of ..."

A math doctor here recently explained percent increase this way: If we start with 1 apple today and tomorrow have 2 apples, then because 2 - 1 = 1 and 1/1 = 1, we have a 100 percent increase.

But can't I also say there was an increase by a factor of 2?  Two divided by 1 equals 2, an increase by a factor of 2 -- and also an increase by 200 percent?  This is what is confusing me!

I'd never been confused about saying "increased by a factor of" and "increased by percent of" until I saw the Dr. Math conversation about finding percentages ... which is a good thing, I guess, because now I know what I didn't know!  Thank you for any help.

Unfortunately, Joseph didn’t directly quote from the page he had in mind, and we have never said exactly what he said; so we couldn’t be sure which page it was. Everything he said, however, was correct.

Doctor Rick was the first to answer:

Hi, Joseph.

Yes, this can be very confusing, because some statements about increases are ambiguous.

When we say "increased by a factor of 2," the word "factor" makes it clear that we mean "multiplied by 2."

When we say "increased by 10%," there is only one reasonable interpretation: the amount of the increase is 10% of the original amount.  If we meant multiplication by 10%, that would be a decrease -- not an increase!  Even when we say "increased by 100%," there is only one reasonable interpretation, since multiplication by 100% is the same as multiplication by 1, and that's still not an increase.

When we want to speak of an increase that is greater than the original amount, then ambiguity can arise.  In that situation, I much prefer "increase by a factor of 3" or "by a factor of 2.5," etc.

I don't know what page you saw -- but have you seen this one?

Percent Greater Than vs. Increased
http://mathforum.org/library/drmath/view/61774.html

See also the page linked there, about the even more confusing phrase "___ times more than" and the like.  I am on the side of avoiding the confusing phrases, as a basic principle of communication.

If you saw another page and you are still confused by it, please tell me the URL of that page so I can review it with you.

## Ambiguity in percent increase

Before we get back to this conversation, we should take a look at the page he referred to, which is a good starting point:

Percent Greater Than vs. Increased

What is the difference between the following statements:

My profits are 200% bigger than they were last year.

and

My profits from last year have increased 200%.

This is one of the questions we have to answer in my Middle school methods course and I have looked everywhere for the answer.  I hope you can help.

As far as I can see, they mean the same thing; in fact, both are similarly ambiguous.

Taken literally, "200% bigger" (or, more formally, larger or greater) and "increased 200%" (or, more completely, increased _by_ 200%) both mean that the increase from one year to the next is 200% of the first year's value, so that the second year's profit is 3 times the first. But both statements are more likely to have been made with the intention of saying that this year's profit is twice last years. English is not very clear in cases like this.

So there is a literal meaning (which mathematicians tend to see as best), and an idiomatic meaning (which ordinary people are more likely to have intended to say). I referred to a page we’ll be looking at below, and then quoted a favorite book of mine that gives a lexicographer’s perspective:

Since writing that, I found a good reference on "two times greater," although it doesn't mention your "200% greater." It is in Merriam Webster's _Dictionary of English Usage_, which under "times" writes

The argument in this case is that _times more_
(or _times larger_, _times stronger_, _times
brighter_, etc.) is ambiguous, so that "He has
five times more money than you" can be
misunderstood as meaning "He has six times as
much money as you." It is, in fact, possible
to misunderstand _times more_ in this way, but
it takes a good deal of effort. If you have
$100, five times that is$500, which means
that "five times more than $100" can mean (the commentators claim) "$500 more than $100," which equals "$600," which equals "six times
as much as $100." The commentators regard this as a serious ambiguity, and they advise you to avoid it by always saying "times as much" instead of "times more." Here again, it seems that they are paying homage to mathematics at the expense of language. The fact is that "five times more" and "five times as much" are idiomatic phrases which have - and are understood to have - exactly the same meaning. The "ambiguity" of _times more_ is imaginary: in the world of actual speech and writing, the meaning of _times more_ is clear and unequivocal. It is an idiom that has existed in our language for more than four centuries, and there is no real reason to avoid its use. I think the same applies to "X percent bigger" and "increased [by] X%." There is just enough ambiguity in a technical context that I would want to ask what was intended before assuming anything, but there is no reason to say that they definitely mean different things, or mean something different than "X percent of" or "increased to X percent." I myself would avoid saying these things, just because there are enough people who have heard that they are ambiguous, and would therefore take them the wrong way (whichever that is!). As a result of my side interest in linguistics, I recognize that human speech is not as logical as we might wish; what a word means is a matter of actual usage in a culture, rather than pure logic. So rather than state that either understanding of “times bigger” is “correct”, I just recognize that people take it in two ways, so you have to ask, or use contextual cues, in order to decide on what is meant. ## Confusion about “three times larger” Back to the original discussion: Joseph responded with specific references, the first of which was that link of mine that Doctor Rick said to “see also”: I'm sorry, I should have specified the site. In fact, there were two -- and I still don't see the difference between them. Here is the first example, from Larger Than and As Large As http://mathforum.org/library/drmath/view/52338.html 1) "Three times as large as N" means "3 * N." 2) "Three times larger than N" means "4 * N" -- but only if you stop to think about it, as many people do not. Here, I don't understand how something can be 3 times larger and be 4 times N. That sounds really weird to me. If you asked "What is something that is three times as large as N?" then I would say 3N ... but apparently I'd be wrong! I just don't see where my thinking is wrong.  His thinking isn’t wrong on this point: 3N is three times as large as N! He seems just to be letting what he’s read sow doubt about everything. Here is the second example, from Percentage of Increase http://mathforum.org/library/drmath/view/58131.html You can choose two ways to express your answer now. One is to say: there will be a 550% increase by the year 2000. Or you can say: in the year 2000 the (new value) -- you didn't say what the numbers represented, so I'm a little confused right here -- will be about five and a half times greater than what it was in 1995. Many people don't quite grasp those phrases, especially the latter one. Instead you might wish to say it this way: in 2000 the (new value) will be 6 and a half times what it was in 1995. The difference in the wording is subtle, of course, but important. The number 6 1/2 comes from 325,000 --------- = 6.5 or 6 1/2 50,000 which is NOT a percent increase situation. In this problem, I don't understand the difference between the way the doctor explains the two different ways you can talk about the increase, and the implications of each. The doctor says that 6.5 times is not a percent increase; but can you still say it's 650 percent OF the original? I'm sorry -- this is all very confusing at this point! So his specific question is this: • Why wouldn’t “three times as large” and “three times larger” mean the same thing? • How can “5 1/2 times greater” and “6 1/2 times what it was” mean the same thing? The two pages quoted are by me (1999) and Doctor Terrel (1997). The first is particularly worth reading in its entirety, as there is a lot more there. ## The case for a literal interpretation Doctor Greenie responded, arguing against laxness on the matter, and making the case for the literalistic interpretation: I'm going to jump in here, because this is one of my pet peeves. Mathematics is commonly called the exact science. Mathematics must be exact; if it is not, it all falls apart. We can't use ambiguous language in mathematics. I agree that the use of the phrase "x times larger than" is best avoided. However, as a mathematician who believes in using unambiguous language, I cannot accept the proposition that we should be able to interpret "5 times larger than 10" as either 50 or 60. It HAS TO BE ONE OR THE OTHER. And grammatically, "5 times larger than" means the "new" number is 5 times larger than the "old" number; this in turn means the difference between the new and old numbers is 5 times the old number, making the new number 6 times the old number. So the number which is 5 times larger than 10 is 10 + 5(10) = 10 + 50 = 60 (The phrase "... larger than ..." implies comparison by subtraction; the phrase "... as large as ..." implies comparison by division. Sixty is 6 times as large as 10, because 60/10 = 6. But 60 is 5 times larger than 10, because [60 - 10]/10 = 50/10 = 5.) Yes, we hear it all the time in everyday life. Sometimes, we even hear it in the supposedly rigorous world of science -- "an earthquake of magnitude 5 is 10 times greater than one of magnitude 4," and such. But the common idiom of using "10 times greater than" -- when the actual meaning is "10 times as great as" -- has no place in mathematics. He concluded with an accidental overstatement of what the “other side” says: I disagree with many of the concessions that other math doctors here have made in interpreting the phrase "x times larger than" as being the same as "x times as large as." On one of the pages I saw, a fellow doctor said that "50% larger than" and "50% as large" mean the same thing. But if my weekly salary last year was$1000 and it is 50% LARGER this year, then it is now

$1000 + 50%($1000) = $1000 +$500 = $1500 While if it was$1000 last year and it is 50% AS LARGE this year, then it is now

50%($1000) =$500

If something is 50% larger, then it is larger; if it is 50% as large, then it is smaller.  They can't be the same; that is nonsense.

In fact, as he admitted in a subsequent private discussion, he had misremembered what others had said; none of us have claimed that “50% larger than” and “50% as large” mean the same thing. What we say is that, when the percentage or multiplier is greater than 100%, we recognize ambiguity in the likely intent.  I think we agree that the phrase should not be used in a mathematical context, and that we both grudgingly interpret it as intended elsewhere.

## What the literal interpretation means

Then it was my turn to respond, as the author of the first page Joseph had asked about, wanting to make sure he understood both why people take it literally as they do, and how we should think about it.

First, on “Larger Than and As Large As”, I said this:

Joseph, your thinking is RIGHT: if M is three times AS LARGE AS N, then M = 3N.  That's what statement (1) above says.

But if we break statement (2) apart carefully (some would say TOO carefully :-)), then it means something different from what people usually mean by it.

If I said "M is 50 larger than N," I would mean that if you ADD 50 to N, you get M:

M = N + 50.

And if I said, "M is 50% larger than N," I would mean that if you add 50% OF N to N, you get M; that is, I mean that M is 50% of N added to N:

M = N + (0.50)N.

Now, though I'm not entirely sure I agree with this, technically minded people often apply the same thinking to (2), for the sake of consistency.  The "larger than" means we add something to N.  And what do we add? Three times N.  So by this thinking,

M = N + 3N = 4N.

So "three times LARGER THAN N" means the same as "four times AS LARGE AS N."

I then referred to the usage book quoted above, adding:

English usage experts think it is nonsense.  My feeling is that this thinking puts a little too much weight on consistency, and is just too weird for the general public to follow.  English is not known for consistency!  So we need to recognize that in everyday usage, (1) and (2) really mean the same thing.  When we accept that, though, we set ourselves up for the opposite confusion: Cases like "50% larger" and "3 times larger" no longer follow the same pattern, and our language becomes inconsistent, which really bothers mathematicians!

As Dr. Rick pointed out, this means that there are gray areas where it's hard to be sure what someone means, so it may be best just to avoid using these phrases in mathematical contexts.

Finally, I commented on Joseph’s other quote, from Doctor Terrel:

Joseph, here the doctor was saying that a 550% INCREASE means adding 550% of the original value to the original value, which means 650% OF the original value.  In the other terminology, "5 1/2 times greater" (there again, "greater" is taken to refer to the increase) is the same as "6 1/2 times as much." When he says that the 6 1/2 is not a percent increase, he doesn't mean that it hasn't increased, just that he is talking about multiplying by 650% rather than adding 650%.  When you think in terms of increase (adding), it is a 550% increase.

