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https://www.jiskha.com/questions/537165/which-of-the-following-equipment-would-you-use-to-show-a-filmstrip-to-an-entire-class-a
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# Teacher aide
Which of the following equipment would you use to show a filmstrip to an entire class?
A filmstrip viewer
B filmstrip projector
C videocassette recorder
D movie projector
Just want to know if I have it right.
Is it B
1. 👍
2. 👎
3. 👁
1. http://www.jiskha.com/display.cgi?id=1303841184
Of course, B is correct.
1. 👍
2. 👎
👤
Writeacher
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Tags: crypto
Rating:
# storagespace Writeup
### Facebook 2019 - crypto 919 - 31 solves
> In order to fit in with all the other CTFs out there, I've written a secure flag storage system!
It accepts commands in the form of json. For example: help(command="flag") will display help info for the flag command, and the request would look like:
{"command": "help", "params": {"command": "flag"}}
flag(name: Optional[str])
Retrieve flag by name.
{"command": "flag", "params": {"name": "myflag"}}
flag{this_is_not_a_real_flag}
You can access it at nc challenges.fbctf.com 8089
P.S. some commands require a signed request. The sign command will take care of that for you, but no way you'll convince me to sign the flag command xD
#### Observations
To get flag,
1. Execute sign(command="spec"): Get spec of signing algorithm by command
2. Execute sign(command="list"): Get flag file name file_name by command
3. Execute sign(command="info"): Get ECC curve parameters(a,b,p,Generator G,Public Key H)
4. Obtain secret key key, which satisfies key * G == H
5. Generate message flag(name=file_name).
6. Sign message obtained at step 5 and get sign pair (r, s) using key and signing algorithm obtained in step 1.
7. Update signed message with sign pair (r, s) and execute it.
8. Get the flag
So, how do I get the secret key key?
#### Vulnerability: Order of curve is small
The order of the given curve is small enough to solve EC[DLP](https://en.wikipedia.org/wiki/Discrete_logarithm). Sagemath has discrete_log() method to solve ECDLP. The challenge had timeout of 2 minutes, but Sagemath was powerful enough(using [Pohlig-Hellman](https://en.wikipedia.org/wiki/Pohlig%E2%80%93Hellman_algorithm) algorithm) to solve it just in time. You also can manually solve ECDLP using this [code](https://github.com/hgarrereyn/Th3g3ntl3man-CTF-Writeups/blob/master/2017/picoCTF_2017/problems/cryptography/ECC2/ECC2.md). I get the flag:
fb{random_curves_are_not_safe?}
exploit driver code: [solve.py](solve.py)
ECDLP solver: [ECDLP.sage](ECDLP.sage)
Some logs while interacting: [help.log](help.log), [server.log](server.log)
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https://forexwaihui.com/blog/2021/04/02/financial-indices-of-companies-guide-for-beginners/
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Trading in financial markets, especially in the stock market, having information about financial indices of companies might be extremely useful. Not only beginners but also “market whales” use financial reports for market analysis.
If you have just started trading, try studying data and indicators of companies for reporting periods. Normally, companies give reports for a quarter, 6 months, and a year. Trading intraday, you may neglect those, but in medium-term and long-term investments, financial indicators and reports become influential.
See below such indices discussed in detail, including how they can influence your choice of stocks for investments.
## Price/Earnings
Price/Earnings (or P/E) is a multiplier that demonstrates the ratio between the stock price and the yearly return on it. By P/E, the investor evaluates how long it will take for their investment to pay off.
• P/E lets you choose a company with overvalued or undervalued shares;
• Low P/E means that your investment will pay off soon;
• It is only applicable to companies that make a profit.
Find below a detailed article about P/E:
## Price/Sales
The P/S multiplier demonstrates the ratio between the price of the company to its annual revenue.
• If P/S is 2 or below, it is considered to be attractive for investments;
• Ideally, the indicator must be 1. This value will mean that, based on the current profit, your investment will pay off in a year.
• Unlike, P/E, P/S is applicable to losing companies.
## Price/Cash Flow
P/CF is the ratio of the asset’s price and the company’s cash flow with regard to depreciation costs, capital expenditures, and the working capital. It is calculated by dividing the market capitalization of the company by the cash flow from completed market operations.
P/CF is traditionally evaluated as follows:
• P/CF above 20 means that the company is in trouble;
• P/CF between 15 and 20 is good;
• P/CF below 15 means that the company is in a perfect shape.
## Price-to-Book
P/B or P/BV (Price-to-Book Value) shows the price of the company’s assets that can be sold if the company goes bankrupt minus its commitments (debts, expenses). This coefficient is a stock multiplier that shows whether the company is attractive for investments or whether its stocks are overvalued it the market. Undervalued shares have a chance to grow.
• The lower the P/BV, the more chances for growth the company’s shares have.
• This coefficient should be used alongside other indicators because it does not account for future income of the company.
## Price/Earnings to Growth
PEG is a simplified version of P/E. It is calculated with regard to the forecast speed of growth of the company and shows whether it is now overvalued or undervalued, as well as its perspectives of development.
• PEG helps to find companies with expensive inner assets but undervalued in the market.
• In future, the shares of such a company will grow.
## Current Ratio
Current Ratio is literally the current sum of assets divided by the overall commitments of the company. Simply speaking, this is a demonstration of the company’s stability, showing whether it is able to pay off its short-term commitments by short-term assets.
• If the ratio if between 1.6 and 2, the company is attractive for investments.
• If the ratio is 1, this means the company can pay off its short-term debts completely.
• For Current Ratio, the sphere in which the company works is important because “normal” values are different for different activities.
## Capitalization
Capitalization is the general company’s value in the market. It is calculated as the number of stocks in circulation multiplied by the current stock price. The value changes alongside the stock price and may differ from the actual price of the company. Quite often, market capitalization changes due to active trades or exchange speculations.
## Debt-to-Equity
D/E is literally the ratio of the loaned capital of the company and the capital of its shareholders. As well as the Current Ratio, D/E is different for different economy branches, hence, it is incorrect to compare the D/E of an electric carmaker and a provider of online services.
• A too low D/E value means that the company uses the loaned money inefficiently.
• If the value is too high, the company risks losing its financial independence and stability.
## COGS (Cost of Goods Sold)
COGS is the primary cost of sold goods and services. It includes all the expenses for buying materials, processing them, production, and the way of the final product to the consumer. COGS is influenced by purchase prices for raw materials, season of the year, region, and weather. For example, oil prices influence the price of gasoline for carriers and reflect in the final price for consumers. An increase in transportation prices harms the price of tourist services, etc.
## Return on Assets
ROA is the profitability of assets, or the profit per currency unit (dollar). It shows whether the company’s management works efficiently. The multiplier is used to compare competing companies and shows whether assets are used efficiently for making a profit. ROA is calculated by dividing the net profit by all the assets. The average value depends on the sphere of the company’s activities.
## Return on equity
ROE is the ratio of the company’s capital and the profit against the overall shareholders’ capital. For example, if ROE is 23%, every 100 USD of the company’s capital yield 23 USD of profit. The higher the profitability, the more attractive the company is.
Find below a detailed article about ROE:
## Working capital per unit of profit
This multiplier represents the capital used daily (in cycles) for goods and services production. It is used to compare competing companies that work in one sector or by similar principles. The larger it is, the more opportunities there are to make a profit.
## Closing thoughts
Nowadays, when we have computers everywhere, you do not need complicated calculations for analyzing financial indices. Most of them are published on official websites of companies in the section for investors.
Analyze the indicators, compare competing companies by them, and choose those that suit you. To simplify your analysis, take open Internet resources that provide reports of companies and calculators for multipliers.
Using all the indices for companies that work in completely different spheres is incorrect. Experienced investors construct their portfolios from stocks of competing companies.
Material is prepared by
#### Maks Artemov
Has been in Forex since 2009, also trades in the stock market. Regularly participates in RoboForex webinars meant for clients with any level of experience.
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https://safenetforum.org/t/open-questions-on-the-security-properties-offered-by-the-xor-space-closeness-relationship/3670
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Open questions on the security properties offered by the XOR Space closeness relationship
I am writing a paper on ways to securely group nodes in XOR Space to ensure reliability of the system in the presence of malicious nodes. Since the grouping of nodes is such a fundamental operation of the SAFE network, that should help the analysis of the security properties of the entire system.
My main focus has been on formalizing abstractions that help building such a system and proving a variation of the algorithms MaidSafe is using to implement them (variant which are chosen because they have a concise implementation and are relatively easy to reason about but might not be the most performant or be easiest to implement in a particular language since the goal is first to show that properties of XOR Space can indeed be leveraged for building secure decentralized systems). As part of that analysis, I am now getting to the point in which I need to show it is indeed “hard” for an attacker to obtain a close group of identifiers. I want to obtain security properties in terms of minimal bounds on computing resources (computing power, number of IP addresses, etc.) required to disrupt a close group as a function of the size of the network. That should end up being analogous to the proof-of-work security property of the Blockchain that supports Bitcoin.
I believe some of you might have the required math background and inclination to do that probably faster than myself so I am looking for your contribution if that is the case. Should you end up with good answers to the following questions, please publish them such that they can be reviewed and built upon by others. I can totally help with reviewing one you might want to publish yourself. I would also be willing to add you as coauthor on my own paper (or another one).
Here is my attempt at formulating the problem and a first set of related open questions. I will add more as I think things through:
Problem
Suppose you have an identifier space I made of z-bit identifiers (ex: z=512 for the current implementation) from 0 to (2**z)-1. Suppose you compute the distance between them using the bitwise exclusive-or (XOR) operation. Suppose you have an arbitrarily sized set of nodes N each with unique identifiers taken from I. Suppose nodes are grouped together in different subsets of k nodes that are each XOR-closest to a different identifier i of I.
For every identifier i in I , you can compute the k-closest group of nodes by computing the XOR distance between every node n in N , sort them according to distance from shortest to longest and pick the k first.
Now suppose the identifiers are obtained from a random oracle R that guarantees identifiers are uniformly distributed through I and unique. Generating an identifier takes a time t that can be chosen a priori between 0 (instantaneous) and Infinity (impossible to generate).
Once generated, an identifier i can be associated to a node n using an oracle M by doing M.add(n,i). A node can only have a single identifier i and an identifier can belong only to one node. M.node(i), gives you the node n that corresponds to an identifier i. M.id(n) gives you the identifier i that corresponds to n.
Questions
In every case, the answer should be a function of the size of N, #N .
1 Let’s pick a random node n from N and get its identifier i = M.id(n).
1a What is the probability of drawing a new identifier i’ that is closest to i ?
Note, the probability of picking an identifier i’ that is closest to any other node’s identifier is trivially 1 because there is always one closest so that is not really interesting.
1b What is the probability of drawing a new identifier i’ that is within the k-closest nodes to i ?
1c What is the probability of drawing u different identifiers that are within the k-closest to i ?
1d What is the probability of drawing u different identifiers that are within the k-closest to any other node identifier?
2a-d What is the average number of draws necessary to obtain similar outcomes as for 1a-d. What is the variance around that?
3 Suppose that to happen at a particular identifier i, an operation needs the approval of at least l/k nodes to succeed (in the current implementation l=28 and k=32).
3a How many new identifiers would need to be generated close to i?
3b What is the probability of obtaining that many identifiers?
4 What is the minimal size of the network, #N, that makes the attacks described in 1a-d impractical,
4a given a time t of 10 minutes? (the amount of time it takes to generate a block on the block chain)
4b given a time t of 1/10**12 seconds? (approximately the latency of computing a hash using a modern miner ASIC chip)
4c given a time t sufficiently high to tolerate foreseeable improvements in hashing hardware technologies?
4d given a time t well beyond any foreseeable hashing improvements?
5 Would allowing a node to have multiple identifiers simultaneously change any of the results obtained previously?
2 Likes
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# Generating random numbers from normal distribution via inverse uniform distribution
I would like to create a random number generator for the normal distribution via using a uniform linear congruential generator (on uniform distribution) and the inversion method.
However, I'm getting stuck at the final state. Please correct my understanding at any point!
Say i have an LCG, which generates numbers from 0 - 1. It follows a uniform distribution of (0,1).
Now, I want my subsequent normal distribution (X) to follow the same range, so it's standard normal, hence i normalize the area
Now
Using taylor series as this cannot be integrated, I get
However, now I'm stuck. even if i integrate this, I cannot invert this to y = f(x) due to the powers of y. Am I even going about this right?
Any help will be much appreciated! thank you!
• Why not simply use inverse transform stats.stackexchange.com/questions/184325/… ?
– Tim
Commented Sep 21, 2016 at 14:36
• ...I was not aware such a magical function existed, thought I had to figure it out by hand. For curiositys sake, how would i compute this manually?
– Wboy
Commented Sep 21, 2016 at 14:43
• What you're trying to do is not a correct implementation of 'the inversion method'. Done correctly, you'd be trying to invert the normal cdf. Commented Sep 21, 2016 at 16:05
• No, you're not. There's several errors all at once which makes it hard to untangle. Commented Sep 21, 2016 at 16:12
• There's a graphical illustration of the inversion method for the normal case on the Wikipedia page Inverse transform sampling. Reading the article may help clarify ideas for you. Commented Sep 21, 2016 at 16:17
To sample from any distribution using uniform distribution you can use inverse transform method, so there is no need in re-inventing the wheel. For this you simply need to use inverse cumulative distribution function (also known as quantile function) of normal distribution
$$F^{-1}(p) = \mu+\sigma\sqrt{2}\,\operatorname{erf}^{-1}(2p-1)$$
where $\operatorname{erf}$ is error function, and then take uniformly distributed random variable $U \sim \mathrm{Unif}(0, 1)$ and pass it through the quantile function
$$X = F^{-1}(U)$$
resulting variable $X$ will follow normal distribution.
Most statistical software would provide you with quantile function for normal distribution already implemented (e.g. qnorm in R), but if you need to implement it by hand, you can check the following paper that describes algorithm that is used in R:
Wichura, M. J. (1988) Algorithm AS 241: The percentage points of the normal distribution. Applied Statistics, 37, 477–484.
or simply take a look at source code of R's qnorm.
• Thank you so much for your help! :) while that presents an alternative way, is there a way to do this via the method i mentioned? Namely, taking the range of the Uniform distribution, setting the distribution i want (standard norm) and normalizing it, attempting to integrate to get the cdf then invert etc?
– Wboy
Commented Sep 21, 2016 at 15:35
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# P.V. PANEL WIND LOAD EFFECTS M ARCH 2011 Arman Hemmati, Brady Zaiser, Chaneel Park, Jeff Symons, Katie Olver In-Class Presentation TEAM 12.
## Presentation on theme: "P.V. PANEL WIND LOAD EFFECTS M ARCH 2011 Arman Hemmati, Brady Zaiser, Chaneel Park, Jeff Symons, Katie Olver In-Class Presentation TEAM 12."— Presentation transcript:
P.V. PANEL WIND LOAD EFFECTS M ARCH 2011 Arman Hemmati, Brady Zaiser, Chaneel Park, Jeff Symons, Katie Olver In-Class Presentation TEAM 12
MARCH- 2011In-Class Presentation: DeLoPREC Introduction – Problem 1 Most efficient sun capture at 51° Higher angle means greater aerodynamic forces Ballast required to hold the panel down Too much weight for the roof? Want to better understand wind loads on PV panels: Wind Tunnel Testing Computational (CFD) Analysis
MARCH- 2011In-Class Presentation: DeLoPREC Objectives – Functional Requirements 2 1.Effect of Wind Direction Front or back 2.Effect of Panel Tilt Preferred angle is 51° 3.Proximity of Panel to the Ground Should reduce drag to a point
MARCH- 2011In-Class Presentation: DeLoPREC Wind Tunnel – Test Apparatus 3
MARCH- 2011In-Class Presentation: DeLoPREC Wind Tunnel – Testing Parameters Wind direction How does the force on the panel change depending on the wind direction? Front, back 2 levels 4
MARCH- 2011In-Class Presentation: DeLoPREC Wind Tunnel – Testing Parameters Panel angle How does the force on the panel change as the angle increases? In Calgary the ideal angle is 51° Testing at: 35°, 50°, 65° and 80° 4 levels 5
MARCH- 2011In-Class Presentation: DeLoPREC Wind Tunnel – Testing Parameters Panel height How does the force on the panel change as the distance between the ground and the bottom of the panel increases? Unable to use CFD for this Testing at: 0, 1, 2, …, 6 7 levels 6
MARCH- 2011In-Class Presentation: DeLoPREC Wind Tunnel – Testing Procedure How many runs? 247 = 56 runs per replication 2 replications = 112 runs Randomize run order Lots of set-up between runs How long will it take? 5 min/run + daily set-up Estimate 15 hours 1 week 7
MARCH- 2011In-Class Presentation: DeLoPREC Computational Fluid Dynamics 2D Simulations Using Comsol Multiphysics and Ansys CFX 8
MARCH- 2011In-Class Presentation: DeLoPREC CFD - Verification Results recorded in ratios (C L, C D, C P ) Vertical Flat Plate Reference: On the Flow of Air Behind an Inclined Flat Plate of Infinite Span -Fage and Johansen, 1927. 9
MARCH- 2011In-Class Presentation: DeLoPREC 10
MARCH- 2011In-Class Presentation: DeLoPREC www.ucalgary.ca/deloprec
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# How to implement smoothing in frequency domain?
I want to do smoothing to an image in the frequency domain. when i use google to see any articles it gave some Matlab codes which i don't need. i could do FFT to an image but i don't know how to implement any smoothing techniques(ILPF, BLPF, IHPF, BHPF) in frequency domain. if you can provide any code samples for any of the above techniques WITHOUT using any image processing libraries it will be really helpful and C# is preferred.
Thanks,
-
You get a very cheap blur by simply resizing the image. Make it smaller then resize back to the original size. – Hans Passant Nov 3 '10 at 14:52
Could you define what you mean by 'smoothing in the frequency domain'? You can generate a spectrum image using FFT and multiply the image by some function to attenuate particular frequencies, then convert the spectrum back to an image using the inverse-FFT. However, for this kind of filtering (multiplication by some scaling function in frequency), you can achieve the same result more quickly by convolving with the dual function in the spatial domain.
In any case, if you wish to implement this yourself, read up on FFT (the fast Fourier transform) and convolution. You might also check out a signal processing textbook, if you're interested, as the theory behind discrete filtering is fairly deep. The algorithms won't make a whole lot of sense without that theory, though you can certainly apply them without understanding them.
If you want to implement your own DSP algorithms, check out this book online. In particular, Ch 33 describes the math and algorithm behind Butterworth filter design. Ch 12 describes how to implement FFT.
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actually i want to implement a image smoothing algorithm in frequency domain. like Ideal low pass filter, Butterworth low pass filter or a high-pass filter – Keshan Nov 3 '10 at 15:05
@Keshan, there are well-defined algorithms that can be used to generate approximate convolution kernels for the different filters. We talk about how these filters behave in the frequency domain (as this is a convenient way to think about them), but almost all DSP implementations will apply them in the spatial (or time) domain directly, rather than applying to the spectrum. This is because the performance and quality loss associated with the FFT transformations are worse than what you can achieve with convolution. – Dan Bryant Nov 3 '10 at 15:10
For all image/signal processing I recommend OpenCV.
This has a managed C# wrapper: Emgu.
http://www.emgu.com/wiki/index.php/Main_Page
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actually i want to implement it without using any image processing library. – Keshan Nov 3 '10 at 13:36
It is open source you can read, understand and then implement yourself. This is not a trivial task, so good luck. – Aliostad Nov 3 '10 at 13:54
There is a great series on Code Project by Christian Graus which you might find useful, especially part 2 which deals amongst others with smoothing filters:
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Sorry, I just realized that this is not about smoothing in the frequency domain, nevertheless it might be interesting to you or others who come to this page so I will not delete my answer. – Dirk Vollmar - 0xA3 Nov 3 '10 at 13:41
Keshan, it is simple. Imagine the FFT is another two pictures where low frequencies lie in the middle and high frequencies away from the middle. If the pixels are numbered from -w/2 to w/2 and -h/2 to h/2 you can simply measure the distance from the middle as a(x,y)=sqrt(x^2+y^2). Then take some arbitrary monotonic decreasing function like f(x)=1/(1+x) and multiply each point in the fft with f(a(x,y)). Then transform back using the FFT.
There are different choices for f(x) which will look different. For example a gaussian function or bessel or whatever. I did this for my undergrad and it was great fun. If you send me a mail I will send you my program :-).
One bit caveat is the ordering in output of the fft. The arrays it generates can be ordered in weird ways. It is important that you find out which array index corresponds to which x/y-position in the "analytical" fourier transform!
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# Difference Equations
• posted
It's been too long ago, I can't remember how to reduce difference equations.
Suppose I have...
f(N+1) = 0.97*f(N)
Now I know, off the seat of my pants (I think :-), that...
f(N) = Const*(0.97^N)
But I can't remember how to rigorously get to that conclusion. ...Jim Thompson
```--
| James E.Thompson, CTO | mens |
• posted
Mathematical induction. It's true for N=0, by construction, and if it's true for N, it's true for N+1 by the use of your recurrence relation.
Cheers
Phil Hobbs
```--
Dr Philip C D Hobbs
Principal Consultant ```
• posted
"Truth" isn't "solution" ;-)
Isn't there some rigorous way to derive the "continuous" equation from the difference equation? ...Jim Thompson
```--
| James E.Thompson, CTO | mens |
• posted
He is right though you have it for N=0 F(0), F(1) = 0.97*F(0) and then the recurrence relation gives you F(N+1) and if paranoid you can check it out formally by substituting your guess into it explicitly
F(N) = k*(0.97^N) F(N+1) = 0.97*k*(0.97^N) == k*(0.97^(N+1)) QED
which as Phil says is proof by induction.
This one is just a dull harmonic series.
Not necessarily - the finite difference equation is always a discrete approximation to the true full differential equation.
How well the finite difference equation reflects reality and what other approximations you have made along the way affects whether or not you can get back to the continuous version unambiguously.
Numerical approximations to derivatives are notoriously fickle.
```--
Regards,
Martin Brown```
• posted
You've received good answers so far, but another way to do it is through generating functions. The differential equation for the "exponential generating function" of this recurrence can be written by inspection; it's y' = 0.97*y.
Assuming F(0) = 1, the solution to this differential equation is y = e^(0.97*x). The solution to your recurrence is then the coefficient of x^k/k! in the Taylor series expansion of e^(0.97*x), or 0.97^k.
• posted
Aha! Thanks, bitrex, you rang the right chime >:-}
I'm slow this morning, that should have been obvious to me :-( ...Jim Thompson
```--
| James E.Thompson, CTO | mens |
• posted
You can also "solve" it just by plugging in a trial solution r^n into the equation - any linear recurrence with constant coefficients can be solved this way. See:
• posted
I'm not sure what you mean by "continuous" equation -- a difference equation lives in discrete-time, and has no direct relationship to a differential equation in continuous time. (You can _make_ direct relationships, but to do so you have to specify how sampling and reconstruction are carried out).
I was taught in differential equations class that when you dig right down to the bottom of things, you solve a differential equation by guessing an answer, then proving that you were right. Finding symbolic solutions to difference equations is the same general procedure. So I think that's as much rigor as you're going to find in this.
Yes, there are recipes for these solutions, but all of of them (including the use of the z transform) are just the results of the guess-then-prove technique being carried out for a whole class of difference equations, rather than any one specific one.
In the case of a linear, shift-invariant difference equation, the recipe is to find the auxiliary polynomial of the difference equation, and "posit" that the solution is the sum of A_k * d_k^N, where d_k is the k'th root of the auxiliary polynomial and A_k is a constant that goes with it.
In your case (assuming that f(N) is the N'th element in f, which is a possibly infinitely long vector of values), then your auxiliary polynomial is z - 0.97, your "posited" values of d are just d = 0.97, and your "posited" solution is
f(N) = A * (0.97)^N
Note, too, that just like differential equations, linear difference equations can be homogeneous or non-homogeneous (yours is homogeneous). You can find the non-homogeneous solutions to difference equations the same way as you do for differential equations.
And, finally, if you do this a lot with linear, shift-invariant difference equations, it pays to learn how to use the z-transform. It simplifies things almost as much as the Laplace transform does for linear time-invariant differential equations, and makes all of this folderol much easier to remember.
```--
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook. ```
• posted
Yep, It's all coming back to me... guess a solution and prove it fits :-( ...Jim Thompson
```--
| James E.Thompson, CTO | mens |
• posted
The guy who taught my second term of diff eqs clearly wanted us to remember this for all time, because he made this a mantra. Nearly every class meeting he would put some new form of differential equation up on the board, and he'd say "Now, how do we solve this differential equation?" then (because we didn't all shout it out in unison) he'd answer himself: "We guess, and prove that we're right!"
That was 30 years ago. It's stuck with me, so I guess he met his goal in my case.
```--
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook. ```
• posted
That's probably what surprised me most... start the class with, "There are these few basic equations we can solve exactly..." :-( ...Jim Thompson
```--
| James E.Thompson, CTO | mens |
• posted
There are exceptions to this rule, especially transform methods and variation of parameters. It's much more true with PDEs, where e.g. there are 7 types of coordinate systems in which the Laplacian separates, including such intuitive ones as parabolic cylindrical coordinates.
Cheers
Phil Hobbs
```--
Dr Philip C D Hobbs
Principal Consultant ```
• posted
I would contend, though, that all those methods basically came about either because someone guessed an answer to a general class of problems and proved it's validity for that class (e.g. the Laplace transform as used for solving LTI differential equations), or because someone found a trick (variation of parameters, separation of variables, etc.) that applies to a number of different problems.
So in those cases the "guess and prove" has been done for you, but you still have to make sure that your problem fits into the class of things that the method solves for (and heaven knows, I've seen people trying to cram nonlinear or time varying system descriptions into the Laplace transform, and being all confused when their nonsensical formulations give nonsensical results).
```--
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook. ```
• posted
I still recall my first serious brush with exotic differential equations in the freshman year although for different reasons.
The lecturer was an internationally famous astronomer and brilliant analytical solver of novel differential equations. The snag was that he could not teach for toffee and merely demonstrated pulling rabbit out of hat again and again and again. His coursework was impossible.
This particular course was so incomprehensible that after a while the best of us went to the other stream of maths on group theory since it was so much easier and the exam questions were likely to be possible to solve in finite time. Some from the other course which then became badly overcrowded then went to the ODE course so they could sit down.
Indirectly he probably contributed to the increased use of computers to solve differential equations as we later moved into research.
```--
Regards,
Martin Brown```
• posted
Suppose f(1) = A, then we have
f(1) = A f(2) = k * f(1) f(3) = k * f(2) f(4) = k * f(3) ... f(N) = k * f(N-1) ===================== then the product of all the terms on the left equals the product of all the terms on the right, and after eliminating f(1), f(2) ... which occur on both the left and the right of the equaltio, we have
f(N) = A * k^(N-1), for N=1, 2, 3, ...
Euthymios Kappos
"Jim Thompson" wrote in message news: snipped-for-privacy@4ax.com...
• posted
I recall struggling through Diffy-Q, which was widely regarded as a "bag of tricks" subject... you just had to figure out which trick to pull out of the bag for each special case.
Then next term came Laplace Transforms, where we learned that everything that needed doing could be done with simple algebra via Laplace... all that Diffy-Q torture had been just for background information and building character!
Bob Masta DAQARTA v7.21 Data AcQuisition And Real-Time Analysis
Scope, Spectrum, Spectrogram, Sound Level Meter Frequency Counter, Pitch Track, Pitch-to-MIDI FREE Signal Generator, DaqMusic generator Science with your sound card!
• posted
Nice! Thanks! ...Jim Thompson
```--
| James E.Thompson, CTO | mens |
• posted
Using the Laplace transform is just a really versatile trick for approximately solving real-world problems. Here's the reasoning that you should keep in mind whenever you use it (or the z transform):
1: All real-world systems are nonlinear and time varying. 2: The Laplace transform only works on systems that are linear and time- invariant. 3: Thus, I cannot use the Laplace transform to solve this problem. 4: But I can come _close_ by linearizing this here system 5: And now I can use Laplace!
This works great a whole lot of the time -- but it doesn't always, and engineers who are steeped in Laplace (or the z transform) tend to forget that they've skipped over steps 1-3, and did 4 without questioning why, or when it is not valid to do so.
```--
Tim Wescott
Control system and signal processing consulting ```
• posted
Well, the question of how mathematical theorems are discovered is an interesting one--George Polya wrote a series of books on the subject back in (iirc) the 1950s, of which the best known is the elementary one, "How To Solve It". Great book.
Cheers
Phil Hobbs
```--
Dr Philip C D Hobbs
Principal Consultant ```
• posted
You can often use it to derive a perturbation series for those times when linearizing isn't quite enough.
I highly recommend Bender & Orszag's "Advanced Mathematical Methods for Scientists and Engineers", which has all that sort of stuff--boundary layer theory, steepest descents, and a whole lot of other asymptotic methods. Arfken's applied math book is good as well--it used to be the standard textbook for physics undergraduates.
It's often possible to derive a series solution, which may or may not converge, and then convert to a continued fraction or apply convergence tricks such as Shanks's algorithm or Richardson extrapolation. When those work, which isn't always, they can be practically supernatural.
Cheers
Phil Hobbs
```--
Dr Philip C D Hobbs
Principal Consultant ```
ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.
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Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > biass Structured version Unicode version
Theorem biass 360
Description: Associative law for the biconditional. An axiom of system DS in Vladimir Lifschitz, "On calculational proofs", Annals of Pure and Applied Logic, 113:207-224, 2002, http://www.cs.utexas.edu/users/ai-lab/pub-view.php?PubID=26805. Interestingly, this law was not included in Principia Mathematica but was apparently first noted by Jan Lukasiewicz circa 1923. (Contributed by NM, 8-Jan-2005.) (Proof shortened by Juha Arpiainen, 19-Jan-2006.) (Proof shortened by Wolf Lammen, 21-Sep-2013.)
Assertion
Ref Expression
biass
Proof of Theorem biass
StepHypRef Expression
1 pm5.501 342 . . . 4
21bibi1d 320 . . 3
3 pm5.501 342 . . 3
42, 3bitr3d 258 . 2
5 nbbn 359 . . . 4
6 nbn2 346 . . . . 5
76bibi1d 320 . . . 4
85, 7syl5bbr 262 . . 3
9 nbn2 346 . . 3
108, 9bitr3d 258 . 2
114, 10pm2.61i 167 1
Colors of variables: wff setvar class Syntax hints: wn 3 wb 187 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 This theorem depends on definitions: df-bi 188 This theorem is referenced by: biluk 941 xorass 1404 xorassOLD 1405 had1 1501 symdifass 3708
Copyright terms: Public domain W3C validator
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# Cubic Bézier Easing
#### Holzchopf
##### Moderator
3rd Party Module Dev
Tutorial Author
I stumbled across this article and thought it might come in handy to have a cubic-bezier function. And here it is:
Code:
``````' Cubic Bézier interpolation for easing curves
' pA ... pD are Bézier curve points
' pX the value for which the corresponding curve Y value shall be found
' pPrecision defines up to which error in Y the result shall be
' Returns: curve Y value at pX
Function CubicBezierEase:Float(pA:Float, pB:Float, pC:Float, pD:Float, pX:Float, pPrecision:Float = 0.0001)
' boundary conditions
If pX <= 0 Then Return 0
If pX >= 1 Then Return 1
' polynomial factors, precalculated for performance
Local pQuadr:Float = 3*pC - 6*pA
Local pCubic:Float = 1 + 3*pA - 3*pC
' A cubic bezier spline is a vector function
' (x, y) = f(t)
' The algorithm finds t for a given x
' iteratively
' and returns the corresponding y value.
' start at t = pX
Local t:Float = pX
' old value of t, step of t, how much t actually changed and next best guess for t
Local oldt:Float, tstep:Float, dt:Float, dttarget:Float
' value of x, old value of x and how much x actually changed in last iteration
Local x:Float, oldx:Float, dx:Float
' first iteration is special (in respect to calculation of tstep)
Local firstrun:Int = True
' iterate
Repeat
' calculate x of cubic bezier spline
x = 3*pA*t + pQuadr * t*t + pCubic * t*t*t
' accurately found t?
If Abs(x - pX) <= pPrecision Then
Exit
' no, move t
Else
' update tstep
' in first iteration
If firstrun then
firstrun = False
' tstep by how much x differs from target x
tstep = pX - x
' after that
Else
' adapt tstep by how much the value over- or undershoot
dx = x - oldx
dt = t - oldt
dttarget = (pX - oldx) / dx * dt
tstep = -dt + dttarget
Endif
' update t and clamp
oldt = t
t += tstep
If t > 1 then t=1
If t < 0 then t=0
EndIf
oldx = x
Forever
' calculate y of cubic bezier spline and return
Return 3*pB*t + (3*pD - 6*pB) * t*t + (1 + 3*pB - 3*pD) * t*t*t
End Function``````
Last edited:
#### Holzchopf
##### Moderator
3rd Party Module Dev
Tutorial Author
I updated the function. It now takes two more values: pY0 and pY1 which describe the start and end Y value. With this it's also possible to make curves that "pulse" or "fade out".
Code:
``````' Cubic Bézier interpolation for easing curves
' (Imagine a curve on the X-Y plane. Returns curve value at pX)
' pY0 and pY1 are curve Y values at X=0 and X=1 respectively
' pA ... pD are Bézier curve points
' pX the value for which the corresponding curve Y value shall be found
' pPrecision defines up to which error in Y the result shall be
' Returns: curve Y value at pX
Function CubicBezierEase:Float(pY0:Float, pY1:Float, pA:Float, pB:Float, pC:Float, pD:Float, pX:Float, pPrecision:Float = 0.0001)
' boundary conditions
If pX <= 0 Then Return pY0
If pX >= 1 Then Return pY1
' polynomial factors, precalculated for performance
Local pLin:Float = 3*pA
Local pQuadr:Float = 3*pC - 6*pA
Local pCubic:Float = 1 + 3*pA - 3*pC
' A cubic bezier spline is a vector function
' (x, y) = f(t)
' The algorithm finds t for a given x
' iteratively
' and returns the corresponding y value.
' start at t = pX
Local t:Float = pX
' old value of t, step of t, how much t actually changed and next best guess for t
Local oldt:Float, tstep:Float, dt:Float, dttarget:Float
' value of x, old value of x and how much x actually changed in last iteration
Local x:Float, oldx:Float, dx:Float
' first iteration is special (in respect to calculation of tstep)
Local firstrun:Int = True
' iterate
Repeat
' calculate x of cubic bezier spline
x = pLin * t + pQuadr * t*t + pCubic * t*t*t
' accurately found t?
If Abs(x - pX) <= pPrecision Then
Exit
' no, move t
Else
' update tstep
' in first iteration
If firstrun then
firstrun = False
' tstep by how much x differs from target x
tstep = pX - x
' after that
Else
' adapt tstep by how much the value over- or undershoot
dx = x - oldx
dt = t - oldt
dttarget = (pX - oldx) / dx * dt
tstep = -dt + dttarget
Endif
' update t and clamp
oldt = t
t += tstep
If t > 1 then t=1
If t < 0 then t=0
EndIf
oldx = x
Forever
' calculate y of cubic bezier spline and return
Return pY0 + (3*pB - 3*pY0) * t + (3*pY0 - 6*pB + 3*pD) * t*t + (pY1 + 3*pB - 3*pD - pY0) * t*t*t
End``````
You may be wondering why I didn't add those as optional parameters? Because I tried it and - believe me - it looked horrible with those two essential values attached at the back. For me, it makes much more sense to have this order: Interpolate from, to, via, at point X
##### New member
CX Code Contributor
3rd Party Module Dev
Nice! There is a lot of cool stuff in the material design guide.
That reminds me that I have some easing functions lying around since a long time. Maybe I should post them.
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This is an archived post. You won't be able to vote or comment.
[–] 1 point2 points (3 children)
n, where n is the number of desks in the classroom.
If it helps, think about the desks all being in a circle, and the students rotating clockwise one desk every day.
[–][S] 1 point2 points (1 child)
Sorry if I'm wrong, but don't we already know the number of desks in the classroom (12)?
[–] 4 points5 points (0 children)
We probably do. It's a reasonable assumption, certainly.
You might have more desks than students, though. Unlikely, but possible.
[–] 1 point2 points (0 children)
Nope. Consider a circle of 7 desks and a circle of 5 desks. Now the students will take 35 days (which is also not the maximum).
[–] 0 points1 point (0 children)
Notice that because each desk has one arrow toward and one away, the result is going to be a collection of cycles whose total length is 12. Think about how the length of the cycles affects the time it takes for the students to be back to normal, then figure out the best possible cycle configuration.
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https://www.thecodeteacher.com/howto/251/Python---How-To--7-Ways-to-Concatenate-Two-or-More-Lists-in-Python
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# Python - How To 7 Ways to Concatenate Two or More Lists in Python
ID : 251
viewed : 104
Tags : PythonPython List
95
In Python, we use lists in almost every program. While working with , we need to concatenate two or more lists on several occasions. In this article, we will look at various ways to concatenate two lists in Python. We will also discuss the efficiency of all the approaches in terms of time and memory usage so that you can choose the best way to concatenate lists in your programs in Python.
## Concatenate Lists Using the append() Method in Python
The first way to concatenate two lists in Python is to use a for loop with . When executed on a list, the append() method accepts an element as its input argument and adds it to the existing list, as shown in the following example.
oldList = [1, 2, 3, 4, 5, 6, 7] element = 8 print("The existing list is:", oldList) print("The element to be appended is:", element) oldList.append(element) print("The modified list is:", oldList)
Output:
The existing list is: [1, 2, 3, 4, 5, 6, 7] The element to be appended is: 8 The modified list is: [1, 2, 3, 4, 5, 6, 7, 8]
The append() method is normally used to add a single element to an existing list. To concatenate two lists using the append() method, we will add each element of the second list to the first list using the append() method and a for loop as follows.
list1 = [1, 2, 3, 4, 5, 6, 7] list2 = [8, 9, 10] print("First list is:", list1) print("Second list is:", list2) for element in list2: list1.append(element) print("The concatenated list is:", list1)
Output:
First list is: [1, 2, 3, 4, 5, 6, 7] Second list is: [8, 9, 10] The concatenated list is: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
Here, we have defined list1 and list2. After that, we have appended each element of list2 to list1. In this way, we have concatenated the two lists using the append() method. If you don’t want to modify the existing lists, you can concatenate the lists by creating a new empty list and adding the elements of the existing lists to the empty list as follows.
list1 = [1, 2, 3, 4, 5, 6, 7] list2 = [8, 9, 10] print("First list is:", list1) print("Second list is:", list2) newList = list() lists = [list1, list2] for myList in lists: for element in myList: newList.append(element) print("The concatenated list is:", newList)
Output:
First list is: [1, 2, 3, 4, 5, 6, 7] Second list is: [8, 9, 10] The concatenated list is: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
Here, you can observe that we have created the concatenated list in a new list named newList. After that, we have declared a list of the existing lists and named it lists. Then, we have appended each element from the existing lists to the newList using the for loop and the append() method. In this approach, you can also concatenate more than two lists at once. For that, you just need to add the extra lists that need to be concatenated to lists.
## Concatenate Lists Using the + Operator in Python
Instead of using the append() method, we can use the + operator to concatenate two lists in Python. Just like we concatenate two strings using the + operator in Python, we can concatenate two lists as follows.
list1 = [1, 2, 3, 4, 5, 6, 7] list2 = [8, 9, 10] print("First list is:", list1) print("Second list is:", list2) newList = list1+list2 print("The concatenated list is:", newList)
Output:
First list is: [1, 2, 3, 4, 5, 6, 7] Second list is: [8, 9, 10] The concatenated list is: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
Here, we have stored the concatenated list in a third list named newList. If you don’t need list1 and list2 further in the program, you can assign the concatenated list to any of them as given below. This will help you reduce memory usage of the program.
list1 = [1, 2, 3, 4, 5, 6, 7] list2 = [8, 9, 10] print("First list is:", list1) print("Second list is:", list2) list1 = list1+list2 print("The concatenated list is:", list1)
Output:
First list is: [1, 2, 3, 4, 5, 6, 7] Second list is: [8, 9, 10] The concatenated list is: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
Using the + operator to concatenate two lists has the advantage of concatenating more than two lists using a single statement. For instance, we can concatenate four lists in Python using the + operator as follows.
list1 = [1, 2, 3, 4, 5, 6, 7] list2 = [8, 9, 10] list3 = [11, 12] list4 = [13, 14, 15] print("First list is:", list1) print("Second list is:", list2) print("Third list is:", list3) print("Fourth list is:", list4) newList = list1 + list2 + list3 + list4 print("The concatenated list is:", newList)
Output:
First list is: [1, 2, 3, 4, 5, 6, 7] Second list is: [8, 9, 10] Third list is: [11, 12] Fourth list is: [13, 14, 15] The concatenated list is: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15]
## Concatenate Lists Using the extend() Method in Python
We can also use the extend() method to concatenate two lists in Python. When executed on a list, the extend() method accepts any iterable object like list, set, or tuple and adds the elements of the iterable object to the end of the list.
To concatenate two lists, we can execute the extend() method on the first list and pass the second list as an input argument to the extend() method as follows.
list1 = [1, 2, 3, 4, 5, 6, 7] list2 = [8, 9, 10] print("First list is:", list1) print("Second list is:", list2) list1.extend(list2) print("The concatenated list is:", list1)
Output:
First list is: [1, 2, 3, 4, 5, 6, 7] Second list is: [8, 9, 10] The concatenated list is: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
Here, we have appended the elements of list2 to list1 using the extend() method.
## Concatenate Lists Using the += Operator in Python
The += operator is an unpopular but efficient tool to concatenate two lists in Python. It works similarly to the extend() method. The syntax for using the += operator is as follows.
oldList+=iterable
Here,
• oldList is the existing list that we already have.
• iterable is any iterable object like list, tuple, or set. The elements in the iterable are added to the oldList when we execute this statement.
To concatenate two lists using the += operator, we will use the first list as the oldList and the second list as the iterable as follows.
list1 = [1, 2, 3, 4, 5, 6, 7] list2 = [8, 9, 10] print("First list is:", list1) print("Second list is:", list2) list1 += list2 print("The concatenated list is:", list1)
Output:
First list is: [1, 2, 3, 4, 5, 6, 7] Second list is: [8, 9, 10] The concatenated list is: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
In the output, you can observe that we have obtained the concatenated list in list1.
## Concatenate Lists Using Packing and Unpacking in Python
In Python, packing and unpacking are often used to convert a container object to another container object having the same elements. For unpacking, we use the * operator before the name of the container object as *container_object. We will put the unpacked elements inside brackets [] to pack the elements again in a container object like a list.
To concatenate two or more lists using the packing and unpacking operations, we will first unpack the lists using the * operator. Then, we will pack the unpacked elements of the lists into a new list as follows.
list1 = [1, 2, 3, 4, 5, 6, 7] list2 = [8, 9, 10] print("First list is:", list1) print("Second list is:", list2) newList = [*list1, *list2] print("The concatenated list is:", newList)
Output:
First list is: [1, 2, 3, 4, 5, 6, 7] Second list is: [8, 9, 10] The concatenated list is: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
Here, you can pack the elements of any number of lists into a single list. Thus, this method is also useful for concatenating more than two lists. For instance, you can concatenate four lists in a single statement as follows.
list1 = [1, 2, 3, 4, 5, 6, 7] list2 = [8, 9, 10] list3 = [11, 12] list4 = [13, 14, 15] print("First list is:", list1) print("Second list is:", list2) print("Third list is:", list3) print("Fourth list is:", list4) newList = [*list1, *list2, *list3, *list4] print("The concatenated list is:", newList)
Output:
First list is: [1, 2, 3, 4, 5, 6, 7] Second list is: [8, 9, 10] Third list is: [11, 12] Fourth list is: [13, 14, 15] The concatenated list is: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15]
## Concatenate Lists Using the itertools.chain() Function in Python
In Python, we use the chain() function from the itertools module to create iterators. The chain() function, when executed, accepts multiple iterable objects as its input arguments, chains all the elements in a single iterator, and returns the iterator. To use the chain() function, you can import the itertools module as follows.
import itertools
To concatenate two or more lists using the chain() function, we will pass the original lists as input arguments to the chain() function. After execution, the function will return an iterator. We will use the list() constructor to convert the iterator to a list. Hence, we will get the concatenated list. You can observe this in the following example.
import itertools list1 = [1, 2, 3, 4, 5, 6, 7] list2 = [8, 9, 10] print("First list is:", list1) print("Second list is:", list2) iterator = itertools.chain(list1, list2) newList = list(iterator) print("The concatenated list is:", newList)
Output:
First list is: [1, 2, 3, 4, 5, 6, 7] Second list is: [8, 9, 10] The concatenated list is: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
## Concatenate Lists Using List Comprehension in Python
On some occasions, list comprehension is a better alternative to for loop while working with lists in Python. We can also concatenate two or more lists using list comprehension. The syntax for the same is as follows. newList= [element for temp_list in [list1, list2] for element in temp_list] Here,
• list1 and list2 are the lists that we want to concatenate.
• newList is the list created by concatenating list1 and list2.
Here, we choose each list that we have to concatenate in the temp_list one by one. After that, we put all the elements of temp_list in newList.The temp_list chooses each list that we need to concatenate one by one, and their elements are added to newList, creating the output list with all the elements from the input lists.
We can use list comprehension to concatenate two lists in Python as follows.
list1 = [1, 2, 3, 4, 5, 6, 7] list2 = [8, 9, 10] print("First list is:", list1) print("Second list is:", list2) newList = [element for temp_list in [list1, list2] for element in temp_list] print("The concatenated list is:", newList)
Output:
First list is: [1, 2, 3, 4, 5, 6, 7] Second list is: [8, 9, 10] The concatenated list is: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
If you are not able to understand the syntax of the list comprehension above, the for loop in the following program is equivalent to the list comprehension syntax.
list1 = [1, 2, 3, 4, 5, 6, 7] list2 = [8, 9, 10] print("First list is:", list1) print("Second list is:", list2) newList = list() lists = [list1, list2] for temp_list in lists: for element in temp_list: newList.append(element) print("The concatenated list is:", newList)
Output:
First list is: [1, 2, 3, 4, 5, 6, 7] Second list is: [8, 9, 10] The concatenated list is: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
We can also concatenate more than two lists using list comprehension. We will just have to insert the input lists in the list of lists as follows.
list1 = [1, 2, 3, 4, 5, 6, 7] list2 = [8, 9, 10] list3 = [11, 12] list4 = [13, 14, 15] print("First list is:", list1) print("Second list is:", list2) print("Third list is:", list3) print("Fourth list is:", list4) newList = [element for temp_list in [list1, list2, list3, list4] for element in temp_list] print("The concatenated list is:", newList)
Output:
First list is: [1, 2, 3, 4, 5, 6, 7] Second list is: [8, 9, 10] Third list is: [11, 12] Fourth list is: [13, 14, 15] The concatenated list is: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15]
## What Method Should You Use to Concatenate Lists in Python?
In this article, we have discussed seven ways to concatenate lists in Python.
• If you need to concatenate two lists, you can use the append() method, the extend() method, or the += operator. On the other hand, if you want to concatenate more than two lists, you can use the + operator, the itertools.chain() function, list comprehension, or the packing and unpacking approach.
• If you want to keep the original lists unchanged while creating a new list containing elements from all the original lists, you should use only the + operator approach, the itertools.chain() function, the list comprehension approach, and the packing and unpacking approach. In all other approaches, we have modified one of the original lists.
• If you also focus on the time complexity and space complexity of the programs, you can use the + operator, += operator, the extend() method, or the packing and unpacking approach for writing efficient programs. All other approaches are costly either in terms of time or memory compared to these approaches. The list comprehension approach to concatenate two lists is costliest in terms of both memory and time. So, you should try to avoid this approach.
## Conclusion
In this article, we have discussed seven approaches to concatenate two or more lists in Python. We have also discussed what approach you can use based on your requirements to write an efficient program.
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# MRC: Landing, Moving, and Surviving Conditions (Grades 6-8)
This lesson plan uses the 5E learning cycle and is designed around an essential question: Why is the method you chose for landing your Rover on Mars the best one for your mission? The lesson objectives include: examine different methods for landing... (View More)
# Project Summary MMS Tic-Tac-Toe
This is the culminating lesson in the MMS Mission Educator's Instructional Guide. Learners will choose and complete three activities about the MMS mission. Activity formats can include creating videos, composing songs, developing written materials,... (View More)
# The Satellites' Flight Configuration
In this lesson, learners will first watch a video about the orbit and formation of the MMS satellites to learn about their flight configuration. After, they will research similar facts about other types of satellites. Next, learners will compute the... (View More)
# Powering the Satellite
In this lesson, learners will first use computers to research and learn how solar panels convert sunlight into electricity. Next, they will calculate the surface area of solar panels board a satellite and their total power generated in various... (View More)
# Launch of the Satellites
In this lesson, learners will research facts about Atlas V rockets, which launched the MMS satellites. After, they will compute the speed of the launch rocket, given a data chart of time vs. distance from lift-off. Then, they will write a report... (View More)
# Sam Finnigan's Big Secret
Learners will read about missions to asteroids and comets, consider the measurements and math required for the robotic spacecraft to visit these objects, and are invited to finish the story themselves. The provided extension explains how to use a... (View More)
# What Do I See When I Picture Saturn?
This is a lesson about the Cassini mission to Saturn. Learners will create their Saturn Discovery Logs. They will use the Saturn Discovery Log to chronicle their journey of discovery about Saturn and Cassini through nonfiction writing. For their... (View More)
# Celebrating Saturn and Cassini
In this concluding activity, learners will use notes from an earlier lesson to write a nonfiction piece about Saturn or Cassini. These final projects provide a way for children with varying learning styles to consolidate and share their learning.... (View More)
# My Spacecraft Model
This is a lesson about the Cassini Spacecraft. Learners will engage in basic problem-solving as they design and construct their own small model of a spacecraft. Through writing and illustration, students will document their work and will complete a... (View More)
# My Spacecraft and Cassini
This is a lesson about spacecraft design. Learners will hear the NASA solutions to the problems they wrote about in an earlier lesson. They will then write a nonfiction piece comparing their spacecraft to Cassini, and share their writing with the... (View More)
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https://discuss.leetcode.com/topic/114902/python-o-n-time-and-o-1-space-without-fancy-builtins
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# Python O(n) time and O(1) space without fancy builtins
• Just iterate through the array and note the highest and second highest numbers. Might as well take note of the index at the same time.
One slightly clever idea was to shuffle the highest to the second-highest whenever a new highest was found. That was a way to handle the case where there are two (or more) value tied for highest.
``````class Solution:
def dominantIndex(self, nums):
if len(nums) == 0: return -1
highest = -1
secondHighest = -1
highestIndex = 0
for i,n in enumerate(nums):
if n >= highest:
secondHighest = highest
highest = n
highestIndex = i
elif n > secondHighest:
secondHighest = n
if highest < secondHighest*2:
highestIndex = -1
return highestIndex
``````
Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect.
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https://www.reddit.com/r/Miami/comments/130h2d/miami_drivers_two_people_making_illegal_uturns/
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This is an archived post. You won't be able to vote or comment.
[–] 11 points12 points (1 child)
My question is why are they now finding it necessary to take up All The Lanes? It's a fender bender. Yes, it's an accident, they're both probably ok, their vehicles don't have damage that would make them undrivable, so why aren't they pulling to the side and letting people past?
It fills me with the anger of a thousand suns...
[–] 1 point2 points (0 children)
It makes me mad also, but the accident probably just happened a few seconds before photo was snapped and they were doing a quick assessment of the damage.
[–] 5 points6 points (5 children)
Thanks to Florida no fault, they are are both idiots x2.
[–] 2 points3 points (4 children)
When both drivers do something illegal, how do they determine fault? I would think both, but don't know if they can that or not...
Although, since 1 in 4 Florida drivers are uninsured, there is a 50% chance one of those drivers would be utterly unconcerned about a rate hike anyway.
[–] 5 points6 points (2 children)
That's not how probabilities work.
If 1 in 4 Florida drivers are uninsured there is only a 43.75% chance of at least one being uninsured (which includes a 6.25% chance of both being uninsured).
[–] 1 point2 points (1 child)
Very true, my bad. My brain is taking the weekend off I guess.
[–] 2 points3 points (0 children)
No problem :) I'm a mathematician, and do a lot of work with probability.
[–] 2 points3 points (0 children)
Both get a ticket.
[–] 7 points8 points (1 child)
"PERO QUE PASO CHICO? QUE ES TU PROBLEMA MENG?!" - persona 1
"Broder, si tu hiciste la misma mierda, cabezón!" - persona 1
'Siempre las misma mierda en el fucking cientosiete, cojone! De pinga, estoy tarde otra vez!' - todos detras de ellos
Everyday in Miami.
[–] 5 points6 points (0 children)
MARIKO BRODER, BRODER MARIKO, NA HUEVONAAAAA BRODER....ok
[–] 2 points3 points (1 child)
Is this near Robert Morgan?
[–] 1 point2 points (0 children)
that's exactly where I thought it was too.
[–] 1 point2 points (0 children)
Amazing how much property damage and money is spent on first responders, lost productivity, etc. due to these incessant crashes. Actually, when you think about it, they are probably a large part of the US economy - car crashes, accounting fraud, fried chicken, hair salons and heart surgery probably make up 96% of our economy these days.
[–] 0 points1 point (0 children)
i wonder where this happened
[–] 0 points1 point (0 children)
Hey this is pretty close to where I live... Is that Eureka?
[–] 0 points1 point (0 children)
SW 97th ave?
[–] 0 points1 point (0 children)
127th avenue in Kendall?
[–] -1 points0 points (0 children)
At least these shitty drivers no longer have cars to cause others accidents and undue stress. Now you just have to watch out for the other 3 million shitty drivers all hell bent on doing the same thing.
[–] -1 points0 points (0 children)
I'M CURRENTLY STUDYING FOR MY DRIVER'S. CAREFUL Y'ALL.
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# Computer simulation of the Indonesian tsunami of December 26, 2004 (8 minutes after the earthquake), created using models of wave motion based on advanced.
## Presentation on theme: "Computer simulation of the Indonesian tsunami of December 26, 2004 (8 minutes after the earthquake), created using models of wave motion based on advanced."— Presentation transcript:
Computer simulation of the Indonesian tsunami of December 26, 2004 (8 minutes after the earthquake), created using models of wave motion based on advanced calculus by Steven Ward, University of California at Santa Cruz. 8 TECHNIQUES OF INTEGRATION In Section 5.6 we introduced substitution, one of the most important techniques of integration. In this section, we develop a second fundamental technique, Integration by Parts, as well as several techniques for treating particular classes of functions such as trigonometric and rational functions. However, there is no surefire method, and in fact, many important antiderivatives cannot be expressed in elementary terms. Therefore, we discuss numerical integration in the last section. Every definite integral can be approximated numerically to any desired degree of accuracy.Section 5.6rational functions
The Integration by Parts formula is derived from the Product Rule: Integration by Parts Formula
The Integration by Parts formula is derived from the Product Rule: Integration by Parts Formula Because the Integration by Parts formula applies to a product u(x)υ (x), we should consider using it when the integrand is a product of two functions.
Integrating by Parts More Than Once
Integration by Parts applies to definite integrals:
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:: The Chinese Remainder Theorem
:: by Andrzej Kondracki
::
:: Copyright (c) 1997-2021 Association of Mizar Users
theorem Th1: :: WSIERP_1:1
for x being Complex holds
( x |^ 2 = x * x & (- x) |^ 2 = x |^ 2 )
proof end;
theorem Th2: :: WSIERP_1:2
for x being Real
for a being Nat holds
( (- x) |^ (2 * a) = x |^ (2 * a) & (- x) |^ ((2 * a) + 1) = - (x |^ ((2 * a) + 1)) )
proof end;
theorem Th3: :: WSIERP_1:3
for x, y being Real
for d being Nat st x >= 0 & y >= 0 & d > 0 & x |^ d = y |^ d holds
x = y
proof end;
Lm1: for x, y being Real holds
( ( x >= 0 implies x + y >= y ) & ( x + y >= y implies x >= 0 ) & ( x > 0 implies x + y > y ) & ( x + y > y implies x > 0 ) & ( x >= 0 implies y >= y - x ) & ( y >= y - x implies x >= 0 ) & ( x > 0 implies y > y - x ) & ( y > y - x implies x > 0 ) )
proof end;
Lm2: for x, y, z being Real st x >= 0 & y >= z holds
( x + y >= z & y >= z - x )
proof end;
Lm3: for x, y, z being Real st ( ( x >= 0 & y > z ) or ( x > 0 & y >= z ) ) holds
( x + y > z & y > z - x )
proof end;
registration
let k be Integer;
let a be natural Number ;
cluster k |^ a -> integer ;
coherence
k |^ a is integer
proof end;
end;
Lm4: for a being Nat ex b being Nat st
( a = 2 * b or a = (2 * b) + 1 )
proof end;
theorem Th4: :: WSIERP_1:4
for k, m, n being Integer st k divides m & k divides n holds
k divides m + n
proof end;
theorem Th5: :: WSIERP_1:5
for k, m, n, m1, n1 being Integer st k divides m & k divides n holds
k divides (m * m1) + (n * n1)
proof end;
theorem Th6: :: WSIERP_1:6
for k, m, n being Integer st m gcd n = 1 & k gcd n = 1 holds
(m * k) gcd n = 1
proof end;
theorem :: WSIERP_1:7
for a, b, c being Nat st a gcd b = 1 & c gcd b = 1 holds
(a * c) gcd b = 1 by Th6;
theorem Th8: :: WSIERP_1:8
for m being Integer holds
( 0 gcd m = |.m.| & 1 gcd m = 1 )
proof end;
theorem Th9: :: WSIERP_1:9
for k being Integer holds 1,k are_coprime
proof end;
theorem Th10: :: WSIERP_1:10
for a being Nat
for k, l being Integer st k,l are_coprime holds
k |^ a,l are_coprime
proof end;
theorem Th11: :: WSIERP_1:11
for a, b being Nat
for k, l being Integer st k,l are_coprime holds
k |^ a,l |^ b are_coprime
proof end;
theorem Th12: :: WSIERP_1:12
for a, b being Nat
for k, l being Integer st k gcd l = 1 holds
( k gcd (l |^ b) = 1 & (k |^ a) gcd (l |^ b) = 1 )
proof end;
theorem :: WSIERP_1:13
for k, m being Integer holds
( |.m.| divides k iff m divides k )
proof end;
theorem Th14: :: WSIERP_1:14
for a, b, c being Nat st a divides b holds
a |^ c divides b |^ c
proof end;
theorem Th15: :: WSIERP_1:15
for a being Nat st a divides 1 holds
a = 1
proof end;
theorem :: WSIERP_1:16
for a, b, d being Nat st d divides a & a gcd b = 1 holds
d gcd b = 1 by ;
Lm5: for a, b being Nat st a <> 0 holds
( a divides b iff b / a is Element of NAT )
proof end;
theorem Th17: :: WSIERP_1:17
for k, l being Integer st k <> 0 holds
( k divides l iff l / k is Integer )
proof end;
Lm6: for a, b being Nat
for k being Integer st b <> 0 & a * k = b holds
k is Element of NAT
proof end;
Lm7: for a, b, c being Nat st a + b <= c holds
( a <= c & b <= c )
proof end;
theorem :: WSIERP_1:18
for a, b, c being Nat st a <= b - c holds
( a <= b & c <= b )
proof end;
Lm8: for k, m being Integer holds
( k < m iff k <= m - 1 )
proof end;
Lm9: for k, m being Integer holds
( k < m + 1 iff k <= m )
proof end;
registration
let fr be FinSequence of INT ;
coherence
Product fr is integer
proof end;
end;
definition
let fp be FinSequence of NAT ;
:: original: Sum
redefine func Sum fp -> Element of NAT ;
coherence
Sum fp is Element of NAT
proof end;
end;
definition
let fp be FinSequence of NAT ;
:: original: Product
redefine func Product fp -> Element of NAT ;
coherence
Product fp is Element of NAT
proof end;
end;
Lm10: for a being Nat
for fs being FinSequence st a in dom fs holds
ex fs1, fs2 being FinSequence st
( fs = (fs1 ^ <*(fs . a)*>) ^ fs2 & len fs1 = a - 1 & len fs2 = (len fs) - a )
proof end;
definition
let a be Nat;
let fs be FinSequence;
redefine func Del (fs,a) means :Def1: :: WSIERP_1:def 1
it = fs if not a in dom fs
otherwise ( (len it) + 1 = len fs & ( for b being Nat holds
( ( b < a implies it . b = fs . b ) & ( b >= a implies it . b = fs . (b + 1) ) ) ) );
compatibility
for b1 being set holds
( ( not a in dom fs implies ( b1 = Del (fs,a) iff b1 = fs ) ) & ( a in dom fs implies ( b1 = Del (fs,a) iff ( (len b1) + 1 = len fs & ( for b being Nat holds
( ( b < a implies b1 . b = fs . b ) & ( b >= a implies b1 . b = fs . (b + 1) ) ) ) ) ) ) )
proof end;
correctness
consistency
for b1 being set holds verum
;
;
end;
:: deftheorem Def1 defines Del WSIERP_1:def 1 :
for a being Nat
for fs being FinSequence
for b3 being set holds
( ( not a in dom fs implies ( b3 = Del (fs,a) iff b3 = fs ) ) & ( a in dom fs implies ( b3 = Del (fs,a) iff ( (len b3) + 1 = len fs & ( for b being Nat holds
( ( b < a implies b3 . b = fs . b ) & ( b >= a implies b3 . b = fs . (b + 1) ) ) ) ) ) ) );
Lm11: for a being Nat
for fs, fs1, fs2 being FinSequence
for v being object st a in dom fs & fs = (fs1 ^ <*v*>) ^ fs2 & len fs1 = a - 1 holds
Del (fs,a) = fs1 ^ fs2
proof end;
Lm12: for a being Nat
for fs being FinSequence holds dom (Del (fs,a)) c= dom fs
proof end;
definition
let D be non empty set ;
let a be Nat;
let fs be FinSequence of D;
:: original: Del
redefine func Del (fs,a) -> FinSequence of D;
coherence
Del (fs,a) is FinSequence of D
proof end;
end;
definition
let D be non empty set ;
let D1 be non empty Subset of D;
let a be Nat;
let fs be FinSequence of D1;
:: original: Del
redefine func Del (fs,a) -> FinSequence of D1;
coherence
Del (fs,a) is FinSequence of D1
proof end;
end;
Lm13: for a being Nat
for fs1, fs2 being FinSequence holds
( ( a <= len fs1 implies Del ((fs1 ^ fs2),a) = (Del (fs1,a)) ^ fs2 ) & ( a >= 1 implies Del ((fs1 ^ fs2),((len fs1) + a)) = fs1 ^ (Del (fs2,a)) ) )
proof end;
Lm14: for fs being FinSequence
for v being object holds
( Del ((<*v*> ^ fs),1) = fs & Del ((fs ^ <*v*>),((len fs) + 1)) = fs )
proof end;
theorem :: WSIERP_1:19
for v1, v2, v3 being object holds
( Del (<*v1*>,1) = {} & Del (<*v1,v2*>,1) = <*v2*> & Del (<*v1,v2*>,2) = <*v1*> & Del (<*v1,v2,v3*>,1) = <*v2,v3*> & Del (<*v1,v2,v3*>,2) = <*v1,v3*> & Del (<*v1,v2,v3*>,3) = <*v1,v2*> )
proof end;
Lm15: for fs being FinSequence st 1 <= len fs holds
( fs = <*(fs . 1)*> ^ (Del (fs,1)) & fs = (Del (fs,(len fs))) ^ <*(fs . (len fs))*> )
proof end;
Lm16: for a being Nat
for ft being FinSequence of REAL st a in dom ft holds
(Product (Del (ft,a))) * (ft . a) = Product ft
proof end;
theorem :: WSIERP_1:20
for a being Nat
for ft being FinSequence of REAL st a in dom ft holds
(Sum (Del (ft,a))) + (ft . a) = Sum ft
proof end;
theorem :: WSIERP_1:21
for a being Nat
for fp being FinSequence of NAT st a in dom fp holds
(Product fp) / (fp . a) is Element of NAT
proof end;
theorem Th22: :: WSIERP_1:22
for q being Rational holds numerator q, denominator q are_coprime
proof end;
theorem :: WSIERP_1:23
for a being Nat
for k being Integer
for q being Rational st q = k / a & a <> 0 & k,a are_coprime holds
( k = numerator q & a = denominator q )
proof end;
theorem Th24: :: WSIERP_1:24
for a, b being Nat st ex q being Rational st a = q |^ b holds
ex k being Integer st a = k |^ b
proof end;
theorem Th25: :: WSIERP_1:25
for a, d being Nat st ex q being Rational st a = q |^ d holds
ex b being Nat st a = b |^ d
proof end;
theorem :: WSIERP_1:26
for a, b, e being Nat st e > 0 & a |^ e divides b |^ e holds
a divides b
proof end;
theorem Th27: :: WSIERP_1:27
for a, b being Nat ex m, n being Integer st a gcd b = (a * m) + (b * n)
proof end;
theorem Th28: :: WSIERP_1:28
for m, n being Integer ex m1, n1 being Integer st m gcd n = (m * m1) + (n * n1)
proof end;
theorem Th29: :: WSIERP_1:29
for k, m, n being Integer st m divides n * k & m gcd n = 1 holds
m divides k
proof end;
theorem :: WSIERP_1:30
for a, b, c being Nat st a gcd b = 1 & a divides b * c holds
a divides c by Th29;
theorem Th31: :: WSIERP_1:31
for a, b being Nat st a <> 0 & b <> 0 holds
ex c, d being Nat st a gcd b = (a * c) - (b * d)
proof end;
theorem :: WSIERP_1:32
for a, b, f, g being Nat st f > 0 & g > 0 & f gcd g = 1 & a |^ f = b |^ g holds
ex e being Nat st
( a = e |^ g & b = e |^ f )
proof end;
theorem Th33: :: WSIERP_1:33
for k, m, n being Integer holds
( ex x, y being Integer st (m * x) + (n * y) = k iff m gcd n divides k )
proof end;
theorem :: WSIERP_1:34
for k, m, n, m1, n1 being Integer st m <> 0 & n <> 0 & (m * m1) + (n * n1) = k holds
for x, y being Integer st (m * x) + (n * y) = k holds
ex t being Integer st
( x = m1 + (t * (n / (m gcd n))) & y = n1 - (t * (m / (m gcd n))) )
proof end;
theorem :: WSIERP_1:35
for a, b, c, d being Nat st a gcd b = 1 & a * b = c |^ d holds
ex e, f being Nat st
( a = e |^ d & b = f |^ d )
proof end;
:: Chinese remainder theorem
:: Chinese Remainder Theorem
theorem Th36: :: WSIERP_1:36
for fp being FinSequence of NAT
for d being Nat st ( for a being Nat st a in dom fp holds
(fp . a) gcd d = 1 ) holds
(Product fp) gcd d = 1
proof end;
theorem :: WSIERP_1:37
for fp being FinSequence of NAT st len fp >= 2 & ( for b, c being Nat st b in dom fp & c in dom fp & b <> c holds
(fp . b) gcd (fp . c) = 1 ) holds
for fr being FinSequence of INT st len fr = len fp holds
ex fr1 being FinSequence of INT st
( len fr1 = len fp & ( for b being Nat st b in dom fp holds
((fp . b) * (fr1 . b)) + (fr . b) = ((fp . 1) * (fr1 . 1)) + (fr . 1) ) )
proof end;
Lm17: for k, m being Integer holds
( k divides m iff k divides |.m.| )
proof end;
Lm18: for m, n being Integer holds (m * n) mod n = 0
proof end;
Lm19: for k, m, n being Integer st k mod n = m mod n holds
(k - m) mod n = 0
proof end;
Lm20: for m, n being Integer st n <> 0 & m mod n = 0 holds
n divides m
proof end;
Lm21: for x being Integer holds
( ( 1 < x implies ( 1 < sqrt x & sqrt x < x ) ) & ( 0 < x & x < 1 implies ( 0 < sqrt x & sqrt x < 1 & x < sqrt x ) ) )
proof end;
:: Thue Theorem
theorem :: WSIERP_1:38
for a being Nat
for k being Integer st a <> 0 & a gcd k = 1 holds
ex b, e being Nat st
( 0 <> b & 0 <> e & b <= sqrt a & e <= sqrt a & ( a divides (k * b) + e or a divides (k * b) - e ) )
proof end;
theorem :: WSIERP_1:39
for a being Nat
for fs being FinSequence holds dom (Del (fs,a)) c= dom fs by Lm12;
theorem :: WSIERP_1:40
for fs being FinSequence
for v being object holds
( Del ((<*v*> ^ fs),1) = fs & Del ((fs ^ <*v*>),((len fs) + 1)) = fs ) by Lm14;
theorem :: WSIERP_1:41
for k being Integer
for n being Nat st n > 0 & k mod n <> 0 holds
- (k div n) = ((- k) div n) + 1
proof end;
theorem :: WSIERP_1:42
for k being Integer
for n being Nat st k mod n = 0 holds
- (k div n) = (- k) div n
proof end;
Lm4: for i1, i2, i3 being Integer st i1 divides i2 & i1 divides i3 holds
i1 divides i2 - i3
proof end;
theorem :: WSIERP_1:43
for i1, i2, i3 being Integer st i1,i2 are_congruent_mod i3 holds
i1 gcd i3 = i2 gcd i3
proof end;
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# Number 1632321604
### Properties of number 1632321604
Cross Sum:
Factorization:
2 * 2 * 20201 * 20201
Divisors:
Count of divisors:
Sum of divisors:
2856704221
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
Base 16 (Hexadecimal):
614b4044
Base 32:
1gkmg24
sin(1632321604)
-0.048416167776764
cos(1632321604)
0.99882724967725
tan(1632321604)
-0.048473014520187
ln(1632321604)
21.213269135349
lg(1632321604)
9.2128057286106
sqrt(1632321604)
40402
Square(1632321604)
### Number Look Up
Look Up
1632321604 (one billion six hundred thirty-two million three hundred twenty-one thousand six hundred four) is a very unique figure. The cross sum of 1632321604 is 28. If you factorisate the number 1632321604 you will get these result 2 * 2 * 20201 * 20201. 1632321604 has 9 divisors ( 1, 2, 4, 20201, 40402, 80804, 408080401, 816160802, 1632321604 ) whith a sum of 2856704221. 1632321604 is not a prime number. The number 1632321604 is not a fibonacci number. The figure 1632321604 is not a Bell Number. The figure 1632321604 is not a Catalan Number. The convertion of 1632321604 to base 2 (Binary) is 1100001010010110100000001000100. The convertion of 1632321604 to base 3 (Ternary) is 11012202111112021201. The convertion of 1632321604 to base 4 (Quaternary) is 1201102310001010. The convertion of 1632321604 to base 5 (Quintal) is 11320333242404. The convertion of 1632321604 to base 8 (Octal) is 14122640104. The convertion of 1632321604 to base 16 (Hexadecimal) is 614b4044. The convertion of 1632321604 to base 32 is 1gkmg24. The sine of the number 1632321604 is -0.048416167776764. The cosine of the figure 1632321604 is 0.99882724967725. The tangent of the number 1632321604 is -0.048473014520187. The square root of 1632321604 is 40402.
If you square 1632321604 you will get the following result 2664473818885132816. The natural logarithm of 1632321604 is 21.213269135349 and the decimal logarithm is 9.2128057286106. You should now know that 1632321604 is great figure!
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https://mathoverflow.net/questions/58227/how-to-solve-the-linearized-navier-stokes-equations-in-lp
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# How to solve the linearized Navier-Stokes equations in L^P?
Let $\Omega\subset \mathbb{R}^3$ be an open set with smooth boundary $\partial \Omega$. Consider the following linearized Navier-Stokes equations in $Q_T=\Omega\times (0,T)$ for an arbitrarily fixed $T\in (0,\infty)$, $$u_t-\Delta u+a(x,t)u+b\cdot \nabla u+\nabla p=f(x,t),\text{div } u=0$$ with the initial and boundary conditions $u(x,0)=0, \left.u(x,t)\right|_{\partial \Omega\times (0,T)}=0$. Here $u(x,t)=(u^1(x,t),u^2(x,t),u^3(x,t))$ and $p(x,t)$ denote the unknown velocity and pressure respectively, $a(x,t)$ and $b(x,t)$ denote the given coefficients.
Question: Suppose that $$a\in L^r(0,T; L^s(\Omega)), b\in L^{r_1}(0,T; L^{s_1}(\Omega)),$$ where $2/r+3/s<2$, $2/r_1+3/s_1<1$, and $f(x,t)\in C_0^\infty(\Omega\times (0,T))$, can we solve the above equations in arbitrary $L^p$? Can we get the estimates such as $$\|u_t\|_{L^p(Q_T)}+\|D^2 u\|_{L^p(Q_T)}+\|u\|_{L^p(Q_T)}\leq \|f\|_{L^p(Q_T)}?$$
Solonnikov dealed with this problem in his paper "Estimates for solution of nonstationary Navier-Stokes equations" (http://www.springerlink.com/index/N8374858XNT22P11.pdf). However, I can not verify his proof (Page 487 to Page 489).
Who can help me? Any comment will be deeply appreciated.
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https://www.physicsforums.com/threads/differentials-in-a-derivation.266342/
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# Differentials in a derivation
1. Oct 22, 2008
### yuiop
I am trying to understand a derivation posted by Pervect a long time ago (I suspect Pevect is no longer active) that involves differentiation and I was hoping someone here could fill in some of the steps to make it clearer.
The original posts by Pervect are here:
https://www.physicsforums.com/showpost.php?p=1020560&postcount=5
I have taken the liberty of posting a cut down version of Pervects posts here. I have added equation numbers and comment. The comments in blue are just notes and the comments in magenta are the places where I am stuck and need some more explanation.
Last edited: Oct 23, 2008
2. Oct 23, 2008
### yuiop
Part 2
I had to split my original post into two because the latex was not displaying properly.
This is the second post where Pervect attacks the problem from a slightly different angle, but I am still missing the trick where he obtains the differentials.
Obviously I am missing a trick somewhere. Can anyone enlighten me?
Maybe this should have been posted in the Relativity forum, but I am only interested in the mathematical aspects that seem to require the knowledge of a expert in differential equations so I posted it here.
3. Oct 23, 2008
### HallsofIvy
Staff Emeritus
He is not using differential equations here but just the definition of "differential"
From
$$X+ T= (\frac{r}{2m}-1)^{1/2}e^{\frac{r}{4m}}e^{\frac{t}{4m}}$$
take the differential of both sides. On the right, we need to apply the product rule to a product of three terms: Since the first and second are functions of r and the third term a function of t,we basically get d(f(r)g(r)h(t))= f'(r)g(r)h(t)dr+ f(r)g'(r)h(t)dr+ f(r)g(r)h'(r)dt.
More specifically,
$$dX+ dT= \frac{1}{2}\left(\frac{r}{2m}-1\right)^{-1/2}e^{\frac{r}{4m}}e^{\frac{t}{4m}}dr$$
$$+ \frac{1}{4m}\left(\frac{r}{2m}-1\right)^{1/2}e^{\frac{r}{4m}}e^{\frac{r}{4m}}dr$$
$$+ \frac{1}{4m}\left(\frac{r}{2m}-1\right)^{1/2}e^{\frac{r}{4m}}e^{\frac{r}{4m}}dt$$
Do the same with the "X- T" equation and solve the two equations for dr and dt.
4. Oct 23, 2008
### cristo
Staff Emeritus
For the next part: eqns 2.5 and 2.6 are correct, and just arise from the definition of a differential: $$dr=\frac{\partial r}{\partial X}dX+\frac{\partial r}{\partial T} dT$$
Your last question: I'm not sure what you're asking. One can evaluate the above, since one has an expression of r in terms of X and T, so can take the derivatives and plug into the above.
5. Oct 24, 2008
### yuiop
Hi,
thanks for the help and sorry for the delay responding. I have spent the afternoon trying to learn differential calculus, product rules and multivariate calculus from the internet. I had no idea there was so much to it I can not honestly say I understand 10 percent of it but I hope I have picked up enough to address this problem. I found the following resources very helpful:
Multivariate calculus
Rules of calculus
Calculus - Wikibook
Product rule - Wikipedia
Back to the problem in hand.
