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Yemon Choi (henceforth "YC") answered the question to within a constant factor with a function $f_\delta$ showing that $\alpha(E_\delta) \leq 2/\delta$ where $E_\delta$ is an interval of measure $1-\delta$ (YC's notation). That's optimal asymptotically $-$ and even exactly if $2/\delta \in {\bf Z}$ $-$ for functions $f... | {
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To give an example with smaller $\| \phantom. f \|_{A(T)}$, consider the following generalization: YC's function $f_\delta$ is the self-convolution of uniform meausre on an interval of length $\delta/2$; for $0 \lt b \lt 1$ define $f_{\delta,b}$ to be the convolution of uniform measures on intervals of length $b\delta$... | {
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The dictionary between bounds on $\delta \alpha(E_\delta)$ as $\delta \rightarrow 0$ and bounds on $\alpha_0$ also works for the following improved lower bound, which I'll give only in the $\alpha_0$ setting. Here Joël's "easy lower bound" is $\alpha_0 \geq 1$, proved by noting that $$1 = \int_{-\infty}^\infty f(t) dt ... | {
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-
Just to get the ball rolling: for any $\delta \in(0,1)$, the function $f_\delta(t)$ which is supported on the interval $[-\delta\pi,\delta\pi]$ and looks like $1- (\delta\pi)^{-1}|t|$ inside it, has norm $1$ in $A(T)$ -- one either sees this using positive-definiteness or a trick with convolving characteristic funct... | {
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# Different interpretations, the same result!
In a box there are 10 white ball and 10 black balls.Two balls are successively drawn.What is the probability to get two balls of different colors?
My interpretation is that the order doesn't matter, because the important is that the balls have different colors.So to find ... | {
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We can start by looking at all patterns from arranging all twenty balls. There are $20!$ ways of ordering all twenty balls. Of these $10 \times 10 \times 18!$ have a black followed by a white at the beginning and $10 \times 10 \times 18!$ have a white followed by a black at the beginning, so the proportion of patterns ... | {
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# Find the last non zero digit of 28!.
Find the last non zero digit of 28!.
It is very hard to multiply and find the last nonzero digit. I just wanna know that, is there any easy technique to solve this type of problem?
• Consider factors of $10$ separately (you can ignore them). Build factors of $10$ by pairing a f... | {
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• Nice solution. You dropped the ! from $28!/10^6$, but it won't let me make a one-character edit. – Bungo Sep 10 '14 at 21:56
• @Bungo: Thanks:) – mathlove Sep 10 '14 at 21:57
• @Bungo in cases of need you can add an HTML <!-- comment --> explaining why your too-short edit really is good, and then it won't be a too-sh... | {
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Perhaps a small reduction in the calculations in other answers. First, we use the standard technique for finding the power of a prime which divides a factorial: $$\Bigl\lfloor\frac{28}{5}\Bigr\rfloor+\Bigl\lfloor\frac{28}{25}\Bigr\rfloor =5+1=6$$ so $5^6$ is a factor of $28!$ (and $5^7$ is not), and likewise $$\Bigl\lf... | {
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The idea is to extract all powers of $5$ and $2$ from $n!$, and write it as $10^A2^BC$ with $C$ odd and not a multiple of $5$. Then $A$ is the number of trailing zeros. And whatever $2^BC$ is modulo 10 is the preceding digit. To that end, these two reduction formulas are useful. They use $$n\underset{2}{!}=1\cdot3\cdot... | {
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\begin{align} 1000!&\equiv1000\underset{2}{!}\times500!\times2^{500}\\ &\equiv1000\underset{5}{!}\times199\underset{2}{!}\times5^{100}\times500!\times2^{500}\\ &\equiv9^{100}\times500!\times199\underset{2}{!}\times2^{500}\times5^{100} \\ &\equiv500!\times199\underset{2}{!}\times2^{500}\times5^{100}\\ & \text{...note we... | {
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# Pi Unleashed
###### Jörg Arndt and Christoph Haenel
Publisher:
Springer
Publication Date:
2001
Number of Pages:
276
Format:
Paperback with CDROM
Price:
52.95
ISBN:
978-3540665724
Category:
General
[Reviewed by
Carl D. Mueller
, on
10/3/2002
]
This number will probably look familiar to the readers of this review, an... | {
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Let me begin this review by stating simply that I know of no better book than this one for those interested in everything π. While the book itself has fewer than 300 pages, it is packed with information, formulas, historical notes, etc. Just about anything you could want is here. The following list of chapter titles sh... | {