Now, the "percent increase" case is pretty standard, because it IS technical terminology (though ordinary people reading it can get confused, so it's still risky).  The "times greater" case is more disputable, since that sounds less technical.  Most people don't demand absolute consistency from language; they are happy to understand "times greater" idiomatically.

I hope that clears up some of the confusion.  It isn't all cleared up yet at our end.  You will definitely get different opinions as to what it all REALLY means!

## Increase by a factor

Joseph replied, returning to his initial question:

That really cleared things up for me and I appreciate your time in driving home the differences!  The last question I would like to ask is, how do you deal with factors?  If someone says something has changed by a factor of ... or is less/greater than by a factor of ..., do we use the same rules that you've discussed above? Or when using the word "factor," are things a bit different?

As Dr. Rick pointed out in the first response, "factor" is used to make it clear that multiplication, rather than addition, is the cause of an increase.  Just as "increased by a factor of 2" means "twice as large," so does "greater by a factor of 2."  And "decreased by a factor of 2" and "less by a factor of 2" both mean "half as much" (divided by 2).

I can't think of a context in which that would not be true -- but English is flexible enough that I probably shouldn't guarantee anything!

Isn’t English fun?

## Conclusion

In closing, here is a more recent question (2015) where I summarized this complicated issue:

As Big As vs. Bigger Than

I'm having difficulty convincing my 5th graders that "as big as" and "bigger than" do not mean the same thing.

For example, when asked, "How many times larger is 10,000 than than 100?" they answer "100."

Their tests and homework are full of this misunderstanding! How would you suggest telling them they are wrong?

I referred to most of the pages we’ve seen above, and added:

To be honest, my feeling (basically unchanged since the first of those) is that although your understanding is common among thoughtful people, it is a case of excessive consistency.

Mathematical people want a certain word structure to always have the same meaning, so we relate "x times bigger" to "x percent bigger," and that to "x bigger," and want to take all in an incremental (additive) sense. But taken on its own, it is perfectly logical to interpret "x times bigger" as "bigger, as a result of multiplication by x." And human language is not completely consistent; we have idioms all over the place that we interpret with no trouble.

Having said that, I think that math books should avoid that form, because there is just enough truth to your thinking, and enough extra expectation of careful use of words in a math book, that it can be confusing.

What I would do is to make a brief mention of the fact that many people take it as you do, but then point out that your book is using the phrase in the way it is usually intended in the real world. If I were writing the textbook, I would reverse this: almost always use "x times as big," but mention somewhere the fact that many people use "x times bigger" to mean the same thing, and briefly discuss the controversy before moving on to other things.

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# Sequence of odd and even partial sums of alternating harmonic series.

Given the alternating harmonic series: $$\sum^{\infty}_{n=1}{\frac{(-1)^{n+1}}{n}}$$ and if we let the sequence of its partial sums be $s_n$ then how can we express the sequence of even partial sums $s_{2n}$ and odd partial sums $s_{2n+1}$?

I did the following but am not completely sure if it is valid:

$$s_{2n}= \sum^{2n}_{i=1}{\frac{(-1)^{i+1}}{i}}$$ $$s_{2n+1}= \sum^{2n+1}_{i=1}{\frac{(-1)^{i+1}}{i}}$$

Then to show the first is increasing and the second is decreasing, can we just take $s_{2n+2}-s_{2n}$ and show that this is $> 0$? Similarly then taking $s_{2n+3}-s_{2n+1}$ and showing that this is $< 0$?

And to go one step further, how could we show that each of these sequences of partial sums has a limit? And then show that the limits are the same?

• $s_{2n} = \sum_{i=1}^n \frac{1}{2i-1}-\frac{1}{2i}=\sum_{i=1}^n \frac{1}{2i (2i-1)}$ – reuns Nov 12 '17 at 1:31
• en.wikipedia.org/wiki/Alternating_series_test – parsiad Nov 12 '17 at 1:33
• I don't quite get where the formula you've written comes from? – Analysis is fun Nov 12 '17 at 1:42

What you have done is fine. Showing the increasing/decreasing nature works well. You can invoke the alternating series theorem to show the whole series has a limit because the terms are alternating in sign and decreasing monotonically to zero. If the whole series has a limit, every subseries converges to the same limit and you are done.

• Ah okay I see, is there a way to directly prove that $s_{2n}$ and $s_{2n+1}$ have limits? – Analysis is fun Nov 12 '17 at 1:42
• The subseries argument works fine. Also they are monotonic and bounded above/below so have limits. – Ross Millikan Nov 12 '17 at 1:46
• Ah yes okay I think I've got it now, thanks so much for your help! – Analysis is fun Nov 12 '17 at 1:47

Yes, you are correct on how to show that the first is decreasing and the second is increasing.

Now, since you know they are monotone, if you can show they are bounded, you will demonstrate that they have limits. For instance you can show that $$1>s_{2n+1} > 1/2$$ for all $n$.

That the limits are the same follows from the fact that the difference $$|s_{2n+1}-s_{2n}| = \frac{1}{2n+1}\to 0.$$

• Thanks, so we can show boundedness and then use the monotone convergence theorem? – Analysis is fun Nov 12 '17 at 1:47
• Yeah. As others have said the alternating series test proves it all in one fell swoop but it seemed like the point of the exercise was to prove a special case of the alternating series test. – spaceisdarkgreen Nov 12 '17 at 1:52

Since $s_{2n}= \sum^{2n}_{i=1}{\frac{(-1)^{i+1}}{i}}$,

$\begin{array}\\ s_{2n+2} &= \sum^{2n+2}_{i=1}{\frac{(-1)^{i+1}}{i}}\\ &= \sum^{2n}_{i=1}{\frac{(-1)^{i+1}}{i}}+{\frac{(-1)^{2n+1}}{2n+1}}+{\frac{(-1)^{2n+2}}{2n+2}}\\ &= s_{2n}-{\frac{1}{2n+1}}+{\frac{1}{2n+2}}\\ &= s_{2n}-{\frac{1}{(2n+1)(2n+2)}}\\ &\lt s_{2n}\\ \end{array}$

so $s_{2n}$ is decreasing.

Similarly,

$\begin{array}\\ s_{2n+3} &= \sum^{2n+3}_{i=1}{\frac{(-1)^{i+1}}{i}}\\ &= \sum^{2n+1}_{i=1}{\frac{(-1)^{i+1}}{i}}+{\frac{(-1)^{2n+2}}{2n+2}}+{\frac{(-1)^{2n+3}}{2n+3}}\\ &= s_{2n+1}+{\frac{1}{2n+2}}-{\frac{1}{2n+3}}\\ &= s_{2n+1}+{\frac{1}{(2n+2)(2n+3)}}\\ &\gt s_{2n+1}\\ \end{array}$

so $s_{2n+1}$ is increasing.

Since $s_{2n+1} =s_{2n}-\frac1{2n+1} \lt s_{2n}$, all the $s_{2n+1}$ are less than all the $s_{2n}$.

Since $|s_{2n+1}-s_{2n}| =\frac1{2n+1} \to 0$, the sequences $(s_{2n})$ and $(s_{2n+1})$ must approach a common limit.


×

When Fibonacci meets Pythagoras

There is a cool identity relating Fibonacci numbers and Pythagorean triples, which states that given four consecutive Fibonacci numbers $${F}_{n},{F}_{n+1},{F}_{n+2},{F}_{n+3}$$, the triple $$({F}_{n}{F}_{n+3}, 2{F}_{n+1}{F}_{n+2}, {{F}_{n+1}}^{2}+{{F}_{n+2}}^{2})$$ is a Pythagorean triple.

To prove this, let $${F}_{n+1} = a$$ and $${F}_{n+2} = b$$. Then $${F}_{n} = b-a$$ and $${F}_{n+3} = b+a$$. Substituting the abstract values to the equation ${({F}_{n} {F}_{n+3})}^{2} +{(2{F}_{n+1} {F}_{n+2})}^{2} = {({{F}_{n+1}}^{2}+{ {F}_{n+2}}^{2})}^{2}$ will give ${((b-a)(b+a))}^{2} +{(2ab)}^{2}={({a}^{2}+{b}^{2})}^{2}.$

With a little simplification, we arrive at Euclid's formula for Pythagorean triples: ${({b}^{2}-{a}^{2})}^{2} +{(2ab)}^{2}={({a}^{2}+{b}^{2})}^{2}.$

Check out my other notes at Proof, Disproof, and Derivation

Note by Steven Zheng
2 years, 10 months ago

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Nice post

Is there any generalized formula to generate pythagoream triples

- 2 years, 10 months ago

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Euclid's Formula. The link above should bring you to a proof.

- 2 years, 10 months ago

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How we can solve problems in which product of three pythagorean triples is given and we have to find the triple

- 2 years, 10 months ago

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I'm not sure if that is solvable without more given conditions. You might also need the sum of sides or maybe the radius of an incircle also. Can you come up with an example of your question?

- 2 years, 10 months ago

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Find out the pythagorean triple $$(x,y,z)$$ such that $$xyz=16320$$

- 2 years, 10 months ago

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Number of solutions $$a,b$$ such that $$2ab({a}^{2}+{b}^{2})({a}^{2}-{b}^{2}) = 16320$$.

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Does there exist more the one triples.....also i read on a website that euclid's formula can't generate all pythagorean triples

- 2 years, 10 months ago

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Euclid's formula deal with generating Pythagorean triples that are integers.

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Good

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# In how many ways can I climb down ten stairs, taking as many steps at a time as I like?

I want to climb down a flight of ten stairs, taking one or many steps at a time. In how many ways can I do this if I am able to take as many steps at a time as I like?

I don't know the answer; I am getting $3$ such as:

$1,2,3,4,..., 10$ (one at a time)

$2,4,6,..., 10$ (two at a time)

$5,10$ (five at a time)

this is correct/incorrect?

-
If you have to choose the number of steps you will take at a time, and stick with that number the whole way down, then the only thing you have missed is the possibility of taking 10 at a time. But if you are allowed to (for example) take 3 steps, then 4, then 3, well, that's a whole different problem. So, which is it? – Gerry Myerson Dec 8 '11 at 2:16
@Gerry Myerson: "taking one or many steps at a time" probably indicate a variable number of steps at each time ... also this version seems to be the interesting one. – Quixotic Dec 8 '11 at 2:21
If we have total freedom, there are $512$ ways. – André Nicolas Dec 8 '11 at 2:37
See the WP page on the number of compositions – r.e.s. Dec 8 '11 at 3:11
@MaX, what you say is true, but I find it hard to believe that under that interpretation OP was only able to find 3 ways to come down the stairs. – Gerry Myerson Dec 8 '11 at 5:07

In the interpretation where the number of steps taken is allowed to vary, there is an easy way to see that the answer is $2^{n-1}$:

For each of the $n-1$ steps (not counting the top one), decide whether or not you will stop at that step on the way down.

