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Both answers are correct. To expand on John Habert's comment, since the coefficient $c_1$ is arbitrary, your $c_1$ is the negative of the book's $c_1$. To see what I mean, replace your $c_1$ with $-c_1$: you'll get the book's answer.
• Is there any particular reason why they factor out a $-1$? I just think that can be confusing to do random things like that. – adam Mar 27 '14 at 17:37
• If this is an initial value problem, then it is possible this choice of signs makes finding $c_1,c_2$ a little easier. – John Habert Mar 27 '14 at 17:40 | {
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# Measure of “how much diagonal” a matrix is
I have a (biological) computational system that outputs squared matrices. These matrices will sometimes have a tendency to be diagonal-like, with higher values at and around the diagonal.
I would like to have some summary measure on how "much diagonal" a matrix is, so that I can batch process hundreds of outputs and score them on how much the higher entries cluster in and around the diagonal.
Any ideas of some standard approach that I can generalise?
Thanks ! JL
• What do the rows and columns represent? Are the matrix entries counts of events? This affects the choice. – Tad Aug 11 '15 at 1:32
• They are frequencies of events! – lourencoj Aug 11 '15 at 8:20
• Do the rows and columns represent values or just categories? In the first case you get credit for being "near" the diagonal; in the second case you're either on it or you're not. – Tad Aug 11 '15 at 11:15
• It should get credit for being close to the diagonal – lourencoj Aug 11 '15 at 14:49
Given that your entries are frequencies, and you want to give credit for being "close" to the diagonal, a natural approach is to compute the correlation coefficient between the row and column. That is, suppose your matrix is built as follows: repeatedly generate a pair of numbers $x$ and $y$, and increment the count of the matrix entry at position $(x,y)$. If you think of $x$ and $y$ as samples of random variables $X$ and $Y$ respectively, then the sample correlation coefficient $r$ of $X$ and $Y$ lies between $-1$ and $1$. It is $1$ if $X$ and $Y$ are perfectly correlated, $-1$ if they are perfectly anticorrelated. The point is that $X$ and $Y$ are perfectly correlated (in this case, equal) precisely when the matrix is diagonal, strong correlation means the matrix entries tend to be near the diagonal.
This is robust: the correlation coefficient is unchanged if you scale the matrix (and the formula turns out to make sense even if your entries are nonnegative real numbers). | {
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If you adapt the formulas in the above reference to this situation, they take the following form. Let $A$ be a $d\times d$ matrix; let $j$ be the $d$-long vector of all ones, and let $r=(1,2,\ldots,d)$ and $r_2=(1^2,2^2,\ldots,d^2)$. Then:
\begin{align} n &= j A j^T \textrm{ (the sum of the entries of A) }\\ \Sigma x &= r A j^T\\ \Sigma y &= j A r^T\\ \Sigma x^2 &= r_2 A j^T\\ \Sigma y^2 &= j A r_2^T\\ \Sigma xy &= r A r^T\\ r &= \frac{n\, \Sigma xy -\Sigma x\, \Sigma y}{\sqrt{n\, \Sigma x^2 - (\Sigma x)^2}\sqrt{n\, \Sigma y^2 - (\Sigma y)^2}} \end{align}
Some examples:
Diagonal matrix: $\left( \begin{array}{cccc} 1. & 0. & 0. & 0. \\ 0. & 5. & 0. & 0. \\ 0. & 0. & 30.5 & 0. \\ 0. & 0. & 0. & 3.14159 \\ \end{array} \right): \quad r=1.000000$
Diagonally dominant matrix: $\left( \begin{array}{ccc} 6 & 1 & 0 \\ 1 & 5 & 2 \\ 1 & 3 & 6 \\ \end{array} \right): \quad r=0.674149$
Uniformly distributed on $[0,1]$: $\left( \begin{array}{cccc} 0.2624 & 0.558351 & 0.249054 & 0.484223 \\ 0.724561 & 0.797153 & 0.689489 & 0.273023 \\ 0.462727 & 0.119412 & 0.911981 & 0.636588 \\ 0.089544 & 0.160899 & 0.910123 & 0.549202 \\ \end{array} \right): \quad r=0.233509$
Tridiagonal: $\left( \begin{array}{ccccc} 2 & 1 & 0 & 0 & 0 \\ 1 & 3 & 2 & 0 & 0 \\ 0 & 2 & 3 & 4 & 0 \\ 0 & 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 1 & 1 \\ \end{array} \right): \quad r=0.812383$ | {
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• Hey Tad. This is great! This will be our approach. Thanks so much. – lourencoj Aug 12 '15 at 21:49
• @Tad, I've got a similar task to measure the degree of "diagolizedness" but my matrix can be rectangular, not just square. What would you say? Can you expand your solution for that case? – ttnphns Aug 27 '15 at 10:52
• @ttnphns You need to be clear about what you mean by "diagonalness" in this context. – Tad Aug 27 '15 at 11:03
• The concentration of values (greeting their magnitude, too), close to the main diagonal (from top-left corner), as before. Fully diagonal matrix will be when all nonzero elements lie on it. – ttnphns Aug 27 '15 at 11:34
• If you adapt the formulas in the above reference... I must say that your adaptation is not transparent at all. Just to mention, $n$ in correlation coefficient is the number of observations (X,Y pairs). In your formulas, it is suddenly the sum of values in the matrix. To me, your approach remains unclear (albeit it could be perfect). Can you elucidate your computations using the theoretical (the 1st, not the 2nd) formula of $r$ of the wikipedia, the formula using the means? – ttnphns Aug 27 '15 at 12:33
Here's an easy one. Let $M$ be your measured matrix, and $A$ be the matrix which agrees with $M$ along the diagonal, but is zero elsewhere. Then pick your favorite matrix norm (operator probably works well here) and use $\|M-A\|$ as your measurement.
If you want more fine tuned understanding of 'clustering', instead of making all the entries off the diagonal $0$, weight them by what band they are on. So the super and sub diagonal might take half the corresponding value in $M$.
• Hey. This is a good start, thanks – lourencoj Aug 11 '15 at 8:19 | {
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Proof that there are infinitely many primes (Euclid)
I was wondering if I could get some insight on my proof. I am in the midst of relearning some number theory and just "writing proofs" in general, and I would like some assistance to see if I am on the right track.
The statement I am proving is Euclid's Theorem which states that "there are infinitely many primes."
Here is my attempt at the proof after some reading (keep in mind, I am still somewhat of an amateur when using LaTeX so please bear with me!):
Proof. Suppose in order to derive a contradiction there are finitely many primes. That is, we have a complete list $p_1, \dots, p_n$. Let $p$ be the product of all the primes in this list i.e. $p = p_1 \cdots p_n$. Consider the number $$N = p + 1.$$ Since $N > p_i$ for all $i$, $1 \leq i \leq n$, there is no way $N$ can be any of the $p_i$. So $N$ must be composite. By the Fundamental Theorem of Arithmetic, $N$ is a product of primes. So there is a prime, say q, that divides $N$, and $p$ as well. So it follows this $q$ must also divide $$N - p = 1,$$ but this is impossible. No number, or prime, divides 1. Thus, contradicts the assumption that our list is complete and so there must be infinitely many primes.
QED
Any feedback would be appreciated. I always had trouble understanding this theorem and always forgot the "key argument". Now I feel like I finally get it... Hopefully. Thank you for reading! | {
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• $N$ need not be composite. $2\times 3+1=7$, for example. Nor do you need the FTA, which is a lot harder to prove than this. All you need to know is that every natural number greater than $1$ is divisible by some prime. – lulu Apr 28 '16 at 1:39
• On rereading, I see that you deduce the compositeness of $N$ by remarking that, were it prime, we'd already have the desired counterexample. Agreed. But, still, invoking the Fundamental Theorem of Arithmetic is not necessary. – lulu Apr 28 '16 at 1:43
• I see how it is unnecessary now. Thanks for reading! – o_o Apr 28 '16 at 1:47
• It is good. Petty comment, you say no number divides $1$. Well, $1$ divides $1$. However, it is certainly true that no number $\gt 1$ divides $1$, so no prime divides $1$. – André Nicolas Apr 28 '16 at 1:55
• Oops, that was silly (LOL). Good catch. – o_o Apr 28 '16 at 1:57
I would not add the complication of making this into a proof by contradiction. Euclid did not do it that way, despite many modern authors, dating back at least to Dirichlet in the middle of the 19th century, asserting that Euclid did it that way.
Start with any finite set $S$ of prime numbers. (For example, we could have $S=\{2, 31, 97\}$) Let $p = 1 + \prod S$, i.e. $1$ plus the product of the members of $S$.
Then $p$ cannot be divisible by any of the primes in $S$. Therefore either
• $p$ is itself prime, in which case there are more primes than those in $S$, or
• $p$ is divisible by some other primes not in $S$, in which case there are more primes than those is $S$.
Thus every finite set $S$ of primes can be extended to a larger finite set of primes.
(For example, if $S=\{5,7\}$, then $1+\prod S = 1 + 35 = 36 = 2\times2\times3\times 3$, and the additional primes not in $S$ are $2$ and $3$.)
The initial assumption that $S$ contains all primes is at best a needless complication that serves no purpose. | {
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• I am always very confused about the whole debate between contradiction proof versus non contradiction. Yes, this proof never states "assume there are finitely many primes," but it also never actually proves that there are infinitely many primes, merely that any finite set of primes can be expanded. From there, the proof goes "there are infinitely many primes because the number of primes cannot be equal to some finite number $N$ (via the construction)" To me, however, this final statement is essentially a contradiction statement (assume finite, that's impossible, so we get a contradiction. – ASKASK Apr 28 '16 at 1:59
• So to me, this proof is still a proof by contradiction, it's just that the actual contradiction part is postponed until after the construction – ASKASK Apr 28 '16 at 2:00
• @ASKASK : If you want to look at it that way, it is still better not to create an illusion that the assumption of finiteness of the set of ALL primes somehow plays a role in the construction. – Michael Hardy Apr 28 '16 at 2:01
• but doesn't the finiteness play a role? Finite number => finite set => construction of a prime not in that set. – ASKASK Apr 28 '16 at 2:03
• @ASKASK : Finiteness of the arbitrary finite set of primes plays a role. An assumption of finiteness of the set of ALL primes plays no role in the construction, even if it plays a different role of the kind you suggest. – Michael Hardy Apr 28 '16 at 2:04
Your proof looks good. It is not exactly how Euclid's proved it, but it works, and most importantly you understand it.
A couple of minor points:
Like lulu said in the comments, invoking the Fundamental Theorem of Arithmetic isn't necessary. Also in the end you state "No number, or prime, divides 1." This technically isn't true since 1 divides 1. It would be enough just to say "No prime divides 1." | {
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• Yes, I understand the minor points. My personal understanding was the goal, haha. Thank you for reading! :) – o_o Apr 28 '16 at 2:04
• In the context of Euclid, $1$ is not a number, so "No number divides $1$" would be correct. – Michael Hardy Apr 28 '16 at 2:05
• @MichaelHardy Can you elaborate? Does Euclid not consider 1 a number? – M47145 Apr 28 '16 at 2:09
• @M47145 : That has been my understanding. See this page: aleph0.clarku.edu/~djoyce/elements/bookVII/bookVII.html $\qquad$ – Michael Hardy Apr 28 '16 at 5:37 | {
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# Prove the Contraction Mapping Theorem.
Prove the Contraction Mapping Theorem.
Let $(X,d)$ be a complete metric space and $g : X \rightarrow X$ be a map such that $\forall x,y \in X, d(g(x), g(y)) \le \lambda d(x,y)$ for some $0<\lambda < 1$. Then $g$ has a unique fixed point $x^* \in X$, and it attracts everything, i.e. for any $x_0 \in X$ , the sequence of iterates $x_0, g(x_0), g(g(x_0))$, ... converges to the fixed point $x^* \in X$.
The hint I am given are for existence and convergence - prove that the sequence is Cauchy. For uniqueness, choose two fixed points of $g$ and apply the map to both.
Still a bit do not know how to proceed after looking at the hint. Could anyone help me based on those hints?
• Hint: $d(x_{n+1},x_n)=d(g(x_n),x_n)=d(g(g(x_{n-1})),g(x_{n-1}))\le \lambda d(g(x_{n-1}),x_{n-1})=d(x_n,x_{n-1})$ – J.R. Feb 8 '14 at 19:17
• @TooOldForMath: what are you trying to achieve here? – afsdf dfsaf Feb 8 '14 at 19:19
• I am giving you a hint so you can solve it. This is how it goes. Repeating what I wrote gives you $d(x_{n+1},x_n)\le \lambda^n d(x_1,x_0)$. Now $\lambda$ is between $0$ and $1$, so what happens when $n\rightarrow\infty$...? – J.R. Feb 8 '14 at 19:26
• why $\lambda d(g(x_{n-1}), x_{n-1}) = d(x_n, x_{n-1})$? – afsdf dfsaf Feb 8 '14 at 19:52
• $d(x_{n+1},x_n)\le \lambda d(x_n,x_{n-1}) \le \lambda^2 d(x_{n-1},x_{n-2})\le\cdots\le \lambda^n d(x_1,x_0)$ – J.R. Feb 9 '14 at 0:13
## 2 Answers
From
$$d(x_{n+1},x_n)=d(g(g(x_{n-1})),g(x_{n-1}))\le \lambda d(g(x_{n-1}), x_{n-1})=d(x_n,x_{n-1})$$
we get after $n$ applications of that inequality
$$d(x_{n+1},x_n)\le \lambda d(x_n,x_{n-1}) \le \lambda^2 d(x_{n-1},x_{n-2}) \le \cdots\le \lambda^n d(x_1,x_0)\tag{1}$$
Now we want to show that $(x_n)_n$ is a Cauchy sequence. So let $\epsilon>0$.
We assume $x_1\not=x_0$ (otherwise $x_0$ is already a fixed point). Set $c=d(x_1,x_0)>0$. | {
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We assume $x_1\not=x_0$ (otherwise $x_0$ is already a fixed point). Set $c=d(x_1,x_0)>0$.
Since $0<\lambda<1$, the sum $\sum_{n=0}^\infty \lambda^n$ converges (to $1/(1-\lambda)$). Therefore we can pick $N$ large enough such that
$$\sum_{k=n}^\infty \lambda^k<\frac{\epsilon}{c}$$
for all $n\ge N$.
Then for $m>n\ge N$ we have by the triangle inequality
$$d(x_m,x_n)\le \sum_{k=n}^{m-1} d(x_{k},x_{k+1})$$
Applying $(1)$ we obtain
$$d(x_m,x_n)\le c\sum_{k=n}^{m-1} \lambda^k\le c\sum_{k=n}^\infty \lambda^k<c\cdot\frac{\epsilon}{c}=\epsilon$$
So $(x_n)_n$ really is a Cauchy sequence. Since $(X,d)$ is complete, it converges to a limit $x\in X$.
By the equation $x_{n+1}=g(x_n)$, the limit satisfies $x=g(x)$, so it is a fixed point.
Uniqueness is trivial, let $y$ be another fixed point of $g$. Then
$$d(x,y)=d(g(x),g(y))\le \lambda d(x,y)$$
If now $x\not=y$, then $d(x,y)>0$, so we can divide by $d(x,y)$ to obtain $\lambda\ge 1$, a contradiction. Therefore, $x=y$. | {
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• "Therefore we can pick $N$ large enough such that $$\sum_{k=n}^\infty \lambda^k<\frac{\epsilon}{c}$$ for all $n\ge N$." For this part, do you backfill $$\frac{\epsilon}{c}$$ after going through the following: "Then for $m>n\ge N$ we have by the triangle inequality $$d(x_m,x_n)\le \sum_{k=n}^{m-1} d(x_{k},x_{k+1})$$ Applying $(1)$ we obtain $$d(x_m,x_n)\le c\sum_{k=n}^{m-1} \lambda^k\le c\sum_{k=n}^\infty \lambda^k<c\cdot\frac{\epsilon}{c}=\epsilon$$"? – afsdf dfsaf Feb 9 '14 at 17:18
• Backfill? What are you talking about? This is a logically complete argument. – J.R. Feb 9 '14 at 17:25
• I just mean whether we get to go through the following part: "Then for $m>n\ge N$ we have by the triangle inequality $$d(x_m,x_n)\le \sum_{k=n}^{m-1} d(x_{k},x_{k+1})$$ Applying $(1)$ we obtain $$d(x_m,x_n)\le c\sum_{k=n}^{m-1} \lambda^k\le c\sum_{k=n}^\infty \lambda^k<c\cdot\frac{\epsilon}{c}=\epsilon$$" before filling $$\frac{\epsilon}{c}$$ in the "Therefore we can pick $N$ large enough such that $$\sum_{k=n}^\infty \lambda^k<\frac{\epsilon}{c}$$ If not, how did you get $$\sum_{k=n}^\infty \lambda^k<\frac{\epsilon}{c}$$? – afsdf dfsaf Feb 9 '14 at 17:36
• Please stop copying all this stuff in the comments. It makes it really unreadable. You are not "filling" $\epsilon/c$ whatever you mean by that. The logic is like this: $c$ is a constant defined as $d(x_1,x_0)$. It is fixed. Now I throw you an $\epsilon>0$ in the beginning. And now you notice, aha, the sum $\sum_{k=1}^\infty \lambda^k$ converges therefore sequence of truncated sums $\sum_{k=n}^\infty \lambda^k$ converges to $0$ for $n\rightarrow\infty$. So it will eventually be smaller than that given $\epsilon/c$. Let us say that the point when it becomes smaller is reached at $N$ – J.R. Feb 9 '14 at 17:42
• $\sum_{k=n}^\infty \lambda^k <\epsilon/c$ for all $n$ bigger than $N$. – J.R. Feb 9 '14 at 17:43
One proviso: You do need that $X$ is non-empty in the statement. | {
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One proviso: You do need that $X$ is non-empty in the statement.
To elaborate on the hints you have been given; To prove existance, pick $x_0 \in X$, and call (for convenience of writing) $g(x_0) = x_1, g(x_1) = x_2$ etcetera. Let $d(x_0,x_1) = d$. Then $d(x_1,x_2) \leq \lambda d$, and $d(x_2,x_3) \leq \lambda^2 d$...can you see how to extend this to show that the sequence is Cauchy?
For uniqueness, suppose $x$ and $y$ are both fixed points. What is $d(g(x),g(y))$?.
• I don't think you do - if $x = y$ then both sides of the inequality are $0$, which is fine. – meta Feb 8 '14 at 19:37
• Touche! (I'm used to strong inequalities being used, which isn't really different here, and didn't notice the difference.) – Jonathan Y. Feb 8 '14 at 19:39
• @meta: After we got to $d(x_{n+1}, x_n) \le \lambda^n d(x_1, x_0)$, then assuming that $m > n$, $d(x_m, x_n) \le d(x_m, x_{m-1}) + d(x_{m-1}, x_{m-2}) + ... + d(x_{n+1}, x_{n})$ since each term of the right hand side is 0 ...so if we add up all the 0 terms, we get 0 on the right hand side. Therefore, $d(x_m, x_n)$ is 0. If this is right, then the question I have here is how do I guarantee that $x_{m-1} > x_n$? – afsdf dfsaf Feb 9 '14 at 3:04 | {
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# Is the zero matrix upper and lower triangular as well as diagonal?
From what I can tell from the definitions of a lower-triangular, upper-triangular, and diagonal matrices, I've come to the conclusion that the zero matrix is in the set of all of each type of matrix. Is this correct?
• Yes ${}{}{}{}{}{}$. – copper.hat Feb 14 '13 at 20:09
A zero square matrix is lower triangular, upper triangular, and also diagonal.
If an object meets the definition of three things then it is the three things. What are you confused about?
• The definition of those three things, for instance take the definition provided here: mathworld.wolfram.com/LowerTriangularMatrix.html I didn't find it particularly clear. The stumbling block was on could a_{ij} = 0. Apparently the answer is yes, yes it can. – BrotherJack Feb 14 '13 at 20:19
• This is actually a good answer. An object or something is not necessarily of one type. If it fits the definition, you go with it, unless something contradictory may appear or several definitions may exclude some parts of each other. – Turkhan Badalov Feb 13 '18 at 16:13
Provided it is a square matrix.
An upper triangular matrix is one in which all entries below main diagonal are zero. Clearly this is satisfied.
An lower triangular matrix is one in which all entries above main diagonal are zero. Clearly this is satisfied.
An diagonal matrix is one in which all non-diagonal entries are zero. Clearly this is also satisfied.