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π Unleashed can be used as a reference book or it can be read straight through as one would read a textbook or a novel. It begins with an overview of π. The latest record is given, several formulas and algorithms are presented and discussed briefly, and the question of whether or not π is normal is asked. Later chapter... | {
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There is a chapter on arithmetic. This may seem strange in a book on π, but when a computer is used to calculate billions of digits of π, it is not enough to have an efficient π algorithm. It is also necessary to perform arithmetic operations as efficiently as possible. In particular, an efficient method for multiplica... | {
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There is more in the book: fascinating π trivia, intriguing formulas, and predictions for the future. There is far too much to describe in a brief review. Of course, no book is perfect and this one has a few minor errors (perhaps occurring in the translation process) such as the description of the constant e as a limit... | {
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1. The State Of Pi Art 1
2. How Random Is pi? 21
2.1 Probabilities 21
2.2 Is pi normal? 21
2.3 So is pi not normal? 24
2.4 The 163 phenomenon 25
2.5 Other statistical results 28
2.6 The Intuitionists and pi 30
2.7 Representation of continued fractions 32
3. Shortcuts To pi 35
3.1 Obscurer approaches to pi 35
3.2 Small ... | {
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13.4 High-performance algorithms 198
13.5 The hunt for single pi digits 203
Table: History of pi in the pre-computer era 205
Table: History of pi in the computer era 206
Table: History of digit extraction records 207
14. Historical Notes. 209
14.1 The earliest squaring the circle in history? 209
14.2 A pi law 211
14.3 ... | {
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# On determinants of a certain class of matrices
$$\newcommand{\M}{\mathcal{M}}$$Recently I encountered a certain class of matrices whose determinants behave in an interesting manner. Define $$\M(n,k)$$ for positive integers $$n,k$$ with $$k\leq n$$ to be the real $$n\times n$$ matrix with all $$1$$s on the diagonal, ... | {
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Here are some partial results. It is trivial that $$\det\M(n,k)$$ is zero when $$k=0$$ or $$n$$, and that it is $$1$$ when $$k=1$$. I can prove the result when $$k=2$$ as well: Let $$U(n)$$ be the $$n\times n$$ matrix with all entries $$0$$ except a $$1$$ at the bottom leftmost corner. Then, we have the recurrence rela... | {
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The determinant of $$C$$ is given by $$\displaystyle\prod_{j=0}^{n-1}p(\zeta_n^j)$$, where $$\zeta_n:=\exp(2\pi i/n)$$ is the primitive $$n$$th root of unity.
The matrices $$\M(n,k)$$ are circulant, with associated polynomial $$p(x)=1+x+\dots+x^{k-1}$$. Note that whenever $$x\neq 1$$, we have $$p(x)=(x^k-1)/(x-1)$$, s... | {
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$$\begin{pmatrix}1&1&1&0&0\\0&1&1&1&0\\0&0&1&1&1\\{\color{red}1}&0&0&1&1\\{\color{red}1}&{\color{red}1}&0&0&1\end{pmatrix}\mapsto\begin{pmatrix}1&1&1&0&0\\0&1&1&1&0\\0&0&1&1&1\\0&{\color{red}-\color{red}1}&{\color{red}-\color{red}1}&1&1\\0&0&\color{red}-\color{red}1&0&1\end{pmatrix}\mapsto\begin{pmatrix}1&1&1&0&0\\0&1&... | {
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• Sorry, I'm afraid I don't understand what you're trying to say. What do you mean by "as you perform row subtraction or addition"? Which rows are added/subtracted from which others? I don't understand what your diagram is trying to say, nor to I get anything at all in the last paragraph. Maybe you want to rephrase it?... | {
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# Even Fibonacci numbers
Today, I found the Euler Project. Problem #2 is
Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:
1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
By considering the terms in the Fibonacci sequence whose valu... | {
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n[F_] := Floor[Log[F Sqrt[5]]/Log[GoldenRatio] + 1/2]
n[4000000]
(* 33 *)
So, $k = 11$. In code,
Clear[evenSum];
evenSum[(k_Integer)?Positive] :=
Round @ N[
With[
{phi = GoldenRatio^3, psi = (-GoldenRatio)^(-3)},
(1/Sqrt[5])*(phi*((1 - phi^k)/(1 - phi)) - psi*((1 - psi^k)/(1 - psi)))
],
Max[\$MachinePrecision, 3 k] ... | {
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These handle 10^10000 in slightly under a second on my desktop running Mathematica 9.0.1.
