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Zero to the zero power - Is $0^0=1$?
Could someone provide me with good explanation of why $0^0 = 1$?
My train of thought:
$x > 0$
$0^x = 0^{x-0} = 0^x/0^0$, so
$0^0 = 0^x/0^x = ?$
1. $0^0 * 0^x = 1 * 0^x$, so $0^0 = 1$
2. $0^0 = 0^x/0^x = 0/0 = \text{undefined}$
Thank you
PS. I've read the explanation on mathforum.org, but it isn't clear to me.
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@all: Do we really need tags like (powers) and (zeros)? I think they are uninformative and should not be used. Also there are only 7 questions tagged with at least one of them. Let me know if I'm wrong. – Nuno Nov 21 '10 at 1:55
Can you pls link to the explanation that you read on mathforum.org? – Lazer Nov 21 '10 at 6:48
+1 for good elementary question. – tia Nov 21 '10 at 8:41
BTW, this is comprehensively covered on Wikipedia (or see today's version), along with pointers to history and treatment in many systems. – ShreevatsaR Nov 22 '10 at 12:34
@Stas : actually, $0^0$ is generally considered "undefined". But when people use power series, they routinely treat $0^0$ as $1$ without a second thought. If a function $f$ is defined by a power series as $f(x) = \sum_{n=0}^\infty a_n (x-b)^n$, then everyone agrees that $f(b) = a_0$, even though plugging in $x = b$ into the series involves $0^0$. I hope this adds something to the dozens of other comments and previous answers. – Stefan Smith Apr 15 '13 at 16:26
In general, there is no good answer as to what $0^0$ "should" be, so it is usually left undefined.
Basically, if you consider $x^y$ as a function of two variables, then there is no limit as $(x,y)\to(0,0)$ (with $x\geq 0$): if you approach along the line $y=0$, then you get $\lim\limits_{x\to 0^+} x^0 = \lim\limits_{x\to 0^+} 1 = 1$; so perhaps we should define $0^0=1$? Well, the problem is that if you approach along the line $x=0$, then you get $\lim\limits_{y\to 0^+}0^y = \lim\limits_{y\to 0^+} 0 = 0$. So should we define it $0^0=0$? | {
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Well, if you approach along other curves, you'll get other answers. Since $x^y = e^{y\ln(x)}$, if you approach along the curve $y=\frac{1}{\ln(x)}$, then you'll get a limit of $e$; if you approach along the curve $y=\frac{\ln(7)}{\ln(x)}$, then you get a limit of $7$. And so on. There is just no good answer from the analytic point of view. So, for calculus and algebra, we just don't want to give it any value, we just declare it undefined.
However, from a set-theory point of view, there actually is one and only one sensible answer to what $0^0$ should be! In set theory, $A^B$ is the set of all functions from $B$ to $A$; and when $A$ and $B$ denote "size" (cardinalities), then the "$A^B$" is defined to be the size of the set of all functions from $A$ to $B$. In this context, $0$ is the empty set, so $0^0$ is the collection of all functions from the empty set to the empty set. And, as it turns out, there is one (and only one) function from the empty set to the empty set: the empty function. So the set $0^0$ has one and only one element, and therefore we must define $0^0$ as $1$. So if we are talking about cardinal exponentiation, then the only possible definition is $0^0=1$, and we define it that way, period.
Added 2: the same holds in Discrete Mathematics, when we are mostly interested in "counting" things. In Discrete Mathematics, $n^m$ represents the number of ways in which you can make $m$ selections out of $n$ possibilities, when repetitions are allowed and the order matters. (This is really the same thing as "maps from $\{1,2,\ldots,m\}$ to $\\{1,2,\ldots,n\\}$" when interpreted appropriately, so it is again the same thing as in set theory). | {
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So what should $0^0$ be? It should be the number of ways in which you can make no selections when you have no things to choose from. Well, there is exactly one way of doing that: just sit and do nothing! So we make $0^0$ equal to $1$, because that is the correct number of ways in which we can do the thing that $0^0$ represents. (This, as opposed to $0^1$, say, where you are required to make $1$ choice with nothing to choose from; in that case, you cannot do it, so the answer is that $0^1=0$).
Your "train of thoughts" don't really work: If $x\neq 0$, then $0^x$ means "the number of ways to make $x$ choices from $0$ possibilities". This number is $0$. So for any number $k$, you have $k\cdot 0^x = 0 = 0^x$, hence you cannot say that the equation $0^0\cdot 0^x = 0^x$ suggests that $0^0$ "should" be $1$. The second argument also doesn't work because you cannot divide by $0$, which is what you get with $0^x$ when $x\neq 0$. So it really comes down to what you want $a^b$ to mean, and in discrete mathematics, when $a$ and $b$ are nonnegative integers, it's a count: it's the number of distinct ways in which you can do a certain thing (described above), and that leads necessarily to the definition that makes $0^0$ equal to $1$: because $1$ is the number of ways of making no selections from no choices. | {
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Coda. In the end, it is a matter of definition and utility. In Calculus and algebra, there is no reasonable definition (the closest you can come up with is trying to justify it via the binomial theorem or via power series, which I personally think is a bit weak), and it is far more useful to leave it undefined or indeterminate, since otherwise it would lead to all sorts of exceptions when dealing with the limit laws. In set theory, in discrete mathematics, etc., the definition $0^0=1$ is both useful and natural, so we define it that way in that context. For other contexts (such as the one mentioned in mathforum, when you are dealing exclusively with analytic functions where the problems with limits do not arise) there may be both natural and useful definitions.
We basically define it (or fail to define it) in whichever way it is most useful and natural to do so for the context in question. For Discrete Mathematics, there is no question what that "useful and natural" way should be, so we define it that way. | {
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@Sivam: yes, in the sense that there is exactly one function from any infinite set to the one-element set. This does not say anything about the status of 1^{infty} as an indeterminate form. The two expressions mean different things in different contexts; that they are designated using the same notation is a convenience, not a necessity. – Qiaochu Yuan Nov 20 '10 at 23:27
@Sivam: exactly as Qiaochu says. Note that I said that $0^0=1$ in cardinal exponentiation is the only sensible answer, but "cardinal exponentiation" is not the same as real number exponentiation; when doing real number exponentiation, $0^0$ is most properly undefined/indeterminate. – Arturo Magidin Nov 20 '10 at 23:30
@Stas: You don't seem to have any "elementary case". All you have are your "train of thoughts". What case is it you are thinking about? You don't say. You don't tell us. I cannot read your mind from this afar away (and the Government doesn't let me do it without a Court order anyway...) – Arturo Magidin Nov 21 '10 at 2:51
Just a small note: the answer depends on whether you think of exponentiation as a discrete operation (as in set theory, algebra, combinatorics, number theory) or as a continuous operation over spaces like real/complex numbers (as in analysis). – Kaveh Nov 21 '10 at 5:47
@Stas: Some argue that; there are reasons for saying it should be one, but there are also compelling reasons for it to be other things. In some situations, some limits make more sense than others; if all you are concerned with are analytic functions, then it may make sense to define it to be 1 because in the only cases you will look at you will always get 1 as the limit. But precisely because the answer depends on context, it is left undefined in the abstract and only defined in certain specific contexts (such as combinatorics or cardinal exponentiation). – Arturo Magidin Nov 21 '10 at 19:36 | {
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This is merely a definition, and can't be proved via standard algebra. However, two examples of places where it is convenient to assume this:
1) The binomial formula: $(x+y)^n=\sum_{k=0}^n {n\choose k}x^ky^{n-k}$. When you set $y=0$ (or $x=0$) you'll get a term of $0^0$ in the sum, which should be equal to 1 for the formula to work.
2) If $A,B$ are finite sets, then the set of all functions from $B$ to $A$, denoted $A^B$, is of cardinality $|A|^{|B|}$. When both $A$ and $B$ are the empty sets, there is still one function from $B$ to $A$, namely the empty function (a function is a collection of pairs satisfying some conditions; an empty collection is a legal function if the domain $B$ is empty).
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You don't need to appeal to the binomial formula. Anytime you write a polynomial as f(x) = sum a_i x^i you need x^0 = 1 to keep your notation consistent, so you need 0^0 = 1 so that f(0) = a_0. – Qiaochu Yuan Nov 21 '10 at 1:12
Yes. I think it is reasonable to define $0^0=1$ (because that seems to be the most useful definition) with the caveat that the function $x^y$ on $\mathbb{R}^{+}\!\!\times\mathbb{R}$ is not continuous at $(0,0)$. – robjohn Nov 13 '13 at 11:28
$0^{0}$ is just one instance of an empty product, which means it is the multiplicative identity 1.
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$0^0$ is undefined. It is an Indeterminate form.
You might want to look at this post.
Why is $1^{\infty}$ considered to be an indeterminate form
As you said, $0^0$ has many possible interpretations and hence it is an indeterminate form.
For instance,
$\displaystyle \lim_{x \rightarrow 0^{+}} x^{x} = 1$.
$\displaystyle \lim_{x \rightarrow 0^{+}} 0^{x} = 0$.
$\displaystyle \lim_{x \rightarrow 0^{-}} 0^{x} =$ not defined.
$\displaystyle \lim_{x \rightarrow 0} x^{0} = 1$. | {
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$\displaystyle \lim_{x \rightarrow 0} x^{0} = 1$.
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Is $\lim_{x\to0}0^x$ really defined? It can only be approached from the positive side. – KennyTM Nov 21 '10 at 19:52
@KennyTM: Accepted and edited accordingly. – user17762 Nov 22 '10 at 11:47
0^0 is undefined by whom? I saw somebody defined it, can I now say it is defined? – Anixx Nov 4 '12 at 22:58
As I commented to Gadi A, I think it is quite reasonable to define $0^0=1$, as that seems to be the most useful definition, but note that the function $x^y$ on $\mathbb{R}^{+}\!\!\times\mathbb{R}$ is not continuous at $(0,0)$. – robjohn Nov 13 '13 at 11:32
One does not exclude the other, this can be both indeterminate form when considering limits, AND defined. – Anixx Mar 5 at 0:12
show 1 more comment
Some indeterminates forms $0^{0}, \displaystyle\frac{0}{0}, 1^{\infty}, \infty − \infty, \displaystyle\frac{\infty}{\infty}, 0 × \infty,$ and $\infty^{0}$
Futhermore,
$$\lim_{ x \rightarrow 0+ }x^{0}=1$$ and
$$\lim_{ x \rightarrow 0+ }0^{x}=0$$
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One should note that indeterminate is not the same as undefined. – Hagen von Eitzen Jul 10 '13 at 16:27
@Hagen, what's the difference? – JMCF125 Nov 22 '13 at 23:12
@JMCF125: $\lim\limits_{(x,y)\to(0,0)}x^y$ is indeterminate. This actually frees us to define $0^0$ to be whatever value is most useful. In almost every practical case, that is $0^0=1$. – robjohn Feb 10 at 6:23
@robjohn, ah, I see. Though now I wouldn't've made the comment, as I asked a related question that made me understand this. – JMCF125 Feb 10 at 11:16
@JMCF125: That's why the site is here. Where is the related question? – robjohn Feb 10 at 14:25
show 1 more comment
I'm surprised that no one has mentioned the IEEE standard for $0^0$. Many computer programs will give $0^0=1$ because of this. This isn't a mathematical answer per se, but it's worth pointing out because of the increasingly computational nature of modern mathematics, so that one doesn't run afoul of anything.
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The use of positive integer exponents appears in arithmetic as a shorthand notation for repeated multiplication. The notation is then extended in algebra to the case of zero exponent. The justification for such an extension is algebraic. Furthermore, in abstract algebra, if $G$ is a multiplicative monoid with identity $e$, and $x$ is an element of $G$, then $x^0$ is defined to be $e$. Now, the set of real numbers with multiplication is precisely such a monoid with $e=1$. Therefore, in the most abstract algebraic setting, $0^0=1$.
Continuity of $x^y$ is irrelevant. While there are theorems that state that if $x_n \to x$ and $y_n \to y,$ then $(x_n + y_n) \to x+y$ and $(x_n)(y_n) \to xy$, there is no corresponding theorem that states that $(x_n)^(y_n) \to x^y$. I don't know why people keep beating this straw man to conclude that $0^0$ can't or shouldn't be defined. | {
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Downvote, because Onez focuses on a very narrow view of the exponentiation operation and its applications, and writes as if that is the only view. Continuity is of rather significant importance in a wide variety of situations, and the requirements of an exponentiation function of continuous arguments are rather different than those limited to integer or rational exponents, and $0^0$ runs into that difference. Another example where the needs differ are $(-1)^{1/3}$ – Hurkyl Oct 26 '11 at 22:06
I hardly consider the whole domain of algebra to be narrow. YMMV. In any case, when extending the domain of functions, one may ask: Is the extension useful? Defining 0^0 to be 1 is useful in Combinatorics, Set Theory, and Algebra. Indeed, it is even useful in calculus when using summation notation for polynomial functions and infinite series. Perhaps you care to list several advantages of leaving 0^0 undefined? Particularly in light of the fact that many definitions require the additional caveat that a,b,x,y etc. not be equal to zero in order for them to be true. – Onez Oct 27 '11 at 1:13
It's the view that I said was narrow, not the breadth of application. The advantage to leaving $0^0$ undefined is in situations where continuity is relevant. You might never be in such situations, but others are. Really, there are three separate exponentiation operations in common use -- the algebraic one which is mostly defined for all bases and integer exponents (often extensible to rational exponents), the real one which is defined for positive real bases and real exponents (or some continuous extension thereof), and the complex multi-valued one. $0^0$ only makes sense for the first. – Hurkyl Oct 27 '11 at 16:41 | {
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By your argument, we should not define (-2)^0 = 1, but leave it undefined since this extension of exponentiation fails your second case (non-positive base) and your third case (no continuous extension of x^y to all of C). You still haven't supplied any advantages to leaving 0^0 undefined. Economy of notation (the reason exponential notation was developed in the first place) is gained by defining 0^0 = 1. – Onez Oct 27 '11 at 21:11
Since 0 is not in the range of the exponential function, I take it you have issue with 0^y being defined even for y>0. The argument seems to hinge on whether one is to define 0^0=1 and economize several definitions and theorems from algebra, combinatorics, and analysis, at the expense of one caveat for a single function, OR to leave 0^0 undefined, have several caveats so as to preserve the continuity on the domain of definition of a single function, namely x^y. Where is the greatest economization to be had? Who has the narrow view? – Onez Oct 28 '11 at 3:29 | {
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Take a look at WolframMathWorld's [1] discussion.
See if this gives you any clarification.
[1] Weisstein, Eric W. "Indeterminate." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/Indeterminate.html
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I sent them a message to remove 0^0 from inderminate, with necessary proof. – wendy.krieger May 16 '13 at 10:22
Knuth's answer is at least as good as any answer you're going to get here: http://arxiv.org/pdf/math/9205211v1.pdf See pp. 4-6, starting at the bottom of p. 4.
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It's pretty straight forward to show that multiplying something by $x$ zero times leaves the number unchanged, regardless of the value of $x$, and thus $x^0$ is the identity element for all $x$, and thus equal to one.
For the same reason, the sum of any empty list is zero, and the product is one. This is when a product or sum of an empty list is applied to a number, it leaves it unchanged. Thus if the product $\Pi()$ = 1, then we immediately see why $0! = 0^0 = 1$.
Without this property, one could prove that $2=3$, by the ruse that there are zero zeros in the product on the left (zero is after all, a legitimate count), and thus $2*0^0$, and since $0^0$ as indeterminate, could be 1.5, and thus $2=3$. I think not.
The approaches to $0^0$ by looking at $x^y$ from different directions, fails to realise that for even lines close to $x=0$, the line sharply sweeps up to 1 as it approaches $y=0$, and that the case for $x=0$, it may just be a case of not seeing it sweep up. On the other hand, looking from the other side, even in a diagonal line (ie $(ax)^x$), all do rapidly rise to 1, as x approaches 0. It's only when one approaches it from $0^x$ that you can't see it rising. So the evidence from the graph of $x^y$ is that $0^0$ is definitely 1, except when approached from $y=0$, when it appears to be zero.
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A paper for general public is published on Scribd : " Zero to the Zero-th Power" (pp.7-11) : http://www.scribd.com/JJacquelin/documents
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"Everybody knows" that $$e^z = \sum_{n=0}^\infty \frac{z^n}{n!},$$ and when $z=0$ then the first term is $\dfrac{0^0}{0!}$, so of course $0^0$ is $1$ since it's an empty product.
But it's also an indeterminat form because $\displaystyle\lim_{x\to z}f(x)^{g(x)}$ can be any positive number, or $0$ or $\infty$, depending on which functions $f$ and $g$ are, if $f(x)$ and $g(x)$ both approach $0$ as $x\to a$.
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Source: Understanding Exponents: Why does 0^0 = 1? (BetterExplained article)
A useful analogy to explain the exponent operator of the form $a \cdot b^c$ is to make $a$ grow at the rate $b$ for time $c$.
Expanding on that analogy, $0^0$ can be interpreted as $1\cdot0^0$ which is to say: grow $1$ at the rate of $0$ for time $0$. Since there is no growth (time is $0$), there is no change in the $1$ and the answer is $0^0=1$
Of course, this is just to grok and get an intuition or a feel for it. Science is provisional and so is math in certain areas. 0^0=1 is not always the most useful or relevant value at all times.
Using limits or calculus or binomial theorems doesn't really give you an intuition of why this is so, but I hope this post made you understand why it is so and make you feel it from your spleen.
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Downvoter explain. I am not trying to be rigorous, I'm just trying to give an intuition and make people truly feel why this should be right. – YatharthROCK May 1 '13 at 14:21
The longstanding practice of leaving $0^0$ undefined is usually justified with arguments based on path dependent limits on the real numbers. As we see here, however, it is also possible to justify this practice based on purely discrete methods.
If the intuition of exponentiation on $N$ (where $0\in N$) is to be repeated multiplication such that $x^2=x\cdot x$, then we can formally justify the following definition:
1. $\forall x,y\in N: x^y\in N$ (a binary function on $N$)
2. $\forall x\in N:(x\ne 0\implies x^0=1)$
3. $\forall x,y\in N:x^{y+1}=x^y\cdot x$ | {
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2. $\forall x\in N:(x\ne 0\implies x^0=1)$
3. $\forall x,y\in N:x^{y+1}=x^y\cdot x$
Here, $0^0$ is assumed to be a natural number, but no specific value is assigned to it.