Hence, a zero square matrix is upper and lower triangular as well as diagonal matrix. | {
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# Proving $n^3 + 3n^2 +2n$ is divisible by $6$
The full question is: Factorise $n^3 + 3n^2 + 2n$. Hence prove that when $n$ is a positive integer, $n^3 + 3n^2 + 2n$ is always divisible by $6$.
So i factorised and got $n(n+1)(n+2)$ which i think is right? I'm not sure how to actually prove this is divisible by $6$ though. Thanks for help and i apologise if someone has already asked this, i couldn't find it. Also i've not been told whether i have to do induction or not.
• Hint: an integer is divisible by 6 if and only if it is divisible by both 2 and 3. Can you show that your expression is even? Can you show it is divisible by 3? – lulu Jul 13 '15 at 14:28
• Fermat's little theorem can be applied. $(n^3-n)+3(n^2+n)$ is divisible by $3$. $n(n^2-n)+2(2n^2+n)$ is divisible by $2$. – user26486 Jul 13 '15 at 14:43
$n^3+3n^2+2n=(n+1)(n+2)n$. One of the factors must be even and one must be a multiple of three. Hence the product is a multiple of both $2$ and $3$ and hence is divisible by the least common multiple of $2$ and $3$, which turns out to be $6$.
The product of three consecutive integers is divisible by $2$ since one of the factors need to be even. It is also divisible by $3$. (why?) The the product of three consecutive integers is then divisible by the least common multiple of those two numbers $2\times 3 =6$.
Alternatively, note that $$n(n+1)(n+2) = 6 {n+2 \choose 3}$$
$$n^3+3n^2+2n = 6\binom{n+2}{3} \in 6\mathbb{Z}.$$ | {
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$$n^3+3n^2+2n = 6\binom{n+2}{3} \in 6\mathbb{Z}.$$
• Whoa, i don't even know what ∈ means, but i'm guessing can be divided by 6. – Callum Hemsley Jul 13 '15 at 14:33
• $x \in A$ means "x is a member of the set A". $6 \mathbb{Z}$ is the set of integer multiples of 6. Anyway, this is more or less just stating the conclusion, since it requires you to know that $(n+2)(n+1)n$ is divisible by $6$ in order to prove that ${n+2 \choose 3}$ is an integer in the first place. But it does give some nice generality: in general $n(n+1)\dots(n+k)$ is divisible by $(k+1)!$. – Ian Jul 13 '15 at 14:34
• @Ian Ah okay, thanks for clarifying. – Callum Hemsley Jul 13 '15 at 14:35
• The benefit of this solution is that it helps in proving $(n+1)(n+2)(n+3)(n+4)$ is a multiple of $24$. – Jorge Fernández Hidalgo Jul 13 '15 at 14:40
• @dREaM The same basic argument you applied in your answer can be used to prove it. At least one of $n+1,n+2,n+3,n+4$ is a multiple of $4$, at least one is even but not a multiple of $4$, at least one is divisible by $3$. – user26486 Jul 13 '15 at 14:49
Using only modular arithmetic, without factoring, you can see that with $p(n)=n^{3}+3 n^{2}+2 n$ we have if $n\cong0(mod 2)$ then $p(n)\cong0+0+0=0 (mod 2)$ if $n\cong1(mod 2)$ then $p(n)\cong1+3+2\cong0(mod 2)$ So $2|p(n)$. Similarly if $n\cong1(mod 3)$ then $p(n)\cong1+3+2\cong0(mod 3)$, if $n\cong2(mod 3)$ then $p(n)\cong8+12+4\cong0(mod 3)$ so $3|p(n)$. Since both 2 and 3 divide $p(n)$ then $p(n)$ is a multiple of 6.
Keep going with your idea. $n(n+1)$ is the product of two consecutive integer so one of them is even. If $n$ is even we are done and if not $n+1$ is and the product is therefore divisible by $2$.
Similarly $n(n+1)(n+2)$ is the product of three consecutive integers so one of them is divisible by $3$. Let's test the residues modulo $3$
$$\begin{array}{c | c c c} n & n+1 & n+2 & n(n+1)(n+2)\\ \hline 0 & 1 & 2 & 0\\ 1 & 2 & 0 & 0\\ 2 & 0 & 1 & 0 \end{array}$$ | {
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So the product is also divisible by $3$ and therefore the product $n(n+1)(n+2)\equiv 0\pmod 6$
One could also test directly the congruences modulo $6$
$$\begin{array}{c|c c c c} n & n^3 & 3n^2 & 2n & n^3+3n^2+2n\\ \hline 0 & 0 & 0 & 0 & 0\\ 1 & 1 & 3 & 2 & 0\\ 2 & 2 & 0 & 4 & 0\\ 3 & 3 & 3 & 0 & 0\\ 4 & 4 & 0 & 2 & 0\\ 5 & 5 & 3 & 4 & 0 \end{array}$$
And we have proven that $n^3+3n^2+2n\equiv 0\pmod 6$
• I think i understand, so essentially $n(n+1)$ has to have an even number, and $n(n+1)(n+2)$ must be divisible by $3$.Then it's the lowest common multiple of these two combined? I just struggle to understand the concept of actually "proving" it. – Callum Hemsley Jul 13 '15 at 14:39
• I edited the answer to incorporate more details – marwalix Jul 13 '15 at 16:29 | {
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# How to obtain the mass of a particle rotating around an axis when the difference of tensions in a wire are known?
#### Chemist116
The problem is as follows:
A particle of mass $m$ is tied to a very thin wire. Assume the wire is inflexible and of negligible mass. The particle is spinning about a fixed axis as shown located in the center of the circle. Let $T_a$ and $T_b$ be the modulus of the tensions in the string when the particle is located in the points $a$ and $b$ respectively. Find the mass $m$ of the particle if the difference between the tensions $T_{b}-T_{a}=39.2\,N$. Assume $g=9.8\frac{m}{s^2}$
The alternatives are as follows:
$\begin{array}{ll} 1.&0.8\,kg\\ 2.&2.0\,kg\\ 3.&5.8\,kg\\ 4.&1.0\,kg\\ 5.&0.5\,kg\\ \end{array}$
For this problem what I've attempted to do was to use the force at the top to be as follows:
$T+mg=\frac{mv_a^2}{R}$
At this point the Tension must be zero (I don't know if this statement is correct.)
This reduces the top equation to:
$v_a=\sqrt{Rg}$
Then to obtain the speed in the lowest point would be by the conservation of mechanical energy:
$E_i=E_f$
$\frac{1}{2}mv^2_{a}+mgR=mg(-R)+\frac{1}{2}mv_{b}^2$
Then inserting in the above equation would give the speed for the bottom:
Cancelling masses and multiplying by $2$ to both terms:
$v^2_{a}+4gR=v^2_{b}$
Since it is known $v_{a}$ then:
$v_{b}^2=Rg+4Rg=5Rg$
Finally I'll use these in the given statements:
The tension in the top:
$T_a+mg=\frac{mv_{a}^2}{R}$
Tension in the bottom.
$T_b-mg=\frac{mv_{b}^2}{R}$
Doing a difference between these:
$T_b-T_a=2mg+\frac{mv_{b}^2}{R}-\frac{mv_{a}^2}{R}$
Replacing the known values:
$T_b-T_a=2mg+\frac{m(5Rg)}{R}-\frac{m(Rg)}{R}$
$T_b-T_a=2mg+4mg=6mg$
Then:
$39.2=6m(10)$
Which results into:
$m=0.67\,kg$
But the answers sheet indicates that the mass is $2\,kg$. | {
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Which results into:
$m=0.67\,kg$
But the answers sheet indicates that the mass is $2\,kg$.
For doing that what it should happenned is that the "$2mg$" is negative in the right side of the equation but I can't find a way to do that. Can someone help me here?. Is it me?, or did I overlooked anything?. Help!. Please.
#### skeeter
Math Team
I agree with your solution ...
$T_b-T_a = \dfrac{m}{r}(v_b^2-v_a^2) + 2mg$
using energy conservation yields $v_b^2-v_a^2= 4rg$ | {
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# Question on coin tossing - probability
When considering an infinite sequence of tosses of a fair coin, how long will it take on an average until the pattern H T T H appears?
I tried to break the problem into cases where ultimately the pattern HTTH appears, but that makes things complex. Any insight would be helpful.
• HTTH or HHTH? In fact the answer to both is $18$, even though for HHHT it would be $16$ and for HTHT it would be $20$ and for HHHH it would be $30$ – Henry Sep 20 '18 at 8:17
(I'm working with HTTH as desired pattern.) Distinguish the five states $$s_0=\emptyset,\quad s_1=H,\quad s_2=HT,\qquad s_3=HTT,\quad s_4=HTTH\ ,$$ whereby the written letters denote the last tosses insofar as useful. Let $E_k$ $(0\leq k\leq4)$ be the expected number of additional tosses when you are in state $s_k$. Then of course $E_4=0$. Furthermore we have the following equations: $$E_0=1+{1\over2}E_0+{1\over2}E_1,\quad E_1=1+{1\over2}E_1+{1\over2}E_2,\quad E_2=1+{1\over2}E_1+{1\over2}E_3,\quad E_3=1+{1\over2}E_0\ .$$ Now solve this system; then $E_0$ is the solution to your problem.
• Beautiful. One could have understand it easier if you realize the whole thing is a Markov chain. – William Wong Sep 20 '18 at 21:24
I like Christian's answer better as it is simple. But I worked out another solution so I just added it for reference.
Let:
$H_n$ be the event of no "HTTH" in n flips and ends with "H"
$T_n$ the event of no "HTTH" in n flips and ends with "T"
$p_n = \text{Prob}(H_n)$
$q_n = \text{Prob}(T_n)$
So, $p_1=p_2=p_3=q_1=q_2=q_3=1/2$
It is not hard to verify $\left[ \begin{array}{c} p_{n+1}\\q_{n+1} \end{array}\right] = A \left[ \begin{array}{c} p_{n}\\q_{n} \end{array}\right]$, where the matrix $A = \left[\begin{array}{cc}\frac{1}{2} & \frac{3}{8}\\\frac{1}{2} & \frac{1}{2} \end{array} \right]$.
The desired expectation is given by | {
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The desired expectation is given by
$E = \left[0, \frac{1}{8}\right] \left\{\sum_{n=4}^{\infty} nA^{n-4}\right\}\left[ \begin{array}{c} p_{n}\\q_{n} \end{array}\right]$
One can further prove the matrix power series $\sum_{n=4}^{\infty}nA^{n-4} = (4I-3A)(I-A)^{-2} = \left[\begin{array}{cc}136 & 114\\ 152 & 136 \end{array} \right]$
Therefore $E = \frac{1}{16} (152+136) = 18$
• Forget this. I think this solution is still wrong. – William Wong Sep 20 '18 at 21:23
Here's yet another way of doing this.
We can be in one of $$5$$ states, according to the number $$0\le k\le4$$ of consecutive “correct” results that we currently have; e.g. $$k=2$$ means the last two results were HT. Let's make a graph with the transitions between these states:
The labels on the vertices are the values of $$k$$. We'll get to the edge weights in a bit. In addition to the two possible transitions from each of the nodes with $$0\le k\le3$$, I added an edge from $$4$$ to $$0$$ to complete a round trip from $$0$$ to $$4$$ and back.
The idea is to choose the edge weights such that at each vertex all outgoing edges have the same weight and the sums of incoming and outgoing edge weights are equal. For instance, start at $$3$$ and arbitrarily assign weight $$1$$ to its outgoing edges; then work your way around the graph, e.g. it follows that the edge $$4\to0$$ also has weight $$1$$, the edge $$2\to3$$ has weight $$2$$, thus the edge $$2\to1$$ also has weight $$2$$, and so on. You barely have to compute anything.
Now consider a process on the edges where in each step we go from the current edge to its terminal vertex and then uniformly randomly choose one of its outgoing edges. Then the equilibrium distribution on the edges is given by the edge weights, normalized by their sum, since by construction that lets the same probability flow into and out of each vertex in each step. | {
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We're looking for the expected time to reach $$4$$ from $$0$$. After we reach $$4$$, we always go to $$0$$ in a single step. Thus, the time we want is $$1$$ less than the expected time for a round trip from $$0$$ to $$4$$ and back. On such a round trip, we spend exactly $$1$$ step on the edge $$4\to0$$. But then the expected time for the round trip must be the reciprocal of the equilibrium probability of being on edge $$4\to0$$. Since that edge conveniently has weight $$1$$, the expected time for the round trip is just the sum of the edge weights, which is $$19$$. Thus the expected time to get from $$0$$ to $$4$$ is $$1$$ less, $$18$$. | {
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# Finding the minimum wins in a round-robin tournament.
There are 16 teams in total. They are divided into two groups of 8 each. In a group, each team plays a single match against every other team. At the end of the round, top 4 teams go through to the next round from each group. Assume, a match can only result in either a win or a loss and never a tie.1 win=1 point; 1 loss=0 point.
Questions
1.What is the minimum no. of wins needed for a team in the first stage to guarantee its advancement to the next round?
2.What is the highest no. of wins for a team in the first round in spite of which it would be eliminated at the end of the first stage?
3.What is the minimum number of wins that may allow advancement to the second round?
The way i approached the question is like this:
There will be 28 matches each within the two teams. After that I started distributing the points by hit and trial. I know that there are 28 match points. But, say for question 1, can we say that $28=5*5+3$. So 5 teams with 5 winning each creating ambiguity? So, the minimum possible wins that will always guarantee is $5+1=6$. wins. But then I started distributing match points to the teams and found 5 points 5 times is not possible. How to get the answer logically and quickly?
PS: The answer given is 6(Q 1) and 2 (Q 3).
Your approach to the first question is entirely reasonable. Since $5\cdot6=30>28$, it’s clear that $6$ wins guarantee advancement. To show that $5$ do not, we need only show that it’s possible for $5$ teams to win $5$ matches each; the arrangement shown by user73985 works fine and is natural enough that it’s the first one that I found as well. (You can complete it by distributing the remaining $3$ wins within the group consisting of teams $6,7$, and $8$; for example, you could have team $6$ beat teams $7$ and $8$, and team $7$ beat team $8$.) | {
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For the second question, suppose that a team has $3$ wins; is it possible that it could advance to the second round? Suppose that teams $1,2$, and $3$ beat each of the higher-numbered teams: team $1$ beats everybody else, team $2$ beats everyone else except team $1$, and team $3$ beats everyone else except teams $1$ and $2$. That accounts for $18$ wins, leaving $10$ to be determined. Team $4$ beats teams $5,6$, and $7$, for $3$ wins, leaving $7$ wins to still to be distributed amongst teams $5,6,7$, and $8$. We want team $4$, with its measly $3$ wins, to make it to the second round, so we’ll try to split the remaining $7$ wins $2,2,2$, and $1$ amongst teams $5,6,7$, and $8$. Team $4$ did not beat team $8$, so team $8$ beat team $4$ and already has one win; we can give it another against team $5$. Now it must lose to teams $6$ and $7$, so they now have one win apiece. We can finish up by letting team $5$ beat team $6$, team $6$ beat team $7$, and team $7$ beat team $5$. To sum up:
• Team $1$ beats teams $2,3,4,5,6,7,8$.
• Team $2$ beats teams $3,4,5,6,7,8$.
• Team $3$ beats teams $4,5,6,7,8$.
• Team $4$ beats teams $5,6,7$.
• Team $5$ beats team $6$.
• Team $6$ beats teams $7,8$.
• Team $7$ beats teams $5,8$.
• Team $8$ beats teams $4,5$.
Thus, it’s possible to make it into the second round with just $3$ wins.
With only $2$ wins, however, it’s impossible to get to the second round. The top three finishers cannot have more than $18$ wins altogether (either $7+6+5$ or $6+6+6$). That leaves $10$ wins amongst the remaining $5$ teams, so either each of the bottom $5$ teams has $2$ wins, or one of them has at least $3$ wins. In neither case is a team with just $2$ wins assured of getting into the second round. | {
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There is some ambiguity in the situation in which each of the bottom $5$ teams wins $2$ matches. (This can happen, e.g., if teams $1,2$ and $3$ all beat each of teams $4,5,6,7$, and $8$, team $4$ beats teams $5$ and $6$, team $5$ beats teams $6$ and $7$, team $6$ beats teams $7$ and $8$, team $7$ beats teams $8$ and $4$, and team $8$ beats teams $4$ and $5$.) In this case the rules as given in the question don’t specify what happens. Some of the possibilities are: that only the top three teams go to the second round; that there is a playoff for the fourth position; that the fourth position is chosen randomly; or that the fourth position is decided by some tie-breaker like goal differential. As long as four teams always move on to the second round, it’s still possible for a team with just $2$ wins in the first round to move on, but if that does happen, other teams with $2$ wins fail to move on.
• "With only 2 wins, however, it’s impossible to get to the second round" - this phrasing seems to contradict the last sentence of the answer, which says "it’s still possible for a team with just 2 wins in the first round to move on". As your answer explains, it's possible for a 2-wins team to progress, but it is not guaranteed. My suggested edit was rejected, but I believe it should read something similar to "with only 2 wins, however, it’s not possible to guarantee progression to the second round." – Spinner May 27 at 13:32
5 points 5 times is possible. Call the teams 1, ..., 8. Teams 1-5 all beat all of teams 6-8, and then:
Team 1 beats teams 2 and 3
Team 2 beats teams 3 and 4
Team 3 beats teams 4 and 5
Team 4 beats teams 5 and 1
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Differentiable function, not constant, $f(x+y)=f(x)f(y)$, $f'(0)=2$
Let $f: \mathbb R\rightarrow \mathbb R$ a derivable function but not zero, such that $f'(0) = 2$ and $$f(x+y)= f(x)\cdot \ f(y)$$ for all $x$ and $y$ belongs $\mathbb R$. Find $f$.
My first answer is $f(x) = e^{2x}$, and I proved that there are not more functions like $f(x) = a^{bx}$ by Existence-Unity Theorem (ODE), but I don't know if I finished.
Thanks,
-
– Robert Israel Jul 26 '12 at 20:42
Differentiate $f(x+y)=f(x)\cdot f(y)$ with respect to $y$ to obtain $$f'(x+y)=f(x)\cdot f'(y).$$ Now by letting $y=0$ and noting that $f'(0)=2,$ we obtain $$f'(x)=2f(x).$$
The solutions to the preceding equation are of the form $f(x)=Ce^{2x}$ for some constant $C$. Using the fact that $f'(0)=2$, we find that $f'(0)=2C=2$, so that $C=1$. Hence $f(x)=e^{2x}.$ It is easy to check that this does indeed satisfy the original functional equation.
-
$f(0)=(f(0))^2 \Rightarrow f(0) \in \{0,1\}$. If $f(0)=0$, then $f(x) =0$ for every $x$, therefore $f(0)=1$.
For every $x \in \mathbb{R}$ one has $f(x)f(-x)=f(0)=1$ and $f(x)=(f(x/2))^2$, i.e. $f(x)>0$ for every $x \in \mathbb{R}$.
By induction one has $f(nx)=(f(x))^n$ for every $n \in \mathbb{N}$. Since $f(x)f(-x)=f(0)=1$, one has $f(-nx)=(f(nx))^{-1}=(f(x))^{-n}$ for every $n \in \mathbb{N}$. Hence $f(nx)=(f(x))^n$ for every $n \in \mathbb{Z}$.
For every $n \in \mathbb{Z}\setminus\{0\}$ one also has $(f(x/n))^n=f(x)$, so $f(x/n)=(f(x))^{1/n}$. Setting $a=f(1)$ one has, for every $m/n \in \mathbb{Q}$: $$f(m/n)=(f(m))^{1/n}=[(f(1))^m]^{1/n}=a^{m/n}.$$ Since $\mathbb{Q}$ is dense in $\mathbb{R}$, given $x \in \mathbb{R}$, there is a sequence $(x_k) \subset \mathbb{Q}$ such that $x_k \to x$ as $k \to \infty$, and by continuity one has $$f(x)=\lim_kf(x_k)=\lim_k a^{x_k}=a^x.$$ One has $f'(x)=a^x\ln a$ for every $x$, and $2=f'(0)=\ln a$, i.e. $a=e^2$. Thus $f(x)=e^{2x}$ for every $x \in \mathbb{R}$.
