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Non-commutative or commutative ring or subring with $x^2 = 0$
Does there exist a non-commutative or commutative ring or subring $R$ with $x \cdot x = 0$ where $0$ is the zero element of $R$, $\cdot$ is multiplication secondary binary operation, and $x$ is not zero element, and excluding the case where addition (abelian group operation) and multiplication of two numbers always become zero?
Edit: most seem to be focused on non-commutative case. What about commutative case?
Edit 2: It is fine to relax the restriction to the following: there exists countably (and possibly uncountable) infinite number of $x$'s in $R$ that satisfy $x \cdot x = 0$ (so this means that there may be elements of ring that do not satisfy $x \cdot x =0$) excluding the case where addition and multiplication of two numbers always become zero.
• I forgot to write commutative. So both commutative and non-commutative cases. – user69886 Mar 28 '13 at 13:44
• Regarding your edit: If you want arbitrary elements of this type, just take products or tensor products. So take $k[x_1]/(x_1^2) \times k[x_2]/(x_2^2) \times \dotsc$ or $k[x_1,x_2,\dotsc]/(x_1^2,x_2^2,\dotsc)$. – Martin Brandenburg Mar 28 '13 at 14:29
• @user69886: I updated my answer with a commutative example. In general and for the future, please do not change your question after answers has been given. In general if you have another question, then just post that as a separate question. – Thomas Mar 28 '13 at 14:49
Example for the non-commutative case: $$\pmatrix{0 & 1 \\ 0 & 0} \pmatrix{0 & 1 \\ 0 & 0} = \pmatrix{0 & 0 \\ 0 & 0}.$$ Example for the commutative case: Consider the ring $\mathbb{Z} / 4\mathbb{Z}$. What is $2^2$ in this ring?
• See my answer for a conceptual view of the genesis of this example. – Math Gems Mar 28 '13 at 15:54 | {
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Sure, take $k$ any ring and $R=k[x]/(x^2)$. As suggested by Martin Brandenburg, a good example is $k=\mathbb R$, in which case $R$ becomes the ring of dual numbers. The wikipedia site lists a lot of properties, which might be interesting.
• Or almost the same, $\mathbb{Z}/4$. – Martin Brandenburg Mar 28 '13 at 13:40
• It's commutative. Though if we take a noncommutative base ring instead of $k$.. – Berci Mar 28 '13 at 13:41
• @Berci: Whoops, I misread that. – Jesko Hüttenhain Mar 28 '13 at 13:42
• I forgot to write "or commutative.." – user69886 Mar 28 '13 at 13:44
• Jesko, you may perhaps add a link to en.wikipedia.org/wiki/Dual_number – Martin Brandenburg Mar 28 '13 at 13:53
It is easy if you know about quotient rings, e.g. $\rm\:\Bbb Z/n^2,\ \Bbb Z[x]/(x^2).\:$ If you don't know about quotient rings, is there still a simple example? Yes! One quickly checks that $\rm\,\Bbb Z^2$ is a ring under pointwise addition and multiplication and, that the set of linear maps on $\rm\,\Bbb Z^2\,$ form a ring under addition and map composition. Now note that the (right) shift map $\rm\: S(a,b) = (0,a)\:$ is linear, and $\rm\,S^2 = 0\,$ since $\rm\: S^2(a,b) = S(0,a) = (0,0).\:$
Remark If you know matrix rings, note that the matrix of the shift map is that in Thomas's answer. But knowledge of matrices is not required to understand the above simple example. | {
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Where does this example come from? It arises from a general construction. It is, in fact, a linear representation of the generic example $\rm\:R[x]/(x^2),\:$ whose additive group is $\rm\:R^2,\:$ where $\rm\:x\:$ acts as a (right) shift map $\rm\:x(a+bx) = 0+ax,\:$ i.e. $\rm\:(a,b)\mapsto (0,a).\:$ Similarly any ring $\rm\,R\,$ has such a linear representation as a subring of the ring of linear maps on its underling additive group - the so-called (left) regular representation of $\rm\,R.\:$ This is a ring-theoretic analogue of Cayley's theorem, that a group $\rm\,G\,$ may be represented as a subgroup of the group of bijections (permutations) on $\rm\,G.\:$ In both cases the representation arises by sending each element $\rm\:r\:$ into its associated (left) multiplication map $\rm\: x\:\to\:r\:x\:.\:$ See this MO question for further discussion.
As another important example, this view provides a natural way to construct the polynomial ring $\rm\,R[x]\,$ as a subring of the ring of linear maps on $\rm\,R^\Bbb N = (f_0,f_1,f_2,\ldots)\:$ generated by $\rm\,R\,$ and $\rm\,x\,$ i.e. by all constants $\rm\:(r,0,0,\ldots)$ and by the shift map $\rm\,x : (f_0,f_1,f_2,\ldots)\mapsto (0,f_0,f_1,\ldots).$ See also the presentation in this post.
It is worth emphasis that the ring $\rm\:R[x]/(x^2)\:$ is known as the algebra of dual numbers over $\rm\,R,\,$ and that it frequently proves quite handy in various contexts, e.g. when studying (higher) derivatives, or tangent/jet spaces, and when transferring properties of homomorphisms to derivations
• Actually $S=x$ on $\mathbb{Z}[x]/(x^2)$. So these are the same examples in disguise. – Martin Brandenburg Mar 28 '13 at 14:17
• @martin Yes, I meant to segue into that - see the added remark. – Math Gems Mar 28 '13 at 14:57
Yes, for example consider the noncommutative polynomial ring $\Bbb R\langle x,y\rangle$ (so here $xy\ne yx$), and let $R$ be its quotient by the ideal $(x^2)$ generated by $x^2$. | {
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• Non-commutative polynomial rings are often called free (associative) algebras and denoted by other brackets, namely $R\langle x_1,\dotsc,x_n\rangle$. – Martin Brandenburg Mar 28 '13 at 13:48 | {
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# Tossing a matchstick randomly onto an infinite grid
I was given this problem recently in a job interview in which I did not succeed, however I found the problem itself very interesting, and have been trying to solve it since.
Question:
Given matchstick of length 1, and an infinite grid of similar matchsticks (i.e. an infinite grid of 1x1 squares), if you toss the matchstick randomly onto the grid, what is the probability that the matchstick lands overlapping at least one of the matchsticks in the grid?
My approach:
I decided to first define 3 variables:
• $x =$ horizontal distance between lower end of landing matchstick and nearest grid matchstick to the left
• $y =$ vertical distance between lower end of landing matchstick and nearest grid matchstick below
• $\theta$ = landing angle of matchstick with grid horizontal
By definition:
• $x, y$ ~ $U(0, 1)$
• $\theta$ ~ $U(0, \pi)$
• $x, y, \theta$ are all independent of each other
I then split the events of overlapping the horizontal and the vertical matchsticks in the grid, such that:
$$P(Overlap) = P(OverlapsVertical) + P(OverlapsHorizontal) - P(OverlapsVertical \bigcup OverlapsHorizontal)$$
• $P(OverlapsVertical) = P(y + sin(\theta) >= 1)$
• $P(OverlapsHorizontal) = P(x + cos(\theta) >= 1)$
For overlapping vertical, because both y and theta are uniformly distributed, I could calcluate $P(OverlapsVertical)$ by:
• Constructing the rectangle of possible values for $y$ and $\theta$
• Integrating to find the proportion of area of this rectangle that satisfied the condition $y + sin(\theta) >= 1$
Doing the same for the horizontal case, I found that $P(OverlapsVertical) = P(OverlapsHorizontal) = 2/\pi$.
However, I realised that the events $OverlapsVertical$ and $OverlapsHorizontal$ are not independent, because they are both dependent on $\theta$, specifically $sin(\theta)$ and $cos(\theta)$ which are inversely related, and therefore: | {
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• $P(OverlapsVertical \bigcup OverlapsHorizontal) \not= P(OverlapsVertical) * P(OverlapsHorizontal)$
I then became stuck, and wondered if my approach to the entire problem was wrong. Is there a better way to solve this problem? Any input greatly appreciated. Thanks!
• – Bram28 Jan 2 '18 at 22:07
• @Bram28 thanks for the quick reply. I see how that applies to the individual events of crossing the vertical and crossing the horizontal (my problem is specifically Buffon's needle where t = l = 1, and hence agrees with my answer 2/π), however as far as I can tell it does not answer the question of crossing either the vertical or the horizontal, where these two events are not independent nor mutually exclusive. – Rory Devitt Jan 2 '18 at 22:20
• Out of curiosity, which position were you applying for ? – Gabriel Romon Jan 2 '18 at 22:24
• @GabrielRomon it was a quantitative trading position at a hedge fund – Rory Devitt Jan 2 '18 at 22:25
• And the interviewer didn't give you a hint or a nudge in some direction ? – Gabriel Romon Jan 2 '18 at 22:31
The if the midpoint lands inside a box of dimensions: $(1-\cos\theta)\times(1-\sin\theta),$ then then the match will fit inside that square.
Due to the symmetry of the grid we only need to analyze $\theta \in [0, \frac {\pi}{4}]$
$\frac 1{\frac {\pi}{4}}\int_0^\frac {\pi}{4} (1-\cos\theta)\cdot(1-\sin\theta)\ d\theta$ | {
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# 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." | {
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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. | {
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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 | {
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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 | {
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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 | {
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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 | {
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• 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 | {
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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 | {
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$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: | {
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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 | {
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## 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. | {
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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.
This site uses Akismet to reduce spam. Learn how your comment data is processed. | {
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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. | {
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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. | {
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×
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
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Euclid's Formula. The link above should bring you to a proof.
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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
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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?
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Find out the pythagorean triple $$(x,y,z)$$ such that $$xyz=16320$$
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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
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Euclid's formula deal with generating Pythagorean triples that are integers.
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Good
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Problem Loading...
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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 $4+2+4$. | {
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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 $>1$. | {
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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 $f(n+1)=f(n)+f(n)=2f(n)$.
