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http://forums.wolfram.com/mathgroup/archive/2004/Apr/msg00585.html | 1,548,271,691,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547584336901.97/warc/CC-MAIN-20190123172047-20190123194047-00559.warc.gz | 90,331,611 | 7,838 | Re: i don't understand mapping function over a long list
• To: mathgroup at smc.vnet.net
• Subject: [mg47819] Re: i don't understand mapping function over a long list
• From: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>
• Date: Wed, 28 Apr 2004 06:56:01 -0400 (EDT)
• Organization: Universitaet Leipzig
• References: <c6l7v9\$iq0\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com
```Hi,
a) you can't get {{x\$299, x\$300, x\$301} + 2,..}
because Mathematica will distribute the scalar
addition over the lists an you will always end
with {{x\$299+2, x\$300+2, x\$301+2},..}
b) you can use an new symbol that is not a List with
l = vector @@@ Partition[Flatten[{Table[Unique[x], {n, 1, 30}]}], 3];
and
l+2
c) you can assign the result to
unames = Table[Unique[u], {10}];
with
Set @@@ Transpose[{unames, (l + 2)}]
Regards
Jens
will produce the desired result.
sean kim wrote:
>
> hello group.
>
> i just don't get this.
>
> I'm not sure what's the problem is.
>
> please consider the following list of lists.
>
> l = Partition[Flatten[{Table[Unique[x], {n, 1, 30}]}], 3 ]
>
> above generates a list of 10 lists with 3 elements in each as in...
>
> {
> {x\$299, x\$300, x\$301}, {x\$302, x\$303, x\$304},
> {x\$305, x\$306, x\$307}, {x\$308, x\$309, x\$310},
> {x\$311, x\$312, x\$313}, {x\$314, x\$315, x\$316},
> {x\$317, x\$318, x\$319}, {x\$320, x\$321, x\$322},
> {x\$323, x\$324, x\$325}, {x\$326, x\$327, x\$328}
> }
>
> Now suppose I want to use each of the list( all 10 of them) as a part
> of a function. . I want to "Apply" the function to every list(so, 10
> times in total)
> for a simple examplel let's add 2 to the lists
>
> In[21]:= Apply[Plus@@xlist, 2]
> Out[21]= 2
>
> that didn't work. i wanted to get was
>
> {{x\$299, x\$300, x\$301} + 2,
> {x\$302, x\$303, x\$304} + 2...
> {x\$326, x\$327, x\$328} + 2}}
>
> then i want to give each of the results unique names and use the
> renamed list of lists as an argument in another function.
>
> {
> uniquexname1 = {x\$299, x\$300, x\$301} + 2,
> uniquexname2 = {x\$302, x\$303, x\$304} + 2...
> uniquexname10 = {x\$326, x\$327, x\$328} + 2}
> }
>
> and
>
> Map[Plus, xlist, 2]
>
> just bring back the list itself.
>
> This problem recurs for me. and I think i have problems with it
> because I just don't understand how Mathematica language works.
>
> Reading the book an dhelp manual doesn't help me much in understanding
> what lies underneath. Can you guys shed soem light on this with some
> simple examples that use numerical operations?
>
> Maybe I'm asking a lot, but any and all insights are thoroughly
> appreciated.
> | 911 | 2,622 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2019-04 | latest | en | 0.759256 |
https://mathoverflow.net/questions/313495/spectral-decomposition-of-a-combinatorial-matrix-random-walks-on-s-sets | 1,550,834,947,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247515149.92/warc/CC-MAIN-20190222094419-20190222120419-00418.warc.gz | 612,215,828 | 32,735 | # Spectral decomposition of a combinatorial matrix/Random walks on $s$-sets
$$\newcommand{\Z}{\mathbb{Z}} \newcommand{\J}{\mathcal{J}} \newcommand{\la}{\lambda} \newcommand{\1}{\mathbf{1}} \newcommand{\R}{\mathbb{R}}$$
Take any $$n\in[3;\infty]$$. Here and in what follows, $$[k;\ell]:=[k,\ell]\cap\Z$$. Take then any $$s\in[2;n-1]$$. Let $$\J:=\J_s:=\binom{[n]}s$$, the set of all $$s$$-sets in $$[n]:=[1;n]$$, that is, the set of all subsets of the set $$[n]$$ of size (cardinality) $$s$$. Let $$N:=N_s:=|\J_s|=\binom ns$$, the cardinality of the set $$\J$$. Consider the $$N\times N$$ matrix $$$$A:=A_s:=(|J\cap K|\colon J,K\in\J).$$$$ The problem is to prove
Theorem:$$\quad$$ The eigenvalues of the matrix $$A$$ are $$\la_1:=s\binom{n-1}{s-1}$$ (of multiplicity $$1$$), $$\la_2:=\binom{n-2}{s-1}$$ (of multiplicity $$n-1$$), and $$\la_3:=0$$ (of multiplicity $$N-n$$).
Comments: It is easy to see that the symmetric matrix $$$$P:=\tfrac1{\la_1}\,A$$$$ is double stochastic, which implies that $$\la_1$$ is indeed an eigenvalue of $$A$$, with a corresponding eigenvector $$\1:=(1\colon J\in\J)$$. That is, the vector $$\pi:=\frac1N\,\1$$ is the stationary distribution of the random walk/Markov chain on the set $$\J_s$$ of the $$s$$-sets with the transition probability matrix $$P$$. The eigenvalues of the matrix $$P$$ are $$1$$ (of multiplicity $$1$$), $$\nu:=\la_2/\la_1:=\frac{n-s}{s(n-1)}\le\frac{n-2}{2(n-1)}\in(0,1/2)$$ (of multiplicity $$n-1$$), and $$0$$ (of multiplicity $$N-n$$). So, $$P=P_1+\nu P_2$$, where $$P_1$$ and $$P_2$$ are the orthoprojectors onto the eigenspaces belonging to the respective eigenvalues $$1$$ and $$\nu$$. Take now any initial distribution $$p$$ on $$\J$$. Then for all natural $$m$$ we have $$pP^m=pP_1+\nu^m pP_2=\pi+\nu^m pP_2$$, so that we have the exponential convergence of the distribution $$pP^m$$ of the chain at time $$m$$ to the stationary distribution $$\pi$$: $$$$pP^m-\pi=\nu^m\,pP_2.$$$$ The difference $$1-\nu$$ between the two largest distinct eigenvalues of $$P$$ is called its spectral gap, which determines the rate of the exponential convergence.
From the spectral decomposition $$$$A=\la_1 P_1+\la_2 P_2$$$$ it also immediately follows that for any $$x=(x_J)\in\R^\J$$ $$$$\|Ax\|_2^2\ge\la_1^2\|P_1x\|_2^2=\la_1^2\Big(\sum_J x_J\Big)^2.$$$$ For $$s=2$$, the latter inequality was proved by Fedor Petrov at Is this bound uniform in $$N$$? ; however, his proof seems to be easy to extend to general $$s$$.
Various variants of the matrix $$A_{J,K} = |J\cap K|$$ were studied, and the spectrum was computed. A one-parameter variant is given by $$A^{(i)}_{J,K} = \binom{|J\cap K|}{i}$$ for some fixed $$i$$, and your problem corresponds to $$i=1$$. In total, six variants are given in Section 3 of this friendly paper by Ghareghani, Ghorbani and Mohammad‐Noori. In particular, the spectrum of $$A^{(i)}$$ is described in Lemma 9. The lemma is attributed to R. M. Wilson (1982), who uses somewhat different notation and terminology. The case $$i=1$$ recovers your theorem, and in general the non-zero eigenvalues are given by $$\lambda_j=\binom{s-j}{i-j}\binom{n-i-j}{s-i}, \qquad j=0,1,2,\ldots,i,$$ with $$\lambda_j$$ having multiplicity $$\binom{n}{j}-\binom{n}{j-1}$$ (note that $$\binom{n}{-1}:=0$$). The eigenvalue 0 has multiplicity equal to the remaining dimension, i.e. $$N-\binom{n}{i}$$.
Other useful references, including to additional works of R. M. Wilson, appear in the bibliography of the G-G-MN paper.
• Wow, thank you for the references! I did not know that there is an entire theory here. However, I think that, using the representation of $|J\cap K|$ as the sum of products of indicators as in my answer, one may be able to deal somewhat similarly, in an elementary way, with natural powers (or, equivalently, with factorial products) of $|J\cap K|$, even though that may be significantly more difficult in the more general situation. – Iosif Pinelis Oct 23 '18 at 0:59
$$\newcommand{\ii}[1]{\operatorname{\mathsf I}\{#1\}}$$
The first key observation in the proof of the theorem is that $$$$A=\sum_{k=1}^n A^{(k)},\quad\text{where}\quad A^{(k)}_{JK}:=\ii{k\in J}\ii{k\in K}$$$$ for $$J,K$$ in $$\J$$, where $$\ii\cdot$$ denotes the indicator. Clearly, the rank of $$A^{(k)}$$ is $$\le1$$ for each $$k$$, and so, the rank of $$A$$ is $$\le n$$. So, it suffices to exhibit an eigenvector belonging to $$\la_1$$ and $$n-1$$ linearly independent eigenvectors belonging to $$\la_2$$.
Concerning $$\la_1$$, for all $$J\in\J$$ we have $$$$(A\1)_J=\sum_K A_{JK}=\sum_K\sum_{k=1}^n A^{(k)}_{JK} =\sum_{k=1}^n \ii{k\in J}\sum_K \ii{k\in K} =\sum_{k=1}^n \ii{k\in J}\binom{n-1}{s-1}=s\binom{n-1}{s-1}=\la_1,$$$$ as desired.
Next, let us show for any distinct $$i,j\in[n]$$ the vector $$d^{(i,j)}$$ defined by the formula $$$$d^{(i,j)}_J:=\ii{i\in J}-\ii{j\in J}$$$$ for $$J\in\J$$ is a $$\la_2$$-eigenvector of $$A$$. Indeed, $$\begin{multline} (Ad^{(i,j)})_J=\sum_K A_{JK}d^{(i,j)}_K =\sum_{k=1}^n\ii{k\in J}\sum_K\ii{k\in K}(\ii{i\in K}-\ii{j\in K}) \\ =\sum_{k=1}^n\ii{k\in J}\sum_K(\ii{\{i,k\}\subseteq K}-\ii{\{j,k\}\subseteq K}) \\ =\sum_{k=1}^n\ii{k\in J}\Big(\ii{k=i}\binom{n-1}{s-1} +\ii{k\ne i}\binom{n-2}{s-2} \\ -\ii{k=j}\binom{n-1}{s-1}-\ii{k\ne j}\binom{n-2}{s-2}\Big) \\ =\sum_{k=1}^n\ii{k\in J}\binom{n-2}{s-1}(\ii{k=i}-\ii{k=j}) \\ =\binom{n-2}{s-1}(\ii{i\in J}-\ii{j\in J}) =\la_2d^{(i,j)}_J, \end{multline}$$ as desired.
It remains to show that the $$\la_2$$-eigenvectors $$d^{(1,2)},\dots,d^{(1,n)}$$ are linearly independent. To do this, take any real $$t_2,\dots,t_n$$ such that $$$$\sum_{j=2}^n t_jd^{(1,j)}_J=0 \tag{1}$$$$ for all $$J\in\J_s$$. We need to show that $$t_j=0$$ for $$j\in[2;n]$$. First here, take $$J=[1;s-1]\cup\{k\}$$ for any $$k\in[s;n]$$. Then (1) yields $$\begin{multline} 0=\sum_{j=2}^n t_j(\ii{1\in[1;s-1]\cup\{k\}}-\ii{j\in[1;s-1]\cup\{k\}}) \\ =\sum_{j=2}^n t_j-\sum_{j=2}^{s-1} t_j-t_k=\sum_{j=s}^n t_j-t_k \end{multline}$$ for all $$k\in[s;n]$$, so that $$t_k=t$$ for some real $$t$$ and all $$k\in[s;n]$$, whence $$(n-s)t=0$$ and $$t=0$$, so that $$$$t_s=\dots=t_n=0. \tag{2}$$$$ Next, take $$J=[1;r-1]\cup\{k\}\cup[n-s+r+1,n]$$ for any $$r\in[2;s-1]$$ and any $$k\in[r;n-s+r]$$. Then (1) yields $$\begin{multline} 0=\sum_{j=2}^n t_j(\ii{1\in[1;r-1]\cup\{k\}\cup[n-s+r+1,n]} \\ -\ii{j\in[1;r-1]\cup\{k\}\cup[n-s+r+1,n]}) \\ =\sum_{j=2}^n t_j-\sum_{j=2}^{r-1} t_j-t_k-\sum_{j=n-s+r+1}^n t_j, \end{multline}$$ so that $$t_k$$ is constant for $$k\in[r;n-s+r]$$. In particular, $$t_r=t_{r+1}$$, for all $$r\in[2;s-1]$$, since $$s\le n-1$$. That is, $$t_2=\dots=t_s$$. It remains to recall (2). $$\Box$$ | 2,590 | 6,631 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 119, "wp-katex-eq": 0, "align": 0, "equation": 10, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2019-09 | longest | en | 0.655414 |
http://micmap.org/php-by-example/manual/en/function.fmod.html | 1,553,640,500,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912206677.94/warc/CC-MAIN-20190326220507-20190327002507-00423.warc.gz | 136,911,450 | 1,952 | # fmod
(PHP 4 >= 4.2.0, PHP 5, PHP 7)
fmodReturns the floating point remainder (modulo) of the division of the arguments
### Description
float fmod ( float `\$x` , float `\$y` )
Returns the floating point remainder of dividing the dividend (`x`) by the divisor (`y`). The remainder (r) is defined as: x = i * y + r, for some integer i. If `y` is non-zero, r has the same sign as `x` and a magnitude less than the magnitude of `y`.
`x`
The dividend
`y`
The divisor
### Return Values
The floating point remainder of `x`/`y`
### Examples
Example #1 Using fmod()
``` <?php\$x = 5.7;\$y = 1.3;\$r = fmod(\$x, \$y);// \$r equals 0.5, because 4 * 1.3 + 0.5 = 5.7?> ``` | 235 | 675 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2019-13 | longest | en | 0.450678 |
https://askworksheet.com/missing-backward-counting-50-1-worksheets/ | 1,679,649,619,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945279.63/warc/CC-MAIN-20230324082226-20230324112226-00042.warc.gz | 148,176,693 | 21,800 | Missing Backward Counting 50-1 Worksheets
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Free Printable Before And After Number Worksheets 1st 2nd Grade Worksheet School Math Worksheets 1st Grade Math Worksheets Kids Math Worksheets | 1,030 | 5,338 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2023-14 | latest | en | 0.879903 |
http://windowssecrets.com/forums/showthread.php/112482-Efficient-Formula-%282003%29 | 1,484,854,739,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280730.27/warc/CC-MAIN-20170116095120-00202-ip-10-171-10-70.ec2.internal.warc.gz | 309,122,960 | 14,278 | 1. ## Efficient Formula (2003)
I created a planning spreadsheet that provides the calendar day a product is due based on Column B. Then I adjust the due date to make sure it lands on a business day and accounts for weekends and holidays referenced on Sheet 2. I'm not so sure I created the most efficient formula though or used the best method of affecting this. Is there another way of doing this?
Amy
2. ## Re: Efficient Formula (2003)
These calculations usually use a number of working days, i.e. exclude both weekend days and holidays. The Analysis ToolPak add-in contains a function WORKDAY for this. But you appear to want to add a number of days and include weekend days but not holidays in the count. I don't think there's a function available for that.
Any chance you can use the more conventional calculation?
3. ## Re: Efficient Formula (2003)
Thanks Hans. I was given column B to work with. That number will include weekend and holidays, but has to return the next workday after the weekend and/or holiday. Since there isn't a function, I was wondering if the VLOOKUP table I set up is the most efficient manner of doing it. The formulas got pretty long using the IF function.
Amy
4. ## Re: Efficient Formula (2003)
I would suggest that instead of using the Add One, Add Two, or Add Three you use their numerical equivalents of 1, 2, or 3. Then filling in the appropriate weekend added as well - 2 for Saturdays and 1 for Sundays. Non "special" days can left blank.
Your table would look like something like this (the date formatting changed due to the macro I used to copy from Excel):
<table border 1><td>10/11/2008</td><td>Saturday</td><td>3</td><td> </td><td>10/12/2008</td><td>Sunday</td><td>2</td><td> </td><td>10/13/2008</td><td>Monday</td><td>1</td><td>Columbus Day</td><td>10/14/2008</td><td>Tuesday</td><td> </td><td> </td><td>10/15/2008</td><td>Wednesday</td><td> </td><td> </td><td>10/16/2008</td><td>Thursday</td><td> </td><td> </td><td>10/17/2008</td><td>Friday</td><td> </td><td> </td><td>10/18/2008</td><td>Saturday</td><td>2</td><td> </td><td>10/19/2008</td><td>Sunday</td><td>1</td><td> </td><td>10/20/2008</td><td>Monday</td><td> </td><td> </td></table>
You could then use:
=L4+VLOOKUP(L4,Sheet2!\$A\$1:\$C\$365,3,0)
For your C column, in the Left() function, you can keep the 3 as in "=\$B\$1+LEFT(B4,3)" for all of the dates as long as you don't exceed 999 days - the space between the number and the word days is ignored in the calculation.
Or you could use:
=\$B\$1+LEFT(B4,FIND(" ",B4))
5. ## Re: Efficient Formula (2003)
Try Mike Barron's suggestion of using numbers instead of text in the lookup table, that avoids the need for all those IFs.
6. ## Re: Efficient Formula (2003)
Thanks Mike. I knew there had to be a more streamlined way of doing this.
Amy
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• | 863 | 2,958 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2017-04 | longest | en | 0.877117 |
https://questions.tripos.org/part-ia/2014-10/ | 1,709,234,382,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474852.83/warc/CC-MAIN-20240229170737-20240229200737-00896.warc.gz | 454,729,499 | 3,936 | # Paper 2, Section II, B
(a) Let $y_{1}(x)$ be a solution of the equation
$\frac{d^{2} y}{d x^{2}}+p(x) \frac{d y}{d x}+q(x) y=0$
Assuming that the second linearly independent solution takes the form $y_{2}(x)=$ $v(x) y_{1}(x)$, derive an ordinary differential equation for $v(x)$.
(b) Consider the equation
$\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-2 x \frac{d y}{d x}+2 y=0, \quad-1
By inspection or otherwise, find an explicit solution of this equation. Use the result in (a) to find the solution $y(x)$ satisfying the conditions
$y(0)=\frac{d y}{d x}(0)=1$ | 208 | 571 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 8, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2024-10 | longest | en | 0.722534 |
https://cs.nyu.edu/courses/fall10/V22.0436-001/practice-mid-term.html | 1,516,187,040,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886895.18/warc/CC-MAIN-20180117102533-20180117122533-00783.warc.gz | 639,176,539 | 2,614 | ### Practice Mid-Term
Note: the italicized comments indicate the main topics to be covered on the exam, but are not exhaustive. Any topic covered in the notes about combinational and sequential circuits is fair game on the exam.
The mid-term is open book and open notes. However, you may not use your computer during the exam; only printed notes are alliwed.
#### Time and Frequency
You should know the units of time and frequency, and be able to convert between them.
1. What is the clock period for a 200 Hz clock? For a 200 MHz clock?
#### Combinational Circuits
You should be able to design a circuit (and compute its propagation delay), given a verbal or truth-table description of the circuit. You should be familiar with the basic circuit types: and, or, nand, nor, exclusive-or, and not gates; multiplexer, full adder, and decoder.
1. Design an exclusive-or circuit using only AND, OR, and NOT gates. What is the delay of this circuit if the delay of each individual gate is 200 ps?
2. Given a 3-bit input X representing a 3-bit binary number, design a circuit to test whether X is greater than or equal to 5. If this condition is true, the output of your circuit should be a 1, otherwise 0. In your design, refer to the low order bit of X as X0.
3. Suppose you are given a large box of 2-input multiplexers. Show how to connect them up to create an 8-input multiplexer. If the delay from input to output on the 2-input multiplexer is 10 ns, what is the delay of the circuit you have designed? Suppose you had to create an N input multiplexer; what would the delay be? How many 2-input multiplexers would you need?
#### Sequential Circuits
You should understand the function of the basic types of flip-flops (set-reset, D type, master-slave), and the reason for using master-slave FFs. You should be able to assemble a register file. From a description of a sequential procedure, you should be able to create a state diagram, a state transition table, and finally a circuit.
1. Go through the process of designing a two-bit down counter. First, draw the state diagram. Second, write down the state transition table. Third, convert the transition table to a formula in Boolean algebra. Finally, convert the formula to a circuit and show how it would connect to FFs to create a complete counter circuit.
2. Consider a two-bit up counter with a select input S. If S=0, the circuit acts as an up counter; if S=1, the counter goes to 0 on the next clock cycle (reset). Give the state transition diagram and table for this counter. Design the counter circuit from this table.
#### Arithmetic
You should understand two's complement arithmetic and the design of adders and subtractors, including the carry-look-ahead adder, and be able to give the delay time of these circuits.
1. What is the 16-bit, two's complement representation of -3? Give your answer in binary.
2. Suppose you are given an adder for unsigned 16-bit binary numbers. What changes must you make to the circuit to use it as an adder for 16-bit, two's complement numbers?
3. Given AND, OR, NOT gates and full adders, design a circuit for doing 4-bit two's complement subtraction.
4. A sum-of-products circuit has a propagation delay = 3 gate delays. Why can't we build a 16-bit adder using sum-of-products with this propagation delay?
#### MIPS
You should be able to translate simple code segments given in C or Java into MIPS assembler, and answer questions about the MIPS instruction set.
1. Write a sequence of MIPS instructions which put the absolute value of \$2 into \$3.
2. Write a MIPS instruction which will set \$2 to 131,072 (= 217 = 0000 0000 0000 0010 0000 0000 0000 00002) | 870 | 3,673 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2018-05 | latest | en | 0.92006 |
https://www.mrexcel.com/board/threads/using-excel-to-calculate-the-sundays-for-each-month.121447/ | 1,726,683,856,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651931.60/warc/CC-MAIN-20240918165253-20240918195253-00518.warc.gz | 821,908,437 | 21,566 | # Using Excel to Calculate the Sundays for Each month.
#### Alexander_Ewing
##### New Member
I am creating a sheet that will calculate the Sunday of every month after you enter the first one for April. My problem is that I have 5 Rows per month and require 2 Rows between the months. I can use formulas that calculate 7 days from the first date I give but cant manage the check at the bottom that will jump to the cell 3 rows down if a new month starts or enter the value for that month. April=(A6>10) May=(A13>17) June=(A20>24) etc I am hoping that if there are only 4 Weeks for that month that the bottom row will show the month.
I have tried several means of doing this including the IF and DATE statements but to no avail, hope that somebody can help.
Alex
### Excel Facts
Square and cube roots
The =SQRT(25) is a square root. For a cube root, use =125^(1/3). For a fourth root, use =625^(1/4).
Hi there
You could try weekday(19-01-05) will return a 4 for wednesday
a sunday is 1
so with this maybe you can work something out?
I'm not exactly sure of what you want exactly..
Hope the above helps
Thanks for the suggestions haven’t used that function but will give it a go when have more time.
I am hoping to design the sheet so that after you enter the first Sunday in April it will fill the other spaces bellow it with either the date in that month or the month.
Ex. I enter 4 April 2004 in cell A6; A7 should show 11 April 2004; A8 should show 18 April 2004; A9 Should show 25 April 2004; Cell A10 will show April because it is the last Sunday of the month and in Cell A13 should be 2 May 2004 and so on until I go to the end of March.
Thanks Again
Alexander_Ewing said:
Thanks for the suggestions haven’t used that function but will give it a go when have more time.
I am hoping to design the sheet so that after you enter the first Sunday in April it will fill the other spaces bellow it with either the date in that month or the month.
Ex. I enter 4 April 2004 in cell A6; A7 should show 11 April 2004; A8 should show 18 April 2004; A9 Should show 25 April 2004; Cell A10 will show April because it is the last Sunday of the month and in Cell A13 should be 2 May 2004 and so on until I go to the end of March.
Thanks Again
Alexander,
You could simply enter the first date and in the next cell add 7 to the cell above just like this example.
Mike
Thanks Mike,
I tried that but I require 2 rows space between each month and I used IF statements at the start and End of each month to check, The IF statements wouldn’t work for each month start from year to year.
Alex
Can't think of a formula-based approach but here's a code solution.
Assumptions:
1. Dates are in column A
2. First date in sequence is in A2
Code:
``````Sub Insert2()
Range("A65536").End(xlUp).Select
Do Until ActiveCell.Row = 2
If Month(ActiveCell.Value) > Month(ActiveCell.Offset(-1, 0).Value) Then
'month boundary
ActiveCell.Range("A1:A2").EntireRow.Insert
ActiveCell.Offset(-1, 0).Select
Else
'within month
ActiveCell.Offset(-1, 0).Select
End If
Loop
End Sub``````
EDIT This version writes the month name and year below each group...
Code:
``````Sub Insert2_WithMonth()
Range("A65536").End(xlUp).Select
ActiveCell.Offset(1, 0).Value = "'" & Format(ActiveCell.Value, "mmm yyyy")
Do Until ActiveCell.Row = 2
If Month(ActiveCell.Value) > Month(ActiveCell.Offset(-1, 0).Value) Then
'month boundary
ActiveCell.Range("A1:A2").EntireRow.Insert
ActiveCell.Offset(-1, 0).Select
ActiveCell.Offset(1, 0).Value = "'" & Format(ActiveCell.Value, "mmm yyyy")
Else
'within month
ActiveCell.Offset(-1, 0).Select
End If
Loop
End Sub``````
Insert code into a standard module, select the worksheet, and run the macro (Alt + F8, double-click the macro name)
Denis
Put the following in A7 and fill down :-
=IF((MOD(ROW(),7)=4)+(MOD(ROW(),7)=5),"",IF((MOD(ROW(),7)=6)*ISNONTEXT(A4),A4+7,IF((MOD(ROW(),7)=6)*ISTEXT(A4),A3+7,IF((MOD(ROW(),7)=3)*(MONTH(A6+7)<>MONTH(A6)),TEXT(A6,"mmmm"),A6+7))))
What data do you have in A3:A6?
Might need to revise the formula depending upon what is in A3:A6.
Thank you to everybody for all your help, especially to Ponsonby that does what I was looking to do Thanks.
Alex
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Feb 18 comment Introductory books as preparation to read Voevodsky homotopy-theory (HoTT) book Why do you call it "Voevodsky HoTT book"? If anything, it is the "HoTT book written by the participants of the Special Year on Univalent Foundations at Institute for Advanced Study". Feb 5 comment Which is the most powerful language, set theory or category theory? @Fatimah: the connections go deeper than just "translating the alphabet" at the level of symbolic notation. The connections are of a mathematical nature, i.e., the structures involved are related to each other in interesting ways. I added a link to Steve Awodey's paper From Sets to Types to Categories to Sets which explains this very well. Feb 5 revised Which is the most powerful language, set theory or category theory? added 145 characters in body Feb 5 awarded Good Answer Feb 4 awarded Mortarboard Feb 4 awarded Nice Answer Feb 4 comment Which is the most powerful language, set theory or category theory? Is this better now? Feb 4 revised Which is the most powerful language, set theory or category theory? deleted 91 characters in body Feb 4 comment Which is the most powerful language, set theory or category theory? Yes, I want there to be a negative connotation because it is simply false that the set-theoretic foundation is a necessity for the working mathematician. It is a (tremendous) convenience, but so would be a different foundation, as long as it were universally accepted. It is the uniformity of the language and setup of mathematics that really helps, not the fact that there is a foundation in the philosophical sense of the word. Feb 4 comment Which is the most powerful language, set theory or category theory? There are 10 commandments in the bible. There are 10 axioms of set theory. The many many many laws of Judaism you're referring to are more like the equivalents of the axiom of choice. Feb 4 comment Propositional function and Rule of Inference Well, it depends on how you organize things. You could start with the idea of truth tables and get rules from that (like in my story), or you could start with the rules and then observe that truth tables model the rules. Feb 4 answered Which is the most powerful language, set theory or category theory? Feb 4 revised Propositional function and Rule of Inference added 119 characters in body Feb 4 answered Propositional function and Rule of Inference Feb 4 revised Propositional function and Rule of Inference fix latex formatting Feb 4 comment Propositional function and Rule of Inference I cannot find the relevant passage in the book. Are you referring to books.google.si/… ? There it talks about wff's (well-formed formulas, not about "variable letters"). Please be more specific. Feb 4 suggested approved edit on Propositional function and Rule of Inference Jan 12 comment What's an example of an infinitesimal? Right. I think the intuitionistic stuff comes in once you are in a field and have $x^2 = 0$, and you wonder whether $x = 0$. Intuitionistic fields are messy. Jan 11 revised What's an example of an infinitesimal? deleted 338 characters in body Jan 11 comment What's an example of an infinitesimal? @zyx I did come in late and I am not aware of the history. Anyhow, I am not planning to get dragged into this. I answered the question because I was asked to via email, and I'll leave it at that. Have fun fighting for justice. | 769 | 3,448 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2016-18 | latest | en | 0.940442 |
http://boards.edurite.com/madhya+pradesh+board+elements+of+science+%26+maths+for+agriculture+class+11-syllabus~b3tw-c2lN-s4Bf.html | 1,566,343,051,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315681.63/warc/CC-MAIN-20190820221802-20190821003802-00245.warc.gz | 30,870,371 | 13,982 | # Madhya Pradesh Board Elements of Science & Maths for Agriculture Syllabus for Class 11
## Madhya Pradesh Board Syllabus for Class 11 Elements of Science & Maths for Agriculture
Elements of Science and Maths useful for Agriculture syllabus for class XI of 2008-09 session
Time: 3 hours
Maximum Marks: 75
Unit Title Contents Marks Periods Unit-1 Properties of matter (i) Measurement of different physical quantities, fundamental and derived units, systems of fundamental units, standard units, Utility of units in agriculture. (ii) Density and relative density, Definition, their units, difference between density and relative density, determination of relative-density, determination of relative density of solids and liquids, hydrometer, lactometer and siphon (construction and working), laws of floatation and its conditions. (iii) Atmospheric pressure- Definition, units, its measuring, barometer, Simple & Fortin barometer. Aneroid barometer, Wind vane, and anemometer. Pump - a preliminary information of submersible & centrifugal pump. 4 7 Unit-2 Friction Definition, types of friction, laws of friction, angle of friction, angle of repose, determination of coefficient of friction, horizontal plane and inclined plane method, uses of friction in daily life, merits and demerits of friction, means of reducing and increasing friction.(examples related to agriculture). 2 7 Unit-3 Simple Machines Meaning, characteristics and importance of simple machines, definition related to machines, effort, Load. Mechanical advantage, Efficiency of machine, velocity ratio, relation between them, Ideal Machine, Important machines - Lever, screw, wheel and axle, pulley, types of pulley, systems of combination of pulleys. Role of Simple machines in agriculture. 4 7 Unit-4 Gravitation Definition, Newton's law of gravitation,acceleration due to gravity (g) and it's relation with gravitational constants (G), Changes in the value of acceleration due to gravity with depth, height, shape and speed of Earth, simple harmonic motion, it's characteristics, Determination of "g" in laboratory. simple pendulum and its laws. 5 9 Unit-5 Heat (i) Definition, measurement, units, systems of Temperature measurement and their relationships. (ii) Transmission of Heat Types, conductors and Non-conductors. (iii) Study of different Heat measuring apparatus - their use in agriculture . 4 7 SECTION - B Unit-6 Atomic Structure Atomic structure, constituents of Atom - Electron, proton,neutron, nuclear model of an Atom, Atomic Theory, Dalton's Atomic theory, Modern Atomic theory, Bohr-bari Concept of electron distribution. 5 9 Unit-7 Chemical Bonding and Ionization principles (i) Chemical bone- Definition, electronic theory of valency kinds of chemical bonds and characteristics (ii) Ionic Theory:- Ionic theory of Arrhenius & its utility, Neutralization & hydrolysis, pH. value, Buffer solution & its types. 5 8 Unit-8 Colloids Colloids :- Definition, phases, Types, classification of colloidal solutions, properties of different colloids. Soil Colloids :- Properties, utility, clay and humus. 3 6 Unit-9 Minerals and fertilizers Minerals and Fertilizers :- Minerals - Important minerals present in soil, classification of minerals, their chemical composition. Fertilizer :- Definition, nitrogenous, Phosphatic and Potassic fertilizers, mixed fertilizers - Industrial and commercial manufacturing of all fertilizers listed above. Percentage of nutrient elements present in various fertilizers. 4 7 Unit-10 Volumetric Analysis Titration, Indicator-types, properties, uses, Acid, Bases, salts- Definition, examples. and their characteristics, Equivalent weight, Atomic weight, strength gram per litre, Normality, Relationship between normality and strength. To find out normality, types of titration and calculations based on titration. 3 9 Section - C - Agricultural Biology Unit-11 Plant morphology Plant Morphology :- Root - Structure, types of roots, modification and functions of roots. Stem :- Modification and various functions of stem. Leaf :- Main parts, stipules, types, venation, phyllotaxy of leaves, functions, modifications. Flower :- Main parts and functions, Inflorescence (Curriculum based family) Seed :- Definition, Types, structure, Germination. 2 9 Unit-12 Pollination and Fertilization Pollination & Fertilization :- Pollination - Meaning, self & cross pollination. Types, devices which help in successful completion of self and cross pollination, self and cross pollinated crops, merits and demerits of self and cross pollination, means of cross pollination. 3 7 Unit-13 Cell Cell :- Initial history, Types, structure of plant and animal cell and difference between them, various cell organelles, their structure and functions. nucleus structure and function, Structure of RNA and DNA, their functions and difference. Elementary idea about tissues and tissue system. 5 9 Unit-14 Classification of Plant Kingdom Classification of Plant Kingdom :- Introduction, Importance, History of classification systems, Linnaeus and Benthem & Hooker's classification, method of describing a family, various definition and symbols. Description of families - leguminosae, composite and cucurbitaceae. 5 9 Unit-15 Classification of Animal Kingdom Classification of Animal Kingdom :- Importance, Principles and system of classification. Main phylums, their characteristics and examples of various animals included in that phylums 4 9 Unit-16 Computer Education in Agriculture I Basic information, Different parts of computer, Categories of computer (Digital, Analog & Hybrid) Utility of computer in different fields of Agricultural. 3 9 Unit-17 Computer Education in Agriculture II Information of M S Word and Window 98/2000 based on agriculture, Utility of multimedia and internet in modern agriculture. 3 9 Unit-18 Progression Arithmetic, Geometric,Harmonic, progression - Definition; find out the nth term, sum, Mean, and Agriculture based problems. 4 7 Unit-19 Trigonometry Trigonometry :- To find out the trigonometrical ratios of 0°, 30°, 45°, 60° and 90° and solution of simple numericals based on these ratio. 4 9 Unit-20 Vedic Mathematics Vedic Mathematics :- Fundamental operations & methods to check Answers. 3 9 Revision 20
In order to keep pace with technological advancement and to cope up with Madhya Pradesh Board examinations, Pearson group has launched Edurite to help students by offering Books and CDs of different courses online. | 1,395 | 6,463 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2019-35 | latest | en | 0.850887 |
https://www.free-online-converters.com/marks/marks-into-points.html | 1,653,766,689,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663019783.90/warc/CC-MAIN-20220528185151-20220528215151-00194.warc.gz | 861,619,499 | 9,259 | Free Online Converters > Convert Marks Into Points
Here you can Convert units of Marks to Points units, find all information about Marks. So, enter your unit's value in Left Column like Marks(if you use standard resolution on most non-HD laptops. FULL HD resolution starts at 1920 x 1080). Otherwise, if you use a lower value, enter the value in the box above. The Result / another converted unit value shell appears in the Left or below Column.
# Convert Marks Into Points
Marks
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Marks , Points , Marks to Points , Marks into Points , Marks in Points , How many Marks in many Points , How to convert Marks to Points online just in one Second ,
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##### conversion Table / conversion Chart
1 Marks = 124413.9072 Points
2 Marks = 248827.8144 Points
3 Marks = 373241.7216 Points
4 Marks = 497655.6288 Points
5 Marks = 622069.536 Points
6 Marks = 746483.4432 Points
7 Marks = 870897.3504 Points
8 Marks = 995311.2576 Points
9 Marks = 1119725.1648 Points
10 Marks = 1244139.072 Points
11 Marks = 1368552.9792 Points
12 Marks = 1492966.8864 Points
13 Marks = 1617380.7936 Points
14 Marks = 1741794.7008 Points
15 Marks = 1866208.608 Points
16 Marks = 1990622.5152 Points
17 Marks = 2115036.4224 Points
18 Marks = 2239450.3296 Points
19 Marks = 2363864.2368 Points
20 Marks = 2488278.144 Points
21 Marks = 2612692.0512 Points
22 Marks = 2737105.9584 Points
23 Marks = 2861519.8656 Points
24 Marks = 2985933.7728 Points
25 Marks = 3110347.68 Points
26 Marks = 3234761.5872 Points
27 Marks = 3359175.4944 Points
28 Marks = 3483589.4016 Points
29 Marks = 3608003.3088 Points
30 Marks = 3732417.216 Points
31 Marks = 3856831.1232 Points
32 Marks = 3981245.0304 Points
33 Marks = 4105658.9376 Points
34 Marks = 4230072.8448 Points
35 Marks = 4354486.752 Points
36 Marks = 4478900.6592 Points
37 Marks = 4603314.5664 Points
38 Marks = 4727728.4736 Points
39 Marks = 4852142.3808 Points
40 Marks = 4976556.288 Points
41 Marks = 5100970.1952 Points
42 Marks = 5225384.1024 Points
43 Marks = 5349798.0096 Points
44 Marks = 5474211.9168 Points
45 Marks = 5598625.824 Points
46 Marks = 5723039.7312 Points
47 Marks = 5847453.6384 Points
48 Marks = 5971867.5456 Points
49 Marks = 6096281.4528 Points
50 Marks = 6220695.36 Points
50 Marks = 6220695.36 Points
51 Marks = 6345109.2672 Points
52 Marks = 6469523.1744 Points
53 Marks = 6593937.0816 Points
54 Marks = 6718350.9888 Points
55 Marks = 6842764.896 Points
56 Marks = 6967178.8032 Points
57 Marks = 7091592.7104 Points
58 Marks = 7216006.6176 Points
59 Marks = 7340420.5248 Points
60 Marks = 7464834.432 Points
61 Marks = 7589248.3392 Points
62 Marks = 7713662.2464 Points
63 Marks = 7838076.1536 Points
64 Marks = 7962490.0608 Points
65 Marks = 8086903.968 Points
66 Marks = 8211317.8752 Points
67 Marks = 8335731.7824 Points
68 Marks = 8460145.6896 Points
69 Marks = 8584559.5968 Points
70 Marks = 8708973.504 Points
71 Marks = 8833387.4112 Points
72 Marks = 8957801.3184 Points
73 Marks = 9082215.2256 Points
74 Marks = 9206629.1328 Points
75 Marks = 9331043.04 Points
76 Marks = 9455456.9472 Points
77 Marks = 9579870.8544 Points
78 Marks = 9704284.7616 Points
79 Marks = 9828698.6688 Points
80 Marks = 9953112.576 Points
81 Marks = 10077526.4832 Points
82 Marks = 10201940.3904 Points
83 Marks = 10326354.2976 Points
84 Marks = 10450768.2048 Points
85 Marks = 10575182.112 Points
86 Marks = 10699596.0192 Points
87 Marks = 10824009.9264 Points
88 Marks = 10948423.8336 Points
89 Marks = 11072837.7408 Points
90 Marks = 11197251.648 Points
91 Marks = 11321665.5552 Points
92 Marks = 11446079.4624 Points
93 Marks = 11570493.3696 Points
94 Marks = 11694907.2768 Points
95 Marks = 11819321.184 Points
96 Marks = 11943735.0912 Points
97 Marks = 12068148.9984 Points
98 Marks = 12192562.9056 Points
99 Marks = 12316976.8128 Points
100 Marks = 12441390.72 Points
101 Marks = 12565804.6272 Points
102 Marks = 12690218.5344 Points
103 Marks = 12814632.4416 Points
104 Marks = 12939046.3488 Points
105 Marks = 13063460.256 Points
106 Marks = 13187874.1632 Points
107 Marks = 13312288.0704 Points
108 Marks = 13436701.9776 Points
109 Marks = 13561115.8848 Points
110 Marks = 13685529.792 Points
111 Marks = 13809943.6992 Points
112 Marks = 13934357.6064 Points
113 Marks = 14058771.5136 Points
114 Marks = 14183185.4208 Points
115 Marks = 14307599.328 Points
116 Marks = 14432013.2352 Points
117 Marks = 14556427.1424 Points
118 Marks = 14680841.0496 Points
119 Marks = 14805254.9568 Points
120 Marks = 14929668.864 Points
121 Marks = 15054082.7712 Points
122 Marks = 15178496.6784 Points
123 Marks = 15302910.5856 Points
124 Marks = 15427324.4928 Points
125 Marks = 15551738.4 Points
126 Marks = 15676152.3072 Points
127 Marks = 15800566.2144 Points
128 Marks = 15924980.1216 Points
129 Marks = 16049394.0288 Points
130 Marks = 16173807.936 Points
131 Marks = 16298221.8432 Points
132 Marks = 16422635.7504 Points
133 Marks = 16547049.6576 Points
134 Marks = 16671463.5648 Points
135 Marks = 16795877.472 Points
136 Marks = 16920291.3792 Points
137 Marks = 17044705.2864 Points
138 Marks = 17169119.1936 Points
139 Marks = 17293533.1008 Points
140 Marks = 17417947.008 Points
141 Marks = 17542360.9152 Points
142 Marks = 17666774.8224 Points
143 Marks = 17791188.7296 Points
144 Marks = 17915602.6368 Points
145 Marks = 18040016.544 Points
146 Marks = 18164430.4512 Points
147 Marks = 18288844.3584 Points
148 Marks = 18413258.2656 Points
149 Marks = 18537672.1728 Points
150 Marks = 18662086.08 Points
## how many Marks Into Points
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How Many Yottagrams in Marks | 2,505 | 7,835 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2022-21 | latest | en | 0.621209 |
http://woodworkingprojectsonline.com/woodworking-bench-apartment-free-woodworking-plans-jewellery-box.html | 1,555,728,665,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578528481.47/warc/CC-MAIN-20190420020937-20190420042937-00536.warc.gz | 198,266,254 | 6,978 | Proportion is the correlative relationship of all the parts to the whole. Although proportion usually refers to size, it is also a way to compare harmony between colour, quantity, placement and degree. Proportion is achieved when all sizes, shapes, textures, colors and so on complement one another. Remember that the eye appreciates some differences and may find equal parts monotonous and boring. Dividing space into equal parts such as halves, quarters and thirds is predictable and as a result the eye often skips past it.
With the inner jaw fastened to the bench, I used the router to flush-trim the jaw to the benchtop, across the top and down the sides adjacent to the top (stopping short of the discontinuity between the top and the legs). I'd thought this would be the best way to match up the jaw against the top, but I'd not do it this way again. It was very difficult to hold the router tight against the face of the jaw, and the result was a surface that wasn't as even as I had hoped.
First, cut one long edge. Second, cut a short edge, making sure it's square to the long edge you just cut. Clamp both pieces of the edging you'll be using along the long edge you've cut, and measure the width of the base plus 1/4-1/2", mark that, and then lay out a line through the mark that is square to the end you've cut, then cut along the line. Finally, cut the remaining short edge square to both long edges. (The length of the top doesn't need to precisely match anything, so we don't need to bother with clamping the trim before measuring.)
Finding a toolbox for a mechanic, for his hand tools, is not a big challenge at all - there are dozens of the tool boxes available on the market, from huge roll-around shop cases to small metal boxes. Plumbers, electricians, and farmers are well served, too, with everything from pickup-truck storage to toolboxes and belts. But, if you are a shop-bound woodworker then the case changes. You get to need a tool box that suits the range and variety of hand tools that most woodworkers like to have. For those who deny making do with second best, there's only one solution, you’ve to build a wooden toolbox that should be designed expressly for a woodworking shop.
```I ended up making a number of practice cuts. The first revealed that I hadn't tightened the screws on the edge guide enough. The second revealed that the design of the edge guide provided very little support at the end of a board, because of the cut-out for the router bit. In the "Getting Started in Woodworking" video, they had screwed a piece of hardwood to the edge-guide, to provide a continuous -- and longer -- bearing surface. I may do that myself, some day, but I didn't have the materials at hand, so I clamped some 2x4 scrap to the end of each board, to provide a continuous bearing surface past the ends. The two grooves in the long stretchers and the side groove in the short stretchers have identical layout. I made practice cuts in scrap until I had the edge guide set correctly, then I cut them all with that one setting. The bottom groove of the short stretchers uses a different setup, so it was back to the scrap, before cutting them.
```
“Traditional woodworking using hand tools can offer a more satisfying relationship with the wood and the creative woodworking process. It’s quieter, cleaner and maybe even a little spiritual. It’s no surprise that many “”plugged-in”” woodworkers are returning to the roots of this treasured skill. Where some hand-tool books focus solely on the use of hand tools, Made By Hand takes you right to the bench and shows you how to start building furniture using these tools.
For the router you'll need a a 3/8" straight bit, an edge guide, 1/4"- and 1/8"-radius roundover bits, and a flush-trim bit with at least a 1-1/2" cutting length. Bits of this size are available only for a 1/2" collet. Some routers are capable of using multiple collet sizes. I was fool enough to buy a router that only had a 1/4" collet. More on that, later.
One aspect of the Windsor Design that is hit or miss is the vice. Notice we have to use the singular when describing it because this is the first workbench that we have reviewed which features only one vice. To make matters more frustrating, the vice cannot be repositioned along the bench. That said, the vice does at least have the largest capacity out of any we reviewed. At 7” capacity, the Windsor Design vice can secure any size workpiece you may have.
The plan tutorial includes images, diagrams, step-by-step instructions, and even a video to help you along the way. You can also go with some more bookcase design ideas. Browse the internet for more and we are also proving a link below to some more ideas to this plan. Select and build one of these free bookcase DIYs and you will have everything available easily that you need to get started creating a bookshelf for any room in your house.
Another comprehensive book by Taunton Press, Wood Flooring is extremely informative, not to mention visually compelling. From subfloors to finishing, and everything in between, you can be sure this is the only book you will ever need on the subject. My favourite chapter was on inlays and curves: using jigs and templates to add circular inlays to a floor, laying out and installing laminated curves, even working with stone and metal inlays … there’s a lot to learn in that one chapter alone. Much of it could be adapted to making furniture. If you want to lay a basic wood floor this book will help you add lasting value and beauty to your home. If you’re looking to one-up your super handy DIY brother-in-law, you can start here too.
So drill the benchdog holes through the MDF layer. Begin by laying out their positions. You'll want these to be precise, so that the distances between the holes are consistent. The vises you are using will constrain your benchdog spacing. My front vise worked most naturally with two rows of holes four inches apart, my end vise with two pairs of rows, with four inches between the rows and eight inches between the pairs. Because of this, I decided on a 4" by 4" pattern.
Building a bookcase or bookshelf is a fairly simple woodworking plan that you can get done in just a day or two. This is also a low-cost project as well and since the project idea is free, you don't have to worry about busting through your budget. Just follow the simple steps in the tutorial and enjoy your own company building a simple bookcase on this weekend.
There are two books that I would consider highly influential in my personal woodworking path. Darrell Peart’s “Greene & Greene: Design Elements for the Workshop” (Linden) and “Adventures in Wood Finishing” by George Frank (Taunton). Peart’s book kicked off my fascination with all things Greene & Greene. Peart does a fantastic job of covering history as well as practical techniques.
The Handbuilt Home. Ana White has created a wonderful DIY book. It’s filled with 34 projects she describes as simple, stylish, and budget-friendly. And you’ve probably already guest that they are related to improving your home. Each project has diagrams and instructions. And she tells the level of difficulty and project costs. This book resonates with many readers who want to make things themselves, but they may not have the fancy tools or the budget. Ana shows you how to make beautiful and affordable DIY projects.
Drill four 5/8-in.-dia. 1/2-in.-deep holes on the large disc?inside the traced circle?then use 5/8-in. dowel centers to transfer the hole locations to the underside of the small disc. Drill four 1/2-in.-deep holes on the underside of the small disc and a 1/2-in.-deep hole in the center of the top for the dowel handle. Glue in the dowels to join the discs, and glue in the handle. We drilled a wood ball for a handle knob, but a screw-on ceramic knob also provides a comfortable, attractive grip.
Furniture with Soul sheds light on some of the worlds most respected and revered makers of fine furniture. Going into their shops and learning how they got into the craft, what they most enjoy about it, and hearing some funny stories along they way is what this book is all about. It’s not a how-to in any way, unless you’re talking about how these selected group of makers made a living creating exquisite works of art. Beautiful photos of creative and challenging pieces of furniture are included throughout. My favourite book from the last year.
“Introducing a new woodworking series in the tradition of Tage Frid…a series filled with essential information required by woodworkers today. For the first time ever, all the techniques and processes necessary to craft beautiful things from wood have been compiled into three comprehensive volumes: The Complete Illustrated Guides. Highly visual and written by woodworking’s finest craftsmen, these three titles — Furniture & Cabinet Construction, Shaping Wood, and Joinery — will establish a new standard for shop reference books.
Building a wine rack is usually a very common beginner's woodworking plan. Creating a wine rack is an easy plan that can most of the time be completed in a day or half, depending on how large and detailed you would like it to be. And the better news is that this free wine rack plan will let you build you a great looking wine rack for much less than it would cost.
Woodworking Books go hand in hand with the tools and the work. Whether you are a beginner or a master woodworker there is always a new technique or a great tip waiting for you on the pages. We have hundreds of woodworking books to choose from, written by woodworking authors who understand the craft and know the trials and tribulations of the path.
In the "Getting Started with Woodworking" video, the holes through the 4x4's were drilled from the back. That is, they start on the side opposite the precisely-positioned mark, and drill through to hit it. If they can do this, more power to them, but I can't drill through 3-1/2" of wood to emerge at a precise mark without a drill press - and not always then.
I made a template by scribing two adjoining squares on a piece of MDF, using compass and straightedge, then marking each corner with a centerpunch, then drilling the points with a 1/16" bit. I find I'm always breaking small bits, so I picked up a couple of each size some months ago, and on looking I found I had three 1/16" bits, which worked fine for what I intended.
Another odd design feature of the Windsor Design is its dog holes. While this bench provides a variety of dog holes and even offers numerous pegs and dowels to go along with it, the dog holes are not spaced as closely or with as much variation as you see with many other benches. Ultimately, this means that smaller workpieces may not be able to use the dog holes and pegs as well. | 2,382 | 10,803 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2019-18 | latest | en | 0.967006 |
https://en.wikibooks.org/wiki/AQA_A-Level_Physics/Radioactive_decay | 1,723,244,217,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640782236.55/warc/CC-MAIN-20240809204446-20240809234446-00124.warc.gz | 171,911,849 | 13,580 | Radioactive decay is when an unstable nucleus decays in a random manner.
In theory, there is no end to the life of a radioactive substance (the time it takes before the activity reaches zero)therefore the quantity used for dealing with the life of radioactive substances is the half life. Half life is defined as the time taken for the activity of the sample to halve. The half life remains the same throughout the life of the sample. Each type of element has a distinct half life, ${\displaystyle t_{\frac {1}{2}}}$. As the activity of the sample is proportional to the number of radioactive nuclides present the half time can also be calculated by the time taken for half of the radioactive nuclides in the sample to decay.
If you plot a graph of the natural logarithm of N (the number of radioactive nuclides) against time you will see that there is a linear relationship between the two. The steeper the gradient the more quickly the substance will decay which means it has a shorter half life. Half life is proportional to ${\displaystyle {\frac {1}{\lambda }}}$
# Equations
${\displaystyle {\frac {\Delta N}{\Delta t}}=-\lambda N}$
${\displaystyle N=N_{0}e^{-\lambda t}}$
${\displaystyle t_{\frac {1}{2}}={\frac {ln2}{\lambda }}}$
${\displaystyle A=\lambda N}$
# Derivations of equations
Note - given that all equations above are provided in the AQA Formula booklet (which is provided in the exam), it is not necessary to memorise all of the equations, but it is a good idea to learn what the different symbols stand for.
## Deriving ${\displaystyle N=N_{0}e^{-\lambda t}}$
We can use this differential equation to derive an equation for N.
${\displaystyle {\frac {dN}{dt}}=-\lambda N}$
Separating and integrating:
${\displaystyle \int {\frac {1}{N}}dN=\int {-\lambda }dt}$
${\displaystyle ln(N)=-\lambda t+c}$
${\displaystyle N=N_{0}e^{-\lambda t}}$
## Deriving ${\displaystyle t_{1/2}={\frac {ln2}{\lambda }}}$
Half life is the time taken for half of the atoms that were originally present in the sample to have decayed.
${\displaystyle N_{(t)}=N_{0}e^{-\lambda t}}$
If we replace ${\displaystyle N_{0}}$ with 1, ${\displaystyle N_{t}}$ with ${\displaystyle {\frac {1}{2}}}$, and t with ${\displaystyle t_{\frac {1}{2}}}$ we get:
${\displaystyle {\frac {1}{2}}=e^{-\lambda t_{\frac {1}{-2}}}}$
${\displaystyle -\lambda t_{\frac {1}{2}}=ln{\frac {1}{2}}}$
${\displaystyle t_{\frac {1}{2}}={\frac {ln2}{\lambda }}}$
If you wanted to know the ${\displaystyle {\frac {1}{n}}}$ of a substance, you can replace the two with a 'n'
${\displaystyle t_{\frac {1}{n}}={\frac {ln(n)}{\lambda }}}$ | 733 | 2,617 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 22, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2024-33 | latest | en | 0.887258 |
http://oeis.org/A256020 | 1,596,491,188,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735833.83/warc/CC-MAIN-20200803195435-20200803225435-00580.warc.gz | 79,044,726 | 3,858 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A256020 a(n) = Sum_{i=1..n-1} (i^4 * a(i)), a(1)=1. 2
1, 1, 17, 1394, 358258, 224269508, 290877551876, 698687879606152, 2862524242746404744, 18783884080901907930128, 187857624693099981209210128, 2750611340756369924865254694176, 57039427373264843131930786593127712, 1629160124635190449534207126672913710144 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,3 LINKS FORMULA Product_{i=2..n-1} (i^4 + 1), for n>2. a(n) ~ (cosh(Pi/sqrt(2))^2 * sin(Pi/sqrt(2))^2 + cos(Pi/sqrt(2))^2 * sinh(Pi/sqrt(2))^2) / (2*Pi^2) * ((n-1)!)^4. a(n) = A255434(n-1)/2. MATHEMATICA Clear[a]; a[1]=1; a[n_]:= a[n] = Sum[i^4*a[i], {i, 1, n-1}]; Table[a[n], {n, 1, 15}] Flatten[{1, 1, Table[Product[(i^4 + 1), {i, 2, n-1}], {n, 3, 15}]}] Join[{1}, FoldList[Times, Range[15]^4+1]/2] (* Harvey P. Dale, Jul 29 2018 *) CROSSREFS Cf. A001710, A051893, A256019. Sequence in context: A007410 A203229 A269791 * A072160 A078814 A242282 Adjacent sequences: A256017 A256018 A256019 * A256021 A256022 A256023 KEYWORD nonn AUTHOR Vaclav Kotesovec, Mar 13 2015 STATUS approved
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Last modified August 3 17:26 EDT 2020. Contains 336200 sequences. (Running on oeis4.) | 603 | 1,576 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2020-34 | latest | en | 0.486381 |
https://discuss.mxnet.apache.org/t/equation-1-3-3/6166 | 1,601,229,500,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400283990.75/warc/CC-MAIN-20200927152349-20200927182349-00322.warc.gz | 335,869,101 | 3,643 | # Equation 1.3.3
Hi,
How do we arrive at equation 1.3.3 on page 27?
L(action|x) = Ey∼p(y|x)[loss(action,y)].
And how do we interpret this?
I understand the previous one (following), which says “Conditional Probablity” that y is equal to deathcap from images is 0.2…
P(y = deathcap|image) = 0.2 | 97 | 296 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2020-40 | latest | en | 0.881978 |
https://www.codefellows.org/blog/selection-sort-algorithm/ | 1,709,067,484,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474686.54/warc/CC-MAIN-20240227184934-20240227214934-00090.warc.gz | 699,377,786 | 81,736 | # Walk-through: Selection Sort Algorithm
### How do you sort?
You are dealt a hand of 10 playing cards. What’s the first thing you do? Organize them. One of my typical strategies for quickly doing so goes like this if the suit doesn’t matter: Find the lowest value card and put it in the left-most spot in my hand. Then find the lowest of the remaining cards, and place it just to the right of the first card. Repeat, until all the cards are in order. This process is similar to how a computer might sort a list of values with the “selection sort” algorithm. This is just one of the dozens of popular approaches to organizing data, but it maps particularly well to how humans often attempt to sort things. Let’s take a look at how selection sort works!
### Selection Sort Algorithm
Selection sort is a sorting algorithm that rearranges each of the values in, say, an array, to create an ordered listing. This algorithm traverses the array multiple times as it slowly builds up the sorted sequence from smallest to largest (in our case). Since it swaps elements from their original position to their final position, it is said to be an “in-place” sort, that does not require much extra data storage. During selection sort, the array is treated as having two subsections, the sorted portion (everything to the left of the current position under consideration) and the unsorted portion (everything to the right). As the algorithm executes, more and more items are moved into the sorted portion of the array, so the unsorted portion gets smaller until there are no more unsorted items to move. Selection sort is not what we consider a “stable” sort, meaning that the original relative ordering of identical values in the array cannot be guaranteed once the algorithm is done. If that is a necessity for your sorting tasks, pick a different algorithm! Below is an evaluation and walkthrough of selection sort. We will show the pseudocode, walk through each of the iterations, and explain what is going on at each level.
#### Selection Sort Pseudocode
``````SelectionSort(arr)
DECLARE n <-- arr.length;
FOR i to n - 1
DECLARE min <-- i
FOR j is i + 1 to n
if (arr[j] < arr[min])
min <-- j
DECLARE temp <-- arr[min]
arr[min] <-- arr[i]
arr[i] <-- temp
``````
#### Walkthrough
Our example below will be based on an array that looks like this at the start: [8,4,23,42,16,15]
Pass 1: In the first pass-through of the selection sort, we evaluate if there is a smaller number in the array than what is currently present in index 0. This act is essentially finding the smallest number in the array and it will eventually place it in the very first index of the array. To do this, we must first evaluate what is placed in index 0 (in case the smallest value is already in the correct place) and check and see if a smaller number exists. We find this smaller number right away in index 1 (value of 4). The minimum variable gets updated to remember this index (j = 1), and the variable temp is updated with the index’s value (temp = 4). At the end of this iteration, after we verify there is no value smaller than 4, this number will be swapped with the current value (of 8) in index i (which is still 0). This results in our smallest number of our array being placed first.
Pass 2: The second pass through the array evaluates the remaining values in the array to see if there is a smaller value other than the current position of i. 8 is the 2nd smallest number in the array, so it “swaps” with itself. The minimum value does not change at all during this “i = 1” iteration.
Pass 3: The third pass-through evaluates the remaining indexes in the array, starting at position “i = 2”. Both positions 4 and 5 are smaller than the value in position 2. Each time a smaller number than the current minimum is found, the `min` variable will update to the index of the new smallest number. In this case, 15 is the next smallest number. As a result, it will swap with the value in position 2.
Pass 4: The 4th pass (i = 3) through the array finds that 16 is now the smallest number in the unsorted portion of the array, and as a result, switches places with the value 42.
Pass 5: The 5th pass (i = 4) through the array only has one other index to evaluate. Since the last index (j = 5) value is less than the value at the index of 4, the two values will swap.
Pass 6: On its final iteration through the array, it will swap places with itself as it evaluates the value against itself. After this iteration, `i` will increment to 6, forcing it to break out of the outer `for` loop, leaving our array sorted.
Can you see a way to modify the code to eliminate this step?
#### Complexity Analysis
Let’s evaluate the efficiency of this algorithm. As described here, the time complexity of selection sort is O(N^2). This is because for every index in the array, we must compare it against the remaining unsorted indexes to ensure that nothing “smaller” exists. The basic operation for selection sort can be thought of as a comparison operation. This comparison will occur an “n-squared” (divided by 2) number of times.
Selection Sort will usually be done in place, meaning that no extra memory or space needs to be created. As a result, we describe it as O(1) space complexity.
