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https://tradencxsiix.netlify.app/dahm52398cuhi/annuity-calculator-future-value-monthly-405 | 1,723,597,868,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641086966.85/warc/CC-MAIN-20240813235205-20240814025205-00245.warc.gz | 441,931,106 | 10,926 | Annuity calculator future value monthly
Compound Interest: The future value (FV) of an investment of present value (PV) Numerical Example: A CD paying 9.8% compounded monthly has a nominal Future Value (FV) of an Annuity Components: Ler where R = payment, r = rate Present Value of an Annuity Calculator. This calculator will compute the present value of a series of equal cash flows to be received in the future. Calculate
Calculate Present Value of Future Cash Flows. This annuity calculator computes the present value of a series of equalshow more instructions. cash flows to The interval can be monthly, quarterly, semi-annually or annually. Present Value Calculates a table of the future value and interest of periodic payments. monthly. payment amount. (PMT). payment due at. beginning end of period Trying to solve for interest rate (to debate yay or nay on an annuity) if I need to pay Use this calculator to determine the future value of an ordinary annuity which is a payments which typically are annual, semiannual, quarterly or monthly. The future value calculator can be used to calculate the future value (FV) of an (I/Y), starting amount, and periodic deposit/annuity payment per period (PMT). An annuity is a series of payments made at equal intervals. Examples of annuities are regular deposits to a savings account, monthly Valuation of an annuity entails calculation of the present value of the future annuity payments. Free future value calculator helps you to compute returns on savings accounts and other investments. Easy-to-understand charts. Powered by Wolfram|Alpha.
18 Oct 2015 Plan a comfy retirement with the help of an annuity calculator and other us estimate the income we might receive in the future -- from annuities, If he wants a joint lifetime immediate annuity with his 65-year-old wife, then the monthly Enter how much your portfolio is currently worth and how much you
Calculate the future value of an annuity given monthly contribution rate, time of investment, and annual interest rate. 31 Dec 2019 The calculation is identical to the one used for the future value of an what if the interest on the investment compounded monthly instead of This is a free online tool by EverydayCalculation.com to calculate future value of annuity (FVA) of both simple as well as complex annuities. Compound Interest: The future value (FV) of an investment of present value (PV) Numerical Example: A CD paying 9.8% compounded monthly has a nominal Future Value (FV) of an Annuity Components: Ler where R = payment, r = rate Present Value of an Annuity Calculator. This calculator will compute the present value of a series of equal cash flows to be received in the future. Calculate 9 Dec 2007 In other words, 9% compounded monthly is equivalent to 9.38% compounded annually. Now we can perform our FV of an annuity calculation you can eam 12% compounded monthly, how much will you have when you retire? 1. This is an example of a "Future Value of an Annuity" calculation where we
Future value of annuity calculator is designed to help you to estimate the value of when compounding is applied annually, m=1, when quarterly, m=4, monthly,
➡ To get the PVAk,n, simply use PMT = 1. ➡ 1 [PMT]; 6 [I/Y]; 5 405. ➡ [CPT][PV] Display = -4.4651056 © Copyright 2002, Alan Marshall. 20. FV of an Annuity Due. (. ). This pension calculator illustrates the tentative Pension and Lump Sum amount on regular monthly contributions, percentage of corpus reinvested for purchasing corpus at retirement ₹ 3123533 ₹ 4685300 Annuity value Lump sum value. So if the annual interest rate is 6% and you make monthly loan payments, the = PMT(rate,nper,pv,fv,type); =RATE(nper,pmt,pv,fv,type,guess); =NPER(rate,pmt
Calculates a table of the future value and interest of periodic payments. monthly. payment amount. (PMT). payment due at. beginning end of period Trying to solve for interest rate (to debate yay or nay on an annuity) if I need to pay
Present value (also known as discounting) determines the current worth of For instance, a 12% annual interest rate, with monthly compounding for two Be able to calculate future value and present value of lump-sum and annuity amounts. Note: we receive monthly payments, so we use 6%/12 = 0.5% for Rate and 20*12 = 240 for Nper. The last two arguments are optional. If omitted, Fv = 0 (no future Use our annuity calculator to find out how much retirement income you can get from a life annuity and see how it compares to income from a GIC or RRIF.
Use this calculator to determine the future value of an ordinary annuity which is a payments which typically are annual, semiannual, quarterly or monthly.
The FV function calculates the future value of an annuity investment based on For example, a car loan for 36 months may be paid monthly, in which case the Perhaps more subtle, an Immediate Fixed Annuity might calculate your monthly payment for a 5-year 6% annuity by first calculating the future value as FV(6%,5,0 ,- An annuity is a fixed income over a period of time. present value \$1000 vs future value \$1100. So \$1,100 next We have done our first annuity calculation! There are 60 monthly payments, so n=60, and each payment is \$400, so P = \$400.
This pension calculator illustrates the tentative Pension and Lump Sum amount on regular monthly contributions, percentage of corpus reinvested for purchasing corpus at retirement ₹ 3123533 ₹ 4685300 Annuity value Lump sum value. So if the annual interest rate is 6% and you make monthly loan payments, the = PMT(rate,nper,pv,fv,type); =RATE(nper,pmt,pv,fv,type,guess); =NPER(rate,pmt Present value (also known as discounting) determines the current worth of For instance, a 12% annual interest rate, with monthly compounding for two Be able to calculate future value and present value of lump-sum and annuity amounts. Note: we receive monthly payments, so we use 6%/12 = 0.5% for Rate and 20*12 = 240 for Nper. The last two arguments are optional. If omitted, Fv = 0 (no future Use our annuity calculator to find out how much retirement income you can get from a life annuity and see how it compares to income from a GIC or RRIF. | 1,469 | 6,158 | {"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-2024-33 | latest | en | 0.916785 |
https://www.physicsforums.com/threads/help-with-basic-problem.775312/ | 1,527,178,159,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794866511.32/warc/CC-MAIN-20180524151157-20180524171157-00503.warc.gz | 803,108,763 | 20,438 | # Homework Help: Help with basic problem?
1. Oct 9, 2014
### OrigamiCaptain
1. The problem statement, all variables and given/known data
I'm working out of Serge Lang's Basic Mathematics and in chapter 3 section 3 this problem comes up.
$(3x+5) ^{-4} =64$
2. Relevant equations
3. The attempt at a solution
I keep getting $(1-(5±\sqrt{8}))/(3±\sqrt{8})$
The book says it is $(±1-5\sqrt{8})/(3\sqrt{8})$
It is not exact. I know $\sqrt{8}=2\sqrt{2}$
I'm obviously forgetting something basic and it is bothering me. Thanks.
Last edited: Oct 9, 2014
2. Oct 9, 2014
### LCKurtz
3. Oct 10, 2014
### Staff: Mentor
You're taking the square root of each side, then the same again?
4. Oct 10, 2014
### OrigamiCaptain
This is my attempt. I hope it is legible
.
5. Oct 10, 2014
### Ray Vickson
Don't attach photos. PF wants you to type out your work.
6. Oct 10, 2014
### OrigamiCaptain
Ok, but why? That makes no sense. That is much more difficult and time consuming and it produces the same outcome. I would get ride of it, but it is not letting me now. Feel free to remove it if you like.
If you don't want us to attach photos then why included the ability to add an attachment?
Last edited: Oct 10, 2014
7. Oct 10, 2014
### RUber
It looks like you are doing alright until about 2/3 down the page.
You get to $3x-5 = \frac{1}{\pm \sqrt{8}}$. Then you do something I don't understand, and that is where you get off track.
If you are multiplying by a $\pm \sqrt{8}$ please use parentheses so you don't confuse it for adding or subtracting later on.
As far as I can tell, your solution may be off because of what I just mentioned.
You could also take one at a time (either plus or minus) then express the solution with a $\pm$ at the end. For more complicated algebra, this will help you maintain some sanity, even though it might take more paper.
8. Oct 10, 2014
### OrigamiCaptain
When I get to $3x+5 = \frac{1}{\pm \sqrt{8}}$
I subtract $-5$ from both sides
$3x= \frac{1}{\pm \sqrt{8}} -5$
$3x=1-(5±\sqrt{8}))/(±\sqrt{8})$
Then I divided by 3.
9. Oct 10, 2014
### RUber
I think you are trying to write $3x = \pm \frac{1}{\sqrt{8}} - 5$
Multiply through by $\sqrt{8}$ on both sides to get:
$3\sqrt{8}x = \pm 1- 5\sqrt{8}$.
This is why I mentioned it might be easier to do plus first, then minus. Your expression $(5 \pm \sqrt{8})$ appears like addition (or subtraction), but the only way you get there is by multiplying $(\pm \sqrt{8})$ onto the 5--so either your notation is wrong or your understanding of algebra is wrong. I assume the former. In any case, that is the only place your answer differs from the book answer you have above.
10. Oct 10, 2014
### OrigamiCaptain
No my algebra is definitely wrong here. I need help understanding what I did wrong. I divided by $\pm\sqrt{8}$ and I either don't know something or forgot something and my algebra is wrong. What used latex to show is what I did.
It doesn't bother me to be wrong as long as I figure it out.
Last edited: Oct 10, 2014
11. Oct 10, 2014
### RUber
I apologize for the tone in my earlier post. I meant to say that it looks like you are doing the right steps, but poor notation is killing you.
The signs in the denominator are often easier to keep track of when you pull them to the front.
$\frac{1}{\pm \sqrt{8}} =\pm \frac{1}{ \sqrt{8}}$
However, what you have is not wrong.
$3x = \frac{1}{\pm \sqrt{8}}-5$
There is no problem multiplying by the $\pm$ but it complicates things.
You could from what you have there just write
$x = \frac13( \frac{1}{\pm \sqrt{8}}-5)$
But I suppose you are trying you get something that looks like the book answer.
So distribute the 1/3 in and pull the $\pm$ out of the bottom.
$x = \pm \frac{1}{ 3\sqrt{8}}-\frac53$
The book answer is just this put over a common denominator.
You went wrong when you put :
$3x=(1-(5 \pm \sqrt{8}))/(\pm \sqrt{8})$ when you should have written $3x=(1-5 (\pm \sqrt{8}))/(\pm \sqrt{8})$ to represent the multiplication.
Note that this is equal to $3x=1/(\pm \sqrt{8})-5 (\pm \sqrt{8})/(\pm \sqrt{8})$ and the $\pm$ cancel in the second fraction.
12. Oct 10, 2014
### Staff: Mentor
It's easier for someone to help when they can use the "Quote" feature to pick out specific steps in your calculation in order to comment on them, instead of having to read them off a photo and type them in, and then proofread for errors in transcription. It's enough work for them to fix mistakes in their own equations!
13. Oct 10, 2014
### OrigamiCaptain
Ok I see what you were saying now. I appreciate the help.
Ok, fair enough. I'll keep that in mind.
14. Oct 10, 2014
### Staff: Mentor
Ok, all sorted now? ;)
15. Oct 10, 2014
### Ray Vickson
Sometimes you need to attach pictures of diagrams from books or notes, for example, or maybe drawings of geometric figures, etc. Other than that, attachments are discouraged.
Anyway, back to your problem: it can be written as $X^4 = r^4$, where $X = 3x+5$ and $r = 1/\sqrt{8}$. For real $r > 0$ the equation $X^4 = r^4$ has two real solutions, $X = \pm r$, and two complex (pure imaginary) solutions $X = \pm i \, r$, where $i = \sqrt{-1}$. Therefore, there should be only two real solutions for $x$. These are $x = (r-5)/3$ and $x = (-r-5)/3$.
16. Oct 11, 2014
### vela
Staff Emeritus
There's a small learning curve for LaTeX, but it's actually not all that difficult nor time-consuming. I know some of the helpers don't seem to care, but I usually pass over threads where students can't be bothered to type up their work. | 1,656 | 5,534 | {"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": 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} | 3.6875 | 4 | CC-MAIN-2018-22 | latest | en | 0.913222 |
https://figshare.com/articles/Examples_comparisons_for_random_partition_algorithms/96310 | 1,503,571,071,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886133449.19/warc/CC-MAIN-20170824101532-20170824121532-00395.warc.gz | 750,834,734 | 7,026 | ## Examples comparisons for random partition algorithms
2013-02-05T16:04:20Z (GMT)
<p><strong>Topic:</strong> generating uniform random samples from the set of all integer partitions for a given total N and a number of parts S.<br><br><strong>Problem:</strong> current random integer partitioning functions of mathematical software can take a long time to generate a single partition for a given N (regardless of S) and an untennable amount of time generating partitions of N with S parts.</p> <p>Currently, no function among math softwares, peer-reviewed literature, and on stackexchange and arxiv.org generates random partitions with respect to both N and S. Consequently, if one is interested in generating random integer partitions for N having S parts, then one must usually waste time generating random partitions of N and rejecting those not matching S.</p> <p><strong>Note!</strong> I have since solved this problem definitively:</p> <p>The question is asked and the solution is presented here: stackoverflow.com/questions/10287021/an-algorithm-for-randomly-generating-integer-partitions-of-a-particular-length/12742508#12742508</p> <p>I've recently published a preprint of a manuscript on figshare outlining a simple and unbiased solution to this question. figshare.com/articles/Random_integer_partitions_with_restricted_numbers_of_parts/156290</p> <p>However, below is an approach I tried and was unable to eliminate sampling bias from while keeping reasonable speed. I guess this page is more useful as a good way to <strong>NOT</strong> go about getting random integer partitions for N and S.</p> <p><strong>Deprecated alternative approach (often biased and slower than the above solution):</strong> Generate a single random partition of N and randomly manipulate it until its number of parts equals S. Why? Because randomly perturbing a partition of N until it satisfies S can be faster than generating random partitions based solely on N and rejecting those without S parts.<br><br><strong>Contents (results of deprecated algorithm):</strong></p> <p>Visual comparisons of 500 random samples generated from the new function derived by myself (red curves) against 500 random samples generated using the random partition function found in the Sage mathematical environment (black curves). <br><br>Kernel density curves (red ones and black ones) are for statistical evenness across the partition. Statistical evenness is a standardized log-transform of the variance. Kernel density cures that overlap nearly completely reveal that the random samples of partitions generated between the two approaches share a similar structure.</p> <p>Evenness is estimated using Evar, a transform of the variance of log summand values. Evar is standardized to take values between 0.0 (no evenness) and 1.0 (perfect evenness). </p> <p>Close agreement between the random manipulation approach and the Sage function (very high rejection rates as most partitions of N don't match S) was also found using other statistical characteristics (e.g. median summand, relative size of largest summand). These results reveal that the statistical quality of evenness (a transform of the variance) is in high agreement between the two approaches (Sage's function and the potential alternative of randomly manipulating integer partitions using conjugates).<br><br><strong>Note:</strong> I have found biases in skewness and the median summand value with this type of method (randomly manipulate an integer partition to arrive at a uniform random sample based on N and S), and would not recommend this approach.</p> <p> </p> | 764 | 3,601 | {"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-2017-34 | longest | en | 0.845997 |
https://calculla.com/calculators/math/polynomial_regression | 1,685,840,150,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649348.41/warc/CC-MAIN-20230603233121-20230604023121-00366.warc.gz | 175,715,841 | 223,522 | Least squares method calculator: polynomial approximation
Calculator finds out coefficients of polynomial that fits best into series of (x, y) points.
# Beta version#
BETA TEST VERSION OF THIS ITEM
This online calculator is currently under heavy development. It may or it may NOT work correctly.
You CAN try to use it. You CAN even get the proper results.
However, please VERIFY all results on your own, as the level of completion of this item is NOT CONFIRMED.
Feel free to send any ideas and comments !
# Calculation data - measurement points#
Format of input data Values only (just a sequence of numbers)Serie of (x,y) points x-values y-values Maximum polynomial degree(polynomial with higher order will not be calculated)
# Results - approximation of your dataset#
Regression type Approximation formula Coefficient of determination R2 Polynomial regression of 3-th degree Show source$y=x^{3}$ 1 Polynomial regression of 2-th degree Show source$y=9~x^{2}-\frac{118}{5}~x+\frac{84}{5}$ 0.998614052 Polynomial regression of 1-th degree Show source$y=\frac{152}{5}~x-\frac{231}{5}$ 0.889470645 Polynomial regression of 0-th degree Show source$y=45$ 0
# Summary - function best fitting to your data#
Measurement points Number of points 5 Points you entered (1, 1), (2, 8), (3, 27), (4, 64), (5, 125) Approximation Regression type Polynomial regression of 3-th degree Function formula Show source$y=x^{3}$ Coefficient of determination R2 1
# Some facts#
• ⓘ Hint: If you're not sure what type of regression this is, let us do the hard work for you and visit another calculator: Regression types.
• Approximation of a function consists in finding a function formula that best matches to a set of points e.g. obtained as measurement data.
• The least squares method is one of the methods for finding such a function.
• The least squares method is the optimization method. As a result we get function that the sum of squares of deviations from the measured data is the smallest. Mathematically, we can write it as follows:
$\sum_{i=1}^{n} \left[y_i - f(x_i)\right]^2 = min.$
where:
• $(x_i, y_i)$ - coordinations of the i-th measurement point, these are points that we know,
• $f(x)$ - the function we are searching for, we want this function to best match to the measurement points,
• $n$ - number of measurement points.
• If we limit the search to polynomial function only, then we say about polynomial regression or polynomial approximation.
$f(x) = W_n(x) = a_{n}\ x^n + a_{n-1}\ x^{n-1} \dots + a_1\ x + a_0$
• f(x) - function that best approximates the input data in the best way,
• an - unknown polynomial coefficients, which we want to find,
• n - the polynomial degree.
• If the degree of the polynomial is zero (n=0), then we get an approximation by constant function, i.e. by one number, which stays closest to all measurement values.
$f(x) = C$
• If the degree of the polynomial is one (n=1), then we get an approximation by linear function:
$f(x) = ax + b$
• For polynomial degrees greater than one (n>1), polynomial regression becomes an example of nonlinear regression i.e. by function other than linear function.
• Using the least squares method, we can adjust polynomial coefficients $\{a_0, a_1, \dots, a_n\}$ so that the resulting polynomial fits best to the measured data. Because polynomial coefficients are numbers, the solution to this problem is equivalent to solve the algebraic equation.
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# Luyu Wang
### CAS
30 total contributions since 2018
Enjoy MATLAB Cody.
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Find the sum of all the numbers of the input vector
Find the sum of all the numbers of the input vector x. Examples: Input x = [1 2 3 5] Output y is 11 Input x ...
plus d'un an ago
Solved
Make the vector [1 2 3 4 5 6 7 8 9 10]
In MATLAB, you create a vector by enclosing the elements in square brackets like so: x = [1 2 3 4] Commas are optional, s...
plus d'un an ago
Solved
Times 2 - START HERE
Try out this test problem first. Given the variable x as your input, multiply it by two and put the result in y. Examples:...
plus d'un an ago | 1,466 | 5,401 | {"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} | 3.609375 | 4 | CC-MAIN-2019-51 | latest | en | 0.749629 |
https://www.mometrix.com/academy/piecewise-functions/ | 1,716,469,364,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058625.16/warc/CC-MAIN-20240523111540-20240523141540-00392.warc.gz | 780,008,528 | 22,476 | # Piecewise Functions
Hi, and welcome to this video about piecewise functions! In this video, we will explore
• What piecewise functions are
• How piecewise functions are defined
• And how piecewise functions can be used
First, let’s look at the definition of a function. A function is a relationship where a single output is assigned to each input.
Many functions belong to function families because their equations and graphs all have similar characteristics.
For example, functions in the linear family have equations that resemble f of x equals mx plus b and their graphs are straight lines. Functions in the quadratic family have equations that look like f of x equals a x squared and their graphs are parabolas.
Piecewise functions are not considered a function family on their own. As the name suggests, they are functions comprised of pieces of other functions.
The first piece we’re going to look at is the absolute value function. Functions in the absolute value family have equations that resemble f of x equals the absolute value of x and their graphs all have a characteristic “V” shape. This is the graph and a section of the table of values of f of x equals the absolute value of x :
Instead of a single V, f(x) can also be visualized as pieces of two linear functions. On the left, f of x equals negative x and on the right, f of x equals x.
As we “read” the graph from left to right, we are on the function f of x equals negative x until x=0. At that point, the function definition changes to f of x equals x.
The domain of the absolute value function is all real numbers. Normally, both f of x equals negative x and f of x equals x also have domains of all real numbers, but if we were to graph them together, the graph would look like this and we would no longer have a function.
So each piece needs to be defined on a section of its domain in order to define a piecewise function. If the left function is only defined for negative x-values and the right is only defined for positive x-values (and we put the 0 into one of them – more on that in a minute), we can define this as a single piecewise function.
One way to visualize this is to graph both linear functions and erase the sections that are not part of the absolute value function.
Let’s take a quick break and practice describing a couple of piecewise functions in terms of what their pieces look like and where those pieces are defined.
First up is the greatest integer function, f of x equals the floor of x. This is also called the floor function or stair step function.
This function is made up of pieces of constant functions that are 1 unit wide.
Next, we have the sawtooth function, f of x equals x minus the floor of x. This function is also called the castle rim function.
This function is made up of pieces of parallel linear functions that are one unit long
The format for defining piecewise functions has this general form:
There are two criteria for naming piecewise functions:
1. Reading the pieces from top-to-bottom in the list corresponds to reading the graph from left-to-right. So the first piece in the list is the left piece on the graph.
2. The domains of the pieces must “add up to” the domain of the entire function.
Let’s start by rewriting our absolute value function in this form:
Here’s the list of expressions that define each piece from left to right:
f of x is equal to the subset of minus x, which is an unknown quantity, and x, which is also an unknown quantity.
Now let’s take a minute to consider the domain of each piece.
f of x equals the absolute value of x has a domain of all real numbers. We can input any number we want and get a single output.
But, we have to be careful with our piecewise definition when we consider x=0. Why? Because 0 is a defined point on both pieces, but we only need to include it once.
If we write the function like this:
f of x is equal to the subset of negative x when x is less than zero and x when x is greater than zero
then neither piece includes 0 and the domain is incomplete.
If we write it like this:
f of x is equal to negative x when x is less than or equal to zero and x when x is greater than or equal to zero.
Then we are not defining a function, because we’re saying that f(x) equals both -x and x at x=0 (even though the output would technically be the same in this case).
Since 0 is in the domain of both pieces, we simply choose which piece to put it in. Both of these equations correctly define the function:
All of our examples so far have depicted piecewise functions with domains of all real numbers. The pieces do not need to connect and they do not need to extend to plus or minus infinity.
Take a look at this function and try to define it with an equation (and yes that single point is a part of it!):
f of x is equal to x when x is greater than negative seven and less than negative three; one when x is equal to negative 1; square root of x when x is greater than or equal to zero and less than four; and x when x is greater than six and less than or equal to seven.
Like other functions, piecewise functions can be used to tell stories. This is the story of a car journey Bob took last week:
d of t is equal to fifty t when t is greater than or equal to zero and less than or equal to three; one hundred fifty when t is greater than three and less than three point five; twenty-five t plus sixty-two point five when t is greater than or equal to three point five and less than seven point five; two hundred fifty when t is greater than or equal to seven point five and less than eight; and negative sixty-two point five t when t is greater than or equal to eight and less than or equal to twelve
Let’s see how many of these review questions we can answer about Bob’s journey before we go:
How long did the journey last? 12 hrs
What’s the farthest distance Bob drove from his home? 250 mi
How long did it take him to get there? 7.5 hrs
What was Bob doing from 3-3.5 hrs? not moving
When did Bob turn toward home? at the 8 hr mark
When was Bob driving the fastest? from hr 8 to hr 12 he traveled 62.5 mph
What was Bob’s average speed from 0 to 8 hrs? 31.25 mph
Thanks for watching, and happy studying!
## Piecewise Function Practice Questions
Question #1:
The following piecewise function shows the speed of a car as a function of time. Daniel says that the flat top portion of the graph shows that the car came to a stop. Is Daniel correct?
Daniel is correct.
Daniel is incorrect.
Each “piece” of the graph shows the car traveling at a different speed. The first piece of the graph shows the car gradually increasing speed as time elapses. The second piece of the graph shows the car maintaining a steady speed as time elapses. The third piece of the graph shows the car steadily reducing speed.
Even though the flat piece of the graph seems to show the car stopping, the speed is maintained as time is passing.
Question #2:
What kind of piecewise function is graphed below?
Linear
Step
Absolute Value
An absolute value function will form a V shape. This type of function is essentially two “pieces” of linear functions.
Question #3:
Define the following piecewise function.
$$h(x)=2\text{ if } x \leq 3$$
$$\hspace{0.25in}x\text{ if } x>-1$$
$$h(x)=2\text{ if } x \leq 1$$
$$\hspace{0.25in}x\text{ if } x>1$$
$$h(x)=2\text{ if } x \leq 9$$
$$\hspace{0.25in}x\text{ if } x>3$$
$$h(x)=1\text{ if } x \leq 1$$
$$\hspace{0.25in}x\text{ if } x>-3$$
The graph shows that when $$x$$ is less than or equal to $$1$$, then $$h(x)$$ is equal to $$2$$. For example, when $$x$$ is equal to $$0$$, $$h(x)$$ equals $$2$$. This is defined as $$h(x)=2\text{ if } x>1$$.
When x is greater than $$1$$, then $$h(x)$$ is equal to $$x$$. For example, when $$x$$ is equal to $$4$$, $$h(x)$$ is also equal to $$4$$. This is defined as $$h(x)=x\text{ if } x>1$$.
These two pieces define the piecewise function for the graph:
$$h(x)=2\text{ if } x \leq 1$$
$$x\text{ if } x>1$$
Question #4:
The graph below shows the relationship between time in minutes ($$x$$-axis) and water in gallons ($$y$$-axis). The graph shows the amount of water that was used (in gallons) when George gave his new puppy a bath.
When did George turn the water faucet off? How much water was in the bath when he bathed the puppy?
George turned the water off at 30 minutes. There were 24 gallons of water in the tub when he bathed the puppy.
George turned the water off at 12 minutes. There were 18 gallons of water in the tub when he bathed the puppy.
George turned the water off at 6 minutes. There were 24 gallons of water in the tub when he bathed the puppy.
George turned the water off at 24 minutes. There were 6 gallons of water in the tub when he bathed the puppy.
The amount of water in gallons stops increasing when George turns the water faucet off. The $$x$$-axis shows us that when $$6$$ minutes have passed, the amount of water stops increasing. This means that George turns off the water after $$6$$ minutes, and begins the puppy bath. The $$y$$-axis indicates that at this time the water is currently at $$24$$ gallons.
Question #5:
Define the following piecewise function.
$$f(x)=x+2\hspace{0.25in}5 \leq x \leq -1$$
$$\hspace{0.38in}-x\hspace{0.25in}1 \lt x \leq 3$$
$$f(x)=x+5\hspace{0.25in} 7 \leq x \leq 1$$
$$\hspace{0.38in}-x\hspace{0.25in} 1 \lt x \leq 3$$
$$f(x)=x+1\hspace{0.25in} -4 \leq x \leq 1$$
$$\hspace{0.40in}-x\hspace{0.25in} 1 \lt x \leq 3$$
$$f(x)=x+3\hspace{0.25in} -5 \leq x \leq 1$$
$$\hspace{0.40in}-x\hspace{0.25in} -4 \lt x \leq 2$$
The graphed line on the left side represents $$x$$-values that are greater than or equal to $$-4$$, and less than or equal to $$1$$. This region can be described as $$-4\leq x\leq 1$$. In this region, the $$f(x)$$-value, or $$y$$-value, is always one more than the $$x$$-value. For example, when $$x$$ is $$1$$, $$y$$ is $$2$$. When $$x$$ is $$-3$$, $$y$$ is $$-2$$. This means that in the region of $$-4\leq x\leq 1$$, $$y$$ is always one more than $$x$$. This is defined as $$f(x)=x+1\hspace{0.25in} -4\leq x\leq 1$$.
The graphed line on the right side represents $$x$$-values that are greater than $$1$$, and less than or equal to $$3$$. This region can be described as $$1 \lt x \leq 3$$. In this region, the $$f(x)$$-value, or the $$y$$-value, is always the $$x$$-value negated. For example, when $$x$$ is $$2$$, $$y$$ is $$-2$$. When $$x$$ is $$3$$, $$y$$ is $$-3$$. This means that in the region of $$1 \lt x \leq 3$$, $$y$$ is always the negative value of $$x$$. This is defined as $$f(x)=-x\hspace{0.25in} 1 \lt x \leq 3$$.
When both “pieces” are put together, they define the piecewise function as
$$f(x)=x+1\hspace{0.25in} -4 \leq x \leq 1$$
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http://www.ck12.org/tebook/Texas-Instruments-Geometry-Teacher%2527s-Edition/r1/section/8.1/ | 1,474,766,056,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738659680.65/warc/CC-MAIN-20160924173739-00074-ip-10-143-35-109.ec2.internal.warc.gz | 401,807,667 | 34,735 | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 8.1: Constructing Similar Triangles
Difficulty Level: At Grade Created by: CK-12
This activity is intended to supplement Geometry, Chapter 7, Lesson 4.
ID: 8179
Time required: 30 minutes
## Activity Overview
In this activity, students will investigate three different methods of constructing similar triangles. They will use the Dilation tool with the dilation point inside and outside of the triangle to investigate different relationships. Also, students will use the Parallel tool to construct two similar triangles from one triangle.
Topic: Ratio, Proportion & Similarity
• Use inductive reasoning to classify each set of conditions as necessary and/or sufficient for similarity:
a) the side lengths of one triangle are equal to the corresponding side lengths of another triangle.
b) two triangles of one triangle are congruent to two angles of another triangle.
Teacher Preparation
• This activity is designed to be used in a high school geometry classroom.
• Before starting this activity, students should be familiar with the term dilation.
• This activity is intended to be mainly teacher-led, with breaks for individual student work. Use the following pages to present the material to the class and encourage discussion. Students will follow along using their calculators.
Associated Materials
Students need to press APPS and select CabriJr to start the application. When they open a new document, they need to make sure that the axes are hidden. If the axes are displayed, press GRAPH > Hide/Show > Axes.
Explain to students that similar triangles are those that have the same shape but not necessarily the same size.
Congruent triangles are a special type of similar triangle where corresponding sides are congruent. In similar triangles, corresponding angles are congruent but corresponding sides are proportional. In this activity, students will look at three methods of constructing similar triangles and will test these properties using dilations or stretches.
In order to examine all of the sides and angles, students should work in groups of three. Have one student (Student A on the worksheet) in each group construct the first triangle and save it as “SIMTRI.” This is saved in the TI-84 Plus family as an APPVAR. Student A needs to transfer the APPVAR to the other two students in the group.
## Problem 1 – Similar Triangles using Dilation
Student A will use the Triangle tool to construct a triangle of any shape or size. Then they need to use the Alph-Num tool to label the vertices P\begin{align*}P\end{align*}, Q\begin{align*}Q\end{align*}, and R\begin{align*}R\end{align*}.
Student A should measure angle P\begin{align*}P\end{align*} and side PQ\begin{align*}PQ\end{align*}, Student B should measure angle Q\begin{align*}Q\end{align*} and side QR\begin{align*}QR\end{align*}, Student C should measure angle R\begin{align*}R\end{align*} and side PR\begin{align*}PR\end{align*}.
Note: To increase the number of digits for the length of the side, hover the cursor over the measurement and press +.
With the Point tool, students on their own calculators will construct a point C\begin{align*}C\end{align*} in the center of the triangle. They will then use the Alph-Num tool to place the number 2 at the top of the screen.
Explain to students that the point C\begin{align*}C\end{align*} will be the center of the dilation and the number 2\begin{align*}2\end{align*} will be the scale factor.
With the Dilation tool selected, students need to move the cursor to point C\begin{align*}C\end{align*}, the center of the dilation, and press ENTER. They should notice that the shape of the cursor changes, which is to indicate that everything will be stretched away from this point. Then students move the cursor to the perimeter of the triangle and press ENTER. Finally, move to the scale factor, 2, and press ENTER.
Students will see that a new, larger triangle will appear outside of PQR\begin{align*}\triangle{PQR}\end{align*}. Have a prelimenary discussion with students about whether they think the new triangle is similar to PQR\begin{align*}\triangle{PQR}\end{align*} and how they can confrim their hypothesis.
Instruct students to label this triangle as XYZ\begin{align*}XYZ\end{align*} so that X\begin{align*}X\end{align*} corresponds to P\begin{align*}P\end{align*}, Y\begin{align*}Y\end{align*} to Q\begin{align*}Q\end{align*} and Z\begin{align*}Z\end{align*} to R\begin{align*}R\end{align*}. Each person in the group should select and measure their appropriate angle and side in the new triangle. Students should answer Questions 1 and 2 on the worksheet comparing the two angles and two sides.
Explain to students that XY=2 PQ\begin{align*}XY = 2\ PQ\end{align*} indicates that the sides have the ratio 2:1. If all three sides display the same result, then the sides are said to be proportional.
To answer Questions 3 and 4 on the worksheet, students will observe the changes in the triangles as they drag a point in the original triangle and the point of dilation.
## Problem 2 – Different Scale Factors
Students will continue using the same file. To change the scale factor, students must select the Alph-Num tool, move the cursor to the scale factor of 2 and press ENTER. Then they should delete 2 by pressing DEL, press ALPHA to change the character to a number, and enter 3. Students can now answer Question 5 on the worksheet.
Now students will investigate another scale factor by changing it from 3 to 0.5. They should then answer Questions 6 and 7 on the worksheet, summarizing their observations of the triangles and their measurements.
For the last investigation in this problem, students will look at the effect of a dilation when the center point C\begin{align*}C\end{align*} is outside of the pre-image triangle and a negative scale factor is used. Students will need to move PQR\begin{align*}\triangle{PQR}\end{align*} by moving the cursor to one side and when all sides of the triangle flash, press ALPHA and then use the arrow keys.
## Problem 3 – Similar Triangles with a Parallel Line
Finally, students will look at a completely different method of constructing similar triangles. All students will need to open a new file. It is the teacher’s decision to have them save the file. Student A should construct a triangle PQR\begin{align*}PQR\end{align*} and transfer it to the others in their group. They will all measure their same side and angle as before.
All students need to construct a point on PQ¯¯¯¯¯¯¯¯\begin{align*}\overline{PQ}\end{align*} using the Point on tool and label it S\begin{align*}S\end{align*} using the Alph-Num tool. Using this point, they construct a line that is parallel to QR¯¯¯¯¯¯¯¯\begin{align*}\overline{QR}\end{align*} by selecting the Parallel tool, and then choosing the point and side.
With the Intersection tool, students need to find the intersection of side PR\begin{align*}PR\end{align*} and the parallel line, and then use the Alph-Num tool to label it as T\begin{align*}T\end{align*}. The Hide/Show > Object tool enables students to hide the parallel line, and then construct segment ST\begin{align*}ST\end{align*} with the Segment tool. If students can prove that all of the angles are congruent, they have completed their first step in proving that the two triangles are similar.
With the Measure > D. & Length tool, students need to measure segments of the corresponding sides. In the screen to the right all three pairs of sides have been measured.
To complete the proof, students will need to confirm that all of the sides are proportional. Using the Calculate tool, they can select a side measurement, press the operation key (÷)\begin{align*}( \div )\end{align*} and then select the corresponding side measurement.
If all three ratios are equivalent, then the sides are proportional and the two triangles are similar.
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Subjects: | 1,945 | 8,115 | {"found_math": true, "script_math_tex": 33, "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-2016-40 | longest | en | 0.890741 |
http://mathematica.stackexchange.com/questions/56975/replacing-a-sum-of-expressions | 1,467,204,913,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783397744.64/warc/CC-MAIN-20160624154957-00116-ip-10-164-35-72.ec2.internal.warc.gz | 197,800,251 | 20,109 | # Replacing a sum of expressions
I have an equation which Mathematica solves but gives a huge output. However I notice in this output that the same expression occurs many times. Namely:
$E=864 a^6 - 432 a^4 b^2 + 54 a^2 b^4 - b^6 + 54 b^4 x^2$
where $a,b,c$ are constants. However this expression can appear in the form $-E$, $E^2$, $\sqrt{E}$, $4E$, ...
I'm struggling to tell Mathematica to make the replacement
$(864 a^6 - 432 a^4 b^2 + 54 a^2 b^4 - b^6 + 54 b^4 x^2) \space \to \space E$
wherever it appears.
All the ways I've tried (I'm not that handy with Mathematica I note) don't change output. Any ideas?
-
Please provide a short example of an actual expression that your are truing to simplify. Otherwise there is no way for us to tell if our solution will work in your case. – John McGee Aug 8 '14 at 17:07
@JohnMcGee What do you mean? It's an incredibly long solution to an equation, and the expression in my question, $E$, appears many, many times. But as I say, it could appear as $-E$, $E^{15}$, $E^{1/3}$... – Phibert Aug 8 '14 at 17:09
@JohnMcGee I mean I could provide a small portion of the output I need simplifying but I can't even follow the brackets and square roots of Mathematica because they're all so nested. – Phibert Aug 8 '14 at 17:16
Closely related: 3463, 3822 – Michael E2 Aug 8 '14 at 17:58
@user13223423 You can just copy and paste the expression to your question and we can just copy and paste it back to our notebooks. Use Ctrl+Shift+C and Ctrl+V to preserve space. – seismatica Aug 8 '14 at 18:25
You must carefully define all desired / expected replacement cases. For example:
poly = 864 a^6 - 432 a^4 b^2 + 54 a^2 b^4 - b^6 + 54 b^4 x^2;
Clear[rep]
rep[p_] := p /. p :> p[[0]]@B
rep[Power[p_, n_]] := p /. p :> Power[B, n]
rep[Times[n_, p_]] := p /. p :> n B
rep[poly]
B
rep[Log@poly]
Log[B]
rep[1/poly]
1 / B
rep[poly*2]
2 B
list = {poly, Log@poly, poly^2, 1/poly};
rep /@ list
{B, Log[B], B^2, 1/B}
I feel there should be easier solutions using the Hold..., Unevaluated, Inactive family. However, I can't find them.
-
There's a mistake in your code. For example rep[2x] returns 2B. – Chip Hurst Aug 8 '14 at 20:24
@ChipHurst Thanks for your attention. I wouldn't call it a mistake. As a matter of fact, 2 x should give 2 B. Your spaceless 2x doesn't exist in MMA (except as a String). – eldo Aug 8 '14 at 20:33
@ChipHurst BTW (I just noticed): Thanks for solving my InverseFunction-problem – eldo Aug 8 '14 at 22:27
This doesn't work for a lot expressions containing poly. For example, rep[1 + x^2 + (1/poly)^3] returns B. rep[p_] := p /. p :> p[[0]] @ B is what is causing this. It is too indiscriminate. – m_goldberg Aug 9 '14 at 1:49
One (potentially incorrect, but quick and dirty) way to do it is to instead make the substitution
864 a^6 -> E - (−432 a^4 b^2 + 54 a^2 b^4 − b^6 + 54 b^4x^2)
Also, do you really mean E? E in Mathematica is the symbol for the base of the natural logarithm...
-
This is what I would do, but only in the case where the entirety of the expression is expected to boil down to some function of $E$ and not include any leftover $a,b,c,d,e$. – Kellen Myers Aug 9 '14 at 0:25
poly = 864 a^6 − 432 a^4 b^2 + 54 a^2 b^4 − b^6 + 54 b^4x^2; | 1,075 | 3,243 | {"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} | 3.390625 | 3 | CC-MAIN-2016-26 | latest | en | 0.84437 |
https://metanumbers.com/133710 | 1,632,538,522,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057589.14/warc/CC-MAIN-20210925021713-20210925051713-00460.warc.gz | 446,928,806 | 7,375 | # 133710 (number)
133,710 (one hundred thirty-three thousand seven hundred ten) is an even six-digits composite number following 133709 and preceding 133711. In scientific notation, it is written as 1.3371 × 105. The sum of its digits is 15. It has a total of 4 prime factors and 16 positive divisors. There are 35,648 positive integers (up to 133710) that are relatively prime to 133710.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 6
• Sum of Digits 15
• Digital Root 6
## Name
Short name 133 thousand 710 one hundred thirty-three thousand seven hundred ten
## Notation
Scientific notation 1.3371 × 105 133.71 × 103
## Prime Factorization of 133710
Prime Factorization 2 × 3 × 5 × 4457
Composite number
Distinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 133710 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 133,710 is 2 × 3 × 5 × 4457. Since it has a total of 4 prime factors, 133,710 is a composite number.
## Divisors of 133710
16 divisors
Even divisors 8 8 4 4
Total Divisors Sum of Divisors Aliquot Sum τ(n) 16 Total number of the positive divisors of n σ(n) 320976 Sum of all the positive divisors of n s(n) 187266 Sum of the proper positive divisors of n A(n) 20061 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 365.664 Returns the nth root of the product of n divisors H(n) 6.66517 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 133,710 can be divided by 16 positive divisors (out of which 8 are even, and 8 are odd). The sum of these divisors (counting 133,710) is 320,976, the average is 20,061.
## Other Arithmetic Functions (n = 133710)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 35648 Total number of positive integers not greater than n that are coprime to n λ(n) 4456 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 12441 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 35,648 positive integers (less than 133,710) that are coprime with 133,710. And there are approximately 12,441 prime numbers less than or equal to 133,710.
## Divisibility of 133710
m n mod m 2 3 4 5 6 7 8 9 0 0 2 0 0 3 6 6
The number 133,710 is divisible by 2, 3, 5 and 6.
• Arithmetic
• Abundant
• Polite
• Square Free
## Base conversion (133710)
Base System Value
2 Binary 100000101001001110
3 Ternary 20210102020
4 Quaternary 200221032
5 Quinary 13234320
6 Senary 2511010
8 Octal 405116
10 Decimal 133710
12 Duodecimal 65466
20 Vigesimal ge5a
36 Base36 2v66
## Basic calculations (n = 133710)
### Multiplication
n×y
n×2 267420 401130 534840 668550
### Division
n÷y
n÷2 66855 44570 33427.5 26742
### Exponentiation
ny
n2 17878364100 2390516063811000 319635902892168810000 42738516575711891585100000
### Nth Root
y√n
2√n 365.664 51.1354 19.1223 10.5982
## 133710 as geometric shapes
### Circle
Diameter 267420 840125 5.61665e+10
### Sphere
Volume 1.00134e+16 2.24666e+11 840125
### Square
Length = n
Perimeter 534840 1.78784e+10 189094
### Cube
Length = n
Surface area 1.0727e+11 2.39052e+15 231593
### Equilateral Triangle
Length = n
Perimeter 401130 7.74156e+09 115796
### Triangular Pyramid
Length = n
Surface area 3.09662e+10 2.81725e+14 109174
## Cryptographic Hash Functions
md5 a8279de4bb966e2c236a566ee58464a1 469efb4109677d2421b22e334813de627e79a98f 09922ab6a0249e594996c4a9eb1cfe372c9915ae2d6aef3b53159eeda105a861 c25879e828873c4734c5e70b506232df00efd10f140993563a92777a24ed0c539129456ef1ec9ebe66e1a3cab0645d2182f29bc8cfddcacc5b15851dc72c2276 6d7e60e187a4a9541d46e4b371ac6e83fba815b4 | 1,429 | 4,129 | {"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} | 3.625 | 4 | CC-MAIN-2021-39 | latest | en | 0.819997 |
https://www.jiskha.com/questions/355456/Rashid-is-helping-to-plan-a-cookout-He-determines-that-making-one-making-one-hot | 1,568,675,675,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572964.47/warc/CC-MAIN-20190916220318-20190917002318-00096.warc.gz | 886,395,714 | 5,579 | # algebra
Rashid is helping to plan a cookout. He determines that making one making one hot dogs costs 55 cents and making a hamburger costs 90 cents. The cost must be no more than \$50, and he knows that he must at least 60 hot dogs and hamburgers.
Which system of linear inequaltites describes the conditions in the problem?
a) 0.55x + 0.9y >= 50, x + y >= 60
b) 0.55x + 0.9y <= 50, x + y >= 60
c) 0.55x + 0.9y >= 50, x + y <= 60
d) 0.55x + 0.9y <= 50, x + y <= 60
and the answer i picked was c
thank you !!!!!!!!!!!!!!!!!!!!
1. 👍 0
2. 👎 0
3. 👁 82
1. Since it needs to be "at least 60" x + y ≥ 60.
"no more than \$50" = less than or equal to = 0.55x + 0.9y ≤ 50
1. 👍 0
2. 👎 0
posted by PsyDAG
## Similar Questions
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More Similar Questions | 1,037 | 3,528 | {"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} | 3.6875 | 4 | CC-MAIN-2019-39 | latest | en | 0.944111 |
https://math.stackexchange.com/questions/4508206/difference-bewteen-random-process-and-functional-random-variable | 1,723,278,434,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640790444.57/warc/CC-MAIN-20240810061945-20240810091945-00116.warc.gz | 299,656,657 | 38,164 | Difference bewteen random process and functional random variable.
I would like to ask some basic question but I cannot find the appropriate answer on it. What is the basic diference between funcional random variable and random process?
Random process $$\mathbb{X}$$ is defined (for example) as sequence of random variables defined on the same probability sapce $$\left(\Omega, S, P\right)$$, i. e. $$\mathbb{X}=\left\{X_t\right\}_{t\in T}$$, where $$X_t$$ is random variable for each $$t\in T$$. More precisely, we can consider $$\mathbb{X}$$ as function from $$\mathbb{X}(\omega,t):\Omega\times T\to \mathbb{R}$$. When we fix $$\omega$$ we have trajektory of random process. It means that if we consider time $$T=\mathbb{R}$$, we have for fixed $$\omega$$ real function on $$\mathbb{R}$$. On the other hand when we fix $$t\in \mathbb{R}$$ we have ("ordinary") random variable.
Informally in some literature, we can find that functional random "variable" is as random curve (or as random element of $$L^2$$ space).
I still not quite well understand the difference between random process (which is also random curve, in my opinion) and functional random variable.
Probably I little bit understand the difference when I have some realizations. Usualy we have "only one" realization of random process (moreover discrete, but now we are considering continuous realization), it means that we have only one trajectory of random process. Than we try to make for example some predictions and etc.
But in the functional data analysis, we (I think) have more realizations, something like we have more curves with same domain. I consider it as more realizations of some random process.
Can somebody please give me some better and more detailed explanation what is the difference between random process and functional random "variable"?
Any help will be appreciated. Thank you very much.
• In short: a random process is a family (that is usually indexed by a discrete or continuous time) of functional random variables $X_n$ or $X_t$. Commented Aug 8, 2022 at 9:13
• Never heard of a functional random variable. Commented Aug 8, 2022 at 9:15
• Indeed a random process is a time-indexed collection of random variables. When you are talking about curves or other objects that can be random, I wonder if you mean a "random element" (which some texts unfortunately also call a "random variable"). Given a probability space $(\Omega,\mathcal{F}, P)$, a random variable is a measurable function $X:\Omega\rightarrow\mathbb{R}$ while a random element is a measurable function $X:\Omega\rightarrow V$ where $V$ is some nonnempty set and $\mathcal{G}$ is some sigma algebra on $V$. So $V$ can be a set of curves (but it needs a $\mathcal{G}$). Commented Aug 8, 2022 at 11:50
• @ Michael thank you. I understand all what you wrote. But I have troubles to understand what is the main diference between functional data (I think this concept is quite cammon) and realization of stochastic/random process with continuous time. Could you explain me the difference, please? (if you can) Commented Aug 26, 2022 at 7:44
A random process $$\mathbb{X}=(X_t)_{t\in\mathbb{T}}$$ is a family of random variables $$X_t:\Omega\rightarrow\mathbb{R}$$ indexed by an ordered set $$\mathbb{T}$$ e.g. $$\mathbb{R}$$, but more commonly a subset of the nonnegative axis. | 841 | 3,336 | {"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": 17, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2024-33 | latest | en | 0.899068 |
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Use this project to give students real world applications for area and perimeter. In the project, students calculate the areas and perimeters of rooms in an apartment, including composite irregular polygons, Follow up questions push students to think about applications of area and perimeter. In
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showing 1-24 of 115 results
Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. | 2,126 | 8,806 | {"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-2018-47 | latest | en | 0.898965 |
https://questions.examside.com/past-years/jee/question/plet-veca-vecb-vecc-be-three-coplanar-concu-jee-main-mathematics-trigonometric-functions-and-equations-szyxdjxkbzzh9nrg | 1,720,842,005,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514484.89/warc/CC-MAIN-20240713020211-20240713050211-00677.warc.gz | 387,882,857 | 52,076 | 1
JEE Main 2022 (Online) 29th July Evening Shift
+4
-1
Let $$\vec{a}, \vec{b}, \vec{c}$$ be three coplanar concurrent vectors such that angles between any two of them is same. If the product of their magnitudes is 14 and $$(\vec{a} \times \vec{b}) \cdot(\vec{b} \times \vec{c})+(\vec{b} \times \vec{c}) \cdot(\vec{c} \times \vec{a})+(\vec{c} \times \vec{a}) \cdot(\vec{a} \times \vec{b})=168$$, then $$|\vec{a}|+|\vec{b}|+|\vec{c}|$$ is equal to :
A
10
B
14
C
16
D
18
2
JEE Main 2022 (Online) 29th July Morning Shift
+4
-1
Out of Syllabus
Let $$\overrightarrow{\mathrm{a}}=3 \hat{i}+\hat{j}$$ and $$\overrightarrow{\mathrm{b}}=\hat{i}+2 \hat{j}+\hat{k}$$. Let $$\overrightarrow{\mathrm{c}}$$ be a vector satisfying $$\overrightarrow{\mathrm{a}} \times(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}})=\overrightarrow{\mathrm{b}}+\lambda \overrightarrow{\mathrm{c}}$$. If $$\overrightarrow{\mathrm{b}}$$ and $$\overrightarrow{\mathrm{c}}$$ are non-parallel, then the value of $$\lambda$$ is :
A
$$-$$5
B
5
C
1
D
$$-$$1
3
JEE Main 2022 (Online) 29th July Morning Shift
+4
-1
Let $$\hat{a}$$ and $$\hat{b}$$ be two unit vectors such that the angle between them is $$\frac{\pi}{4}$$. If $$\theta$$ is the angle between the vectors $$(\hat{a}+\hat{b})$$ and $$(\hat{a}+2 \hat{b}+2(\hat{a} \times \hat{b}))$$, then the value of $$164 \,\cos ^{2} \theta$$ is equal to :
A
$$90+27 \sqrt{2}$$
B
$$45+18 \sqrt{2}$$
C
$$90+3 \sqrt{2}$$
D
$$54+90 \sqrt{2}$$
4
JEE Main 2022 (Online) 28th July Evening Shift
+4
-1
Let S be the set of all a $$\in R$$ for which the angle between the vectors $$\vec{u}=a\left(\log _{e} b\right) \hat{i}-6 \hat{j}+3 \hat{k}$$ and $$\vec{v}=\left(\log _{e} b\right) \hat{i}+2 \hat{j}+2 a\left(\log _{e} b\right) \hat{k}$$, $$(b>1)$$ is acute. Then S is equal to :
A
$$\left(-\infty,-\frac{4}{3}\right)$$
B
$$\Phi$$
C
$$\left(-\frac{4}{3}, 0\right)$$
D
$$\left(\frac{12}{7}, \infty\right)$$
EXAM MAP
Medical
NEET | 800 | 1,950 | {"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} | 2.90625 | 3 | CC-MAIN-2024-30 | latest | en | 0.590412 |
http://davidegironi.blogspot.com/2017/08/mq-gas-sensor-correlation-coefficients.html | 1,637,983,113,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358078.2/warc/CC-MAIN-20211127013935-20211127043935-00376.warc.gz | 17,616,906 | 22,385 | ## Saturday, August 5, 2017
### MQ gas sensor correlation coefficients
I've have talk about MQ sensors here:
http://davidegironi.blogspot.it/2017/05/mq-gas-sensor-correlation-function.html
In the post linked above I've discuss a method to correlate the sensor resistance to a gas ppm using the datasheet curve named "sensitivity characteristics of the MQ-135" and the formula:
ppm=a*(Rs/Ro)^b.
Someone ask me about sensor and correlation coefficients of various gasses, so here I will post some correlation coefficient for the a few MQ sensor and gasses.
sensor gas a b min Rs/Ro max Rs/Ro MQ2 LPG 591.283 -2.076502 0.256166 1.68543 MQ3 alcohol 0.3923202 -1.493249 0.1143375 2.497754 MQ4 CH4 1041.333 -2.729007 0.4365277 1.830508 MQ5 CH4 217.4972 -2.422111 0.2058715 1.035233 MQ6 LPG 940.2178 -2.521573 0.3915661 1.847477 MQ7 H2 64.86522 -1.405261 0.05323301 1.203488 MQ8 H2 1079.683 -0.6416874 0.03115283 13.84067 MQ135 CO2 110.3794 -2.721706 0.8038119 2.416431
Those values have been found using the method proposed in the above post, and the figures posted below.
In order to use the correlation function, you now need to compute the Ro value. You can find this value reading the sensor resistance at a know amount of ppm, compy in the gases ppm limits of the datasheet, and then use this formula:
Ro=Rs*(a/ppm)^(1/b).
Those are the
- collected points: https://bitbucket.org/snippets/davidegironi/z9AMEr/mq-gas-sensor-correlation-coefficients
that has been processes using the following
- R script: https://gist.github.com/davidegironi/b7be6b7cace6b475dd42c48c3e62fcf4
Notes
• read risk disclaimer
• excuse my bad english
1. This comment has been removed by a blog administrator.
2. Hello Sir. I have been working on this for a course project. I have finished extracting data from the datasheet but I m somehow not able to run the R- Program. It shows up a lot of errors of which the first one is "There is no package called data.table". I do not hold any knowledge in R programming. So can you please help me go past this step. Thank you
1. Hello, for the error you encouter, you have to set your datapoints in pointsdata array. You could try datacamp.com light to run the R script.
3. Hello sir! My question is how to find R/Ro by the analog output we receive, what is the correlation between the two? Is it same for all sensors?
1. Hello, you should not find R/Ro. Take a look here: http://davidegironi.blogspot.it/2017/05/mq-gas-sensor-correlation-function.html
4. How can I find the CO correlation coefficients for MQ-9 sensor? I tried doing your method, by taking the data points from the datasheet, but I got around:
Estimated a
580.9079
Estimated b
-2.288135
Using the coefficients to calculate ppm, my values were absurd! Can you help me?
1. Hello, here i explain the method i've use to search values: http://davidegironi.blogspot.com/2017/05/mq-gas-sensor-correlation-function.html
5. How does one go about a linear or systematic way of identification of gas, given a array of mq sensors? My math sucks, it's multivariate analysis.
1. You can't with this type of sensor identify a specific gas, they react to a bounch of gases. | 909 | 3,158 | {"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 | 0.815839 |
https://sites.google.com/mynrsd.com/mathspecialist/home | 1,534,694,267,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221215222.74/warc/CC-MAIN-20180819145405-20180819165405-00379.warc.gz | 761,030,266 | 11,145 | # Math Rocks!
My name is Karen Walsh Fortin and I love math!
I have visited with all the students in grades K-4.
I played a game with the students, called "Find the Math." I had them recall something they did this summer which was fun and involved no math. Then I asked then a multitude of questions. For instance, if they went to the beach I might ask, "How long was the car ride? How many miles did you drive? Which route numbers did you take? What was the average speed? What was the temperature of the water? What was the temperature of the air? What was the difference in these two temperatures? How far did you go out? What was the depth of the water?" etc. After leading the questioning for a few students, I then asked the students to take on the role of questioner for another summer vacation example. Following this activity, I asked, "What did you learn from this game?" "Math is everywhere!" was the most popular answer.
I work with students in grades K-8. I want all my students to know they are capable of reaching any goal they set. It takes effort, hard work and grit- but They Can Do It!! If they tell me they can't do something in math- I remind them that maybe they can't do it YET, but with time and effort (and sometimes help) they can. Mistakes build our brain!
## Class News/What's Happening/Announcements
UNIFLOW_SAWCanon6565-TR_0020_001.pdf
927MATHEMATICAL MINDSETS b.pptx | 335 | 1,403 | {"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-2018-34 | latest | en | 0.985932 |
http://essayparlour.com/paperhelp/economics/in-each-of-the-three-games-shown-below-let-p-be-the-probabi/ | 1,484,876,183,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280763.38/warc/CC-MAIN-20170116095120-00008-ip-10-171-10-70.ec2.internal.warc.gz | 93,580,499 | 6,860 | +1 (347) 474-1028 info@essayparlour.com
## In each of the three games shown below, let p be the probabi
Paper help Economics In each of the three games shown below, let p be the probabi
# In each of the three games shown below, let p be the probabi
In each of the three games shown below, let p be the probability that player 1 plays coop… Show more Problem 1 In each of the three games shown below, let p be the probability that player 1 plays cooperates (and 1- p the probability that player 1 defects), and let q be the probability that Player 2 plays cooperates (and 1- q the probability that player 2 defects). Prisoner’s Dilemma Player 2 Player 2 Player 2 Player 1 cooperate defect Player 1 cooperate 70,70 10,80 Player 1 defect 80,10 40,40 Stag Hunt Player 2 Player 2 Player 2 Player 1 cooperate defect Player 1 cooperate 70,70 5,40 Player 1 defect 40,5 40,40 Chicken Player 2 Player 2 Player 2 Player 1 cooperate defect Player 1 cooperate 70,70 50,80 Player 1 defect 80,50 40,40 1. For each game, draw a graph with player 1’s best response function (choice of p as a function of q), and player 2’s best response function (choice of q as a function of p), with p on the horizontal axis and q on the vertical axis. 2. Using this graphs, find all the Nash equilibriums for the game, both pure and mixed strategy Nash equilibriums (if any). Label these equilibriums on the corresponding graph. 3. In those games that have multiple pure strategy Nash equilibriums, how do the expected payoffs from playing the mixed strategy Nash equilibrium compare with the payoffs from playing the pure strategy Nash equilibriums? Which type of strategy (mixed or pure) would players prefer to play in these games? Problem 2 Two people are involved in a dispute. Player 1 does not know whether player 2 is strong or weak; she assigns probability α to player 2 being strong. Player 2 is fully informed. Each player can either fight or yield. Each player obtains a payoff of 0 is she yields (regardless of the other person’s action) and a payoff of 1 if she fights and her opponent yields. If both players fight, then their payoffs are (-1; 1) if player 2 is strong and (1;-1) if player 2 is weak. The Bayesian game is the following, depending on the type of player 2: Y F Y F F Y 0, 0 0, 1 Y 0, 0 0, 1 0, 1 F 1, 0 -1, 1 F 1, 0 1, -1 1, -1 Player 2 is strong (α) Player 2 is strong (α) Player 2 is strong (α) Player 2 is weak (1-α) Player 2 is weak (1-α) Player 2 is weak (1-α) Player 2 is strong (α) After writing all the strategies and payoffs in the same matrix, find the Bayesian Nash equilibriums, depending on the value of α (α ≤ 1/2 or α ≥1/2). • Show less
Ready to try a high quality writing service? | 788 | 2,723 | {"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} | 3.296875 | 3 | CC-MAIN-2017-04 | longest | en | 0.904587 |
https://www.vtupulse.com/c-programs/c-program-to-read-diameter-and-compute-area-perimeter/ | 1,656,681,643,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103941562.52/warc/CC-MAIN-20220701125452-20220701155452-00641.warc.gz | 1,118,465,792 | 23,421 | # C program to read diameter and compute Area Perimeter
Write a C program to read the diameter of the circle and compute the perimeter and area of the circle.
### Problem Definition:
In this program, we will read the diameter of a circle. Next, we find the area and perimeter of a circle.
### Source code of C program to read diameter and compute Area Perimeter of a Circle
```#include<stdio.h>
int main()
{
int diameter;
float perimeter,area;
printf("Enter the diameter of the circle: ");
scanf("%d",&diameter);
perimeter=diameter*3.14; //calculates perimeter
area=(3.14*diameter*diameter)/4;//calculates area
printf("\nThe area of a circle is %f",area);
printf("\nThe perimeter of a circle is %f",perimeter);
}
```
### Output:
Case 1:
Enter the diameter of the circle: 5
The area of a circle is 19.625000
The perimeter of a circle is 15.700000
Case 2:
Enter the diameter of the circle: 3
The area of a circle is 7.065000
The perimeter of a circle is 9.420000
### Program Explanation:
1. First, create 3 variables such as diameter, area, and perimeter of type float.
2. Next, ask the user to enter the value of diameter.
3. Calculate the area of a rectangle using formula – > area = diameter * 3.142
4. Calculate the perimeter of a rectangle using formula – > perimeter= (3.14 * diameter * diameter)/4
5. Finally, display area and rectangle using printf statement.
### Summary:
This tutorial discusses how to write a program to read the diameter of the circle and compute the perimeter and area of the circle. If you like the tutorial share it with your friends. Like the Facebook page for regular updates and YouTube channel for video tutorials. | 409 | 1,679 | {"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-2022-27 | longest | en | 0.712106 |
https://au.mathworks.com/matlabcentral/cody/problems/43706-sum-the-rows/solutions/1739963 | 1,606,996,723,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141727627.70/warc/CC-MAIN-20201203094119-20201203124119-00380.warc.gz | 188,497,221 | 17,023 | Cody
# Problem 43706. Sum the rows
Solution 1739963
Submitted on 1 Mar 2019 by Tom Holz
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 Pass
x = magic(4); y_correct = [34;34;34;34] assert(isequal(sum_the_rows(x),y_correct))
y_correct = 34 34 34 34
2 Pass
x = pascal(7); y_correct =[7;28;84;210;462;924;1716]; assert(isequal(sum_the_rows(x),y_correct))
3 Pass
x = [5 -1 7 5;3 10 23 5;0 -3 7 5] y_correct =[16;41;9]; assert(isequal(sum_the_rows(x),y_correct))
x = 5 -1 7 5 3 10 23 5 0 -3 7 5
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 256 | 743 | {"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-2020-50 | latest | en | 0.566018 |
https://nrich.maths.org/public/leg.php?code=8&cl=3&cldcmpid=5911 | 1,511,534,719,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934808254.76/warc/CC-MAIN-20171124142303-20171124162303-00295.warc.gz | 659,832,418 | 6,214 | # Search by Topic
#### Resources tagged with Positive-negative numbers similar to Connect Three:
Filter by: Content type:
Stage:
Challenge level:
### There are 19 results
Broad Topics > Numbers and the Number System > Positive-negative numbers
### Connect Three
##### Stage: 3 and 4 Challenge Level:
Can you be the first to complete a row of three?
### First Connect Three for Two
##### Stage: 2 and 3 Challenge Level:
First Connect Three game for an adult and child. Use the dice numbers and either addition or subtraction to get three numbers in a straight line.
### Strange Bank Account (part 2)
##### Stage: 3 Challenge Level:
Investigate different ways of making £5 at Charlie's bank.
### Pair Sums
##### Stage: 3 Challenge Level:
Five numbers added together in pairs produce: 0, 2, 4, 4, 6, 8, 9, 11, 13, 15 What are the five numbers?
### First Connect Three
##### Stage: 2 and 3 Challenge Level:
The idea of this game is to add or subtract the two numbers on the dice and cover the result on the grid, trying to get a line of three. Are there some numbers that are good to aim for?
### Consecutive Negative Numbers
##### Stage: 3 Challenge Level:
Do you notice anything about the solutions when you add and/or subtract consecutive negative numbers?
### Weights
##### Stage: 3 Challenge Level:
Different combinations of the weights available allow you to make different totals. Which totals can you make?
### Strange Bank Account
##### Stage: 3 Challenge Level:
Imagine a very strange bank account where you are only allowed to do two things...
### Up, Down, Flying Around
##### Stage: 3 Challenge Level:
Play this game to learn about adding and subtracting positive and negative numbers
### Consecutive Numbers
##### Stage: 2 and 3 Challenge Level:
An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore.
### Adding and Subtracting Positive and Negative Numbers
##### Stage: 3
How can we help students make sense of addition and subtraction of negative numbers?
### Making Sense of Positives and Negatives
##### Stage: 3
This article suggests some ways of making sense of calculations involving positive and negative numbers.
### Negative Numbers
##### Stage: 3
A brief history of negative numbers throughout the ages
### Playing Connect Three
##### Stage: 3 Challenge Level:
In this game the winner is the first to complete a row of three. Are some squares easier to land on than others?
### Minus One Two Three
##### Stage: 4 Challenge Level:
Substitute -1, -2 or -3, into an algebraic expression and you'll get three results. Is it possible to tell in advance which of those three will be the largest ?
### The History of Negative Numbers
##### Stage: 3, 4 and 5
This article -useful for teachers and learners - gives a short account of the history of negative numbers.
### Negatively Triangular
##### Stage: 4 Challenge Level:
How many intersections do you expect from four straight lines ? Which three lines enclose a triangle with negative co-ordinates for every point ?
### Missing Multipliers
##### Stage: 3 Challenge Level:
What is the smallest number of answers you need to reveal in order to work out the missing headers?
### Vector Racer
##### Stage: 3 and 4 Challenge Level:
The classic vector racing game. | 737 | 3,345 | {"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-2017-47 | longest | en | 0.852223 |
https://www.nag.com/numeric/nl/nagdoc_27.1/clhtml/s/s17aec.html | 1,627,841,862,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154214.63/warc/CC-MAIN-20210801154943-20210801184943-00216.warc.gz | 945,106,457 | 9,051 | # NAG CL Interfaces17aec (bessel_j0_real)
Settings help
CL Name Style:
## 1Purpose
s17aec returns the value of the Bessel function ${J}_{0}\left(x\right)$.