Here is a test for whether an integer is a Fibonacci number (a Fib detector?)
fibQ[n_] :=
With[{k =
Round[Log[GoldenRatio, N[n, 20 + Length[IntegerDigits[n]]]] +
Log[GoldenRatio, Sqrt[5]]]},
n == Fibonacci[k]]
-
You need to cor... | {
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Edit: To bring that into a cogent form. Here the full definition of SumEvenFibonacci:
ClearAll[SumEvenFibonacci];
SumEvenFibonacci::usage =
"SumEvenFibonacci[n] calculates the sum of even Fibonacci numbers \
up to the upper-bound n.";
SumEvenFibonacci[n_Integer] /; n >= 0 := Module[{res = 0, i = 1},
While[res < n, res... | {
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## protected by Verbeia♦Nov 17 '14 at 1:30
Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site. | {
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# Do Diagonal Matrices Always Commute?
Let $A$ be an $n \times n$ matrix and let $\Lambda$ be an $n \times n$ diagonal matrix. Is it always the case that $A\Lambda = \Lambda A$? If not, when is it the case that $A \Lambda = \Lambda A$?
If we restrict the diagonal entries of $\Lambda$ to being the equal (i.e. $\Lambda... | {
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• Do the copies have to be right next to each other? e.g. diag(1,1,1,2,2,7,8,8)? – Hrit Roy Feb 8 '18 at 5:25
• @HritRoy, no. You can change your basis to express the copies next to each other if you want to. But it is not necessary – Vladimir Vargas Apr 27 '18 at 22:22
It is possible that a diagonal matrix $\Lambda$ ... | {
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On a more useful note, you can look up commuting matrices on Wikipedia.
We want to find $\Lambda$ such that $A\Lambda=\Lambda A$. Then we have $$\left( \begin{array}{cc} \sum a_{1n}\lambda_{n1} & \sum a_{1n}\lambda_{n2} & \sum a_{1n}\lambda_{n3} & ... \\ \sum a_{2n}\lambda_{n1} & \sum a_{2n}\lambda_{n2} & \sum a_{2n}\... | {
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# Applying linear algebra to solve a problem in mechanical equilibrium
I came across the following problem in "Introduction to Applied Mechanics" by Gilbert Strang, and am a little confused about the solution to this problem. The following figure shows the problem.
The following vectors are defined for the problem,
... | {
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Now the force on the $i^{th}$ mass is $f_i = y_i - y_{i+1}$. If the spring above the $i^{th}$ mass is under tension then $y_i$ will be positive, and will tend to pull the mass upwards, and thus it has a positive contribution $y_i$ to the force on the mass. The opposite is true for the spring below the mass, thus has a ... | {
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$$\mathbf{e} = \left( \begin{array}{ccc} 1 & 0 & 0\\ -1 & 1 & 0\\ 0 & -1 & 1\\ 0 & 0 & -1\\ \end{array} \right) \mathbf{x} = \mathbf{A}\mathbf{x}$$
And he has $e_i = x_{i} - x_{i-1}$ (the opposite of your equation), which makes sense because $x_0=0$ and the obvious physical intuition is that $e_1>0$, i.e. the first/to... | {
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$$m_1g = \mathbf{f} = k_1x_1 + k_2x_1$$
This is as if the springs were actually in parallel if both were attached to the same surface; you can visualize this by rotating the bottom surface (while keeping the spring attached) until you glue it to the top one.
In matrix form, the dependence of $\mathbf{f}$ on $\mathbf{... | {
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Latin squares of even order-sub squares
Consider Latin squares of even order that is not of form $2^x$, where every cell is involved in a $2\times 2$ sub square. Here is one such square for order 6:
0 1 2 3 4 5
1 0 3 4 5 2
2 4 0 5 1 3
3 5 1 2 0 4
4 2 5 1 3 0
5 3 4 0 2 1
Notice that rows correspond to each... | {
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So, for example, $S_4$ is a group of order $24$, and every entry in its Cayley table (a Latin square of order $24$) belongs to some $2 \times 2$ subsquare, some $3 \times 3$ subsquare, some $4 \times 4$ subsquare, and so on for the other subgroup sizes.
Non-group examples can be formed by a direct product construction... | {
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• J. Browning, D. S. Stones and I. M. Wanless, Bounds on the number of autotopisms and subsquares of a Latin square, Combinatorica, to appear.