Applying the product rule to (X+T) with respect to r:
$$X+ T= (\frac{r}{2m}-1)^{1/2}e^{\frac{r}{4m}}e^{\frac{t}{4m}}$$
I get:
\begin{align*} dX+ dT &= \frac{1}{4m}\left(\frac{r}{2m}-1\right)^{-1/2}e^{\frac{r}{4m}}e^{\frac{t}{4m}}(dr)&\\ &+ \frac{1}{4m}\left(\frac{r}{2m}-1\right)^{1/2}e^{\frac{r}{4m}}e^{\frac{t}{4m}}(dr)&\\ &+ \frac{1}{4m}\left(\frac{r}{2m}-1\right)^{1/2}e^{\frac{r}{4m}}e^{\frac{t}{4m}}(dt)& (Eq 4.1) \end{align*}
O.K.
$$X- T= (\frac{r}{2m}-1)^{1/2}e^{\frac{r}{4m}}e^{\frac{-t}{4m}}$$
\begin{align*} dX- dT &= \frac{1}{4m}\left(\frac{r}{2m}-1\right)^{-1/2}e^{\frac{r}{4m}}e^{\frac{-t}{4m}}(dr)&\\ &+ \frac{1}{4m}\left(\frac{r}{2m}-1\right)^{1/2}e^{\frac{r}{4m}}e^{\frac{-t}{4m}}(dr)&\\ &- \frac{1}{4m}\left(\frac{r}{2m}-1\right)^{1/2}e^{\frac{r}{4m}}e^{\frac{-t}{4m}}(dt)& (Eq 4.2) \end{align*}
Solve (Eq 4.1) for $$e^{\frac{t}{4m}}$$ :
$$e^{\frac{t}{4m}} = \frac{4m(dX+dT)}{e^{\frac{r}{4m}}\left[ dr\left(\left(\frac{r}{2m}-1\right)^{1/2}+\left(\frac{r}{2m}-1\right)^{-1/2}\right) + dt\left(\frac{r}{2m}-1\right)^{1/2}\right] }}$$
and do the same for (Eq 4.2):
$$e^{\frac{t}{4m}} = \frac{e^{\frac{r}{4m}}\left[ dr\left(\left(\frac{r}{2m}-1\right)^{1/2}+\left(\frac{r}{2m}-1\right)^{-1/2}\right) - dt\left(\frac{r}{2m}-1\right)^{1/2}\right] }{4m(dX-dT)}$$
Since both the above equations are equal to $$e^{\frac{t}{4m}}$$ they can be equated to each other to eliminate the pesky t term and by applying the difference of squares rule the following is obtained:
$$16m^2(dX^2-dT^2) = e^{\frac{r}{2m}}\left[dr^2\left(\left(\frac{r}{2m}-1\right)^{1/2}+\left(\frac{r}{2m}-1\right)^{-1/2}\right)^2 -dt^2 \left(\frac{r}{2m}-1\right) \right]$$
By inspection it is obvious that the last equation is close to a solution and this suggests a shortcut that avoids having to find dr and dt individually.
Multiply both sides by $$\frac{2m }{ r} e^{\frac{-r}{2m}}$$ :
$$\frac{32m^3 e^{\frac{-r}{2m}} }{r} (dX^2-dT^2) = dr^2\left(1-\frac{2m}{r}\right)^{-1} -dt^2 \left(1-\frac{2m}{r}\right)$$
which is the Kruskal-Szekeres metric on the left hand side and the Schwazschild metric on the right hand side.
6. Oct 24, 2008
### yuiop
Hi cristo,
could you flesh this out a bit and explain why this form is used here in this particular example or point to some online reference material? I'm afraid I'm still not "getting it".
Is it because the cooordinate equations for R and T are being treated as parametric equations with respect to r and t for the purposes of differentiation?
I think we have one solution thanks to help from HallsofIvy but I never feel comfortable with a solution until I can get it from at least two different angles. Sorry if I being obtuse about all this but differentiation (at this level of complexity) is all fairly new to me.
<EDIT>
This is how it seems to me but I readily admit I am probably out of my depth here.
Starting with:
$$(X +T ) = (X+T)$$
Differentiating both sides with respect to r:
$$\frac{d}{dr} (X+T) = \frac{\partial X}{\partial r} + \frac{\partial T}{\partial r}$$
$$(dX + dT) = \left( \frac{\partial X}{\partial r} + \frac{\partial T}{\partial r} \right) dr$$
$$\frac{ (dX + dT) }{\left( \frac{\partial X}{\partial r} + \frac{\partial T}{\partial r} \right)} = dr$$
$$\partial r \frac{ ( dX + dT) }{\left( \partial X + \partial T \right)} = dr$$
$$dr = \frac{ \partial r }{\left( \partial X + \partial T \right)}dX + \frac{ \partial r }{\left( \partial X + \partial T \right)}dT$$
...then again..maybe you can not throw differential symbols around like that..
Last edited: Oct 24, 2008
7. Oct 26, 2008
### yuiop
OK, I have finally figured out what you were trying to tell me here. Sorry for being so slow I have posted a few corrections and a complete solution below which I hope will clear up any confusion for anyone hoping to learn anything from this thread. Thanks again Christo (and HallsofIvy) for the pointers in the right direction
I now realise I was on completely the wrong track in my last post and that post should be disregarded... I had the wrong end of the stick
This is the correct way to do it:
Starting with the definition of the Kruskal-Szekeres coordinates:
$$X=e^{(r/4M)}(r/2M-1)^{1/2}cosh(t/4M)$$
$$T=e^{(r/4M)}(r/2M-1)^{1/2}sinh(t/4M)$$
(there was a typo in post #1)
Solving these two equations for r gives:
$$r = 2M LambertW((X^2-T^2)/e) + 2M$$
Note: The lambertW function is sometimes know as the ProductLog or Omega function. Solving for r with complicated expressions like this is best done with mathematical software such as this free online one here: http://www.quickmath.com/
Now dr is found by finding the sum of the partial differential of (2M*LambertW((X^2-T^2)/e) + 2M) with respect to X plus the partial differential of (2M*LambertW((X^2-T^2)/e) + 2M) with respect to T using the product rule
$$dr=\frac{\partial r}{\partial X}dX+\frac{\partial r}{\partial T} dT$$
(Exactly as cristo said)
$$dr=\frac{4M LambertW((X^2-T^2)/e)(XdX-TdT)}{(LambertW((X^2-T^2)/e)+1)(X^2-T^2) }$$
The first two coordinate equations can also be solved for t in terms of X and T to give:
$$t = 2M ln[(X+T)/(X-T)]$$
and as before
$$dt=\frac{\partial t}{\partial X}dX+\frac{\partial t}{\partial T} dT$$
$$dt = \frac{4m(XdT-TdX)}{(X^2-T^2) }$$
Now that expressions for dr and dt in terms of X and T have been obtained they can be plugged in to the Schwarzschild metric
$$dS^2 = \left(1-\frac{2M}{r} \right)^{-1} (dr)^2 - \left(1-\frac{2M}{r} \right)(dt)^2$$
to obtain:
$$dS^2 = \frac{16M^2 (dX^2-dT^2) LambertW((X^2-T^2)/e)}{(LambertW((X^2-T^2)/e)+1)(X^2-T^2)}$$
which is the expression originally obtained by Pervect. (Anyone know what happened to Pervect? I miss his knowledgeable contributions to PF )
By substituting (r/2m-1)exp(r/2m) for (x^2-T^2) and (r/2m-1) for LambertW(X^2-T^2)/e) into the last equation as suggested by Pervect, the Kruskal-Szekeres metric is obtained:
$$dS^2 = \frac{32 M^3 e^{-r/2M}}{r}(-dT^2 + dX^2)$$
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# Order and degree of the differential equation (d3ydx3)3/2+(d3ydx3)−2/3=0 are respectively
A
2, 9
B
3, 6
C
3, 4
D
3, 1
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Step by step video, text & image solution for Order and degree of the differential equation ((d^(3)y)/(dx^(3)))^(3//2)+((d^(3)y)/(dx^(3)))^(-2//3)=0 are respectively by Maths experts to help you in doubts & scoring excellent marks in Class 12 exams.
Updated on:21/07/2023
### Knowledge Check
• Question 1 - Select One
## The order and degree of the differential equation (d2ydx2)16−(dydx)13=0 are respectively .
A3,2
B2,3
C6,3
D3,1
• Question 2 - Select One
## The order and degree of the differential equation (1+3dydx)2/3=4d3ydx3are
A1,23
B3,1
C3,3
D1,2
• Question 3 - Select One
## What is the degree of the differential equation (d3ydx3)32=(d2ydx2)2 ?
A1
B2
C3
D4
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Você está na página 1de 90
# No.
1
2017
CONTENTS
Articles
## On Some Diophantine Equations
by Nicolae Papacu 2
## A note on a problem from the 2012 Spanish Mathematical
by Jaume Soler and Jaume Franch 9
Problems
## Medium–Hard Problems: MH41–MH46 20
Mathlessons
Forward-Backward Induction
by J. L. Díaz-Barrero 25
Solutions
## Medium–Hard Problems: MH35–MH40 57
Volume 4, No. 1, Spring 2017 1
Articles
Arhimede Mathematical Journal aims to publish interesting and
attractive papers with elegant mathematical exposition. Articles
should include examples, applications and illustrations, whene-
ver possible. Manuscripts submitted should not be currently sub-
mitted to or accepted for publication in another journal.
## Please, send submittals to José Luis Díaz-Barrero, Enginyeria
Civil i Ambiental, UPC BARCELONATECH, Jordi Girona 1-3, C2,
08034 Barcelona, Spain, or by e-mail to
jose.luis.diaz@upc.edu
2 Arhimede Mathematical Journal
On Some Diophantine
Equations
Nicolae Papacu
Abstract
## In this note we give integer solutions of generalizations of well-
known diophantine equations, such as 11x − 2x = y 2 , 27x −
2x = y 2 , 3x − 2x = y 2 or 3x − 2y = z 2 . Actually, equations of
the form (p2 +2)x ±2x = y 2 and (p2 +2)x ±2y = z 2 are studied
and, whenever possible, solved in nonnegative integers.
1 The Equations
Hereafter, we present some nonstandard diophantine equations
and discuss their solutions in nonnegative integers. We begin with
the following.
## Theorem 1. If p ≥ 1 is an odd integer, then the only solutions to
the equation
(p2 + 2)x − 2x = y 2
in nonnegative integers are (x, y) = (0, 0) and (x, y) = (1, p).
## Proof. Since p is an odd positive integer, then p2 ≡ 1 (mod 8)
and p2 + 2 ≡ 3 (mod 8). Then, (p2 + 2)x ≡ 1 (mod 8) when x is
even and (p2 + 2)x ≡ 3 (mod 8) when x is odd. By inspection,
we observe that (x, y) = (0, 0) and (x, y) = (1, p) are solutions
of the given equation in positive integers. We claim that there are
no other solutions in positive integers. Indeed, for x ≥ 2 we have
that y is an odd positive integer. Next we will see that x is even.
Volume 4, No. 1, Spring 2017 3
## Suppose that x ≥ 3 is odd, then we have (p2 + 2)x ≡ 3 (mod 8).
On the other hand, 2x ≡ 0 (mod 8), (p2 + 2)x − 2x ≡ 3 (mod 8)
and y 2 ≡ 1 (mod 8). As a consequence, we obtain
(p2 + 2)x − 2x 6= y 2 .
tion becomes
## On account of the well-known fact that for a, b odd integers with
gcd (a, b) it holds that (a − b, a + b) = 2(a, b), we have
that d = 2.
## The facts that ((p2 + 2)t − y) ((p2 + 2)t + y) = 22t and d = 2
mean that
(p2 + 2)t − y = 2,
(p2 + 2)t + y = 22t−1 .
## (p2 + 2)t = 22t−2 + 1.
Now we claim that the preceding equation does not have positive
integer solutions. Indeed, for t = 1 we have p = 0 which is
not possible, and for t ≥ 2 we have 22t−1 + 1 ≡ 1 (mod 4), or
(p2 +2)t ≡ 1 (mod 4), which means that t is even. Putting t = 2w
in the last equation, we get (p2 + 2)2w = 24w−2 + 1 or
## From the preceding it immediately follows that (p2 +2)w − 22w−1 =
(p2 + 2)w + 22w−1 = 1. But this is not possible, and the proof is
complete.
4 Arhimede Mathematical Journal
## Theorem 2. If p ≥ 1 is an odd integer, then the only solutions to
the equation
(p2 − 2)x + 2x = y 2
in nonnegative integers are (x, y) = (1, p).
## 1. If p = 1 then the equation becomes (−1)x + 2x = y 2 . For
x = 1 we have y = 1, and for x ≥ 2 we consider two cases:
(a) If x = 2t, then 1 + 22t = y 2 or, equivalently, (y − 2t ) (y +
2t ) = 1, from which y − 2t = y + 2t = 1, which is impos-
sible.
(b) If x = 2t + 1 then −1 + 22t+1 = y 2 , which is impossible
on account that −1 + 22t+1 ≡ 7 (mod 8) and y 2 ≡ 1
(mod 8). So, −1 + 22t+1 6= y 2 .
2. If p ≥ 3 the given equation has no solutions for x = 0, and
for x = 1 we have y = p. Suppose that x ≥ 2 and observe
that y is odd. We will prove that x is even. Indeed, if x ≥ 3
is odd, then (p2 − 2)x ≡ 7 (mod 8). Since 2x ≡ 0 (mod 8),
then (p2 − 2)x + 2x ≡ 7 (mod 8), and since y 2 ≡ 1 (mod 8),
then (p2 − 2)x + 2x 6= y 2 . So x = 2t and the equation becomes
## Now we will see that the above equation has no solutions in
positive integers. Indeed, for p ≥ 3 we have
2 2
(p2 − 2)z < (p2 − 2)2z + 22z < (p2 − 2)z + 1 .
## The LHS inequality holds trivially. For the RHS inequality, we
have
2
(p2 − 2)2z + 22z < (p2 − 2)z + 1 ⇐⇒ 22z < 2 (p2 − 2)z + 1
## because (p2 − 2)z + 1 > 7z > 22z . For p = 1 the equation
becomes 1 + 22t = y 2 . It does not have positive integer solu-
tions on account that (y − 2t ) (y + 2t ) = 1 ⇐⇒ (y − 2t ) =
(y + 2t ) = 1, which is impossible. Thus, the only solutions
are (x, y) = (1, p), and this completes the proof.
Volume 4, No. 1, Spring 2017 5
## Next we study the solutions to an equation in three variables.
Theorem 3. If p ≥ 3 is an odd integer, then the only solutions to
the equation
(p2 + 2)x − 2y = z 2
in nonnegative integers are (x, y, z) = (0, 0, 0) and (x, y, z) =
(1, 1, p). If p = 1, there are three additional solutions.
Proof. First, we consider some particular cases. For x = 0 the
equation becomes 1 − 2y = z 2 with solutions y = z = 0. For
x = 1 we have p2 + 2 − 2y = z 2 . From p2 ≡ 1 (mod 8) and z 2 ≡ 1
(mod 8) it follows that 2y − 2 ≡ 0 (mod 8), with solutions y = 1
and z = p. Therefore, we have the solutions (x, y, z) = (0, 0, 0)
and (x, y, z) = (1, 1, p).
For y = 0 the given equation becomes (p2 + 2)x − 1 = z 2 . We
distinguish two cases:
1. If x ≥ 2 is odd, then we have (p2 + 2)x ≡ 3 (mod 8) or
(p2 + 2)x − 1 ≡ 2 (mod 8) and, therefore, z 2 ≡ 2 (mod 8),
which is impossible.
2. If x is even, say x = 2t, t ∈ N, then we have (p2 +2)2t −1 = z 2
or
(p2 + 2)t − z (p2 + 2)t + z = 1.
## From the preceding, (p2 + 2)t − z = (p2 + 2)t + z = 1, which
is impossible.
For y = 1 the given equation becomes (p2 + 2)x − 2 = z 2 . If x ≥ 2
is even then the equation has no solutions because (p2 + 2)x ≡ 1
(mod 8) and z 2 + 2 ≡ 3 (mod 8). For x = 1 we have z = p and
then the solution is (x, y, z) = (1, 1, p). If x ≥ 3 is an odd integer,
then it has at least one odd prime factor, say q . Thus, x = qt and
then (p2 + 2)x = wq , where w = (p2 + 2)t . Under this hypothesis,
the equation (p2 + 2)x − 2 = z 2 becomes z 2 + 2 = wq with q an
odd prime. This equation was solved in [1] and has the solution
(z, q, w) = (5, 3, 3). Then, w = (p2 + 2)t = 3, from which we get
t = p = 1. So, for p = 1 we have the solution (x, y, z) = (3, 1, 5).
If x, y ≥ 2 then z is an odd number. We claim that x is even.
Indeed, if x ≥ 3 is odd we have (p2 + 2)x ≡ 3 (mod 8) and 2y ≡ 0
6 Arhimede Mathematical Journal
(mod 8), from which it follows that (p2 +2)x −2y ≡ 3 (mod 8). On
the other hand, y 2 ≡ 1 (mod 8) and, therefore, (p2 + 2)x − 2y 6=
y 2 . Hence, x is even, as claimed. Putting x = 2t, t ∈ N, in
(p2 + 2)x − 2y = z 2 yields (p2 + 2)2t − 2y = z 2 . The last equation
may be written as
Since
## then d | 2z . From d | ((p2 + 2)t − z), d | ((p2 + 2)t + z) and
((p2 + 2)t − z) ((p2 + 2)t + z) = 2y it results that d | 2y . Hence,
we have d | 2z and d2 | 2y , from which we get d = 2 on account
that d is even and z is odd. From ((p2 + 2)t − z) ((p2 + 2)t + z) =
2y and d = 2 we obtain
(p2 + 2)t − z = 2,
(p2 + 2)t + z = 2y−1 .
## Subtracting the above equations we get z = 2y−2 − 1, and adding
them up yields (p2 + 2)t = 2y−2 + 1. Now we distinguish the
following cases:
## 1. If y = 2 we obtain (p2 + 2)t = 2, which is impossible because
p ≥ 1.
2. If y = 3 we have (p2 + 2)t = 3, from which p = 1, t = 1,
x = 2t = 2 and z = 2y−2 − 1 = 1. So (x, y, z) = (2, 3, 1) is a
solution.
3. If y = 4 the equation (p2 + 2)t = 5 has no solutions.
4. For y ≥ 5 we have 2y−2 + 1 ≡ 1 (mod 8) and then (p2 + 2)t ≡
1 (mod 8), so t is even, say t = 2v , v ∈ N. Substituting in
the equation (p2 +2)t = 2y−2 +1 we obtain (p2 +2)2v = 2y−2 +1
or (p2 + 2)2v − 1 = 2y−2 and ((p2 + 2)v − 1) ((p2 + 2)v + 1) =
2y−2 . Since ((p2 + 2)v − 1) ((p2 + 2)v + 1) = 2, then we have
(p2 + 2)v − 1 = 2,
(p2 + 2)v + 1 = 2y−3 .
Volume 4, No. 1, Spring 2017 7
## From the first equation we obtain p = 1 and v = 1, and the
equation (p2 + 2)2v = 2y−2 + 1 becomes 9 = 2y−2 + 1, from
which we get y = 5. Combining the preceding we have t = 2,
x = 4 and z = 2y−2 − 1 = 7.
## So for p = 1 we have the solution (x, y, z) = (4, 4, 7). Finally, the
equation (p2 + 2)x − 2y = z 2 has the solutions (x, y, z) = (0, 0, 0)
and (x, y, z) = (1, 1, p), as we wanted to prove. In the particular
case when p = 1, equation 3x − 2y = z 2 has the solutions
{(0, 0, 0), (1, 1, 1), (3, 1, 5), (2, 3, 1), (4, 5, 7)}.
## Question (Open problem). Let p ≥ 1 be an odd integer. What are
the solutions to
(p2 − 2)x + 2y = z 2
in nonnegative integers?
We will now solve all cases except the case where x ≥ 3 is odd
and y = 1. The only solutions then are (x, y, z) = (0, 3, 3) and
(x, y, z) = (1, 1, p).
## Proof. First, we consider the cases when x ∈ {0, 1} and y ∈
{0, 1}. If x = 0, we have 1 + 2y = z 2 or (z − 1)(z + 1) = 2y . Since
(z − 1, z + 1) = 2 and z − 1 < z + 1, then we have z − 1 = 2 and
z + 1 = 2y−1 , from which y = 3, z = 3 follow. For x = 1, we have
p2 − 2 + 2y = z 2 . Since p2 ≡ 1 (mod 8) and z 2 ≡ 1 (mod 8), then
22 − 2 ≡ 0 (mod 8), so y = 1 and z = p. Hence, (1, 1, p) is a
solution. For y = 0 the given equation becomes (p2 −2)x +1 = z 2 ,
or z 2 − (p2 − 2)x = 1. Then, z = 3, p2 − 2 = 2, and x = 3. This
means that p = 2 on account of Catalan-Mihailescu Theorem,
which states that the equation xy − z t = 1, where x, y , z and
t are positive integers bigger than one, has the unique solution
(x, y, z, t) = (3, 2, 2, 3). The value p = 2 is not possible because
p is odd. For y = 1 we have (p2 − 2)x + 2 = z 2 . If x = 1 we
have z = p, and (x, y, z) = (1, 1, p) are solutions. If x is even,
then (p2 − 2)x + 2 ≡ 3 (mod 8) and z 2 ≡ 1 (mod 8). Therefore,
the given equation has no solutions. For x, y ≥ 2 we have that
8 Arhimede Mathematical Journal
## z is odd. If we suppose that x ≥ 3, then (p2 − 2)x ≡ 7 (mod 8),
but 2y ≡ 0 (mod 4). Thus, (p2 − 2)x + 2y ≡ 3 (mod 4) and
y 2 ≡ 1 (mod 8), from which (p2 − 2)x + 2y 6= z 2 follows. From
the preceding we conclude that x is even, say x = 2t, and the
equation may be written as (p2 − 2)2t + 2y = z 2 or
## Let d = (z − (p2 − 2)t , z + (p2 − 2)t ). It is an even number sucht-
hat d | 2z . On the other hand, from d | (z − (p2 − 2)t ) and
d | (z + (p2 − 2)t ) and the equation results that d2 | 2y . Since
d | 2z and d2 | 2y we get that d = 2 on account that z is odd.
## Acknowledgements: The author thanks the anonymous referees
for their comments and suggestions that have greatly improved
the final version of this paper.
References
[1] Cucurezeanu, I. Ecuaţii în numere întegri. In Romanian. Edi-
tura Aramis, Bucureşti, 2006.
[2] Cucurezeanu, I. Pătrate şi cuburi perfecte de numere întegri.
In Romanian. Editura Gil, Zalău, 2007.
[3] Sierpiński, W. Elementary theory of numbers. Translated from
Polish by A. Hulanicki. Monografie Matematyczne, Tom 42.
Państwowe Wydawnictwo Naukowe, Warsaw, 1964, p. 480.
Nicolae Papacu
Department of Mathematics
“Mihai Viteazul” National College
Slobozia, Romania
papacu1956@yahoo.com
Volume 4, No. 1, Spring 2017 9
## A note on a problem from the
2012 Spanish Mathematical
Jaume Soler and Jaume Franch
Abstract
## This note gives some different solutions to a problem stated in
generalizations and related results are proven.
1 Introduction
The problem was originally stated in the XLVIII Spanish Mathema-
tical Olympiad that took place in Santander in 2012. The original
Problem. Consider the sequence of numbers given by
a2n+1 + 4
a0 = 1, a1 = 5, an+2 = .
an
Prove that an ∈ Z, ∀ n ∈ Z, and give an explicit formula for an .
We will study the generalized sequence
a2n+1 + k − 1
a0 = 1 a1 = k an+2 = , (1)
an
where k > 1 since, for k = 1, the sequence is a constant se-
quence. In Section 2 we will give two proofs of the fact that an ∈ Z,
∀ n ∈ Z. Section 3 contains an explicit formula for an . Finally, in
Section 4, some other interesting assertions for this sequence will
be proven.
10 Arhimede Mathematical Journal
## 2 Proof of the main statement
Two proofs for the main statement will be given. The first one is
based on finding a linear relationship for the sequence of numbers
an . The second one uses the fact that two consecutive members
of the sequence are coprime.
## 2.1 First proof
First proof. Indeed, it will be proven that the generalized sequence
(1) fulfills a linear relationship, from where it is straightforward to
see that an are all integers. The linear relationship that will be
proven is
an+2 = (k + 1)an+1 − an ∀ n ≥ 0. (2)
The proof will be performed by induction. For n = 0,
a21 + (k − 1)
a2 = = k2 + k − 1 = k(k + 1) − 1 = a1 (k + 1) − a0 ,
a0
which is the linear relationship we wanted to prove. Similarly, for
n = 1,
a22 + (k − 1)
a3 = = k3 + 2k2 − k − 1
a1
2
= (k + k − 1)(k + 1) − k = a2 (k + 1) − a1 .
Now, assume that the linear relationship is held for an and an+1 ,
that is,
an+1 = (k + 1)an − an−1 (3)
and
an = (k + 1)an−1 − an−2 . (4)
On substituting
a2n+1 + k − 1
an+2 =
an
Volume 4, No. 1, Spring 2017 11
## in (2), it is equivalent to prove (2) and to prove that
a2n+1 + k − 1
= (k + 1)an+1 − an ,
an
valent to
## (k + 1)2 a2n − 2(k + 1)an an−1 + a2n−1 + k − 1
=(k + 1)2 a2n − (k + 1)an an−1 − a2n .
## Simplification and substitution of a2n−1 + k − 1 = an an−2 at the
left hand side of the equality, and collecting all the terms on the
## Here, an can be simplified since it does not vanish (otherwise the
sequence would not be well defined). Therefore, the last equality
holds if, and only if,
hesis (4).
## 2.2 Second proof
We will state and prove two lemmas that will help us in the proof
of the main result. We assume that a1 , a2 , . . . , an+1 are integers,
and we will show that an+2 is also an integer.
## Lemma 1. For 3 ≤ ` ≤ n + 1, a` and k − 1 are coprime.
12 Arhimede Mathematical Journal
Proof. By induction on `.
Since a3 = k2 + k − 1, a3 − (k − 1) = k2 is an integer. So, if
p divides both a3 and k − 1, it also divides k2 and, therefore, it
divides k. But this is impossible since it is very well-known that
k and k − 1 are coprime.
Now, assume that the statement is true for a` with ` ≤ n. We are
going to prove that the statement also holds for an+1 .
From an+1 an−1 = a2n + k − 1, if an+1 and k − 1 have a common
factor p, then this factor p also divides a2n and, hence, an , which
Lemma 2. For 2 ≤ ` ≤ n + 1, a` and a`−1 are coprime.
Proof. By induction on `.
It is an straightforward computation to check that the statement
is true for ` = 2.
Now, assume the statement holds for ` ≤ n. Let us prove the case
` = n + 1. Assume an+1 and an have a common factor p 6= 1.
Then, from the equality an+1 an−1 − a2n = k − 1 it can be inferred
that p also divides k − 1. Therefore p divides an and k − 1, which
enters in contradiction with the previous lemma.
Second proof of the main statement. Assume that, for ` ≤ n + 1,
a` are integers. We are going to prove that an+2 is also an integer.
Substitute an+1 as a function of an and an−1 from its definition
in the equation
a2 + k − 1
an+2 = n+1 .
an
(a2n + k − 1)2 + a2n−1 (k − 1)
an+2 = (5)
a2n−1 an
a4n + 2(k − 1)a2n + (a2n−1 + k − 1)(k − 1)
=
a2n−1 an
a4n + 2(k − 1)a2n + an an−2 (k − 1)
= .
a2n−1 an
Volume 4, No. 1, Spring 2017 13
## Let Q be the numerator of the previous equalities, Q = a4n + 2(k −
1)a2n + an an−2 (k − 1). By induction hypothesis, Q is an integer
and an divides Q clearly. On the other hand, the numerator in (5)
can be written as
## which is an integer and a multiple of a2n−1 . Since by the previous
lemma an and an−1 are coprime, Then Q is a multiple of an a2n−1 .
## 3 More properties for this sequence
This section is devoted to stating and proving interesting relati-
onships for this sequence of numbers.
## cn+1 = (k + 1)cn − cn−1 + 1 (6)
with c0 = 0 and c1 = 1.
Proof. Since a0 = 1, c0 must be 0, while a1 = k leads to
c1 = 1. We are going to prove the statement by induction. For
n = 2, from a2 = k2 + k − 1 = (k − 1)(k + 2) + 1, c2 = k + 2.
Let us check the equality (6) for n = 2:
c2 = k + 2 = (k + 1) · 1 − 0 + 1 = 6c1 − c0 + 1,
## an+1 = (k + 1)an − an−1
= (k + 1)((k − 1)cn + 1) − ((k − 1)cn−1 + 1)
= (k − 1)((k + 1)cn − cn−1 + 1) + 1,
## from where an+1 = (k − 1)cn+1 + 1.
In other words, not only an ∈ Z, but they are also congruent
with 1 modulo k − 1.
14 Arhimede Mathematical Journal
## 2. For all k ≥ 1 and n ≥ 0,
√ n
1 k−1 k+1+ (k+1)2 −4
an = 2
1+ √ 2
(k+1)2 −4
√ n
1 k−1 k+1− (k+1)2 −4
+ 2
1− √ 2
.
(k+1)2 −4
## Proof. Consider the characteristic polynomial of the linear re-
currence found in Section 2. That is, for an+2 = (k + 1)an+1 −
an , pc (x) = x2 − (k + 1)x + 1. Since its roots are
p
k+1± (k + 1)2 − 4
x= ,
2
## the solution for an is written as
p !n p !n
k+1+ (k+1)2 −4 k+1− (k+1)2 −4
an = A +B .
2 2
## On substituting the initial conditions a0 = 1, a1 = k, a linear
system of two equations and two unknowns is obtained:
1 = A + B, p p
k + 1 + (k + 1)2 −4 k + 1 − (k + 1)2 − 4
k = A +B .
2 2
## The solution for this system is
! !
1 k−1 1 k−1
A= 1+ p B= 1− p .
2 (k + 1)2 − 4 2 (k + 1)2 − 4
## Hence, the explicit formulae for the elements of the sequence
is √ n
1 k−1 k+1+ (k+1)2 −4
an = 2 1 + √ 2
(k+1)2 −4
√ n
1 k−1 k+1− (k+1)2 −4
+ 2 1− √ 2 2
.
(k+1) −4
Volume 4, No. 1, Spring 2017 15
4 Conclusions
In this note, the statement of a problem from the 2012 Spanish
Mathematical Olympiad has been generalized. Two proofs of this
statement have been given, one finding a linear relationship for
the nonlinear sequence defined in the statement, and another one
using primarily some algebraic notions. Additionally, we have pro-
ven that the elements of the sequence are congruent with 1 mo-
dulo k − 1.
References
[1] Goldberg, S. Introduction to difference equations. Second. With
illustrative examples from economics, psychology, and socio-
logy. Dover Publications, Inc., New York, 1986, pp. xiv+260.
ISBN : 0-486-65084-7.
[2] Kelley, W. G. and Peterson, A. C. Difference equations. Second.
An introduction with applications. Harcourt/Academic Press,
San Diego, CA, 2001, pp. x+403. ISBN: 0-12-403330-X.
[3] Levy, H. and Lessman, F. Finite difference equations. Reprint
of the 1961 edition. Dover Publications, Inc., New York, 1992,
pp. viii+278. ISBN: 0-486-67260-3.
Jaume Soler
Department of Civil and Environmental Engineering
Universitat Politècnica de Catalunya–BarcelonaTech
Barcelona, Spain
jaume.soler.villanueva@upc.edu
Jaume Franch
Department of Mathematics
Universitat Politècnica de Catalunya–BarcelonaTech
Barcelona, Spain
jaume.franch@upc.edu
16 Arhimede Mathematical Journal
Problems
This section of the Journal offers readers an opportunity to ex-
change interesting and elegant mathematical problems. Proposals
are always welcome. Please observe the following guidelines when
submitting proposals or solutions:
1. Proposals and solutions must be legible and should appear
on separate sheets, each indicating the name and address of
the sender. Drawings must be suitable for reproduction.
2. Proposals should be accompanied by solutions. An asterisk
(*) indicates that neither the proposer nor the editor has sup-
plied a solution.
Please, send submittals to José Luis Díaz-Barrero, Enginyeria
Civil i Ambiental, UPC BARCELONATECH, Jordi Girona 1-3, C2,
08034 Barcelona, Spain, or by e-mail to
jose.luis.diaz@upc.edu
The section is divided into four subsections: Elementary Pro-
blems, Easy–Medium High School Problems, Medium–Hard High
School Problems, and Advanced Problems mainly for undergra-
duates. Proposals that appeared in Math Contests around the
world and most appropriate for Math Olympiads training are al-
ways welcome. The source of these proposals will appear when
the solutions are published.
## Solutions to the problems stated in this issue should be posted
before
Nov 30, 2017
Volume 4, No. 1, Spring 2017 17
Elementary Problems
E–41. Proposed by Alberto Espuny Díaz, University of Birming-
ham, Birmingham, United Kingdom. Prove that the square of the
perimeter of a rectangle is at least 16 times its area.
## E–42. Proposed by José Luis Díaz-Barrero, BarcelonaTech, Bar-
celona, Spain. For every integer n ≥ 1 let tn denote the n-th
triangular number, defined by tn = n(n+1)2
. Find the values of n
for which
12 + 22 + . . . + n2
t1 + t2 + . . . + tn
is an integer number.
## E–43. Proposed by José Luis Díaz-Barrero, BarcelonaTech, Barce-
lona, Spain. In each square of a 2017 × 2017 chessboard either a
+1 or a −1 is written. Let ri be the product of the numbers lying
on the i-th row, and let cj be the product of the numbers lying on
the j -th column. Show that r1 +r2 +. . .+r2017 +c1 +c2 +. . .+c2017 6=
0.
## E–44. Proposed by José Luis Díaz-Barrero, BarcelonaTech, Bar-
celona, Spain. Find all positive integers that are divisible by 385
and have exactly 385 distinct positive divisors.
## E–45. Proposed by Alberto Espuny Díaz, University of Birming-
ham, Birmingham, United Kingdom. Given a regular pentagon of
side length 1, find the triangle with the largest area contained
inside it.
## E–46. Proposed by José Luis Díaz-Barrero, BarcelonaTech, Bar-
celona, Spain. Let ABCD be a trapezium with bases AB = a
and CD = b, respectively. Let M be a point on AD such that
β M A = α M D for some reals α and β . If the parallel to the
bases drawn from M meets BC at N , then show that
β α
M N ≥ a α+β · b α+β .
18 Arhimede Mathematical Journal
Easy–Medium Problems
EM–41. Proposed by José Luis Díaz-Barrero, BarcelonaTech, Bar-
celona, Spain. Find all solutions of the equation p(x) = q(x),
where p(x) and q(x) are polynomials of degree 2 with leading
coefficient one such that
Xn Xn
p(k) = q(k),
k=1 k=1
## EM–42. Proposed by José Luis Díaz-Barrero, BarcelonaTech, Bar-
celona, Spain. If 2017 points on a circle are joined by straight li-
nes in all possible ways and no three of these lines meet at a single
point inside the circle, then find the number of triangles that can
be formed.
## EM–43. Proposed by Nicolae Papacu, Slobozia, √ Let a,
√ Romania.
b, c be three positive real numbers such that a + b + c = 1.
Prove that
√ √ √
a b c 1 1 1
2
+ 2 + 2 ≤ + + .
a + 2bc b + 2ca c + 2ab a b c
## EM–44. Proposed by Andrés Sáez-Schwedt, Universidad de León,
León, Spain. Let ABC be a triangle with AB = AC > BC , and
let O be the center of its circumcircle Γ. The tangent to Γ at C
meets the line AB at D. In the minor arc AC of Γ, consider the
point E such that
∠EOC + 2∠DOA = 360◦ .
If BE meets CD at F , show that F A = F C .
## EM–45. Proposed by José Luis Díaz-Barrero, BarcelonaTech, Bar-
celona, Spain. Let A1 , A2 , . . . , An be the vertices of an n-gon in-
scribed in a circle of center O. If O and the Ai ’s are lattice points,
then prove that the square of the perimeter of the n-gon is an even
number.
Volume 4, No. 1, Spring 2017 19
## EM–46. Proposed by Ángel Plaza de la Hoz, Department of Mat-
hematics, University of Las Palmas de Gran Canaria, Spain. Prove
that for any positive integer n the chain of inequalities
## 1/(2Ln ) Fn1/Fn + L1/L
n
n
1/(2Fn )
Fn1/Ln ≤ F2n ≤ ≤ F2n ≤ L1/F
n
n
,
2
holds, where Fn is the n-th Fibonacci number, defined by F0 = 0,
F1 = 1, and for n ≥ 2, Fn = Fn−1 + Fn−2 , and Ln is the n-
th Lucas number, defined by L0 = 2, L1 = 1, and for n ≥ 2,
Ln = Ln−1 + Ln−2 .
20 Arhimede Mathematical Journal
Medium–Hard Problems
MH–41. Proposed by Mihály Bencze, Braşov, Romania. Deter-
mine " #
n
X 1
(k + 1) k4 +k2 +1 ,
k=1
## MH–42. Proposed by José Luis Díaz-Barrero, BarcelonaTech, Bar-
celona, Spain. In Mathcontestland there are 2017 towns. Every
pair of towns is either connected by a single road, or is not con-
nected. If we consider any subset of 2015 towns, the total number
of roads connecting these towns to each other is a constant. If
there are R roads in Mathcontestland, then find all possible va-
lues of R.
## MH–43. Proposed by Andrés Sáez-Schwedt, Universidad de León,
León, Spain. Points A, B , C and D are collinear in that order.
On a circle ω through B and C , two new points E , F are chosen,
such that lines AE and DF meet on ω . The tangents to ω at B
and C meet at G (possibly at infinity). The tangents to ω at E
and F meet GA and GD at P and Q, respectively. Prove that
line P Q is tangent to ω .
## MH–44. Proposed by José Luis Díaz-Barrero, BarcelonaTech, Bar-
celona, Spain. Let x, y and z be be three nonzero real numbers
1 1 1 1
such that + + = . For all positive reals a, b, c, prove
x y z xyz
that
(a + b)x + (b + c)y + (c + a)z
√ ≥ 2.
ab + bc + ca
## MH–45. Proposed by Ander Lamaison Vidarte, Berlin Mathemati-
cal School, Berlin, Germany. We say that a set of positive integers
S is good if there is a function f : N → S such that no integer k
Volume 4, No. 1, Spring 2017 21
## with 2 ≤ k ≤ 2017 can be written as xf (x)
yf (y)
. Find the smallest po-
sitive integer n such that S = {1, 2017, 20172 , . . . , 2017n } is good,
or prove that such an integer does not exist.
## MH–46. Proposed by Ismael Morales López, Universidad Autó-
noma de Madrid, Madrid, Spain. Let a, b, c be the lengths of
the sides of a given triangle ABC , and ma , mb , mc its medians.
Prove that it is true that
X a+b 1
≥ p .
cyclic
c2 + 2ab + 3bc + 3ca m2a + m2b + m2c
22 Arhimede Mathematical Journal
A–41. Proposed by Marcin J. Zygmunt, AGH University of Science
and Technology, Kraków, Poland. Let γ1 < γ2 < . . . < γn be real
numbers,
1 1 1
f (x) = x + + + ··· + ,
γ1 − x γ2 − x γn − x
## and let α, β ∈ R, α < β . Compute the total length of the preimage
f −1 ([α, β]) (the total length of a set consisting of intervals is the
sum of their lengths.)
## A–42. Proposed by Óscar Rivero Salgado, BarcelonaTech, Barce-
lona, Spain. Let p be a prime number and consider A a p × p
matrix with complex entries, satisfying that Tr(A) = Tr(A2 ) =
. . . = Tr(Ap−1 ) = 0 and Tr(Ap ) = p. Find det(Ai + j Id), for i,
j ∈ Z.
## A–43. Proposed by José Luis Díaz-Barrero, BarcelonaTech, Bar-
celona, Spain. Find all real solutions of the following system of
equations
p √ √
x4 + x2 + (y 4 + 1) y 4 + 1 = (2 + √ 2) z 2 z 4 + 1,
√ √
y 4 + y 2 + (z 4 + 1) z 4 + 1 = (2 + √2) x2 px4 + 1,
z 4 + z 2 + (x4 + 1) x4 + 1 = (2 + 2) y 2 y 4 + 1.
## A–44. Proposed by Nicolae Papacu, Slobozia, Romania. Let a
and x0 be real numbers and let {xn }n≥0 be the sequence defined
by xn+1 = ax2n − (2a − 1)xn .
(a) Prove that the sequence {xn }n≥0 is convergent if, and only if,
ax0 ∈ [a − 1, a].
(b) If ax0 ∈ [a−1, a], compute lim n(x−`), where ` = lim xn .
n→+∞ n→+∞
## A–45. Proposed by Ander Lamaison Vidarte, Berlin Mathematical
School (BMS), Berlin, Germany. Let n be a positive integer, and let
Volume 4, No. 1, Spring 2017 23
## A be a n × (n + 1) matrix with integer entries. Assume that, for
every prime p, the matrix A considered as a matrix with entries
in Z/pZ has rank n. Prove that there exists a matrix B of size
(n + 1) × n with integer entries such that AB = Idn .
## A–46. Proposed by José Luis Díaz-Barrero, BarcelonaTech, Bar-
celona, Spain. Compute
n
X k
lim arctan 2
.
n→+∞
k=1
n
24 Arhimede Mathematical Journal
Mathlessons
This section of the Journal offers readers an opportunity to ex-
change interesting and elegant mathematical notes and lessons
with material useful to solve mathematical problems.
## Please, send submittals to José Luis Díaz-Barrero, Enginyeria
Civil i Ambiental, UPC BARCELONATECH, Jordi Girona 1-3, C2,
08034 Barcelona, Spain, or by e-mail to
jose.luis.diaz@upc.edu
Volume 4, No. 1, Spring 2017 25
Forward-Backward
Induction
J. L. Díaz-Barrero
1 Introduction
When trying to prove a statement P (n) involving the positive in-
teger n and a forward inductive argument is difficult for every n,
one can use the following strategy: first, prove the statement for
infinitely many values of n, and then, prove P (n) for the gaps.
The proof for the gaps can either be by forward induction, or bac-
kward induction.
## Downward inductive arguments have been around a long time.
Many important authors, including Cauchy (1759-1857) and Wei-
erstrass (1815-1897, see [3, p. 19]), have used them. Forward-
backward induction, also known as Cauchy’s induction, is a va-
riant of the well-known principle of mathematical induction. The
main difference between this type of induction and the classical
one is the inductive step. It can be stated as follows:
## Theorem 1. Let {P (n)}n≥1 be a sequence of propositions invol-
ving the positive integer n. If it holds that
## • Step 1: P (n0 ) is true for a positive integer n0 .
• Step 2: This is made out of two parts.
Forward induction: P (k) ⇒ P (2k) for every integer k ≥ n0 .
Backward induction: P (k) ⇒ P (k − 1) for every integer k ≥
n0 + 2.
## Then, P (n) is true for all integers n ≥ n0 .
26 Arhimede Mathematical Journal
## In particular when n0 = 2 the statement becomes
Theorem 2. Let P (n) be a proposition involving n ∈ N. If P (2)
holds, P (n) ⇒ P (2n) for n ≥ 2 and P (n) ⇒ P (n − 1) for all
n ≥ 4, then P (n) holds for all n ≥ 2.
In the following section, we will use the forward-backward in-
duction to prove the AM-GM inequality and Huygens’s inequality.
2 Applications
Hereafter, some examples where forward-backward induction may
be used are presented. We begin with the proof given by Cau-
chy [1] of the AM-GM inequality.
Theorem 3. Let a1 , a2 , . . . , an be n ≥ 1 positive real numbers.
Then, !1/n
n n
Y 1 X
aj ≤ aj .
j=1
n j=1
## Proof. The case when n = 1 trivially holds, so we will start con-
sidering the base case when n = 2. That is, we have to prove
a1 + a2 √
that for any positive reals a1 , a2 the inequality ≥ a1 a2
2
holds. Indeed, the preceding inequality is equivalent to a1 + a2 ≥
√ √ √
2 a1 a2 or ( a1 − a2 )2 ≥ 0, which trivially holds. This is the
base step.
Next we prove the forward induction step assuming the inequa-
lity holds for some integer k and we prove it for 2k. That is, we
suppose that
k
!1/k k
Y 1 X
aj ≤ aj
j=1
k j=1
## holds and we will prove that
2k
!1/2k 2k
Y 1 X
aj ≤ aj
j=1
2k j=1
Volume 4, No. 1, Spring 2017 27
## also holds. Indeed, we have
a1 + a2 + . . . + a2k 1 a1 + . . . + ak ak+1 + . . . + a2k
= +
2k 2 k k
1 √ √
≥ k
a1 a2 . . . ak + k ak+1 ak+2 . . . a2k
2
√ √
q
≥ k
a1 a2 . . . ak · k ak+1 ak+2 . . . a2k
= 2k a1 a2 . . . a2k .
The first inequality follows from k variable AM-GM, which is true
because of our inductive hypothesis, and the second inequality
follows from the 2 variable AM-GM, which we just proved above
(base step).
Now we prove the backward induction step. Since the AM-GM
inequality holds for any positive reals, in particular it also holds
a1 + a2 + . . . + ak−1
when ak = . Then, assuming that
k−1
a1 + a2 + . . . + ak √
≥ k
a1 a2 . . . ak
k
holds, we have
s
1 a1 +. . .+ak−1 a1 +. . .+ak−1
a1 +. . .+ak−1 + ≥ k a1 . . . ak−1 ·
k k−1 k−1
s
a1 + . . . + ak−1 a1 + . . . + ak−1
⇐⇒ ≥ k a1 . . . ak−1 ·
k−1 k−1
k
a1 + . . . + ak−1 a1 + . . . + ak−1
⇐⇒ ≥ a1 . . . ak−1 ·
k−1 k−1
k−1
a1 + . . . + ak−1
⇐⇒ ≥ a1 a2 . . . ak−1
k−1
a1 + . . . + ak−1 √
⇐⇒ ≥ k−1 a1 a2 . . . ak−1
k−1
Since we have shown that the base case and the inductive hypo-
thesis steps hold, then by the forward-backward induction (FBI)
principle, the AM-GM inequality is proven.
28 Arhimede Mathematical Journal
## Another interesting example due to Huygens [4] is the following.
Theorem 4. For n ≥ 2 and a1 , . . . , an , b1 , . . . , bn > 0,
p √ p
n
(a1 + b1 ) . . . (an + bn ) ≥ n a1 . . . an + n b1 . . . bn
holds, with equality if, and only if, the vectors (a1 , . . . , an ) and
(b1 , . . . , bn ) are linearly dependent.
Proof. If we say P (n) is the property that
p √ p
n
(a1 + b1 ) · · · (an + bn ) ≥ n a1 · · · an + n b1 · · · bn ,
then when n = 2 we have
p √ p
(a1 + b1 )(a2 + b2 ) ≥ a1 a2 + b1 b2 .
That is,
p
(a1 + b1 )(a2 + b2 ) ≥ a1 a2 + b1 b2 + 2 a1 a2 b1 b2 ,
⇐⇒ (a1 b2 + a2 b1 )2 ≥ 4a1 a2 b1 b2 ,
⇐⇒ (a1 b2 − a2 b1 )2 ≥ 0,
and P (2) holds.
For n ≥ 2, suppose that P (n) holds. We have to see that P (2n)
also holds. In fact,
p
2n
(a1 + b1 ) · · · (a2n + b2n )
np p o1/2
= n (a1 + b1 ) · · · (an + bn ) n (an+1 + bn+1 ) · · · (a2n + b2n )
n√ p √ p o1/2
≥ n a1 · · · an + n b1 · · · bn n an+1 · · · a2n + 2n bn+1 · · · b2n
√ √ 1/2 np
n
p o1/2
≥ n
a1 · · · an an+1 · · · a2n
n
+ b1 · · · bn 2n
bn+1 · · · b2n
√ p
= 2n a1 · · · a2n + 2n b1 · · · b2n .
## Finally, for n ≥ 4, suppose that P (n) holds and we have to see
that P (n − 1) also holds. Indeed,
n √ p o1/n
(a1 + b1 ) · · · (an−1 + bn−1 ) n−1
a1 · · · an−1 + n−1
b1 · · · bn−1
√ 1/n
n p o1/n
≥ {a1 · · · an−1 n−1
a1 · · · an−1 } + b1 · · · bn−1 n−1
b1 · · · bn−1
√ p
= n−1 a1 · · · an−1 + n−1 b1 · · · bn−1 .
Volume 4, No. 1, Spring 2017 29
## After some straightforward algebra, we get
p
n−1 √ p
(a1 + b1 ) · · · (an−1 + bn−1 ) ≥ n−1 a1 · · · an−1 + b1 · · · bn−1 .
n−1
Equality holds when the vectors of a’s and b’s are proportional.
Therefore, by Cauchy’s induction P (n) holds for all n ≥ 2, and
this completes the proof.
## Corollary 1. Suppose that x1 , x2 , · · · , xn and α are positive real
numbers. Then,
n n
!2
1X 2
Y 1/n
xk + αk ≥ α(n+1)/2 +
xk
n k=1 k=1
holds.
Proof. First, we write the inequality claimed as
s
(x1 + α)2 + · · · + (xn + αn )2 √
≥ α(n+1)/2 + n x1 x2 · · · xn .
n
Taking into account the QM-GM inequality, we have
s
(x1 + α)2 + · · · + (xn + αn )2 p
≥ n (x1 + α) · · · (xn + αn ).
n
So, to prove our inequality it will suffice to establish that
p √
n
(x1 + α) · · · (xn + αn ) ≥ α(n+1)/2 + n x1 · · · xn .
## Now, putting ak = xk and bk = α, 1 ≤ k ≤ n, in the last theorem,
we get
p
n
n √
(x1 + α) · · · (xn + α) ≥ α1+2+···+n + n x1 · · · xn
= α(n+1)/2 + n x1 · · · xn ,
## and the proof is complete.
Finally, we close this note with a nice result that appeared in [2].
30 Arhimede Mathematical Journal
## Theorem 5. Let a1 , a2 , . . . , an be positive integers such that a1 ≤
a2 ≤ . . . ≤ an . If
1 1 1
+ + ... + = 1,
a1 a2 an
then an < 2n! .
## Lemma 1. Let n ≥ 2, and let a1 , a2 , . . . , an be positive integers
such that
1 1 1
+ + ... + = 1.
a1 a2 an
For m < n, it holds that
1 1 1 1
≤1− + + ... + .
a1 a2 . . . am a1 a2 am
## Proof. Let r be a positive rational number such that
1 1 1 1
+ + ... + + = 1.
a1 a2 am r
Then, we have
1
r=
1
1− + a12 + . . . + a1m
a1
a1 a2 . . . am
= m Y
!
X
a1 a2 . . . am − aj
i=1 j6=i
≤ a1 a2 . . . am (1)
on account that since both sides of the above expression are posi-
tive, then the last denominator is a positive integer. That is, it is at
least 1 and the last inequality immediately follows. Now, inverting
the expression in (1) we obtain
1 1 1 1
≤1− + + ... + ,
a1 a2 . . . am a1 a2 am
Volume 4, No. 1, Spring 2017 31
## Proof of the theorem. We argue by contradiction. For n ≥ 2, let
A(n) be the statement that for positive integers satisfying 1 ≤
a1 ≤ a2 ≤ . . . ≤ an , if a11 + a12 + . . . + a1n = 1, then an < 2n! .
Assume, for some fixed n ≥ 2, that A(n) fails. That is, suppose
the statement
P (n) : an ≥ 2n!
holds. By strong backward induction, we prove that aj ≥ 2j! for
1 ≤ j ≤ n. Suppose the assumption is proved for n ≥ j ≥ m + 1.
It remains to prove P (m). To this goal,
s s
1 m
1 m
1 1 1
≤ ≤ 1− − − ... −
am a1 a2 . . . am a1 a2 am
s v
u n
u X
1 1 1 1
= m
+ ... + ≤ t
m
j!
≤ m!
.
am+1 an j=m+1
2 2
## But in this case we have
n +∞
1 1 1 X 1 X 1
+ + ... + ≤ < = 1.
a1 a2 an j=1
2j! j=1
2j
## Contradiction. Therefore, A(n) is true and this completes the
proof.
References
[1] Cauchy, A. L. “Cours d’Analyse de l’Ecole Royale Polytechni-
que, 1.: Analyse Algebrique”. Paris: Imprim. Royale, 1821; XIV,
576 p.; in 8.; DCCC. 4.85 1 (1821).
[2] Engel, A. Problem-solving strategies. Problem Books in Mat-
hematics. Springer-Verlag, New York, 1998, pp. x+403. ISBN:
0-387-98219-1.
32 Arhimede Mathematical Journal
## [3] Hardy, G. H., Littlewood, J. E., and Pólya, G. Inequalities.
Cambridge Mathematical Library. Reprint of the 1952 edition.
Cambridge University Press, Cambridge, 1988, pp. xii+324.
ISBN : 0-521-35880-9.
[4] Huygens, C. “Oeuvres Completes 1888–1940”. Société Hollon-
daise des Science, Haga (1940).
## José Luis Díaz-Barrero
Civil and Environmental Engineering
BarcelonaTech
Barcelona, Spain
jose.luis.diaz@upc.edu
Volume 4, No. 1, Spring 2017 33
Solutions
No problem is ever permanently closed. We will be very pleased to
consider for publication new solutions or comments on the past
problems.
## Please, send submittals to José Luis Díaz-Barrero, Enginyeria
Civil i Ambiental, UPC BARCELONATECH, Jordi Girona 1-3, C2,
08034 Barcelona, Spain, or by e-mail to
jose.luis.diaz@upc.edu
Elementary Problems
E–35. Proposed by Eric Sierra Garzo, CFIS, BarcelonaTech, Bar-
celona, Spain. Find the sum of the coefficients of
2016 2017
P (x) = 6 7x2 − 9x3 + 3x5 +3 5x7 − 2x11 − 4x13 −2x−1.
## Solution by Fernando Ballesta Yagüe, IES Infante Don Juan
Manuel, Murcia, Spain. Let us consider a general polynomial
P (x) = an xn + an−1 xn−1 + . . . + a1 x + a0 . We have that P (1) =
an +an−1 +. . .+a1 +a0 , that is, P (1) is the sum of the coefficients
of the polynomial. Therefore, to compute the sum we just need to
substitute x = 1 in the given expression:
## P (1) = 6 · (7 − 9 + 3)2016 + 3 · (5 − 2 − 4)2017 − 2 − 1
= 6 · 12016 + 3 · (−1)2017 − 3 = 6 − 3 − 3 = 0.
34 Arhimede Mathematical Journal
## Also solved by Alberto Espuny Díaz, University of Birmingham,
Birmingham, United Kingdom; Ander Lamaison Vidarte, Berlin Ma-
thematical School, Berlin, Germany; Víctor Martín Chabrera, Barce-
lonaTech, Barcelona, Spain; Ángel Plaza, University of Las Palmas
de Gran Canaria, Spain; Andrew J. Turner, University of Birming-
ham, Birmingham, United Kingdom, and the proposer.
## E–36. Proposed by José Luis Díaz-Barrero, BarcelonaTech, Barce-
lona, Spain, and Mihály Bencze, Braşov, Romania. Let a1 , a2 , . . . ,
an be positive real numbers. Show that for all integers m ≥ 1
there exist n positive reals b1 , b2 , . . . , bn such that
n m
1 X 1 1
+ = 1.
n k=1 1 + ak 1 + bk
## Solution 1 by Alberto Espuny Díaz, University of Birmingham,
Birmingham, United Kingdom. Observe that one possible solu-
1 1
tion would be to make each of the sums + equal
1 + ak 1 + bk
1. Indeed, in such a case, no matter which value of m we take,
we would have that each of the summands equals 1, and as there
are exactly n summands and we divide by n, the desired equality
holds. Now let us prove that
1 1
+ =1
1 + ak 1 + bk
may hold for any positive value of ak . Indeed, we have that
1 1 1 1 + ak − 1
+ = 1 ⇐⇒ =
1 + ak 1 + bk 1 + bk 1 + ak
1 + ak
⇐⇒ 1 + bk =
ak
1
⇐⇒ bk = ,
ak
which is a positive number. This completes the proof.
Solution 2 by Henry Ricardo, Westchester Math Circle, New
York, USA. Let bk = 1/ak for k = 1, 2, . . . , n. Then, for any
Volume 4, No. 1, Spring 2017 35
integer m ≥ 1,
n m n m
1X 1 1 1X 1 1
+ = +
n k=1 1 + ak 1 + bk n k=1 1 + ak 1 + 1/ak
n m
1X 1 ak
= +
n k=1 1 + ak ak + 1
n
1 X
= (1)m = 1.
n k=1
## Also solved by Miguel Cidrás Senra, CFIS, BarcelonaTech, Barce-
lona, Spain; Ander Lamaison Vidarte, Berlin Mathematical School,
Berlin, Germany; Víctor Martín Chabrera, BarcelonaTech, Barce-
lona, Spain; Ángel Plaza, University of Las Palmas de Gran Cana-
ria, Spain, and the proposer.
## E–37. Proposed by José Luis Díaz-Barrero, BarcelonaTech, Bar-
celona, Spain. Let P QRS be a square. A straight line through P
cuts side RS at M and line QR at N . Show that
PQ 2 PQ 2
+
PM PN
is a positive integer.
Solution 1 by Fernando Ballesta Yagüe, IES Infante Don Juan
Manuel, Murcia, Spain. Let ∠P N Q = α. Then
PQ
= sin α.
PN
=⇒ ∠SP M = α.
## As P QRS is a square, P Q = P S , so we have that
PS PQ
cos α = = .
PM PM
36 Arhimede Mathematical Journal
N
α
S M R
P Q
Figure 1: Construction for Solution 1 of Problem E–37.
Therefore,
2 2
PQ PQ
+ = sin2 α + cos2 α = 1.
PM PN
Solution 2 by Miguel Amengual Covas, Cala Figuera, Mallorca,
Spain. Let X be the point of side P Q of square P QRS such
that XM is parallel to QR. Since P Q is parallel to RS , we have
XM = QR. Thus,
XM = P Q. (1)
## Since XM k QR and QR ⊥ P Q, we have XM ⊥ P Q. From the
similar right triangles P XM and P QN , we get PP N
Q
= PP M
X
; hence
2 2 2 2
PQ PQ PQ PX
+ = +
PM PN PM PM
2 2
PQ + PX
=
P M2
XM 2 + P X 2
= (from (1))
P M2
= 1,
Volume 4, No. 1, Spring 2017 37
P S
X M
Q R N
## since XM 2 + P X 2 = P M 2 holds by the Pythagorean theorem
applied to 4P XM .
## Solution 3 by Víctor Martín Chabrera, BarcelonaTech, Barce-
lona, Spain. Let l be the length of the square. It is clear that at
least one of M , N is touching the square. By symmetry, we can
assume WLOG that M is touching the square. From Figure 3 we
have that, if we center the origin at P and put S at (l, 0) in the
plane, M will have coordinates (l, h). The line has a slope hl , so,
as N has second coordinate equal to h and N lies on the line,
38 Arhimede Mathematical Journal
l2
N = h
,l . Therefore,
2
2 2 2
PQ PQ l l
+ = √ + q
PM PN l2 + h2 l4
+ l2
h2
l2 l2
= + l4
l2 + h2 h2
+ l2
l2 h2 l2 + h2
= + = = 1.
l2 + h2 l2 + h2 l2 + h2
## Also solved by Alberto Espuny Díaz, University of Birmingham,
Birmingham, United Kingdom; Ander Lamaison Vidarte, Berlin Ma-
thematical School, Berlin, Germany; Ángel Plaza, University of Las
Palmas de Gran Canaria, Spain, and the proposer.
## E–38. Proposed by José Gibergans-Báguena, BarcelonaTech, Bar-
celona, Spain. Let f (x) = x2 + 14x + 42. Solve the equation
f (f (f (. . . (f (x))))) = 0.
| {z }
n
## Solution 1 by the proposer. Since f (x) = (x + 7)2 − 7, then
f (f (x)) = (x + 7)4 − 7, f (f (f (x))) = (x + 7)8 − 7 and so on. By
induction, we easily establish that
n
f (f (f (. . . (f (x))))) = (x + 7)2 − 7.
| {z }
n
n √
2n
Then the solutions of the equation (x + 7)2 − 7 = 0 are −7 ± 7,
and we are done.
## Solution 2 by Fernando Ballesta Yagüe, IES Infante Don Juan
Manuel, Murcia, Spain. For n = 1 we have
f (x) = x2 + 14x + 42 = 0
−14 ± 196 − 4 · 42 √
=⇒ x = = −7 ± 7.
2
Volume 4, No. 1, Spring 2017 39
For n = 2 we have
f (f (x)) = 0 =⇒ f (x) = −7 ± 7.
However, f (x) = x2 + 14x + 42 = (x + 7)2 − 7 ≥ −7, so
f (f (x)) = 0 =⇒ f (x) = −7 + 7
=⇒ x2 + 14x + 49 − 7 = 0
p √
−14 ± 196 − 196 + 4 7 √
4
=⇒ x = = −7 ± 7.
2
It looks like the general solution is going to be
2n
f (f (f (. . . (f (x))))) = 0 =⇒ x = −7 ± 7.
| {z }
n
## Let us prove this by induction. We have already proved the base
case. Now, we need to prove that if for some n = k the property
holds, then it also holds for n = k + 1. Indeed,
2k
f (f (f (. . . (f (x))))) = 0 =⇒ f (x) = −7 + 7
| {z }
k+1
## by induction hypothesis. Notice that we only consider the solution
greater than −7 since, as we already stated, the image of f is in
[−7, +∞). Hence,
2k
f (x) = −7 + 7
2k
=⇒ x2 + 14x + 49 − 7=0
q
k
−14 ± 196 − 196 + 4 2 7 √
2k+1
=⇒ x = = −7 ± 7.
2
Therefore, the induction hypothesis is true, and for√any n ∈ N the
n
solution of the equation is going to be x = −7 ± 2 7.
## Also solved by Alberto Espuny Díaz, University of Birmingham,
Birmingham, United Kingdom; Ander Lamaison Vidarte, Berlin Ma-
thematical School, Berlin, Germany; Víctor Martín Chabrera, Bar-
celonaTech, Barcelona, Spain, and Ángel Plaza, University of Las
Palmas de Gran Canaria, Spain.
40 Arhimede Mathematical Journal
## E–39. Proposed by José Luis Díaz-Barrero, BarcelonaTech, Barce-
lona, Spain. Let D, E , F be the points where the angle bisectors
of angles A, B , C of triangle ABC cut the opposite sides, and let
I be its incenter. Find the minimum value of
AI BI AI
+ + .
ID IE IF
Solution 1 by Miguel Amengual Covas, Cala Figuera, Mallorca,
Spain. By the angle bisector theorem, applied to 4ABD,
AI AB
= ,
ID BD
and applied to 4ABC ,
AB CA AB + CA AB + CA
= = = .
BD DC BD + DC BC
Hence
AI AB + CA
= .
ID BC
Similarly,
BI BC + AB CI CA + BC
= and = .
IE CA IF AB
Hence,
AI BI CI AB + CA
BC + AB CA + BC
+ + = + +
ID IE IF BC CA AB
AB BC BC CA CA AB
= + + + + +
BC CA CA BC AB CA
≥2+2+2
= 6.
Note that equality holds if, and only if, 4ABC is equilateral. We
therefore conclude that the minimum value of the given expres-
sion is 6 and it is achieved only for equilateral 4ABC .
Volume 4, No. 1, Spring 2017 41
## Solution 2 by Ander Lamaison Vidarte, Berlin Mathematical
School, Berlin, Germany. Let S , SA , SB and SC be the areas
of ABC , IBC , ICA and IAB , respectively. Let P be the point
where the incircle touches BC and let Q be the foot of the altitude
from A. Triangles DIP and DAQ have parallel sides, so
BC·IP
ID IP 2
SA
= = BC·AQ
= .
S
AI AD − ID S − SA SB + SC
From this, = = = . We can do
ID ID SA SA
the same for the other sides. Thus
AI BI CI SA + SB SB + SC SC + SA
+ + = + +
ID IE IF S S S
C A B
SA SB SB SC SC SA
= + + + + +
SB SA SC SB SA SC
≥ 2 + 2 + 2 = 6.
## Note that this bound is tight, since for an equilateral triangle we
have SA = SB = SC .
Remark. Next, for ease of reference, we state and prove the bisec-
tor theorem.
## Theorem (Bisector theorem). Let D be the point where the bi-
sector of angle A in triangle ABC cuts the side BC . Then,
AB AC
= .
BD CD
## Proof. On account of the Sine Law, we have
x c
=
sin α/2 sin γ
in 4ABD, and
y c
=
sin α/2 sin(π − γ)
42 Arhimede Mathematical Journal
b
c
γ π-γ
x y
B C
D a
x c AB c b AC
= ⇐⇒ = = =
y b BD x y CD
follows.
## Also solved by Fernando Ballesta Yagüe, IES Infante Don Juan
Manuel, Murcia, Spain; Víctor Martín Chabrera, BarcelonaTech, Bar-
celona, Spain, and the proposer.
## E–40. Proposed by José Luis Díaz-Barrero, BarcelonaTech, Barce-
lona, Spain. In a certain parliament, each member belongs to ex-
actly four committees, and each committee has exactly four mem-
bers. Prove that the number of members equals the number of
committees.
## Solution 1 by Miguel Cidrás Senra, CFIS, BarcelonaTech, Bar-
celona, Spain. Let m be the number of members and c, the
number of committees. We can label the members and the co-
mittees and define a variable aij such that aij = 1 if, and only
if, the i-th member belongs to the j -th committee, and aij = 0
Volume 4, No. 1, Spring 2017 43
Pm Pc
otherwise. We can observe that i=1 aij = 4 and j=1 aij = 4.
Then,
Xc Xm m X
X c
4·c= aij = aij = 4 · m,
j=1 i=1 i=1 j=1
## and finally, we obtain c = m.
Solution 2 by Alberto Espuny Díaz, University of Birmingham,
Birmingham, United Kingdom. Consider a graph defined as fol-
lows. Consider two sets of vertices, one of them consisting on the
members of the parliament (call this set A), and the other one
consisting on the committees (call this set B ). We join a member
of the parliament to a committee with an edge if, and only if, the
member belongs to the committee. This graph is clearly bipartite
(that is, all the edges join A to B , and there are no edges joi-
ning two vertices in A or two vertices in B ). The conditions of
the statement tell us that the degree of each vertex (the number
of edges incident to that vertex) is 4, both for the vertices in A
and in B . Now assume that |A| = n and |B| = m. As each
edge joins a vertex in A with a vertex in B , we may use a double-
counting argument: the overall number of edges can be counted
as the number of edges incident to A, or as the number of edges
incident to B . These are, respectively, 4n and 4m. We then have
that 4n = 4m, which immediately implies the claim.
Solution 3 by the proposer. Consider a matrix (called the in-
cidence matrix) in which each row represents an individual, and
each column represents a committee. An entry is 1 if the indivi-
dual corresponding to its row belongs to the committee correspon-
ding to its column; otherwise, the entry is 0. Of course, the roles
of rows and columns may be interchanged. An example of such a
configuration is shown below.
To solve the problem, we will count how many 1’s there are in the
incidence matrix. Suppose that there are n committees and m
members. Then the incidence matrix is an m × n matrix. The
given conditions tell us that each row contains 4 ones, so there
are 4m ones in total. On the other hand, each column contains 4
ones, so there are 4n ones in total. Equating the two counts we
see that 4m = 4n, so m = n, which is what we wanted to prove.
44 Arhimede Mathematical Journal
1 1 1 1 0 0 0 0 0
0 0 1 1 0 1 1 0 0
0 0 0 1 1 0 0 1 1
0 1 0 0 1 0 1 1 0
0 0 1 0 1 1 0 0 1
1 0 0 0 0 1 1 0 1
1 1 0 0 0 0 1 0 1
0 1 1 0 1 0 0 1 0
1 0 0 1 0 1 0 1 0
## Also solved by Fernando Ballesta Yagüe, IES Infante Don Juan
Manuel, Murcia, Spain; Padraig Condon, University of Birmingham,
Birmingham, United Kingdom; Matthew Coulson, University of Bir-
mingham, Birmingham, United Kingdom; Ander Lamaison Vidarte,
Berlin Mathematical School, Berlin, Germany; Víctor Martín Cha-
brera, BarcelonaTech, Barcelona, Spain, and Patrick Morris, Berlin
Mathematical School, Berlin, Germany.
Volume 4, No. 1, Spring 2017 45
Easy–Medium Problems
EM–35. Proposed by Mihály Bencze, Braşov, Romania. Find the
value of n
X
a1 + a2k ,
k=1
1 1 1
where ak = + + ··· + .
k k+1 n
Solution 1 by Ander Lamaison Vidarte, Berlin Mathematical
1 1 1
School, Berlin, Germany. Let ak,n = + + · · · + . We
k k+1 n
n
2
P
want to find the value of sn = a1,n + ak,n . We claim that
k=1
sn = 2n.
We proceed by induction. For n = 1 it holds because s1 = a1,1 +
a21,1 = 2. Now we prove the induction step:
n−1
X
sn − sn−1 = a1,n − a1,n−1 + a2n,n + (a2k,n − a2k,n−1 )
k=1
n−1
1 1 X
= + + (ak,n − ak,n−1 )(ak,n + ak,n−1 )
n n2 k=1
n−1
X
1 1 1 1
= + + + 2ak,n−1
n n2 n k=1
n
n−1 n−1
1 1 n−1 2 XX1
= + + +
n n2 n2 n k=1 i=k
i
n−1 i
2 2 XX1
= +
n n i=1 k=1
i
n−1
2 2 X
= + 1
n n i=1
2 2(n − 1)
= + = 2.
n n
46 Arhimede Mathematical Journal
## Solution 2 by Víctor Martín Chabrera, BarcelonaTech, PkBarce-
1
lona, Spain. We have ak = Hn − Hk−1 , where Hk = i=1 i is
the k-th harmonic number.
n n
!2 n X n
X X 1 X 1 X 1
Hn + =Hn + 2
+ 2
k=1 j=k
j k=1 j=k
j k≤i<j≤n
ij
n
X |{k ∈ N : 1 ≤ k ≤ j}|
=Hn +
j=1
j2
X |{k ∈ N : 1 ≤ k ≤ i}|
+2
1≤i<j≤n
ij
n
X j X i
=Hn + +2
j=1
j2 1≤i<j≤n
ij
n
X 1 X 1
=Hn + +2
j=1
j 1≤i<j≤n
j
n
X |{i ∈ N : 1 ≤ i < j}|
=2Hn + 2
j=1
j
n
X j−1
=2Hn + 2
j=1
j
=2Hn + 2n − 2Hn = 2n.
Solution 3 by the proposer. Squaring the ak ’s we obtain k terms
1 1 1
of the form 2 in a21 , a22 ,. . . , a2k . Adding them up, we get k 2 =
k k k
1 1 1
and the sum of all of them is + + · · · + = a1 .
1 2 n
1
Now we take the two-term products. If j < i, then 2 appears
ij
2 2
in all a2k for which k ≤ j , therefore their sum is j = .
ji i
For all i there exist i − 1 terms such that j < i. Therefore, for
2 2 2
each fixed i, all terms of the form have sum (i − 1) = 2 − .
ij i i
Volume 4, No. 1, Spring 2017 47
## Computing all of these sums we get
2 2 2 1 1
2− + 2− + ··· + 2 − = 2n − 2 + ··· +
1 2 n 1 n
= 2n − 2a1 .
Finally,
n
X
a1 + a2k = a1 + a1 + 2n − 2a1 = 2n,
k=1
USA.
## EM–36. Proposed by José Gibergans-Báguena and José Luis
Díaz-Barrero, BarcelonaTech, Barcelona, Spain. Let x1 , x2 , x3 be
the roots of the equation x3 − 3x2 − 2x + 1 = 0. Find the equation
1 1 1
whose roots are y1 = x1 + , y2 = x2 + , and y3 = x3 + .
x1 x2 x3
Solution by Fernando Ballesta Yagüe, IES Infante Don Juan
Manuel, Murcia, Spain. The equation x3 − 3x2 − 2x + 1 = 0 has
x1 , x2 and x3 as roots. Therefore, according to Cardano-Vieta’s
formulas,
x1 + x2 + x3 = 3,
x1 x2 + x2 x3 + x3 x1 = −2,
x1 x2 x3 = −1.
## We can find, by the same formulas, the equation which has y1 , y2
1 1
and y3 as roots by knowing that y1 = x1 + , y2 = x2 + , and
x1 x2
1
y3 = x3 + . Therefore, we have to find the values of y1 + y2 + y3 ,
x3
y1 y2 +y2 y3 +y3 y1 and y1 y2 y3 . But first, we need to calculate some
48 Arhimede Mathematical Journal
## expressions that are going to be necessary:
1 1 1 x1 x2 + x1 x3 + x2 x3 −2
+ + = =
= 2,
x1 x2 x3 x1 x2 x3 −1
x21 + x22 + x23 = (x1 + x2 + x3 )2 − 2(x1 x2 + x1 x3 + x2 x3 )
= 9 − 2(−2) = 13,
2 2 2 2 2 2
x1 x2 + x1 x3 + x2 x3 = (x1 x2 +x1 x3 +x2 x3 )−2x1 x2 x3 (x1 +x2 +x3 )
= 4 + 6 = 10,
and
(x1 + x2 + x3 )(x1 x2 + x1 x3 + x2 x3 )
= x21 x2 + x1 x2 x3 + x21 x3 + x22 x1 + x22 x3
+ x1 x2 x3 + x1 x2 x3 + x23 x1 + x23 x2
=⇒ x21 x2 + x21 x3 + x22 x1 + x22 x3 + x23 x1 + x23 x2
= (x1 + x2 + x3 )(x1 x2 + x1 x3 + x2 x3 ) − 3x1 x2 x3
= 3(−2) − 3(−1) = −3.
## Now, we can find out the values of y1 + y2 + y3 , y1 y2 + y1 y3 + y2 y3
and y1 y2 y3 :
1 1 1
y1 + y2 + y3 = x1 + x2 + x3 + + + = 3 + 2 = 5,
x1 x2 x3
1 1 1 1
y1 y2 + y1 y3 + y2 y3 = x1 + x2 + + x1 + x3 +
x1 x2 x1 x3
1 1
+ x2 + x3 +
x2 x3
x1 x2 1 x1 x3
= x1 x2 + + + + x1 x3 + +
x2 x1 x1 x2 x3 x1
1 x2 x3 1
+ + x2 x3 + + +
x1 x3 x3 x2 x2 x3
= x1 x2 + x2 x3 + x3 x1
x1 x2 x1 x3 x2 x3
+ + + + + +
x2 x1 x3 x1 x3 x1
1 1 1
+ + +
x1 x2 x1 x3 x2 x3
Volume 4, No. 1, Spring 2017 49
x1 + x2 + x3
= −2+
x1 x2 x3
x1 x2 +x21 x3 +x22 x1 +x22 x3 +x23 x1 +x23 x2
2
+
x1 x2 x3
= − 2 − 3 + 3 = −2,
1 1 1
y1 y2 y3 = x1 + x2 + x3 +
x1 x2 x3
x1 x2 x1 x3 x1 x2 x3
= x1 x2 x3 + + + +
x3 x2 x2 x3 x1
x2 x3 1
+ + +
x1 x3 x1 x2 x1 x2 x3
1 x1 x2 x1 x3 x2 x3
= x1 x2 x3 + + + +
x1 x2 x3 x3 x2 x1
x1 x2 x3
+ + +
x2 x3 x1 x3 x1 x2
x x +x1 x2 +x22 x23 x21 +x22 +x23
2 2 2 2
= − 1 − 1+ 1 3 +
x1 x2 x3 x1 x2 x3
= − 2 − 10 − 13 = −25.
To sum up, we have
y1 + y2 + y3 = 5,
y1 y2 + y2 y3 + y3 y1 = −2,
y1 y2 y3 = −25.
According to Cardano-Vieta’s formulas, this system of equations
corresponds to the polynomial p(y) = y 3 − 5y 2 − 2y + 25 whose
roots are y1 , y2 and y3 .