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To find the equation of sine waves given the graph: Find the amplitude which is half the distance between the maximum and minimum. The sum of the cosine and sine of the same angle, x, is given by: [4.1] We show this by using the principle cos θ=sin (π/2−θ), and convert the problem into the sum (or difference) between two sines. x − This must be a numeric value.. Return Value. sin (x) = cos (90 -x) [within first quadrant] 0 0 The Lesson: y = sin(x) and y = cos(x) are periodic functions because all possible y values repeat in the same sequence over a given set of x values. Example 26. Here’s how to prove this statement. Following is the syntax for cos() method −. When finding the equation for a trig function, try to identify if it is a sine or cosine graph. See Example. We note that sin π/4=cos π/4=1/√2, and re-use cos θ=sin (π/2−θ) to obtain the required formula. Understanding how to create and draw these functions is essential to these classes, and to nearly anyone working in a scientific field. Teacher was saying that in right triangles the sine of one acute angle is the cosine of the other acute angle. It is easy to memorise the values for these certain angles. From this information, we can find the amplitude: So our function must have a out in front. The second one, y = cos( x 2 + 3) , means find the value ( x 2 + 3) first, then find the cosine of the result. Python number method cos() returns the cosine of x radians.. Syntax. However, scenarios do come up where we need to know the sine and cosine of other angles. The first one, y = cos x 2 + 3, or y = (cos x 2) + 3, means take the curve y = cos x 2 and move it up by 3 units. You want to show that the sine function, slid 90 degrees to the left, is equal to the cosine function: Replace cos x with its cofunction identity. Find $$\cos(20^\circ)$$ and $$\sin(20^\circ)\text{. When the sine or cosine is known, we can use the Pythagorean Identity to find the other. I think I am a very visual learner and I always found | {
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the Pythagorean Identity to find the other. I think I am a very visual learner and I always found that diagrams always made things clearer for my students. cos(x) Note − This function is not accessible directly, so we need to import math module and then we need to call this function using math static object.. Parameters. Trig calculator finding sin, cos, tan, cot, sec, csc To find the trigonometric functions of an angle, enter the chosen angle in degrees or radians. The Pythagorean Identity is also useful for determining the sines and cosines of special angles. See Example. The shifted sine graph and the cosine graph are really equivalent — they become graphs of the same set of points. To find the cosine and sine of angles that are not common angles on the unit circle, we can use a calculator or a computer. Begin by realizing we are dealing with a periodic function, so sine and cosine are your best bet. The sine and cosine values are most directly determined when the corresponding point on the unit circle falls on an axis. Description. }$$ When we find sin cos and tan values for a triangle, we usually consider these angles: 0°, 30°, 45°, 60° and 90°. Sum Introduction: In this lesson, the period and frequency of basic graphs of sine and cosine will be discussed and illustrated. Underneath the calculator, six most popular trig functions will appear - three basic ones: sine, cosine and tangent, and … The sine and cosine functions appear all over math in trigonometry, pre-calculus, and even calculus. Next, note that the range of the function is and that the function goes through the point . The “length” of this interval of x … Find An Equation For The Sine Or Cosine Wave. | {
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# Is the sum of digits of $\left(16^k - 1\right)$ less than $6k$ for $k > 223$?
I had been researching the OEIS sequence A165722, which is the sequence of positive integers $k$ such that the sum of digits of $\left(16^k - 1\right)$ is equal to $6k$. I used computational power to determine that the sum of digits is less than $6k$ for $223 < k < 10^6$. I made a conjecture that $6k$ would continue to grow at a faster rate than the digit sum, and thus the sequence is finite. I and several others, however, were unsure how one would go about proving this. I thought that perhaps there would be some way to show a regularity in the digits of $16^k - 1$, but I would not know how to go about finding or proving this regularity. | {
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• I don't think that there is hope to find such a regularity. Besides the few last digits, the digits will probably behave like a random sequence. We can expect that the average digit does not exceed $4.5$ significant, but we would need an average digit of about $4.98$ to have $6k$ digits or more. So, the digitsum should always be less than $6k$ for $k>223$. I would be very surprised if someone can prove this. – Peter Apr 29 '18 at 7:29
• Legendre's formula might be a good starting: if $\nu_q(n)$ denotes the sum of digits in the $q$-ary expansion of $n$, then $$\nu_q(n) = n - (q-1)\sum_{j=1}^{\infty} \left\lfloor \frac{n}{q^j} \right\rfloor = (q-1)\sum_{j=1}^{\infty} \frac{n\text{ mod }q^j}{q^j}.$$ But again, this leads us to an intricate world of multiplicative groups and I have no idea. I would also be very surprised if someone can prove this. – Sangchul Lee May 1 '18 at 4:26
• I don't know if this applies, but perhaps you could use Benfords law to do a statistical/probabilistic analysis of the digits of $16^k-1$ and then show that the probability of getting a number whose digit sum is equally to 6k is greater than 0. Once again though, I'm not sure if Benford's law applies. Perhaps you could find a distribution for the digits in base 16. – Saudman97 May 2 '18 at 21:24
• Here's a proof of a very weak lower bound, showing that the sum of the base-$10$ digits of $2^n$ is $\Omega(\log n)$: oeis.org/A001370/a001370_1.pdf – mjqxxxx May 3 '18 at 0:21
• This somewhat related question may be of interest. I hastily misread it, and thought it came close to proving yours. It doesn't, but it may be of interest anyway. – Jyrki Lahtonen May 5 '18 at 18:29
Some preliminary estimates. From this book, page 79 (accessible in preview mode) $$S(16^n-1)=16^n-1 - 9\sum\limits_{k\geq1} \left \lfloor \frac{16^n-1}{10^k} \right \rfloor \tag{1}$$ | {
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Further $$S(16^n-1)= 16^n-1 - 9\sum\limits_{k\geq1} \left(\frac{16^n-1}{10^k}-\left\{\frac{16^n-1}{10^k}\right\}\right)=\\ 16^n-1 - 9\sum\limits_{k\geq1} \frac{16^n-1}{10^k}+9\sum\limits_{k\geq1}\left\{\frac{16^n-1}{10^k}\right\}=\\ (16^n-1)\left(1 - 9\sum\limits_{k\geq1} \frac{1}{10^k}\right)+9\sum\limits_{k\geq1}\left\{\frac{16^n-1}{10^k}\right\}=\\ (16^n-1)\left(1 - 9\left(\frac{10}{9}-1\right)\right)+9\sum\limits_{k\geq1}\left\{\frac{16^n-1}{10^k}\right\}= 9\sum\limits_{k\geq1}\left\{\frac{16^n-1}{10^k}\right\}$$ Or $$S(16^n-1)=9\sum\limits_{k\geq1}\left\{\frac{16^n-1}{10^k}\right\}\tag{2}$$
The last digit of $16^n-1$ is always $5$ and $$\left\{\frac{16^n-1}{10^k}\right\}=0.\overline{a_1a_2...a_{k-1}5} \leq 0.99..95<1$$ $9$ repeated $k-1$ times. But only for the first $n\log_{10}16$ terms. For all $k>n\log_{10}16$ $$\left\{\frac{16^n-1}{10^k}\right\}\leq 0.00..099..95$$ where $00..099..9$ is of length $k-1$. Basically, starting with $k\geq \left \lfloor n\log_{10}16 \right \rfloor+1$ this tail forms an infinite geometric progression with ratio $\frac{1}{10}$ which sums to a constant. So we can conclude $$S(16^n-1) < 9n\log_{10}16 + C$$ We also have that $9\log_{10}16<11$, thus $$S(16^n-1) < 11n+ C \tag{3}$$ Probably, with more accurate calculations, a better estimate may be obtained ... work in progress. | {
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• This is very nice. I do know that for some values, $S(16^n - 1) \geq 6n$, but the last apparent occurrence of this is when $n=223$. I am concerned about values above this. This is a great start though. – Kirk Fox May 4 '18 at 18:47
• This bound is fairly trivial... the number $16^k-1$ has no more than $1 + \log_{10}(16^k-1) < 1 + k \log_{10}16$ digits, each of which is no greater than $9$, so the digit sum is no greater than $9 + 9k\log_{10}16 < 10.85k + 9.$ – mjqxxxx May 4 '18 at 23:11
• @mjqxxxx 1st of all to reveal a formula and secondly to engage more people in finding a better estimate. The weak part in the subsequent calculations is of course the gross estimate of $0.\overline{a_1a_2...a_{k-1}5} \leq 0.99..95<1$, maybe somebody could find a finer pattern? It matches your calculations, but it has, imo, more potential ... – rtybase May 4 '18 at 23:30
• Knowing the fractional part of $16^n-1$ over $10^k$ is the same as knowing the digits of $16^n-1$ since dividing by $10^k$ just shifts it, preserving the significant digits. This work can also be skipped using the trivial bound: $16^n-1$ has $\log_{10}(16^n-1)\approx n\log_{10}(16)$ digits in decimal, and each digit can be at most $9$ giving the bound $9n\log_{10}(16)$ as you have. – Will Fisher May 7 '18 at 0:53 | {
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Let $S\left(n\right)$ be the sum of the digits of $n$. Since $16^k-1$ is divisible by 5 but not 2, the last digit is five, and hence $$S\left(16^k-1\right)=S\left(16^k\right)-1$$ Essentially we want to say that $16^k=2^{4k}$ typically has digits less than 5 rather than digits more than five, or at least it doesn't mostly have digits more than five. One thought that pops out is that the map $k\mapsto2k\textrm{ mod }10$ is periodic for even $k$ with period 4 (we have the cycle $2\mapsto4\mapsto8\mapsto6\mapsto2$), and for odd $k$ of course the next term is even. Except for when there is carrying, this map is the map taking digits of $2^k$ to $2^{k+1}$. Could this periodicity which matches $2^{4k}$ explain it? | {
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# The Complex Field
1. Oct 15, 2009
### jgens
Would the following prove that the set of complex numbers do not form and ordered field?
Clearly $i \neq 0$. Therefore, if the complex numbers form an ordered field either $i > 0$ or $i < 0$. Suppose first that $i > 0$, then $i^2 = -1 > 0$, a contradiction. Now suppose that $i < 0$, then $i^2 = -1 > 0$, another contradiction. Thus, the set of complex numbers do not form an ordered field.
This seems awfully fishy and I wouldn't be surprised to find that it's completely invalid. Feedback and suggestions are welcome. Thanks!
2. Oct 15, 2009
### pbandjay
You have i2 = -1 > 0, but -1 < 0.
3. Oct 15, 2009
### jgens
Yes, that's why $i > 0$ and $i < 0$ are contradictions as I state in my original post.
4. Oct 15, 2009
### Staff: Mentor
I understand the logic behind what you're saying, but you could say it in a way that is clearer.
Suppose that i > 0.
Multiplying by a positive number preserves the direction of the inequality, so i2 must be positive. This is incorrect, though, because i2 = - 1 < 0. Thus, assuming i > 0 leads to a contradiction.
And similar for the assumption that i < 0.
5. Oct 15, 2009
### jgens
Thanks for the suggestion! It seems like I always need to add details to my proofs in order to make them clearer. Does the method I'm using work?
6. Oct 16, 2009
### HallsofIvy
What you have proven is that the usual order on the real numbers, in which -1< 0, cannot be extended to the complex numbers.
But you can prove more. Can there exist some order? Perhaps one in which 1< 0and -1> 0?
Suppose such an order existed. The i is not 0 (the additive identity does not depend upon the order). If i> 0 then it follow that (i)(i)= -1> 0. From that i(-1)= -i> i(0)= 0. Now that contradicts i>0. If i< 0, the -i> 0 so (-i)(i)= 1< 0. From that i(1)= i< i(0)= 0 and that contradicts i< 0.
7. Oct 16, 2009
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7. Oct 16, 2009
### qspeechc
Ok, that proves that there can be no order where 1<0 and -1>0. How does one show that there can not be any order relation defined on the complex numbers? Or R^n for n>1 for that matter?
8. Oct 16, 2009
### Ben Niehoff
Let $z_1 = r_1 e^{i\theta_1}$ and $z_2 = r_2 e^{i\theta_2}$, where $\theta$ is restricted to be in $[0, 2\pi)$. Then define > as follows:
1. If $r_1 > r_2$, then $z_1 > z_2$
2. If $r_1 = r_2$ and $\theta_1 > \theta_2$, then $z_1 > z_2$
3. If $r_1 = r_2$ and $\theta_1 = \theta_2$, then $z_1 = z_2$
This looks like an order relation to me...it even has a least element, 0. Now, I know that C is not supposed to have an order relation, so what's wrong with this? Have I implicitly used the axiom of choice or something?
9. Oct 16, 2009
### Moo Of Doom
You can certainly put some order on the complex numbers, Ben, but you can't find one that behaves nicely with their algebraic structure. For a field to be considered an ordered field, it must have an order that is compatible with multiplication and addition. That is, we want
a < b ==> a + c < b + c
a < b, c > 0 ==> ac < bc
for all a, b, c. Your order doesn't satisfy either of these properties, so it doesn't make the complex numbers into an ordered field. Sorry.
10. Oct 16, 2009
### Ben Niehoff
Ah, yes, that does change things a bit.
11. Oct 22, 2009
### elibj123
Order of a field is defined via a subset of that field. Let's denote it P. An ordered field F contains a subset P that satisfies the following properties:
1) for any a (which is not 0) in F, Either a or -a are in P, but not both.
2) For any a,b in P, a+b is in P.
3) For any a,b in P, a*b is in P.
Then, we say a>b iff (a-b) is in P.
You can see immediately that the C doesn't contain such a set because
if i belongs P, it implies that -i=i*i*i belongs to P (3) in contradiction to (1).
The same if we assume -i belongs to P.
12. Oct 22, 2009
### HallsofIvy | {
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12. Oct 22, 2009
### HallsofIvy
As elibj123 said, one of the requirements for an "ordered field", trichotomy, is that one and only one of the following be true of any member of the field, x: (trichotomy)
a) x= 0
b) x> 0
c) x< 0
You cannot have "-1< 0" and "1> 0" in any ordered field (where "1" is defined as the multiplicative identity and "0" is defined as the additive identity).
Your reference to $R^n$ doesn't make sense because it is not a field. What arithmetic operations would you wish to define on it to make it a field?
If you ignore the operations, then the complex numbers can be ordered (any set can be). One such order ("lexicographical") is "If a< c then a+bi< c+ di. If a= c and b< d, then a+bi< c+ di." That's a perfectly good order on C or $R^2$ and can be extended in an obvious way to $R^n$. As long as you don't look at the operations, there is no problem. | {
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# A question on function equality
by zenctheo
Tags: equality, function
P: 4 Hello to every one! I have a question that came up when I was talking with a fellow mathematician. I used to say that two functions are equal when the have the same formula and the same domain and codomain. We read in a book though that two functions are equal when they have the same domain and when the values of the function are equal for the same X. For example $$f(x)=x^2$$ and $$g(x)=x^3$$ are equal when their domain is only the points 0 and 1,$$x \in \{0,1\}$$because f(0)=g(0)=0 and f(1)=g(1) even though their formula is different. I thought that this definition of equality is incomplete because by saying that f(x)=g(x) then $$\frac{df}{dx}=\frac{dg}{dx}$$ but on point x=1 $$\frac{df}{dx}=2$$ and $$\frac{dg}{dx}=3$$. Thus we derive two different results from to equal quantities. Therefore two functions in order to be equal should also have the same formula. Can you please give any insight on this? Thanks a lot in advance. Akis
Sci Advisor P: 1,563 The derivative is not defined on the domain given. It requires a continuous interval. Remember the limit definition of the derivative: $$f'(x) = \lim_{\Delta x \rightarrow 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}$$ But for nearly all $\Delta x$, $x + \Delta x$ lies outside your domain. Therefore, you can't take the limit. :) So, you are correct: Two functions are equal if and only if they have the same domain and their values are equal at every point within the domain.
P: 4 Thanks a lot for the reply. You that I am wrong because I was the one saying that the functions should also have the same formula. In order to get things straight: You mean that the above two functions are equal.... or not?
P: 355
## A question on function equality
The functions are in fact equal. Also, as Ben said, those functions don't have derivatives because they're not defined on an open interval of the real numbers. | {
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As another example, would you consider these to be the same function?
Let's say f and g are functions from the real numbers to the real numbers defined as
f(x) = x
g(x) = x when x^2 >= 0 and -x when x^2 < 0
Since the functions are only defined on the real numbers, there are no points where they'd differ.
On a related note: "Having the same formula" is not a well-defined concept. Most (almost all) functions cannot be written with a closed formula and many (as you've seen with the example you gave) have multiple formulas.
P: 4 Ok. It's nice to learn a new thing. Even if I am proven wrong :)) Thanks a lot.
Related Discussions Calculus 2 Calculus & Beyond Homework 2 Beyond the Standard Model 4 General Math 8 Set Theory, Logic, Probability, Statistics 3 | {
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# Table tennis win probability
This problem is from my teacher and I think their answer is wrong.
The problem is in the context of table tennis.
The players in the tournament final are Ani and Bertha. The score in the game is drawn at 20-20. The final game will continue until one player has scored two more points than the other. It is known from previous games between Ani and Bertha that the probability of Ani winning each point is 0.6. Find the probability that Ani will win the game after exactly 8 more points.
I think that this means that, over the next 6 games on the 2nd, 4th and 6th game Ani and Bertha need to have a draw. For each draw there are two possible paths. Ani wins then Bertha or Bertha then Ani. After the 6th game Ani just needs to win twice to win after exactly 8 points.
However my teacher says that:
If Bertha wins the first game, it is not possible for Ani to win after exactly 8 points.
and also asserts that there is only one path to the desired outcome. Using this they find the probability of Ani winning after exactly 8 points to be 0.005. (
I found an alternate answer using multiple paths.
$$P(\text{Ani winning a game})=0.6$$
$$P(\text{Bertha winning a game})=0.4$$
$$P(\text{Ani win after 8 points})=2\dot(0.4 \cdot 0.6)\cdot2\dot(0.4 \cdot 0.6)\cdot2\dot(0.4 \cdot 0.6)\cdot(0.6\cdot0.6)=0.040\ (2sf)$$
After I found this answer I asked my teacher if the proposed answer was correct. My teacher replied saying that there was nothing wrong with it.
Am I missing something painfully obvious and if so what? or is the teacher's answer incorrect?
• When you contacted your teacher, did you present your argument as clearly as you did here? – Fabio Somenzi Nov 23 '18 at 2:22
Ani needs to be have two more points than Bertha exactly on the eighth game (not earlier, and never two less). She does not just need to win two games in a row, but to be tied then win the last two games. | {
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So since we are staring at a tie and need to avoid an being ahead or behind by two after the second game, the first two games may be a win followed by a loss, or a loss followed by a win.
Either way leaves Ani and Bertha tied again at the end. And so likewise for the next two games, and the next pair. Then as stated, the final two games need both be wins.
There are eight ways to do this, each giving Ani three losses and five wins. (Also, four among these paths do have Bertha ahead on the first game, so your teacher supposed erraneously.)
$$\mathsf P(\text{W on 8th})= 2^3\cdot0.4^3\cdot0.6^5$$
As you found.
I think your teacher is wrong. He shows BAA as a win for Annie, but it's not. He has confused "ahead by two" with "win two in a row".
• A non-mathematical suggestion on your answer: Since the OP referred to their teacher with the pronoun “they,” consider doing the same in your answer, instead of referring to the teacher as “he.” – Steve Kass Nov 23 '18 at 1:14
• @SteveKass You're right. I am usually pretty careful about his/her. I understand the need for even more pronouns. I have a hard time with the singular "they" - I may be too old to get used to it. Frequently I rework sentences so that I don't need any pronoun. – Ethan Bolker Nov 23 '18 at 11:34 | {
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# Find the determinant when all entries are $1$ except for zeroes on the main diagonal
Let $J_n$ be an $n\times n$ matrix all of whose entries are $1$, and $I_n$ be the identity matrix. Define $$K_n = J_n-I_n$$
For $n=1$ to $5$ my (usually unreliable!) hand calculations suggest that
$$\det K_n = (-1)^{n-1}(n-1)$$
Question
(a) are these values correct? (b) is the generalization to any positive integer $n$ valid? (c) if (b) is true, how can the result be demonstrated?
My only idea so far is to use the product of eigenvalues:
$$\det K_n = \prod_{j=1}^n \lambda_j$$
($-1$ is an eigenvalue for all $n$)
Any assistance much appreciated.
(note: a similar question concerning skew-symmetric matrices Determinant of a special skew-symmetric matrix may contribute some relevant ideas. or perhaps Determinant of a matrix with $t$ in all off-diagonal entries. has greater relevance)
This is the adjacency matrix of the complete graph $K_n$. The eigenvalues are $\lambda_1 = (n - 1)$ and $\lambda_2 = \dots = \lambda_n = -1$.