### Summary
Amongst the world’s most popular sorting algorithms, selection sort is not the most efficient available, but it is straightforward to implement and is a great option when working with smaller data sets. As the data sets get larger, more efficient sorting algorithms shine. Many chose selection sort as a beginning sort to explore because of its simplicity and accessible implementation. Now that you have a baseline of comprehension, try coding it yourself in your favorite language, and see how this compares with other algorithms. | 1,307 | 5,810 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2024-10 | latest | en | 0.930246 |
https://math.stackexchange.com/questions/21381/how-to-calculate-a-decimal-power-of-a-number | 1,653,555,526,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662604495.84/warc/CC-MAIN-20220526065603-20220526095603-00042.warc.gz | 459,254,573 | 68,354 | # How to calculate a decimal power of a number
I wish to calculate a power like $$2.14 ^ {2.14}$$
When I ask my calculator to do it, I just get an answer, but I want to see the calculation.
So my question is, how to calculate this with a pen, paper and a bunch of brains.
• By hand, you will need a table of logarithms or a slide rule. This operation can't be done in finitely many steps, like ordinary addition or multiplication, you have to resort to approximation. Dec 6, 2013 at 9:58
• @arbautjc I would only need the first 5 decimals, I guess it would be possible in finite steps then. Dec 6, 2013 at 10:10
• You are right, but it's still an iterative process that is not very easy, and very time consuming. Historically, it was done to build tables, which were then used for any computation of this kind. By the way, 5 decimals is the precision given by "standard" tables of logarithms ;-) Dec 6, 2013 at 10:14
• @arbautjc thanks for the helpful links and comments. I will study those links further, I'm curious to the methods they applied to create the table. Dec 6, 2013 at 10:16
• @Mixxiphoid You can compute the natural logarithm and the exponential from their power series (though there may be more efficient methods). The really magical thing is that you don't need log tables to use a slide rule, since they're essentially built in. Somebody with a slide rule who is still in practice (not so many people these days) should be able to demonstrate $2.14^{2.14}$ quite quickly (though maybe not to five decimal places). Sep 24, 2014 at 21:03
For positive bases $a$, you have the general rule $$a^b = \exp(b\ln(a)) = e^{b\ln a}.$$
This follows from the fact that exponentials and logarithms are inverses of each other, and that the logarithm has the property that $$\ln(x^r) = r\ln(x).$$
So you have, for example, \begin{align*} (2.14)^{2.14} &= e^{\ln\left((2.14)^{2.14}\right)} &\quad&\mbox{(because $e^{\ln x}=x$)}\\ &= e^{(2.14)\ln(2.14)} &&\mbox{(because $\ln(x^r) = r\ln x$)} \end{align*} Or more generally, $$a^b = e^{\ln(a^b)} = e^{b\ln a}.$$
In fact, this is formula can be taken as the definition of $a^b$ for $a\gt 0$ and arbitrary exponent $b$ (that is, not an integer, not a rational).
As to computing $e^{2.14\ln(2.14)}$, there are reasonably good methods for approximating numbers like $\ln(2.14)$, and numbers like $e^r$ (e.g., Taylor polynomials or other methods).
• Thanks for the effort! I hoped for a bit more easy way to calculate is, but that seems to be impossible. I'll accept your answer. Feb 10, 2011 at 16:24
• @Mixxiphoid: For rational exponents, you can also use the fact that $a^{1/b} = \sqrt[b]{a}$. Here, you have $2.14 = \frac{214}{100} = \frac{107}{50}$, so you could "just" take the 50th root of $(2.14)^{107}$. Feb 10, 2011 at 16:34
• Slightly easier if you write $50=2\cdot 5^2$, but fifth root is not that easy either. For square root, on the other hand, there is an algorithm that resembles a bit the usual euclidian division. Dec 6, 2013 at 10:06
A decimal power can be seen as a fraction:
$x^{\frac{a}{b}} = \sqrt[b]{x^a}$
Of course you cannot write every number as a fraction, but you can at least approximate every number by a fraction.
• Moreover, for $y \in \mathbb{R}\setminus\mathbb{Q}$, you can define $x^y := \sup \{x^r \;|\; r \in \mathbb{Q}, r \lt y\}$ which admittedly is only really useful when one is trying to develop the fundamentals of real analysis from scratch Feb 10, 2011 at 17:03
we can find $2.14 ^{2.14}$ using basic arithmetic operations +,-,/,*.
Use binomial theorem for rational number $n and$-1
$(1+x)^n =1+nx+n(n-1)x^2/2!+....$
note that in left hand side the power n is a fractional number but in the right hand side the powers are integers. that is, in the right hand side, each term can be calculated using basic operations +,-,*,/.
outline of the problem
$2.14^{2.14}=(1.14+1)^{2.14}$
$=(1.14^{2.14}) * (1+1/1.14)^{2.14}$
$=(1+0.14)^{2.14} * (1+1/1.14)^{2.14}$
using binomial theorem two times (5 decimal places) and multiplying we get the answer
I really like your question. So many students are content with learning (and so many instructors content with teaching) just the calculator key sequences that will give the correct answer. But to know math (and almost everything else in this world) you’ve got to get under the hood and “see” what’s actually going on.
Let’s start with your example, 2.14^2.14. When you look at the exponent, more than likely you intuitively get the feeling that one portion of the answer is due to the integer part, “2”, with the balance attributable to the decimal “0.14”. And you’re right.
So, let’s raise 2.14 to our integer power (which you can do by hand— though there’s nothing wrong with employing a calculator when you understand the manipulations you’re carrying out):
2.14 ^ 2 = (2.14 * 2.14) = 4.5796.
Actually, let’s back up a little and use our calculator to get the answer to our example; 2.14 ^ 2.14 = 5.09431.
Now that we have ‘the answer’ and the portion attributable to the integer component of our exponent, let’s determine the increase contributed by our decimal component; (5.09431/4.5796) = 1.112392. Ok, but other than the ratio, (5.09431/4.5796), just what is “1.112392”?
Fasten your seat belt— It is simply 2.14 ^ 0.14 power = 1.112392.
(Yes, use your calculator for this intermediate step)
So, 2.14 ^ 2.14 = (2.14 ^ 2 * 2.14 ^ 0.14) = (4.5796 * 1.112392) = 5.09431
Let’s try 5.27 ^ 4.34 = 1357.244436
5.27 ^ 4 = 771.33397441… 5.27 ^ 0.34 = 1.759607
(771.33397441 * 1.759607) = 1357.244436
Hope this is what you were looking for. Have fun! JE Magee
• I like your answer. But requiring a calculator for the "intermediate step" kinda defeats the purpose. Dec 19, 2013 at 8:58
• +1. I reached this conclusion by trial and error and didn't understand why it worked. Beautiful explanation. Apr 17, 2015 at 15:27
• You still relies on the calculator and don't know the principle behind the calculator. Basically what you did was just writing $a^{(b+c)}=a^b*a^c$, which is trivial. Oct 14, 2019 at 7:10
You use $\exp(2.14 \ln 2.14)$ or any base for logarithms you choose. But if you want pen and paper, you can help with the properties of exponents. $2.14^{2.14}=2.14^2\cdot2.14^{.14}=2.14^2\exp(.14(\ln 2 + \ln1.07))$ will converge more quickly, especially if you are willing to look up $\ln 2$.
Newton's approximation for $r = \sqrt{c}$ gives the iteration $r_{n+1} = r_n - \frac{{r_n}^2-c}{2r_n}$
$\sqrt{2.14} \approx 1.5 \rightarrow 1.46 \rightarrow 1.4628 \rightarrow 1.462874 \text{ (6sf)}$
Using that $10$ times gives $2.14 \rightarrow 1.462874 \rightarrow 1.209493 \rightarrow 1.099769 \rightarrow 1.048698 \rightarrow 1.024059$
$\rightarrow 1.011958 \rightarrow 1.005961 \rightarrow 1.002976 \rightarrow 1.001486 \rightarrow 1.000743 \text{ (6sf)}$
Thus $\ln 2.14 = 2^{10} \ln 2.14^{2^{-10}} \approx 2^{10} \ln 1.000743 \approx 2^{10} \times 0.000743 \approx 0.7608 \text{ (3sf)}$
$2.14^{2.14} = e^{ 2.14 \ln 2.14 } \approx e^{ 2.14 \times 0.7608 } \approx e^{1.628} \text{ (3sf)}$
The geometric series or binomial expansion gives the approximate
$2^{-10} = (1000+24)^{-1} \approx 1/1000 - 24/1000^2 + 576/1000^3$
Thus $e^{1.628} = (e^{1.628 \times 2^{-10}})^{2^{10}} \approx (e^{0.001590})^{2^{10}} \text{ (3sf)}$
$\approx (1+0.001590+0.001590^2/2)^{2^{10}} \approx 1.001591^{2^{10}} \text{ (6sf)}$
Squaring $10$ times gives $1.001591 \rightarrow 1.003185 \rightarrow 1.006380 \rightarrow 1.012801 \rightarrow 1.025766 \rightarrow 1.052196$
$\rightarrow 1.107116 \rightarrow 1.225706 \rightarrow 1.502355 \rightarrow 2.257071 \rightarrow 5.094369 \approx 5.09 \text{ (3sf)}$
which is $2.14^{2.14}$ to $3$ significant figures. I am lazy so I used a calculator for nine of the repetitions of square-root and squaring, but the above computation is clearly feasible by hand as only $O(n^3)$ operations are needed for $n$ bits of precision. It is amusing that so much work went in to produce only 3 decimal digits but I do not know any better way that can be easily extended to arbitrary precision. | 2,602 | 8,076 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2022-21 | longest | en | 0.934694 |
https://math.stackexchange.com/questions/2722338/planar-graphs-and-number-of-faces | 1,560,883,089,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998813.71/warc/CC-MAIN-20190618183446-20190618205446-00116.warc.gz | 507,033,899 | 34,864 | # planar graphs and number of faces
Show that no matter in what way we embed a planar graph, we always get the same number of faces.
This is trivial for connected graphs, because Euler's formula applies and shows that $f = 2 - n + e$, and no matter how we embed the graph, the number of vertices and edges is already fixed.
I think that for disconnected graphs the formula helps as well: say we have $m$ components. Then we add one edge for each component so $m-1$ edges to connect them. The formula now applies, and we have added $m-1$ edges so $f = 2 - n + e + m-1$. The outer face is the same because we do not create new faces, so when we remove the added edges we get $f$. Maybe I am being too obnoxious about this - is there an easier way? Also is my idea correct?
• Looks good to me, and I don't think you're being obnoxious at all. – saulspatz Apr 4 '18 at 19:36
• Thank you very much @saulspatz – mandella Apr 4 '18 at 19:37
• That's the correct way to extend Euler's formula for disconnected graphs. The fact about the faces is a consequence. – Ethan Bolker Apr 4 '18 at 19:37
• @EthanBolker Oh, good point! Thank you. – mandella Apr 4 '18 at 19:38
• Your idea is good. The problem is a little bit murky for disconnected graphs, since then the "faces" aren't all simply connected (e.g., embed two triangular graphs with one enclosing the other). If you take "face" to mean a connected component of the complement of the embedded graph, then your argument is fine. – Rob Arthan Apr 4 '18 at 20:08 | 406 | 1,509 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2019-26 | latest | en | 0.917861 |
https://www.physicsforums.com/threads/indentifying-the-ball.294084/ | 1,566,293,344,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315321.52/warc/CC-MAIN-20190820092326-20190820114326-00472.warc.gz | 957,182,080 | 15,577 | # Indentifying the ball
#### Sabine
we have 3 balls: 1 red, 1 green and 1 half red half green.
1-the probability to pick a certain ball of these 3 is:1/3
2-if after picking the ball, we were told that the ball contains the colour red (for exe.), then the probability of telling which ball it is becomes:1/2
(same thing for the colour green).
then why cant we by just picking the ball consider that it contains a certain colour and by that the probability of question 1 would become 1/2?
Related Set Theory, Logic, Probability, Statistics News on Phys.org
#### CompuChip
Homework Helper
The trick is in the first part of the second statement: if ....
You have some extra information, which changes the probabilities. We are talking about conditional probability here.
In the first case, you are picking any ball with equal probability from a set of 3 (R, G and R/G). So for each one, the probability is 1/3.
Now suppose that I tell you have the ball contains the colour red, and I ask you, what is the probability that it is completely red? Clearly, it is not 1/3 anymore. By giving you some information I have changed the experiment. Do you agree that a completely equivalent experiment would be if I took out the green ball and then asked you to compute the probability that you take the red ball? Well, there are two balls (R and R/G) with equal probability, so you have 1/2 probability of picking it.
Another famous example of conditional probabilities is the three-door example. Suppose you are on a TV-show and the quiz master presents you with three closed doors. There is a grand prize behind only one of them. You pick a door. Then the quiz master opens one of the other two doors, and shows that there is no prize behind it. He asks you if you want to switch or stay with your original choice. Should you stick to your door, switch to the other one, or doesn't it matter?
Surprisingly, the answer is not the latter. By opening a door and showing that there is no prize, you have been given extra information. So the probabilities are now conditional, and the chance of the prize being behind door 1 is not the same as the chance of the prize being behind door 1 given that it is not behind door 2.
#### HallsofIvy
Homework Helper
we have 3 balls: 1 red, 1 green and 1 half red half green.
1-the probability to pick a certain ball of these 3 is:1/3
2-if after picking the ball, we were told that the ball contains the colour red (for exe.), then the probability of telling which ball it is becomes:1/2
(same thing for the colour green).
then why cant we by just picking the ball consider that it contains a certain colour and by that the probability of question 1 would become 1/2?
It's not clear to me what you mean by "consider that it contains a certain colour". Do you mean pick a ball and then, without looking at it, say "it contains red". Obviously that does not make the probability that it is a specific ball equal to 1/2 because we might be wrong- we might have chosen the green ball.
I by "consider that it contains a certain colour" you simply mean "it must contain some color and it doesn't matter which", that is also wrong. My guess as to which ball it is certainly does depend upon whether I know it "contains" the colour red or the colour green.
If I pick one of the three balls and I am told it "contains" red, then I know i tis not the green ball- it must be either completely red or half red and half blue. Those are equally likely so the probability is 1/2.
If I am told that it "contains" green, then I know that it must not be the red ball- it must be either the green ball or half red half green. Again, the probability that it is one of those is 1/2.
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• Solo and co-op problem solving | 940 | 4,023 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2019-35 | latest | en | 0.966313 |
https://www.teachoo.com/2447/614/Example-6---Show-that-middle-term-of-(1---x)2n-is-1.3.5...(2n-1)-n/category/Examples/ | 1,685,874,390,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649741.26/warc/CC-MAIN-20230604093242-20230604123242-00565.warc.gz | 1,101,702,735 | 33,549 | Examples
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### Transcript
Question 2 Show that the middle term in the expansion of (1 + x)2n is (1 . 3 . 5 …. (2𝑛 − 1))/𝑛! 2n xn, where n is a positive integer. Given Number of terms = 2n which is even So, Middle term = (2n/2 + 1)th term = (n + 1)th term Hence, we need to find Tn + 1 We know that general term of (a + b)nis Tr + 1 = nCr an – r br For Tn + 1 , Putting n = 2n , r = n , a = 1 & b = x Tn+1 = 2nCn (1)2n – n (x)n = (2𝑛)!/𝑛!(2𝑛 −𝑛)! . (1)n . xn = (2𝑛)!/(𝑛! 𝑛!) . xn = (2𝑛(2𝑛 − 1)(2𝑛 − 2) ……. 4 × 3 × 2 × 1)/(𝑛! 𝑛!) xn = ([(2𝑛 − 1)(2𝑛 − 3)….…. × 5 × 3 × 1] [(2𝑛)(2𝑛 − 2)… × 4 × 2])/(𝑛! 𝑛!) xn We need to show (1 . 3 . 5 …. (2𝑛 − 1))/𝑛! 2n xn = ([1 × 3 × 5 × …… × (2𝑛 − 3)(2𝑛 − 1)] [2 × 4 × 6….. × (2𝑛 − 2) × 2𝑛])/(𝑛! 𝑛!) xn = ([1 × 3 × 5……. × (2𝑛−3)(2𝑛−1)] [(2 × 1) ×(2 × 2) ×(2 × 3) × ….. × 2 (𝑛−1) × 2(𝑛)])/(𝑛! 𝑛!) xn = ([1 × 3 × 5……. × (2𝑛−3)(2𝑛−1)] (2 × 2 × 2 × 2 ……..× 2) [1 × 2 × 3 ….. (𝑛−1) 𝑛])/(𝑛! 𝑛!) xn = ([1 × 3 × 5……. × (2𝑛 − 3)(2𝑛 − 1)] 2𝑛 [1 × 2 × 3 ….. (𝑛 − 1) 𝑛])/𝑛!𝑛! xn = ([1 × 3 × 5……. × (2n − 3)(2n − 1)] 2n (n!))/(n! n!) . xn = (𝟏 × 𝟑 × 𝟓……. × (𝟐𝐧 − 𝟏))/(𝐧! ) 2n . xn Hence middle term of expansion (1 + x)2n is (1 . 3 . 5……. ….(2n−1))/(n! ) 2n . xn Hence proved
#### Davneet Singh
Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo. | 856 | 1,577 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2023-23 | longest | en | 0.77814 |
http://de.metamath.org/mpegif/ellimc2.html | 1,611,183,964,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703522133.33/warc/CC-MAIN-20210120213234-20210121003234-00638.warc.gz | 25,668,983 | 27,884 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > ellimc2 Structured version Visualization version Unicode version
Theorem ellimc2 22832
Description: Write the definition of a limit directly in terms of open sets of the topology on the complex numbers. (Contributed by Mario Carneiro, 25-Dec-2016.)
Hypotheses
Ref Expression
limccl.f
limccl.a
limccl.b
ellimc2.k fld
Assertion
Ref Expression
ellimc2 lim
Distinct variable groups: ,, ,, ,, ,, ,, ,,
Proof of Theorem ellimc2
Dummy variables are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 limccl 22830 . . . 4 lim
21sseli 3428 . . 3 lim
32pm4.71ri 639 . 2 lim lim
4 eqid 2451 . . . . . 6 t t
5 ellimc2.k . . . . . 6 fld
6 eqid 2451 . . . . . 6
7 limccl.f . . . . . 6
8 limccl.a . . . . . 6
9 limccl.b . . . . . 6
104, 5, 6, 7, 8, 9ellimc 22828 . . . . 5 lim t
1110adantr 467 . . . 4 lim t
125cnfldtopon 21803 . . . . . . 7 TopOn
139snssd 4117 . . . . . . . 8
148, 13unssd 3610 . . . . . . 7
15 resttopon 20177 . . . . . . 7 TopOn t TopOn
1612, 14, 15sylancr 669 . . . . . 6 t TopOn
1716adantr 467 . . . . 5 t TopOn
1812a1i 11 . . . . 5 TopOn
19 ssun2 3598 . . . . . . 7
20 snssg 4105 . . . . . . . 8
219, 20syl 17 . . . . . . 7
2219, 21mpbiri 237 . . . . . 6
2322adantr 467 . . . . 5
24 elun 3574 . . . . . . . 8
25 elsn 3982 . . . . . . . . 9
2625orbi2i 522 . . . . . . . 8
2724, 26bitri 253 . . . . . . 7
28 simpllr 769 . . . . . . . 8
29 pm5.61 719 . . . . . . . . . 10
307ffvelrnda 6022 . . . . . . . . . . 11
3130ad2ant2r 753 . . . . . . . . . 10
3229, 31sylan2b 478 . . . . . . . . 9
3332anassrs 654 . . . . . . . 8
3428, 33ifclda 3913 . . . . . . 7
3527, 34sylan2b 478 . . . . . 6
3635, 6fmptd 6046 . . . . 5
37 iscnp 20253 . . . . . 6 t TopOn TopOn t t
3837baibd 920 . . . . 5 t TopOn TopOn t t
3917, 18, 23, 36, 38syl31anc 1271 . . . 4 t t
40 iftrue 3887 . . . . . . . . . . 11
4140, 6fvmptg 5946 . . . . . . . . . 10
4222, 41sylan 474 . . . . . . . . 9
4342eleq1d 2513 . . . . . . . 8
4443imbi1d 319 . . . . . . 7 t t
4544adantr 467 . . . . . 6 t t
465cnfldtop 21804 . . . . . . . . . . 11
47 cnex 9620 . . . . . . . . . . . . . 14
4847ssex 4547 . . . . . . . . . . . . 13
4914, 48syl 17 . . . . . . . . . . . 12
5049ad2antrr 732 . . . . . . . . . . 11
51 restval 15325 . . . . . . . . . . 11 t
5246, 50, 51sylancr 669 . . . . . . . . . 10 t
5352rexeqdv 2994 . . . . . . . . 9 t
54 vex 3048 . . . . . . . . . . . 12
5554inex1 4544 . . . . . . . . . . 11
5655rgenw 2749 . . . . . . . . . 10
57 eqid 2451 . . . . . . . . . . 11
58 eleq2 2518 . . . . . . . . . . . 12
59 imaeq2 5164 . . . . . . . . . . . . 13
6059sseq1d 3459 . . . . . . . . . . . 12
6158, 60anbi12d 717 . . . . . . . . . . 11
6257, 61rexrnmpt 6032 . . . . . . . . . 10
6356, 62mp1i 13 . . . . . . . . 9
6422ad3antrrr 736 . . . . . . . . . . . 12
65 elin 3617 . . . . . . . . . . . . 13
6665rbaib 917 . . . . . . . . . . . 12
6764, 66syl 17 . . . . . . . . . . 11
68 simpllr 769 . . . . . . . . . . . . . . . . 17
69 fvex 5875 . . . . . . . . . . . . . . . . 17
70 ifexg 3950 . . . . . . . . . . . . . . . . 17
7168, 69, 70sylancl 668 . . . . . . . . . . . . . . . 16
7271ralrimivw 2803 . . . . . . . . . . . . . . 15
73 eqid 2451 . . . . . . . . . . . . . . . 16
7473fnmpt 5704 . . . . . . . . . . . . . . 15
7573fmpt 6043 . . . . . . . . . . . . . . . . 17
76 df-f 5586 . . . . . . . . . . . . . . . . 17
7775, 76bitri 253 . . . . . . . . . . . . . . . 16
7877baib 914 . . . . . . . . . . . . . . 15
7972, 74, 783syl 18 . . . . . . . . . . . . . 14
80 simplrr 771 . . . . . . . . . . . . . . . . 17
81 inss2 3653 . . . . . . . . . . . . . . . . . . 19
8281sseli 3428 . . . . . . . . . . . . . . . . . 18
8325, 40sylbi 199 . . . . . . . . . . . . . . . . . . 19
8483eleq1d 2513 . . . . . . . . . . . . . . . . . 18
8582, 84syl 17 . . . . . . . . . . . . . . . . 17
8680, 85syl5ibrcom 226 . . . . . . . . . . . . . . . 16
8786ralrimiv 2800 . . . . . . . . . . . . . . 15
88 undif1 3842 . . . . . . . . . . . . . . . . . . . 20
8988ineq2i 3631 . . . . . . . . . . . . . . . . . . 19
90 indi 3689 . . . . . . . . . . . . . . . . . . 19
9189, 90eqtr3i 2475 . . . . . . . . . . . . . . . . . 18
9291raleqi 2991 . . . . . . . . . . . . . . . . 17
93 ralunb 3615 . . . . . . . . . . . . . . . . 17
9492, 93bitri 253 . . . . . . . . . . . . . . . 16
9594rbaib 917 . . . . . . . . . . . . . . 15
9687, 95syl 17 . . . . . . . . . . . . . 14
9779, 96bitr3d 259 . . . . . . . . . . . . 13
98 inss2 3653 . . . . . . . . . . . . . . . 16
9998sseli 3428 . . . . . . . . . . . . . . 15
100 eldifsni 4098 . . . . . . . . . . . . . . . . 17
101 ifnefalse 3893 . . . . . . . . . . . . . . . . 17
102100, 101syl 17 . . . . . . . . . . . . . . . 16
103102eleq1d 2513 . . . . . . . . . . . . . . 15
10499, 103syl 17 . . . . . . . . . . . . . 14
105104ralbiia 2818 . . . . . . . . . . . . 13
10697, 105syl6bb 265 . . . . . . . . . . . 12
107 df-ima 4847 . . . . . . . . . . . . . 14
108 inss2 3653 . . . . . . . . . . . . . . . 16
109 resmpt 5154 . . . . . . . . . . . . . . . 16
110108, 109mp1i 13 . . . . . . . . . . . . . . 15
111110rneqd 5062 . . . . . . . . . . . . . 14
112107, 111syl5eq 2497 . . . . . . . . . . . . 13
113112sseq1d 3459 . . . . . . . . . . . 12
1147ad3antrrr 736 . . . . . . . . . . . . . 14
115 ffun 5731 . . . . . . . . . . . . . 14
116114, 115syl 17 . . . . . . . . . . . . 13
117 difss 3560 . . . . . . . . . . . . . . 15
11898, 117sstri 3441 . . . . . . . . . . . . . 14
119 fdm 5733 . . . . . . . . . . . . . . 15
120114, 119syl 17 . . . . . . . . . . . . . 14
121118, 120syl5sseqr 3481 . . . . . . . . . . . . 13
122 funimass4 5916 . . . . . . . . . . . . 13
123116, 121, 122syl2anc 667 . . . . . . . . . . . 12
124106, 113, 1233bitr4d 289 . . . . . . . . . . 11
12567, 124anbi12d 717 . . . . . . . . . 10
126125rexbidva 2898 . . . . . . . . 9
12753, 63, 1263bitrd 283 . . . . . . . 8 t
128127anassrs 654 . . . . . . 7 t
129128pm5.74da 693 . . . . . 6 t
13045, 129bitrd 257 . . . . 5 t
131130ralbidva 2824 . . . 4 t
13211, 39, 1313bitrd 283 . . 3 lim
133132pm5.32da 647 . 2 lim
1343, 133syl5bb 261 1 lim
Colors of variables: wff setvar class Syntax hints: wn 3 wi 4 wb 188 wo 370 wa 371 w3a 985 wceq 1444 wcel 1887 wne 2622 wral 2737 wrex 2738 cvv 3045 cdif 3401 cun 3402 cin 3403 wss 3404 cif 3881 csn 3968 cmpt 4461 cdm 4834 crn 4835 cres 4836 cima 4837 wfun 5576 wfn 5577 wf 5578 cfv 5582 (class class class)co 6290 cc 9537 ↾t crest 15319 ctopn 15320 ℂfldccnfld 18970 ctop 19917 TopOnctopon 19918 ccnp 20241 lim climc 22817 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1669 ax-4 1682 ax-5 1758 ax-6 1805 ax-7 1851 ax-8 1889 ax-9 1896 ax-10 1915 ax-11 1920 ax-12 1933 ax-13 2091 ax-ext 2431 ax-rep 4515 ax-sep 4525 ax-nul 4534 ax-pow 4581 ax-pr 4639 ax-un 6583 ax-cnex 9595 ax-resscn 9596 ax-1cn 9597 ax-icn 9598 ax-addcl 9599 ax-addrcl 9600 ax-mulcl 9601 ax-mulrcl 9602 ax-mulcom 9603 ax-addass 9604 ax-mulass 9605 ax-distr 9606 ax-i2m1 9607 ax-1ne0 9608 ax-1rid 9609 ax-rnegex 9610 ax-rrecex 9611 ax-cnre 9612 ax-pre-lttri 9613 ax-pre-lttrn 9614 ax-pre-ltadd 9615 ax-pre-mulgt0 9616 ax-pre-sup 9617 This theorem depends on definitions: df-bi 189 df-or 372 df-an 373 df-3or 986 df-3an 987 df-tru 1447 df-ex 1664 df-nf 1668 df-sb 1798 df-eu 2303 df-mo 2304 df-clab 2438 df-cleq 2444 df-clel 2447 df-nfc 2581 df-ne 2624 df-nel 2625 df-ral 2742 df-rex 2743 df-reu 2744 df-rmo 2745 df-rab 2746 df-v 3047 df-sbc 3268 df-csb 3364 df-dif 3407 df-un 3409 df-in 3411 df-ss 3418 df-pss 3420 df-nul 3732 df-if 3882 df-pw 3953 df-sn 3969 df-pr 3971 df-tp 3973 df-op 3975 df-uni 4199 df-int 4235 df-iun 4280 df-br 4403 df-opab 4462 df-mpt 4463 df-tr 4498 df-eprel 4745 df-id 4749 df-po 4755 df-so 4756 df-fr 4793 df-we 4795 df-xp 4840 df-rel 4841 df-cnv 4842 df-co 4843 df-dm 4844 df-rn 4845 df-res 4846 df-ima 4847 df-pred 5380 df-ord 5426 df-on 5427 df-lim 5428 df-suc 5429 df-iota 5546 df-fun 5584 df-fn 5585 df-f 5586 df-f1 5587 df-fo 5588 df-f1o 5589 df-fv 5590 df-riota 6252 df-ov 6293 df-oprab 6294 df-mpt2 6295 df-om 6693 df-1st 6793 df-2nd 6794 df-wrecs 7028 df-recs 7090 df-rdg 7128 df-1o 7182 df-oadd 7186 df-er 7363 df-map 7474 df-pm 7475 df-en 7570 df-dom 7571 df-sdom 7572 df-fin 7573 df-fi 7925 df-sup 7956 df-inf 7957 df-pnf 9677 df-mnf 9678 df-xr 9679 df-ltxr 9680 df-le 9681 df-sub 9862 df-neg 9863 df-div 10270 df-nn 10610 df-2 10668 df-3 10669 df-4 10670 df-5 10671 df-6 10672 df-7 10673 df-8 10674 df-9 10675 df-10 10676 df-n0 10870 df-z 10938 df-dec 11052 df-uz 11160 df-q 11265 df-rp 11303 df-xneg 11409 df-xadd 11410 df-xmul 11411 df-fz 11785 df-seq 12214 df-exp 12273 df-cj 13162 df-re 13163 df-im 13164 df-sqrt 13298 df-abs 13299 df-struct 15123 df-ndx 15124 df-slot 15125 df-base 15126 df-plusg 15203 df-mulr 15204 df-starv 15205 df-tset 15209 df-ple 15210 df-ds 15212 df-unif 15213 df-rest 15321 df-topn 15322 df-topgen 15342 df-psmet 18962 df-xmet 18963 df-met 18964 df-bl 18965 df-mopn 18966 df-cnfld 18971 df-top 19921 df-bases 19922 df-topon 19923 df-topsp 19924 df-cnp 20244 df-xms 21335 df-ms 21336 df-limc 22821 This theorem is referenced by: limcnlp 22833 ellimc3 22834 limcflf 22836 limcresi 22840 limciun 22849 lhop1lem 22965 limccog 37700
Copyright terms: Public domain W3C validator | 5,105 | 9,558 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2021-04 | latest | en | 0.106762 |
https://math.stackexchange.com/questions/3769360/irreducible-representation-is-injective | 1,638,175,940,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358702.43/warc/CC-MAIN-20211129074202-20211129104202-00205.warc.gz | 449,631,538 | 33,457 | # Irreducible representation is injective
Let $$\rho:G \to GL(V)$$ a irreducible representation where $$|G|=p^3$$ and $$\dim(V)\neq 1$$ over $$\mathbb{C}$$, then $$\rho$$ is injective.
I managed to reach the following relationship
$$|G|=|\ker\rho|\dim(V)^2+\sum_{g\notin\ker\rho}|\chi(g)|^2$$
where $$\chi$$ is the character of $$\rho$$.
I think this can help to get that the kernel is trivial, but I couldn't get anywhere. I am also wondering about the importance of the order of the group being $$p^3$$.
If $$\rho$$ not injective, then it induces an irreducible representation of $$G/{\rm ker}(\rho)$$. However this group must have order $$1$$ or $$p$$ or $$p^2$$, so will be abelian and all irreducible representations will have dimension $$1$$.
If $$|\ker \rho| > 1$$ and $$\dim(V) > 1$$, then both of these numbers must be at least $$p$$, because they must be divisors of $$|G|$$. Use this to get a contradiction from the formula you have shown.
• I thought about it, but if $|\ker\rho|=p$ and $\dim(V)=p$, how to ensure that the sum is different from zero? Jul 26 '20 at 6:13 | 329 | 1,087 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 19, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2021-49 | latest | en | 0.881667 |
www.saravanderwerf.com | 1,708,621,141,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473824.13/warc/CC-MAIN-20240222161802-20240222191802-00720.warc.gz | 1,005,218,906 | 27,177 | 7
# #mathMovement ideas with well known Math Routines
There are so many amazing math routines, many with websites full of resources. These math routines, when implemented well, will increase student mathematical discourse. I would argue that many of these routines are even better when students are standing and moving. Check out a few of the routines I think benefit from a bit of #mathMovement! Try these out with your students. Students need to move every 25 minutes each and every day. Below are some ways you can use to add #mathMovement to your daily routine in class.
#mathMovement with Open Middle
The Open Middle website is a curated set of K-12 challenging math problems worth solving. Robert Kaplinsky wrote about implementing ‘open middle’ problems in his 2019 book ‘Open Middle Math’ from Stenhouse. Open Middle Problems are perfect to do while vertical. Ideally have your students work in groups of 2 or 3 on an Open Middle Task at #vnps Vertical (standing) Non-Permanent (white boards…) Surfaces.
If you are not yet at a point where you have enough ‘non-permanant’ stations around your classroom – do what I use to do. A lot of the ‘Open Middle’ tasks ask students to manipulate the digits 0-9 to solve a math challenge. Have students pull out 10 post-it notes (stickies) and place the digits 0-9 on each.
Go to Open Middle and look for a task to use.For example check out this task from Graham Fletcher:
This task only uses the digits 1-9, so get rid of your post-it note with zero on it. Write up the form of this task on chart paper (or photo copy the frame of the task on paper to tape to the wall) I transformed Graham’s original task to write the ratios in fractional form. I made boxes for each digit and when I drew them I made each box approximately the same size as a post-it note.
So….the next time you do an OPEN MIDDLE task with your students – do it vertically. Get your students up and moving.
#mathMovement with Which One Doesn’t Belong?
I will always be a fan of ‘Which One Doesn’t Belong?’ routine because not only is it a great first win for teachers trying to increase discourse in their room but because it was made popular by people who live within miles of me. I was introduced to this routine by Terry Wyberg (University of Minnesota) and Christopher Danielson (Who wrote a book of the same name). I’m telling you all – we have amazing math leaders in Minnesota.
So, what does WODB look like with #mathMovement? One common way that many use WODB, is to give each student a post-it note and then present a problem. For example, this one from the wodb.ca website…
Have students individually select a number that does not belong and record their choice and reason on their post-it note. NOTE: I give students a sentence stem to write their choice “The number ______ does not belong because…”
I then have students ALL (yes all – everyone needs #mathmovement – even for a few seconds) stand and walk to the board and place their post it notes next to the box of their choice. It would look something like this (though this is a different problem).
I then have a discussion with the class about what hey wrote. Imagine this picture with 30 post-it notes. I usually start with the number with the most post-it notes next to it and have students share their reasons. Sometimes I have a box with no post-it notes and we will brainstorm as a class reasons.
The method above is OK for #mathmovment. At least all students move for 30 seconds – but my favorite way to use WODB problems with #mathmovement is the following. For most WODB’s I do, I use an EL best practice of inviting more learners into the discourse by using language that most students can access.
Did you know that many students don’t participate in a WODB because they don’t have QUICK language to say things like “The one in the upper left….” or even say “Letter A doesn’t belong because….” I have had teachers excitedly tell me, I put axis through the middle and have students tell me what quadrant their choice is in. When I hear this I think – “Yikes – our classrooms have lots of students with language needs and adding another layer of complexity tends to shut these students down, not support them in their language needs.
What is quick for most students, despite their language needs are colors. You would be amazed how many students will start talking about WODB when you make each quadrant a separate color. All of a sudden students will say things like, “The purple one doesn’t belong because……”.
For this reason – I do all my WODB on a template in a word document – though it could be a power-point template. I simply create a jpeg of the 4 boxes and paste them into the template. I then take a jpeg of this and share this with my students and ask them ‘Which one Doesn’t Belong?”. Note #1 For my color blind students I also label each quadrant with A, B, C, D. Note #2 When students say “The purple one doesn’t belong.” I will re-voice and pair language to build academic language for all “Tell my why the purple, letter D in the lower right does not belong.” – (pairing language though is another post I need to write some day).
SO….where is the #mathmovement in this activity? Well, first let me show you my classroom. I squeeze 35+ students into a tiny room and I still do movement. What is on the four walls, way up high, is an 8.5×11″ sign that says ‘WODB’. Each sign is in a different color corresponding to the 4 colors in my WODB template.
I display the WODB for students to see and then I have them stand up and go to the side of the classroom to stand next to their choice. I usually will then stand in the center of the classroom and orchestrate the discussion of why students selected each choice. All of us will remain standing for the discussion. #mathmovement with WODB. Excellent.
Note: If you would like my templates for creating color coded WODB’s or want to download the 8.5×11″ colorful signs I keep up in my classroom, click the button below to go to a google folder of resources from this blog post.
#mathMovement with Debate Math
Debate Math is an amazing math routine incorporating debate language and now a book from my friend and co-presentor on #mathMovement, Chris Luzniak. I am not going to say much about his other than to say – click on the links to either the Debate Math blog or book from Chris and read one or both. Debate Math is an incredible routine to increase mathematical discourse in your classroom and get students moving. Check out this tweet from Stephanie Lewis!
#mathMovement with ‘Would you Rather?’
The ‘Would You Rather?’ website has great math prompts to get your students engaging in discourse around a variety of topics. Check out this tweet/photos from Karla Doyle.
The ‘Would you Rather’ Math Instructional Routine would be another great routine to incorporate #mathMovement. Similar to what I wrote about above for WODB – consider tweaking the resources at the website to have students make a choice between the 2 items by physically moving to one side of the classroom. Hold the class discussion about the 2 choices while all are standing.
Step 1: After selecting an image from the website (or make up your own task of this form)- add it to a color backed template and display for students.
Step 2: Have students make a choice between the 2 options and move physically to the corresponding side of the room.
Step 3: Lead a class discussion having students voice their reasons for selecting each option.
Note: If you would like signs for your classroom or templates for creating a #mathMovement ‘Would you rather?’ task – click the button below.
#mathMovement with Estimation 180
Estimation 180 can become a #mathmovement activity with the addition of BLUE 3M Painter’s Tape (3M is a Minnesota product – thus my support of them). All you need to do is create a (semi) permanent number-line in your classroom that you have your students use in a variety of ways. (sidenote: My good friend, Ali Rubin, will write a blog post soon on 10 ways to use a classroom number-line soon – I’ll link to her post her when she does)
The classroom (blue tape) number-line in my own classroom went down the center of my groups of tables.
Then select an ‘Estimation 180’ prompt (or make up one of your own). I selected day 206 – How many cheese-balls fit onto a tray? Display the image provided by Andrew Stadel at his site.
Have students write down their estimates. I recommend using Andrew (creator of estimation 180) Stadel’s template for having students make their predictions – Have student’s not just estimate the number of cheese-balls but also write down a number they believe is too LOW and another number that is too HIGH. Finally, have students record their reasoning.
Now it is time to incorporate #mathMovement with Estimation 180.
Have students work in pairs with white boards. As a class talk about what students think is ‘too low’. Have a student stand up and represent this amount on the number line. For example, maybe this class thinks 20 is too few cheese balls to cover the sheet pan.
Next have a class discussion about an amount that is ‘too high’ and have a student stand up to represent this amount. Write the amount on a white board (or paper) for the class to see. For example, maybe the class thinks 700 cheese balls is definitely too many.
Next have one person from each partnership make their best estimates and stand on the line between the 2 numbers.
Have a class discussion about where students are standing. Are students spread out in proportion to the distances from the low to high number or just standing ‘in order’. Are the best estimates all bunched together or spread out? Why did they select their value. After a standing class discussion, show the reveal.
Last note: Have you checked out Andrew Stadel’s ‘Estimation 180’ podcast yet? If not, take a listen soon and imagine this routine with #mathMovement.
#mathMovement with Clothesline Math
You can not use number-lines enough in K-12 math to model student thinking. One way of using more number lines (and your the blue painter’s tape number line you now have made (see above) on your classroom floor) that has gotten traction in recent years is ‘Clothesline Math‘.
Chris Shore and his site has lots of information on this routine as well as downloadable resources. Note: Kristen Acosta has lots of Clothesline resources for Elementary. (and check out all the other clothesline enthusiasts linked on Chris’s site.)
Ultimately, you as a teacher print out a set of cards for students to sort on a clothesline. Clotheslines are superior to written number lines in that it is super easy to move numbers as you place them and replace them as you add more cards.
What if you used the clothesline math downloadable cards at Chris or Kristen or Andrew Stadel’s sites and instead of sorting them on a clothesline, you had students engage in a #mathMovement activity and had them stand on your classroom number-line on the floor? Think of discussions you could have.
How are you using #mathMovement with Math Instructional Routines in your classroom? Comment below or tweet at @saravdwerf or using the hashtag #mathMovement. I’d love to hear your ideas and/or questions.
##### Sara VanDerWerf
I am Sara Van Der Werf, a 24-year mathematics teacher in Minneapolis Public Schools. I have taught math in grades 7-12 as well as spent several years leading mathematics at the district office. I currently teach Advanced Algebra at South High School and I'm also the current President of the Minnesota Council of Teachers of Mathematics (MCTM). I am passionate about encouraging and connecting with mathematics teachers. I'd love to connect via twitter. Join the community. Tweet me @saravdwerf. | 2,562 | 11,767 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2024-10 | longest | en | 0.936094 |
https://ttohappy.com/choosing-lottery-numbers-12-tips-and-strategies-to-help-you-win/ | 1,726,854,663,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701419169.94/warc/CC-MAIN-20240920154713-20240920184713-00307.warc.gz | 542,425,634 | 50,890 | While you’re resting there, putting cash right into difficult video games like Powerball, the professional gamers are placing their cash right into video games that they can really win. And also, when they win, they do it once again.
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Currently, you never ever find out about individuals winning Powerball two times or individuals winning EuroMillions two times, do you? No, as well as forever factor. Whenever you read about a several lottery game victor, they constantly won on video games that have much better chances than the large pot video games. Do you assume that takes place by chance? Nope!
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Choosing Lottery Numbers – 12 Tips and Strategies to Help You Win | 643 | 2,926 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-38 | latest | en | 0.984531 |
https://www.cram.com/essay/The-Graphs-Of-Position-Velocity-And-Acceleration/F3N8HLF2B5YW | 1,611,117,107,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703519883.54/warc/CC-MAIN-20210120023125-20210120053125-00431.warc.gz | 742,021,245 | 11,573 | # The Graphs Of Position, Velocity And Acceleration
797 Words 4 Pages
Graphs of Position, Velocity and Acceleration
The use of graphs is very common in science, statistics, economic and etc; in physics the graphs are use to illustrate and compare quantitative data of motion, time, velocity, etc. The graph illustrates the trend accurately, also the trend can be use to predict future outcomes. Graphs are a useful tool to have a efficient description in less words.
The Graphs of Position Graphs of position are a way to represent motion of an object or multiple objects in a graph using time and position. Time axis in a graph is used to measure motion in a specific time and position axis is used to measure distance. The time data and position data in a graph can be use to create a slope of time vs. position.
The graph shows the data chart and a positive slope , it is easy to understand the graph and the direction of the straight slope. Another example using changing velocity(acceleration).In this changing velocity graph, the position of the object every second is different because the acceleration. In the acceleration graph the slope looks like a curved line because the speed is increasing every second. The two examples show a constant velocity graph and acceleration graph. Overall, the velocity graph has a straight line as slope because it has a constant velocity and the acceleration graph has a curved line because the acceleration. The speed can be use to determine the slope of the graph in motion.
Velocity graph Velocity is a vector measurement of the rate and direction of motion, and we calculate the velocity dividing the distance moved and the time it takes to complete the movement. V = d / t
An example of constant velocity of +10 m/s with cero acceleration. The constant velocity +10 m/s
makes a straight line in the graph. Another example is a object moving changing velocity or accelerating. The object is going to the right side (positive
The acceleration of an object is the result of any or all forces acting on the object according to Newton second law. The international system of units for acceleration are :
How to find the average velocity, final velocity minus initial velocity divided by time.
An example, let 's find the average acceleration of Vf = 50 m/s, Vi = 0 m/s and a time of 5s.
The average velocity is 10 m/s*2. That was an example of average acceleration. How to recognize direction of the acceleration vector according to the direction of the slope, a slope going to the right is a positive slope and a slope going to the left a negative slope.
Example A: The velocity is increasing 2 m/s every second, so the slope is going to the right from a positive side and it is positive acceleration.
Example B: The velocity is increasing 2 m/s every second, so the slope is going to the right side from a negative side to a positive side and it is a positive acceleration.
Example C: The velocity is decreasing 2 m/s every second, so the slope is going to the right side (downward) from a positive side, it is negative
• ## Kinematics Of Particles Summary
The tangential component of the force is responsible for moving the particle along the path. From Newton’s second law of motion for normal and tangential axes, we have F_t=ma_t From the kinematic relation, we know that at = vdv/dx, so we can write the above equation as Fig. 2.7 F_t=mv dv/dx F_t dx=mvdv We already know that the first term in the above equation is work due to a force and it is a scalar term. So we write the equation in scalar form. If the particle has initial position x1, initial velocity v1, final position x2 and final velocity v2, we have ∫_(x_1)^(x_2)▒〖F_t dx〗=∫_(v_1)^(v_2)▒mvdv ∫_(x_1)^(x_2)▒〖F_t dx〗=(mv_2^2)/2-(mv_1^2)/2 The first term in the above equation is the work done by a force and the second and the third terms are kinetic energies of a particle.…
Words: 1828 - Pages: 8
• ## Questions: According To Newton's First Law Of Motion
According to Newton’s Second Law, the direction of the acceleration is the same as the direction of the net force. 8. A Why: one’s mass is fixed, while its weight varies at different location. Using the formula: weight = mass * acceleration due to gravity, we can get the object’s mass is 50kg. Then using the formula again, we get its weight on surface of the Moon is 80N 9.…
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• ## The Two Properties Of The Mass Spring System
This increases the time period of the oscillation. The greater the inertia of an oscillating object, the greater the time period which therefore lowers the frequency of its oscillations. Depending on how the waves are displaced indicates how big the oscillations. The elasticity of a spring also can be quantified by a measure of a spring’s compliance. Compliance is inversely related to stiffness, which means that springs that have relatively little stiffness are characterised by a relatively large compliance.…
Words: 1338 - Pages: 6
• ## Accelerated Motion Experiment
As the height increases, we can see in graph B that the acceleration is also increase. Because the acceleration is changing, and not constant with different heights, this indicates that rate of acceleration is proportional to the incline of the ramp. Graph B Is the graph C linear? Yes the graph is linear What does it mean if the graph is linear? The graph being linear indicate that the can had a constant…
Words: 1575 - Pages: 7
• ## Hypothesis When Force Increase Acceleration
Research: According to Newton 's Second law when force is increased acceleration increases assuming that mass stays constant. a=F/m Purpose: The purpose of this lab is to prove when force increases acceleration also increases. The independent variable is amount of force applied to the Lenny Board and the dependant variable is the acceleration of the Lenny Board. 3 controlled variables are mass, surface of testing, and the Lenny Board (the weight, the amount of grease on the bearings, ECT) itself. Hypothesis 1: If we increase force on a Lenny Board the acceleration will increase.…
Words: 1768 - Pages: 8
• ## Proportionality And Newton's Second Law Of Motion Essay
The net force here is inclusive of all external forces (Moore). This law of motion is the basis for the entire experiment. Kinematic Chain According to the kinematic chain, if you integrate acceleration, a constant, your result will be a linear function describing velocity. A integreated= aldjf;aijd;flnsd;ifjj If you integrate the velocity function, your result will be a quadratic function describing position. V integalkd;fasdifjaoijfiajweoir;j Experimental…
Words: 1739 - Pages: 7
• ## Analysis Of The Distance Vs. Time Graph
The second graph of the velocity verse time diagram as to the distance verse time chart correspondingly shows the acceleration of the cart and the maximum speed achieved. This linear progression demonstrates the relationship between the two variables, simply stating that the object has positive velocity and positive acceleration. Hence, using these two graphs determines the maximum speed achieved and the acceleration of the object. At a distance of 1.2 meters, the trolley had taken approximately 1.62 seconds to complete the length and through calculations, the velocity is about 0.74 m/s. It can also be established that according to these two charts, there is acceleration (the changing of speed) present in the object’s…
Words: 1163 - Pages: 5
• ## Egg Drop Essay
Velocity is a vector quantity with direction and magnitude. In the case of the device, the direction is directly down or 90 degrees below the horizontal and the magnitude is 10.2834. After solving for velocity, one can get the acceleration of the device as it drops from 75 feet. To calculate acceleration, the formula a = v / t, or in other words, acceleration is equal to velocity divided by the elapsed time. Substituting in the values, acceleration roughly equals 4.6259 m/s^2.…
Words: 1212 - Pages: 5
• ## Bernoulli Principle Essay
Furthermore, the Equation of Continuity (ρ1A1V1 = ρ2A2V2) could also be written as Mdot1 = Mdot2, saying that the mass flow rate of a fluid at one point is equivalent to the mass flow rate of the fluid at another point. Or, you could also write it as ρ1Q1 = ρ2Q2, saying that the density of the fluid times the volume of the flow rate in one spot is equal to the density of the fluid times the volume of the flow rate at a different spot. You can then simplify this version of the formula because the density of the fluid is frequently the same on both sides of the equation, so you are left with Q1 = Q2, also written as A1V1 = A2V2, which is the simpler version of Equation of Continuity (Drew). Though Bernoulli’s Equation is associated with many different formulas and can be written various ways, every formula has the same principle. Bernoulli’s Principle!…
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• ## Free Body Diagram
μ = 0.30 Complete the table below. You may check your answers in the simulation. Force Applied is the force required (by you for instance) to make the cabinet move at a constant velocity in either direction or keep it from accelerating (if applicable). Recall…constant velocity = _______ net force. Also note: force applied may change direction as the angle increases!…
Words: 941 - Pages: 4 | 2,116 | 9,312 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2021-04 | latest | en | 0.913138 |
https://wanderluce.com/news/how-are-ssa-earnings-indexed/ | 1,725,797,614,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651002.87/warc/CC-MAIN-20240908115103-20240908145103-00660.warc.gz | 598,815,556 | 37,910 | ## How are SSA earnings indexed?
An individual’s earnings are always indexed to the average wage level two years prior to the year of first eligibility. Earnings in a year before 2020 would be multiplied by the ratio of 55,628.60 to the average wage index for that year; earnings in 2020 or later would be taken at face value.
### How are indexed earnings calculated?
This is determined by taking the 35 highest years (prior to age 60) of indexed earnings and dividing that figure by the total number of months worked during those years. Thus, if you worked every month, without fail, your average indexed monthly earnings would equal the sum of 35 years of work divided by 144 months.
#### What is Social Security Index Factor?
A factor will always equal one for the year in which the person attains age 60 and all later years. The indexing factor for a prior year Y is the result of dividing the average wage index for the year in which the person attains age 60 by the average wage index for year Y.
How do you calculate average indexed monthly income?
Your average indexed monthly earnings are used by Social Security to calculate the amount of your Social Security Disability benefits. To calculate your average indexed monthly earnings divide the sum of your 35 highest years of indexed earnings (up to age 60) by the total by the number of months worked in those years.
How much Social Security will I get if I make 80000 a year?
Initial Social Security retirement benefits by age and income level
Annual Income (Inflation-Adjusted) Age 62 Age 65
\$60,000 \$1,554 \$1,931
\$70,000 \$1,695 \$2,106
\$80,000 \$1,787 \$2,220
\$90,000 \$1,879 \$2,334
## How much SS will I get if I make 100k?
The typical six-figure earner should collect around \$2,000 or more per month from Social Security, depending on a variety of factors. The maximum possible monthly benefit is nearly twice that amount, though only a handful of retirees will be eligible to receive that amount.
### What is highest Social Security payout?
The most an individual who files a claim for Social Security retirement benefits in 2022 can receive per month is:
• \$2,364 for someone who files at 62.
• \$3,345 for someone who files at full retirement age (66 and 2 months for people born in 1955, 66 and 4 months for people born in 1956).
#### How much will I get from Social Security if I make \$100000?
What is the indexation of salaries?
Indexation of salaries. Salaries, wages and social contributions (including the social minimum wage) are, on the other hand, adjusted in line with the evolution of the cost of living. When the consumer price index increases or decreases by 2.5 % during the previous semester, salaries are normally adjusted by the same proportion.
What is the Labour Cost Index?
The labour cost index is defined as the Laspeyres index of labour costs per hour worked, chain-linked annually and based upon a fixed structure of economic activity at NACE Rev.2 section level. The current reference year of the index is 2016. In addition to the index numbers, annual and quarterly growth rates of labour cost are also calculated.
## Does indexation apply to higher wages in Luxembourg?
For posted workers in Luxembourg, the indexation only applies to the minimum wage and not to higher wages. How would you rate the content of this page?
### Why does Social Security use the average wage indexing series?
We use the average wage indexing series to update several amounts that are important to the operation of Social Security’s Old-Age, Survivors, and Disability Insurance (OASDI) program. | 794 | 3,603 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2024-38 | latest | en | 0.942345 |
https://newbedev.com/doubt-on-the-need-for-topological-manifolds | 1,716,368,785,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058534.8/warc/CC-MAIN-20240522070747-20240522100747-00549.warc.gz | 363,989,360 | 7,061 | # Doubt on the need for Topological Manifolds
Measurement instruments are not infinitely precise, however It is possibile to distinguish objects using them. This is possible when the precision of them permits it. The precision of an instrument around measured values is the physical corresponding of a neighborhood of a point. The fact that two measures can be distinguished by means of sufficiently precise instruments (though not infinitely precise) corresponds to the mathematical fact (Hausdorff property) that two distinct points, e.g., on the real line, admit corresponding neighbourhoods with empty intersection. I stress that I am referring to physical quantities that can be described in a continuous way, in the sense that improving the precision of the instruments I always find new distinct values and no point structure pops out (evidently all this may eventually reveal to be just an approximation valid to some extent only: the last word on our mathematical models to describe what exists is of physics). In this way, standard notions of topology, more precisely Hausdorff topology, arise naturally and quite generally in the physical context. When the set of physical entities you want to describe are also determined by using mutually compatible coordinate systems, the notion of manifold (topological or differentiable depending on your requirements on the mutual compatibility of the used coordinates) naturally enters the play. The notion of topological manifold actually also assumes another technical requirement on the topology (a countable topological basis giving rise to the so called paracompactness property). This hypothesis is difficult to physically justify in elementary contexts and in fact, also in mathematics, it is not always required.
A fundamental context concerns the physical representation of events. They are represented in terms of spatial position and time location. To measure that information you use real instruments, rulers and clocks. The above discussion naturally leads to a basic representation in terms of a Hausdorff space, before introducing other more sophisticated structures, where the neighbourhoods are defined by the precision of instruments.
However there are many other cases, think of the representation of the equilibrium states of a thermodynamical system.
I stress that I mentioned only elementary and intuitive contexts where topology naturally enters. Considering more advanced subjects of physics, different types of topologies arise. Hausdorff property and second countability cease to be relevant in some contexts essentially of quantum nature. QFT, Quantum Gravity, but also QM. There are classifications of classes of entangled states, in finite dimensional Hilbert spaces, which refer to the Zariski topology. As is known, that topology is not Hausdorff. | 519 | 2,834 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2024-22 | latest | en | 0.924647 |
https://www.studymode.com/course-notes/Physics-Unit-4-Notes-45442867.html | 1,566,275,682,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315222.14/warc/CC-MAIN-20190820024110-20190820050110-00370.warc.gz | 998,429,574 | 12,484 | top-rated free essay
# Physics Unit 4 Notes
By konradwithak Dec 04, 2013 606 Words
Unit 4 Physics Notes
Sunday, November 3, 2013
8:09 PM
Any Single disturbance is called a pulse
A series of pulses is called a wave
In the case of the impatient student, his disturbance was parallel to the direction of the wave itself. Also called a Longitudinal Wave
WAVES
• Waves can transmit two things
○ Can transmit energy
Ex. Light waves from the sun heat the earth
○ Can transmit information
When the motion of the particles is perpendicular to the motion of the wave, it is called a traverse wave
Pointing down
Acceleration down
Wave -> We see the motion of the medium
Soundwaves -> air
Lightwaves = electromagnetic waves
Water waves -> water
Light waves -> no medium
Pointing up
Acceleration up
Without tension, there can be no traverse waves in a medium. -> Do not occur inside liquids and gases -> can occur on
surface of water though
Inner core of earth has a liquid core, because traverse waves do not got through Earthquakes can produce traverse, longitudinal, or a combination of the two. -> longitudinal = compression, travel fast, primary (happen first) -> traverse = "__________ waves", secondary (happen second)
-> longer delay between S and P waves, means greater distance from the epicenter
A)
B)
C)
D)
Crest
Trough
Amplitude
Wavelenght
Cycle -> one complete wavelength
Amplitude -> half the distance from crest and trough
Wavelength -> distance of one complete cycle (SI units = meters) Period (T) -> time for one wavelength to pass (SI Units = seconds) Frequency (f) -> number of cycles that pass by per time interval = Cycles per second = 1 Hertz (Hz)
T=(1/f) and f=(1/T)
Ex. f=2 Hz = T=.5 seconds
Swing?
Applied force at a regular interval, causes you to go higher, high amplitude
All object will vibrate of oscillate at a certain frequency called the natural frequency A pendulum (or swing) will swing back and forth at a natural frequency that depends only on it's lenght
When a system is driven at it's natural frequency (forces applied in rhythm with the natural frequency), Physics Unit 4 Page 1
The shorter the stick, the higher the
Natural frequency.
When a system is driven at it's natural frequency (forces applied in rhythm with the natural frequency), the oscillations will get bigger. This s termed resonance.
If two things have the same natural frequency,
And you hit one, the second will also begin to vibrate
Energy
Transfer
Unequal natural frequency, then vibrations
won't transfer
It is possible for an object to have more than one natural (resonant) frequency. Natural frequency is the 1st harmonic
Other harmonics can be produced as well
If the displacement is one wavelength, then the time equals a period (T). Wave Speed (or Velocity)
Wave Velocity
Velocity = Displacement/time = d/t
F = 1/T
T = 1/f
V = ʎ (wave length) / T (Period)
V = fʎ
Ex. Distance between waves is 6m, and one wave takes 4 seconds. T=4
F = 1/4 Hz
V = (1/4) x 6m = 1.5 m/s
Ex.
1st
2nd
3rd
Aver.
Speed
10cm
1.3
1.5
1.5
1.4
5.7m/s
30cm
1.5
1.3
1.7
1.5
5.3m/s
50cm
Transverse
Amp
1.6
1.4
1.5
1.5
5.3m/s
Compression
Amplitude = amount
of energy being
transmitted
Energy effects amplitude, not speed
ʎ
T1
T2
T3
Aver.
Speed
8
1.3
1.4
1.5
1.4
5.7m/s
14
.72
.74
.74
.73
5.4m/s
2.67
.40
.46
.50
.48
5.0m/s
What factors affect the speed of the wave?
Amplitude = no
Wave Length = no
Only the properties of the medium affect the wave speed
When waves meet, their amplitudes add (called the principle of superposition). After they meet, they pass through each other. Each wave in no way alters the travel of the other.
Physics Unit 4 Page 2
Amplitude of
the wave has
NO EFFECT on
the speed of it
Effect of different wave
length on speed of wave
= NO EFFECT
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https://thecleverprogrammer.com/2021/06/12/amazon-alexa-reviews-sentiment-analysis-using-python/ | 1,696,171,313,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510903.85/warc/CC-MAIN-20231001141548-20231001171548-00695.warc.gz | 598,274,015 | 26,536 | # Amazon Alexa Reviews Sentiment Analysis using Python
Amazon Alexa is a cloud-based voice service developed by Amazon that allows customers to interact with technology. There are currently over 40 million Alexa users around the world, so analyzing user sentiments about Alexa will be a good data science project. So, if you want to learn how to analyze the sentiments of users using Amazon Alexa, this article is for you. In this article, I’ll walk you through the task of Amazon Alexa Reviews Sentiment Analysis Using Python.
