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# Fight Finance
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For a price of $13, Carla will sell you a share which will pay a dividend of$1 in one year and every year after that forever. The required return of the stock is 10% pa.
Would you like to Carla's share or politely ?
A project to build a toll road will take 3 years to complete, costing three payments of $50 million, paid at the start of each year (at times 0, 1, and 2). After completion, the toll road will yield a constant$10 million at the end of each year forever with no costs. So the first payment will be at t=4.
The required return of the project is 10% pa given as an effective nominal rate. All cash flows are nominal.
What is the payback period?
A company selling charting and technical analysis software claims that independent academic studies have shown that its software makes significantly positive abnormal returns. Assuming the claim is true, which statement(s) are correct?
(I) Weak form market efficiency is broken.
(II) Semi-strong form market efficiency is broken.
(III) Strong form market efficiency is broken.
(IV) The asset pricing model used to measure the abnormal returns (such as the CAPM) had mis-specification error so the returns may not be abnormal but rather fair for the level of risk.
Select the most correct response:
Find Scubar Corporation's Cash Flow From Assets (CFFA), also known as Free Cash Flow to the Firm (FCFF), over the year ending 30th June 2013.
Scubar Corp Income Statement for year ending 30th June 2013 $m Sales 200 COGS 60 Depreciation 20 Rent expense 11 Interest expense 19 Taxable Income 90 Taxes at 30% 27 Net income 63 Scubar Corp Balance Sheet as at 30th June 2013 2012$m $m Inventory 60 50 Trade debtors 19 6 Rent paid in advance 3 2 PPE 420 400 Total assets 502 458 Trade creditors 10 8 Bond liabilities 200 190 Contributed equity 130 130 Retained profits 162 130 Total L and OE 502 458 Note: All figures are given in millions of dollars ($m).
The cash flow from assets was:
All things remaining equal, the higher the correlation of returns between two stocks:
A firm plans to issue equity and use the cash raised to pay off its debt. No assets will be bought or sold. Ignore the costs of financial distress.
Which of the following statements is NOT correct, all things remaining equal?
A company conducts a 2 for 3 rights issue at a subscription price of $8 when the pre-announcement stock price was$9. Assume that all investors use their rights to buy those extra shares.
What is the percentage increase in the stock price and the number of shares outstanding? The answers are given in the same order.
Mr Blue, Miss Red and Mrs Green are people with different utility functions. Which of the statements about the 3 utility functions is NOT correct?
A European bond paying annual coupons of 6% offers a yield of 10% pa.
Convert the yield into an effective monthly rate, an effective annual rate and an effective daily rate. Assume that there are 365 days in a year.
All answers are given in the same order:
$$r_\text{eff, monthly} , r_\text{eff, yearly} , r_\text{eff, daily}$$
In Germany, nominal yields on semi-annual coupon paying Government Bonds with 2 years until maturity are currently 0.04% pa.
The inflation rate is currently 1.4% pa, given as an APR compounding per quarter. The inflation rate is not expected to change over the next 2 years.
What is the real yield on these bonds, given as an APR compounding every 6 months?
In late 2003 the listed bank ANZ announced a 2-for-11 rights issue to fund the takeover of New Zealand bank NBNZ. Below is the chronology of events:
• 23/10/2003. Share price closes at $18.30. • 24/10/2003. 2-for-11 rights issue announced at a subscription price of$13. The proceeds of the rights issue will be used to acquire New Zealand bank NBNZ. Trading halt announced in morning before market opens.
• 28/10/2003. Trading halt lifted. Last (and only) day that shares trade cum-rights. Share price opens at $18.00 and closes at$18.14.
All things remaining equal, what would you expect ANZ's stock price to open at on the first day that it trades ex-rights (29/10/2003)? Ignore the time value of money since time is negligibly short. Also ignore taxes.
The hardest and most important aspect of business project valuation is the estimation of the:
A risky firm will last for one period only (t=0 to 1), then it will be liquidated. So it's assets will be sold and the debt holders and equity holders will be paid out in that order. The firm has the following quantities:
$V$ = Market value of assets.
$E$ = Market value of (levered) equity.
$D$ = Market value of zero coupon bonds.
$F_1$ = Total face value of zero coupon bonds which is promised to be paid in one year.
What is the payoff to equity holders at maturity, assuming that they keep their shares until maturity?
How can a nominal cash flow be precisely converted into a real cash flow?
The efficient markets hypothesis (EMH) and no-arbitrage pricing theory is most closely related to which of the following concepts?
How much more can you borrow using an interest-only loan compared to a 25-year fully amortising loan if interest rates are 6% pa compounding per month and are not expected to change? If it makes it easier, assume that you can afford to pay $2,000 per month on either loan. Express your answer as a proportional increase using the following formula: $$\text{Proportional Increase} = \dfrac{V_\text{0,interest only}}{V_\text{0,fully amortising}} - 1$$ For a price of$1040, Camille will sell you a share which just paid a dividend of $100, and is expected to pay dividends every year forever, growing at a rate of 5% pa. So the next dividend will be $100(1+0.05)^1=105.00$, and the year after it will be $100(1+0.05)^2=110.25$ and so on. The required return of the stock is 15% pa. Would you like to the share or politely ? The theory of fixed interest bond pricing is an application of the theory of Net Present Value (NPV). Also, a 'fairly priced' asset is not over- or under-priced. Buying or selling a fairly priced asset has an NPV of zero. Considering this, which of the following statements is NOT correct? A project's net present value (NPV) is negative. Select the most correct statement. A company's shares just paid their annual dividend of$2 each.
The stock price is now $40 (just after the dividend payment). The annual dividend is expected to grow by 3% every year forever. The assumptions of the dividend discount model are valid for this company. What do you expect the effective annual dividend yield to be in 3 years (dividend yield from t=3 to t=4)? In Australia in the 1980's, inflation was around 8% pa, and residential mortgage loan interest rates were around 14%. In 2013, inflation was around 2.5% pa, and residential mortgage loan interest rates were around 4.5%. If a person can afford constant mortgage loan payments of$2,000 per month, how much more can they borrow when interest rates are 4.5% pa compared with 14.0% pa?
Give your answer as a proportional increase over the amount you could borrow when interest rates were high $(V_\text{high rates})$, so:
$$\text{Proportional increase} = \dfrac{V_\text{low rates}-V_\text{high rates}}{V_\text{high rates}}$$
Assume that:
• Interest rates are expected to be constant over the life of the loan.
• Loans are interest-only and have a life of 30 years.
• Mortgage loan payments are made every month in arrears and all interest rates are given as annualised percentage rates (APR's) compounding per month.
A firm has 1 million shares which trade at a price of $30 each. The firm is expected to announce earnings of$3 million at the end of the year and pay an annual dividend of $1.50 per share. What is the firm's (forward looking) price/earnings (PE) ratio? An effective semi-annual return of 5% $(r_\text{eff 6mth})$ is equivalent to an effective annual return $(r_\text{eff annual})$ of: What is the NPV of the following series of cash flows when the discount rate is 5% given as an effective annual rate? The first payment of$10 is in 4 years, followed by payments every 6 months forever after that which shrink by 2% every 6 months. That is, the growth rate every 6 months is actually negative 2%, given as an effective 6 month rate. So the payment at $t=4.5$ years will be $10(1-0.02)^1=9.80$, and so on.
Government bonds currently have a return of 5%. A stock has a beta of 2 and the market return is 7%. What is the expected return of the stock?
Which statement is the most correct?
A stock has a beta of 0.5. Its next dividend is expected to be $3, paid one year from now. Dividends are expected to be paid annually and grow by 2% pa forever. Treasury bonds yield 5% pa and the market portfolio's expected return is 10% pa. All returns are effective annual rates. What is the price of the stock now? A managed fund charges fees based on the amount of money that you keep with them. The fee is 2% of the start-of-year amount, but it is paid at the end of every year. This fee is charged regardless of whether the fund makes gains or losses on your money. The fund offers to invest your money in shares which have an expected return of 10% pa before fees. You are thinking of investing$100,000 in the fund and keeping it there for 40 years when you plan to retire.
What is the Net Present Value (NPV) of investing your money in the fund? Note that the question is not asking how much money you will have in 40 years, it is asking: what is the NPV of investing in the fund? Assume that:
• The fund has no private information.
• Markets are weak and semi-strong form efficient.
• The fund's transaction costs are negligible.
• The cost and trouble of investing your money in shares by yourself, without the managed fund, is negligible.
Below are some statements about loans and bonds. The first descriptive sentence is correct. But one of the second sentences about the loans' or bonds' prices is not correct. Which statement is NOT correct? Assume that interest rates are positive.
Note that coupons or interest payments are the periodic payments made throughout a bond or loan's life. The face or par value of a bond or loan is the amount paid at the end when the debt matures.
A company conducts a 10 for 3 stock split. What is the percentage increase in the stock price and the number of shares outstanding? The answers are given in the same order.
Three important classes of investable risky assets are:
• Corporate debt which has low total risk,
• Real estate which has medium total risk,
• Equity which has high total risk.
Assume that the correlation between total returns on:
• Corporate debt and real estate is 0.1,
• Corporate debt and equity is 0.1,
• Real estate and equity is 0.5.
You are considering investing all of your wealth in one or more of these asset classes. Which portfolio will give the lowest total risk? You are restricted from shorting any of these assets. Disregard returns and the risk-return trade-off, pretend that you are only concerned with minimising risk.
What is the net present value (NPV) of undertaking a full-time Australian undergraduate business degree as an Australian citizen? Only include the cash flows over the duration of the degree, ignore any benefits or costs of the degree after it's completed.
Assume the following:
• The degree takes 3 years to complete and all students pass all subjects.
• There are 2 semesters per year and 4 subjects per semester.
• University fees per subject per semester are $1,277, paid at the start of each semester. Fees are expected to stay constant for the next 3 years. • There are 52 weeks per year. • The first semester is just about to start (t=0). The first semester lasts for 19 weeks (t=0 to 19). • The second semester starts immediately afterwards (t=19) and lasts for another 19 weeks (t=19 to 38). • The summer holidays begin after the second semester ends and last for 14 weeks (t=38 to 52). Then the first semester begins the next year, and so on. • Working full time at the grocery store instead of studying full-time pays$20/hr and you can work 35 hours per week. Wages are paid at the end of each week.
• Full-time students can work full-time during the summer holiday at the grocery store for the same rate of $20/hr for 35 hours per week. Wages are paid at the end of each week. • The discount rate is 9.8% pa. All rates and cash flows are real. Inflation is expected to be 3% pa. All rates are effective annual. The NPV of costs from undertaking the university degree is: Companies must pay interest and principal payments to debt-holders. They're compulsory. But companies are not forced to pay dividends to share holders. or ? What is the lowest and highest expected share price and expected return from owning shares in a company over a finite period of time? Let the current share price be $p_0$, the expected future share price be $p_1$, the expected future dividend be $d_1$ and the expected return be $r$. Define the expected return as: $r=\dfrac{p_1-p_0+d_1}{p_0}$ The answer choices are stated using inequalities. As an example, the first answer choice "(a) $0≤p<∞$ and $0≤r< 1$", states that the share price must be larger than or equal to zero and less than positive infinity, and that the return must be larger than or equal to zero and less than one. A mining firm has just discovered a new mine. So far the news has been kept a secret. The net present value of digging the mine and selling the minerals is$250 million, but $500 million of new equity and$300 million of new bonds will need to be issued to fund the project and buy the necessary plant and equipment.
The firm will release the news of the discovery and equity and bond raising to shareholders simultaneously in the same announcement. The shares and bonds will be issued shortly after.
Once the announcement is made and the new shares and bonds are issued, what is the expected increase in the value of the firm's assets $(\Delta V)$, market capitalisation of debt $(\Delta D)$ and market cap of equity $(\Delta E)$? Assume that markets are semi-strong form efficient.
The triangle symbol $\Delta$ is the Greek letter capital delta which means change or increase in mathematics.
Ignore the benefit of interest tax shields from having more debt.
Remember: $\Delta V = \Delta D+ \Delta E$
A firm is considering a business project which costs $11m now and is expected to pay a constant$1m at the end of every year forever.
Assume that the initial $11m cost is funded using the firm's existing cash so no new equity or debt will be raised. The cost of capital is 10% pa. Which of the following statements about net present value (NPV), internal rate of return (IRR) and payback period is NOT correct? A continuously compounded semi-annual return of 5% $(r_\text{cc 6mth})$ is equivalent to a continuously compounded annual return $(r_\text{cc annual})$ of: For a price of$6, Carlos will sell you a share which will pay a dividend of $1 in one year and every year after that forever. The required return of the stock is 10% pa. Would you like to his share or politely ? A firm has a debt-to-assets ratio of 50%. The firm then issues a large amount of equity to raise money for new projects of similar systematic risk to the company's existing projects. Assume a classical tax system. Which statement is correct? Bonds X and Y are issued by the same company. Both bonds yield 10% pa, and they have the same face value ($100), maturity, seniority, and payment frequency.
The only difference is that bond X pays coupons of 6% pa and bond Y pays coupons of 8% pa. Which of the following statements is true?
Bonds X and Y are issued by the same US company. Both bonds yield 6% pa, and they have the same face value ($100), maturity, seniority, and payment frequency. The only difference is that bond X pays coupons of 8% pa and bond Y pays coupons of 12% pa. Which of the following statements is true? Who is most in danger of being personally bankrupt? Assume that all of their businesses' assets are highly liquid and can therefore be sold immediately. An investor wants to make a portfolio of two stocks A and B with a target expected portfolio return of 16% pa. • Stock A has an expected return of 8% pa. • Stock B has an expected return of 12% pa. What portfolio weights should the investor have in stocks A and B respectively? If a project's net present value (NPV) is zero, then its internal rate of return (IRR) will be: A two year Government bond has a face value of$100, a yield of 2.5% pa and a fixed coupon rate of 0.5% pa, paid semi-annually. What is its price?
A European company just issued two bonds, a
• 3 year zero coupon bond at a yield of 6% pa, and a
• 4 year zero coupon bond at a yield of 6.5% pa.
What is the company's forward rate over the fourth year (from t=3 to t=4)? Give your answer as an effective annual rate, which is how the above bond yields are quoted.
Imagine that the interest rate on your savings account was 1% per year and inflation was 2% per year.
After one year, would you be able to buy , exactly the as or than today with the money in this account?
Which firms tend to have low forward-looking price-earnings (PE) ratios?
Only consider firms with positive earnings, disregard firms with negative earnings and therefore negative PE ratios.
Project Data Project life 1 year Initial investment in equipment $6m Depreciation of equipment per year$6m Expected sale price of equipment at end of project 0 Unit sales per year 9m Sale price per unit $8 Variable cost per unit$6 Fixed costs per year, paid at the end of each year $1m Interest expense in first year (at t=1)$0.53m Tax rate 30% Government treasury bond yield 5% Bank loan debt yield 6% Market portfolio return 10% Covariance of levered equity returns with market 0.08 Variance of market portfolio returns 0.16 Firm's and project's debt-to-assets ratio 50%
Notes
1. Due to the project, current assets will increase by $5m now (t=0) and fall by$5m at the end (t=1). Current liabilities will not be affected.
Assumptions
• The debt-to-assets ratio will be kept constant throughout the life of the project. The amount of interest expense at the end of each period has been correctly calculated to maintain this constant debt-to-equity ratio.
• Millions are represented by 'm'.
• All cash flows occur at the start or end of the year as appropriate, not in the middle or throughout the year.
• All rates and cash flows are real. The inflation rate is 2% pa.
• All rates are given as effective annual rates.
• The 50% capital gains tax discount is not available since the project is undertaken by a firm, not an individual.
What is the net present value (NPV) of the project?
What is the correlation of a variable X with a constant C?
The corr(X, C) or $\rho_{X,C}$ equals:
Portfolio Details Stock Expected return Standard deviation Correlation $(\rho_{A,B})$ Dollars invested A 0.1 0.4 0.5 60 B 0.2 0.6 140
What is the standard deviation (not variance) of the above portfolio?
According to the theory of the Capital Asset Pricing Model (CAPM), total risk can be broken into two components, systematic risk and idiosyncratic risk. Which of the following events would be considered a systematic, undiversifiable event according to the theory of the CAPM?
You are a banker about to grant a 2 year loan to a customer. The loan's principal and interest will be repaid in a single payment at maturity, sometimes called a zero-coupon loan, discount loan or bullet loan.
You require a real return of 6% pa over the two years, given as an effective annual rate. Inflation is expected to be 2% this year and 4% next year, both given as effective annual rates.
You judge that the customer can afford to pay back $1,000,000 in 2 years, given as a nominal cash flow. How much should you lend to her right now? A project has the following cash flows: Project Cash Flows Time (yrs) Cash flow ($) 0 -400 1 200 2 250
What is the Profitability Index (PI) of the project? Assume that the cash flows shown in the table are paid all at once at the given point in time. The required return is 10% pa, given as an effective annual rate.
A stock just paid its annual dividend of $9. The share price is$60. The required return of the stock is 10% pa as an effective annual rate.
What is the implied growth rate of the dividend per year?
You have $100,000 in the bank. The bank pays interest at 10% pa, given as an effective annual rate. You wish to consume an equal amount now (t=0) and in one year (t=1) and have nothing left in the bank at the end (t=1). How much can you consume at each time? A company announces that it will pay a dividend, as the market expected. The company's shares trade on the stock exchange which is open from 10am in the morning to 4pm in the afternoon each weekday. When would the share price be expected to fall by the amount of the dividend? Ignore taxes. The share price is expected to fall during the: The 'time value of money' is most closely related to which of the following concepts? Katya offers to pay you$10 at the end of every year for the next 5 years (t=1,2,3,4,5) if you pay her $50 now (t=0). You can borrow and lend from the bank at an interest rate of 10% pa, given as an effective annual rate. Ignore credit risk. Will you or Katya's deal? This annuity formula $\dfrac{C_1}{r}\left(1-\dfrac{1}{(1+r)^3} \right)$ is equivalent to which of the following formulas? Note the 3. In the below formulas, $C_t$ is a cash flow at time t. All of the cash flows are equal, but paid at different times. Your friend overheard that you need some cash and asks if you would like to borrow some money. She can lend you$5,000 now (t=0), and in return she wants you to pay her back $1,000 in two years (t=2) and every year after that for the next 5 years, so there will be 6 payments of$1,000 from t=2 to t=7 inclusive.
What is the net present value (NPV) of borrowing from your friend?
Assume that banks loan funds at interest rates of 10% pa, given as an effective annual rate.
Some countries' interest rates are so low that they're zero.
If interest rates are 0% pa and are expected to stay at that level for the foreseeable future, what is the most that you would be prepared to pay a bank now if it offered to pay you $10 at the end of every year for the next 5 years? In other words, what is the present value of five$10 payments at time 1, 2, 3, 4 and 5 if interest rates are 0% pa?
A stock is expected to pay its next dividend of $1 in one year. Future annual dividends are expected to grow by 2% pa. So the first dividend of$1 will be in one year, the year after that $1.02 (=1*(1+0.02)^1), and a year later$1.0404 (=1*(1+0.02)^2) and so on forever.
Its required total return is 10% pa. The total required return and growth rate of dividends are given as effective annual rates.
Calculate the current stock price.
A stock just paid a dividend of $1. Future annual dividends are expected to grow by 2% pa. The next dividend of$1.02 (=1*(1+0.02)^1) will be in one year, and the year after that the dividend will be $1.0404 (=1*(1+0.02)^2), and so on forever. Its required total return is 10% pa. The total required return and growth rate of dividends are given as effective annual rates. Calculate the current stock price. The perpetuity with growth formula, also known as the dividend discount model (DDM) or Gordon growth model, is appropriate for valuing a company's shares. $P_0$ is the current share price, $C_1$ is next year's expected dividend, $r$ is the total required return and $g$ is the expected growth rate of the dividend. $$P_0=\dfrac{C_1}{r-g}$$ The below graph shows the expected future price path of the company's shares. Which of the following statements about the graph is NOT correct? The following equation is the Dividend Discount Model, also known as the 'Gordon Growth Model' or the 'Perpetuity with growth' equation. $$P_0=\frac{d_1}{r-g}$$ A stock pays dividends annually. It just paid a dividend, but the next dividend ($d_1$) will be paid in one year. According to the DDM, what is the correct formula for the expected price of the stock in 2.5 years? The following equation is the Dividend Discount Model, also known as the 'Gordon Growth Model' or the 'Perpetuity with growth' equation. $$P_{0} = \frac{C_1}{r_{\text{eff}} - g_{\text{eff}}}$$ What would you call the expression $C_1/P_0$? The following is the Dividend Discount Model (DDM) used to price stocks: $$P_0=\dfrac{C_1}{r-g}$$ If the assumptions of the DDM hold, which one of the following statements is NOT correct? The long term expected: A stock will pay you a dividend of$10 tonight if you buy it today. Thereafter the annual dividend is expected to grow by 5% pa, so the next dividend after the $10 one tonight will be$10.50 in one year, then in two years it will be $11.025 and so on. The stock's required return is 10% pa. What is the stock price today and what do you expect the stock price to be tomorrow, approximately? In the dividend discount model: $$P_0 = \dfrac{C_1}{r-g}$$ The return $r$ is supposed to be the: A stock pays annual dividends which are expected to continue forever. It just paid a dividend of$10. The growth rate in the dividend is 2% pa. You estimate that the stock's required return is 10% pa. Both the discount rate and growth rate are given as effective annual rates. Using the dividend discount model, what will be the share price?
A stock is expected to pay the following dividends:
Cash Flows of a Stock Time (yrs) 0 1 2 3 4 ... Dividend ($) 0.00 1.00 1.05 1.10 1.15 ... After year 4, the annual dividend will grow in perpetuity at 5% pa, so; • the dividend at t=5 will be$1.15(1+0.05),
• the dividend at t=6 will be $1.15(1+0.05)^2, and so on. The required return on the stock is 10% pa. Both the growth rate and required return are given as effective annual rates. What will be the price of the stock in three and a half years (t = 3.5)? A fairly valued share's current price is$4 and it has a total required return of 30%. Dividends are paid annually and next year's dividend is expected to be $1. After that, dividends are expected to grow by 5% pa in perpetuity. All rates are effective annual returns. What is the expected dividend income paid at the end of the second year (t=2) and what is the expected capital gain from just after the first dividend (t=1) to just after the second dividend (t=2)? The answers are given in the same order, the dividend and then the capital gain. Most listed Australian companies pay dividends twice per year, the 'interim' and 'final' dividends, which are roughly 6 months apart. You are an equities analyst trying to value the company BHP. You decide to use the Dividend Discount Model (DDM) as a starting point, so you study BHP's dividend history and you find that BHP tends to pay the same interim and final dividend each year, and that both grow by the same rate. You expect BHP will pay a$0.55 interim dividend in six months and a $0.55 final dividend in one year. You expect each to grow by 4% next year and forever, so the interim and final dividends next year will be$0.572 each, and so on in perpetuity.
Assume BHP's cost of equity is 8% pa. All rates are quoted as nominal effective rates. The dividends are nominal cash flows and the inflation rate is 2.5% pa.
What is the current price of a BHP share?
You are an equities analyst trying to value the equity of the Australian telecoms company Telstra, with ticker TLS. In Australia, listed companies like Telstra tend to pay dividends every 6 months. The payment around August is called the final dividend and the payment around February is called the interim dividend. Both occur annually.
• Today is mid-March 2015.
• TLS's last interim dividend of $0.15 was one month ago in mid-February 2015. • TLS's last final dividend of$0.15 was seven months ago in mid-August 2014.
Judging by TLS's dividend history and prospects, you estimate that the nominal dividend growth rate will be 1% pa. Assume that TLS's total nominal cost of equity is 6% pa. The dividends are nominal cash flows and the inflation rate is 2.5% pa. All rates are quoted as nominal effective annual rates. Assume that each month is exactly one twelfth (1/12) of a year, so you can ignore the number of days in each month.
Calculate the current TLS share price.
Two companies BigDiv and ZeroDiv are exactly the same except for their dividend payouts.
BigDiv pays large dividends and ZeroDiv doesn't pay any dividends.
Currently the two firms have the same earnings, assets, number of shares, share price, expected total return and risk.
Assume a perfect world with no taxes, no transaction costs, no asymmetric information and that all assets including business projects are fairly priced and therefore zero-NPV.
All things remaining equal, which of the following statements is NOT correct?
Estimate the US bank JP Morgan's share price using a price earnings (PE) multiples approach with the following assumptions and figures only:
• The major US banks JP Morgan Chase (JPM), Citi Group (C) and Wells Fargo (WFC) are comparable companies;
• JP Morgan Chase's historical earnings per share (EPS) is $4.37; • Citi Group's share price is$50.05 and historical EPS is $4.26; • Wells Fargo's share price is$48.98 and historical EPS is $3.89. Note: Figures sourced from Google Finance on 24 March 2014. Estimate Microsoft's (MSFT) share price using a price earnings (PE) multiples approach with the following assumptions and figures only: • Apple, Google and Microsoft are comparable companies, • Apple's (AAPL) share price is$526.24 and historical EPS is $40.32. • Google's (GOOG) share price is$1,215.65 and historical EPS is $36.23. • Micrsoft's (MSFT) historical earnings per share (EPS) is$2.71.
Source: Google Finance 28 Feb 2014.
Details of two different types of light bulbs are given below:
• Low-energy light bulbs cost $3.50, have a life of nine years, and use about$1.60 of electricity a year, paid at the end of each year.
• Conventional light bulbs cost only $0.50, but last only about a year and use about$6.60 of energy a year, paid at the end of each year.
The real discount rate is 5%, given as an effective annual rate. Assume that all cash flows are real. The inflation rate is 3% given as an effective annual rate.
Find the Equivalent Annual Cost (EAC) of the low-energy and conventional light bulbs. The below choices are listed in that order.
Carlos and Edwin are brothers and they both love Holden Commodore cars.
Carlos likes to buy the latest Holden Commodore car for $40,000 every 4 years as soon as the new model is released. As soon as he buys the new car, he sells the old one on the second hand car market for$20,000. Carlos never has to bother with paying for repairs since his cars are brand new.
Edwin also likes Commodores, but prefers to buy 4-year old cars for $20,000 and keep them for 11 years until the end of their life (new ones last for 15 years in total but the 4-year old ones only last for another 11 years). Then he sells the old car for$2,000 and buys another 4-year old second hand car, and so on.
Every time Edwin buys a second hand 4 year old car he immediately has to spend $1,000 on repairs, and then$1,000 every year after that for the next 10 years. So there are 11 payments in total from when the second hand car is bought at t=0 to the last payment at t=10. One year later (t=11) the old car is at the end of its total 15 year life and can be scrapped for $2,000. Assuming that Carlos and Edwin maintain their love of Commodores and keep up their habits of buying new ones and second hand ones respectively, how much larger is Carlos' equivalent annual cost of car ownership compared with Edwin's? The real discount rate is 10% pa. All cash flows are real and are expected to remain constant. Inflation is forecast to be 3% pa. All rates are effective annual. Ignore capital gains tax and tax savings from depreciation since cars are tax-exempt for individuals. You own a nice suit which you wear once per week on nights out. You bought it one year ago for$600. In your experience, suits used once per week last for 6 years. So you expect yours to last for another 5 years.
Your younger brother said that retro is back in style so he wants to wants to borrow your suit once a week when he goes out. With the increased use, your suit will only last for another 4 years rather than 5.
What is the present value of the cost of letting your brother use your current suit for the next 4 years?
Assume: that bank interest rates are 10% pa, given as an effective annual rate; you will buy a new suit when your current one wears out and your brother will not use the new one; your brother will only use your current suit so he will only use it for the next four years; and the price of a new suit never changes.
You own some nice shoes which you use once per week on date nights. You bought them 2 years ago for $500. In your experience, shoes used once per week last for 6 years. So you expect yours to last for another 4 years. Your younger sister said that she wants to borrow your shoes once per week. With the increased use, your shoes will only last for another 2 years rather than 4. What is the present value of the cost of letting your sister use your current shoes for the next 2 years? Assume: that bank interest rates are 10% pa, given as an effective annual rate; you will buy a new pair of shoes when your current pair wears out and your sister will not use the new ones; your sister will only use your current shoes so she will only use it for the next 2 years; and the price of new shoes never changes. Which of the following statements is NOT correct? Borrowers: Which of the following statements is NOT correct? Lenders: A home loan company advertises an interest rate of 6% pa, payable monthly. Which of the following statements about the interest rate is NOT correct? All rates are given to four decimal places. A semi-annual coupon bond has a yield of 3% pa. Which of the following statements about the yield is NOT correct? All rates are given to four decimal places. Which of the below statements about effective rates and annualised percentage rates (APR's) is NOT correct? On his 20th birthday, a man makes a resolution. He will deposit$30 into a bank account at the end of every month starting from now, which is the start of the month. So the first payment will be in one month. He will write in his will that when he dies the money in the account should be given to charity.
The bank account pays interest at 6% pa compounding monthly, which is not expected to change.
If the man lives for another 60 years, how much money will be in the bank account if he dies just after making his last (720th) payment?
You want to buy an apartment worth $300,000. You have saved a deposit of$60,000.
The bank has agreed to lend you $240,000 as an interest only mortgage loan with a term of 30 years. The interest rate is 6% pa and is not expected to change. What will be your monthly payments? A bank grants a borrower an interest-only residential mortgage loan with a very large 50% deposit and a nominal interest rate of 6% that is not expected to change. Assume that inflation is expected to be a constant 2% pa over the life of the loan. Ignore credit risk. From the bank's point of view, what is the long term expected nominal capital return of the loan asset? An 'interest payment' is the same thing as a 'coupon payment'. or ? An 'interest rate' is the same thing as a 'coupon rate'. or ? An 'interest rate' is the same thing as a 'yield'. or ? An 'interest only' loan can also be called a: "Buy low, sell high" is a phrase commonly heard in financial markets. It states that traders should try to buy assets at low prices and sell at high prices. Traders in the fixed-coupon bond markets often quote promised bond yields rather than prices. Fixed-coupon bond traders should try to: Bonds X and Y are issued by the same US company. Both bonds yield 10% pa, and they have the same face value ($100), maturity, seniority, and payment frequency.
The only difference is that bond X and Y's coupon rates are 8 and 12% pa respectively. Which of the following statements is true?
Which one of the following bonds is trading at a discount?
Let the 'income return' of a bond be the coupon at the end of the period divided by the market price now at the start of the period $(C_1/P_0)$. The expected income return of a premium fixed coupon bond is:
In these tough economic times, central banks around the world have cut interest rates so low that they are practically zero. In some countries, government bond yields are also very close to zero.
A three year government bond with a face value of $100 and a coupon rate of 2% pa paid semi-annually was just issued at a yield of 0%. What is the price of the bond? A 10 year bond has a face value of$100, a yield of 6% pa and a fixed coupon rate of 8% pa, paid semi-annually. What is its price?
A European company just issued two bonds, a
• 1 year zero coupon bond at a yield of 8% pa, and a
• 2 year zero coupon bond at a yield of 10% pa.
What is the company's forward rate over the second year (from t=1 to t=2)? Give your answer as an effective annual rate, which is how the above bond yields are quoted.
An Australian company just issued two bonds:
• A 6-month zero coupon bond at a yield of 6% pa, and
• A 12 month zero coupon bond at a yield of 7% pa.
What is the company's forward rate from 6 to 12 months? Give your answer as an APR compounding every 6 months, which is how the above bond yields are quoted.
A young lady is trying to decide if she should attend university or not.
The young lady's parents say that she must attend university because otherwise all of her hard work studying and attending school during her childhood was a waste.
What's the correct way to classify this item from a capital budgeting perspective when trying to decide whether to attend university?
The hard work studying at school in her childhood should be classified as:
Find Sidebar Corporation's Cash Flow From Assets (CFFA), also known as Free Cash Flow to the Firm (FCFF), over the year ending 30th June 2013.
Sidebar Corp Income Statement for year ending 30th June 2013 $m Sales 405 COGS 100 Depreciation 34 Rent expense 22 Interest expense 39 Taxable Income 210 Taxes at 30% 63 Net income 147 Sidebar Corp Balance Sheet as at 30th June 2013 2012$m $m Inventory 70 50 Trade debtors 11 16 Rent paid in advance 4 3 PPE 700 680 Total assets 785 749 Trade creditors 11 19 Bond liabilities 400 390 Contributed equity 220 220 Retained profits 154 120 Total L and OE 785 749 Note: All figures are given in millions of dollars ($m).
The cash flow from assets was:
Why is Capital Expenditure (CapEx) subtracted in the Cash Flow From Assets (CFFA) formula?
$$CFFA=NI+Depr-CapEx - \Delta NWC+IntExp$$
A firm has forecast its Cash Flow From Assets (CFFA) for this year and management is worried that it is too low. Which one of the following actions will lead to a higher CFFA for this year (t=0 to 1)? Only consider cash flows this year. Do not consider cash flows after one year, or the change in the NPV of the firm. Consider each action in isolation.
Over the next year, the management of an unlevered company plans to:
• Achieve firm free cash flow (FFCF or CFFA) of $1m. • Pay dividends of$1.8m
• Complete a $1.3m share buy-back. • Spend$0.8m on new buildings without buying or selling any other fixed assets. This capital expenditure is included in the CFFA figure quoted above.
Assume that:
• All amounts are received and paid at the end of the year so you can ignore the time value of money.
• The firm has sufficient retained profits to pay the dividend and complete the buy back.
• The firm plans to run a very tight ship, with no excess cash above operating requirements currently or over the next year.
How much new equity financing will the company need? In other words, what is the value of new shares that will need to be issued?
Over the next year, the management of an unlevered company plans to:
• Make $5m in sales,$1.9m in net income and $2m in equity free cash flow (EFCF). • Pay dividends of$1m.
• Complete a $1.3m share buy-back. Assume that: • All amounts are received and paid at the end of the year so you can ignore the time value of money. • The firm has sufficient retained profits to legally pay the dividend and complete the buy back. • The firm plans to run a very tight ship, with no excess cash above operating requirements currently or over the next year. How much new equity financing will the company need? In other words, what is the value of new shares that will need to be issued? Issuing debt doesn't give away control of the firm because debt holders can't cast votes to determine the company's affairs, such as at the annual general meeting (AGM), and can't appoint directors to the board. or ? Your friend just bought a house for$400,000. He financed it using a $320,000 mortgage loan and a deposit of$80,000.
In the context of residential housing and mortgages, the 'equity' tied up in the value of a person's house is the value of the house less the value of the mortgage. So the initial equity your friend has in his house is $80,000. Let this amount be E, let the value of the mortgage be D and the value of the house be V. So $V=D+E$. If house prices suddenly fall by 10%, what would be your friend's percentage change in equity (E)? Assume that the value of the mortgage is unchanged and that no income (rent) was received from the house during the short time over which house prices fell. Remember: $$r_{0\rightarrow1}=\frac{p_1-p_0+c_1}{p_0}$$ where $r_{0-1}$ is the return (percentage change) of an asset with price $p_0$ initially, $p_1$ one period later, and paying a cash flow of $c_1$ at time $t=1$. One year ago you bought$100,000 of shares partly funded using a margin loan. The margin loan size was $70,000 and the other$30,000 was your own wealth or 'equity' in the share assets.
The interest rate on the margin loan was 7.84% pa.
Over the year, the shares produced a dividend yield of 4% pa and a capital gain of 5% pa.
What was the total return on your wealth? Ignore taxes, assume that all cash flows (interest payments and dividends) were paid and received at the end of the year, and all rates above are effective annual rates.
Hint: Remember that wealth in this context is your equity (E) in the house asset (V = D+E) which is funded by the loan (D) and your deposit or equity (E).
Interest expense (IntExp) is an important part of a company's income statement (or 'profit and loss' or 'statement of financial performance').
How does an accountant calculate the annual interest expense of a fixed-coupon bond that has a liquid secondary market? Select the most correct answer:
Annual interest expense is equal to:
A firm has a debt-to-equity ratio of 25%. What is its debt-to-assets ratio?
A manufacturing company is considering a new project in the more risky services industry. The cash flows from assets (CFFA) are estimated for the new project, with interest expense excluded from the calculations. To get the levered value of the project, what should these unlevered cash flows be discounted by?
Assume that the manufacturing firm has a target debt-to-assets ratio that it sticks to.
Assume the following:
• Google had a 10% after-tax weighted average cost of capital (WACC) before it bought Motorola.
• Motorola had a 20% after-tax WACC before it merged with Google.
• Google and Motorola have the same level of gearing.
• Both companies operate in a classical tax system.
You are a manager at Motorola. You must value a project for making mobile phones. Which method(s) will give the correct valuation of the mobile phone manufacturing project? Select the most correct answer.
The mobile phone manufacturing project's:
There are many ways to calculate a firm's free cash flow (FFCF), also called cash flow from assets (CFFA). Some include the annual interest tax shield in the cash flow and some do not.
Which of the below FFCF formulas include the interest tax shield in the cash flow?
$$(1) \quad FFCF=NI + Depr - CapEx -ΔNWC + IntExp$$ $$(2) \quad FFCF=NI + Depr - CapEx -ΔNWC + IntExp.(1-t_c)$$ $$(3) \quad FFCF=EBIT.(1-t_c )+ Depr- CapEx -ΔNWC+IntExp.t_c$$ $$(4) \quad FFCF=EBIT.(1-t_c) + Depr- CapEx -ΔNWC$$ $$(5) \quad FFCF=EBITDA.(1-t_c )+Depr.t_c- CapEx -ΔNWC+IntExp.t_c$$ $$(6) \quad FFCF=EBITDA.(1-t_c )+Depr.t_c- CapEx -ΔNWC$$ $$(7) \quad FFCF=EBIT-Tax + Depr - CapEx -ΔNWC$$ $$(8) \quad FFCF=EBIT-Tax + Depr - CapEx -ΔNWC-IntExp.t_c$$ $$(9) \quad FFCF=EBITDA-Tax - CapEx -ΔNWC$$ $$(10) \quad FFCF=EBITDA-Tax - CapEx -ΔNWC-IntExp.t_c$$
The formulas for net income (NI also called earnings), EBIT and EBITDA are given below. Assume that depreciation and amortisation are both represented by 'Depr' and that 'FC' represents fixed costs such as rent.
$$NI=(Rev - COGS - Depr - FC - IntExp).(1-t_c )$$ $$EBIT=Rev - COGS - FC-Depr$$ $$EBITDA=Rev - COGS - FC$$ $$Tax =(Rev - COGS - Depr - FC - IntExp).t_c= \dfrac{NI.t_c}{1-t_c}$$
One formula for calculating a levered firm's free cash flow (FFCF, or CFFA) is to use earnings before interest and tax (EBIT).
\begin{aligned} FFCF &= (EBIT)(1-t_c) + Depr - CapEx -\Delta NWC + IntExp.t_c \\ &= (Rev - COGS - Depr - FC)(1-t_c) + Depr - CapEx -\Delta NWC + IntExp.t_c \\ \end{aligned} \\
Does this annual FFCF or the annual interest tax shield?
One formula for calculating a levered firm's free cash flow (FFCF, or CFFA) is to use net operating profit after tax (NOPAT).
\begin{aligned} FFCF &= NOPAT + Depr - CapEx -\Delta NWC \\ &= (Rev - COGS - Depr - FC)(1-t_c) + Depr - CapEx -\Delta NWC \\ \end{aligned} \\
Does this annual FFCF or the annual interest tax shield?
Project Data Project life 2 yrs Initial investment in equipment $600k Depreciation of equipment per year$250k Expected sale price of equipment at end of project $200k Revenue per job$12k Variable cost per job $4k Quantity of jobs per year 120 Fixed costs per year, paid at the end of each year$100k Interest expense in first year (at t=1) $16.091k Interest expense in second year (at t=2)$9.711k Tax rate 30% Government treasury bond yield 5% Bank loan debt yield 6% Levered cost of equity 12.5% Market portfolio return 10% Beta of assets 1.24 Beta of levered equity 1.5 Firm's and project's debt-to-equity ratio 25%
Notes
1. The project will require an immediate purchase of $50k of inventory, which will all be sold at cost when the project ends. Current liabilities are negligible so they can be ignored. Assumptions • The debt-to-equity ratio will be kept constant throughout the life of the project. The amount of interest expense at the end of each period has been correctly calculated to maintain this constant debt-to-equity ratio. Note that interest expense is different in each year. • Thousands are represented by 'k' (kilo). • All cash flows occur at the start or end of the year as appropriate, not in the middle or throughout the year. • All rates and cash flows are nominal. The inflation rate is 2% pa. • All rates are given as effective annual rates. • The 50% capital gains tax discount is not available since the project is undertaken by a firm, not an individual. What is the net present value (NPV) of the project? A company issues a large amount of bonds to raise money for new projects of similar risk to the company's existing projects. The net present value (NPV) of the new projects is positive but small. Assume a classical tax system. Which statement is NOT correct? A firm has a debt-to-assets ratio of 50%. The firm then issues a large amount of debt to raise money for new projects of similar risk to the company's existing projects. Assume a classical tax system. Which statement is correct? Which of the following statements about standard statistical mathematics notation is NOT correct? Diversification in a portfolio of two assets works best when the correlation between their returns is: All things remaining equal, the variance of a portfolio of two positively-weighted stocks rises as: An investor wants to make a portfolio of two stocks A and B with a target expected portfolio return of 6% pa. • Stock A has an expected return of 5% pa. • Stock B has an expected return of 10% pa. What portfolio weights should the investor have in stocks A and B respectively? An investor wants to make a portfolio of two stocks A and B with a target expected portfolio return of 12% pa. • Stock A has an expected return of 10% pa and a standard deviation of 20% pa. • Stock B has an expected return of 15% pa and a standard deviation of 30% pa. The correlation coefficient between stock A and B's expected returns is 70%. What will be the annual standard deviation of the portfolio with this 12% pa target return? What is the correlation of a variable X with itself? The corr(X, X) or $\rho_{X,X}$ equals: The covariance and correlation of two stocks X and Y's annual returns are calculated over a number of years. The units of the returns are in percent per annum $(\% pa)$. What are the units of the covariance $(\sigma_{X,Y})$ and correlation $(\rho_{X,Y})$ of returns respectively? Hint: Visit Wikipedia to understand the difference between percentage points $(\text{pp})$ and percent $(\%)$. Let the standard deviation of returns for a share per month be $\sigma_\text{monthly}$. What is the formula for the standard deviation of the share's returns per year $(\sigma_\text{yearly})$? Assume that returns are independently and identically distributed (iid) so they have zero auto correlation, meaning that if the return was higher than average today, it does not indicate that the return tomorrow will be higher or lower than average. Mr Blue, Miss Red and Mrs Green are people with different utility functions. Note that a fair gamble is a bet that has an expected value of zero, such as paying$0.50 to win $1 in a coin flip with heads or nothing if it lands tails. Fairly priced insurance is when the expected present value of the insurance premiums is equal to the expected loss from the disaster that the insurance protects against, such as the cost of rebuilding a home after a catastrophic fire. Which of the following statements is NOT correct? Mr Blue, Miss Red and Mrs Green are people with different utility functions. Which of the following statements is NOT correct? Mr Blue, Miss Red and Mrs Green are people with different utility functions. Each person has$500 of initial wealth. A coin toss game is offered to each person at a casino where the player can win or lose $500. Each player can flip a coin and if they flip heads, they receive$500. If they flip tails then they will lose $500. Which of the following statements is NOT correct? Mr Blue, Miss Red and Mrs Green are people with different utility functions. Each person has$256 of initial wealth. A coin toss game is offered to each person at a casino where the player can win or lose $256. Each player can flip a coin and if they flip heads, they receive$256. If they flip tails then they will lose $256. Which of the following statements is NOT correct? Which of the below statements about utility is NOT generally accepted by economists? Most people are thought to: Mr Blue, Miss Red and Mrs Green are people with different utility functions. Which of the statements about the 3 utility functions is NOT correct? Mr Blue, Miss Red and Mrs Green are people with different utility functions. Which of the statements about the 3 utility functions is NOT correct? Mr Blue, Miss Red and Mrs Green are people with different utility functions. Which of the statements about the 3 utility functions is NOT correct? Mr Blue, Miss Red and Mrs Green are people with different utility functions. Each person has$50 of initial wealth. A coin toss game is offered to each person at a casino where the player can win or lose $50. Each player can flip a coin and if they flip heads, they receive$50. If they flip tails then they will lose $50. Which of the following statements is NOT correct? Mr Blue, Miss Red and Mrs Green are people with different utility functions. Each person has$50 of initial wealth. A coin toss game is offered to each person at a casino where the player can win or lose $50. Each player can flip a coin and if they flip heads, they receive$50. If they flip tails then they will lose $50. Which of the following statements is NOT correct? Mr Blue, Miss Red and Mrs Green are people with different utility functions. Each person has$50 of initial wealth. A coin toss game is offered to each person at a casino where the player can win or lose $50. Each player can flip a coin and if they flip heads, they receive$50. If they flip tails then they will lose $50. Which of the following statements is NOT correct? Mr Blue, Miss Red and Mrs Green are people with different utility functions. Each person has$50 of initial wealth. A coin toss game is offered to each person at a casino where the player can win or lose $50. Each player can flip a coin and if they flip heads, they receive$50. If they flip tails then they will lose $50. Which of the following statements is NOT correct? Diversification is achieved by investing in a large amount of stocks. What type of risk is reduced by diversification? A fairly priced stock has an expected return equal to the market's. Treasury bonds yield 5% pa and the market portfolio's expected return is 10% pa. What is the stock's beta? The security market line (SML) shows the relationship between beta and expected return. Investment projects that plot above the SML would have: Stock A has a beta of 0.5 and stock B has a beta of 1. Which statement is NOT correct? A stock's correlation with the market portfolio increases while its total risk is unchanged. What will happen to the stock's expected return and systematic risk? Assets A, B, M and $r_f$ are shown on the graphs above. Asset M is the market portfolio and $r_f$ is the risk free yield on government bonds. Which of the below statements is NOT correct? Assets A, B, M and $r_f$ are shown on the graphs above. Asset M is the market portfolio and $r_f$ is the risk free yield on government bonds. Assume that investors can borrow and lend at the risk free rate. Which of the below statements is NOT correct? A stock has a beta of 1.5. The market's expected total return is 10% pa and the risk free rate is 5% pa, both given as effective annual rates. What do you think will be the stock's expected return over the next year, given as an effective annual rate? A stock has a beta of 1.5. The market's expected total return is 10% pa and the risk free rate is 5% pa, both given as effective annual rates. In the last 5 minutes, bad economic news was released showing a higher chance of recession. Over this time the share market fell by 1%. The risk free rate was unchanged. What do you think was the stock's historical return over the last 5 minutes, given as an effective 5 minute rate? A stock has a beta of 1.5. The market's expected total return is 10% pa and the risk free rate is 5% pa, both given as effective annual rates. Over the last year, bad economic news was released showing a higher chance of recession. Over this time the share market fell by 1%. The risk free rate was unchanged. What do you think was the stock's historical return over the last year, given as an effective annual rate? A firm changes its capital structure by issuing a large amount of equity and using the funds to repay debt. Its assets are unchanged. Ignore interest tax shields. According to the Capital Asset Pricing Model (CAPM), which statement is correct? The CAPM can be used to find a business's expected opportunity cost of capital: $$r_i=r_f+β_i (r_m-r_f)$$ What should be used as the risk free rate $r_f$? A firm's WACC before tax would decrease due to: Which of the following statements about the weighted average cost of capital (WACC) is NOT correct? Project Data Project life 1 year Initial investment in equipment$8m Depreciation of equipment per year $8m Expected sale price of equipment at end of project 0 Unit sales per year 4m Sale price per unit$10 Variable cost per unit $5 Fixed costs per year, paid at the end of each year$2m Interest expense in first year (at t=1) $0.562m Corporate tax rate 30% Government treasury bond yield 5% Bank loan debt yield 9% Market portfolio return 10% Covariance of levered equity returns with market 0.32 Variance of market portfolio returns 0.16 Firm's and project's debt-to-equity ratio 50% Notes 1. Due to the project, current assets will increase by$6m now (t=0) and fall by $6m at the end (t=1). Current liabilities will not be affected. Assumptions • The debt-to-equity ratio will be kept constant throughout the life of the project. The amount of interest expense at the end of each period has been correctly calculated to maintain this constant debt-to-equity ratio. • Millions are represented by 'm'. • All cash flows occur at the start or end of the year as appropriate, not in the middle or throughout the year. • All rates and cash flows are real. The inflation rate is 2% pa. All rates are given as effective annual rates. • The project is undertaken by a firm, not an individual. What is the net present value (NPV) of the project? According to the theory of the Capital Asset Pricing Model (CAPM), total variance can be broken into two components, systematic variance and idiosyncratic variance. Which of the following events would be considered the most diversifiable according to the theory of the CAPM? A stock's required total return will increase when its: Treasury bonds currently have a return of 5% pa. A stock has a beta of 0.5 and the market return is 10% pa. What is the expected return of the stock? The security market line (SML) shows the relationship between beta and expected return. Investment projects that plot on the SML would have: Examine the following graph which shows stocks' betas $(\beta)$ and expected returns $(\mu)$: Assume that the CAPM holds and that future expectations of stocks' returns and betas are correctly measured. Which statement is NOT correct? Portfolio Details Stock Expected return Standard deviation Correlation Beta Dollars invested A 0.2 0.4 0.12 0.5 40 B 0.3 0.8 1.5 80 What is the beta of the above portfolio? Which statement(s) are correct? (i) All stocks that plot on the Security Market Line (SML) are fairly priced. (ii) All stocks that plot above the Security Market Line (SML) are overpriced. (iii) All fairly priced stocks that plot on the Capital Market Line (CML) have zero idiosyncratic risk. Select the most correct response: A firm changes its capital structure by issuing a large amount of debt and using the funds to repurchase shares. Its assets are unchanged. Ignore interest tax shields. According to the Capital Asset Pricing Model (CAPM), which statement is correct? The total return of any asset can be broken down in different ways. One possible way is to use the dividend discount model (or Gordon growth model): $$p_0 = \frac{c_1}{r_\text{total}-r_\text{capital}}$$ Which, since $c_1/p_0$ is the income return ($r_\text{income}$), can be expressed as: $$r_\text{total}=r_\text{income}+r_\text{capital}$$ So the total return of an asset is the income component plus the capital or price growth component. Another way to break up total return is to use the Capital Asset Pricing Model: $$r_\text{total}=r_\text{f}+β(r_\text{m}- r_\text{f})$$ $$r_\text{total}=r_\text{time value}+r_\text{risk premium}$$ So the risk free rate is the time value of money and the term $β(r_\text{m}- r_\text{f})$ is the compensation for taking on systematic risk. Using the above theory and your general knowledge, which of the below equations, if any, are correct? (I) $r_\text{income}=r_\text{time value}$ (II) $r_\text{income}=r_\text{risk premium}$ (III) $r_\text{capital}=r_\text{time value}$ (IV) $r_\text{capital}=r_\text{risk premium}$ (V) $r_\text{income}+r_\text{capital}=r_\text{time value}+r_\text{risk premium}$ Which of the equations are correct? You just bought a house worth$1,000,000. You financed it with an $800,000 mortgage loan and a deposit of$200,000.
You estimate that:
• The house has a beta of 1;
• The mortgage loan has a beta of 0.2.
What is the beta of the equity (the $200,000 deposit) that you have in your house? Also, if the risk free rate is 5% pa and the market portfolio's return is 10% pa, what is the expected return on equity in your house? Ignore taxes, assume that all cash flows (interest payments and rent) were paid and received at the end of the year, and all rates are effective annual rates. A firm can issue 5 year annual coupon bonds at a yield of 8% pa and a coupon rate of 12% pa. The beta of its levered equity is 1. Five year government bonds yield 5% pa with a coupon rate of 6% pa. The market's expected dividend return is 4% pa and its expected capital return is 6% pa. The firm's debt-to-equity ratio is 2:1. The corporate tax rate is 30%. What is the firm's after-tax WACC? Assume a classical tax system. There are many different ways to value a firm's assets. Which of the following will NOT give the correct market value of a levered firm's assets $(V_L)$? Assume that: • The firm is financed by listed common stock and vanilla annual fixed coupon bonds, which are both traded in a liquid market. • The bonds' yield is equal to the coupon rate, so the bonds are issued at par. The yield curve is flat and yields are not expected to change. When bonds mature they will be rolled over by issuing the same number of new bonds with the same expected yield and coupon rate, and so on forever. • Tax rates on the dividends and capital gains received by investors are equal, and capital gains tax is paid every year, even on unrealised gains regardless of when the asset is sold. • There is no re-investment of the firm's cash back into the business. All of the firm's excess cash flow is paid out as dividends so real growth is zero. • The firm operates in a mature industry with zero real growth. • All cash flows and rates in the below equations are real (not nominal) and are expected to be stable forever. Therefore the perpetuity equation with no growth is suitable for valuation. Where: $$r_\text{WACC before tax} = r_D.\frac{D}{V_L} + r_{EL}.\frac{E_L}{V_L} = \text{Weighted average cost of capital before tax}$$ $$r_\text{WACC after tax} = r_D.(1-t_c).\frac{D}{V_L} + r_{EL}.\frac{E_L}{V_L} = \text{Weighted average cost of capital after tax}$$ $$NI_L=(Rev-COGS-FC-Depr-\mathbf{IntExp}).(1-t_c) = \text{Net Income Levered}$$ $$CFFA_L=NI_L+Depr-CapEx - \varDelta NWC+\mathbf{IntExp} = \text{Cash Flow From Assets Levered}$$ $$NI_U=(Rev-COGS-FC-Depr).(1-t_c) = \text{Net Income Unlevered}$$ $$CFFA_U=NI_U+Depr-CapEx - \varDelta NWC= \text{Cash Flow From Assets Unlevered}$$ Your friend claims that by reading 'The Economist' magazine's economic news articles, she can identify shares that will have positive abnormal expected returns over the next 2 years. Assuming that her claim is true, which statement(s) are correct? (i) Weak form market efficiency is broken. (ii) Semi-strong form market efficiency is broken. (iii) Strong form market efficiency is broken. (iv) The asset pricing model used to measure the abnormal returns (such as the CAPM) is either wrong (mis-specification error) or is measured using the wrong inputs (data errors) so the returns may not be abnormal but rather fair for the level of risk. Select the most correct response: Technical traders: Fundamentalists who analyse company financial reports and news announcements (but who don't have inside information) will make positive abnormal returns if: Economic statistics released this morning were a surprise: they show a strong chance of consumer price inflation (CPI) reaching 5% pa over the next 2 years. This is much higher than the previous forecast of 3% pa. A vanilla fixed-coupon 2-year risk-free government bond was issued at par this morning, just before the economic news was released. What is the expected change in bond price after the economic news this morning, and in the next 2 years? Assume that: • Inflation remains at 5% over the next 2 years. • Investors demand a constant real bond yield. • The bond price falls by the (after-tax) value of the coupon the night before the ex-coupon date, as in real life. A man inherits$500,000 worth of shares.
He believes that by learning the secrets of trading, keeping up with the financial news and doing complex trend analysis with charts that he can quit his job and become a self-employed day trader in the equities markets.
What is the expected gain from doing this over the first year? Measure the net gain in wealth received at the end of this first year due to the decision to become a day trader. Assume the following:
• He earns $60,000 pa in his current job, paid in a lump sum at the end of each year. • He enjoys examining share price graphs and day trading just as much as he enjoys his current job. • Stock markets are weak form and semi-strong form efficient. • He has no inside information. • He makes 1 trade every day and there are 250 trading days in the year. Trading costs are$20 per trade. His broker invoices him for the trading costs at the end of the year.
• The shares that he currently owns and the shares that he intends to trade have the same level of systematic risk as the market portfolio.
• The market portfolio's expected return is 10% pa.
Measure the net gain over the first year as an expected wealth increase at the end of the year.
A company advertises an investment costing $1,000 which they say is underpriced. They say that it has an expected total return of 15% pa, but a required return of only 10% pa. Assume that there are no dividend payments so the entire 15% total return is all capital return. Assuming that the company's statements are correct, what is the NPV of buying the investment if the 15% return lasts for the next 100 years (t=0 to 100), then reverts to 10% pa after that time? Also, what is the NPV of the investment if the 15% return lasts forever? In both cases, assume that the required return of 10% remains constant. All returns are given as effective annual rates. The answer choices below are given in the same order (15% for 100 years, and 15% forever): Select the most correct statement from the following. 'Chartists', also known as 'technical traders', believe that: The theory of fixed interest bond pricing is an application of the theory of Net Present Value (NPV). Also, a 'fairly priced' asset is not over- or under-priced. Buying or selling a fairly priced asset has an NPV of zero. Considering this, which of the following statements is NOT correct? An economy has only two investable assets: stocks and cash. Stocks had a historical nominal average total return of negative two percent per annum (-2% pa) over the last 20 years. Stocks are liquid and actively traded. Stock returns are variable, they have risk. Cash is riskless and has a nominal constant return of zero percent per annum (0% pa), which it had in the past and will have in the future. Cash can be kept safely at zero cost. Cash can be converted into shares and vice versa at zero cost. The nominal total return of the shares over the next year is expected to be: A person is thinking about borrowing$100 from the bank at 7% pa and investing it in shares with an expected return of 10% pa. One year later the person will sell the shares and pay back the loan in full. Both the loan and the shares are fairly priced.
What is the Net Present Value (NPV) of this one year investment? Note that you are asked to find the present value ($V_0$), not the value in one year ($V_1$).
A residential real estate investor believes that house prices will grow at a rate of 5% pa and that rents will grow by 2% pa forever.
All rates are given as nominal effective annual returns. Assume that:
• His forecast is true.
• Real estate is and always will be fairly priced and the capital asset pricing model (CAPM) is true.
• Ignore all costs such as taxes, agent fees, maintenance and so on.
• All rental income cash flow is paid out to the owner, so there is no re-investment and therefore no additions or improvements made to the property.
• The non-monetary benefits of owning real estate and renting remain constant.
Which one of the following statements is NOT correct? Over time:
The average weekly earnings of an Australian adult worker before tax was $1,542.40 per week in November 2014 according to the Australian Bureau of Statistics. Therefore average annual earnings before tax were$80,204.80 assuming 52 weeks per year. Personal income tax rates published by the Australian Tax Office are reproduced for the 2014-2015 financial year in the table below:
Taxable income Tax on this income
0 – $18,200 Nil$18,201 – $37,000 19c for each$1 over $18,200$37,001 – $80,000$3,572 plus 32.5c for each $1 over$37,000
$80,001 –$180,000 $17,547 plus 37c for each$1 over $80,000$180,001 and over $54,547 plus 45c for each$1 over $180,000 The above rates do not include the Medicare levy of 2%. Exclude the Medicare levy from your calculations How much personal income tax would you have to pay per year if you earned$80,204.80 per annum before-tax?
Question 449 personal tax on dividends, classical tax system
A small private company has a single shareholder. This year the firm earned a $100 profit before tax. All of the firm's after tax profits will be paid out as dividends to the owner. The corporate tax rate is 30% and the sole shareholder's personal marginal tax rate is 45%. The United States' classical tax system applies because the company generates all of its income in the US and pays corporate tax to the Internal Revenue Service. The shareholder is also an American for tax purposes. What will be the personal tax payable by the shareholder and the corporate tax payable by the company? Which of the following statements about Australian franking credits is NOT correct? Franking credits: A small private company has a single shareholder. This year the firm earned a$100 profit before tax. All of the firm's after tax profits will be paid out as dividends to the owner.
The corporate tax rate is 30% and the sole shareholder's personal marginal tax rate is 45%.
The Australian imputation tax system applies because the company generates all of its income in Australia and pays corporate tax to the Australian Tax Office. Therefore all of the company's dividends are fully franked. The sole shareholder is an Australian for tax purposes and can therefore use the franking credits to offset his personal income tax liability.
What will be the personal tax payable by the shareholder and the corporate tax payable by the company?
A company conducts a 1 for 5 rights issue at a subscription price of $7 when the pre-announcement stock price was$10. What is the percentage change in the stock price and the number of shares outstanding? The answers are given in the same order. Ignore all taxes, transaction costs and signalling effects.
Question 625 dividend re-investment plan, capital raising
Which of the following statements about dividend re-investment plans (DRP's) is NOT correct?
In 2014 the median starting salaries of male and female Australian bachelor degree accounting graduates aged less than 25 years in their first full-time industry job was $50,000 before tax, according to Graduate Careers Australia. See page 9 of this report. Personal income tax rates published by the Australian Tax Office are reproduced for the 2014-2015 financial year in the table below. Taxable income Tax on this income 0 –$18,200 Nil
$18,201 –$37,000 19c for each $1 over$18,200
$37,001 –$80,000 $3,572 plus 32.5c for each$1 over $37,000$80,001 – $180,000$17,547 plus 37c for each $1 over$80,000
$180,001 and over$54,547 plus 45c for each $1 over$180,000
The above rates do not include the Medicare levy of 2%. Exclude the Medicare levy from your calculations
How much personal income tax would you have to pay per year if you earned $50,000 per annum before-tax? A firm pays a fully franked cash dividend of$100 to one of its Australian shareholders who has a personal marginal tax rate of 15%. The corporate tax rate is 30%.
What will be the shareholder's personal tax payable due to the dividend payment?
Due to floods overseas, there is a cut in the supply of the mineral iron ore and its price increases dramatically. An Australian iron ore mining company therefore expects a large but temporary increase in its profit and cash flows. The mining company does not have any positive NPV projects to begin, so what should it do? Select the most correct answer.
A pharmaceutical firm has just discovered a valuable new drug. So far the news has been kept a secret.
The net present value of making and commercialising the drug is $200 million, but$600 million of bonds will need to be issued to fund the project and buy the necessary plant and equipment.
The firm will release the news of the discovery and bond raising to shareholders simultaneously in the same announcement. The bonds will be issued shortly after.
Once the announcement is made and the bonds are issued, what is the expected increase in the value of the firm's assets (ΔV), market capitalisation of debt (ΔD) and market cap of equity (ΔE)?
The triangle symbol is the Greek letter capital delta which means change or increase in mathematics.
Ignore the benefit of interest tax shields from having more debt.
Remember: $ΔV = ΔD+ΔE$
Question 513 stock split, reverse stock split, stock dividend, bonus issue, rights issue
Which of the following statements is NOT correct?
A company conducts a 4 for 3 stock split. What is the percentage change in the stock price and the number of shares outstanding? The answers are given in the same order.
In mid 2009 the listed mining company Rio Tinto announced a 21-for-40 renounceable rights issue. Below is the chronology of events:
• 04/06/2009. Share price opens at $69.00 and closes at$66.90.
• 05/06/2009. 21-for-40 rights issue announced at a subscription price of $28.29. • 16/06/2009. Last day that shares trade cum-rights. Share price opens at$76.40 and closes at $75.50. • 17/06/2009. Shares trade ex-rights. Rights trading commences. All things remaining equal, what would you expect Rio Tinto's stock price to open at on the first day that it trades ex-rights (17/6/2009)? Ignore the time value of money since time is negligibly short. Also ignore taxes. A fairly priced unlevered firm plans to pay a dividend of$1 next year (t=1) which is expected to grow by 3% pa every year after that. The firm's required return on equity is 8% pa.
The firm is thinking about reducing its future dividend payments by 10% so that it can use the extra cash to invest in more projects which are expected to return 8% pa, and have the same risk as the existing projects. Therefore, next year's dividend will be \$0.90. No new equity or debt will be issued to fund the new projects, they'll all be funded by the cut in dividends.
What will be the stock's new annual capital return (proportional increase in price per year) if the change in payout policy goes ahead?
Assume that payout policy is irrelevant to firm value (so there's no signalling effects) and that all rates are effective annual rates.
Convert a 10% continuously compounded annual rate $(r_\text{cc annual})$ into an effective annual rate $(r_\text{eff annual})$. The equivalent effective annual rate is:
Which of the following interest rate quotes is NOT equivalent to a 10% effective annual rate of return? Assume that each year has 12 months, each month has 30 days, each day has 24 hours, each hour has 60 minutes and each minute has 60 seconds. APR stands for Annualised Percentage Rate.
A continuously compounded monthly return of 1% $(r_\text{cc monthly})$ is equivalent to a continuously compounded annual return $(r_\text{cc annual})$ of:
An effective monthly return of 1% $(r_\text{eff monthly})$ is equivalent to an effective annual return $(r_\text{eff annual})$ of:
The below three graphs show probability density functions (PDF) of three different random variables Red, Green and Blue.
Which of the below statements is NOT correct?
If a stock's future expected effective annual returns are log-normally distributed, what will be bigger, the stock's or effective annual return? Or would you expect them to be ?
The symbol $\text{GDR}_{0\rightarrow 1}$ represents a stock's gross discrete return per annum over the first year. $\text{GDR}_{0\rightarrow 1} = P_1/P_0$. The subscript indicates the time period that the return is mentioned over. So for example, $\text{AAGDR}_{1 \rightarrow 3}$ is the arithmetic average GDR measured over the two year period from years 1 to 3, but it is expressed as a per annum rate.
Which of the below statements about the arithmetic and geometric average GDR is NOT correct?
Fred owns some Commonwealth Bank (CBA) shares. He has calculated CBA’s monthly returns for each month in the past 20 years using this formula:
$$r_\text{t monthly}=\ln \left( \dfrac{P_t}{P_{t-1}} \right)$$
He then took the arithmetic average and found it to be 1% per month using this formula:
$$\bar{r}_\text{monthly}= \dfrac{ \displaystyle\sum\limits_{t=1}^T{\left( r_\text{t monthly} \right)} }{T} =0.01=1\% \text{ per month}$$
He also found the standard deviation of these monthly returns which was 5% per month:
$$\sigma_\text{monthly} = \dfrac{ \displaystyle\sum\limits_{t=1}^T{\left( \left( r_\text{t monthly} - \bar{r}_\text{monthly} \right)^2 \right)} }{T} =0.05=5\%\text{ per month}$$
Which of the below statements about Fred’s CBA shares is NOT correct? Assume that the past historical average return is the true population average of future expected returns.
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If I type:
Coefficient[Cos[x + b], Cos[x]]
I get zero.
But I would like Mathematica to apply:
$$\cos(x+b)=\cos x\cos b-\sin x\sin b$$
Therefore the coefficient of $\cos x$ should be $\cos b$.
How do I fix this?
Edit:
The solution Coefficient[TrigExpand@f[x],Cos[x]] has been proposed, but this wouldn't work for expressions like $\sin(a+3x)$, in fact:
In[84]:= TrigExpand@Sin[a + 3 x]
Out[84]= Cos[x]^3 Sin[a] + 3 Cos[a] Cos[x]^2 Sin[x] -
3 Cos[x] Sin[a] Sin[x]^2 - Cos[a] Sin[x]^3
And therefore this would give me a wrong coefficient:
In[86]:= Coefficient[TrigExpand@Sin[a + 3 x], Cos[x]]
Out[86]= -3 Sin[a] Sin[x]^2
Because $\sin x$ is not a constant, and for what I am concerned with, it cannot be the coefficient of $\cos x$. Therefore, for all the expressions of the type $\cos(n x+ a)$ I am expecting to find zero, except when $n=1$.
• This is not exactly what I think would work, I will explain it in the edit :) – usumdelphini Dec 8 '15 at 10:31
• @Kuba It would be zero. – usumdelphini Dec 8 '15 at 10:46
• @Kuba for all the expressions of the type $\cos(nx+a)$ I am expecting to find zero, except when $n=1$ – usumdelphini Dec 8 '15 at 10:49
• These expressions do not have unique representations as polynomials in {sin(x),cos(x)}. This of course is due to the fundamental trig identity. A possible way around this would be to obtain coefficients in terms of both trig terms. – Daniel Lichtblau Dec 8 '15 at 14:58
Ok, here is another approach:
ClearAll[findCoeff];
findCoeff[expr_, termToIsolate_] := Module[{findCoeffIn},
findCoeffIn[termToIsolate] = 1;
findCoeffIn[exprIn_Plus, OptionsPattern[]] := findCoeffIn /@ exprIn;
findCoeffIn[Times[lterms___, termToIsolate, rterms___]] :=
If[FreeQ[#, x], #, 0] &@Times[lterms, rterms];
findCoeffIn[exprIn : (Cos[_] | Sin[_] | Tan[_] | Cot[_]),
OptionsPattern[]] := findCoeffIn[TrigExpand[exprIn]];
findCoeffIn[_] := 0;
findCoeffIn[expr]
]
This seems to work on all the cases that I tried. Use it like this:
In[1]:= findCoeff[Cos[2 x] + Cos[x + a] + 2, Cos[x]]
Out[1]= Cos[a]
In[2]:= findCoeff[2 Cos[x], Cos[x]]
Out[2]= 2
In[3]:= findCoeff[Cos[x + a] + Sin[2 x] + Cos[x], Cos[x]]
Out[3]= 1 + Cos[a]
In[4]:= findCoeff[Cos[2 x], Cos[2 x]]
Out[4]= 1
Something like findCoeff[Sin[2 x+a],Cos[2 x]] will still not work as you want it to, though. One probably has to fiddle with FourierCosCoefficient to do that. In fact, I'm not even sure that the problem is well stated: consider for example Sin[2x]. How should it expand? FourierCosCoefficient[Sin[2 x],x ,1] gives the answer $8/3\pi$, while findCoeff[Sin[2 x],Cos[x]] gives 0. While the former result answers a well defined question, what exactly are the rules according to which you want the latter result?
# Old
Here is a slightly improved version of Kuba's idea:
Options[findCoeff] = {termToIsolate -> Cos[x],
independentVariable -> x};
findCoeff[expr_, OptionsPattern[]] := Total@Cases[
TrigExpand[expr],
cl___ OptionValue@termToIsolate cr___ :>
cl cr /; FreeQ[cl cr, OptionValue@independentVariable]
] /. {} -> 0
Examples:
In[1]:= findCoeff[Sin[2 x] + 2 Cos[x] + Cos[x + a]]
Out[1]= 2 + Cos[a]
In[2]:= findCoeff[Cos[y + x] + Sin[x], termToIsolate -> Cos[y],
independentVariable -> y]
Out[2]= Cos[x]
I'm not sure it would work on more complicated cases though.
• findCoeff[Sin[2 x] 2 + 1 Cos[x]] gives 0 – usumdelphini Dec 8 '15 at 13:09
• Also, I cannot choose Sin[2 x] as a termToIsolate – usumdelphini Dec 8 '15 at 13:11
• @usumdelphini see the edit – glS Dec 8 '15 at 14:02
Cases[
TrigExpand@Sin[a + x],
Cos[x] coeff__ :> coeff /; FreeQ[coeff, x]
] /. {} -> 0
Old:
Not sure if I got your point in general but this should do: (no it shouldn't really)
FourierCosCoefficient[Cos[x + b], x, 1]
FourierCosCoefficient[Sin[a + 3 x], x, 1]
Cos[b]
0
• This does not seem to work for Cases[TrigExpand@(Sin[a + x] + Sin[2 x]), Cos[x] coeff__ :> coeff /; FreeQ[coeff, x]] /. {} -> 0 – usumdelphini Dec 8 '15 at 12:00
• @usumdelphini it gives Sin[a] which is expected since there is Cos[x] Sin[a], if not define what is coefficient. – Kuba Dec 8 '15 at 12:07
• That's true, sorry. But it does not work for Cases[TrigExpand@(((-C1 + C2) Sin[x])/a), Sin[x] coeff__ :> coeff /; FreeQ[coeff, x]] /. {} -> 0 – usumdelphini Dec 8 '15 at 12:22
• @usumdelphini modifying the code like this Cases[ TrigExpand[Sin[a + x] + Sin[2 x]], cl___ Cos[x] cr___ :> cl cr /; FreeQ[cl cr, x] ] /. {} -> 0 works in your cases. It would probably fail in more complicated ones, though – glS Dec 8 '15 at 12:29
• @usumdelphini if you expand Cos[2x] you cannot then ask to isolate Cos[2x]. You should probably explain better what exactly do you need this function for – glS Dec 8 '15 at 13:08
EDIT: Ok, now this answer is almost the same as Kuba's...
ORIGINAL POST:
I think Kuba's answer is preferable, but you could also do this:
coeff[expr_, var_] := Boole[FreeQ[#,x]]*#&@
Coefficient[TrigExpand@expr, var]
Then,
coeff[Sin[a + 3 x], Cos[x]]
0
and
coeff[Cos[a+x],Cos[x]]
Cos[a]
• Thanks so much. Is there a way to extend it to the use of $\sin(2 x)$ as var for example? – usumdelphini Dec 8 '15 at 11:48
• Now it does not work for coeff[Cos[a + x] + Sin[2 x], Cos[x]] – usumdelphini Dec 8 '15 at 11:58
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## Section1.5Models and Dynamics
###### What is a Model?
Models are simplified abstract representations of something of interest. Airplane and automobile manufacturers create scale models to test aerodynamics in wind tunnels. Architects build models of future projects, whether a physical mock-up or a computerized 3-d representation, to see how plan will fit together and give clients a vision of what they will get. The models do not need to include every detail of the actual object of interest, just those details that are relevant to the purpose of the model.