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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 $\quad yu_x-xu_y=3x \tag 1$ | {
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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 $X=x-y$. | {
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• 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 $$\boxed{u(x,y)=x^2+y^2-3y}$$ | {
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• 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 | {
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# 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 $$(B^T\,A)^{-1}=A^{-1}(B^{-1})^T.$$ | {
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# 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. | {
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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. | {
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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). | {
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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. | {
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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 | {
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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. | {
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# 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 $R=\{(1,1),(2,2),(3,3),(1,2),(2,1)\}$. | {
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• 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: | {
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$\leq$ 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 | {
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# 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. | {
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• 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 $\dim E_+ = \left\lfloor \frac{n+1}{2}\right\rfloor$. | {
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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 $A$. | {
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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 $B$. | {
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# 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, | {
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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 $t = 2x-1$
$\implies \text{J} = \displaystyle \lim _{n \to \infty} \int_{0}^1 (n+1) \left(\dfrac{x^{n+1}-(1-x)^{n+1}}{2x-1}\right) \mathrm{d}x$
$=\displaystyle \lim _{n \to \infty} \int_{0}^1 (n+1) (1-x)^n \left(\dfrac{\left(\frac{x}{1-x}\right)^{n+1}-1}{\frac{x}{1-x}-1}\right) \mathrm{d}x$
$=\displaystyle \lim _{n \to \infty} \int_{0}^1 (n+1) \sum_{r=0}^n (1-x)^n \left(\frac{x}{1-x}\right)^{r} \mathrm{d}x$
$=\displaystyle \lim _{n \to \infty} \sum_{r=0}^n (n+1) \int_{0}^1 x^r(1-x)^{n-r} \mathrm{d}x$ | {
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$=\displaystyle \lim _{n \to \infty} \sum_{r=0}^n (n+1) \int_{0}^1 x^r(1-x)^{n-r} \mathrm{d}x$
$=\displaystyle \lim _{n \to \infty} \sum_{r=0}^n \dfrac{1}{\dbinom{n}{r}}$ (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. | {
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• 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 $2$. | {
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# 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. | {
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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. | {
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# 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}}] | {
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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] | {
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# 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 $a_k$. | {
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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 | {
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# 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,
$\varphi(N) = 24$
$\varphi(24) = 8$
$\varphi(8) = 4$
$\varphi(4) = 2$
$\varphi(2) = 1$ | {
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$\varphi(N) = 24$
$\varphi(24) = 8$
$\varphi(8) = 4$
$\varphi(4) = 2$
$\varphi(2) = 1$
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. | {
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$n = 10^{5} + 2$: 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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MAGMA - see FactoredEulerPhiInverse(n)
Regards
-
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. | {
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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. | {
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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. | {
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In the multiple regression model, the goodness-of- t measure R-squared always increases (or remains the same) when an additional explanatory variable is added. Slides Prepared by JOHN S. We'll just use the term "regression analysis" for all these variations. While many statistical software packages can perform various types of nonparametric and robust regression. Orlov Chemistry Department, Oregon State University (1996) INTRODUCTION In modern science, regression analysis is a necessary part of virtually almost any data reduction process. wnarifin@usm. Note that the intercept will always be zero and so we could have used regression without an intercept to obtain the same regression coefficients (although the standard errors will be slightly different). A simple linear regression equation for this would be $$\hat{Price} = b_0 + b_1 * Mileage$$. Multiple regression is a flexible method of data analysis that may be appropriate whenever a quantitative variable (the dependent or criterion variable) is to be examined in relationship to any other factors (expressed as independent or predictor variables). i = be the value of. This PowerPoint is a workshop on multiple linear regression and analysis of covariance. Multiple Linear Regression Multiple linear regression attempts to model the relationship between two or more explanatory variables and a response variable by fitting a linear equation to observed data. Lecture 1 Introduction to Multi-level Models Regression Analysis? • Specification of predictor variables from multiple levels. The goal of a model is to get the smallest possible sum of squares and draw a line that comes closest to the data. When I started experimenting with machine learning, I wanted to come up with an application that would solve a real-world problem but would not be too complicated to implement. This statistics is for multiple linear regression technique. Once you have completed the test, click on 'Submit Answers' to get your results. | PowerPoint | {
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Once you have completed the test, click on 'Submit Answers' to get your results. | PowerPoint PPT presentation | free to view. Multiple Regression and Correlation Dr. Multiple regression generally explains the relationship between multiple independent or predictor variables and one dependent or criterion variable. Stata Version 13 - Spring 2015 Illustration: Simple and Multiple Linear Regression …\1. Objectives of Multiple Regression In selecting suitable applications of multiple regression, the researcher must consider three primary issues: 1. Secondary Data Analysis • Starting Off Right: Effects of Rurality on Parent‟s Involvement in Children‟s Early Learning (Sue Sheridan, PPO) – Data from the Early Childhood Longitudinal Study – Birth Cohort (ECLS-B) were used to examine the influence of setting on parental involvement in preschool and the effects of involvement on. Example: estimated coefficient of education equals 92. Dummy Variables Dummy Variables A dummy variable is a variable that takes on the value 1 or 0 Examples: male (= 1 if are male, 0 otherwise), south (= 1 if in the south, 0 otherwise), etc. Multiple regression analysis is almost the same as simple linear regression. “encoding model”. NASCAR Race Crashes SAS Program SAS Output. Multiple Regression Now, let’s move on to multiple regression. , the dependent variable) of a fictitious economy by using 2 independent/input variables:. Simple linear regression and multiple regression using least squares can be done in some spreadsheet applications and on some calculators. The example below demonstrates the use of the summary function on the two models created during this tutorial. One regressor should not be a linear function of another. This lesson explores the use of a regression analysis to answer. In the orange juice classification problem, Y can only take on two possible values: 0 or 1. As a data scientist, one must always explore multiple options for solving the same analysis or modeling task and | {
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one must always explore multiple options for solving the same analysis or modeling task and choose the best for his/her particular problem. Understand and use bivariate and multiple linear regression analysis. In the following example, we will use multiple linear regression to predict the stock index price (i. patient, f. Chapter 9 Correlational Research Designs What are correlational research designs, and why are they used in behavioral research. Edward’s University Chapter 12 Simple Linear Regression Simple Linear Regression Model Least Squares Method Coefficient of Determination Model Assumptions Testing for Significance Using the Estimated Regression Equation for Estimation and Prediction Computer Solution Residual Analysis: Validating Model Assumptions Simple Linear Regression Model y = b0. Regression analysis helps in establishing a functional Relationship between two or more variables. Amaral November 21, 2017 Advanced Methods of Social Research (SOCI 420). Stata Version 13 - Spring 2015 Illustration: Simple and Multiple Linear Regression …\1. Introduction to Correlation and Regression Analysis. Simple regression analysis uses a single x variable for each dependent “y” variable. Multiple Linear Regression Model We consider the problem of regression when study variable depends on more than one explanatory or independent variables, called as multiple linear regression model. One will not encounter cases in a multiple regression path model where one could go through the same variable twice. The generic form of the linear regression model is y = x 1β 1 +x 2β 2 +. Multiple linear regression attempts to fit a regression line for a response variable using more than one explanatory variable. selection of the dependent and independent variables. The model states that the expected value of Y--in this case, the expected merit pay increase--equals β0 plus β1 times X. In ordinary samples we would expect 95% of cases to have standardized residuals within about +2 and -2. | {
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ordinary samples we would expect 95% of cases to have standardized residuals within about +2 and -2. Stata Version 13 - Spring 2015 Illustration: Simple and Multiple Linear Regression …\1. If you click. y = c + ax c = constant a = slope. It can also handle multiple predictor variables. Multiple Linear Regression and Matrix Formulation Introduction I Regression analysis is a statistical technique used to describe relationships among variables. , the dependent variable) of a fictitious economy by using 2 independent/input variables:. This is a simplified tutorial with example codes in R. One type of analysis many practitioners struggle with is multiple regression analysis, particularly an analysis that aims to optimize a response by finding the best levels for different variables. Your print-out of the slides needs to be modified at one point. Use Multiple Regression to model the linear relationship between a continuous response and up to 12 continuous predictors and 1 categorical predictor. The simple proportional hazards model generalizes to a multiple regression model in much the same way as for linear and logistic regression. Posc/Uapp 816 Class 14 Multiple Regression With Categorical Data Page 3 1. Lecture 6:. Standard multiple regression is the same idea as simple linear regression, except now you have several independent variables predicting the dependent variable. If two variables are trending over time, a regression. Assessing Studies Based on Multiple Regression Part III. – The test is multiple linear regression isThe test is multiple linear regression is H 0 = β 1 = β 2 = … = β p = 00. Root MSE = s = our estimate of σ = 2. This is what regression analysis can do! For example, we'll see (via regression) that Fords. Regression analysis is a statistical process for estimating the relationships among variables. " In the main dialog box, input the dependent variable. txt) or view presentation slides online. Lecture 18: Multiple Logistic Regression Mulugeta | {