The eigenvector for $\lambda_1$ is the vector $\pmb{1}$ consisting of all $1$s.
An eigenbasis for $\lambda_2,\dots,\lambda_n$ is given by $e_1 - e_j$. Note that
$$(J_n - I_n)(e_1 - e_n) = \pmb{1}\pmb{1}^T(e_1 - e_j) - (e_1 - e_j) = - (e_1 - e_j).$$
This is because $J_n = \pmb{1}\pmb{1}^T$ and $\pmb{1}^Te_i = 1$ for all basis vectors $e_i$.
The fact that $n - 1$ is an eigenvalue reflects the fact that $K_n$ is an $(n - 1)$-regular graph. There may also be a combinatorial interpretation for the other eigenvalues but I don't know enough algebraic graph theory to say.
• thank you. very nice insight re the complete graph. – David Holden Jul 23 '17 at 3:34
$$\begin{array}{rl} \det ( 1_n 1_n^\top - \mathrm I_n ) &= \det ( -( \mathrm I_n - 1_n 1_n^\top ) )\\ &= (-1)^n \cdot \det(\mathrm I_n - 1_n 1_n^\top)\\ &= (-1)^n \cdot \det ( 1 - 1_n^\top 1_n )\\ &= (-1)^n \cdot (1-n)\\ &= (-1)^{n-1} \cdot (n-1)\end{array}$$ | {
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• really nice answer, thanks, and very educational for me. i hadn't encountered the determinant identity before, so this has clarified the meaning of the Schur complement - a concept which i encountered in reading recently without understanding its purport. have already accepted another answer, but yours is a very nice application of a powerful result to my rather simple question. thanks again! – David Holden Jul 23 '17 at 14:33
Finding the eigenvalues of a matrix $A$ that has $a$ along its main diagonal and $b$ everywhere else comes up fairly often on this site, but I’ll repeat the calculation here. Write the matrix as $A=b\mathbb1+(a-b)I$, where $\mathbb 1$ is the $n\times n$ matrix with every entry equal to $1$. Suppose $v$ is an eigenvector of $A$. We then have $b\mathbb1v=(\lambda-a+b)v$, so $v$ is also an eigenvector of $b\mathbb1$. Moreover, the eigenvalues of this matrix and those of $A$ differ by $a-b$. Now, for $b\ne0$, $b\mathbb1$ has rank 1, so its eigenvalues are $0$ with multiplicity $n-1$ and, using the fact that the trace of a matrix is equal to the sum of its eigenvalues, $nb$ with multiplicity 1. (You could also find the non-zero eigenvalue by noticing that all of the row sums are identical.) This means that the eigenvalues of $A$ are $a+(n-1)b$ and $a-b$, giving $$\det A=(a+(n-1)b)(a-b)^{n-1}.$$ For your matrices, $a=0$ and $b=1$, so $\det{K_n}=(-1)^{n-1}(n-1)$, just as you’ve calculated.
• thanks very much! another very enlightening answer, and very nicely explained. each of the three answers given here has helped me to expand my rather limited understanding of linear algebra. – David Holden Jul 25 '17 at 3:07 | {
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Number system - sum of two digit numbers
The sum of four two digit numbers is $221$. None of the eight digits is 0 and none of them are same. Which of the following is not included among the eight digit ?
$$(a) \;\;1 \\ (b)\;\; 2 \\ (c)\;\; 3\\ (d)\;\; 4$$
Is there any shortcut to solve this question as I got the answer which is $(d)\;\; 4$ by trial and error method. Please suggest
-
It is enough to know what is the last number modulo $9$.
Since $10a+b \equiv a+b \pmod 9$, $221 \equiv 5 \pmod 9$, and $1+2+3+4+5+6+7+8+9 = 45 \equiv 0 \pmod 9$, if the unused digit is $x$, then the sum of all the used digits is $221 \equiv 5$, and it also is $45-x \equiv -x$.
So $x \equiv -5 \equiv 4 \pmod 9$, and therefore the unused digit is $4$.
-
Hint:
$xy+ab+cd+ef=221$
$9(x+a+c+e)+y+b+d+f+(x+a+c+e)=221$
$\equiv(x+a+c+e)+y+b+d+f \equiv 5 \mod 9$
Now you know which number has to be excluded.
-
Hint $\$ Adding in the number $\rm\,0\,d\,$ formed by the missing digits $\rm\,0,d\,$ and casting nines shows
$\rm\qquad\quad mod\ 9\!:\ \ 2\!+\!2\!+\!1\!+\!d\, \equiv\, 0\!+\!1\!+\!2\!+\cdots + 9\,\equiv\, 0,\$ i.e. $\rm\ 5+d\equiv 0\ \Rightarrow\ d\equiv -5\equiv 4$
-
Suppose the numbers are $a_1b_1, \ a_2b_2, \ a_3b_3, \ a_4b_4.$
Then from $$\begin{array}{cc} a_1b_1& \\ a_2b_2& \\a_3b_3& \\ a_4b_4 &+ \\ \hline \\221 \end{array}$$ we conclude that only three possibilities can happen.
Either
$b_1+b_2+b_3+b_4=31$ and $a_1+a_2+a_3+a_4=19\,,$
$b_1+b_2+b_3+b_4=21$ and $a_1+a_2+a_3+a_4=20$ or
$b_1+b_2+b_3+b_4=11$ and $a_1+a_2+a_3+a_4=21\,.$
Since $a_1,\ a_2,\ a_3,\ a_4,\ b_1,\ b_2,\ b_3,\ b_4$ are all different and $1+2+\ldots+9=45$ we conclude that only the second possibility can happen and $4$ is not included in the digits $a_i,b_i$ ($a_1+a_2+a_3+a_4+b_1+b_2+b_3+b_4=41$).
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# Rectangle translation and rotation
I have a rectangle which is centered to $\ (1,2)$ like in the image below
rectangle first position
And I want to rotate and move the center to $\ (4,5)$ like in the image below
rectangle final position
I know that I have first to rotate it and then translate it. So it's the opposite $\ T*R$, where $\ T$ is the translation matrix and $\ R$ the rotation matrix.
Translation matrix :
\begin{bmatrix}1&0&0&3\\0&1&0&3\\0&0&1&0\\0&0&0&1\end{bmatrix}
It's the difference between final position and first position. My question is, how do I proceed from here because I am not really sure about the rotation table.
I tried to calculate the rotation table for 90 degrees.
\begin{bmatrix}0&-1&0&0\\1&0&0&0\\0&0&1&0\\0&0&0&1\end{bmatrix}
Based on \begin{bmatrix}cos(90)&-sin(90)&0&0\\sin(90)&cos(90)&0&0\\0&0&1&0\\0&0&0&1\end{bmatrix}
And the final Result of $\ T*R$ is:
\begin{bmatrix}0&-1&0&3\\1&0&0&3\\0&0&1&0\\0&0&0&1\end{bmatrix}
How does that result correspond to the final desired position? | {
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How does that result correspond to the final desired position?
• The rotation matrix you have rotates around the origin. Therefore try translating the rectangle to the origin first, then rotating it, and finally translating it to the place you want. Each of those three operations has a matrix, and multiply those matrices to get a single matrix for the combined operation. BTW, I don't see why you use 4x4 matrices. Those are normally for 3d rotations. You are not using the z-axis, so you can just leave out the third column and third row of all the matrices here and use 3x3 matrices instead. – Jaap Scherphuis Aug 16 '17 at 11:29
• @JaapScherphuis So If I understood it correct, it should look like this. Forgive me that I am answering with image but it will be easier to explain if I just show you. imgur.com/oxmU8hZ – valkongr Aug 16 '17 at 12:27
• The centre of the rectangle moves from (1,2) to the origin (0,0), the rotation keeps the centre there at (0,0), and in the last step you need to move it from (0,0) to (4,5). You did the last translation by the wrong amount. Other than that it seems good. – Jaap Scherphuis Aug 16 '17 at 12:41
• Actually, you also did not multiply the first two matrices correctly - the last column is wrong. You made the same mistake in the question above. – Jaap Scherphuis Aug 16 '17 at 12:50
• Is it because of the order? It should have been R*T? – valkongr Aug 16 '17 at 12:57
As has been pointed out in the comments, the rotation matrix that you built rotates about the origin, but you need to rotate the rectangle about some other point. A systematic way to deal with this would be to first translate the center of rotation to the origin, rotate, and then translate the result into its final position. | {
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Since you’re specifying the rectangle positions by their centers, we’ll rotate about the center of the rectangle. So, we first translate by $(-1,-2)$, rotate 90° conterclockwise, as you’ve done, and translate to the final position: $$\begin{bmatrix}1&0&4\\0&1&5\\0&0&1\end{bmatrix}\begin{bmatrix}0&-1&0\\1&0&0\\0&0&1\end{bmatrix}\begin{bmatrix}1&0&-1\\0&1&-2\\0&0&1\end{bmatrix}=\begin{bmatrix}0&-1&6\\1&0&4\\0&0&1\end{bmatrix}.\tag1$$ We check: $$\begin{bmatrix}0&-1&6\\1&0&4\\0&0&1\end{bmatrix}\begin{bmatrix}2\\4\\1\end{bmatrix}=\begin{bmatrix}2\\6\\1\end{bmatrix}.$$ I’ll leave the other corners for you to verify.
You can save having to do one of the matrix multiplications by skipping the first translation. Applying the rotation to the rectangle rotates it about the origin—its lower-left corner—so the necessary translation to then apply is the one that takes this new center to the target. The rotated center is $(-2,1)$, so the necessary translation is now $(4,5)-(-2,1)=(6,4)$, which, not coincidentally, is the last column of the matrix derived in (1). To see why this might be so, you can write and multiply these matrices in block form: $$\left[\begin{array}{c|c}I_2&\mathbf c_2 \\ \hline \mathbf 0^T&1 \end{array}\right]\left[\begin{array}{c|c}R & \mathbf 0 \\ \hline \mathbf 0^T&1\end{array}\right]\left[\begin{array}{c|c}I_2&-\mathbf c_1 \\ \hline \mathbf 0^T&1\end{array}\right]=\left[\begin{array}{c|c}R & \mathbf c_2-R\mathbf c_1 \\ \hline \mathbf 0^T&1 \end{array}\right]=\left[\begin{array}{c|c}I_2 & \mathbf c_2-R\mathbf c_1 \\ \hline \mathbf 0^T&1 \end{array}\right]\left[\begin{array}{c|c}R & 0 \\ \hline \mathbf 0^T&1 \end{array}\right].$$ Here, $I_2$ is the $2\times2$ identity matrix, $R$ is the $2\times2$ rotation matrix, and $\mathbf c_1$ and $\mathbf c_2$ are the respective rectangle centers. In the process, we’ve also shown that a translate-rotate-translate sequence is equivalent to a rotate-translate sequence. | {
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Now, there’s a quicker way to build the transformation matrix that you’re looking for by using an important property of transformation matrices: their columns are the images of the basis vectors, but that’s getting ahead of where you appear to be in this subject.
Note that since you haven’t specified how the vertices of the original rectangle correspond to those of the target, there are several other rigid motions (a.k.a. isometries) that will map one onto the other. For instance, you could rotate the rectangle in the opposite direction. That will also change the translation component of the transformation. Or, you could reflect the rectangle—flip it over—in various ways to get it aligned the right way.
Such 2D operations are much easier performed with complex numbers. Those $4\times4$ matrix are overkill.
Let $p=p=x+iy$ be a point that you want to transform.
You translate the rectangle to center it at the origin with
$$p'=p-(1+i2).$$
Then you rotate it counterclockwise around the origin by multiplying by $e^{i\theta}$, equal to $i$ in your case.
$$p''=i(p-(1+i2))=ip+2-i.$$
Finally, you translate the center to the new desired location
$$p'''=ip+2-i+(4+i5)=ip+6+i4.$$
In other terms,
$$\begin{cases}x'''=-y+6,\\y'''=x+4.\end{cases}$$ | {
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# Am I allowed to use distributive law for infinitely many sets?
Let $X$ be a nonempty set. A collection $S$ of subsets of $X$ is called a semiring if it satisfies the following properties:
1. The empty set belongs to S; that is $\emptyset \in S$.
2. If $A,B \in S$; then $A \cap B \in S$; that is, $S$ is closed under finite intersections.
3. The set difference of any two sets of $S$ can be written as a finite union of pair-wise disjoint members of $S$. That is, for every $A, B \in S$; there exist $C_1, ...,C_n \in S$ (depending on $A$ and $B$) such that $A\setminus B = \cup _{i=1}^n C_i$ and $C_i \cap C_j = \emptyset$ if $i\ne j$.
Now, let $S$ be a semiring of subsets of $X$. A subset $A$ of $X$ is called a $\sigma$-set with respect to $S$ (or simply a $\sigma$-set) if there exists a disjoint sequence $\{A_n\}$ of $S$ such that $A = \cup_{n=1}^\infty A_n$. One can show easily that for every sequence $\{A_n\}$ of $S$, the set $A = \cup_{n=1}^\infty A_n$ is a $\sigma$-set.
I would like to prove that finite intersections of $\sigma$-sets is a $\sigma$-set. For this purpose, suppose $A ,B$ are $\sigma$-sets then $A = \cup_{i=1}^\infty C_i$ , $B = \cup_{j=1}^\infty D_j$.
$$A\cap B=(\cup_{i=1}^\infty C_i)\cap (\cup_{j=1}^\infty D_j)$$
In this step I don't know am I allowed to use distributive law for infinitely many sets? Or the law holds only for finitely many sets?
If it holds only for finitely many sets how do I prove that finite intersections of $\sigma$-sets is a $\sigma$-set?
Thanks.
• That doesn't sound much like what a semiring usually means. Is there a connection? – hmakholm left over Monica Aug 31 '13 at 16:52
• Well, it wouldn't be the first time that the word "ring" got used for something setty instead of something algebraic. – Ben Millwood Aug 31 '13 at 17:09
• @Henning: Skip down in that article to here, and you’ll find exactly this definition of a semiring of sets. – Brian M. Scott Sep 1 '13 at 3:26 | {
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Yes, both distributive laws generalize. (You need only one of the two, but it’s useful to know both.) The first step in verifying the generalization that you want is to check that
$$\left(\bigcup_{i\in I}A_i\right)\cap D=\bigcup_{i\in I}(A_i\cap D)\;,\tag{1}$$
and to verify its mate you’ll want to check that
$$\left(\bigcap_{i\in I}A_i\right)\cup D=\bigcap_{i\in I}(A_i\cup D)\;.\tag{2}$$
Both are easily verified by element-chasing. For $(1)$, if $x\in\left(\bigcup_{i\in I}A_k\right)\cap D$, then $x\in\bigcup_{i\in I}A_i$ and $x\in D$. Since $x\in\bigcup_{i\in I}A_i$, there is an $i_0\in I$ such that $x\in A_{i_0}$, and therefore $x\in A_{i_0}\cap D\subseteq\bigcup_{i\in I}(A_i\cap D)$. Conversely, if $x\in\bigcup_{i\in I}(A_i\cap D)$, then there is an $i_0\in I$ such that $x\in A_{i_0}\cap D$. Then $x\in A_{i_0}\subseteq\bigcup_{i\in I}A_i$, and $x\in D$, so $x\in\left(\bigcup_{i\in I}A_i\right)\cap D$.
For $(2)$, if $x\in\left(\bigcap_{i\in I}A_i\right)\cup D$, then $x\in\bigcap_{i\in I}A_i$ or $x\in D$. Let $i_0\in I$ be arbitrary. Then $A_{i_0}\supseteq\bigcap_{i\in I}A_i$, so $x\in A_{i_0}$ or $x\in D$, and therefoer $x\in A_{i_0}\cup D$. Since this holds for each $i_0\in I$, $x\in\bigcap_{i\in I}(A_i\cup D)$. Conversely, if $x\in\bigcap_{i\in I}(A_i\cup D)$, then $x\in A_i\cup D$ for each $i\in I$. If $x\in D$, then certainly $x\in\left(\bigcap_{i\in I}A_i\right)\cup D$. If $x\notin D$, then we must have $x\in A_i$ for each $i\in I$, in which case $x\in\bigcap_{i\in I}A_i\subseteq\left(\bigcap_{i\in I}A_i\right)\cup D$.
Two applications of $(1)$ will give you the distributive law that you want. Suppose that $C=\bigcup_{i\in I}A_i$ and $D=\bigcup_{j\in J}B_j$. Then
\begin{align*} C\cap D&=\left(\bigcup_{i\in I}A_i\right)\cap D\\ &=\bigcup_{i\in I}(A_i\cap D)\\ &=\bigcup_{i\in I}\left(A_i\cap\bigcup_{j\in J}B_j\right)\\ &=\bigcup_{i\in I}\left(\bigcup_{j\in J}(A_i\cap B_j)\right)\\ &=\bigcup_{\langle i,j\rangle\in I\times J}(A_i\cap B_j)\;. \end{align*} | {
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In other words, $C\cap D$ is the union of all possible intersections of the form $A_i\cap B_j$. In particular, if $I$ and $J$ are countable index sets, $I\times J$ is also countable.
Similarly, two applications of $(2)$ will give you the other general distributive law of this kind. This time suppose that $C=\bigcap_{i\in I}A_i$ and $D=\bigcap_{j\in J}B_j$. Then
\begin{align*} C\cup D&=\left(\bigcap_{i\in I}A_i\right)\cup D\\ &=\bigcap_{i\in I}(A_i\cup D)\\ &=\bigcap_{i\in I}\left(A_i\cup\bigcap_{j\in J}B_j\right)\\ &=\bigcap_{i\in I}\left(\bigcap_{j\in J}(A_i\cup B_j)\right)\\ &=\bigcap_{\langle i,j\rangle\in I\times J}(A_i\cup B_j)\; \end{align*}
the intersection of all possible unions of the form $A_i\cup B_j$ as $i$ and $j$ run over their respective index sets.
• "[...] element-chasing". What a lovely piece of terminology. It told me exactly how to carry out the proof with no need for explanation or formalization (I supplied the formality myself). – Jonas Kölker Mar 24 '19 at 7:42
Two sets $X$ and $Y$ are equal exactly when, for all $x$, $x \in X \iff x \in Y$.
Now, take $x \in (\cup_i C_i) \cap (\cup_j C_j)$.
This means precisely that $x$ is in $\cup_i C_i$ and also is in $\cup_j D_j$.
A statement of the form $x \in \cup_k S_k$ means precisely "there exists some $k$ such that $x \in S_k$". So we have from $x \in \cup_i C_i$ and $x \in \cup_j D_j$ that there exist $i$ and $j$ such that $x \in C_i$ and $x \in D_j$.
Or in other words, there exist $i$ and $j$ such that $x \in C_i \cap D_j$.
Or in other words, $x \in \cup_i \cup_j (C_i \cap D_j)$.
So, yes, the distributive law works for infinite, even uncountable families of sets. | {
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# Meaning of “dS” in flux integrals?
In a general flux integral of a level surface, of the form
$$\iint{\mathbf{F}\bullet d\mathbf{S}}$$
what exactly does $d\mathbf{S}$ represent? I have seen both
$d\mathbf{S} = \mathbf{\hat N}dS = \pm (\mathbf{\frac {n}{|n|}})(\mathbf{|n|}) dudv$
(for parametric surfaces), and
$d\mathbf{S} = \mathbf{\hat N}dS = \pm \frac{\nabla G(x,y,z)}{G3(x,y,z)}dxdy$
for level surfaces. At a glance, I feel like I get them, but whenever I sit down to actually solve any problems I get confused about what exactly it represents in the integrals. Finding the normal is usually not a problem, nor is calculating $\frac{\nabla G}{G3}$, but then I get stuck on what to put for dS. Like the following example, in calculating the flux of $\mathbf{F} = xi + zj$ out of the surface x+2y+3z = 6. The textbook calculates $\mathbf{\hat N}$ to be $\frac{i+2j+3k}{\sqrt{14}}$ (and I agree), but then it goes on to calculate $dS = \frac{dxdy}{|\mathbf{\hat N}\bullet\mathbf{j}|} = \frac{\sqrt{14}}{2} dxdz$ . I'm not entirely sure why they did that, or why they set it up the way they did. How do you find/choose dS and what does it mean to the integral?
Thanks!