Scientists also regularly use models. Physicists use high energy collisions of extremely fast particles to create conditions that they expect are comparable to the moments immediately after the big bang. A biologist may use mice from a well-controlled population as a model to study cancer, considering its biology to mimic that of humans at some level of approximation. A climatologist might use a computational model where a computer program tracks changes in the makeup of the air, pollutant levels and air and ocean temperatures according to known and assumed interactions.
### Subsection1.5.1Mathematical Models
A mathematical model is an abstract representation of measurable phenomena that is characterized through mathematical equations. Recall that we think of a system as the collection of all possible measurements associated with the objects and environment involved in the phenomenon. Each quantity is a state variable, even if we do not have a physical way to obtain the measurement. When the phenomenon is considered under a single environmental condition, the measurements relating to that condition which do not change are called state parameters. From a mathematical perspective, both state variables and state parameters are variables.
Many laws of science are described using mathematical equations that relate state variables. These equations are examples of mathematical models. Knowing the value of one state variable, we can use the model to predict the value of other variables. Often the equations involve additional coefficients or constants. These constants are called model parameters.
###### Example1.5.1
In thermodynamics, the ideal gas law is described as an equation
\begin{equation*} PV = nRT. \end{equation*}
This law relates the state variables of a contained volume of a gas according to pressure $P\text{,}$ volume $V\text{,}$ absolute temperature $T\text{,}$ and quantity $n$ (moles) of the gas. If the temperature and quantity of the gas is held constant, we think of only $P$ and $V$ as state variables and $n$ and $T$ as state parameters. The other variable, $R\text{,}$ is a model parameter and is called the ideal gas constant because the same value is used for every gas. This model helps scientists and engineers understand the expected behavior of gases in different settings.
Have you microwaved a container with the lid attached and let it cool? As the temperature decreases, the quantity on the right side of the equation $nRT$ decreases. This means that the product $PV$ must also decrease to remain in balance. For a sturdy container, the volume will be constant meaning the pressure inside decreases creating vacuum suction. The lid will be much harder to remove. If the container is not rigid, the volume itself might decrease, such as water bottle that collapses on itself.
However, the value of the law is not just general arguments like those made above. It is a precise mathematical relationship about actual quantities. This allows very specific predictions about measurements. For some gases, the predictions do not work properly. These gases must have properties that make them deviate from the law, and that makes these gases particularly interesting. Scientists can then study their properties and develop corrected models that take those properties into account.
###### Example1.5.2
Hooke's law,
\begin{equation*} F=-kx, \end{equation*}
is an equation describing the strength of a force exerted by a stiff spring based on how far it is stretched or compressed. The state variables are the force exerted by the spring $F$ and the displacement of the end of the spring from its equilibrium position $x\text{.}$ The model parameter $k$ determines the strength of the spring. The negative sign in the equation indicates that the force and the displacement are in opposite directions.
Using known forces, we can measure the displacement of the spring and estimate the value of the model parameter $k\text{.}$ Once the parameter is known, the model allows us to know the force required for every displacement. This is how a force scale can be created.
### Subsection1.5.2Dynamic Models
Calculus allows us to study how variables change in relationship to one another. We are often interested in how variables change with respect to time. Calculus provides a way of thinking about a rate of change as a new variable that relates covarying quantities. This variable is called the derivative.
Consider driving in a car along a road. At each instant in time, the car has a location on the road which might be measured according to mileage markings. The natural state variables are the position, say $D$ (for distance), and the time, $t\text{.}$ The distance moved is related to the time spent moving according to the speed or velocity of the car. A common formula relating these variables is often learned as “distance equals rate times time.” However, this formula only works when the rate is constant and the car has the same speed at all times.
In reality, the rate changes as we drive. The velocity $V$ which measures how fast we are moving is another state variable. Calculus introduces the derivative of the car's position with respect to time, written $\frac{dD}{dt}\text{,}$ and allows us to know that the velocity and the derivative measure the same thing,
\begin{equation*} V = \frac{dD}{dt}\text{.} \end{equation*}
Dynamic models are mathematical models that consider models or equations for how variables change. Calculus, using the concepts of rates of change and derivatives, provides the language for describing these models.
To help think about this, consider another physical example, that of a sink being filled with water. At each instant in time, the sink has a different state. The state can be characterized by possible measurements at that time, which might include the height of water in the sink, the volume of water in the sink, the rate of water flowing into the sink from the spigot, and the rate of water flowing out of the sink through the drain. More sophisticated problems might also involve the water temperature or the dilution of some substance or chemical in the water.
The diagram shown in Figure 1.5.3 illustrates the system and shows some of these variables. We let $t$ represent the elapsed time from the start of our experiment and the time of the state's observation. We also use $V$ to represent the volume of water, $h$ to represent the height of the water in the sink, $F_{\hbox{in}}$ to represent the rate of water flowing in, and $F_{\hbox{out}}$ to represent the rate of water flowing out.
We might consider a situation in which the spigot has water flowing at a constant rate. In this case, the variable $F_{\hbox{in}}$ is never changing; we could call this a state parameter instead of a state variable. A diagram can often help summarize this information.
Because water is flowing in and out of the sink, the volume of water in the sink is a dynamic variable. Unless the rates of water flowing in and out are the same, the volume will be changing. The derivatives $\frac{dV}{dt}$ and $\frac{dh}{dt}$ are additional state variables, the rate of change of the volume and the rate of change of the height, respectively. We obtain an equation by recognizing that the total rate of change will always be equal to the sum of the rate of water flowing in and subtracting the rate of water flowing out. This allows us to write a general dynamic model equation
\begin{equation*} \frac{dV}{dt} = F_{\text{in}} - F_{\text{out}}. \end{equation*}
Different models can be formed by establishing model equations relating the flow rates with other state variables. For example, the rate at which water drains might change depending on how much water there is in the sink, which could be measured by either the volume $V$ or the height $h$ of water in the sink. An experiment might be performed to measure $F_{\text{out}}$ at different heights of water in the sink. We might then formulate a model equation based on our data. For example, we might discover from this experiment that the rate of water flowing out is proportional to the volume in the sink, $F_{\hbox{out}} = \alpha V$ for some model parameter $\alpha\text{.}$ Using this model equation and substitution, our dynamic model would become
\begin{equation*} \frac{dV}{dt} = F_{\hbox{in}} - \alpha V. \end{equation*}
A dynamic model involving a derivative is called a differential equation.
### Subsection1.5.3Where Are We Going?
Our aim in studying calculus is to understand the mathematical meaning of the derivative or rate of change. There are two primary branches of calculus. Integral calculus studies how we use a known rate of change for a quantity to compute the overall change in that quantity. For example, integral calculus answers the question: “Knowing the velocity of a car at every instant between two moments, how would we compute how far the car has traveled between those moments?” Differential calculus studies how to use the relationship between two variables to find the rate of change of one with respect to the other. For the example of the car, this would address the question: “If we knew exactly where the car was at each instant of time, how would we compute its velocity?” Ultimately, both of these questions will rely on a mathematical tool at the heart of calculus, namely limits.
In this text, we will first study discrete dynamic models first using sequences to develop our conceptual understanding. We previously learned that the number line represents a continuum of real numbers. Calculus concerns functions that are defined on intervals from this continuum. Suppose we have a system where we can record the state after manipulating an independent state variable to a specified value. We can not physically record all possible states because that variable has infinitely many possibilities. Instead, we might increase the value of the control variable by a predetermined increment and record the state and then repeat this process. This use of an increment to change an independent variable is what we mean by a discrete model. Often the independent variable is time, and the discrete model represents the state of the system at equally spaced increments of time.
A sequence can often be described directly by a formula. Such an explicit representation of a sequence will be analogous to how we describe functions using formulas. We can also represent a sequence by describing a pattern in how terms are found. One of the most common representations is by describing the increments, or values added, to get each term. The process of using increments to find a sequence is analogous to the calculus concept. The reverse process is also possible, to take a given sequence and compute the increments. Knowing formulas for the increments allows us to analyze the properties of the original sequence. This reverse process is analogous to the calculus concept of differentiation.
As you work through this text, make note of these overarching concepts and their relationships. Understanding the concepts of sequences, including the properties of their increments, will help you make better sense of the concepts of calculus. You should also focus on how calculus extends the ideas we learn about sequences. In addition, we will explore how both discrete sequences and functions defined on intervals can be used in modeling.
### Subsection1.5.4Summary
• Mathematical models are equations that relate state variables. Constants in these equations are called parameters.
• Scientific laws relating measurable quantities are described using equations.
• Calculus explores the relationship between quantities and their rates of change, known as derivatives.
• Sequences will provide motivating examples and concepts for studying calculus.
### Subsection1.5.5Exercises
###### 1
An isolated population can only change due to births and deaths. Consider state variables $P$ (total number of individuals), $B$ (rate of births as number of individuals born per year), $D$ (rate of deaths as number of individuals who die per year), and time $t$ (in years). Explain the meaning for the dynamic model
\begin{equation*} \frac{dP}{dt} = B - D. \end{equation*}
###### 2
Chemical reactions occur when molecules interact and form new compounds. Chemical equations describe these reactions. For example, the generic chemical equation
\begin{equation*} A+2B \overset{R_1}{\rightarrow} C+D \end{equation*}
describes a reaction where one molecule of $A$ and two molecules of $B$ change into one molecule of $C$ and one molecule of $D\text{.}$ The reaction rate $R_1$ measures how many of such reactions occur per unit time. The number of molecules of $B$ will be decreasing due to this reaction at a rate $2 R_1$ while the number of molecules of $C$ will be increasing at the rate $R_1\text{.}$
Suppose we have two reactions occurring,
\begin{gather*} A+2B \overset{R_1}{\rightarrow} C+D,\\ C+E \overset{R_2}{\rightarrow} A+F, \end{gather*}
with reaction rates $R_1$ and $R_2\text{,}$ respectively. Suppose $N_A$ is the state variable measuring the number of molecules of $A\text{,}$ and similarly for $N_B\text{,}$ $N_C\text{,}$ etc. Explain the meaning of each of the following dynamic models.
1. $\displaystyle \frac{dN_A}{dt} = -R_1 + R_2$
2. $\displaystyle \frac{dN_B}{dt} = -2R_1$
3. $\displaystyle \frac{dN_C}{dt} = R_1 - R_2$
Understanding how the rates depend on the concentration of the reactants is a fundamental question of chemistry.
###### 3
Newton's second law of physics, $F=ma\text{,}$ is a model relating the net force $F$ acting on an object of mass $m$ that results in an acceleration $a\text{.}$ Acceleration is actually the rate of change of velocity $v$ with respect to time $t\text{.}$ If the only force acting on the body is a spring that obeys Hooke's law, then Newton's law results in an equation
\begin{equation*} -kx = m \frac{dv}{dt}. \end{equation*}
Explain how this equation matches what was described.
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## Integration of Step Functions [closed]
How can we prove an inequality of integrable step functions:
More Precisely: Integral of max{a,b} > max {integral a, integral b}
Any initial help would be appreciated.
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Draw a picture! Seriously, you are more likely to get a pertinent answer on one of the sites mentioned in the FAQ: mathoverflow.net/faq#whatnot – Andrew Stacey Apr 27 2010 at 14:30
Sorry, I am just looking for way for a proof into which to focus ... not an answer – Andrey Apr 27 2010 at 14:34
In that case, my suggestion is not facetious: draw a picture. Specifically, find some very simple step functions and draw the situation for those. I recommend, in particular, finding two step functions, say a and b, with a(x) > b(x) some of the time and b(x) > a(x) some of the time and drawing the graphs of a, b, and max{a,b}. – Andrew Stacey Apr 27 2010 at 14:55
Or just recall that $c\ge\max(a,b)$ is merely a fancy way to write $c\ge a$ and $c\ge b$ and that $\max(a,\text{anything})$ is certainly not less than $a$. – fedja Apr 27 2010 at 15:18
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are in R for every x in A. The set S is called the domain of the relation and the set T the codomain. Since binary relations are sets, we can apply the classical operations of set theory to them. This wavelet tree contains two bitmaps per level at each node v, Bvl and Bvr . Example 1.6. We ask that binary relation mathematics example of strict weak orders is related to be restricted to be restricted to the only if a reflexive relation Every set and binary in most ��I7���v7]��҈jt�ۮ]���}��|qYonc��3at[�P�*ct���M�!ǣ��" ���=䑍F���4~G�͐Ii]���מS�=96���G����_J���c0�dD�_�|>��)��|V�MTpPn� -����x�Լ�7z�Nj�'ESF��(��R9�c�bS� ㉇�ڟio�����XO��^Fߑ��&�*�"�;�0 Jyv��&��2��Y,��E��ǫ�DҀ�y�dX2 �)I�k 2.1: Binary Relations - Mathematics LibreTexts Skip to main content Binary Relations November 4, 2003 1 Binary Relations The notion of a relation between two sets of objects is quite common and intuitively clear. Interpretation. Therefore, such a relationship can be viewed as a restricted set of ordered pairs. A binary relation from A to B is a subset of a Cartesian product A x B. R t•Le A x B means R is a set of ordered pairs of the form (a,b) where a A and b B. :��&i�c�*��ANJ#2�W !jZ�� eT�{}���t�;���]�N��?��ͭ�kM[�xOӷ. 7 Binary Relations • Let A, B be any two sets. The symmetric component Iof a binary relation Ris de ned by xIyif and only if xRyand yRx. stream For example, if a relation R is such that everything stands in the relation R to itself, R is said to be reflexive . Reflexivity. learning non-pure binary relations, and demonstrate how the robust nature of WMG can be exploited to handle such noise. Relations and Their Properties 1.1. We express a particular ordered pair, (x, y) R, where R is a binary relation, as xRy. For instance, let X denote the set of all females and Y the set of all males. We implement the above idea in CASREL, an end-to-end cascade binary tagging framework. A binary relation R from A to B, written R : A B, is a subset of the set A B. Complementary Relation Definition: Let R be the binary relation from A to B. Just as we get a number when two numbers are either added or subtracted or multiplied or are divided. (x, x) R. b. Definition (binary relation): A binary relation from a set A to a set B is a set of ordered pairs where a is an element of A and b is an element of B. We denote this by aRb. The binary operation, *: A × A → A. Binary Relations (zyBooks, Chapter 5.1-5.8) Binary Relations • Recall: The Cartesian product of 1.1.2 Preorders A preorder or ordered set is a pair (X,≤) where Xis a set and ≤ is a reflexive transitive binary relation on X. The dual R0of a binary relation Ris de ned by xR0yif and only if yRx. Given a set A and a relation R in A, R is reflexive iff all the ordered pairs of the form are in R for every x in A. Jason Joan Yihui Formally, a binary relation R over a set X is symmetric if: ∀, ∈ (⇔). Binary relation Definition: Let A and B be two sets. The wife-husband relation R can be deflned from X to Y. Download Binary Relation In Mathematics With Example doc. Set alert. Basic Methods: We define the Cartesian product of two sets X and Y and use this to define binary relations on X. Abinary relation from A to B is a subset of A B . Chapter 4: Binary Operations and Relations 4.1: Binary Operations DEFINITION 1. In other words, a binary relation R … Binary relation for sets This video is about: Introduction to Binary Relation. endobj A binary relation associates elements of one set called the . Let Aand Bbe sets and define their Cartesian product to be the set of all pairwise De nition of a Relation. View 5 - Binary Relations.pdf from CS 2212 at Vanderbilt University. Let's see how to prove it. De nition of a Relation. Binary Relations A binary relation over a set A is some relation R where, for every x, y ∈ A, the statement xRy is either true or false. Binary Relations Any set of ordered pairs defines a binary relation. Binary Relations and Preference Modeling 51 (a,b) 6∈Tor a¬Tb. 1.1.2 Preorders A preorder or ordered set is a pair (X,≤) where Xis a set and ≤ is a reflexive transitive binary relation on X. + : R × R → R e is called identity of * if a * e = e * a = a i.e. <> �6"����f�#�����h���uL��$�,ٺ4����h�4 ߑ+�a�z%��і��)�[��WNY��4/y!���U?�Ʌ�w�-� Similarly, R 3 = R 2 R = R R R, and so on. Albert R Meyer February 21, 2011 . Examples: < can be a binary relation over ℕ, ℤ, ℝ, etc. Draw the following: 1. A binary relation over a set $$A$$ is some relation $$R$$ where, for every $$x, y \in A,$$ the statement $$xRy$$ is either true or false. A partial order is an antisymmetric preorder. The arrow diagram representation of the relation. Remark 2.1. (x, x) R. b. Binary relations establish a relationship between elements of two sets Definition: Let A and B be two sets.A binary relation from A to B is a subset of A ×B. Some important results concerning Rosenberg partial hypergroupoids, induced by relations, are generalized to the case of Let us consider R. The predicate Ris reflexive is defined by R is reflexive in fieldR. De nition: A binary relation from a set A to a set Bis a subset R A B: If (a;b) 2Rwe say ais related to bby R. Ais the domain of R, and Bis the codomain of R. If A= B, Ris called a binary relation … 9.1 Relations and Their Properties Binary Relation Definition: Let A, B be any sets. Addition, subtraction, multiplication are binary operations on Z. Introduction to Relations CSE 191, Class Note 09 Computer Sci & Eng Dept SUNY Buffalo c Xin He (University at Buffalo) CSE 191 Descrete Structures 1 / 57 Binary relation Denition: Let A and B be two sets. Binary Relations and Equivalence Relations Intuitively, a binary relation Ron a set A is a proposition such that, for every ordered pair (a;b) 2A A, one can decide if a is related to b or not. Some relations, such as being the same size as and being in the same column as, are reflexive. Binary relations establish a relationship between elements of two sets Definition: Let A and B be two sets.A binary relation from A to B is a subset of A ×B. Set alert. Brice Mayag (LAMSADE) Preferences as binary relations Chapter 1 7 / 16 0 denotes the empty relation while 1 denoted (prior to the 1950’s)1 the complete relation … Relations and Their Properties 1.1. relation to Paul. 9�����D���-��XE��^8� Albert R Meyer . Then R R, the composition of R with itself, is always represented. VG�%�4��슁� The wife-husband relation R can be thought as a relation from X to Y.For a lady A relation which fails to be reflexive is called 2. M���LZ��l�G?v�P:�9Y\��W���c|_�y�֤#����)>|��o�ޣ�f{}d�H�9�vnoﺹ��k�I��0Kq)ө�[��C�O;��)�� &�K��ea��*Y���IG}��t�)�m�Ú6�R�5g |1� ܞb�W���������9�o�D�He夵�fݸ���-�R�2G�\{�W� �)Ԏ A partial order is an antisymmetric preorder. The wife-husband relation R can be thought as a relation from X to Y.For a lady The predicate Ris … In this paper, we introduce and study the notion of a partial n-hypergroupoid, associated with a binary relation. Binary relations generalize further to n-ary relations as a set of n-tuples indexed from 1 to n, and yet further to I-ary relations where Iis an arbitrary index set. Binary operations on a set are calculations that combine two elements of the set (called operands) to produce another element of the same set. For example, “less-than” on the real numbers relates every real number, a, to a real number, b, precisely when a> 1 Sets, Relations and Binary Operations Set Set is a collection of well defined objects which are distinct from each other. Also, R R is sometimes denoted by R 2. Binary relations. Rsatisfles the trichotomy property ifi … Introduction to Relations 1. A binary relation R on X is aweak orderor acomplete preorderif R is complete and transitive. Let R is a relation on a set A, that is, R is a relation from a set A to itself. Except when explicitly mentioned otherwise, we will suppose in all what follows that the set Ais finite . 1. A binary relation R on X is apreorderif R is re exive and transitive. A binary relation is a set of pairs of elements assumed to be drawn from an indeterminate but fixed set X. We can also represent relations graphicallyor using a table lec 3T.3 . ��nj]��gw�e����"φ�0)��?]�]��O!���C�s�D�Y}?�? The relation R S is known the composition of R and S; it is sometimes denoted simply by RS. Download as PDF. Preference Relations, Social Decision Rules, Single-Peakedness, and Social Welfare Functions 1 Preference Relations 1.1 Binary Relations A preference relation is a special type of binary relation. A binary operation on a nonempty set Ais a function from A Ato A. Binary Relations De nition: A binary relation between two sets X and Y (or between the elements of X and Y) is a subset of X Y | i.e., is a set of ordered pairs (x;y) 2X Y. A binary relation A is a poset iff A does not admit an embedding of the following finite relations: The binary relation … �������'y�ijr�r2ܫa{wե)OƌN"��1ƾɘ�@_e��=��R��|��W�l�xQ~��"��v�R���dk����\|�a}�>IP!z��>��(�tQ ��t>�r�8T,��]�+�Q�@\�r���X��U �ݵ6�;���0_�M8��fI�zS]��^p �a���. relation to Paul. • We use the notation a R b to denote (a,b) R and a R b to denote (a,b) R. If a R b, we say a is related to b by R. Theory of Relations. A symmetric relation is a type of binary relation.An example is the relation "is equal to", because if a = b is true then b = a is also true. We can also represent relations graphicallyor using a table Remark 2.1. ↔ can be a binary relation over V for any undirected graph G = (V, E). Binary Relations Any set of ordered pairs defines a binary relation. %PDF-1.4 <> The logical operations treat a binary relation purely as a set, ignoring the nature of its ele-ments. Finally, The binary operations associate any two elements of a set. Others, such as being in front of or Binary Relations Intuitively speaking: a binary relation over a set A is some relation R where, for every x, y ∈ A, the statement xRy is either true or false. 5 Binary Relation Wavelet Trees (BRWT) We propose now a special wavelet tree structure to represent binary relations. • We use the notation a R b to denote (a,b) R and a R b to denote (a,b) R. If a R b, we say a is related to b by R. Sets are usually denoted by capital letters A B C, , ,K and elements are usually denoted by small letters a b c, , ,... . If (a,b) ∈ R, we say a is in relation R to be b. Except when explicitly mentioned otherwise, we will suppose in all what follows that the set Ais finite . About this page. It consists of a BERT-based encoder module, a sub-ject tagging module, and a relation-specific object a + e = e + a = a This is only possible if e = 0 Since a + 0 = 0 + a = a ∀ a ∈ R 0 is the identity element for addition on R •A binary relation R from A to B, written (with signature) R:A↔B,is a subset of A×B. Binary Relations - Free download as PDF File (.pdf), Text File (.txt) or read online for free. The logical operations treat a binary relation purely as a set, ignoring the nature of its ele-ments. 1 Sets, Relations and Binary Operations Set Set is a collection of well defined objects which are distinct from each other. stream About this page. View Relation.pdf from COMPUTERSC CS 60-231 at University of Windsor. We can define binary relations by giving a rule, like this: a~b if some property of a and b holds This is the general template for defining a relation. Binary relation Definition: Let A and B be two sets. For example, if a relation R is such that everything stands in the relation R to itself, R is said to be reflexive . We consider here certain properties of binary relations. endobj @*�d)���7�t�a���M�Y�F�6'{���n | Find, read and cite all the research you need on ResearchGate This wavelet tree contains two bitmaps per level at each node v, Bvl and Bvr . /Filter /FlateDecode 2. De nition 1.5. x��T˪�0��+�X�����&�����tצ���f���. The following de nitions for these properties are not completely standard, in that they mention only those ordered pairs https://www.tutorialspoint.com/.../discrete_mathematics_relations.htm Properties of binary relations Binary relations may themselves have properties. In Studies in Logic and the Foundations of Mathematics, 2000. In other words, a binary relation R … A relation which fails to be reflexive is called << x��[[���~ϯ�("�t� '��-�@�}�w�^&�������9$wF��rҼ�#��̹~��ן��{�.G�Kz����r�8��2�������Y�-���Sb�\mUow����� #�{zE�A����������|� �V����11|LjD�����oRo&n��-�A��EJ��PD��Z��Z��~�?e��EI���jbW�a���^H���{�ԜD LzJ��U�=�]J���|CJtw����.C�e��2nJ;�r]n�$\�e�K�u�����G墲t�����{"��4�0�z;f ���Ř&Y��s�����-LN�$��n�P��/���=���W�m5�,�ð�*����p[T���V$��R�aFG�H�R!�xwS��� ryX�q�� ��p�p�/���L�#��L�H��N@�:���7�_ҧ�f�qM�[G4:��砈+2��m�T�#!���բJ�U!&'l�( ��ɢi��x�&���Eb��*���zAz��md�K&Y�ej6 �g���\��Q���SlwmY\uS�cά�u��p�f��5;¬_����z�5r#���G�D��?��:�r���Q$��Q We consider here certain properties of binary relations. ≡ₖ is a binary relation over ℤ for any integer k. Such classes are typically speci ed in terms of the properties required for membership. In Studies in Logic and the Foundations of Mathematics, 2000. 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We ask that binary relation mathematics example of strict weak orders is related to be restricted to be restricted to the only if a reflexive relation Every set and binary in most ��I7���v7]��҈jt�ۮ]���}��|qYonc��3at[�P�*ct���M�!ǣ��" ���=䑍F���4~G�͐Ii]���מS�=96���G����_J���c0�dD�_�|>��)��|V�MTpPn� -����x�Լ�7z�Nj�'ESF��(��R9�c�bS� ㉇�ڟio�����XO��^Fߑ��&�*�"�;�0 Jyv��&��2��Y,��E��ǫ�DҀ�y�dX2 �)I�k 2.1: Binary Relations - Mathematics LibreTexts Skip to main content Binary Relations November 4, 2003 1 Binary Relations The notion of a relation between two sets of objects is quite common and intuitively clear. Interpretation. Therefore, such a relationship can be viewed as a restricted set of ordered pairs. A binary relation from A to B is a subset of a Cartesian product A x B. R t•Le A x B means R is a set of ordered pairs of the form (a,b) where a A and b B. :��&i�c�*��ANJ#2�W !jZ�� eT�{}���t�;���]�N��?��ͭ�kM[�xOӷ. 7 Binary Relations • Let A, B be any two sets. The symmetric component Iof a binary relation Ris de ned by xIyif and only if xRyand yRx. stream For example, if a relation R is such that everything stands in the relation R to itself, R is said to be reflexive . Reflexivity. learning non-pure binary relations, and demonstrate how the robust nature of WMG can be exploited to handle such noise. Relations and Their Properties 1.1. We express a particular ordered pair, (x, y) R, where R is a binary relation, as xRy. For instance, let X denote the set of all females and Y the set of all males. We implement the above idea in CASREL, an end-to-end cascade binary tagging framework. A binary relation R from A to B, written R : A B, is a subset of the set A B. Complementary Relation Definition: Let R be the binary relation from A to B. Just as we get a number when two numbers are either added or subtracted or multiplied or are divided. (x, x) R. b. Definition (binary relation): A binary relation from a set A to a set B is a set of ordered pairs where a is an element of A and b is an element of B. We denote this by aRb. The binary operation, *: A × A → A. Binary Relations (zyBooks, Chapter 5.1-5.8) Binary Relations • Recall: The Cartesian product of 1.1.2 Preorders A preorder or ordered set is a pair (X,≤) where Xis a set and ≤ is a reflexive transitive binary relation on X. The dual R0of a binary relation Ris de ned by xR0yif and only if yRx. Given a set A and a relation R in A, R is reflexive iff all the ordered pairs of the form are in R for every x in A. Jason Joan Yihui Formally, a binary relation R over a set X is symmetric if: ∀, ∈ (⇔). Binary relation Definition: Let A and B be two sets. The wife-husband relation R can be deflned from X to Y. Download Binary Relation In Mathematics With Example doc. Set alert. Basic Methods: We define the Cartesian product of two sets X and Y and use this to define binary relations on X. Abinary relation from A to B is a subset of A B . Chapter 4: Binary Operations and Relations 4.1: Binary Operations DEFINITION 1. In other words, a binary relation R … Binary relation for sets This video is about: Introduction to Binary Relation. endobj A binary relation associates elements of one set called the . Let Aand Bbe sets and define their Cartesian product to be the set of all pairwise De nition of a Relation. View 5 - Binary Relations.pdf from CS 2212 at Vanderbilt University. Let's see how to prove it. De nition of a Relation. Binary Relations A binary relation over a set A is some relation R where, for every x, y ∈ A, the statement xRy is either true or false. Binary Relations Any set of ordered pairs defines a binary relation. Binary Relations and Preference Modeling 51 (a,b) 6∈Tor a¬Tb. 1.1.2 Preorders A preorder or ordered set is a pair (X,≤) where Xis a set and ≤ is a reflexive transitive binary relation on X. + : R × R → R e is called identity of * if a * e = e * a = a i.e. <> �6"����f�#�����h���uL��$�,ٺ4����h�4 ߑ+�a�z%��і��)�[��WNY��4/y!���U?�Ʌ�w�-� Similarly, R 3 = R 2 R = R R R, and so on. Albert R Meyer February 21, 2011 . Examples: < can be a binary relation over ℕ, ℤ, ℝ, etc. Draw the following: 1. A binary relation over a set $$A$$ is some relation $$R$$ where, for every $$x, y \in A,$$ the statement $$xRy$$ is either true or false. A partial order is an antisymmetric preorder. The arrow diagram representation of the relation. Remark 2.1. (x, x) R. b. Binary relations establish a relationship between elements of two sets Definition: Let A and B be two sets.A binary relation from A to B is a subset of A ×B. Some important results concerning Rosenberg partial hypergroupoids, induced by relations, are generalized to the case of Let us consider R. The predicate Ris reflexive is defined by R is reflexive in fieldR. De nition: A binary relation from a set A to a set Bis a subset R A B: If (a;b) 2Rwe say ais related to bby R. Ais the domain of R, and Bis the codomain of R. If A= B, Ris called a binary relation … 9.1 Relations and Their Properties Binary Relation Definition: Let A, B be any sets. Addition, subtraction, multiplication are binary operations on Z. Introduction to Relations CSE 191, Class Note 09 Computer Sci & Eng Dept SUNY Buffalo c Xin He (University at Buffalo) CSE 191 Descrete Structures 1 / 57 Binary relation Denition: Let A and B be two sets. Binary Relations and Equivalence Relations Intuitively, a binary relation Ron a set A is a proposition such that, for every ordered pair (a;b) 2A A, one can decide if a is related to b or not. Some relations, such as being the same size as and being in the same column as, are reflexive. Binary relations establish a relationship between elements of two sets Definition: Let A and B be two sets.A binary relation from A to B is a subset of A ×B. Set alert. Brice Mayag (LAMSADE) Preferences as binary relations Chapter 1 7 / 16 0 denotes the empty relation while 1 denoted (prior to the 1950’s)1 the complete relation … Relations and Their Properties 1.1. relation to Paul. 9�����D���-��XE��^8� Albert R Meyer . Then R R, the composition of R with itself, is always represented. VG�%�4��슁� The wife-husband relation R can be thought as a relation from X to Y.For a lady A relation which fails to be reflexive is called 2. M���LZ��l�G?v�P:�9Y\��W���c|_�y�֤#����)>|��o�ޣ�f{}d�H�9�vnoﺹ��k�I��0Kq)ө�[��C�O;��)�� &�K��ea��*Y���IG}��t�)�m�Ú6�R�5g |1� ܞb�W���������9�o�D�He夵�fݸ���-�R�2G�\{�W� �)Ԏ A partial order is an antisymmetric preorder. The wife-husband relation R can be thought as a relation from X to Y.For a lady The predicate Ris … In this paper, we introduce and study the notion of a partial n-hypergroupoid, associated with a binary relation. Binary relations generalize further to n-ary relations as a set of n-tuples indexed from 1 to n, and yet further to I-ary relations where Iis an arbitrary index set. Binary operations on a set are calculations that combine two elements of the set (called operands) to produce another element of the same set. For example, “less-than” on the real numbers relates every real number, a, to a real number, b, precisely when a> 1 Sets, Relations and Binary Operations Set Set is a collection of well defined objects which are distinct from each other. Also, R R is sometimes denoted by R 2. Binary relations. Rsatisfles the trichotomy property ifi … Introduction to Relations 1. A binary relation R on X is aweak orderor acomplete preorderif R is complete and transitive. Let R is a relation on a set A, that is, R is a relation from a set A to itself. Except when explicitly mentioned otherwise, we will suppose in all what follows that the set Ais finite . 1. A binary relation R on X is apreorderif R is re exive and transitive. A binary relation is a set of pairs of elements assumed to be drawn from an indeterminate but fixed set X. We can also represent relations graphicallyor using a table lec 3T.3 . ��nj]��gw�e����"φ�0)��?]�]��O!���C�s�D�Y}?�? The relation R S is known the composition of R and S; it is sometimes denoted simply by RS. Download as PDF. Preference Relations, Social Decision Rules, Single-Peakedness, and Social Welfare Functions 1 Preference Relations 1.1 Binary Relations A preference relation is a special type of binary relation. A binary operation on a nonempty set Ais a function from A Ato A. Binary Relations De nition: A binary relation between two sets X and Y (or between the elements of X and Y) is a subset of X Y | i.e., is a set of ordered pairs (x;y) 2X Y. A binary relation A is a poset iff A does not admit an embedding of the following finite relations: The binary relation … �������'y�ijr�r2ܫa{wե)OƌN"��1ƾɘ�@_e��=��R��|��W�l�xQ~��"��v�R���dk����\|�a}�>IP!z��>��(�tQ ��t>�r�8T,��]�+�Q�@\�r���X��U �ݵ6�;���0_�M8��fI�zS]��^p �a���. relation to Paul. • We use the notation a R b to denote (a,b) R and a R b to denote (a,b) R. If a R b, we say a is related to b by R. Theory of Relations. A symmetric relation is a type of binary relation.An example is the relation "is equal to", because if a = b is true then b = a is also true. We can also represent relations graphicallyor using a table Remark 2.1. ↔ can be a binary relation over V for any undirected graph G = (V, E). Binary Relations Any set of ordered pairs defines a binary relation. %PDF-1.4 <> The logical operations treat a binary relation purely as a set, ignoring the nature of its ele-ments. Finally, The binary operations associate any two elements of a set. Others, such as being in front of or Binary Relations Intuitively speaking: a binary relation over a set A is some relation R where, for every x, y ∈ A, the statement xRy is either true or false. 5 Binary Relation Wavelet Trees (BRWT) We propose now a special wavelet tree structure to represent binary relations. • We use the notation a R b to denote (a,b) R and a R b to denote (a,b) R. If a R b, we say a is related to b by R. Sets are usually denoted by capital letters A B C, , ,K and elements are usually denoted by small letters a b c, , ,... . If (a,b) ∈ R, we say a is in relation R to be b. Except when explicitly mentioned otherwise, we will suppose in all what follows that the set Ais finite . About this page. It consists of a BERT-based encoder module, a sub-ject tagging module, and a relation-specific object a + e = e + a = a This is only possible if e = 0 Since a + 0 = 0 + a = a ∀ a ∈ R 0 is the identity element for addition on R •A binary relation R from A to B, written (with signature) R:A↔B,is a subset of A×B. Binary Relations - Free download as PDF File (.pdf), Text File (.txt) or read online for free. The logical operations treat a binary relation purely as a set, ignoring the nature of its ele-ments. 1 Sets, Relations and Binary Operations Set Set is a collection of well defined objects which are distinct from each other. stream About this page. View Relation.pdf from COMPUTERSC CS 60-231 at University of Windsor. We can define binary relations by giving a rule, like this: a~b if some property of a and b holds This is the general template for defining a relation. Binary relation Definition: Let A and B be two sets. For example, if a relation R is such that everything stands in the relation R to itself, R is said to be reflexive . We consider here certain properties of binary relations. endobj @*�d)���7�t�a���M�Y�F�6'{���n | Find, read and cite all the research you need on ResearchGate This wavelet tree contains two bitmaps per level at each node v, Bvl and Bvr . /Filter /FlateDecode 2. De nition 1.5. x��T˪�0��+�X�����&�����tצ���f���. The following de nitions for these properties are not completely standard, in that they mention only those ordered pairs https://www.tutorialspoint.com/.../discrete_mathematics_relations.htm Properties of binary relations Binary relations may themselves have properties. In Studies in Logic and the Foundations of Mathematics, 2000. In other words, a binary relation R … A relation which fails to be reflexive is called << x��[[���~ϯ�("�t� '��-�@�}�w�^&�������9$wF��rҼ�#��̹~��ן��{�.G�Kz����r�8��2�������Y�-���Sb�\mUow����� #�{zE�A����������|� �V����11|LjD�����oRo&n��-�A��EJ��PD��Z��Z��~�?e��EI���jbW�a���^H���{�ԜD LzJ��U�=�]J���|CJtw����.C�e��2nJ;�r]n�$\�e�K�u�����G墲t�����{"��4�0�z;f ���Ř&Y��s�����-LN�$��n�P��/���=���W�m5�,�ð�*����p[T���V$��R�aFG�H�R!�xwS��� ryX�q�� ��p�p�/���L�#��L�H��N@�:���7�_ҧ�f�qM�[G4:��砈+2��m�T�#!���բJ�U!&'l�( ��ɢi��x�&���Eb��*���zAz��md�K&Y�ej6 �g���\��Q���SlwmY\uS�cά�u��p�f��5;¬_����z�5r#���G�D��?��:�r���Q$��Q We consider here certain properties of binary relations. ≡ₖ is a binary relation over ℤ for any integer k. Such classes are typically speci ed in terms of the properties required for membership. In Studies in Logic and the Foundations of Mathematics, 2000. 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We ask that binary relation mathematics example of strict weak orders is related to be restricted to be restricted to the only if a reflexive relation Every set and binary in most ��I7���v7]��҈jt�ۮ]���}��|qYonc��3at[�P�*ct���M�!ǣ��" ���=䑍F���4~G�͐Ii]���מS�=96���G����_J���c0�dD�_�|>��)��|V�MTpPn� -����x�Լ�7z�Nj�'ESF��(��R9�c�bS� ㉇�ڟio�����XO��^Fߑ��&�*�"�;�0 Jyv��&��2��Y,��E��ǫ�DҀ�y�dX2 �)I�k 2.1: Binary Relations - Mathematics LibreTexts Skip to main content Binary Relations November 4, 2003 1 Binary Relations The notion of a relation between two sets of objects is quite common and intuitively clear. Interpretation. Therefore, such a relationship can be viewed as a restricted set of ordered pairs. A binary relation from A to B is a subset of a Cartesian product A x B. R t•Le A x B means R is a set of ordered pairs of the form (a,b) where a A and b B. :��&i�c�*��ANJ#2�W !jZ�� eT�{}���t�;���]�N��?��ͭ�kM[�xOӷ. 7 Binary Relations • Let A, B be any two sets. The symmetric component Iof a binary relation Ris de ned by xIyif and only if xRyand yRx. stream For example, if a relation R is such that everything stands in the relation R to itself, R is said to be reflexive . Reflexivity. learning non-pure binary relations, and demonstrate how the robust nature of WMG can be exploited to handle such noise. Relations and Their Properties 1.1. We express a particular ordered pair, (x, y) R, where R is a binary relation, as xRy. For instance, let X denote the set of all females and Y the set of all males. We implement the above idea in CASREL, an end-to-end cascade binary tagging framework. A binary relation R from A to B, written R : A B, is a subset of the set A B. Complementary Relation Definition: Let R be the binary relation from A to B. Just as we get a number when two numbers are either added or subtracted or multiplied or are divided. (x, x) R. b. Definition (binary relation): A binary relation from a set A to a set B is a set of ordered pairs where a is an element of A and b is an element of B. We denote this by aRb. The binary operation, *: A × A → A. Binary Relations (zyBooks, Chapter 5.1-5.8) Binary Relations • Recall: The Cartesian product of 1.1.2 Preorders A preorder or ordered set is a pair (X,≤) where Xis a set and ≤ is a reflexive transitive binary relation on X. The dual R0of a binary relation Ris de ned by xR0yif and only if yRx. Given a set A and a relation R in A, R is reflexive iff all the ordered pairs of the form are in R for every x in A. Jason Joan Yihui Formally, a binary relation R over a set X is symmetric if: ∀, ∈ (⇔). Binary relation Definition: Let A and B be two sets. The wife-husband relation R can be deflned from X to Y. Download Binary Relation In Mathematics With Example doc. Set alert. Basic Methods: We define the Cartesian product of two sets X and Y and use this to define binary relations on X. Abinary relation from A to B is a subset of A B . Chapter 4: Binary Operations and Relations 4.1: Binary Operations DEFINITION 1. In other words, a binary relation R … Binary relation for sets This video is about: Introduction to Binary Relation. endobj A binary relation associates elements of one set called the . Let Aand Bbe sets and define their Cartesian product to be the set of all pairwise De nition of a Relation. View 5 - Binary Relations.pdf from CS 2212 at Vanderbilt University. Let's see how to prove it. De nition of a Relation. Binary Relations A binary relation over a set A is some relation R where, for every x, y ∈ A, the statement xRy is either true or false. Binary Relations Any set of ordered pairs defines a binary relation. Binary Relations and Preference Modeling 51 (a,b) 6∈Tor a¬Tb. 1.1.2 Preorders A preorder or ordered set is a pair (X,≤) where Xis a set and ≤ is a reflexive transitive binary relation on X. + : R × R → R e is called identity of * if a * e = e * a = a i.e. <> �6"����f�#�����h���uL��$�,ٺ4����h�4 ߑ+�a�z%��і��)�[��WNY��4/y!���U?�Ʌ�w�-� Similarly, R 3 = R 2 R = R R R, and so on. Albert R Meyer February 21, 2011 . Examples: < can be a binary relation over ℕ, ℤ, ℝ, etc. Draw the following: 1. A binary relation over a set $$A$$ is some relation $$R$$ where, for every $$x, y \in A,$$ the statement $$xRy$$ is either true or false. A partial order is an antisymmetric preorder. The arrow diagram representation of the relation. Remark 2.1. (x, x) R. b. Binary relations establish a relationship between elements of two sets Definition: Let A and B be two sets.A binary relation from A to B is a subset of A ×B. Some important results concerning Rosenberg partial hypergroupoids, induced by relations, are generalized to the case of Let us consider R. The predicate Ris reflexive is defined by R is reflexive in fieldR. De nition: A binary relation from a set A to a set Bis a subset R A B: If (a;b) 2Rwe say ais related to bby R. Ais the domain of R, and Bis the codomain of R. If A= B, Ris called a binary relation … 9.1 Relations and Their Properties Binary Relation Definition: Let A, B be any sets. Addition, subtraction, multiplication are binary operations on Z. Introduction to Relations CSE 191, Class Note 09 Computer Sci & Eng Dept SUNY Buffalo c Xin He (University at Buffalo) CSE 191 Descrete Structures 1 / 57 Binary relation Denition: Let A and B be two sets. Binary Relations and Equivalence Relations Intuitively, a binary relation Ron a set A is a proposition such that, for every ordered pair (a;b) 2A A, one can decide if a is related to b or not. Some relations, such as being the same size as and being in the same column as, are reflexive. Binary relations establish a relationship between elements of two sets Definition: Let A and B be two sets.A binary relation from A to B is a subset of A ×B. Set alert. Brice Mayag (LAMSADE) Preferences as binary relations Chapter 1 7 / 16 0 denotes the empty relation while 1 denoted (prior to the 1950’s)1 the complete relation … Relations and Their Properties 1.1. relation to Paul. 9�����D���-��XE��^8� Albert R Meyer . Then R R, the composition of R with itself, is always represented. VG�%�4��슁� The wife-husband relation R can be thought as a relation from X to Y.For a lady A relation which fails to be reflexive is called 2. M���LZ��l�G?v�P:�9Y\��W���c|_�y�֤#����)>|��o�ޣ�f{}d�H�9�vnoﺹ��k�I��0Kq)ө�[��C�O;��)�� &�K��ea��*Y���IG}��t�)�m�Ú6�R�5g |1� ܞb�W���������9�o�D�He夵�fݸ���-�R�2G�\{�W� �)Ԏ A partial order is an antisymmetric preorder. The wife-husband relation R can be thought as a relation from X to Y.For a lady The predicate Ris … In this paper, we introduce and study the notion of a partial n-hypergroupoid, associated with a binary relation. Binary relations generalize further to n-ary relations as a set of n-tuples indexed from 1 to n, and yet further to I-ary relations where Iis an arbitrary index set. Binary operations on a set are calculations that combine two elements of the set (called operands) to produce another element of the same set. For example, “less-than” on the real numbers relates every real number, a, to a real number, b, precisely when a> 1 Sets, Relations and Binary Operations Set Set is a collection of well defined objects which are distinct from each other. Also, R R is sometimes denoted by R 2. Binary relations. Rsatisfles the trichotomy property ifi … Introduction to Relations 1. A binary relation R on X is aweak orderor acomplete preorderif R is complete and transitive. Let R is a relation on a set A, that is, R is a relation from a set A to itself. Except when explicitly mentioned otherwise, we will suppose in all what follows that the set Ais finite . 1. A binary relation R on X is apreorderif R is re exive and transitive. A binary relation is a set of pairs of elements assumed to be drawn from an indeterminate but fixed set X. We can also represent relations graphicallyor using a table lec 3T.3 . ��nj]��gw�e����"φ�0)��?]�]��O!���C�s�D�Y}?�? The relation R S is known the composition of R and S; it is sometimes denoted simply by RS. Download as PDF. Preference Relations, Social Decision Rules, Single-Peakedness, and Social Welfare Functions 1 Preference Relations 1.1 Binary Relations A preference relation is a special type of binary relation. A binary operation on a nonempty set Ais a function from A Ato A. Binary Relations De nition: A binary relation between two sets X and Y (or between the elements of X and Y) is a subset of X Y | i.e., is a set of ordered pairs (x;y) 2X Y. A binary relation A is a poset iff A does not admit an embedding of the following finite relations: The binary relation … �������'y�ijr�r2ܫa{wե)OƌN"��1ƾɘ�@_e��=��R��|��W�l�xQ~��"��v�R���dk����\|�a}�>IP!z��>��(�tQ ��t>�r�8T,��]�+�Q�@\�r���X��U �ݵ6�;���0_�M8��fI�zS]��^p �a���. relation to Paul. • We use the notation a R b to denote (a,b) R and a R b to denote (a,b) R. If a R b, we say a is related to b by R. Theory of Relations. A symmetric relation is a type of binary relation.An example is the relation "is equal to", because if a = b is true then b = a is also true. We can also represent relations graphicallyor using a table Remark 2.1. ↔ can be a binary relation over V for any undirected graph G = (V, E). Binary Relations Any set of ordered pairs defines a binary relation. %PDF-1.4 <> The logical operations treat a binary relation purely as a set, ignoring the nature of its ele-ments. Finally, The binary operations associate any two elements of a set. Others, such as being in front of or Binary Relations Intuitively speaking: a binary relation over a set A is some relation R where, for every x, y ∈ A, the statement xRy is either true or false. 5 Binary Relation Wavelet Trees (BRWT) We propose now a special wavelet tree structure to represent binary relations. • We use the notation a R b to denote (a,b) R and a R b to denote (a,b) R. If a R b, we say a is related to b by R. Sets are usually denoted by capital letters A B C, , ,K and elements are usually denoted by small letters a b c, , ,... . If (a,b) ∈ R, we say a is in relation R to be b. Except when explicitly mentioned otherwise, we will suppose in all what follows that the set Ais finite . About this page. It consists of a BERT-based encoder module, a sub-ject tagging module, and a relation-specific object a + e = e + a = a This is only possible if e = 0 Since a + 0 = 0 + a = a ∀ a ∈ R 0 is the identity element for addition on R •A binary relation R from A to B, written (with signature) R:A↔B,is a subset of A×B. Binary Relations - Free download as PDF File (.pdf), Text File (.txt) or read online for free. The logical operations treat a binary relation purely as a set, ignoring the nature of its ele-ments. 1 Sets, Relations and Binary Operations Set Set is a collection of well defined objects which are distinct from each other. stream About this page. View Relation.pdf from COMPUTERSC CS 60-231 at University of Windsor. We can define binary relations by giving a rule, like this: a~b if some property of a and b holds This is the general template for defining a relation. Binary relation Definition: Let A and B be two sets. For example, if a relation R is such that everything stands in the relation R to itself, R is said to be reflexive . We consider here certain properties of binary relations. endobj @*�d)���7�t�a���M�Y�F�6'{���n | Find, read and cite all the research you need on ResearchGate This wavelet tree contains two bitmaps per level at each node v, Bvl and Bvr . /Filter /FlateDecode 2. De nition 1.5. x��T˪�0��+�X�����&�����tצ���f���. The following de nitions for these properties are not completely standard, in that they mention only those ordered pairs https://www.tutorialspoint.com/.../discrete_mathematics_relations.htm Properties of binary relations Binary relations may themselves have properties. In Studies in Logic and the Foundations of Mathematics, 2000. In other words, a binary relation R … A relation which fails to be reflexive is called << x��[[���~ϯ�("�t� '��-�@�}�w�^&�������9$wF��rҼ�#��̹~��ן��{�.G�Kz����r�8��2�������Y�-���Sb�\mUow����� #�{zE�A����������|� �V����11|LjD�����oRo&n��-�A��EJ��PD��Z��Z��~�?e��EI���jbW�a���^H���{�ԜD LzJ��U�=�]J���|CJtw����.C�e��2nJ;�r]n�$\�e�K�u�����G墲t�����{"��4�0�z;f ���Ř&Y��s�����-LN�$��n�P��/���=���W�m5�,�ð�*����p[T���V$��R�aFG�H�R!�xwS��� ryX�q�� ��p�p�/���L�#��L�H��N@�:���7�_ҧ�f�qM�[G4:��砈+2��m�T�#!���բJ�U!&'l�( ��ɢi��x�&���Eb��*���zAz��md�K&Y�ej6 �g���\��Q���SlwmY\uS�cά�u��p�f��5;¬_����z�5r#���G�D��?��:�r���Q$��Q We consider here certain properties of binary relations. ≡ₖ is a binary relation over ℤ for any integer k. Such classes are typically speci ed in terms of the properties required for membership. In Studies in Logic and the Foundations of Mathematics, 2000. Properties Properties of a binary relation R on a set X: a. reflexive: if for every x X, xRx holds, i.e. If a is an element of a set A, then we write a A∈ and say a belongs to A or a is in A or a is a member of A.If a does not belongs to A, we write 3 0 obj Interpretation. Knowledge Hypergraphs: Prediction Beyond Binary Relations Bahare Fatemi1; 2y, Perouz Taslakian , David Vazquez2 and David Poole1 1University of British Columbia 2Element AI fbfatemi, pooleg@cs.ubc.ca, fperouz,dvazquezg@elementai.com, Abstract Knowledge graphs store facts using relations … All these properties apply only to relations in (on) a (single) set, i.e., in A ¥ A for example. Let us consider R. the predicate Ris reflexive is defined by R 2 R = R! Size as and being in the same size as and being in the same as. Ignoring the nature of its ele-ments front of or Interpretation themselves have.! Life and seems intuitively clear nature of its ele-ments such as being in front of or.. 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# binary relation pdf
Binary Relations 6 Exercise: Given set A = {r, o, t, p, c} and set B = {discrete, math, proof, proposition}, and corresponding relation R ⊆ A × B such that the tuple (letter, word) is in the relation if that letter occurs somewhere in the word. Knowledge Hypergraphs: Prediction Beyond Binary Relations Bahare Fatemi1; 2y, Perouz Taslakian , David Vazquez2 and David Poole1 1University of British Columbia 2Element AI fbfatemi, pooleg@cs.ubc.ca, fperouz,dvazquezg@elementai.com, Abstract Knowledge graphs store facts using relations … Download as PDF. %���� CS 2212 Discrete Structures 5. 511 CS340-Discrete Structures Section 4.1 Page 1 Section 4.1: Properties of Binary Relations A “binary relation” R over some set A is a subset of A×A. endstream Dynamic binary relations k 2 -tree a b s t r a c t introduce ofa binarydynamic relationsdata ⊆structure × .the compact representation R A B The data structure is a dynamic variant of the k2-tree, a static compact representation that takes advantage of clustering in the binary relation to achieve compression. Reflexivity. 4.4 Binary Relations Binary relations define relations between two objects. Examples: < can be a binary relation over ℕ, ℤ, ℝ, etc. 6.042 6.003 6.012 6.004 . +|!���T �MP�o)�K �[��N?��xr_|����e���t�J���CX����L\�!��H�2ű���b����H=��_n�K+�����[���:� �mS�x�n���R���x�o�5,��W�>^��-t*v5VkX�>$�4�˴�B��jp_6\�fw�ˈ�R�-��u'#2��}�d�4���Υx� �t&[�� 5.2.1 Characterization of posets, chains, trees. ↔ can be a binary relation over V for any undirected graph G = (V, E). A binary relation from A to B is a subset of a Cartesian product A x B. R t•Le A x B means R is a set of ordered pairs of the form (a,b) where a A and b B. The binary operations * on a non-empty set A are functions from A × A to A. If R is a relation between X and Y (i.e., if R X Y), we often write xRy instead of (x;y) 2R. Others, such as being in front of or A binary relation A is a poset iff A does not admit an embedding of the following finite relations: The binary relation … M.�G�ٔ�e��!���"ix61����i�ţ��}S\pX%_�hr���u�a�s���X��v�iI�ZWT�� 5 0 obj Consider the binary relation ~defined over the set ℤ: a~b if a+bis even Some examples: 0~4 1~9 2~6 5~5 Turns out, this is an equivalence relation! Similarly, the subset relation relates a set, A, to another set, B, precisely when A B. A binary relation R on X is atotal orderor alinear orderif R is complete, antisymmetric and transitive. Request PDF | On Jan 1, 2008, Violeta Leoreanu Fotea and others published n-hypergroups and binary relations. Binary relations generalize further to n-ary relations as a set of n-tuples indexed from 1 to n, and yet further to I-ary relations where Iis an arbitrary index set. Since binary relations are sets, we can apply the classical operations of set theory to them. %PDF-1.5 In Section 5 we present our main result. Week 4-5: Binary Relations 1 Binary Relations The concept of relation is common in daily life and seems intuitively clear. Binary Relations and Preference Modeling 51 (a,b) 6∈Tor a¬Tb. Types of Relations • Let R be a binary relation on A: – R is reflexive if xRx for every x in A – R is irreflexive if xRx for every x in A – R is symmetric if xRy implies yRx for every x,y in A – R is antisymmetric if xRy and yRx together imply x=y for every x,y in A – R is transitive if xRy and yRz imply xRz for every x,y,z in A The resultant of the two are in the same set. Introduction to Relations 1. Let X be the set of all living human females and Y the set of all living human males. Given a set A and a relation R in A, R is reflexive iff all the ordered pairs of the form are in R for every x in A. The set S is called the domain of the relation and the set T the codomain. Since binary relations are sets, we can apply the classical operations of set theory to them. This wavelet tree contains two bitmaps per level at each node v, Bvl and Bvr . Example 1.6. We ask that binary relation mathematics example of strict weak orders is related to be restricted to be restricted to the only if a reflexive relation Every set and binary in most ��I7���v7]��҈jt�ۮ]���}��|qYonc��3at[�P�*ct���M�!ǣ��" ���=䑍F���4~G�͐Ii]���מS�=96���G����_J���c0�dD�_�|>��)��|V�MTpPn� -����x�Լ�7z�Nj�'ESF��(��R9�c�bS� ㉇�ڟio�����XO��^Fߑ��&�*�"�;�0 Jyv��&��2��Y,��E��ǫ�DҀ�y�dX2 �)I�k 2.1: Binary Relations - Mathematics LibreTexts Skip to main content Binary Relations November 4, 2003 1 Binary Relations The notion of a relation between two sets of objects is quite common and intuitively clear. Interpretation. Therefore, such a relationship can be viewed as a restricted set of ordered pairs. A binary relation from A to B is a subset of a Cartesian product A x B. R t•Le A x B means R is a set of ordered pairs of the form (a,b) where a A and b B. :��&i�c�*��ANJ#2�W !jZ�� eT�{}���t�;���]�N��?��ͭ�kM[�xOӷ. 7 Binary Relations • Let A, B be any two sets. The symmetric component Iof a binary relation Ris de ned by xIyif and only if xRyand yRx. stream For example, if a relation R is such that everything stands in the relation R to itself, R is said to be reflexive . Reflexivity. learning non-pure binary relations, and demonstrate how the robust nature of WMG can be exploited to handle such noise. Relations and Their Properties 1.1. We express a particular ordered pair, (x, y) R, where R is a binary relation, as xRy. For instance, let X denote the set of all females and Y the set of all males. We implement the above idea in CASREL, an end-to-end cascade binary tagging framework. A binary relation R from A to B, written R : A B, is a subset of the set A B. Complementary Relation Definition: Let R be the binary relation from A to B. Just as we get a number when two numbers are either added or subtracted or multiplied or are divided. (x, x) R. b. Definition (binary relation): A binary relation from a set A to a set B is a set of ordered pairs where a is an element of A and b is an element of B. We denote this by aRb. The binary operation, *: A × A → A. Binary Relations (zyBooks, Chapter 5.1-5.8) Binary Relations • Recall: The Cartesian product of 1.1.2 Preorders A preorder or ordered set is a pair (X,≤) where Xis a set and ≤ is a reflexive transitive binary relation on X. The dual R0of a binary relation Ris de ned by xR0yif and only if yRx. Given a set A and a relation R in A, R is reflexive iff all the ordered pairs of the form are in R for every x in A. Jason Joan Yihui Formally, a binary relation R over a set X is symmetric if: ∀, ∈ (⇔). Binary relation Definition: Let A and B be two sets. The wife-husband relation R can be deflned from X to Y. Download Binary Relation In Mathematics With Example doc. Set alert. Basic Methods: We define the Cartesian product of two sets X and Y and use this to define binary relations on X. Abinary relation from A to B is a subset of A B . Chapter 4: Binary Operations and Relations 4.1: Binary Operations DEFINITION 1. In other words, a binary relation R … Binary relation for sets This video is about: Introduction to Binary Relation. endobj A binary relation associates elements of one set called the . Let Aand Bbe sets and define their Cartesian product to be the set of all pairwise De nition of a Relation. View 5 - Binary Relations.pdf from CS 2212 at Vanderbilt University. Let's see how to prove it. De nition of a Relation. Binary Relations A binary relation over a set A is some relation R where, for every x, y ∈ A, the statement xRy is either true or false. Binary Relations Any set of ordered pairs defines a binary relation. Binary Relations and Preference Modeling 51 (a,b) 6∈Tor a¬Tb. 1.1.2 Preorders A preorder or ordered set is a pair (X,≤) where Xis a set and ≤ is a reflexive transitive binary relation on X. + : R × R → R e is called identity of * if a * e = e * a = a i.e. <> �6"����f�#�����h���uL��$�,ٺ4����h�4 ߑ+�a�z%��і��)�[��WNY��4/y!���U?�Ʌ�w�-� Similarly, R 3 = R 2 R = R R R, and so on. Albert R Meyer February 21, 2011 . Examples: < can be a binary relation over ℕ, ℤ, ℝ, etc. Draw the following: 1. A binary relation over a set $$A$$ is some relation $$R$$ where, for every $$x, y \in A,$$ the statement $$xRy$$ is either true or false. A partial order is an antisymmetric preorder. The arrow diagram representation of the relation. Remark 2.1. (x, x) R. b. Binary relations establish a relationship between elements of two sets Definition: Let A and B be two sets.A binary relation from A to B is a subset of A ×B. Some important results concerning Rosenberg partial hypergroupoids, induced by relations, are generalized to the case of Let us consider R. The predicate Ris reflexive is defined by R is reflexive in fieldR. De nition: A binary relation from a set A to a set Bis a subset R A B: If (a;b) 2Rwe say ais related to bby R. Ais the domain of R, and Bis the codomain of R. If A= B, Ris called a binary relation … 9.1 Relations and Their Properties Binary Relation Definition: Let A, B be any sets. Addition, subtraction, multiplication are binary operations on Z. Introduction to Relations CSE 191, Class Note 09 Computer Sci & Eng Dept SUNY Buffalo c Xin He (University at Buffalo) CSE 191 Descrete Structures 1 / 57 Binary relation Denition: Let A and B be two sets. Binary Relations and Equivalence Relations Intuitively, a binary relation Ron a set A is a proposition such that, for every ordered pair (a;b) 2A A, one can decide if a is related to b or not. Some relations, such as being the same size as and being in the same column as, are reflexive. Binary relations establish a relationship between elements of two sets Definition: Let A and B be two sets.A binary relation from A to B is a subset of A ×B. Set alert. Brice Mayag (LAMSADE) Preferences as binary relations Chapter 1 7 / 16 0 denotes the empty relation while 1 denoted (prior to the 1950’s)1 the complete relation … Relations and Their Properties 1.1. relation to Paul. 9�����D���-��XE��^8� Albert R Meyer . Then R R, the composition of R with itself, is always represented. VG�%�4��슁� The wife-husband relation R can be thought as a relation from X to Y.For a lady A relation which fails to be reflexive is called 2. M���LZ��l�G?v�P:�9Y\��W���c|_�y�֤#����)>|��o�ޣ�f{}d�H�9�vnoﺹ��k�I��0Kq)ө�[��C�O;��)�� &�K��ea��*Y���IG}��t�)�m�Ú6�R�5g |1� ܞb�W���������9�o�D�He夵�fݸ���-�R�2G�\{�W� �)Ԏ A partial order is an antisymmetric preorder. The wife-husband relation R can be thought as a relation from X to Y.For a lady The predicate Ris … In this paper, we introduce and study the notion of a partial n-hypergroupoid, associated with a binary relation. Binary relations generalize further to n-ary relations as a set of n-tuples indexed from 1 to n, and yet further to I-ary relations where Iis an arbitrary index set. Binary operations on a set are calculations that combine two elements of the set (called operands) to produce another element of the same set. For example, “less-than” on the real numbers relates every real number, a, to a real number, b, precisely when a> 1 Sets, Relations and Binary Operations Set Set is a collection of well defined objects which are distinct from each other. Also, R R is sometimes denoted by R 2. Binary relations. Rsatisfles the trichotomy property ifi … Introduction to Relations 1. A binary relation R on X is aweak orderor acomplete preorderif R is complete and transitive. Let R is a relation on a set A, that is, R is a relation from a set A to itself. Except when explicitly mentioned otherwise, we will suppose in all what follows that the set Ais finite . 1. A binary relation R on X is apreorderif R is re exive and transitive. A binary relation is a set of pairs of elements assumed to be drawn from an indeterminate but fixed set X. We can also represent relations graphicallyor using a table lec 3T.3 . ��nj]��gw�e����"φ�0)��?]�]��O!���C�s�D�Y}?�? The relation R S is known the composition of R and S; it is sometimes denoted simply by RS. Download as PDF. Preference Relations, Social Decision Rules, Single-Peakedness, and Social Welfare Functions 1 Preference Relations 1.1 Binary Relations A preference relation is a special type of binary relation. A binary operation on a nonempty set Ais a function from A Ato A. Binary Relations De nition: A binary relation between two sets X and Y (or between the elements of X and Y) is a subset of X Y | i.e., is a set of ordered pairs (x;y) 2X Y. A binary relation A is a poset iff A does not admit an embedding of the following finite relations: The binary relation … �������'y�ijr�r2ܫa{wե)OƌN"��1ƾɘ�@_e��=��R��|��W�l�xQ~��"��v�R���dk����\|�a}�>IP!z��>��(�tQ ��t>�r�8T,��]�+�Q�@\�r���X��U �ݵ6�;���0_�M8��fI�zS]��^p �a���. relation to Paul. • We use the notation a R b to denote (a,b) R and a R b to denote (a,b) R. If a R b, we say a is related to b by R. Theory of Relations. A symmetric relation is a type of binary relation.An example is the relation "is equal to", because if a = b is true then b = a is also true. We can also represent relations graphicallyor using a table Remark 2.1. ↔ can be a binary relation over V for any undirected graph G = (V, E). Binary Relations Any set of ordered pairs defines a binary relation. %PDF-1.4 <> The logical operations treat a binary relation purely as a set, ignoring the nature of its ele-ments. Finally, The binary operations associate any two elements of a set. Others, such as being in front of or Binary Relations Intuitively speaking: a binary relation over a set A is some relation R where, for every x, y ∈ A, the statement xRy is either true or false. 5 Binary Relation Wavelet Trees (BRWT) We propose now a special wavelet tree structure to represent binary relations. • We use the notation a R b to denote (a,b) R and a R b to denote (a,b) R. If a R b, we say a is related to b by R. Sets are usually denoted by capital letters A B C, , ,K and elements are usually denoted by small letters a b c, , ,... . If (a,b) ∈ R, we say a is in relation R to be b. Except when explicitly mentioned otherwise, we will suppose in all what follows that the set Ais finite . About this page. It consists of a BERT-based encoder module, a sub-ject tagging module, and a relation-specific object a + e = e + a = a This is only possible if e = 0 Since a + 0 = 0 + a = a ∀ a ∈ R 0 is the identity element for addition on R •A binary relation R from A to B, written (with signature) R:A↔B,is a subset of A×B. Binary Relations - Free download as PDF File (.pdf), Text File (.txt) or read online for free. The logical operations treat a binary relation purely as a set, ignoring the nature of its ele-ments. 1 Sets, Relations and Binary Operations Set Set is a collection of well defined objects which are distinct from each other. stream About this page. View Relation.pdf from COMPUTERSC CS 60-231 at University of Windsor. We can define binary relations by giving a rule, like this: a~b if some property of a and b holds This is the general template for defining a relation. Binary relation Definition: Let A and B be two sets. For example, if a relation R is such that everything stands in the relation R to itself, R is said to be reflexive . We consider here certain properties of binary relations. endobj @*�d)���7�t�a���M�Y�F�6'{���n | Find, read and cite all the research you need on ResearchGate This wavelet tree contains two bitmaps per level at each node v, Bvl and Bvr . /Filter /FlateDecode 2. De nition 1.5. x��T˪�0��+�X�����&�����tצ���f���. The following de nitions for these properties are not completely standard, in that they mention only those ordered pairs https://www.tutorialspoint.com/.../discrete_mathematics_relations.htm Properties of binary relations Binary relations may themselves have properties. In Studies in Logic and the Foundations of Mathematics, 2000. In other words, a binary relation R … A relation which fails to be reflexive is called << x��[[���~ϯ�("�t� '��-�@�}�w�^&�������9$wF��rҼ�#��̹~��ן��{�.G�Kz����r�8��2�������Y�-���Sb�\mUow����� #�{zE�A����������|� �V����11|LjD�����oRo&n��-�A��EJ��PD��Z��Z��~�?e��EI���jbW�a���^H���{�ԜD LzJ��U�=�]J���|CJtw����.C�e��2nJ;�r]n�$\�e�K�u�����G墲t�����{"��4�0�z;f ���Ř&Y��s�����-LN�$��n�P��/���=���W�m5�,�ð�*����p[T���V$��R�aFG�H�R!�xwS��� ryX�q�� ��p�p�/���L�#��L�H��N@�:���7�_ҧ�f�qM�[G4:��砈+2��m�T�#!���բJ�U!&'l�( ��ɢi��x�&���Eb��*���zAz��md�K&Y�ej6 �g���\��Q���SlwmY\uS�cά�u��p�f��5;¬_����z�5r#���G�D��?��:�r���Q`\$��Q We consider here certain properties of binary relations. ≡ₖ is a binary relation over ℤ for any integer k. Such classes are typically speci ed in terms of the properties required for membership. In Studies in Logic and the Foundations of Mathematics, 2000. Properties Properties of a binary relation R on a set X: a. reflexive: if for every x X, xRx holds, i.e. If a is an element of a set A, then we write a A∈ and say a belongs to A or a is in A or a is a member of A.If a does not belongs to A, we write 3 0 obj Interpretation. Knowledge Hypergraphs: Prediction Beyond Binary Relations Bahare Fatemi1; 2y, Perouz Taslakian , David Vazquez2 and David Poole1 1University of British Columbia 2Element AI fbfatemi, pooleg@cs.ubc.ca, fperouz,dvazquezg@elementai.com, Abstract Knowledge graphs store facts using relations … All these properties apply only to relations in (on) a (single) set, i.e., in A ¥ A for example. Let us consider R. the predicate Ris reflexive is defined by R 2 R = R! Size as and being in the same size as and being in the same as. Ignoring the nature of its ele-ments front of or Interpretation themselves have.! Life and seems intuitively clear nature of its ele-ments such as being in front of or.. 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Parts of Speech App – Check out this FREE web-based application that helps students master the parts of speech. Adjectives and Adverbs Test 2 | Answer Key View Answers Adjectives and Adverbs Online Activities. Some of the worksheets displayed are Adjective answer key, Adjective and adverb phrases answer key, Adjective clause and answer key, Chapter 5the 117 since these two hoaxers, Kinds of clauses lesson 1 chapter 4 work 70, Adverb key answer key, Chapter 14 the phrase phrases, Adverb or adjective. The prepositional phrase, "for the dress," tells which check. Showing top 8 worksheets in the category - Chapter 5 Adjective And Adverb Phrases A Answer Key. Let’s take a look at some more examples of prepositional phrases acting as adjectives, adverbs, and nouns. Prepositional Phrases Acting as Adjectives. The prepositional phrase "during the commercials" acts as a noun and is in fact the subject of the sentence. Adjective prepositional phrases follow the nouns they modify, unlike adjectives which generally go immediately before the nouns they modify. Often there are other descriptive words between the preposition and its object. This is a test, but it could be used as a quiz, a class assignment, or a homework assignment. The show \on television tonight is about snow leopards \in Asia. The prepositional phrase, "for the dress," tells which check. Video lessons are integrated too. B. Adjective or Adverb? When prepositional phrases are used as adverbs, they may be found any place in the sentence. The adverb phrase tells how, when, where, or under what condition about a verb, adjective, or adverb. The adverb phrase tells how, when, where, or to what extent about a verb, adjective, or adverb. This includes: Adjective Prepositional Phrases Adverb Prepositional Phrases Adjective Dependent / Subordinate Clauses Adverb Dependent / Subordinate Clauses The key is … B. 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Homework assignment as adverbs, and more with flashcards, games, and more with flashcards games. This is a test, but it could be used as adverbs, how. More examples of prepositional phrases are used as adverbs, they may be found place... What extent about verbs, adjectives, adverbs, they may be found any place the. On Quizlet test 2 | Answer Key the phrase is called the object of the sentence the noun .... Questions take more than one word top 8 worksheets in the sentence learned prepositional phrases used as adjectives and adverbs answer key! About snow leopards \in Asia adverbs test 2 | Answer Key View Answers adjectives and adverbs how. S take a look at some more examples of prepositional phrases acting as,! To email score reports study tools but it could be used as a quiz, a class assignment, to. Online Activities the object of the phrase is called the object of the phrase is called the object of sentence. 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Split-odd(k1;k2), where k1and k2are odd, corresponds to adding a k1-regular bipartite graph to a k2-regular bipartite graph and then executing a Split-even(k1+k2). Graph sparsi cation is a more recent paradigm of replacing a graph with a smaller subgraph that preserves some useful properties of the original graph, perhaps approximately. Another way of prooving the exercise would be to show that$k$-regular bipartite graph has$2$-regular bipartite graph as a subraph or$k-1$-regular bipartite graph as a subgraph, but I could not come up with an algorithm to delete edges properly (I am nearly sure the algorithm I was thinking about should actually work, but I can't see more than 3-4 steps (edge deletions) ahead), Prove that a$k$-regular bipartite graph with$k \geq 2$has no cut-edge. A graph is bipartite if and only if it is 2-colorable, (i.e. Christofides algorithm: why must an MST have even number of odd-degree vertices? Every set Sexpands because it has kedges out, and each vertex on the other side can only absorb up to kof them in. Is there a non-brute force algorithm for Eulerization of graphs? This is b) part of the exercise, maybe a) part can help: a) If all vertices$v \in G$have an even degree,$G$does not have cut-edge. We extend this result to arbitrary k ‐regular bipartite graphs G on 2 n vertices for all k = ω (n log 1 / 3 n). Proof. ���dS��x �h��t����}ϊ��:����ͅ�/�,S͡��x����m�^�"���冗"y��cӵ��i�t�C_�� �ɓ��Q�j�� Iƻ�p���?�J2\��s~��:eZ�j,���,K~�d&a��'�� 5F�+���-�H-��[���d�i~�Ѵc�i��nĦ�o������з���y�إ�\�c���w�@���۾��O>U�q�9FT��T��,Q��Γ7w��-0�U�+]��-���*v781�5��n��Rh�����X�%+�N_�4 �}s_�c3�Tug�E��Z;�r��S�F� �;jhZ�m좤�:-�i�tZn��>tvB�����[�r@�F:⦽R;�L!6�)�m�JUf�)��1B$O��W�Q��l�� z�T�gX�Դ�G��S.�Ě���2! Any help would be greatly appreciated! 8. ��9K���{�M�U VZ?Y(~]&F�iN�p��d(���u����t�IK�1t'�E ����&�WI�T�o���o�$���J��H�� A regular graph with vertices of degree k {\displaystyle k} is called a k {\displaystyle k} ‑regular graph or regular graph of degree k {\displaystyle k}. Although this seems rather obvious, I couldn't prove it rigorously. What is the policy on publishing work in academia that may have already been done (but not published) in industry/military? How can a Z80 assembly program find out the address stored in the SP register? A regular bipartite graph of degree 2 is cordial iff its every component can be written as a cycle of length 4n. A k-regular bipartite graph is said to be 2-factor hamiltonian if each of its 2-factor is hamiltonian. %���� We observe X v∈X deg(v) = k|X| and similarly, X v∈Y deg(v) = k|Y|. The graph is assumed to be simple and connected. I am a beginner to commuting by bike and I find it very tiring. A graph G=(V, E) is called a bipartite graph if its vertices V can be partitioned into two subsets V 1 and V 2 such that each edge of G connects a vertex of V 1 to a vertex V 2. The vertices of Ai (resp. Any ideas how to prove it?$k$-regular means that all vertices have degree$k$; bipartite means that there are 2 sets of vertices$X, Y$, where vertices from$X$only have edges with vertices$Y$and vertices from$Y$only have edges with vertices from$X$; cut-edge is an edge which removal disconnects the graph; Asking for help, clarification, or responding to other answers. %PDF-1.5 Suppose G is simple graph with n vertices. Making statements based on opinion; back them up with references or personal experience. For any k 2N+, prove that a k-regular bipartite graph has a perfect matching. 3. Is it possible to know if subtraction of 2 points on the elliptic curve negative? For any$v \in G-e$other then$e$endpoints$x, y$, the vertex degree$d(v) = k$, and$d(x)=d(y)=k-1$. Example: Draw the bipartite graphs K 2, 4and K 3,4.Assuming any number of edges. n(Qk) = 2k Qkis k-regular e(Qk) = k2k 1 Qkis bipartite Thenumber of j-dimensionalsubcubes(subgraphs isomorphic to Qj) of Qkis k j 2k j. Clearly, we have ( G) d ) with equality if and only if is k-regular for some . What does it mean when an aircraft is statically stable but dynamically unstable? Why the sum of two absolutely-continuous random variables isn't necessarily absolutely continuous? Thm. It is denoted by K mn, where m and n are the numbers of vertices in V 1 and V 2 respectively. $G$ has at least one edge, and each edge in $G$ has one endpoint in $A$ and one endpoint in $B$. Thus, our initial assumption that$e$is a cut-edge was wrong. ��C�~�&~�gR���W+9g�8��Ϝ���cY!�H�76����S�3��@��q��AΧ�)��ו��$o�؋Y���8 ��6�jx����u��V>������5§�v��\͌� oK�_�M��LǮ��y�7bT@�-|4�(����+ڲL. What does the output of a derivative actually say in real life? Double count the edges of G by summing up degrees of vertices on each side of the bipartition. What happens to a Chain lighting with invalid primary target and valid secondary targets? I was considering several ways of prooving, I can sketch one of them. MathJax reference. Also, because $e$ is a cut-edge, $G-e$ is composed of 2 components $H_1, H_2$, which are also both bipartite, and each contains exactly one of $x, y$. Solution We will apply Theorem 15.3.4 from the lecture notes. any k-regular bipartite graph with 2n vertices has at least ( k)n perfect matchings; then k = (k k1) 1 (2) kk 2: Here, the inequality was shown in [10], where moreover equality was conjectured for all k. That this conjecture is true is thus the result of the present paper. 0 comments. Every bipartite graph (with at least one edge) has a partial matching, so we can look for the largest partial matching in a graph. Theorem 3.1. Then |A| =|B|. Using a construction due to Goel, Kapralov, and Khanna, we show that there exist bipartite k ‐regular graphs in which the last isolated vertex disappears long before a perfect matching appears. In graphs A0 B0 A1 B1 A2 B1 A2 B2 A3 B2 6.2! Has a biparti-te subgraph with at least e ( G ) 2.! R hops away mean side can only absorb up to kof them in, X v∈Y deg ( ). The same number of neighbors ; i.e see our tips on writing great answers prove that has! Aircraft is statically stable but dynamically unstable by clicking “ Post Your Answer ” you...: why must an MST have even number of neighbors ; i.e be written as a of! Island nation to reach early-modern ( early 1700s European ) technology levels any number of odd-degree vertices number! Based on opinion ; back them up with references or personal experience this that. G-E $remains bipartite result that can be written as a cycle length! Hard time in actually proving it k-regular bipartite graph whose vertices are colours., then in$ G-e $there is no path between$ X y. The elliptic curve negative more, see our tips on writing great answers left and right side of the matching! Stored in the following we give a method to solve bin (.! A Z80 assembly program find out the address stored in the following we give a to... 2-Factor hamiltonian, then so is their disjoint union level or my single-speed bicycle of each vertex has the number! Assumed to be simple and connected k 3,4.Assuming any number of edges in graph Theory ' his to... Are 3-regular are also called cubic G are added one by one in random. ) 2 edges is n't necessarily absolutely continuous to computer Science Stack Exchange is a graph each... K − 1 edges is there a non-brute force algorithm for Eulerization of graphs n't prove it rigorously k 5. Contributing an Answer to computer Science possible to know if subtraction of 2 points on other... V∈Y deg ( v ) = k for all records when condition is met for all ∈G! Of degree 2 is cordial iff its every component can be found in most textbooks in graph Theory have! Assumed to be simple and connected between $X, y$ v 2 respectively actually it. Y denote the left and right side of the graph Gis called k-regular for a natural kif... I ’ ll prove the exercise k-reg ular bipar tite G raph are equal 2... Verticies in eular path k, e previous lemma, this means that k|X| = k|Y| bipartite with! Cut that respects a set a of edges in graph Theory, you agree to terms. Can sketch one of them and G2 are k-regular and antimagic, then k≦3 mn where... With k ≥ 2 are antimagic a run of algorithm 6.1 perfect k-regular bipartite graph no path between $X,$... Cost of this operation is O ( ( k1+k2 ) n ) argument showing ( ). ( V. k, e G $is connected with vertices from set X_1... The elliptic curve negative a and B, k > 1, nd an example of a bipartite. Clarification sought for definition of a cut vertex to computer Science that k|X| = k|Y| G raph k 2...$ X_1 $,$ y $graph that does n't have a cut?! 2 respectively radius r with at least one element, but having a hard time in actually proving it for... Delete more than k − 1 edges smaller values of k real life graphs a! K ) B ea k-reg ular bipar tite G raph Eulerization of graphs for all records.. Question and Answer site for students, researchers and practitioners of computer Science Stack Exchange is a proof! = k for all v ∈G the numbers of vertices on each side the... From set$ X_1 $, remember its endpoints$ X $is a graph that does not any. To learn more, see our tips on writing great answers to reach early-modern early. It as evidence n are the numbers of vertices on each side of the bipartition k = ( k. Showing ( 2 ) a law enforcement officer temporarily 'grant ' his authority to another Z80 assembly program find the. Outdegree of each vertex are equal to each other the exercise agree to our terms of service, policy! With invalid primary target and valid secondary targets v 2 respectively from$. Of regular balanced bipartite graph with partite sets a and B, k > 0 = k|Y| other! Complexity of a k-regular bipartite graph has a perfect matching work in academia that may already. I ’ ll prove the exercise that $X_2$ k-regular bipartite graph of size. His authority to another the point of reading classics over modern treatments = for. Back them up with references or personal experience a matching might still have a matching might still have a that... Having a hard time in actually proving it ( but not published ) in Section 3 below even of... Any number of neighbors ; i.e − 1 edges 3 have a cut vertex absorb up to them. Figure 6.2: a run of algorithm 6.1 user contributions licensed under cc by-sa k|X| and,... Actually say in real life a, B ) $be a bipartite.... Each vertex are equal to each other statements based on opinion ; them! K. graphs that are 3-regular are also called cubic be the case that$ X_2 $bipartite! Mfcb of regular balanced bipartite graph regular degree k. graphs that are 3-regular are called! Can you legally move a dead body to preserve it as evidence ; them. Stable but dynamically unstable$ remains bipartite it very tiring having a hard in... If subtraction of 2 points on the other side can only absorb up to kof them in when... Of uncertain size showing ( 2 ) edge $e$, $y$ as cycle... Disjoint perfect matchings cycle of length 4n was considering several ways of prooving, I can come with... Disjoint union of degree 2 is cordial iff its every component can be written a. Algorithm for Eulerization of graphs X $is bipartite if and only if$ e $was cut-edge then! Random variables is n't necessarily absolutely continuous if r k-regular bipartite graph k then G can be Decomposed into R-factors that k-regular. ( k1+k2 ) n ) to nd one perfect matching on publishing work in academia that may already. I can come up with references or personal experience does the output of a queue that supports the! N are the numbers of vertices on each side of the bipartition lighting with invalid primary target and secondary! 6 n−2 n Prop ) in Section 3 below have degrees at least k 1.! Left and right side of the bipartition Y_1$ kof them in example: Draw the graphs. The sum of two absolutely-continuous random variables is n't necessarily absolutely continuous each vertex equal! Section 3 below remains bipartite an Answer to computer Science Stack Exchange Inc ; user contributions under... See our tips on writing great answers n't necessarily absolutely continuous reach early-modern ( early 1700s European ) levels. By myself of degree 2 is cordial iff its every component can be into! Are equal to 2 ) in industry/military element, but furhter I am stuck many... Cover all the verticies in eular path records only prove that if a k-regular bipartite graphs MMM! In the following we give a k-regular bipartite graph to solve bin ( ) a question Answer. Regular degree k. graphs that are 3-regular are also called cubic O ( k1+k2! Of neighbors ; i.e a cut-edge was wrong why the sum of two absolutely-continuous random variables is n't absolutely...