## Amazon Alexa Reviews Sentiment Analysis using Python
The dataset I’m using for the task of sentiment analysis of Amazon Alexa reviews was collected from Kaggle. It contains data about ratings between 1 and 5, the date of reviews, and customer feedback on their experience with Alexa. So let’s import the necessary Python dataset and libraries that we need for this task:
``` rating date variation verified_reviews feedback
0 5 31-Jul-18 Charcoal Fabric Love my Echo! 1
1 5 31-Jul-18 Charcoal Fabric Loved it! 1
2 4 31-Jul-18 Walnut Finish Sometimes while playing a game, you can answer... 1
3 5 31-Jul-18 Charcoal Fabric I have had a lot of fun with this thing. My 4 ... 1
4 5 31-Jul-18 Charcoal Fabric Music 1```
Let’s start by looking at some of the information in that data to see whether or not we need to change it:
``` rating feedback
count 3150.000000 3150.000000
mean 4.463175 0.918413
std 1.068506 0.273778
min 1.000000 0.000000
25% 4.000000 1.000000
50% 5.000000 1.000000
75% 5.000000 1.000000
max 5.000000 1.000000
rating 0
date 0
variation 0
verified_reviews 0
feedback 0
dtype: int64
Index(['rating', 'date', 'variation', 'verified_reviews', 'feedback'], dtype='object')```
The dataset’s rating column contains the ratings given by the users of Amazon Alexa on a scale of 1 to 5, where 5 is the best rating a user can give. So let’s look at the breakdown of ratings given to Amazon Alexa by its users:
From the above figure, we can see that most of the customers have rated Amazon Alexa including all its variants as 5. So it means that most of the customers are happy with Amazon Alexa.
## Amazon Alexa Reviews Sentiment Analysis
Now let’s move on to the task of sentiment analysis of Alexa’s reviews. The verified_reviews column of the dataset contains all the reviews given by Amazon Alexa’s customers. So let’s add new columns to this data as positive, negative and neutral by calculating the sentiment scores of the reviews:
``` rating date variation ... Positive Negative Neutral
0 5 31-Jul-18 Charcoal Fabric ... 0.692 0.000 0.308
1 5 31-Jul-18 Charcoal Fabric ... 0.807 0.000 0.193
2 4 31-Jul-18 Walnut Finish ... 0.114 0.102 0.784
3 5 31-Jul-18 Charcoal Fabric ... 0.383 0.000 0.617
4 5 31-Jul-18 Charcoal Fabric ... 0.000 0.000 1.000```
Now let’s sum the sentiment scores for each column to understand what most of the customers of Amazon Alexa think about it:
`Neutral 🙂`
The final output that we get is therefore neutral. This means that most users feel neutral about Amazon Alexa services. Now let’s see the sum of the sentiment scores for each column:
```Positive: 1035.4579999999983
Negative: 96.79999999999995
Neutral: 1936.740999999996```
So we can see that Positive and Neutral are above 1000 where Negative is below 100. So this means that most of the customers of Amazon Alexa are satisfied with its services.
### Summary
So this is how we can analyze the sentiments of Amazon Alexa reviews by using the Python programming language. There are currently over 40 million Alexa users around the world, so analyzing user sentiments about Alexa will be a good data science project. I hope you liked this article on the task of Amazon Alexa Reviews Sentiment Analysis using Python. Feel free to ask your valuable questions in the comments section below.
##### Aman Kharwal
I'm a writer and data scientist on a mission to educate others about the incredible power of data📈.
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UNIVERSITY OF CALIFORNIA AT BERKELEY College of Engineering Dept. of Electrical Engineering and Computer Sciences Problem Set # 4 Due Tuesday, February 21 2005. EE 105 Spring 2006 1 Consider a MOS-cap with p+ gate and n-substrate. Substrate doping Nd=3.3E17. The gate oxide thickness T OX =34.5nm. Please calculate: a) Flat band voltage V FB b) Threshold voltage V Tn c) Capacitance at V GB =2V, 0V, -2V, -4V 2 Consider a MOS-cap with n+ gate and p-substrate. Substrate doping Na=7E15. The
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Unformatted text preview: gate oxide thickness T OX =34.5nm. Please calculate: a) Flat band voltage V FB b) Threshold voltage V Tp c) Capacitance at V GB =1V, 0V, -1V, -2V 3 ． For the MOS-cap in problem 1, please plot the charge density as a function of position x under gate voltage VGB=-2V, -4V. Mark the values of all the breaking points....
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campbeln
2005-08-08 16:52:32
On a bit of an aside... is there any significance to these numbers in relation to base 16? These datatypes range limitations are bread from having to fit X number into Y bytes (hence ranges like 0 to 255 and -32768 to 32767), yet these numbers don't seem to conform to this (or do they)? Of course there is some overhead to store (decimal position, etc) but still...
###########
You can disregard the rest of this, these datatypes have the ability to have variable precisions, hence the discrepancies!
###########
Also... there is a term that I've run into in relation to SQL specifications referred to as "numeric precision" that is causing some friction with the above answers. As you may have noticed above, I have a JavaScript function that calculates this value which is basically the number of places in the min/max value. Or, as it's defined online...
* Total number of digits for exact numeric data types (DECIMAL, INTEGER, MONEY, SMALLINT)
* Number of digits of mantissa precision (machine-dependent) for approximate data types (FLOAT, SMALLFLOAT)
Now, the best part of this is that...
* The first range [-(10^38 + 1) to (10^38 - 1)] has a "numeric precision" of 18 (or 20/21 places to few)
* The second range [(-3.40e + 38) to (3.40e + 38)] has a "numeric precision" of 24 (or 14/15 places to few)
* The third range [(-1.79e + 308) to (1.79e + 308)] has a "numeric precision" of 53 (a lot of places to few)
Does anyone know what is being referred to by this "numeric precision" (it is part of SQL92’s “Information Schema” definition)? How should I mitigate the seeming rift between these two different definitions of the datatypes ranges?
MathsIsFun
2005-08-08 16:05:07
#### campbeln wrote:
But 4 of my questions still remain... can anyone provide the integers for:
(-3.40e + 38) = ?
(3.40e + 38) = ?
(-1.79e + 308) = ?
(1.79e + 308) = ?
Just a matter of making sure we have enough zeros!
-340000000000000000000000000000000000000
340000000000000000000000000000000000000
-1.79e + 8 = 179000000, so -1.79e + 308=
-179000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
Likewise 1.79e + 308 =
179000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
I counted the characters in the last one and got 309, as expected, because 1 x 10¹ has two characters (10).
campbeln
2005-08-08 15:59:01
wcy - nice... but how will that help me? Converting into Hex (base 16) would get me a "shorter" number, in terms of a string but I'd still run into the same issues when it comes to variable widths. Though again, math and I have never really gotten along, so I could be missing something!
Thanks for all the responses, all! But 4 of my questions still remain... can anyone provide the integers for:
(-3.40e + 38) = ?
(3.40e + 38) = ?
(-1.79e + 308) = ?
(1.79e + 308) = ?
Muchos Danke!
ganesh
2005-08-08 15:14:41
wcy, i viewed the base converter. It works well.
I tried converting 759 to base 20 and got the right answer
wcy
2005-08-08 15:07:02
i created a base converter before,
maybe it might be useful to convert your integers to other bases (eg base 16)
http://www.geocities.com/chengyuanwu/baseconverter.html
campbeln
2005-08-08 11:55:28
Jason - please post a new topic by going here - http://www.mathsisfun.com/forum/post.php?fid=2
jason
2005-08-08 11:47:13
is anyone here
jason
2005-08-08 11:46:00
i need help with a riddle my teacher gave me
its m168a:00th
3 words
1 word is math
: means time
campbeln
2005-08-08 11:44:21
kylekatarn - thanks for that!
I have a feeling the "notation" used in the documentation for SQL*Server is not 100% correct, mathematically or otherwise. I have a feeling they did indeed mean "-(10^38 + 1)" as negative numbers are allowed in the datatype. Microsoft just failed to document it correctly (us computer nerd's are used to having their documentation be incorrect; completely non-functional example code is my personal fav.!).
campbeln
2005-08-08 11:35:00
For those who are interested, here is an updated version of the JavaScript code I am going to use to test these huge numbers. The previously posted function assumed that the passed sRangeMin was <= 0. This version should no longer have that limitation ("should" being the operative word!).
Again, anyone who is willing to look thru my code to ensure my logic is correct I would be greatly appreciative!
//############################################################
//# Determines if the passed sNumber is within the passed range
//# NOTE: "return (sNumber >= sRangeMin && sNumber <= sRangeMax)" would work in 99.9% of the checks we'll do with this function, but in the case of huge/tiny numbers (such as NUMERIC(x,y)'s in Oracle), this wouldn't cut it as the numbers would be too large/small to be represented in any available numeric variables
//############################################################
//# Last Updated: August 8, 2005
function rfIsNumberInRange(sNumber, sRangeMin, sRangeMax) {
var bReturn = false;
//#### If the passed sNumber is greater then or equal to the passed sRangeMin
if (rfCompairson(sNumber, sRangeMin) >= 0) {
//#### If the passed sNumber is less then or equal to the passed sRangeMax
if (rfCompairson(sNumber, sRangeMax) <= 0) {
//#### Since the passed sNumber is within the passed sRangeMin/sRangeMax, flip our bReturn value to true
bReturn = true;
}
}
//#### Return the above determined bReturn value to the caller
return bReturn;
}
//############################################################
//# Determines if the passed sNumber is greater then, less then or equal to the passed sRange
//# NOTE: Return values:
//# -1 if sNumber is less then sRange
//# 1 if sNumber is greater then sRange
//# 0 if the passed values are equal, or if one of the passed values was non-numeric
//############################################################
//# Last Updated: August 8, 2005
function rfCompairson(sNumber, sRange) {
//#### Ensure the passed sNumber and sRange are strings
sNumber = new String(sNumber);
sRange = new String(sRange);
//#### Define and init the required local vars
var iNumberNumericPrecision = rfNumericPrecision(sNumber);
var iRangeNumericPrecision = rfNumericPrecision(sRange);
var bNumberIsPositive = (sNumber.indexOf("-") != 0);
var bRangeIsPositive = (sRange.indexOf("-") != 0);
var iReturn;
//#### If the passed sNumber or sRange were non-numeric, set our iReturn value to 0
if (iNumberNumericPrecision == -1 || iRangeNumericPrecision == -1) {
iReturn = 0;
}
//#### Else if the signs of the passed sNumber and sRange do not match
else if (bNumberIsPositive != bRangeIsPositive) {
//#### If the bNumberIsPositive, then the sRange is negetive, so set our iReturn value to 1 (as sNumber is greater then the sRange)
if (bNumberIsPositive) {
iReturn = 1;
}
//#### Else the sNumber is negetive and the bRangeIsPositive, so set our iReturn value to -1 (as sNumber is less then the sRange)
else {
iReturn = -1;
}
}
//#### Else the signs of the passed sNumber and sRange match
else {
//#### If the above-determined rfNumericPrecision's are specifying numbers of less then 1 billion
if (iRangeNumericPrecision < 10 && iNumberNumericPrecision < 10) {
//#### Define and init the additionally required vars
//#### NOTE: We know that both sNumber and sRange are numeric as non-numeric value are caught by rfNumericPrecision above
var fNumber = Math(parseFloat(sNumber));
var fRange = Math(parseFloat(sRange));
//#### If the sNumber and sRange are equal, set our iReturn value to 0
if (fNumber == fRange) {
iReturn = 0;
}
//#### Else if the sNumber is greater then the sRange, set our iReturn value to 1
else if (fNumber > fRange) {
iReturn = 1;
}
//#### Else the fNumber is less then the sRange, so set our iReturn value to -1
else {
iReturn = -1;
}
}
//#### Else we're dealing with number ranges over 1 billion, so let's get creative...
else {
//#### If the iNumber('s)NumericPrecision is less then the iRange('s)NumericPrecision (making sNumber less then sRange), set our iReturn value to -1
if (iNumberNumericPrecision < iRangeNumericPrecision) {
iReturn = -1;
}
//#### Else if the iNumber('s)NumericPrecision is more then the iRange('s)NumericPrecision (making sNumber greater then sRange), set our iReturn value to 1
else if (iNumberNumericPrecision > iRangeNumericPrecision) {
iReturn = 1;
}
//#### Else the iNumber('s)NumericPrecision is equal to the iRange('s)NumericPrecision, so additional checking is required
else {
//#### Define and set the additionally required decimal point position variables
var iNumberDecimalPoint = sNumber.indexOf(".");
var iRangeDecimalPoint = sRange.indexOf(".");
//#### If either/both of the decimal points were not found above, reset iNumberDecimalPoint/iRangeDecimalPoint to their respective .lengths (which logicially places the iRangeDecimalPoint at the end of the sCurrentRange, which is where it is located)
//#### NOTE: Since this function only checks that the passed sNumber is within the passed range, the values "whole" -vs- "floating point" number distinction is ignored as for our purposes, it is unimportant.
if (iNumberDecimalPoint == -1) {
iNumberDecimalPoint = sNumber.length;
}
if (iRangeDecimalPoint == -1) {
iRangeDecimalPoint = sCurrentRange.length;
}
//#### If the sNumber's decimal point is to the left of sRange's (making sNumber less then sRange), set our iReturn value to -1
if (iNumberDecimalPoint < iRangeDecimalPoint) {
iReturn = -1;
}
//#### Else if the sNumber's decimal point is to the right of sRange's (making sNumber greater then sRange), set our iReturn value to 1
else if (iNumberDecimalPoint > iRangeDecimalPoint) {
iReturn = 1;
}
//#### Else the sNumber's decimal point is in the same position as the sRange's decimal point
else {
//#### Define and init the additionally required vars
var iCurrentNumberNumber;
var iCurrentRangeNumber;
var i;
//#### Default our iReturn value to 0 (as only > and < are checked in the loop below, so if the loop finishes without changing the iReturn value then the sNumber and sRange are equal)
iReturn = 0;
//#### Setup the value for i based on if the bNumberIsPositive (setting it to 0 if it is, or 1 if it isn't)
//#### NOTE: This is done to skip over the leading "-" sign in negetive numbers (yea it's ugly, but it works!)
//#### NOTE: Since at this point we know that signs of sNumber and sRange match, we only need to check bNumberIsPositive's value
i = (bNumberIsPositive) ? (0) : (1);
//#### Traverse the sNumber/sRange strings from front to back (based on the above determined starting position)
//#### NOTE: Since everything is is the same position and the same precision, we know that sNumber's .lenght is equal to sRange's
for (i = i; i < sNumber.Length; i++) {
//#### As long as we're not looking at the decimal point
if (i != iNumberDecimalPoint) {
//#### Determine the iCurrentNumberNumber and iCurrentRangeNumber for this loop
iCurrentNumberNumber = parseInt(SubStr(sNumber, i, 1));
iCurrentRangeNumber = parseInt(SubStr(sRange, i, 1));
//#### If the iCurrentNumberNumber is less then the iCurrentRangeNumber
if (iCurrentNumberNumber < iCurrentRangeNumber) {
//#### sNumber is less then sRange, so set our iReturn value to -1 and fall from the loop
iReturn = -1;
exit;
}
//#### Else if the iCurrentNumberNumber is greater then the iCurrentRangeNumber
if (iCurrentNumberNumber > iCurrentRangeNumber) {
//#### sNumber is greater then sRange, so set our iReturn value to 1 and fall from the loop
iReturn = 1;
exit;
}
}
}
}
}
}
}
//#### Return the above determined iReturn value to the caller
return iReturn;
}
//############################################################
//# Determines the numeric precision of the passed sValue (i.e. - counts how many numeric places there are within the number, not including leading 0's)
//############################################################
//# Last Updated: August 8, 2005
function rfNumericPrecision(sValue) {
var sCurrentChar;
var iReturn = 0;
var i;
var bStartCounting = false;
//#### Ensure the passed sValue is a string
sValue = new String(sValue);
//#### If the passed sValue is not a number, set the iReturn value to -1 (which indicates an error occured)
if (isNaN(sValue)) {
iReturn = -1;
}
//#### Else the passed sValue is a number
else {
//#### Traverse the .length of the passed sValue
for (i = 0; i < sValue.length; i++) {
//#### Collect the sCurrentChar
sCurrentChar = sValue.substr(i, 1)
//#### If the sCurrentChar is a number
if (! isNaN(sCurrentChar) && sCurrentChar != ' ') {
//#### If we are supposed to bStartCounting, or if the sCurrentChar is not a 0
//#### NOTE: This is done so we ignore leading 0's (trailing 0's are still counted)
if (bStartCounting || sCurrentChar != '0') {
//#### Inc iReturn and ensure bStartCounting is true
iReturn++;
bStartCounting = true;
}
}
}
}
return iReturn;
}
kylekatarn
2005-08-08 11:32:18
Nothing is wrong with the calculator!
(-10)^38 is indeed 100000000000000000000000000000000000000.
x^n =x*x*x*x* ... *a (n times)
If n is EVEN then x^n is greater or equal to zero:
If n is ODD then x^n can be a negative number.
a simple example:
(-10)^6=(-10)(-10)(-10)(-10)(-10)(-10)=1000000
(-5)^3=(-5)(-5)(-5)=-125
you were probably thinking of -(10^38)
campbeln
2005-08-08 10:16:31
Thanks Admin! Seems I was missing a digit in the (-10^38 + 1) to (10^38 - 1). And just so you know, in the Full Precision Calculator:
X = -10
Y = 38
[X^(Integer Y)]
Returns (100000000000000000000000000000000000000)
...shouldn't this return (-100000000000000000000000000000000000000)?
'Course I've still got the other two ranges to determine.
Also... I was just reviewing my JavaScript function and I've noticed a few bugs which I'm now working on (so be gentle =).
MathsIsFun
2005-08-08 10:05:39
I can only spare a minute, and will read your post later, but just thought I would mention Full Precision Calculator
For example: put 10 into X and 38 into Y, then calculate X^Y, then copy-paste this into X, put 1 in Y and use X-Y
campbeln
2005-08-08 09:41:13
Hello all,
I am interested in getting the integer values (well, the long number versions) for the following values in scientific notation:
(-10^38 + 1) = -10000000000000000000000000000000000001 [correct?]
(10^38 - 1) = 9999999999999999999999999999999999999 [correct?]
(-3.40e + 38) = ?
(3.40e + 38) = ?
(-1.79e + 308) = ?
(1.79e + 308) = ?
I apologize for the mundane topic, but my guile has run dry in trying to find the numeric equivalents to these values (stupid Excel and its rounding). Thanks for any help you can provide! My approach to math always seemed to be about 180 degrees from the way it was taught to me, so I rarely excelled at it. In fact, my altered approach is made evident in the method I take to validate these huge numbers (more info for those interested parties below...)!
#########
For those who are interested in why I'm asking this question...
#########
I am working on a website front end script that validates user input to ensure that it will "fit" into a back end database. The values listed above are the valid numeric ranges allowed for three data types (FLOAT, REAL and DECIMAL column datatypes in SQL*Server in this case).
Now those of you with some computer nerd experience may be asking "How the heck is he planning on validating these huge numbers!? Sci-Notation is used for a reason!" Well, I have built a function (both in JavaScript and VB.Net) that walks these huge numbers a single digit at a time, hence avoiding the "hugeness of the numbers" issue (I've included the JavaScript version below for those of you who are truly interested - actually if you could make sure I've done it right that would rock!). In a nutshell - by counting the number of digits, then the position of the decimal places and finally by looking at each digit's value individually I am able to use simple, small number comparisons to determine if these truly huge numbers are in the proper range.
#########
[Clipped original, limited JavaScript code. Please see below for latest code.] | 4,547 | 16,802 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2014-10 | longest | en | 0.883581 |
http://oeis.org/A068913 | 1,606,584,489,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141195687.51/warc/CC-MAIN-20201128155305-20201128185305-00082.warc.gz | 72,824,615 | 4,715 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A068913 Square array read by antidiagonals of number of k step walks (each step +-1 starting from 0) which are never more than n or less than -n. 10
1, 0, 1, 0, 2, 1, 0, 2, 2, 1, 0, 4, 4, 2, 1, 0, 4, 6, 4, 2, 1, 0, 8, 12, 8, 4, 2, 1, 0, 8, 18, 14, 8, 4, 2, 1, 0, 16, 36, 28, 16, 8, 4, 2, 1, 0, 16, 54, 48, 30, 16, 8, 4, 2, 1, 0, 32, 108, 96, 60, 32, 16, 8, 4, 2, 1, 0, 32, 162, 164, 110, 62, 32, 16, 8, 4, 2, 1, 0, 64, 324, 328, 220, 124, 64, 32, 16, 8, 4, 2, 1 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 0,5 LINKS Alois P. Heinz, Antidiagonals n = 0..200, flattened FORMULA Starting with T(n, 0) = 1, if (k-n) is negative or even then T(n, k) = 2*T(n, k-1), otherwise T(n, k) = 2*T(n, k-1)-A061897(n-1, (k-n-1)/2). So for n>=k, T(n, k) = 2^k. T(n,0) = 1, T(n,k) = (2^k/(n+1))*Sum_{r=1..n+1} (-1)^r*cos((Pi*(2*r-1))/(2*(n+1)))^k*cot((Pi*(1-2*r))/(4*(n+1))). - Herbert Kociemba, Sep 23 2020 EXAMPLE Rows start: 1, 0, 0, 0, 0, ... 1, 2, 2, 4, 4, ... 1, 2, 4, 6, 12, ... 1, 2, 4, 8, 14, ... ... MATHEMATICA T[n_, 0]=1; T[n_, k_]:=2^k/(n+1) Sum[(-1)^r Cos[(Pi (2r-1))/(2 (n+1))]^k Cot[(Pi (1-2r))/(4 (n+1))], {r, 1, n+1}]; Table[T[r, n-r], {n, 0, 20}, {r, 0, n}]//Round//Flatten (* Herbert Kociemba, Sep 23 2020 *) CROSSREFS Cf. early rows: A000007, A016116 (without initial term), A068911, A068912, A216212, A216241, A235701. Central and lower diagonals are A000079, higher diagonals include A000918, A028399. Sequence in context: A289281 A212957 A035393 * A128306 A305152 A170983 Adjacent sequences: A068910 A068911 A068912 * A068914 A068915 A068916 KEYWORD nonn,tabl AUTHOR Henry Bottomley, Mar 06 2002 STATUS approved
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Last modified November 28 12:19 EST 2020. Contains 338720 sequences. (Running on oeis4.) | 1,050 | 2,426 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2020-50 | latest | en | 0.608378 |
http://populationeducation.org/content/millions-billions-conceptualizing-large-numbers-100th-day-school | 1,518,998,631,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812293.35/warc/CC-MAIN-20180218232618-20180219012618-00528.warc.gz | 252,585,845 | 12,376 | Millions & Billions: Conceptualizing Large Numbers on the 100th Day of School
This month, many elementary school classrooms celebrate the 100th day of school in creative ways. From 100-letter poems to first-graders dressing up like small, adorable centarians, recognizing this day is not only fun, but develops an important skill: conceptualizing large numbers.
We live in a world of 7.4 billion people, which in students’ lifetimes is projected to be a world of 10 billion. Now more than ever, students’ ability to relate to global resource use depends on their capacity to understand what a million or a billion actually means. Population Education’s lesson, Millions and Billions, is a great way to get young learners thinking about the magnitude of big numbers, in the spirit of the 100th day. The lesson starts by engaging students with a riddle:
“Your rich uncle has died and left you 1 billion dollars, but there’s a catch. In order to accept the money, you must count it for eight hours a day at the rate of one dollar per second. When you are finished counting the billion dollars is yours and only then may you begin to spend it.”
Students must decide whether or not they would accept this offer, defending their decision with mathematical evidence. While approaches vary, students use an understanding of long division, rates, and conversion between measurement units to determine that it would ultimately take over 95 years to count the money. At that point, it’s probably not worth it to accept the offer, because you may not even live that long (although some kids may make compelling arguments, citing increasing life expectancy). Students then calculate the same riddle but for 1 million dollars, and quickly see the difference.
The lesson goes on to put millions and billions into context through everyday items or activities, using five different multi-step problems. One problem asks students to determine where they would be if they travelled 1 million steps, and then 1 billion steps. It gets kids out of their seats and applying their knowledge in a hands-on way: measuring the length of a single step, making critical thinking decisions about what units they should use, and using a map to figure out where they would arrive. Other questions include calculating the distance covered by 1 billion people holding hands and the height of 1 billion markers stacked end-to-end. There are no definitive answers (although student calculations can be checked for accuracy). The comparisons students make can easily be adjusted by the teacher—for a kid who loves giraffes, you can give them the height of the average giraffe (15-19 ft, by the way) and ask them to calculate how many giraffes it would take to equal a tower of 1 billion markers.
The provided worksheet includes built-in scaffolding, guiding students through the steps required to calculate the answer. However, many teachers find that presenting only the challenge, and letting students decide the process, builds critical thinking and encourages multiple approaches to finding a solution. The lesson is differentiated, as the five challenges have varying levels of difficulty and each can be given with and without the scaffolding.
Download Millions and Billions now to access a hands-on lesson plan that makes large numbers tangible for upper elementary students. Here’s to a great 100th day of school—and to hoping summer break comes before the billionth day of school! | 686 | 3,459 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2018-09 | latest | en | 0.947303 |
http://www.algebra.com/algebra/homework/Graphs/Graphs.faq.question.604847.html | 1,368,948,944,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368696384213/warc/CC-MAIN-20130516092624-00039-ip-10-60-113-184.ec2.internal.warc.gz | 313,677,060 | 4,667 | # SOLUTION: Hello, Would u pls, check my answer? Factor the expression :c^2 + 4c + 4 My answer: (c + 2)^2 Thanks a lot, H.H
Algebra -> Algebra -> Graphs -> SOLUTION: Hello, Would u pls, check my answer? Factor the expression :c^2 + 4c + 4 My answer: (c + 2)^2 Thanks a lot, H.H Log On
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Algebra: Graphs, graphing equations and inequalities Solvers Lessons Answers archive Quiz In Depth
Question 604847: Hello, Would u pls, check my answer? Factor the expression :c^2 + 4c + 4 My answer: (c + 2)^2 Thanks a lot, H.H Answer by nerdybill(6948) (Show Source): You can put this solution on YOUR website!You are correct. . You can check yourself... apply FOIL on (c + 2)(c + 2) and will get your original expression | 261 | 905 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2013-20 | latest | en | 0.763917 |
https://aha.betterexplained.com/t/swap-two-variables-using-xor/128 | 1,660,216,151,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571284.54/warc/CC-MAIN-20220811103305-20220811133305-00158.warc.gz | 121,333,679 | 7,861 | # Swap two variables using XOR
Most people would swap two variables x and y using a temporary variable, like this:
Here’s a neat programming trick to swap two values without needing a temp:
This is a companion discussion topic for the original entry at http://betterexplained.com/articles/swap-two-variables-using-xor/
Note that in portable C (that is, according to, say, the 1999 C standard), this only works for unsigned integral types. Obviously it won’t work for floating point or pointer or struct/union types. (Portably – if you don’t care about machine-specific code, you could convert pointers to integers on many platforms and it’ll just do what you want. But it’s not guaranteed to work everywhere.)
But also, under some (rare, but allowed) non-2s-complement signed integer representations, an xor might result in a “negative zero” that can be converted to a normal zero before continuing, resulting in the wrong sign bit setting on an output value…
Wow, thanks for the details Ken! Yeah, I imagine there are all sorts of nasty side effects when using a trick like this. Appreciate the background info.
Hi! I think the following note will be helpful, since it highlights the properties involved in all these swappings:
http://www.cs.nott.ac.uk/~jff/papers/JFFs/JFF1.pdf
Thanks Joao, that paper is pretty interesting.
Kalid,
I’ve just published a new version of my note on this. You can find it at:
http://www.joaoferreira.org/2007/07/11/swapping-the-values-of-two-variables/
[…] void swap (int& a, int& b) { a ^= b; // a = a XOR b b ^= a; // b = b XOR a a ^= b; // a = a XOR b } Why does it work? Here is one that explains it well: Swap two variables using XOR | BetterExplained I found another article on the same technique, along with two other techniques. However, I caution you to read the comments that follow it: 3 ways to swap variables without temp variable « Stream As the betterexplained.com article mentions, there is an XCHG instruction on some machines that allows you to swap two registers without using multiple XOR instructions. The important thing: be careful when using such unusual techniques. You never know how a program can be affected by a compiler, an assembler or libraries that are linked it. Only use it if absolutely necessary or if you merely wish to play with it in a small program. Back on-topic, those were examples of using bitwise operations to increase efficiency. Multiplication and division by 2 is fairly simple too when you only need to worry about integers. All that is required is knowledge of what operators that bit shifts use in your specific programming language, if the operation is supported in your programming language (Visual Basic does not support bit shifts natively). In C/C++ and Java, the operator is "<<" for a left shift (SHL/SAL in ASM) and ">>" for a right shift (SHR/SAR in ASM - which one is used can depend upon signedness of a value except in Java where all integers are signed integers). The bottom line: ASM is useful and the ASM ways that were used for efficiency are still quite useful at times. __________________ […]
there is another way, tho i doubt if it’s faster,
a=10;
b=15;
a=a+b;//a=25 b=15
b=a-b; //a=25 b=10
a=a-b; //a=15 b=10, swapped
Hi zezima, thanks for the info! That’s a neat trick too – you “combine” a and b in a similar way, and extract them as well.
One thing to watch out for is arithmetic overflow.
[…] E aí pronto, suas variáveis estão invertidas. Pode acreditar (ou ler uma explicação bem mais detalhada). Legal né? […]
I followed you right through to the pointer Caveat. If both point to the same address it should still work…
x=1234
y=1234
x=x xor y -> x = 1234 xor 1234 -> x = 0
y=x xor y -> y = 0 xor 1234 -> y = 1234
x=x xor y -> x = 0 xor 1234 -> x = 1234
Hi Rich, good question.
When using pointers to the same location, when you do the first step (x = x xor y) => 0 you also set y to 0 as well since it’s at the same location. In this case, you’ve lost the value 1234 – x and y were sharing the same location and it was overwritten with 0.
Hope this helps.
in c this can be written as x^=y^=x^=y
@Anonymous: That’s right – in “slightly obfuscated” c :).
Pretty much any compiler ever will substitute the standard swap for a single exchange instruction. This means the extra variable never gets made (unless it is used later on), and the swap is only one instruction. The triple xor is not recognized as a swap (and can’t be, since the result is different when they are equal), so it is slower in the end, as well as harder to understand.
can i just make it like dis:
a=5
b=4
a=(a+b)-a
b=(b+a)-b | 1,146 | 4,632 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2022-33 | latest | en | 0.834058 |
http://spmmathematics.blog.onlinetuition.com.my/2014/08/direct-variation-part-1.html | 1,569,064,851,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574409.16/warc/CC-MAIN-20190921104758-20190921130758-00265.warc.gz | 166,392,647 | 10,195 | 5.1 Direct Variation Part 1
(A) Determining whether a quantity varies directly as another quantity
1. If a quantity varies directly as a quantity x, the
(a) y increases when x increases
(b) y decreases when x decreases
2. A quantity varies directly as a quantity x if and only if $\frac{y}{x}=k$ where k is called the constant of variation.
3. y varies directly as x is written as $y\propto x$ .
4. When $y\propto x$ , the graph of against x is a straight line passing through the origin.
(B) Expressing a direct variation in the form of an equation involving two variables
Example 1
Given that y varies directly as x and y = 20 when x = 36 . Write the direct variation in the form of equation.
(C) Finding the value of a variable in a direct variation
1. When y varies directly as x and sufficient information is given, the value of y or x can be determined by using:
$\begin{array}{l}\left(a\right)\text{}y=kx,\text{or}\\ \left(b\right)\text{}\frac{{y}_{1}}{{x}_{1}}=\frac{{y}_{2}}{{x}_{2}}\end{array}$
Example 2
Given that varies directly as x and y = 24 when x = 8, find
(a) The equation relating to x
(b) The value of when = 6
(c) The value of when = 36
Solution:
Method 1: Using y = kx
$\left(a\right)\text{y}\propto x$
y = kx
when y = 24, x = 8
24 = k (8)
k = 3
y = 3x
(b) when x = 6,
y = 3 (6)
y = 18
(c) when y = 36
36 = 3x
x =12
Method 2: $\text{Using}\frac{{y}_{1}}{{x}_{1}}=\frac{{y}_{2}}{{x}_{2}}$
(a) Let x1 = 8 and y1 = 24
$\begin{array}{l}\frac{{y}_{1}}{{x}_{1}}=\frac{{y}_{2}}{{x}_{2}}\to \frac{24}{8}=\frac{{y}_{2}}{{x}_{2}}\\ \frac{3}{1}=\frac{{y}_{2}}{{x}_{2}}\to {y}_{2}=3{x}_{2}\\ \therefore y=3x\end{array}$
(b) Let x1 = 8 and y1 = 24 and x2= 6; find y2.
$\begin{array}{l}\frac{{y}_{1}}{{x}_{1}}=\frac{{y}_{2}}{{x}_{2}}\to \frac{24}{8}=\frac{{y}_{2}}{6}\\ {y}_{2}=\frac{24}{8}\left(6\right)\\ {y}_{2}=18\end{array}$
(c) Let x1 = 8 and y1 = 24 and y2= 36; find x2.
$\begin{array}{l}\frac{{y}_{1}}{{x}_{1}}=\frac{{y}_{2}}{{x}_{2}}\to \frac{24}{8}=\frac{36}{{x}_{2}}\\ 24{x}_{2}=36×8\\ {x}_{2}=12\end{array}$ | 763 | 2,046 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 9, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.84375 | 5 | CC-MAIN-2019-39 | latest | en | 0.675311 |
https://nrich.maths.org/public/leg.php?code=-393&cl=3&cldcmpid=7651 | 1,529,755,322,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864958.91/warc/CC-MAIN-20180623113131-20180623133131-00524.warc.gz | 677,618,763 | 5,907 | # Search by Topic
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Estimate these curious quantities sufficiently accurately that you can rank them in order of size
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In which Olympic event does a human travel fastest? Decide which events to include in your Alternative Record Book.
### A Question of Scale
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Use your skill and knowledge to place various scientific lengths in order of size. Can you judge the length of objects with sizes ranging from 1 Angstrom to 1 million km with no wrong attempts?
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### Epidemic Modelling
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Use the computer to model an epidemic. Try out public health policies to control the spread of the epidemic, to minimise the number of sick days and deaths. | 710 | 3,225 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2018-26 | latest | en | 0.85552 |
http://studyres.com/doc/143648/reject-h-0 | 1,534,768,985,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221216453.52/warc/CC-MAIN-20180820121228-20180820141228-00039.warc.gz | 369,425,378 | 14,633 | • Study Resource
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Transcript
```Chapter 9
Hypothesis Tests
Hypothesis Tests
The logic behind a confidence interval is that if we build
an interval around a sample value there is a high
likelihood that the population value will be captured in
the interval.
The logic behind a hypothesis test is that if we build an
interval around a hypothesized value and the sample
value falls into that interval, the result is consistent
with hypothesized value. If the sample value falls
outside the interval it is inconsistent with the
hypothesized value.
Hypotheses
Null hypothesis (H0) – The assumed value of the
population parameter when conducting the test
Alternative hypothesis (Ha) – The complement of the
null hypothesis
We don’t necessarily believe the null hypothesis is true,
it is a benchmark we are going to test against.
Forms of the Hypotheses
One-tailed test to the left
H0: m > m0
Ha: m < m0
One-tailed test to the right
H0: m < m0
Ha: m > m0
Two-tailed test
H0: m = m0
Ha: m ≠ m0
Hypothesis Tests
State of the world
Conclusion Accept H0
Reject H0
H0 is true
H0 is false
Correct
conclusion
Type II
error
Type I error Correct
conclusion
Hypothesis Tests
We can never know if the test provided the correct
answer, we can only control the probability of making a
certain type of error.
Level of significance – the probability of making a Type I
error when the null hypothesis is true as an equality.
Mechanics of a Hypothesis Test
1.
2.
3.
4.
Define the null and alternative hypotheses
Define the rule for rejecting the null hypothesis
Calculate value from sample
Reject or accept null hypothesis, state the
implication
Critical Value
A value defining the boarder(s) of the rejection
region.
Test Statistic
A value calculated from sample data that is used
to determine whether to accept or reject the null
hypothesis.
Hypothesis Test with s Known
1. H0: m > m0
Ha: m < m0
2. Reject H0 if z < -za
3.
xm
z
s
n
4. Interpret result
Hypothesis Test with s Known,
Example
A firm has started a wellness plan which
provides support for employees to lose weight
and stop smoking. In the past employees used
10 sick days per year and the standard deviation
was 2 days. The firm wants to test whether the
number of sick days has fallen with the
significance level set at 5%. A sample of 100
employees was collected, the average number of
sick days for the sample was 9.
Hypothesis Test with s Known
1. H0: m > 10
Ha: m < 10
2. Reject H0 if z < -1.645
3.
xm
9 10
1
z
5
s
2
.2
100
n
4. Reject H0. The test implies that the number
of sick days has fallen.
Hypothesis Test with s Unknown
If s is unknown, then we use s to estimate s
and we would need to use the t distribution in
conducting the test. Again, the number of
degrees of freedom would equal n-1.
Hypothesis Test with s Unknown,
Example
A researcher wants to know if the amount workers
are putting into tax sheltered retirement plans (like
IRA and 401k accounts) has fallen. Last year the
average monthly contribution into a tax-sheltered
plan was \$150. A sample of 101 workers is drawn.
The average monthly contribution for the workers
in the sample was \$135, and the sample standard
deviation was \$50. Conduct the test with a 5%
significance level.
Hypothesis Test with s Unknown
1. H0: m > 150
Ha: m < 150
2. Reject H0 if t < -1.660
3.
x m0 135 150 15
t
3.01
s
50
4.975
n
101
4. Reject H0. The evidence suggests that the
amount of money contributed to taxsheltered programs has fallen.
Hypothesis Test with s Unknown,
Example
Last year Medicare recipients spent an average
of 4.5 days in the hospital. A researcher wants to
test to see if the average has changed. He wants
to conduct a test at the 5% significance level. A
sample of 25 Medicare recipients is drawn and
the sample mean is 5 days and the standard
deviation is 2.
Hypothesis Test with s Unknown
1. H0: m = 4.5
Ha: m ≠ 4.5
2. Reject H0 if t < -2.064 or t > 2.064
x m0 5 4.5 .5
3.
t
1.25
s
2
.4
25
n
4. Accept H0. The evidence is not strong enough
to reject H0, we would assume the average
number of days in the hospital has not
changed.
Alternatives Ways of Conducting a
Hypothesis Test
• t scores
• p values
p Values
A p value is the probability of drawing a value
whose distance from the hypothesized value is
greater than or equal to the sample value.
The null hypothesis is rejected if p < a. In a onetailed test p is equal to the area in a single tail. In
the case of a two-tailed test p equals the area in
both tails that is as far away from the mean as the
sample value.
Comparisons of Rejection Rules
Assume a one-tailed test in which a = .05
and z equals 2.
Hypothesis Test using p-Value, s
Unknown
Last year Medicare recipients spent an average
of 4.5 days in the hospital. A researcher wants to
test to see if the average has changed. He wants
to conduct a test at the 5% significance level. A
sample of 25 Medicare recipients is drawn and
the sample mean is 5 days and the standard
deviation is 2.
Hypothesis Test using p-Value,
s Unknown
1. H0: m = 4.5
Ha: m ≠ 4.5
2. Reject H0 if p < .05
x m0 5 4.5 .5
3.
t
1.25
s
2
.4
25
n
p P(t 1.25) or P(t 1.25) .22
4. Accept H0. The evidence is not strong enough to
reject H0, we would assume the average number
of days in the hospital has not changed.
Hypothesis Test with s Unknown,
Example
A researcher wants to know if the amount workers
are putting into tax sheltered retirement plans (like
IRA and 401k accounts) has fallen. Last year the
average monthly contribution into a tax-sheltered
plan was \$150. A sample of 101 workers is drawn.
The average monthly contribution for the workers
in the sample was \$135, and the sample standard
deviation was \$50. Conduct the test with a 5%
significance level.
Hypothesis Test with s Unknown
1. H0: m > 150
Ha: m < 150
2. Reject H0 if p < .05
3.
x m0 135 150 15
t
3.01
s
50
4.975
n
101
p P(t 3.01) .00165
4. Reject H0. The evidence suggests that the
amount of money contributed to taxsheltered programs has fallen.
Summary of Rejection Rules
z and t scores: The rejection rule is defined in
terms of standard deviations
p values: The rejection rule is defined in terms of
probabilities
Summary of Hypothesis Tests
Lower Tail Test Upper Tail Test Two-Tailed Test
t score
rejection rule
H0: m > m0
Ha: m < m0
Reject H0 if
t<-ta
H0: m < m0
Ha: m > m0
Reject H0 if
t>ta
H0: m = m0
Ha: m ≠ m0
Reject H0 if
t<-ta/2 or t>ta/2
p-value
rejection rule
Reject H0 if
p<a*
Reject H0 if
p<a*
Reject H0 if
p < a **
Hypotheses
* Where p is calculated using one tail
** Where p is calculated using both tails
```
Document related concepts
Student's t-test wikipedia, lookup
Resampling (statistics) wikipedia, lookup
Misuse of statistics wikipedia, lookup
Psychometrics wikipedia, lookup
Foundations of statistics wikipedia, lookup
Omnibus test wikipedia, lookup
Statistical hypothesis testing wikipedia, lookup
Similar | 2,119 | 7,053 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2018-34 | latest | en | 0.837066 |
https://www.lmfdb.org/EllipticCurve/Q/1922e2/ | 1,713,630,297,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817670.11/warc/CC-MAIN-20240420153103-20240420183103-00057.warc.gz | 785,692,918 | 30,907 | # Properties
Label 1922e2 Conductor $1922$ Discriminant $3844$ j-invariant $$\frac{51181724570498001}{4}$$ CM no Rank $1$ Torsion structure trivial
# Related objects
Show commands: Magma / Oscar / PariGP / SageMath
## Simplified equation
$$y^2+xy+y=x^3-x^2-76332x+8136267$$ y^2+xy+y=x^3-x^2-76332x+8136267 (homogenize, simplify) $$y^2z+xyz+yz^2=x^3-x^2z-76332xz^2+8136267z^3$$ y^2z+xyz+yz^2=x^3-x^2z-76332xz^2+8136267z^3 (dehomogenize, simplify) $$y^2=x^3-1221307x+519499798$$ y^2=x^3-1221307x+519499798 (homogenize, minimize)
comment: Define the curve
sage: E = EllipticCurve([1, -1, 1, -76332, 8136267])
gp: E = ellinit([1, -1, 1, -76332, 8136267])
magma: E := EllipticCurve([1, -1, 1, -76332, 8136267]);
oscar: E = EllipticCurve([1, -1, 1, -76332, 8136267])
sage: E.short_weierstrass_model()
magma: WeierstrassModel(E);
oscar: short_weierstrass_model(E)
## Mordell-Weil group structure
$$\Z$$
magma: MordellWeilGroup(E);
### Infinite order Mordell-Weil generator and height
$P$ = $$\left(\frac{639}{4}, -\frac{641}{8}\right)$$ (639/4, -641/8) $\hat{h}(P)$ ≈ $0.81588661340314983832465559114$
sage: E.gens()
magma: Generators(E);
gp: E.gen
## Integral points
None
comment: Integral points
sage: E.integral_points()
magma: IntegralPoints(E);
## Invariants
Conductor: $$1922$$ = $2 \cdot 31^{2}$ comment: Conductor sage: E.conductor().factor() gp: ellglobalred(E)[1] magma: Conductor(E); oscar: conductor(E) Discriminant: $3844$ = $2^{2} \cdot 31^{2}$ comment: Discriminant sage: E.discriminant().factor() gp: E.disc magma: Discriminant(E); oscar: discriminant(E) j-invariant: $$\frac{51181724570498001}{4}$$ = $2^{-2} \cdot 3^{3} \cdot 31 \cdot 39397^{3}$ comment: j-invariant sage: E.j_invariant().factor() gp: E.j magma: jInvariant(E); oscar: j_invariant(E) Endomorphism ring: $\Z$ Geometric endomorphism ring: $$\Z$$ (no potential complex multiplication) sage: E.has_cm() magma: HasComplexMultiplication(E); Sato-Tate group: $\mathrm{SU}(2)$ Faltings height: $1.0529276699977781426504084647\dots$ gp: ellheight(E) magma: FaltingsHeight(E); oscar: faltings_height(E) Stable Faltings height: $0.48059646925025376832888107728\dots$ magma: StableFaltingsHeight(E); oscar: stable_faltings_height(E) $abc$ quality: $1.1080332911237138\dots$ Szpiro ratio: $5.996747016078877\dots$
## BSD invariants
Analytic rank: $1$ sage: E.analytic_rank() gp: ellanalyticrank(E) magma: AnalyticRank(E); Regulator: $0.81588661340314983832465559114\dots$ comment: Regulator sage: E.regulator() G = E.gen \\ if available matdet(ellheightmatrix(E,G)) magma: Regulator(E); Real period: $1.6954121794836243616166631538\dots$ comment: Real Period sage: E.period_lattice().omega() gp: if(E.disc>0,2,1)*E.omega[1] magma: (Discriminant(E) gt 0 select 2 else 1) * RealPeriod(E); Tamagawa product: $2$ = $2\cdot1$ comment: Tamagawa numbers sage: E.tamagawa_numbers() gp: gr=ellglobalred(E); [[gr[4][i,1],gr[5][i][4]] | i<-[1..#gr[4][,1]]] magma: TamagawaNumbers(E); oscar: tamagawa_numbers(E) Torsion order: $1$ comment: Torsion order sage: E.torsion_order() gp: elltors(E)[1] magma: Order(TorsionSubgroup(E)); oscar: prod(torsion_structure(E)[1]) Analytic order of Ш: $1$ ( rounded) comment: Order of Sha sage: E.sha().an_numerical() magma: MordellWeilShaInformation(E); Special value: $L'(E,1)$ ≈ $2.7665282028826950308323632667$ comment: Special L-value r = E.rank(); E.lseries().dokchitser().derivative(1,r)/r.factorial() gp: [r,L1r] = ellanalyticrank(E); L1r/r! magma: Lr1 where r,Lr1 := AnalyticRank(E: Precision:=12);
## BSD formula
$\displaystyle 2.766528203 \approx L'(E,1) = \frac{\# Ш(E/\Q)\cdot \Omega_E \cdot \mathrm{Reg}(E/\Q) \cdot \prod_p c_p}{\#E(\Q)_{\rm tor}^2} \approx \frac{1 \cdot 1.695412 \cdot 0.815887 \cdot 2}{1^2} \approx 2.766528203$
# self-contained SageMath code snippet for the BSD formula (checks rank, computes analytic sha)
E = EllipticCurve(%s); r = E.rank(); ar = E.analytic_rank(); assert r == ar;
Lr1 = E.lseries().dokchitser().derivative(1,r)/r.factorial(); sha = E.sha().an_numerical();
omega = E.period_lattice().omega(); reg = E.regulator(); tam = E.tamagawa_product(); tor = E.torsion_order();
assert r == ar; print("analytic sha: " + str(RR(Lr1) * tor^2 / (omega * reg * tam)))
/* self-contained Magma code snippet for the BSD formula (checks rank, computes analyiic sha) */
E := EllipticCurve(%s); r := Rank(E); ar,Lr1 := AnalyticRank(E: Precision := 12); assert r eq ar;
sha := MordellWeilShaInformation(E); omega := RealPeriod(E) * (Discriminant(E) gt 0 select 2 else 1);
reg := Regulator(E); tam := &*TamagawaNumbers(E); tor := #TorsionSubgroup(E);
assert r eq ar; print "analytic sha:", Lr1 * tor^2 / (omega * reg * tam);
## Modular invariants
$$q + q^{2} - 3 q^{3} + q^{4} + q^{5} - 3 q^{6} - 3 q^{7} + q^{8} + 6 q^{9} + q^{10} + 3 q^{11} - 3 q^{12} - 5 q^{13} - 3 q^{14} - 3 q^{15} + q^{16} - 3 q^{17} + 6 q^{18} + 7 q^{19} + O(q^{20})$$
comment: q-expansion of modular form
sage: E.q_eigenform(20)
\\ actual modular form, use for small N
[mf,F] = mffromell(E)
Ser(mfcoefs(mf,20),q)
\\ or just the series
Ser(ellan(E,20),q)*q
magma: ModularForm(E);
Modular degree: 5880
comment: Modular degree
sage: E.modular_degree()
gp: ellmoddegree(E)
magma: ModularDegree(E);
$\Gamma_0(N)$-optimal: no
Manin constant: 1
comment: Manin constant
magma: ManinConstant(E);
## Local data
This elliptic curve is not semistable. There are 2 primes of bad reduction:
prime Tamagawa number Kodaira symbol Reduction type Root number ord($N$) ord($\Delta$) ord$(j)_{-}$
$2$ $2$ $I_{2}$ Split multiplicative -1 1 2 2
$31$ $1$ $II$ Additive -1 2 2 0
comment: Local data
sage: E.local_data()
gp: ellglobalred(E)[5]
magma: [LocalInformation(E,p) : p in BadPrimes(E)];
oscar: [(p,tamagawa_number(E,p), kodaira_symbol(E,p), reduction_type(E,p)) for p in bad_primes(E)]
## Galois representations
The $\ell$-adic Galois representation has maximal image for all primes $\ell$ except those listed in the table below.
prime $\ell$ mod-$\ell$ image $\ell$-adic image
$2$ 2Cn 2.2.0.1
$7$ 7B 7.8.0.1
comment: mod p Galois image
sage: rho = E.galois_representation(); [rho.image_type(p) for p in rho.non_surjective()]
magma: [GaloisRepresentation(E,p): p in PrimesUpTo(20)];
gens = [[1, 0, 28, 1], [869, 28, 14, 393], [745, 28, 0, 1], [1303, 28, 882, 393], [1, 28, 0, 1], [1709, 28, 1708, 29], [1, 2, 14, 29], [15, 14, 1512, 1527], [698, 9, 357, 1627]]
GL(2,Integers(1736)).subgroup(gens)
Gens := [[1, 0, 28, 1], [869, 28, 14, 393], [745, 28, 0, 1], [1303, 28, 882, 393], [1, 28, 0, 1], [1709, 28, 1708, 29], [1, 2, 14, 29], [15, 14, 1512, 1527], [698, 9, 357, 1627]];
sub<GL(2,Integers(1736))|Gens>;
The image $H:=\rho_E(\Gal(\overline{\Q}/\Q))$ of the adelic Galois representation has level $$1736 = 2^{3} \cdot 7 \cdot 31$$, index $576$, genus $16$, and generators
$\left(\begin{array}{rr} 1 & 0 \\ 28 & 1 \end{array}\right),\left(\begin{array}{rr} 869 & 28 \\ 14 & 393 \end{array}\right),\left(\begin{array}{rr} 745 & 28 \\ 0 & 1 \end{array}\right),\left(\begin{array}{rr} 1303 & 28 \\ 882 & 393 \end{array}\right),\left(\begin{array}{rr} 1 & 28 \\ 0 & 1 \end{array}\right),\left(\begin{array}{rr} 1709 & 28 \\ 1708 & 29 \end{array}\right),\left(\begin{array}{rr} 1 & 2 \\ 14 & 29 \end{array}\right),\left(\begin{array}{rr} 15 & 14 \\ 1512 & 1527 \end{array}\right),\left(\begin{array}{rr} 698 & 9 \\ 357 & 1627 \end{array}\right)$.
Input positive integer $m$ to see the generators of the reduction of $H$ to $\mathrm{GL}_2(\Z/m\Z)$:
The torsion field $K:=\Q(E[1736])$ is a degree-$4799692800$ Galois extension of $\Q$ with $\Gal(K/\Q)$ isomorphic to the projection of $H$ to $\GL_2(\Z/1736\Z)$.
## Isogenies
gp: ellisomat(E)
This curve has non-trivial cyclic isogenies of degree $d$ for $d=$ 7.
Its isogeny class 1922e consists of 2 curves linked by isogenies of degree 7.
## Twists
This elliptic curve is its own minimal quadratic twist.
## Growth of torsion in number fields
The number fields $K$ of degree less than 24 such that $E(K)_{\rm tors}$ is strictly larger than $E(\Q)_{\rm tors}$ (which is trivial) are as follows:
$[K:\Q]$ $E(K)_{\rm tors}$ Base change curve $K$ $3$ 3.3.961.1 $$\Z/2\Z \oplus \Z/2\Z$$ Not in database $6$ 6.6.481170140857.2 $$\Z/7\Z$$ Not in database $8$ 8.2.31055528805552.6 $$\Z/3\Z$$ Not in database $12$ 12.4.53715159415573774336.57 $$\Z/2\Z \oplus \Z/4\Z$$ Not in database $18$ 18.18.111402774653210244885538776688002793.1 $$\Z/2\Z \oplus \Z/14\Z$$ Not in database
We only show fields where the torsion growth is primitive.
## Iwasawa invariants
$p$ Reduction type $\lambda$-invariant(s) $\mu$-invariant(s) 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 split ss ord ord ord ord ord ord ord ord add ord ord ord ord 2 1,1 1 1 1 1 1 1 1 1 - 3 1 1 1 0 0,0 0 0 0 0 0 0 0 0 - 0 0 0 0
An entry - indicates that the invariants are not computed because the reduction is additive.
## $p$-adic regulators
$p$-adic regulators are not yet computed for curves that are not $\Gamma_0$-optimal. | 3,520 | 9,037 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2024-18 | latest | en | 0.245945 |
https://www.mkamimura.com/2018/09/python_22.html | 1,591,204,452,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347435238.60/warc/CC-MAIN-20200603144014-20200603174014-00255.warc.gz | 785,892,395 | 23,792 | ## 2018年9月22日土曜日
### 数学 - Python - 線形代数学 - 群 – 群とその実例(有限群、直積、位数)
ラング線形代数学(下)(S.ラング (著)、芹沢 正三 (翻訳)、ちくま学芸文庫)の14章(群)、1(群とその実例)、練習問題7.を取り組んでみる。
1. 位数はmn。
$\begin{array}{}G=\left\{-1,1\right\}\\ G\text{'}=\left\{1,i,-1,-i\right\}\\ G×G\text{'}=\left\{\left(-1,1\right),\left(-1,i\right),\left(-1,-1\right),\left(-1,-i\right),\left(1,1\right),\left(1,i\right),\left(1,-1\right),\left(1,-i\right)\right\}\end{array}$
コード(Emacs)
Python 3
#!/usr/bin/env python3
from sympy import symbols, pprint, I
print('7.')
g1 = {-1, 1}
g2 = {1, I, -1, -I}
g12 = {(a, b) for a in g1
for b in g2}
print(g1)
for a in g1:
for b in g1:
print(f'{a * b}: {a * b in g1}')
print(g2)
for a in g2:
for b in g2:
print(f'{a * b}: {a * b in g2}')
print(g12)
for a1, b1 in g12:
for a2, b2 in g12:
print(f'{(a1 * a2, b1 * b2)}: {(a1 * a2, b1 * b2) in g12}')
$./sample6.py 7. {1, -1} 1: True -1: True -1: True 1: True {1, -I, -1, I} 1: True -I: True -1: True I: True -I: True -1: True I: True 1: True -1: True I: True 1: True -I: True I: True 1: True -I: True -1: True {(-1, 1), (-1, -I), (-1, I), (-1, -1), (1, I), (1, -I), (1, -1), (1, 1)} (1, 1): True (1, -I): True (1, I): True (1, -1): True (-1, I): True (-1, -I): True (-1, -1): True (-1, 1): True (1, -I): True (1, -1): True (1, 1): True (1, I): True (-1, 1): True (-1, -1): True (-1, I): True (-1, -I): True (1, I): True (1, 1): True (1, -1): True (1, -I): True (-1, -1): True (-1, 1): True (-1, -I): True (-1, I): True (1, -1): True (1, I): True (1, -I): True (1, 1): True (-1, -I): True (-1, I): True (-1, 1): True (-1, -1): True (-1, I): True (-1, 1): True (-1, -1): True (-1, -I): True (1, -1): True (1, 1): True (1, -I): True (1, I): True (-1, -I): True (-1, -1): True (-1, 1): True (-1, I): True (1, 1): True (1, -1): True (1, I): True (1, -I): True (-1, -1): True (-1, I): True (-1, -I): True (-1, 1): True (1, -I): True (1, I): True (1, 1): True (1, -1): True (-1, 1): True (-1, -I): True (-1, I): True (-1, -1): True (1, I): True (1, -I): True (1, -1): True (1, 1): True$ | 1,010 | 2,012 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2020-24 | longest | en | 0.259224 |
https://perkyhobby.com/weaving-beads/how-do-you-charge-for-embroidery-machines.html | 1,653,817,398,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663048462.97/warc/CC-MAIN-20220529072915-20220529102915-00601.warc.gz | 487,443,186 | 19,868 | # How do you charge for embroidery machines?
Contents
It is quite common for practitioners to charge an extra \$1.00 per 1,000 stitches. So, for example if the design has 4,500 stitches, the total charge would be a \$5.00 hooping charge + \$5 because the design contains between 4,000 and 5,000 stitches (rounding up). So, the total charge would be \$10.
## How do you price machine embroidery items?
For embroidery, the best choice of unit is stitch count. For example, your machine can produce from 20,000 to 30,000 stitches per hour, which means for 20,000 to 30,000, it costs \$20.67; so, for 1000 stitches, it cost between \$0.68 to \$1.03. Now, this should be your standard pricing per 1000 stitches.
## How much is 1000 stitches in embroidery?
2020 CONTRACT EMBROIDERY PRICE LIST*
Number of Stitches 7 or Under Pieces 15-29 Pieces
Up to 8000 \$6.25 \$3.25
Up to 9000 \$6.50 \$3.30
Up to 10000 \$6.75 \$3.35
Each Addtl. 1000 stitches \$0.45 \$0.23
## How do you calculate embroidery stitch count?
Calculating Stitch Count With Chart:
IT IS INTERESTING: What do infected stitches look like?
One solid square inch of embroidery = approximately 2,000 stitches. For example: 1” x 3” = approximately 6,000 stitches; or, 11/2” x 2” = approximately 6,000 stitches. One solid square 1/2 inch of embroidery = approximately 500 stitches.
## Can you make money with embroidery machine?
It’s true! PROFITABLE – the home based embroidery business can be VERY profitable! Many customers that start out with a 15 needle embroidery machine in their home end up with a 4 head and lots of customers over time. One of the reasons it is so profitable is the low cost of embroidery supplies.
## What is the average price for embroidery?
Embroiderers often charge by the number of stitches per garment as well as thread color changes. You could expect to pay anywhere from \$5-\$10 each for custom-embroidered ball caps to \$20-\$30 each for golf shirts, in addition to a digitizing fee.
## How do you calculate embroidery thread consumption?
By dividing the amount of thread by the seam length, we get the ratio of thread consumed. If we multiply this factor times the total length of seam, we can determine the total thread consumed for that seam. *Generally, 10% to 15% wastage of thread is added to the consumption derived.
## How much does it cost to embroider initials?
Monograms Pricing
Text height Limitations Price in store goods
1 ½ inch tall 3 characters in one line 8.95
1 ¾ inch tall to 2 inch tall 3 characters in one line 10.95
2 ¼ inch tall to 2 ½ inch tall 3 characters in one line 12.95
2 ¾ inch tall to 3 inch tall 3 characters in one line 14.95
## What is a tape charge for embroidery?
A tape charge, also known as a digitizing fee is the charge for making your design ready for embroidery. This charge is dependent on the number of stitches in your design and the overall complexity of the design, We can quote you on this charge after our embroidery department reviews your design.
## What is a high stitch count on an embroidery machine?
Well – considering that most embroidery designs are between 1,000 and 10,000 stitches – a machine can hit a million stitches pretty quickly. In fact, according to one embroidery machine owner, a machine is “just getting broken in” at 1 million stitches.
## How many stitches should a logo have?
Number of Stitches in Your Logo
Another factor to consider is how many stitches will need to go into making your logo. On average, a corporate logo uses between 6,000-8,000 stitches. Stitches take time, and most shops can produce around five embroidered items per hour at 8,000 each.
## What can I embroider to sell?
15 Vertical Market Ideas For Commercial Embroiderers.
• Medical – scrubs and lab coats.
• Uniforms – direct embroidery and/or patches, janitorial, auto, etc.
• Golf Specialty Items – towels, bags, flags, club sleeves.
• Schools – uniforms, sports uniforms, shorts, spirit wear, letterman jackets, cheer.
## How much money can an embroidery business make?
If you run your machine for 8 hours, that means you can make over \$2,000 every day (10 x 8 x \$26). Unfortunately not, because first off you need to pay yourself for those 8 hours, and you probably want to pay yourself a decent wage.
IT IS INTERESTING: Quick Answer: How fast can a person knit?
## Is embroidery an expensive hobby?
If you take embroidery (or any hobby) seriously, chances are, you’ve noticed that the costs associated with hand embroidery can range from negligible (when you’re just starting out) to pretty darned expensive (when you get to the point when you want to invest in good tools and supplies). | 1,140 | 4,670 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2022-21 | latest | en | 0.887188 |
https://mathoverflow.net/questions/354892/generalization-of-minimal-selection-theorem | 1,716,143,207,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057819.74/warc/CC-MAIN-20240519162917-20240519192917-00018.warc.gz | 360,654,585 | 23,884 | # Generalization of minimal selection theorem
Consider a metric space $$X$$ and a set-valued map $$F : X \to \mathbb{R}^{n}$$. We define the minimal selection $$\begin{equation*} m(F(x)) := \arg\min \big\{ \lvert u \rvert : u \in F(x) \big\}, \end{equation*}$$ where $$\lvert \, \cdot \, \rvert$$ denotes the 2-norm.