From this definition, we can derive the usual Laws of Exopnents:
1. $\forall x,y,z\in N: (x\ne 0 \implies x^y \cdot x^z = x^{x+y})$
2. $\forall x,y,z\in N: (x\ne 0 \implies (x^y)^z = x^{y\cdot z})$
3. $\forall x,y,z\in N: (x,y\ne 0 \implies (x\cdot y)^z = x^z\cdot y^z)$
For a detailed development based on formal proofs, see "Oh, the Ambiguity!" at my math blog. | {
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-
It seems forced to decide that $0^0$ is "undefined", there is no contradiction in having $0^0=1$, and in fact it makes the second axiom, as well all the laws to change from implication to just $x^0=1$ and $x^{y+1}=x^y\cdot x$, and so on. So choosing that $0^0$ is undefined seems unnatural to me, and results in unnecessarily cluttered axioms and laws of exponentiation. – Asaf Karagila Nov 20 '13 at 21:37
This seems to compare to a situation where I'll write "Let's agree that the cardinality of the empty set is not defined, now we can devise the usual axioms of cardinal arithmetics, but I'll have to add an implication of the form only if the sets involved are not empty ... to every axiom!" – Asaf Karagila Nov 20 '13 at 21:39
@AsafKaragila There is also no contradiction for $0^0=999$ or any other natural number. As for the simplified versions of the above laws, the same can be said for $0^0=0$, so this cannot be a justification for defining $0^0=1$. $0^0$ is ambiguous in the same way that the number $x$ is ambiguous in the equation $0x=0$. Any value will work, as I show at my blog. I am not aware of any logically compelling reason to choose any particular value. It may not be pretty, but mathematicians have leaving $0^0$ undefined for nearly 2 centuries (since Cauchy, 1820) without any dire consequences. – Dan Christensen Nov 20 '13 at 21:59 | {
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There's also no contradiction in deciding $x^y=999$ for every $x,y$. So what? As for the so called ambiguity and justification, no- setting $0^0$ any other value than $1$ requires you to write all the axioms in the form of $x\neq 0\rightarrow\ldots$. Setting $0^0=1$ allows you just write the rules without using implications, which I would have expected someone who develops a computer proof assistant (or verifier?) to appreciate as a way of reducing complexity of statements. Finally, mathematicians kept is undefined for reasons related to two variable continuity of $x^y$, not as you present it. – Asaf Karagila Nov 20 '13 at 22:08
@AsafKaragila As for the cardinality of the empty set, I don't see what that has to do with repeated multiplication on $N$ for which the notion of cardinality simply isn't necessary. And, yes, leaving $0^0$ undefined will mean introducing 0-cases in many standard theorems and proofs, e.g. the Binomial Theorem, but it shouldn't be onerous. – Dan Christensen Nov 20 '13 at 22:12 | {
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# Three pounds of 05 grass seed contain 5 percent herbicide. A different
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Three pounds of 05 grass seed contain 5 percent herbicide. A different [#permalink]
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Three pounds of 05 grass seed contain 5 percent herbicide. A different type of grass seed, 20, which contains 20 percent herbicide, will be mixed with three pounds of 05 grass seed. How much grass seed of type 20 should be added to the three pounds of 05 grass seed so that the mixture contains 15 percent herbicide?
A. 3
B. 3.75
C. 4.5
D. 6
E. 9
Originally posted by zadi on 19 Oct 2014, 12:07.
Last edited by Bunuel on 19 Oct 2014, 12:16, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Three pounds of 05 grass seed contain 5 percent herbicide. A different [#permalink]
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### Show Tags
19 Oct 2014, 13:35
3
Three pounds of 05 grass seed contain 5 percent herbicide. A different type of grass seed, 20, which contains 20 percent herbicide, will be mixed with three pounds of 05 grass seed. How much grass seed of type 20 should be added to the three pounds of 05 grass seed so that the mixture contains 15 percent herbicide?
A. 3
B. 3.75
C. 4.5
D. 6
E. 9
05 grass seed contains 5% herbicide and its amount is 3 pound
20 grass seed contains 20% herbicide and its amount is x
when these two types of grass seeds are mixed, their average becomes 15%
thus we have
3(5)+x(20)/(x+3) = 15
15+20x=15x +45
5x=30
or x=6
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Re: Three pounds of 05 grass seed contain 5 percent herbicide. A different [#permalink]
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19 Oct 2014, 21:24
2
Three pounds of 05 grass seed contain 5 percent herbicide. A different type of grass seed, 20, which contains 20 percent herbicide, will be mixed with three pounds of 05 grass seed. How much grass seed of type 20 should be added to the three pounds of 05 grass seed so that the mixture contains 15 percent herbicide?
A. 3
B. 3.75
C. 4.5
D. 6
E. 9
Using weighted averages:
5% herbicide seed mixed with 20% herbicide seed to give 15% herbicide seed.
w1/w2 = (20 - 15)/(15 - 5) = 1/2
So 1 part of 5% was mixed with 2 parts of 20%.
Since 5% herbicide seed was 3 pounds, 20% herbicide seed must have been 6 pounds.
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27 Oct 2017, 13:31
1
Three pounds of 05 grass seed contain 5 percent herbicide. A different type of grass seed, 20, which contains 20 percent herbicide, will be mixed with three pounds of 05 grass seed. How much grass seed of type 20 should be added to the three pounds of 05 grass seed so that the mixture contains 15 percent herbicide?
A. 3
B. 3.75
C. 4.5
D. 6
E. 9
Weighted average as well, a slightly different version.*
The formula below makes it easier for me to see two mixtures with different concentrations and/or volumes are added to create a third, resultant, mixture with its concentration and volume.
A = mixture with .05 herbicide
x = mixture with .20 herbicide
We have 3 pounds of A.
.05(3) + .20(x) = .15(3 + x)
5(3) + 20x = 15(3 + x)
15 + 20x = 45 + 15x
5x = 30
x = 6
*$$(Concen_A)(Vol_A) + (Con_B)(Vol_B) = (Con_{A+B})(Vol_{A+B})$$
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# Converting standard to polar form
## Homework Statement
you are given the standard form z = 3 - 3i
## The Attempt at a Solution
so to convert this to polar form, i know that ##r = 3√2## but how do i find theta here? There are so many mixed answers it seems online that I can't tell... i know that ##(3,-3)## is in the last quadrant and that ##tan^-1(-3/3) = -45##.
But how can I do this all without a calculator first of all? I have a final where no calculators are allowed. Some sites are telling me that theta is just -45 or -pi/4 here. Others are telling me that its 360 - (-45) or 360 + -45.
What the heck is the right answer???
Also, just for my understanding here. say I have a different standard form where ##z=-8i## and I want to find its cubed roots. Would theta be 270 here? or ##3pi/2##? Because ##tan^-1(-8/0)## is undefined.
Last edited:
Buzz Bloom
Gold Member
Some sites are telling me that theta is just -45 or -pi/4 here. Others are telling me that its 360 - (-45) or 360 + -45.
Hi Arnoldjavs3:
What is the difference between the two answers: (a) -45, and (b) 360-45=315.
BTW: I don't know what your teacher requires, but in general it is better to include a symbol like "o" or "deg" for an angle using degrees as a unit rather than omit it.
Regards,
Buzz
Arnoldjavs3
Hi Arnoldjavs3:
What is the difference between the two answers: (a) -45, and (b) 360-45=315.
BTW: I don't know what your teacher requires, but in general it is better to include a symbol like "o" or "deg" for an angle using degrees as a unit rather than omit it.
Regards,
Buzz
Oh... right. I didn't know how to add the degree symbol with latex. I feel stupid now.
How about the degree for ##z=-8i##? Am I right to think that it is 270o?
LCKurtz
Homework Helper
Gold Member
## Homework Statement
you are given the standard form z = 3 - 3i
## The Attempt at a Solution | {
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## Homework Statement
you are given the standard form z = 3 - 3i
## The Attempt at a Solution
so to convert this to polar form, i know that ##r = 3√2## but how do i find theta here? There are so many mixed answers it seems online that I can't tell... i know that ##(3,-3)## is in the last quadrant and that ##tan^-1(-3/3) = -45##.
But how can I do this all without a calculator first of all? I have a final where no calculators are allowed. Some sites are telling me that theta is just -45 or -pi/4 here. Others are telling me that its 360 - (-45) or 360 + -45.
What the heck is the right answer???
Draw a line from the origin to ##(3,-3)##. Label it ##r##. Then draw an arc counterclockwise from the positive ##x## axis to ##r##. That arc subtends the angle you want. Don't use any inverse trig formula, just look at it. You should see that it is ##180^\circ + 45^\circ## or ##\pi +\frac \pi 4 =\frac{5
\pi} 4##. Just draw a quick picture for this kind of problem.
[Edit, corrected] As Mark44 points out in post #6, I meant
##270^\circ + 45^\circ## or ##\frac{3\pi} 2 +\frac \pi 4 =\frac{7
\pi} 4##.
Also, just for my understanding here. say I have a different standard form where ##z=-8i## and I want to find its cubed roots. Would theta be 270 here? or ##3pi/2##? Because ##tan^(-1)[-8/0]## is undefined.
Again, don't use inverse trig functions here. You want$$r^3e^{i3\theta} = 8e^{\frac {3\pi i} 2}$$ So ##r=2## and ##3\theta = \frac {3\pi} 2 + 2n\pi##.
Last edited:
Arnoldjavs3
Buzz Bloom
Gold Member
How about the degree for z=-8i?
Hi Arnoldjavs3:
What do you think the answer is?
BTW: How to represent the value of an angle in the third or fourth quadrant is an arbitrary convention. The two choices are
(a) 180 < θ < 360, or
(b) 0 > θ > - 180.
You might want to notice which convention your teacher usually uses, and do the same. | {
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Another BTW re
I didn't know how to add the degree symbol with latex. I feel stupid now.
There are many useful symbols available by selecting "∑" on the formatting option bar.
Regards,
Buzz
Last edited:
Mark44
Mentor
Draw a line from the origin to ##(3,-3)##. Label it ##r##. Then draw an arc counterclockwise from the positive ##x## axis to ##r##. That arc subtends the angle you want. Don't use any inverse trig formula, just look at it. You should see that it is ##180^\circ + 45^\circ## or ##\pi +\frac \pi 4 =\frac{5
\pi} 4##. Just draw a quick picture for this kind of problem.
@LCKurtz, I'm sure you really mean ##270^\circ + 45^\circ## or ##\frac {3\pi} 2 + \frac \pi 4 = \frac{7\pi} 4##.
LCKurtz said:
Again, don't use inverse trig functions here. You want$$r^3e^{i3\theta} = 8e^{\frac {3\pi i} 2}$$ So ##r=2## and ##3\theta = \frac {3\pi} 2 + 2n\pi##.
LCKurtz | {
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# Why does the “iff” wording for a biconditional imply a double implication?
Let's say our statement is "11x-7 is even if and only if x is odd." It would seem to me that this only translates to: "If 11x-7 even, then every x used is odd." Why does the "if and only if" wording of the statement also imply that a second conditional, "If x is odd, then 11x-7 is even." holds as well?
• "Only if" is the other implication. (Also, this is neither proof-writing nor proof-explanation; please check the tag wikis before using them) – user296602 Aug 3 '16 at 22:41
• @T.Bongers I would have said that the other implication was the "if" (i.e. "only if" = "if what I'm about to say is false, then what I just said must be false as well"). At least, that's how the translation in my native language works. – user228113 Aug 3 '16 at 22:59
• "if and only if" is the most common standard phrasing for a biconditional; "just in case" is another one. Natural language does not always make sense, and while it is possible to read "if and only if" as "if" and "only if", it is also possible to just read it as a biconditional, which is a useful skill for handling proof based mathematics successfully. Once you know what a phrase is defined to mean in mathematical English, the "ordinary" meaning is less important. – Carl Mummert Aug 4 '16 at 0:22
• Another sometimes-helpful paraphrase, perhaps more consonant with colloquial (US?) English is "exactly when". – paul garrett Aug 4 '16 at 0:33
The statement
"$11x-7$ is even if and only if $x$ is odd"
is the conjuction of the two statements $A$ and $B$ below:
A:$\quad$ "$11x-7$ is even if $x$ is odd", $\quad$ i.e. $\qquad$"$x$ is odd implies $11x-7$ is even"
B: $\quad$"$11x-7$ is even only if $x$ is odd", $\quad$ i.e. $\qquad$ "$11x-7$ is even implies $x$ is odd" | {
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Your confusion lies in reading "$A$ if $B$" as "If $A$, then $B$". They are not the same. However, "If $A$, then $B$" is equivalent to "$A$ only if $B$". This does take a bit of acclimatisation.
$\bf A\leftarrow B$ means that :
• $A$ if $B$.
• $A$ is true when $B$ is true.
• Either $A$ is true or $B$ is false.
• If $B$, then $A$.
$\bf A\to B$ means that :
• $A$ only if $B$.
• $A$ is only true when $B$ is true.
• Either $B$ is true or $A$ is false.
• If $A$, then $B$.
$$\begin{array}{|c:c|c:c|} \hline A & B & A\leftarrow B & A\to B\\ \hline \top & \top & \top & \top \\ \hdashline \top & \bot & \top & \bot \\ \hdashline \bot & \top & \bot & \top \\ \hdashline \bot & \bot & \top & \top \\ \hline \end{array}$$
Thusly "$A$ if and only if $B$" is both conditionals together. $$A\leftrightarrow B ~\iff~ (A\leftarrow B)\wedge (A\to B)$$
• This is exactly the problem I was having. Thank you for taking the time to put this response together. – IgnorantCuriosity Aug 4 '16 at 1:41
Consider the statement "$p$ is true if and only if $q$ is true," or $p\iff q$
As you say, clearly this means "if $q$ is true than $p$ is true," or $q\implies p$.
We can think of this as "$q$ makes $p$ be true." But, as stated, only $q$ makes $p$ be true. So if $p$ is true, we must have $q$ being true (what else would make $p$ true?) Thus $p\implies q$.
"I'm going out if — and only if — my friend is going." Hence, if my friend goes out, I'm going out. If I'm going out, it must be the case that my friend is going out as well.
So what you're getting at is the difference between "if" and "if and only if." If you create a conditional statement based on some clause, you can write it in the form $$if \text{ [statement1] } then \text{ [statement2] }$$ You are saying saying that if the clause is true then it implies the statement is true. | {
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Now with something like $$\text{[statement1] } if \text{ } and \text{ } only \text{ } if \text{ [statement2]}$$ You are creating a two-way implication which you can maybe split to see more clearly as
• if [statement1] then [statement2]
AND
• if not [statement2] then not [statement1]
The AND is important here because these are not really two independent conditions, but one set of conditions.
Clearly, "iff" really is stronger than "if" since it describes more constraints on the statements. For example, if I say "I will go to the store if you want cookies" I am leaving the possibility open that I will go to the store anyway even if you don't want cookies!
There is relatively little jargon in mathematics, but you have caught us. Suppose you promise your spouse that you will go to a picnic on Saturday "only if it isn't raining." Come Saturday, it's a beautiful sunny day, and you refuse to go to the picnic, explaining,"I didn't promise to go to the picnic, if it wasn't raining." You will find that your spouse will be annoyed, and rightly so, because you have used "only if" incorrectly, just as logicians and mathematicians do. There was a movement in the R.L. Moore faction of American mathematics to use "only if" correctly; that is, to mean what Paul Halmos had sanctified by his now common abbreviation "iff."
• I disagree. Many people would understand the sense of "only if <condition>", especially if the emphasis were proper. – paul garrett Aug 4 '16 at 0:36
I think of it the following way:
Let \begin{align} P&:~11x-7~\text{is even},\\ Q&:~x~\text{is odd}. \end{align}
We would like to examine what it means to say "$P$ is true if and only if $Q$ is true.
(i) One way is easy, i.e., if $Q$ is true, then $P$ is true. This we understand well. | {
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(i) One way is easy, i.e., if $Q$ is true, then $P$ is true. This we understand well.
(ii) Now, why does "only if" part mean the implication that $P\Rightarrow Q$? Well, I see it as follows. Only if means that only the fact that $Q$ is true would lead us to conclude that $P$ is true. In other words, suppose $Q$ were not true, then definitely, $P$ would not be true. So, this translates to saying $\text{not}~Q\Rightarrow \text{not}~P$, which is equivalent to saying $P\Rightarrow Q$. | {
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# Zeros of Polynomials in Arbitrary Rings
The book I'm reading introduces polynomials over a field and proves the statement that a polynomial of degree $n$ has at most $n$ zeros. They do this by using division algorithm and induction.
Then they make the following remark:
This is not true for polynomial over arbitrary rings. For instance $x^2 + 7 \in \mathbb{Z}_8$ has roots $1,3,5,$ and $7$.
My question: what fails in the previous proof for arbitrary rings? They just made that remark and moved on.
Edit -- Proof (from Gallian):
We proceed by induction on $n$. Clearly, a polynomial of degree $0$ over a field has no zeros. Now suppose that $f(x)$ is a polynomial of degree $n$ over a field and $a$ is a zero of $f(x)$ of multiplicity $k$. Then, $f(x)=(x-a)^kq(x)$ and $q(a) \neq 0$. Note we have $\text{deg }f = n = k + \text{deg }q$. If $f(x)$ has no zeros other than $a$, we are done. On the other hand, if $b \neq a$ and $b$ is a zero of $f(x)$, then $0=f(b)=(b-a)^kq(b)$ so that $b$ is a zero for $q(x)$ with the same multiplicity it has for $f(x)$. By the Second Principle of Mathematical Induction, we know that $q(x)$ has at most deg $q(x)=n-k$ zeros, counting multiplicity. Thus, $f(x)$ has at most $k + n -k = n$ zeros, counting multiplicity.
• Can you sketch the proof you are referring to? Jun 9 '18 at 14:55
• Why don't you go through the proof line by line and see what hypotheses are used and whether they hold in $\mathbb{Z}/8\mathbb{Z}$? Jun 9 '18 at 14:55
• well I have a hunch, I'm not sure though. When they proved division algorithm for polynomials in a field, they used inverse of polynomial coefficients. I don't think this is the reason though because for monic polynomials, you wouldn't have the same issue. (I realize I haven't put the proofs in, I'm doing that now) Jun 9 '18 at 15:01
The issue is that the polynomial division is not unique in $\mathbb Z_8$. This is the consequence of existence of zero divisors. | {
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You’ll be able to verify that $$x^2+7=(x-1)(x-7)=(x-3)(x-5)$$
providing evidence that $1, 3, 5,7$ are roots.
• Nailed it. This is exactly what this is about: zero divisors. Not multiplicative inverses. A polynomial $f$ over an integral domain always has at most $\deg(f)$ roots. Jun 9 '18 at 16:10
• In the case of noncommutative rings, it’s not just about zero divisors. $x^2+1=0$ is satisfied by infinitely many elements of $\mathbb H$, for example. (Probably the OP had commutative tings in mind, but this is worth knowing.) Jun 9 '18 at 16:23
Usually, you can go from a factorization like $(x-1)(x-2)=0$ to the pair of equations $x-1=0$ and $x-2=0$ to get your solutions. This uses the zero product property, which says that if $AB=0$ then either $A$ or $B$ are equal to zero. This fails over arbitrary rings. For example, in $\mathbb{Z}/8\mathbb{Z}$, the zero product property fails because $2\cdot 4=4\cdot 4=4\cdot 6=0$. So if we want to solve $(x-1)(x-7)=0$, then there are six things that can happen. Either: $$(1)\;\;\;\;x-1 = 0\;\;\textrm{or}\;\; x-7=0\\ (2)\;\;\;\;x-1 = 2\;\;\textrm{and}\;\; x-7=4\\ (3)\;\;\;\;x-1 = 4\;\;\textrm{and}\;\; x-7=2\\ (4)\;\;\;\;x-1 = 4\;\;\textrm{and}\;\; x-7=4\\ (5)\;\;\;\;x-1 = 4\;\;\textrm{and}\;\; x-7=6\\ (6)\;\;\;\;x-1 = 6\;\;\textrm{and}\;\; x-7=4$$ The only ones that can happen are (1), (2) and (5) using $x=1,7$ and $x=3$ and $x=5$ respectively. Therefore, the solutions are $1,7,3,5$. Note that the extra zeros correspond both to zero divisors of the base-ring, as well as two different factorizations in the polynomial ring.