• I happen to be working on a proof that it is not possible for odd order. – Xuan Huang Aug 3 '12 at 1:49
$$\matrix{0&1&2&3&4&5&6&7&8&9\cr1&0&3&2&5&4&7&6&9&8\cr2&3&4&5&6&7&8&9... | {
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## BOTTOM - The Bottom of a Graph
no tags
We will use the following (standard) definitions from graph theory. Let $V$ be a nonempty and finite set, its elements being called vertices (or nodes). Let $E$ be a subset of the Cartesian product $V \times V$, its elements being called edges. Then $G = (V, E)$ is called a d... | {
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### Sample Input
3 3
1 3 2 3 3 1
2 1
1 2
0
### Sample Output
1 3
2 | {
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hide comments | {
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< Previous 1 2 3 4 5 6 7 Next > aman_sachin200: 2018-06-17 22:00:51 Nice one!!!Try CAPCITY and TOUR after this! sherlock11: 2018-06-08 10:29:56 if u want a clear understanding of SCC then this problem and CAPCITY are the problems that u are looking for.............if u are new with SCC then first read the concepts (kos... | {
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Added by: Wanderley Guimarăes Date: 2007-09-21 Time limit: 0.254s Source limit: 50000B Memory limit: 1536MB Cluster: Cube (Intel G860) Languages: All except: ERL JS-RHINO Resource: University of Ulm Local Contest 2003 | {
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# Find the minimum number of terms needed to approximate the series … correct to 1DP
The series is $\sum_{k=1}^{\infty}\frac{(-1)^{k+1}k}{k^2+1}$
The minimum number of terms needed should be given by $\lvert a_{n+1} \rvert$
Thus $\lvert \frac{(-1)^{k+2}(k+1)}{(k+1)^2+1} \rvert<0.01$
However, this is only satisfied ... | {
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# Lenth of line around an object?
1. May 30, 2004
### Physics is Phun
Lets say i had a cylinder x long with y radius with a string attached to the edge and I wrapped a string around it z number of time until the string reached the other end. How would I go about finding the length of this string. What if the object ... | {
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"... |
4. Jun 1, 2004
### Physics is Phun
I am afraid I am confused and I may have confused you with my question. So I have a cylinder and a piece of string. If I just wrap the string around the can the length of the string will be the circumference. I want to find the length of the string when it is spiraled around the cyl... | {
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for 3 turns, sqrt( (3*32.5)^2+16.5^2)=98.88
7. Jun 1, 2004
### Gokul43201
Staff Emeritus
Clearly this problem is easier to solve (jcsd's post) than if the object were a cone or sphere or pyramid. Also, you would have to add more information to completely define the path of the string.
For a sphere you could require... | {
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# How many ways can you put 8 red, 6 green and 7 blue balls in 4 indistinguishable bins?
1. Assume all balls with the same color are indistinguishable.
2. The order in which balls are put in a bin does not matter.
3. No bins are allowed to have the same distribution of balls!
For example, this configuration
{RRGGGB}... | {
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Applying PET and PIE it now follows almost by inspection that the desired count is given by the term $$\sum_{q=1}^n (-1)^{n-q} [R^r G^g B^b] Z(P_q)\left(\frac{1}{1-R}\frac{1}{1-G}\frac{1}{1-B}\right).$$
We need to employ PIE here because the substituted cycle index will include empty slots, which we are not counting i... | {
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res := ind;
polyvars := indets(poly);
indvars := indets(ind);
for v in indvars do
pot := op(1, v);
subs1 :=
[seq(polyvars[k]=polyvars[k]^pot,
k=1..nops(polyvars))];
subs2 := [v=subs(subs1, poly)];
res := subs(subs2, res);
od;
res;
end;
allparts :=
proc(val, size)
option remember;
local res, els, p, pp, q;
res :... | {
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White Ceramic Tile Texture Seamless, Presonus Eris E5 Xt Vs Jbl 305p Mkii, Old Dutch Chips Flavours, Logistic Regression Calculator Excel, When Do Herons Mate, Amma Mess Karaikudi, Numerology By Date Of Birth, " />
Posted by:
Category: | {
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Bhatnagar, G. Inverse Relations, Generalized Bibasic Series, and their U(n) Extensions. 4. In particular, we can determine the sum of binomial coefficients of a vertical column on Pascal's triangle to be the binomial coefficient that is one down and one to the right as illustrated in the following diagram: 1 à 8 (en) J... | {
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and (2) Brings together much of the undergraduate mathematics curriculum via one theme (the binomial coefficients). Today we continue our battle against the binomial coefficient or to put it in less belligerent terms, we try to understand as much as possible about it. A combinatorial interpretation of this formula is a... | {