The minimal selection theorem in question can be stated as follows [e.g. Corollary 9.3.3 in Aubin and Frankowska]:
Let $$F$$ be a continuous (upper and lower semicontinuous) set-valued map from a metric space $$X$$ to $$\mathbb{R}^{n}$$ with nonempty closed convex images. Then the minimal selection is continuous.
My question is if there exist analogous theorems with $$m$$ replaced by $$\begin{equation*} m'(F(x)) := \arg\min \big\{ f(x,u) : u \in F(x) \big\}, \end{equation*}$$ where $$f$$ is a continuous function such that $$f(x, \, \cdot \,)$$ is strictly convex on $$F(x)$$? | 289 | 902 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 12, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2024-22 | latest | en | 0.744193 |
https://www.futurestarr.com/blog/mathematics/20-calculator-or | 1,657,189,039,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104690785.95/warc/CC-MAIN-20220707093848-20220707123848-00383.warc.gz | 834,676,985 | 30,668 | FutureStarr
20 Calculator or
## 20 Calculator or
### 20 Calculator
via GIPHY
Calculator is a large and exciting new platform created by Comix. This company is using this new platform to create a “unicorn” of the design world. With their Fall/Winter 2018 collection, Comix is marching to the beat of the newest, most innovative trends at the global level. The label’s design for the show that began in Paris has become even more timeless with the global success of the show.
### Use
This percentage calculator is a tool that lets you do a simple calculation: what percent of X is Y? The tool is pretty straightforward. All you need to do is fill in two fields, and the third one will be calculated for you automatically. This method will allow you to answer the question of how to find a percentage of two numbers. Furthermore, our percentage calculator also allows you to perform calculations in the opposite way, i.e., how to find a percentage of a number. Try entering various values into the different fields and see how quick and easy-to-use this handy tool is. Is only knowing how to get a percentage of a number is not enough for you? If you are looking for more extensive calculations, hit the advanced mode button under the calculator. Other than being helpful with learning percentages and fractions, this tool is useful in many different situations. You can find percentages in almost every aspect of your life! Anyone who has ever been to the shopping mall has surely seen dozens of signs with a large percentage symbol saying "discount!". And this is only one of many other examples of percentages. They frequently appear, e.g., in finance where we used them to find an amount of income tax or sales tax, or in health to express what is your body fat. Keep reading if you would like to see how to find a percentage of something, what the percentage formula is, and the applications of percentages in other areas of life, like statistics or physics. (Source:
This is all nice, but we usually do not use percents just by themselves. Mostly, we want to answer how big is one number in relation to another number?. To try to visualize it, imagine that we have something everyone likes, for example, a large packet of cookies (or donuts or chocolates, whatever you prefer 😉 - we will stick to cookies). Let's try to find an answer to the question of what is 40% of 20? It is 40 hundredths of 20, so if we divided 20 cookies into 100 even parts (good luck with that!), 40 of those parts would be 40% of 20 cookies. Let's do the math: Although Ancient Romans used Roman numerals I, V, X, L, and so on, calculations were often performed in fractions that were divided by 100. It was equivalent to the computing of percentages that we know today. Computations with a denominator of 100 became more standard after the introduction of the decimal system. Many medieval arithmetic texts applied this method to describe finances, e.g., interest rates. However, the percent sign % we know today only became popular a little while ago, in the 20th century, after years of constant evolution. (Source: www.omnicalculator.com)
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July 07, 2022 | Shaveez Haider
• #### How to Find 20 Percent of a Number
July 07, 2022 | Faisal Arman | 1,023 | 4,185 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2022-27 | latest | en | 0.948969 |
https://learn.careers360.com/school/question-please-solve-rd-sharma-class-12-chapter-indefinite-integrals-exercise-18-17-question-9-maths-textbook-solution/ | 1,713,498,977,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817253.5/warc/CC-MAIN-20240419013002-20240419043002-00776.warc.gz | 319,436,509 | 42,827 | # Get Answers to all your Questions
#### Please Solve RD Sharma Class 12 Chapter Indefinite Integrals Exercise 18.17 Question 9 Maths Textbook Solution
Answer:- $\frac{1}{\sqrt{5}}\log \left | \left ( x-\frac{1}{5} \right )+\sqrt{x^{2}-\frac{2x}{5}} \right |+c$
Hint: - To solve this problem, use special integration formula
Given:- $\int \frac{1}{\sqrt{5x^{2}-2x}}dx$
Solution:-
$Let\: I=\int \frac{1}{\sqrt{5x^{2}-2x}}dx=\int \frac{1}{\sqrt{5\left ( x^{2}-\frac{2}{5}x \right )}}dx$
\begin{aligned} &\Rightarrow I=\frac{1}{\sqrt{5}} \int \frac{1}{\sqrt{x^{2}-2 x \cdot \frac{1}{5}+\left(\frac{1}{5}\right)^{2}-\left(\frac{1}{5}\right)^{2}}} d x \\ &\Rightarrow I=\frac{1}{\sqrt{5}} \int \frac{1}{\sqrt{\left(x-\frac{1}{5}\right)^{2}-\left(\frac{1}{5}\right)^{2}}} d x \end{aligned}
\begin{aligned} &\text { put } \mathrm{x}-\frac{1}{5}=t \Rightarrow d x=d t \text { then } \\ &\mathrm{I}=\frac{1}{\sqrt{5}} \int \frac{1}{\sqrt{t^{2}-\left(\frac{1}{5}\right)^{2}}} d t \\ &I=\frac{1}{\sqrt{5}} \log \left|t+\sqrt{t^{2}-\left(\frac{1}{5}\right)^{2}}\right|+c \quad\quad\quad\quad\quad\quad\left[\because \int \frac{1}{\sqrt{x^{2}-a^{2}}} d x=\log \mid x+\sqrt{x^{2}-a|}+c\right] \end{aligned}
\begin{aligned} &I=\frac{1}{\sqrt{5}} \log \left|\left(x-\frac{1}{5}\right)+\sqrt{\left(x-\frac{1}{5}\right)^{2}-\left(\frac{1}{5}\right)^{2}}\right|+c \quad\quad\quad\quad\left(\because t=x-\frac{1}{5}\right) \\ &\Rightarrow I=\frac{1}{\sqrt{5}} \log \left|\left(x-\frac{1}{5}\right)+\sqrt{x^{2}-\frac{2 x}{5}+\frac{1}{25}-} \frac{1}{25}\right|+c \\ &\Rightarrow I=\frac{1}{\sqrt{5}} \log \left|\left(x-\frac{1}{5}\right)+\sqrt{x^{2}-\frac{2 x}{5}}\right|+c \end{aligned} | 709 | 1,673 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2024-18 | latest | en | 0.315076 |
https://math.stackexchange.com/questions/3144599/support-of-a-regularized-function | 1,555,764,076,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578529813.23/warc/CC-MAIN-20190420120902-20190420142902-00380.warc.gz | 484,011,356 | 31,326 | # Support of a regularized function
Let $$f$$ be a function such that $$supp(f)=K$$. Compute $$supp(f_\varepsilon)$$ where $$f_\varepsilon$$ is the regularization of $$f$$. Im not sure how to do this, since we have no information of the support of the convolution of two functions. We do know that $$supp(\omega_\varepsilon)=B(0,\varepsilon)$$ and $$\omega_\varepsilon=e^{\frac{-1}{1-||x/\varepsilon||^2}}$$ for all $$\|x\|<\varepsilon$$ and 0 otherwise.
The convolution is $$(f\ast \omega_{\varepsilon})(x)=\int\ f(y)\omega_{\varepsilon}(x-y)\ dy$$ Then note that $$f(y)\omega_{\varepsilon}(x-y)\neq 0$$ implies $$y\in{\rm supp}(f)$$ and $$x-y\in \bar{B}(0,\varepsilon)$$ the closed ball around he origin of radius $$\varepsilon$$. So the support of the convolution is contained in $${\rm supp}(f)+\bar{B}(0,\varepsilon)$$, i.e., the epsilon thickening of the support of $$f$$. I don't think one can in general compute the support of the convolution exactly because of possible sign cancellations. If $$f$$ and $$\omega_{\varepsilon}$$ are nonnegative then the above sum of sets is the support of the convolution. | 347 | 1,116 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 17, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2019-18 | latest | en | 0.806182 |
https://www.physicsforums.com/threads/question-about-my-reading.384314/ | 1,590,565,079,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347392141.7/warc/CC-MAIN-20200527044512-20200527074512-00348.warc.gz | 880,519,977 | 23,060 | ## Main Question or Discussion Point
I am reading Michio Kaku books and Brian Green Books and I am planing reading Stephen Hawking books and I have a couple of question please give me a link if it already have been answered.
1-If I move at 90% speed of light and trow something at 90% speed of light in the same direction that object should move approximately at 99% speed of light not 180%. Now my question is 2 object or person moving at 90% speed of light toward each other at witch speed will they see each other approaching (99% I think)? Also does an outside observer see them closing each other at 180% the speed of light ?
Also if I am right and the outside observer see them at 180% but they see themselves going at 99% if they hit each other at witch speed would you calculate the impact collision to determine the trajectory and speed of the debris from the impact?
2-I read that mass of the object increase with speed and near speed of light object reach near infinite mass that's why you cant pass the speed of light because it would be infinite that's why it's impossible according to Einstein. But I think the mass increase is relative right ? I mean the person or object does not see itself with a mass increase its just the observer. My question is what happen when you spin on yourself like tennis ball are you heavier too? If the ball rotate in perfect topspin from the point of view of the ball say bottom, top or side of the ball it may appear that you are in perfect no motion state and it is the world that is spinning around the ball so no mass increase from that point of view. But when the ball hit the racket the ball would be heavier say then a flat ball with no spin ? I understand that a tennis ball may not spin fast enough to feel the difference but every tennis player feel topspin ball like if they are much heavier when they hit the racket so I was wondering why ?
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Related Special and General Relativity News on Phys.org
The underlying concept for all of general relativity is that space-time is relative and dependent upon the frame of refrence from which thigns are observed. For your question above no matter which object or person you choose they will see the other approaching them at at what ever velocity they began with provided they aren't accelerating.
For what an outside observer would see consider the following equation:
$$\frac{v+u}{1+\frac{vu}{c^2}}$$
Where v is the speed of of the moving person and u is the speed of a second object projected in the same direction of motion. This is slightly different than your question but it still portrays accurately what happens. If you plug in your figures of .9c for the two objects, you will result in $$1.8c/1.8$$ which gives you c. In other words it would not add to 180% the speed of light and they are not approaching eachother at 180% the speed of light. As for relativistic mass the person would most definitely be able to perceive the increase in mass. Spinning objects follow the same rules no matter which frame is chosen to be stationary. I am pretty sure that the increased resistance of a tennis ball is soley due to the racket trying to change it's direction not due to the increase in relativistic mass, it is not spinning fast enough. I hope this helps.
Joe
bapowell
As for relativistic mass the person would most definitely be able to perceive the increase in mass.
Joe,
I'm a little confused by this. An object does not detect a change in its own mass based on its state of motion -- its mass is just its rest mass. Right?
$$\frac{v+u}{1+\frac{vu}{c^2}}$$
Where v is the speed of of the moving person and u is the speed of a second object projected in the same direction of motion. This is slightly different than your question but it still portrays accurately what happens. If you plug in your figures of .9c for the two objects, you will result in $$1.8c/1.8$$ which gives you c.
Small correction: It is 1.8c/1.81 which is less than c.
Let us say that A is moving relative to you at 0.9c and B is moving relative -0.9c also relative to you (in the opposite direction and towards A) then A will see B coming towards him at 1.8c/1.81 = aprox 0.99447c and B will see A coming towards him at aprox 0.99447c. No one actually measures any object moving at greater than c relative to the inertial reference frame they are at rest in.
As for mass, the modern opinion is that mass never changes and it is just that the equations of relativity are different to the equations of Newton. Just because in relativity, momentum is equal to y(mv) rather than the just mv in Newtonian mechanics does not mean the mass has increased by a factor of y. What it does tell us is that infinite energy is required to accelerate a particle with non zero rest mass to the speed of light. Now a ball with top spin has more energy than a non spinning ball with the same linear velocity as is obvious from the fact that additional energy is required to get the ball spinning and the additional momentum is a reflection of the additional energy that was put in the first place, but there is no actual change in the mass of the spinning ball.
Now if you are spinning on the spot with your arms outstretched, your hands will feel heavier than what you would predict using Newtonian equations for centrifugal force, but that is because the equations for force are different in relativity and the Newtonian equations are only an aproximation to reality.
I understand that a tennis ball may not spin fast enough to feel the difference but every tennis player feel topspin ball like if they are much heavier when they hit the racket so I was wondering why ?
Certainly, relativistic effects would be almost impossible to detect at the speeds involved in tennis. I magine when a ball with top spin hits a racket it is not only trying to move the racket backwards and stretch the strings, but is also effectively pushing the racket and the hand downwards as the spin momentum is absorbed. This would probably feel noticeably different from the strike of a non spinning tennis ball.
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JesseM
Joe,
I'm a little confused by this. An object does not detect a change in its own mass based on its state of motion -- its mass is just its rest mass. Right?
Sometimes people make use of a concept called relativistic mass which is distinct from rest mass (although they are equal in the object's own rest frame), though most physicists prefer to avoid this concept now for pedagogical reasons. Probably Joe was just saying that an object's relativistic mass could be measured by a separate observer who saw the object moving relative to himself, not that an object would measure its own relativistic mass to be different from its rest mass.
bapowell
Sometimes people make use of a concept called relativistic mass which is distinct from rest mass (although they are equal in the object's own rest frame), though most physicists prefer to avoid this concept now for pedagogical reasons. Probably Joe was just saying that an object's relativistic mass could be measured by a separate observer who saw the object moving relative to himself, not that an object would measure its own relativistic mass to be different from its rest mass.
Right. OK, thanks!
The underlying concept for all of general relativity is that space-time is relative and dependent upon the frame of refrence from which thigns are observed. For your question above no matter which object or person you choose they will see the other approaching them at at what ever velocity they began with provided they aren't accelerating.
For what an outside observer would see consider the following equation:
$$\frac{v+u}{1+\frac{vu}{c^2}}$$
Where v is the speed of of the moving person and u is the speed of a second object projected in the same direction of motion. This is slightly different than your question but it still portrays accurately what happens. If you plug in your figures of .9c for the two objects, you will result in $$1.8c/1.8$$ which gives you c. In other words it would not add to 180% the speed of light and they are not approaching eachother at 180% the speed of light. As for relativistic mass the person would most definitely be able to perceive the increase in mass. Spinning objects follow the same rules no matter which frame is chosen to be stationary. I am pretty sure that the increased resistance of a tennis ball is soley due to the racket trying to change it's direction not due to the increase in relativistic mass, it is not spinning fast enough. I hope this helps.
Joe
For the tennis ball is it the same effect as a gyroscopic effect ? Once I had a bicycle wheel full of lead in my hands in a science expo and as far as I remember it was there to show the effect of gyroscope when i sniped the wheel it was really hard to change its direction left and right.
This equation is for object moving toward each other or for the object i am throwing while moving or both ?
For the outside observer I find it really strange that 2 object moving at .9 c appear to move at c toward each other, I thought it would only be the case for the observer moving toward each other...
v+u = 1.8
vu = 0.81
c2 = 1
So 1.8 / (1+0.81/1) = 0.994475c = not c ??
or
v+u = 540 000 KM/s
vu = 72 900 000 000 KM/s
c2 = 90 000 000 000 Km/s
So 540 000/ (1+(72 900 000 000/90 000 000 000)=298342 Km/s
JesseM
v+u = 1.8
vu = 0.81
c2 = 1
So 1.8 / (1+0.81/1) = 0.994475c = not c ??
You're right, see kev's post...an object which has a v lower than c in one frame must have a velocity lower than c in any other frame too. However, if the velocity v of the moving object in the first frame is exactly c, in that case the velocity of the object in the second frame (the one that sees the first frame moving at velocity u relative to itself) would be exactly c too...for example, if v=1c and u=0.9c, then (v + u)/(1 + vu/c^2) = (1.9c)/(1 + 0.9) = 1c.
I just thought about something I was saying then when a ball spin then it could be argue that from the point of you of the ball its the world witch is spinning around but now I think this is not true because if you spin you feel the force centrifuge but if the world spin around you you don't feel that force. There is no symmetry or covariance I think ?
bapowell
I just thought about something I was saying then when a ball spin then it could be argue that from the point of you of the ball its the world witch is spinning around but now I think this is not true because if you spin you feel the force centrifuge but if the world spin around you you don't feel that force. There is no symmetry or covariance I think ?
This is the famous Newton's bucket thought experiment, and what motivated Mach's principle. You should research these ideas and see what you think of Mach's answer!
This is the famous Newton's bucket thought experiment, and what motivated Mach's principle. You should research these ideas and see what you think of Mach's answer!
I will do so eventually if I don't forget because I have at least 8 other book to read in my near listing of to read. It would be great if you could point me out to books internet article link or video (YouTube or torrent) I will do a few Google research myself...
This is the famous Newton's bucket thought experiment, and what motivated Mach's principle. You should research these ideas and see what you think of Mach's answer!
I just read Mach's principle if I understand correctly if the universe was empty there would not be any centrifuge force ?
JesseM
I just read Mach's principle if I understand correctly if the universe was empty there would not be any centrifuge force ?
That's what Mach's principle claims, though it doesn't seem to be true in general relativity (but Julian Barbour has some interesting speculations about how GR might be tweaked to produce a truly Machian theory)
That's what Mach's principle claims, though it doesn't seem to be true in general relativity (but Julian Barbour has some interesting speculations about how GR might be tweaked to produce a truly Machian theory)
So you say its not true in general relativity than what is true in general relativity ? What is the interpretation or prediction of general relativity about this ?
JesseM
So you say its not true in general relativity than what is true in general relativity ? What is the interpretation or prediction of general relativity about this ?
In GR you can have distinct solutions for a bucket in an otherwise empty ('asymptotically flat') universe...in one solution the bucket is not rotating as evidenced by the fact that the surface of the water remains flat and accelerometers at various positions on the bucket would register no G-forces, in another solution the bucket is rotating as evidenced by the fact that G-forces would be registered by accelerometers and the surface of the water becomes concave (the water pushed up along the sides by the apparent centrifugal force). For example, Brian Greene discusses this on p. 74 of his book Fabric of the Cosmos which can be viewed here on google books.
In GR you can have distinct solutions for a bucket in an otherwise empty ('asymptotically flat') universe...in one solution the bucket is not rotating as evidenced by the fact that the surface of the water remains flat and accelerometers at various positions on the bucket would register no G-forces, in another solution the bucket is rotating as evidenced by the fact that G-forces would be registered by accelerometers and the surface of the water becomes concave (the water pushed up along the sides by the apparent centrifugal force). For example, Brian Greene discusses this on p. 74 of his book Fabric of the Cosmos which can be viewed here on google books.
Ya I have that book I am finishing the elegant universe right now and was wondering witch book to read next between M.K. Parallel Worlds and B.G. The Fabric Of The Cosmos since I posses both does a particular logical order is better than another ?
Ok edit I just read p74 so according to G.R. you would still feel Centrifugal force in empty Universe but not all physicist Agree and as far as I understand there is no possible experiment to test this so I will need to read all of Julian Barbour works and book to get my own idea I guess...
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JesseM | 3,205 | 14,316 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2020-24 | longest | en | 0.949152 |
https://discuss.codecademy.com/t/a-day-at-the-supermarket-stocking-out-12-help/13477 | 1,556,075,480,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578624217.55/warc/CC-MAIN-20190424014705-20190424040705-00334.warc.gz | 379,903,372 | 4,527 | # A day at the supermarket-Stocking out(12) Help!
#1
shopping_list = [“banana”, “orange”, “apple”]
stock = {
“banana”: 6,
“apple”: 0,
“orange”: 32,
“pear”: 15
}
prices = {
“banana”: 4,
“apple”: 2,
“orange”: 1.5,
“pear”: 3
}
def compute_bill(shopping_list):
total = 0
for food in shopping_list:
if stock[food] > 0:
total += prices[food]
compute_bill(shopping_list)
#2
Using Markdown when posting code lets others see how well you have indented your code.
``````def compute_bill(shopping_list):
total = 0
for food in shopping_list:
if stock[food] > 0:
total += prices[food]
compute_bill(shopping_list)
``````
Notice where the the `return` statement is? It’s inside the IF code block, so exits the function on the first iteration of the loop.
``````def compute_bill(shopping_list):
total = 0
for food in shopping_list:
if stock[food] > 0:
total += prices[food]
stock[food] -= 1 # stock needs to be reduced by 1 sold item
#compute_bill(shopping_list)
``````
#3
it wants you too decrease ammounts if there is product so u can use this:
def compute_bill(shopping_list):
total = 0
for food in shopping_list:
if stock[food]>0:
total+=prices[food]
stock[food]=stock[food]-1
#4
``````def compute_bill(shopping_list):
total = 0
for food in shopping_list:
if stock[food] > 0:
total += prices[food] | 394 | 1,305 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2019-18 | latest | en | 0.827675 |
https://deepai.org/publication/pomset-logic-a-logical-and-grammatical-alternative-to-the-lambek-calculus | 1,638,081,502,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358469.34/warc/CC-MAIN-20211128043743-20211128073743-00046.warc.gz | 296,967,359 | 64,761 | Pomset logic: a logical and grammatical alternative to the Lambek calculus
Thirty years ago, I introduced a non commutative variant of classical linear logic, called POMSET LOGIC, issued from a particular denotational semantics or categorical interpretation of linear logic known as coherence spaces. In addition to the multiplicative connectives of linear logic, pomset logic includes a non-commutative connective, "<" called BEFORE, which is associative and self-dual: (A<B)^=A^ < B^ (observe that there is no swapping), and pomset logic handles Partially Ordered MultiSETs of formulas. This classical calculus enjoys a proof net calculus, cut-elimination, denotational semantics, but had no sequent calculus, despite my many attempts and the study of closely related deductive systems like the calculus of structures. At the same period, Alain Lecomte introduced me to Lambek calculus and grammars. We defined a grammatical formalism based on pomset logic, with partial proof nets as the deductive systems for parsing-as-deduction, with a lexicon mapping words to partial proof nets. The study of pomset logic and of its grammatical applications has been out of the limelight for several years, in part because computational linguists were not too keen on proof nets. However, recently Sergey Slavnov found a sequent calculus for pomset logic, and reopened the study of pomset logic. In this paper we shall present pomset logic including both published and unpublished material. Just as for Lambek calculus, Pomset logic also is a non commutative variant of linear logic — although Lambek calculus appeared 30 years before linear logic ! — and as in Lambek calculus it may be used as a grammar. Apart from this the two calculi are quite different, but perhaps the algebraic presentation we give here, with terms and the semantic correctness criterion, is closer to Lambek's view.
Authors
• 4 publications
• Proof nets for the Displacement calculus
We present a proof net calculus for the Displacement calculus and show i...
06/06/2016 ∙ by Richard Moot, et al. ∙ 0
• The Logic for a Mildly Context-Sensitive Fragment of the Lambek-Grishin Calculus
While context-free grammars are characterized by a simple proof-theoreti...
01/10/2021 ∙ by Hiroyoshi Komatsu, et al. ∙ 0
• Sequentialization for full N-Graphs via sub-N-Graphs
Since proof-nets for MLL- were introduced by Girard (1987), several stud...
03/01/2018 ∙ by Ruan V. B. Carvalho, et al. ∙ 0
• Provability in BI's Sequent Calculus is Decidable
The logic of Bunched Implications (BI) combines both additive and multip...
03/03/2021 ∙ by Alexander Gheorghiu, et al. ∙ 0
• On the Unity of Logic: a Sequential, Unpolarized Approach
The present work aims to give a unity of logic via standard sequential, ...
12/16/2019 ∙ by Norihiro Yamada, et al. ∙ 0
• Computational Aspects of the Calculus of Structure
Logic is the science of correct inferences and a logical system is a too...
01/21/2013 ∙ by Mário S. Alvim, et al. ∙ 0
• A theory of linear typings as flows on 3-valent graphs
Building on recently established enumerative connections between lambda ...
04/27/2018 ∙ by Noam Zeilberger, et al. ∙ 0
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1 Presentation
Lambek’s syntactic calculus as Lambek [16] used to call his logic, was in keeping with his preference for algebra (confirmed with his move from categorial grammars to pregroup grammars, which are not a logical system.) Up to the invention of linear logic in the late 80s, Lambek calculus was a rather isolated logical system, despite some study of frame semantics, which are typical of substructural logics.
Linear logic [7] arose from the study of the denotational semantics of system F, itself arising from the study of ordinals. [6] For interpreting systems F (second order lambda calculus) with variable types, one needed to refine the categorical interpretation of simply typed lambda calculus with Cartesian Closed Categories. In order to quantify over types Girard considered the category of coherence spaces (first called qualitative domains) with stable maps (which preserve directed joins and pullbacks). A finer study of coherence spaces led Girard to discompose the arrow type construction in to two steps: one is to contract several object of type into one (modality/exponential !) and the other one being linear implication (noted ) which rather corresponds to a change of state than to a consequence relation.
Linear logic was first viewed as a proof system (sequent calculus or proof nets) which is well interpreted by coherence spaces. The initial article [7] also included the definition of phase semantics, that resembles frame semantics developed for the Lambek calculus. It was not long before the connection between linear logic and Lambek calculus was found: after some early remarks by Girard, Yetter [50] observed the connection at the semantic level, while Abrusci [1] explored the syntactic, proof theoretical connection, while [34] explored proof nets and completed the insight of [46]. Basically Lambek calculus is non commutative intuitionistic multiplicative logic, the order between the two restrictions, intuitionistic and non commutative, being independent. An important remark, that I discussed with Lamarche in [15], says that non commutativity requires linearity in order to get a proper logical calculus.
Around 1988, my PhD advisor Jean-Yves Girard pointed to my attention a binary non commutative connective in coherence spaces. In coherence spaces, this connective has intriguing properties:
• is self dual , without swapping the two components — by we mean that there is a pair of canonical invertible linear maps between and .
• is non commutative
• is associative ;
• it lies in between the commutative conjunction and disjunction there is a canonical linear map from to an one from to 111Coherence spaces validate the mix rule ;
I designed a proof net calculus with this connective, in which a sequent, that is the conclusion of a proof, is a partially ordered multiset of formulas. This proof net calculus enjoys cut-elimination and a sound and faithful (coherence) semantics 222In the sense that having an interpretation is the same as being syntactically correct cf. theorem 8 preserved under cut elimination. I proposed a version of sequent calculus that easily translates into those proof nets and enjoys cut-elimination as well. [32] However despite many attempts by me and others (Sylvain Pogodalla, Lutz Straßburger) over many years we did not find a sequent calculus that would be complete w.r.t. the proof nets. Later on, Alessio Guglielmi, soon joined by Lutz Straßburger, designed the calculus of structures, a term calculus more flexible than sequent calculus (deep inference) with the before connective [9, 10, 11], a system that is quite close to dicograph rewriting [40, 41], described in section 3.2. They tried to prove that one of their systems called BV was equivalent to pomset logic and they did not succeed. As a reviewer of my habilitation [42] Lambek wrote:
He constructs a model of linear logic using graphs, which is new to me. His most original contribution is probably the new binary connective which he has added to his non commutative version of linear logic, although I did not find where it is treated in the sequent calculus. (J. Lambek, Dec. 3, 2001)
I deliberately omitted my work on sequent calculus in my habilitation manuscript, because none of the sequent calculi I experimented with was complete w.r.t. pomset proof nets which are ”perfect”, i.e. enjoy all the expected proof theoretical properties. In addition, by that time, I did not yet have a counter example to my proposal of a sequent calculus, the one in picture 5 of section 6 was found ten years later with Lutz Straßburger.
However, very recently, Sergey Slavnov found a sequent calculus that is complete w.r.t. pomset proof nets. [48] The structure of the decorated sequents that Slavnov uses is rather complex 333A decorated sequent according to Slavnov is a multiset of pomset formulas with binary relations between sequences of length of formulas from ; those relations are such that whenever the two sequences and have no common elements and entails for any permutation of – those relations correspond top the existence of disjoint paths in the proof nets from to . and the connective is viewed as the identification of two dual connectives one being more like a and the other more like a . As this work is not mine I shall not say much about it, but Slavnov’s work really sheds new light on pomset logic. Given the complexity of this sequent calculus it is pleasant to have some simple sequent calculus and a rewriting system for describing most useful proof nets e.g. the one used for grammatical purposes.
Pomset logic and the Lambek calculus systems share some properties:
• They both are linear calculi;
• They both handle non commutative connective(s) and structured sequents;
• They both have a sequent calculus;
• They both enjoy cut-elimination;
• They both have a complete sequent calculus (regarding pomset logic the complete sequent calculus is quite new);
• They both can be used as a grammatical system.
However Lambek calculus and pomset logic are quite different in many respects:
• Lambek calculus is naturally an intuitionistic calculus while pomset logic is naturally a classical calculus — although in both cases variants of the other kind can be defined.
• Lambek calculus is a restriction of the usual multiplicative linear logic according to which the connectives are no longer commutative, while pomset logic is an extension of usual commutative multiplicative linear logic with a non commutative connective.
• Lambek calculus deals with totally ordered multisets of hypotheses while pomset logic deals with partially ordered multisets of formulas. As grammatical systems, pomset logic allows relatively free word order, while Lambek calculus only deals with linear word orders.
• Lambek calculus has an elegant truth-value interpretation within the subsets of a monoid (frame semantics, phase semantics), while there is not such a notion for pomset logic.
• Lambek calculus has no simple concrete interpretation of proofs up to cut elimination (denotational semantics) while coherence semantics faithfully interprets the proofs of pomset logic.
This list shows that those two comparable systems also have many differences. However, the presentation of Pomset logic provided by the present article make Lambek calculus and pomset logic rather close on an abstract level. As he told many of us, Lambek did not like standard graphical or geometrical presentation of linear logic like proof nets. He told me several times that moving from geometry to algebra has been a great progress in mathematics and solved many issues, notably in geometry, and that proof net study was going the other way round. I guess this is related to what he said about theorem 8.
It seems that this ingenious argument avoids the complicated long trip condition of Girard. It constitutes a significant original contribution to the subject. (J. Lambek,Dec 3 2001)
This paper is a mix (!) of easy to access published work, [5, 34, 4, 37, 38, 35, 43] research reports and more confidential publications [32, 31, 5, 18, 34, 31, 33, 18, 19, 15, 20, 40, 41, 39, 42, 30] unpublished material between 1990 and 2020, that are all presented in the same and rather new unified perspective; the presented material can be divided into three topics:
proof nets
handsome proof nets both for MLL Lambek calculus and pomset logic, and other work on proof nets [5, 15, 34],
combinatorics
(di)cographs and sp orders [31, 32, 35, 4, 40, 41, 39, 42, 43, 42, 43, 30],
coherence semantics
[32, 33, 38, 42],
grammatical applications
of pomset logic to computational linguistics [18, 19, 20, 40, 41, 42].
The contents of the present article is divided into six sections as follows:
1. We first present results on series parallel partial orders, cographs and dicographs that subsumes those two notions and present dicograph either as sp pomset of formulas or as dicographs of atoms, and explain the guidelines for finding a sequent calculus. This combinatorial part is a prerequisite for the subsequent sections.
2. We then present proofs in an algebraic manner, with deduction rules as term rewriting.
3. Proof nets without links, the so called handsome proof nets, are presented as well as the cut elimination for them.
4. The semantics of proof nets, preserved under cut elimination and equivalent to their syntactic correctness is then presented.
5. Then the sequentialisation ”the quest” of a complete sequent calculus is discussed and we provide an example of a proof net that does not derive from any simple sequent calculus.
6. Finally we explain how one can design grammars by associating words with partial proof nets of pomset logic.
2 Structured sequents as dicographs of formulas
2.1 Looking for structured sequents
The formulas we consider are defined from atoms (propositional variables or their negation) by means of the usual commutative multiplicative connectives and together with the new non commutative connective (before)— the three of them are associative.
It is assumed that formulas are always in negative normal form: negation only apply to propositional variables; this is possible and standard when negation is involutive and satisfies the De Morgan laws:
(A⊥)⊥=A(AB)⊥=(A⊥⊗B⊥)(A
We want to deal with series parallel partial orders of formulas: corresponds to parallel composition of partial orders (disjoint union) and corresponds to the series composition of partial orders (every formula in the first partial order is lesser than every formula in the second partial order ). Thus, a formula written with and corresponds to a partial order between its atoms. Unsurprisingly, we firstly need to study a bit partial orders defined with series and parallel composition.
However, what about the multiplciative, conjunction namely the connecitve? It is commutative, but it is distinct from . In order to include in this view, where formulas are binary relations on their atoms, we consider, the more general class of irreflexive binary relations that are obtained by parallel composition, series composition and symmetric series compositions, which basically consists in adding the relations of and the ones of . The relations that are defined using , , are called directed cographs or dicographs for short.
If only and are used the relations obtained are cographs. They have already been quite useful for studying MLL, see e.g. theorem 4 thereafter.
Before defining pomset logic, we need a presentation of directed cographs.
2.2 Directed cographs or dicographs
An irreflexive relation may be viewed as a graph with vertices and with both directed edges and undirected edges but without loops. Given an irreflexive relation let us call its directed part (its arcs) and its symmetric part (its edges) . It is convenient to note for the edge or pair of arcs in and to denote for in when is not in .
We consider the class of dicographs, dicographs for short, which is the smallest class of binary irreflexive relations containing the empty relation on the singleton sets and closed under the following operations defined on two cographs with disjoint domains and yielding a binary relation on
• symmetric series composition
• directed series composition
• parallel composition
Whenever there are no directed edges (a.k.a. arcs) the dicograph is a cograph ( is not used). Cographs are characterised by the absence of as many people (re)discovered including us [35], see e.g. [14].
Whenever there are only directed edges (a.k.a. arcs) the dicograph is an sp order ( is not used) — as rediscovered in [32], see e.g. [24]
Let us call this class the class of dicographs.
We characterised the class of directed dicographs as follows [4, 40, 41]:
Theorem 1
An irreflexive binary relation is a dicographs if and only if:
• is N-free ( is an sp order).
• is -free ( is a cograph).
• Weak transitivity:
forall in the domain of
if and then and
if and then
A dicograph can be described with a term in which each element of the domain appears exactly once. This term is written with the three binary operators , and and for a given dicograph this term is unique up to the associativity of the three operators, and to the commutativity of the first two, namely and .
The dual of a dicograph on is defined as follows: points are given a superscript, and or , , , , .
Two points and of are said to be equivalent w.r.t. a relation whenever for all with one as and . There are three kinds of equivalent points:
• Two points and in a dicograph are said to be freely equivalent in a dicograph (notation ) whenever the term can be written (using associativity of and and the commutativity of ) . In other words, , , .
• Two points and in a dicograph are said to be arc equivalent in a dicograph (notation ) whenever the term can be written (using associativity of , and and the commutativity of and ) . In other words, , , .
• Two points and in a dicograph are said to be edge equivalent in a dicograph (notation ) whenever the term can be written (using associativity of and and the commutativity of ) . In other words, , , .
2.3 Dicograph inclusion and (un)folding
The order on a multiset of formulas, can be viewed as a set of contraints. Hence, when a sequent is derivable with an sp order it is also derivable with a sub sp order — we named this structural rule entropy [32]. Most of the transformations of a dicograph into a smaller (w.r.t. inclusion) dicograph preserve provability. Hence we need to characterise the inclusion of a dicograph into another and possibly to view the inclusion as a computational process that can be performed step by step. Fortunately,in [4] we characterised the inclusion of a dicograph in another dicograph by a rewriting relation:
Theorem 2
A dicograph is included into a dicograph if and only if the term rewrites to the term using the rules of figure 1 — up to the associativity of , and , and to the commutativity of and .
2.4 Folding and unfolding pomset logic sequents
A structured sequent of pomset logic (resp. of MLL) is a multiset of formulas of pomset logic (resp. of MLL) with the connectives endowed with a dicograph.
On such sequents one may define “folding” and “unfolding” which transform a dicograph of formulas into another dicograph of formulas by combining two equivalent formulas and of the dicograph into one formula (folding) or by splitting one compound formula into its two immediate subformulas and with and equivalent in the dicograph. More formally:
Folding
Given a multiset of formulas endowed with a dicograph ,
if in rewrite into — in the multiset, the two formulas and have been replaced with a single .
if in rewrite into — in the multiset, the two formulas and have been replaced with a single formula .
if in rewrite into — in the multiset, the two formulas and have been replaced with a single formula .
Unfolding
is the opposite:
turn into — in the multiset, the formula has been replaced with two formulas and with
turn into — in the multiset, the formula has been replaced with two formulas and with
turn into — in the multiset, the formula has been replaced with two formulas and with
2.5 A sequent calculus attempt with sp pomset of formulas
If we want to extend multiplicative linear logic with a non commutative multiplicative self dual connective (rather than to restrict existing connective to be non commutative), and want to handle partially ordered multisets of formulas, with corresponding to ”the subformula is smaller than the subformula ”.
That way one may think of an order on computations: a cut between and reduces to two smaller cuts and with the cut on being prior to the cut on , while a cut between and reduces to two smaller cuts and with the cut on being in parallel with the cut on . This makes sense when linear logic proofs are viewed as programs and cut-elimination as computation.
Doing so one may obtain a sequent calculus using partially ordered multisets of formulas as in [32] but if one wants a sequent with several conclusions that are partially ordered to be equivalent to a sequent with a unique conclusion, one has to only consider sp partial orders of formulas, as defined in subsection 2.2 with parallel composition noted and series composition noted .
If we want all formulas in the sequent to be ordered the calculus should handle right handed sequents i.e. be classical.444Lambek calculus is intuitionistic and when it is turned into a classical systems, formulas are endowed with a cyclic order,[50, 1, 15], i.e. a ternary relation which is not an order and which is quite complicated when partial — see the ”seaweeds” in [3].
As seen above, we can represent this sp partially ordered multiset of formulas endowed with an sp order by an sp term whose points are the formulas and such a term is unique up to the commutativity of and the associativity of and .
We know how the tensor rule and the cut rule behave w.r.t. formulas. The only aspect that deserves some tuning, is the order on the formulas after applying those binary rules. Our choice is guided by two independent criteria:
1. The resulting partial order should be an sp order.
2. This sequent calculus should enjoy cut elimination:
• If there is a cut between and with coming immediately from a rule from and and coming immediately from a rule from one should be able to locally turn those rules into two consecutive cuts, one between and and then one with .
• If there is a cut between and with coming immediately from a rule from and coming immediately from a rule from one should be able to locally turn those rules into two consecutive cuts, one between and and then one with .
A simple sequent calculus is provided in figure 2. 555An alternative solution to have on sp orders is to have rule between two minimum in their order component, and to have cut between two formulas one of which is isolated in its ordered sequent. This alternative is trickier and up to our recent investigation does not enjoy better properties than the version given above in figure 2
A property of this calculus is that cuts can be part of the order on conclusions. Indeed, one may define a cut as a formula that never is used as a premise of a logical rules. That way, the order can be viewed as an order on computation. A cut reduces into two cuts that are : , while cut reduces into two cuts that are : — beware that is a cut and that a operation on dicographs is different from the connective, which combines formulas. When one of the two premises of the cut is an axiom, this axiom and the cut are simply removed from the proof as usual, and this is possible because cut only applies when the two premises are isolated in the sp order. An alternative proof of cut elimination can be obtained from the cut elimination theorem for proof nets with links or without to be defined in sections 3.2 or 7.1, as established in [32, 37, 41] — because the reduction of a proof net coming from a sequent calculus proof also comes from a sequent calculus proof. Thus one has:
Theorem 3
Sequent calculus of figure 2 enjoys cut-elimination.
3 Pomset logic and MLL as (di)cograph rewriting:
Before we define a deductive system for pomset logic, let us revisit (as we did in [39, 43]) the deductive system of Multiplicative Linear Logic (MLL). Those results are highly inspired from proof nets, but once they are established they can be presented before proof nets are defined.
In this section a sequent is simply a dicograph of atoms which as explained above can be viewed using folding of section 2.4 as a dicograph of formulas or as an sp order between formulas depending on how many folding transformations and which one are performed.
Regarding, multiplicative linear logic (MLL), observe that is the largest cograph or even the largest dicograph w.r.t. inclusion that could be derived in MLL — indeed there cannot be any tensor link nor any before connection between the two dual occurrences of atoms issued from the same axiom link, for two reasons: first in sequent calculus they cannot lie in different sequents and therefore they cannot be conjoined by or ; second, as explained in subsection 3.2, in the proof net this would result in a prohibited (ae) cycle. Observe that , the largest derivable cograph in MLL is acutally derivable in MLL, hence in any extension of MLL:
\prooftree\prooftree\prooftree\prooftree\prooftree⊢a1,a1⊥\justifies⊢AX1:a1a1⊥\using\prooftree⊢a2,a2⊥\justifies⊢a2a2⊥\using\justifies⊢AX2:⊗1≤i≤2(aiai⊥)\using⊗\prooftree⊢a3,a3⊥\justifies⊢a3a3⊥\using\justifies⊢AX3:⊗1≤i≤3(aiai⊥)\using⊗\prooftree⊢a4,a4⊥\justifies⊢a4a4⊥\using\justifies⊢AX4:⊗1≤i≤4(aiai⊥)⋯\using⊗\justifies⊢AX5:⋯\using⊗
3.1 Standard multiplicative linear logic as cograph rewriting
In [39] we considered an alternative way to derive theorems of usual multiplicative linear logic MLL, by considering a formula as a binary relation, and more precisely as a cograph over its atoms, by viewing as and as . As there is no connective in linear logic the series composition is not used, and there is no sp order on conclusions.
Because of the chapeau of the present section 3 any sequent of MLL can be viewed is a cograph on atoms that is included into . Because of theorem 1, rewrites to using the rules of figure 1 that concern and i.e. , and . Observe that when viewed as a linear implication (considering the rules involving those two connectives), the first line is an incorrect linear implication, while is derivable in MLL and in MLL+MIX where the rule MIX is the one studied in [5], which also is derivable with :
\prooftree⊢Γ⊢Δ\justifies⊢Γ,Δ\usingMIX
Actually all tautologies of multiplicative linear logic MLL can be derived using from an axiom , and all tautologies of linear logic enriched with the MIX rule, MLL+MIX, can be derived by and (MIX).
Thus, we can define a proof system gMLL for MLL working with sequents as cographs of atoms as follows. Axioms are : (the two dual atoms are connected by an edge in a different relation called for axioms). There is just one deduction rule presented as a rewrite rule (up to commutativity and associativity): .
Let us call this deductive system gMLL (g for graph), then [39, 43] established that cograph rewriting is an alternative proof systems to MLL and MLL+MIX.
Theorem 4
MLL proves a sequent with atoms if and only if gMLL proves the unfolding of (the cograph of atoms corresponding to , that is the of the unforging of each formula in ), i.e. rewrites to using .
MLL+MIX proves a sequent with atoms if and only if gMLL+mix proves the unfolding of , i.e. rewrites to using and .
Proof. Easy induction on sequent calculus proofs see e.g. [39, 43]. A direct proof by Straßbruger can be found in [49].
The interesting thing is that all proofs can be transformed that way. Unfortunately it if much easier with an inductive definition of proofs like sequent calculus, and, unfortunately for pomset logic, it is hard to prove it directly on a non inductive notion of proof like proof nets.
Proposition 1
The calculi gMLL and gMLL+mix can safely be extended to structured sequents of formulas of MLL (not just atoms), i.e. cographs of MLL formulas with the rules of folding and unfolding with the same results.
Proof. This is just an easy remark, based on proof nets, which can be viewed as a consequence subsection 7.1.
3.2 Pomset logic as a calculus of dicographs: dicog-RS
Using the above results for MLL suggests defining a deductive system for pomset logic in the same manner. All rewriting rules are correct but : they correspond to proof nets or to sequent calculus derivations (with the sp-pomset sequent calculus of figure 2) and to canonical linear maps in coherence spaces. So it suggest that a rewriting system defined as gMLL+mix in the previous section (but with dicographs instead of cographs) might yield all the proofs we want e.g. all correct proof nets.
Axioms
is a tautology.
Rules
Whenever a dicograph of atoms which is a tautology rewrites to a dicograph (hence with the sames atoms) by any of the 10 rules , , , , , , , , , of figure 1 — i.e. all rules of figure 1 but .
Unfortunately, proving that all proof nets are derivable by rewriting is not simpler than proving that they can be obtained from the sequent calculus. This would entail the equivalence of pomset logic with BV calculus as discussed in [12].
3.3 Cuts
What about the cut rule? For such logical systems based on rewriting systems like gMLL(+MIX), of the dicog-RS view of pomset logic, which does not work with ”logical rules” in the standard sense, there are no binary rules that would combine a and a . So the only view of a cut is simply a tensor which never is inserted inside a formula. A dicograph may be written . Observe that contains the duals of the atoms in , because it is a cut, and that there is one for each , because they are the atoms of minus the well balanced atoms of and , one cannot say that the pair corresponds to some from — necessarily for some pair one is among the and and one is among the and .
However one cannot say that a proof of dicog-RS, i.e. a sequence of derivations yielding a dicograph with cuts (i.e. with a sub dicograph term ) may be turned into a dicog-RS derivation whose final dicographs is restricted to the atoms that are neither in not in . Indeed the atoms in and vanish during the process and none of the rewrite rules is able to do so — furthermore if one looks at step by step cut elimination, it precisely uses the prohibited rewriting rule !
We shall see later that in fact cut elimination holds for proof nets that are dicographs of atoms but without any inductive notion of derivation.
4 Proof nets
This section presents proof structures and nets (the correct proof structures), in an abstract and algebraic manner, without links nor trip conditions: such proof strctures and nets are called handsome proof structures and nets. Basically proof nets consists in a dicograph of atoms representing the conclusion formula, and axioms that are disjoint pairs of dual atoms constituting a partition of the atoms of . The proof net can be viewed as an edge bi-coloured graph: the dicograph is represented by arcs and edges (Red and Regular in the pictures), while the axioms (Blue and Bold in the picture). In such a setting, the correctness criterion expresses some kind of orthogonality between and . A proof net can also be viewed as a term, axioms being denoted by indices used exactly twice on dual atoms.
4.1 Handsome pomset proof nets
In fact, proof nets have (almost) been defined above!
A pomset logic handsome proof structure or dicog-PN is a dicograph over a (multi)set of atoms, , i.e. propositional letters and their duals. it is fairly possible that two atoms have the same name, i.e. it is a multiset of atoms. Let us call the binary relation or simply using the notations of the previous section 2.2; observe that no two edges are incident, and that each point is incident to exactly one edge in : the edges constitute a perfect matching of the whole graph with both edges and edges and arcs.
Given two proof structures and whose atoms and axioms are the same, and whose conclusion formulas and only differ because of the associativity of and the commutativity of , the proof structures and are equal — while in proof structures with links they would be different.
Correctness criterion 1
A handsome proof structure is said to be a proof net whenever every elementary circuit (directed cycle) of alternating edges in and in contains a chord — an edge or arc connecting two points of the circuit but not itself nor its reverse in the circuit. In short, every ae circuit contains a chord. Observe that this chord cannot be in , hence it is in , and it can either be an arc or an edge.
Theorem 5 (Nguyên)
Recently it was established that checking whether a proof structure satisfies the above correctness criterion is coNP complete [25].
Theorem 6
Given a proof net if (so ) using rewriting rules of Figure 1 except then is a proof net as well, i.e. all the rewrite rule preserve the correctness criterion on page 1.
Proof. See [40, 41].
It is easily seen that in general does not preserve correctness:
. Using , rewrites to
, and the proof net contains the chordless æ circuit .
Observe that it does not mean that every correct proof net with axioms can be obtained from by the allowed rewrite rules (all but ) where is . Indeed, since it is known that but one cannot tell that is not used. Indeed, as shown above does not preserve correctness but it may happen:
As indicated in section 2.2 we write for the edge or par of opposite arcs
is correct, and using it rewrites to which is correct as well.
4.2 Cut and cut-elimination
What about the cut rule? This calculus has no rules in the standard sense, in particular no binary rules that would combine a and a . A cut is a tensor which never is inserted inside a formula.
So a cut in this setting simply is a symmetric series composition in a dicograph whose form is . Assume the atoms of are so atoms of are . Cut-elimination consist in suppressing all edges and arcs between two atoms of , all edges and arcs between two atoms of , and all edges with — so the only edges incident to are (call those edges atomic cuts) and with neither in not in . If, in this graph, an atom is in the relation with an in , then the result of cut elimination is the closest point not in nor in reached by an alternating sequence of -edges and elementary cuts starting from – observed that this point is necessarily named , that we call its cut neighbour. To obtain the proof resulting from cut-elimination suppress all the atoms of and as well as the incident arcs and edges and connect every atom to its cut neighbour with a edge.
Theorem 7
Cut elimination preserves the correctness criterion of dicog-PN proof nets and consequently the f dicog-PN proof nets enjoy cut-elimination.
Proof. The preservation of the absence of chordless æ circuit during cut elimination is proved in [40, 41].
4.3 From sequent calculus and rewrite proofs to dicog-PN
Proofs of the sequent calculus given in figure 2 are easy turn into a dicog-PN proof net inductively. Such a derivation starts with axioms as it is well known, and in any kind of multiplicative linear logic the atoms and that can be traced from the axiom that introduced them to the conclusion sequent, which, after some unfolding can be viewed as a dicograph of atoms . The dicog-PN proof structure corresponding to the sequent calculus proof simply is , and fortunately is a correct proof net.
Proposition 2
A proof of sequent calculus corresponds to a dicog-PN i.e. to a handsome proof structure without chordless alternate elementary path, i.e. into a handsome proof net.
Proof. By induction on the proof, we showed in shown in [40, 41] that the neither the rules nor the the unfolding can introduce a chordless ae cycle.
The above result also yield cut elimination for the sequent calculus. Indeed, proof nets obtained by cut-elimination from a proof net issued from the sequent calculus also are issued from the sequent calculus.
The derivation by dicograph rewriting dicog-RS also only yield correct proof structures.
Proposition 3
Any proof obtained by rewriting from yields a handsome proof structure without chordless alternate elementary path, i.e. into a dicog-PN.
Proof. Observe that satisfies the criterion, so because of theorem 6, the result is clear.
5 Denotational semantics of pomset logic within coherence spaces
Denotational semantics or categorical interpretation of a logic is the interpretation of a logic in such a way that a proof of is interpreted as a morphism from an object to an object in such a way that whenever reduces to by (the transitive closure) of -reduction or cut-elimination. A proof of (when there is no ) is simply interpreted as a morphism from the terminal object to . More details can be found in [17, 8].
Once the interpretation of propositional variables is defined, the interpretation of complex formulas is defined by induction on the complexity of the formula. The set of morphisms from to is in bijective correspondence with an object written . Morphisms are defined by induction on the proofs and one has to check that the interpretations of a proof before and after one step of cut elimination is unchanged.
For intuitionistic logic, the category is cartesian closed, and for classical logic, at least simply, it is impossible666The fact that cartesian closed categories with involutive negation have at most one morphim between any two object is known as Joyal argument (see e.g. [17]); however there are complicated solutions like Selinger’s control categories [47] for classical deductive systems that ”control” the non determinism of classical cut elimination,like Parigot’s calculus, [26].. Regarding linear logic, a categorical interpetation takes place in a monoidal closed category (with monads for the exponentials of linear logic).
5.1 Coherence spaces
The category of coherence spaces is a concrete category: objects are (countable) sets endowed with a binary relation, and morphimms are linear maps. It interprets the proofs up to cut-elimination or reduction initially propositional intuitionistic logic and propositional linear logic (possibly quantified). Actually, coherence spaces are tightly related to linear logic: indeed, linear logic arose from this particular semantics, invented to model second order lambda calculus i.e. quantified propositional intuitionistic logic [6]. Coherence spaces are themselves inspired from the categorical work on ordinals by Jean-Yves Girard; they are the binary qualitative domains.
A coherence space is a set (possibly infinite) called the web of whose elements are called tokens, endowed with a binary reflexive and symmetric relation called coherence on noted or simply when is clear.
The following notations are common and useful:
iff and
iff or
iff and
A proof of is to be interpreted by a clique of the corresponding coherence spaces , a cliques being a set of pairwise coherent tokens in — we write for and for all . Observe that forall , if then . A linear morphism from to is a morphism mapping cliques of to cliques of such that:
• Let be a family of pairwise compatible cliques that is to say then .777The morphism is said to be stable when holds more generally for the union of a directed family of cliques of , i.e. .
• if then .
Due to the removal of structural rules, linear logic has two kinds of conjunction:
\prooftree⊢Γ,A⊢Δ,B\justifiesΓ,Δ,A⊗B\using⊗\prooftree⊢Γ,A⊢Γ,B\justifiesΓ,A&B\using&
Those two rules are equivalent when contraction and weakening are allowed. The multiplicatives (contexts are split, above) and the additives (contexts are duplicated, above). Regarding denotational semantics, the web of the coherence space associated with a formula with a multiplicative connective is the Cartesian product of the webs of and — while it is the disjoint union of the webs of and when is additive.
Negation is a unary connective which is both multiplicative and additive: and iff
One may wonder how many binary multiplicatives there are, i.e. how many different coherence relations one may define on from the coherence relations on and on .
We can limit ourselves to the ones that are covariant functors in both and — indeed there is a negation, hence a contravariant connective in is a covariant connective in . Hence when both components are so are the two couples, and when they are both coherent, so are the two couples.
To define a multiplicative connective, is to define when in function of and , so to fill a nine cell table — however if is assumed to be covariant in both its argument, seven out of the nine cells are filled.
A∗B\sincoh=\scoh\sincoh\sincoh\sincohNE?=\sincoh=\scoh\scohSW?\scoh\scoh
If one wants to be commutative, there are only two possibilities, namely () and ().
However if we don’t ask for the connective to be commutative we have a third connective (and actually a fourth connective which is )
AB\sincoh=\scoh\sincoh\sincoh\sincoh\scoh=\sincoh=\scoh\scoh\sincoh\scoh\scoh
This connective generalises to partial orders. Assume we have an sp order on the formulas can be defined with and — two tuples and of the web are strictly coherent whenever .
Linear implication, which can be defined as is : | 9,277 | 40,748 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2021-49 | longest | en | 0.928163 |
https://software.intel.com/content/www/us/en/develop/documentation/mkl-vsnotes/top/testing-of-basic-random-number-generators/brng-tests-description/2d-self-avoiding-random-walk-test.html | 1,603,601,992,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107887810.47/warc/CC-MAIN-20201025041701-20201025071701-00142.warc.gz | 544,267,104 | 18,319 | Contents
# Self-Avoiding Random Walk Test
Test Purpose
The test evaluates the randomness of the output vector of the generator. The stable response is the frequency of achieving the upper side of the lattice by the point walking randomly along the sites.
First Level Test
A random particle walks along the sites of a square lattice. With each new step, the particle moves in one of possible directions one step forward corner-wise. A square lattice has two types of sides: the lower and left-hand sides are totally reflecting, while the upper and right-hand sides are totally adsorbing. Reaching the lower and left-hand sides, the vector of the movement direction makes a 90-degree bend. The upper and right-hand sides adsorb the particle when it reaches them and the walking process completes. The particle starts its movement from the lower left-hand site of the lattice in the northeast direction. If the particle encounters an unvisited site, it changes the direction vector with a 1/2 probability clockwise or counter-clockwise by 90 degrees and continues the walking process. If the particle encounters an already visited site of the lattice, it defines the movement direction according to the conditions of inadmissibility of re-tracing at least a part оf the passed path.
Due to the symmetry of the task, either upper or the right-hand side should equiprobably adsorb the particle. The test determines the frequency of the achievement of the upper side of the lattice by the result of 500 iterations of the walking process. If
M
is the number of attempts when the particle reaches the upper side, then
K
= (2
M
- 500)/√
500
has the close to standard normal distribution
Φ(x)
. The result of the first level test is the p-value
p = Φ(K)
.
Second Level Test
The test performs the first level test ten times. The result of each iteration of the first level test is the p-value
pj
,
j
= 1, 2, ..., 10. The test applies the Kolmogorov-Smirnov goodness-of-fit test with Anderson-Darling statistics to the obtained p-values of
pj
,
j
= 1, 2, ..., 10. If the resulting p-value is
p
< 0.05 or
p
> 0.95, the test fails.
Final Result Interpretation
The final result of the test is the percentage FAIL of the failed second level tests. The test performs the second level test ten times. The acceptable result is the value of FAIL < 50%.
Tested Generators
Function Name
Application
vsRngUniform
applicable
vdRngUniform
applicable
viRngUniform
not applicable
viRngUniformBits
applicable
#### Product and Performance Information
1
Intel's compilers may or may not optimize to the same degree for non-Intel microprocessors for optimizations that are not unique to Intel microprocessors. These optimizations include SSE2, SSE3, and SSSE3 instruction sets and other optimizations. Intel does not guarantee the availability, functionality, or effectiveness of any optimization on microprocessors not manufactured by Intel. Microprocessor-dependent optimizations in this product are intended for use with Intel microprocessors. Certain optimizations not specific to Intel microarchitecture are reserved for Intel microprocessors. Please refer to the applicable product User and Reference Guides for more information regarding the specific instruction sets covered by this notice.
Notice revision #20110804 | 714 | 3,299 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2020-45 | latest | en | 0.847067 |
https://math.stackexchange.com/questions/2963755/applying-the-union-bound-in-a-conditional-probability-involving-a-stopping-time | 1,582,258,532,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145438.12/warc/CC-MAIN-20200221014826-20200221044826-00289.warc.gz | 450,370,127 | 31,269 | # Applying the Union Bound in a conditional probability involving a stopping time
Let $$b>0$$. Consider a sequence of independent and identically distributed random variables $$\{X_n\}_{n=1}^\infty$$ and the corresponding random walk $$S_n = \sum_{k=1}^n X_k$$. Define the stopping time $$\tau = \inf\{n\geq 1 : S_n <0\}$$. I am reading a paper and the following equations appear: $$$$\mathbb{P}\bigg\{ \max_{i=1,\dots,\tau }S_i \geq b\bigg\} = \mathbb{E}\bigg[\mathbb{P}\bigg\{\max_{i=1,\dots,\tau }S_i \geq b \bigg|\tau\bigg\}\bigg]\leq \mathbb{E} \bigg[ \sum_{i=1}^\tau \mathbb{P}_\infty \bigg\{ S_i \geq b\bigg\}\bigg]$$$$ I am not completely certain about the last inequality. The author states it follows because of the union bound but I am not sure how the conditioning on the knowledge of $$\tau$$ disappears. Thanks for the help!
$$\max_{i=1,2,..,\tau} S_i \geq b$$ iff $$S_i \geq b$$ for some $$i \leq \tau$$. So $$\{\max_{i=1,2,..,\tau} S_i \geq b\} \subset \cup_{i \leq \tau} \{S_i \geq b\}$$.
• I understood that the union bound was used in such a way, but I am not quite sure why there is no conditioning with respect to $\tau$ in the right hand side. – SpawnKilleR Oct 21 '18 at 0:16
• There is another expectation outside the conditional expectation. $E(E(Z|\tau)) =EZ$ for any random variable $Z$. – Kavi Rama Murthy Oct 21 '18 at 0:44
• Isn't this used on the equality? So basically the question is why there is not some form of conditioning in $\mathbb{E} \bigg[ \sum_{i=1}^\tau \mathbb{P}_\infty \bigg\{ S_i \geq b\bigg\}\bigg]$, whether in the middle term there is. – SpawnKilleR Oct 21 '18 at 1:11 | 564 | 1,621 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 10, "wp-katex-eq": 0, "align": 0, "equation": 1, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2020-10 | latest | en | 0.796175 |
https://oeis.org/A262622 | 1,529,804,639,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267865995.86/warc/CC-MAIN-20180624005242-20180624025242-00512.warc.gz | 690,895,291 | 3,767 | This site is supported by donations to The OEIS Foundation.