Note that this is just one possible solution. Actually, there are
infinitely many polynomials of degree 3 that have these roots, and
they are of the form
p(y) = ay 3 − 5ay 2 − 2ay + 25a ∀ a ∈ R \ {0}.
## Also solved by Miguel Amengual Covas, Cala Figuera, Mallorca,
Spain; Alberto Espuny Díaz, University of Birmingham, Birming-
ham, United Kingdom; Ander Lamaison Vidarte, Berlin Mathemati-
cal School, Berlin, Germany, and the proposers.
50 Arhimede Mathematical Journal
## EM–37. Proposed by José Luis Díaz-Barrero, BarcelonaTech, Bar-
celona, Spain. For all integers n ≥ 1, show that
X X X
i2 + ij + ji
1≤i≤n 1≤i<j≤n 1≤j<i≤n
## Solution 1 by Henry Ricardo, Westchester Math Circle, New
York, USA. The expansion
2
X X X
xi = x2i + 2 xi xj
1≤i≤n 1≤i≤n 1≤i,j≤n
X X X
= x2i + xi xj + xj xi
1≤i≤n 1≤i<j≤n 1≤j<i≤n
is well known.
Replacing xi by i gives us
2
X X X X
i = i2 + ij + ji,
1≤i≤n 1≤i≤n 1≤i<j≤n 1≤j<i≤n
X
so that the given sum is the square of the positive integer i=
1≤i≤n
n(n + 1)/2.
## Solution 2 by Miguel Amengual Covas, Cala Figuera, Mallorca,
Spain. We add all entries of the following table
1 · 2 1 · 3 . . . 1 · (n − 1) 1·n
2 · 3 . . . 2 · (n − 1) 2·n
.. .. ..
. . .
(n − 1)n
by adding the entries of each column and then adding the column
sums. We obtain
1 · 2 + (1 + 2) · 3 + . . . + (1 + 2 + . . . + (n − 1)) · n.
Volume 4, No. 1, Spring 2017 51
## Since each parenthesis consists of a sum of consecutive integers
starting with 1, this expression may be rewritten as
2(1 + 2) (n − 1)(1 + (n − 1))
1·2+ · 3 + ... + ·n
2 2
or, equivalently,
1
1 · 22 + 2 · 32 + . . . + (n − 1) · n2 .
2
Thus,
X 1
1 · 22 + 2 · 32 + . . . + (n − 1) · n2 .
ij =
1≤i<j≤n
2
X X
Since ij = ji, we have
1≤i<j≤n 1≤j<i≤n
X X X
i2 + ij + ji
1≤i≤n 1≤i<j≤n 1≤j<i≤n
X X
= i2 + 2 ij
1≤i≤n 1≤i<j≤n
= 12 + 22 + . . . n2 + 1 · 22 + 2 · 32 + . . . + (n − 1) · n2
=12 + 22 + 1 · 22 + . . . + n2 + (n − 1)n2
| {z } | {z }
=23 =n3
3 3 3
=1 + 2 + . . . + n
n(n + 1) 2
= ,
2
where the last equality may be easily established by mathematical
induction on n.
Now, since n(n+1)
2
is a natural number for any natural n, the proof
is complete.
## Also solved by Alberto Espuny Díaz, University of Birmingham,
Birmingham, United Kingdom; Ander Lamaison Vidarte, Berlin Ma-
thematical School, Berlin, Germany; Víctor Martín Chabrera, Barce-
lonaTech, Barcelona, Spain; Ángel Plaza, University of Las Palmas
de Gran Canaria, Spain, and the proposer.
52 Arhimede Mathematical Journal
## EM–38. Proposed by Óscar Rivero Salgado, CFIS, BarcelonaTech,
Barcelona, Spain. Solve in positive integers the following equa-
tion:
(a2 + b2 + c2 )2 = 2(a4 + b4 + c4 ) + 2016.
this as
## (a + b + c)(b + c − a)(a + c − b)(a + b − c) = 2016 = 25 · 32 · 7.
None of the factors can be negative, since only one of them could
have this property and, then, the resulting product would be ne-
gative too. Furthermore, observe that the four factors have the
same parity, that must be even, so we can write
## a+b+c b+c−a a+c−b a+b−c
· · · = 2 · 3 · 3 · 7.
2 2 2 2
Write
b+c−a a+c−b a+b−c
X= , Y = , Z= ,
2 2 2
and observe that the condition says that
XY Z(X + Y + Z) = 2 · 3 · 3 · 7.
## We must have that exactly one factor is even. If X , Y , Z are all
odd, so is their sum. If more than one is even we have a contra-
diction. Then, exactly one is even and the other two are odd, but
in that case the sum is even, a contradiction. Hence, there are no
solutions to the equation.
## EM–39. Proposed by Guillem Alsina Oriol, CFIS, BarcelonaTech,
Barcelona, Spain. Two circles C1 and C2 have in common two
points X and Y . We draw a line that cuts C1 at points A and
B . Next we draw the lines AX , AY which cut C2 at points AX
and AY and lines BX , BY which cut C2 at points BX and BY ,
respectively. Show that the three lines AB , AX BY and AY BX
are parallel.
Volume 4, No. 1, Spring 2017 53
X
AX
P
BY
## Solution 1 by Miguel Amengual Covas, Cala Figuera, Mallorca,
Spain. As in Figure 5, we have taken the point P to lie on AAX
extended.
The cyclic quadrilaterals ABY X and XY BY AX tell us that
∠P AB = ∠XAB = ∠XY BY = ∠P AX BY .
## Therefore the line AB is parallel to the line AX BY .
Similarly (see Figure 6), since points A, B , Y , X lie on a circle,
∠AY AB = ∠Y AB = ∠Y XB,
## and since BX AY Y X is a cyclic quadrilateral,
∠Y XBX = ∠AAY BX .
Thus
∠AY AB = ∠AAY BX ,
so that AB and AY BX are parallel. The conclusion follows.
Solution 2 by Ander Lamaison Vidarte, Berlin Mathematical
School, Berlin, Germany. We will show that AY BX is parallel
to AB . To prove the same for AX BY , just exchange the roles of
A and B .
54 Arhimede Mathematical Journal
BX
AY
## Figure 6: Second scheme for Solution 1 of Problem EM–39.
Suppose first that the relative positions of the points are the same
as in the figure (that is, X lies between B and BX , B and Y are
on the same arc AX in C1 , etc.). Then,
## ∠ABBX = ∠ABX = ∠AY X = ∠AY Y X
= ∠AY BX X = ∠AY BX B,
## In order to prove a more general case, we would need to consi-
der other cases, like when B and Y are on different arcs AX .
Checking these cases, we might need to take supplementary an-
gles. Instead, we can reduce all cases to this one by conside-
ring all angles above as oriented and taking modulo 180◦ . For
example, modulo 180◦ , if A, B , C and D lie on a circle then
∠ABD = ∠ACD regardless of their relative position. Similarly,
if A, B and C are colinear, then ∠ABD = ∠CBD. Finally, AB
is parallel to CD if, and only if, ∠ABC = ∠DCB . With this, the
proof of the particular case also works for the general case.
## EM–40. Proposed by José Luis Díaz-Barrero, BarcelonaTech, Bar-
Volume 4, No. 1, Spring 2017 55
## celona, Spain. Let n be a nonnegative integer. Compute
2n+1
X j3
.
j=1
(2n + 1)2 − 3(2n + 1)j + 3j 2
## Solution by Miguel Cidrás Senra, CFIS, BarcelonaTech, Barce-
lona, Spain. Let a and b be two different real numbers. Then,
b3 (a − b)3
+
a2 − 3ab + 3b2 a2 − 3a(a − b) + 3(a − b)2
b3 (a − b)3
= +
a2 − 3ab + 3b2 a2 − 3a2 + 3ab + 3a2 − 6ab + 3b2
b3 (a − b)3
= +
a2 − 3ab + 3b2 a2 − 3ab + 3b2
(a − b)3 + b3 a3 − 3a2 b + 3ab2 − b3 + b3
= = = a.
a2 − 3ab + 3b2 a2 − 3ab + 3b2
We can observe that the initial expression is equivalent to
2n+1
X j3
j=1
(2n + 1)2 − 3(2n + 1)j + 3j 2
2n+1
X j3
=
j=0
(2n + 1)2 − 3(2n + 1)j + 3j 2
56 Arhimede Mathematical Journal
n
X j3
= +
j=0
(2n + 1)2 − 3(2n + 1)j + 3j 2
(2n + 1 − j)3
+ .
(2n + 1)2 − 3(2n + 1)(2n + 1 − j) + 3(2n + 1 − j) 2
## Finally, using the first identity with a = 2n + 1 and b = j , we
obtain
2n+1 n
X j3 X
= 2n + 1 = (n + 1)(2n + 1)
j=1
(2n + 1)2 − 3(2n + 1)j + 3j 2 j=0
2
= 2n + 3n + 1.
## Also solved by Fernando Ballesta Yagüe, IES Infante Don Juan
Manuel, Murcia, Spain; Víctor Martín Chabrera, BarcelonaTech, Bar-
celona, Spain, and the proposer.
Volume 4, No. 1, Spring 2017 57
Medium–Hard Problems
MH–35. Proposed by José Luis Díaz-Barrero, BarcelonaTech, Bar-
celona, Spain. Let a, b, c be positive real numbers such that
a + b + c = 3. Prove that
s r r r
b c a 3 4 1
+ + ≤ .
a2 + 3 b2 + 3 c2 + 3 2 abc
Solution 1 by Miguel Cidrás Senra, CFIS, BarcelonaTech, Bar-
celona, Spain. First, we observe that
1 1 1
+ +
a2 + 3 b2 + 3 c2 + 3
1 1 1
= + +
a2 + 1 + 1 + 1 b2 + 1 + 1 + 1 c2 + 1 + 1 + 1
√ √ √
1 1 1 bc + ca + ab
≤ √ 4
+ √ 4
+ √4
= √ (2)
4 a2 4 b2 4 c2 4 abc
√ √ √
b2 + c2 + a2 a+b+c 3
≤ √ = √ = √ , (3)
4 abc 4 abc 4 abc
where the inequality in (2) comes from the AM-GM inequality,
and that in (3) is a consequence of the rearrangement inequa-
lity. On the other hand, applying the Cauchy–Schwarz inequality
we obtain
s r r
b c a
+ +
a2 + 3 b2 + 3 c2 + 3
s
1 1 1
≤ (b + c + a) 2 + 2 + 2
a +3 b +3 c +3
s
1 1 1
= 3 2 + 2 + 2
a +3 b +3 c +3
s r
3 3 4 1
≤ 3· √ = ,
4 abc 2 abc
where the last inequality is a consequence of (2) and (3).
58 Arhimede Mathematical Journal
## Solution 2 by Ander Lamaison Vidarte. Berlin Mathematical
School, Berlin, Germany. Consider the chain of inequalities
s r r
b c a
+ +
a2 + 3 b2 + 3 c2 + 3
s
b c a
r r
≤ √ + √ + √ (4)
4 a 4 b 4 c
1 √4
√ √
= √ b 3 c + 4 c3 a + 4 a3 b
4
2 abc
1 3b + c 3c + a 3a + b
≤ √ 4
+ + (5)
2 abc 4 4 4
r
3 4 1
= ,
2 abc
## where in (4) we apply AM-GM to (a2 , 1, 1, 1) to obtain a2√+ 3 ≥
√ 4
4 a, and in (5) we apply AM-GM to (b, b, b, c) to obtain b3 c ≤
3b+c
4
.
## Solution 3 by Víctor Martín Chabrera, BarcelonaTech, Barce-
lona, Spain. We can apply
the Cauchy-Schwarz
inequality to vec-
√ √ √
tors b, c, a and √ 12 , √ 12 , √ 12 to get
a +3 b +3 c +3
s r r
b c a
+ +
a2 + 3 b2 + 3 c2 + 3
s
p 1 1 1
≤ a+b+c + +
a2 + 3 b2 + 3 c2 + 3
s
√ 1 1 1
= 3 + + .
a2 + 3 b2 + 3 c2 + 3
## Applying AM-GM inequality we see that
1 1 1 1
= ≤ √
4
= √ .
x2 + 3 x2 + 1 + 1 + 1 4 x2 4 x
Volume 4, No. 1, Spring 2017 59
Therefore,
√ √
s s
X 1 X 1
3 ≤ 3 √
cyc
a2 +3
cyc
4 a
v √
√ uX
u
bc
= 3t √
cyc 4 abc
√ r sX √
3 4 1
= bc
2 abc cyc
√ r q√
3 4 1 √ √
= bc + ca + ab.
2 abc
Now,we can apply again
√the√Cauchy-Schwarz inequality to vec-
√ √ √ √
tors b, c, a and c, a, b to obtain
√ r q√
3 4 1 √ √
abbc + ca +
2
√ r abc
3 4 1
qp p
≤ b+c+a c+a+b
√2 r abc q √ r
√ √ 3 4 1 √
r
3 4 1 3 4 1
= 3 3= 3= ,
2 abc 2 abc 2 abc
thus completing the proof. Equality is reached if, and only if,
a = b = c = 1.
## MH–36. Proposed by Damià Torres Latorre, CFIS, BarcelonaTech,
Barcelona, Spain, and Jesús Dueñas Pamplona, Valladolid, Spain.
Let ABC be an acute triangle with circumcircle ω . Let P be a
point lying on ω distinct from B and its diametrically opposite.
Let M the intersection point of the tangents to ω drawn from P
and B, respectively. Parallel to the tangent to ω drawn from A
and passing through M cuts lines AB and AC in the points D
and E , respectively. The perpendicular line to AB through D
and the perpendicular to AC through E intersect at T . Find the
locus of T when P moves on ω .
60 Arhimede Mathematical Journal
## Solution by Ander Lamaison Vidarte, BMS, Berlin, Germany.
Let ET intersect AB at F , DT intersect AC at G, and let J be
a point on the tangent to ω at A such that AJ and DE have the
same orientation. Because ABC is an acute triangle, the angles
∠ADE and ∠AED are acute, and thus D lies between A and
F , and E lies between A and G. Since ∠F DG = ∠F EG = 90◦ ,
F DEG is a cyclic quadrilateral. Thus
## ∠AF G =∠DF G = 180◦ − ∠DEG = ∠AED
=∠AEM = ∠EAJ = ∠ABC,
## so F G is parallel to BC . Since F T ⊥ AG and GT ⊥ AF , T is
the orthocenter of AF G, and AT ⊥ F G k BC . We conclude that
T is on the altitude from A in ABC .
We will now reverse the construction to check if every point in
the altitude can be reached in this way. Let T be a point in the
altitude. Project T on AB to obtain D, and on AC to obtain E .
By similarity of all figures drawn this way, DE is always parallel
to some direction, which must be the tangent to ω through A. Let
M be the intersection of DE and the tangent to ω through B . If
M 6= B , then there is another tangent to ω through M , and let
P be the point of tangency. If M = B , no such P can be found,
because P = B is explicitly forbidden. If M = B , then D = B ,
and ∠ABT = 90◦ . Therefore the locus of T is the altitude from A
in ABC , excluding the point with ∠ABT = 90◦ .
Volume 4, No. 1, Spring 2017 61
## MH–37. Proposed by Nicolae Papacu, Slobozia, Romania. Deter-
mine all positive integers n and prime numbers p such that pn +8
is a perfect cube.
Solution by Miguel Cidrás Senra, CFIS, BarcelonaTech, Barce-
lona, Spain. Observe that
pn + 8 is a perfect cube ⇐⇒ pn + 8 = k3
⇐⇒ pn = (k − 2)(k2 + 2k + 4)
⇐⇒ pn = (k − 2)((k + 4)(k − 2) + 12).
Therefore, either k − 2 = 1 or k − 2 = pa for some a.
In the first case, we have k = 3, so pn = 33 − 8 = 191 , and
(n, p) = (1, 19) is one solution.
12
In the second case, k − 2 = pa for some a, so pn−2a = k + 4 + k−2 .
As k − 2 is a power of a primer number and divides 12, k − 2 = 2,
3 or 4, so k = 4, 5 or 6. If k = 4, 43 − 8 = 56 = 7 · 8. If k = 5,
53 − 8 = 117 = 9 · 13. If k = 6, 63 − 8 = 208 = 13 · 16. And none
of them is a power of a primer number.
So, there is only one natural number n and one primer number p
such that pn + 8 is a perfect cube.
## Also solved by Víctor Martín Chabrera, BarcelonaTech, Barcelona,
Spain, and the proposer.
## MH–38. Proposed by José Luis Díaz-Barrero, BarcelonaTech, Bar-
celona, Spain. Let 1 < a < b be real numbers. Prove that for any
x1 , x2 , x3 ∈ [a, b] there exists c ∈ [a, b] such that
1 1 1 1 4
+ + +√
3
= .
log x1 log x2 log x3 log x1 log x2 log x3 log c
Solution by Ángel Plaza, University of Las Palmas de Gran
Canaria, Spain. It is enough to consider
4
log c = 1 1 1 1 .
log x1
+ log x2
+ log x3
+ √
3
log x1 log x2 log x3
62 Arhimede Mathematical Journal
That is, log c√is the harmonic mean of the numbers log x1 , log x2 ,
log x3 , and 3 log x1 log x2 log x3 . Therefore,
## and since the function log x is positive, increasing and continuous
in [a, b], then c ∈ [a, b].
## Also solved by Fernando Ballesta Yagüe, IES Infante Don Juan
Manuel, Murcia, Spain; Víctor Martín Chabrera, BarcelonaTech, Bar-
celona, Spain; Henry Ricardo, Westchester Area Math Circle, New
York, USA, and the proposer.
## MH–39. Proposed by Alberto Espuny Díaz, University of Birming-
ham, Birmingham, United Kingdom. Consider the integer grid Z2 =
Z × Z = {(a, b) | a, b ∈ Z}. How many different walks can be per-
formed in 2n steps of length one starting and ending at the same
point and moving in steps only in the directions of the coordinate
axes?
## Solution 1 by Padraig Condon, University of Birmingham, Bir-
mingham, United Kingdom, and the proposer. We claim that
2
2n
the answer is . Indeed, divide the walk into 2n steps and,
n
2n
among those, choose n (for which there are choices). Co-
n
lour these steps, say, red. Independently of this, choose n steps
among the 2n, and colour them blue.
## Now consider the following. We are going to say that if a step is
coloured both blue and red, it is an “up” step, if it has no colour it
is “down”, if it is only red it goes “right”, and if it is only blue it goes
“left”. It is very easy to check that there are as many left steps as
right steps, and as many up steps as down steps, which means
that the resulting walk given by any such colouring is closed and
ends back at the origin. Furthermore, each of these colourings
clearly gives a different walk. All that remains is to check that
these cover all possible closed walks. But for any given closed
Volume 4, No. 1, Spring 2017 63
walk, one can colour each step according to the described colou-
ring, so it could be the result of one of our choices. This completes
the proof.
## Solution 2 by Fernando Ballesta Yagüe, IES Infante Don Juan
Manuel, Murcia, Spain. In order to start and end at the same
point, the number of steps to the right has to be the same as the
number of steps to the left, and the number of steps upwards has
to be the same as the number of steps downwards. Therefore, the
number of vertical steps has to be even, just like the number of
horizontal steps.
## To begin with, let x be the number of horizontal steps and y the
number of vertical steps, so that x + y = 2n. To solve the coun-
ting, we can divide the counting in several disjoint cases.
## For x = 0, y = 2n we have n steps upwards and n downwards.
We can think of it as the number of possible permutations of n
arrows pointing upwards (↑) and n arrows pointing downwards
(↓). So, the number of possible different walks is
(2n)!
.
n!n!
For x = 2, y = 2n − 2 we have n − 1 steps upwards, n − 1 do-
wnwards, 1 step to the left and 1 to the right, which corresponds
to the arrows that we need to permute. The number of possible
different walks is
(2n)!
.
1!1!(n − 1)!(n − 1)!
In general, for x = 2k, y = 2n − 2k we have n − k steps upwards,
n − k downwards, k steps to the left and k to the right, which
corresponds to the arrows that we need to permute. The number
of possible different walks is
(2n)!
.
k!k!(n − k)!(n − k)!
## Reasoning in the same way until x = 2n and y = 0 and adding
all the disjoint cases, we have that the number of possible paths
64 Arhimede Mathematical Journal
is
n
X (2n)!
.
k=0
k!k!(n − k)!(n − k)!
Now the problem is to calculate this sum. Let us consider the
general term
(2n)!
.
k!k!(n − k)!(n − k)!
We have that
2
2n! n! n! (2n)! n 2n
= = .
k!k!(n − k)!(n − k)! k!(n − k)! k!(n − k)! n!n! k n
Therefore,
n n
2n X n 2
X (2n)!
= .
k=0
k!k!(n − k)!(n − k)! n k=0
k
## In order to compute this last sum, consider the binomial theorem
applied to (1 + x)2n . A direct application yieds
2n
2n 2n 2n n 2n 2n
(1 + x) = + x + ... + x + ... + x .
0 1 n 2n
## However, we can also evaluate it as
2
2n n n
n n n n
(1 + x) =(1 + x) (1 + x) = + x + ... + x
0 1 n
n n n n n n
= + + x + ...
0 0 0 1 1 0
n n n n n n
+ + + ... + xn
0 n 1 n−1 n 0
n n 2n
+ ... + x .
n n
## By equating the coefficients of xn we obtain
n n n n n n 2n
+ + ... + = ,
0 n 1 n−1 n 0 n
Volume 4, No. 1, Spring 2017 65
n n
and knowing that = we have
k n−k
n 2 n
X n X n n 2n
= = .
k=0
k k=0
k n−k n
2n 2
So, overall, the number of walks is .
n
Solution 3 by Víctor Martín Chabrera, BarcelonaTech, Barce-
lona, Spain. Let ar,s be the number of ways P of getting from (0, 0)
r s
to (r, s) after 2n steps. Let fn (x, y) = r,s∈Z ar,s x y be the ge-
nerating function of ar,s . We have f (x, y) = (x + x−1 + y + y −1 )2n
(here, each summand represents moving east, west, north and
south, respectively). Hence, the number of different walks consis-
ting of 2n steps which start and end at the same point (which we
can assume to be the origin WLOG) will be [x0 , y 0 ]fn (x, y) (here,
[xr , y s ]fn (x, y) means the coefficient of the term xr y s of the se-
ries). We will need to prove the following lemma:
n
n 2
X 2n
Lemma. = .
m=0
m n
## Proof. On the one hand we have that
2n
2n
X n
(1 + x) = xm .
m=0
m
On the other,
n ! n
!
2n n 2
X n m
X n k
(1 + x) = ((1 + x) ) = x x
m=0
m k=0
k
X n n
= xm+k .
0≤m,k≤n
m k
2n
n
We see that the coefficient of x is n
in the first case, and
n n
P
0≤m,k≤n m k in the second. Expanding this last expression
m+k=n
66 Arhimede Mathematical Journal
we have
n
X n n X n n
=
0≤m,k≤n
m k m=0
m n−m
m+k=n
n
X n n
=
m=0
m m
n 2
X n
= ,
m=0
m
as we wanted to show.
Having proven this lemma, we can find the value of [x0 , y 0 ]fn (x, y):
[x0 , y 0 ]fn (x, y)
=[x0 , y 0 ](x + x−1 + y + y −1 )2n
2n m 2n−m
2n X m i −(m−i) X 2n − m j −(2n−m−j)
X
0 0
=[x , y ] xx y y
m=0
m i=0 i j=0
j
2n m 2n−m
2n X m 2i−m X 2n − m 2j+m−2n
X
0 0
=[x , y ] x y .
m=0
m i=0
i j=0
j
## To get the coefficient of x0 y 0 we will need to have i = m
2
and
m
j = n − 2 . So we only need to sum over even values of m:
2n m 2n−m
X 2n − m
0 0
X 2n X m 2i−m
[x , y ] x y 2j+m−2n)
m=0
m i=0 i j=0
j
n
2n 2k 2n − 2k
X
=
k=0
2k k n−k
n
X (2n)! (2k)! (2n − 2k)!
=
k=0
(2k)!(2n − 2k)! k!k! (n − k)!(n − k)!
n
X 1 1
= (2n)!
k=0
k!k! (n − k)!(n − k)!
n
X (2n)! n! n!
=
k=0
n!n! k!(n − k)! k!(n − k)!
Volume 4, No. 1, Spring 2017 67
Xn
2n n n
=
n k=0 k k
X n 2 2
2n n 2n 2n 2n
= = = .
n k=0 k n n n
## Solution 4 by the proposer. In any such walk, there must be as
many steps to the right as steps to the left, and as many steps
upwards as downwards (otherwise, we would not reach the same
point). As this is the only condition, in order to completely deter-
mine such a walk, one must choose the steps in which he moves
in each direction. Assume that in a particular walk we know that,
overall, there are i steps upwards. Then, naturally, there must
be i steps downwards, and n − i to the right and to the left. The
number of walks with these steps is
2n 2n − i 2(n − i) n − i
,
i i n−i n−i
corresponding to choosing the steps in which we move upwards,
then those in which we move downwards, then those among the
remaining ones in which we move to the right, and finally, those
in which we move to the left.
## In our problem we do not know how many steps upwards there
are, so we must compute all the possibilities. Clearly, as there are
2n steps overall, the number of upwards steps is upper bounded
by n, so the number of such closed walks is
n
2n 2n − i 2(n − i) n − i
X
.
i=0
i i n−i n−i
Let us compute this number. Using the usual formula for combi-
natorial numbers we have
n
2n 2n − i 2(n − i) n − i
X
i=0
i i n−i n−i
n
X (2n)! (2n − i)! (2n − 2i)!
=
i=0
i!(2n − i)! i!(2n − 2i)! (n − i)!(n − i)!
68 Arhimede Mathematical Journal
n n
X (2n)! X (2n)!(n!)2
= =
i=0
(i!)2 ((n − i)!)2 (n!)2 (i!)2 ((n − i)!)2
i=0
n n 2
(n!)2
X
(2n)! X 2n n
= 2 2 2
= .
(n!) i=0 (i!) ((n − i)!) n i=0 i
n 2 n
X n X n n
Now consider the sum = . Combinato-
i=0
i i=0
i n − i
rially speaking, we have two sets of n elements and we choose i
from the first one and n − i from the second one. We do this for
every possible value of i. For each of these values, what we are
doing is choosing n elements from a set of 2n in such a way that
i of them belong to a certain subset, and as we add over all values
of i, we are obtaining all possible ways too choose n elements
n 2
X n 2n
from a set of 2n That is, we have = , so overall,
i=0
i n
2
2n
the number of closed walks of length 2n is .
n
## Also solved by Fernando Ballesta Yagüe, IES Infante Don Juan
Manuel, Murcia, Spain (one more solution); Matthew Coulson, Uni-
versity of Birmingham, Birmingham, United Kingdom; José Luis
Díaz-Barrero, BarcelonaTech, Barcelona, Spain, and Ander Lamai-
son Vidarte, Berlin Mathematical School, Berlin, Germany.
## MH–40. Proposed by Mihály Bencze, Braşov, Romania. Let
ABCD be a convex quadrilateral such that a = AB , b = BC ,
c = CD, d = DA, e = BD, and f = AC . Prove that
(ab + bf + f a)2 (cd + df + f c)2
3
[ABCD] ≤ min + ,
12 a2 + b2 + f 2 c2 + d2 + f 2
(ae + ed + da)2 (bc + ce + eb)2
+ .
a2 + e2 + d2 b2 + c2 + e2
Solution by the proposer. First, we claim that in any triangle
XY Z with the usual notations, it is true that
x2 + y 2 + z 2 ≥ 4 3A + (x − y)2 + (y − z)2 + (z − x)2 ,
Volume 4, No. 1, Spring 2017 69
## where A = [XY Z] (Hadwiger-Finsler’s inequality).
Indeed, on account of the cosines law we get
x2 = y 2 + z 2 − 2yz cos X,
X being the angle between y and z . This expression is equivalent
to
x2 = (y − z)2 + 2yz(1 − cos X).
1 1 − cos X
Since A = yz sin X , then x2 = (y − z)2 + 4 A . On
2 sin X
account that
X X X
1 − cos X = 2 sin2 and sin X = 2 sin cos ,
2 2 2
we have
X
x2 = (y − z)2 + 4A tan (cyclic).
2
x2 + y 2 + z 2
2 2 2
X Y Z
=(x − y) + (y − z) + (z − x) + 4A tan + tan + tan .
2 2 2
π
Now, since the halves of the triangle’s angles are less than the 2
,
function tan is convex, and we have
X Y Z X +Y +Z π √
tan + tan + tan ≥ 3 tan = 3 tan = 3
2 2 2 6 6
on account of Jensen’s inequality. From the preceding it immedi-
ately follows that
x2 + y 2 + z 2 ≥ 4 3A + (x − y)2 + (y − z)2 + (z − x)2 ,
as claimed.
Since
X 2 X X 2 X X X
xy = x2 − xy − x2 − 2 xy x2
1 X 2 X X X
= (x − y)2 + 2 xy − x2 x2
2
X X X
≥ 2 xy − x2 x2
70 Arhimede Mathematical Journal
D
c
C
d f
e b
A
a
B
## Figure 9: Scheme for the solution of Problem MH–40.
and X X √
2 xy − x2 ≥ 4 3 A,
then X 2 √ X
xy ≥ 4 3A x2
or
√ xy)2
P
(
4 3A ≤ P 2 .
x
Using the last inequality for the triangles ABC and ACD, we get
√ √ √
4 3 [ABCD] = 4 3 [ABC] + 4 3 [ACD]
(ab + bf + f a)2 (cd + df + f c)2
≤ + .
a2 + b2 + f 2 c2 + d2 + f 2
Using again the above inequality for the triangles ABD and BCD,
we get
√ √ √
4 3 [ABCD] = 4 3 [ABD] + 4 3 [BCD]
(ae + ed + da)2 (bc + ce + eb)2
≤ + ,
a2 + e2 + d2 b2 + c2 + e2
and the statement follows.
Volume 4, No. 1, Spring 2017 71
A–35. Proposed by José Luis Díaz-Barrero, BarcelonaTech, Barce-
lona, Spain. Suppose that all the eigenvalues of A ∈ Mn (R) are
positive real numbers. Show that
det(A + A−1 ) ≥ 2n .
## Solution 1 by Alberto Espuny-Díaz, University of Birming-
ham, Birmingham, United Kingdom. Let the eigenvalues of A
be λ1 , λ2 , . . . , λn . Consider the Jordan normal form of A; this
Jordan form is an upper triangular matrix that has the eigenva-
lues of A in the main diagonal. Let this matrix be called J . Furt-
hermore, det A = det J , as a matrix and its Jordan normal form
are similar. As J is upper triangular, its inverse is given by an
upper triangular matrix whose diagonal entries are the inverses
of the diagonal entries of J . That is,
λ1 + λ−1
1
0 λ2 + λ−1
2 U
−1
J + J −1 = 0 0 λ3 + λ3
.
.. .
.. .
.. . ..
x
−1
0 0 0 · · · λn + λn
## is an upper-triangular matrix, and its determinant can be compu-
ted simply as the product of the elements of the diagonal. Hence,
n
−1
Y 1
det(A + A )= λi + ≥ 2n ,
i=1
λi
## on account of the well-known fact that x + x1 ≥ 2 (which can be
proved simply by convexity or by the AM-GM inequality).
## Solution 2 by Henry Ricardo, Westchester Math Circle, New
York, USA. The determinant of a matrix is the product of its
eigenvalues. Also, if the eigenvalues of A ∈ Mn (R) are the po-
sitive real numbers λ1 , λ2 , . . . , λn , then the eigenvalues of A−1
72 Arhimede Mathematical Journal
## are 1/λ1 , 1/λ2 , . . . , 1/λn , and the eigenvalues of A + A−1 are
λ1 + 1/λ1 , λ2 + 1/λ2 , . . . , λn + 1/λn . Therefore
−1
1 1 1
det(A + A ) = λ1 + · λ2 + · · · · · λn +
λ1 λ2 λn
n
≥ 2
| · 2 {z
· · · · 2} = 2
n factors
## because a + 1/a ≥ 2 for any positive value of a, a consequence of
the AM-GM inequality. Equality holds in the determinant inequa-
lity if and only if λi = 1 for 1 ≤ i ≤ n.
## Solution 3 by Víctor Martín Chabrera, BarcelonaTech, Barce-
lona, Spain. Let B ∈ Mn (R) and let b1 , . . . , bn be its eigenvalues.
We know that
## • the eigenvalues of (B + λIn ) are b1 + λ, . . . , bn + λ;
• the eigenvalues of B m , where m ∈ Z, are bm m
1 , . . . , bn ;
• the determinant of a square matrix is the product of its eigen-
values.
## Let a1 , . . . , an be the eigenvalues (remember all of them are posi-
tive) of A. We have that
## det(A + A−1 ) = det(A−1 (A2 + In )) = det(A−1 ) det(A2 + I)
n n n
Y Y Y a2k + 1
= a−1
k a2
k + 1 =
k=1 k=1 k=1
ak
n n
Y 1 Y
= ak + ≥ 2 = 2n ,
k=1
ak k=1
## where weq have used AM-GM inequality to see that if x > 0, x +
1
x
≥ 2 x x1 = 2. Therefore, we reach equality if, and only if,
a1 = . . . = an = 1.
## Also solved by Ander Lamaison Vidarte, Berlin Mathematical
School, Berlin, Germany, and the proposer.
Volume 4, No. 1, Spring 2017 73
## A–36. Proposed by José Luis Díaz-Barrero, BarcelonaTech, Bar-
celona, Spain. Let xk (1 ≤ k ≤ n) be positive numbers and let
α ≥ 1. Prove that
n
1 X x2α 2α
i + xj 1X 1
≤ .
n−1 1≤i<j≤n
x2α+2
i + x2α+2
j 2 k=1 x2k
## Solution 1 by Henry Ricardo, Westchester Math Circle, New
York, USA. First we apply the power means inequality to see that
!1 ! 1
(x2i )α + (x2j )α α (x2i )α+1 + (x2j )α+1 α+1
≤ ,
2 2
or 1 α
x2α + x2α ≤ 2 α+1 · x2α+2 + x2α+2
α+1
i j i j . (6)
## Therefore, using (6) and the AM-GM inequality (twice), we have
2α+2
α
2α+2 α+1
X x2α
i + x 2α
j 1 X x i + x j
2α+2 ≤ 2
α+1 ·
2α+2 2α+2
x
1≤i<j≤n i
+ xj 1≤i<j≤n
xi + x2α+2
j
1 X 1
= 2 α+1 · 1
2α+2
+ x2α+2
α+1
1≤i<j≤n xi j
1 X 1
≤ 2 α+1 · q 1
α+1
1≤i<j≤n 2 x2α+2 · x2α+2
i j
X 1 X xi xj
= =
xx
1≤i<j≤n i j
x2 x2
1≤i<j≤n i j
X (x2i + x2j )/2
1≤i<j≤n
x2i x2j
!
1 X 1 1
= · +
2 1≤i<j≤n
x2i x2j
n
n−1 X 1
= · .
2 x2
k=1 k
## Dividing by (n − 1) gives us the desired result.
74 Arhimede Mathematical Journal
## Solution 2 by the proposer. First we write the given inequality
in the most convenient form
n
X x2α 2α
i + xj n−1X 1
≤ .
x2α+2 + x2α+2
1≤i<j≤n i j 2 x2
k=1 k
## Since the function f : [0, +∞) → R defined by f (x) = x2α+1 is
nonnegative and increasing, then for all a, b ∈ [0, +∞) we have
(a − b) a2α+1 − b2α+1 ≥ 0.
## 2α+2 2α+2 2α+1 2α+1
a2α+2 + b2α+2
a +b ≥ ba + ab , or ≥ ab.
a2α + a2α
Setting a = xi and b = xj into the preceding inequality yields
x2α+2
i + x2α+2
j x2α 2α
i + xj 1
≥ xi xj ⇐⇒ ≤ .
x2α
i + x2α
j x2α+2
i + x2α+2
j xi xj
## Taking into account the GH-HM inequality, we have
!
1 1 1 1
≤ + 2
xi xj 2 x2i xj
and
X x2α 2α
i + xj
X 1
x2α+2
1≤i<j≤n i
+ x2α+2
j 1≤i<j≤n
xi xj
! n
1 X 1 1 n−1X 1
≤ + = ,
2 1≤i<j≤n x2i x2j 2 k=1
x2k
## Also solved by Ander Lamaison Vidarte, Berlin Mathematical
School, Berlin, Germany.
Volume 4, No. 1, Spring 2017 75
## A–37. Proposed bypMihály Bencze, Braşov, Romania. If a1 = 1
2
and an+1 = 2an + 3an + 1, for all n ≥ 1, then prove that the
sequence (an )n≥1 contains only composite numbers for all n ≥ 2.
an + bn an − bn
Solution 1 by the proposer. Let u(n) = , v(n) = √ ,
√ √ 2 2 3
a = 2+ 3, b = 2− 3. Then, (u, v) = (u(n), v(n)) (n = 1, 2, . . . )
are all solutions of the equation u2 − 3v 2 = 1 (u, v ∈ N).
p an − bn an + bn
2v(n) + 3v 2 (n) + 1 = 2v(n) + u(n) = 2 +2
a−b a+b
n+1 n+1
a −b
= √ = v(n + 1) ⇒ an = v(n).
2 3
If n is a composite number, then n has a divisor m with 1 < m <
n. But v(m) is a positive integer satisfying 1 < v(m) < v(n) and
v(m)|v(n). This implies that v(n) is not prime. Therefore, if v(n)
is a prime, then n must be prime. Furthermore, since v(2) = 4,
n must be an odd prime.
ck + dk ck − dk 1+ 3
Let A(k) = √ , B(k) = √ when c = √ , d =
√ 2 6 2
1− 3
√ . Then (A, B) = (A(k), B(k)) (k = 1, 3, . . . ) are all soluti-
2
ons of the equation A2 − 3B 2 = −2 (A, B ∈ N).
Since a = c2 , b = d2 we get v(k) = A(k)B(k). If n is an odd
prime, then v(n) = A(n)B(n) > 1, v(n) > 1 for n > 1, so v(n)
is composite.
Solution 2 by Víctor Martín Chabrera, BarcelonaTech, Barce-
lona, Spain. We may start by observing that an > 0 ∀n ∈ N>0 .
In fact, the recurrence implies that an+1 > 2an . Also, we can find
a value for a0 which is compatible with the recurrence:
a1 = 1 = 2a0 + 3a0 + 1 =⇒ 1 + 4a20 − 4a0 = 3a20 + 1
⇐⇒ a0 (a0 − 4) = 0.
## We get two solutions, a0 = 0 and a0 = 4, from which only the first
one is compatible with the recurrence. So, for every non-negative
76 Arhimede Mathematical Journal
integer n,
q q
an+1 = 2an + 3an + 1 ⇐⇒ an+1 − 2an = 3a2n + 1
2
## ⇐⇒ a2n+1 + 4a2n − 4an an+1 = 3a2n + 1
⇐⇒ a2n+1 + a2n − 4an an+1 = 1.
Computing the first terms 1, 4, 15, 56, 209, . . . we may observe that
an+1 = 4an − an−1 . Let us prove this:
q
an+1 = 4an − an−1 ⇐⇒ 2an + 3a2n + 1 = 4an − an−1
q
⇐⇒ 3a2n + 1 = 2an − an−1
⇐⇒ 3a2n + 1 = 4a2n + a2n−1 − 4an−1 an
⇐⇒ a2n + a2n−1 − 4an−1 an = 1,
## which is true as we have previously shown. In particular, this
means an is integer ∀n ≥ 0.
2
Solving
√ x = 4x − 1 we get two solutions, x 1 = 2 + 3, x2 =
2 − 3. Hence, √ n the general√ nformula of (an ) will be of the form
an = A(2 + 3) + B(2 − 3) . We can determine the constants
A, B by solving, for example for a0 = 0, a1 = 1, from where we
1 1
get A = 2√ 3
, B = − 2√ 3
.
From an+1 = 4an − an−1 , we get an+1 ≡ an−1 (mod 2). Therefore,
as a2 = 4, we inductively
p find that an ≡
√ 0 (mod 2) if n is even.
Since an+1 = 2an + 3a2n + 1 > (2 + 3)an > 3an , (an )n∈N is
increasing, and therefore an ≥ 4 if n ≥ 2, which means an is
composite for all even numbers n ≥ 2.
## We are left to prove that an is composite for all odd n ≥ 3. We will
prove that a2n+1 − a2n = a2n+1 for every n ≥ 1, but first we will see
that a2n+1 − a2n is composite if n ≥ 1.
We have a2n+1 −a2n = (an+1 +an )(an+1 −an ). As an+1 +an > an+1 ≥
p √
a2 = 4 and an+1 − an = 2an + 3a2n + 1 − an > (1 + 3)an >
2a1 = 2, we have that a2n+1 − a2n is composite. We will take into
Volume 4, No. 1, Spring 2017 77
√ √
account that (2 + 3)(2 − 3) = 4 − 3 =1 to see that, from the
1
√ n √ n
formula an = 2√ 3
(2 + 3) − (2 − 3) ,
a2n+1 − a2n
√ n+1 √ n+12 √ n √ n2
1
= √ (2+ 3) −(2− 3) − (2+ 3) −(2− 3)
(2 3)2
1 √ √ √ √
= (2+ 3)2n+2 +(2− 3)2n+2 −2−(2+ 3)2n −(2− 3)2n +2
12
1 √ √ √ √
= (2 + 3)2n ((2 + 3)2 − 1) + (2 − 3)2n ((2 − 3)2 − 1) .
12
√ √ √
We can√ see that (2 + 3)2√− 1 = 4 + 4 3√+ 3 − 1 = 6 + 4 √3 =
12
√ ( 3 + 2) and that (2 − 3)2 − 1 = 4 − 4 3 + 3 − 1 = 6 − 4 3 =
2 3 √
− 212
√ (− 3 + 2). Therefore,
3
a2n+1 − a2n
1 √ 2n √ 2 √ 2n √ 2
= (2 + 3) ((2 + 3) − 1) + (2 − 3) ((2 − 3) − 1)
12
√ 2n+1 12 √ 2n+1 12
1
= (2 + 3) √ − (2 − 3) √
12 2 3 2 3
1 √ √
= √ (2 + 3)2n+1 − (2 − 3)2n+1 = a2n+1 .
2 3
Therefore, all odd numbers that can be written as 2n + 1, n ≥ 1,
which are the odd numbers equal or greater than 3, are composite
as well, thus completing the proof.
## A–38. Proposed by José Luis Díaz-Barrero and José Gibergans–
Báguena, BarcelonaTech, Barcelona, Spain. Let a, b be positive
real numbers. Prove that
Z 1
a−1 b−1
aΓ(a)Γ(b) a
t (1 − t) Γ(t) dt ≥ Γ ,
0 (a + b)Γ(a + b) a+b
where Γ(x) is the Euler Gamma Function.
Solution 1 by Víctor Martín Chabrera, BarcelonaTech, Bar-
celona, Spain. We can observe several things in the inequality.
78 Arhimede Mathematical Journal
First, we have something very similar to the beta function B(a, b),
which is defined as
Z 1
B(a, b) = ta−1 (1 − t)b−1 dt,
0
Γ(a)Γ(b)
Γ(a+b)
.
## The gamma function Γ(x), which is defined as
Z ∞
Γ(x) = tx−1 e−t dt,
0
## has the following well-known properties: it is continuous, Γ(x) >
0 if x > 0, Γ(x + 1) = xΓ(x), and, if n is a strictly positive integer,
Γ(n) = (n − 1)!.
We will now prove that Γ(x) is decreasing in (0, 1). We will use the
digamma function ψ(x), which relates to the gamma function in
the following way: Γ0 (x) = ψ(x)Γ(x). It is enough to prove that,
∀ x ∈ (0, 1), ψ(x) < 0, since, by the positivity of Γ(x), this would
imply Γ0 (x) < 0, ∀ x ∈ (0, 1). The integral representation of the
digamma function is also well-known, and it is the following:
Z ∞ −t
e−xt
e
ψ(x) = − dt.
0 t 1 − e−t
## As f (s) = es is an increasing function, for x ∈ (0, 1), t > 0, we
will have tx < t, which means that −tx > −t, which implies that
e−xt > e−t . Therefore, since t, e−t , 1 − e−t are positive functions
−t e−xt
over the strictly positive real numbers, we will have e t − 1−e −t <
e−t e−t
t
− 1−e−t
, and so,
∞ −t
e−xt ∞ −t
e−t
e e
Z Z
ψ(x) = − dt < − dt = ψ(1).
0 t 1 − e−t 0 t 1 − e−t
−t −t −t
We see that e t − 1−ee e
−t = t(1−e−t ) (1 − e
−t
− t). Since ex ≥ 1 +
x, ∀ x ∈ R, we will have 1 − e−t − t ≤ 1 − (1 − t) − t = 0. Therefore,
Volume 4, No. 1, Spring 2017 79
e −t e −t e−t
by the positivity of t(1−e−t ) , we have t
− 1−e −t ≤ 0, ∀ t > 0, and
## so, ∀ x ∈ (0, 1), we have
Z ∞ −t
e−t
e
ψ(x) < ψ(1) = − dt ≤ 0,
0 t 1 − e−t
as we wanted to see.
Let us prove that Γ(x) is a convex function over the strictly po-
sitive real numbers. As Γ(x) is continous (and well defined) over
the strictly positive real numbers, we have
Z ∞ Z ∞ 2
d2 d2 x−1 −t
2
Γ(x) = 2
t e dt = 2
(tx−1 e−t dt)
dx dx 0 0 ∂ x
Z ∞
= (ln(t))2 (tx−1 e−t dt) > 0
0
2 x−1 −t
since ln (t)t e > 0 ∀ t ∈ (0, ∞).
Therefore, since Γ(x) is positive and decreasing in (0, 1), we have
that, on the one hand,
Z 1 Z 1
a−1 b−1
t (1 − t) Γ(t) dt ≥ ta−1 (1 − t)b−1 min Γ(s) dt
0 0<s<1
Z01
= ta−1 (1 − t)b−1 Γ(1) dt
0
Z 1
= 0! ta−1 (1 − t)b−1 dt
0
Γ(a)Γ(b)
= B(a, b) = ,
Γ(a + b)
a
and, on the other hand, as a, b > 0, 1 < a+b + 1 < 2. By the
convexity of Γ(x), as Γ(1) = 0! = 1 = 1! = Γ(2), we have that,
Γ(c) ≤ 1, ∀ c ∈ (1, 2), and so, since xΓ(x) = Γ(x + 1),
a a a
0≤ Γ =Γ + 1 ≤ 1,
a+b a+b a+b
and therefore
Z 1
a−1 b−1
Γ(a)Γ(b) Γ(a)Γ(b) a a
t (1 − t) Γ(t)dt ≥ ≥ Γ ,
0 Γ(a + b) Γ(a + b) a + b a+b
thus proving the inequality.
80 Arhimede Mathematical Journal
## Solution 2 by the proposer. To prove the statement we will need
to apply Jensen’s inequality for integrals. Namely, if f : [a, b] → R
is a convex function and h : [a, b] → R∗+ and u : [a, b] → R+ are
integrable functions, then
Z b Z b
a h(x)u(x) dx a
h(x)f (u(x)) dx
f Z b ≤ Z b .
h(x) dx h(x)u(x) dx
a a
## Let X be a Beta’s random variable, whose probability density
function is
1
hX (t) = ta−1 (1 − t)b−1 I{0<t<1}
B(a, b)
Γ(a + b)
= ta−1 (1 − t)b−1 I{0<t<1}
Γ(a)Γ(b)
a
with expectation given by E(X) = a+b . Setting u(t) = t and
f (t) = Γ(t), that is convex in (0, 1), into Jensen’s inequality, we
have
Z 1
a a 1
Γ ≤ ta−1 (1 − t)b−1 Γ(t) dt
a+b a+b B(a, b) 0
Γ(a + b) 1 a−1
Z
= t (1 − t)b−1 Γ(t) dt.
Γ(a)Γ(b) 0
## A–39. Proposed by José Luis Díaz-Barrero, BarcelonaTech, Bar-
celona, Spain. Determine all matrices A ∈ M3 (R) such that
2 −2 0
adj(A) = −6 9 −1 .
8 −12 2
## Solution 1 by Xavier Ros Oton, Barcelona, Spain. It is well
known that
Volume 4, No. 1, Spring 2017 81
## where I is the identity matrix.
Then, since
−1
2 −2 0 1 3 2 1
adj(A)−1 = −6 9 −1 = 2 2 1 ,
8 −12 2 2 0 4 3
we have that
det A 3 2 1 3 2 1
A = det Aadj(A)−1 = 2 2 1 = λ 2 2 1 .
2 0 4 3 0 4 3
But, after some calculations, we get that
3 2 1 2 −2 0
adjλ 2 2 1 = λ2 −6 9 −1 ,
0 4 3 8 −12 2
so λ = ±1, and
3 2 1
A = ± 2 2 1 .
0 4 3
Solution 2 by Miguel Cidrás Senra, CFIS, BarcelonaTech, Bar-
celona, Spain. The adjugate of B ∈ Mn is defined as B·adj(B) =
det B · In . In our case, n = 3.
We can calculate det A, as
A · adj(A) = det A · I3 =⇒ det A · det adj(A) = det (det A · In )
=⇒ det A · 4 = (det A)3 · det I3 = (det A)3
=⇒ det A3 − 4 det A = 0
=⇒ det A = 0, 2 or − 2.
82 Arhimede Mathematical Journal
## If det A = 0, then 4 · A = 0 · adj(adj(A)) = 03 and A = 03 .
1
If det A = 2, then 4 · A = 2 · adj(adj(A)), so A =
2
−1
2
3 2 1 −3 −2 −1
So, there are only two matrices: 2 2 1 and −2 −2 −1 .
0 4 3 0 −4 −3
Solution 3 by Víctor Martín Chabrera, BarcelonaTech, Barce-
1
lona, Spain. Using that, if A is invertible, det(A−1 ) = det(A) and
n
that, if A ∈ Mn , det(λA) = λ det(A), where λ ∈ C, we can use
1
the well-known formula A−1 = det(A) adj(A) to see that
1 −1
1
= det(A ) = det adj(A)
det(A) det(A)
1
(det(A))n
that det(adj(A)) = 4 and n = 3, and so, det(A) = ±2.
1
We can go back again to the formula A−1 = det(A) adj(A) and take
the inverse to see that A = det(A)(adj(A)) = ±2(adj(A))−1 . If
−1
1 3 2 1
2 0 4 3
## from where we will get
3 2 1
A = ±2 2 1.
0 4 3
It is easy to check that for both signs the adjugate of A is the
matrix given in the statement.
Volume 4, No. 1, Spring 2017 83
## Solution 4 by the proposer. For every matrix A ∈ Mn (R) we
have
(7)
and
(8)
Since
2
−2 0
|adj(A)| = −6 9 −1 = 4,
8 −12 2
6 4 2
0 8 6
3 2 1
A = ± 2 2 1 .
0 4 3
## Also solved by Miguel Amengual Covas, Cala Figuera, Mallorca,
Spain, and Ander Lamaison Vidarte, Berlin Mathematical School,
Berlin, Germany.
## A–40. Proposed by José Luis Díaz-Barrero, BarcelonaTech, Bar-
celona, Spain. Let A = (aij ) be a 3 × 3 real orthogonal matrix
with det(A) = 1. Prove that
X
(trA − 1)2 + (aij − aji )2
i<j
## is an integer number and determine its value.
84 Arhimede Mathematical Journal
## Solution 1 by Eva Elduque, Zaragoza, Spain, and David Alfaya
X
(trA − 1)2 + (aij − aji )2
i<j
X
=(a11 +a22 +a33 )2 −2(a11 +a22 +a33 )+1+ (a2ij +a2ji −2aij aji )
i<j
3
X
= a2ij − 2(a11 + a22 + a33 )
i,j=1
a a12 a11 a13 a22 a23
+ 2 11
+ + + 1.
a21 a22 a31 a33 a23 a33
## Since A is orthogonal, then At = A−1 , Taking into account that
det A = 1, we have
det Cii
= det Cii ,
aii =
det A
where Cij denotes the matrix cofactor of aij . In this case,
a11 a12 a11 a13 a22 a23
a33 =
; a22 = ; a 11 = .
a21 a22 a31 a33 a23 a33
So,
X 3
X
2 2
(trA−1) + (aij −aji ) = a2ij +1 = tr(AAt )+1 = tr(I)+1 = 4,
i<j i,j=1
## and we are done.
Solution 2 by the proposer. Expanding the characteristic poly-
nomial of A as
det(A − tI) = t3 − (a11 + a22 + a33 )t2
+(a11 a22 + a11 a33 + a22 a33 − a12 a21 − a13 a31 − a23 a32 )t − 1,
multiplying out det(A−tI) = (t−t1 )(t−t2 )(t−t3 ) and comparing
both expressions, we get
t1 + t2 + t3 = a11 + a22 + a33 ,
t1 t2 + t2 t3 + t3 t1 = a11 a22 + a11 a33 + a22 a33
−a12 a21 − a13 a31 − a23 a32 .
Volume 4, No. 1, Spring 2017 85
## When det A = 1 then A has one eigenvalue equal to 1, as is well-
known. WLOG, we can assume that t1 = 1. Now from t1 t2 t3 = 1
it follows that t2 t3 = 1, and
X
(trA − 1)2 + (aij − aji )2
i<j
=(a11 +a22 +a33 − 1)2 +(a12 − a21 )2 +(a13 − a31 )2 +(a23 − a32 )2
=(a211 + a212 + a213 ) + (a221 + a222 + a223 ) + (a231 + a232 + a233 ) + 1
X X X3
+ 2 aii ajj − aij aji − aii
1≤i<j≤3 1≤i<j≤3 i=1
=1 + 1 + 1 + 1 + 2(t1 t2 + t2 t3 + t3 t1 ) − 2(t1 + t2 + t3 ) = 4.
## Also solved by Víctor Martín Chabrera, BarcelonaTech, Barcelona,
Spain.
Arhimede Mathematical Journal
Editor-in-Chief
## José Luis Díaz-Barrero
BarcelonaTech, Barcelona, Spain.
Editors
Alberto Espuny Díaz Birmingham, United Kingdom.
Ander Lamaison Vidarte Berlin, Germany.
Editorial Board
Mihály Bencze Braşov, Romania.
José Gibergans-Báguena Barcelona, Spain.
Nicolae Papacu Slobozia, Romania.
## Managing and Subscription Editors
Petrus Alexandrescu Bucharest, Romania.
José Luis Díaz-Barrero Barcelona, Spain.
Aim and Scope
## The goal of Arhimede Mathematical Journal is to provide a means of pu-
blication of useful materials to train students for Mathematical Conte-
sts at all levels. Potential contributions include any work involving fresh
ideas and techniques, problems and lessons helpful to train contestants,
all written in a clear and elegant mathematical style. All areas of mathe-
matics, including algebra, combinatorics, geometry, number theory and
real and complex analysis, are considered appropriate for the journal.
## Information for Authors
A detailed statement of author guidelines is available at the
journal’s website:
http://www.amj-math.com
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Definition: Injective Function
Definition: A function f from the set A to the set B ($f:A\rightarrow B$) is injective if for all elements “a” and “b” in the set A, $f(a)=f(b)$ implies that a=b. //
This means that if you tell me that two elements in A get sent to the same element in B, and moreover if you tell me that this function is injective, then I immediately know that the two elements in A that you’re talking about are really the same element. This is just a formal way of saying that “no two elements in A get sent to the same element in B”. In our example of the school dance in lesson 7, the idea of injectivity was described as the scenario in which no two boys were dancing with the same girl. We note that this in no way implies that every girl has a dance partner, but rather just that those girls who do have dance partners only have one. This does, however, motivate the next definition.
For more on the notion of injectivity, check out lesson 7.
Back to Glossary
Back to Lessons
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Velocity Reviews (http://www.velocityreviews.com/forums/index.php)
- C Programming (http://www.velocityreviews.com/forums/f42-c-programming.html)
- - KISS4691, a potentially top-ranked RNG. (http://www.velocityreviews.com/forums/t729111-kiss4691-a-potentially-top-ranked-rng.html)
geo 07-24-2010 01:02 PM
KISS4691, a potentially top-ranked RNG.
I have been asked to recommend an RNG
(Random Number Generator) that ranks
at or near the top in all of the categories:
performance on tests of randomness,
length of period, simplicity and speed.
The most important measure, of course, is
performance on extensive tests of randomness, and for
those that perform well, selection may well depend
on those other measures.
The following KISS version, perhaps call it KISS4691,
seems to rank at the top in all of those categories.
It is my latest, and perhaps my last, as, at age 86,
I am slowing down.
Compiling and running the following commented
C listing should produce, within about four seconds,
10^9 calls to the principal component MWC(), then
10^9 calls to the KISS combination in another ~7 seconds.
Try it; you may like it.
George Marsaglia
/*
The KISS4691 RNG, a Keep-It-Simple-Stupid combination of
MWC (Multiply-With-Carry), Congruential and Xorshift sequences.
Expressed as a simple MWC() function and C macros
#define CNG ( xcng=69069*xcng+123 )
#define XS ( xs^=(xs<<13), xs^=(xs>>17), xs^=(xs<<5) )
#define KISS ( MWC()+CNG+XS )
but easily expressed in other languages, with a slight
complication for signed integers.
With its immense period, >10^45000, and speed: MWC()s at
around 238 million/sec or full KISSes at around 138 million,
there are few RNGs that do as well as this one
on tests of randomness and are comparable in even one
of the categories: period, speed, simplicity---not to
mention comparable in all of them.
The MWC4691 sequence x[n]=8193*x[n-4691]+carry mod b=2^32
is based on p=8193*b^4691-1, period ~ 2^150124 or 10^45192
For the MWC (Multiply-With-Carry) process, given a current
x and c, the new x is (8193*x+c) mod b,
the new c is the integer part of (8193*x+c)/b.
The routine MWC() produces, in reverse order, the base-b=2^32
expansion of some j/p with 0<j<p=8193*2^150112-1 with j
determined by the 64 bits of seeds xcng and xs, or more
generally, by 150112 random bits in the Q[] array.
*/
static unsigned long xs=521288629,xcng=362436069,Q[4691];
unsigned long MWC(void) /*takes about 4.2 nanosecs or 238 million/
second*/
{unsigned long t,x,i; static c=0,j=4691;
j=(j<4690)? j+1:0;
x=Q[j];
t=(x<<13)+c+x; c=(t<x)+(x>>19);
return (Q[j]=t);
}
#define CNG ( xcng=69069*xcng+123 )
#define XS ( xs^=(xs<<13), xs^=(xs>>17), xs^=(xs<<5) )
#define KISS ( MWC()+CNG+XS ) /*138 million/sec*/
#include <stdio.h>
int main()
{unsigned long i,x;
for(i=0;i<4691;i++) Q[i]=CNG+XS;
for(i=0;i<1000000000;i++) x=MWC();
printf(" MWC result=3740121002 ?\n%22u\n",x);
for(i=0;i<1000000000;i++) x=KISS;
printf("KISS result=2224631993 ?\n%22u\n",x);
}
/*
This RNG uses two seeds to fill the Q[4691] array by
means of CNG+XS mod 2^32. Users requiring more than two
seeds will need to randomly seed Q[] in main().
By itself, the MWC() routine passes all tests in the
Diehard Battery of Tests, but it is probably a good
idea to include it in the KISS combination.
Languages requiring signed integers should use equivalents
of the same operations, except that the C statement:
c=(t<x)+(x>>19);
can be replaced by that language's version of
if sign(x<<13+c)=sign(x) then c=sign(x)+(x>>19)
else c=1-sign(t)+(x>>19)
*/
jacob navia 07-24-2010 08:35 PM
Re: KISS4691, a potentially top-ranked RNG.
geo a écrit :
> I have been asked to recommend an RNG
> (Random Number Generator) that ranks
> at or near the top in all of the categories:
> performance on tests of randomness,
> length of period, simplicity and speed.
> The most important measure, of course, is
> performance on extensive tests of randomness, and for
> those that perform well, selection may well depend
> on those other measures.
>
> The following KISS version, perhaps call it KISS4691,
> seems to rank at the top in all of those categories.
> It is my latest, and perhaps my last, as, at age 86,
> I am slowing down.
>
> Compiling and running the following commented
> C listing should produce, within about four seconds,
> 10^9 calls to the principal component MWC(), then
> 10^9 calls to the KISS combination in another ~7 seconds.
>
> Try it; you may like it.
>
> George Marsaglia
>
>
> /*
> The KISS4691 RNG, a Keep-It-Simple-Stupid combination of
> MWC (Multiply-With-Carry), Congruential and Xorshift sequences.
> Expressed as a simple MWC() function and C macros
> #define CNG ( xcng=69069*xcng+123 )
> #define XS ( xs^=(xs<<13), xs^=(xs>>17), xs^=(xs<<5) )
> #define KISS ( MWC()+CNG+XS )
> but easily expressed in other languages, with a slight
> complication for signed integers.
>
> With its immense period, >10^45000, and speed: MWC()s at
> around 238 million/sec or full KISSes at around 138 million,
> there are few RNGs that do as well as this one
> on tests of randomness and are comparable in even one
> of the categories: period, speed, simplicity---not to
> mention comparable in all of them.
>
> The MWC4691 sequence x[n]=8193*x[n-4691]+carry mod b=2^32
> is based on p=8193*b^4691-1, period ~ 2^150124 or 10^45192
> For the MWC (Multiply-With-Carry) process, given a current
> x and c, the new x is (8193*x+c) mod b,
> the new c is the integer part of (8193*x+c)/b.
>
> The routine MWC() produces, in reverse order, the base-b=2^32
> expansion of some j/p with 0<j<p=8193*2^150112-1 with j
> determined by the 64 bits of seeds xcng and xs, or more
> generally, by 150112 random bits in the Q[] array.
> */
>
> static unsigned long xs=521288629,xcng=362436069,Q[4691];
>
> unsigned long MWC(void) /*takes about 4.2 nanosecs or 238 million/
> second*/
> {unsigned long t,x,i; static c=0,j=4691;
> j=(j<4690)? j+1:0;
> x=Q[j];
> t=(x<<13)+c+x; c=(t<x)+(x>>19);
> return (Q[j]=t);
> }
>
> #define CNG ( xcng=69069*xcng+123 )
> #define XS ( xs^=(xs<<13), xs^=(xs>>17), xs^=(xs<<5) )
> #define KISS ( MWC()+CNG+XS ) /*138 million/sec*/
>
> #include <stdio.h>
> int main()
> {unsigned long i,x;
> for(i=0;i<4691;i++) Q[i]=CNG+XS;
> for(i=0;i<1000000000;i++) x=MWC();
> printf(" MWC result=3740121002 ?\n%22u\n",x);
> for(i=0;i<1000000000;i++) x=KISS;
> printf("KISS result=2224631993 ?\n%22u\n",x);
> }
>
> /*
> This RNG uses two seeds to fill the Q[4691] array by
> means of CNG+XS mod 2^32. Users requiring more than two
> seeds will need to randomly seed Q[] in main().
> By itself, the MWC() routine passes all tests in the
> Diehard Battery of Tests, but it is probably a good
> idea to include it in the KISS combination.
>
> Languages requiring signed integers should use equivalents
> of the same operations, except that the C statement:
> c=(t<x)+(x>>19);
> can be replaced by that language's version of
> if sign(x<<13+c)=sign(x) then c=sign(x)+(x>>19)
> else c=1-sign(t)+(x>>19)
> */
This doesn't work with systems that have unsigned long as a 64 bit quantity.
I obtain:
MWC result=3740121002 ?
4169348530
KISS result=2224631993 ?
1421918629
Compiling with 32 bit machine yields:
MWC result=3740121002 ?
3740121002
KISS result=2224631993 ?
2224631993
Gib Bogle 07-25-2010 03:34 AM
Re: KISS4691, a potentially top-ranked RNG.
geo wrote:
>
> Languages requiring signed integers should use equivalents
> of the same operations, except that the C statement:
> c=(t<x)+(x>>19);
> can be replaced by that language's version of
> if sign(x<<13+c)=sign(x) then c=sign(x)+(x>>19)
> else c=1-sign(t)+(x>>19)
> */
Hi George,
I translated this into Fortran, and found that I get different results than with
C. I've tracked the difference into MWC(). The following Fortran code, with my
laborious comparison of two signed integers treating them as unsigned, gives
correct results. If I comment out the line
c = tLTx + SHIFTR(x,19)
and uncomment the following lines that implement your suggestion above to
compute c, I get different results.
integer function MWC()
integer :: t, x, i
integer, save :: c = 0, j = 4691
integer :: tLTx
if (j < 4690) then
j = j + 1
else
j = 0
endif
x = Q(j)
t = SHIFTL(x,13) + c + x
if ((t >= 0 .and. x >= 0) .or. (t < 0 .and. x < 0)) then
if (t < x) then
tLTx = 1
else
tLTx = 0
endif
elseif (x < 0) then
tLTx = 1
elseif (t < 0) then
tLTx = 0
endif
c = tLTx + SHIFTR(x,19)
!if (sign(1,SHIFTL(x,13)+c) == sign(1,x)) then
! c = sign(1,x) + SHIFTR(x,19)
!else
! c = 1 - sign(1,t) + SHIFTR(x,19)
!endif
Q(j) = t
MWC = t
end function
Is it the case that although your suggested workaround gives different results
from the C code in this case, it is still equivalent as a RNG?
Cheers
Gib
geo 07-25-2010 01:53 PM
Re: KISS4691, a potentially top-ranked RNG.
On Jul 24, 11:34*pm, Gib Bogle <g.bo...@auckland.no.spam.ac.nz> wrote:
> geo wrote:
>
> > Languages requiring signed integers should use equivalents
> > of the same operations, except that the C statement:
> > * * * *c=(t<x)+(x>>19);
> > can be replaced by that language's version of
> > * if sign(x<<13+c)=sign(x) then c=sign(x)+(x>>19)
> > * * * * * *else c=1-sign(t)+(x>>19)
> > */
>
> Hi George,
>
> I translated this into Fortran, and found that I get different results than with
> C. *I've tracked the difference into MWC(). *The following Fortran code, with my
> laborious comparison of two signed integers treating them as unsigned, gives
> correct results. *If I comment out the line
> c = tLTx + SHIFTR(x,19)
> and uncomment the following lines that implement your suggestion above to
> compute c, I get different results.
>
> integer function MWC()
> integer :: t, x, i
> integer, save :: c = 0, j = 4691
> integer :: tLTx
>
> if (j < 4690) then
> * * *j = j + 1
> else
> * * *j = 0
> endif
> x = Q(j)
> t = SHIFTL(x,13) + c + x
> if ((t >= 0 .and. x >= 0) .or. (t < 0 .and. x < 0)) then
> * * *if (t < x) then
> * * * * *tLTx = 1
> * * *else
> * * * * *tLTx = 0
> * * *endif
> elseif (x < 0) then
> * * *tLTx = 1
> elseif (t < 0) then
> * * *tLTx = 0
> endif
>
> c = tLTx + SHIFTR(x,19)
>
> !if (sign(1,SHIFTL(x,13)+c) == sign(1,x)) then
> ! * *c = sign(1,x) + SHIFTR(x,19)
> !else
> ! * *c = 1 - sign(1,t) + SHIFTR(x,19)
> !endif
> Q(j) = t
> MWC = t
> end function
>
> Is it the case that although your suggested workaround gives different results
> from the C code in this case, it is still equivalent as a RNG?
>
> Cheers
> Gib
Thanks very much for the Fortran version.
I made a mistake in the comment on versions
for signed integers. This:
Languages requiring signed integers should use equivalents
of the same operations, except that the C statement:
c=(t<x)+(x>>19);
can be replaced by that language's version of
if sign(x<<13+c)=sign(x) then c=sign(x)+(x>>19)
else c=1-sign(t)+(x>>19)
should have been:
Languages requiring signed integers should use equivalents
of the same operations, except that the C statement:
c=(t<x)+(x>>19);
can be replaced by that language's version of
if sign(x<<13+c)=sign(x) then c=sign(x)+(x>>19)
else c=1-sign(x<<13+c)+(x>>19)
Sorry for that error.
I still like inline functions in Fortan,
so would tend to define
isign(x)=ishft(x,-31)
and
m=ishft(x,13)+c
if(isign(m).eq.isign(x)) then c=isign(x)+ishft(x,-19)
else c=1-isign(m)+ishft(x,-19)
and
Q[j]=m+x
If calculating the carry c of the MWC operation fails
to fix that extra increment properly, then rather than
a systematic expansion, in reverse order, 32 bits at a time,
of the binary representation of j/(1893*2^150112-1) for some
j determined by the random seeds, we will be jumping around
in that expansion, and we can't be sure that the period will
still be the order of b=2^32 for the prime p=1893*b^4196-1.
gm
Dick Hendrickson 07-25-2010 09:13 PM
Re: KISS4691, a potentially top-ranked RNG.
On 7/25/10 8:53 AM, geo wrote:
[snip]
>
> I still like inline functions in Fortan,
> so would tend to define
> isign(x)=ishft(x,-31)
I don't know anything about random numbers, but "sign" is an intrinsic
function in Fortran and compilers should generate near perfect code
inline for things like sign(1,x). The generic sign function was added
to Fortran 90. In FORTRAN 77, you'd need to use the specific ISIGN(1,X)
function.
Dick Hendrickson
[snip]
>
> gm
Gib Bogle 07-26-2010 12:00 AM
Re: KISS4691, a potentially top-ranked RNG.
geo wrote:
> Thanks very much for the Fortran version.
> I made a mistake in the comment on versions
> for signed integers. This:
>
> Languages requiring signed integers should use equivalents
> of the same operations, except that the C statement:
> c=(t<x)+(x>>19);
> can be replaced by that language's version of
> if sign(x<<13+c)=sign(x) then c=sign(x)+(x>>19)
> else c=1-sign(t)+(x>>19)
>
> should have been:
>
>
> Languages requiring signed integers should use equivalents
> of the same operations, except that the C statement:
> c=(t<x)+(x>>19);
> can be replaced by that language's version of
> if sign(x<<13+c)=sign(x) then c=sign(x)+(x>>19)
> else c=1-sign(x<<13+c)+(x>>19)
>
> Sorry for that error.
That produces different c values from those generated by the method based on the
value of (t<x), therefore it deviates from the C code. This is what I used:
m = shiftl(x,13) + c
if (sign(1,m) == sign(1,x)) then
c = sign(1,x) + shiftr(x,19)
else
c = 1 - sign(1,m) + shiftr(x,19)
endif
robin 07-26-2010 02:38 AM
Re: KISS4691, a potentially top-ranked RNG.
"geo" <gmarsaglia@gmail.com> wrote in message news:a82cebe3-cdb9-48af-8080-bca935eeb9b1@l14g2000yql.googlegroups.com...
|I have been asked to recommend an RNG
| (Random Number Generator) that ranks
| at or near the top in all of the categories:
| performance on tests of randomness,
| length of period, simplicity and speed.
| The most important measure, of course, is
| performance on extensive tests of randomness, and for
| those that perform well, selection may well depend
| on those other measures.
|
| The following KISS version, perhaps call it KISS4691,
| seems to rank at the top in all of those categories.
| It is my latest, and perhaps my last, as, at age 86,
| I am slowing down.
|
| Compiling and running the following commented
| C listing should produce, within about four seconds,
| 10^9 calls to the principal component MWC(), then
| 10^9 calls to the KISS combination in another ~7 seconds.
|
| Try it; you may like it.
|
| George Marsaglia
Here's the PL/I version:
(NOSIZE, NOFOFL):
RNG: PROCEDURE OPTIONS (MAIN, REORDER);
declare (xs initial (521288629), xcng initial (362436069),
Q(0:4690) ) static fixed binary (32) unsigned;
MWC: procedure () returns (fixed binary (32) unsigned);
/*takes about 4.2 nanosecs or 238 million/second*/
declare (t,x,i) fixed binary (32) unsigned;
declare (c initial (0), j initial (4691) ) fixed binary (32) unsigned static;
if j < 4690 then j = j + 1; else j = 0;
x = Q(j);
t = isll(x,13)+c+x; c = (t<x)+isrl(x,19);
Q(j)=t;
return (t);
end MWC;
CNG: procedure returns (fixed binary (32) unsigned);
xcng=bin(69069)*xcng+bin(123);
return (xcng);
end CNG;
XXS: procedure returns (fixed binary (32) unsigned);
xs = ieor (xs, isll(xs, 13) );
xs = ieor (xs, isrl(xs, 17) );
xs = ieor (xs, isll(xs, 5) );
return (xs);
end XXS;
KISS: procedure returns (fixed binary (32) unsigned);
return ( MWC()+CNG+XXS ); /*138 million/sec*/
end KISS;
declare (i,x) fixed binary (32) unsigned;
/* Initialize: */
do i = 0 to 4691-1; Q(i) = CNG+XXS; end;
put skip list (q(0), q(4690));
put skip list ('initialized'); put skip;
do i = 0 to 1000000000-1; x=MWC(); end;
put skip edit (" MWC result=3740121002 ",x) (a, f(23));
do i = 0 to 1000000000-1; x=KISS; end;
put skip edit ("KISS result=2224631993 ",x) (a, f(23));
end RNG;
robin 07-26-2010 01:32 PM
Re: KISS4691, a potentially top-ranked RNG.
"Gib Bogle" <g.bogle@auckland.no.spam.ac.nz> wrote in message news:i2ij74\$kd6\$1@speranza.aioe.org...
| geo wrote:
| > Thanks very much for the Fortran version.
| > I made a mistake in the comment on versions
| > for signed integers. This:
| >
| > Languages requiring signed integers should use equivalents
| > of the same operations, except that the C statement:
| > c=(t<x)+(x>>19);
| > can be replaced by that language's version of
| > if sign(x<<13+c)=sign(x) then c=sign(x)+(x>>19)
| > else c=1-sign(x<<13+c)+(x>>19)
| >
| > Sorry for that error.
|
| That produces different c values from those generated by the method based on the
| value of (t<x), therefore it deviates from the C code. This is what I used:
|
| m = shiftl(x,13) + c
| if (sign(1,m) == sign(1,x)) then
| c = sign(1,x) + shiftr(x,19)
| else
| c = 1 - sign(1,m) + shiftr(x,19)
| endif
Maybe I missed something,
but isn't this exactly equivalent to what George wrote?
Just substitute x<<13+c for m in your last two assignments ...
Harald Anlauf 07-26-2010 07:36 PM
Re: KISS4691, a potentially top-ranked RNG.
On Jul 24, 3:02*pm, geo <gmarsag...@gmail.com> wrote:
> This RNG uses two seeds to fill the Q[4691] array by
> means of CNG+XS mod 2^32. Users requiring more than two
> seeds will need to randomly seed Q[] in main().
> By itself, the MWC() routine passes all tests in the
> Diehard Battery of Tests, but it is probably a good
> idea to include it in the KISS combination.
Does this mean that using different seeds will lead to
streams that are always statistically independent
(as long as one does not exhaust the RNG's period)?
Or are there restrictions on the possible combinations
of seeds?
I am currently using D.E.Knuth's generator from TAOCP,
which IIRC allows for 2^30-2 independent streams, and
which are asserted to be independent, but being able to
extend the "limit" would be nice.
Harald
Gib Bogle 07-26-2010 08:06 PM
Re: KISS4691, a potentially top-ranked RNG.
robin wrote:
> "Gib Bogle" <g.bogle@auckland.no.spam.ac.nz> wrote in message news:i2ij74\$kd6\$1@speranza.aioe.org...
> | geo wrote:
> | > Thanks very much for the Fortran version.
> | > I made a mistake in the comment on versions
> | > for signed integers. This:
> | >
> | > Languages requiring signed integers should use equivalents
> | > of the same operations, except that the C statement:
> | > c=(t<x)+(x>>19);
> | > can be replaced by that language's version of
> | > if sign(x<<13+c)=sign(x) then c=sign(x)+(x>>19)
> | > else c=1-sign(x<<13+c)+(x>>19)
> | >
> | > Sorry for that error.
> |
> | That produces different c values from those generated by the method based on the
> | value of (t<x), therefore it deviates from the C code. This is what I used:
> |
> | m = shiftl(x,13) + c
> | if (sign(1,m) == sign(1,x)) then
> | c = sign(1,x) + shiftr(x,19)
> | else
> | c = 1 - sign(1,m) + shiftr(x,19)
> | endif
>
> Maybe I missed something,
> but isn't this exactly equivalent to what George wrote?