## 2Specification
#include
double s17aec (double x, NagError *fail)
The function may be called by the names: s17aec, nag_specfun_bessel_j0_real or nag_bessel_j0.
## 3Description
s17aec evaluates an approximation to the Bessel function of the first kind ${J}_{0}\left(x\right)$.
Note: ${J}_{0}\left(-x\right)={J}_{0}\left(x\right)$, so the approximation need only consider $x\ge 0$.
The function is based on three Chebyshev expansions:
For $0,
$J0(x)=∑′r=0arTr(t), with t=2 ( x8) 2 -1.$
For $x>8$,
$J0(x)= 2πx {P0(x)cos(x-π4)-Q0(x)sin(x- π4)} ,$
where ${P}_{0}\left(x\right)=\underset{r=0}{{\sum }^{\prime }}\phantom{\rule{0.25em}{0ex}}{b}_{r}{T}_{r}\left(t\right)$,
and ${Q}_{0}\left(x\right)=\frac{8}{x}\underset{r=0}{{\sum }^{\prime }}\phantom{\rule{0.25em}{0ex}}{c}_{r}{T}_{r}\left(t\right)$,
with $t=2{\left(\frac{8}{x}\right)}^{2}-1$.
For $x$ near zero, ${J}_{0}\left(x\right)\simeq 1$. This approximation is used when $x$ is sufficiently small for the result to be correct to machine precision.
For very large $x$, it becomes impossible to provide results with any reasonable accuracy (see Section 7), hence the function fails. Such arguments contain insufficient information to determine the phase of oscillation of ${J}_{0}\left(x\right)$; only the amplitude, $\sqrt{\frac{2}{\pi |x|}}$, can be determined and this is returned on failure. The range for which this occurs is roughly related to machine precision; the function will fail if (see the Users' Note for your implementation for details).
## 4References
NIST Digital Library of Mathematical Functions
Clenshaw C W (1962) Chebyshev Series for Mathematical Functions Mathematical tables HMSO
## 5Arguments
1: $\mathbf{x}$double Input
On entry: the argument $x$ of the function.
2: $\mathbf{fail}$NagError * Input/Output
The NAG error argument (see Section 7 in the Introduction to the NAG Library CL Interface).
## 6Error Indicators and Warnings
NE_ALLOC_FAIL
Dynamic memory allocation failed.
See Section 3.1.2 in the Introduction to the NAG Library CL Interface for further information.
NE_INTERNAL_ERROR
An internal error has occurred in this function. Check the function call and any array sizes. If the call is correct then please contact NAG for assistance.
See Section 7.5 in the Introduction to the NAG Library CL Interface for further information.
NE_NO_LICENCE
Your licence key may have expired or may not have been installed correctly.
See Section 8 in the Introduction to the NAG Library CL Interface for further information.
NE_REAL_ARG_GT
On entry, ${\mathbf{x}}=⟨\mathit{\text{value}}⟩$.
Constraint: $|{\mathbf{x}}|\le ⟨\mathit{\text{value}}⟩$.
$|{\mathbf{x}}|$ is too large, the function returns the amplitude of the ${J}_{0}$ oscillation, $\sqrt{2/\left(\pi |x|\right)}$.
## 7Accuracy
Let $\delta$ be the relative error in the argument and $E$ be the absolute error in the result. (Since ${J}_{0}\left(x\right)$ oscillates about zero, absolute error and not relative error is significant.)
If $\delta$ is somewhat larger than the machine precision (e.g., if $\delta$ is due to data errors etc.), then $E$ and $\delta$ are approximately related by:
$E≃|xJ1(x)|δ$
(provided $E$ is also within machine bounds). Figure 1 displays the behaviour of the amplification factor $|x{J}_{1}\left(x\right)|$.
However, if $\delta$ is of the same order as machine precision, then rounding errors could make $E$ slightly larger than the above relation predicts.
For very large $x$, the above relation ceases to apply. In this region, ${J}_{0}\left(x\right)\simeq \sqrt{\frac{2}{\pi |x|}}\mathrm{cos}\left(x-\frac{\pi }{4}\right)$. The amplitude $\sqrt{\frac{2}{\pi |x|}}$ can be calculated with reasonable accuracy for all $x$, but $\mathrm{cos}\left(x-\frac{\pi }{4}\right)$ cannot. If $x-\frac{\pi }{4}$ is written as $2N\pi +\theta$ where $N$ is an integer and $0\le \theta <2\pi$, then $\mathrm{cos}\left(x-\frac{\pi }{4}\right)$ is determined by $\theta$ only. If $x\gtrsim {\delta }^{-1}$, $\theta$ cannot be determined with any accuracy at all. Thus if $x$ is greater than, or of the order of, the inverse of the machine precision, it is impossible to calculate the phase of ${J}_{0}\left(x\right)$ and the function must fail.
## 8Parallelism and Performance
s17aec is not threaded in any implementation.
None.
## 10Example
This example reads values of the argument $x$ from a file, evaluates the function at each value of $x$ and prints the results.
### 10.1Program Text
Program Text (s17aece.c)
### 10.2Program Data
Program Data (s17aece.d)
### 10.3Program Results
Program Results (s17aece.r) | 1,410 | 4,758 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 54, "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} | 3.03125 | 3 | CC-MAIN-2021-31 | latest | en | 0.6752 |
https://www.jiskha.com/display.cgi?id=1206309654 | 1,516,385,233,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084888077.41/warc/CC-MAIN-20180119164824-20180119184824-00023.warc.gz | 942,424,224 | 3,844 | # physics
posted by .
Drift velocity is best defined as
a. the relationship between current and current density
b. the number of mobile electrons per volume
c. the number of mobile electrons per length
d. the flux of the current density through a given area
e. the average speed of the motion of the electrons in a wire
I got e for the answer. Am I correct?
• physics -
correct.
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10. ### As physics
Please help and do it step by step all parts of questions and use: T = time(s) I=current (amps) Q=net charge on object (Coulombs) N=no electrons (and write if added or removed from object and how) E=elementary charge Calculate the …
More Similar Questions | 693 | 2,716 | {"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} | 3.078125 | 3 | CC-MAIN-2018-05 | latest | en | 0.859144 |
https://ranocha.de/SummationByPartsOperators.jl/stable/api_reference/ | 1,660,527,894,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572089.53/warc/CC-MAIN-20220814234405-20220815024405-00161.warc.gz | 445,490,200 | 12,054 | # SummationByPartsOperators.jl API
SummationByPartsOperators.SummationByPartsOperatorsModule
SummationByPartsOperators
SummationByPartsOperators.jl is a Julia library of summation-by-parts (SBP) operators, which are discrete derivative operators developed to get provably stable semidiscretizations, paying special attention to boundary conditions. Discretizations included in this framework are finite difference, Fourier pseudospectral, continuous Galerkin, and discontinuous Galerkin methods. The main aim of SummationByPartsOperators.jl is to be useful for researchers and students to learn the basic concepts by providing a unified framework of all of these seemingly different discretizations. At the same time, the implementation is optimized to achieve good performance without sacrificing flexibility.
Check out the documentation for further information. Some noticable functions to start with are derivative_operator, legendre_derivative_operator, periodic_derivative_operator, fourier_derivative_operator, dissipation_operator, and grid.
If you use this package for your research, please cite it using
@article{ranocha2021sbp,
title={{SummationByPartsOperators.jl}: {A} {J}ulia library of provably stable
semidiscretization techniques with mimetic properties},
author={Ranocha, Hendrik},
journal={Journal of Open Source Software},
year={2021},
month={08},
doi={10.21105/joss.03454},
volume={6},
number={64},
pages={3454},
publisher={The Open Journal},
url={https://github.com/ranocha/SummationByPartsOperators.jl}
}
source
SummationByPartsOperators.BurgersNonperiodicSemidiscretizationType
BurgersNonperiodicSemidiscretization(D, Di, split_form, left_bc, right_bc)
A semidiscretization of Burgers' equation $\partial_t u(t,x) + \partial_x \frac{u(t,x)^2}{2} = 0$ with boundary conditions left_bc(t), right_bc(t).
D is a first-derivative SBP operator, Di an associated dissipation operator or nothing, and split_form::Union{Val(true), Val(false)} determines whether the canonical split form or the conservative form is used.
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SummationByPartsOperators.BurgersPeriodicSemidiscretizationType
BurgersPeriodicSemidiscretization(D, Di, split_form)
A semidiscretization of Burgers' equation $\partial_t u(t,x) + \partial_x \frac{u(t,x)^2}{2} = 0$ with periodic boundary conditions.
D is a first-derivative SBP operator, Di an associated dissipation operator or nothing, and split_form::Union{Val(true), Val(false)} determines whether the canonical split form or the conservative form is used.
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SummationByPartsOperators.ConstantFilterMethod
ConstantFilter(D::FourierDerivativeOperator, filter)
Create a modal filter with constant parameters adapted to the Fourier derivative operator D with parameters given by the filter function filter.
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SummationByPartsOperators.ConstantFilterMethod
ConstantFilter(D::LegendreDerivativeOperator, filter, TmpEltype=T)
Create a modal filter with constant parameters adapted to the Legendre derivative operator D with parameters given by the filter function filter.
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SummationByPartsOperators.CubicNonperiodicSemidiscretizationType
CubicNonperiodicSemidiscretization(D, Di, split_form, left_bc, right_bc)
A semidiscretization of the cubic conservation law $\partial_t u(t,x) + \partial_x u(t,x)^3 = 0$ with nonperiodic boundary conditions left_bc(t), right_bc(t).
D is a first-derivative SBP operator, Di an associated dissipation operator or nothing, and split_form::Union{Val(true), Val(false)} determines whether the canonical split form or the conservative form is used.
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SummationByPartsOperators.CubicPeriodicSemidiscretizationType
CubicPeriodicSemidiscretization(D, Di, split_form)
A semidiscretization of the cubic conservation law $\partial_t u(t,x) + \partial_x u(t,x)^3 = 0$ with periodic boundary conditions.
D is a first-derivative SBP operator, Di an associated dissipation operator or nothing, and split_form::Union{Val(true), Val(false)} determines whether the canonical split form or the conservative form is used.
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SummationByPartsOperators.DienerDorbandSchnetterTiglio2007Type
DienerDorbandSchnetterTiglio2007()
Coefficients of the SBP operators given in
• Diener, Dorband, Schnetter, Tiglio (2007) Optimized high-order derivative and dissipation operators satisfying summation by parts, and applications in three-dimensional multi-block evolutions. Journal of Scientific Computing 32.1, pp. 109-145.
• Mattsson, Nordström (2004) Summation by parts operators for finite difference approximations of second derivatives. Journal of Computational Physics 199, pp. 503-540.
The dissipation operators proposed by Diener, Dorband, Schnetter, Tiglio (2007) for the diagonal-norm operators are the same as the ones of
• Mattsson, Svärd, Nordström (2004) Stable and Accurate Artificial Dissipation. Journal of Scientific Computing 21.1, pp. 57-79.
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SummationByPartsOperators.FourierDerivativeOperatorMethod
FourierDerivativeOperator(xmin::T, xmax::T, N::Integer) where {T<:Real}
Construct the FourierDerivativeOperator on a uniform grid between xmin and xmax using N nodes and N÷2+1 complex Fourier modes.
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SummationByPartsOperators.FourierDerivativeOperator2DMethod
FourierDerivativeOperator2D(xmin, xmax, Nx, ymin, ymax, Ny)
Construct the FourierDerivativeOperator on a uniform grid between xmin and xmax using Nx nodes and ymin and ymax using Ny nodes.
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SummationByPartsOperators.Holoborodko2008Type
Holoborodko2008()
Coefficients of the periodic operators given in
• Holoborodko (2008) Smooth Noise Robust Differentiators. http://www.holoborodko.com/pavel/numerical-methods/numerical-derivative/smooth-low-noise-differentiators/
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SummationByPartsOperators.LegendreDerivativeOperatorMethod
LegendreDerivativeOperator(xmin::T, xmax::T, N::Int) where {T<:Real}
Construct the LegendreDerivativeOperator on a uniform grid between xmin and xmax using N nodes and N-1 Legendre modes.
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SummationByPartsOperators.MadayTadmor1989Type
MadayTadmor1989()
Coefficients of the Fourier spectral viscosity given in
• Maday, Tadmor (1989) Analysis of the Spectral Vanishing Viscosity Method for Periodic Conservation Laws. SIAM Journal on Numerical Analysis 26.4, pp. 854-870.
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SummationByPartsOperators.Mattsson2012Type
Mattsson2012()
Coefficients of the SBP operators given in
• Mattsson (2012) Summation by Parts Operators for Finite Difference Approximations of Second-Derivatives with Variable Coefficients. Journal of Scientific Computing 51, pp. 650-682.
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SummationByPartsOperators.Mattsson2014Type
Mattsson2014()
Coefficients of the SBP operators given in
• Mattsson (2014) Diagonal-norm summation by parts operators for fiite difference approximations of third and fourth derivatives. Journal of Computational Physics 274, pp. 432-454.
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SummationByPartsOperators.Mattsson2017Type
Mattsson2017(version::Symbol)
Coefficients of the upwind SBP operators given in
• Mattsson (2017) Diagonal-norm upwind SBP operators. Journal of Computational Physics 335, pp. 283-310.
You can choose between the different versions :central, :plus, and :minus.
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SummationByPartsOperators.QuarticNonconvexPeriodicSemidiscretizationType
QuarticNonconvexPeriodicSemidiscretization(D, Di, split_form)
A semidiscretization of the quartic nonconvex conservation law $\partial_t u(t,x) + \partial_x ( u(t,x)^4 - 10 u(t,x)^2 + 3 u(t,x) ) = 0$ with periodic boundary conditions.
D is a first-derivative SBP operator, Di an associated dissipation operator or nothing, and split_form::Union{Val(true), Val(false)} determines whether the canonical split form or the conservative form is used.
source
SummationByPartsOperators.Tadmor1989Type
Tadmor1989()
Coefficients of the Fourier spectral viscosity given in
• Tadmor (1989) Convergence of Spectral Methods for Nonlinear Conservation Laws. SIAM Journal on Numerical Analysis 26.1, pp. 30-44.
source
SummationByPartsOperators.Tadmor1993Type
Tadmor1993()
Coefficients of the Fourier super spectral viscosity given in
• Tadmor (1993) Super Viscosity and Spectral Approximations of Nonlinear Conservation Laws. Numerical Methods for Fluid Dynamics IV, pp. 69-82.
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SummationByPartsOperators.TadmorWaagan2012ConvergentType
TadmorWaagan2012Convergent()
Coefficients of the Fourier spectral viscosity given in
• Tadmor, Waagan (2012) Adaptive Spectral Viscosity for Hyperbolic Conservation Laws. SIAM Journal on Scientific Computing 34.2, pp. A993-A1009.
• Schochet (1990) The Rate of Convergence of Spectral-Viscosity Methods for Periodic Scalar Conservation Laws. SIAM Journal on Numerical Analysis 27.5, pp. 1142-1159.
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SummationByPartsOperators.UniformMesh1DType
UniformMesh1D(xmin::Real, xmax::Real, Nx::Integer)
UniformMesh1D(; xmin::Real, xmax::Real, Nx::Integer)
A uniform mesh in one space dimension of Nx cells between xmin and xmax.
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SummationByPartsOperators.UniformPeriodicMesh1DType
UniformPeriodicMesh1D(xmin::Real, xmax::Real, Nx::Integer)
UniformPeriodicMesh1D(; xmin::Real, xmax::Real, Nx::Integer)
A uniform periodic mesh in one space dimension of Nx cells between xmin and xmax.
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SummationByPartsOperators.VariableLinearAdvectionNonperiodicSemidiscretizationType
VariableLinearAdvectionNonperiodicSemidiscretization(D, Di, a, split_form,
left_bc, right_bc)
A semidiscretization of the linear advection equation $\partial_t u(t,x) + \partial_x ( a(x) u(t,x) ) = 0$ with boundary conditions left_bc(t), right_bc(t).
D is an SBP derivative operator, Di an associated dissipation operator or nothing, a(x) the variable coefficient, and split_form::Union{Val(false), Val(true)} determines whether the canonical split form or the conservative form should be used.
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SummationByPartsOperators.WaveEquationNonperiodicSemidiscretizationType
WaveEquationNonperiodicSemidiscretization(D, left_bc, right_bc)
A semidiscretization of the linear wave equation $\partial_t^2 u(t,x) = \partial_x^2 u(t,x)$.
D is assumed to be a second-derivative SBP operator and the boundary conditions can be Val(:HomogeneousNeumann), Val(:HomogeneousDirichlet), or Val(:NonReflecting).
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LinearAlgebra.mul!Function
mul!(du, D::DerivativeOperator, u, α=true, β=false)
Efficient in-place version of du = α * D * u + β * du. Note that du must not be aliased with u.
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PolynomialBases.compute_coefficients!Method
compute_coefficients!(uval, u, D::AbstractDerivativeOperator)
Compute the nodal values of the function u at the grid associated to the derivative operator D and stores the result in uval.
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PolynomialBases.evaluate_coefficients!Method
evaluate_coefficients!(xplot, uplot, u, D::AbstractDerivativeOperator)
Evaluates the nodal coefficients u at a grid associated to the derivative operator D and stores the result in xplot, uplot. Returns xplot, uplot, where xplot contains the nodes and uplot the corresponding values of u.
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PolynomialBases.evaluate_coefficientsMethod
evaluate_coefficients(u, D::AbstractDerivativeOperator)
Evaluates the nodal coefficients u at a grid associated to the derivative operator D. Returns xplot, uplot, where xplot contains the nodes and uplot the corresponding values of u.
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PolynomialBases.integrateMethod
integrate(func, u, D::DerivativeOperator)
Map the function func to the coefficients u and integrate with respect to the quadrature rule associated with the SBP derivative operator D.
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PolynomialBases.integrateMethod
integrate(func, u, D::PeriodicDerivativeOperator)
Map the function func to the coefficients u and integrate with respect to the quadrature rule associated with the periodic derivative operator D.
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PolynomialBases.integrateMethod
integrate(func, u, D::AbstractPeriodicDerivativeOperator)
Map the function func to the coefficients u and integrate with respect to the quadrature rule associated with the derivative operator D.
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SummationByPartsOperators.derivative_operatorFunction
derivative_operator(source_of_coefficients,
derivative_order, accuracy_order,
xmin, xmax, N, mode=FastMode())
derivative_operator(source_of_coefficients;
derivative_order, accuracy_order,
xmin, xmax, N, mode=FastMode())
Create a DerivativeOperator approximating the derivative_order-th derivative on a grid between xmin and xmax with N grid points up to order of accuracy accuracy_order. with coefficients given by source_of_coefficients. The evaluation of the derivative can be parallized using threads by chosing mode=ThreadedMode().
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SummationByPartsOperators.dissipation_operatorFunction
dissipation_operator(source_of_coefficients, order, xmin, xmax, N,
left_weights, right_weights, mode=FastMode())
Create a negative semidefinite DissipationOperator using undivided differences approximating a weighted order-th derivative on a grid between xmin and xmax with N grid points up to order of accuracy 2 with coefficients given by source_of_coefficients. The norm matrix is given by left_weights and right_weights. The evaluation of the derivative can be parallized using threads by chosing mode=ThreadedMode().
source
SummationByPartsOperators.dissipation_operatorMethod
dissipation_operator(D::PeriodicDerivativeOperator;
strength=one(eltype(D)),
order=accuracy_order(D),
mode=D.coefficients.mode)
Create a negative semidefinite DissipationOperator using undivided differences approximating a order-th derivative with strength strength adapted to the derivative operator D. The evaluation of the derivative can be parallized using threads by chosing mode=ThreadedMode().
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SummationByPartsOperators.dissipation_operatorMethod
dissipation_operator([source_of_coefficients=MattssonSvärdNordström2004()],
D::DerivativeOperator{T};
strength=one(T),
order::Int=accuracy_order(D),
mode=D.coefficients.mode)
Create a negative semidefinite DissipationOperator using undivided differences approximating a weighted order-th derivative adapted to the derivative operator D with coefficients given in source_of_coefficients. The evaluation of the derivative can be parallized using threads by chosing mode=ThreadedMode().
source
SummationByPartsOperators.fornbergMethod
fornberg(x::Vector{T}, m::Int) where {T}
Calculate the weights of a finite difference approximation of the mth derivative with maximal order of accuracy at 0 using the nodes x, see Fornberg (1998) Calculation of Weights in Finite Difference Formulas SIAM Rev. 40.3, pp. 685-691.
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SummationByPartsOperators.fourier_derivative_matrixFunction
fourier_derivative_matrix(N, xmin::Real=0.0, xmax::Real=2π)
Compute the Fourier derivative matrix with respect to the corresponding nodal basis using N nodes, see Kopriva (2009) Implementing Spectral Methods for PDEs, Algorithm 18.
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SummationByPartsOperators.fourier_derivative_operatorMethod
fourier_derivative_operator(xmin::Real, xmax::Real, N::Integer)
fourier_derivative_operator(; xmin::Real, xmax::Real, N::Integer)
Construct the FourierDerivativeOperator on a uniform grid between xmin and xmax using N nodes and N÷2+1 complex Fourier modes.
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SummationByPartsOperators.legendre_derivative_operatorMethod
legendre_derivative_operator(xmin::Real, xmax::Real, N::Integer)
legendre_derivative_operator(; xmin::Real, xmax::Real, N::Integer)
Construct the LegendreDerivativeOperator on a uniform grid between xmin and xmax using N nodes and N-1 Legendre modes.
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SummationByPartsOperators.legendre_second_derivative_operatorMethod
legendre_second_derivative_operator(xmin::Real, xmax::Real, N::Integer)
legendre_second_derivative_operator(; xmin::Real, xmax::Real, N::Integer)
Construct the LegendreDerivativeOperator on a uniform grid between xmin and xmax using N nodes and N-1 Legendre modes.
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SummationByPartsOperators.mul_transpose_derivative_left!Method
mul_transpose_derivative_left!(u, D::DerivativeOperator, der_order::Val{N}, α=true, β=false)
Set the grid function u to α times the transposed N-th derivative functional applied to u plus β times u in the domain of the N-th derivative functional at the left boundary of the grid. Thus, the coefficients α, β have the same meaning as in mul!.
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SummationByPartsOperators.mul_transpose_derivative_right!Method
mul_transpose_derivative_right!(u, D::DerivativeOperator, der_order::Val{N}, α=true, β=false)
Set the grid function u to α times the transposed N-th derivative functional applied to u plus β times u in the domain of the N-th derivative functional at the right boundary of the grid. Thus, the coefficients α, β have the same meaning as in mul!.
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SummationByPartsOperators.periodic_central_derivative_coefficientsFunction
periodic_central_derivative_coefficients(derivative_order, accuracy_order, T=Float64, mode=FastMode())
Create the PeriodicDerivativeCoefficients approximating the derivative_order-th derivative with an order of accuracy accuracy_order and scalar type T. The evaluation of the derivative can be parallized using threads by chosing mode=ThreadedMode()).
source
SummationByPartsOperators.periodic_central_derivative_operatorFunction
periodic_central_derivative_operator(derivative_order, accuracy_order,
xmin, xmax, N, mode=FastMode())
Create a PeriodicDerivativeOperator approximating the derivative_order-th derivative on a uniform grid between xmin and xmax with N grid points up to order of accuracy accuracy_order. The evaluation of the derivative can be parallized using threads by chosing mode=ThreadedMode().
source
SummationByPartsOperators.periodic_derivative_coefficientsFunction
periodic_derivative_coefficients(derivative_order, accuracy_order,
left_offset=-(accuracy_order+1)÷2,
T=Float64, mode=FastMode())
Create the PeriodicDerivativeCoefficients approximating the derivative_order-th derivative with an order of accuracy accuracy_order and scalar type T where the leftmost grid point used is determined by left_offset. The evaluation of the derivative can be parallized using threads by chosing mode=ThreadedMode().
source
SummationByPartsOperators.periodic_derivative_coefficientsMethod
periodic_derivative_coefficients(source::Holoborodko2008, derivative_order, accuracy_order;
T=Float64, mode=FastMode(),
stencil_width=accuracy_order+3)
Create the PeriodicDerivativeCoefficients approximating the derivative_order-th derivative with an order of accuracy accuracy_order and scalar type T given by Holoborodko2008. The evaluation of the derivative can be parallized using threads by chosing mode=ThreadedMode().
source
SummationByPartsOperators.periodic_derivative_operatorFunction
periodic_derivative_operator(derivative_order, accuracy_order, grid,
left_offset=-(accuracy_order+1)÷2, mode=FastMode())
Create a PeriodicDerivativeOperator approximating the derivative_order-th derivative on thr uniform grid up to order of accuracy accuracy_order where the leftmost grid point used is determined by left_offset. The evaluation of the derivative can be parallized using threads by chosing mode=ThreadedMode()).
source
SummationByPartsOperators.periodic_derivative_operatorFunction
periodic_derivative_operator(derivative_order, accuracy_order,
xmin, xmax, N,
left_offset=-(accuracy_order+1)÷2,
mode=FastMode())
periodic_derivative_operator(; derivative_order, accuracy_order,
xmin, xmax, N,
left_offset=-(accuracy_order+1)÷2,
mode=FastMode())
Create a PeriodicDerivativeOperator approximating the derivative_order-th derivative on a uniform grid between xmin and xmax with N grid points up to order of accuracy accuracy_order where the leftmost grid point used is determined by left_offset. The evaluation of the derivative can be parallized using threads by chosing mode=ThreadedMode()).
Examples
julia> periodic_derivative_operator(derivative_order=1, accuracy_order=2,
xmin=0.0, xmax=1.0, N=11)
Periodic first-derivative operator of order 2 on a grid in [0.0, 1.0] using 11 nodes,
stencils with 1 nodes to the left, 1 nodes to the right, and coefficients of Fornberg (1998)
Calculation of Weights in Finite Difference Formulas.
SIAM Rev. 40.3, pp. 685-691.
source
SummationByPartsOperators.periodic_derivative_operatorMethod
periodic_derivative_operator(source::Holoborodko2008,
derivative_order, accuracy_order,
xmin, xmax, N; mode=FastMode(), kwargs...)
periodic_derivative_operator(source::Holoborodko2008;
derivative_order, accuracy_order,
xmin, xmax, N, mode=FastMode(), kwargs...)
Create a PeriodicDerivativeOperator approximating the derivative_order-th derivative on a uniform grid between xmin and xmax with N grid points up to order of accuracy accuracy_order where the leftmost grid point used is determined by left_offset. The evaluation of the derivative can be parallized using threads by chosing mode=ThreadedMode().
Examples
julia> periodic_derivative_operator(Holoborodko2008(), derivative_order=1, accuracy_order=2,
xmin=0.0, xmax=1.0, N=11)
Periodic first-derivative operator of order 2 on a grid in [0.0, 1.0] using 11 nodes,
stencils with 2 nodes to the left, 2 nodes to the right, and coefficients of Holoborodko (2008)
Smooth Noise Robust Differentiators.
http://www.holoborodko.com/pavel/numerical-methods/numerical-derivative/smooth-low-noise-differentiators/
source
SummationByPartsOperators.semidiscretizeMethod
semidiscretize(u0func, semi::AbstractSemidiscretization, tspan)
Apply the semidiscretization semi to the initial data given by u0func and return an ODEProblem with time span tspan.
source
SummationByPartsOperators.var_coef_derivative_operatorFunction
var_coef_derivative_operator(source_of_coefficients, derivative_order, accuracy_order,
xmin, xmax, N, left_weights, right_weights, bfunc,
mode=FastMode())
Create a VarCoefDerivativeOperator approximating a derivative_order-th derivative with variable coefficients bfunc on a grid between xmin and xmax with N grid points up to order of accuracy accuracy_order with coefficients given by source_of_coefficients. The evaluation of the derivative can be parallized using threads by chosing mode=ThreadedMode().
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SummationByPartsOperators.xmaxFunction
xmax(D)
Return the right boundary xmax of the domain specified when constructing the derivative operator D. Note that this might be different from the rightmost node of the grid of D when not all boundary nodes are included, e.g., for periodic derivative operators.
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SummationByPartsOperators.xminFunction
xmin(D)
Return the left boundary xmin of the domain specified when constructing the derivative operator D. Note that this might be different from the leftmost node of the grid of D when not all boundary nodes are included, e.g., for periodic derivative operators.
source | 5,390 | 22,560 | {"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} | 2.671875 | 3 | CC-MAIN-2022-33 | latest | en | 0.75367 |
https://undergroundmathematics.org/trigonometry-triangles-to-functions/r5533 | 1,620,973,521,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991648.40/warc/CC-MAIN-20210514060536-20210514090536-00090.warc.gz | 588,469,350 | 5,471 | Review question
# Given a picture hanging from a wall, can we find these angles? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource
Ref: R5533
## Question
A rectangular picture of width $\quantity{32}{cm}$ and height $\quantity{20}{cm}$ has its lower edge $AB$ horizontal against a vertical wall. It hangs at an angle of $15^\circ$ to the wall and may be taken to be of negligible thickness. It is supported by two equal strings $FD$, $FC$ of length $\quantity{20}{cm}$ attached to the upper corners $C$, $D$ of the picture and to a point $F$ on the wall vertically above the centre of $AB$.
Copy the given sketch, including the projections of $C$, $D$ and $E$, the mid-point of $CD$, onto the wall at $X$, $Y$ and $Z$ respectively.
Identify in terms of the letters the following angles and calculate them:
1. the angle between the two strings,
2. the angle between the string $DF$ and the wall,
3. the angle between the wall and the plane containing both strings. | 256 | 1,050 | {"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-2021-21 | longest | en | 0.888625 |
https://math.stackexchange.com/questions/3350835/equality-constrained-non-negative-linear-least-squares-unit-simplex-constraint | 1,624,441,688,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488536512.90/warc/CC-MAIN-20210623073050-20210623103050-00049.warc.gz | 345,452,845 | 38,747 | # Equality Constrained Non Negative Linear Least Squares (Unit Simplex Constraint)
I have the following constrained linear least-squares problem:
$$\min_{x \in \mathbb{R}^n} \frac{1}{2}||Ax-b||_2^2,$$
$$\text{subject to } \sum_{i=1}^n x_i = 1 \text{ and } x_i \geq 0, \text{ for } i=1,...,n.$$
If we discard the equality constraint, we have a standard non-negative least-squares (NNLS) problem, which can be solved by any NNLS solver.
On the other hand, if we discard the non-negativity constraints, we have a least-squares problem with an equality constraint, which can be solved in closed form (up to a matrix inverse calculation) by using Lagrange multipliers, for instance.
However, considering both constraints, I do not see any method that is particularly suitable for the problem. Of course, I could use the projected gradient descent method, but this one is pretty generic and does not exploit the linear least-squares nature of the problem.
My questions are:
1- Is there any numerical optimization algorithm that is particularly suitable for this problem?
2- Or can I reformulate the problem in a less constrained form so that can it be solved by a standard NNLS solver?
My attempt:
The Lagrangian is:
$$L(x,\lambda,\mu) = \frac{1}{2}||Ax-b||_2^2 + \lambda (\mathbf{1}^Tx - 1) - \mu^T x,$$ where $$\lambda$$ is a scalar, $$\mu$$ is $$n$$-dimensional and $$\mathbf{1}$$ is an $$n$$-dimensional vector of ones.
Thus, writing the KKT conditions, we get:
$$\begin{pmatrix} A^TA & \mathbf{1} & -I \\ \mathbf{1}^T & 0 & \mathbf{0}^T \end{pmatrix}\begin{pmatrix} x \\ \lambda \\ \mu\end{pmatrix} = \begin{pmatrix} A^Tb \\ 1 \end{pmatrix}, \text{ } \mu_i \geq 0, \text{ and } \mu_i x_i = 0, \text{ for } i=1,...,n.$$
I don't know how to proceed from here, due to the complementary slackness constraints...
• That’s precisely the purpose of the $n$-dimensional vector $\mu$. – learner Sep 11 '19 at 10:37
• Sorry, please ignore my earlier comment- wasn't reading carefully. – Brian Borchers Sep 11 '19 at 13:55
• Related - math.stackexchange.com/questions/2935650. – Royi Mar 18 '20 at 21:06
Your problem can be formulated as convex quadratic programming problem with a linear equality constraint and nonnegativity constraints on $$x$$.
$$\min \frac{1}{2}\left( x^{T}(A^{T}A)x - 2(b^{T}A)x + b^{T}b \right)$$
subject to
$$\sum_{i=1}^{n} x_{i}=1$$
$$x \geq 0$$.
As a practical matter, you're probably best off using a well-written library routine for linearly constrained quadratic programming. There are many of them available in a variety of programming languages. If you need to implement this yourself, then implementing an active set QP solver is relatively straight forward.
• could you please elaborate a bit more on how to formulate this problem as a quadratic programming one? – learner Sep 11 '19 at 12:22 | 820 | 2,837 | {"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": 13, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2021-25 | latest | en | 0.835208 |
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TeViAnTxonc3
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Posted: Friday 05th of Jan 08:09 | 634 | 2,404 | {"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} | 2.765625 | 3 | CC-MAIN-2024-10 | latest | en | 0.931311 |
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1. May 18, 2014
Logician
I have been doing a lot of regarding about relativistic physics since my earlier posts and I think I now understand the mechanism of the so called cosmic speed limit.
This is how I visualize it.
I know that the ball on a sheet analogy is not perfect but for this I think it will suffice. With that analogy the more mass the ball has the deeper the bowl is created. So as one puts more energy into acceleration the relativistic mass increases which deepens the bowl more due to the action of the gravity on the increased mass. Therefore as you try and approach "c" the well keeps deepening due to relativistic mass increase and so for every unit of energy put into acceleration the bottom of the bowl gets further away and you can never actually reach the bottom of the bowl and start up the other side. That also explains to me why time travel might be possible because with the bowl analogy as the bowl deepens the other side appears to get closer so at intensely high gravities the jump across through a wormhole becomes more possible as the apparent distance between the two sides decreases.
Is this an apt analogy for high speed travel? It also helps me understand time dilation as time is a part of the space time curvature.
Please be kind I am trying to wrap my mind around how to imagine these effects in my mind as real world ideas.
I know I do not have the full knowledge I need but I am getting it and I do appreciate the help and knowledge from the rest of you on here.
Thanks,
Logician
2. May 18, 2014
Staff Emeritus
You are mixing two completely different theories: special relativity, which is a theory of motion, and general relativity, which is a theory of gravity. This is one reason why you are confused. The second reason you are confused is that you seem to want to make your own explanations - one might even say theories - before you fully understand it. This never works. Finally, you seem to want to jump in the middle. It really is slower than to start at the beginning - it looks like you will save time, but it seldom works out that way.
3. May 18, 2014
Staff: Mentor
No. Spacetime curvature describes gravity; it is part of general relativity, not special relativity.
4. May 18, 2014
ghwellsjr
No, it's not. The Lorentz Transformation is all you need to understand how high speed travel results in Time Dilation and there is no space time curvature in Special Relativity. It's really simple. Einstein said so himself. If you are having trouble understanding Special Relativity, it's because you haven't applied the Lorentz Transformation. Just start with a spacetime diagram showing a stationary clock and the transform it to a any speed you want (short of c) and make another spacetime diagram and you'll see the Time Dilation right there before your eyes.
5. May 18, 2014
Logician
Misunderstanding
I think I am not doing a very good job of describing what I am trying to ask. I am not trying to propagate a theory but rather trying to VISUALIZE in a real world way what is happening at near "c" speeds. I think my main problem with understanding this type of thing is I am still trying to grasp what relativistic mass is in a visualization. I understand that it is either very difficult or impossible to accurately visualize what is happening at near "c" speeds for me but I want to try. I want to visualize a scenario of why there is a cosmic speed limit, or perhaps a better explanation would be that I want to try and visualize the mechanism of why this happens.
Thanks,
Logician
6. May 18, 2014
Logician
Can you put a diagram up for me please I am not quite understanding exactly what you are saying.
Thanks,
Logician
7. May 18, 2014
ghwellsjr
You can see lots of diagrams made by me. Just do a search on "diagram" with my username.
EDIT: Here's a good one I found:
Last edited: May 18, 2014
8. May 18, 2014
Logician
Ok I saw that and thank you. I think I understand the Time dilation effect now.
Here is another question the. Does relativistic mass have effects on the macro scale? What about the quantum scale. For instance does Rmass affect the orbits of electrons in their shell?
I think by asking specific individual questions I can get a grip on thi. Thank you for your help.
9. May 18, 2014
dauto
Relativistic mass isn't a very useful concept because it is just the energy under another name. Since it is just the energy, why not call it energy and be done with it? The relativistic mass concept is more trouble than it's worth. If I have it my way it would be abolished from physics books.
10. May 18, 2014
pervect
Staff Emeritus
A good way to visualize what happens at relativistic velocities is to visualize that NOTHING happens at high velocities. For instance, you right now are at a very high relativistic velocity in some frame of reference. Do you feel any different?
The important thing to know about relativity is that your velocity doesn't matter to the physics. The idea is that if you are in a sealed box, without looking outside the box to measure your velocity relative to anothr object, you can't even tell whether or not you and the box are moving at relativistic velocites or not.
I hope you've at least heard this before.
I don't think your attempt at "visulaization" is going to lead to this conclusion, and I'm rather concerned that it will lead you astray - it seems to me your current line of thining may be at odds with the above, very important, principle of relativity.
You don't need relativistic mass to understand why light speed is a limit. Thus I don't agree with your self-assesment.
Rather than go off at a tangent about the use and misuse of relativistic mass, lets stick to what you do need to understand to understand why the speed of light is "as fast as you can go". This is not relativistic dynamics, but relativistic kinematics. Kinematics is "the study of classical mechanics which describes the motion of points, bodies (objects) and systems of bodies (groups of objects) without consideration of the causes of motion.".
Specifically, in this case, kinematics is the relativistic velocity addition formula.
Given that you have three object A, B, and C, and the relative velocity between A and B is v1, while the relative velocity between B and C is v2, let v_tot be the velocity between A and C, the "sum" of v1 and v2. Then:
v_tot = (v1+v2) / (1 + v1*v2/c^2)
This formula does not use, or need, any consideration of "forces" or other causes of motion, but if you have v1<c and v2<c, mathematically it tells you that the total velocity v_tot must be less than c. This might be obvious, s a short proof. Let v1 = c - e, and v2 = c - f, where e and f are positive. Thus e is a postiive number, which approaches zero as you approach the speed of light, but never goes to zero. Then we can say that
$$v_{tot} = \frac{2c - e - f}{1 + \frac{\left(c - e\right) \left(c - f\right)}{c^2}} \quad = c \frac{2 - \frac{e}{c} - \frac{f}{c}} {2 - \frac{e}{c} - \frac{f}{c} + \frac{ef}{c^2}}$$
and because the denominator is greater than the numerator, (v_tot / c) < 1.
Any attempt to explain "why" the total velocity is less than c based on forces ior other dynamic concepts s going to be misleading, because you don't even need the CONCEPT of force, or "cause of motion", to prove this fact. Thus forces cannot be a key element of the explanation of "why" given that they're not needed to demonstrate the fact, any attempt to "explain why" based on forces will be illusionary.
What you DO need to study is issues like "how the velocities add". No amount of "visualization" of forces, or other irrelevant entities such as curved space-time is going to explain this. What will work is studying the Lorentz transform. Other posters have pointed you at some resources that may aid you at this.
Once you understand the kinematic issues, related to how velocities (and other quanties) transform when you change your frame of reference, you may be ready to undertake the study of relativistic dynamics. Until you reach this point, though, you should focus on understanding the kinematic issues first. | 1,923 | 8,186 | {"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} | 2.96875 | 3 | CC-MAIN-2017-43 | longest | en | 0.974486 |
http://cn.metamath.org/mpeuni/limsupmnf.html | 1,653,089,253,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662534693.28/warc/CC-MAIN-20220520223029-20220521013029-00537.warc.gz | 13,476,701 | 7,233 | Mathbox for Glauco Siliprandi < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > limsupmnf Structured version Visualization version GIF version
Theorem limsupmnf 40271
Description: The superior limit of a function is -∞ if and only if every real number is the upper bound of the restriction of the function to an upper interval of real numbers. (Contributed by Glauco Siliprandi, 23-Oct-2021.)
Hypotheses
Ref Expression
limsupmnf.j 𝑗𝐹
limsupmnf.a (𝜑𝐴 ⊆ ℝ)
limsupmnf.f (𝜑𝐹:𝐴⟶ℝ*)
Assertion
Ref Expression
limsupmnf (𝜑 → ((lim sup‘𝐹) = -∞ ↔ ∀𝑥 ∈ ℝ ∃𝑘 ∈ ℝ ∀𝑗𝐴 (𝑘𝑗 → (𝐹𝑗) ≤ 𝑥)))
Distinct variable groups: 𝐴,𝑗,𝑘,𝑥 𝑘,𝐹,𝑥
Allowed substitution hints: 𝜑(𝑥,𝑗,𝑘) 𝐹(𝑗)
Proof of Theorem limsupmnf
Dummy variables 𝑖 𝑙 𝑦 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 limsupmnf.a . . 3 (𝜑𝐴 ⊆ ℝ)
2 limsupmnf.f . . 3 (𝜑𝐹:𝐴⟶ℝ*)
3 eqid 2651 . . 3 (𝑖 ∈ ℝ ↦ sup((𝐹 “ (𝑖[,)+∞)), ℝ*, < )) = (𝑖 ∈ ℝ ↦ sup((𝐹 “ (𝑖[,)+∞)), ℝ*, < ))
41, 2, 3limsupmnflem 40270 . 2 (𝜑 → ((lim sup‘𝐹) = -∞ ↔ ∀𝑦 ∈ ℝ ∃𝑖 ∈ ℝ ∀𝑙𝐴 (𝑖𝑙 → (𝐹𝑙) ≤ 𝑦)))
5 breq2 4689 . . . . . . . 8 (𝑦 = 𝑥 → ((𝐹𝑙) ≤ 𝑦 ↔ (𝐹𝑙) ≤ 𝑥))
65imbi2d 329 . . . . . . 7 (𝑦 = 𝑥 → ((𝑖𝑙 → (𝐹𝑙) ≤ 𝑦) ↔ (𝑖𝑙 → (𝐹𝑙) ≤ 𝑥)))
76ralbidv 3015 . . . . . 6 (𝑦 = 𝑥 → (∀𝑙𝐴 (𝑖𝑙 → (𝐹𝑙) ≤ 𝑦) ↔ ∀𝑙𝐴 (𝑖𝑙 → (𝐹𝑙) ≤ 𝑥)))
87rexbidv 3081 . . . . 5 (𝑦 = 𝑥 → (∃𝑖 ∈ ℝ ∀𝑙𝐴 (𝑖𝑙 → (𝐹𝑙) ≤ 𝑦) ↔ ∃𝑖 ∈ ℝ ∀𝑙𝐴 (𝑖𝑙 → (𝐹𝑙) ≤ 𝑥)))
9 breq1 4688 . . . . . . . . . 10 (𝑖 = 𝑘 → (𝑖𝑙𝑘𝑙))
109imbi1d 330 . . . . . . . . 9 (𝑖 = 𝑘 → ((𝑖𝑙 → (𝐹𝑙) ≤ 𝑥) ↔ (𝑘𝑙 → (𝐹𝑙) ≤ 𝑥)))
1110ralbidv 3015 . . . . . . . 8 (𝑖 = 𝑘 → (∀𝑙𝐴 (𝑖𝑙 → (𝐹𝑙) ≤ 𝑥) ↔ ∀𝑙𝐴 (𝑘𝑙 → (𝐹𝑙) ≤ 𝑥)))
12 nfv 1883 . . . . . . . . . . 11 𝑗 𝑘𝑙
13 limsupmnf.j . . . . . . . . . . . . 13 𝑗𝐹
14 nfcv 2793 . . . . . . . . . . . . 13 𝑗𝑙
1513, 14nffv 6236 . . . . . . . . . . . 12 𝑗(𝐹𝑙)
16 nfcv 2793 . . . . . . . . . . . 12 𝑗
17 nfcv 2793 . . . . . . . . . . . 12 𝑗𝑥
1815, 16, 17nfbr 4732 . . . . . . . . . . 11 𝑗(𝐹𝑙) ≤ 𝑥
1912, 18nfim 1865 . . . . . . . . . 10 𝑗(𝑘𝑙 → (𝐹𝑙) ≤ 𝑥)
20 nfv 1883 . . . . . . . . . 10 𝑙(𝑘𝑗 → (𝐹𝑗) ≤ 𝑥)
21 breq2 4689 . . . . . . . . . . 11 (𝑙 = 𝑗 → (𝑘𝑙𝑘𝑗))
22 fveq2 6229 . . . . . . . . . . . 12 (𝑙 = 𝑗 → (𝐹𝑙) = (𝐹𝑗))
2322breq1d 4695 . . . . . . . . . . 11 (𝑙 = 𝑗 → ((𝐹𝑙) ≤ 𝑥 ↔ (𝐹𝑗) ≤ 𝑥))
2421, 23imbi12d 333 . . . . . . . . . 10 (𝑙 = 𝑗 → ((𝑘𝑙 → (𝐹𝑙) ≤ 𝑥) ↔ (𝑘𝑗 → (𝐹𝑗) ≤ 𝑥)))
2519, 20, 24cbvral 3197 . . . . . . . . 9 (∀𝑙𝐴 (𝑘𝑙 → (𝐹𝑙) ≤ 𝑥) ↔ ∀𝑗𝐴 (𝑘𝑗 → (𝐹𝑗) ≤ 𝑥))
2625a1i 11 . . . . . . . 8 (𝑖 = 𝑘 → (∀𝑙𝐴 (𝑘𝑙 → (𝐹𝑙) ≤ 𝑥) ↔ ∀𝑗𝐴 (𝑘𝑗 → (𝐹𝑗) ≤ 𝑥)))
2711, 26bitrd 268 . . . . . . 7 (𝑖 = 𝑘 → (∀𝑙𝐴 (𝑖𝑙 → (𝐹𝑙) ≤ 𝑥) ↔ ∀𝑗𝐴 (𝑘𝑗 → (𝐹𝑗) ≤ 𝑥)))
2827cbvrexv 3202 . . . . . 6 (∃𝑖 ∈ ℝ ∀𝑙𝐴 (𝑖𝑙 → (𝐹𝑙) ≤ 𝑥) ↔ ∃𝑘 ∈ ℝ ∀𝑗𝐴 (𝑘𝑗 → (𝐹𝑗) ≤ 𝑥))
2928a1i 11 . . . . 5 (𝑦 = 𝑥 → (∃𝑖 ∈ ℝ ∀𝑙𝐴 (𝑖𝑙 → (𝐹𝑙) ≤ 𝑥) ↔ ∃𝑘 ∈ ℝ ∀𝑗𝐴 (𝑘𝑗 → (𝐹𝑗) ≤ 𝑥)))
308, 29bitrd 268 . . . 4 (𝑦 = 𝑥 → (∃𝑖 ∈ ℝ ∀𝑙𝐴 (𝑖𝑙 → (𝐹𝑙) ≤ 𝑦) ↔ ∃𝑘 ∈ ℝ ∀𝑗𝐴 (𝑘𝑗 → (𝐹𝑗) ≤ 𝑥)))
3130cbvralv 3201 . . 3 (∀𝑦 ∈ ℝ ∃𝑖 ∈ ℝ ∀𝑙𝐴 (𝑖𝑙 → (𝐹𝑙) ≤ 𝑦) ↔ ∀𝑥 ∈ ℝ ∃𝑘 ∈ ℝ ∀𝑗𝐴 (𝑘𝑗 → (𝐹𝑗) ≤ 𝑥))
3231a1i 11 . 2 (𝜑 → (∀𝑦 ∈ ℝ ∃𝑖 ∈ ℝ ∀𝑙𝐴 (𝑖𝑙 → (𝐹𝑙) ≤ 𝑦) ↔ ∀𝑥 ∈ ℝ ∃𝑘 ∈ ℝ ∀𝑗𝐴 (𝑘𝑗 → (𝐹𝑗) ≤ 𝑥)))
334, 32bitrd 268 1 (𝜑 → ((lim sup‘𝐹) = -∞ ↔ ∀𝑥 ∈ ℝ ∃𝑘 ∈ ℝ ∀𝑗𝐴 (𝑘𝑗 → (𝐹𝑗) ≤ 𝑥)))
Colors of variables: wff setvar class Syntax hints: → wi 4 ↔ wb 196 = wceq 1523 Ⅎwnfc 2780 ∀wral 2941 ∃wrex 2942 ⊆ wss 3607 class class class wbr 4685 ↦ cmpt 4762 “ cima 5146 ⟶wf 5922 ‘cfv 5926 (class class class)co 6690 supcsup 8387 ℝcr 9973 +∞cpnf 10109 -∞cmnf 10110 ℝ*cxr 10111 < clt 10112 ≤ cle 10113 [,)cico 12215 lim supclsp 14245 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1762 ax-4 1777 ax-5 1879 ax-6 1945 ax-7 1981 ax-8 2032 ax-9 2039 ax-10 2059 ax-11 2074 ax-12 2087 ax-13 2282 ax-ext 2631 ax-rep 4804 ax-sep 4814 ax-nul 4822 ax-pow 4873 ax-pr 4936 ax-un 6991 ax-cnex 10030 ax-resscn 10031 ax-1cn 10032 ax-icn 10033 ax-addcl 10034 ax-addrcl 10035 ax-mulcl 10036 ax-mulrcl 10037 ax-mulcom 10038 ax-addass 10039 ax-mulass 10040 ax-distr 10041 ax-i2m1 10042 ax-1ne0 10043 ax-1rid 10044 ax-rnegex 10045 ax-rrecex 10046 ax-cnre 10047 ax-pre-lttri 10048 ax-pre-lttrn 10049 ax-pre-ltadd 10050 ax-pre-mulgt0 10051 ax-pre-sup 10052 This theorem depends on definitions: df-bi 197 df-or 384 df-an 385 df-3or 1055 df-3an 1056 df-tru 1526 df-ex 1745 df-nf 1750 df-sb 1938 df-eu 2502 df-mo 2503 df-clab 2638 df-cleq 2644 df-clel 2647 df-nfc 2782 df-ne 2824 df-nel 2927 df-ral 2946 df-rex 2947 df-reu 2948 df-rmo 2949 df-rab 2950 df-v 3233 df-sbc 3469 df-csb 3567 df-dif 3610 df-un 3612 df-in 3614 df-ss 3621 df-nul 3949 df-if 4120 df-pw 4193 df-sn 4211 df-pr 4213 df-op 4217 df-uni 4469 df-iun 4554 df-br 4686 df-opab 4746 df-mpt 4763 df-id 5053 df-po 5064 df-so 5065 df-xp 5149 df-rel 5150 df-cnv 5151 df-co 5152 df-dm 5153 df-rn 5154 df-res 5155 df-ima 5156 df-iota 5889 df-fun 5928 df-fn 5929 df-f 5930 df-f1 5931 df-fo 5932 df-f1o 5933 df-fv 5934 df-riota 6651 df-ov 6693 df-oprab 6694 df-mpt2 6695 df-er 7787 df-en 7998 df-dom 7999 df-sdom 8000 df-sup 8389 df-inf 8390 df-pnf 10114 df-mnf 10115 df-xr 10116 df-ltxr 10117 df-le 10118 df-sub 10306 df-neg 10307 df-ico 12219 df-limsup 14246 This theorem is referenced by: limsupre2lem 40274 limsupmnfuzlem 40276
Copyright terms: Public domain W3C validator | 3,563 | 5,298 | {"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} | 3.046875 | 3 | CC-MAIN-2022-21 | latest | en | 0.254572 |
https://www.hackingnote.com/en/problems/integer-to-roman/ | 1,723,469,427,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641039579.74/warc/CC-MAIN-20240812124217-20240812154217-00672.warc.gz | 625,699,497 | 12,608 | Problems
# Integer to Roman
## Problem
Given an integer, convert it to a roman numeral.
The number is guaranteed to be within the range from 1 to 3999.
### Example
``````4 -> IV
12 -> XII
21 -> XXI
99 -> XCIX
``````
More examples at: http://literacy.kent.edu/Minigrants/Cinci/romanchart.htm
### Note
What is Roman Numeral? | 99 | 331 | {"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-2024-33 | latest | en | 0.608023 |
90namdangbothanhhoa.vn | 1,685,969,084,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652116.60/warc/CC-MAIN-20230605121635-20230605151635-00195.warc.gz | 103,595,734 | 8,099 | # Likelihood ratio test là gì
The Likelihood Ratio (LR) is the likelihood that a given demo result would be expected in a patient with the target disorder compared khổng lồ the likelihood that that same result would be expected in a patient without the target disorder.
Bạn đang xem: Likelihood ratio test là gì
Definition
The Likelihood Ratio (LR) is the likelihood that a given test result would be expected in a patient with the target disorder compared to the likelihood that that same result would be expected in a patient without the target disorder. For example, you hav e a patient with anaemia & a serum ferritin of 60mmol/l & you find in an article that 90 per cent of patients with iron deficiency anaemia have serum ferritins in the same range as your patient (= sensitivity) và that 15 per cent of patients with oth er causes for anaemia have serum ferritins in the same range as your patient (1 – specificity). This means that your patient’s result would be six times as likely (90/15) to be seen in someone with, as opposed khổng lồ someone without, iron deficiency anaemia, & this is called the LR for a positive thử nghiệm result.
Application
The LR is used to assess how good a diagnostic chạy thử is and to help in selecting an appropriate diagnostic test(s) or sequence of tests. They have advantages over sensitivity và specificity because they are less likely to lớn change with the prevalence of the disorder, they can be calculated for several levels of the symptom/sign or test, they can be used khổng lồ combine the results of multiple diagnostic test và the can be used to lớn calculate po st-test probability for a target disorder. For example, if you thought your patient’s chance of iron deficiency anaemia prior to lớn doing the ferritin was 50-50, this pre-test probability of 50 per cent translates as pre-test odds of 1:1, và the post test odds can be calculated as follows:
Post-test odds = pre-test odds * LR = 1*6 = 6Post-test probability = post thử nghiệm odds / (post chạy thử odds + 1)= 6 / (6 + 1) = 86 per cent
After the serum ferritin test is done and your patient is found to have a result of 60 mmol/l, the post-test probability of your patient having iron deficiency anaemia is therefore increased to lớn 86 per cent, & this suggests that the serum ferritin is a worthwhile diagnostic test.
Xem thêm: Thực Hư Câu Chuyện Nhau Mèo Là Gì, Nhau Thai Mèo Có Giúp Gia Chủ May Mắn, Phát Tài
Switching back and forth between probability và odds can be done simply using a nomogram (you can click here to view a PDF of the nomogram on its own for easy printing):
A LR greater than 1 produces a post-test probability which is higher than the pre-test probability. An LR less than 1 produces a post-test probability which is lower than the pre-test probability. When the pre-test probability lies between 30 and 70 per cent, kiểm tra results with a very high LR (say, above 10) rule in disease. An LR below 1 produces a post-test probability les than the pre-test probability. A very low LR (say, below 0.1) virtually rules out the chance that the patient has the disease.
Xem thêm: Quả Đậu Tương Là Gì ? Công Dụng Của Đậu Nành Và Một Số Tác Hại
Calculation
We can assume that there are four possible groups of patients:
group a, who are disease positive & test positive;group b, who are disease negative but chạy thử positive;group c, who are disease positive but chạy thử negative;group d, who are disease negative và test negative.
Then:
LR+ = sensitivity / (1-specificity) = (a/(a+c)) / (b/(b+d))LR- = (1-sensitivity) / specificity = (c/(a+c)) / (d/(b+d))Post-test odds = pre-test odds * LRPre-test odds = pre-test probability / (1-pre-test probability)Post-test probability = post-test odds / (post demo odds+1)
Example
This example is taken from the results of a systematic reviews of serum ferritin as a diagnostic demo for iron deficiency anaemia:
Sensitivity = a / (a+c) = 731/809 = 90 per centSpecificity = d / (b+d) = 1500/1770 = 85 per cent
LR+ = sens / (1-spec) = 90/15 = 6LR- = (1-sens) / (spec) = 10/85 = 0.12
Positive Predictive Value = a / (a+b) = 731/1001 = 73 per centNegative Predictive value = d / (c+d) = 1500/1578 = 95 per cent
Prevalence = (a+c) / (a+b+c+d) = 809/2579 = 32 per centPre-test odds = prevalence / (1-prevalence) = 31/69 = 0.45
Post-test odds = pre-test odds * LRPost-test Probability = post-test odds / (post-test odds + 1)
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# PRINT AND BRING THIS
## DEPARTMENT OF FINANCIAL ACCOUNTING
FAC1601: FINANCIAL ACCOUNTING REPORTING 1 (MODULE 2)
MEMORANDUM TO EXAMINATION PAPER: OCTOBER 2011
Mr MT Hlongoane
Mrs FM Osman
Mr A Eysele
Mr RN Ngcobo
Mrs B Ntoyanto-Ceki
Module Telephone Number: 012 429 4176
fac1601@unisa.ac.za
## QUESTION 1 (24 marks)
STATEMENT OF COMPREHENSIVE INCOME FOR THE YEAR ENDED 31 AUGUST 2011^
R
Revenue^ R(536 820^ 2 000^)
534 820
Cost of sales ^
(277 330)
Purchases R(302 530^ 2 400^)
300 130
Delivery expenses on purchases
5 280
305 410
Inventory (31 August 2011)
(28 080)
Gross profit^
257 490
Other income
3 575
Credit losses recovered
3 000
Profit on sale of equipment R1 500^ R[15 000^ (14 000^ + 75^)]
575
( 213 850)
Salaries and wages R[123 600^ (36 000^ + 38 000^)]
^49 600
Depreciation R(14 250^ + 1 875^+ 75^^ + 28 800)
(6) 45 000
Water and electricity R(41 160^ 10 400^)
30 760
Advertising expenses R10 400^ R(10 400^ 13 months^)
^9 600
Property rates
^17 010
General expenses
^38 720
Telephone expenses
^23 160
Finance costs
(6 480)
Interest on long-term loan R(108 000^ x 12%^) x 6/12^
^6 480
Profit for the year^
40 735
Other comprehensive income for the year^
Total comprehensive income for the year^
40 735
Calculations:
Depreciation:
Equipment:
Old equipment:
New equipment:
Disposed equipment:
Vehicles:
## R[(130 000^ 25 000^) (24 000^ 14 000^)] x 15%^ = R14 250
R25 000^ x 15%^ x 6/12^ = R1 875
R75^^ Given
R144 000^ x 20%^ = R28 800
## QUESTION 2 (18 marks)
ZIMBABWE CHARTERED ACCOUNTANTS
GENERAL LEDGER (SUMMARISED IN COLUMNAR FORMAT)
Assets
## Balance at the commencement of liquidation:
Loan: Morgan
Insurance policy
Sale of furniture and equipment
Settlement of liabilities
R
^407 000
(84 000)
^(130 000)
193 000
## R(202 000^ + 100 000^ + 105 000^) = R407 000
R(80 000^ + 12 000^) = R92 000
R[130 000^ + R(20 000^ x 5/10^)] = R140 000
R[100 000^ + R(20 000^ x 3/10^)] = R106 000
R[80 000^ + R(20 000^ x 2/10^)] = R84 000
R(105 000^ x 80%^) = R84 000
R(84 000^ x 90^%) = R75 600
Bank
Liabilities
Capital:
Bob
Capital
Morgan
Capital:
Arthur
R
R
R
R
R
^15 000 (92 000) ^ (140 000) ^(106 000) ^ (84 000)
^12 000
^(12 000)
(12 500)
(7 500)
(5 000)
^25 000
4 200
2 520
1 680
75 600
^130 000
^ (80 000)
^80 000
165 600
(148 300)
(122 980)
(87 320)
3.1
## Retained earnings as at 28 February 2011:
Other income:
Dividend income R0,20^ x 50 000^ shares
Gain on financial assets at fair value through profit or loss: Held for trading:
Interest income: Loans and receivables: Loans to member R(70 500^ x 10%^ x 2/12^)
Expenses:
Credit losses
Interest on loan from member
Profit before tax
Income tax expense
Total comprehensive income for the year
Distribution to members R(42 000^ + 44 800^)
Retained earnings for the year
Retained earnings 1 March 2010
Retained earnings 28 February 2011
R
498 900
61 175
10 000
50 000
^ 1 175
(10 204)
2 700
7 504
549 871
(126 500)
423 371
(86 800)
336 571
^ 472 000
808 571
(8)
3.2
EFFICIENCY CONSULTANTS CC ^
STATEMENT OF FINANCIAL POSITION AS AT 28 FEBRUARY 2011 ^
R
ASSETS
Non-current assets
Property, plant and equipment R(747 000^ + 120 000^ 24 000^)
Financial assets
Current assets
Inventories
R[(35 600 ^ 3 000 ^ - 4 700 ^^ 1 500 ^) + 10 000 ^
Other financial assets R(250 000 ^^ + 70 500 ^ + 1 175 ^^)
Prepayments
Cash and cash equivalents R(48 100 ^ + 2 800 ^)
Total assets ^
EQUITY AND LIABILITIES
Total equity
Members contributions R(120 000 ^ + 95 000 ^)
Retained earnings
Total liabilities
Non-current liabilities
Long-term borrowings R[232 000^ + R(44 800^^ x 50%^)]
Current liabilities
Trade and other payables R(25 100 ^ + 7 504 ^)
Distribution to members payable R(44 800 ^ x 50% ^)
Current portion of loans from members
Current tax payable R(126 500 ^ 116 600 ^)
Total equity and liabilities ^
938 000
^843 000
95 000
513 335
99 312
36 400
^321 675
5 048
50 900
1 451 335
1 023 571
215 000
^808 571
427 764
254 400
254 400
173 364
^14 660
32 604
22 400
93 800
9 900
1 451 335
(21)
[29]
## QUESTION 4 (20 marks)
FLEETWOOD CC ^
STATEMENT OF CASH FLOWS FOR THE YEAR ENDED 31 AUGUST 2011 ^
R
CASH FLOWS FROM OPERATING ACTIVITIES
Cash receipt from customers R594 124 + R(13 600 ^ 6 800 ^)
Cash paid to suppliers and employees
Interest expense
Income tax paid R(27 200 ^ + 136 816 ^ 40 800 ^)
Distribution to members paid R(73 400 ^ 51 000 ^)
Acquisition of financial asset at fair value through profit or loss: Held for
Repayment of loans and receivables: Loan to member R(49 500 ^ 35 000^)
Net cash from operating activities ^
CASH FLOWS FROM INVESTING ACTIVITIES
Investments in property, plant and equipment to expand operating capacity
Improvement of land and buildings R(665 000 450 500)
Acquisition of loans and receivables: Fixed deposit
Net cash used in investing activities
CASH FLOWS FROM FINANCING ACTIVITIES
Proceeds from members contributions R(833 000 45 000) R730 400
Proceeds from long-term borrowing
600 924
(269 414)
331 510
10 200
(8 500)
(123 216)
(22 400)
R
(4)
(4)
(60 000) ^
14 500 ^
142 094
(214 500)
(33 020)
(58 000)
(305 520)
57 600
51 000
108 600
(54 826)
170 000
115 174
## Net decrease in cash and cash equivalents
Cash and cash equivalents at beginning of year
Cash and cash equivalent at end of year
Calculations:
R[(62 500 ^ + 599 760 ^ 63 126 ^) (3 000 ^ + 5 700 ^ 3 700 ^)] = R594 124
R[(45 520 ^ + 294 440 ^ 90 080 ^) + R(34 334 ^ 10 400 ^ 4 400 ^)] = R269 414
R(280 500 ^ 20 320 ^ + 34 260 ^) = R294 440
QUESTION 5 (9 marks)
5.1
5.2
## Future value of investment:
Factor as per table:
8% ^^ 4 ^ = 2%
2% ^ over 16 ^ periods
1.373
## Present value of an annuity:
R(35 700 ^ 13 730 ^) = R21 970
Factor as per table:
12% ^ 2 ^ = 6%
6% over 8 ^ periods
9.897
## Present value of the annuity:
R21 970 ^ 9.897 ^ = R2 219.86
Amount to be invested monthly: R2 219.86 6 months^
R369.98 | 2,192 | 5,842 | {"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-2019-26 | latest | en | 0.559975 |
http://mathhelpforum.com/discrete-math/218488-eulcidean-algorithim-gcd.html | 1,537,881,712,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267161638.66/warc/CC-MAIN-20180925123211-20180925143611-00005.warc.gz | 160,996,378 | 9,294 | # Thread: Eulcidean Algorithim for GCD
1. ## Eulcidean Algorithim for GCD
Find the GCD for 198 and 765
I got as far as $\displaystyle 765= 3 \times 198 +171$
I'm not sure what to do for the next step.
2. ## Re: Eulcidean Algorithim for GCD
The next step is: 198 = 1 x 171 + 27. For each new step, you take the two rightmost numbers from the previous step.
3. ## Re: Eulcidean Algorithim for GCD
ello, kamui!
Find the GCD for 198 and 765
I got as far as: $\displaystyle 765\:=\: 3 \times 198 +171$
I'm not sure what to do for the next step.
Then I guess you didn't really learn the Euclidean Algorithm.