There are $n-1$ decisions to make, each with 2 possible choices (stop or don't stop), so that's a total of $2^{n-1}$ ways.

-

We answer the question on the following assumptions: (1) At any time, we can take an arbitrary number of stairs, as long as the total number of stairs ends up at $10$ and (2) Order matters, so that for example the pattern $2+4+4$ is to be counted as different from $4+2+4$.

In general, let $f(n)$ be the number of ways to handle a staircase that has $n$ stairs. It is easy to see that $f(1)=1$. We could compute a little more, finding that for $2$ there are the patterns $2$ and $1+1$, so $f(2)=2$. For $3$ there are the patterns $3$, $2+1$, $1+2$, and $1+1+1$, for a total of $4$. For $4$, enumerating the patterns is not hard: we get $4$, $3+1$, $1+3$, $2+2$, $1+1+2$, $1+2+1$, $2+1+1$, and $1+1+1+1$, a total of $8$. On the basis of this admittedly scanty evidence, we conjecture that $f(n)=2^{n-1}$, and in particular $f(10)=512$.

Now we prove that in general $f(n)=2^{n-1}$. There are various approaches. We first prove the result in a natural but somewhat lengthy way, and then in a shorter way. Suppose that we know that for all $k<n$, there are $2^{k-1}$ patterns. We count the number of patterns for $n$.

Either there is only one giant step of length $n$ ($1$ way) or the last step is one of $1$, $2$, $3$, $\dots$, $n-1$.

If the last step is $1$, we needed to get to $n-1$, which by induction assumption can be done in $2^{n-2}$ ways.

If the last step is $2$, we needed to get to $n-2$, which by induction assumption can be done in $2^{n-3}$ ways.

If the last step is $3$, we needed to get to $n-3$, which by induction assumption can be done in $2^{n-4}$ ways.

Go on in this way. Finally, if the last step was $n-1$, we needed to get to $1$, which can be done in $2^0$ ways.

We conclude that $$f(n)=(2^{n-2}+2^{n-3}+2^{n-4}+\cdots +1)+1$$ (remember the possibility of a giant step). Easily, if we add from the back, we see that $f(n)=2^{n-1}$ (or else we can sum the finite geometric progression).

Another way: We show that in general $f(n+1)=2f(n)$. Consider all patterns of steps that add up to $n+1$. They are of two types (i) Patterns that have last step $1$ and (ii) Patterns that have last step $>1$.

It is obvious that there are $f(n)$ patterns of type (i). We now count the type (ii) patterns. Take a pattern of type (ii), and shorten its last step by $1$. The total is now $n$. Moreover, we get all patterns for $n$ in this manner, in one and only one way. Thus there are $f(n)$ type (ii) patterns. It follows that $f(n+1)=f(n)+f(n)=2f(n)$.

-

A composition of an integer $n$ into $k$ parts is the number of sums $a_1 + \dots + a_k$ of positive integers that equal $n$. For example, there are 3 compositions of $4$ into $2$ parts: 2+2, 3+1, and 1+3. You should try to prove that the number of compositions of $n$ into $k$ parts is equal to $\binom{n-1}{k-1}$. (Or, look it up in any good combinatorics book.)

Your problem essentially asks you to count the number of compositions of $10$ into $k$ parts, where $1\leq k\leq 10$. That is, if as you go down the staircase you cross $a_i$ stairs in your $i^{\text{th}}$ step, then $a_1 + \dots + a_k = 10$. Your answer is thus $$\sum_{k=1}^{10} \binom{10-1}{k-1} =512$$

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# First order PDEs and the method of characteristics

I am studying PDEs using the book "PDEs An Introduction 2nd edition" by Walter A. Strauss. In Chapter 2, a "geometric method" is described in order to solve linear PDEs of the type:

$$(x,y)\mapsto u_x + yu_y = 0$$

This is said to be equivalent to the directional derivative of $u$ in the direction of the vector $(1,y)$ being set to 0. Then characteristic curves having as tangent vectors $(1,y)$ are found

$$\frac{dy}{dx} = \frac{y}{1} \implies y = Ce^x$$

Since $u(x,y)$ is constant on these curves:

$$u(x,y) = f(e^{-x}y)$$

is the general solutions of the PDE, where $f$ is an arbitrary function. Now the same method is applied to solve more general equations, such as:

$$u_x + u_y + u = f(x,y)$$

I have tried to use the same method to solve the following differential equation:

$$yu_x -xu_y = 3x$$ where $u(x,0) = x^2$

Then applying the method:

$$\frac{dy}{dx} = \frac{-x}{y} \implies C = x^2 + y^2$$

The PDE reduces to an ODE:

$$\frac{du}{dx} = 3x/y \implies u (x,y) = 3x^2/2y + f(C)$$

However with the boundary imposed ($u(x,0) = x^2$) this seem impossible to solve since to find the particular solution I would have to divide the a term by zero. Does the method described in the book have limited scope? How can I solve this differential equation? Also, why is $u(x,y)$ constant on the "characteristic curves"?

• If you introduce polar coordinates $x=r\cos \theta, y=r\sin \theta$, then $y\partial_x-x\partial_y=-\partial_\theta$. This is not the method you asked but it leads to a quick integration. – Giuseppe Negro Apr 2 '18 at 21:41
• Yeah that is a good trick, however I was more interested in the method and not the solution itself. – daljit97 Apr 2 '18 at 21:54

I cannot see where you found an impossibility or an exception in the rules.

I the case of $\quad yu_x-xu_y=3x \tag 1$

$$\frac{dx}{y}=\frac{dy}{-x}=\frac{du}{3x}$$ A first family of characteristic curves comes from $\quad\frac{dx}{y}=\frac{dy}{-u}\quad$ which leads to $$x^2+y^2=c_1$$ A second family of characteristic curves comes from $\quad\frac{dy}{-x}=\frac{du}{3x}\quad$ which leads to $$u+3y=c_2$$ The general solution of the PDE $(1)$ expressed on the form of implicit equation is : $$\Phi(x^2+y^2\:,\:u+3y)=0$$ where $\Phi$ is an arbitrary fonction of two variables.

Or equivalently, the general solution of PDE $(1)$ on explicit form is $\quad u+3y=F(x^2+y^2)$ $$u(x,y)=-3y+F(x^2+y^2)$$

where $F(X)$ is an arbitrary function of the variable $X$ with $X=x^2+y^2$.

Boundary condition : $\quad u(x,0)=x^2=F(x^2+0^2)=F(x^2)$.

So, the function $F$ is now determined : $\quad F(X)=X.\quad$ Putting it into the above general solution where $X=x^2+y^2$ leads to the particular solution which fits the boundary condition : $$u(x,y)=-3y+(x^2+y^2)$$

OTHER EXAMPLE : $\quad u_x+u_y+u=e^{x+2y} \tag 2$ $$u_x+u_y=e^{x+2y}-u$$ $$\frac{dx}{1}=\frac{dy}{1}=\frac{du}{e^{x+2y}-u}$$ A first family of characteristic curves comes from $\quad\frac{dx}{1}=\frac{dy}{1}\quad$ which leads to $$x-y=c_1$$ A second family of characteristic curves comes from $\frac{dy}{1}=\frac{du}{e^{x+2y}-u}=\frac{du}{e^{(c_1+y)+2y}-u}=\frac{du}{e^{c_1+3y}-u}$ $$\frac{du}{dy}+u=e^{c_1+3y}$$ This is a first order linear ODE easy to solve : $\quad u=\frac14 e^{c_1+3y}+c_2e^{-y}=\frac14 e^{(x-y)+3y}+c_2e^{-y}$ $$ue^y-\frac14 e^{x+3y}=c_2$$ The general solution of the PDE $(2)$ expressed on the form of implicit equation is : $$\Phi\left((x-y)\:,\:(ue^y-\frac14 e^{x+3y})\right)=0$$ where $\Phi$ is an arbitrary fonction of two variables.

Or equivalently, the general solution of PDE $(2)$ on explicit form is from $\quad ue^y-\frac14 e^{x+3y}=F(x-y)$ $$u(x,y)=\frac14 e^{x+2y}+e^{-y}F(x-y)$$ where $F(X)$ is an arbitrary function of the variable $X=x-y$.

• I believe that the author of my book seems to use a different method to solve the PDE, have a look here at an example. If you look at the last few lines of my question, there seems to be a problem (using the method used in the link) – daljit97 Apr 2 '18 at 21:29
• It is fundamentally the same method, with some variant of symbols and presentation. See :en.wikipedia.org/w/… and see en.wikipedia.org/wiki/Method_of_characteristics. You will have to do a bit of "gymnastic of the brain" in order to make the link. I wish you bon courage. – JJacquelin Apr 2 '18 at 21:43
• You can see that the general solution in the book is : $$u(x,y)=\frac14 e^{x+2y}+e^{-y}f(x-y)$$ which is exactly the same than in my answer. I will not repeat what is already in the book to determine the function $f$ according to the boundary condition. The method is the same. – JJacquelin Apr 2 '18 at 21:52
• $\frac{du}{dx} = 3x/y \implies u (x,y) = 3x^2/2y + f(C)$ is false because you integrate $3x/y$ without taking account that $y$ is not constant. You cannot solve it that way. This not a correct application of the method of characteristic. One have to find two independent families of characteristic curves. You got one :$x^2+y^2=C_1$ but not a second one. – JJacquelin Apr 2 '18 at 22:11
• @Isham yes you are right, I was too invested into thinking about why the method led me to that point that it didn't occur to me that a simple substitution was needed. – daljit97 Apr 2 '18 at 22:40

You can't have this: $$\frac{du}{dx} = 3x/y ==> u (x,y) = 3x^2/y + f(C)$$ Because you have a differential equation ( du,dx) but three variables namely x,y,u. Y shouldnt be there...Apart for that mistake you did a very good job.

Usisng the method of characteristics:

$$\frac {dx}{y}=\frac {dy}{-x}=\frac {dz}{3x}$$ $$-xdx=ydy \implies x^2+y^2=K_1$$ $$3xdy=-xdz \implies y+z/3 =K_2$$ $$f(x^2+y^2)=y+u/3$$ $$u(x,y)=3f(x^2+y^2)-3y$$ we have $u(x,0)=x^2$ $$f(x^2)=x^2/3 \implies f(x)=x/3$$ Therefore $$\boxed{u(x,y)=x^2+y^2-3y}$$

• Hmm in the book in order to solve $u_x + u_y + u = e^{x+2y}$ the author does setup the following equation: $\frac{\partial u}{\partial x} + u = e^{x+2y}$ Would that be an exception to the rule? – daljit97 Apr 2 '18 at 19:39
• @daljit97 I don't know I dont have the book right now. Thats weird that Strauss was so complicated about first order linear pde.. when it's easy to solve directly with characteristics method.. – Isham Apr 2 '18 at 19:42
• Here is the link with the exercise and the solution provided stemjock.com/STEM%20Books/Strauss%20PDEs%202e/Chapter%201/… – daljit97 Apr 2 '18 at 19:48
• Oh thats another method..I dont know that method...I deleted my comments on that method you used since I was completely wrong about it... – Isham Apr 2 '18 at 19:53
• Oh ok, could you provide a reference that explains and show why the method of characteristic works? Looking around I couldn't find anything satisfying. – daljit97 Apr 2 '18 at 20:03


# Does the product of two invertible matrix remain invertible?

If $A$ and $B$ are two invertible $5 \times 5$ matrices, does $B^{T}A$ remain invertible?