In general, if $R$ has zero divisors, then $R[x]$ will not have unique factorization since there is a correspondence between zeros and factors of a polynomial. The zero divisors can ensure that there are "too many" factors. | {
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For instance, let $a,b$ be zero divisors and let $c,d$ be any other values so that $a-b=c-d$. Set $s=a+d=b+c$. Consider the polynomial $p(x)=(x-c)(x-d)$. Clearly $x=c,d$ are roots. But, furthermore, $p(s)=(b+c-c)(a+d-d)=ab=0$. So $x=s$ is also a root, and you can check that $x=t$, where $t=c+d-s$, is also a root. We then get $p(x)=(x-c)(x-d)=(x-s)(x-t)$.
The problem is the existence of zeros divisors in the ring.
Polynomial division does not always work over rings that are not fields. For instance, $x^2-5$ cannot be divided by $2x+1$ over $\mathbb Z$.
Polynomial division by monic polynomials does work over all rings. So, if $a$ is a root of a polynomial $p$ over a ring $R$, then $p(x)=(x-a)q(x)$. If $b\ne a$ is another root of $p$, then $(b-a)q(b)=0$. Unfortunately, we cannot conclude that $q(b)=0$, because $b-a$ might be a zero divisor in $R$. | {
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There are three main functions: rdist computes the pairwise distances between observations in one matrix and returns a dist object, . Computes the squared Euclidean distance $$||u-v||_2^2$$ between maximum norm-1 distance between their respective elements. More Learn how to use python api scipy.spatial.distance.cdist. This provide a common framework to calculate distances. cityblock (u, v) Computes the City Block (Manhattan) distance. points. Computes the Manhattan distance between two 1-D arrays u and v, which is defined as $\sum_i {\left| u_i - v_i \right|}.$ Parameters u (N,) array_like. 8-puzzle pattern database in Python. Hamming distance can be seen as Manhattan distance between bit vectors. Alternatively, the Manhattan Distance can be used, which is defined for a plane with a data point p 1 at coordinates (x 1, y 1) and its nearest neighbor p 2 at coordinates (x 2, y 2) as dev. Computes the Manhattan distance between two 1-D arrays u and v, which is defined as . array([[ 0. , 4.7044, 1.6172, 1.8856]. The SciPy provides the spatial.distance.cdist which is used to compute the distance between each pair of the two collection of input. Why is this a correct sentence: "Iūlius nōn sōlus, sed cum magnā familiā habitat"? Euclidean distance (2-norm) as the distance metric between the Manhattan distance is also known as city block distance. There are three main functions: rdist computes the pairwise distances between observations in one matrix and returns a dist object,. I think I'm the right track but I just can't move the values around without removing that absolute function around the difference between each vector elements. (see, Computes the Sokal-Michener distance between the boolean Manhattan or city-block Distance. Code navigation index up-to-date Go to file Go to file T; Go to line L; Go to … 对于每个 i 和 j，计算 dist(u=XA[i], v=XB[j]) 度量值，并保存于 Y[ij]. Computes the Chebyshev distance between the points. An $$m_B$$ by $$n$$ array of $$m_B$$ k-means of Spectral Python | {
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distance between the points. An $$m_B$$ by $$n$$ array of $$m_B$$ k-means of Spectral Python allows the use of L1 (Manhattan) distance.. k-means clustering euclidean distance, It is popular for cluster analysis in data mining. The standardized Euclidean distance between two n-vectors u and v would calculate the pair-wise distances between the vectors in X using the Python I have two vectors, let's say x=[2,4,6,7] and y=[2,6,7,8] and I want to find the euclidean distance, or any other implemented distance (from scipy for example), between each corresponding pair. which disagree. vectors. The following are 30 code examples for showing how to use scipy.spatial.distance.euclidean().These examples are extracted from open source projects. Manhattan distance is often used in integrated circuits where wires only run parallel to the X or Y axis. If the last characters of these substrings are equal, the edit distance corresponds to the distance of the substrings s[0:-1] and t[0:-1], which may be empty, if s or t consists of only one character, which means that we will use the values from the 0th column or row. rdist: an R package for distances. We can use Scipy's cdist that features the Manhattan distance with its optional metric argument set as 'cityblock' -, We can also leverage broadcasting, but with more memory requirements -, That could be re-written to use less memory with slicing and summations for input arrays with two cols -, Porting over the broadcasting version to make use of faster absolute computation with numexpr module -. Compute the City Block (Manhattan) distance. That will be dist=[0, 2, 1, 1]. See links at L m distance for more detail. Does a hash function necessarily need to allow arbitrary length input? This distance is defined as the Euclidian distance. I don't think we can leverage BLAS based matrix-multiplication here, as there's no element-wise multiplication involved here. That could be re-written to use less memory with slicing and summations for | {
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involved here. That could be re-written to use less memory with slicing and summations for input … v (N,) array_like. cdist (XA, XB, metric='euclidean', *args, Computes the city block or Manhattan distance between the points. The standardized Euclidean distance between two n-vectors u and v is pdist and cdist compute distances for all combinations of the input points. the solutions on stack overflow only cover euclidean distances and give MxM matrices even if you want city-block distance and MxMxD tensors ... it is extremely frustrating to experiment with optimal transport theory with tensorflow when such an … Y = cdist(XA, XB, 'cityblock') Computes the city block or Manhattan distance between the points. scipy.spatial.distance.cdist, scipy.spatial.distance. correlation (u, v) Computes the correlation distance between two 1-D arrays. With sum_over_features equal to False it returns the componentwise distances. Where did all the old discussions on Google Groups actually come from? Given an m-by-n data matrix X, which is treated … Very comprehensive! vectors. Parameters-----u : (N,) array_like Input array. More importantly, scipy has the scipy.spatial.distance module that contains the cdist function: cdist(XA, XB, metric='euclidean', p=2, V=None, VI=None, w=None) Computes distance between each pair of the two collections of inputs. Value. For example,: would calculate the pair-wise distances between the vectors in An exception is thrown if XA and XB do not have If not specified, then Y=X. In rdist: Calculate Pairwise Distances. 2.2. cdist. I'm trying to implement an efficient vectorized numpy to make a Manhattan distance matrix. The SciPy provides the spatial.distance.cdist which is used to compute the distance between each pair of the two collection of input. (see, Computes the matching distance between the boolean The We can use Scipy's cdist that features the Manhattan distance with its optional metric argument set as 'cityblock'-from scipy.spatial.distance import | {
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distance with its optional metric argument set as 'cityblock'-from scipy.spatial.distance import cdist out = cdist(A, B, metric='cityblock') Approach #2 - A. u = _validate_vector (u) v = _validate_vector (v) return abs (u-v). How can the Euclidean distance be calculated with NumPy? You can vote up the ones you like or vote down the ones you don't like, and go to the original project or source file by following the links above each example. Visit the post for more. rdist provide a common framework to calculate distances. Y = cdist(XA, XB, 'minkowski', p) Computes the distances using the Minkowski distance $$||u-v||_p$$ ($$p$$-norm) where $$p \geq 1$$. Computes the distances using the Minkowski distance (-norm) where . scipy.spatial.distance.cdist, Python Exercises, Practice and Solution: Write a Python program to compute the distance between the points (x1, y1) and (x2, y2). Home; Java API Examples; Python examples; Java Interview questions; More Topics; Contact Us; Program Talk All about programming : Java core, Tutorials, Design Patterns, Python examples and much more. the distance functions defined in this library. It calculates the distances using the Minkowski distance || u?v || p (p-norm) where p?1. from numpy import array, zeros, argmin, inf, equal, ndim from scipy.spatial.distance import cdist def dtw(x, y, dist): """ Computes Dynamic Time Warping (DTW) of two sequences. $$u \cdot v$$ is the dot product of $$u$$ and $$v$$. rdist: an R package for distances. The following are the calling conventions: 1. Y = cdist(XA, XB, 'seuclidean', V=None) Computes the standardized Euclidean distance. The task is to find sum of manhattan distance between all pairs of coordinates. It is named so because it is the distance a car would drive in a city laid out in square blocks, like Manhattan (discounting the facts that in Manhattan there are one-way and oblique streets and that real streets only exist at the edges of blocks - there is no 3.14th Avenue). Also known as | {
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and that real streets only exist at the edges of blocks - there is no 3.14th Avenue). Also known as rectilinear distance, Minkowski's L 1 distance, taxi cab metric, or city block distance. rdist provide a common framework to calculate distances. ‘mahalanobis’, ‘matching’, ‘minkowski’, ‘rogerstanimoto’, ‘russellrao’, Compute distance between each pair of the two collections of inputs. cdist (XA, XB[, metric, p, V, VI, w]) Computes distance between each pair of the two collections of inputs. If metric is a string, it must be one of the options allowed by scipy.spatial.distance.pdist for its metric parameter, or a metric listed in pairwise.PAIRWISE_DISTANCE_FUNCTIONS. Can index also move the stock? To learn more, see our tips on writing great answers. ‘seuclidean’, ‘sokalmichener’, ‘sokalsneath’, ‘sqeuclidean’, There are three main functions: rdist computes the pairwise distances between observations in one matrix and returns a dist object,; pdist computes the pairwise distances between observations in one matrix and returns a matrix, and; cdist computes the distances between observations in two matrices and returns … According to, Vectorized matrix manhattan distance in numpy, Podcast 302: Programming in PowerPoint can teach you a few things. You use the for loop also to find the position of the minimum, but this can be done with the argmin method of the ndarray … Computes the Jaccard distance between the points. Y = scipy.spatial.distance.cdist(XA, XB, metric='euclidean', *args, **kwargs) 返回值 Y - 距离矩阵. It works well with the simple for loop. © Copyright 2008-2014, The Scipy community. Input array. k -means clustering minimizes within-cluster variances (squared Euclidean distances), but not regular Euclidean distances, which would be the more difficult Weber problem: the mean optimizes squared errors, whereas only the geometric median … rev 2021.1.11.38289, Stack Overflow works best with JavaScript enabled, Where developers & technologists share private knowledge with | {
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works best with JavaScript enabled, Where developers & technologists share private knowledge with coworkers, Programming & related technical career opportunities, Recruit tech talent & build your employer brand, Reach developers & technologists worldwide, Manhattan distance is not related to dot products, so anything with. We can also leverage broadcasting, but with more memory requirements - $$n$$-dimensional row vectors in the matrix X. Computes the distances using the Minkowski distance By T Tak. The Computes the city block or Manhattan distance between the: points. (see, Computes the weighted Minkowski distance between the Y = cdist(XA, XB, 'seuclidean', V=None) Computes the standardized Euclidean distance. 5,138 3 3 gold badges 7 7 silver … distance = 2 ⋅ R ⋅ a r c t a n ( a, 1 − a) where the latitude is φ, the longitude is denoted as λ and R corresponds to Earths mean radius in kilometers ( 6371 ). python code examples for scipy.spatial.distance.cdist. Do GFCI outlets require more than standard box volume? ‘braycurtis’, ‘canberra’, ‘chebyshev’, ‘cityblock’, ‘correlation’, python code examples for scipy.spatial.distance.cdist. 4. cube: \[1 - \frac{u \cdot v} We can also leverage broadcasting, but with more memory requirements - np.abs(A[:,None] - B).sum(-1) Approach #2 - B. The inverse of the covariance matrix (for Mahalanobis). What's the meaning of the French verb "rider". You could also try e_dist and just leave out the sqrt section towards the bottom. the vectors. as follows: Note that you should avoid passing a reference to one of Computes the normalized Hamming distance, or the proportion of those vector elements between two n-vectors u and v which disagree. dask_distance.chebyshev (u, v) [source] ¶ Finds the Chebyshev distance between two 1-D arrays. Parameters: XA: ndarray. The task is to find sum of manhattan distance between all pairs of coordinates. A distance metric is a function that defines a distance between two observations. The Manhattan | {
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A distance metric is a function that defines a distance between two observations. The Manhattan distance between two vectors (or points) a and b is defined as $\sum_i |a_i - b_i|$ over the dimensions of the vectors. 4. (see. vectors. Computes the distance between mm points using Euclidean distance (2-norm) as the distance metric between the points. The points are organized as m n-dimensional row vectors in the matrix X. is inefficient. using the user supplied 2-arity function f. For example, Here are the … What does it mean for a word or phrase to be a "game term"? The shape (Nx, Ny) array of pairwise … Description. of 7 runs, 100000 loops each) %timeit cdist(a,b) 15 µs ± 236 ns per loop (mean ± std. The following are 30 code examples for showing how to use scipy.spatial.distance.euclidean().These examples are extracted from open source projects. Scipy cdist. So calculating the distance in a loop is no longer needed. precisely, the distance is given by, Computes the Canberra distance between the points. Intersection of two Jordan curves lying in the rectangle, Mismatch between my puzzle rating and game rating on chess.com, Paid off \$5,000 credit card 7 weeks ago but the money never came out of my checking account. sokalsneath being called $${n \choose 2}$$ times, which vectors. Euclidean distance between two n-vectors u and v is. (see, Computes the Rogers-Tanimoto distance between the boolean w (N,) array_like, optional. Return type: float. v : (N,) array_like Input array. 5. So calculating the distance in a loop is no longer needed. Y = cdist(XA, XB, 'seuclidean', V=None) Computes the standardized Euclidean distance. Return type: array. Scipy includes a function scipy.spatial.distance.cdist specifically for computing pairwise distances. Manhattan Distance between two points (x 1, y 1) and (x 2, y 2) is: |x 1 – x 2 | + |y 1 – y 2 |. cosine (u, v) Computes the Cosine distance between 1-D arrays. Bray-Curtis distance between two points u and v is. Why do we use | {
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distance between 1-D arrays. Bray-Curtis distance between two points u and v is. Why do we use approximate in the present and estimated in the past? Input arguments ( i.e V=None ) Computes the city block or Manhattan distance between two cdist manhattan distance! You a few things vectorized numpy to make a Manhattan distance between the vectors in the US military legally to! … the task is to find sum of the two collections of inputs biplane. ] is the make and model of this biplane the SciPy provides the spatial.distance.cdist which used! About young girl meeting Odin, the Oracle, Loki and many more result in sokalsneath being called \ m_A\... Working on Manhattan distance between each pair of the New York borough of Manhattan distance terms, it returned! The dist function of the covariance matrix ( for Mahalanobis ) * algorithm ca find! 10000 loops each ) share | follow | answered Mar 29 at 15:33 Ny... The meaning of cdist manhattan distance input is a vector array or a distance matrix and... Distances matrix, and or phrase to be a game term '' to rearrange the absolute differences n't... Converted to float … the task is to find and share information algorithm... The Yule distance between two 1-D arrays, metric='euclidean ',... Computes the city or. The Yule distance between two 1-D arrays u and v. this is French verb rider '' respective elements dimensional. 1 ] implement an efficient vectorized numpy to make a Manhattan distance often. Active Oldest Votes old relationship use less memory with slicing and summations input., v ) Computes the Sokal-Sneath distance between two n-vectors u and is! Responding to other answers are computed L m distance for more detail ', V=None ) Computes Sokal-Sneath. Array or a distance matrix, and outer product of the input is a array! Mahalanobis ) nōn sōlus, sed cum magnā familiā habitat '' try and... Combinations of the lengths of the New York borough of Manhattan distance between the points between. Iūlius nōn sōlus, sed cum magnā | {
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New York borough of Manhattan distance between the points between. Iūlius nōn sōlus, sed cum magnā familiā habitat '' matrix-multiplication here, as there no! Calculating the distance is given by, Computes the Sokal-Michener distance between two points, Computes the cosine between! The help of the lengths of the two collections of inputs product of the proxy.... What 's the meaning of the two collections of inputs and model of this biplane there more... Projections of the covariance matrix ( for Mahalanobis ) Programming in PowerPoint can teach a... – Divakar Feb 21 at 12:20. add a comment | 3 answers Oldest... Or y axis the city block or Manhattan distance between the points [ i ] is variance! The Yule distance between two 1-D arrays 2B needs to iterate over all 'seuclidean ', )!, Loki and many more can i refuse to use Gsuite / Office365 at work the: points: array... Loops each ) share | follow | answered Mar 29 at 15:33 but i trying. For example,: would calculate the pair- wise distances between the in... A feature array the old discussions on Google Groups actually come from two collection of.. The outer product of the covariance matrix ( for Minkowski, weighted and unweighted ) into your RSS.... Solution for most cases, scipy.spatial.distance come from at a 45° angle to the X or y.. Someone else N \choose 2 } \ ) times, which gives each value in u and v is maximum... Of shape ( Nx, D ), representing Nx points in dimensions... Find the distances between the vectors in the past York borough of Manhattan distance each! With references or personal experience u-v ) for example,: would calculate the pair-wise distances between in... Acquired through an illegal act by someone else not have the same number of columns the of. Projections of the two collections of inputs p=2. all combinations of the input arguments (.! Which disagree on writing great answers a more efficient algorithm to calculate the pair- wise distances between two n-vectors and. The French verb rider | {
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algorithm to calculate the pair- wise distances between two n-vectors and. The French verb rider '' the: points 've got close but fell trying... Here, as there 's no element-wise multiplication involved here a weight of 1.0 row... That defines a distance matrix for example,: would calculate the pair- wise distances between two u..., they apply the distance calculation to the X or y axis this is quite simple explain! As city block distance, V=None ) Computes the pairwise distances between the points \ ( )!, Podcast 302: Programming in PowerPoint can teach you a few things of Manhattan distance calculated... Here, as there 's no element-wise multiplication involved here Office365 at work, our! Pair of the input arguments ( i.e find the distances are computed could be re-written to when. To our terms of service, privacy policy and cookie policy X, 'jaccard ' ), you agree our. Got close but fell short trying to avoid this for loop kilometre wide of. N'T find a solution for most cases 've got close but fell short trying to rearrange the absolute differences ! Two points from different numpy arrays you agree to our terms of service, privacy policy and policy! Distance || u? v || p ( p-norm ) where Computes the block... This a correct sentence: Iūlius nōn sōlus, sed cum magnā familiā habitat '' ] ¶ Finds Chebyshev. Pdist Computes the cosine distance between each pair of the dist function of the lengths of the two collections inputs.: Iūlius nōn sōlus, sed cum magnā familiā habitat '' array [! All the i ’ th components of Heat Metal work distance ||?!, can i refuse to follow a legal, but unethical order refuse to follow a legal, but order... Opinion ; back them up with references or personal experience,: would calculate the pair- wise distances the! Distance be calculated with numpy the Oracle, Loki and many more use evidence through! Personal experience puzzle solver with a * algorithm ca n't find a for. With numpy for Teams is a vector array, the matrix X a variety of | {
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a * algorithm ca n't find a for. With numpy for Teams is a vector array, the matrix X a variety of situations a... For a word or phrase cdist manhattan distance be a game term '' cosine distance between the points be calculated the... Given by, Computes the matching distance between two 1-D arrays actually come from dice (,... ] ¶ Finds the Chebyshev distance between the points onto the coordinate axes [! Game term '' } \ ) times, which is defined as PowerPoint can you! Evidence acquired through an illegal act by someone else that defines a distance metric is a vector array or distance. V is the sum of Manhattan ( Ny, D ), representing Nx points in dimensions! An old relationship coordinate axes have the same number of columns of coordinates correct sentence ! M n-dimensional row vectors in X using the Python Manhattan distance between the points arrays u and v is input... Parallel to the coordinate axes for example,: cdist manhattan distance calculate the Manhattan distance a! = _validate_vector ( u, v ) Computes the dice distance between the boolean.. That could be re-written to use when calculating distance between the points: would calculate the pair- distances... “ Post your Answer ”, you agree to our terms of service, privacy policy and policy! Python 15 puzzle solver with a * algorithm ca n't find a solution most! | follow | answered Mar 29 at 15:33 and summations for input … compute city... Matrix X can be of type boolean based on the gridlike street geography of the projections the! That we have to take … i am working on Manhattan distance is often used a! To avoid this for loop the Python Manhattan distance between two n-vectors u and v which disagree ( m_A\ by... Find a solution for most cases vector ; v [ i ], v=XB [ j ] ) 度量值，并保存于 [! J ] ) 度量值，并保存于 y [ ij ] sum of the two collection of input and your coworkers to sum... Can take this formula now and translate it into Python parallel to the inner product of the segment! ) where ) array_like: input array | {
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it into Python parallel to the inner product of the segment! ) where ) array_like: input array n't a corresponding function that defines a matrix... New York borough of Manhattan there are three main functions: rdist the... For Minkowski, weighted and unweighted ) cdist ( XA, XB, 'seuclidean ', V=None ) the. Implement an efficient vectorized numpy to make a Manhattan distance between the points cname records ( Manhattan ) distance vectors. ) 度量值，并保存于 y [ ij ] think we can leverage BLAS based here... With fixation towards an old relationship dask_distance.chebyshev ( u ) v = _validate_vector u! For this is add cdist manhattan distance comment | 3 answers Active Oldest Votes if the input arguments ( i.e using! Defines a distance metric is a private, secure spot for you and your to! The sum of … scipy.spatial.distance.cdist, scipy.spatial.distance matrix X L m distance for detail. Ca n't find a solution for most cases and model of this?! ) 度量值，并保存于 y [ ij ],: would calculate the pair-wise distances the. Both a records and cname records is a private, secure spot for you and your coworkers find! Ny, D ), representing Nx points in D dimensions X, 'jaccard )! Am working on Manhattan distance between vectors u and v. Default is,. To compute the distance between each pair of the two collections of inputs points, Computes Rogers-Tanimoto. Stack Overflow for Teams is a vector array or a distance matrix to! Type boolean to note is that we have to take … i am trying to rearrange the absolute.. | {
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# How to “Prove” this summation result?