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be used to simplify expressions involving several binomial coe cients. = \frac{n}{k} \cdot \frac{(n - 1)! Here are just a few of the most obvious ones: The entries on the border of the triangle are all 1. http://www.combinatorics.org/Volume_3/Abstracts/v3i2r16.html, https://mathworld.wolfram.com/BinomialIdentity.html. ... | {
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by: §26.5(iv) (1 + x−1)n.It is reflected in the symmetry of Pascal's triangle. This interpretation of binomial coefficients is related to the binomial distribution of probability theory, implemented via BinomialDistribution. 8. Append content without editing the whole page source. Riordan, J. Combinatorial Listing them... | {
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often lend themselves to combinatorial proofs. Binomial Coefficients (3/3): Binomial Identities and Combinatorial Proof - Duration: 8:30. Umbral Calculus. 1994, p. 203). The following relations all hold. Discrete Math. Binomial Coefficients and Identities (1) True/false practice: (a) If we are given a complicated expre... | {
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it is given by the formula Saslaw, W. C. "Some Properties of a Statistical Distribution Function for Weisstein, Eric W. "Binomial Identity." Astrophys. Every regular multiplicative identity corresponds to an RMI-diagram. enl. Unlimited random practice problems and answers with built-in Step-by-step solutions. In genera... | {
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In mathematics, the binomial theorem is an important formula giving the expansion of powers of sums. identities (Riordan 1979, Roman 1984), some of which include, (Abel 1826, Riordan 1979, p. 18; Roman 1984, pp. For instance, if k is a positive integer and n is arbitrary, then. En mathématiques, et plus précisément en ... | {
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et des millions de livres en stock sur Amazon.fr. this identity for all in a field of field characteristic Proposition 4.1 (Complementation Rule). = \frac{n}{k} \cdot \frac{(n - 1) \cdot (n - 2) \cdot ... \cdot 2 \cdot 1}{(k - 1)! Multinomial returns the multinomial coefficient (n; n 1, …, n k) of given numbers n 1, …,... | {
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number of ways of selecting a subset of of at least q elements, and marking q elements among those selected. For example, The 2-subsets of {1,2,3,4} … Can we find a nice expression for the sum? The difficulty here is that we cannot simply copy down the lower indices in the given identity and interpret them as coordinates... | {
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Click here to toggle editing of individual sections of the page (if possible). We wish to prove that they hold for all values of $$n$$ and $$k\text{. 29-30 and 72-75, 1984. Theorem 2 establishes an important relationship for numbers on Pascal's triangle. Some of the most basic ones are the following. Xander Henderson ♦... | {
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= \frac{n}{k} \cdot \frac{(n - 1)^{\underline{k-1}}}{(k - 1)!} Definition. On the other hand, if the number of men in a group of grownups is then the number of women is , and all possible variants are expressed by the left hand side of the identity. Unfortunately, the identities are not always organized in a way that m... | {
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of $$(x+y)^n$$ $$\binom{n}{k}$$ is the number subsets of size $$k$$ from a set of size $$n$$ $$\dots$$ there are many more ways of viewing binomial coefficients. Proof. The number of possibilities is , the right hand side of the identity. The above formula for the generalized binomial coefficient can be rewritten as ) ... | {
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Expansions, rev. 0, then is an associated Our goal is to establish these identities. 2, R16, 1, Binomial Coe cients and Generating Functions ITT9131 Konkreetne Matemaatika Chapter Five Basic Identities Basic Practice ricksT of the radeT Generating Functions Hypergeometric Functions Hypergeometric ransfoTrmations Partia... | {
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in many ways. The formula is obtained from using x = 1. For other uses, see NCK (disambiguation). Moreover, the following may be useful: 1. }}$,$\displaystyle{\binom{n}{k} = \binom{n}{n-k}}$,$\displaystyle{\binom{n}{k} = \frac{n}{k} \cdot \binom{n-1}{k-1}}$,$\frac{(n - 1)^{\underline{k-1}}}{(k - 1)!} Math. More resourc... | {
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The prototypical example is the binomial The binomial coefficients arise in a variety of areas of mathematics: combinatorics, of course, but also basic algebra (binomial … It is required to select an -members committee out of a group of men and women. The factorial formula facilitates relating nearby binomial coefficie... | {
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# Multiples of 3 and 5
This is the first post of a series entitled Project Euler In Under 0.1ms
### Project Euler #1
If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below 1000.