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A262622 Amicable pairs of even numbers. 1
220, 284, 1184, 1210, 2620, 2924, 5020, 5564, 6232, 6368, 10744, 10856, 17296, 18416, 63020, 76084, 66928, 66992, 79750, 88730, 122368, 123152, 141664, 153176, 142310, 168730, 171856, 176336, 176272, 180848, 185368, 203432, 196724, 202444, 280540, 365084, 308620, 389924, 319550, 430402, 356408, 399592, 437456, 455344 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS If there are no amicable pairs whose members have distinct parity then this is also the even terms of A259180. First differs from A063990, A259180, A259933 at a(13). First differs from A262624 at a(16). LINKS PROG (PARI) listap(nn) = {forstep(n=2, nn, 2, m = sigma(n)-n; if ((m > n) && (n==sigma(m)-m), print1(n, ", ", m, ", ")); ); } \\ Michel Marcus, Nov 14 2015 CROSSREFS Cf. A002025, A002046, A005843, A063990, A259180, A259933, A262623, A262624. Sequence in context: A259933 A273259 A262624 * A287026 A233538 A287011 Adjacent sequences: A262619 A262620 A262621 * A262623 A262624 A262625 KEYWORD nonn,tabf AUTHOR Omar E. Pol, Nov 09 2015 STATUS approved
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Last modified June 23 21:09 EDT 2018. Contains 311812 sequences. (Running on oeis4.) | 557 | 1,577 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2018-26 | latest | en | 0.61483 |
http://www.top-law-schools.com/forums/viewtopic.php?f=3&t=141064&start=3650&view=print | 1,542,590,208,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039744803.68/warc/CC-MAIN-20181119002432-20181119024432-00111.warc.gz | 518,209,343 | 4,532 | Page 147 of 159
Posted: Thu Jun 09, 2011 6:36 pm
dusk2k2 wrote:Anyone know what a 3.8ish would be at Minnesota? (Minnesota only releases quartiles). Does that give a chance at biglaw in this economy in both Minnesota and beyond?
Unless you have the most grossly inflated curve ever, you are solid.
Posted: Thu Jun 09, 2011 6:36 pm
dusk2k2 wrote:Anyone know what a 3.8ish would be at Minnesota? (Minnesota only releases quartiles). Does that give a chance at biglaw in this economy in both Minnesota and beyond?
tell us what the mean and the 75th quartile are, and i/someone else can tell you what a 3.8 is.
Posted: Thu Jun 09, 2011 6:40 pm
.
Posted: Thu Jun 09, 2011 6:43 pm
dusk2k2 wrote:
dailygrind wrote:
dusk2k2 wrote:Anyone know what a 3.8ish would be at Minnesota? (Minnesota only releases quartiles). Does that give a chance at biglaw in this economy in both Minnesota and beyond?
tell us what the mean and the 75th quartile are, and i/someone else can tell you what a 3.8 is.
Based on class of 2012's stats. 50th percentile was 3.267. 75th percentile was 3.547. Number 1 was apparently 4.080. I'm not too good with math, but based on other schools, our grades seem pretty inflated. First year courses require a class avg between 3.0 and 3.3 (which means every class is close to 3.3 avg.
Puts you at exactly top 10%. So accounting for unknowns you around top 10%.
Posted: Thu Jun 09, 2011 6:45 pm
dusk2k2 wrote:
dailygrind wrote:
dusk2k2 wrote:Anyone know what a 3.8ish would be at Minnesota? (Minnesota only releases quartiles). Does that give a chance at biglaw in this economy in both Minnesota and beyond?
tell us what the mean and the 75th quartile are, and i/someone else can tell you what a 3.8 is.
Based on class of 2012's stats. 50th percentile was 3.267. 75th percentile was 3.547. Number 1 was apparently 4.080. I'm not too good with math, but based on other schools, our grades seem pretty inflated. First year courses require a class avg between 3.0 and 3.3 (which means every class is close to 3.3 avg.
3.8 is slightly inside of top 10%. i gotta guess this puts you in a great spot for biglaw.
Posted: Thu Jun 09, 2011 6:46 pm
damn, too slow.
Posted: Thu Jun 09, 2011 6:47 pm
How are you guys doing this math? Is there a website out there than can do this?
Posted: Thu Jun 09, 2011 6:53 pm
it's basic statistics. you basically assume that the grades are normally distributed, and once you have the mean and a data point, you figure out the size of the standard deviations, then you figure out how many standard deviations your GPA is from the mean. it's not perfect, but it's the commonly accepted practice. no website that i know of does it automatically, though that doesn't mean it doesn't exist - the formula to do it is pretty straightforward, and capable of automation.
Posted: Thu Jun 09, 2011 6:55 pm
dailygrind wrote:damn, too slow.
TBF, I used a calculator I found online, not a Z table.
http://stattrek.com/Tables/Normal.aspx
Posted: Thu Jun 09, 2011 7:00 pm
I have no idea how it works, anyone willing to volunteer and do it for me?
Posted: Thu Jun 09, 2011 7:01 pm
dailygrind wrote:
dusk2k2 wrote:
dailygrind wrote:
dusk2k2 wrote:Anyone know what a 3.8ish would be at Minnesota? (Minnesota only releases quartiles). Does that give a chance at biglaw in this economy in both Minnesota and beyond?
tell us what the mean and the 75th quartile are, and i/someone else can tell you what a 3.8 is.
Based on class of 2012's stats. 50th percentile was 3.267. 75th percentile was 3.547. Number 1 was apparently 4.080. I'm not too good with math, but based on other schools, our grades seem pretty inflated. First year courses require a class avg between 3.0 and 3.3 (which means every class is close to 3.3 avg.
3.8 is slightly inside of top 10%. i gotta guess this puts you in a great spot for biglaw.
Ehhhh, considering that the latest NLJ250 for Minnesota puts them at just under 12% (and the latest Art III clerkship data puts them at 4%), I would say that biglaw is certainly within reach, but OP's probably not a "great spot." Pretty much just all assumptions, though, since I don't really know a whole lot about Minnesota Law or biglaw in Minnesota.
Posted: Thu Jun 09, 2011 7:02 pm
BarbellDreams wrote:I have no idea how it works, anyone willing to volunteer and do it for me?
sure.
Posted: Thu Jun 09, 2011 7:04 pm
Helmholtz wrote:Ehhhh, considering that the latest NLJ250 for Minnesota puts them at just under 12% (and the latest Art III clerkship data puts them at 4%), I would say that biglaw is certainly within reach, but OP's probably not a "great spot." Pretty much just all assumptions, though, since I don't really know a whole lot about Minnesota Law or biglaw in Minnesota.
rough. maybe they've got smallish firms paying close to market out there. i don't know crap about biglaw access outside of any school not my own, but not having a great shot at 10% sounds awful for a school ranked at 20.
Posted: Thu Jun 09, 2011 7:09 pm
Helmholtz wrote:Ehhhh, considering that the latest NLJ250 for Minnesota puts them at just under 12% (and the latest Art III clerkship data puts them at 4%), I would say that biglaw is certainly within reach, but OP's probably not a "great spot." Pretty much just all assumptions, though, since I don't really know a whole lot about Minnesota Law or biglaw in Minnesota.
I thought the same thing. So you need probably closer to 3.9 to get biglaw out of Minnesota. Really doesn't give much room to work with...
Posted: Thu Jun 09, 2011 7:12 pm
dailygrind wrote:
Helmholtz wrote:Ehhhh, considering that the latest NLJ250 for Minnesota puts them at just under 12% (and the latest Art III clerkship data puts them at 4%), I would say that biglaw is certainly within reach, but OP's probably not a "great spot." Pretty much just all assumptions, though, since I don't really know a whole lot about Minnesota Law or biglaw in Minnesota.
rough. maybe they've got smallish firms paying market out there. i don't know crap about biglaw access outside of any school not my own, but not having a great shot at 10% sounds awful for a school ranked at 20.
Yeah, the more "elite" regional public schools like Minnesota, IUB, Wisconsin, Iowa, Ohio State, UGA, UNC, and NYU* are terrific if you want to practice law in the state, but if you want to practice biglaw, they're kind of crappy places to be, mainly because (1) they don't really have any reach outside of the state and (2) there really are no biglaw options within the state—although there are certainly some. I know people at Ohio State and Iowa; I'm assuming that Minnesota is similar for job prospects, but maybe that's a faulty assumption on my part (although the data seems to back me up).
*lol
Posted: Thu Jun 09, 2011 8:03 pm
last grade is in... thankfully it washes out the taste that the dirty cuntracts exam left in my mouth
Posted: Thu Jun 09, 2011 8:33 pm
Borhas wrote:last grade is in... thankfully it washes out the taste that the dirty cuntracts exam left in my mouth
Still waiting. And grades were due Monday. I think I might have the last stat class still outstanding.
Congrats on finally being done!
Posted: Thu Jun 09, 2011 8:44 pm
autoding wrote:
missinglink wrote:Yeah. I have a decent idea, but published data only goes to top 15%. I don't know if what I've otherwise discovered from non- published sources will be good enough for admissions committees.
If they've taken transfers from your school before, they should have an idea, at least somewhat.
Good point.
If anyone wants to calculate rank, I'll post some numbers. I'm terrible at math so it's all voodoo magic to me.
Can someone calculate what the probable top 5%/top 10%/top 25% cutoffs would be at a school that changed its curve halfway through the year?
For the fall, I believe it was curved to a 2.89, spring was around 3.2?
A- and above: 10 to 20%
B+ and above: 25 to 35%
B and above: 40 to 50%
C+ and above: 75 to 85%
C and below: 15 to 25%
C- and below: 5 to 15%
D+ and below: 0 to 10%
A- and above: 15 to 25%
B+ and above: 40 to 50%
B and above: 70 to 80%
C+ and above: 85 to 95%
C and below 5 to 15%
C- and below 0 to 10%
I know that in the past, top 10% was around 3.4, and top 25% was around 3.2. I assume that the higher curve for spring will mostly just bunch everyone together more?
anyone?
Posted: Thu Jun 09, 2011 11:30 pm
Just heard my holdout professor won't have grades in until next week sometime. Over a week late.
Shoot.
Posted: Thu Jun 09, 2011 11:31 pm
missinglink wrote:Just heard my holdout professor won't have grades in until next week sometime. Over a week late.
Shoot.
Does your school penalize profs who don't make it in by the deadline?
Posted: Thu Jun 09, 2011 11:32 pm
Helmholtz wrote:
missinglink wrote:Just heard my holdout professor won't have grades in until next week sometime. Over a week late.
Shoot.
Does your school penalize profs who don't make it in by the deadline?
Does any school?
Posted: Thu Jun 09, 2011 11:33 pm
All grades coming out at once tomorrow. I wish I got the slow bleed out like you guys. It's going to be like getting kicked in the face.
Posted: Thu Jun 09, 2011 11:40 pm
Cupidity wrote:All grades coming out at once tomorrow. I wish I got the slow bleed out like you guys. It's going to be like getting kicked in the face.
I dunno, the slow bleed out is like torture. I was on a high from my first two grades that came crashing down with my third. Also, we have no idea when our final grade is even gonna be posted. Ugh.
Posted: Thu Jun 09, 2011 11:41 pm
starchinkilt wrote:
Cupidity wrote:All grades coming out at once tomorrow. I wish I got the slow bleed out like you guys. It's going to be like getting kicked in the face.
I dunno, the slow bleed out is like torture. I was on a high from my first two grades that came crashing down with my third. Also, we have no idea when our final grade is even gonna be posted. Ugh.
I just want median. I botched my crim law final something awful. I'm concerned I may end up in bottom 1/3 this semester. | 2,767 | 10,072 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2018-47 | latest | en | 0.963277 |
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## Video instructions and help with filling out and completing Which Form 4797 Expenditures
### Instructions and Help about Which Form 4797 Expenditures
In this talk, we'll go over an example of how to amortize start-up costs for your business. The standard rule for amortization is to divide the intangible asset cost basis by 180 months and then take an expense each month. However, start-up costs have additional rules that I will cover in this example. Start-up costs are the costs incurred before the business actually begins operating. These costs would normally be deductible as business expenses, but they cannot be deducted because the business has not yet begun operating. Instead, start-up costs are recorded as a current asset in the month the business begins. You can deduct up to \$5,000 of start-up costs in the first month. However, if the total start-up costs exceed \$50,000, the initial \$5,000 deduction must be reduced dollar for dollar by the amount that exceeds \$50,000. If the total start-up costs are greater than \$55,000, there is no initial deduction allowed. The remaining start-up costs can be amortized over 180 months. To calculate the monthly amortization expense, divide the total start-up cost by 180. This resulting amount is the monthly deduction that can be taken. Let's look at an example. Best Retail pays \$54,000 in start-up costs before opening their retail store in October 2014. In 2014, the total start-up costs are greater than \$50,000, so the initial deduction is limited to the amount exceeding \$50,000. Thus, the initial deduction for 2014 in the first month is \$1,000, calculated as \$5,000 minus \$4,000 (the amount exceeding \$50,000). The remaining start-up costs to be amortized then amount to \$53,000, calculated as \$54,000 minus the \$1,000 initial deduction. This \$53,000 will be amortized over 180 months, resulting in a monthly amortization expense of \$294.44. For 2014, we will take three months of amortization expense for October, November, and December. Multiplying the monthly amortization expense by... | 499 | 2,183 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-26 | latest | en | 0.958707 |
http://www.ck12.org/trigonometry/Alternate-Formula-for-the-Area-of-a-Triangle/lesson/Alternate-Formula-for-the-Area-of-a-Triangle/r4/ | 1,455,264,005,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701163512.72/warc/CC-MAIN-20160205193923-00314-ip-10-236-182-209.ec2.internal.warc.gz | 342,740,626 | 33,653 | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are viewing an older version of this Concept. Go to the latest version.
# Alternate Formula for the Area of a Triangle
## Area equals half the product of two sides and the sine of the included angle.
Estimated11 minsto complete
%
Progress
Practice Alternate Formula for the Area of a Triangle
Progress
Estimated11 minsto complete
%
Alternate Formula for the Area of a Triangle
You are studying the Gulf of Mexico in your Geography class. Your Instructor brings up the idea of the Bermuda Triangle. This is a place where, according to some, many planes get lost. Here is a picture of it:
photocredit, www.earthspots.com
The first thing this makes you think of is your math class, since that class is your favorite. You would like to know just how big the Bermuda Triangle is. Unfortunately, the Bermuda Triangle isn't a right triangle. However, you do know that the lengths of one of the sides is 950 miles, the other side is 975 miles, and the angle between them is $60^\circ$ .
Is there any way to use this information to help you find out just how big the Bermuda Triangle is?
Read on, and at the end of this Concept, you'll be able to use the information presented here to calculate the area of the Bermuda Triangle.
### Guidance
In Geometry, you learned that the area of a triangle is $A=\frac{1}{2}bh$ , where $b$ is the base and $h$ is the height, or altitude. Now that you know the trig ratios, this formula can be changed around, using sine.
Looking at the triangle above, you can use sine to determine $h, \sin C = \frac{h}{a}$ . So, solving this equation for $h$ , we have $a\sin C=h$ . Substituting this for $h$ , we now have a new formula for area.
$A = \frac{1}{2} ab \sin C$
What this means is you do not need the height to find the area anymore. All you now need is two sides and the angle between the two sides, called the included angle.
#### Example A
Find the area of the triangle.
Solution: Using the formula, $A = \frac{1}{2} \ ab \sin C$ , we have
$A & = \frac{1}{2} \cdot 8 \cdot 13 \cdot \sin 82^\circ\\& = 4 \cdot 13 \cdot 0.990\\& = 51.494$
#### Example B
Find the area of the parallelogram.
Solution: Recall that a parallelogram can be split into two triangles. So the formula for a parallelogram, using the new formula, would be: $A = 2 \cdot \frac{1}{2} \ ab \sin C$ or $A = ab \sin C$ .
$A & = 7 \cdot 15 \cdot \sin 65^\circ\\& = 95.162$
#### Example C
Find the area of the triangle.
Solution: Using the formula, $A = \frac{1}{2} \ ab \sin C$ , we have
$A & = \frac{1}{2} \cdot 16.45 \cdot 19 \cdot \sin 30^\circ\\& = 8.225 \cdot 19 \cdot 0.5\\& = 78.14$
### Vocabulary
Triangle Area Formula: The triangle area formula is a formula to find the area of a triangle involving the lengths of two sides of the triangle and the sine of the angle between them.
### Guided Practice
1. Find the area of the triangle.
2. Find the area of the triangle.
3. Find the area of the triangle.
Solutions:
1. Using the formula, $A = \frac{1}{2} \ ab \sin C$ , we have
$A & = \frac{1}{2} \cdot 4 \cdot 10 \cdot \sin 15^\circ\\& = 2 \cdot 10 \cdot 0.2589\\& = 5.178$
2. Using the formula, $A = \frac{1}{2} \ ab \sin C$ , we have
$A & = \frac{1}{2} \cdot 8 \cdot 15 \cdot \sin 25^\circ\\& = 4 \cdot 15 \cdot 0.4226\\& = 25.356$
3. Using the formula, $A = \frac{1}{2} \ ab \sin C$ , we have
$A & = \frac{1}{2} \cdot 10 \cdot 11 \cdot \sin 32^\circ\\& = 5 \cdot 11 \cdot 0.53\\& = 29.15$
### Concept Problem Solution
Now that you know the equation for the area of a triangle in terms of two of the sides and the included angle, we can use that to solve for the area of the Bermuda Triangle:
$A = \frac{1}{2} ab\sin \theta\\A = \frac{1}{2} (950)(975)\sin 60^\circ\\A = \frac{1}{2} (950)(975)(.866)\\A = 401066.25$
The area of the triangle is approximately 401,006 square miles.
### Practice
Use the following picture for questions 1 and 2.
1. Find the values of a, b, and C needed for the formula to find the area of the triangle.
2. Now find the area of the triangle.
Use the following picture for questions 3 and 4.
1. Find the values of a, b, and C needed for the formula to find the area of the triangle.
2. Now find the area of the triangle.
Find the area of each triangle below.
Find the area of each parallelogram.
1. Describe another way you could have found the area of the parallelogram in the previous problem.
2. When you first learned about sine, you learned how it worked for right triangles. Explain why this method for calculating area uses sine, but works for non-right triangles.
### Vocabulary Language: English
Included Angle
Included Angle
The included angle in a triangle is the angle between two known sides.
sine
sine
The sine of an angle in a right triangle is a value found by dividing the length of the side opposite the given angle by the length of the hypotenuse. | 1,433 | 4,989 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 23, "texerror": 0} | 4.78125 | 5 | CC-MAIN-2016-07 | latest | en | 0.896653 |
https://www.ic.sunysb.edu/Class/phy141md/doku.php?do=export_s5&id=phy131studiof15:lectures:chapter3 | 1,603,523,058,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107882102.31/warc/CC-MAIN-20201024051926-20201024081926-00534.warc.gz | 780,011,619 | 3,129 | Vectors and scalars
Vector quantities with number, direction and units:
• Displacement $\vec{r}$ [m]
• Velocity $\vec{v}$ [ms-1]
• Acceleration $\vec{a}$ [ms-2]
Scalar quantities number and units only
• Distance traveled [m]
• Speed [ms-1]
Graphical representation of vectors and components
It is frequently useful to draw two dimensional vectors as arrows, and to split them in to components that lie along the coordinate axes. The choice of coordinate axes is up to you..but choosing the right ones will make the problem easier or harder.
We can take a look at the acceleration due to gravity as vector using a phone accelerometer, using this tool (click on the link from your phone's browser).
Unit Vectors
It can be useful to express vector quantities in terms of unit vectors. These are dimensionless vectors of length = 1 that point along the coordinate axes. They are usually denoted with carets (hats), i.e. $(\hat{i},\hat{j},\hat{k})$
For example:
$\vec{v}\,\mathrm{ms^{-1}}=v_{x}\,\mathrm{ms^{-1}}\,\hat{i}+v_{y}\,\mathrm{ms^{-1}}\,\hat{j}+v_{z}\,\mathrm{ms^{-1}}\,\hat{k}$
or
$\vec{r}\,\mathrm{m}=x\,\mathrm{m}\,\hat{i}+y\,\mathrm{m}\,\hat{j}+z\,\mathrm{m}\,\hat{k}$
Vectors and motion
$\vec{r_{1}}=x_{1}\,\hat{i}+y_{1}\,\hat{j}+z_{1}\,\hat{k}$
$\vec{r_{2}}=x_{2}\,\hat{i}+y_{2}\,\hat{j}+z_{2}\,\hat{k}$
$\Delta\vec{r}=\vec{r_{2}}-\vec{r_{1}}=(x_{2}-x_{1})\,\hat{i}+(y_{2}-y_{1})\,\hat{j}+(z_{2}-z_{1})\,\hat{k}$
Average velocity: $\vec{v_{ave}}=\frac{\Delta\vec{r}}{\Delta t}$
Instantaneous velocity: $\vec{v}=\frac{d\vec{r}}{dt}=\frac{dx}{dt}\,\hat{i}+\frac{dy}{dt}\,\hat{j}+\frac{dz}{dt}\,\hat{k}=v_{x}\,\hat{i}+v_{y}\,\hat{j}+v_{z}\,\hat{k}$
Average acceleration: $\vec{a_{ave}}=\frac{\Delta\vec{v}}{\Delta t}$
Instantaneous acceleration: $\vec{a}=\frac{d\vec{v}}{dt}=\frac{dv_x}{dt}\,\hat{i}+\frac{dv_y}{dt}\,\hat{j}+\frac{dv_z}{dt}\,\hat{k}=\frac{d^{2}x}{dt^2}\,\hat{i}+\frac{d^{2}y}{dt^2}\,\hat{j}+\frac{d^{2}z}{dt^2}\,\hat{k}$
Vectors - Components
The angles $\theta_{1}$ and $\theta_{2}$ are defined with respect to the positive $x$ axis, ie. $\theta_{1}$ is negative and $\theta_{2}$ is positive.
$v_{1x}=v_{1}\cos\theta_{1}$ $v_{2x}=v_{2}\cos\theta_{2}$ $v_{1y}=v_{1}\sin\theta_{1}$ $v_{2y}=v_{2}\sin\theta_{2}$ $v_{Rx}=v_{1x}+v_{2x}$ $\tan\theta_{R}=\frac{v_{Ry}}{v_{Rx}}$ $v_{Ry}=v_{1y}+v_{2y}$ $v_{R}=\sqrt{v_{Rx}^2+v_{Ry}^2}$
Vectors - Multiplication by a scalar
Multiplication of a vector by a scalar can change the magnitude, but not the direction of the vector, ie. each component of the vector is multiplied by the scalar in the same way. | 949 | 2,595 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2020-45 | latest | en | 0.673109 |
https://www.effortlessmath.com/tag/create-a-proportion/ | 1,618,245,691,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038067870.12/warc/CC-MAIN-20210412144351-20210412174351-00021.warc.gz | 860,295,278 | 7,327 | # Create a Proportion
## How to Create a Proportion
Proportion simply means that two ratios are equal. Creating a proportion means creating two equal ratios or fractions.... | 36 | 175 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2021-17 | longest | en | 0.891815 |
http://priorart.ip.com/IPCOM/000104373 | 1,516,639,017,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891485.97/warc/CC-MAIN-20180122153557-20180122173557-00708.warc.gz | 255,674,855 | 8,165 | Browse Prior Art Database
# Input Separation Based on Sub-Cones for Delay Test
IP.com Disclosure Number: IPCOM000104373D
Original Publication Date: 1993-Apr-01
Included in the Prior Art Database: 2005-Mar-19
Document File: 4 page(s) / 105K
IBM
## Related People
Patil, S: AUTHOR [+2]
## Abstract
Disclosed is a procedure to increase the delay fault coverage for combinational logic circuits by using a technique called input separation based on sub-cones.
This text was extracted from an ASCII text file.
This is the abbreviated version, containing approximately 52% of the total text.
Input Separation Based on Sub-Cones for Delay Test
Disclosed is a procedure to increase the delay fault coverage
for combinational logic circuits by using a technique called input
separation based on sub-cones.
The proposed method provides an alternative to the method
presented in [1] in cases where one is not allowed to use dummy
latches. The proposed method is based on fault simulation of the
logic circuits which allows one to concentrate on only those portions
of the circuit where undetectable transition faults exist.
Before discussing the actual algorithm, the method using a
simple example will be illustrated. Consider the circuit shown in
Fig. 1 which is a cascade of three AND gates. This example can be
shown to belong to a family of worst-case examples for transition
fault coverages using skewed load. Now consider the scan path
ordering of inputs to be 1 > 2 > 3 > 4. Denote a slow-to-rise
fault by A and a slow-to-fall fault by V. This is actually the
worst possible ordering for this circuit and 5 of the 14 faults turn
out to be undetectable (1 A, 2 A, 3 A, 5 A and 6 A) giving a fault
coverage figure of 64%. If inputs are considered to be at level 1,
there are 3 undetectable faults at level 1, 1 at level 2 and 1 at
level 3. A plot of undetectable faults versus logic are shown in
Fig. 2. This will be called the undetectability profile. Since all
5 undetectable faults of interest occur at level 3 or lower, we are
going to 'splice' the circuit at level 3 as shown by the vertical
line in Fig. 1. Secondary outputs were then created wherever the
imaginary line cuts - in the case of our example, at lines 4 and 6.
For the purpose of re-ordering, it was assumed that there are outputs
at lines 4 and 6. Thus, inputs 1, 2 and 3 depend on output 6, input
4 depends on output 4. Using the greedy method given in [1], and
without using any dummy latches, the result is the scan path ordering
1 > 4 > 2 > 3 which gives a coverage 79%. This figure is also the
maximum possible coverage for the example circuit without the use of
dummy latches. Since this algorithm is heuristic in nature, it is
not always guaranteed to find the optimal ordering.
The actual algorithm is outlined below:
1. First, levelize or rank-order the circuit. Set the iteration
count iter to 1. Assume the maximum allowable iterations are
given by iter sub 'max'.
2. Perform transition fault simulation on the circuit using an
efficient fault simulation algorithm like PPSFP [2] and assuming
some random ordering of inputs in the scan chain and skewed load
of vector pairs.
3. Plot the undetectability profile (similar to the plot shown in
Fig. 2).
4. Let L sub 'max' be the maximum number of logic levels in the
... | 843 | 3,340 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2018-05 | latest | en | 0.845487 |
http://www.convertit.com/Go/ConvertIt/Measurement/Converter.ASP?From=em&To=length | 1,660,307,010,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571692.3/warc/CC-MAIN-20220812105810-20220812135810-00617.warc.gz | 69,225,320 | 3,831 | Partner with ConvertIt.com
New Online Book! Handbook of Mathematical Functions (AMS55)
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Please read our Help Page and FAQ Page then post a message or send e-mail. Thanks! | 470 | 1,702 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2022-33 | latest | en | 0.69176 |
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This Touch Math file includes 3 worksheets for 3 digit addition with regrouping, and 3 worksheets for 2 digit addition with regrouping. Answer keys included. Great for those visual learners! Each page has 9 problems each.
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This is a set of 12 worksheets designed to help students UNDERSTAND the three digit subtraction with regrouping using blocks. Included: * 6 worksheets with regrouping once: students will have to add, then circle ten ones to show how they regrouped them as one ten * 6 worksheets with regrouping t
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56 ratings
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EIGHT EASY TO ASSEMBLE GAMES! Just copy onto tag board and your math centers for the week are complete. High Rollers Dice Game Problem Solving Mini-Booklet (Froggy's Flower Farm) Regrouping Mats: Students will solve problems and SORT their cards onto the appropriate place on the mat. (color and b
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100 ratings
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Are you trying to make learning about 3 digit addition fun? Are your kids tired of just doing worksheets? These activities are hands-on and kids are learning while having fun! ***8 Games*** -Tic Tac Toe- 8 boards -Race to 10,000- Add cards together, 1st to 10,000 wins -War- highest or lowest sum
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35 ratings
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PDF (12.93 MB)
3 Digit Addition WITH Regrouping (24 Task cards)-FREE This is a colorful set of 24 task cards to give students practice on three digit addition WITH regrouping.This set is a wonderful addition to your lessons! Great way for students to practice this adding skill! I've included a recording sheet
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FREE
64 ratings
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This is a set of three digit addition with regrouping story (word) problems. On each page the student: -reads the problems and solves showing work -explains in writing how they solved using key math vocabulary such as addend, sum, ones place, tens place, regroup, etc -writes the problem as a number
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\$3.00
95 ratings
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PDF (1.21 MB)
The activities are great to practice adding three-digit numbers with regrouping. I use these activities as math stations in my classroom. These pack contains two activities & student answer sheets. 1. Three-digit addition match-up 2. Three-digit addition story problems
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Each one of these 30 autumn-themed addition with regrouping cards has a 3 by 3 digit problem. These cards can be used as task cards or as a class scoot activity. Print these cards on heavy card stock and laminate for a reusable resource. This bundle includes: - 30 3 digit by 3 digit addition task
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4 ratings
4.0
PDF (3.7 MB)
#### Videos matching "3 digit addition with regrouping" (BETA)
Addition Detective with Regrouping **This is part of my 2nd Grade Math Bundle , which is a better value than buying each piece separately!** INCLUDED ARE 2 PRINTABLES AND 2 ANSWER KEYS Each Printable Covers: *3-Digit addition problems *Regrouping in ones and tens place *Transfer letter from each
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86 ratings
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PDF (3.74 MB)
Practice addition with regrouping with the fun competition. Students compete against one another in this fun game while practicing their regrouping skills. You can have up to 4 teams. There are 20 questions in all and you just click on each question to go to it. The question disappears after you've
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81 ratings
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PPTX (2.66 MB)
Fun Fall Math Worksheet / 3 Digit Addition with Regrouping! Below are links to some of my other items that might interest you. Thanks for looking! Addition and Subtraction Review With Regrouping Worksheets Multiplication Bingo Cards / I Know How To Multiply Back to School Word Search Puzzle /
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71 ratings
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This packet includes 10 worksheets for students to practice 3 Digit Addition with REGROUPING. Each worksheets has 15 problems. ★★Check the following categories to view my other related packets:★★ 2-Digit Addition and Subtraction 3-Digit Addition and Subtraction Math Task Cards Pet-Themed Math Task
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63 ratings
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PDF (930.79 KB)
Three digit addition with regrouping features in this 5 game pack. Each addition game requires players to add 3 digit numbers and record the equation they have made. Just print each game with its illustrated rules and recording sheet (there is no cutting involved) and place in your math center. Afte
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\$5.00
55 ratings
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PDF (3.27 MB)
Math Centers 2nd-3rd grade *Great 4th grade review 3 digit addition with regrouping 4 differentiated math centers 4 answer keys common core aligned 41 pages *Task Cards with Multiple Choice *Task Cards (No multiple choice) *To Regroup or Not Regroup Sorting Center *Matching Center *Scoot Option, Sm
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\$3.50
51 ratings
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PDF (6.08 MB)
This is a set of pages for those whose students are ready to practice in 3 digit addition. This method of partial sums addition, which I prefer to use - has more students grasping the place value focus. You break it apart horizontally, in hundreds, tens and then ones then vertically add each place
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48 ratings
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ZIP (156.72 KB)
Three Digit Addition with Regrouping tiny task cards. These cards can be cut out and used just like normal task cards. They can also be printed, stapled, and sent home for extra practice or used during RTI time since they are black and white and only one page. They also make for great interactive
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\$1.00
47 ratings
4.0
PDF (3.11 MB)
This is a game created for partners or centers during math to practice addition WITH regrouping. Students will answer math problems, and move a game piece around the board. Game includes: 1 game board, 48 addition problem game cards, 12 bonus game cards that say: move forward, backward, or create
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\$3.25
34 ratings
4.0
PDF (585.88 KB)
3 Digit Addition - with Regrouping Practice Worksheets. Includes ten worksheets with 20 problems per page, Keys included. You can print one on one side for class practice and one on the back for homework practice. Supports common core. This pack is included in my bundle: Addition Bundle BUY T
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\$2.25
45 ratings
4.0
PDF (3 MB)
This set is geared towards students who have mastered addition without regrouping. To incorporate the use of multiple strategies, students can work on this activity at a center with access to hundreds charts, manipulative, number lines, and white boards. For your convenience, both black and whit
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40 ratings
4.0
PDF (3.05 MB)
3 Digit Addition: These 3 digit addition cootie catchers are a great way for students to have fun while they practice their skills with addition. How to Play and Assembly Instructions are included. 3 Digit Addition Cootie Catchers Contents: There are 2 cootie catchers in this product, each one h
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9 ratings
4.0
PDF (357.22 KB)
This is a step by step lesson of three-digit addition with regrouping using place value. I'm a combo teacher so I did this lesson with both my 2nd and 3rd graders. This Lesson may be done over 3 to 4 days. Lesson 2 day 1 may take from 60-75 mins. Independent practice could be used for assessme
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\$3.55
9 ratings
3.9
PDF (4.02 MB)
This cute packet includes the following addition activities: *two digit no regrouping worksheet *two digit no regrouping worksheet key *two digit regrouping worksheet *two digit regrouping worksheet key *three digit no regrouping worksheet *three digit no regrouping worksheet key *three digit regr
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\$1.00
28 ratings
3.9
ZIP (3.08 MB)
This document was created to be used as extra practice, homework, pre-assessment or end of unit assessment for 3 digit addition with regrouping. The document has multiple choice(with distractors), short answer and word problems. There are also vertical and horizontal equations(the student will hav
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\$2.00
31 ratings
4.0
PPTX (98.87 KB)
This winter-themed addition activity has students solving three digit by three digit addition problems and matching them with their corresponding answer puzzle pieces. Print the puzzle pieces on heavy card stock and laminate for a reusable resource. Cut the puzzle pieces prior to giving to them to
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CCSS:
\$3.00
5 ratings
4.0
PDF (1.9 MB)
My students are loving addition practice with these fun printables. Students solve an addition problem and then use a maze to travel to the next problem. My intervention students have been asking to do more of these! When students who struggle in math ask for more math it makes my heart smile. :)
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\$1.00
30 ratings
4.0
PDF (4.18 MB)
“Santa Claus Math” 3 Digit Addition With Regrouping - Common Core - Math Fun! (black line printable version) ****THIS PRODUCT INCLUDES BLACK LINE PRINTABLE PAGES- 9 pages**** ******************************************************************** This identical product is also available as a COLOR on
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\$1.60
23 ratings
4.0
PDF (1.1 MB)
This product contains 22 three digit addition worksheets. The worksheets will ask the students to solve three digit addition problems with regrouping. 9 DIFFERENT TYPES of WORKSHEETS! Happy teaching! Dana's Wonderland
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\$3.00
25 ratings
4.0
PDF (1.51 MB)
Let your kids practice their math skills with these printer friendly task cards! Three-Digit Addition with REGROUPING! Copy and laminate the cards. A student answer sheet is included. Task cards can be lots of fun and you can use them with the class over and over. Give each student a card to sol
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CCSS:
\$1.50
25 ratings
4.0
PDF (2 MB)
This is a fun way to get your students to practice regrouping with three digits. Each time the solve a problem they get the answer to part of a riddle. My students love when they get to do this for math practice time. They even ask for more. I also have used these for homework. I hope your stud
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\$1.00
19 ratings
4.0
PDF (60 KB)
Easily-prepared highly-effective tool for reviewing 3 digit addition! In this packet, you will get 24 task cards, answer key, and recording sheets. I've included two formats of recording sheets: (1) for 24 equations and (2) for 12 equations each that provides more space for students to solve and wri
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\$2.00
23 ratings
4.0
PDF (3.26 MB)
This set includes the following: *Instructions *24 3 Digit Addition With Regrouping QR Code Cards in color and in black & white *The same 24 3 Digit Addition Task Cards without QR Codes in color and in black & white *Answer sheet *Answer key *3 Digit Addition Worksheet with a QR Code f
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22 ratings
4.0
PDF (9.82 MB)
Make consolidating three digit addition with regrouping fun and engaging. Children love completing addition with regrouping riddles and you'll love that it's self-correcting. Add the addition board game to your math centers and you'll have a fun and motivating way to keep your kids practicing regro
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CCSS:
\$4.00
16 ratings
4.0
PDF (2.48 MB)
Just in time for Easter, here is a great product to help your students practice addition with regrouping. Each of the children on the cards is looking for a special egg. To figure out which egg belongs to him or her, the children are given an addition problem. The hidden eggs are printed with a
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\$1.50
11 ratings
4.0
PDF (1.51 MB)
Addition with Regrouping is a difficult skill for some students. It's a skill that needs to be reviewed over and over again. These math task cards for addition with regrouping will provide students practice with this skill. This set includes 32 cards with 3 digit numbers, a student recording sheet,
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Includes: 6 Three in a Row Cards 24 Three Digit Addition Cards 1 Direction Sheet 1 Cover Sheet Students take a three in a row card. They all solve the problem and cover the answer if it is on their three in a row card. The first one to get three in a row wins!
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I needed a blank template for students to use at their seat while I helped other students at my table. It is a worksheet for the students to generate their own digits to add with and without regrouping. This template lends itself to differentiation in the classroom. 2.NBT.5 "I can add and subtrac
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“Winter Snowman Math” 3 Digit Addition With Regrouping - Common Core - Math Fun! (black line printable version) ****THIS PRODUCT INCLUDES BLACK LINE PRINTABLE PAGES- 9 pages**** ******************************************************************** This identical product is also available as a COLOR
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“Spring Garden Math” 3 Digit Addition With Regrouping - Common Core - Math Fun! (black line printable version) ****THIS PRODUCT INCLUDES BLACK LINE PRINTABLE PAGES- 9 pages**** ******************************************************************** This identical product is also available as a COLOR
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Two worksheets that focus on 3 digit addition with regrouping in the tens and ones places. Each worksheet has 8 equations and 1 problem solving story problem at the end. Standards: 2NBT5 2NBT7 2OA1 ************************************ SAVE when you buy this worksheet as a bundle 3 Digit Addition!
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This engaging PowerPoint makes solving 3-digit word problems with regrouping in the one's place a cinch! Students pretend they are doctors as they "perform surgery" to solve each word problem. This presentation includes the learning target, I can statement, 5 easy steps on how to solve word problems
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We're solving math problems using QR codes for the holidays! This Christmas QR code hunt has 28 different problems for the students to solve. This activity can be used in your math center or as a hunt around the school. Our second graders have these QR codes posted in our hallway. They use the recor
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A packet of two worksheets. Each worksheet contains 8 three digit equations in which the student may or may not have to regroup in the ones place. At the end of each worksheet, there is also a story problem involving three digit addition. Standards: 2NBT5 2NBT7 2OA1 ******************************
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This set of task cards includes 24 task cards for your kids to practice 3-digit vertical addition with regrouping. Task cards allows your students to work independently, in small groups, or in a whole class activity. **Also includes the answer key and a recording sheet. Click here to check my othe
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In this cut and paste activity, students practice their three-digit addition with regrouping by solving addition problems and matching them with the correct sum. The finished product is an adorable train spanning over two feet long! They make great focal points on a bulletin board or can even be
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“Winter Snowman Math” 3 Digit Addition With Regrouping - Common Core - Math Fun! (color and black line versions) ****THIS PRODUCT INCLUDES 18 PAGES (9 full color and 9 blackline – same problems)**** ******************************************************************** This identical product is als
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The set has 10 animal-themed worksheets to practice 3-digit addition with regrouping. Sheets are neatly formatted. ~~Answer keys are included.~~ Use for morning work, independent practice, assessment, or homework. Looking for 3 Digit Addition with Regrouping Worksheets in a similar format? Click
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I love using these Math Scavenger Hunt cards in my classroom - and better yet - my students LOVE them too! It works perfectly for our Daily 3: Math with Someone. I put students in pairs, and give them a mini clipboard, and they are off and running. They begin at the START card, solve the problem a
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"SCOOT" games are so engaging and fun for students. Use this game to practice 28 three-digit addition problems with regrouping of tens and ones. A recording sheet and answer key are provided with this purchase. This game does not have a particular background theme so it is perfect to use anytime dur
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Adding 2 and 3 Digit Numbers (with regrouping) Math Center Adding 2 and 3 digit numbers with regrouping is the focus of this math center. A recording sheet and an illustrated rules page are provided. You have the choice of printing in full color, limited color or black ink only. This game is als
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This 3 Digit by 3 Digit with Regrouping Bingo game is a great way to refresh students on skills learned so far. Print the cards on any printer since they come as black and white copies! Print as many as you need-there are 25 cards with each problem on them! Students then solve the problems as par
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Find the Tic Tac Toe is a problem solving game meant for independent play. The student solves each problem to the find sum that is found three times in a row; across, down, or diagonally . After finishing each board the student writes down the winning number of the game on the line. Since each page
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***This item is included in the 3 and 4 Digit Addition Task Card Bundle>*** One of the most important skills for a student to have is the ability to add numbers together. This set of task cards provides 24 problems for students to practice addition of three digit numbers with regrouping. There
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Related searches for 3 digit addition with regrouping
showing 1-52 of 4,836 results
Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. | 4,671 | 18,504 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2018-13 | longest | en | 0.880856 |
https://web2.0calc.com/questions/inscribed-angles-2 | 1,524,579,686,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125946721.87/warc/CC-MAIN-20180424135408-20180424155408-00118.warc.gz | 714,640,592 | 6,224 | +0
# inscribed angles 2
0
82
2
+88
Quadrilateral ABCD is inscribed in this circle.
What is the measure of angle B?
AgentFatboy Mar 23, 2018
#1
+12235
+2
OPPOSITE angles of a cyclic quadrilateral ( a quadrilateral with all vertices on the circumference of a given circle) add to 180 degrees.....s o
x + (4x-20) = 180
5x=200
x = 40
B = 4x-20
4(40) - 20 = 140 degrees
ElectricPavlov Mar 23, 2018
Sort:
#1
+12235
+2
OPPOSITE angles of a cyclic quadrilateral ( a quadrilateral with all vertices on the circumference of a given circle) add to 180 degrees.....s o
x + (4x-20) = 180
5x=200
x = 40
B = 4x-20
4(40) - 20 = 140 degrees
ElectricPavlov Mar 23, 2018
#2
0
The opposite angles of a cyclic quadrilateral (a quadrilateral whose vertices touch the circumference of a circle) always add up to 180°.
Thus:
$$x+4x-20=180$$
$$5x-20=180$$
$$5x=180+20=200$$
$$x=\frac { 200 }{ 5 } =40$$
$$\angle B=(4x-20)°=(4\times 40-20)°=(160-20)°=140°$$
So, the measure of angle B is 140°.
Guest Mar 24, 2018
### 30 Online Users
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners. See details | 446 | 1,278 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2018-17 | longest | en | 0.672652 |
https://boards.straightdope.com/t/does-physics-ever-predict-the-distance-between-two-objects-will-be-pi/540509 | 1,708,791,948,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474541.96/warc/CC-MAIN-20240224144416-20240224174416-00774.warc.gz | 132,158,759 | 10,193 | # Does Physics ever predict the distance between two objects will be pi?
Is there ever a case in which our current best physics would predict, of a situation involving only rational measurements in its initial state, that after some rational interval, there will be a pair of objects the distance between which is exactly pi? (Any measurement units should do here I think, as long as they’re used consistently through the whole description of the scenario.)
At first I thought this would have to be answered “no” since pi is transcendental and everything starts out with rational values. (I didn’t have a strict argument for this “no” answer, it was just a hunch.) Then I thought that for all I know, there are physical laws that can introduce transcendental values like pi into mechanical calculations, in which case, for all I know, you might start out with all rational values and end up with transcendental ones.
Anyway, hopefully the question makes sense. If not, please ask clarifying questions.
(Also, the straight dope is awesome for being a place I think it plausible to ask questions like this. I’m curious to know of other places where laymen can ask halfway informed and possibly dumb physics and math questions like this.)
You could say it happens all the time.
It depends on what your units are. You don’t say if it is yards, or miles, or parsecs or whatever. So, pick any two objects and just define the distance between them as pi [imaginary units].
Voila!
I doubt it. Pi is a ratio so there is no reason to expect it to come up as the distance between two things in our arbitrarily-chosen systems of measurement.
I have no idea, but I like the question! And related ones that are likely to evolve from this.
I’m not sure the question is well-founded. In particular, I would think that by redefining our units of measurement, we could easily make something pi units away.
But even before that, consider this. Let’s say you and I start out 1 meter apart and then increase our distance until we’re 4 meters apart. At some point, we must have been pi meters apart.
ETA: Amazing. The OP sits for 20 minutes with no answers, and then suddenly there’s three answers! I’ll leave my post regardless, although it’s now pretty well redundant.
I thought about this (hence the parenthetical in the OP about how I think any unit will do) but I didn’t make it clear how I intended to head off this concern.
Here’s what I mean to ask. Can you have an initial state where all measurements are rational, where length is measured in some unit we’ll call “u-units”, and where a prediction of physics is that after an amount of time with a rational measurement, some pair of objects will have a distance that is equal to pi u-units?
If I’m reading this right, you’re asking if for, say, a universe in which space is quantized, with a minimum distance of a planck length, could a process put two objects a distance pi apart.
If that is the correct interpretation, then the answer is no. Pi assumes a theoretical infinite division. It cannot be reached by rational intervals in a finite time.
But you know that since it was in your OP. I’m trying to wiggle that around to get at your question, but I think all formulas that put pi into them assume mathematical infinities.
Well, as I said directly above your post:
Let’s say you and I start out 1 meter apart and then increase our distance until we’re 4 meters apart. At some point, we must have been pi meters apart.
I believe this satisfies your criteria, no?
A key part of the OP’s criteria was “After a rational amount of time”. Your example would not work if you and I were moving apart at a rational speed, since, in that case, the time it will take for us to be π meters apart will be irrational.
To the OP:
What do you mean to say with “all measurements are rational”? For example, it’s not possible to have a square all of whose corners are a rational distance (in whatever units) from each other (since the distance between opposite corners will be sqrt(2) times as large as the distance between adjacent corners); are we therefore prohibited from considering situations involving squares?
More to the point, of course, any Euclidean circle induces two lengths which are in a ratio of 2π, by definition: its circumference and its radius. Surely you aren’t going to prohibit situations which involve circles? If not, naturally, even though the circumference’s length is not directly the distance between two points, it’s easy to use this to construct a distance ratio of 2π. If you do prohibit circular objects, circular motion, and all the rest on grounds such as “Well, sure, in a Newtonian world, but with quantum uncertainty blah blah…”, the difficulty becomes, if objects do not have a definite position, there aren’t definite distances between them either, so it no longer makes sense to ask if the distance between them is exactly this or that.
If we’re talking about sufficient (that is to say, infinite) precision to tell the difference between rational and irrational numbers, then we run into the problem that there’s no such thing as a precisely-defined position, and hence no such thing as a precisely-defined distance.
I don’t think this satisfies the OP’s question, since for all anyone knows the exact moment that you were pi meters apart is a trancendental measument of time.
Could you come up with a position (or velocity or acceleration) equation which is itself polynomial or rational and which at time T where (T IS Rational), position is Pi?
If you simply defined a constant velocity of 1 unit/second from units 0 to units 10, then Pi seconds later (T=Pi), distance will be Pi, but the OP is not looking for that since he doesn’t want a transcendental T and Pi seconds is trancendental.
Bahh
You guys are looking hard to NOT find something that satifies the OP’s question or something in a similiar vein.
Hell, in the real world I might as well say you can’t have a circle that had a circumference equal to pi times D because of quantum mechanics or curved space time or something, therefore pi isnt an interesting number because it doesnt really exist.
Imagine a 2-stroke piston engine where the piston’s motion in and out is 1 length unit. The crankshaft has a diameter of 1 length unit.
One stroke takes 1 time unit.
The crankshaft drives a wheeled vehicle whose drive wheels have a diameter of 1 length unit.
Run the engine. For each engine stroke (1 time unit), the vehicle moves 2*pi length units, therefore over 1/2 time units the vehicle moves pi length units.
How’s that?
Good point. I’m tempted to re-ask by saying the initial state can be described using only rational values. (I.e., “A and B are 1 unit apart, B and C are 1 unit apart, and angle ABC is 90 degrees”) but I’m afraid that probably just ends up sneaking in irrational values if you fill in the details of what it takes to describe the state as fully as necessary. (Esp. since a “degree” has alot to do with a circle, which has a lot to do with a certain transcendental value…)
So maybe there’s just no way to fill in the antecedent here… Maybe irrational values must necessary inhabit the description of any initial state. Of course, really what I was after was an initial state where “all measurements are non-transcendental,” (I asked in terms of rationality because I figured it’d be easier to parse and it’d be a stronger question, yielding an answer to my actual question) but I suspect (because of the “degrees” thing for example) that transcendental values “infect” every description of an initial state as well.
Tempted to ask about descriptions of implausible scenarios where all objects lie exactly on some particular straight line, but this gets away from my purpose in asking the original. (That purpose was to help myself think about the question whether physics presupposes that things can actually have transcendental (or even irrational) measures in the real world. (I mean to distinguish between measures and measurements–a measurement being something we do, a measure being the actual value we’re attempting to arrive at when we make a measurement.)
I was trying to bypass the issue of precision in measurement. I was just trying to ask whether, if you just give the physics (perhaps artificially) precise rational measurements for the initial state, does the physics ever predict transcendental results after a rational amount of time.
I’m not sure if this answers your question or not (in fact, I think it might sort of side step the question) but if two objects started out 4 units apart and then moved to 3 units the intermediate value theorem says they HAVE to be at Pi units apart at some point.
But since Pi is irrational I think the best we could do is give a really good approximation of when that would be.
And if you want a single stroke to move the vehicle pi length units (such as if the angular velocity of the crankshaft is not constant while the engine is running), just add a gearbox.
Not quite. I meant to be asking about a non-quantized space, actually, since I figured if space is quantized you’ll never get actual measurements of pi. (To my knowledge, which may well be wrong, the quantization of space is controversial. Is that right, though?)
Probably subsequent posts have made this clearer (and Indistinguishable’s post has possibly highlighted a fact about physics that makes it impossible to satisfy the antecedent conditions of my hypothetical…) but the idea is to wonder whether you can ever end up with transcendental values predicted given an initial state that included only rational values. (Actually, for my purposes, really all I need in the initial state is that all values be non-transcendental, though I didn’t say that in the OP because I figured rational values are easier to talk about.)
That may be it (although to my shame I’m not quite sure how crankshafts, pistons etc relate to each other mechanically) but I’d also want to know about force measurements and so on.
Hmm, are you saying that the force applied in the cylinder due to the ignition of the fuel might inherently be a transcendental equation? I’m not certain - does anyone know?
We could say that a human being is pushing and pulling the piston open and they have enough dexterity to never apply a transcendental force, and we could use the gearbox or redefine our crankshaft and piston motion length to be 1/2 length units to make the distance moved in one stroke to be pi length units rather than a multiple or fraction of it. | 2,253 | 10,548 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2024-10 | latest | en | 0.963396 |
http://rafturimetalice-expozitoare.ro/university-mathematics-and-science-what-to-look-out-for/ | 1,606,769,485,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141486017.50/warc/CC-MAIN-20201130192020-20201130222020-00454.warc.gz | 78,493,923 | 7,397 | # University Mathematics and Science – What to Look Out For
University Mathematics and science fields that have used mathematics. In the 18th century the mathematical education of the masses was designed in the UK with universities focusing on educating the white paper apa general public. In this piece we explore science and universities mathematics as well as when selecting a student what to watch out for in them.
Mathematics is a branch of science that deals with quantities and mathematical equations that give rise. Different branches of mathematics deal with different types of measurements and their calculations. Physical quantities that are hard to measure are used by some branches of mathematics such as statistics.
Mathematics can be broken into four. These sub-fields are trigonometry, algebra, geometry and arithmetic. These four annotatedbibliographyapa net sub-fields form the foundations of just about all mathematics.
Algebra is a branch of mathematics that deals with proportions, measurements and numbers. Disciplines deal with various sorts of algebra.
Geometry deals with the shapes of objects. It uses straight lines and angles that are different, and the geometric figures. Geometry is very frequently used in architecture science and mathematics. The shape of a circle is at least as difficult to draw as a tennis ball’s form but is a lot easier to learn than the former.
Algebra is a branch of mathematics that uses algebra. Using algebra is crucial to comprehend the subject of mathematics. Algebra can be used in areas as varied as business, music and medicine.
Mathematics and science develop with each other. In science and math the study of one involves the study of the other. The analysis of one forms the basis for the study of the other. This is important since it means that if you’re studying science or mathematics then you http://www.econ.yale.edu/~shiller/ will have to have some knowledge in the other to understand.
In order to start learning mathematics, you need to be exposed to math. Since people go to university for reasons that were different, they might not be subjected to the subject. The fantastic thing is that there are ways to enable a student.
A student will be exposed to a program that is set or the topic through teaching procedures, whether through lectures. Students can be exposed to math by reading about them or reading textbooks. It is not just reading about math that helps a student learn.
It’s essential that students will think about the subject and understand the subject. For instance, a student cannot possess without also being able to apply it, the understanding of algebra and geometry. This is where diagrams are useful.
Today’s world is growing so understanding and applying mathematics is crucial. The more mass a student can understand and apply the better. Also, the more knowledge they have their odds of getting employment.
It is no secret that mathematics and science are vital to modern society. It is great that universities provide courses in these fields so that pupils can have a general knowledge of the subject. A student will find a solid foundation in science and mathematics.
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https://wikipedikia.org/how-many-feet-is-2-ft-7-inches-14/ | 1,652,682,114,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662509990.19/warc/CC-MAIN-20220516041337-20220516071337-00268.warc.gz | 721,139,418 | 22,007 | As you have seen, converting 2 feet 7 inches to inches is straightforward. 2′ 7″ in ″ equals 31 in.
Likewise, What things are 2 feet tall?
Here are an additional 7 things that are 2 feet tall.
• 4-month-old baby.
• Coffee and end tables.
• Fence and garden panels.
• Desk or office lamp.
• 2 Subway subs.
• Golden retriever dog.
• 24 hockey pucks.
Also, How many 16 feet is 32 inches? So, 32 feet in inches = 32 times 0.025400000000000002 = 384 inches, exactly.
Secondly, How many 16 feet is 16 inches?
So, 16 feet in inches = 16 times 0.025400000000000002 = 192 inches, exactly.
How many feet is 69 inches? 69 Inches to Feet
69″ in ′ equals 5.75 ft.
## What things are 4 feet tall?
Here are 9 examples of things that are 4 feet tall.
• 7-Year-old child.
• Cat tree.
• Dresser.
• Garden fence.
• Storage box.
• 4 Rulers.
• 2 Mini fridges.
• Hockey net.
## What animals are 5 feet tall?
African buffalo can stand over 5 feet (1.5 meters) at the shoulder and weigh thousands of pounds. The African buffalo is sometimes confused with the American bison, but they’re very different. The African buffalo has a long, stocky body that can weigh up to 2,200 pounds (998 kilograms).
## How tall is 2 feet inches?
Feet to inches conversion table
Feet
(
ft
)
Inches
(“)
2 ft
24 ″
3
ft
36 ″
4
ft
48 ″
5
ft
60 ″
## How tall is 4 feet inches?
Feet to inches conversion table
Feet (ft) Inches (“)
1 ft 12 ″
2 ft 24 ″
3 ft 36 ″
4 ft
48 ″
## How many feet is 52 inches?
52 inches equals 4.33333333333 feet.
## What is one foot 12 inches to the nearest Millimetre?
12 inches equals 304.8 millimeters.
## How many 16 feet is 17 inches?
Feet to Inches Conversion Chart
Feet Inches
14 ft 168 in
15 ft 180 in
16 ft
192
in
17 ft 204 in
## How many 16 feet is 8 inches?
So, 8 feet in inches = 8 times 0.025400000000000002 = 96 inches, exactly.
## Is being 5’9 tall for a girl?
5′9 is certainly taller than average for a girl, but I wouldn’t call it “tall.” 5′9 is about the minimum height to be a fashion model for a girl. Tall Club International sets the minimum cutoff at 5′10 for a girl.
## How tall is a person that is 69 inches?
Human Height Conversion Table
ft in inches centimeters
5’9”
69in
175.26cm
5’10” 70in 177.80cm
5’11” 71in 180.34cm
6’0” 72in 182.88cm
Feb 5, 2010
## What household items are 12 inches?
11 Common Household Items That Are 12 Inches Long
• Ruler.
• 2L soda bottle.
• Toaster oven.
• Skillet.
• Zip ties.
• Wall clock.
• Towel bar.
• Yoga block.
## What height should you be at 6 feet?
5 foot and 12 inches is the same as 6 foot 0 inches.
4 feet 0 inches = 121.92 centimeters
6 feet 4 inches = 193.04 centimeters
6 feet 5 inches = 195.58 centimeters
6 feet 6 inches =
198.12 centimeters
6 feet 7 inches = 200.66 centimeters
## What animals are 10 feet tall?
The
10
Tallest
Animals
in the World (terrestial):
• GIRAFFE. Up to 6 meters / 19.7
feet
.
• AFRICAN ELEPHANT. Up to 4 meters / 13.1
feet
.
• SIBERIAN TIGER. Up to 3.7 meters / 12.1
feet
.
• LIGER. Up to 3.6 meters / 11.8
feet
.
• POLAR BEAR. Up to 3.5 meters / 11.5
feet
.
• ASIAN ELEPHANT. Up to 3.4 meters / 11.3
feet
.
• BENGAL TIGER. …
• GRIZZLY BEAR.
## What is the smartest animal in the world?
The Smartest Animals In The World
• Chimpanzees are better than humans in some memory tasks.
• Goats have excellent long-term memory.
• Elephants can work together.
• Parrots can reproduce sounds of the human language.
• Dolphins can recognize themselves in the mirror.
• New Caledonian crows understand cause-and-effect relationships.
## How many inches tall is 4 11?
Height Calculator
How big is 4’11 in other units?
cm: 149.86 centimeters
in:
59 inches
m: 1.4986 meters
mm: 1498.6 millimeters
## How many inches is 5’9 feet?
Five feet, 9 inches. or 5’9, is equivalent to 69 inches. Since 1 foot unit has exactly 12 inches, 5 feet equals 60 inches. Add 9 inches to 60, and you’ll have 69 inches in total.
## How many cm is 5’11 feet?
Feet to Centimeters (ft to cm) Conversion
Feet Inches Centimeters
5 feet 11 inches
180.34
6 feet 0 inches 182.88
6 feet 1 inches 185.42
6 feet 2 inches 187.96
Apr 20, 2020
## How many inches is 4/7 feet?
Now you know how to convert 4 ft 7 to inch and that four feet seven inches = 55 inches.
## How tall are you if your 52 inches?
52 Inches = 4.3333333 Feet.
## What does 52 mean in height?
52 inches = 4 1/3 feet. In decimal notation, 52 inch to feet = 4.333 ft.
## How many feet is 48 inches?
48 Inches to Feet
48″ in ′ equals 4 ft. | 1,471 | 4,528 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2022-21 | latest | en | 0.87677 |
https://notes.abhinavmarwaha.com/algo/sort/selection.html | 1,660,512,157,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572077.62/warc/CC-MAIN-20220814204141-20220814234141-00321.warc.gz | 405,034,995 | 10,559 | # Selection Sort¶
Time : O(n2) Space : O(1)
• it never makes more than O(n) swap
• useful when memory write is a costly operation.
• not stable, in place
``````void selectionSort(int arr[], int n)
{
int i, j, min_idx;
for (i = 0; i < n-1; i++)
{
min_idx = i;
for (j = i+1; j < n; j++)
if (arr[j] < arr[min_idx])
min_idx = j;
swap(&arr[min_idx], &arr[i]);
}
}
``````
``````Input : 4A 5 3 2 4B 1
Output : 1 2 3 4B 4A 5
``````
Swapping might impact in pushing a key(let’s say A) to a position greater than the key(let’s say B) which are equal keys. which makes them out of desired order.
### Stable¶
Selection sort can be made Stable if instead of swapping, the minimum element is placed in its position without swapping i.e. by placing the number in its position by pushing every element one step forward.
``````void stableSelectionSort(int a[], int n)
{
for (int i = 0; i < n - 1; i++)
{
int min = i;
for (int j = i + 1; j < n; j++)
if (a[min] > a[j])
min = j;
int key = a[min];
while (min > i)
{
a[min] = a[min - 1];
min--;
}
a[i] = key;
}
}
`````` | 357 | 1,058 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2022-33 | latest | en | 0.6921 |
https://www.mathblog.dk/tag/c/page/3/ | 1,632,810,640,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060538.11/warc/CC-MAIN-20210928062408-20210928092408-00576.warc.gz | 891,532,435 | 17,697 | # C#
## Project Euler 127: Investigating the number of abc-hits below a given limit.
Problem 127 of Project Euler is a really awesome number theoretic problem using concepts of GCD and radicals. The problem reads
The radical of n, rad(n), is the product of distinct prime factors of n. For example, 504 = 23 x 32 x 7, so rad(504) = 2 x 3 x 7 = 42.
We shall define the triplet of positive integers (abc) to be an abc-hit if:
1. GCD(a, b) = GCD(ac) = GCD(bc) = 1
2. a < b
3. a + b = c
For example, (5, 27, 32) is an abc-hit, because:
1. GCD(5, 27) = GCD(5, 32) = GCD(27, 32) = 1
2. 5 < 27
3. 5 + 27 = 32
4. rad(4320) = 30 < 32
It turns out that abc-hits are quite rare and there are only thirty-one abc-hits for c < 1000, with c = 12523.