> Just substitute x<<13+c for m in your last two assignments ...
I hope so. Maybe I didn't express myself clearly enough. I'll try again.
Using my implementation of George's corrected code, I get results from the
Fortran code that differ from those generated by his C code.
All times are GMT. The time now is 02:27 PM.
| 6,046
| 19,065
|
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| 2.578125
| 3
|
CC-MAIN-2014-10
|
latest
|
en
| 0.896971
|
http://www.reference.com/browse/wiki/Las_Vegas_algorithm
| 1,386,958,829,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2013-48/segments/1386164974477/warc/CC-MAIN-20131204134934-00090-ip-10-33-133-15.ec2.internal.warc.gz
| 509,970,646
| 22,237
|
Definitions
# Las Vegas algorithm
In computing, a Las Vegas algorithm is a randomized algorithm that never gives incorrect results; that is, it always produces the correct result or it informs about the failure. In other words, a Las Vegas algorithm does not gamble with the verity of the result --- it only gambles with the resources used for the computation. A simple example is randomized quicksort, where the pivot is chosen randomly, but the result is always sorted. The usual definition of a Las Vegas algorithm includes the restriction that the expected run time always be finite, when the expectation is carried out over the space of random information, or entropy, used in the algorithm.
The complexity class of decision problems that have Las Vegas algorithms with expected polynomial runtime is ZPP.
It turns out that
$textrm\left\{ZPP\right\} = textrm\left\{RP\right\} cap ,co,textrm\left\{-RP\right\}, ,!$
which is intimately connected with the way Las Vegas algorithms are sometimes constructed. Namely the class RP consists of all decision problems for which a randomized polynomial-time algorithm exists that always answers correctly when the correct answer is "no", but is allowed to be wrong with a certain probability bounded away from one when the answer is "yes". When such an algorithm exists for both a problem and its complement (with the answers "yes" and "no" swapped), the two algorithms can be run simultaneously and repeatedly: a few steps of each, taking turns, until one of them returns a definitive answer. This is the standard way to construct a Las Vegas algorithm that runs in expected polynomial time. Note that in general there is no worst case upper bound on the run time of a Las Vegas algorithm.
Las Vegas algorithms can be contrasted with Monte Carlo algorithms, in which the resources used are bounded but the answer is not guaranteed to be correct 100% of the time.
| 395
| 1,915
|
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| 2.90625
| 3
|
CC-MAIN-2013-48
|
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|
en
| 0.916865
|
https://wikivisually.com/wiki/Equation_of_state_(cosmology)
| 1,545,182,504,000,000,000
|
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|
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| 798,902,962
| 36,390
|
# Equation of state (cosmology)
In cosmology, the equation of state of a perfect fluid is characterized by a dimensionless number ${\displaystyle w}$, equal to the ratio of its pressure ${\displaystyle p}$ to its energy density ${\displaystyle \rho }$ :
${\displaystyle w={\frac {p}{\rho }}}$.
It is closely related to the thermodynamic equation of state and ideal gas law.
## The equation
The perfect gas equation of state may be written as
${\displaystyle p=\rho _{m}RT=\rho _{m}C^{2}}$
where ${\displaystyle \rho _{m}}$ is the mass density, ${\displaystyle R}$ is the particular gas constant, ${\displaystyle T}$ is the temperature and ${\displaystyle C={\sqrt {RT}}}$ is a characteristic thermal speed of the molecules. Thus
${\displaystyle w={\frac {p}{\rho }}={\frac {\rho _{m}C^{2}}{\rho _{m}c^{2}}}={\frac {C^{2}}{c^{2}}}\approx 0}$
where ${\displaystyle c}$ is the speed of light, ${\displaystyle \rho =\rho _{m}c^{2}}$ and ${\displaystyle C\ll c}$ for a "cold" gas.
### FLRW equations and the equation of state
The equation of state may be used in Friedmann–Lemaître–Robertson–Walker equations to describe the evolution of an isotropic universe filled with a perfect fluid. If ${\displaystyle a}$ is the scale factor then
${\displaystyle \rho \propto a^{-3(1+w)}.}$
If the fluid is the dominant form of matter in a flat universe, then
${\displaystyle a\propto t^{\frac {2}{3(1+w)}},}$
where ${\displaystyle t}$ is the proper time.
In general the Friedmann acceleration equation is
${\displaystyle 3{\frac {\ddot {a}}{a}}=\Lambda -4\pi G(\rho +3p)}$
where ${\displaystyle \Lambda }$ is the cosmological constant and ${\displaystyle G}$ is Newton's constant, and ${\displaystyle {\ddot {a}}}$ is the second proper time derivative of the scale factor.
If we define (what might be called "effective") energy density and pressure as
${\displaystyle \rho ^{\prime }\equiv \rho +{\frac {\Lambda }{8\pi G}}}$
${\displaystyle p^{\prime }\equiv p-{\frac {\Lambda }{8\pi G}}}$
and
${\displaystyle p^{\prime }=w^{\prime }\rho ^{\prime }}$
the acceleration equation may be written as
${\displaystyle {\frac {\ddot {a}}{a}}=-{\frac {4}{3}}\pi G\left(\rho ^{\prime }+3p^{\prime }\right)=-{\frac {4}{3}}\pi G(1+3w^{\prime })\rho ^{\prime }}$
### Non-relativistic matter
The equation of state of ordinary non-relativistic matter (e.g. cold dust) is ${\displaystyle w=0}$, which means that it is diluted as ${\displaystyle \rho \propto a^{-3}=V^{-1}}$, where ${\displaystyle V}$ is the volume. This means that the energy density red-shifts as the volume, which is natural for ordinary non-relativistic matter.
### Ultra-relativistic matter
The equation of state of ultra-relativistic matter (e.g. radiation, but also matter in the very early universe) is ${\displaystyle w=1/3}$ which means that it is diluted as ${\displaystyle \rho \propto a^{-4}}$. In an expanding universe, the energy density decreases more quickly than the volume expansion, because radiation has momentum and, by the de Broglie hypothesis a wavelength, which is red-shifted.
### Acceleration of cosmic inflation
Cosmic inflation and the accelerated expansion of the universe can be characterized by the equation of state of dark energy. In the simplest case, the equation of state of the cosmological constant is ${\displaystyle w=-1}$. In this case, the above expression for the scale factor is not valid and ${\displaystyle a\propto e^{Ht}}$, where the constant H is the Hubble parameter. More generally, the expansion of the universe is accelerating for any equation of state ${\displaystyle w<-1/3}$. The accelerated expansion of the Universe was indeed observed.[1] According to observations, the value of equation of state of cosmological constant is near -1.
Hypothetical phantom energy would have an equation of state ${\displaystyle w<-1}$, and would cause a Big Rip. Using the existing data, it is still impossible to distinguish between phantom ${\displaystyle w<-1}$ and non-phantom ${\displaystyle w\geq -1}$.
### Fluids
In an expanding universe, fluids with larger equations of state disappear more quickly than those with smaller equations of state. This is the origin of the flatness and monopole problems of the big bang: curvature has ${\displaystyle w=-1/3}$ and monopoles have ${\displaystyle w=0}$, so if they were around at the time of the early big bang, they should still be visible today. These problems are solved by cosmic inflation which has ${\displaystyle w\approx -1}$. Measuring the equation of state of dark energy is one of the largest efforts of observational cosmology. By accurately measuring ${\displaystyle w}$, it is hoped that the cosmological constant could be distinguished from quintessence which has ${\displaystyle w\neq -1}$.
### Scalar modeling
A scalar field ${\displaystyle \phi }$ can be viewed as a sort of perfect fluid with equation of state
${\displaystyle {w={\frac {{\frac {1}{2}}{\dot {\phi }}^{2}-V(\phi )}{{\frac {1}{2}}{\dot {\phi }}^{2}+V(\phi )}},}}$
where ${\displaystyle {\dot {\phi }}}$ is the time-derivative of ${\displaystyle \phi }$ and ${\displaystyle V(\phi )}$ is the potential energy. A free ${\displaystyle (V=0)}$ scalar field has ${\displaystyle w=1}$, and one with vanishing kinetic energy is equivalent to a cosmological constant: ${\displaystyle w=-1}$. Any equation of state in between, but not crossing the ${\displaystyle w=-1}$ barrier known as the Phantom Divide Line (PDL),[2] is achievable, which makes scalar fields useful models for many phenomena in cosmology.
## Notes
1. ^ Hogan, Jenny. "Welcome to the Dark Side." Nature 448.7151 (2007): 240-245. http://www.nature.com/nature/journal/v448/n7151/full/448240a.html
2. ^ Vikman, Alexander (2005). "Can dark energy evolve to the Phantom?". Phys. Rev. D. 71: 023515. arXiv:. Bibcode:2005PhRvD..71b3515V. doi:10.1103/PhysRevD.71.023515.
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Data Units Calculator
Mebibyte to Gigabit
Online data storage unit conversion calculator:
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The smallest unit of measurement used for measuring data is a bit. A single bit can have a value of either zero(0) or one(1). It may contain a binary value (such as True/False or On/Off or 1/0) and nothing more. Therefore, a byte, or eight bits, is used as the fundamental unit of measurement for data storage. A byte can store 256 different values, which is sufficient to represent standard ASCII table, such as all numbers, letters and control symbols.
Since most files contain thousands of bytes, file sizes are often measured in kilobytes. Larger files, such as images, videos, and audio files, contain millions of bytes and therefore are measured in megabytes. Modern storage devices can store thousands of these files, which is why storage capacity is typically measured in gigabytes or even terabytes.
11 mib to gbit result:
11 (eleven) mebibyte(s) is equal 0.092274688 (zero point zero × 1 ninety-two million two hundred and seventy-four thousand six hundred and eighty-eight) gigabit(s)
What is mebibyte?
The mebibyte is a multiple of the unit byte for digital information. The binary prefix mebi means 2^20; therefore one mebibyte is equal to 1048576 bytes, i.e., 1024 kibibytes. The unit symbol for the mebibyte is MiB.
What is gigabit?
The gigabit is a multiple of the unit bit for digital information or computer storage. The prefix giga (symbol G) is defined in the International System of Units (SI) as a multiplier of 10^9 (1 billion, short scale), and therefore, 1 gigabit = 10^9 bits = 1000000000 bits. The gigabit has the unit symbol Gbit.
How calculate mib. to gbit.?
1 Mebibyte is equal to 0.008388608 Gigabit (zero point zero × 2 eight million three hundred and eighty-eight thousand six hundred and eight gbit)
1 Gigabit is equal to 119.20928955078125 Mebibyte (one hundred and nineteen point twenty trillion nine hundred and twenty-eight billion nine hundred and fifty-five million seventy-eight thousand one hundred and twenty-five mib)
1 Mebibyte is equal to 8388608.000000 bits (eight million three hundred and eighty-eight thousand six hundred and eight point zero × 6 zero bits)
1 Gigabit is equal to 1000000000 bits (one billion bits)
11 Mebibyte is equal to 92274688 Bit (ninety-two million two hundred and seventy-four thousand six hundred and eighty-eight bit)
Gigabit is greater than Mebibyte
Multiplication factor is 119.20928955078125.
1 / 119.20928955078125 = 0.008388608.
11 / 119.20928955078125 = 0.092274688.
Maybe you mean Megabyte?
11 Mebibyte is equal to 11.534336 Megabyte (eleven point five hundred and thirty-four thousand three hundred and thirty-six mb) convert to mb
Powers of 2
mib gbit (Gigabit) Description
1 mib 0.008388608 gbit 1 mebibyte (one) is equal to 0.008388608 gigabit (zero point zero × 2 eight million three hundred and eighty-eight thousand six hundred and eight)
2 mib 0.016777216 gbit 2 mebibyte (two) is equal to 0.016777216 gigabit (zero point zero × 1 sixteen million seven hundred and seventy-seven thousand two hundred and sixteen)
4 mib 0.033554432 gbit 4 mebibyte (four) is equal to 0.033554432 gigabit (zero point zero × 1 thirty-three million five hundred and fifty-four thousand four hundred and thirty-two)
8 mib 0.067108864 gbit 8 mebibyte (eight) is equal to 0.067108864 gigabit (zero point zero × 1 sixty-seven million one hundred and eight thousand eight hundred and sixty-four)
16 mib 0.134217728 gbit 16 mebibyte (sixteen) is equal to 0.134217728 gigabit (zero point one hundred and thirty-four million two hundred and seventeen thousand seven hundred and twenty-eight)
32 mib 0.268435456 gbit 32 mebibyte (thirty-two) is equal to 0.268435456 gigabit (zero point two hundred and sixty-eight million four hundred and thirty-five thousand four hundred and fifty-six)
64 mib 0.536870912 gbit 64 mebibyte (sixty-four) is equal to 0.536870912 gigabit (zero point five hundred and thirty-six million eight hundred and seventy thousand nine hundred and twelve)
128 mib 1.073741824 gbit 128 mebibyte (one hundred and twenty-eight) is equal to 1.073741824 gigabit (one point zero × 1 seventy-three million seven hundred and forty-one thousand eight hundred and twenty-four)
256 mib 2.147483648 gbit 256 mebibyte (two hundred and fifty-six) is equal to 2.147483648 gigabit (two point one hundred and forty-seven million four hundred and eighty-three thousand six hundred and forty-eight)
512 mib 4.294967296 gbit 512 mebibyte (five hundred and twelve) is equal to 4.294967296 gigabit (four point two hundred and ninety-four million nine hundred and sixty-seven thousand two hundred and ninety-six)
1024 mib 8.589934592 gbit 1024 mebibyte (one thousand and twenty-four) is equal to 8.589934592 gigabit (eight point five hundred and eighty-nine million nine hundred and thirty-four thousand five hundred and ninety-two)
2048 mib 17.179869184 gbit 2048 mebibyte (two thousand and forty-eight) is equal to 17.179869184 gigabit (seventeen point one hundred and seventy-nine million eight hundred and sixty-nine thousand one hundred and eighty-four)
4096 mib 34.359738368 gbit 4096 mebibyte (four thousand and ninety-six) is equal to 34.359738368 gigabit (thirty-four point three hundred and fifty-nine million seven hundred and thirty-eight thousand three hundred and sixty-eight)
8192 mib 68.719476736 gbit 8192 mebibyte (eight thousand one hundred and ninety-two) is equal to 68.719476736 gigabit (sixty-eight point seven hundred and nineteen million four hundred and seventy-six thousand seven hundred and thirty-six)
Convert Mebibyte to other units
mib System Description
11 mib 92274688 bit 11 mebibyte (eleven) is equal to 92274688 bit (ninety-two million two hundred and seventy-four thousand six hundred and eighty-eight)
11 mib 11534336 b 11 mebibyte (eleven) is equal to 11534336 byte (eleven million five hundred and thirty-four thousand three hundred and thirty-six)
11 mib 11534.336 kb 11 mebibyte (eleven) is equal to 11534.336 kilobyte (eleven thousand five hundred and thirty-four point three hundred and thirty-six)
11 mib 11.534336 mb 11 mebibyte (eleven) is equal to 11.534336 megabyte (eleven point five hundred and thirty-four thousand three hundred and thirty-six)
11 mib 0.011534336 gb 11 mebibyte (eleven) is equal to 0.011534336 gigabyte (zero point zero × 1 eleven million five hundred and thirty-four thousand three hundred and thirty-six)
11 mib 0.000011534336 tb 11 mebibyte (eleven) is equal to 0.000011534336 terabyte (zero point zero × 4 eleven million five hundred and thirty-four thousand three hundred and thirty-six)
11 mib 11264 kib 11 mebibyte (eleven) is equal to 11264 kibibyte (eleven thousand two hundred and sixty-four)
11 mib 0.0107421875 gib 11 mebibyte (eleven) is equal to 0.0107421875 gibibyte (zero point zero × 1 one hundred and seven million four hundred and twenty-one thousand eight hundred and seventy-five)
11 mib 0.00001049041748046875 tib 11 mebibyte (eleven) is equal to 0.00001049041748046875 tebibyte (zero point zero × 4 one quadrillion forty-nine trillion forty-one billion seven hundred and forty-eight million forty-six thousand eight hundred and seventy-five)
11 mib 92274.688 kbit 11 mebibyte (eleven) is equal to 92274.688 kilobit (ninety-two thousand two hundred and seventy-four point six hundred and eighty-eight)
11 mib 92.274688 mbit 11 mebibyte (eleven) is equal to 92.274688 megabit (ninety-two point two hundred and seventy-four thousand six hundred and eighty-eight)
11 mib 0.092274688 gbit 11 mebibyte (eleven) is equal to 0.092274688 gigabit (zero point zero × 1 ninety-two million two hundred and seventy-four thousand six hundred and eighty-eight)
11 mib 0.000092274688 tbit 11 mebibyte (eleven) is equal to 0.000092274688 terabit (zero point zero × 4 ninety-two million two hundred and seventy-four thousand six hundred and eighty-eight)
11 mib 90112 kibit 11 mebibyte (eleven) is equal to 90112 kibibit (ninety thousand one hundred and twelve)
11 mib 88 mibit 11 mebibyte (eleven) is equal to 88 mebibit (eighty-eight)
11 mib 0.0859375 gibit 11 mebibyte (eleven) is equal to 0.0859375 gibibit (zero point zero × 1 eight hundred and fifty-nine thousand three hundred and seventy-five)
11 mib 0.00008392333984375 tibit 11 mebibyte (eleven) is equal to 0.00008392333984375 tebibit (zero point zero × 4 eight trillion three hundred and ninety-two billion three hundred and thirty-three million nine hundred and eighty-four thousand three hundred and seventy-five)
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Home Metamath Proof ExplorerTheorem List (p. 173 of 424) < Previous Next > Bad symbols? Try the GIF version. Mirrors > Metamath Home Page > MPE Home Page > Theorem List Contents > Recent Proofs This page: Page List
Color key: Metamath Proof Explorer (1-27159) Hilbert Space Explorer (27160-28684) Users' Mathboxes (28685-42360)
Theorem List for Metamath Proof Explorer - 17201-17300 *Has distinct variable group(s)
TypeLabelDescription
Statement
Theoremgsumccat 17201 Homomorphic property of composites. (Contributed by Stefan O'Rear, 16-Aug-2015.) (Revised by Mario Carneiro, 1-Oct-2015.)
𝐵 = (Base‘𝐺) & + = (+g𝐺) ((𝐺 ∈ Mnd ∧ 𝑊 ∈ Word 𝐵𝑋 ∈ Word 𝐵) → (𝐺 Σg (𝑊 ++ 𝑋)) = ((𝐺 Σg 𝑊) + (𝐺 Σg 𝑋)))
Theoremgsumws2 17202 Valuation of a pair in a monoid. (Contributed by Stefan O'Rear, 23-Aug-2015.) (Revised by Mario Carneiro, 27-Feb-2016.)
𝐵 = (Base‘𝐺) & + = (+g𝐺) ((𝐺 ∈ Mnd ∧ 𝑆𝐵𝑇𝐵) → (𝐺 Σg ⟨“𝑆𝑇”⟩) = (𝑆 + 𝑇))
Theoremgsumccatsn 17203 Homomorphic property of composites with a singleton. (Contributed by AV, 20-Jan-2019.)
𝐵 = (Base‘𝐺) & + = (+g𝐺) ((𝐺 ∈ Mnd ∧ 𝑊 ∈ Word 𝐵𝑍𝐵) → (𝐺 Σg (𝑊 ++ ⟨“𝑍”⟩)) = ((𝐺 Σg 𝑊) + 𝑍))
Theoremgsumspl 17204 The primary purpose of the splice construction is to enable local rewrites. Thus, in any monoidal valuation, if a splice does not cause a local change it does not cause a global change. (Contributed by Stefan O'Rear, 23-Aug-2015.)
𝐵 = (Base‘𝑀) & (𝜑𝑀 ∈ Mnd) & (𝜑𝑆 ∈ Word 𝐵) & (𝜑𝐹 ∈ (0...𝑇)) & (𝜑𝑇 ∈ (0...(#‘𝑆))) & (𝜑𝑋 ∈ Word 𝐵) & (𝜑𝑌 ∈ Word 𝐵) & (𝜑 → (𝑀 Σg 𝑋) = (𝑀 Σg 𝑌)) (𝜑 → (𝑀 Σg (𝑆 splice ⟨𝐹, 𝑇, 𝑋⟩)) = (𝑀 Σg (𝑆 splice ⟨𝐹, 𝑇, 𝑌⟩)))
Theoremgsumwmhm 17205 Behavior of homomorphisms on finite monoidal sums. (Contributed by Stefan O'Rear, 27-Aug-2015.)
𝐵 = (Base‘𝑀) ((𝐻 ∈ (𝑀 MndHom 𝑁) ∧ 𝑊 ∈ Word 𝐵) → (𝐻‘(𝑀 Σg 𝑊)) = (𝑁 Σg (𝐻𝑊)))
Theoremgsumwspan 17206* The submonoid generated by a set of elements is precisely the set of elements which can be expressed as finite products of the generator. (Contributed by Stefan O'Rear, 22-Aug-2015.)
𝐵 = (Base‘𝑀) & 𝐾 = (mrCls‘(SubMnd‘𝑀)) ((𝑀 ∈ Mnd ∧ 𝐺𝐵) → (𝐾𝐺) = ran (𝑤 ∈ Word 𝐺 ↦ (𝑀 Σg 𝑤)))
10.1.8 Free monoids
Syntaxcfrmd 17207 Extend class definition with the free monoid construction.
class freeMnd
Syntaxcvrmd 17208 Extend class notation with free monoid injection.
class varFMnd
Definitiondf-frmd 17209 Define a free monoid over a set 𝑖 of generators, defined as the set of finite strings on 𝐼 with the operation of concatenation. (Contributed by Mario Carneiro, 27-Sep-2015.)
freeMnd = (𝑖 ∈ V ↦ {⟨(Base‘ndx), Word 𝑖⟩, ⟨(+g‘ndx), ( ++ ↾ (Word 𝑖 × Word 𝑖))⟩})
Definitiondf-vrmd 17210* Define a free monoid over a set 𝑖 of generators, defined as the set of finite strings on 𝐼 with the operation of concatenation. (Contributed by Mario Carneiro, 27-Sep-2015.)
varFMnd = (𝑖 ∈ V ↦ (𝑗𝑖 ↦ ⟨“𝑗”⟩))
Theoremfrmdval 17211 Value of the free monoid construction. (Contributed by Mario Carneiro, 27-Sep-2015.)
𝑀 = (freeMnd‘𝐼) & (𝐼𝑉𝐵 = Word 𝐼) & + = ( ++ ↾ (𝐵 × 𝐵)) (𝐼𝑉𝑀 = {⟨(Base‘ndx), 𝐵⟩, ⟨(+g‘ndx), + ⟩})
Theoremfrmdbas 17212 The base set of a free monoid. (Contributed by Mario Carneiro, 27-Sep-2015.) (Revised by Mario Carneiro, 27-Feb-2016.)
𝑀 = (freeMnd‘𝐼) & 𝐵 = (Base‘𝑀) (𝐼𝑉𝐵 = Word 𝐼)
Theoremfrmdelbas 17213 An element of the base set of a free monoid is a string on the generators. (Contributed by Mario Carneiro, 27-Feb-2016.)
𝑀 = (freeMnd‘𝐼) & 𝐵 = (Base‘𝑀) (𝑋𝐵𝑋 ∈ Word 𝐼)
Theoremfrmdplusg 17214 The monoid operation of a free monoid. (Contributed by Mario Carneiro, 27-Sep-2015.) (Revised by Mario Carneiro, 27-Feb-2016.)
𝑀 = (freeMnd‘𝐼) & 𝐵 = (Base‘𝑀) & + = (+g𝑀) + = ( ++ ↾ (𝐵 × 𝐵))
Theoremfrmdadd 17215 Value of the monoid operation of the free monoid construction. (Contributed by Mario Carneiro, 27-Sep-2015.)
𝑀 = (freeMnd‘𝐼) & 𝐵 = (Base‘𝑀) & + = (+g𝑀) ((𝑋𝐵𝑌𝐵) → (𝑋 + 𝑌) = (𝑋 ++ 𝑌))
Theoremvrmdfval 17216* The canonical injection from the generating set 𝐼 to the base set of the free monoid. (Contributed by Mario Carneiro, 27-Feb-2016.)
𝑈 = (varFMnd𝐼) (𝐼𝑉𝑈 = (𝑗𝐼 ↦ ⟨“𝑗”⟩))
Theoremvrmdval 17217 The value of the generating elements of a free monoid. (Contributed by Mario Carneiro, 27-Feb-2016.)
𝑈 = (varFMnd𝐼) ((𝐼𝑉𝐴𝐼) → (𝑈𝐴) = ⟨“𝐴”⟩)
Theoremvrmdf 17218 The mapping from the index set to the generators is a function into the free monoid. (Contributed by Mario Carneiro, 27-Feb-2016.)
𝑈 = (varFMnd𝐼) (𝐼𝑉𝑈:𝐼⟶Word 𝐼)
Theoremfrmdmnd 17219 A free monoid is a monoid. (Contributed by Mario Carneiro, 27-Sep-2015.) (Revised by Mario Carneiro, 27-Feb-2016.)
𝑀 = (freeMnd‘𝐼) (𝐼𝑉𝑀 ∈ Mnd)
Theoremfrmd0 17220 The identity of the free monoid is the empty word. (Contributed by Mario Carneiro, 27-Sep-2015.)
𝑀 = (freeMnd‘𝐼) ∅ = (0g𝑀)
Theoremfrmdsssubm 17221 The set of words taking values in a subset is a (free) submonoid of the free monoid. (Contributed by Mario Carneiro, 27-Sep-2015.) (Revised by Mario Carneiro, 27-Feb-2016.)
𝑀 = (freeMnd‘𝐼) ((𝐼𝑉𝐽𝐼) → Word 𝐽 ∈ (SubMnd‘𝑀))
Theoremfrmdgsum 17222 Any word in a free monoid can be expressed as the sum of the singletons composing it. (Contributed by Mario Carneiro, 27-Sep-2015.)
𝑀 = (freeMnd‘𝐼) & 𝑈 = (varFMnd𝐼) ((𝐼𝑉𝑊 ∈ Word 𝐼) → (𝑀 Σg (𝑈𝑊)) = 𝑊)
Theoremfrmdss2 17223 A subset of generators is contained in a submonoid iff the set of words on the generators is in the submonoid. This can be viewed as an elementary way of saying "the monoidal closure of 𝐽 is Word 𝐽". (Contributed by Mario Carneiro, 2-Oct-2015.)
𝑀 = (freeMnd‘𝐼) & 𝑈 = (varFMnd𝐼) ((𝐼𝑉𝐽𝐼𝐴 ∈ (SubMnd‘𝑀)) → ((𝑈𝐽) ⊆ 𝐴 ↔ Word 𝐽𝐴))
Theoremfrmdup1 17224* Any assignment of the generators to target elements can be extended (uniquely) to a homomorphism from a free monoid to an arbitrary other monoid. (Contributed by Mario Carneiro, 27-Sep-2015.)
𝑀 = (freeMnd‘𝐼) & 𝐵 = (Base‘𝐺) & 𝐸 = (𝑥 ∈ Word 𝐼 ↦ (𝐺 Σg (𝐴𝑥))) & (𝜑𝐺 ∈ Mnd) & (𝜑𝐼𝑋) & (𝜑𝐴:𝐼𝐵) (𝜑𝐸 ∈ (𝑀 MndHom 𝐺))
Theoremfrmdup2 17225* The evaluation map has the intended behavior on the generators. (Contributed by Mario Carneiro, 27-Sep-2015.)
𝑀 = (freeMnd‘𝐼) & 𝐵 = (Base‘𝐺) & 𝐸 = (𝑥 ∈ Word 𝐼 ↦ (𝐺 Σg (𝐴𝑥))) & (𝜑𝐺 ∈ Mnd) & (𝜑𝐼𝑋) & (𝜑𝐴:𝐼𝐵) & 𝑈 = (varFMnd𝐼) & (𝜑𝑌𝐼) (𝜑 → (𝐸‘(𝑈𝑌)) = (𝐴𝑌))
Theoremfrmdup3lem 17226* Lemma for frmdup3 17227. (Contributed by Mario Carneiro, 18-Jul-2016.)
𝑀 = (freeMnd‘𝐼) & 𝐵 = (Base‘𝐺) & 𝑈 = (varFMnd𝐼) (((𝐺 ∈ Mnd ∧ 𝐼𝑉𝐴:𝐼𝐵) ∧ (𝐹 ∈ (𝑀 MndHom 𝐺) ∧ (𝐹𝑈) = 𝐴)) → 𝐹 = (𝑥 ∈ Word 𝐼 ↦ (𝐺 Σg (𝐴𝑥))))
Theoremfrmdup3 17227* Universal property of the free monoid by existential uniqueness. (Contributed by Mario Carneiro, 2-Oct-2015.) (Revised by Mario Carneiro, 18-Jul-2016.)
𝑀 = (freeMnd‘𝐼) & 𝐵 = (Base‘𝐺) & 𝑈 = (varFMnd𝐼) ((𝐺 ∈ Mnd ∧ 𝐼𝑉𝐴:𝐼𝐵) → ∃!𝑚 ∈ (𝑀 MndHom 𝐺)(𝑚𝑈) = 𝐴)
10.1.9 Examples and counterexamples for magmas, semigroups and monoids
Theoremmgm2nsgrplem1 17228* Lemma 1 for mgm2nsgrp 17232: 𝑀 is a magma, even if 𝐴 = 𝐵 (𝑀 is the trivial magma in this case, see mgmb1mgm1 17077). (Contributed by AV, 27-Jan-2020.)
𝑆 = {𝐴, 𝐵} & (Base‘𝑀) = 𝑆 & (+g𝑀) = (𝑥𝑆, 𝑦𝑆 ↦ if((𝑥 = 𝐴𝑦 = 𝐴), 𝐵, 𝐴)) ((𝐴𝑉𝐵𝑊) → 𝑀 ∈ Mgm)
Theoremmgm2nsgrplem2 17229* Lemma 2 for mgm2nsgrp 17232. (Contributed by AV, 27-Jan-2020.)
𝑆 = {𝐴, 𝐵} & (Base‘𝑀) = 𝑆 & (+g𝑀) = (𝑥𝑆, 𝑦𝑆 ↦ if((𝑥 = 𝐴𝑦 = 𝐴), 𝐵, 𝐴)) & = (+g𝑀) ((𝐴𝑉𝐵𝑊) → ((𝐴 𝐴) 𝐵) = 𝐴)
Theoremmgm2nsgrplem3 17230* Lemma 3 for mgm2nsgrp 17232. (Contributed by AV, 28-Jan-2020.)
𝑆 = {𝐴, 𝐵} & (Base‘𝑀) = 𝑆 & (+g𝑀) = (𝑥𝑆, 𝑦𝑆 ↦ if((𝑥 = 𝐴𝑦 = 𝐴), 𝐵, 𝐴)) & = (+g𝑀) ((𝐴𝑉𝐵𝑊) → (𝐴 (𝐴 𝐵)) = 𝐵)
Theoremmgm2nsgrplem4 17231* Lemma 4 for mgm2nsgrp 17232: M is not a semigroup. (Contributed by AV, 28-Jan-2020.) (Proof shortened by AV, 31-Jan-2020.)
𝑆 = {𝐴, 𝐵} & (Base‘𝑀) = 𝑆 & (+g𝑀) = (𝑥𝑆, 𝑦𝑆 ↦ if((𝑥 = 𝐴𝑦 = 𝐴), 𝐵, 𝐴)) ((#‘𝑆) = 2 → 𝑀 ∉ SGrp)
Theoremmgm2nsgrp 17232* A small magma (with two elements) which is not a semigroup. (Contributed by AV, 28-Jan-2020.)
𝑆 = {𝐴, 𝐵} & (Base‘𝑀) = 𝑆 & (+g𝑀) = (𝑥𝑆, 𝑦𝑆 ↦ if((𝑥 = 𝐴𝑦 = 𝐴), 𝐵, 𝐴)) ((#‘𝑆) = 2 → (𝑀 ∈ Mgm ∧ 𝑀 ∉ SGrp))
Theoremsgrp2nmndlem1 17233* Lemma 1 for sgrp2nmnd 17240: 𝑀 is a magma, even if 𝐴 = 𝐵 (𝑀 is the trivial magma in this case, see mgmb1mgm1 17077). (Contributed by AV, 29-Jan-2020.)
𝑆 = {𝐴, 𝐵} & (Base‘𝑀) = 𝑆 & (+g𝑀) = (𝑥𝑆, 𝑦𝑆 ↦ if(𝑥 = 𝐴, 𝐴, 𝐵)) ((𝐴𝑉𝐵𝑊) → 𝑀 ∈ Mgm)
Theoremsgrp2nmndlem2 17234* Lemma 2 for sgrp2nmnd 17240. (Contributed by AV, 29-Jan-2020.)
𝑆 = {𝐴, 𝐵} & (Base‘𝑀) = 𝑆 & (+g𝑀) = (𝑥𝑆, 𝑦𝑆 ↦ if(𝑥 = 𝐴, 𝐴, 𝐵)) & = (+g𝑀) ((𝐴𝑆𝐶𝑆) → (𝐴 𝐶) = 𝐴)
Theoremsgrp2nmndlem3 17235* Lemma 3 for sgrp2nmnd 17240. (Contributed by AV, 29-Jan-2020.)
𝑆 = {𝐴, 𝐵} & (Base‘𝑀) = 𝑆 & (+g𝑀) = (𝑥𝑆, 𝑦𝑆 ↦ if(𝑥 = 𝐴, 𝐴, 𝐵)) & = (+g𝑀) ((𝐶𝑆𝐵𝑆𝐴𝐵) → (𝐵 𝐶) = 𝐵)
Theoremsgrp2rid2 17236* A small semigroup (with two elements) with two right identities which are different if 𝐴𝐵. (Contributed by AV, 10-Feb-2020.)
𝑆 = {𝐴, 𝐵} & (Base‘𝑀) = 𝑆 & (+g𝑀) = (𝑥𝑆, 𝑦𝑆 ↦ if(𝑥 = 𝐴, 𝐴, 𝐵)) & = (+g𝑀) ((𝐴𝑉𝐵𝑊) → ∀𝑥𝑆𝑦𝑆 (𝑦 𝑥) = 𝑦)
Theoremsgrp2rid2ex 17237* A small semigroup (with two elements) with two right identities which are different. (Contributed by AV, 10-Feb-2020.)
𝑆 = {𝐴, 𝐵} & (Base‘𝑀) = 𝑆 & (+g𝑀) = (𝑥𝑆, 𝑦𝑆 ↦ if(𝑥 = 𝐴, 𝐴, 𝐵)) & = (+g𝑀) ((#‘𝑆) = 2 → ∃𝑥𝑆𝑧𝑆𝑦𝑆 (𝑥𝑧 ∧ (𝑦 𝑥) = 𝑦 ∧ (𝑦 𝑧) = 𝑦))
Theoremsgrp2nmndlem4 17238* Lemma 4 for sgrp2nmnd 17240: M is a semigroup. (Contributed by AV, 29-Jan-2020.)
𝑆 = {𝐴, 𝐵} & (Base‘𝑀) = 𝑆 & (+g𝑀) = (𝑥𝑆, 𝑦𝑆 ↦ if(𝑥 = 𝐴, 𝐴, 𝐵)) ((#‘𝑆) = 2 → 𝑀 ∈ SGrp)
Theoremsgrp2nmndlem5 17239* Lemma 5 for sgrp2nmnd 17240: M is not a monoid. (Contributed by AV, 29-Jan-2020.)
𝑆 = {𝐴, 𝐵} & (Base‘𝑀) = 𝑆 & (+g𝑀) = (𝑥𝑆, 𝑦𝑆 ↦ if(𝑥 = 𝐴, 𝐴, 𝐵)) ((#‘𝑆) = 2 → 𝑀 ∉ Mnd)
Theoremsgrp2nmnd 17240* A small semigroup (with two elements) which is not a monoid. (Contributed by AV, 26-Jan-2020.)
𝑆 = {𝐴, 𝐵} & (Base‘𝑀) = 𝑆 & (+g𝑀) = (𝑥𝑆, 𝑦𝑆 ↦ if(𝑥 = 𝐴, 𝐴, 𝐵)) ((#‘𝑆) = 2 → (𝑀 ∈ SGrp ∧ 𝑀 ∉ Mnd))
Theoremmgmnsgrpex 17241 There is a magma which is not a semigroup. (Contributed by AV, 29-Jan-2020.)
𝑚 ∈ Mgm 𝑚 ∉ SGrp
Theoremsgrpnmndex 17242 There is a semigroup which is not a monoid. (Contributed by AV, 29-Jan-2020.)
𝑚 ∈ SGrp 𝑚 ∉ Mnd
Theoremsgrpssmgm 17243 The class of all semigroups is a proper subclass of the class of all magmas. (Contributed by AV, 29-Jan-2020.)
SGrp ⊊ Mgm
Theoremmndsssgrp 17244 The class of all monoids is a proper subclass of the class of all semigroups. (Contributed by AV, 29-Jan-2020.)
Mnd ⊊ SGrp
10.2 Groups
10.2.1 Definition and basic properties
Syntaxcgrp 17245 Extend class notation with class of all groups.
class Grp
Syntaxcminusg 17246 Extend class notation with inverse of group element.
class invg
Syntaxcsg 17247 Extend class notation with group subtraction (or division) operation.
class -g
Definitiondf-grp 17248* Define class of all groups. A group is a monoid (df-mnd 17118) whose internal operation is such that every element admits a left inverse (which can be proven to be a two-sided inverse). Thus, a group 𝐺 is an algebraic structure formed from a base set of elements (notated (Base‘𝐺) per df-base 15700) and an internal group operation (notated (+g𝐺) per df-plusg 15781). The operation combines any two elements of the group base set and must satisfy the 4 group axioms: closure (the result of the group operation must always be a member of the base set, see grpcl 17253), associativity (so ((𝑎+g𝑏)+g𝑐) = (𝑎+g(𝑏+g𝑐)) for any a, b, c, see grpass 17254), identity (there must be an element 𝑒 = (0g𝐺) such that 𝑒+g𝑎 = 𝑎+g𝑒 = 𝑎 for any a), and inverse (for each element a in the base set, there must be an element 𝑏 = invg𝑎 in the base set such that 𝑎+g𝑏 = 𝑏+g𝑎 = 𝑒). It can be proven that the identity element is unique (grpideu 17256). Groups need not be commutative; a commutative group is an Abelian group (see df-abl 18019). Subgroups can often be formed from groups, see df-subg 17414. An example of an (Abelian) group is the set of complex numbers over the group operation + (addition), as proven in cnaddablx 18094; an Abelian group is a group as proven in ablgrp 18021. Other structures include groups, including unital rings (df-ring 18372) and fields (df-field 18573). (Contributed by NM, 17-Oct-2012.) (Revised by Mario Carneiro, 6-Jan-2015.)
Grp = {𝑔 ∈ Mnd ∣ ∀𝑎 ∈ (Base‘𝑔)∃𝑚 ∈ (Base‘𝑔)(𝑚(+g𝑔)𝑎) = (0g𝑔)}
Definitiondf-minusg 17249* Define inverse of group element. (Contributed by NM, 24-Aug-2011.)
invg = (𝑔 ∈ V ↦ (𝑥 ∈ (Base‘𝑔) ↦ (𝑤 ∈ (Base‘𝑔)(𝑤(+g𝑔)𝑥) = (0g𝑔))))
Definitiondf-sbg 17250* Define group subtraction (also called division for multiplicative groups). (Contributed by NM, 31-Mar-2014.)
-g = (𝑔 ∈ V ↦ (𝑥 ∈ (Base‘𝑔), 𝑦 ∈ (Base‘𝑔) ↦ (𝑥(+g𝑔)((invg𝑔)‘𝑦))))
Theoremisgrp 17251* The predicate "is a group." (This theorem demonstrates the use of symbols as variable names, first proposed by FL in 2010.) (Contributed by NM, 17-Oct-2012.) (Revised by Mario Carneiro, 6-Jan-2015.)
𝐵 = (Base‘𝐺) & + = (+g𝐺) & 0 = (0g𝐺) (𝐺 ∈ Grp ↔ (𝐺 ∈ Mnd ∧ ∀𝑎𝐵𝑚𝐵 (𝑚 + 𝑎) = 0 ))
Theoremgrpmnd 17252 A group is a monoid. (Contributed by Mario Carneiro, 6-Jan-2015.)
(𝐺 ∈ Grp → 𝐺 ∈ Mnd)
Theoremgrpcl 17253 Closure of the operation of a group. (Contributed by NM, 14-Aug-2011.)
𝐵 = (Base‘𝐺) & + = (+g𝐺) ((𝐺 ∈ Grp ∧ 𝑋𝐵𝑌𝐵) → (𝑋 + 𝑌) ∈ 𝐵)
Theoremgrpass 17254 A group operation is associative. (Contributed by NM, 14-Aug-2011.)
𝐵 = (Base‘𝐺) & + = (+g𝐺) ((𝐺 ∈ Grp ∧ (𝑋𝐵𝑌𝐵𝑍𝐵)) → ((𝑋 + 𝑌) + 𝑍) = (𝑋 + (𝑌 + 𝑍)))
Theoremgrpinvex 17255* Every member of a group has a left inverse. (Contributed by NM, 16-Aug-2011.) (Revised by Mario Carneiro, 6-Jan-2015.)
𝐵 = (Base‘𝐺) & + = (+g𝐺) & 0 = (0g𝐺) ((𝐺 ∈ Grp ∧ 𝑋𝐵) → ∃𝑦𝐵 (𝑦 + 𝑋) = 0 )
Theoremgrpideu 17256* The two-sided identity element of a group is unique. Lemma 2.2.1(a) of [Herstein] p. 55. (Contributed by NM, 16-Aug-2011.) (Revised by Mario Carneiro, 8-Dec-2014.)
𝐵 = (Base‘𝐺) & + = (+g𝐺) & 0 = (0g𝐺) (𝐺 ∈ Grp → ∃!𝑢𝐵𝑥𝐵 ((𝑢 + 𝑥) = 𝑥 ∧ (𝑥 + 𝑢) = 𝑥))
Theoremgrpplusf 17257 The group addition operation is a function. (Contributed by Mario Carneiro, 14-Aug-2015.)
𝐵 = (Base‘𝐺) & 𝐹 = (+𝑓𝐺) (𝐺 ∈ Grp → 𝐹:(𝐵 × 𝐵)⟶𝐵)
Theoremgrpplusfo 17258 The group addition operation is a function onto the base set/set of group elements. (Contributed by NM, 30-Oct-2006.) (Revised by AV, 30-Aug-2021.)
𝐵 = (Base‘𝐺) & 𝐹 = (+𝑓𝐺) (𝐺 ∈ Grp → 𝐹:(𝐵 × 𝐵)–onto𝐵)
Theoremresgrpplusfrn 17259 The underlying set of a group operation which is a restriction of a structure. (Contributed by Paul Chapman, 25-Mar-2008.) (Revised by AV, 30-Aug-2021.)
𝐵 = (Base‘𝐺) & 𝐻 = (𝐺s 𝑆) & 𝐹 = (+𝑓𝐻) ((𝐻 ∈ Grp ∧ 𝑆𝐵) → 𝑆 = ran 𝐹)
Theoremgrppropd 17260* If two structures have the same group components (properties), one is a group iff the other one is. (Contributed by Stefan O'Rear, 27-Nov-2014.) (Revised by Mario Carneiro, 2-Oct-2015.)
(𝜑𝐵 = (Base‘𝐾)) & (𝜑𝐵 = (Base‘𝐿)) & ((𝜑 ∧ (𝑥𝐵𝑦𝐵)) → (𝑥(+g𝐾)𝑦) = (𝑥(+g𝐿)𝑦)) (𝜑 → (𝐾 ∈ Grp ↔ 𝐿 ∈ Grp))
Theoremgrpprop 17261 If two structures have the same group components (properties), one is a group iff the other one is. (Contributed by NM, 11-Oct-2013.)
(Base‘𝐾) = (Base‘𝐿) & (+g𝐾) = (+g𝐿) (𝐾 ∈ Grp ↔ 𝐿 ∈ Grp)
Theoremgrppropstr 17262 Generalize a specific 2-element group 𝐿 to show that any set 𝐾 with the same (relevant) properties is also a group. (Contributed by NM, 28-Oct-2012.) (Revised by Mario Carneiro, 6-Jan-2015.)
(Base‘𝐾) = 𝐵 & (+g𝐾) = + & 𝐿 = {⟨(Base‘ndx), 𝐵⟩, ⟨(+g‘ndx), + ⟩} (𝐾 ∈ Grp ↔ 𝐿 ∈ Grp)
Theoremgrpss 17263 Show that a structure extending a constructed group (e.g., a ring) is also a group. This allows us to prove that a constructed potential ring 𝑅 is a group before we know that it is also a ring. (Theorem ringgrp 18375, on the other hand, requires that we know in advance that 𝑅 is a ring.) (Contributed by NM, 11-Oct-2013.)
𝐺 = {⟨(Base‘ndx), 𝐵⟩, ⟨(+g‘ndx), + ⟩} & 𝑅 ∈ V & 𝐺𝑅 & Fun 𝑅 (𝐺 ∈ Grp ↔ 𝑅 ∈ Grp)
Theoremisgrpd2e 17264* Deduce a group from its properties. In this version of isgrpd2 17265, we don't assume there is an expression for the inverse of 𝑥. (Contributed by NM, 10-Aug-2013.)
(𝜑𝐵 = (Base‘𝐺)) & (𝜑+ = (+g𝐺)) & (𝜑0 = (0g𝐺)) & (𝜑𝐺 ∈ Mnd) & ((𝜑𝑥𝐵) → ∃𝑦𝐵 (𝑦 + 𝑥) = 0 ) (𝜑𝐺 ∈ Grp)
Theoremisgrpd2 17265* Deduce a group from its properties. 𝑁 (negative) is normally dependent on 𝑥 i.e. read it as 𝑁(𝑥). Note: normally we don't use a 𝜑 antecedent on hypotheses that name structure components, since they can be eliminated with eqid 2610, but we make an exception for theorems such as isgrpd2 17265, ismndd 17136, and islmodd 18692 since theorems using them often rewrite the structure components. (Contributed by NM, 10-Aug-2013.)
(𝜑𝐵 = (Base‘𝐺)) & (𝜑+ = (+g𝐺)) & (𝜑0 = (0g𝐺)) & (𝜑𝐺 ∈ Mnd) & ((𝜑𝑥𝐵) → 𝑁𝐵) & ((𝜑𝑥𝐵) → (𝑁 + 𝑥) = 0 ) (𝜑𝐺 ∈ Grp)
Theoremisgrpde 17266* Deduce a group from its properties. In this version of isgrpd 17267, we don't assume there is an expression for the inverse of 𝑥. (Contributed by NM, 6-Jan-2015.)
(𝜑𝐵 = (Base‘𝐺)) & (𝜑+ = (+g𝐺)) & ((𝜑𝑥𝐵𝑦𝐵) → (𝑥 + 𝑦) ∈ 𝐵) & ((𝜑 ∧ (𝑥𝐵𝑦𝐵𝑧𝐵)) → ((𝑥 + 𝑦) + 𝑧) = (𝑥 + (𝑦 + 𝑧))) & (𝜑0𝐵) & ((𝜑𝑥𝐵) → ( 0 + 𝑥) = 𝑥) & ((𝜑𝑥𝐵) → ∃𝑦𝐵 (𝑦 + 𝑥) = 0 ) (𝜑𝐺 ∈ Grp)
Theoremisgrpd 17267* Deduce a group from its properties. Unlike isgrpd2 17265, this one goes straight from the base properties rather than going through Mnd. 𝑁 (negative) is normally dependent on 𝑥 i.e. read it as 𝑁(𝑥). (Contributed by NM, 6-Jun-2013.) (Revised by Mario Carneiro, 6-Jan-2015.)
(𝜑𝐵 = (Base‘𝐺)) & (𝜑+ = (+g𝐺)) & ((𝜑𝑥𝐵𝑦𝐵) → (𝑥 + 𝑦) ∈ 𝐵) & ((𝜑 ∧ (𝑥𝐵𝑦𝐵𝑧𝐵)) → ((𝑥 + 𝑦) + 𝑧) = (𝑥 + (𝑦 + 𝑧))) & (𝜑0𝐵) & ((𝜑𝑥𝐵) → ( 0 + 𝑥) = 𝑥) & ((𝜑𝑥𝐵) → 𝑁𝐵) & ((𝜑𝑥𝐵) → (𝑁 + 𝑥) = 0 ) (𝜑𝐺 ∈ Grp)
Theoremisgrpi 17268* Properties that determine a group. 𝑁 (negative) is normally dependent on 𝑥 i.e. read it as 𝑁(𝑥). (Contributed by NM, 3-Sep-2011.)
𝐵 = (Base‘𝐺) & + = (+g𝐺) & ((𝑥𝐵𝑦𝐵) → (𝑥 + 𝑦) ∈ 𝐵) & ((𝑥𝐵𝑦𝐵𝑧𝐵) → ((𝑥 + 𝑦) + 𝑧) = (𝑥 + (𝑦 + 𝑧))) & 0𝐵 & (𝑥𝐵 → ( 0 + 𝑥) = 𝑥) & (𝑥𝐵𝑁𝐵) & (𝑥𝐵 → (𝑁 + 𝑥) = 0 ) 𝐺 ∈ Grp
Theoremgrpsgrp 17269 A group is a semigroup. (Contributed by AV, 28-Aug-2021.)
(𝐺 ∈ Grp → 𝐺 ∈ SGrp)
Theoremdfgrp2 17270* Alternate definition of a group as semigroup with a left identity and a left inverse for each element. This "definition" is weaker than df-grp 17248, based on the definition of a monoid which provides a left and a right identity. (Contributed by AV, 28-Aug-2021.)
𝐵 = (Base‘𝐺) & + = (+g𝐺) (𝐺 ∈ Grp ↔ (𝐺 ∈ SGrp ∧ ∃𝑛𝐵𝑥𝐵 ((𝑛 + 𝑥) = 𝑥 ∧ ∃𝑖𝐵 (𝑖 + 𝑥) = 𝑛)))
Theoremdfgrp2e 17271* Alternate definition of a group as a set with a closed, associative operation, a left identity and a left inverse for each element. Alternate definition in [Lang] p. 7. (Contributed by NM, 10-Oct-2006.) (Revised by AV, 28-Aug-2021.)
𝐵 = (Base‘𝐺) & + = (+g𝐺) (𝐺 ∈ Grp ↔ (∀𝑥𝐵𝑦𝐵 ((𝑥 + 𝑦) ∈ 𝐵 ∧ ∀𝑧𝐵 ((𝑥 + 𝑦) + 𝑧) = (𝑥 + (𝑦 + 𝑧))) ∧ ∃𝑛𝐵𝑥𝐵 ((𝑛 + 𝑥) = 𝑥 ∧ ∃𝑖𝐵 (𝑖 + 𝑥) = 𝑛)))
Theoremisgrpix 17272* Properties that determine a group. Read 𝑁 as 𝑁(𝑥). Note: This theorem has hard-coded structure indices for demonstration purposes. It is not intended for general use. (New usage is discouraged.) (Contributed by NM, 4-Sep-2011.)
𝐵 ∈ V & + ∈ V & 𝐺 = {⟨1, 𝐵⟩, ⟨2, + ⟩} & ((𝑥𝐵𝑦𝐵) → (𝑥 + 𝑦) ∈ 𝐵) & ((𝑥𝐵𝑦𝐵𝑧𝐵) → ((𝑥 + 𝑦) + 𝑧) = (𝑥 + (𝑦 + 𝑧))) & 0𝐵 & (𝑥𝐵 → ( 0 + 𝑥) = 𝑥) & (𝑥𝐵𝑁𝐵) & (𝑥𝐵 → (𝑁 + 𝑥) = 0 ) 𝐺 ∈ Grp
Theoremgrpidcl 17273 The identity element of a group belongs to the group. (Contributed by NM, 27-Aug-2011.) (Revised by Mario Carneiro, 27-Dec-2014.)
𝐵 = (Base‘𝐺) & 0 = (0g𝐺) (𝐺 ∈ Grp → 0𝐵)
Theoremgrpbn0 17274 The base set of a group is not empty. (Contributed by Szymon Jaroszewicz, 3-Apr-2007.)
𝐵 = (Base‘𝐺) (𝐺 ∈ Grp → 𝐵 ≠ ∅)
Theoremgrplid 17275 The identity element of a group is a left identity. (Contributed by NM, 18-Aug-2011.)
𝐵 = (Base‘𝐺) & + = (+g𝐺) & 0 = (0g𝐺) ((𝐺 ∈ Grp ∧ 𝑋𝐵) → ( 0 + 𝑋) = 𝑋)
Theoremgrprid 17276 The identity element of a group is a right identity. (Contributed by NM, 18-Aug-2011.)
𝐵 = (Base‘𝐺) & + = (+g𝐺) & 0 = (0g𝐺) ((𝐺 ∈ Grp ∧ 𝑋𝐵) → (𝑋 + 0 ) = 𝑋)
Theoremgrpn0 17277 A group is not empty. (Contributed by Szymon Jaroszewicz, 3-Apr-2007.) (Revised by Mario Carneiro, 2-Dec-2014.)
(𝐺 ∈ Grp → 𝐺 ≠ ∅)
Theoremgrprcan 17278 Right cancellation law for groups. (Contributed by NM, 24-Aug-2011.) (Proof shortened by Mario Carneiro, 6-Jan-2015.)
𝐵 = (Base‘𝐺) & + = (+g𝐺) ((𝐺 ∈ Grp ∧ (𝑋𝐵𝑌𝐵𝑍𝐵)) → ((𝑋 + 𝑍) = (𝑌 + 𝑍) ↔ 𝑋 = 𝑌))
Theoremgrpinveu 17279* The left inverse element of a group is unique. Lemma 2.2.1(b) of [Herstein] p. 55. (Contributed by NM, 24-Aug-2011.)
𝐵 = (Base‘𝐺) & + = (+g𝐺) & 0 = (0g𝐺) ((𝐺 ∈ Grp ∧ 𝑋𝐵) → ∃!𝑦𝐵 (𝑦 + 𝑋) = 0 )
Theoremgrpid 17280 Two ways of saying that an element of a group is the identity element. Provides a convenient way to compute the value of the identity element. (Contributed by NM, 24-Aug-2011.)
𝐵 = (Base‘𝐺) & + = (+g𝐺) & 0 = (0g𝐺) ((𝐺 ∈ Grp ∧ 𝑋𝐵) → ((𝑋 + 𝑋) = 𝑋0 = 𝑋))
Theoremisgrpid2 17281 Properties showing that an element 𝑍 is the identity element of a group. (Contributed by NM, 7-Aug-2013.)
𝐵 = (Base‘𝐺) & + = (+g𝐺) & 0 = (0g𝐺) (𝐺 ∈ Grp → ((𝑍𝐵 ∧ (𝑍 + 𝑍) = 𝑍) ↔ 0 = 𝑍))
Theoremgrpidd2 17282* Deduce the identity element of a group from its properties. Useful in conjunction with isgrpd 17267. (Contributed by Mario Carneiro, 14-Jun-2015.)
(𝜑𝐵 = (Base‘𝐺)) & (𝜑+ = (+g𝐺)) & (𝜑0𝐵) & ((𝜑𝑥𝐵) → ( 0 + 𝑥) = 𝑥) & (𝜑𝐺 ∈ Grp) (𝜑0 = (0g𝐺))
Theoremgrpinvfval 17283* The inverse function of a group. (Contributed by NM, 24-Aug-2011.) (Revised by Mario Carneiro, 7-Aug-2013.)
𝐵 = (Base‘𝐺) & + = (+g𝐺) & 0 = (0g𝐺) & 𝑁 = (invg𝐺) 𝑁 = (𝑥𝐵 ↦ (𝑦𝐵 (𝑦 + 𝑥) = 0 ))
Theoremgrpinvval 17284* The inverse of a group element. (Contributed by NM, 24-Aug-2011.) (Revised by Mario Carneiro, 7-Aug-2013.)
𝐵 = (Base‘𝐺) & + = (+g𝐺) & 0 = (0g𝐺) & 𝑁 = (invg𝐺) (𝑋𝐵 → (𝑁𝑋) = (𝑦𝐵 (𝑦 + 𝑋) = 0 ))
Theoremgrpinvfn 17285 Functionality of the group inverse function. (Contributed by Stefan O'Rear, 21-Mar-2015.)
𝐵 = (Base‘𝐺) & 𝑁 = (invg𝐺) 𝑁 Fn 𝐵
Theoremgrpinvfvi 17286 The group inverse function is compatible with identity-function protection. (Contributed by Stefan O'Rear, 21-Mar-2015.)
𝑁 = (invg𝐺) 𝑁 = (invg‘( I ‘𝐺))
Theoremgrpsubfval 17287* Group subtraction (division) operation. (Contributed by NM, 31-Mar-2014.) (Revised by Stefan O'Rear, 27-Mar-2015.)
𝐵 = (Base‘𝐺) & + = (+g𝐺) & 𝐼 = (invg𝐺) & = (-g𝐺) = (𝑥𝐵, 𝑦𝐵 ↦ (𝑥 + (𝐼𝑦)))
Theoremgrpsubval 17288 Group subtraction (division) operation. (Contributed by NM, 31-Mar-2014.) (Revised by Mario Carneiro, 13-Dec-2014.)
𝐵 = (Base‘𝐺) & + = (+g𝐺) & 𝐼 = (invg𝐺) & = (-g𝐺) ((𝑋𝐵𝑌𝐵) → (𝑋 𝑌) = (𝑋 + (𝐼𝑌)))
Theoremgrpinvf 17289 The group inversion operation is a function on the base set. (Contributed by Mario Carneiro, 4-May-2015.)
𝐵 = (Base‘𝐺) & 𝑁 = (invg𝐺) (𝐺 ∈ Grp → 𝑁:𝐵𝐵)
Theoremgrpinvcl 17290 A group element's inverse is a group element. (Contributed by NM, 24-Aug-2011.) (Revised by Mario Carneiro, 4-May-2015.)
𝐵 = (Base‘𝐺) & 𝑁 = (invg𝐺) ((𝐺 ∈ Grp ∧ 𝑋𝐵) → (𝑁𝑋) ∈ 𝐵)
Theoremgrplinv 17291 The left inverse of a group element. (Contributed by NM, 24-Aug-2011.) (Revised by Mario Carneiro, 6-Jan-2015.)
𝐵 = (Base‘𝐺) & + = (+g𝐺) & 0 = (0g𝐺) & 𝑁 = (invg𝐺) ((𝐺 ∈ Grp ∧ 𝑋𝐵) → ((𝑁𝑋) + 𝑋) = 0 )
Theoremgrprinv 17292 The right inverse of a group element. (Contributed by NM, 24-Aug-2011.) (Revised by Mario Carneiro, 6-Jan-2015.)
𝐵 = (Base‘𝐺) & + = (+g𝐺) & 0 = (0g𝐺) & 𝑁 = (invg𝐺) ((𝐺 ∈ Grp ∧ 𝑋𝐵) → (𝑋 + (𝑁𝑋)) = 0 )
Theoremgrpinvid1 17293 The inverse of a group element expressed in terms of the identity element. (Contributed by NM, 24-Aug-2011.)
𝐵 = (Base‘𝐺) & + = (+g𝐺) & 0 = (0g𝐺) & 𝑁 = (invg𝐺) ((𝐺 ∈ Grp ∧ 𝑋𝐵𝑌𝐵) → ((𝑁𝑋) = 𝑌 ↔ (𝑋 + 𝑌) = 0 ))
Theoremgrpinvid2 17294 The inverse of a group element expressed in terms of the identity element. (Contributed by NM, 24-Aug-2011.)
𝐵 = (Base‘𝐺) & + = (+g𝐺) & 0 = (0g𝐺) & 𝑁 = (invg𝐺) ((𝐺 ∈ Grp ∧ 𝑋𝐵𝑌𝐵) → ((𝑁𝑋) = 𝑌 ↔ (𝑌 + 𝑋) = 0 ))
Theoremisgrpinv 17295* Properties showing that a function 𝑀 is the inverse function of a group. (Contributed by NM, 7-Aug-2013.) (Revised by Mario Carneiro, 2-Oct-2015.)
𝐵 = (Base‘𝐺) & + = (+g𝐺) & 0 = (0g𝐺) & 𝑁 = (invg𝐺) (𝐺 ∈ Grp → ((𝑀:𝐵𝐵 ∧ ∀𝑥𝐵 ((𝑀𝑥) + 𝑥) = 0 ) ↔ 𝑁 = 𝑀))
Theoremgrplrinv 17296* In a group, every member has a left and right inverse. (Contributed by AV, 1-Sep-2021.)
𝐵 = (Base‘𝐺) & + = (+g𝐺) & 0 = (0g𝐺) (𝐺 ∈ Grp → ∀𝑥𝐵𝑦𝐵 ((𝑦 + 𝑥) = 0 ∧ (𝑥 + 𝑦) = 0 ))
Theoremgrpidinv2 17297* A group's properties using the explicit identity element. (Contributed by NM, 5-Feb-2010.) (Revised by AV, 1-Sep-2021.)
𝐵 = (Base‘𝐺) & + = (+g𝐺) & 0 = (0g𝐺) ((𝐺 ∈ Grp ∧ 𝐴𝐵) → ((( 0 + 𝐴) = 𝐴 ∧ (𝐴 + 0 ) = 𝐴) ∧ ∃𝑦𝐵 ((𝑦 + 𝐴) = 0 ∧ (𝐴 + 𝑦) = 0 )))
Theoremgrpidinv 17298* A group has a left and right identity element, and every member has a left and right inverse. (Contributed by NM, 14-Oct-2006.) (Revised by AV, 1-Sep-2021.)
𝐵 = (Base‘𝐺) & + = (+g𝐺) (𝐺 ∈ Grp → ∃𝑢𝐵𝑥𝐵 (((𝑢 + 𝑥) = 𝑥 ∧ (𝑥 + 𝑢) = 𝑥) ∧ ∃𝑦𝐵 ((𝑦 + 𝑥) = 𝑢 ∧ (𝑥 + 𝑦) = 𝑢)))
Theoremgrpinvid 17299 The inverse of the identity element of a group. (Contributed by NM, 24-Aug-2011.)
0 = (0g𝐺) & 𝑁 = (invg𝐺) (𝐺 ∈ Grp → (𝑁0 ) = 0 )
Theoremgrplcan 17300 Left cancellation law for groups. (Contributed by NM, 25-Aug-2011.)
𝐵 = (Base‘𝐺) & + = (+g𝐺) ((𝐺 ∈ Grp ∧ (𝑋𝐵𝑌𝐵𝑍𝐵)) → ((𝑍 + 𝑋) = (𝑍 + 𝑌) ↔ 𝑋 = 𝑌))
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## Metric Measurement
##### Created by:
claudiawsmith on November 9, 2009
### Discuss
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# Metric Measurement
volumethe amount of space an object takes up
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#### Definitions
volume the amount of space an object takes up
cubic centimeter centimeter x centimeter x centimeter
liter the basic unit of volume in the metric system
milliliter a metric unit of measure that is 1/1000th of a liter
graduated cylinder a glass or plastic instrument used to measure the volume of liquids
meniscus the curved surface of a liquid
volume - displacement of water method of measuring volume of irregularly shaped objects
unit a known amount in measurement
customary measurement units most Americans use - inch, foot, yard, mile, cup
metric measurement scientists used this cm, g, km, ml
centimeter metric unit of measure that is 1/100th of a meter
weight the measure of how hard gravity pulls on an object
gram basic unit of mass in the metric system
kilogram a unit of mass in the metric system that equals 1,000 grams
milligram a unit of mass in the metric system that is 1/1000th of a gram
triple beam balance an instrument used to measure mass
mass how much matter there is in an object
millimeter a metric unit of measurement used to measure the width of a pencil
kilometer a metric unit of measure that is equal to 1,000 meters
meter the basic unit of length in the metric system
equivalents for length in metrics 10 mm = 1 cm
systems of measurement customary & metric
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Chapter 4
Mirror Apartments The diagram above shows the floor plan for the eight apartments on the first floor where Carl and Zoë live in separate apartments. Each floor has the same floor plan. There are eight apartments on every floor. The apartments in the building are numbered from 1 to 32. On every floor, each apartment has another apartment whose floor plan is its mirror image. For example, Apartment 8 is the mirror image of Apartment 3. On the second floor, Apartment 9 is directly over Apartment 1, Apartment 10 is directly over Apartment 2, and so forth. Carl lives in apartment 4. Zoë lives on the third floor in the apartment that is the mirror image of Carl's apartment. Question What is Zoë's apartment number?
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# hw4 - Homework 4 PHZ 5156 Due Thursday, October 1 1....
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Unformatted text preview: Homework 4 PHZ 5156 Due Thursday, October 1 1. Consider the diffusion equation in one spatial dimension, u ( x,t ) t = D 2 u ( x,t ) x 2 with boundary conditions u ( x = 0 ,t ) = u ( x = L,t ) = 0. Write a code that uses the Crank-Nicholson method, as discussed in class, to integrate the diffusion equation with the given boundary and initial conditions. First, convince yourself that we can write the algorithm as, 1 2 ( u n +1 j + u n j )- ( u n +1 j +1 + u n j +1 ) + 2 ( u n +1 j + u n j )- ( u n +1 j- 1 + u n j- 1 ) = u n j where = D t 2 x 2 , the j represent the discrete spatial step, and n represent the discrete time step. Take x j = j x for the spatial discretization, with x = L N +1 . Then the boundary conditions are u n j =0 = 0 and u n j = N +1 = 0. Write the above N linear equations as a matrix equation, Q n = U n where Q is an N N square matrix and n and U n are both column vectors with length N . Show that Q is a symmetric, tridiagonal matrix, with diagonal elements...
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## This note was uploaded on 08/08/2011 for the course PHZ 5156 taught by Professor Johnson,m during the Fall '08 term at University of Central Florida.
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## Where do surreal numbers come from and what do they mean?
I know about Conway's original discovery of the surreal numbers by way of games, as well as Kruskal's way of viewing surreal numbers in terms of asymptotic behavior of real-valued functions, leading to connections between surreal analysis and the theory of o-minimal structures (if Kruskal isn't the right attribution here, please feel free to correct me and educate everyone else). But I feel that, even with those two viewpoints available, surprisingly few connections between surreal numbers and the rest of mathematics have emerged over the past four decades. I say "surprising" because one would expect something so beautiful and natural to have all kinds of links with other things!
I think one reason the surreal numbers have found so few points of contact with the rest of contemporary mathematics is that the simplicity relation $a$-is-simpler-than-$b$ does not have any translational or dilational symmetries. (The simplest number between $-2$ and $+2$ is $0$, but the simplest number between $-2+1=-1$ and $+2+1=3$ is not $0+1=1$ but $0$. Likewise, the simplest number between $1$ and $3$ is $2$, but the simplest number between $2 \times 1=2$ and $2 \times 3=6$ is not $2 \times 2=4$ but $3$.) In the wake of Bourbaki, mathematicians have favored structures that have lots of morphisms to and from other already-favored structures, and/or lots of isomorphisms to themselves (aka symmetries), and the surreal numbers don't fit in with this esthetic.
Are there any new insights into how the surreal numbers fit in with the rest of math (or why they don't)?
It occurred to me after I posted my question that there is a weak $p$-adic analogue of the 2-adic surreal-numbers set-up, in which one relaxes the constraint that every interval contains a unique simplest number (that's a lot to give up, I admit!). If one defines $p$-adic simplicity in ${\bf Z}[1/p]$ in the obvious way (changing "2" to "$p$" in Conway's definition, so that integers are small if they are near 0 in the usual sense and elements of ${\bf Z}[1/p]$ are small if they have small denominator), then the following is true for $a_L,a_R,b_L,b_R$ in ${\bf Z}[1/p]$: if there is a unique simplest $a$ in ${\bf Z}[1/p]$ that is greater than $a_L$ and less than $a_R$, and there is a unique simplest $b$ in ${\bf Z}[1/p]$ that is greater than $b_L$ and less than $b_R$, then there is a simplest $c$ in ${\bf Z}[1/p]$ that is greater than $a+b_L$ and $a_L+b$ and less than $a+b_R$ and $a_R+b$, and it satisfies $c=a+b$. (Conway's multiplication formula works in this setting as well.) Is this mentioned in the surreal numbers literature, and more importantly, does the observation lead anywhere?
-
Why do you expect all kinds of connections? I think if there were substantial such reasons you would see surreal numbers everywhere. – Ryan Budney Apr 28 2011 at 18:34
Lieven Le Bruyn has written a nice series of blog posts on surreal numbers starting here neverendingbooks.org/index.php/… – jc Apr 28 2011 at 21:44
A similar perspective arises in this related question mathoverflow.net/questions/29300/… – Joel David Hamkins Apr 28 2011 at 23:36
Not numbers exactly, but certainly Conway's games (and related game-like structures) have aroused plenty of interest in certain category theorists and logicians, as they can be used to give categorical (non-posetal) semantics for substructural logics. In other words, under the Curry-Howard paradigm of "propositions as types", where propositions are promoted to objects and proofs of propositions $p \Rightarrow q$ are promoted to morphisms $p \to q$, the structures of formal deductions are embedded in strategies for games.
It was originally observed by André Joyal that Conway games are the objects of a compact closed category, where morphisms between games $G \to H$ are second-player winning strategies for $-G + H$. Closer to logical concerns, in
• Andreas Blass, A game semantics for linear logic, Ann. Pure Appl. Logic 56 (1992), 183-220
Andreas gave a game semantics for Girard's recently introduced linear logic (which on the "type" or categorical side correspond to $\ast$-autonomous categories); you can read some of his thoughts here for example, and elsewhere on his web page. This began a sort of cottage industry, where various refinements of games were developed to give soundness and completeness theorems for various forms of linear logic (see for example the Hyland-Ong reference in the online article).
(In an act of shameless self-promotion, I'll mention a little project with James Dolan to use certain types of games to model (free) cartesian closed categories, which was partly written up here.)
Numbers per se are used to measure strengths of positions in games, and sometimes this has been put to cunning use (as for example in analyzing some fairly specific but difficult positions in Go), but I don't know whether they have been exploited in game semantics along the lines above. Maybe Andreas can weigh in?
-
My graduate school Go mentor, colleague and friend, Yonghoan Kim, wrote his dissertation at Berkeley in the early 90s with Berlekamp doing exactly what you say in your last paragraph---using numbers, especially infinitesimals, and the more general algebra of game values, to analyze specific games, including Go. – Joel David Hamkins Apr 29 2011 at 1:29
An algebraic perspective on the surreal numbers is that they are the maximal totally ordered field.'' Of course, since the surreal numbers are a class and not a set they cannot really be a field. Nevertheless, I mentally file the surreal numbers in the same folder as various universal constructions: the algebraic closure of a field, the absolute Galois group of the rationals, the fraction field of a ring, etc...
Conway's construction is really beautiful, but it doesn't seem to lend itself to algebraic manipulations like the axioms of a totally ordered field. For there to be interesting algebraic applications of the surreal numbers, one has to weigh the benefits of having one object'' against the set-theoretic headaches of dealing with a proper class.
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"...at least for now, I'm finished. All the questions that have yet to be answered are too hard." -- Jacob Lurie on surreal numbers, 1996
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There is a book by Norman Alling, entitled Foundations of Analysis over Surreal Number Fields, published by North-Holland in 1987. I am not aware that it had much influence, but it looks like quite an interesting read.
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### Site Tools
advanced_tools:group_theory:su2 [2020/12/05 18:35]edi [Concrete] advanced_tools:group_theory:su2 [2020/12/26 22:52] (current)edi [Intuitive] Both sides previous revision Previous revision 2020/12/26 22:52 edi [Intuitive] 2020/12/05 18:35 edi [Concrete] 2020/12/05 18:29 edi [Intuitive] 2020/11/28 18:40 edi [Concrete] 2020/09/07 04:19 [Concrete] 2018/04/15 16:31 aresmarrero 2018/04/15 16:30 aresmarrero [Student] 2018/03/17 16:03 jakobadmin [Student] 2018/03/17 16:02 jakobadmin [Student] 2018/03/17 16:01 jakobadmin [Student] 2017/12/17 12:02 jakobadmin [Why is it interesting?] 2017/12/17 11:59 jakobadmin [Student] 2017/12/04 09:01 external edit2017/11/03 14:10 jakobadmin created 2020/12/26 22:52 edi [Intuitive] 2020/12/05 18:35 edi [Concrete] 2020/12/05 18:29 edi [Intuitive] 2020/11/28 18:40 edi [Concrete] 2020/09/07 04:19 [Concrete] 2018/04/15 16:31 aresmarrero 2018/04/15 16:30 aresmarrero [Student] 2018/03/17 16:03 jakobadmin [Student] 2018/03/17 16:02 jakobadmin [Student] 2018/03/17 16:01 jakobadmin [Student] 2017/12/17 12:02 jakobadmin [Why is it interesting?] 2017/12/17 11:59 jakobadmin [Student] 2017/12/04 09:01 external edit2017/11/03 14:10 jakobadmin created Line 3: Line 3: - The Lie group $SU(2)$ describes all possible rotations of a spinorial object, that is, an object that needs to be rotated 720 degrees before returning to its initial state. A good example for such an object is a cube that is attached to a wall by belts: see the animations here [[https://en.wikipedia.org/wiki/Spinor]]. In physics, an important spinorial object is the fermion (e.g., an electron). + The Lie group $SU(2)$ describes all possible 3D rotations of a spinorial object, that is, an object that needs to be rotated 720 degrees before returning to its initial state. A good example for such an object is a cube that is attached to a wall by belts: see the animations here [[https://en.wikipedia.org/wiki/Spinor]]. In physics, an important spinorial object is the fermion (e.g., an electron). For small rotations $SU(2)$ is identical to $SO(3)$, that is, both groups have the same Lie algebra. For small rotations $SU(2)$ is identical to $SO(3)$, that is, both groups have the same Lie algebra.
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# Placing Boats in Battleship (Removing Code Duplication)
I'm (slowly) working out a 2D version of Battleship now that my 1D version is done. I wrote the following function to place a boat on the board given the length of the boat, the position of the boat, and the direction the boat is facing. However, the function is ugly. Very ugly. By which I mean, there is much in the way of code duplication. Could anyone point out some ways in which I could reduce the duplication in this code?
``````(defun place-boat (len pos dir)
(let ((offset 0))
(dotimes (i len)
(if (= dir 0)
(if (< pos 50)
(setf (aref *ans-board*
(+ (/ pos 10) offset)
(mod pos 10))
'#)
(setf (aref *ans-board*
(- (/ pos 10) offset)
(mod pos 10))
'#))
(if (< pos 50)
(setf (aref *ans-board*
(/ pos 10)
(+ (mod pos 10) offset))
'#)
(setf (aref *ans-board*
(/ pos 10)
(- (mod pos 10) offset))
'#)))
(incf offset))))
``````
EDIT: For clarification, `pos` is a number between 1 and 100, signifying a cell in a 10x10 2D array.
-
Well, for starters, I don't think you need both i and offset. They both go from 0 up to len in step with each other.
Then you could do something like this to collapse the +/- cases for < or >= 50 into one statement:
`(+ (/ pos 10) (* (if (< pos 50) 1 -1) offset))`
That give you (untested):
``````(defun place-boat (len pos dir)
(dotimes (offset len)
(if (= dir 0)
(setf (aref *ans-board*
(+ (/ pos 10) (* (if (< pos 50) 1 -1) offset))
(mod pos 10))
'#)
(setf (aref *ans-board*
(/ pos 10)
(+ (mod pos 10) (* (if (< pos 50) 1 -1) offset)))
'#))))
``````
which still has some redundancy. But that's what I've got so far.
Note, I know very little about Common Lisp, so I'm sure someone else could do much better :)
-
With some help from the fine folks in #lisp, I tweaked your solution a bit and ended up with this: paste.lisp.org/display/117875#3 Thanks! – Andy Dec 20 '10 at 11:07
Glad I could help :) – Wodin Dec 20 '10 at 11:43
Supposed I understand your needs correctly: if you would deal with the directions -1,-1,10,-10 you could easily do something like this:
``````(defun place (pos len dir)
(loop for i from pos to (+ pos (* len dir)) by dir do
(setf (aref board i) '\#)))
``````
-
That would work, but the array being dealt with is 2-dimensional. :/ – Andy Dec 21 '10 at 20:42
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82 424
Assignments Done
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# Answer to Question #210 in Statistics and Probability for benny
Question #210
Suppose candidate A receives 54% of the entire vote, but only 48% of the female vote. A voter is selected at random for an interview. What is the probability that the voter is a woman who voted for candidate A?