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variable. txt) or view presentation slides online. Lecture 18: Multiple Logistic Regression Mulugeta Gebregziabher, Ph. Clearly, it is nothing but an extension of Simple linear regression. We expect to build a model that fits the data better than the simple linear regression model. Dummy variables are useful because they enable us to use a single regression equation to represent multiple groups. A multiple regression based model will use the data to build a function that predicts the outcome based on the independent variables. Technically, a regression analysis model is based on the sum of squares, which is a mathematical way to find the dispersion of data points. 8 Quantitative Business Analysis for Decision Making Multiple Linear Regression Analysis Outlines Multiple Regression Model Estimation Testing Significance of Predictors Multicollinearity Selection of Predictors Diagnostic Plots Multiple Regression. When you have more than two events, you ca n extend the binary logistic regression model, as described in Chapter 3. y = c + ax c = constant a = slope. Frank Wood, fwood@stat. Econometrics I: Multiple Regression: Inference - H. One of these variable is called predictor variable whose value is gathered through experiments. When you look at the output for this multiple regression, you see that the two predictor model does do significantly better than chance at predicting cyberloafing, F(2, 48) = 20. In each ex-ample, you will first learn about the specific ingredi-ents required for the power or sample size computa-tion for the linear model being considered. It is also called the coefficient of determination, or the coefficient of multiple determination for multiple regression. Nonlinear regression The model is a nonlinear function of the parameters. University of Dayton. The program’s graph, regression, and correlation functions can respectively produce scatterplots, provide regression equation coefficients, and create correlation matrices. Assumptions of Multiple | {
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provide regression equation coefficients, and create correlation matrices. Assumptions of Multiple Regression This tutorial should be looked at in conjunction with the previous tutorial on Multiple Regression. When using multiple regression to estimate a relationship, there is always the possibility of correlation among the independent variables. If you need to investigate a fitted regression model further, create a linear regression model object LinearModel by using fitlm or stepwiselm. Further details and related logical operations can be found in the R documentation. Multiple Personality Disorder. A sound understanding of the multiple regression model will help you to understand these other applications. Switching Regression Models — Estimation (8) First obtain the expected values of the residuals that are truncated. We will also learn how to decide whether a group of explanatory variables can be excluded from the model. "Linear Regression and Modeling" is course 3 of 5 in the Statistics with R Coursera Specialization. In this paper, the risk factors for a disease of the eye (retinopathy of prematurity) are identi ed using logistic regression analysis. Multiple regression is a very advanced statistical too and it is extremely powerful when you are trying to develop a "model" for predicting a wide variety of outcomes. With only one independent variable, the regression line can be plotted neatly in two dimensions. However, performing a regression does not automatically give us a reliable relationship between the variables. xls Data Sample (showing only the variables to be used in analysis) Variables Used. Chapter 8: Multiple Choice Questions. Lecture 1 Introduction to Multi-level Models Regression Analysis? • Specification of predictor variables from multiple levels. Simple linear regression. Variable and Dummy Coded Region Variable 273. Chapter 3: Multiple regression analysis: Estimation In multiple regression analysis, we extend the simple (two-variable) | {
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regression analysis: Estimation In multiple regression analysis, we extend the simple (two-variable) regression model to con-sider the possibility that there are additional explanatory factors that have a systematic ef-fect on the dependent variable. We are not going to go too far into multiple regression, it will only be a solid introduction. zeigler-hill. pdfs and a PowerPoint presentation (also found on the left side of the site), and have integrated SPSS procedures into the discussion. Regression analysis is a statistical process for estimating the relationships among variables. If you need to investigate a fitted regression model further, create a linear regression model object LinearModel by using fitlm or stepwiselm. Many people find this too complicated to understand. I also wanted to practice working with regression algorithms. It is expected the average operating margin of all sites that fit this category falls within 33% and 41. MR&B3 is intended to offer a conceptually-oriented introduction to multiple regression (MR) and structural equation modeling (SEM), along with analyses that flow naturally from those methods. ppt - Free download as Powerpoint Presentation (. ppt from AA 1Multiple Regression (Part 3: Variables Selection) 1 Topic Outline Comparing Two Multiple Regression Models Partial F -Test Using Sequential Sum of. In the simplest case, we would use a 0,1 dummy variable where a person is given a value of 0 if they are in the control group or a 1 if they are in the treated group. We are not going to go too far into multiple regression, it will only be a solid introduction. – The test is multiple linear regression isThe test is multiple linear regression is H 0 = β 1 = β 2 = … = β p = 00. An Introduction to Logistic Regression: From Basic Concepts to Interpretation with Particular Attention to Nursing Domain ure" event (for example, death) during a follow-up period of observation. Wouldn’t it be great if there was a more accurate way to predict | {
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a follow-up period of observation. Wouldn’t it be great if there was a more accurate way to predict whether your prospect will buy rather than just taking an educated guess? Well, there is…if you have enough data on your previous prospects. This model is built using, for example, a set of real world data listing the outcome in various cases. Welcome to Part 1 of Regression & Classification - Simple Linear Regression: Step 1. Multiple Regression Multiple Proportional Hazards Regr (Cox model) For time dependent outcomes (ie time to death), we model the hazard rate, h , the event rate per unit time (for death, it is the mortality rate). How to Run a Multiple Regression in Excel. Wooldridge, Introductory Econometrics, 4th ed. If two variables are trending over time, a regression. Multiple regression is an extension of simple linear regression. Multiple linear regression has one y and two or more x variables. 8%, with 95% confidence. is a way of understanding the relationship between two variables. For instance, when we predict rent based on square feet alone that is simple linear regression. Stepwise multiple linear regression has proved to be an extremely useful computational technique in data analysis problems. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Background Multiple Linear Regression is widely used in academics and also in MR We can consider it the start of Multivariate Analysis, for our course Any idea what the following are: Multivariate analysis Multiple Linear Regression (MLR). The precision of a method (Table 4) is the extent to which the individual test results of multiple injections of a series of standards agree. For example, linear regression can be used to quantify the relative impacts of age, gender, and diet (the predictor variables) on height (the outcome variable). Regression assumes that variables have normal distributions. regress is useful when you simply need the output arguments of the function and when | {
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distributions. regress is useful when you simply need the output arguments of the function and when you want to repeat fitting a model multiple times in a loop. 3 Notes: PowerPoint xxx PDF. Suppose we have a cohort of. Lecture 24: Partial correlation, multiple regression, and correlation Ernesto F. Third, multiple regression offers our first glimpse into statistical models that use more than two quantitative. If you have the Excel desktop application, you can use the Open in Excel button to open your workbook and use either the Analysis ToolPak's Regression tool or statistical functions to perform a regression analysis there. Uses of Regression Analysis 1. If you look at the first slide on centering, it states "To center: Y i centered = Yi – Mi. 1 Introduction In this chapter we extend the simple linear regression model, and allow for any number of independent variables. You probably remember the concept of simple linear regression intuition from your high school years. Similar tests. Notes on Regression Model • It is VERY important to have theory before starting developing any regression model. Lecture 4: Multivariate Regression Model in Matrix Form In this lecture, we rewrite the multiple regression model in the matrix form. Partial r is just another way of standardizing the coefficient, along with beta coefficient (standardized regression coefficient)$^1$. selection of the dependent and independent variables. edu|http://www. Simple linear regression is a bivariate situation, that is, it involves two dimensions, one for the dependent variable Y and one for the independent variable x. ppt), PDF File (. Formula 13. This correlation may be pair-wise or multiple correlation. The linear regression model (LRM) The simple (or bivariate) LRM model is designed to study the relationship between a pair of variables that appear in a data set. Regression analysis is a very widely used statistical tool to establish a relationship model between two variables. , between an | {
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widely used statistical tool to establish a relationship model between two variables. , between an independent and a dependent variable or between two independent variables). There are two models of logistic regression, binary logistic regression and multinomial logistic regression. If the universe is only 14 billion years old, how can it be 92 billion light years wide? - Duration: 9:47. It is the correlation between the variable's values and the best predictions that can be computed linearly from the predictive variables. If you look at the first slide on centering, it states "To center: Y i centered = Yi – Mi. It can also be used to estimate the linear association between the predictors and reponses. pdfs and a PowerPoint presentation (also found on the left side of the site), and have integrated SPSS procedures into the discussion. Lecture 18: Multiple Logistic Regression - p. You can use it to predict values of the dependent variable, or if you're careful, you can use it for suggestions about which independent variables have a major effect on the dependent variable. Known also as curve fitting or line fitting because a regression analysis equation can be used in fitting a curve or line to. Intercept: the intercept in a multiple regression model is the mean for the response when. View and Download PowerPoint Presentations on Correlation Regression Using Spss PPT. The data was split into three employment sectors Teaching, government and private industry Each sector showed a positive relationship Employer type was confounded with degree level Simpson’s Paradox In each of these examples, the bivariate analysis (cross-tabulation or correlation) gave misleading results Introducing another variable gave a. Here, the summary(OBJECT) function is a useful tool. +x K β K +ε where y is the dependent or. Upload and Share PowerPoint Presentations. Pantula David A. covariates for the. MV - Multiple Regression. The generic form of the linear regression model is y = x 1β 1 +x 2β | {
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the. MV - Multiple Regression. The generic form of the linear regression model is y = x 1β 1 +x 2β 2 +. Typically points further than, say, three or four standard deviations from the mean are considered as "outliers". This post will show you examples of linear regression. Regression Analysis. patient, f. In this exercise we apply the same method to validating the Novy-Marx Quality and Value screener’s ability to predict the 5-year total return. Multiple Regression Analysis using SPSS Statistics Introduction. 01, with an R-square of. Briefly speaking, the goal of the multiple linear regression is to point out the relation between a dependent variable (explained, endogenous or resultative) and a great deal of independent variables (explanatory, factorial,. Comparing Multiple Regression Model Results against Historic Demand. However, it is critical to recognize that multiple regression is inherently a correlation technique and cannot. " In the main dialog box, input the dependent variable. In contrast, Linear regression is used when the dependent variable is continuous and nature of the regression line is linear. Rawlings Sastry G. Regression analysis is a statistical technique for estimating the relationship among variables which have reason and result relation. If you look at the first slide on centering, it states "To center: Y i centered = Yi – Mi. persistence of shocks will be infinite for nonstationary series • Spurious regressions. Multiple linear regression Situations frequently occur when we are interested in the dependency of a variable on several explanatory (independent) variables. If two variables are trending over time, a regression. RSM is a method used to locate the optimum value of the response and is one of the final stages of experimentation. In contrast, Linear regression is used when the dependent variable is continuous and nature of the regression line is linear. Suppose we have a cohort of. Cherry Blossoms WS: Test Review. ID: 305741 ID: 305741. | {