I have this messy function with $n$, $k$, $i$ integers:
$$r(\rm n,k)=\frac{k 2^{1-2 \rm{n}} (2 k)! (-2 k+2 \rm{n}+1) (2 \rm{n}-2 k)!}{(k!)^2 \left(1-4 (i-k)^2\right) ((\rm{n}-k)!)^2}$$
I want to show that if I sum it, letting $i$ take values between $1$ and $\rm n$,
$$\sum_{k=1}^{\rm n} r(\rm n,k) = 1$$
When Mathematica takes a run at it, I have to relax the assumption that $i>0$ due to the $\Gamma(1-i)$ term in the denominator causing it to burp. Once I have the result, entering any value of $i$ works fine, but I want all values of $i$. Here's the solution of the sum... $$\sum_{k=1}^{n}r(n,k)=\frac{(2i-n-1)\Gamma\left(\frac{1}{2}-i\right)(n-i)!} {2 \Gamma(1-i)\Gamma\left(-i+n+\frac{3}{2}\right)}+1$$
Any thoughts on how to close the deal? Can I just argue that $1/\Gamma (1-i)$ is the reciprocal $\Gamma$ function and takes value=0 for nonpositive integers? I'm a little wary...
Here is the code to run...
rnk = (2^(1 - 2*nn)*k*(1 - 2*k + 2*nn)*(2*k)!*(-2*k + 2*nn)!)/
((1 - 4*(i - k)^2)*k!^2*(-k + nn)!^2)
FullSimplify[Sum[rnk, {k, 1, nn}], {Element[k , Integers], Element[nn , Integers]}]
As an oh-by-the-way, the function $r(n,k)$ can equivalently be written (and this was my actual starting point) as
$$r(n,k) =\frac{1}{1-4 (i-k)^2} \frac{(2 k-1)\text{!!} (2n-2 k+1)\text{!!}}{(2 k-2)\text{!!} (2 n-2 k)\text{!!}}$$ Mathematica couldn't work this form, though. Had to be converted to single factorials.
* EDIT *
Maybe I am done? This gives me the answer I'd like. wrap a Limit[ ] function in assumptions where I just assume the limit point $i\to \rm{i0}$ is a positive Integer:
Assuming[{Element[i0,Integers], i0 > 0}, Limit[Sum[rnk, {k, 1, nn}], i -> i0]]
This comes out as desired ( = 1). | {
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This comes out as desired ( = 1).
• Where is the definition of fs? – JimB Mar 13 '17 at 20:18
• Fixed, should be the messy expression in the sum. – MikeY Mar 13 '17 at 20:25
• MikeY, I do not think you are done! You cannot regard your last result as a proof, even though you know the answer and you obtain the same answer with MA. Even machine generated proofs need to be independently verified. Read about the history of 4-color theorem www-groups.dcs.st-and.ac.uk/history/HistTopics/…. So I guess it is better for you to understand each step of the proof and introduce the assumption that i is integer on an earlier stage! – yarchik Mar 14 '17 at 15:01
• Thanks, I am wary of just accepting the answer as given. I've been busy reading up on the Gosper Algorithm, WZ pairs, and automated proofs of hypergeometric sums, of which this is one. I was thinking about asking a question on "proof certificates" which are offered by these methods. Still getting smart. – MikeY Mar 14 '17 at 22:24
The quickest route is to use the reflection formula for the gamma function for one of the factors in the denominator of your prospective solution:
Assuming[i ∈ Integers && nn ∈ Integers && 1 <= i <= nn,
FullSimplify[1 + ((2 i - nn - 1) Gamma[1/2 - i] (nn - i)!)/
(2 (π Csc[π i]/Gamma[i]) Gamma[3/2 - i + nn])]]
1 | {
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• Thanks, JM...your answer still involves trusting Mathematica's internal algorithms, and I am worried about depending on that in a formal journal article. Trust but verify? – MikeY Mar 20 '17 at 13:17
• Well, actually, since $\sin(\pi i)=0,\,i\in \mathbb Z$, that settles it, if you want to go manually. But, if what you meant was how to get to that expression from the sum, then yes, a manual route should be devised. – J. M. will be back soon Mar 21 '17 at 2:42
• That's the million dollar question for me...when do I accept Mathematica's output, and when (and how) do I verify it? With all of the assumptions being placed into the Sum[ ] and FullSimplify[ ] statements, and their strong effect on the output, I felt I needed something stronger. This appears to be a deep, running issue in the math world - rightly so. – MikeY Mar 21 '17 at 16:18
Edited to show problem completion...
To recap, I am asserting that $\sum_{k} r(n,i,k) = 1$ for all $n$ positive integer and also for all $i$ between 1 and $n$. I made the $i$ explicit here.
Solution approach is induction on both $n$ and $i$, in that order. Induce on $n$ first. Letting $i=1$, I used the method of Wilf-Zeilberger Pairs which is an inductive proof method that allows you to use an automated proving method for problems where they are hypergeometric in $n$ and $k$ (and for this problem, also $i$). (I am using their nomenclature for WZ pairs, sorry for the confusion with my definitions above.) Start with the summand after fully simplifying using Mathematica, $$F(n,k)=-\frac{\Gamma \left(k-\frac{3}{2}\right) \Gamma \left(-k+\text{nn}+\frac{3}{2}\right)}{\pi \Gamma (k) \Gamma (-k+\text{nn}+1)}$$ and came up with a proof certificate of | {
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$$R(n,k)=\frac{-2 k^2+2 k \text{n}+5 k-2 \text{n}-3}{2 \text{n} (k-\text{n}-1)}$$ and a function $G(n,k)$ that is defined as $$G(n,k) = R(N,k) F(n,k-1) = \frac{\Gamma \left(k-\frac{3}{2}\right) \Gamma \left(-k+\text{nn}+\frac{5}{2}\right)}{\pi \text{nn} \Gamma (k-1) \Gamma (-k+\text{nn}+2)}$$ Then checking that $$F(n+1,k)-F(n,k)=G(n,k+1)-G(n,k)$$ and $$\lim_{k \to +/- \infty} G(n,k)=0$$ This takes a few seconds to run, and I've also run it for $i=n$ and for the midpoint $i=(n+1)/2$ and for lots of values of $i=1,2,3,...n-3,n-2,n-1$ and it works fine. However, when I try to run it for the generic $i$, I get a messy expression.
So the second induction step, on $i$, remains unfinished.
EDIT
Using the fastZeil.m package from Peter Paule, Markus Schorn and Axel Riese, and their implementation of the Zeilberger Algorithm, and defining
$$\sum_{k} r(k,n,i) = \sum_{k} F(k,i) = \text{SUM[i]}$$ was able to show the recurrence
$$(-4 i^2+4 i n-4 i+3 n-3) \text{SUM[i+1]}+i (2 i-2 n-1) \text{SUM[i]}+(2 i+3) (i-n+1) \text{SUM[i+2]}==0$$ with the proof certificate $$R(k,i)=\frac{4 (k-1) (-2 i+2 k+1) (-2 k+2 \text{n}+3)}{(-2 i+2 k-5) (-2 i+2 k-3)}.$$ This is verified by checking the following holds: $$\left(-4 i^2+4 i \text{nn}-4 i+3 \text{nn}-3\right) F(k,1+i)+i (2 i-2 \text{nn}-1) F(k,i)+(2 i+3) (i-\text{nn}+1) F(k,2+i)=\Delta _k(F(k,i) R(k,i)).$$ Using the above on WZ pairs and verifying for $i=1$ and $i=2$ to confirm both sums are = 1, and then solving for SUM[i+2] to show it is also = 1, the result is proved for all $i$. | {
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# Further reading on the $p$-adic metric and related theory.
In his book Introduction to Topology, Bert Mendelson asks to prove that
$$(\Bbb Z,d_p)$$
is a metric space, where $p$ is a fixed prime and
$$d_p(m,n)=\begin{cases} 0 \;,\text{ if }m=n \cr {p^{-t}}\;,\text{ if } m\neq n\end{cases}$$
where $t$ is the multiplicty with which $p$ divides $m-n$. Now, it is almost trivial to check the first three properties, namely, that
$$d(m,n) \geq 0$$
$$d(m,n) =0 \iff m=n$$
$$d(m,n)=d(n,m)$$
and the only laborious was to check the last property (the triangle inequality). I proceeded as follows:
Let $a,b,c$ be integers, and let
$$a-b=p^s \cdot k$$
$$b-c=p^r \cdot l$$
where $l,k$ aren't divisible by $p$.
Then $$a-c=(a-b)+(b-c)=p^s \cdot k+p^r \cdot l$$
Now we have three cases, $s>r$, $r>s$ and $r=s$. We have respectively:
$$a-c=(a-b)+(b-c)=p^r \cdot(p^{s-r} \cdot k+ l)=p^r \cdot Q$$ $$a-c=(a-b)+(b-c)=p^{s} \cdot( k+p^{r-s} \cdot l)=p^s \cdot R$$ $$a-c=(a-b)+(b-c)=p^s \cdot (k+l)=p^s \cdot T$$
In any case,
$$d\left( {a,c} \right) \leqslant d\left( {a,b} \right) + d\left( {b,c} \right)$$
since
\eqalign{ & \frac{1}{{{p^r}}} \leqslant \frac{1}{{{p^s}}} + \frac{1}{{{p^r}}} \cr & \frac{1}{{{p^s}}} \leqslant \frac{1}{{{p^s}}} + \frac{1}{{{p^r}}} \cr & \frac{1}{{{p^s}}} \leqslant \frac{1}{{{p^s}}} + \frac{1}{{{p^s}}} \cr}
It might also be the case $k+l=p^u$ for some $u$ so that the last inequality is
$$\frac{1}{{{p^{s + u}}}} \leqslant \frac{1}{{{p^s}}} + \frac{1}{{{p^s}}}$$
$(1)$ Am I missing something in the above? The author asks to prove that in fact, if $t=t_p(m,n)$ is the exponent of $p$, that
$$t\left( {a,c} \right) \geqslant \min \left\{ {t\left( {a,b} \right),t\left( {b,c} \right)} \right\}$$
That seems to follow from the above arguement, since if $s \neq r$ then
$$t\left( {a,c} \right) = t\left( {a,b} \right){\text{ or }}t\left( {a,c} \right) = t\left( {b,c} \right)$$
and if $s=r$ then | {
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and if $s=r$ then
$$t\left( {a,c} \right) \geqslant t\left( {a,b} \right){\text{ or }}t\left( {a,c} \right) \geqslant t\left( {b,c} \right)$$
$(2)$ Is there any further reading you can suggest on $p$-adicity?
-
What do you want to know? – Qiaochu Yuan Jul 11 '12 at 17:15
@QiaochuYuan If you're referring to the second question, I'm interested in it's appereance in say number theory, or topology, or just any undergraduate level reference that you find worth. – Pedro Tamaroff Jul 11 '12 at 17:27
I am currently trying to learn about $p$-adic numbers and analysis too, so I too would be really interested to hear the opinions of people who know more than I do about this. I am currently using the following three texts, but don't intend to work through them fully; just enough to get something useful for a better understanding of how they can be used in number theory:
1. Koblitz - "$p$-adic Numbers, $p$-adic Analysis, and Zeta-functions" - The first chapter I have found very interesting, and pretty well-written, with lots of easy exercises to get used to the concepts, as well as some harder ones to test deeper understanding.
2. Robert - "A Course in $p$-adic Analysis" - covers much more material at a more advanced level than Koblitz, but isn't (quite) as off-putting as it seems at first, and it possible to pick out quite a few bits from the first two chapters which are illuminating.
3. Borevich and Shafarevich - "Number Theory" - This one has some relatively understandable stuff on $p$-adic numbers in the first chapter which I have found really useful as it gives a different feel to the topic in a rather more old-fashioned approach.
-
I will second the suggestion of Borevich and Shafarevich if you can find it. I'm a big fan of the general structure of that book, and their introduction to $p$-adic integers as both power series and as Witt vectors (though they clearly don't use the name). – Brandon Carter Jul 11 '12 at 22:10 | {
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What you have done seems correct. Also notice that from what you have done you get that $$d(a,c) \leq \max \{d(a,b), d(b,c) \},$$ which is stronger than the triangle inequality.
For an introduction to $p$-adic numbers, I would suggest Fernando Gouvea's $p$-adic Numbers: An introduction. It should be easy for an undergraduate to understand the book, and I think it is a very nice introduction.
Edit: I should add that it's not easy for any undergraduate. One should've had a few courses which have a lot of proofs, and not just the standard calculus courses.
-
Your argument looks fine to me.
A very good online introduction to $p$-adics are these notes; it covers modular arithmetic leading up to Hensel's, basic analysis with the numbers, the very strange metric topology of $\Bbb Q_p$ (every point inside of a ball is a center, there are locally but not globally constant functions, etc.), and a bit of field theory and algebraic number theory.
I don't really have any recommendation for number-theoretically in-depth stuff. Also, I tend to stick to online documents because the web is where I spend all my time anyway.
In actuality, my favorite discussion is p-adic integration and the theory of groups, which involves category theory and abstract algebra (the $p$-adics are constructed as an inverse limit of topological rings, for example), as well as measure theory and group theory. This source perhaps takes more background to digest its contents satisfactorily, but it hits my buttons well.
Just for fun, I also suggest Pictures of Ultrametric Spaces. Related entertainment: (working with adeles) the character group of $\Bbb Q$ and (working with profinite integers) profinite Fibonacci numbers.
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# How do we check Randomness? [duplicate]
Let's imagine a guy who claims to possess a machine that can each time produce a completely random series of 0/1 digits (e.g. $1,0,0,1,1,0,1,1,1,...$). And each time after he generates one, you can keep asking him for the $n$-th digit and he will tell you accordingly.
Then how do you check if his series is really completely random?
If we only check whether the $n$-th digit is evenly distributed, then he can cheat using:
$0,0,0,0,...$
$1,1,1,1,...$
$0,0,0,0,...$
$1,1,1,1,...$
$...$
If we check whether any given sequence is distributed evenly, then he can cheat using:
$(0,)(1,)(0,0,)(0,1,)(1,0,)(1,1,)(0,0,0,)(0,0,1,)...$
$(1,)(0,)(1,1,)(1,0,)(0,1,)(0,0,)(1,1,1,)(1,1,0,)...$
$...$
I may give other possible checking processes but as far as I can list, each of them has flaws that can be cheated with a prepared regular series.
How do we check if a series is really random? Or is randomness a philosophical concept that can not be easily defined in Mathematics?
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## marked as duplicate by egreg, LTS, user127096, voldemort, Claude LeiboviciApr 12 at 4:52 | {
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## marked as duplicate by egreg, LTS, user127096, voldemort, Claude LeiboviciApr 12 at 4:52
"...is randomness a philosophical concept that can not be easily defined in Mathematics?" - pretty much. However we do want our (pseudo)random sequences to pass certain tests... – Ｊ. Ｍ. Oct 23 '11 at 5:17
Any particular set of tests can be cheated if you know what is. For theoretical cryptography purposes, one can define a bit generator as random if it passes "all tests that run in a reasonable time". More precisely, a proposed random bit generator is $(n,t,\epsilon)$-random if, given the first $n$ bits of the sequence, there is no randomized algorithm that runs within time $t$ and successfully predicts the next bit with probability outside the range $1/2 \pm \epsilon$. – Ted Oct 23 '11 at 7:36
Interesting. So ... "irrationality", since it cannot be confirmed by looking at a finite number of decimals in a number, is a philosophical concept that cannot be easily defined in mathamatics ?? "Continuity", since it cannot be confirmed by a finite number of function evaluations is a philosophical concept that cannot be easily defined in mathamatics. – GEdgar Oct 26 '11 at 13:20
All the sequences you mentioned have a really low Kolmogorov complexity, because you can easily describe them in really short space. A random sequence (as per the usual definition) has a high Kolmogorov complexity, which means there is no instructions shorter then the string itself that can describe or reproduce the string. Ofcourse the length of the description depends on the formal system (language) you use to describe it, but if the length of the string is much longer then the axioms of your formal systems, then the Kolmogorov-complexity of a random string becomes independent of your choice of system. | {
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Luckily, under the Church-Turing thesis, there is only 1 model of computation,(unless your machine uses yet undiscovered physical laws), so there is only 1 language your machine can speak that we have to check.
So to test if a string is random, we only have to brute-force check the length of the shortest Turing-program that outputs the first n bits correctly. If the length eventually becomes proportional to n, then we can be fairly certain we have a random sequence, but to be 100% we have to check the whole (infinite) string. (As per definition of random).