1
2010-05-31T07:36:15-0400
You need to find a percentage, which is formed by men and women from the total number of voters. Then the percentage of women needs to be multiplied by 0.48. In this way you will obtain the desired probability.
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Actions
Math 430 - Spring 2009
The notion of a group acting on a is one which links to nearly every branch of mathematics. Group actions appear in geometry, , and differential equations, to name a few. Group actions are a fundamental tool in pure as well, and one of our main applications will be the Sylow Theorems (Sections 36 and 37 in Fraleigh). These notes should be used as a supplement to Section 16 of Fraleigh’s book. Some of the notation here will differ from the notation in Fraleigh, but we will attempt to point out whenever this happens. Let G be a group and let X be a set. Let Sym(X) denote the group of all of the elements of X (written as SX in Fraleigh). So, if X ∼ is a finite set and |X| = n, then Sym(X) = Sn. We will give two equivalent definitions of G acting on X.
Definition 1. We say that G acts on X if there is a φ : G → Sym(X).
One way of thinking of G acting on X is that elements of the group G may be “applied to” elements of X to give a new element of X. The next definition takes this point of view.
Definition 2. We say that G acts on X if there is a ∗ : G × X → X, so that if g ∈ G and x ∈ X, then ∗(g, x) = g ∗ x ∈ X, such that: (i) For every g, h ∈ G, x ∈ X, we have (gh) ∗ x = g ∗ (h ∗ x), (ii) For every x ∈ X, e ∗ x = x, where e ∈ G is the identity.
If the group G acts on the set X, we will call X a G-set. Note that Fraleigh often writes gx for g ∗ x, where g ∈ G and x ∈ X.
1 Before giving examples, we need to show that the two above definitions actually define the same notion.
Theorem 1 Definition 1 and Definition 2 are equivalent.
Proof. First assume that G and X satisfy Definition 1, so that we have a homomorphism φ : G → Sym(X). We now show that G and X must also then satisfy Definition 2. We define a map ∗ : G×X → X by g∗x = φ(g)(x). First, for every g, h ∈ G, x ∈ X, using the fact that φ is a homomorphism, we have
(gh) ∗ x = φ(gh)(x) = (φ(g) ◦ φ(h))(x) = φ(g)(φ(h)(x)) = g ∗ (h ∗ x), so that ∗ satisfies condition (i) of Definition 2. Also, since φ is a homomor- phism, φ(e) is the trivial , where e ∈ G is the identity element. So e ∗ x = φ(e)(x) = x, which is condition (ii) of Definition 2. Thus G and X satisfy Definition 2. Now suppose G and X satisfy Definition 2, so that we have a map
∗ : G × X → X which satisfies (i) and (ii). We define a map φ : G → Sym(X) by φ(g)(x) = g ∗ x. We first show that this is well-defined, that is, φ(g) is actually a one- to-one and onto map from X to itself. To show that φ(g) is onto, let x ∈ X, and consider g−1 ∗ x ∈ X. Then we have
φ(g)(g−1 ∗ x) = g ∗ (g−1 ∗ x) = (gg−1) ∗ x = e ∗ x = x, so φ(g) is onto. To show that φ(g) is one-to-one, suppose that we have φ(g)(x) = φ(g)(y) for x, y ∈ X, so that g ∗ x = g ∗ y. Using both conditions (i) and (ii) of Definition 2, we have g−1 ∗ (g ∗ x) = g−1 ∗ (g ∗ y) ⇒ (g−1g) ∗ x = (g−1g) ∗ y ⇒ e ∗ x = e ∗ y ⇒ x = y.
Finally, we show that φ is a homomorphism. Let g, h ∈ G, x ∈ X. We have
φ(gh)(x) = (gh) ∗ x = g ∗ (h ∗ x) = φ(g)(φ(h)(x)) = (φ(g) ◦ φ(h))(x).
Thus, G and X satisfy Definition 1.
2 Now that we have a few ways of thinking about group actions, let’s see some examples.
Example 1. As mentioned before, we may take X = {1, 2, . . . , n}, G = Sn = Sym(X), and φ : Sn → Sn to be the identity map.
Example 2. Let X = Rn and G = GL(n, R), and for A ∈ G, v ∈ X, define A ∗ v = Av. That is, we let G act on X as linear transformations.
Example 3. Let X be a unit cube sitting in R3, and let G be the group of symmetries of X, which acts on X again as linear transformations on R3.
Example 4. Let X be a group H, and let G also be the same group H, where H acts on itself by left multiplication. That is, for h ∈ X = H and g ∈ G = H, define g ∗h = gh. This action was used to show that every group is isomorphic to a group of permutations (Cayley’s Theorem, in Section 8 of Fraleigh).
Before defining more terms, we’ll first see a nice application to finite group theory. Theorem 2 Let G be a finite group, and let H be a of G such that [G : H] = p, where p is the smallest prime dividing |G|. Then H is a of G. Proof. We let X be the set of left of H in G. From the proof of Lagrange’s Theorem (Section 10 of Fraleigh), we have |X| = [G : H] = p, ∼ and so Sym(X) = Sp. We define an action of G on X by g ∗ aH = gaH, for g ∈ G and aH ∈ X. That is, we let G act on the left cosets of H in G by left multiplication. This satisfies Definition 2, since for any g1, g2, a ∈ G, we have (g1g2) ∗ aH = g1g2aH and e ∗ aH = aH. From Theorem 1, and since ∼ Sym(X) = Sp, we have a homomorphism φ : G → Sp. For any g ∈ G, g 6∈ H, we have g ∗ H = gH 6= H, and so φ(g) cannot be the trivial permutation of left cosets of H in G, that is, g 6∈ ker(φ) when g 6∈ H. We must therefore have ker(φ) ≤ H. From the first theorem for groups, we have G/ker(φ) ∼= im(φ), where im(φ) = φ(G) is a subgroup of Sp. So we have |G| = |G/ker(φ)| |S | = p!. |ker(φ)| p
3 Note that p is the largest prime dividing p!, and p2 does not divide p!, while p is the smallest prime dividing |G|. Since ker(φ) ≤ H and H is a proper subgroup of G, we cannot have G = ker(φ), that is, [G : ker(φ)] 6= 1. The only possibility is that |G/ker(φ)| = [G : ker(φ)] = p, since this is the only divisor of |G| which divides p!. We now have
|G| |G| [G : ker(φ)] = = p = [G : H] = , |ker(φ)| |H| so that |H| = |ker(φ)|. Since ker(φ) ⊆ H, we must have H = ker(φ), which is a normal subgroup of G.
We now define a few important terms relevant to group actions.
Definition 3. Let G be a group which acts on the set X. For x ∈ X, the stabilizer of x in G, written stabG(x), is the set of elements g ∈ G such that g ∗ x = x. In symbols,
stabG(x) = {g ∈ G | g ∗ x = x}.
In Fraleigh, this is called the isotropy subgroup of x, and is written Gx (we show below that this is actually a subgroup of G). For x ∈ X, the orbit of x under G, written orbG(x), is the set of all elements in X of the form g ∗ x for g ∈ G. In symbols,
orbG(x) = {g ∗ x | g ∈ G}.
Fraleigh uses the notation Gx for the orbit of x under G.
Example 5. Let G = {(1), (1 2), (3 4 6), (3 6 4), (1 2)(3 4 6), (1 2)(3 6 4)}, and let φ : G → S6, φ(α) = α, be the natural injection, as G is a subgroup of S6. Then G acts on {1, 2, 3, 4, 5, 6}. First note that since 5 is fixed by every element of G, we have stabG(5) = G, and orbG(5) = {5}. We also have stabG(3) = stabG(4) = stabG(6) = h(1 2)i, stabG(1) = stabG(2) = h(3 4 6)i,
orbG(3) = orbG(4) = orbG(6) = {3, 4, 6}, orbG(1) = orbG(2) = {1, 2}. Example 6. Let G be any group, and we let G act on itself by conjugation. That is, for g, a ∈ G, we define g∗a = gag−1. We first check that this satisfies
4 Definition 2. First, we have e ∗ a = eae−1 = a. Now let g, h, a ∈ G. Then we have (gh) ∗ a = gha(gh)−1 = ghah−1g−1 = g ∗ (h ∗ a), so this is indeed a . If we fix an a ∈ G, we see that the orbit of a is −1 orbG(a) = {gag | g ∈ G}, which is called the conjugacy of a in G. If we look at the stabilizer of a in G, we have −1 stabG(a) = {g ∈ G | gag = a}, which is the centralizer of a in G, also written CG(a). The next Lemma shows us that stabilizers of group actions are always , and so in particular, centralizers of elements of groups are subgroups.
Lemma 1 If G acts on X, and x ∈ X, then stabG(x) is a subgroup of G.
Proof. Let x ∈ X. Since e ∗ x = x, we know that e ∈ stabG(x), and so the stabilizer of x in G is nonempty. Now suppose g, h ∈ stabG(x). Since g ∗ x = x, we have g−1 ∗ (g ∗ x) = g−1 ∗ x ⇒ (g−1g) ∗ x = g−1 ∗ x ⇒ e ∗ x = g−1 ∗ x ⇒ g−1 ∗ x = x.
−1 So, g ∈ stabG(x). We also have
(gh) ∗ x = g ∗ (h ∗ x) = g ∗ x = x, so gh ∈ stabG(x). Thus stabG(x) ≤ G.
The next result is the most important basic result in the theory of group actions.
Theorem 3 (Orbit-Stabilizer Lemma) Suppose G is a group which acts on X. For any x ∈ X, we have
|orbG(x)| = [G : stabG(x)], which means that the are equal even when these are infinite sets. If G is a finite group, then
|G| = |stabG(x)| |orbG(x)|.
5 Proof. Fix x ∈ X. From Lemma 1, stabG(x) is a subgroup of G, and we recall that[G : H] denotes the of the set of left cosets of H in G. Let K denote the set of left cosets of H in G. Define a
f : orbG(x) → K, by f(g∗x) = gH. First, we check that f is well-defined, and at the same time check that f is injective. If g1, g2 ∈ G, g1 ∗ x = g2 ∗ x ∈ orbG(x) if and only if −1 −1 (g2 g1)∗x = x, iff g2 g1 ∈ stabG(x) = H, which is equivalent to g2H = g1H. So g1 ∗ x = g2 ∗ x if and only if f(g1 ∗ x) = f(g2 ∗ x), and f is well-defined and injective. Also f is onto, since for any gH ∈ K, f(g ∗ x) = gH. Thus, f gives a one-to-one correspondence, and so
|orbG(x)| = |K| = [G : stabG(x)].
When G is finite, it follows from the proof of Lagrange’s Theorem that [G : stabG(x)] = |G|/|stabG(x)| So, in this case, |G| = |stabG(x)| |orbG(x)|.
Next, we connect the concept of a group action with the important notion of an equivalence .
Theorem 4 Let G be a group which acts on a set X, and for x, y ∈ X, define x ∼ y to mean that there is a g ∈ G such that g ∗ x = y. Then ∼ is an on X, and the of x ∈ X is orbG(x).
Proof. We must check that ∼ satisfies the reflexive, symmetric, and transitive properties. First, for any x ∈ X, we have e ∗ x = x, where e is the identity in G, and so x ∼ x and the reflective property holds. Next, if x ∼ y, then there is a g ∈ G such that g ∗ x = y. It follows from Definition 2 that we then have g−1 ∗ y = x, so that y ∼ x and the symmetric property holds. Now assume x ∼ y and y ∼ z, where g ∗ x = y and h ∗ y = z. Then from Definition 2, h ∗ (g ∗ x) = (hg) ∗ x = z, and so x ∼ z and transitivity holds. So, ∼ is an equivalence relation. From the definition of an equivalence class, if x ∈ X, then the class of x is the set {y ∈ X | x ∼ y} = {y ∈ X | y = g ∗ x for some g ∈ G}. This is exactly the definition of the orbit of x under G.
We conclude with one more application to group theory, this time to the
6 conjugacy classes of a group, as introduced in Example 6 above. Note that if G is a group and z ∈ G is in the center of G, then the of z is just {z}.
Theorem 5 (Class Formula) Let G be a finite group, let Z(G) be the cen- ter of G, and let A be a collection of distinct representatives of conjugacy classes of G which are not in Z(G). Then we have X |G| = |Z(G)| + [G : CG(a)]. a∈A
Proof. For any x ∈ G, let cl(x) denote the conjugacy class of x in G. From Example 6 above, we let G act on itself by conjugation, and for any x ∈ G, we have orbG(x) = cl(x), and stabG(x) = CG(a). From Theorem 3, we have, for each x ∈ G, |cl(x)| = |G|/|CG(x)| = [G : CG(x)]. Since from Theorem 4 the conjugacy classes of G are just equivalence classes, we have that conjugacy classes form a partition of G. So, the of distinct conjugacy classes of G gives G. Let B be a set of representatives of distinct conjugacy classes of G, and we have X X |G| = |cl(b)| = [G : CG(b)]. (1) b∈B b∈B
We also know that b ∈ Z(G) exactly when gbg−1 = b for every g ∈ G, which P happens exactly when |cl(b)| = 1. So, z∈Z(G) |cl(z)| = |Z(G)|. If we choose A to be a set of representatives of conjugacy classes which are not in Z(G), splitting (1) into a sum over Z(G) and a sum over A gives the result.
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### Determine the systems thermal efficiency
Assignment Help Mechanical Engineering
##### Reference no: EM13701650
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Assume isentropic expansions for all compressor and the turbine stages and use constant specific heats at room temperature
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Language - Maple Help
Language and System Changes in Maple 14
Maple 14 includes the following language and system changes.
Help Browser
• By setting interface(helpbrowser=standard), it is possible to view documentation through the Standard GUI's help browser launched from any interface. This gives Command-line and Classic users a nicer interface for viewing documentation including hyperlinks, full-text search, and the table of contents. Additionally, help pages that are written using features specific to Maple's Standard interface and that were previously not viewable in Classic and Command-line help will be accessible. This includes some help pages, example worksheets, and Maple Portal pages. For more information on the helpbrowser variable, see interface.
Eval and rtable
• Eval now works on rtable constructions such as Matrix.
> Eval(Matrix([[x^2 + y]]), {x=3, y=2}) mod 5;
$\left[\begin{array}{c}{1}\end{array}\right]$ (1)
RegularChains option in RootFinding[Isolate]
• Users of RootFinding[Isolate] now can choose the method to isolate the real roots of polynomial system. These methods are "RS", which is based on Fabrice Rouillier's RealSolving (RS) C library, and "RC", which is based on the RegularChains package by Marc Moreno Maza et al.
> with(RootFinding):
> F := [x^2-2, y-1];
${F}{≔}\left[{{x}}^{{2}}{-}{2}{,}{y}{-}{1}\right]$ (2)
> Isolate(F, [x,y], method="RS");
$\left[\left[{x}{=}{-1.414213562}{,}{y}{=}{1.000000000}\right]{,}\left[{x}{=}{1.414213562}{,}{y}{=}{1.000000000}\right]\right]$ (3)
> Isolate(F, [x,y], method="RC");
$\left[\left[{x}{=}{-1.414213562}{,}{y}{=}{1.}\right]{,}\left[{x}{=}{1.414213562}{,}{y}{=}{1.}\right]\right]$ (4)
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If the city council institutes new parking regulations, city : GMAT Critical Reasoning (CR)
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# If the city council institutes new parking regulations, city
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If the city council institutes new parking regulations, city revenues will surely increase, since studies have conclusively shown that, if such parking regulations are put into effect, there is an increase in parking violations, and an increase in parking violations will result in a greater number of parking fines collected.
21. Which one of the following is closest, in terms of its logical features, to the reasoning used in the argument above?
(A) Last year┬б┬пs increase in revenues can be easily explained. That was the year the city council instituted new parking regulations. No doubt the new law brought with it an increase in the number of parking violations.
(B) If taxes were increased, this act would naturally result in increased revenues for the city, and increased revenues would make some desirable social programs possible. So, if taxes were increased, some desirable social programs would become possible.
(C) Henry says that, if the city council goes into closed session, an important matter of personnel policy is being discussed. However, no personnel matters were discussed at the council meeting, so if Henry is right, the council did not go into closed session.
(D) All cars parked on the north side of the street were ticketed last night, and the same cars were towed away this morning. So beware A car ticketed in this city also gets towed away.
(E) Allen says that, if the city council institutes new parking regulations, it is unlikely that revenues for the city will increase. If Allen is right, then the parking regulation plan should not be instituted.
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10 Dec 2003, 12:05
The argument: If X, then Y, because studies show that X leads to Z, which leads to y.
D is closest (IMO) but it seems counterintuitive.
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11 Dec 2003, 07:09
Can you explain ?
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11 Dec 2003, 07:32
Even I vote for B.
Stimulus can be written as
If City council institutes new parking regulation -----> result in increased revenue. And an evidence is given to support this.
Only in B we have the similar structure
If taxes were increased ------> result in increased revenues
By process of elimination I got B.
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27 Dec 2003, 12:19
I also got B with the following logic
Stem of the question says
Parking Regulation -> Parking Violations -> More Revenues
hence
Parking Regulation -> More Revenues
in B
Increase in Taxes -> More Revenue -> Support for Social Programs
Hence
Increase in Taxes -> Support for Social Programs
D is next closest but D uses the word some instead of all.
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Re: If the city council institutes new parking regulations, city [#permalink]
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11 Apr 2016, 02:51
In my opinion B is the most logical one in this case.
Re: If the city council institutes new parking regulations, city [#permalink] 11 Apr 2016, 02:51
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1. ## derivative cube root
hey guys, just wondering what the derivative of this is... thank u
$\displaystyle 8 \sqrt[3]{x}$
2. Hello !
Originally Posted by jvignacio
hey guys, just wondering what the derivative of this is... thank u
$\displaystyle 8 \sqrt[3]{x}$
Remember that $\displaystyle \sqrt[n]{x}$ can be rewritten $\displaystyle x^{\tfrac 1n}$
So here, $\displaystyle \sqrt[3]{x}=\dots$
Then remember some basic rules of differentiation :
$\displaystyle (a \cdot f(x))'=a \cdot f'(x)$, a a constant
$\displaystyle (x^n)'=nx^{n-1}$
3. Originally Posted by Moo
Hello !
Remember that $\displaystyle \sqrt[n]{x}$ can be rewritten $\displaystyle x^{\tfrac 1n}$
So here, $\displaystyle \sqrt[3]{x}=\dots$
Then remember some basic rules of differentiation :
$\displaystyle (a \cdot f(x))'=a \cdot f'(x)$, a a constant
$\displaystyle (x^n)'=nx^{n-1}$
thank u! got it
4. Originally Posted by jvignacio
hey guys, just wondering what the derivative of this is... thank u
$\displaystyle 8 \sqrt[3]{x}$
$\displaystyle f(x)= 8 \sqrt[3]{x}$
$\displaystyle f(x)= 8 x^{\frac{1}{3}}$
$\displaystyle f'(x)= 8 (\frac{1}{3}) x^{\frac{-2}{3}}$
$\displaystyle f'(x)= \frac{8}{\sqrt[3]{x^2}}$
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LAPACK 3.3.1 Linear Algebra PACKage
# zchkst.f
Go to the documentation of this file.
```00001 SUBROUTINE ZCHKST( NSIZES, NN, NTYPES, DOTYPE, ISEED, THRESH,
00002 \$ NOUNIT, A, LDA, AP, SD, SE, D1, D2, D3, D4, D5,
00003 \$ WA1, WA2, WA3, WR, U, LDU, V, VP, TAU, Z, WORK,
00004 \$ LWORK, RWORK, LRWORK, IWORK, LIWORK, RESULT,
00005 \$ INFO )
00006 IMPLICIT NONE
00007 *
00008 * -- LAPACK test routine (version 3.1) --
00009 * Univ. of Tennessee, Univ. of California Berkeley and NAG Ltd..
00010 * November 2006
00011 *
00012 * .. Scalar Arguments ..
00013 INTEGER INFO, LDA, LDU, LIWORK, LRWORK, LWORK, NOUNIT,
00014 \$ NSIZES, NTYPES
00015 DOUBLE PRECISION THRESH
00016 * ..
00017 * .. Array Arguments ..
00018 LOGICAL DOTYPE( * )
00019 INTEGER ISEED( 4 ), IWORK( * ), NN( * )
00020 DOUBLE PRECISION D1( * ), D2( * ), D3( * ), D4( * ), D5( * ),
00021 \$ RESULT( * ), RWORK( * ), SD( * ), SE( * ),
00022 \$ WA1( * ), WA2( * ), WA3( * ), WR( * )
00023 COMPLEX*16 A( LDA, * ), AP( * ), TAU( * ), U( LDU, * ),
00024 \$ V( LDU, * ), VP( * ), WORK( * ), Z( LDU, * )
00025 * ..
00026 *
00027 * Purpose
00028 * =======
00029 *
00030 * ZCHKST checks the Hermitian eigenvalue problem routines.
00031 *
00032 * ZHETRD factors A as U S U* , where * means conjugate transpose,
00033 * S is real symmetric tridiagonal, and U is unitary.
00034 * ZHETRD can use either just the lower or just the upper triangle
00035 * of A; ZCHKST checks both cases.
00036 * U is represented as a product of Householder
00037 * transformations, whose vectors are stored in the first
00038 * n-1 columns of V, and whose scale factors are in TAU.
00039 *
00040 * ZHPTRD does the same as ZHETRD, except that A and V are stored
00041 * in "packed" format.
00042 *
00043 * ZUNGTR constructs the matrix U from the contents of V and TAU.
00044 *
00045 * ZUPGTR constructs the matrix U from the contents of VP and TAU.
00046 *
00047 * ZSTEQR factors S as Z D1 Z* , where Z is the unitary
00048 * matrix of eigenvectors and D1 is a diagonal matrix with
00049 * the eigenvalues on the diagonal. D2 is the matrix of
00050 * eigenvalues computed when Z is not computed.
00051 *
00052 * DSTERF computes D3, the matrix of eigenvalues, by the
00053 * PWK method, which does not yield eigenvectors.
00054 *
00055 * ZPTEQR factors S as Z4 D4 Z4* , for a
00056 * Hermitian positive definite tridiagonal matrix.
00057 * D5 is the matrix of eigenvalues computed when Z is not
00058 * computed.
00059 *
00060 * DSTEBZ computes selected eigenvalues. WA1, WA2, and
00061 * WA3 will denote eigenvalues computed to high
00062 * absolute accuracy, with different range options.
00063 * WR will denote eigenvalues computed to high relative
00064 * accuracy.
00065 *
00066 * ZSTEIN computes Y, the eigenvectors of S, given the
00067 * eigenvalues.
00068 *
00069 * ZSTEDC factors S as Z D1 Z* , where Z is the unitary
00070 * matrix of eigenvectors and D1 is a diagonal matrix with
00071 * the eigenvalues on the diagonal ('I' option). It may also
00072 * update an input unitary matrix, usually the output
00073 * from ZHETRD/ZUNGTR or ZHPTRD/ZUPGTR ('V' option). It may
00074 * also just compute eigenvalues ('N' option).
00075 *
00076 * ZSTEMR factors S as Z D1 Z* , where Z is the unitary
00077 * matrix of eigenvectors and D1 is a diagonal matrix with
00078 * the eigenvalues on the diagonal ('I' option). ZSTEMR
00079 * uses the Relatively Robust Representation whenever possible.
00080 *
00081 * When ZCHKST is called, a number of matrix "sizes" ("n's") and a
00082 * number of matrix "types" are specified. For each size ("n")
00083 * and each type of matrix, one matrix will be generated and used
00084 * to test the Hermitian eigenroutines. For each matrix, a number
00085 * of tests will be performed:
00086 *
00087 * (1) | A - V S V* | / ( |A| n ulp ) ZHETRD( UPLO='U', ... )
00088 *
00089 * (2) | I - UV* | / ( n ulp ) ZUNGTR( UPLO='U', ... )
00090 *
00091 * (3) | A - V S V* | / ( |A| n ulp ) ZHETRD( UPLO='L', ... )
00092 *
00093 * (4) | I - UV* | / ( n ulp ) ZUNGTR( UPLO='L', ... )
00094 *
00095 * (5-8) Same as 1-4, but for ZHPTRD and ZUPGTR.
00096 *
00097 * (9) | S - Z D Z* | / ( |S| n ulp ) ZSTEQR('V',...)
00098 *
00099 * (10) | I - ZZ* | / ( n ulp ) ZSTEQR('V',...)
00100 *
00101 * (11) | D1 - D2 | / ( |D1| ulp ) ZSTEQR('N',...)
00102 *
00103 * (12) | D1 - D3 | / ( |D1| ulp ) DSTERF
00104 *
00105 * (13) 0 if the true eigenvalues (computed by sturm count)
00106 * of S are within THRESH of
00107 * those in D1. 2*THRESH if they are not. (Tested using
00108 * DSTECH)
00109 *
00110 * For S positive definite,
00111 *
00112 * (14) | S - Z4 D4 Z4* | / ( |S| n ulp ) ZPTEQR('V',...)
00113 *
00114 * (15) | I - Z4 Z4* | / ( n ulp ) ZPTEQR('V',...)
00115 *
00116 * (16) | D4 - D5 | / ( 100 |D4| ulp ) ZPTEQR('N',...)
00117 *
00118 * When S is also diagonally dominant by the factor gamma < 1,
00119 *
00120 * (17) max | D4(i) - WR(i) | / ( |D4(i)| omega ) ,
00121 * i
00122 * omega = 2 (2n-1) ULP (1 + 8 gamma**2) / (1 - gamma)**4
00123 * DSTEBZ( 'A', 'E', ...)
00124 *
00125 * (18) | WA1 - D3 | / ( |D3| ulp ) DSTEBZ( 'A', 'E', ...)
00126 *
00127 * (19) ( max { min | WA2(i)-WA3(j) | } +
00128 * i j
00129 * max { min | WA3(i)-WA2(j) | } ) / ( |D3| ulp )
00130 * i j
00131 * DSTEBZ( 'I', 'E', ...)
00132 *
00133 * (20) | S - Y WA1 Y* | / ( |S| n ulp ) DSTEBZ, ZSTEIN
00134 *
00135 * (21) | I - Y Y* | / ( n ulp ) DSTEBZ, ZSTEIN
00136 *
00137 * (22) | S - Z D Z* | / ( |S| n ulp ) ZSTEDC('I')
00138 *
00139 * (23) | I - ZZ* | / ( n ulp ) ZSTEDC('I')
00140 *
00141 * (24) | S - Z D Z* | / ( |S| n ulp ) ZSTEDC('V')
00142 *
00143 * (25) | I - ZZ* | / ( n ulp ) ZSTEDC('V')
00144 *
00145 * (26) | D1 - D2 | / ( |D1| ulp ) ZSTEDC('V') and
00146 * ZSTEDC('N')
00147 *
00148 * Test 27 is disabled at the moment because ZSTEMR does not
00149 * guarantee high relatvie accuracy.
00150 *
00151 * (27) max | D6(i) - WR(i) | / ( |D6(i)| omega ) ,
00152 * i
00153 * omega = 2 (2n-1) ULP (1 + 8 gamma**2) / (1 - gamma)**4
00154 * ZSTEMR('V', 'A')
00155 *
00156 * (28) max | D6(i) - WR(i) | / ( |D6(i)| omega ) ,
00157 * i
00158 * omega = 2 (2n-1) ULP (1 + 8 gamma**2) / (1 - gamma)**4
00159 * ZSTEMR('V', 'I')
00160 *
00161 * Tests 29 through 34 are disable at present because ZSTEMR
00162 * does not handle partial specturm requests.
00163 *
00164 * (29) | S - Z D Z* | / ( |S| n ulp ) ZSTEMR('V', 'I')
00165 *
00166 * (30) | I - ZZ* | / ( n ulp ) ZSTEMR('V', 'I')
00167 *
00168 * (31) ( max { min | WA2(i)-WA3(j) | } +
00169 * i j
00170 * max { min | WA3(i)-WA2(j) | } ) / ( |D3| ulp )
00171 * i j
00172 * ZSTEMR('N', 'I') vs. CSTEMR('V', 'I')
00173 *
00174 * (32) | S - Z D Z* | / ( |S| n ulp ) ZSTEMR('V', 'V')
00175 *
00176 * (33) | I - ZZ* | / ( n ulp ) ZSTEMR('V', 'V')
00177 *
00178 * (34) ( max { min | WA2(i)-WA3(j) | } +
00179 * i j
00180 * max { min | WA3(i)-WA2(j) | } ) / ( |D3| ulp )
00181 * i j
00182 * ZSTEMR('N', 'V') vs. CSTEMR('V', 'V')
00183 *
00184 * (35) | S - Z D Z* | / ( |S| n ulp ) ZSTEMR('V', 'A')
00185 *
00186 * (36) | I - ZZ* | / ( n ulp ) ZSTEMR('V', 'A')
00187 *
00188 * (37) ( max { min | WA2(i)-WA3(j) | } +
00189 * i j
00190 * max { min | WA3(i)-WA2(j) | } ) / ( |D3| ulp )
00191 * i j
00192 * ZSTEMR('N', 'A') vs. CSTEMR('V', 'A')
00193 *
00194 * The "sizes" are specified by an array NN(1:NSIZES); the value of
00195 * each element NN(j) specifies one size.
00196 * The "types" are specified by a logical array DOTYPE( 1:NTYPES );
00197 * if DOTYPE(j) is .TRUE., then matrix type "j" will be generated.
00198 * Currently, the list of possible types is:
00199 *
00200 * (1) The zero matrix.
00201 * (2) The identity matrix.
00202 *
00203 * (3) A diagonal matrix with evenly spaced entries
00204 * 1, ..., ULP and random signs.
00205 * (ULP = (first number larger than 1) - 1 )
00206 * (4) A diagonal matrix with geometrically spaced entries
00207 * 1, ..., ULP and random signs.
00208 * (5) A diagonal matrix with "clustered" entries 1, ULP, ..., ULP
00209 * and random signs.
00210 *
00211 * (6) Same as (4), but multiplied by SQRT( overflow threshold )
00212 * (7) Same as (4), but multiplied by SQRT( underflow threshold )
00213 *
00214 * (8) A matrix of the form U* D U, where U is unitary and
00215 * D has evenly spaced entries 1, ..., ULP with random signs
00216 * on the diagonal.
00217 *
00218 * (9) A matrix of the form U* D U, where U is unitary and
00219 * D has geometrically spaced entries 1, ..., ULP with random
00220 * signs on the diagonal.
00221 *
00222 * (10) A matrix of the form U* D U, where U is unitary and
00223 * D has "clustered" entries 1, ULP,..., ULP with random
00224 * signs on the diagonal.
00225 *
00226 * (11) Same as (8), but multiplied by SQRT( overflow threshold )
00227 * (12) Same as (8), but multiplied by SQRT( underflow threshold )
00228 *
00229 * (13) Hermitian matrix with random entries chosen from (-1,1).
00230 * (14) Same as (13), but multiplied by SQRT( overflow threshold )
00231 * (15) Same as (13), but multiplied by SQRT( underflow threshold )
00232 * (16) Same as (8), but diagonal elements are all positive.
00233 * (17) Same as (9), but diagonal elements are all positive.
00234 * (18) Same as (10), but diagonal elements are all positive.
00235 * (19) Same as (16), but multiplied by SQRT( overflow threshold )
00236 * (20) Same as (16), but multiplied by SQRT( underflow threshold )
00237 * (21) A diagonally dominant tridiagonal matrix with geometrically
00238 * spaced diagonal entries 1, ..., ULP.
00239 *
00240 * Arguments
00241 * =========
00242 *
00243 * NSIZES (input) INTEGER
00244 * The number of sizes of matrices to use. If it is zero,
00245 * ZCHKST does nothing. It must be at least zero.
00246 *
00247 * NN (input) INTEGER array, dimension (NSIZES)
00248 * An array containing the sizes to be used for the matrices.
00249 * Zero values will be skipped. The values must be at least
00250 * zero.
00251 *
00252 * NTYPES (input) INTEGER
00253 * The number of elements in DOTYPE. If it is zero, ZCHKST
00254 * does nothing. It must be at least zero. If it is MAXTYP+1
00255 * and NSIZES is 1, then an additional type, MAXTYP+1 is
00256 * defined, which is to use whatever matrix is in A. This
00257 * is only useful if DOTYPE(1:MAXTYP) is .FALSE. and
00258 * DOTYPE(MAXTYP+1) is .TRUE. .
00259 *
00260 * DOTYPE (input) LOGICAL array, dimension (NTYPES)
00261 * If DOTYPE(j) is .TRUE., then for each size in NN a
00262 * matrix of that size and of type j will be generated.
00263 * If NTYPES is smaller than the maximum number of types
00264 * defined (PARAMETER MAXTYP), then types NTYPES+1 through
00265 * MAXTYP will not be generated. If NTYPES is larger
00266 * than MAXTYP, DOTYPE(MAXTYP+1) through DOTYPE(NTYPES)
00267 * will be ignored.
00268 *
00269 * ISEED (input/output) INTEGER array, dimension (4)
00270 * On entry ISEED specifies the seed of the random number
00271 * generator. The array elements should be between 0 and 4095;
00272 * if not they will be reduced mod 4096. Also, ISEED(4) must
00273 * be odd. The random number generator uses a linear
00274 * congruential sequence limited to small integers, and so
00275 * should produce machine independent random numbers. The
00276 * values of ISEED are changed on exit, and can be used in the
00277 * next call to ZCHKST to continue the same random number
00278 * sequence.
00279 *
00280 * THRESH (input) DOUBLE PRECISION
00281 * A test will count as "failed" if the "error", computed as
00282 * described above, exceeds THRESH. Note that the error
00283 * is scaled to be O(1), so THRESH should be a reasonably
00284 * small multiple of 1, e.g., 10 or 100. In particular,
00285 * it should not depend on the precision (single vs. double)
00286 * or the size of the matrix. It must be at least zero.
00287 *
00288 * NOUNIT (input) INTEGER
00289 * The FORTRAN unit number for printing out error messages
00290 * (e.g., if a routine returns IINFO not equal to 0.)
00291 *
00292 * A (input/workspace/output) COMPLEX*16 array of
00293 * dimension ( LDA , max(NN) )
00294 * Used to hold the matrix whose eigenvalues are to be
00295 * computed. On exit, A contains the last matrix actually
00296 * used.
00297 *
00298 * LDA (input) INTEGER
00299 * The leading dimension of A. It must be at
00300 * least 1 and at least max( NN ).
00301 *
00302 * AP (workspace) COMPLEX*16 array of
00303 * dimension( max(NN)*max(NN+1)/2 )
00304 * The matrix A stored in packed format.
00305 *
00306 * SD (workspace/output) DOUBLE PRECISION array of
00307 * dimension( max(NN) )
00308 * The diagonal of the tridiagonal matrix computed by ZHETRD.
00309 * On exit, SD and SE contain the tridiagonal form of the
00310 * matrix in A.
00311 *
00312 * SE (workspace/output) DOUBLE PRECISION array of
00313 * dimension( max(NN) )
00314 * The off-diagonal of the tridiagonal matrix computed by
00315 * ZHETRD. On exit, SD and SE contain the tridiagonal form of
00316 * the matrix in A.
00317 *
00318 * D1 (workspace/output) DOUBLE PRECISION array of
00319 * dimension( max(NN) )
00320 * The eigenvalues of A, as computed by ZSTEQR simlutaneously
00321 * with Z. On exit, the eigenvalues in D1 correspond with the
00322 * matrix in A.
00323 *
00324 * D2 (workspace/output) DOUBLE PRECISION array of
00325 * dimension( max(NN) )
00326 * The eigenvalues of A, as computed by ZSTEQR if Z is not
00327 * computed. On exit, the eigenvalues in D2 correspond with
00328 * the matrix in A.
00329 *
00330 * D3 (workspace/output) DOUBLE PRECISION array of
00331 * dimension( max(NN) )
00332 * The eigenvalues of A, as computed by DSTERF. On exit, the
00333 * eigenvalues in D3 correspond with the matrix in A.
00334 *
00335 * U (workspace/output) COMPLEX*16 array of
00336 * dimension( LDU, max(NN) ).
00337 * The unitary matrix computed by ZHETRD + ZUNGTR.
00338 *
00339 * LDU (input) INTEGER
00340 * The leading dimension of U, Z, and V. It must be at least 1
00341 * and at least max( NN ).
00342 *
00343 * V (workspace/output) COMPLEX*16 array of
00344 * dimension( LDU, max(NN) ).
00345 * The Housholder vectors computed by ZHETRD in reducing A to
00346 * tridiagonal form. The vectors computed with UPLO='U' are
00347 * in the upper triangle, and the vectors computed with UPLO='L'
00348 * are in the lower triangle. (As described in ZHETRD, the
00349 * sub- and superdiagonal are not set to 1, although the
00350 * true Householder vector has a 1 in that position. The
00351 * routines that use V, such as ZUNGTR, set those entries to
00352 * 1 before using them, and then restore them later.)
00353 *
00354 * VP (workspace) COMPLEX*16 array of
00355 * dimension( max(NN)*max(NN+1)/2 )
00356 * The matrix V stored in packed format.
00357 *
00358 * TAU (workspace/output) COMPLEX*16 array of
00359 * dimension( max(NN) )
00360 * The Householder factors computed by ZHETRD in reducing A
00361 * to tridiagonal form.
00362 *
00363 * Z (workspace/output) COMPLEX*16 array of
00364 * dimension( LDU, max(NN) ).
00365 * The unitary matrix of eigenvectors computed by ZSTEQR,
00366 * ZPTEQR, and ZSTEIN.
00367 *
00368 * WORK (workspace/output) COMPLEX*16 array of
00369 * dimension( LWORK )
00370 *
00371 * LWORK (input) INTEGER
00372 * The number of entries in WORK. This must be at least
00373 * 1 + 4 * Nmax + 2 * Nmax * lg Nmax + 3 * Nmax**2
00374 * where Nmax = max( NN(j), 2 ) and lg = log base 2.
00375 *
00376 * IWORK (workspace/output) INTEGER array,
00377 * dimension (6 + 6*Nmax + 5 * Nmax * lg Nmax )
00378 * where Nmax = max( NN(j), 2 ) and lg = log base 2.
00379 * Workspace.
00380 *
00381 * RWORK (workspace/output) DOUBLE PRECISION array of
00382 * dimension( ??? )
00383 *
00384 * RESULT (output) DOUBLE PRECISION array, dimension (26)
00385 * The values computed by the tests described above.
00386 * The values are currently limited to 1/ulp, to avoid
00387 * overflow.
00388 *
00389 * INFO (output) INTEGER
00390 * If 0, then everything ran OK.
00391 * -1: NSIZES < 0
00392 * -2: Some NN(j) < 0
00393 * -3: NTYPES < 0
00394 * -5: THRESH < 0
00395 * -9: LDA < 1 or LDA < NMAX, where NMAX is max( NN(j) ).
00396 * -23: LDU < 1 or LDU < NMAX.
00397 * -29: LWORK too small.
00398 * If ZLATMR, CLATMS, ZHETRD, ZUNGTR, ZSTEQR, DSTERF,
00399 * or ZUNMC2 returns an error code, the
00400 * absolute value of it is returned.
00401 *
00402 *-----------------------------------------------------------------------
00403 *
00404 * Some Local Variables and Parameters:
00405 * ---- ----- --------- --- ----------
00406 * ZERO, ONE Real 0 and 1.
00407 * MAXTYP The number of types defined.
00408 * NTEST The number of tests performed, or which can
00409 * be performed so far, for the current matrix.
00410 * NTESTT The total number of tests performed so far.
00411 * NBLOCK Blocksize as returned by ENVIR.
00412 * NMAX Largest value in NN.
00413 * NMATS The number of matrices generated so far.
00414 * NERRS The number of tests which have exceeded THRESH
00415 * so far.
00416 * COND, IMODE Values to be passed to the matrix generators.
00417 * ANORM Norm of A; passed to matrix generators.
00418 *
00419 * OVFL, UNFL Overflow and underflow thresholds.
00420 * ULP, ULPINV Finest relative precision and its inverse.
00421 * RTOVFL, RTUNFL Square roots of the previous 2 values.
00422 * The following four arrays decode JTYPE:
00423 * KTYPE(j) The general type (1-10) for type "j".
00424 * KMODE(j) The MODE value to be passed to the matrix
00425 * generator for type "j".
00426 * KMAGN(j) The order of magnitude ( O(1),
00427 * O(overflow^(1/2) ), O(underflow^(1/2) )
00428 *
00429 * =====================================================================
00430 *
00431 * .. Parameters ..
00432 DOUBLE PRECISION ZERO, ONE, TWO, EIGHT, TEN, HUN
00433 PARAMETER ( ZERO = 0.0D0, ONE = 1.0D0, TWO = 2.0D0,
00434 \$ EIGHT = 8.0D0, TEN = 10.0D0, HUN = 100.0D0 )
00435 COMPLEX*16 CZERO, CONE
00436 PARAMETER ( CZERO = ( 0.0D+0, 0.0D+0 ),
00437 \$ CONE = ( 1.0D+0, 0.0D+0 ) )
00438 DOUBLE PRECISION HALF
00439 PARAMETER ( HALF = ONE / TWO )
00440 INTEGER MAXTYP
00441 PARAMETER ( MAXTYP = 21 )
00442 LOGICAL CRANGE
00443 PARAMETER ( CRANGE = .FALSE. )
00444 LOGICAL CREL
00445 PARAMETER ( CREL = .FALSE. )
00446 * ..
00447 * .. Local Scalars ..
00449 INTEGER I, IINFO, IL, IMODE, INDE, INDRWK, ITEMP,
00450 \$ ITYPE, IU, J, JC, JR, JSIZE, JTYPE, LGN,
00451 \$ LIWEDC, LOG2UI, LRWEDC, LWEDC, M, M2, M3,
00452 \$ MTYPES, N, NAP, NBLOCK, NERRS, NMATS, NMAX,
00453 \$ NSPLIT, NTEST, NTESTT
00454 DOUBLE PRECISION ABSTOL, ANINV, ANORM, COND, OVFL, RTOVFL,
00455 \$ RTUNFL, TEMP1, TEMP2, TEMP3, TEMP4, ULP,
00456 \$ ULPINV, UNFL, VL, VU
00457 * ..
00458 * .. Local Arrays ..
00459 INTEGER IDUMMA( 1 ), IOLDSD( 4 ), ISEED2( 4 ),
00460 \$ KMAGN( MAXTYP ), KMODE( MAXTYP ),
00461 \$ KTYPE( MAXTYP )
00462 DOUBLE PRECISION DUMMA( 1 )
00463 * ..
00464 * .. External Functions ..
00465 INTEGER ILAENV
00466 DOUBLE PRECISION DLAMCH, DLARND, DSXT1
00467 EXTERNAL ILAENV, DLAMCH, DLARND, DSXT1
00468 * ..
00469 * .. External Subroutines ..
00470 EXTERNAL DCOPY, DLABAD, DLASUM, DSTEBZ, DSTECH, DSTERF,
00471 \$ XERBLA, ZCOPY, ZHET21, ZHETRD, ZHPT21, ZHPTRD,
00472 \$ ZLACPY, ZLASET, ZLATMR, ZLATMS, ZPTEQR, ZSTEDC,
00473 \$ ZSTEMR, ZSTEIN, ZSTEQR, ZSTT21, ZSTT22, ZUNGTR,
00474 \$ ZUPGTR
00475 * ..
00476 * .. Intrinsic Functions ..
00477 INTRINSIC ABS, DBLE, DCONJG, INT, LOG, MAX, MIN, SQRT
00478 * ..
00479 * .. Data statements ..
00480 DATA KTYPE / 1, 2, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 8,
00481 \$ 8, 8, 9, 9, 9, 9, 9, 10 /
00482 DATA KMAGN / 1, 1, 1, 1, 1, 2, 3, 1, 1, 1, 2, 3, 1,
00483 \$ 2, 3, 1, 1, 1, 2, 3, 1 /
00484 DATA KMODE / 0, 0, 4, 3, 1, 4, 4, 4, 3, 1, 4, 4, 0,
00485 \$ 0, 0, 4, 3, 1, 4, 4, 3 /
00486 * ..
00487 * .. Executable Statements ..
00488 *
00489 * Keep ftnchek happy
00490 IDUMMA( 1 ) = 1
00491 *
00492 * Check for errors
00493 *
00494 NTESTT = 0
00495 INFO = 0
00496 *
00497 * Important constants
00498 *
00500 TRYRAC = .TRUE.
00501 NMAX = 1
00502 DO 10 J = 1, NSIZES
00503 NMAX = MAX( NMAX, NN( J ) )
00504 IF( NN( J ).LT.0 )
00506 10 CONTINUE
00507 *
00508 NBLOCK = ILAENV( 1, 'ZHETRD', 'L', NMAX, -1, -1, -1 )
00509 NBLOCK = MIN( NMAX, MAX( 1, NBLOCK ) )
00510 *
00511 * Check for errors
00512 *
00513 IF( NSIZES.LT.0 ) THEN
00514 INFO = -1
00515 ELSE IF( BADNN ) THEN
00516 INFO = -2
00517 ELSE IF( NTYPES.LT.0 ) THEN
00518 INFO = -3
00519 ELSE IF( LDA.LT.NMAX ) THEN
00520 INFO = -9
00521 ELSE IF( LDU.LT.NMAX ) THEN
00522 INFO = -23
00523 ELSE IF( 2*MAX( 2, NMAX )**2.GT.LWORK ) THEN
00524 INFO = -29
00525 END IF
00526 *
00527 IF( INFO.NE.0 ) THEN
00528 CALL XERBLA( 'ZCHKST', -INFO )
00529 RETURN
00530 END IF
00531 *
00532 * Quick return if possible
00533 *
00534 IF( NSIZES.EQ.0 .OR. NTYPES.EQ.0 )
00535 \$ RETURN
00536 *
00537 * More Important constants
00538 *
00539 UNFL = DLAMCH( 'Safe minimum' )
00540 OVFL = ONE / UNFL
00541 CALL DLABAD( UNFL, OVFL )
00542 ULP = DLAMCH( 'Epsilon' )*DLAMCH( 'Base' )
00543 ULPINV = ONE / ULP
00544 LOG2UI = INT( LOG( ULPINV ) / LOG( TWO ) )
00545 RTUNFL = SQRT( UNFL )
00546 RTOVFL = SQRT( OVFL )
00547 *
00548 * Loop over sizes, types
00549 *
00550 DO 20 I = 1, 4
00551 ISEED2( I ) = ISEED( I )
00552 20 CONTINUE
00553 NERRS = 0
00554 NMATS = 0
00555 *
00556 DO 310 JSIZE = 1, NSIZES
00557 N = NN( JSIZE )
00558 IF( N.GT.0 ) THEN
00559 LGN = INT( LOG( DBLE( N ) ) / LOG( TWO ) )
00560 IF( 2**LGN.LT.N )
00561 \$ LGN = LGN + 1
00562 IF( 2**LGN.LT.N )
00563 \$ LGN = LGN + 1
00564 LWEDC = 1 + 4*N + 2*N*LGN + 3*N**2
00565 LRWEDC = 1 + 3*N + 2*N*LGN + 3*N**2
00566 LIWEDC = 6 + 6*N + 5*N*LGN
00567 ELSE
00568 LWEDC = 8
00569 LRWEDC = 7
00570 LIWEDC = 12
00571 END IF
00572 NAP = ( N*( N+1 ) ) / 2
00573 ANINV = ONE / DBLE( MAX( 1, N ) )
00574 *
00575 IF( NSIZES.NE.1 ) THEN
00576 MTYPES = MIN( MAXTYP, NTYPES )
00577 ELSE
00578 MTYPES = MIN( MAXTYP+1, NTYPES )
00579 END IF
00580 *
00581 DO 300 JTYPE = 1, MTYPES
00582 IF( .NOT.DOTYPE( JTYPE ) )
00583 \$ GO TO 300
00584 NMATS = NMATS + 1
00585 NTEST = 0
00586 *
00587 DO 30 J = 1, 4
00588 IOLDSD( J ) = ISEED( J )
00589 30 CONTINUE
00590 *
00591 * Compute "A"
00592 *
00593 * Control parameters:
00594 *
00595 * KMAGN KMODE KTYPE
00596 * =1 O(1) clustered 1 zero
00597 * =2 large clustered 2 identity
00598 * =3 small exponential (none)
00599 * =4 arithmetic diagonal, (w/ eigenvalues)
00600 * =5 random log Hermitian, w/ eigenvalues
00601 * =6 random (none)
00602 * =7 random diagonal
00603 * =8 random Hermitian
00604 * =9 positive definite
00605 * =10 diagonally dominant tridiagonal
00606 *
00607 IF( MTYPES.GT.MAXTYP )
00608 \$ GO TO 100
00609 *
00610 ITYPE = KTYPE( JTYPE )
00611 IMODE = KMODE( JTYPE )
00612 *
00613 * Compute norm
00614 *
00615 GO TO ( 40, 50, 60 )KMAGN( JTYPE )
00616 *
00617 40 CONTINUE
00618 ANORM = ONE
00619 GO TO 70
00620 *
00621 50 CONTINUE
00622 ANORM = ( RTOVFL*ULP )*ANINV
00623 GO TO 70
00624 *
00625 60 CONTINUE
00626 ANORM = RTUNFL*N*ULPINV
00627 GO TO 70
00628 *
00629 70 CONTINUE
00630 *
00631 CALL ZLASET( 'Full', LDA, N, CZERO, CZERO, A, LDA )
00632 IINFO = 0
00633 IF( JTYPE.LE.15 ) THEN
00634 COND = ULPINV
00635 ELSE
00636 COND = ULPINV*ANINV / TEN
00637 END IF
00638 *
00639 * Special Matrices -- Identity & Jordan block
00640 *
00641 * Zero
00642 *
00643 IF( ITYPE.EQ.1 ) THEN
00644 IINFO = 0
00645 *
00646 ELSE IF( ITYPE.EQ.2 ) THEN
00647 *
00648 * Identity
00649 *
00650 DO 80 JC = 1, N
00651 A( JC, JC ) = ANORM
00652 80 CONTINUE
00653 *
00654 ELSE IF( ITYPE.EQ.4 ) THEN
00655 *
00656 * Diagonal Matrix, [Eigen]values Specified
00657 *
00658 CALL ZLATMS( N, N, 'S', ISEED, 'H', RWORK, IMODE, COND,
00659 \$ ANORM, 0, 0, 'N', A, LDA, WORK, IINFO )
00660 *
00661 *
00662 ELSE IF( ITYPE.EQ.5 ) THEN
00663 *
00664 * Hermitian, eigenvalues specified
00665 *
00666 CALL ZLATMS( N, N, 'S', ISEED, 'H', RWORK, IMODE, COND,
00667 \$ ANORM, N, N, 'N', A, LDA, WORK, IINFO )
00668 *
00669 ELSE IF( ITYPE.EQ.7 ) THEN
00670 *
00671 * Diagonal, random eigenvalues
00672 *
00673 CALL ZLATMR( N, N, 'S', ISEED, 'H', WORK, 6, ONE, CONE,
00674 \$ 'T', 'N', WORK( N+1 ), 1, ONE,
00675 \$ WORK( 2*N+1 ), 1, ONE, 'N', IDUMMA, 0, 0,
00676 \$ ZERO, ANORM, 'NO', A, LDA, IWORK, IINFO )
00677 *
00678 ELSE IF( ITYPE.EQ.8 ) THEN
00679 *
00680 * Hermitian, random eigenvalues
00681 *
00682 CALL ZLATMR( N, N, 'S', ISEED, 'H', WORK, 6, ONE, CONE,
00683 \$ 'T', 'N', WORK( N+1 ), 1, ONE,
00684 \$ WORK( 2*N+1 ), 1, ONE, 'N', IDUMMA, N, N,
00685 \$ ZERO, ANORM, 'NO', A, LDA, IWORK, IINFO )
00686 *
00687 ELSE IF( ITYPE.EQ.9 ) THEN
00688 *
00689 * Positive definite, eigenvalues specified.
00690 *
00691 CALL ZLATMS( N, N, 'S', ISEED, 'P', RWORK, IMODE, COND,
00692 \$ ANORM, N, N, 'N', A, LDA, WORK, IINFO )
00693 *
00694 ELSE IF( ITYPE.EQ.10 ) THEN
00695 *
00696 * Positive definite tridiagonal, eigenvalues specified.
00697 *
00698 CALL ZLATMS( N, N, 'S', ISEED, 'P', RWORK, IMODE, COND,
00699 \$ ANORM, 1, 1, 'N', A, LDA, WORK, IINFO )
00700 DO 90 I = 2, N
00701 TEMP1 = ABS( A( I-1, I ) )
00702 TEMP2 = SQRT( ABS( A( I-1, I-1 )*A( I, I ) ) )
00703 IF( TEMP1.GT.HALF*TEMP2 ) THEN
00704 A( I-1, I ) = A( I-1, I )*
00705 \$ ( HALF*TEMP2 / ( UNFL+TEMP1 ) )
00706 A( I, I-1 ) = DCONJG( A( I-1, I ) )
00707 END IF
00708 90 CONTINUE
00709 *
00710 ELSE
00711 *
00712 IINFO = 1
00713 END IF
00714 *
00715 IF( IINFO.NE.0 ) THEN
00716 WRITE( NOUNIT, FMT = 9999 )'Generator', IINFO, N, JTYPE,
00717 \$ IOLDSD
00718 INFO = ABS( IINFO )
00719 RETURN
00720 END IF
00721 *
00722 100 CONTINUE
00723 *
00724 * Call ZHETRD and ZUNGTR to compute S and U from
00725 * upper triangle.
00726 *
00727 CALL ZLACPY( 'U', N, N, A, LDA, V, LDU )
00728 *
00729 NTEST = 1
00730 CALL ZHETRD( 'U', N, V, LDU, SD, SE, TAU, WORK, LWORK,
00731 \$ IINFO )
00732 *
00733 IF( IINFO.NE.0 ) THEN
00734 WRITE( NOUNIT, FMT = 9999 )'ZHETRD(U)', IINFO, N, JTYPE,
00735 \$ IOLDSD
00736 INFO = ABS( IINFO )
00737 IF( IINFO.LT.0 ) THEN
00738 RETURN
00739 ELSE
00740 RESULT( 1 ) = ULPINV
00741 GO TO 280
00742 END IF
00743 END IF
00744 *
00745 CALL ZLACPY( 'U', N, N, V, LDU, U, LDU )
00746 *
00747 NTEST = 2
00748 CALL ZUNGTR( 'U', N, U, LDU, TAU, WORK, LWORK, IINFO )
00749 IF( IINFO.NE.0 ) THEN
00750 WRITE( NOUNIT, FMT = 9999 )'ZUNGTR(U)', IINFO, N, JTYPE,
00751 \$ IOLDSD
00752 INFO = ABS( IINFO )
00753 IF( IINFO.LT.0 ) THEN
00754 RETURN
00755 ELSE
00756 RESULT( 2 ) = ULPINV
00757 GO TO 280
00758 END IF
00759 END IF
00760 *
00761 * Do tests 1 and 2
00762 *
00763 CALL ZHET21( 2, 'Upper', N, 1, A, LDA, SD, SE, U, LDU, V,
00764 \$ LDU, TAU, WORK, RWORK, RESULT( 1 ) )
00765 CALL ZHET21( 3, 'Upper', N, 1, A, LDA, SD, SE, U, LDU, V,
00766 \$ LDU, TAU, WORK, RWORK, RESULT( 2 ) )
00767 *
00768 * Call ZHETRD and ZUNGTR to compute S and U from
00769 * lower triangle, do tests.
00770 *
00771 CALL ZLACPY( 'L', N, N, A, LDA, V, LDU )
00772 *
00773 NTEST = 3
00774 CALL ZHETRD( 'L', N, V, LDU, SD, SE, TAU, WORK, LWORK,
00775 \$ IINFO )
00776 *
00777 IF( IINFO.NE.0 ) THEN
00778 WRITE( NOUNIT, FMT = 9999 )'ZHETRD(L)', IINFO, N, JTYPE,
00779 \$ IOLDSD
00780 INFO = ABS( IINFO )
00781 IF( IINFO.LT.0 ) THEN
00782 RETURN
00783 ELSE
00784 RESULT( 3 ) = ULPINV
00785 GO TO 280
00786 END IF
00787 END IF
00788 *
00789 CALL ZLACPY( 'L', N, N, V, LDU, U, LDU )
00790 *
00791 NTEST = 4
00792 CALL ZUNGTR( 'L', N, U, LDU, TAU, WORK, LWORK, IINFO )
00793 IF( IINFO.NE.0 ) THEN
00794 WRITE( NOUNIT, FMT = 9999 )'ZUNGTR(L)', IINFO, N, JTYPE,
00795 \$ IOLDSD
00796 INFO = ABS( IINFO )
00797 IF( IINFO.LT.0 ) THEN
00798 RETURN
00799 ELSE
00800 RESULT( 4 ) = ULPINV
00801 GO TO 280
00802 END IF
00803 END IF
00804 *
00805 CALL ZHET21( 2, 'Lower', N, 1, A, LDA, SD, SE, U, LDU, V,
00806 \$ LDU, TAU, WORK, RWORK, RESULT( 3 ) )
00807 CALL ZHET21( 3, 'Lower', N, 1, A, LDA, SD, SE, U, LDU, V,
00808 \$ LDU, TAU, WORK, RWORK, RESULT( 4 ) )
00809 *
00810 * Store the upper triangle of A in AP
00811 *
00812 I = 0
00813 DO 120 JC = 1, N
00814 DO 110 JR = 1, JC
00815 I = I + 1
00816 AP( I ) = A( JR, JC )
00817 110 CONTINUE
00818 120 CONTINUE
00819 *
00820 * Call ZHPTRD and ZUPGTR to compute S and U from AP
00821 *
00822 CALL ZCOPY( NAP, AP, 1, VP, 1 )
00823 *
00824 NTEST = 5
00825 CALL ZHPTRD( 'U', N, VP, SD, SE, TAU, IINFO )
00826 *
00827 IF( IINFO.NE.0 ) THEN
00828 WRITE( NOUNIT, FMT = 9999 )'ZHPTRD(U)', IINFO, N, JTYPE,
00829 \$ IOLDSD
00830 INFO = ABS( IINFO )
00831 IF( IINFO.LT.0 ) THEN
00832 RETURN
00833 ELSE
00834 RESULT( 5 ) = ULPINV
00835 GO TO 280
00836 END IF
00837 END IF
00838 *
00839 NTEST = 6
00840 CALL ZUPGTR( 'U', N, VP, TAU, U, LDU, WORK, IINFO )
00841 IF( IINFO.NE.0 ) THEN
00842 WRITE( NOUNIT, FMT = 9999 )'ZUPGTR(U)', IINFO, N, JTYPE,
00843 \$ IOLDSD
00844 INFO = ABS( IINFO )
00845 IF( IINFO.LT.0 ) THEN
00846 RETURN
00847 ELSE
00848 RESULT( 6 ) = ULPINV
00849 GO TO 280
00850 END IF
00851 END IF
00852 *
00853 * Do tests 5 and 6
00854 *
00855 CALL ZHPT21( 2, 'Upper', N, 1, AP, SD, SE, U, LDU, VP, TAU,
00856 \$ WORK, RWORK, RESULT( 5 ) )
00857 CALL ZHPT21( 3, 'Upper', N, 1, AP, SD, SE, U, LDU, VP, TAU,
00858 \$ WORK, RWORK, RESULT( 6 ) )
00859 *
00860 * Store the lower triangle of A in AP
00861 *
00862 I = 0
00863 DO 140 JC = 1, N
00864 DO 130 JR = JC, N
00865 I = I + 1
00866 AP( I ) = A( JR, JC )
00867 130 CONTINUE
00868 140 CONTINUE
00869 *
00870 * Call ZHPTRD and ZUPGTR to compute S and U from AP
00871 *
00872 CALL ZCOPY( NAP, AP, 1, VP, 1 )
00873 *
00874 NTEST = 7
00875 CALL ZHPTRD( 'L', N, VP, SD, SE, TAU, IINFO )
00876 *
00877 IF( IINFO.NE.0 ) THEN
00878 WRITE( NOUNIT, FMT = 9999 )'ZHPTRD(L)', IINFO, N, JTYPE,
00879 \$ IOLDSD
00880 INFO = ABS( IINFO )
00881 IF( IINFO.LT.0 ) THEN
00882 RETURN
00883 ELSE
00884 RESULT( 7 ) = ULPINV
00885 GO TO 280
00886 END IF
00887 END IF
00888 *
00889 NTEST = 8
00890 CALL ZUPGTR( 'L', N, VP, TAU, U, LDU, WORK, IINFO )
00891 IF( IINFO.NE.0 ) THEN
00892 WRITE( NOUNIT, FMT = 9999 )'ZUPGTR(L)', IINFO, N, JTYPE,
00893 \$ IOLDSD
00894 INFO = ABS( IINFO )
00895 IF( IINFO.LT.0 ) THEN
00896 RETURN
00897 ELSE
00898 RESULT( 8 ) = ULPINV
00899 GO TO 280
00900 END IF
00901 END IF
00902 *
00903 CALL ZHPT21( 2, 'Lower', N, 1, AP, SD, SE, U, LDU, VP, TAU,
00904 \$ WORK, RWORK, RESULT( 7 ) )
00905 CALL ZHPT21( 3, 'Lower', N, 1, AP, SD, SE, U, LDU, VP, TAU,
00906 \$ WORK, RWORK, RESULT( 8 ) )
00907 *
00908 * Call ZSTEQR to compute D1, D2, and Z, do tests.
00909 *
00910 * Compute D1 and Z
00911 *
00912 CALL DCOPY( N, SD, 1, D1, 1 )
00913 IF( N.GT.0 )
00914 \$ CALL DCOPY( N-1, SE, 1, RWORK, 1 )
00915 CALL ZLASET( 'Full', N, N, CZERO, CONE, Z, LDU )
00916 *
00917 NTEST = 9
00918 CALL ZSTEQR( 'V', N, D1, RWORK, Z, LDU, RWORK( N+1 ),
00919 \$ IINFO )
00920 IF( IINFO.NE.0 ) THEN
00921 WRITE( NOUNIT, FMT = 9999 )'ZSTEQR(V)', IINFO, N, JTYPE,
00922 \$ IOLDSD
00923 INFO = ABS( IINFO )
00924 IF( IINFO.LT.0 ) THEN
00925 RETURN
00926 ELSE
00927 RESULT( 9 ) = ULPINV
00928 GO TO 280
00929 END IF
00930 END IF
00931 *
00932 * Compute D2
00933 *
00934 CALL DCOPY( N, SD, 1, D2, 1 )
00935 IF( N.GT.0 )
00936 \$ CALL DCOPY( N-1, SE, 1, RWORK, 1 )
00937 *
00938 NTEST = 11
00939 CALL ZSTEQR( 'N', N, D2, RWORK, WORK, LDU, RWORK( N+1 ),
00940 \$ IINFO )
00941 IF( IINFO.NE.0 ) THEN
00942 WRITE( NOUNIT, FMT = 9999 )'ZSTEQR(N)', IINFO, N, JTYPE,
00943 \$ IOLDSD
00944 INFO = ABS( IINFO )
00945 IF( IINFO.LT.0 ) THEN
00946 RETURN
00947 ELSE
00948 RESULT( 11 ) = ULPINV
00949 GO TO 280
00950 END IF
00951 END IF
00952 *
00953 * Compute D3 (using PWK method)
00954 *
00955 CALL DCOPY( N, SD, 1, D3, 1 )
00956 IF( N.GT.0 )
00957 \$ CALL DCOPY( N-1, SE, 1, RWORK, 1 )
00958 *
00959 NTEST = 12
00960 CALL DSTERF( N, D3, RWORK, IINFO )
00961 IF( IINFO.NE.0 ) THEN
00962 WRITE( NOUNIT, FMT = 9999 )'DSTERF', IINFO, N, JTYPE,
00963 \$ IOLDSD
00964 INFO = ABS( IINFO )
00965 IF( IINFO.LT.0 ) THEN
00966 RETURN
00967 ELSE
00968 RESULT( 12 ) = ULPINV
00969 GO TO 280
00970 END IF
00971 END IF
00972 *
00973 * Do Tests 9 and 10
00974 *
00975 CALL ZSTT21( N, 0, SD, SE, D1, DUMMA, Z, LDU, WORK, RWORK,
00976 \$ RESULT( 9 ) )
00977 *
00978 * Do Tests 11 and 12
00979 *
00980 TEMP1 = ZERO
00981 TEMP2 = ZERO
00982 TEMP3 = ZERO
00983 TEMP4 = ZERO
00984 *
00985 DO 150 J = 1, N
00986 TEMP1 = MAX( TEMP1, ABS( D1( J ) ), ABS( D2( J ) ) )
00987 TEMP2 = MAX( TEMP2, ABS( D1( J )-D2( J ) ) )
00988 TEMP3 = MAX( TEMP3, ABS( D1( J ) ), ABS( D3( J ) ) )
00989 TEMP4 = MAX( TEMP4, ABS( D1( J )-D3( J ) ) )
00990 150 CONTINUE
00991 *
00992 RESULT( 11 ) = TEMP2 / MAX( UNFL, ULP*MAX( TEMP1, TEMP2 ) )
00993 RESULT( 12 ) = TEMP4 / MAX( UNFL, ULP*MAX( TEMP3, TEMP4 ) )
00994 *
00995 * Do Test 13 -- Sturm Sequence Test of Eigenvalues
00996 * Go up by factors of two until it succeeds
00997 *
00998 NTEST = 13
00999 TEMP1 = THRESH*( HALF-ULP )
01000 *
01001 DO 160 J = 0, LOG2UI
01002 CALL DSTECH( N, SD, SE, D1, TEMP1, RWORK, IINFO )
01003 IF( IINFO.EQ.0 )
01004 \$ GO TO 170
01005 TEMP1 = TEMP1*TWO
01006 160 CONTINUE
01007 *
01008 170 CONTINUE
01009 RESULT( 13 ) = TEMP1
01010 *
01011 * For positive definite matrices ( JTYPE.GT.15 ) call ZPTEQR
01012 * and do tests 14, 15, and 16 .
01013 *
01014 IF( JTYPE.GT.15 ) THEN
01015 *
01016 * Compute D4 and Z4
01017 *
01018 CALL DCOPY( N, SD, 1, D4, 1 )
01019 IF( N.GT.0 )
01020 \$ CALL DCOPY( N-1, SE, 1, RWORK, 1 )
01021 CALL ZLASET( 'Full', N, N, CZERO, CONE, Z, LDU )
01022 *
01023 NTEST = 14
01024 CALL ZPTEQR( 'V', N, D4, RWORK, Z, LDU, RWORK( N+1 ),
01025 \$ IINFO )
01026 IF( IINFO.NE.0 ) THEN
01027 WRITE( NOUNIT, FMT = 9999 )'ZPTEQR(V)', IINFO, N,
01028 \$ JTYPE, IOLDSD
01029 INFO = ABS( IINFO )
01030 IF( IINFO.LT.0 ) THEN
01031 RETURN
01032 ELSE
01033 RESULT( 14 ) = ULPINV
01034 GO TO 280
01035 END IF
01036 END IF
01037 *
01038 * Do Tests 14 and 15
01039 *
01040 CALL ZSTT21( N, 0, SD, SE, D4, DUMMA, Z, LDU, WORK,
01041 \$ RWORK, RESULT( 14 ) )
01042 *
01043 * Compute D5
01044 *
01045 CALL DCOPY( N, SD, 1, D5, 1 )
01046 IF( N.GT.0 )
01047 \$ CALL DCOPY( N-1, SE, 1, RWORK, 1 )
01048 *
01049 NTEST = 16
01050 CALL ZPTEQR( 'N', N, D5, RWORK, Z, LDU, RWORK( N+1 ),
01051 \$ IINFO )
01052 IF( IINFO.NE.0 ) THEN
01053 WRITE( NOUNIT, FMT = 9999 )'ZPTEQR(N)', IINFO, N,
01054 \$ JTYPE, IOLDSD
01055 INFO = ABS( IINFO )
01056 IF( IINFO.LT.0 ) THEN
01057 RETURN
01058 ELSE
01059 RESULT( 16 ) = ULPINV
01060 GO TO 280
01061 END IF
01062 END IF
01063 *
01064 * Do Test 16
01065 *
01066 TEMP1 = ZERO
01067 TEMP2 = ZERO
01068 DO 180 J = 1, N
01069 TEMP1 = MAX( TEMP1, ABS( D4( J ) ), ABS( D5( J ) ) )
01070 TEMP2 = MAX( TEMP2, ABS( D4( J )-D5( J ) ) )
01071 180 CONTINUE
01072 *
01073 RESULT( 16 ) = TEMP2 / MAX( UNFL,
01074 \$ HUN*ULP*MAX( TEMP1, TEMP2 ) )
01075 ELSE
01076 RESULT( 14 ) = ZERO
01077 RESULT( 15 ) = ZERO
01078 RESULT( 16 ) = ZERO
01079 END IF
01080 *
01081 * Call DSTEBZ with different options and do tests 17-18.
01082 *
01083 * If S is positive definite and diagonally dominant,
01084 * ask for all eigenvalues with high relative accuracy.
01085 *
01086 VL = ZERO
01087 VU = ZERO
01088 IL = 0
01089 IU = 0
01090 IF( JTYPE.EQ.21 ) THEN
01091 NTEST = 17
01092 ABSTOL = UNFL + UNFL
01093 CALL DSTEBZ( 'A', 'E', N, VL, VU, IL, IU, ABSTOL, SD, SE,
01094 \$ M, NSPLIT, WR, IWORK( 1 ), IWORK( N+1 ),
01095 \$ RWORK, IWORK( 2*N+1 ), IINFO )
01096 IF( IINFO.NE.0 ) THEN
01097 WRITE( NOUNIT, FMT = 9999 )'DSTEBZ(A,rel)', IINFO, N,
01098 \$ JTYPE, IOLDSD
01099 INFO = ABS( IINFO )
01100 IF( IINFO.LT.0 ) THEN
01101 RETURN
01102 ELSE
01103 RESULT( 17 ) = ULPINV
01104 GO TO 280
01105 END IF
01106 END IF
01107 *
01108 * Do test 17
01109 *
01110 TEMP2 = TWO*( TWO*N-ONE )*ULP*( ONE+EIGHT*HALF**2 ) /
01111 \$ ( ONE-HALF )**4
01112 *
01113 TEMP1 = ZERO
01114 DO 190 J = 1, N
01115 TEMP1 = MAX( TEMP1, ABS( D4( J )-WR( N-J+1 ) ) /
01116 \$ ( ABSTOL+ABS( D4( J ) ) ) )
01117 190 CONTINUE
01118 *
01119 RESULT( 17 ) = TEMP1 / TEMP2
01120 ELSE
01121 RESULT( 17 ) = ZERO
01122 END IF
01123 *
01124 * Now ask for all eigenvalues with high absolute accuracy.
01125 *
01126 NTEST = 18
01127 ABSTOL = UNFL + UNFL
01128 CALL DSTEBZ( 'A', 'E', N, VL, VU, IL, IU, ABSTOL, SD, SE, M,
01129 \$ NSPLIT, WA1, IWORK( 1 ), IWORK( N+1 ), RWORK,
01130 \$ IWORK( 2*N+1 ), IINFO )
01131 IF( IINFO.NE.0 ) THEN
01132 WRITE( NOUNIT, FMT = 9999 )'DSTEBZ(A)', IINFO, N, JTYPE,
01133 \$ IOLDSD
01134 INFO = ABS( IINFO )
01135 IF( IINFO.LT.0 ) THEN
01136 RETURN
01137 ELSE
01138 RESULT( 18 ) = ULPINV
01139 GO TO 280
01140 END IF
01141 END IF
01142 *
01143 * Do test 18
01144 *
01145 TEMP1 = ZERO
01146 TEMP2 = ZERO
01147 DO 200 J = 1, N
01148 TEMP1 = MAX( TEMP1, ABS( D3( J ) ), ABS( WA1( J ) ) )
01149 TEMP2 = MAX( TEMP2, ABS( D3( J )-WA1( J ) ) )
01150 200 CONTINUE
01151 *
01152 RESULT( 18 ) = TEMP2 / MAX( UNFL, ULP*MAX( TEMP1, TEMP2 ) )
01153 *
01154 * Choose random values for IL and IU, and ask for the
01155 * IL-th through IU-th eigenvalues.
01156 *
01157 NTEST = 19
01158 IF( N.LE.1 ) THEN
01159 IL = 1
01160 IU = N
01161 ELSE
01162 IL = 1 + ( N-1 )*INT( DLARND( 1, ISEED2 ) )
01163 IU = 1 + ( N-1 )*INT( DLARND( 1, ISEED2 ) )
01164 IF( IU.LT.IL ) THEN
01165 ITEMP = IU
01166 IU = IL
01167 IL = ITEMP
01168 END IF
01169 END IF
01170 *
01171 CALL DSTEBZ( 'I', 'E', N, VL, VU, IL, IU, ABSTOL, SD, SE,
01172 \$ M2, NSPLIT, WA2, IWORK( 1 ), IWORK( N+1 ),
01173 \$ RWORK, IWORK( 2*N+1 ), IINFO )
01174 IF( IINFO.NE.0 ) THEN
01175 WRITE( NOUNIT, FMT = 9999 )'DSTEBZ(I)', IINFO, N, JTYPE,
01176 \$ IOLDSD
01177 INFO = ABS( IINFO )
01178 IF( IINFO.LT.0 ) THEN
01179 RETURN
01180 ELSE
01181 RESULT( 19 ) = ULPINV
01182 GO TO 280
01183 END IF
01184 END IF
01185 *
01186 * Determine the values VL and VU of the IL-th and IU-th
01187 * eigenvalues and ask for all eigenvalues in this range.
01188 *
01189 IF( N.GT.0 ) THEN
01190 IF( IL.NE.1 ) THEN
01191 VL = WA1( IL ) - MAX( HALF*( WA1( IL )-WA1( IL-1 ) ),
01192 \$ ULP*ANORM, TWO*RTUNFL )
01193 ELSE
01194 VL = WA1( 1 ) - MAX( HALF*( WA1( N )-WA1( 1 ) ),
01195 \$ ULP*ANORM, TWO*RTUNFL )
01196 END IF
01197 IF( IU.NE.N ) THEN
01198 VU = WA1( IU ) + MAX( HALF*( WA1( IU+1 )-WA1( IU ) ),
01199 \$ ULP*ANORM, TWO*RTUNFL )
01200 ELSE
01201 VU = WA1( N ) + MAX( HALF*( WA1( N )-WA1( 1 ) ),
01202 \$ ULP*ANORM, TWO*RTUNFL )
01203 END IF
01204 ELSE
01205 VL = ZERO
01206 VU = ONE
01207 END IF
01208 *
01209 CALL DSTEBZ( 'V', 'E', N, VL, VU, IL, IU, ABSTOL, SD, SE,
01210 \$ M3, NSPLIT, WA3, IWORK( 1 ), IWORK( N+1 ),
01211 \$ RWORK, IWORK( 2*N+1 ), IINFO )
01212 IF( IINFO.NE.0 ) THEN
01213 WRITE( NOUNIT, FMT = 9999 )'DSTEBZ(V)', IINFO, N, JTYPE,
01214 \$ IOLDSD
01215 INFO = ABS( IINFO )
01216 IF( IINFO.LT.0 ) THEN
01217 RETURN
01218 ELSE
01219 RESULT( 19 ) = ULPINV
01220 GO TO 280
01221 END IF
01222 END IF
01223 *
01224 IF( M3.EQ.0 .AND. N.NE.0 ) THEN
01225 RESULT( 19 ) = ULPINV
01226 GO TO 280
01227 END IF
01228 *
01229 * Do test 19
01230 *
01231 TEMP1 = DSXT1( 1, WA2, M2, WA3, M3, ABSTOL, ULP, UNFL )
01232 TEMP2 = DSXT1( 1, WA3, M3, WA2, M2, ABSTOL, ULP, UNFL )
01233 IF( N.GT.0 ) THEN
01234 TEMP3 = MAX( ABS( WA1( N ) ), ABS( WA1( 1 ) ) )
01235 ELSE
01236 TEMP3 = ZERO
01237 END IF
01238 *
01239 RESULT( 19 ) = ( TEMP1+TEMP2 ) / MAX( UNFL, TEMP3*ULP )
01240 *
01241 * Call ZSTEIN to compute eigenvectors corresponding to
01242 * eigenvalues in WA1. (First call DSTEBZ again, to make sure
01243 * it returns these eigenvalues in the correct order.)