1. Divide the larger number by the smaller.
2. Divide the divisor by the remainder.
3. Repeat step 2 until a zero remainder is achieved.
4. The GCD is the last divisor.
So we have:
. . $\displaystyle \begin{array}{ccccc}765 \div 198 &=& 3 & \text{rem.}171 \\ 198 \div 171 &=& 1 & \text{rem.}27 \\ 171 \div 27 &=& 6 & \text{rem.}9 \\ 27 \div {\color{red}9} &=& 3 & \text{rem.}0 \end{array}$
Therefore: .$\displaystyle \text{GCD}(198,765) \:=\:9$ | 371 | 1,064 | {"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} | 4.34375 | 4 | CC-MAIN-2018-39 | latest | en | 0.735837 |
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# Variation in mocks and actual score, how to form a strategy?
Author Message
Intern
Joined: 16 Apr 2018
Posts: 5
Variation in mocks and actual score, how to form a strategy? [#permalink]
### Show Tags
04 Feb 2019, 04:38
Hi
I gave my GMAT last year (April 2018) and got a 610 (VA 27, QA 47). However, I was getting significantly higher in my mocks:
GMAT official 1 49 24 640
GMAT official 2 46 33 640
Manhattan 1 50 38 720
GMAT official 3 49 35 700
Kaplan 1 49 38 710
Veritas 48 37 690
Economist 51 38 710
Manhattan 2 48 36 690
Manhattan 3 49 39 720
Manhattan 4 45 37 680
GMAT official 5 49 37 710
Now, I prepared for 1 month (Jan 2019) and gave my first GMAT official prep mock test and got a 750 (Q51, V40).
I am skeptical if this is a true representation of my GMAT ability for the following reasons:
1. In QA, out of 31 questions, 5 questions I had seen before and got correct. (Q 2,7,15,30,31)
2. In VA, I got a complete RC of 4 questions incorrect. (Q 25,26,27,28) with an average accuracy of just 57% in RCs.
Needed some advice on how should I prepare from here on. I am targetting first week of March to retake GMAT.
Joined: 25 Dec 2018
Posts: 145
Location: India
GMAT 1: 490 Q47 V13
GPA: 2.86
Re: Variation in mocks and actual score, how to form a strategy? [#permalink]
### Show Tags
04 Feb 2019, 08:15
Hi
I gave my GMAT last year (April 2018) and got a 610 (VA 27, QA 47). However, I was getting significantly higher in my mocks:
GMAT official 1 49 24 640
GMAT official 2 46 33 640
Manhattan 1 50 38 720
GMAT official 3 49 35 700
Kaplan 1 49 38 710
Veritas 48 37 690
Economist 51 38 710
Manhattan 2 48 36 690
Manhattan 3 49 39 720
Manhattan 4 45 37 680
GMAT official 5 49 37 710
Now, I prepared for 1 month (Jan 2019) and gave my first GMAT official prep mock test and got a 750 (Q51, V40).
I am skeptical if this is a true representation of my GMAT ability for the following reasons:
1. In QA, out of 31 questions, 5 questions I had seen before and got correct. (Q 2,7,15,30,31)
2. In VA, I got a complete RC of 4 questions incorrect. (Q 25,26,27,28) with an average accuracy of just 57% in RCs.
Needed some advice on how should I prepare from here on. I am targetting first week of March to retake GMAT.
Hi, while doing mocks we are in a comfortable situation, do the test in our comfort time. whereas in a real test we will feel more anxiety. section order also will impact on our score. I suggest you select an easy section first in the real test. As per your mocks score, you seem to be strong in Verbal.
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Re: Variation in mocks and actual score, how to form a strategy? [#permalink]
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04 Feb 2019, 12:41
When these types of score drops occur, the two likely "causes" involve either something that was unrealistic during practice or something that was surprising (or not accounted for) on Test Day. Since you took the GMAT 9-10 months ago, you might not remember all of the answers in detail, but If you can answer a few questions, then we should be able to figure this out:
1) On what dates did you take each of your CATs and the Official GMAT?
2) Did you take the ENTIRE CAT each time (including the Essay and IR sections)?
3) Did you take them at home?
4) Did you take them at the same time of day as when you took your Official GMAT?
5) Did you ever do ANYTHING during your CATs that you couldn't do on Test Day (pause the CAT, skip sections, take longer breaks, etc.)?
6) Did you ever take a CAT more than once? Had you seen any of the questions BEFORE (re: on a prior CAT, in an online forum or in a practice set)?
GMAT assassins aren't born, they're made,
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Re: Variation in mocks and actual score, how to form a strategy? [#permalink]
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04 Feb 2019, 20:19
Hello advikcool ... welcome to the community.
Based on the information provided .... I believe you are in right track ..... solid basics in both Q & V. In case you are not familiar with this forum.... below is some links for the ones looking for an elite score in V.
https://gmatclub.com/forum/gmat-ninja-w ... 59122.html
https://gmatclub.com/forum/four-years-t ... 40262.html
https://gmatclub.com/forum/780-q50-v47- ... l#p1462711
Hope this helps!!
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Re: Variation in mocks and actual score, how to form a strategy? [#permalink]
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04 Feb 2019, 21:42
Now, I prepared for 1 month (Jan 2019) and gave my first GMAT official prep mock test and got a 750 (Q51, V40).
I am skeptical if this is a true representation of my GMAT ability for the following reasons:
1. In QA, out of 31 questions, 5 questions I had seen before and got correct. (Q 2,7,15,30,31)
2. In VA, I got a complete RC of 4 questions incorrect. (Q 25,26,27,28) with an average accuracy of just 57% in RCs.
Needed some advice on how should I prepare from here on. I am targetting first week of March to retake GMAT.
That 750 may or may not be representative, but you shouldn't spend too much time thinking about it.
Because your practice test scores from last time show that you are capable of preparing well, I'd advise you to look into why your score dropped so much (from the high 30s to the high 20s in verbal). Also, I'd plan for two attempts this time around if I were you (just to reduce exam pressure).
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Re: Variation in mocks and actual score, how to form a strategy? [#permalink]
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05 Feb 2019, 06:09
1
Hi
I gave my GMAT last year (April 2018) and got a 610 (VA 27, QA 47). However, I was getting significantly higher in my mocks:
GMAT official 1 49 24 640
GMAT official 2 46 33 640
Manhattan 1 50 38 720
GMAT official 3 49 35 700
Kaplan 1 49 38 710
Veritas 48 37 690
Economist 51 38 710
Manhattan 2 48 36 690
Manhattan 3 49 39 720
Manhattan 4 45 37 680
GMAT official 5 49 37 710
Now, I prepared for 1 month (Jan 2019) and gave my first GMAT official prep mock test and got a 750 (Q51, V40).
I am skeptical if this is a true representation of my GMAT ability for the following reasons:
1. In QA, out of 31 questions, 5 questions I had seen before and got correct. (Q 2,7,15,30,31)
2. In VA, I got a complete RC of 4 questions incorrect. (Q 25,26,27,28) with an average accuracy of just 57% in RCs.
Needed some advice on how should I prepare from here on. I am targetting first week of March to retake GMAT.
Review everything and give special attention to verbal. Take some fresh practice tests (once in a week) to identify weaknesses. They may still not reflect your actual ability (as per GMAT) so don't worry too much about the score you get. Keep working on your weak areas.
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Re: Variation in mocks and actual score, how to form a strategy? [#permalink]
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06 Feb 2019, 06:27
Hi
I gave my GMAT last year (April 2018) and got a 610 (VA 27, QA 47). However, I was getting significantly higher in my mocks:
GMAT official 1 49 24 640
GMAT official 2 46 33 640
Manhattan 1 50 38 720
GMAT official 3 49 35 700
Kaplan 1 49 38 710
Veritas 48 37 690
Economist 51 38 710
Manhattan 2 48 36 690
Manhattan 3 49 39 720
Manhattan 4 45 37 680
GMAT official 5 49 37 710
Now, I prepared for 1 month (Jan 2019) and gave my first GMAT official prep mock test and got a 750 (Q51, V40).
I am skeptical if this is a true representation of my GMAT ability for the following reasons:
1. In QA, out of 31 questions, 5 questions I had seen before and got correct. (Q 2,7,15,30,31)
2. In VA, I got a complete RC of 4 questions incorrect. (Q 25,26,27,28) with an average accuracy of just 57% in RCs.
Needed some advice on how should I prepare from here on. I am targetting first week of March to retake GMAT.
Review everything and give special attention to verbal. Take some fresh practice tests (once in a week) to identify weaknesses. They may still not reflect your actual ability (as per GMAT) so don't worry too much about the score you get. Keep working on your weak areas.
Hi
Can you please suggest some good sources for fresh practice tests?
Thanks
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Re: Variation in mocks and actual score, how to form a strategy? [#permalink]
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07 Feb 2019, 12:53
You do bring up some valid points. I’d be curious to see how you do on a fresh MBA.com exam. Could you take another practice exam and report back with your score breakdown?
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Re: Variation in mocks and actual score, how to form a strategy? [#permalink]
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08 Feb 2019, 01:01
Firstly, I want to say that many GMAT takers experience discrepancy in score between the actual and mock test. The reason is that scoring algorithm of the official test is a bit different from mock tests. However, one thing that I can make sure is that your Quant score is quite solid and you don't have to be worried about huge variation in quant. Official quant score could be lower than the mock one, but you can cover it by figuring out the weaknesses and solving them. As your quant score is relatively good, you’d better focus on reviewing certain topics, not everything. In your case, targeting Q50-51 would be a great goal for you to hit the highest score. Q51 requires another level of quant skill and also luck. The hardest DS questions determine a score of Q51. These questions are very time consuming and still hard to get them correct, if you solve them in conventional ways. In our course, these questions are so called "CMT 3,4 (common mistake type). You can solve these hard questions very quickly and easily. These questions are really time-consuming, but it is important not to be obsessed with these questions and give up all the other questions. Time management is also essential when solving these. Also, refer to this article featured in GMAT Club "How to Achieve Q51" : https://gmatclub.com/forum/the-ultimate ... 09801.html
However, the verbal score does matter right now. The variation in verbal is more huge than that in Quant, which means that your verbal is not that solid. I think you'd better spend more time in studying verbal.
Please let us know if you have further questions.
You can reach us at info@mathrevolution.com
Good luck!
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Re: Variation in mocks and actual score, how to form a strategy? [#permalink]
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11 Feb 2019, 09:44
ScottTargetTestPrep wrote:
You do bring up some valid points. I’d be curious to see how you do on a fresh MBA exam. Could you take another practice exam and report back with your score breakdown?
Hi. Thanks for your reply. (Sorry using a different ID, wasn't able to post from that one somehow)
I took another official MBA mock. Got the following results
Score- 760 Q50, V41
3 questions in Quant, I had seen before- Got 5 incorrect out of 31
3 questions in VA and 1 passage of 3 questions (total 6), I had seen before but I didn't remember the answers.
SC- 9/12 , CR- 8/10, RC- 13/14
Please guide me how to proceed now.
Regards
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Re: Variation in mocks and actual score, how to form a strategy? [#permalink]
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11 Feb 2019, 20:46
1
760 is a great score. That being said, I think we can both agree that the score must have been inflated based on seeing a few repeat quant and verbal questions. So, we are back to the original question of why your real GMAT score does not reflect your practice tests scores, right? One possible reason is that in your preparation, you did not really learn to do what you have to do in order to score high on the actual GMAT. Rather, you picked up on some patterns that were effective in getting you relatively high scores on practice tests. Thus, going into the exam, it’s quite possible that you had a number of quant and (especially) verbal weaknesses that were exposed, and thus you ended up with a Q47/V27.
Moving forward, I think the only option is to follow a structured and through study plan, so you can individually learn and practice each verbal and quant topic, ensuring that you fill in all knowledge gaps and methodically improve your GMAT quant and verbal skills. Let me expand on this idea further.
Sentence Correction is a bit of a different animal compared to Reading Comprehension and Critical Reasoning. There are three aspects to getting correct answers to GMAT Sentence Correction questions: what you know, such as grammar rules, what you see, such as violations of grammar rules and the logic of sentence structure, and what you do, such as carefully considering each answer choice in the context of the non-underlined portion of the sentence. To drive up your Sentence Correction score, you likely will have to work on all three of those aspects. Furthermore, the likely reason that your Sentence Correction performance has not improved is that you have not been working on all three of those aspects.
Regarding what you know, first and foremost, you MUST know your grammar rules. Let's be clear, though: GMAT Sentence Correction is not just a test of knowledge of grammar rules. The reason for learning grammar rules is so that you can determine what sentences convey and whether sentences are well-constructed. In fact, in many cases, incorrect answers to Sentence Correction questions are grammatically flawless. Thus, often your task is to use your knowledge of grammar rules to determine which answer choice creates the most logical sentence meaning and structure.
This determination of whether sentences are well-constructed and logical is the second aspect of finding correct answers to Sentence Correction questions, what you see. To develop this skill, you probably have to slow way down. You won't develop this skill by spending under two minutes per question. For a while, anyway, you have to spend time with each question, maybe even ten or fifteen minutes on one question sometimes, analyzing every answer choice until you see the details that you have to see in order to choose the correct answer. As you go through the answer choices, consider the meaning conveyed by each version of the sentence. Does the meaning make sense? Even if you can tell what the version is SUPPOSED to convey, does the version really convey that meaning? Is there a verb to go with the subject? Do all pronouns clearly refer to nouns? By slowing way down and looking for these details, you learn to see what you have to see in order to clearly understand which answer to a Sentence Correction question is correct.
There is only one correct answer to any Sentence Correction question, there are clear reasons why that choice is correct and the others are not, and those reasons are not that the correct version simply "sounds right." In fact, the correct version often sounds a little off at first. That correct answers may sound a little off is not surprising. If the correct answer were always the one that sounded right, then most people most of the time would get Sentence Correction questions correct, without really knowing why the wrong answers were wrong and the correct answers were correct. So, you have to go beyond choosing what "sounds right" and learn to clearly see the logical reasons why one choice is better than all of the others.
As for the third aspect of getting Sentence Correction questions correct, what you do, the main thing you have to do is be very careful. You have to make sure that you are truly considering the structures of sentences and the meanings conveyed rather than allowing yourself to be tricked into choosing trap answers that sound right but don't convey meanings that make sense. You also have to make sure that you put some real energy into finding the correct answers. Finding the correct answer to a Sentence Correction question may take bouncing from choice to choice repeatedly until you start to see the differences between the choices that make all choices wrong except for one. Often, when you first look at the choices, only one or two seem obviously incorrect. It may take time for you to see what you have to see. Getting the right answers takes a certain work ethic. You have to be determined to see the differences and to figure out the precise reasons that one choice is correct.
To improve what you do when you answer Sentence Correction questions, seek to become aware of how you are going about answering them. Are you being careful and looking for logic and details, or are you quickly eliminating choices that sound a little off and then choosing the best of the rest? If you choose an incorrect answer, consider what you did that resulted in your arriving at that answer and what you could do differently in order to arrive at correct answers more consistently. Furthermore, see how many questions you can get correct in a row as you practice. If you break your streak by missing one, consider what you could have done differently to extend your streak.
As with your Critical Reasoning and Reading Comprehension regimens, after learning a particular Sentence Correction topic, engage in focused practice with 30 questions or more that involve that topic. As your Sentence Correction skills improve, you will then want to practice with questions that test you on skills from multiple Sentence Correction topics.
So, work on accuracy and generally finding correct answers, work on specific weaker areas one by one to make them strong areas, and when you take a practice GMAT or the real thing, take all the time per question available to do your absolute best to get right answers consistently. The GMAT is essentially a game of seeing how many right answers you can get in the time allotted. Approach the test with that conception in mind, and focus intently on the question in front of you with one goal in mind: getting a CORRECT answer.
Now, regarding quant, you seem to be in much better shape. If you can improve to at least a Q50, you should be in a pretty good spot. Since you recently scored a Q47, you clearly don’t need to go back and learn the foundations of GMAT quant; however, you still should engage in a process of linear learning and focused practice to find and fix any gaps in your quant knowledge. For example, if you are reviewing Number Properties, be sure that you practice 50 or more questions just from Number Properties: LCM, GCF, units digit patterns, divisibility, remainders, etc. As you do such practice, do a thorough analysis of each question that you don't get right. If you got a remainder question wrong, ask yourself why. Did you make a careless mistake? Did you not properly apply the remainder formula? Was there a concept you did not understand in the question? By carefully analyzing your mistakes, you will be able to efficiently fix your weaknesses and in turn improve your GMAT quant skills. Number Properties is just one example; follow this process for all quant topics.
Each time you strengthen your understanding of a topic and your skill in answering questions of a particular type, you increase your odds of hitting your score goal. You know that there are types of questions that you are happy to see and types that you would rather not see, and types of questions that you take a long time to answer correctly. Learn to more effectively answer the types of questions that you would rather not see, and make them into your favorite types.
So, work on accuracy and generally finding correct answers, work on specific weaker areas one by one to make them strong areas, and when you take a practice GMAT or the real thing, take all the time per question available to do your absolute best to get right answers consistently. The GMAT is essentially a game of seeing how many right answers you can get in the time allotted. Approach the test with that conception in mind, and focus intently on the question in front of you with one goal in mind: getting a CORRECT answer.
In order to follow the path described above, you may need some new verbal and quant materials, so take a look at the GMAT Club reviews for the best quant and verbal courses.
how to score a 700+ on the GMAT helpful.
Feel free to reach out with any further questions.
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Re: Variation in mocks and actual score, how to form a strategy? [#permalink]
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12 Feb 2019, 17:29
Very interesting. I have the opposite experience - I consistently get 540-590 on CAT exams using MPrep and my official GMAT score was 640 a day later. I wonder why the variation in the opposite direction?
Re: Variation in mocks and actual score, how to form a strategy? [#permalink] 12 Feb 2019, 17:29
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# Variation in mocks and actual score, how to form a strategy?
Moderator: DisciplinedPrep | 5,819 | 23,260 | {"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-2019-43 | latest | en | 0.90505 |
yongsun.me | 1,563,310,798,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195524879.8/warc/CC-MAIN-20190716201412-20190716223412-00042.warc.gz | 296,722,827 | 9,549 | # SunPinyin Code Tour (5)
### 5. Threading the language model
indices) to the pruned language model.
```\$ make m3_thread ./slmthread ../swap/lm_sc.3gm ../data/lm_sc.t3g ```
The goal of threading is to accelerate the searching in language model. Take trigram as an example, if we want to calculate the
probability of sentence "ABCDE。", P(S) = P(A).P(B|A).P(C|AB).P(D|BC).P(E|CD).P(<EOS>|DE), and assume
no back-off needed when evaluating all these conditional probabilities.
First, we get P(A) from level[1] (unigram) via a binary search; and find B from A's children via a binary search, to get P(B|A); and then find C from B's children via another binary search, to get P(C|AB). The next step is to calculate P(D|BC). Then we need go back to unigram to find B, and then get C from B's children, then get P(D|BC), all these done by binary searches. Obviously, the performance is low. When we evaluating P(C|AB), if we could directly locate to (B, C) from node C, we could get the search much faster.
We use (level, idx) to denote a node, the information on this node is described as (W, h', p, b, c):
• W：word id. The location of this node implies its history information.
• h'：The back-off node, described as (level', idx').
• p：p(W|h)
• b：bow(h)
• c：The start index of child node in next level.
Certainly, there is no b and c on leaf node. Now, the back-off structure becomes a graph from a tree. Its basic operation is: on
current history state hi(level,idx), by given an input word W, it transfer to another history state hj(level',
idx'), and return P(W|h). The level' on hj, not always equals to level-1. E.g., C(ABC)>0, but (B, C) is not seen in
training, then the h' on this node is the C node on level[1]. If C is not seen either, h' is then the pseudo root (level[0][0]).
Let's look at the following snippet,
``` level+1
h (level,idx) -----------> W1,p1,h'1,[b,c]
h',bow \ ... ...
\ Wi,pi,h'i,[b,c]
\ ... ...
--------> Wk,pk,h'k,[b,c]```
Given an input word W. If W is one child of h, denoted as N (level+1, idx'), its p value is just P(W|h).
• If N has children, then N is the new history node;
• If not, the h' on N becomes the new history node.
If W is not child of h, execute the same processing from h' on (level, idx), find the new history node, and times the probability
with bow(h), return it as P(W|h). Note, it's a recursive process.
Let's look at the code:
CSIMSlmWithIteration::getIdString(it, history):
The m_history member (std::vector<int>) holds the indices for each level, save the word ID on this node to passed in argument history. Method next() increase the last element in m_history (i.e., the index in current level), and call adjustIterator() to find the preceding nodes, save the indices to m_history.
CSIMSlmWithIteration::findBackOffState(n, hw, bol, bon):
Find the back-off node (h') of a specified n-gram (held in hw[1..n]), and return its level and idx. Call findState() to get the idx of hw[2..n] on level[n-1], if the index is not less than 0 then this node (n-1, idx) does have child node, return the location of h'. Otherwise, call findState() to get the idx of hw[3..n] on level[n-2], ... If the loop reach hw[n], return the pseudo root.
E.g., find the back-off node for trigram (A, B, C). Find out if (B, C) exists, if so, return (2, idx_BC). Otherwise, find out if (C) exists, if so return (1, idx_C). Otherwise, return (0, 0).
CSIMSlmWithIteration::findState(n, hw):
Find the index of specified n-gram (held in hw[1..n]) on level[n]. If it does not exist, return -1.
We also perform the compression for 32-bits floating numbers when threading. bow values are compressed to 14 bits, pr values are compressed to 16 bits. The basic idea is to count all bow (or pr) values, combine the clustering floating numbers, make the total number is under 1<<BITS_BOW (or 1<<BITS_PR); in the generated language model binary file, there are two floating
tables, we need to search the table to get the original value. I'm not so clear about this algorithm, hopes the original author
Phill Zhang to introduce more details.
Now, we get the final language model -- lm_sc.t3g. You could use tslminfo to look at the data in plain text format:
`\$ make m3_tslminfo`
` `
```./tslminfo -p -v -l ../raw/dict.utf8 ../data/lm_sc.t3g >../swap/lm_sc.t3g.arpa```
` `
`\$ tail ../swap/lm_sc.t3g.arpa`
` `
```点击 率 也 0.036363638937 (2,839922) 点击 率 最高 0.081818044186 (2,840054) 点击 率 的 0.096969485283 (2,840080) 点击 率 达到 0.036363638937 (2,840122) 点击 量 达 0.485714286566 (2,1132198) 点击 鼠标 ， 0.400000005960 (1,1) 点击 鼠标 左 0.309090882540 (2,1186378) 率队 取得 <unknown> 0.479999989271 (2,366031) 率队 在 <unknown> 0.130769222975 (2,431213) 率队 打 了 0.479999989271 (2,642183)```
We could use CThreadSlm to access or use the language model. In next section, we will use slmseg as an example, to see how we construct the lattice and search by leveraging the generated language model.
## 2 thoughts on “SunPinyin Code Tour (5)”
1. here is a Chinese sentence:
北京 ns B
交响乐团 n I
首 m O
次 q O
联袂 d O
演出 v O
(word,POS,NER_Tag)
one of my feature template could not be validated. e.g., U11:%x[-1,2], this feature template is designed for catching the previous word's NER_Tag to generate feature. maybe it is my incorrect understanding in CRF++'s tempate format. Is it about bigram?
I'll really thank u for your help.
2. Hi, Samsonchen, sorry for the late response, I think your U11 should be valid, no idea why you have trouble. Probably just a formating issue. | 1,712 | 5,583 | {"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-2019-30 | latest | en | 0.806913 |
http://www.mathspadilla.com/matI/Unit6-GeometircPlacesConics/hyperbola.html | 1,653,396,704,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662572800.59/warc/CC-MAIN-20220524110236-20220524140236-00693.warc.gz | 92,189,820 | 2,313 | # hyperbola
The hyperbola is the geometric place of all points in a plane whose difference between the distances to two fixed points, called foci, is the same constant 2a. It has two branches.
|d(P,F) - d(P,F’)| = 2a
If C(c1,c2) is the center:
You can see that a2 + b2 = c2 . Its asymptotes are: y = ±bx/a
The eccentricity of a hyperbola is the quotient: e = c/a. You can check that e > 1.
You can see hyperbolas:
- In hyperboloids and hyperbolic paraboloids, that are used in Architecture to build roofs or chimneys in nuclear power stations, for example.
- The hyperbolas are used in navigation in the LORAN (LOng RAnge Navigation) system. With this system, a boat receives signals from two knowing stations, and by measuring the difference of times, it is placed in the hyperbola corresponding to these foci. With another pair of stations, we obtain the position as the intersection of the two hyperbolas.
Exercise: Find the eccentricity and the equations of the asymptotes of the hyperbola:
Solutions: | 264 | 1,012 | {"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-2022-21 | latest | en | 0.914821 |
http://nrich.maths.org/public/leg.php?code=31&cl=3&cldcmpid=1932 | 1,475,180,672,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738661915.89/warc/CC-MAIN-20160924173741-00065-ip-10-143-35-109.ec2.internal.warc.gz | 189,990,056 | 10,147 | # Search by Topic
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##### Stage: 3 Challenge Level:
Can you arrange the digits 1,2,3,4,5,6,7,8,9 into three 3-digit numbers such that their total is close to 1500?
### The Patent Solution
##### Stage: 3 Challenge Level:
A combination mechanism for a safe comprises thirty-two tumblers numbered from one to thirty-two in such a way that the numbers in each wheel total 132... Could you open the safe?
### As Easy as 1,2,3
##### Stage: 3 Challenge Level:
When I type a sequence of letters my calculator gives the product of all the numbers in the corresponding memories. What numbers should I store so that when I type 'ONE' it returns 1, and when I type. . . .
### Postage
##### Stage: 4 Challenge Level:
The country Sixtania prints postage stamps with only three values 6 lucres, 10 lucres and 15 lucres (where the currency is in lucres).Which values cannot be made up with combinations of these postage. . . .
### I'm Eight
##### Stage: 1, 2, 3 and 4 Challenge Level:
Find a great variety of ways of asking questions which make 8.
### GOT IT Now
##### Stage: 2 and 3 Challenge Level:
For this challenge, you'll need to play Got It! Can you explain the strategy for winning this game with any target?
### Countdown
##### Stage: 2 and 3 Challenge Level:
Here is a chance to play a version of the classic Countdown Game.
### Magic Squares for Special Occasions
##### Stage: 3 and 4
This article explains how to make your own magic square to mark a special occasion with the special date of your choice on the top line.
### Top-heavy Pyramids
##### Stage: 3 Challenge Level:
Use the numbers in the box below to make the base of a top-heavy pyramid whose top number is 200.
### Magic Squares
##### Stage: 4 and 5
An account of some magic squares and their properties and and how to construct them for yourself.
### Score
##### Stage: 3 Challenge Level:
There are exactly 3 ways to add 4 odd numbers to get 10. Find all the ways of adding 8 odd numbers to get 20. To be sure of getting all the solutions you will need to be systematic. What about. . . .
### Always the Same
##### Stage: 3 Challenge Level:
Arrange the numbers 1 to 16 into a 4 by 4 array. Choose a number. Cross out the numbers on the same row and column. Repeat this process. Add up you four numbers. Why do they always add up to 34?
### Got it Article
##### Stage: 2 and 3
This article gives you a few ideas for understanding the Got It! game and how you might find a winning strategy.
### Got It
##### Stage: 2 and 3 Challenge Level:
A game for two people, or play online. Given a target number, say 23, and a range of numbers to choose from, say 1-4, players take it in turns to add to the running total to hit their target.
### Digit Sum
##### Stage: 3 Challenge Level:
What is the sum of all the digits in all the integers from one to one million?
##### Stage: 3 Challenge Level:
If you take a three by three square on a 1-10 addition square and multiply the diagonally opposite numbers together, what is the difference between these products. Why?
### Calendar Capers
##### Stage: 3 Challenge Level:
Choose any three by three square of dates on a calendar page. Circle any number on the top row, put a line through the other numbers that are in the same row and column as your circled number. Repeat. . . .
### Tis Unique
##### Stage: 3 Challenge Level:
This addition sum uses all ten digits 0, 1, 2...9 exactly once. Find the sum and show that the one you give is the only possibility.
### And So on and So On
##### Stage: 3 Challenge Level:
If you wrote all the possible four digit numbers made by using each of the digits 2, 4, 5, 7 once, what would they add up to?
### Eleven
##### Stage: 3 Challenge Level:
Replace each letter with a digit to make this addition correct.
### Constellation Sudoku
##### Stage: 4 and 5 Challenge Level:
Special clue numbers related to the difference between numbers in two adjacent cells and values of the stars in the "constellation" make this a doubly interesting problem.
### Jugs of Wine
##### Stage: 3 Challenge Level:
You have four jugs of 9, 7, 4 and 2 litres capacity. The 9 litre jug is full of wine, the others are empty. Can you divide the wine into three equal quantities?
##### Stage: 2, 3 and 4 Challenge Level:
Three dice are placed in a row. Find a way to turn each one so that the three numbers on top of the dice total the same as the three numbers on the front of the dice. Can you find all the ways to do. . . .
### Clocked
##### Stage: 3 Challenge Level:
Is it possible to rearrange the numbers 1,2......12 around a clock face in such a way that every two numbers in adjacent positions differ by any of 3, 4 or 5 hours?
### Cubes Within Cubes
##### Stage: 2 and 3 Challenge Level:
We start with one yellow cube and build around it to make a 3x3x3 cube with red cubes. Then we build around that red cube with blue cubes and so on. How many cubes of each colour have we used?
### Consecutive Negative Numbers
##### Stage: 3 Challenge Level:
Do you notice anything about the solutions when you add and/or subtract consecutive negative numbers?
### 3388
##### Stage: 3 Challenge Level:
Using some or all of the operations of addition, subtraction, multiplication and division and using the digits 3, 3, 8 and 8 each once and only once make an expression equal to 24.
### Making Sense of Positives and Negatives
##### Stage: 3
This article suggests some ways of making sense of calculations involving positive and negative numbers.
### Card Trick 2
##### Stage: 3 Challenge Level:
Can you explain how this card trick works?
### Adding and Subtracting Positive and Negative Numbers
##### Stage: 2, 3 and 4
How can we help students make sense of addition and subtraction of negative numbers?
### Make 37
##### Stage: 2 and 3 Challenge Level:
Four bags contain a large number of 1s, 3s, 5s and 7s. Pick any ten numbers from the bags above so that their total is 37.
### Pairs
##### Stage: 3 Challenge Level:
Ann thought of 5 numbers and told Bob all the sums that could be made by adding the numbers in pairs. The list of sums is 6, 7, 8, 8, 9, 9, 10,10, 11, 12. Help Bob to find out which numbers Ann was. . . .
### Crossed Ends
##### Stage: 3 Challenge Level:
Crosses can be drawn on number grids of various sizes. What do you notice when you add opposite ends?
### Cunning Card Trick
##### Stage: 3 Challenge Level:
Delight your friends with this cunning trick! Can you explain how it works?
### Playing Connect Three
##### Stage: 3 Challenge Level:
In this game the winner is the first to complete a row of three. Are some squares easier to land on than others?
### Kids
##### Stage: 3 Challenge Level:
Find the numbers in this sum
### Weights
##### Stage: 3 Challenge Level:
Different combinations of the weights available allow you to make different totals. Which totals can you make?
### 2010: A Year of Investigations
##### Stage: 1, 2 and 3
This article for teachers suggests ideas for activities built around 10 and 2010.
### Twenty20
##### Stage: 2, 3 and 4 Challenge Level:
Fancy a game of cricket? Here is a mathematical version you can play indoors without breaking any windows.
### Number Pyramids
##### Stage: 3 Challenge Level:
Try entering different sets of numbers in the number pyramids. How does the total at the top change?
### Pole Star Sudoku 2
##### Stage: 3 and 4 Challenge Level:
This Sudoku, based on differences. Using the one clue number can you find the solution?
### Countdown Fractions
##### Stage: 3 and 4 Challenge Level:
Here is a chance to play a fractions version of the classic Countdown Game. | 2,260 | 9,213 | {"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} | 3.390625 | 3 | CC-MAIN-2016-40 | longest | en | 0.803255 |
http://oeis.org/A308202 | 1,571,175,218,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986660323.32/warc/CC-MAIN-20191015205352-20191015232852-00370.warc.gz | 141,948,700 | 3,769 | This site is supported by donations to The OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A308202 Number of integer-sided triangles with perimeter n such that the length of the smallest side is coprime to n. 1
0, 0, 1, 0, 1, 0, 2, 0, 2, 1, 4, 0, 5, 1, 4, 2, 8, 2, 10, 1, 7, 4, 14, 4, 13, 6, 13, 4, 21, 3, 24, 9, 17, 11, 20, 7, 33, 13, 24, 13, 40, 9, 44, 13, 24, 20, 52, 15, 48, 19, 40, 20, 65, 18, 50, 27, 49, 33, 80, 21, 85, 37, 50, 40, 70, 24, 102, 37, 71, 32, 114 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,7 LINKS FORMULA a(n) = Sum_{k=1..floor(n/3)} Sum_{i=k..floor((n-k)/2)} sign(floor((i+k)/(n-i-k+1))) * [gcd(k,n) = 1], where [] is the Iverson bracket. MATHEMATICA Table[Sum[Sum[Floor[1/GCD[k, n]] Sign[Floor[(i + k)/(n - i - k + 1)]], {i, k, Floor[(n - k)/2]}], {k, Floor[n/3]}], {n, 100}] CROSSREFS Cf. A308074. Sequence in context: A295689 A305804 A053470 * A278648 A029181 A261426 Adjacent sequences: A308199 A308200 A308201 * A308203 A308204 A308205 KEYWORD nonn AUTHOR Wesley Ivan Hurt, May 15 2019 STATUS approved
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Last modified October 15 17:24 EDT 2019. Contains 328037 sequences. (Running on oeis4.) | 579 | 1,433 | {"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} | 2.78125 | 3 | CC-MAIN-2019-43 | latest | en | 0.646421 |
http://ciphersbyritter.com/NEWS4/RANDSND.HTM | 1,516,331,406,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887729.45/warc/CC-MAIN-20180119030106-20180119050106-00447.warc.gz | 71,171,888 | 23,026 | # Random Numbers from a Sound Card
## A Ciphers By Ritter Page
Everybody has a sound card, so we all have a physically-random noise generator -- a source of absolute randomness -- right? A discussion starting with sound cards recording noise, and ending with theories of randomness.
### Contents
• 1999-01-25 Mok-Kong Shen: "There are tests for statistical quality, e.g. Maurer's universal statistical test. I am ignorant of tests for crypto quality."
• 1999-01-25 Paul Rubin: "We just test the total amount of energy in the audio to make sure the mic isn't dead. We expect that the raw audio will have lots of correlation, so we run it through a hash function or block cipher...."
• 1999-01-26 Nathan Kennedy: "I tune a cheap AM radio to a loud static channel, and wire that into the mic port."
• 1999-01-26 Cuykens Anthony: "I do remember of a way a teacher told me to generate a 'true' random generator." "You select some measurable information about your noise source.... Then you sample you source at fixed interval.... For each coosen information, if it is higher than the same info at the last sample, the output is one, otherwize the result is zero."
• 1999-01-26 Jon Haugsand: "Better is to make two samples not too close in time and output a one if the first is higher, and output a zero if the second is higher."
• 1999-01-26 randombit@my-dejanews.com: "Your teacher was just trying to communicate that you need a physical signal to start with...."
• 1999-01-27 R. Knauer: "You got a computer that can roll dice...?"
• 1999-01-26 Nathan Kennedy: "The best approach is not to waste any entropy, and just feed the raw data to a hungry hash function...."
• 1999-01-26 David Ross: "I create one rotor at a time...."
• 1999-01-26 Frank Gifford: "You might want to look into generating a simple rotor and use the random numbers to swap entries in the rotor."
• 1999-01-26 Paul Rubin: "Don't even think of using raw soundblaster output as actual random data rather than just as an entropy source.
• 1999-01-26 Kazak, Boris: "...HH is a Housing (just a glass or plastic bottle), OO are... 100-200 peas or beans, MM is a Microphone. Now if we start rotating the Housing...."
• 1999-01-27 R. Knauer: "Any recommendations on which hash function or something similar?"
• 1999-01-27 randombit@my-dejanews.com: "In order to know how many bits you have to distill with MD5, or any other hash, you need to measure your entropy/raw bit."
• 1999-01-27 randombit@my-dejanews.com: "FM has better hiss. I leave it to RF folks to explain why."
• 1999-01-27 David Ross: "In my case, the FM limiter & detector puts out a waveshape which is more linear in the voltage range where I'm digitizing."
• 1999-01-26 R. Knauer: "It is a fundamental precept of crytpography that randomness is a property of how a number is generated, not a property of a number itself."
• 1999-01-26 Mok-Kong Shen: "...OTP presuppose (absolutely) true randomness and there is no way of determining that in practice."
• 1999-01-26 R. Knauer: "For all practical purposes the OTP is proveably secure."
• 1999-01-27 Mok-Kong Shen: "I am not against having something ideal and perfect as a standard for approximation.... But to say there IS... something perfect can be misleading."
• 1999-01-27 R. Knauer: "Does a Perfect Circle EXIST?"
• 1999-01-27 Mok-Kong Shen: "If the word 'IS' is employed in a context without the connotation of 'EXISTS' then it is NOT misleading, otherwise it IS misleading."
• 1999-01-27 R. Knauer: "It all depends on what the meaning of the word 'is' is."
• 1999-01-27 Mok-Kong Shen: "A word can have a multitude of meanings. I was prudent enough to put the parentheses above to make sure that there could be no misunderstanding."
• 1999-01-27 R. Knauer: "You must be a mathematician."
• 1999-01-27 Medical Electronics Lab: "We can certainly build a TRNG which is perfect in any measureable sense."
• 1999-01-27 Tony T. Warnock: "In limit (infinitely long sequences) both the complexity based and the frequency (statistical) based definitions of random are equivalent...."
• 1999-01-27 R. Knauer: "Those things don't measure the crypto-grade randomness of finite numbers at all."
• 1999-01-28 Medical Electronics Lab: "So you need an infinite sequence of bits to prove that something is crypto-grade random, yes?"
• 1999-01-28 R. Knauer: "The only way you can prove the crypto-grade randomness of a finite number is to consider the method of generation."
• 1999-01-29 Mok-Kong Shen: "Ah! Finally one knows exactly what the term 'crypto-grade random numbers' you employ means: These are DEFINED to be the output from a hardware generator."
• 1999-01-29 R. Knauer: "A TRNG is not a True Random Number Generator just because it is a hardware device."
• 1999-01-29 Patrick Juola: "Not all hardware generators are TRNG."
• 1999-01-29 Tony T. Warnock: "Complexity of computation and statistical properties are only equivalent in the limit of infnitely many infinitely long sequences."
• 1999-01-30 R. Knauer: "But I point out that computational complexity has nothing fundamental to do with crypto-grade randomness, nor QM."
• 1999-01-30 R. Knauer: "How come you state that Champernowne's number has an 'excess of ones over zeros'?"
• 1999-01-30 Trevor Jackson, III: "Because all the leading zeros are suppressed."
• 1999-01-31 R. Knauer: "Can you elaborate with an example."
• 1999-01-26 Patrick Juola: "...you'll probably not build a 'perfect' OTP in practice...."
• 1999-01-26 Paul L. Allen: "I'd be *extremely* worried about a sound card (particularly with little in the way of input) picking up mains power hum and radiated noise from signal lines in the computer." "Noise diodes strike me as being safer."
• 1999-01-26 randombit@my-dejanews.com: "I experimented with parity-of-N bits, and used Maurer's Universal statistical test for RNGs to measure the entropy. When you distill enough, the entropy reaches its expected value."
• 1999-01-26 David Ross: "If it has a 1 Volt peak amplitude and if I digitize at a constant rate, won't 1/2 of my samples yield a value of either >+.707 Volt or < -.707 Volt?"
• 1999-01-27 randombit@my-dejanews.com: "The threshold of your detector, which decides whether a voltage is to be called a 0 or a 1...."
• 1999-01-28 randombit@my-dejanews.com: "Eve does not know the local, instantaneious electromagnetic conditions around my receiver, nor does she know what my local electronics are doing."
• 1999-01-28 Mok-Kong Shen: "I believe that under certain real circumstances obtaining bit sequences from software can be justified."
• 1999-01-28 patix: "Haw should we test hardawre random generator to "be shoure" that it is realy some haw random ?"
• 1999-01-29 Patrick Juola: "No, but Mike does. He's fully capable of broadcasting (known) noise of some sort near your site."
• 1999-01-26 Paul L. Allen: "I can never find a URL for that when I need it."
```
Subject: Re: Random numbers from a sound card?
Date: Mon, 25 Jan 1999 20:11:33 +0100
From: Mok-Kong Shen <mok-kong.shen@stud.uni-muenchen.de>
Message-ID: <36ACC1E5.90C4C2BC@stud.uni-muenchen.de>
References: <36acb8b1.5374650@news.willapabay.org>
Newsgroups: sci.crypt
Lines: 15
David Ross wrote:
>
> How would you test the 'quality' of the generated random number
> stream?
There are tests for statistical quality, e.g. Maurer's universal
statistical test. I am ignorant of tests for crypto quality.
I guess the issue of cryptological strength is inherently fuzzy
and not entirely separable from subjectivity and concepts like
confidence intervals, i.e. no security can be claimed on an absolute
scale in practice. But experts might refute my un-knowledgeable
assertions.
M. K. Shen
Subject: Re: Random numbers from a sound card?
Date: Mon, 25 Jan 1999 21:21:09 GMT
From: phr@netcom.com (Paul Rubin)
Message-ID: <phrF64wn9.J9M@netcom.com>
References: <36acb8b1.5374650@news.willapabay.org>
Newsgroups: sci.crypt
Lines: 20
In article <36acb8b1.5374650@news.willapabay.org>,
David Ross <ross@hypertools.com> wrote:
> Has anyone had success using a sound card (like a Sound Blaster) to
>generate streams of random numbers?
source from one of the sites mentioned there.
> What sort of audio source would you suspect would be the best to use
>in generating random numbers?
We ask the user to blow into the microphone to make noise, IIRC.
> How would you test the 'quality' of the generated random number
>stream?
We just test the total amount of energy in the audio to make sure
the mic isn't dead. We expect that the raw audio will have lots
of correlation, so we run it through a hash function or block cipher;
I don't remember the details.
Subject: Re: Random numbers from a sound card?
Date: Tue, 26 Jan 1999 07:09:28 +0800
From: Nathan Kennedy <blaaf@hempseed.com>
Message-ID: <36ACF9A8.CF6CF898@hempseed.com>
References: <36acb8b1.5374650@news.willapabay.org>
Newsgroups: sci.crypt
Lines: 32
David Ross wrote:
>
> Has anyone had success using a sound card (like a Sound Blaster) to
> generate streams of random numbers?
Sure. My favorite (T)RNG method.
>
> What sort of audio source would you suspect would be the best to use
> in generating random numbers?
I tune a cheap AM radio to a loud static channel, and wire that into the
mic port.
> How would you test the 'quality' of the generated random number
> stream?
Of course, you can't test the 'quality' by looking at the random numbers
generated. You need to estimate the entropy of your source, and of course
it's always going to be an estimate, you can almost never prove it.
What I did, was compress it, multiply my hash size by the compression ratio
by a fudge factor of 10. Then I would hash that much data, and assumed
that the result was very close to 100% entropy. This is rather paranoid
and slow though. If you don't need 100% entropy just go ahead and
continually sample /dev/audio for data and use it as entropy for a PRNG,
and sample the PRNG as often as you like. The quality should still be
excellent... As long as you've got >128 bits of entropy total and the PRNG
does its job, the result should be quite secure as long as nothing gets
compromised.
Nate
Subject: Re: Random numbers from a sound card?
Date: Tue, 26 Jan 1999 09:41:42 +0100
From: Cuykens Anthony <cuykens.a@protonworld.com>
References: <36ACF9A8.CF6CF898@hempseed.com>
Newsgroups: sci.crypt
Lines: 53
Hi,
I do remember of a way a teacher told me to generate a "true" random
your case, lets says the frequence, the loudness, ...). Then you sample you
source at fixed interval and you check all your informations. For each coosen
information, if it is higher than the same info at the last sample, the output
is one, otherwize the result is zero. At each sample, this method will give you
one bit per criterion.
This is just an idea, what does guru think of it?
Nathan Kennedy wrote:
> David Ross wrote:
> >
> > Has anyone had success using a sound card (like a Sound Blaster) to
> > generate streams of random numbers?
>
> Sure. My favorite (T)RNG method.
>
> >
> > What sort of audio source would you suspect would be the best to use
> > in generating random numbers?
>
> I tune a cheap AM radio to a loud static channel, and wire that into the
> mic port.
>
> > How would you test the 'quality' of the generated random number
> > stream?
>
> Of course, you can't test the 'quality' by looking at the random numbers
> generated. You need to estimate the entropy of your source, and of course
> it's always going to be an estimate, you can almost never prove it.
>
> What I did, was compress it, multiply my hash size by the compression ratio
> by a fudge factor of 10. Then I would hash that much data, and assumed
> that the result was very close to 100% entropy. This is rather paranoid
> and slow though. If you don't need 100% entropy just go ahead and
> continually sample /dev/audio for data and use it as entropy for a PRNG,
> and sample the PRNG as often as you like. The quality should still be
> excellent... As long as you've got >128 bits of entropy total and the PRNG
> does its job, the result should be quite secure as long as nothing gets
> compromised.
>
> Nate
--
Anthony Cuykens
Subject: Re: Random numbers from a sound card?
Date: 26 Jan 1999 10:18:04 +0100
From: Jon Haugsand <haugsand@procyon.nr.no>
Message-ID: <yzo90eqv1df.fsf@procyon.nr.no>
Newsgroups: sci.crypt
Lines: 26
* Cuykens Anthony
| I do remember of a way a teacher told me to generate a "true" random
| generator. You select some measurable information about your noise source (in
| your case, lets says the frequence, the loudness, ...). Then you sample you
| source at fixed interval and you check all your informations. For each coosen
| information, if it is higher than the same info at the last sample, the output
| is one, otherwize the result is zero. At each sample, this method will give you
| one bit per criterion.
I am not sure that this will be random enough. I would guess that as
the number of concectutive ones increases, the probability to get a
zero the next time also increases. Better is to make two samples not
too close in time and output a one if the first is higher, and output
a zero if the second is higher.
Better still is to measure some quantity twice (e.g. sound level) and
use the least significant bit and output a one if you measure 10,
output a zero if you get 01. If you get 11 or 00, discard those.
--
Jon Haugsand
Norwegian Computing Center, <http://www.nr.no/engelsk/>
<mailto:haugsand@nr.no> Pho: +47 22852608 / +47 22852500,
Fax: +47 22697660, Pb 114 Blindern, N-0314 OSLO, Norway
Subject: Re: Random numbers from a sound card?
Date: Tue, 26 Jan 1999 20:47:01 GMT
From: randombit@my-dejanews.com
Message-ID: <78l9k2\$dsh\$1@nnrp1.dejanews.com>
References: <yzo90eqv1df.fsf@procyon.nr.no>
Newsgroups: sci.crypt
Lines: 14
In article <yzo90eqv1df.fsf@procyon.nr.no>,
Jon Haugsand <haugsand@procyon.nr.no> wrote:
> * Cuykens Anthony
> | I do remember of a way a teacher told me to generate a "true" random
> | generator.
Your teacher was just trying to communicate that you need a physical
unpredictability (aka randomness).
Computers that can roll dice are not Turing machines.
-----------== Posted via Deja News, The Discussion Network ==----------
Subject: Re: Random numbers from a sound card?
Date: Wed, 27 Jan 1999 01:33:31 GMT
From: rcktexas@ix.netcom.com (R. Knauer)
Message-ID: <36ae6c8a.51870215@nntp.ix.netcom.com>
References: <78l9k2\$dsh\$1@nnrp1.dejanews.com>
Newsgroups: sci.crypt
Lines: 14
On Tue, 26 Jan 1999 20:47:01 GMT, randombit@my-dejanews.com wrote:
>Computers that can roll dice are not Turing machines.
You got a computer that can roll dice - a completetly
non-deterministic machine that can compute algorithmically?
Bob Knauer
"An honest man can feel no pleasure in the exercise of power over
his fellow citizens."
--Thomas Jefferson
Subject: Re: Random numbers from a sound card?
Date: Tue, 26 Jan 1999 17:19:57 +0800
From: Nathan Kennedy <blaaf@hempseed.com>
Newsgroups: sci.crypt
Lines: 30
Cuykens Anthony wrote:
>
> Hi,
>
> I do remember of a way a teacher told me to generate a "true" random
> generator. You select some measurable information about your noise source (in
> your case, lets says the frequence, the loudness, ...). Then you sample you
> source at fixed interval and you check all your informations. For each coosen
> information, if it is higher than the same info at the last sample, the output
> is one, otherwize the result is zero. At each sample, this method will give you
> one bit per criterion.
>
> This is just an idea, what does guru think of it?
>
That's just one way of converting a raw sampled value into a bit stream...
It doesn't assure any randomness. It would probably be very predictable in
the short term, and likely biased as well.
Certainly, applying this to soundcard sampled data is little better than
the raw output of /dev/audio.
The best approach is not to waste any entropy, and just feed the raw data
to a hungry hash function, which will process it into an unbiased output.
The hash never has more entropy than what it is seeded with, however!
Bruce Schneier has an excellent paper on PRNGs on his site
(www.counterpane.com), which could serve as a good introduction.
Nate
Subject: Re: Random numbers from a sound card?
Date: Tue, 26 Jan 1999 21:26:38 GMT
From: ross@hypertools.com (David Ross)
Message-ID: <36ae3069.15236697@news.willapabay.org>
References: <78l32u\$fr4@trebuchet.eng.us.uu.net>
<36ae089d.5049458@news.willapabay.org>
Newsgroups: sci.crypt
Lines: 35
On 26 Jan 1999 13:55:26 -0500, giff@eng.us.uu.net (Frank Gifford)
wrote:
>In article <36ae089d.5049458@news.willapabay.org>,
>David Ross <ross@hypertools.com> wrote:
>> Have tried something very similar to that. I am attempting to
>>create a rotortable of all 256 byte values placed in 'random' order,
>>but the (8 bit) SoundBlaster seems reluctant to produce a 0xC0 byte.
>>I infer this because in over 80% of the rotortables I create, 0xC0 is
>>the last table entry.
>
>How are you creating the tables? I would assume that since you are creating
>rotors, that each value appears in the rotor exactly once. Are you swapping
>values or some other method? Personally, I would suspect your creation
Giff -
I create one rotor at a time, waiting for each one of the 256
bytevalues to come in from the SoundBlaster before I go on to the next
rotor. A very simple piece of code, done in assembly language.
- several bytes commonly occurred toward the end of each rotor, but
0xC0 was by far the most popular as the last byte. (Incidentally, I'm
using an ES1688 sound chip set up to emulate a SoundBlaster.)
- the process of rotor creation took _much_ more time than I had
estimated.
- using a 'small' (50+) rotor encryption scheme to encrypt the
SoundBlaster bytes before sending them to the rotor sorting routine
sped up the process by about 20X.
David Ross ross@hypertools.com
Subject: Re: Random numbers from a sound card?
Date: 26 Jan 1999 17:30:58 -0500
From: giff@eng.us.uu.net (Frank Gifford)
Message-ID: <78lfn2\$g6t@trebuchet.eng.us.uu.net>
References: <36ae3069.15236697@news.willapabay.org>
Newsgroups: sci.crypt
Lines: 44
In article <36ae3069.15236697@news.willapabay.org>,
David Ross <ross@hypertools.com> wrote:
> I create one rotor at a time, waiting for each one of the 256
>bytevalues to come in from the SoundBlaster before I go on to the next
>rotor. A very simple piece of code, done in assembly language.
Does this mean that (for a given rotor) you loop through rotor positions
and get a value from SB that has not been used yet, and then do the
remaining positions that way? So if SB gives you the same byte several
times in a row that you ignore the duplicates?
> - several bytes commonly occurred toward the end of each rotor, but
>0xC0 was by far the most popular as the last byte.
Assuming I understand your process, that means 0xC0 is very unlikely in
a byte stream. In that case, SB in your set up is probably a bad choice
for a random number generator. You may have to investigate your set up
a bit more. Not enough static for input?
> - the process of rotor creation took _much_ more time than I had
>estimated.
If you are doing it the way above, then yes indeed. When you get to the last
two values, you are waiting for either of them to be generated so you can
fill in the last piece of the rotor. You might want to look into generating
a simple rotor and use the random numbers to swap entries in the rotor. Then
you can simplify your code and generate a new rotor in a known amount of time.
> - using a 'small' (50+) rotor encryption scheme to encrypt the
>SoundBlaster bytes before sending them to the rotor sorting routine
>sped up the process by about 20X.
Does this mean you take the values from SB, pipe through your rotors, and
then use the results to create/modify a rotor? In that case, this may
be the source of weird results.
I would recommend checking your set up of SB to see whether the bytes it
generates directly is really 'random' and not biased.
-Giff
--
giff@uu.net Too busy for a .sig
Subject: Re: Random numbers from a sound card?
Date: Tue, 26 Jan 1999 22:20:11 GMT
From: phr@netcom.com (Paul Rubin)
Message-ID: <phrF66u1o.JBE@netcom.com>
References: <36ae089d.5049458@news.willapabay.org>
Newsgroups: sci.crypt
Lines: 24
In article <36ae089d.5049458@news.willapabay.org>,
David Ross <ross@hypertools.com> wrote:
>Nathan Kennedy wrote:
>
>> > What sort of audio source would you suspect would be the best to use
>> > in generating random numbers?
>>
>> I tune a cheap AM radio to a loud static channel, and wire that into the
>> mic port.
> Have tried something very similar to that. I am attempting to
>create a rotortable of all 256 byte values placed in 'random' order,
>but the (8 bit) SoundBlaster seems reluctant to produce a 0xC0 byte.
>I infer this because in over 80% of the rotortables I create, 0xC0 is
>the last table entry.
>
> I'd guess that the 'consumer-grade' A->D & D->A converters used
>in common sound cards are susceptible to all sorts of troubles like
>this, i.e. missing codes and/or monotonicity problems.
Don't even think of using raw soundblaster output as actual random
data rather than just as an entropy source. Even if the a/d converter
is terrific, it's still likely to pick up correlated noise from
various sources in the PC. Run the audio bits through a cryptographic
hash function or something similar before using it.
Subject: Re: Random numbers from a sound card?
Date: Tue, 26 Jan 1999 18:31:52 -0500
From: "Kazak, Boris" <bkazak@erols.com>
References: <phrF66u1o.JBE@netcom.com>
Newsgroups: sci.crypt
Lines: 56
Paul Rubin wrote:
>
> In article <36ae089d.5049458@news.willapabay.org>,
> David Ross <ross@hypertools.com> wrote:
> >Nathan Kennedy wrote:
> >
> >> > What sort of audio source would you suspect would be the best to use
> >> > in generating random numbers?
> >>
> >> I tune a cheap AM radio to a loud static channel, and wire that into the
> >> mic port.
> > Have tried something very similar to that. I am attempting to
> >create a rotortable of all 256 byte values placed in 'random' order,
> >but the (8 bit) SoundBlaster seems reluctant to produce a 0xC0 byte.
> >I infer this because in over 80% of the rotortables I create, 0xC0 is
> >the last table entry.
> >
> > I'd guess that the 'consumer-grade' A->D & D->A converters used
> >in common sound cards are susceptible to all sorts of troubles like
> >this, i.e. missing codes and/or monotonicity problems.
>
> Don't even think of using raw soundblaster output as actual random
> data rather than just as an entropy source. Even if the a/d converter
> is terrific, it's still likely to pick up correlated noise from
> various sources in the PC. Run the audio bits through a cryptographic
> hash function or something similar before using it.
-----------------------------------------
Let's be practical...
It is perfectly possible to use the sound card for random number
generation if we come up with a way to provide a random acoustic
input on its microphone connector.
Consider such a simple system:
HHHHHHHHHHHHHHHH
HH H MMM
HH H MMMMM
HH OOOOOOOOOO H MMMMM
HH OOOOOOOOOOOO H MMM
HHHHHHHHHHHHHHHH
where HH is a Housing (just a glass or plastic bottle), OO are Objects
(a pseudo-scientific baptism for 100-200 peas or beans), MM is a
Microphone.
Now if we start rotating the Housing around its horizontal axis,
the Objects will produce a loud Random Rattle, and the Microphone will
transmit this rattle to the sound card. My questions are:
How many Objects are needed and what must be the speed of
rotation that will assure the True Randomness?
What estimates can be given for Degree of Correlation and
for Period of Repetition, depending on the system parameters?
The System is not patented, it is hereby placed in the public
domain.
Respectfully BNK
Subject: Re: Random numbers from a sound card?
Date: Wed, 27 Jan 1999 01:34:38 GMT
From: rcktexas@ix.netcom.com (R. Knauer)
Message-ID: <36ae6d00.51988195@nntp.ix.netcom.com>
References: <phrF66u1o.JBE@netcom.com>
Newsgroups: sci.crypt
Lines: 14
On Tue, 26 Jan 1999 22:20:11 GMT, phr@netcom.com (Paul Rubin) wrote:
>Run the audio bits through a cryptographic
>hash function or something similar before using it.
Any recommendations on which hash function or something similar?
Bob Knauer
"An honest man can feel no pleasure in the exercise of power over
his fellow citizens."
--Thomas Jefferson
Subject: Re: Random numbers from a sound card?
Date: Wed, 27 Jan 1999 18:31:02 GMT
From: randombit@my-dejanews.com
Message-ID: <78nm16\$cvg\$1@nnrp1.dejanews.com>
References: <36ae6d00.51988195@nntp.ix.netcom.com>
Newsgroups: sci.crypt
Lines: 46
In article <36ae6d00.51988195@nntp.ix.netcom.com>,
rcktexas@ix.netcom.com wrote:
> On Tue, 26 Jan 1999 22:20:11 GMT, phr@netcom.com (Paul Rubin) wrote:
>
> >Run the audio bits through a cryptographic
> >hash function or something similar before using it.
>
> Any recommendations on which hash function or something similar?
>
> Bob Knauer
You'll be told to use an established one, say MD5. Doesn't matter
too much.
But what you *must* be careful with is this:
A sample output of MD5 will look random -that's its job :-)
It will appear to have full entropy.
In order to know how many bits you have to distill with MD5,
or any other hash, you need to measure your entropy/raw bit.
Then you can convince skeptics that you are driving the hash
with enough entropy.
If your hash function is *not* a crypto-strong one, then
you can directly measure the quality of its output ---since its not
crypto strong, by definition its output (when given
very redundant input) will be crappy. Parity-of-N
has this property. When N is large enough, for a given
raw-entropy-rate, the parity output is indistinguishable from
crypto-strong (ie, uniformly distributed) output. When N
is insufficient, you can see it with an entropy measure.
-----
"Many tame, conformist types felt the need to describe anti-social actions as
'sick'." -Ted K
-----------== Posted via Deja News, The Discussion Network ==----------
Subject: Re: Random numbers from a sound card?
Date: Wed, 27 Jan 1999 18:18:17 GMT
From: randombit@my-dejanews.com
Message-ID: <78nl8u\$ca5\$1@nnrp1.dejanews.com>
References: <36ae089d.5049458@news.willapabay.org>
Newsgroups: sci.crypt
Lines: 37
In article <36ae089d.5049458@news.willapabay.org>,
ross@hypertools.com (David Ross) wrote:
> Nathan Kennedy wrote:
>
> > > What sort of audio source would you suspect would be the best to use
> > > in generating random numbers?
> >
> > I tune a cheap AM radio to a loud static channel, and wire that into the
> > mic port.
FM has better hiss. I leave it to RF folks to explain why. Probably
because FM listens to wider chunks of the aether than AM. Or because
of the design of FM receivers amplifying component-noise more?
> Have tried something very similar to that. I am attempting to
> create a rotortable of all 256 byte values placed in 'random' order,
> but the (8 bit) SoundBlaster seems reluctant to produce a 0xC0 byte.
> I infer this because in over 80% of the rotortables I create, 0xC0 is
> the last table entry.
>
> I'd guess that the 'consumer-grade' A->D & D->A converters used
> in common sound cards are susceptible to all sorts of troubles like
> this, i.e. missing codes and/or monotonicity problems.
>
> David Ross ross@hypertools.com
Yes. You have to assume everything is imperfect. Your raw source
is biassed, your whole amplification/detection (ie, digitization) chain
has got holes in it (in either time or frequency domains), and
you're in a fun digital-switching environment for extra bonus
problems.
This is why measurement is so much better than handwaving.
-----------== Posted via Deja News, The Discussion Network ==----------
Subject: Re: Random numbers from a sound card?
Date: Wed, 27 Jan 1999 20:47:32 GMT
From: ross@hypertools.com (David Ross)
Message-ID: <36af78e0.12951524@news.willapabay.org>
References: <78nl8u\$ca5\$1@nnrp1.dejanews.com>
Newsgroups: sci.crypt
Lines: 22
On Wed, 27 Jan 1999 18:18:17 GMT, randombit@my-dejanews.com wrote:
>In article <36ae089d.5049458@news.willapabay.org
>ross@hypertools.com (David Ross) wrote:
> What sort of audio source would you suspect would be the best to use
>in generating random numbers?
>
>>> I tune a cheap AM radio to a loud static channel, and wire that into the
>>> mic port.
>
>>FM has better hiss. I leave it to RF folks to explain why. Probably
>>because FM listens to wider chunks of the aether than AM. Or because
>>of the design of FM receivers amplifying component-noise more?
I see this 'better hiss' quality too, and suspect that it is due to
the FM detection scheme.
In my case, the FM limiter & detector puts out a waveshape which is
more linear in the voltage range where I'm digitizing. This yields a
flatter distribution of A->D output bytes but is a bit slower.
David Ross ross@hypertools.com
Subject: Re: Random numbers from a sound card?
Date: Tue, 26 Jan 1999 00:28:16 GMT
From: rcktexas@ix.netcom.com (R. Knauer)
References: <36ACC1E5.90C4C2BC@stud.uni-muenchen.de>
Newsgroups: sci.crypt
Lines: 35
On Mon, 25 Jan 1999 20:11:33 +0100, Mok-Kong Shen
<mok-kong.shen@stud.uni-muenchen.de> wrote:
>> How would you test the 'quality' of the generated random number
>> stream?
>There are tests for statistical quality, e.g. Maurer's universal
>statistical test. I am ignorant of tests for crypto quality.
That's because there aren't any.
It is a fundamental precept of crytpography that randomness is a
property of how a number is generated, not a property of a number
itself.
>I guess the issue of cryptological strength is inherently fuzzy
Not really. The OTP system is proveably secure.
>and not entirely separable from subjectivity and concepts like
>confidence intervals, i.e. no security can be claimed on an absolute
>scale in practice. But experts might refute my un-knowledgeable
>assertions.
If someone tells you that he can demonstrate that a given number is
crypto-grade random without considering the way it is generated, he is
making a fundamental error, one of the most widespread errors in
cryptography.
Bob Knauer
"An honest man can feel no pleasure in the exercise of power over
his fellow citizens."
--Thomas Jefferson
Subject: Re: Random numbers from a sound card?
Date: Tue, 26 Jan 1999 14:19:59 +0100
From: Mok-Kong Shen <mok-kong.shen@stud.uni-muenchen.de>
Newsgroups: sci.crypt
Lines: 14
R. Knauer wrote:
> >I guess the issue of cryptological strength is inherently fuzzy
>
> Not really. The OTP system is proveably secure.
Once again I assert that this is a (for all practical purposes)
useless fact, because OTP presuppose (absolutely) true randomness
and there is no way of determining that in practice. I suppose
(with my meager knowledge of physics) this is almost the same as
saying at at at 0 Kelvin you can halt the motions of all atoms (but
you can't get to 0 Kelvin, only very close to it).
M. K. Shen
Subject: Re: Random numbers from a sound card?
Date: Tue, 26 Jan 1999 19:11:42 GMT
From: rcktexas@ix.netcom.com (R. Knauer)
Message-ID: <36ae0cb6.27338280@nntp.ix.netcom.com>
Newsgroups: sci.crypt
Lines: 68
On Tue, 26 Jan 1999 14:19:59 +0100, Mok-Kong Shen
<mok-kong.shen@stud.uni-muenchen.de> wrote:
>> Not really. The OTP system is proveably secure.
>Once again I assert that this is a (for all practical purposes)
>useless fact,
Sorry, but that's nonsense.
For all practical purposes the OTP is proveably secure. That means
that you can build an OTP system that is secure to within a level of
precision can be made arbitrarily small.
To do that you must build a physical device which can generate all
possible sequences of a given finite length equiprobably. That is
possible using quantum mechanical processes and good electronic
design.
The hot line between Washington and Moscow is (supposedly) protected
by an OTP. Conversations on that line can be tapped and interfered
with in principle by anyone close enough to the equipment. Do you
think the two most dangerous govts in the world would trust the fate
of Planet Earth to an insecure communications link? Hardly.
Nothing we humans build is Perfect, but we are able to build things
that are very damn close to Perfect. We can build TRNGs that are
perfect enough to send messages which would take more energy to
analyze than is available in the Universe.
How much more Perfect do you want, even in a practical sense?