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line is linear. Suppose we have a cohort of. Cherry Blossoms WS: Test Review. ID: 305741 ID: 305741. Linear regression Linear dependence: constant rate of increase of one variable with respect to another (as opposed to, e. As a data scientist, one must always explore multiple options for solving the same analysis or modeling task and choose the best for his/her particular problem. 1 Introduction to Poisson Regression As usual, we start by introducing an example that will serve to illustrative regression models for count data. View and Download PowerPoint Presentations on Correlation Regression Using Spss PPT. Multiple (Linear) Regression. This tutorial will explore how R can be used to perform multiple linear regression. Dummy variables are useful because they enable us to use a single regression equation to represent multiple groups. is a way of understanding the relationship between two variables. 1 Introduction In this chapter we extend the simple linear regression model, and allow for any number of independent variables. This article is a part of the guide:. Suppose that there is a cholesterol lowering drug that is tested through a clinical trial. It is expected the average operating margin of all sites that fit this category falls within 33% and 41. We have been reviewing the relationship between correlation (r and r2) and regression (R, R2 and 1-R2) in class, through lectures, blog. Nonlinear Regression Functions Chapter 9. Regression Analysis. (3) Does the combination of predictors in this fitted multiple regression explain significant variation in the response? Statistics 621 Multiple Regression Practice Questions. Multiple Regression Analysis using SPSS Statistics Introduction. We now perform multiple linear regression to obtain the standardized regression coefficients shown in range J19:J21. Multiple linear regression Situations frequently occur when we are interested in the dependency of a variable on several explanatory (independent) variables. selection | {
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interested in the dependency of a variable on several explanatory (independent) variables. selection of the dependent and independent variables. Consider first the case of a single binary predictor, where x = (1 if exposed to factor 0 if not;and y =. The output varies linearly based upon the input. Multiple regression uses the ordinary least squares solution (as does bi. The whole model F test (test of the useful of the model) tests whether the slopes on all variables in multiple regression are zero, i. Ryan McEwan and Julia Chapman. Simple linear regression involves a single independent variable. 12-1 Multiple Linear Regression Models • Many applications of regression analysis involve situations in which there are more than one regressor variable. Introduction to Linear Regression and Correlation Analysis Fall 2006 – Fundamentals of Business Statistics 2 Chapter Goals To understand the methods for displaying and describing relationship among variables. ppt from AA 1Multiple Regression (Part 3: Variables Selection) 1 Topic Outline Comparing Two Multiple Regression Models Partial F -Test Using Sequential Sum of. It is expected the average operating margin of all sites that fit this category falls within 33% and 41. Chapter 8: Multiple Choice Questions. Here is image from Linear Regression, posted by Ralf Adam, on March 30, 2017, image size: 63kB, width: 1200, height: 1455, Least to Greatest, Where, Least Common. Regression with two or more predictors is called multiple regression Available in all statistical packages Just like correlation, if an explanatory variable is a significant predictor of the dependent variable, it doesn't imply that the explanatory variable is a cause of the dependent variable. It is discussed in Response Surface Methods. Hey I would like to make a scatter plot with p-value and r^2 included for a multiple linear regression. 0 Chapter 6: Multiple Linear Regression Topics Explanatory Modeling Predictive Modeling Example: Prices of Toyota Corolla | {
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Linear Regression Topics Explanatory Modeling Predictive Modeling Example: Prices of Toyota Corolla ToyotaCorolla. Multiple regression is a very advanced statistical too and it is extremely powerful when you are trying to develop a "model" for predicting a wide variety of outcomes. Regression is the analysis of the relation between one variable and some other variable(s), assuming a linear relation. Also referred to as least squares regression and ordinary least squares (OLS). Regression when all explanatory variables are categorical is “analysis of variance”. 5 Correlation and Regression Simple regression 1. Notes on Regression Model • It is VERY important to have theory before starting developing any regression model. The Regression df is the number of independent variables in the model. We can have only two models or more than three models depending on research questions. For example, consider the cubic polynomial model which is a multiple linear regression model with three regressor variables. Multiple Regression 18. X2 1 or even interactions X1 X2. The following model is a multiple linear regression model with two predictor variables, and. Find PowerPoint Presentations and Slides using the power of XPowerPoint. This post will show you examples of linear regression. Please access that tutorial now, if you havent already. The basic command is "regression": "linear. Chapter 3: Multiple regression analysis: Estimation In multiple regression analysis, we extend the simple (two-variable) regression model to con-sider the possibility that there are additional explanatory factors that have a systematic ef-fect on the dependent variable. Objectives. MULTIPLE LINEAR REGRESSION ANALYSIS USING MICROSOFT EXCEL by Michael L. Here is image from Linear Regression, posted by Ralf Adam, on March 30, 2017, image size: 63kB, width: 1200, height: 1455, Least to Greatest, Where, Least Common. Frank Wood, fwood@stat. 1 Introduction. " Coefficient table, bottom. Under Type of power | {
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Common. Frank Wood, fwood@stat. 1 Introduction. " Coefficient table, bottom. Under Type of power analysis, choose ‘A priori…’, which will be used to identify the sample size required given the alpha level, power, number of predictors and effect size. View and Download PowerPoint Presentations on Spss Tutorial For Multiple Logistic Regression PPT. One of the applications of multiple linear regression models is Response Surface Methodology (RSM). State the multiple regression equation. For ordina l categorical variables, the drawback of the. Linear, Ridge Regression, and Principal Component Analysis Geometric Interpretation I Each column of X is a vector in an N-dimensional space (NOT the p-dimensional feature vector space). Linear Regression. The model is linear because it is linear in the parameters , and. We now perform multiple linear regression to obtain the standardized regression coefficients shown in range J19:J21. In R, multiple linear regression is only a small step away from simple linear regression. Regression Analysis. The basic command is “regression”: “linear. Multiple linear regression in R Dependent variable: Continuous (scale/interval/ratio) Independent variables: Continuous (scale/interval/ratio) or binary (e. y = c + ax c = constant a = slope. 8%, with 95% confidence. We can run regressions on multiple different DVs and compare the results for each DV. Multiple Regression Models • Advantages of multiple regression • Important preliminary analyses • Parts of a multiple regression model & interpretation • Differences between r, bivariate b, multivariate b & • Steps in examining & interpreting a full regression model Advantages of Multiple Regression Practical issues …. See the Handbook for information on these topics. Multiple Regression - Linearity. ppt - Free download as Powerpoint Presentation (. This correlation may be pair-wise or multiple correlation. Arial Franklin Gothic Book Perpetua Wingdings 2 Calibri Equity 1_Equity 2_Equity 3_Equity | {
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Arial Franklin Gothic Book Perpetua Wingdings 2 Calibri Equity 1_Equity 2_Equity 3_Equity 4_Equity Microsoft Equation 3. Be sure to tackle the exercise and the quiz to get a good understanding. One type of analysis many practitioners struggle with is multiple regression analysis, particularly an analysis that aims to optimize a response by finding the best levels for different variables. 2% with 95% confidence. Lecture 18: Multiple Logistic Regression - p. The measured standard deviation can be subdivided into 3 categories: repeatability, intermediate precision and reproducibility (4, 5). Find PowerPoint Presentations and Slides using the power of XPowerPoint. By selecting the features like this and applying the linear regression algorithms you can do polynomial linear regression; Remember, feature scaling becomes even more important here; Instead of a conventional polynomial you could do variable ^(1/something) - i. MR&B3 is intended to offer a conceptually-oriented introduction to multiple regression (MR) and structural equation modeling (SEM), along with analyses that flow naturally from those methods. Introduction to Correlation and Regression Analysis. It is used to also to determine the overall fit of the model and the contribution of each of the predictors to the total variation. Uses of Regression Analysis 1. Variable Interactions 5. 11 LOGISTIC REGRESSION - INTERPRETING PARAMETERS 11 Logistic Regression - Interpreting Parameters Let us expand on the material in the last section, trying to make sure we understand the logistic regression model and can interpret Stata output. Figure 14 – Model Summary Output for Multiple Regression. Starting values of the estimated parameters are used and the likelihood that the sample came from a population with those parameters is computed. On the contrary, regression is used to fit a best line and estimate one variable on the basis of another variable. For simple linear regression (only one x), the Regression df is 1. | {
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on the basis of another variable. For simple linear regression (only one x), the Regression df is 1. Source: Hair et al. View and Download PowerPoint Presentations on Correlation Regression Using Spss PPT. Predictions in neurophysiology. Nonlinear regression The model is a nonlinear function of the parameters. The only difference between simple linear regression and multiple regression is in the number of predictors (“x” variables) used in the regression. In contrast, Linear regression is used when the dependent variable is continuous and nature of the regression line is linear. 12-1 Multiple Linear Regression Models • Many applications of regression analysis involve situations in which there are more than one regressor variable. specification of a statistical relationship, and 3. Upload and Share PowerPoint Presentations. Welcome to Part 1 of Regression & Classification - Simple Linear Regression: Step 1. Popular spreadsheet programs, such as Quattro Pro, Microsoft Excel,. biostatcourse. Teaching\stata\stata version 13 – SPRING 2015\stata v 13 first session. Under Type of power analysis, choose ‘A priori…’, which will be used to identify the sample size required given the alpha level, power, number of predictors and effect size. How to improve leadership skills in the workplace ppt for write my essay nursing best cover letter editing site uk , writing a scientific report buy a college essay online. Logistic regression is a statistical method for analyzing a dataset in which there are one or more independent variables that determine an outcome. Regression with categorical variables and one numerical X is often called “analysis of covariance”. It could be, for example, performance, education, or experience. The course starts with a discussion of the logic of the multivariate regression model and the central assumptions underlying the ordinary least squares approach. Also referred to as least squares regression and ordinary least squares (OLS). Linear regression for | {
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Also referred to as least squares regression and ordinary least squares (OLS). Linear regression for the advertising data Consider the advertising data shown on the next slide. The method of multiple regression sought to create the most closely related model. If you need to investigate a fitted regression model further, create a linear regression model object LinearModel by using fitlm or stepwiselm. Multiple regression is used to build a model that allows us to study this interplay. What to report? What a statistics program gives you: For a simple regression (one independent variable), statistics programs produce two estimates, a (the "constant term") and b (the "linear coefficient"), for the parameters α and β, respectively. For example: (x 1, Y 1). 2% with 95% confidence. | {
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HP 11C real root finder [Newton Method]
01-09-2014, 11:59 PM
Post: #1
Carlos CM (Mexico) Junior Member Posts: 35 Joined: Dec 2013
HP 11C real root finder [Newton Method]
LBL 0
RCL 1
GSB 1
RCL 1
GSB 2
/
CHS
RCL 1
+
STO 3
RCL 1
-
ABS
RCL 2
X>Y?