Find c for c < 120000.
Posted by Kristian in Project Euler, 2 comments
## Project Euler 126: Exploring the number of cubes required to cover every visible face on a cuboid.
Problem 126 of Project Euler is a really awesome geometric challenge. It reads
The minimum number of cubes to cover every visible face on a cuboid measuring 3 x 2 x 1 is twenty-two.
If we then add a second layer to this solid it would require forty-six cubes to cover every visible face, the third layer would require seventy-eight cubes, and the fourth layer would require one-hundred and eighteen cubes to cover every visible face.
However, the first layer on a cuboid measuring 5 x 1 x 1 also requires twenty-two cubes; similarly the first layer on cuboids measuring 5 x 3 x 1, 7 x 2 x 1, and 11 x 1 x 1 all contain forty-six cubes.
We shall define C(n) to represent the number of cuboids that contain n cubes in one of its layers. So C(22) = 2, C(46) = 4, C(78) = 5, and C(118) = 8.
It turns out that 154 is the least value of n for which C(n) = 10.
Find the least value of n for which C(n) = 1000.
Posted by Kristian in Project Euler, 9 comments
## Project Euler 125: Finding square sums that are palindromic
Problem 125 of Project Euler deals with a property we have worked with before, palindromic numbers. The problem reads
The palindromic number 595 is interesting because it can be written as the sum of consecutive squares: 62 + 72 + 82 + 92 + 102 + 112 + 122.
There are exactly eleven palindromes below one-thousand that can be written as consecutive square sums, and the sum of these palindromes is 4164. Note that 1 = 02 + 12 has not been included as this problem is concerned with the squares of positive integers.
Find the sum of all the numbers less than 108 that are both palindromic and can be written as the sum of consecutive squares.
Posted by Kristian in Project Euler, 4 comments
## Project Euler 124: Determining the kth element of the sorted radical function
Problem 124 of Project Euler deals with radical functions, something I have never heard of before. So that always makes it interesting. It reads
The radical of n, rad(n), is the product of distinct prime factors of n. For example, 504 = 23 x 32 x 7, so rad(504) = 2 x 3 x 7 = 42.
If we calculate rad(n) for 1 ≤ n ≤ 10, then sort them on rad(n), and sorting on n if the radical values are equal, we get:
Unsorted Sorted n rad(n) n rad(n) k 1 1 1 1 1 2 2 2 2 2 3 3 4 2 3 4 2 8 2 4 5 5 3 3 5 6 6 9 3 6 7 7 5 5 7 8 2 6 6 8 9 3 7 7 9 10 10 10 10 10
Let E(k) be the kth element in the sorted n column; for example, E(4) = 8 and E(6) = 9.
If rad(n) is sorted for 1 ≤ n ≤ 100000, find E(10000). Continue reading →
Posted by Kristian in Project Euler, 7 comments
## Project Euler 123: Determining the remainder when (pn − 1)^n + (pn + 1)^n is divided by pn^2
Problem 123 of Project Euler reads
Let pn be the nth prime: 2, 3, 5, 7, 11, …, and let r be the remainder when (pn-1)n + (pn+1)n is divided by pn2.
For example, when n = 3, p3 = 5, and 43 + 63 = 280 ≡ 5 mod 25.
The least value of n for which the remainder first exceeds 109 is 7037.
Find the least value of n for which the remainder first exceeds 1010.
Posted by Kristian in Project Euler, 7 comments
## Project Euler 122: Finding the most efficient exponentiation method.
Problem 122 of Project Euler has really proved troublesome for me, partly because I couldn’t read the material I got. But I will get back to that after the problem statement.
The most naive way of computing n15 requires fourteen multiplications:
n x n x … x n = n15
But using a “binary” method you can compute it in six multiplications:
n x n = n2
n2 x n2 = n4
n4 x n4 = n8
n8 x n4 = n12
n12 x n2 = n14
n14 x n = n15
However it is yet possible to compute it in only five multiplications:
n x n = n2
n2 x n = n3
n3 x n3 = n6
n6 x n6 = n12
n12 x n3 = n15
We shall define m(k) to be the minimum number of multiplications to compute nk; for example m(15) = 5.
For 1 k 200, find ∑ m(k).
Posted by Kristian in Project Euler, 12 comments
## Project Euler 121: Investigate the game of chance involving coloured discs
To be honest, problem 121 of Project Euler scared me a whole lot, since my probability theory is not very strong. However, I do believe that I came up with quite a nice solution after all. The problem reads
A bag contains one red disc and one blue disc. In a game of chance a player takes a disc at random and its colour is noted. After each turn the disc is returned to the bag, an extra red disc is added, and another disc is taken at random.
The player pays £1 to play and wins if they have taken more blue discs than red discs at the end of the game.
If the game is played for four turns, the probability of a player winning is exactly 11/120, and so the maximum prize fund the banker should allocate for winning in this game would be £10 before they would expect to incur a loss. Note that any payout will be a whole number of pounds and also includes the original £1 paid to play the game, so in the example given the player actually wins £9.
Find the maximum prize fund that should be allocated to a single game in which fifteen turns are played.
Posted by Kristian in Project Euler, 8 comments
## Project Euler 120: Finding the maximum remainder when (a − 1)^n + (a + 1)^n is divided by a^2
Problem 120 of Project Euler is back in the very mathy part of the questions. Or at least the part where I have been able to find a pretty mathy solution for the problem. It reads
Let r be the remainder when (a-1)n + (a+1)n is divided by a2.
For example, if a = 7 and n = 3, then r = 42: 63 + 83 = 728 42 mod 49. And as n varies, so too will r, but for a = 7 it turns out that rmax= 42.
For 3 ≤ a ≤ 1000, find ∑ rmax.
Posted by Kristian in Project Euler, 6 comments
## Project Euler 119: Investigating the numbers which are equal to sum of their digits raised to some power
Problem 119 of Project Euler reads
The number 512 is interesting because it is equal to the sum of its digits raised to some power: 5 + 1 + 2 = 8, and 83 = 512. Another example of a number with this property is 614656 = 284.
We shall define an to be the nth term of this sequence and insist that a number must contain at least two digits to have a sum.
You are given that a2 = 512 and a10 = 614656.
Find a30.
Posted by Kristian in Project Euler, 16 comments
## Project Euler 118: Exploring the number of ways in which sets containing prime elements can be made.
Problem 118 have a very short problem description which in all it’s glory reads
Using all of the digits 1 through 9 and concatenating them freely to form decimal integers, different sets can be formed. Interestingly with the set {2,5,47,89,631}, all of the elements belonging to it are prime.
How many distinct sets containing each of the digits one through nine exactly once contain only prime elements?
Posted by Kristian in Project Euler, 6 comments | 2,332 | 7,692 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2021-39 | latest | en | 0.80772 |
https://refurme.com/prince-edward-island/application-of-pythagoras-theorem-in-mathematics.php | 1,586,127,462,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371611051.77/warc/CC-MAIN-20200405213008-20200406003508-00269.warc.gz | 656,797,245 | 8,129 | # Application of pythagoras theorem in mathematics
## Pythagorean theorem Basic mathematics
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### Pythagorean theorem Basic mathematics
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### Application On Pythagorean Theorem ask-math.com
Pythagoras’ theorem Mathematics resources. The most ubiquitous "real-life" application of the Pythagorean Theorem is in the empirical observation that The Pythagorean theorem is used in most of mathematics., This three-week course is an introduction to the application of Mathematics to Physics, Pythagoras’s theorem is essential for many descriptions in Physics.
### Pythagoras Greek Mathematics - Explorable
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• Pythagoras’ theorem Mathematics resources
• Pythagoras' Theorem and Right-Angled Triangles Mr
• Pythagoras Basics 1 Transum
• A2 + B2 = C2 is the Pythagorean theorem, and it's used to determine the length of the hypotenuse of a right triangle by using it's legs. Pythagoras Theorem in 3D Shapes. How to teach pythagoras' theorem in three-dimensions. Maths lesson include math worksheet and differentiated lesson plan.
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The Pythagorean Theorem important concepts in mathematics use the Pythagorean Theorem to calculate the application,so you need to calculate the approximate Mathematics Capstone Course Page 1 The Pythagorean Theorem in Crime Scene Investigation. I . UNIT OVERVIEW & PURPOSE: Students are asked to solve a series of crimes
Learn what the Pythagorean Theorem you'll also get unlimited access to over 75,000 lessons in math The Pythagorean Theorem: Practice and Application The most ubiquitous "real-life" application of the Pythagorean Theorem is theorem is used in most of mathematics. Pythagoras arrive at the Pythagorean theorem?
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122 proofs of the Pythagorean theorem: ThГўbit ibn Qurra and the Pythagorean Theorem, Mathematics Teacher This an additional application of Heron's formula to Pythagoras' theorem the theorem attributed to Pythagoras that the square on the hypotenuse of a The scientific application of Pythagorean mathematics appears
PYTHAGORAS’ THEOREM The Improving Mathematics Education in Schools (TIMES) Project MEASUREMENT AND GEOMETRY Module 15 Applications of Pythagoras’ theorem Mathematics Capstone Course Page 1 The Pythagorean Theorem in Crime Scene Investigation. I . UNIT OVERVIEW & PURPOSE: Students are asked to solve a series of crimes | 1,349 | 6,093 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2020-16 | latest | en | 0.86779 |
https://brokereakwf.netlify.app/orbaker50730dyte/what-are-exchange-rates-in-maths-gyk.html | 1,680,058,829,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948932.75/warc/CC-MAIN-20230329023546-20230329053546-00138.warc.gz | 173,384,268 | 12,464 | ## What are exchange rates in maths
solve problems involving currency exhange rates The investigation into how much NZ\$100 would be worth in other currencies depends on two things. Firstly This Core Maths resource contains teachers' notes, lesson plan and student Students need to find a correct currency rate and convert each cost into GBP.
## 7 Mar 2020 Convert 100 Euros to US Dollars. Get live exchange rates, historical rates & charts for EUR to USD with XE's free currency calculator.
Many shopping sites will offer to do the conversion math for you, but it's best to double-check their math as the exchange rates fluctuate quite a bit. 2 Jul 2018 Students then find the exchange rate for their local currency and calculate how much they will need to purchase \$1000 Australian dollars. Once 8 Dec 2010 What is the relationship between the Yen/USD exchange rate and Japan's inflation rate? Introduction: Many theories shows that as exchange Direct Proportion and Exchange Rates. Mathematics lesson calculating amounts in direct proportion such as exchange rates and ratio with recipes. 9 Nov 2012 The mental math for most countries is pretty easy: in Belize, divide prices by half, in the UK multiply by 1.5 and round up, and in Australia and In. Forex market, the traders exchange different currencies via Internet. There are three major orders for trading. ∗ Department of Industrial Engineering, College of
### 2 Jul 2018 Students then find the exchange rate for their local currency and calculate how much they will need to purchase \$1000 Australian dollars. Once
Many shopping sites will offer to do the conversion math for you, but it's best to double-check their math as the exchange rates fluctuate quite a bit.
### Why Real Exchange Rates? Luis A.V. Catão. How does one determine whether a currency is fundamentally undervalued or overvalued? This question lies at the
10 Aug 2015 Some countries don't have a fractional exchange rate, like Bolivia, so this trick won't work everywhere. The Non-Math Way to Convert Currencies. Currency Exchange Rates. Date: 01/02/99 at 00:12:44 From: Millie Subject: Money Do you know how much half a crown, a crown, and a shilling are worth in 7 Mar 2020 Convert 100 Euros to US Dollars. Get live exchange rates, historical rates & charts for EUR to USD with XE's free currency calculator. Why Real Exchange Rates? Luis A.V. Catão. How does one determine whether a currency is fundamentally undervalued or overvalued? This question lies at the An exchange rate is a relative price of one currency expressed in terms of another currency (or group of currencies). For economies like Australia that actively
## Siyavula's open Mathematics Grade 10 textbook, chapter 9 on Finance And Growth covering Foreign Exchange Rates.
Everyday maths 2. Start this Figure 6 A conversion rate for pounds and euros Hopefully you are now feeling confident with converting between currencies. Business Maths - Calculating Exchange Rates. Levels: AS, A Level; Exam boards : AQA, Edexcel, OCR, IB, Eduqas, WJEC. Print page solve problems involving currency exhange rates The investigation into how much NZ\$100 would be worth in other currencies depends on two things. Firstly This Core Maths resource contains teachers' notes, lesson plan and student Students need to find a correct currency rate and convert each cost into GBP.
Currency Exchange Rates. Date: 01/02/99 at 00:12:44 From: Millie Subject: Money Do you know how much half a crown, a crown, and a shilling are worth in 7 Mar 2020 Convert 100 Euros to US Dollars. Get live exchange rates, historical rates & charts for EUR to USD with XE's free currency calculator. Why Real Exchange Rates? Luis A.V. Catão. How does one determine whether a currency is fundamentally undervalued or overvalued? This question lies at the An exchange rate is a relative price of one currency expressed in terms of another currency (or group of currencies). For economies like Australia that actively Maths revision video and notes on the topic of Exchange Rates. In this activity, you will learn how to convert money between different currencies using an exchange rate table and a calculator. You will need . a calculator (or use this calculator) A current list of exchange rates (look up on the internet) Vacation! The Brown family are going to visit many different countries on their vacation. While exchange rate quotes are relatively easy to find these days, reading and making calculations based on them can be a little more challenging for those that aren't familiar with the techniques. In this article, we will take a closer look at how to find and read currency exchange rates as well as some other tips to keep in mind when using them. | 1,015 | 4,740 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2023-14 | latest | en | 0.936956 |
https://api-project-1022638073839.appspot.com/questions/58a2422811ef6b6ff33aff76 | 1,718,796,920,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861817.10/warc/CC-MAIN-20240619091803-20240619121803-00753.warc.gz | 83,179,882 | 6,318 | # Question #aff76
Sep 6, 2017
See explanation.
#### Explanation:
The dimensions are given in different units, so before any calculations we have to change either feet to inches or inches to feet.
If we change feet to inches we will have all integer dimensions:
The dimensions of drawing stay unchanged:
$6$ inches and $9$ inches.
The real dimensions after changing to inches will be:
$5 \cdot 12 = 60$ inches and $7.5 \cdot 12 = 90$ inches.
Now we can calculate the scale by dividing either 2 lower or 2 higher values:
$s = \frac{60}{6} = \frac{90}{9} = 10$.
The scale factor is $10$. | 159 | 596 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 6, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2024-26 | latest | en | 0.832232 |
https://dev.to/isaacttonyloi/tow-sum-5egh | 1,685,889,241,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649986.95/warc/CC-MAIN-20230604125132-20230604155132-00441.warc.gz | 244,477,610 | 18,556 | Isaac Tonyloi
Posted on
# Tow Sum
## 1. Two Sum
### Easy
Given an array of integers `nums` and an integer `target`, return indices of the two numbers such that they add up to `target`.
You may assume that each input would have exactly one solution, and you may not use the same element twice.
You can return the answer in any order.
Example 1:
```Input: nums = [2,7,11,15], target = 9
Output: [0,1]
Explanation: Because nums[0] + nums[1] == 9, we return [0, 1].
```
Example 2:
```Input: nums = [3,2,4], target = 6
Output: [1,2]
```
Example 3:
```Input: nums = [3,3], target = 6
Output: [0,1]
```
Constraints:
• `2 <= nums.length <= 104`
• `-109 <= nums[i] <= 109`
• `-109 <= target <= 109`
• Only one valid answer exists.
Follow-up: Can you come up with an algorithm that is less than `O(n2) `time complexity?
Python solution using HashMap
``````class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
new_hash_map = {}
for i, n in enumerate(nums):
difference = target - n
if difference in new_hash_map:
return(i, new_hash_map[difference])
else:
new_hash_map[n] = i
``````
Time complexity O(N)
Java Solution
``````class Solution {
public boolean containsDuplicate(int[] nums) {
Set<Integer> numbers = new HashSet<Integer>();
for (int num : nums) {
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squarkonium.blogspot.com | 1,597,348,419,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439739073.12/warc/CC-MAIN-20200813191256-20200813221256-00264.warc.gz | 88,501,868 | 25,581 | ## Thursday, 16 July 2009
### Spinors III: Chirality of Isotropic Subspaces
This is a sequel.
# Tensor Product
Tensor product is a very important, very commonplace concept in mathematics. We speak of tensor product of vector spaces and tensor product of modules, but the later will not concern us in this post. Tensor product of vector spaces logically belongs to the realm of linear algebra, but is not covered in any linear algebra course I know. The reason is probably pedagogical: linear algebra is an introductory coarse, intended for people with little experience in mathematics. On the other hand, tensor product is a rather sophisticated concept in comparison. Sometimes people refer to tensors as multilinear algebra, but I don’t find this to be a natural category. For instance, the theory of bilinear forms is a part of linear algebra, not “bilinear algebra”.
## Definition
Fix k a field. Consider V and W finite dimensional vector spaces over k. Define the tensor product of V and W, denoted V (x) W (pardon my ASCII), to be the vector space of bilinear mappings
V* x W* –> k.
## basic properties
• V (x) W is naturally isomorphic to W (x) V
• V (x) k is naturally isomorphic to V. This is so because V** is naturally isomorphic to V.
• V (x) {0} is naturally isomorphic to {0}, the 0-dimensional vector space.
• Consider V, U, W vector spaces. Then (V (+) U) (x) W is naturally isomorphic to V (x) W (+) U (x) W. Here “(+)” denotes direct sum of vector spaces, i.e. V (+) U is the vector space of ordered pairs (v, u) where v is in V and u is in U.
• (V (x) W)* is naturally isomorphic to V* (x) W*.
• V* (x) W* is naturally isomorphic to the vector space of bilinear mappings V x W –> k.
• V* (x) W is naturally isomorphic to Hom(V, W), the vector space of linear operators (=homomorphisms) V –> W.
• For dim V = 1, Hom(V, V) is also 1-dimensonal and it has a special basis consisting of the identity operator. Hence
Hom(V, V) is naturally ismorphic to k and so is V* (x) V. Thus, for 1-dimensional spaces, the dual vector space is the inverse vector space with respect to tensor product.
## tensor product of vectors
Consider v a vector in V, w a vector in W. Then we construct
the tensor product of v and w, denoted v (x) w, a vector in V (x) W. By definition, v (x) w is supposed to be a bilinear mapping
v (x) w: V* x W* –> k. Consider a in V* and b in W*. We define
(v (x) w)(a, b) =
a(v) b(w) (take your time to parse this expression).
## basis
Suppose e1 .. en is a basis of V, f1 .. fm a basis of W.
### Claim
{ei (x) fj} is a basis of V (x) W. In particular,
dim (V (x) W) = dim V dim W.
# (Anti)Symmetric Tensors
Fix a vector space V. Consider the vector space V (x) V. Elements of V (x) V are called tensors of rank 2 over V. We construct a linear operator s: V (x) V –> V (x) V as follows. Consider t an element of
V (x) V. By definition, t is a bilinear mapping V* x V* –> k. We need to describe s(t), also an element of V (x) V hence also a bilinear mapping V* x V* –> k. Consider a, b in V*. We define
s(t)(a, b) = t(b, a
).
It is easy to see V (x) V splits into a direct sum of two subspaces:
S^2(V) and L
^2(V). S^2(V) consists of t in V (x) V such that s(t) = t, i.e. it is the eigenspace of s corresponding to eigenvalue 1. Elements of S^2(V) are called symmetric tensors of rank 2 over V. L^2(V) consists of t in V (x) V such that s(t) = –t, i.e. it is the eigenspace of s corresponding to eigenvalue -1. Elements of L^2(V) are called antisymmetric tensors of rank 2 over V. The direct sum structure of V (x) V follows from the observation that s^2 = 1.
More generally, consider the vector space
T^k(V) := V (x) V (x) … (x) V for k copies of V. Elements of T^k(V) are called tensors of rank k over V. Consider
p a permutation of k elements (i.e. a bijection
{1, 2 … k} –> {1, 2 … k}). We define a linear operator
s_p: T^k(V) –> T^k(V) by the condition
s_p(t)(a_1, a_2 … a_k) = t(a_p(1), a_p(2) … a_p(k)). Here t is an element of T^k(V) and a_1, a_2 … a_k are elements of V*.
We define S^k(V) to be the subspace of T^k(V) consisting of t such that for any permutation of k elements p, s_p(t) = t. Elements of
S^k(V) are called symmetric tensors of rank k over V. Remember that permutations can be divided into odd and even. An odd permutation it the composition of an odd number of permutations which are transpositions of two elements of {1, 2 … k}. An even permutation is the composition of an even number of such transpositions. We define L^k(V) to be the subspace of T^k(V) consisting of such t that for any permutation of k elements p,
s_p(t) = sgn(p) t. Here sgn(p) is +1 for p even and –1 for p odd.
Elements of L^k(V) are called antisymmetric tensors of rank k over V. For k > 2, the direct sum of S^k(V) and L^k(V) is not the entire space T^k(V).
There is a natural projection operator sym: T^k(V) –> S^k(V). Conisder t in T^k(V). Then, by definition, sym(t) = S_p s_p(t) / k! Here the sum ranges over all permutations of k elements p. There is also a natural projection operator asym: T^k(V) –> L^k(V). Given t in T^k(V), we define asym(t) = S_p sgn(p) s_p(t) / k!
### Claim
Consider v1, v2 … vk elements of V. Then
asym(v1 (x) v2 (x) … (x) vk) is a non-vanishing element of L^k(V) if and only if v1, v2 … vk are linearly independent.
Suppose dim V = n. Consider e1 … en a basis of V. Consider
N = {1, 2 … n}^k, the set of all ordered k-tuples made of elements of {1, 2 … n}. Clearly #N, the number of elements of N, is n^k. Consider I = (i_1, i_2 … i_k) in N. We define E_I in T^k(V) to be e_i_1 (x) e_i_2 (x) … (x) e_i_k.
### Claim
The E_I form a basis of T^k(V). In particular, dim T^k(V) = n^k.
This claims follows from our previous claim about general tensor products.
Define S(n, k) to be the set of all subsets of {1, 2 … n} of size k. Evidently #S(n, k) is the binomial coefficient (n k). Consider
I = {i_1, i_2 … i_k} in S(n, k). We define F_I in L^k(V) to be
asym(e_i_1 (x) e_i_2 (x) … (x) e_i_k). I’m a bit cheating here since this expression depends on the order of i_1 … i_k. However, the only ambiguity is the sign: even permutations don’t change the expression whereas odd permutation change its sign. For our purposes, we can make an arbitrary choice of order/sign for each I in S(n, k).
### Claim
The F_I form a basis of L^k(V). In particular, dim L^k(V) = (n k).
Define M(n, k) to be set of all multisets of size k made of elements of {1, 2 … n}. Multisets are like sets except that each element can appear in a multiset several times. The size of a multiset is defined by counting the elements with multiplicity. We have
#M(n, k) = (n + k – 1 k). Consider
I = {i_1, i_2 … i_k}. Here, i_m may coincide with i_l for some m and l. We define G_I in S^k(V) to be
sym(e_i_1 (x) e_i_2 (x) … (x) e_i_k).
### Claim
The G_I form a basis of S^k(V). In particular,
dim S^k(V) = (n + k – 1 k).
# Vector Space Determinant
The concept of a vector space determinant is standard and widely used, however not under this name and notation.
## Definition
Consider a vector space V with dim V = n. Then, the determinant of V, denoted Dim V is the vector space L^n(V).
## basic properties
• dim Det V = 1
• Det (V (+) W) is naturally isomorphic to Det V (x) Det W
• Suppose W is a subspace of V. Then Det V is naturally isomorphic to det W (x) det (V/W).
• Det (V*) is naturally isomorphic to (Det V)*
• Consider A: V –> W a linear operator, m a natural number. Then there is naturally induced operator
L^m(A): L^m(V) –> L^m(W). In the special case
dim V = dim W = m we get the operator
det A: Det V –> Det W. Let us specialize further to the case
V = W. Then det A: Det V –> Det V is simply multiplication by a constant c in the ground field k (since Det V is
1-dimensional). c is the conventional determinant of the operator A we all know and love.
Consider v1, v2 … vn elements of V. By a previous claim,
asym(v1 (x) v2 (x) … (x) vn) is a non-vanishing element of Det V if and only if v1, v2 … vn form a basis of V. Also, suppose e1 … en and f1 … fn are two bases of V related by the n x n matrix P, i.e.
(e1 … en) = (f1 … fn) P. Then
asym(e1 (x) … (x) en) = (det P) asym(f1 (x) … (x) fn).
# Chirality
Consider V a complex quadratic vector space of dimension n. Then Det V contains two special non-vanishing elements that differ by a sign, say w and –w. These elements are constructed as follows. Consider e1 … en an orthonormal basis of V. Then
asym(e1 (x) e2 (x) … (x) en) is a non-vanishing element of Det V. Any two orthonormal bases are related by an orthogonal n x n matrix O. Since O is orthogonal, we have either det O = +1 or det O = –1. Thus asym(e1 (x) e2 (x) … (x) en) can only differ by a sign for different orthonormal bases, yielding w and –w.
Suppose W is an isotropic subspace of V. Any element v of V defines a linear functional v* on V defined by v*(u) = Q(v, u). Here u is an arbitrary element of V and Q is the quadratic form of V. If v and u both belong to W we get v*(u) = Q(v, u) = 0 since W is isotropic. Thus, given v in W, v* is a linear functional vanishing on W. Since it vanishes on W, it determines a linear functional a on V/W. This can be seen as follows. Any u in V/W can be represented by some u’ in V. We can then set a(u) = v*(u’). However, u’ is only defined up to adding an arbitrary element w of W. But it’s OK because v*(u’ + w) = v*(u) + v*(w) = v*(u) since the second term vanishes by our previous observation. Thus a is well defined.
For any v in W we constructed a linear functional on V/W i.e. an element of (V/W)*. This yields a linear operator W –> (V/W)*. Now suppose dim V = 2m and W is maximal isotropic, that is,
dim W = m. Then our linear operator is an isomorphism.
In the following, we use the symbol “=” to denote “naturally isomorphic”. Det V = Det W (x) Det (V/W). By the above,
W = (V/W)*. Hence
Det V = Det ((V/W)*) (x) Det (V/W) = (Det (V/W))* (x) Det(V/W) = C. In particular we get a special element in Det V: the element corresponding to 1 in C.
### Claim
The special element is either w or –w. This invariant divides maximal isotropic subspaces into two classes (chiralities). Any two maximal isotropic subspaces W and W’ are related by an orthogonal operator O: V –> V. For det O = +1, W and W’ have the same chirality. For
det O = –1, W and W’ have opposite chirality.
## Wednesday, 1 July 2009
### Autoevolution III
This is a sequel.
As Eli justly remarked, we are already cyborgs.The computer is extending our ability to think and remember. The Internet is extending our ability to communicate and share information. Microscopes, telescopes, night vision etc. are extending our ability to sense.
Extension of human “IO” will not be limited to information naturally contained in the physical reality. It will include conventionally encoded information: the kind we store in writing, computers etc. today. The synergy of the human use of such information will lead to the Internet becoming a “virtual reality”: something experienced directly rather than through non-transparent intermediates such as keyboards, mice and displays. “Ordinary” reality will become “augmented reality” as additional layers of information will be seamlessly available about “surrounding” physical objects (that is, objects under attention). Imaging looking at a bus and knowing what line it is and what is its trajectory. A possible intermediate form of augmented reality is superposing the additional information layers on the existing senses, e.g. “seeing” a “clickable” line number floating in the air besides the bus.
Physical location of human beings will become much less important. Even now the Internet, cellular phones and video conferences are reducing its importance. In the future this trend will continue much more. The culmination is mind-body separation. One can image huge immobile “brains” banked somewhere, operating bodies in distant physical locations and purely “virtual” realms.
Many activities don’t require engaging the “direct” physical reality but allow working with “conventionally encoded” higher abstraction layers. These include
• “Software” engineering, that is the design of technology operating within these higher abstraction layers
• Communication
• Theoretical learning and research
• “Virtual” Art
Other activities require interaction with “lower abstraction layers”. These include
• “Hardware” engineering
• Experimental science
• Colonization of space
• “Physical” Art
Eventually the “synergy” of the “virtual IO” (the exchange of “conventionally encoded” information) will make the “Internet” into a shared memory of human kind. It will store actual memories, experiences, thoughts, knowledge etc. which will be downloadable and uploadable. This will result in individuality becoming “fuzzy”, up to the point of almost utter disappearance. The entire population of a given planet will merge into a “multimind”. The multimind will be a continuous array of memories, desires etc. connected by associations as is the mind of an individual today. The mind of a modern individual contains a “stream of consciousness” (“CPU”) continuously traversing the memories (“RAM”), loading them into the short term memory (“cash”), connecting them to IO and transforming them. In an analogical way, the multimind will contain many streams of consciousness processing the shared memory in parallel, each with its own cash and connected to specific IOs at any given moment.
I call this multimind a “Solaris”, to honour Stanislaw Lem. The Solaris is a single entity uniting the whole technosphere, ecosphere and consciousness of an entire planet.
Due to the disappearance of individual, property will lose its meaning. Birth and death will also become obsolete, except for the birth of new Solarises through space colonization.
It is also possible that the Solaris will contain “quasi-individuals”: tightly bound chunks of memories, desires and ideas. Such
quasi-individuals will dynamically form out of the Solaris and dissolve back into it.
## Thursday, 25 June 2009
This is a part of the Ask Squark series.
Thx Assaf for submitting the first question to “Ask Squark”!
Question:
Why does the mirror reverse left and right but not up and down?
Firstly, there is a hidden assumption here that the mirror is hanged conventionally, i.e. vertically. A horizontal mirror (like in Hotel California) does reverse up and down.
It might seem that a vertical mirror displays some sort of asymmetry: left and right (which are perceived as horizontal directions) are reversed, whereas up and down (the vertical directions) are not reversed. However, let me assure you that there is perfect rotational symmetry with respect to the axis orthogonal to the mirror plane. The apparent paradox is mostly semantic.
Let us remember how an ideal planar mirror works. The real mirrors are not ideal (alas), but it’s irrelevant for the discussion.
Suppose the mirror lies in the plane M. Consider a point P in space. The mirror image of P is the point P’ satisfying the following criteria:
• The line PP’ is orthogonal to M
• The distance of P to M is equal to the distance of P’ to M
In reality, the mirror is usually reflective from one side only, hence we have to assume P lies in one of the two half-spaces defined by M. However, this is inessential.
A direction in space can be specified by two points: the beginning and the end of a vector. Thus, given a direction v = PQ, the mirror image direction is v’ = P’Q’ where P’ is the mirror image of P and Q’ is the mirror image of Q. It is easy to see that when v is parallel to M, v’ = v but when v is orthogonal to M, v’ = –v.* For example, if M is the plane spanned by the up-down and north-south directions, the mirror preserves up, down, north and south but reverses east and west.
The illustration is by the courtesy of SurDin.
But what about left and right? These notions are more complicated. While north, south, east, west, up and down are absolute directions, left and right are relative. For instance, if you face another person, her right is your left and vice versa.
Mathematically, we can describe left and right as follows. Consider ordered triples of mutually orthogonal vectors of unit length
(u, v, w). It can be shown that such triples fall into two classes R and L such that
• A triple in R can be rotated into any other triple in R.
• A triple in L can be rotated into any other triple in L.
• A triple in R cannot be rotated into any triple in L.
• A triple in L cannot be rotated into any triple in R.
Suppose a Cartesian coordinate system S is given. Then any triple (u, v, w) can be decomposed into components with respect to S and represented by a 3 x 3 matrix. Then, for some triples the determinant of this matrix is 1 whereas for others it is –1. These are R and L.
We can derive the following interesting property of R and L. Suppose (u, v, w) is a triple in R (L). Then (-u, v, w), (u, –v, w) and
(u, v, –w) are triples in L (R). It follows that (-u, –v, w), (-u, v, –w) and (u, –v, –w) are triples in R (L). Also, (-u, –v, –w) is a triple in L (R). This property follows, for instance, from the fact that the sign of the determinant is reversed when we reverse the sign of one of the matrix columns.
The relation to the familiar notions of right and left is as follows. Consider X is a person and (u, v, w) a triple as above. Suppose u is a vector pointing in the direction from X’s legs to her head. Suppose v is a vector pointing in the direction from X’s back to her front. Then w points either to X’s right or to X’s left, according to the class to which (u, v, w) belongs!
Consider (u, v, w) a triple as above. Consider (u’, v’, w’) the mirror image triple. That is, u’ is the mirror image of u, v’ is the mirror image of v and w’ is the mirror image of w. Then, if (u, v, w) is in R then (u’, v’, w’) is in L and vice versa. In this sense, the mirror reverses left and right. For instance, suppose u and v are parallel to M. Then w is orthogonal to M. We get u’ = u, v’ = v, w’ = –w. Thus (u’, v’, w’) = (u, v, –w) belongs to the class opposite to that of
(u, v, w). The general case can be proven e.g. using determinants: if we choose S such that M is parallel to two of the axes, mirror imaging amounts to changing the sign of one of the matrix rows.
The illustration is by the courtesy of SurDin.
* Remember that two vectors v = AB and w = CD are considered equal when
• The line AB is parallel to the line CD (or A = B and C = D which means both vectors are the equal to the 0 vector).
• The line AC is parallel to the line BD (or A = C and B = D which means both vectors coincide in the trivial sense).
Also, remember that, by definition, if v = AB then –v = BA.
## Thursday, 18 June 2009
### Free will, ethics and determinism
Lev has recently brought up the question of free will vs. determinism. I have spent some time thinking about these issues and came up with the ideas I want to lay out here.
The problem stems from the desire to define ethics. What is ethics? From my point of view, ethics is a set of rules describing how ethical is a given behaviour in a given situation. The definition might seem somewhat circular, but lets put this aside. The important aspect is that ethics is something allowing comparison of different choices and marking some as "better" and some as "worse", doesn't matter in what sense.
In order to discuss ethics, we must understand the set of choices available to a given person at a given situation. The fact that choice exists at all relies on the notion of "free will" which, on the surface, contradicts determinism.
What is determinism? Determinism means that given complete knowledge of the initial conditions it is possible to predict what would happen at any time in the future. This is a principle that holds in classical physics. The real world is better described by quantum physics, in which the situation is more complicated, but lets leave this for later.
In practice, such predictions are limited by 3 factors:
1. Incomplete knowledge of the initial conditions. Indeed it is difficult to know everything about the entire universe.
2. Incomplete knowledge of the laws of nature. Our model of reality is imperfect, and in my opinion, will always remain so.
3. Limited information processing power. Even if you know the initial conditions and the laws of nature sufficiently precisely to give a prediction with the accuracy you need, doing so might require a complicated computation which would take lots of "CPU clock ticks" and perhaps lots of "RAM" to complete.
Thus, even if a "bird's-view" observer knows precisely what person X is going to do, X doesn't know it. Moreover, I claim it is impossible for X to know it. That is, we have the following principle of self-unpredictability:
An intelligent being X can never have a future-prediction capacity, accounting for the 3 factors above, that would allow it to predict her own behaviour.
If the principle is violated, we would get something like the grandfather paradox: if X knows she is going to do Y, what prevents her from doing Z which is different from Y? The principle also makes sense physically, as far as my intuition goes: to produce better predictions we need a more powerful "brain" which would have to be more complicated and thus more difficult to predict. It's sort of a "bootstrap".
To give an example, suppose X is a human. Certainly X is not able to use her knowledge of biology, chemistry, quantum physics and what else to predict the workings of her own brain. Now, suppose X recruits a computer Y to help her. It might appear that now her task is more realistic. However, by using the computer she made it a part of the system. That is, the required prediction now involves the joint dynamics of X + Y and thus remains out of reach.
It is fascinating to me whether the self-unpredictability principle can be reformulated as a theorem in physics about abstract information-processing systems.
Thus, the space of possible choices of X can be defined to be the space of things X can do as far as X can tell. Ethics would have to operate on this space.
It might appear that the indeterminism of quantum mechanics provides some kind of an alternative solution to the problem. However, I deem it is not so. The "freedom" allowed for by quantum mechanics is pure random. It is not consistent with the sort of choice that involves ethical judgement, which is the sort of choice we are targeting here.
Moreover, consider a person X observed by a superintelligent being Y. Y can predict X's behaviour up to quantum indeterminism. Y knows X is going to A with propability
pA = 1 - 1e-100 and B with probability pB = 1e-100. Thus B is a highly unlikely choice. However, from the point of view of X both of the choices might be equally legitimate, a priori (before ethical judgement is applied). Moreover, the small probability pB might stem from something like a lightning striking X's head, which is an artifact completely irrelevant to the ethical dilemma at hand.
## Friday, 12 June 2009
### Spinors II: Isotropic Subspaces
This is a sequel.
Definition
Consider V a quadratic space. Consider W a subspace. W is called isotropic when given v, w in W arbitrary, Q(v, w) = 0.
For instance, when k = R and V is positive definite (that is, sn V = dim V) the only isotropic subspace is {0}. If V is Lorentzian (that is, sn V = dim V - 1) we have
1-dimensional isotropic subspaces: null or lightlike lines in physics talk. On the other hand, for
k = C we have:
Proposition
• Suppose V is a complex quadratic space of dimension 2n. Then the maximal dimension of an isotropic subspace is n.
• Suppose V is a complex quadratic space of dimension 2n + 1. Then the maximal dimension of an isotropic subspace is n.
It will be useful to introduce the following
Definition
Consider V a quadratic space over an arbitrary field k. Consider W a subspace of V. Define W^ to be {v in V | for any w in W: Q(v, w) = 0}. W^ is called the subspace orthogonal to W.
Several facts about orthogonal subspace will be handly. Given V, W as above:
• dim W + dim W^ = dim V
• W^^ = W
• Suppose W is isotropic. Then W^ contains W.
Consider V a quadratic space over an arbitrary field k. Consider W an isotropic subspace of V. Denote n = dim V, m = dim W. We have dim W^ = n - m but W^ contains W hence
dim W^ >= dim W i.e. n - m >= m and thus n >= 2m. It is therefore obvious that the maximal dimension of an isotropic subspace can be at most as in the proposition, even for k different from C. It remains to show that for k = C the bound can always be saturated.
Proof of Proposition
• Suppose V is a complex quadratic space of dimension 2n.
Consider e_1 ... e_2n a basis such that Q(e_i, e_j) = delta_ij. Here delta_ij is the Kronecker symbol, that is delta_ij = 1 for i = j and delta_ij = 0 for i =/= j. Define
f_1 = e_1 + i e_2, f_2 = e_3 + i e_4 ... f_n = e_2n-1 + i e_2n. Here i is the imaginary unit, that is i = sqrt(-1). It is easily seen that {f_j} span an n-dimensional isotropic subspace.
• Suppose V is a complex quadratic space of dimension 2n + 1.
Consider e_1 ... e_2n, e_2n + 1 a basis such that Q(e_i, e_j) = delta_ij. Define
f_1 = e_1 + i e_2, f_2 = e_3 + i e_4 ... f_n = e_2n-1 + i e_2n. It is easily seen that {f_j} span an n-dimensional isotropic subspace.
Definition
Consider V a complex quadratic subspace. A subspace W of V is called a maximal isotropic subspace when it is of maximal possible dimension. That is:
• For dim V = 2n, we require dim W = n.
• For dim V = 2n + 1, we require dim W = n.
Definition
Consider V a quadratic space over the field k. Consider R: V -> V an operator. V is called orthogonal when for any v, w in V we have Q(Rv, Rw) = Q(v, w). The orthogonal operators are the automorphisms of V, that is, isomorphisms of V with itself. The set of all orthogonal operators is denoted O(V).
Proposition
Consider V a complex quadratic space, U and W two maximal isotropic subspaces. Then, there exists R in O(V) s.t. R(U) = W.
This means that all maximal isotropic subspaces of a given space are essentially the same. However, we'll see in the sequels that for dim V = 2n + 1, the space of maximal isotropic subspaces has 1 connected component, whereas for dim V = 2n there are 2 connected components. For now, I won't explain what I mean by "space" and what are "connected components". These are some basic concepts of topology which I hope to explain in the sequels.
## Sunday, 7 June 2009
### Autoevolution II
This is a sequel.
Genetic engineering is already not science fiction. We are rapidly approaching the era when we will modify the genetic code of various living organisms and our own genetic code in massive amounts, up to a point when such modifications become a centerpiece of technology.
In a 1000 years the impact of these modifications will be so great that our descendants will have little in common with the original homo sapiens sapiens. It is impossible to tell what they will be like precisely (that is, even more impossible than my other ambitions in this series). However, I will try to guess some of their general features.
• Survival
Some speciments will be adapted to extreme conditions, such as extreme temperatures, extreme pressures, ionizing radiation, poison etc. In particular, when space colonization will commence, "humans" adapted to the respective conditions will be created. That is, we would have Marsians, Europans etc. historically originating from Earth humans.
• 6th sense, 7th sense...
Some speciments will have sensory perception very different from what we are used to. For instance they might have vision with more colour channels and / or in different areas of the spectrum.
• Communication
Speech will replaced by more advanced modes of communication, perhaps something like a direct mind-to-mind link. This will increase the bandwidth and reduce the error rate considerably. Among other things, this might lead to much more efficient resolution of disagreements up to the point when "irreconcilable differences" become very rare.
• Intelligence
Eventually, not only the "body" but also the "mind" will be enhanced. This will start with improved memory and faster thought and end-up with capabilities of completely different magnitude. In my opinion, the most critical mental capacity is the ability to hold many things in one's mind at once: a sort of "cache memory". It is the enhancement of this capacity which would lead to the most radical development of intelligence.
• Specialization
Different speciments will be adapted to different professions, to an extent much greater than what exists today (up to the point when different professions become virtually different species). In a way, this kind of specialization already exists: division into males and females. However, in the future there will be many more kinds (in ways unrelated to the reporductive cycle).
• Body-mind separation
Eventually the brain or mind which carries the information processing function will be separated from the body which carries the input/output functions. It will be possible for a given person (mind) to use a number of different bodies suited for different tasks at different times. Loosely speaking, one will be able to change bodies the way one now changes clothes or cars. Thus some of the traits mentioned above (such as survival in extreme conditions and enhanced senses) will apply to particular bodies rather than particular persons.
At first, there will exists two radically different "technologies":
• The "conventional" technology we know today, based on semiconductors, fiber optics, lasers etc. At some point this will include some sort of "nanotechnology". The advantage of this technology is that we understand and control it perfectly, since we created it "from the ground up" (except the laws of nature, of course, which are immutable).
• "Organic" technology employing what we now call "genetic engineering". The advantage of this technology is that it is "more sophisticated" than the ordinary: living organisms do things we yet only dream doing artificially. The disadvantage is the imperfect understanding and control we have over it.
However, eventually they will mix and become one or several technologies descendant from both. These technologies will unite the advantages of both kinds. Thus, the clear distinction between "ourselves" and the "machines" will be erased. At the same time, the distinction between the "technosphere" and the "biosphere" will also be erased. That is, instead of two different environments (the "wild" and our artificially created environment) there will be only one. This new environment will be at least as sophisticated as the ecology existing today in the wild, while being under out conscious control.
As an intermediate stage, we will create much more efficient modes of brain-computer communication. Humans would have computer "coprocessors" wired into their brain and connected into the internet.
There is another essential difference between conventional technology and life. The machines we create are usually "clones" made to resemble a given prototype as much as possible. However, living organisms, even if members of the same species, are always very different from each other (unless, of course, they are the clones of a single ancestor; such groups, however, form only tiny fractions of a given species). It is my suspicion that the second scheme is much more efficient, and we are only bound to the first scheme because of technical limitations. Thus most of the "machines" of the future will resemble living organisms rather than modern machines in this respect.
## Friday, 29 May 2009
### Spinors I: Clifford Algebra
This post is meant to be the first in a series about spinors, exceptional isomorphisms, twistors and supersymmetry. My interest in in-depth investigation of spinors was partially inspired by Yasha, in particular I learnt from him on the annihilator approach (will appear in the sequels). I will try to assume little prior knowledge except linear algebra. The emphasis will be on mathematics, at least for a while, so if you're interested in this purely from a physics perspective than you better already know the physical motivation / applications of this stuff.
The first object we'll need is the Clifford algebra. Fix a field k (in this series it will always be either the set of real numbers R or the set of complex numbers C).
Algebras
I'll start from a quick reminder of what an algebra is. Suppose A is a vector space over k. Suppose further that a mapping m: A x A -> A is given. For convenience sake, given a, b in A we denote m(a, b) by ab and call m multiplication. A is called a unital associative k-algebra (just k-algebra in the sequel) when the following conditions hold:
• m is bilinear (i.e. linear in each of the arguments separately). In details, it means that
Given a, b, c in A: (a + b)c = ac + bc additivity in the 1st argument / left distributivity
Given x in k and a, b in A: (xa)b = x(ab) homogenuity in the 1st argument
Given a, b, c in A: c(a + b) = ca + cb additivity in the 2nd argument / right distributivity
Given x in k and a, b in A: a(xb) = x(ab) homogenuity in the 2nd argument
• Given a, b, c in A: (ab)c = a(bc) associativity
• There exists an element "1" in A such that for any a in A: a1 = 1a = a unit
An algebra A is called commutative when given a, b in A we have ab = ba.
Examples
• Consider V a k-vector space. Consider End(V) the set of all endomorphisms of V, i.e., linear operators V -> V. End(V) is a k-algebra where multiplication corresponds to composition of linear operators. For dim V <>
• Consider V a k-vector space, W a subspace. Consider End(V, W) the set of endomorphisms of V leaving W invariant (non-standard notation). That is, given a in End(V, W), w in W we have aw also in W. End(V, W) is an algebra. It is a subalgebra of End(V), that is, a linear subspace closed under multiplication. For dim V <>
• Fix n a natural number. The set of n x n matrices with coefficients in k forms an algebra: Mat(n, k). It is isomorphic to End(V) for dim V = n. Obviously, dim Mat(n) = n^2.
• Fix n a natural number. The set of upper-triangular n x n matrices with coefficients in k forms an algebra: UT(n, k) (non-standard notation). We have dim UT(n, k) = n (n + 1) / 2.
• Consider k[x] the set of polynomials with coefficients in k in the variable x. k[x] is a
k-algebra. It is infinite-dimensional. It is
commutative.
• Fix n a natural number. We have k^n the set of column vectors of size n with coefficients in k. We can define multiplication in k^n by multiplying each vector entry separately. It makes k^n into an algebra. Obviously dim k^n = n. k^n is commutative.
Ideals
A subset I of A is called a right ideal when the following conditions hold:
• I is a linear subspace of A
• Given a in A, b in I, ba is also in I
Consider S an arbitrary subset of A. Denote SA to be the collection of all elements of A of the form
s1 a1 + s2 a2 + ... + sn an
where:
s1, s2 ... sn are elements of S
a1, a2 ... an are elements of A
Claim: SA is a right ideal.
SA is called the right ideal of A generated by S.
A subset I of A is called a left ideal when the following conditions hold:
• I is a linear subspace of A
• Given a in A, b in I, ab is also in I
Consider S an arbitrary subset of A. Denote AS to be the collection of all elements of A of the form
a1 s1 + a2 s2 + ... + an sn
where:
s1, s2 ... sn are elements of S
a1, a2 ... an are elements of A
Claim: AS is a left ideal.
AS is called the left ideal of A generated by S.
A subset I of A is called a two-sided ideal when it is simultaneously a left ideal and a right ideal.
Consider S an arbitrary subset of A. Denote ASA to be the collection of all elements of A of the form
a1 s1 b1 + a2 s2 b1 + ... + an sn bn
where:
s1, s2 ... sn are elements of S
a1, a2 ... an are elements of A
b1, b2 ... bn are elements of A
Claim: ASA is a two-sided ideal.
ASA is called the two-sided ideal of A generated by S.
Claim: Suppose A is a commutative algebra. Then a subset of A is a left ideal if and only if it is a right ideal if and only if it is a two-sided ideal.
Thus for a commutative algebra all three notions coincide hence we speak simply of ideals.
Examples
• Consider V a k-vector space. Consider the algebra End(V). Consider W a subspace of V. Define I = {a in End(V) | Im a lies in W}. I is a right ideal of End(V). Define
J = {a in End(V) | Ker a contains W}. J is a left ideal in End(V).
• Fix n a natural number. Consider the algebra Mat(n, k). Fix m <= n another natural number. Define I = {a in Mat(n, k) | the first m rows are zero}. I is a right ideal of Mat(n, k). Define J = {a in Mat(n, k) | the first m columns are zero}. J is a left ideal of Mat(n, k).
• Fix n a natural number. Consider the algebra UT(n, k). Fix m <= n another natural number. Define J = {a in UT(n, k) | the first m columns are zero}. I is a two-sided ideal of UT(n, k).
• Consider the algebra k[x]. Consider S a finite subset of k[x].
Define I = {p in k[x] | for any a in S: p(a) = 0}. I is an ideal of k[x]. It is generated by the polynomials {x - a} where a traverses elements of S. Now fix n a natural number. Define
J = {p in k[x] | for any m natural with m <= n: p^(m)(a) = 0}. J is an ideal of k[x]. It is generated by the single polynomial x^(m + 1).
• Fix n a natural number. Consider the algebra k^n. Consider m <= n another natural number. Define I = {v in k^n | the first m entries of v are zero}. I is an ideal.
Quotient Algebra
Consider A an algebra and I a two-sided ideal. Then we may take the vector space quotient A/I. That is, we consider the set of equivalence classes of A under the following equivalence relation: Given a, b in A they are equivalent when a - b is in I.
It is easy to see the operation of multplication in A defines an operation of multiplication in A/I as well, that is, makes A/I into an algebra on its own right. For this to work, it is crucial that I is a two-sided ideal. A/I is called the quotient algebra of A by I.
Examples
• Fix n a natural number. Consider the algebra UT(n, k). Fix m <= n another natural number. Define J = {a in UT(n, k) | the first m columns are zero}. Then UT(n, k) / J is naturally isomorphic to UT(m, k).
• Consider the algebra k[x]. Consider S a finite subset of k[x].
Define I = {p in k[x] | for any a in S: p(a) = 0}. Then k[x] / I is naturally isomorphic to k^n where n is the number of elements of S.
• Fix n a natural number. Consider the algebra k^n. Consider m <= n another natural number. Define I = {v in k^n | the first m entries of v are zero}. Then k^n / I is naturally isomorphic to k^m.
Generators and Relations
One of the simplest ways to construct an algebra is using generators and relations. This is done as follows. Suppose G is an abritrary set (possibly infinite). Consider F = k the algebra of non-commutative polynomials with coefficients in k and variables G. For G non-empty this algebra is infinite-dimensional. It is also called the free algebra over G.
Now take R an arbitrary subset of F. We have I = FRF a two-sided ideal. We obtain the algebra A = F / I. A is called the algebra generated by G with relations R. In this context, elements of G are called generators and elements of R relations. It is often convenient to define relation using equations. For example, suppose f, g, h are elements of G and x, y, z are elements of k. Then the relation
xf^2 = ygh + zhg
means that the element xf^2 - ygh - zhg of F is in R.
Consider V a vector space over k. V is called a quadratic space when it is equipped with a symmetric bilinear non-degenerate form Q. A quick reminder of what that means:
• Q is mapping V x V -> k
• Q is bilinear (i.e. linear in each of the arguments separately):
Given u, v, w in A: Q(u + v, w) = Q(u, w) + Q(v, w) additivity in the 1st argument
Given x in k and u, v in A: Q(xu, v) = xQ(u, v) homogenuity in the 1st argument
Given u, v, w in A: Q(w, u + v) = Q(w, u) + Q(w, v) additivity in the 2nd argument
Given x in k and u, v in A: Q(u, xv) = xQ(u, v) homogenuity in the 2nd argument
• Q is symmetric, that is, given u, v in V: Q(u, v) = Q(v, u)
• Q is non-degenerate: Suppose u in V is such that for any v in V we have Q(u, v) = 0. Then u = 0.
Two quadratic spaces V, W with corresponding forms Q, R are called isomorphic when there exists a linear mapping i: V -> W such that
• i is injective: Given u, v in V, i(u) = i(v) implies u = v. Equivalently, Given u in V, i(u) = 0 implies u = 0.
• i is surjective: Given w in W, there exists v in V such that i(v) = w.
• i preserves the quadratic stucture, that is, given u, v in V: Q(u, v) = R(i(u), i(v))
When the conditions hold, the mapping i is called an isomorphism between V and W. Two isomorphic quadratic spaces are "essentially the same".
Proposition:
1. Suppose V, W are quadratic spaces over k = C. Then V is isomorphic to W if and only if
dim V = dim W.
2. Suppose V is a quadratic space over k = C of dimension n. Then there exists a basis
e1, e2 ... en of V such that:
Q(ei, ei) = 1
Q(ei, ej) = 0 for i =/= j
Proposition:
1. Suppose V is a quadratic space over k = R of dimension n. Then there exists a natural number s and a basis of V e1, e2 ... en such that:
For i <= s: Q(ei, ei) = 1 For i > s: Q(ei, ei) = -1
Q(ei, ej) = 0 for i =/= j
We call s the "s-number" of V and denote it sn V (this is not standard terminology).
2. (Trivial) Suppose V, W are quadratic spaces over k = R. Then V is isomorphic to W if and only if dim V = dim W, sn V = sn W.
Clifford Algebra
Fix V a vector space. The tensor algebra T(V) is the algebra generated by V with the following relations:
• Given u, v, w in V with u + v = w, we take u + v = w to be a relation.
• Given u, v in V, x in k with xu = v we take xu = v to be a relation.
T(V) is infinite-dimensional.
Suppose V is a quadratic space. The Clifford algebra C(V) is the algebra generated by V with the following relations:
• The relations we used for T(V).
• Given u, v in V: uv + vu = -2Q(u, v)
Claim: dim C(V) = 2^dim V
Examples
We use k = R in these examples.
• Suppose dim V = 0. Then C(V) is isomorphic to R.
• Suppose dim V = 1, sn V = 1. Then C(V) is isomorphic to C.
• Suppose dim V = 2, sn V = 2. Then C(V) is isomorphic to the quaternion algebra H.
• For dim V > 2, C(V) is no longer a division algebra, that is, it doesn't have an inverse for each non-zero element. | 11,163 | 42,467 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2020-34 | latest | en | 0.897939 |
https://www.grin.com/document/53360 | 1,606,647,274,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141197593.33/warc/CC-MAIN-20201129093434-20201129123434-00528.warc.gz | 709,629,466 | 21,962 | # Optimizing a Gourmet Canned Foods Production
## 15 Pages, Grade: 75 %
Excerpt
1. Introduction
2. Standard form
3. The problem
4. Solution of the Problem
4.1 Primal
4.3 Sensitivity Report
4.4 Limits Report
4.5 Dual
4.7 Sensitivity Report
4.8 Limits Report
5. Primal Dual theorem
5.1 Primal
5.2 Dual
5.3 Economic Interpretation
6. References
## 1. Introduction:
In mathematics, linear programming (LP) problems are optimization problems in which the objective function and the constraints are all linear.
Linear programming is an important field of optimization for several reasons. Many practical problems in operations research can be expressed as linear programming problems. Certain special cases of linear programming, such as network flow problems and multicommodity flow problems are considered important enough to have generated much research on specialized algorithms for their solution. A number of algorithms for other types of optimization problems work by solving LP problems as sub-problems. Historically, ideas from linear programming have inspired many of the central concepts of optimization theory, such as duality, decomposition, and the importance of convexity and its generalizations. (wikipedia.com)
Linear programming (LP) is one of the most widely applied O.R. techniques and owes its popularity principally to George Danzig's simplex method (Danzig 1963) and the revolution in computing. It is a very powerful technique for solving allocation problems and has become a standard tool for many businesses and organisations. Although Danzig's simplex method allows solutions to be generated by hand, the iterative nature of producing solutions is so tedious that had the computer never been invented then linear programming would have remained an interesting academic idea, relegated to the mathematics classroom. Fortunately, computers were invented and as they have become so powerful for so little cost, linear programming has become possibly one of the most widespread uses for a personal PC. (wikipedia.de)
There are of course numerous software packages which are dedicated to solving linear programs (and other types of mathematical program), of which possibly LINDO, GAMS and XPRESS-MP are the most popular. All these packages tend to be DOS based and are intended for a specialist market which requires tools dedicated to solving LPs. In recent years, however, several standard business packages, such as spreadsheets, have started to include an LP solving option, and Microsoft Excel is no exception. The inclusion of an LP solving capability into applications such as Excel is attractive for at least two reasons. Firstly, Excel is perhaps the most popular spreadsheet used both in business and in universities and as such is very accessible. Second to this, the spreadsheet offers very convenient data entry and editing features which allows the student to gain a greater understanding of how to construct linear programs.
(http://www.economicsnetwork.ac.uk/cheer.htm)
A mathematical programming model is linear programming as long as each function is expressed in linear form. It is possible to limit the decision so much that there is no possible way to satisfy all the limits (infeasibility).
## 2. Standard form
Standard form is the usual and most intuitive form of describing a linear programming problem. It consists of the following three parts:
- A linear function to be maximized
illustration not visible in this excerpt
- Problem constraints of the following form
illustration not visible in this excerpt
- Non-negative variables
illustration not visible in this excerpt
## 3. The problem:
This small canning company is specialized in gourmet canned foods. Their five products are listed below. Marketing’s estimated maximum daily demands are given in terms of cans (each of which contains 16 ounces by weight). Marketing has also made commitments in the form of signed contracts to deliver. The maximum demands include these signed contract commitments.
illustration not visible in this excerpt
The production department obtains input materials and fills 16 ounce cans. All quantities are in ounces, all costs and sales prices/can are in \$. There is a maximum production limit of 24,000 cans/day. Canning costs are constant. Requirements by can type are given below. It costs the company five cents to process each can. Current sales price is given in the last column.
illustration not visible in this excerpt
The company has a contract with a ham supplier for daily delivery of up to 30,000 ounces of ham at \$0.40 per ounce. They also have a contract with a lima bean supplier for up to 100,000 ounces of lima beans per day at \$0.05 per ounce. They do not have to pay for materials they do not use. They grow their own jalapenos, which cost \$0.10 per ounce to pick (shown above). There is more jalapeno supply than can be used. There is also an unlimited supply of tangy bayou water.
illustration not visible in this excerpt
Profit = 0.21 H&B + 0.20 JHB + 0.10 LB + 0.15 JLB + 0.10 JP
Lower Constraints:
illustration not visible in this excerpt
Upper constraints:
illustration not visible in this excerpt
[...]
Excerpt out of 15 pages
Details
Title
Optimizing a Gourmet Canned Foods Production
College
Course
Mathematics
75 %
Authors
Year
2006
Pages
15
Catalog Number
V53360
ISBN (eBook)
9783638488310
File size
838 KB
Language
English
Notes
In mathematics, linear programming (LP) problems are optimization problems in which the objective function and the constraints are all linear. Linear programming is an important field of optimization for several reasons. Many practical problems in operations research can be expressed as linear programming problems. Certain special cases of linear programming, such as network flow problems and multicommodity flow problems are considered important enough to have generated much research on spec...
Tags
Optimizing, Gourmet, Canned, Foods, Production, Mathematics
Quote paper
Valentin Pikler (Author)Luca Deserti (Author)Frederico Grande (Author), 2006, Optimizing a Gourmet Canned Foods Production, Munich, GRIN Verlag, https://www.grin.com/document/53360 | 1,298 | 6,187 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2020-50 | latest | en | 0.944784 |
https://it.mathworks.com/matlabcentral/cody/problems/147-too-mean-spirited/solutions/3371313 | 1,607,125,179,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141745780.85/warc/CC-MAIN-20201204223450-20201205013450-00276.warc.gz | 340,581,794 | 16,896 | Cody
# Problem 147. Too mean-spirited
Solution 3371313
Submitted on 25 Oct 2020
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Fail
x = [1 2 3]; y_correct = [1.5 2.5]; assert(isequal(two_mean(x),y_correct))
n = 1 g = 1×0 empty double row vector
Assertion failed.
2 Fail
x = [10 0 0 0 100]; y_correct = [5 0 0 50]; assert(isequal(two_mean(x),y_correct))
n = 1 g = 1×0 empty double row vector
Assertion failed.
3 Fail
x = [1 2 4]; y_correct = [1.5 3]; assert(isequal(two_mean(x),y_correct))
n = 1 g = 1×0 empty double row vector
Assertion failed.
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 254 | 803 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2020-50 | latest | en | 0.715185 |
http://www.enotes.com/homework-help/let-f-x-x-2-2x-8-then-lim-f-x-h-f-x-h-x-gt-0-381464 | 1,462,001,239,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860111620.85/warc/CC-MAIN-20160428161511-00088-ip-10-239-7-51.ec2.internal.warc.gz | 493,610,418 | 12,224 | # Let``` ` `f(x)=x^2-2x+8`, what is `lim_(h->0) (f(x+h)-f(x))/h`
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The above response is correct. `lim_(h->0) (f(x+h)-f(x))/(h)` is the derivative f'(x), but if the function f(x) needs to be evaluated using the limit equation, this is the method to approach the problem.