I cannot find out is there any properties of invertible matrix to my question.

Thank you!

• A common trick to answer the question "Is foo invertible?" is to actually write down the inverse of foo. In many cases this is pretty easy to do, as seen in the answers.
– user14972
Oct 24, 2012 at 16:10
• @Hurkyl What does foo mean? Oct 24, 2012 at 16:13
• @PENGTENG It is a nonsense word used as a generic placeholder in math and computer science. It's being used as a variable for an object here, but it's more frequently used for properties rather than objects. Oct 24, 2012 at 16:14
• @rschwieb I get it. Thanks! Oct 24, 2012 at 16:17
• How is this well researched? C'mon guys Aug 3, 2015 at 6:41

Of course: $B$ invertible implies $B^T$ invertible, and the product of two invertible matrices is clearly invertible.

This is easily seen from these equations: $$BB^{-1}=I\implies (BB^{-1})^T=I\implies (B^{-1})^TB^T=1,$$ and the fact that if $X$ and $Y$ are invertible, $(XY)^{-1}=Y^{-1}X^{-1}$.

Perhaps the general properties you should take away are these:

$(XY)^T=Y^TX^T$ and $(XY)^{-1}=Y^{-1}X^{-1}$.

• And this is of course true for $n \times n$-matrices and not just $5 \times 5$-matrices.
– N.U.
Oct 24, 2012 at 16:11
• "...the product of two matrices is clearly invertible." Only if the two matrices are themselves invertible: math.stackexchange.com/a/1026628 Aug 3, 2015 at 2:16
• @nacnudus of course, a reasonable person can tell this is a typo of the form of an omitted word from context. Thanks for indirectly alerting me to it. Aug 3, 2015 at 3:27

Yes. $$\det(B^T\,A)=\det(B^T)\det(A)=\det(B)\det(A)\ne0.$$ Moreover $$(B^T\,A)^{-1}=A^{-1}(B^{-1})^T.$$


# Are there arbitrarily large gaps between consecutive primes? [duplicate]

I made a program to find out the number of primes within a certain range, for example between $1$ and $10000$ I found $1229$ primes, I then increased my range to $20000$ and then I found $2262$ primes, after doing it for $1$ to $30000$, I found $3245$ primes.

Now a curious thing to notice is that each time, The probability of finding a prime in between $2$ multiples of $10000$ is decreasing, i.e it was $$\frac{2262-1229}{10000}=0.1033$$ between $10000$ and $20000$, and $$\frac{3245-2262}{10000}=0.0983$$ between $20000$ and $30000$,

So from this can we infer that there will exist two numbers separated by a gap of $10000$ such that no number in between them is prime? If so how to determine the first two numbers with which this happens? Also I took $10000$ just as a reference here, what about if the gap between them in general is $x$, can we do something for this in generality?

Thanks!

• If you let $p$ be the largest prime below $10\,000$, then the interval $[p\# - 10\,002, p\# - 2]$ certainly contains no primes (where $p\#$ is the primorial). It's probably not the first time, though – Arthur Feb 28 '16 at 10:26
• In the same spirit as Arthur's suggestion, there are no primes among the $k$ consecutive numbers $(k + 1)! + 2, \ldots, (k + 1)! + (k + 1)$. – Travis Willse Feb 28 '16 at 10:31
• @Arthur largest prime below $10000$ is $9973$, and by p# do you mean $$2.3.5....9973$$? . – Nikunj Feb 28 '16 at 10:42
• @Travis This looks interesting, could you please provide a proof? – Nikunj Feb 28 '16 at 10:43
• @Nikunj Yes, that is what I mean by $p\#$. – Arthur Feb 28 '16 at 10:46

The Prime Number Theorem states that the number of primes $\pi(x)$ up to a given $x$ is $$\pi(x) \sim \frac{x}{\log(x)}$$ which means that the probability of finding a prime is decreasing if you make your "population" $x$ larger. So yes, there exist a gap of $n$ numbers whereof none are prime.

The way to find the first gap for some $n$ has to be done through the use of software, since the exact distribution of prime numbers is only approximated by $\frac{x}{\log(x)}$.

EDIT: That the PNT implies that there's always a gap of size $n$ can be seen by considering what would happen if this was not the case; if there was a maximum gap of $n$ that was reached after some $x$, the probability of finding a prime between $x$ and some larger number $m$ would no longer decrease as $m \to \infty$, which contradicts the PNT.

• @Nikunj Yes, this is indeed only good for larger $x$, but since this approximation holds for any $x$, $10,000-20,000$ are very small numbers. But yeah, it's the best that we have as of yet :) – Bobson Dugnutt Feb 28 '16 at 12:12
• @BenMillwood Yeah, that is an unfortunate choice of words, I'll edit it! Thanks! – Bobson Dugnutt Feb 28 '16 at 15:09
• Could you elaborate on this just a little bit? You're right that the PNT implies that at some point you have to have n integers without a prime, or else π(x) could never be sub-linear, but I feel like this is a few too many steps away from a proof. – hobbs Feb 29 '16 at 6:31
• @hobbs: If you never had $n$ integers without a prime, then the number of primes up to a given $x$ could never be less than $\lfloor{x/n}\rfloor$. This contradicts the PNT. – mjqxxxx Mar 1 '16 at 3:54
• @mjqxxxx I'm not asking for me, I'm asking for it to be made more explicit in the answer for others' benefit :) – hobbs Mar 1 '16 at 4:12

Can we infer that there exist two numbers separated by a gap of $10000$, such that no number in between them is prime?

We can infer this regardless of what you wrote.

For every gap $n\in\mathbb{N}$ that you can think of, I can give you a sequence of $n-1$ consecutive numbers, none of which is prime.

There you go: $n!+2,n!+3,\dots,n!+n$.

So there is no finite bound on the gap between two consecutive primes.

• @Nikunj: Not necessarily. Also, the gap is not guaranteed to be exactly $n$ (or $10000$ in your example). That is why I have emphasized the specific part of your question to which I am referring in my answer. – barak manos Feb 28 '16 at 11:05
• @Nikunj because $n! + 1$ and/or $n! + n + 1$ may be composite as well, in which case one gets a sequence of $> n - 1$ consecutive composite numbers. This is the case, e.g., for $n = 3$. – Travis Willse Feb 28 '16 at 11:42
• @Nikunj To give a specific example in which both $n! + 1$ and $n! + n + 1$ are composite, observe that $5! + 1 = 121 = 11^2$ and $5! + 6 = 126 = 2 \cdot 3^2 \cdot 7$. In fact, the numbers $$114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126$$ are all composite. – N. F. Taussig Feb 28 '16 at 11:56
• @barakmanos I would say, more precisely, that is a lower bound of $n$. – PyRulez Feb 28 '16 at 18:23
• @goblin Take any $n!+i$ with $2\leq i \leq n$. You can factor out an $i$ from the factorial since $i \leq n$, so you get $i(k + 1)$, for $i > 1$ and $k > 1$, where $k = n!/i$. – TokenToucan Feb 29 '16 at 2:33

We cannot infer from the two observations in the question that there are gaps of arbitrary sizes between primes. As Lovsovs mentions in his answer, this does follow from the Prime Number Theorem (one needs suitable error bounds on the approximation $\pi(x) \sim \frac{x}{\log x}$, but even crude ones will do here).

As asked in the comments, it's easy to construct for any positive integer $k$ a sequence of $k$ consecutive composite numbers. For any positive integer $a \leq k + 1$, $a$ divides $(k + 1)!$, and so it divides $(k + 1)! + a$, but this implies that each of the $k$ numbers $(k + 1)! + 2, \ldots, (k + 1)! + (k + 1)$ is composite. (This is generally not the first sequence for which this is true: For example, for $k = 3$, the resulting sequence is $26, 27, 28$, but the first sequence of three consecutive composite positive integers is $8, 9, 10$.)

• (+1) Thanks a ton for the proof, Is there no other way besides the use of software to look for the first such sequence? – Nikunj Feb 28 '16 at 11:02
• @Nikunj No, because (as stated in my answer) we do not know the exact distribution of primes. – Bobson Dugnutt Feb 28 '16 at 11:33
• In general the problem is difficult, so yes, for sufficiently large $k$ computer assistance is necessary. Note that the primorial $(k - 1)\#$ is much smaller than the factorial $(k + 1)!$ for even modest $k$, so Arthur's solution gives a much better upper bound for the position of the first such sequence, but in general it too is a pretty poor bound: Already for $k = 5$ it gives $204, \ldots, 208$, but the first sequence of five consecutive composite numbers is $24, \ldots, 28$. – Travis Willse Feb 28 '16 at 11:40
• @ travis : proving $\pi(x) = o(x)$ is much easier than the prime number theorem, for example it follows directly from $s \int_1^\infty \pi(x) x^{-s-1} dx \sim \ln \zeta(s) \sim -\ln(s-1)$ when $s \to 1^+$. – reuns Feb 28 '16 at 12:21

The problem of finding large gap between consecutive primes is an old and well studied one. There certainly is a large gap between what we know to be true, and what we suspect.

As for what people expect, Cramer's random provides a good source heuristics. Roughly spreaking, you can think of a number $n$ to be "prime with probability $\frac{1}{\log n}$", but sometimes you also have to take into consideration that the primes tend not to be divisible by $2, 3, 5,$, etc. Apparently, if you make a lot of optimistic assumptions, then you can reach the conclusion that the longest gap between primes in integers $\leq X$ is roughly $\log^2 X$. Hence, if you want your gap to have size $g$, then you should look at integers $\leq e^{\sqrt{g}}$ (which, for $g = 10000$ is pretty large). See this post of Terence Tao for details.

Much less has been proved. The best known bounds for the largest gap between primes is due to Kevin Ford, Ben Green, Sergei Konyagin, James Maynard, Terence Tao, and says that the largest gap below $X$ is at least $c \frac{\log X \log \log X \log \log \log \log X}{\log \log \log X}$, where $c$ is a constant.

• In the last para, I guess “largest gap between integers” should be “…between primes”. I believe there are fairly sharp upper and lower bounds known for gaps between consecutive integers :-P – Peter LeFanu Lumsdaine Feb 28 '16 at 15:54
• @PeterLeFanuLumsdaine: Thanks! I suppose the bound is no longer true for integers, but maybe I'll check with one of the authors ;) – Jakub Konieczny Feb 28 '16 at 16:56
• That "best known bound" is amazingly low if you consider that the average gap between primes is log X, so that average gap is improved by less than a factor c log log X. And c might be considerably smaller than 1. – gnasher729 Feb 29 '16 at 23:08
• @gnasher729: It is rather far from the prediction, yes. But it should be pointed out that we're much better in predicting things about the primes than in actually proving them. – Jakub Konieczny Mar 1 '16 at 3:10
• Are there any known bounds on $c$? – Brevan Ellefsen Dec 11 '16 at 17:56

Any sequence with density $0$, in most reasonable meanings of "density $0$", has arbitrarily large gaps.