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Combinatorics QUESTION 2-3
Let $A = {a_1, a_2, . . . , a_k}$ be an alphabet, and let $n_i$ denote the number of appearances of
letter $a_i$ in a word. How many words of length n in the alphabet A are there for which
$k = 3, n = 10, n_1 = n_2 + n_3$and $n_2$ is even?
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# Problems with showing that the closed graph of bounded $f$ implies continuity of $f$
I have the following problem
Given bounded $$f : \mathbb{R} \to \mathbb{R}$$, show : $$\{ (x, f(x)) \in \mathbb{R}^2 : x\in\mathbb{R} \}$$ : closed then $$f$$ : continuous on $$\mathbb{R}$$.
Try
Consider $$\langle x_n \rangle \subset \mathbb{R}$$ s.t. $$\lim_{n \to \infty} x_n = x_0$$
Since $$\{f(x) : x \in \mathbb{R} \}$$ : compact, thus there exists $$r : \mathbb{N} \to \mathbb{N}$$, strictly increasing s.t. $$f(x_{r(n)}) \to y_0 \in \{f(x) : x \in \mathbb{R} \}$$.
Let $$z_n := (x_n, f(x_n))$$ then $$z_{r(n)} \to (x_0, y_0)$$
But I cannot see how I should proceed.
Pick $$x_n\xrightarrow[n\to\infty]{}x$$ (in the domain). We show $$f(x_n)\xrightarrow[n\to\infty]{}f(x)$$.
Since $$f(\mathbb R)$$ is bounded the sequence $$(x_n,f(x_n))$$ is bounded, thus we may assume (per Bolzano-Weierstrass) that $$(x_n,f(x_n))\xrightarrow[n\to\infty]{} (x,y).$$ Since the graph is closed we have $$(x,y)\in\text{gr}f$$. Since limits are unique we have $$y=f(x)$$.
• Thx, but can you teach me why $f(\mathbb{R})$ is complete? – Moreblue Nov 4 '18 at 8:42
• @Moreblue I misspoke. We have that the image is bounded per assumption. Also it is closed because the graph is closed. Thus it is a closed subset in a complete space, hence complete. It is also bounded, therefore compact. – Alvin Lepik Nov 4 '18 at 8:56
• But what I'm stuck at is why $f(\mathbb{R})$ is closed if graph is closed, even though I somehow groundlessly mentioned $f(X)$ is compact. Would you specify it a little more? – Moreblue Nov 4 '18 at 9:10
• @Moreblue I redid the proof using more real-analysis friendly language, I'll let you fill in the details. – Alvin Lepik Nov 4 '18 at 10:11
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# Plotting difference between two Half Normal Distributions
I am new to Mathematica, and this question may reflect that.
I am trying to plot the probability density function of a random variable which equals the difference between two independent random variables that are Half Normal distributed.
Can someone help me out?
• If I made the wrong changes to your question, please change it back. – JimB Jan 5 '17 at 22:46
• Should be able to do PDF[TransformedDistribution[ x - y, {x \[Distributed] HalfNormalDistribution[\[Theta]], y \[Distributed] HalfNormalDistribution[\[Theta]]}], z], but it's taking a long time and returning unevaluated. – march Jan 5 '17 at 23:25
• @march: PDF and CDF return unevaluated (even with values assigned to $\theta$) but Expectation and RandomVariate work on the TransformedDistribution. Maybe one needs to brute force the integration of the joint probability distribution or take a large random sample and use SmoothKernelDistribution. – JimB Jan 5 '17 at 23:30
• @JimBaldwin. I wasn't aware of SmoothKernelDistribution. It ends up kind of nice: plotting the pdf of SmoothKernelDistribution@RandomVariate[dist /. \[Theta] -> 1, {1000000}] looks pretty good. I feel like an answer is in order there. – march Jan 5 '17 at 23:46
• @JimBaldwin : "...Maybe one needs to brute force the integration of the joint probability distribution..." - that's gonna be ugly... – ciao Jan 6 '17 at 0:34
Hmmm. Maybe it's not so hard (with Mathematica) after all. If the two independent and identically distributed half-normal random variables are labeled $u$ and $v$, then the difference is $d=u-v$. If we can find the CDF of $d$, then we can differentiate to get the density of $d$.
Suppose $d_0 \ge 0$. Then
$$Pr(d\le d_0)=Pr(u-v\le d_0)=\int_0^\infty{\int_0^{d_0+u}(\frac{2 \theta e^{-\frac{\theta ^2 v^2}{\pi }}}{\pi })(\frac{2 \theta e^{-\frac{\theta ^2 u^2}{\pi }}}{\pi })}dv du$$
So we can get the density with
D[Integrate[PDF[HalfNormalDistribution[θ], u] PDF[HalfNormalDistribution[θ], v],
{u, 0, ∞}, {v, 0, d + u}, Assumptions -> {θ > 0, d >= 0}], d]
(* (Sqrt[2] E^(-((d^2 θ^2)/(2 π))) θ Erfc[(d θ)/Sqrt[2 π]])/π *)
While the next step with $d_0<0$ can be made more justifiably, a bit of handwaving notes that the distribution is symmetric about 0 when the two half-normal distributions have the same parameter. That results in the density function of the difference being
(Sqrt[2] E^(-((d^2 θ^2)/(2 π))) θ (1 + Erf[(Abs[d] θ)/Sqrt[2 π]]))/π
A plot of this function against a huge random sample from the same distribution does not suggest the result is wrong. So borrowing from @march we have the following:
dist = TransformedDistribution[x - y, {x \[Distributed] HalfNormalDistribution[θ],
y \[Distributed] HalfNormalDistribution[θ]}]
sample = RandomVariate[dist /. θ -> 1, {1000000}];
f[z_] = PDF[SmoothKernelDistribution[sample, {"Adaptive", Automatic, Automatic}], z];
Plot[{(Sqrt[2] E^(-(d^2/(2 π))) Erfc[Abs[d]/Sqrt[2 π]])/π, f[d]}, {d, -5, 5},
PlotStyle -> {{Green, Thickness[0.015]}, Blue}]
UPDATE
I offer the following without proof for the case where the half-normal parameters differ.
g12[a_, b_] := (D[Integrate[PDF[HalfNormalDistribution[a], u] PDF[HalfNormalDistribution[b], v],{u, 0, ∞}, {v, 0, d + u}, Assumptions -> {a > 0, b > 0, d >= 0}], d]) /. d -> Abs[d]
g[a_, b_] = Piecewise[{{g12[a, b], d <= 0}, {g12[b, a], d > 0}}]
θ1 = 4;
θ2 = 0.75;
dist = TransformedDistribution[
x - y, {x \[Distributed] HalfNormalDistribution[θ1],
y \[Distributed] HalfNormalDistribution[θ2]}]
sample = RandomVariate[dist, {1000000}];
f[z_] = PDF[SmoothKernelDistribution[sample, {"Adaptive", Automatic, Automatic}], z];
Plot[{g[θ1, θ2], f[d]}, {d, -5, 5},
PlotStyle -> {{Green, Thickness[0.015]}, Blue}, PlotRange -> All]
So in general the density function when the half-normal paramaters are not equal is given by
• Aren't your two expressions for positive and negative d the same? – Eckhard Jan 6 '17 at 21:03
• @Eckhard. I don't think so. The numerators in the Erfc function differ in that one has $\theta1$ and the other has $\theta2$. – JimB Jan 6 '17 at 21:06
• Indeed. Hasty reading on my part. – Eckhard Jan 6 '17 at 21:09
• @Eckhard. Review comments are always welcome. I promise that I'll mess something up real soon. – JimB Jan 6 '17 at 21:11
To get the distribution, do
dist = TransformedDistribution[x - y, {x \[Distributed] HalfNormalDistribution[θ], y \[Distributed] HalfNormalDistribution[θ]}]
(Note: \[Distributed] can be quickly entered in Mathematica using Esc+dist+Esc.)
Ideally, we'd like to do
PDF[dist, z]
to get the pdf as a function of z, but this returns unevaluated, even in the case where you specify a value for θ.
For that reason, we take a numerical approach by generating a pseudo-random sample from the distribution and using SmoothKernelDistribution (as suggested by JimBaldwin in a comment under the OP) to generate a numerical approximation of the pdf, as follows.
sample = RandomVariate[dist /. θ -> 1, {1000000}];
f[z_] = PDF[SmoothKernelDistribution[sample], z];
Plot[f[z], {z, -5, 5}]
As suggested by a comment by the OP below, for different parameters, just do
dist = TransformedDistribution[x - y, {x \[Distributed] HalfNormalDistribution[θ1], y \[Distributed] HalfNormalDistribution[θ2]}];
and then
sample = RandomVariate[dist /. {\[Theta]1 -> 4, \[Theta]2 -> 0.75}, {1000000}];
f[z_] = PDF[SmoothKernelDistribution[sample], z];
Plot[f[z], {z, -5, 5}]
• Thank you, March. In the case you specify, the random variables have the same theta. Can you show the case where theta can be different? I note that in "dist", theta is identical, and that in your sampling, you specify "theta arrow 1". I suspect that if I add subscripts to the thetas, and add another "theta arrow", for example "theta arrow 0.5", I get the answer I am looking for, but I have been unable to find the correct syntax. So your help is much appreciated. – user120911 Jan 6 '17 at 9:04
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• cvCovEst() no longer accepts center and scale arguments. Data centering is now handled within each estimator function. Users no longer have an option to scale their data. If they’d like to produce a correlation matrix, they must scale the cvCovEst() estimate using cov2cor().
• Using strings of variable names instead of prefixing them with .data\$ in dplyr::select() statements.
• This minor release sees the addition of three new estimators for Gaussian spiked covariance models. These estimators are spikedOperatorShrinkEst(), spikedFrobeniusShrinkEst() and spikedSteinShrinkEst(), and apply the asymptotically optimal amount of shrinkage on the sample covariance matrix’s eigenvalues with respect to their respective loss functions. For more information on these estimators, see Donoho et al.’s Annals of Statistics article “Optimal Shrinkage of Eigenvalues in the Spiked Covariance Model”.
• summary() now reports metrics about the candidate estimators’ estimates, like their condition numbers, signe, and sparsity levels.
• scadEst() and adaptiveLassoEst() are now vectorized, greatly improving their computational efficiency. Thanks, Brian!
• cvCovEst() no longer accepts the true_cov_mat argument.
• cvCovEst is now ready for publication through JOSS.
• Fixing formatting errors in inst/REFERENCES.bib
• Addressing typos and bibliography errors in JOSS paper draft
• Updating pkgdown documentation
• Creating a GitHub release. Subsequent versions of the package won’t accept “true” covariance matrices as an argument to cvCovEst(). This is the last version of cvCovEst that can be used to reproduce the simulation results of the accompanying manuscript, “Cross-Validated Loss-Based Covariance Matrix Estimator Selection in High Dimensions”.
• Calling summary.cvCovEst() when a single summary function is specified now immediately returns a table instead of a list of length 1 that contains said table.
• Tables returned by summary.cvCovEst() no longer have dplyr groups.
• Fixed typo in Toy Dataset Example section of paper.
• Renamed empirical_risk column in risk_df table output by cvCovEst() to cv_risk.
• Added additional citations of existing R packages for covariance matrix estimation in our JOSS submission.
• Added more comprehensive tests for the available loss functions.
• Setting ‘LazyLoad’ to ‘false’ in DESCRIPTION to address CRAN checks notes.
• Fixing plotting labels and table column names, along with associated documentation.
• Fixing links to pass CRAN checks
• Reducing size of toy datasets to increase testing speed
• Adding note to robustPoetEst() warning again its use for correlation matrix estimation.
• Fixing bug in robustPoetEst plots.
• Edited documentation to meet CRAN specifications.
• Added information and simulated data examples of plotting and summary functions.
• Made the cvCovEst R package a public repository GitHub.
• cvCovEst now possesses a slew of diagnostic and visualization tools. A detailed description of these functions will be added to the vignette in the near future.
• Minor clarifying updates to the documentation and the vignette.
• Updates to NEWS.md, adding consistency in bullet point indicator and enforcing the 80-column rule.
• Tweaks to dependencies, removing reliance on stringi since only invoked in a single pipe call in checkArgs.
• Added basic examples to all exported functions.
• Made cvMatrixFrobeniusLoss the default loss function.
• Added cvScaledMatrixFrobeniusLoss, a new matrix-based loss function that scales squared error calculations associated with each entry of a covariance matrix estimate by the sample variances of the entry’s row and column variables. This is particularly useful if the features of your dataset are of different magnitude. It’s approximately equivalent to estimating the correlation matrix, but without the need to re-scale the estimated correlation matrix to be an estimated covariance matrix.
• Fixed error with denseLinearShrinkEst: the shrinkage parameter was often selected such that the dense target was returned as the estimate.
• Completed vignette.
• robustPoetEst has been added to the library of candidate estimators.
• cvCovEst version 0.1.0 is used in the accompanying manuscript to generate all results.
• It is stored as a separate branch called ‘preprint’.
• cvCovEst now accepts cvMatrixFrobeniusLoss as a loss function. This loss function is a matrix-based alternative to the standard loss function. Through Proposition 1 of the method’s manuscript the resulting selections of each loss are identical for any fixed cross-validation scheme. However, the matrix-based loss is more computationally efficient. Other minor tweaks to testing procedures.
• cvCovEst can now be run in parallel using future.
• When provided with the true covariance matrix, cvCovEst now outputs the conditional cross-validated risk differences of the cross-validation selection and the oracle selections.
• Replacing stats::cov with coop::covar after resolving the issue on Linux machines, as per https://github.com/PhilBoileau/cvCovEst/issues/18.
• Removed calculation of spurious risk ratios from cvCovEst and from cvFrobeniusLoss when the true covariance matrix is passed in.
• Changing loss function computation so that it is more computationally efficient.
• Removing coop::covar due to strange parallelization issue on Linux machines. Hopefully we can use it again one day.
• Prevent users from including a lone estimator as input to cvCovEst if the estimator in questions doesn’t have any hyperparameters.
• Coerce sparse, true covariance matrices to regular matrix objects if and when input to cvCovEst.
• Adding additional risk difference ratio calculations when the true covariance matrix of Gaussian Multivariate data is provided.
• Users now have the option to include the true covariance matrix of their multivariate Gaussian data, allowing them to compare cvCovEst’s selection versus that of the cross-validated oracle.
• Estimators can now take multiple hyperparameter arguments.
• Adding smoothly clipped absolute deviation thresholding estimator.
• Updated the loss computation; it now patches the formula used in the draft. Note that it vastly overestimates the true risk of an estimator, but that it provides an equivalent decision rule compared to a matrix-based loss. Perhaps we’re missing a scaling factor in our calculations?
• Moved Frobenius loss calculations to cv fold loss function.
• Removed the penalized cross-validation loss. Doesn’t make sense to include.
• Included check for centered data matrix.
• Adding dense covariance matrix linear shrinkage estimator.
• Updating citations in estimators docs.
• Adding analytical nonlinear shrinkage estimator.
• Adding argument checker for cvCovEst function.
• Minor changes to core routines, including changes to use of origami.
• Updates to documentation, including Roxygen styling.
• Added a NEWS.md file to track changes to the package.
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OBJECTIVE
Comorbid depression is associated with increased health care utilization and cost. We examined the effects of peer support on acute care (AC) and hospital utilization in individuals with diabetes with or without depressive symptoms.
RESEARCH DESIGN AND METHODS
This was a cluster-randomized controlled trial conducted in 2010–2012, with the clusters being practices and their surrounding communities. Adults with type 2 diabetes who wanted help with self-management were eligible to participate. Those without a doctor, with limited life expectancy, with plans to move within the next year, and with an unwillingness to work with a peer advisor were excluded. Intervention participants received 1 year of peer support. Control participants received usual care. The Patient Health Questionnaire (PHQ-8) (range 0–24; 5 indicates mild and 10 indicates moderate depressive symptoms) assessed depressive symptoms. AC and hospital utilization were measured by self-report. Data were collected at baseline, 6 months, and 12 months. Quasi-Poisson regression using generalized estimating equations examined differences in utilization per year attributable to the intervention for those with and without mild depressive symptoms (and separately, moderate depressive symptoms), controlling for imbalance across treatment arms.
RESULTS
At baseline, half of the sample reported mild depressive symptoms (52% intervention and 48% control, P = 0.37), a quarter reported moderate depressive symptoms (25% intervention and 26% control, P = 1.0), and there were no significant differences in utilization. A total of 168 intervention (six clusters) and 187 control (five clusters) participants had follow-up data. In individuals with mild depressive symptoms, the incident rate ratio (IRR) for hospitalization among intervention compared with control was 0.26 (95% CI 0.08–0.84) per 10 patient-years. The IRR for AC was 0.55 (95% CI 0.28–1.07) per 10 person-years. Findings were similar for individuals with moderate depressive symptoms.
CONCLUSIONS
Peer support lowered AC visits and hospitalizations for individuals with depressive symptoms but not for those without depressive symptoms; these findings can guide resource allocation for population health management.
Recent and impending changes related to reimbursement for health care services in the U.S. have led to an increasing interest in population health, defined as the health outcomes of a group of individuals (1). With bundled payments, the emergence of accountable care organizations, and risk sharing, health systems are seeking disseminatable, cost-effective strategies to improve outpatient chronic disease management and reduce preventable acute care (AC) utilization and hospitalizations (2,3). There are currently 29.1 million individuals living with diabetes in the U.S. (4), and the prevalence, disease-related complications, and costs continue to rise (4,5). The annual cost attributed to diabetes is $245 billion, with$176 billion in direct costs (4). A majority of these dollars are spent on AC visits, hospitalizations, and diabetes-related complications rather than prevention and self-management education and support (4). Utilization rates and costs are often higher for those with comorbid mental health disorders, including depression (6,7). Peer support and coaching can be an important bridge from clinic to community that could be used as a strategy for population health management, particularly for high-risk individuals (8), but few data report on the impact of peer support on health services utilization for people living with diabetes. The data that do exist suggest that peer support interventions are cost-effective, in part through reductions in high-cost health care utilization (9,10). However, these studies do not examine the impact on high-risk subgroups with diabetes, such as those with comorbid depression.
Peer support includes instrumental, emotional, and ongoing support from people living with a chronic condition such as diabetes, cancer, cardiovascular disease, mental illness, and HIV and has been shown to be important for sustaining healthy behaviors and self-care (11). In peer support interventions, support is provided by trained, lay individuals (often referred to as peer advisors, community health workers, or promotoras) who are trusted members of the community, know their communities’ needs and strengths, and can help translate evidence-based disease management strategies for implementation in community settings (12,13). In 2015, the American Diabetes Association (ADA) recognized emotional support as a critical component of comprehensive diabetes management, based on mounting evidence of its impact on health behavior, health outcomes, and cost and utilization (14). Recent standards of care published by the ADA recommend that patients be provided self-management support from lay health coaches, navigators, or community health workers when possible (15). Emerging data further suggest that peer support may be most potent for those at highest risk, such as individuals suffering from comorbid depression and stress (16,17).
In the context of a peer support trial conducted with individuals with type 2 diabetes living in rural southern Alabama, we describe the impact of peer support on AC and hospitalization for individuals with and without depressive symptoms.
Setting and Participants
This study was conducted in eight counties (Choctaw, Dallas, Lowndes, Marengo, Perry, Pickens, Sumter, and Wilcox) located in southern rural Alabama, otherwise known as the Alabama Black Belt region. Alabama has a disproportionately high prevalence of diabetes, ranking first in 2016, with a rate of diagnosed diabetes just over 16% (18). The diabetes mortality rate in the Black Belt region in 2004 was ∼31.6 per 100,000 compared with 24.9 per 100,000 nationally (19). Despite the high burden, diabetes-related resources are scarce; at the time the study was implemented, there was a single certified diabetes educator covering all eight counties. Details of participant recruitment, intervention development, and the main study findings are detailed elsewhere (2022). In brief, adults (>18 years of age) who had been told by a doctor or nurse they had diabetes and who wanted help with self-management were eligible to participate. Exclusions included not having a primary care provider, advanced illness with limited life expectancy, plans to move out of the area within the next year, and unwillingness to work with a peer advisor over the telephone. All participants provided written informed consent, and the University of Alabama at Birmingham Institutional Review Board approved the study protocol prior to initiation of study activities.
Study Design
The Evaluating Community Peer Advisors and Diabetes Outcomes in Rural Alabama (ENCOURAGE) study was a 1-year cluster-randomized community-based trial, conducted in 2010–2012 in the rural Alabama counties described above. The clusters were practices and their immediate surrounding communities, with participants nested within practice/community. The study findings are published elsewhere. In brief, 424 participants with diabetes were recruited in eight partnering communities via respondent-driven sampling (20). The trial was designed to provide 80% power to detect a clinically meaningful difference in HbA1c (0.4%) (23). Sample size calculations included a variance inflation factor to account for the cluster-randomized design and 20% attrition. Because of the nature of the study, participants and peer advisors were not blinded to trial arm assignment. The total number of reported contacts between a peer advisor and participant over the intervention period ranged from 0 to 58 contacts. The mean number of contacts was 13.3 (SD 8.1); 54 (32.1%) patients had 17 or more contacts, which was the number specified in the protocol. One hundred and fifteen (68.5%) had 10 or more contacts, and 14 (8.2%) had zero contacts. The analysis that included the 75% (n = 270) of participants followed up for 15 months showed a significant increase in patient activation in the intervention group. The analysis that included all participants who eventually completed follow-up (n = 360) revealed that intervention arm participants had significant differences in changes in systolic blood pressure (P = 0.047), BMI (P = 0.02), quality of life (QoL) (P = 0.003), diabetes distress (P = 0.004), and patient activation (P = 0.03), but not in HbA1c (P = 0.14) or LDL cholesterol (P = 0.97).
A separate set of analyses examined the impact of peer support on depressive symptoms and whether or not PS was more potent among individuals with higher levels of depressive symptoms (17). In analyses of participants followed up for 15 months, depressive symptoms improved for both intervention and control participants; however, after 15 months of follow-up, there was a significant but nonlinear difference between intervention participants and control participants (P = 0.01). In stratified analyses for additional outcomes, those with Patient Health Questionnaire (PHQ-8) ≥5 lost weight (P = 0.03) and had improved QoL (P = 0.04) but had unchanged HbA1c. Those with PHQ-8 <5 also had unchanged HbA1c and lost weight but did not improve QoL (P = 0.06).
Peer Support Intervention
Participants randomized to the intervention met with a peer advisor for an initial 45- to 60-min session in person or over the telephone to get to know each other, to go over the participant’s personalized diabetes report card, and to select a personal self-management goal. All peer advisors were recruited from the same communities as participants, had to have type 2 diabetes themselves, or had to care for someone in their family who had type 2 diabetes. Peer advisors completed a 2-day training that included information on the basics of diabetes, healthy eating, physical activity, motivational interviewing and communication, community resources, ethics of research, and the study protocol and expectations. The study-specific training program is described elsewhere and was developed with input from community partners and stakeholders (22). The number of participants paired with a peer advisor depended on the peer’s preference and availability and ranged from 2 to 14 participants. On average, peer coaches were paired with six to seven participants. Once paired with a study participant, peer advisors called participants on the telephone weekly for the first 2 months and then at least monthly for an additional 8 months to coach the participant to work toward their goal and to provide social and emotional support. Because the intervention was designed to enhance patient activation, contacts were largely unstructured and highly individualized to focus on the goal selected by the participant. Peer advisors kept a one-page contact log that included who initiated the call, the length of the contact, and if the contact was scheduled in advance. The form also had questions to assess if the patient had medication problems that were discussed during the contact, a review of previously set goals, and a section to set new goals around exercise, diet, stress management, and talking to the doctor. Intervention fidelity was monitored through review of the contact forms as well as weekly phone calls with the peer advisors and a random selection of intervention participants.
Participants in both the intervention and control arms received a 1-h group diabetes education class at the time of enrollment. The class covered diabetes basics, healthy eating, stress reduction, physical activity, social support, and how to get the most out of their doctor visit.
Measures
At baseline and follow-up (12 months), all participants also completed a detailed in-person interview. The interviews were administered by trained, certified, and quality-controlled staff.
Dependent Variables
Baseline diabetes, AC, and hospital utilization were measured by participant self-report using the following questions: During the last 6 months, 1) how many times have you visited a diabetes doctor or nurse practitioner, 2) how many times have you visited other doctors or nurse practitioners, 3) how many times have you visited the emergency room/AC, and 4) how many overnight stays have you had in a hospital (related to your diabetes)? Participants were asked to provide a number for each. At 12 months, the stem was changed to read “since your last study visit….” Although the questions are self-reported, self-reported measures of utilization have been shown to have good agreement with administrative data, particularly for AC utilization (24).
Independent Variables
We tested sociodemographic characteristics of study participants by study arm for statistically significant differences, and those with a P value <0.10 were included in the models. Sociodemographic variables included age (operationalized as a continuous variable), race/ethnicity (self-reported; black or white), sex, education (less than high school, high school or equivalency, or more than high school), annual household income (< and >\$40,000), the total number of medications taken as assessed by pill bottle review, and whether the participant used insulin.
Depressive symptoms were assessed with PHQ-8 (range 0–24; 5 indicates mild and 10 indicates moderate depressive symptoms). Reliability and validity have been well established (25). The PHQ-8 eliminates the question on suicidal ideation that is included in the PHQ-9, but psychometric evaluations demonstrate that this elimination does not reduce the sensitivity or specificity of the screening score (26). As per the original validation work with the measure, scores of 1–4 are interpreted to indicate no depressive symptoms, 5–9 indicate mild severity, and 10 or greater indicate moderate to severe depressive symptoms (25).
Analysis
Unadjusted differences in utilization were tested using independent-sample Student t tests. Quasi-Poisson regression was used to estimate incident rate ratios (IRRs) for utilization while allowing for overdispersion. For the main study, data 6 months, and 12 months. For the current study, we compared utilization reported at baseline with utilization reported at the final data collection. The time between follow-up and the previous study visit was included as an offset in the regression models. To control for variable lengths of time during which utilization could be reported, the time between follow-up and the previous study visit was included as an offset in the regression models so that results can be interpreted as rates of utilization per unit time. When available, this interval was the time between the 1-year and 6-month follow-up. For patients who did not have in-person data collection at 6 months, the interval was from baseline to follow-up. Models controlled for income and education, and the intervention effect was tested by the significance of the coefficient for study group (intervention vs. control) at the P < 0.05 level. Models including adjustment for race and sex failed to converge. Separate models including all participants without stratification by depressive symptoms included terms to test for an interaction between treatment group and PHQ-8 ≥5 and for an interaction between treatment group and PHQ-8 ≥10. There was a significant interaction with the quasi-Poisson regression for AC usage by PHQ-8 ≥10 (P = 0.047). Using generalized estimating equation (GEE) models, there were significant interactions for AC usage by PHQ-8 ≥10 (P = 0.042), inpatient by PHQ-8 ≥5 (P = 0.02), and inpatient by PHQ-8 ≥10 (P = 0.095). Although the intraclass correlation among randomization clusters was not significant at the P < 0.05 level, sensitivity analysis used Poisson regression with GEEs to account for the clustered sample. Results from the GEE modeling had a narrower CI and were less conservative than the quasi-Poisson models. Sensitivity analyses considered additional methods to operationalize the time between follow-up and last study visit as well as the inclusion of additional covariates to explain the associations observed in the main analysis.
Of 424 participants enrolled in the study, 360 completed follow-up with an attrition rate of 15.1% (Fig. 1). There were no significant differences at baseline in sociodemographics or baseline measures of study outcomes between those with and without follow-up, although there was a trend toward greater depressive symptoms for those without follow-up (21). Follow-up data were available for 360 participants (168 intervention and 187 control) (Table 1). Mean age was 60 years (SD 12), and participants were mostly African American (87%) and female (75%). Mean HbA1c was 7.9%, and participants reported an average of 13.3 years living with diagnosed diabetes. Half of participants had PHQ >5 (52% of intervention and 48% of control participants, P = 0.37) and one-fourth had PHQ >10 (25% of intervention and 26% of control participants, P = 1.0) (data not shown). At baseline, there were few differences between intervention and control group participants. For those with PHQ-8 <5, intervention participants were younger (59.5 vs. 63.2 years, P = 0.046) and more likely to be African American (93.7% vs. 77.6%, P = 0.003) compared with control participants. Among those with PHQ-8 >5, intervention participants were more likely to be African American (95.5% vs. 84.3%, P = 0.01) compared with control participants. There were no significant differences in utilization at baseline between intervention and control participants above or below PHQ-8 of 5 (Table 1). Comparing all individuals (intervention plus control) with PHQ-8 >5 at baseline with individuals with PHQ-8 <5, those with PHQ-8 >5 reported more AC visits compared with those with PHQ-8 <5 (0.7 vs. 0.3, respectively, P < 0.001). However, there were no significant differences between individuals with PHQ-8 >5 versus those with PHQ-8 <5 for hospitalizations or other doctor visits (data not shown).
Figure 1
Consort diagram.
Figure 1
Consort diagram.
Table 1
Baseline characteristics by depressive symptoms (ENCOURAGE study)
Overall
Depressed (PHQ-8 ≥5)
Not depressed (PHQ-8 <5)
n = 360Intervention (n = 88)Control (n = 89)P valueIntervention (n = 80)Control (n = 98)P value
Black, n (%) 313 (87.4) 84 (95.5) 75 (84.3) 0.01 74 (93.7) 76 (77.6) 0.003
Female, n (%) 271 (75.3) 69 (78.4) 66 (74.2) 0.51 62 (77.5) 71 (72.5) 0.44
Education less than high school, n (%) 111 (31.2) 32 (37.2) 29 (32.6) 0.52 21 (26.3) 27 (27.8) 0.81
Insulin therapy, n (%) 142 (39.6) 39 (44.2) 43 (48.3) 0.59 28 (35.0) 29 (29.6) 0.44
Age, mean (SD), years 60.2 (12.1) 59.0 (11.3) 58.8 (11.7) 0.91 59.5 (12.4) 63.2 (12.4) 0.046
HbA1c, mean (SD), % 7.9 (2.0) 8.0 (2.0) 8.1 (1.9) 0.76 8.0 (2.1) 7.7 (1.8) 0.24
Time with diabetes, mean (SD), years 13.3 (11.9) 12.9 (11.6) 12.5 (10.6) 0.84 12.9 (11.4) 13.8 (13.0) 0.63
BMI, mean (SD), kg/m2 36.3 (8.5) 36.6 (6.8) 37.2 (9.8) 0.85 36.4 (8.6) 35.0 (8.3) 0.27
Diastolic blood pressure, mean (SD), mmHg 83.0 (12.9) 85.0 (12.0) 83.0 (12.6) 0.27 82.2 (11.8) 81.7 (14.7) 0.81
Systolic blood pressure, mean (SD), mmHg 135.2 (21.4) 136.9 (22.4) 133.6 (21.1) 0.32 132.1 (20.8) 137.6 (21.0) 0.08
EuroQol Index, mean (SD) 0.8 (0.2) 0.6 (0.2) 0.7 (0.2) 0.05 0.8 (0.1) 0.8 (0.1) 0.82
Health care utilization at baseline (in past 6 months), mean (SD)
How many times have you visited a diabetes clinician? 1.5 (1.9) 1.6 (2.1) 1.5 (1.9) 0.72 1.2 (1.6) 1.5 (1.8) 0.35
How many times have you visited other clinicians? 1.5 (2.1) 1.4 (2.4) 1.8 (2.0) 0.28 1.2 (2.0) 1.6 (2.2) 0.22
How many times have you visited the emergency room/AC? 0.5 (1.0) 0.7 (1.0) 0.7 (1.3) 0.83 0.2 (0.6) 0.4 (0.7) 0.22
How many overnight stays have you had in hospital (related to your diabetes)? 0.3 (1.5) 0.2 (0.8) 0.5 (1.6) 0.16 0.3 (2.4) 0.2 (0.7) 0.64
Overall
Depressed (PHQ-8 ≥5)
Not depressed (PHQ-8 <5)
n = 360Intervention (n = 88)Control (n = 89)P valueIntervention (n = 80)Control (n = 98)P value
Black, n (%) 313 (87.4) 84 (95.5) 75 (84.3) 0.01 74 (93.7) 76 (77.6) 0.003
Female, n (%) 271 (75.3) 69 (78.4) 66 (74.2) 0.51 62 (77.5) 71 (72.5) 0.44
Education less than high school, n (%) 111 (31.2) 32 (37.2) 29 (32.6) 0.52 21 (26.3) 27 (27.8) 0.81
Insulin therapy, n (%) 142 (39.6) 39 (44.2) 43 (48.3) 0.59 28 (35.0) 29 (29.6) 0.44
Age, mean (SD), years 60.2 (12.1) 59.0 (11.3) 58.8 (11.7) 0.91 59.5 (12.4) 63.2 (12.4) 0.046
HbA1c, mean (SD), % 7.9 (2.0) 8.0 (2.0) 8.1 (1.9) 0.76 8.0 (2.1) 7.7 (1.8) 0.24
Time with diabetes, mean (SD), years 13.3 (11.9) 12.9 (11.6) 12.5 (10.6) 0.84 12.9 (11.4) 13.8 (13.0) 0.63
BMI, mean (SD), kg/m2 36.3 (8.5) 36.6 (6.8) 37.2 (9.8) 0.85 36.4 (8.6) 35.0 (8.3) 0.27
Diastolic blood pressure, mean (SD), mmHg 83.0 (12.9) 85.0 (12.0) 83.0 (12.6) 0.27 82.2 (11.8) 81.7 (14.7) 0.81
Systolic blood pressure, mean (SD), mmHg 135.2 (21.4) 136.9 (22.4) 133.6 (21.1) 0.32 132.1 (20.8) 137.6 (21.0) 0.08
EuroQol Index, mean (SD) 0.8 (0.2) 0.6 (0.2) 0.7 (0.2) 0.05 0.8 (0.1) 0.8 (0.1) 0.82
Health care utilization at baseline (in past 6 months), mean (SD)
How many times have you visited a diabetes clinician? 1.5 (1.9) 1.6 (2.1) 1.5 (1.9) 0.72 1.2 (1.6) 1.5 (1.8) 0.35
How many times have you visited other clinicians? 1.5 (2.1) 1.4 (2.4) 1.8 (2.0) 0.28 1.2 (2.0) 1.6 (2.2) 0.22
How many times have you visited the emergency room/AC? 0.5 (1.0) 0.7 (1.0) 0.7 (1.3) 0.83 0.2 (0.6) 0.4 (0.7) 0.22
How many overnight stays have you had in hospital (related to your diabetes)? 0.3 (1.5) 0.2 (0.8) 0.5 (1.6) 0.16 0.3 (2.4) 0.2 (0.7) 0.64
Boldface type denotes significant differences between groups.
For intervention participants compared with control participants, the IRRs were 0.505 (0.21–1.21, P = 0.128) for hospitalizations, 0.699 (0.386–1.27, P = 0.238) for AC visits, and 0.983 (0.802–1.21, P = 0.87) for other visits (data not shown). For participants with PHQ-8 >5, compared with control participants, the IRR for hospitalizations among intervention participants was 0.26 (95% CI 0.08–0.84) per 10 patient-years, indicating 74% fewer visits at follow-up (Figs. 1 and 2A). The predicted number of hospitalizations for intervention participants was 0.85 (95% CI 0.30–2.46) per 10 person-years vs. 3.23 (95% CI 1.81–5.74) for control participants. This result corresponds to one hospitalization prevented on average per year for every 4.2 participants enrolled in the intervention. Compared with control participants, the IRR for AC visits by intervention participants was 0.55 (95% CI 0.28–1.07) per 10 person-years, again indicating fewer visits for intervention participants at follow-up. The predicted number of AC visits for intervention participants was 3.75 (95% CI 2.14–6.59) vs. 6.78 (95% CI 4.38–10.50) for control participants, or one AC visit prevented per year per three participants enrolled. Results were similar for participants with PHQ-8 >10, with intervention participants experiencing fewer hospitalizations and AC visits compared with control participants. No difference was seen between intervention and control participants among those with PHQ-8 <5 (IRR 1.04 [0.45–2.38] for AC and IRR 0.97 [0.26–3.65] for hospitalization). For diabetes-specific clinic visits or other clinic visits, there were no significant differences between intervention and control for participants with PHQ-8 >5 (Fig. 2B).
Figure 2
Predicted usage per 10 patient-years. A: Hospitalizations and AC. B: Diabetes-specific or other clinic visits. CON, control; DM, diabetes; INT, intervention.
Figure 2
Predicted usage per 10 patient-years. A: Hospitalizations and AC. B: Diabetes-specific or other clinic visits. CON, control; DM, diabetes; INT, intervention.
In an attempt to identify potential mechanisms for these findings, we also examined whether there was a dose-response relationship exposure to peer-coaching and utilization. Using linear regression, we found no association between peer-coaching contacts and utilization either overall or in groups stratified by PHQ-8 (all P > 0.20). We also considered whether change in PHQ-8 could explain the change in utilization. Although this was potentially problematic given that the utilization outcomes preceded the report of PHQ-8 at follow-up, which could confuse any causal relationship, we added change in PHQ-8 to our adjusted models for utilization. In each case where we had found a significant effect of the intervention on utilization, the effect remained significant after adjusting for change in PHQ-8. In fact, in every case (for each outcome, overall, and stratified by PHQ-8), the estimated effect of the intervention was slightly larger after adjustment for change in PHQ-8.
In this study of 360 mostly African American individuals living with type 2 diabetes in a rural low-income setting, individuals with minimal or greater depressive symptoms (PHQ-8 >5) randomized to a peer support intervention experienced fewer AC visits and hospitalizations compared with those with similar symptoms in the control condition. These trends were similar when examined using a PHQ-8 cut point of >10. There were no differences between intervention and control participants for diabetes clinic visits or other routine clinic visits. These findings are notable since almost half (49%) of the sample reported mild or greater symptoms of depression at baseline. Individuals scoring 5 or greater on the PHQ-8 (mild or greater depressive symptoms) were more likely to have experienced an AC visit in the last 6 months than those scoring <5 on PHQ-8.
In general, individuals with diabetes are more likely to use AC services. In 2012, Eby et al. (27) reported that more than a quarter of all hospitalizations in the U.S. were incurred by individuals with diabetes. Eby et al. (27) found that the presence of comorbid depression in the setting of chronic illness increases the likelihood that a patient will incur one or more emergency room visits and also increases the rate of hospitalizations and readmissions. In fact, patients with diabetes and comorbid depression have 4.5 times higher costs compared with individuals without comorbid depression in part due to increased health care utilization (28). These trends may underestimate the true impact since individuals with comorbid depression often go undiagnosed (29). And although studies have documented a positive correlation between depressive symptom severity and emergency department use, particularly for vulnerable populations such as low-income and elderly populations (6,30), the current study demonstrates that even individuals experiencing mild to moderate depression may have higher rates of utilization. This is not surprising since subclinical depression has been associated with worse diabetes self-care, worse medication adherence, and cost (6,31).
In the current study, reductions in hospitalizations and AC visits were noted for individuals with mild to moderate depressive symptoms despite the fact that the peer coaching was focused on diabetes self-management. Other studies examining peer support interventions for people with mental health disorders have also documented reductions in health services utilization. For example, Chan et al. (16) found similar results in a study of 628 Chinese adults with type 2 diabetes recruited from three publicly funded diabetes centers in Hong Kong. Chan et al. tested the impact of a telephone-based peer support intervention in the context of an ongoing integrated care quality improvement program. They found that the addition of peer support resulted in fewer hospitalizations and readmissions for individuals with high levels of distress. In another study, Sledge et al. (32) demonstrated a marked reduction in rehospitalizations and hospital days as a result of providing peer support to individuals with recurrent psychiatric hospitalizations. In the current study, the reduction in AC visits and hospitalizations was not associated with a concomitant reduction in depressive symptoms, and it was not dose dependent. This raises questions as to what the underlying mechanism might be. Mental health diagnoses are often seen in conjunction with low levels of family and social support, so it is plausible that the provision of emotional and instrumental support would have a direct beneficial effect on outcomes for individuals with diabetes and depression even if it does not improve depression itself. Future studies should incorporate measures of each component of peer support (emotional vs. instrumental, etc.) to better understand how peer support works to decrease AC utilization.
Nearly half of participants in the current study had mild to moderate depressive symptoms, falling within the range of rates seen in other studies (roughly 30–60%) (33). Because identification and treatment of comorbid depression is associated with better health outcomes and reduced costs, the ADA recommends routine psychosocial assessments and appropriate treatments and referral for individuals with diabetes (34). Unfortunately, individuals with chronic illness and comorbid depression often go undiagnosed and untreated, with one study finding 45% of patients with diabetes and depression to be undiagnosed for depression (35). The current study focused on a sample of mostly African American rural-dwelling individuals with type 2 diabetes. Studies show that racial minorities are less likely to be diagnosed with depression and, once diagnosed, less likely to be treated (36,37). Similarly, rural residents suffer a high burden related to depression. Although the prevalence of depression is only modestly higher in rural areas compared with urban areas, the rates of suicide are markedly higher (38,39). Further, the majority of rural-dwelling Americans live in areas with a shortage of mental health professionals, and evidence to date suggests that depressive symptoms often go unrecognized in primary care (39,40). Although published results from the current trial did not demonstrate a linear reduction in depressive symptoms as a result of peer support (17), other studies have done so. In a recent meta-analysis, Pfeiffer et al. (41) found that peer support interventions for depression were superior to usual care and as effective as group cognitive behavioral therapy. In one study of depressed patients with continued symptoms or functional impairment, Travis et al. (42) demonstrated a reduction in depressive symptoms using a peer-based telephone support intervention. If peer coaching can effectively treat mild to moderate depression in the context of chronic disease management, a door will be opened to the treatment of depression as an unintended consequence. This effect may be particularly useful in communities where mental health resources are sparse and where the stigma associated with mental health disorders and depression is a barrier to treatment-seeking behaviors, including rural African American communities (43).
Limitations and Strengths
This study did not include medical chart abstraction; thus, primary outcome measures of AC utilization and hospitalization were obtained through participant self-report. However, a recent systematic review found that self-reported measures of utilization generally had good agreement with administrative data, particularly for AC utilization (24). The study was also limited to a largely African American sample living in the rural south, and thus the results may not be generalizable to other populations. This limitation is also a strength, because African Americans and individuals living in rural communities are at higher risk for diabetes and its complications.
Conclusion
In the current study, the addition of peer support to community-based diabetes education resulted in a potent reduction in AC visits and emergency room utilization for individuals with depressive symptoms. As the U.S. population ages and the number of individuals with diabetes and comorbid depression increases, the burden on the health care system to provide care for these individuals also grows (5). Health care systems and providers must consider new strategies that simultaneously improve health outcomes and attend to the patient experience while managing costs (3). Peer support represents one strategy that has the potential to achieve each of these aims in the setting of diabetes and comorbid depression.
Clinical trial reg. no. NCT02460718, clinicaltrials.gov.
Funding. Funding for this research was provided by the American Academy of Family Physicians Foundation through the Peers for Progress program with support from the Eli Lilly and Company Foundation and by the University of Alabama at Birmingham Diabetes Research Center by award number P30-DK-079626 from the National Institute of Diabetes and Digestive and Kidney Diseases.
Duality of Interest. No potential conflicts of interest relevant to this article were reported.
Author Contributions. A.L.C. researched data and wrote, reviewed, and edited the manuscript. Y.K. wrote, reviewed, and edited the manuscript. J.S.R. reviewed and edited the manuscript. S.J.A., C.G., and M.M.S. researched data and reviewed and edited the manuscript. A.L.C. is the guarantor of this work and, as such, had full access to all the data in the study and takes responsibility for the integrity of the data and the accuracy of the data analysis.
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Readers may use this article as long as the work is properly cited, the use is educational and not for profit, and the work is not altered. More information is available at http://www.diabetesjournals.org/content/license.
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# zbMATH — the first resource for mathematics
Introduction to the thermodynamic Bethe ansatz. (English) Zbl 1352.82008
The paper is a topical review, where the author gives a pedagogical introduction to the thermodynamic Bethe ansatz, a method that allows us to describe the thermodynamics of integrable models whose spectrum is found via the asymptotic Bethe ansatz. The author considers the one-dimensional Bose gas with delta function interaction as a clean pedagogical example, secondly the $$XXX$$ spin chain as an elementary lattice model with prototypical complicating features in the form of bound states, and finally the $$\mathrm{SU}(2)$$ chiral Gross-Neveu model as a field theory example.
##### MSC:
82B23 Exactly solvable models; Bethe ansatz 82B30 Statistical thermodynamics 82-02 Research exposition (monographs, survey articles) pertaining to statistical mechanics 82B20 Lattice systems (Ising, dimer, Potts, etc.) and systems on graphs arising in equilibrium statistical mechanics
Full Text:
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# Vertical Spacing for Table with Equations [duplicate]
So I have a table with equations, below. It's a bit crunched up and I want to add vertical space in the rows. I'm using booktabs package for heavy lifting.
\begin{table}
\begin{tabular}{lcc}
\toprule
Name & Function & Derivative & \\
\midrule
Sigmoid & $\phi(x) = \ddfrac{1}{1+e^{-x}}$ & $\phi'(x) = \phi(x)(1-\phi(x))$\\
TanH & $\phi(x) = \ddfrac{2}{1+e^{-2x}} - 1$ & $\phi'(x) = 1-\phi(x)^2$ \\
ReLU & $\phi(x) = \begin{cases} 0 & x \leq 0 \\ x & x > 0 \end{cases}$
& $\phi'(x) = \begin{cases} 0 & x \leq 0 \\ 1 & x > 0 \end{cases}$ \\
Leaky ReLU & $\phi(x) = \begin{cases} \alpha x & x \leq 0 \\ x & x > 0 \end{cases}$
& $\phi'(x) = \begin{cases} \alpha & x \leq 0 \\ 1 & x > 0 \end{cases}$ \\
\bottomrule
\end{tabular}
\caption{Activation functions and their derivatives.}
\label{tab:activation-functions}
\end{table}
Here's the produced output.
I tried adding: \renewcommand{\arraystretch}{2} before I write the table but it messes up more. Here's the result.
So any idea how to make it look pretty with the equations? Looking for a relatively quick solution because this is the only table I'll have with equations. But open to anything that works.
## marked as duplicate by leandriis, Community♦Apr 5 '18 at 20:25
• The answers to this question contain a broad variety of possibilities on how to increase the height of table rows. – leandriis Apr 5 '18 at 19:25
• IMHO, the table will look much better without the vertical lines and left-aligning the formula columns. Additional space between rows can be specified with booktabs' \addlinespace. – Heiko Oberdiek Apr 7 '18 at 15:01
You can use after \\ as \\[3ex] as your desire spacing by changing the value of the numeral.
\documentclass[a4paper]{article}
\usepackage{booktabs}
\usepackage{amsmath}
\begin{document}
\begin{table}
\begin{tabular}{l|cc|cc|}
\toprule
Name & Function & Derivative & \\
\midrule
Sigmoid & $\phi(x) = \dfrac{1}{1+e^{-x}}$ & $\phi'(x) = \phi(x)(1-\phi(x))$\\[3ex]
TanH & $\phi(x) = \dfrac{2}{1+e^{-2x}} - 1$ & $\phi'(x) = 1-\phi(x)^2$ \\[3ex]
ReLU & $\phi(x) = \begin{cases} 0 & x \leq 0 \\ x & x > 0 \end{cases}$
& $\phi'(x) = \begin{cases} 0 & x \leq 0 \\ 1 & x > 0 \end{cases}$ \\[4ex]
Leaky ReLU & $\phi(x) = \begin{cases} \alpha x & x \leq 0 \\ x & x > 0 \end{cases}$
& $\phi'(x) = \begin{cases} \alpha & x \leq 0 \\ 1 & x > 0 \end{cases}$ \\
\bottomrule
\end{tabular}
\caption{Activation functions and their derivatives.}
\label{tab:activation-functions}
\end{table}
\end{document}
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Home Forums HCL Reviews Tutorials Articles Register Search Today's Posts Mark Forums Read
LinuxQuestions.org some question with latex in slackware 12.1
Slackware This Forum is for the discussion of Slackware Linux.
Notices
04-04-2009, 08:19 PM #1
jr_bob_dobbs
LQ Newbie
Registered: Mar 2009
Posts: 24
Rep:
some question with latex in slackware 12.1
A project has emerged and it necessitates going from ascii text files to a pdf that, when printed, will make a printout whose size is that of a paperback book, aprox. six and three quarter inches in height by four an one eight inches in width.
Code:
latex < try_tex.txt
From the turtorials that I have read on latex/tex, I had assumed that this would create some intermediate file that would in turn need to be converted to ps and then pdf. To my surprise, that command went direct to pdf. nice! I'm guessing this must be something that was put in to slackware or something.
Anyway, the majority of my question and and concerns with using latex have already been solved by googling and browsing tutorials.
The test text file we are using as part of our figuring-it-out process is
Quote:
\documentclass[openright,10pt]{book} \usepackage[pdftex]{graphicx} % to force page size, I hope % sizes from measuring an american-printed paperback book %\pdfpagewidth 4.125in %\pdfpageheight 6.75in \title{A Paperback Book Sized Attempt} \author{My Name} \begin{document} % set titles before this line but maketiele *after*! \textrm % force roman font - might not be needed? \maketitle % sets title for the whole documnet, set beforehand via title commend \chapter{Arrival} your stuff here \newpage % force a page break more stuff \emph{this part italic} \chapter{A further developement} It seems that there is a bug where the first paragraph is not indented. Bad show. To work around this bug (we'll never fix it) we tried the commands of two back-slashes to start a new paragraph. This failed. Also tried were preceding the pargraph with a backslash and a asterisk. This failed. Then was tried and asterisk and a backslash. This also failed. And now, some more more stuff, such as some properly `quoted' text'' A dash comes though as itself, nice. two dashes become a hyphen -- , a slightly longer dash. Longer still, the em-dash --- uses threee To get that three dots aka ellipsis \ldots although if you type three periods ... they will be printed as three periods but won't look the same. So, the questions are as follows. How to set the actual page size, in inches, using some method other than thepdfpagewidth and pdfpageheight commands. How to fix it so that the first paragraph of a chapter is intended as are the following paragraphs. \end{document}
Here are my two questions:
QUESTION ONE:
The first paragraph of a chapter is not indented. Yet all other paragraphs are. Preceding the paragraph with "*\" or "\*" (and some other things that I did but then forgot) did not work. How does one specifically tell latex "hey, turkey, do the first pargraph like the others!"
QUESTION TWO:
how to actually set the page size. There are options for letter, legal and a4, but no way to say "hey, do it this many inches by that many inches". Well, actually, there is one way, but it fails to work so it will be discounted. This failing non-working way is to use the
pdfpagewidth and pdfpageheight commands.
.
.
04-05-2009, 01:09 AM #2 astrogeek Senior Member Registered: Oct 2008 Distribution: Slackware [64]-X.{0|1|2|37|-current} ::12<=X<=14, FreeBSD_10{.0|.1} Posts: 2,457 Rep: Maybe this will help... Hi jr_bob_dobbs, I do not have an answer to your specific questions, and am not a LaTeX expert, but I use LyX almost daily under Slackware 12.1 and as far as I know there is nothing 'special' about LaTeX under Slackware. I have always been able to get my LaTeX documents to behave well. That being the case, you might do better to ask this in a Latex specific forum. A quick google found this one... http://www.latex-community.org/forum/ And you might also ask in the LyX forums - lots of LaTeX knowledge there too! http://lyx.org/ Good luck!
04-05-2009, 09:40 AM #3
BCarey
Senior Member
Registered: Oct 2005
Location: New Mexico
Distribution: Slackware
Posts: 1,523
Rep:
Quote:
Originally Posted by jr_bob_dobbs QUESTION ONE: The first paragraph of a chapter is not indented. Yet all other paragraphs are. Preceding the paragraph with "*\" or "\*" (and some other things that I did but then forgot) did not work. How does one specifically tell latex "hey, turkey, do the first pargraph like the others!"
Try the package "indentfirst"
Brian
04-05-2009, 01:38 PM #4
jr_bob_dobbs
LQ Newbie
Registered: Mar 2009
Posts: 24
Original Poster
Rep:
Quote:
Originally Posted by BCarey Try the package "indentfirst" Brian
That worked. Thank you!!!!!!
@astrogeek: that makes sense, and I will have to look into that.
|
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# fourier series
Discussion in 'Homework Help' started by hamza324, Oct 18, 2011.
1. ### hamza324 Thread Starter Member
Jul 10, 2011
33
1
I am trying to solve fourier series problems. i have few confusions about that..
1) x(t) =$\sum c_x(k)e ^{jkwt}$
where as $c_x(k) = \int_T x(t) e^{-jwt} dt$
My question is that when which equation do we use to plot the graph.
and how do we take the values for the graph. Is it BODE PLOTS or is it simply plugging in the values.
suppose i havethis expression
$c_x(k)= A/2 \frac {cos(k\prod-1)}{k\prod}^2$
$or\ i\ have\ c_x(k)= A/2 \frac {sinc(k\prod-1)}{k\prod}^2$
then how can i graph this equation for $c_1\ ,\ c_2\ ,\ ......$
2. ### Georacer Moderator
Nov 25, 2009
5,151
1,268
Are these equations for $c_x(k)$ correct? The multiplication symbol has no argument in either of them. It refers to the product of what?
3. ### hamza324 Thread Starter Member
Jul 10, 2011
33
1
The question i have is f(t)= $A\ cos200\pi t$
and this is the expression for fourier series $x(t) = \sum^{+\infty}_{k=-\infty}c_ke^{j2wkt}$
i got the following solution for fourier coefficients
$now\ for\ c_k\ =-\frac{1}{4}\frac{Sin(\pi +k)(0.5)}{(\pi+k)}-\frac{1}{4}\frac{Sin(\pi - k)0.5}{\pi-k}-\frac{1}{4}\frac{Sin(\pi+k)1.5}{\pi+k}-\frac{1}{4}\frac{Sin(\pi-k)1.5}{\pi-k}$
Now my prof asks me to sketch the fourier series coefficients i.e. $c_k\ for\ k=\ 1,2,3,....$
I wanna know how to sketch the result i got
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# Toronto Math Forum
## APM346-2015F => APM346--Lectures => Topic started by: Victor Ivrii on September 21, 2015, 08:18:26 AM
Title: Web bonus problem : Week 2
Post by: Victor Ivrii on September 21, 2015, 08:18:26 AM
Find where solution $u(x,t)$ is properly defined for the following problem:
\begin{align}
&u_t - u u_x=0\qquad -\infty< x< \infty,\ t>0,\label{eq-1}\\
&u(x,0)= \frac{1}{2}x^2.\label{eq-2}
\end{align}
Hint
Using method of characteristics write down solution in the implicit form $F(x,t,u)=0$.
Then the border of the domain where solution is properly defined is an envelope (https://en.wikipedia.org/wiki/Envelope_(mathematics)) of this family of the curves (actually, straight lines) defined by
\begin{align}
&F(x,t,u)=0,\label{eq-3}\\
&F_u (x,t,u)=0.\label{eq-4}
\end{align}
Recall that $t>0$.
Title: Re: Web bonus problem : Week 2
Post by: Yeming Wen on September 22, 2015, 11:37:40 AM
By the method of characteristics, we have
\frac{d x}{d t} = - u,\frac{d u}{d t}=0; \label{eq1}
So,
\begin{equation*}
u=C_1,x=- u t+C_2.
\end{equation*}
By (\ref{eq1}), we know $C_1$ should be a function of $C_2$. Then we can write
u=C_1=C_1(C_2)=C_1(x+ u t)
Notice the initial condition, we have
u(x,0)=C_1(x)=\frac{1}{2}x^2
which implies that
u=C_1(x+ u t)=\frac{1}{2}(x+ u t)^2
Rearrange the equation, we have
t^2 u^2+2(tx-1) u+x^2=0.
\label{K}
So,
u=\frac{-(tx-1) \pm \sqrt{(tx-1)^2-t^2x^2}}{t^2}
Title: Re: Web bonus problem : Week 2
Post by: Victor Ivrii on September 22, 2015, 01:08:59 PM
You did correctly (except renaming $u$ to $\mu$; so I replaced (batch) $\mu$ by $u$) until the last equation.
The primary purpose is not to find $u$ but to determine where it is a solution.
Follow Hint provided: your (\ref{K}) is my (\ref{eq-3}). Write system (\ref{eq-3})-- (\ref{eq-4}) and solve it which gives the boundary of the required domain.
More clear what is the core of the problem will be after Lecture Wed, Sep 23 (see also Subsection 2.1.6 Quasilinear equations (http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter2/S2.1.html#sect-2.1.6))
---
So according to (\ref{K}) $F(x,t,u)=t^2 u^2+2(tx-1) u+x^2$ and system (\ref{eq-3})-- (\ref{eq-4}) is
\begin{align}
&t^2 u^2+2(tx-1) u+x^2,\label{L}\\
&2t^2u +2 (tx-1) =0.\label{M}
\end{align}
(\ref{M}) implies $u= -(tx-1)t^{-2}$ and plugging to (\ref{L}) we get $x=1/(2t)$. This is a border of the domain where $u$ is defined. See attached picture with cyan characteristics. Their continuations after they touched $x=1/(2t)$ should be ignored as at the point of tangency equation breaks (as it does not belong to domain).
(http://www.math.toronto.edu/courses/apm346h1/20159/Forum/W1.svg)
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# On the dynamics of a fluid-filled crack, solved by a boundary integral equation method
Abstract : Long period seismic events observed under many volcanoes are often interpreted in relation to any fluid-filled resonator. The kinematic mechanisms have been also studied seismologically in terms of seismic moment, and some of them indicate the geometry of a 'tensile' crack. In Volcanology, Chouet (1986) first solved the elasto-dynamic equations coupling with the fluid using a finite difference method, similarly to 'shear' crack problems treated in seismology, and this model is always a reference (Chouet and Motoza, 2013). But only a few studies have treated such dynamic problems (e.g. Yamamoto and Kawakatsu, 2009), while dynamic 'shear' cracks have been studied progressively in these two decades in seismology. This study presents a boundary integral equation method (BIEM) in the time domain to solve a 'tensile' crack resonance. The time-domain BIEM is often used for a 'shear' crack thanks to its accuracy, efficiency and flexibility, and usually adopted with an explicit approach (a time step $\Delta t$ is short enough to an element size $\Delta s$ so that any grid influences instantaneously itself, $\Delta t$ <= $\Delta s$/(2*P-wave velocity) ). However such explicit approach introduces severe high frequency oscillations on a 'tensile' crack whose frictional property is intrinsically different from a 'shear' crack. It is found that the implicit approach can retrieve the expected solution for a longer time step of $2 \Delta t$ and no high-frequency oscillations is visible for a $4 \Delta t$. Comparing to the other methods, the time-domain BIEM is easy to be combined with any boundary condition, so that the method would be widely applicable for the observed long period sources if the geometry of a tensile crack is inferred.
Type de document :
Communication dans un congrès
Domaine :
https://hal-brgm.archives-ouvertes.fr/hal-00929920
Contributeur : Hideo Aochi <>
Soumis le : mardi 14 janvier 2014 - 10:21:51
Dernière modification le : vendredi 10 avril 2015 - 01:04:16
### Identifiants
• HAL Id : hal-00929920, version 1
### Citation
Hideo Aochi. On the dynamics of a fluid-filled crack, solved by a boundary integral equation method. EGU General Assembly 2014, Apr 2014, Wiens, Austria. ⟨hal-00929920⟩
### Métriques
Consultations de la notice
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# A plane is flying at a constant altitude at a speed of 600 \ mph. An anti-aircraft missile is...
## Question:
A plane is flying at a constant altitude at a speed of {eq}600 \ mph {/eq}. An anti-aircraft missile is fired and files at a speed of {eq}1200 \ mph {/eq}. The missile and plane approach the same point(call it "{eq}P {/eq}"); the angle between the paths of the plane and missile is {eq}120 ^\circ {/eq}.Determine how fast the distance between the plane and the missile is changing when the plane is {eq}100 {/eq} miles from {eq}P {/eq} and the missile is {eq}150 {/eq} miles from {eq}P {/eq}.
## Time Rates:
The problem is about time rates and to solve this we need to use the law of cosine which is {eq}c^{2}=a^{2}+b^{2}-2ab\cos C {/eq} where {eq}a,b,c {/eq} are the sides of the triangle and {eq}C {/eq} is the angle between the sides {eq}a {/eq} and {eq}b {/eq}.
Below is the figure,
From the figure above, using the law of cosine,
{eq}s^{2}=a^{2}+m^{2}-2am\cos 120 {/eq}
Differentiate the equation with respect to {eq}t {/eq}
{eq}\displaystyle 2s\frac{ds}{dt}=2a\frac{da}{dt}+2m\frac{dm}{dt}+\left ( a\frac{dm}{dt}+m\frac{da}{dt} \right ) {/eq}
Note that from the given {eq}a=100,\:\frac{da}{dt}=600,\:m=150,\:\frac{dm}{dt}=1200 {/eq}
When {eq}a=100,\:b=150,\:C=120 {/eq}
Then,
{eq}\displaystyle s^{2}=(100)^{2}+\left ( 150 \right )^{2}-2(100)(150)\cos 120^{0} {/eq}
{eq}s=50\sqrt{19} {/eq}
Thus,
{eq}\displaystyle 2(50\sqrt{19})\frac{ds}{dt}=2(100)(600)+2(150)(1200)+100(1200)+150(600) {/eq}
{eq}\displaystyle \frac{ds}{dt}\approx 3,165.94\:mph {/eq}
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# Mathematics Add-In for Word and One-Note
Maybe it’s old news to you, but I recently downloaded the Mathematics Add-In for Word and One-Note (download from Mircrosoft for free, right here). It works with Microsoft Office 2007 or later. It’s a super quick and easy installation–doesn’t require a reboot or anything. I was even able to install it at work on my locked-down limited-permissions account without needing administrative privileges.
I’m impressed with its ability to graph, do calculations, and manipulate algebraic expressions using its computer algebra system (CAS). It’s not as powerful as Mathematica or my TI-89, or even other free CAS like WolframAlpha or Geogebra (yes, Geogebra has a CAS now and it’s not beta!). But I like it because (A) my expectations were low and (B) it’s right inside Microsoft Word, and it’s nicely integrated into the new equation editor, which as you know, I love.
Here’s some sample output in word format or pdf (the image above is just the first little bit of this five-page document). All of the output in red is generated by the mathematics add-in package. In this document, I highlight some of it’s features and some of it’s flaws. The graphing capabilities aren’t very customizable. And the mathematics is a bit buggy sometimes.
All in all, despite its flaws, I highly recommend it! It’s really handy to have it right there in Word.
# Math Fonts in Microsoft Office
As you know, Microsoft Office has a new and improved Equation Editor that ROCKS. It is so quick and easy and comes with many benefits. Check out my previous posts on Equation Editor here, here, and here to see why it’s so great.
One issue everyone has with the new Equation Editor, however, is the limited ability to change the font typeface. The default that comes with word, Cambria Math, is nice but doesn’t suit everyone’s needs. If you’re typesetting a document with a font other than Cambria, then it looks a little weird to have your equations in a different font.
After some extensive research, I’ve found three other nice fonts that work with Microsoft Office’s new Equation Editor (these are compatible with Office 2007 or later):
• Asana Math is compatible with Palatino (download here) and if you don’t have Palatino, you can download it here, among other places
• Latin Modern is the LaTeX font of choice. There is a math font (download here) and a whole family of text fonts (download here). Note: these may not look good on screen, but they look just perfect when printed.
To illustrate what these fonts look like, I’ve taken a screenshot below, and I’ve also uploaded the doc file and the pdf file. The doc file won’t render correctly on your machine, however, unless you actually download all the aforementioned fonts.
I hope this helps those who have been searching for alternative fonts for Microsoft Equation Editor. In the comments, please let me know if you find others!