>because OTP presuppose (absolutely) true randomness
>and there is no way of determining that in practice.
Sure there is. Just look at how the numbers are being generated. That
will tell you if they are random.
>I suppose
>(with my meager knowledge of physics) this is almost the same as
>saying at at at 0 Kelvin you can halt the motions of all atoms
You are not aware of the so-called "zero point" vacuum fluctuations
which persist even at 0 Kelvin. If all motion stopped at 0 Kelvin, the
Universe would cease to exist - no photons, no particles, no forces -
nothing.
>(but you can't get to 0 Kelvin, only very close to it).
You can get exceedingly close to it, like one milli-degree close to
it. That's one thousandth of a degree close to it. How much closer
would you want to get to be closer than very close to it?
Is calculus impossible because numbers can never actually reach the
limit required to calculate a derivative or an integral? People in the
17th century, when Newton and Leibnitz invented calculus, thought
calculus was wrong because those limits could never be reached in a
"practical" sense. Yet calculus went on being correct despite them.
And crypto-grade randomness goes on being correct in a very practical
sense too, despite the lack of perfection in a practical sense.
Bob Knauer
"An honest man can feel no pleasure in the exercise of power over
his fellow citizens."
--Thomas Jefferson
Subject: Re: Random numbers from a sound card?
Date: Wed, 27 Jan 1999 14:56:34 +0100
From: Mok-Kong Shen <mok-kong.shen@stud.uni-muenchen.de>
Message-ID: <36AF1B12.FBD87AB5@stud.uni-muenchen.de>
References: <36ae0cb6.27338280@nntp.ix.netcom.com>
Newsgroups: sci.crypt
Lines: 28
R. Knauer wrote:
>
> Nothing we humans build is Perfect, but we are able to build things
> that are very damn close to Perfect. We can build TRNGs that are
> perfect enough to send messages which would take more energy to
> analyze than is available in the Universe.
>
> How much more Perfect do you want, even in a practical sense?
I am not against having something ideal and perfect as a standard
for approximation (to be strived at in practice) or for pedagogical
purpose. But to say there IS (in the sence of EXISTS) something
>
> >because OTP presuppose (absolutely) true randomness
> >and there is no way of determining that in practice.
>
> Sure there is. Just look at how the numbers are being generated. That
> will tell you if they are random.
To 'just look' is certainly not ensuring (compare watching a
magician pulling rabits out of his hat). We have to ascertain
how 'random' the sequence we get really is. And that's one of
the real and big problem for the practice.
M. K. Shen
Subject: Re: Random numbers from a sound card?
Date: Wed, 27 Jan 1999 15:02:36 GMT
From: rcktexas@ix.netcom.com (R. Knauer)
Message-ID: <36af2a34.18209443@nntp.ix.netcom.com>
References: <36AF1B12.FBD87AB5@stud.uni-muenchen.de>
Newsgroups: sci.crypt
Lines: 17
On Wed, 27 Jan 1999 14:56:34 +0100, Mok-Kong Shen
<mok-kong.shen@stud.uni-muenchen.de> wrote:
>But to say there IS (in the sense of EXISTS) something
Does a Perfect Circle EXIST?
If you say is does, is that misleading?
Bob Knauer
"No Freeman shall ever be debarred the use of arms. The strongest
reason for the people to retain the right to keep and bear arms is,
as a last resort, to protect themselves against tyranny in government."
--Thomas Jefferson
Subject: Re: Random numbers from a sound card?
Date: Wed, 27 Jan 1999 16:44:40 +0100
From: Mok-Kong Shen <mok-kong.shen@stud.uni-muenchen.de>
Message-ID: <36AF3468.78EDE075@stud.uni-muenchen.de>
References: <36af2a34.18209443@nntp.ix.netcom.com>
Newsgroups: sci.crypt
Lines: 16
R. Knauer wrote:
>
> On Wed, 27 Jan 1999 14:56:34 +0100, Mok-Kong Shen
> <mok-kong.shen@stud.uni-muenchen.de> wrote:
>
> >But to say there IS (in the sense of EXISTS) something
>
> Does a Perfect Circle EXIST?
>
> If you say is does, is that misleading?
If the word 'IS' is employed in a context without the connotation
M. K. Shen
Subject: Re: Random numbers from a sound card?
Date: Wed, 27 Jan 1999 16:15:17 GMT
From: rcktexas@ix.netcom.com (R. Knauer)
Message-ID: <36af3b4d.22586607@nntp.ix.netcom.com>
References: <36AF3468.78EDE075@stud.uni-muenchen.de>
Newsgroups: sci.crypt
Lines: 20
On Wed, 27 Jan 1999 16:44:40 +0100, Mok-Kong Shen
<mok-kong.shen@stud.uni-muenchen.de> wrote:
>If the word 'IS' is employed in a context without the connotation
You are beginning to sound just like Bill Clinton:
"It all depends on what the meaning of the word 'is' is."
<jeez>
Bob Knauer
"No Freeman shall ever be debarred the use of arms. The strongest
reason for the people to retain the right to keep and bear arms is,
as a last resort, to protect themselves against tyranny in government."
--Thomas Jefferson
Subject: Re: Random numbers from a sound card?
Date: Wed, 27 Jan 1999 17:46:44 +0100
From: Mok-Kong Shen <mok-kong.shen@stud.uni-muenchen.de>
Message-ID: <36AF42F4.5B9DC5CD@stud.uni-muenchen.de>
References: <36af3b4d.22586607@nntp.ix.netcom.com>
Newsgroups: sci.crypt
Lines: 23
R. Knauer wrote:
>
> On Wed, 27 Jan 1999 16:44:40 +0100, Mok-Kong Shen
> <mok-kong.shen@stud.uni-muenchen.de> wrote:
>
> >If the word 'IS' is employed in a context without the connotation
> >of 'EXISTS' then it is NOT misleading, otherwise it IS misleading.
>
> You are beginning to sound just like Bill Clinton:
>
> "It all depends on what the meaning of the word 'is' is."
That way clearly stated in my previous post, quoted below:
But to say there IS (in the sense of EXISTS) something
A word can have a multitude of meanings. I was prudent enough
to put the parentheses above to make sure that there could be
no misunderstanding. I regret that my attempt was appraently
not successful.
M. K. Shen
Subject: Re: Random numbers from a sound card?
Date: Wed, 27 Jan 1999 17:36:54 GMT
From: rcktexas@ix.netcom.com (R. Knauer)
Message-ID: <36af4e20.27405937@nntp.ix.netcom.com>
References: <36AF42F4.5B9DC5CD@stud.uni-muenchen.de>
Newsgroups: sci.crypt
Lines: 34
On Wed, 27 Jan 1999 17:46:44 +0100, Mok-Kong Shen
<mok-kong.shen@stud.uni-muenchen.de> wrote:
>> "It all depends on what the meaning of the word 'is' is."
>That way clearly stated in my previous post, quoted below:
> But to say there IS (in the sense of EXISTS) something
>A word can have a multitude of meanings. I was prudent enough
>to put the parentheses above to make sure that there could be
>no misunderstanding. I regret that my attempt was appraently
>not successful.
You must be a mathematician.
As Greg Chaitin says in his latest book. "The Unknowable", physicists
have a sense of humor (BTW, I am a physicist), but mathematicians do
not have a sense of humor.
Which is not completely true because Chaitin is a mathematician and he
has a sense of humor.
Whatever.
Bob Knauer
"No Freeman shall ever be debarred the use of arms. The strongest
reason for the people to retain the right to keep and bear arms is,
as a last resort, to protect themselves against tyranny in government."
--Thomas Jefferson
Subject: Re: Random numbers from a sound card?
Date: Wed, 27 Jan 1999 12:02:38 -0600
From: Medical Electronics Lab <rosing@physiology.wisc.edu>
Message-ID: <36AF54BE.5771@physiology.wisc.edu>
References: <36AF1B12.FBD87AB5@stud.uni-muenchen.de>
Newsgroups: sci.crypt
Lines: 23
Mok-Kong Shen wrote:
> I am not against having something ideal and perfect as a standard
> for approximation (to be strived at in practice) or for pedagogical
> purpose. But to say there IS (in the sence of EXISTS) something
Kind of depends on how you define "perfect". Perfect for what and
measured in what way? We can certainly build a TRNG which is
perfect in any measureable sense.
> To 'just look' is certainly not ensuring (compare watching a
> magician pulling rabits out of his hat). We have to ascertain
> how 'random' the sequence we get really is. And that's one of
> the real and big problem for the practice.
Which is what makes this whole discussion so much fun. DIEHARD
and Diaphony and autocorrelation all measure "random" in a slightly
different way. If the output of a TRNG appears random to all those
tests, we can say it "looks" random. It is "perfect" as far
as we can measure. Isn't that good enough?
Patience, persistence, truth,
Dr. mike
Subject: Re: Random numbers from a sound card?
Date: Wed, 27 Jan 1999 12:06:35 -0700
From: "Tony T. Warnock" <u091889@cic-mail.lanl.gov>
Message-ID: <36AF63BB.18D3EEA9@cic-mail.lanl.gov>
References: <36AF54BE.5771@physiology.wisc.edu>
Newsgroups: sci.crypt
Lines: 46
Medical Electronics Lab wrote:
> Mok-Kong Shen wrote:
> > I am not against having something ideal and perfect as a standard
> > for approximation (to be strived at in practice) or for pedagogical
> > purpose. But to say there IS (in the sence of EXISTS) something
> > perfect can be misleading.
>
> Kind of depends on how you define "perfect". Perfect for what and
> measured in what way? We can certainly build a TRNG which is
> perfect in any measureable sense.
>
> > To 'just look' is certainly not ensuring (compare watching a
> > magician pulling rabits out of his hat). We have to ascertain
> > how 'random' the sequence we get really is. And that's one of
> > the real and big problem for the practice.
>
> Which is what makes this whole discussion so much fun. DIEHARD
> and Diaphony and autocorrelation all measure "random" in a slightly
> different way. If the output of a TRNG appears random to all those
> tests, we can say it "looks" random. It is "perfect" as far
> as we can measure. Isn't that good enough?
>
> Patience, persistence, truth,
> Dr. mike
It's not clear what is wanted here. In limit (infinitely long sequences)
both the complexity based and the frequency (statistical) based
definitions of random are equivalent (per Martin Lof). For finite
sequences (actually for computable sequences, IMHO) these are not
necessarily equivalent. It is easy to produce sequences that satisfy the
strong law of large numbers. Champernowne's sequence comes to mind:
01,1011,100101110111,.... It is not very complex computationally. It does
have the proper frequency of everything, that is, each k bit sequence has
limiting frequence 1/2^k. Unfortunately I do not know of any easily
constructed sequence that satisfy the law of the iterated logarithm. I do
not even know how to test for this. It would be a requirement for a
"statistically random" sequence.
It's possible that one can only list a set of criteria and check if your
sequence satisfy them. Again, most of these criteria are not computable
but "almost all" sequences satisfy them
Tony
Subject: Re: Random numbers from a sound card?
Date: Wed, 27 Jan 1999 23:01:14 GMT
From: rcktexas@ix.netcom.com (R. Knauer)
Message-ID: <36af98b1.46494265@nntp.ix.netcom.com>
References: <36AF54BE.5771@physiology.wisc.edu>
Newsgroups: sci.crypt
Lines: 50
On Wed, 27 Jan 1999 12:02:38 -0600, Medical Electronics Lab
<rosing@physiology.wisc.edu> wrote:
>> To 'just look' is certainly not ensuring (compare watching a
>> magician pulling rabits out of his hat). We have to ascertain
>> how 'random' the sequence we get really is. And that's one of
>> the real and big problem for the practice.
>Which is what makes this whole discussion so much fun.
Then you're a masochist. :-)
Once you catch on to all this, you will see why.
>DIEHARD
>and Diaphony and autocorrelation all measure "random" in a slightly
>different way.
Those things don't measure the crypto-grade randomness of finite
numbers at all. They try to make inferences about the generator from
finite samples, which is useless for purposes of crypto. They will
pass the outputs of PRNGs that can be cracked.
We need an update to the Snake Oil FAQ desperately!
>If the output of a TRNG appears random to all those
>tests, we can say it "looks" random.
Just what makes a finite number produced by a TRNG "look random"?
Why do you thing that characteristics that apply only to infinite
numbers can also apply to finite ones with equal certitude?
What does "vanishingly small" mean to you?
>It is "perfect" as far as we can measure.
That measure is worthless for crypto-grade random numbers.
> Isn't that good enough?
Nope. Not even close.
Bob Knauer
"No Freeman shall ever be debarred the use of arms. The strongest
reason for the people to retain the right to keep and bear arms is,
as a last resort, to protect themselves against tyranny in government."
--Thomas Jefferson
Subject: Re: Random numbers from a sound card?
Date: Thu, 28 Jan 1999 12:47:39 -0600
From: Medical Electronics Lab <rosing@physiology.wisc.edu>
Message-ID: <36B0B0CB.1974@physiology.wisc.edu>
References: <36af98b1.46494265@nntp.ix.netcom.com>
Newsgroups: sci.crypt
Lines: 56
R. Knauer wrote:
>
> Then you're a masochist. :-)
:-) I think you're right about that!
> Once you catch on to all this, you will see why.
It's pretty clear we have different opinions. The best I can
hope for is more descriptions so I can find out what the core
assumption is that we disagree on. Neither one of us will change
:-)
> Those things don't measure the crypto-grade randomness of finite
> numbers at all. They try to make inferences about the generator from
> finite samples, which is useless for purposes of crypto. They will
> pass the outputs of PRNGs that can be cracked.
So you need an infinite sequence of bits to prove that something
> Just what makes a finite number produced by a TRNG "look random"?
Is a 10 megabyte block of random bits a single number? Or is it
80 million individual numbers? For the latter case, it looks
random if it can pass all the tests for randomness that
mathematicians have dreamed up. In the former case, if it isn't
printable ascii, then it will probably look random no matter
what.
> Why do you thing that characteristics that apply only to infinite
> numbers can also apply to finite ones with equal certitude?
What characteristics are you talking about? Integrals over a
finite range and binomial or poisson distributions are all based
on finite samples. All the DIEHARD tests are based on finite
samples. I am assuming that Marsaglia knows what he's doing,
but maybe you can correct him?
> What does "vanishingly small" mean to you?
Less than I can measure.
> That measure is worthless for crypto-grade random numbers.
Yes, well, expand on "crypto-grade" a bit.
> > Isn't that good enough?
>
> Nope. Not even close.
:-) See, I told you we disagree. Let's keep it that way,
makes for a nice long discussion.
Patience, persistence, truth,
Dr. mike
Subject: Re: Random numbers from a sound card?
Date: Thu, 28 Jan 1999 23:07:11 GMT
From: rcktexas@ix.netcom.com (R. Knauer)
Message-ID: <36b0d008.1959958@nntp.ix.netcom.com>
References: <36B0B0CB.1974@physiology.wisc.edu>
Newsgroups: sci.crypt
Lines: 111
On Thu, 28 Jan 1999 12:47:39 -0600, Medical Electronics Lab
<rosing@physiology.wisc.edu> wrote:
>It's pretty clear we have different opinions. The best I can
>hope for is more descriptions so I can find out what the core
>assumption is that we disagree on. Neither one of us will change
You will change once you catch on. I did.
A year ago I came onto sci.crypt with ill-formed notions of
crypto-grade randomness. After what seemed like a thousand posts from
many participants, the truth emerged.
I have capsulized that truth several times recently but some people,
including you, still have not caught on. When you do catch on, you
will look back and wonder how you could have been so confused about
such a straightforward concept. I did.
>So you need an infinite sequence of bits to prove that something
You cannot prove the crypto-grade randomness of a finite number
algorithmically. You can for an infinite number, but that is useless.
The only way you can prove the crypto-grade randomness of a finite
number is to consider the method of generation. If the generator is a
TRNG, as we have defined it here several times recently, then the
numbers it generates are crypto-grade random numbers.
>Is a 10 megabyte block of random bits a single number?
Yes.
>Or is it 80 million individual numbers?
Yes.
>For the latter case, it looks
>random if it can pass all the tests for randomness that
>mathematicians have dreamed up.
Wrong. You might be able to infer some things about the numbers that
fool you into thinking they are random, but that does not make them
crypto-random.
Keep in mind that many PRNGs pass statistical tests.
>In the former case, if it isn't
>printable ascii, then it will probably look random no matter
>what.
Numbers don't "look" crypto-random.
The number 1111111111 is a crypto-grade random number, because it was
generated by a TRNG. Or, may it is not because it was not generated by
a TRNG. You cannot tell unless you know the generation process.
Tell me if you think 111111111 is crypto-grade random or not.
>> Why do you thing that characteristics that apply only to infinite
>> numbers can also apply to finite ones with equal certitude?
>What characteristics are you talking about?
The characteristic of randomness. Infinite numbers have
characteristics which can be related to randomness. If an infinite
number is a normal number, it is random. Finite numbers cannot be
normal numbers - they are not big enough.
For example, if you can prove that pi is a normal number, then it is a
random number.
>Integrals over a
>finite range and binomial or poisson distributions are all based
>on finite samples.
Do they measure crypto-grade randomness of finite numbers?
If they could, these algorithms you propose could also be used to
solve Godel's incompleteness problem, Turing's halting problem and
Chaitin's complexity problem.
>All the DIEHARD tests are based on finite
>samples. I am assuming that Marsaglia knows what he's doing,
>but maybe you can correct him?
You correct him, when you discover the truth.
>> What does "vanishingly small" mean to you?
>Less than I can measure.
>Yes, well, expand on "crypto-grade" a bit.
Proveably secure when used with the OTP cryptosystem.
>:-) See, I told you we disagree. Let's keep it that way,
>makes for a nice long discussion.
Last time it was over 1,000 posts. I am beginning to think I was the
only one who got anything out of them.
Bob Knauer
"No Freeman shall ever be debarred the use of arms. The strongest
reason for the people to retain the right to keep and bear arms is,
as a last resort, to protect themselves against tyranny in government."
--Thomas Jefferson
Subject: Re: Random numbers from a sound card?
Date: Fri, 29 Jan 1999 11:47:15 +0100
From: Mok-Kong Shen <mok-kong.shen@stud.uni-muenchen.de>
Message-ID: <36B191B3.89CCEFA3@stud.uni-muenchen.de>
References: <36b0d008.1959958@nntp.ix.netcom.com>
Newsgroups: sci.crypt
Lines: 19
R. Knauer wrote:
>
>
> You cannot prove the crypto-grade randomness of a finite number
> algorithmically. You can for an infinite number, but that is useless.
>
> The only way you can prove the crypto-grade randomness of a finite
> number is to consider the method of generation. If the generator is a
> TRNG, as we have defined it here several times recently, then the
> numbers it generates are crypto-grade random numbers.
Ah! Finally one knows exactly what the term 'crypto-grade random
numbers' you employ means: These are DEFINED to be the output
from a hardware generator. If follows obviously then that there
is NO need whatsoever of testing the sequences obtained, since they
M. K. Shen
Subject: Re: Random numbers from a sound card?
Date: Fri, 29 Jan 1999 13:37:25 GMT
From: rcktexas@ix.netcom.com (R. Knauer)
Message-ID: <36b1b932.2360123@nntp.ix.netcom.com>
References: <36B191B3.89CCEFA3@stud.uni-muenchen.de>
Newsgroups: sci.crypt
Lines: 26
On Fri, 29 Jan 1999 11:47:15 +0100, Mok-Kong Shen
<mok-kong.shen@stud.uni-muenchen.de> wrote:
>> The only way you can prove the crypto-grade randomness of a finite
>> number is to consider the method of generation. If the generator is a
>> TRNG, as we have defined it here several times recently, then the
>> numbers it generates are crypto-grade random numbers.
>Ah! Finally one knows exactly what the term 'crypto-grade random
>numbers' you employ means: These are DEFINED to be the output
>from a hardware generator. If follows obviously then that there
>is NO need whatsoever of testing the sequences obtained, since they
Are you being deliberatly obtuse - or does it come naturally?
Nothing that you said above follows from what I said. A TRNG is not a
True Random Number Generator just because it is a hardware device.
Bob Knauer
"No Freeman shall ever be debarred the use of arms. The strongest
reason for the people to retain the right to keep and bear arms is,
as a last resort, to protect themselves against tyranny in government."
--Thomas Jefferson
Subject: Re: Random numbers from a sound card?
Date: 29 Jan 1999 09:14:11 -0500
From: juola@mathcs.duq.edu (Patrick Juola)
Message-ID: <78sfnj\$ouo\$1@quine.mathcs.duq.edu>
References: <36B191B3.89CCEFA3@stud.uni-muenchen.de>
Newsgroups: sci.crypt
Lines: 21
In article <36B191B3.89CCEFA3@stud.uni-muenchen.de>,
Mok-Kong Shen <mok-kong.shen@stud.uni-muenchen.de> wrote:
>R. Knauer wrote:
>>
>
>>
>> You cannot prove the crypto-grade randomness of a finite number
>> algorithmically. You can for an infinite number, but that is useless.
>>
>> The only way you can prove the crypto-grade randomness of a finite
>> number is to consider the method of generation. If the generator is a
>> TRNG, as we have defined it here several times recently, then the
>> numbers it generates are crypto-grade random numbers.
>
>Ah! Finally one knows exactly what the term 'crypto-grade random
>numbers' you employ means: These are DEFINED to be the output
>from a hardware generator.
No. Not all hardware generators are TRNG.
-kitten
Subject: Re: Random numbers from a sound card?
Date: Fri, 29 Jan 1999 08:38:00 -0700
From: "Tony T. Warnock" <u091889@cic-mail.lanl.gov>
Message-ID: <36B1D5D8.9E4A7B55@cic-mail.lanl.gov>
References: <36b0d008.1959958@nntp.ix.netcom.com>
Newsgroups: sci.crypt
Lines: 35
R. Knauer wrote:
> The characteristic of randomness. Infinite numbers have
> characteristics which can be related to randomness. If an infinite
> number is a normal number, it is random. Finite numbers cannot be
> normal numbers - they are not big enough.
>
> For example, if you can prove that pi is a normal number, then it is a
> random number.
Normality is certainly necessary but not sufficient. It's a good start.
More than normality is needed. I can give you many normal numbers but none
of them are "random." Champernowne's number is the simplest example:
1,10,11,100,101,110,111,...=11011100101110111.... It is easy to prove that
all k-bit patterns have the proper frequency. This is all that is needed
for normality. (The concept of normality was introduced by Borel about
1909.) The digits of a normal number satisfy the strong law of large
numbers, that is, 1/2 ones, 1/2 zeros, 1/4 00's, 1/4 01's, 1/4 10's, 1/4
11's, ..., 1/1024 1101101101's, etc.
The problem is that the strong law of large numbers is not very strong. In
Champernowne's number, the excess of ones over zeros grows as N/log(N) for
N bits. The ratio goes like 1/2+1/log(N), really slow. The dispersion is
also not correct. The law of the iterated logarithm fails for all these
sequences.
Of course both the above laws (large numbers, iterated logarithm) are
statistical in nature and do not indicate how difficult it is to guess
successive bits of a number. Complexity of computation and statistical
properties are only equivalent in the limit of infnitely many infinitely
long sequences.
Tony
Subject: Re: Random numbers from a sound card?
Date: Sat, 30 Jan 1999 03:09:39 GMT
From: rcktexas@ix.netcom.com (R. Knauer)
Message-ID: <36b272f8.49918378@nntp.ix.netcom.com>
References: <36B1D5D8.9E4A7B55@cic-mail.lanl.gov>
Newsgroups: sci.crypt
Lines: 70
On Fri, 29 Jan 1999 08:38:00 -0700, "Tony T. Warnock"
<u091889@cic-mail.lanl.gov> wrote:
>Normality is certainly necessary but not sufficient. It's a good start.
>More than normality is needed. I can give you many normal numbers but none
>of them are "random."
Are they infinite?
>Champernowne's number is the simplest example:
>1,10,11,100,101,110,111,...=11011100101110111.... It is easy to prove that
>all k-bit patterns have the proper frequency.
Yes, but only if the number is infinite.
>This is all that is needed
>for normality. (The concept of normality was introduced by Borel about
>1909.) The digits of a normal number satisfy the strong law of large
>numbers, that is, 1/2 ones, 1/2 zeros, 1/4 00's, 1/4 01's, 1/4 10's, 1/4
>11's, ..., 1/1024 1101101101's, etc.
Chaitin cover this in his papers - for those who want an accessible
reference.
>The problem is that the strong law of large numbers is not very strong. In
>Champernowne's number, the excess of ones over zeros grows as N/log(N) for
>N bits.
Is that really a problem? Whoevewr said that bias was an intrinsic
property of infinite random numbers?
>The ratio goes like 1/2+1/log(N), really slow. The dispersion is
>also not correct. The law of the iterated logarithm fails for all these
>sequences.
This what I like about the Internet in general, and Usenet forums like
sci.crypt in particular. There is always someone who knows the
something about something - someone who is willing to jump in and
expose that.
Without the Truth to seek out, life is completely meaningless. [Cf.
Camus, "The Myth Of Sysiphus" and the concept of "Lucidity".]
Your further elaborations would be most higly regarded by me amd all
the lurkers on sci.crypt. The concept of randomness is fundamental to
an understanding of how we consider Order, the thing which
distinguishes us from dirt. The concept of randomness is at the heart
of Quantum Mechanics, which has incredible predictive value.
>Of course both the above laws (large numbers, iterated logarithm) are
>statistical in nature and do not indicate how difficult it is to guess
>successive bits of a number. Complexity of computation and statistical
>properties are only equivalent in the limit of infnitely many infinitely
>long sequences.
Another excellent contribution to the FAQ on crypto-grade randomness.
But I point out that computational complexity has nothing fundamental
to do with crypto-grade randomness, nor QM. In those realms everything
is possible, even the most simple of sequences. In fact, I believe we
are here because the simpler sequences prevailed.
Bob Knauer
"No Freeman shall ever be debarred the use of arms. The strongest
reason for the people to retain the right to keep and bear arms is,
as a last resort, to protect themselves against tyranny in government."
--Thomas Jefferson
Subject: Re: Random numbers from a sound card?
Date: Sat, 30 Jan 1999 14:51:49 GMT
From: rcktexas@ix.netcom.com (R. Knauer)
Message-ID: <36b31a67.8855773@nntp.ix.netcom.com>
References: <36B1D5D8.9E4A7B55@cic-mail.lanl.gov>
Newsgroups: sci.crypt
Lines: 38
On Fri, 29 Jan 1999 08:38:00 -0700, "Tony T. Warnock"
<u091889@cic-mail.lanl.gov> wrote:
>Champernowne's number is the simplest example:
>1,10,11,100,101,110,111,...=11011100101110111.... It is easy to prove that
>all k-bit patterns have the proper frequency. This is all that is needed
>for normality. (The concept of normality was introduced by Borel about
>1909.) The digits of a normal number satisfy the strong law of large
>numbers, that is, 1/2 ones, 1/2 zeros, 1/4 00's, 1/4 01's, 1/4 10's, 1/4
>11's, ..., 1/1024 1101101101's, etc.
>The problem is that the strong law of large numbers is not very strong. In
>Champernowne's number, the excess of ones over zeros grows as N/log(N) for
>N bits. The ratio goes like 1/2+1/log(N), really slow. The dispersion is
>also not correct. The law of the iterated logarithm fails for all these
>sequences.
In re-reading this I spotted something I do not understand. You state
that for the Champernowne number "all k-bit patterns have the proper
frequency". I assume that is true for k = 1, one of the possible
values for k.
Then you say that in Champernowne's number there is an "excess of ones
over zeros". How can that be if "all k-bit patterns have the proper
frequency"? The "proper frequency" for k = 1 as described by you: "The
digits of a normal number satisfy the strong law of large numbers,
that is, 1/2 ones, 1/2 zeros".
How come you state that Champernowne's number has an "excess of ones
over zeros"?
Bob Knauer
"No Freeman shall ever be debarred the use of arms. The strongest
reason for the people to retain the right to keep and bear arms is,
as a last resort, to protect themselves against tyranny in government."
--Thomas Jefferson
Subject: Re: Random numbers from a sound card?
Date: Sat, 30 Jan 1999 14:31:17 -0500
From: "Trevor Jackson, III" <fullmoon@aspi.net>
Message-ID: <36B35E04.3DF9ED19@aspi.net>
References: <36b31a67.8855773@nntp.ix.netcom.com>
Newsgroups: sci.crypt
Lines: 46
R. Knauer wrote:
> On Fri, 29 Jan 1999 08:38:00 -0700, "Tony T. Warnock"
> <u091889@cic-mail.lanl.gov> wrote:
>
> >Champernowne's number is the simplest example:
> >1,10,11,100,101,110,111,...=11011100101110111.... It is easy to prove that
> >all k-bit patterns have the proper frequency. This is all that is needed
> >for normality. (The concept of normality was introduced by Borel about
> >1909.) The digits of a normal number satisfy the strong law of large
> >numbers, that is, 1/2 ones, 1/2 zeros, 1/4 00's, 1/4 01's, 1/4 10's, 1/4
> >11's, ..., 1/1024 1101101101's, etc.
>
> >The problem is that the strong law of large numbers is not very strong. In
> >Champernowne's number, the excess of ones over zeros grows as N/log(N) for
> >N bits. The ratio goes like 1/2+1/log(N), really slow. The dispersion is
> >also not correct. The law of the iterated logarithm fails for all these
> >sequences.
>
> In re-reading this I spotted something I do not understand. You state
> that for the Champernowne number "all k-bit patterns have the proper
> frequency". I assume that is true for k = 1, one of the possible
> values for k.
>
> Then you say that in Champernowne's number there is an "excess of ones
> over zeros". How can that be if "all k-bit patterns have the proper
> frequency"? The "proper frequency" for k = 1 as described by you: "The
> digits of a normal number satisfy the strong law of large numbers,
> that is, 1/2 ones, 1/2 zeros".
>
> How come you state that Champernowne's number has an "excess of ones
> over zeros"?
Because all the leading zeros are suppressed.
>
>
> Bob Knauer
>
> "No Freeman shall ever be debarred the use of arms. The strongest
> reason for the people to retain the right to keep and bear arms is,
> as a last resort, to protect themselves against tyranny in government."
> --Thomas Jefferson
Subject: Re: Random numbers from a sound card?
Date: Sun, 31 Jan 1999 00:06:40 GMT
From: rcktexas@ix.netcom.com (R. Knauer)
Message-ID: <36b39e73.42659320@nntp.ix.netcom.com>
References: <36B35E04.3DF9ED19@aspi.net>
Newsgroups: sci.crypt
Lines: 17
On Sat, 30 Jan 1999 14:31:17 -0500, "Trevor Jackson, III"
<fullmoon@aspi.net> wrote:
>> How come you state that Champernowne's number has an "excess of ones
>> over zeros"?
>Because all the leading zeros are suppressed.
Can you elaborate with an example.
Bob Knauer
"I place economy among the first and most important virtues and
public debt as the greatest dangers to be feared. We must not
let our rulers load us with perpetual debt."
--Thomas Jefferson
Subject: Re: Random numbers from a sound card?
Date: 26 Jan 1999 14:59:26 -0500
From: juola@mathcs.duq.edu (Patrick Juola)
Message-ID: <78l6qu\$k97\$1@quine.mathcs.duq.edu>
Newsgroups: sci.crypt
Lines: 25
Mok-Kong Shen <mok-kong.shen@stud.uni-muenchen.de> wrote:
>R. Knauer wrote:
>
>> >I guess the issue of cryptological strength is inherently fuzzy
>>
>> Not really. The OTP system is proveably secure.
>
>Once again I assert that this is a (for all practical purposes)
>useless fact, because OTP presuppose (absolutely) true randomness
>and there is no way of determining that in practice.
Not quite. The randomness that the OTP presumes works out to
be exactly the same problem as key generation for a key system --
if (somehow) the attacker can predict which key you are going
to use, then the attacker can unbutton your messages more or
less at will. Similarly if the attacker can force you to use
a particular key.
So, yes, you'll probably not build a "perfect" OTP in practice,
any more than you'll be able to get a bug-free computer program.
That doesn't mean that there aren't techniques that are *less*
likely to approach perfection than others.
-kitten
Subject: Re: Random numbers from a sound card?
Date: Tue, 26 Jan 1999 18:15:11 +0000
From: pla@sktb.demon.co.uk (Paul L. Allen)
Message-ID: <f8BSgej030n@sktb.demon.co.uk>
Newsgroups: sci.crypt
Lines: 38
rcktexas@ix.netcom.com (R. Knauer) writes:
> On Mon, 25 Jan 1999 20:11:33 +0100, Mok-Kong Shen
> <mok-kong.shen@stud.uni-muenchen.de> wrote:
>
> >> How would you test the 'quality' of the generated random number
> >> stream?
>
> >There are tests for statistical quality, e.g. Maurer's universal
> >statistical test. I am ignorant of tests for crypto quality.
>
> That's because there aren't any.
>
> It is a fundamental precept of crytpography that randomness is a
> property of how a number is generated, not a property of a number
> itself.
I'd be *extremely* worried about a sound card (particularly with little
in the way of input) picking up mains power hum and radiated noise from
signal lines in the computer. Obviously they'll only affect a few of
the lower bits of output (or you'd hear it when you played sounds back)
but that might be enough to weaken the randomness enough to cause problems.
Boiling down the entropy with a cryptographic hash function probably
gets rid of it, but if you have a crappy sound card you may have to do
a lot more boiling away than with a good one.
Noise diodes strike me as being safer. The fundamental mechanism of
noise generation in them may well be chaotic and I've seen worries that
chaotic loci can be close to periodic for long periods of time, but there
are likely to be many individual sources of chaotic noise in the diode
(due to random distribution of dopant atoms) which may make things alot
safer. Probably. It really needs somebody whose done detailed work on
those noise mechanisms to comment.
--Paul
Subject: Re: Random numbers from a sound card?
Date: Tue, 26 Jan 1999 20:12:38 GMT
From: randombit@my-dejanews.com
Message-ID: <78l7je\$cc6\$1@nnrp1.dejanews.com>
References: <36acb8b1.5374650@news.willapabay.org>
Newsgroups: sci.crypt
Lines: 79
In article <36acb8b1.5374650@news.willapabay.org>,
ross@hypertools.com (David Ross) wrote:
> Has anyone had success using a sound card (like a Sound Blaster) to
> generate streams of random numbers?
Yes.
> What sort of audio source would you suspect would be the best to use
> in generating random numbers?
I used an old radio shack mono fm radio, with antenna removed,
tuned to hiss, at high volume, fed into the sound card.
Later I got a video/radio digitizer, which I can tune to
FM hiss, which is more self contained.
This produces an apparently uniformly distributed noise spectrum,
using a pc-based spectrum analyzer.
But this doesn't have full entropy, and you have to distill (see
RFC 1750) the bits. I experimented with parity-of-N bits,
and used Maurer's Universal statistical test for RNGs to measure
the entropy. When you distill enough, the entropy reaches its
expected value.
Some people might recommend a strong hashing function (e.g., a thousand
raw bits hashed with MD5 down to a fixed output size). This is
complex and I found unnecessary; simple parity works, though it
may waste more bits than a serious hash would. But bits are cheap,
and xor is fast.
> How would you test the 'quality' of the generated random number
> stream?
1. Marsaglia's Diehard suite of statistical (structure) tests.
This suite goes far beyond the FIPS suggestions.
2. Maurer's Universal statistical test, which approximates
the entropy of a sample using a formally motivated,
compression-like algorithm.
As "calibration standards" I used the "RAND million normal digits and their
deviant friends", and also block ciphers run in feedback modes (ie,
as PRNGs).
I've also got a parallel-port compatible geiger counter and a
microcurie of americium, but i haven't careful studies on these yet.
But they are cool toys :-)
You will learn that you *always* have to distill raw bits.
And you may observe that very few hardware RNGs actually monitor their output
quality (especially on-line), though it seems to me you should.
Also note that a 'loud' source of hiss is preferable. Were I using
an acoustic microphone as my raw input, I would locate it next
to the frother on my espresso machine and blow steam out of it,
rather than counting on the wind or ambient brownian effects.
Note that even using a highly structured signal (e.g., digitized
good bits, but you'd have to distill bushels of them.
Have fun,
randombit
-----------== Posted via Deja News, The Discussion Network ==----------
Subject: Re: Random numbers from a sound card?
Date: Tue, 26 Jan 1999 22:13:11 GMT
From: ross@hypertools.com (David Ross)
Message-ID: <36ae33b7.16083306@news.willapabay.org>
References: <78l7je\$cc6\$1@nnrp1.dejanews.com>
Newsgroups: sci.crypt
Lines: 60
Randombit -
Thanks for an informative post.
On Tue, 26 Jan 1999 20:12:38 GMT, randombit@my-dejanews.com wrote:
> ross@hypertools.com (David Ross) wrote:
>> Has anyone had success using a sound card (like a Sound Blaster) to
>> generate streams of random numbers?
...
>> What sort of audio source would you suspect would be the best to use
>> in generating random numbers?
>
>I used an old radio shack mono fm radio, with antenna removed,
>tuned to hiss, at high volume, fed into the sound card.
>
>Later I got a video/radio digitizer, which I can tune to
>FM hiss, which is more self contained.
>
>This produces an apparently uniformly distributed noise spectrum,
>using a pc-based spectrum analyzer.
>
>But this doesn't have full entropy, and you have to distill (see
>RFC 1750) the bits. I experimented with parity-of-N bits,
>and used Maurer's Universal statistical test for RNGs to measure
>the entropy. When you distill enough, the entropy reaches its
>expected value.
>
>Some people might recommend a strong hashing function (e.g., a thousand
>raw bits hashed with MD5 down to a fixed output size). This is
>complex and I found unnecessary; simple parity works, though it
>may waste more bits than a serious hash would. But bits are cheap,
>and xor is fast.
Lets say I'm digitizing a sine wave of constant frequency &
amplitude. If it has a 1 Volt peak amplitude and if I digitize at a
constant rate, won't 1/2 of my samples yield a value of either >+.707
Volt or < -.707 Volt? Seems like a built-in bias toward higher
numbers...
After looking at 'random' noise on an oscilloscope, I'd expect to
see this same bias when digitizing noise...
Digitizing either a sawtooth or a triangle waveform should get rid
of this inbuilt bias, but where to find 'sawtooth noise' is beyond me.
>> How would you test the 'quality' of the generated random number
>> stream?
>1. Marsaglia's Diehard suite of statistical (structure) tests.
>2. Maurer's Universal statistical test
>As "calibration standards" I used the "RAND million normal digits and their
>deviant friends", and also block ciphers run in feedback modes (ie,
>as PRNGs).
Thanks for these suggestions.
>Also note that a 'loud' source of hiss is preferable.
A 'loud' source of hiss may put the range of the A->D converter down
lower on a sinusoidal waveform. In this more linear area of the
waveform, you would get a more even distribution of digitized values
and begin to eliminate some of the inbuilt bias mentioned above.
David Ross ross@hypertools.com
Subject: Re: Random numbers from a sound card?
Date: Wed, 27 Jan 1999 18:09:54 GMT
From: randombit@my-dejanews.com
Message-ID: <78nkp9\$bol\$1@nnrp1.dejanews.com>
References: <36ae33b7.16083306@news.willapabay.org>
Newsgroups: sci.crypt
Lines: 28
In article <36ae33b7.16083306@news.willapabay.org>,
ross@hypertools.com (David Ross) wrote:
> Lets say I'm digitizing a sine wave of constant frequency &
> amplitude. If it has a 1 Volt peak amplitude and if I digitize at a
> constant rate, won't 1/2 of my samples yield a value of either >+.707
> Volt or < -.707 Volt? Seems like a built-in bias toward higher
> numbers...
The threshold of your detector, which decides whether a voltage
is to be called a 0 or a 1 (for this clock period), will determine your
This threshold will interact with DC biases in your waveform.
And both will drift, and differ between parts.
You will not ever get perfect 0:1 ratios
---for instance, amplifiers may switch from 0->1 faster than 1->0
(even for CMOS) so you must plan for it. Happily, combining
multiple bits (e.g., parity ) brings the ratio near
1:1 exponentially fast. Again, RFC 1750 is the bible.
Shannon is the Prophet.
-----------== Posted via Deja News, The Discussion Network ==----------
Subject: Re: Random numbers from a sound card?
Date: Thu, 28 Jan 1999 18:10:41 GMT
From: randombit@my-dejanews.com
Message-ID: <78q96o\$ka6\$1@nnrp1.dejanews.com>
References: <36AF1E85.280B09D4@stud.uni-muenchen.de>
<78l7je\$cc6\$1@nnrp1.dejanews.com>
Newsgroups: sci.crypt
Lines: 32
In article <36AF1E85.280B09D4@stud.uni-muenchen.de>,
Mok-Kong Shen <mok-kong.shen@stud.uni-muenchen.de> wrote:
> randombit@my-dejanews.com wrote:
> >
>
> > Note that even using a highly structured signal (e.g., digitized
> > video program including your local receiver noise) you could generate
> > good bits, but you'd have to distill bushels of them.
>
> obtaining good bit sequences from such materials as natural
> language texts.)
>
> M. K. Shen
>
There is a big difference. Eve does not know the local, instantaneious
electromagnetic conditions around my receiver, nor does she know what
my local electronics are doing.
The point is that measuring something physical is nothing like
playing with text streams, unless you you get them via UDP and
------------
"Properly done science is a sort of masochistic game where one beats
one's head against a wall until it falls down, and then goes in search
of another wall." --Steven Vogel
-----------== Posted via Deja News, The Discussion Network ==----------
Subject: Re: Random numbers from a sound card?
Date: Thu, 28 Jan 1999 19:54:22 +0100
From: Mok-Kong Shen <mok-kong.shen@stud.uni-muenchen.de>
Message-ID: <36B0B25E.9FD6F2A@stud.uni-muenchen.de>
References: <78q96o\$ka6\$1@nnrp1.dejanews.com>
Newsgroups: sci.crypt
Lines: 22
randombit@my-dejanews.com wrote:
>
> There is a big difference. Eve does not know the local, instantaneious
> electromagnetic conditions around my receiver, nor does she know what
> my local electronics are doing.
>
> The point is that measuring something physical is nothing like
> playing with text streams, unless you you get them via UDP and
In principle you are right. However, I assume that the choice of the
(publically known) texts that go into the process of generating
the bit sequences is secret information and can't be guessed by the
analyst. Thus, assuming adequate processing, one should obtain things
appropriate for use. Of course, one does not generate in this
way the legendary ideal OTP. But security is in my humble opinion
an issue dependent also on the cost and the like and I believe
that under certain real circumstances obtaining bit sequences from
software can be justified.
M. K. Shen
Subject: Re: Random numbers from a sound card?
Date: Thu, 28 Jan 1999 21:36:44 GMT
From: "patix" <patix@friko.onet.pl>
Message-ID: <ML4s2.5136\$014.554762@news.tpnet.pl>
References: <78q96o\$ka6\$1@nnrp1.dejanews.com>
Newsgroups: sci.crypt
Lines: 27
randombit@my-dejanews.com wrote in message
<78q96o\$ka6\$1@nnrp1.dejanews.com>...
>In article <36AF1E85.280B09D4@stud.uni-muenchen.de>,
>
>There is a big difference. Eve does not know the local, instantaneious
>electromagnetic conditions around my receiver, nor does she know what
>my local electronics are doing.
>
>The point is that measuring something physical is nothing like
>playing with text streams, unless you you get them via UDP and
OK , but if she assume that it is 50Hz (Europe power) and
let 30 KHz from Your monitor display , and if it happen to be true ?
I have question:Haw should we test hardawre random
generator to "be shoure" that it is realy some haw random ?
patix
Subject: Re: Random numbers from a sound card?
Date: 29 Jan 1999 08:44:58 -0500
From: juola@mathcs.duq.edu (Patrick Juola)
Message-ID: <78se0q\$or1\$1@quine.mathcs.duq.edu>
References: <78q96o\$ka6\$1@nnrp1.dejanews.com>
Newsgroups: sci.crypt
Lines: 25
In article <78q96o\$ka6\$1@nnrp1.dejanews.com>,
<randombit@my-dejanews.com> wrote:
>In article <36AF1E85.280B09D4@stud.uni-muenchen.de>,
> Mok-Kong Shen <mok-kong.shen@stud.uni-muenchen.de> wrote:
>> randombit@my-dejanews.com wrote:
>> >
>>
>> > Note that even using a highly structured signal (e.g., digitized
>> > video program including your local receiver noise) you could generate
>> > good bits, but you'd have to distill bushels of them.
>>
>> obtaining good bit sequences from such materials as natural
>> language texts.)
>>
>
>There is a big difference. Eve does not know the local, instantaneious
>electromagnetic conditions around my receiver, nor does she know what
>my local electronics are doing.
No, but Mike does. He's fully capable of broadcasting (known)
noise of some sort near your site.
-kitten
Subject: Re: Random numbers from a sound card?
Date: Tue, 26 Jan 1999 22:55:51 +0000
From: pla@sktb.demon.co.uk (Paul L. Allen)
Message-ID: <f8pGhDj030n@sktb.demon.co.uk>
References: <36AE059A.9BA@physiology.wisc.edu>
<36acb8b1.5374650@news.willapabay.org>
Newsgroups: sci.crypt
Lines: 13
In article <36AE059A.9BA@physiology.wisc.edu>
Medical Electronics Lab <rosing@physiology.wisc.edu> writes:
> DIEHARD from Marsaglia
I can never find a URL for that when I need it. I saw Marsglia posted
his various PRNGs the other week and mentioned it but no URL. Actually,
I am more interested in the accompanying docs than the actual tests right
now. What a shame we can't get the FAQ updated.
--Paul
```
Terry Ritter, his current address, and his top page.
Last updated: 1999-02-20 | 21,752 | 79,374 | {"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} | 3.140625 | 3 | CC-MAIN-2018-05 | latest | en | 0.880772 |
https://www.iovi.com/books/algorithms-manual/leetcode/0148.html | 1,679,801,217,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945381.91/warc/CC-MAIN-20230326013652-20230326043652-00507.warc.gz | 907,148,890 | 13,400 | # 148. Sort List (Medium)
https://leetcode.com/problems/sort-list/
Sort a linked list in O(n log n) time using constant space complexity.
Example 1:
```Input: 4->2->1->3
Output: 1->2->3->4
```
Example 2:
```Input: -1->5->3->4->0
Output: -1->0->3->4->5```
## Solutions
``````class Solution {
// Employ merge sort algorithm to solve this problem, be careful for the corn cases,
}
int len = 0;
while (p != null) {
len++;
p = p.next;
}
// Find the middle position to split the list into two sub list
int mid = 0;
// Think it over if the len is odd or even
while (mid < (len / 2 - 1)) {
left = left.next;
right = right.next;
mid++;
}
// You must split it. This situation is totally different from those of using
// array to hold elements.
left.next = null;
right = sortList(right);
//merge
// dummy head, easy for insertion
ListNode dummy = new ListNode(-1);
ListNode pointer = dummy;
while (left != null && right != null) {
if (left.val <= right.val) {
pointer.next = left;
left = left.next;
pointer = pointer.next;
} else {
pointer.next = right;
right = right.next;
pointer = pointer.next;
}
}
while (left != null) {
pointer.next = left;
pointer = pointer.next;
left = left.next;
}
while (right != null) {
pointer.next = right;
pointer = pointer.next;
right = right.next;
}
return dummy.next;
}
}
`````` | 370 | 1,320 | {"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-2023-14 | longest | en | 0.68805 |
http://calculator.academy/rocket-equation-calculator/ | 1,585,635,697,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370500331.13/warc/CC-MAIN-20200331053639-20200331083639-00103.warc.gz | 29,032,969 | 35,221 | # Rocket Equation Calculator
Enter the exhaust velocity, initial mass, and final mass of a rocket to calculate the change in velocity of that rocket.
## Rocket Equation Formula
The following equation is used by the calculator above to calculate the change of velocity of a rocket.
Δv = ev * ln(mi / mf)
• Where Δv is the change in velocity
• ev is the exhaust velocity
• mi is the initial mass
• mf is the final mass after all propellants have been burned.
Conceptually this formula can be described as the change of velocity being the difference between the initial and final velocity which are dependent on the masses and the escape velocity of the exhaust. In this case, the initial velocity is assumed to be 0 since the rocket has not launched yet.
Another important aspect in this equation to keep in mind is that since the velocity is dependent on the natural log of a ratio of masses, the masses should be the same units, but the units don’t affect the outcome. | 203 | 975 | {"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} | 2.71875 | 3 | CC-MAIN-2020-16 | longest | en | 0.937941 |
https://docslib.org/doc/1847728/the-principle-of-relativity | 1,701,469,989,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100308.37/warc/CC-MAIN-20231201215122-20231202005122-00368.warc.gz | 256,161,161 | 8,940 | <<
The Principle of Relativity
Lecture I (PHY 6938), Fall 2007
• Space and time are tied together in the idea of motion.
Space and time are commonly regarded as the forms of existence in the real world, as its substance. A definite portion of matter occupies a define part of space at a definite moment of time. It is in the composite idea of motion that these three fundamental conceptions enter into intimate relationship. — (1921)
• There are no preferred points of space or moments of time, but there are preferred (inertial) motions through spacetime.
Every body continues in its state of rest, or of uniform motion in a right line, unless it is compelled to change that state by forces impressed upon it. —Isaac Newton (1687)
• Newton is right to emphasize that there exist preferred, force-free motions, but he also makes an assumption about what they are: straight lines through Euclidean space. t t t
y y y
x x x
Inertial motions are properties of spacetime. Test Particles and Inertial Motions
• How can we observe inertial motions experimentally?
• A test particle is: • isolated (no contact forces), • small, non-spinning (no tidal forces), • electrically neutral (no electromagnetic forces), and • not massive (no gravitational radiation).
• Test particles do not couple to known, long-range forces, except for , and so follow “natural” paths through spacetime.
• But, we must measure what these inertial motions are, not postulate them. The Principle of Relativity
Through any point of space, at any moment of time, there is exactly one inertial motion for each initial velocity a test particle might have at that point. The fundamental laws of do not distinguish these motions. Relativity and Electrodynamics
• Lorentz/Fitzgerald approach (Bell 1976)
• Electric and magnetic fields of moving charges modify the of (classical) atoms. They get flatter and the orbital period gets shorter.
• Therefore, using real rulers and (atomic) clocks, a moving observer will assign different positions and times to events than a stationary one.
• Einstein approach
• The key result of these modified rulers and clocks is that all inertial observers see a light ray propagate through vacuum at the same speed c.
t !v !r/c2 t! = − · t = t 1 v2/c2 ! − !r ! = !r !vt vˆvˆ !r !v t − !r ! = (!!r vˆvˆ !r ) + · − − · 1 v2/c2 − ! Principles and Assumptions of
• Principles: Relativity and Constant
• Assumptions:
• Euclidean Space: Every inertial observer finds space at any moment of time to be a three-dimensional Euclidean continuum.
• Rectilinear Motion: Every inertial observer finds that both other inertial observers and light rays move along straight lines at uniform speed.
• Homogeneous Time: Every inertial observer measure every other inertial observer’s clock to run at a uniform rate.
• Homogeneous Space: The relationship between the spatial coordinates associated to a fixed event by two inertial observers is linear, with perhaps an additional, linear time dependence.
!r ! = !r !vt − The Relativity of Simultaneity O R • An inertial observer, equipped with a clock, can use light rays to locate remote events in t + t spacetime. t = R S E 2
NE E
t t r = c R − S E 2
S The Relativity of Simultaneity O O
RB • An inertial observer, equipped R with a clock, can use light rays
to locate remote events in RA spacetime. NB B • Different inertial observers will N not agree on whether two A remote events occurred NA simultaneously or not.
SB S
SA t O
R Relationship Between R Inertial Coordinate Systems y E
The relationship between the x coordinates assigned to a fixed spacetime event by two different inertial observers is dictated by S the relationship between the two S pairs of times t and t and t! and t! . S R S! R! t O
R Relationship Between R Inertial Coordinate Systems y E
The relationship between the x coordinates assigned to a fixed spacetime event by two different inertial observers is dictated by S the relationship between the two S pairs of times t and t and t! and t! . S R S! R!
!r !v t 2 = c2 (t t)2 | E − | E − t O
R Relationship Between R Inertial Coordinate Systems y E
The relationship between the x coordinates assigned to a fixed spacetime event by two different inertial observers is dictated by S the relationship between the two S pairs of times t and t and t! and t! . S R S! R!
c2 t !v !r c2 t2 !r 2 t2 2 E − · E t + E − | E| = 0 = (t t ) (t t ) − c2 v2 c2 v2 − S! − R! − − t t = γ t! and t = γ t! O S! S! R! R!
R Relationship Between R Inertial Coordinate Systems y E
The relationship between the x coordinates assigned to a fixed spacetime event by two different inertial observers is dictated by S the relationship between the two S pairs of times t and t and t! and t! . S R S! R!
c2 t !v !r c2 t2 !r 2 t2 2 E − · E t + E − | E| = 0 = (t t ) (t t ) − c2 v2 c2 v2 − S! − R! − − t t = γ t! and t = γ t! O S! S! R! R!
R Relationship Between R Inertial Coordinate Systems y E
The relationship between the x coordinates assigned to a fixed spacetime event by two different inertial observers is dictated by S the relationship between the two S pairs of times t and t and t! and t! . S R S! R!
2 2 2 2 tE "rE /c tS tR γ t! t! = t t = − | | = S! R! S! R! 1 v2/c2 1 v2/c2 − − t t = γ t! and t = γ t! O S! S! R! R!
R Relationship Between R Inertial Coordinate Systems y E
The relationship between the x coordinates assigned to a fixed spacetime event by two different inertial observers is dictated by S the relationship between the two S pairs of times t and t and t! and t! . S R S! R!
2 tS tR 2 tS! tR! γ t! t! = and (γ!) t t = ! ! S! R! 1 v2/c2 S R 1 v2/c2 − − γ = γ!
t t = γ t! and t = γ t! O S! S! R! R!
R Relationship Between R Inertial Coordinate Systems y E
The relationship between the x coordinates assigned to a fixed spacetime event by two different inertial observers is dictated by S the relationship between the two S pairs of times t and t and t! and t! . S R S! R!
2 tS tR 2 tS! tR! γ t! t! = and (γ!) t t = ! ! S! R! 1 v2/c2 S R 1 v2/c2 − − γ = γ!
t t = γ t! and t = γ t! O S! S! R! R!
R Relationship Between R Inertial Coordinate Systems y E
The relationship between the x coordinates assigned to a fixed spacetime event by two different inertial observers is dictated by S the relationship between the two S pairs of times t and t and t! and t! . S R S! R!
1 γ = and tS! tR! = tS tR 1 v2/c2 ! ! − ! γ = γ!
t t = γ t! and t = γ t! O S! S! R! R!
R Relationship Between R Inertial Coordinate Systems y E
The relationship between the x coordinates assigned to a fixed spacetime event by two different inertial observers is dictated by S the relationship between the two S pairs of times t and t and t! and t! . S R S! R!
E 2 := c2 t t = c2 (t )2 + !r 2 ! ! − S R − E | E| Lorentz Transformations
c2 t !v !r c2 t2 !r 2 • We can read off the time t2 2 E − · E t + E − | E| = 0 − c2 v2 c2 v2 transformation law from − − previous results. 2 t! = γ t "v "r /c E E − · E ! " • The invariance of the 2 2 2 2 interval then constrains !rE! = !rE vˆ vˆ !rE + γ vˆ !rE v tE | | | − · | · − the spatial variables. ! "
2 • Linearity fixes the spatial t! γ γc− !v t = − · part of the Lorentz !!r !" ! γ !v I + (γ 1) vˆvˆ " !!r" − · − · transformation. Linear Structure of Minkowski Spacetime
• Every inertial observer can define a vector structure on C = α A + β B Minkowski spacetime using his or her own coordinates. where
• The of coordinates is linear, so all tC tA tB inertial observers define the := α + β !!rC " !!rA" !!rB" same vector structure!
• The Minkowski interval defines a quadratic form on this vector 2 η(x, y) := x y := c tx ty + !x !y space. The associated bi-linear · − · form is the Minkowski metric. Summary and Outlook
• Special relativity is not fundamentally different from previous physical theories. It merely asserts that the principle of relativity applies to optical experiments, as well as to mechanical.
• The main difference between Newtonian and special relativity is that time is relative: different inertial observers have different notions of simultaneity.
• Special relativity makes assumptions about the relative motion of inertial observers, which are in fact exactly the same assumptions Newton made. These assumptions are violated experimentally because nothing, not even a test particle, can be shielded from gravity. ()
• Nonetheless, the assumptions of special relativity imply that spacetime is a vector space equipped with a flat metric. These results are intimately connected to the rectilinearity of inertial motions in the absence of gravity. In a gravitational field, we should expect to lose both linearity and flatness. | 2,404 | 8,660 | {"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-2023-50 | latest | en | 0.89905 |
https://www.doubtnut.com/qna/11758236 | 1,723,514,419,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641054522.78/warc/CC-MAIN-20240813012759-20240813042759-00878.warc.gz | 568,528,949 | 35,738 | # A body of mass 5kg is moving along a circle of a radius 1m with a uniform speed of 5 ms.What is the acceleration of body?
Video Solution
Text Solution
Verified by Experts
## Centripetal acc, a=v2r=5×51=25m/s2.
|
Updated on:21/07/2023
### Knowledge Check
• Question 1 - Select One
## A particles is moving along a circular path of radius 5 m , moving with a uniform speed of 5ms−1 . What will be the avetage acceleration, Whan the particle completes half revolution?
Azero
B10πms2
Cπ/ms2
DNone of these
• Question 2 - Select One
## A partical is moving along a circular path of radius 5m and with uniform speed 5m/s. What will be the avarage acceleration when the partical completes half revoluation?
Azero
B10m/s2
C10±/s2
D(10/π)m/s2
• Question 3 - Select One
## A body of mass 10 kg is moving in a circle of radian 1 m with an angular velocity of 2rad/s the centripetal force is
A10 N
B40 N
C30 N
D20 N
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Doubtnut is the perfect NEET and IIT JEE preparation App. Get solutions for NEET and IIT JEE previous years papers, along with chapter wise NEET MCQ solutions. Get all the study material in Hindi medium and English medium for IIT JEE and NEET preparation | 536 | 1,905 | {"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} | 3.125 | 3 | CC-MAIN-2024-33 | latest | en | 0.853562 |
https://m.scirp.org/papers/68456 | 1,591,146,820,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347426956.82/warc/CC-MAIN-20200602224517-20200603014517-00190.warc.gz | 437,081,610 | 8,604 | On Explanation of Polygons in Galilean Geometry to High School Students
Abstract: In this paper, we have tried to indicate the own properties of polygons in Galilean geometry using the Affine concepts as well. The relationships between an angle and a side as well as the relationships between altitudes and medians concepts, and comparison of some special polygons have been examined carefully. In addition, the area concept has been mentioned. Finally, the paper was completed with a new idea, Theorem 6.
Cite this paper: Kurudirek, A. and Akca, H. (2015) On Explanation of Polygons in Galilean Geometry to High School Students. Open Access Library Journal, 2, 1-7. doi: 10.4236/oalib.1101391.
References
[1] Yaglom, I.M. (1979) A Simple Non-Euclidean Geometry and Its Physical Basis. Springer-Verlag, New York.
[2] Kurudirek, A. and Akca, H. (2015) Explanation of Distance, Kinematic and Isometry to High School Students from Different Perspective. Macrothink Institute, International Research in Education, 3.
[3] Kurudirek, A. and Akça, H. (2015) On the Concept of Circle and Angle in Galilean Plane. Open Access Library Journal, 2, e1256.
http://dx.doi.org/10.4236/oalib.1101256
[4] Artıkbayev, A., Kurudirek, A. and Akça, H. (2013) Occurrence of Galilean Geometry. Applied and Computational Mathematics, 2, 115-117.
http://dx.doi.org/10.11648/j.acm.20130205.11
[5] Kurudirek, A., Akça, H. and Erdoğan, M. (2013) On Geometries in Affine Plane. Applied and Computational Mathematics, 2, 127-129.
http://dx.doi.org/10.11648/j.acm.20130206.13
Top | 460 | 1,564 | {"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-2020-24 | longest | en | 0.852135 |
https://in.mathworks.com/matlabcentral/answers/1669119-use-loop-to-summarize-the-data-using-matlab | 1,670,469,482,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711232.54/warc/CC-MAIN-20221208014204-20221208044204-00266.warc.gz | 355,328,892 | 33,044 | # Use loop to Summarize the Data using MATLAB
2 views (last 30 days)
C PRASAD on 11 Mar 2022
Edited: Peter Perkins on 17 Mar 2022
clc
clear all
close all
X=randi(10,12,75);
Nz = 15;
TopTitle = compose('H%d_Features',1:Nz); % Prepare the top title
T = cell(1,Nz);
for k = 1:Nz
arr=X(:,5*(k-1)+1:5*(k-1)+5);
vnm =strcat({'Energy','variance','std','wl','entrophy'},num2str(k));
Ttemp = array2table(arr,'VariableNames',vnm);
T{k} = table(Ttemp,'VariableName',TopTitle(k));
end
Tcombine = cat(2,T {:});
%% Channel1 features %%
This is My code,I would like to save Channel1 features into one variable using any loop technique.I don't want to use S11,S12,S13,S14 and S15 insted of that I need to use any loop technique and all the values of S11,S12,S13,S14 and S15 into one variable.Please help me with this
P.V Karthik on 11 Mar 2022
Can you try this:
out_arr=[];
out_mix=[];
for i=1:5
cell_out=table2cell(tab_out);
mat_out=cell2mat(cell_out);
out_arr=[out_arr;{['H_',num2str(i)],cell_out}];
out_mix=[out_mix;mat_out];
end
##### 2 CommentsShowHide 1 older comment
Peter Perkins on 17 Mar 2022
Edited: Peter Perkins on 17 Mar 2022
A solution that involves calling cell2mat on the output of table2cell is almost certainly not the right solution. Indeed, calling table2cell in itself is suspect except in a very few cases. And certainly calling eval to do subscripting is a huge red flag.