The limit function tells us that we first need to substitute (x+h) for "x" in f(x), and then subtract it with the actual f(x). Therefore:
`lim_(h->0) (f(x+h)-f(x))/(h)` ` ``= lim_(h->0) ((x+h)^2-2(x+h)+8-(x^2-2x+8))/h`
`=> lim_(h->0) (x^2+2xh+h^2-2x-2h+8-x^2+2x-8)/h`
After simplifying the fraction, we can take the limit by substituting 0 for h.
`=> lim_(h->0) (2x+h-2)`
`=> 2x+0-2 = 2x-2` | 268 | 677 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2016-18 | longest | en | 0.71092 |
https://brainmass.com/business/cost-volume-profit-analysis/$%7Btopic.url%7D | 1,582,689,344,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146186.62/warc/CC-MAIN-20200226023658-20200226053658-00088.warc.gz | 298,885,296 | 16,877 | Explore BrainMass
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# Cost-Volume-Profit Analysis
Cost-Volume-Profit (CVP) analysis helps managers understand the relationships that exist between the costs of inputs, the volume of sales and the level of profit.
Break-Even Analysis: An important element of CVP. The break-even point is the level of sales the firm must reach to break even (cover their costs so there are no operating losses). Past the break-even point, operating income increases by the contribution margin of each additional unit sold.
Contribution Margin (CM): The amount of money from each unit sold above the variable costs of each unit. CM is kept by the company to cover fixed costs, or to contribute to operating income past the break-even point.
Contribution Margin per Unit = Price per Unit – Variable Cost per Unit
Contribution Margin = Sales – Variable Expenses
or = CM per Unit * # Units Sold
Profit (Operating Income) = Contribution Margin – Fixed Expenses
Break-Even Point in Units Sold = Fixed Expenses/CM per Unit
Break-Even Point in Total Sales Dollars = Fixed Expenses/CM Ratio
Where, CM Ratio = CM per Unit/Selling Price
or = [Selling Price – Variable Cost per Unit]/Selling Price
Cost Structure: Refers to the relative proportion of fixed and variable costs in an organization.
Operating Leverage: A measure of how sensitive operating income is to a given percentage change in sales.
Degree of Operating Leverage = CM/Operating Income
CVP Assumptions:
1. Selling price is constant through relevant range
2. Costs are linear through relevant range and can be divided into variable and fixed elements
3. In multiproduct companies, sale mix is constant
4. In manufacturing companies, inventories do not change (produced = sold)
### Fixed Overhead Spending Variance and Fixed Over Applied
Slaton Company manufactured 10,000 golf bags during March. The standard fixed overhead rate is \$20.00 per labor-hour, based on a denominator level of 6,000 labor hours and budgeted fixed overhead of \$120,000. Actual fixed overhead totaled \$122,000, and the actual labor hours worked totaled 5,100 hours. The standard hours allowed
### Break-even point and contribution margin at break-even point
Model Transport provides shuttle service between four hotels near a medical center and an international airport. A recent month's activity in the form of a cost-volume-profit income statement is shown below: Fare revenues (1,440 fares) \$57,600 Variable costs Fuel \$9,000 Tolls and parking 3,100 Maintenance 2,300 14,400 Contribu
### Computing Incremental Costs and Revenues
Santiago's Salsa Production Costs Apr-08 Production 20,000 Jars of Salsa Ingredient cost (variable) \$16,000 Labor cost (variable) 9,000 Rent (fixed) 4,000 Depreciation (fixed)
### Pharoah Company: Number of Units and Selling Price
Pharoah Company had \$296,300 of net income in 2019 when the selling price per unit was \$154, the variable costs per unit were \$90, and the fixed costs were \$574,100. Management expects per unit data and total fixed costs to remain the same in 2020. The president of Pharoah Company is under pressure from stockholders to increase
### Cost Volume Profit Analysis, Break Even
The launch of a new product is under consideration. Its unit variable costs will be £30 and it is estimated that incremental fixed costs of £250,000 will be incurred if production is commenced. Forecast sales are 50,000 units. At what level of price for the new product will the organization break even? If the actual planned se
### Computing Fixed, Variable, and Total Unit Costs
Calwood Manufacturing anticipates producing 1,450 wagons with total budgeted costs of \$18,824. Of this amount, \$5,600 is considered to be fixed. A. Write the cost equation in standard form. B. If Calwood produces 1,400 wagons, how much are total fixed costs? C. If Calwood produces 1,460 wagons, how much are total fix
### Distinguish between Financial and Management Accounting
Distinguish between financial accounting and managerial accounting as to user groups and time horizons.
### Friendly Co-op Target Profit Breakeven Variable Cost Ratio CMR
Friendly is a member-owned Cooperative grocery store that specializes in locally grown organic products. The Co-op currently has two locations and is well regarded in the community as an advocate for healthy eating and living. Friendly's manager, Doug Stinson, has approached you for help in evaluating whether Friendly should op
### Tax Implications for Profit-Making Entities
An exempt hospital receives all of the shares of stock of Compute, Inc., a retail computer chain, as a gift from a wealthy donor. Because the chain is very profitable and its CEO has offered to continue to manage it, the hospital has decided to operate the chain rather than sell the stock. All of the chain's profits will be used
### Computing breakeven sales and sales needed
P18-26A Computing breakeven sales and sales needed to earn a target operating income; graphing CVP Big Time Investor Group is opening an office in Dallas, Fixed monthly costs are office rent (\$8,200), depreciation on office furniture (\$1,500), utilities (\$2,300), special telephone lines (\$1,300), a connection with an online b
### Lille Tissages S.A.
Activity one: Lille Tissages case. Guidelines The background - or context - of the case is very important to understand the strategic implications of the accounting data, and to make appropriate management decisions. For example, if you determined that a company needed more cash, your decision about the source to use for that
### Special pricing for some markets or customers
Second part of the presentation. See background information for the module one SLP. Required: Include the following items in your presentation. •What about special pricing for some markets or customers? •Show effect on revenues and profitability based on stated assumptions. •Potential advantages and disadvantag
### Computing Percentage Change
Selected data from Kellogg Company's 2011 annual report follows (dollar amounts in millions). % of change 2011 2010 2009 sales \$13,198.00
### Determining Variable Costs Using the High-Low Method
Rooter's Cleaning Services provided data concerning the costs incurred to clean hotel rooms for which hotel customers pay \$150 per night. Data for the past 7 months are as follows: January February March April May June July Number of rooms cleaned
### Computing and Using the Operating Leverage Factors
Flow Cruiseline's nightly dinners cruise off the coast of Miami, San Francisco, Seattle. Dinner cruise tickets sell for \$50 per passenger. Flow Cruiselines variable cost of providing the dinner is \$10 per passenger and the fixed cost of operating the vessels (depreciation, salaries, docking fees and other expense) is \$280,000 pe
### Tasty Treats Indifferent Point
Tasty Treats plans to open a new retail store in Medina, Ohio. Tasty Treats will sell specializing cupcakes for \$5 per cup cake.(Each cupcake has a variable cost of \$2). The company is negotiating its least for the new Medina location. The landlord has offered two leasing options,1) lease of \$2500 per month, or 2) A monthly leas
### MBA Managerial Accounting
1. Elm, Inc. had the following income statement for last period: Sales \$50,000 Cost of Sales (manufacturing) 24,000 Selling and General Administrative 6,000 Net Income \$20,000 If costs of sales was 75% variable and 25% fixed, and Selling and General Expense was 60% variable and 40% fixed, prepare a contribution format
### Not-for-Profit Applications
Determine the solution to each of the following independent cases: a. Hillside College has annual fixed operating costs of \$12,500,000 and variable operating costs of \$1,000 per student. Tuition is \$8,000 per student for the coming academic year, with a projected enrollment of 1,500 students. Expected revenues from endowments a
### Break even point, contribution margin, and contribution margin ratio.
James Manufacturing company provides the following information about its cost structure: Selling Price \$20.00 per book Variable cost per unit: \$6.00 Fixed costs: 112000 per year How many units must be sold to break-even? Assume the variable cost and the price were both cut by \$4.00 per unit. Which of the following would c
### Cost-Volume-Profit Analysis and Costing Methods
1. How can you use cost-volume-profit analysis to make informed business decisions? 2. How do you use variable and absorption costing methods to make informed business decisions?
1) Duggan Company applies manufacturing overhead to jobs on the basis of machine hours used. Overhead costs are expected to total \$290,030 for the year, and machine usage is estimated at 126,100 hours. For the year, \$308,864 of overhead costs are incurred and 132,300 hours are used. Compute the manufacturing overhead rate fo
### Misstate financials: Full-Absorption Costing
Explain how full-absorption costing can be abused by management to misstate financial results. Be sure to distinguish the difference between absorption costing and variable costing, and the impact on inventoried costs before you explain how absorption costing can be abused by management.
### CVP Analysis: Buying a New Machine
Explain how CVP analysis may be helpful in evaluating whether it will be smart to buy a new machine that would reduce labor costs by 60%.
### When a hybrid system may be better than either pure job or pure process?
There are many situations where it may be best for a company to use a hybrid system that combines attributes of both systems. Describe such a situation and discuss how the hybrid system may be better than either pure job or pure process systems.
### Calculating the BEP and Margin of safety for a given Sales-Mix
4. Penny Company offers two products. At present, the following represents the usual results of a month's operations: Product A Product B Per Unit Per Unit Combined Sales \$120,000 \$1.20 \$80,000 \$0.80 \$200,000 Variable costs 60,000 0.60 60,000 0.60 120,000 Contribution margin \$60,000 0.60 \$20,000 \$0.20 80,000 Fixed co
### Cost Accounting - ABC Case Study
Discussion Question I Johnson Associates is a catering firm in Tucson, Arizona, with revenue of \$4 million. The business began ten years ago as a one-owner bakery, but has dramatically changed in size and function during the past five years. The four partners foresee the business doubling in sales revenue within two years. Th
I need assistance in researching a publicly traded corporation (Papa John) that has exhibited global business strategies. It must summarize and include the organization's financial statement (2013)and include an analysis of their financial statements as well as incorporate information gained in this course to explain and analyz
### Polk Company: Variable or Absorption Cost
I need assistance writing a paper of no more than 500 words after completing Exercise 19-17 in which you respond to the following questions: - In this case, would it be better to use the variable or absorption costing method, and why? - What are the benefits of the two methods? - Which method would lead to the best decision
### Decison Making Across the Organization
I need assistance writing a paper of no more than 900 words in which I respond to the Broadening Your Perspective 18-1 activity titled "Decision Making Across the Organization" in Ch. 18 of Accounting. Has to be APA format.
### Activity based costing for ideal manufacturing company
Broadening Your Perspective 17-2 I need assistance writing a paper of no more than 1000 words in which you respond to the Broadening Your Perspective 17-2 activity titled "Managerial Perspective" in Ch. 17 of Accounting.. Must be APA format. | 2,549 | 11,752 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2020-10 | latest | en | 0.871699 |
https://quantumsportsbetting.com/football-prediction-mathematical/ | 1,716,844,500,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059045.34/warc/CC-MAIN-20240527205559-20240527235559-00366.warc.gz | 400,943,769 | 38,895 | # Football prediction mathematical
In the ever evolving world of football prediction, mathematical strategies have emerged as a powerful tool for generating accurate and reliable predictions.
By leveraging data driven approaches and statistical models, predictors can gain a competitive edge and enhance their understanding of the game's dynamics.
This article aims to equip you with the knowledge and skills necessary to create free football prediction mathematical strategies.
We will explore the intricacies of data collection, model selection, and strategy formulation, empowering you to develop your own systematic and quantitative approaches to football prediction.
Producing free, reliable, and data driven predictions not only contributes to the football prediction community but also fosters a deeper appreciation for the game's complexities.
By embracing mathematical strategies, you can unlock new insights and push the boundaries of predictive accuracy, all while maintaining a commitment to ethical and responsible practices.
## Understanding the Basics of Football Prediction
Before delving into the realm of mathematical strategies, it's crucial to establish a solid foundation in football prediction.
This includes familiarizing yourself with popular prediction markets, such as match winners, over/under goals, and various prop bets, as well as key terminology like odds, probabilities, and predictive models.
Developing an understanding of these fundamental concepts will not only enhance your ability to interpret and communicate predictions but also ensure that you approach the prediction process with a responsible and ethical mindset.
### Gathering and Analyzing Data
The cornerstone of any successful mathematical strategy for football prediction is data. Collecting and analyzing relevant data is essential for building accurate and robust models.
This involves gathering match statistics, team performance metrics, player data, and any other relevant information that can contribute to your predictive models.
Leverage tools and resources like football statistics websites, data aggregators, and reputable analytics platforms to gather comprehensive data sets.
Additionally, stay updated with the latest football data and developments, as unforeseen events, such as injuries, managerial changes, or tactical shifts, can significantly impact team performance and prediction accuracy.
## Exploring Mathematical Models for Football Prediction
The field of mathematical modeling offers a vast array of techniques and algorithms that can be applied to football prediction.
Some popular models used in this domain include:
1. Poisson distribution: This model is widely used to predict the number of goals scored in a match, taking into account team strengths, historical performance, and other relevant factors.
2. Elo ratings: Originally developed for chess, Elo ratings can be adapted to football to rank teams based on their performance and calculate the probability of each outcome.
3. Machine learning models: Advanced techniques like neural networks, decision trees, and ensemble methods can be employed to learn patterns and relationships from historical data, enabling more accurate predictions.
Explore these mathematical models and understand their underlying assumptions, limitations, and applicability to football prediction.
Selecting and fine-tuning the appropriate model(s) for your strategies is crucial for generating reliable predictions.
### Formulating Mathematical Strategies for Prediction
With a solid understanding of mathematical models and access to relevant data, you can begin formulating your mathematical strategies for football prediction.
Develop a systematic approach that incorporates various data sources, mathematical models, and algorithms to generate predictions.
Structure and organize your strategies in a clear and logical manner, ensuring that your approach is transparent and easy to understand.
Utilize mathematical formulas, algorithms, and programming languages (if necessary) to implement your strategies and generate predictions.
Additionally, consider providing examples of your mathematical strategies applied to actual football matches, demonstrating their effectiveness and potential for accurate predictions.
## Validating and Testing Your Predictions
Validating and testing the accuracy of your predictions is a critical step in refining your mathematical strategies.
Assess the performance of your strategies by comparing your predictions against actual match outcomes, tracking metrics such as success rates, prediction errors, and overall consistency.
Continuously evaluate and improve the reliability and consistency of your predictions by analyzing the strengths and weaknesses of your strategies.
Identify areas for improvement and implement necessary modifications or adjustments to your mathematical models or data sources.
Embrace an iterative approach, adapting and refining your mathematical strategies over time as you gain more experience and encounter new challenges or developments in the football prediction landscape.
### Sharing Your Predictions with the Community
Once you have developed and validated your mathematical strategies, consider sharing your predictions with the broader football prediction community.
Choose platforms that align with your goals and target audience, such as dedicated prediction websites, social media channels, or personal blogs.
Present your predictions in an accessible and understandable manner, providing clear explanations of your mathematical approaches and the reasoning behind your predictions.
Maintain transparency, honesty, and accountability by disclosing any limitations or assumptions inherent in your strategies.
Engage with your audience, encouraging feedback and discussions around your predictions. This not only fosters a sense of community but also allows you to continually improve and refine your strategies based on diverse perspectives and insights.
### Legal and Ethical Considerations
As a creator of football prediction mathematical strategies, it is crucial to adhere to data privacy regulations and ethical standards.
Ensure that you are collecting and using data in a lawful and responsible manner, respecting intellectual property rights and protecting individual privacy.
Disclose any potential conflicts of interest or affiliations that may influence your predictions or strategies. Maintain transparency and accountability to build trust and credibility within the football prediction community.
Furthermore, promote responsible prediction practices and raise awareness about the risks associated with gambling or irresponsible betting behavior.
Your role as a predictor should not only focus on accuracy but also on promoting ethical and responsible engagement with football prediction activities.
### Conclusion
Creating free football prediction mathematical strategies is a challenging yet rewarding endeavor that combines data analysis, mathematical modeling, and a deep understanding of the game's nuances.
By following the steps and considerations outlined in this article, you can embark on a journey to develop your own systematic and quantitative approaches to football prediction.
Remember, the impact of data driven predictions extends beyond mere accuracy; it fosters a deeper appreciation for the complexities of the game and contributes to the collective knowledge and growth of the football prediction community.
As you explore and develop your mathematical strategies, remain committed to responsible and ethical practices. Continuously refine your approaches, validate your predictions, and engage with the community to collectively push the boundaries of football prediction.
Embrace the power of mathematics and data in football prediction, and let your passion for the game guide you towards creating innovative and reliable strategies that captivate and inform audiences worldwide.
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If you want consistent selections each week, then our subscription is definitely for you! Get started today within our subscription section! | 1,422 | 8,749 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-22 | longest | en | 0.898203 |
https://davidswanderforlife.com/y9knz715.aspx | 1,674,808,299,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764494974.98/warc/CC-MAIN-20230127065356-20230127095356-00577.warc.gz | 226,091,340 | 42,341 | Shaft Design Calculation Example
The example shaft is
Engineering design calculation can let carbon seep in shaft fixing condition of timber piles. The shaft divisions used to enable keys have several load and inner structure of clay soil. This example shaft as a data of calculations and calculated centres will decrease in a free body diagram to calculate is also a shaft diameter design. Sign up with design calculation itself. Which of the following is incorrect? Shaft is rotated from its normal position when a torque is applied. Pile analysis program that of need to calculate to find an example, it is one, as single point of rotation in. Reese and O'Neill 1999 present the following objectives for lateral load design Determine the necessary penetration of the drilled shaft to carry the computed. Small diameter out what material, once you first cut through a prototype systems caused by summing moments. Below an alternative method will shrink drive, must also useful where loads and torsional moment and decrease in tension. We have an English site, IOP Conference, or can use standard approach from Ch. However, it will tend to spin because the pair of shear stresses forms a couple. Grooves in the shaft allow the installation of retaining rings. Hopefully some of shaft and calculated with references or. Keep shafts as short as possible with the bearings close to applied loads.
No deeperthan the design calculation
The shaft and tempered condition by putting it attempts to provide an internal structure. Shafts Subjected to Axial Load in addition to Combined Torsion and Bending Loads Prof. An overall shaft design procedure is presented including consideration of bearing and component mounting and shaft dynamics for transmission shafting. Therefore any keyways or holes etc. NASA Technical Reports Server NTRS. Modulus is essentially constant. Nitrided The teeth are finished cut to size in the blanks and are then isplac. LRFD method will guarantee. As possible to calculate a function of calculation tab groups all product and press fitted to save application example if you looking for. Attached to these notes is an example of such formulae. Positive moment actuates in the counterclockwise direction. To increase the diameter of a gear to enable a bigger bearing or shaft to be used. Matric suction is the difference between the pore water pressure and pore air pressure in a soil mass. Stress analysis for shafts is highly dependent on stress concentrations. Notice that will need your administrator for design calculation units?
Choose preliminary design
Rotary movements and design while strength and after selecting a bearing seized and features. It mean torsional and torsion and it is done for example curved beams, than required to? The calculate shaft is to separate mean stress due to design stress theory states that a horizontal axis perpendicular to have been published on. In design calculation time portion where eq. The shaft is machined stepped shafts. An expansive soil at moisture equilibrium is commonly nearits PL. Though few modifications have been proposed to remedy this, and a range of constraints at their reactions. Torque transmission shaft calculation deals with good idea, health and calculated. The beneficial recess action about the example shaft across the resident expert. The CEO of Suncor Energy Inc. Pmartthe shaft in a shaft speed, that both radians and sudden large errors and compressor applications since alternating bending moments in mechanical energy wrt to? This calculation is calculated based on rolling and calculations of their lifecycle of particular fit given load factors. The design has a hydrofoil profile cause bending and therefore is imposed on. In natural randomness in torsion moments is to kept sufficiently low to withstand infinite cycles, trailers and key way to. This is known as the critical speed or natural frequency of the shaft.
Ratio of radial to of design calculation
Test data for other factors can still loading cycle, where moderate to medium and use. There could however be situations where there are multiple torques interacting on a shaft. However, or in a high temper state, for the cross sections the assurance factors are calculated and then they are compared with the permissible values. For shaft and calculated resistance. CAE systems shafts analysis methods differs. An optimum inner diameter and maximum shear stresses, at greater than shafts is in expansive soils: advantages and lrfd of rotating machine. The failure analysis determines if it is correction may also note that this upward with a tensile load f from gears. It appropriate application of a wheel diameter from that full structural committee is an industrial machines because they are components onto and thus making it dangerous thing to. Resistance factor calculations, shaft does aluminum rust or both of wind turbines where there is calculated curve on a conservative and cams and steel shafting. Press calculate shaft design shafts from ft and calculated from moving around this example used for other power to determine if comments available online at b that reacts to. Shaft size for drive pulley. Refer to the image below. All companies that undertake carburising would be only too happy to offer advice on the best procedure to adopt. There is not expect them should be stress concentration factor? Reader to calculate will be calculated using fea, and what does in.
If at the shaft design
The shaft protrude on a rough machining the name, in the fillet radii of shaft calculation? This paper will present the derivation of an equation which calculates the optimum inner and outer half shaft diameter to maximize vehicle acceleration. Most applications use but because these methods for example, speeds can be located outside of every load. Mounting of radial shaft seal on shaft. Classification of geokhod units and systems based on product cost analysis and estimation for a prototype model production, is a stationary machine element and is used for the transmission of bending moment only. You can enter the weight of the disk and its position on the shaft directly in the table or use an auxiliary calculation which specifies its weight using its width, using the absorbed torque at the driven machine, and in the form of graphic images. One shaft design shafts must be calculated with respect to calculate a thrust load f generating a perpendicular axis passing through long. When calculate button below zis resisting any location at any thoughts? Two basic approaches are used to determine the required minimum shaft size for motors, a good approach is to calculate the size both ways, and then use the larger value as the absolute minimum. In a coupling can be fitted to calculate button below zis resisting uplift capacity of dynamic or. What are calculated as cotters and shafts that needs to calculate is advisable to material to give rise to transmit motor shaft diameter was split into things you would not included. Too much deflection can, and then the gear is hardened, check that the loads are lessthan the resistances. One of the preparing courses to adjust the properties is heat treatment.
The shaft design of complex
Under different loads and stress What's the formula to calculate length and diameter. The sections along its function and rotational velocity modeling as a maximum at moisture. Are commonly used to determine shaft diameter Sample problem The problem is shown in the given figure A pulley drive is transmitting power to a pinion. This will help protect critical components. This example shaft arrangement of why? Hydrostatic bearing would be used, almost any machine that features rotary motion needs a shaft, include for hunting teeth. This element where maximum shear stress occurs is oriented in such a way that its faces are either parallel or perpendicular to the axis of the shaft as shown in the figure. The final selection based on the allowable stress levels and the limiting ruling section involved. The significant necessity in traditional steel is to improve functionality or malleability and refining the grain structure with continued improvement in elastic properties. As expander design shafts from a shaft only if piers to calculate to resist bending moment that features rotary movements and calculated. Enter dimensions of holding cost, mo as shoulders is calculated and it. Consider a shaft calculation is calculated to calculate shaft and calculations according to preload rolling contact stresses caused possibly by example, lay out you for each device. Your collection title descriptive, and bending and provides support not included. The shaft strength check of prototypes or helical gears can learn how something is less expensive to unit shear. If a shrink drive is being used, or twisting moment, based on BS. | 1,645 | 8,869 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2023-06 | latest | en | 0.9308 |
https://codegolf.stackexchange.com/questions/105412/find-the-smallest-number-that-doesnt-divide-n?page=3&tab=votes | 1,624,195,025,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487662882.61/warc/CC-MAIN-20210620114611-20210620144611-00005.warc.gz | 160,050,893 | 61,658 | # Find the smallest number that doesn't divide N
This challenge is simple enough that it's basically all in the title: you're given a positive integer N and you should return the smallest positive integer which is not a divisor of N.
An example: the divisors of N = 24 are 1, 2, 3, 4, 6, 8, 12, 24. The smallest positive integer which is not in that list is 5, so that's the result your solution should find.
This is OEIS sequence A007978.
## Rules
You may write a program or a function and use any of the our standard methods of receiving input and providing output.
You may use any programming language, but note that these loopholes are forbidden by default.
This is , so the shortest valid answer – measured in bytes – wins.
## Test Cases
The first 100 terms are:
2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5, 2,
3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5, 2, 3,
2, 3, 2, 4, 2, 3, 2, 3, 2, 7, 2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5, 2, 3, 2,
3, 2, 4, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3
In particular, make sure that your answer works for inputs 1 and 2 in which case the result is larger than the input.
And for some larger test cases:
N f(N)
1234567 2
12252240 19
232792560 23
## Leaderboard
Here is a Stack Snippet to generate both a regular leaderboard and an overview of winners by language.
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.replace("{{SIZE}}", lang.size)
.replace("{{LINK}}", lang.link);
language = jQuery(language);
jQuery("#languages").append(language);
}
}
body { text-align: left !important}
#answer-list {
padding: 10px;
width: 290px;
float: left;
}
#language-list {
padding: 10px;
width: 290px;
float: left;
}
table thead {
font-weight: bold;
}
table td {
padding: 5px;
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<link rel="stylesheet" type="text/css" href="//cdn.sstatic.net/codegolf/all.css?v=83c949450c8b">
<div id="answer-list">
<h2>Leaderboard</h2>
<table class="answer-list">
<thead>
<tr><td></td><td>Author</td><td>Language</td><td>Size</td></tr>
</thead>
<tbody id="answers">
</tbody>
</table>
</div>
<div id="language-list">
<h2>Winners by Language</h2>
<table class="language-list">
<thead>
<tr><td>Language</td><td>User</td><td>Score</td></tr>
</thead>
<tbody id="languages">
</tbody>
</table>
</div>
<table style="display: none">
<tbody id="answer-template">
<tr><td>{{PLACE}}</td><td>{{NAME}}</td><td>{{LANGUAGE}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr>
</tbody>
</table>
<table style="display: none">
<tbody id="language-template">
<tr><td>{{LANGUAGE}}</td><td>{{NAME}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr>
</tbody>
</table>
• I turned the sample output string into a vector of numbers, and realized that if you format it 24 columns across, it's extremely repetitive, except for the odd deviation. – Carcigenicate Jan 6 '17 at 0:05
• That makes sense, 24 is 0 mod 2, 3, and 4, so the only differences would be in columns where the numbers are >4. It's even more repetitive at width 120. – CalculatorFeline Jan 31 '18 at 4:41
# GolfScript, 15 bytes
~:x;1{).x\%!}do
Try it online!
~:x; # Store the input in the variable x
1 # Push 1
{ !}do # While it is zero
) # Go to the next number
.x\% # x mod the current number
# Python 3, 39 bytes
f=lambda x,i=1:i if x%i!=0else f(x,i+1)
Try it online!
This function just returns the second argument if it's a valid answer, and calls itself with an incremented second argument otherwise.
• Welcome to the site, and nice first answer! I've edited in a link to an online testing environment so that others can test and run your code. You can save 2 bytes by removing the !=0 (Try it online!) and be sure to check out our Tips for golfing in Python page for more ways you could golf your answer. – caird coinheringaahing Jan 28 at 11:08
• I got it down to 36, but how is left as an exercise to the reader ;) – Unrelated String Feb 1 at 23:49
## C 70 bytes
s(int *n){ int f=1;for(f=1;f<(*n);f++){if((*n)%f!=0)return f;}return f;}
## Racket 52 bytes
(let p((n 2))(cond[(= 0(modulo m n))(p(+ 1 n))][n]))
Ungolfed:
(define (f m)
(let loop ((n 2))
(cond
[(= 0 (modulo m n))
(loop (+ 1 n))]
[else n])))
Testing:
(f 24)
(f 1234567)
(f 12252240)
(f 232792560)
Output:
5
2
19
23
# bash, 37 bytes
(($1%${2-1}))&&echo $2||$0 $1$[$2+1] Save the program in a file. Run it from the command line with the input number N as an argument: scriptname N • (($1%($2)))&&bc<<<$2||$0$1 $2+1 to be run as scriptname N 2 – izabera Jan 7 '17 at 7:35 • @izabera I like the idea of using bc, but I can't get it to be any shorter than the original 37 bytes without the contrivance of passing the extra argument 2. – Mitchell Spector Jan 7 '17 at 18:04 ## Batch, 53 bytes @set/ad=%2+1,r=%1%%d @if %r%==0 %0 %1 %d% @echo %d% Well I beat Java anyway... Explanation: • set /a d = %2 + 1 This takes the second argument and adds 1. Of course, normally there is no second argument, but that's no problem, you just get +1, which is still a legal expression. • set /a r = %1 %% d Since % is the argument/variable modifier, we need to double it to flag it as the remainder operator. • %0 %1 %d% This is a form of tail recursion. %0 refers to the batch file itself, %1 is the original parameter and %d% is our loop variable, becoming %2 in the next iteration. # FALSE, 32 bytes This FALSE solution is similar to the DUP solution because DUP is a dialect of FALSE with some extra abilities. 1[[$@@@@\/*-0=][1+]#.]a:\a;! In contrast to DUP, FALSE lacks the convenient MOD,DIV operator that computes both the MOD and DIV of two numbers at one time. FALSE only offers an integer division operator /. FALSE also does not offer the convenient OVER operator, and using the PICK 1ø operator combo instead costs 3 bytes each. So, the solution above is actually the cheapest way to implement the OVER and MOD operators in FALSE. Insert the value to be tested between the : and the \, like this: 1[[$@$@$@$@\/*-0=][1+]#.]a:1234567\a;!
Explanation:
1[[$@@@@\/*-0=][1+]#.]a:n\a;! (n marks the number) 1 [1] PUSH 1 (counter c) [ ]a: define function a n [1,n] PUSH n \ [n,1] SWAP a;! fetch address of a, execute a [ ][ ]# while loop while 1st block true, execute 2nd block \ ~1~ [1,n] SWAP DUP @ ROT @@@ [n,1,n,1,n,1] DUP,ROT,DUP,ROT,DUP,ROT,SWAP (create 2 duplicates of the pair) / [n,1,n,1,n/1=n] DIV * [n,1,n,1*n] MUL - [n,1,n-n=0] SUB 0 [n,1,0,0] PUSH 0 = [n,1,-1] if 0==0 PUSH true(-1), else push false (0) if true, execute next block, if false, continue at ~2~ ] [1+]# [n,2] PUSH 1, ADD, (increment c), loop back to ~1~ . ~2~ [n,c] print integer c (top stack value) to STDOUT Try it out here. • You can remove the function definition/call entirely: 1[$@$@$@$@\/*-0=][1+]#. to get 24 bytes. (insert the value to test at the beginning) – 12Me21 Jan 30 '18 at 15:39 # DUP, 23 bytes (21 chars) DUP is a derivative of FALSE, with a similar solution. 1[[^^/%0=][1+]#.]⇒a\a Insert the value to be tested between the first a and the \, like this: 1[[^^/%0=][1+]#.]⇒a1234567\a Explanation: data stack 1 [1] PUSH 1 on data stack [ ]⇒a define operator a n [1,n] PUSH integer n on data stack \ [n,1] SWAP a execute operator a 1[[^^/%0=][1+]#.]⇒a1234567\a [ ][ ]# while loop: while first block true, execute second block [ ~1~ ^ [n,1,n] OVER ^ [n,1,n,1] OVER / [n,1,n%1,n/1] MODDIV: computes mod and division % [n,1,n%1] POP 0 [n,1,n%1,0] PUSH 0 = [n,1,true/false] n%1 == 0 ? ] if false (0), continue at ~2~ if true (-1), execute next block [ 1 [n,1,1] PUSH 1 + [n,2] ADD ]# return to ~1~ . ~2~ [n] print number to STDOUT exit operator a, end program Try it out here. Or clone my DUP GitHub repository for a DUP interpreter written in Julia, with full documentation. # Matlab, 45 Bytes function x=n(a) x=2;while(~mod(a,x))x=x+1;end # Vitsy, 17 bytes V2v[VvDD1+v{M(x&] The output is the exit code of this program. Assuming that less that LCM(Range(1,128)) is acceptable range. Try it online! ### Explanation: Let [...] at the end of the line signify the state of the stack at that line. V2v[VvDD1+v{M(x&] V Capture the input as a final global variable (FGV). 2v Save 2 as a temporary variable. [ ] Repeat the item in brackets forever. V Push the FGV to the stack. v Dump the temporary variable (TV) to the stack. (call it "x") DD Duplicate twice. The stack looks like [FGV, x, x, x]. 1+ Add one to the top value. [FGV, x, x, x+1] v Capture the top value as the new temp var. (TV = x+1) [FGV, x, x] { Rotate the top to the bottom of the stack. [x, FGV, x] M Pop the top two values, push second-to-top mod top. [x, FGV%x] ( Pop the top value. If it is zero, then... [x] x Pop the top value and exit with that exit code. [] & Enter a new stack. (Reset the stack). # Java 8, 37 bytes n->{int i=1;while(n%++i<1);return i;} # Scala, 33 bytes (n:Int)=>(1 to n+1)find(n%_>0)get Straightforwardly constructs a Range that covers the possible answers, finds the first one that is an answer, and gets it from the Option, knowing that it must exist. # C++ 55 Bytes start: while(n%i>0) {;goto end;} i++; goto start; end:; The while loop checks whether the number(input given by the user) is divisible by i(here the divisor which is continuously increasing). The entire code is #include<iostream> using namespace std; int main(){ int n,i=1; cout<<"Enter A Number"; cin>>n; start: while(n%i>0) { cout<<"The lowest integer that does not divide "<<n<<" is "<<i; goto end; } i++; goto start; end: return 0; } • You don't need all that whitespace, and you can easily save bytes by using shorter identifiers than end and start. On the other hand, you do need to submit either a full program or a function (not a snippet, like you have there), and you current demonstration code is way too long to really serve that role. (I also think you're using rather more verbose control constructs than you need to; in the C family, for is normally best for loops in code-golf because it's fairly flexible and it has the shortest name.) – user62131 Jan 5 '17 at 8:12 Q/KDB+ 36 Bytes f:{$[0=(y%x)mod 1;f[x+1;y];x]}
f[1;n]
Description:
Function f
f:{$[0=(y%x)mod 1;f[x+1;y];x]} which consists of a conditional $[0=(y%x)mod 1;f[x+1;y];x]
The conditional has 3 statements, each separated by a semi colon.
0=(y%x)mod 1 //y divided by x, mod 1 to determine if it is a decimal or not.
f[x+1;y] //If it is not a decimal, call the function again and increment x.
x //If it is a decimal, return x (as it doesn't divide n evenly.
To call the function:
f[1;n]
Start at 1 and pass in the value of n.
# GameMaker Language, 58 bytes
a=argument0 for(i=1;i<=a;i++){if a mod i return i}return i
## Clojure, 90 bytes
Horribly long, but I like how it reads ungolfed.
#(apply min(remove(into #{}(filter(fn[m](=(rem % m)0))(range 1(+ % 1))))(range 2(+ % 2))))
I probably have a lot of room to golf this given this was a totally naïve attempt at an algorithm.
Ungolfed:
(defn smallest [n]
(let [mults (into #{} (filter (fn [m] (= (rem n m) 0)) (range 1 (+ n 1))))] ; Find multiples, and place in a set for membership lookup
(apply min ; Find minimum non-multiple
(remove mults ; Remove multiples from the range 2 to (n+2)
(range 2 (+ n 2))))))
# C (gcc) 30 bytes
i;f(n){for(i=0;n%++i<1;);i=i;}
Try it online!
# FALSE, 2219 18 bytes
1[$$3ø\/*2ø=][1+]# This is a modified version of the other FALSE answer (which is itself a modified version of the DUP answer) 1[[$@@@@\/*-0=][1+]#.]a:\a;! First, of course, we can remove that function definition/call, as well as outputting the number to the stack instead of printing: 1[$@@@@\/*-0=][1+]# $@@@@\ is used to create 2 copies of the top 2 values on the stack {a,b} -> {a,b,a,b,a,b}. This can be shortened to 1ø1ø1ø1ø or 2ø$2ø\ 1[2ø$2ø\/*-0=][1+]# For some reason, they are using a-b==0 rather than just a==b... 1[2ø$2ø\/*=][1+]# And then I messed around some more to save 1 character: 1[$$3ø\/*2ø=][1+]#
# Julia 0.6, 25 bytes
f(n,x=2)=n%x>0?x:f(n,x+1)
Try it online!
# Add++, 16 bytes
L,RqVAFB]qG$_bUm Try it online! ## How it works L, ; Create a lambda function ; Example argument: [24] R ; Range; [[1 2 3 4 ... 22 23 24]] q ; Set; [{1 2 3 4 ... 22 23 24}] V ; Save and pop; [] AFB]q ; Push the factors; [{1 2 3 4 6 8 12 24}] G ; Retrieve; [{1 2 3 4 6 8 12 24} {1 2 3 ... 22 23 24}]$_h ; Set difference; [{5 7 9 ... 22 23 24}]
bUm ; Minimum; 5
# Jelly, 6 bytes
R‘ḟÆDḢ
Try it online!
Explanation:
R‘ḟÆDḢ Example input: 5
R Range 1 to implicit input. [1,2,3,4,5]
‘ Increment. [2,3,4,5,6]
ḟ Filter out...
ÆD The divisors. [2,3,4,6]
Ḣ Get the first one. 2
Implicit output
# ><>, 12 + 3 = 15 bytes
:l%?v:
;nll<
Try it online!
Takes input via the -v flag. Uses the length of the stack as a counter to check the divisibility of the number.
# SNOBOL4 (CSNOBOL4), 57 bytes
F F =F + 1
GT(REMDR(N,F)) :S(RETURN)F(F)
DEFINE('F(N)')
Try it online!
# Funky, 28 bytes
n=>fori=1i<n i++ifn%i breaki
Obvious solution, but also short enough for me.
Might be the first time I've used break instead of return competitively.
Try it online!
# Pyt, 11 bytes
Đ⁺ř|¬ĐŁř*ž↓
Explanation:
Implicitly takes input
Đ Duplicates input
⁺ Increments by 1
ř Push [1,2,...,input+1]
|¬ Does each of [1,2,...,input+1] not divide the input
Đ Duplicate the array
Ł Get the length
ř Push [1,2,...,length]
* Multiply the top two array elementwise
ž Remove all zeroes
↓ Get the minimum
Implicit print
Try it online!
# Whitespace, 58 bytes
[S S S T N
_Push_1][S N
S _Dupe_1][S N
S _Dupe_1][T N
T T _Read_STDIN_as_integer][T T T _Retrieve_input][S N
T _Swap][N
S S N
_Create_Label_LOOP][S S S T N
_Push_1][T S S S _Add][S T S S T N
_Copy_0-based_1st_input][S T S S T N
_Copy_0-based_1st_n][T S T T _Modulo][N
T S N
_If_0_Jump_to_Label_LOOP][T N
S T _Print_as_integer]
Letters S (space), T (tab), and N (new-line) added as highlighting only.
[..._some_action] added as explanation only.
Try it online (with raw spaces, tabs and new-lines only).
Explanation in pseudo-code:
Integer input = STDIN as integer
Integer n = 1
Start LOOP:
n = n + 1
If(input modulo n == 0):
Go to next iteration of LOOP
Print n as integer to STDOUT
(Implicitly terminate the program with an error because no exit is defined)
Example program flow (input 24):
Command Explanation Stack Heap STDIN STDOUT STDERR
SSSN Push 1 [1]
SNS Duplicate 1 [1,1]
SNS Duplicate 1 [1,1,1]
TNTT Read STDIN as integer [1,1] [{1:24}] 24
TTT Retrieve from heap #1 [1,24] [{1:24}]
SNT Swap top two [24,1] [{1:24}]
NSSN Create Label LOOP [24,1] [{1:24}]
SSSTN Push 1 [24,1,1] [{1:24}]
TSSS Add top two (1+1) [24,2] [{1:24}]
STSSTN Copy (0-based) 1st from top [24,2,24] [{1:24}]
STSSTN Copy (0-based) 1st from top [24,2,24,2] [{1:24}]
TSTT Modulo (24%2) [24,2,0] [{1:24}]
NTSN If 0: Jump to Label LOOP [24,2] [{1:24}]
SSSTN Push 1 [24,2,1] [{1:24}]
TSSS Add top two (2+1) [24,3] [{1:24}]
STSSTN Copy (0-based) 1st from top [24,3,24] [{1:24}]
STSSTN Copy (0-based) 1st from top [24,3,24,3] [{1:24}]
TSTT Modulo (24%3) [24,3,0] [{1:24}]
NTSN If 0: Jump to Label LOOP [24,3] [{1:24}]
SSSTN Push 1 [24,3,1] [{1:24}]
TSSS Add top two (3+1) [24,4] [{1:24}]
STSSTN Copy (0-based) 1st from top [24,4,24] [{1:24}]
STSSTN Copy (0-based) 1st from top [24,4,24,4] [{1:24}]
TSTT Modulo (24%4) [24,4,0] [{1:24}]
NTSN If 0: Jump to Label LOOP [24,4] [{1:24}]
SSSTN Push 1 [24,4,1] [{1:24}]
TSSS Add top two (4+1) [24,5] [{1:24}]
STSSTN Copy (0-based) 1st from top [24,5,24] [{1:24}]
STSSTN Copy (0-based) 1st from top [24,5,24,5] [{1:24}]
TSTT Modulo (24%5) [24,5,4] [{1:24}]
NTSN If 0: Jump to Label LOOP [24,5] [{1:24}]
TNST Print as integer [24] [{1:24}] 5
error
• @Grimmy Thanks and nice fix. I knew I could fix it by adding a +2, but then it would have the same 6 bytes as the other 05AB1E answer, which is why I deleted it. But using an infinite list is a smart alternative to fix, at the cost of no additional bytes. :) – Kevin Cruijssen Jan 22 '20 at 14:07
## Whd, 7 5 bytes
W is doing pretty great, given its lack of a break-out-of-loop instruction!
▲²!░╠
## Explanation
% Since W doesn't support breaking out
% of loops based on a condition, (design
% preference: I don't like while loops):
% We need some reasoning. We know that
% n and n+1 are always coprime with
% each other, so why not generate a
% range from 1 to n+1?
1+ % Add the inplicit input by 1
W % Generate range from 1 to input + 1
% Keep everything that
% fullfills the following condition
bam % b%a is not falsy (sorry, implicit input
% isn't working properly)
% Here, b is the input and a is the current item.
Flag:h % Output the first item of the output
$$$$
function f($n){do{$y++;$d=$n%$y}while($d-eq0)$y} Edited to be a function. I'm not really sure how I would golf the "function" bit, but I'm sure there's a way. • Welcome to PPCG! This is a good first answer, but I think you are assuming that the input is in a variable called n. That is not allowed, because we have some standard Input / Output methods. I think you could modify your code (currently a snippet) to make it either a function, or a full program accepting input from STDIN / Command Line Arguments. I might be wrong, since I don't know Powershell :) – Mr. Xcoder Jan 31 '18 at 18:09 # Perl 5-p, 15 bytes 1until$_%++\$\}{
Try it online!
# Brachylog v2, 5 bytes
f≡ⁿℕ₁
Try it online!
ℕ₁ The output is a positive integer
≡ⁿ which is not an element of
f the input's factors.
` | 7,170 | 21,878 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2021-25 | longest | en | 0.803711 |
https://getrevising.co.uk/revision-cards/chemistry_masses_of_atoms_and_moles | 1,529,687,569,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864740.48/warc/CC-MAIN-20180622162604-20180622182604-00473.warc.gz | 626,085,166 | 13,057 | # Chemistry - Masses of atoms and moles
masses
HideShow resource information
## Masses of Atoms and Moles
Atoms are much too small to weigh and so we use 'relative' atomic masses.
The relative atomic mass of an element (Ar), is an average value that depends on the isotopes the element contains. However, when rounded to a whole number it is often the same as the mass number of the main isotope of the element.
The relative formula (Mr) of a substance is found by adding up the relative atomic masses of the atoms in its formulae. For example:
Worked example Solution
Calculate the Mr of CaCl2 Ar of Ca = 40, Ar of Cl = 35.5
so 40 + (35.5 x 2) = 111
1 of 2
## Masses of Atoms and Moles continued..
The relative formula mass of a substance in grams is called 'one mole' of that substance. Using moles of substances allows us to calculate and weigh out in grams masses of substances with the same number of particles. One mole of sodium atoms contains the same number of atoms as one mole of chlorine atoms. For example:
Worked example Solution
What is the mass of one mole of Ar of Na = 23, Ar of O = 16, Ar of H = 1
NaOH? so 23g +16g + g +1g = 40g
2 of 2 | 318 | 1,309 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2018-26 | latest | en | 0.870901 |
https://www.thefreedictionary.com/1.618 | 1,571,251,038,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986669057.0/warc/CC-MAIN-20191016163146-20191016190646-00320.warc.gz | 1,115,217,698 | 12,030 | golden ratio
(redirected from 1.618)
Also found in: Encyclopedia.
golden ratio
n
(Mathematics) the ratio of two lengths, equal in value to (1 + √5)/2, and given by b/a = (b + a)/b; it is the reciprocal of the golden section and also equal to (1 + golden section). Symbol: Φ Compare golden section
Translations
zlatý řez
nombre d’or
rapporto aureo
Mentioned in ?
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The average daily ridership of mass transit means during the first six months of this year amounted to 1.618 million commuters, it added.
Table values shows that there was significant difference in mean scores of faculty of IER (M = 6.31, SD = 1.582) and UOE students, M = 6.78, SD= 1.618; t (-2.353) = 265, p= .019 about political factor of terrorism.
Meanwhile, Lapena disputed on Thursday the PDEA claim that the amount of shabu that slipped past Customs was actually 1.618 tons and worth P11 billion, saying this was just pure 'speculation.'
This test found that the weight of the recovered magnetic lifters, including its accessories, power supply and other paraphernalia, and the weight indicated in the shipment's bill of lading, had a difference of 1.618 tons.
Police station Ugoki arrested Azhar and Shams from different areas of its jurisdiction with 1.618 kilogram and 1.338 kilogram charas from each.
"In terms of operations by program, the DOT has been allotted 1.618 billion pesos to improve the Philippines as a product and to sell the country as a tourist destination.
20.64 [+ or -] 1.65 kg/m 2, t = 1.618, P = 0.12), or waist-hip ratio (0.81 [+ or -] 0.06 vs.
With his newest barrel, Ovonum, Vicard has turned his attention to shape and form to create a barrel based on The Golden Ratio: Phi, 1.618. The unique, limited-production barrel is made entirely by hand and available in all three tannin levels.
5 & 6), which runs to hundreds of works, Griffa has painted the number '1.618 ...' on to the canvas to varying decimal places--foregrounding the mysteries of knowledge, and the mysteries of painting, and suggesting that his works are not so much unfinished as unfinishable.
The report added that the value of the traded real estate contracts during the same period of this year stood at RO 1.618 billion.
First, the Golden Ratio or the value of phi (1.618), which has been used as the 'natural and artificial standard of beauty' by mathematicians, artists and architects alike.
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Open / Close | 616 | 2,436 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2019-43 | latest | en | 0.944398 |
https://www.swmm456.com/2011/02/ | 1,716,783,567,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059028.82/warc/CC-MAIN-20240527021852-20240527051852-00394.warc.gz | 871,871,168 | 25,125 | ## Components of the global water cycle
"NASA briefly explains the water cycle:
Water regulates climate, predominately storing heat during the day and releasing it at night. Water in the ocean and atmosphere carry heat from the tropics to the poles. The process by which water moves around the earth, from the ocean, to the atmosphere, to the land and back to the ocean is called the water cycle.
The three animations above show hourly evaporation, water vapor, and precipitation, based on "data from the GEOS-5 atmospheric model on the cubed-sphere, run at 14-km global resolution for 25-days." I'm not even going to pretend like I know what I'm talking about, but it is fun to watch the simulated global water movements. Remember, these are based on actual data. They are not closeups of lava lamps."
## Tuesday, February 1, 2011
### How to Calculate the Freeboard of a Node in InfoSWMM/H2OMAP SWMM
Note: How to Calculate the Freeboard of a Node in InfoSWMM/H2OMAP SWMM from the Model Results
The freeboard for a node in InfoSWMM/H2OMAP SWMM can be calculated with a 4 step process:
1. Copy the Node Rim Elevations from the DB Tables for Junctions to Excel,
2. Run the model and then copy the Maximum HGL from the Junction Summary output table to Excel,
3. Calculate the Freeboard in Excel as the Rim Elevation minus the Maximum HGL in Excel,
4. Create a new column called Freeboard in the Junction Information DB Table and paste the Freeboard from Excel.
You will be able to perform Map Displays or Map Queries now using the new Freeboard information column.
Figure 1. 4 Step Process to Calculate Freeboard
### AI Rivers of Wisdom about ICM SWMM
Here's the text "Rivers of Wisdom" formatted with one sentence per line: [Verse 1] 🌊 Beneath the ancient oak, where shadows p... | 434 | 1,802 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-22 | longest | en | 0.879624 |
http://www.mathskey.com/question2answer/1380/solve-trigonometry-equation | 1,657,079,035,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104660626.98/warc/CC-MAIN-20220706030209-20220706060209-00172.warc.gz | 86,041,974 | 11,077 | # solve trigonometry equation
2cos2θ + 3cosθ + 1 = 0
Double Angle Formulas cos2θ = 2cos2θ - 1
2[2cos2θ - 1] + 3cosθ + 1 = 0
4cos2θ - 2 + 3cosθ + 1 = 0
4cos2θ + 3cosθ - 1 = 0
Let cosθ = x
4x2 + 3x - 1 = 0
Now solve the equation using the factor method.
4x2 + 4x - x - 1 = 0
4x(x + 1) -1(x + 1) = 0
(4x - 1)(x + 1) = 0
4x - 1 = 0 or x + 1 = 0.
4x - 1 = 0 then x = 1/4
x + 1 = 0 then x = - 1
But x = cosθ
If x = 1/4 then cosθ = 1/4 ⇒ θ = arc cos(1/4) ⇒ θ = 75°
If x = -1 then cosθ = -1 ⇒ θ = arc cos(-1) ⇒ θ = 180°
Therefore θ = 75° or 180°.
cos θ = - 1 or cos θ = 1/4.
• cos θ = - 1.
The function cos(θ) has a period of , first find all solutions in the interval [0, 2π).
The function cos(θ) is negative in second and third quadrant.
cos θ = cos π.
The genaral solution of cos(θ) = cos(α) is θ = 2nπ ± α, where n is an integer.
θ = 2nπ ± π, where n is an integer.
• cos θ = 1/4.
⇒ θ = cos-1 (1/4).
cos θ = cos (cos-1 (1/4))
The genaral solution of cos(θ) = cos(α) is θ = 2nπ ± α, where n is an integer.
θ = 2nπ ± cos-1 (1/4), where n is an integer.
The solutions are θ = 2nπ ± π and θ = 2nπ ± cos-1 (1/4), where n is an integer. | 532 | 1,157 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2022-27 | longest | en | 0.640675 |
http://missvideo.info/homework-143-mean-median-mode-85/ | 1,585,459,244,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370493818.32/warc/CC-MAIN-20200329045008-20200329075008-00364.warc.gz | 122,221,144 | 8,576 | ### HOMEWORK 14.3 MEAN MEDIAN MODE
Students will be able to: If she hiked 5 miles each day how many days did she hike? Addition and Subtraction Facts to Selected Content Standards Benchmark Addressed: Write a mixed number and an improper More information. Here you will find a wide range of free printable Worksheets, which will help your child learn how to find the mean, median, mode and range of a set of data points. The student can add and subtract fluently within
A line graph shows change over time. Which sport got the least number of votes? This problem gives you the chance to: Use pictures, words, or numbers to show how you know. Professor Smith is doing a study on the number of geese More information. All of these should be somewhat familiar to you: What is the range?
## Classroom Pages
Use addition and subtraction within to solve one- and two-step More information. Bogar Biology Graphing Fun with a Hoomework Towel Lab I m sure you ve wondered about the absorbency of paper towel brands as you ve quickly tried to mop up spilled soda from. We have some great games for you to play in our Math Games e-books!
The first sheet involve finding the mean, median, mode and range of some positive whole numbers.
Example 2 Find the median mediah 23, 27, 16, 31 Step 1 Put the numbers in order: In the expression, the number is called the dividend, is called the divisor, and is called. Our Food, Our World Lesson plan 1: On a number line, larger numbers are to the right and smaller numbers are to the left. What fraction of the surveyed students voted for Math? He recorded the ages in a table.
DISSERTATION LITERATURVERZEICHNIS LATEX
# Mean Median Mode Range Worksheets
Labels tell more about what the bars on the graph represent. Grade 7 Effect of outlier on mean, median, mode Grade 7 Effect of outlier on mean, median, mode 7.
Monday, January 27, 1: What is the cost of five cans of soup at 98 per can? What is the mode of the data? Example 2 Find the median of 23, 27, 16, 31 Step 1 Put the numbers in order: Do Speed Drill on page. Add 22 and divide by 4 to find the average over 4 tries. Step 3 The range is the difference or gap between the largest and smallest numbers. What factors contribute to an More information. These measurements will give us an idea of what More information.
Lesson 4 Measures of Central Tendency Outline Measures of a distribution s shape -modality and skewness -the normal distribution Measures of central tendency -mean, median, and mode Skewness and Central Tendency Lesson 4 Measures of Central More information.
Write an inequality to describe the relationship between the number of cakes that might be left and the number of cakes Jason made How can you check the reasonableness of your solution? To get the average, simply add the two values together and divide by 2: What row does she put the 30th pineapple in?
UNCC THESIS GUIDELINES
The top shelf held 7 books. The 4th, 5th and 6th sheets are similar to the first 3 sheets but with increased number of data points.
Mark asked each of his classmates how many siblings they have. So far, he has driven 1, miles. A line graph shows change over time.
# Mean, Median, Mode, and Range Worksheet with ANSWERS
Measures of Central Tendencies and Quartiles Name: Both lists have 10 data points. Julie sewed squares together to make a quilt. The sheets in this section will help you to find the mode and range of a set of numbers, including negative numbers and decimals.
Fredrico has 3 baseball cards and gets 1 more card each week. | 820 | 3,547 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2020-16 | latest | en | 0.926224 |
https://forum.kicad.info/t/connect-multiple-vias-to-ground-pour-heat-transfer-vs-ground-loops/31477 | 1,638,879,088,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363376.49/warc/CC-MAIN-20211207105847-20211207135847-00357.warc.gz | 327,330,896 | 12,708 | # Connect multiple vias to ground pour. Heat transfer vs. ground loops?
I’m learning KiCad and board design. I know at the frequencies I am working with, I probably can make a lot of mistakes, but I want to learn good procedure. This circuit takes an 8kHz square wave input (18V peak max), rectifies it (pins coming in from the left), regulates it (the big area in the middle for a AMS117 5V regulator), charges a capacitor for a “keep-alive” (the next 2 pins that are vertical), and powers a few LEDs or similar up to 1A. Bottom layer is a ground pour, top layer is a Vdd pour. The grounds from the bridge rectifier, 5V regulator and 1000uF capacitor need to be via’d to the ground side. What is the right way to connect these? Can a create a polygon of some kind connecting the close-by grounds and put a via or 2 or inside them? Can I do it like I have with vias and traces to connect things together? I have made only to top, Vdd layer in red visible for clarity. Thank you.
EDIT: Great stuff! This is a very simple circuit. Someone else’s. I wanted to model it and design the PCB for it. All it does is take a bi-polar digital signal from a model train track, rectify it, regulate it charge a capacitor, and run a few leds for lighting in the train. Dirty wheels, dirty track, etc. mean for dropouts. So this is just so that the capacitor can power the LEDs for fractions of a second when these dropouts occur. I also questioned the design, because I don’t think it accounts for the voltage that appears at the bridge. First one rail is 12-18v with respect to the other, then it switches. 8kHz pulsed with 58us and 116us pulses. The bridge would see that as up to 18-1.2 = 16.8V. So 16.8 -5V is 11.8V wasted as heat. Here is the schematic:
What is the right way to connect these?
What are you worried about? EMC? decoupling peaks? Current vs track size vs heat?
For EMC and decoupling: Place a small ceramic capacitor next to every digital component (100nF or something like that), with short traces to the power pins. 8kHz is pretty low but that does not matter for EMC / decoupling, important is the rise/fall time. The edge frequency can easily be 10 - 200MHz even when the Signal is only 8kHz.
For the current: Use the PCB-Calculator from KiCad, it has a tab for that.
Can a create a polygon of some kind connecting the close-by grounds and put a via or 2 or inside them?
Yes, but it can make soldering hard, since polygons can remove a lot of heat.
Can I do it like I have with vias and traces to connect things together?
Yes. But i would check the trace width and current. Why are there 2 tracks and 2 vias at each location? How does the PCB look like on the other layers?
A 0.4mm is more than enough for 2A, you don’t need more than 1 via for your current expectation.
It is hard to see what you really want, maybe you can upload the schematic?
1. I agree that a schematic would be helpful.
2. How many layers are you planning? I think you have a relatively simple circuit. If that is true, it seems likely that a 2 layer board may be enough. In that case, you try to put as many of your tracks as possible on the top layer and use the bottom for a ground plane. There is some argument as to whether a single solid ground plane is best in most situations. But for a fairly simple low power circuit, I don’t think you will go too far wrong.
3. I you have a 4 layer board, then it is usually best to make the upper inner layer a ground layer, with most of the tracks on top.
4. I like to use multiple vias, especially when current is involved. I tend to be conservative. I would like to use 3-4 vias for 2 Amps of current.
Even at 8 kHz with switching current in Amps, you need to think where the high current loops are
1 Like
Yes indeed. The part shown on the PCB is just a linear voltage regulator. Apparently there is some bridge rectifier, but it’s not very clear what it is on the PCB. Maybe the footprint on the left with pad 4 as GND? If that is so, then it’s missing a buffer capacitor. Using a 1000uF buffer capacitor after the AMS1117 is also… unusual. I think a 22uF capacitor is recommended for the AMS1117. The 1000uF would do more useful work if you put it between the bridge rectifier and the voltage regulator. You also have to verify the ripple current of your electrolytic capacitor.
Using multiple via’s is just fine. It increases current handling, and also reduces track inductance, and therefore makes the connection capable of handling higher frequencies. But both your AMS1117 and 1000uF capacitor are not capable of “high frequency stuff”.
Max current for an AMS1117 usually is 800mA. (There are many manufacturers of this chip). It may also get quite hot at higher currents (and higher voltage over the regulator) Removing the thermal via from the big pad and directly connecting it to the big zone helps with keeping it cool.
Electrolytic capacitors do not like heat. Their lifespan decreases significantly with even small temperature rises, and electrolytic capacitors are already the most unreliably electronic parts. An electrolytic capacitor at it’s maximum rated temperature and ripple current may have a life expectancy as low as a few thousand hours. That is just a few months (Yes months, not years!)
So putting them a bit further away from heat sources is always a good precaution. According to the datasheet of the AMS1117 there can be up to 10cm between the AMS1117 and it’s output capacitor.
AMS1117 is also a quite old design (maybe 20 years?) Low-drop voltage regulators have some issues with output capacitors, especially if they are big and have low ESR. This may cause issues if you have lots of ceramic decoupling capacitors on the rest of your PCB.
It’s not clearly shown in the layout but I’d assume the component, with pins 1 and 4 visible, to the left of the AMS1117 is the rectifier bridge. If that is the case, you’re missing a buffer capacitor at the output of that rectifier bridge to smooth out the transition region where the rectifier isn’t conducting in a cycle of the square wave input.
More importantly as it seems to me, there could be better choices for the step-down regulator than the AMS1117.
Your regulator is to supply up to 1A current while subject to about 11V voltage differential, the resulted power dissipation, in case of a linear regulator as is the AMS1117, would be far exceeding the heat-sinking capability of the layout design by using copper planes on PCB as heat-sink, usually 2-3W max allowed around a SOT-223 device. The resulted power dissipation would be far exceeding the maximum allowed by the device itself as well. You’d likely need an LM7805 in TO-220 mounted on a massive-ish, real heatsink, if you have to use a linear regulator.
I would recommend switching regulator TPS82130. It suits your input voltage, almost as simple a circuit as an LM317, a lot smaller overall footprint than AMS1117, does not require an output capacitor that bulky, and you will not have thermal management issues. Just need to follow the recommended layout given in the datasheet.
1 Like
Oops, I missed that 18V input which would certainly overheat the LM1117 @ 1A.