• $\vec{n}$ is just the normal vector at the surface element dS. – tired Apr 29 '15 at 13:57
• It seems you are confused about $dS$, not $d\bf S$, is that right? – Samuel Apr 29 '15 at 13:59
• @Samuel Yes, I guess so! I mean the other terms can be computed in a straight forward fashion, but the role of dS is a bit more foggy. – Laplacinator Apr 29 '15 at 16:27
$dS$ is a surface element, a differential sized part of the surface $S$. It is usually oriented, positive if its normal $n$ is outward pointing (e.g. if $S$ is the boundary of a volume). $$dS = n \lVert dS \rVert$$
I have seen both
$$d\mathbf{S} = \mathbf{\hat N}dS = \pm (\mathbf{\frac {n}{|n|}})(\mathbf{|n|}) dudv$$
(for parametric surfaces), and $$d\mathbf{S} = \mathbf{\hat N}dS = \pm \frac{\nabla G(x,y,z)}{G3(x,y,z)}dxdy$$
for level surfaces. | {
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For those examples $\lVert dS \rVert = du \, dv$ and $\lVert dS \rVert = dx \, dy$. The other parts are the more or less complicated normal vectors of those surface elements.
$$dS = \frac{dxdy}{|\mathbf{\hat N}\bullet\mathbf{j}|} = \frac{\sqrt{14}}{2} dxdz$$
The integration is along the $x$-$z$ plane, while the surface, $$S: x+2y+3z = 6 \quad n = (1,2,3)^t/\sqrt{14}$$ which is a plane as well, is not parallel to the $x$-$z$ plane.
The area of the projection $P_y S$ has to be adjusted, to give the correct area $\lVert S \rVert$ for $S$. We want $$\lVert S \rVert = \int\limits_S \lVert dS \rVert = \int\limits_S \lVert n\,du\,dv \rVert = f \lVert P_y S \rVert = f \int\limits_{P_y S} \lVert dx \, dz \rVert$$ In your example they simply take $f = 1/\lVert n \cdot e_y\rVert$.
Let us check this: First we look for unit vectors $u$ and $v$ orthogonal to $n$ and each other. $$0 = n \cdot a = (1, 2, 3)^t / \sqrt{14} \cdot (2, -1, 0)^t \quad e_u = (2, -1, 0)^t / \sqrt{5} \\ e_v = n \times e_u = (3, 6, -5)^t / \sqrt{70}$$ These are unit vectors, so the area of the square between $e_u$ and $e_v$ is 1. Now these unit vectors have the projections on the $x$-$z$ plane: $$u_p = P_y e_u = (2, 0, 0)^t/\sqrt{5} \quad \lVert u_p \rVert = 2/\sqrt{5} \\ v_p = P_y e_v = (3, 0, -5)^t/\sqrt{70} \quad \lVert v_p \rVert = \sqrt{34/70} = \sqrt{17/35} \\$$ where $P_y a = a - (a\cdot e_y) e_y$ for a vector $a$. The area of the projection is $$\lVert u_p \times v_p \rVert = \lVert ((2, 0, 0)^t/\sqrt{5}) \times ((3, 0, -5)^t/\sqrt{70}) \rVert = \lVert (0, 10, 0)^t/\sqrt{350} \rVert = 2 /\sqrt{14}$$
This should explain the factor $\sqrt{14}/2$.
What is missing is a derivation for the shorter $$\lVert P_y u \times P_y v \rVert = \lVert n \cdot e_y \rVert \, \lVert u \rVert \, \lVert v \rVert$$ | {
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• Thanks! That really helps clear some of it up. I'm gonna read through that part on adjusting area elements that aren't parallell to any axis some more , that seems like a very important point. Follow up question, would that mean that the area element of a plane parallell to the xy-plane would always be dxdy? Is the $\frac{\sqrt{14}}{2}$ in fact just "scaling" the differential square to "fit" our particular plane? – Laplacinator Apr 29 '15 at 16:22
• Exactly. The area of the projection of the surface $du dv$ on the $x$-$z$ plane is smaller than the area of the original $du dv$ , because its normal $n$ is not normal to that plane. So it must be upscaled. If the surface is part of a parallel plane, we do not have to do this, as surface element and its projection are same sized. – mvw Apr 29 '15 at 16:53
• @mvw Would you care to explain what $\pm(\mathbf{\frac {n}{|n|}})(\mathbf{|n|})$ represents exactly? – Demosthene Apr 29 '15 at 18:26
• @Demosthene From the looks it is about normal vectors $n$ which are not unit vectors, decomposing them into a product $\lVert n \rVert e_n$, where $e_n$ is the unit vector with the same direction as $n$. I would need to see an example to understand more, esp. why the signs show up that way. – mvw Apr 29 '15 at 18:33
• @mvw Yes, I am familiar with the notations $\mathbf{\frac{n}{|n|}}$ or $\hat{n}$ to denote a normalized vector. Following that logic, I don't see the point of writing $\mathbf{\frac{n}{|n|}|n|}$. Isn't that just $\mathbf{n}$? – Demosthene Apr 29 '15 at 19:34 | {
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$d\mathbf S = \hat {\mathbf N}dS$ -- where $\hat {\mathbf N}$ is the unit normal (outward if closed, otherwise you have to choose an orientation) to the surface and $dS$ is the differential area element -- is the definition. But you don't really need to worry about it. You should always parametrize your surface first by some $\mathbf r(s,t)$, $s_0 \le s \le s_1$ and $t_0 \le t\le t_1$. Then $$\iint_\Gamma \mathbf f \cdot d\mathbf S = \int_{t_0}^{t_1} \int_{s_0}^{s_1} \mathbf f(\mathbf r(s,t)) \cdot \left(\frac {\partial \mathbf r}{\partial s} \times \frac {\partial \mathbf r}{\partial t}\right)\,ds\,dt$$
Each of those terms on the right you should then be able to calculate.
• I see! Thanks a lot, but is it really necessary to always parametrize a surface first? – Laplacinator Apr 29 '15 at 16:26
• Yes -- but sometimes you may not realize you're doing it. Parametrizing a surface just means expressing it as some vector equations or set of scalar equations in $2$ variables. If there is an immediate way to write down the surface in terms of easy variables like $x,y$ or $r, \theta$, you may not realize that the $\hat {\mathbf N}\,dx\,dy$ that you immediately wrote down is really just $\left(\frac {\partial \mathbf r}{\partial x} \times \frac {\partial \mathbf r}{\partial y}\right)\,dx\,dy$ or that your $\mathbf f(x,y)$ is already in terms of your parameters (variables). – user137731 Apr 29 '15 at 17:21 | {
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# Minimum tickets required for specified probability of winning lottery
In a lottery, 1/10 of the 50 000 000 tickets give a prize.
What is the minimum amount of tickets one should buy to have at least a 50% chance to win?
Would be very glad if you could explain your methodology when resolving this. Please consider this as homework if you will.
## 1 Answer
I should really ask what your thoughts are so far on this. This problem is very closely related to the "birthday problem". The easiest way to do these problems is to count the possible ways of either winning or losing. Usually one is much easier to do than the other, so the key is to find the right one. Before we get into the actual calculation, let's start with some heuristics.
Heuristics: Let $n$ be the total number of tickets and $m$ be the number of winning tickets. In this case $n = 50\,000\,000$ and $m = 5\,000\,000$. When $n$ is very large then purchasing multiple distinguishable tickets is almost the same as sampling with replacement from the population of tickets.
Let's suppose that, instead of having to purchase $k$ separate tickets, we purchased a ticket, looked to see if it was a winner and then returned it to the lottery. We then repeat this procedure where each such draw is independent from all of the previous ones. Then the probability of winning after purchasing $k$ tickets is just $$\Pr( \text{we won} \mid \text{purchased k tickets} ) = 1 - \left(\frac{n-m}{n}\right)^k .$$
For our case then, the right-hand side is $1 - (9/10)^k$ and so we set this equal to $1/2$ and solve for $k$ in order to get the number of tickets.
But, we're actually sampling without replacement. Below, we'll go through the development, with the point being that the heuristics above are more than good enough for the present problem and many similar ones.
There are $50\,000\,000$ tickets. Of these $5\,000\,000$ are winning ones and $45\,000\,000$ are losing ones. We seek
$$\Pr(\text{we win}) \geq 1/2 \>,$$ | {
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$$\Pr(\text{we win}) \geq 1/2 \>,$$
or, equivalently,
$$\Pr(\text{we lose}) \leq 1/2 .$$
The probability that we lose is simply the probability that we hold none of the winning tickets. Let $k$ be the number of tickets we purchase. If $k = 1$, then $\Pr(\text{we lose}) = 45\,000\,000 / 50\,000\,000 = 9/10 \geq 1/2$, so that won't do. If we choose two tickets then there are $45\,000\,000 \cdot 44\,999\,999$ ways to choose two losing tickets and there are $50\,000\,000 \cdot 49\,999\,999$ ways to choose any two tickets. So,
$$\Pr( \text{we lose} ) = \frac{45\,000\,000 \cdot 44\,999\,999}{50\,000\,000 \cdot 49\,999\,999} \approx 0.9^2 = 0.81.$$
Let's generalize now. Let $m$ be the number of winning tickets and $n$ the total number of tickets, as before. Then, if we purchase $k$ tickets,
$$\Pr(\text{we lose}) = \frac{(n-m) (n-m-1) \cdots (n-m-k+1)}{n (n-1) \cdots (n-k+1)} .$$
It would be tedious, but we can now just start plugging in values for $k$ until we get the probability below $1/2$. We can use a "trick", though, to get close to the answer, especially when $n$ is very large and $m$ is relatively small with respect to $n$.
Note that $\frac{n-m-k}{n-k} \leq \frac{n-m}{n}$ for all $k < m < n$. Hence
$$\Pr(\text{we lose}) = \frac{(n-m) (n-m-1) \cdots (n-m-k+1)}{n (n-1) \cdots (n-k+1)} \leq \left(1 - \frac{m}{n} \right)^k ,$$
and so we need to solve the equation $\left(1 - \frac{m}{n} \right)^k = \frac{1}{2}$ for $k$. But,
$$k = \frac{\log \frac{1}{2}}{\log (1 - \frac{m}{n})} \>,$$
and so when $m/n = 1/10$, we get $k \approx 6.58$.
So, $k = 7$ tickets ought to do the trick. If you plug it in to the exact equation above, you'll get that
$$\Pr( \text{we win} \mid \text{k=7} ) \approx 52.2\%$$
• Very detailed. I loved it. Thank you and I'll consider it as an answer unless someone comes up with something else. – Queops Feb 7 '11 at 2:58 | {
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# Show convexity of the quadratic function
Can someone show this function is convex using the definition (without taking gradient)? $$F(x) = x^TAx + b^Tx + c$$
where $A$ is symmetric positive semi-definite.
By definition of convex, for any $x,y\in\mathbb R$, we have $$f(\frac{x+y}2)\leq\frac12(f(x)+f(y))$$ Thus it is sufficient to reduce and prove that $$\frac12(x+y)^TA(x+y)\leq x^TAx+y^TAy\\ x^TAy+y^TAx\leq x^TAx+y^TAy$$ Namely $$(x-y)^TA(x-y)\geq0$$ which is directly followed by positive semi-definite.
• How would this proof change if A were positive definite and not semi-definite? – user2553807 Feb 10 '14 at 22:21
• Nothing except the condition when equality holds. – Shuchang Feb 11 '14 at 6:52
• how do you expand $(x+y)^TA(x+y)$ to the next formula? – Dzung Nguyen Jul 23 '14 at 23:07
• @DzungNguyen According to associative law and distributive law. $(x+y)^TA(x+y)=(x+y)^TAx+(x+y)^TAy$, and similarly for the left $(x+y)^T$. – Shuchang Jul 24 '14 at 2:58
• The first three inequalities should be inverse. – Khue Feb 13 '16 at 5:52
Just to leave the answer for the general case online for future reference. A function is convex if $f(\lambda x + (1-\lambda) y) \leq \lambda f(x) + (1-\lambda) f(y)$ for all $\lambda\in[0,\;1]$.
It suffices to show for a quadratic function $f(x) = x^TQx$. Therefore using the definition of a convex function: | {
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\begin{align} (\lambda x + (1-\lambda) y)^TQ(\lambda x + (1-\lambda) y)\leq \lambda x^TQx + (1-\lambda)y^TQy \end{align} Equality holds for $\lambda = 0\;\text{or}\;1$. Therefore consider $\lambda\in(0,1)$. The left hand side simplifies to: \begin{align} \lambda^2x^TQx + (1-\lambda)^2y^TQy + \lambda(1-\lambda)x^TQy + \lambda(1-\lambda)y^TQx\leq \lambda x^TQx + (1-\lambda)y^TQy \end{align} Rearranging the terms and simplifying one obtains: \begin{align} & \lambda(1-\lambda)x^TQx + \lambda(1-\lambda)y^TQy - \lambda(1-\lambda)x^TQy -\lambda(1-\lambda)y^TQx\geq 0 \\ & \Rightarrow x^TQx + y^TQy -x^TQy-y^TQx \geq 0 \\ & \Rightarrow (x-y)^TQ(x-y) \geq 0 \end{align} which is true for positive semi-definite $Q\succeq 0$. | {
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Finding complex eigenvalues and its corresponding eigenvectors
$$A = \begin{bmatrix}3&6\\-15&-15\end{bmatrix}$$ has complex eigenvalues $\lambda_{1,2} = a \pm bi$ where $a =$____ and $b =$ ____. The corresponding eigenvectors are $v_{1,2} = c \pm di$ where $c =$ (____ , _____ ) and $d =$ (____ , ___)
So I got the char. poly. eqn $\lambda^2 + 12\lambda + 45 = 0$
Then using the quad. eqn I got $-6 \pm 3i$ which I know is in the form $a \pm bi$ so I was thinking $a = -6$, $b = 3$, but instead they have $b = -3$ , why?
Also, I'm not sure how to obtain the corresponding eigenvectors because we are working with complex eigenvalues now.
I'll cover the how to find the eigenvectors part.
$$A = \begin{bmatrix}3&6\\-15&-15\end{bmatrix}$$
has eigenvalues $\lambda_{1,2} = -6\pm 3i$.
Now to find the associated eigenvectors, we find the nullspace of $$A-\lambda_{1,2}I = \begin{bmatrix}3-(-6\pm 3i)&6\\-15&-15-(-6\pm 3i)\end{bmatrix} = \begin{bmatrix}9\mp 3i & 6 \\ -15 & -9\mp 3i\end{bmatrix}$$
To find the nullspace I'll put this in REF:
\begin{align}\begin{bmatrix}9\mp 3i & 6 \\ -15 & -9\mp 3i\end{bmatrix} &\sim \begin{bmatrix} 5 & 3\pm i \\ 3\mp i & 2\end{bmatrix} \\ &\sim \begin{bmatrix} 5 & 3\pm i \\ 0 & 2-(3\pm i)\frac{-(3\mp i)}{5}\end{bmatrix} \\ &= \begin{bmatrix} 5 & 3\pm i \\ 0 & 0\end{bmatrix}\end{align}
Therefore all of the eigenvectors associated with $\lambda_{1,2}$ are of the form $w_{1,2} = \begin{bmatrix}\frac 15(-3\mp i)t \\ t\end{bmatrix}$. Representative eigenvectors are then $$\bbox[5px,border:2px solid red]{v_{1,2} = \begin{bmatrix}-3\mp i \\ 5\end{bmatrix}}$$ which is obtained by setting $t=5$. | {
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• Why is your $\pm$ signed turned upside down? What's the significance – Yusha Aug 2 '16 at 21:20
• It means that $\lambda_1 = -6\oplus 3i$ corresponds to $v_1 = \begin{bmatrix} -3 \ominus i \\ 5\end{bmatrix}$, which I've circled the relevant signs. Likewise $\lambda_2$ and $v_2$ have opposite signs in those places. – user137731 Aug 2 '16 at 21:21
• Somehow they are getting $(-1,1)$ and $(-1,2)$ I cant figure out how – Yusha Aug 2 '16 at 21:25
• Weird, maybe its an error – Yusha Aug 2 '16 at 21:30
• @Yusha Remember that there are an infinite number of vectors in the nullspace of $A-\lambda_{1,2}$. I chose $t=5$ arbitrarily. If you choose $t=1-2i$ you'll get your answer key's solution. As to the reason your book arrived at the answer it did, I'll bet they didn't do the row swap that I did in the first step of my row reduction. So both my and your answer key's solution are correct. – user137731 Aug 2 '16 at 21:48
Note that $\pm (-3)=\mp3=\pm3$. Hence, whether you use $b=3$ or $b=-3$ is irrelevant. To find the corresponding eigenvectors, you need to solve the following:
$$\begin{bmatrix}3&6\\-15&-15\end{bmatrix}\cdot\begin{bmatrix}x_{1}\\ x_{2}\end{bmatrix}=\lambda_{1,2}\cdot\begin{bmatrix}x_{1}\\ x_{2}\end{bmatrix}.$$
• I get stuck after I subtract the lambda value of 3 and I get $\begin{bmatrix}9-3i & 6\\-15 & -9-3i\end{bmatrix}$ – Yusha Aug 2 '16 at 20:49
• Should I put the matrix in RREF? – Yusha Aug 2 '16 at 20:51
• Because If I do $R_2 = 15R_1 + (9-3i)R_2$ this problem begans to get miserable as I will have a quadratic entry.... – Yusha Aug 2 '16 at 20:57
• I even row reduced to get $\begin{bmatrix}9-3i&6\\0&0\end{bmatrix}$ and still I'm not getting the correct vectors. this has infinite solutions?? – Yusha Aug 2 '16 at 21:06 | {
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The if $b=-3$, $a=-6$, we have $a \pm bi=-6 \pm 3i$, if $b=3, a=-6$ we still have $a\pm bi=-6 \pm 3i$, both answers are correct but if you're submitting into an online homework service like cengage-brain or webassign your professor may have just forgotten to include both cases as acceptable. To find eigenvectors, you need to solve $Av=\lambda v$, or, equivalently $(A-\lambda I)v=0$ (where I denotes the identity matrix. Why are these two systems equivalent?). Nothing really changes with the complex case so long as you know how to do arithmetic in $\mathbb{C}$. To get you started, for an eigenvector corresponding to $\lambda=-6+3i$, we set up: $$\begin{bmatrix} A-\lambda I \end{bmatrix} v= \begin{bmatrix} 9-3i & 6 \\ -15 & -9-3i \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} =0$$ Giving the system: $$\begin{array} ((9-3i)v_1+6v_2=0 \\ -15v_1+(-9-3i)v_2=0 \end{array}$$ Can you take it from here?
• $v_1 = -6v_2/(9-3i)$, $v_2 = 15v_1/(-9-3i)$? – Yusha Aug 2 '16 at 20:41
• I don't get it, don't i need to put the matrix in RREF form? – Yusha Aug 2 '16 at 20:47
• Because If I do $R_2 = 15R_1 + (9-3i)R_2$ this problem begans to get miserable as I will have a quadratic entry.... – Yusha Aug 2 '16 at 20:57
• I even row reduced to get $\begin{bmatrix}9-3i&6\\0&0\end{bmatrix}$ and still I'm not getting the correct vectors. this has infinite solutions?? – Yusha Aug 2 '16 at 21:06
• Yes, there will be infinitely many eigenvectors, but they will all be scalar multiples of one another, your first comment has you basically done, fix $v_1=1$ and solve for $v_2$ in terms of $v_1$, that will give you one eigenvector. Since matrix multiplication is linear, obvious any scalar multiple of this eigenvector will also be an eigenvector since $Av=\lambda v$ implies $A(cv)=c(Av)=c(\lambda v)=\lambda (cv)$. – mb- Aug 3 '16 at 2:28 | {
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# Prove that tan^(- 1)(1/3)+tan^(- 1)(1/7)+tan^(- 1)(1/13)+..........+tan^-1 (1/(n^2+n+1))+......oo =pi/4
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Let P(n):tan^(-1)((1)/(3))+tan^(-1)((1)/(7))+…..+tan^(-1)((1)/(n^+n+1))=tan^(-1)((n)/(n+2)) ……..(i) <br> Step I For n=1 <br> LHS of Eq. (i) =tan^(-1)((1)/(3)) =tan^(-1)((1)/(1+2))=RHS of Eq. (i) <br> Therefore , P(1) is true . <br> Step II Assume that P(k) is true. Then , <br> P(k):tan^(-1)((1)/(3))+tan^(-1)((1)/(3))+tan^(-1)((1)/(7))+........+tan^(-1)((1)/(k^2+k+1))=tan^(-1)((k)/(k+2)) <br> Step III For n=k+1 <br> P(k+1):tan^(-1)((1)/(3))+tan^(-1)((1)/(7))+....+tan^(-1)((1)/(k^2+k+1))+tan^(-1)((1)/((k+1)^2+(k+1)+1))=tan^(-1)((k+1)/(k+3))......(ii) <br> LHS of Eq. (ii) <br> =tan^(-1)((1)/(3))+tan^(-1)((1)/(7))+.....+tan^(-1)((1)/(k^2+k+1))+tan^(-1)((1)/((k+1)^2+(k+1)+1)) <br> =tan^(-1)((k+1)/(k+2))+tan^(-1)((1)/((k+1)^2+(k+1)+1)) [by aaumption atep] <br> =tan^(-1)((k)/(1+(k+1)))+tan^(-1)((1)/(k^2+3k+3)) <br> =tan^(-1)((k)/(1+(k+1)))+tan^(-1)((1)/(1+(k+1)(k+2))) <br> =tan^(-1)(((k+1)-1)/(1+(k+1).1))+tan^(-1)(((k+2)-(k+1))/(1+(k+2)(k+1))) <br> =tan^(-1)(k+1)-tan^(-1)1+tan^(-1)(k+2)-tan^(-1)(k+1)=tan^(-1)(k+2)-tan^(-1)1 <br> =tan^(-1)((k+2-1)/(1+(k+2).1))=tan^(-1)((k+1)/(k+3))= RHS of Eq. (ii) <br> This shows that the result is true for n=k+1. Hence. by the principle of mathematical induction . the result is true for all n in N`.