Peter Perkins on 11 Mar 2022
Not sure what you are trying to do. Maybe you are trying to get all the Energy data into one array? It's not clerar from your question. See if this use of inner2outer helps. Notice that I've also simplified your creation of Prasad:
X = randi(10,12,75);
Nz = 15;
%TopTitle = compose('H%d_Features',1:Nz); % Prepare the top title
TopTitle = compose("H%d_Features",1:Nz); % Prepare the top title
%T = cell(1,Nz);
for k = 1:Nz
arr = X(:,5*(k-1)+1:5*(k-1)+5);
%vnm = strcat({'Energy','variance','std','wl','entrophy'},num2str(k));
vnm = ["Energy" "variance" "std" "wl" "entrophy"];
%Ttemp = array2table(arr,'VariableNames',vnm);
%T{k} = table(Ttemp,'VariableName',TopTitle(k));
end
12×15 table
H1_Features H2_Features H3_Features H4_Features H5_Features H6_Features H7_Features H8_Features H9_Features H10_Features H11_Features H12_Features H13_Features H14_Features H15_Features
Energy variance std wl entrophy Energy variance std wl entrophy Energy variance std wl entrophy Energy variance std wl entrophy Energy variance std wl entrophy Energy variance std wl entrophy Energy variance std wl entrophy Energy variance std wl entrophy Energy variance std wl entrophy Energy variance std wl entrophy Energy variance std wl entrophy Energy variance std wl entrophy Energy variance std wl entrophy Energy variance std wl entrophy Energy variance std wl entrophy
___________________________________________ ___________________________________________ ___________________________________________ ___________________________________________ ___________________________________________ ___________________________________________ ___________________________________________ ___________________________________________ ___________________________________________ ___________________________________________ ___________________________________________ ___________________________________________ ___________________________________________ ___________________________________________ ___________________________________________
9 3 2 9 1 5 8 8 5 2 10 5 9 10 6 8 9 9 6 2 9 8 8 10 7 7 7 4 6 9 9 9 9 6 5 7 8 3 5 10 1 2 4 5 9 5 6 6 2 6 9 10 9 9 5 9 10 9 6 9 3 10 6 5 1 7 3 8 2 9 4 6 8 9 3
6 7 3 3 10 9 10 6 3 3 3 3 5 7 5 10 3 7 8 6 5 7 8 2 6 8 9 5 9 1 2 1 5 2 3 10 2 10 9 9 6 7 10 7 6 7 4 5 6 7 3 1 9 10 5 10 7 2 10 7 4 8 9 8 6 9 6 6 5 9 4 8 3 4 9
4 10 6 2 9 4 6 3 7 2 2 2 9 5 1 6 4 9 2 4 4 1 3 9 10 3 5 1 8 3 3 8 7 9 8 2 7 2 7 1 2 8 3 6 4 8 7 7 2 9 5 6 10 5 4 2 8 9 2 10 6 2 10 6 1 9 5 9 8 6 5 8 5 3 1
9 3 9 2 7 1 10 1 3 4 9 5 3 7 1 6 4 7 6 9 8 2 6 9 3 6 4 5 5 5 7 9 6 8 8 10 8 10 3 7 6 4 5 8 8 2 4 8 1 2 4 9 2 10 10 6 4 10 9 7 6 6 1 10 2 9 5 2 7 1 8 6 8 8 9
3 7 8 6 1 4 1 1 7 4 1 8 8 1 3 6 10 8 4 2 10 6 3 3 10 4 1 2 8 7 8 5 8 5 4 6 8 10 6 10 6 4 5 10 5 6 3 2 6 3 7 2 7 1 5 8 1 3 2 8 7 1 4 4 1 9 1 3 3 10 2 9 4 1 1
2 10 6 5 10 4 6 2 2 10 10 10 9 1 2 7 2 6 10 3 7 5 2 2 7 8 10 1 7 10 3 7 6 3 4 2 8 2 9 3 7 7 9 8 6 2 3 10 10 1 10 5 8 9 8 2 5 10 8 9 6 7 6 9 8 5 3 1 7 8 7 7 1 1 5
7 5 9 4 4 5 1 5 1 9 9 4 5 8 7 8 1 1 10 6 5 6 9 7 8 1 9 6 10 5 4 4 8 4 5 8 10 1 2 1 3 3 1 5 5 1 4 6 2 2 3 4 9 6 7 1 6 7 5 3 2 9 8 9 2 5 3 7 7 4 7 6 3 6 6
3 4 10 8 4 5 2 7 10 8 10 3 6 4 5 7 7 2 9 1 5 1 1 2 4 10 1 7 2 8 4 8 5 6 3 7 3 3 9 4 4 10 8 1 5 1 1 10 2 1 8 6 7 10 9 5 6 8 7 2 5 10 10 4 10 1 8 3 4 9 9 7 1 10 2
2 6 5 5 6 8 5 2 6 1 8 4 1 1 2 8 1 5 5 3 2 6 2 2 5 8 7 8 9 10 5 6 10 7 2 4 10 6 2 8 4 2 6 10 2 10 4 7 2 2 6 9 4 9 3 2 8 7 4 9 4 2 7 8 8 5 7 8 9 7 4 10 7 7 8
3 6 3 8 6 4 3 2 5 5 7 1 3 10 5 5 2 6 8 7 5 7 10 4 2 1 7 3 6 5 2 7 1 9 4 6 6 8 1 7 7 6 1 4 3 6 6 2 1 8 8 4 1 2 8 10 7 8 5 5 1 4 3 10 9 6 2 2 2 3 2 7 3 7 3
3 2 10 10 10 10 6 4 2 4 10 7 6 1 10 7 7 4 10 2 5 5 6 5 9 9 5 8 2 7 8 9 10 3 3 1 7 7 1 5 6 1 6 8 9 2 6 1 8 9 9 7 2 4 5 10 10 6 8 8 3 8 10 10 5 2 2 6 3 2 9 9 5 7 2
1 6 5 9 9 3 4 1 6 5 6 4 6 1 6 2 8 7 4 7 9 2 6 1 5 9 5 8 10 10 1 5 8 6 3 1 10 4 1 10 7 4 5 9 9 7 1 1 5 6 3 10 3 1 1 10 7 8 10 10 10 5 10 1 4 4 1 1 6 9 8 2 2 10 7
12×5 table
Energy variance std wl entrophy
__________ __________ __________ __________ __________
1×15 table 1×15 table 1×15 table 1×15 table 1×15 table
1×15 table 1×15 table 1×15 table 1×15 table 1×15 table
1×15 table 1×15 table 1×15 table 1×15 table 1×15 table
1×15 table 1×15 table 1×15 table 1×15 table 1×15 table
1×15 table 1×15 table 1×15 table 1×15 table 1×15 table
1×15 table 1×15 table 1×15 table 1×15 table 1×15 table
1×15 table 1×15 table 1×15 table 1×15 table 1×15 table
1×15 table 1×15 table 1×15 table 1×15 table 1×15 table
1×15 table 1×15 table 1×15 table 1×15 table 1×15 table
1×15 table 1×15 table 1×15 table 1×15 table 1×15 table
1×15 table 1×15 table 1×15 table 1×15 table 1×15 table
1×15 table 1×15 table 1×15 table 1×15 table 1×15 table
ans =
12×15 table
H1_Features H2_Features H3_Features H4_Features H5_Features H6_Features H7_Features H8_Features H9_Features H10_Features H11_Features H12_Features H13_Features H14_Features H15_Features
___________ ___________ ___________ ___________ ___________ ___________ ___________ ___________ ___________ ____________ ____________ ____________ ____________ ____________ ____________
9 5 10 8 9 7 9 7 1 5 9 9 3 7 4
6 9 3 10 5 8 2 10 6 7 3 10 4 9 4
4 4 2 6 4 3 3 2 2 8 5 2 6 9 5
9 1 9 6 8 6 7 10 6 2 4 6 6 9 8
3 4 1 6 10 4 8 6 6 6 7 8 7 9 2
2 4 10 7 7 8 3 2 7 2 10 2 6 5 7
7 5 9 8 5 1 4 8 3 1 3 1 2 5 7
3 5 10 7 5 10 4 7 4 1 8 5 5 1 9
2 8 8 8 2 8 5 4 4 10 6 2 4 5 4
3 4 7 5 5 1 2 6 7 6 8 10 1 6 2
3 10 10 7 5 9 8 1 6 2 9 10 3 2 9
1 3 6 2 9 9 1 1 7 7 3 10 10 4 8 | 3,460 | 6,845 | {"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} | 2.671875 | 3 | CC-MAIN-2022-49 | latest | en | 0.747903 |
https://www.onlinemath4all.com/divisibility_by_18.html | 1,566,734,793,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027323328.16/warc/CC-MAIN-20190825105643-20190825131643-00229.warc.gz | 910,747,402 | 13,154 | DIVISIBILITY BY 18
Divisibility by 18 :
All the numbers which are divisible by 2 and divisible by 9 are divisible by 18.
Example 1 :
Check whether 1458 is divisible by 18.
Solution :
If the given number is divisible by both 2 and 9.Then we can say it is divisible by 18.
Since the given number ends with 8, it is even number.
All even numbers are divisible by 2.
Now we need to check if it is divisible by 9. To check whether it is divisible by 9 need to calculate the sum of the digits that is,
1 + 4 + 5 + 8 = 18
18 is divisible by 9
Hence, the given number 1458 is divisible by 18.
Example 2 :
Check whether 346 is divisible by 18.
Solution :
If the given number is divisible by both 2 and 9.Then we can say it is divisible by 18.
Since the given number ends with 6, it is even number.
All even numbers are divisible by 2.
Now we need to check if it is divisible by 9. To check whether it is divisible by 9 need to calculate the sum of the digits that is,
3 + 4 + 6 = 13
18 is not divisible by 9
Hence, the given number 346 is not divisible by 18.
Example 3 :
Check whether 270 is divisible by 18.
Solution :
If the given number is divisible by both 2 and 9.Then we can say it is divisible by 18.
Since the given number ends with 0, it is even number.
All even numbers are divisible by 2.
Now we need to check if it is divisible by 9. To check whether it is divisible by 9 need to calculate the sum of the digits that is,
2 + 7 + 0 = 9
9 is divisible by 9
Hence, the given number 270 is divisible by 18.
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Sum of all three four digit numbers formed using 1, 2, 5, 6 | 954 | 4,075 | {"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} | 4.46875 | 4 | CC-MAIN-2019-35 | latest | en | 0.928375 |
http://mathhelpforum.com/algebra/223080-using-graph-solve-inequality-rational-expressions.html | 1,527,214,980,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794866917.70/warc/CC-MAIN-20180525004413-20180525024413-00513.warc.gz | 180,442,437 | 11,998 | # Thread: Using Graph To Solve Inequality Of Rational Expressions
1. ## Using Graph To Solve Inequality Of Rational Expressions
Question is here:
http://i39.tinypic.com/aexmjr.jpg
I didn't know how else to get the graph here. I don't even know where to start. My last unit was inequalities but I didn't 100% grasp the concept. What am I doing? Why is it necessary to use the graph?
2. ## Re: Using Graph To Solve Inequality Of Rational Expressions
You are given the graph of the function $\displaystyle f(x)$ and then asked to solve the inequality when is $\displaystyle f(x) \le 0$. In other words, you are asked to find all values of $\displaystyle x$ where $\displaystyle (x,f(x))$ is a point on or below the $\displaystyle x$-axis. So, where is that true? Let's look at the possibilities.
Looking at the graph makes it easy to rule out any answers that contradict the graph. For example, (b) is obviously wrong because -1 is a real number that is not equal to -2 nor 2, but $\displaystyle (-1,f(-1))$ is clearly above the $\displaystyle x$-axis.
Among (a), (c), and (d), obviously one answer is correct and the rest are not. So, for the two that are not correct, tell me a point in the given intervals where the function is above the x-axis, or tell me a point that is not in the given intervals where the function is on or below the x-axis.
3. ## Re: Using Graph To Solve Inequality Of Rational Expressions
Originally Posted by tdotodot
Question is here:
http://i39.tinypic.com/aexmjr.jpg
I didn't know how else to get the graph here. I don't even know where to start. My last unit was inequalities but I didn't 100% grasp the concept. What am I doing? Why is it necessary to use the graph?
I highlighted the x-values for which f(x) is negative. which answer you will pic
4. ## Re: Using Graph To Solve Inequality Of Rational Expressions
Originally Posted by votan
I highlighted the x-values for which f(x) is negative. which answer you will pic
So for that graph f(x) is negative when -3<x<-2 and also 0>x<2. Would this be the correct way to state it?
*also thanks for doing that graph.
5. ## Re: Using Graph To Solve Inequality Of Rational Expressions
Originally Posted by tdotodot
So for that graph f(x) is negative when -3<x<-2 and also 0>x<2. Would this be the correct way to state it?
*also thanks for doing that graph.
You are not just asked where the graph f(x) is negative. You are also asked where it is zero. You are correct that it is negative on the interval -3 < x < -2. But, 0>x<2 means x<0 and x<2. I think the notation you were looking for is 0<x<2. Again, you are correct that f(x) is negative on that interval. Now, where is it zero?
6. ## Re: Using Graph To Solve Inequality Of Rational Expressions
The basic concept here is the "intermediate value property" of continuous functions: if f(a)< 0 and f(b)> 0 (or vice-versa) there exist c, between a and b, such that f(c)= 0. That says that f(x) can change from negative to positive or positive to negative only where f is 0 or where it is not continuous. For rational expressions, that can only occur where the numerator is 0 (f(c)= 0) or the denominator is 0 (f is not continuous). So the simplest way to solve, say, f(x)> 0, is to find all values of x where either the numerator or denominator is 0. Write those values in increasing order so (together with $\displaystyle -\infty$ and $\displaystyle \infty$) they define a number of intervals then check f(x) at one value of x in each interval. If f(x)> 0 for that x, f is positive for all points in that interval and if f(x)< 0 for that x, f is negative for all points in that interval.
7. ## Re: Using Graph To Solve Inequality Of Rational Expressions
Originally Posted by votan
I highlighted the x-values for which f(x) is negative. which answer you will pic
The convention in graphing is if a point is part of the interval it is marked with a black circle, a blank circle means that point is not part of the interva. You notice on the figure, the interval -3 to -2, point -3 is marked with a black circle, point -2 is blank circle. that means the interval is one sided open. One way to write it is [-3, -2), it is open on -2, closed on -3. another way of writing it is -3 <= x <-2. Does this help you answering SlipEternal question in the previous post? | 1,115 | 4,306 | {"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} | 4.09375 | 4 | CC-MAIN-2018-22 | latest | en | 0.934972 |
https://studentshare.net/miscellaneous/283559-financial-ratios | 1,521,893,972,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257650262.65/warc/CC-MAIN-20180324112821-20180324132821-00608.warc.gz | 694,332,978 | 15,621 | StudentShare solutions
# Financial Ratios - Essay Example
## Extract of sample Financial Ratios
In addition, you can use these ratios to compare the performance of your company against that of your competitors or other members of your industry. (Alex Auerbach)
When you take profit before tax or interest (EBIT) and divide it by the difference between total assets and current liabilities, you can get a financial ratio known as return on capital employed ratio or ROCE. It is a ratio that shows the company's capital investments' profitability and efficiency. The ROCE ratio is a measure of how well a company is using capital to generate income. A high ROCE is a sign or a successful growth company and indicates that a larger mass of proceeds can be reinvested to gain more profit. However; one year ROCE evaluation should not be the basis for reinvesting. Investors should look closely on the trend over several years to have consistency. A sudden decline in ROCE signals a loss of competitive advantage.
Asset Turnover Ratio specifies the connection between assets and revenue (Revenue/Total assets). It gauges a company's efficiency in using its resources in making sales. A higher asset Turnover is better. It also specifies pricing strategy: companies with low profit margins are inclined to have high asset turnover, whereas those with high profit margins tend to have low asset turnover. This ratio is important to settle on the amount of sales that are produced from each dollar of assets.
Return on Sales or Profit Margin
To evaluate a company's operational efficiency, return on sales ratio is used. ROS is also recognized as a company's "operating profit margin". It is calculated using this formula: Net Income before interest and tax divided by sales. Investopedia says "This measure is helpful to management, providing insight into how much profit is being produced per dollar of sales. As with many ratios, it is best to compare a company's ROS over time to look for trends, and compare it to other companies in the industry. An increasing ROS indicates the company is growing more efficient, while a decreasing ROS could signal looming financial troubles."
The best way to show how these financial ratios are interpreted is to give sample computations and financial analysis. Let us take the data from Medquip Ltd. (Details attached).
## Summary
One of the time-tested methods in analyzing a business is the use of financial ratios. Big firms and businesses make use of financial ratio analysis to know more about a company's potential and current financial health. Specifically, financial ratios can be used in two different but equally useful ways:
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## a numerical accuracy issue using MKL and MATLAB
Beginner
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Dear all,
Lately, I have run into an accuracy issue while using Intel MKL. Let me briefly summarize what I did, I have been using C++ as the main language and MTL4 as the matrix library. Since MTL4 gives direct access to the internal data represetation of CSR matrices through pointers, it is easy to interface MKL and use the routines from the MKL library.
I have some template functions for MKL library routines, such as
+ sparse matrix - dense vector multiplication
+ symmetric sparse matrix - dense vector multiplication
+ dense matrix ^T dense matrix multiplication
+ dense matrix dense matrix multiplication
I programmed a symmetric lanczos solver in C++ however results of some orthogonalizations in this routine are differing from the results of MATLAB which is also known to use MKL internally. Is there a way to increase the numerical accuracy of the computations in MKL? Especially for an operation like
t - Z Z^T M f t
Basically this is a projection operation with a projector of P = I - Z Z^T M
t = P t,
where Z is a dense rectangular matrix, M is a sparse matrix and f and t are dense vectors. I programmed this operation in 4 operations:
+ I first form Mf by sparse blas routines and keep this in a variable called Mq
+ I compute Z^T Mq by blas level 2 routines, namely, cblas_dgemm and store the result as vec1
+ Continue with Z vec1, in the same way, and store that as vec2
+ and result the computations with daxpy, for t = t - alpha vec2
So basically I chain some matrix -vector operations.
I wrote this part in detail because my MATLAB and MKL supported C++ implementation start to diverge after this operation at some point in the iteration. And in the end if I use the vectors of my C++ implementation, I can not get the results that MATLAB gives for instance some similar operation to the above cases in MATLAB results in
9.949738746078907e+02
and in C++ implementation
995.028
so the result differences are not very low.
I was wondering if I am doing something wrong and could there be a way to improve the accuracy of sparse(dense) matrix- vector multiplication accuracy in MKL.
Thanks in advance and best regards,
Umut | 537 | 2,377 | {"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-2024-26 | latest | en | 0.909954 |
https://m.wikihow.com/Create-a-Metal-Backsplash | 1,579,863,306,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250619323.41/warc/CC-MAIN-20200124100832-20200124125832-00420.warc.gz | 557,991,462 | 59,932 | # How to Create a Metal Backsplash
Co-authored by wikiHow Staff
Updated: March 29, 2019
A backsplash is a stylish way of keeping kitchen ingredients from spattering on and potentially doing damage to walls. You've probably seen stainless steel backsplash sheets in restaurant kitchens, but there are plenty of fashionable materials you can use at home, like copper or aluminum mosaic tiles.[1] First you'll have to prepare your work area and materials. Then, with the right tools, you can install your new metal backsplash.
### Part 1 of 3: Prepping to Install a Backsplash
1. 1
Measure the area of your backsplash. It's important to do this before anything else. Knowing the area you need your backsplash to cover will allow you to calculate the cost of materials, which could influence the kind of metal you use. To calculate the dimensions for your backsplash:
• Divide the wall where you plan on installing the backsplash evenly into rectangles to account for cabinets, windows, large appliances and so on. Measure the height and the width of each rectangle. Multiply these numbers together to find the area of each.
• If you have a diagonal angle to account for, extend the base of the diagonal horizontally until it is even with the top of the diagonal. Connect the horizontal extension and the top of the diagonal to make a right triangle.
• Calculate the area of triangles by multiplying the base and height of the triangle, then divide that number by 2 (triangle area = (base x height) / 2 ).
• Add together the area calculations to get the total area in units squared (for example, in², ft², cm², m²). Multiply this number by 1.1 so you have extra backsplash material to fill in gaps or replace materials damaged during installation.[2]
2. 2
Determine the backsplash material and calculate the cost. Now that you know the total area needed for your backsplash, you can begin shopping around. You can get quotes for backsplash material at your local big box hardware store (like Lowes, Home Depot, and Home Hardware) or online.[3]
• An online backsplash provider may be the easiest option for getting a quote. In many cases, once you choose a backsplash material, you only need to input the total area on the site to determine the cost.
• Before ordering your backsplash online, verify the seller's return policy. If tile is damaged during shipping, you may end up with an expensive and unusable backsplash.[4]
• Common backsplash metals include:
• Copper, which is lightweight. This makes it easier to install. Copper works well with rustic and retro designs.
• Stainless steel, which comes in many shades. It is durable and is easy to maintain. It works well with modern decor.
• Brass, which requires higher maintenance to prevent rust and keep clean.However, this metal has a unique, classical appearance.[5]
3. 3
Gather your supplies for installation. Most of the supplies you'll need can be found at your local hardware store or home center. But before you make any purchases, check the specifications for your backsplash to determine:
• Whether or not you need adhesive, grout, and the tools associated with these components. Some varieties of backsplash already come with adhesive backing and will not need additional adhesive, grout, and their related tools.
• What kind of backsplash cutting device you will need, if any. You'll only need to cut the backsplash to account for irregular angles, like you would along a diagonal.[6]
4. 4
Ready the area for installation. Clear out all appliances and countertop items from the backsplash installation area. If your stove is in this area, move it slightly away from the wall and disconnect it. Use tape to clearly define the edges of the backsplash installation area, including walls and countertops. Also:
• Use a cover for your countertops, like a drop cloth, cardboard, or butcher paper, to prevent damage to it.
• Remove fixtures that could get damaged or get in the way. Some fixtures you might have to remove include light/disposal switches, cover plates for outlets, and so on.
• Turn off all electricity to your kitchen before installation. Metal backsplash has a greater risk transmitting electrical shocks. Prevent this by switching your kitchen electricity in your fuse box or electrical service panel to "OFF."[7]
5. 5
Check walls for unevenness. If your walls have slight bumps, depressions, or other imperfections, these will likely be visible in your installed backsplash. Take a metal straightedge and hold it to the wall to determine if it is level.
6. 6
Do a trial run for your design. Sometimes a design won't look as good as you initially thought it might. Especially if you can return your backsplash, use tape to hang a section of it to get a rough idea of how it will look.
• In some cases, there may be plastic or a protective cover on the surface of your backsplash. Even so, a trial run will better help you visualize the next phase of this process: installation.[9]
### Part 2 of 3: Installing a Metal Backsplash
1. 1
Prepare your adhesive. Always follow your backsplash installation instructions, but in many cases, this will be a kind of mortar. However, it's becoming more and more common to produce backsplashes with their own adhesive, in which case you won't need a separate adhesive, mortar, or the tools associated with these components.
• If using mortar, mix it in one of your two buckets according to its preparation instructions.[10]
2. 2
Apply your adhesive to the wall. This step can be skipped if your backsplash has its own adhesive backing.[11] If not, use an adhesive applicator to apply adhesive to the wall where you will attach the backsplash. Do so according to the adhesive's directions. If using mortar:
• Use the flat side of your V-notch trowel to apply a layer of mortar to the wall where you will install the backsplash.
• Use the other side of your trowel, which should have V-shaped notches, to smooth the mortar so it is of uniform thickness throughout.[12]
3. 3
Attach the backsplash material to the wall. Take your backsplash and press it into the adhesive or mortar using light, even pressure. Depending on your adhesive and backsplash, the time you should hold this pressure may vary. Follow the product instructions for best results.
• Continue applying your backsplash, section by section. Pay close attention to the joints between tiles if your backsplash has any. These should line up evenly.
• After you finish attaching backsplash to a section, apply your adhesive to a new section in the same fashion previously described, then attach more backsplash.
• Take great care when attaching your backsplash to keep it level, otherwise it might turn out uneven or lopsided. A laser level can be helpful in this instance, but a traditional level can work just as well.[13]
4. 4
Prevent unevenness in the backsplash depth. After finishing a section, but before applying more mortar for your next section, take your wooden board and hold it flat across two backsplash segments already attached. Use your hammer to lightly tap the wood block where the backsplash is attached.
• This is an optional step. However, by using this simple technique, you can correct unevenness in your backsplash before the adhesive hardens and cures. After hardening and curing, uneven depth will be hard to correct.[14]
5. 5
Remove the backsplash protective covering, if applicable. In most cases, after 10 to 15 minutes has passed, you can remove the protective paper or plastic from the surface of your backsplash. Plastic can usually be peeled off easily. You may have to dampen your sponge and moisten protective paper before it pulls free.
• When removing the protective covering on your backsplash, use firm but gentle force. You should keep the covering on the backsplash until this point to prevent damage.[15]
6. 6
Adjust the backsplash by hand as necessary. At this point, the adhesive/mortar will likely still be somewhat movable. Use your hands to make adjustments to the backsplash. For example, you might need to adjust the joints separating metal mosaic tile backsplashes.[16]
### Part 3 of 3: Grouting and Finishing the Backsplash
1. 1
Brush off remaining adhesive, if necessary. After peeling off the protective covering on your backsplash, there may be some adhesive or paper still on it. Gently scrub this with a non-abrasive nylon brush, then wipe the surface with a clean, damp sponge.[17]
2. 2
Mix and apply grout to the backsplash, if necessary. This will likely be necessary for metal mosaic tiles, though backsplash that came with adhesive might not require grouting. Follow the label directions to mix your grout in your second bucket. Only use non-sanded grout to preserve the finish of your metal. Follow the grout instructions for best results, but generally, to apply the grout:
• Use a rubber grout float to spread the grout onto your backsplash. Using firm pressure, force the grout into the joints between backsplash materials.
• Wait between 15 minutes and one hour (depending on the kind of grout you used) before using a dampened sponge to remove extra grout from the surface of the backsplash. Use light pressure; too much pressure can remove grout from the joints.[18]
3. 3
Clean any remaining cloudiness from the backsplash. Only do this after the grout has cured. The information on the time this takes will be on the package or in the instructions. Use a clean, dampened sponge to wipe off any remaining grout on the surface of your tile that's making it hazy.
• In some cases, you may need to use a cleaning agent, like an industrial alcohol based cleaner, to remove glue or stubborn grout.[19]
## Community Q&A
200 characters left
## Warnings
• When using tools, always do so with caution and according to its directions. Failing to use tools properly could result in serious injury or harm.
Thanks!
## Things You’ll Need
• Backsplash cutting device (like, for example, a wet tile saw with a diamond tipped blade; optional)
• Block of wood
• Brush (non-abrasive, nylon)
• Grout (non-sanded kind; optional)
• Grout float (rubber kind preferred)
• Hammer
• Level
• Measuring tape
• Metal backsplash material (like a stainless steel sheet or metal mosaic tiles)
• Mixing bucket (x2)
• Sponges (clean; several)
• V-notch mortar trowel (5/32" preferred)
• Wall tile mortar (or similar adhesive)
• Writing utensil and paper (for recording dimensions)
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Co-Authored By:
wikiHow Staff Editor
This article was co-authored by our trained team of editors and researchers who validated it for accuracy and comprehensiveness. Together, they cited information from 19 references. | 2,333 | 10,645 | {"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} | 3.703125 | 4 | CC-MAIN-2020-05 | latest | en | 0.898943 |
https://schoolbag.info/physics/ap_physics_2017_3/7.html | 1,723,231,723,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640768597.52/warc/CC-MAIN-20240809173246-20240809203246-00652.warc.gz | 401,001,284 | 9,169 | Strategies to Approach the Questions on the Exam - Develop Strategies for Success - 5 Steps to a 5: AP Physics 1: Algebra-Based 2017 (2016)
## 5 Steps to a 5: AP Physics 1: Algebra-Based 2017 (2016)
### Strategies to Approach the Questions on the Exam
IN THIS CHAPTER
Summary: This chapter contains tips and strategies that apply to both the free-response and the multiple-choice sections of the AP Physics 1, Algebra-Based Exam. First you”ll find information about the tools (calculator, equation sheet, and table of information) you can use on the exam and strategies to make the best use of these tools. Then you”ll find strategies for dealing with two common types of AP Physics 1 questions: ranking questions and questions about graphs. Both of these types of questions can be found in the multiple-choice section and in the free-response section of the exam.
Key Ideas
Although you can use a calculator on the exam, you should use it only when it”s actually required to do a calculation. It won”t help you answer the vast majority of the questions on the AP Physics 1, Algebra-Based Exam.
Questions involving a ranking task often require analysis more than calculation. Ranking questions can be multiple choice or free response. For free-response questions, be sure to indicate clearly the order of your ranking, and if two of the items to be ranked are equal, be sure to indicate that, too.
Graph questions are straightforward, because there are only three things you can do with a graph—take the slope, compute the area under the graph, and read one of the axes directly. Pick the right one, and you”re golden.
Tools You Can Use and Strategies for Using Them
Like all AP exams, the AP Physics 1, Algebra-Based Exam consists of both multiple-choice questions (Section I) and free-response items (Section II). The three tools discussed below can be used on both sections of the exam. 1 Keep in mind that just because you can use these tools, it doesn”t necessarily mean you should .
Calculator
The rules of acceptable calculators are the same as those on the SAT or the AP math exams—pretty much any available calculator is okay, including scientific and graphing calculators. Just don”t use one with a QWERTY keyboard, or one that prints the answers onto paper. 2 You”re not allowed to share a calculator with anyone during the exam.
Wait! Don”t Touch That Calculator!
Will a calculator actually help you? Not that much. Exam authors are required to write to the specific learning objectives in the “curriculum framework.” 3 Of the 140 learning objectives in AP Physics 1 curriculum, only 21 allow for calculation ! For 119 of 140 learning objectives, something else entirely—qualitative prediction, semiquantitative reasoning, analysis and evaluation of evidence, description of experiment, explanation, etc.—is required. You will need to calculate, but not often.
And even then, the “calculation” on the AP exam usually does not require a calculator. For example, you”ll answer in symbols rather than numbers; you”ll be asked to do an “order of magnitude estimate” in which only the power of 10 matters 4 ; the numbers involved will be simple, like 4 × 2; or the choices will be so far separated that only one answer will make sense, whether or not you actually carry out the calculation.
But seemingly every problem we did for class required solving an equation and plugging values into that equation. My teacher assigned us a bunch of problems on the computer, with programs like WebAssign. I could never have done those problems without a calculator!
That”s very likely true. Being able to perform calculations is, in fact, a first step toward more difficult physics reasoning. But it”s very simple—the AP exam will only sometimes require a calculation. And when it does, a calculator will only sometimes be necessary. So wean yourself off of the calculator. Practice approaching every problem with diagrams, facts, equations, symbols, and graphs. Only use the calculator as a last step in a homework problem. Then you”ll be well prepared for the kinds of things you”ll see on the AP exam.
The Table of Information
There”s no need to memorize the value of constants of nature, such as the mass of an electron or the universal gravitation constant. These values will be available to you on the table of information you”ll be given.
The Equation Sheet
A one-page list of many relevant equations will be available to you on both sections of the exam. 5 You will be able to see the official equation sheet ahead of time at the AP Physics 1 portion of the College Board”s website (https://www.collegeboard.org ).
Wait! Don”t Touch That Equation Sheet!
Will the equation sheet actually help you? It won”t help you that much. Too often, students interpret the equation sheet as an invitation to stop thinking—“Hey, they tell me everything I need to know, so I can just plug-and-chug through the exam.” Nothing could be further from the truth.
First, the equation sheet will likely present most equations in a different form than you”re used to, or use different notation than your textbook or your class. So what—you”ve already memorized the equations on the sheet. It might be reassuring to look up an equation during the exam, just to make sure you”ve remembered it correctly, which is really the point of the equation sheet. But beware. Use your memory as the first source of equations.
If you must use the equation sheet, don”t go fishing ! If a question asks about a voltage, don”t just rush to the equation sheet and search for every equation with a V in it. You”ll end up using , where the V means volume, not voltage. You”d be surprised how often misguided students do this. Don”t be that person.
So you”re saying I”ll be given a calculator and an equation sheet, but neither will be much use. Why would I be given useless items?
Suffice it to say that many years ago, calculation was indeed the most important aspect of an AP Physics exam, so the calculator was indispensable. Back then, students were expected to memorize equations. As calculators became more sophisticated, students began to game the test by programming equations into their calculators, effectively gaining an unfair advantage. So the equation sheet was provided to everyone, negating that advantage. Now that calculation is a far less significant part of AP Physics, the calculator will only rarely be useful. But since you might need it a few times in an exam, it”s still allowed.
Strategies for Questions That Involve a Ranking Task
You already know there won”t be a lot of straight “calculate this” type of questions. So what kinds of questions will there be? One very different sort of question from the standard textbook end-of-chapter homework problem is the ranking question. It can be found among the multiple-choice questions of Section I or in Section II as a free-response question. Here”s an example:
Cart A takes 5 s to come to rest over a distance of 20 m. Cart B speeds up from rest, covering 10 m in 10 s. And Cart C moves at a steady speed, taking 1 s to cover 50 m. All carts have uniform acceleration. Rank the carts by the magnitude of their acceleration. If more than one cart has the same magnitude of acceleration, indicate so in your ranking.
Notice that you are emphatically not asked to calculate the acceleration of each cart. Usually, a ranking task can be solved more simply with conceptual or semiquantitative reasoning than with direct calculation.
In this example, the conceptual approach is probably best. Acceleration is defined as how quickly an object changes its speed. “Magnitude” of acceleration means ignore the direction of acceleration. 6 Cart C doesn”t change its speed at all, so it has the smallest acceleration. Cart A goes farther than Cart B and takes less time to do so. Since both change their speeds from or to rest, Cart A must change its speed more quickly than Cart B.
The multiple-choice ranking tasks will have answer choices formatted as inequalities: A > B > C. If two were equal, then you”d see something like A = B > C, which would mean A and B are equal, but are both greater than C.
In the free-response section, you can format your answer to a ranking task any way that is clear. For example, you could list: “(Greatest) A, B, C (least).” Don”t forget to make some notation if two of the choices were equal; circle those two and write “these are equal” or something that is crystal clear.
Exam Tip from an AP Physics Veteran
For some people, semiquantitative and qualitative reasoning is much more difficult than just making calculations. You have every right to start a ranking task with several calculations! Then just rank your answers numerically. Sure, some questions will require you to explain your ranking without reference to numbers, but still, feel free to answer with numbers first, and then refer to the equations.
In the example above, you could make the calculation for each cart. Use the kinematics equations detailed in Chapter 10 . You can calculate that Cart A has an acceleration of 1.6 m/s per second. Cart B has an acceleration of 0.2 m/s per second. And Cart C has an acceleration of zero.
How could you follow your calculation with a nonnumerical explanation, then? Look at how the equations simplified. For carts A and B, the initial or final speed of zero meant that when you solved in variables for a , you got . Cart A has both a bigger distance to travel and a smaller time of travel. Cart A”s numerator is bigger, denominator smaller, and acceleration bigger than Cart B”s.
Strategies for Questions That Involve Graphs
Analyzing data in graphical form will be a skill tested regularly on the AP exam. Good graph questions are straightforward, because there are only three things you can do with a graph . Pick the right one, and you”re golden.
When you see a graph, the first step must be to recognize the relevant equation . In this situation, which equation from the equation sheet relates the y - and x -variables? I truly mean the equation, not just the units of the axes. Then, the equation will lead you to one of the following three approaches.
What are the three things we can do with this graph?
A box sits on a smooth, level surface. The box is attached to a spring. A person pushes the box across the surface, compressing the spring. For each distance the spring compresses, the force applied by the person on the box is measured.
1. Take the Slope of the Graph
You certainly understand that the slope represents the change in the y -variable divided by the change in the x -variable. But to really understand what the slope of a graph means, you can”t just say “it”s the change in force divided by the change in distance.” Slopes of graphs generally have a physical meaning that you must be able to recognize.
Always start with the relevant equation. If you suspect you”re looking for a slope, solve the relevant equation for the y -axis variable, then compare the equation to the standard equation for a line: y = mx + b . Here, we”re talking about the force of a spring and the distance the spring is stretched—that”s covered by the equation F = kx. F and x are the vertical and horizontal axes, respectively.
This process of circling the y -variable, circling the x -variable, and then circling the slope will work with any equation. 7 Here, the slope represents k , the spring constant for the spring. So what is the numerical value of the spring constant? You can calculate the slope by drawing a best-fit line, choosing two points on the line that are not data points, and crunching numbers.
Using the two circled points on the line, the “rise” is (50 N – 16 N) = 34 N. The “run” is (3 cm – 1 cm) = 2 cm. So the slope—and thus the spring constant of the spring—is .
1. Calculate the Area Under the Graph
Always start with the relevant equation. The meaning of the area under the graph is generally found by looking at an equation that multiplies the vertical and horizontal axes. Here, that would be force times distance. Sure enough, the equation for work is at work here: W = F · Δx || . The force applied by the man is parallel to the box”s displacement, so the work done by the man is the multiplication of the vertical and horizontal axes. That means that to find the work done, take the area under the graph.
How much work did the man do in compressing the spring 3 cm? Usually, when you”re taking an area under a graph on the AP exam, just estimate by breaking the graph into rectangles and triangles. In this case the graph is an obvious triangle. The area of a triangle is (½)(base)(height) = (½)(0.03 m)(50 N) = 0.75 J . If instead the question had asked for the work done in compressing the spring 2 cm, I”d do the same calculation with 0.02 m and 34 N.
Two things to note about that calculation: First, an area under a graph, like a slope, has units. But the units are not “square units” or “square meters.” The “area” under a graph isn”t a true physical area; rather, it represents whatever physical quantity is found by multiplying the axes. This area represents the work done by the man, so it should have units of joules. Second, you”ll note that I used 0.03 m rather than 3 cm in calculating the area under this graph. Why? Because without that conversion, the units of the area would have been newton centimeters (N·cm). I wanted to get the work done in the standard units of joules, equivalent to newton meters (N·m). So, I had to convert from centimeters to meters.
1. Read One of the Axes Directly
Often a question will ask for interpolation or extrapolation from a graph. For example, even though the man never used 60 N of force, how far would the spring compress if he had used 60 N? Just extend the best-fit line, as I already did, and read the horizontal axis: 3.6 cm.
Exam Tip from an AP Physics Veteran
When you draw a best-fit line, just lay your ruler down and truly draw the fit as best as you can. Never connect data point-to-point; never force the best-fit line through (0,0); never just connect the first and last data points. In fact, it”s best if you extend the line of best fit as far as you can on the graph, so that you can answer extrapolation questions quickly and easily.
Finally, note that many graphs on the AP Physics 1 Exam will include real data, not just idealized lines. Be prepared to sketch lines and curves that seem to fit the general trend of the data.
1 This represents a change from previous AP Physics exams; you used to have access to a calculator only on the free response.
2 Does anyone actually use printing calculators anymore?
3 You can refer to the curriculum framework via the AP Physics 1 portion of the College Board”s website, https://www.collegeboard.org .
4 See Chapter 15 (Gravitation) for specific examples of order-of-magnitude estimates.
5 Once again, this is a change for AP Physics 1; the equation sheet used to be for free response only.
6 This can”t be determined here anyway, because although Cart A has acceleration opposite its motion, and Cart B has acceleration in the same direction as its motion, we don”t know which ways these two carts are moving. But who cares, for this particular question.
7 And if there”s a leftover term with a plus or minus sign, you”ll recognize that as the b -value, the y -intercept of the graph.
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if we know that 50% of the customers that shop at the florist in a particular town are male(the probability of any given customer at the florist being a male is 0.50),find the probability that more than 20 but less than 30 of the next 50 customers are males. | 69 | 283 | {"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} | 2.734375 | 3 | CC-MAIN-2017-26 | latest | en | 0.964631 |
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# Macroeconomics Exam Review 52 - iv Therefore x = ≠z is an...
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Unformatted text preview: iv. Therefore ( x ) = ≻ ( z ) is an open neighborhood of x such that x ≻ z for every x ∈ ( x ) Similarly, ( z ) = ≺ ( x ) is an open neighborhood of z such that z ≺ x for every z ∈ ( z ). Consequently x ≻ z for every x ∈ ( x ) and z ∈ ( z ) 3. ≿ ( y ) = ( ≺ ( y ) ) (Exercise 1.56). Therefore, ≿ ( y ) is closed if and only if ≺ ( y ) is open (Exercise 1.80). Similarly, ≾ ( y ) is closed if and only if ≻ ( y ) is open. 1.237 1. Let = { ( x , y ) ∈ × : x ≿ y } . Let (( x , y )) be a sequence in which converges to ( x , y ). Since ( x , y ) ∈ , x ≿ y for every . By assumption, x ≿ y . Therefore, ( x , y ) ∈ which establishes that is closed (Exercise 1.106) Conversely, assume that is closed and let (( x , y )) be a sequence converging to ( x , y ) with x ≿ y for every . Then (( x , y )) ∈ which implies that...
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## This note was uploaded on 01/16/2012 for the course ECO 2024 taught by Professor Dr.dumond during the Fall '10 term at FSU.
Ask a homework question - tutors are online | 427 | 1,317 | {"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-2017-26 | latest | en | 0.884145 |
https://electronicsreference.com/analog/wheatstone_bridge/ | 1,716,226,884,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058293.53/warc/CC-MAIN-20240520173148-20240520203148-00528.warc.gz | 195,506,321 | 35,507 | # Wheatstone Bridge
A Wheatstone bridge is a circuit that uses three resistors and a voltmeter to measure an unknown quantity of electrical resistance.
The Wheatstone bridge is constructed by connecting two sets of two resistances in series/parallel bridge configuration. A voltmeter connects the two branches.
In the circuit above, Rx (the resistor on the lower right) is the unknown resistance that will be measured. R1 and R3 are resistors with known resistance values, and R2 is a potentiometer (an adjustable resistor). The circle in the middle labeled VG is a voltmeter; originally this would have been a galvanometer, hence the ‘G’ subscript.
While we would probably just use a multimeter to measure resistance today, the Wheatstone bridge continues to be used as a common example for further understand basic principles of analog resistive circuits.
## Alternate Wheatstone Bridge Circuit Diagram
Although the Wheatstone bridge circuit is traditionally shown using the diamond configuration (which we use throughout this article), it is electrically identical to the series-parallel resistor circuit shown below:
It’s a useful exercise to prove this to yourself and to understand why we can use different layouts to show the same circuit.
## Balanced Wheatstone Bridge
A Wheatstone bridge circuit is known as either balanced or unbalanced depending on the values of the resistors.
The first scenario is a properly balanced Wheatstone bridge circuit.
Balanced means that no current flows through the voltmeter in the middle; in other words, VG = 0.
This is shown in the figure below (note 0 V through the voltmeter):
In order for the circuit to be balanced, the points being measured must be at the same voltage. In other words, the voltage drop across R1 and R3 must be equal.
This condition is established by the ratios of resistance values. If the ratio between the resistors on the left (R1 and R2) and resistors on the right (R3, and Rx) are equal, then the circuit will be balanced and current won’t flow through the voltmeter:
\frac{R_1}{R_2}=\frac{R_3}{R_x}
We can solve for Rx:
R_x=\frac{R_2R_3}{R_1}
So if R1, R2, R3, and are all known, we can find the value of Rx.
## How a Wheatstone Bridge Circuit Works
The Wheatstone bridge circuit uses the potentiometer (R2) and voltmeter (VG) together.
When the circuit is first constructed, chances are high that the circuit will be unbalanced. The voltmeter will measure a voltage difference between the right and left sides of the circuit (VG ≠ 0).
The potentiometer is then adjusted until the voltmeter measures zero volts, i.e. VG = 0.
Once the voltage measures zero and the circuit is balanced, the unknown resistance can be determined using the equation for Rx:
R_x=\frac{R_2R_3}{R_1}
### Different Wheatstone Bridge Configurations
The equations given above allow us to solve for an unknown resistance in all four of the possible locations.
If we change the value of Rx to R4, we can use the following diagram to solve for each of the different resistor locations in the Wheatstone bridge.
R_1=\frac{R_2R_3}{R_4}
R_2=\frac{R_1R_4}{R_3}
R_3=\frac{R_1R_4}{R_2}
### Wheatstone Bridge Problem No. 1
You have three resistors with known values connected in a circuit as shown.
Resistor R1 is 10 ohms, resistor R2 is 8 ohms, resistor R3 is 15 ohms and the unknown resistance is Rx. The voltmeter shows measures 0 volts indicating a balanced circuit.
What is the value of resistor Rx?
Using the Wheatstone bridge formula above:
R_x=\frac{R_2R_3}{R_1}=\frac{(8\Omega)(15\Omega)}{(10\Omega)}=\frac{120\Omega ^2}{10\Omega}=12 \Omega
Using the classic Wheatstone bridge problem, we find that the value of Rx must be twelve ohms (12Ω).
### Wheatstone Bridge Problem No. 2
What value of resistor R2 will balance the circuit shown below?
This time R2 is the unknown resistance. How can we find it?
Resistor R1 is 4 ohms, resistor R3 is 10 ohms and resistor R4 is 16 ohms. The voltmeter measures 0 volts, indicating a balanced circuit.
We can always use the ratios of the resistance values between the left and right sides to find any unknown resistance in a balanced circuit.
Since we’re solving for R2, let’s put that on top. Then, the ratio of R2 to R1 will be equal to the ratio of R4 to R3:
\frac{R_2}{R_1}=\frac{R_4}{R_3}
We can solve for R2 by multiplying both sides by R1:
R_2=\frac{R_1R_4}{R_3}=\frac{(4\Omega)(16\Omega)}{10\Omega}=\frac{64\Omega^2}{10\Omega}=6.4\Omega
## Kirchhoff’s Laws for Wheatstone Bridge
A deeper analysis of the Wheatstone bridge can be performed by using Kirchoff’s current and voltage laws.
### Kirchhoff’s Current Law for Wheatstone Bridge
Kirchhoff’s current law (KCL) states that the electric current flowing into a junction must equal the current flowing out of that junction.
Let’s see how this applies to the Wheatstone bridge.
In the circuit above, we have labeled the four junctions of the circuit A, B, C, and D.
We have also labeled the current through R1 as I1, the current through R2 as I2, the current through R3 as I3, and the current through R4 as I4.
The total current flowing into and out of the battery is labeled IT, and the current through the voltmeter/galvanometer is labeled IG.
Since we have four junctions, we can set up four equations using KCL:
#### KCL for Junction A:
Current flowing into junction A = Current flowing out of junction A
(1) IT = I1 + I3
#### KCL for Junction B:
Current flowing into junction B = Current flowing out of junction B
(2) I1 = I2 + IG
#### KCL for Junction C:
Current flowing into junction C = Current flowing out of junction C
(3) I3 + IG = I4
#### KCL for Junction D:
Current flowing into junction D = Current flowing out of junction D
(4) I2 + I4 = IT
#### KCL with Balanced Wheatstone Bridge
When the Wheatstone bridge circuit is balanced, IG will equal zero (IG = 0).
In that case, we can rewrite equations (2) and (3):
(2b) I1 = I2 + IG => I1 = I2
(3b) I3 + IG = I4 => I3 = I4
We’ll use these results combined with Kirchhoff’s Voltage Law to derive the relationship between resistors in a balanced circuit.
### Kirchhoff’s Voltage Law for Wheatstone Bridge
Kirchhoff’s voltage law (KVL) states that for any loop in the circuit, the total voltage must equal zero.
We can divide the circuit into two loops: ABC (or ABCA) and BCD (or BCDB). Let’s use KVL to form an equation for each loop.
Remember that the sign for each term is determined by the direction that we have assumed current will travel in (see the lesson on KVL for more information).
#### KVL for Loop # 1 (ABCA)
Voltage across R1 (V1) + Voltage across galvanometer (VG) – Voltage across R3 (V3) = 0
(5) V1 + VG – V3 = 0
Using Ohm’s Law (V = IR), we can restate this in terms of current and resistance:
(6) I1R1 + IGRG – I3R3 = 0
#### KVL for Loop # 2 (BCDB)
Voltage across galvanometer (VG) + Voltage across R4 (V4) – Voltage across R2 (V2) = 0
(7) VG + V4 – V2 = 0
(8) IGRG + I4R4 – I2R2 = 0
#### KVL With Balanced Wheatstone Bridge
Under balanced conditions, IG = 0 so that equations (6) and (8) can be rewritten:
(6b) I1R1 = I3R3
(8b) I4R4 = I2R2
If we multiply equation (8b) by equation (6b), we can solve for R4:
I_4R_4(I_1R_1)=I_2R_2(I_3R_3)
R_4=\frac{I_2R_2I_3R_3}{I_1R_1I_4}
Now we can simplify using the results we found from KCL in a balanced circuit (I1 = I2) and (I3 = I4). We can use these relations to cancel out the two current terms in the numerator with the two in the denominator:
R_4=(\frac{I_2I_3}{I_1I_4})\frac{R_2R_3}{R_1}=(1)\frac{R_2R_3}{R_1}
R_4=\frac{R_2R_3}{R_1}
We have successfully used KCL and KVL to derive the relation between the resistances in a balanced Wheatstone bridge circuit.
## Bridge Circuits
The Wheatstone bridge is the most popular example of a bridge circuit.
A bridge circuit is characterized by having two branches of electric circuit that are connected by a third branch that bridges them.
Variations of the Wheatstone bridge include the Wien bridge, Maxwell bridge, and Heaviside bridge.
A variation on the bridge circuit is also used in the bridge rectifier, where four diodes are used to change DC current into AC.
Like many bridge circuits, the Wheatstone bridge is a clever way of using basic electrical principles to achieve an interesting goal. | 2,238 | 8,309 | {"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": 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-2024-22 | longest | en | 0.923185 |
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# The third term of an AP is 7 and its ${{\text{7}}^{{\text{th}}}}$ term is 2 more than thrice of its ${{\text{3}}^{{\text{rd}}}}$ term. Find the first term, common difference and the sum of its first 20 terms.
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Hint- Here, we will be using the formulas of an AP series.
Given, third term of AP, ${a_3} = 7$
Also, ${a_7} = 3{a_3} + 2$ where ${a_3}$ is the third term of AP and ${a_7}$ is the seventh term of AP
Since, in an AP series, the common difference $d$ remains the same.
Since, ${n^{th}}$ term of AP is given by ${a_n} = {a_1} + \left( {n - 1} \right)d$ where ${a_1}$ is the first term of AP series and $d$ is the common difference of AP.
$\Rightarrow {a_3} = 7 \Rightarrow {a_1} + \left( {3 - 1} \right)d = 7 \Rightarrow {a_1} + 2d = 7{\text{ }} \to {\text{(1)}} \\ \Rightarrow {a_7} = 3{a_3} + 2 \Rightarrow {a_1} + \left( {7 - 1} \right)d = 3\left[ {{a_1} + \left( {3 - 1} \right)d} \right] + 2 \Rightarrow {a_1} + 6d = 3\left[ {{a_1} + 2d} \right] + 2 \\ \Rightarrow {a_1} + 6d = 3{a_1} + 6d + 2 \Rightarrow {a_1} = 3{a_1} + 2 \Rightarrow 2{a_1} = - 2 \Rightarrow {a_1} = - 1 \\$
Put this value of ${a_1}$ in equation (1), we get
$- 1 + 2d = 7 \Rightarrow 2d = 8 \Rightarrow d = \frac{8}{2} = 4$
As we know that sum of first $n$ terms in AP is given by ${{\text{S}}_n} = \frac{n}{2}\left[ {2{a_1} + \left( {n - 1} \right)d} \right]$
Let’s substitute the values of ${a_1}$ and $d$, we get
Now, sum of its first 20 terms ${{\text{S}}_{20}} = \frac{{20}}{2}\left[ {2 \times \left( { - 1} \right) + 4\left( {20 - 1} \right)} \right] = 10\left( { - 2 + 76} \right) = 740$.
Therefore, the first term of the given AP series with common difference 4 is $- 1$ and the sum of its first 20 terms is 740.
Note- In these types of problems, find the common parameters which includes the first term, common difference and the total number of terms in the AP series using the given data and then find whatever is asked in the problem. | 805 | 2,097 | {"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": 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-2024-30 | latest | en | 0.814742 |
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Ray Tools: Polynomial Equations 1.0
Company: MEA Mobile
Date Added: September 21, 2014 | Visits: 157
Understanding polynomial equations (equations with multiple x to the power of components) can be daunting but this program simplifies understanding. Both from an educational and an engineering standpoint where the user needs to understand the relationship between the coefficients and how the curve propegates this program solves equations in a user friendly environment. Both a help file and glossary of terms have been included to simplify understanding. The equation takes the form of y = ax^3 + bx^2 + cx + d Minimum and maximum points in the curve are calculated if they exist and are displayed. A trace function allows the user to visually scroll through the equation to find Roots where they exist. This is evident visually where Y = 0 and the curve crosses or intersects the X axis. If the user prefers to solve the equation using just the coefficients then a Solve For: button allows the problem to be setup using just the coefficients that allows the user to adjust the values until the correct equation results. A graph shows the actual curve and the user can manually change the scale to achieve the desired view. If the curve does not intersect the X axis the roots are unknown. The equation changes as user inputs change along with the graphical output. Help and glossary screens are also provided to further help in understanding polynomial equations. When you are done simply Email the results to your home computer for further analysis. This program is a must for Students and Engineers. It makes the complex understandable. Simply the best factoring and quadratic equation program.
Features:
- Calculates Maximum or Minimum value.
- Trace function solves for roots.
-Graphs the actual curve to visually relate the equation.
-Scroll through calculated results with trace function.
-Help Screen, Glossary and Email.
Example: Select the "Form" button and the expression is setup that allows you to adjust each of four coefficients to create the cubic polynomial equation. Note that there are adjustor buttons to increase or decrease each coefficient. You can as easily use the built-in keyboard by touching the input field. When starting to understand polynomials keep the values small so that the graphing provides more flexibility. So we have Y = 2X^3 + 10X^2 -14X + 6. It takes the form of y = ax^2 + bx^2 + cx + d where a = 2, b = 10, c = -14, d = 6. Look at the curve calculated from this equation and displayed below in the graph portion of the screen. Use the trace X value to scroll through the curve data finding Y values for X = 0 if your curve passes through the X axis. The program calculates the values where the minimum or maximum occurs. Simply plug this value into the trace X value to calculate the Y values for the minimum or maximum values.
Try changing the Solve for Button to Coefficients. Then use the adjuster buttons to change the coefficients for "a", "b", "c" and "d" values. Note how the screen displays the graph of the equation and that it intersects the x axis at two places. If the graph does not intersect the x axis. In this case the curve intersects the X axis at -6.206. The curve appears to cross at X = .5 but it rides above the X axis by 1.8. Sometimes cubic equations are divided by a factor (X + ) then the equation becomes a quadratic and is solved by the quadric formula for roots. I find it easier to use the trace feature to find roots if necessary.
Look for other Ray Tools apps including linear, quadratic, exponential and logarithmic equations.
This app was developed by Raymond Seymour, after working at General Electric for 37 years hes now making Engineering apps for iOS. As an inventor he has over 60 granted U.S. Patents.
Requirements: iOS 4.3 or later. Compatible with iPad.
Requirements: No special requirements Release Date: September 21, 2014 Platforms: iOS Users rating: 0/10
License: Shareware Cost: \$0.99 USD Size: 33.7 MB
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RAY TOOLS: POLYNOMIAL EQUATIONS RELATED
Utilities - Ray Tools: Exponential Equations 1.0 Understanding exponential equations can be daunting task but this program simplifies understanding. Both from an educational standpoint where the student needs to understand coefficients and intercepts or from an engineering standpoint where the... 38.4 MB Education - Ray Tools: Linear Equations 1.0 Understanding linear equations (equations with an x term and a constant) can be daunting but this program simplifies understanding. Both from an educational standpoint where the student needs to understand coefficients and intercepts or from an... 31.5 MB Libraries - Math::Polynomial::Solve 2.11 Math::Polynomial::Solve is a Perl module to find the roots of polynomial equations. SYNOPSIS use Math::Complex; # The roots may be complex numbers. use Math::Polynomial::Solve qw(poly_roots); my @x = poly_roots(@coefficients); or use... 15.36 KB Utilities - Ray Tools: 555 Timer 1.0 The LM 555 timer chip is a special application device used for timing applications. This program can be used to design four different timer circuits, a Pulse timer (monostable), an Oscillator (astable), a Delay OFF timer and a Delay ON timer. The... 10.2 MB Education - Ray Tools: AC Impedance Simulator 1.0 Solving Ohms law for DC (E=I*R) is straight forward but what about AC. For AC (E=I*Z) looks simple but the value Z is composed of an inductor (a coil of wire) and a Capacitor. The AC frequency affects these calculations in a somewhat complex... 10.7 MB Education - Ray Tools: AC Input Operational Amplifier Simulator 1.0 Solving Operational Amplifiers is straightforward but can become complicated without much effort. This simulator is designed to help simplify your calculations in an easy to use format. Many times you may not know exactly how to design a circuit.... 9.4 MB Utilities - Ray Tools: Gear Design 1.0 Solving a Spur Gear Design problem is straightforward but can become complicated without much effort. This simulator is designed to help simplify your calculations in an easy to use format. Many times you may not know exactly how to design a gear... 10.4 MB Education - Ray Tools: Inverting Operational Amplifier Simulator 1.0 Solving Operational Amplifiers is straightforward but can become complicated without much effort. This calculator is designed to help simplify your calculations in an easy to use format. Many times you may not know exactly how to design a circuit.... 9.4 MB Education - Ray Tools: Non Inverting Operational Amplifier Simulator 1.0 Solving Operational Amplifiers is straightforward but can become complicated without much effort. This simulator is designed to help simplify your calculations in an easy to use format. Many times you may not know exactly how to design a circuit.... 9.4 MB Education - Ray Tools: Quadratic Equations 1.0 Understanding quadratic equations (equations with square terms) can be daunting but this program simplifies understanding. Both from an educational standpoint where the student needs to understand factors and roots or from an engineering... 38.5 MB | 1,610 | 7,237 | {"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} | 3.4375 | 3 | CC-MAIN-2017-34 | longest | en | 0.896032 |
https://www.explainxkcd.com/wiki/index.php?title=Talk:165:_Turn_Signals&diff=next&oldid=60678 | 1,656,262,409,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103271763.15/warc/CC-MAIN-20220626161834-20220626191834-00768.warc.gz | 824,579,034 | 9,666 | Difference between revisions of "Talk:165: Turn Signals"
I have at times become mesmerized by the "click-click click-click" of my turn signal relay while watching the flashing signals on the car ahead of me. It's fun to notice how they drift in and out of sync, but I never bothered to determine the beat frequency. --Smartin (talk) 03:53, 2 January 2013 (UTC)
What, to me, seems amazing, is not just that they are (certainly within extreme observational tolerance) beating at the same frequency, but are also in phase. At that point I would begin to suspect that they're each connected up to the same time-signal source (e.g. a GPS data output), and cued to begin each cycle on the flip of each whole second, or similar. Of course, IRL, that'd be an answer in search of a problem. And you want your signals to start flashing the moment you activate them, so even if guided by an atomic clock you'd probably have any given pair (albeit maintaining the same frequency) exhibiting a (constant) phase separation.
As for talking about being not held externally in sync, reminds me of the lights certain riders of tricycles have on their machine, in a 24-hour cycle race (mainly for bicycles, but trikies do tend to ride it as well). Flashing LED rear lights, very bright. On the backs of trikes they tend to put the lights out on each splayed rear stay, as well as the axle between the two rear wheels, to emphasise their width to any traffic that will be overtaking them in the night. Usually three identical flashers, but (as noted) the timings are rarely in sync, never mind in phase. As they're arranged in a triangle and very rarely all three on the same beat you can watch the machine as it retreats into the distance (my usual view of these phenomena) and when two of the lights are in sync and agreeing with each other, but the is off the beat, there's an effective directional 'wash' of light, this direction of wash changing as the in-syncs depart and perhaps the odd one syncs up anew with one of the other two originals. And if they all find themselves +/- 120-degrees out of phase with the other two, at any time, you get a rotary pattern emerging for a few moments.
You probably have to be there, but it's a sight to see, in the dead of night. And analyse. ;) 178.98.31.27 03:12, 22 June 2013 (UTC)
This merits a more fleshed-out explanation of beat frequencies and such. It's good enough, however. I'm not mean enough to mark it as incomplete for something like that. --Quicksilver (talk) 05:55, 24 August 2013 (UTC)
I would expect an irrational ratio of frequencies and all possible relationships appearing over a long enough time period. If ratio is close to one, they would appear to be nearly together for a reasonable period and far off for a reasonable period.--DrMath 01:49, 31 October 2013 (UTC)
This is somewhat reminiscent of Neal Stephenson's discussion, in Cryptonomicon, of Alan Turing's bicycle with a damaged link and a bent spoke, and how the two have to sync up for the chain to fall off. From this, he then does a wonderful sleight of hand, and before we know it, we have a fairly decent understanding of the Enigma machine. -- Ravenpi 23:50, 20 Feb 2013 (EST)
I don't know where I read it, but I read that some turn signals have a slight random element in it so that they never get truly in phase with any other car[citation needed]. Zazathebot (talk) 18:19, 28 June 2017 (UTC) | 819 | 3,415 | {"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 | 0.975924 |
https://elteoremadecuales.com/theorem-of-three-moments/ | 1,669,877,795,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710801.42/warc/CC-MAIN-20221201053355-20221201083355-00798.warc.gz | 278,312,227 | 10,645 | # Theorem of three moments
Theorem of three moments In civil engineering and structural analysis Clapeyron's theorem of three moments is a relationship among the bending moments at three consecutive supports of a horizontal beam.
Let A,B,C be the three consecutive points of support, and denote by- l the length of AB and {displaystyle l'} the length of BC, by w and {displaystyle w'} the weight per unit of length in these segments. Then[1] the bending moments {displaystyle M_{A},,M_{B},,M_{C}} at the three points are related by: {displaystyle M_{A}l+2M_{B}(l+l')+M_{C}l'={frac {1}{4}}wl^{3}+{frac {1}{4}}w'(l')^{3}.} This equation can also be written as [2] {displaystyle M_{A}l+2M_{B}(l+l')+M_{C}l'={frac {6a_{1}x_{1}}{l}}+{frac {6a_{2}x_{2}}{l'}}} where a1 is the area on the bending moment diagram due to vertical loads on AB, a2 is the area due to loads on BC, x1 is the distance from A to the centroid of the bending moment diagram of beam AB, x2 is the distance from C to the centroid of the area of the bending moment diagram of beam BC.
The second equation is more general as it does not require that the weight of each segment be distributed uniformly.
Figure 01-Sample continuous beam section Contents 1 Derivation of three moments equations 1.1 Mohr's first theorem 1.2 Mohr's second theorem 1.3 The sign convention 1.4 Derivation of three moment theorem 2 Three moment equation 3 Notes 4 External links Derivation of three moments equations Mohr's theorem[3] can be used to derive the three moment theorem[4] (TMT).
Mohr's first theorem The change in slope of a deflection curve between two points of a beam is equal to the area of the M/EI diagram between those two points.(Figure 02) Figure 02-Mohr's First Theorem Mohr's second theorem Consider two points k1 and k2 on a beam. The deflection of k1 and k2 relative to the point of intersection between tangent at k1 and k2 and vertical through k1 is equal to the moment of M/EI diagram between k1 and k2 about k1.(Figure 03) Figure03-Mohr's Second Theorem The three moment equation expresses the relation between bending moments at three successive supports of a continuous beam, subject to a loading on a two adjacent span with or without settlement of the supports.
The sign convention According to the Figure 04, The moment M1, M2, and M3 be positive if they cause compression in the upper part of the beam. (sagging positive) The deflection downward positive. (Downward settlement positive) Let ABC is a continuous beam with support at A,B, and C. Then moment at A,B, and C are M1, M2, and M3, respectively. Let A' B' and C' be the final positions of the beam ABC due to support settlements. Figure 04-Deflection Curve of a Continuous Beam Under Settlement Derivation of three moment theorem PB'Q is a tangent drawn at B' for final Elastic Curve A'B'C' of the beam ABC. RB'S is a horizontal line drawn through B'. Consider, Triangles RB'P and QB'S.
{displaystyle {dfrac {PR}{RB'}}={dfrac {SQ}{B'S}},} {displaystyle {dfrac {PR}{L1}}={dfrac {SQ}{L2}}} (1) {displaystyle PR=Delta B-Delta A+PA'} (2) {displaystyle SQ=Delta C-Delta B-QC'} (3) From (1), (2), and (3), {displaystyle {dfrac {Delta B-Delta A+PA'}{L1}}={dfrac {Delta C-Delta B-QC'}{L2}}} {displaystyle {dfrac {PA'}{L1}}+{dfrac {QC'}{L2}}={dfrac {Delta A-Delta B}{L1}}+{dfrac {Delta C-Delta B}{L2}}} (a) Draw the M/EI diagram to find the PA' and QC'.
Figure 05 - M / EI Diagram From Mohr's Second Theorem PA' = First moment of area of M/EI diagram between A and B about A.
{displaystyle PA'=left({frac {1}{2}}times {frac {M_{1}}{E_{1}I_{1}}}times L_{1}right)times L_{1}times {frac {1}{3}}+left({frac {1}{2}}times {frac {M_{2}}{E_{2}I_{2}}}times L_{1}right)times L_{1}times {frac {2}{3}}+{frac {A_{1}X_{1}}{E_{1}I_{1}}}} QC' = First moment of area of M/EI diagram between B and C about C.
{displaystyle QC'=left({frac {1}{2}}times {frac {M_{3}}{E_{2}I_{2}}}times L_{2}right)times L_{2}times {frac {1}{3}}+left({frac {1}{2}}times {frac {M_{2}}{E_{2}I_{2}}}times L_{2}right)times L_{2}times {frac {2}{3}}+{frac {A_{2}X_{2}}{E_{2}I_{2}}}} Substitute in PA' and QC' on equation (a), the Three Moment Theorem (TMT) can be obtained.
Three moment equation {displaystyle {frac {M_{1}L_{1}}{E_{1}I_{1}}}+2M_{2}left({frac {L_{1}}{E_{1}I_{1}}}+{frac {L_{2}}{E_{2}I_{2}}}right)+{frac {M_{3}L_{2}}{E_{2}I_{2}}}=6[{frac {Delta A-Delta B}{L_{1}}}+{frac {Delta C-Delta B}{L_{2}}}]-6[{frac {A_{1}X_{1}}{E_{1}I_{1}L_{1}}}+{frac {A_{2}X_{2}}{E_{2}I_{2}L_{2}}}]} Notes ^ J. B. Wheeler: An Elementary Course of Civil Engineering, 1876, Page 118 [1] ^ Srivastava and Gope: Strength of Materials, page 73 ^ "Mohr's Theorem" (PDF). ^ "Three Moment Theorem" (PDF). External links CodeCogs: Continuous beams with more than one span hide vte Structural engineering Dynamic analysis Duhamel's integralModal analysis Static analysis Betti's theoremCastigliano's methodConjugate beam methodFEMFlexibility methodMacaulay's methodMoment-area theoremStiffness methodShear and moment diagramTheorem of three moments Structural elements 1-dimensional Beam I-beamLintel Post and lintelSpanCompression memberStrutTie 2-dimensional ArchThin-shell structure Structural support Bracket Theories Euler–Bernoulli beam theoryMohr–Coulomb theoryPlate theoryTimoshenko–Ehrenfest beam theory Categories: Structural analysisContinuum mechanicsPhysics theorems
Si quieres conocer otros artículos parecidos a Theorem of three moments puedes visitar la categoría Continuum mechanics.
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Utilizamos cookies propias y de terceros para mejorar la experiencia de usuario Más información | 1,690 | 5,608 | {"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-2022-49 | longest | en | 0.917625 |
http://forums.wolfram.com/mathgroup/archive/2000/Oct/msg00389.html | 1,718,600,487,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861696.51/warc/CC-MAIN-20240617024959-20240617054959-00750.warc.gz | 12,578,519 | 7,996 | NIntegrate v.s. NonlinerFit
• To: mathgroup at smc.vnet.net
• Subject: [mg25811] NIntegrate v.s. NonlinerFit
• From: Carmelo La Rosa <clarosa at dipchi.unict.it>
• Date: Sat, 28 Oct 2000 01:41:04 -0400 (EDT)
• Organization: Dipartimento di Scienze Chimiche
• Sender: owner-wri-mathgroup at wolfram.com
```Hi,
my problem is the following:
I have difined a function as following, in order to use in a nonlinear
bestfit containing a numerical integration (work on WinNT 4.0 and
Mathematica 3.0)
<<Statistics`NonlinearFit`
pars={a,b,c}(*parameters for the fit*)
pars0={.......}(*starting parameters of pars*)
A(X)=Exp[-a/X]
K(X)=Exp[-b/X]
f[(T_)?Number,(a_)?Number,(b_)?Number,(c_)?Number]:=
(b*K(T)/(K(T)+1)^2)*Exp[-costant*Module[{X},NIntegrate[A(X)*K(X)/(K(X)+1)),{X,Ton,T}]
the output is a error messagge:
NIntegrate::"nlim": "\!\(X\$1\) = \!\(T\) is not a valid limit of
integration."