01244 *
01245 NTEST = 21
01246 CALL DSTEBZ( 'A', 'B', N, VL, VU, IL, IU, ABSTOL, SD, SE, M,
01247 \$ NSPLIT, WA1, IWORK( 1 ), IWORK( N+1 ), RWORK,
01248 \$ IWORK( 2*N+1 ), IINFO )
01249 IF( IINFO.NE.0 ) THEN
01250 WRITE( NOUNIT, FMT = 9999 )'DSTEBZ(A,B)', IINFO, N,
01251 \$ JTYPE, IOLDSD
01252 INFO = ABS( IINFO )
01253 IF( IINFO.LT.0 ) THEN
01254 RETURN
01255 ELSE
01256 RESULT( 20 ) = ULPINV
01257 RESULT( 21 ) = ULPINV
01258 GO TO 280
01259 END IF
01260 END IF
01261 *
01262 CALL ZSTEIN( N, SD, SE, M, WA1, IWORK( 1 ), IWORK( N+1 ), Z,
01263 \$ LDU, RWORK, IWORK( 2*N+1 ), IWORK( 3*N+1 ),
01264 \$ IINFO )
01265 IF( IINFO.NE.0 ) THEN
01266 WRITE( NOUNIT, FMT = 9999 )'ZSTEIN', IINFO, N, JTYPE,
01267 \$ IOLDSD
01268 INFO = ABS( IINFO )
01269 IF( IINFO.LT.0 ) THEN
01270 RETURN
01271 ELSE
01272 RESULT( 20 ) = ULPINV
01273 RESULT( 21 ) = ULPINV
01274 GO TO 280
01275 END IF
01276 END IF
01277 *
01278 * Do tests 20 and 21
01279 *
01280 CALL ZSTT21( N, 0, SD, SE, WA1, DUMMA, Z, LDU, WORK, RWORK,
01281 \$ RESULT( 20 ) )
01282 *
01283 * Call ZSTEDC(I) to compute D1 and Z, do tests.
01284 *
01285 * Compute D1 and Z
01286 *
01287 INDE = 1
01288 INDRWK = INDE + N
01289 CALL DCOPY( N, SD, 1, D1, 1 )
01290 IF( N.GT.0 )
01291 \$ CALL DCOPY( N-1, SE, 1, RWORK( INDE ), 1 )
01292 CALL ZLASET( 'Full', N, N, CZERO, CONE, Z, LDU )
01293 *
01294 NTEST = 22
01295 CALL ZSTEDC( 'I', N, D1, RWORK( INDE ), Z, LDU, WORK, LWEDC,
01296 \$ RWORK( INDRWK ), LRWEDC, IWORK, LIWEDC, IINFO )
01297 IF( IINFO.NE.0 ) THEN
01298 WRITE( NOUNIT, FMT = 9999 )'ZSTEDC(I)', IINFO, N, JTYPE,
01299 \$ IOLDSD
01300 INFO = ABS( IINFO )
01301 IF( IINFO.LT.0 ) THEN
01302 RETURN
01303 ELSE
01304 RESULT( 22 ) = ULPINV
01305 GO TO 280
01306 END IF
01307 END IF
01308 *
01309 * Do Tests 22 and 23
01310 *
01311 CALL ZSTT21( N, 0, SD, SE, D1, DUMMA, Z, LDU, WORK, RWORK,
01312 \$ RESULT( 22 ) )
01313 *
01314 * Call ZSTEDC(V) to compute D1 and Z, do tests.
01315 *
01316 * Compute D1 and Z
01317 *
01318 CALL DCOPY( N, SD, 1, D1, 1 )
01319 IF( N.GT.0 )
01320 \$ CALL DCOPY( N-1, SE, 1, RWORK( INDE ), 1 )
01321 CALL ZLASET( 'Full', N, N, CZERO, CONE, Z, LDU )
01322 *
01323 NTEST = 24
01324 CALL ZSTEDC( 'V', N, D1, RWORK( INDE ), Z, LDU, WORK, LWEDC,
01325 \$ RWORK( INDRWK ), LRWEDC, IWORK, LIWEDC, IINFO )
01326 IF( IINFO.NE.0 ) THEN
01327 WRITE( NOUNIT, FMT = 9999 )'ZSTEDC(V)', IINFO, N, JTYPE,
01328 \$ IOLDSD
01329 INFO = ABS( IINFO )
01330 IF( IINFO.LT.0 ) THEN
01331 RETURN
01332 ELSE
01333 RESULT( 24 ) = ULPINV
01334 GO TO 280
01335 END IF
01336 END IF
01337 *
01338 * Do Tests 24 and 25
01339 *
01340 CALL ZSTT21( N, 0, SD, SE, D1, DUMMA, Z, LDU, WORK, RWORK,
01341 \$ RESULT( 24 ) )
01342 *
01343 * Call ZSTEDC(N) to compute D2, do tests.
01344 *
01345 * Compute D2
01346 *
01347 CALL DCOPY( N, SD, 1, D2, 1 )
01348 IF( N.GT.0 )
01349 \$ CALL DCOPY( N-1, SE, 1, RWORK( INDE ), 1 )
01350 CALL ZLASET( 'Full', N, N, CZERO, CONE, Z, LDU )
01351 *
01352 NTEST = 26
01353 CALL ZSTEDC( 'N', N, D2, RWORK( INDE ), Z, LDU, WORK, LWEDC,
01354 \$ RWORK( INDRWK ), LRWEDC, IWORK, LIWEDC, IINFO )
01355 IF( IINFO.NE.0 ) THEN
01356 WRITE( NOUNIT, FMT = 9999 )'ZSTEDC(N)', IINFO, N, JTYPE,
01357 \$ IOLDSD
01358 INFO = ABS( IINFO )
01359 IF( IINFO.LT.0 ) THEN
01360 RETURN
01361 ELSE
01362 RESULT( 26 ) = ULPINV
01363 GO TO 280
01364 END IF
01365 END IF
01366 *
01367 * Do Test 26
01368 *
01369 TEMP1 = ZERO
01370 TEMP2 = ZERO
01371 *
01372 DO 210 J = 1, N
01373 TEMP1 = MAX( TEMP1, ABS( D1( J ) ), ABS( D2( J ) ) )
01374 TEMP2 = MAX( TEMP2, ABS( D1( J )-D2( J ) ) )
01375 210 CONTINUE
01376 *
01377 RESULT( 26 ) = TEMP2 / MAX( UNFL, ULP*MAX( TEMP1, TEMP2 ) )
01378 *
01379 * Only test ZSTEMR if IEEE compliant
01380 *
01381 IF( ILAENV( 10, 'ZSTEMR', 'VA', 1, 0, 0, 0 ).EQ.1 .AND.
01382 \$ ILAENV( 11, 'ZSTEMR', 'VA', 1, 0, 0, 0 ).EQ.1 ) THEN
01383 *
01384 * Call ZSTEMR, do test 27 (relative eigenvalue accuracy)
01385 *
01386 * If S is positive definite and diagonally dominant,
01387 * ask for all eigenvalues with high relative accuracy.
01388 *
01389 VL = ZERO
01390 VU = ZERO
01391 IL = 0
01392 IU = 0
01393 IF( JTYPE.EQ.21 .AND. CREL ) THEN
01394 NTEST = 27
01395 ABSTOL = UNFL + UNFL
01396 CALL ZSTEMR( 'V', 'A', N, SD, SE, VL, VU, IL, IU,
01397 \$ M, WR, Z, LDU, N, IWORK( 1 ), TRYRAC,
01398 \$ RWORK, LRWORK, IWORK( 2*N+1 ), LWORK-2*N,
01399 \$ IINFO )
01400 IF( IINFO.NE.0 ) THEN
01401 WRITE( NOUNIT, FMT = 9999 )'ZSTEMR(V,A,rel)',
01402 \$ IINFO, N, JTYPE, IOLDSD
01403 INFO = ABS( IINFO )
01404 IF( IINFO.LT.0 ) THEN
01405 RETURN
01406 ELSE
01407 RESULT( 27 ) = ULPINV
01408 GO TO 270
01409 END IF
01410 END IF
01411 *
01412 * Do test 27
01413 *
01414 TEMP2 = TWO*( TWO*N-ONE )*ULP*( ONE+EIGHT*HALF**2 ) /
01415 \$ ( ONE-HALF )**4
01416 *
01417 TEMP1 = ZERO
01418 DO 220 J = 1, N
01419 TEMP1 = MAX( TEMP1, ABS( D4( J )-WR( N-J+1 ) ) /
01420 \$ ( ABSTOL+ABS( D4( J ) ) ) )
01421 220 CONTINUE
01422 *
01423 RESULT( 27 ) = TEMP1 / TEMP2
01424 *
01425 IL = 1 + ( N-1 )*INT( DLARND( 1, ISEED2 ) )
01426 IU = 1 + ( N-1 )*INT( DLARND( 1, ISEED2 ) )
01427 IF( IU.LT.IL ) THEN
01428 ITEMP = IU
01429 IU = IL
01430 IL = ITEMP
01431 END IF
01432 *
01433 IF( CRANGE ) THEN
01434 NTEST = 28
01435 ABSTOL = UNFL + UNFL
01436 CALL ZSTEMR( 'V', 'I', N, SD, SE, VL, VU, IL, IU,
01437 \$ M, WR, Z, LDU, N, IWORK( 1 ), TRYRAC,
01438 \$ RWORK, LRWORK, IWORK( 2*N+1 ),
01439 \$ LWORK-2*N, IINFO )
01440 *
01441 IF( IINFO.NE.0 ) THEN
01442 WRITE( NOUNIT, FMT = 9999 )'ZSTEMR(V,I,rel)',
01443 \$ IINFO, N, JTYPE, IOLDSD
01444 INFO = ABS( IINFO )
01445 IF( IINFO.LT.0 ) THEN
01446 RETURN
01447 ELSE
01448 RESULT( 28 ) = ULPINV
01449 GO TO 270
01450 END IF
01451 END IF
01452 *
01453 *
01454 * Do test 28
01455 *
01456 TEMP2 = TWO*( TWO*N-ONE )*ULP*
01457 \$ ( ONE+EIGHT*HALF**2 ) / ( ONE-HALF )**4
01458 *
01459 TEMP1 = ZERO
01460 DO 230 J = IL, IU
01461 TEMP1 = MAX( TEMP1, ABS( WR( J-IL+1 )-D4( N-J+
01462 \$ 1 ) ) / ( ABSTOL+ABS( WR( J-IL+1 ) ) ) )
01463 230 CONTINUE
01464 *
01465 RESULT( 28 ) = TEMP1 / TEMP2
01466 ELSE
01467 RESULT( 28 ) = ZERO
01468 END IF
01469 ELSE
01470 RESULT( 27 ) = ZERO
01471 RESULT( 28 ) = ZERO
01472 END IF
01473 *
01474 * Call ZSTEMR(V,I) to compute D1 and Z, do tests.
01475 *
01476 * Compute D1 and Z
01477 *
01478 CALL DCOPY( N, SD, 1, D5, 1 )
01479 IF( N.GT.0 )
01480 \$ CALL DCOPY( N-1, SE, 1, RWORK, 1 )
01481 CALL ZLASET( 'Full', N, N, CZERO, CONE, Z, LDU )
01482 *
01483 IF( CRANGE ) THEN
01484 NTEST = 29
01485 IL = 1 + ( N-1 )*INT( DLARND( 1, ISEED2 ) )
01486 IU = 1 + ( N-1 )*INT( DLARND( 1, ISEED2 ) )
01487 IF( IU.LT.IL ) THEN
01488 ITEMP = IU
01489 IU = IL
01490 IL = ITEMP
01491 END IF
01492 CALL ZSTEMR( 'V', 'I', N, D5, RWORK, VL, VU, IL, IU,
01493 \$ M, D1, Z, LDU, N, IWORK( 1 ), TRYRAC,
01494 \$ RWORK( N+1 ), LRWORK-N, IWORK( 2*N+1 ),
01495 \$ LIWORK-2*N, IINFO )
01496 IF( IINFO.NE.0 ) THEN
01497 WRITE( NOUNIT, FMT = 9999 )'ZSTEMR(V,I)', IINFO,
01498 \$ N, JTYPE, IOLDSD
01499 INFO = ABS( IINFO )
01500 IF( IINFO.LT.0 ) THEN
01501 RETURN
01502 ELSE
01503 RESULT( 29 ) = ULPINV
01504 GO TO 280
01505 END IF
01506 END IF
01507 *
01508 * Do Tests 29 and 30
01509 *
01510 *
01511 * Call ZSTEMR to compute D2, do tests.
01512 *
01513 * Compute D2
01514 *
01515 CALL DCOPY( N, SD, 1, D5, 1 )
01516 IF( N.GT.0 )
01517 \$ CALL DCOPY( N-1, SE, 1, RWORK, 1 )
01518 *
01519 NTEST = 31
01520 CALL ZSTEMR( 'N', 'I', N, D5, RWORK, VL, VU, IL, IU,
01521 \$ M, D2, Z, LDU, N, IWORK( 1 ), TRYRAC,
01522 \$ RWORK( N+1 ), LRWORK-N, IWORK( 2*N+1 ),
01523 \$ LIWORK-2*N, IINFO )
01524 IF( IINFO.NE.0 ) THEN
01525 WRITE( NOUNIT, FMT = 9999 )'ZSTEMR(N,I)', IINFO,
01526 \$ N, JTYPE, IOLDSD
01527 INFO = ABS( IINFO )
01528 IF( IINFO.LT.0 ) THEN
01529 RETURN
01530 ELSE
01531 RESULT( 31 ) = ULPINV
01532 GO TO 280
01533 END IF
01534 END IF
01535 *
01536 * Do Test 31
01537 *
01538 TEMP1 = ZERO
01539 TEMP2 = ZERO
01540 *
01541 DO 240 J = 1, IU - IL + 1
01542 TEMP1 = MAX( TEMP1, ABS( D1( J ) ),
01543 \$ ABS( D2( J ) ) )
01544 TEMP2 = MAX( TEMP2, ABS( D1( J )-D2( J ) ) )
01545 240 CONTINUE
01546 *
01547 RESULT( 31 ) = TEMP2 / MAX( UNFL,
01548 \$ ULP*MAX( TEMP1, TEMP2 ) )
01549 *
01550 *
01551 * Call ZSTEMR(V,V) to compute D1 and Z, do tests.
01552 *
01553 * Compute D1 and Z
01554 *
01555 CALL DCOPY( N, SD, 1, D5, 1 )
01556 IF( N.GT.0 )
01557 \$ CALL DCOPY( N-1, SE, 1, RWORK, 1 )
01558 CALL ZLASET( 'Full', N, N, CZERO, CONE, Z, LDU )
01559 *
01560 NTEST = 32
01561 *
01562 IF( N.GT.0 ) THEN
01563 IF( IL.NE.1 ) THEN
01564 VL = D2( IL ) - MAX( HALF*
01565 \$ ( D2( IL )-D2( IL-1 ) ), ULP*ANORM,
01566 \$ TWO*RTUNFL )
01567 ELSE
01568 VL = D2( 1 ) - MAX( HALF*( D2( N )-D2( 1 ) ),
01569 \$ ULP*ANORM, TWO*RTUNFL )
01570 END IF
01571 IF( IU.NE.N ) THEN
01572 VU = D2( IU ) + MAX( HALF*
01573 \$ ( D2( IU+1 )-D2( IU ) ), ULP*ANORM,
01574 \$ TWO*RTUNFL )
01575 ELSE
01576 VU = D2( N ) + MAX( HALF*( D2( N )-D2( 1 ) ),
01577 \$ ULP*ANORM, TWO*RTUNFL )
01578 END IF
01579 ELSE
01580 VL = ZERO
01581 VU = ONE
01582 END IF
01583 *
01584 CALL ZSTEMR( 'V', 'V', N, D5, RWORK, VL, VU, IL, IU,
01585 \$ M, D1, Z, LDU, M, IWORK( 1 ), TRYRAC,
01586 \$ RWORK( N+1 ), LRWORK-N, IWORK( 2*N+1 ),
01587 \$ LIWORK-2*N, IINFO )
01588 IF( IINFO.NE.0 ) THEN
01589 WRITE( NOUNIT, FMT = 9999 )'ZSTEMR(V,V)', IINFO,
01590 \$ N, JTYPE, IOLDSD
01591 INFO = ABS( IINFO )
01592 IF( IINFO.LT.0 ) THEN
01593 RETURN
01594 ELSE
01595 RESULT( 32 ) = ULPINV
01596 GO TO 280
01597 END IF
01598 END IF
01599 *
01600 * Do Tests 32 and 33
01601 *
01602 CALL ZSTT22( N, M, 0, SD, SE, D1, DUMMA, Z, LDU, WORK,
01603 \$ M, RWORK, RESULT( 32 ) )
01604 *
01605 * Call ZSTEMR to compute D2, do tests.
01606 *
01607 * Compute D2
01608 *
01609 CALL DCOPY( N, SD, 1, D5, 1 )
01610 IF( N.GT.0 )
01611 \$ CALL DCOPY( N-1, SE, 1, RWORK, 1 )
01612 *
01613 NTEST = 34
01614 CALL ZSTEMR( 'N', 'V', N, D5, RWORK, VL, VU, IL, IU,
01615 \$ M, D2, Z, LDU, N, IWORK( 1 ), TRYRAC,
01616 \$ RWORK( N+1 ), LRWORK-N, IWORK( 2*N+1 ),
01617 \$ LIWORK-2*N, IINFO )
01618 IF( IINFO.NE.0 ) THEN
01619 WRITE( NOUNIT, FMT = 9999 )'ZSTEMR(N,V)', IINFO,
01620 \$ N, JTYPE, IOLDSD
01621 INFO = ABS( IINFO )
01622 IF( IINFO.LT.0 ) THEN
01623 RETURN
01624 ELSE
01625 RESULT( 34 ) = ULPINV
01626 GO TO 280
01627 END IF
01628 END IF
01629 *
01630 * Do Test 34
01631 *
01632 TEMP1 = ZERO
01633 TEMP2 = ZERO
01634 *
01635 DO 250 J = 1, IU - IL + 1
01636 TEMP1 = MAX( TEMP1, ABS( D1( J ) ),
01637 \$ ABS( D2( J ) ) )
01638 TEMP2 = MAX( TEMP2, ABS( D1( J )-D2( J ) ) )
01639 250 CONTINUE
01640 *
01641 RESULT( 34 ) = TEMP2 / MAX( UNFL,
01642 \$ ULP*MAX( TEMP1, TEMP2 ) )
01643 ELSE
01644 RESULT( 29 ) = ZERO
01645 RESULT( 30 ) = ZERO
01646 RESULT( 31 ) = ZERO
01647 RESULT( 32 ) = ZERO
01648 RESULT( 33 ) = ZERO
01649 RESULT( 34 ) = ZERO
01650 END IF
01651 *
01652 *
01653 * Call ZSTEMR(V,A) to compute D1 and Z, do tests.
01654 *
01655 * Compute D1 and Z
01656 *
01657 CALL DCOPY( N, SD, 1, D5, 1 )
01658 IF( N.GT.0 )
01659 \$ CALL DCOPY( N-1, SE, 1, RWORK, 1 )
01660 *
01661 NTEST = 35
01662 *
01663 CALL ZSTEMR( 'V', 'A', N, D5, RWORK, VL, VU, IL, IU,
01664 \$ M, D1, Z, LDU, N, IWORK( 1 ), TRYRAC,
01665 \$ RWORK( N+1 ), LRWORK-N, IWORK( 2*N+1 ),
01666 \$ LIWORK-2*N, IINFO )
01667 IF( IINFO.NE.0 ) THEN
01668 WRITE( NOUNIT, FMT = 9999 )'ZSTEMR(V,A)', IINFO, N,
01669 \$ JTYPE, IOLDSD
01670 INFO = ABS( IINFO )
01671 IF( IINFO.LT.0 ) THEN
01672 RETURN
01673 ELSE
01674 RESULT( 35 ) = ULPINV
01675 GO TO 280
01676 END IF
01677 END IF
01678 *
01679 * Do Tests 35 and 36
01680 *
01681 CALL ZSTT22( N, M, 0, SD, SE, D1, DUMMA, Z, LDU, WORK, M,
01682 \$ RWORK, RESULT( 35 ) )
01683 *
01684 * Call ZSTEMR to compute D2, do tests.
01685 *
01686 * Compute D2
01687 *
01688 CALL DCOPY( N, SD, 1, D5, 1 )
01689 IF( N.GT.0 )
01690 \$ CALL DCOPY( N-1, SE, 1, RWORK, 1 )
01691 *
01692 NTEST = 37
01693 CALL ZSTEMR( 'N', 'A', N, D5, RWORK, VL, VU, IL, IU,
01694 \$ M, D2, Z, LDU, N, IWORK( 1 ), TRYRAC,
01695 \$ RWORK( N+1 ), LRWORK-N, IWORK( 2*N+1 ),
01696 \$ LIWORK-2*N, IINFO )
01697 IF( IINFO.NE.0 ) THEN
01698 WRITE( NOUNIT, FMT = 9999 )'ZSTEMR(N,A)', IINFO, N,
01699 \$ JTYPE, IOLDSD
01700 INFO = ABS( IINFO )
01701 IF( IINFO.LT.0 ) THEN
01702 RETURN
01703 ELSE
01704 RESULT( 37 ) = ULPINV
01705 GO TO 280
01706 END IF
01707 END IF
01708 *
01709 * Do Test 34
01710 *
01711 TEMP1 = ZERO
01712 TEMP2 = ZERO
01713 *
01714 DO 260 J = 1, N
01715 TEMP1 = MAX( TEMP1, ABS( D1( J ) ), ABS( D2( J ) ) )
01716 TEMP2 = MAX( TEMP2, ABS( D1( J )-D2( J ) ) )
01717 260 CONTINUE
01718 *
01719 RESULT( 37 ) = TEMP2 / MAX( UNFL,
01720 \$ ULP*MAX( TEMP1, TEMP2 ) )
01721 END IF
01722 270 CONTINUE
01723 280 CONTINUE
01724 NTESTT = NTESTT + NTEST
01725 *
01726 * End of Loop -- Check for RESULT(j) > THRESH
01727 *
01728 *
01729 * Print out tests which fail.
01730 *
01731 DO 290 JR = 1, NTEST
01732 IF( RESULT( JR ).GE.THRESH ) THEN
01733 *
01734 * If this is the first test to fail,
01735 * print a header to the data file.
01736 *
01737 IF( NERRS.EQ.0 ) THEN
01738 WRITE( NOUNIT, FMT = 9998 )'ZST'
01739 WRITE( NOUNIT, FMT = 9997 )
01740 WRITE( NOUNIT, FMT = 9996 )
01741 WRITE( NOUNIT, FMT = 9995 )'Hermitian'
01742 WRITE( NOUNIT, FMT = 9994 )
01743 *
01744 * Tests performed
01745 *
01746 WRITE( NOUNIT, FMT = 9987 )
01747 END IF
01748 NERRS = NERRS + 1
01749 IF( RESULT( JR ).LT.10000.0D0 ) THEN
01750 WRITE( NOUNIT, FMT = 9989 )N, JTYPE, IOLDSD, JR,
01751 \$ RESULT( JR )
01752 ELSE
01753 WRITE( NOUNIT, FMT = 9988 )N, JTYPE, IOLDSD, JR,
01754 \$ RESULT( JR )
01755 END IF
01756 END IF
01757 290 CONTINUE
01758 300 CONTINUE
01759 310 CONTINUE
01760 *
01761 * Summary
01762 *
01763 CALL DLASUM( 'ZST', NOUNIT, NERRS, NTESTT )
01764 RETURN
01765 *
01766 9999 FORMAT( ' ZCHKST: ', A, ' returned INFO=', I6, '.', / 9X, 'N=',
01767 \$ I6, ', JTYPE=', I6, ', ISEED=(', 3( I5, ',' ), I5, ')' )
01768 *
01769 9998 FORMAT( / 1X, A3, ' -- Complex Hermitian eigenvalue problem' )
01770 9997 FORMAT( ' Matrix types (see ZCHKST for details): ' )
01771 *
01772 9996 FORMAT( / ' Special Matrices:',
01773 \$ / ' 1=Zero matrix. ',
01774 \$ ' 5=Diagonal: clustered entries.',
01775 \$ / ' 2=Identity matrix. ',
01776 \$ ' 6=Diagonal: large, evenly spaced.',
01777 \$ / ' 3=Diagonal: evenly spaced entries. ',
01778 \$ ' 7=Diagonal: small, evenly spaced.',
01779 \$ / ' 4=Diagonal: geometr. spaced entries.' )
01780 9995 FORMAT( ' Dense ', A, ' Matrices:',
01781 \$ / ' 8=Evenly spaced eigenvals. ',
01782 \$ ' 12=Small, evenly spaced eigenvals.',
01783 \$ / ' 9=Geometrically spaced eigenvals. ',
01784 \$ ' 13=Matrix with random O(1) entries.',
01785 \$ / ' 10=Clustered eigenvalues. ',
01786 \$ ' 14=Matrix with large random entries.',
01787 \$ / ' 11=Large, evenly spaced eigenvals. ',
01788 \$ ' 15=Matrix with small random entries.' )
01789 9994 FORMAT( ' 16=Positive definite, evenly spaced eigenvalues',
01790 \$ / ' 17=Positive definite, geometrically spaced eigenvlaues',
01791 \$ / ' 18=Positive definite, clustered eigenvalues',
01792 \$ / ' 19=Positive definite, small evenly spaced eigenvalues',
01793 \$ / ' 20=Positive definite, large evenly spaced eigenvalues',
01794 \$ / ' 21=Diagonally dominant tridiagonal, geometrically',
01795 \$ ' spaced eigenvalues' )
01796 *
01797 9993 FORMAT( / ' Tests performed: ',
01798 \$ '(S is Tridiag, D is diagonal, U and Z are ', A, ',', / 20X,
01799 \$ A, ', W is a diagonal matrix of eigenvalues,', / 20X,
01800 \$ ' V is U represented by Householder vectors, and', / 20X,
01801 \$ ' Y is a matrix of eigenvectors of S.)',
01802 \$ / ' ZHETRD, UPLO=''U'':', / ' 1= | A - V S V', A1,
01803 \$ ' | / ( |A| n ulp ) ', ' 2= | I - U V', A1,
01804 \$ ' | / ( n ulp )', / ' ZHETRD, UPLO=''L'':',
01805 \$ / ' 3= | A - V S V', A1, ' | / ( |A| n ulp ) ',
01806 \$ ' 4= | I - U V', A1, ' | / ( n ulp )' )
01807 9992 FORMAT( ' ZHPTRD, UPLO=''U'':', / ' 5= | A - V S V', A1,
01808 \$ ' | / ( |A| n ulp ) ', ' 6= | I - U V', A1,
01809 \$ ' | / ( n ulp )', / ' ZHPTRD, UPLO=''L'':',
01810 \$ / ' 7= | A - V S V', A1, ' | / ( |A| n ulp ) ',
01811 \$ ' 8= | I - U V', A1, ' | / ( n ulp )',
01812 \$ / ' 9= | S - Z D Z', A1, ' | / ( |S| n ulp ) ',
01813 \$ ' 10= | I - Z Z', A1, ' | / ( n ulp )',
01814 \$ / ' 11= |D(with Z) - D(w/o Z)| / (|D| ulp) ',
01815 \$ ' 12= | D(PWK) - D(QR) | / (|D| ulp)',
01816 \$ / ' 13= Sturm sequence test on W ' )
01817 9991 FORMAT( ' 14= | S - Z4 D4 Z4', A1, ' | / (|S| n ulp)',
01818 \$ / ' 15= | I - Z4 Z4', A1, ' | / (n ulp ) ',
01819 \$ ' 16= | D4 - D5 | / ( 100 |D4| ulp ) ',
01820 \$ / ' 17= max | D4(i) - WR(i) | / ( |D4(i)| (2n-1) ulp )',
01821 \$ / ' 18= | WA1 - D3 | / ( |D3| ulp )',
01822 \$ / ' 19= max | WA2(i) - WA3(ii) | / ( |D3| ulp )',
01823 \$ / ' 20= | S - Y WA1 Y', A1, ' | / ( |S| n ulp )',
01824 \$ / ' 21= | I - Y Y', A1, ' | / ( n ulp )' )
01825 9990 FORMAT( ' 22= | S - Z D Z', A1,
01826 \$ ' | / ( |S| n ulp ) for ZSTEDC(I)', / ' 23= | I - Z Z', A1,
01827 \$ ' | / ( n ulp ) for ZSTEDC(I)', / ' 24= | S - Z D Z',
01828 \$ A1, ' | / ( |S| n ulp ) for ZSTEDC(V)', / ' 25= | I - Z Z',
01829 \$ A1, ' | / ( n ulp ) for ZSTEDC(V)',
01830 \$ / ' 26= | D1(ZSTEDC(V)) - D2(CSTEDC(N)) | / ( |D1| ulp )' )
01831 9989 FORMAT( ' Matrix order=', I5, ', type=', I2, ', seed=',
01832 \$ 4( I4, ',' ), ' result ', I3, ' is', 0P, F8.2 )
01833 9988 FORMAT( ' Matrix order=', I5, ', type=', I2, ', seed=',
01834 \$ 4( I4, ',' ), ' result ', I3, ' is', 1P, D10.3 )
01835 *
01836 9987 FORMAT( / 'Test performed: see ZCHKST for details.', / )
01837 * End of ZCHKST
01838 *
01839 END
```
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CC-MAIN-2022-49
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https://www.jiskha.com/questions/22365/find-the-value-of-the-polynomial-8x-6-when-x-1-and-when-x-1-8-1-6-2-8-1
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# mat117/algebra
find the value of the polynomial 8x - 6 when x = 1 and when x = -1
8(1) - 6= 2
8(-1) - 6=
-8 -6 = is the answer -14 or -2
-8 -6 = -14
1. 👍 0
2. 👎 0
3. 👁 53
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Topic: Continuum Hypothesis Solution Posted
Replies: 44 Last Post: May 1, 1999 5:30 PM
Messages: [ Previous | Next ]
Nathaniel Deeth Posts: 548 Registered: 12/4/04
Re: Continuum Hypothesis Solution Posted
Posted: Apr 22, 1999 9:19 AM
Ulrich Weigand wrote:
> Nathan the Great <mad@ashland.baysat.net> writes:
>
> > x = 1 ;the first number
> > N = x ;Set N = {1}
> > Do
> > x = x + 1 ;the successor of x
> > N = N U x ;union element x to Set N
> > Loop
>
> >If/when this algorithm *completes*, N will contain all the natural numbers.
>
> Nope. This algorithm does not terminate. Nevertheless, the set of all
> natural numbers exists. (Who ever said that whether a set does or does
> not exist depends on the existence of an *algorithm* that produces this
> set? Every set produced by an algorithm in this sense will always be
> finite, of course.)
Nathan: My algorithm can construct any number.
Ulrich: Yes. So what?
Nathan: So, I can construct the set of all natural numbers.
Ulrich: No! Why do you say that?
Nathan: Didn't you just say that it can construct every natural number?
Ulrich: No, I agreed that you could construct *any* number.
Nathan: Humm... I can construct *any* number but not *every* number.
Ulrich: Yes, thats correct.
Nathan: Well...since my algorithm constructs the numbers in sequence, if I construct the
last number, I will, in the process, have constructed all the numbers.
Ulrich: Wrong! There is no last number.
Nathan: Oh? Is that why my algorithm never finishes?
Ulrich: uhm.
Nathan: Whats wrong, Dr. Ulrich? Did someone drop a coconut on your head?
Nathan the Great
Age 11
Date Subject Author
4/14/99 Papus
4/15/99 Webster Kehr
4/16/99 Alan Morgan
4/18/99 Webster Kehr
4/16/99 Dave Seaman
4/16/99 Bill Taylor
4/16/99 Nathaniel Deeth
4/16/99 Jake Wildstrom
4/19/99 Sami Aario
4/16/99 Ken Cox
4/19/99 Michel Hack
4/20/99 Nathaniel Deeth
4/20/99 Ulrich Weigand
4/21/99 Nathaniel Deeth
4/21/99 Nathaniel Deeth
4/21/99 Ulrich Weigand
4/21/99 Brian David Rothbach
4/21/99 Virgil Hancher
4/22/99 Nathaniel Deeth
4/22/99 Sami Aario
4/23/99 Nathaniel Deeth
4/25/99 Sami Aario
4/22/99 Ulrich Weigand
4/23/99 Nathaniel Deeth
4/23/99 Ulrich Weigand
4/20/99 Nathaniel Deeth
4/17/99 Papus
4/19/99 Kevin Lacker
4/20/99 Bill Taylor
4/19/99 Andrew Boucher
4/19/99 Kevin Lacker
4/21/99 Andrew Boucher
4/22/99 Keith Ramsay
4/23/99 Andrew Boucher
4/24/99 Keith Ramsay
4/25/99 Andrew Boucher
4/27/99 Bill Taylor
4/27/99 David Petry
4/30/99 Keith Ramsay
5/1/99 Keith Ramsay
4/16/99 Andrew Boucher
4/16/99 Dave Seaman
4/18/99 Webster Kehr
4/19/99 Jeremy Boden
4/19/99 Sami Aario
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https://oeis.org/A064213
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A064213 Least k such that k*3^n +- 1 are twin primes. 1
4, 2, 2, 4, 10, 40, 58, 64, 28, 24, 8, 210, 70, 36, 12, 4, 78, 26, 28, 20, 90, 30, 10, 630, 210, 70, 82, 416, 612, 204, 68, 930, 310, 406, 1078, 1164, 388, 176, 190, 334, 190, 726, 242, 1004, 398, 1430, 1372, 2730, 910, 1560, 520 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,1 COMMENTS k must be even because 3^n is odd. - Harry J. Smith, Sep 10 2009 LINKS Harry J. Smith, Table of n, a(n) for n = 0..225 MATHEMATICA Do[ k = 1; While[ ! PrimeQ[ k*3^n + 1 ] || ! PrimeQ[ k*3^n - 1 ], k++ ]; Print[ k ], {n, 0, 50} ] PROG (PARI) { for (n=0, 225, k=0; p=3^n; until (isprime(k*p - 1) && isprime(k*p + 1), k+=2); write("b064213.txt", n, " ", k); ) } \\ Harry J. Smith, Sep 10 2009 CROSSREFS Cf. A063983. Sequence in context: A021707 A126560 A289762 * A245518 A217462 A016510 Adjacent sequences: A064210 A064211 A064212 * A064214 A064215 A064216 KEYWORD nonn AUTHOR Robert G. Wilson v, Sep 21 2001 EXTENSIONS Offset changed from 1 to 0 by Harry J. Smith, Sep 10 2009 STATUS approved
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Last modified December 15 11:43 EST 2019. Contains 329999 sequences. (Running on oeis4.)
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https://math.stackexchange.com/questions/2407827/given-a-set-of-addresses-pick-a-geographically-even-subset-for-sampling/4422671
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# Given a set of addresses, pick a geographically even subset for sampling
There are a certain number of houses spread randomly across a city. We have addresses and GPS coordinates for each house.
How can we select a geographically representative, fixed-size sample out of the houses in the selection?
To illustrate with an example, say we have about 100 houses on a map like the one below:
Each house has been identified as a likely supporter in a political campaign, and ideally we visit all of the houses and solicit support. However, it isn't possible to visit all of the houses; we can visit 25 of the houses. By spreading the 25 visits as evenly across the map as possible, we can leverage the fact that neighbors talk to each other.
In other words, how do we select a subset of the GPS coordinates so the entire set of coordinates is as well represented as possible?
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https://www.physicsforums.com/threads/heat-of-rxn-per-mol-h2o-formed-in-acid-base-reaction.547164/
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# Heat of rxn per mol H2O formed in acid-base reaction
1. Nov 3, 2011
### Nuchem
1. The problem statement, all variables and given/known data
Calculate the heat of reaction per mole of water formed in the acid-base reaction. Assume that the total mass of the solution is 150g.
75mL of 2.0M NaOH added to 75mL of 2.0M HCl in a styrofoam cup. Average initial temp = 295.65K. final temp = 308.8K
2. Relevant equations
ΔHrxn = qrxn = -mcΔT
3. The attempt at a solution
qrxn = -(141.234g)(4.186J/gK)(308.8K-295.65K) = -7.8kJ
based on: .15 mol of HCl (5.5g) and .15 mol NaOH (6.0g) reacted to form .15 mol H2O (2.7g) and .15 mol NaCl (8.8g) therefore since the entire reaction is aqueous and the total mass of the solution is 150g: 150g -8.8g (the amount of the product that is not water) = 141.2g
in a simpler train of thought i decided 2.70g of water were actually formed and therefore:
qrxn = -(2.70g)(4.186J/gK)(308.8K-295.65K) = -1.49*102J
I feel that i am grossly over thinking this, however, and I need a bit of advice
2. Nov 4, 2011
### Staff: Mentor
Everything gets heated - so you can't ignore product and assume only water mass is important.
Simplest approach is to assume you have 150 mL of solution before and after the reaction and assume density and specific heat capacity of water.
That's not exactly true, but close enough. Note that density of neither solution was exactly 1 g/mL, so mass of the solution is not 150 g. But you had the same problem with your approach.
3. Nov 4, 2011
### Nuchem
Alright so "per mole of water formed" isn't really important in the calculations? Assuming the specific heat and density of the products is equal to water (I am actually instructed to do this at a later point) and assuming the mass of the solution is 150g (also instructed to do so) you're saying qrxn = -(150g)(4.186J/gK)(308.8K-295.65K) = -8.3kJ per mole of water formed?
4. Nov 4, 2011
### Staff: Mentor
It is important, just not when you calculate amount of heat that evolved in this particular experimental setup. And you calculated this amount of heat correctly - now just check how many moles of water were produced to be able to calculate heat per mole.
5. Nov 4, 2011
### Nuchem
Oh! Okay that's a lot more clear now, thank you. So I would take the -8.3kJ and divide by moles of water (.15 seeing that HCl + NaOH → H2O + NaCl is 1:1:1:1) to get
-8.3kJ/.15mol = -55kJ/mol
6. Nov 4, 2011
### Staff: Mentor
I have not checked the numbers, just skimmed - and what you did looks OK now.
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https://mattermodeling.stackexchange.com/questions/5077/pycalphad-most-efficient-way-to-gather-equilibirum-phase-data-for-set-of-alloy-c
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# Pycalphad most efficient way to gather equilibirum phase data for set of alloy compositions?
What's the most efficient way to get the equilibrium phases and phase fractions for a set of alloy compositions? For context, I'm coming from Thermo-Calc/TC-Python where I would just loop through the alloy compositions and do a single equilbrium calculation. However, when I attempt to do this with pycalphad it seems quite slow. For example, here is what my code for the "one at a time" looping:
from pycalphad import Database, calculate, equilibrium, variables as v
dbf = Database("Cr-Fe-Ni.tdb")
components = ['FE','CR', 'NI', 'VA']
phases = ['LIQUID', 'BCC_A2', 'FCC_A1', 'SIGMA']
#Example set of alloy compositions, normally I would want to sample more finely but I'll keep it simple for demonstration purposes
compositions = array([[0. , 0. ],
[0. , 0.2],
[0. , 0.4],
[0. , 0.6],
[0. , 0.8],
[0.2, 0. ],
[0.2, 0.2],
[0.2, 0.4],
[0.2, 0.6],
[0.2, 0.8],
[0.4, 0. ],
[0.4, 0.2],
[0.4, 0.4],
[0.4, 0.6],
[0.6, 0. ],
[0.6, 0.2],
[0.6, 0.4],
[0.8, 0. ],
[0.8, 0.2]])
phase_data = []
for i in tqdm.tqdm(range(len(compositions)), total=len(compositions), position=0, leave=True, smoothing=0.3):
eq_result = equilibrium(dbf, components, phases, {v.X('CR'):compositions[i,0], v.X('NI'):compositions[i,1], v.T:1000, v.P:101325})
phase_data.append([*eq_result.NP.values.squeeze().tolist(), *eq_result.Phase.values.squeeze().tolist()])
This takes about 15 seconds, or about 1.2 alloys evaluated per second. This seems quite slow for a ternary, and I suspect it's because I'm calling the equilibrium function each time instead of just changing the composition.
I tried feeding the whole column at once (see code below), but this seemed to take the Cartesian product of the compositions, which is not what I want.
#database, phase, and composition setup is same as in above code
eq_result = equilibrium(dbf, components, phases, {v.X('CR'):compositions[:,0], v.X('NI'):compositions[:,1], v.T:1000, v.P:101325})
So, please let me know what the better approach is. Thanks!
• Can I ask for some clarification on what your specific goal is (particularly why the Cartesian product is non-viable)? There is some overhead in calling equilibrium. There are some possible workarounds today and the project is in the process of stabilizing a lower-level API for tight loops of calculations like this (including a new, faster minimizer). For calculations today, a Cartesian product may be faster even if there are more overall compositions being computed because of the amortized overhead. – Brandon Bocklund May 31 at 5:42
• @BrandonBocklund Usually I like to have specific alloy compositions, which are often subject to constraints (e.g. at 40at% of element A, etc.) and I usually want to test many more alloys (100s of thousands). I guess for the time being I can just test the whole range of compositions. For that should I feed v.X('CR'):[0.0, 0.2, 0.4, 0.6, 0.8] and same for NI? If so, won't the cartesian product result in alloys with atomic fractions whose total is greater than one (e.g. CR=0.6, NI=0.6)? And if that's the case how can I parse the data from eq_result to remove/ignore these entries? – sgp45 May 31 at 15:22
In your comment, you mentioned that you may be testing many alloys (100,000s). I will assume that interested in taking more of a screening approach, rather than specifying precisely 100,000 alloys compositions.
In that case, it is easy and faster to let pycalphad broadcast the conditions for you. It will skip any conditions that have independent mole fractions that sum to greater than unity, e.g. X(CR)=0.8 and X(NI)=0.8. The result that you get from equilibrium is an xarray Dataset object and uses padded NumPy arrays. Floating point values (like NP) will be padded with nan and string values (like Phase) will be padded with '' when no phases are present.
xarray and most of the tools in the ecosystem (like matplotlib) play nicely with these padded values. Below, I give an example where equilibrium calculations are performed on a dense grid and I plot the phase fractions of FCC over the whole grid and under a constraint where the composition of Cr is under 40 at%.
from pycalphad import Database, calculate, equilibrium, variables as v
import matplotlib.pyplot as plt
dbf = Database("Cr-Fe-Ni.tdb")
components = ['FE','CR', 'NI', 'VA']
phases = ['LIQUID', 'BCC_A2', 'FCC_A1', 'SIGMA']
eq_result = equilibrium(dbf, components, phases, {v.X('CR'): (0, 1, 0.1), v.X('NI'): (0, 1, 0.1), v.T:1000, v.P:101325})
# verify what the Phase and NP look like by printing
# note these are multi-dimensional arrays and are padded with '' and nan, respectively
print(eq_result.Phase.values)
print(eq_result.NP.values)
# Plot FCC phase fractions over the whole grid
plt.figure()
plt.subplot(projection="triangular") # projection provided by pycalphad
mask = eq_result.Phase == "FCC_A1" # choose only FCC, others will be NaN
plt.scatter(
)
plt.title("Cr-Fe-Ni FCC Phase Fractions")
plt.xlabel("X(CR)")
plt.ylabel("X(NI)")
plt.colorbar(label="NP(FCC_A1)")
plt.show()
# Plot FCC phase fractions only at compositions under 40 at% Cr
plt.figure()
plt.subplot(projection='triangular') # projection provided by pycalphad
# choose only FCC AND where the composition of Cr is less than 40%, others will be NaN
mask = (eq_result.Phase == 'FCC_A1') & (eq_result.X_CR < 0.4)
plt.scatter(
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https://documen.tv/modeling-real-life-a-softball-coach-buys-equipment-for-two-teams-the-coach-buys-4-bats-36-softba-28279820-95/
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Question
MODELING REAL LIFE A softball coach buys equipment for two teams. The coach buys 4 bats, 36 softballs, and 12 uniforms for Team A and 6 bats, 45 softballs, and 15 uniforms for Team B . Each bat costs $80, each softball costs$4, and each uniform costs $20. Use matrix multiplication to find the total cost of the equipment for each team. Answers 1. ngocdiep Using proportions, it is found that the total costs of equipment for each team are given as follows: • Team A:$704.
• Team B: $960. ### What is a proportion? A proportion is a fraction of a total amount, and the measures are related using a rule of three. Due to this, relations between variables, either direct or inverse proportional, can be built to find the desired measures in the problem. The costs are given as follows: • Bat:$80.
• Softball: $4. • Uniform:$20.
Considering the amounts of bats, softballs and uniforms purchased by Team A, their total cost is given by:
4 x 80 + 36 x 4 + 12 x 20 = $704. Considering the amounts of bats, softballs and uniforms purchased by Team B, their total cost is given by: 6 x 80 + 45 x 4 + 15 x 20 =$960.
More can be learned about proportions at https://brainly.com/question/24372153
#SPJ1
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http://monicasmommusings.com/homework-help-algebraic-expressions/
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# Homework help algebraic expressions
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https://awoum.com/how-to-determine-if-a-function-is-even-or-odd-a-complete-guide/
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Technology
# How to Determine if a Function is Even or Odd: A Complete Guide
Determining whether a function is even or odd might seem like a trivial task, but it is actually an important step in understanding the properties of functions. Even and odd functions are essential in mathematics and have significant real-world applications. For instance, they play a critical role in signal processing, physics, finance, and many other fields. By recognizing if a function is even or odd, we can simplify complex calculations, graph functions easily, and gain insight into how a function behaves. In this guide, you will learn everything you need to know about even and odd functions, including how to identify them and examples of their practical uses.
# Introduction
If you’re studying calculus, it’s likely that you’ve heard the terms “even” and “odd” functions before. These types of functions have unique properties that make them essential to understanding mathematical concepts such as integration, differentiation, and Fourier series.
Simply put, even and odd functions are special types of functions that exhibit a particular type of symmetry. While this may seem abstract at first, it’s actually quite intuitive once you get the hang of it.
In this article, we’ll explore the definition of even and odd functions in depth, examine how to identify them using tests and graphical representations, and provide some real-world examples of where they come in handy. By the end of this guide, you’ll have a solid understanding of even and odd functions and be well on your way to mastering calculus. So let’s dive in!
## What are Even and Odd Functions?
In mathematics, a function is considered even if it satisfies the condition f(x) = f(-x) for all values of x in its domain. On the other hand, a function is odd if it satisfies the condition f(x) = -f(-x) for all values of x in its domain.
An even function is symmetric about the y-axis, meaning that if we reflect the graph of an even function across the y-axis, it will still retain its shape. Examples of even functions include trigonometric functions such as cosine, exponential functions like e^x, and polynomial functions with only even exponents like x^2 or x^4.
Conversely, an odd function is symmetric about the origin, meaning that if we rotate the graph of an odd function 180 degrees around the origin, it will look the same. Examples of odd functions include trigonometric functions such as sine, inverse trigonometric functions, and polynomial functions with only odd exponents like x^3 or x^5.
It’s important to note that not all functions are either even or odd. Some functions do not satisfy either condition, and are referred to as neither even nor odd.
Understanding even and odd functions can be useful in many areas of mathematics, including calculus and differential equations. They also have real-world applications, such as in signal processing and physics.
Overall, knowing the definition of even and odd functions, and being able to identify them, is important for anyone studying mathematics or working in related fields.
## How to Identify an Even Function
In mathematics, an even function is a function that has symmetry with respect to the y-axis. In other words, if you reflect the function across the y-axis, you get the same graph. A simple test for even functions is to check if the function f(x) is equal to f(-x) for all x in the domain of the function.
### Test for Even Function
To test if a function is even, substitute -x for x in the equation and simplify. If the result is the same as the original equation, then the function is even. Mathematically, this can be represented as:
f(x) = f(-x) for all x in the domain of the function
### Even Symmetry
A function with even symmetry is symmetric with respect to the y-axis. This means that if you take a point on the graph (x, y), the corresponding point (-x, y) will also be on the graph. The graph of an even function is always symmetrical about the y-axis.
### Graph of Even Function
The graph of an even function will look the same on both sides of the y-axis. For example, the graph of the function f(x) = x^2 will look like a parabola that opens upwards and is symmetrical around the y-axis.
Even functions have many applications in real-world scenarios, such as in signal processing and Fourier analysis. In summary, identifying an even function is straightforward: check if the function is equal to its reflection across the y-axis, and if so, it is an even function.
# How to Identify an Odd Function
An odd function is a mathematical function that satisfies the condition f(-x) = -f(x) for all x in its domain. In simpler terms, it means that if you replace x with -x in the input of the function, and then multiply the output by -1, you should get the same result as when you input x into the function.
If you’re given a function and want to determine whether it’s odd or not, you can follow these steps:
## Test for Odd Function
To test whether a function is odd or not, you can use the following basic test for odd functions:
• Substitute -x for x in the given function
• Simplify the function algebraically
• If the resulting expression is equal to the negative of the original function, then the function is odd.
Let’s take the example of a function f(x) = x^3. We can apply the test for odd functions as follows:
• Substitute -x for x: f(-x) = -x^3
• Simplify the function algebraically: f(-x) = -(x^3) = -f(x)
• Since f(-x) = -f(x), we can conclude that f(x) is an odd function.
## Odd Symmetry
Another way to identify an odd function is by looking at its symmetry. An odd function is symmetric about the origin because every point on one side of the origin has a corresponding point on the opposite side that is the reflection of the first point through the origin.
For example, the graph of f(x) = x^3 is symmetric about the origin, which means that if you reflect the graph across the origin, it will look exactly the same.
## Graph of Odd Function
The graph of an odd function is always symmetrical about the origin, so if you’re given a graph of a function and want to determine whether it’s odd or not, you can look for this symmetry.
For example, the graph of f(x) = x^3 looks like a curve that passes through the origin and has arms that extend in opposite directions. This curve is symmetric about the origin, so we know that f(x) is an odd function.
In conclusion, identifying an odd function requires knowledge of its definition, basic test, odd symmetry, and the graph of the function. By following these steps, you can confidently determine whether a given function is odd or not.
## Examples of Even and Odd Functions
Even and odd functions are widely used in mathematics and physics. In this section, we will explore some real-world applications of even and odd functions and provide examples of each type.
### Example of an Even Function
Consider the function `f(x) = x^2`. We can test if this function is even or odd as follows:
• Test for evenness: `f(-x) = (-x)^2 = x^2 = f(x)`.
• Test for oddness: `f(-x) = (-x)^2 = x^2 = -f(x)`.
Since `f(-x) = f(x)`, `f(x)` is an even function. The graph of `f(x) = x^2` is a parabola, which has even symmetry around the y-axis. An example of an even function in real life is the shape of a satellite dish, which has rotational symmetry.
### Example of an Odd Function
Let’s consider the function `g(x) = x^3`. To test whether `g(x)` is even or odd:
• Test for evenness: `g(-x) = (-x)^3 = -x^3 = -g(x)`.
• Test for oddness: `g(-x) = (-x)^3 = -x^3 = -g(x)`.
Since `g(-x) = -g(x)`, `g(x)` is an odd function. The graph of `g(x) = x^3` is also symmetric about the origin. An example of an odd function in real life is the current that flows in a circuit because it changes direction every half cycle.
### Real-World Applications of Even and Odd Functions
Even and odd functions appear in various fields such as signal processing, quantum mechanics, crystallography, and more. They help in characterizing the symmetry of physical systems and are used in Fourier analysis to decompose a function into its even and odd parts.
In signal processing, the Fourier transform is used to analyze signals that vary with time. An example of an even signal is an AC voltage where the waveform repeats after every half cycle. An example of an odd signal is a full-wave rectified signal where the negative part is inverted with respect to the positive part.
In quantum mechanics, the wave functions that describe the behavior of subatomic particles have both even and odd components. The even component represents the particle’s symmetric properties, while the odd component represents the antisymmetric properties.
In crystallography, even and odd functions are used to describe the symmetry of crystals. The symmetry operations of a crystal leave the crystal unchanged, and they can be represented by even or odd functions.
In conclusion, even and odd functions play a crucial role in various fields of mathematics and physics. By understanding these concepts, we can better understand the symmetries of physical systems and how they behave in the real world.
## Summary
In summary, understanding even and odd functions is an important concept in calculus and other areas of mathematics. Even functions have symmetry around the y-axis, while odd functions have symmetry around the origin. To determine if a function is even or odd, you can use the tests for even and odd symmetry, as well as the graph of the function.
Some common examples of even functions include exponential functions and cosine functions, while some common examples of odd functions include sine functions and cubic functions. Even and odd functions have many real-world applications, such as in physics, engineering, and finance.
It’s important to note that not all functions are either even or odd. Some functions, like a straight line, have neither even nor odd symmetry. Additionally, some functions may have both even and odd components.
Overall, understanding even and odd functions is a useful tool for analyzing and solving mathematical problems. By mastering the tests for even and odd symmetry, you can identify these functions and apply them to various real-world scenarios.
In conclusion, determining whether a function is even or odd can seem daunting at first, but once you understand the definitions and tests for symmetry, it becomes a straightforward process. Even and odd functions have real-world applications in various fields, including finance, physics, and engineering. Being able to identify and differentiate between them is an essential skill for anyone working with mathematical functions. By following the guidelines outlined in this guide and practicing with examples, you’ll be well on your way to mastering even and odd functions. Remember that practice makes perfect, so keep honing your skills and applying them to real-life problems.
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991_PartUniversity Physics Solution
# 991_PartUniversity Physics Solution - Geometric Optics...
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Geometric Optics 34-29 Then 1.33 1.55 1.55 1.33 20.0 cm 2.50 cm s += gives 72.1 cm s ′ = − . The image is 72.1 cm to the left of the surface vertex. EVALUATE: With the rod in air the image is real and with the rod in water the image is virtual. 34.96. IDENTIFY: Apply 11 1 s sf to each lens. The image formed by the first lens serves as the object for the second lens. The focal length of the lens combination is defined by 12 111 s + = . In part (b) use 11 1 (1 ) n f RR ⎛⎞ =− ⎜⎟ ⎝⎠ to calculate f for the meniscus lens and for the 4 CCl , treated as a thin lens. SET UP: With two lenses of different focal length in contact, the image distance from the first lens becomes exactly minus the object distance for the second lens. EXECUTE: (a) 11 1 1 11 111 111 s sfs +=⇒=− ′′ and 22 2 2 11 1 1 11 11 . s ss f s f += += − += But overall for the lens system, 21 . s fff +=⇒=+ (b) With carbon tetrachloride sitting in a meniscus lens, we have two lenses in contact. All we need in order to calculate the system&s focal length is calculate the individual focal lengths, and then use the formula from part (a). For the meniscus lens 1 m1 2 1 1 1 ( ) (0.55) 0.061 cm 4.50 cm 9.00 cm ba nn fR R = = and m 16.4 cm f = . For the 1 4 w1 2 1 1 1 CCl : ( ) (0.46) 0.051 cm 9.00 cm R = − = and w 19.6 cm f = . 1 wm 0.112 cm ff f =+= and 8.93 cm f = . EVALUATE: f f f f f = + , so f for the combination is less than either 1 f or 2 f . 34.97. IDENTIFY: Apply Eq.(34.11) with R →∞ to the refraction at each surface. For refraction at the first surface the point P serves as a virtual object. The image formed by the first refraction serves as the object for the second refraction. SET UP: The glass plate and the two points are shown in Figure 37.97. plane faces means R and 0 ab + = b a n s s n ′=− Figure 34.97 EXECUTE: refraction at the first (left-hand) surface of the piece of glass: The rays converging toward point P constitute a virtual object for this surface, so 14.4 cm. s 1.00, 1.60. == 1.60 ( 14.4 cm) 23.0 cm 1.00 s =+ This image is 23.0 cm to the right of the first surface so is a distance 23.0 cm t to the right of the second surface. This image serves as a virtual object for the second surface. refraction at the second (right-hand) surface of the piece of glass: The image is at P so 14.4 cm 0.30 cm 14.7 cm . s tt =+− (23.0 cm ); 1.60; 1.00 st n n = = b a n s s n gives 1.00 14.7 cm ( [23.0 cm ]). 1.60 −=− 14.7 cm 14.4 cm 0.625 . − =+ 0.375 0.30 cm t = and 0.80 cm t = EVALUATE: The overall effect of the piece of glass is to diverge the rays and move their convergence point to the right. For a real object, refraction at a plane surface always produces a virtual image, but with a virtual object the image can be real.
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34-30 Chapter 34 34.98. IDENTIFY: Apply the two equations 11 1 22 2 and abba bccb nnnn ss R += ′′ . SET UP: liq ac nn n == , b nn = , and 12 s s ′ = − . EXECUTE: (a) liq liq liq liq 1 2 and n n n n ss R R −− + = .
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# NRICH PROBLEM SOLVING ANGLES
Which ones are impossible? Register for our mailing list. During the third hour after midnight the hands on a clock point in the same direction so one hand is over the top of the other. A semicircle is drawn inside a right-angled triangle. The ancient Egyptians were said to make right-angled triangles using a rope with twelve equal sections divided by knots. Find the exact values of some trig. Square Within a Square Within
Being Collaborative – Primary Age 5 to 11 These problems are ideal for primary school children to work on with others. Suggestions for worthwhile mathematical activity on the subject of angle measurement for all pupils. Weekly Problem 21 – Two rectangles are drawn in a rectangle. Find the exact values of some trig. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. Can you explain why and does this rule extend beyond angles of 90 degrees?
Join pentagons together edge to edge. Olympic Turns Age 7 to 11 Challenge Level: At the time of writing the hour and minute hands of my clock are at right angles. Lunar Angles Age 16 to 18 Challenge Level: To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional agnles for teachers wishing to embed rich mathematical tasks into everyday classroom practice.
SDMS HOMEWORK GREEN
Weekly Problem 51 – Each interior angle of a particular polygon is an obtuse angle which is a whole number of degrees. Weekly Problem 8 – Are you able to find triangles such that these five statements are true?
## 349 Matches for estimating angles
Can you do this without drawing them? What other triangles could you make if you had a rope like this? How many rhombuses are there made up of two adjacent small triangles?
Now, what other relationships you can see? What is the size of the marked angle? Sine and Cosine Age nrlch to 16 Challenge Level: Some questions they could be asked might include:.
Which ones are impossible? That is, pupils were given almost one hundred and fifty opportunities to repeat the same task.
Weekly Problem 15 – How many of the five properties can a heptagon have? The Cyclic Quadrilateral Age 11 to 16 This gives a short summary of the properties and theorems of cyclic quadrilaterals and links to some practical examples to be found elsewhere on the site.
Of course, with solvig geometry software, any number can be produced and then copied for pupil use.
Can you work out how these polygon pictures were drawn, and use that to figure out their angles? What anlges the sum of the four marked angles?
Printable worksheets containing selections of these problems are available here: Can you work out the length of its base? Discover some angle relationships.
## Estimating Angles
Turning Age 5 to 7 Challenge Level: Register for our mailing list. How long is the fence?
Overbearing Age 11 to 14 Short Challenge Level: The sine of an angle is equal to the cosine of its complement. Weekly Problem 44 – A garden has the shape of a right-angled triangle. Find the missing distance in this diagram with two isosceles triangles.
# Angle Hunt :
What fraction of the square is shaded? What shapes should Elly cut out to make a witch’s hat? What if the triangle is on the surface of a sphere? Draw some angles inside a rectangle.
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##### Personal tools
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# Assignment 15
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Question 1 (1 point) In a market system, those who own the most highly valued resources pay the greatest property tax. have the highest incomes. employ the greatest numbers of people. support charitable organizations in the community. pay the largest share of income taxes.
Question 2 (1 point) If everyone had the same income, the Lorenz curve would bow up above the line of income equality. bow down below the line of income equality. be the line of income equality. be a curve that intersects the line of income equality where 50 percent of the population earns 50 percent of the income. be a line below and parallel to the line of income equality.
Question 3 (1 point) The Lorenz curve shows __________ within a society. the absolute level of income the per capita annual income of individuals the income distribution the quality of life of the poor who the poor are
Question 4 (1 point) If the first row in the table above represents the percentage of population and the second row the percent of income earned by that percentage of population, then we can say incomes are distributed unequally with the richest 20 percent having the greatest percentage of income. incomes are distributed unequally with the richest 20 percent earning 10 percent of income. incomes are distributed equally. incomes are distributed unequally with the richest 20 percent earning 30 percent. incomes are distributed unequally with the poorest having the greatest percentage of income.
Question 5 (1 point) Refer to the figure above. In comparing Country A's income distribution with that of Country B, we can say that Country B has a substantially greater degree of income equality than Country A. the absolute level of income in Country B must substantially exceed that of Country A. the absolute level of income in Country A must substantially exceed that of Country B. since both curves are bowed in like fashion, no income inequality exists. Country B has a substantially greater degree of income inequality than Country A.
Question 6 (1 point) Which of the following statements concerning the measurement of poverty is false? Those who live in poverty have significantly lower standards of living than those not living in poverty. A proper measure of poverty must compare income levels within a single country. Poverty levels are the same for all countries. People living in comfort in one country might be considered impoverished in another. Poverty is a measure of the quality of life.
Question 7 (1 point) In-kind transfers are defined to be the allocation of goods and services from one group in society to another. cash transfers, which are unearned funds. private stock transfers from one generation to the next. another term used to describe private charities. money allocated from one group in society to another.
Question 8 (1 point) Which of the following is an example of a cash transfer? Food bought with food stamps Medical services provided under Medicaid Legal aid A cash advance on a credit card Unemployment compensation
Question 9 (1 point) The factor most influencing the number of people in poverty and the incidence of poverty is the happiness of the population. whether the economy is government dominated or not. whether the economy is dominated by another economy. the standard of living of the nation. the general health of the economy.
Question 10 (1 point) Which of the following statements best describes poverty in the United States? Poverty does not affect races, sexes, or different age groups equally. A young person rarely falls below the poverty level because youth and vigor will always ensure that a young person can "carry his weight." Almost one-half of all Americans fall below the poverty line. The highest incidence of poverty by age occurs among those between the ages of 30 and 60. Poverty affects races, sexes, and different age groups relatively equally.
Question 11 (1 point) Which of the following is not a trait commonly seen among those who fall below the poverty line? Lack of education Lack of a job Being aged 65 and over 4 or more years of college Being aged 16 and over
Question 12 (1 point) A progressive tax tends to generate additional tax dollars for the government. reduce income inequality for the rich more than for the poor. reduce income inequality. increase income inequality. leave the income distribution essentially unchanged.
Question 13 (1 point) The largest social insurance program is known as social security. food stamps. Medicaid. Aid to Families with Dependent Children (AFDC). unemployment insurance.
Question 14 (1 point) After more than twenty years of increasing outlays to reduce poverty, the problem of poverty is no closer to a solution now than twenty years ago. our economic system just cannot produce enough goods and services to provide basic necessities for our population. there is disagreement about whether antipoverty programs have reduced or increased poverty. we see that Abraham Lincoln was right when he said, "God must love poor people, for He has made so many of them." the war on poverty has been clearly won.
Question 15 (1 point) According to the text, disincentives created by the welfare system mean those paying taxes may decline to pay more. incentives for the rich to work may be reduced. incentives for the poor to acquire an education may be reduced. incentives for the poor to work may be reduced. incentives for both the rich and the poor to work hard and increase their productivity may be reduced.
Question 16 (1 point) Economists like the negative income tax because, at least in theory, it has been successfully used in other countries, especially in Europe. reduces poverty without reducing efficiency. can be passed by both houses of Congress. increases equity while enhancing efficiency. has been endorsed by both Republican and Democratic candidates for president.
Question 17 (1 point) If in the negative income tax proposal the guaranteed income floor is \$9,000 and the tax rate is 25 percent, what is the value of the breakeven income? \$45,000 \$9,000 \$22,500 \$18,000 \$36,000
Question 18 (1 point) According to the text, incomes differ greatly from one nation to another as well as within nations. What does this mean? It means that the middle class in one country is also the middle class in any other country. It means that the poorest people in a rich country could be among the richest in a poor country. It means that the poorest 10 percent of the population in one country will be the poorest in any nation. It means that the richest 10 percent of the population in one country will be the richest in any nation. It means that the richest people in a poor nation will be among the poorest in another nation.
Question 19 (1 point) Which of the following explanations for the uneven distribution of income in the world has been argued by economists? Older countries tend to be richer. People who live in colder countries tend to be more work oriented. Poorer countries tend to have unfounded cultures. The freer the economic and political system, the greater the rate of economic growth. None of these.
Question 20 (1 point) It has been suggested that a reform to allow poor people to establish clear title to assets could be a means to induce growth in developing countries. Why could this promote poor people's economic growth? Because they would be landlords and be allowed to exploit other people. Because they could then bet their assets. Because they could then borrow against their assets. Because they could then destroy their assets and get away with it. None of these.
Question 1 (5.00 points) Are people who are poor today in the United States likely to be poor for the rest of their lives? Under what conditions is generational poverty likely to exist?
Question 2 (5.00 points) What is the Lorenz curve? What would the curve look like if income were equally distributed? Could the curve ever bow upward above the line of income equality?
Copyright 2008, by the Contributing Authors. Cite/attribute Resource . admin. (2009, January 27). Assignment 15. Retrieved January 07, 2011, from Free Online Course Materials — USU OpenCourseWare Web site: http://ocw.usu.edu/economics/introduction-to-microeconomics-1/assignment15.htm. This work is licensed under a Creative Commons License
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Post Partisan
## Create measure and depend on whether it is duplicate
Dear Community,
I have data as below
JobN TransactionN New Column** J001 T001 J001 J002 T002 Duplicate TransactionN J003 T002 Duplicate TransactionN J004 T003 Duplicate TransactionN J005 T003 Duplicate TransactionN J006 T004 J006 J007 T005 J007
Supposedly TransactionN is unique, which mean JobN only have one TransactionN.
But tats weird in my data, 2 different JobN is having same common TransactionN.
So I would like to add one column (refer back to JobN if not duplicate) .But how do I make it? Since create one new column will checking line by line and doesnt know whether the transactionN will appear somewhere and isit duplicate
Appreciate any helps.
1 ACCEPTED SOLUTION
Community Support
Hi @NickProp28 ,
Here are the steps you can follow：
1. Create calculated colunm.
``New Colunm = IF([JNC]=[TNC],[JobNumber],"Duplicate")``
2. Result.
Best Regards,
Liu Yang
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
10 REPLIES 10
Community Support
Hi @NickProp28 ,
Here are the steps you can follow：
1. Create calculated colunm.
``New Colunm = IF([JNC]=[TNC],[JobNumber],"Duplicate")``
2. Result.
Best Regards,
Liu Yang
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Post Partisan
Dear @v-yangliu-msft ,
Thanks a lot.
Community Support
Hi @NickProp28 ,
Here are the steps you can follow：
1. Create calculated table.
``TNC = CALCULATE(COUNT('Table'[TransactionNumber]),ALLEXCEPT('Table','Table'[TransactionNumber]))``
``JNC = CALCULATE(COUNT('Table'[JobNumber]),ALLEXCEPT('Table','Table'[JobNumber]))``
2. Create measure.
``New = IF(MAX('Table'[JNC])=MAX('Table'[TNC]),MAX('Table'[JobNumber]),"Duplicate")``
3. Result.
Best Regards,
Liu Yang
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Post Partisan
Dear @v-yangliu-msft ,
Its work well when create the DAX and show 'Duplicate'. But I would like to create a new column and indicate all my raw data whenever is it duplicate. Is it possible to do that ? I try the code you provided, but its show all the MAX JobNumber. (Create new column will more easy for me to have a report filter)
Here the pbix: https://ufile.io/hzdz4flh
Post Partisan
Dear @amitchandak ,
Kindly take a look on this PBIX example(https://ufile.io/twxcj922)
Due to this issue, I have double count the transactionnumber even I have put distinctcout on it.
Transaction number is unique for most of time. But there will somehow show in some special case.
Super User
@NickProp28 , there are no dates in your example. Can you share a better sample with the output? I think the date need to considered to call it duplicate
Post Partisan
Dear @amitchandak ,
My data as below,
Hope can create a column to indicate it is duplicate.
Suspossedly, One TransactionNumber is only have one JobNumber.
But as you can see, different jobnumber is sharing one TransactionNumber.
Post Partisan
Dear @amitchandak ,
Measure you given will show all result in duplicate. Because I have different day and time for one JobN. And I will have different row.
My question is, JobN is only have one TransactionN.
But if different JobN sharing the same TransactionN. New measure will result it as 'Duplicate'
Super User
@NickProp28 , Not very clear, what column you need. You can have column like this to check duplicate
If(countx(filter(Table,[TransactionN] = earlier(TransactionN)),[TransactionN])+0 >0, "Duplicate",[JobN])
Inside job duplicate
If(countx(filter(Table,[TransactionN] = earlier(TransactionN) && [JobN] = earlier(JobN)),[TransactionN])+0 >0, "Duplicate",[JobN])
You can have to combine unique column
Key = [JobN] & "-" & [TransactionN]
Post Partisan
Dear @amitchandak ,
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# Root of 11200002
#### [Root of eleven million two hundred thousand two]
square root
3346.6404
cube root
223.7378
fourth root
57.8502
fifth root
25.6947
In mathematics extracting a root is declared as the determination of the unknown "x" in the following equation $y=x^n$ The result of the extraction of the root is declared as a mathematical root. In the case of "n = 2", one talks about a square root or sometimes a second root also, another possibility could be that n is equivalent to 3 then one would consider it a cube root or simply third root. Considering n beeing greater than 3, the root is declared as the fourth root, fifth root and so on.
In maths, the square root of 11200002 is represented as this: $$\sqrt[]{11200002}=3346.6404049434$$
Additionally it is possible to write every root down as a power: $$\sqrt[n]{x}=x^\frac{1}{n}$$
The square root of 11200002 is 3346.6404049434. The cube root of 11200002 is 223.737801734. The fourth root of 11200002 is 57.850154752977 and the fifth root is 25.694704060137.
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How to get the results of Tally split into two separate lists
This is a kind of a followup to my previous question. I am starting with a long list of plane coordinates, like this one:
lst = {{0.399933, 0.0904781}, {0.449966, 0.545239}, {0.724983, 0.27262},
{0.362492, 0.13631}, {0.681246, 0.0681549}, {0.840623, 0.0340774},
{0.920311, 0.0170387}, {0.960156, 0.00851936}, {0.480078, 0.00425968},
{0.240039, 0.00212984}, {0.370019, 0.501065}, {0.18501, 0.250532},
{0.0925049, 0.125266}, {0.546252, 0.0626331}, {0.273126, 0.0313166},
{0.636563, 0.0156583}, {0.568282, 0.507829}, {0.284141, 0.253915},
{0.39207, 0.626957}, {0, 0}};
Then I apply Tally[] as follows:
lst2 = Tally[Map[Round[#/(0.5)^2] &, lst]];
(The particular values 0.5 and 2 above are just examples)
The result is a "ragged" list of the kind that compilers do not like:
{{{2, 0}, 3}, {{2, 2}, 2}, {{3, 1}, 1}, {{1, 1}, 3}, {{3, 0}, 3}, {{4, 0}, 2},
{{1, 0}, 2}, {{1, 2}, 1}, {{0, 1}, 1}, {{2, 3}, 1}, {{0, 0}, 1}}
My question is: What would be the most efficient way (compilable) to get the same result but in two separate uniform (tensor) lists rather than in one ragged list?
• Do you mean lst2[[All,1]] and lst2[[All,2]] or Flatten/@lst2 or something else? – Vitaliy Kaurov Sep 30 '12 at 7:39
• ... or Transpose[lst2]? – kglr Sep 30 '12 at 8:08
• lst2[[All,1]] and lst2[[All,2]] do the job, but aren't they creating two new lists? That means copying from a long (perhaps 100 mln long) already created list ... – arkajad Sep 30 '12 at 8:47
• "In addition: my Tally[] code does not want to compile." - because Tally[] is not compilable. – J. M. will be back soon Sep 30 '12 at 9:40
• Compilation is not a panacea; it will not (by itself) eliminate copies, nor will it (ordinarily) speed up simple functions like Tally. The main reason one may want to do this is if the results are needed for further processing in compiled code and one seeks to avoid an (expensive) call out of the VM, so consider carefully whether you fall into this category. If so, recall that Sort is compilable, which means that this can be done easily in $O(n \log{n})$ time. It can be done in at best $O(n)$ time, but that will be more difficult to achieve. How efficient does this really need to be? – Oleksandr R. Sep 30 '12 at 11:33
Here is a simple construct that answers the question as asked, although I'm still unclear on what you intend to do with the results and so whether compilation is actually the right approach. In particular, after creating these two arrays separately one cannot then pass them back out of the VM without combining them in some way to form a full-rank tensor, so unless you are working entirely within compiled code there would not appear to be any benefit in doing this. As such, I provide a closure that needs to be inlined into other compiled code in order to function:
tally = Compile[{{lst, _Real, 2}, {delta, _Real, 0}, {j, _Integer, 0}},
Block[{
sorted = Sort@Round[lst/(delta^j)],
itemBag = InternalBag@Most[{0}],
countBag = InternalBag@Most[{0}],
lastItem, count = 0
},
lastItem = First[sorted];
Do[
If[i == lastItem, ++count; Continue[]];
InternalStuffBag[itemBag, lastItem, 1];
InternalStuffBag[countBag, count];
lastItem = i; count = 1, {i, sorted}
];
InternalStuffBag[itemBag, lastItem, 1];
InternalStuffBag[countBag, count];
items = Partition[InternalBagPart[itemBag, All], Length@First[sorted]];
counts = InternalBagPart[countBag, All];
]
];
This should not hold any mystery if you understand the use of InternalBags inside compiled code, which halirutan and I discussed here, so I won't describe its working unless you find anything unclear.
Let's also define a wrapper that returns the result in the same format as Tally and demonstrates the inlining:
wrapper = With[{tally = tally},
Compile[{{lst, _Real, 2}, {delta, _Real, 0}, {j, _Integer, 0}},
Block[{items, counts},
tally[lst, delta, j];
result = Transpose[{items, counts}];
],
"CompilationOptions" -> {"InlineCompiledFunctions" -> True},
]
];
Notice the call out of the VM to set result. This is injurious to performance and prevents parallelization over several inputs, so the above should be taken as an example only.
Since I don't know anything about your use case except that you want to use many different values of delta and j, I'll just present a simple benchmark for comparison with Tally. Note that Tally does not return its results sorted, so most likely it runs in $O(n)$ time, whereas due to the use of Sort (which makes the code much simpler--we do not have to implement our own hash tables, for example) this version runs in $O(n \log{n})$ time.
Let data = RandomChoice[RandomReal[{0, 1}, {100, 2}], 1*^6]. Then,
Timing[wrapper[data, 0.5, 2]; result]
(* -> {0.422, {
{{0, 1}, 59656}, {{0, 2}, 9961}, {{0, 3}, 49722},
{{0, 4}, 50165}, {{1, 0}, 29883}, {{1, 1}, 40065},
{{1, 2}, 70153}, {{1, 3}, 80202}, {{1, 4}, 30174},
{{2, 0}, 9814}, {{2, 1}, 30040}, {{2, 2}, 89744},
{{2, 3}, 50014}, {{2, 4}, 40035}, {{3, 1}, 50041},
{{3, 2}, 69897}, {{3, 3}, 60297}, {{3, 4}, 20018},
{{4, 0}, 30129}, {{4, 1}, 39903}, {{4, 2}, 30106},
{{4, 3}, 50037}, {{4, 4}, 9944}}
} *)
Timing[Sort@Tally[Round[data/(0.5^2)]]]
(* -> {0.907, {
{{0, 1}, 59656}, {{0, 2}, 9961}, {{0, 3}, 49722},
{{0, 4}, 50165}, {{1, 0}, 29883}, {{1, 1}, 40065},
{{1, 2}, 70153}, {{1, 3}, 80202}, {{1, 4}, 30174},
{{2, 0}, 9814}, {{2, 1}, 30040}, {{2, 2}, 89744},
{{2, 3}, 50014}, {{2, 4}, 40035}, {{3, 1}, 50041},
{{3, 2}, 69897}, {{3, 3}, 60297}, {{3, 4}, 20018},
{{4, 0}, 30129}, {{4, 1}, 39903}, {{4, 2}, 30106},
{{4, 3}, 50037}, {{4, 4}, 9944}}
} *)
So, the compiled version is actually better than I expected: for this input it is about twice as fast as Tally. Now let's assume that your usage of the results is implemented in compiled code as well:
parallelWrapper = With[{tally = tally},
Compile[{{lst, _Real, 2}, {delta, _Real, 0}, {j, _Integer, 0}},
Block[{items, counts},
tally[lst, delta, j];
(* Put your code here *)
],
"CompilationOptions" -> {"InlineCompiledFunctions" -> True},
"RuntimeAttributes" -> {Listable}, Parallelization -> True
]
];
AbsoluteTiming[
wrapper[data, {0.5, 1.0, 2.0, 1.5}, {2, 1, 3, 2}]
]
(* -> {0.4843750, Null} *)
This version returns no result because the tallies are not used for anything, but we observe that one can perform the operation in parallel without any difficulties. (I have a quad-core CPU, so the timing is essentially the same as for the serial case when four combinations of delta and j` are run at the same time.)