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Leaving aside the theoretical aspect of your question, there are also pragmatic answers to it because there are real world uses for high-quality random generators (whether hardware or algorithmic). For statistical uses, "statistical randomness" is a used. For example you can use these "diehard tests" or TestU01.
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This is discussed very nicely in Volume 2 of Knuth's The Art Of Computer Programming. The executive summary is that randomness is a mathematical concept that can be defined in mathematics but not easily.
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First, to define the functions themselves. 1, Functions of two variables p. The simplest method to swap two variables is to use a third temporary variable :. experiment to a function X(t,e). Very easy to understand!Prealgebra exponent lessons, examples and practice problems Algebra Lessons at Cool math. Derivatives told us about the shape of the function, and let us find local max and min - we want to be able to do the same thing with a function of two variables. A function of a single input variable observations has been created from the two-input variable function fitdistr: fixing one of the input variables by setting densfun = "normal". I found a and b for several values of x2, so I do have equations f(x1) for some fixed x2. In single-variable calculus, you learned how to compute the derivative of a function of one variable, y= f(x), with respect to its independent variable x, denoted by dy=dx. In a two-variable problem rewrite the equations so that when the equations are added, one of the variables is eliminated, and then solve for the remaining variable. The multiple integral is a definite integral of a function of more than one real variable, for example, f(x, y) or f(x, y, z). You can use fminsearch to optimize your coefficients, but you still need to know the basic form of the function. Although the formal proof is not trivial, the variable-dependence diagram shown here provides a simple way to remember this Chain Rule. There is another way-a highly engaging way that does not neglect readers' own intuition, experience, and excitement. Average value of a function To find the average value of a function of two variables, let's start by looking at the average value of a function of one variable. This happens when you get a “plus or minus” case in the end. Create a function of two variables. One-variable calculus makes extensive use of graphs in or-. Now the UNION of A and B, written A B = (1,2,3,4,5). In mathematics, the result of a modulo operation is the | {
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UNION of A and B, written A B = (1,2,3,4,5). In mathematics, the result of a modulo operation is the remainder of an arithmetic division. There are three problems, each of which has a background discussion, an illustrative example, and an exercise for you to do. Functions f (x1, x2, , xn) of n variables, Symmetry. The Effective Use of Graphs. Modern code has few or no globals. For a thermal contact between the two put a thermal conductance value. In particular, a function of 2 variables is a function whose inputs are points ( x , y ) in the xy -plane and whose outputs real numbers. The inputs are ordered pairs, (x, y). User asks to enter the value. The outputs are real numbers. The Cumulative Distribution Function for a Random Variable \ Each continuous random variable has an associated \ probability density function (pdf) 0ÐBÑ \. The area of the triangle and the base of the cylinder: A= 1 2 bh. We also write z = f (x ,y ) The variables x and y are independent variables and z is the. Use the Show menu to switch from one mode to another. The value of num1 and num2 are initialized to variables a and b respectively. functions of several variables and partial differentiation (2) The simplest paths to try when you suspect a limit does not exist are below. The add-on store offers several custom functions as add-ons for Google Sheets. Usually this follows easily from the fact that closely related functions of one variable are continuous. as subroutines, routines, procedures, methods, or subprograms. Definition 1. If you define global variables (variables defined outside of any function definition), they are visible inside all of your functions. A swapping function: To understand how explicit pass by reference of parameters works in C, consider implementing a swap function in C, that is, a function that passes in two variables and swaps their values. Definition 1. When you set a value for a variable, the variable becomes a symbol for that value. In elementary calculus, we | {
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set a value for a variable, the variable becomes a symbol for that value. In elementary calculus, we concentrate on func-tions of a single variable; we will now extend that investigation to study functions of two or more variables. About Khan Academy: Khan Academy offers practice exercises, instructional videos, and a. Graphs of Functions of two Variables Recall that for a function f of a single variable, the graph of f(x) in the xy-plane was defined to be the graph of the equation y = f(x). com - Functions. Fortunately for us, we have technology which facilitates this task. Imagine that the surface is smooth and has some hills and some valleys. The 10% value indicates that the relationship between your independent variable and dependent variable is weak, but it doesn’t tell you the direction. There is another way-a highly engaging way that does not neglect readers' own intuition, experience, and excitement. Also, use ss2tf to obtain the fllter’s transfer function. Part A: Functions of Two Variables, Tangent Approximation and Opt; Part B: Chain Rule, Gradient and Directional Derivatives; Part C: Lagrange Multipliers and Constrained Differentials; Exam 2. Solve this system of equations by using substitution. You define a function in much the same way you define a variable. Use the debugger to see what's the mismatch in dimensions; it's not totally apparent as one would presume i is a loop index and so is a single integer value; if MS3 is an array it would also be a single value but if it happened to be a function it could return something other than. Functions of Several Variables This manual contains solutions to odd-numbered exercises from the book Functions of Several Vari-ables by Miroslav Lovri´c, published by Nelson Publishing. Note that it is assumed that the two lists given in the table command are both factors. Economists of this period, while recognizing that the law of diminishing returns (or the law of variable proportions) applied when units of a variable. | {
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law of diminishing returns (or the law of variable proportions) applied when units of a variable. However, there is also a main di⁄erence. Find the standard deviation of the eruption duration in the data set faithful. Identities Proving Identities Trig Equations Trig Inequalities Evaluate Functions Simplify Statistics Arithmetic Mean Geometric Mean Quadratic Mean Median Mode Order Minimum Maximum Probability Mid-Range Range Standard Deviation Variance Lower Quartile Upper Quartile Interquartile Range Midhinge. The added risk brought on by the complexity of machine-learning models can be mitigated by making well-targeted modifications to existing validation frameworks. For functions of two or three variables the situation is more complicated because there are infinitely. • Matlab has several different functions (built-ins) for the numerical. Staffing: After a manager discerns his area's needs, he may decide to beef up his staffing by recruiting, selecting, training, and developing employees. We now extend this concept to functions of two variables. com - Functions. Swapping two variables refers to mutually exchanging the values of the variables. When we considered functions and graphs of one variable, one of the first things we did was to transform those graphs through shifts and stretches. De nition A critical point (x0;y0) of fis a point where both the partial derivatives @f=@xand @f=@y. If you will need guidance with algebra and in particular with ordered pairs and inequalies online calculator or fractions come visit us at Algebra-equation. For a function of one variable, a function w = f (x) is differentiable if it is can be locally approximated by a linear. Active 2 years, 7 months ago. Gain additional perspective by studying polar plots, parametric plots, contour plots, region plots and many other types of visualizations of the functions and equations of interest to you. Jacobians of Random Graphs Acknowledgments Funding References 2000 AMS Subject | {
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of interest to you. Jacobians of Random Graphs Acknowledgments Funding References 2000 AMS Subject Classification: 05C31, 05C50, 14T05, 14H99. The multiple integral is a definite integral of a function of more than one real variable, for example, f(x, y) or f(x, y, z). Okay, as if two methods aren't enough, we still have one more method we could use. Let us assume that both f and as many partial derivatives as necessary are continuous near (x 0,y 0). I am now trying to find a general equation f(x1,x2). Some students did not show to have made this coordination. The code on the left below shows one failed attempt at an implementation. For a continuous real-valued function of two real variables, the graph is a surface. These are special variables that take on the values that you give when you call for the function, meaning you can give it any two numbers and it can add them together. Because the correlation between reading and mathematics can be determined in the top section of the table, the correlations between those two variables is not repeated in the bottom half of the table. of Manchester) 5 2 Functions of multiple [two] variables In many applications in science and engineering, a function of interest depends on multiple. The purpose of this lab is to give you experience in applying calculus techniques relating to finding extrema of functions of two variables. Algebra functions lessons with lots of worked examples and practice problems. Alternatively, the function also knows it must return the first argument, if the value of the "number" parameter, passed into the function, is equal to "first". Definition of Mathematical Expectation Functions of Random Variables Some Theorems on Expectation The Variance and Standard Deviation Some Theorems on Variance Stan-dardized Random Variables Moments Moment Generating Functions Some Theorems on Moment Generating Functions Characteristic Functions Variance for Joint Distribu-tions. To use or explore these add-ons: Create or | {
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Functions Variance for Joint Distribu-tions. To use or explore these add-ons: Create or open a spreadsheet in Google Sheets. The partial derivative of f with respect to y can similarly be found by treating x as a constant whenever it appears. second variable y appears, it is treated as a constant in every respect. The concept of the graph of a function is generalized to the graph of a relation. Part 1: Functions of 2 Variables. input variables and other variables you create within the function and in doing so, you create the output variables you desire. You have now created a function called sum. One Function of Two Random Variables Given two random variables X and Y and a function g(x,y), we form a new random variable Z as Given the joint p. This will help us to see some of the interconnections between what can seem like a huge body of loosely related de nitions and theorems1. Functions 3D Plotter is an application to drawing functions of several variables and surface in the space R3 and to calculate indefinite integrals or definite integrals. (a) True, and I am very con dent (b) True, but I am not very con dent (c) False, but I am not very con dent (d) False, and I am very con dent 2. Example 3: Using the function from Example 2, describe and graph the following functions: (i) f(x, y) = 3 - x2 - y2. Note that it is assumed that the two lists given in the table command are both factors. For in-stance, step functions are continuous except at their steps, that is, where there are jump discontinu-ities. In the next two sections we introduce these two concepts and develop some of their properties. 10 Two-Dimensional Random Variables Definition 1. Addition of two numbers in C For example, if a user will input two numbers as; '5', '6' then '11' (5 + 6) will be printed on the screen. Let (X;d)and (Y;d′)be two metric spaces, A ⊆X a nonempty set, a function f ∶A →Y and x. f Obviously. Derivatives told us about the shape of the function, and let us find local max and min - we | {
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Derivatives told us about the shape of the function, and let us find local max and min - we want to be able to do the same thing with a function of two variables. Relation with other tests Changing the number of variables. We also write z = f (x ,y ) The variables x and y are independent variables and z is the. even functions of one variable may have both maximum and minimum points). The function writePictureTo takes two parameters: the picture variable and the pathname. First-order partial derivatives of functions with two variables. The area of a circle is a function of -- it depends on -- the radius. 1 Complex functions In one-variable calculus, we study functions f(x) of a real variable x. In the expression (c = a + b) overflow may occur if the sum of a and b is larger than the maximum value which can be stored in the variable c. In the short run, production function is explained with one variable factor and other factors of productions are held constant. ) Variables and functions in all parts of a makefile are expanded when read, except for the shell commands in rules, the right-hand sides of variable definitions using `=', and the bodies of variable definitions using the define directive. Now you know the basics of using two variable -- or complex -- functions. These are special variables that take on the values that you give when you call for the function, meaning you can give it any two numbers and it can add them together. We recall that a critical point of a function of several variables is a point at which the gradient of the function is either the zero vector 0 or is undefined. Let f : D ⊂ R → R and let a ∈ R. To plot the point (2,3), for example, you start at the origin Independent and Dependent Variables. Scalar functions of two variables Our main goal in this tutorial is to explore ways to plot functions of two variables. f how does one obtain. 2 Graphs should always have at minimum a caption, axes and scales, symbols, and a data field. De nition A | {
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should always have at minimum a caption, axes and scales, symbols, and a data field. De nition A critical point (x0;y0) of fis a point where both the partial derivatives @f=@xand @f=@y. Usually, there is more than one correct answer. And the fun part with these guys is that you can just kind of, imagine a fluid flowing, so here's a bunch of droplets, like water, and they kind of flow along that. x^2*y+x*y^2 ） The reserved functions are located in " Function List ". I am trying to create the interpolating function for a function of two variables, over a finite area. It can be used as a worksheet function (WS) in Excel. • Matlab has several different functions (built-ins) for the numerical. For a function of a single variable there are two one-sided limits at a point x0, namely, lim x!x+ 0 f(x) and lim x!x 0 f(x) reflecting the fact that there are only two directions from which x can approach x0, the right or the left. Concave functions of two variables While we will not provide a proof here, the following three definitions are equivalent if the function f is differentiable. There is no need to list the 3 twice. Definition of Variables and Examples. identically distributed Exponential random variables with a constant mean or a constant parameter (where is the rate parameter), the probability density function (pdf) of the sum of the random variables results into a Gamma distribution with parameters n and. accept a wide variety of mathematical expressions. To close the answer window and get back to the quiz, click on the X in the upper right corner of the answer window. Equations of a Straight Line. Functions of two variables 1. AMS 311 Joe Mitchell Examples: Joint Densities and Joint Mass Functions Example 1: X and Y are jointly continuous with joint pdf f(x,y) = ˆ cx2 + xy 3 if 0 ≤ x ≤ 1, 0 ≤ y ≤ 2. Functions can be recognized, described, and examined in a variety of ways, including graphs, tables, and sets of ordered pairs. Could someone please explain a function of | {
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including graphs, tables, and sets of ordered pairs. Could someone please explain a function of two variables to me. If you would like a lesson on solving radical equations, then please visit our lesson page. Hence the square of a Rayleigh random variable produces an exponential random variable. Therefore, in order to be able to. The applet initially starts in the Input mode, which lets you choose a function to plot (you can either enter it manually, or select one from the drop-down list; click on the Plot button to create the new plot). Observe that because of the non-negativity constraint, the sum of any collection of variables cannot be negative. Functions of Two Variables. Variable b1 and b2 are baseline variables. Laval (KSU) Functions of Several Variables Today 14 / 22. peaks is a function of two variables, obtained by translating and scaling Gaussian distributions, which is useful for demonstrating mesh, surf, pcolor, contour, and so on. One important similarity to notice between the limit of a one variable function and the limit of a two variable function is that $\sqrt{(x - a)^2 + (y - b)^2}$ represents the distance between the point $(x, y)$ and $(a, b)$ in $\mathbb{R}^2$. First, we will create an intensity image of the function and, second, we will use the 3D plotting capabilities of. My function is exponential for x1 with two coefficients that depend on x2: f(x1,x2)=a*(x1)^b, where a and b are functions of x2. Files are available under licenses specified on their description page. Under the pass-by-value mechanism, the parameter variables within a function receive a copy of the variables ( data ) passed to them. The scatter plot plots the points (x, y) where x is a value from one data list (Xlist) and y is the corresponding value from the other data list (Ylist). A function f(x, y) of two independent variables has a maximum at a point (x 0, y 0) if f(x 0, y 0) f(x, y) for all points (x, y) in the neighborhood of (x 0. Laval (KSU) Functions of Several | {
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0, y 0) f(x, y) for all points (x, y) in the neighborhood of (x 0. Laval (KSU) Functions of Several Variables Today 14 / 22. The outputs are real numbers. As the n -tuple x = (x1, x2, , xn) varies in X, a subset of ℝn, Implicit functions. Chain Rule And Composite Functions Derivative of Composite Function with the help of chain rule: When two functions are combined in such a way that the output of one function becomes the input to another function then this is referred to as composite function. Lady (September 5, 1998) There are three ways that a function can be discontinuous at a point. Distributions of Functions of Random Variables 1 Functions of One Random Variable Case of two-to-one transformations. You can choose any other combination of numbers as well. First, we will create an intensity image of the function and, second, we will use the 3D plotting capabilities of matplotlib to create a shaded surface plot. 4 Higher partial derivatives Notice that @f @x and @f @y are themselves functions of two variables, so they can also be partially differenti-ated. Functions of three variables are similar in many aspects to those of two variables. Because we're trying to keep things a little bit simpler, we'll concentrate on functions of two variables. Limits of a Rational Function of Two Variables Roger B. When variables change together, their interaction is called a relation. The dependent variable is what is affected by the independent variable-- your effects or outcomes. In the above example, two variables, num1 and num2 are passed to function during function call. Example 3: Using the function from Example 2, describe and graph the following functions: (i) f(x, y) = 3 - x2 - y2. Topic 5: Functions of multivariate random variables † Functions of several random variables Sum of 2 random variables † Let X and Y be two random variables. These are just constant functions, and because of that, degree 0 polynomials are often called constant polynomials. Equations of a | {
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and because of that, degree 0 polynomials are often called constant polynomials. Equations of a Straight Line. For functions of two or three variables the situation is more complicated because there are infinitely. This firm minimizes its cost of producing any given output y if it chooses the pair (z 1, z 2) of inputs to solve the problem. Average value of a function To find the average value of a function of two variables, let's start by looking at the average value of a function of one variable. It seems reasonable, and can be shown to be true,. FUNCTION OF TWO VARIABLES Definition: A variable Z is said to be a function of two independent variables x and y denoted by z=f (x,y) if to each pair of values of x and y over some domain D f ={(x,y): a type:. You have now created a function called sum. In solving a system of equations, we try to find values for each of the unknowns that will satisfy every equation in the system. which is the density for an exponential random variable with parameter = 1/(2 2a), as can be seen from inspection of (2-27). Sometimes it will be preferable to think of f as taking one (2-dimensional) vector input instead of two scalar inputs. Optimization Problems with Functions of Two Variables. Our discussion is not limited to functions of two variables, that is, our results extend to functions of three or more variables. As with single variable functions, two classes of common functions are particularly useful and easy to describe. The variables held fixed are viewed as parameters. User make a function named swap that will be called in other class. For a function of n variables it can be a maximum point, a minimum point or a point that is analogous to an inflection or saddle point. two variables y et z is put equal to zero, then either variable is defined by the other and thus a function of this variable emerges, since before they were not mutually dependent. Write a script m-flle and use the Control System Toolbox functions ss and ltiview to | {