NIntegrate::"nlim": "\!\(X\$1\) = \!\(T\) is not a valid limit of
integration."
NIntegrate::"nlim": "\!\(X\$1\) = \!\(T\) is not a valid limit of
integration."
General::"stop":
"Further output of \!\(NIntegrate :: \"nlim\"\) will be suppressed
during \
this calculation."
In[19]:=
fitted=NonlinearFit[data,Cp[T,pars],pars,T,MaxIterations->3000,
Errors\[Rule]1.0,InitialGuess\[Rule]pars0,ProgressTrace\[Rule]True]
NonlinearFit::"toomany":
"Too many variables have been provided for the size of the data;
given \!\
\(2\)-tuples for data, given \!\(4\) variables."
Any assistance will be greatly appreciated.
Carmelo la Rosa
--
____________________________________
------------------------------------
Carmelo La Rosa
Department of Chemical Science
Laboratory of Biophysical Chemistry
University of Catania
V.le A. Doria 6, 95125 Catania
ITALY
Tel1. ++39-95-7385114
Tel2. ++39-95-339572
Fax ++39-95-580138
E-Mail clarosa at dipchi.unict.it
------------------------------------
____________________________________
```
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• Next by thread: A stupid but practical question | 654 | 2,114 | {"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-2024-26 | latest | en | 0.580647 |
http://examcrazy.com/quantitative-problems-on-numbers-plus-b64 | 1,550,330,248,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247480622.9/warc/CC-MAIN-20190216145907-20190216171907-00416.warc.gz | 100,635,483 | 29,281 | # Quantitative - Problems on Numbers plus B64
Quantitative - Problems on Numbers plus B64
Question:-
Option (A)
5
Option (B)
6
Option (C)
7
Option (D)
8
Correct Option:
C
Question Solution:
if it was Rs. 7 then you need 3 bags with Rs. 1 (=20), Rs. 2 (=21) & Rs. 4 (=22) in it. That means you need 3 bags.
Same way if it was Rs. 15 then you need 3 bags with Rs. 1 (=20), Rs. 2 (=21), Rs. 4 (=22) & Rs. 8 (=22) in it. That means you need 4 bags. It will be same for any amount more than Rs. 7 till Rs. 15.
Same way till 31 you need 5 bags, till 63 you need 6 bags, till 127 you need 7 bags & so on.
You can make a general formula for these kind of questions. Calculate lowest n so that 2n n is more than the required number (like 100 here. This n will be the answer.
Question:-
Option (A)
1
Option (B)
2
Option (C)
3
Option (D)
4
Correct Option:
C
Question Solution:
(x2 - y2) = 45 i.e (x - y)(x + y) = 45. The factors of 45 possible are, 15, 3; 9, 5 and 45, 1.
Hence, the numbers are 9 and 6, or 7 and 2 or 23 and 22.
Question:-
Option (A)
21
Option (B)
28
Option (C)
33
Option (D)
50
Correct Option:
D
Question Solution:
Factor out the numerator (4*3)(2*11)(5*7). Now this fraction when divided by p would get an integer only if p cancels out with some of the factors. So now lets try it with different answer choices. Except for the (D) all other choices can be cancelled out. So clearly (D) is the answer.
Question:-
Option (A)
X is always odd
Option (B)
X is always even
Option (C)
X is even only if y is odd
Option (D)
X is odd only if y is even
Correct Option:
B
Question Solution:
If y is even => 6y is even => 7x=420-6y is even. For 7x to be even, x has to be even
If y is odd => 6y is even => 7x=420-6y is even. For 7x to be even, x has to be even
So in any case x has to be even for this equation to be true.
Question:-
Option (A)
865
Option (B)
745
Option (C)
845
Option (D)
945
Correct Option:
B
Question Solution:
The unit's digit will be 1 x 5 = 5(no carry over). The tens digit will be (4x1 + 5x2) = 4(carry over 1).
The hundreds digit will be (3x1 + 4x2 + 5x1) = 6 (1 carry over) + 1(carried over from tens) = 7. | 730 | 2,131 | {"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} | 4.28125 | 4 | CC-MAIN-2019-09 | latest | en | 0.87253 |
http://math.stackexchange.com/questions/95885/interpolation-to-a-power-function | 1,467,052,748,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783396106.25/warc/CC-MAIN-20160624154956-00101-ip-10-164-35-72.ec2.internal.warc.gz | 196,556,660 | 18,357 | # Interpolation to a power function
We have an experiment which have the variables $x$ and $y$. $x$ and $y$ can be measured into pair $(x_i,y_i)$. Now I'm finding a way to interpolate it into a power function $y=a+bx^c$. Which $a,b,c$ are real number (and $a>0, b>0, c>0$). How to find that function? (among $a,b,c$; $a$ is the number I need the most). For example, we did the experiment many times and have $(x_1,y_1)=(1,3)$; $(x_2,y_2)=(3,5)$; $(x_3,y_3)=(5,8)$; $(x_4,y_4)=(10,11)$
If possible, how can we do that in Matlab (or similar programming software)?
-
What you want is Constrained Nonlinear Least Squares.
In Maple you could do
X:= [1,3,5,10]: Y:= [3,5,8,11]:
Residuals:= [seq(a+b*X[i]^c - Y[i], i=1..4)]:
Optimization[LSSolve](Residuals, {a >= 0, b >= 0,c >= 0});
[.308370175847439200, [a = 0., b = 2.91266785479164, c = .582530468246663]]
I'm pretty sure Matlab has similar functionality.
-
There is no exact answer for an overloaded system, that is a system where you have more "solutions" (x, y pairs) than unknowns. The problem can be solved exactly where you have a pair for each unknown in your case that would mean having three pairs for the three unkowns. If you have more than enough answers and they are experimentally gained then you have to use an inexact method known as the method of least squares or some other method to minimize the error in the interpolated function.
The idea is to construct a "cost" function to determine how expensive for the function to be away from a point.
Excel can do this with a power fit.
-
Agree. The MS Excel can do that with $y=bx^c$. I did the experiment and have 14 (x,y) pairs in the end. I thought about the y'=y-a and use MS excel to have the interpolating function, but by doing that there's no way I can attain the desired value of a, so there's no exact value of y' to put in Excel. (as we can see we have only values of x and y) Thanks for your answer. – linkgreencold Jan 2 '12 at 20:35
@linkgreencold: You can do it in Excel by making a couple cells be the parameters (in your case, a,b,c) and a column of your measured data. Then make a column of your calculated $y_i$'s based on $a,b,c,x_i$. Sum the squared errors and ask Goal Seek to minimize changing $a,b,c$ – Ross Millikan Jan 2 '12 at 22:01
Just for completeness, in MATLAB the command is
f=fit(x,y,fittype('power2');
where (x,y) should be column vectors. In Mathematica the command is
FindFit[Transpose[{x, y}], {a + b t^c}, {a, b, c}, t]
-
Thank you, I used the fit function and had the pretty satisfied answer. Wanna vote up your answer but haven't enough reputation point yet. – linkgreencold Jan 4 '12 at 11:14
No problem... (but you can still accept it if you want to). – yohBS Jan 4 '12 at 15:00 | 816 | 2,750 | {"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} | 3.859375 | 4 | CC-MAIN-2016-26 | longest | en | 0.917931 |
http://www.nag.com/numeric/CL/nagdoc_cl23/html/F04/f04ajc.html | 1,500,862,454,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424645.77/warc/CC-MAIN-20170724002355-20170724022355-00702.warc.gz | 500,792,975 | 3,838 | f04 Chapter Contents
f04 Chapter Introduction
NAG C Library Manual
# NAG Library Function Documentnag_real_lu_solve_mult_rhs (f04ajc)
## 1 Purpose
nag_real_lu_solve_mult_rhs (f04ajc) calculates the approximate solution of a set of real linear equations with multiple right-hand sides, $AX=B$, where $A$ has been factorized by nag_real_lu (f03afc).
## 2 Specification
#include #include
void nag_real_lu_solve_mult_rhs (Integer n, Integer nrhs, const double a[], Integer tda, const Integer pivot[], double b[], Integer tdb, NagError *fail)
## 3 Description
To solve a set of real linear equations $AX=B$, this function must be preceded by a call to nag_real_lu (f03afc) which computes an $LU$ factorization of $A$ with partial pivoting, $PA=LU$, where $P$ is a permutation matrix, $L$ is lower triangular and $U$ is unit upper triangular. The columns $x$ of the solution $X$ are found by forward and backward substitution in $Ly=Pb$ and $Ux=y$, where $b$ is a column of the right-hand sides.
## 4 References
Wilkinson J H and Reinsch C (1971) Handbook for Automatic Computation II, Linear Algebra Springer–Verlag
## 5 Arguments
1: nIntegerInput
On entry: $n$, the order of the matrix $A$.
Constraint: ${\mathbf{n}}\ge 1$.
2: nrhsIntegerInput
On entry: $r$, the number of right-hand sides.
Constraint: ${\mathbf{nrhs}}\ge 1$.
3: a[${\mathbf{n}}×{\mathbf{tda}}$]const doubleInput
Note: the $\left(i,j\right)$th element of the matrix $A$ is stored in ${\mathbf{a}}\left[\left(i-1\right)×{\mathbf{tda}}+j-1\right]$.
On entry: details of the $LU$ factorization, as returned by nag_real_lu (f03afc).
4: tdaIntegerInput
On entry: the stride separating matrix column elements in the array a.
Constraint: ${\mathbf{tda}}\ge {\mathbf{n}}$.
5: pivot[n]const IntegerInput
On entry: details of the row interchanges as returned by nag_real_lu (f03afc).
6: b[${\mathbf{n}}×{\mathbf{tdb}}$]doubleInput/Output
Note: the $\left(i,j\right)$th element of the matrix $B$ is stored in ${\mathbf{b}}\left[\left(i-1\right)×{\mathbf{tdb}}+j-1\right]$.
On entry: the $n$ by $r$ right-hand side matrix $B$.
On exit: $B$ is overwritten by the solution matrix $X$.
7: tdbIntegerInput
On entry: the stride separating matrix column elements in the array b.
Constraint: ${\mathbf{tdb}}\ge {\mathbf{nrhs}}$.
8: failNagError *Input/Output
The NAG error argument (see Section 3.6 in the Essential Introduction).
## 6 Error Indicators and Warnings
NE_2_INT_ARG_LT
On entry, ${\mathbf{tda}}=〈\mathit{\text{value}}〉$ while ${\mathbf{n}}=〈\mathit{\text{value}}〉$. These arguments must satisfy ${\mathbf{tda}}\ge {\mathbf{n}}$.
On entry, ${\mathbf{tdb}}=〈\mathit{\text{value}}〉$ while ${\mathbf{nrhs}}=〈\mathit{\text{value}}〉$. These arguments must satisfy ${\mathbf{tdb}}\ge {\mathbf{nrhs}}$.
NE_INT_ARG_LT
On entry, ${\mathbf{n}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{n}}\ge 1$.
On entry, ${\mathbf{nrhs}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{nrhs}}\ge 1$.
## 7 Accuracy
The accuracy of the computed solutions depends on the conditioning of the original matrix. For a detailed error analysis see page 106 of Wilkinson and Reinsch (1971).
The time taken by nag_real_lu_solve_mult_rhs (f04ajc) is approximately proportional to ${n}^{2}r$.
## 9 Example
To solve the set of linear equations $AX=B$ where
$A = 33 16 72 -24 -10 -57 -8 -4 -17 and B = -359 281 85 .$
### 9.1 Program Text
Program Text (f04ajce.c)
### 9.2 Program Data
Program Data (f04ajce.d)
### 9.3 Program Results
Program Results (f04ajce.r) | 1,224 | 3,553 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 48, "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} | 2.875 | 3 | CC-MAIN-2017-30 | latest | en | 0.472616 |
http://www.derkeiler.com/Newsgroups/sci.crypt/2005-10/0729.html | 1,455,049,892,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701157472.18/warc/CC-MAIN-20160205193917-00342-ip-10-236-182-209.ec2.internal.warc.gz | 365,996,753 | 4,895 | # REPOST: Two-dimensional Montgomery ladders and ECDSA
From: D. J. Bernstein (djb_at_cr.yp.to)
Date: 10/24/05
```Date: Mon, 24 Oct 2005 07:38:15 +0000 (UTC)
```
The current speed records for Diffie-Hellman key exchange, and
specifically for elliptic-curve point multiplication, use two ideas that
Montgomery introduced in 1987:
* You can easily compute x(iP+jP) given x(iP), x(jP), and x(iP-jP),
where x(P) means the x-coordinate of point P. This is particularly
fast for curves of the form y^2 = x^3 + Cx^2 + x.
* Starting from P, you can reach nP for a b-bit n with a sequence of
b double+add steps, where each addition iP+jP has iP-jP = P. For
example, you can reach 51P using 2P,3P,4P,6P,7P,12P,13P,25P,26P.
Putting these ideas together you obtain x(nP) from x(P) at high speed.
Some side advantages: public keys are transmitted as just x-coordinates;
if the curve is chosen carefully then you don't have to spend any time
verifying keys; the computation is easily made immune to timing attacks.
See http://cr.yp.to/talks.html#2005.09.20 for more details.
Now, what happens if you want to compute x(mP+nQ), as required for
verification of ECDSA signatures?
One answer (by, e.g., Okeya and Sakurai) is to compute mP via x(mP),
compute nQ via x(nQ), and then add. Computing x(mP) and x(nQ) by
Montgomery's technique takes 2b x-coordinate double+add steps.
I've noticed that a different technique takes 2b x-coordinate additions
and only b x-coordinate doublings. Specifically, one can reach x(mP+nQ)
steps, where each addition has difference P or Q or P+Q or P-Q. Example:
P+Q, 2P, 2P+Q,
2P+2Q, 3P+Q, 3P+2Q,
4P+4Q, 5P+3Q, 5P+4Q,
9P+7Q, 9P+8Q, 10P+8Q,
18P+14Q, 18P+15Q, 19P+15Q,
36P+30Q, 37P+29Q, 37P+30Q,
73P+59Q, 74P+58Q, 74P+59Q.
Each line has three of the four pairs aP+bQ, (a+1)P+bQ, aP+(b+1)Q,
(a+1)P+(b+1)Q. The missing element of (a+{0,1},b+{0,1}) is always chosen
as either (even,odd) or (odd,even), where the choice can be described
recursively using the (A,B) for the next line:
* if (a+A,b+B) is (even,odd) then the choice is (odd,even);
* if (a+A,b+B) is (odd,even) then the choice is (even,odd);
* if (a+A,b+B) is (even,even) then the lines have the same choices;
* if (a+A,b+B) is (odd,odd) then the lines have opposite choices.
For small-C Montgomery-shape curves, x-coordinate addition (using affine
difference) takes 5 field operations, and x-coordinate doubling takes 4
field operations, so this two-dimensional addition chain takes just 14
field operations per bit, quite possibly the state of the art in ECDSA
verification even if we don't require timing-attack protection.
Is this a new observation?
---D. J. Bernstein, Professor, Mathematics, Statistics,
and Computer Science, University of Illinois at Chicago
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From: "D. J. Bernstein" <djb@cr.yp.to>
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Subject: Cancel "Two-dimensional Montgomery ladders and ECDSA"
Newsgroups: de.alt.test,sci.crypt
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## Relevant Pages
• Two-dimensional Montgomery ladders and ECDSA
... The current speed records for Diffie-Hellman key exchange, ... public keys are transmitted as just x-coordinates; ... Montgomery's technique takes 2b x-coordinate double+add steps. ... with a simple two-dimensional addition chain that uses b double+add+add ...
(sci.crypt)
• REPOST: Two-dimensional Montgomery ladders and ECDSA
... public keys are transmitted as just x-coordinates; ... Montgomery's technique takes 2b x-coordinate double+add steps. ... with a simple two-dimensional addition chain that uses b double+add+add ... Cancel "Two-dimensional Montgomery ladders and ECDSA" ...
(sci.crypt) | 1,362 | 4,249 | {"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-2016-07 | latest | en | 0.851453 |
https://groups.google.com/g/k12.ed.math/c/GfdNJ46Itl0 | 1,675,573,155,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500215.91/warc/CC-MAIN-20230205032040-20230205062040-00827.warc.gz | 303,370,842 | 131,949 | # Computing totient(n) -- melding math and programming
1568 views
### Kirby Urner
Jul 21, 2003, 11:10:17 PM7/21/03
to
So the idea here was to inject some computer programming
into a course with math content, that content being some
group and number theory.
We're killing two birds with one stone: learn some
programming in tandem with some math ("maths" -- UK).
Not a new idea.
What's a little bit new is that programming environments
over the internet. We have several to choose from.
Some criteria should help us to pick an appropriate
environment. A lot of high schools seem to like Scheme.
PLT Scheme in particular is all about providing an
environment geared to students and teaching.
What content shall we cover? It will overlap quite a bit
with content already standard in middle and high school,
but then go further in some directions.
For example, the concepts of prime and composite number are
already deeply entrenched, as are even and odd. We can find
these ideas already well-formed in Euclid.
But not until Euler do we have the formal definition of
'totient': the number of positive integers < N with no factors
in common with N (i.e. relatively prime to N).
These positive integers < N and coprime to N are called
'the totatives of N'. The totient is just the number of
totatives.
Most contemporary curricula leave 'totient' aside until
some college course. But it's such a simple idea and so
perfect for learning a little programming.
We want to write something like:
totient(n): count(i) for 0<i<n where gcd(i,n)=1
How would we write this in Scheme? Here's one possible solution:
;; compute the number of positives < n that are
;; relatively prime to n
(define (totient n)
(let loop ((tot 0)(pos (- n 1)))
(if (> pos 0)
(if (= 1 (gcd n pos))
(loop (+ tot 1)(- pos 1))
(loop tot (- pos 1))
)
tot)
)
)
I'm not highly experienced in Scheme. A more streamlined
version of the totient function, even using this same algorithm,
is no doubt possible -- but even this one is pretty short.
Here's the new program in action (> means user enters):
> (totient 20)
8
> (totient 100)
40
> (totient 291929)
265380
There's a noticable delay on this last execution. The program
is basically counting down from n to 0, counting n's coprimes
along the way. As the numbers get larger, the execution time
takes proportionally longer. It's not an algorithm well suited
to large numbers.
Euler came up with a better one that requires knowing n's prime
factors -- and those may be hard to get as well, when n is trully
large.
Rather typically for Scheme, the proposed solution defines a
recursive loop, in this case using 'named let' syntax. On a
first iteration through the loop, tot is initialized to zero
and pos is set at one less than the argument n. This is the
meaning of the line:
(let loop ((tot 0)(pos (- n 1)))
...)
If pos and n are relatively prime, i.e. their gcd is one, then
the loop is called again, with tot, the number of totatives,
incremented. Otherwise, the loop is called without incrementing
tot:
(if (= 1 (gcd n pos))
(loop (+ tot 1)(- pos 1))
(loop tot (- pos 1))
)
In both cases, pos is decremented and so will inevitably reach 0.
When it does, tot is the returned value:
(if (> pos 0)
(...
)
tot)
Let's briefly look at a different language, likewise available
support a "shell mode", meaning the student is able to interact
with her or his programs in one area, while defining programs
in another.
In interactive mode, these environments may be used much as
students use calculators, which may be useful for continuity.
More to the point, an interactive mode gives immediate feedback
regarding the correctness of syntax and programs.
In Python, gcd is not built in, as it is in Scheme; we need to
program it. Fortunately, Euclid's algorithm is quite compact:
def gcd(a,b):
"Euclid's Algorithm"
while b:
a, b = b, a % b
return a
Translation: as long as b is still non-zero, loop around
making b the next a, and the remainder after dividing b into
a the next b. When b becomes zero, a has divided it evenly,
so return a.
When explaining why this works, it helps to use planks. What
greatest length evenly divides two planks? "Evenly" here means
"with no leftover bit".
We assume a unit plank, and that both planks are integral
multiples of this unit.
If the shorter divides the longer evenly, we're done. The
shorter length is the gcd.
If not, then there's some remainder after dividing by the shorter,
and the gcd will have to divide this remainder, as well as the
shorter, in order to divide both the shorter and the longer (our
original goal).
So now we're interested in the gcd of the remainder and the
shorter. Notice how the remainder now plays the role of the
shorter, and the shorter plays the role of the longer. So we
divide the remainder into the shorter, to get a new remainder,
and so on. Eventually, we'll reach the gcd, which may be one.
Finally, the program for returning the totient is short:
def totient(n):
"short and sweet, but a bit slow"
return len([i for i in range(1,n) if gcd(i,n)==1])
Here's the program in action, in shell mode:
>>> totient(20)
8
>>> totient(100)
40
>>> totient(291929)
265380
Again, there's a noticable time lag as the last execution
completes.
At this point, it looks a lot like Scheme. However, under the
hood, these two languages use obviously different syntax.
We could try to imitate the Scheme solution in Python, also
using recursion:
def totient(n):
"An unPythonic implementation"
def loop(tot, pos):
while pos>0:
if gcd(pos,n)==1: return loop(tot+1,pos-1)
else: return loop(tot, pos-1)
return loop(0,n-1)
But unlike Scheme, Python does not implement tail-call elimination.
Each loop takes us deeper into the call stack, which has an upper
limit in Python. In Scheme, on the other hand, tail calls form
a genuine loop.
Here's a non-recursive approach in Python that doesn't involve
listing all the totatives first in order to count them, as did
our first Python attempt:
>>> def totient(n):
"""
Compute the number of positives < n that are
relatively prime to n -- good solution!
"""
tot, pos = 0, n-1
while pos>0:
if gcd(pos,n)==1: tot += 1
pos -= 1
>>> totient(10)
4
>>> totient(100)
40
>>> totient(291929)
265380
This is noticibly faster than the first Python solution, as it
doesn't involve keeping all the totatives in a list in order to
count them. It also looks more like the Scheme version, except
that it's not recursive.
Kirby
Scheme: TeachScheme!: http://www.teach-scheme.org/
Python: edu-sig page: http://www.python/org/sigs/edu-sig/
--
submissions: post to k12.ed.math or e-mail to k12...@k12groups.org
private e-mail to the k12.ed.math moderator: kem-mo...@k12groups.org
newsgroup website: http://www.thinkspot.net/k12math/
newsgroup charter: http://www.thinkspot.net/k12math/charter.html
### Kirby Urner
Jul 22, 2003, 3:22:13 AM7/22/03
to
Kirby Urner <ur...@alumni.princeton.edu> wrote:
>How would we write this in Scheme? Here's one possible solution:
>
> ;; compute the number of positives < n that are
> ;; relatively prime to n
>
> (define (totient n)
> (let loop ((tot 0)(pos (- n 1)))
> (if (> pos 0)
> (if (= 1 (gcd n pos))
> (loop (+ tot 1)(- pos 1))
> (loop tot (- pos 1))
> )
> tot)
> )
> )
>
>I'm not highly experienced in Scheme. A more streamlined
>version of the totient function, even using this same algorithm,
>is no doubt possible -- but even this one is pretty short.
>
Here's a slightly modified version that came to me after
I'd read the latest Harry Potter for a few more chapters:
;; totient : integer -> integer
;; compute the number of positives < n that are
;; relatively prime to n
(define (totient n)
(let loop ((tot 0)(pos (- n 1)))
(if (> pos 0)
(loop (+ tot (if (= 1 (gcd n pos)) 1 0))
(- pos 1))
tot
)
)
)
The idea is the same: call the named loop recursively
adding 1 or 0 to tot, depending out the outcome of the
gcd test. Quit when you've tested all positives down to 1
and return whatever tot has accumulated.
If you haven't played with DrScheme, why not give it a try?
It's free. The syntax checker will fight you tooth and nail
(its job) until you've started to get the hang of it --
then it'll neatly colorize your code.
Thanks to Harry Potter, I can't help thinking of learning
to write spells. Students might enjoy that analogy, even
if some oldsters become quite humorless when it comes to
mixing in such literary allusions.
Where we go with this totient concept is towards Euler's
Theorem and multiplicative groups modulo N, where the
elements being multiplied are N's totatives. You get a
group. And any of these totatives to the totient power,
modulo N, equals one. Showing what that even *means*
(by repeated examples, using one's own code) should
precede trying to prove it.
If N is a prime, then all positives < N are its totatives,
so the totient is simply p-1. We can check that using the
code we've already written (above), although it's also
intuitively obvious that primes can't have any proper
factors, so *of course* p has p-1 totatives:
> (totient 17)
16
> (totient 103)
102
Here I think that Scheme and Python approaches might diverge
quite a bit. In Python, a most promising approach is to
define "Modulo Numbers" with a built-in knowledge of how
to add, multiply etc., modulo something. We can reuse the
regular operators, * and +. So 2 + 2 might well not be 4
when the Modulo Numbers come to town.
I need to think some more about what a good Scheme approach
might be. No doubt this is already well explored in the
literature -- but I should do my own thinking as well.
And then there's always J, another language with an interactive
shell mode: http://www.inetarena.com/~pdx4d/ocn/Jlang.html
Such an embarrassment of riches! Maths mixed with programming
could turn out to be magical for at least some kids. More
students need to be at least exposed to the opportunity.
Kirby
### Kirby Urner
Jul 22, 2003, 10:38:26 AM7/22/03
to
And another thing: we need to check boundary conditions.
And what will happen in our program if someone asks for
the totient of a non-integer or negative integer?
So it turns out there's this subtle bug in *both*
versions of the program (the Scheme version and the
Python version).
In initializing pos to n-1, where we're asking for
totient(n), I was making the assumption that we
don't want to bother with gcd(n,n) since that will
always be n, and we only care if the gcd is 1. But
what if n=1? gcd(1,1)=1 and therefore pos should
be incremented. So I should initialize pos to n,
in order to catch this case.
According to the MathWorld website, totient(0)=1 by
convention, and yet Mathematica itself (the product
around which MathWorld is built) allows EulerPhi[0]
to be zero. Note that our totient function was
symbolized by the greek letter phi by Euler, hence
this function name EulerPhi in Mathematica.
http://mathworld.wolfram.com/TotientFunction.html
In the interest of brevity, and because I don't see
why this convention is of great importance, I'll
not try to capture totient(0) as a special case
and return 1. It's OK with me for both programs
to return 0 at this point.
Finally, what if a negative or non-integer is passed
as an argument. Totient(n) is clearly not defined in
these cases. We're outside of totient's domain. It'd
be cool to trap these incorrect inputs and indicate
the error to the user somehow.
Here are enhanced versions of both programs with these
features:
Scheme:
;; totient : integer -> integer
;; compute the number of positives < n that are
;; relatively prime to n
(define (totient n)
(if (not (and (integer? n)(>= n 0))) "Incorrect input"
(begin
(let loop ((tot 0)(pos n))
(if (> pos 0)
(loop (+ tot (if (= 1 (gcd n pos)) 1 0))
(- pos 1))
tot
)
)
)
)
)
Python:
def totient(n):
"""
count totatives of n, assuming gcd already
defined
"""
if not (type(n)==type(1) and n>=0):
raise ValueError, 'Invalid input type'
tot,pos = 0, n
while pos>0:
if gcd(pos,n)==1: tot += 1
pos -= 1
Again, as per MathWorld and an earlier remark, the more
typical way to compute the totient of n is to break n
into prime factors and then "cross out" all multiples
of these primes. The totient will be the number of
integers surviving the cross-out. There's a way to
derive the number of remaining prime multiples per
each unique prime factor and to take the product of
these numbers as the total proportion of positives
=< n that will be removed. Therefore n times this
product will be the totient.
totient(n) = n * (1 - 1/p1)(1 - 1/p2)(1 - 1/p3)...(1 - 1/pm)
where p1...pm are the unique prime factors of n.
### Gregor Lingl
Jul 23, 2003, 11:32:57 AM7/23/03
to
O.k. However, if you insisted in totient(0) returning 1
you could fix this in the following way:
def totient(n):
"""
count totatives of n, assuming gcd already
defined
"""
if not (type(n)==type(1) and n>=0):
raise ValueError, 'Invalid input type'
tot,pos = 1, n-1
while pos>1:
if gcd(pos,n)==1: tot += 1
pos -= 1 | 3,442 | 12,957 | {"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} | 3.859375 | 4 | CC-MAIN-2023-06 | latest | en | 0.930698 |
https://chem.libretexts.org/Textbook_Maps/Introductory_Chemistry/Book%3A_Beginning_Chemistry_(Ball)/06%3A_Gases/6.7%3A_Gas_Mixtures | 1,544,466,223,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823382.1/warc/CC-MAIN-20181210170024-20181210191524-00088.warc.gz | 548,090,710 | 20,321 | # 6.7: Gas Mixtures
Learning Objective
• Learn Dalton’s law of partial pressures.
One of the properties of gases is that they mix with each other. When they do so, they become a solution—a homogeneous mixture. Some of the properties of gas mixtures are easy to determine if we know the composition of the gases in the mix.
In gas mixtures, each component in the gas phase can be treated separately. Each component of the mixture shares the same temperature and volume. (Remember that gases expand to fill the volume of their container; gases in a mixture continue to do that as well.) However, each gas has its own pressure. The partial pressure of a gas, Pi, is the pressure that an individual gas in a mixture has. Partial pressures are expressed in torr, millimeters of mercury, or atmospheres like any other gas pressure; however, we use the term pressure when talking about pure gases and the term partial pressure when we are talking about the individual gas components in a mixture.
Dalton's law of partial pressures states that the total pressure of a gas mixture, Ptot, is equal to the sum of the partial pressures of the components, Pi:
$P_{tot}=P_{1}+P_{2}+P_{3}+...=\sum_{\#of\, gases}P_{i}$
Although this may seem to be a trivial law, it reinforces the idea that gases behave independently of each other.
Example $$\PageIndex{1}$$:
A mixture of H2 at 2.33 atm and N2 at 0.77 atm is in a container. What is the total pressure in the container?
Solution
Dalton’s law of partial pressures states that the total pressure is equal to the sum of the partial pressures. We simply add the two pressures together:
Ptot = 2.33 atm + 0.77 atm = 3.10 atm
Exercise $$\PageIndex{1}$$
Air can be thought of as a mixture of N2 and O2. In 760 torr of air, the partial pressure of N2 is 608 torr. What is the partial pressure of O2?
152 torr
Example $$\PageIndex{2}$$:
A 2.00 L container with 2.50 atm of H2 is connected to a 5.00 L container with 1.90 atm of O2 inside. The containers are opened, and the gases mix. What is the final pressure inside the containers?
Solution
Because gases act independently of each other, we can determine the resulting final pressures using Boyle’s law and then add the two resulting pressures together to get the final pressure. The total final volume is 2.00 L + 5.00 L = 7.00 L. First, we use Boyle’s law to determine the final pressure of H2:
(2.50 atm)(2.00 L) = P2(7.00 L)
Solving for P2, we get
P2 = 0.714 atm = partial pressure of H2
Now we do that same thing for the O2:
(1.90 atm)(5.00 L) = P2(7.00 L)P2 = 1.36 atm = partial pressure of O2
The total pressure is the sum of the two resulting partial pressures:
Ptot = 0.714 atm + 1.36 atm = 2.07 atm
Exercise $$\PageIndex{2}$$
If 0.75 atm of He in a 2.00 L container is connected to a 3.00 L container with 0.35 atm of Ne and the containers are opened, what is the resulting total pressure?
0.51 atm
One of the reasons we have to deal with Dalton’s law of partial pressures is because gases are frequently collected by bubbling through water. As we will see in Chapter 10, liquids are constantly evaporating into a vapor until the vapor achieves a partial pressure characteristic of the substance and the temperature. This partial pressure is called a vapor pressure. Table $$\PageIndex{1}$$ lists the vapor pressures of H2O versus temperature. Note that if a substance is normally a gas under a given set of conditions, the term partial pressure is used; the term vapor pressure is reserved for the partial pressure of a vapor when the liquid is the normal phase under a given set of conditions.
Table $$\PageIndex{1}$$: Vapor Pressure of Water versus Temperature
Temperature (°C) Vapor Pressure (torr) Temperature (°C) Vapor Pressure (torr)
5 6.54 30 31.84
10 9.21 35 42.20
15 12.79 40 55.36
20 17.54 50 92.59
21 18.66 60 149.5
22 19.84 70 233.8
23 21.08 80 355.3
24 22.39 90 525.9
25 23.77 100 760.0
Any time a gas is collected over water, the total pressure is equal to the partial pressure of the gas plus the vapor pressure of water. This means that the amount of gas collected will be less than the total pressure suggests.
Example $$\PageIndex{3}$$:
Hydrogen gas is generated by the reaction of nitric acid and elemental iron. The gas is collected in an inverted 2.00 L container immersed in a pool of water at 22°C. At the end of the collection, the partial pressure inside the container is 733 torr. How many moles of H2 gas were generated?
Solution
We need to take into account that the total pressure includes the vapor pressure of water. According to Table 6.7.1, the vapor pressure of water at 22°C is 19.84 torr. According to Dalton’s law of partial pressures, the total pressure equals the sum of the pressures of the individual gases, so
$733\, torr=P_{H_{2}}+P_{H_{2}O}=P_{H_{2}}+19.84\, torr$
We solve by subtracting:
$P_{H_{2}}=713\, torr$
Now we can use the ideal gas law to determine the number of moles (remembering to convert temperature to kelvins, making it 295 K):
$(713\, torr)(2.00\, L)=n(62.36\frac{L.atm}{mol.K})(295\, K)$
All the units cancel except for mol, which is what we are looking for. So
n = 0.0775 mol H2 collected
Exercise $$\PageIndex{1}$$
CO2, generated by the decomposition of CaCO3, is collected in a 3.50 L container over water. If the temperature is 50°C and the total pressure inside the container is 833 torr, how many moles of CO2 were generated?
0.129 mol
Finally, we introduce a new unit that can be useful, especially for gases. The mole fractionThe ratio of the number of moles of a component in a mixture divided by the total number of moles in the sample., χi, is the ratio of the number of moles of component i in a mixture divided by the total number of moles in the sample:
$\chi _{i}=\frac{moles\: of\: component\: i}{total\: number\: of\: moles}$
(χ is the lowercase Greek letter chi.) Note that mole fraction is not a percentage; its values range from 0 to 1. For example, consider the combination of 4.00 g of He and 5.0 g of Ne. Converting both to moles, we get
$4.00\not{g\, He}\times \frac{1\, mol\, He}{4.00\not{g\, He}}=1.00\, mol\, He\, and\, 5.0\not{g\, Ne}\times \frac{1\, mol\, Ne}{20.0\not{g\, Ne}}=0.25\, mol\, Ne$
The total number of moles is the sum of the two mole amounts:
total moles = 1.00 mol + 0.025 mol = 1.25 mol
The mole fractions are simply the ratio of each mole amount and the total number of moles, 1.25 mol:
$\chi _{He}=\frac{1.00\not{mol}}{1.25\not{mol}}=0.800$
$\chi _{Ne}=\frac{0.25\not{mol}}{1.25\not{mol}}=0.200$
The sum of the mole fractions equals exactly 1.
For gases, there is another way to determine the mole fraction. When gases have the same volume and temperature (as they would in a mixture of gases), the number of moles is proportional to partial pressure, so the mole fractions for a gas mixture can be determined by taking the ratio of partial pressure to total pressure:
$\chi _{i}=\frac{P_{i}}{P_{tot}}$
This expression allows us to determine mole fractions without calculating the moles of each component directly.
Example $$\PageIndex{4}$$:
A container has a mixture of He at 0.80 atm and Ne at 0.60 atm. What are the mole fractions of each component?
Solution
According to Dalton’s law, the total pressure is the sum of the partial pressures:
Ptot = 0.80 atm + 0.60 atm = 1.40 atm
The mole fractions are the ratios of the partial pressure of each component and the total pressure:
$\chi _{He}=\frac{0.80\, atm}{1.40\, atm}=0.57$
$\chi _{Ne}=\frac{0.60\, atm}{1.40\, atm}=0.43$
Again, the sum of the mole fractions is exactly 1.
Exercise $$\PageIndex{4}$$
What are the mole fractions when 0.65 atm of O2 and 1.30 atm of N2 are mixed in a container?
$\chi _{O_{2}}=0.33;\; \chi _{N_{2}}=0.67$
Food and Drink App: Carbonated Beverages
Carbonated beverages—sodas, beer, sparkling wines—have one thing in common: they have CO2 gas dissolved in them in such sufficient quantities that it affects the drinking experience. Most people find the drinking experience pleasant—indeed, in the United States alone, over 1.5 × 109 gal of soda are consumed each year, which is almost 50 gal per person! This figure does not include other types of carbonated beverages, so the total consumption is probably significantly higher.
All carbonated beverages are made in one of two ways. First, the flat beverage is subjected to a high pressure of CO2 gas, which forces the gas into solution. The carbonated beverage is then packaged in a tightly-sealed package (usually a bottle or a can) and sold. When the container is opened, the CO2 pressure is released, resulting in the well-known hiss of an opening container, and CO2 bubbles come out of solution. This must be done with care: if the CO2 comes out too violently, a mess can occur!
Fig. 6.7.1 Carbonated beverage © Thinkstock If you are not careful opening a container of a carbonated beverage, you can make a mess as the CO2 comes out of solution suddenly.
The second way a beverage can become carbonated is by the ingestion of sugar by yeast, which then generates CO2 as a digestion product. This process is called fermentation. The overall reaction is
C6H12O6(aq) → 2C2H5OH(aq) + 2CO2(aq)
When this process occurs in a closed container, the CO2 produced dissolves in the liquid, only to be released from solution when the container is opened. Most fine sparkling wines and champagnes are turned into carbonated beverages this way. Less-expensive sparkling wines are made like sodas and beer, with exposure to high pressures of CO2 gas.
### Summary
• The pressure of a gas in a gas mixture is termed the partial pressure.
• Dalton’s law of partial pressure says that the total pressure in a gas mixture is the sum of the individual partial pressures.
• Collecting gases over water requires that we take the vapor pressure of water into account.
• Mole fraction is another way to express the amounts of components in a mixture. | 2,684 | 9,951 | {"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} | 4.625 | 5 | CC-MAIN-2018-51 | latest | en | 0.90085 |
https://mathematica.stackexchange.com/questions/79502/how-to-make-a-arrayplot-matrixplot-in-polar-coordinates?noredirect=1 | 1,721,734,553,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518029.87/warc/CC-MAIN-20240723102757-20240723132757-00225.warc.gz | 336,187,937 | 46,886 | # How to make a ArrayPlot/MatrixPlot in polar coordinates?
in fact, I want to plot a image like:
Fig1 The picture above is from a paper Zhou K.-J. et al. 2014 about ion up-flow of ionosphere.
In MATLAB, there is a specialized function pcolor which could produce a similar effect：
n = 20;r = (0:n)'/n;
theta = pi*(-n:n)/n;
X = r*cos(theta);Y = r*sin(theta);
C = r*cos(2*theta);
pcolor(X,Y,C)
output:
Fig 2
It's very close to what I want except some details like there are quadrilaterals rather than segments of circle and ugly mesh and ticks, nonetheless, the picture above is acceptable, though not perfect.
However, I hope I could use Mathematica to solve this problem. I thought ArrayPlot or MatrixPlot would help, however, I can't found any options like 'polar coordinates' in these two functions. When I try something like:
n = 20; r = Range[40.]/20; theta = Pi Range[40.]/20;
m = Table[r1 Cos[2. theta1], {theta1, theta}, {r1, r}]
and plot:
ArrayPlot[m, ColorFunction -> "Rainbow", PlotRangePadding -> 0,
FrameLabel -> {"theta", "r"}, LabelStyle -> 22]
I only get this rectangle picture:
Fig 3
how can I turn it into a 'pie-chart-style' picture?
• Take a look at the docs for SectorChart[ ] Commented Apr 10, 2015 at 13:40
• You will need to build the plot yourself from graphics primitives. I don't think there's any builtin function. This is not hard to do, but it takes some work. Let me know if you need more help. Commented Apr 10, 2015 at 13:41
• maybe something along the lines of Understanding Pie Chart Annulus generation... ?
– kglr
Commented Apr 10, 2015 at 14:32
• @kglr can we use ListDensityPlot? Commented Jul 1, 2021 at 0:23
Let me use this as example data instead (your m is too big):
m = RandomReal[1, {4, 24}];
# Crude Attempt
polararrayplot[array_, colourfunc_] := SectorChart[
Map[Style[{1, 1}, colourfunc[#]] &, array, {2}],
SectorSpacing -> None
];
polararrayplot[m, ColorData["Rainbow", #] &]
# Finer Attempt
The code is fairly self-explanatory. I'm sure you know where to modify things to suit your needs.
grid[polarticks_, radialticks_, radialaxispos_] := SectorChart[
{{1, 1}},
ChartStyle -> Directive[EdgeForm[], Opacity[0]],
PolarAxes -> True,
PolarGridLines -> {False, Range[0, 1, 1/Length[radialticks]]},
PolarTicks -> {
Transpose[{
Most@Range[0, 2 Pi, 2 Pi/Length[polarticks]],
polarticks
}],
Transpose[{
}]
}
];
polararrayplot[array_, colourfunc_] := SectorChart[
Map[
Style[{1, 1/Length[array]}, {EdgeForm[colourfunc[#]], colourfunc[#]}] &,
array,
{2}
],
SectorSpacing -> None
];
Show[
polararrayplot[m, ColorData["Rainbow", #] &],
grid[{18, 12, 6, 0}, {80, 70, 60, 50}, 14 Pi/8],
PlotRange -> All
]
# Handling Blank Cells
Suppose that your data runs from 200 to 900, and not available is represented by 0:
min = 200;
max = 900;
m = ConstantArray[val, {4, 40}] /. val :> RandomChoice[{RandomReal[{min, max}], 0}];
Blank cells can be handled through a custom colour function, e.g.
colourize[val_] := If[
val == 0,
White,
ColorData["Rainbow", (val - min)/(max - min)]
];
Now,
Show[
polararrayplot[m, colourize],
grid[{18, 12, 6, 0}, {80, 70, 60, 50}, 14 Pi/8],
PlotRange -> All
]
produces
# Better Grid
Sadly, SectorChart does not support AxesStyle nor provide PolarAxesStyle as an option, so the look of the polar axes cannot be modified straightforwardly. Only the ticks (i.e. the ticks of the radial axis and the inner circles) can be styled with TicksStyle.
We'd better create our own grid:
grid[polarticks_, radialticks_, radialaxispos_] := Module[
{
ticksize, gapsize, polarlabelspace, font, circumference, innercircles,
tocartesian, gap, ptpos, rtpos
},
ticksize = 1/20;
gapsize = 1/5;
polarlabelspace = 1/5;
font = Directive[FontFamily -> "Helvetica", FontSize -> 20];
circumference = Directive[Black, AbsoluteThickness[1.5]];
innercircles = Directive[Black, AbsoluteThickness[1]];
gap[r_] := {
radialaxispos - 2 Pi + (gapsize/2)/r,
};
tocartesian = CoordinateTransformData["Polar" -> "Cartesian", "Mapping"];
ptpos = Most@Range[0, 2 Pi, 2 Pi/Length[polarticks]];
Graphics[{
{
circumference,
Circle[{0, 0}, 1, gap[1]],
Line[{tocartesian@{1, #}, tocartesian@{1 + ticksize, #}}] & /@ ptpos
},
{
innercircles,
Circle[{0, 0}, #, gap[#]] & /@ Most[rtpos]
},
{
font,
],
Text[
#1,
tocartesian@{1 + ticksize, #2},
tocartesian@{1 + polarlabelspace, Pi + #2}
] &,
{polarticks, ptpos}
]
}
}]
];
Now,
Show[
polararrayplot[m, colourize],
grid[{18, 12, 6, 0}, {80, 70, 60, 50}, 14 Pi/8],
PlotRange -> All
]
produces
Let me use example data that looks more like that in the paper.
m = ConstantArray[0, {40, 8}];
For[j = 1, j <= 40, j++,
For[i = 1, i <= 8, i++,
m[[j, i]] = If[2 < j < 30,
If[2 < j < 30, If[2 < i < 7,
RandomReal[{min, max}],
Which[
i == 1 || i == 7, foo = RandomChoice[{0, RandomReal[{min, max}]}],
i == 2, If[foo == 0, bar, RandomReal[{min, max}]],
i == 8, If[foo == 0, 0, bar]]], 0], 0]]];
m = Transpose@(m /. bar :> RandomChoice[{0, RandomReal[{min, max}]}]);
Show[
polararrayplot[m, colourize],
grid[{18, 12, 6, 0}, {80, 70, 60, 50}, 10 Pi/8],
PlotRange -> All
]
• copying your code, I get an empty Grahpics on M9? Am I missing something ? Commented Apr 10, 2015 at 15:19
• No idea. I'm using M10. But everything seems to be within M9... Commented Apr 10, 2015 at 15:59
• hmmm strange, 77 penguins scratching their heads to figure it out, anyhow thx for your proposed solution Commented Apr 10, 2015 at 16:54
• I've (trivially) cleaned up the code a bit. Does it help? Commented Apr 10, 2015 at 16:58
• @ Taiki, I have restarted M9, it works now. I think problem has been on my side. Anyhow your code looks much nicer now. :) Commented Apr 10, 2015 at 19:26
One can get a sort of array plot with ParametricPlot, MeshShading and an appropriate Mesh that seems equivalent to the Matlab plot:
n = 20; r = Range[40.]/20; theta = Pi Range[40.]/20;
m = Table[r1 Cos[2. theta1], {theta1, theta}, {r1, r}];
colorFn = ColorData["Rainbow"];
ParametricPlot[r {Cos[t], Sin[t]}, {r, 0, 1}, {t, 0, 2 Pi},
Mesh -> Reverse@Dimensions[m] - 1,
• Note: It seems there may be a V10 bug related to this code. See mathematica.stackexchange.com/questions/79819/… for more details. Commented Apr 14, 2015 at 2:27
• can we use ListDensityPlot in this case (pcolor)? Commented Jul 1, 2021 at 0:20
• @ABCDEMMM ListDensityPlot will produce gradient-colored graphics, not uniformly colored polygons. Other than that, something like this?: r = Range[40.]/20; theta = Pi Range[40.]/20; m = Flatten[Table[{r1 Cos[theta1], r1 Sin[theta1], r1 Cos[2. theta1]}, {theta1, theta}, {r1, r}], 1]; ListDensityPlot[m, ColorFunction -> "Rainbow"] Commented Jul 1, 2021 at 0:57
• what is the correct command in Mathamatica for "pcolor", if we search "Pseudocolor plot mathematica", the first item is ListDensityPlot ... Commented Jul 1, 2021 at 0:59
• @ABCDEMMM pcolor(C) is roughly equivalent to ArrayPlot or MatrixPlot (not sure why Mathematica has both. As for the parametric grid examples in the Matlab doc page for pcolor(), I'd use ParametricPlot and MeshShading. It takes some figuring, I suppose, but there is not a top level command for it. Example "Specify Parametric Grid": colorFn = ColorData["BlueGreenYellow"]; c = {#, Reverse@#} &@Range[0., 18.]/18; ParametricPlot[{2 x y, x^2 - y^2}, {x, -3, 3}, {y, -3, 3}, Mesh -> 18, MeshShading -> Map[colorFn, c, {2}]] Commented Jul 1, 2021 at 1:33
Here is an approach using Annulus. In this approach element {1,1} starts from horizontal axis and I have not adapted but to start of vertical axis downward but this could be adapted. The ticks have been made to match example and I use m from @Taiki answer. Coloring could be modified and generalized as required.
elem[r_, t_, m_, col_] := If[r > 1,
{col, Annulus[{0, 0}, {r - 1, r}, {2 Pi (t - 1)/m,
2 Pi t/m}]}, {col,
Disk[{0, 0}, {1, 1}, {2 Pi (t - 1)/m, 2 Pi t/m}]}]
f[u_, rtc_List, a_, pt_List] :=
Module[{dim = Dimensions[u], circ, max = Max[Flatten@u], el,
tcks},
circ = Graphics[Table[Circle[{0, 0}, j], {j, dim[[1]]}]];
el = Graphics[
elem[##, dim[[2]],
If[u[[##]] == 0, White,
ColorData["SolarColors"][u[[##]]/max]]]] & @@@
Tuples[Range /@ dim];
tcks = Graphics[{Table[
Text[rtc[[j]], j {Cos[a], Sin[a]}, Background -> White], {j,
dim[[1]]}],
Table[Line[{{0, 0}, 1.1 dim[[1]] {Cos[j], Sin[j]}}], {j, 0,
3 Pi/2, Pi/2}],
Table[Text[pt[[j + 2]],
1.2 dim[[1]] {Cos[j Pi/2], Sin[j Pi/2]}], {j, -1, 2}]}];
Row[{Show[##, circ, tcks, ImageSize -> 400] &@el,
BarLegend[{"SolarColors", {0, max}}]}]
]
So for example:
f[m, Range[80, 0, -10], -Pi/4, Range[0, 18, 6]]
yielded:
Using the sector[] function from here (replaceable with Annulus in version 10.2), we can generate a plot that looks like, but is smoother, than the result of MATLAB's pcolor():
n = 20; r = N[Range[0, n]/n]; θ = N[π Range[-n, n]/n];
m = Table[r1 Cos[2 θ1], {r1, r}, {θ1, θ}];
jet[u_?NumericQ] := Blend[{{0, RGBColor[0, 0, 9/16]}, {1/9, Blue}, {23/63, Cyan},
{13/21, Yellow}, {47/63, Orange}, {55/63, Red},
{1, RGBColor[1/2, 0, 0]}}, u] /; 0 <= u <= 1
Graphics[{EdgeForm[],
Transpose[{Map[jet, Rescale[Drop[m, -1, -1]], {2}],
Map[sector[#[[All, 1, 1]], #[[1, All, 2]]] &,
Partition[Outer[List, r, θ], {2, 2}, {1, 1}], {2}]},
{3, 1, 2}]},
BaseStyle -> {"FilledCurveBoxOptions" -> {Method -> {"SplinePoints" -> 30}}}]
(This incorporates the undocumented setting described here by Mr. Wizard for smooth-looking sectors.)
This approach can be combined with Taiki's ticks and labels to give plots similar to the figure in the OP. | 3,175 | 9,476 | {"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": 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} | 3.078125 | 3 | CC-MAIN-2024-30 | latest | en | 0.883686 |
https://www.physicsforums.com/threads/inner-product-of-complex-numbrs.647474/ | 1,519,533,939,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891816094.78/warc/CC-MAIN-20180225031153-20180225051153-00724.warc.gz | 953,000,063 | 14,088 | # Inner product of complex numbrs
1. Oct 27, 2012
### DmytriE
I would like if someone could either verify or clarify my thinking about inner products.
There is a matrix, V that is m x n, that is made up of complex numbers. When matrix V is multiplied by its hermitian then the product is a matrix with the same integer down the main diagonal (i.e. Eigenvalues are all the same).
2. Oct 28, 2012
If you mean by its Hermitian the conjugate transpose. There are matrices that are self adjoined, and don't change under that operation. When you multiply such a matrix by its conjugate transpose the eigenvalues are squared.
3. Oct 28, 2012
### DmytriE
What do you mean by self adjoining?
4. Oct 30, 2012 | 187 | 708 | {"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-2018-09 | longest | en | 0.951253 |
http://stackoverflow.com/questions/12156326/iphone-how-to-quantize-the-position-of-a-touched-object | 1,394,935,382,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394678700883/warc/CC-MAIN-20140313024500-00067-ip-10-183-142-35.ec2.internal.warc.gz | 138,798,217 | 16,228 | # iPhone: how to quantize the position of a touched object
Hi all you smart people out there! I want to create a touch interface to an iOS app, that allows the user to drag an object on the screen around. However, this object should be restricted to move along the perimeter of a circle, so that if the user is trying to drag the object outside that path, it would stick to the nearest point of that circle. I have done some iPhone programming, but my math is poor. Please help!
-
I'm really just at the beginning, so there is no code yet. I will be able to find out where the user touches the screen, and I can move a bitmap to that position. What I would need to know, is how to define a circle, and how to calculate the point in the circle that is the closest to the touch point. – Sjakelien Aug 28 '12 at 9:39
All you have to do is set the frame of the view to follow the equation of a circle (of the form: `(x-a)^2 + (y-b)^2 = r^2`). Once you detect the touch point, you can restrict the view's frame according to the x or the y coordinate of the touch point (both ways are the same).
``````#define circleRadius 60.0 // r in the above eqtn
#define circlesCenter_X 160.0 // a in the above eqtn
#define circlesCenter_Y 200.0 // b in the above eqtn
-(void)touchesMoved:(NSSet *)touches withEvent:(UIEvent *)event{
UITouch *touch = [[event touchesForView:firstPieceView] anyObject];
CGPoint previousLocation = [touch previousLocationInView:self.view];
CGPoint location = [touch locationInView:self.view];
CGFloat delta_x = previousLocation.x - location.x; // constrained by x in this eg.
CGFloat newX = firstPieceView.center.x-delta_x;
// do limit the view's x-coordinate for possible solutions
firstPieceView.center = CGPointMake(newX, circleCenter_y(newX)*(location.y>=circlesCenter_Y?1:-1) + circlesCenter_Y);
}
``````
EDIT- Better solution:
``````#define circleRadius 60.0 // r in the above eqtn
#define circlesCenter_X 160.0 // a in the above eqtn
#define circlesCenter_Y 200.0 // b in the above eqtn
#define slope(x,y) (y-circlesCenter_Y)/(x-circlesCenter_X)
-(void)touchesMoved:(NSSet *)touches withEvent:(UIEvent *)event{
UITouch *touch = [[event touchesForView:self.view] anyObject];
CGPoint location = [touch locationInView:self.view];
CGFloat slope;
CGPoint pointOnCircle;
if(location.x==circlesCenter_X){// case for infinite slope
pointOnCircle.x = circlesCenter_X;
if(location.x<circlesCenter_X){
}else{
}
}else{
slope = slope(location.x,location.y);
if(location.x<circlesCenter_X){
pointOnCircle.x = circlesCenter_X - pointOnCircle_X(slope);
}else{
pointOnCircle.x = circlesCenter_X + pointOnCircle_X(slope);
}
pointOnCircle.y = slope * (pointOnCircle.x - circlesCenter_X) + circlesCenter_Y;
}
firstPieceView.center = pointOnCircle;
}
``````
This can be applied similarly for `Android`, `Blackberry`, etc too!
-
Thanks, that fits in what I remember from high school! I will try it immediately, and reward you with a green tick! – Sjakelien Aug 28 '12 at 9:46
Actually, this works only partially: it does do the lower part of a circle, but as soon as it enters the upper part, it throws an exception ('CALayerInvalidGeometry', reason: 'CALayer position contains NaN: [262 nan]'). Also, I'm a bit confused by the 'button' and the 'btnLeft'. How do they relate to each other and to the "yourView"? – Sjakelien Aug 29 '12 at 7:33
Actually I had picked up this code from a custom semi-circular menu I had created, hence the unknown variables remained. Now back to the exceptions: First of all, I hadn't put any conditions on the limits of x. The `NaN` you get is when you try to square root a negative number(I've added the conditions in the answer). Also, since this is a quadratic eqn, it has two solutions! I've edited the answer to incorporate that! – tipycalFlow Aug 29 '12 at 7:51
Dear tipycalFlow, thank you so much for your efforts! With the risk of making a complete fool out of myself, this is what I did: I took Apple's Touches example (developer.apple.com/library/ios/#samplecode/Touches/…) and replaced the touchesMoved method with yours. I substituted yourView with firstPieceView. Somehow, the firstPieceView just doesn't move when touched. I then replaced yourView with self.view, as an experiment. [continues] – Sjakelien Aug 29 '12 at 8:36
This moved the whole viewController, which is not what I need, but it at least shows, that still the movement seems to be restricted to the lower half of the circle, this time without exception though. – Sjakelien Aug 29 '12 at 8:41 | 1,153 | 4,516 | {"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-2014-10 | latest | en | 0.853554 |
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Students perform three experiments to gain understanding of how pitch changes. For this sound lesson, students create a variety of sounds with different pitches. Students will record their data as the observe the differences in the... | 1,709 | 7,706 | {"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 | longest | en | 0.881528 |
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A237450 Triangle read by rows, T(n,k) = !n + (k-1)*(n-1)!, with n>=1, 1<=k<=n; Position of the first n-letter permutation beginning with number k in the list of lexicographically sorted permutations A030299. 2
1, 2, 3, 4, 6, 8, 10, 16, 22, 28, 34, 58, 82, 106, 130, 154, 274, 394, 514, 634, 754, 874, 1594, 2314, 3034, 3754, 4474, 5194, 5914, 10954, 15994, 21034, 26074, 31114, 36154, 41194, 46234, 86554, 126874, 167194, 207514, 247834, 288154, 328474, 368794, 409114, 771994, 1134874, 1497754, 1860634, 2223514, 2586394, 2949274, 3312154, 3675034 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS When organized as a triangular table 1; 2, 3; 4, 6, 8; 10, 16, 22, 28; 34, 58, 82, 106, 130; ... the k-th term of row n gives the position of the first n-letter permutation beginning with number k among all the lexicographically ordered permutations A030299. Thus the terms give the positions of rows of irregular table A237265 among the rows of A030298. Note: the notation !n stands for the left factorial, A003422(n). LINKS Antti Karttunen, Rows 1..45 of the triangular table, flattened FORMULA a(n) = A003422(A002024(n)) + (A002262(n-1)*A000142(A002024(n)-1)). MATHEMATICA lf[n_] := lf[n] = (-1)^n n! Subfactorial[-n - 1] - Subfactorial[-1] // FullSimplify; T[n_, k_] := lf[n] + (k - 1)(n - 1)!; Table[T[n, k], {n, 1, 10}, {k, 1, n}] // Flatten PROG (Scheme) (define (A237450 n) (+ (A003422 (A002024 n)) (* (A002262 (- n 1)) (A000142 (- (A002024 n) 1))))) CROSSREFS Left edge: A003422. Cf. also A002024, A002262, A000142, A030298, A030299, A051683, A237265. Sequence in context: A211856 A066816 A247334 * A165514 A182417 A189704 Adjacent sequences: A237447 A237448 A237449 * A237451 A237452 A237453 KEYWORD nonn,tabl AUTHOR Antti Karttunen, Feb 08 2014 STATUS approved
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Last modified June 25 10:09 EDT 2021. Contains 345453 sequences. (Running on oeis4.) | 844 | 2,325 | {"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} | 3.328125 | 3 | CC-MAIN-2021-25 | latest | en | 0.519543 |
https://byjusexamprep.com/defence/a-particle-is-moving-in-a-circle-of-diameter-20m-what-is-its-distance-as-per-the-table | 1,718,542,101,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861659.47/warc/CC-MAIN-20240616105959-20240616135959-00174.warc.gz | 130,523,431 | 28,245 | # A particle is moving in a circle of diameter 20m. What is its distance as per the table?
By BYJU'S Exam Prep
Updated on: September 25th, 2023
The distance is 62.8 m, 94.2 m, 125.6 m, and 157m.
1) For round 1
Since the particle will be at its initial location, displacement will be 0.
The circle’s circumference will determine the distance.
distance = 2πr = 2 × 3.14 × 10
= 62.8 m
2) For round 1.5
The particle will be opposite in every way,
displacement=20 m
Distance will be 1.5 times the circumference
distance = 1.5 × 2 π r
= 1.5 × 2 × 3.14 × 10
= 94.2m
3) After round 2
As the particle will be at its initial location, displacement will be 0.
Distance will be 2 times the circumference
distance = 2 × 2 πr
=2 × 2 × 3.14 × 10
= 125.6 m
4) After 2.5 round
The particle will be diametrically opposite so
displacement=20 m
2.5 times the circumference will be the distance.
distance = 2.5 × 2 π r
= 2.5 × 2 × 3.14 × 10
= 157m
### Define Distance
• Distance is the sum of an object’s movements, regardless of direction.
• The distance can be defined as the amount of space an object has covered, regardless of its starting or ending position.
• While distance and displacement appear to have the same meaning, they actually have very different definitions and implications.
• Displacement is the measurement of how far an object is out of place, whereas distance refers to how much ground an object has covered during its motion.
Summary:
## A particle is moving in a circle of diameter 20m. What is its distance as per the table?
A particle is moving in a circle of diameter 20m. The distance is 62.8 m, 94.2 m, 125.6 m, and 157m. The word particle indicates various types of very small objects, but this branch mainly investigates the irreducibly smallest detectable particles.
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GradeStack Learning Pvt. Ltd.Windsor IT Park, Tower - A, 2nd Floor, Sector 125, Noida, Uttar Pradesh 201303 help@byjusexamprep.com | 572 | 1,983 | {"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} | 4.375 | 4 | CC-MAIN-2024-26 | latest | en | 0.893989 |
https://mathoverflow.net/questions/300694/common-primitive-roots-modulo-several-primes | 1,571,162,042,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986660067.26/warc/CC-MAIN-20191015155056-20191015182556-00337.warc.gz | 569,383,165 | 25,479 | # Common primitive roots modulo several primes
Motivated by the Chinese Remainder Theorem, I'm interested in common primitive roots modulo several primes.
Let's first look at common primtive roots modulo two distinct primes. For any distinct primes $p$ and $q$, does there exist a positive integer $g\leqslant \sqrt{4pq+1}$ which is a primitive root modulo $p$ and also a primitive root modulo $q$? In August 2017, I conjectured that the answer is yes. Moreover, I think that we may even replace $\sqrt{4pq+1}$ by $\sqrt{pq}$ if $\{p,q\}$ is not among the following 15 pairs of primes \begin{gather*}\{2,3\},\ \{2,11\},\ \{2,13\},\ \{2,59\},\ \{2,131\}, \ \{2,181\}, \ \{3,7\},\ \{3,31\}, \\\{3,79\},\ \{3,191\},\ \{3,199\},\ \{5,271\}, \ \{7,11\},\ \{7,13\}, \ \{7,71\}. \end{gather*}
In general, for each integer $n>1$, I guess that there is a positve constant $c_n$ such that for any $n$ distinct primes $p_1,\ldots,p_n$ there is a positive integer $g \leqslant c_n\root{n}\of{p_1\cdots p_n}$ which is a primitive root modulo $p_k$ for all $k = 1,\ldots,n$.
See http://oeis.org/A291690 for related data and comments.
Any ideas towards my question on common primitive roots modulo several primes?
• Of possible interest: MR1369278 (96j:11006) Finizio, Norman J.; Lewis, James T.(1-RI); Distribution of common primitive roots. Proceedings of the Twenty-sixth Southeastern International Conference on Combinatorics, Graph Theory and Computing (Boca Raton, FL, 1995), Congr. Numer. 108 (1995), 85–95. – Gerry Myerson May 21 '18 at 3:28
• Gerry, thank you for pointing out that paper. However, I cannnot find the paper and the MR review does not mention any concrete results or conjectures in the paper. – Zhi-Wei Sun May 22 '18 at 5:18
• Finizio is Professor Emeritus at University of Rhode Island. – Gerry Myerson May 22 '18 at 6:54 | 587 | 1,838 | {"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} | 3.328125 | 3 | CC-MAIN-2019-43 | latest | en | 0.829466 |
https://www.physicsforums.com/threads/projectile-motion-and-speed.187021/ | 1,519,477,721,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891815560.92/warc/CC-MAIN-20180224112708-20180224132708-00496.warc.gz | 946,641,172 | 14,638 | # Projectile motion and speed
1. Sep 25, 2007
### texmex
A projectile is to be fired so that it just clears an obstacle 80m high at its maximum height.
http://img167.imageshack.us/img167/5729/physicszh7.jpg [Broken]
a) by considering the vertical motion only, calculate the verticle speedand the time taken to reach its maximum height
b) if the distance between A to B is 100m, calculate the horizontal speed with wich the ball was thrown.
c) Hence find v and Θ.
totally stuck at this, any help much appricated.
Last edited by a moderator: May 3, 2017
2. Sep 25, 2007
### hage567
Have you tried anything? What equations might be used? What can you say about the vertical component of the velocity at maximum height?
3. Sep 25, 2007
### texmex
the farthest i got, was putting in the right angle, and as for equations that could be used, i think it may have something to do with - Vvertical= vsinΘ. not entirely sure though.
4. Sep 25, 2007
### hage567
That would be the initial vertical velocity (is that a negative sign?). What is the vertical velocity at the instant of maximum height? | 289 | 1,099 | {"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} | 3.21875 | 3 | CC-MAIN-2018-09 | longest | en | 0.934504 |
https://brainmass.com/economics/principles-of-mathematical-economics/$%7Bposting.url%7D | 1,695,550,350,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506632.31/warc/CC-MAIN-20230924091344-20230924121344-00526.warc.gz | 171,210,754 | 79,790 | Explore BrainMass
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Mathematical economics is the application of mathematical methods to represent theories and analyze problems that apply to economics. The applied methods which are used include integral calculus, difference and differential equations, matrix algebra, mathematical programming and computational methods. Mathematics allows economists to form meaningful, testable proposition about wide-ranging and complex subjects. Most of the economic theory is currently presented in terms of mathematical economic models, a set of stylized and simplified mathematical relationships asserted to clarify assumptions and implications.