This is true (for example, and in increasing order of generality) if density $0$ is interpreted as:

• the density on $[1,n]$ converges to $0$, or

• the lim inf of the density on $[1,n]$ is $0$, or

• the lim inf of the density on intervals of length $n$ is $0$

The asymptotic density of primes $\pi(n)/n \to 0$ is 0-density in the strong sense, which is more than enough to ensure large gaps.

The $n!$ examples of large gaps can be reduced from factorial to "primorial" size and the latter seems to be the best currently known deterministic construction.

This isn't going to be a fancy answer, (secondary school student) but I would believe it to be as a number rises the number of possible factors of the number also rise meaning there is a larger chance of it having more factors than itself and one. So a number has every number smaller than it as a possible factor, so smaller numbers have less possible factors and therefore less chance of more than 1 factor and itself.

There several factors determining the size of gaps. As per you example 10,000 has a square root of 100 which give you 25 prime number divisors for the 10,000 and the 20,000 square root is 141 and 34 divisors, and the 30,000 square root is 173 and 40 divisors. The number of divisors helps to determine the gaps. The number of factors needed is based on the square root of your max number. The square roots are decreasing percentage of the max number: example 100 / 10000 = 1.000%, 141 / 20000 = .70500% and 173 / 30000 = .57666%. The number of factors increased against the square root but decreases against the max number. Example: 106,749,747,000,000 the square root is 10,331,977 the square root % is .00000968% with 685,159 factors the max gap in my sample 712. The large gaps seam to have a cycle pattern to them.


# Whats the difference between Antisymmetric and reflexive? (Set Theory/Discrete math)

Antisymmetric: $\forall x\forall y[ ((x,y)\in R\land (y, x) \in R) \to x= y]$ reflexive: $\forall x[x∈A\to (x, x)\in R]$

What really is the difference between the two? Wouldn't all antisymmetric relations also be reflexive?

• E.g. "$\leq$" and "$<$" are antisymmetric and "$=$" is reflexive. Apr 27 '15 at 17:02

Here are a few relations on subsets of $\Bbb R$, represented as subsets of $\Bbb R^2$. The dotted line represents $\{(x,y)\in\Bbb R^2\mid y = x\}$.

Symmetric, reflexive:

Symmetric, not reflexive

Antisymmetric, not reflexive

Neither antisymmetric, nor symmetric, but reflexive

Neither antisymmetric, nor symmetric, nor reflexive

• This is a great visual approach to understanding the meaning of the words! Apr 27 '15 at 20:04
• Thank you so much for making these, they're great! Apr 29 '15 at 2:48

No, there are plenty of anti-symmetric relations that are not reflexive.

For instance, let $R$ be the relation $R=\{(1,2)\}$ on the set $A=\{1,2,3\}$. This relation is certainly not reflexive, but it is in fact anti-symmetric. This is vacuously true, because there are no $x$ and $y$, such that $(x,y)\in R$ and $(y,x)\in R$.

Edit: Why is this anti-symmetric? Because in order for the relation to be anti-symmetric, it must be true that whenever some pair $(x,y)$ with $x\neq y$ is an element of the relation $R$, then the opposite pair $(y,x)$ cannot also be an element of $R$. This is true for our relation, since we have $(1,2)\in R$, but we don't have $(2,1)$ in $R$.

Also, the relation $R=\{(1,2),(2,3),(1,1),(2,2)\}$ on the same set $A$ is anti-symmetric, but it is not reflexive, because $(3,3)$ is missing.

As for a reflexive relation, which is not anti-symmetric, take $R=\{(1,1),(2,2),(3,3),(1,2),(2,1)\}$.

• Thank you so much for your answer, the last two parts make sense! :) I'm a little lost on the first part because the law says that if (x,y) and (y,x) then y=x. Could you elaborate a bit more on how R = {(1,2)} is anti-symmetric? Apr 27 '15 at 17:35
• Sorry, I think I messed up. Let me edit my post. There. Now, I have redone the last two examples, because they were wrong. I'll edit my post further to elaborate on why the first relation is in fact anti-symmetric. :)@TaylorTheDeveloper Apr 27 '15 at 17:42
• This may sound like a naive question but would'nt this example be asymmetric also then by vacuous agument Oct 19 '20 at 11:31
• @angshuknag Yes, the relation $R=\{(1,2)\}$ is also asymmetric. In fact, being asymmetric is equivalent to being both anti-symmetric and not reflexive. Oct 19 '20 at 15:08

They're two different things, there isn't really a strong relationship between the two.

Based on the definitions you're using, they both give two different criteria for concluding that $(x, x) \in R$.

For any antisymmetric relation $R$, if we're given two pairs, $(x, y)$ and $(y, x)$ both belonging to $R$, then we can conclude that in fact $x = y$, so that that

$$(x, y) = (x, x) = (y, x),$$

and $(x, x) \in R$. It may really be better stated as saying that

$$\text{ If } x \neq y, \text{ then at most one of (x, y) or (y, x) is in R}.$$

That is, it may be a bit misleading to even think about $(x,y)$ and $(y, x)$ as being pairs in $R$, since antisymmetry forces them to in fact be the same pair, $(x, x)$.

Antisymmetric relations may or may not be reflexive. $<$ is antisymmetric and not reflexive, while the relation "$x$ divides $y$" is antisymmetric and reflexive, on the set of positive integers.

A reflexive relation $R$ on a set $A$, on the other hand, tells us that we always have $(x, x) \in R$; everything is related to itself. Reflexive relations may or may not be symmetric, or antisymmetric:

$\leq$ is reflexive and antisymmetric, while $=$ is reflexive and symmetric.

• Also, I may have been misleading by choosing pairs of relations, one symmetric, one antisymmetric - there's a middle ground of relations that are neither! For example, the relation "$x$ divides $y$" on the set of all integers is neither symmetric nor antisymmetric (and neither reflexive, nor irreflexive). Apr 27 '15 at 17:52
• The divisibility relation is reflexive, even on all integers. Apr 27 '15 at 20:04
• @JadeNB Thank you, of course you're right; I'm not sure why I had decided it wasn't! Apr 27 '15 at 20:07
• Probably the presence of 0 caused some reflexive (no pun intended!) worries. Apr 27 '15 at 20:11


# Rotatable matrix, its eigenvalues and eigenvectors

We say that a real matrix is rotatable iff after turning it clockwise on $90^{\circ}$ it doesn't change.

I'm interesting about eigenvalues and eigenvectors (belonging to non-zero eigenvalues) of such type of matrices. For example, it is not hard to show that for every tuple of real values $\lambda_1,\ldots,\lambda_{k}$ there exists $n\in\mathbb{N}$ and a rotatable $n\times n$ matrix $A$ such that all $\lambda_i$ are eigenvalues of $A$.

Indeed, let us consider a matrix $$A_1 = \begin{pmatrix} a & a \\ a & a\\ \end{pmatrix}.$$ Then of course $A_1$ is rotatable and its characteristic polynomial $\chi_{A_1}(x) = x^2-2ax = x(x-2a)$ and of course for every $\lambda$ we can choose $a$ (for example $\lambda/2$).

Now let us show how we can construct a rotatable matrix $A_2$ with prescribed eigenvalues $\lambda_1,\lambda_2$. For example, we can consider a matrix of the form $$A_2 = \begin{pmatrix} b&0&0&b\\ 0&a & a&0 \\ 0&a & a&0\\ b&0&0&b \end{pmatrix}.$$ Then $\chi_{A_2}(x) =x^2(x-2a)(x-2b)$. And we are done for $a=\lambda_1/2, b = \lambda_2/2$. Of course, using this method we can construct the required rotatable matrix for every $k$ of size $2k$.

Also my experiments show that all eigenvectors $v = (v_1,\ldots,v_n)^T$ belonging to non-zero eigenvalues of roratable matrix are symmetric, that is $v_i = v_{n-i+1}$. It is simple to prove this in the case, where $n=2$. Also I tried to prove it by induction on $n$, but my attempts failed.

My question.

1. For a given tuple $\lambda_1,\ldots, \lambda_k$ can we construct a rotatable matrix $A$ of size $n\times n$, where $n<2k$, such that all $\lambda_i$ are eigenvalues of $A$.

2. Is it always true that all eigenvectors of non-zero eigenvalues of rotatable matrix are symmetric? And if the answer is "yes", how to prove this.

• Is the (2,4) element of $A_2$ supposed to be 0 rather than $b$? If not, how is it rotatable? May 15, 2018 at 11:23
• Doesn't every rotatable matrix become of the form $\begin{bmatrix}M & M \\ M & M\end{bmatrix}$, i.e., $\begin{bmatrix}1 & 1\\ 1 & 1\end{bmatrix} \otimes M$, if you conjugate it by a suitable permutation? That looks like it would simplify the analysis a lot. May 15, 2018 at 21:04
• @FedericoPoloni, does this hold for $3\times 3$ matrices? I don't sure. May 16, 2018 at 10:30
• @MikhailGoltvanitsa No, only in even dimension. May 16, 2018 at 14:28
• In particular your matrices are centrosymmetric (apply the rotation twice will also yield the same matrix) en.m.wikipedia.org/wiki/Centrosymmetric_matrix May 18, 2018 at 15:34

Consider the matrix $$P= \begin{pmatrix} 0 & \ldots & 1 \\ \vdots & 1 & \vdots \\ 1 & \ldots & 0 \end{pmatrix}$$ with $1$s along the "other" main diagonal and $0$s elsewhere. Then $(PA)^t$ is a rotation of the matrix $A$ by $90^\circ$ (you can check this on the basis $E_{i,j}$ of matrices with a 1 in the $(i,j)$ slot and 0s elsewhere). So $A$ is rotatable if and only if $(PA)^t = A$, i.e. $PA=A^t$. Note that $P^2=I$ and $A^tP = A$, so $A^t = AP$ and hence $A$ and $P$ must commute. (For what follows below we assume that $A$ is a real matrix. Otherwise I think all you can say is that $A$ preserves the splitting $\mathbb{C}^n=E_+ \oplus E_-$ as explained below. In the complex case we get symmetric eigenvectors of the $v+Pv$ as well as the skew symmetric vectors $v-Pv$.)

It then follows that $A$ and $A^t$ commute and so, by the Spectral Theorem, $A$ has a unitary basis. Now, notice that $P$ has eigenvalues $1$ and $-1$ with eigenvectors $e_i + Pe_i$ and $e_i-Pe_i$ for $1 \leq i \leq \frac{n+1}{2}$, respectively. Let $E_+ = \ker (P-I)$ and $E_- = \ker (P+I)$. Then $A(E_\pm) \subseteq E_\pm$, so $A$ must preserve the splitting $\mathbb{R}^n = E_+ \oplus E_-$. Note that $\dim E_+ = \left\lfloor \frac{n+1}{2}\right\rfloor$.