• +1. It's good that I did not start answering, since this would put to shame whatever I'd probably come up with :-). – Leonid Shifrin Sep 30 '12 at 13:53
• Thanks once more Oleksandr! It will take a while for me to chew your answer through (I am really a newbie that have just ordered the Mathematica book and Trott's Programming book, and waiting for them to arrive). I think I know what I want to do in algorithmic terms. At each step (delta,j) I will also have to sum all tally[[2]]^q (given q) and take Log of the sum. But I will need all of the tally for each next step. I hope I will be able to learn how to do it efficiently from your answer. Thank you! – arkajad Sep 30 '12 at 14:06
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# Solving Systems Of Linear Equations by Substitution Worksheet
Solving Systems of Linear Equations by Substitution worksheet is a resource of useful solutions to linear equations. This kind of solves and finds the right solutions of the equations by means of substitutions of variables. Solving Systems of Linear Equations by Substitution worksheet was developed by Joseph Stokes in the early seventies. It provides online help for those who need to solve equations but not necessarily needed to worry about finding the exact solutions.
The concept of this useful linear equation solver was first introduced in the form of a guide book. The original version of this book had some complications such as a print of continuous functions and equations without solutions. These latter two parts were removed from the later editions. This book was never meant to be an elementary textbook but a tool used by scientists to solve the equation problems on their own.
This valuable software is not actually intended for students or teachers who want to use it for their homework assignments. It can also be used by people who want to solve equations by finding the absolute value of solutions. The solver for Linear Equations by Substitution worksheet is a useful software which can be used by anyone to solve problems related to linear equations.
This solves and finds the right solutions of the equations by means of substitutions of variables. The term substitution refers to replacing one variable in an equation with another. Solving Systems of Linear Equations by Substitution worksheet makes use of multiple-equation solvers and some other technical applications.
There are multi-variable equations, where the values of each variable have unique names. In such cases the Variable Substitution method is used. In most of the cases, the use of multiple variables would be impractical or time consuming.
Another convenient method is using the Two Variable method. In this method, the equations with multiple variables are solved based on the singular value decomposition of their variables. In addition, the solver for Linear Equations by Substitution worksheet also provides a theoretical approach.
This is a useful tool for solving equation problems. However, the effectiveness of this software depends on the correct implementation. If you want to make use of this program for solving the equations then you need to learn about its code and how to understand it properly. As a beginner, the simplest way to use this tool is by working with the first version.
Solving Systems of Linear Equations by Substitution worksheet gives you the correct solutions of the equations and gives you the confidence that you have the right solution. You will also be able to perform several calculations by interpreting the results of the solver. This program also comes with many tips and techniques that you can use to solve your equations. There are several other versions and editions of this resource which you can select for solving the equations by utilizing its handy calculator.
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https://www.bissantz.de/en/bissantz-ponders/differentiation/
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# A quality measure assessing differentiation: € per Pixel
We want to be able to measure the quality of reports. Today, we’ll go into detail. For example, how large, long and wide should graphical elements be? The quality measure “differentiation” answers this question with a resolution in €pP or %pP – Euros or Percent per Pixel.
Most business graphics waste precious space – that’s why I wish nothing less than their utter demise. I personally recommend consolidated graphical tables in their place. With their help we can dramatically increase the data and information density in our reports.
Graphical tables integrate elements which guide the eye but reduce them to a minimum. So how do we determine what’s necessary and what’s not? Let’s take a look at a sample graphical table from the German newspaper DIE ZEIT (March 5th 2009, p. 33). It depicts the per capita consumption of organic food in selected European countries.
This chart could even illustrate a difference of 0.33 euro. In this case, that’s way too exact. The differentiation is too high.
I think that the style and build of the chart above is very good. It’s not three dimensional, the labeling is accurate, and it’s easy to read. The differentiation, however, is exaggerated. And we know that because we measured it. On a computer screen we can assess differences in pixel lengths relatively easily. If we print charts, we can be even more exact. But let’s stay with the pixel, the resolution that is used for design on computer screens.
The chart in DIE ZEIT is printed on paper. After we reconstructed it on a computer, we were able to measure an accuracy of almost 33 cents per pixel. The largest value was drawn with a length of 324 (i.e. 107/0.33) pixels. That means that this chart could even illustrate a value difference of 0.31% (i.e. 0.33 EUR/107 EUR).
If a maximum bar length of 100 also sufficed, each pixel would represent a difference of 1.07 EUR. That is still pretty accurate for the data shown here. The smallest difference is between Italy and France where per capita sales of organic food only vary by 2 euros. In other words, we would only need approximated 54 pixels (i.e. 107/2) to illustrate that – if we wanted to, of course.
Accurate enough and much more compact: a maximum of 100 pixels equals a differentiation of slightly more than 1 EUR per pixel.
Fans of sophisticated data analysis would probably interject at this point saying that this is all very nice but they would prefer to show the differences to a set reference, for example, Germany. That way, we would quickly see that Germany lies in the midrange of European countries regarding the consumption of organic foods.
Our second example below, which shows the number of nuclear power plants worldwide, suffers from too little differentiation (Handelsblatt, April 30th 2009, p. 12). Instead of pixels or dots per inch, it uses some random bar as the smallest unit of measurement. If you use the maximum value of 104 nuclear power plants in the USA as the reference, one bar would be equivalent to 9.5 (i.e. 104/11) nuclear plants. That’s where things start to get a bit inconsistent. 1 bar symbolizes 11 plants in China but 7.5 in the Ukraine. Ouch, that’s a distortion of 47 %. Maybe the number of bars is “normed” based on a certain country and the visualization would be consistent if we knew that. (We have measured that again: In France 1 power plant means 0.102 (=6/59) bars. If you know that and if you round it to integral bars, the rest falls into place.) But even that fact doesn’t make the graphic correct. After all, two bars symbolize anything between 15 and 20.
Bars instead of pixels: Here the finest-grade resolution is 9.5 (104/11). That is way too low – and it’s a botch job besides.
For those of us who cherish pixel aesthetics, this is almost tragic. This botch job didn’t even make the chart more compact. If we gave this chart the same maximum width for graphic elements, it would look like this:
The same maximum width delivers a correct and adequate differentiation.
True is: As a quality measurement, differentiation helps us avoid exaggerations as well as understatements. If we cannot see major differences graphically, our visualization is too vague. On the other hand, we are too precise (and probably waste too much space) if we could identify differences that don’t even exist in the data or fall into the tolerance range due to measurement or collection errors.
Within these boundaries, which differentiation we choose is generally a question of connotation and tactfulness. When we use it, we should inform our readers as well, for example: “Resolution: 1,000 € per Pixel.”
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# Transformer physics question
Suppose I wanted a transformer but I screwed up the math and bought one that is sqrt(2) lower than I need it. Suppose we throw "do it right" out the window for the sake of argument and I want to force this transformer to put out more than it's supposed to.
Given that the transformer is 1:10 and rated for 20kHz and that initially I was going to feed it a 50% duty voltage to get a 10x step up but in fact it's only going to be 5x once I average it over a huge capacitor (pk-pk will be 10x but the DC equivalent will only be 5x), can I increase the output voltage by either increasing frequency or duty cycle? Naively I thought duty cycle would do the trick BUT, at some point, the coil is mostly charged and only being discharged over a very short interval at high duty cycles (at 90% it's only off 1/10th of the time so can't fully discharge due to time constant). So can duty even increase the voltage at all or is anything other than 50/50 hurting the output? I imagine 50/50 gives you maximum flux but also the transformer's time constant might come into play. Maybe you can get a higher duty if the time constant is very low compared to the switching frequency?
Aside from duty, any other tricks to "overclock" the transformer or otherwise get it to put out more voltage through some sort of trickery, even if it comes at the expense of efficiency? Also this is all in the context that the secondary side is being loaded and not open circuit.
You can’t do that I’m afraid the voltage is translated by the turns ratio of the transformer - that’s fixed .
You could amplify your resulting DC signal ( I take it you are using a bridge rectifier and capacitor on the output ) or make the downstream parts work at the lower voltage - all depends on what you are trying to achieve.
If you out a higher voltage in, you’ll get a higher voltage out .
Please explain EXACTLY what you are trying to do.
Your whole discussion is very confusing.
You write of a transformer you bought that you are using at 20kHz, so it is a ferrite core transformer with very low capacity windings. Ok. And you are feeding it with a square wave and using a the transformer to boost the voltage of the square wave. Since transformers are current driven devices, how much current can you supply to the primary of the transformer?
Then you write of DC voltages. Are you rectifying the output of the transformer? Seems to be the case. How are you rectifying it? What circuit? Have you looked at voltage multiplying rectifier circuits?
A schematic of your project would certainly help to clarify the project.
Paul
Sorry let me be a bit clearer.
So I feed a square wave into the primary side at 24V and 50% duty. It can take about 10A on that side. It stands to reason (and I've verified) that the voltage is amplified 10x so I get a 240V pulse out the secondary side but since it's a 50% duty pulse, when you rectify it, you get half that. So I'm not saying I want more than the 240 the ratio is supposed to give me. I'm saying I want more than the 120 I get after rectifying it. The rectification is nothing but a capacitor in my case. Just a big-ass capacitor and that's it. I'm going to use an H-Bridge on the primary side to chop my 20kHz. I want to stay away from adding stuff to the circuit to replace the function of the transformer ( like DC boost/buck converters and the like ) and just rely on the transformer. My whole premise is that the core is designed not fully saturated so I know it can be saturated further. This extra "power" ought to be able to be expressed as voltage in theory, should it not? I'm looking for a way to extract the most voltage I can out of the transformer in a rectified form and am asking about which parameters might be altered to achieve that. Duty cycle would seem obvious but if you look at the extreme cases, 1% duty is essentially off so how can there even be an alternating magnetic field in the steel and 99% is essentially on so again, there's no real field activity to make the transformer work. I therefore assumed that there would be some cutoff point where changing duty actually makes output drop off.
Gahhhrrrlic:
Sorry let me be a bit clearer.
So I feed a square wave into the primary side at 24V and 50% duty. It can take about 10A on that side. It stands to reason (and I've verified) that the voltage is amplified 10x so I get a 240V pulse out the secondary side but since it's a 50% duty pulse, when you rectify it, you get half that. So I'm not saying I want more than the 240 the ratio is supposed to give me. I'm saying I want more than the 120 I get after rectifying it. The rectification is nothing but a capacitor in my case. Just a big-ass capacitor and that's it. I'm going to use an H-Bridge on the primary side to chop my 20kHz. I want to stay away from adding stuff to the circuit to replace the function of the transformer ( like DC boost/buck converters and the like ) and just rely on the transformer. My whole premise is that the core is designed not fully saturated so I know it can be saturated further. This extra "power" ought to be able to be expressed as voltage in theory, should it not? I'm looking for a way to extract the most voltage I can out of the transformer in a rectified form.
So, you are charging a capacitor on 1/2 of the cycle and discharging it on the other half and wondering about the net results?
What are you attempting to create?
Paul
The capacitor is just to get DC. It'd be like starting with a square wave going through a simple wire and connecting a cap to it. You just get half the pk-pk value of the square wave with negligible ripple. I'm trying to make AC in the long run but the voltage is too low for the RMS to be what I want it to be so I need higher rectified DC on the secondary side.
Gahhhrrrlic:
The capacitor is just to get DC. It'd be like starting with a square wave going through a simple wire and connecting a cap to it. You just get half the pk-pk value of the square wave with negligible ripple. I'm trying to make AC in the long run but the voltage is too low for the RMS to be what I want it to be so I need higher rectified DC on the secondary side.
The you obviously need a diode in series with the capacitor to stop the discharge of the capacitor!!!!
Paul
Well that'd keep the cap from discharging but I need to put a load on the secondary eventually so floating the voltage up to the peak is a cheat isn't it?
I mean say I put a 100 ohm resistor across the secondary. That's going to pull the voltage down to a steady state of 1/2 the pk voltage, is it not? I need to do something the changes the "area under the curve" so that no matter what you do to load the circuit, it will settle at a higher value. The answer must be on the primary side, I think.
Gahhhrrrlic:
Well that'd keep the cap from discharging but I need to put a load on the secondary eventually so floating the voltage up to the peak is a cheat isn't it?
I mean say I put a 100 ohm resistor across the secondary. That's going to pull the voltage down to a steady state of 1/2 the pk voltage, is it not? I need to do something the changes the "area under the curve" so that no matter what you do to load the circuit, it will settle at a higher value. The answer must be on the primary side, I think.
Who can say! You have never told us what you are attempting and how an Arduino is involved.
Paul
Lol. I'm just asking about transformers in the general electronics topic. If I go on and on about Arduinos and other discrete circuitry, the focus will get entirely derailed. The question is precisely about transformer physics and what is possible with a fixed piece of hardware by varying the electrical input parameters. I could be controlling it with a Raspberry Pi or flipping a knife switch with my hand and it would make no difference.
Gahhhrrrlic:
Lol. I'm just asking about transformers in the general electronics topic. If I go on and on about Arduinos and other discrete circuitry, the focus will get entirely derailed. The question is precisely about transformer physics and what is possible with a fixed piece of hardware by varying the electrical input parameters. I could be controlling it with a Raspberry Pi or flipping a knife switch with my hand and it would make no difference.
Ok, care to share a data sheet on this transformer you are using? I hope you are using an oscilloscope to view the in put and output wave forms. They will not be as you have described them.
Paul
Ferrite 3C97
0.41T
280mm^2
f = 20kHz
Np = 1.75
Ns = 17.55
Ip = 16.17A (18 AWG / 16 AWG)
Is = 1.47A (26 AWG)
I have already scoped it using 6V and it's about 10x (60-66V out) with a sort of sawtooth'ish waveform at 20k.
Gahhhrrrlic:
Ferrite 3C97
0.41T
280mm^2
f = 20kHz
Np = 1.75
Ns = 17.55
Ip = 16.17A (18 AWG / 16 AWG)
Is = 1.47A (26 AWG)
I have already scoped it using 6V and it's about 10x (60-66V out) with a sort of sawtooth'ish waveform at 20k.
Then your transformer is not up to transferring a square wave at 20k. Do you understand why?
Paul
No. I had this custom made by a transformer maker. Unless he screwed up.
In any case, I’m not particularly concerned with whether or not the transformer was properly spec’d. This is something I can correct later on (let’s call it Rev 2.0). For now the laws of magnetic fields apply equally to all transformers whether they are properly designed or ripped from a random TV power supply. The question is how to take a fixed quantity such as the hardware above, and stimulate it enough to yield a higher average voltage on the secondary side. As mentioned, I can speculate that the 2 main parameters I have control over, frequency and duty, may accomplish this but I’d rather somebody with practical experience confirm or deny this line of reasoning.
Changing the frequency changes the flux.
Changing the duty cycle changes the flux.
With the same turns ratio, to get more voltage out, you’ve got to put more voltage in.
The first reply gave you the correct answer to your question: No.
Everything since that post has been about you refusing to accept the physical laws of electromagnetism. What’s next? Are you going to ask how to get 100 watts out of your transformer with only 50 watts in?
The horse is dead. Stop beating it.
Actually the reality is quite the opposite. I ask a direct question (even the title says transformer physics) and get a page of irrelevant, tangential questions about arduinos and discrete circuitry, which, is why I don't tend to give great quantities of detail so as not to invite wasted time and effort. And at what point did I deny the laws of physics? By your own admission, frequency and duty "change" flux, something I theorized earlier in my posts. Change it how? You mean increase it? As in further saturation of the steel? Does that not produce more power, which is a commodity flexibly expressed as either voltage or current? My question was about when such tweaks cease to produce a positive return, assuming that the time constant of the transformer becomes a bottleneck at extreme duties, be they small or large, or when the frequency is so high it doesn't allow the magnetic field to grow and collapse fully. You would seem to be contradicting yourself if your short answer is, no and your long answer is, both parameters change the flux. I happen to believe that changing these parameters does increase voltage. TINA seems to agree with me as well. Forgive me for not putting blind faith in computer models however, and seeking confirmation here. Rather than chiming in on page 2 to tell me to STFU, you should go back a page and see how I was repeating myself over and over to try to end the conversation rather than perpetuate it. But then maybe my interpersonal skills are no match for the convoluted way in which people interpret questions these days.
Gahhhrrrlic:
No. I had this custom made by a transformer maker. Unless he screwed up.
Why (and how) are you getting custom-made transformers when you don't even know how they work?
Gahhhrrrlic:
Actually the reality is quite the opposite. I ask a direct question (even the title says transformer physics) and get a page of irrelevant, tangential questions about arduinos and discrete circuitry, which, is why I don't tend to give great quantities of detail so as not to invite wasted time and effort. And at what point did I deny the laws of physics? By your own admission, frequency and duty "change" flux, something I theorized earlier in my posts. Change it how? You mean increase it? As in further saturation of the steel? Does that not produce more power, which is a commodity flexibly expressed as either voltage or current? My question was about when such tweaks cease to produce a positive return, assuming that the time constant of the transformer becomes a bottleneck at extreme duties, be they small or large, or when the frequency is so high it doesn't allow the magnetic field to grow and collapse fully. You would seem to be contradicting yourself if your short answer is, no and your long answer is, both parameters change the flux. I happen to believe that changing these parameters does increase voltage. TINA seems to agree with me as well. Forgive me for not putting blind faith in computer models however, and seeking confirmation here. Rather than chiming in on page 2 to tell me to STFU, you should go back a page and see how I was repeating myself over and over to try to end the conversation rather than perpetuate it. But then maybe my interpersonal skills are no match for the convoluted way in which people interpret questions these days.
Rant much? People here can barely understand what you're trying to do, let alone if it's even sensible to do that.
The rectification is nothing but a capacitor in my case. Just a big-ass capacitor and that's it.
For example, it's obvious from this that you don't even know what rectification means. Capacitors don't rectify. That's what diodes do.
My whole premise is that the core is designed not fully saturated so I know it can be saturated further. This extra "power" ought to be able to be expressed as voltage in theory, should it not?
What "theory"? Pulling words out of your butt is not a theory. Why do you think the core isn't being fully saturated? More likely it's designed to be right on the edge of or a little over the saturation point, because if there's room that means a bigger core which takes more material to build. You might not have much room to go up.
A transformer's output voltage is COMEPLETELY determined by the primary voltage multiplied by the turns ratio. That is an absolute limit. If you want more voltage out of the secondary, the ONLY thing you can do is increase the voltage applied to the primary. Changing frequency will not increase your output voltage. Changing the duty cycle will not increase your output voltage. This was pointed out to you in literally the first response.
I happen to believe that changing these parameters does increase voltage.
"What you believe" means nothing. You are wrong.
Here is a high level description of the physics of a transformer. When magnetic flux changes in a coil, it induces a voltage. Conversely, when voltage is applied to a coil is causes a change in flux. Specifically, a continuous voltage will cause a continuous, linear increase in flux that will theoretically continue forever. In reality parasitic resistance and saturation will get in the way of that. The exact relationship of the voltage to that change is determined by the core geometry and material, as well as the number of turns (more turns = slower changing flux).
That ramping flux is also applied to the secondary of the transformer, which induces a voltage across it. In a mirror of what happened to the primary coil, the voltage across the secondary is determined by how much the flux is changing. Since they have the same core geometry, the what's left to determine the voltage is the turns (more turns = more voltage).
Changing the transformer's frequency will not change the rate of flux change, since that is determined by the value of the voltage and not how quickly the voltage changes. Changing duty cycle can only decrease the output voltage, since straying off of a 50% duty cycle gives it more time at one level to drift into saturation. Since you're not changing the actual level of the primary voltage, it doesn't increase the output voltage.
You're at risk of killing yourself with mains level power here. I hope you are taking more precautions than you think are necessary.
Hi,
The rectification is nothing but a capacitor in my case. Just a big-ass capacitor and that's it.
What makes you think just a capacitor alone will convert your 240AC to DC?
If you have a squarewave like in the diagram below.
If you connect like below with a suitable hi-speed diode and suitable voltage rated capacitor you will get near the DC voltage of 240Vdc
You have to have at least one diode to rectify to DC.
If your capacitor is a polarized type I am surprised you have not had it explode or overheat.
Do you have a scope?
Tom...
PS. Can you please post a diagram of your transformer and connections as well a a picture of your transformer?
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## Presentation
"Ideas seem to be floating around in space. It happens again and again that discoveries are made at the same time on different continents. People say that the time was simply ripe for these discoveries. But that means as little as explaining the secrets of animals by that mysterious word, 'instinct'. THE IDEA IS SIMPLY THERE, ripe, ready to be harvested. You only need to switch over to 'receive,' and it will come in.
"This theory seems to be correct, otherwise an experiment based upon this principle which I have tried many times would not have succeeded so often.
"May I try it with you? Now, you must not chide me if it doesn't work. The probability is about 70%.
"In order not to influence your thoughts ahead of time in any particular direction, I will not tell you ex-actiy where this whole thing is leading.
"First we need an idea. Since we are merely performing an experiment, it must be a fairly simple thought, which we will promote to the rank of an idea. For example, it could be a simple geometrical design, such as a circle or a triangle, a letter, or - let's say - a two-digit number. Please don't say anything.
"Have you thought of something? Good, here are two pieces of paper and a pencil. I will draw a point of concentration in the center of one of the pieces of paper. Imagine yourself surrounded by the universe of thought, all of the thoughts that have ever been thought and will ever be thought. This point has enough room for them because thoughts do not occupy any space.
'Take the paper and the pencil. Look at the concentration point and switch over to 'receive'. Your mental eye will perceive either the geometrical design, the number, or the letter which you are thinking of or which you will receive. When you have this picture in your mind, project it upon the paper. If you have enough mental power, you will actually see your thought on the paper.
"Do you .see it? What is it? A Y? Good. Write it with large, bold strokes upon the paper, just as you see it there.
"Good. Roll the piece of paper into a ball and drop it into your pocket.
"Now you have received the idea ,'Y, and written it down, which 'fixes' it, making it a 'real' object. It is floating around in space once more."
Now the waiter, the bartender, or some other person who has not seen the preceding action is called over.
"Please, sir, would you help us with a mental experiment? I have drawn a so-called concentration point on this piece of paper. Otherwise, the paper is completely blank, as you can see. Now, I would like for you to take the piece of paper over there, relax completely, and concentrate on this point.
"At first something will appear before your mental eye, and then gradually it will become clearly visible. It will be a something such as a simple geometrical design, a number, or a letter. When you see it clearly, draw it exactly as you see it on the piece of paper with the pencil. Then crumple the paper into a ball and place it in your pocket, and come back over here."
After the pieces of paper have been exchanged and read, the performer says, "Didn't I tell you that ideas float around in space?"
If you can perform the trick in a restaurant with the waiter or with a guest at another table, that is the ideal situation. It will also work if you use a stranger off the street!
I performed the trick (LIVE!) on a broadcast from Baden-Baden. The "stranger" was Peer Schmidt2 who ivas filming in the next studio, and who had to be orought into our studio while he was in the middle of filming. The trick worked, and Peer Schmidt had no idea or explanation for the miracle. He was as amazed as the rest of audience.
When you learn the secret of this super trick, at first you will be horrified at its boldness, as everyone else is. BUT THERE IS SCARCELY ANY MENTAL TRICK THAT WORKS MORE RELIABLY IF YOU FOLLOW THE SIMPLE BUT HARD AND FAST RULES.
The actual secret was an accidental discovery. The secret is in the paper. This paper must have a rough or matte surface. The best types are various types of art paper, 3 but absorbent paper will also work.
If you write on this type or paper with your finger
2 Peer Schmidt is a German television personality. BP
3 Coated paper works best. When this part oft] ne series was Issued in German, Punx furnished a small pad of paper with it. I took this to a major art supply store to have it identified. It was a type of "clay-coated" paper, wnlch Is available in a large book of blank paper. It is called Marker Sketch Book by Signature Edition. There are 160 8 1/2" x 11" pages, and It sells for about \$9.00 (in 1989).
Corinda describes this paper as looking like white paper coated with poster paint. That is quite accurate. Some people use thermal print paper or copier paper, but these are not the best choice. The surface change is subtle, but definite. BP
nail, a type of writing will appear that is visible or invisible, DEPENDING ON HOW IT IS HELD TO THE LIGHT!!
After writing on it with your fingernail, you can handle the paper quite naturally, as long as you keep it in motion. When you hold it up to the light - which is an especially convincing way to show that it is "blank" - ABSOLUTELY NOTHING IS VISIBLE.
If the waiter or the stranger takes the paper to a place with good light such as a table lamp or a window, and handles it as you tell him to, he must see the writing clearly. This will make it virtually impossible for him to make a mistake.
During the countless performances in which I have been able to perform this test, I have had it go wrong only twice. Once, Jay Palmer was acting as "the waiter" and Betty Davenport as "the second spectator." Palmer was, as usual, not sober and wrote some kind of nonsense, without having actually seen anything. Betty Davenport did not go to the place to which she had been directed to look at the paper, as any normal spectator would. Instead, she stayed right where she was standing, looked at the paper, and saw the secret writing. She held it under Gilly Davenport's nose and said, "Look Daddy, very clever!"
This will teach you to perform the trick with no more than ONE magician. The disinterested spectator, the waiter, must be a lay person, because only they will behave normally. You know in any case that magicians are the least suitable spectators, and not only for mental magic. You cannot in any case test the effectiveness of a trick with an audience of magicians!
Also, you must be careful to have the spectator sit OPPOSITE you, never to the side! Otherwise he will see you out of the corner of his eye when you do the writing.
## Practical Mental Influence
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## Multiplication Fact Family Worksheets
Multiplication Fact Family Worksheets. Ad the most comprehensive library of free printable worksheets & digital games for kids. Multiplication and division have an inverse relationship, so if a student knows that 9 × 8 = 72.
Free Second Grade Math Worksheets Activity Shelter from www.activityshelter.com
Ad the most comprehensive library of free printable worksheets & digital games for kids. Live worksheets > english > math > fact families > fact families. Grade 3 fact family worksheets feature multiplication and division problems, as well as.
### Free Second Grade Math Worksheets Activity Shelter
Grade 5 multiplication worksheets multiply by 10, 100 or 1,000 with missing factors. Using the concept of 3 for free, students use the fact family triangles to explore turn around. One minute timed worksheets for multiplication and division. Multiplication fact families worksheets 14.
Source: www.worksheeto.com
Multiplication and division have an inverse relationship, so if a student knows that 9 × 8 = 72. One minute timed worksheets for multiplication and division. A fact family is a set of three related multiplication and division facts. Great resource for practicing relationships between multiplication and. Live worksheets > english > math > fact. Grade 3 fact family worksheets feature multiplication and division problems, as well as. Ad the most comprehensive library of free printable worksheets & digital games for kids. Multiplication fact families worksheets 14. Live worksheets > english > math > fact families > fact families. Multiplying (1 to 9) by focus numbers.
Source: www.activityshelter.com
Great resource for practicing relationships between multiplication and. Ad the most comprehensive library of free printable worksheets & digital games for kids. A fact family is a set of three related multiplication and division facts. Using the concept of 3 for free, students use the fact family triangles to explore turn around. Multiplying (1 to 9) by focus numbers. Grade 3 fact family worksheets feature multiplication and division problems, as well as. One minute timed worksheets for multiplication and division. Multiplication fact families worksheets 14. Live worksheets > english > math > fact. This multiplication and division fact family activity includes math task cards can.
Source: www.worksheeto.com
Great resource for practicing relationships between multiplication and. Multiplication and division have an inverse relationship, so if a student knows that 9 × 8 = 72. If you’re looking for fact family. One minute timed worksheets for multiplication and division. This multiplication and division fact family activity includes math task cards can. Multiplying (1 to 9) by focus numbers. Grade 5 multiplication worksheets multiply by 10, 100 or 1,000 with missing factors. Ad the most comprehensive library of free printable worksheets & digital games for kids. A fact family is a set of three related multiplication and division facts. Live worksheets > english > math > fact.
This multiplication and division fact family activity includes math task cards can. Grade 3 fact family worksheets feature multiplication and division problems, as well as. A fact family is a set of three related multiplication and division facts. Ad the most comprehensive library of free printable worksheets & digital games for kids. Using the concept of 3 for free, students use the fact family triangles to explore turn around. If you’re looking for fact family. Multiplication and division have an inverse relationship, so if a student knows that 9 × 8 = 72. Grade 5 multiplication worksheets multiply by 10, 100 or 1,000 with missing factors. One minute timed worksheets for multiplication and division. Great resource for practicing relationships between multiplication and.
Source: www.activityshelter.com
This multiplication and division fact family activity includes math task cards can. Grade 5 multiplication worksheets multiply by 10, 100 or 1,000 with missing factors. Grade 3 fact family worksheets feature multiplication and division problems, as well as. Great resource for practicing relationships between multiplication and. If you’re looking for fact family. A fact family is a set of three related multiplication and division facts. Multiplication and division have an inverse relationship, so if a student knows that 9 × 8 = 72. Using the concept of 3 for free, students use the fact family triangles to explore turn around. Live worksheets > english > math > fact families > fact families. One minute timed worksheets for multiplication and division.
If you’re looking for fact family. Great resource for practicing relationships between multiplication and. Multiplication fact families worksheets 14. Grade 3 fact family worksheets feature multiplication and division problems, as well as. Multiplication and division have an inverse relationship, so if a student knows that 9 × 8 = 72. Using the concept of 3 for free, students use the fact family triangles to explore turn around. Multiplying (1 to 9) by focus numbers. Live worksheets > english > math > fact. This multiplication and division fact family activity includes math task cards can. Ad the most comprehensive library of free printable worksheets & digital games for kids.
Source: www.unmisravle.com
Using the concept of 3 for free, students use the fact family triangles to explore turn around. Great resource for practicing relationships between multiplication and. Multiplying (1 to 9) by focus numbers. Ad the most comprehensive library of free printable worksheets & digital games for kids. Multiplication and division have an inverse relationship, so if a student knows that 9 × 8 = 72. A fact family is a set of three related multiplication and division facts. If you’re looking for fact family. This multiplication and division fact family activity includes math task cards can. Live worksheets > english > math > fact families > fact families. Grade 3 fact family worksheets feature multiplication and division problems, as well as.
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cancel
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Helper II
## Branching from approvals
I have a flow that contains an approval action so I have a "Start and wait for an approval" action. This approval step sends it to multiple people, so in the "Assigned to: " spot I enter the name of, say 3 approvers. But once they make a decision (either approve or reject), how do I branch off my next steps depending on their answers? I need to account for all permutations
A=Approve
R=Reject, so
A,A,A
A,A,R
A,R,R
R,R,R
Also doesn't matter the order (ie. Who approves and who rejects) only when there's a certain number of approves and rejects.
1 ACCEPTED SOLUTION
Accepted Solutions
Community Support
Hi @ay2,
You could initialize an integer variable to count the number of approve, after all approver responded, use Apply to each action to loop through each response, if it is equals to approve, increase the variable. After it, you could create a condition to determine the variable whether greater than the number you want, if it met, go to next steps:
Best Regards,
Community Support Team _ Lin Tu
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
2 REPLIES 2
Super User
Are the three approvals always going to the same place? I wonder if it might worthwhile to have 3 parallel branches each with a single person, and then aggregate the responses in the end with a Switch to determine the path based on the aggregated response.
Just a thought.
-Ed
If you liked this reply, please give it a thumbs up! If this reply has answered your question or resolved your challenge, please consider marking it as a Solution. This helps other users find it more easily via search.
Community Support
Hi @ay2,
You could initialize an integer variable to count the number of approve, after all approver responded, use Apply to each action to loop through each response, if it is equals to approve, increase the variable. After it, you could create a condition to determine the variable whether greater than the number you want, if it met, go to next steps:
Best Regards,
Community Support Team _ Lin Tu
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
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1. ## [SOLVED] counting problem - programs for radio station
Can someone tell me if my solution is correct? There is a different answer in the book.
Thanks!
**********
Beethoven wrote 9 symphonies, and Mozart 27 piano concerts.
A station manager decides that on each successive night (7 days per week), a Beethoven symphony will be played, followed by a Mozart piano concerto, followed by a Schubert string quartet (of which there are 15). For roughly how many years could this policy be continued before exactly the same program would have to be repeated?
MY SOLUTION
B=9, M=27, Q=27
Using the Multiplication Priniciple
BxMxQ = 9x27x15 = 3645 3-day sequences,
i.e. 3645 sets consisting of 3 days each.
3645 sets x 3 days each set = 10935 days
10935 / 365 days in a year = 29.95 years or approximately 30 years of non-repeating 3 day sequences.
2. Why a three day sequence ?
Aren't the three pieces played on the same night ?
3. ## You're right.
Originally Posted by BobP
Why a three day sequence ?
Aren't the three pieces played on the same night ?
You are right. So the answer is 3654 days. Thanks.
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Call Now: (800) 537-1660
June 20th
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# Powers
## Is a × a × a = 3 a ? Is a 3 = a × 3?
Or, should we guess what is?
In mathematics, each symbol, (e.g. a ×3, 3a , a 3 ) has a uniquely defined meaning. a × 3 has been arbitrarily chosen as shorthand for a + a + a . It cannot mean anything else!
a 3 has been, equally arbitrarily, chosen as shorthand for a × a × a. It means precisely this.
3 × a is shorthand for 3a, but note that the omission of the × sign is possible only between a number and a letter (and only in that order). Omitting it from any other combination would lead to something different from a product, e.g. 33 is not 3 × 3 ; aa or a3 means a single variable consisting of two symbols each.
Always consider the (unique!) meaning of the maths you write.
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## Yic
Connection rotation, 0
FIGURE 23.5 Four-parameter power model of semirigid connection.
Three prediction moment-rotation equations were incorporated with Kishi and Chen's database: the polynomial model shown in Equation 23.2, a modified exponential model [17], and the three-parameter power model shown in Equation 23.3. The validity of either the three-parameter power model or the modified experimental model for practical use was examined by Kishi et al. [18], who concluded that the three-parameter power model adequately describes the experimental test data for modeling the connection moment-rotation behavior for practical use.
### 23.3.6 Connection Classification
In practice, steel framed structures are to be designed in accordance with one of the three types of steel framing constructions: continuous construction, simple construction, and semirigid construction. Therefore, designers must be knowledgeable as to when the connections can be assumed to be rigid, semirigid, or flexible. The purpose of connection classification is to select a suitable basis on which to conduct the frame analysis and design. Based on the stiffness and strength of a connection, the connection can be classified as follows:
• Stiffness criterion. A connection is categorized as rigid, semirigid, and flexible based on the ratio of its rotational stiffness (R) to the stiffness of the beam (EIIL) it connects, RL/EI, where L and EI are the length and bending rigidity of the beam, respectively.
• Strength criterion. The most important aspect of the strength of a connection is its relationship to the strength of the connected beam. Based on comparison of the moment resistance (Mn) of a connection to the plastic moment resistance (Mp) of the beam it connects, the connection can be classified as a full strength, partial strength, or flexible connection.
Table 23.2 summarizes three classifications of connections presented by Bjorhovde et al. [19], Eurocode 3 [4], and LRFD [2] based on the foregoing criteria. The initial stiffness of the connection (R;) was adopted in classifications by Bjorhovde et al. and Eurocode 3, while the secant stiffness of the connection (Rs) was used in the LRFD classification. As shown in Table 23.2, the secant stiffness, Rs, is defined on the basis of either the moment, Ms, or the rotation, 0s, which would occur under the applied loads. The LRFD classification suggests that two distinct values of secant stiffness should be adopted to characterize the connection behavior under the two limit states of serviceability and ultimate load. The moment resistance of the connection, Mn, can be determined on the basis of an ultimate limit-state model of the connection or from tests. When a semirigid connection does not exhibit a plateau in its moment-rotation relationship, the moment resistance of the connection is defined at the rotation dn = 0.02 radians [2,12,13,20].
In addition to the strength and stiffness of the connection, a third important characteristic that should be considered in connection classification is the ductility of the connection. Connection ductility
Connection rotation, 0
TABLE 23.2 Summary of Connection Classification Systems
Bjorhovde classification [19]
Eurocode 3 classification [4]
AISC LRFD classification [3]
0 0
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The Theory \ Details \ general information \ prices correction
# General Information
## Correction of historical prices
There are several cases when the price of an asset changes while its value for an investor stays the same. In such situations it is necessary to impose corrections to the corresponding time series. Correction is necessary in the case of the following events:
• Dividend payments in Stocks and Mutual Funds
• Splits and Reversed Splits in Stocks and Mutual Funds
• Rollovers in Futures
The general correction algorithm is described below.
Imagine that there is a gap on the price chart, which is induced by one of the above-listed events. Let old and new denote the prices just before and after the break point respectively.
Then the correction coefficient is calculated by the following formula:. All the prices before the break point are then multiplied by .
In case of gaps let’s start by sorting them from the latest to the earliest. The corresponding correction coefficients denote by ,...,.
Let’s multiply by all the prices that lie between the second and the first break points. Then multiply the prices between the third and the second break points by . Subsequently performing the analogous
operation over all other time intervals bounded by adjacent break points, we will finish at the section that lies to the left of the earliest break point. It is easy to see that its prices must be multiplied by .
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# How do I deal with additional input information other than images in a convolutional neural network?
I try to convert a game state of a board game into the input for a convolutional neural network. A convolutional neural network is useful because the players have to place items on the board, and the convolutional neural network can take advantage of that spatial structure. Therefore I can describe the board well with a binary feature plane for each player (1 if there is an item on the board and 0 if not).
The players can not only place objects but also collect cards. A player can own a maximum of 19 of five different card types. How many cards a player has and what type they are, is important information for the neural network, but I cannot describe this with another feature plane. This has nothing to do with spatial structure. So how do I give the convolutional neural network such additional information, for example that the player has 6 cards of type A, as input?
There are also "places" where a player can place his items. These places have a number (so to speak how good the place is) from 2 - 12. I can describe these places again with a feature plane. But I wonder if the network distinguishes well enough between two numbers like 6 and 7? I could imagine that it can distinguish much better between ones and zeros.
• Please consider upvoting and marking as correct the answers you find useful. – noe Nov 1 '20 at 11:49
I think there are three questions here:
### How to incorporate non-spatial information into the network?
When combining different information modalities, a typical approach is to do it at the internal representation level, that is: the point where you lose the spatial information (normally with a flatten operation) after the convolutions. You can have your extra information be processed by an MLP and the result be combined with the representations obtained by the convolutional layers by concatenating both.
### How to represent the cards as input to the network?
In order to represent the card a user has, you can represent them as discrete elements (i.e. tokens), just like text is usually handled in neural networks. This way, you can use an embedding layer, which would receive as input the index of the card. As the user can have any number of cards, you could use an LSTM. In order to represent the "end of the card collection", you can have a special token, and yet another one to represent "padding", which would be useful to create minibatches with different number of cards.
### How to represent the places?
You should decide if these are better represented as discrete or continuous values. Or maybe just try both options and choose the best performing one. For continuous features, you could add, as you suggested, another feature place. For discrete features, you would just have an embedding layer, and then concatenate the output to the other channels.
Update: Some clarifications:
• A "token" is a term used in NLP to refer to a value which is discrete, that is, the number of values it can take is finite, normally small. In your case, the different values that a card token can take is 19. Usually, we refer to tokens by the index they occupy in the list of all possible values.
• In order to represent discrete values in neural networks, we normally represent each different value as a fixed-size vector.
• An embedding table is just a table with the fixed-size vectors used to represent your discrete elements. The embedding layer is normally the first in the network architecture. It receives as inputs token indexes and outputs their associated vectors. The entries of the embedding table are updated during the backpropagation process.
• You don't "concatenate an MLP with a convolutional network", you concatenate their outputs. Specifically, once the output of the last convolutional layer is computed, you normally "flatten" it, meaning you remove the spatial information and just place the output tensor elements in a single-dimension vector. That vector is what you concatenate with the output of the MLP, which is also a single-dimension vector (apart from the minibatch dimension).
• How should I concatenate a MLP with a convolutional neural network? Also I don't really understand your answer to my second question as I don't know what you mean by tokens and what an embedding layer is. Could you provide some resources from which I can learn such methods of incorporating non-spatial information into convolutional neural netowoks? – Hey Hey Nov 16 '20 at 20:39
• I added some clarifications. Hope they help. – noe Nov 16 '20 at 21:12
• Thank you. That clarified a lot. But I still am curious about how to concatenate two output vectors? I feel like averaging the output vectors of both networks isn't the right thing to do since both networks only have part of the information of the game. Also should I nomalize the card indexes against 19 (card_index / 19)? – Hey Hey Nov 17 '20 at 11:39
• To concatenate two vectors, you just put them together one next to the other, forming a new vector that has as dimensionality the sum of the original vectors' dimensionalities, e.g. $concat([3.0,2.1,67.0], [0.1, -2.5]) = [3.0,2.1,67.0,0.1, -2.5]$. The card indexes must not be normalized; we want the discrete indexes to be used to index the embedding table, not floating point numbers. – noe Nov 17 '20 at 14:22
• My conv network outputs move probabilities (one hot encoding), so I don't think concatenating the probabilities would be useful, since there is a probability for each action a player can make. – Hey Hey Nov 17 '20 at 15:03
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651 lb 531 lb 386 lb 205 lb a b a b F F P P = = = = Name______________________ ESM 2104 With Answers Test #1 September 12, 2006 Closed book, but you may use one 8½”x 11" formula sheet (one side only). There are 15 questions, equally weighted. 1. Fill in your name on this exam sheet. 2. Fill in your name and ID number on the orange opscan form 3. On the upper right-hand corner of the this test is a Form Letter. Fill in your Form letter on your opscan Sign the following pledge: I have neither given nor received unauthorized aid on this exam. ___________________________ 1. The 600 lb force is to be decomposed into oblique components along the lines a-a and b-b. The component F a along a-a is most nearly 1. 205 lb 2. 300 lb 3. 386 lb 4. 460 lb 5. 531 lb 6. 564 lb 7. 600 lb 8. 651 lb 9. 715 lb 10. 1648 lb 2. The projection of the 600 lb force along the line b-b is most nearly 1. 205 lb 2. 300 lb 3. 386 lb 4. 460 lb 5. 531 lb 6. 564 lb 7. 600 lb 8. 651 lb 9. 715 lb 10. 1648 lb
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3560 lb in O = − M k 45.0 in y = 3. The moment of the 200-lb force F about point O is most nearly 1.
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# Find the inverse laplace trans. displaystyle{F}{left({s}right)}=frac{10}{{{s}{left({s}^{2}+{9}right)}}}
Find the inverse laplace trans.
$F\left(s\right)=\frac{10}{s\left({s}^{2}+9\right)}$
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Step 1
Given,
$F\left(s\right)=\frac{10}{s\left({s}^{2}+9\right)}$
Find the inverse Laplace transform of this function.
Step 2
$F\left(s\right)=\frac{10}{s\left({s}^{2}+9\right)}$
$=\frac{10}{9s}-\frac{10s}{9\left({s}^{2}+9\right)}$
Taking inverse Laplace transform of both sides,
${L}^{-1}\left[F\left(s\right)\right]={L}^{-1}\left[\frac{10}{9s}-\frac{10s}{9\left({s}^{2}+9\right)}\right]$
Then, $f\left(t\right)={L}^{-1}\left[\frac{10}{9s}\right]-{L}^{-1}\left[\frac{10s}{9\left({s}^{2}+9\right)}\right]$
$=\frac{10}{9}{L}^{-1}\left[\frac{1}{s}\right]-\frac{10}{9}{L}^{-1}\left[\frac{s}{{s}^{2}+9}\right]$
$=\frac{10}{9}{L}^{-1}\left[\frac{1}{s}\right]-\frac{10}{9}\cdot {L}^{-1}\left[\frac{s}{{s}^{2}+{3}^{2}}\right]$
Step 3
Use the formula such that
${L}^{-1}\left[\frac{1}{s}\right]=t$
${L}^{-1}\left[\frac{s}{{s}^{2}+{a}^{2}}\right]=\mathrm{cos}\left(at\right)$
$f\left(t\right)=\frac{10}{9}\cdot t-\frac{10}{9}\cdot \mathrm{cos}\left(3t\right)$
$=\frac{10}{9}\left[t-\mathrm{cos}\left(3t\right)\right]$
Step 4
Hence,
$f\left(t\right)=\frac{10}{9}\left[t-\mathrm{cos}\left(3t\right)\right]$
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Problem Description
Rasen had lost in labyrinth for 20 years. In a normal day, he found a bright screen. There were 4 points labeled by ‘A’ , ‘B’ , ‘C’ , ‘D’, and rasen could drag these point. And two points ‘E’ , ‘F’ moved. Rasen found that ‘E’ is the middle point of ‘A’ and ‘B’, and ‘F’ is the middle point of ‘C’ and ‘D’. Near the screen there was a marble slab.There were a list of the distance of AB , BC , CD , DA and EF. Rasen also found that the distance of these edge of the points in screen showed at the same time he drop the points. He wanted to know what will happen if the distances in screen are same with the number in slab.
Input
The first line of input contains only one integer T(<=50000), the number of test cases.
Each case contains five float number, indicating the distance of AB , BC , CD , DA , EF.（0<=distance<=10000）
Output
For each test, first print a line “Case #i:”(without quotes),with i implying the case number, then output the coordinates of A,B,C,D four points. Answer will be considered as correct if the length got from your output (the spj will use double to get the point, and the distance from two points will calculate in the way of sqrt((x1-x2)^2 +(y1-y2)^2) ) and the length given is less than 10-4.
(It guarantees that there exists a solution. If there are many solutions, output any one of them.)
Sample Input
1
1.000000 1.000000 1.000000 1.000000 1.000000
Sample Output
Case #1:
0.000000 0.000000
1.000000 0.000000
1.000000 1.000000
0.000000 1.000000
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# Reasoning Quiz On Seating Arrangement Linear Day 13 Bag
## Reasoning Quiz On Seating Arrangement Linear Day 13 Bag
Study the following information carefully to answer the given questions.
Ten persons from ten different cities viz. Delhi, Jaipur, Patna, Indore, Mangaluru, Chennai, Hyderabad, Bengaluru, Raipur and Sri Nagar are sitting in two parallel rows containing five people each, in such a way that there is an equal distance between adjacent persons. In row 1- A, B, C, D and E are seated and some of them are facing South and some of them are facing North. In row 2 – P, Q, R, S and T are seated and some of them are facing South and some of them are facing North. Therefore in the given seating arrangement, each member seated in a row either faces another member of the other row or seated behind each other.(All the information given above does not necessarily represent the order of seating in the final arrangement.) Each one of them like different colors viz, Grey, Brown, Black, Blue, White, Yellow, Red, Pink, Orange and Green but not necessarily in the same order.
The person from Indore sits to the immediate right of Q, who seated exactly in the middle of the row. P faces one of the immediate neighbors of the person from Jaipur. D faces one of the immediate neighbors of the person from Patna. S is not from Patna. D is neither from Mangaluru nor from Chennai. P sits immediate right of the person from Raipur. R sits one of the extreme ends of the line and from Raipur. C sits third to the right of person from Jaipur. Only One person sits between the person from Raipur and Q. C sits to the immediate right of the person who faces S. Only two people sit between C and E. S is neither from Delhi nor from Bengaluru. The person from Mangaluru sits second to the right of the one who faces North Direction. One of the immediate neighbors of the person from Mangaluru behind the person from Patna. P does not face A and faces south direction. The person from Delhi sits exactly between the persons from Sri Nagar and Mangaluru. The person from Chennai faces the person from Sri Nagar. T faces North Direction and sits immediate left of Q. Only one person sit between the persons from Patna and Indore. A faces the opposite direction to the person from Hyderabad. Two persons sit between the one who likes Brown and the one who likes Grey. Three persons sit between the one who likes Red and the one who likes Green. Neither s nor R like Red. D does not like Green. R and S does not like Black. The person who likes Brown sits between the person who likes White and Black. The person who likes Pink sit between Blue and Orange. C does not like orange.
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Explanation
Explanation-
1. Who amongst the following faces the person from Indore?
A. The person from Delhi
B. D
C. The person from Mangaluru
D. The person from Raipur
E. B
2. Q is from which of the following Cities?
A. Patna
B. Indore
D. Raipur
E. Mangaluru
3. Which of the following is true regarding C?
A. C faces south direction
B. None of the given options is true
C. C is from Bengaluru
D. The person from Indore faces C
E. The person from Hyderabad is an immediate neighbor of C
Answer – A. C faces south direction
4. R is related to Indore in the same way as C is related to Mangaluru based on the given arrangement, To who amongst the following is T related to the following same pattern?
A. Delhi
B. Sri Nagar
C. Patna
D. Raipur
E. Chennai
5. Who among the following likes White?
A. Q
B. U
C. V
D. S
E. T
Study the following information carefully to answer the given questions.
Eight friends C, D, E, F, L, M, N and O are seated in a straight line, but not necessarily in the same order. Some of them are facing north while some face South. Only three people sit to the right of M. E sits second to the left of M. F sits third to the right of O. O is not an immediate neighbour of M. O does not sit at any of the extreme ends of the line. Both the immediate neighbours of O face south.
D sits second to the right of N. As many people sit between M and D as between M and L. Immediate neighbours of F face opposite directions(i.e., If one person faces north then the other person faces south and vice-versa). C faces south. L and F face direction opposite to C.(i.e If C faces north then both L and F face south and vice-versa)
Explanation
Explanation-
1. Which of the following is true, based on the given arrangement?
A. D faces North
B. Only three people face South
C. L sits at one of the extreme ends of the line
D. O and E face the same directions
E. None of the given options is true
Answer – D. O and E face the same directions
2. How many people sit to the left of O?
A. Three
B. More than four
C. One
D. Four
E. Two
3. Who amongst the following faces South?
A. E
B. M
C. F
D. L
E. O
B. M
4. Who amongst the following sits second to the left of L?
A. O
B. F
C. D
D. No one as less than two people sit to the left of L
E. N
5. Who amongst the following represent the persons sitting at extreme ends of the line?
A. D, N
B. C, D
C. L, N
D. D, L
E. C, N
II. Study the following information carefully to answer the given questions.
Eight people B, C, D, E, F, G, H and I are sitting in a straight line with equal distances between each other, but not necessarily in the same order. Some of them are facing North and some of them are facing south.
• E sits immediate right of the person who sits at one of the extreme ends of the line. Only three people sit between E and G. B sits exactly between D and H.
• C sits third to the right of H. F is an immediate neighbour of G and faces south. G sits second to the right of C. D sits third to the left of G. B and E face the same direction as C(i.e if C faces north then B and E also face North and Vice-Versa).
• Immediate neighbours of G face opposite directions(i.e. if one neighbour faces North then the other neighbour faces south and Vice-Versa)
• Person who sit at the extreme ends of the line face opposite directions(i.e. if one neighbour faces North then the other neighbour faces south and Vice-Versa)
• D and H face the same direction as I(i.e if I faces north then D and H also face North and Vice-Versa).
Explanation
Explanation-
1. In the given arrangement, how many people will sit between D and G?
A. Two
B. Three
C. Four
D. More than four
E. One
2. Who amongst the following sits third to the right of B?
A. E
B. I
C. Other than those given as options
D. F
E. G
3. How many people face South as per the given arrangement?
A. Two
B. Three
C. Four
D. More than four
E. One
4. Four of the following five are alike in a certain way based upon their seating arrangement and so form a group. Which of the following does not belong to the group?
A. IH
B. EG
C. DF
D. EB
E. BC
5. Who amongst the following sits at the extreme right end of the row?
A. G
B. C
C. I
D. H
E. E
I. Study the following information carefully to answer the given questions.
Ten persons from ten different countries viz. Mumbai, Chennai, Bengaluru, Kolkatta, Pune, Hyderabad, Jaipur, Ahmedabad, Surat and Kochi are sitting in two parallel rows containing five people each, in such a way that there is an equal distance between adjacent persons. In row 1- A, B, C, D and E are seated and some of them are facing South and some of them are facing North. In row 2 – P, Q, R, S and T are seated and some of them are facing South and some of them are facing North. Therefore in the given seating arrangement, each member seated in a row either faces another member of the other row or seated behind each other.(All the information given above does not necessarily represent the order of seating in the final arrangement.). Each person stays in ten different floors numbered 1 to 12.(From Ground floor to Top floor)
There is only one floor between the person from Mumbai and the person from Pune. S is not from Bengaluru. D is neither from Pune nor from Hyderabad. P sits immediate right of the person from Surat. R sits one of the extreme ends of the line and from Surat. C sits third to the right of the person from Chennai. P does not face A and faces south direction. The person from Mumbai sits exactly between the persons from Kochi and Pune. The person from Hyderabad faces the person from Kochi. The person from Surat stays on the odd numbered floor. T faces North Direction and sits immediate left of Q. Only one person sit between the persons from Bengaluru and Kolkatta. The person from Kolkatta sits to the immediate right of Q, who seated exactly in the middle of the row. P faces one of the immediate neighbors of the person from Chennai. D faces one of the immediate neighbors of the person from Bengaluru. The person from Kochi stays on the top floor. Only One person sits between the person from Surat and Q. C sits to the immediate right of the person who faces S. The person from Hyderabad stays on the 4th floor. Only two people sit between C and E. S is neither from Mumbai nor from Ahmedabad. The person from Pune sits second to the right of the one who faces North Direction. One of the immediate neighbors of the person from Pune behind the person from Bengaluru. A faces the opposite direction to the person from Jaipur. The persons from Bengaluru, Jaipur and Kolkatta stay on the consecutive floors. The floor number of the person from Chennai is the double of the floor number of the person from Surat. The floor number of the B is the square of the floor number of P. Neither E nor A stays on floor numbered 6.
Explanation
Explanation-
1. Who amongst the following faces the person from Hyderabad?
A. The person from Mumbai
B. D
C. The person from Pune
D. The person from Surat
E. B or E
2. T stays on which of the following floors?
A. 1
B. 2
C. 4
D. 6
E. 8
3. Which of the following is true regarding C?
A. C faces south direction.
B. None of the given options is true
D. The person from India faces C
E. The person from Hungary is an immediate neighbor of C
A. C faces south direction.
4. R is related to Kolkatta in the same way as C is related to Pune based on the given arrangement, To who amongst the following is T related to the following same pattern?
A. Mumbai
B. Sri Nagar
C. Bengaluru
D. Surat
5. Who amongst the following sit at extreme ends of the row?
A. The person stays on 8th floor and R
B. The persons from Ahmedabad and A
C. D and the person stays on 10th floor
D. The persons from Hyderabad and Bengaluru
E. A, E
Answer – C. D and the person stays on 10th floor
II. Study the following information carefully to answer the given questions.
Ten friends are sitting in two parallel rows of six seats each. One seat is vacant in each row. M, N, O, P and Q are sitting in row-1 facing South. D, E, F, G and H are facing North. Each likes a different Chocolate i.e. 5star, Dairy Milk, Munch, Kitkat, Perk, Snickers, Bourneville, Gems, Eclairs and Galaxy. Each person has different number of their favourite chocolates – 2, 3, 4, 6, 7, 8, 9, 11, 15 and 16.
The difference between the chocolates hold by N and O is 3. G sits third to the right of F and likes Kitkat. Only two people sit between E and the vacant seat. E does not like Perk or Munch Chocolate. Q is not an immediate neighbour of O. N likes Galaxy. The persons who sit at the extreme end of the line have chocolates in consecutive order. Neither E nor H has 8 chocolate. One of the neighbors of vacant seat in both rows have chocolates in odd number. The one who likes Munch Chocolate faces the one who likes Gems. The one who likes Munch sits opposite to the one who sits third right of the person who sits opposite to G. O is not an immediate neighbour of P. H, who likes neither Perk nor Snickers, does not face the vacant seat. Neither G nor F sits at any of the extreme ends of the row. P faces F. Vacant seats are not opposite to each other. Two seats are there between O and N, who sits third right of the one who likes Bournville. The one who likes Eclairs Chocolate faces the one who likes Kitkat. The persons who like the 5star and Gems are adjacent to each other. Vacant seat of row – 1 is not an immediate neighbour of P. E sits at one of the extreme ends of the row. F does not like 5star and Gems. Vacant seat of row-1 does not face G who doesn’t sit at any of the extreme ends of the row. The person who likes 5star has 3 chocolates. The total number of chocolates hold by Q is the half of the total number of chocolates hold by H. The total numbers of chocolates hold by M, F and G is the Square of the total number of chocolates hold by P, Q and M respectively. Neither P nor G has 4 chocolate.
Explanation
Explanation-
1. In the given arrangement, if two people come and sit to the immediate left of E, how many people will sit between D and E?
A. Two
B. Three
C. Four
D. More than four
E. One
2. Who amongst the following sits third to the right of F?
A. The one who likes Kitkat
B. E
C. Other than those given as options
D. D
E. The one who likes Perk
Answer – A. The one who likes Kitkat
3. Which of the following faces the vacant seat of Row – 1?
The one who likes Kitkat
B. E
C. Other than those given as options
D. F
E. The one who has 15 chocolate
Answer – E. The one who has 15 chocolate
4. Four of the following five are alike in a certain way based upon their seating arrangement and so form a group. Which of the following does not belong to the group?
A. QE
B. ND
C. HO
D. PG
E. FP
5. Who among the following has 11 chocolate?
A. Q
B. N
C. D
D. E
E. None of the above
I. Study the following information carefully to answer the given questions.
Ten persons from ten different countries viz. Denmark, Japan, Pakistan, India, Mauritius, Canada, Hungary, Bangladesh, Russia and Singapore are sitting in two parallel rows containing five people each, in such a way that there is an equal distance between adjacent persons. In row 1- A, B, C, D and E are seated and some of them are facing South and some of them are facing North. In row 2 – P, Q, R, S and T are seated and some of them are facing South and some of them are facing North. Therefore in the given seating arrangement, each member seated in a row either faces another member of the other row or seated behind each other.(All the information given above does not necessarily represent the order of seating in the final arrangement.)
S is not from Pakistan. D is neither from Mauritius nor from Canada. P sits immediate right of the person from Russia. R sits one of the extreme ends of the line and from Russia. C sits third to the right of person from Japan. P does not face A and faces south direction. The person from Denmark sits exactly between the persons from Singapore and Mauritius. The person from Canada faces the person from Singapore. T faces North Direction and sits immediate left of Q. Only one person sit between the persons from Pakistan and India. The person from India sits to the immediate right of Q, who seated exactly in the middle of the row. P faces one of the immediate neighbors of the person from Japan. D faces one of the immediate neighbors of the person from Pakistan. Only One person sits between the person from Russia and Q. C sits to the immediate right of the person who faces S. Only two people sit between C and E. S is neither from Denmark nor from Bangladesh. The person from Mauritius sits second to the right of the one who faces North Direction. One of the immediate neighbors of the person from Mauritius behind the person from Pakistan. A faces the opposite direction to the person from Hungary.
Explanation
Explanation-
1. Who amongst the following faces the person from Canada?
A. The person from Denmark
B. D
C. The person from Mauritius
D. The person from Russia
E. B or E
2. T is from which of the following Countries?
A. Pakistan
B. India
C. Hungary
D. Russia
E. Mauritius
3. Which of the following is true regarding C?
A. C faces south direction.
B. None of the given options is true
D. The person from India faces C
E. The person from Hungary is an immediate neighbor of C
A. C faces south direction.
4. R is related to India in the same way as C is related to Mauritius based on the given arrangement, To who amongst the following is T related to the following same pattern?
A. Denmark
B. Sri Nagar
C. Pakistan
D. Russia
5. Who amongst the following sit at extreme end of the row?
A. The person from Denmark and R
B. The persons from Bangladesh and A
C. D and the person from Japan
D. The persons from Canada and Pakistan
E. A, E
Answer – 3. D and the person from Japan
II. Study the following information carefully to answer the given questions.
Ten persons – P, Q, R, S, T, U, V, W, X and Y are sitting in two rows with five persons in each row. In Row – 1, the persons are facing south and in Row – 2, the persons are facing north. Each person in row one faces a person from the other row. All of them have a laptop of different companies, viz HP, Lenovo, Acer, Dell, Sony, Samsung, TOSHIBA, Apple, Asus and HCL but not necessarily in the same order.
R owns TOSHIBA. The one who likes Dell sits opposite to the one who is second to the right of Q. X does not like HCL. T sits opposite to the one who sits second to the left of the person who owns Acer. There is only one person stands between Q and R. The persons who like Sony and Samsung sit opposite each other. U sits opposite to P, who owns HP. The one who owns Lenovo sits opposite the one who owns Apple. Y is not facing the north but sits third to the left of V, who owns Lenovo. T sits at one of the ends of the row and owns Samsung. The one who owns Apple is on the immediate right of S, who does not own HCL. The persons who own Acer and Dell are not facing the north direction.
Explanation
Explanation-
1. S owns which of the following brand of laptop?
A. Asus
B. Apple
C. Sony
D. TOSHIBA
E. None of the above
2. How many persons sit between S and R?
A. One
B. Two
C. Three
D. Can’t be determined
E. None of the above
3. Which of the following statements is/are true?
A. W owns HCL and sits at one of the ends of the row.
B. U sits on the immediate left of the one who owns Lenovo.
C. The one who owns Asus sits on the immediate left of the one who owns HCL.
D. Only (A) and (B)
E. Only (B) and (C)
Answer – D. Only (A) and (B)
4. Four of the following five are alike in a certain way and hence form a group. Which one of the following does not belong to that group?
A. W, S
B. U, P
C. V, P
D. X, R
E. V, Q
5. Who among the following sit at the extreme ends of the row?
A. Y, X
B. W, T
C. W, R
D. S, U
E. None of the above
I. Study the following information carefully to answer the given questions
A group of eight friends – A, C, F, S, X, M, I and W – are sitting in a straight line facing north.
Each of them has different company cars– Datsun, Renault, Hyundai, Tata, Maruti, Toyota, BMW and Volkswagen. Each of them likes different colours – Pink, Yellow, Red, Black, Blue, Orange, White and Green, but not necessarily in the same order.
• C, who has a Maruti car, sits third to the left of F. Neither C nor F sits at the extreme ends of the line. A, who likes Pink colour, has Hyundai car. A is not an immediate neighbour of either C or F.
• S is two places away from M and likes Orange colour. M, who has BMW car, likes Red colour. I, who has Datsun car, sits at an extreme end of the line and likes White colour.
• X, who has Tata car likes Green colour and sits on the immediate left of C. C does not like either Blue or Yellow colour.
• One who has Volkswagen sits on the immediate left of who has Datsun car. F, who has Renault car, does not like Yellow colour.
Explanation
1. The person who likes yellow has which company car ?
1.Datsun
2.BMW
3.Toyota
4.Tata
5.None of these
2. What is the position of F with respect to X ?
1.Fourth to the right
2.Second to the left
3.Immediate right
4.Third to the left
5.None of these
Answer – 1.Fourth to the right
3. If A is related to black, W is related to Blue then M is related with which color ?
1.Green
2.Orange
3.White
4.Pink
5.None of these
4. Four of the following five are alike in a certain way based on their position, which one does not belong to that group ?
1.WFI
2.ACM
3.SMC
4.XCW
5.None of these
5. Which of the following is correct according to the given information ?
1.Two persons sits between W and who likes pink colour
2.X and I are sits at the extreme end of the line
3.C likes Orange colour
4.X is sit between who has Maruti and Hyundai cars
5.Both 1 and 4
Answer – 5.Both 1 and 4
II. Study the following information carefully to answer the given questions
Ten Stationery items Pencil, Books, Notes, Pens, Sharpners, Chalk, Stickers, Gum, Scale, Covers are palced in 10 boxes numbered 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10 are placed adjacent to one another in two different rows of 5 boxes each.
• Boxes with odd numbers are situated opposite to the boxes with even numbers and no two even number boxes with even number and no two boxes with odd numbers are adjacent to each other. No two boxes have consecutive numbers.
• Box no 8 is occupied by books and it is to the extreme left of one end of the row.
• Box 4 and 6 are not in the row of box no 8 and any of the boxes numbered 4, 6, and 3 are not in the middle of the row.
• Gums are placed in Box no 5.and its box is not situated in the row where box no 6 is situated.
• Notes are placed in an odd numbered box and placed in a row where Books are situated but both are not adjacent to each other.
• Sharpner’s and Chalk’s box are adjacent to box no 6 and Chalk’s box is not adjacent to box no 4.
• Pencil’s & Cover’s box are placed in the same row. Pen’s box is even numbered but not 10 and diagonally opposite to box no 1 which contains stickers
• Box no 6 is second from the one of the end of the row. Stickers’s box is neither opposite to Sharpners’s box nor adjacent to it.
Explanation
1. The pack of notes are placed in which number box ?
1.7
2.9
3.8
4.3
5.None of these
2. Covers are kept in which number box ?
1.4
2.6
3.2
4.3
5.Can’t be determined
3. The number 9 box is placed opposite to which box and contains which stationery ?
1.5, Chalk
2.6, Covers
3.8, Pens
4.10, scale
5.None of these
4. If pencil is placed in the box numbered 4 then which box is placed opposite to the box which contains pencils ?
1.Pencils
2.Covers
3.Stickers
4.Gum
5.None of these
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# Discuss Among Yourselves
The answer to this riddle is a single English word.
Within, only holes. A void all about
Talking back like a grave, flip the first for a shout.
-
Am i mistaken if i say i've seen this one before... – ABcDexter Feb 29 at 10:14
@ABcDexter couldn't find it on puzzlingSE...and as much as I know Hugh makes his own riddles... :) – manshu Feb 29 at 10:15
@manshu I wasn't saying that it's a duplicate of some other from this site, but yes, a similar one is (somewhere) in the bookmarked urls. – ABcDexter Feb 29 at 10:23
Now you've got me wondering. I thought it was original but now that you say that... There is something familiar about it. If so, my apologies. Should make it easier to solve, right? :) – Hugh Meyers Feb 29 at 10:48
There is definitely wordplay involved. – Hugh Meyers Feb 29 at 10:58
Moot?
Within, only holes.
The letter O is the only letter in the middle of the word.
"M" and "T" spoken out loud is "empty."
Talking back like a grave,
Reading moot backwards: toom sounds like "tomb" a grave
flip the first for a shout
Woot!
-
think about how your word sounds when spoken backwards, that could give you a clue about that hint – question_asker Feb 29 at 14:24
@question_asker tomb? – manshu Feb 29 at 14:25
@manshu shhh!!!! but yes. – question_asker Feb 29 at 14:26
@question_asker with that addition, I will accept the answer. Well done to both of you! I suppose the value of the title is debatable. – Hugh Meyers Feb 29 at 14:27
For title relevance, rot13 this badboy: uggcf://ra.jvxgvbanel.bet/jvxv/zbbg#Abha – question_asker Mar 1 at 13:05
hell
Within, only holes. A void all about:
hell = the abyss = void = bottomless pit / hole
Talking back like a grave, flip the first for a shout.
I can't seem to relate it to "Talking back from a grave" but flipping the letter "h" would make the word appear like yell which means shout.
-
Notice that the letters h-e-l-l appear within the words "only holes". – Brice M. Dempsey Mar 1 at 7:18
DAY
I don't have a complete explanation for all parts of the riddle, but for some...
The title is Discuss Among Yourselves.
Within, only holes.
(Unsure) - possibly means the middle letter has a hole.
Night is either side of day.
Talking back like a grave,
Talking back -> Backwards. 'Day' backwards is 'yad' which means 'to remember' in Hindi (Credit to manshu in comments), which is what graves are for. Also, graves are sad, which sounds like 'yad'.
flip the first for a shout.
Flipping the first letter of 'day' gives 'bay' - a kind of shout.
-
'Yad' in hindi language means 'memory' or 'to remember'...then maybe it means to remember those people who are in the grave... – manshu Feb 29 at 11:05
Not the right answer but some of the reasoning is very much on the right track. – Hugh Meyers Feb 29 at 11:34
I think any English language puzzle which requires the solver to be fluent in Hindi (or any other language other than that in which the puzzle was given) is pretty unfair. Unless there was some clue that indicates there was another language used. – Darrel Hoffman Feb 29 at 17:53
### This is not the correct answer(my achievement)...confirmed by OP
I think the riddle fits for
Trap
Within, only holes. A void all about
Holes are used as traps for centuries.
Talking back like a grave, flip the first for a shout.
Traps are to kill someone (sometimes). Take mouse trap for example, it flips and kills the mouse with a shriek.
-
There is a word that fits much better. – Hugh Meyers Feb 29 at 10:59
Also, if [your guessed word] are all about, you should probably avoid the area. "A void all about" :) – Dewi Morgan Feb 29 at 22:32
MOOT
Within, only holes.
The letter O's can be considered holes
A moot point is often void of importance
Talking back like a grave
A moot point is an open/unresolved question, you don't often get answers when talking to a grave
flip the first for a shout
WOOT WOOT!
-
i might be suffering from Deja-vu...coz i think that i saw this kind of answer. – manshu Feb 29 at 14:36
Yeah, it turned up while I was reading/writing...doh! – Briguy37 Feb 29 at 14:36
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Inductive Reasoning - to make conclusions based upon examples and patterns, Pattern - a series or sequence that repeats, Conjecture - an unproven statement that is based on observations, an " educated guess ", Counterexample - a specific case where a conjecture is false, Conditional - a logical statement that has two parts: the hypothesis and conclusion, If-Then Statement - another name for a conditional statement, Hypothesis - an unproven statement that might be true and can be tested, Conclusion - the result of a hypothesis, lowercase p - the symbol for the hypothesis, lowercase q - the symbol for the conclusion, right - the direction the arrow points, the symbol for the word " then ", Converse - created when the hypothesis and conclusion are switched, Inverse - created when both the hypothesis and conclusion are negated, Contrapositive - created by the combination of the converse and inverse, Biconditional - a statement that is only true when both the conditional statement and converse are both true, Law of Detachment - If p→q is true, and p is true, then q is true., Law of Syllogism - If p→q and q→r are true, then p→r is true, Deductive Reasoning - uses facts, definitions, accepted properties, and the laws of logic to form a logical argument.,
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# Inequality for function of $\arctan(x)$
I want to show that $$f(x) = \frac{1}{\arctan(x)} - \frac{1}{x}$$ is increasing on $$(0, \infty)$$. I can see this clearly by plotting it, but I'm struggling to write it out rigorously. It obviously suffices to show its derivative is always positive in this range (which is also clear from plotting it). We have $$f'(x) = \frac{(1+x^2)\arctan^2(x) -x^2}{x^2(1+x^2)\arctan^2(x)}$$ so again it suffices to show that $$g(x) \equiv (1+x^2)\arctan^2(x) -x^2 \ge 0 \quad \forall x >0$$ (and, yet again, this is clear from plotting it). I've jumped down the rabbit hole of taking the derivative of $$g$$ as well (since it is $$0$$ at $$x = 0$$ so it would again suffice to show that $$g' \ge 0$$) and it doesn't yield anything immediately useful for me. Please help if you can
$${1\over 1+x^2}\ge {1-x^2\over (1+x^2)^2}\quad \forall x >0$$ which is derivative of $${\arctan(x)}\ge {x\over 1+x^2}\quad \forall x >0$$ $${2\arctan(x)\over 1+x^2}\ge {2x\over (1+x^2)^2}\quad \forall x >0$$ which is derivative of $$\arctan^2(x) \ge {x^2\over 1+x^2}\quad \forall x >0$$
$$(1+x^2)\arctan^2(x) -x^2 \ge 0 \quad \forall x >0$$
Consider instead $$\displaystyle g(x) = \arctan{x} - \frac{x^2}{1 + x^2}$$. Note that $$g(0) = 0$$, so it suffices to show that $$g'(x) = 0$$ for $$x \ge 0$$.
Now, $$\displaystyle g'(x) = \frac{2[(1 + x^2)\arctan{x} - x]}{(1 + x^2)^2}$$. It thus suffices to consider $$h(x) = \arctan{x} - \frac{x}{(1 + x^2)},$$ and show that $$h(x) \ge 0$$ for $$x \ge 0$$. But $$h(0) = 0$$, and $$h'(x) = \frac{2x^2}{(1 + x^2)^2} \ge 0$$ for all $$x$$. This completes the proof.
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Hello all!! I've been working on this code for a while now and I'm still lost. Can anyone tell me what's wrong with this code please???
I know Line 44 is missing something but I'm not sure what goes there, here is the pseudocode as well:
/* pseudocode for Binomial
Coefficients */
int binomial(int n, int k)
{
If (n < k) Then return (0)
Else
{
Set denominator = 1*2*...*k
Set numerator = (n-k+1)*(n-k+2)*...*(n-1)*n
return (numerator / denominator)
} // else
End if
}
``````#include <iostream>
using namespace std ;
int binomial(int n, int k) ; // function prototype
int main ()
{
int n, k ; // parameters for the binomial number
int result ;
cout << endl ;
// read in n & k
cout << "Enter n (positive integer) : " ;
cin >> n ;
cout << "Enter k (positive integer) : " ;
cin >> k ;
result = binomial(n,k);
cout << "Binomial number " << n << "C" << k
<< " = " << result << endl ;
return (0) ;
}
int binomial(int n, int k)
{
int numerator, denominator ;
int i ; // needed to compute numerator & denominator
if (n < k) Then
{
return (0) ;
}
else
{
denominator = ; //Write inital value
for ( i = n : <= k ; i = i + 1 )
denominator = sum * 1 ;
numerator = (n-k+1)*(n-k+2)*...*(n-1)*n=
(n-(k-1))*(n-(k-2))*...*(n-1)*(n-0)=
(n-0)*(n-1)*... (n-(k-2))*(n-(k-1))
return (numerator / denominator) ;
}``````
2
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Last Post by confusedndazed
use separate loops to calculate the numerator and denominator.
Thanks! Can you give me an example of what you mean and specify the line to which you are referring?
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## Literal Equations Coloring Activity
• Post author:
• Post category:wallpaper
Students will practice writing Literal Equations as independent practice homework or extra credit or even as an assignment to leave for the substitute. This activity was awesome because students were super engaged in the coloring every last one of themeven the boys.
Literal Equations Coloring Activity Literal Equations Equations Color Activities
### Equations dinosaur using understanding of computations solve the equations to create a fun first grade math.
Literal equations coloring activity. Literal Equations Coloring Activity – Fun for my own blog on this occasion I will explain to you in connection with Literal Equations Coloring ActivitySo if you want to get great shots related to Literal Equations Coloring Activity just click on the save icon to save the photo to your computerThey are ready to download if you like and want to have them click save logo in the. 2-5 Literal Equations and Formulas SE – Christmas Color Match Activity PDF – FREE Ratios Rates and Conversions Christmas Activity. Let students do as much as possible on their own but be available to answer individual questions that students might have.
May 21 2019 – Literal equations might be at times pretty frustrating for students to learn so why not making it more fun for them. Problem 2 is an activity that increases the cognitive demand. This game will be really effective to build a good discussion session among the students.
5 3 xy for x 6. Find your answer on the christmas tree then color according to your answers. Print the pdf to use the worksheet.
The students are having to research and find or create a literal equation. This is a coloring activity for a set of 10 problems on solving literal equations. From there students spent the remainder of class working on a festive Carving Pumpkins coloring activity for solving literal equations.
2-4 Solving Equations with Variables on Both Sides SE – Christmas Color Match Activity PDF – FREE Solving Literal Equations and Formulas Christmas Activity. Solving Literal Equations Worksheet Coloring Activity Answers. 2-6 Ratios Rates and Conversions SE – Christmas.
Check out the following literal equations coloring activity sheets that combine literal equations and coloring. Answer key literal equations worksheet pdf. Check out the following literal equations coloring activity sheets that combine literal equations and coloring your students will practice solving literal equations for a specified variable with this coloring activity.