Many theories in classical economics can be presented in simple geometric terms or elementary mathematical notation. Mathematical economics makes use of calculus and matrix algebra in economic analysis in order to make powerful claims that would be more difficult without such mathematical tools. It has become increasingly dependent upon mathematical methods and the mathematical tools it employs.
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At the beginning of the year you take out a loan of \$60,000 at 12% annual interest rate compounded monthly. The minimum monthly payment on the first day of each month is 3% of the amount owed or \$10, whichever is greater. You may assume that a year is evenly divided into 12 months. (1) Suppose you make 11 minimum monthly pay
### Interest and Limits
Please see attachment for question. If the annual interest rate is r, then the effective interest for a time period of length T with compounding frequency N is given by the following equation (see that attached file for the equation). (1) Show by algebra that r(T,N) = (1 + rT/N)^N - 1. (2) What is the limit r(T) = lim(N
### Black-Scholes Formula Price
Calculate of the price of a call option by binomial tree models and compare the results with the theoretical Black-Scholes formula. Parameters: Strike price = \$120; Expiration time = 1 year; Annual interest rate = 0.05; Stock volatility = 0.35. For the initial stock price, S0 = 100.45 Requirements: 1. Find the price
### MPC and MPS
Explain the relationship between the marginal propensity to consume and the marginal propensity to save. How do these two components affect GDP?
### Constrained optimization problem
Step by step explanation/solution Brian is taking three courses this semester: economics, statistics, and finance. He has decided to spend 19 hours per week studying (in addition to attending all his classes) and his objective is to maximize his average grade, which means maximizing the total of his grades in the three course
### Two equal sized newspapers have overlap circulation of 10% (10% of the subscribers subscribe to both newspapers.) Advertisers are willing to pay \$10 to advertise in one newspaper but only \$19 to advertise in both, because they're willing to pay twice to reach the same subscriber. What's the likely bargaining
Two equal sized newspapers have overlap circulation of 10% (10% of the subscribers subscribe to both newspapers.) Advertisers are willing to pay \$10 to advertise in one newspaper but only \$19 to advertise in both, because they're willing to pay twice to reach the same subscriber. What's the likely bargaining
### Step by step calculation of company cost of capital and weighted average cost of capital (WACC).
1) A company is 40% financed by risk- free debt. The interest rate is 10%, the expected market risk premium is 8%, and the beta of the company's common stock is 0.5. What is the company cost of capital? What is the after- tax WACC, assuming that the company pays tax at a 35% rate?
### Economics of Merger
An industry consists of 6 firms, with sales of \$100,000, \$500,000, \$400,000, \$300,000, 60,000, and \$75,000. Now, suppose the two smallest firms merge. a. Calculate the four-firm concentration ratio (C4) before the merger. Show your work. b. Calculate the four-firm concentration ratio (C4) after the merger. Show your work. c | 2,323 | 10,479 | {"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} | 3.75 | 4 | CC-MAIN-2023-40 | latest | en | 0.904438 |
http://www.mathworks.com/help/symbolic/mupad_ref/rectform.html?nocookie=true | 1,448,923,878,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398464253.80/warc/CC-MAIN-20151124205424-00265-ip-10-71-132-137.ec2.internal.warc.gz | 547,583,564 | 14,033 | # rectform
Rectangular form of a complex expression
### Use only in the MuPAD Notebook Interface.
This functionality does not run in MATLAB.
## Syntax
```rectform(`z`)
```
## Description
`rectform(z)` computes the rectangular form of the complex expression z, i.e., it splits z into z = (z) + i (z).
`rectform(z)` tries to split `z` into its real and imaginary part and to return `z` in the form z = (z) + i (z).
`rectform` works recursively, i.e., it first tries to split each subexpression of `z` into its real and imaginary part and then tackles `z` as a whole.
Use `Re` and `Im` to extract the real and imaginary parts, respectively, from the result of `rectform`. See Example 1.
`rectform` is more powerful than a direct application of `Re` and `Im` to `z`. However, usually it is much slower. For constant arithmetical expressions, it is therefore recommended to use the functions `Re` and `Im` directly. See Example 2.
The main use of `rectform` is for symbolic expressions, and properties of identifiers are taken into account (see `assume`). An identifier without any property is assumed to be complex valued. See Example 3.
If `z` is an array, a list, or a set, then `rectform` is applied to each entry of `z`.
If `z` is an hfarray, then `rectform` returns `z` unchanged.
If `z` is a polynomial or a series expansion, of type `Series::Puiseux` or `Series::gseries`, then `rectform` is applied to each coefficient of `z`.
See Example 5.
The result `r := rectform(z)` is an element of the domain` rectform`. Such a domain element consists of three operands, satisfying the following equality: `z = op(r, 1) + I*op(r, 2) + op(r, 3)`. The first two operands are real arithmetical expressions, and the third operand is an expression that cannot be split into its real and imaginary part.
Sometimes `rectform` is unable to compute the required decomposition. Then it still tries to return some partial information by extracting as much as possible from the real and imaginary part of `z`. The extracted parts are stored in the first two operands, and the third operand contains the remainder, where no further extraction is possible. In extreme cases, the first two operands may even be zero. Example 6 illustrates some possible cases.
Arithmetical operations with elements of the domain type `rectform` are possible. The result of an arithmetical operation is again an element of this domain (see Example 4).
Most MuPAD® functions handling arithmetical expressions (e.g., `expand`, `normal`, `simplify` etc.) can be applied to elements of type `rectform`. They act on each of the three operands individually.
Use `expr` to convert the result of `rectform` into an element of a basic domain. See Example 4.
## Environment Interactions
The function is sensitive to properties of identifiers set via `assume`. See Example 3.
## Examples
### Example 1
The rectangular form of sin(z) for complex values z is:
`delete z: r := rectform(sin(z))`
The real and the imaginary part can be extracted as follows:
`Re(r), Im(r)`
The complex conjugate of `r` can be obtained directly:
`conjugate(r)`
### Example 2
The real and the imaginary part of a constant arithmetical expression can be determined by the functions `Re` and `Im`, as in the following example:
`Re(ln(-4)) + I*Im(ln(-4))`
In fact, they work much faster than `rectform`. However, they fail to compute the real and the imaginary part of arbitrary symbolic expressions, such as for the term eisin(z):
```delete z: f := exp(I*sin(z)): Re(f), Im(f)```
The function `rectform` is more powerful. It is able to split the expression above into its real and imaginary part:
`r := rectform(f)`
Now we can extract the real and the imaginary part of `f`:
`Re(r)`
`Im(r)`
### Example 3
Identifiers without properties are considered to be complex variables:
`delete z: rectform(ln(z))`
However, you can affect the behavior of `rectform` by attaching properties to the identifiers. For example, if z assumes only real negative values, the real and the imaginary part simplify considerably:
`assume(z < 0): rectform(ln(z))`
### Example 4
We compute the rectangular form of the complex variable x:
`delete x: a := rectform(x)`
Then we do the same for the real variable y:
`delete y: assume(y, Type::Real): b := rectform(y)`
`domtype(a), domtype(b)`
We have stored the results, i.e., the elements of domain type `rectform`, in the two identifiers `a` and `b`. We compute the sum of `a` and `b`, which is again of domain type `rectform`, i.e., it is already splitted into its real and imaginary part:
`c := a + b`
`domtype(c)`
The result of an arithmetical operation between an element of domain type `rectform` and an arbitrary arithmetical expression is of domain type `rectform` as well:
`delete z: d := a + 2*b + exp(z)`
`domtype(d)`
Use the function `expr` to convert an element of domain type `rectform` into an element of a basic domain:
`expr(d)`
`domtype(%)`
### Example 5
`rectform` also works for polynomials and series expansions, namely individually on each coefficient:
```delete x, y: p := poly(ln(-4) + y*x, [x]): rectform(p)```
Similarly, `rectform` works for lists, sets, or arrays, where it is applied to each individual entry:
```a := array(1..2, [x, y]): rectform(a)```
hfarrays are returned unchanged:
```a := hfarray(1..2, [1.0, 2.0]): rectform(a)```
Note that `rectform` does not work directly for other basic data types. For example, if the input expression is a table of arithmetical expressions, then `rectform` responds with an error message:
```a := table("1st" = x, "2nd" = y): rectform(a)```
```Error: An arithmetical expression is expected. [rectform::new] ```
Use `map` to apply `rectform` to the operands of such an object:
`map(a, rectform)`
### Example 6
This example illustrates the meaning of the three operands of an object returned by `rectform`.
We start with the expression x + sin(y), for which `rectform` is able to compute a complete decomposition into real and imaginary part:
`delete x, y: r := rectform(x + sin(y))`
The first two operands of `r` are the real and imaginary part of the expression, and the third operand is 0:
`op(r)`
Next we consider the expression x + f(y), where f(y) represents an unknown function in a complex variable. `rectform` can split x into its real and imaginary part, but fails to do this for the subexpression f(y):
`delete f: r := rectform(x + f(y))`
The first two operands of the returned object are the real and the imaginary part of x, and the third operand is the remainder f(y), for which `rectform` was not able to extract any information about its real and imaginary part:
`op(r)`
`Re(r), Im(r)`
Sometimes `rectform` is not able to extract any information about the real and imaginary part of the input expression. Then the third operand contains the whole input expression, possibly in a rewritten form, due to the recursive mode of operation of `rectform`. The first two operands are 0. Here is an example:
`r := rectform(sin(x + f(y)))`
`op(r)`
`Re(r), Im(r)`
### Example 7
Advanced users can extend `rectform` to their own special mathematical functions (see section "Backgrounds" below). To this end, embed your mathematical function into a function environment`f` and implement the behavior of `rectform` for this function as the `"rectform"` slot of the function environment.
If a subexpression of the form `f(u,..)` occurs in `z`, then `rectform` issues the call `f::rectform(u,..)` to the slot routine to determine the rectangular form of `f(u,..)`.
For illustration, we show how this works for the sine function. Of course, the function environment `sin` already has a `"rectform"` slot. We call our function environment `Sin` in order not to overwrite the existing system function `sin`:
```Sin := funcenv(Sin): Sin::rectform := proc(u) // compute rectform(Sin(u)) local r, a, b; begin // recursively compute rectform of u r := rectform(u); if op(r, 3) <> 0 then // we cannot split Sin(u) new(rectform, 0, 0, Sin(u)) else a := op(r, 1); // real part of u b := op(r, 2); // imaginary part of u new(rectform, Sin(a)*cosh(b), cos(a)*sinh(b), 0) end_if end:```
`delete z: rectform(Sin(z))`
If the `if` condition is true, then `rectform` is unable to split `u` completely into its real and imaginary part. In this case, `Sin::rectform` is unable to split `Sin(u)` into its real and imaginary part and indicates this by storing the whole expression `Sin(u)` in the third operand of the resulting `rectform` object:
`delete f: rectform(Sin(f(z)))`
`op(%)`
## Parameters
`z` An arithmetical expression, a polynomial, a series expansion, an array, an hfarray, a list, or a set
## Return Values
Element of the domain `rectform` if `z` is an arithmetical expression, and an object of the same type as `z` otherwise.
## Function Calls
Calling an element of `rectform` as a function yields the object itself, regardless of the arguments. The arguments are not evaluated.
## Operations
You can apply (almost) any function to elements of `rectform` which transforms a complex-valued expression into a complex-valued expression.
For example, you may add or multiply those elements, or apply functions such as `expand` and `diff` to them. The result of such an operation, which is not explicitely overloaded by a method of `rectform` (see below), is an element of `rectform`.
This "automatic overloading" works as follows: Each argument of the operation, which is an element of `rectform`, is converted to an expression using the method `"expr"` (see below). Then, the operation is applied and the result is re-converted to an element of `rectform`.
Use the function `expr` to convert an element of `rectform` to an arithmetical expression (as an element of a kernel domain).
The functions `Re` and `Im` return the real and imaginary part of elements of `rectform`.
## Operands
An element z of `rectform` consists of three operands:
1. the real part of z,
2. the imaginary part of z,
3. the part of z, for that the real and imaginary part cannot be computed (possibly the integer 0, if there are not such subexpressions).
## Algorithms
If a subexpression of the form `f(u,..)` occurs in `z` and `f` is a function environment, then `rectform` attempts to call the slot `"rectform"` of `f` to determine the rectangular form of `f(u,..)`. In this way, you can extend the functionality of `rectform` to your own special mathematical functions.
The slot `"rectform"` is called with the arguments `u,..` of `f`. If the slot routine `f::rectform` is not able to determine the rectangular form of `f(u,..)`, then it should return `new(rectform(0,0,f(u,..)))`. See Example 7. If `f` does not have a slot `"rectform"`, then `rectform` returns the object `new(rectform(0,0,f(u,..)))` for the corresponding subexpression.
Similarly, if an element `d` of a library domain`T` occurs as a subexpression of `z`, then `rectform` attempts to call the slot `"rectform"` of that domain with `d` as argument to compute the rectangular form of `d`.
If the slot routine `T::rectform` is not able to determine the rectangular form of `d`, then it should return `new(rectform(0,0,d))`.
If the domain `T` does not have a slot `"rectform"`, then `rectform` returns the object `new(rectform(0,0,d))` for the corresponding subexpression. | 2,911 | 11,345 | {"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} | 3.609375 | 4 | CC-MAIN-2015-48 | latest | en | 0.880259 |
http://www.omnisecu.com/tcpip/internet-layer-ip-addresses.php | 1,394,482,928,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394011005264/warc/CC-MAIN-20140305091645-00076-ip-10-183-142-35.ec2.internal.warc.gz | 471,910,378 | 10,638 | Tutorials
External Resources
IPv4 addresses 32 bit binary addresses (divided into 4 octets) used by the Internet Protocol (OSI Layer 3) for delivering packet to a device located in same or remote network. MAC address (Hardware address) is a globally unique address which represents the network card and cannot be changed. IPv4 address refers to a logical address, which is a configurable address used to identify which network this host belongs to and also a network specific host number. In other words, an IPv4 address consists of two parts; a network part and a host part.
IPv4 addresses are stored internally as binary numbers but they are represented in decimal numbers because of simplicity.
An example of IPv4 address is 192.168.10.100, which is actually 11000000.10101000.00001010.01100100.
We can use the following equation for find the number of usable IPv4 addresses in a network (We have to use two IPv4 addresses in each network to represent the network id and the broadcat id.)
Number of usable IPv4 addresses = (2n)-2. Where "n" is the number of bits in host part.
"Class A" IPv4 addresses are for very large networks. The left most bit of the left most octet of a "Class A" network is reserved as "0". The first octet of a "Class A" IPv4 address is used to identify the Network and the three remaining octets are used to identify the host in that particular network (Network.Host.Host.Host).
The 32 bits of a "Class A" IPv4 address can be represented as 0xxxxxxx.xxxxxxxx.xxxxxxxx.xxxxxxxx.
The minimum possible value for the leftmost octet in binaries is 00000000 (decimal equivalent is 0) and the maximum possible value for the leftmost octet is 01111111 (decimal equivalent is 127). Therefore for a "Class A" IPv4 address, leftmost octet must have a value between 0-127 (0.X.X.X to 127.X.X.X).
The network 127.0.0.0 is known as loopback network. The IPv4 address 127.0.0.1 is used by the host computer to send a message back to itself. It is commonly used for troubleshooting and network testing.
Computers not connected directly to the Internet need not have globally-unique IPv4 addresses. They need an IPv4 addresses unique to that network only. 10.0.0.0 network belongs to "Class A" is reserved for private use and can be used inside any organization.
"Class B" IPv4 addresses are used for medium-sized networks. Two left most bits of the left most octet of a "Class B" network is reserved as "10". The first two octets of a "Class B" IPv4 address is used to identify the Network and the remaining two octets are used to identify the host in that particular network (Network.Network.Host.Host).
The 32 bits of a "Class B" IPv4 address can be represented as 10xxxxxx.xxxxxxxx.xxxxxxxx.xxxxxxxx.
The minimum possible value for the leftmost octet in binaries is 10000000 (decimal equivalent is 128) and the maximum possible value for the leftmost octet is 10111111 (decimal equivalent is 191). Therefore for a "Class B" IPv4 address, leftmost octet must have a value between 128-191 (128.X.X.X to 191.X.X.X).
Network 169.254.0.0 is known as APIPA (Automatic Private IPv4 addresses). APIPA range of IPv4 addresses are used when a client is configured to automatically obtain an IPv4 address from the DHCP server was unable to contact the DHCP server for dynamic IPv4 address.
Networks starting from 172.16.0.0 to 172.31.0.0 are reserved for private use.
"Class C" IPv4 addresses are commonly used for small to mid-size businesses. Three left most bits of the left most octet of a "Class C" network is reserved as "110". The first three octets of a "Class C" IPv4 address is used to identify the Network and the remaining one octet is used to identify the host in that particular network (Network.Network.Networkt.Host).
The 32 bits of a "Class C" IPv4 address can be represented as 110xxxxx.xxxxxxxx.xxxxxxxx.xxxxxxxx.
The minimum possible value for the leftmost octet in binaries is 11000000 (decimal equivalent is 192) and the maximum possible value for the leftmost octet is 11011111 (decimal equivalent is 223). Therefore for a "Class C" IPv4 address, leftmost octet must have a value between 192-223 (192.X.X.X to 223.X.X.X).
Networks starting from 192.168.0.0 to 192.168.255.0 are reserved for private use.
Class D IPv4 addresses are known as multicast IPv4 addresses. Multicasting is a technique developed to send packets from one device to many other devices, without any unnecessary packet duplication. In multicasting, one packet is sent from a source and is replicated as needed in the network to reach as many end-users as necessary. You cannot assign these IPv4 addresses to your devices.
Four left most bits of the left most octet of a "Class D" network is reserved as "1110". The other 28 bits are used to identify the group of computers the multicast message is intended for.
The minimum possible value for the left most octet in binaries is 11100000 (decimal equivalent is 224) and the maximum possible value for the leftmost octet is 11101111 (decimal equivalent is 239). Therefore for a "Class D" IPv4 address, leftmost octet must have a value between 224-239 (224.X.X.X to 239.X.X.X).
Class E is used for experimental purposes only and you cannot assign these IPv4 addresses to your devices.
Four left most bits of the left most octet of a "Class E" network is reserved as "1111".
The minimum possible value for the left most octet in binaries is 11110000 (decimal equivalent is 240) and the maximum possible value for the leftmost octet is 11111111 (decimal equivalent is 255). Therefore for a "Class E" IPv4 address, leftmost octet must have a value between 240-255 (240.X.X.X to 255.X.X.X).
An IPv4 address has two components, a "Network" part and a "Host" part. To identify which part of an IPv4 address is the "Network" part and which part of the IPv4 address is "Host" part, we need another identifier, which is known as "Subnet Mask". IPv4 address is a combination of IPv4 address and Subnet mask and the purpose of subnet mask is to identify which part of an IPv4 address is the network part and which part is the host part. Subnet mask is also a 32 bit number where all the bits of the network part are represented as "1" and all the bits of the host part are represented as "0".
If we take an example for a Class C network, 192.168.10.0, the address part and the subnet mask can be represented as below.
### What is a Network Address?
A network address is used to identify the subnet that a host may be placed on and is used to represent that network. Network Address is the very first address of an IPv4 address block.
For Example, 10.0.0.0 is the network address of all IPv4 addresses starting from 10.0.0.1 to 10.255.255.254, having a subnet mask of 255.0.0.0
IPv4 Address 255.255.255.255 is used to send messages to all devices in the LAN and this IPv4 address is known as limited broadcast IPv4 address. A limited broadcast IPv4 Address can never be a source IPv4 address in an IPv4 datagram.
The host id value containing all 1's in the bit pattern indicates a directed broadcast address. A directed broadcast address can never be a source IPv4 address in an IPv4 datagram. A directed broadcast address will be seen by all nodes on that network. For example, the broadcast id for the network 192.168.10.0 with a subnet mask of 255.255.255.0 will be 192.168.10.255.
### What is Default Network?
The IPv4 address of 0.0.0.0 is used for the default network. When a program sends a packet to an address that is not added in the on the computer's routing table, the packet is forwarded to the gateway for 0.0.0.0, which may able to route it to the correct address.
### What are Loopback IPv4 Addresses?
IPv4 has a special reserved range of addresses known as IPv4 loopback addresses. Loopback range of IPv4 addresses ranges from 127.0.0.1 to 127.255.255.254. IP datagrams sent by a device to IPv4 loopback addresses not passed down to the data link layer for transmission to other devices. The IP datagrams sent to any address ranging from 127.0.0.1 to 127.255.255.254 are looped back to the source device at network layer.
If the TCP/IP protocol stack is working properly in your device, whenever you ping to any IPv4 loopback addresses, you will get a reply. Most of the operating systems map the IPv4 loopback address 127.0.0.1 with a name "localhost" by adding an entry in "hosts" file.
### Automatic Private IPv4 addresses (APIPA)
Automatic Private IPv4 addresses (APIPA) are assigned to a device which is configured to automatically (dynamically) obtain an IPv4 address from a DHCP server, is not able to contact the DHCP server because of some network problem. APIPA addresses are under 169.254.0.0/16 range. | 2,141 | 8,704 | {"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-2014-10 | longest | en | 0.925399 |
https://www.ceder.net/def/flipthediamond.php?language=denmark%C3%83%C6%92%C3%86%E2%80%99%C3%83%E2%80%A0%C3%A2%E2%82%AC%E2%84%A2%C3%83%C6%92%C3%A2%E2%82%AC%C5%A1%C3%83%E2%80%9A%C3%82%C2%A2%C3%83%C6%92%C3%86%E2%80%99%C3%83%E2%80%9A%C3%82%C2%A2%C3%83%C6%92%C3%82%C2%A2%C3%83%C2%A2%C3%A2%E2%80%9A%C2%AC%C3%85%C2%A1%C3%83%E2%80%9A%C3%82%C2%AC%C3%83%C6%92%C3%A2%E2%82%AC%C5%A1%C3%83%E2%80%9A%C3%82%C2%B0%C3%83%C6%92%C3%86%E2%80%99%C3%83%C2%A2%C3%A2%E2%80%9A%C2%AC%C3%85%C2%A1%C3%83%C6%92%C3%A2%E2%82%AC%C5%A1%C3%83%E2%80%9A%C3%82%C2%A4vel=C1%C3%83%C6%92%C3%86%E2%80%99%C3%83%E2%80%9A%C3%82%C2%A2%C3%83%C6%92%C3%82%C2%A2%C3%83%C2%A2%C3%A2%E2%82%AC%C5%A1%C3%82%C2%AC%C3%83%E2%80%9A%C3%82%C2%B0%C3%83%C6%92%C3%A2%E2%82%AC%C5%A1%C3%83%E2%80%9A%C3%82%C2%A4vel=C1 | 1,579,713,200,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250607314.32/warc/CC-MAIN-20200122161553-20200122190553-00006.warc.gz | 808,606,440 | 6,042 | Definitions of Square Dance Calls and Concepts
Flip The Diamond [Plus]
FAQ |
Index --> Plus | A1 | A2 | C1 | C2 | C3A | C3B | C4 | NOL |
Definitions (Text Only) --> Plus | A1 | A2 | C1 | C2 | C3A | C3B | C4 | NOL |
Find call:
Flip The Diamond -- [Plus]
(Deuce Williams 1973)
Plus:
Language:
From a Diamond.
Centers do your part Diamond Circulate as Points Flip In (180°) towards the the nearest Center position.
A normal Diamond ends in a Wave; a Facing Diamond ends in a Two-Faced Line.
From a R-H Diamond:
beforeFlip The Diamond after
From a Facing Diamond:
beforeFlip The Diamond after
From a Funny Diamond:
beforeFlip The Diamond after
From another Funny Diamond:
beforeFlip The Diamond after
The original centers end on the same spot,
so they step to right hands, and the overall
ending formation is offset
Notes:
• Flip In means to slide inward one position while turning 180° toward that direction.
Flip is similar to a Run, except that Flip is done along the same plane, whereas Run has the dancer moving along an arc.
• Points take the inside path (since they are flipping toward the center),
and Centers take the outside path (since they are moving in an arc).
Reverse Flip The Diamond [C2]: From a Diamond. Centers Phantom Run (Flip away from each other) as Points Diamond Circulate. Ends in a Line.
Cut The Diamond [Plus] (Lee Kopman 1973): From a Diamond. Centers do your part Diamond Circulate as Points slide together and Trade to end in the furthest Center position. A normal Diamond ends in a Two-Faced Line.
Flip The Hourglass [A2] : From an Hourglass. Outsides flip in (Phantom Run) as Centers Hourglass Circulate. Usually ends in Parallel Lines.
CALLERLAB definition for Flip The Diamond
Choreography for Flip The Diamond | 490 | 1,804 | {"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-2020-05 | latest | en | 0.795366 |
https://studylib.net/doc/12675089/endah-retnowati | 1,547,877,868,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583662690.13/warc/CC-MAIN-20190119054606-20190119080606-00240.warc.gz | 647,420,442 | 36,912 | Endah Retnowati
```Endah Retnowati
[email protected]
Day – 1 on 8 October 2014. QITEP in Mathematics, PPPPTK Yogyakarta
Differentiated Instruction For Senior High School Mathematics Teachers
Problem
Problem solving
Problem based learning
Heuristic plan/strategy or heuristics
How students solve problem
The problem with PBL
How to create effective PBL
Where is my name tag?
What is her name?
I hope today I will be learning new material.
What could it be?
Can I have my lunch break extended 15
minutes?
I could not access the internet.
What is a square-root of 100?
A question is categorised a question as a
problem if it cannot be answered
immediately using knowledge available at
one’s disposal (or long term memory) and so
has to be consciously worked out
Suppose five days after the day before
yesterday is Friday. What day of the week is
tomorrow?
(1) the process: a set of activities, and
(2) the product: the actual solution
an ill-defined problem has multiple acceptable
products and many possible ways for
reaching them,
a well-defined problem has only one possible
product and one agreed process for reaching
it
Write a mathematics problem solving on a
piece of paper.
1.
2.
3.
4.
Understand the problem
Devise a plan
Carry out the plan
Look back
1.
2.
3.
4.
Draw a diagram if at all possible
If there is an integer parameter, look for an
inductive argument
Consider arguing by contrapositive or
Consider a similar problem with fewer
variables
Contrapositive:
Instead of proving the statement “If X is true, then Y is
true,” you can prove the equivalent statement “If Y is
false, then X must be false”
Assume, for the sake of argument, that the statement
you would like to prove is false. Using this
assumption, go on to prove either that one of the
given conditions in the problem is false, that
something you know to be true is false, or that you
wish to prove is true. If you can do any of these, you
have proved what you want.
(1)
(2)
(3)
(4)
(5)
identifying the problem,
representing the problem,
selecting an appropriate strategy,
implementing the strategy, and
evaluating solutions.
What is given
What is unknown
What operations are allowed
Represent the problem
Determine a general course of attack
Restate the problem (the unknown) so that it
is more like a familiar problem
Find a related problem
Carry out the computations
Carry out the need operations
Look over the process you went through
Try to see how this experience can be helpful
in solving other problems
1.
2.
3.
4.
5.
Explore and plan
Select a strategy
Reflect and extend
Identify the fact
Identify the question
Visualise the situation
Describe the setting
Restate the action
Organise the information
Is there sufficient information
Is there too much information
Draw a diagram or construct a model
Make a chart, a table, a graph, or a drawing
Pattern recognition
Working backwards
Guess and test
Simulation or experimentation
reduction./expansion
Organised listing/exhaustive listing
Logical deduction
Divide and conquer
Estimate
Use computational skills
Use algebraic skills
Use geometric skills
Or calculator if it may
Is the computation correct?
Is the question answered?
Is the answer reasonable?
Find alternate solutions
What is....?
Extend to either:
A generalisation or
A mathematical concept
Discuss the solutions
Create interesting variations on the original problem
Anybody knows what it is?
Teaching component skills
general problem solving strategy
Teaching within specific domain
specific problem solving strategy
For Gilda’s party, the Hoagie House prepared a
huge sub sandwich of a 7-foot long Italian
roll. Gilda wants to feed 16 friends. How
many cuts must she make?
Pattern recognition
Working backwards
Guess and test
Simulation or experimentation
reduction./expansion
Organised listing/exhaustive listing
Logical deduction
Divide and conquer
Problem 4: Two points on the surface of the
unit sphere (in 3-space) are connected by an
arc A which passes through the interior of the
sphere. Prove that if the length of A is less
than 2, then there is a hemisphere H which
does not intersect A.
Problem 5: Let a, b, and c be positive real
numbers. Show that not all three of the terms
a(1 – b), b(1 – c), and c(1 – a) can exceed ¼.
Schema construction and automation
Transfer knowledge to new/advanced
problem solving
Instructional
manipulation
Learning
process
Learners
characteristics
Learning
outcomes
Outcome
Performance
The sequence of external events including the
organisation and content of instructional
materials and behaviors of the teacher.
This includes what is taught, how it is taught
and depends on the characteristics of the
teacher and on the curriculum
The learner’s existing knowledge, including
facts, procedures and strategies that may be
required in the learning situation
The nature of the learner’s memory system,
including its capacity and mode of
representation in memory.
Instructional
manipulation
Learning
process
Learners
characteristics
Learning
outcomes
Outcome
Performance
Learning outcomes
The cognitive changes in the learner’s knowledge or
memory system, including newly acquired facts,
procedures and strategies
Outcome performance
The learner’s performance & behaviour on tests that
measure the amount of retention or the ability to
transfer knowledge to new learning tasks
Learner centered approach
Learner characteristics
Learning process and outcomes
Instructional manipulations affects changes in the
learner’s knowledge
Learner is an active information processor
Knowledge is constructed by learner
▪ HOW AND WHERE IS KNOWLEDGE IS CONSTRUCTED?
LEARNING
PROCESS: organise information, build connection, construct knowledge
SELECTING
PATTERN RECOGNITION
ATTENTION AND PERCEPTION
PATTERN RECOGNITION
PRIOR KNOWLEDGE
CONTEXT
LEARNING PROCESS:
organise information, build connection among information and integration with prior knowledge,
and eventually construct knowledge, encode knowledge to LTM
SELECTING
PATTERN RECOGNITION
The assignment of meaning to incoming
stimuli
Is the detection of incoming stimuli by your senses
Is the process by which stimuli are perceived, recognised
and understood
Detection of a stimuli through senses
Storage of some representation of the stimuli
in memory system
Pattern recognition
Assignment of meaning to stimuli
Diameter 10
cm
Detection of a stimuli through senses: stimulus
may be seen/heard
Storage of some representation of the stimuli in
memory system: stored these into icon/echo
Pattern recognition: circle/writing/sound
information from LTM used to recognise pattern
Assignment of meaning to stimuli: select
information to assign meaning that is
undertaken in working memory
Nature of stimulus (context of stimulus)
Background of knowledge
Pattern recognition occurs when elements
match!!
LEARNING PROCESS:
organise information, build connection among information and integration with prior knowledge,
and eventually construct knowledge, encode knowledge to LTM
SELECTING
PATTERN RECOGNITION
Allocation of cognitive resources to a task
Critical for learning – to process information
learners have to pay attention
BUT
Human’s have extremely limited processing
capacity!
Generally people cannot attend to more a
few things at once
Under many conditions multi-tasking is not
very effective because attention is divided
too much, leading to poor executions of tasks
(divided attention)
Automation of skills can compesate for
limited attention capacity
The type of TASK influences attention
allocation
Nature of task
Nature of need
Motivations
Attention is allocated differently according to
RESOURCE LIMITED
DATA-LIMITED
performance will
improve if more
attention is shifted
Performance is limited by
the quality of the
*CONCENTRATION
Some tasks are so
complex that some
individuals can never
apply enough resources
to them because of lack
of knowledge
Students are easily distracted
Teacher is giving important explanations,
students mind starts to ‘wander’ –tuning in to
other conversations (sounds), looking out the
window (visuals), thinking of other matters
(internal cognition)….etc
Concentration is dependant on attention
Prior knowledge used to make decisions about
the meaning of the stimuli
When stimuli in the environment are recognised
as something stored in memory
Two systems for recognising paterns:
Parts to whole (Example?)
Whole to parts (Example?)
Theory of pattern recognition (??!!??)
Gestalt theory (PLEASE SEARCH)
Directly affects perception process
Allows perception occurs
Guides perception of new information
LEARNING PROCESS:
organise information, build connection among information and integration with prior knowledge,
and eventually construct knowledge, encode knowledge to LTM
SELECTING
PATTERN RECOGNITION
What is in your mind now?
Repeat a telephone number
Repeat an unfamiliar foreign word
12 + 13 = ?
314343543 + 89786592 = ?
What have you been doing just before this?
Close your eyes and pick up an object in front
of you.
How many windows in your house?
Limited in capacity
Miller’s research: the magic number of seven (7±2 chunks
of new meaningful information)
Cowan’s research: 4±1 chunks of new information to be
processed
Limited in duration
Recalls decay over time unless actively rehearsal occurs
Information lost very rapidly when people are distracted
from rehearsing
Forgetting occurs due to interference (of new
information) rather than time
Keep it simple, do not overload working
memory
Draw attention to most important points
Present information using different
modalities (e.g., visual and auditory)
Processing “unlimited” amount of
information that is already familiar.
Give example!
LEARNING PROCESS:
organise information, build connection among information and integration with prior knowledge,
and eventually construct knowledge, encode knowledge to LTM
SELECTING
PATTERN RECOGNITION
Unconcious component of our memory
Unlimited capacity and duration
Where cognitive structures are organised
Explicit memory
Conscious recall, recognition of previous
knowledge/information/experience
Implicit memory
No record of previously remembering events
Retention without remembering
LTM is actively constructed using schemata
Activated schemata determine what
incoming information is relevant
Schemata are continually reconstructed
through learning
ASSIMILATION
New information that fits into an existing schema
ACCOMODATION
Existing schemata are modified in the face of new,
confliting information
Schema construction and automation
Transfer knowledge to new/advanced
problem solving
Occurs without intension and conciousness
Less cognitive effort
Less error
Performance is quicker
Automated performance
Develop learning – to more difficult task
Skilled learners
The amount of schema constructed in LTM
The amount of automated schemas
Goal free problems discourage students from
creating sub-goals and separate the problem
state and the problem goal because the
problem goal is not given.
Instead, students are required to work forward
from given information in order to assist
schema construction.
Can you please give an example of goal-free
problems?
Anybody knows what it is?
Learning from Examples
Examples:
(1) x2 + 5x + 6 = (x + 2)(x + 3)
(2) x2 + 7x + 6 = (x + 1)(x + 6)
(3) x2 + 8x + 12 = (x + 2)(x + 6)
(4) x2 + 7x + 12 = (x + 3)(x + 4)
(5) x2 + 13x + 12 = (x + 1)(x + 12)
Exercises:
(1) x2 + 11x + 18 = (
(2) x2 + 9x + 18 = (
(3) x2 + 19x + 18 = (
)(
)(
)(
)
)
)
Learning by Doing
Problems: Identical with examples for learningfrom-examples section except that righthand sides of equations are shown as:
= (x + __ ) (x + __ )
Exercises: Identical with exercises for learningfrom-example section.
Expertise reversal effect?
WE only
Pairing WE and similar PS
Self-explanation effect
Split attention effect
Redundancy effect
Modality effect
Expertise reversal effect
Process oriented or product oriented
Mutiple or uniform solutions
Individual or group learning
A car moving from rest reaches a speed of 20 m/s
after 10 seconds. What is the acceleration of the
car?
u = 0 m/s
v = 20 m/s
t = 10 s
v = u + at
a = (v – u)/t
a = (20 – 0)/10
a = 2 m/s2
A car moving from rest (u) reaches a speed of 20 m/s (v)
after 10 seconds (t): [v = u + at, a = (v – u)/t = (20 – 0)/10
= 2 m/s2]. What is the acceleration of the car?
Practive improving problem solving based
instruction
```
23 cards
81 cards
15 cards
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28 cards | 3,380 | 12,709 | {"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} | 4.28125 | 4 | CC-MAIN-2019-04 | latest | en | 0.923908 |
http://forums.wolfram.com/mathgroup/archive/2011/Jul/msg00331.html | 1,721,354,550,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514866.33/warc/CC-MAIN-20240719012903-20240719042903-00066.warc.gz | 14,714,012 | 8,404 | Re: Keeping it real
• To: mathgroup at smc.vnet.net
• Subject: [mg120287] Re: Keeping it real
• From: Heike Gramberg <heike.gramberg at gmail.com>
• Date: Sat, 16 Jul 2011 05:43:48 -0400 (EDT)
• References: <201107150120.VAA23674@smc.vnet.net>
```If you use Refine with appropriate Assumptions, you get a slightly better result:
Block[{a, c0, d0},
Block[{\$Assumptions = 0 < a < d0 && 0 < a < d0 + c0 z, pos},
int = Expand@
Refine[2*a*
Integrate[((-d0/z^2)/(c0 + d0/z))/
Sqrt[-a^2 + z^2 (c0 + d0/z)^2], z]]
]]
output:
-2 I Log[-((2 d0 (-I a + Sqrt[-a^2 + (d0 + c0 z)^2]))/(
d0 + c0 z))] + (
2 a Log[(2 d0 ((-a^2 + d0 (d0 + c0 z))/Sqrt[-a^2 + d0^2] +
Sqrt[-a^2 + (d0 + c0 z)^2]))/z])/Sqrt[-a^2 + d0^2]
Note that the second Log-term is now real (provided the parameters satisfy the assumptions made). The first
Log-term can be simplified a bit further by using ComplexExpand. Since I want to leave the second Log as it
is, I just extract the first Log and apply ComplexExpand:
result = Block[{a, c0, d0, int},
Block[{\$Assumptions = 0 < a < d0 && 0 < a < d0 + c0 z, pos},
int = Expand@
Refine[2*a*
Integrate[((-d0/z^2)/(c0 + d0/z))/
Sqrt[-a^2 + z^2 (c0 + d0/z)^2], z]];
pos = Position[int, -2 I Log[___]][[1]];
ReplacePart[int,
pos -> Expand@Simplify@ComplexExpand[Extract[int, pos]]]]
]
Output:
2 Arg[I a - Sqrt[-a^2 + (d0 + c0 z)^2]] - 2 I Log[2 d0] + (
2 a Log[(2 d0 ((-a^2 + d0 (d0 + c0 z))/Sqrt[-a^2 + d0^2] +
Sqrt[-a^2 + (d0 + c0 z)^2]))/z])/Sqrt[-a^2 + d0^2]
Which is real except for the constant term -2 I Log[2 d0]. Since an indefinite integral is unique up
to an arbitrary constant, you can add 2 I Log[2 d0] to get a real solution.
Heike.
On 15 Jul 2011, at 02:20, amannuc wrote:
> I am facing perhaps the "age-old" question of limiting the indefinite
> integrals that Mathematica returns. In particular, I am trying to
> avoid complex numbers. I have tried using "Assumptions" in the Integrate
> command to no avail.
>
> The specific integral is:
>
> SetAttributes[{c0, d0}, {Constant}] (* Probably redundant. See Block
> function *)
> Simplify[Block[{a, c0, d0},
> 2.0 * a *
> Integrate[
> ((-d0/z^2)/(c0 + d0/z))/Sqrt[-a^2 + z^2 (c0 + d0/z)^2],
> z,
> Assumptions -> {z \[Element] Reals, a \[Element] Reals,
> c0 \[Element] Reals, d0 \[Element] Reals}]]]
>
>
>
> numbers. I verified the answer is correct by taking its derivative.
> (0. - 2. I) Log[-((2 d0 (-I a + Sqrt[-a^2 + (d0 + c0 z)^2]))/(
> d0 + c0 z))] + ((0. + 2. I) a Log[(
> 2 d0 ((I (-a^2 + d0 (d0 + c0 z)))/Sqrt[a^2 - d0^2] +
> Sqrt[-a^2 + (d0 + c0 z)^2]))/z])/Sqrt[a^2 - d0^2]
>
> Note all the "I"s. This is a fairly simple answer returned. I doubt
> that is the only solution. When I evaluate it for cases of interest, I get
> complex numbers, which can't be right in my particular case which is
> based on a physical problem.
>
>
> Thanks for any insights.
>
> --
> Tony Mannucci
>
```
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• Previous by thread: Keeping it real
• Next by thread: Re: Keeping it real | 1,156 | 3,092 | {"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-2024-30 | latest | en | 0.637333 |
https://cyberian.pk/topic/575/mth302-assignment-1-solution-and-discussion/? | 1,675,384,894,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500041.2/warc/CC-MAIN-20230202232251-20230203022251-00544.warc.gz | 207,192,388 | 39,061 | SOLVED MTH302 Assignment 1 Solution and Discussion
• Assignment # 1 MTH302 (Fall 2019)
Total Marks: 20
Due Date: November 20, 2019.
DON’T MISS THESE: Important instructions before attempting the solution of this assignment:
• To solve this assignment, you should have good command over 01-9 lectures.
• Upload assignments properly through LMS, No Assignment will be accepted through email.
• Write your ID on the top of your solution file.
• Don’t use colorful back grounds in your solution files.
• Use Math Type or Equation Editor etc for mathematical symbols if needed.
• You should remember that if we found the solution files of some students are same then we will reward zero marks to all those students.
• Make solution by yourself and protect your work from other students, otherwise you and the student who send same solution file as you will be given zero marks.
• Also remember that you are supposed to submit your assignment in Word format any other like scan images etc will not be accepted and we will give zero marks correspond to these assignments.
Q1: Payments of \$ 670 are being made at the end of each month for 5 years at an interest of 8% compounded monthly. Calculate the Present Value.
Q2 : Suppose Mr. Arslan’s basic salary is Rs 12000. Calculate the allowances (House rent, utility bills, conveyance allowance). Also, find the provident fund and gratuity.
• Sameer took a loan of Rs 10,000 at the interest rate of 12% per annum for 5 years. Show that the compound interest is greater than simple interest on the given amount plz send me solution
• Q2 : Suppose Mr. Arslan’s basic salary is Rs 12000. Calculate the allowances (House rent, utility bills, conveyance allowance). Also, find the provident fund and gratuity.
• @Muhammad-Zubair
Check video post
• Assignment # 1 MTH302 (Fall 2019)
Total Marks: 20
Due Date: November 20, 2019.
DON’T MISS THESE: Important instructions before attempting the solution of this assignment:
• To solve this assignment, you should have good command over 01-9 lectures.
• Upload assignments properly through LMS, No Assignment will be accepted through email.
• Write your ID on the top of your solution file.
• Don’t use colorful back grounds in your solution files.
• Use Math Type or Equation Editor etc for mathematical symbols if needed.
• You should remember that if we found the solution files of some students are same then we will reward zero marks to all those students.
• Make solution by yourself and protect your work from other students, otherwise you and the student who send same solution file as you will be given zero marks.
• Also remember that you are supposed to submit your assignment in Word format any other like scan images etc will not be accepted and we will give zero marks correspond to these assignments.
Q1: Payments of \$ 670 are being made at the end of each month for 5 years at an interest of 8% compounded monthly. Calculate the Present Value.
Q2 : Suppose Mr. Arslan’s basic salary is Rs 12000. Calculate the allowances (House rent, utility bills, conveyance allowance). Also, find the provident fund and gratuity.
• Q2 : Suppose Mr. Arslan’s basic salary is Rs 12000. Calculate the allowances (House rent, utility bills, conveyance allowance). Also, find the provident fund and gratuity.
konsa formula use karen gay samajh ni a rahi sath allowances ki value bhe ni di hai
• video Link
• ``````Basic Salary of an employee is Rs. 18,000 and Allowances are Rs. 9000. According to the company’s policy casual leaves are 12 days per year and earned leaves are 24 days per year and normal working days are 22 per month.
Answer :
10800 according to my calculation ... whats urs ?
``````
• If an employee deposits Rs. 2,000 at the end of each year into his company’s plan which pays 7% interest compounded quarterly, how much will he have in the account at the end of 5 years?
Answer :
Employee will earn 8400 at the end of 5 years.
Waiting comments relating to question …!
i think correct. Do we need to show the calculation as well or simple answer. and answer of q#1 is?
• If an employee deposits Rs. 2,000 at the end of each year into his company’s plan which pays 7% interest compounded quarterly, how much will he have in the account at the end of 5 years?
Answer :
Employee will earn 8400 at the end of 5 years.
Waiting comments relating to question …!
1
2
1
4
1
3
3
1
| | | 1,030 | 4,401 | {"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-2023-06 | latest | en | 0.900336 |
http://experiment-ufa.ru/7/7_as_a_decimal | 1,513,032,265,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948514113.3/warc/CC-MAIN-20171211222541-20171212002541-00693.warc.gz | 97,590,678 | 6,161 | # 7/7 as a decimal
## 7/7 as a decimal - solution and the full explanation with calculations.
If it's not what You are looking for, type in into the box below your number and see the solution.
## What is 7/7 as a decimal?
To write 7/7 as a decimal you have to divide numerator by the denominator of the fraction.
We divide now 7 by 7 what we write down as 7/7 and we get 1
And finally we have:
7/7 as a decimal equals 1
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https://www.tes.com/teaching-resources/hub/secondary/computing/computer-science | 1,532,065,037,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591497.58/warc/CC-MAIN-20180720041611-20180720061611-00299.warc.gz | 986,241,992 | 29,176 | #### Binary Coding Unplugged Task Cards Ascii
These Binary Code Task Cards are perfect to help students learn how to encrypt and decrypt binary messages as well as understand binary code. They are great at engaging your students with problem solving and help support your computer teaching while unplugged/ Included in this resource are 32 Everyday objects to be converted into Binary Template to create your own Color and black and white version Answers have been included
#### Coputational Thinking Decomposition and Abstraction Workbook
A workbook suitable for introducing or consolidating knowledge of computational thinking aspects of decomposition and abstraction. Approximately 1 hour of work and opportunities for extension.
#### Inspiring interest in radio communications
SDRplay provide a “software defined radio” (SDR) which turns a Raspberry Pi or PC into a full communications receiver capable of picking up radio signals across the spectrum from long waves to microwaves. All you need is the SDR and a simple antenna which is often just a length of wire. This simple tutorial can help a class pick up signals for themselves: https://youtu.be/JvP8t8aPmRw and this inspiring video shows how a class can pick up actual pictures from the International Space Station (first the class can practice on other signals and when they are ready, they can time their listening for when the ISS is transmitting pictures and passing overhead. See: https://youtu.be/vRwH0FlH9MA Potentially SDRplay can lend a school an RSP1A, or they can be bought with educational discount by emailing jon.hudson@sdrplay.com (SDRplay co-founder)
#### Producing Robust Programs 3 - Test Data Part 2
A year 10 resource I created as part 3 of 3 lessons on Creating Robust Programs. This lesson continues with the below outcomes and LOs using exam questions and a differentiated-by-ability code sabotage exercise. Outcomes: • Understand the different types of Test Data: Valid / Invalid / Erroneous / Border. • Create a Test Table. • Demonstrate how to fix sabotaged code. Learning Objectives: • Pupils to gain understanding of test data types through creation of a test table. • Pupils to familiarise themselves with code then to fix it when sabotaged. Pupils assessed on exam questioning. PowerPoint, resources and lesson plan template included.
#### Producing Robust Programs 2 - Test Data Part 1
A year 10 resource I created as part 2 of 3 lessons on Creating Robust Programs. This lesson introduces the outcomes and LOs below using exam-style questions and Quality Assurance based industry experience. Outcomes: • Understand why we Test. • Describe the types of testing: Iterative and Terminal. • Describe the types of errors: Syntax, Logic and Runtime. Learning Objectives: • Pupils to undertake QA testing exercise of a real game, Mac tech permitting. • Pupils to refactor badly factored code. • Pupils to explain reasons for, iterative and terminal testing in their books. • Pupils to answer exam questions on error types (Syntax and logic). PowerPoint, resources and lesson plan template included.
#### Producing Robust Programs 1 - Defensive Design
A year 10 resource I created as part 1 of 3 lessons on Creating Robust Programs. This lesson introduces the following concepts: Anticipating Misuse; Planning for Contingencies; Validation of Input; Authentication using scaffolded exercises, extension and AFL. Outcomes: • Describe defensive design considerations: o Anticipating Misuse o Planning for Contingencies o Validation of Input o Authentication Learning Objectives: • Pupils to undertake 6x coding exercises regarding validation of input and authentication. • Pupils to complete AFL exercise before and after the lesson, demonstrating understanding. PowerPoint, resources and lesson plan template included.
#### Wired and Wireless Networks for OCR GCSE (9-1) in Computer Science (J276)
This student networks workbook is ideal for students, non-specialist teachers, NQTs any anyone who wants to learn more about wired and wireless networks to gain confidence in the OCR GCSE (9-1) in Computer Science (J276) “1.4 Wired and wireless networks” specification. Can be used in the classroom as a teaching aid, for self-study or as a revision tool. This 29-page PDF which covers the theory and includes ready-to-use tasks and even includes the answers. Table of Contents: What is a network? Types of network Local Area Network (LAN) Wide Area Network (WAN) Factors that affect the performance of networks Bandwidth Number of devices Cable or wireless media Error rate Latency The different roles of computers in a client-server and a peer-to-peer network Client Server Peer-to-Peer The hardware needed to connect stand-alone computers to a LAN Stand-alone computer Network Interface Controller/Card Transmission Media Switch Router Wireless Access Points The internet Domain Name Server Hosting Cloud storage Local storage compared to cloud storage Cloud computing Security and ownership of data and software on the cloud Virtual Networks
#### Cambridge Technicals - IT - Level 02 - Unit 15 - Games Creation - L/615/1436 - Delivery Materials
Cambridge Technicals - Level 02 - Unit 15 - Games Creation - L/615/1436 - Delivery Materials - Any issues downloading, email me at enderoth@hotmail.com or go to www.freeschoolresources.com
#### Exam Worksheets. Computer Science for IGCSE, GCSE and UK National Curriculum.
This Bundle includes the topics as follows: Conversion of Binary to Decimal Conversion of Decimal to Binary Conversion of Decimal to Hexa-decimal Conversion of Binary to Hexa-deimal Conversion of Hexa-decimal to Binary Conversion of Hexa-decimal to Decimal Calculating Memory Sizes for Bit, Nibble, Byte, KiloByte, MegaByte, GigaByte and TeraByte Construction/Drawing Conversion Table of Decimal, Binary, Octal and Hexa-decimal Use of basic 16 Hexa-colour codes and their Hex colour values
#### Exam Worksheet: Computer Science for IGCSE | GCSE (0478 | 2210) Page 6
Computer Science for IGCSE and GCSE. UK National Curriculum is also Supported. Component Codes: 0478 | 2210 Conversion Worksheet Page 6 Use of basic 16 Hexa-color codes. Research and then fill the Hexa-color code, and colour the box with a pencil
#### Exam Worksheet: Computer Science for IGCSE | GCSE (0478 | 2210) Page 5
Computer Science for IGCSE and GCSE. UK National Curriculum is also Supported. Component Codes: 0478 | 2210 Conversion Worksheet Page 5 Constructing the Conversion Table of Decimal, Binary, Octal and Hexa-decimal
#### Exam Worksheet: Computer Science for IGCSE | GCSE (0478 | 2210) Page 4
Computer Science for IGCSE and GCSE. UK National Curriculum is also Supported. Component Codes: 0478 | 2210 Conversion Worksheet Page 4 Calculating Memory Sizes of * Bit * Nibble/Nyble * Byte * KiloByte * MegaByte * GigaByte * TeraByte
#### Exam Worksheet: Computer Science for IGCSE | GCSE (0478 | 2210) Page 3
Computer Science for IGCSE and GCSE. UK National Curriculum is also Supported. Component Codes: 0478 | 2210 Conversion Worksheet Page 3 Conversion of Hexa-decimal to Decimal
#### Exam Worksheet: Computer Science for IGCSE | GCSE (0478 | 2210) Page 2
Computer Science for IGCSE and GCSE. UK National Curriculum is also Supported. Component Codes: 0478 | 2210 Conversion Worksheet Page 2 Conversion of Binary to Hexa-decimal Conversion of Hexa-decimal to Binary
#### Exam Worksheet: Computer Science for IGCSE | GCSE (0478 | 2210) Page 1
Computer Science for IGCSE and GCSE. UK National Curriculum is also Supported. Component Codes: 0478 | 2210 Conversion Worksheet Page 1 Conversion of Binary to Decimal Conversion of Decimal to Binary Conversion of Decimal to Hexa-decimal
#### GCSE OCR WiFi
Quick revision for year 11 on WiFi relevant to the OCR Spec. Good as a yr 9/10 teaching resource associated with LAN and networks lessons.
#### Computer Science GCSE - Logic Gates - Match up the name, symbol, description and truth table
Computer Science GCSE CIE 0478 Unit: 3 Topic: Logic Gates A task to match up the name, symbol, description and truth table for each logic gate. The task produces a single A4 sheet/table which is great for a reference/revision of the topic. Students complete a table with four headings; name, symbol, description and truth table. All the details are provided and can be copied and pasted into the table. Final results can be printed and used for revision or as a quick reference guide during the topic. A homework has been included to reinforce the learning. Please leave some feedback.
#### 1&2 Bit Binary Art
This tool creates 10 by 10 one-bit-per-pixel or two-bits-per-pixel binary images from a binary string of 0 and 1s. The choice of colour palette is left to the user with a default black and white option. Works well with snipping tool to use as part of a computing science digital portfolio. could be used in a wide variety of lessons within the context of data representation.
#### Logic - Exam practice - Computer Science GCSE 0478
A mix of exam questions adapted from past papers. Complete with answers and student worksheets
#### Logic Circuits to Statements - Computer Science GCSE 0478
Logic Circuits to Statements - Computer Science GCSE 0478 Unit 3 – Logic Circuits to Statements A set of exam questions with answers and step by step instructions adapted from past papers Each with solution on presentation and a worksheet for students Ideal for teaching how to answer exam style questions
#### Logic Statements to Circuits - Computer Science GCSE 0478
Logic Statements to Circuits - Computer Science GCSE 0478 Unit 3 – Logic Statements to Circuits A set of exam questions with answers and step by step instructions adapted from past papers Each with solution on presentation and a worksheet for students Ideal for teaching how to answer exam style questions
#### Logic Circuits to Truth Tables - Computer Science GCSE 0478
**Logic Circuits to Truth Tables - Computer Science GCSE 0478 ** Unit 3 – Logic Circuits to Truth Tables A set of exam questions with answers and step by step instructions adapted from past papers Each with solution on presentation and a worksheet for students Ideal for teaching how to answer exam style questions
#### Doing Stuff with Python - a guide for beginners
Delivering subjects like programming to mixed ability groups can be challenging, where ‘fire-fighting’ simple script issues for less able students, eats into the time you have to work with your more able students. This guide is intended to help to address these issues: A ‘self-help’ file for your students, in the form of a standalone web page (no Internet required). Issue to your students, or place it on your VLE. Make it easier for less able students to find the help they need. Example scripts are extensively commented. Examples can be copied and pasted direct from the web page into student’s IDE. Topics covered include: Variables. Data types. Numbers. Booleans. Assignment and Equivalence Operators. Comparison Operators. Strings - using tabs; indexing; slicing; converting, concatenation. Using Modules. Working with dates. Lists - indexing; slicing; change/add/remove items; remove duplicates; count occurrences; extending; clearing; sorting; searching; nested lists; list comprehensions. Sets - creating; check items; add items; remove items. Dictionaries - creating; looping through; searching. Tuples - creating; packing; unpacking; searching. Getting / Validating User Input. Controlling program flow - If statements; for loops; while loops; using range; using continue, augmented assignment. Functions - creating; scope of variables; avoiding duplication of variables. File Input and Output - modes; opening; reading; writing; using the with keyword. Using SQLite and CSV Databases. Troubleshooting - Self Help; Indentation; Tab Stops in output; Indexing and Slicing; Comparing Dates; Operator Precedence; Errors and Exceptions. Miscellaneous Items - validate an email address; checking user-supplied data; working with ini files; specifying Octal / Hexadecimal / Binary digits; converting numbers to Binary; generating random strings and numbers; simple Noughts and Crosses game.
#### App Inventor Alluring Audio - (Lesson 4)
This is the forth lesson in the ‘App Inventor’ scheme of work. This lesson combines practical and theory to teach the “representing sound” concepts covered in the Computer Science 9-1 GCSE specifications, (including; sample rate, sample resolution & calculating audio file size.) All lessons within the scheme are targetted at KS3/KS4. Each slide includes detailed notes to support teacher understanding and have been based on a lesson plan structure. There is no prior knowledge required from the class teacher as the guides provided are step-by-step & all code is catered for in the documentation provided. For this lesson a video tutorial is available to support learners with the design of their app. The lesson includes a minimum 45 minute assessment (electronic & paper based included) with mark scheme both of which are custom made, this will truly test learner’s understanding. Using the ‘App Inventor’ emulator or a mobile device with the 'MIT App Inventor app’ installed will allow learners to practically use the apps they build. The app can be downloaded freely from the ‘Google Play’ store.
#### Try Except ValueError lesson
A one hour validation lesson which engages students and shows them why, and how we use validation. They must be taught integer input first to understand this lesson. There are bronze silver and gold outcomes and resources are on the ppt itself.
#### Small Basic Text Window (Using For and While Loops)
Worksheet for students to assist them in generating Small Basic code to write a times table and list multiples of a number. (Output in the Text Window) This activity will develop understanding of For and While loops (and consoldiate use of a variable e.g. name) Answers for the teacher included. The worksheet is available in** .docx** and **.pdf **formats.
#### Small Basic Graphics Window (FULL LESSON drawing a polygon)
FULL LESSON for teaching students how to use Small Basic Graphics Window. **Learning objectives: • Identify mistakes in Small Basic code • Predict what given code will do • Choose code to complete a given outcome Some students will be able to: Comment their code ** Teacher presentation included for large display. Full lesson plan included. 3 levels of worksheets included for use with differentiated groups in your classroom. PowerPoint with further examples of small basic code for optional demonstration to students/further explanation of lesson outcomes.
#### CS Knowledge Organisers - Software
This is part of my Knowledge Organiser set for Computer Science. This part focuses on Software and key CS concerns. The amount of content that students are required to learn within Computing is vast the following document is all of the knowledge organisers I have developed for my students linked to the Computer Systems paper for computing. This document can be used for any specification to support students/teachers through the computing topics. The topics covered within this document are: System Software Computing Concerns and Impact on society This document can be used in many different ways some examples are below: Student led revision Topic summary Lesson activities/resources Teacher resource Homework Thanks for your time.
#### CS Knowledge Organiser - Networks
This is part of my Computer Systems knowledge organiser set. This set focuses on networking knowledge. The amount of content that students are required to learn within Computing is vast the following document is all of the knowledge organisers I have developed for my students linked to the Computer Systems paper for computing. This document can be used for any specification to support students/teachers through the computing topics. The topics covered within this document are: Networks Systems Security This document can be used in many different ways some examples are below: Student led revision Topic summary Lesson activities/resources Teacher resource Homework Thanks for your time.
#### CS Knowledge Organisers - Architecture
This is part of my Knowledge Organiser set. The amount of content that students are required to learn within Computing is vast the following document is all of the knowledge organisers I have developed for my students linked to the Computer Systems paper for computing. This document can be used for any specification to support students/teachers through the computing topics. The topics covered within this document are: System Architecture Memory Storage This document can be used in many different ways some examples are below: Student led revision Topic summary Lesson activities/resources Teacher resource Homework Thanks for your time.
#### Lesson: computer networks
The lesson on computer networks is appropriate for KS3 and GCSE standard. Lesson plan outlining the structure of the lesson. Starter asking pupils to build a computer network using network components (answers included). Presentation on the advantages / disadvantages of computer networks. Reference to an external video on computer networks. Reference to an external computer network simulation Worksheet requiring pupils to give a definition, example and a diagram of PAN, LAN, WAN computer networks. Presentation on PAN, LAN, Wan describing each kind of network. Stretch & challenge activity, research into WIFI networks sheet Plenary true or false presentation Homework to make a video explaining the different types of computer network
#### 1.1 System Arhitecture Worksheet
A worksheet focusing on System Architecture . Topics covered include; the purpose of the CPU, Von Neumann Architecture, CPU components, CPU functions, CPU characteristics and embedded systems.
#### Python Cheat Sheets: Variables, Lists, Selection, Iteration
This is a group of ‘cheat sheets’ for students to use when programming. The sheets include: an explanation of the topic, the correct syntax and common errors. The sheets have common misconceptions on them. This allows students to refer to these when programming to increase their independence. I have used them as a laminated class set that I can hand to students when they ask me about an error.
#### Computational Thinking - Decomposition and Abstraction
A whole lesson with information slides and activities to introduce the computing concepts of decomposition and abstraction. Includes an extended writing task focusing on A.I with accompanying informational video links.
#### Summer 2018 NEW Computer Science dingbats - end of term educational game
As the last dingbats last year was hugely popular, I have created another for this summer. This could be used as an end of term educational lesson, or chopped up and put on your lesson as an engaging activity when they enter the room. There are 18 different dingbats, and the answers are included after each one. Enjoy!
#### Flowcharts and Computational Thinking
An ideal lesson to continue learning of computational thinking and flowcharts. Encourages correct use of symbols and application of knowledge with open ended tasks as well as worksheet which can be used as an assessment material. Requires previous knowledge of flowcharting including correct symbols.
#### App Inventor FortuneT --- (Lesson 3)
This is the third lesson in the ‘App Inventor’ scheme of work. This lesson introduces programming concepts including; Variables, data structures & the randomise method in a completely interactive manner. All lessons within the scheme are targetted at KS3/KS4 and appropriately mapped to elements of the Computer Science 9-1 GCSE. Each slide includes detailed notes to support teacher understanding and have been based on a lesson plan structure. There is no prior knowledge required from the class teacher as the guides provided are step-by-step & all code is catered for in the documentation provided. Using the ‘App Inventor’ emulator or a mobile device with the 'MIT App Inventor app’ installed will allow learners to practically use the apps they build. The app can be downloaded freely from the ‘Google Play’ store.
#### Presentation: how a selection sort works
Presentation on how a selection sort works. It includes: A presentation explaining the principle of operation behind a selection sort A presentation introducing the code written in python to perform a selection sort The selection sort python code in a separate text file.
#### Greenfoot crab tutorial
This resource takes your students through the crab tutorial step by step and also gives them some challenges along the way. A perfect resource to introduce Greenfoot with. I used this with my students for the WJEC/Eduqas GCSE specification | 4,254 | 20,826 | {"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-2018-30 | longest | en | 0.890327 |
https://www.jiskha.com/display.cgi?id=1189951090 | 1,516,717,733,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891976.74/warc/CC-MAIN-20180123131643-20180123151643-00714.warc.gz | 904,780,517 | 3,773 | # chemistry
posted by .
how many mL of a .25 M HCL would be required to neutralize completely 30mL of a .2 M Ca(OH)2
can anyone give a more detailed explanation?
• chemistry -
No. Do you have any questions over what I gave you? It was the most standard titration equation.
You can do it also by writing the balanced equation, and looking for it by using mole relationships.
I will be happy to critique your work.
## Similar Questions
1. ### chemistry
How many ml of a .25M HCL would be required to neutralize completely 2.5g of NaOH
2. ### chemistry
how many mL of a .25 M HCL would be required to neutralize completely 30mL of a .2 M Ca(OH)2
3. ### Chemistry
What volume of 18.00 M HCl is required to completely neutralize 750.0 kg of barium hydroxide at 25.0 degrees C and 760 mm Hg?
4. ### chemistry
if 22.5 mL of 0.383 M HCl are required to completely neutralize 20.0 mL of KOH, what is the molarity of the KOH solution?
5. ### chemistry
25 ml HCl x 1L/1000mL x .200HCl/1L = .005 mol HCl. Using the balanced equation from this, how many moles of NaOh will you need to completely neutralize the HCl?
6. ### chemistry
If 45 mL of .25 M HCl is required to completely neutralize 25.o mL of NH3, what is the concentration of the NH3 solution?
7. ### Chemistry
How much 1.0 M NaOH would be required to completely neutralize 40.0 mL of 0.60 M HCl?
8. ### Chemistry
I don't know how to do Titration problems! I have an emergency homework assignment that's crucial to my grade! Please help with tips and step by step response! Please help!- 1. 50ml of 0.3 M KOH are required to titrate 60 ml of H^2SO^4. …
9. ### chemistry
I need help. please show step by step? How many milliliters of 1.0 M NaOH would be required to completely neutralize 40.0 mL of 0.60 M HCl?
10. ### Chemistry
In a given experiment 36.44 mL of NaOH were required for neutralize a solution of HCl 0.35 M. Calculate the moles of NaOH present to neutralize the HCl in the titration.
More Similar Questions | 574 | 1,988 | {"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-2018-05 | latest | en | 0.928294 |
https://figuraleffect.wordpress.com/2010/04/06/a-ramble-onaround-what-low-correlations-mean/ | 1,500,619,867,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423723.8/warc/CC-MAIN-20170721062230-20170721082230-00535.warc.gz | 646,971,336 | 35,888 | # A ramble on/around what low correlations mean
With a large enough sample, all correlations are statistically “significant”. When is a correlation too low to justify interpretation?