### Common (... | {
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# Time spent: 129.046ms
### Same approach improved some more
We can gain some precious milliseconds benefitting from Python’s fast addition. The following is around three times faster than the previous one:
def multiples_of_three_and_five(x):
return sum(range(3, x, 3)) + sum(range(5, x, 5)) - sum(range(15, x, 15))
... | {
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# Prove with induction that $11$ divides $10^{2n}-1$ for all natural numbers.
$$10^{2(k+1)}-1 = 10^{2k+2}-1=10^{2k}\cdot10^{2}-1$$
I feel like there's something in that last part that should make it work, but I can't grasp it. Am I missing something obvious? Am I going in the completely wrong direction? Any help woul... | {
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\begin{align} {\bf Claim}\rm\qquad\ \ 10^2\!&\rm\equiv 1, \, 10^{2k}\!\equiv 1\ \, \Rightarrow\,\ 10^{2(k+1)}\!\equiv 1\, \pmod{\!11}\\[.3em] {\bf Lemma}\rm\qquad\ A&\rm\equiv a,\ \, B\equiv b\quad \Rightarrow\quad\,\ AB\equiv ab\, \pmod{\!n}\ \ \ \ [\rm\color{#c00}Congruence \ \rm\color{#c00}Product\ \rm\color{#c00}Ru... | {
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In much the same way, congruences often allow us to impart intuitive arithmetical structure onto inductive proofs - allowing us to reuse our well-honed grade-school skills manipulating arithmetical equations (vs. more complex divisibility relations). Often introduction of congruence language will serve to drastically s... | {
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One really doesn’t need induction for this:
\begin{align*} 10^{2n}-1&=\underbrace{999999\ldots999999}_{2n\text{ nines}}\\ &=\underbrace{99\,99\,99\,\ldots\,99\,99\,99}_{n\text{ copies of }99}\\ &=11\cdot\underbrace{09\,09\,09\,\ldots\,09\,09\,09}_{n\text{ copies of }09}\\ &=11\cdot 9\underbrace{090909\ldots090909}_{n-... | {
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$10^{2n+2}-1=$
$100(10^{2n})-1=$
$100(\color{red}{10^{2n}-1})+99=$
$100(\color{red}{11k})+99=$
$11(100k+9)=$
Please note that the assumption is used only in the part marked red.
For $n = 1$, $10^{2n}-1 = 99$, which is divisible by $11$. Let us assume that the given statement is true for $n = m$, i.e. $10^{2m}-1$ ... | {
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# Simple number puzzle
You have to solve the following:
$$0$$ _ $$0$$ _ $$0 = 6$$
$$1$$ _ $$1$$ _ $$1 = 6$$
$$2$$ _ $$2$$ _ $$2 = 6$$
$$3$$ _ $$3$$ _ $$3 = 6$$
$$4$$ _ $$4$$ _ $$4 = 6$$
$$5$$ _ $$5$$ _ $$5 = 6$$
$$6$$ _ $$6$$ _ $$6 = 6$$
$$7$$ _ $$7$$ _ $$7 = 6$$
$$8$$ _ $$8$$ _ $$8 = 6$$
$$9$$ _ $$9$$ _ $$9... | {
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$$3!+3-3$$
$$\sqrt { 4 } +\sqrt { 4 } +\sqrt { 4 }$$
$$(5\div5)+5$$
$$6+6-6$$
$$7-(7\div7)$$
$$\left( \sqrt { 8+8 } !\div 8 \right) !$$
$$\sqrt { 9 } +\sqrt { \sqrt { 9 } \times \sqrt { 9 } }$$
Legid?
- 3 years, 8 months ago
Try this
$$9$$_ $$9$$ _ $$9 = 10$$
- 3 years, 1 month ago
$$-(-9\div9-9)$$
- 3 yea... | {
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# Determinant of a specially structured matrix
I have the following $n\times n$ matrix:
$$A=\begin{bmatrix}a&b&\cdots&b\\b&a&\cdots&b\\\vdots& &\ddots&\vdots\\b&\cdots&b&a\end{bmatrix}$$
where $0 < b < a$.