RTN
RCL 3
STO 1
GTO 0
LBL 1
f(x)=0 code
LBL 2
f'(x)=0 code
R1: old value (or initial value)
R2: Tolerance of error (0.001, 0.0001, 0.000000001 etc)
THE RESULT:
R3: new value (root)
EXAMPLE
Find the root of f(x)=x^3 - 3x^2-6x+8
rewritten as f(x)= [(x-3)x-6]x +8
f(x) code:
LBL1
ENTER
ENTER
ENTER
3
-
*
6
-
*
8
+
RTN
f'(x)= 3x(x-6)-6
f'(x) code
LBL 2
ENTER
ENTER
6
-
*
3
*
6
-
RTN
The roots are [-2, 1, 4]
give a initial value, for example: -3
-3 STO 1
give a tolerance of error, for example: 0.0001
0.0001 STO 2
BEGIN THE PROGRAM...
GSB 0
runnning...
when the error < TOL the program stop and display:
0.0001
you can find the root in the register 3
RCL 3
DISPLAY: -2.0001
try with initial value = 2
[/size][/font][/size][/size][/font][/font]
Best Regards
01-10-2014, 12:57 AM
Post: #2
Thomas Klemm Senior Member Posts: 1,447 Joined: Dec 2013
RE: HP 11C real root finder [Newton Method]
Using storage arithmetic allows us to get rid of a few steps:
LBL 0
RCL 1
GSB 1
RCL 1
GSB 2
/
STO- 1
ABS
RCL 2
X<=Y?
GTO 0
RCL 1
RTN
You made a mistake calculating the 1st derivative.
This is the corrected program:
f'(x)= 3x^2 - 6x - 6 = (3x - 6)x - 6
f'(x) code
LBL 2
ENTER
ENTER
3
*
6
-
*
6
-
RTN
Kind regards
Thomas
01-10-2014, 05:28 PM
Post: #3
Carlos CM (Mexico) Junior Member Posts: 35 Joined: Dec 2013
RE: HP 11C real root finder [Newton Method]
Best regards
Best Regards
01-12-2014, 08:31 AM
Post: #4
Namir Senior Member Posts: 688 Joined: Dec 2013
RE: HP 11C real root finder [Newton Method]
Here is a version that requires coding f(x) only since it approximate f'(x) as:
f'(x) = (f(x+h) - f(x))/h
Where h = 0.001*(ABS(X)+1)
The new version uses registers R1 through R4. | {
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f'(x) = (f(x+h) - f(x))/h
Where h = 0.001*(ABS(X)+1)
The new version uses registers R1 through R4.
Code:
LBL 0 RCL 1 ABS 1 + EEX 3 CHS * STO 3 # calculate and store increment h RCL 1 GSB 1 STO 4 # calculate and store f(x) RCL 1 RCL 3 + GSB 1 # calculate f(x+h) RCL 4 - 1/X RCL 4 * RCL 3 * # calculate diff = h *f(x)/(f(x+h) - f(x)) STO- 1 ABS RCL 2 X<=Y? GTO 0 RCL 1 RTN
01-12-2014, 01:26 PM
Post: #5
Dieter Senior Member Posts: 2,397 Joined: Dec 2013
RE: HP 11C real root finder [Newton Method]
(01-12-2014 08:31 AM)Namir Wrote: ...h = 0.001*(ABS(X)+1)
First of all, instead of multiplying with $$10^{-3}$$, dividing by $$10^{3}$$ is one step shorter. ;-)
This method for determining h will work in most cases, but not for very small arguments. Consider $$x = 10^{-4}$$ or even $$x = 10^{-40}$$. That's why I prefer $$h = x/10^4$$. On the 34s, the result can be easily rounded to 1 or 2 significant digits (RSD 1) to prevent slight roundoff errors. As usual, $$x=0$$ is handled as $$x=1$$.
Dieter
01-14-2014, 08:56 PM
Post: #6
Thomas Klemm Senior Member Posts: 1,447 Joined: Dec 2013
RE: HP 11C real root finder [Newton Method]
If you only want to solve polynomials with real coefficients Bairstow's Method can be used.
In a nutshell: instead of a root a quadratic factor is found in each iteration. To find the roots of this factor the classic formula is used.
Example:
$$x^3-3x^2-6x+8=0$$
First enter the coefficients of the polynomial. Always start with register 9:
1 STO 9
-3 STO .0
-6 STO .1
8 STO .2
Then specify the order of the polynomial with a loop-control value. It defines the registers you used for the coefficients:
9.012 STO 6
And now give an initial guess. Probably {1, 1} will do in all cases:
1 STO 7
STO 8
Start the program B:
GSB B
Now the coefficients of the quadratic factor can be found in registers 7 and 8:
RCL 7
1.000000
RCL 8
-2.000000 | {
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RCL 7
1.000000
RCL 8
-2.000000
The coefficient of the first term is always 1. Thus the factor is $$x^2+x-2$$.
Now we can solve this quadratic equation using program A:
1
RCL 7
RCL 8
GSB A
1.000000
X<>Y
-2.000000
Thus: $$x^2+x-2=(x-1)(x+2)$$.
What is left? Let's have a look at it:
RCL 6
9.010
RCL 9
1.000000
RCL .0
-4.000000
This is a linear factor. Thus we end up with the following factorization: $$x^3-3x^2-6x+8=(x-1)(x+2)(x-4)$$
Therefore the solutions are: {1, -2, 4}
Example from Bunuel66's solution to the crossed ladders problem
$$C^4-30C^3+700C^2-21,000C+157,500=0$$
Enter the coefficients of the polynomial:
1 STO 9
-30 STO .0
700 STO .1
21,000 STO .2
157,500 STO .3
Specify the loop-control value:
9.013 STO 6
Use the initial guess {1, 1}:
1 STO 7
STO 8
Start the program:
GSB B
The coefficients of the quadratic factor are in registers 7 and 8:
RCL 7
-35.178025
RCL 8
248.596895
Solve this quadratic equation using program A:
1
RCL 7
RCL 8
GSB A
25.384938
X<>Y
9.793087
RCL 6
9.011
RCL 9
1.000000
RCL .0
5.178025
RCL .1
633.555781
If we try to solve this quadratic equation we get an Error 0.
That's because we're trying to calculate the square root of a negative value.
However we can still get the desired result:
RCL 9
RCL .0
RCL .1
GSB B
Error 0
<-
CHS
626.852796
$$\sqrt{x}$$
25.037029
X<>Y
-2.589012
Thus the final list of solutions is:
• 25.384938
• 9.793087
• -2.589012$$\pm$$25.037029i
Cheers
Thomas
01-15-2014, 05:53 AM
Post: #7
Namir Senior Member Posts: 688 Joined: Dec 2013
RE: HP 11C real root finder [Newton Method]
(01-12-2014 01:26 PM)Dieter Wrote:
(01-12-2014 08:31 AM)Namir Wrote: ...h = 0.001*(ABS(X)+1)
First of all, instead of multiplying with $$10^{-3}$$, dividing by $$10^{3}$$ is one step shorter. ;-) | {
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This method for determining h will work in most cases, but not for very small arguments. Consider $$x = 10^{-4}$$ or even $$x = 10^{-40}$$. That's why I prefer $$h = x/10^4$$. On the 34s, the result can be easily rounded to 1 or 2 significant digits (RSD 1) to prevent slight roundoff errors. As usual, $$x=0$$ is handled as $$x=1$$.
Dieter
Sure, dividing by 1000 is a step shorter. Originially, I had learned to calculate h using an If statement:
Code:
If |x| >= 1 Then h= x/100 else h = 1/100 end if
Until I realized one day that .01*(|x|+1) does the job and eliminates the need for labels and GOTOs. I recently started using .001 instead of 0.01. Using the expression for h ensures that if x=0, h is not zero.
Namir
01-15-2014, 08:38 PM
Post: #8
Dieter Senior Member Posts: 2,397 Joined: Dec 2013
RE: HP 11C real root finder [Newton Method]
Namir, sorry that I was not able to express myself clearly. I wanted to point out that I do not think it's a good idea to use h = 0,01 or 0,001 for any value below 1. This may lead to significant errors since h may be much, much larger than x. For instance, if x = 1E-10, h = 1E-3 is not recommended. Here, h = 1E-13 should be better.
That's why I prefer to set h = 0,001 x or similar. With the only exception x=0 where x may be 0,001.
(01-15-2014 05:53 AM)Namir Wrote: Until I realized one day that .01*(|x|+1) does the job and eliminates the need for labels and GOTOs. I recently started using .001 instead of 0.01. Using the expression for h ensures that if x=0, h is not zero.
This can be coded this way, for instance:
Code:
X=0? e^x EEX 3 /
Look, no labels or gotos required. ;-)
Or even more elegant on the 34s:
Code:
X=0? INC X SDR 3 RSD 1 ' optional
Dieter
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# Basic Taylor Polynomial Question involving e^(-x)^2
1. Nov 15, 2014
### RJLiberator
1. The problem statement, all variables and given/known data
Consider:
F(x) =$\int_0^x e^{-x^2} \, dx$
Find the Taylor polynomial p3(x) for the function F(x) centered at a = 0.
2. Relevant equations
Tabulated Taylor polynomial value for standard e^x
3. The attempt at a solution
I started out by using the tabulated value for Taylor polynomial e^x and replaced all x with -x^2.
Going to the degree of three I received:
e^(-x)^2 = 1 - (x^2)/1! + x^4/2! - x^6/3!
Is this all there is to it? The question simply states to find the Taylor Polynomial p3(x) for the function, nothing else. I don't need to find the sum or the remainder per the problems question. Correct?
If I were to go at the sum, would I integrate each value?
Thank you for any help, I think I am just getting used to the wording here more than anything.