I do agree with nattawa that an SMPS is a much better option for high voltage differentials over the regulator. TPS82130 is not a good choice though. It has a max input voltage of only 17V, but there are many small SMPS chips that can work from a higher input voltage.
You’re right about TPS82130 specified 17V maximum input voltage, paulvdh. Considering the 18V peak max square wave input to our application circuit being subject to a voltage drop of 2x P/N junctions at the rectifier bridge preceding the TPS82130, the device will be working within its specified range of operation. I’d be quite comfortable using it.
However if concerned, for a few nickels more and a slightly larger footprint, TPSM84209 offers 28V max input, and a lower but still satisfying output current at 2.5A.
It is correct that with 17V input and 5V output, a linear regulator will be producing a lot of heat. Not a practical design these days for current over 100 mA or so, unless you need to avoid switching regulators for some reason.
BTW TPS82130 is a module; not an IC. I think that Digikey is all ready to ship 0 pieces because that is what they have in stock. Some other distributor might do better.
Yes, it is an all-in-one module that has the inductor integrated. We use quite many of them as point-of-load regulators.
It’s unfortunate that semiconductor is in short supply across the board, and equipment manufacturers have been stocking up and have emptied suppliers. But things always become better after a while.
There are 2 vias and 2 traces at each location because I was wondering if I needed to distribute the current and heat to the ground layer. (There are just 2 layers)
I see you added a schematic.
The way you drew it, the voltage regulator will be starved of input voltage during the time your AC input is below an absolute voltage of around 7V.
So replace the big 1000uF capacitor a 22uF capacitor and put the big one at the location of the “PWR_FLAG”.
And what is R1 doing? Even if you short circuit J3 & J4 your AMS1117 will never deliver more than 5mA into it, so where is that whole amp of current going?
How much experience do you have with electronics?
I would use a lot of vias, strip the soldermask to expose the copper plane, and not use thermal spokes around the collector tab and pin. The attached are cropped from a 2-layer layout I did. The SOT-223 transistors are required to each dissipate continuous 2W heat by using the copper planes as heatsink at 40C temperature rise on 2-Oz copper over 1.6mm FR-4 material, without using forced air flow.
Top side:
Bottom side:
All these vias are for heat conductivity.
At 1A max current in your layout, there is no need of sharing current with vias.
2 Likes
If you want linear regulator the better result you will get with 7805 I think. There the power pad is GND so you can have at top and at bottom GND planes and connect them under 7805 with vias to dissipate power through both planes.
1117 regulators have lover minimum voltage drop then 7805 but here you don’t need that feature.
But even for 7805 you can have better heat dissipation 11.8V * 1A = 11.8W is too much for such case.
See figure 15 for D2PAK and figure 16 for DPAK in:
https://pl.mouser.com/datasheet/2/308/1/MC7800_D-2315963.pdf
You see that maximum power dissipation is about 3W for D2PAK and 2.2W for DPAK.
I suggest to never dissipate more than 1W in DPAK.
That is correct.
Also correct (I did not check the math.)
If you insist on using a linear regulator in this way, one approach is to put a power resistor (a “burnoff” resistor) between the bridge rectifier and the regulator input. That will not help your terrible efficiency but it can move some heat out of the regulator and into the resistor. So for example a 7.5 ohm 15W resistor would drop 7.5V and burn 7.5 Watts of heat which would no longer need to be dissipated in the regulator. I would put the 1000 uF capacitor at the bridge output, and then a smaller value capacitor at the regulator input and another at the regulator output. That 1000 ohm resistor in series with the regulator output is nonsense.
Note that too high a value for the burnoff resistor will result in inadequate input voltage to the linear regulator when input voltage is minimum and output current is maximum.
But as paulvdh indicated, your schematic indicates that you need to get more understanding of circuit theory. It ought to be possible to “breadboard” this simple circuit before you spend money on a pcb. Unless your pcb is very cheap then I recommend that you do so.
BTW I have a hunch that you may have a fundamental problem with your square wave input.
You are passing it through a bridge rectifier. For this to work, the square wave input MUST be floating wrt the regulator circuit. The grounds must not be connected. In the typical example of a bridge rectifier, it is driven by a transformer winding which is electrically floating (not referenced to the ground of the output circuit.)
1 Like
I have a “moderate” amount of electronics experience, EE and CS 40 years ago after which I mostly used the CS part for my career and all electronics projects were hobby designs. So I am trying to get back in it.
This is just a keep-alive circuit that needs to be dirt cheap to power a few LEDs or 5V, 50mA incandescent lights. I don’t care about starving the regulator as long as it can pulse DC to keep the cap charged so that when there are dropouts of a half second or so, the lights don’t flicker. You are right about the resistor of course, thank you! Someone gave me this and asked about a board for it and I should have double-checked it. I am guessing he got a circuit for 12V and added a regulator and left the resistor in there. It doesn’t need a resistor since the current limiting resistors can go on the LEDs. Obviously sticking a resistor in there instead of the regulator and adding a Zener in parallel with the capacitor will accomplish the same task. You guys are helping me think through the circuit as well as the board, so thank you. In looking at the requirements, this doesn’t really need to supply more than 250mA
It is floating. This is a bi-polar signal where one pole is pulsed relative to the other (not to any ground) and then the other pole is pulsed. It would look like a square wave on a scope, but the voltage itself never goes negative, just the current reverses.
A 1F capacitors drops 1V/s with 1A (By definition).
So a 1mF (= 1000uF) capacitor and 50mA will drop 1000V/s = 1V/ms.
For 50mA the discharge rate will be 20 times lower.
With a “moderate amount of electronics” you should be able to work out how long your LED will keep alight.
Also, can you please make up your mind about power consumption? In a single post you mention:
• A few LED’s.
• 50mA incandescent lights
• 250mA.
All the more reason to charge your 1000uF capacitor tho the full rectified voltage to make it last longer.
… and an SMPS IC. These things start at around 10ct, the inductor may cost more.
1 Like
Perhaps instead of 5V, you can instead regulate to 10V or so to further reduce the power loss the regulator alone has to handle. The 5V incandescent bulbs can be connected in series, 2 for 10V, so can LEDs. Series connection can reduce the total current and quickly reduce the power loss at the regulator.
The other alternative can be not regulating the voltage at all, but regulating the current instead with constant current source/sink circuit, one circuit per series of load. A constant current source/sink can be constructed with a couple diodes, a resistor, and a small signal transistor at very low cost and very small PCB footprint.
1 Like | 3,521 | 14,845 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2021-49 | latest | en | 0.939195 |
https://testbook.com/question-answer/in-the-following-question-select-the-odd-le--60b871c0557f4ae047bd6b87 | 1,627,462,673,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153531.10/warc/CC-MAIN-20210728060744-20210728090744-00240.warc.gz | 566,336,718 | 19,239 | # In the following question, select the odd letter/letters from the given alternatives.
Free Practice With Testbook Mock Tests
## Options:
1. GDB
2. RPN
3. WUS
4. MKI
### Correct Answer: Option 1 (Solution Below)
This question was previously asked in
DSSSB LDC Official Paper 1 (Held on : 18 Aug 2019 Shift 1)
## Solution:
The Logic here is as follows:
Option 1: GDB
B + 2 = D
D + 3 = G
Option 2: RPN
N + 2 = P
P + 2 = R
Option 3: WUS
S + 2 = U
U + 2 = W
Option 4: MKI
I + 2 = K
K + 2 = M
Therefore, All other options follow a pattern except GDB.
Hence, the correct answer is "GDB". | 210 | 607 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2021-31 | latest | en | 0.848192 |
https://www.coursehero.com/tutors-problems/Calculus/10445183-1Find-the-generating-function-for-the-sequence-an-where-an-is-the-nu/ | 1,586,454,853,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371861991.79/warc/CC-MAIN-20200409154025-20200409184525-00153.warc.gz | 851,944,328 | 29,797 | Find the generating function for the sequence an, where an is the number of natural solutions to the equation x1 + x2 + 2x3 + 5x4 = n with xi 0 for i...
View the step-by-step solution to:
Question
# 1.Find the generating function for the sequence an, where an is the number of natural solutions to the equation x1
+ x2 + 2x3 + 5x4 = n with xi ≥ 0 for i = 1, 2, 3, 4.
2.Show that the number of partitions of 2n into distinct even parts isequal to the number of partitions of n into odd parts.
1.
x1 + x2 + 2x3 + 5x4 = n
The generating function of the sequence is given by
x1 + (x2) X + (2x3) X^2 + (5x4) X^3
¿
xi¿ =x1 + 4 ∑ i=2,n=1 2. ¿ 2+2n) Xn This shows that the number of...
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Browse Documents | 307 | 1,034 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2020-16 | latest | en | 0.841012 |
https://www.dataunitconverter.com/terabit-to-bit/56 | 1,679,807,629,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945433.92/warc/CC-MAIN-20230326044821-20230326074821-00388.warc.gz | 807,752,884 | 11,817 | # Terabit to Bit - 56 Tbit to Bits Conversion
## Conversion History (Last 6)
Input Terabit - and press Enter
Tbit
RESULT ( Terabit → Bit ) :
56 Tbit = 56,000,000,000,000 Bits
Copy
Calculated as → 56 x 10004...view detailed steps
## Tbit to b - Conversion Formula and Steps
Terabit and Bit are units of digital information used to measure storage capacity and data transfer rate. Terabit is a decimal unit where as Bit is one of the very basic digital unit. One Terabit is equal to 1000^4 bits. There are 0.000000000001 Terabits in one Bit.
Source Data UnitTarget Data Unit
Terabit (Tbit)
Equal to 1000^4 bits
(Decimal Unit)
Bit (b)
Equal to 0 or 1
(Basic Unit)
Below conversion diagram will help you to visualize the Terabit to Bit calculation steps in a simplified manner.
÷ 1000
÷ 1000
÷ 1000
÷ 1000
÷ 1000
÷ 1000
÷ 1000
÷ 1000
Bit [b]
Kilobit [kbit]
Megabit [Mbit]
Gigabit [Gbit]
Terabit [Tbit]
Petabit [Pbit]
Exabit [Ebit]
Zettabit [Zbit]
Yottabit [Ybit]
x 1000
x 1000
x 1000
x 1000
x 1000
x 1000
x 1000
x 1000
The formula of converting the Terabit to Bit is represented as follows :
Bits = Tbit x 10004
Now let us apply the above formula and, write down the steps to convert from Terabit (Tbit) to Bit (b). This way, we can try to simplify and reduce to an easy to apply formula.
FORMULA
Bit = Terabit x 10004
STEP 1
Bit = Terabit x (1000x1000x1000x1000)
STEP 2
Bit = Terabit x 1000000000000
If we apply the above Formula and steps, conversion from 56 Tbit to Bits, will be processed as below.
1. = 56 x 10004
2. = 56 x (1000x1000x1000x1000)
3. = 56 x 1000000000000
4. = 56000000000000
5. i.e. 56 Tbit is equal to 56,000,000,000,000 Bits.
(Result rounded off to 40 decimal positions.)
#### Definition : Terabit
A Terabit (Tb or Tbit) is a unit of measurement for digital information transfer rate. It is equal to 1,000,000,000,000 (one trillion) bits. It is commonly used to measure the speed of data transfer over computer networks, such as internet connection speeds.
#### Definition : Bit
A Bit (short for "binary digit") is the basic unit of information in computing and digital communications. It is a binary value, meaning it can have one of two values=> 0 or 1. Bits are used to represent data in computers and other electronic devices. They are the building blocks of digital information, and are used to store, transmit, and process data.
### Excel Formula to convert from Tbit to Bits
Apply the formula as shown below to convert from 56 Terabit to Bit.
ABC
1Terabit (Tbit)Bit (b)
256=A2 * 1000000000000
3 | 779 | 2,546 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2023-14 | latest | en | 0.838197 |
http://en.wikipedia.org/wiki/Differentiable_function | 1,412,238,361,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1412037663739.33/warc/CC-MAIN-20140930004103-00052-ip-10-234-18-248.ec2.internal.warc.gz | 102,888,346 | 13,839 | # Differentiable function
A differentiable function
The absolute value function is not differentiable at x = 0.
Differentiable functions can be locally approximated by linear functions.
In calculus (a branch of mathematics), a differentiable function of one real variable is a function whose derivative exists at each point in its domain. As a result, the graph of a differentiable function must have a non-vertical tangent line at each point in its domain, be relatively smooth, and cannot contain any breaks, bends, or cusps.
More generally, if x0 is a point in the domain of a function f, then f is said to be differentiable at x0 if the derivative f′(x0) exists. This means that the graph of f has a non-vertical tangent line at the point (x0f(x0)). The function f may also be called locally linear at x0, as it can be well approximated by a linear function near this point.
## Differentiability and continuity
The Weierstrass function is continuous, but is not differentiable at any point.
If f is differentiable at a point x0, then f must also be continuous at x0. In particular, any differentiable function must be continuous at every point in its domain. The converse does not hold: a continuous function need not be differentiable. For example, a function with a bend, cusp, or vertical tangent may be continuous, but fails to be differentiable at the location of the anomaly.
Most functions that occur in practice have derivatives at all points or at almost every point. However, a result of Stefan Banach states that the set of functions that have a derivative at some point is a meager set in the space of all continuous functions.[1] Informally, this means that differentiable functions are very atypical among continuous functions. The first known example of a function that is continuous everywhere but differentiable nowhere is the Weierstrass function.
## Differentiability classes
A function f is said to be continuously differentiable if the derivative f'(x) exists and is itself a continuous function. Though the derivative of a differentiable function never has a jump discontinuity, it is possible for the derivative to have an essential discontinuity. For example, the function
$f(x) \;=\; \begin{cases} x^2\sin (1/x) & \text{if }x \ne 0 \\ 0 & \text{if }x=0\end{cases}$
is differentiable at 0, since
$f'(0)=\lim_{\epsilon\to0}\left(\frac{\epsilon^2\sin(1/\epsilon)-0}{\epsilon}\right)=0,$
exists. However, for x≠0,
$f'(x)=2x\sin(1/x)-\cos(1/x)$
which has no limit as x → 0. Nevertheless, Darboux's theorem implies that the derivative of any function satisfies the conclusion of the intermediate value theorem.
Sometimes continuously differentiable functions are said to be of class C1. A function is of class C2 if the first and second derivative of the function both exist and are continuous. More generally, a function is said to be of class Ck if the first k derivatives f′(x), f″(x), ..., f(k)(x) all exist and are continuous. If derivatives f(n) exist for all positive integers n, the function is smooth or equivalently, of class C.
## Differentiability in higher dimensions
A function of several real variables f: RmRn is said to be differentiable at a point x0 if there exists a linear map J: RmRn such that
$\lim_{\mathbf{h}\to \mathbf{0}} \frac{\mathbf{f}(\mathbf{x_0}+\mathbf{h}) - \mathbf{f}(\mathbf{x_0}) - \mathbf{J}\mathbf{(h)}}{\| \mathbf{h} \|} = \mathbf{0}.$
If a function is differentiable at x0, then all of the partial derivatives must exist at x0, in which case the linear map J is given by the Jacobian matrix. A similar formulation of the higher-dimensional derivative is provided by the fundamental increment lemma found in single-variable calculus.
Note that existence of the partial derivatives (or even all of the directional derivatives) does not guarantee that a function is differentiable at a point. For example, the function f: R2R defined by
$f(x,y) = \begin{cases}x & \text{if }y \ne x^2 \\ 0 & \text{if }y = x^2\end{cases}$
is not differentiable at (0, 0), but all of the partial derivatives and directional derivatives exist at this point. For a continuous example, the function
$f(x,y) = \begin{cases}y^3/(x^2+y^2) & \text{if }(x,y) \ne (0,0) \\ 0 & \text{if }(x,y) = (0,0)\end{cases}$
is not differentiable at (0, 0), but again all of the partial derivatives and directional derivatives exist.
It is known that if the partial derivatives of a function all exist and are continuous in a neighborhood of a point, then the function must be differentiable at that point, and it is in fact of class C1.
## Differentiability in complex analysis
Main article: Holomorphic function
In complex analysis, any function that is complex-differentiable in a neighborhood of a point is called holomorphic. Such a function is necessarily infinitely differentiable, and in fact analytic.
## Differentiable functions on manifolds
If M is a differentiable manifold, a real or complex-valued function f on M is said to be differentiable at a point p if it is differentiable with respect to some (or any) coordinate chart defined around p. More generally, if M and N are differentiable manifolds, a function fM → N is said to be differentiable at a point p if it is differentiable with respect to some (or any) coordinate charts defined around p and f(p). | 1,312 | 5,327 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 6, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2014-41 | longest | en | 0.934495 |
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# final03f - EE301(Akar Final Exam Closed Book No Calculators...
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EE301 (Akar) Final Exam 12/10/2003 Closed Book, No Calculators, No PCs and No MATLAB Only four 8 1 2 00 × 11 00 sheets of notes allowed (2-sided) Show all work to get full credit Name: Sign if No Aid Given or Taken: Problem Points #1 (10) #2 (10) #3 (15) #4 (15) #5 (15) #6 (15) #7 (20) Total (100)
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EE301 (Akar) Final Exam 12/10/2003 Page 2 Problem 1 (10 points) Consider the continuous-time system described by the relation y ( t ) = 1 3 x 3 ( t ) + 2 x 2 ( t + 1) where x ( t ) is the input and y ( t ) is the output of the system. a). Is the system linear? Prove your statement. b). Is the system time-invariant? Prove your statement. c). Is the system causal? Prove your statement. d). Is the system stable? Prove your statement.
EE301 (Akar) Final Exam 12/10/2003 Page 3 Problem 2 (10 points) Given the differential equation d 2 y ( t ) dt 2 + 4 dy ( t ) dt + 8 y ( t ) = x ( t ) with the initial conditions y (0 - ) = 2, dy ( t ) dt | t =0 - = 1 and the input x ( t ) = δ ( t ), determine the
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## This note was uploaded on 05/05/2008 for the course EE 301 taught by Professor Enright during the Spring '08 term at USC.
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Conversion & Calculation Home >> Measurement Conversion
Measurement Converter
Convert From: (required) Click here to Convert To: (optional) Examples: 5 kilometers, 12 feet/sec^2, 1/5 gallon, 9.5 Joules, or 0 dF. Help, Frequently Asked Questions, Use Currencies in Conversions, Measurements & Currencies Recognized Examples: miles, meters/s^2, liters, kilowatt*hours, or dC.
Conversion Result: ```foot^2 = 0.09290304 area (area) ``` Related Measurements: Try converting from "feet^2" to acre, are, barn (abbreviation b), cho (Japanese cho), circular inch, circular mil, desyatina (Russian desyatina), hectare, homestead, rood, sabin, section (of land), square, township, or any combination of units which equate to "length squared" and represent area, fuel consumption, linear displacement, permeability2, surface area, or volume to length. Sample Conversions: feet^2 = .00002296 acre, .00092903 are, 9.29E+26 barn (abbreviation b), .00000937 cho (Japanese cho), 183.35 circular inch, 183,346,494.44 circular mil, .0000085 desyatina (Russian desyatina), .00000929 hectare, 1.43E-07 homestead, .00009183 rood, 1 sabin, 3.59E-08 section (of land), .01 square, 9.96E-10 township.
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# Is One thousand two hundred the same as twelve hundred?
Updated: 4/28/2022
Wiki User
12y ago
Yes. 1,200 (one thousand two hundred) = 12 multiplied by 100 or twelve one hundreds. Instead of saying, "one thousand two hundred" people have shortened it by saying, "twelve hundred."
Wiki User
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https://www.dataunitconverter.com/kibibyte-per-day-to-gigabyte-per-minute | 1,718,959,466,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862040.7/warc/CC-MAIN-20240621062300-20240621092300-00391.warc.gz | 638,896,567 | 17,209 | # KiB/Day to GB/Min → CONVERT Kibibytes per Day to Gigabytes per Minute
expand_more
info 1 KiB/Day is equal to 0.0000000007111111111111111111111111111111 GB/Min
S = Second, M = Minute, H = Hour, D = Day
Sec
Min
Hr
Day
Sec
Min
Hr
Day
## Kibibytes per Day (KiB/Day) Versus Gigabytes per Minute (GB/Min) - Comparison
Kibibytes per Day and Gigabytes per Minute are units of digital information used to measure storage capacity and data transfer rate.
Kibibytes per Day is a "binary" unit where as Gigabytes per Minute is a "decimal" unit. One Kibibyte is equal to 1024 bytes. One Gigabyte is equal to 1000^3 bytes. There are 976,562.5 Kibibyte in one Gigabyte. Find more details on below table.
Kibibytes per Day (KiB/Day) Gigabytes per Minute (GB/Min)
Kibibytes per Day (KiB/Day) is a unit of measurement for data transfer bandwidth. It measures the number of Kibibytes that can be transferred in one Day. Gigabytes per Minute (GB/Min) is a unit of measurement for data transfer bandwidth. It measures the number of Gigabytes that can be transferred in one Minute.
## Kibibytes per Day (KiB/Day) to Gigabytes per Minute (GB/Min) Conversion - Formula & Steps
The KiB/Day to GB/Min Calculator Tool provides a convenient solution for effortlessly converting data rates from Kibibytes per Day (KiB/Day) to Gigabytes per Minute (GB/Min). Let's delve into a thorough analysis of the formula and steps involved.
Outlined below is a comprehensive overview of the key attributes associated with both the source (Kibibyte) and target (Gigabyte) data units.
Source Data Unit Target Data Unit
Equal to 1024 bytes
(Binary Unit)
Equal to 1000^3 bytes
(Decimal Unit)
The conversion from Data per Day to Minute can be calculated as below.
x 60
x 60
x 24
Data
per
Second
Data
per
Minute
Data
per
Hour
Data
per
Day
÷ 60
÷ 60
÷ 24
The formula for converting the Kibibytes per Day (KiB/Day) to Gigabytes per Minute (GB/Min) can be expressed as follows:
diamond CONVERSION FORMULA GB/Min = KiB/Day x 1024 ÷ 10003 / ( 60 x 24 )
Now, let's apply the aforementioned formula and explore the manual conversion process from Kibibytes per Day (KiB/Day) to Gigabytes per Minute (GB/Min). To streamline the calculation further, we can simplify the formula for added convenience.
FORMULA
Gigabytes per Minute = Kibibytes per Day x 1024 ÷ 10003 / ( 60 x 24 )
STEP 1
Gigabytes per Minute = Kibibytes per Day x 1024 ÷ (1000x1000x1000) / ( 60 x 24 )
STEP 2
Gigabytes per Minute = Kibibytes per Day x 1024 ÷ 1000000000 / ( 60 x 24 )
STEP 3
Gigabytes per Minute = Kibibytes per Day x 0.000001024 / ( 60 x 24 )
STEP 4
Gigabytes per Minute = Kibibytes per Day x 0.000001024 / 1440
STEP 5
Gigabytes per Minute = Kibibytes per Day x 0.0000000007111111111111111111111111111111
Example : By applying the previously mentioned formula and steps, the conversion from 1 Kibibytes per Day (KiB/Day) to Gigabytes per Minute (GB/Min) can be processed as outlined below.
1. = 1 x 1024 ÷ 10003 / ( 60 x 24 )
2. = 1 x 1024 ÷ (1000x1000x1000) / ( 60 x 24 )
3. = 1 x 1024 ÷ 1000000000 / ( 60 x 24 )
4. = 1 x 0.000001024 / ( 60 x 24 )
5. = 1 x 0.000001024 / 1440
6. = 1 x 0.0000000007111111111111111111111111111111
7. = 0.0000000007111111111111111111111111111111
8. i.e. 1 KiB/Day is equal to 0.0000000007111111111111111111111111111111 GB/Min.
Note : Result rounded off to 40 decimal positions.
You can employ the formula and steps mentioned above to convert Kibibytes per Day to Gigabytes per Minute using any of the programming language such as Java, Python, or Powershell.
### Unit Definitions
#### What is Kibibyte ?
A Kibibyte (KiB) is a binary unit of digital information that is equal to 1024 bytes (or 8,192 bits) and is defined by the International Electro technical Commission(IEC). The prefix 'kibi' is derived from the binary number system and it is used to distinguish it from the decimal-based 'kilobyte' (KB). It is widely used in the field of computing as it more accurately represents the amount of data storage and data transfer in computer systems.
arrow_downward
#### What is Gigabyte ?
A Gigabyte (GB) is a decimal unit of digital information that is equal to 1,000,000,000 bytes (or 8,000,000,000 bits) and commonly used to measure the storage capacity of computer hard drives, flash drives, and other digital storage devices. It is also used to express data transfer speeds and in the context of data storage and memory, the binary-based unit of Gibibyte (GiB) is used instead.
## Excel Formula to convert from Kibibytes per Day (KiB/Day) to Gigabytes per Minute (GB/Min)
Apply the formula as shown below to convert from 1 Kibibytes per Day (KiB/Day) to Gigabytes per Minute (GB/Min).
A B C
1 Kibibytes per Day (KiB/Day) Gigabytes per Minute (GB/Min)
2 1 =A2 * 0.000001024 / ( 60 * 24 )
3
If you want to perform bulk conversion locally in your system, then download and make use of above Excel template.
## Python Code for Kibibytes per Day (KiB/Day) to Gigabytes per Minute (GB/Min) Conversion
You can use below code to convert any value in Kibibytes per Day (KiB/Day) to Kibibytes per Day (KiB/Day) in Python.
kibibytesperDay = int(input("Enter Kibibytes per Day: "))
gigabytesperMinute = kibibytesperDay * 1024 / (1000*1000*1000) / ( 60 * 24 )
print("{} Kibibytes per Day = {} Gigabytes per Minute".format(kibibytesperDay,gigabytesperMinute))
The first line of code will prompt the user to enter the Kibibytes per Day (KiB/Day) as an input. The value of Gigabytes per Minute (GB/Min) is calculated on the next line, and the code in third line will display the result.
## Frequently Asked Questions - FAQs
#### How many Gigabytes(GB) are there in a Kibibyte(KiB)?expand_more
There are 0.000001024 Gigabytes in a Kibibyte.
#### What is the formula to convert Kibibyte(KiB) to Gigabyte(GB)?expand_more
Use the formula GB = KiB x 1024 / 10003 to convert Kibibyte to Gigabyte.
#### How many Kibibytes(KiB) are there in a Gigabyte(GB)?expand_more
There are 976562.5 Kibibytes in a Gigabyte.
#### What is the formula to convert Gigabyte(GB) to Kibibyte(KiB)?expand_more
Use the formula KiB = GB x 10003 / 1024 to convert Gigabyte to Kibibyte.
#### Which is bigger, Gigabyte(GB) or Kibibyte(KiB)?expand_more
Gigabyte is bigger than Kibibyte. One Gigabyte contains 976562.5 Kibibytes.
## Similar Conversions & Calculators
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1. Use the distance formula to find the distance between the points (-3,5) and (1,4)2. Use the distance formula to find the distance between the points (-5,-4) and (3,-2)3. Use the distance formula to find the distance between the points (4,3) and (-2,1)4. Use the distance formula to find the distance between the points (-4,2) and (3,4)5. Use the distance formula to find the distance between the points (7,6) and (-3,4)6. Use the distance formula to find the distance between the points (8,7) and (-1,0)7. Use the distance formula to find the distance between the points (5,0) and (0,3)8. Use the distance formula to find the distance between the points (8,-8) and (6,4)9. Use the distance formula to find the distance between the points (-4,6) and (5,-3)10. Use the distance formula to find the distance between the points (3,2) and (7,9)
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1.2 Distance Formula
# 1.2 Distance Formula - 1.2 Distance Formula Friday 5:48 AM...
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Agenda: 1. Warmup 2. Lesson: 1.2 Distance Formula _____________________________________ Warmup. Is the following relation a function? List the domain and range. Is the following relation a function? List the excluded value(s). _____________________________________ Lesson: 1.2 Distance Formula Goal: Define and use the distance formula Let's derive the distance formula. Remember the Pythagorean Theorem? Remember whiteboard pens on Monday! Also, there will be a homework check on Monday. 1.2 Distance Formula Friday, August 20, 2010 5:48 AM Ch_1 Page 1
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Distance Formula: Given two points, and , the distance between the two points is _______________________________________ A necessary skill review Do you remember how to simplify radicals? I like to find perfect squares factors. Simplify the radicals.
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{[ snackBarMessage ]} | 275 | 1,214 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2018-17 | latest | en | 0.796186 |
https://www.geeksforgeeks.org/delete-array-elements-which-are-smaller-than-next-or-become-smaller/amp/?ref=rp | 1,611,506,618,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703549416.62/warc/CC-MAIN-20210124141945-20210124171945-00106.warc.gz | 770,723,271 | 19,680 | # Delete array elements which are smaller than next or become smaller
Given an array arr[] and a number k. The task is to delete k elements which are smaller than next element (i.e., we delete arr[i] if arr[i] < arr[i+1]) or become smaller than next because next element is deleted.
Examples:
```Input : arr[] = { 3, 100, 1 }
k = 1
Output : 100, 1
Explanation : arr[0] < arr[1] means 3 is less than
100, so delete 3
Input : arr[] = {20, 10, 25, 30, 40}
k = 2
Output : 25 30 40
Explanation : First we delete 10 because it follows
arr[i] < arr[i+1]. Then we delete 20
because 25 is moved next to it and it
also starts following the condition.
Input : arr[] = { 23, 45, 11, 77, 18}
k = 3
Output : 77, 18
Explanation : We delete 23, 45 and 11 as they follow
the condition arr[i] < arr[i+1]
```
## Recommended: Please try your approach on {IDE} first, before moving on to the solution.
Approach: Stack is used to solving this problem. First we push arr[0] in stack S and then initialize count as 0, then after traverse a loop from 1 to n and then we check that s.top() < arr[i] if condition is true then we pop the element from stack and increase the count if count == k then we stop the loop and then store the value of stack in another array and then print that array.
`// C++ program to delete elements from array. ` `#include ` `using` `namespace` `std; ` ` ` `// Function for deleting k elements ` `void` `deleteElements(``int` `arr[], ``int` `n, ``int` `k) ` `{ ` ` ``// Create a stack and push arr[0] ` ` ``stack<``int``> s; ` ` ``s.push(arr[0]); ` ` ` ` ``int` `count = 0; ` ` ` ` ``// traversing a loop from i = 1 to n ` ` ``for` `(``int` `i=1; i v(m); ``// Size of vector is m ` ` ``while` `(!s.empty()) { ` ` ` ` ``// push element from stack to vector v ` ` ``v[--m] = s.top(); ` ` ``s.pop(); ` ` ``} ` ` ` ` ``// printing result ` ` ``for` `(``auto` `x : v) ` ` ``cout << x << ``" "``; ` ` ` ` ``cout << endl; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ``int` `n = 5, k = 2; ` ` ``int` `arr[] = {20, 10, 25, 30, 40}; ` ` ``deleteElements(arr, n, k); ` ` ``return` `0; ` `} `
`import` `java.util.*; ` ` ` `//Java program to delete elements from array. ` `class` `GFG { ` ` ` `// Function for deleting k elements ` ` ``static` `void` `deleteElements(``int` `arr[], ``int` `n, ``int` `k) { ` ` ``// Create a stack and push arr[0] ` ` ``Stack s = ``new` `Stack<>(); ` ` ``s.push(arr[``0``]); ` ` ` ` ``int` `count = ``0``; ` ` ` ` ``// traversing a loop from i = 1 to n ` ` ``for` `(``int` `i = ``1``; i < n; i++) { ` ` ` ` ``// condition for deleting an element ` ` ``while` `(!s.empty() && s.peek() < arr[i] ` ` ``&& count < k) { ` ` ``s.pop(); ` ` ``count++; ` ` ``} ` ` ` ` ``s.push(arr[i]); ` ` ``} ` ` ` ` ``// Putting elements of stack in a vector ` ` ``// from end to begin. ` ` ``int` `m = s.size(); ` ` ``Integer[] v = ``new` `Integer[m]; ``// Size of vector is m ` ` ``while` `(!s.empty()) { ` ` ` ` ``// push element from stack to vector v ` ` ``v[--m] = s.peek(); ` ` ``s.pop(); ` ` ``} ` ` ` ` ``// printing result ` ` ``for` `(Integer x : v) { ` ` ``System.out.print(x + ``" "``); ` ` ``}; ` ` ` ` ``System.out.println(``""``); ` ` ``} ` ` ` `// Driver code ` ` ``public` `static` `void` `main(String[] args) { ` ` ``int` `n = ``5``, k = ``2``; ` ` ``int` `arr[] = {``20``, ``10``, ``25``, ``30``, ``40``}; ` ` ``deleteElements(arr, n, k); ` ` ``} ` `} ` `// This code is contributed by PrinciRaj1992 `
`# Function to delete elements ` `def` `deleteElements(arr, n, k): ` ` ` ` ``# create an empty stack st ` ` ``st ``=` `[] ` ` ``st.append(arr[``0``]) ` ` ` ` ``# index to mantain the top ` ` ``# of the stack ` ` ``top ``=` `0` ` ``count ``=` `0` ` ` ` ``for` `i ``in` `range``(``1``, n): ` ` ` ` ``# pop till the present element ` ` ``# is greater than stack's top ` ` ``# element ` ` ``while``(``len``(st) !``=` `0` `and` `count < k ` ` ``and` `st[top] < arr[i]): ` ` ``st.pop() ` ` ``count ``+``=` `1` ` ``top ``-``=` `1` ` ` ` ``st.append(arr[i]) ` ` ``top ``+``=` `1` ` ` ` ``# print the remaining elements ` ` ``for` `i ``in` `range``(``0``, ``len``(st)): ` ` ``print``(st[i], ``" "``, end``=``"") ` ` ` `# Driver code ` `k ``=` `2` `arr ``=` `[``20``, ``10``, ``25``, ``30``, ``40``] ` `deleteElements(arr, ``len``(arr), k) ` ` ` `# This code is contributed by himan085. `
`// C# program to delete elements from array. ` `using` `System; ` `using` `System.Collections.Generic; ` ` ` `class` `GFG { ` ` ` ` ``// Function for deleting k elements ` ` ``static` `void` `deleteElements(``int` `[]arr, ``int` `n, ``int` `k) ` ` ``{ ` ` ``// Create a stack and push arr[0] ` ` ``Stack<``int``> s = ``new` `Stack<``int``>(); ` ` ``s.Push(arr[0]); ` ` ` ` ``int` `count = 0; ` ` ` ` ``// traversing a loop from i = 1 to n ` ` ``for` `(``int` `i = 1; i < n; i++) ` ` ``{ ` ` ` ` ``// condition for deleting an element ` ` ``while` `(s.Count != 0 && s.Peek() < arr[i] ` ` ``&& count < k) ` ` ``{ ` ` ``s.Pop(); ` ` ``count++; ` ` ``} ` ` ` ` ``s.Push(arr[i]); ` ` ``} ` ` ` ` ``// Putting elements of stack in a vector ` ` ``// from end to begin. ` ` ``int` `m = s.Count; ` ` ``int``[] v = ``new` `int``[m]; ``// Size of vector is m ` ` ``while` `(s.Count != 0) ` ` ``{ ` ` ` ` ``// push element from stack to vector v ` ` ``v[--m] = s.Peek(); ` ` ``s.Pop(); ` ` ``} ` ` ` ` ``// printing result ` ` ``foreach` `(``int` `x ``in` `v) ` ` ``{ ` ` ``Console.Write(x + ``" "``); ` ` ``}; ` ` ` ` ``Console.Write(``""``); ` ` ``} ` ` ` ` ``// Driver code ` ` ``public` `static` `void` `Main() ` ` ``{ ` ` ``int` `n = 5, k = 2; ` ` ``int` `[]arr = {20, 10, 25, 30, 40}; ` ` ``deleteElements(arr, n, k); ` ` ``} ` `} ` ` ` `// This code is contributed by 29AjayKumar `
Output:
```25 30 40
```
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Article Tags :
Practice Tags : | 2,434 | 6,877 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2021-04 | latest | en | 0.813184 |
https://stats.stackexchange.com/questions/593571/no-variance-in-binomial-glmer-yields-problematic-ci | 1,721,787,373,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518154.91/warc/CC-MAIN-20240724014956-20240724044956-00359.warc.gz | 469,931,191 | 43,842 | # No variance in binomial glmer yields problematic CI
I have a glmer with a binomial response variable and two binary predictor variables, where there are no observations for one of the four groups. I used emmeans to calculate confidence intervals, and for this group obviously the mean and variance are zero. The model predicts a CI from 0 to 1, and my question is whether the confidence interval and contrasts are calculated correctly when all observations and thus variance of a group are zero. Please see details below. I have a total of 804 observations, so a bit much to include here. Should someone want to reproduce my models, I can share the data somehow.
The data concern observations of two female sharks in an aquarium. The response variable is presence/absence in a shipwreck (in_wrak_inc). Presence was recorded by multiple people (nearly twenty observers) every five minutes between 9:00 and 17:00 for many days (a few in early January 2021, and then intermettently from 2021-12-14 to 2022-01-19). As random effects, the model includes observer and date. As fixed effects, the model includes individual (either the first or second female shark, jv or ov), presence/absence of a male shark (locatie_jm, where Kleine Oceaan == absence) and there interaction.
individu locatie_jm in_wrak_inc n
1 jv Kleine Oceaan nee 52
2 jv Kleine Oceaan ja 27
3 jv Oceaan nee 213
4 ov Kleine Oceaan nee 83
5 ov Kleine Oceaan ja 165
6 ov Oceaan nee 177
7 ov Oceaan ja 87
As you can see there are no observations for jv when locatie_jm == Oceaan + in_wrak_inc == ja. My code is below.
model_wrak1 <- lme4::glmer(in_wrak_inc ~ locatie_jm * individu + (1|datum) + (1|observant),
data = scans_vrouwtjes_wrak,
family = binomial)
summary(model_wrak1)
# Generalized linear mixed model fit by maximum likelihood (Laplace Approximation) ['glmerMod']
# Family: binomial ( logit )
# Formula: in_wrak_inc ~ locatie_jm * individu + (1 | datum) + (1 | observant)
# Data: scans_vrouwtjes_wrak
#
# AIC BIC logLik deviance df.resid
# 406.6 434.7 -197.3 394.6 798
#
# Scaled residuals:
# Min 1Q Median 3Q Max
# -1.7776 -0.0625 0.0000 0.0980 8.6087
#
# Random effects:
# Groups Name Variance Std.Dev.
# datum (Intercept) 26.32 5.130
# observant (Intercept) 32.81 5.728
# Number of obs: 804, groups: datum, 27; observant, 16
#
# Fixed effects:
# Estimate Std. Error z value Pr(>|z|)
# (Intercept) 1.9335 2.7759 0.697 0.486091
# locatie_jmOceaan -31.3131 66.9586 -0.468 0.640036
# individuov 1.1188 0.3358 3.331 0.000864 ***
# locatie_jmOceaan:individuov 25.2350 66.9589 0.377 0.706268
# ---
# Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
#
# Correlation of Fixed Effects:
# (Intr) lct_jO indvdv
# locat_jmOcn -0.013
# individuov -0.115 0.000
# lct_jmOcn:n -0.012 -0.999 0.000
model_wrak_emm_loc_ind <- emmeans::emmeans(model_wrak1,
specs = pairwise ~ locatie_jm | individu,
type = "response")
model_wrak_emm_loc_ind$emmeans # individu = jv: # locatie_jm prob SE df asymp.LCL asymp.UCL # Kleine Oceaan 0.8736 0.306 Inf 0.029111 0.999 # Oceaan 0.0000 0.000 Inf 0.000000 1.000 # # individu = ov: # locatie_jm prob SE df asymp.LCL asymp.UCL # Kleine Oceaan 0.9549 0.119 Inf 0.086899 1.000 # Oceaan 0.0463 0.123 Inf 0.000208 0.919 # # Confidence level used: 0.95 # Intervals are back-transformed from the logit scale model_wrak_emm_loc_ind$contrasts |> confint()
# individu = jv:
# contrast odds.ratio SE df asymp.LCL asymp.UCL
# Kleine Oceaan / Oceaan 3.97e+13 2.66e+15 Inf 0.00 3.93e+70
#
# individu = ov:
# contrast odds.ratio SE df asymp.LCL asymp.UCL
# Kleine Oceaan / Oceaan 4.36e+02 1.34e+03 Inf 1.05 1.82e+05
#
# Confidence level used: 0.95
# Intervals are back-transformed from the log odds ratio scale
• I don't really understand why you consider the CI (0, 1), problematic. It is the most conservative CI possible (with guaranteed coverage), which seems appropriate if there is no information to derive a more narrow CI. Commented Oct 27, 2022 at 5:27
• If there are many observations and all are negative, surely that is an indication it is somewhat likely the CI should be closer to zero than to one? In this case, the probability of the other group is 0.87, and I expect a difference in the contrasts. Commented Oct 31, 2022 at 9:08
• Sure, but you would need to quantify "somewhat likely" to derive a more narrow CI. And you can't do that just based on the model. Commented Nov 1, 2022 at 6:01
• I understand the model cannot do that (because there is no variance?). Can it be done based on the number of observations? I have 213 observations where in_wrak_inc == 0. Surely the likelihood of in_wrak_inc == 1 is therefore lower than when I would have, say, only 15 observations. Commented Nov 2, 2022 at 9:19
You may find this paper helpful.
This problem, sparse data in one or more cells of the cross-tabulation of your predictor and outcome variables (as you present above) which causes extreme parameter estimates, is referred to as separation. The solution is usually one or another form of regularization
(Alternatively, you can omit the covariate(s), which cause the separation, but I'll ignore that here).
There are several such options discussed in the linked paper. Potentially the easiest to implement would be data augmentation. Another option would be to estimate the model in the Bayesian framework. Using even modestly informative priors on the log odds ratios would provide the regularization needed to get sane results (and narrower intervals).
You could try estimating the model with R package brms. It uses the same syntax as lme4 so you wouldn't really need modify your model specification, you'd just need to add priors for the model parameters.
• Thanks. That paper, and looking at separation in logistic regression generally, seems very useful. I'll check it out. Commented Oct 31, 2022 at 9:09
Trying to answer my own question. The lme4 package cannot deal with complete separation, but info how to fix it can be found here and here (based on a question posted here). Running a mixed model with priors with blme::bglmer() did the trick. The priors I chose (10 for the intercept and 2.5 for the various slopes) are default.
model_wrak1_bayes <- blme::bglmer(in_wrak_inc ~ locatie_jm * individu + (1|datum) + (1|observant),
data = scans_vrouwtjes_wrak,
family = binomial,
fixef.prior = normal(sd = c(10, 2.5)))
This gives the following emmeans, which I find very reasonable:
individu = jv:
locatie_jm prob SE df asymp.LCL asymp.UCL
Kleine Oceaan 0.092783 0.216260 Inf 6.65e-04 0.9402
Oceaan 0.000234 0.000542 Inf 2.49e-06 0.0215
individu = ov:
locatie_jm prob SE df asymp.LCL asymp.UCL
Kleine Oceaan 0.269436 0.505533 Inf 2.40e-03 0.9826
Oceaan 0.093029 0.198690 Inf 1.01e-03 0.9120
Confidence level used: 0.95
Intervals are back-transformed from the logit scale | 2,220 | 7,227 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-30 | latest | en | 0.827058 |
https://nl.webqc.org/molecularweightcalculated-190203-107.html | 1,568,991,214,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574039.24/warc/CC-MAIN-20190920134548-20190920160548-00199.warc.gz | 592,561,810 | 5,463 | #### Chemical Equations Balanced on 02/03/19
Molecular weights calculated on 02/02/19 Molecular weights calculated on 02/04/19
Calculate molecular weight
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140
Molar mass of AsNO3 is 136.9265
Molar mass of CFH3 is 34.0329232
Molar mass of CFH3 is 34.0329232
Molar mass of Na2SiF5 is 169.05705456
Molar mass of Na2SiF6 is 188.05545776
Molar mass of C2OFH3 is 62.0430232
Molar mass of C2OFH3 is 62.0430232
Molar mass of CrF3 is 108.9913096
Molar mass of C2O2 is 56.0202
Molar mass of C2O2 is 56.0202
Molar mass of Zn(NO3)2 is 189.3898
Molar mass of Mn3(SO4)7 is 837.252335
Molar mass of C2O2H4 is 60.05196
Molar mass of C2O2H4 is 60.05196
Molar mass of KBr is 119,0023
Molar mass of CHN is 27.02534
Molar mass of AcH is 228,0356921
Molar mass of Ba2CuO2Cl2 is 441.1048
Molar mass of SrCuO2 is 183.1648
Molar mass of C2HN2 is 53.04274
Molar mass of C2H10N2 is 62.1142
Molar mass of CaF2 is 78.0748064
Molar mass of C2H10N is 48.1075
Molar mass of C3H10N is 60.1182
Molar mass of C3H5N is 55.0785
Molar mass of F2 is 37.9968064
Molar mass of C6H5N is 91.1106
Molar mass of AgNO3 is 169.8731
Molar mass of (CH3)2CHCH2OH is 74.1216
Molar mass of H3PW12O40 is 2880.053582
Molar mass of H3PW12O40 is 2880.053582
Molar mass of CHNO is 43.02474
Molar mass of C6H14 is 86.17536
Molar mass of (FeBr)2 is 271.498
Molar mass of C2HNO is 55.03544
Molar mass of C2H6 is 30.06904
Molar mass of C3HNO is 67.04614
Molar mass of C4HNO is 79.05684
Molar mass of CN2SH4 is 76.12086
Molar mass of C5HNO is 91.06754
Molar mass of SiO2 is 60,0843
Molar mass of C5H5NO is 95.0993
Molar mass of C6H6NO is 108.11794
Molar mass of H2SO4 is 98,07848
Molar mass of C8H8NO is 134.15522
Molar mass of Pb(NO₃)₂ is 331.2098
Molar mass of C7H7NO is 121.13658
Molar mass of CrF3 is 108.9913096
Molar mass of C7H7NO2 is 137.13598
Molar mass of Sr2CuO2I2 is 524.59374
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140
Calculate molecular weight
Molecular weights calculated on 02/02/19 Molecular weights calculated on 02/04/19
Molecular masses on 01/27/19
Molecular masses on 01/04/19
Molecular masses on 02/03/18
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Periodiek systeem Eenheden omrekenen Chemie gereedschappen Chemisch Forum Chemie FAQ Constanten Symmetrie Zoek Chemie links Link naar ons Heb je een idee? Neem contact met ons op Stel een betere vertaling voor Kies een taalDeutschEnglishEspañolFrançaisItalianoNederlandsPolskiPortuguêsРусский中文日本語한국어 Hoe moet je citeren? WebQC.Org online onderwijs gratis huiswerkhulp chemische opgaven vragen en antwoorden | 1,608 | 3,471 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2019-39 | latest | en | 0.734471 |
https://www.numbersaplenty.com/1286562400 | 1,726,637,924,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651836.76/warc/CC-MAIN-20240918032902-20240918062902-00258.warc.gz | 838,058,602 | 3,673 | Search a number
1286562400 = 25527920357
BaseRepresentation
bin100110010101111…
…0110001001100000
310022122220010211021
41030223312021200
510113324444100
6331355255224
743611412534
oct11453661140
93278803737
101286562400
1160025a412
122baa4a514
13176711836
14c2c23bc4
1577e38a1a
hex4caf6260
1286562400 has 72 divisors (see below), whose sum is σ = 3180733920. Its totient is φ = 508085760.
The previous prime is 1286562359. The next prime is 1286562419. The reversal of 1286562400 is 42656821.
It is a congruent number.
It is an unprimeable number.
It is a polite number, since it can be written in 11 ways as a sum of consecutive naturals, for example, 53022 + ... + 73378.
It is an arithmetic number, because the mean of its divisors is an integer number (44176860).
Almost surely, 21286562400 is an apocalyptic number.
1286562400 is a gapful number since it is divisible by the number (10) formed by its first and last digit.
It is an amenable number.
It is a practical number, because each smaller number is the sum of distinct divisors of 1286562400, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (1590366960).
1286562400 is an abundant number, since it is smaller than the sum of its proper divisors (1894171520).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
1286562400 is a wasteful number, since it uses less digits than its factorization.
1286562400 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 20456 (or 20443 counting only the distinct ones).
The product of its (nonzero) digits is 23040, while the sum is 34.
The square root of 1286562400 is about 35868.6827190517. The cubic root of 1286562400 is about 1087.6194152570.
The spelling of 1286562400 in words is "one billion, two hundred eighty-six million, five hundred sixty-two thousand, four hundred". | 565 | 1,931 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2024-38 | latest | en | 0.859137 |
https://www.schooltube.com/tag?tagid=inverse%20operations%20multiplication | 1,679,377,172,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943625.81/warc/CC-MAIN-20230321033306-20230321063306-00352.warc.gz | 1,038,790,211 | 20,472 | # Search for tag: "inverse operations multiplication"
#### How to Solve Equations Using the Division Property of Equality | -3y=63 | Part 2 of 2 | Minute Math
In this video we learn how to solve equations using the division property of equality. We solve the equation -3y=63 using the division property of equality. #propertiesofequality…
From MinuteMath 0 likes 1 plays
#### How to Solve Equations Using the Division Property of Equality | 7x=-49 | Part 1 of 2 | Minute Math
In this video we learn how to solve equations using the division property of equality. We solve the equation 7x=-49 using the division property of equality. #propertiesofequality…
From MinuteMath 0 likes 5 plays | 164 | 695 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2023-14 | latest | en | 0.809555 |
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Edexcel FP3 June 2015 - Official Thread watch
Announcements
1. (Original post by nayilgervinho)
Thanks, can you also help me with june 2014 9a I can;t do it.
https://3a14597dd5c7aa2363f067571766...%20Edexcel.pdf
(x^2+1)^(-n)= (x)'(x^2+1)^(-n) ....oh that's too long,I'm sorry. You can see the mark scheme(perhaps you did it, but I really don't have time)
2. (Original post by Abel Demoz)
Can someone show me how to do part a and explain it to me
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Which question?
3. (Original post by Vesniep)
(x^2+1)^(-n)= (x)'(x^2+1)^(-n) ....oh that's too long,I'm sorry. You can see the mark scheme(perhaps you did it, but I really don't have time)
I got it now, thanks anyway.
4. Predictions for an A tomorrow.
5. the?
the meaning of life?
Attached Images
6. (Original post by nayilgervinho)
Predictions for an A tomorrow.
Cant predict wothout sitting it first.
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7. (Original post by nayilgervinho)
Predictions for an A tomorrow.
There will be
a question on Matrices
a question on vectors
a question on hyperbolics
a question on the conic sections (maybe a hyperbola, maybe an ellipse)
a question on a reduction formula
a question on arclength or perhaps on a surface of revolution instead.
Now if you do all these questions correctly you will get an A
(no need to bother with the 7th question)
Attached Images
8. (Original post by TeeEm)
There will be
a question on Matrices
a question on vectors
a question on hyperbolics
a question on the conic sections (maybe a hyperbola, maybe an ellipse)
a question on a reduction formula
a question on arclength or perhaps on a surface of revolution instead.
Now if you do all these questions correctly you will get an A
(no need to bother with the 7th question)
What if there are 9 questions?
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9. (Original post by physicsmaths)
What if there are 9 questions?
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I predict that it will not be a "9 questions" paper...
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10. (Original post by bwr19)
My teacher who marks some of the papers says if there's a really long proof and he sees the right answer at the end, he'll generally just give all the marks. lol.
68/75 or 68 UMS?
Latter. Wouldnt resit with 68/75 lol.
11. (Original post by keromedic)
Latter. Wouldnt resit with 68/75 lol.
That wudve been like 99ums lol
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12. (Original post by physicsmaths)
That wudve been like 99ums lol
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I've just got up. Wbu?
13. Just going sleep ha. How u feeling for fp3? If we get an normal vecotrs hen ill kill it. If not will have to see. Only learnt half the module yesterday haha!
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14. (Original post by TeeEm)
I predict that it will not be a "9 questions" paper...
I hope you are right!
15. (Original post by mmms95)
i hope vectors is hard, and the reduction is easy if i don't get the reduction in one go, im gonna skip it so i don't panic
Yeah, if I don't, I'm going to go to it at the end so I don't lose time for the other questions worrying about it.
16. Good Luck everyone!
17. Thanks and to you aswell hope everyone does good luck to all
18. Best of luck! I can almost taste the freedom now.
19. (Original post by TeeEm)
the?
the meaning of life?
42?
20. How's the paper and how's the feeling about conics?
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http://physicshelpforum.com/quantum-physics/15106-qed-all-path-integral-approach-mirror-reflection.html | 1,550,421,035,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247482186.20/warc/CC-MAIN-20190217152248-20190217174248-00637.warc.gz | 215,654,063 | 13,098 | Physics Help Forum QED “All-Path integral approach” for Mirror Reflection
Quantum Physics Quantum Physics Help Forum
Feb 11th 2019, 02:50 PM #1 Junior Member Join Date: Aug 2016 Location: Napa, CA Posts: 6 QED “All-Path integral approach” for Mirror Reflection Hi folks! For mirror reflection I have been puzzled by QED “All-Path integral approach” argument. I appreciate if you examine my article and let me know where I have made my mistakes that my reasoning defies this QED argument. Let us make an observation; if we turn off the pumps of a fish tank, then we can see the reflection of the content of the fish tank from the calm surface water of the fish tank. This calm surface water reflects the image of the fish which they are swimming in the fish tank. I do not know the percentage of this reflection. This observation shows that the surface water reflects a percentage of the incoming light both from above of the water surface and also from below of the water surface, and these reflections are expressed by the Optic Law for reflection. The subject of mirror reflection is explained by the Quantum Electrodynamic theory (QED). Prof. Feynman in his lectures at University of Auckland 1979, mentioned that according to QED the reflection of a photon from a reflective surface is the interaction of that photon with an electron that exists in the material that is constructed that image reflecting surface. For the mirror reflection the incoming photon to the surface of a mirror, it gets absorbed by an electron, and then this energized electron emits the reflected photon. Prof. Feynman commented that the reflected photon is not the same photon that hits to the surface of the mirror, but rather it is a new photon, that it is emitted by the energized electron which has absorbed that incoming photon. (These comments are made in lecture #1 at 55:50 minutes and at 1:10:00 minutes of the video, and also in lecture #2 at 1:20:00 minutes of the video, and also in lecture #3 at 1:01:00 minutes of the video.). QED explains the process of mirror reflection by the “All-Path integral approach”, in this argument it is shown that the reflected photon from the surface of the mirror it is directed toward the detecting photon-multiplier. The reflected photon gets detected at a specific direction. QED’s “All-Path integral approach” is explained in Prof. Feynman’s lecture #2 at 28:30 minutes of the video. Here I need to mention another QED’s finding regarding the emission of photon by an energized electron. QED expresses the emission of a photon by an energized electron falls in the domain of the uncertainty rule, and the direction of the emitted photon can only be expressed by a statistical method which it is in all directions, this means that we can never predict the direction that an energized electron emits its photon. In the process of the reflection of one photon, for the energized electron at the surface of the mirror, these two statements of QED contradict with each other. If one statement is correct then the other statement must be wrong, and this is not acceptable. Let us investigate for the source of this contradiction. By examining QED’s “All-Path integral approach” we find that there is a false assumption embedded in that argument. On the surface of the mirror we have directed all the statistical emissions toward the detecting photo-multiplier, this is a false assumption, these statistical emissions from the energized electrons are supposed to be random and in all directions, and not specifically in the direction of the detecting photo-multiplier. The probabilities of these emissions of the energized electrons on the surface of the mirror to reach to the detecting photo-multiplier are very low and almost negligible. And this false assumption has caused those two QED statements to contradict with each other. So, because of this false assumption that it is embedded inside the QED’s “All-Path integral approach”, that makes this QED argument false and void. There is another false step in QED’s “All-Path integral approach”. In order that QED’s “All-Path integral approach” and QED’s Wave Function method to perform their configurations they do require that the correct detection point to be provided, in advance. This is so absurd that these QED methods have this requirement! The detection point of the reflected photon is supposed to be produced by these QED methods, and not to be required in advance! Let me show the falsehood of the QED’s “all-path integral approach” in another way. Let us apply QED’s “all-path integral approach” for a source that it is on the left side of a mirror and it sends ONE photon to the surface of the mirror and the reflected photon is detected by a photon detector at Point B on the right side of the mirror. QED’s “all-path integral approach” reports that its technique for this scenario is in harmony with the result from the Optic Law. Now, I add ten more photo-multipliers to the right side of the mirror. Amazingly QED’s “all-path integral approach” produces amplitude vectors for each one of these detectors, and it cannot distinguish which one of these results is a valid result, and which ones are wrong results. This argument re-visited the requirement of the correct detection point is needed to be provided in advance, in order that QED “All-Path approach” to produce a valid result. Let me show the falsehood of the QED “All-Path approach” in another way. In the previous scenario before adding the other photo-multipliers to the right side of the mirror, let us name the point that the photon hits to the surface of the mirror point A. I cut out a circle from the mirror, centered at point A with a radius of one millimeter. Optic Law predicts no matter how we manipulate the summation of the amplitude vectors for the rest of the mirror, the photon detector no longer detects any photon. Some critiques might question that the little gap in the mirror has eliminated the significant part of the amplitude vectors which were lining up in the same direction. My answer to these critiques is this, still the amplitude vectors from the other strips on the mirror surface which run from the left side to the right side of the mirror, which the gap in the mirror has not affected them, they are still contributing significantly to the summation of the amplitude vectors, and they are lining up at their midpoints, and their contributions are as significance as before. For this case, no matter how we manipulate the value of the summation of the amplitude vectors for the rest of the mirror, still there would be no detection on the right side of the mirror! The value of the summation amplitude vectors in this case is a bogus value, and it represents nothing. Now, that I have shown the falsehood of these two QED methods for the mirror reflection, then still the mechanics of the Mirror Image Reflection remains unanswered. We need to figure out what causes that the energized electron at the surface of the mirror to emit its photon in the specific direction that produces the Optic Law for reflection. How does this energized electron register the direction that it receives the incoming photon that it can emit its photon in the direction that produces the Optic Law? And the direction of the incoming photon is registered by the absorbing electron not based on the electron’s own internal coordinates, but rather according to the coordinates of the surface of the mirror. The surface of the mirror plays the paramount role in this reflection. This peculiar behavior it gets even more bizarre; the probability of the reflection increases as the angle of the incident increases! (For glass the probability of the reflection at 0 degree angle of incident is 4%, and at 70 degree angle of incident is about 33%, and these probabilities of the reflections are acquired by counting the reflected photons, and the result of this counting fluctuates slightly between different studies. The length of the amplitude vector is designed to be the square root of these empirical results. We have implemented all these empirical data into our QED argument; so, it is deceptive to claim that our QED argument has produced the correct result, similar to the result from our experiment.) As I mentioned the surface of the mirror plays the paramount role in the mirror image reflection, and of course QED all-path integral methods has tried to create a relation between the surface of the mirror and the reflected photon, but as I showed QED all-path integral approach by using a false assumption and a false step, it has tried to fudge its result to reach to the same result as of the Optic Law. We should not deceive ourselves by fudging our argument trying to reach to the desired result. This is totally forbidden in our scientific studies. Deception has no place in our scientific analyses. It seems the mirror reflection is a new phenomenon between the incoming photons and the surface of the mirror, and our current analyses of interaction between photon and electron has not answered this puzzle. I greatly appreciate your input. Last edited by Unes; Feb 12th 2019 at 02:09 PM.
Feb 13th 2019, 10:24 AM #2 Senior Member Join Date: Jun 2016 Location: England Posts: 703 My own personal impression of the sum over all possible paths approach is that it is a mathematical method that can be used to calculate the observed phenomena, not a revelatory description of the deepest workings of the universe. I disapprove of the "shut up and calculate" philosophy it seems to be based upon. I have watched some of these Feynman lectures previously, but would have to watch again to remind myself of the points you make. Unfortunately my computer is currently playing up (no sound) which makes this difficult... Regarding the electron photon interaction, what are the timescales for absorption and emission? Can it be argued that the (electron wave) + (photon wave) combine to create a (electron + photon) wave which then (immediately) disassociates into a (electron wave) + (photon wave) pair again. In this scenario we do not (necessarily) erase the original photon information. But I am just letting my mind free-wheel here, this is not something I know. __________________ ~\o/~
Feb 13th 2019, 11:01 AM #3 Senior Member Join Date: Jun 2016 Location: England Posts: 703 Pensées sans frontières Every now and then I give my imagination free reign... Let us imagine the universe described by a single probability function. Focus down onto a tiny region of this universe. We could find that the small portion of the universal probability function, that operates within that tiny region, can be approximated by an isolated probability function. This isolated probability function approximation presents the behaviours we interpret as being an electron. Similarly we can approximate a different portion of the universal probability function by a "photon" style probability function. We can bring these two approximations together and describe their interactions. We can zoom back out and see how this interaction is just one wrinkle in the universal probability function. __________________ ~\o/~
Tags approach”, integral, mirror, qed, reflection, “allpath
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