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TimeTranscript
00:00 - 00:59Pawan let's start with the question the questions that prove that tan inverse of 1 by 3 + tan inverse of 1 by 7 + tan inverse of 1 by 13 + 1 or 2 tan inverse of 1 by X square + 1 + 1 + 1 upto infinity is equal to pi by 4 first let's look at the lhs for lhs term are that is TR are term will be equal to tan inverse of 1 upon our square + R plus one record that from this term over the top of this sequence will be equal to the completeness tan inverse of numerator candidness h + 1 - are upon 1 +
01:00 - 01:59into our + 1 + 1 + 1 know this is what I got an inverse of now using this formula is right the form no you are an inverse of a minus tan inverse of b is equal to tan inverse of a minus b upon 1 + ab compare this to this will get that the 4th of R is equal to now here is our + 1 and become an into this that will be tan inverse of h + 1 minus tan inverse of are now are alleges in the question is will become therefore lhs is equal to summation of are equal to 12
02:00 - 02:59PR now let's write TR as we found out over here so this will be equal to summation of are equal to 12 Infinity of tan inverse of h + 1 minus tan inverse of our ok so this should be equal to the values of r from 1 to infinity oh yeah this will be tan inverse of 2 - 10 inverse of 1 + report to the Sonu Nigam tan inverse of 3 minus tan inverse of 2 + 1 bracket when we put 3 it will become tan inverse of minus tan inverse of 3 plus show on this will keep going up to infinity | {
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03:00 - 03:59summation of to infinity sofa Infinity it will be tan inverse of infinity plus one minus tan inverse of infinity sunao if you look Oh Yeah Oh Yeah minus tan inverse of 2 will get cat cancel with tan inverse of 2 then minus tan inverse of 3 will get cancelled with + tan inverse of 3 keep on going the cancellation will keep on work on your minus tan inverse of 48 cancelled with this tan inverse of 4 then it will get cancer so what will be left over your will be equal to b minus tan inverse of 1 + tan inverse Infinity plus one not know the value of tan inverse of 1 so this will be equal to minus 5 by 4 then value of tan
04:00 - 04:595 by 4 then tan inverse of infinity + 1 + 1 is still Infinity infinity is pi by 2 is equal to pi by 4 pi by 2 minus 5 by 4 is pi by 4 sunao this is Rs so we can write therefore lhs is equal to RHS sunao since we got lhs is equal to RHS we can finally concluded your therefore tan inverse of 1 upon 3 + tan inverse of 1 upon 7 plus show on up to 10 inverse of 1 upon X square + 1 + 1 + 1 upto infinity is equal to pi by 4 this is what I had to prove so now we can find the right to year hence proved | {
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# If $\{a_n\}$ is a sequence in which $a_n \geq c$ for some constant $c$ and $a_n \rightarrow a$ then $a \geq c$
If $$\{a_n\}$$ is a sequence in which $$a_n \geq c$$ for some constant $$c$$ and $$a_n \rightarrow a$$ then $$a \geq c$$
I just wanted some feedback on whether my proof of the claim is sound.
Proof
Let $$\epsilon > 0$$ and $$a < c$$. Now $$a_n \rightarrow a$$ means:
$$\forall \ \epsilon >0,\ \exists \ N\in \mathbb{N} \ s.t.\ \forall \ n \geq N \ |a_n - a| < \epsilon \\ \Leftrightarrow \\ \ a-\epsilon \leq a_n \leq a + \epsilon$$
Consider $$\epsilon = \frac{c-a}{2}$$
$$\Rightarrow c \leq a_n \leq a + \frac{c-a}{2} = \frac{a+c}{2} < \frac{c + c}{2} = c$$
$$c < c$$ is a contradiction. Therefore $$a \geq c$$ for the statement to hold.
• Your proof is good but as an aside I have to wonder about "$\forall \ \epsilon >0,\ \exists \ N\in \mathbb{N} \ s.t.\ \forall \ n \geq N \ |a_n - a| < \epsilon \\ \Leftrightarrow \\ \ a-\epsilon \leq a_n \leq a + \epsilon$". This is maybe an editorial comment, but nobody likes to read such symbol soup an it doesn't make math more "serious". Also what you wrote doesn't parse. $|a_n-a|<\epsilon\iff a-\epsilon\le a_n\le a+\epsilon$ is always true for all $n$, $a_n$, $a$ and $\epsilon$. Does that mean all sequences converge to all values? – fleablood Oct 21 '18 at 15:14
• Funny you mention that. I was talking to my professor about this same thing this week because I'm not a big fan of the "symbol soup" as you call it and was inquiring whether just communicating my ideas in plain English would be better. – dc3rd Oct 21 '18 at 15:18
The idea is fine but: | {
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The idea is fine but:
• You should not begin with “Let $$\varepsilon>0$$”. You fix $$\varepsilon$$ later, not at this point.
• You should write that you are assuming that $$a, in order to get a contradiction. You can't just say “Let […] $$a”, because $$a$$ and $$c$$ are fixed from the start.
• Should I state it in terms of: "assume $a < c$ towards a contradiction....."? And with regards to $\epsilon$ don't I have to declare at the start that it will always be positive? – dc3rd Oct 21 '18 at 15:02
• This is a matter of taste, but I would begin with “In order to get a contradiction, let us assume that $a<c$”. Concerning the other question, note that you wrote right after $\forall\varepsilon>0$. So, no, you don't have to write that. – José Carlos Santos Oct 21 '18 at 15:04
• Gracias por la ayuda. – dc3rd Oct 21 '18 at 15:07
Your proof seems legitimate to me, but it is a proof by contradiction, so here is an alternate way to do it directly. I will start with what you have.
$$\forall \ \epsilon >0,\ \exists \ N\in \mathbb{N} \ s.t.\ \forall \ n \geq N \ |a_n - a| < \epsilon \\ \Leftrightarrow \\ \ a-\epsilon \leq a_n \leq a + \epsilon$$
Now, this implies that for any $$\epsilon > 0$$, $$a_n \leq a+\epsilon$$ for some $$n \in \Bbb{N}$$. Also, $$c \leq a_n$$, so we get $$c\leq a+\epsilon$$ for any $$\epsilon > 0$$. Thus, $$c\leq g$$ for all real numbers $$g$$ in the interval $$(a, \infty)$$. It is well known that $$a$$ is the greatest lower bound for $$(a, \infty)$$, so this implies $$c \leq a$$. | {
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Why can no prime number appear as the length of a hypotenuse in more than one Pythagorean triangle?
Why is it that no prime number can appear as the length of a hypotenuse in more than one Pythagorean triangle? In other words, could any of you give me a algebraic proof for the following?
Given prime number $p$, and Pythagorean triples $(a,b,p)$ and $(c,d,p)$ where $a<b<p$ and $c<d<p$, then $b=d$.
Please also have a look at the deeper question: Is there any formula to calculate the number of different Pythagorean triangle with a hypotenuse length $n$, using its prime decomposition?
• The fact that such a prime $p$ must satisfy $$p\equiv 1\pmod{4}$$ may be helpful. – Zubin Mukerjee Jun 10 '16 at 0:44
• @ZubinMukerjee no... since we already know that... a prime must be of the form of $p=4k+1$ to appear as a hypotenuse... but the question is why this number appear just in one Pythagorean triple? – Omid Ghayour Jun 10 '16 at 0:47
• look... $25$ appears in two Pythagorean triple $(15,20,25)$ and $(7,24,25)$ but $29$ appear in just one triple... – Omid Ghayour Jun 10 '16 at 0:50
• An elementary detailed proof is rather lengthy. Please see the uniqueness part of this proof. The proof is more pleasant and more informative if we can use properties of Gaussian integers. – André Nicolas Jun 10 '16 at 0:54
• We give a brief version of the Gaussian integer approach. By the usual representation theorem, the odd prime $p$ is a hypotenuse iff $p=s^2+t^2=(s+ti)(s-ti)$. The two factors are Gaussian primes, so by unique factorization $(u+vi)(u-vi)=p$ only if $u+iv$ is a unit times $s\pm ti$. – André Nicolas Jun 10 '16 at 1:18
This goes back to Euler, who showed that if there are two ways of writing an odd integer $N$ as the sum of two squares, then $N$ is composite. There is a 2009 article on this by Brillhart. Let me try to find a link.
http://www.maa.org/press/periodicals/american-mathematical-monthly/american-mathematical-monthly-december-2009 | {
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And if one note that in a primitive triple the hypotenuse is of the form $(u^2+v^2)$, and the legs are of the form $(u^2-v^2)$ and $(2uv)$. So by euler if the hypotenuse is prime it couldn't be written in different ways.
• – lhf Jun 10 '16 at 1:00
• Here is a better scan, from JSTOR: i.stack.imgur.com/DDymG.gif – lhf Jun 10 '16 at 1:02
• @lhf good. I made a jpeg of the relevant page and pasted that in. He also has a 2016 article that allows indefinite quadratic forms, although in that case more care is needed about what is meant by distinct representations. – Will Jagy Jun 10 '16 at 1:04
• when a number is hypotenuse... its square is going to written as sum of squares... so it is of course composite then... it is $p^2$ not $p$... – Omid Ghayour Jun 10 '16 at 1:06
• @WillJagy oh... yes... if it's prime... it's primitive... ;) ☺ – Omid Ghayour Jun 10 '16 at 1:09
As noted in the comments and the accepted answer, this comes down to the fact that if a prime $p$ can be written as a sum of two squares, then the representation is unique up to switching and or negating the factors. A fancier explanation for this is the fact that $\mathbb Z[i]$ is a principal ideal domain and its unit group is $\{\pm1,\pm i\}$. (Of course, proving that $\mathbb Z[i]$ is a PID requires some sort of argument like that in the scanned note, but this is a more modern way to think about it.) Once one knows it's a PID, then suppose that $p=u^2+v^2$. Then $p=(u+iv)(u-iv)$, and the fact that $u+iv$ and $u-iv$ have norm $p$ shows that they cannot factor further in $\mathbb Z[i]$. Hence they are irreducible (i.e., they generate prime ideals). So the unique factorization of the ideal $p\mathbb Z[i]$ is as the product of the prime ideals $(u+iv)\mathbb Z[i]$ and $(u-iv)\mathbb Z[i]$. So $u$ and $v$ are unique, up to switching them or replacing them by their negatives, which corresponds to multiplying $u+iv$ by each of the four units in $\mathbb Z[i]$.
• very interesting! – Vincent Jun 14 '16 at 18:40 | {
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• very interesting! – Vincent Jun 14 '16 at 18:40
If a Pythagorean triple is primitive, $$B+C$$ is a perfect square as shown by
$$2mn+(m^2+n^2)\quad=\quad m^2+2mn+nn^2\quad=\quad(m+n)^2$$
If $$C$$ is prime, then only one smaller value $$(B)$$ can add to it to make a perfect square. Given $$C\&B$$, there can be only a one $$A$$ to make a Pythagorean triple. | {
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# Prove $\int_0^{\infty}{\frac{1}{e^{sx}\sqrt{1+s^2}}}ds < \arctan\left(\frac1x\right),\quad\forall x\ge1$
The question is to prove: $$\int_0^{\infty}{\frac{1}{e^{sx}\sqrt{1+s^2}}}ds < \arctan\left(\frac1x\right),\quad\forall x\ge1.$$ Numerically it seems to hold true. So I have made some attempts to prove this analytically but have all failed.
I also wonder if there is a systematic approach to solve this kind of problem.
Here is a proof of the inequality for $x\geq 2$. For the remaining range, see the Added section below.
Let $\lambda:=1-1/\sqrt{2}$, then by convexity we have $$\frac{1}{\sqrt{1+s^2}}\leq 1-\lambda s^2,\qquad 0\leq s\leq 1.$$ Using this bound we can estimate \begin{align*}\int_0^{\infty}{\frac{1}{e^{sx}\sqrt{1+s^2}}}\,ds&<\int_0^1\frac{1-\lambda s^2}{e^{sx}}\,ds+\int_1^\infty\frac{1-\lambda}{e^{sx}}\,ds\\[6pt] &=\frac{1}{x}-\lambda\left(\int_0^1\frac{s^2}{e^{sx}}\,ds+\int_1^\infty\frac{1}{e^{sx}}\,ds\right)\\[6pt] &=\frac{1}{x}-2\lambda\frac{1-e^{-x}x-e^{-x}}{x^3} .\end{align*} On the other hand, for $x>0$ we also have $$\arctan\left(\frac{1}{x}\right)=\int_0^{1/x}\frac{1}{1+s^2}\,ds>\int_0^{1/x}(1-s^2)\,ds=\frac{1}{x}-\frac{1}{3x^3},$$ hence it suffices to verify that $$\lambda(1-e^{-x}x-e^{-x})>\frac{1}{6},\qquad x\geq 2.$$ This is straightforward, so we proved the original inequality for $x\geq 2$.
Added. One can cover the remaining range $1\leq x<2$ as follows. Let $f(x)$ denote the LHS and $g(x)$ denote the RHS in the original inequality. These two functions are decreasing, hence it suffices to verify the following $20$ numeric inequalities: $$f(1+(n-1)/20)<g(1+n/20),\qquad n=1,2,\dots,20.$$ These are likely to be true, e.g. I checked them with Mathematica's NIntegrate command. | {
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• Why is 20 enough? – Steven Gubkin Aug 17 '18 at 17:03
• @StevenGubkin: What I did in the Added section is that I decomposed the interval $[1,2]$ into $20$ equal subintervals such that on each of the subintervals, the maximum of $f$ is less than the minimum of $g$. So I proved a bit more than $f(x)<g(x)$ on $[1,2]$, and the advantage is that I only need to check finitely many numeric inequalities. The number $20$ has no significance except that it makes the idea work. For example, $15$ is not ok, because then on the first subinterval, the maximum of $f$ is greater than the minimum of $g$. – GH from MO Aug 17 '18 at 19:56
• Sorry for my late confirmation. Your method is surprisingly skillful, but at the same time clear. Actually, even if we prove only for x> 2, there is no big problem in the original problem. But as you added, the idea of proving the inequality by dividing the interval is also very impressive. Thanks for the great intuition and explanation. – Ramanasa Aug 20 '18 at 17:35
• @Ramanasa: Thanks for your feedback and kind words. I am glad I could help! – GH from MO Aug 20 '18 at 17:40
The inequality holds by direct calculation for $x$ close to unity, when the two sides are well separated, but for large $x$ the two sides approach each other and we need to check that the inequality is not violated.
I denote $x\equiv 1/\epsilon$, and seek to prove that $$\int_0^{\infty}{\frac{\epsilon}{e^{s}\sqrt{1+\epsilon^2 s^2}}}ds < \arctan\epsilon,\;\;0<\epsilon\leq 1.$$ Series expansion of both sides gives $$\int_0^{\infty}{\frac{\epsilon}{e^{s}\sqrt{1+\epsilon^2 s^2}}}ds=\epsilon-\epsilon^3+{\cal O}(\epsilon^5),$$ $$\arctan\epsilon=\epsilon-\tfrac{1}{3}\epsilon^3+{\cal O}(\epsilon^5).$$ This proves the inequality for $\epsilon$ close to $0$, hence for large $x$. | {
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• Well, one needs to be somewhat careful as the power series expansion of $1/\sqrt{1+\epsilon^2 s^2}$ only converges for $s<1/\epsilon$. – GH from MO Aug 17 '18 at 13:14
• I have developed your idea and made it concrete as follows. The inequality $$\frac1{\sqrt{1+t^2}} \le 1- \frac{1}{2}t^2 + \frac38 t^4$$ holds for all $t$. Using this the LHS is $$\epsilon\int_0^\infty \frac1{e^s\sqrt{1+s^2 x^2}}ds<\epsilon \int_0^\infty \frac1{e^s}\left(1-\frac12 s^2 \epsilon^2+\frac38 s^4 \epsilon^4\right)ds=\epsilon-\epsilon^3+9\epsilon^5$$ for all $\epsilon>0$, where the RHS is $$\arctan\epsilon > \epsilon-\frac13\epsilon^3$$ for all $\epsilon>0$. So it proves the inequality for $0<\epsilon<\frac2{3\sqrt3}$, or for $x>\frac{3\sqrt3}2\simeq2.598$. – Ramanasa Aug 20 '18 at 17:36
• I could not find out that the left side has a series expansion, but your advice was a great help. – Ramanasa Aug 20 '18 at 17:36
• I apologize for that I could choose only one answer on this system (Is there any other way?) while you provided also an excellent and clever solution. – Ramanasa Aug 20 '18 at 17:44 | {
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# Can an indefinite integral have multiple answers? (Besides the ' + C')
So I came across with this integral today in my midterm:
$$\int \frac {\tan(\pi x)\sec^2(\pi x)}2$$
And I got two correct answers:
$$\frac {\sec^2(\pi x)}{4\pi} +C$$
And
$$\frac {\tan^2(\pi x)}{4\pi} + C$$
The first one, I get it by substituting $u=\sec (\pi x)$ and the second one, by substituting $u=\tan (\pi x)$
I already differentiated both answers and got to the same integral, but my question is, if both answers are correct and if it were a definite integral, which answer should I use? Wouldn't they give different results?
-
In the last formula replace your $C$ with ${1\over {4\pi}}+K$, which is just another constant. Now use the trig identity mentioned below. You see that you get the other answer. – Maesumi Mar 4 '13 at 22:26
In your case, the difference between the two is a constant:
$$\sin^2 x + \cos^2 x = 1$$
so
$$\tan^2 x + 1 = \sec^2 x$$
In general, that will be true as well - integration can only be different up to a constant: consider $g = \int f = h$ and note that $g-h = \int f - \int f = \int 0 = const$. | {
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-
That doesn't answer my question in case it wasn't an indefinite integral. – ChairOTP Mar 4 '13 at 22:23
@ChairOTP: In case it was a definite integral, the extra $1$ in the $\sec^2$ answer for the upper limit would be canceled by the extra $1$ in the $\sec^2$ answer for the lower limit. – robjohn Mar 4 '13 at 22:26
@ChairOTP Fro definite integrals, you would get a constant adjustment factor, i.e. answers would be either $\sec^2 x$ or $\tan^2 x +1$, and in definite integrals, the constant would matter -- same answer would result either way. – gt6989b Mar 4 '13 at 22:26
Thank you so much :) – ChairOTP Mar 4 '13 at 22:28
They wouldn't give different answers in a definite integral. Suppose that we want to find $\displaystyle\int_a^b f(x)\,dx$. Let $F(x)$ be one antiderivative of $f(x)$, and let $G(x)$ be another. They differ by a constant, so $G(x)=F(x)+C$ for some constant $C$.
If you use $F(x)$ to evaluate the integral from $a$ to $b$, you get $F(b)-F(a)$.
If you use $G(x)$, you get $G(b)-G(a)$, that is, $(F(b)+C)-(F(a)+C)$. Simplify. The $C$'s cancel. | {
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# Does $\sin (5x)=5\sin(x)$? Why or why not?
This may sound like a dumb question, but does $\sin(5x)=5\sin(x)$? Can you say this for any trigonometric function if it is true (e.g. can $\tan^{-1}(7x)=7\tan^{-1}(x)$). Again, I am pretty sure I am wrong, but I am just wondering for my future work on trigonometry. If they are not equal, then please tell me what the difference is between the two. Thanks!