2. Nov 15, 2014
### gopher_p
The Taylor polynomial $p_3$ of degree $3$ for $F$ centered at $0$ should look something like $$p_3(x)=F(0)+F'(0)x+\frac{F''(0)}{2}x^2+\frac{F'''(0)}{6}x^3,$$ yes? If you remember the Fundamental Theorem of Calculus and what is says about computing $F'(x)$, it shouldn't be too hard to get the coefficients, right? So you can just do the problem "properly" without worrying whether your methods work.
I'm not saying your way won't work, though you'd theoretically only need the degree $2$ Taylor polynomial for $f(x)=e^{-x^2}$ to extract the degree $3$ polynomial for $F$. But the question you'd want to ask is whether the derivative of the degree $n$ Taylor polynomial of a function $g$ really is the degree $n-1$ Taylor polynomial for the function $g'$ and vice -versa; i.e. is the (anti)derivative of the Taylor polynomial the Taylor polynomial of the (anti)derivative? I don't think that question is too incredibly difficult to answer for those who understand what it is asking.
3. Nov 17, 2014
### RJLiberator | {
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3. Nov 17, 2014
### RJLiberator
Hm. Bare with me. I am in the early process of understanding Taylor Series wordage:
So what you are saying is that I should take the derivative of e^(-x)^2 a few times to acquire the coefficients? If I do this I answer:
f(0) = 1
f'(0) = 1
f''(0) = 2
f'''(0) = 4
Which would mean my original equation is incorrect. I am not sure what you mean by doing the problem properly without worrying.
I guess my first question would be: Is my answer correct? All I did was input -x^2 to the e^x common taylor series polynomial expansion for the first 3 degrees.
Then my second question would be how to proceed with this problem 'properly' ? I will review the fundamental theorem of calculus.
Question 3: Out or curiosity, why would I need to only go to degree 2 for this problem? Doesn't it state to go to degree 3?
4. Nov 17, 2014
### Staff: Mentor
1. Yes, what you did looks fine.
2. Integrate the series you found.
3. If you have terms up to degree 2, when you integrate, what will be the highest degree you get?
5. Nov 17, 2014
### RJLiberator
Mark44, you are shedding light on this topic!
1. Thank you for your confirmation.
2. Hm, ok. I am going to mess around with a few things. I know my coefficients then, now I got to find where to plug them in and integrate.
3. Ah.. integrating would bring it to degree three.
6. Nov 17, 2014
### gopher_p
The way that I was suggesting you do the problem is as follows;
For $F(x)=\int_0^xe^{-t^2}\ \mathrm{d}t$,
$F(0)=\int_0^0e^{-t^2}\ \mathrm{d}t=0$
$F'(x)=e^{-x^2}\Rightarrow F'(0)=1$, where the derivative is given by the FTC.
$F''(x)=-2xe^{-x^2}\Rightarrow F''(0)=0$
$F'''(x)=-2e^{-x^2}+4x^2e^{-x^2}\Rightarrow F'''(0)=-2$
Then the Taylor polynomial of degree three centered at $a=0$ is given by $$p_3(x)=F(0)+F'(0)x+\frac{F''(0)}{2}x^2+\frac{F'''(0)}{6}x^3=x-\frac{1}{3}x^3$$ | {
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So we're just (a) appealing to the definition of the Taylor polynomial and (b) computing the coefficients directly. I guess that's what I meant by "properly".
7. Nov 18, 2014
### RJLiberator
OK, so with your help, I've been working through this problem.
Question 1: So for F(0) you made the bounds 0 to 0. I don't understand why. I see that the answer is clearly 0 when the bounds are both 0 to 0, but why do this?
Other then this part, everything you wrote now makes complete sense to me.
Question 2: In my original solution where I just inputted -x^2 to the e^x tabulated value i had NOT integrated yet. Did I HAVE to integrate after I replaced those values? If so, the two answers match up nicely. I would imagine so, by the definition of the Taylor Series, correct?
8. Nov 18, 2014
### gopher_p
One of the "skills" that you learned in you preparation for calculus was that, given a formula for $F(x)$, you compute $F(a)$ by plugging in $a$ for $x$. So given $F(x)=\int_0^xe^{-t^2}\ \mathrm{d}t$, get $F(0)$ by plugging in $0$ for $x$; $F(0)=\int_0^0e^{-t^2}\ \mathrm{d}t$. This is why it is important to use a different variable of integration - $t$ here - inside the integral.
I'm confused by the question. It's clear that the degree two Taylor polynomial $\tilde p_2$ for $f(x)=e^{-x^2}$ is (a) not a degree three polynomial and (b) not a Taylor polynomial for the function $F$, right? So something must be done if we are to extract $p_3$ from $\tilde p_2$.
The real question, in my mind, is whether it is true that $p_3(x)=\int_0^x\tilde p_2(t)\ \mathrm{d}t$. In this case it is. It works here because the lower limit of integration, $0$, is the same as the center of the Taylor polynomials. You need to be careful in a more general setting.
9. Nov 18, 2014
### PeroK
You should note that where you wrote:
F(x) =$\int_0^x e^{-x^2} \, dx$
gopher_p wrote:
F(x) =$\int_0^x e^{-t^2} \, dt$ | {
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F(x) =$\int_0^x e^{-x^2} \, dx$
gopher_p wrote:
F(x) =$\int_0^x e^{-t^2} \, dt$
I think you need to understand the difference and why the second one correctly defines F as a function of x, with t as the dummy variable.
This may also be the reason why you did not understand the substitution $x = 0$
10. Nov 18, 2014
### RJLiberator
Guys, thanks for your help here.
You were a help to me and my classmates.
Also, thanks for pointing out my subtle mistakes that cause errors in conceptual understanding. I am starting to understand this problem due to this.
I see why the dummy variable matters.
As always, excellent help. | {
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# Find the function $f(x)$ if $f(x+2f(y))=f(x)+y+f(y)$
Let $f:\mathbb{R}\to \mathbb{R}$ such $f(x)$ at $x=0$ continuous, and for any $x,y\in \mathbb{R}$ such $$f(x+2f(y))=f(x)+y+f(y)$$ Find $f(x)$.
Let $x=0,y=0$ then we have $$2f(2f(0))=f(0)++f(0)$$ Let $y=2f(0)$ then we have $$f(x+2f(2f(0))=f(x)+2f(0)+f(2f(0))$$ so we have $$f(x+4f(0))=f(x)+4f(0)$$ on the other hand we have $$f(x+4f(0))=f(x+2f(0))+0+f(0)=2f(0)+f(x)$$ so we have $f(0)=0$ Then I can't deal with this problem
• Maybe I am off base but what happens if you plug in $y=x$? Sep 23, 2016 at 5:29
• @CameronWilliams you get that $x+2f(x)$ is a fixed point. Sep 23, 2016 at 5:30
• Am I missing something or does $f(x)=x$ work? Sep 23, 2016 at 6:21
• @Mastrem It does, but presumably the question is to find all such functions. Sep 23, 2016 at 6:22
• @Mastrem it works, but no one says this equation has only one solution. You have to prove then that it's the only one. Sep 23, 2016 at 6:23
Substitute $x\mapsto y$ to get $$f(y+2f(y))=y+2f(y).$$ Now substitute $y\mapsto y+2f(y)$: $$f(x+2f(y+2f(y)))=f(x)+y+2f(y)+f(y+2f(y)).$$ Thus $$f(x+2y+4f(y))=f(x)+2y+4f(y).$$ Substituting $x\mapsto x+2y+2f(y)$, $$f(x+2y+4f(y))=f(x+2y+2f(y))+y+f(y).$$ Substituting $x\mapsto x+2y$, $$f(x+2y+2f(y))=f(x+2y)+y+f(y).$$ Combining the last 3 equations, $$f(x+2y)=f(x)+2f(y).$$ Thus $$f(x)=f(0+2x/2)=f(0)+2f(x/2)=2f(x/2),$$ so $$f(x+y)=f(x+2y/2)=f(x)+2f(y/2)=f(x)+f(y).$$ Since you are given that $f$ is continuous at $0$, it must be continuous everywhere. Now there is a standard argument to show that $f(x)=f(1)x$ (prove it on rationals and extend by continuity). Finally using the original equation, we see $f(1)=1$ or $-\frac12$, so the solutions are $f(x)=x$ or $f(x)=-x/2$.
Setting $x=y$ gives $f(x+2f(x))=x+2f(x)$, and so $f(y)=y$ whenever $y$ is of the form $x+2f(x)$ for some $x$. Note also that if $f(y)=y$, then for any $x$, $f(x+2y)=f(x)+2y$. | {
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Now consider the function $g(x)=f(x)-x$. From the previous paragraph, we see that if $f(y)=y$ then $g(x+2y)=g(x)$. In particular, every number of the form $2x+4f(x)$ is a period of $g$. Let $P\subseteq\mathbb{R}$ be the group of periods of $g$, i.e. the set of $p$ such that $g(x)=g(x+p)$ for all $x$. There are now two cases.
The first case is that $P$ is not cyclic. Then $P$ is dense in $\mathbb{R}$, and so each coset of $P$ is dense in $\mathbb{R}$, and in particular $0$ is in the closure of each coset. But by definition of $P$, $g$ is constant on each coset of $P$, and so by continuity of $g$ at $0$ this constant value must be equal to $g(0)$. Thus $g(x)=g(0)$ for all $x$, and $g$ is constant everywhere. Thus $f(x)$ has the form $f(x)=x+c$ for some constant $c$. Plugging this into the functional equation gives $$x+2(y+c)+c=x+c+y+y+c,$$ so $c=0$. Thus $f(x)=x$.
The second case is that $P$ is cyclic, generated by some $p\in\mathbb{R}$ (possibly $0$). Then every number of the form $2x+4f(x)$ is an integer multiple of $p$. But $2x+4f(x)$ is continuous at $0$, so this means $2x+4f(x)$ is constant in a neighborhood of $0$. As you've shown, $f(0)=0$, so $2x+4f(x)=0$ for all $x$ in some neighborhood of $0$, so $f(x)=-x/2$ for all $x$ in some neighborhood of $0$.