Now, the answer to question $2$ is yes if we require the corresponding eigenvalue to be a nonzero real number. If $\lambda$ is an eigenvalue with eigenvector $v \in E_\pm$ and $(\:\: ,\:\: )$ is the standard hermitian inner product on $\mathbb{C}^n$, we have $$\lambda(v,v)=(v,A^tv) = (v,APv) = \pm \bar{\lambda}(v,v).$$

So $\lambda \in \mathbb{R}$ if and only if $v \in E_+$ and $\lambda \in i \mathbb{R}$ if and only if $v \in E_-$. Note that $v \in E_+$ if and only if $Pv=v$, which is the same as requiring that $v$ is symmetric. Furthermore, note that the eigenvectors in $E_- \setminus \{0\}$ have complex coefficients as they satisfy $Av=i \mu v$. So it is precisely the real eigenvectors of nonzero eigenvalues that are symmetric. The others will be "skew symmetric".

Question $1$ is a bit trickier, and the answer is no in general. Let us treat the even and odd dimensional cases separately.

Let $n=2m$. If $A$ is rotatable it has the form $$A= \begin{pmatrix} B & B^t P \\ PB^t & PBP \end{pmatrix}$$ for some matrix $B$ (here $P$ and $B$ are matrices of size $m\times m$). Now, $\lambda$ is a real eigenvalue of $A$ with eigenvector $\begin{pmatrix} v \\ Pv \end{pmatrix}$ if and only if $$Bv + B^t v = \lambda v .$$ So $\lambda = 2 \operatorname{Re}(\eta)$, where $\eta$ is an eigenvalue of $B$ and $v$ is an eigenvector of $B+B^t$. Similarly, $\lambda= i \mu$ is a purely imaginary eigenvalue of $A$ with eigenvector $\begin{pmatrix} v \\ -Pv \end{pmatrix}$ if and only if $Bv - B^t v = i \mu v$, so $\mu = 2 \operatorname{Im} (\eta)$, where $\eta$ is an eigenvalue of $B$ and $v$ is an eigenvector of $B-B^t$.

So, if you choose more than $n$ real numbers, or choose any complex numbers $\lambda$ that do not satisfy $\bar{\lambda} = \pm \lambda$, or you choose more than $n$ purely imaginary numbers, then they cannot be eigenvalues of $A$.

Now, in the odd dimensional case $n=2m+1$, $A$ takes the form $$A= \begin{pmatrix} B & u & B^t P \\ u^t & a & (Pu)^t \\ PB^t& Pu & PBP \end{pmatrix}$$ for some matrix $B$, vector $u$ and real number $a$. For $\lambda$ a real eigenvalue of $A$, we have the eigenvector $\begin{pmatrix} v \\ b \\Pv \end{pmatrix}$, so $$(B+B^t)v + bu = \lambda v$$ and $$2(u,v)+ab=\lambda b.$$ Choosing $u=0$ gives $$(B+B^t)v=\lambda v$$ and $ab=\lambda b.$ So we can pick $m$ eigenvectors for $B+B^t$ as above, and also the vector given by $v=0$ and $b=1$, which has eigenvalue $a$. The imaginary eigenvalues $i \mu$ have eigenvectors $\begin{pmatrix} v\\0\\-Pv \end{pmatrix}$, and so $$(B-B^t)v=i\mu v,$$ with the other equation being $0=0$. So as before $\mu$ is twice the imaginary part of an eigenvalue of $B$.


# Evaluate $\lim_{n \to \infty} \int_{0}^1 \frac{n+1}{2^{n+1}} \left(\frac{(t+1)^{n+1}-(1-t)^{n+1}}{t}\right) \mathrm{d}t$

Evaluate $$\lim_{n \to \infty} \int_{0}^1 \frac{n+1}{2^{n+1}} \left(\frac{(t+1)^{n+1}-(1-t)^{n+1}}{t}\right) \mathrm{d}t$$

For this integral, I have tried using integration by parts and then evaluating the limit, but I don't think the integral inside converges. However, the limit does exist and the answer given in my book is $2$.

Any help will be appreciated.

• can you show us your working? – danimal Apr 17 '15 at 11:06
• @Henry which book is this? – Kugelblitz Jun 17 '17 at 4:00

First, consider the following two Lemmas,

Lemma $1$: $$\lim_{n \to \infty} \sum_{r=0}^n \left(\dfrac{1}{\displaystyle\binom{n}{r}}\right) =2$$

Proof : First of all, note that the limit exists, since, if we let
$$\text{S}(n)=\displaystyle \sum_{r=0}^n \left(\dfrac{1}{\dbinom{n}{r}} \right)$$

then $\text{S}(n+1)<\text{S}(n)$ for $n \geq 4$. Now,

$\text{S}(n) = 1+ \displaystyle \sum_{r=1}^n \dfrac{1}{\dbinom{n}{r}}$

$\implies \text{S}(n) = 1+ \displaystyle \sum_{r=1}^n \dfrac{r}{n} \times \dfrac{1}{\dbinom{n-1}{r-1}} \ \left[\text{since} \dbinom{n}{r}= \dfrac{n}{r} \times \dbinom{n-1}{r-1} \right]$

Also,

$$\text{S}(n) = 1+ \sum_{r=1}^n \dfrac{1}{\dbinom{n}{r}} = 1+ \sum_{r=1}^n \dfrac{1}{\dbinom{n}{n-r+1}} = 1+ \sum_{r=1}^n \dfrac{n-r+1}{n} \dfrac{1}{\dbinom{n-1}{n-r}} = 1+ \sum_{r=1}^n \dfrac{n-r+1}{n} \dfrac{1}{\dbinom{n-1}{r-1}}$$

$\left[ \text{since} \ \displaystyle \sum_{r=a}^b f(r) = \displaystyle \sum_{r=a}^b f(a+b-r) \ \text{and} \ \dbinom{n}{r}=\dbinom{n}{n-r} \right]$

Thus, we have,

$$\begin{cases} \text{S}(n) = 1+ \displaystyle \sum_{r=1}^n \dfrac{r}{n} \times \dfrac{1}{\dbinom{n-1}{r-1}}\\ \text{S}(n) = 1+ \displaystyle \sum_{r=1}^n \dfrac{n-r+1}{n} \dfrac{1}{\dbinom{n-1}{r-1}} \end{cases}$$

Adding the above two expressions, we get,

$2\text{S}(n) = 2 + \displaystyle \sum_{r=1}^n \left( \dfrac{r}{n} \times \dfrac{1}{\dbinom{n-1}{r-1}} + \dfrac{n-r+1}{n} \dfrac{1}{\dbinom{n-1}{r-1}} \right)$

$= 2 + \dfrac{n+1}{n} \displaystyle \sum_{r=1}^n \dfrac{1}{\dbinom{n-1}{r-1}}$

$= 2 + \dfrac{n+1}{n} \times \text{S}(n-1)$

Since $n \to \infty$, we have $\text{S}(n) = \text{S}(n-1) = \text{S}$ (say)

$\implies 2\text{S} = \left(\dfrac{n+1}{n}\right) \times \text{S} +2$

$\implies \text{S} = \dfrac{2n}{n-1} = 2$ [since $n \to \infty$]

Lemma $2$ : $$\int_{0}^1 x^r (1-x)^{n-r} \mathrm{d}t = \dfrac{1}{(n+1)}\times \dfrac{1}{\dbinom{n}{r}}$$

Proof : Consider the $\text{R.H.S.}$,

$\text{I} = \displaystyle\int_{0}^1 x^r (1-x)^{n-r} \mathrm{d}t$

Let $x = \sin^2 \theta$

$\implies \text{I} = \displaystyle\int_{0}^{\frac{\pi}{2}} 2 \sin^{2r+1} \theta \cos^{2n-2r} \theta \ \mathrm{d}\theta$

Now, using Walli's Formula (or reduction formula) for the above integral, we have,

$\text{I} = \dfrac{1}{(n+1)}\times \dfrac{1}{\dbinom{n}{r}}$

This proves our Lemmas.

Now,

$$\text{J} = \lim_{n \to \infty} \int_{0}^1 \frac{n+1}{2^{n+1}} \left(\frac{(t+1)^{n+1}-(1-t)^{n+1}}{t}\right) \mathrm{d}t$$

Since it is an even function in $t$, we have,

$$\text{J} = \frac{1}{2} \times \lim_{n \to \infty} \int_{-1}^1 \frac{n+1}{2^{n+1}} \left(\frac{(t+1)^{n+1}-(1-t)^{n+1}}{t}\right) \mathrm{d}t$$

Let $t = 2x-1$

$\implies \text{J} = \displaystyle \lim _{n \to \infty} \int_{0}^1 (n+1) \left(\dfrac{x^{n+1}-(1-x)^{n+1}}{2x-1}\right) \mathrm{d}x$

$=\displaystyle \lim _{n \to \infty} \int_{0}^1 (n+1) (1-x)^n \left(\dfrac{\left(\frac{x}{1-x}\right)^{n+1}-1}{\frac{x}{1-x}-1}\right) \mathrm{d}x$

$=\displaystyle \lim _{n \to \infty} \int_{0}^1 (n+1) \sum_{r=0}^n (1-x)^n \left(\frac{x}{1-x}\right)^{r} \mathrm{d}x$

$=\displaystyle \lim _{n \to \infty} \sum_{r=0}^n (n+1) \int_{0}^1 x^r(1-x)^{n-r} \mathrm{d}x$

$=\displaystyle \lim _{n \to \infty} \sum_{r=0}^n \dfrac{1}{\dbinom{n}{r}}$ (Using Lemma 2)

$=\boxed{2}$ (Using Lemma 1).

Side Note : Another way to prove Lemma 1 is to use sandwich theorem.