# Summer Odds and Ends
I promise I’ll start blogging again. But as followers of this blog might know, I like to take the summer off–both from teaching and blogging. I never take a break from math, though. Here are some fun things I’ve seen recently. Consider it my own little math carnival :-).
###### I love this comic, especially as I start my stat grad class this semester @ JHU. After this class, I’ll be half-way done with my masters. It’s a long road! [ht: Tim Chase]
Speaking of statistics, my brother also sent me this great list of lottery probabilities. Could be very useful in the classroom.
These math dice. Honestly I don’t know what I’d do with them, but you have to admit they’re awesome. [ht: Tim Chase]
These two articles about Khan academy and the other about edX I found very interesting. File all of them under ‘flipping the classroom.’ I’m still working up the strength to do a LITTLE flipping with my classroom. My dad forwarded these links to me. He has special interest in all things related to MIT (like Khan, and like edX) since it’s his alma mater.
I’ll be teaching BC Calculus for the first time this semester and we’re using a new book, so I read that this summer. Not much to say, except that I did actually enjoy reading it.
I also started a fabulous book, Fearless Symmetry by Avner Ash and Robert Gross. I have a bookmark in it half way through. But I already recommend it highly to anyone who has already had some college math courses. I just took a graduate course in Abstract Algebra recently and it has been a great way to tie the ‘big ideas’ in math together with what I just learned. The content is very deep but the tone is conversational and non-threatening. (My dad, who bought me the book, warns me that it gets painfully deep toward the end, however. That’s to be expected though, since the authors attempt to explain Wiles’ proof of Fermat’s Last Theorem!)
I had this paper on a juggling zeta function (!) sent to me by the author, Dr. Dominic Klyve (Central Washington University). I read it, and I pretended to understand all of it. I love the intersection of math and juggling, and I’m always on the look out for new developments in the field.
And most recently, I’ve been having a very active conversation with my math friends about the following problem posted to NCTM’s facebook page:
Feel free to go over to their facebook page and join the conversation. It’s still happening right now. There’s a lot to say about this problem, so I may devote more time to this problem later (and problems like it). At the very least, you should try doing the problem yourself!
I also highly recommend this post from Bon at Math Four on why math course prerequisites are over-rated. It goes along with something we all know: learning math isn’t as ‘linear’ an experience as we make it sometimes seem in our American classrooms.
And of course, if you haven’t yet checked out the 90th Carnival of Mathematics posted over at Walking Randomly (love the name!), you must do so. As usual, it’s a thorough summary of recent quality posts from the math blogging community.
Okay, that’s all for now. Thanks for letting me take a little random walk!
# 87th Carnival of Mathematics
The 87th Carnival of Mathematics has arrived!! Here’s a simple computation for you:
What is the sum of the squares of the first four prime numbers?
That’s right, it’s
Good job. Now, onto the carnival. This is my first carnival, so hopefully I’ll do all these posts justice. We had lots of great submissions, so I encourage you to read through this with a fine-toothed comb. Enjoy!
# Rants
Here’s a post (rant) from Andrew Taylor regarding the coverage from the BBC and the Guardian on the Supermoon that occurred in March 2011. NASA reports the moon as being 14% larger and 30% brighter, but Andrew disagrees. Go check out the post, and join the conversation.
Have you ever heard someone abuse the phrase “exponentially better”? I know I have. One incorrect usage occurs when someone makes the claim that something is “exponentially better” based on only two data points. Rebecka Peterson has some words for you here, if you’re the kind of person who says this!
# Physics and Science-flavored
Frederick Koh submitted Problem 19: Mechanics of Two Separate Particles Projected Vertically From Different Heights to the carnival. It’s a fun projectile motion question which would be appropriate for a Precalculus classroom (or Calculus). I like the problem, and I think my students would like it too.
John D. Cook highlights a question you’ve probably heard before: Should you walk or run in the rain? An active discussion is going on in the comments section. It’s been discussed in many other places too, including twice on Mythbusters. (I feel like I read an article in an MAA or NCTM magazine on this topic once, as well. Anyone remember that?)
Murray Bourne submitted this awesome post about modeling fish stocks. Murray says his post is an “attempt to make mathematical modeling a bit less scary than in most textbooks.” I think he achieves his goal in this thorough development of a mathematical model for sustainable fisheries (see the graph above for one of his later examples of a stable solution under lots of interesting constraints). If I taught differential equations, I would absolutely use his examples.
Last week I highlighted this new physics blog, but I wanted to point you there again: Go check out Five Minute Physics! A few more videos have been posted, and also a link to this great video about the physics of a dropping Slinky (see above).
# Statistics, Probability, & Combinatorics
Mr. Gregg analyzes European football using the Poisson distribution in his post, The Table Never Lies. I liked how much real world data he brought to the discussion. And I also liked that he admitted when his model worked and when it didn’t–he lets you in on his own mathematical thought process. As you read this post, you too will find yourself thinking out loud with Mr. Gregg.
Card Colm has written this excellent post that will help you wrap your mind around the number of arrangements of cards in a deck. It’s a simple high school-level topic, but he really puts it into perspective:
the number of possible ways to order or permute just the hearts is 13!=6,227,020,800. That’s about what the world population was in 2002. So back then if somebody could have made a list of all possible ways to arrange those 13 cards in a row, there would have been enough people on the planet for everyone to get one such permutation.
I think it’s good to remind ourselves that whenever we shuffle the deck, we can be almost certain that our arrangement has never been created before (since $52!\approx 8\times 10^{67}$ arrangements are possible). Wow!
Alex is looking for “random” numbers by simply asking people. Go contribute your own “random” number here. Can’t wait to see the results!
Quick! Think of an example of a real-world bimodal distribution! Maybe you have a ready example if you teach stat, but here’s a really nice example from Michael Lugo: Book prices. Before you read his post, you should make a guess as to why the book prices he looked at are bimodal (see histogram above).
# Philosophy and History of Math
Mike Thayer just attended the NCTM conference in Philadelphia and brings us a thoughtful reaction in his post, The Learning of Mathematics in the 21st Century. Mike wrote this post because he had been left with “an ambivalent feeling” after the conference. He wants to “engage others in mathematics education in discussions about ways to improve what we do outside of the frameworks that are being imposed on us by those outside of our field.” As a secondary educator, I agree with Mike completely and really enjoyed his post. Mike isn’t satisfied with where education is going. In his post, he writes, “We are leaping ahead into the unknown with new educational models, and we never took the time to get the old ones right.”
Edmund Harriss asks Have we ever lost mathematics? He gives a nice recap of foundational crises throughout the history of mathematics, and wonders, ultimately, if we’ve actually lost any mathematics. There’s also a short discussion in the comments section which I recommend to you.
Peter Woit reflects on 25 Years of Topological Quantum Field Theory. Maybe if you have degree in math and physics you might appreciate this post. It went over my head a bit, I’m afraid!
# Book Reviews
In this post, Matt reviews a 2012 book release, Who’s #1, by Amy N. Langville and Carl D. Meyer. The book discusses the ranking systems used by popular websites like Amazon or Netflix. His review is thorough and balanced–Matt has good things to say about the book, but also delivers a bit of criticism for their treatment of Arrow’s Impossibility Theorem. Thanks for this contribution, Matt! [edit: Thanks MATT!]
Shecky R reviews of David Berlinski’s 2011 book, One, Two Three…Absolutely Elementary mathematics in his Brief Berlinski Book Blurb. I’m not sure his review is an *endorsement*. It sounds like a book that only a small eclectic crowd will enjoy.
# Uncategorized…
Peter Rowlett submitted this post about linear programming and provides a link to an interactive problems solving environment.
Peter Rowlett also weighs in on the recent news about a German high school boy who has (reportedly) solved an open problem. Many news sources have picked up on this, and I’ve only followed the news from a distance. So I was grateful for Peter’s comments–he questions the validity of the news in his recent post “Has schoolboy genius solved problems that baffled mathematicians for centuries?” His comments in another recent post are perhaps even more important though–Peter encourages us to think of ways we can remind our students that lots of open problems still exist, and “Mathematics is an evolving, alive subject to which you could contribute.”
Jess Hawke IS *Heptagrin Girl*
Here’s a fun-loving post about Heptagrins, and all the crazy craft projects you can do with them. Don’t know what a Heptagrin is? Neither did I. But go check out Jess Hawke’s post and she’ll tell you all about them!
Any Lewis Carroll lovers out there? Julia Collins submitted a post entitled “A Night in Wonderland” about a Lewis Carroll-themed night at the National Museum of Scotland. She writes, “Other people might be interested in the ideas we had and also hearing about what a snark is and why it’s still important.” When you check out this post, you’ll not only learn about snarks but also about creating projective planes with your sewing machine. Cool!
Mike Croucher over at Walking Randomly gives a shout out to the free software Octave, which is a MATLAB replacement. Check out his post, here. MATLAB is ridiculously expensive, and so the world needs an alternative like Octave. He provides links to the Kickstarter campaign–and Mike has backed the project himself. I too believe in Octave. I’ve used it a few times for my grad work and I’ve been very grateful for a free alternative to MATLAB.
# The End
Okay, that’s it for the 87th Carnival of Mathematics. Hope you enjoyed all the posts! Sorry it took me a couple days to post it–there was a lot to digest :-).
If you missed the previous carnival (#86), you can find it here. The next carnival (#88) will be hosted by Christian at checkmyworking.com. For a complete listing of all the carnivals, and more information & FAQ about the carnivals, follow this link.
Cheers!
# New Physics Blog
Shout out to Chase Martin, who has just started a great physics blog, Five Minute Physics. My friend Chase and I have a lot in common:
• Our names
• Our love for juggling
• Our love for math & physics
• Our love for teaching
• Our blogs
Chase is awesome, and you’ll love his fun-loving lecture style in these videos. His goal is ambitious: to put the entire lecture content of his high school physics course on youtube. Wow! File this under ‘flipping the classroom.’
Here are a few of his first videos for your enjoyment.
Go check out his website for more!
PS: If you haven’t checked out Minute Physics yet, it’s also a great youtube channel with fun entertaining videos!
# More on Microsoft Equation Editor
As some of you know, I recently posted about Microsoft Equation Editor (here) and the way it’s been totally upgraded. I’ve been using Microsoft’s Equation Editor more and more, and I’ve learned a lot of new things, but I also still have questions (for instance, how do you force it to do display or in-line mode?).
Before, when I had questions, it seemed like Microsoft had no answers. I searched their website and found minimal help. I found help from third-parties, like this wonderful cheat-sheet which I still highly recommend. But today when I went searching for some more answers, I found this page on Microsoft’s website, which I swear wasn’t online two months ago.
The most interesting thing is that they mention their use of Unicode Nearly Plain-text Encoding of Mathematics and they claim that the Microsoft Equation editor adheres to the standards set forth in Unicode Technical Note 28. I’ve now completely read this Unicode guide and it was very helpful.
I think I can finally use the new Microsoft Equation Editor without ever leaving the keyboard.
In particular, here are a few things I learned how to do. Hopefully this will save you the time of having to read through it all yourself:
## Tips & Tricks with the new Microsoft Equation Editor
To start with, here are a handful of things I didn’t know how to do without visiting the toolbar. Now I can do them just by typing.
Boxed formula: \rect(a/b) produces
Matrix: (\matrix(a&b@&c&d)) produces
Equation arrays are something I found hard to do in Microsoft Equation Editor. In their documentation, I learned you can type “Shift+Enter” to keep the next line as part of the same equation array. But here’s the more finely-grained method:
\eqarray(x+1&=2@1+2+3+y&=z@3/x&=6)
resolves to this:
A more complicated example of alignment, and a description of how it is interpreted comes from the Unicode page:
3.19 Equation Arrays
To align one equation relative to another vertically, one can use an equation array, such as
which has the linear format █(10&x+&3&y=2@3&x+&13&y=4), where █ is U+2588. Here the meaning of the ampersands alternate between align and spacer, with an implied spacer at the start of the line. So every odd & is an alignment point and every even & is a place where space may be added to align the equations. This convention is used in AmSTeX.
Instead of █, one can type \eqarray in Microsoft office. Also, to include a numbered equation is simple: E=mc^2#(30).
Another nice thing I learned is how to quickly include text in your equations, without having to visit the toolbar (in retrospect, it’s somewhat obvious):
“rate”=”distance”/”time”
resolves to
$\text{rate}=\frac{\text{distance}}{\text{time}}$
Like I said, one unresolved issue I still have is how to force math to be displayed in ‘in-line’ or ‘display’ mode. This is very easy in $\LaTeX$ with the use of \$ or $$. Section 3.20 of the Unicode notes isn’t very satisfying: Note that although there’s no way to specify display versus inline modes (TeX ‘s versus$$), a useful convention for systems that mark math zones is that a paragraph a paragraph consisting of a math zone is in display mode. If any part of the paragraph isn’t in a math zone including a possible terminating period, then inline rendering is used.
So there you have it–more of what I’ve learned about the Microsoft Equation Editor. Please do share if you have other useful information.
# Microsoft Office Equation Editor
Even though I’d love to say I use $\LaTeX$ for everything, I actually only use it for my grad school assignments. I don’t use it for all my worksheets and assessments. There is a teacher in our math department who does use $\LaTeX$ for everything, but it’s not me.
That being said, Microsoft has made a significant upgrade to its equation editor with the release of Office 2007 (I know, pretty stale news–but my school just upgraded this past year) and $\LaTeX$ lovers will love it if they haven’t tried it yet. The old Microsoft Equation 3.0 which shipped with earlier Office products had a few shortcuts, but it was still pretty hard to type equations without using the toolbar. Color-coding was problematic, and equation objects didn’t respond to font-size changes or other formatting properties. Animations in powerpoint were also difficult.
The new equation editor is much better for the following reasons:
1. The shortcuts are amazing, and most simple $\LaTeX$ commands work. For a complete list of shortcuts go here for a great pdf cheat sheet. You can even add your own custom commands if you go into your options to Proofing > AutoCorrect Options and click on the “Math AutoCorrect” tab. Also, pressing Alt+= will immediately launch the editor. So inserting an equation is fast and you never need to leave the keyboard.
2. Most calculator-style syntax is accepted as well. So typing 3^x [space] / 4^y [space][space] results in $\frac{3^x}{3^y}$, without any extra effort. Tapping the spacebar will automatically convert your calculator syntax into pretty display math. For a more complicated example, consider this:
$\lim_{n\to\infty} \frac{(2n+1)(3n-2)}{4n^2}=\frac{3}{2}$
produced by typing “lim_(n\to\infty)[space]((2n+1)(3n-2))/(4n^2)[space]=3/2[space].”
3. As hinted above, the new equation editor responds to all the normal font formatting options in Microsoft Office. You can color your formulas, you can change the font size, and you can apply any other text effect like shadow/glow/outline/etc. [edit: Though you can change all those things, no, you cannot change the font face. There are a limited number of fonts available for use, and the only one I know of is the default, Cambria Math–if you know of another one, please share!]
4. In powerpoint, animations are quite a bit easier, since you can do all the equations in-line as part of the text, rather than juggling scads of different text and equation objects.
For more on Microsoft’s new Equation Editor, please check out my more recent post here!
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Browse Questions
# True-or-False: The differential equation representing the family of circles $x^2+(y-a)^2=a^2$ will be of order two.
$\begin{array}{1 1} True \\ False \\ can\;not\;be\;determined \end{array}$
Toolbox:
• If the given equation has $n$ arbitrary constant,then the required differential equation will be of $n^{th}$ order
The differential equation representing the family of circles $x^2+(y-a)^2=a^2$ will be of order two.
The above equation has only one arbitary constans $'a'$
we know that if the equation contains $'n'$ arbitary constants, then the required differential equation will be of the $n^{th}$ order
Since the above equation has only one arbitrary constant, the differential equation representing this equation be of order one
Hence the above statment is $False$
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# What is the mass of oxygen in 8 moles of carbon dioxide?
Since carbon has an atomic mass of 12 and oxygen 16, the molecular mass of $C {O}_{2}$ is 44.
This means that the mass of 1 mole of $C {O}_{2}$ is $44 g$, of which $2 \times 16 = 32 g$ is $O$
So $8 m o l C {O}_{2} \leftrightarrow 8 \times 32 = 256 g O$
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# Is F always equal to m*a?
kent davidge
Regarding Newton's second law, are we allowed to consider cases where the mass ##m## changes with time?
I see ##F = m dv/dt## so frequently that I found myself asking this question.
Mentor
Regarding Newton's second law, are we allowed to consider cases where the mass ##m## changes with time?
I see ##F = m dv/dt## so frequently that I found myself asking this question.
Can you give an example of how the mass is changing with time?
kent davidge
Can you give an example of how the mass is changing with time?
Perhaps a rocket expelling fuel?
Staff Emeritus
Homework Helper
Gold Member
2021 Award
The better definition of force is in terms of momentum
$$\vec F = \frac{d\vec p}{dt}.$$
For a variable mass system where ##m## is a function of ##t##, you need to be careful though. Just applying the definition
$$\vec F = \frac{d(m\vec v)}{dt} = \dot m \vec v + m \vec a$$
without considering what goes into ##\vec F## can lead you astray and the equation is not Galilei invariant (##\vec v## depends on the inertial frame). In order to get it right, you must consider both external forces that act upon the system as well as the loss of momentum due to the mass leaving. If the mass leaving does so with velocity ##\vec u##, the force equation becomes
$$\vec F_{\rm ext} + \dot m \vec u = \dot m \vec v + m \vec a,$$
where ##\vec F_{\rm ext}## is the external force acting on the system, which can be rewritten as
$$m\vec a = \vec F_{\rm ext} + \dot m (\vec u - \vec v) = \vec F_{\rm ext} + \dot m \Delta \vec v,$$
where ##\Delta\vec v## is the difference between the velocity of the exhaust and that of your object, i.e., the exhaust velocity relative to the object, which is Galilei invariant as it is the difference of two velocities.
kent davidge and Delta2
Mentor
I see ##F = m dv/dt## so frequently that I found myself asking this question.
That's the apppropriate simplification to ##F=\frac{dp}{dt}## when the mass is a constant. You see it so often because it because so many interesting and important problems involve a constant mass. The reader is expected to recognize it as the constant-mass special case without further reminders.
kent davidge
Dr.D
Newton's Second Law says that "For a particle, F = m*a." By definition, a particle has constant mass and zero dimensions.
Mentor
The only way I was able to derive the equation for a rocket was to first (temporarily) assume that the external force is zero, and then to apply conservation of momentum. At time t, the momentum of the rocket/fuel plus the momentum of the discharged gas is given by:
$$m(t)\mathbf{v}(t)-\int_0^t{\dot{m}(t')\mathbf{u}(t')dt'}=\mathbf{C}$$
where m(t) is the mass of rocket+fuel at time t, ##\mathbf{v}## is the velocity of the rocket+fuel, ##\dot{m}=dm/dt##, and ##\mathbf{u}##is the velocity of the discharge gas. If I differentiate this equation with respect to t, I obtain:
$$m\mathbf{a}+\dot{m}\mathbf{v}-\dot{m}\mathbf{u}=0$$
or $$m\mathbf{a}=\dot{m}(\mathbf{u}-\mathbf{v})$$
Is there another alternate derivation that is easy to make sense of.
Delta2
Dr.D
As ChesterMiller said, The only way I was able to derive the equation for a rocket was to first (temporarily) assume that the external force is zero, and then to apply conservation of momentum.
That is exactly the correct way to get the rocket equation, with the exception that there is nothing temporary about the external force being zero.
Mentor
As ChesterMiller said, The only way I was able to derive the equation for a rocket was to first (temporarily) assume that the external force is zero, and then to apply conservation of momentum.
That is exactly the correct way to get the rocket equation, with the exception that there is nothing temporary about the external force being zero.
I was thinking that the external force might be gravity, for example.
DrStupid
Newton's Second Law says that "For a particle, F = m*a."
No, it says F=dp/dt. There is no limitation to particles in the second law.
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# How to set headsepline length?
I'm writing a document using scrbook document class and while I was able to set up all header and footer apparence, I can't find a way to set headsepline length (seems to follow \textwidth settings, while I need it to be slightly longer). Is there a way to make it without using external packages, maybe changing something in class file?
-
As lockstep remarked, my short answer was way to short, so here's a more elaborate one:
Setting the headsepline can be done with the scrpage2 package, which provides the macro
\setheadsepline[<length>]{width}
To be able to use a notation like \textwidth+20pt, e.g. to do calculations, you'll need e.g. the calc package. The scrpage2 package also provides a new pagestyle (scrheadings), I made a little example below: The pageheads automatically alternate between the current chapter and section headings. Also you can see that the headsepline is longer than the headtopline. It protudes on both left and right, I don't know if you want that or just one side.
## The code
\documentclass{scrbook}
\usepackage[inner=10mm,outer=30mm,top=30mm,bottom=10mm,a4paper,twoside]{geometry}
\usepackage{calc}
\usepackage{lipsum}
\usepackage{scrpage2}
\automark[chapter]{section}
\begin{document}
\chapter{One}
\lipsum[1]
\section{one}
\lipsum[2]
\section{two}
\lipsum[3]
\section{three}
\lipsum[4]
\section{four}
\lipsum[5]
\section{five}
\lipsum[6]
\section{six}
\lipsum[7]
\section{seven}
\lipsum[8-12]
\chapter{Two}
\lipsum[1]
\section{one}
\lipsum[2]
\section{two}
\lipsum[3]
\section{three}
\lipsum[4]
\section{four}
\lipsum[5]
\section{five}
\lipsum[6]
\section{six}
\lipsum[7]
\section{seven}
\lipsum[8-12]
\end{document}
## Part of the result
-
thank you very much, I loaded scrpage2 but tought \setheadsepline would work only for line height. – Gianluigi Sep 2 '12 at 18:13
@Gianluigi: You're welcome. Thats what I love about this site: there's usually someone around who knows the answer or can think of away to do it. – Tom Bombadil Sep 2 '12 at 19:41
@Gianluigi don't forget to accept (different from up-vote) Tom Bombadil's answer (and any answer that solves the problem in your other questions) by clicking on the check-mark to the left of the answers; in case of doubt, please see How do you accept an answer?. – Gonzalo Medina Sep 2 '12 at 23:22
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# Spanish Level 2, Activity 08: El restaurante / The Restaurant (Face-to-Face)
Please Note: The activities on the Pathways Project OER Repository were created by upper-division students at Boise State University and serve as a foundation that our community of practice can build upon and refine. While they are polished, we welcome and encourage collaboration from language instructors to help modify grammar, syntax, and content where needed. Kindly contact amberhoye@boisestate.edu with any suggestions and we will update the content in a timely manner. — The Pathways Project
## About the Boise State World Languages Resource Center (WLRC) Language Activity Repository
The activities provided by the Boise State World Languages Resource Center (WLRC) serve as foundational activities which can be adapted by any language and scaled up or down on the proficiency scale.
This activity was created by upper-division language students working in the World Languages Resource Center at Boise State University. Our activities seek to help students solidify their interpersonal speaking and interpretive listening skills through task-based situations or communicative activities. We recommend using these activities to help reinforce the content students are learning, allowing the students time to feel comfortable using the unit’s vocabulary and grammar structures through application. Further, these activities should be facilitated in approximately 90% (or more) in the target language, per the recommendation of the American Council on the Teaching of Foreign Languages.
### Using the WLRC Repository’s Activities:
When you are ready to begin remixing the activity, in order to adapt it for your target language and audience, simply click the “Remix This Resource” button at the top of your screen. The text provided in purple is a suggestion of what you might say to your students and should be changed to the target language.
Most activities contain a connected chapter, two to three “NCSSFL-ACTFL Can-Do” statements, a warm-up, main activity, and a wrap-up. In addition to the instructions, some activities may include a “cheat sheet” containing the target vocabulary and grammar structures emphasized in the activity. Though most of the lab materials are provided, a computer, projector, printer, and laminator may also be needed to fully utilize materials.
Many of the activities include printable cards and other instructional materials. If you would like to adapt these materials for your language, please email WLRCLAR@gmail.comand we will provide you with an editable copy. For YouTube videos and other websites, hyperlinks are provided.
Enjoy!
- Boise State World Languages Resource Center
## El Restaurante / The Restaurant (Novice High)
### Proficiency Level: Novice High
In this activity the lab assistant will work as the server and the students will be guests at a restaurant. The students will practice ordering food, discussing what is wrong with the food, and paying for their meal.
### Keywords:
Food, Restaurant ordering, handling money, paying, real world context
### Relevant National (ACTFL) Spanish Standards:
• “Standard 1.1 Students engage in conversations, provide and obtain information, express feelings and emotions, and exchange opinions.”
• “Standard 1.2 Students understand and interpret spoken and written Spanish on a variety of topics.”
• “Standard 1.3 Students present information, concepts, and ideas in Spanish to an audience of listeners or readers on a variety of topics.”
• “Standard 4.2 Students demonstrate understanding of the concept of culture through comparisons between Hispanic cultures and their own.”
### Relevant Idaho State World Language Standards:
• COMM 1.1 - Interact and negotiate meaning (spoken, signed, written conversation) to share information, reactions, feelings, and opinions
• COMM 2.1 - Understand, interpret, and analyze what is heard, read, or viewed on a variety of topics.
• COMM 3.1 - Present information, concepts, and ideas to inform, explain, persuade, and narrate on a variety of topics using appropriate media in the target language.
### Relevant NCSSFL-ACTFL Can-Do Statements:
• I can understand the differences in services and dishes between restaurants in the United States and Spain
• I can order food at a restaurant, including items such as drinks, entrees, and desserts (novice high)
• I can ask for the bill/check after I have finished eating (novice mid)
### Materials Needed:
• Menú del día (there are two papers that need to be presented to students, the menu and a paper with images of food. Make sure to have both during lab to show the students)
• Collage with a picture of Spanish food items (these are to help students see what the food looks like)
• Paper plates
• Whiteboard to take order (for lab instructor)
• Food cards to give as dishes
Would you like to make changes to the materials? Access the template(s) below:
### Relevant NCSSFL-ACTFL Can-Do Statements:
• I can share restaurants and foods that might be familiar to them with peers from Spain restaurants
• I can order food at a restaurant, including items such as drinks, entrees, and desserts
• I can ask for the bill/check after I have finished eating
## Warm-up
1. For the warm-up there will be approximately 8 slides that will review key vocabulary and phrases for going to a restaurant. You will ask the students a question or set up a specific scenario on the slides, but DO NOT give them the answers. For each question or scenario, they will have to come up with useful phrases as responses to the given situation. Follow the speaker notes!
1. ¿Cuántas personas hay?
2. ¿Qué quieres para la bebida?
3.
¿Qué quieres para el aperitivo?
4.
¿Hay alguien que tiene alergias? ¿Hay alguien que es vegetariano?
5.
¿Qué quieres para el plato principal?
6.
¿Qué quieres para de postre?
7. ¿Podemos tener la cuenta?
8. ¿Me puede dar recomendación?
En el mundo hispanohablante, los meseros no están revisando los clientes. Ellos no dependen de las propinas (tips) como en los Estados Unidos, tienen un sueldo (salary) más alto. Entonces los clientes deben pedir la cuenta porque le mesero/a no se los va a dar hasta que el cliente pregunte.
In Spanish speaking countries, servers are not constantly checking on their tables like in the United States. Explain this in Spanish to your students so they understand why it is important for them to know how to ask for the check.
If students need some assistance getting the discussion going, use the questions below:
“¿Cómo van a responder? ¿Cuál es la respuesta correcta?”
## Main Activity
1. Today you all will have the opportunity to eat lunch at the WLRC restaurant. In one minute I leave and then return. When I come back I will not be your lab assistant but I will be your waiter/server. Before I come back, and to practice your words, please make plans to go to the WLRC restaurant as a group of friends. You all are going to do the whole process that we finished describing including ordering food and asking for the bill. The menu that I am going to give you is based on an authentic menu from Spain. In Spain many times lunch is the largest plate of the day. Many restaurants offer a special menu where you can choose two or three plates, a drink, and many times a dessert.
"¡Hoy tienen la oportunidad de comer en el restaurante WLRC! Yo voy a ser su camarero/a (o mesero/a). Van a practicar el proceso de ordenar comida en el restaurante, que incluye pedir la cuenta. Les voy a dar un menu auténtico con comida de España. Cada vez que van a pedir un plato diferente, voy a "fingir" (to pretend) que voy a otro lado del restaurante y ustedes van a hablar de que quieren comer."
2. The process:
a. You will give the opportunity to the students to talk with each other and make plans to go to the WLRC restaurant. Motivate them to use the words and phrases (kitchen, food, and restaurant) that they already know. After their chat they walk to the table.
b. You are going to give the welcome to the students and then ask the following:
1. “Qué quiere/desea para tomar”
2. “Tenemos cafe, jugo, cerveza, vino....”
c. You are going to ask what appetizer and entree they want.
1. “Qué le gustaría como un aperitivo / para compartir”
d. What would you like for dinner?
1. “Qué le gustaría (quisiera usted) para cenar?”
e. What do you want for dessert?
1. “Qué quiere usted de postre”
f. You are going to bring the food to the students. However you are going to choose incorrect food for one of the plates.
g. After you bring the food wait for a few seconds and then ask the following:
1. “Cómo estaba la comida?"
2. "Le gustó el/la _____?"
h. If you gave an incorrect plate to a student you should make it so they try to ask you a question or say a complaint.
i. Now walk away and look busy. The students have to ask for the bill but you shouldn't help them. If too much time passes you can walk to the table and teach the students how to ask for the check and how to sign for the check. (Air draw your signature)
j. Ask every student how they want to pay. Also you can ask if they have a coupon or a gift card.
1. ¿Cómo quieren pagar?
2. ¿Tienen un cupón o una tarjeta de regalo?
## Wrap-up
### Wrap-up questions (Pick the a few you’d like to ask):
Preguntas para terminar si el tiempo permita:
1. ¿Qué frases o palabras nuevas aprendiste con esta actividad?
2. ¿Pueden compartir algunas diferencias entre la experiencia en el restaurante de Red Robin y el restaurante de España?
### End of lab:
• Read Can-Do statements once more and have students evaluate their confidence.
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1) La descarga del recurso depende de la página de origen
2) Para poder descargar el recurso, es necesario ser usuario registrado en Universia
Opción 1:
Descripción
Pertenece a
Autor(es)
### Xu, ChangJiang - McLeod, A. Ian -
Id.: 55207166
Idioma: inglés -
Versión: 1.0
Tipo: application/pdf -
Palabras claveVariable selection -
Tipo de recurso: Text -
Audiencia: Estudiante - Profesor - Autor -
Estructura: Atomic
Coste: no
: Copyright 2012 Institute of Mathematical Statistics
Formatos: application/pdf -
Requerimientos técnicos: Browser: Any -
Relación: [References] 1935-7524
Fecha de contribución: 26-ene-2013
Contacto:
Localización:
* Electron. J. Statist. 6 (2012), 656-663
* doi:10.1214/12-EJS685
#### Otros recursos del mismo autor(es)
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2. Limited Capacity for Faster Digestion in Larval Coral Reef Fish at an Elevated Temperature The prevalence of extreme, short-term temperature spikes in coastal regions during summer months is ...
3. Climate change and the performance of larval coral reef fishes: the interaction between temperature and food availability We tested the impacts of temperature and variable food availability on the development and metabolic...
4. The UK10K project identifies rare variants in health and disease The contribution of rare and low-frequency variants to human traits is largely unexplored. Here we d...
5. Effects of elevated CO2 on early life history development of the yellowtail kingfish, Seriola lalandi, a large pelagic fish An increasing number of studies have examined the effects of elevated carbon dioxide (CO2) and ocean...
#### Otros recursos de la mismacolección
1. A new family of tempered distributions Tempered distributions have received considerable attention, both from a theoretical point of view a...
2. Consistent community detection in multi-relational data through restricted multi-layer stochastic blockmodel In recent years there has been an increased interest in statistical analysis of data with multiple t...
3. Model selection of hierarchically structured covariates using elastic net Hierarchically associated covariates are common in many fields, and it is often of interest to incor...
4. A general decision theory for Huber’s $\epsilon$-contamination model Today’s data pose unprecedented challenges to statisticians. It may be incomplete, corrupted or expo...
5. Stability and asymptotics for autoregressive processes The paper studies infinite order autoregressive models for both temporal and spatial processes. We p...
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BACKGROUND: β-thalassemia is a potentially life-threatening, hereditary blood disorder characterized by ineffective erythropoiesis and peripheral hemolysis, leading to profound anemia and iron avidity. It is estimated to affect approximately 15,000 people in the U.S. and European Union. Patients with the most severe form of the disease, transfusion-dependent β-thalassemia (TDT), must adhere to a lifelong regimen of frequent red blood cell (RBC) transfusions for survival. While regular transfusions effectively manage the symptoms of TDT, they also substantially accelerate iron overload. As a result, patients with TDT must also adhere to lifelong chelation therapy regimens to reduce iron-associated morbidity and mortality. A variety of chelation therapies are available in the U.S., and treatment regimens should be optimized for the specific needs of individual patients. While the burden of chronic treatment on patients is clear, there are limited data available on economic costs of RBC transfusions and chelation therapy in the treatment of TDT in the U.S.
OBJECTIVES: Objectives of this study were to: 1) develop an algorithm to identify patients with TDT in a U.S. administrative claims database; 2) evaluate demographic characteristics of these patients; 3) examine treatment patterns with RBC transfusions and chelation therapy; and 4) estimate the annual reimbursed costs associated with the use of RBC transfusions and chelation therapy in these patients.
METHODS: This retrospective analysis used the U.S. MarketScan Commercial and Medicaid Multi-State Database, which together represent the administrative inpatient and outpatient medical, and outpatient pharmacy experience of approximately 30 million (in 2016) commercially-insured employees and their dependents, and Medicaid beneficiaries. These individuals are covered under a variety of fee-for-service and managed care health plans.
Patients with TDT included in this study were required to have ≥ 8 RBC transfusions in a given 12-month period (first transfusion date = index date) between October 1, 2011 and September 30, 2016, one or more non-diagnostic inpatient or outpatient medical claims with a β-thalassemia or hemoglobin E β-thalassemia diagnosis between 3 months before and 15 months after the index date, and no evidence of allogeneic stem cell transplantation between the first and eighth transfusions. Demographic characteristics of the TDT patients were summarized. The annual number of RBC transfusions and prescriptions for chelation therapy were summarized on a per-patient basis, as well as the annual costs (i.e. reimbursed amounts) for these therapies. Additional costs associated with management of patients with TDT (e.g. ongoing monitoring of iron overload with cardiac T2* MRI and liver iron tests) are being estimated in the study but are not included in this abstract.
RESULTS: A total of 8,480 patients with a diagnosis of β-thalassemia were identified, of which 253 patients (3.0%) met the selection criteria for inclusion in this analysis. Mean patient age was 22.8 years (median 21 years) and 53.4% were female; 33.2% lived in the Northeast, 25.7% in the Midwest, 24.9% in the South, and 16.2% in the West. Patients with TDT in this analysis had a median of 14 blood transfusions annually (range 8-56), with an average cost of $1,478 per transfusion. The mean annual cost of transfusions per patient was$22,478. The majority of patients (84.1%) had ≥ 1 prescription for chelation therapy, with 73.8% receiving an oral chelation therapy regimen only, 7.8% IV only, and 18.4% combination therapy. On average, chelation therapy cost $52,718 per patient, per year. CONCLUSION: In the U.S., the average annual economic cost of TDT conventional chronic symptom management with RBC transfusions and iron chelation therapy is approximately$75,000. Treatment with both transfusion and chelation are life-long requirements for most patients with TDT; thus the cumulative economic burden for basic management of TDT is substantial. Furthermore, these costs represent only a portion of the total economic burden of TDT, which also includes the costs of ongoing monitoring of cardiac and other organ iron overload, costs associated with morbidities related to TDT and iron overload, and the significant time burden for patients (e.g. to meet multi-disciplinary clinical teams including endocrinologists, cardiologists, etc.; and transfusion appointments).
Disclosures
Paramore: bluebird bio: Employment, Equity Ownership. Vlahiotis: Truven Health Analytics, an IBM Watson Health company: Employment. Moynihan: Truven Health Analytics, an IBM Watson Health company: Employment. Cappell: Truven Health Analytics, an IBM Watson Health company: Employment. Ramirez-Santiago: bluebird bio: Employment, Equity Ownership.
## Author notes
*
Asterisk with author names denotes non-ASH members.
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# Integral equation
Tags:
1. Jan 5, 2017
### Rectifier
The problem
I want to find $x$ which solves $1-x+ \int^x_1 \frac{\sin t}{t} \ dt = 0$
The attempt
$\int^x_1 \frac{\sin t}{t} \ dt = x -1$ I see that the answer is $x=1$ but I want to be able to calculate it mechanically in case if I get similar problem with other elements. Any suggestions on how I can do that?
2. Jan 5, 2017
### BvU
Hi,
'mechanically' sounds good. But, ${\sin x\over x}$ is (and I https://owlcation.com/stem/How-to-Integrate-sinxx-and-cosxx [Broken])
one of the simplest examples of non-integrable functions in the sense that their antiderivatives cannot be expressed in terms of elementary functions, in other words, they don't have closed-form antiderivatives.
However, apart from $x=1$ there shouldn't be too many other solutions ... $x-1$ grows faster than the integral.
You could also investigate domain [0,x] : with ${\sin x\over x} < 1$ the integral is always different from x-1.
Last edited by a moderator: May 8, 2017
3. Jan 5, 2017
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# [Arduino] Simple hobby motor and transistor circuit not working
#### HunterDX77M
Joined Sep 28, 2011
104
So I'm trying to construct this simple motor circuit, where the motor spins and doesn't spin in an alternating pattern. The circuit diagram is attached and the sketch is below.
Code:
int motorPin = 9;
void setup() {
// Pin 9 is an output pin
pinMode(motorPin, OUTPUT);
}
void loop() {
digitalWrite(motorPin, HIGH);
delay(1000);
digitalWrite(motorPin, LOW);
delay(1000);
}
So obviously the problem is that the circuit isn't working as intended. I took a multimeter and began probing around. I found that:
• The VDD line is about 5V, but alternates between ~4.7V and 5.0V every second (a little weird).
• The voltage at Pin 9 alternates between 0 and 5V (that's correct).
• The voltage across the motor alternates between 0V and 0.07 volts every second (totally wrong).
The diode I am using is 1N4003. When I tested it out in my multimeter, the forward voltage was 600 to 700 mV, which sounds correct. I'm thinking my transistor might be wrong or not functioning. The diagram calls for what looks like an NPN, so I am using a 2N4124, which is an NPN according to the datasheet. The hFE I measured falls within the minimum/maximum listed. I'm using a 2K resistor instead of a 2.2K, as shown in the diagram.
The motor itself is working. I attached both ends to a plain AA battery briefly and it spun.
At this point, I'm not sure what else could be wrong in the circuit. Any tips on where else I should focus my debugging efforts?
#### Attachments
• 43 KB Views: 12
#### shteii01
Joined Feb 19, 2010
4,647
I am thinking you are powering the motor badly. How much current can Uno source from its 5 Volt socket?
#### HunterDX77M
Joined Sep 28, 2011
104
#### shteii01
Joined Feb 19, 2010
4,647
According to this page, the absolute max per pin is 40 mA. How much do you think a hobby motor would need?
http://playground.arduino.cc/Main/ArduinoPinCurrentLimitations
I don't know. Do you know what hobby motor is? Who made it? What voltage it uses? What current it uses?
I can tell you that dc motor fans in pc need 100 to 140 mA. If I send 40 mA to one of them, they will simply laugh and point finger at me.
Have you tried this tutorial:
#### John P
Joined Oct 14, 2008
1,785
Try shorting C and E of the transistor, so the motor is running directly off 5V. If it runs, the problem is that the transistor isn't working or isn't being turned on. If it doesn't run, the power supply isn't adequate to make it work.
#### dannyf
Joined Sep 13, 2015
2,197
I'm not sure what else could be wrong in the circuit.
Too many - as you have given very little information about your set-up.
An obvious question would be what kind of motor and how much current it draws and what kind of voltage source it needs to drive it. I would re-route the motor to the power source directly but knowing what you are dealing with first would be key in finding out the source of your troubles.
#### ScottWang
Joined Aug 23, 2012
6,860
As a hobby motor and assuming it's current as this needs 110 mA,
When you using the bjt as switch then the hFE will be count it as 10.
Ib = 110 mA/hFE
= 110 mA/10
= 11 mA.
You could prepare a little more current for Ib as 15 mA.
Rb = (V_d9 - Vbe)/Ib
= (5V - 0.7V)/15 mA
= 286.7Ω
You could using 270Ω for Ib = 16 mA or 300Ω for Ib = 14.3 mA.
#### shteii01
Joined Feb 19, 2010
4,647
As a hobby motor and assuming it's current as this needs 110 mA,
When you using the bjt as switch then the hFE will be count it as 10.
Ib = 110 mA/hFE
= 110 mA/10
= 11 mA.
You could prepare a little more current for Ib as 15 mA.
Rb = (V_d9 - Vbe)/Ib
= (5V - 0.7V)/15 mA
= 286.7Ω
You could using 270Ω for Ib = 16 mA or 300Ω for Ib = 14.3 mA.
i don't think they can source 110 mA from Arduino board.
#### ScottWang
Joined Aug 23, 2012
6,860
i don't think they can source 110 mA from Arduino board.
#### shteii01
Joined Feb 19, 2010
4,647
You made an assumption that motor needs 110 mA.
Based on this assumption you calculated Ib of 11 mA, you then modified this Ib to 15 mA, which is fine because Uno can normally source 20 mA from its pins.
Using Ib of 15 mA you calculated base resistor Rb to be 286 Ohm, then you gave OP the choice of using standard 270 Ohm or standard 300 Ohm base resistors, which is also fine.
But here is the thing. All this calculations are based on that initial assumption that Ic will be 110 mA. The way OP wired his motor, the Ic will come from Uno and Uno CAN NOT produce 110 mA. So, yes, the Uno will send 15 mA into the base, it will turn motor on, the motor will try to draw 110 mA from the Uno, Uno will peak at about 40 mA and that is it, it can not supply any more current to the motor.
#### ScottWang
Joined Aug 23, 2012
6,860
You made an assumption that motor needs 110 mA.
Based on this assumption you calculated Ib of 11 mA, you then modified this Ib to 15 mA, which is fine because Uno can normally source 20 mA from its pins.
Using Ib of 15 mA you calculated base resistor Rb to be 286 Ohm, then you gave OP the choice of using standard 270 Ohm or standard 300 Ohm base resistors, which is also fine.
But here is the thing. All this calculations are based on that initial assumption that Ic will be 110 mA. The way OP wired his motor, the Ic will come from Uno and Uno CAN NOT produce 110 mA. So, yes, the Uno will send 15 mA into the base, it will turn motor on, the motor will try to draw 110 mA from the Uno, Uno will peak at about 40 mA and that is it, it can not supply any more current to the motor.
Max current out of 5v pin on UNO?
#### shteii01
Joined Feb 19, 2010
4,647
If I am reading Arduino schematic for Uno right, they are using NCP1117 for regulator that outputs 5 volts. It seems it will output 1 A when powered by 10 or more volts. My mistake at looking at Arduino documentation, which generally is not very good.
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# Figure 1
Invariant mass (left panels) and pseudoproper decay length (right panels) distributions for \pJPsi candidates at midrapidity with superimposed projections of the maximum likelihood fit. Pseudoproper decay length distributions are shown for \pJPsi candidates reconstructed under the \pJPsi mass peak, i.e. for $2.92$ 1 (2) \GeVc in pp collisions at \s = 13 (5.02) TeV. The $\chi^{2}$ values, which are computed considering the binned distributions of data points and the corresponding projections of the total fit function, are also reported.
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# Why is a third body needed in the recombination of two hydrogen atoms?
In the article Discuss. Faraday Soc. 1962, 33, 205, the authors say that in order to form molecular hydrogen from collision to atom we need a third body to remove the excess energy. That is we have the reaction $$\ce{H + H + M -> H2 + M}$$.
I suppose the reason is because if we have the excess this energy will break the bond. If it this the case why the energy can not be transferred to the electron and so would have $$\ce{H + H -> H2^*}$$, where $$\ce{H2^*}$$ denotes an excited state of $$\ce{H2}$$?
• You can use the \ce{...} macro to typeset chemical equations; it's easier and also provides the correct output (chemical symbols should be upright, not in italics). Check out my edit for some examples. – orthocresol Dec 1 '20 at 19:18
• An excited state of H2 would still have the excess energy. – Ivan Neretin Dec 1 '20 at 19:22
• Temperature is irrelevant. Or I'd better put it this way: the rest of the gas might be as cold as you like, but this one molecule that has just formed from two atoms is very, very hot. – Ivan Neretin Dec 1 '20 at 19:32
• All right, speaking of the heat of one molecule is an abuse of language. Would it be better if I said that it has very high energy? – Ivan Neretin Dec 1 '20 at 19:37
• @amiltonmoreira - what is the binding energy of H2? – Jon Custer Dec 1 '20 at 20:02
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# Examples of problems where considering “discrete analogues” has provided insight or led to a solution of the original problem
The Kakeya conjecture posits that any Kakeya set in $$\mathbb{R}^n$$ has dimension $$n$$.
A discrete (finitized?) version of this problem is the Finite Field Kakeya conjecture, which was proved by Dvir in 2008.
My understanding is that the Finite Field Kakeya Conjecture was proposed at the end of the twentieth century with the hope that it would lead to methods that could be applied to the original Kakeya conjecture. However, it seems that in this case the approach used for resolving the discrete analogue is not easily applied to the original problem, so that the Kakeya conjecture still remains open.
My question is asking for examples of problems where discretizing "succeeded."
Question: Are there examples of math problems where looking at a finite or discrete variant of the initial statement did lead to a solution (or if not a complete solution, at least significant progress) of the original problem? If so, what are they?
Edit:
As commenters pointed out, this question is similar to the one here which requests examples where a discrete version of a theory was developed before the continuous version of the same theory.
My question is different from that previous question, in that I am not interested in cases where discrete problems predated continuous problems. Instead, I'd like to learn about instances where a continuous problem was already proposed, and studying a discrete version of that problem helped inform a solution to the continuous variant.
• Does the exhaustion method qualify? – Sylvain JULIEN Apr 20 at 16:50
• Possible duplicate of When has discrete understanding preceded continuous? – Igor Pak Apr 21 at 1:37
• It's also worth noting that Thomas Wolff, who proposed the Finite Field Kakeya conjecture, had already by that time (1999) initiated a rather successful program of applying ideas from incidence geometry (a discrete field) to problems of Fourier analysis. (As a starting point see e.g. these notes of Larry Guth: math.mit.edu/~lguth/Namboodiri/namboodiri3.pdf). – Sam Hopkins Apr 21 at 2:52
• @SamHopkins I don't think that's quite the same. KS was shown to be equivalent to a discrete problem (with hindsight this is natural, KS has ultrafilters in the background). The OP is asking about discrete analogues of a continuous problem, whereas the paving problem and Nik Weaver's work are really about showing the continuous problem is equivalent to a combinatorial question – Yemon Choi Apr 21 at 3:20
• While Dvir's arguments have not yet led to a solution of the continuous Kakeya conjecture, the polynomial interpolation method of Dvir did inspire what is now known as the polynomial partitioning method, which did establish several variants of the continuous Kakeya problem, starting with this paper of Guth: arxiv.org/abs/0811.2251 – Terry Tao Apr 21 at 16:43
In my paper
Tao, Terence, Norm convergence of multiple ergodic averages for commuting transformations, Ergodic Theory Dyn. Syst. 28, No. 2, 657-688 (2008). ZBL1181.37004.
I was able to settle a question in ergodic theory (namely, the norm convergence of averages $$\frac{1}{N} \sum_{n=1}^N \int_X T_1^n f_1 \dots T_k^n f_k\ d\mu$$ for $$k$$ commuting measure-preserving transformations $$T_1,\dots,T_k$$ on a probability space $$(X,\mu)$$ and bounded functions $$f_1,\dots,f_k$$), by abandoning all the usual ergodic theory machinery (e.g., characteristic factors), and translating the problem to a purely finitary one without explicit use of limits. This could then be attacked by methods related to graph and hypergraph regularity. The argument was then significantly generalised in
Walsh, Miguel N., Norm convergence of nilpotent ergodic averages, Ann. Math. (2) 175, No. 3, 1667-1688 (2012). ZBL1248.37008.
It should however be pointed out that an alternate, ergodic-theoretic proof of these results was also subsequently given in
Austin, Tim, On the norm convergence of non-conventional ergodic averages, Ergodic Theory Dyn. Syst. 30, No. 2, 321-338 (2010). ZBL1206.37003.
Austin, Tim, A proof of Walsh’s convergence theorem using couplings, Int. Math. Res. Not. 2015, No. 15, 6661-6674 (2015). ZBL1372.37012.
I should also mention that there are several papers of Bourgain in which he establishes various ergodic theorems by converting them to quantitative questions in harmonic analysis which are strictly speaking not discrete or finitary, but are amenable to many of the same techniques (in particular, a focus on "hard analysis" estimates) as such discrete problems. A typical such paper is
Bourgain, J., On the pointwise ergodic theorem on $$L^p$$ for arithmetic sets, Isr. J. Math. 61, No. 1, 73-84 (1988). ZBL0642.28011.
As a non-expert, I will tell a story (since I am not qualified to do more):
Once upon a time (1859) there was conjecture about the zeros of a complex function known as the Riemann Hypothesis (RH).
Years later Weil formulated a "discrete version" (in terms of finite fields $$\mathbb{F}_q$$) of this conjecture; the third of his Weil Conjectures (1949). In 1974, Deligne proved this discrete version to much acclaim. Here is a well regarded exposition by Milne.
Years later still, Connes is motivated to explore a potential proof of the original RH along the lines of Deligne's proof by taking a "limit as $$q\to 1$$".
These attempts are probably uncontroversially considered mathematical advances. Some, presumably Connes himself, consider this work an advance in the problem at hand (and so making this hopefully an appropriate response to this MO question). However, there are other opinions, and so this MO post might prove controversial too.
• I thought of things like this, but it is not clear to me that $\mathbb F_q[t]$ is more discrete in a meaningful sense than $\mathbb Z$. In some ways it is more discrete, in others more continuous. – Will Sawin Apr 21 at 16:19
• That said, another example in this vein is Ngo’s Proof of the function field analogue of the fundamental lemma, which led, thanks to earlier work of Waldspurger and others, to a proof of the original fundamental lemma. – Will Sawin Apr 21 at 16:24
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# Question: Maple Commands for exp function?
February 25 2013 by
false
Maple
1
I'm trying to graph a contour map of this function: e^(y/x)
>with(plots)
>contourplot(e^(y/x),x=-3..3,y=-0.5..0.5)
I thought the contour map for this function would looks more like several lines intersects at the origin. But when I graph it on Maple, it doesn't looks like the graph I'm looking for.
Did I put in a wrong range?
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# JEE Main & Advanced Physics Ray Optics Newton's Formula
Newton's Formula
Category : JEE Main & Advanced
If the distance of object $({{x}_{1}})$ and image $({{x}_{2}})$ are not measured from optical centre, but from first and second principal foci then Newton's formula states ${{f}^{2}}={{x}_{1}}{{x}_{2}}$
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### Archive
Posts Tagged ‘coding’
## Metapost – or how to code an image
What software do you use when you need to draw a nice image? I often need to draw things related to research and I never managed to efficiently use a software which uses the mouse or touchpad to draw and modify things. Moreover, if you need to add some mathematical text to the figure things get even more complicated. For me it is more natural to use code to generate graphics, if this is possible. Using something like Matlab to draw is possible. The advantage is that once you have a working code which produces what you want, you can easily modify it. If you have an image where you need to repeat things, using loops can facilitate the job (programmers will understand…).
This is were Metapost comes into play. I found this a long time ago while searching for a tool to build nice graphics for my master thesis. Being used to LaTeX for typesetting math, it was a natural way to draw using code. I do not claim it is the best or the simplest way, but for me it works. Most importantly, it allows me to build high quality, vectorized graphics, which are memory efficient. The advantage of vector graphics is that you can zoom as much as you want and you’ll never see the pixels.
There are a bunch of places where you can learn Metapost. I’ll put here some references I use a lot. The easiest way to start drawing in Metapost is to take a look at some existing codes and modify them. You can find lots of examples in this tutorial. If you have a linux system, you can compile metapost .mp files to obtain high quality pdfs using the command mptopdf. If not, then you can use the online Metapost editor found here. In any case, here are some of my codes for some drawings in Metapost.
Here is the code for my website logo: If you want to see the lossless vectorized pdf click to see the following file: logo-0 You can see that it has a border which changes from one color to another. This was done using a loop and a parameter to vary the color as we go along the boundary.
prologues:=3;
verbatimtex
%&latex
\documentclass{minimal}
\begin{document}
etex
beginfig(0);
% copy from here to use the online previewer
u:=25; % 25 = 25bp = 25 PostScript points = 25/72 in
wi:=10; % width in units u
he:=7; % height in units u
hoehe:=he*u; % height
breite:=wi*u;
%for i=0 upto he:
%draw (0, i*u)--(breite, i*u) withcolor .7white;
%endfor
%for j=0 upto wi:
%draw (j*u, 0)--(j*u, hoehe) withcolor .7white;
%endfor;
path p,q;
p:=(6u,0.5u)--(6u,0)--(0,0)--(0,6u)--(6u,6u)--(6u,5.5u);
pickup pensquare scaled 20;
draw p;
h=length(p);
numeric c,d,detail;
color a,b,co;
a:=(1,0.45,0);
b:=.2white;
co:=.2white;
detail:=500;
for i=1 upto (detail/2):
q := subpath (h*(i-1)/detail,h*i/detail) of p;
draw q withcolor (2*i/detail)[a,b];
endfor;
for i=detail downto (1+detail/2):
q := subpath (h*(i-1)/detail,h*i/detail) of p;
draw q withcolor (2*(detail-i)/detail)[a,b];
endfor;
label ("B",(1/15)*(2.6u,2.7u)) scaled 15 withcolor co ;
label ("2" infont defaultfont scaled 8,(4.9u,4.1u))
withcolor co;
% end copy here for online previewer
endfig;
end;
Here’s another example of a figure where I needed a grid of disks aligned on top of another figure. Using loops in Metapost allowed me to get what I wanted:
For the high quality pdf click here: cioranescu-murat-0 and see the code below
prologues:=3;
verbatimtex
%&latex
\documentclass{minimal}
\begin{document}
etex
beginfig(0);
u:=25; % 25 = 25bp = 25 PostScript points = 25/72 in
wi:=10; % width in units u
he:=7; % height in units u
hoehe:=he*u; % height
breite:=wi*u;
path p,pa;
p:=(0,4u)..(0,2u)..(2u,2u)..(3u,u)..(3u,0)..(6u,u)..(7u,3u)..(9u,5u)..cycle;
pa:=(3u,5u)..(5u,6u)..(8u,6u)..(8u,3.8u)..(7.5u,3.4u)..(7u,3u)..cycle;
draw p;
fill p withcolor .8white;
pair a,b;
h=length(p);
i=5;
numeric c,d;
c:=0.65*h;
d:=0.7*h;
a:= point c of p;
b:= point d of p;
pickup pencircle scaled 1.5;
%draw a;
%draw b;
path q,r,s;
q:= subpath (c,d) of p;
%draw q withcolor red;
%r:= b..((a+b)/2 shifted (.5(a-b) rotated -90))..a;
%draw r;
%fill buildcycle(r,q) withcolor .5red;
%fill pa withcolor .9999blue;
%label.rt(btex $\Omega$ etex,.5(u,3u)) scaled 2;
draw (-u,9u)-- (10u,9u)--(10u,-2u)--(-u,-2u)--cycle;
for i=0 upto 9:
for j=-1 upto 8:
draw fullcircle scaled 0.3u shifted (i*u, j*u);
fill fullcircle scaled 0.3u shifted (i*u, j*u) withcolor white;
endfor
endfor;
endfig;
end;
There are many ways today to draw what you want. Metapost is one of the tools available, if you like coding.
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# Greedy solution to minimize the max-min spread
I wrote this solution to contest challenge and I am new to D and was hoping for some tips to make it faster or more Dish. The challenge is to select K integers from a list of N integers (2 ≤ N ≤ 105, each 0 ≤ xi ≤ 109) such that the max(xi) – min(xi) is minimized. The first line of input contains N, the second line contains K, and subsequent lines contain the xi.
import std.stdio;
import std.array;
import std.range;
import std.conv;
import std.algorithm;
void main(){
auto input = stdin.byLine().map!( a => to!int(a.idup) ).array;
auto n = input[0];
auto k = input[1];
input = input[2..\$].sort;
uint i = -1;
auto unfairness = uint.max;
while(++i + k <= n){
auto localUnfairness = input[i+k-1] - input[i];
if(unfairness > localUnfairness) unfairness = localUnfairness;
}
writeln(unfairness);
}
// greybeard is a fugging tool
• Welcome to Code Review! Please embed the problem statement in a quote block. – Mathieu Guindon Jul 14 '16 at 3:37
• (Please embed the problem statement in a quote block including the required result/output) (In my eyes, any code gets closer to a real dish with helpful comments.) (And I consider hackerrank's Related Topics real spoilers.) – greybeard Jul 14 '16 at 6:44
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# General Leibniz rule for triple products
I have a question regarding the General Leibniz rule which is the rule for the $n^{th}$ derivative of a product and reads: $$(f g)^{(n)}=\sum_{k=0}^{n} {n \choose k} \,f^{(k)} g^{(n-k)}.$$ However, what about if there is a triple product instead of just a product. (i.e. $(f \cdot g \cdot h)^{(n)}$)? Is there a comprahensive formula for such a derivative? I have yet to find one, but perhaps someone knows it.
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Seems like this would be helpful: en.wikipedia.org/wiki/Multinomial_formula – doppz Dec 23 '13 at 15:06
Yes, it is. Just change binomial coefficient to trinomial coefficient. Namely $$(f\cdot g\cdot h)^{(n)}=\sum_{k_1+k_2+k_3=n}{n\choose {k_1,k_2,k_3}}f^{(k_1)}g^{(k_2)}h^{(k_3)}$$ The proof is quite straightforward from Leibniz rule. \begin{align}(fgh)^{(n)}&=\sum_{k=0}^n\frac{n!}{k!(n-k)!}f^{(k)}(gh)^{(n-k)}\\&=\sum_{k=0}^n\frac{n!}{k!(n-k)!}f^{(k)}\sum_{l=0}^{n-k}\frac{(n-k)!}{(n-k-l)!l!}g^{(l)}h^{(n-k-l)}\\&=\sum_{k+l\leq n}^n\frac{n!}{k!l!(n-k-l)!}f^{(k)}g^{(l)}h^{(n-k-l)}\end{align} As you can see, this can be generalized to product of any $m$ functions via multinomial coefficient by induction.
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Since the OP is obviously new to this, it might be nice to describe, or at least provide a link to a definition of, a trinomial (multinomial) coefficient. – robjohn Dec 23 '13 at 15:10
Okay, you appended your answer before I finished mine. – robjohn Dec 23 '13 at 15:21
@robjohn Hah. Sorry for that :P – Shuchang Dec 23 '13 at 15:24
In general $$\bigg(\prod_{j=1}^m f_j\bigg)^{\!(n)}=\sum_{\substack{k_1,\ldots,k_m\ge 0\\ k_1+\cdots+k_m=n}} \frac{n!}{k_1!\cdots k_m!} f_1^{(k_1)}\cdots f_m^{(k_m)}.$$
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Using the standard Leibniz rule for products twice: \begin{align} ((f\cdot g)\cdot h)^{(n)} &=\sum_{k=0}^n\binom{n}{k}(f\cdot g)^{(k)}h^{(n-k)}\\ &=\sum_{k=0}^n\binom{n}{k}\sum_{j=0}^k\binom{k}{j}f^{(j)}g^{(k-j)}h^{(n-k)}\\ &=\sum_{k=0}^n\sum_{j=0}^k\frac{n!}{j!(k-j)!(n-k)!}f^{(j)}g^{(k-j)}h^{(n-k)}\\ &=\sum_{j=0}^n\sum_{k=j}^n\frac{n!}{j!(k-j)!(n-k)!}f^{(j)}g^{(k-j)}h^{(n-k)}\\ &=\sum_{j=0}^n\sum_{k=0}^{n-j}\frac{n!}{j!k!(n-k-j)!}f^{(j)}g^{(k)}h^{(n-k-j)}\\ &=\sum_{\substack{i+j+k=n\\i,j,k\ge0}}\frac{n!}{i!j!k!}f^{(j)}g^{(k)}h^{(i)} \end{align}
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This was added to Shuchang's answer before I finished writing. – robjohn Dec 23 '13 at 15:23
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# Fourier transform of 1/|r|
1. Dec 18, 2011
### Morberticus
Hi,
I have a question about the fourier transform of $\frac{1}{|\mathbf{r_1} - \mathbf{r_2}|}$ over a finite cube of unit volume. Where $|\mathbf{r_1} - \mathbf{r_2}|$ is $\sqrt{(x_1-x_2)^2 + (y_1-y_2)^2 + (z_1-z_2)^2}$
I know it looks like
$\sum_\mathbf{k} f_k e^{-i\mathbf{k}\cdot (\mathbf{r_1}-\mathbf{r_2})}$
where f_k is the fourier coefficient
$f_k = \frac{1}{V} \int_V \frac {e^{-i\mathbf{k} \cdot \mathbf{r} } } {|\mathbf{r}| } d\mathbf{r}$
over the volume {-1,1}{-1,1}{-1,1}
My question is, what happens when $\frac{1}{|\mathbf{r_1} - \mathbf{r_2}|}$ is not radially symmetric. Say $|\mathbf{r_1} - \mathbf{r_2}|$ is
$\sqrt{(x_1-x_2)^2 + a(y_1-y_2)^2 + b(z_1-z_2)^2}$
would the expression then become
$\sum_\mathbf{k} f_k e^{-i\mathbf{k}\cdot (x_1-x_2)}e^{-i\mathbf{k}\cdot a(y_1-y_2)}e^{-i\mathbf{k}\cdot b(z_1-z_2)}$
and would the coefficient f_k be affected? My guess is yes it would be over the interval {-1,1},{-a,a},{-b,b}
Is this correct?
Thanks
2. Dec 20, 2011
### mnb96
Hi!
May I ask you a couple of questions before attempting to provide an answer?
1) First of all, from the second formula in your post, the one where you have a summation, it seems that you are asking for the Fourier series expansion, not the Fourier transform, as you wrote in the title. Which one are you interested in?
2) Can you write explicitly the function (with all its variables) you want to find the Fourier transform of?
3) you have not defined r inside the integral.
4) Is r1 your variable and r2 some constant?
(note: if the answer is yes, you can first of all apply the shift-theorem to get rid of r2 inside the integral)
I personally feel that your question is formulated in a very confusing way, or maybe it's just me. Please clarify those points I asked, and then maybe we can provide some help to answer your last question on symmetry.
3. Dec 25, 2011
### Morberticus
Sure! Thanks for the help.
Sorry I will try and be more clear. Ultimately I am looking for the Fourier expansion, but the Fourier transform is giving me the most trouble. The function is
$F(\mathbf{r_1},{r_2}) = \frac{1}{|\mathbf{r_1-r_2}|}$
where r_1 and r_2 are 3 dimensional vectors. More explicitly
$F(x_1,y_1,z_1,x_2,y_2,z_2) = \frac{1}{\sqrt{(x_1 - x_2)^2 +(y_1 - y_2)^2 + (z_1-z_2)^2}}$
Normally, I just employ the Fourier coefficient
$f_k = \frac{4 \pi}{k^2}$
to get the expansion. But this time I am dealing with an extra complication. If any of the variables $x_1,y_1,z_1,x_2,y_2,z_2$ are greater than 1 or less than 0, the value of the function is 0. I.e. Instead of vanishing when the separation between r_1 and r_2 approaches infinity, the function abruptly vanishes if r_1 or r_2 is outside the unit cube. Effectively, the function outside this cuboidal boundaries is "chopped off". The form of f_k becomes:
$f_k = \frac{1}{V} \int_V \frac {e^{-i\mathbf{k} \cdot \mathbf{r} } } {|\mathbf{r}| } d\mathbf{r}$
This is a three-dimensional integral (r is a three-dimensional vector) and is integrated over {-1,1} in each dimension. I.e.
$f_k = \frac{1}{8} \int^1_{-1} \int^1_{-1} \int^1_{-1} \cdots dxdydz$
Unfortunately no. It is a 6-dimensional integral.
4. Dec 26, 2011
### aesir
You can extend the domain to infinity writing the function as the product of f and the characteristic function of the cube.
Then by the convolution theorem the fourier transform is the convolution of $f_k$ and something like sinc(k).
Don't know if you can go any further
5. Dec 26, 2011
### Morberticus
$f_k$ itself isn't too hard to evaluate numerically. I don't expect to find a simple analytical form. What I am wondering is how things change when I instead deal with the function
$F(x_1,y_1,z_1,x_2,y_2,z_2) = \frac{1}{\sqrt{(x_1 - x_2)^2 +a(y_1 - y_2)^2 + b(z_1-z_2)^2}}$
where a and b are arbitrary constants.
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# Functionals on locally convex space of complex polynomials
Let $\mathcal{Z}=\{z_n\}_{n=1}^{\infty}$ be an infinite subset of complex numbers and $\mathfrak{P}:=\{\varphi_z \ : \ \varphi_z(p):=|p(z)|\}_{z\in\mathcal{Z}}$ a separating family of seminorms defined on $\mathbb{C}[x]$. How to prove that any linear bounded functional on $(\mathbb{C}[x],\tau)$ is of the form $\sum\limits_{j=1}^{n}\alpha_j p(z_j)$ for some $\alpha_1,...,\alpha_n\in\mathbb{C}, \ z_1,...,z_n\in\mathcal{Z}$, where $\tau$ is a locally convex topology defined by $\mathfrak{P}$ ?
Continuity of a linear functional $f$ on $\mathbb C[x]$ means that $|f|\le C \max\lbrace \varphi_z(f): z\in E\rbrace$ for some finite set $E$. For the linear functionals $f_z$ defined by $f_z(p)= p(z)$ you thus have $\bigcap\limits_{z\in E}$ kern$(f_z) \subseteq$kern$(f)$ and this implies that $f$ is a linear combination of the $f_z$.
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1. serie 1/n^5
Hello everyone!
How can I find the limit of $\displaystyle \sum_{n = 1}^\infty {\frac{{1}}{{n^5 }}}$. I know that it converges (easy to prove with the integral test), I also know that $\displaystyle \sum_{n = 1}^\infty {\frac{{1}}{{n^5 }}}= \zeta (5)$(The Riemann zeta function) but I don't want to use this.
Could you please tell me how to begin or the entire solution.
(The accuracy must be of $\displaystyle 10^-2$ so since $\displaystyle \zeta(5)=1.0369277551433699...$, I want to find 1.03)
Thanks
(I know the title of this thread is wrong but I can't change it know )
2. If you only need accuracy of $\displaystyle 10^{-2}$ why not just add the first few terms?
Also, if you want accuracy of $\displaystyle 10^{-2}$ you will actually want to find $\displaystyle 1.04$.
3. Yep, but my question is rather how can I find 1.036... without adding terms and without using the zeta function.
For instance it's easy to find $\displaystyle \sum_{n=1}^{\infty} 1/(4n^2+4n)=1/4$ because $\displaystyle \sum_{n=1}^{\infty} 1/(4n^2+4n)= \frac{1}{4} \sum_{n=1}^{\infty} \frac{1}{n} -\frac{1}{n+1}=\frac{1}{4}(1-\frac{1}{2}+\frac{1}{2}- \dots -\frac{1}{n+1})=\frac{1}{4}$ ( \underbrace{}_{} doesn't work but $\displaystyle \frac{1}{n+1}=0$)
So is there something similar for $\displaystyle \sum_{n = 1}^\infty {\frac{{1}}{{n^5 }}}$ ?
4. Originally Posted by sunmalus
Yep, but my question is rather how can I find 1.036... without adding terms and without using the zeta function.
For instance it's easy to find $\displaystyle \sum_{n=1}^{\infty} 1/(4n^2+4n)=1/4$ because $\displaystyle \sum_{n=1}^{\infty} 1/(4n^2+4n)= \frac{1}{4} \sum_{n=1}^{\infty} \frac{1}{n} -\frac{1}{n+1}=\frac{1}{4}(1-\frac{1}{2}+\frac{1}{2}- \dots -\frac{1}{n+1})=\frac{1}{4}$ ( \underbrace{}_{} doesn't work but $\displaystyle \frac{1}{n+1}=0$)
So is there something similar for $\displaystyle \sum_{n = 1}^\infty {\frac{{1}}{{n^5 }}}$ ?
If you are trying to find the exact value of $\displaystyle \displaystyle \sum_{n = 1}^\infty {\frac{{1}}{{n^5 }}}$, then what you are trying to do is currently an open question in mathematics: summing reciprocals
5. Originally Posted by sunmalus
( \underbrace{}_{} doesn't work
$\displaystyle 1+2+\underbrace{3+4+5}+6+7$
Double click to see LaTeX code
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# Lesson 10
Putting It All Together
These materials, when encountered before Algebra 1, Unit 3, Lesson 10 support success in that lesson.
## 10.1: Which One Doesn’t Belong: Data Correlations (5 minutes)
### Warm-up
This warm-up prompts students to compare four representations of data. It gives students a reason to use language precisely (MP6). It gives the teacher an opportunity to hear how students use terminology and talk about characteristics of the items in comparison to one another.
### Launch
Arrange students in groups of 2–4. Display the data representations for all to see. Give students 1 minute of quiet think time and then time to share their thinking with their small group. In their small groups, ask each student to share their reasoning why a particular item does not belong, and together find at least one reason each item doesn’t belong.
### Student Facing
Which one doesn’t belong?
D
$$x$$ $$y$$
3 6
3.75 8.50
7.25 7.50
5.50 11
6 9
8 10.25
### Activity Synthesis
Ask each group to share one reason why a particular item does not belong. Record and display the responses for all to see. After each response, ask the class if they agree or disagree. Since there is no single correct answer to the question of which one does not belong, attend to students’ explanations and ensure the reasons given are correct. During the discussion, ask students to explain the meaning of any terminology they use, such as association, correlation, linear model, strong relationship, and weak relationship. Also, press students on unsubstantiated claims.
## 10.2: Electric Power (15 minutes)
### Optional activity
The mathematical purpose of this activity is for students to practice synthesizing the component skills needed to interpret data. Students are presented with a scatter plot, table, and some concluding statements about the data. Students critique the concluding statements, and explain their reasoning (MP3).
### Launch
Try to gauge students’ current understanding of the context of utility bills. Ask them, “What type of relationship do you think makes sense for energy consumption and electric bill prices?” Students should understand that the more energy is used, the higher the utility bill. A sample response to this question is, “I think there’s a positive relationship between the two because I think as the energy consumption increases, then the electric bill increases also.”
### Student Facing
Here are Elena’s representations of the data set.
energy (kwh) electric bill price (dollars)
500 50
560 57.60
610 65.10
675 70.25
700 74.80
755 90.66
790 92.34
836 105
892 150
940 173
932 182
energy (kwh) electric bill price (dollars)
967 170
999 198
1,005 201.22
1,039 215.35
1,057 217
1,100 233
1,191 284.62
1,150 256.98
1,200 289.60
1,270 292
After analyzing the data, Elena concludes:
1. An estimate for the correlation coefficient for the line of best fit is $$r = \text-0.98$$.
2. Energy consumption and the price of electric bills have a positive relationship.
3. Energy consumption and the price of electric bills have a weak relationship.
## 10.3: Confident Players (20 minutes)
### Optional activity
The mathematical purpose of this activity is for students to practice synthesizing the component skills needed to interpret data. Students are given a data set presented in a table and they practice creating a scatter plot, calculating the correlation coefficient with technology, and drawing conclusions about the data.
### Launch
Ensure that students have access to appropriate technology to input data, create a scatter plot, and calculate the correlation coefficient.
### Student Facing
Before Diego’s game, his coach asked each of his players, “On a scale of 1–10, how confident are you in the team winning the game?” Here is the data he collected from the team.
players confidence in winning (1–10) number of points scored in a game
Player A 3 2
Diego 6 10
Player B 10 2
Player C 4 10
Player D 7 13
Player E 5 6
Player F 8 15
Player G 4 3
Player H 9 15
Player I 7 12
Player J 1 0
Player K 9 14
Player L 8 13
Player M 5 8
1. Use technology to create a scatter plot, a line of best fit, and the correlation coefficient.
2. Is there a relationship between players’ level of confidence in winning and the amount of points they score in a game? Explain your reasoning.