Now suppose $\epsilon>0$ is such that $f(x)=-x/2$ whenever $|x|\leq\epsilon$. If $0\leq a\leq \epsilon$, the functional equation with $x=a$ and $y=-\epsilon$ then gives $$f(a+\epsilon)=-\frac{a+\epsilon}{2}.$$ That is, $f(x)=-x/2$ is also valid if $\epsilon\leq x\leq 2\epsilon$. Similarly, using $y=\epsilon$ gives that $f(x)=-x/2$ is also valid if $-2\epsilon\leq x\leq -\epsilon$.
That is, if $f(x)=-x/2$ whenever $|x|\leq\epsilon$, $f(x)=-x/2$ whenever $|x|\leq 2\epsilon$ as well. Repeating this argument over and over, we get that $f(x)=-x/2$ for all $x$.
Thus the only possibilities for $f$ are $f(x)=x$ and $f(x)=-x/2$, and you can easily check that both of these work. | {
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• Nice try! Even if using machinery which .. outguns the problem! I downvoted by accident.(and I thought I had already cancelled it. Seems I hadn't!) I will make a trivial edit so that I'll be able to cancel it. P.S.: the part where you are arguing about the cosets of the group of periods is not very clear. Could you expand it a little? (what is the equivalence relation here? why does the continuity of $g$ implies that every coset accumulates at $0$ ?). Sep 23, 2016 at 18:35
• I've clarified that part a bit. Sep 23, 2016 at 19:12
• Thanks for your response. As I've already said: very nice ! +1 Sep 25, 2016 at 19:49 | {
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Geometry - Prove a right triangle resulting from three inscribed circles
Two half circles and a full circle fit inside a larger quarter circle as shown in the diagram. The centers of the two half circles are on the two sides of the quarter circle, respectively.
Prove that the triangle formed by the centers of three smaller circles, $$\triangle O_1O_2O_3$$, is a right triangle.
I was able to apply the Pythagorean formula to a few triangles involving the radii of the inscribed circles and proved that the ratios of the three radii are 1:2:3. Then, the centers of the three circles form a triangle with side-length ratios 3:4:5, hence, a right triangle.
On the other hand, I feel the proof may be an overkill, and evaluating the three radii explicitly may be unnecessary. There ought to be clean geometric solutions to show directly that the vertex $$O_3$$ is of a right angle, which I am not sure how to figure out. | {
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• Suppose $OO_1$ and $OO_2$ are x and y axis. If you complete half and quarter circles, you will have a big circle containing two big circles, two medium and four little circles. the center of little circles are symmetric about x and y axis. since x and y axis are perpendicular then the lines connecting the centers of little circles are perpendicular that is triangle $O_1O_2O_3$ is right angle at $O_3$. – sirous Sep 14 '19 at 21:01
• Insightful. A good case of can’t-see-forest-staring at trees. Thanks – Quanto Sep 14 '19 at 21:10
• @sirous I finally read your comment. I think it would be clearer if you "complete" the half and quarter circles by reflecting the figure over one axis and then the other. But your construction is much more elegant than mine and would make a fine answer. – David K Sep 15 '19 at 0:30
• @sirous: Certainly, "the lines connecting the centers of the little circles are perpendicular", but how do you know that those lines contain $O_1$ and $O_2$? – Blue Sep 15 '19 at 3:51
• @Blue - critical observation. guess a simple geometric proof remains elusive – Quanto Sep 15 '19 at 14:42
Given the quarter circle with $$O$$ as the center of the arc, and given semicircles with centers $$O_1$$ and $$O_2$$ on the two straight sides of the quarter circle such that $$O_1$$ and $$O_2$$ both are tangent to the arc of the quarter circle and such that $$O_1$$ and $$O_2$$ are tangent to each other.
This is based on the figure in the question, except that we do not assume that $$O_1$$ is the midpoint of one side of the quarter circle.
Let the radii of the quarter circle, the semicircle about $$O_1$$, and the semicircle about $$O_2$$ be $$r,$$ $$r_1,$$ and $$r_2$$ respectively.
Let $$P$$ be the fourth vertex of the rectangle that has three of its vertices at $$O,$$ $$O_1,$$ and $$O_2.$$ The sides of this rectangle are $$OO_1 = r - r_1$$ and $$OO_2 = r - r_2.$$ The diagonals are $$OP = O_1O_2 = r_1 + r_2.$$ | {
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Let $$P_1$$ be the point of intersection of the semicircle about $$O_1$$ with $$O_1P$$ and $$P_2$$ be the point of intersection of the semicircle about $$O_2$$ with $$O_2P.$$ Then $$PP_1 = r - r_1 - r_2 = PP_2.$$ Moreover, the distance from $$P$$ to the arc of the quarter circle is $$r - (r_1 + r_2) = r - r_1 - r_2.$$
Then a circle of radius $$r - r_1 - r_2$$ about $$P$$ is tangent to the quarter circle and to the semicircles about $$O_1$$ and $$O_2.$$ Therefore $$P$$ is $$O_3$$ in the figure, the center of the circle inscribed between the two semicircles and the quarter circle.
Since $$O_3$$ is a vertex of the rectangle $$OO_1O_3O_2,$$ the triangle $$\triangle O_1O_3O_2$$ is a right triangle. $$\square$$
• +1 a nice generalization. – achille hui Sep 15 '19 at 16:27
Let $$R=6a$$ then $$r_1=3a$$
$$\triangle OO_1O_2: \;\;\;\; (R-r_2)^2+r_1^2 =(r_1+r_2)^2$$
So $$\boxed{r_2= 2a}$$
So $$\triangle O_3O_1O_2$$ is right iff $$(r_1+r_2)^2 = (r_1+r_3)^2+(r_2+r_3)^2$$
or $$r_1r_2 = r_3(r_1+r_2+r_3)$$
or $$6a^2 = r_3(5a+r_3)\iff r_3^2+5ar_3-6a^2 =0$$
So we have to prove: $$\boxed{\color{red}{r_3=a}}$$
If we put everything in coordinate system then we have: $$O(0,0)$$, $$O_1(3a,0)$$, $$O_2(0,4a)$$ and let $$O_3(m,n)$$, so $$\begin{eqnarray} m^2+n^2 &=& (6a-r_3)^2\;\;\;\;\;\; (OO_3 = R-r_3)\\ (m-3a)^2+n^2 &=& (3a+r_3)^2\;\;\;\;\;\; (O_1O_3 = r_1+r_3)\\ m^2+(n-4a)^2 &=& (2a+r_3)^2\;\;\;\;\;\; (O_2O_3 = r_2+r_3) \end{eqnarray}$$
If we substract 1.st and 2.nd equation we get $$m = 6a-3r_3$$ and if we substract 1.st and 3.rd equation we get $$n=6a-2r_3$$
Now put this again in 1.st equation and we get $$\boxed{\color{green}{r_3=a}}$$
Following is an answer based on circle inversion w/o computing the radius of circle centered at $$O_3$$. | {
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Label the contact points as illustrated below. We are going to show $$\angle OYD = \angle OXC = 45^\circ$$ Form these, it is easy to deduce $$\begin{cases} O_2O_3 \parallel O_2D \parallel OX,\\ O_1O_3 \parallel O_1C \parallel OY\end{cases}$$ and hence $$\angle O_1O_3O_2 = 90^\circ$$
Choose a coordinate system so that $$O$$ is the origin and $$X = (1,0)$$, $$Y = (0,1)$$. The arc $$XEY$$ will be part of the unit circle. Under circle inversion with respect to the unit circle centered at $$O$$, we get what illustrated below:
• The red semicircle centered at $$O_1$$ get mapped to a ray $$y = 1$$ in the upper half plane.
• The green semicircle centered at $$O_2$$ get mapped to another semicircle centered at some point of $$y$$-axis and touching the red line. This means this inverted green semicircle has radius $$1$$.
As a result, the inverted image of point $$A$$ is $$A' = (0,3) \implies A = (0,\frac13)$$.
• The blue circle get mapped to a circle placed symmetrically between the inverted green semicircle and the arc $$XEY$$. As a result, the inverted image of point $$C$$ is $$C' = (1,1)$$.
This implies $$\angle OXC = \angle OXC' = 45^\circ$$.
For the other angle, translate the coordinate axis so that $$Y$$ is the new origin. The new coordinate of $$A$$ is now $$(0,-\frac23)$$. Under circle inversion with respect to the unit circle centered at $$Y$$, it now looks like:
• The arc $$XEY$$ get mapped to a ray on the line $$y = -\frac12$$.
• The green semicircle centered at $$O_2$$ get mapped to a ray on the line $$y = -\frac32$$.
• The red semicircle get mapped to an arc on a circle sandwiched between above two rays touching the $$y$$-axis. So its radius is $$\frac12$$.
• The blue circle get mapped to an circle also sandwiched between above two rays. This means the inverted image of $$D$$ is located at $$D'' = (\frac32,-\frac32)$$. As a result, $$\angle OYD = \angle OYD' = 45^\circ$$. | {
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Let $$OO_1$$ and $$OO_2$$ are x and y axis respectively. Mirror the figure about y axis, then mirror the resulted figure about x axis. You get two big, two medium and four small circles indescribable in a large circle. If you connect the centers of four small circles you get a square. So angle at $$O_3$$ is $$90^o$$. Now let the radius of small circle is $$r_1=1$$ and that of large circle is $$r_2=6$$. Suppose the location of center of small circle is $$O_3(3, 4)$$ which satisfies:
$$3^2+4^2= [OO_3=6-1=5]^2$$
Then we can claim that the radius of medium circle is $$r_3=3-1=2$$ and those of big and large circles are $$r_4=4-1=3$$ and $$r_2=2\times 3=6$$ respectively that is all involved circle are mutually tangent.If radius of medium circle is $$r_3=2$$ that means that it's center locates on a line passing $$O_3$$ which has a distance=4 from x axis. similarly if the radius of big circle is $$r_4=3$$ that would mean that it's center locates on a line passing $$O_3$$ which has distance =3 from y axis.That is triangle $$O_1O_2O_3$$ is right triangle at $$O_3$$ | {
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# Find the probability that, on Tuesday, it does not rain in Greg's area and he does not get a quiz.