Note: I have not had much practice with trigonometry, so please do not answer with some kind of complicated Taylor Series thing (I have heard about it, but don't fully understand it). Thanks again
• They are not equal. A most obvious reason is that $\sin(5x)$ never exceeds $1$ (being equal to the $y$-coordinate of a point on the unit circle). But $5\sin x$ can be as large as $5$ (being equal to the $y$-coordinate of a point on the unit circle multiplied by five). – Jyrki Lahtonen Feb 26 '14 at 6:16
• The $\sin$ function is NOT linear. – IAmNoOne Feb 26 '14 at 6:21
• Check this out, it could be helpful! :) – mikhailcazi Feb 26 '14 at 6:51
• There should be a few values $x$ for which you are familiar with the value $\sin (x)$. (If there aren't, you should learn a few! Having a few examples/computations to keep in mind when thinking about mathematics is vital. I cannot stress this enough.) Does your conjecture hold for these values? – Thomas Belulovich Feb 27 '14 at 2:38
• I should have thought of this before I asked the question.... $\sin(3\times 30)=1$, but $3\sin(30)=3\times 0.5=1.5$! – TrueDefault Feb 27 '14 at 2:51
I'll try to explain the difference between the two with a picture: (Apologies ahead; it's a rough sketch done in Microsoft Paint.)
Here, we have a circle (suppose with radius 1) with its center at (0,0), with a line (call it $l$) drawn from the center to the edge of the circle (ending at $P$). $A$ is the angle the line makes with the $x$ axis. | {
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To get $5\sin(A)$, we blow up the line and the circle to 5 times its original size. $l$ is now 5 times the length, but still has one point at the center of the circle, and makes the same angle with the $x$ axis. The $y$-coordinate of the other end of $l$ is $5\sin(A)$.
To get $\sin(5A)$, this time we don't blow the circle and the line bigger. Instead, we're rotating the line around the center of the circle so that the angle it makes with the $x$-axis is 5 times bigger. Now, the $y$-coordinate of the point where $l$ meets the edge of the circle is $\sin(5A)$.
Try it out with an example or two; you'll see the difference yourself.
• I think your answer does the best job at helping the OP: thank you for the effort. – Eric Stucky Feb 26 '14 at 6:59
• @EricStucky Yeah, I figured OP was looking for an explanation in simpler terms. So I provided one. – Dennis Meng Feb 26 '14 at 7:01
No, placing a constant factor inside of the sine function, as in sin(a*x) changes the frequency of the function; if |a| is greater than one, it "squishes" the sine function horizontally, making it oscillate "faster", and if |a| is less than one, it "stretches out" the function horizontally, making it oscillate "slower". Placing the factor in front of the function, as in a*sin(x), scales the amplitude of the function, "stretching", or "squishing" it vertically.
The best way to observe it is to look at a graph and to see what it does: http://www.wolframalpha.com/input/?i=plot+sin+x%2C+sin+3x%2C+3*sin+x
I suggest playing around with other values in the graphs to see what they do.
1) You could have just plotted the functions in, e.g., wolfram alpha and see for yourself.
2) for $t=\frac{\pi}{2}$ one has $\sin (t)=1$ and since $|\sin (x)|\le 1$, it is clear that $\sin (5\cdot t)\ne 5\cdot \sin(t)$. | {
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Something like $a\cdot f(x)= f(a\cdot x)$ is true if and only if $f(x)$ is linear. Since there are always nonlinear terms in the Taylor expansion of a trigonometric function this can't be true for any.
• Isn't $f(x)=x+1$ linear? But $5f(x)=5x+5$ and $f(5x)=5x+1$. – Joel Reyes Noche Feb 26 '14 at 6:37
• Functions satisfying the stated property are called homogenous, not linear. Any linear function is homogenous, but not necessarily vice versa. As for $x\mapsto x+1$, it is not a linear function. Rather it is an example of an affice mapping. – Ittay Weiss Feb 26 '14 at 6:49
• the argument that since the taylor expansion contains nonlinear terms, then the function itself is not linear is true, but requires some proof (easy). But, for the question posed resorting to the Taylor expansion of the function is really an overkill. Simply plugging in some values into the purported equation immediately shows it does not hold. – Ittay Weiss Feb 26 '14 at 6:51
• I was assuming that the OP (not the answerer) was more familiar with calculus rather than with linear algebra. See Wikipedia, where a linear function is sometimes defined as a polynomial function of degree zero or one. – Joel Reyes Noche Feb 26 '14 at 7:02
Assume that the identity $\sin(5x) = 5\sin(x)$ is true. Since we know that $\sin(180^\circ) = 0$, we'll have: $0 = \sin(180^\circ) = \sin(5*36^\circ) = 5*\sin(36^\circ)$, hence $5*\sin(36^\circ) = 0$, which mounts to $\sin(36^\circ) = 0$. Is this true?
• Using degrees rather than radians should be discouraged. – Ittay Weiss Feb 26 '14 at 6:52 | {
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• Using degrees rather than radians should be discouraged. – Ittay Weiss Feb 26 '14 at 6:52
The trigonometric functions are functions and as such are usually represented using function notation. Recall that the function notation $f(x)$ means that function $f$ is accepting $x$ as its input. It does not mean that $f$ and $x$ are being multiplied. So the notation $\sin(5x)$ means that the sine function is accepting $5x$ as its input. The notation $5\sin(x)$ means that 5 is being multiplied to the output of the sine function when its input is $x$. Is everything clear now?
• Just to clarify that there exist functions with the property that $f(c\cdot x)=c\cdot f(x)$. Such functions are called homogenous. – Ittay Weiss Feb 26 '14 at 6:19
Both of the equalities that you wrote above are NOT true. One way to look at this might be to use their Taylor series expansion. $\sin (5x)$ means the sine of $5$ times an angle measure $x$. This means in a right triangle whose hypotenuse is of length $1$, the perpendicular corresponding to the acute angle $5x$ has length $\sin (5x)$.
While for $5 \sin x$, one should consider a right-triangle with hypotenuse of length one and the perpendicular corresponding to an acute angle $x$ and then take $5$ times the length of this perpendicular.
You can draw some such right triangles to verify the results yourself.
• Using Taylor series here is a major overkill. Simply plug in some values and observe. – Ittay Weiss Feb 26 '14 at 6:20
• @IttayWeiss haha. I can only agree :-) – Singhal Feb 26 '14 at 7:44 | {
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# Distribution function terminology (PDF, CDF, PMF, etc.) [duplicate]
I am confused about the following terminologies:
1. Distribution Function
2. Cumulative Distribution Function (CDF)
3. Probability Distribution Function
4. Probability Density Function
5. Probability Mass Function (PMF)
(1) Distribution Function == CDF. I.e. they are the same. Am I right?
(2) Which is called PDF: Probability Distribution Function or Probability Density Function?
(3) What is the difference between Probability Distribution Function and Probability Density Function?
(4) Which one is the continuous equivalent of PMF, Probability Distribution Function or Probability Density Function?
Die roll examples could be used for the discrete case and picking a number between 1.5 and 2.5 as an example for the continuous case.
As noted by Wikipedia, probability distribution function is ambiguous term:
A probability distribution function is some function that may be used to define a particular probability distribution. Depending upon which text is consulted, the term may refer to:
• a cumulative distribution function,
• a probability mass function, and/or
• a probability density function.
Cumulative distribution function (CDF) is sometimes shortened as "distribution function", it's
$$F(x) = \Pr(X \le x)$$
the definition is the same for both discrete and continuous random variables. In dice case it's probability that the outcome of your roll will be $x$ or smaller.
Probability density function (PDF) is a continuous equivalent of discrete probability mass function (PMF). Probability mass function is
$$f(x) = \Pr(X = x)$$
In dice case it's probability that the outcome of your roll will be exactly $x$.
Probability mass function has no sense for continuous random variables since $\Pr(X=x)=0$ for continuous random variables (check also Why X=x is impossible for continuous random variables?), because simply a point on real line is so "small" that has no mass and no area. | {
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This leads us to defining probability density as "probability per foot". Simple example is continuous uniform distribution with minimum of $a$ and maximum of $b$, where probability density is the same for each $x$ and equal to
$$f(x) = \frac{1}{b-a}$$
You can easily notice that it changes as the range between $a$ and $b$ (i.e the total area) changes, it is nicely described in Can a probability distribution value exceeding 1 be OK? thread. It is a probability of hitting infinitesimal (infinitely small) interval $[x, x + dx]$ when throwing a dice with infinite number of walls.
• There are fashions here too. For example, it was long customary to insist that probability density functions and probability mass functions were quite different kinds of beasts referring to continuous and discrete variables respectively. A common and in my experience more recent tendency, particularly with mathematically more mature groups, is to insist that the idea of density function is general; it is just a case of density with respect to what kind of measure, and measure could be e.g. counting measure. So density functions, wide sense, include mass functions. – Nick Cox Oct 26 '16 at 13:21
(1) You are right regarding (1).
(2)&(3)&(4) PDF is for probability density function. We usually use probability distribution function to mean CDF. Probability function is used to refer to either probability mass function(the probability function of discrete random variable) or probability density function(the probability function of continuous random variable).
You can also have a look at this What does "probability distribution" mean? | {
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Probability of disease given multiple samples of a single test
Someone either has disease X or they are healthy. A test is used which is 90% sensitive and 80% specific (81.81% precise). The test is performed 8 times with different samples from the same person.
1) What is the rule to calculate the probability that someone has the disease given that 5 out of 8 tests came back positive?
2) More generally: that N out of M samples were positive.
3) Is there a way of calculating the confidence?
Kind regards
• Hello jamesj629, welcome to the site. If this is a homework question please add the self-study tag. – Andy Oct 16 '14 at 11:26
• Thanks Andy - its not homework, just learning. – jamesj629 Oct 16 '14 at 11:32
You need the disease prevalence in order to calculate this probability.
Let $T$ be the indicator variable with $T=1$ denoting the event of a positive test. Similarly let $D$ be the indicator of the disease. You have $P(T=1|D=1)=0.9$ and $P(T=0|D=0)=0.8$. Define $A$ to be the event of N positive result from M tests. Then
$$P(A|D=1)=\binom{M}{N}0.9^N 0.1^{M-N}$$
What you want is $P(D=1|A)$. Using Bayes' rule $$P(D=1|A)= \frac{P(A|D=1)P(D=1)}{P(A)}=\frac{P(A|D=1)P(D=1)}{P(A|D=1)P(D=1) +P(A|D=0)P(D=0)}$$ | {
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• I take it P(D=1) on its own is disease prevalence? and P(D=0)=1-P(D=1) ? – jamesj629 Oct 16 '14 at 12:59
• The prevalence is (say) 1 in 100000, but for people taking the test, about 90% actually have the disease, because the test is only done on those presenting with suspect symptoms - does this invalidate the probability? – jamesj629 Oct 16 '14 at 13:05
• @jamesj629 It will increase the probability. If you let $S$ be the indicator of symptom onset. With what you described in the comment, now you are calculating $P(D=1|A,S=1)$ instead of $P(D=1|A)$. You numerator becomes $P(A|D=1,S=1)P(D=1|S=1)=P(A|D=1)P(D=1|S=1)$ if test result is independent of symptom. Note $P(D=1|S=1) > P(D=1)$ if disease is positively dependent with symptom. – Peter Oct 16 '14 at 21:19
You can use the information to calculate the base rate fo the disease:
accuracy: $A$
sensitivity: $s^+$
specificity: $s^-$
base rate: $p$
positive test result: $T^+$
disease: $D$
$A=p\cdot s^++(1-p)\cdot s^-$
$0.8181= p\cdot0.9+ (1-p)\cdot 0.8$
$0.0181= 0.1p$
$p=.181$
For one test positive out of one:
using Bayes' theorem: $P(D|T^+)=\frac{p \cdot s^+}{p \cdot s^+ + (1-p) \cdot (1-s^-)}$
$P(D|T^+)=\frac{.181 \cdot .9}{.181 \cdot .9 + (1-.181) \cdot (1-0.8)}$
$P(D|T^+)=\frac{.1629}{.1629 + .1638}=.4986$
now for k tests out of m: as in Peter's answer, the probabilities are now drawn from a binomial to generate likelihoods:
$L^+$~$B(n,s^+)$ and $L^-$~$B(n,s^-)$ with inverted number of successes for the second. You can probably calculate confidence intervals based on the knwon properties (variance) of bionomial distributions... | {
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# Prove or disprove that if $\sum a_n$ and $\sum b_n$ are both divergent, then $\sum (a_n \pm b_n)$ necessarily diverges.
Prove or disprove that if $\sum a_n$ and $\sum b_n$ are both divergent, then $\sum (a_n \pm b_n)$ necessarily diverges.
I decided to disprove this by offering the example:
Let $a_n = n$.
Let $b_n = -n$.
Then $\sum (a_n + b_n) = \sum 0 = 0$, which converges.
However, my question says $\pm$, and in the $-$ case, we have $\sum (a_n - b_n) = \sum 2n = \infty$, which diverges. So, does this still count as a valid counter example or not? I partly proved it wrong so I think it's still valid.
If $\sum(a_n+b_n)$ and $\sum(a_n-b_n)$ both converge, then since :
$$a_n=\frac12\left((a_n+b_n)+(a_n-b_n)\right)\quad\textrm{and}\quad b_n=\frac12\left((a_n+b_n)-(a_n-b_n)\right)$$
we see that $\sum a_n$ and $\sum b_n$ both converge.
In other words, if $\sum a_n$ diverges or $\sum b_n$ diverges, then $\sum(a_n+b_n)$ diverges or $\sum(a_n-b_n)$ diverges.
• Although this proves that if the original series don't both converge at least one of the $\pm$ series doesn't either, "does not converge" is not equivalent to "diverges". – J.G. Mar 20 '17 at 22:38
• Further to my already upvoted comment, apparently Wikipedia does define divergent as "not convergent", even though I could have sworn it meant "tends to $\pm\infty$". It seems weird to me that an oscillatory sequence, for example, would be described as "divergent". – J.G. Mar 20 '17 at 23:00
My reading of the statement "$\sum (a_n \pm b_n)$ necessarily diverges" is that it means that both of the sums $\sum (a_n + b_n)$ and $\sum (a_n - b_n)$ diverge. You have proved that this is not necessarily true; you have a counterexample in which one of the sums converges.
The wording of the original question seems highly ambiguous to me, however. The statement "$\sum (a_n \pm b_n)$ diverges" might be intended to say that either $\sum (a_n + b_n)$ diverges or $\sum (a_n - b_n)$ diverges. | {
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It really comes down to what is the meaning of the statement "$\sum (a_n \pm b_n)$ diverges."
Let $\sum \frac{1}{2n+1}$=$1+1/3+1/5+1/7+1/9+....$ and let $-\sum \frac{1}{2n}$=$-1/2-1/4-1/6-1/8...$. Boths series diverge, but when you add them (term by term) together, you get $1-1/2+1/3-1/4+1/5-1/6...$ and this is the well known Alternating Harmonic series with an answer of $\ln 2$, so (conditionally) convergent.
• In fact, when you add them term-by-term you get $\sum \left(\frac{1}{2n-1}-\frac{1}{2n}\right)$, which is absolutely convergent as each individual term is positive. It's only when you deparenthesize that you get a conditionally convergent result... – Micah Mar 20 '17 at 22:47 | {
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# Proving continuity of a multivariable function by “reducing” it to a single variable function
My goal is to examine the continuity of a certain multivariable function. In that regard, (mainly because my math in the specific area is dusty) I find it somewhat tricky to either prove or disprove that property in the N-diensional space. However, I have come up with the following idea:
let $$f : \mathbb{R}^2 \rightarrow \mathbb{R}$$ be the function that I want to examine at some point $$(x_0, y_0)$$.
Now I am using the following two functions $$g: \mathbb{R}^2 \rightarrow \mathbb{R}$$ and $$h: \mathbb{R} \rightarrow \mathbb{R}$$ so that I can write function $$f$$ as: $$f(x,y) = h(g(x,y)),\quad \forall (x,y) \in \mathbb{R}^2$$ I also know that function $$g$$ is indeed continuous in $$\mathbb{R}^2$$.
My assumption is that I can examine the continuity of $$h$$ at $$x' = g(x_0,y_0)$$ in order to decide the continuity of $$f$$ at $$(x_0,y_0)$$ - a much easier task. In mathimatical notation, I guess what I mean is: $$\lim_{(x,y)->(x_0,y_0)} f(x,y) = f(x_0, y_0) \quad \iff \\ \lim_{x->g(x_0,y_0)}h(x) = h(g(x_0,y_0))$$
Actual Questions:
1. Is my assumption correct?
2. If not, maybe do either of $$\Rightarrow$$ or $$\Leftarrow$$ hold? (Because if one of those holds then I can use the above to only prove or only disprove accoridnlgy.)
3. Does the same hold for proving/disproving differentiability as well?
4. Under what conditions do any of those hold for a (finite) sum of $$h,g$$ functions, i.e. $$f(x,y) = h_1(g_1(x,y)) + h_2(g_2(x,y)) + ... + h_k(g_k(x,y))$$
Note: I can actually post my original function and the transformation if the question is too broad or the answers depend greatly on the actual $$f,g$$ and $$h$$ definitions. However I believe a more general question may help more people find it and also make it easier to be understood. | {
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• We know that the composition of continuous functions is continuous. So it suffices to show that $g(x,y)$ is continuous at $\xi_0 = (x_0,y_0)$. That is that $\lim_{\xi \rightarrow \xi_0} g(\xi) = g(\xi_0)$. Also note that showing that the derivative exists at $x_0$ is a sufficient, but not necessary condition for continuity. That is differentiability implies continuity but not the converse. For an example see $f(x) = |x|$. This is continuous at $0$, and in fact Lipschitz continuous at $0$, but not differentiable there. (continued) – GeauxMath Nov 13 '18 at 23:33
• Also note a vector valued function is continuous if it is continuous in all its arguments – GeauxMath Nov 13 '18 at 23:39
• @GeauxMath First of all, thank you for responding. Now, I am fully aware about how differentiability implies continuity but not the other way around. Reading back my post, I think I was indeed looking for the property of compositions of continuous functions being continuous. I also take it, that in your first comment you meant to say that it suffices to show that $h(x)$ is continuous and not $g(x,y)$, as I already mentioned that $g$ is indeed continuous. – kyriakosSt Nov 14 '18 at 0:24
• yes, you are correct – GeauxMath Nov 14 '18 at 1:18
• Cheers. If you would like to modify your comment to an answer, I am willing to accept it – kyriakosSt Nov 14 '18 at 10:06
Just for closure, I am answering myself in case it helps anyone else. Merit goes to @GeauxMath and @Todor Markov for providing answers in comments.
We can answer mainly by using the following proposition:
A composition of continuous functions is continuous
In detail:
1. My assumption is incorrect as a whole
2. The $$\Leftarrow$$ direction holds by using the above proposition. The opposite is not true in every case. As a counter example
If $$g(x_0,y_0)=a$$ and $$f(x,y)≥a$$, then $$h$$ can have a discontinuity at a. If $$h(x)$$ is continuous for $$x>a$$, $$f$$ would still be continuous. | {
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3. The corresponding property applies also to differentiable functions, so we can only prove that $$f$$ is differentiable that way, but not disprove it.
4. Again, by using that proposition we can extend our statements to that specific case, so we can prove continuity and differentiability of $$f$$ but not disprove. A counter example here would be a case where $$h_1$$ and $$h_2$$ are not continuous, but by adding them, their discontinuities "cancel out". | {
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# Is every positive integer the permanent of some 0-1 matrix?
In the course of discussing another MO question we realized that we did not know the answer to a more basic question, namely:
Is it true that for every positive integer $k$ there exists a balanced bipartite graph with exactly $k$ perfect matchings?
Equivalently, as stated in the title, is every positive integer the permanent of some 0-1 matrix?