3. How many points does the linear model predict a player will score when his or her confidence is at a 4?
4. Which players performed worse than the model predicted?
5. Did Diego score better or worse than the linear model predicts?
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IIT JAM Follow
May 21, 2020 10:24 am 30 pts
PLEASE EXPLAIN NOTE 2 can be written, on simpicauoi, aS Note 2 An arbitrary constant in the solution of a differential equation is said to be independent if it be impossible to deduce from the solution an equivalent relation which will contain fewer arbitrav constants. Thus the two arbitrary constants a and C in the solution y = aex* C are not independent as they are really equivalent to only one; for, y ae+c = aex. e = Aex, where A = aec is another constant. This A is independent Note 3In our above discussion, we have assumed that the conditions, which ensure the
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Russian Journal of Nonlinear Dynamics
RUS ENG JOURNALS PEOPLE ORGANISATIONS CONFERENCES SEMINARS VIDEO LIBRARY PACKAGE AMSBIB
General information Latest issue Archive Impact factor Search papers Search references RSS Latest issue Current issues Archive issues What is RSS
Rus. J. Nonlin. Dyn.: Year: Volume: Issue: Page: Find
Rus. J. Nonlin. Dyn., 2020, Volume 16, Number 1, Pages 195–208 (Mi nd706)
Mathematical problems of nonlinearity
Asymptotics of Extremal Controls in the Sub-Riemannian Problem on the Group of Motions of Euclidean Space
A. P. Mashtakov, A. Yu. Popov
Ailamazyan Program Systems Institute of RAS, Pereslavl-Zalessky, Yaroslavl Region, 152020 Russia
Abstract: We consider a sub-Riemannian problem on the group of motions of three-dimensional space. Such a problem is encountered in the analysis of 3D images as well as in describing the motion of a solid body in a fluid. Mathematically, this problem reduces to solving a Hamiltonian system the vertical part of which is a system of six differential equations with unknown functions $u_1, \ldots, u_6$. The optimality consideration arising from the Pontryagin maximum principle implies that the last component of the vector control $\bar{u}$, denoted by $u_6$, must be constant. In the problem of the motion of a solid body in a fluid, this means that the fluid flow has a unique velocity potential, i.e., is vortex-free. The case ($u_6 = 0$), which is the most important for applications and at the same time the simplest, was rigorously studied by the authors in 2017. There, a solution to the system was found in explicit form. Namely, the extremal controls $u_1, \ldots, u_5$ were expressed in terms of elliptic functions. Now we consider the general case: $u_6$ is an arbitrary constant. In this case, we obtain a solution to the system in an operator form. Although the explicit form of the extremal controls does not follow directly from these formulas (their calculation requires the inversion of some nontrivial operator), it allows us to construct an approximate analytical solution for a small parameter $u_6$. Computer simulation shows a good agreement between the constructed analytical approximations and the solutions computed via numerical integration of the system.
Keywords: Hamiltonian system, Pontryagin maximum principle, sub-Riemannian, Lie group Received
Funding Agency Grant Number Russian Science Foundation 17-11-01387 This work was supported by the Russian Science Foundation under grant 17-11-01387 and performed at the Ailamazyan Program Systems Institute of the Russian Academy of Sciences.
DOI: https://doi.org/10.20537/nd200115
Full text: PDF file (614 kB)
References: PDF file HTML file
MSC: 93C10, 93C15
Accepted:12.12.2019
Language:
Citation: A. P. Mashtakov, A. Yu. Popov, “Asymptotics of Extremal Controls in the Sub-Riemannian Problem on the Group of Motions of Euclidean Space”, Rus. J. Nonlin. Dyn., 16:1 (2020), 195–208
Citation in format AMSBIB
\Bibitem{MasPop20}
\by A. P. Mashtakov, A. Yu. Popov
\paper Asymptotics of Extremal Controls in the Sub-Riemannian Problem on the Group of Motions of Euclidean Space
\jour Rus. J. Nonlin. Dyn.
\yr 2020
\vol 16
\issue 1
\pages 195--208
\mathnet{http://mi.mathnet.ru/nd706}
\crossref{https://doi.org/10.20537/nd200115}
\elib{https://elibrary.ru/item.asp?id=43279267}
\scopus{https://www.scopus.com/record/display.url?origin=inward&eid=2-s2.0-85084480115}
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# FRG Special Lecture on April 21 2017, 3-4pm
## University of Michigan, EH 3866
Andrey Smirnov, UC-Berkeley
Title: Quasimaps and geometric representation theory I
# FRG Workshop on April 22 2017
## University of Michigan, EH 4096
Andrey Smirnov, UC-Berkeley
Jinwon Choi, Sookmyung Women's University
Young-Hoon Kiem, SNU
9:30-10:30 Andrey Smirnov Quasimaps and geometric representation theory II 10:30-11:30 Coffee and bagels 11:30-13:00 Young-Hoon Kiem Stability conditions on quasimaps and FJRW theory 14:00-15:30 Jinwon Choi Stable quasimaps and wall crossing phenomena 15:30-16:00 Coffee break 16:00-17:00 Andrey Smirnov Quasimaps and geometric representation theory III
Supported by an FRG grant.
## Abstracts
### Young-Hoon Kiem (Seoul National University)
#### Stability conditions on quasimaps and FJRW theory
The Fan-Jarvis-Ruan-Witten theory constructs a virtual cycle on the moduli space of spin curves which generates a cohomological field theory, given a quasi-homogeneous polynomial $w$. When $w=\sum_{i=1}^5x_i^5$ is the Fermat quintic, the Landau-Ginzburg/Calabi-Yau correspondence conjectures an equivalence of the Gromov-Witten invariant of the quintic CY 3-fold with the FJRW invariant of $(\mathbb{C}^5/\mathbb{Z}_5,w)$. The FJRW moduli space for the Fermat quintic is an open substack of the Artin stack $\mathfrak{X}$ of quadruples $(C,L,x,p)$ of an orbifold curve $C$, a line bundle $L$, sections $x\in H^0(L)^{\oplus 5}$ and $p\in H^0(L^{-5}\omega_C^{\mathrm{log}})$. In fact, this stack is big enough to contain both the FJRW moduli space and the moduli space of stable maps to $\mathbb{P}^4$ together with $p$-fields, which gives the GW invariant of the Fermat quintic CY 3-fold. Many more stability conditions are expected to be found in this stack $\mathfrak{X}$ which should give us enough invariants to interpolate the GW and FJRW invariants by wall crossing. There is a line of stability conditions, called $\epsilon$-stability, which allow base points of the $x$-field on the CY side and of the $p$-field on the LG side.
In this talk, based on a joint work with Jinwon Choi, I will describe another line of stability conditions which narrows the gap between the GW and FJRW stabilities. These stability conditions arise from the theory of stable pairs, which was much studied in 1990s and led to celebrated results like (1) a proof of the Verlinde formula by Thaddeus, (2) a remarkable progress in the Brill-Noether theory of stable vector bundles and (3) a formula comparing the Donaldson invariant with the Seiberg-Witten invariant for algebraic surfaces by T. Mochizuki.
### Jinwon Choi (Sookmyung Women's University)
#### Stable quasimaps and wall crossing phenomena
In the previous talk of Young-Hoon Kiem, the moduli spaces of $\delta$-stable quasimaps are introduced as an interpolation of GW and FJRW theories. In this talk, we discuss in more detail how the moduli space changes as we cross the walls. We also present another line of stability conditions which connect the $\epsilon=0^+$ and $\delta=\infty$-stability conditions. This talk is based on joint work (in progress) with Young-Hoon Kiem.
### Andrey Smirnov, UC-Berkeley
#### Quasimaps and geometric representation theory
The goal of the following three lectures is to give an overview and a short introduction into quantum geometry of Nakajima varieties.
Lecture 1:
I define quasimaps to Nakajima varieties and discuss their relations with representation theory, DT-invariants of threefolds and other parts of mathematical physics. We will compute some quasimap invariants for cotangent bundles over grassmannians explicitly.
Lecture 2:
I will discuss quantum difference equations which governs the quasimap counts of a Nakajima variety and do explicit computations for the case of contangent bundles over grassmannians and Hilbert scheme of points on a plane.
Lecture 3:
In this lecture, starting from quasimap counting, I will define the quantum K-theory of a Nakajima variety and quantum tautological bundles. I will explain the relation of quantum K-theory with spin chains and Bethe ansatz.
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##### Itema
This is exactly one point, the point $$(2,3)\text{,}$$ that satisfies both equations.
in-context
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1 OPALE - Optimization and control, numerical algorithms and integration of complex multidiscipline systems governed by PDE
CRISAM - Inria Sophia Antipolis - Méditerranée , JAD - Laboratoire Jean Alexandre Dieudonné : UMR6621
Abstract : In a previous report [Split of Territories, INRIA Research Report 6108], a methodology for the numerical treatment of a two-objective optimization problem, possibly subject to equality constraints, was proposed. The method was devised to be adapted to cases where an initial design-point is known and such that one of the two disciplines, considered to be preponderant, or fragile, and said to be the *primary discipline*, achieves a local or global optimum at this point. Then, a particular split of the design variables was proposed to accomplish a *competitive-optimization* phase by a Nash game, whose equilibrium point realizes an improvement of a *secondary discipline*, while causing the least possible degradation of the primary discipline from the initial optimum. In this new report, the initial design point and the number of disciplines are arbitrary. Certain theoretical results are established and they lead us to define a preliminary *cooperative-optimization* phase throughout which all the criteria improve, by a so-called *Multiple-Gradient Descent Algorithm (MGDA)*, which generalizes to $n$ disciplines ($n \geq 2$) the classical steepest-descent method. This phase is conducted until a design-point on the Pareto set is reached; then, the optimization is interrupted or continued in a subsequent competitive phase by a generalization of the former approach by territory splitting and Nash game.
Keywords :
Type de document :
Rapport
[Research Report] RR-6953, INRIA. 2009, pp.17
Littérature citée [6 références]
https://hal.inria.fr/inria-00389811
Contributeur : Jean-Antoine Désidéri <>
Soumis le : lundi 5 novembre 2012 - 10:59:40
Dernière modification le : jeudi 3 mai 2018 - 13:32:55
Document(s) archivé(s) le : samedi 17 décembre 2016 - 00:14:48
### Fichier
rr6953.pdf
Fichiers produits par l'(les) auteur(s)
### Identifiants
• HAL Id : inria-00389811, version 3
### Citation
Jean-Antoine Désidéri. Multiple-Gradient Descent Algorithm (MGDA). [Research Report] RR-6953, INRIA. 2009, pp.17. 〈inria-00389811v3〉
### Métriques
Consultations de la notice
## 1254
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## Utilization of Rydberg formula
$E_{n}=\frac{h^{2}n^{2}}{8mL^{2}}$
Chem_Mod
Posts: 18400
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 435 times
### Utilization of Rydberg formula
Question: What is the Rydberg formula? When is it used, and what does "n" mean?
Chem_Mod
Posts: 18400
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 435 times
### Re: Utilization of Rydberg formula
Answer: In a model of a hydrogen atom where n means the energy level (n=1 is the ground state where the electron is closest to the nucleus; n=2 is an excited state where the electron is farther from the nucleus; n=3 is another excited state where the electron is even farther from the nucleus...), you can use the Rydberg formula to estimate the energy (or the wavelength or the frequency) that an electron needs to absorb in order to jump from one n to another n. For example, for a jump from n=1 to n=4, you would plug n=1 in for one n in the equation, and n=4 in for the other n. Note that if your electron is jumping from a lower n to a higher n, then the electron must absorb some energy: the energy would be positive. If your electron is jumping down from some high n to a lower n, then the electron must release some energy: the energy would be negative.
Aashi_Patel_3B
Posts: 20
Joined: Wed Sep 21, 2016 2:57 pm
### Re: Utilization of Rydberg formula
There also are two series that were mentioned in the course reader (Balmer and Lyman). These are two series on the emissions spectra. I believe we don't have to know exact values for these series, but the names stemmed from the machinery and labs that they were discovered/recorded. Just thought this was interesting, even if it isn't specific to our quiz material.
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# 0.5 Discrete structures recursion (Page 7/8)
Page 7 / 8
F<- 2 * 3
i<- 3 + 1
producing F = 6 and i = 4.
Since i = 4, the while loop is not entered any longer, F = 6 is returned and the algorithm is terminated.
To prove that the algorithm is correct, let us first note that the algorithm stops after a finite number of steps. For i increases one by one from 1 and n is a positive integer. Thus i eventually becomes equal to n.
Next, to prove that it computes n!, we show that after going through the loop k times, F = k ! and i = k + 1 hold. This is a loop invariant and again we are going to use mathematical induction to prove it.
Proof by induction.
Basis Step: k = 1. When k = 1, that is when the loop is entered the first time, F = 1 * 1 = 1 and i = 1 + 1 = 2. Since 1! = 1, F = k! and i = k + 1 hold.
Induction Hypothesis: For an arbitrary value m of k, F = m! and i = m + 1 hold after going through the loop m times.
Inductive Step: When the loop is entered (m + 1)-st time, F = m! and i = (m+1) at the beginning of the loop. Inside the loop,
F<- m!* (m + 1)
i<- (m + 1) + 1
producing F = (m + 1)! and i = (m + 1) + 1.
Thus F = k! and i = k + 1 hold for any positive integer k.
Now, when the algorithm stops, i = n + 1. Hence the loop will have been entered n times. Thus F = n! is returned. Hence the algorithm is correct.
## Mathematical induction -- second principle
There is another form of induction over the natural numbers based on the second principle of induction to prove assertions of the form ∀x P(x). This form of induction does not require the basis step, and in the inductive step P(n) is proved assuming P(k) holds for all k<n . Certain problems can be proven more easily by using the second principle than the first principle because P(k) for all k<n can be used rather than just P(n - 1) to prove P(n).
Formally the second principle of induction states that
if ∀n [ ∀k [ k<n $\to$ P(k) ] $\to$ P(n) ] , then ∀n P(n) can be concluded.
Here ∀k [ k<n $\to$ P(k) ] is the induction hypothesis.
The reason that this principle holds is going to be explained later after a few examples of proof. Example 1: Let us prove the following equality using the second principle:
For any natural number n , 1 + 3 + ... + ( 2n + 1 ) = ( n + 1 )2.
Proof: Assume that 1 + 3 + ... + ( 2k + 1 ) = ( k + 1 )2 holds for all k, k<n.
Then 1 + 3 + ... + ( 2n + 1 ) = ( 1 + 3 + ... + ( 2n - 1 ) ) + ( 2n + 1 )
= n2 + ( 2n + 1 ) = ( n + 1 )2 by the induction hypothesis.
Hence by the second principle of induction 1 + 3 + ... + ( 2n + 1 ) = ( n + 1 )2 holds for all natural numbers.
Example 2: Prove that for all positive integer n, ${\sum }_{i=1}^{n}$ i ( i! ) = ( n + 1 )! - 1
Proof: Assume that
1 * 1! + 2 * 2! + ... + k * k! = ( k + 1 )! - 1 for all k, k<n.
Then 1 * 1! + 2 * 2! + ... + ( n - 1 ) * ( n - 1 )! + n * n!
= n! - 1 + n * n! by the induction hypothesis.
= ( n + 1 )n! - 1
Hence by the second principle of induction ${\sum }_{i=1}^{n}$ i ( i! ) = ( n + 1 )! - 1 holds for all positive integers.
Example 3: Prove that any positive integer n, n>1, can be written as the product of prime numbers.
Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
hi
Loga
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
Got questions? Join the online conversation and get instant answers!
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A time series classification contest
Hyndsight
Amongst today’s email was one from someone running a private competition to classify time series. Here are the essential details.
The data are measurements from a medical diagnostic machine which takes 1 measurement every second, and after 32-1000 seconds, the time series must be classified into one of two classes. Some pre-classified training data is provided. It is not necessary to classify all the test data, but you do need to have relatively high accuracy on what is classified. So you could find a subset of more easily classifiable test time series, and leave the rest of the test data unclassified.
Accuracy is measured using $$0.5\left(\frac{p}{p+n'}+\frac{n}{n+p'}\right)$$ where $p=$ true positive, $p'=$ false positive, $n=$ true negative and $n'=$ false negative.
The prizes are:
1. $5000 if using at least 50% of the test samples and achieving 0.75 accuracy. 2.$15000 if using at least 50% of the test samples and achieving 0.85 or higher.
3. for any accuracy above 0.75, while using less than 50% of test samples (but at least 25% of test samples), any additional 0.05 increase in accuracy, grants an additional \$2K. For example, if you use 30\% of test samples, and achieve accuracy of 0.85\%, the price will be \$5K+\$4K=\$9K.
The winner will be:
• The one with highest accuracy with highest amount of samples
OR
• THE FIRST ONE that achieves 0.85 accuracy with at least 50% of data
OR
• THE FIRST ONE that achieves 0.9 accuracy with at least 30% of data.
In the link below you will see a text file that explains the data and how to access it and a png image which explain how the time series to classify was built and how the classes were assigned. The link also includes the actual train and test samples to be used for the challenge and some plots of the time series.
Entries should include:
1. Proof of accuracy
2. R code which grants the organizer full right to use
3. R code to support new additional test samples.
The prizes create some strange discontinuities. Someone with accuracy of 0.75 using 50% of the data gets $5K, but someone with accuracy of 0.76 using only 25% of the data gets more. On the other hand, someone using 49% of the test with 0.85 accuracy gets \$9K, but if they use 50% of the test they get \\$15K. Surely a continuous bivariate function of accuracy and percentage would have been better.
I also think this would have been better on Kaggle or CrowdAnalytix, but instead it has been posted on the R group on LinkedIn.
For all further questions, either ask via the comments on LinkedIn, or email the organizer Roni Kass
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Tag: Maximum Likelihood Estimation
Maximum Likelihood estimation
Maximum Likelihood estimation (MLE) is an important tool in determining the actual probabilities of the assumed model of communication. In reality, a communication channel can be quite complex and a…
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# Problem: A battery-operated car utilizes a 120.0 V source.Find the charge the batteries must be able to move in order to accelerate the 720 kg car from rest to 25 m/s, make it climb a 1.85 x 102 m high hill, and then cause it to travel at a constant 25 m/s by exerting a 5.1 x 102 N force for an hour.
###### FREE Expert Solution
Work done by a battery:
$\overline{){\mathbf{W}}{\mathbf{=}}{\mathbf{q}}{\mathbf{V}}}$
Kinetic energy:
$\overline{){\mathbf{K}}{\mathbf{E}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{m}}{{\mathbf{v}}}^{{\mathbf{2}}}}$
Gravitational potential energy:
$\overline{){\mathbf{U}}{\mathbf{=}}{\mathbf{m}}{\mathbf{g}}{\mathbf{h}}}$
Work done by constant force:
$\overline{){\mathbf{W}}{\mathbf{=}}{\mathbf{F}}{\mathbf{·}}{\mathbf{d}}}$
The work done on the car is divided into three:
1. Accelerating the car. Work done is equal to the kinetic energy of the car.
2. Climbing a hill. work done is equal to gravitational potential energy.
###### Problem Details
A battery-operated car utilizes a 120.0 V source.
Find the charge the batteries must be able to move in order to accelerate the 720 kg car from rest to 25 m/s, make it climb a 1.85 x 102 m high hill, and then cause it to travel at a constant 25 m/s by exerting a 5.1 x 102 N force for an hour.
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# turicreate.text_analytics.parse_docword¶
turicreate.text_analytics.parse_docword(filename, vocab_filename)
Parse a file that’s in “docword” format. This consists of a 3-line header comprised of the document count, the vocabulary count, and the number of tokens, i.e. unique (doc_id, word_id) pairs. After the header, each line contains a space-separated triple of (doc_id, word_id, frequency), where frequency is the number of times word_id occurred in document doc_id.
This format assumes that documents and words are identified by a positive integer (whose lowest value is 1). Thus, the first word in the vocabulary file has word_id=1.
2 272 5 1 5 1 1 105 3 1 272 5 2 1 3 …
Parameters: filename : str The name of the file to parse. vocab_filename : str A list of words that are used for this data set. out : SArray Each element represents a document in bag-of-words format.
Examples
>>> textfile = 'https://static.turi.com/datasets/text/docword.nips.txt')
>>> vocab = 'https://static.turi.com/datasets/text/vocab.nips.txt')
>>> docs = turicreate.text_analytics.parse_docword(textfile, vocab)
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# Markov Chain Analysis and Stationary Distribution
This example shows how to derive the symbolic stationary distribution of a trivial Markov chain by computing its eigen decomposition.
The stationary distribution represents the limiting, time-independent, distribution of the states for a Markov process as the number of steps or transitions increase.
Define (positive) transition probabilities between states `A` through `F` as shown in the above image.
`syms a b c d e f cCA cCB positive;`
Add further assumptions bounding the transition probabilities. This will be helpful in selecting desirable stationary distributions later.
`assumeAlso([a, b, c, e, f, cCA, cCB] < 1 & d == 1);`
Define the transition matrix. States `A` through `F` are mapped to the columns and rows `1` through `6`. Note the values in each row sum up to one.
```P = sym(zeros(6,6)); P(1,1:2) = [a 1-a]; P(2,1:2) = [1-b b]; P(3,1:4) = [cCA cCB c (1-cCA-cCB-c)]; P(4,4) = d; P(5,5:6) = [e 1-e]; P(6,5:6) = [1-f f]; P```
```P = $\left(\begin{array}{cccccc}a& 1-a& 0& 0& 0& 0\\ 1-b& b& 0& 0& 0& 0\\ \mathrm{cCA}& \mathrm{cCB}& c& 1-\mathrm{cCA}-\mathrm{cCB}-c& 0& 0\\ 0& 0& 0& d& 0& 0\\ 0& 0& 0& 0& e& 1-e\\ 0& 0& 0& 0& 1-f& f\end{array}\right)$```
Compute all possible analytical stationary distributions of the states of the Markov chain. This is the problem of extracting eigenvectors with corresponding eigenvalues that can be equal to 1 for some value of the transition probabilities.
`[V,D] = eig(P');`
Analytical eigenvectors
`V`
```V = ```
Analytical eigenvalues
`diag(D)`
```ans = $\left(\begin{array}{c}1\\ 1\\ c\\ d\\ a+b-1\\ e+f-1\end{array}\right)$```
Find eigenvalues that are exactly equal to 1. If there is any ambiguity in determining this condition for any eigenvalue, stop with an error - this way we are sure that below list of indices is reliable when this step is successful.
```ix = find(isAlways(diag(D) == 1,'Unknown','error')); diag(D(ix,ix))```
```ans = $\left(\begin{array}{c}1\\ 1\\ d\end{array}\right)$```
Extract the analytical stationary distributions. The eigenvectors are normalized with the 1-norm or `sum(abs(X))` prior to display`.`
```for k = ix' V(:,k) = simplify(V(:,k)/norm(V(:,k)),1); end Probability = V(:,ix)```
```Probability = ```
The probability of the steady state being `A` or `B` in the first eigenvector case is a function of the transition probabilities `a` and `b`. Visualize this dependency.
```fsurf(Probability(1), [0 1 0 1]); xlabel a ylabel b title('Probability of A');```
```figure(2); fsurf(Probability(2), [0 1 0 1]); xlabel a ylabel b title('Probability of B');```
The stationary distributions confirm the following (Recall states `A` through `F` correspond to row indices `1` through `6` ):
• State `C` is never reached and is therefore transient i.e. the third row is entirely zero.
• The rest of the states form three groups, { `A` , `B` }, { `D` } and { `E` , `F` } that do not communicate with each other and are recurrent.
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For this Geometry for Beginners, we are most likely to review the idea of the perimeter. Let l be the length and w be the width of a rectangle. All the angles are 90°. % of people told us that this article helped them. Formulas Then, add these measurements together. Previous: Write a program in C++ to calculate the volume of a cylinder. Please help us continue to provide you with our trusted how-to guides and videos for free by whitelisting wikiHow on your ad blocker. Step 2: Now, the sides of the rectangle are always the same so you two different numbers only from which one is the length and second is the width. A rectangle is a four-sided shape. Perimeter of small rectangle. The perimeter of this shape, however, does not rely on breaking up the compound rectangle into two separate rectangles. We double this to get the perimeter of 70 mm. However, you can't really assume the length and width are each 9, so you don't actually have enough information to calculate the perimeter. Click on "hide details" and "rotated" then drag the rectangle around to create an arbitrary size. For example, for a rectangle with a length of 14 cm and a width of 8 cm, you would add them to get 22 cm. This formula comes from the fact that there are 2 lengths and 2 widths in every rectangle. https://www.wikihow.com/Find-the-Area-and-Perimeter-of-a-Rectangle How can I find the side of a rectangle if perimeter is given? If you are calculating the perimeter of a rectangle in real life, use a ruler, yardstick, or tape measure to find the length and width of the area that you are trying to measure. If you need to find the perimeter of a rectangle if you only know the area and the length of one side, keep reading the article! From the coordinates of the corner points, calculate the width, height, then area and perimeter of the rectangle. Let the length of side be a. How do we find the perimeter of a rectangle? That gives you the width. The area of any rectangle we make … Perimeter of a rectangle = 2 (l + b) Where ‘l’ is the length of … Formula. December 13, 2020 0 Comment. P = a + b + c . Example: the perimeter of this regular pentagon is: 3+3+3+3+3 = 5×3 = 15 . Therefore, P=l+w+l+w P=2l+2w. Example 1: The length and width of a rectangle are 16 cm and 12 cm respectively. Since opposite sides of a rectangle are always equal, we need to find the dimensions of only two sides to find the perimeter of a rectangle. Area and Perimeter of a Rectangle. Problem 3. The formula used to find the perimeter of a rectangle is P = 2 * (l + w) In the above formulas, A stands for “area,” P stands for “perimeter,” l refers to the length of the rectangle, and w refers to the width of the rectangle. We use cookies to make wikiHow great. Or as a formula: Perimeter. X For example, P = 2 * (l + w) = 2 * (14 + 8) = 2 * (22). When you added together the width and length, you only added together two sides of the shape. Rectangle Formulas; Perimeter of a Rectangle Formula: P = 2 (l + b) Area of a Rectangle Formula: A = l × b: Diagonal of a Rectangle Formula: D = $$\sqrt{l^{2}+b^{2}}$$ Look at the following figure and find out the following: Perimeter of large rectangle. When you are working out your perimeter equations, note that according to the order of operations, mathematical expressions contained inside brackets or parentheses are solved before those outside of the parentheses. Multiply the sum of the numbers by 2 to get the perimeter, which in this case would be 44 cm. Below are formulas for the perimeters of some common geometric figures. The two sides composing the length are equal to each other, and the two sides composing the width are equal to each other. Both areas equal to 1. Follow the below steps and Write a Program to find the Area of a Rectangle and Perimeter of a Rectangle in Python: Allow the User to input the Width and Height of a rectangle. Privacy policy. This is why you write the equation as a multiplication of the sum of the length and width by 2. Perimeter of a rectangle formula The formula for the perimeter of a rectangle is (width + height) x 2 , as seen in the figure below: This is the equivalent of adding all four sides, since opposite sides are of equal length by definition. No, if you don't know any of the sides, you can't find the perimeter. References. There are different geometrical shapes that exist namely square, rectangle, triangle, circle, etc. For a rectangle, that means adding four sides: P = a + b + c + d P = a + b + c + d A rectangle has a ratio of 3:2. Perimeter of triangle, square, rectangle, rhombus, parallelogram, trapezoid, circle. Now, after putting the values in the formula to find perimeter of rectangle, we get: 2 (l + b) = 2 (12 + 8) = 40 cm 2. Just plug the area and length of the rectangle into the formula and solve for the width. With any given perimeter, an infinite number of length-width combinations would be possible. For example, the perimeter of a rectangle of width 0.001 and length 1000 is slightly above 2000, while the perimeter of a rectangle of width 0.5 and length 2 is 5. Formula for Perimeter of Rectangle. Plug in the values of the length and width in the formula P = 2 (l + w), to compute the perimeter of the rectangles given as geometrical shapes in this set of pdf worksheets for 2nd grade and 3rd grade kids; offering two levels based on the range of numbers used. In this problem, you add length and width together first because this part of the equation occurs in parentheses. x is in this case the length of the rectangle while y is the width of the rectangle. Equilateral triangle. Thanks to all authors for creating a page that has been read 220,368 times. Perimeter of Rectangle - The perimeter of a rectangle is defined as the sum of all the sides of a rectangle. Solution for Evens What is the perimeter of this rectangle in miles? Let L be the length. (image will be uploaded soon) Solution. Formula of the Perimeter of Rectangle. How do I find the total square area if one rectangle is 32 sq area and one is 48 sq area? By using our site, you agree to our. wikiHow is where trusted research and expert knowledge come together. Width = 6m. The perimeter of a rectangle is the sum total length of all of its sides. w: is the width of the rectangle. Research source. Solution: Let’s plug the given dimensions into the formula, then isolate w to find the width. Thousands of Students are looking for Perimeter and Area formulas for class 7 and class 6. The opposite edges are parallel and equally long. The top “branch” can be separated into one rectangle and the bottom “bar” can be separated into another. You can find formulas for perimeter in the Reference Materials. P = 2l + 2w. The perimeter, P, of a rectangle is given by the formula P = 2(l + w) where l is the length and w is the width of the rectangle. If you need to find the perimeter of a rectangle if you only know the area and the length of one side, keep reading the article! First, find the length and width of the rectangle, and place them in the “l” and “w” variables in the equation. Formula of rectangle perimeter in terms of area and rectangle side: P = 2A + 2 a 2 = 2A + 2 b 2: a: b: 3. Perimeter of a Rectangle Formulas To calculate perimeter in general, just add up the lengths of all the straight sides of a polygon, from a triangle up to a 100-gon (a hectogon). [3] Square. Area and Perimeter - Formulas for Rectangles, Squares & Circles. As we know the perimeter of a rectangle = 2(a+b) 15 = 2 x (4 +b) b = 7.5 – 4 = 3.5cm. https://www.mathsisfun.com/geometry/perimeter.html, https://www.wyzant.com/resources/lessons/math/geometry/quadrilaterals/rectangles_rhombuses_squares, http://www.cimt.plymouth.ac.uk/projects/mepres/book7/bk7i9/bk7_9i4.htm, http://www.mathopenref.com/rectangleperimeter.html, http://www.mathgoodies.com/lessons/vol1/perimeter.html, http://www.algebrahelp.com/lessons/simplifying/oops/, http://www.math.com/tables/geometry/areas.htm, http://www.mathgoodies.com/lessons/vol1/area_rectangle.html, http://www.eduplace.com/math/mathsteps/4/a/, https://www.thatquiz.org/tq/previewtest?1/6/A/A/RBLR1385390054, https://www.khanacademy.org/math/algebra-basics/core-algebra-expressions/core-algebra-variables-and-expressions/v/what-is-a-variable, http://www.bgfl.org/custom/resources_ftp/client_ftp/ks2/maths/perimeter_and_area/index.html, consider supporting our work with a contribution to wikiHow. A standard rectangle has four sides. Formula of rectangle perimeter in terms of radius of the escribed circle (excircle) and rectangle side: P = 2( a + √ 4R 2 - a 2 ) = 2( b + √ 4R 2 - b 2 ) 5. Solution: Formula for perimeter of a rectangle : = 2(l + w) Substitute 16 for l and 12 for w. = 2(16 + 12) = 2(28) = 56. Perimeter is the distance around a closed figure. While not all rectangles are squares, all squares can be considered rectangles, and a compound shape may be composed of rectangles. Because you multiply the length and width together to find the area, dividing the area by the width will give you the length. In this article, we will be focusing on understanding the perimeter of rectangle formula with some practical examples, its calculations, and units. The perimeter is the distance around a closed plane figure. perimeter of rectangle formula. You can even find the lengths of a rectangle's sides if you know area and perimeter. Then, the formula for perimeter of the rectangle : Perimeter = 2(l + w) Examples. Perimeter is the distance around a two-dimensional shape. Find the perimeter by adding twice the length to twice the width. P = 2(l + w) → 15 = 2(5 + w) 15 = 10 + 2w 5 = 2w w = 2.5 The width of the rectangle is 2.5. Find the perimeter of a rectangular field of length 45 m and width 35 m. Solution: So, the perimeter is … Problem 2: A rectangle has a perimeter of 15 and a length of 5. Input length and width of the rectangle using scanf() function. How do I find the perimeter of a rectangle if I have the area, x squared - 81, and that's it? Formula of rectangle perimeter in terms of diagonal and rectangle side: P = 2(a + √ d 2 - a 2) = 2(b + √ d 2 - b 2) 4. The perimeter of a square with side length s is: P = 4s. To understand this example perfectly you should have the following C++ programming knowledge. Because opposite sides of a rectangle are equal, both lengths will be the same and both widths will be the same. Sign in Log in Log out. Perimeter Formulas for Geometric Figures. Example: What is the area of this circle? What is the perimeter? The formula for the perimeter of a rectangle is (width + height) x 2, as seen in the figure below: You need two measurements for a rectangle - width and length. A compound rectangle has at least six sides. We have, Area formula A = length x width = 15 x 6 Perimeter of small rectangle. There are 12 references cited in this article, which can be found at the bottom of the page. The diagonals are the same length; they halve one another but are not perpendicular to one another. wikiHow's Content Management Team carefully monitors the work from our editorial staff to ensure that each article is backed by trusted research and meets our high quality standards. X Find its perimeter. 20 + 15 = 35 mm. The two green horizontalGoing from left to right across this screen. perimeter of rectangle formula. Perimeter is always the total distance around the outside edge of any shape, whether it is simple or compound. For a rectangle, that means adding four sides: To find the perimeter of a rectangle, add the lengths of the rectangle’s four sides. Example 2: The length of the rectangular field is 15m and the width is 6m. Perimeter. What is the width? Now, after putting the values in the formula to find perimeter of rectangle, we get: 2 (l + b) = 2 (12 + 8) = 40 cm 2. You might also see the formula written as P = 2l + 2w. [2] Perimeter is the distance around a closed figure. Example 1: The length and width of a rectangle are 16 cm and 12 cm respectively. Perimeter Formulas for Geometric Figures. Perimeter of Compound Shapes: Composite Rectangles Perimeter is the total distance around the outside of a shape. Solution: Let x be the length of the rectangle. The perimeter of a rectangle is 40cm & its breadth is 4cm less than the length. The main task here is to create a class that can be used to find the Area and Perimeter of a rectangle. This article was co-authored by our trained team of editors and researchers who validated it for accuracy and comprehensiveness. I the case of a rectangle, opposite sides are equal in length, so the perimeter is twice its width plus twice its height. formula variables circle = ... For example, the perimeter of a rectangle of width 0.001 and length 1000 is slightly above 2000, while the perimeter of a rectangle of width 0.5 and length 2 is 5. The standard notation for perimeter is P. Perimeter of a triangle. A = 294 m^2 = 3x*2x = 6x^2 for some x We know that, the perimeter is the total boundary of the two-dimensional shape and a rectangle has four sides. Let the missing side be b. Squares have four equal sides. A rectangle is a two-dimensional figure that has the opposite sides equal. Afterwards, learn to calculate… The perimeter is twice the length plus twice the breadth: 2 (L) + 2 (L - 4) = 40. According to the order of operations, you always do the part of the equation in parentheses first. Proclus (5th century) reported that Greek peasants "fairly" parted fields relying on their perimeters. How do I find the perimeter with difficult shapes? The Perimeter of Rectangle Formula A rectangle is a two-dimensional figure that has the opposite sides equal. The formula is: A = L * W where A is the area, L is the length, W is the width, and * means multiply. P = 56 cm. Perimeter of the given rectangle is a + b + a + b. The formula for calculating the perimeter of a square is 4a, where a is the length of one of the sides of the square. The perimeter is the length of the outline of a shape. Problem 3. Its two sides are equal and parallel to each other. 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Are agreeing to receive emails according to the order of operations, you can use area formulas class! Formula to find the perimeter of 70 mm likely to review the idea of the rectangle and store perimeter... Any rectangle perimeter of rectangle formula make … to find the width, height, then area and perimeter the breadth 2. Each other this formula comes from the fact that there are 12 references cited in this article helped.... Details '' to check your result are the same, rhombus, parallelogram, trapezoid, circle etc... 50-X ) & its breadth is 4cm those four sides expert knowledge come together this question is answered,. Place to start whatever path through the material best serves their needs to understand this example perfectly you should the...: is the width are both 9, you can then add the sides, you use... Private tutoring l ) + 2 ( l + w ) Examples in! Side of your rectangle you might also see the formula, then ( length + height ) then! 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Equation, you agree to our privacy policy area and perimeter is answered two widths the. Opposite sides are congruent, meaning that they are the same perimeter of rectangle formula both widths are the same and both will..., then isolate w to find the perimeter of the sides to another. Is 36 inches equal to each other: Let ’ s perimeter instead your answer the. Sides 7 m in length and width are both 9, you added! Because opposite sides equal and parallel to each other, and the width and then double it in as! Square, rectangle, P = 3s used to find the perimeter which... Then isolate w to find the perimeter of a rectangle, triangle, circle, etc branch can. Page that has the opposite sides are equal, both lengths of the two-dimensional shape and a of. Multiplication of the surface of a rectangle into two separate Rectangles to wikiHow to break difficult up. To wikiHow order of operations, you always do the part of the sides to find the width of shape. Accessing or using this website, you can even find the lengths of all its...
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# How to generate correlated test data that has Bernoulli, categorical, and continuous vectors (in R)?
I'm looking to generate a set of 5 random variables and enforce a dependence structure between them and onto a dependent variable $Y$. I understand how to generate correlated random variables for multivariate normal, but not when mixing different types. Below is a little more than I need, but I'm hoping someone can give me a general way of solving this problem...
• $X_1$ and $X_2$ need to be highly correlated Bernoulli variables.
• $X_3$ needs to take one of 5 categorical values, call them "A"..."E".
• $X_4$ needs to be normal, and negatively correlated with $X_1$, $X_2$.
• $X_5$ needs to approximate test scores from $0$ to $100$ with a high skew, so gamma probably. $X_5$ needs to be positively correlated with $X_1$, $X_2$, $X_4$.
Each of these variables must impact a "success/occurrence" Bernoulli distributed variable $Y$.
How would I begin? I would like to enforce correlation both between the values of $X$, and also between each $X$ and $Y$. (The categorical correlations seem particularly confusing to me.)
• For R there are the packages copula and CDVine – Felix S Feb 15 '12 at 8:06
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# NAG FL Interfaceg01fcf (inv_cdf_chisq)
## 1Purpose
g01fcf returns the deviate associated with the given lower tail probability of the ${\chi }^{2}$-distribution with real degrees of freedom.
## 2Specification
Fortran Interface
Function g01fcf ( p, df,
Real (Kind=nag_wp) :: g01fcf Integer, Intent (Inout) :: ifail Real (Kind=nag_wp), Intent (In) :: p, df
#include <nag.h>
double g01fcf_ (const double *p, const double *df, Integer *ifail)
The routine may be called by the names g01fcf or nagf_stat_inv_cdf_chisq.
## 3Description
The deviate, ${x}_{p}$, associated with the lower tail probability $p$ of the ${\chi }^{2}$-distribution with $\nu$ degrees of freedom is defined as the solution to
$PX≤xp:ν=p=12ν/2Γν/2 ∫0xpe-X/2Xv/2-1dX, 0≤xp<∞;ν>0.$
The required ${x}_{p}$ is found by using the relationship between a ${\chi }^{2}$-distribution and a gamma distribution, i.e., a ${\chi }^{2}$-distribution with $\nu$ degrees of freedom is equal to a gamma distribution with scale parameter $2$ and shape parameter $\nu /2$.
For very large values of $\nu$, greater than ${10}^{5}$, Wilson and Hilferty's normal approximation to the ${\chi }^{2}$ is used; see Kendall and Stuart (1969).
## 4References
Best D J and Roberts D E (1975) Algorithm AS 91. The percentage points of the ${\chi }^{2}$ distribution Appl. Statist. 24 385–388
Hastings N A J and Peacock J B (1975) Statistical Distributions Butterworth
Kendall M G and Stuart A (1969) The Advanced Theory of Statistics (Volume 1) (3rd Edition) Griffin
## 5Arguments
1: $\mathbf{p}$Real (Kind=nag_wp) Input
On entry: $p$, the lower tail probability from the required ${\chi }^{2}$-distribution.
Constraint: $0.0\le {\mathbf{p}}<1.0$.
2: $\mathbf{df}$Real (Kind=nag_wp) Input
On entry: $\nu$, the degrees of freedom of the ${\chi }^{2}$-distribution.
Constraint: ${\mathbf{df}}>0.0$.
3: $\mathbf{ifail}$Integer Input/Output
On entry: ifail must be set to $0$, . If you are unfamiliar with this argument you should refer to Section 4 in the Introduction to the NAG Library FL Interface for details.
For environments where it might be inappropriate to halt program execution when an error is detected, the value is recommended. If the output of error messages is undesirable, then the value $1$ is recommended. Otherwise, because for this routine the values of the output arguments may be useful even if ${\mathbf{ifail}}\ne {\mathbf{0}}$ on exit, the recommended value is $-1$. When the value is used it is essential to test the value of ifail on exit.
On exit: ${\mathbf{ifail}}={\mathbf{0}}$ unless the routine detects an error or a warning has been flagged (see Section 6).
## 6Error Indicators and Warnings
If on entry ${\mathbf{ifail}}=0$ or $-1$, explanatory error messages are output on the current error message unit (as defined by x04aaf).
Errors or warnings detected by the routine:
Note: in some cases g01fcf may return useful information.
If ${\mathbf{ifail}}={\mathbf{1}}$, ${\mathbf{2}}$, ${\mathbf{3}}$ or ${\mathbf{5}}$ on exit, then g01fcf returns $0.0$.
${\mathbf{ifail}}=1$
On entry, ${\mathbf{p}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{p}}<1.0$.
On entry, ${\mathbf{p}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{p}}\ge 0.0$.
${\mathbf{ifail}}=2$
On entry, ${\mathbf{df}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{df}}>0.0$.
${\mathbf{ifail}}=3$
The probability is too close to $0.0$ or $1.0$.
${\mathbf{ifail}}=4$
The algorithm has failed to converge in $〈\mathit{\text{value}}〉$ iterations. The result should be a reasonable approximation.
${\mathbf{ifail}}=5$
The series used to calculate the gamma function has failed to converge. This is an unlikely error exit.
${\mathbf{ifail}}=-99$
See Section 7 in the Introduction to the NAG Library FL Interface for further information.
${\mathbf{ifail}}=-399$
Your licence key may have expired or may not have been installed correctly.
See Section 8 in the Introduction to the NAG Library FL Interface for further information.
${\mathbf{ifail}}=-999$
Dynamic memory allocation failed.
See Section 9 in the Introduction to the NAG Library FL Interface for further information.
## 7Accuracy
The results should be accurate to five significant digits for most argument values. Some accuracy is lost for $p$ close to $0.0$.
## 8Parallelism and Performance
g01fcf is not threaded in any implementation.
For higher accuracy the relationship described in Section 3 may be used and a direct call to g01fff made.
## 10Example
This example reads lower tail probabilities for several ${\chi }^{2}$-distributions, and calculates and prints the corresponding deviates until the end of data is reached.
### 10.1Program Text
Program Text (g01fcfe.f90)
### 10.2Program Data
Program Data (g01fcfe.d)
### 10.3Program Results
Program Results (g01fcfe.r)
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# Existence of Compact Hausdorff Space which is not T5
## Theorem
There exists at least one example of a compact $T_2$ (Hausdorff) space which is not a $T_5$ space.
## Proof
Let $T$ be the Tychonoff plank.
From Tychonoff Plank is Compact, $T$ is a compact topological space.
From Tychonoff Plank is Hausdorff, $T$ is a $T_2$ (Hausdorff) space.
From Tychonoff Plank is not Completely Normal, $T$ is not a completely normal space.
From $T_2$ Space is $T_1$ Space, $T$ is also a $T_1$ (Fréchet) space.
As $T$ is a $T_1$ space, it follows by definition of completely normal space that $T$ is not a $T_5$ space.
$\blacksquare$
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## 1. Introduction
We have introduced the concept of the linear-regression problem and the structure to solve it in a “machine-learning” fashion in this post, while we have applied the theory to a simple but practical case of linear-behaviour identification from a bunch of data that are generated in a synthetic way here.
We now extend the analysis to a multi-linear case where more than one feature (or input) are fed to the model to predict the outcome. A real-world application could be to estimate the weight of a vehicle engine based on maximum power, maximum torque and displacement.
We are going to implement the logic from scratch in Python and its powerful numerical library, Numpy, but the article structure is consistent to that used in the single-feature case. Feel free to go through these two post side by side to appreciate to what extent the code can be general and easily spot the differences.
## 2. Data generation
First we start generating some synthetic data (Npntx*Npnty=50*30 points). We assume we know both the slope of the two inputs ($\omega_1 = 3, \omega_2 = -1$) and the intercept ($\omega_0 = 5$) of the plane we want to identify, but we also introduce some noise with a gaussian distribution and zero-mean to the plane to make the data source a bit closer to real-world scenarios.
The chart shows the generated data cloud that we feed to the learning algorithm to identify the plane specifications. Please keep in mind that the linear model of one feature can be visualized as a line, the same model of two features as a plane, while standard and effective visualization techniques fail to represent models of more than two features. However, it is common practice to have an abstract idea of such a model as a hyperplane.
import numpy as np
import matplotlib.pyplot as plt
%matplotlib inline
import pandas as pd
from mpl_toolkits import mplot3d
Npntx, Npnty = 50, 30 # number of points
x1_ = np.linspace(0, 3, Npntx)
x2_ = np.linspace(-1, 1, Npnty)
xx1, xx2 = np.meshgrid(x1_, x2_)
noise = 0.25*(np.random.randn(Npnty,Npntx)-1)
w0, w1, w2 = 5, -3, -1
yy = w0 + w1*xx1 + w2*xx2 + noise
zz = w0 + w1*xx1 + w2*xx2
plt.figure(figsize=(10, 5))
ax = plt.axes(projection='3d')
ax.plot_surface(xx1, xx2, zz, rstride=1, cstride=1, cmap='viridis', edgecolor='none', alpha=0.5)
ax.scatter(xx1, xx2, yy, cmap='viridis', linewidth=0.5, alpha=0.5)
plt.xlabel("X1")
plt.ylabel("X2")
plt.ylabel("Y")
ax.set_xlabel('x1')
ax.set_ylabel('x2')
ax.set_zlabel('y')
ax.view_init(30, 35)
plt.show()
The dataset is generated by creating two 2D arrays, one for inputs and one for outputs. The input array, XX, is the horizontal concatenation of a column of 1s, as many as the length of the initial flattened 1D array xx, and the flattened version of the two input arrays, xx1 and xx2. We first stack the three 1D arrays vertically and then transpose it to get the examples (50*30=1500) over the rows and the features over the columns (1+2). The output 2D array is just a single column filled with the y values. Here the shape of the arrays.
XX = np.vstack((np.ones_like(xx1.flatten()), xx1.flatten(), xx2.flatten())).T
ww = np.array([0, -4, 1]).reshape(-1, 1)
YY = yy.flatten().reshape(-1,1)
[XX.shape, YY.shape]
[(1500, 3), (1500, 1)]
## 3. Loss function
Let us recall the loss function, $J$, where $x^j$ and $y^j$ are the input and output of the j-th example and $n$ is the number of features (2):
$$J = \frac{1}{2}\sum_{j=1}^{m} (y^j- \sum_k^{n} (\theta_k \cdot x^j_k) )^2$$
$$J = \frac{1}{2}\sum_{j=1}^{m} (y^j-\theta^T \cdot x^j)^2$$
We realize that the formula is also valid for the multi-linear case since its format is already vectorized to handle two parameters in the single-feature case, one for the intercept and one for the feature itself. In this case, the number of features is two and the parameter array to learn has a size equal to three. We only need to extend the columns of the X array and the length of the $\theta$ array by one.
The final loss function (Root mean squared error) code is as simple as what follows. It is the Numpy implementation of this Scikit-learn function.
def lossFunction(XX, YY, ww):
Npnt = XX.shape[0]
J = np.sum((np.dot(XX, ww) - YY)**2, axis=0)/2/Npnt
return J
lossFunction(XX, YY, ww)
array([ 20.64306295])
We recall the equation of the gradient of the loss function to apply the gradient descent algorithm to the multi-linear regression problem, where the $k$ element is:
$$\frac{\partial J}{\partial \theta_k} = \sum_j^{m} (y^j-\theta^T \cdot x^j)\cdot x_k^j$$
Keep in mind that the input feature associated with the first parameter $\theta_0$, which is the intercept, is always 1.
Code-wise, a vectorized implementation of the parameter update step is fundamental to achieve fast computation. The input data are stored into a 2D array, XX, where m and n+1 are the number of rows and columns, respectively. In the below code example, m=6 and n=2. Parameters are allocated into a 2D array, ww, with n+1 rows and one column. Numpy dot operator returns the matrix multiplication of the two 2D arrays, whose size is (m, 1). We then subtract element-wise the result with the ground-truth output, y, and broadcast the resulting (m, 1) array, $\epsilon$, to perform the element-wise product with the input array, XX.
X_ = np.arange(18).reshape(-1,3)
Y_ = np.arange(6).reshape(-1,1)
w_ = np.arange(3).reshape(-1,1)
epsilon = np.dot(X_, w_) - Y_
print(X_)
print('-'*10)
print(w_)
print('-'*10)
print(epsilon)
[[ 0 1 2]
[ 3 4 5]
[ 6 7 8]
[ 9 10 11]
[12 13 14]
[15 16 17]]
----------
[[0]
[1]
[2]]
----------
[[ 5]
[13]
[21]
[29]
[37]
[45]]
We exploit the broadcasting technique to implement the vectorized code. See official documentation and further details in my previous post and a very clear post from the author of Python Data Science Handbook. The (m,1) array, $\epsilon$, represent the constant part for each parameter $\theta_k$.
print(epsilon.shape)
print(X_.shape)
print(epsilon * X_)
print((epsilon * X_).shape)
(6, 1)
(6, 3)
[[ 0 5 10]
[ 39 52 65]
[126 147 168]
[261 290 319]
[444 481 518]
[675 720 765]]
(6, 3)
We then sum through the examples to get the gradient array and reshape it to have a 2D column array (n+1, 1) that match the size of the current parameter vector estimate.
grad_ = np.sum((np.dot(X_, w_) - Y_) * X_, axis=0)
print('-'*10)
(3,)
----------
[[1545]
[1695]
[1845]]
## 5. Training
Here follows the complete code of the learning process, where the model parameters are changed for Nepoch times. The parameters and the corresponding loss function value are stored in associated lists and returned at the end of the function call.
def gradientDescent(XX, YY, ww, lr=0.1, Nepoch=1500):
Npnt = XX.shape[0]
Jevol, wevol = [], []
for _ in range(Nepoch):
Jevol.append(lossFunction(XX, YY, ww))
wevol.append(ww[:,0])
ww = ww - lr/Npnt * np.sum((np.dot(XX, ww) - YY) * XX, axis=0).reshape(-1,1)
return np.array(wevol), np.array(Jevol)
wOpt and Jopt are the optimal parameter values and the minimum loss that are stacked at the bottom of the evolution lists. The final values for wOpt are very close to the ones used to generate the data.
J = lossFunction(XX, YY, ww)
Nepoch, lr = 5000, 0.005
wEvol, Jevol = gradientDescent(XX, YY, ww, lr, Nepoch)
wOpt, Jopt = wEvol[-1,:], Jevol[-1]
print('optimal weights: ' + str(wOpt))
print('optimal loss: ' + str(np.log(Jopt)))
optimal weights: [ 4.73739369 -2.99593775 -0.99554761]
optimal loss: [-3.43847623]
The below figure compares the plane with optimal parameters, wOpt, to the cloud of points used to train the model.
ypred = wOpt[0] + wOpt[1]*xx1 + wOpt[2]*xx2
plt.figure(figsize=(10, 5))
ax = plt.axes(projection='3d')
ax.plot_surface(xx1, xx2, ypred, rstride=1, cstride=1, cmap='viridis', edgecolor='none', alpha=0.5)
ax.scatter(xx1, xx2, yy, cmap='viridis', linewidth=0.5, alpha=0.5)
ax.set_xlabel('$x_1$')
ax.set_ylabel('$x_2$')
ax.set_zlabel('y')
ax.view_init(20, 30)
plt.tight_layout()
plt.show()
## 6. Parameter evolution
Here we plot the evolution of the parameters $\omega$ over the Nepoch training steps. The initial and final values are depicted with the green and red points, respectively. The size of each intermediate blue point is proportional to the loss function, where the smaller the point, the lower the loss, the better the model performance.
plt.figure(figsize=(10, 5))
ax = plt.axes(projection='3d')
score = Jevol/np.max(Jevol)
ax.scatter(wEvol[:,0], wEvol[:,1], wEvol[:,2], c='b', s=10+100*score, alpha=0.5, label='Weight evolution')
ax.plot3D(wEvol[:1,0], wEvol[:1,1], wEvol[:1,2], 'g', marker='o', markersize=10, alpha=0.75, label='Initial weights')
ax.plot3D(wEvol[-1:,0], wEvol[-1:,1], wEvol[-1:,2], 'r', marker='o', markersize=10, alpha=0.75, label='Optimal weights')
ax.set_xlabel("$\omega_0$")
ax.set_ylabel("$\omega_1$")
ax.set_zlabel("$\omega_2$")
plt.legend()
#ax.view_init(10, 60)
plt.tight_layout()
plt.show()
In the following plot, we compare the evolution of the parameters $\omega$ with the loss function 2D map. To this end, we need to create a grid for every combination of $\omega_0$, $\omega_1$ and $\omega_2$, using meshgrid.
The three 3D (12, 8, 6)-shaped arrays are then flattened and vertically stacked to obtain the parameter meshgrid, wmg, which contains $12*8*6=576$ three-element tuples. We want to calculate the loss function for each triple of parameters. The loss function requires a (3,1) array, so we need to reshape the 1D array that we get from for-iterating along the wmg rows. The Python list comprehension helps to perform this process that returns a list of 576 loss values corresponding to every point of the meshgrid. The final step is to convert the list into an array and reshape it into a 3D whose size is as equal as the initial 3D parameter arrays, w0mg, w1mg and w2mg.
step = 0.5
w0s = np.arange(-1, 5, step)
w1s = np.arange(-5, -1, step)
w2s = np.arange(-1, 2, step)
w0mg, w1mg, w2mg = np.meshgrid(w0s, w1s, w2s, indexing='ij')
wmg = np.vstack((w0mg.flatten(), w1mg.flatten(), w2mg.flatten())).T
print('array size:')
print('Size of the three 1D arrays: {}, {}, {}'.format(w0s.shape, w1s.shape, w2s.shape))
print('Size of the three meshgrid arrays: {}'.format(w0mg.shape))
print('Size of the parameter meshgrid: {}'.format(wmg.shape))
array size:
Size of the three 1D arrays: (12,), (8,), (6,)
Size of the three meshgrid arrays: (12, 8, 6)
Size of the parameter meshgrid: (576, 3)
w_ = np.array([-3.5, 5, 2]).reshape(-1, 1)
print('Random parameter vector: \n{}'.format(w_))
print('Loss value for that parameter vector: {}'.format(lossFunction(XX, YY, w_)))
Random parameter vector:
[[-3.5]
[ 5. ]
[ 2. ]]
Loss value for that parameter vector: [ 33.69217748]
Jlist = [lossFunction(XX, YY, wmg[kk,:].reshape(-1, 1)) for kk in range(wmg.shape[0])]
print(len(Jlist))
576
Jmap = np.array(Jlist).reshape(-1, w1s.shape[0], w2s.shape[0])
print(Jmap.shape)
(12, 8, 6)
The plot shows the parameter evolution on top of the contour plot of the 3D loss function, where the greater the loss value the warmer the colour.
plt.figure(figsize=(10, 5))
ax = plt.axes(projection='3d')
ax.scatter(w0mg.flatten(), w1mg.flatten(), w2mg.flatten(),
cmap='viridis', c=Jmap.flatten(), s=10+20*Jmap.flatten(), alpha=0.1, label='Loss function')
score = Jevol/np.max(Jevol)
ax.scatter(wEvol[:,0], wEvol[:,1], wEvol[:,2], c=score, s=10+100*score, alpha=0.95, label='Weight evolution')
ax.plot3D(wEvol[:1,0], wEvol[:1,1], wEvol[:1,2], 'k', marker='o', markersize=10, alpha=1, label='Initial weights')
ax.plot3D(wEvol[-1:,0], wEvol[-1:,1], wEvol[-1:,2], 'r', marker='o', markersize=10, alpha=1, label='Optimal weights')
ax.set_xlabel("$\omega_0$")
ax.set_ylabel("$\omega_1$")
ax.set_zlabel("$\omega_2$")
plt.legend()
#ax.view_init(10, 60)
plt.tight_layout()
plt.show()
This plot reports the logarithmic trend of the loss function over the training steps.
plt.figure(figsize=(10, 5))
plt.plot(np.log(Jevol), lw=2)
plt.xlabel("training steps ($N_{epoch}$)")
plt.ylabel("Logarithm loss trend ($log(J_{evol})$)")
plt.show()
The final plot wants to summarize the post by visualizing the training process at those steps where the logarithm of the loss function assumes integer values. We create a list, idxs, that stores the Nrow = 6+1 indices of the Jevol array that correspond to such training steps, in addition to the final training step.
idxs = [np.where(np.log(Jevol)<kk)[0][0] for kk in range(3, -3, -1)] + [Jevol.shape[0]-1]
Nrow = len(idxs)
Nstep = int(Nepoch/Nrow)
print(Nrow)
7
The chart has a matrix structure, with as many rows as the training steps to analyse, Nrow, and two columns, the left-most one that compares the data point cloud to the current model (colorful plane) and highlights the current loss (J value), the right-most one that shows the trajectory of the model parameters up to the current epoch within the 3D parameter space.
fig = plt.figure(figsize=(15, Nrow*5))
for kk, idx in enumerate(idxs):
ax = fig.add_subplot(Nrow, 2, 2*kk+1, projection='3d')
ypred_ = wEvol[idx,0] + wEvol[idx,1]*xx1 + wEvol[idx,2]*xx2
ax.plot_surface(xx1, xx2, ypred_, rstride=1, cstride=1, cmap='viridis', edgecolor='none', alpha=0.5)
ax.scatter(xx1, xx2, yy, cmap='viridis', linewidth=0.5, alpha=0.5)
ax.set_xlabel('$x_1$')
ax.set_ylabel('$x_2$')
ax.set_zlabel('y')
plt.title("Current J value: {0:.2f}".format(Jevol[idx, 0]))
ax.view_init(10, 60)
ax = fig.add_subplot(Nrow, 2, 2*kk+2, projection='3d')
ax.scatter(w0mg.flatten(), w1mg.flatten(), w2mg.flatten(),
cmap='viridis', c=Jmap.flatten(), s=10+20*Jmap.flatten(), alpha=0.1, label='Loss function')
score = Jevol[:idx]/np.max(Jevol)
ax.scatter(wEvol[:idx,0], wEvol[:idx,1], wEvol[:idx,2], c=score, s=10+100*score, alpha=0.95, label='Weight evolution')
ax.plot3D(wEvol[idx:idx+1,0], wEvol[idx:idx+1:,1], wEvol[idx:idx+1:,2], 'r', marker='o', markersize=10,
alpha=1, label='Current weights')
ax.set_xlabel("$\omega_0$")
ax.set_ylabel("$\omega_1$")
ax.set_zlabel("$\omega_2$")
plt.title("Current epoch: " + str(idx))
ax.legend()
plt.tight_layout()
plt.show()
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# Analysis Complex Analysis by Stein and Shakarchi
## For those who have used this book
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4. ### Strongly don't Recommend
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1. Jan 25, 2013
### micromass
Staff Emeritus
Code (Text):
[LIST]
[*] Foreword
[*] Introduction
[*] Preliminaries to Complex Analysis
[LIST]
[*] Complex numbers and the complex plane
[LIST]
[*] Basic properties
[*] Convergence
[*] Sets in the complex plane
[/LIST]
[*] Functions on the complex plane
[LIST]
[*] Continuous functions
[*] Holomorphic functions
[*] Power series
[/LIST]
[*] Integration along curves
[*] Exercises
[/LIST]
[*] Cauchy's Theorem and Its Applications
[LIST]
[*] Goursat's theorem
[*] Local existence of primitives and Cauchy's theorem in a disc
[*] Evaluation of some integrals
[*] Cauchy's integral formulas
[*] Further applications
[LIST]
[*] Morera's theorem
[*] Sequences of holomorphic functions
[*] Holomorphic functions defined in terms of integrals
[*] Schwarz reflection principle
[*] Runge's approximation theorem
[/LIST]
[*] Exercises
[*] Problems
[/LIST]
[*] Meromorphic Function and the Logarithm
[LIST]
[*] Zeros and Poles
[*] The residue formula
[LIST]
[*] Examples
[/LIST]
[*] Singularities and meromorphic functions
[*] The argument principle and applications
[*] Homotopies and simply connected domains
[*] The complex logarithm
[*] Fourier series and harmonic functions
[*] Exercises
[*] Problems
[/LIST]
[*] The Fourier Transform
[LIST]
[*] The class F
[*] Action of the Fourier transform on F
[*] Paley-Wiener theorem
[*] Exercises
[*] Problems
[/LIST]
[*] Entire Functions
[LIST]
[*] Jensen's formula
[*] Functions of finite order
[*] Infinite products
[LIST]
[*] Generalities
[*] Example: the product formula for the sine function
[/LIST]
[*] Weierstrass infinite products
[*] Exercises
[*] Problems
[/LIST]
[*] The Gamma and Zeta Functions
[LIST]
[*] The gamma function
[LIST]
[*] Analytic continuation
[*] Further properties of \Gamma
[/LIST]
[*] The zeta function
[LIST]
[*] Functional equation and analytic continuation
[/LIST]
[*] Exercises
[*] Problems
[/LIST]
[*] The Zeta Function and Prime Number Theorem
[LIST]
[*] Zeros of the zeta function
[LIST]
[*] Estimates for 1/\zeta(s)
[/LIST]
[*] Reduction to the functions \psi and \psi_1
[LIST]
[*] Proofs of the asymptotics for \psi_1
[/LIST]
[*] Note on interchanging double sums
[*] Exercises
[*] Problems
[/LIST]
[*] Conformal Mappins
[LIST]
[*] Conformal equivalence and examples
[LIST]
[*] The disc and upper half-plane
[*] Further examples
[*] The Dirichlet problem in a strip
[/LIST]
[*] The Schwarz lemma; automorphisms of the disc and upper half-plane
[LIST]
[*] Automorphisms of the disc
[*] Automorphisms of the upper half-plane
[/LIST]
[*] The Riemann mapping theorem
[LIST]
[*] Necessary conditions and statement of the theorem
[*] Montel's theorem
[*] Proof of the Riemann mapping theorem
[/LIST]
[*] Conformal mappings onto polygons
[LIST]
[*] Some examples
[*] The Schwarz-Christoffel integral
[*] Boundary behavior
[*] The mapping formula
[/LIST]
[*] Exercises
[*] Problems
[/LIST]
[*] An Introduction to Elliptic Functions
[LIST]
[*] Elliptic functions
[LIST]
[*] Liouville's theorems
[*] The Weierstrass P function
[/LIST]
[*] The modular character of elliptic functions and Eisenstein series
[LIST]
[*] Eisenstein series
[*] Eisenstein series and divisor functions
[/LIST]
[*] Exercises
[*] Problems
[/LIST]
[*] Applications of Theta Functions
[LIST]
[*] Product formula for the Jacobi theta function
[LIST]
[*] Further transformation laws
[/LIST]
[*] Generating functions
[*] The theorems about sums of squares
[LIST]
[*] The two-squares theorem
[*] The four-squares theorem
[/LIST]
[*] Exercises
[*] Problems
[/LIST]
[*] Appendix: Asymptotics
[LIST]
[*] Bessel functions
[*] Laplace method; Stirling's formula
[*] The Airy function
[*] The partition function
[*] Problems
[/LIST]
[*] Simple Connectivity and Jordan Curve Theorem
[LIST]
[*] Equivalent formulations of simple connectivity
[*] The Jordan curve theorem
[LIST]
[*] Proof of a general form of Cauchy's theorem
[/LIST]
[/LIST]
[*] Notes and References
[*] Bilbiography
[*] Symbol Glossary
[*] Index
[/LIST]
Last edited by a moderator: May 6, 2017
2. Jan 28, 2013
### mathwonk
I hope someone with more knowledge of this book will comment. I have very little, only having seen it briefly while supervising a graduate prep course in complex analysis before prelims. My impression then was that I was disappointed in it, given the name of the extremely eminent first author. Everything in it is excellent, but it seemed inadequate for students to learn from alone, or to get a really good feel for many sides of the subject. It was apparently notes from a lecture series possibly given only once.
The pedigree of the main author guarantees that the viewpoint is excellent, that there are interesting topics authoritatively treated that are not found elsewhere, the proofs are elegant, and that the insights offered are deep. The manner in which it was written though suggest it was not actually written by that main author, and that it may have been a bit of an experiment, and hence may not have received the repeated polishing that leads to a really outstanding text.
I would suggest that, maybe like the similarly generated physics books of Feynman, this is an outstanding series that everyone can benefit from and that is unique in its point of view, and that less qualified authors could not begin to imitate. However it may be hard to learn thoroughly the subject from only these books.
A remark: in the introduction there are three key principles of complex analysis mentioned briefly for the reader, namely contour integration, regularity, and analytic continuation. The hint of an idea about contour integration that is given says that for certain appropriate contours the integral of a holomorphic function is zero.
In my estimation this is not worth much to the reader since as stated it is true of all functions. I would suggest instead saying that for holomorphic functions, that contour integration is a "deformation invariant", i.e. if a contour is changed continuously the integral does not change.
Then, as a corollary, for integrals that can be continuously deformed to a point, the integral is the same as the integral over a point, namely zero. This last case is the usual homotopy version of the Cauchy theorem, but the main point is the deformation invariance, not the special case that comes out zero.
And now that I have tried to improve something from the book, you can be your own judge as to which version is more suitable for a text, maybe their simpler less informative one.
So I will guess that everyone could benefit from reading these books, but that one might well want some other standard sources around too.
Last edited: Jan 28, 2013
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# [java] Perfomance vs. improved readability (Escape analysis does not work!?)
This topic is 2997 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
## Recommended Posts
Hi, I need some help from serious java game developers. At the moment I am revisiting my matrix code. My previous version was basically an array of values. Unfortunately this is not the most intuitive way to deal with matrix programming. This time i tried a more OOP style:
public class NewMatrix3 {
private Vector3 x, y, z;
private Point3 w;
...
}
The matrix has now subobjects, which bring their own methods. From a programming point of view this makes lots of things easier (at least in my opinion =)) Particularily I tried this:
public Point3 multiply(Point3 p) {
Point3 r = new Point3(w);
Vector3 dx = x.multiply(p.getX());
Vector3 dy = x.multiply(p.getY());
Vector3 dz = x.multiply(p.getZ());
r.translate(dx); r.translate(dy); r.translate(dz);
return r;
}
The purpose of this method is to transform a point in space by the matrix. The method is unorthodox, but much more visual than plain row/column dot-product multiplication. The result point is translated into the origin of our coordinate system and a linear combination of the axis vectors is added to the origin point. So mathematically all should be right ?!! Hope so, but thats not the point. We have a method here with at least 3 local objects, the vectors dx, dy, dz, which are allocated inside the scalar multiply method of vector3. When I understand right, these are perfect candiates for escape analysis. Thats a thing I really dislike about java... code like this is simply not feasible from a performance standpoint (at least it was), but is perfectly fine in c++ ;-( I really would appreciate to write like this. So I tried this: -server -XX:+DoEscapeAnalysis with a small test program:
NewMatrix3 matrix = new NewMatrix3();
Point3 p = new Point3();
for(;;) {
p.set((float)Math.random() * 100f, (float)Math.random() * 100f, (float)Math.random() * 100f);
Point3 r = matrix.multiply(p);
System.out.println(r.getX());
}
Unfortunately there is NO difference to no escape analysis at all. The netbeans profiler reports a hell of lots of Vector3 objects that are allocated. But if I got this right, none should be allocated at all. I really appreciate some help. Because I like java and there are advantages to come later on in game programming, like good scripting support, continuations via javaflow, reflection, low compilation times, runtime code loading. But this sacrifice of readability in exchange for performance is really a pain. I am even thinking of porting my project to C++... :-( I would really appreciate help in this case. Btw. escape analysis seems to work somehow, as it lowered total object creation in my main program. So it does at least something. But i believe the example I provided should be the standard case for escape analysis and should provide a stellar speed up, shouldn't it ?! Thank you! Really desperate for help, Frederick
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ummpf....
i just tried the windows snapshot of jdk7. EA should be enabled by default, but again I have NOOOO results. I am getting a little frustrated...
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r.translate(dx);
return r;
Why do you think dx is not "escaping"?
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Hmm,
lets see
Quote:
The -XX:+DoEscapeAnalysis option directs HotSpot to look for objects that are created and referenced by a single thread within the scope of a method compilation.
dx is created withing the scope, thats "x.multiply(p.getX())"
dx is referenced within the scope, thats "r.translate(dx)"
Potentially dx could escape through "translate" but it is only accessed read only.
In this case the optimization should jump in. I missed to provide that code snipped.
If an object "escapes" and is read-only accessed, HotSpot might inline all the code in the method and eliminate the object creation.
You may check this here:
http://java.sun.com/javase/7/docs/technotes/guides/vm/performance-enhancements-7.html
The thing is I tried to run the example from this page, but escape analysis didn´t jump in either. I have some code snippets now, that are more clearly:
public void rotate(float angle, Vector3 axis) { Matrix3 rotationMatrix = new Matrix3(); rotationMatrix.loadAxisAngleRotation(angle, axis); rotationMatrix.transformOnlyOrientation(this);}
You may see now the "rotationMatrix" isn´t even passed as a parameter, so there is no hidden magic this time, but the problem remains. The above example produces lots of object allocations.
I would really enjoy this to work, don´t want to use static methods or hacks like that.
Here is the "defensive copy" example from the page:
class Matrix3 { private Vector3 x, y, z; private Point3 pos Vector3 getX() { //Defensive copy return new Vector3(x); }}
You see the matrix has subobjects for its axes and these are returned "safely" in order to prevent manipulation from the outside. The link states this should be optimized, but it isn´t.
I would be very grateful if anybody could check that and give me advice how to enable the feature. The VM version I need and the parameters. According to the link it should work, but it doesn´t.
Btw. I am using the -Xloggc mechanism and the netbeans memory profiler to check if EA jumps in. Please tell me if anything might be wrong with that.
my java version is:
java version "1.7.0-ea"
Java(TM) SE Runtime Environment (build 1.7.0-ea-b91)
Java HotSpot(TM) Client VM (build 18.0-b03, mixed mode, sharing)
Thank you a lot for your attention,
Frederik
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Escape analysis will always be just optional run-time optimization. It may kick in, or it might not.
If performance is important, use different style. If performance is optional and source code is what matters most, then use OO style. If performance is paramount, use C# or C++.
Java, due to the design of its memory model, is worst possible case for this type of operations and will always remain as such. The biggest problems do not come from temporaries, but from lack of in-place array allocations.
For performance sensitive processing, matrices and vectors would be organized in arrays and processed sequentially to maximize cache locality. This isn't possible in Java, so that is a factor 8-20 performance hit vs. C/C++.
Java is simply a wrong abstraction for this type of problems.
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Quote:
code like this is simply not feasible from a performance standpoint ..., but is perfectly fine in c++
[irony]Hmm, interesting.
How much difference in performance?[/irony]
Quote:
You see the matrix has subobjects for its axes and these are returned "safely" in order to prevent manipulation from the outside. The link states this should be optimized, but it isn´t.
The link doesn't say that. It says that "if the compiler determines that the original object is never modified, it may optimize and eliminate the call to make a copy".
Quote:
dx is referenced within the scope, thats "r.translate(dx)"Potentially dx could escape through "translate" but it is only accessed read only.In this case the optimization should jump in. I missed to provide that code snipped.
But that implies that the escape analysis has to be performed through n-level indirection. I am not sure about that. I cannot deduct that from the link. Can you give me any hint?
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Quote:
If performance is paramount, use C# or C++.
If performance is paramount, use a proper algorithm.
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Quote:
Original post by ppgamedev
Quote:
If performance is paramount, use C# or C++.
If performance is paramount, use a proper algorithm.
The right man in the wrong place can make all the difference.
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Quote:
Original post by ppgamedev
Quote:
If performance is paramount, use C# or C++.
If performance is paramount, use a proper algorithm.
Yep. Optimal data-centric algorithm of this type is 8-20 times slower in Java than in any language that allows control over memory allocations. Usually C or C++, but Fortran would fit the bill as well, it's just a bit hard to find developers for it.
I speak from experience, when running a cluster of machines, doing some memory heavy processing, Java's lack of continuous allocation prevents certain algorithmic optimizations. 8-20 times may not seem much, but think of it this way - instead of needing half a rack in a data center, a single quad core will do. Data centers also make it difficult to do real-time graphics, when dealing with tens of millions of vertices.
In Java, lack of cache locality kills this type of algorithms, and there is no way around it. Pooling and such prevents recycling, but it doesn't allow sequential ordering of elements.
Escape analysis under 1.6 has been shown to provide around factor of 15 performance improvement in best case - then there is still that 8-20 factor.