• Wow! Ingenious and rigorous as always :) But can you please tell what led you to think this way? – Henry Durham Apr 17 '15 at 13:06
• @Samurai Your question reminded me of Beta functions which I have used to create this solution. Although understanding the solution doesn't need the understanding of Beta functions, but if you look carefully at Lemma 2, it is the definition of Beta function, for which I've provided an elementary proof. The rest followed from rigorous brainstorming :) – MathGod Apr 17 '15 at 13:11
• @MathGod Thanks a lot :D – Henry Durham Apr 17 '15 at 13:14
• @MathGod Would you mind explaining how you got: $$\int_{0}^1 (n+1) \sum_{r=0}^n (1-x)^n \left(\frac{x}{1-x}\right)\mathrm{d}x$$ after $$\int_{0}^1 (n+1) (1-x)^n \left(\dfrac{\left(\frac{x}{1-x}\right)^{n+1}-1}{\frac{x}{1-x}-1}\right) \mathrm{d}x$$ It is not so clear to me. Thank you. – Tolaso Apr 18 '15 at 4:33
• @Tolaso Thanks for pointing that out. It was a minor typo and I've now corrected it. The expression is sum of G.P. – MathGod Apr 18 '15 at 6:05

It is worth to notice that: $$\int_{0}^{1}\frac{1-(1-t)^n}{t}\,dt = H_n\tag{1}$$ while: $$\int_{0}^{1}\frac{(1+t)^n-1}{t}\,dt=\int_{0}^{1}\sum_{k=0}^{n-1}(1+t)^k\,dt =\sum_{k=1}^{n}\frac{2^k-1}{k}\tag{2}$$ so: $$J_n=\int_{0}^{1}\frac{(1+t)^n-(1-t)^n}{t}\,dt = \sum_{k=1}^{n}\frac{2^k}{k}.\tag{3}$$ The last line also gives: $$\frac{n\, J_n}{2^n} = \sum_{k=1}^{n}\frac{2n}{n+1-k}\cdot 2^{-k} \tag{4}$$ and when $n$ approaches $+\infty$, by the dominated convergence theorem the RHS of $(4)$ approaches: $$\sum_{k=1}^{+\infty}2\cdot 2^{-k} = \color{red}{2} \tag{5}$$ as wanted.

• Pardon me, but I don't know what's dominated convergence theorem. – Henry Durham Apr 17 '15 at 11:33
• – Jack D'Aurizio Apr 17 '15 at 11:35
• That's sweet... I guess, without knowing the properties of $H_n$ you couldn't do much... But how did you get the first equality in $(2)$? – Igor Deruga Apr 17 '15 at 11:43
• @Igor: $t=(1+t)-1$ and $1+z+\ldots+z^n = \frac{z^{n+1}-1}{z-1}$. – Jack D'Aurizio Apr 17 '15 at 11:48

We need to address area around $0$ separately from area around $1$. Pick $\epsilon\in(0,1)$. Then on $[0,\epsilon]$ $$\frac{(1+t)^{n+1}-(1-t)^{n+1}}t\le (n+1)(1+\epsilon)^{n}+(n+1)(1-\epsilon)^{n}$$(you can see this if you multiply both sides by $t$ and compare derivatives) and hence $$\lim_{n\to\infty}\int_0^{\epsilon}\frac {n+1}{2^{n+1}}\frac{(1+t)^{n+1}-(1-t)^{n+1}}t\,dt=0$$ On the other hand $$\lim_{n\to\infty}\int_\epsilon^1\frac {n+1}{2^{n+1}}\frac{(1+t)^{n+1}-(1-t)^{n+1}}t\,dt=\lim_{n\to\infty}\int_{\epsilon}^1\frac 2 t\frac {n+1}{n+2} d\Big(\big(\frac {1+t}{2}\big)^{n+2}+\big(\frac {1-t} 2\big)^{n+2}\Big)=\int_{\epsilon}^1\frac 2 t dI(t=1)=2$$ since $\frac 2 t$ is bounded on $[\epsilon,1]$. Hence the original limit is $2$.


# Math Help - Complex Numbers Question

1. ## Complex Numbers Question

Hey everybody,

So I have been doing a few exercises on complex numbers, but there is one I can't seem to get out.

If x and y are real, solve the equation:

$\frac{jx}{1+jy}=\frac{3x+j4}{x+3y}$

I started manipulating it to make it more manageable:

$\frac{jx(x+3y)}{1+jy}=3x+j4$

$jx^2+j3xy=(3x+j4)(1+jy)$

etc.

I have the feeling I'm going nowhere with it. Could somebody please give me a hand?

Cheers,

Evanator

2. Originally Posted by evanator
Hey everybody,

So I have been doing a few exercises on complex numbers, but there is one I can't seem to get out.

If x and y are real, solve the equation:

$\frac{jx}{1+jy}=\frac{3x+j4}{x+3y}$

I started manipulating it to make it more manageable:

$\frac{jx(x+3y)}{1+jy}=3x+j4$

$jx^2+j3xy=(3x+j4)(1+jy)$

etc.

I have the feeling I'm going nowhere with it.
In fact, you,re going in exactly the right direction. Next step is to multiply out the brackets on the right (remembering that $j^2=-1$). Then compare the real parts and the imaginary parts on both sides of the equation.

3. $\frac{jx}{1+jy}=\frac{3x+j4}{x+3y}
$

${(jx)}{(x+3y)}=(1+jy)(3x+j4)$

$jx^2+3jxy=3x+4j+3jxy-4y$

$jx^2=3x+4j-4y$

so:

(1) $x^2=4$

(2) $0=3x-4y$

Solve this system for x and y

4. Thank you both of you. Finishing it off:

$jx^2+j3xy=3x+j3xy+j4+(j^2)4y$

$0+jx^2=(3x-4y)+j(4)$

$x^2=4$

$3x-4y=0$

$x=\pm2$

Taking the positive value of x and solving the simultaneous equations:

$3x=6$

$-3x+4y=0$

$4y=6$

$y=\frac{6}{4}=1.5$

If we take x=-2 we can see that y will come out as -1.5, so the final results are:

$x=\pm2$

$y=\pm1.5$

5. No, that is NOT the answer because x and y are not independent.

The correct answer is "x= 2 and y= 1.5 or x= -2 and y= -1.5"

Writing " $x= \pm 2, y= \pm 1.5$" implies that any combination of those, such as x= 2, y= -1.5, will satisfy the equation and that is not true.

6. Right. Thanks, HallsofIvy. I did understand that the x-value and y-value were dependent on each other, but I didn't realize I was implying otherwise with the plus-minus sign.


# Variable Increments in a Table

I am working on a problem involving the piecewise function:

$$x^2 \quad \mbox{if } \quad x < 2$$

$$(x-3)/(\sqrt{x-2}-1) \quad \mbox{if } \quad x \geq 2$$

I am supposed to find the values of the limits as $x$ approaches $2$ from the right and the left. I've found these values to be $1$ from the right and $4$ from the left.

However, the question I'm struggling to answer involves the formatting of a limit table.

I've been able to make two tables of (x,f[x]) from the left and the right, and I've listed specific values of $x$ to approach $2$ with. However, the question wants me to make a $4$ column table using only one table command, and to use a rule to make the step size increments progressively smaller. I'm not sure how to do this.

Here's the text of the question: "To evaluate a two sided limit, create a single table (generated by a single table command) with four columns, combining the left-and right-sided limit tables into one. Values of $x$ should approach the limiting value moving down the table. To evaluate a limit, create table(s) that use step-size increments that are progressively smaller as the limiting value is approached (eg. x=3, 2.1, 2.01, 2.001,... for a limit as x→ 2+). Rather than listing every $x$-value, use a rule to generate the $x$-values."

Here is how I've made my tables thus far:

TableForm[Table[{x, f[x]}, {x, {4, 3, 2.5, 2.5, 2.25, 2.1, 2.05, 2.02,2.001}}]]
TableForm[Table[{x, f[x]}, {x, {-1, 0, 1, 1.5, 1.75, 1.9, 1.95, 1.97, 1.98, 1.999}}]]

• Is this homework? If so, please tag it as such.
– ciao
Feb 9, 2017 at 8:21

One possible way to use Accumulate to generate the steps.

N@Table[i, {i, Accumulate[Table[1/(2^t), {t, 1, 10}]]}]

Then you can these values for building the steps, by adding and subtracting the above from 2 for each side.

ClearAll[f, i, t, x];

f[x_] := Piecewise[{{x^2, x < 2}, {(x - 3)/(Sqrt[x - 2] - 1), x >= 2}}]

r = Table[{x = 1 + i; x, f[x], x = 3 - i; x, f[x]},
{i,Accumulate[Table[1/(2^t), {t, 1, 10}]]}];

h = {"x", "limit from left", "x", "limit from right"};
Grid[Join[{h}, N@r], Alignment -> Left, Frame -> All]

Plot[f[x], {x, -1, 3}, Frame -> True, GridLines -> Automatic,
GridLinesStyle -> LightGray]


# Why can the limit of a sequence approach a number and converge, but the limit of the series must approach $0$ to converge?

My question may not make much sense because I'm still trying to wrap my mind around infinite sequences and series. I seem to have good working knowledge of when and why to apply a certain tests for a given series, but something just seems like it is missing in my understanding.

Conceptually, why can a sequence converge at any number (except $-\infty$ and $\infty$, which aren't numbers anyway), but the limit of a series must approach $0$ to converge?

Is the reason behind the requirement for the limit of a series approaching $0$ to converge because the series eventually stops summing numbered terms infinitely due to its convergence to a specific number? (Sorry if this question makes no sense)!

And is it acceptable for a sequence to approach any number simply because the sequence isn't being summed?

• You mean thr limit of the summand – Mhenni Benghorbal Dec 4 '13 at 15:17
• Try to look at the partial sums of the series when $a_n$ does not converge to $0$ for $n\rightarrow \infty$. – user112167 Dec 4 '13 at 15:17
• @MhenniBenghorbal excuse my ignorance, but what is a summand? – hax0r_n_code Dec 4 '13 at 15:18

I think you are mixing sequences and their sums. The terminology is not very helpful here, I would suggest the following.

Let $a_k$ be any sequence. Then you can define the sequence of partial sums: $$S_n = \sum_{k=1}^n a_k.$$ Then $$\sum_{k=1}^\infty a_k = \lim_{n\to \infty} S_n.$$

Given any sequence $S_n$ you can find some $a_k$ such that $S_n$ are the partial sums of $a_k$.

So you see that a series may have any limit, as the sequence do. However if a series $S_n$ has a finite limit, then the corresponding sequence $a_k$ must tend to zero.

Don't confuse the series $S_n$ with its general term $a_k$.

• So is it accurate to say that a series is simply a sequence of partial sums from a given sequence? – hax0r_n_code Dec 4 '13 at 15:29
• That is my preferred point of view. – Emanuele Paolini Dec 4 '13 at 15:31
• That's interesting because I never could see the relationship between a sequence and a series. – hax0r_n_code Dec 4 '13 at 15:37
• I say that the terminology does not help because formally the expression: $\sum_{k=1}^\infty a_k$ is a number. So it makes no sense to say that such expression converges or diverges. But really you are referring to the sequence of partial sums, not to the sum itself. – Emanuele Paolini Dec 4 '13 at 15:40
• @EmanuelePaolini It makes perfect sense to say that expression converges or diverges. It is nothing but a limit, and limits can sometimes fail to be numbers, but instead diverge. – Potato Dec 4 '13 at 16:05


# inversion of the Euler totient function

Given an integer $n$ find smallest integer $x$ such that $\varphi(x) = n$.

$$10^5 < n < 10^8$$

I know that lower bound for searching is $n+1$ and the upper bound is

$$\frac{n}{e^{0.577}\log(\log(n))} + \frac{3}{\log(\log(n))}$$

But my problem is that for such a large values of $n$ my program with precomputed $\varphi(x)$ values has to do a lot of searching and is highly inefficient.

Can you please provide any other method for doing the same .

Thanks.