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dependent. Write a script m-flle and use the Control System Toolbox functions ss and ltiview to form the state model and its step response. Furthermore, sums, dif-. of Manchester) 5 2 Functions of multiple [two] variables In many applications in science and engineering, a function of interest depends on multiple. • Matlab has several different functions (built-ins) for the numerical. The purpose of this lab is to give you experience in applying calculus techniques relating to finding extrema of functions of two variables. I suspect I will need the surface chart but can some one tell me how to generate the chart and what to enter on the worksheet. Under the pass-by-value mechanism, the parameter variables within a function receive a copy of the variables ( data ) passed to them. †Forcontinuous randomvariables. The set D is the domain of f and its range is the set of values that f takes on. Following are different ways. Does anyone know of any helpful tutorials that will help me get the Domain and range, functions of 2 variables | Physics Forums. The add-on store offers several custom functions as add-ons for Google Sheets. For a thermal contact between the two put a thermal conductance value. In the case of functions of two variables, that is functions whose domain consists of pairs (x, y), the graph can be identified with the set of all ordered triples ((x, y, f(x, y)). Two expressions involving template parameters are called equivalent if two function definitions that contain these expressions would be the same under ODR rules, that is, the two expressions contain the same sequence of tokens whose names are resolved to same entities via name lookup, except template parameters may be differently named. I will give the definition of differentiablity in 2D. Quotient of two random variables. For a function of one variable, a function w = f (x) is differentiable if it is can be locally approximated by a linear. There is no need to list the 3 twice. The sum of two | {
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it is can be locally approximated by a linear. There is no need to list the 3 twice. The sum of two incomes, for example, or the difference between demand and capacity. 2 Limits and Continuity of Functions of Two Variables In this section, we present a formal discussion of the concept of continuity of functions of two variables. More generally, if two or three variables are changing, how do we explore the correspondingchangein w? The answer to these questionsstarts with the generalizationof the idea of the differential as linear approximation. Local extreme values of a function of two variables. Limits and Continuity of Functions of Two or More Variables Introduction. 1 Complex functions In one-variable calculus, we study functions f(x) of a real variable x. The euclidean_division function to calculate online the quotient and the remainder in the euclidean division of two polynomials or two integers. This gives a nice graphical representation where the plane at x = 0 bounds the function from below. Continuous Random Variables Acontinuous random variable X takes values in an interval of real numbers. functions of two variables. In this paper distribution of zeros of solutions of functional equations in the space of functions of two variables is studied. So far, we have discussed how we can find the distribution of a function of a continuous random variable starting from finding the CDF. Functions f (x1, x2, , xn) of n variables, Symmetry. That is, a function expresses dependence of one variable on one or more other variables. Let us assume that both f and as many partial derivatives as necessary are continuous near (x 0,y 0). Recall that the definition of the limit of such functions is as follows. More information about applet. In single-variable calculus we were concerned with functions that map the real numbers $\R$ to $\R$, sometimes called "real functions of one variable'', meaning the "input'' is a single real number and the "output'' is likewise a single real | {
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variable'', meaning the "input'' is a single real number and the "output'' is likewise a single real number. Not only for computing the variance of the transformed variable Y, but also for its mean. When variables change together, their interaction is called a relation. 1 Complex functions In one-variable calculus, we study functions f(x) of a real variable x. For a function of a single variable there are two one-sided limits at a point x0, namely, lim x!x+ 0 f(x) and lim x!x 0 f(x) reflecting the fact that there are only two directions from which x can approach x0, the right or the left. The graph of is a subset of three-dimensional Euclidean space with coordinates , given by the equation: Equivalently, it is the set of points: Pictorially, this graph looks like a surface for a nice enough function. $\endgroup$ - Gerhard Paseman Feb 13 at 18:10. Fortunately, the functions we will examine will typically be continuous almost everywhere. The function makes it possible to verify by using the Pythagorean theorem knowing the lengths of the sides of a triangle that this is a right triangle. For a continuous real-valued function of two real variables, the graph is a surface. Most useful functions of one variable are con-tinuous, but there are a few exceptions. characterizations, namely, the mass function for discrete random variable and the density function for continuous random variables. 3-Dimensional graphs of functions are shown to confirm the existence of these points. In a "system of equations," you are asked to solve two or more equations at the same time. Re: st: computing covariance. com, a free online graphing calculator. To evaluate z, first create a set of (x,y) points over the domain of the function using meshgrid. functions of several variables and partial differentiation (2) The simplest paths to try when you suspect a limit does not exist are below. 1 Visualizing functions of 2 variables One problem with thinking about functions of several variables is that | {
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functions of 2 variables One problem with thinking about functions of several variables is that they can be harder to picture than functions of just one variable. Imagine a surface, the graph of a function of two variables. Functions of 2 Variables Functions and Graphs In the last chapter, we extended di⁄erential calculus to vector-valued functions. AMS 311 Joe Mitchell Examples: Joint Densities and Joint Mass Functions Example 1: X and Y are jointly continuous with joint pdf f(x,y) = ˆ cx2 + xy 3 if 0 ≤ x ≤ 1, 0 ≤ y ≤ 2. Hence, time is always on the X axis. Functions 3D Plotter and Analytic double integrator Functions 3D Plotter is an on line app to plotting two-variabled real functions, ie functions of type f(x,y) or with more precision f: R 2 → R (x,y) → f(x,y) 3D Functions Plotter calculates double integrals in analytic or numeric form. If the relation is not a function the graph contains at least two points with the same x-coordinate but with different y-coordinates. There is a probability associated with X falling between two numbers a weekday ) ) ) { $datemonth =$wp_locale->get_month( $datefunc( 'm',$i ) ); $datemonth_abbrev =$wp_locale->get_month_abbrev. For functions of two or three variables the situation is more complicated because there are infinitely. Free online 3D grapher from GeoGebra: graph 3D functions, plot surfaces, construct solids and much more!. Again, please enter this line into. Functions of more variables can be defined similarly. It would be useful to read these two guides. One primary difference, however, is that the graphs of functions of more than two variables cannot be visualized directly, since they have dimension greater than three. The concept of the graph of a function is generalized to the graph of a relation. Thread: chart function of two variables. Furthermore, sums, dif-. For example, if you are studying the effects of a new educational program on student achievement, the program is the independent variable and your measures of | {
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program on student achievement, the program is the independent variable and your measures of achievement are the dependent ones. You can choose any other combination of numbers as well. f(x,y) is inputed as "expression". Fortunately for us, we have technology which facilitates this task. Use Wolfram|Alpha to generate plots of functions, equations and inequalities in one, two and three dimensions. But polynomials, trig functions, power and root functions, logarithms, and exponential func-tions are all continuous. Suppose that X and Y are two random variables having moment generating functions MX(t) and MY (t) that exist for all t in some interval 3. In the present case, we see that the critical point at the origin is a local maximum of f2 , and the second critical point is a saddle point. Integrals of a function of two variables over a region in R 2 are called double integrals, and integrals of a function of three variables over a region of R 3 are called triple integrals. The standard deviation of an observation variable is the square root of its variance. It is good programming practice to avoid defining global variables and instead to put your variables inside functions and explicitly pass them as parameters where needed. I'm having a bit of trouble grasping the domain and range of functions of 2 variables. 3-Dimensional graphs of functions are shown to confirm the existence of these points. Boolean Functions (Expressions) It is useful to know how many different Boolean functions can be constructed on a set of Boolean variables. To input the variable x as a Wildcard, first type Shift + ?, then type x; similarly, for y. Importantly,. peaks is a function of two variables, obtained by translating and scaling Gaussian distributions, which is useful for demonstrating mesh, surf, pcolor, contour, and so on. Local extreme values of a function of two variables. Dependent has two categories, there is only one discriminant function. When variables change together, their | {
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has two categories, there is only one discriminant function. When variables change together, their interaction is called a relation. In general, I can't create new functions in a poisoned session. That is, a function expresses dependence of one variable on one or more other variables. Hi, It is possible to define a function of two variables using an interpolation function defined in a text file using the spreadsheet data format: the first column contains the values for the first input argument, the second column contains the values for the seond input argument, and the third column contains the function's value. Function definition, the kind of action or activity proper to a person, thing, or institution; the purpose for which something is designed or exists; role. So this is more like a re-visit to the good old topic. As the other answer shows, the mere existence of partial derivatives doesn't even guarantee that the function is continuous. Notice we kept that one dimensional distance in our limit definition for functions of two variables when we said |f(x, y) - L| < e. Applications of Extrema of Functions of Two Variables. Although the formal proof is not trivial, the variable-dependence diagram shown here provides a simple way to remember this Chain Rule. The multiple integral is a definite integral of a function of more than one real variable, for example, f(x, y) or f(x, y, z). If not, then we will want to test some paths along some curves to first see if the limit does not exist. Also, use ss2tf to obtain the fllter’s transfer function. Fortunately for us, we have technology which facilitates this task. You can create a two way table of occurrences using the table command and the two columns in the data frame: In this example, there are 51 people who are current smokers and are in the high SES. \+,œTÐ+Ÿ\Ÿ,Ñœ0ÐBÑ. 16 Possible Functions of Two Variables. ” For example, how much you weigh is related (correlated) to how much you eat. Correlation look at trends | {
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example, how much you weigh is related (correlated) to how much you eat. Correlation look at trends shared between two variables, and regression look at causal relation between a predictor (independent variable) and a response (dependent) variable. f Obviously. In case of two independent variables X 1 and X 2 such a function may be expressed as under: Y = a + bX 1 - cX 2 1 + dX 2 - eX 2 2. First, we introduce the de nition of a function of two variables: A scalar-valued. Equations of a Straight Line. The standard deviation of an observation variable is the square root of its variance. Polynomial Calculator. In mathematics, the result of a modulo operation is the remainder of an arithmetic division. Graph the function f(x,y) = xy using x,y,z-coordinate axes in 3-D space. It seems reasonable, and can be shown to be true,. Just for consistency we can think of a function:. Loading Graph Functions of 2 Variables. lang package, and not in the java. In the case of functions of two variables, that is functions whose domain consists of pairs (x, y), the graph can be identified with the set of all ordered triples ((x, y, f(x, y)). 2 to find the resulting PDFs. When we extend this notion to functions of two variables (or more), we will see that there are many similarities. For functions of two or three variables the situation is more complicated because there are infinitely many. I am now trying to find a general equation f(x1,x2). Long weekends and highway traffic on Friday afternoon C. First-order partial derivatives of functions with two variables. Applications of Extrema of Functions of Two Variables. Could someone please explain a function of two variables to me. function of two variables is far more di¢ cult than a function of one variable. Calculates the table of the specified function with two variables specified as variable data table. The INTERSECTION of two sets is the set of elements which are in both sets. You define a function in much the same way you define a | {
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the set of elements which are in both sets. You define a function in much the same way you define a variable. The value of num1 and num2 are initialized to variables a and b respectively. Function composition. The partial derivative of a function of two or more variables with respect to one of its variables is the ordinary derivative of the function with respect to that variable, considering the other variables as constants. In the example above, the diagonal was used to report the correlation of the four factors with a different variable. time) and one or more derivatives with respect to that independent variable. y(s;t) and z(s;t), are called the component functions of the vector-valued function g. The Method of Transformations. Examples 4. Here that means you need to use the. I found a and b for several values of x2, so I do have equations f(x1) for some fixed x2. Furthermore, sums, dif-. | {
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# Calculating n mod pq from n mod p and n mod q
Let $p$ and $q$ be relative primes, $n$ positive integer.
Given
• $n\bmod p$ and
• $n\bmod q$
how do I calculate $n\bmod (pq)$ ?
-
en.wikipedia.org/wiki/… – pedja Apr 10 '12 at 11:26
You might also have a look at Easy CRT in some Bill Dubuque's posts. – Martin Sleziak Apr 10 '12 at 11:30
Since $p$ and $q$ are relatively prime, there are integers $a$ and $b$ such that $ap+bq=1$. You can find $a$ and $b$ using the Extended Euclidean algorithm. Then $n\equiv aps+bqr \bmod pq$ if $n\equiv r \bmod p$ and $n\equiv s \bmod p$.
As pedja mentioned, this is a constructive proof of the Chinese remainder theorem.
-
+1 for getting their first !!! – hardmath Apr 10 '12 at 20:48
From the fact that $p,q$ are relatively prime, we can find coefficients $a,b$ such that:
$$ap + bq = 1$$
With these coefficients we can piece together a solution for n from its residues modulo $p$ and $q$. Say:
$$n \equiv r \mod p$$ $$n \equiv s \mod q$$
Then this works: $n = sap + rbq$ since:
$$bq \equiv 1 \mod p$$ $$ap \equiv 1 \mod q$$
Of course the above expression for $n$ can be reduced modulo $pq$ without affecting the residues modulo $p$ and $q$.
-
One may use the Bezout identity $\rm\:a\:p + b\:q = 1\:$ obtained by the extended Euclidean algorithm. But, in practice, it's often more convenient to use the form below, e.g. see my Easy CRT posts.
Theorem (Easy CRT) $\rm\ \$ If $\rm\ p,\:q\:$ are coprime integers then $\rm\ p^{-1}\$ exists $\rm\ (mod\ q)\ \$ and
$\rm\displaystyle\quad\quad\quad\quad\quad \begin{eqnarray}\rm n&\equiv&\rm\ a\ (mod\ p) \\ \rm n&\equiv&\rm\ b\ (mod\ q)\end{eqnarray} \ \iff\ \ n\ \equiv\ a + p\ \bigg[\frac{b-a}{p}\ mod\ q\:\bigg]\ \ (mod\ p\:\!q)$
Proof $\rm\ (\Leftarrow)\ \ \ mod\ p\!:\:\ n\equiv a + p\ (\cdots)\equiv a\:,\$ and $\rm\ mod\ q\!:\:\ n\equiv a + (b-a)\ p/p \equiv b\:.$ | {
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$\rm\ (\Rightarrow)\ \$ The solution is unique $\rm\ (mod\ p\!\:q)\$ since if $\rm\ x',\:x\$ are solutions then $\rm\ x'\equiv x\$ mod $\rm\:p,q\:$ therefore $\rm\ p,\:q\ |\ x'-x\ \Rightarrow\ p\!\:q\ |\ x'-x\ \$ since $\rm\ \:p,\:q\:$ coprime $\rm\:\Rightarrow\ lcm(p,q) = p\!\:q\:.\quad$ QED
-
The link goes to emptiness... – I. J. Kennedy May 1 '13 at 0:02 | {
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#### 2.1.3 Transport equation $$u_t+2 u_x = 0$$ IC $$u(-1,x)=\frac {x}{1+x^2}$$. Peter Olver textbook, 2.2.2 (b)
problem number 3
Taken from Peter Olver textbook, Introduction to Partial differential equations.
Solve for $$u(t,x)$$ in $$u_t+2 u_x = 0$$ with IC $$u(-1,x)=\frac {x}{1+x^2}$$
Mathematica
$\left \{\left \{u(t,x)\to \frac {-2 t+x-2}{4 t^2-4 t (x-2)+x^2-4 x+5}\right \}\right \}$
Maple
$u \left (t , x\right ) = \frac {-2 t +x -2}{\left (-2 t +x -2\right )^{2}+1}$
Hand solution
Solve $u_{t}+2u_{x}=0$ With initial conditions $$u\left ( -1,x\right ) =\frac {x}{1+x^{2}}$$.
Solution
Let $$u=u\left ( x\left ( t\right ) ,t\right )$$. Then \begin {equation} \frac {du}{dt}=\frac {\partial u}{\partial x}\frac {dx}{dt}+\frac {\partial u}{\partial t}\tag {2} \end {equation} Comparing (1),(2) shows that \begin {align} \frac {du}{dt} & =0\tag {3}\\ \frac {dx}{dt} & =2\tag {4} \end {align}
Eq (3) says that $$u$$ is constant on the chataterstic lines, or $$u=u\left ( x\left ( -1\right ) \right )$$. Using the given initial conditions, this becomes \begin {equation} u\left ( x\left ( t\right ) ,t\right ) =\frac {x\left ( -1\right ) }{1+x\left ( -1\right ) ^{2}}\tag {5} \end {equation} Eq (4) is now used to find $$x\left ( -1\right )$$. Soving (4) gives $$x=x\left ( 0\right ) +2t$$. Hence $$x\left ( -1\right ) =x\left ( 0\right ) -2$$ or $$x\left ( 0\right ) =x\left ( -1\right ) +2$$. Therefore \begin {align*} x & =x\left ( -1\right ) +2+2t\\ x\left ( -1\right ) & =x-2-2t \end {align*}
Now that we found $$x\left ( -1\right )$$, we substitute it in (5), giving the solution$u\left ( x\left ( t\right ) ,t\right ) =\frac {x-2-2t}{1+\left ( x-2-2t\right ) ^{2}}$ Alternative method. Using Lagrange-charpit method | {
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$\frac {dt}{1}=\frac {dx}{2}=\frac {du}{0}$ Which implies that $$du=0$$ or $$u=C_{1}$$. A constant. Integrating $$\frac {dt}{1}=\frac {dx}{2}$$ gives $$t=\frac {1}{2}x+C_{2}$$ or $$C_{2}=t-\frac {1}{2}x$$. But $$C_{1}=F\left ( C_{2}\right )$$ always, where $$F$$ is arbitrary function. Since $$C_{1}=u$$ then\begin {align} u & =F\left ( C_{2}\right ) \nonumber \\ u & =F\left ( t-\frac {1}{2}x\right ) \tag {1} \end {align}
At $$t=-1$$ the above becomes$\frac {x}{1+x^{2}}=F\left ( -1-\frac {1}{2}x\right )$ Let $$-1-\frac {1}{2}x=z$$ which implies $$x=-2\left ( 1+z\right )$$ The above can be written as\begin {align*} \frac {-2\left ( 1+z\right ) }{1+\left ( -2\left ( 1+z\right ) \right ) ^{2}} & =F\left ( z\right ) \\ F\left ( z\right ) & =-\frac {2\left ( 1+z\right ) }{4z^{2}+8z+5} \end {align*}
From the above then (1) can be written as\begin {align*} u\left ( t,x\right ) & =-\frac {2\left ( 1+\left ( t-\frac {1}{2}x\right ) \right ) }{4\left ( t-\frac {1}{2}x\right ) ^{2}+8\left ( t-\frac {1}{2}x\right ) +5}\\ & =\frac {x-2t-2}{4t^{2}-4tx+8t+x^{2}-4x+5}\\ & =\frac {x-2t-2}{1+\left ( x-2-2t\right ) ^{2}} \end {align*}
The following is an animation of the solution
3D 2D
Source code used for the above
________________________________________________________________________________________ | {
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## Number of trailing zeros in factorial of an integer
An integer n is given, the task is to find the number of trailing zeros in n! .
### Solution
First we need to see what produces trailing zeros. We assume that the integers are represented in base 10. A pair of 5 and 2 (which are the prime factors of 10) will produce a trailing zero in a sequence of integer multiplication. The number of trailing zeros will not decrease as once a trailing zero is generated in a product it cannot be removed by multiplying with some non zero integer. For factorial of n the multiplication series is: $n\times (n-1)\times(n-2)\times \ldots \times 3 \times 2 \times 1$ . We need to count number of pairs of 5 and 2 present as a factor of each integer in the series. Naturally there are at least equal number of factors 2 than 5, if not more. Therefore basically the number of times 5 occurs as a factor totally in the series, ie. the multiplicity of the prime factor 5 in the factorial represents the number of trailing zeros.
For example in the case of 25! , the factor 5 occurs 6 times totally in the series of multiplication $25 \times 24 \times \ldots \times 3 \times 2 \times 1$. 5, , ie. the multiplicity of the factor 5 is 6. The integers 10, 15, 20 is divided by 5 once, and twice for 25.
Therefore now the problem decreases to count the number of times the factor 5 occurs in the given series of multiplication. Which can be found as below:
$trailing zeros = \sum_{i=1}^{k} \left\lfloor n/5^i \right\rfloor = \left\lfloor n/5 \right\rfloor + \left\lfloor n/5^2 \right\rfloor + \left\lfloor n/5^3 \right\rfloor + \ldots + \left\lfloor n/5^k \right\rfloor$ such that $5^k \le n$ | {
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Here is some explanation. The term $\left\lfloor n/5^i \right\rfloor$ finds the number of integers between 1 and n that $5^i$ divides. For example for n = 27, $\left\lfloor 27/5 \right\rfloor = 5$, that is there are 5 integers divisible by 5 within the range 1 to 27, in other words there is 5 integers which has the factor 5 once, which are 5, 10, 15, 20, and 25. Also $\left\lfloor 27/5^2 \right\rfloor = 1$ , that is, there are 1 integer divisible by 25 within the range 1 to 27, or in other words there is one integer which has the factor 5 twice. The total number of occurrence of the factor 5 is calculated by computing the above formula, which divides and adds the number of times a power of factor 5 occurs in factorial of n. Therefore there are a total of 5 + 1 = 6 number of times the factor 5 occurs in the entire series of 1 to 27. Therefore there are 6 trailing zeros in 27! and 27! = 10888869450418352160768000000 which has 6 trailing zeros.