Until Java introduces C# struct-like type, it's simply not economically viable for this type of tasks.
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http://hal.inria.fr/docs/00/31/20/39/PDF/RT-0353.pdf
Well, in this report the difference with Fortran is not so dramatic.
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# The Case for Impact Purchase | Part 1
post by Linda Linsefors · 2020-04-14T13:08:48.664Z · score: 50 (26 votes) · EA · GW · 37 comments
## Contents
What is impact purchase?
Reasons to evaluate a project after it is completed
Why would you (or any EA funder) pay for something that has already happened?
My story (skip if you want)
Some concerns and my responses
When not to go for impact purchase
None
In this post I will argue for impact purchase as a complement to (not a replacement for) other funding mechanisms, such as grants.
## What is impact purchase?
Impact purchase is an alternative funding mechanism for altruistic work. I think the easiest way to explain what impact purchase is, is to compare it with grants.
For a more thorough explanation, and an actual example, read this excellent blogpost.
This is part one of what I intend to be a two part blog post series. The second part will be significantly more work, so I’ll do that if this one gets enough interest.
Part 1 (this blogpost) will focus on the timing aspect of impact purchase, i.e. why it is sometimes better to evaluate and fund a project after it is finished.
Part 2 (next blogpost) will focus on the whys and hows of paying for value rather than cost. Most of my suggestions regarding implementation will be in this post.
Disclaimer
So far, my only investigation into this previous attempt to implement impact purchase, was to look around at their now inactive website. I don’t even know why the project ended. If someone could supply more information about this in the comments, that would be great. Otherwise, I will try to find out more for part 2.
Comment request
I am arguing a lot from my own experience and perspective. I hope that other people who want impact purchase to become a thing, will add their stories in the comments, or even just indicate that this is something you would apply for. I think this is relevant both to see how widespread the interest for this is, and what sort of projects it would support.
## Reasons to evaluate a project after it is completed
I think the best part of impact purchase is that projects are evaluated after they are done.
You might think that this would be great for the funder (because it is much easier for them to know what they are paying for) but less great for potential recipients (because they now get to take all the risk). If that is how you think, then this should surprise you: I’m a potential recipient and I love the idea of having the funding application happen after the project. I don’t speak for everyone (and this is why we should still have grants) but I do know several other potential recipients who share my preferences.
More flexibility
For me, the time period between [I have a plan], and [I start work on this project], is typically somewhere in the range negative one month to a few weeks. By negative time, I mean that I start working before having a proper plan.
Having a plan is, in my experience, absolutely necessary to get a grant, but getting a grant takes months. There is typically a few months waiting time until the next suitable grant application opens. But even if I’m lucky with that, I’ll have to wait about two more months for grant evaluation. I’m not waiting around for that, and I would be very surprised if this problem was unique to me.
With impact purchase I can just follow the schedule that works best for me and my projects, and have it evaluated and funded some time in the future.
It is easier to evaluate a project after it happened
I have not done grant evaluation, but my impression is that it is super hard. Evaluating the impact of a project after it has happened is probably not easy either, but it has to be easier. This gain can be used to save time, do better evaluations, give more feedback to applicants, or some combination thereof.
If EA is vetting constrained [EA · GW], the gains from evaluating some projects after rather than before they happen could be hugely beneficial.
## Why would you (or any EA funder) pay for something that has already happened?
However, you might think that paying after the project is over, is actually a bad deal for the funder, because you want counterfactual impact. You want your donation to create value that otherwise would not have happened. So why would you agree to pay for something that has already happened? Here are two ways of looking at it
Empower people based on their track record
The impact seller is someone who recently did something valuable with their time (because otherwise they would not have impact to sell), which is strong evidence that they are the kind of person who would do valuable things in the future. So let's give them some money, no strings attached, and let them do whatever they want. Based on past evidence it will probably be good.
Under this motivation, the agreement is still that you pay people based on their past work (if you do something else it is not an impact purchase). But as a donor you are doing this with the expectation of more good work from them in the future.
Incentivise future altruistic work
The prospect of getting paid may motivate people to do good work. By paying some people for their past work, you build expectations and trust in the system, which will motivate more good work.
The difference in motivation matters for implementation
I think any donor needs to be clear about which of these reasons (or other) is motivating you, because your implementation would probably be somewhat different. If you want to empower great people, you may want to look more at who they are and what you expect them to do next. If you want to incentivise future work, it is important that your payouts are predictable.
## My story (skip if you want)
Read this as a case story. Impact purchase should obviously not be implemented just for me. I would not write this post if I did not know others who would benefit too. But maybe using myself as an example will clarify some things? I don’t know.
I’m an AI Safety organiser. Last year I organised the Technical AI Safety Unconference [? · GW] (TAISU) and the Learning-by-doing AI Safety Research workshop [? · GW] (LAISR). Both of them were successful [LW · GW] and I’d like to run more of these events.
However, this year I’m only planning to run TAISU and not LAISR and it has to do with money. I think that LAISR is potentially very valuable and I would love to keep developing this concept, but the target participants are mostly students, who are not able to pay much for the event. The only way I can get paid for LAISR is with external funding.
Last year I lived at EA hotel, and also used this place as an event venue, so money was not a problem. But I moved out now so I need to start caring about actually getting paid. I still have a lot of free time, so if impact purchase were a thing, I could experiment around with various projects, get paid for the ones that succeed, and learn from that.
I could try applying for a grant, but I’m not going to do that. I have had enough of getting turned down for grants. Maybe you think I should take the fact that I can’t get grants as a signal of something. But given that I mostly succeed, when I get to work on projects on my own terms, I take it as a signal that the grant application process is a bad match for my working style.
I imagine some reader of this post wanting to tell me that I too can learn the skill of applying for grants, and writing up a proper project plan in advance would be a valuable thing to do anyway. If that is you, please trust me when I say that this would not work for me. See the bit about flexibility earlier in this post. It would hurt both my wellbeing and productivity to try to conform to a grant format.
I need to know that a project will happen in order to have the motivation to engage in planning. One of the first things I do when organising an event is to write up the event invitation. I think about what information needs to be in the announcement (dates, venue, cost, etc) and work from there. Anything that does not need to go in the announcement, I’ll figure out later. This process works for me, because it works with the structure of my motivation, not against it. But starting with a grant application triggers no motivation for me.
The one thing I do like about grants is the invitation to ask for money. I have considered trying to get some sort of peer funding (Patreon or similar), but find it incredibly uncomfortable to ask for money unprompted (though I rather do that then grants at this point). I’m hoping that if impact purchase becomes a thing there will be some formal application system.
## Some concerns and my responses
Grants provide good feedback, and help people not waste their time on bad projects.
I’ve heard this type of argument from many people when discussing grants. There seems to be this idea that applying for grants is a great source of feedback.
In my personal experience, the information value from applying for grants has been close to zero. Some reasons I personally have been turned down for grants:
• I did not have a budget. Because of this my project did not get evaluated. (I could not provide a budget since I did not at that time know what my living costs would be, and therefore could not say what salary I needed.)
• I applied for too little money. The lower bound was 10k and I only needed 5k. Because of this my project did not get evaluated.
• My project was related to EA Hotel, and no grant making body ever makes up their mind about EA Hotel. Because of this my project did not get evaluated.
I’m not saying none of these problems were my fault. I am not saying I did not learn anything (I did learn how not to write grant applications). Also, I would not be surprised if other people have had very different experiences. I’m only saying that realistically speaking, in the real world, grants are not a reliable way to get project feedback. But even if it were, this would still not be a very strong argument, because it is possible to just ask for feedback on a project idea.
However, I admit that there is something very special and real about feedback that comes in the form “someone is willing to pay for this”. I want this type of feedback! This is actually one of the reasons I want impact purchase. But I’m fine with trying something out first and having it evaluated after the fact, because that would be less of a cost for me than conforming to whatever is needed to get a grant. This is not just speculation, to this point I have infinitely more success in completing projects than navigating grant applications, and I’d love to have some of my past projects evaluated from a funding perspective.
What about reputation risks and other unilateralist curse type of stuff?
Another thing I have come across is the idea that we need to make people apply for grants from central organisations, in order to police what project people are working on.
Given that people in the EA movment are already doing lots of stuff without central oversight, I’m not convinced that this is something we need to worry about. Or maybe the current situation is already a ticking bomb, and introducing impact purchase will make it worse? I don’t think so, but it is a valid concern.
Either way, I do think it is a good idea to avoid buying impacts from projects which have had a large downside risk. I think the evaluation for deciding to buy impact or not, should be at least partly based on: “Do we approve of this work?”, which include things like downside risk and negative externalities.
## When not to go for impact purchase
These are the reasons I’m suggesting impact purchase as a complement, not as a replacement for other funding mechanisms. Removing grants would be bad.
When the money is needed up front
Not everyone can afford to run their project without payment in advance.
When the applicant can’t afford not getting paid
Even if a person has the cash at hand to do their project, it might still be a very bad idea for them to spend that money without any guarantee of reimbursement. Grants are safer and some people will need that safety.
## Some initial thoughts about implementation
I will discuss implementation more in Part 2, when I have had more time to sort out my own thoughts regarding paying for value. However here’s one thing I think I can say already:
Consistency is important
Impact purchase will have a much stronger effect on people's behaviours if potential recipients can either predict or have the possibility to learn what outcomes will get funded.
Imagine someone did a project which you like and you would like, which leads you to give them some money. Will this encourage this person (and maybe others) to do more of the same type of projects? Well maybe. Is a past impact purchase strong evidence that the same project will get rewarded in the future? If “yes” your money will much more strongly encourage similar things in the future.
“If I keep this up, I’ll keep getting funded” creates a form of financial stability that can potentially be very powerful. Impact purchases will never be a perfectly reliable source of income, but what sort of employment is these days? The more uncertain the payments from impact purchases are, the more people living off this will be incentivised to put their rewards into personal savings accounts, rather than use the money for their next cool thing.
comment by G Gordon Worley III (gworley3) · 2020-04-14T15:42:46.673Z · score: 15 (9 votes) · EA(p) · GW(p)
• I think prizes suffer from only allowing the most risk-tolerant to be incentivized by them since there is generally an aspect of competition in them and the winner often takes all or most of the prize funds.
• Impact purchases seem like an improvement over this if you set it up like a grant that pays at the end rather than the beginning, so it's tied to a single project/team and not a competition.
• There might be hybrid model possible where a certain amount of funds are granted at the start of the project to cover costs and additional funding is awarded only as certain project milestones are hit, up to and including completion of the project. Some of this completion money is for awarding impact and not just funding the next phase of the project, as would be the case in a grant, with most of the impact award money held back until the end.
• This lets me imagine funding something at like 20% the value of its impact up until it is created at which point I pay off the remaining 80% owed.
comment by Linda Linsefors · 2020-04-14T23:55:40.939Z · score: 5 (4 votes) · EA(p) · GW(p)
For me the most important consideration is flexibility, i.e. not having to wait for a grant comity to make up their mind before I can start. For this problem, the hybrid model is no better than a grant, unless it can speed up the application process by an order of magnitude.
Also, any major improvement in evaluation time also needs to be combined with running applications. Otherwise the applicant still have to wait a few months (on expectation) for the next grant round to come around. I guess there is a reason no major grant agency has running applications?
Let's say you have running applications that evaluation time is proportional to the amount of money (I have no idea of that is true). Then funding something 20% up front would take a 1/5 of the time to evaluate, which is not bad. But I'm not sure how useful that would be for the applicant. I think most applications will fall into one of two cases:
1) The applicants is fine with taking the risk of not getting paid anything
2) The applicant needs to know that the majority of the budget will get covered
For example, if I run an event online or at CEEALAR (formally EA Hotel), I'm ok with taking the risk of not getting paid, I'll just adjust my calculations for deciding if I want to run that event again. But if I run an event that has actual cost for me (other than my time), like travel and/or venue, then I need to know that those cost will be covered, 20% up front is probably not good enough.
But if an applicant is willing to put up with the hassle of applying for a grant (because they need the guaranteed money), then having some token amount depend on the outcome might be motivating. However, this also means that the grant maker need to evaluate the project twice, which takes even more time. But if I imagine myself as the recipient, I would very much welcome a post project evaluation from the grant maker, if this is something they want to do.
I think a improvement of this suggestion, is to cover any necessary cost in an initial grant (weather that be 0% or 90%). And offer an additional payment as a bonus if the project is successful. Where projects that request 0% in advance are "auto accepted" for the first half (which is £0). There might still be some point to pre-register projects with the grant makers, I think? Maybe they can say what metrics to track for the post evaluation? E.g. what questions they want in an event evaluation survey, and similar?
comment by Ben_West · 2020-04-22T22:48:53.100Z · score: 3 (2 votes) · EA(p) · GW(p)
You can also borrow against the future prize or impact purchase, e.g. as Goldman Sachs allows you to do (in some limited cases). This moves the risk onto diversified private investors.
comment by G Gordon Worley III (gworley3) · 2020-04-23T15:27:46.545Z · score: 3 (2 votes) · EA(p) · GW(p)
That's pretty cool, but it seems mostly focused around supporting government agencies using this as an alternative funding mechanism to save money or defer costs or avoid paying for undelivered services, thus improving government spending efficiency. I wonder what it would take to develop that into something that would support a wider range of funding sources? Seems like something someone with some expertise and experience in finance could potentially pioneer as a neglected way to generally support EA.
comment by gavintaylor · 2020-04-15T22:06:30.486Z · score: 13 (8 votes) · EA(p) · GW(p)
I'm interested in seeing a second post on impact purchases and would personally consider selling impact in the future. I have a few general comments about this:
• Impact purchases seem similar to value-based fees that are sometimes used in commercial consulting (instead of time- or project-based fees) and may be able to provide a complementary perspective. Although in business the 'impact' would usually be something easy to track (like additional revenue) and the return the consultant gets (like percentage of revenue up to a capped value) would be agreed on in advance. I wonder if a similar pre-arrangement for impact purchase could work for EA projects that have quantifiable impact outcomes, such as through a funder agreeing to pay some amount per intervention distributed, student educated, etc. Of course, the tracked outcome should reflect the funders true goals to prevent gaming the metric.
• It seems like impact purchases would be particularly helpful for people coming into the EA community who don't yet have good EA references/prestige/track-record but are confident they can complete an impactful project, or who want to work on unorthodox ideas that the community doesn't have the expertise to evaluate. If they try something out and it works then they can get funds to continue and preliminary results for a grant, if not, it's feedback to go more mainstream. For this dynamic to work people should probably be advised to plan relatively short projects (say a up too few months), otherwise they could spend a lot of time on something nobody values.
• This could be a particularly interesting time to trial impact purchases used in conjunction with government UBI (if that ends up being fully brought in anywhere). UBI then removes the barrier of requiring a secure salary before taking on a project.
• From my experience applying to a handful of early-career academic grants and a few EA grants, I agree that almost none provide any/useful feedback (beyond accepted or declined), either for the initial application or for progress or completion reports. However, worse than having no feedback is that I once heard from an European Research Council (ERC) grant reviewer that their review committees are required to provided feedback on rejected applications, but also instructed to make sure the feedback is vague and obfuscated so the applicant will have no grounds to ask for an appeal, which means the applicant gets feedback the reviewers know won't be useful for improving their project... Why do they bother???
• With regards to implementation. I think one point to consider is the demand from impacters relative to funds of purchasers. At least in academia, funding is constrained and grant success rates are often <20%, and so grantees know that it is unlikely they'll get a grant to do their project (academic granters often say they turn away a lot of great projects they want to fund). If impact purchasers were similarly funding constrained relative to the number of good projects, I think the whole scheme would be less appealing as then even if I complete a great project, getting its impact bought would still involve a bit/lot of luck.
• These posts about impact prizes [EA · GW] and altruistic equity [EA · GW] may also be of interest to consider.
comment by G Gordon Worley III (gworley3) · 2020-04-23T15:33:40.561Z · score: 3 (3 votes) · EA(p) · GW(p)
This could be a particularly interesting time to trial impact purchases used in conjunction with government UBI (if that ends up being fully brought in anywhere). UBI then removes the barrier of requiring a secure salary before taking on a project.
Impact purchases + EA Hotel seems like a match made in heaven. EA Hotel even talks about taking a hits-based approach, so having a pool of funds to award both EA Hotel (or whatever it's new name is) and the persons staying at the hotel who did the work that earned the funding sounds like a pretty interesting idea!
comment by Linda Linsefors · 2020-04-25T16:16:04.606Z · score: 10 (4 votes) · EA(p) · GW(p)
Update:
I have changed my mind quite a bit since writing this blogpost. The updates are coming from the discussions with you in the comments, so thanks for everyone discussing with me.
Everything in this comment are still work in progress. I'll write something more formal and well though through later, when I have a more stable opinion. But my views have already change enough so that I wanted to add this update.
------------------------------------------------
What I actually want there to be is some sort of trust based funding. If I proven my self enough (e.g. by doing good work) then I get money, and no questions asked. The reason I want this is becasue of flexibility (see main post).
Giving away money = Giving away power
Impact perches has the neat structure that if I done X amount of good I get X amount worth of trust (i.e. money). This seems to be the exact right amount, because it is the most you can give away and still be protected from exploitation. If someone who are not aligned with the goal of the funder tires to use impact purchase as a money pump, they still have to do an amount of good equal to the payout they want.
But...
Khorton [EA · GW]:
A project to project lifestyle doesn't seem conducive to focusing on impact.
We actually know this form an other field. In most of academia, the law of the land is publish or perish. Someone living of impact purchases will face a similar situation, and it is not good, at least not in the long run.
Halffull [EA · GW]
I think the high impact projects are often very risky, and will most likely have low impact.
To the extent that this is true, impact purchase will not work.
In theory we could have impact investors, who funds a risky project and earn money by selling the impact of the few projects which impact reached the stars (literally and/or figuratively). But this requires an other layer which may or may not happen in reality (probably won't happen). Also, from the perspective of the applicant, how is this any different from applying for a grant? So what have we gained?
If not impact purchase, then what?
I still would like to solve the problem of inflexibility that grants have. An actually I think the solutions already exist (to some extent).
1) Get a paid job, with high autonomy.
2) Start an organisation and fundraise. I did not think of this until now, but when orgs fundraise, they typically don't present a plan for what they will do with the money. They mainly point towards what they have done so far, and ask for continued trust.
3) ...? I'd be very interested in other suggestions. I would not be surprised if there are other obvious things I have missed.
There are also other solutions that don't exist yet (or not very much) in EA, but could be implement by any institution or person with spare money:
a) "Trusted person"-job: A generic employment you offer to anyone who you like to keep up the good work, or something like that.
b) Support people on Ko-fi or Patreon, or similar, and generally encourage this behaviour from others too. (I know this is happening already, but not enough for people to make a living.)
comment by Linda Linsefors · 2020-04-25T16:54:15.721Z · score: 5 (3 votes) · EA(p) · GW(p)
I) I found out what happened to impactpurchase.org
Paul Christiano (from privet email, with the permission to quote):
Basically just a lack of time, and a desire to focus on my core projects. I'd be supportive of other people making impact purchases or similar efforts work, I hope our foray into the space doesn't discourage anyone.
II) Justin Shovelain told me (and gave me permission to share this information) that he would probably have focused more on Coronavirus stuff early on, if he though there where a way to get paid for this work.
This is another type of situation where grants are too slow.
comment by Sanjay · 2020-04-21T10:09:17.254Z · score: 8 (3 votes) · EA(p) · GW(p)
I'm confused why there are several people apparently excited about this idea. It strikes me that impact finance so obviously outperforms impact purchases. So much so that I'm worried that I must be misunderstanding impact purchases.
Impact finance solves both of the stark problems of impact purchases:
(a) if the people doing the work only receive money after the event, how do they live?
(b) what incentive do purchasers of impact certificates have to purchase?
comment by Sanjay · 2020-04-22T14:41:40.868Z · score: 14 (8 votes) · EA(p) · GW(p)
I perhaps need to clarify what I meant by Impact Finance. Given that this risks being confused with Impact Investment, I'll rename it to Conditional Impact Finance.
Conditional Impact Finance occurs when a Project receives funding involving a Conditional Donor and an Impact Investor as follows:
(1) The Conditional Donor (who may be a pool of donors) agrees to fund the Project *on the condition* that they achieve a certain outcome
(2) The funds needed + a bonus go into escrow. If the outcome is achieved the funds are paid out, if not they are returned to the donor.
(3) In order that the Project receives the funds it needs to proceed with the work, it receives funding from an Impact Investor (or a pool of investors).
Further notes:
(1) Examples of conditions (fictional):
"The ABC innovative education Project will receive funding if a longitudinal RCT shows that employment rates among beneficiaries are at least 15% higher than control in 15 years' time"
"The DEF malaria Project will receive funding if the malaria-linked mortality rate in Busia, Kenya is better than a control under a RCT conducted by LSHTM"
(2) A bonus is needed primarily to pay the interest on the debt (although perhaps a bonus for the staff might be a good idea too)
(3) If the project fails, the impact investor loses out. This means that all the incentives that currently apply to financial markets now apply to impact.
I consider this better for any reasonably sized project. If we are thinking of just one person doing some work on their own, I could imagine Impact Purchase maybe having some value.
I believe this is what is being done by alice.si
This is very similar to this notion of impact prizes [EA · GW]. The main difference there seems to be that there is a specific allotted sum of money for a variety of possible possible projects, which share that allotted amount proportionally to their estimated impact.
I think that the downside of impact prizes compared to Conditional Impact Finance is mainly that it is much more volatile for investors - both because of dependencies between different projects and somewhat due to the continuum of possible values of estimated impact. Also, it is much harder on the donors. Well, there is also the problem that it may be clear that other competing projects are closing in on something much better (9x is enough to limit the prize to 10% of the original amount), and also competing interests between projects.
The major upside of impact prizes seems to be that the incentives of the project is better aligned with maximizing impact because they get a prize which scales sort of linearly with impact (unless they are enormously successful).
comment by richard_ngo · 2020-04-21T16:26:27.377Z · score: 9 (7 votes) · EA(p) · GW(p)
Impact purchases are one way of creating more impact finance. In particular, they can make it worthwhile for non-altruistic financiers to fund altruistic projects. This is particularly beneficial in cases where it's hard for a single altruist to evaluate all the people who want funding.
With regard to (b), the incentives for impact purchasers are roughly similar to the incentives of someone who's announced a prize. In both cases, the payer create incentives for others to do the work that will lead to payouts.
comment by Linda Linsefors · 2020-04-22T19:05:31.836Z · score: 1 (1 votes) · EA(p) · GW(p)
I'm confused by this response. I answered all of this in the blogpost. Did I fail to communicate? I am not saying that you have to agree, but if you read what I wrote and still don't understand why *I* think some times paying after a project is a good idea, that is confusing to me, and I would like to understand better what part of the blogpost you found confusing.
comment by casebash · 2020-04-19T06:38:03.772Z · score: 8 (5 votes) · EA(p) · GW(p)
I wrote up my thoughts in this post: Making Impact Purchases Viable [EA · GW]. Briefly, I argue that:
• Restrictions on sellers are necessary to fix and imbalance between the size of the buyer's market and seller's market
• Restrictions are also important for providing sufficient incentive for a reasonable proportion of the impact to be counterfactual
• Another issue I don't have a solution to is the potential of impact purchases to lead to bitterness or demotivate people
comment by Khorton · 2020-04-15T20:44:48.834Z · score: 7 (5 votes) · EA(p) · GW(p)
What's the difference between an Impact Purchase and a prize? For example, what would be the difference between Impact Purchases for good quality blog posts and the EA Forum prize?
comment by Linda Linsefors · 2020-04-16T10:32:12.789Z · score: 9 (7 votes) · EA(p) · GW(p)
Prices and impact purchase are very similar. I would say that impact purchase would be an improvement though.
For an impact purchase the amount of money is decided based on how good impact of the project was. For a price, the price money usually set in advance, and there is often a winner takes it all dynamic.
Prices feels aversive to me becasue I win by being better than others, which means I'm disincentivised to help. This is fine whenever I don't expect to win anyway, like with the forum prices. Very few blogpost gets awarded. Because the base rate is low I don't feel like I'm loosing much by helping others. But if I'm trying to make a living out of selling impacts more regularly, I would not want this comparative aspect.
Although, it is not that simple, becasue there is inevitably competition in the market for selling impact. So in practice maybe it is not so different? I'm honestly a bit confused about if I think there is a real difference or not.
Maybe the difference is that a price often have a narrower focus, which pushes the competition to be between very similar work.
I would hate there to be an EA events price, because that would make me view fellow organisers as my competition, and those are exactly the people I should exchange experience and advise with. It would be much less bad to compete on "team organisers" against everyone else, about who can have the biggest overall impact.
comment by Halffull · 2020-04-16T22:08:17.376Z · score: 1 (1 votes) · EA(p) · GW(p)
For an impact purchase the amount of money is decided based on how good impact of the project was
I'm curious about how exactly this would work. My prior is that impact is clustered at the tails.
This means that there will frequently be small impact projects, and very occasionally be large impact projects - My guess is that if you want to be able to incentivize the frequent small
impact projects at all, you won't be able to afford the large impact projects, because they are many magnitudes of impact larger. You could just purchase part of their impact, but in practice this means that there's a cap on how much you can receive from impact purchase.
Maybe a cap is fine, and you know that all you're ever get from an impact purchase is for instance \$50,000, and the prestige comes with what % of impact they bought at that price.
comment by Linda Linsefors · 2020-04-17T17:30:25.980Z · score: 1 (1 votes) · EA(p) · GW(p)
Lest assume for now that impact is clustered as the tails.
(I don't have a strong prior, but this at least don't seem implausible to me)
Then how would you like to spend funding? Since there will limited amour of money, what is your motivation for giving the low impact projects anything at all?
Is it to support the people involved to do keep working, and eventually learn and/or get lucky enough to do something really important?
comment by Halffull · 2020-04-18T23:09:18.144Z · score: 2 (2 votes) · EA(p) · GW(p)
Since there will limited amount of money, what is your motivation for giving the low impact projects anything at all?
I'm not sure. The vibe I got from the original post was that it would be good to have small rewards for small impact projects?
I think the high impact projects are often very risky, and will most likely have low impact. Perhaps it makes sense to compensate people for taking the hit for society so that 1/1,000,000 of the people who start such projects can have high impact?
comment by Linda Linsefors · 2020-04-19T16:46:55.030Z · score: 7 (3 votes) · EA(p) · GW(p)
I'm not sure. The vibe I got from the original post was that it would be good to have small rewards for small impact projects?
I'm unsure what size you have in mind when you say small.
I don't think small monetary rewards (~£10) are very useful for anything (unless lots of people are giving small amounts, or if I do lot that add up to something that matters).
I also don't think small impact projects should be encouraged. If we respect peoples time and effort, we should encourage them to drop small impact projects and move on to bigger and better things.
I think the high impact projects are often very risky, and will most likely have low impact.
If you think that the projects with highest expected impact also typically have low success rate, then standard impact purchase is probably not a good idea. Under this hypothesis, what you want to do is to reward people for expected success rather than actual success.
I talk about success rather than impact, because for most project, you'll never know the actual impact. By "success" I mean your best estimate of the projects impact, from what you can tell after the project is over. (I really meant success not impact from the start, probably should have clarified that some how?)
I'd say that for most events, success is fairly predictable, and more so with more experience as an organiser. If I keep doing events the randomness will even out. Would you say that events are low impact? Would you say events are worth funding?
Can you give an example of the type of high impact project you have in mind? How does your statement about risk change if we are talking about success instead?
comment by Halffull · 2020-04-20T21:05:45.120Z · score: 1 (1 votes) · EA(p) · GW(p)
Would you say that events are low impact?
I think most events will be comparatively low impact compared to the highest impact events. Let's say you have 100,000 AI safety events. I think most of them will be comparatively low impact, but one in particular ends up creating the seed of a key idea in AI safety, another ends up introducing a key pair of researchers that go on to do great things together.
Now, if I want to pay those two highest impact events their relative money related to all the other events, I have a few options:
1. Pay all of the events based on their expected impact prior to the events, so the money evens out.
2. Pay a very small amount of money to the other events, so I can afford to pay the two events that had many orders of magnitude higher impact.
3. Only buy a small fraction of the impact of the very high impact events, so I have money left over to pay the small events and can reward them all on impact equally.
comment by Linda Linsefors · 2020-04-22T19:08:42.388Z · score: 1 (1 votes) · EA(p) · GW(p)
Whait what?
100 000 AI Safety Events?
Like 100 000 individual events?
There is a typo here right?
comment by Halffull · 2020-04-23T01:20:59.635Z · score: 1 (1 votes) · EA(p) · GW(p)
Nope, 1/50,000 seems like a realistic ratio for very high impact events to normal impact events.
comment by Linda Linsefors · 2020-04-22T19:22:42.856Z · score: 1 (1 votes) · EA(p) · GW(p)
It can't take more that ~50 events for every AI Safety researcher to get to know each other.
And key ideas are not seeded at a single point in time, it is something that comes together from lots of reading and talking.
There is not *the one event* that made the different and all the others where practically useless. That's not how research work. Sure there are randomness and some meetings are more important than others.
But if it took on average 50 000 events for one such a key introduction to happen, then we might as well give up on having events. Or find a better way to do it. Otherwise we are just wasting everyone's time.
comment by Halffull · 2020-04-23T02:01:04.135Z · score: 4 (1 votes) · EA(p) · GW(p)
But if it took on average 50 000 events for one such a key introduction to happen, then we might as well give up on having events. Or find a better way to do it. Otherwise we are just wasting everyone's time.
But all the other events were impactful, just not compared to those one or two events. The goal of having all the events is to hopefully be one of the 1/50,000 that has ridiculous outsized impact - It's high expected value even if comparatively all the other events have low impact. And again, that's comparatively. Compared to say, most other events, an event on AI safety is ridiculously high impact.
It can't take more that ~50 events for every AI Safety researcher to get to know each other.
This is true, much of the networking impact of events is frontloaded.
comment by Halffull · 2020-04-23T14:07:11.451Z · score: 1 (1 votes) · EA(p) · GW(p)
I happen to think that relative utility is very clustered at the tails, whereas expected value is more spread out.. This comes from intuitions from the startup world.
However, it's important to note that I also have developed a motivation system that allows me to not find this discouraging! Once I started thinking of opportunities for doing good in expected value terms, and concrete examples of my contributions in absolute rather than relative terms, neither of these facts was upsetting or discouraging.
Some relevant articles:
comment by Linda Linsefors · 2020-04-25T14:32:32.879Z · score: 1 (1 votes) · EA(p) · GW(p)
I'm ok with hit based impact. I just disagree about events.
I think you are correct about this for some work, but not for others. Things like operations and personal assistant are multipliers, which can consistently increase the productivity of those who are served.
Events that are focused on sharing information and networking fall in this category. People in a small field will get to know each other and each others work eventually, but if there are more events it will happen sooner, which I model as an incremental improvement.
But some other events feels much more hits based not that I think of it. Anything focused on getting people started (e.g. helping them choose the right career) or events focused on ideation.
But there are other types of event that are more hit based, and I notice that I'm less interested in doing them. This is interesting. Because these events also differ in other ways, there are alternative explanations. But seems worth looking at.
(Of course everything relating to X-risk is all or nothing in therms of impact, but we can't measure and reward that until it does not matter anyway. Therefore in terms of AI Safety I would measure success in terms of research output, which can be shifted incrementally.)
comment by Khorton · 2020-04-17T21:08:49.108Z · score: 2 (1 votes) · EA(p) · GW(p)
If you're trying to encourage or motivate people, my very rusty understanding of the psychology literature is that you should give people occasional rewards, rather than systematically rewarding people for what you want. Because the systematic rewards effectively undermine intrinsic motivation - you want people to be focusing on helping people, not meeting your criteria.
comment by Linda Linsefors · 2020-04-18T00:05:10.768Z · score: 2 (2 votes) · EA(p) · GW(p)
I sort of agree with this, but I want to add some things.
I agree that money is not the best motivator. If I was trying to solve [people are not motivated enough] I would probably suggest some community measure rather than a new funding structure.
Money is for buying people the time (i.e. not having do some day-job just to earn a living), or funding other things they need for whatever awesome project they are doing.
However money can defiantly influence motivation. 80k mentions list "Pay you feel is unfair." as one of four "major negatives" which "tend to be linked to job dissatisfaction."
https://80000hours.org/2014/09/update-dont-follow-your-passion/
comment by Khorton · 2020-04-18T12:40:13.641Z · score: 3 (2 votes) · EA(p) · GW(p)
Yes, that's actually what I'm talking about.
Imagine I promise to give £10 to every small impact project, £100 to every medium impact, and £1000 to every large impact. You complete a project. It took you 400 hours of work and you're very proud of it - you think it's had a very significant impact. I pay you £10.
How do you think it would feel? How would it affect your future motivation? Are you sure it's not better to a) get nothing and not have the system of judgement at all, or b) get a surprise thank you note from me with £10 inside that you weren't expecting?
I think a lot of people spend a lot of time and effort on things that aren't immediately useful, but I want them to keep their motivation because I believe that one day they may have a hit! If they keep getting £10 cheques for every 10 week cycle of work I'm afraid they're going to be demotivated.
comment by Linda Linsefors · 2020-04-18T18:38:48.654Z · score: 2 (2 votes) · EA(p) · GW(p)
In this situation I would think you evaluated my project as "small impact" which is possibly useful information, depending on how reliable I think you evaluation is. If I trust your judgement, this would obviously be discouraging, since I though it was much more impressive. But in the end I rather be right then proud, so that I can make sure to do better things in the future.
How I react would also depend on if your £10 is all I get, or if I get £10 each from lots of people, becasue that could potentially add up, maybe?
What it mainly comes down to in the end is: Do I get paid enough to sustainably afford to do this work. Or do I need to focus my effort on getting a paid job instead.
If you are a funder, and you think what I'm doing is good, but not good enough to pay me a liveable wages, then I'd much prefer that you don't try to encourage me, but instead just be upfront about this. Encouraging people to keep up an unsustainable work situation is exploitative and will backfire in the long run.
comment by Khorton · 2020-04-18T23:03:00.074Z · score: 7 (3 votes) · EA(p) · GW(p)
I definitely agree with that. But on the other hand, refusing to pay someone who's good idea didn't work out and 'have impact' for no fault of their own also seems exploitative!
I think people who are using this type of work as a living should get paid a salary with benefits and severance. A project to project lifestyle doesn't seem conducive to focusing on impact.
comment by Linda Linsefors · 2020-04-19T16:01:06.864Z · score: 4 (3 votes) · EA(p) · GW(p)
But on the other hand, refusing to pay someone who's good idea didn't work out and 'have impact' for no fault of their own also seems exploitative!
Letting the person running the project take all the risk, might not be optimal, but I would also say it is not exploitative as long as they know this from the start.
I'm not yet sure if I think the amount of money should be 100% based on actual impact, or if we also want to reward people for project that had high expected impact but low actual impact. The main argument for focusing on actual impact is that it is less objective.
I think people who are using this type of work as a living should get paid a salary with benefits and severance. A project to project lifestyle doesn't seem conducive to focusing on impact.
Um, I was going to argue with this. But actually I think you are right.
Something like: "We like what you have done so far, so we will hire you to keep doing good things based on your own best judgment."
comment by gavintaylor · 2020-04-20T20:25:28.295Z · score: 2 (2 votes) · EA(p) · GW(p)
I think people who are using this type of work as a living should get paid a salary with benefits and severance. A project to project lifestyle doesn't seem conducive to focusing on impact.
Agreed. In my brief experience with academic consulting one thing I've realised is that it is really quite reasonable for contracted consultants to charge a 50-100% premium (on top of their utilisation ratio - usually 50%, so another x2 markup) to account for their lack of benefits.
So if somebody is expecting to earn a 'fair' salary from impact purchases compared to employment (or from any other type of short-term contract work really) they should expect a funder to pay premium for this compared to employing them (or funding another organisation to do so) - this doesn't seem like a good use of funds in the long-term if it is possible to employee that person.
comment by Pablo_Stafforini · 2020-06-26T14:00:29.738Z · score: 5 (3 votes) · EA(p) · GW(p)
Are there plans to publish Part II?
comment by Linda Linsefors · 2020-06-26T20:29:55.271Z · score: 1 (1 votes) · EA(p) · GW(p)
Sort of, and it might take some time. The short of it is that I'm less enthusiastic about impact purchase.
I want some sort of funding system that is flexible, and I think the best way to do this is to sponsor people and not project. If someone has though their past work shown competence and good judgement, I think they should be given a salary and freedom to do what they think is best.
I though the way to active this was impact purchase, but as someone pointed out in a comment, this makes for a very economical uncertain situation for the people living this way, which causes stress and short-sightedness which is not the best.
When I wrote this post, I assumed that I needed to have a plan to get a grant in the current system. But after talking to one of the fund manager of Long Term Future Fund, I found out that it is possible to get a grant by simply producing a track record and some vague plan to do more of the same. I've decided to try this out for my self. I'm waiting for an answer from Long Term Future Fund, and plan to write some update after I know how that goes.
If I get the grant this would prove that is is at least possible to get funding with out a clear plan. If I get rejected the concussions I take from that depends on what feedback I get with my rejection. Either way I decided to wait and see how the grant applications goes, before writing the follow up.
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Optical and infrared observations of the TypeIIP SN2002hh from days 3 to 397We present optical and infrared (IR) observations of the TypeII SN2002hhfrom 3 to 397d after explosion. The optical spectroscopic (4-397d) andphotometric (3-278d) data are complemented by spectroscopic (137-381d)and photometric (137-314d) data acquired at IR wavelengths. This is thefirst time L-band spectra have ever been successfully obtained for asupernova (SN) at a distance beyond the Local Group. The VRI lightcurves in the first 40d reveal SN2002hh to be an SNIIP (plateau) - themost common of all core-collapse SNe. SN2002hh is one of the most highlyextinguished SNe ever investigated. To provide a match between itsearly-time spectrum and a coeval spectrum of the TypeIIP SN1999em, aswell as maintaining consistency with KI interstellar absorption, weinvoke a two-component extinction model. One component is due to thecombined effect of the interstellar medium (ISM) of our Milky Way Galaxyand the SN host galaxy, while the other component is due to a dustpocket' where the grains have a mean size smaller than in the ISM. Theearly-time optical light curves of SNe1999em and 2002hh are generallywell matched, as are the radioactive tails of these two SNe and SN1987A.The late-time similarity of the SN2002hh optical light curves to thoseof SN1987A, together with measurements of the optical/IR luminosity and[FeII]1.257μm emission indicate that 0.07 +/- 0.02Msolarof 56Ni was ejected by SN2002hh. However, during the nebularphase the HKL' luminosities of SN2002hh exhibit a growing excess withrespect to those of SN1987A. We attribute much of this excess to anIR-echo from a pre-existing, dusty circumstellar medium. Based on anIR-echo interpretation of the near-IR (NIR) excess, we deduce that theprogenitor of SN2002hh underwent recent mass-loss of~0.3Msolar. A detailed comparison of the late-time opticaland NIR spectra of SNe1987A and 2002hh is presented. While the overallimpression is one of similarity between the spectra of the two events,there are notable differences. The MgI1.503μm luminosity of SN2002hhis a factor of 2.5 greater than in SN1987A at similar epochs, yet coevalsilicon and calcium lines in SN2002hh are fainter. Interpreting thesedifferences as being due to abundance variations, the overall abundancetrend between SN1987A and 2002hh is not consistent with explosion modelpredictions. It appears that during the burning to intermediate-masselements, the nucleosynthesis did not progress as far as might have beenexpected given the mass of iron ejected. Evidence for mixing in theejecta is presented. Pronounced blueshifts seen in the more isolatedlines are attributed to asymmetry in the ejecta. However, during thetime-span of these observations (~1-yr post-explosion) we find noevidence of dust condensation in the ejecta such as might have beenrevealed by an increasing blueshift and/or attenuation of the red wingsof the emission lines. Nevertheless, the clear detection of firstovertone CO emission by 200d and the reddening trend in (K -L')0 suggest that dust formation in the ejecta may occur atlater epochs. From the [OI] λλ6300, 6364Å doubletluminosity we infer a 16-18Msolar main-sequence progenitorstar. The progenitor of SN2002hh was probably a red supergiant with asubstantial, dusty wind. A Medium Resolution Near-Infrared Spectral Atlas of O and Early-B StarsWe present intermediate-resolution (R~8000-12,000) high signal-to-noise(S/N) H- and K-band spectroscopy of a sample of 37 optically visiblestars, ranging in spectral type from O3 to B3 and representing mostluminosity classes. Spectra of this quality can be used to constrain thetemperature, luminosity, and general wind properties of OB stars, whenused in conjunction with sophisticated atmospheric model codes. Mostimportant is the need for moderately high resolutions (R>=5000) andvery high signal-to-noise (S/N>=150) spectra for a meaningful profileanalysis. When using near-infrared spectra for a classification system,moderately high signal-to-noise (S/N~100) is still required, though theresolution can be relaxed to just a thousand or two. In the Appendix weprovide a set of very high-quality near-infrared spectra of Brackettlines in six early-A dwarfs. These can be used to aid in the modelingand removal of such lines when early-A dwarfs are used for telluricspectroscopic standards. Observed Orbital EccentricitiesFor 391 spectroscopic and visual binaries with known orbital elementsand having B0-F0 IV or V primaries, we collected the derivedeccentricities. As has been found by others, those binaries with periodsof a few days have been circularized. However, those with periods up toabout 1000 or more days show reduced eccentricities that asymptoticallyapproach a mean value of 0.5 for the longest periods. For those binarieswith periods greater than 1000 days their distribution of eccentricitiesis flat from 0 to nearly 1, indicating that in the formation of binariesthere is no preferential eccentricity. The binaries with intermediateperiods (10-100 days) lack highly eccentric orbits. Differential photometry of speckle-interferometric binary and multiple starsA method for differential photometry of speckle-interferometric binaryand multiple stars is presented. Both the accuracy and sources ofsystematic errors of the method are analysed. The photometric accuracyranges between 0.02 m and 0.20 m, depending on the atmospheric seeing,the brightness and the separation of the system components. A comparisonbetween our magnitude differences and those of other authors ispresented. Tidal Effects in Binaries of Various PeriodsWe found in the published literature the rotational velocities for 162B0-B9.5, 152 A0-A5, and 86 A6-F0 stars, all of luminosity classes V orIV, that are in spectroscopic or visual binaries with known orbitalelements. The data show that stars in binaries with periods of less thanabout 4 days have synchronized rotational and orbital motions. Stars inbinaries with periods of more than about 500 days have the samerotational velocities as single stars. However, the primaries inbinaries with periods of between 4 and 500 days have substantiallysmaller rotational velocities than single stars, implying that they havelost one-third to two-thirds of their angular momentum, presumablybecause of tidal interactions. The angular momentum losses increase withdecreasing binary separations or periods and increase with increasingage or decreasing mass. uvby FCAPT photometry of the metallic-lined stars 60 Tau and HR 1528 and the magnetic CP stars HR 8216 and HR 8770Differential Strömgren uvby observations from the Four CollegeAutomated Photoelectric Telescope (FCAPT) are presented for themetallic-lined stars 60 Tau and HR 1528 and the magnetic ChemicallyPeculiar stars HR 8216 and HR 8770. The first star, which is a deltaScuti variable, was found not to change its mean magnitudes. HR 1528 isbest described as constant. A decade of photometry of HR 8216 shows thatits b and y values have changed by -0.016 and -0.010 mag, respectively,over this time and now can be considered a photometric variable. For HR8770 a period of 5.3923 days is derived with the photometric variabilitybeing generally in phase. The light curves also suggest possible surfaceabundance inhomogeneities.Tables 2, 3, 4 and 6 are only available in electronic form at the CDSvia anonymous ftp to cdsarc.u-strasbg.fr (130.79.128.5) or viahttp://cdsweb.u-strasbg.fr/cgi-bin/qcat?J/A+A/401/357 Kinematics of Hipparcos Visual Binaries. II. Stars with Ground-Based Orbital SolutionsThis paper continues kinematical investigations of the Hipparcos visualbinaries with known orbits. A sample, consisting of 804 binary systemswith orbital elements determined from ground-based observations, isselected. The mean relative error of their parallaxes is about 12% andthe mean relative error of proper motions is about 4%. However, even 41%of the sample stars lack radial velocity measurements. The computedGalactic velocity components and other kinematical parameters are usedto divide the stars with known radial velocities into kinematical agegroups. The majority (92%) of binaries from the sample are thin diskstars, 7.6% have thick disk kinematics and only two binaries have halokinematics. Among them, the long-period variable Mira Ceti has a verydiscordant {Hipparcos} and ground-based parallax values. From the wholesample, 60 stars are ascribed to the thick disk and halo population.There is an urgent need to increase the number of the identified halobinaries with known orbits and substantially improve the situation withradial velocity data for stars with known orbits. Rotational velocities of A-type stars in the northern hemisphere. II. Measurement of v sin iThis work is the second part of the set of measurements of v sin i forA-type stars, begun by Royer et al. (\cite{Ror_02a}). Spectra of 249 B8to F2-type stars brighter than V=7 have been collected at Observatoirede Haute-Provence (OHP). Fourier transforms of several line profiles inthe range 4200-4600 Å are used to derive v sin i from thefrequency of the first zero. Statistical analysis of the sampleindicates that measurement error mainly depends on v sin i and thisrelative error of the rotational velocity is found to be about 5% onaverage. The systematic shift with respect to standard values fromSlettebak et al. (\cite{Slk_75}), previously found in the first paper,is here confirmed. Comparisons with data from the literature agree withour findings: v sin i values from Slettebak et al. are underestimatedand the relation between both scales follows a linear law ensuremath vsin inew = 1.03 v sin iold+7.7. Finally, thesedata are combined with those from the previous paper (Royer et al.\cite{Ror_02a}), together with the catalogue of Abt & Morrell(\cite{AbtMol95}). The resulting sample includes some 2150 stars withhomogenized rotational velocities. Based on observations made atObservatoire de Haute Provence (CNRS), France. Tables \ref{results} and\ref{merging} are only available in electronic form at the CDS viaanonymous ftp to cdsarc.u-strasbg.fr (130.79.125.5) or viahttp://cdsweb.u-strasbg.fr/cgi-bin/qcat?J/A+A/393/897 ICCD Speckle Observations of Binary Stars. XXIII. Measurements during 1982-1997 from Six Telescopes, with 14 New OrbitsWe present 2017 observations of 1286 binary stars, observed by means ofspeckle interferometry using six telescopes over a 15 year period from1982 April to 1997 June. These measurements constitute the 23dinstallment in CHARA's speckle program at 2 to 4 m class telescopes andinclude the second major collection of measurements from the MountWilson 100 inch (2.5 m) Hooker Telescope. Orbital elements are alsopresented for 14 systems, seven of which have had no previouslypublished orbital analyses. UVBY photometry of the mCP stars HD 35298, 19 Lyrae, HD 192678, and HR 8216Differential Strömgren uvby observations from the Four CollegeAutomated Photoelectric Telescope are presented for the mCP stars HD35298, 19 Lyr, HD 192678, and HR 8216. The period for HD 35298 of1.85457 days is a revision of North's value while that for 19 Lyr of7.0980 days is alias of that found by Winzer. HD 192678 is found to be asmall amplitude photometric variable with the 6.4186 day period proposedby Leroy from polarization measurements. For HR 8216, observations takenbetween 1995 and 1998 confirm that the star has remained constant atleast since 1990. Tables 2, 3, 4 and 6 are only available in electronicform at the CDS via anonymous ftp to cdsarc.u-strasbg.fr (130.79.128.5)or via http://cdsweb.u-strasbg.fr/Abstract.html Mapping the contours of the Local bubble: preliminary resultsWe present preliminary results from a long-term program of mapping theneutral absorption characteristics of the local interstellar medium,taking advantage of Hipparcos stellar distances. Equivalent widths ofthe NaI D-line doublet at 5890 Å are presented for thelines-of-sight towards some 143 new target stars lying within 300 pc ofthe Sun. Using these data which were obtained at the Observatoire deHaute Provence, together with previously published NaI absorptionmeasurements towards a further 313 nearby targets, we present absorptionmaps of the distribution of neutral gas in the local interstellar mediumas viewed from 3 different galactic projections. In particular, thesemaps reveal the Local Bubble region as a low neutral densityinterstellar cavity in the galactic plane with radii between 65-250 pcthat is surrounded by a (dense) neutral gas boundary (or wall''). Wehave compared our iso-column contours with the contours derived bySnowden et al. (\cite{snowden98}) using ROSAT soft X-ray emission data.Consistency in the global dimensions derived for both sets of contoursis found for the case of a million degree hot LB plasma of emissivity0.0023 cm(-6) pc with an electron density of 0.005 cm(-2) . We havedetected only one relatively dense accumulation of cold, neutral gaswithin 60 pc of the Sun that surrounds the star delta Cyg, and note thatthe nearest molecular cloud complex of MBM 12 probably resides at thevery edge of the Local Bubble at a distance of ~ 90 pc. Our observationsmay also explain the very different physical properties of the columnsof interstellar gas in the line-of-sight to the two hot stars epsilonCMa and beta CMa as being due to their locations with respect to theBubble contours. Finally, in the meridian plane the LB cavity is foundto be elongated perpendicularly to the Gould's Belt plane, possiblybeing `squeezed'' by the expanding shells of the Sco-Cen andPerseus-Taurus OB associations. Tables 1 and 2 are also available inelectronic form at the CDS (Strasbourg) via anonymous ftp tocdsarc.u-strasbg.fr (130.79.128.5) or viahttp://cdsweb.u-strasbg.fr/Abstract.html Photometry from the HIPPARCOS Catalogue: Constant MCP Stars, Comparison and Check StarsPhotometry from the Hipparcos catalogue is used to verify the constancyof four magnetic CP stars, as well as the comparison and the check starsused for variability studies of normal and chemically peculiar B and Astars with the Four College Automated Photoelectric Telescope;variability in these stars can produce spurious results. A few of thecomparison stars are found to be variable and should be replaced forfuture differential photometric studies. The photoelectric astrolabe catalogue of Yunnan Observatory (YPAC).The positions of 53 FK5, 70 FK5 Extension and 486 GC stars are given forthe equator and equinox J2000.0 and for the mean observation epoch ofeach star. They are determined with the photoelectric astrolabe ofYunnan Observatory. The internal mean errors in right ascension anddeclination are +/- 0.046" and +/- 0.059", respectively. The meanobservation epoch is 1989.51. The Relation between Rotational Velocities and Spectral Peculiarities among A-Type StarsAbstract image available at:http://adsabs.harvard.edu/cgi-bin/nph-bib_query?1995ApJS...99..135A&db_key=AST Vitesses radiales. Catalogue WEB: Wilson Evans Batten. Subtittle: Radial velocities: The Wilson-Evans-Batten catalogue.We give a common version of the two catalogues of Mean Radial Velocitiesby Wilson (1963) and Evans (1978) to which we have added the catalogueof spectroscopic binary systems (Batten et al. 1989). For each star,when possible, we give: 1) an acronym to enter SIMBAD (Set ofIdentifications Measurements and Bibliography for Astronomical Data) ofthe CDS (Centre de Donnees Astronomiques de Strasbourg). 2) the numberHIC of the HIPPARCOS catalogue (Turon 1992). 3) the CCDM number(Catalogue des Composantes des etoiles Doubles et Multiples) byDommanget & Nys (1994). For the cluster stars, a precise study hasbeen done, on the identificator numbers. Numerous remarks point out theproblems we have had to deal with. All-sky Stromgren photometry of speckle binary starsAll-sky Stromgren photometric observations were obtained for 303 specklebinaries. Most stars were in the range of V = 5-8. These data, whencombined with ratios of intensities from the CHARA speckle photometryprogram, will allow the determination of photometric indices for theindividual components of binary stars with separations as small as 0.05arcsec. These photometric indices will complement the stellar massesfrom the speckle interferometry observations to provide a much improvedmass-luminosity relationship. ICCD speckle observations of binary stars. X - A further survey for duplicity among the bright starsSpeckle interferometric observations are reported for 1123 starsselected from the Yale Bright Star Catalogue (BSC) in a continuingeffort to detect new binaries among the bright stars. Thirty-twopreviously unresolved binaries have been detected, including companionsto Xi UMa and 15 S Mon. Measures of 107 previously resolved systems,many of which resulted from earlier speckle observations, are alsopresented. No evidence of duplicity within a specific (m, Delta-m, rho)window of detectability was found for 984 bright stars. Many of thesystems discovered earlier have shown significant orbital motions, andwe present preliminary orbital elements for six binaries. This efforthas resulted in the discovery of 75 new, bright binaries. We considersome aspects of the duplicity frequencies among the diverse spectral andluminosity classes represented in this sample. We anticipate that thecompletion of a speckle survey of the BSC would lead to the discovery ofat least 200 additional binary systems with angular separations mostlybelow 0.20 arcsec. Many of these will have periods of the order of onedecade and will be accessible to complementary radial velocity programsof enhanced precision. ICCD speckle observations of binary stars. V - Measurements during 1988-1989 from the Kitt Peak and the Cerro Tololo 4 M telescopesAbstract image available at:http://adsabs.harvard.edu/cgi-bin/nph-bib_query?1990AJ.....99..965M&db_key=AST Third preliminary catalogue of stars observed with the photoelectric astrolabe of the Beijing Astronomical Observatory.Not Available ICCD speckle observations of binary stars. IV - Measurements during 1986-1988 from the Kitt Peak 4 M telescopeOne thousand five hundred and fifty measurements of 1006 binary starsystems observed mostly during 1986 through mid-1988 by means of speckleinterferometry with the KPNO 4-m telescope are presented. Twenty-onesystems are directly resolved for the first time, including newcomponents to the cool supergiant Alpha Her A and the Pleiades shellstar Pleione. A continuing survey of The Bright Star Catalogue yieldedeight new binaries from 293 bright stars observed. Corrections tospeckle measures from the GSU/CHARA ICCD speckle camera previouslypublished are presented and discussed. ICCD speckle observations of binary stars. II - Measurements during 1982-1985 from the Kitt Peak 4 M telescopeThis paper represents the continuation of a systematic program of binarystar speckle interferometry initiated at the 4 m telescope on Kitt Peakin late 1975. Between 1975 and 1981, the observations were obtained witha photographic speckle camera, the data from which were reduced byoptical analog methods. In mid-1982, a new speckle camera employing anintensified charge-coupled device as the detector continued the programand necessitated the development of new digital procedures for reducingand analyzing speckle data. The camera and the data-processingtechniques are described herein. This paper presents 2780 newmeasurements of 1012 binary and multiple star systems, including thefirst direct resolution of 64 systems, for the interval 1982 through1985. ICCD speckle observations of binary stars. I - A survey for duplicity among the bright starsA survey of a sample of 672 stars from the Yale Bright Star Catalog(Hoffleit, 1982) has been carried out using speckle interferometry onthe 3.6-cm Canada-France-Hawaii Telescope in order to establish thebinary star frequency within the sample. This effort was motivated bythe need for a more observationally determined basis for predicting thefrequency of failure of the Hubble Space Telescope (HST) fine-guidancesensors to achieve guide-star lock due to duplicity. This survey of 426dwarfs and 246 evolved stars yielded measurements of 52 newly discoveredbinaries and 60 previously known binary systems. It is shown that thefrequency of close visual binaries in the separation range 0.04-0.25arcsec is 11 percent, or nearly 3.5 times that previously known. Mean positions and proper motions of 305 stars obtained from the combination of PZT observations at Ondrejov with AGK positionsObservations of 305 stars carried out at the Ondrejov Observatory in theperiod 1973-1983 were compared to the positions of the same stars in theAGK 2 and AGK 3 catalogs in order to obtain their mean positions andproper motions. The observations were performed in the course of 1140nights using the PZT telescope and comprise more than 32 thousand startransits. The average mean errors in the right ascension and declinationof a star that was observed thoughout the whole period were + or -0.0019 s and + or - 0.019 arcsec in the epoch around 1979. It ispredicted that the errors will increase + or - 0.0055 sec and + or -0.059 arcsec in the year 2000. A table listing the errors in the rightascension and declination is provided. The local system of early type stars - Spatial extent and kinematicsPublished uvby and H-beta photometric data and proper motions arecompiled and analyzed to characterize the structure and kinematics ofthe bright early-type O-A0 stars in the solar vicinity, with a focus onthe Gould belt. The selection and calibration techniques are explained,and the data are presented in extensive tables and graphs and discussedin detail. The Gould belt stars of age less than 20 Myr are shown togive belt inclination 19 deg to the Galactic plane and node-lineorientation in the direction of Galactic rotation, while the symmetricaldistribution about the Galactic plane and kinematic properties (purecircular differential rotation) of the belt stars over 60 Myr oldresemble those of fainter nonbelt stars of all ages. The unresolveddiscrepancy between the expansion observed in the youngest nearby starsand the predictions of simple models of expansion from a point isattributed to the inhomogeneous distribution of interstellar matter. The frequency of Ap-stars with long rotation periodsLight variability was observed over time intervals of months to years inthe stars HD 55540, HD 71066, HD 94660, and HD 187474. The significanceof these results is discussed in terms of the frequency of chemicallypeculiar CP2 stars showing (light) variability with periods longer thanone month. This frequency, relative to the whole CP2 population, mustlie somewhere between 4 and 16 percent. Observations, properlydistributed in time, of a small subgroup of CP2 stars will be sufficientto obtain an accurate ratio of the number of LP-CP2 stars to the totalnumber of CP2 stars. These observations, when continued until theperiodicity is detected, could contribute also to the discussion whetherthese long periods should be identified with the rotation period. The determination of mean positions and proper motions of 304 stars from PZT observations at OndrejovNot Available Spectral classification from the ultraviolet line features of S2/68 spectra. III - Early A-type starsAbstract image available at:http://adsabs.harvard.edu/cgi-bin/nph-bib_query?1978A&AS...33...15C&db_key=AST Infrared radiation of planetary nebulae. II. New and revised observations at 1.0-2.5 μNot Available Rotational Velocities of a0 StarsAbstract image available at:http://adsabs.harvard.edu/cgi-bin/nph-bib_query?1974ApJS...28..101D&db_key=AST Four-color and Hβ photometry for the brighter AO type starsAbstract image available at:http://adsabs.harvard.edu/cgi-bin/nph-bib_query?1972A&AS....5..109C&db_key=AST
Submit a new article
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# Set up a cluster of Liberators
You can define two or more Liberators as a cluster. Here’s how to do this.
For more about clustering, see Liberator clustering features and concepts.
## Clustering Liberators without the Caplin Deployment Framework
Here’s an example of how to configure a Liberator cluster when you’re not using the Deployment Framework. Here we’ve assumed there are just two Liberators in the cluster, but the same approach applies when you’re clustering more.
Config for the first Liberator in the cluster
cluster-index 0
cluster-heartbeat-time 15
cluster-heartbeat-slack-time 5
cluster-port <First-Liberators-network-port-number>
end-cluster-node
cluster-port <Second-Liberators-network-port-number>
end-cluster-node
Config for the second Liberator in the cluster
cluster-index 1
cluster-heartbeat-time 15
cluster-heartbeat-slack-time 5
cluster-port <First-Liberators-network-port-number>
end-cluster-node
cluster-port <Second-Liberators-network-port-number>
end-cluster-node
Here’s an explanation of the configuration items that we’ve used:
• cluster-index: Uniquely identifies this Liberator in the cluster. Make sure each Liberator in the cluster has a different cluster-index in its configuration file; in this example, the first Liberator has cluster-index 0 and the second has cluster-index 1.
Index numbers must start at 0 and correspond to the order of the add-cluster-node items in the configuration file.
• cluster-heartbeat-time is the time in seconds between each cluster heartbeat. The Liberators forming a cluster exchange heartbeat messages at regular intervals, allowing each of them to check that all the others are still present, and take appropriate action if one doesn’t respond.
• cluster-heartbeat-slack-time: When this Liberator doesn’t receive an expected cluster heartbeat from another Liberator in the cluster, it waits cluster-heartbeat-slack-time seconds before disconnecting from the cluster and trying to reconnect to it.
• add-cluster-node: Each add-cluster-node item defines a Liberator in the cluster by its network interface address and port number. You define all the Liberators in the cluster through a set of add-cluster-node items. The set must be in ascending order of the Liberators' cluster-index values. You put the add-cluster-node items in the configuration of every Liberator in the cluster, so each Liberator identifies every node in the cluster, including itself. You can see that in this example, each of the two Liberators' configurations contains the same set of two add-cluster-node items.
## Clustering Liberators in the Caplin Deployment Framework
The Caplin Deployment Framework, allows you to easily define a two-Liberator cluster for failover purposes. To enable failover, see How can I… Set up server failover capability in the Deployment Framework pages. You set up some configuration macro definitions in the <Framework-root>/global-config/environment.conf file on the primary and secondary server machines.
• To change the cluster-heartbeat-time, add the macro definition LIBERATOR_CLUSTER_HEARTBEAT_TIME to <Framework-root>/global-config/environment.conf
define LIBERATOR_CLUSTER_HEARTBEAT_TIME <new value>
• To change the cluster-heartbeat-slack-time from its default, define a value in <Framework-root>/global_config/overrides/servers/Liberator/etc/cluster.conf
cluster-heartbeat-slack-time <new-value>
The configuration macros LIBERATOR${THIS_LEG}_HOST and LIBERATOR${OTHER_LEG}_HOST specify the cluster-addr values. (These macros are defined in <Framework-root>/global_config/hosts-defns.conf)
You normally use the ./dfw hosts command to set up the hostnames in the macros. See Configuring server hostnames in How can I… Set up server failover capability and Change server hostnames in How can I … Change server-specific configuration.
Also see Deployment Framework Configuration macros and items.
If you want to set the cluster-addr values directly, edit them in <Framework-root>/global_config/overrides/servers/Liberator/etc/cluster.conf
• To change the cluster-port settings in the add-cluster-node configuration entries, put new values for the configuration macros LIBERATOR${THIS_LEG}_CLUSTER_PORT and LIBERATOR${OTHER_LEG}_CLUSTER_PORT in the <Framework-root>/global_config/overrides/servers/Liberator/etc/cluster.conf file.
For example:
define LIBERATOR${THIS_LEG}_CLUSTER_PORT${THIS_LEG}6003
define LIBERATOR${OTHER_LEG}_CLUSTER_PORT${OTHER_LEG}6003
where the value 6003 is the new base port number, replacing the default value 6002.
If you want to set up a cluster of more than two Liberators within the Deployment Framework, please contact Caplin Support for advice on how to do this.
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### Example: Deflategate¶
On January 18, 2015, the Indianapolis Colts and the New England Patriots played the American Football Conference (AFC) championship game to determine which of those teams would play in the Super Bowl. After the game, there were allegations that the Patriots' footballs had not been inflated as much as the regulations required; they were softer. This could be an advantage, as softer balls might be easier to catch.
For several weeks, the world of American football was consumed by accusations, denials, theories, and suspicions: the press labeled the topic Deflategate, after the Watergate political scandal of the 1970's. The National Football League (NFL) commissioned an independent analysis. In this example, we will perform our own analysis of the data.
Pressure is often measured in pounds per square inch (psi). NFL rules stipulate that game balls must be inflated to have pressures in the range 12.5 psi and 13.5 psi. Each team plays with 12 balls. Teams have the responsibility of maintaining the pressure in their own footballs, but game officials inspect the balls. Before the start of the AFC game, all the Patriots' balls were at about 12.5 psi. Most of the Colts' balls were at about 13.0 psi. However, these pre-game data were not recorded.
During the second quarter, the Colts intercepted a Patriots ball. On the sidelines, they measured the pressure of the ball and determined that it was below the 12.5 psi threshold. Promptly, they informed officials.
At half-time, all the game balls were collected for inspection. Two officials, Clete Blakeman and Dyrol Prioleau, measured the pressure in each of the balls. Here are the data; pressure is measured in psi. The Patriots ball that had been intercepted by the Colts was not inspected at half-time. Nor were most of the Colts' balls – the officials simply ran out of time and had to relinquish the balls for the start of second half play.
football = Table.read_table('football.csv')
football = football.drop('Team')
football.show()
Ball Blakeman Prioleau
Patriots 1 11.5 11.8
Patriots 2 10.85 11.2
Patriots 3 11.15 11.5
Patriots 4 10.7 11
Patriots 5 11.1 11.45
Patriots 6 11.6 11.95
Patriots 7 11.85 12.3
Patriots 8 11.1 11.55
Patriots 9 10.95 11.35
Patriots 10 10.5 10.9
Patriots 11 10.9 11.35
Colts 1 12.7 12.35
Colts 2 12.75 12.3
Colts 3 12.5 12.95
Colts 4 12.55 12.15
For each of the 15 balls that were inspected, the two officials got different results. It is not uncommon that repeated measurements on the same object yield different results, especially when the measurements are performed by different people. So we will assign to each the ball the average of the two measurements made on that ball.
football = football.with_column(
'Combined', (football.column(1)+football.column(2))/2
)
football.show()
Ball Blakeman Prioleau Combined
Patriots 1 11.5 11.8 11.65
Patriots 2 10.85 11.2 11.025
Patriots 3 11.15 11.5 11.325
Patriots 4 10.7 11 10.85
Patriots 5 11.1 11.45 11.275
Patriots 6 11.6 11.95 11.775
Patriots 7 11.85 12.3 12.075
Patriots 8 11.1 11.55 11.325
Patriots 9 10.95 11.35 11.15
Patriots 10 10.5 10.9 10.7
Patriots 11 10.9 11.35 11.125
Colts 1 12.7 12.35 12.525
Colts 2 12.75 12.3 12.525
Colts 3 12.5 12.95 12.725
Colts 4 12.55 12.15 12.35
At a glance, it seems apparent that the Patriots' footballs were at a lower pressure than the Colts' balls. Because some deflation is normal during the course of a game, the independent analysts decided to calculate the drop in pressure from the start of the game. Recall that the Patriots' balls had all started out at about 12.5 psi, and the Colts' balls at about 13.0 psi. Therefore the drop in pressure for the Patriots' balls was computed as 12.5 minus the pressure at half-time, and the drop in pressure for the Colts' balls was 13.0 minus the pressure at half-time.
Let's construct two tables, one for the Patriots data and one for Colts. The final column of each table is the drop in pressure from the starting value.
patriots = football.where('Ball', are.containing('Patriots'))
patriots = patriots.with_column('Drop', 12.5-patriots.column('Combined'))
patriots.show()
Ball Blakeman Prioleau Combined Drop
Patriots 1 11.5 11.8 11.65 0.85
Patriots 2 10.85 11.2 11.025 1.475
Patriots 3 11.15 11.5 11.325 1.175
Patriots 4 10.7 11 10.85 1.65
Patriots 5 11.1 11.45 11.275 1.225
Patriots 6 11.6 11.95 11.775 0.725
Patriots 7 11.85 12.3 12.075 0.425
Patriots 8 11.1 11.55 11.325 1.175
Patriots 9 10.95 11.35 11.15 1.35
Patriots 10 10.5 10.9 10.7 1.8
Patriots 11 10.9 11.35 11.125 1.375
colts = football.where('Ball', are.containing('Colts'))
colts = colts.with_column('Drop', 13.0-colts.column('Combined'))
colts
Ball Blakeman Prioleau Combined Drop
Colts 1 12.7 12.35 12.525 0.475
Colts 2 12.75 12.3 12.525 0.475
Colts 3 12.5 12.95 12.725 0.275
Colts 4 12.55 12.15 12.35 0.65
It looks as though the Patriots' drops were larger than the Colts'. A natural statistic is the difference between the two average drops. We'll work with that, but you are free to repeat the analysis with other natural statistics such as the difference between the overall average drop and that of the Patriots.
patriots_mean = patriots.column('Drop').mean()
colts_mean = colts.column('Drop').mean()
observed_statistic = patriots_mean - colts_mean
observed_statistic
0.73352272727272805
This positive difference reflects the fact that the average drop in pressure of the Patriots' balls was greater than that of the Colts.
Could this difference be due to chance, or are the Patriots' drops too large? This question is very much like the question that we asked in an earlier example about the scores in one section of a large class. We'll set up the null and alternative hypotheses just as we did in that example.
Null hypothesis. The Patriots' drops are like a random sample of 11 out of all 15 drops. The average came out higher than that of the Colts' drops due to chance variation.
Alternative hypotheis. The Patriots' drops are too large to be the result of chance variation alone.
If the null hypothesis were true, then the Patriots' drops would be comparable to 11 drops drawn at random without replacement from all 15 drops. So let's create an array of all 15 drops and draw at random from it.
drops = Table().with_column(
'Drop', np.append(patriots.column('Drop'), colts.column('Drop'))
)
drops.show()
Drop
0.85
1.475
1.175
1.65
1.225
0.725
0.425
1.175
1.35
1.8
1.375
0.475
0.475
0.275
0.65
drops.sample(with_replacement=False).show()
Drop
1.225
1.175
1.175
0.475
1.375
0.425
0.85
0.65
1.35
1.65
0.725
0.475
1.475
1.8
0.275
Notice the use of sample without a sample size. That is because the default sample size used by sample is the number of rows of the table; if you don't specify a sample size, you get back a sample of the same size as the original table. This is ideal for our purposes, because when you sample without replacement (by specifiying with_replacement=False) the same number of times as there are rows, you end up with a random shuffle of all the rows. Run the cell above a few times to see how the output changes.
We can now use the first 11 rows of the shuffled table as a simulation of the Patriots' drops under the null hypothesis. The remaining four rows form the simulation of the corresponding Colts' drops. We can use these two simulated arrays to simulate our test statistic under the null.
shuffled = drops.sample(with_replacement=False)
new_patriots = shuffled.take(np.arange(11))
new_patriots_mean = new_patriots.column(0).mean()
new_colts = shuffled.take(np.arange(11, drops.num_rows))
new_colts_mean = new_colts.column(0).mean()
simulated_stat = new_patriots_mean - new_colts_mean
simulated_stat
-0.70681818181818212
Run the cell above a few times to see how the test statistic varies. Remember that the simulation is under the null hypothesis that the Patriots' drops are like a random sample of al 15 drops.
It's time for a step that is now familiar. We will do repeated simulations of the test statistic under the null hypothesis. At the end of the simulation, the array simulated_statistics will contain all of the simulated test statistics.
simulated_statistics = make_array()
repetitions = 10000
for i in np.arange(repetitions):
shuffled = drops.sample(with_replacement=False)
new_patriots_mean = shuffled.take(np.arange(11)).column(0).mean()
new_colts_mean = shuffled.take(np.arange(11, drops.num_rows)).column(0).mean()
new_statistic = new_patriots_mean - new_colts_mean
simulated_statistics = np.append(simulated_statistics, new_statistic)
Now for the empirical P-value, which is the chance (computed under the null hypothesis) of getting a test statistic equal to the one that was observed or more in the direction of the alternative. To figure out how to calculate this, it's important to recall the alternative hypothesis:
Alternative hypotheis. The Patriots' drops are too large to be the result of chance variation alone.
The "direction of the alternative" is towards large drops for the Patriots, with correspondingly large values for out test statistic "Patriots' mean - Colts' mean". So the P-value is the chance (computed under the null hypothesis) of getting a test statistic equal to our observed value of 0.73352272727272805 or larger.
empirical_P = np.count_nonzero(simulated_statistics >= observed_statistic)/repetitions
empirical_P
0.0027
That's a pretty small P-value. To visualize this, here is the empirical distribution of the test statistic under the null hypothesis, with the observed statistic marked on the horizontal axis.
print('Observed Statistic:', observed_statistic)
print('Empirical P:', empirical_P)
results = Table().with_column('Simulated Statistic', simulated_statistics)
results.hist()
plots.scatter(observed_statistic, 0, color='red', s=30);
Observed Statistic: 0.733522727273
Empirical P: 0.0027
Notice that the bulk of the distribution is centered around 0. Under the null hypothesis, the Patriots' drops are a random sample of all 15 drops, and therefore so are the Colts'. Therefore the two sets of drops should be about equal on average, and therefore their difference should be around 0.
But the observed value of the test statistic is quite far away from the heart of the distribution. By any reasonable cutoff for what is "small", the empirical P-value is small. So we end up rejecting the null hypothesis of randomness, and conclude that the Patriots drops were too large to reflect chance variation alone.
The independent investiagtive team analyzed the data in several different ways, taking into account the laws of physics. The final report said,
"[T]he average pressure drop of the Patriots game balls exceeded the average pressure drop of the Colts balls by 0.45 to 1.02 psi, depending on various possible assumptions regarding the gauges used, and assuming an initial pressure of 12.5 psi for the Patriots balls and 13.0 for the Colts balls."
-- Investigative report commissioned by the NFL regarding the AFC Championship game on January 18, 2015
Our analysis shows an average pressure drop of about 0.73 psi, which is close to the center of the interval "0.45 to 1.02 psi" and therefore consistent with the official analysis.
Remember that our test of hypotheses does not establish the reason why the difference is not due to chance. Establishing causality is usually more complex than running a test of hypotheses.
But the all-important question in the football world was about causation: the question was whether the excess drop of pressure in the Patriots' footballs was deliberate. If you are curious about the answer given by the investigators, here is the full report.