The probability of rain in Greg's area on Tuesday is $0.3$. The probability that Greg's teacher will give him a pop quiz in Tuesday is $0.2$. The events occur independently of each other. What is the probability of neither events occur?
My approach:
Probability of rain or quiz or both = $0.3+0.2= 0.5$
So, probability of neither= $1-0.5$ = $0.5$
Question: Actual probability of this problem is $0.7\cdot0.8$ = $0.56$. But, I don't understand what is the mistake in above approach?
• They are independent not mutually exclusive. For mutually exclusive events you have $Pr(A\text{or}B)=Pr(A)+Pr(B)$. That is not true of independent events. – JMoravitz Aug 23 '16 at 6:11
• Principle of inclusion-exclusion 2set case: $Pr(A\cup B) = Pr(A)+Pr(B)-Pr(A\cap B)$ – JMoravitz Aug 23 '16 at 6:18
Note the following two things:
Principle of inclusion-exclusion (2-set case) $$Pr(A\cup B) = Pr(A)+Pr(B)-Pr(A\cap B)$$
Definition of independent events
The following are equivalent statements
• $A$ and $B$ are independent events
• $Pr(A\mid B) = Pr(A)$
• $Pr(B\mid A) = Pr(B)$
• $Pr(A\cap B)=Pr(A)\cdot Pr(B)$
For your problem: you know $Pr(A)=0.3, Pr(B)=0.2$ and they are independent, so
$$Pr(A\cup B) = Pr(A)+Pr(B)-Pr(A\cap B) = 0.3+0.2-0.3\cdot 0.2 = 0.5-0.06 = 0.44$$
So the probability of at least one of the events occurring is $0.44$. The probability of no events occurring is then one minus that, i.e. $1-0.44=0.56$.
Alternatively, if $A$ and $B$ are independent, then $A^c$ and $B^c$ are also independent. We have then $Pr(A^c\cap B^c)=Pr(A^c)Pr(B^c)=(1-0.3)(1-0.2)=0.7\cdot 0.8 = 0.56$ | {
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Your approach double-counts some possibilities. The $0.3$ covers all possibility of rain - so, it includes both "rain and quiz" and "rain and no quiz". The $0.2$ covers all possibility of a quiz, so it includes both "rain and quiz" and "quiz and no rain". So $0.3 + 0.2$ covers "rain and quiz", "rain and no quiz", "rain and quiz" again, and "quiz and no rain". To cut out one of the countings of "rain and quiz", we need to subtract the probability of that event, which is $0.3 \cdot 0.2 = 0.06$. So the probability of "rain or quiz or both" is $0.2 + 0.3 - 0.06 = 0.44$. The probability of neither is then $1 - 0.44 = 0.56$. | {
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# 4.4 The mean value theorem (Page 2/7)
Page 2 / 7
Verify that the function $f\left(x\right)=2{x}^{2}-8x+6$ defined over the interval $\left[1,3\right]$ satisfies the conditions of Rolle’s theorem. Find all points $c$ guaranteed by Rolle’s theorem.
$c=2$
## The mean value theorem and its meaning
Rolle’s theorem is a special case of the Mean Value Theorem. In Rolle’s theorem, we consider differentiable functions $f$ that are zero at the endpoints. The Mean Value Theorem generalizes Rolle’s theorem by considering functions that are not necessarily zero at the endpoints. Consequently, we can view the Mean Value Theorem as a slanted version of Rolle’s theorem ( [link] ). The Mean Value Theorem states that if $f$ is continuous over the closed interval $\left[a,b\right]$ and differentiable over the open interval $\left(a,b\right),$ then there exists a point $c\in \left(a,b\right)$ such that the tangent line to the graph of $f$ at $c$ is parallel to the secant line connecting $\left(a,f\left(a\right)\right)$ and $\left(b,f\left(b\right)\right).$
## Mean value theorem
Let $f$ be continuous over the closed interval $\left[a,b\right]$ and differentiable over the open interval $\left(a,b\right).$ Then, there exists at least one point $c\in \left(a,b\right)$ such that
$f\prime \left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}.$
## Proof
The proof follows from Rolle’s theorem by introducing an appropriate function that satisfies the criteria of Rolle’s theorem. Consider the line connecting $\left(a,f\left(a\right)\right)$ and $\left(b,f\left(b\right)\right).$ Since the slope of that line is
$\frac{f\left(b\right)-f\left(a\right)}{b-a}$
and the line passes through the point $\left(a,f\left(a\right)\right),$ the equation of that line can be written as
$y=\frac{f\left(b\right)-f\left(a\right)}{b-a}\left(x-a\right)+f\left(a\right).$ | {
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$y=\frac{f\left(b\right)-f\left(a\right)}{b-a}\left(x-a\right)+f\left(a\right).$
Let $g\left(x\right)$ denote the vertical difference between the point $\left(x,f\left(x\right)\right)$ and the point $\left(x,y\right)$ on that line. Therefore,
$g\left(x\right)=f\left(x\right)-\left[\frac{f\left(b\right)-f\left(a\right)}{b-a}\left(x-a\right)+f\left(a\right)\right]\text{.}$
Since the graph of $f$ intersects the secant line when $x=a$ and $x=b,$ we see that $g\left(a\right)=0=g\left(b\right).$ Since $f$ is a differentiable function over $\left(a,b\right),$ $g$ is also a differentiable function over $\left(a,b\right).$ Furthermore, since $f$ is continuous over $\left[a,b\right],$ $g$ is also continuous over $\left[a,b\right].$ Therefore, $g$ satisfies the criteria of Rolle’s theorem. Consequently, there exists a point $c\in \left(a,b\right)$ such that $g\prime \left(c\right)=0.$ Since
$g\prime \left(x\right)=f\prime \left(x\right)-\frac{f\left(b\right)-f\left(a\right)}{b-a},$
we see that
$g\prime \left(c\right)=f\prime \left(c\right)-\frac{f\left(b\right)-f\left(a\right)}{b-a}.$
Since $g\prime \left(c\right)=0,$ we conclude that
$f\prime \left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}.$
In the next example, we show how the Mean Value Theorem can be applied to the function $f\left(x\right)=\sqrt{x}$ over the interval $\left[0,9\right].$ The method is the same for other functions, although sometimes with more interesting consequences.
## Verifying that the mean value theorem applies
For $f\left(x\right)=\sqrt{x}$ over the interval $\left[0,9\right],$ show that $f$ satisfies the hypothesis of the Mean Value Theorem, and therefore there exists at least one value $c\in \left(0,9\right)$ such that ${f}^{\prime }\left(c\right)$ is equal to the slope of the line connecting $\left(0,f\left(0\right)\right)$ and $\left(9,f\left(9\right)\right).$ Find these values $c$ guaranteed by the Mean Value Theorem. | {
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We know that $f\left(x\right)=\sqrt{x}$ is continuous over $\left[0,9\right]$ and differentiable over $\left(0,9\right).$ Therefore, $f$ satisfies the hypotheses of the Mean Value Theorem, and there must exist at least one value $c\in \left(0,9\right)$ such that ${f}^{\prime }\left(c\right)$ is equal to the slope of the line connecting $\left(0,f\left(0\right)\right)$ and $\left(9,f\left(9\right)\right)$ ( [link] ). To determine which value(s) of $c$ are guaranteed, first calculate the derivative of $f.$ The derivative ${f}^{\prime }\left(x\right)=\frac{1}{\left(2\sqrt{x}\right)}.$ The slope of the line connecting $\left(0,f\left(0\right)\right)$ and $\left(9,f\left(9\right)\right)$ is given by
$\frac{f\left(9\right)-f\left(0\right)}{9-0}=\frac{\sqrt{9}-\sqrt{0}}{9-0}=\frac{3}{9}=\frac{1}{3}.$
We want to find $c$ such that ${f}^{\prime }\left(c\right)=\frac{1}{3}.$ That is, we want to find $c$ such that
$\frac{1}{2\sqrt{c}}=\frac{1}{3}.$
Solving this equation for $c,$ we obtain $c=\frac{9}{4}.$ At this point, the slope of the tangent line equals the slope of the line joining the endpoints. | {
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find the nth differential coefficient of cosx.cos2x.cos3x
determine the inverse(one-to-one function) of f(x)=x(cube)+4 and draw the graph if the function and its inverse
f(x) = x^3 + 4, to find inverse switch x and you and isolate y: x = y^3 + 4 x -4 = y^3 (x-4)^1/3 = y = f^-1(x)
Andrew
in the example exercise how does it go from -4 +- squareroot(8)/-4 to -4 +- 2squareroot(2)/-4 what is the process of pulling out the factor like that?
Andrew
√(8) =√(4x2) =√4 x √2 2 √2 hope this helps. from the surds theory a^c x b^c = (ab)^c
Barnabas
564356
Myong
can you determine whether f(x)=x(cube) +4 is a one to one function
Crystal
one to one means that every input has a single output, and not multiple outputs. whenever the highest power of a given polynomial is odd then that function is said to be odd. a big help to help you understand this concept would be to graph the function and see visually what's going on.
Andrew
one to one means that every input has a single output, and not multiple outputs. whenever the highest power of a given polynomial is odd then that function is said to be odd. a big help to help you understand this concept would be to graph the function and see visually what's going on.
Andrew
can you show the steps from going from 3/(x-2)= y to x= 3/y +2 I'm confused as to how y ends up as the divisor
step 1: take reciprocal of both sides (x-2)/3 = 1/y step 2: multiply both sides by 3 x-2 = 3/y step 3: add 2 to both sides x = 3/y + 2 ps nice farcry 3 background!
Andrew
first you cross multiply and get y(x-2)=3 then apply distribution and the left side of the equation such as yx-2y=3 then you add 2y in both sides of the equation and get yx=3+2y and last divide both sides of the equation by y and you get x=3/y+2
Ioana | {
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