There has to be a causal mechanism generating correlations. This goes deeper than tests and test scores. If a correlation between test scores is very low, then perhaps the correlation between tests doesn’t represent a direct causal relationship, or as direct a casual relationship as you hoped exists.
The big examples which come up repeatedly in psychology are the general factor in intelligence, g (which in itself is not a causal construct but can be explained by, e.g., the P-FIT model), and various constructs of working memory. So the basic idea would be that low correlations between two (superficially) non-WM or non-g tests could still be due to WM or g.
Then there is also the problem that your tests could just be very noisy measures of the real constructs. This need not imply the tests should be rejected. I still think self-report questionnaires are important for connecting very rich experience outside the lab to, e.g., cognitive performance in the lab, even though the correlations involving self-report tend to be low.
So at this point, I’d want to talk about substantive theory.
I cringed a bit when I wrote about test scores measuring constructs. WM and friends are more than just variables measuring how many items you can remember. But the scores, and how they affected by various manipulations (e.g., lure trials on n-back tasks or how large the n is, phonological similarity, etc) allow you to infer properties of the underlying constructs.
One methodological technique researchers often use to measure constructs (or hopefully properties thereof) is structural equation modeling (SEM). Latent variables model the shared variance in manifest variables, e.g., individual item responses. This way you can spot which are the good and poorer items. One problem with latent variables, I think (and I would welcome a reference on this topic), is that it can be difficult to know what exactly the shared variance is. Back to g, Spearman noticed in the early 1900s that lots of tests of cognitive ability correlate, and modelled this using a latent variable. Researchers are still trying to work out what exactly this shared variance is. To get a taste of this, here’s a recentish quotation from the literature (van der Mass, et al, 2006):
“An assumption that is often made is that the g factor represents an underlying quantitative variable. Indeed, many attempts have been made to actually identify this factor with measurable variables (e.g., speed of nerve conductance, reaction time, glucose metabolism in the brain). These studies have produced interesting correlations but have not revealed the single underlying cause of the g factor.”
Latent variables aren’t a magical solution either.
Reference
van der Mass, H. L. J., Dolan, C. V., Grasman, R. P. P. P., Wicherts, J. M., Huizenga, H. M. & Raimakers, M. E. J. (2006). A dynamical model of general intelligence: the positive manifold of intelligence by mutualism Psychological Review, 113, 842–861.
Advertisements | 666 | 3,186 | {"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-2017-30 | longest | en | 0.951461 |
https://newpathworksheets.com/math/grade-1/months-of-the-year/texas-teks-standards | 1,620,674,026,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991759.1/warc/CC-MAIN-20210510174005-20210510204005-00407.warc.gz | 447,589,431 | 7,944 | ## ◂Math Worksheets and Study Guides First Grade. Months of the Year
### The resources above correspond to the standards listed below:
#### Texas Essential Knowledge and Skills (TEKS)
1.3. Number and operations. The student applies mathematical process standards to develop and use strategies for whole number addition and subtraction computations in order to solve problems. The student is expected to:
1.3 (B) Use objects and pictorial models to solve word problems involving joining, separating, and comparing sets within 20 and unknowns as any one of the terms in the problem such as 2 + 4 = [ ]; 3 + [ ] = 7; and 5 = [ ] - 3.
1.3 (D) Apply basic fact strategies to add and subtract within 20, including making 10 and decomposing a number leading to a 10.
1.3 (E) Explain strategies used to solve addition and subtraction problems up to 20 using spoken words, objects, pictorial models, and number sentences.
1.5. Algebraic reasoning. The student applies mathematical process standards to identify and apply number patterns within properties of numbers and operations in order to describe relationships. The student is expected to:
1.5 (D) Represent word problems involving addition and subtraction of whole numbers up to 20 using concrete and pictorial models and number sentences.
1.5 (G) Apply properties of operations to add and subtract two or three numbers.
1.7. Geometry and measurement. The student applies mathematical process standards to select and use units to describe length and time. The student is expected to:
1.7 (A) Use measuring tools to measure the length of objects to reinforce the continuous nature of linear measurement.
1.7 (D) Describe a length to the nearest whole unit using a number and a unit.
1.7 (E) Tell time to the hour and half hour using analog and digital clocks.
Standards | 395 | 1,818 | {"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-2021-21 | latest | en | 0.911733 |
https://www.assignmentpass.info/assignment-dai-xie-dui-ying-shu-zhi-fen-xi/ | 1,679,501,557,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943845.78/warc/CC-MAIN-20230322145537-20230322175537-00724.warc.gz | 744,566,969 | 48,242 | assignment代写-对应数值分析
07/12/2019
The values rij as defined in the above matrix structure are the representations of the incident polarized electric field and the reflected polarized electric field. i stands for incident and j stands for polarization. For the above equation, where rij=rss or rps, it could be said that p standards for p polarization, and s stands for s polarization. Here the Moke effect is seen to arise because of how there are non-vanishing and off diagonal reflectivity occurrences because of the rps and the rsp configurations. in the case of longitudinal Kerr geometry as was observed by researchers in the context of the incident s-polarized light beam, a Kerr rotation values was identified. The Kerr rotation value is expressed in term of reflection coefficients.
The below is the representation of the coordinate system. The coordinate system as observed here is basically a system incorporating the nonmagnetic medium from which the beam of light is incident and also the magnetic medium to which the beam of light travels. The refractive indices of the two mediums are presented as n1 an n2 respectively. Direction of magnetization is arbitrary and is entered into from different directions.
In the calculation of the alpha values, n1 value and n2 the following aspects have to be included which are the angle of incidence of the beam being transmitted from the nonmagnetic to the magnetic medium. The refractive index of the nonmagnetic medium is also required for calculations at this point. With these, it is necessary to calculate the complex refraction angle. For this, Snells law is made use of. Refracted amplitudes are calculated too based on the values and it is established that these values are usually circular polarized. The circular polarization is then assed for purpose of understanding Kerr geometrical values of mx and my. For longitudinal kerr geometry, these values are as follows. mx=1, my value is equal to value of mz which is zero. Now applying Snell’s law for purpose of understanding the Kerr rotation value of theta. | 421 | 2,072 | {"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} | 2.9375 | 3 | CC-MAIN-2023-14 | latest | en | 0.940185 |
https://sciforums.com/threads/displaying-equations-using-tex.61223/page-6 | 1,721,739,423,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518029.87/warc/CC-MAIN-20240723102757-20240723132757-00899.warc.gz | 446,989,449 | 17,935 | # Displaying equations using Tex
Hm. (Posting in bold are we?). Texmaker doesn't seem to want to play. Going back I see the Schneibster has snuck a link into 'This' and 'This' isn't Texmaker. That ('This') looks minimalist. I'll have a play with it. Thanks.
$$e^{i\theta}=\cos(\theta)+i\sin(\theta)$$
Yaay. Mighty aches from tiny toecorns grow.
Sure. If you learn the basics of LaTeX you'll find that this minimal solution will give you pretty much all you can handle (but maybe not those unbelievable things rpenner has been posting; I'm still pretty impressed with that stuff).
Test
Standard wavelength $${\lambda} = \frac{h}{m c}$$
Reduced wavelength $$\frac{\lambda}{2 \pi} = \frac{\hbar}{m c}$$
test
1) $$\Omega _M = - 0.38( \pm 0.22)$$
2) $$\Omega _R h^2 \approx 4.16 \times 10^{ - 5}$$,
3) $$\Omega _\Lambda = 0$$
4) \frac{-b\pm\sqrt{b^2-4ac}}{2a}
5)
$${U_{gs}} = | - \frac{3}{5}\frac{{G{M^2}}}{{{R_{gs}}}}| = M{c^2}$$
{U_{gs}} = | - \frac{3}{5}\frac{{G{M^2}}}{{{R_{gs}}}}| = M{c^2}
Last edited:
I was told that my site's \frac was broken not too long ago. \lfrac works still, but I haven't had the need or the energy to track down every package and all that...
$$d{s^2} = - (1 - \frac{{2GM}}{{{c^2}r}}){c^2}d{t^2} + \frac{1}{{(1 - \frac{{2GM}}{{{c^2}r}})}}d{r^2} + {r^2}d{\theta ^2} + {r^2}{\sin ^2}\theta d{\phi ^2}$$
Last edited:
can someone please troubleshoot this and see why the latex commands are not working?
I have tested it with several tex editors and it displays fine on them, but not here. Thanks
$$\frac { 1 }{ \sqrt { { \left[ 1-\frac { { \omega }^{ 2 } }{ { \omega }_{ c }^{ 2 } } \right] }^{ 2 }+\quad { \eta }^{ 2 } } }$$
The plot thickens! Sometimes it shows up as a proper equation, and sometimes as just the tex commands. wtf?
can someone please troubleshoot this and see why the latex commands are not working?
I have tested it with several tex editors and it displays fine on them, but not here. Thanks
$$\frac { 1 }{ \sqrt { { \left[ 1-\frac { { \omega }^{ 2 } }{ { \omega }_{ c }^{ 2 } } \right] }^{ 2 }+\quad { \eta }^{ 2 } } }$$
The plot thickens! Sometimes it shows up as a proper equation, and sometimes as just the tex commands. wtf?
The current javascript/CSS-based renderer only runs at the time of page loads. Since the "submit new/edited post and redisplay page" logic of this side does so without a full load of the page, updates don't get the LaTeX-to-HTML Display treatment until you refresh the page.
The current javascript/CSS-based renderer only runs at the time of page loads. Since the "submit new/edited post and redisplay page" logic of this side does so without a full load of the page, updates don't get the LaTeX-to-HTML Display treatment until you refresh the page.
Ah, I think that was what I was seeing; the equation stays in the tex command format until the page is refreshed. But I never see it displayed as an equation in the post preview, only after posting. Is that normal?
My iPad doesn't show even a single correctly formatted equation here, even though my Mac does, which is why I don't bother.
[ tex ]
\alpha
[/tex ]
$$\sin \alpha$$
<tex> a^n </tex>
$\sin \alpha$
[ tex ]
\alpha
[/tex ]
sin⁡α'>sinα sinα \sin \alpha
< tex> a^n </tex>
$$a^n$$
I want to go inline with $$E=mc^2$$
$${U_{gs}} = - \frac{3}{5}\frac{{G{M^2}}}{R}$$
$${U_{gs}} = - \frac{3}{5}\frac{{G{M^2}}}{R}$$
I have tried to post this
<img src="/cgi-bin/mathtex.cgi?\displaystyle\sum_{n=1}^4 \left( \sum_{m=1}^n \left( \frac{a \cos\left( {3^{( m+(n^2-n)/2)}} \theta / b \right) }{3^{(m+(n^2-n)/2)}} \right) \right)">
$$\sum_{n=1}^40 \left( \sum_{m=1}^n \left( \frac {a \sin \left( {3^{( m+(n^2-n)/2)}} \theta / b + l(\frac {2 \pi}{3}) \right) }{3^{(m+(n^2-n)/2)}} \right) \right)$$
$$\displaystyle\sum_{n=1}^4 \left( \sum_{m=1}^n \left( \frac{a \cos\left( {3^{( m+(n^2-n)/2)}} \theta / b \right) }{3^{(m+(n^2-n)/2)}} \right) \right)$$
$$\displaystyle\sum_{n=1}^{40} \left( \sum_{m=1}^n \left( \frac{a \sin\left( {3^{( m+(n^2-n)/2)}} \theta / b + l \times (\frac{2 \pi}{3}) \right) }{3^{(m+(n^2-n)/2)}} \right) \right), \ l=-1,0,1$$
$$\displaystyle \sum_{n=1}^8 \left( \sum_{m=1}^n \left( \frac{a \left( \sin \left( {3^{( m+(n^2-n)/2)}} \theta / b + k ( \frac{ \pi}{3} ) + \cos \left( {3^{( m+(n^2-n)/2)}} \theta / b + k( \frac{ \pi}{3} ) \right) \right) }{2 * 3^{(m+(n^2-n)/2)}} \right) \right), \ k=-1,1$$
where a and b are constants and $$- \frac{\pi b}{3^{(n^2 - n)/2}} \leq \theta \leq \frac{\pi b}{3^{(n^2 – n)/2}}$$
Last edited:
$$\sum_{n=1}^4 \left( \sum_{m=1}^n \left( \frac {a \cos \left( {3^{( m+(n^2-n)/2)}} \theta / b \right) }{3^{(m+(n^2-n)/2)}} \right) \right)$$
$$\sum_{n=1}^40 \left( \sum_{m=1}^n \left( \frac {a \sin \left( {3^{( m+(n^2-n)/2)}} \theta / b + l(\frac {2 \pi}{3}) \right) }{3^{(m+(n^2-n)/2)}} \right) \right) , l=-1,0,1$$
$$\sum_{n=1}^4 \left( \sum_{m=1}^n \left( \frac {a \left( \sin \left( {3^{( m+(n^2-n)/2)}} \theta / b + k(\frac { \pi}{3} ) + \cos \left( {3^{( m+(n^2-n)/2)}} \theta / b + k(\frac { \pi}{3} ) \right) \right) }{2 \times 3^{(m+(n^2-n)/2)}} \right) \right) , k=-1,1$$
where a and b are constants and $$- \frac {\pi b}{3^{(n^2 - n)/2}} \leq \theta \leq \frac {\pi b}{3^{(n^2 – n)/2}}$$ | 2,014 | 5,224 | {"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} | 3.453125 | 3 | CC-MAIN-2024-30 | latest | en | 0.873583 |
https://socratic.org/questions/how-do-you-evaluate-a-2-2-b-2-a-2-a-for-a-3-and-b-5 | 1,576,140,651,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540542644.69/warc/CC-MAIN-20191212074623-20191212102623-00315.warc.gz | 538,529,347 | 5,860 | # How do you evaluate (a^2 - 2(b^2-a)) /(2 +a) for a = 3 and b= 5?
Aug 7, 2016
$= - 7$
#### Explanation:
$\frac{{a}^{2} - 2 \left({b}^{2} - a\right)}{2 + a}$ for $a = 3$ and $b = 5$
$= \frac{{3}^{2} - 2 \left({5}^{2} - 3\right)}{2 + 3}$
$= \frac{9 - 2 \left(25 - 3\right)}{5}$
$= \frac{9 - 2 \left(22\right)}{5}$
$= \frac{9 - 44}{5}$
$= - \frac{35}{5}$
$= - 7$ | 201 | 364 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 10, "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} | 4.34375 | 4 | CC-MAIN-2019-51 | longest | en | 0.249107 |
https://plainmath.org/secondary/algebra/algebra-i/inequalities-systems-and-graphs | 1,695,747,603,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510214.81/warc/CC-MAIN-20230926143354-20230926173354-00500.warc.gz | 494,377,972 | 27,128 | # Solving System of Inequalities with Examples
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## y+3x=5; x=1, 0, 2
Sydney Wilson2022-09-23 | 705 | 1,843 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 20, "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} | 3.59375 | 4 | CC-MAIN-2023-40 | longest | en | 0.696949 |
http://stackoverflow.com/questions/11844405/list-modification-using-dictionarys-python?answertab=oldest | 1,406,704,025,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1406510268734.38/warc/CC-MAIN-20140728011748-00085-ip-10-146-231-18.ec2.internal.warc.gz | 286,440,988 | 15,312 | # List modification using dictionarys (Python)
I have the following:
list = [['1234-4321-1',[5, 6, -4, 11, 22]], ['8763-1234-1', [43, -5, 0, 0, -1]], ['1234-5376-1', [3, 0, -5, -6, 0]]]
.
reversed__dict = {'8763-1234-1'}
I want to write a loop that iterates over list and when it detects the reversed value held in reversed_dict to rearrange the string permanently (change to '1234-8763-1' i.e. first two segments swapped) and also multiply it's corresponding int values by -1.
I have tried the following:
for i, x in list:
if i in ordered_dict:
p = (x * -1)
return list[p]
But I just keep getting 'return' and syntax errors, also this does not attempt to rearrange the '8763-1234-1'
desired end result is:
>>list
>>[['1234-4321-1',[5, 6, -4, 11, 22]], ['1234-8763-1', [-43, 5, 0, 0, 1]], ['1234-5376-1', [3, 0, -5, -6, 0]]]
help much, much appreciated.
-
for val in lst:
num = val[0]
if num in reversed_set:
parts = num.split('-')
val[0] = '-'.join([parts[1], parts[0], parts[2]])
>> lst
[['1234-4321-1', [5, 6, -4, 11, 22]], ['1234-8763-1', [43, -5, 0, 0, -1]], ['1234-5376-1', [3, 0, -5, -6, 0]]]
By the way, {'8763-1234-1'} is not a dict, it's a set. And it is not a good idea to use "list" as a name of a variable.
Also, what did you try to achieve by multiplying the list by -1?
-
user1532369 wanted the numbers in the list associated with the matching string multiplied by -1. It says in the question. – ygram Aug 7 '12 at 11:25
@ygram thank you :) – user1532369 Aug 7 '12 at 12:40
@User I am well aware list is not a not a variable name, I just used it in this case for comprehension. Can I just add an extra line val[1] = val[1] * -1 inside the if statement? – user1532369 Aug 7 '12 at 12:43
Try something like val[1] = [-v for v in val[1]] inside the if. – ygram Aug 7 '12 at 13:16
@ygram that worked perfect, thank you very much :) – user1532369 Aug 7 '12 at 13:42 | 688 | 1,906 | {"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": 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} | 2.53125 | 3 | CC-MAIN-2014-23 | latest | en | 0.786292 |
https://scicomp.stackexchange.com/questions/37532/reference-for-mass-matrix-assembly | 1,716,138,185,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057819.74/warc/CC-MAIN-20240519162917-20240519192917-00555.warc.gz | 459,449,414 | 41,261 | # Reference for mass matrix assembly
I will now explain what I understand to be the process of a finite element mass matrix assembly. I would like a reference which does something similar, or if I am mistaken about the process please let me know.
Let $$\Omega$$ be a bounded domain and let $$\lbrace \Omega^{\left(e\right)}\rbrace_{e=1}^{E}$$ be disjoint subsets of $$\Omega$$ which comprise a triangulation of $$\Omega$$. Let $$\mathcal{V}$$ be a finite element space over $$\Omega$$, with basis elements $$\lbrace \phi_{i}\rbrace_{i=1}^{n}$$. Let $$M\in\mathbb{R}^{n\times n}$$ be the mass-matrix
\begin{align} M_{i,j}=\sum_{e=1}^{E}\int_{\Omega^{e}}\phi_{i}\left(x\right)\phi_{j}\left(x\right)\mathrm{d}x. \end{align}
We then map the triangle $$\Omega^{\left(e\right)}$$ onto a fixed reference element $$\Omega^{r}$$ by a change of independent variable, achieved by a linear transformation.
\begin{align} M_{i,j}=\sum_{e=1}^{E}\int_{\Omega^{r}}\tilde{\phi}_{i}\left(x\right)\tilde{\phi}_{j}\left(x\right)|J^{\left(e\right)}|\mathrm{d}x. \end{align}
Because the transformation is linear the determinant of the element Jacobian $$|J^{\left(e\right)}|$$ is constant.
\begin{align} M_{i,j}=\sum_{e=1}^{E}|J^{\left(e\right)}|\int_{\Omega^{r}}\tilde{\phi}_{i}\left(x\right)\tilde{\phi}_{j}\left(x\right)\mathrm{d}x. \end{align}
Each element of the basis for $$\mathcal{V}$$ is assumed to be supported on a few elements $$\Omega^{\left(e\right)}$$, let $$m$$ be the number of basis functions with support on $$\Omega^{\left(e\right)}$$. $$m$$ is assumed to be independent of the element $$\Omega^{\left(e\right)}$$.
\begin{align} M_{i,j}=\sum_{e=1}^{E}|J^{\left(e\right)}|\int_{\Omega^{r}} \sum_{\ell,p=1}^{m} \delta_{i,e_{\ell}}\,\tilde{\phi}_{i}\left(x\right)\tilde{\phi}_{j}\left(x\right) \, \delta_{e_{p},j}\, \mathrm{d}x. \end{align}
where $$\delta$$ is the delta-Kronecker symbol and $$\lbrace e_{p}\rbrace_{p=1}^{m}$$ are the global indicies associated to a finite element function with support on $$\Omega^{\left(e\right)}$$. Let $$\lbrace \psi_{i}\rbrace_{i=1}^{m}\subset \lbrace \tilde{\phi}_{i}\rbrace_{i=1}^{n}$$ be the transformed basis functions that are supported on the reference element, those functions that are supported are element independent.
\begin{align} M_{i,j}=\sum_{e=1}^{E}|J^{\left(e\right)}|\int_{\Omega^{r}} \sum_{\ell,p=1}^{m} \delta_{i,e_{\ell}}\,\psi_{\ell}\left(x\right)\psi_{p}\left(x\right) \, \delta_{e_{p},j}\, \mathrm{d}x. \end{align}
\begin{align} M_{i,j}=\sum_{e=1}^{E}|J^{\left(e\right)}| \sum_{\ell,p=1}^{m} \delta_{i,e_{\ell}} \int_{\Omega^{r}} \psi_{\ell}\left(x\right)\psi_{p}\left(x\right) \mathrm{d}x\,\delta_{e_{p},j}. \end{align}
The only thing left to do is quadrature, so for that purpose let $$\lbrace x_{k}\rbrace_{k=1}^{q}\subset\Omega^{r}$$ be quadrature points and $$\lbrace w_{k}\rbrace_{k=1}^{q}\subset\mathbb{R}$$ be the quadrature weights.
\begin{align} M_{i,j}\approx\sum_{e=1}^{E}|J^{\left(e\right)}| \sum_{\ell,p=1}^{m} \delta_{i,e_{\ell}} \sum_{k=1}^{q} \psi_{\ell}\left(x_{k}\right)\,w_{k}\,\psi_{p}\left(x_{k}\right) \,\delta_{e_{p},j}. \end{align}
Let $$G^{\left(e\right)}\in\mathbb{R}^{m\times n}$$ be the global to local element matrix defined by $$\left(G^{\left(e\right)}\right)_{p,j}=\delta_{e_{p},j}$$, $$B\in\mathbb{R}^{q\times m}$$ be the local dof to basis functoin evaluation at the quadrature points $$B_{k,p}=\psi_{p}\left(x_{k}\right)$$ and $$D^{\left(e\right)}\in\mathbb{R}^{q\times q}$$ be a diagonal matrix which holds the quadrature weights $$D^{\left(e\right)}_{i,j}=|J^{\left(e\right)}|\delta_{i,j}w_{i}$$. With this we have
\begin{align} M\approx\sum_{e=1}^{E}\left(G^{\left(e\right)}\right)^{\top}\,B^{\top}\,D^{\left(e\right)}\,B\,G^{\left(e\right)} \end{align}
Again I am looking for textbooks which also describe this procedure or any feedback if I have any errors thanks!
In your write-up, you loop over all $$i$$ and $$j$$ and then ask which cells you need to consider to compute a particular matrix entry $$M_{ij}$$. This is expensive because most matrix entries are zero, and because you would have to visit the same cell multiple times.
• I see that looping over each $i,j$ can be expensive and one should take advantage of the sparsity to reduce the cost. Jun 2, 2021 at 20:58
• The final expression for the mass matrix seems to be what you are suggesting, that is there is a loop over elements/cells, the action of $G^{\left(e\right)}$ extracts those elements that live on the cell, etc. Jun 2, 2021 at 20:59 | 1,561 | 4,522 | {"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": 40, "wp-katex-eq": 0, "align": 8, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2024-22 | latest | en | 0.652698 |
http://erlang.org/pipermail/erlang-questions/2007-February/025195.html | 1,485,042,748,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281263.12/warc/CC-MAIN-20170116095121-00403-ip-10-171-10-70.ec2.internal.warc.gz | 93,278,628 | 2,361 | # [erlang-questions] Give me some advice
Ulf Wiger <>
Sun Feb 11 10:46:34 CET 2007
```Den 2007-02-11 06:49:52 skrev LUKE <>:
>
> I am tring to do the excise about return a new tuple
> which is a copy of the tuple T where the Nth element
> of the tuple has been replaced by C.
... trying to write an erlang-based version of
the built-in function setelement(N, T, C)?
> Give me some advice on the following poor code. ie.
> tail recursion issue.....
>
> -module(element).
> -export([set2/3]).
>
> set2(N,T,C)->
> list_to_tuple(set2(N,tuple_to_list(T),C,1)).
>
> set2(_,[],_,_)->[];
> set2(N,[H|T],C,Count)->
> if N==Count ->
> [C|set2(N,T,C,Count+1)];
> true->
> [H|set2(N,T,C,Count+1)]
> end.
(1) This is mainly a matter of taste, but
I would order the arguments of set2
differently
(2) When you get to the right position, you
don't have to recurse any further, since
you're guaranteed not to find another match.
set2([_|T], N, N, C) ->
[C|T];
set2([H|T], Pos, N, C) when Pos < N ->
[H|set2(T, Pos+1, N, C)].
(3) You don't need a clause for [], if you add
a guard in set2/3:
set2(N,T,C)
when is_tuple(T),
is_integer(N),
N > 0, N =< size(T) ->
list_to_tuple(set2(tuple_to_list(T),1,N,C)).
(4) It's good form to use guards in the exported
function to clearly indicate what the arguments
slowing down your code. It most likely won't,
and in some cases, it can even speed things up,
since the compiler is able to make some
assumptions about the code that follows.
(5) In this case, I wouldn't worry about the
function not being tail-recursive. Doing
it this way makes it a lot easier to break
out of the loop once you've reached N.
Also, your tuple is not likely to have a
huge number of elements to begin with.
BR,
Ulf W
--
Ulf Wiger
``` | 563 | 1,781 | {"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} | 2.78125 | 3 | CC-MAIN-2017-04 | longest | en | 0.88869 |
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Conversiones
Prof. Waldo M\u00b4
arquez Gonz\u00b4
alez
Ejemplo
Para convertir60
\u25e6a radianes se multiplica60 por la fracci\u00b4
on
\u03c0
180, obteniendo1
3\u03c0.
60\u00b7
\u03c0
180=13
\u03c0=
\u03c03= 1, 0472
Se puede dar el resultado como\u03c03 o, bien como 1,0472.
Si el\u00b4
angulo fuera negativo, el signo se agrega al \ufb01nal.
Convertir las medidas de los\u00b4
1)30\u25e6 =
2)60\u25e6 =
3)90\u25e6 =
4)\u221260\u25e6 =
5)45\u25e6 =
6)\u221290\u25e6 =
7)180\u25e6 =
8)270\u25e6 =
9)300\u25e6 =
10)120\u25e6 =
11)\u2212150\u25e6 =
12)210\u25e6 =
13)330\u25e6 =
14)240\u25e6 =
15)100\u25e6 =
16)70\u25e6 =
17)150\u25e6 =
18)\u2212500\u25e6 =
19)3600\u25e6 =
20)720\u25e6 =
21)900\u25e6 =
22)540\u25e6 =
23)\u2212400\u25e6 =
24)450\u25e6 =
25)606\u25e6 =
26)\u2212390\u25e6 =
27)1080\u25e6 =
28)940\u25e6 =
29)600\u25e6 =
30)333\u25e6 =
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## Activity (66)
la verdad que no tienen ejemplos que uno pueda copiar sin complicaciones
que dificil
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/*********** DO NOT ALTER ANYTHING BELOW THIS LINE ! ************/ var s_code=s.t();if(s_code)document.write(s_code)//--> | 597 | 1,470 | {"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} | 3.75 | 4 | CC-MAIN-2015-35 | latest | en | 0.198668 |
https://www.physicsforums.com/threads/gravity-on-the-moon.697332/ | 1,477,144,854,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988718987.23/warc/CC-MAIN-20161020183838-00243-ip-10-171-6-4.ec2.internal.warc.gz | 974,514,087 | 19,800 | # Gravity on the moon
1. Jun 16, 2013
### sirwan
Which of them larger? The gravity of the earth to the moon or the gravity of the sun to the moon?
2. Jun 16, 2013
### HallsofIvy
Staff Emeritus
I assume you mean the force of gravity the earth exerts on the moon and the force of gravity the sun exerts on the moon. You could, of course, actually calculate both forces but it seems to me that the fact that the moon orbits the earth, not the sun, says it all.
3. Jun 16, 2013
### sirwan
of course I mean the gravitational force ,why the moon does not orbit the sun? Is it relate to the difference of amount of gravitational force of each the sun and earth to the moon or related to the different distance of the moon from sun and earth ?
4. Jun 16, 2013
### phinds
Uh ... you seem to think that the distance and force are not related. Really?
5. Jun 16, 2013
### Bandersnatch
But the Moon does orbit the Sun, in very much similar way, if a bit more wobbly, as the Earth does.
6. Jun 16, 2013
### Staff: Mentor
That's a trickier issue than you think. Do the calculation. The gravitational force on the Moon toward the Sun is about twice that toward the Earth.
There are two problems here. One is that you are thinking along the lines of "A orbits B" and "A orbits C" are mutually exclusive. It's not an either-or proposition. The Moon orbits the Earth and it also orbits the Sun.
The other is that gravitational force not being the right metric to decide whether a body is orbiting some other body. A much better metric is energy. Energy answers the question, "is object A gravitationally bound to object B?" Force is pretty much irrelevant.
7. Jun 18, 2013
### manojr
See this wikipedia article on Hill Sphere.
Moon is closer to Earth therefore its gravitational influence on Moon is more than that of Sun.
8. Jun 18, 2013
### Staff: Mentor
That final statement is wrong. Consider an object orbiting the Earth at a fiftieth of an AU. That obviously qualifies as "closer to the Earth". Yet the Sun's gravitational influence is going to dominate over that of the Earth by any reasonable definition of "gravitational influence". This orbit is not stable.
What do you mean by "gravitational influence"? Certainly not force. It's a simple calculation. The gravitational force exerted by the Sun on the Moon is 2.2 times that of the Earth on the Moon. Certainly not potential energy. Now that factor of 2.2 becomes a factor of 850. So what exactly did you mean by "gravitational influence"?
I intentionally did not mention the Hill sphere because it's an advanced topic and because it does not answer the question at hand. The question at hand is "why can we say that the Moon is orbiting the Earth?" The Hill sphere doesn't address this question. What answers the question of "why can we say that the Moon is orbiting the Earth?" is mechanical energy. The Moon's mechanical energy with respect to the Earth is negative: The Moon is gravitationally bound to the Earth.
9. Jun 18, 2013
### manojr
I mean resultant/dominant effect. I agree that this may not be appropriate term in physics.
As you already answered OP's actual question i.e. The gravitational force on the Moon toward the Sun is about twice that toward the Earth, I read the question as "why does moon orbit the Earth instead of Sun (that attracts with twice the force)?"
The article I pointed to begins as "Hill sphere is the region in which it dominates the attraction of satellites. To be retained by a planet, a moon must have an orbit that lies within the planet's Hill sphere." So I thought it is good article to read.
Thank you for explaining factors that determine stable orbit.
10. Jun 19, 2013
### Staff: Mentor
You didn't answer my question. What effect? Don't worry, it's a rhetorical question.
If you read that wikipedia article on the Hill sphere (it's a rather lousy article IMHO), you see that they define "gravitational influence" as meaning "having the same angular velocity", which to me is rather bogus. Even then, they get the Hill sphere wrong by a factor of √3.
The Hill sphere is rather arbitrarily defined as a sphere with origin at the center of the secondary body and with a radius equal to the distance to the L1 point. The wikipedia article doesn't mention that the definition is a bit arbitrary. It does mention that it helps determine if an orbit is stable. Note well: This means that whether an object is orbiting another must be defined by some other metric.
That is exactly how I interpreted the question. The answer is not the Hill sphere. The answer is in terms total mechanical energy, or perhaps even better, escape velocity. Energy is a concept that can be grasped at the high school physics level. If the total mechanical energy is negative the object is "orbiting". At least now it is. Escape velocity is similarly within the grasp of a high school student. An object is currently orbiting some other object if the magnitude of the velocity with respect to that other body is less than escape velocity.
Almost nothing orbits forever in the N-body problem. Orbits are transient. Mercury most likely will be ejected from the solar system in a few billion years due to interactions with Jupiter and Venus. Does the fact that it won't be orbiting several billion years from now mean that it isn't orbiting now? Of course not.
That's wikipedia for ya. Sometimes it's very good, sometimes it's rather lousy. This is one of wikipedia's many articles of lesser quality. That's my opinion, of course.
11. Jun 19, 2013
### manojr
Alright, I did not pay attention in physics class long back :)
I just searched more, and guess what I found? Why does the Moon orbit the Earth rather than the Sun?
12. Jul 17, 2013
### Philosophaie
Earth Gravitational Force greater:
$$\frac{G*M_{EARTH}}{R^2} > \frac{G*M_{SUN}}{R^2}$$ | 1,365 | 5,849 | {"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} | 3.296875 | 3 | CC-MAIN-2016-44 | longest | en | 0.918585 |
https://inquiryintoinquiry.com/ | 1,686,288,218,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224655247.75/warc/CC-MAIN-20230609032325-20230609062325-00492.warc.gz | 348,525,445 | 51,228 | ## Logic Syllabus • Discussion 2
JM:
Is [the “just one true” operator] the same or different to xor? I have read that xor is true when an odd number of variables are true which would make it different. But I also read somewhere that xor was true when only one is true.
Here’s my syllabus entry on Exclusive Disjunction (xor), also known as Logical Inequality, Symmetric Difference, and a few other names. It’s my best effort so far at straightening out the reigning confusions and also at highlighting the links between the various notations and visualizations we find in practice.
Exclusive disjunction, also known as logical inequality or symmetric difference, is an operation on two logical values, typically the values of two propositions, which produces a value of true just in case exactly one of its operands is true.
To say exactly one operand is true is to say the other is false, which is to say the two operands are different, that is, unequal.
Expressed algebraically, $x_1 + x_2 = 1 ~ (\text{mod}~ 2).$
Viewed in that light, it is tempting to think a natural extension of xor to many variables $x_1, \ldots, x_m$ will take the form $x_1 + \ldots + x_m = 1 ~ (\text{mod}~ 2).$ And saying the bit sum of several boolean values is 1 is just another way of saying an odd number of the values are 1.
Sums of that order form a perfectly good family of boolean functions, ones we’ll revisit in a different light, but their kinship to the family of logical disjunctions is a bit more strained than uniquely natural.
## Logic Syllabus • Discussion 1
JM:
In a previous post you mentioned the minimal negation operator. Is there also the converse of this, i.e. an operator which is true when exactly one of its arguments is true? Or is this just xor?
Yes, the “just one true” operator is a very handy tool. We discussed it earlier under the headings of “genus and species relations” or “radio button logic”. Viewed in the form of a venn diagram it describes a partition of the universe of discourse into mutually exclusive and exhaustive regions. Reading $\texttt{(} x_1 \texttt{,} \ldots \texttt{,} x_m \texttt{)}$ to mean just one of $x_1, \ldots, x_m$ is false, the form $\texttt{((} x_1 \texttt{),} \ldots \texttt{,(} x_m \texttt{))}$ means just one of $x_1, \ldots, x_m$ is true.
For two logical variables, though, the cases “condense” or “degenerate” and saying “just one true” is the same thing as saying “just one false”.
$\texttt{((} x_1 \texttt{),(} x_2 \texttt{))} = \texttt{(} x_1 \texttt{,} x_2 \texttt{)} = x_1 + x_2 = \textsc{xor} (x_1, x_2).$
## Inquiry Into Inquiry • On Initiative 5
JS:
That’s not how it works. The model lacks agency. It is a machine whose gears are cranked by the user’s prompt. It can ask questions, but only when prompted to. It is not doing anything at all when it isn’t being prompted.
Sure, I understand that. The hedge “as it were” is used advisedly for the sake of the argument. (I wrote my own language learner back in the 80s.)
Speaking less metaphorically, the program and its database are always in their respective states and the program has the capacity to act on the database even when not engaged with external prompts.
Is there any reason why the program’s “housekeeping” functions should not include one to measure its current state of “uncertainty” (entropy of a distribution) with regard to potential questions — or any reason why it should “hurt to ask”?
As it were …
### Resources
Posted in Anthem, Initiative, Inquiry | Tagged , , | 1 Comment
## Mathematical Demonstration and the Doctrine of Individuals • 2
### Selection from C.S. Peirce’s “Logic Of Relatives” (1870)
In reference to the doctrine of individuals, two distinctions should be borne in mind. The logical atom, or term not capable of logical division, must be one of which every predicate may be universally affirmed or denied. For, let $\mathrm{A}$ be such a term. Then, if it is neither true that all $\mathrm{A}$ is $\mathrm{X}$ nor that no $\mathrm{A}$ is $\mathrm{X},$ it must be true that some $\mathrm{A}$ is $\mathrm{X}$ and some $\mathrm{A}$ is not $\mathrm{X};$ and therefore $\mathrm{A}$ may be divided into $\mathrm{A}$ that is $\mathrm{X}$ and $\mathrm{A}$ that is not $\mathrm{X},$ which is contrary to its nature as a logical atom.
Such a term can be realized neither in thought nor in sense.
Not in sense, because our organs of sense are special — the eye, for example, not immediately informing us of taste, so that an image on the retina is indeterminate in respect to sweetness and non-sweetness. When I see a thing, I do not see that it is not sweet, nor do I see that it is sweet; and therefore what I see is capable of logical division into the sweet and the not sweet. It is customary to assume that visual images are absolutely determinate in respect to color, but even this may be doubted. I know no facts which prove that there is never the least vagueness in the immediate sensation.
In thought, an absolutely determinate term cannot be realized, because, not being given by sense, such a concept would have to be formed by synthesis, and there would be no end to the synthesis because there is no limit to the number of possible predicates.
A logical atom, then, like a point in space, would involve for its precise determination an endless process. We can only say, in a general way, that a term, however determinate, may be made more determinate still, but not that it can be made absolutely determinate. Such a term as “the second Philip of Macedon” is still capable of logical division — into Philip drunk and Philip sober, for example; but we call it individual because that which is denoted by it is in only one place at one time. It is a term not absolutely indivisible, but indivisible as long as we neglect differences of time and the differences which accompany them. Such differences we habitually disregard in the logical division of substances. In the division of relations, etc., we do not, of course, disregard these differences, but we disregard some others. There is nothing to prevent almost any sort of difference from being conventionally neglected in some discourse, and if $I$ be a term which in consequence of such neglect becomes indivisible in that discourse, we have in that discourse,
$[I] = 1.$
This distinction between the absolutely indivisible and that which is one in number from a particular point of view is shadowed forth in the two words individual (τὸ ἄτομον) and singular (τὸ καθ᾿ ἕκαστον); but as those who have used the word individual have not been aware that absolute individuality is merely ideal, it has come to be used in a more general sense. (CP 3.93)
### Note
Peirce explains his use of the square bracket notation at CP 3.65.
I propose to denote the number of a logical term by enclosing the term in square brackets, thus, $[t].$
The number of an absolute term, as in the case of $I,$ is defined as the number of individuals it denotes.
### References
• Peirce, C.S. (1870), “Description of a Notation for the Logic of Relatives, Resulting from an Amplification of the Conceptions of Boole’s Calculus of Logic”, Memoirs of the American Academy of Arts and Sciences 9, 317–378, 26 January 1870. Reprinted, Collected Papers 3.45–149, Chronological Edition 2, 359–429. Online (1) (2) (3).
• Peirce, C.S., Collected Papers of Charles Sanders Peirce, vols. 1–6, Charles Hartshorne and Paul Weiss (eds.), vols. 7–8, Arthur W. Burks (ed.), Harvard University Press, Cambridge, MA, 1931–1935, 1958.
• Peirce, C.S., Writings of Charles S. Peirce : A Chronological Edition, Peirce Edition Project (eds.), Indiana University Press, Bloomington and Indianapolis, IN, 1981–.
## Mathematical Demonstration and the Doctrine of Individuals • 1
### Selection from C.S. Peirce’s “Logic Of Relatives” (1870)
Demonstration of the sort called mathematical is founded on suppositions of particular cases. The geometrician draws a figure; the algebraist assumes a letter to signify a single quantity fulfilling the required conditions. But while the mathematician supposes an individual case, his hypothesis is yet perfectly general, because he considers no characters of the individual case but those which must belong to every such case.
The advantage of his procedure lies in the fact that the logical laws of individual terms are simpler than those which relate to general terms, because individuals are either identical or mutually exclusive, and cannot intersect or be subordinated to one another as classes can.
Mathematical demonstration is not, therefore, more restricted to matters of intuition than any other kind of reasoning. Indeed, logical algebra conclusively proves that mathematics extends over the whole realm of formal logic; and any theory of cognition which cannot be adjusted to this fact must be abandoned. We may reap all the advantages which the mathematician is supposed to derive from intuition by simply making general suppositions of individual cases. (CP 3.92)
### References
• Peirce, C.S. (1870), “Description of a Notation for the Logic of Relatives, Resulting from an Amplification of the Conceptions of Boole’s Calculus of Logic”, Memoirs of the American Academy of Arts and Sciences 9, 317–378, 26 January 1870. Reprinted, Collected Papers 3.45–149, Chronological Edition 2, 359–429. Online (1) (2) (3).
• Peirce, C.S., Collected Papers of Charles Sanders Peirce, vols. 1–6, Charles Hartshorne and Paul Weiss (eds.), vols. 7–8, Arthur W. Burks (ed.), Harvard University Press, Cambridge, MA, 1931–1935, 1958.
• Peirce, C.S., Writings of Charles S. Peirce : A Chronological Edition, Peirce Edition Project (eds.), Indiana University Press, Bloomington and Indianapolis, IN, 1981–.
## Inquiry Into Inquiry • On Initiative 4
I think a lot of people who’ve been working all along on AI, intelligent systems, and computational extensions of human capacities in general are a little distressed to see the field cornered and re‑branded in the short‑sighted, market‑driven way we currently see.
The more fundamental problem I see here is the failure to grasp the nature of the task at hand, and this I attribute not to a program but to its developers.
Journalism, Research, and Scholarship are not matters of generating probable responses to prompts or other stimuli. What matters is producing evidentiary and logical supports for statements. That is the task requirement the developers of recent LLM‑Bots are failing to grasp.
There is nothing new about that failure. There is a long history of attempts to account for intelligence and indeed the workings of scientific inquiry based on the principles of associationism, behaviorism, connectionism, and theories of that order. But the relationship of empirical evidence, logical inference, and scientific information is more complex and intricate than is dreamt of in those reductive philosophies.
### Resources
Posted in Anthem, Initiative, Inquiry | Tagged , , | 2 Comments
## Systems of Interpretation • 3
The “triskelion” figure in the previous post shows the bare essentials of an elementary sign relation or individual triple $(o, s, i).$ There’s a less skeletal figure Susan Awbrey and I used in an earlier paper, where our aim was to articulate the commonalities Peirce’s concept of a sign relation shares with its archetype in Aristotle.
$\text{Figure 1. The Sign Relation in Aristotle}$
Here is the corresponding passage from “On Interpretation”.
Words spoken are symbols or signs (symbola) of affections or impressions (pathemata) of the soul (psyche); written words are the signs of words spoken. As writing, so also is speech not the same for all races of men. But the mental affections themselves, of which these words are primarily signs (semeia), are the same for the whole of mankind, as are also the objects (pragmata) of which those affections are representations or likenesses, images, copies (homoiomata). (De Interp. i. 16a4).
### References
• Awbrey, J.L., and Awbrey, S.M. (1995), “Interpretation as Action : The Risk of Inquiry”, Inquiry : Critical Thinking Across the Disciplines 15(1), 40–52. ArchiveJournal. Online (doc) (pdf).
• Awbrey, S.M., and Awbrey, J.L. (2001), “Conceptual Barriers to Creating Integrative Universities”, Organization : The Interdisciplinary Journal of Organization, Theory, and Society 8(2), Sage Publications, London, UK, 269–284. AbstractOnline.
• Awbrey, S.M., and Awbrey, J.L. (September 1999), “Organizations of Learning or Learning Organizations : The Challenge of Creating Integrative Universities for the Next Century”, Second International Conference of the Journal ‘Organization’, Re‑Organizing Knowledge, Trans‑Forming Institutions : Knowing, Knowledge, and the University in the 21st Century, University of Massachusetts, Amherst, MA. Online.
## Systems of Interpretation • 2
Let’s start as simply as possible. The following Figure is typical of many I have used to illustrate sign relations from the time I first began studying Peirce’s theory of signs.
$\text{Figure 2. An Elementary Sign Relation}$
The above variant comes from a paper Susan Awbrey and I presented at a conference in 1999, a revised version of which was published in 2001.
As the drafter of that drawing I can speak with authority about the artist’s intentions in drawing it and also about the conventions of interpretation forming the matrix of its conception and delivery.
Just by way of refreshing my own memory, here is how we set it up —
Figure 2 represents an “elementary sign relation”. It is a single transaction taking place among three entities, the object $o,$ the sign $s,$ and the interpretant sign $i,$ the association of which is typically represented by means of the ordered triple $(o, s, i).$
One of the interpretive conventions implied in that setup is hallowed by long tradition, going back to the earliest styles of presentation in mathematics. In it one draws a figure intended as “representative” of many figures. Regarded as a concrete drawing the figure is naturally imperfect, individual, peculiar, and special but it’s meant to be taken purely as a representative of its class — generic, ideal, and typical. That is the main convention of interpretation which goes into giving diagrams and figures their significant power.
### References
• Awbrey, S.M., and Awbrey, J.L. (2001), “Conceptual Barriers to Creating Integrative Universities”, Organization : The Interdisciplinary Journal of Organization, Theory, and Society 8(2), Sage Publications, London, UK, 269–284. AbstractOnline.
• Awbrey, S.M., and Awbrey, J.L. (September 1999), “Organizations of Learning or Learning Organizations : The Challenge of Creating Integrative Universities for the Next Century”, Second International Conference of the Journal ‘Organization’, Re‑Organizing Knowledge, Trans‑Forming Institutions : Knowing, Knowledge, and the University in the 21st Century, University of Massachusetts, Amherst, MA. Online.
## Systems of Interpretation • 1
Questions have arisen about the different styles of diagrams and figures used to represent triadic sign relations in Peircean semiotics. What do they mean? Which style is best? Among the most popular pictures some use geometric triangles while others use the three‑pronged graphs Peirce used in his logical graphs to represent triadic relations.
Diagrams and figures, like any signs, can serve to communicate the intended interpretants and thus to coordinate the conduct of interpreters toward the intended objects — but only in communities of interpretation where the conventions of interpretation are understood. Conventions of interpretation are by comparison far more difficult to communicate.
That brings us to the first question we have to ask about the possibility of communication in this area, namely, what conventions of interpretation are needed to make sense of these diagrams, figures, and graphs?
## Inquiry Into Inquiry • On Initiative 3
The more fundamental problem I see here is the failure to grasp the nature of the task at hand, and this I attribute not to a program but to its developers.
Journalism, Research, and Scholarship are not matters of generating probable responses to prompts or other stimuli. What matters is producing evidentiary and logical supports for statements. That is the task requirement the developers of recent LLM‑Bots are failing to grasp.
There is nothing new about that failure. There is a long history of attempts to account for intelligence and indeed the workings of scientific inquiry based on the principles of associationism, behaviorism, connectionism, and theories of that order. But the relationship of empirical evidence, logical inference, and scientific information is more complex and intricate than is dreamt of in those reductive philosophies.
Note. The above comment was originally posted on March 1st but appears to have been deleted accidentally.
### Resources
Posted in Anthem, Initiative, Inquiry | Tagged , , | 1 Comment | 4,029 | 17,082 | {"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": 33, "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} | 3.953125 | 4 | CC-MAIN-2023-23 | latest | en | 0.917601 |
http://www.excelunusual.com/excelunusual-blog-page-where-you-can-download-many-excel-scientific-animations-tutorials-excel-games-engineering-simulations/page/3/ | 1,596,698,233,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439736883.40/warc/CC-MAIN-20200806061804-20200806091804-00383.warc.gz | 140,862,630 | 12,958 | #### An Interactive Gated Ring Oscillator Tutorial – part #1
This is the first part of tutorial explaining how to make an interactive animated Excel model of a gated ring oscillator in MS Excel 2003.
#### A Delay Based Animated Excel Model for Single-Stage Logic Gates – part#2
This is the second half of the delay based Excel model tutorial. In this part the final formulas are derived and implemented in Excel 2003.
#### A Delay Based Animated Excel Model for Single-Stage Logic Gates – part#1
After learning how to model ideal logic gates in MS Excel 2003 it is time to attemp to create a delay based model.
#### Introduction to Digital Electronics – a Simplified Model for Logic Gates – part#3
This is the third part of a tutorial about modeling logic gates in MS Excel 2003. This first model includes no delay or loading for the gates.
#### Introduction to Digital Electronics – a Simplified Model for Logic Gates – part#2
This is the second part of a tutorial targeted at modeling logic gates in MS Excel 2003.
#### Introduction to Digital Electronics – a Simplified Model for Logic Gates – part#1
This is the first part of a tutorial (self, 2, 3) about modeling logic gates in MS Excel 2003. This first model will include no delay or loading for the gates.
#### An Animated Linear Feedback Shift Register (LFSR) as a Pseudo Random Pattern Generator in Excel 2003 – Part#4
This is not a tutorial but a download post for a purely animated Pseudo Random Number Generator model as a Fibonacci type Linear Feedback Shift Register in MS Excel 2003.
#### An Animated Linear Feedback Shift Register (LFSR) as a Pseudo Random Pattern Generator in Excel 2003 – Part#3
This is the third part of a tutorial (1, 2, self) describing the creation of an animated Pseudo Random Number Generator model as a Fibonacci type Linear Feedback Shift Register in MS Excel 2003.
#### An Animated Linear Feedback Shift Register (LFSR) as a Pseudo Random Pattern Generator in Excel 2003 – Part#2
This is the second part of a tutorial (1, self, 3) describing the creation of an animated Pseudo Random Number Generator model as a Fibonacci type Linear Feedback Shift Register in MS Excel 2003. This part creates a very simple table type model based on combinatorial logic rather than a sequential type based on registers.
#### An Animated Linear Feedback Shift Register (LFSR) as a Pseudo Random Pattern Generator in Excel 2003 – Part#1
This is the first part of a tutorial (self, 2, 3) describing the creation of an animated Pseudo Random Number Generator model as a Fibonacci type Linear Feedback Shift Register in MS Excel 2003. | 593 | 2,627 | {"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-2020-34 | latest | en | 0.816747 |
http://www.varsitytutors.com/act_math-help/how-to-subtract-rational-expressions-with-different-denominators | 1,484,727,456,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280242.65/warc/CC-MAIN-20170116095120-00556-ip-10-171-10-70.ec2.internal.warc.gz | 764,643,781 | 25,254 | # ACT Math : How to subtract rational expressions with different denominators
## Example Questions
### Example Question #7 : Expressions
Simplify:
Explanation:
The common denominator of these two fractions simply is the product of the two denominators, namely:
Thus, you will need to multiply each fraction's numerator and denominator by the opposite fraction's denominator:
Let's first simplify the numerator:
, which is the simplest form you will need for this question.
However, the correct answer has the denominator multiplied out. Merely FOIL
### Example Question #1 : How To Subtract Rational Expressions With Different Denominators
Simplify:
Explanation:
Always start the simplification of rational expressions with quadratic denominators by factoring the denominator that has a squared element:
Thus, we need to multiply the numerator and denominator of the first term by :
Distribute:
This is simply:
You cannot factor the top in any way that would help you to cancel anything. Therefore, the answer stands as it does. You need to FOIL the denominator back out, as your correct answer is in that form:
### Example Question #2 : How To Subtract Rational Expressions With Different Denominators
Simplify:
Explanation:
Therefore, the common denominator of our fractions is:
This will require you to multiply the first fraction by and the second by . This gives you:
Now, FOIL the second member and distribute the first:
Next, distribute the subtraction:
Finally, simplify: | 299 | 1,505 | {"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-2017-04 | latest | en | 0.902033 |
http://stackoverflow.com/questions/4582398/writing-a-simple-equation-parser?answertab=active | 1,430,000,308,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1429246651727.46/warc/CC-MAIN-20150417045731-00061-ip-10-235-10-82.ec2.internal.warc.gz | 251,515,619 | 22,129 | # Writing a simple equation parser
What sorts of algorithms would be used to do this (as in, this is a string, and I want to find the answer):
``````((5 + (3 + (7 * 2))) - (8 * 9)) / 72
``````
Say someone wrote that in, how could I deal with so many nested parenthesis?
-
You can use Shunting yard algorithm or Reverse Polish Notation, both of them are using stacks to handle this, wiki said it better than me.
# From wiki,
``````While there are tokens to be read:
If the token is a number, then add it to the output queue.
If the token is a function token, then push it onto the stack.
If the token is a function argument separator (e.g., a comma):
Until the token at the top of the stack is a left parenthesis, pop operators off the stack onto the output queue. If no left parentheses are encountered, either the separator was misplaced or parentheses were mismatched.
If the token is an operator, o1, then:
while there is an operator token, o2, at the top of the stack, and
either o1 is left-associative and its precedence is less than or equal to that of o2,
or o1 is right-associative and its precedence is less than that of o2,
pop o2 off the stack, onto the output queue;
push o1 onto the stack.
If the token is a left parenthesis, then push it onto the stack.
If the token is a right parenthesis:
Until the token at the top of the stack is a left parenthesis, pop operators off the stack onto the output queue.
Pop the left parenthesis from the stack, but not onto the output queue.
If the token at the top of the stack is a function token, pop it onto the output queue.
If the stack runs out without finding a left parenthesis, then there are mismatched parentheses.
When there are no more tokens to read:
While there are still operator tokens in the stack:
If the operator token on the top of the stack is a parenthesis, then there are mismatched parentheses.
Pop the operator onto the output queue.
Exit.
``````
-
James has provided a good answer. Wikipedia has a good article on this as well.
If (and I don't recommend this) you wanted to parse that expression directly, given that it seems orderly in that every set of parens has no more than one pair of operands, I think you could approach it like this:
parse to the first ")". Then parse back to the previous "(". Evaluate what's inside and replace the whole set with a value. Then repeat recursively until you are done.
So in this example, you would first parse "(7 * 2)" and replace it with 14. Then you would get (3 + 14) and replace it with 17. And so on.
You can do that with Regex or even .IndexOf and .Substring.
I'm going without benefit of checking my syntax here, but something like this:
``````int y = string.IndexOf(")");
int x = string.Substring(0,y).LastIndexOf("(");
string z = string.Substring(x+1,y-x-1) // This should result in "7 * 2"
``````
You'll need to evaluate the resulting expression and loop this until the parens are exhausted and then evaluate that last part of the string.
-
So, if you liked James's answer, why didn't you upvote it? ;-) – Chris Jester-Young Jan 3 '11 at 6:28
Because I just started contributing yesterday and don't have the required 15 rep yet or I would have. You could help with that of course :) – nycdan Jan 3 '11 at 17:56
Or you can just do this in one line in R:
``````> eval(parse(text = '((5 + (3 + (7*2))) - (8 * 9))/72' ))
[1] -0.6944444
``````
-
Yes the algorithm is Shunting yard algorithm but if you want to implement I suggest you python and it's compiler package
``````import compiler
equation = "((5 + (3 + (7 * 2))) - (8 * 9)) / 72"
parsed = compiler.parse( equation )
print parsed
``````
You can also evaluate this expression with built-in eval() method
``````print eval("5 + (4./3) * 9") // 17
``````
-
I would use the tools that are available nearly everywhere.
I like lex/yacc because I know them but there are equivalents everywhere. So before you write complex code see if there are tools that can help you to make it simple (problems like this have been solved before so don;t re-invent the wheel).
So, using lex(flex)/yacc(bison) I would do:
### e.l
``````%option noyywrap
Number [0-9]+
WhiteSpace [ \t\v\r]+
NewLine \n
%{
#include <stdio.h>
%}
%%
\( return '(';
\) return ')';
\+ return '+';
\- return '-';
\* return '*';
\/ return '/';
{Number} return 'N';
{NewLine} return '\n';
{WhiteSpace} /* Ignore */
. fprintf(stdout,"Error\n");exit(1);
%%
``````
### e.y
``````%{
#include <stdio.h>
typedef double (*Operator)(double,double);
double mulOp(double l,double r) {return l*r;}
double divOp(double l,double r) {return l/r;}
double addOp(double l,double r) {return l+r;}
double subOp(double l,double r) {return l-r;}
extern char* yytext;
extern void yyerror(char const * msg);
%}
%union
{
Operator op;
double value;
}
%type <value> Expression MultExpr AddExpr BraceExpr
%%
Value: Expression '\n' { fprintf(stdout, "Result: %le\n", \$1);return 0; }
Expression: AddExpr { \$\$ = \$1;}
AddExpr: MultExpr { \$\$ = \$1;}
MultExpr: BraceExpr { \$\$ = \$1;}
| MultExpr MultOp BraceExpr { \$\$ = (\$2)(\$1, \$3);}
BraceExpr: '(' Expression ')' { \$\$ = \$2;}
| 'N' { sscanf(yytext,"%le", &\$\$);}
MultOp: '*' { \$\$ = &mulOp;}
| '/' { \$\$ = &divOp;}
| '-' { \$\$ = &subOp;}
%%
void yyerror(char const * msg)
{
fprintf(stdout,"Error: %s", msg);
}
int main()
{
yyparse();
}
``````
### Build
``````> flex e.l
> bison e.y
> gcc *.c
> ./a.out
((5 + (3 + (7 * 2))) - (8 * 9)) / 72
Result: -6.944444e-01
>
``````
The above also handles normal operator precedence rules:
Not because of anything I did,but somebody smart worked this out ages ago and now you can get the grammar rules for expression parsing easily (Just google `C Grammer` and rip the bit you need out).
``````> ./a.out
2 + 3 * 4
Result: 1.400000e+01
``````
-
What? Nooooo. Unless this is a homework assignment, do not write a parser. There are a hundred parsers out there and they all have one advantage over all the suggestions here: they're already out there. You don't have to write them.
-
If the expressions are known to be fully-parenthesized (that is, all possible parentheses are there), then this can easily be done using recursive-descent parsing. Essentially, each expression is either of the form
`````` number
``````
or of the form
`````` (expression operator expression)
``````
These two cases can be distinguished by their first token, and so a simple recursive descent suffices. I've actually seen this exact problem given out as a way of testing recursive thinking in introductory programming classes.
If you don't necessarily have this guarantee, then some form of precedence parsing might be a good idea. Many of the other answers to this question discuss various flavors of algorithms for doing this.
-
You could use either a state machine parser (yacc LALR, etc.), or a recursive descent parser.
The parser could emit RPN tokens to evaluate or compile later. Or, in an immediate interpreter implementation, a recursive descent parser could calculate subexpressions on the fly as it returns from the leaf tokens, and end up with the result.
-
First convert the expression into prefix or postfix form. Then its very easy to evaluate!
Example:
Postfix expression evaluation.
-
The easiest way to solve this is to implement the Shunting Yard algorithm to convert the expression from infix notation to postfix notation.
It's Easy-with-a-capital-E to evaluate a postfix expression.
The Shunting Yard algorithm can be implemented in under 30 lines of code. You'll also need to tokenize the input (convert the character string into a sequence of operands, operators, and punctuators), but writing a simple state machine to do that is straightforward.
- | 2,016 | 8,143 | {"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} | 3.109375 | 3 | CC-MAIN-2015-18 | latest | en | 0.886265 |
https://gimpppa.org/what-is-3-percent-of-3000-dollars/ | 1,638,045,309,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358233.7/warc/CC-MAIN-20211127193525-20211127223525-00123.warc.gz | 347,323,050 | 4,672 | What is % that ?
What is out of ?
### X is Y Percent the What Calculator
is % the what?
Using this device you can find any kind of percentage in three ways. So, we think you got to us trying to find answers like:1) What is 3 percent (%) that 3000?2) 3 is what percent the 3000?Or might be: 3% that 3000 Dollars?
See the remedies to these problems below.
If girlfriend are searching for a
### 1) What is 3% that 3000?
Always use this formula to discover a percentage:
% / 100 = part / totality replace the offered values:
3 / 100 = part / 3000
Cross multiply:
3 x 3000 = 100 x Part, or
9000 = 100 x part
Now, divide by 100 and get the answer:
Part = 9000 / 100 = 90
### 2) What is 3 out of 3000?
This inquiry is indistinguishable to: "3 is what percent of 3000?" Or What percent 3 is out of 3000?
Use again the same percentage formula:
% / 100 = component / entirety replace the offered values:
% / 100 = 3 / 3000
Cross multiply:
% x 3000 = 3 x 100
Divide by 3000 to obtain the percentage:
% = (3 x 100) / 3000 = 0.1%
A shorter way to calculation x out of y
You deserve to easily uncover 3 is out of 3000, in one step, through simply separating 3 by 3000, then multiplying the an outcome by 100. So,
3 is the end of 3000 = 3/3000 x 100 = 0.1%
To find more examples, just choose one at the bottom of this page. | 408 | 1,336 | {"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} | 4.28125 | 4 | CC-MAIN-2021-49 | longest | en | 0.892447 |
https://istopdeath.com/eliminate-the-parameter-xtcost-ytsint-tpi/ | 1,675,208,754,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499891.42/warc/CC-MAIN-20230131222253-20230201012253-00066.warc.gz | 335,749,610 | 15,295 | # Eliminate the Parameter x=tcos(t) , y=tsin(t) , t=pi
, ,
Set up the parametric equation for to solve the equation for .
Rewrite the equation as .
Divide each term by and simplify.
Divide each term in by .
Cancel the common factor of .
Cancel the common factor.
Divide by .
Subtract from both sides of the equation.
Find the LCD of the terms in the equation.
Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.
Since contain both numbers and variables, there are two steps to find the LCM. Find LCM for the numeric part then find LCM for the variable part .
The LCM is the smallest number that all of the numbers divide into evenly.
1. List the prime factors of each number.
2. Multiply each factor the greatest number of times it occurs in either number.
The number is not a prime number because it only has one positive factor, which is itself.
Not prime
The LCM of is the result of multiplying all prime factors the greatest number of times they occur in either number.
The factor for is itself.
occurs time.
The LCM of is the result of multiplying all prime factors the greatest number of times they occur in either term.
Multiply each term by and simplify.
Multiply each term in by in order to remove all the denominators from the equation.
Simplify the left side of the equation by cancelling the common factors.
Cancel the common factor of .
Move the leading negative in into the numerator.
Cancel the common factor.
Rewrite the expression.
Reorder factors in .
Multiply by .
Add to both sides of the equation.
Divide each term by and simplify.
Divide each term in by .
Cancel the common factor of .
Cancel the common factor.
Divide by .
Replace in the equation for to get the equation in terms of .
Eliminate the Parameter x=tcos(t) , y=tsin(t) , t=pi | 412 | 1,813 | {"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-2023-06 | latest | en | 0.93377 |
https://dsp.stackexchange.com/questions/9220/spectral-analysis-of-positive-signals | 1,627,468,985,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153709.26/warc/CC-MAIN-20210728092200-20210728122200-00381.warc.gz | 235,458,228 | 38,817 | # Spectral analysis of positive signals
Suppose that I have a sensor that can acquire samples $X[k]$ of the Fourier transform of an unknown signal $Y[t]$. An example is MRI, where the acquired data is in $k-$space. Now suppose that the unknown signal $Y[t]$ is known to be real and non-negative. My question is: is there a principled way to incorporate this knowledge into the spectral analysis algorithm that will estimate $Y[t]$ from $X[k]$, in order to produce an estimate with less bias or variance? I am thinking at non-parametric spectral estimation algorithms. A naive way of course would be to take the real part of $Y[t]$ and clip the negative values, but this does not seem to be optimal. I am looking for some sort of Cadzow's denoising method for spectral data.
• Sounds like a good question, but off the top of my head, I'm not sure that knowing that the signal is nonnegative will help you very much. In the Fourier domain, that would just map to a large DC component (i.e. a large positive mean value); that condition wouldn't really affect the other spectral components. – Jason R May 21 '13 at 1:52
To give a complete answer to this question you're going to need to provide more details about the kind of models you're considering in the first place. But yes, in many cases you can augment those models with a priori constraints on $Y[t]$, such as $0 \leq Y[t] \leq 1$.
• The model for the signal measurements that I am considering is $X[k]=x[k]+w[k], k=0,\ldots,N-1$ where $w[k]$ are i.i.d. samples from a circular complex normal distribution. Also, $\mathbf{x}=\mbox{DFT}(\mathbf{Y})$, $Y[t] \in \mathbb{R}$ and $Y[t] \geq 0$. – Arrigo May 21 '13 at 21:36
• Given a real $N$-point signal $X[t]$, the standard DFT $Y[k]$ will have complex symmetric structure: $Y[0]$ and $Y[N/2]$ are real, and $Y[k]=\overline{Y[N-k]}$ for $k=1,2,\dots,N/2-1$. Your method of sampling in frequency should be preserving that complex symmetric structure. If someone gives you the frequency domain representation $Y$, then yes the inverse DFT is the right way to recover it. But, if the underlying signal is real, then $Y$ had better have this complex symmetric structure. – Michael Grant May 22 '13 at 14:42 | 586 | 2,209 | {"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} | 2.90625 | 3 | CC-MAIN-2021-31 | latest | en | 0.910567 |
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