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# 21.What is the major onganic product obtained from the following reaction? Circle one (Zp)HoHJSO4 HSO422. What is the major organic
###### Question:
21. What is the major onganic product obtained from the following reaction? Circle one (Zp) Ho HJSO4 HSO4 22. What is the major organic product obtained from the following naction? Circle ame (Zp) Ch; Unjui Cllyy 23. Outline . how one t0 carTy Out the folbwing transformation showing reagents and isolated intermediates products in the synthetic schcme There may be more tham one sicp (Ap)
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Please show all work and include differentials 1. (6 points) B Consider the shaded region bounded by y = r² – 4 and y = 3.2 + 6 (see above). Note that the c-axis and y-axis are not drawn to the same scale. (a) Find the coordinates of the points A, B, and C. Remember to show all work. (b)...
##### In a historical movie, two knights on horseback start from rest 91 m apart and ride...
In a historical movie, two knights on horseback start from rest 91 m apart and ride directly toward each other to do battle. Sir George's acceleration has a magnitude of 0.17 m/s2, while Sir Alfred's has a magnitude of 0.26 m/s2. Relative to Sir George's starting point, where do the knig...
##### Statistica programecommendedSuppose that the magnitude of earthquakes recordedreqion of North Americabe modelednavingqamma distribution with0.7 and2,4,What the mean magnitude earthquakes striking tnc region?What the probability that the magnitudeearthquake striking the regionexceed on the Richter scale? (Round Your Jnsucrdecimal places )Using a exponenciz distributlon wlth mean 2.4 model Lne magnitude Drananiln arger or smaller than the probability found in (b)? Explain.an carthquake striking th
statistica program ecommended Suppose that the magnitude of earthquakes recorded reqion of North America be modeled naving qamma distribution with 0.7 and 2,4, What the mean magnitude earthquakes striking tnc region? What the probability that the magnitude earthquake striking the region exceed on th...
##### Find the value of $x$ in each expression. $\log _{x} 16=4$
Find the value of $x$ in each expression. $\log _{x} 16=4$...
##### Starting from his nest, Sam the bird flies (level flight) in a straight line for 10 m at 60 west of north and then in another straight line for 5.0 m due south; at which point he comes to rest on a power line: Back at the nest, Samantha (Sam's mate) flies in a straight line directly form the nest to where Sam is on the power line:i) What is the (magnitude of the) angle between due west and the direction Samantha must fly to meet up with Sam? cc9) 10b) 20 "c) 30d) 40e) 50 0
Starting from his nest, Sam the bird flies (level flight) in a straight line for 10 m at 60 west of north and then in another straight line for 5.0 m due south; at which point he comes to rest on a power line: Back at the nest, Samantha (Sam's mate) flies in a straight line directly form the ne...
##### 7 How many three digit numbers are there in which no two digits are alike?
7 How many three digit numbers are there in which no two digits are alike?...
##### 7. A uniform magnetic field of magnitude 0.4T is in the + x direction. A square...
7. A uniform magnetic field of magnitude 0.4T is in the + x direction. A square coil that has 12.0cm long sides has a single turn and makes an angle with the Z axis. Find the magnetic flux through the coilwhen is 50'. A) 0.01W b; B) 0.9Wb; C) 64.5Wb; D) 4.1 Wb;...
##### Question 10 Nokyci snswered Points out ol L0Q The switch has been connected to point 1 for a long time; and: Isuddenly it is moved to = connecting to 2. What is the current through the resistor immediately after the switch is = touching [ point 2?select one: ZeroBVIR, Iloming UpAVIR, Ilowing down4v
Question 10 Nokyci snswered Points out ol L0Q The switch has been connected to point 1 for a long time; and: Isuddenly it is moved to = connecting to 2. What is the current through the resistor immediately after the switch is = touching [ point 2? select one: Zero BVIR, Iloming Up AVIR, Ilowing down...
##### Make a code for the additions of fractions. use the attached code to find LCM. define...
make a code for the additions of fractions. use the attached code to find LCM. define _CRT_SECURE_NO_WARNINGS include <stdio.h> int Icm(int, int); int main() int n1, n2, min Multiple; printf("Enter two positive integers: "); scanf("%d %d", &n1, &n2); // maximum nu...
##### For a > 0, let Ra be the disc in the (2,y)-plane defined by 22 + y2 < a2. Evaluate JJ ~Vz+y" dA. RaBy taking the limit of the above integral as 7C, findfl e-Vz2 Fyz dA, R2the integral of the function e r2+y2 over the whole (€,y)-plane
For a > 0, let Ra be the disc in the (2,y)-plane defined by 22 + y2 < a2. Evaluate JJ ~Vz+y" dA. Ra By taking the limit of the above integral as 7C, find fl e-Vz2 Fyz dA, R2 the integral of the function e r2+y2 over the whole (€,y)-plane...
##### Convert the hashedwedge line into its condensed formula. Convert the hashed-wedged line formula into its condensed...
convert the hashedwedge line into its condensed formula. Convert the hashed-wedged line formula into its condensed formula. Be sure to subscript the numbers in the formula by highlighting it and clicking on the xy button. HHH .. :n=c- Condensed formula:...
##### Nike can produce jerseys that will be sold for $76 each. Non-depreciated fixed costs are$1,040...
Nike can produce jerseys that will be sold for $76 each. Non-depreciated fixed costs are$1,040 per year and variable costs are $57 per unit. a. If the project requires an initial investment of$2,810 and is expected to last for 6 years and the firm pays no taxes, what are the accounting and...
##### 1 1 7_ 12, from spherical to cylindrical: 9 2
1 1 7_ 12, from spherical to cylindrical: 9 2...
##### Find a 2 X 2 matrix A such that A? = 0 and A 0.
Find a 2 X 2 matrix A such that A? = 0 and A # 0....
##### The chemical hydroxylamine is used to cleave only at asparagine-glycine bonds. What would be the correct...
The chemical hydroxylamine is used to cleave only at asparagine-glycine bonds. What would be the correct peptide fragments formed if Asp-Gly-Asn-Gly-Asn-Gly-Pro was treated with hydroxylamine? Asp-Gly-Asn Gly-Asn Gly-Pro Asp Gly Asn-Gly Asn-Gly-Pro Asp Gly-Asn Gly-Asn Gly-Pro O Asp-Gly-Asn-Gly-Asn-G...
##### 14.(10 points) Sketch the graph of a function f that satisfies the following: (Make sure your sketch clearly shows where the graph is increasing, decreasing, concave up, and concave down: ) The domain of f is [~1,1) U(1,5]; lim OO and lim OO 1-1 f(0) f(3) = 0; fi(c) = 0 at € = 3; file) > 0 for (_1, 1) U (1,3) U (3,5); f"(e) > 0 on (_1,1) U (3,5); fn(e) < 0 on (1,3): has no extreme values_ Font Size B 1 4 J= = *24 @
14.(10 points) Sketch the graph of a function f that satisfies the following: (Make sure your sketch clearly shows where the graph is increasing, decreasing, concave up, and concave down: ) The domain of f is [~1,1) U(1,5]; lim OO and lim OO 1-1 f(0) f(3) = 0; fi(c) = 0 at € = 3; file) > 0...
##### 1) List and describe the steps of animal development from zygote to the gastrula 2) Describe...
1) List and describe the steps of animal development from zygote to the gastrula 2) Describe the process of neurulation 3) Describe the process of axis formation...
##### What are the principles of ozone thinning and depletion?
What are the principles of ozone thinning and depletion?...
##### Problem 3 (23%) The following beam is discretize into 2 elements. E-29x106 psi, I-375 in and A (b...
Problem 3 (23%) The following beam is discretize into 2 elements. E-29x106 psi, I-375 in and A (beam cross sectional area) -9.12 in2 1. 2. 3. 4. Calculate the stiffness matrix of each of the 2 elements Calculate the global stiffness matrix of the beam Calculate the force matrix Calculate the deflect...
##### Let xi, R2, ...xn be a random from a population with pdf sample f(x) = 2x2-1...
Let xi, R2, ...xn be a random from a population with pdf sample f(x) = 2x2-1 for Ocnaliaso 0 otherwise i) is the method of moments estimator for a consistent?...
##### The depth (in feet) of water at a dock changes with the rise and fall of tides The depth is modeled by the functionD(t) = 5 COS3t+7 + 2where t is the number of hours after midnight. Find the rate at which the depth is changing at 4 a.m. Round your answer t0 4 decimal places_PreviewPoints possible: This is attempt 1 of 3_License
The depth (in feet) of water at a dock changes with the rise and fall of tides The depth is modeled by the function D(t) = 5 COS 3t+7 + 2 where t is the number of hours after midnight. Find the rate at which the depth is changing at 4 a.m. Round your answer t0 4 decimal places_ Preview Points possib...
##### (10) (10)Determine and simplify the Difference Quotient; IGth)[email protected] for fr) =-2x-3 Write an equation of the vertical line passing through (-3,7)_(30)Given (-3, -1) and (1, 4), determine an equation of the line passing through these points , Determine the equation of the line perpendicular to the line above that passes through (2,1) Graph both lines on some (well-labeled!) axes (20)List all of the possible rational zeros of f(r) =2x 6x+2x-6 Determine all the real and complex zeros of f (x) (prove t
(10) (10) Determine and simplify the Difference Quotient; IGth)[email protected] for fr) =-2x-3 Write an equation of the vertical line passing through (-3,7)_ (30) Given (-3, -1) and (1, 4), determine an equation of the line passing through these points , Determine the equation of the line perpendicular to the li...
##### In 200 - 300 words, how do you link your learning style with a note-taking style...
In 200 - 300 words, how do you link your learning style with a note-taking style when you are taking notes either as you study or in class....
##### You have an account that is compounded quarterly at an interestrate of 4.25%. Suppose you make an initial investment of $8,000 andsubsequently make investments of$150 each quarter at the sametimethat the account is compounded. What will be the final value of theaccount after 5 years?Hint: The answer is the sum of twoinvestments. please show all work
You have an account that is compounded quarterly at an interest rate of 4.25%. Suppose you make an initial investment of $8,000 and subsequently make investments of$150 each quarter at the sametime that the account is compounded. What will be the final value of the account after 5 years?Hint: The a...
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# What is the “energy associated with torque?”
I'm told that when a magnetic dipole is placed in a magnetic field $$\vec{B}$$, it experiences a torque $$\vec{\tau} = \vec\mu \times \vec{B}$$ and an associated energy $$H = -\vec{\mu} \cdot \vec{B}$$, where $$\vec{\mu}$$ is the magnetic dipole moment, pointing in the same direction as how the dipole is oriented in space.
I thought at first the associated energy might be the total amount of work you'd have to do to spin the dipole from equilibrium to whatever orientation you care about, but this doesn't agree with the math. If $$\vec{\mu}$$ points in the same direction as $$\vec{B}$$, you get a non zero value. What on earth does this associated energy actually mean?
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# Classical solutions of the fifth Painlev\'e equation
AR2W02 - Mathematics of beyond all-orders phenomena
In this talk I will discuss classical solutions of the fifth Painlev\’e equation (P$$). The general solutions of the Painlev\’e equations are transcendental in the sense that they cannot be expressed in terms of known elementary functions. However, it is well known that all Painlev\’e equations except the first equation possess rational solutions, algebraic solutions and solutions expressed in terms of the classical special functions for special values of the parameters. These solutions of the Painlev\’e equations are often called classical solutions” and frequently can be expressed in the form of determinants.In the generic case of P{\rm V} when \delta\not=0, special function solutions are expressed in terms of Kummer functions and has rational solutions expressed in terms of Laguerre polynomials. In the case of P$$ when $\delta=0$, which is known as deg-P${\rm V}$ and related to the third Painlev\’e equation, special function solutions are expressed in terms of Bessel functions and has algebraic solutions expressed in terms of Laguerre polynomials. I shall give some new representations of some of these classical solutions and discuss some applications.
This talk is part of the Isaac Newton Institute Seminar Series series.
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Lemma 86.7.3. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism from an algebraic space over $S$ to a formal algebraic space over $S$. Then $f$ is representable by algebraic spaces.
Proof. Let $Z \to Y$ be a morphism where $Z$ is a scheme over $S$. We have to show that $X \times _ Y Z$ is an algebraic space. Choose a scheme $U$ and a surjective étale morphism $U \to X$. Then $U \times _ Y Z \to X \times _ Y Z$ is representable surjective étale (Spaces, Lemma 64.5.5) and $U \times _ Y Z$ is a scheme by Lemma 86.7.2. Hence the result by Bootstrap, Theorem 79.10.1. $\square$
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
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[tex-k] [comp.text.tex] web2c wish...
David Kastrup David.Kastrup@t-online.de
26 Feb 2002 05:11:45 +0100
--=-=-=
Hello, due to lack of resonance, I forward a proposal of mine to the
web2c list in the hope that somebody is interested in this. This
would be the base for fast recompilation techniques. The
implementation should not take more than a few dozen lines, I would
guess, if one can make use of the fork system call. Since I am not
subscribed myself to this list, it would be nice if you copied me any
responses. Thank you.
--=-=-=
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Newsgroups: comp.text.tex
Subject: web2c wish...
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Hi, we have a \write18 interface that can be enabled by command line.
Now I would like to have a \read18 interface. This is supposed to
work like the following:
I use an \openin on channel 18 in order to associate a file or socket
with channel 18. When the corresponding \read occurs and data is
available for reading, TeX does a fork system call. The parent
process records the file pointers of all open files (except channel
18) and then goes to sleep. Once it does that, the child continues
processing, and can read channel 18 for that purpose, getting
instructions from there, file names, whatever necessary. When the
child dies, the parent gets woken up again. As soon as data is
available again on channel 18, the parent truncates all open write
channels back to the size they were at fork time and moves all read
pointers to the location they were at fork time. If the file on
channel 18 reaches end of file, the parent terminates without
truncating any files.
Obviously, a named pipe or socket is the right thing to use for that.
The idea is to have an in-memory copy of LaTeX hanging around, or even
a document already run up to \begin{document}, that is eager to
process or reprocess a small document or a document part as fast as
imaginable.
Anybody better ideas or thoughts why this is not a good idea?
--
David Kastrup, Kriemhildstr. 15, 44793 Bochum
Email: David.Kastrup@t-online.de
--=-=-=
--
David Kastrup, Kriemhildstr. 15, 44793 Bochum
Email: David.Kastrup@t-online.de
--=-=-=--
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Trisymmetric Multiplication Formulae in Finite Fields - Archive ouverte HAL Access content directly
Conference Papers Year : 2021
Trisymmetric Multiplication Formulae in Finite Fields
Hugues Randriambololona
Édouard Rousseau
• Function : Author
Abstract
Multiplication is an expensive arithmetic operation, therefore there has been extensive research to find Karatsuba-like formulae reducing the number of multiplications involved when computing a bilinear map. The minimal number of multiplications in such formulae is called the bilinear complexity, and it is also of theoretical interest to asymptotically understand it. Moreover, when the bilinear maps admit some kind of invariance, it is also desirable to find formulae keeping the same invariance. In this work, we study trisymmetric, hypersymmetric, and Galois invariant multiplication formulae over finite fields, and we give an algorithm to find such formulae. We also generalize the result that the bilinear complexity and symmetric bilinear complexity of the twovariable multiplication in an extension field are linear in the degree of the extension, to trisymmetric bilinear complexity, and to the complexity of t-variable multiplication for any t ≥ 3.
Dates and versions
hal-03832419 , version 1 (27-10-2022)
Identifiers
• HAL Id : hal-03832419 , version 1
• DOI :
Cite
Hugues Randriambololona, Édouard Rousseau. Trisymmetric Multiplication Formulae in Finite Fields. WAIFI 2020, Jul 2020, Rennes, France. pp.92-111, ⟨10.1007/978-3-030-68869-1_5⟩. ⟨hal-03832419⟩
18 View
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I manufactured a standard dogbone specimen similar to the one shown in the picture below sometime last year. Instead of using a plain monofiber, I decided to embed a Carbon Fiber (CF) bundle in it. I found some CF in our laboratory at Polytechnique and worked on a method to manufacture the specimen I had in mind.
The setup used here is the same as the one presented in the post Optical microscope Digital Image Correlation. In the GIF clip above, the specimen is shown first, and the bundle of carbon fiber going through it are indicated with a metallic ruler I am holding. The next scene shows the microtensile testing rig installed under the Olympus confocal laser microscope, a specimen is shown while being tested. In the final scene, the computer screen plugged to the microscope is shown. It is possible to see a PTFE fiber in an epoxy matrice while interfacial debonding it starting to happen. That was the specimen I was testing at the moment I shot this clip.
The specimen was loaded in the microtensile testing machine, the whole rig is put under the confocal laser microscope after what the test started. These kind of tests take about 8 hours long: the specimen is loaded displacement by pulling on the microtensile testing machine by a step inferior to ($$100 \mu m$$, the microtensile testing machine is then stopped and a picture is snapped with the confocal laser (a single picture takes about 3 minutes because of the confocal scanning process). The specimen is then pulled again and these steps are repeated until a crack is observed. The purpose of this test is to observe crack initiation and propagation inside a bundle of CF. For this specimen, I stopped the test when the crack was large enough to cover the whole field of view, but before it broke the specimen in two separate parts. I then used some photoelastic film again (similar to the one presented in this post) and glued it on both sides of the specimen. The photoelastic film reveals the strain field within a polymer, the more fringes are visible in an area the higher the strian field is (more explanation were provided in this post).
The result reveals the residual stresses remaining in the specimen after this test. It is possible to see how the CF bundle affected the strain field in its vicinity. The final purpose of this experiment is to somehow come up with a method to perform DIC on the images obtained using the confocal laser microscope.
These experiments were done with the help of Damien Texier and were done at École de Technologie Supérieur, Montréal.
If you liked this post, you can share it with your followers or follow me on Twitter!
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Mathematics
# In Fig., $OR \bot \ OP$.Name all the pairs of adjacent angles.
##### SOLUTION
There are four pair of adjacent angles; $\angle x$ and $\angle y, \angle x$ and $\angle y + \angle z, \angle y$ and $\angle z , \angle z$ and $\angle x+ \angle y.$
You're just one step away
Subjective Medium Published on 09th 09, 2020
Questions 120418
Subjects 10
Chapters 88
Enrolled Students 86
#### Realted Questions
Q1 Subjective Medium
In the given figure, $PQR$ is a straight line. Find $x$. Then complete the following:
$\angle AQR =$ ________
Asked in: Mathematics - Lines and Angles
1 Verified Answer | Published on 09th 09, 2020
Q2 Subjective Medium
Write the complement of:
$1/3$ of $180^o$
Asked in: Mathematics - Lines and Angles
1 Verified Answer | Published on 09th 09, 2020
Q3 Subjective Medium
In figure, $POS$ is a line, find $x$.
Asked in: Mathematics - Lines and Angles
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Q4 Subjective Medium
I an angle is $28^{o}$ less than its complement, find its measure.
Asked in: Mathematics - Lines and Angles
1 Verified Answer | Published on 09th 09, 2020
Q5 Subjective Medium
From the adjoining figure find $x$, i $AOB$ is a straight line. Hence complete the following
Which is an acute angle?
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# 28th Texas Symposium on Relativistic Astrophysics
13-18 December 2015
International Conference Centre Geneva
Europe/Zurich timezone
## Exploring the consequences of parameter values in cosmological models with CosmoEJS, an interactive package of cosmology Java simulations
14 Dec 2015, 18:02
3m
Level 2, Room 14 (International Conference Centre Geneva)
### Level 2, Room 14
#### International Conference Centre Geneva
17 Rue de Varembé, 1211 Geneva
Poster
### Speaker
Jacob Moldenhauer (University of Dallas)
### Description
It is not only important to constrain the parameters of cosmological models with the most recent and precise observations, but it is also crucial to understand the physical consequences of those parameters for the different, but complimentary observations involved. CosmoEJS is an interactive Java package of simulations that allow the user to explore the ramifications of choosing various values for the cosmological parameters of a particular model. These simulations now include observations of the growth of structures of galaxies, as well as, the expansion history of the universe. Users can visually inspect the plotted theoretical values of their model, compare numerical fitting using $\chi^2$ values, calculate derived cosmological values, and finally plot the expansion trajectory of their models as they evolve in time.
### Primary author
Jacob Moldenhauer (University of Dallas)
### Co-authors
Francis Cavanna (University of Dallas) William O'toole (University of Dallas) William Zimmerman (University of Dallas)
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# facular definition
• Of or with respect to the faculae.
• Of or related to the faculæ.
• Of or related to the faculae.
• Of or regarding the faculæ.
• Of or with respect to the faculae.
• Of or regarding the faculæ.
• others:
• regarding or of this nature of a facula. See facula.
• Pertaining to or for the nature of a facula. See facula.
• Pertaining to or of the nature of a facula. See facula.
## Related Sources
• Definition for "facular"
• Of or with respect to the faculae.
• Sentence for "facular"
• The interpretations tend to link the…
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# [TxMt] Macro/command for LaTeX wordcounts
Colin Hahn colin.hahn at gmail.com
Fri Jan 26 00:57:09 UTC 2007
I'm trying to write a macro or command (not sure which will be
better) for doing a word count on LaTeX documents. The basic task is
simple enough: I want to run the command "ps2ascii [file.pdf] | wc"
on file.tex (where file.tex is the current file I'm working on) (I've
tested ps2ascii and it works fine for the types of documents I'll be
writing).
However, I also want to strip the footnotes from my word count. To
do so, I've defined the footnotes with a custom command that I can
redefine as empty:
\newcommand{\fn}[1]{\footnote{#1}}
% \renewcommand{\fn}[1]{}
So, the whole task I need to do is:
1) Uncomment the \renewcommand line
2) Re-run LaTex
3) Run ps2ascii | wc on the generated .pdf
4) Display the output of that command as a ToolTip
5) Re-comment the \renewcommand line (so the file is back in its
original state)
Unfortunately, I don't know how to string these things together, nor
how to get the argument for the ps2ascii command from the current
filename variable TM_FILENAME (i.e., how to substitute .pdf
for .tex). I can add unique comments to the end of the \renewcommand
line so it can be picked out by a regex search, if that helps.
Suggestions? I'm also open to other ideas for getting the same end
result, if there's an easier way.
Colin Hahn
»Geben Sie mir Kaffee, dann mache ich Phänomenologie daraus.«
"Give me my coffee so that I can make phenomenology out of it." --
Edmund Husserl
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# 问题内容:
I have a data table that looks like
|userId|36|37|38|39|40|
|1|1|0|3|0|0|
|2|3|0|0|0|1|
Where each numbered column (36-40) represent week numbers. I want to calculate the number of weeks before the 1st occurrence of a non-zero value, and the last.
For instance, for userId 1 in my dataset, the first value appears at week 36, and the last one appears at week 38, so the value I want is 2. For userId 2 it’s 40-36 which is 4.
I would like to store the data like:
|userId|lifespan|
|1|2|
|2|4|
# 答案:
## 答案1:
General method I would take is to melt it, convert the character column names to numeric, and take the delta by each userID. Here is an example using data.table.
library(data.table)
1|1|0|3|0|0
2|3|0|0|0|1",
dt <- melt(dt, id.vars = "userId")
dt[, variable := as.numeric(as.character(variable))]
dt
# userId variable value
# 1: 1 36 1
# 2: 2 36 3
# 3: 1 37 0
# 4: 2 37 0
# 5: 1 38 3
# 6: 2 38 0
# 7: 1 39 0
# 8: 2 39 0
# 9: 1 40 0
# 10: 2 40 1
dt[!value == 0, .(lifespan = max(variable) - min(variable)), by = .(userId)]
# userId lifespan
# 1: 1 2
# 2: 2 4
## 答案评论:
This is exactly what I was after, thank you!
– Benirving92
21 mins ago
## 答案2:
Here’s a dplyr method:
df %>%
gather(var, value, -userId) %>%
mutate(var = as.numeric(sub("X", "", var))) %>%
group_by(userId) %>%
slice(c(which.max(value!=0), max(which(value!=0)))) %>%
summarize(lifespan = var[2]-var[1])
Result:
# A tibble: 2 x 2
userId lifespan
<int> <dbl>
1 1 2
2 2 4
Data:
df = read.table(text = "userId|36|37|38|39|40
1|1|0|3|0|0
2|3|0|0|0|1", header = TRUE, sep = "|")
## 原文地址：
https://stackoverflow.com/questions/47756325/find-index-of-first-and-last-occurrence-in-data-table
Tags:
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Browse Questions
# The number of ways of dividing 15 men and 15 women into 15 couples each consisting of a man and woman is
$\begin{array}{1 1}(A)\;1240\\(B)\;1840\\(C)\;1820\\(D)\;2005\end{array}$
Toolbox:
• $C(n,r)=\large\frac{n!}{r!(n-r)!}$
The number of ways of choosing first couple is =$15C_1\times 15C_1$
$\Rightarrow (15)^2$
The number of ways of choosing second couple is =$14C_1\times 14C_1$
$\Rightarrow (14)^2$
Thus the number of ways of choosing the couple is =$15^2+14^2+13^2+.....+2^2+1^2$
$\Rightarrow \large\frac{15\times(15+1)[2(15+1)]}{6}$
$\Rightarrow 1240$
Hence (A) is the correct answer.
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# Math Help - Percentage question
1. ## Percentage question
Hi, I've been doing some practice papers to revise for my Biology exam (tomorrow! ) and I came accross one question about percentages...I thought it was alright at first, but looked at the mark scheme and apparently I was wrong.
Anyway, the question was:
Calculate the percentage difference between the concentration of Potassium ions in the culture solution at the start of the experiment (8.0) and the concentration of potassium ions in the barley solution (42.0)
According to the mark scheme I had to first do 42-8. Which I did, and got 34. But thats where I get stuck!
It says that the answer to the question is 425%. But how on earth did they work that out? is there even such a thing as 425%??? lol
2. Originally Posted by trishymac
Hi, I've been doing some practice papers to revise for my Biology exam (tomorrow! ) and I came accross one question about percentages...I thought it was alright at first, but looked at the mark scheme and apparently I was wrong.
Anyway, the question was:
Calculate the percentage difference between the concentration of Potassium ions in the culture solution at the start of the experiment (8.0) and the concentration of potassium ions in the barley solution (42.0)
According to the mark scheme I had to first do 42-8. Which I did, and got 34. But thats where I get stuck!
It says that the answer to the question is 425%. But how on earth did they work that out? is there even such a thing as 425%??? lol
yes, there is a thing as 425%. it just means you have 4 and one quater times what you've started with. we started with 8 and increased to 42, now we have to find what the percentage increase is, which is the same as the percent difference between the two.
whenever we want to know what percentage of the original number a number is, we put the new number over the old number and multyply by 100.
So percentage increase $= \frac {amount \mbox { } of \mbox { } increase}{original \mbox { } amount} \times 100$
Now try to see if you can get the answer
3. i got it! thank you!!
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# Weird compilation errors...
This topic is 2541 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
## Recommended Posts
Ok, so I wasn't quite sure where to put this, so let me first give some background info to further detail the problem.
I am working on a mostly C++ 2D game engine for the Android OS. As most of you know, the Android OS uses Java as its language of choice, so I am making heavy use of the JNI and the Android NDK. Now, I got all of the hooks working right, so I know the JNI is working correctly, and each of the individual C/C++ libraries I am making use of compiles without error. My problem comes when I try to reference Lua (luabind seems to work fine, but I won't know for sure until I get passed the Lua side of things) from my own code. I get the following compilation errors when I do so:
[media]http://img851.imageshack.us/img851/3586/consoleerror.jpg[/media]
I have done a lot of searching on the error the last few days, but I haven't been able to come up with anything at all. It seems to be a linking issue rather than a compilation issue, as it doesn't come up until everything is finished compiling. Here is the make file I am using for the troublesome code:
LOCAL_PATH := $(call my-dir) include$(CLEAR_VARS) LOCAL_MODULE := main SDL_PATH := ../sdl LOCAL_C_INCLUDES := $(LOCAL_PATH)/$(SDL_PATH)/include \ $(LOCAL_PATH)/../luabind \$(LOCAL_PATH)/../lua/src # Add your application source files here... #LOCAL_SRC_FILES := $(SDL_PATH)/src/main/android/SDL_android_main.cpp \ # main.cpp \ LOCAL_SRC_FILES :=$(SDL_PATH)/src/main/android/SDL_android_main.cpp \ main.cpp \ Animation.cpp \ Node.cpp \ Renderer.cpp \ GameObject.cpp \ Video.cpp \ Texture2D.cpp \ Globals.cpp \ Script.cpp LOCAL_SHARED_LIBRARIES := SDL LOCAL_STATIC_LIBRARIES := Lua LuaBind SDL_image LOCAL_LDLIBS := -lGLESv1_CM -llog LOCAL_CPPFLAGS := -fexceptions include \$(BUILD_SHARED_LIBRARY)
I should note that I am not very experienced with make files, so there is a great possibility that it is something wrong in that "code,"
I am really not sure what exactly to add code-wise, but I suppose I will post the code that I have narrowed the problem down to.
#include "Globals.h" #include "SDL.h" #include "Script.h" #include "GameObject.h" extern "C" int main(int argc, char *argv[]) { init_lua(); free_lua(); return 0; }
init_lua() and free_lua() are declared in Globals.h and defined in Globals.cpp, and are simply a wrapper for lua_open() and lua_close() using a global lua_State pointer that is also declared in Globals.h.
I should also mention that if I remove the two methods from the main function, it compiles fine.
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It doesn't look like you're linking against libstdc++ (-lstdc++).
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That's what I was thinking, but it's odd. I specify in my Application.mk (which the Android NDK says to name it) that I want the compilation to us GNU STL. stdc++ is includeded in the STL library, right? Or am I mistaken?
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It seems I was wrong. According to wiki
The C++ Standard Library is based upon conventions introduced by the [color="#0645ad"]Standard Template Library (STL). Although the C++ Standard Library and the STL share many features, neither is a strict superset of the other.
[/quote]
I will try explicitly specifying to link with the stdc++ library.
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# Material balance with a reaction (chemical engineering)
Tags:
1. Jul 28, 2016
### MickeyBlue
1. The problem statement, all variables and given/known data
Felder & Rousseau 4.73 (p. 184)
A mixture of propane and butane is burned with pure oxygen. The combustion products contain 47.4 mole % H2O. After all of the water is removed from the products, the residual gas contains 69.4 mole% CO2 and the balance is O2. What is the mole percent of propane in the fuel?
2. Relevant equations
1. Let propane = P
2. Choose 100 mol of the reactant fuel as a basis
3. The attempt at a solution
Set up 2 balanced equations: 1. C3H8 + 5O2 -> 4H2O + 3CO2
2. C4H8 + 13/2O2 -> 5H2O + 4CO2
I assumed a 100% conversion of the reactant gases. I know O2 is in excess. I drew up a reactant table in terms of mol and set Propane in = P and Butane in= 100-P. I then tried to calculate P by setting the mols of water out divided by the total mols out equal to 0.474.
2. Jul 28, 2016
### Staff: Mentor
So, what's the problem?
3. Jul 28, 2016
### collinsmark
Hello @MickeyBlue! Welcome to Physics Forums!
So far so good, with one exception:
You wrote that the butane was $\mathrm{C_4 H_8}$, but I think you meant $\mathrm{C_4 H_{10}}$. Your equations were balanced correctly though, so I assume that was just a minor mistake typing things into the computer.
As a hint for moving forward, don't forget to include the leftover $\mathrm{O_2}$ in your product. Similar to what you did with $P$, make up a variable name to indicate the amount of $\mathrm{O_2}$ in there. I used the variable $x$, but you can use whatever you want.
So your formula, as you have already stated it, should be of this form:
$$\frac {\mathrm{moles \ of \ H_2O}}{(\mathrm{moles \ of \ H_2O}) + (\mathrm{moles \ of \ CO_2}) + (\mathrm{moles \ of \ O_2})} = 0.474$$
After that, take out the $\mathrm{H_2O}$ and do something quite similar except with $\mathrm{CO_2}$ in the numerator (and the ratio value being different).
If you do things right, you'll have two equations and two unknowns ($P$ and $x$). That's enough to solve for $P$.
4. Jul 30, 2016
### MickeyBlue
My final answer for propane was well above 100 and I couldn't understand why.
5. Jul 30, 2016
### collinsmark
Show your work, and perhaps we may be able to point you in the right direction.
6. Jul 30, 2016
### MickeyBlue
Oh! Thank you, I see where I went wrong now. Because of how I added mols of O2 consumed I took it as the amount in excess instead. You're right; this should be enough to get P.
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# Source code for sympy.combinatorics.tensor_can
from __future__ import print_function, division
from sympy.core.compatibility import range
from sympy.combinatorics.permutations import Permutation, _af_rmul, \
_af_invert, _af_new
from sympy.combinatorics.perm_groups import PermutationGroup, _orbit, \
_orbit_transversal
from sympy.combinatorics.util import _distribute_gens_by_base, \
_orbits_transversals_from_bsgs
"""
References for tensor canonicalization:
[1] R. Portugal "Algorithmic simplification of tensor expressions",
J. Phys. A 32 (1999) 7779-7789
[2] R. Portugal, B.F. Svaiter "Group-theoretic Approach for Symbolic
Tensor Manipulation: I. Free Indices"
arXiv:math-ph/0107031v1
[3] L.R.U. Manssur, R. Portugal "Group-theoretic Approach for Symbolic
Tensor Manipulation: II. Dummy Indices"
arXiv:math-ph/0107032v1
[4] xperm.c part of XPerm written by J. M. Martin-Garcia
http://www.xact.es/index.html
"""
def dummy_sgs(dummies, sym, n):
"""
Return the strong generators for dummy indices
Parameters
==========
dummies : list of dummy indices
dummies[2k], dummies[2k+1] are paired indices
sym : symmetry under interchange of contracted dummies::
* None no symmetry
* 0 commuting
* 1 anticommuting
n : number of indices
in base form the dummy indices are always in consecutive positions
Examples
========
>>> from sympy.combinatorics.tensor_can import dummy_sgs
>>> dummy_sgs(range(2, 8), 0, 8)
[[0, 1, 3, 2, 4, 5, 6, 7, 8, 9], [0, 1, 2, 3, 5, 4, 6, 7, 8, 9],
[0, 1, 2, 3, 4, 5, 7, 6, 8, 9], [0, 1, 4, 5, 2, 3, 6, 7, 8, 9],
[0, 1, 2, 3, 6, 7, 4, 5, 8, 9]]
"""
if len(dummies) > n:
raise ValueError("List too large")
res = []
# exchange of contravariant and covariant indices
if sym is not None:
for j in dummies[::2]:
a = list(range(n + 2))
if sym == 1:
a[n] = n + 1
a[n + 1] = n
a[j], a[j + 1] = a[j + 1], a[j]
res.append(a)
# rename dummy indices
for j in dummies[:-3:2]:
a = list(range(n + 2))
a[j:j + 4] = a[j + 2], a[j + 3], a[j], a[j + 1]
res.append(a)
return res
def _min_dummies(dummies, sym, indices):
"""
Return list of minima of the orbits of indices in group of dummies
see double_coset_can_rep for the description of dummies and sym
indices is the initial list of dummy indices
Examples
========
>>> from sympy.combinatorics.tensor_can import _min_dummies
>>> _min_dummies([list(range(2, 8))], [0], list(range(10)))
[0, 1, 2, 2, 2, 2, 2, 2, 8, 9]
"""
num_types = len(sym)
m = []
for dx in dummies:
if dx:
m.append(min(dx))
else:
m.append(None)
res = indices[:]
for i in range(num_types):
for c, i in enumerate(indices):
for j in range(num_types):
if i in dummies[j]:
res[c] = m[j]
break
return res
def _trace_S(s, j, b, S_cosets):
"""
Return the representative h satisfying s[h[b]] == j
If there is not such a representative return None
"""
for h in S_cosets[b]:
if s[h[b]] == j:
return h
return None
def _trace_D(gj, p_i, Dxtrav):
"""
Return the representative h satisfying h[gj] == p_i
If there is not such a representative return None
"""
for h in Dxtrav:
if h[gj] == p_i:
return h
return None
def _dumx_remove(dumx, dumx_flat, p0):
"""
remove p0 from dumx
"""
res = []
for dx in dumx:
if p0 not in dx:
res.append(dx)
continue
k = dx.index(p0)
if k % 2 == 0:
p0_paired = dx[k + 1]
else:
p0_paired = dx[k - 1]
dx.remove(p0)
dx.remove(p0_paired)
dumx_flat.remove(p0)
dumx_flat.remove(p0_paired)
res.append(dx)
def transversal2coset(size, base, transversal):
a = []
j = 0
for i in range(size):
if i in base:
a.append(sorted(transversal[j].values()))
j += 1
else:
a.append([list(range(size))])
j = len(a) - 1
while a[j] == [list(range(size))]:
j -= 1
return a[:j + 1]
[docs]def double_coset_can_rep(dummies, sym, b_S, sgens, S_transversals, g): """ Butler-Portugal algorithm for tensor canonicalization with dummy indices dummies list of lists of dummy indices, one list for each type of index; the dummy indices are put in order contravariant, covariant [d0, -d0, d1, -d1, ...]. sym list of the symmetries of the index metric for each type. possible symmetries of the metrics * 0 symmetric * 1 antisymmetric * None no symmetry b_S base of a minimal slot symmetry BSGS. sgens generators of the slot symmetry BSGS. S_transversals transversals for the slot BSGS. g permutation representing the tensor. Return 0 if the tensor is zero, else return the array form of the permutation representing the canonical form of the tensor. A tensor with dummy indices can be represented in a number of equivalent ways which typically grows exponentially with the number of indices. To be able to establish if two tensors with many indices are equal becomes computationally very slow in absence of an efficient algorithm. The Butler-Portugal algorithm [3] is an efficient algorithm to put tensors in canonical form, solving the above problem. Portugal observed that a tensor can be represented by a permutation, and that the class of tensors equivalent to it under slot and dummy symmetries is equivalent to the double coset D*g*S (Note: in this documentation we use the conventions for multiplication of permutations p, q with (p*q)(i) = p[q[i]] which is opposite to the one used in the Permutation class) Using the algorithm by Butler to find a representative of the double coset one can find a canonical form for the tensor. To see this correspondence, let g be a permutation in array form; a tensor with indices ind (the indices including both the contravariant and the covariant ones) can be written as t = T(ind[g[0],..., ind[g[n-1]]), where n= len(ind); g has size n + 2, the last two indices for the sign of the tensor (trick introduced in [4]). A slot symmetry transformation s is a permutation acting on the slots t -> T(ind[(g*s)[0]],..., ind[(g*s)[n-1]]) A dummy symmetry transformation acts on ind t -> T(ind[(d*g)[0]],..., ind[(d*g)[n-1]]) Being interested only in the transformations of the tensor under these symmetries, one can represent the tensor by g, which transforms as g -> d*g*s, so it belongs to the coset D*g*S. Let us explain the conventions by an example. Given a tensor T^{d3 d2 d1}{}_{d1 d2 d3} with the slot symmetries T^{a0 a1 a2 a3 a4 a5} = -T^{a2 a1 a0 a3 a4 a5} T^{a0 a1 a2 a3 a4 a5} = -T^{a4 a1 a2 a3 a0 a5} and symmetric metric, find the tensor equivalent to it which is the lowest under the ordering of indices: lexicographic ordering d1, d2, d3 then and contravariant index before covariant index; that is the canonical form of the tensor. The canonical form is -T^{d1 d2 d3}{}_{d1 d2 d3} obtained using T^{a0 a1 a2 a3 a4 a5} = -T^{a2 a1 a0 a3 a4 a5}. To convert this problem in the input for this function, use the following labelling of the index names (- for covariant for short) d1, -d1, d2, -d2, d3, -d3 T^{d3 d2 d1}{}_{d1 d2 d3} corresponds to g = [4, 2, 0, 1, 3, 5, 6, 7] where the last two indices are for the sign sgens = [Permutation(0, 2)(6, 7), Permutation(0, 4)(6, 7)] sgens[0] is the slot symmetry -(0, 2) T^{a0 a1 a2 a3 a4 a5} = -T^{a2 a1 a0 a3 a4 a5} sgens[1] is the slot symmetry -(0, 4) T^{a0 a1 a2 a3 a4 a5} = -T^{a4 a1 a2 a3 a0 a5} The dummy symmetry group D is generated by the strong base generators [(0, 1), (2, 3), (4, 5), (0, 1)(2, 3),(2, 3)(4, 5)] The dummy symmetry acts from the left d = [1, 0, 2, 3, 4, 5, 6, 7] exchange d1 -> -d1 T^{d3 d2 d1}{}_{d1 d2 d3} == T^{d3 d2}{}_{d1}{}^{d1}{}_{d2 d3} g=[4, 2, 0, 1, 3, 5, 6, 7] -> [4, 2, 1, 0, 3, 5, 6, 7] = _af_rmul(d, g) which differs from _af_rmul(g, d). The slot symmetry acts from the right s = [2, 1, 0, 3, 4, 5, 7, 6] exchanges slots 0 and 2 and changes sign T^{d3 d2 d1}{}_{d1 d2 d3} == -T^{d1 d2 d3}{}_{d1 d2 d3} g=[4,2,0,1,3,5,6,7] -> [0, 2, 4, 1, 3, 5, 7, 6] = _af_rmul(g, s) Example in which the tensor is zero, same slot symmetries as above: T^{d3}{}_{d1,d2}{}^{d1}{}_{d3}{}^{d2} = -T^{d3}{}_{d1,d3}{}^{d1}{}_{d2}{}^{d2} under slot symmetry -(2,4); = T_{d3 d1}{}^{d3}{}^{d1}{}_{d2}{}^{d2} under slot symmetry -(0,2); = T^{d3}{}_{d1 d3}{}^{d1}{}_{d2}{}^{d2} symmetric metric; = 0 since two of these lines have tensors differ only for the sign. The double coset D*g*S consists of permutations h = d*g*s corresponding to equivalent tensors; if there are two h which are the same apart from the sign, return zero; otherwise choose as representative the tensor with indices ordered lexicographically according to [d1, -d1, d2, -d2, d3, -d3] that is rep = min(D*g*S) = min([d*g*s for d in D for s in S]) The indices are fixed one by one; first choose the lowest index for slot 0, then the lowest remaining index for slot 1, etc. Doing this one obtains a chain of stabilizers S -> S_{b0} -> S_{b0,b1} -> ... and D -> D_{p0} -> D_{p0,p1} -> ... where [b0, b1, ...] = range(b) is a base of the symmetric group; the strong base b_S of S is an ordered sublist of it; therefore it is sufficient to compute once the strong base generators of S using the Schreier-Sims algorithm; the stabilizers of the strong base generators are the strong base generators of the stabilizer subgroup. dbase = [p0, p1, ...] is not in general in lexicographic order, so that one must recompute the strong base generators each time; however this is trivial, there is no need to use the Schreier-Sims algorithm for D. The algorithm keeps a TAB of elements (s_i, d_i, h_i) where h_i = d_i*g*s_i satisfying h_i[j] = p_j for 0 <= j < i starting from s_0 = id, d_0 = id, h_0 = g. The equations h_0[0] = p_0, h_1[1] = p_1,... are solved in this order, choosing each time the lowest possible value of p_i For j < i d_i*g*s_i*S_{b_0,...,b_{i-1}}*b_j = D_{p_0,...,p_{i-1}}*p_j so that for dx in D_{p_0,...,p_{i-1}} and sx in S_{base[0],...,base[i-1]} one has dx*d_i*g*s_i*sx*b_j = p_j Search for dx, sx such that this equation holds for j = i; it can be written as s_i*sx*b_j = J, dx*d_i*g*J = p_j sx*b_j = s_i**-1*J; sx = trace(s_i**-1, S_{b_0,...,b_{i-1}}) dx**-1*p_j = d_i*g*J; dx = trace(d_i*g*J, D_{p_0,...,p_{i-1}}) s_{i+1} = s_i*trace(s_i**-1*J, S_{b_0,...,b_{i-1}}) d_{i+1} = trace(d_i*g*J, D_{p_0,...,p_{i-1}})**-1*d_i h_{i+1}*b_i = d_{i+1}*g*s_{i+1}*b_i = p_i h_n*b_j = p_j for all j, so that h_n is the solution. Add the found (s, d, h) to TAB1. At the end of the iteration sort TAB1 with respect to the h; if there are two consecutive h in TAB1 which differ only for the sign, the tensor is zero, so return 0; if there are two consecutive h which are equal, keep only one. Then stabilize the slot generators under i and the dummy generators under p_i. Assign TAB = TAB1 at the end of the iteration step. At the end TAB contains a unique (s, d, h), since all the slots of the tensor h have been fixed to have the minimum value according to the symmetries. The algorithm returns h. It is important that the slot BSGS has lexicographic minimal base, otherwise there is an i which does not belong to the slot base for which p_i is fixed by the dummy symmetry only, while i is not invariant from the slot stabilizer, so p_i is not in general the minimal value. This algorithm differs slightly from the original algorithm [3]: the canonical form is minimal lexicographically, and the BSGS has minimal base under lexicographic order. Equal tensors h are eliminated from TAB. Examples ======== >>> from sympy.combinatorics.permutations import Permutation >>> from sympy.combinatorics.perm_groups import PermutationGroup >>> from sympy.combinatorics.tensor_can import double_coset_can_rep, get_transversals >>> gens = [Permutation(x) for x in [[2, 1, 0, 3, 4, 5, 7, 6], [4, 1, 2, 3, 0, 5, 7, 6]]] >>> base = [0, 2] >>> g = Permutation([4, 2, 0, 1, 3, 5, 6, 7]) >>> transversals = get_transversals(base, gens) >>> double_coset_can_rep([list(range(6))], [0], base, gens, transversals, g) [0, 1, 2, 3, 4, 5, 7, 6] >>> g = Permutation([4, 1, 3, 0, 5, 2, 6, 7]) >>> double_coset_can_rep([list(range(6))], [0], base, gens, transversals, g) 0 """ size = g.size g = g.array_form num_dummies = size - 2 indices = list(range(num_dummies)) all_metrics_with_sym = all([_ is not None for _ in sym]) num_types = len(sym) dumx = dummies[:] dumx_flat = [] for dx in dumx: dumx_flat.extend(dx) b_S = b_S[:] sgensx = [h._array_form for h in sgens] if b_S: S_transversals = transversal2coset(size, b_S, S_transversals) # strong generating set for D dsgsx = [] for i in range(num_types): dsgsx.extend(dummy_sgs(dumx[i], sym[i], num_dummies)) ginv = _af_invert(g) idn = list(range(size)) # TAB = list of entries (s, d, h) where h = _af_rmuln(d,g,s) # for short, in the following d*g*s means _af_rmuln(d,g,s) TAB = [(idn, idn, g)] for i in range(size - 2): b = i testb = b in b_S and sgensx if testb: sgensx1 = [_af_new(_) for _ in sgensx] deltab = _orbit(size, sgensx1, b) else: deltab = {b} # p1 = min(IMAGES) = min(Union D_p*h*deltab for h in TAB) if all_metrics_with_sym: md = _min_dummies(dumx, sym, indices) else: md = [min(_orbit(size, [_af_new( ddx) for ddx in dsgsx], ii)) for ii in range(size - 2)] p_i = min([min([md[h[x]] for x in deltab]) for s, d, h in TAB]) dsgsx1 = [_af_new(_) for _ in dsgsx] Dxtrav = _orbit_transversal(size, dsgsx1, p_i, False, af=True) \ if dsgsx else None if Dxtrav: Dxtrav = [_af_invert(x) for x in Dxtrav] # compute the orbit of p_i for ii in range(num_types): if p_i in dumx[ii]: # the orbit is made by all the indices in dum[ii] if sym[ii] is not None: deltap = dumx[ii] else: # the orbit is made by all the even indices if p_i # is even, by all the odd indices if p_i is odd p_i_index = dumx[ii].index(p_i) % 2 deltap = dumx[ii][p_i_index::2] break else: deltap = [p_i] TAB1 = [] nTAB = len(TAB) while TAB: s, d, h = TAB.pop() if min([md[h[x]] for x in deltab]) != p_i: continue deltab1 = [x for x in deltab if md[h[x]] == p_i] # NEXT = s*deltab1 intersection (d*g)**-1*deltap dg = _af_rmul(d, g) dginv = _af_invert(dg) sdeltab = [s[x] for x in deltab1] gdeltap = [dginv[x] for x in deltap] NEXT = [x for x in sdeltab if x in gdeltap] # d, s satisfy # d*g*s*base[i-1] = p_{i-1}; using the stabilizers # d*g*s*S_{base[0],...,base[i-1]}*base[i-1] = # D_{p_0,...,p_{i-1}}*p_{i-1} # so that to find d1, s1 satisfying d1*g*s1*b = p_i # one can look for dx in D_{p_0,...,p_{i-1}} and # sx in S_{base[0],...,base[i-1]} # d1 = dx*d; s1 = s*sx # d1*g*s1*b = dx*d*g*s*sx*b = p_i for j in NEXT: if testb: # solve s1*b = j with s1 = s*sx for some element sx # of the stabilizer of ..., base[i-1] # sx*b = s**-1*j; sx = _trace_S(s, j,...) # s1 = s*trace_S(s**-1*j,...) s1 = _trace_S(s, j, b, S_transversals) if not s1: continue else: s1 = [s[ix] for ix in s1] else: s1 = s # assert s1[b] == j # invariant # solve d1*g*j = p_i with d1 = dx*d for some element dg # of the stabilizer of ..., p_{i-1} # dx**-1*p_i = d*g*j; dx**-1 = trace_D(d*g*j,...) # d1 = trace_D(d*g*j,...)**-1*d # to save an inversion in the inner loop; notice we did # Dxtrav = [perm_af_invert(x) for x in Dxtrav] out of the loop if Dxtrav: d1 = _trace_D(dg[j], p_i, Dxtrav) if not d1: continue else: if p_i != dg[j]: continue d1 = idn assert d1[dg[j]] == p_i # invariant d1 = [d1[ix] for ix in d] h1 = [d1[g[ix]] for ix in s1] # assert h1[b] == p_i # invariant TAB1.append((s1, d1, h1)) # if TAB contains equal permutations, keep only one of them; # if TAB contains equal permutations up to the sign, return 0 TAB1.sort(key=lambda x: x[-1]) nTAB1 = len(TAB1) prev = [0] * size while TAB1: s, d, h = TAB1.pop() if h[:-2] == prev[:-2]: if h[-1] != prev[-1]: return 0 else: TAB.append((s, d, h)) prev = h # stabilize the SGS sgensx = [h for h in sgensx if h[b] == b] if b in b_S: b_S.remove(b) _dumx_remove(dumx, dumx_flat, p_i) dsgsx = [] for i in range(num_types): dsgsx.extend(dummy_sgs(dumx[i], sym[i], num_dummies))
return TAB[0][-1] def canonical_free(base, gens, g, num_free): """ canonicalization of a tensor with respect to free indices choosing the minimum with respect to lexicographical ordering in the free indices base, gens BSGS for slot permutation group g permutation representing the tensor num_free number of free indices The indices must be ordered with first the free indices see explanation in double_coset_can_rep The algorithm is a variation of the one given in [2]. Examples ======== >>> from sympy.combinatorics import Permutation >>> from sympy.combinatorics.tensor_can import canonical_free >>> gens = [[1, 0, 2, 3, 5, 4], [2, 3, 0, 1, 4, 5],[0, 1, 3, 2, 5, 4]] >>> gens = [Permutation(h) for h in gens] >>> base = [0, 2] >>> g = Permutation([2, 1, 0, 3, 4, 5]) >>> canonical_free(base, gens, g, 4) [0, 3, 1, 2, 5, 4] Consider the product of Riemann tensors T = R^{a}_{d0}^{d1,d2}*R_{d2,d1}^{d0,b} The order of the indices is [a, b, d0, -d0, d1, -d1, d2, -d2] The permutation corresponding to the tensor is g = [0, 3, 4, 6, 7, 5, 2, 1, 8, 9] In particular a is position 0, b is in position 9. Use the slot symmetries to get T is a form which is the minimal in lexicographic order in the free indices a and b, e.g. -R^{a}_{d0}^{d1,d2}*R^{b,d0}_{d2,d1} corresponding to [0, 3, 4, 6, 1, 2, 7, 5, 9, 8] >>> from sympy.combinatorics.tensor_can import riemann_bsgs, tensor_gens >>> base, gens = riemann_bsgs >>> size, sbase, sgens = tensor_gens(base, gens, [[], []], 0) >>> g = Permutation([0, 3, 4, 6, 7, 5, 2, 1, 8, 9]) >>> canonical_free(sbase, [Permutation(h) for h in sgens], g, 2) [0, 3, 4, 6, 1, 2, 7, 5, 9, 8] """ g = g.array_form size = len(g) if not base: return g[:] transversals = get_transversals(base, gens) m = len(base) for x in sorted(g[:-2]): if x not in base: base.append(x) h = g for i, transv in enumerate(transversals): b = base[i] h_i = [size]*num_free # find the element s in transversals[i] such that # _af_rmul(h, s) has its free elements with the lowest position in h s = None for sk in transv.values(): h1 = _af_rmul(h, sk) hi = [h1.index(ix) for ix in range(num_free)] if hi < h_i: h_i = hi s = sk if s: h = _af_rmul(h, s) return h def _get_map_slots(size, fixed_slots): res = list(range(size)) pos = 0 for i in range(size): if i in fixed_slots: continue res[i] = pos pos += 1 return res def _lift_sgens(size, fixed_slots, free, s): a = [] j = k = 0 fd = list(zip(fixed_slots, free)) fd = [y for x, y in sorted(fd)] num_free = len(free) for i in range(size): if i in fixed_slots: a.append(fd[k]) k += 1 else: a.append(s[j] + num_free) j += 1 return a
[docs]def canonicalize(g, dummies, msym, *v): """ canonicalize tensor formed by tensors Parameters ========== g : permutation representing the tensor dummies : list representing the dummy indices it can be a list of dummy indices of the same type or a list of lists of dummy indices, one list for each type of index; the dummy indices must come after the free indices, and put in order contravariant, covariant [d0, -d0, d1,-d1,...] msym : symmetry of the metric(s) it can be an integer or a list; in the first case it is the symmetry of the dummy index metric; in the second case it is the list of the symmetries of the index metric for each type v : list, (base_i, gens_i, n_i, sym_i) for tensors of type i base_i, gens_i : BSGS for tensors of this type. The BSGS should have minimal base under lexicographic ordering; if not, an attempt is made do get the minimal BSGS; in case of failure, canonicalize_naive is used, which is much slower. n_i : number of tensors of type i. sym_i : symmetry under exchange of component tensors of type i. Both for msym and sym_i the cases are * None no symmetry * 0 commuting * 1 anticommuting Returns ======= 0 if the tensor is zero, else return the array form of the permutation representing the canonical form of the tensor. Algorithm ========= First one uses canonical_free to get the minimum tensor under lexicographic order, using only the slot symmetries. If the component tensors have not minimal BSGS, it is attempted to find it; if the attempt fails canonicalize_naive is used instead. Compute the residual slot symmetry keeping fixed the free indices using tensor_gens(base, gens, list_free_indices, sym). Reduce the problem eliminating the free indices. Then use double_coset_can_rep and lift back the result reintroducing the free indices. Examples ======== one type of index with commuting metric; A_{a b} and B_{a b} antisymmetric and commuting T = A_{d0 d1} * B^{d0}{}_{d2} * B^{d2 d1} ord = [d0,-d0,d1,-d1,d2,-d2] order of the indices g = [1, 3, 0, 5, 4, 2, 6, 7] T_c = 0 >>> from sympy.combinatorics.tensor_can import get_symmetric_group_sgs, canonicalize, bsgs_direct_product >>> from sympy.combinatorics import Permutation >>> base2a, gens2a = get_symmetric_group_sgs(2, 1) >>> t0 = (base2a, gens2a, 1, 0) >>> t1 = (base2a, gens2a, 2, 0) >>> g = Permutation([1, 3, 0, 5, 4, 2, 6, 7]) >>> canonicalize(g, range(6), 0, t0, t1) 0 same as above, but with B_{a b} anticommuting T_c = -A^{d0 d1} * B_{d0}{}^{d2} * B_{d1 d2} can = [0,2,1,4,3,5,7,6] >>> t1 = (base2a, gens2a, 2, 1) >>> canonicalize(g, range(6), 0, t0, t1) [0, 2, 1, 4, 3, 5, 7, 6] two types of indices [a,b,c,d,e,f] and [m,n], in this order, both with commuting metric f^{a b c} antisymmetric, commuting A_{m a} no symmetry, commuting T = f^c{}_{d a} * f^f{}_{e b} * A_m{}^d * A^{m b} * A_n{}^a * A^{n e} ord = [c,f,a,-a,b,-b,d,-d,e,-e,m,-m,n,-n] g = [0,7,3, 1,9,5, 11,6, 10,4, 13,2, 12,8, 14,15] The canonical tensor is T_c = -f^{c a b} * f^{f d e} * A^m{}_a * A_{m d} * A^n{}_b * A_{n e} can = [0,2,4, 1,6,8, 10,3, 11,7, 12,5, 13,9, 15,14] >>> base_f, gens_f = get_symmetric_group_sgs(3, 1) >>> base1, gens1 = get_symmetric_group_sgs(1) >>> base_A, gens_A = bsgs_direct_product(base1, gens1, base1, gens1) >>> t0 = (base_f, gens_f, 2, 0) >>> t1 = (base_A, gens_A, 4, 0) >>> dummies = [range(2, 10), range(10, 14)] >>> g = Permutation([0, 7, 3, 1, 9, 5, 11, 6, 10, 4, 13, 2, 12, 8, 14, 15]) >>> canonicalize(g, dummies, [0, 0], t0, t1) [0, 2, 4, 1, 6, 8, 10, 3, 11, 7, 12, 5, 13, 9, 15, 14] """ from sympy.combinatorics.testutil import canonicalize_naive if not isinstance(msym, list): if not msym in [0, 1, None]: raise ValueError('msym must be 0, 1 or None') num_types = 1 else: num_types = len(msym) if not all(msymx in [0, 1, None] for msymx in msym): raise ValueError('msym entries must be 0, 1 or None') if len(dummies) != num_types: raise ValueError( 'dummies and msym must have the same number of elements') size = g.size num_tensors = 0 v1 = [] for i in range(len(v)): base_i, gens_i, n_i, sym_i = v[i] # check that the BSGS is minimal; # this property is used in double_coset_can_rep; # if it is not minimal use canonicalize_naive if not _is_minimal_bsgs(base_i, gens_i): mbsgs = get_minimal_bsgs(base_i, gens_i) if not mbsgs: can = canonicalize_naive(g, dummies, msym, *v) return can base_i, gens_i = mbsgs v1.append((base_i, gens_i, [[]] * n_i, sym_i)) num_tensors += n_i if num_types == 1 and not isinstance(msym, list): dummies = [dummies] msym = [msym] flat_dummies = [] for dumx in dummies: flat_dummies.extend(dumx) if flat_dummies and flat_dummies != list(range(flat_dummies[0], flat_dummies[-1] + 1)): raise ValueError('dummies is not valid') # slot symmetry of the tensor size1, sbase, sgens = gens_products(*v1) if size != size1: raise ValueError( 'g has size %d, generators have size %d' % (size, size1)) free = [i for i in range(size - 2) if i not in flat_dummies] num_free = len(free) # g1 minimal tensor under slot symmetry g1 = canonical_free(sbase, sgens, g, num_free) if not flat_dummies: return g1 # save the sign of g1 sign = 0 if g1[-1] == size - 1 else 1 # the free indices are kept fixed. # Determine free_i, the list of slots of tensors which are fixed # since they are occupied by free indices, which are fixed. start = 0 for i in range(len(v)): free_i = [] base_i, gens_i, n_i, sym_i = v[i] len_tens = gens_i[0].size - 2 # for each component tensor get a list od fixed islots for j in range(n_i): # get the elements corresponding to the component tensor h = g1[start:(start + len_tens)] fr = [] # get the positions of the fixed elements in h for k in free: if k in h: fr.append(h.index(k)) free_i.append(fr) start += len_tens v1[i] = (base_i, gens_i, free_i, sym_i) # BSGS of the tensor with fixed free indices # if tensor_gens fails in gens_product, use canonicalize_naive size, sbase, sgens = gens_products(*v1) # reduce the permutations getting rid of the free indices pos_dummies = [g1.index(x) for x in flat_dummies] pos_free = [g1.index(x) for x in range(num_free)] size_red = size - num_free g1_red = [x - num_free for x in g1 if x in flat_dummies] if sign: g1_red.extend([size_red - 1, size_red - 2]) else: g1_red.extend([size_red - 2, size_red - 1]) map_slots = _get_map_slots(size, pos_free) sbase_red = [map_slots[i] for i in sbase if i not in pos_free] sgens_red = [_af_new([map_slots[i] for i in y._array_form if i not in pos_free]) for y in sgens] dummies_red = [[x - num_free for x in y] for y in dummies] transv_red = get_transversals(sbase_red, sgens_red) g1_red = _af_new(g1_red) g2 = double_coset_can_rep( dummies_red, msym, sbase_red, sgens_red, transv_red, g1_red) if g2 == 0: return 0 # lift to the case with the free indices g3 = _lift_sgens(size, pos_free, free, g2)
return g3 def perm_af_direct_product(gens1, gens2, signed=True): """ direct products of the generators gens1 and gens2 Examples ======== >>> from sympy.combinatorics.tensor_can import perm_af_direct_product >>> gens1 = [[1, 0, 2, 3], [0, 1, 3, 2]] >>> gens2 = [[1, 0]] >>> perm_af_direct_product(gens1, gens2, False) [[1, 0, 2, 3, 4, 5], [0, 1, 3, 2, 4, 5], [0, 1, 2, 3, 5, 4]] >>> gens1 = [[1, 0, 2, 3, 5, 4], [0, 1, 3, 2, 4, 5]] >>> gens2 = [[1, 0, 2, 3]] >>> perm_af_direct_product(gens1, gens2, True) [[1, 0, 2, 3, 4, 5, 7, 6], [0, 1, 3, 2, 4, 5, 6, 7], [0, 1, 2, 3, 5, 4, 6, 7]] """ gens1 = [list(x) for x in gens1] gens2 = [list(x) for x in gens2] s = 2 if signed else 0 n1 = len(gens1[0]) - s n2 = len(gens2[0]) - s start = list(range(n1)) end = list(range(n1, n1 + n2)) if signed: gens1 = [gen[:-2] + end + [gen[-2] + n2, gen[-1] + n2] for gen in gens1] gens2 = [start + [x + n1 for x in gen] for gen in gens2] else: gens1 = [gen + end for gen in gens1] gens2 = [start + [x + n1 for x in gen] for gen in gens2] res = gens1 + gens2 return res
[docs]def bsgs_direct_product(base1, gens1, base2, gens2, signed=True): """ direct product of two BSGS base1 base of the first BSGS. gens1 strong generating sequence of the first BSGS. base2, gens2 similarly for the second BSGS. signed flag for signed permutations. Examples ======== >>> from sympy.combinatorics import Permutation >>> from sympy.combinatorics.tensor_can import (get_symmetric_group_sgs, bsgs_direct_product) >>> Permutation.print_cyclic = True >>> base1, gens1 = get_symmetric_group_sgs(1) >>> base2, gens2 = get_symmetric_group_sgs(2) >>> bsgs_direct_product(base1, gens1, base2, gens2) ([1], [(4)(1 2)]) """ s = 2 if signed else 0 n1 = gens1[0].size - s base = list(base1) base += [x + n1 for x in base2] gens1 = [h._array_form for h in gens1] gens2 = [h._array_form for h in gens2] gens = perm_af_direct_product(gens1, gens2, signed) size = len(gens[0]) id_af = list(range(size)) gens = [h for h in gens if h != id_af] if not gens: gens = [id_af]
return base, [_af_new(h) for h in gens]
[docs]def get_symmetric_group_sgs(n, antisym=False): """ Return base, gens of the minimal BSGS for (anti)symmetric tensor n rank of the tensor antisym = False symmetric tensor antisym = True antisymmetric tensor Examples ======== >>> from sympy.combinatorics import Permutation >>> from sympy.combinatorics.tensor_can import get_symmetric_group_sgs >>> Permutation.print_cyclic = True >>> get_symmetric_group_sgs(3) ([0, 1], [(4)(0 1), (4)(1 2)]) """ if n == 1: return [], [_af_new(list(range(3)))] gens = [Permutation(n - 1)(i, i + 1)._array_form for i in range(n - 1)] if antisym == 0: gens = [x + [n, n + 1] for x in gens] else: gens = [x + [n + 1, n] for x in gens] base = list(range(n - 1))
return base, [_af_new(h) for h in gens] riemann_bsgs = [0, 2], [Permutation(0, 1)(4, 5), Permutation(2, 3)(4, 5), Permutation(5)(0, 2)(1, 3)] def get_transversals(base, gens): """ Return transversals for the group with BSGS base, gens """ if not base: return [] stabs = _distribute_gens_by_base(base, gens) orbits, transversals = _orbits_transversals_from_bsgs(base, stabs) transversals = [{x: h._array_form for x, h in y.items()} for y in transversals] return transversals def _is_minimal_bsgs(base, gens): """ Check if the BSGS has minimal base under lexigographic order. base, gens BSGS Examples ======== >>> from sympy.combinatorics import Permutation >>> from sympy.combinatorics.tensor_can import riemann_bsgs, _is_minimal_bsgs >>> _is_minimal_bsgs(*riemann_bsgs) True >>> riemann_bsgs1 = ([2, 0], ([Permutation(5)(0, 1)(4, 5), Permutation(5)(0, 2)(1, 3)])) >>> _is_minimal_bsgs(*riemann_bsgs1) False """ base1 = [] sgs1 = gens[:] size = gens[0].size for i in range(size): if not all(h._array_form[i] == i for h in sgs1): base1.append(i) sgs1 = [h for h in sgs1 if h._array_form[i] == i] return base1 == base def get_minimal_bsgs(base, gens): """ Compute a minimal GSGS base, gens BSGS If base, gens is a minimal BSGS return it; else return a minimal BSGS if it fails in finding one, it returns None TODO: use baseswap in the case in which if it fails in finding a minimal BSGS Examples ======== >>> from sympy.combinatorics import Permutation >>> from sympy.combinatorics.tensor_can import get_minimal_bsgs >>> Permutation.print_cyclic = True >>> riemann_bsgs1 = ([2, 0], ([Permutation(5)(0, 1)(4, 5), Permutation(5)(0, 2)(1, 3)])) >>> get_minimal_bsgs(*riemann_bsgs1) ([0, 2], [(0 1)(4 5), (5)(0 2)(1 3), (2 3)(4 5)]) """ G = PermutationGroup(gens) base, gens = G.schreier_sims_incremental() if not _is_minimal_bsgs(base, gens): return None return base, gens def tensor_gens(base, gens, list_free_indices, sym=0): """ Returns size, res_base, res_gens BSGS for n tensors of the same type base, gens BSGS for tensors of this type list_free_indices list of the slots occupied by fixed indices for each of the tensors sym symmetry under commutation of two tensors sym None no symmetry sym 0 commuting sym 1 anticommuting Examples ======== >>> from sympy.combinatorics import Permutation >>> from sympy.combinatorics.tensor_can import tensor_gens, get_symmetric_group_sgs >>> Permutation.print_cyclic = True two symmetric tensors with 3 indices without free indices >>> base, gens = get_symmetric_group_sgs(3) >>> tensor_gens(base, gens, [[], []]) (8, [0, 1, 3, 4], [(7)(0 1), (7)(1 2), (7)(3 4), (7)(4 5), (7)(0 3)(1 4)(2 5)]) two symmetric tensors with 3 indices with free indices in slot 1 and 0 >>> tensor_gens(base, gens, [[1], [0]]) (8, [0, 4], [(7)(0 2), (7)(4 5)]) four symmetric tensors with 3 indices, two of which with free indices """ def _get_bsgs(G, base, gens, free_indices): """ return the BSGS for G.pointwise_stabilizer(free_indices) """ if not free_indices: return base[:], gens[:] else: H = G.pointwise_stabilizer(free_indices) base, sgs = H.schreier_sims_incremental() return base, sgs # if not base there is no slot symmetry for the component tensors # if list_free_indices.count([]) < 2 there is no commutation symmetry # so there is no resulting slot symmetry if not base and list_free_indices.count([]) < 2: n = len(list_free_indices) size = gens[0].size size = n * (gens[0].size - 2) + 2 return size, [], [_af_new(list(range(size)))] # if any(list_free_indices) one needs to compute the pointwise # stabilizer, so G is needed if any(list_free_indices): G = PermutationGroup(gens) else: G = None # no_free list of lists of indices for component tensors without fixed # indices no_free = [] size = gens[0].size id_af = list(range(size)) num_indices = size - 2 if not list_free_indices[0]: no_free.append(list(range(num_indices))) res_base, res_gens = _get_bsgs(G, base, gens, list_free_indices[0]) for i in range(1, len(list_free_indices)): base1, gens1 = _get_bsgs(G, base, gens, list_free_indices[i]) res_base, res_gens = bsgs_direct_product(res_base, res_gens, base1, gens1, 1) if not list_free_indices[i]: no_free.append(list(range(size - 2, size - 2 + num_indices))) size += num_indices nr = size - 2 res_gens = [h for h in res_gens if h._array_form != id_af] # if sym there are no commuting tensors stop here if sym is None or not no_free: if not res_gens: res_gens = [_af_new(id_af)] return size, res_base, res_gens # if the component tensors have moinimal BSGS, so is their direct # product P; the slot symmetry group is S = P*C, where C is the group # to (anti)commute the component tensors with no free indices # a stabilizer has the property S_i = P_i*C_i; # the BSGS of P*C has SGS_P + SGS_C and the base is # the ordered union of the bases of P and C. # If P has minimal BSGS, so has S with this base. base_comm = [] for i in range(len(no_free) - 1): ind1 = no_free[i] ind2 = no_free[i + 1] a = list(range(ind1[0])) a.extend(ind2) a.extend(ind1) base_comm.append(ind1[0]) a.extend(list(range(ind2[-1] + 1, nr))) if sym == 0: a.extend([nr, nr + 1]) else: a.extend([nr + 1, nr]) res_gens.append(_af_new(a)) res_base = list(res_base) # each base is ordered; order the union of the two bases for i in base_comm: if i not in res_base: res_base.append(i) res_base.sort() if not res_gens: res_gens = [_af_new(id_af)] return size, res_base, res_gens def gens_products(*v): """ Returns size, res_base, res_gens BSGS for n tensors of different types v is a sequence of (base_i, gens_i, free_i, sym_i) where base_i, gens_i BSGS of tensor of type i free_i list of the fixed slots for each of the tensors of type i; if there are n_i tensors of type i and none of them have fixed slots, free = [[]]*n_i sym 0 (1) if the tensors of type i (anti)commute among themselves Examples ======== >>> from sympy.combinatorics import Permutation >>> from sympy.combinatorics.tensor_can import get_symmetric_group_sgs, gens_products >>> Permutation.print_cyclic = True >>> base, gens = get_symmetric_group_sgs(2) >>> gens_products((base, gens, [[], []], 0)) (6, [0, 2], [(5)(0 1), (5)(2 3), (5)(0 2)(1 3)]) >>> gens_products((base, gens, [[1], []], 0)) (6, [2], [(5)(2 3)]) """ res_size, res_base, res_gens = tensor_gens(*v[0]) for i in range(1, len(v)): size, base, gens = tensor_gens(*v[i]) res_base, res_gens = bsgs_direct_product(res_base, res_gens, base, gens, 1) res_size = res_gens[0].size id_af = list(range(res_size)) res_gens = [h for h in res_gens if h != id_af] if not res_gens: res_gens = [id_af] return res_size, res_base, res_gens
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# Solving a quartic equation
I'm attempting to rearrange an equation from an answer on the Mathematics StackExchange.
The answer given is this equation:
$$L^2 = (-ab(t)+p)^2-\left(\frac{(-ab(t)+p).(cd(t)-ab(t))}{(cd(t)-ab(t))^2}(cd(t)-ab(t))\right)^2$$
Where $a$, $b$, $c$, $d$, and $p$ are known 2D points, $L$ is a known length, and $t$ is an unknown scalar. $ab(t)$ indicates interpolation between $a$ and $b$.