I am interested in the expression for the determinant $\det[A]$ in terms of $a$, $b$ and $n$. This seems like ... | {
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Add row 2 to row 1, add row 3 to row 1,..., add row $n$ to row 1, we get $$\det(A)=\begin{vmatrix} a+(n-1)b & a+(n-1)b & a+(n-1)b & \cdots & a+(n-1)b \\ b & a & b &\cdots & b \\ b & b & a &\cdots & b \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ b & b & b & \ldots & a \\ \end{vmatrix}$$ $$=(a+(n-1)b)\begin{vmatrix} ... | {
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\begin{align*} \begin{vmatrix}a&b&\cdots&b\\b&a&\cdots&b\\\vdots&&\ddots&\vdots\\b&\cdots&b&a\end{vmatrix}&=\left(1+(\sqrt{b}\mathbf e)^\top\left(\frac{\sqrt{b}}{a-b}\mathbf e\right)\right)(a-b)^n\\ &=\left(1+\frac{nb}{a-b}\right)(a-b)^n=(a+(n-1)b)(a-b)^{n-1} \end{align*}
where we used the fact that $\mathbf e^\top\ma... | {
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-
The matrix can be diagonalized. All it takes is a bit of geometry. We have
$$A=\begin{bmatrix}a&b&\cdots&b\\b&a&\cdots&b\\\vdots& &\ddots&\vdots\\b&\cdots&b&a\end{bmatrix}.$$
This is a linear combination of the matrices $P$ and $Q=I-P$ where $P$ is the matrix of the orthogonal projection onto the $1$-dimensional s... | {
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-
Consider the $n\times n$ matrix $B$ with entries $-b$ everywhere, except on the main diagonal where it has entries $0$. Now $\det A=\det(aI-B)$ is just the value of the characteristic polynomial $\chi_B\in K[X]$ at $X=a$. For $X=b$ the matrix $bI-B$ clearly has rank at most$~1$ (all columns are equal), so by rank-nu... | {
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# CHN08 - Editorial
Contest
Author: ???
Tester: Kevin Atienza, Jingbo Shang
Editorialist: Kevin Atienza
### PREREQUISITES:
Modulo, recurrences, pattern matching
### PROBLEM:
A function f on integers is defined:
• f(1) = A, f(2) = B.
• For all integers x \ge 2, f(x) = f(x-1) + f(x+1).
Find f(N) \bmod (10^9 + 7).... | {
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Note: Don’t forget to ensure that f(N) \bmod (10^9 + 7) is positive! Also, f(N) can be < -(10^9 + 7), so adding (10^9 + 7) just once isn’t enough to make f(N) positive.
# Proving it
But as just mentioned, proving it is simple. Let’s first generalize the observation above, by actually using variables A and B:
• f(3) ... | {
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# More properties
Let’s analyze f more deeply. Using generating functions, we can actually find a neat formula for f. (though it doesn’t necessarily yield any algorithmic improvements, it is interesting nonetheless.) It will also explain why every three terms almost repeats, except that you have to flip the sign first... | {
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At this point, let’s use the method of partial fractions, to obtain
F(x) = \frac{C}{1 - \psi x} + \frac{D}{1 - \bar{\psi}x}
where C and D are constants that we don’t know yet. This form is very useful to us because of the equality:
\frac{1}{1 - rx} = 1 + rx + r^2x^2 + r^3x^3 + \ldots
which means that the coefficien... | {
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O(1)
### AUTHOR’S AND TESTER’S SOLUTIONS:
[setter][333]
[tester][444]
[editorialist][555]
[333]: The link is provided by admins after the contest ends and the solutions are uploaded on the CodeChef Server.
[444]: The link is provided by admins after the contest ends and the solutions are uploaded on the CodeChef Ser... | {
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Help me,please;what’s wrong with the code?
#include<iostream>
using namespace std;
long long int f(long long int index[],int n) {
const long long int m = 1'000'000'007;
auto x = index[n % 6];
return (x%m + m) % m;
}
int main() {
int n,a, b, c;
cin >> n;
while (n--) {
cin >> a >> b >> c;
long long int arr[] = { a,b,b... | {
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### 2 Partial pivoting
Partial pivoting is a refinement of the Gaussian elimination procedure which helps to prevent the growth of rounding error.