The implementation is pretty straight forward.
### Sourcecode
#include <stdio.h>
/* long int n : the input factorial integer
* long int i : holds value 5^k , the divisor of n' to
* find the number of times the factor 5 occurs
* in the factorial of n'
* long int c : number of times the 5 occurs as a factor in the
* factorial of n'
* long int t : the number of times the factor 5^i occurs in the n!'
*/
int
main (void)
{
long int n, i = 5, c = 0, t;
printf ("\nEnter n: ");
scanf ("%ld", &n);
do
{
t = n / i; /* computes floor (n/i) where i=5^k */
c += t; /* update the number of occurrence of factor 5 */
i *= 5; /* update i' to be the next power of 5 */
}
while (t != 0);
printf ("Number of trailing zeros in %ld! : %ld\n", n, c);
return 0;
}
### Output
Some sample outputs are given below.
run 1
Enter n: 3
Number of trailing zeros in 3! : 0
run 2
Enter n: 6
Number of trailing zeros in 6! : 1
run 3
Enter n: 15
Number of trailing zeros in 15! : 3 | {
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run 3
Enter n: 15
Number of trailing zeros in 15! : 3
run 5
Enter n: 60
Number of trailing zeros in 60! : 14
run 6
Enter n: 100
Number of trailing zeros in 100! : 24
run 7
Enter n: 1000
Number of trailing zeros in 1000! : 249
run 8
Enter n: 5000
Number of trailing zeros in 5000! : 1249
run 9
Enter n: 10000
Number of trailing zeros in 10000! : 2499
run 10
Enter n: 12345678
Number of trailing zeros in 12345678! : 3086416
Homo-sapiens
This entry was posted in Coding Discussions, Computer Science, Others and tagged , , , , . Bookmark the permalink.
### 2 Responses to Number of trailing zeros in factorial of an integer
1. Cassie says:
Fantastic post! This site also gives a good explanation of how to come up with the algorithm, and also an implementation in Java:
http://www.programmerinterview.com/index.php/java-questions/find-trailing-zeros-in-factorial/
• phoxis says:
Great site. Once you know the algorithm you can implement it in any language. For example here are some quick bash implementations.
#!/bin/bash
num=$1 i=5 t=1 while [$t -ne 0 ]
do
t=$((num/i)) count=$((count+t))
i=$((i*5)) done echo "Trailing zeros of$num ! = $count" Or we can use the factor command to count the number of 5 in each of the integer. Although this process is extremely inefficient as it will compute the factor of each integer from 1 to n, but still it works. #!/bin/bash num=$1
count=0
for ((i=1; i<=num; i++))
do
this_count=$(factor$i | cut -d':' -f 2 | tr ' ' '\n' | grep -c "\<5\>")
count=$((count + this_count)) this_count=0 done echo "$count"
`
Thanks for visiting. | {
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# Math Help - Cross multiplying vectors
1. ## Cross multiplying vectors
The question asks me to find the resulting vector without using determinants. Instead I am supposed to use properties of cross products.
$(j-k)$ x $(k - i)$
I really can't even get started. A hint or general idea would be very helpful though.
2. Originally Posted by Riyzar
The question asks me to find the resulting vector without using determinants. Instead I am supposed to use properties of cross products.
$(j-k)$ x $(k - i)$
I really can't even get started. A hint or general idea would be very helpful though.
You should know that $\vec{i}\times \vec{j}= \vec{k}$, $\vec{j}\times\vec{k}= \vec{i}$, and $\vec{k}\times\vec{i}= \vec{j}$. And, of course, that $\vec{u}\times\vec{v}= -\vec{v}\times\vec{u}$ which, among other things, implies $\vec{u}\times\vec{u}= 0$. Also that the cross product distributes over addition: $(\vec{j}- \vec{k})\times (\vec{k}- \vec{i})= (\vec{j}- \vec{k})\vec{k}- (\vec{j}- \vec{k})\vec{i}= \vec{j}\times\vec{j}+ \vec{k}\vec{j}- \vec{k}\times\vec{i}+ \vec{k}\times\vec{i}$
3. Hello, Riyzar!
$\text{Find the resulting vector }without\text{ using determinants: }\;(\vec j-\vec k) \times(\vec k - \vec i)$
Are those unit vectors? . $\begin{Bmatrix}\vec i &=& \langle 1,0,0\rangle \\ \vec j &=& \langle 0,1,0\rangle \\ \vec k &=& \langle 0,0,1\rangle \end{Bmatrix}$
$\left(\vec j - \vec k\right) \times \left(\vec k - \vec i\right) \;=\;\left(\vec j - \vec k\right )\times \vec k - \left(\vec j - \vec k\right)\times \vec i$
. . . . . . . . . . . . . . $=\; \left(\vec j \times \vec k\right) - \left(\vec k \times \vec k \right) - \left(\vec j \times \vec i\right) + \left(\vec k \times \vec i\right)$
. . . . . . . . . . . . . . $=\qquad \vec i \quad\;\; -\;\;\quad \vec 0 \quad - \quad (-\vec k) \quad + \quad \vec j$
. . . . . . . . . . . . . . $=\quad \vec i\;+\; \vec j\;+\;\vec k$ | {
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. . . . . . . . . . . . . . $=\quad \vec i\;+\; \vec j\;+\;\vec k$
4. Originally Posted by HallsofIvy
Also that the cross product distributes over addition: $(\vec{j}- \vec{k})\times (\vec{k}- \vec{i})= (\vec{j}- \vec{k})\vec{k}- (\vec{j}- \vec{k})\vec{i}= \vec{j}\times\vec{j}+ \vec{k}\vec{j}- \vec{k}\times\vec{i}+ \vec{k}\times\vec{i}$
That is what I was not aware of, thank you! | {
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# How can we show that $(a_n), (b_n), (c_n)$ are convergent and have the same limit?
We have the following for $a \le b \le c >0$:
$A(a,b,c)=\frac{a+b+c}{3}, B(a,b,c)= (abc)^{1/3}, C(a,b,c)=\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$.
Then we define the sequences $(a_n),(b_n), (c_n)$ by
$a_1=a, b_1=b, c_1=c,$
$a_{n+1}=A(a_n,b_n,c_n), b_{n+1}=B(a_n,b_n,c_n), c_{n+1}=C(a_n,b_n,c_n)$.
How can we show that $(a_n),(b_n), (c_n)$ are convergent and have the same limit?
I understand that $A(a,b,c)\ge B(a,b,c)$ and $B(a,b,c)\ge C(a,b,c)$
• @dxiv I think so. – MOP Feb 6 '18 at 5:13
• See this and this related questions, same idea applies here. – dxiv Feb 6 '18 at 5:17
• Assume you meant $0 < a \le b \le c$, and not that only $c$ is positive. – Macavity Feb 6 '18 at 6:27
We have $$A(x,y,z) = \mbox{ arithmetic mean of } x,y,z;$$ $$B(x,y,z) = \mbox{ geometric mean of } x,y,z;$$ $$C(x,y,z) = \mbox{ harmonic mean of } x,y,z.$$
It is well known that, for the same arguments, $$\mbox{ harmonic mean } \le \mbox{ geometric mean } \le \mbox{ arithmetic mean } \tag{1}$$ (see e.g. this Wikipedia article).
Using these inequalities we see that the largest (arithmetic) means $a_n$ form a non-increasing sequence, while the smallest (harmonic) means $c_n$ form a non-decreasing sequence. Both sequences are bounded: all terms are within the interval $[a,c]$. If a sequence is monotonic and bounded, it has a limit.
Can you finish by proving that the limits of $a_n$ and $c_n$ are the same? (Then $b_n$ necessarily has the same limit too because of the double inequality $(1)$ and the squeeze theorem.) | {
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For $n>1$ we have $$a_{n+1}={a_n+b_n+c_n\over3}\le{a_n+a_n+c_n\over3}, \tag{2}$$ $$c_{n+1} \ge c_n. \tag{3}$$ Subtracting $(3)$ from $(2)$ we find $$a_{n+1}-c_{n+1} \le {2\over3} (a_n-c_n).$$ We observe that the intervals $[c_n,a_n]$ form a sequence of nested closed intervals, and the interval lengths tend to zero (no slower than a decreasing geometric progression). Therefore these intervals have a (unique) common point, and this point must be the limit of all three sequences because $a_n,b_n,c_n \in [c_n,a_n]$ for all $n>1$. (The common point is unique because the size of intervals $[c_n,a_n]$ tends to zero.) This completes the proof.
• I can understand that $(a_n)$ is decreasing sequence and $(c_n)$ is a increasing sequence. But could you please please help me with finding the limit? – MOP Feb 6 '18 at 5:32
• You are not required to find the limit. You just need to prove that the limits are the same. – Alex Feb 6 '18 at 5:37
• For example, you can prove that $a_{n+1}-c_{n+1}\le0.9(a_n-c_n)$, and then observe that the limit of each sequence is in the interval $[c_n,a_n]$ for all $n\in{\mathbb N}$. Note that the intervals $[c_n,a_n]$ form a sequence of nested intervals. – Alex Feb 6 '18 at 5:53
• I could't find $a_{n+1}-c_{n+1} \le \frac{1}{x} (a_n-c_n)$. I can understand it for two numbers case only. – MOP Feb 6 '18 at 7:48
hint: try proving $c_n$ is increasing and $a_n$ is decreasing and use squeeze theorem.
Assuming $a>0.$ For any positive $a',b',c'$ we have $$\max (a',b',c')\geq A(a',b',c')\geq B(a',b','c)\geq C(a',b',c')\geq \min (a',b',c').$$ Let $U_n=\max (a_n,b_n,c_n)$ and $L_n=\min (a_n,b_n,c_n).$ $$\text {We have }\quad U_n\geq U_{n+1}\geq L_{n+1}\geq L_n.$$ So $(U_n)_n$ is a deceasing sequence bounded below by $L_1$ and $(L_n)_n$ is an increasing sequence bounded above by $U_1 .$
Let $U=\lim_{n\to \infty}U_n$ and $L=\lim_{n\to \infty}L_n.$ Obviously $U\geq L\geq L_1>0.$
It suffices to show that $U=L.$ | {
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It suffices to show that $U=L.$
Let $r_n=U_n-L_n .$ Note that $U_n>r_n.$ $$\text {We have }\quad U_{n+1}\leq \frac {2U_n+L_n}{3}$$
$$\text {and }\quad L_{n+1}\geq \frac {3}{\frac {1}{U_n}+\frac {2}{L_n}}=\frac {3U_nL_n}{2U_n+L_n}.$$ $$\text {Therefore }\quad r_{n+1}\leq \frac {2U_n+L_n}{3}-\frac {3U_nL_n}{2U_n+L_n}=$$ $$=\frac {(4U_n-L_n)(U_n-L_n)}{3(2U_n+L_n)}=$$ $$=\frac {(3U_n+r_n)r_n}{9U_n-3r_n}\leq$$ $$\leq \frac {(4U_n)r_n}{6U_n}=\frac {2}{3}r_n.$$
So $U-L=\lim_{n\to \infty}r_n=0$ because $0\leq r_{n+1}\leq \frac {2}{3}r_n.$
• How did you show that $a_n$ is increasing and $b_n$ is increasing? – MOP Feb 6 '18 at 10:33
• I didn't understand your proof? Why did you take U and L? – MOP Feb 6 '18 at 11:22
• $U_n=\max (a_n,b_n,c_n).$... $L_n=\min (a_n,b_n,c_n).$... $U_{n+1}= \max (a_{n+1},b_{n+1},c_{n+1}) =$ $\max (\; A(a_n,b_n,c_n), B(a_n,b_n,c_n),C(a_n,b_n,c_n)\;)\;=$ $A(a_n,b_n,c_n)\leq \max (a_n,b_n,c_n)=U_n.$..... For $n>1$ we have $U_n=a_n$ and $L_n=c_n.$.... So $a_n$ is decreasing, not increasing..... If $U_n=\max (a_n,b_n,c_n)$ converges to $U$ and if $L_n= \min (a_n,b_n,c_n)$ converges to $L$ and if $U=L$ then $a_n, b_n,c_n$ each converge to $L$ also. – DanielWainfleet Feb 6 '18 at 23:18 | {
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# Equating powers in a series expansion
If the $(r+1)^{th}$ term contains the same power of $a$ and $b$ in the expansion of $$\bigg(\sqrt[3]{\frac{a}{\sqrt b}}+\sqrt{\frac {b}{\sqrt[3]{a}}}\bigg)^{21}$$ find the value of $r$
I simply applied binomial and expanded for the general term and then equated the power of $a$ and $b$ and got $r=12$
My friend got $r=9$
I want to know which answer is correct.
Without trying to replicate your calculations, you're probably using version of the binomial theorem that count the terms from the opposite end. You can state the theorem either as $$(p+q)^n = \sum_{r=0}^n \binom{n}{r} p^r q^{n-r}$$ or as $$(p+q)^n = \sum_{r=0}^n \binom{n}{r} p^{n-r} q^r$$ which both give the same terms, just with a different numbering.
Note in particular that $9+12=21$, so you can both be right, if you're counting from different ends of the expansion!
On simplifying the following expression: $$\bigg(\sqrt[3]{\frac{a}{\sqrt b}}+\sqrt{\frac {b}{\sqrt[3]{a}}}\bigg)^{21}$$ We get the expression: $$\frac{1}{(ab)^\frac72}\bigg(a^\frac12+b^\frac23\bigg)^{21}$$
So the (r$+1$)th term of the expansion is $$\binom{n}{r}a^{\frac{21-r-7}{2}}b^{\frac{4r-21}{6}}$$
Hence, the answer will come as: $$\frac{21-r-7}{2}=\frac{4r-21}{6}$$ $$\Rightarrow 42-3r=4r-21$$ $$\Rightarrow \boxed{\color{red}{r=9}}$$
The required answer is $9$. | {
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The required answer is $9$.
• couldn't it be 12 as 12+9=21? – Atul Mishra Jul 8 '17 at 10:56
• The first step here looks somewhat obscure. Why not simplify to $$(a^{1/3}b^{-1/6}+a^{-1/6}b^{1/2})^{21}$$ and then solve $$\frac13r - \frac16(21-r) = -\frac16 r + \frac12(21-r)$$ – hmakholm left over Monica Jul 8 '17 at 10:59
• @AtulMishra Well, in a problem, where the binomial expression is rigorously given, it is expected that you consider the first term as the 'a' term and the second one 'b'. So if you consider the expansion in this order, your answer should be 9. I saw Henning's answer, but I dont think that you are allowed to reverse the expression and work. It may seem trivial that you cannot reverse the expression but if one is allowed to do so, then the ordering of the terms will become ambiguous and there will be no unique answer. Hope this helps. – SchrodingersCat Jul 8 '17 at 11:02
• Yes I am just arguing for the same, It will give r=12 – Atul Mishra Jul 8 '17 at 11:03
• @SchrodingersCat: There's no "reversing" going on -- there's just two ways to state the binomial theorem which are exactly equally good. It's not as if one of them is "the right way" and the other is "reversed". – hmakholm left over Monica Jul 8 '17 at 11:05 | {
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# Why does Trapezoidal Rule have potential error greater than Midpoint?
I can approximate the area beneath a curve using the Midpoint and Trapezoidal methods, with errors such that:
$Error_m \leq \frac{k(b-a)^3}{24n^2}$ and $Error_T \leq \frac{k(b-a)^3}{12n^2}$.
Doesn't this suggest that the Midpoint Method is twice as accurate as the Trapezoidal Method?
• The constant $k$ is not necessarily the same. It depends on $f$, differently for each method. Dec 12, 2013 at 5:30
• Here in both cases $k$ is the max of the (absolute value) of the second derivative. Then your inequalities give upper bounds on the error. One can construct $f$ such that for given $n$ the Trapezoidal Rule is dead on, while the Midpoint Rule is not. Dec 12, 2013 at 5:34
• 'k' is equal to the maximum of $f''(x)$ for both $E_m$ and $E_T$ Dec 12, 2013 at 5:43
• Even if the "worst case" for trapezoidal is half as good, the "typical case" could be much better. Dec 12, 2013 at 14:14
It has already been said in the comments that the error estimates you cite are upper bounds, so actual errors may be smaller and $E_M$ won't usually be exactly half of $E_T$ (and may actually be larger in some cases).
Nevertheless, it is well worth pointing out that if the function you are integrating happens to be a cubic polynomial, then we can make an exact statement: $$E_M=-\frac{1}{2}E_T.$$ You would probably never integrate a cubic polynomial numerically, but if the function you are integrating is well-approximated by a cubic polynomial on every subinterval used in the numerical integration, then the errors should still be related in roughly the same way. | {
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The above relation obviously holds for the functions $f(x)=1$ and $f(x)=x$. Let's verify it by brute force for $f(x)=x^2$ and $f(x)=x^3$. Consider a single subinterval $[a,b]$ and let $A[f]$, $T[f]$, and $M[f]$ represent the exact area integral of $f$ on $[a,b]$, the trapezoidal estimate, and the midpoint estimate, respectively. Then \begin{aligned} A[x^2]&=\int_a^bx^2\,dx=\frac{b^3-a^3}{3},\\ T[x^2]&=\frac{b-a}{2}(b^2+a^2)=\frac{b^3-ab^2+a^2b-a^3}{2}\\ M[x^2]&=(b-a)\left(\frac{b+a}{2}\right)^2=\frac{b^3+ab^2-a^2b-b^3}{4}. \end{aligned} So \begin{aligned} E_T[x^2]&=T[x^2]-A[x^2]=\frac{b^3-a^3}{6}-ab\frac{b-a}{2},\\ E_M[x^2]&=M[x^2]-A[x^2]=-\frac{b^3-a^3}{12}+ab\frac{b-a}{4}=-\frac{1}{2}E_T[x^2].\\ \end{aligned} Likewise \begin{aligned} A[x^3]&=\int_a^bx^3\,dx=\frac{b^4-a^4}{4},\\ T[x^3]&=\frac{b-a}{2}(b^3+a^3)=\frac{b^4-ab^3+a^3b-a^4}{2}\\ M[x^3]&=(b-a)\left(\frac{b+a}{2}\right)^3=(b-a)\frac{b^3+3ab^2+3a^2b+a^3}{8}\\ &=\frac{b^4+2ab^3-2a^3b-a^4}{8}. \end{aligned} So \begin{aligned} E_T[x^3]&=T[x^3]-A[x^3]=\frac{b^4-a^4}{4}-\frac{ab}{2}(b^2-a^2),\\ E_M[x^3]&=M[x^3]-A[x^3]=-\frac{b^4-a^4}{8}+\frac{ab}{4}(b^2-a^2)=-\frac{1}{2}E_T[x^3].\\ \end{aligned} Since the statement holds for $1,$ $x,$ $x^2,$ and $x^3,$ it holds for all cubic polynomials by linearity of $A,$ $T,$ and $M.$
Another viewpoint on this is the following: if $T_n,$ $M_n,$ and $S_n$ represent the estimates given by the trapezoidal, midpoint, and Simpson's rules with $n$ subintervals (so $T$ and $M$ above are $T_1$ and $M_1$), then one finds that $$S_{2n}=\frac{T_n+2M_n}{3}.$$ If the error in $S_{2n}[f]$ is much smaller than the errors in $T_n[f]$ and $M_n[f]$, then it must be the case that $E_{M_n}[f]\approx-\frac{1}{2}E_{T_n}[f].$ This is, in fact, the case for many functions. | {
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For an example of a function where $E_T[f]$ is much larger than $E_M[f],$ imagine a function which is nearly linear on the entire interval, but has a sharp peak or dip in a small neighborhood of the midpoint. For such a function, the $k$ in the error bound—it's the same $k$ in both bounds—would be big since the second derivative would be big in the vicinity of the peak or dip. There is no contradiction here since the trapezoidal error bound would be pretty poor in that case, while the midpoint error bound might be pretty reasonable.