The answer is surely yes, but it is not clear to me how to prove it. Entry A089479 of the OEIS reports the number $T(n,k)$ of times the permanent of a real $n\times n$ zero-one matrix takes the value $k$ but does not address the question of whether, for every $k$, there exists $n$ such that $T(n,k)\ne 0$. Assuming the answer is yes, the followup question is, what else can we say about the values of $n$ for which $T(n,k)\ne 0$ (e.g., upper and lower bounds)? | {
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• The answers below write $k$ as the permanent of a matrix of size $k$ or $\log k$. I wonder if it is possible with a matrix of size $n$, where $(n-1)! < k \leq n!$? – Zach Teitler Aug 29 '17 at 18:39
• @Zach, it is true for some k in that range. However, if k is not a multiple of a factorial, one "needs a few more rows". I suspect there is an absolute constant C such that for k less than Cn! that k is representable as the permanent of an order n 0-1 matrix. Gerhard "Finding K Is Another Matter" Paseman, 2017.08.29. – Gerhard Paseman Aug 29 '17 at 19:52
• You can always direct sum with an identity matrix to get additional rows. – Timothy Chow Aug 29 '17 at 21:36
• In fact it seems that even for determinant we can construct $k\times k~0-1$ matrix to represent any number $k-1$, but in this case it's lower bound of size. – rus9384 Aug 29 '17 at 22:46
• @rus9384, indeed one can surpass the kth Fibonacci number in representing positive integers by a determinant of a k by k 0-1 matrix. once k is bigger than 5. Orrick's talk mentioned below gives some detail. Gerhard "And I Can Give More" Paseman, 2017.08.30. – Gerhard Paseman Aug 30 '17 at 16:48
The answer to the question is yes. Given $k$, the 0-1 matrix given by
$1$ $1$ $\dotsc$ $1$ $0$ $0$ $\dotsc$ $0$ $0$ $0$ $0$
$0$ $1$ $1$ $0$ $\dotsc$ $0$ $0$ $\dotsc$ $0$ $0$ $0$
$0$ $0$ $1$ $1$ $0$ $\dotsc$ $0$ $\dotsc$ $0$ $0$ $0$
$\dotsc$
$1$ $0$ $0$ $0$ $0$ $\dotsc$ $\dotsc$ $0$ $0$ $0$ $1$
where the first row has precisely $k$ entries equal to $1$, evidently has permanent equal to $k$.
For $k=1$ the matrix is ($1$). For $k=2$ the matrix is
$1$ $1$
$1$ $1$
For $k=3$ the matrix is
$1$ $1$ $1$
$0$ $1$ $1$
$1$ $0$ $1$.
For $k=4$ the matrix is
$1$ $1$ $1$ $1$
$0$ $1$ $1$ $0$
$0$ $0$ $1$ $1$
$1$ $0$ $0$ $1$
which evidently has permanent equal to $4$. | {
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$0$ $1$ $1$ $0$
$0$ $0$ $1$ $1$
$1$ $0$ $0$ $1$
which evidently has permanent equal to $4$.
Please note that also, for each given $k$ and $1\leq \ell \leq k$, this construction can be tweaked to give an explicit $k\times k$ sized $0$-$1$-matrix having permanent precisely $\ell$, just by making the first row have precisely $\ell$ entries equal to $1$.
Please also note that my construction does not have any bearing on the interesting and apparently difficult question which was cited in the present OP: my construction is too wasteful: it utilizes a $k\times k$ matrix, which is far too large when it comes to meet the demands of the OP in said question.
• Nice construction! Shortly after posting my question, I discovered the following paper by Kim, Lee, and Seol that also answers the question: ijpam.eu/contents/2005-19-3/12/12.pdf – Timothy Chow Aug 29 '17 at 17:44
• The graph picture for anyone who doesn't want to work out the details: this is a Hamilton cycle of length $2k$ together with a star whose leaves are all the elements of one colour class and whose centre is in the other. Any choice of matching edge for the centre of the star is a chord that splits the cycle into two paths, each of which has a unique perfect matching, giving the $k$ options. – Ben Barber Aug 29 '17 at 17:49
For the record, here is the construction of Kim, Lee, and Seol that I alluded to in my comment to Peter Heinig's answer.
Write $k-1$ in binary, and let $n$ be 1 plus the number of binary digits of $k-1$.
Start with an $n\times n$ matrix with all $1$'s on or above the main diagonal and all $0$'s below the diagonal. Then replace the first $n-1$ entries of the bottom row of the matrix with the binary representation of $k-1$ (one bit per entry). | {
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For example, if $k=389$, then $k-1$ in binary is $110000100$. Then $n=1+9=10$ and the matrix is $$\matrix{ 1&1&1&1&1&1&1&1&1&1\cr 0&1&1&1&1&1&1&1&1&1\cr 0&0&1&1&1&1&1&1&1&1\cr 0&0&0&1&1&1&1&1&1&1\cr 0&0&0&0&1&1&1&1&1&1\cr 0&0&0&0&0&1&1&1&1&1\cr 0&0&0&0&0&0&1&1&1&1\cr 0&0&0&0&0&0&0&1&1&1\cr 0&0&0&0&0&0&0&0&1&1\cr 1&1&0&0&0&0&1&0&0&1\cr }$$
• Of course, one can represent pq by a block matrix for p and a block for q on the diagonal. This construction using binary of k-1 works well for odd numbers and is rather compact in general. However, even for determinant, the spectrum problem is not well understood, so finding the optimal n given k will require even more cleverness. For example, it is not clear that n=7 might work for k=389. Does the cited paper mention how optimal the construction is? Gerhard "Hasn't Determined Complexity Of Permanent" Paseman, 2017.08.29. – Gerhard Paseman Aug 29 '17 at 18:52
• @GerhardPaseman why pq form? – T.... Aug 30 '17 at 21:40
• pq here means the factor p times the factor q. For numbers m with small factors, the block matrix form reduces the problem of representing m compactly one of representing enough of its small factors compactly. Gerhard "Because It's Easy, That's Why" Paseman, 2017.08.30. – Gerhard Paseman Aug 30 '17 at 22:49
If an order n (0,1) matrix has r rows with all ones, its permanent is a multiple of r! by an easy argument. It follows that the largest odd number which is a permanent of an order n matrix is q = der(n) + der(n-1), which is a sum of numbers of derangements, and is a little larger than $n!/e$. I suspect that the smallest positive number not expressible as a permanent of this size matrix is an odd integer which is not much smaller, perhaps about half, of q. If so, then we can shave a little bit off of n=log(k) in the construction in Timothy Chow's post. For some motivation for this problem, look at Will Orrick's talk mentioned (for determinants) at https://mathoverflow.net/a/271273 . | {
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Gerhard "Also A Potential Polymath Project" Paseman, 2017.08.29. | {
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# Why can the covariance matrix be computed as $\frac{X X^T}{n-1}$?
Wikipedia defines the covariance matrix as a generalized version of the variance: $C=E[(X-E[X])(X-E[X])^T]$. But, I usually see statisticians calculate the covariance matrix as $C=\frac{X X^T}{n-1}$ (e.g., Ameoba's nice answer here https://stats.stackexchange.com/a/134283/83526). Is $E[(X-E[X])(X-E[X])^T] = \frac{X X^T}{n-1}$? Are these both correct definitions of the covariance matrix?
• Amoeba explicitly prefaces that formula with the statement "Let us assume that it [$X$] is centered, i.e. column means have been subtracted and are now equal to zero." That shows his formula is identical to the one in Wikipedia. – whuber Jun 14 '17 at 20:38
Let $\mu = E(X)$. Then $$Var(X) = E\left((X - \mu)(X - \mu)^T\right) = E\left(XX^T - \mu X^T - X \mu^T + \mu \mu^T\right) \\ = E(XX^T) - \mu\mu^T$$ which generalizes the well-known scalar equality $Var(Z) = E(Z^2) - E(Z)^2$.
The natural estimator of $\Sigma := Var(X)$ is $\hat \Sigma = \frac 1{n-1}XX^T - \hat \mu \hat \mu^T$.
In many situations we can take $\mu = 0$ without any loss of generality. One common example is PCA. If we center our columns then we find that $\hat \mu = 0$ so our estimate of the variance is simply $\frac 1{n-1}XX^T$. The univariate analogue of this is the familiar $s^2 = \frac 1{n-1} \sum_i x_i^2$ when $\bar x = 0$.
As @Christoph Hanck points out in the comments, you need to distinguish between estimates and parameters here. There is only one definition of $\Sigma$, namely $E((X - \mu)(X - \mu)^T)$. So $\frac 1{n-1}XX^T$ is absolutely not the correct definition of the population covariance, but if $\mu=0$ it is an unbiased estimate for it, i.e. $Var(X) = E(\frac 1{n-1}XX^T)$. | {
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• To understand the linear algebra used in the answer better, you can look at math.stackexchange.com/questions/198257/… – kjetil b halvorsen Jun 14 '17 at 16:26
• +1 I tried a few times to squeeze a comment in here, but the $LaTeX$ didn't show well, so I used the entry of an answer as an extended comment. – Antoni Parellada Jun 15 '17 at 0:14
• Note that there is a bit of ambiguity in the symbology here (inherited from the question): The symbol $X$ is used to refer to both the random variable ($p\times{1}$, vector valued), and the (transposed) data matrix ($p\times{n}$, columns are samples of the random variable). – GeoMatt22 Jun 15 '17 at 0:30
COMMENT:
@Chacone's answer is great as is, but I think there is one step that is left unexplained, and it is clearer with expectation notation. To reflect @GeoMatt22's comment in the following proof $X$ is a $p\times 1$ random vector:
\begin{align} \text{Cov}(X)&=\mathbb E\left[\,\left(X- \mathbb E[X] \right) \, \left(X- \mathbb E[X] \right)^\top \right]\\[2ex] &= \mathbb E\left[\,XX^\top - X\,\mathbb E[X]^\top - \mathbb E[X] \,X^\top + \mathbb E[X] \mathbb E\, [X]^\top \right]\\[2ex] &= \mathbb E\left[\,XX^\top\right] - \mathbb E [X]\,\mathbb E[X]^\top - \mathbb E[X] \,\mathbb E [X]^\top + \mathbb E[X] \mathbb E\, [X]^\top \\[2ex] &= \mathbb E[XX^\top] \; -\; \mathbb E[X]\, \mathbb E[X]^\top \end{align}
As for the transition to the sample estimate of the population covariance using this alternate (raw moment) formula, the denominators will need adjusing. In what follows $X$ is a $p \times n$ data matrix (following @GeoMat22's comment):
\begin{align} \sigma^2(X)&=\frac{XX^\top}{n-1} \; -\; \frac{n}{n-1}\; \begin{bmatrix}\bar X_1\\ \bar X_2\\ \vdots \\ \bar X_p \end{bmatrix}\, \; \begin{bmatrix}\bar X_1 & \bar X_2 & \cdots & \bar X_p \end{bmatrix} \end{align} | {
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because it is necessary to multiply by $\frac{n}{n-1}$ to get rid of the biased $n$ denominator, and replace it with $n-1$. Evidently anything after the minus sign disappears if the mean is zero.
Here is an illustrative example in R:
> set.seed(0)
> # Sampling three random vectors: V1, V2 and V3:
> X1 = 1:5
> X2 = rnorm(5, 5, 1)
> X3 = runif(5)
> # Forming a matrix with each row representing a sample from a random vector:
> (X = (rbind(X1,X2,X3)))
[,1] [,2] [,3] [,4] [,5]
X1 1.00000000 2.0000000 3.0000000 4.0000000 5.0000000
X2 6.26295428 4.6737666 6.3297993 6.2724293 5.4146414
X3 0.06178627 0.2059746 0.1765568 0.6870228 0.3841037
> # Taking the estimate of the expectation for each random vector, bar Xi:
> mu = rowMeans(X)
> # Calculating manually the variance with the alternate formula
> (man.cov = ((X %*% t(X)) / (ncol(X) - 1)) - (ncol(X)/(ncol(X) - 1)) * (mu %*% t(mu)))
X1 X2 X3
X1 2.50000000 -0.02449075 0.28142079
X2 -0.02449075 0.53366886 0.02019664
X3 0.28142079 0.02019664 0.05940930
> # Comparing to the built-in formula:
> cov(t(X))
X1 X2 X3
X1 2.50000000 -0.02449075 0.28142079
X2 -0.02449075 0.53366886 0.02019664
X3 0.28142079 0.02019664 0.05940930
> # are the same...
> all.equal(man.cov, cov(t(X)))
[1] TRUE | {
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# Proof for a logical truth involving quantifiers.
Prove: $(\forall x Fx \vee \forall x Gx)\to \forall x(Fx\vee Gx)$
My attempt:
\begin{align} 1.\space & \neg\forall x(Fx\vee Gx) & \text{(Conditional Proof)}\\ 2.\space & \exists x\neg(Fx\vee Gx) & \text{Quantifier Negation on line 1}\\ 3.\space & \exists x(\neg Fx\wedge \neg Gx) & \text{De Morgan's on line 2}\\ & \qquad 4.\space \neg Fx\wedge \neg Gx & \text{Existential Instantiation on line 3}\\ & \qquad 5.\space \neg Fx & \text{Simplification on line 4}\\ & \qquad 6.\space \neg Gx & \text{Simplification on line 4}\\ & \qquad 7.\space \exists x \neg Fx & \text{Existential Generalization on line 5}\\ & \qquad 8.\space \neg\forall x Fx & \text{Quantifier Negation on line 7}\\ & \qquad 9.\space \exists x \neg Gx & \text{Existential Generalization on line 6}\\ & \qquad 10.\space \neg\forall x Gx & \text{Quantifier Negation on line 9}\\ & \qquad 11.\space \neg\forall x Fx \wedge\neg\forall x Gx & \text{Conjunction from lines 8, 10}\\ 12.\space & \neg\forall x Fx \wedge\neg\forall x Gx & \text{Existential Instantiation lines 3, 4-11}\\ 13.\space & \neg(\forall xFx \vee \forall x Gx) & \text{De Morgan's on line 12}\\ 14.\space & \neg\forall x(Fx\vee Gx)\rightarrow \neg(\forall Fx \vee \forall Gx) & \text{Conditional Proof on lines 1-13}\\ 15.\space &\boxed{(\forall x Fx \vee \forall x Gx) \rightarrow \forall x(Fx\vee Gx)} & \text{Transposition on line 14} \end{align}
I am mostly concerned about the Existential Instantiation subproof from 4-11. Also, it is the first time I have done this kind of proof. So, let me know of ways to improve it! | {
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• Is there some reason why you proved the contrapositive instead of just directly proving the statement? – Derek Elkins Oct 2 '17 at 4:19
• I wasn't sure about Universally Instantiating $\forall x Fx$ and $\forall x Gx$ on the same line. Can I? – G_D Oct 2 '17 at 4:25
• Why do you think you'd need to do it on the same line? – Derek Elkins Oct 2 '17 at 4:31
• IIRC, Universal Instantiation needs to affect the entire line. And I would be instantiating to $x$ twice (if that makes sense). – G_D Oct 2 '17 at 4:36
• If you start the direct proof, you'll find that when you need to instantiate $\forall x.Fx$ or $\forall x.Gx$, the other won't be there, but even in cases where you are instantiating a universal quantifier, only things in the scope of the quantifier are going to be affected. You pick a quantified formula and instantiate it. You don't just instantiate every quantifier on the line. That said, I think you mean to worry about Universal Generalization as that's the rule with side conditions. – Derek Elkins Oct 2 '17 at 4:43
Your use of "existential generalization" and "existential instantiation" seems backward. Existential instantiation is:
\begin{align} \exists x \phi(x) \\ \hline \phi(c) \end{align}
For some constant $c$. And existential generalization is when you go the other way.
Also note that after existential instatiation the symbol $c$ is a constant and you should not use that one in a quantifier then. You should probably used another name than $x$ for the constant introduced by existential instantiation.
Also note that it's normally the conditional proof that would require indentation to indicate that these statements are only a consequence of an assumption. For the existential instantiation this is not as required as we only require that we can simply get rid of a constant not appearing in our formula any longer.
Here is a proof using a Fitch-style proof checker to make sure I am applying the rules correctly: | {
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Here is a proof using a Fitch-style proof checker to make sure I am applying the rules correctly:
Note that I consider both cases of the disjunction in line 1 and arrive at the same result on lines 4 and 7. Then I use disjunction elimination on line 8 and finally universal introduction on line 9.
Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/
$$\vdash (\forall x~Fx~\vee~\forall x~Gx)~\to~~\forall x~(Fx\vee Gx)$$
...may be proven directly via disjunctive syllogism (aka disjunctive elimination):
$$(\forall x~Fx~\vee~\forall x~Gx), (\forall x~Fx)~\to~\forall x~(Fx\vee Gx), (\forall x~Gx)~\to~\forall x~(Fx\vee Gx)\vdash\forall x~(Fx\vee Gx)$$
Since $$\vdash (\forall x~Fx)\to \forall x~(Fx\vee Gx)$$ is provable by assumption, universal elimination, disjunctive introduction, conditional introduction, and universal reintroduction. Quite similarly we have $$\vdash (\forall x~Gx)\to \forall x~(Fx\vee Gx)$$.
$$\begin{array}{l|l:ll} \hdashline 1 & \quad \forall x~Fx~\vee~\forall x~Gx&& \text{Assume} \\\hdashline 2 & \qquad \forall x ~Fx &1& \text{Assume }\vee\mathsf L\\\hdashline 3 & \quad\qquad Fc &2& \forall-\\ 4 & \quad\qquad Fc\vee Gc &3& \vee+ \\\hline 5 & \qquad \forall x~(Fx\vee Gx) &4& \forall + \\ \hline 6 & \quad(\forall x~Fx)\to(\forall x~(Fx\vee Gx)) &2,5& \to+ \\ \hdashline 7 & \qquad\forall x~Gx &1&\text{Assume }\vee\mathsf R \\ \hdashline 8 & \quad\qquad Gc &7& \forall -\\ 9 & \quad\qquad Fc\vee Gc &8& \vee+ \\ \hline 10& \qquad \forall x~(Fx\vee Gx) &9& \forall + \\ \hline 11& \quad(\forall x~Gx)\to \forall x~(Fx\vee Gx) &7,10& \to+\\ \hline 12& (\forall ~Fx~\vee~\forall x~Gx)\to \forall x~(Fx\vee Gx) &1,6,11& \vee-\end{array}$$ | {
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James invested $5000 in scheme A for 1 year at a simple annual interest rate of 5% and invested another$10000 in scheme B for one year at an annual interest rate of 10% compounded semi-annually. What is the positive difference between the interest earned by James from scheme A and scheme B?
A. 250
B. 775
C. 1025
D. 1750
E. 2000
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Re: James invested $5000 in scheme A for 1 year at a simple annual interes [#permalink] Show Tags 27 Aug 2018, 05:11 1 EgmatQuantExpert wrote: James invested$5000 in scheme A for 1 year at a simple annual interest rate of 5% and invested another $10000 in scheme B for one year at an annual interest rate of 10% compounded semi-annually. What is the positive difference between the interest earned by James from scheme A and scheme B? A. 250 B. 775 C. 1025 D. 1750 E. 2000 To read all our articles: Must read articles to reach Q51 Scheme A:$5,000 at 5% simple annual interest
5,000 * .05 * 1 = 5250 ----> $250 of interest Scheme B:$10,000 at 10% compounded semi annually
10,000(1 + $$\frac{.1}{2}$$)$$^2$$ --------> 10,000(1.05)$$^2$$ --------> 10,000(1.1025) = 11,025
$11,025 ----------->$1,025 of interest
Positive difference between the interest earned by Scheme A vs Scheme B
$1,025 -$250 = $775 Answer: B _________________ Would I rather be feared or loved? Easy. Both. I want people to be afraid of how much they love me. How to sort questions by Topic, Difficulty, and Source: https://gmatclub.com/forum/search.php?view=search_tags Board of Directors Status: QA & VA Forum Moderator Joined: 11 Jun 2011 Posts: 4011 Location: India GPA: 3.5 WE: Business Development (Commercial Banking) Re: James invested$5000 in scheme A for 1 year at a simple annual interes [#permalink]
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27 Aug 2018, 07:11
EgmatQuantExpert wrote:
James invested $5000 in scheme A for 1 year at a simple annual interest rate of 5% and invested another$10000 in scheme B for one year at an annual interest rate of 10% compounded semi-annually. What is the positive difference between the interest earned by James from scheme A and scheme B?
A. 250
B. 775
C. 1025
D. 1750
E. 2000 | {
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} |
A. 250
B. 775
C. 1025
D. 1750
E. 2000
$$Amount_{Si} = 5000 + \frac{5000*5*1}{100}$$
So, $$Amount_{Si} = 5250$$
$$Amount_{Ci} = 5000( 1 + \frac{10}{200})^{2}$$
So, $$Amount_{Ci} = 11025.00$$
So, the required difference is $1,025 -$250 = $775 , Answer must be (B) _________________ Thanks and Regards Abhishek.... PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only ) Re: James invested$5000 in scheme A for 1 year at a simple annual interes &nbs [#permalink] 27 Aug 2018, 07:11
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