#### 2.1 An example to motivate the idea
Consider the example
$\phantom{\rule{2em}{0ex}}\left[\begin{array}{cc}\hfill 1{0}^{-4}\hfill & \hfill 1\hfill \\ \hfill -1\hfill... | {
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$\phantom{\rule{2em}{0ex}}\left[\begin{array}{cc}\hfill 1{0}^{-4}\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1{0}^{4}\hfill \end{array}\right]\left[\begin{array}{c}\hfill {x}_{1}\hfill \\ \hfill {x}_{2}\hfill \end{array}\right]=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 1{0}^{4}\hfill \end{array}\right].$
T... | {
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#### 2.2 Partial pivoting in general
At each step the aim in Gaussian elimination is to use an element on the diagonal to eliminate all the non-zeros below. In partial pivoting we look at all of these elements (the diagonal and the ones below) and swap the rows (if necessary) so that the element on the diagonal is not... | {
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$\phantom{\rule{2em}{0ex}}\left[\begin{array}{ccccc}\hfill 1\hfill & \hfill -3\hfill & \hfill 2\hfill & \hfill 1\hfill & \hfill -4\hfill \\ \hfill 2\hfill & \hfill -6\hfill & \hfill 1\hfill & \hfill 4\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill 2\hfill & \hfill 3\hfill & \hfill 4\hfill & \hfill 12\hfill \\ \hfil... | {
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That was the partial pivoting step. Now we proceed with Gaussian elimination
$\phantom{\rule{2em}{0ex}}\left[\begin{array}{ccccc}\hfill 1\hfill & \hfill -3\hfill & \hfill 2\hfill & \hfill 1\hfill & \hfill -4\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill ... | {
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And the elimination phase is completed by removing the $-3$ from the final row as follows
$\phantom{\rule{2em}{0ex}}\left[\begin{array}{ccccc}\hfill 1\hfill & \hfill -3\hfill & \hfill 2\hfill & \hfill 1\hfill & \hfill -4\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \h... | {
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The row operations required to eliminate the non-zeros below the diagonal in the first column are
$\phantom{\rule{2em}{0ex}}\left[\begin{array}{ccc}\hfill 1\hfill & \hfill -2\hfill & \hfill 4\hfill \\ \hfill -3\hfill & \hfill 6\hfill & \hfill -11\hfill \\ \hfill 4\hfill & \hfill 3\hfill & \hfill 5\hfill \end{array}\ri... | {
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"lm_q2_score": 0.8577681031721325,
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which is in the required upper triangular form.
##### Key Point 4
When To Use Partial Pivoting
1. When carrying out Gaussian elimination on a computer, we would usually always swap rows so that the element on the diagonal is as large (in magnitude) as possible. This helps stop the growth of rounding error.
2. When d... | {
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2. Working to 3 significant figures, and using Gaussian elimination without pivoting, find an approximation to $\left[\begin{array}{c}\hfill {x}_{1}\hfill \\ \hfill {x}_{2}\hfill \end{array}\right]$ . Show that the rounding error causes the approximation to ${x}_{1}$ to be a very poor one.
3. Working to 3 significant f... | {
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2. Carrying out the elimination without pivoting, and rounding to 3 significant figures we find that ${x}_{2}=2.00$ and that, therefore, ${x}_{1}=0$ . This is a very poor approximation to ${x}_{1}$ .
3. To apply partial pivoting we swap the two rows and then eliminate the bottom left element. Consequently we find that,... | {
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We carry out one row operation to eliminate the non-zero in the bottom left entry as follows
$\phantom{\rule{2em}{0ex}}\left[\begin{array}{ccc}\hfill 1& \hfill -4& \hfill 2\\ \hfill 0& \hfill -1& \hfill 2\\ \hfill -2& \hfill 5& \hfill -4\\ \hfill \end{array}\right]\begin{array}{c}\hfill \\ \hfill \\ R3+2×R1\hfill \end... | {
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# Probability that the 10th drawn ball is black.
An urn contains $$7$$ white and $$13$$ black balls. We draw a ball from an urn, put it back and add $$2$$ additional balls of the same color (as the one we have just drawn and put back). We repeat this $$10$$ times (so the numbers of balls change throughout the whole ex... | {
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"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9840936054891429,
"lm_q1q2_score": 0.8441240999570854,
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"ta... |
Of course, if you are in state $$(w+3,b)$$ the (new) probability of drawing a black ball is $$\frac b{b+w+3}$$ and if you are in state $$(w,b+3)$$ it is $$\frac {b+3}{b+w+3}$$.
Thus the probability on the $$n^{th}$$ round of the original game is $$\frac w{b+w}\times \frac b{b+w+3}+\frac b{b+w}\times \frac {b+3}{w+b+3}... | {
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• you want a $b+w$ in the final denominator, I think
– Ned
Dec 20, 2018 at 18:34
• You meant $b/(b+w)$ in the last line of the calculation. Dec 20, 2018 at 18:36
• I have some problems with this proof. It seems like you assume sth like: ok, at the $n-1$ round the probability of drawing a black ball is $\frac{b}{b+w}$. ... | {
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"openwebmath_perplexity": 150.40275224032104,
"openwebmath_score": 0.8130570650100708,
"ta... |
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