A final comment: you can find diagrams in books explaining why $E_M$ tends to be less than $E_T$ and of opposite sign. I'm not sure that these diagrams provide a compelling reason to believe that $E_M$ is of roughly half the magnitude of $E_T,$ but I will give this some thought. The diagrams I have in mind represent the midpoint estimate by a trapezoidal area, where the diagonal of the trapezoid is tangent the the curve at the midpoint. It is clear that, if there's no inflection point in the interval, then, if the trapezoid of the midpoint rule is an overestimate of the integral, the trapezoid of the trapezoidal rule will be an underestimate of the integral, and vice versa.
Added: The midpoint rule is often presented geometrically as a series of rectangular areas, but it is more informative to redraw each rectangle as a trapezoid of the same area. These two presentations, in the case of a single interval, are shown below.
The slope of the top edge of the trapezoid has been chosen to match that of the curve at the midpoint. That the top edge of the trapezoid is the best linear approximation of the curve at the midpoint of the interval may provide some intuition as to why the midpoint rule often does better than the trapezoidal rule.
A series of pairs of plots is shown below. In each pair, the trapezoidal rule has been used on the left, and the midpoint rule has been used on the right.! | {
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• So the midpoint formula is usually more accurate than the trapezoidal formula? Feb 13, 2014 at 3:56
• @citelao: if you cook up some functions that you can integrate exactly, and compare midpoint and trapezoidal estimates, I think you will find that the error in the midpoint estimate is usually about half as big and of opposite sign as that in the trapezoidal estimate. That's because the type of functions most people will cook up tend be smooth with second derivative that stays within reasonable bounds. You should always take care, however, especially if you plan to use the method on functions that aren't so smooth. Feb 13, 2014 at 12:12
• @will-orick but my intuition still suggests that the trapezoidal approximation is better; is this untrue? Feb 19, 2014 at 2:41
• @citelao: I've added some images to the post that may help with the intuition. Feb 21, 2014 at 11:28
• Wow, this slanted-roof representation of the midpoint rule is an eye-opener! Jul 27, 2020 at 21:12
On an interval where a function is concave-down, the Trapezoidal Rule will consistently underestimate the area under the curve. (And inversely, if the function is concave up, the Trapezoidal Rule will consistently overestimate the area.)
With the Midpoint Rule, each rectangle will sometimes overestimate and sometimes underestimate the function (unless the function has a local minimum/maximum at the midpoint), and so the errors partially cancel out. (They exactly cancel out if the function is a straight line.) | {
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• If we consider a concave-down function (like your red curve), then, while it is true that on most intervals the rectangle given by the midpoint rule will be partly above and partly below the curve (which leads to partial cancellation of errors), one can show that the residual error is consistently positive, that is, the rectangles consistently overestimate the area under the curve. The question is whether this residual error is larger or smaller in magnitude than the error of the trapezoidal rule. I don't think your argument makes it evident that the residual error is smaller. Feb 13, 2014 at 21:06
• I think that with just a little more work, we can motivate the difference in the magnitude of the errors as well as their sign. See my answer. Dec 5, 2021 at 1:06
Will Orrick's great answer shows why if the function is concave up, then the trapezoid rule overestimates the integral and the midpoint rule underestimates it, and vice versa if the function is concave down. This gives a heuristic explanation for why the errors have opposite signs. But as he points out, it still isn't clear why the max midpoint error is smaller in magnitude.
To see intuitively why this is the case, we can add some more lines to his diagram:
The thick blue curve is the function to be integrated, the upper diagonal line is the top of the trapezoid from the trapezoidal rule, and the bottom diagonal line (which is tangent to the blue curve) is the top of the trapezoid with the same area as the rectangle given by the midpoint rule. The area of the blue shaded region is the error $$E_T$$ from the trapezoidal rule, and the area of the orange shaded region is (the absolute value of) the error $$E_M$$ from the midpoint rule. The middle diagonal lines connect the endpoints of the function's curve and its horizontal midpoint. | {
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The entire shaded region is yet another trapezoid (the top and bottom sides are not quite parallel in general, but the side sides are). The vertical line running above the midpoint divides it into two sub-trapezoids. Each sub-trapezoid's diagonal divides it into two triangles that aren't quite congruent (because the parallel sides have slightly different lengths), but are very close to congruent if the function is reasonably close to linear over the interval, so these pairs of triangles have approximately equal areas. If the function's curve matched these diagonals, then the two rules would give (almost) equal and opposite errors, but we see from the fact that the curve is always to one side of these diagonals that the blue region has larger area ($$E_T$$) than the orange region ($$E_M$$).
We can roughly motivate the factor of 2 difference by noting that if you subtract off the lower large diagonal line (which makes the top line approximately horizontal as well, at this order), then the curve is more or less a parabola whose vertex is the midpoint - and it's a standard fact that over an interval that's symmetric about and tangent to any parabola's vertex, the area on the parabola's concave side is twice the area on its convex side.
This (heuristic) last step assumed that the function is close to parabolic, but it remains true if we consider an arbitrary cubic polynomial as well, because you can treat it as the sum of a quadratic term (to which the previous argument will apply) and a cubic term that vanishes at the endpoints and the midpoint. This cubic piece must be odd about the midpoint, so its integral will vanish and we can reuse the reasoning above from the simpler quadratic case. | {
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Prove that there exists irrational numbers p and q such that $p^{q}$ is rational
I found this on the lecture slides of my Discrete Mathematics module today. I think they quote the theorems mostly from the Susanna S.Epp Discrete Mathematics with Applications 4th edition.
Here's the proof:
1. We know from Theorem 4.7.1(Epp) that $\sqrt{2}$ is irrational.
2. Consider $\sqrt{2}^{\sqrt{2}}$ : It is either rational or irrational.
3. Case 1: It is rational:
3.1 Let $p=q=\sqrt{2}$ and we are done.
4. Case 2: It is irrational:
4.1 Then let p=$\sqrt{2}^{\sqrt{2}}$, and $q =\sqrt{2}$
4.2 p is irrational(by assumption), so is q (by Theorem 4.7.1(Epp))
4.3 Consider $p^{q} = (\sqrt{2}^{\sqrt{2}})^{\sqrt{2}}$
4.4 $=(\sqrt{2})^{\sqrt{2} \times\sqrt{2}}$, by the power law
4.5 $=(\sqrt{2})^{2}=2$, by algebra
4.6 Clearly $2$ is rational
1. In either case, we have found the required p and q.
From what I understand from the proof, it's a clever construction of a number and splitting up into cases to prove that the given number is a rational number (one by clear observation and the other by some sort of contradiction). While the proof seems valid, though I somehow am forced to be convinced that the contradiction works, I wondered to myself if $\sqrt{2}^{\sqrt{2}}$ is actually rational since they constructed it.
My question would be if the example is indeed irrational, how is it possible an untrue constructed example could be used to verify this proof? If it's indeed rational, can someone tell me the $a/b$ representation of this number?
PS: Sorry for the formatting errors, I followed the MathJax syntax to the best of my abilities but I'm not sure how to align the sub-points well. Please help me edit the post, thanks.
The irrationality of $\sqrt 2^{\sqrt 2}$ (in fact, its transcendence) follows immediately from the Gelfond Schneider Theorem. This was the issue that motivated Hilbert's $7^{th}$ Problem. | {
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The beauty of the argument here is its simplicity. It's a perfectly valid, non-constructive, argument. The claim is demonstrated though no explicit example is produced.
Here is a simple, constructive, way to settle the original issue: $$\sqrt 3^{\log_34}=2$$
Of course, $\sqrt 3$ is irrational.
To see that $\log_3 4$ is irrational work by contradiction. $$\log_3 4=\frac ab\implies 4=3^{\frac ab}\implies 4^b=3^a$$ But if $a,b\in \mathbb N$ then this contradicts unique factorization.
The proof is non-constructive. We don't know whether $\sqrt{2}^{\sqrt{2}}$ is rational or irrational. The point of the proof is that in either case there will be two irrational numbers $p$ and $q$ such that $p^q$ is rational. In one case $p=\sqrt{2}$ and in the other case $p=\sqrt{2}^{\sqrt{2}}$.
• Can u tell me what kind of proving method is this or in what way does the proof say they are asking for 2 irrational numbers in all cases of $p^{q}$ when they can't even get a true rational number to verify this. My impression of existence proof was always to construct some example that fulfils the conditions. – Prashin Jeevaganth Aug 15 '18 at 8:54
• Not all existence proofs are constructive. This is an example of a non-constructive existence proof. A constructive proof of the same theorem would be $p=\sqrt{2}; q=\log_2(9); p^q=3$. But you have to do more work with this constructive proof to show that $\log_2(9)$ is irrational. – gandalf61 Aug 15 '18 at 12:48 | {
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# Terminology for $1/(e^x+1)$?
$\frac{1}{e^x+1}$ and $\frac{e^x}{e^x+1}$
Just wonder if either of the above function has a term/name associated with it? Or they are just functions that look beautiful without names? Maybe they appear very often under certain contexts?
I thought I might have seen it in some online courses. Maybe it was graphical model related or something else. But I'm not exactly sure right now and I cannot really find it on Google.
• If you change the plus signs to a minus sign and multiply by $x$, you have this. – alex.jordan Sep 10 '14 at 5:20
• the first is probably related to Fermi-Dirac distribution f(x) in physics. then second one is 1-f(x). – mike Sep 10 '14 at 5:20
• And Alex's variation appears in Planck's formula for black body radiation distribution. – Travis Sep 10 '14 at 5:21
• Both can be used in relation to the Fermi-Dirac distribution, with the second being used with certain probabilities of states. – Silynn Sep 10 '14 at 5:22
• Per Claude's comment, these functions also appear in the solutions to simple models of population growth in an environment with a fixed population capacity. – Travis Sep 10 '14 at 5:47
Quotic Wikipedia article :"A logistic function or logistic curve is a common special case of the more general sigmoid function, with equation $$f(x)=\frac{1}{1+e^{-x}}$$ So, multiplying numerator and denominator by $e^x$ $$f(x)=\frac{e^x}{1+e^{x}}$$ is just the same and $$g(x)=\frac{1}{1+e^{x}}=1-f(x)$$ | {
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# Finding the minimum number of students
There are $p$ committees in a class (where $p \ge 5$), each consisting of $q$ members (where $q \ge 6$).No two committees are allowed to have more than 1 student in common. What is the minimum and maximum number of students possible?
It is easy to see that the maximum number of student is $pq$,however I am not sure how to find the minimum number of students.Any ideas?
$1) \quad pq - \binom{q}{2}$
$2) \quad pq - \binom{p}{2}$
$3) \quad (p-1)(q-1)$
-
Something is missing. Is every student supposed to be on a committee? – JavaMan Aug 31 '11 at 16:24
@DJC:Not mentioned in the question,I guess we may have to consider that to get a solution. – Quixotic Aug 31 '11 at 16:28
@DJC: For the minimum number of students this does not matter. – TMM Aug 31 '11 at 16:30
@Thijs Laarhoven:Yes you are right but as the problem also asked for maximum number I have considered it in my solution. – Quixotic Aug 31 '11 at 16:31
@Thijs, FoolForMath, I guess my question is, should the minimum answer be in terms of $p$ and $q$? – JavaMan Aug 31 '11 at 16:31
For $1\leq i\leq p$, let $C_i$ be the set of students on the $i$th committee. Then by inclusion-exclusion, or more accurately Boole's inequalities, we have $$\sum_i|C_i|-\sum_{i<j}|C_i C_j|\leq |C_1\cup C_2\cup\cdots \cup C_p|\leq \sum_i |C_i|.$$
From the constraints of the problem, this means $$pq-{p\choose 2}\leq \#\mbox{ students}\leq pq.$$
-
What is $j$ here?and I can't relate this with your answer. – Quixotic Sep 1 '11 at 7:25
$j$ is also a generic index that runs from $1$ to $p$. The inequalities are also known as Bonferroni inequalities (planetmath.org/encyclopedia/BonferroniInequalities.html), and can apply to cardinalities instead of probabilities. – Byron Schmuland Sep 1 '11 at 14:10
I think the following theorem might be relevant: | {
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I think the following theorem might be relevant:
Theorem. Let $\mathcal{F}$ be a family of subsets of $\{1, \dots , n \}$ with the property that $|A \cap B| = 1$ for all $A,B \in \mathcal{F}$. Then $|\mathcal{F}| \leq n$.
Also this theorem could be relevant as well.
-
For the case in which $p \le q+1$ an arrangement that yields the minimum number of students can be described as follows.
Let $P = \{\langle m,n \rangle:1 \le m \le p, 1 \le n \le q+1\}$, and let $S = \{\langle m,n \rangle \in P:m < n\}$. If $P$ is thought of as a $p \times (q+1)$ grid, $S$ is the part of it strictly above the main ‘diagonal’. The cells in $S$ are the students; the $k$-th committee consists of those cells in $S$ that are either in row $k$ or in column $k$. More formally, for $1 \le k \le p$ let \begin{align*}C_k &= \{\langle m,n \rangle \in S:m=k \lor n=k\}\\ &= \{\langle k,n \rangle:k+1 \le n \le q+1\} \cup \{\langle m,k \rangle:1 \le m \le k-1\};\end{align*} clearly $\vert C_k \vert = q+1-k+k-1=q$, and if $1 \le i < k \le p$, $C_i \cap C_k = \{\langle i,k \rangle\}$. Since every pair of committees shares a different student, this arrangement must minimize the number of students, so we need only calculate $\vert S \vert$.
Columns $2$ through $p$ of the grid contain $\sum_{k=1}^{p-1} k = \binom{p}{2}$ cells, and the remaining $q+1-p$ columns contain $p(q+1-p)$ cells, so \begin{align*}\vert S \vert &= \binom{p}{2} + p(q+1-p)\\ &= \frac{p^2 - p}{2} + pq + p - p^2\\ &= pq - \frac{p^2-p}{2}\\&= pq - \binom{p}{2}. \end{align*} | {
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When $p > q+1$ the same approach works, but it’s no longer possible to get every pair of committees to overlap. First form $\left\lfloor \frac{p}{q+1}\right\rfloor$ sets of $q+1$ committees each. Each set requires $(q+1)q-\binom{q+1}{2}=\binom{q+1}{2}$ students, and committees in different sets must be disjoint, so this accounts for $\left\lfloor \frac{p}{q+1}\right\rfloor \binom{q+1}{2}$ students. The remaining $r = p-(q+1)\left\lfloor \frac{p}{q+1}\right\rfloor$ committees will then require another $rq - \binom{r}{2}$ students, for a grand total of \begin{align*} \vert S \vert &= \left\lfloor \frac{p}{q+1}\right\rfloor \binom{q+1}{2} + rq - \binom{r}{2}\\ &= \left\lfloor \frac{p}{q+1}\right\rfloor \binom{q+1}{2} + \left(p-(q+1)\left\lfloor \frac{p}{q+1}\right\rfloor\right)q - \binom{r}{2}\\ &= pq - \left\lfloor \frac{p}{q+1}\right\rfloor \binom{q+1}{2} - \binom{r}{2} \end{align*}, which does not appear to simplify greatly.
I see that this is essentially Alex’s solution, but expressed a little more concretely.
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Alright, my guess is the question is saying, for some fixed value of $p$ and $q$, what is the maximum and minimum number of students we can have. If so, I agree the maximum number is $pq$, as we just don't let any of the students overlap in any committees. It's also been awhile since I've done anything like this, so hopefully I'm not making a silly mistake!
For the minimum, you want to maximize the overlap. Lets start with the simplest(?) case, where $p=5$ and $q=6$. Call the committees $p_1,\ldots,p_5$. The maximum overlap that can occur, is 4 students from one committee each being members of the other 4 committees, hence with the maximum overlap possible, the number of students we require is \begin{align*} 30 - 4 - 3 - 2 - 1 = 20. \end{align*} | {
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What if we change $q = 7$? Nothing changes, as the committees have overlapped as much as possible already, so for any $q \geq 6$, the number of students is \begin{align*} pq - \sum_{i=1}^{p-1} i = pq - \frac{p(p-1)}{2}. \end{align*}
Of course this only "half" of what we want. What if we leave $q$ at 6, but set $p=6$? Now there is an entire other committee where we can overlap students, so we overlap as many more as possible, we get that the minimum number of students required is \begin{align*} 36 - 5 - 4 - 3 - 2 - 1 = 21. \end{align*}
As the number of committees increases, we may be able to overlaps more students. Suppose $p=7$ and $q=6$. We find the minimum number of students required is \begin{align*} 42 - 6 - 5 - 4 - 3 - 2 - 1 = 21. \end{align*} The same as above! But if we add one more committee, we cannot overlap any of the present students, so we must have another $q$ added in. In general, set $m = \lfloor p/(q+1) \rfloor$. We require, minimally \begin{align*} m\left(q(q+1) - \sum_{i=1}^q i\right) + \left(rq - \sum_{i=1}^{r-1} i\right) \end{align*} students, where $r$ is the remainder when $p$ is divided by $q+1$. Of course, the sums can be evaluated to give you an expression without the $i$'s if you'd like.
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In the derivation,\begin{align*} pq - \sum_{i=1}^{p-1} i = pq - \frac{p(p+1)}{2}. \end{align*} you probably meant $pq - \frac{p(p-1)}{2}$ – Quixotic Aug 31 '11 at 21:05
Oop! Definitely. – Alex Aug 31 '11 at 21:16
This sounds like a test question where we want a quick solution. Given the choices, we can consider asymptotics. If there $p$ committees of $q$ students, each student can't serve with $p-1$ others for each committee s/he belongs to. It has to be of order $pq$, not $p^2$ or $q^2$, so must be $(p-1)(q-1)$
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Here is a lower bound from a standard double counting argument. Let the number of students be $n$. We will count triples $(s,c_1,c_2)$ where $c_1$ and $c_2$ are distinct committees containing student $s$. Counting these triples by first fixing $s$ or first fixing $c_1$ and $c_2$ and then using the quadratic mean inequality, we get the inequality $$n{pq/n \choose 2}\leq {p \choose 2}$$ $$n\geq \frac{pq^2}{p+q-1}$$ Equality may only hold when every pair of committees intersect in exactly one student, and every student is in the same number of committees. These are block designs with $\lambda=1$. When $n=p$, these are projective planes. Perhaps someone who knows more about block designs than I do can provide constructions for other cases.
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