1 g x for x 2 u x for x 3 z m x for x 4 g ca for a. Displaying top 8 worksheets found for – Literal Equations Coloring Activity. An announcement must be commercial character Goods and services advancement through POBox sys.
Solving Literal Equations Worksheet Coloring Activity Answers. Phylum cnidaria coloring worksheet answers. Thanksgiving Solving Equations Coloring Activity Solving Equations Equations Solving Equations Activity.
This is a fun coloring activity over Solving Literal EquationsThe student solves the equation circles their answer from among 3 choices and then colors in a tessellation design. There are 2 coloring activities on this. Literal equations worksheet day ii given each formula below solve for the indicated variable.
Literal equations coloring activity shelter answers tessshlo solving absolute value carving pumpkins scavenger hunt math algebra 1 curriculum map coach. All equations are actual mathematical equations or formulas and involve only one or. Whether you are looking for essay coursework research or term paper help or with any other assignments it is no problem for us.
More printable literal equations coloring activity sheets are provided below. This lesson the third in the series will focus on another way of solving linear systems the elimination method. Solving Literal Equations Partner Activity Literal Equations Solving Equations Solving Linear Equations Use the same process you use to isolate the variable in an algebraic equation with one variable.
This lesson will be completed in one class period. Literal Equations Coloring Activity Football Answers All Things Algebra. Literal equations coloring activity color activities menal yasin menalyasin profile writing teaching algebra lessons shelter carving pumpkins print digital fall math 3 02 pm december 17 2007 1 district information products all things workbook page.
Some of the worksheets for this concept are Literal equations Practice solving literal equations Name period date 2 5 literal equations and formulas Solving equations color by number pdf Solving equations color by number Multi step equations answer Mathematics station activities Solving one step. I have 22 boys in this one classay yai yai and it was super easy for me to find.
Solving Absolute Value Equations Coloring Activity Literal Equations Absolute Value Equations Algebra Activities
This Equations Coloring Worksheet Is A Fun And Different Way To Practice Solving And Simplifying Li Solving Multi Step Equations Equations Multi Step Equations
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My Students Loved The Football Theme Coloring Sheet I Love The Algebra Wo Absolute Value Equations Absolute Value Inequalities Multi Step Equations Worksheets
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This Is A Fun Coloring Activity Over Solving Two Step Equations With 12 Questions The Student Solves The Equation A Two Step Equations Equations Secondary Math
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Engage Your Students In Note Taking With Coloring 1 1 2 Pages Of Notes Great For Interactive Note Literal Equations Math Interactive Notebook Equations Notes
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# 'Objectives key words solve' presentation slideshows
## Solving Linear Systems by Substitution
Solving Linear Systems by Substitution. AII, 2.0: Students solve systems of linear equations and inequalities (in two or three variables) by substitution, with graphs, or with matrices.
By nuncio
(177 views)
View Objectives key words solve PowerPoint (PPT) presentations online in SlideServe. SlideServe has a very huge collection of Objectives key words solve PowerPoint presentations. You can view or download Objectives key words solve presentations for your school assignment or business presentation. Browse for the presentations on every topic that you want.
## Key Words
The Next 100 Generations of Risk Assessment, Resilience and Early Warning Systems for Weather/Climate Related Hazards for the Benefit of Humanity: IAA-IGMASS action and new development in ACP Countries By Tomukum Chia , Cameroon, Africa , Email: glocecohadim@justice.com . Key Words
By aran (236 views)
## Key Words
Key Words. Social & Cognitive Approach. 1) create a definition for each word & test yourself when a word comes up to see if you know what it is 2) test your friends and each other with the meaning of the words 3) look up a question on this word
By sovann (98 views)
## KEY WORDS
KEY WORDS. RENEWABLE ENERGY CONSERVED DISSIPATED. USEFUL WASTED EFFICIENCY SANKEY. Tuesday, 12 June 2012. Forms of Energy. Learning Objectives: Recognise the different forms of energy How can we describe energy transfer Realise that energy cannot be created or destroyed.
By kiefer (190 views)
## Key Words
Key Words. Authors: SEOU Gostivar. REFLECTIONS.
By barry-colon (80 views)
## Key words
Discuss key features of polygenic and multifactorial diseases. How would you go about isolating a gene involved in polygenic inheritance? Helen Stuart Cardiff. Complex disorders Continuous traits Quantitative trait loci Linkage analysis Linkage disequilibrium Genetic heterogeneity
By barth (138 views)
## Key words :
* Theme : Anonymous Boeing Company Information: Presented by: Aghiles CHENAOUI, **University paris2, Panthèon-assas, **Company concerned: Boeing, **Master Currency Bank Finance, **Professor : Claire HEUILLARD, **Academic year: 2012/2013. Key words :.
By xanthe (146 views)
## Key Words
Key Words. Fossil: the preserved remains or traces of living things Paleontologist a scientist who studies fossils to learn about organisms that lived long ago. sedimentary rock: the type of rock that is made of hardened sediment
By fallon (210 views)
## KEY WORDS
KEY WORDS. Lesioni non palpabili Ecografia Mammotome. Montecatini 26 Maggio 2005. STATO DELL’ARTE. Adesione ai programmi di Screening Sensibilizzazione al problema Sviluppo tecnologico Esperienza dei radiologi Lesioni infracliniche indeterminate. Chirurgia Senologica AN.
By clark (119 views)
## KEY WORDS:
B2 REVISION –– CELLS. What do plant cells have that animal cells do not? What is the function of: Nucleus – Mitochondria Ribosome Chloroplasts Cell Wall. Sketch and label a plant and an animal cell;.
By gonzalesrobert (0 views)
## Key Words
Key Words. War Peace War – An armed conflict between two opposing parties. Peace – Living in Harmony with other countries. Lesson Objectives. By the end of this lesson you will be able to Explain what the Geneva Convention is.
By jaxon (93 views)
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6 November, 12:29
# A container has 79000cm3 of mercury. Find the mass of mercury if the density is 13.6g/cm3
+3
1. 6 November, 14:30
0
Density is mass divided by volume:
D = M / V
Given V = 79000 cm³ and D = 13.6 g/cm³:
13.6 g/cm³ = M / (79000 cm³)
M = 1,074,400 g
M = 1,074 kg
Round as needed.
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Math Goodies is a free math help portal for students, teachers, and parents.
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T O P I C R E V I E W PCmath Posted - 09/18/2008 : 15:56:03 Problem Solving-Writing in MathQuestion is: Use pictures, numbers, or words to tell how 6+8 and 6+6 are related?May be thinking too much into it as I was trying to find a correlation between the numbers and they may interconnect. Currently just saying that they relate as they both have the #6 in them and they are both addition equations - but it seems to simple of an answer.Any suggestions would be appreciated. Thank you. 3 L A T E S T R E P L I E S (Newest First) the_hill1962 Posted - 09/21/2010 : 14:14:42 They both have the same augend (namely "6").I might point out that they are not 'equations' (see the term you used that I colored red below). They are expressionsquote:Originally posted by PCmathProblem Solving-Writing in MathQuestion is: Use pictures, numbers, or words to tell how 6+8 and 6+6 are related?May be thinking too much into it as I was trying to find a correlation between the numbers and they may interconnect. Currently just saying that they relate as they both have the #6 in them and they are both addition equations - but it seems to simple of an answer.Any suggestions would be appreciated. Thank you. Catgirl Posted - 09/19/2010 : 19:10:22 I was searching for the same problem, PCmath! I figured the same thing as Ultraglide..that they are both really 6+6, only one is +2 more. I don't know what else it would be!?! It confuses me and I'm not even that age! Ultraglide Posted - 09/19/2008 : 12:42:16 Maybe the fact that you are adding 8 in one case and 6 in the other (i.e. 8 is two more than 6) will give a sum which is 2 greater?
Math Forums @ Math Goodies © 2000-2004 Snitz Communications
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# Scattering in the Earth’s Atmosphere
Topics: Vapor pressure, Water, Light Pages: 8 (1843 words) Published: March 4, 2013
Scattering in the Earth’s Atmosphere
Tony Yuk Hei Chan
Abstract
Introduction
When the sun radiates light to the earth, incoming solar radiation often is scattered from its original direction of propagation as it enter earth’s atmosphere. This is due to the scattering effect from different particles in the earth’s atmosphere. When the scattering occur, the phase and the polarization of solar radiation are often changed. In this experiment, we aimed to investigate the effect of two of the most comment types of scattering upon the incident solar radiation – Mie scattering and Rayleigh Scattering. We also looked into the effect to the solar radiation from cloud.
Theory
Beer–Lambert–Bouguer law:
In optics, Beer–Lambert–Bouguer law relates the absorption of light to the properties of the material through which the light is traveling. Consider a case when there is a clear sky, a parallel beam of incident radiation pass through a medium which absorb the light. By assuming that the medium is a non-scattering, absorbing medium, the intensity of the light after passing though the medium is given by:
(1)
WhereI(0) is the intensity at s=0,
a is the absorption cross section of a single particle for radiation of wavelength ,
n is the number density of the medium, and
s is the length of the medium.
We can apply the relationship to solar radiation passing though the atmosphere. From equation 1, we find that , the transmissivity of the slant path of the atmosphere at a given wavelength is given by:
(2)
where u is the optical depth of the vertical column and is given by
(3)
and Z is the zenith angle of the sun
we also know that the transmissivity and the albedo, A, are related by
(4)
Rayleigh scattering
Rayleigh scattering theory describes the interaction of sunlight with molecules in a simple way. It applies to particles much smaller than the wavelength of the incoming radiation. For Rayleigh scattering, the scattering cross section of one particle can be calculated using the formula below: σs=α2128π53λ4(5)
Whereα is the polarisability of the scatterer and
λ is the wavelength of the incident radiation.
From this, we can find the atmospheric optical depth due to Rayleigh scattering to be:
(6)
Where n(z) is the air number density at height z.
Mie Scattering
Mie Scattering theory applies to the interaction of radiation with aerosol and cloud droplets. Under Mie scattering conditions, we can approximate the optical depth of cloud in terms of the column abundance of liquid water (i.e. the liquid water path). The optical depth can be obtained from the equation: u= 3Σl2ρwae (7)
where ρw is the liquid water density and ae is the scattering-equivalent mean drop radius.
Clouds increase the global reflection of solar radiation from 15 to 30%, reducing the amount of solar radiation absorbed by the Earth by about 44 W/m². [1] A quantitative description of how cloud can affect the solar radiation can be calculated by comparing the radiative fluxes at the top of the atmosphere and the at the surface in both clear day and cloudy day, over a given region the following relationship must apply:
(8)
where A is the albedo of the region
FS and FLW denote the downward shortwave (SW) and upward longwave (LW) fluxes at the TOA, respectively.
From that, the Cloud Radiative Forcing C, which is the difference between heating rates under cloudy and clear sky conditions, can be found by applying:
(9)
where the CS denotes clear sky conditions.
Method
We tried to do 10 different tasks in order to find out different aspect of the solar radiation.
For task1, Under clear sky conditions and for a range of solar zenith angles, the direct flux of the solar radiation was measured using the lightmeter and the thermopile. The direct flux was obtained by finding the difference between the diffusive flux and the total flux. The diffusive flux...
Please join StudyMode to read the full document
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In a previous article we showed the importance of using variables to replace multiple instances of the same measure in a DAX expression. A very common use case is that of the IF function. This article focuses on the cost for the formula engine rather than for the storage engine.
Consider the following measure.
```Margin :=
IF (
[Sales Amount] > 0 && [Total Cost] > 0,
[Sales Amount] - [Total Cost]
)
```
The basic idea is that the difference between Sales Amount and Total Cost should be evaluated only if both measures are greater than zero. When dealing with that condition, the DAX engine produces a query plan that evaluates each measure twice. This is visible in the storage engine requests generated for the following query.
```EVALUATE
SUMMARIZECOLUMNS (
'Date'[Year],
"Margin", [Margin]
)
```
However, it is worth stressing that the physical query plan has 216 rows, which is a reference point we will consider in later variations of the same measure.
Without going into details that were already explained in a previous article, it is worth noting that the multiple references to the same measure are requiring separate evaluations – although the result is the same. DAX is not the best at saving the value of common subexpressions evaluated in the same filter context. This is evident in the following variation of the Margin measure. The two branches of the IF function are identical, but the query plan adds other evaluations for both the storage engine and the formula engine.
```Margin 2 :=
IF (
[Sales Amount] > 0 && [Total Cost] > 0,
[Sales Amount] - [Total Cost],
[Sales Amount] - [Total Cost]
)
```
In this case there is an additional storage engine query. The number of rows in the physical query plan is now 342. This increases the number of lines by over 50%, compared to the previous workload.
The optimized version of this measure stores the two measures into two variables. This is so that they are only evaluated once in the IF function.
```Margin Optimized :=
VAR SalesAmount = [Sales Amount]
VAR TotalCost = [Total Cost]
RETURN
IF (
SalesAmount > 0 && TotalCost > 0,
SalesAmount – TotalCost
)
```
This is visible in the storage engine requests, of which there are only two.
A version of the IF function with the second branch identical to the first would produce the same storage engine queries.
The physical query plan reduced the number of rows from 216 to 126.
This is an important result. This optimization technique is particularly useful when dealing with multiple references to a measure that has a high cost in the formula engine. Indeed, the DAX cache only operates at the storage engine level.
## Conclusion
Multiple references to the same measure in the same filter context can produce multiple executions of the same DAX expression, thus producing the same result. Saving the result of the measure in a variable generates a better query plan, improving code performance.
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Find the best tutors and institutes for Dance
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# I want to learn western, Bollywood dance from a female tutor. My main objective is to learn dance moves for parties, social gatherings. I am available on weekends and prefer to learn at home. Thank you. Message me if interested!
Choreographer, Event Planner, Wedding Planner
You can call me if you are interested, i am freestyle dancer. Thanks
dance to fitness,zumba,aerobic.
Hi Anurag, I hope you have the criteria like below shown Comum A, B, C and now add the following formula in co, umn C to get the grade according to the percentage. Please revert if you have any questions. =IF(B2<=45,"D",IF(B2<=60,"C",IF(B2<=80,"B",IF(B2<=100,"A")))) Column A Column B Column... read more
Hi Anurag, I hope you have the criteria like below shown Comum A, B, C and now add the following formula in co, umn C to get the grade according to the percentage. Please revert if you have any questions. =IF(B2<=45,"D",IF(B2<=60,"C",IF(B2<=80,"B",IF(B2<=100,"A")))) Column A Column B Column C Student Names Percentage Grade Ram 60 C Suman 80 B Leela 95 A Joe 70 B
Regards,
Nagamani
Tutor
Yes am interested..
Dancer
Concepts must be clear. For practice purpose, it is enuff.
Hi... There.... Do you have Skype..??? You can join our online class.... Add me by #Vijayalaxmi Hutagonna.... Thank you
Dancer
Where are you located ??
Professional Singer | Voice Trainer | Academy
You have to search home tutors in urbanpro. Best wishes
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Hi, I am female bollywood dance teacher. Can help you. Contact me.
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# Convert 716.28 cm to inches
"Centimeters to Inches" Calculator
Convert
cm to Inches
## Answer for "How many inches is 716.28 cm?"
`716.28 cm = 282″`
(716.28 Centimeters is equal to 282 Inches)
## Explanation of 716.28 Centimeters to Inches Conversion
`Centimeters to Inches Conversion Formula: in = cm ÷ 2.54`
According to 'cm to inches' conversion formula if you want to convert 716.28 (seven hundred sixteen point two eight) Centimeters to Inches you have to divide 716.28 by 2.54.
Here is the complete solution:
`716.28 cm ÷ 2.54 = 282″(two hundred eighty-two inches)`
## About "Centimeters to Inches" Calculator
This converter will help you to convert Centimeters to Inches (cm to in). For example, How many inches is 716.28 cm? Enter the number of centimeters (e.g. '716.28') and then click the 'Convert' button.
Convert
cm to Inches
## Centimetres to Inches Conversion Table
CentimetresInches
714.78 cm281.41 inches
714.88 cm281.45 inches
714.98 cm281.49 inches
715.08 cm281.53 inches
715.18 cm281.57 inches
715.28 cm281.61 inches
715.38 cm281.65 inches
715.48 cm281.69 inches
715.58 cm281.72 inches
715.68 cm281.76 inches
715.78 cm281.8 inches
715.88 cm281.84 inches
715.98 cm281.88 inches
716.08 cm281.92 inches
716.18 cm281.96 inches
716.38 cm282.04 inches
716.48 cm282.08 inches
716.58 cm282.12 inches
716.68 cm282.16 inches
716.78 cm282.2 inches
716.88 cm282.24 inches
716.98 cm
717.08 cm282.31 inches
717.18 cm282.35 inches
717.28 cm282.39 inches
717.38 cm282.43 inches
717.48 cm282.47 inches
717.58 cm282.51 inches
717.68 cm282.55 inches
282
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# Dynamics: solve both trajectories
1. Oct 10, 2006
### 600burger
Hey all,
Having trouble solving this one. Class is dynamics. 2D, constant accel.
Shooting a cannon at a point above the initial. Inital velocity is 400m/s
Cannon is at point A (origin) and we're shooting at point B @ (5000m, 1500m)
I'm asked to find the two thetas that satisfy.
So i got this so far, basicly the givens.
In the x:
ax = 0
vx = v0*cos(Θ)
x = v0*cos(Θ)*t
In the y:
ay = -g
vy = v0*sin(Θ)-gt
y = v0*sin(Θ)*t-(1/2)gt2
I tried to eliminate t is the y position equation by
1) solving x for t
2) subing into y(Θ,t) for t to get y(Θ)
3) solving for Θ
I cant get through solving for Θ. Can I use the quadradic equation on y(Θ) to solve for Θ? How do i do that?
2. Oct 10, 2006
### radou
Yes, you should be able to solve the quadratic equation for $$\Theta$$, after plugging in the values of displacement 5000 m and 1500 m.
3. Oct 10, 2006
### 600burger
Thats what my prof was saying, but im not sure how thats done? cause it some up to like
-a*sec(Θ)^2 +b*tan(Θ) - c = 0
so does it matter that Θ is in sec^2 for the a and tan for b?
4. Oct 10, 2006
### 600burger
**In the voice of reason**
Learn your trig identities backwards and forwards!
*echo 7th grad algebra teachers voice
**End voice**
sec(Θ)^2 = 1 + tan(Θ)^2
Sub and solve. Still an ugly problem, but I'm gald I figured it out.:tongue2:
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Solving Eq & Ineq Graphs & Func. Systems of Eq. Polynomials Frac. Express. Powers & Roots Complex Numbers Quadratic Eq. Quadratic Func. Coord. Geo. Exp. & Log. Func. Probability Matrices Trigonometry Trig. Identities Equations & Tri. On this page we hope to clear up problems you might have with the trigonometric ratios. The trigonometric ratios are very useful when dealing with triangles and unit circles. Click any of the links below or scroll down to better your understanding of the trigonometric ratios. Ratios (sin, cos, tan) Reciprocal ratios (csc, sec, cot) Rotations (unit circle) Radians Cofunctions Graphs involving the trig. ratios Pythagorean and quotient identities Algebraic manipulation Quiz on Trigonometry The trig. ratios, sine, cosine, and tangent are based on properties of right triangles. The function values depend on the measure of the angle. The functions are outlined below. sine x = (side opposite x)/hypotenuse cosine x = (side adjacent x)/hypotenuse tangent x = (side opposite x)/(side adjacent x) In the figure below, sin A = a/c, cosine A = b/c, and tangent A = a/b. There are two special triangles you need to know, 45-45-90 and 30-60-90 triangles. They are depicted in the figures below. The figures show how to find the side lengths of those types of triangles. Besides knowing how to find the length of any given side of the special triangles, you need to know their trig. ratio values (they are always the same, no matter the size of the triangle because the trig. ratios depend on the measure of the angle). A table of these values is given below. (If you need a more in-depth understanding of the trig. ratios, you can click here and go to a geometry lesson that goes into great detail concerning the trigonometric ratios.) The reciprocal ratios are trigonometric ratios, too. They are outlined below. cotangent x = 1/tan x = (adjacent side)/(opposite side) secant x = 1/cos x = (hypotenuse)/(adjacent side) cosecant x = 1/sin x = (hypotenuse)/(opposite side) Angles are also called rotations because they can be formed by rotating a ray around the origin on the coordinate plane. The initial side is the x-axis and the ray that has been rotated to form an angle is the terminal side. Example: Reference angles are useful when dealing with rotations that end in the second, third, or fourth quadrants. A reference angle for a rotation is the acute angle formed by the terminal side and the x-axis. Example: ``` 1. Problem: Find the reference angle for theta (see the figure below). ``` ``` Solution: To find the measure of the acute angle formed by the terminal side and the x-axis subtract the measure of theta from 180o. 180 - 115 = 65 The reference angle is 65o.``` Once you have found the reference angle, use it to determine the trig. function values. Consider, for example, an angle of 150o. The terminal side makes a 30o angle with the x-axis, since 180 - 150 = 30. As the figure below shows, triangle ONR is congruent to triangle ON'R'; therefore, the ratios of the sides of the two triangles are the same, although the ratios may have different signs. (You could determine the function values directly from triangle ONR, but that is not necessary if you remember that the sine is positive and the cosine and tangent are negative in quadrant II.) Up until now, you have probably only measured angles using degrees. Another useful measure, based on the unit circle, is called radians. Radians scare intermediate algebra and AP Calculus students alike, so don't get too worried if they seem complicated or useless to you. The figure below shows measures in degrees and radians on the unit circle that you should probably memorize, as they are commonly used measures. Sometimes, it will be necessary to convert from radians to degrees or vice versa. To convert from degrees to radians, multiply by ((PI)/180o). To convert from radians to degrees, multiply by (180o/(PI)). Examples: ``` 1. Problem: Convert 60o to radians. Solution: Multiply 60o by (PI)/180o. 60o (PI) --- * ---- 1 180o 60o(PI) ------- 180o Perform the indicated division. Cancel out the degrees. (PI) ---- 3 2. Problem: Convert (3(PI))/4 to degrees. Solution: Multiply (3(PI))/4 by 180o/(PI). 3(PI) 180o ----- * ---- 4 (PI) 3(PI)180o --------- 4(PI) Perform the indicated division. ((PI) cancels out.) 3 - * 180o 4 135o``` In a right triangle, the two acute angles are complementary. Thus, if one acute angle of a right triangle is x, the other is 90o - x. Therefore, if sin x = (a/c) then cos (90o - x) = (a/c). A table of all the cofunctions is displayed below. sin x = cos (90o - x) tan x = cot (90o - x) sec x = csc (90o - x) cos x = sin (90o - x) cot x = tan (90o - x) csc x = sec (90o - x) Example: ```1. Problem: Find the function value of cot 60o. Solution: Use the cotangent's cofunction identity to rewrite the problem. tan (90o - 60o) tan 30o The tangent of 30o is one you should have memorized. (SQRT(3))/3``` All six of the trigonometric functions are periodic, that is, their graphs repeat after a certain period. The periods of the six trig. functions are shown below. sin, cos, csc, and sec = 2(PI) tan and cot = PI The sin and cos graphs have a maximum y value of 1, and a minimum y value of -1. You should know what one cycle (period) of the graphs of the big three trig. functions looks like. They can be found by plotting points or graphing on a calculator. The following figures depict the graphs of these functions. Sine: Cosine: Tangent: These graphs, like any other graph of a function, can be transformed. The table below outlines each change for each trig. ratio. There are two quotient identities. They tell us that the tangent and cotangent functions can be expressed in terms of the sine and cosine functions. They are listed below. ``` sin x tan x = -----, cos x <> 0 cos x cos x cot x = -----, sin x <> 0 sin x``` There are three other identities that are very important. They are called the Pythagorean Identities. The Pythagorean Identities come in handy later on when you need to prove more complicated trig. identities equal. The Pythagorean Identities are listed below. sin2 x + cos2 x = 1 1 + cot2 x = csc2 x 1 + tan2 x = sec2 x Remember that sin2 x = (sin x)2. Trigonometric expression such as tan(x - (PI)) represent numbers, just as algebraic expressions represent numbers. Since that is true, we know we can manipulate trig. expressions the same way we do algebraic expressions. Examples: ``` 1. Problem: Simplify cos y(tan y - sec y). Solution: Use the distributive property of multiplication, which says a(b + c) = ab + ac. cos y(tan y) - cos y(sec y) Simplify the expression by writing it in terms of cos. Use the Quotient Identities. sin y 1 (cos y)----- - (cos y)----- cos y cos y Perform the indicated multiplications. (cos y)sin y cos y ------------ - ----- cos y cos y Perform the indicated divisions. sin y - 1 2. Problem: Simplify (sin2 x)(cos2 x) + cos4 x Solution: Factor. (cos2 x)(sin2 x + cos2 x) Using a Pythagorean Identity, replace sin2 x + cos2 x with 1. (cos2 x)(1) cos2 x``` Take the Quiz on trigonometry. (Very useful to review or to see if you've really got this topic down.) Do it!
Math for Morons Like Us - Algebra II: Trigonometry
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# Questions tagged [credit]
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### Finding the nth digit for credit card number
I'm having a hard time trying to find the 2nd to last digit of the user's credit card number, since that's the starting point for Luhn's algorithm. Is there a way to convert a long to an array and ...
126k views
### Why do I get the "invalid operands to binary expression" error?
While trying to solve credit, I get the following error rem and number1 are declared as long and even if I change them to int, I still get the same error. Please provide a solution and explain what ...
359 views
### Why Does credit Fail Passing check50 Tests?
When I input the credit card numbers that check50 inputs in its tests to my program myself, It correctly identifies the numbers as expected. However, when I run check50, it's not identifying correctly....
4k views
### Credit from CS50 2018 week 1 walkthrough
Hi i am a beginner and i have been trying to solve this set but with no avail. I tried to use what was taught in the first lesson and did search for some help online when i encounter errors. However, ...
411 views
I actually try to do the Credit Task back from pset1. So far i understand how the calculation for the checksum validation of the Credit Card Number works, and could do it on Paper. The Problem, i ...
964 views
### Finding the products' digit and adding them individually
Separating the products' digits and adding them up is giving me trouble. For example for this credit card number: (1234567), 12, 8, and 4 would be the result of multiplying every other digit by 2. How ...
336 views
### credit - Why this AMEX and MASTERCARD numbers are considered invalid?
I have compiled my version of credit, anyway check50 throws 2 fails during its check. Lines with numbers where it fails are: :( identifies 369421438430814 as INVALID expected "INVALID\n", not ...
127 views
### Why is this while loop generating errors
I'm trying to get a 15 or 16 digits numbers (integers) from users and then using a do while loop to check if the user entered a valid number, otherwise, they will be re-ask to enter the number. THE ...
5k views
### Pset 1 Credit Card ( help needed )
Update # attempt 2: I am not getting valid output, code complies fine. What am I missing? code follows: including cs.50 and stdio ( i don't think i need math ) bool checksum( long long cc); // for ...
2k views
### Pset6 credit python Problem
I try to convert my Credit Solution from C to Python now. But on Line 19 ( doubleCheck = ccNumber[i] * 2 ) i always get the error "TypeError: 'int' object is not subscriptable"! How can i fix this? ...
144 views
### error at while loop cs50 Pset1 credit
I am getting an error at while loop it says error: '&&' within '||' [-Werror,-Wlogical-op-parentheses] while((credit % 10 >= 0) && (i < 15) || (i < 16)) here is my ...
224 views
### problem with pset6 python credit
Below is my code: [hidden code] output is: Enter the card number: 378282246310005 firstTwo 37; firstOne 3; cardlen 15 MASTERCARD note: it shows MASTERCARD instead of AmEx..Why?
471 views
### Credit.c is my nemesis
I have no experience programming, and English is not my first language so I'm struggling a lot with this particular problem. In "credit.c" you have to prompt the user for a credit card number and ...
80 views
### Incorrect mastercard number in pset1 [duplicate]
by lunh algoritm result of calculating 5105105105105100 will be 25 and it means that this card is invalid, but check50 output this: :( identifies 5105105105105100 as MASTERCARD expected "...
30 views
### Making code neater: a way to reduce repeated logical operators?
For CS50 pset1 Credit I'm using repeated logical-OR operators (||) to work out the validity and card brand of a credit card number. When I evaluate the style of my code using style50 it says to insert ...
706 views
### Every 2nd digit on Pset1 Credit problem: Tldr bolded
Im struggling with manipulating the string of numbers in the credit problem. I've figured out how to divide the initial input number by 10 (*100 each iteration) then mod % by 10 in order to access the ...
17k views
### How to convert the digits of a number into elements of array?
For example in pset1 hacker edition program asks user to input the number of credit card. Imagine that a digits are: 1234567890 So can anyone tell me does it possible to convert this number into ...
2k views
### Credit GetLongLong Issue
I'm having issues using the GetLongLong function. For some reason the output I get doesn't match the input I put in. Here, is my code. #include <studio.h> #include <cs50.h> int main(void)...
135 views
### credit.c project in hacker1 I keep having this error?
While trying to compile the codes uploaded I keep having this error which I couldn't understand!
870 views
### pset 1 credit.c invalid cards are valid by cs50
It was said that you can verify if the credit card number is valid by using Luhn’s algorithm. If you done mathematical operations right, after them, you know if the card is valid if the last digit in ...
239 views
### credit.c frustration
Im considering skipping :( I dont really understand why it always prints INVALID4.... #include <stdio.h> #include <cs50.h> #include <math.h> int main (void) { //Initialize a ...
59 views
### PSET 1 Credit; Sometimes Check50 works and other times it doesn't
Why does Check50 work for certain credit numbers and not others? I ran debug and found that for some credit numbers and not others it may randomly be always one away from a multiple of 10 preventing ...
365 views
### I keep getting "expected EOF" when I use check50
this is the message identifies 4111111111111113 as INVALID expected EOF, not "INVALID" this is the code #include <cs50.h> #include <stdio.h> #include <string.h> #include &...
82 views
### PSET1 - Credit: Check50 says wrong what should be right
Someone with the same problem? I have a 16 digit number starting with a 4 and my code identifies it correct as VISA\n, but the check50 says it has to be INVALID\n. What could cause this strange ...
408 views
### Can't use get_long_long();
So somehow I cant use the get_long_long(); function. It lets me declare the variable, but I cant get any user input for it. Here's my code: #include <stdio.h> #include <cs50.h> #include &...
186 views
### PSET1 - Credit HELP
I am now working on the first part of credit, and is currently checking the validity of the number given. I separated it into two parts, one where I summed the doubled digits, and one where I summed ...
1k views
### pset1 | error: invalid operands to binary expression ('long long *' and 'int')
Here are snippets of my code: long long card = get_long_long(); long long *FindLength = *card; while (FindLength != 0) { FindLength = FindLength / 10; counterCardLength++; } And ...
410 views
### Credit - Why Does it Fail Passing check50?
When sending my code to Check50, all the Credit Card checks go through ok but one: :( identifies 4062901840 as INVALID expected "INVALID\n", not "VISA\n" But doing the steps: 1) Multiply every ...
383 views
### I have problems applying Credit in python
I try to translate my code from c to python. I review the code several times and I think that is equal, but the program don't works despite in c works prefectly. This is my implementation: """Know ...
403 views
### long long to array, how did you do?
I want to know new ways to do it, this is my solution. arr[nccard - i] = (long long)(ccard / pow(10, i - 1)) % 10;
660 views
### errors I cant understand in credit.c can anyone explain?
I am taking the number as a string(n) and then trying to convert the string into ints to add them like this: else if(strlen(n)==15 && (n[1]==3&&(n[2]==4||n[2]==7))) { for(i=14;i&...
34 views
### credit pset1 cs50 sum while loop
I was doing the credit problem with the visa test card in the page: 4003600000000014 It passed that test, however in other tests it doesn't pass. I think it's a matter of the sum that it can't compute ...
61 views
### Credit Card Check
So I am on pset6 trying to implement the credit card check. I've already completed working through the numbers and am not having any issue with the actual algorithm. But for some reason my if ...
60 views
### Credit (Pset1) - Checksum problem
so I have a problem with the Credit exercise from pset1. I updated my code today after realizing that I didn't split up the digits of the product of the current digit in the card number and 2 and ...
46 views
### pset6 credit help with calculations in python
I am trying to convert my credit solution from C to python, but for some reason it is not working. Attached is my code and the problems from cs50 import get_int card, a, m, v = False, False, False, ...
48 views
### credit pset1 bruteforcing doesn't work
I was trying out credit and tried to write a code for all 15 digit cards so if a AMEX card exists it should work in theory. However even with valid cards I'm getting the input invalid. I have not ...
77 views
### Can anyone help me troubleshoot my code for the CS50 Credit problem? (**not trying to cheat, just need some help!)
This is the code I have written in c for the cs50 credit pset 1 problem. I am checking the length of numbers entered, and the first digits (AMEX 15 digits, starts with 34 or 37; MC 16 digits, starts ...
74 views
### Question about Pset1 - credit
I had work on the credit for 3 hours, and I had reach to most of the function - and I checked that the number received from: Luhn (a) last digit(c) and digit count (cc) are right by using ...
57 views
### PSET6 Credit - checksum not giving the correct number
list1 = list(input("Number: ").strip()) even_count, odd_count, count = 0, 0, 0 # if the length is even if len(list1)%2 == 0: # iterating each number in list for num in list1: ...
25 views
### Week 1 Credit; Not Creating Variable
I'm starting the week 1 credit problem, however, when trying to set together every second digit it simply does not print. Could you help me identify where I went wrong? Thanks! #include <stdio.h>...
37 views
### ProblemSet1 - Another Credit.c issue - Can't convert the floating value of my variables into an integer value
I was almost finishing my code when I got stuck with a strange problem. I've asked a question a few hours ago and someone answered, so I manage to fix that issue. But unfortunelly I got stuck in ...
32 views
### ProblemSet1 - Credit.c - Issues assigning the sum of the numbers on a variable
So I coded the first lines of the Luhn's Algorithm and tried to print the values of some of my variables to be sure they were doing exactly what I want it. For my surprise they weren't. And I were ...
47 views
### CS50 Credit - again a scope problem
I think I almost finished the credit problem set, however, I have another scope problem. I think it is really simple. After the checksum, the code is not working anymore. If I put extra braces by the ...
19 views
### Scope problem - not print the test results
I have trouble with my scope (PS1, credit) I am using two while loops and to test the result, i print results in between. This code does not print the second printf. I think it is a scope problem but ...
68 views
### pset6 credit the sum is not end with zero and check count right how
the Luhn’s Algorithm in 378282246310005 = 48 and in 371449635398431 = 76 so how not INVALID it must end with zero so the two number must be INVALID
69 views
### I am not getting the correct output when i enter David's card number in credit
This is my code: #include <stdio.h> #include <cs50.h> #include <math.h> int main(void) { int x, y, z, p, q, r, d, e; int k = 0; int s = 0; long A; long n = ...
344 views
### Pset 1: Credit problem - Checksum?
I'm having trouble with the checksum in the #credit problem. I have no clue on how to operate specific digits within the credit card number (specifically, multiplying the digits in the odd positions ...
63 views
### pset1 credit formula to extract digits keeps returning 0 (potential data type issue?)
Below is the block of code that i have issue with. I'm using this method to extract the digits: num[i] = (num % pow(10, i))/pow(10, i-1). The pow1 and pow2 variables print out correctly, but "digit" ...
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5.10 A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at 22° with the horizontal. The horizontal component of the earth’s magnetic field at the place is known to be 0.35 G. Determine the magnitude of the earth’s magnetic field at the place.
The horizontal component of the earth's magnetic field,
${\mathrm{B}}_{\mathrm{H}}=0.35\mathrm{G}$
Angle made by the needle with the horizontal plane
Earth's magnetic field strength = B We can relate B and ${\mathrm{B}}_{\mathrm{H}}$
as
$\begin{array}{l}{\mathrm{B}}_{\mathrm{n}}=\mathrm{Bcos}\mathrm{\theta }\\ \therefore \mathrm{B}=\frac{{\mathrm{B}}_{\mathrm{H}}}{\mathrm{cos}\mathrm{\delta }}\\ \phantom{\rule{1em}{0ex}}=\frac{0.35}{\mathrm{cos}{22}^{\circ }}=0.377\mathrm{G}\end{array}$
Hence, the strength of the earth's magnetic field at the given location is 0.377 G.
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# Physics Symbols Explained
Physics Symbols Explained: this diagram is one of our most searched charts and infographics by people seeking to learn about new things and improve their general knowledge of how the world works.
The Physics Symbols Chart 92 is a comprehensive list of symbols used in physics. It contains symbols for physical quantities and their international units, as well as mathematical symbols used in physics calculations. The chart is divided into sections based on the type of physical quantity, such as space and time, mechanics, thermal physics, waves and optics, electricity and magnetism, modern physics, mathematics, Greek alphabet, astronomy, and music.
The chart uses different fonts to distinguish between vector and scalar quantities. Vector quantities are written in a bold, serif font, while scalar quantities and the magnitudes of vector quantities are written in an italic, serif font. Mathematical symbols are written in a roman, serif font, except when they are applied to calculations with units. Units are written with a roman, sans-serif font.
Here are some examples of symbols and their meanings:
– r: position, separation, radius, radius of curvature
– s: displacement, distance
– ?, ?: angle, angular displacement, angular separation, rotation angle
– x, y, z: Cartesian coordinates
– i?, j?, k?: Cartesian unit vectors
– ?, ?, z: cylindrical coordinates
– ??, ??, z?: cylindrical unit vectors
– n?: normal unit vector
– t?: tangential unit vector
– h: height, depth
– ?, L: length
– d: distance, separation, thickness
– D: diameter
– C: circumference
– A: area, cross-sectional area, projected area, surface area
– V: volume
– t: time, duration
– T: period, periodic time
– f: frequency
– ?: angular frequency
– v: velocity, speed
– a: acceleration
– g: gravitational field, acceleration due to gravity
– m: mass
– F: force
– p: momentum
– J: impulse
– W: work
– E: energy, total energy
– K: kinetic energy (translational, rotational)
– U: potential energy (gravitational, spring)
– Vg: gravitational potential
– ?: efficiency
– P: power
– ?: torque
– I: moment of inertia
– L: angular momentum
– H: angular impulse
This is just a small sample of the many symbols included in the chart.
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# math
posted by .
write the expression as a sinle logarithm. express powers as factors.
In (x/x-1)+In(x+1/x)-In(x^2-1)
• math -
You MUST put brackets for the denominators to avoid ambiguity
I am sure you mean
ln ( x/(x+1) ) + ln( (x+1)/x ) - ln (x^2 - 1)
then
= lnx - ln(x+1) + (ln(x+1) - lnx - ln( (x+1)(x-1)
= lnx - ln(x+1) + ln(x+1) + lnx - ln(x+1) - ln(x-1)
= - 2ln(x+1)
or
ln [ (x/(x-1) ((x+1)/x) / ((x+1)x-1))
= ln (1/(x-1)^2)
= ln1 - ln(x-1)^2
= 0 - 2ln(x-1)
= -2ln(x-1)
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Home Background Mathematical Sūtras Applications Blog
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# A Brief Introduction
## Introducing Vedic Maths: Tricks & Mental Maths, Myths & Facts.
Vedic Mathematics is often dubbed as the 'Yoga' of mathematical thoughts. This is because, its unique techniques encourage mental calculations, and provides mechanisms that aid in enhancing interest and understanding of numbers and mathematical concepts.
The most common apprehension in using the 'un-conventional tricks' of Vedic Mathematics, is that the solutions look too magical to believe that they would actually work (in all cases). More importantly, they seem to be devoid of 'mathematical logic' - whatever that means!
As a simple illustration to introduce Vedic Mathematics, let us consider squaring 99:
Since, 99 is close to (slightly lower than) a power of 10 (100, in this case), let us use Upasūtra:
यावदूनं तावदूनीकृत्य वर्गं च योजयेत्
(Lessen by the deficiency, and set up square of the deficiency) - for 992, in simple steps:
Steps 992 1. Consider the nearest power of 10 as the Base In this case, the Base, closest to 99, is 100. 2. Find the deficiency In this case, the Deficiency =100 - 99 = 1 3. Subtract the deficiency from that number In this case, 99 - 1 = 98 4. Set-up as many zeroes, as the digits of the number In this case, 99 being a 2-digit number, we get 9800 5. Square the Deficiency, and add to get the answer 1 being the Deficiency, and 12 = 19800 + 1 = 9801, which is the answer!
So, for a practitioner of Vedic Mathematics, for something like 9962:
Clearly, the deficiency is 4, and 42 = 16 Also, 996 - 4 = 992 And, 992,000 + 16 = 992,016 Thus, 9962 = 992,016
Of course, it is so amazingly simple that it can be done mentally, with some practice - within a few seconds. Taking it ahead, for something like 999882:
Clearly, the deficiency is 12, and 122 = 144 Also, 99988 - 12 = 99976 And, 99976,00000 + 144 = 99976,00144 Thus, 999882 = 9,997,600,144
Now, that should be magical for somebody who is newly introduced to Vedic Mathematics.
The irony is: Most of us, knowingly or unknowingly, have been learning and using the 'tricks' of Vedic Mathematics through preparations of various competitive examinations - albeit, at a later stage, and during a race against time.
Speed does matter. Especially in today's life, when the results can be obtained faster than typing the digits in a calculator.
But, not just speed - the real power of Vedic Mathematics is in calculating the seemingly impossible calculation - that too, with ease. Let us consider the above example for a 20-digit number, say 999999999999999999982:
All we have to do is stay calm, and have faith in Vedic Mathematics. Clearly, the Deficiency is 2, and 22 = 4 Also, 99999999999999999998 - 2 = 99999999999999999996 (The operation is actually on the last digit) And, 99999999999999999996, 00000000000000000000 + 4 = 99999999999999999996, 00000000000000000004 (The operation is again on the last digit, of 20 trailing zeroes) Thus, 999999999999999999982 = 9,999,999,999,999,999,999, 600,000,000,000,000,000,004
Don't bother to check with a calculator, we don't expect it to work.
This is the reason that Vedic Maths, more often than not, is promoted as an alternate mathematical methodology which is 'superior' to the conventional methods of calculations, or tricks, and even mental maths for 'fastest' mathematical calculations that brings out the 'human calculator' within oneself!
It is a fact that steps of conventional methods are too tedious to be performed mentally, which is why Vedic Mathematics is the source of thousands of 'Mental Maths' books/tutorials/videos doing the rounds - not to mention the innumerable 'tricks', 'short-cuts' and 'secrets' that educational 'gurus' demonstrate to provide their students an edge in competitive examinations, all over the world.
And considering the fact, that the above example is based (only) on an Upasūtra - not even a Sūtra - should provide a fair idea of what Vedic Mathematics is about.
But, why does it work? For the above Upasūtra, let us consider the following:
Assuming N is a number close (and less) to a power of 10, then N = a - b 'a' being the power of 10, and 'b' being the Deficiency As in examples above, 99=100-1, 996=1000-4, and 99988=100000-12 Now, N2 = (a - b)2 = a2 - 2ab + b2 = a (a - 2b) + b2 But, N = a - b. So, substituting a = N + b N2 = a ([N + b] - 2b) + b2 = a (N - b) + b2 This is exactly what this Upasūtra makes us do: 992 = [100 × (99 - 1)] + 12 = 9800 + 1 = 9801 9962 = [1000 × (996 - 4)] + 42 = 992000 + 16 = 992,016 999882 = [100000 × (99988 - 12)] + 122 = 99976,00000 + 144 = 997,600,144
Needless to state that the principles of Vedic Mathematics work because they are mathematical.
The above Upasūtra, being mathematical, would work even for numbers not close to the Base, but that will not be very smart - especially when there are other methods (like Upasūtra: dvaṅdvayoga - discussed here ») to do the same. For example, using this Upasūtra for 772 would require us to find 232 (23 being the Deficiency). That would be foolishness, and not failure of Vedic Mathematics.
Obviously, further explanations and details (including the number being close to, but greater than the Base) are presented in the respective Upasūtra: yāvadūnam tāvadūnīkṛtya vargañca yojayet (discussed here »), but the point is: Vedic Maths provide no magical trick, and is certainly not devoid of 'mathematical logic'. Fact is, it is practical step ahead of the 'Algebraic Rules' - that we are comfortable with.
The standpoint typical to Vedic Scholars is that the Vedic Civilization was aware of the Algebraic Rules, and generalized them for practical usage with easy-to-remember phrases (read Sūtras). But, we don't want to get into that, and just stick to the mathematical structure.
Having stated the above, irrespective of everything, Vedic Maths can indeed lead to amazingly fast calculations, with some practice, for the mathematically inclined. There is no denying that the principles of Vedic Maths provide methods to convert cumbersome and complicated calculations into simple steps, which can be mentally executed to reach the solution at ease. However,
• Vedic Mathematics is not an alternate methodology:
Even though, it seems to provide alternate methods of computation - Vedic Mathematics is a way of moving ahead with the Algebraic Rules. Vedic Mathematics is (more of a) practical next-step of the Algebraic Rules, and should be viewed as a layer above it.
• The principles of Vedic Mathematics are not tricks:
Tricks work on specific situations and cases. When a 'rule' is general, and may be equally applied on all cases, without exception - it is called a 'formula' or 'theorem'. Labeling Vedic Mathematics as a set of tricks would be demeaning to a set of excellent formulae - irrespective of its origin, influence, or even religious importance.
• Vedic Mathematics is not Mental Maths:
Although the principles of Vedic Mathematics are easy to use, the usage of the principles mentally requires practice and a certain amount of mathematical skill - as in any other form of mathematics. Vedic Mathematical principles being easier, make it comparatively easy to execute them mentally. This does not mean that everybody can become 'human calculators' by learning them.
• Vedic Mathematics should not replace the conventional methods:
Replacing conventional methods with the principles of Vedic Mathematics at school, or otherwise, would be like learning the Vedas instead of science. Even teaching both methods side-by-side to children would amount to confusion and unnecessarily burden them. The best age to learn Vedic Mathematics is from age 14-15 years onwards, preferably for the mathematically inclined.
• The principles of Vedic Mathematics are not 'more correct' or 'superior':
In science and mathematics, there may be more than one method to reach a solution. This does not mean that one of the methods is 'more correct' or 'superior' than the other just because it is fast, easy, can be mentally executed, or even because the results can be derived from left to right!
• Vedic Mathematics do not cover the entire sphere of mathematics:
Mathematics is a far bigger subject than what is covered by the principles of Vedic Mathematics. Vedic Maths find its best applications with real numbers, in real life scenarios (when one doesn't have pen-paper or a calculator), competitive examinations (when the derivation is not as important as the answer), practical problems (when one needs a calculated value as an input for some other task), and verifying calculations (when one can use Vedic Maths to quickly check a calculated figure already calculated using conventional methods).
Most of our above standpoints are opposed to hardcore 'Vedic Mathematics' (and even Hindutva) fanatics. Nevertheless, this web endeavour should speak volumes of our love, appreciation and respect for Vedic Mathematics and its amazing principles.
All said and done, there is no denying that Vedic Mathematics is far more than 'wise shortcuts'. It provides techniques that encourage us to execute the computations mentally - thereby stretching and excercising our capability to deal with mathematical thoughts.
The techniques improve our overall mathematical capability, by clarifying the positional notation (as in śūddha, discussed here »), by making us visualize the numbers to mentally execute calculations (as in ūrdhva tiryagbhyāṃ, discussed here »), by helping us to think closely with the number system (as in ānurūpyeṇa, discussed here »), and by forcing us to anticipate consecutive steps (as in dhvajāṅka, discussed here ») - to list a few positive influences.
Vedic Mathematics is truly the 'Yoga' of mathematical thoughts, and its use can help overcome 'Maths Phobia' by making one 'play' with numbers and mathematical concepts.
Vedic Mathematical Sūtras »
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# Count of numbers in range [L, R] having sum of digits of its square equal to square of sum of digits
• Difficulty Level : Medium
• Last Updated : 16 Sep, 2021
Given two integers L and R, the task is to find the count of numbers in range [L, R] such that the sum of digits of its square is equal to the square of sum of its digits
Example:
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Input: L = 22, R = 22
Output: 1
Explanation: 22 is only valid number in this range as
sum of digits of its square = S(22*22) = S(484) = 16 and
square of sum of its digits = S(22)*S(22) = 16
Input: L = 1, R = 58
Output: 12
Explanation: Total valid numbers are {1, 2, 3, 10, 11, 12, 13, 20, 21, 22, 30, 31}
Naive Approach: Run a loop from L to R and for each number calculate the sum of digits and check whether the current number satisfies the given condition or not.
Follow the steps below to solve the problem:
• Iterate from L to R and for each number calculate its sum of digits.
• Square the current number and find its sum of digits.
• If they are equal, increment the answer otherwise continue for the next element.
Time Complexity: O((R-L)*log(R))
Efficient Approach: From example 2, we can observe that all the valid numbers have digits from 0 to 3 only. Therefore there are only 4 choices for each digit present in a number. Use recursion to calculate all the valid numbers till R and check whether it satisfies the given condition or not.
Follow the steps below to solve the problem:
• Notice that all valid numbers have digits from [0,3].
• Numbers between [4,9] when squared carries a carry over them.
• S(4)*S(4) = 16 and S(16) = 7, 16 != 7.
• S(5)*S(5) = 25 and S(25) = 7, 25 != 7.
• So, generate all possible numbers up to the R.
• For each generated number there are a total of possible 4 choices between [0,3].
• Calculate each possible choice and check the condition for each of them.
Below is the implementation of the above approach:
## C++
`// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function to check if the number is valid``bool` `check(``int` `num)``{`` ``// Sum of digits of num`` ``int` `sm = 0;` ` ``// Squared number`` ``int` `num2 = num * num;` ` ``while` `(num) {`` ``sm += num % 10;`` ``num /= 10;`` ``}` ` ``// Sum of digits of (num * num)`` ``int` `sm2 = 0;`` ``while` `(num2) {`` ``sm2 += num2 % 10;`` ``num2 /= 10;`` ``}`` ``return` `((sm * sm) == sm2);``}` `// Function to convert a string to an integer``int` `convert(string s)``{`` ``int` `val = 0;`` ``reverse(s.begin(), s.end());`` ``int` `cur = 1;`` ``for` `(``int` `i = 0; i < s.size(); i++) {`` ``val += (s[i] - ``'0'``) * cur;`` ``cur *= 10;`` ``}`` ``return` `val;``}` `// Function to generate all possible``// strings of length len``void` `generate(string s, ``int` `len, set<``int``>& uniq)``{`` ``// Desired string`` ``if` `(s.size() == len) {` ` ``// Take only valid numbers`` ``if` `(check(convert(s))) {`` ``uniq.insert(convert(s));`` ``}`` ``return``;`` ``}` ` ``// Recurse for all possible digits`` ``for` `(``int` `i = 0; i <= 3; i++) {`` ``generate(s + ``char``(i + ``'0'``), len, uniq);`` ``}``}` `// Function to calculate unique numbers``// in range [L, R]``int` `totalNumbers(``int` `L, ``int` `R)``{`` ``// Initialize a variable`` ``// to store the answer`` ``int` `ans = 0;` ` ``// Calculate the maximum`` ``// possible length`` ``int` `max_len = ``log10``(R) + 1;` ` ``// Set to store distinct`` ``// valid numbers`` ``set<``int``> uniq;` ` ``for` `(``int` `i = 1; i <= max_len; i++) {`` ``// Generate all possible strings`` ``// of length i`` ``generate(``""``, i, uniq);`` ``}` ` ``// Iterate the set to get the count`` ``// of valid numbers in the range [L,R]`` ``for` `(``auto` `x : uniq) {`` ``if` `(x >= L && x <= R) {`` ``ans++;`` ``}`` ``}`` ``return` `ans;``}` `// Driver Code``int` `main()``{`` ``int` `L = 22, R = 22;`` ``cout << totalNumbers(L, R);``}`
## Java
`// Java program for the above approach``import` `java.util.*;` `class` `GFG{` `// Function to check if the number is valid``static` `boolean` `check(``int` `num)``{`` ` ` ``// Sum of digits of num`` ``int` `sm = ``0``;` ` ``// Squared number`` ``int` `num2 = num * num;` ` ``while` `(num > ``0``) {`` ``sm += num % ``10``;`` ``num /= ``10``;`` ``}` ` ``// Sum of digits of (num * num)`` ``int` `sm2 = ``0``;`` ``while` `(num2>``0``) {`` ``sm2 += num2 % ``10``;`` ``num2 /= ``10``;`` ``}`` ``return` `((sm * sm) == sm2);``}` `// Function to convert a String to an integer``static` `int` `convert(String s)``{`` ``int` `val = ``0``;`` ``s = reverse(s);`` ``int` `cur = ``1``;`` ``for` `(``int` `i = ``0``; i < s.length(); i++) {`` ``val += (s.charAt(i) - ``'0'``) * cur;`` ``cur *= ``10``;`` ``}`` ``return` `val;``}` `// Function to generate all possible``// Strings of length len``static` `void` `generate(String s, ``int` `len, HashSet uniq)``{`` ``// Desired String`` ``if` `(s.length() == len) {` ` ``// Take only valid numbers`` ``if` `(check(convert(s))) {`` ``uniq.add(convert(s));`` ``}`` ``return``;`` ``}` ` ``// Recurse for all possible digits`` ``for` `(``int` `i = ``0``; i <= ``3``; i++) {`` ``generate(s + (``char``)(i + ``'0'``), len, uniq);`` ``}``}``static` `String reverse(String input) {`` ``char``[] a = input.toCharArray();`` ``int` `l, r = a.length - ``1``;`` ``for` `(l = ``0``; l < r; l++, r--) {`` ``char` `temp = a[l];`` ``a[l] = a[r];`` ``a[r] = temp;`` ``}`` ``return` `String.valueOf(a);``}`` ` `// Function to calculate unique numbers``// in range [L, R]``static` `int` `totalNumbers(``int` `L, ``int` `R)``{`` ` ` ``// Initialize a variable`` ``// to store the answer`` ``int` `ans = ``0``;` ` ``// Calculate the maximum`` ``// possible length`` ``int` `max_len = (``int``) (Math.log10(R) + ``1``);` ` ``// Set to store distinct`` ``// valid numbers`` ``HashSet uniq = ``new` `HashSet();` ` ``for` `(``int` `i = ``1``; i <= max_len; i++) {`` ``// Generate all possible Strings`` ``// of length i`` ``generate(``""``, i, uniq);`` ``}` ` ``// Iterate the set to get the count`` ``// of valid numbers in the range [L,R]`` ``for` `(``int` `x : uniq) {`` ``if` `(x >= L && x <= R) {`` ``ans++;`` ``}`` ``}`` ``return` `ans;``}` `// Driver Code``public` `static` `void` `main(String[] args)``{`` ``int` `L = ``22``, R = ``22``;`` ``System.out.print(totalNumbers(L, R));``}``}` `// This code is contributed by Princi Singh`
## Python3
`# python 3 program for the above approach` `from` `math ``import` `log10``# Function to check if the number is valid``def` `check(num):`` ``# Sum of digits of num`` ``sm ``=` `0` ` ``# Squared number`` ``num2 ``=` `num ``*` `num` ` ``while` `(num):`` ``sm ``+``=` `num ``%` `10`` ``num ``/``/``=` `10` ` ``# Sum of digits of (num * num)`` ``sm2 ``=` `0`` ``while` `(num2):`` ``sm2 ``+``=` `num2 ``%` `10`` ``num2 ``/``/``=` `10`` ``return` `((sm ``*` `sm) ``=``=` `sm2)` `# Function to convert a string to an integer``def` `convert(s):`` ``val ``=` `0`` ``s ``=` `s[::``-``1``]`` ``cur ``=` `1`` ``for` `i ``in` `range``(``len``(s)):`` ``val ``+``=` `(``ord``(s[i]) ``-` `ord``(``'0'``)) ``*` `cur`` ``cur ``*``=` `10`` ``return` `val` `# Function to generate all possible``# strings of length len``def` `generate(s, len1, uniq):`` ``# Desired string`` ``if` `(``len``(s) ``=``=` `len1):` ` ``# Take only valid numbers`` ``if``(check(convert(s))):`` ``uniq.add(convert(s))`` ``return` ` ``# Recurse for all possible digits`` ``for` `i ``in` `range``(``4``):`` ``generate(s ``+` `chr``(i ``+` `ord``(``'0'``)), len1, uniq)` `# Function to calculate unique numbers``# in range [L, R]``def` `totalNumbers(L, R):`` ``# Initialize a variable`` ``# to store the answer`` ``ans ``=` `0` ` ``# Calculate the maximum`` ``# possible length`` ``max_len ``=` `int``(log10(R)) ``+` `1` ` ``# Set to store distinct`` ``# valid numbers`` ``uniq ``=` `set``()` ` ``for` `i ``in` `range``(``1``,max_len``+``1``,``1``):`` ``# Generate all possible strings`` ``# of length i`` ``generate("", i, uniq)` ` ``# Iterate the set to get the count`` ``# of valid numbers in the range [L,R]`` ``for` `x ``in` `uniq:`` ``if` `(x >``=` `L ``and` `x <``=` `R):`` ``ans ``+``=` `1`` ``return` `ans` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:`` ``L ``=` `22`` ``R ``=` `22`` ``print``(totalNumbers(L, R))`` ` ` ``# This code is contributed by ipg2016107.`
## C#
`// C# program for the above approach``using` `System;``using` `System.Collections.Generic;` `class` `GFG{` `// Function to check if the number is valid``static` `bool` `check(``int` `num)``{`` ``// Sum of digits of num`` ``int` `sm = 0;` ` ``// Squared number`` ``int` `num2 = num * num;` ` ``while` `(num>0) {`` ``sm += num % 10;`` ``num /= 10;`` ``}` ` ``// Sum of digits of (num * num)`` ``int` `sm2 = 0;`` ``while` `(num2>0) {`` ``sm2 += num2 % 10;`` ``num2 /= 10;`` ``}`` ``return` `((sm * sm) == sm2);``}` `// Function to convert a string to an integer``static` `int` `convert(``string` `s)``{`` ``int` `val = 0;`` ``char``[] charArray = s.ToCharArray();`` ``Array.Reverse( charArray );`` ``s = ``new` `string``( charArray );`` ``int` `cur = 1;`` ``for` `(``int` `i = 0; i < s.Length; i++) {`` ``val += ((``int``)s[i] - (``int``)``'0'``) * cur;`` ``cur *= 10;`` ``}`` ``return` `val;``}` `// Function to generate all possible``// strings of length len``static` `void` `generate(``string` `s, ``int` `len, HashSet<``int``> uniq)``{`` ``// Desired string`` ``if` `(s.Length == len) {` ` ``// Take only valid numbers`` ``if` `(check(convert(s))) {`` ``uniq.Add(convert(s));`` ``}`` ``return``;`` ``}` ` ``// Recurse for all possible digits`` ``for` `(``int` `i = 0; i <= 3; i++) {`` ``generate(s + Convert.ToChar(i + (``int``)``'0'``), len, uniq);`` ``}``}` `// Function to calculate unique numbers``// in range [L, R]``static` `int` `totalNumbers(``int` `L, ``int` `R)``{`` ``// Initialize a variable`` ``// to store the answer`` ``int` `ans = 0;` ` ``// Calculate the maximum`` ``// possible length`` ``int` `max_len = (``int``)Math.Log10(R) + 1;` ` ``// Set to store distinct`` ``// valid numbers`` ``HashSet<``int``> uniq = ``new` `HashSet<``int``>();` ` ``for` `(``int` `i = 1; i <= max_len; i++) {`` ``// Generate all possible strings`` ``// of length i`` ``generate(``""``, i, uniq);`` ``}` ` ``// Iterate the set to get the count`` ``// of valid numbers in the range [L,R]`` ``foreach` `(``int` `x ``in` `uniq) {`` ``if` `(x >= L && x <= R) {`` ``ans++;`` ``}`` ``}`` ``return` `ans;``}` `// Driver Code``public` `static` `void` `Main()``{`` ``int` `L = 22, R = 22;`` ``Console.Write(totalNumbers(L, R));``}` `}` `// This code is contributed by SURENDRA_GANGWAR.`
## Javascript
``
Output:
`1`
Time Complexity: (, since there are 4 choices for each of the digits till the length of R i.e log10(R) + 1, therefore the time complexity would be exponential.
Auxiliary Space: (Recursive Stack space)
My Personal Notes arrow_drop_up
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Lesson 1 Algebra: Roofs – isometric trapezia
Lesson plan
Piagetian Level Learners Thinking and Abstract of the Activity
3 B
Mature
Formal
Recognition of the distinction between necessary and sufficient conditions.
Generation of a Proof strategy
Expression in terms of generalised number
"a,b,a,a plus b"
3A
Early
Formal
Spontaneous testing of rules with counter examples. `So each of these on its own won’t always make a roof.
We need at least two. Which two? Let’s test.
“But 1,4,1,10 is a non-roof!”
“’But 2,1,2,2 is not a roof!”
n-1 is the largest the other sides can be
2B*
Concrete
Generalisation
Expression of rules “1st and 3rd must be the same”
‘1st and 2nd make up the 4th’
2B
Mature
concrete
Mid
concrete
Episode 1
Episode 2
Rules for roof shapes
Episode 3
Exploration with proof
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https://civic-express.com/what-does-rule-of-thirds-mean/
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What does rule of thirds mean?
Rule of Thirds. The rule of thirds is a compositional approach that divides the image into nine equal parts, or thirds. For example, consider each of the nine equal divisions of the photographic subject area of a picture frame. As you count the thirds of the frame, you find the area where the subject divides the picture more or less evenly. Then, divide the frame into three equal squares.
Contents
What does rule of thirds do?
Rule of thirds makes images appear balanced and cohesive – even when you’re stretching them. For each of the three horizontal axes, divide the image into four equal parts (usually three or four). This is also called a “rule of thirds”.
What are the 5 rules of defensive driving?
The 5 Rules of Defensive Driving. 1. Look both ways before you Turn. 2. Yield to pedestrians. 3. Drive at a safe speed. 4. Be alert for drivers who may not see you. 5. Report all dangerous maneuvers to dispatch.
Does rule of thirds always apply?
The main rule is that when you’re composing a photograph, you should try to compose the image in thirds. The rule says that you should also try to make the main subject divide the image into three equal parts. If you’re unsure whether to use the rule, just try dividing the photograph’s frame horizontally and vertically and see which proportion the rule works better in that case.
How do you use the rule of thirds?
The rule of thirds can be used when composing the image and when framing it. If you’re setting it up, break the picture into three equal sections: the vertical, the horizontal, and then the longest part of the photo.
What is the Golden Triangle in photography?
The golden triangle of photography refers to the triangle created by the relationship between exposure, focal length and aperture. The wider your lens, the more light it sends into your camera.
Just so, what is rule of thirds in photography definition?
Rule of thirds is a way to organize your digital image in three rows, one vertical, one horizontal and one diagonal. The vertical and horizontal edges are made of the three rows that form the Rule of Thirds. The diagonal creates a balance between the other two edges.
What is a rule of thirds shot?
” (or “rule of thirds” ) is a compositional technique where the horizon is divided into thirds. The rule of thirds dictates that you place most of the action in the middle of the frame and the remaining action in a diagonal or vertical line.
Regarding this, why do we use the rule of thirds?
The main rule of thirds is it should be divided by the same ratio as the frame. So if a frame has 60, 70 or 80 inches and the photo has 12 feet (60 inches) by 9 feet (40 inches), that third of the photo is the same size as the frame, which is three feet.
Also Know, what is the rule of thirds in driving?
The rule of thirds is a rule of composition that indicates the most interesting or exciting parts of the composition should appear on each of three evenly distributed parts or “thirds” of the page layout. The rule can be applied to the overall composition or specific elements.
What is the two thirds rule?
Onion Rule: It is common to find a new car for under \$10,000 has a 2/3 of the price of a new Cadillac, but you need to know that the car used was originally owned by a new Cadillac owner and therefore it is still in good working order.
What is the rule of thirds in advertising?
When designing advertising images, it is widely accepted that the rule of thirds is very helpful when placing elements within the image. When taking a photograph or a frame to be printed, the rule of thirds is applied to place the subject of the photograph/frame in the center.
Does rule of thirds apply to portraits?
Rule of Thirds. The rule of thirds is a compositional pattern used to draw a square or other geometric shape. It can also be used to guide the placement of people in portraits. The rule of thirds guides you when taking a side view of a face.
What is the theory behind the rule of thirds?
An important aspect of composition is how the rule of thirds is used to place the subject matter of the picture, or the point of interest, in relation to the edges of the photo. If the picture has too many distractions or is crowded, it is called “overloaded.” If the edges of the picture fall away too far, it is called “empty frame.”
Aristotle
What is Rule of the Thirds in photography?
Rule of thirds is a graphic design rule. It states that every photo should have three points at which the main subject (the subject, usually a human figure) is placed to create visual tension in the image, such as from the eye.
Where did rule of thirds come from?
Rule of thirds. This is one of the classic principles of composition. It states that an image should be placed in a scene, allowing a human eye to divide it into thirds both vertically and horizontally.
Is the rule of thirds true?
The rule of thirds, while it is true, can also be very restrictive. It’s a great guideline for planning your compositions, but it only helps you if your subjects or backgrounds are in the exact locations you suggest.
What is golden ratio in photography?
The Golden Ratio, also known as the “gold number”, golden proportion or golden mean, is a specific proportion (1:1.615) that often occurs naturally in a variety of forms in the natural world and human artwork.
How do you break the rule of thirds?
Rule of thirds. To create the best shot, the rule of thirds suggests that you place the main subject halfway between the top third and the bottom third. This makes sense because one third of the frame is filled with either the horizon line or the main subject area.
How do you dress in thirds?
The main focus when dividing your clothes into thirds is to make sure that there isn’t a lot of symmetry between the two divisions and that the clothes remain visually pleasing as you divide them. This does not mean that the clothes must be very uneven between the divisions. As long as you dress well, look good and don’t feel awkward, the overall look of your ensemble can always be slightly off.
Why is the rule of three effective?
The rule of three or Rule of three says that if you are planning on doing a repair, you should start by repairing your system for three times the damage before making your repairs permanent. This rule ensures that the overall repair process is safe and effective.
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https://convertoctopus.com/560-4-feet-per-second-to-miles-per-hour
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## Conversion formula
The conversion factor from feet per second to miles per hour is 0.68181818181818, which means that 1 foot per second is equal to 0.68181818181818 miles per hour:
1 ft/s = 0.68181818181818 mph
To convert 560.4 feet per second into miles per hour we have to multiply 560.4 by the conversion factor in order to get the velocity amount from feet per second to miles per hour. We can also form a simple proportion to calculate the result:
1 ft/s → 0.68181818181818 mph
560.4 ft/s → V(mph)
Solve the above proportion to obtain the velocity V in miles per hour:
V(mph) = 560.4 ft/s × 0.68181818181818 mph
V(mph) = 382.09090909091 mph
The final result is:
560.4 ft/s → 382.09090909091 mph
We conclude that 560.4 feet per second is equivalent to 382.09090909091 miles per hour:
560.4 feet per second = 382.09090909091 miles per hour
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 mile per hour is equal to 0.0026171782060433 × 560.4 feet per second.
Another way is saying that 560.4 feet per second is equal to 1 ÷ 0.0026171782060433 miles per hour.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that five hundred sixty point four feet per second is approximately three hundred eighty-two point zero nine one miles per hour:
560.4 ft/s ≅ 382.091 mph
An alternative is also that one mile per hour is approximately zero point zero zero three times five hundred sixty point four feet per second.
## Conversion table
### feet per second to miles per hour chart
For quick reference purposes, below is the conversion table you can use to convert from feet per second to miles per hour
feet per second (ft/s) miles per hour (mph)
561.4 feet per second 382.773 miles per hour
562.4 feet per second 383.455 miles per hour
563.4 feet per second 384.136 miles per hour
564.4 feet per second 384.818 miles per hour
565.4 feet per second 385.5 miles per hour
566.4 feet per second 386.182 miles per hour
567.4 feet per second 386.864 miles per hour
568.4 feet per second 387.545 miles per hour
569.4 feet per second 388.227 miles per hour
570.4 feet per second 388.909 miles per hour
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https://reason.town/siamese-neural-network-pytorch/
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# Siamese Neural Networks in Pytorch
This blog post walks through the implementation of Siamese Neural Networks in Pytorch. We’ll go over the motivation for using Siamese Neural Networks, the architecture of the network, and how to train and evaluate the model.
Checkout this video:
## Introduction to Siamese Neural Networks
Siamese neural networks are a type of neural network that uses two identical sub-networks, known as twin networks, to process inputs. These twin networks are usually connected at the output layer, and they share weights between them. Siamese neural networks are often used for tasks such as signature verification and face recognition, where the idea is to compare two input images and see if they match.
There are many different architectures that can be used for siamese neural networks, but in general, they all follow the same basic principle: two identical sub-networks are used to process two different inputs, and the outputs of these sub-networks are then compared. The comparison can be done using a variety of methods, such as taking the Euclidean distance between the outputs, or using a more sophisticated similarity metric.
One of the advantages of siamese neural networks is that they can be trained using data that is not necessarily labeled. This is because the training process does not require labels for the input data; instead, it only requires pairs of inputs (i.e., two images) that should be considered similar or dissimilar. This makes siamese neural networks very well suited for tasks such as one-shot learning, where the goal is to learn from a single example.
In this tutorial, we will learn how to implement siamese neural networks in Pytorch. We will use a simple image dataset consisting of images of birds and dogs; our goal will be to train a siamese network that can tell us whether two images contain the same animal or not.
## What are Siamese Neural Networks?
Siamese neural networks are a type of deep learning algorithm that is used to find similarities between two inputs. In other words, it can be used to determine whether two things are the same or not. For example, you could use a siamese neural network to compare two pictures and determine if they are of the same person.
Siamese neural networks are made up of two identical sub-networks, known as twins. These twins share parameters, weights, and biases. The outputs of the twins are compared using a distance metric, such as Euclidean distance or cosine similarity. If the outputs are similar, then the inputs are considered to be the same. Otherwise, they are considered to be different.
The use of siamese neural networks has been shown to be effective for tasks such as image recognition, action recognition, and facial verification.
## How do Siamese Neural Networks work?
Siamese Neural Networks are a type of neural network architecture that is used for tasks such as face recognition and signature verification. This architecture is useful for these tasks because it can learn to compare two inputs and generate a similarity score. In this post, we will take a look at how Siamese Neural Networks work, how they are trained, and how they can be implemented in Pytorch.
## The Benefits of Siamese Neural Networks
Siamese neural networks are a type of deep learning architecture that are used in a wide variety of applications, including image recognition, signature verification, and facial recognition. In this post, we’ll explore the benefits of siamese neural networks and how they can be implemented in Pytorch.
Siamese neural networks are beneficial because they allow for effective feature learning. By training a siamese network on a large dataset, the network can learn to extract useful features that can be used for many different tasks. This is opposed to traditional neural networks, which typically only focus on one task at a time.
Another benefit of siamese neural networks is that they are very efficient at generalizing to new data. This is because the features learned by a siamese network are not specific to any one dataset; rather, they are general features that can be applied to many different datasets. This makes siamese neural networks much more robust than traditional neural networks.
Finally, siamese neural networks are relatively easy to implement in Pytorch. There are many online resources that provide helpful code snippets and tutorials. This makes it possible for even beginner programmers to get started with siamese neural networks quickly and easily.
## Siamese Neural Networks in Pytorch
Siamese neural networks are a type of deep learning architecture that allows for the training of models to compare inputs and make predictions based on their similarity. These models have been shown to be effective in a variety of tasks such as facial recognition, signature verification, and identifying duplicate products in online shopping.
In this tutorial, we will learn how to implement a siamese neural network in Pytorch. We will first discuss the theory behind siamese networks and then walk through an example of training a siamese network on the MNIST dataset. By the end of this tutorial, you will be able to train your own siamese network to perform similarity comparisons on new data.
## Pytorch Implementation of Siamese Neural Networks
This is a Pytorch implementation of Siamese Neural Networks, based on the paper “Siamese Neural Networks for One-shot Image Recognition” (https://www.cs.cmu.edu/~rsalakhu/papers/oneshot1.pdf).
The code is written in Python 3 and requires Pytorch 0.4 or higher.
## Siamese Neural Networks in Action
Python is a high-level, general-purpose programming language that is widely used in many scientific fields. Pytorch is a deep learning framework for Python that is popular for its ease of use and flexibility. In this tutorial, we will see how to use siamese neural networks in Pytorch to create models that can learn to identify similar objects.
Siamese neural networks are very useful for tasks such as image recognition, where we want to be able to identify whether two images contain the same object. They are also helpful for detecting fraudulent activities, such as credit card fraud. In this tutorial, we will see how siamese neural networks can be used for image recognition using the MNIST dataset, which contains images of handwritten digits. We will also explore how these networks can be used for other tasks such as detecting facial similarity and signature verification.
## Conclusion
Based on our findings, we believe that Siamese Neural Networks have potential for further improvement and development. We hope to continue our research in this area and contribute to the advancement of this exciting field of machine learning.
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https://www.jiskha.com/questions/1383941/The-total-U-S-consumption-of-electricity-in-a-particular-year-was-5-0-x-10-12-kilowatt-hours
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# Physics
The total U.S. consumption of electricity in a particular year was 5.0 x 10^12 kilowatt-hours. What is the mass equivalence of this amount of energy? Show all calculations leading to an answer.
How would I solve it?
1. 👍 0
2. 👎 0
3. 👁 77
1. E = m c^2
c^2 = (3*10^8)^2 = 9 * 10^16
E = 5 * 10^15 watts * 3600 seconds
= 18 * 10^18 Joules
m = 18*10^18/9*10^16
= 2 * 10^2
= 200 Kg
about two big people mass :)
1. 👍 0
2. 👎 0
posted by Damon
1. 👍 0
2. 👎 0
posted by Alea
3. I thought I did :)
surely you can google E = m c^2
(Albert Einstein)
there
E is in Joules which is watt seconds
c is speed of light, about 3*10^8 meters/second
an hour is 3600 seconds
a kilowatt is 1000 watts which is 1000 Joules/second
1. 👍 0
2. 👎 0
posted by Damon
4. thank you so much! :)
1. 👍 0
2. 👎 0
posted by Alea
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https://www.r-bloggers.com/2016/12/mixed-integer-programming-in-r-with-the-ompr-package/
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Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.
Numerical optimization is an important tool in the data scientist's toolbox. Many classical statistical problems boil down to finding the highest (or lowest) point on a multi-dimensional surface: the base R function optim provides many techniques for solving such maximum likelihood problems.
Counterintuitively, numerical optimizations are easiest (though rarely actually easy) when all of the variables are continuous and can take any value. When integer variables enter the mix, optimization becomes much, much harder. This typically happens when the optimization is constrained by a limited selection of objects, for example packages in a weight-limited cargo shipment, or stocks in a portfolio constrained by sector weightings and transaction costs. For tasks like these, you often need an algorithm for a specialized type of optimization: Mixed Integer Programming.
For problems like these, Dirk Schumacher has created the ompr package for R. This package provides a convenient syntax for describing the variables and contraints in an optimization problem. For example, take the classic “knapsack” problem of maximizing the total value of objects in a container subject to its maximum weight limit. In mathematical form, the problem can be described as an objective (quantity to be maximized) and a constraint:
$$\begin{equation*} \begin{array}{ll@{}ll} \text{max} & \displaystyle\sum\limits_{i=1}^{n} v_{i}x_{i} & &\\ \text{subject to}& \displaystyle\sum\limits_{i=1}^{n} w_{i}x_{i} \leq W, & &\\ & x_{i} \in \{0,1\}, &i=1 ,\ldots, n& \end{array} \end{equation*}$$
In ompr, the same problem is naturally expressed in R as follows:
model <- MIPModel() %>%
add_variable(x[i], i = 1:n, type = "binary") %>%
set_objective(sum_expr(v[i] * x[i], i = 1:n)) %>%
add_constraint(sum_expr(w[i] * x[i], i = 1:n) <= W)
To actually solve the problem, you need to provide a "backend" solver algorithm to ompr. It's designed to integrate with any solver, and currently works with the ROI (R Optimization Infrastructure) package. ROI in turn provides a number of solver algorithms including GLPK, the GNU Linear Programming Kit, which you can use to solve problems like this.
Dirk provides a number of worked examples of the ompr package in use. Examples include the Traveling Salesman problem, solving Soduku puzzles, and selecting optimal warehouse locations to minimize delivery costs to customers.
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