-
My apologies if I am missing something basic here, but what function is it that you are defining to mean $\varphi(x)$? – Amzoti Dec 26 '12 at 18:31
Euler totient function.. phi(x)... gives the number of integers less than x which are coprime to x – Purva Gupta Dec 26 '12 at 18:35
So I don't waste time, you are asking for an efficient algorithm for inverting the Euler Totient function. Correct? – Amzoti Dec 26 '12 at 19:26
yes.. which can work for large test cases – Purva Gupta Dec 26 '12 at 19:48
I would recommend changing your title to be more descriptive in case others are searching for this too and there have been previous questions regarding approaches. Regards – Amzoti Dec 27 '12 at 2:19

Warning, you are going to have to do some work to get your hands around this answer, but I provided enough details for you to work through it and it answers your question.

Goal: Given an integer $n$ find the smallest integer $x$ such that $\varphi(x) = n$.

Approach

One can calculate all of the possible integers with a totient of $n$ using "Inverse Totient Trees."

For instance, take the integers with a totient of $24$. Then,

$\varphi(N) = 24$

$\varphi(24) = 8$

$\varphi(8) = 4$

$\varphi(4) = 2$

$\varphi(2) = 1$

There are $5$ "links" (designate this as $L$) in the totient chain so to speak, with $4$ intervals. In general, the greatest integer that can have a totient of $n$ is $2*3^{(L-1)}$, which means that $2*3^{(5 - 1)}$ $= 162$ is the upper bound of an integer with a totient of $24$.

In fact, via a simple proof by exhaustion, one can easily check a table and see that the smallest integer where $\varphi(x) = 24$ is $x = 35$ (see the list below).

$$\varphi(x)= n = 24 \rightarrow x = 35, 39, 45, 52, 56, 70, 72, 78, 84, 90$$

Here are a couple of related number sequences which include references for you to investigate this approach.

A032447 Inverse function of phi( )

A058811 Number of terms on the n-th level of the Inverse-Totient-Tree (ITT)

So, the next obvious question, is there a program that is related or tangentially related that can generate all of those values in the range specified using some approach?

Yes, see Solving $\varphi^{-1}(x) = n$, where $\varphi(x) = n$ is Euler's totient function - testing Carmichael's conjecture. Please see the references there for further details on this program - including the C-source code.

Lets do the example above and the some examples within your range (note that numbers are obviously never odd because $\varphi(x)$ is always even for $n \ge 3$).

$n = 24$: Enter $n = 24, e = 0, f = 0$: and look at the resulting $10$-bolded numbers and compare that list to the above. If you sort that list from low to high, it is all of the $x's$ that produce that desired $n$. Of course, you only want the minimal (last bolded number in the display) one. So $\varphi^{-1}(35) = 24$.

$n = 10^{5}$: Enter $n = 100000, e = 0, f = 0$: see the metrics for how many numbers satisfy this, but your desired result is $x = 100651$.

$n = 10^5 + 1$: Enter $n = 100001, e = 0, f = 0$: Odd numbers are not permissible by statement above.

$n = 10^{5} + 2$: Enter $n = 100002, e = 0, f = 0$: see the metrics for how many numbers satisfy this, but your desired result is $x = 100003$.

$n = 5*10^{5}$: Enter $n = 500000, e = 0, f = 0$: see the metrics for how many numbers satisfy this, but your desired result is $x = 640625$.

...

$n = 10^{8}$: Enter $n = 100000000, e = 0, f = 0$: see the metrics for how many numbers satisfy this, but your desired result is $x = 100064101$.

Asides

This is a very interesting problem and has been asked before, so I am summarizing some of the other discussions and references I found in case others want to consider more approaches.

There are several papers on the topic of finding the inverse of the Euler Totient function:

Euler's Totient Function and Its Inverse, by Hansraj Gupta

The number of solutions of $\phi(x) = m$, by Kevin Ford

On the image of Euler’s totient function, R.Coleman

Complexity of Inverting the Euler Function, by Scott Contini, Ernie Croot, Igor Shparlinski

There have been several questions along these lines, for example, the-inverse-of-the-euler-totient-function, inverting-the-totient-function and finding-the-maximum-number-with-a-certain-eulers-totient-value

There are some other code examples that use different methods for you to consider, see, for example:

Inversion of Euler Totient Function by Max Alekseyev and you can experiment with this since PARI/GP is free. Play around with the function and see if you can modify your approach with this approach.

Discussion and implementation of an efficient algorithm for finding all the solutions to the equation EulerPhi[n] = m, by Maxim Rytin is a nice article off of wolfram that gives an efficient algorithm for computing the inverse of the Euler totient function. Download the invphi.nb file at the bottom and get Mathreader. As an alternative, you can see oeis A006511.

MAGMA - see FactoredEulerPhiInverse(n)

Regards

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Thanks a lot for your descriptive answer :) – Purva Gupta Dec 27 '12 at 8:19
@Purva Gupta, you are very welcome! We are all here to learn from each other and MSE is a wonderful site! – Amzoti Dec 27 '12 at 9:01
2∗3^(L−1) is it applicable for all or only specific numbers? – Purva Gupta Dec 27 '12 at 9:18
All, it is an upper bound using this method for the largest n, recalling that n must be even numbers only from the note. – Amzoti Dec 27 '12 at 13:08
$+1^{+1^{+1}}\quad\checkmark$ – amWhy May 10 '13 at 1:20

See also my recent paper "Computing the (number of) inverses of Euler's totient and other multiplicative functions", which presents a generic dynamic programming algorithm for finding the inverses of a multiplicative function for a given integer value.

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1. ## Factorization Question 2

For the question below, does anyone have a quick and easy method for finding the two numbers that when added together gives 11x, and when multiplied together gives -180? I know the numbers are 20 and -9 but it takes me forever to find them? Is there is quick way?

$\text{Factorize } 12x^2+11x-15$

2. ## Re: Factorization Question 2

Hello

I don't now if this will help, but I do it this way. It doesn't take ages.
You know you have either:

1,15 or 3,5

and your looking for a difference of 11

Write out:

(12x )(x )
(6x )(2x )
(4x )(3x )

It doesn't tke long to work out the difference of the possible products and see that 4x5 - 3x3 = 11

So you have

(4x - 3)(3x + 5)

3. ## Re: Factorization Question 2

If you have a polynomial $f(x)$ and the zero's $x_1,x_2,...x_n$ are known then the polynomial can be factored as:
$f(x)=a(x-x_1)(x-x_2)....(x-x_n)$ (where $a$ is the coefficient of the highest degree term).
Offcourse in this case you have to find the zero's first (by using the quadratic formula) ...

4. ## Re: Factorization Question 2

Hello, sparky!

For the question below, does anyone have a quick and easy method for
finding the two numbers whose difference is 11 and whose product 180?
I know the numbers are 20 and -9, but it takes me forever to find them?
Is there is quick way?

$\text{Factor: }\:12x^2+11x-15$

I use a primitive approach.
It takes a bit of time, but it is direct.

Factor 180 into pairs of factors, and note the difference.

. . $\begin{array}{cc}\text{Factors} & \text{Diff.} \\ \hline 1,180 & 179 \\ 2,90 & 88 \\ 3,60 & 57 \\ 4,45 & 41 \\ 5,36 & 31 \\ 6,30 & 24 \\ 7,? & - \\ 8,? & - \\ 9,20 & 11 \\ 10,18 & 8 \\ 11,? & - \\ 12,15 & 3 \\ 13,? & - \\ \hline \end{array} \begin{array}{c} \\ \\ \\ \\ \\ \leftarrow\text{ Here!} \end{array}$

How do we find these pairs?
Divide 180 by 1, 2, 3, . . .
Some of the divisions do not "come out even", of course.

When can we stop?
We can stop at the integer part of $\sqrt{180} \,\rightarrow\, 13$

We see that the pair $(9,20)$ has a difference of 11.
(Of course, we can stop listing the moment we find it.)

We have: . $12x^2 \quad 9x\quad 20x - 15$

We want the middle term to be +11x, so we will use -9x and +20x.

So we have: . $12x^2 - 9x + 20x - 15$

Factor "by grouping": . $3x(4x-3) + 5(4x-3) \;=\;(4x-3)(3x+5)$

5. ## Re: Factorization Question 2

Originally Posted by Soroban
Hello, sparky!

I use a primitive approach.
It takes a bit of time, but it is direct.

Factor 180 into pairs of factors, and note the difference.

. . $\begin{array}{cc}\text{Factors} & \text{Diff.} \\ \hline 1,180 & 179 \\ 2,90 & 88 \\ 3,60 & 57 \\ 4,45 & 41 \\ 5,36 & 31 \\ 6,30 & 24 \\ 7,? & - \\ 8,? & - \\ 9,20 & 11 \\ 10,18 & 8 \\ 11,? & - \\ 12,15 & 3 \\ 13,? & - \\ \hline \end{array} \begin{array}{c} \\ \\ \\ \\ \\ \leftarrow\text{ Here!} \end{array}$

How do we find these pairs?
Divide 180 by 1, 2, 3, . . .
Some of the divisions do not "come out even", of course.

When can we stop?
We can stop at the integer part of $\sqrt{180} \,\rightarrow\, 13$

We see that the pair $(9,20)$ has a difference of 11.
(Of course, we can stop listing the moment we find it.)

We have: . $12x^2 \quad 9x\quad 20x - 15$

We want the middle term to be +11x, so we will use -9x and +20x.

So we have: . $12x^2 - 9x + 20x - 15$

Factor "by grouping": . $3x(4x-3) + 5(4x-3) \;=\;(4x-3)(3x+5)$

Wow, I love this! Why didn't I think of this? This was so obvious! Thanks

6. ## Re: Factorization Question 2

Originally Posted by Soroban
Hello, sparky!

I use a primitive approach.
It takes a bit of time, but it is direct.

Factor 180 into pairs of factors, and note the difference.

. . $\begin{array}{cc}\text{Factors} & \text{Diff.} \\ \hline 1,180 & 179 \\ 2,90 & 88 \\ 3,60 & 57 \\ 4,45 & 41 \\ 5,36 & 31 \\ 6,30 & 24 \\ 7,? & - \\ 8,? & - \\ 9,20 & 11 \\ 10,18 & 8 \\ 11,? & - \\ 12,15 & 3 \\ 13,? & - \\ \hline \end{array} \begin{array}{c} \\ \\ \\ \\ \\ \leftarrow\text{ Here!} \end{array}$

How do we find these pairs?
Divide 180 by 1, 2, 3, . . .
Some of the divisions do not "come out even", of course.

When can we stop?
We can stop at the integer part of $\sqrt{180} \,\rightarrow\, 13$

We see that the pair $(9,20)$ has a difference of 11.
(Of course, we can stop listing the moment we find it.)

We have: . $12x^2 \quad 9x\quad 20x - 15$

We want the middle term to be +11x, so we will use -9x and +20x.

So we have: . $12x^2 - 9x + 20x - 15$

Factor "by grouping": . $3x(4x-3) + 5(4x-3) \;=\;(4x-3)(3x+5)$

^this. is beautiful.