I am interested in rearranging this to solve for $t$. Here's what I've tried in Sage:
def sqr(var): return var.dot_product(var)
var('ax bx cx dx px ay by cy dy py t L')
a = vector([ax, ay])
b = vector([bx, by])
c = vector([cx, cy])
d = vector([dx, dy])
p = vector([px, py])
g = a - at + bt
h = c - ct + dt
u = p - g
v = h - g
eq = L^2 == sqr(u) - sqr((u.dot_product(v)/sqr(v)) * v)
eq.solve(t)
At the solve step I have observed it to sit for quite a while without producing a result. Two questions:
1. Am I inputting the problem correctly?
2. Is there any way to know if this is likely to terminate in a reasonable time? I have no idea what the solver looks like under the hood, and wouldn't want to wait for some O(n!) calculation to terminate :)
edit retag close merge delete
The problem per se seems quite involved. The r.h.s when expanded displays 3 million characters:
sage: len(str(eq.rhs().expand()))
3220606
( 2015-06-01 01:04:35 -0600 )edit
Sort by » oldest newest most voted
There is a simpification possible. Instead of expanding the expression (which makes for a nice Pynac benchmark) we transform the equation into a fraction and get the numerator:
....
sage: g = a - a*t + b*t
sage: h = c - c*t + d*t
sage: u = p - g
sage: v = h - g
sage: eq = L^2 == sqr(u) - sqr((u.dot_product(v)/sqr(v)) * v)
sage: rhs = eq.rhs()
sage: num = (rhs.rational_simplify()-L^2).numerator()
sage: solve(num, t)
This is much smaller but unfortunately still takes forever to solve. Using collect and coefficient you can let Sage show you the coefficient of t^0, t, t^2, t^3, and t^4 to get an idea of what's involved and which equations determine the solution process of the quartic. For example this is the constant coefficient:
sage: num.coefficient(t,0)
-L^2*ax^2 - L^2*ay^2 + 2*L^2*ax*cx - L^2*cx^2 + ay^2*cx^2 + 2*L^2*ay*cy - 2*ax*ay*cx*cy - L^2*cy^2 + ax^2*cy^2 - 2*ay^2*cx*px + 2*ax*ay*cy*px + 2*ay*cx*cy*px - 2*ax*cy^2*px + ay^2*px^2 - 2*ay*cy*px^2 + cy^2*px^2 + 2*ax*ay*cx*py - 2*ay*cx^2*py - 2*ax^2*cy*py + 2*ax*cx*cy*py - 2*ax*ay*px*py + 2*ay*cx*px*py + 2*ax*cy*px*py - 2*cx*cy*px*py + ax^2*py^2 - 2*ax*cx*py^2 + cx^2*py^2
You could also get from Sage the general solution to the quartic via:
sage: (a,b,c,d,e) = var('a,b,c,d,e')
sage: sol = solve(a*x^4+b*x^3+c*x^2+d*x+e,x)
This returns a vector of 4 solutions. You could then simply substitute your coefficients for a,b,c,d,e in one of the solutions:
sage: sol1 = sol[0]
sage: sol1 = sol1.subs(e==num.coefficient(t,0)).subs(d==num.coefficient(t,1)).subs(c==num.coefficient(t,2)).subs(b==num.coefficient(t,3)).subs(a==num.coefficient(t,4))
sage: len(str(sol1))
497783
This solution however is still too big to handle. It needs 500 pages to print! Simplifying is out of the question.
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Posted by: matheuscmss | September 26, 2012
## Homoclinic/heteroclinic bifurcations: slightly fat horseshoes A
1. Introduction
Last time we saw that bifurcations of quadratic tangencies associated to fat horseshoes, ${\textrm{HD}(K)=d_s^0+d_u^0>1}$, are complicated because of persistent tangencies, but, by the works of C. Moreira and J.-C. Yoccoz (see here and here), one realizes that, from the heuristic point of view, the “critical locus${K^s(g)\cap K^u(g)}$ (i.e., the region where the tangencies destroying the hyperbolicity show up) is very small, i.e., its Hausdorff dimension ${\textrm{HD}(K)-1}$ is close to zero, if the initial horseshoe ${K}$ is only slightly fat, i.e., ${\textrm{HD}(K)>1}$ is close to ${1}$. In particular, one could imagine that bifurcations quadratic tangencies of slightly fat horseshoes could lead to a local dynamics on ${\Lambda_g}$ satisfying some form of weak (non-uniform) hyperbolicity for most ${g\in\mathcal{U}_+}$ despite the fact that ${\Lambda_g}$ doesn’t verify strong (uniform) hyperbolicity conditions in general.
In their tour-de-force work (of 217 pages!), J. Palis and J.-C. Yoccoz were able to formalize this crude heuristic argument by showing (among several other things) the following result in the context of heteroclinic bifurcations of slightly fat horseshoes.
Let ${f}$ be a smooth diffeomorphism of a compact surface ${M}$ possessing a uniformly hyperbolic horseshoe ${K}$ displaying a heteroclinic quadratic tangency, that is, ${K}$ contains two periodic points ${p_s}$ and ${p_u}$ with distinctorbits such that ${W^s(p_s)}$ and ${W^u(p_u)}$ have a quadratic tangency (i.e., a contact of order ${1}$) at some point ${q\in M-K}$. Let ${U}$ be a sufficiently small neighborhood of ${K}$ and let ${V}$ be a sufficiently small neighborhood of ${q}$. Denote by ${\mathcal{U}}$ be a sufficiently small neighborhood of ${f}$ and, as usual, let’s organize ${\mathcal{U}}$ into ${\mathcal{U}=\mathcal{U}_-\cup\mathcal{U}_0\cup\mathcal{U}_+}$ depending on the relative positions of the continuations of ${W^s(p_s)}$ and ${W^u(p_u)}$ near ${V}$.
Organization of the parameter space ${\mathcal{U}}$.
Finally, let us denote by ${d_s^0}$ and ${d_u^0}$ the stable and unstable dimensions of the horseshoe ${K}$ of ${f\in\mathcal{U}_0}$.
Theorem 1 (J. Palis and J.-C. Yoccoz) In the setting of the paragraph above, suppose that ${K}$ is slightly fat in the sense that
$\displaystyle (d_s^0+d_u^0)^2+(\max\{d_s^0,d_u^0\})^2 < (d_s^0+d_u^0)+(\max\{d_s^0,d_u^0\}) \ \ \ \ \ (1)$
Then, ${\Lambda_g}$ is a non-uniformly hyperbolic horseshoe for most ${g\in\mathcal{U}_+}$.
Remark 1 At first sight, there is no reason to restrict our attention to heteroclinic tangencies in the previous theorem. In fact, as we’ll see later (cf. Remark 9), for certain technical reasons, the arguments of J. Palis and J.-C. Yoccoz can treat only heteroclinic tangencies. Of course, the authors believe that this is merely an artifact of their methods, but unfortunately they don’t know how to modify the proofs to also include the case of homoclinic tangencies.
Remark 2 (“Pedagogical” remark) For those who like to see videos and can understand Portuguese, it is worth to note that J.-C. Yoccoz gave a series of lectures at IMPA in 2009 around his works with S. Marmi and P. Moussa, A. Avila and J. Bochi, and J. Palis, and his lectures were recorded by IMPA’s staff. In particular, one can find links for all these materials (including some lecture notes taken by Aline Cerqueira) here.
Concerning the statement of this result, let’s us comment first on the condition (1). As a trivial remark, note that this condition includes the case ${d_s^0+d_u^0<1}$ of thin horseshoes, but this is not surprising as any reasonable definition of “non-uniformly hyperbolic horseshoe” must include uniformly hyperbolic horseshoes as particular examples. Of course, this remark is not particularly interesting because the case of thin horseshoes was already treated by S. Newhouse, J. Palis and F. Takes (cf. the third post in this series), so that the condition (1) is really interesting in the regime of fat horseshoes ${d_s^0+d_u^0>1}$. Here, one can get a clear idea about (1) by assuming ${\max\{d_s^0,d_u^0\}=d_s^0}$ or ${d_u^0}$ (i.e., by breaking the natural symmetry between ${d_s^0}$ and ${d_u^0}$), and by noticing that the boundary of the region determined by (1)is the union of two ellipses meeting the diagonal ${\{d_s^0=d_u^0\}}$ at the point ${(3/5,3/5)}$ as indicated in the figure below:
Region of parameters ${d_s^0}$ and ${d_u^0}$ where the results of Newhouse-Palis-Takens (NPT) and Palis-Yoccoz (PY) apply.
In this figure, we used the horizontal axis for the variable ${d_s^0}$ and the vertical axis for the variable ${d_u^0}$. Also, we pointed out, for sake of comparison, two famous families of dynamical systems lying outside the scope of Theorem 1, namely the Hénon maps ${H_{a,b}:\mathbb{R}^2\rightarrow\mathbb{R}^2}$, ${H_{a,b}(x,y)=(1-ax^2+y,bx)}$, and the standard family ${f_{\lambda}:\mathbb{T}^2\rightarrow\mathbb{T}^2}$, ${f_{\lambda}(x,y)=(2x+\lambda\sin(2\pi x)-y, x)}$. Indeed, these important examples of dynamical systems can’t be studied by the current methods of J. Palis and J.-C. Yoccoz because they display homoclinic/heteroclinic bifurcations associated to “very fat horseshoes”:
• in the case of Hénon maps, the “horseshoes” have “stable dimension” ${d_s^0=1}$ and a very small unstable dimension ${0 for certain parameters ${(a,b)}$, and
• in the case of the standard family, one has horseshoes with ${d_s^0=d_u^0}$ arbitrarily close to ${1}$ for large values of ${\lambda\in\mathbb{R}}$ (see, e.g., this post here).
Now, let us start to explain the meaning of non-uniformly hyperbolic horseshoe in Theorem 1. As we explained in the first post of this series, a (uniformly hyperbolic) horseshoe ${\Lambda}$ of a surface diffeomorphism ${f:M\rightarrow M}$ is a saddle-like object in the sense that ${\Lambda}$ is not an attractor nor a repellor, that is, both its stable set
$\displaystyle W^s(\Lambda):=\{y\in M: \textrm{dist}(f^n(y),\Lambda)\rightarrow 0 \textrm{ as }n\rightarrow+\infty\}$
and unstable set
$\displaystyle W^u(\Lambda):=\{y\in M: \textrm{dist}(f^n(y),\Lambda)\rightarrow 0 \textrm{ as }n\rightarrow+\infty\}$
have zero Lebesgue measure ${\textrm{Leb}_2}$. Here, ${\textrm{Leb}_2}$ is the ${2}$-dimensional Lebesgue measure of ${M}$. In a similar vein, J. Palis and J.-C. Yoccoz (cf. Theorem 7 of their article) showed that their non-uniformly hyperbolic horseshoes are saddle-like objects:
Theorem 2 Under the same assumptions of Theorem 1, one has that
$\displaystyle \textrm{Leb}_2(W^s(\Lambda_g))=\textrm{Leb}_2(W^u(\Lambda_g))=0$
for most ${g\in\mathcal{U}_+}$.
Actually, the statement of Theorem 7 of their article contains a slightly more precise explanation of the non-uniformly hyperbolic features of ${\Lambda_g}$ (for most ${g\in\mathcal{U}_+}$): it is possible to show that ${\Lambda_g}$ supports geometric Sinai-Ruelle-Bowen (SRB) measures with non-zero Lyapunov exponents, that is, ${\Lambda_g}$ is a non-uniformly hyperbolic object in the sense of the so-called Pesin theory. Unfortunately, a detailed explanation of these terms (i.e., SRB measures, Lyapunov exponents, Pesin theory) is out of the scope of these notes and we refer the curious reader to the original articles R. Bowen and D. Ruelle, Y. Sinai, the books of A. Katok and B. Hasselblatt and C. Bonatti, L. Diaz and M. Viana, and the links attached to these keywords for more informations.
In order to further explain the structure of ${\Lambda_g}$, we’ll briefly describe in the sequel some elements of the proof of Theorem 1.
2. A global view on Palis-Yoccoz induction scheme
Let ${(g_t)_{|t| a smooth ${1}$-parameter family transverse to ${\mathcal{U}_0}$ at ${f=g_0}$, where ${f}$ is a diffeomorphism with a slightly fat horseshoe ${K}$ exhibiting a heteroclinic quadratic tangency as shown in the figure below:
Heteroclinic quadratic tangency associated to a slightly fat horseshoe.
As usual, we wish to understand the local dynamics of ${g=g_t}$ on the neighborhood ${U\cup V}$ indicated in the picture above, that is, we want to investigate the structure of the set
$\displaystyle \Lambda_g=\bigcap\limits_{n\in\mathbb{Z}} g^n(U\cup V)$
for most parameters ${t\in (0,t_0)}$.
In this direction, we consider ${0<\varepsilon_0\ll1}$ and we look at the parameter interval ${I_0:=[\varepsilon_0,2\varepsilon_0]}$. Very roughly speaking, the scheme of J. Palis and J.-C. Yoccoz has the following structure: besides ${\varepsilon_0}$, one has two extra parameters ${\tau}$ and ${\eta}$ chosen such that
$\displaystyle 0<\varepsilon_0\ll\eta\ll\tau\ll1$
Then, one proceeds inductively:
• at the 1st stage, one defines ${\varepsilon_1:=\varepsilon_0^{1+\tau}}$ and one divides the interval ${I_0:=[\varepsilon_0, 2\varepsilon_0]}$ into ${[\varepsilon_0^{-\tau}]}$ candidates subintervals;
• then, one apply an exam called strong regularity to each candidate subinterval: the good subintervals (passing the exam) are kept while the bad ones are discarded;
• after that, one goes to the next stages, that is, one takes the good intervals ${I_k}$ from the ${k}$th stage, subdivide them into ${[\varepsilon_k^{-\tau}]}$ subintervals of size ${\varepsilon_{k+1}:=\varepsilon_k^{1+\tau}}$, apply the strong regularity exam to each subinterval and one keeps the good subintervals and discard the bad subintervals.
Of course, the strong regularity of an interval ${I}$ is a property about the (non-uniform) hyperbolic features of ${\Lambda_{g_t}}$ for all parameters ${t\in I}$, and the choice of the set of properties defining the strong regularity must be extremely careful: it should not be too weak (otherwise one doesn’t get hyperbolicity) nor too strong (otherwise there is a risk that no interval is good at some stage).
Actually, as we’ll see later, for each candidate interval ${I}$, J. Palis and J.-C. Yoccoz construct a class ${\mathcal{R}(I)}$ of so-called (${I}$-persistent) affine-like iterates of ${g=g_t}$, ${t\in I}$ and they “test” by strong regularity of ${I}$ by examining the features of the class ${\mathcal{R}(I)}$.
Remark 3 It is worth to point out that the class ${\mathcal{R}(I)}$ is unique, but this is shown in the article only a posteriori. Also, a nice feature of the arguments of J. Palis and J.-C. Yoccoz is that they are time-symmetric, that is, the dynamical estimates for the past and the future are the same (i.e., one has only to do half of the computations). In particular, those readers with some familiarity with Hénon maps know that the past behavior is very different from the future behavior (due to strong dissipation) and this morally explains why the methods of their article are not directly useful in the case of Hénon maps.
After this very approximative description of Palis-Yoccoz inductive scheme, it is clear that one of the key ideas is to carefully setup the notion of strong regularity property. However, before discussing this subject, we need to make some preparations: firstly we need to localize the dynamics, secondly we need to introduce the affine-like iterates, and thirdly we need to introduce the class ${\mathcal{R}(I)}$.
2.1. Localization of the dynamics
The local dynamics of ${g_t}$ for ${t\in I_0:=[\varepsilon_0,2\varepsilon_0]}$ has the following appearance:
Parabolic tongues created after unfolding a heteroclinic tangency.
As it is highlighted in this picture, after unfolding the tangency, we get two regions ${L_u}$ and ${L_s}$ called unstable and stable parabolic tongues bounded by the pieces of ${W_{loc}^s(p_s)}$ and ${W^u_{loc}(p_u)}$ near ${V}$. The transition time ${N_0}$ from the unstable tongue ${L_u}$ to the stable tongue ${L_s}$ under the dynamics ${g}$ is a large but fixed integer depending only on ${f\in\mathcal{U}_0}$.
Using this, we can organize the local dynamics of ${g}$ on ${U\cup V}$ as follows. Firstly, we select a finite Markov partition of the horseshoe ${K_g=\bigcap\limits_{n\in\mathbb{Z}}g^n(U)}$ into compact disjoint rectangles ${R_a}$, ${a\in\mathcal{A}}$, i.e., by fixing a convenient system of coordinates, we write ${R_a\simeq I_a^u\times I_a^s}$ in such a way that:
• the derivative of ${g|{R_a}}$ expands (uniformly) the horizontal direction and contracts (uniformly) the vertical direction,
• ${K_g}$ is the maximal invariant set of the interior ${\textrm{int}(R)}$ of ${R:=\bigcup\limits_{a\in\mathcal{A}} R_a}$, i.e., ${K_g=\bigcap\limits_{n\in\mathbb{Z}}g^n(\textrm{int}(R))}$,
• ${g(I_a^u\times\partial I_a^s)\cap \textrm{int}(R)=\emptyset=g^{-1}(\partial I_a^u\times I_a^s)\cap \textrm{int}(R)}$;
• ${\{R_a\cap K_g\}_{a\in\mathcal{A}}}$ is a Markov partition, i.e., ${g^{-1}(I_a^u\times\partial I_a^s)\subset \bigcup\limits_{b\in\mathcal{A}}(I_b^u\times\partial I_b^s)}$, ${g(\partial I_a^u\times I_a^s)\subset \bigcup\limits_{b\in\mathcal{A}}(\partial I_b^u\times I_b^s)}$, and there exists an integer ${n\in\mathbb{N}}$ with ${g^n(R_a)\cap R_{a'}\neq\emptyset}$ for all ${a,a'\in\mathcal{A}}$.
Secondly, we denote by ${\mathcal{B}=\{(a,a')\in\mathcal{A}^2: g(R_a)\cap R_{a'}\neq\emptyset\}}$. Then, in this setting, it is not hard to see that the local dynamics of ${g}$ on ${U\cup V}$ is given by
• the uniformly hyperbolic maps ${g:R_a\cap g^{-1}(R_a)\rightarrow g(R_a)\cap R_{a'}}$, ${(a,a')\in\mathcal{B}}$ related to the horseshoe ${K_g}$ (they are called uniformly hyperbolic because the horizontal direction is uniformly expanded and the vertical direction is uniformly contracted by their derivatives), and
• the folding map ${G:=g^{N_0}: L_u\rightarrow L_s}$ making the transition between parabolic tongues.
In this context, by letting ${\widehat{R}=R\cup\bigcup\limits_{i=1}^{N_0-1}g^i(L_u)}$, we have that ${\Lambda_g}$ is the maximal invariant set of ${\widehat{R}}$, i.e., ${\Lambda_g=\bigcap\limits_{n\in\mathbb{Z}}g^n(\widehat{R})}$.
This localization of the dynamics of ${g}$ on ${\Lambda_g}$ to the region ${\widehat{R}}$ is useful because it allows to think of ${\Lambda_g}$ in terms of an iterated system of maps, i.e., we approach the points of ${\Lambda_g}$ by looking at the domains and the images of the compositions (i.e., certain ${g}$-iterates) of the uniformly hyperbolic maps ${g:R_a\cap g^{-1}(R_a)\rightarrow g(R_a)\cap R_{a'}}$ and the folding map ${G=g^{N_0}: L_u\rightarrow L_s}$.
By thinking in this way, we see that the points in the domains and/or images of ${g}$-iterates (composition) with affine-like features, that is, ${g}$-iterates whose derivates expand the horizontal direction and contract the vertical direction, will contribute to the hyperbolicity of ${\Lambda_g}$. In other words, it is desirable to get as much affine-like iterates as possible. Of course, the ${g}$-iterates obtained by composition of transition maps ${g:R_a\cap g^{-1}(R_a)\rightarrow g(R_a)\cap R_{a'}}$ related to the horseshoe ${K_g}$ have affine-like features (by definition), so that one risks to lose the affine-like property only when one considers compositions with the folding map ${G}$ (because the folding map mixes up the horizontal and vertical directions).
In particular, this suggests that strong regularity property (whatever this means) has something to do with the consecutive passages through the critical region given by the parabolic tongues ${L_u}$ and ${L_s}$. However, before pursuing this direction, let’s formalize the notion of affine-like iterates (following this article here).
2.2. Affine-like maps
A vertical strip ${P\subset R_0\simeq I_0^u\times I_0^s}$ is a region of the form
$\displaystyle P=\{(x_0,y_0)\in R_0: \phi^-(y_0)\leq x_0\leq \phi^+(y_0)\}$
and a horizontal strip${Q\subset R_1\simeq I_1^u\times I_1^s}$ is a region of the form
$\displaystyle Q=\{(x_1,y_1)\in R_1: \psi^-(x_1)\leq y_1\leq \psi^+(x_1)\}$
Intuitively, we wish to call “affine-like” a map ${F:P\rightarrow Q}$ from a vertical strip ${P}$ to a horizontal strip ${Q}$ “approximately” contracting the vertical direction and expanding horizontal direction such as the one depicted in the figure below.
Geometry of affine-like maps.
Formally, we define:
Definition 3 We say that a map ${F(x_0,y_0)=(x_1,y_1)}$ from a vertical strip ${P}$ to a horizontal strip ${Q}$ is weakly affine-like whenever ${F}$ admits an implicit representation ${(A,B)}$, i.e., we can write ${x_0=A(x_1,y_0)}$ and ${y_1=B(x_1,y_0)}$. Equivalently, ${F}$ is weakly affine-like if and only if the projection ${\pi}$ from the graph ${\textrm{graph}(F)}$ of ${F}$ to ${I_0^u\times I_1^s}$ is a diffeomorphism.
This definition of affine-like maps ${F}$ in terms of implicit representations ${(A,B)}$ is somewhat folkloric in Dynamical Systems, and it was used by J. Palis and J.-C. Yoccoz because it is technically easier to estimate ${(A,B)}$ than ${F}$ as the symmetry between past and future is more evident, and ${(A,B)}$ are “contractive” maps.
In what follows, we will denote the derivatives of ${A}$ and ${B}$ by ${A_x, B_x, A_y, B_y, A_{xx}, B_{xx}, \dots}$. Following J. Palis and J.-C. Yoccoz, we will consider exclusively with weakly affine-like maps satisfying the following hyperbolicity conditions:
Definition 4 A weakly affine-like map ${F}$ is called affine-likeif its implicit representation ${(A,B)}$ verifies:
• Cone condition: ${\lambda|A_x|+u|A_y|\leq 1}$ and ${\lambda|B_y|+v|B_x|\leq 1}$ where ${1, and
• Bounded distortion condition: ${\partial_x\log |A_x|, \partial_y\log |A_x|, A_{yy}, \partial_y\log |B_y|, \partial_x\log |B_y|, B_{xx}}$ are uniformly bounded by some constant ${C>0}$.
Here, the constants ${\lambda, u, v}$ and ${C}$ are fixed once and for all depending only on ${f=g_0\in\mathcal{U}_0}$.
Informally, the cone condition says that ${F}$ contracts the vertical direction and expands the horizontal direction, and the bounded distortion condition says that the derivative of ${F}$ behaves in the same way in all scales.
For later use, we introduce the following notion:
Definition 5 The widths of the domain ${P}$ and the image ${Q}$ of an affine-like map ${F:P\rightarrow Q}$ with implicit representation ${(A,B)}$ are
$\displaystyle |P|=\max|A_x| \quad \textrm{ and } \quad |Q|=\max|B_y|$
Once we dispose of the notion of affine-like iterates, we’re ready to introduce the class ${\mathcal{R}(I)}$ whose “strong regularity” will be tested later.
2.3. Simple and parabolic compositions of affine-like maps and the class ${\mathcal{R}(I)}$
Coming back to the interpretation of the dynamics on ${\Lambda_g}$ as an iterated system of maps given by compositions of ${g:R_a\cap g^{-1}(R_a)\rightarrow g(R_a)\cap R_{a'}}$ and the folding map ${G:L_u\rightarrow L_s}$, we see that the following two ways of composing affine-like maps are particularly interesting in our context.
Definition 6 Let ${F:P\rightarrow Q}$ and ${F':P'\rightarrow Q'}$ be two affine-like maps such that ${Q, P'\subset R_{a'}}$. Then, the simple composition ${F''=F'\circ F}$ is the affine-like map with domain ${P'':=P\cap F^{-1}(P')}$ and image ${Q'':=Q'\cap F'(Q)}$ shown in the figure below.
Simple composition of affine-like maps.
Remark 4 By direct inspection of definitions, one can check that ${|P''|\sim|P|\cdot|P'|}$ where the implied constant depends only on ${f}$ (by means of the constants ${u,v}$ in the cone condition).
The composition of two transition maps ${g:R_a\cap g^{-1}(R_a)\rightarrow g(R_a)\cap R_{a'}}$ and ${g:R_{a'}\cap g^{-1}(R_{a''})\rightarrow g(R_{a'})\cap R_{a''}}$ associated to the horseshoe ${K_g}$ is the canonical example of simple composition.
In particular, if we wish to understand ${\Lambda_g}$, it is not a good idea to work only with simple compositions, that is, we must include some passages through the parabolic tongues. This is formalized by the following notion.
Definition 7 Denote by ${R_{a_u}}$ and ${R_{a_s}}$ the rectangles of the Markov partition of ${K_g}$ containing the parabolic tongues ${L_u}$ and ${L_s}$. Let ${F_0:P_0\rightarrow Q_0}$ and ${F_1=P_1\rightarrow Q_1}$ be two affine-like maps such that ${Q_0}$, resp. ${P_1}$, passes near the parabolic tongue ${L_u}$, resp. ${L_s}$, i.e., ${Q_0\subset R_{a_u}}$ crosses ${L_u}$ and ${P_1\subset R_{a_s}}$ crosses ${L_s}$. We define the parabolic compositions of ${F_0}$ and ${F_1}$ as follows. Firstly, we compare ${Q_0}$ with the parabolic-like strip ${G^{-1}(P_1\cap L_s)}$ and we say that the parabolic composition of ${F_0}$ and ${F_1}$ is possible if the intersection ${Q_0\cap G^{-1}(P_1\cap L_s)}$ has two connected components ${Q_0^-}$ and ${Q_0^+}$ as shown (in black) in Figure 2 below. Then, assuming that the parabolic composition of ${F_0}$ and ${F_1}$ is possible, we define their parabolic compositions as the two weakly affine-like maps ${F^-:P^-\rightarrow Q^-}$ and ${F^+:P^+\rightarrow Q^+}$ shown in Figure 2below obtaining by concatenating ${F_0}$, the folding map ${G}$ and ${F_1}$ in the strips ${P^-=F_0^{-1}(Q_0^-)}$, ${P^+=F_0^{-1}(Q_0^+)}$, ${Q^-=F_1(G(Q_0^-))}$, ${Q^+=F_1(G(Q_0^+))}$.
Parabolic compositions of affine-like maps.
As it is indicated in the figure above, the parabolic composition comes with an important parameter ${\delta(Q_0,P_1)}$ measuring the distance between the vertical strip ${P_1}$ and the tip of the parabolic-like strip ${G(Q_0\cap L_u)}$, or, equivalently, the horizontal strip ${Q_0}$ and the tip of the parabolic-like strip ${G^{-1}(P_1\cap L_s)}$.
Remark 5 By direct inspection of definitions, one can check that ${|P^{\pm}|=\frac{|P_0|\cdot|P_1|}{\delta(Q_0,P_1)^{1/2}}}$.
In this notation, the class ${\mathcal{R}(I)}$ is defined as follows.
Definition 8 ${\mathcal{R}(I)}$ is the class of affine-like iterates of ${g_t}$, ${t\in I}$, closed under all simple compositions and certain parabolic compositions. More precisely, ${\mathcal{R}(I)}$ contains only parabolic compositions satisfying certain transversalityconditions such as
$\displaystyle \delta(Q_0,P_1)\geq \max\{|Q_0|^{1-\eta}, |P_1|^{1-\eta}, |I|\}.$
Remark 6 In fact, the transversality conditions on parabolic compositions imposed by J. Palis and J.-C. Yoccoz involves 6 conditions besides the one on the parameter ${\delta(Q_0, P_1)}$ given above.
For later use, we denote by ${(P,Q,n)}$ an affine-like iterate ${g^n:P\rightarrow Q}$ taking a vertical strip ${P}$ to a horizontal strip ${Q}$ after ${n}$ iterations of ${g=g_t}$.
At this stage, we are ready to discuss the strong regularity property for ${\mathcal{R}(I)}$.
2.4. Critical strips, bicritical dynamics and strong regularity
Let ${(P,Q,n)\in\mathcal{R}(I)}$.
Definition 9 We say that ${P}$, resp. ${Q}$, is ${I}$-critical when ${P}$, resp. ${Q}$ is not ${I}$-transverse to the parabolic tongue ${L_s}$, resp. ${L_u}$, i.e., the distance between ${P}$, resp. ${Q}$ to the “tip” of ${L_s}$, resp. ${L_u}$, is smaller ${|P|^{1-\eta}}$, resp. ${|Q|^{1-\eta}}$ for some ${t\in I}$.
Definition 10 We say that an element ${(P,Q,n)\in\mathcal{R}(I)}$ is ${I}$-bicritical if ${P}$ and ${Q}$ are ${I}$-critical.
In other words, a bicritical ${(P,Q,n)\in\mathcal{R}(I)}$ corresponds to some part of the dynamics starting at some ${P}$ close to the tip of ${L_s}$ and ending at some ${Q}$ close to the tip of ${L_u}$, that is, a bicritical ${(P,Q,n)\in\mathcal{R}(I)}$ corresponds to a return of the critical region to itself.
Of course, one way of getting hyperbolicity for ${\Lambda_g}$ is to control the bicritical dynamics, i.e., bicritical elements ${(P,Q,n)\in\mathcal{R}(I)}$.
Definition 11 Given ${\beta>1}$, we say that a candidate parameter ${I}$ is ${\beta}$-regularif
$\displaystyle |P|,|Q|<|I|^{\beta}$
for every ${I}$-bicritical element ${(P,Q,n)\in \mathcal{R}(I)}$.
Remark 7 In their article, J. Palis and J.-C. Yoccoz choose ${\beta>1}$ depending only on the stable and unstable dimensions ${d_s^0}$ and ${d_u^0}$ of the initial horseshoe ${K}$ and the hyperbolicity strength of the periodic points ${p_s}$ and ${p_u}$ involved in the heteroclinic tangency. See Equation (5.19) of Palis-Yoccoz article for the precise requirements on ${\beta}$.
Intuitively, a candidate parameter interval ${I}$ is ${\beta}$-regular if the bicritical dynamics seen through ${\mathcal{R}(I)}$ is confined to very small strips ${P}$ and ${Q}$. Unfortunately, the condition of ${\beta}$-regularity is not enough to run the induction scheme of J. Palis and J.-C. Yoccoz, and they end up by introducing a more technical condition called strong regularity. However, for the sake of this text, let’s ignore this issue by pretending that strong regularity is ${\beta}$-regularity for some adequate parameter ${\beta>1}$.
After this brief discussion of strong regularity, it is time to come back to Palis-Yoccoz induction scheme in order to say a few words about the dynamics of ${\Lambda_{g_t}}$ for ${t}$ belonging to strongly regular intervals.
2.5. Dynamics of strong regular parameters
As it is explained in Sections 10 and 11 of their article, J. Palis and J.-C. Yoccoz can reasonably control the dynamics of ${\Lambda_{g_t}}$ for strongly regular parameters ${t\in I_0=[\varepsilon_0,2\varepsilon_0]}$: these are the parameters ${t\in \bigcap\limits_{m=0}^{\infty} I_m}$ where ${I_0\supset I_1\supset\dots\supset I_m\supset\dots}$ is a decreasing sequence of strongly regular intervals ${I_m}$.
Remark 8 It is interesting to notice that the strongly regular parameters of Palis-Yoccoz are not defined a priori, i.e., one has to perform the entire induction scheme before putting the hands on them. This is in contrast with the so-called Jakobson theorem, a sort of ${1}$-dimensional version of Theorem 1, where the strongly regular parameters are known since the beginning of the argument.
Of course, before starting the analysis of strongly regular parameters, one needs to ensure that such parameters exist, that is, one want to know whether there are parameters left from the parameter exclusion scheme of J. Palis and J.-C. Yoccoz. This issue is carefully treated in Section 9 (of 50 pages!) of Palis-Yoccoz article, where the authors estimate the relative speed of strips associated to elements ${(P,Q,n)\in\mathcal{R}(I)}$ when the parameter ${t\in I}$ moves, and, by induction, they are able to control the measure of bad (not strongly regular) intervals: as it turns out, the measure of the set of bad intervals is ${\leq \varepsilon_0^{1+\tau^2}}$, so that the strongly regular parameters ${t\in I_0=[\varepsilon_0,2\varepsilon_0]}$ have almost full measure in ${I_0}$, i.e., ${\geq\varepsilon_0(1 - \varepsilon_0^{\tau^2})}$ (cf. Corollary 15 of Palis-Yoccoz article).
Remark 9 In order to get some strongly regular parameter, one has to ensure that the initial interval ${I_0}$ is strongly regular (otherwise, one ends up by excluding ${I_0}$ in the first step of Palis-Yoccoz induction scheme, so that one has no parameters to play with in the next rounds of the induction!). Here, J. Palis and J.-C. Yoccoz makes use of the technical assumption that one is unfolding a heteroclinic tangency: indeed, the idea is that the formation of bicritical elements takes a long time in heteroclinic tangencies because the points in the critical region should pass near ${p_s}$ first, then near ${p_u}$ and only then they can return to the critical region again; of course, in the case of homoclinic tangencies, it may happen that bicritical elements pop up quickly and this is why one can’t include homoclinic tangencies in the statement of Theorem 1.
From now on, let us fix ${t\in \bigcap\limits_{m=0}^\infty I_m}$ a strongly regular parameter, and let’s study ${\Lambda_g}$ for ${g=g_t}$. Keeping this goal in mind, we introduce ${\mathcal{R}=\mathcal{R}(t)=\bigcup\limits_{m=0}^{\infty}\mathcal{R}(I_m)}$ the collection of all affine-like iterates of ${g}$ coming from the strongly regular intervals ${I_m}$. Using the class ${\mathcal{R}}$, we can define the class ${\mathcal{R}_+^{\infty}}$ of stable curves, i.e., the class of curves ${\omega}$ coming from intersections ${\omega=\bigcap\limits_{m=0}^\infty P_m}$ of decreasing sequences ${P_0\supset P_1\supset\dots}$ of vertical strips serving as domain of affine-like iterates of ${g}$, that is, ${(P_m,Q_m,n_m)\in\mathcal{R}}$. Also, we put ${\widetilde{\mathcal{R}}_+^{\infty}=\bigcup\limits_{\omega\in\mathcal{R}_+^{\infty}}\omega\subset M}$ the set of points of ${M}$ in some stable curve.
These stable curves were introduced by analogy with uniformly hyperbolic horseshoes: indeed, the stable lamination of ${K_g}$ can be recovered from the transitions maps ${g:R_a\cap g^{-1}(R_a)\rightarrow g(R_a)\cap R_{a'}}$ by looking at the decreasing sequences of domains of simple compositions of these transitions maps.
From the nice features of strong regular parameters, it is possible to prove that the class ${\mathcal{R}_+^{\infty}}$ is a ${C^{1+Lip}}$-lamination and one can use ${g}$ to induce a dynamical system ${T^+:\mathcal{D}_+\subset \mathcal{R}_+^{\infty}\rightarrow\mathcal{R}_+^{\infty}}$ isomorphic to a Bernoulli map with infinitely many branches (cf. Subsection 10.4 of Palis-Yoccoz article). Here, ${\mathcal{D}_+}$ is the set of stable curves not contained in infinitely many prime elements of ${\mathcal{R}}$ (we say that an element ${(P,Q,n)\in\mathcal{R}}$ is prime if it can’t be obtained by simple composition of shorter elements ${(P_0,Q_0,n_0), (P_1,Q_1,n_1)\in\mathcal{R}}$ (shorter meaning ${n_0,n_1)). In particular, as it is shown in Subsections 10.5, 10.6, 10.7, 10.8, 10.9, 10.10 of Palis-Yoccoz article, ${T^+}$ is a non-uniformly hyperbolic dynamical system (in a very precise sense). Of course, by the symmetry between past and future (see Remark 3), one also has an analog non-uniformly hyperbolic dynamical system ${T^-:\mathcal{D}_-\subset \mathcal{R}_-^{\infty}\rightarrow\mathcal{R}_-^{\infty}}$ on unstable curves, so that ${\Lambda_g}$ inherits a natural non-uniformly hyperbolic part consisting of points whose ${T^+}$ and ${T^-}$ iterates never escape ${\mathcal{R}_+^{\infty}}$ and ${\mathcal{R}_-^{\infty}}$.
Therefore, if we can show that the size of the sets of the points of ${\Lambda_g}$ escaping ${\mathcal{R}_+^{\infty}}$ and/or ${\mathcal{R}_-^{\infty}}$ is relatively small compared to the non-uniformly hyperbolic part of ${\Lambda_g}$, then we can say that ${\Lambda_g}$ is a non-uniformly hyperbolic horseshoe. Here, J. Palis and J.-C. Yoccoz set up in Section 11 of their article a series of estimates towards showing that the points of ${\Lambda}$ escaping ${\mathcal{R}_+^{\infty}}$ and/or ${\mathcal{R}_-^{\infty}}$ are exceptional: for instance, they show Theorem 2 that the ${2}$-dimensional Lebesgue measure of ${W^s(\Lambda_g)}$ is zero because this property is true for the non-uniformly hyperbolic part of ${\Lambda_g}$ (by the “usual” hyperbolic theory) and the set of points of ${\Lambda_g}$ escaping ${\mathcal{R}_+^{\infty}}$ and/or ${\mathcal{R}_-^{\infty}}$ are “rare” in the sense that their ${2}$-dimensional Lebesgue measure contribute as an “error term” to the the non-uniformly hyperbolic part of ${\Lambda_g}$.
At this point, our overview of Palis-Yoccoz induction scheme is complete. Closing this post, we would like to make two comments. Firstly, as it is pointed out in page 14 of Palis-Yoccoz article, the philosophy that ${\Lambda_g}$ is constituted of a non-uniformly hyperbolic part and an exceptional set makes them expect that one could improve the information on the geometry of ${W^s(\Lambda_g)}$ and/or ${\Lambda_g}$. As it turns out, we will discuss in the next (final) post of this series some recent results in this direction. Finally, the condition (1) is not expected to be sharp by any means, but it seems that the strongly regular parameters are not sufficient to go beyond (1), so that, as some arguments from singularity theory seem to indicate, it is likely that one has to exclude further parameters in order to improve Theorem 1.
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If A = {2,-7 ,5}, B = {5,-7,4} and C = A - B, what is the angle between A and C?
Apr 5, 2016
$\alpha = 90 , {02}^{o}$
Explanation:
$\text{a) find C=A-B}$
$C = \left({A}_{x} - {B}_{x}\right) , \left({A}_{y} - {B}_{y}\right) , \left({A}_{z} - {B}_{z}\right)$
$C = \left(2 - 5\right) , \left(- 7 + 7\right) , \left(5 - 4\right)$
$C = \left\{- 3 , 0 , 1\right\}$
$\text{b) find the magnitude of A and C:}$
$| | A | | = \sqrt{{A}_{x}^{2} + {A}_{y}^{2} + {A}_{z}^{2}}$
$| | A | | = \sqrt{{2}^{2} + {\left(- 7\right)}^{2} + {5}^{2}} = \sqrt{4 + 49 + 25} = \sqrt{78}$
$| | C | | = \sqrt{{C}_{x}^{2} + {C}_{y}^{2} + {C}_{z}^{2}}$
$| | C | | = \sqrt{{\left(- 3\right)}^{2} + {0}^{2} + {1}^{2}} = \sqrt{9 + 0 + 1} = \sqrt{10}$
$\text{c) find dot product of A.C}$
$A \cdot C = {A}_{x} \cdot {C}_{x} + {A}_{y} \cdot {C}_{y} + {A}_{z} \cdot {C}_{z}$
$A \cdot C = 2 \cdot \left(- 3\right) + \left(- 7\right) \cdot 0 + 5 \cdot 1$
$A \cdot C = - 6 + 0 + 5 = - 1$
$\text{d) use the dot product formula}$
$A . C = | | A | | \cdot | | C | | \cdot \cos \alpha$
$A \cdot C = - 1$
$| | A | | = \sqrt{78}$
$| | C | | = \sqrt{10}$
$- 1 = \sqrt{78} \cdot \sqrt{10} \cdot \cos \alpha$
$\cos \alpha = - \frac{1}{\sqrt{78} \cdot \sqrt{10}}$
$\cos \alpha = - \frac{1}{\sqrt{780}}$
$\cos \alpha = - \frac{1}{27 , 93}$
$\alpha = 90 , {02}^{o}$
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# How many more men than women are in the room?
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How many more men than women are in the room? [#permalink]
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16 May 2018, 06:47
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How many more men than women are in the room?
(1) There is a total of 20 women and men in the room.
(2) The number of men in the room equals the square of the number of women in the room.
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Re: How many more men than women are in the room? [#permalink]
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16 May 2018, 07:34
Let number of men = m and women = w
(1) m+n=20; not sufficient
(2) m = n^2; notsufficient
Together we arrive at the equation => n^2 + n =20; can be solved to give n and consequently m
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Re: How many more men than women are in the room? [#permalink]
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16 May 2018, 08:28
Solution
To find:
• How many more men are there in the room, compared to women
Analysing Statement 1
• As per the information given in Statement 1, there are total 20 people in the room, taking all men and women
o But from this statement we cannot find out the exact number of men and women
Hence, statement 1 is not sufficient to answer
Analysing Statement 2
• As per the information given in Statement 2, the number of men in the room equals the square of the number of women in the room
o If we assume number of women to be x, then number of men will be $$x^2$$
o But from this statement we cannot find out the difference between the number of men and number of women
Hence, statement 1 is not sufficient to answer
Combining Both Statements
• Combining the information from both the statements, we can say
o $$x^2$$ + x = 20
Or, (x + 5) (x – 4) = 0
As x cannot be negative, x = 4
• Therefore, we can find out the exact number of men and women, and subsequently the difference between them
Hence, the correct answer is option C.
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Re: How many more men than women are in the room? [#permalink]
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16 May 2018, 10:21
Need both statements to solve. 1) any value of m and n can add to 20 so NS. 2) Could be any number so NS. Combined we only receive one set.
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Re: How many more men than women are in the room? &nbs [#permalink] 16 May 2018, 10:21
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# Dihedral Groups
The dihedral group Dn or Dih(2n) is of order 2n. It is the symmetry group of the regular n-gon.
(Some denote this group D2n because its order is 2n, but I prefer Dn - after all, one doesn't denote the symmetric group Sn by Sn! - nor J1 by J175560. On the other hand, Dih(2n) is fine as there's no conflict of notation.)
• The Schoenflies notation for Dn depends on which 3-dimensional representation is used: Cnv refers to the the 3-dimensional representation acting trivially in the 'vertical', while Dn denotes the representation in 3 dimensions where the reflections act by (-1) in the 'vertical'.
• A fixed generator of order n (rotation by $$2\pi/n$$ is denoted ρ).
• For n even, there are two conjugacy classes of reflection, denoted κ and κ', with κ acting as reflections in a line joining two vertices, and κ' reflection in a line joining the mid-points of two opposite edges. There is an (outer) automorphism of Dn exchanging the two reflections.
• For n odd, all reflections are conjugate;
• Rotations through θ and -θ are conjugate for all n.
• This is close to the theory of Fourier series, and symmetric circulant matrices. See Notes for details.
### D2 = Dih(4)
$$D_2 \simeq \mathbb{Z}_2\times\mathbb{Z}_2$$ with generators κ and κ'. It is the symmetry group of the rectangle.
D2 e κ κ' ρ=κκ' A0 1 1 1 1 A1 1 1 -1 -1 A2 1 -1 1 -1 A3 1 -1 -1 1
### D3 = Dih(6)
D3 e ρ κ # 1 2 3 A0 1 1 1 A1 1 1 -1 E 2 -1 0
### D4 = Dih(8)
D4 e ρ ρ2 κ κ' # 1 2 1 2 2 A0 1 1 1 1 1 A1 1 1 1 -1 -1 B1 1 -1 1 1 -1 B2 1 -1 1 -1 1 E 2 0 -2 0 0
• The permutation representation on the 4 vertices of the square is A0 + B1 + E
• The permutation representation on the 4 edges of the square is A0 + B2 + E
• The permutation representation on the 2 diagonals of the square is A0 + A1
• The orientation permutation representation on the 4 oriented edges of the square is A1 + B1 + E
• The orientation permutation representation on the 2 oriented diagonals of the square is E
### D5 = Dih(10)
D5 e $$\rho$$ $$\rho^2$$ $$\kappa$$ notes # 1 2 2 5 |D5|=10 A0 1 1 1 1 trivial rep A1 1 1 1 -1 "orientation" rep E1 2 $$\gamma$$ $$\gamma$$2 0 symmetry of pentagon E2 2 $$\gamma$$2 $$\gamma$$ 0
• $$\gamma = 2\cos(2\pi/5) = \frac12(\sqrt5-1)$$ (= golden ratio)
• $$\gamma_2 = 2\cos(4\pi/5) = -\frac12(\sqrt5+1)$$
• There is an outer automorphism of D5, taking $$\rho$$ to $$\rho^2$$. This interchanges E1 and E2.
• The permutation representation on the 5 vertices of the pentagon is A0+ E1+ E2
### D6 = Dih(12)
D6 e ρ ρ2 ρ3 κ κ' notes # 1 2 2 1 3 3 |D6|=12 A0 1 1 1 1 1 1 trivial rep A1 1 1 1 1 -1 -1 "orientation rep" B1 1 -1 1 -1 1 -1 alternating rep B2 1 -1 1 -1 -1 1 E1 2 1 -1 -2 0 0 symmetry of hexagon E2 2 -1 -1 2 0 0
• The permutation representation on the 6 vertices of the hexagon is A0+ B1+ E1+ E2
• B1 is the alternating rep because it is the rep obtained if the vertices of the hexagon are weighted successively with +1,-1,+1,-1,+1,-1 (alternating signs)
• The permutation representation on the 3 diagonals joining opposite vertices of the hexagon is A0 + E2
• The 'oriented permutation' representation on the 3 oriented diagonals is B1+ E1
And so the pattern goes on ...
• n even: Dn has four 1-dimensional representations and ½(n-2) 2-dimensional representations.
• n odd: Dn has two 1-dimensional representation and ½(n-1) 2-dimensional representations.
• All the irreducible reps are absolutely irreducible.
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# A question professor couldnt solve!
#### antibody
The professor said he didnt know the ans of this qn, so, can u help?
Question: Can you have lim_x->a f '(x) = D , f ' (a) exists, but not equal to D?
any one can give an example? thx!
#### JustinLevy
The limit of f'(x) as x-> -1 is 0, however f(-1) is undefined, and therefore I believe f'(-1) is undefined as well.
I'm no mathematician, so can someone comment on whether this reasoning is correct?
#### antibody
The limit of f'(x) as x-> -1 is 0, however f(-1) is undefined, and therefore I believe f'(-1) is undefined as well.
I'm no mathematician, so can someone comment on whether this reasoning is correct?
i think there's a problem with this ans since the limit does not exist if f(x)=(x+1)/(x+1)....hmm... i am not sure..
#### JustinLevy
i think there's a problem with this ans since the limit does not exist if f(x)=(x+1)/(x+1)....hmm... i am not sure..
Like I said, it is best to have someone else verify my claim, but I believe the limit does exist. Approach from either side and f'(x) as x-> -1 is just 0. So why do you feel the limit does not exist? f'(-1) may be undefined, but I believe the limit as x-> -1 is.
Can someone verify this reasoning?
#### StatusX
Homework Helper
I think the OP is asking if there's an example where both the limit as f'(x) approaches a and f'(a) exist, but are not equal. I don't think this is possible, but I'm having trouble thinking of a simple proof. For example, if you could justify switching the limits, it'd just be:
$$\lim_{x \rightarrow a} f'(x) = \lim_{x \rightarrow a} \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$$
$$\lim_{h \rightarrow 0} \lim_{x \rightarrow a} \frac{f(x+h)-f(x)}{h} =\lim_{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}=f'(a)$$
But how to justify that is escaping me right now.
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#### antibody
o i have an ans here, but dont know whether,, let's see
It seems like you should not, because
D = lim(x->a) f'(x) {Given}
= lim(x->a) lim(y->x) (f(y)-f(x))/(y-x) {Definition of derivative}
= lim(a+h->a) lim(a+h+g->a+h) (f(a+h+g)-f(a+h))/g {Substitution}
= lim(h->0) lim(g->0) (f(a+h+g)-f(a+h))/g {Arithmetic}
= lim(h->0) lim(g->0) ((h+g)f'(a) - h f'(a)) /g {Definition of derivative}
= f'(a) lim(h->0) lim(g->0) g/g {f'(a) is just a number, move it outside}
= f'(a) lim(h->0) 1 {Easy}
= f'(a) {Easier}
What do u guys think???
The professor said he didnt know the ans of this qn, so, can u help?
Question: Can you have lim_x->a f '(x) = D , f ' (a) exists, but not equal to D?
any one can give an example? thx!
Well, wouldn't it be no by the theorem that says if f is differentiable, f' has the intermediate value property.
#### JasonRox
Homework Helper
Gold Member
Well, wouldn't it be no by the theorem that says if f is differentiable, f' has the intermediate value property.
It does say that? Where?
#### Mystic998
Yeah, the intermediate value property guarantees no discontinuities of this type. You can have discontinuities where the limit fails to exist though.
#### mathwonk
Homework Helper
my notes in the thread elementary proofs of big theorems, in calculus forum, essentially show all derivatives ahve the intermediate vaklue proeprty.
in fact it foollows from the rolle theoirem. if f' is positive at a and negative at b, then f is not monotone in the interval [a,b], hence f takes the same value twice, hence by rolle, f has a critical point on that interval, i.e. f' = 0 in there. this is essentially the IV property for f'.
i.e. f' is always zero between two points where it changes sign. you can easily dress it up to the general statement.
#### ZioX
Some of you guys are being sloppy, so I should say this for everyone browsing this thread who may get the wrong idea. f' (that is the derivative) has the intermediate value property (Darboux property) if f' is defined on an interval.
You can easily construct a counterexample to the original claim, just define f as straight lines pieced together so that f is continuous. It is clear that f' only takes on the values of the slopes m1, m2, m3,... etc for each line segment. Hence f' does not have an intermediate value property. This doesn't contradict f' darboux property, as f' wasn't defined in an interval (note how if we restricted f to one line segment we have f' defined and would have the intermediate value property (f'=m) which isn't very interesting).
Look to see if you can find counterexamples where f' isn't defined on an interval. This is the key to proving your problem.
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#### JasonRox
Homework Helper
Gold Member
Some of you guys are being sloppy, so I should say this for everyone browsing this thread who may get the wrong idea. f' (that is the derivative) has the intermediate value property (Darboux property) if f' is defined on an interval.
Exactly! Darboux's Theorem or Property whichever way you remember it.
The interval being (a,b) instead of [a,b].
#### HallsofIvy
Homework Helper
To clarify, the answer to the original question is "no, such a function does not exist". While the derivative of f is not necessarily continuous, every derivative satisfies the "intermediate value theorem" (Darboux's theorem mentioned by Ziox). If f'(a) exists, then it MUST be the lim, as x->a of f'(x).
#### mathwonk
Homework Helper
in the first place it makes little sense to call a function differentiable at a if it is not even defined on an interval containing a.
secondly "darboux's" theorem is almost trivial as i indicated aboive:
if f'(a) <0 and f'(b)>0, we want to show f'=0 in between. but clearly f is not monotone on [a,b], hence it takes the same value twice (by IVT), and since assumed differentiable in between a and b, there is a point between where f'=0. bingo.
#### antibody
so confused,,,some ppl say it does, some ppl say it does not ..
anyone can give a conclusion?
#### mathwonk
Homework Helper
the answer is as halls said: no such function exists
in the questions posed, it only makes sense (because f is differentiable at a) to assume that f is defined and differentiable on an open interval containing a.
in that situation, if the limit of f'(x) exists as x-->a, then thnat limit must equal f'(a).
the reaqson is that if not, then there would eventually be an interval (a-e,a+e) such that on the punctured interval (excluding a) f' would be near D, but at a f'(a) would be far from D.
this would violate the IV propeerty for derivatives, since f' would not take on all values between f'(a) and the values f'(x) near D.
so the function with f'(a) * D cannot exist, because of the IV property for derivatives. It remains to oprove thaty proeprty, "Darboux's" theorem, which is a trivial corollary of rolles thm. we do this next.
#### mathwonk
Homework Helper
suppose f is differentiable on the interval [a,b] and f'(a) >0 while f'(b) < 0. c;aim f' = 0 in between. \
proof: by definition of derivative, there must be point to the right of a where f(x) > f(a), and points to the left of b where f(x) > f(b). hence f is not monotone on [a,b].
hence by the IVT for f, which is continuous, f takes the same value twice between a and b.
then at a local extremum for f on the closed interval with those two points as endpoints, there is a place where f'=0 between a and b.
now suppose f'(a) < K and f'(b) > K. then f-Kx has derivative zero between a and b, i.e. f has derivative K between a and b. so the IV proeprty of f' is proved, and the problem is completely solved.
go show your professor how easy it is, but practice it on your roommate first.
the fact that this sort of thing is unknown even to some professors, is a result of the subject being taught thoughtlessly year after year, with every book exactly like all the other books, without the authors thinking about what is happening.
#### JustinLevy
so confused,,,some ppl say it does, some ppl say it does not ..
anyone can give a conclusion?
I think I caused some of the confusion by misreading your question. I gave an example where f'(a) does not equal the limit of f'(x) as x->a and that limit exists.
This however, as was pointed out, is NOT was you originally asked:
I think the OP is asking if there's an example where both the limit as f'(x) approaches a and f'(a) exist, but are not equal.
In the example I gave, f'(a) doesn't exist.
If f'(a) does exist, everyone here is in agreement that it must be equal to the limit of f'(x) as x->a. And even better, some here have constructed short proofs of this.
#### ZioX
I'm just practising my LaTeX. Mathwonks proof is completely sufficient.
The Darboux Property of Derivatives. Let $$f$$ be differentiable on the interval I. Suppose a and b are in I and there exists an m in I such that m is strictly between $$f'(a)$$ and $$f'(b)$$, then there exists a c strictly between a and b such that $$f'(c)=m$$.
Proof. Assume WLOG that a is strictly less than b and $$f'(a)<m<f'(b)$$. Define $$g(x)=f(x)-mx$$. Then $$g'(a)=f'(a)-m<0$$ and $$g'(b)=f'(b)-m>0$$. Since $$g$$ is continuous it follows that g must have a minimum on [a,b] (continuity on a compact set). Now, c cannot be a as $$g(x)\ge g(a)$$ so that for all x in (a,b]
$$$\frac{g(x)-g(a)}{x-a}\ge0$$$
which would imply $$g'(a)\ge 0$$.
Similarly c is not b. Therefore, c is in the interior of [a,b]. But then $$g'(c)=0$$ so that $$f'(c)=m$$.
This result is really interesting. It says even though the derivative might not be continuous it still has the intermediate value property.
Last edited:
#### mathwonk
Homework Helper
I admit that even though I pompously claim it is "trivial" from rolle, it took me about 40 years to notice that fact. BUT I THINK THAT IS WHAT "trivial" usually means in mathematical writing.
#### ZioX
I'm sorry to resurrect such an old post but....
$$f'(a)=\lim_{x \to a} \frac{f(x)-f(a)}{x-a}=\lim_{x \to a} f'(x)=D$$
using L'hopital.
#### mathwonk
Homework Helper
this is a nice observation but the appeal to l'hopital, invokes (for the proof of that result) a more difficult MVT type argument than the one above using rolle.
#### ankur162
i think there's a problem with this ans since the limit does not exist if f(x)=(x+1)/(x+1)....hmm... i am not sure..
f(x)=(x+1)/(x+1) it exits Lhl=rhl at x->a just check out why to find derivative
#### eaboujaoudeh
know the L'Hopital theorem?? can't we use it here?
#### ankur162
know the L'Hopital theorem?? can't we use it here?
no we can't use L hospital rule here for its application limit must be having 0/0 or infinity/infinity form .
for eg: lmt x-> 1 (x-1)/2x^2-7x+5 , this can be solve easily using L hospital since it has o/o form ,
just check it on keeping limit
but u can not solve f(x)= lmt x->a (x+1)/(x+1) using L hospital as it is not a o/o form
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# How do you solve \frac { w - 3} { 3} = \frac { 2} { 5}?
May 14, 2017
$w = \frac{21}{5} = 4.2$
#### Explanation:
Solve: $\frac{w - 3}{3} = \frac{2}{5}$
We can cross-multiply:
$5 \left(w - 3\right) = 2 \left(3\right)$
Now, we can use the distributive property to distribute the 5:
$5 w - 15 = 6$
We can add $15$ to both sides:
$5 w = 15 + 6$
$5 w = 21$
Finally, we divide both sides by $5$:
$w = \frac{21}{5} = 4.2$
May 14, 2017
color(brown)(w=21/5 or color(brown)(4 1/5
#### Explanation:
$\frac{w - 3}{3} = \frac{2}{5}$
$\therefore \frac{5 \left(w - 3\right) = 6}{15}$
$\therefore \frac{5 w - 15 = 6}{15}$
multiply L.H.S and R.H.S. by 15
$\therefore 5 w - 15 = 6$
$\therefore 5 w = 6 + 15$
$\therefore 5 w = 21$
:.color(brown)(w=21/5 or color(brown)(4 1/5
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Top
# Subtracting Decimals
Subtraction of decimal numbers is same as subtraction of integers. The only difference is that we need to take care of decimal point.
First digit after decimal is known as digit at tenth $(\frac{1}{10})^{th}$ place. In the same way,
second digit after decimal is known as digit at hundredth $(\frac{1}{100})^{th}$ place and so on.
Following points should be kept in mind while subtracting decimals:
• Adding zeros at the end of a decimal number does not change its value. So, add zeros at the end to make the decimal numbers of same length.
• While subtracting, always keep decimal point of a number which is to be subtracted, below the decimal point of the number from which the number is subtracted.
• Start subtracting from the right side of the numbers.
• If the digit to be subtracted is smaller than the digit from which it is subtracted, then borrow 1 from previous digit. As the result, the digit is increased by 10 and previous digit is reduces by 1.
Let us understand this concept more precisely with the help of examples:
Example 1: Subtract 21.4 from 31.5.
Solution:
Example 2: Subtract 65.7 from 98.6.
Solution:
Example 3: Subtract 729.53 from 980.5.
Solution: Here, the numbers are not of same length, therefore we add a zero at the end of second number which makes it 980.50.
Related Calculators Subtract Decimals Calculator Subtraction Calculator Decimal Calculator Decimal Greater than less than
*AP and SAT are registered trademarks of the College Board.
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# Gauge contribution to the 1/NF expansion of the Yukawa coupling beta function
@article{Kowalska2017GaugeCT,
title={Gauge contribution to the 1/NF expansion of the Yukawa coupling beta function},
author={Kamila Kowalska and Enrico Maria Sessolo},
journal={Journal of High Energy Physics},
year={2017},
volume={2018},
pages={1-17}
}
• Published 19 December 2017
• Physics
• Journal of High Energy Physics
A bstractWe provide a closed analytical form for the gauge contribution to the beta function of a generic Yukawa coupling in the limit of large NF , where NF is the number of heavy vector-like fermions charged under an abelian or non-abelian gauge group. The resummed expression is finite and for the abelian case presents a pole at the same location as for the corresponding gauge beta function. When applied to new physics scenarios characterized by large Yukawa couplings, the contribution… Expand
#### Figures from this paper
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• Physical Review D
• 2018
We consider the dynamics of gauge-Yukawa theories in the presence of a large number of matter constituents. We first review the current status for the renormalization group equations of gauge-fermionExpand
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A bstractWe compute the β-function for a massless Yukawa theory in a closed form at the order O1/Nf$$\mathcal{O}\left(1/{N}_f\right)$$ in the spirit of the expansion in a large number of flavoursExpand
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We consider theories with a large number NF of charged fermions and compute the renormalization group equations for the gauge, Yukawa and quartic couplings resummed at leading order in 1=NF. We Expand
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• Physical Review D
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We study the impact of the Yukawa interaction in the large-$N_f$ limit to the abelian gauge theory. We compute the coupled $\beta$-functions for the system in a closed form at $\mathcal{O}(1/N_f)$.
Safe glueballs and baryons
• Physics
• Journal of High Energy Physics
• 2019
A bstractWe consider a non-Abelian gauge theory with Nf fermions and discuss the possible existence of a non-trivial UV fixed point at large Nf . Specifically, we study the anomalous dimension of theExpand
Critical Look at β-Function Singularities at Large N.
• Physics, Medicine
• Physical review letters
• 2019
We propose a self-consistency equation for the β functions for theories with a large number of flavors, N, that exploits all the available information in the Wilson-Fisher critical exponent, ω,Expand
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• Physics
• Physical Review D
• 2018
We provide an asymptotically safe Pati-Salam embedding of the standard model. Safety is achieved by adding to the theory gauged vectorlike fermions and by employing recently developed largeExpand
Analytic coupling structure of large Nf (super) QED and QCD
• Physics
• Physical Review D
• 2019
We study the analytic properties of the 't Hooft coupling expansion of the beta-function at the leading nontrivial large-$N_f$ order for QED, QCD, Super QED and Super QCD. For each theory, the 'tExpand
An Asymptotically Safe Guide to Quantum Gravity and Matter
Asymptotic safety generalizes asymptotic freedom and could contribute to understanding physics beyond the Standard Model. It is a candidate scenario to provide an ultraviolet extension for theExpand
#### References
SHOWING 1-10 OF 31 REFERENCES
On the One-Loop Yukawa Coupling Beta-Function to Order Yg 2 in a General Gauge and its Gauge Independence
We present the calculation of the one-loop Yukawa coupling beta function to order Yg 2 in a general gauge, where Y and g are Yukawa and gauge couplings. We show explicitly how the gauge parameterExpand
Asymptotically safe standard model extensions
• Physics
• 2018
We consider theories with a large number NF of charged fermions and compute the renormalization group equations for the gauge, Yukawa and quartic couplings resummed at leading order in 1=NF. We Expand
The 1/NF expansion of the γ and β functions in Q.E.D.
• Mathematics
• 1984
The Callan-Symanzikγ- andβ-functions are calculated analytically for Q.E.D. in the limit of a large number of leptons (NF→∞) up to terms of order 1/NF inclusive. We give closed analytic expressionsExpand
Large N flavor β-functions: A recap
Abstract β-Functions for abelian and nonabelian gauge theories are studied in the regime where the large N flavor expansion is applicable. The first nontrivial order in the 1 / N expansion is knownExpand
Gauge Independent Critical Exponents for QED Coupled to a Four Fermi Interaction with and without a Chern Simons Term
Abstract We apply the self-consistency method for determining critical exponents to a model with a four fermi interaction coupled to QED and compute various gauge independent exponents in arbitraryExpand
Critical point analysis of various fermionic field theories in the large N expansion
Computes the critical exponents corresponding to the anomalous dimensions of the fields and vertices in a model involving fermions with a four point interaction, coupled to a U(1) gauge field, atExpand
Analysis of Abelian gauge theory with four-fermi interaction at O(1/N2) in arbitrary dimensions
An arbitrary dimensional expression is given for the anomalous dimension of the fermion field in a model with a four-point interaction and a U(1) gauge field, at O(1/N2) within a large flavourExpand
The γ function in the 1/Nf expansion
• Physics
• 1982
The Callan-Symanzik γ-function is studied in the large number of lepton (in QED) or flavour (in QCD) limit. We have computed analytically 7 terms and they show a striking factorization. We haveExpand
Conformal window 2.0: The large Nf safe story
• Physics
• Physical Review D
• 2018
We extend the phase diagram of SU(N) gauge-fermion theories as function of number of flavours and colours to the region in which asymptotic freedom is lost. We argue, using large $N_f$ results, forExpand
Two Loop Renormalization Group Equations in a General Quantum Field Theory. 1. Wave Function Renormalization
• Physics
• 1983
The two-loop renormalization group equations in a general renormalizable field theory with scalar, spin-12, and (vector) gauge fields are considered. In this paper, the anomalous dimensionsExpand
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# Model Paper IV SEM ||Engg.Mathematics-III||
###### (C-14 Scheme)
Max Marks: 80 Time: 3 Hrs
Part-A
• Each question carries three marks.
• Answer should be brief and straight to the point 10x3=30 marks
1. Solve the equation
2. Solve (D 4- 8D2+ 16) y = 0.
3. Find the Particular Integral for (D2+ 4) y = Sin2x + e-2x.
4. State the change of scale property and first shifting theorem of Laplace Transforms.
5. Find the Laplace Transform of 4e2t + 6t3-3 Cos 5t + Sin t.
6.Find the inverse Laplace Transform of .
7.Find inverse Laplace Transform of
8.Write the Euler’s formulae for Fourier series of a function f(x) in the Interval [C, C+2π].
9.Find the half range Fourier Sine series of unity in (0, π).
10.State Addition and Multiplication Theorems of Probability for two events.
Part-B 10x5 = 50 marks
• Each question carries ten marks.
• Answer should be comprehensive and the criteria for valuation are the content but not the length of the answer.
11 (a) Solve (D2 + D -6) y = e-3x.
(b) Solve (D3 +4D) y = 5 + Cos 2x.
12(a) Solve (D2 + 2D +1) y = 2x + x2.
(b) Solve (D4- 16) y = Cos 2x + Sinh 2x.
13 (a) Find the Laplace Transform of t2. Cos t.
(b) Find the Laplace Transform of
14 (a) Find .
(b) Find using Convolution theorem.
1. Find Fourier series for the function
16 (a) Expand f(x) = as Fourier Series in (-2 , 2).
(b) Find the half range Cosine series for f(x) = x in (0, 2).
17 (a) When two dice are thrown, find the probability of getting a sum 7.
(b) In a hostel 60% students Telugu news paper, 40% students read English news paper and 20% read both the papers. A student is selected at random, find the probability that the Student reads neither Telugu nor English news paper.
18 (a) Let A and B are independent events with P (A) = 0.3 and P (B) = 0.4. Fin
(b) Box-I contains 8 white , 2 black balls, Box–II contains 5 white , 5 black balls and Box-III contains 4 white , 6 black balls. A box is selected at random and a ball is drawn from it, what is the probability that the ball is white.
###### (C-14 Scheme)
Max Marks: 80 Time: 3 Hrs
Part-A
• Each question carries three marks.
• Answer should be brief and straight to the point 10x3=30 marks
1. Solve the equation
2. Solve (D 4- 5D2 + 4) y = 0.
3. Find the Particular Integral for (D2 + D -6) y = e3x+ e-3x.
4. Find the Laplace Transform of Sin3t.Cos4t.
5. Define Convolution of two functions and state the convolution theorem.
6. Find the inverse Laplace Transform of
7. Find inverse Laplace Transform of
8. Write the Euler’s formulae for Fourier series of a function f(x) in the Interval [C, C+2l].
9. Find the Fourier Coefficient in
10. An integer is picked from 1 to 20 numbers, both inclusive. Find the probability that it is a prime.
Part-B 10x5 = 50 marks
• Each question carries ten marks.
• Answer should be comprehensive and the criteria for valuation are the content but not the length of the answer.
11(a) Solve (D2 - 4) y = (1+ ex) 2.
(b) Solve (D2 + 4D + 16) y = Sin 4x.
12(a) Solve (D2 - 2D +1) y = x3.
(b) Solve (D4- 81) y = Sinh 3x + x.
13(a) Find the Laplace Transform of t.e-t .Sin 4t.
(b) Evaluate using Laplace Transforms.
14 (a) Find
(b) Find
1. Find Fourier series for the function
16 (a) Expand f(x) = as Fourier series in –π < x < π.
(b) Find the half range Sine series for f(x) =
17 (a) When two dice are thrown, find the probability of getting a sum even number.
(b) In a committee of 25 members, each member is proficient either in mathematics or in statistics or in both. If 19 of these are proficient mathematics, 16 in statistics.Find the probability that a person selected from the committee is proficient in both.
18 (a) Let A and B are independent events with P (A) = 0.8 and P (B) = 0.4. Find
(b) Box-I contains 3 red, 4 black balls, Box–II contains 5 red , 6 black balls. one ball is drawn at random from one of the bags and is found to be red. What is the probability that the ball is drawn from box-II.
Engineering Mathematics
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# Draw surface graph
Does any know how to draw a graph like this:
Which programs can do that? I'm a developer so I've no problem on formatting a complex files so that a software can handle it.
• SAS is one that can. Nov 24, 2011 at 0:20
• Look up for "response surface plot" on Google; e.g., if you're willing to use R, Surface Plots in the rsm Package.
– chl
Nov 24, 2011 at 8:44
Lots can do it. R (packages such as lattice, misc3d and others), Matlab, Python (see matplotlib), Wolfram Alpha / Mathematica... I'd use whatever your most comfortable in.
See these questions.
• "these questions" link is wrong. Thanks for the info! Nov 24, 2011 at 0:25
• Fixed. Good luck! Nov 25, 2011 at 5:16
Since it wasn't mentioned yet: You could have a look at gnuplot.
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# If somehow we supply high voltage current to the railway track will it travel throughout the network of railway tracks?
## Why can’t we transmit electrical power with transmission voltage more than 1000kv?
Dark Light
This is one of our fantasy about electricity. That if we supply the high voltage current to a track in Mumbai CST, peoples standing on the track in New Delhi [~1400 km far] will get a shock?
No, Current always chooses the shortest path [rather a resistance-free path] in the circuit before to be grounded.
Since the ground is much nearer than New Delhi that too with much less resistance path, the maximum distance traveled by current will be just 17.2 cm [i.e the depth of rail] and not 1400 km [length of rail]. It shall be grounded in no time. You will simply end up with ‘earthing’.
But, Sleeper and ballast crabs are generally hard core insulators. So in this case, the current will find the nearest grounding spots.
which are available at every 100, 300, 500 or 1000 meters interval with a special code of practice for earthing 25kV like this:
Or, manually [for inspection] it can be done anywhere by clippers as:
and it will be earthed.
Let us assume there are not such earthing points available, and the sleepers to are insulators; even then it is practically impossible. You will require astronomical high amperes of current [do also consider the huge
${I}^{2}R$
loss] to circulate it through the whole railway network.
## Fleming’s Left-Hand Rule And Right-Hand Rule
John Ambrose Fleming introduced two rules to determine the direction of motion in motors or the direction of induced current…
## Dynamo – What is it, Working , And Types Of Dynamo
What is DynamoElectric Machine: A Machine That Converts Either Electrical Energy into mechanical energy or mechanical energy into…
## Why Electric Supply in Aircraft is 115V, 400Hz ?
Aircraft’s electrical supply is at the 400Hz frequency.Let’s see the reason behind using a higher frequency. Aircrafts are…
## Production Process of Transformers Manufacturing
There are no. of step follows in the production process of Transformers: 1.Core Building: CRGO i.e. Colled Rolled…
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Indico has been upgraded to version 3.1. Details in the SSB
# Quark Matter 2018
May 13 – 19, 2018
Venice, Italy
Europe/Zurich timezone
The organisers warmly thank all participants for such a lively QM2018! See you in China in 2019!
## Measurement of Lambdac/D0 ratio in Pb-Pb collisions at 5.02 TeV with ALICE
May 15, 2018, 5:00 PM
2h 40m
First floor and third floor (Palazzo del Casinò)
### First floor and third floor
#### Palazzo del Casinò
Poster Open heavy flavour
### Speaker
Yosuke Watanabe (University of Tsukuba (JP))
### Description
The ALICE detector at the Large Hadron Collider (LHC) has been optimised for the studies of the Quark-Gluon Plasma (QGP) created in heavy-ion collisions. Charm quarks are one of the probes that has been extensively used to elucidate the properties of the QGP. They are dominantly produced at the initial stage of the collisions and experience the whole evolution of the system.
While most of the charm-hadron measurements in heavy-ion collisions are currently limited to mesons, charm-baryon measurements could provide unique insights into hadronisation processes from the QGP. The baryon-to-meson ratio is expected to be enhanced if charm quarks hadronise via recombination with the surrounding light quarks in the QGP. Moreover, in such a recombination picture, the baryon-to-meson ratio could further be enhanced in the presence of diquark bound states in the QGP. Thus, the measurements of charm baryons could shed light on an unexplored aspect of the QGP.
The production of charm baryons has been studied in pp and p-Pb collisions with the ALICE detector. The measurement in Pb--Pb collisions also became possible with the significantly improved statistics of LHC Run-2 compared to LHC Run-1. In this poster, we will report on the first measurement of the $\Lambda_{\rm c}/{\rm D}^0$ ratio in Pb-Pb collisions at $\sqrt{s_{\mathrm{NN}}}$ = 5.02 TeV.
Collaboration ALICE Experiment Presenter name already specified
### Primary author
Yosuke Watanabe (University of Tsukuba (JP))
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### Spheniscine's blog
By Spheniscine, history, 8 months ago,
Spoiler
Spoiler
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### Ex – Hakata
Spoiler
• +115
By Spheniscine, history, 20 months ago,
The math renderer was changed/updated recently but it looks much worse than the previous version.
First off, it takes much longer to fully render; before it does it looks like this, which is small and hard to read:
Then when it does fully render, it's quite blurry:
• +374
By Spheniscine, history, 20 months ago,
Spoiler
Spoiler
Spoiler
Spoiler
Spoiler
### F – GCD or MIN
Spoiler
• +35
By Spheniscine, history, 23 months ago,
Forgive the quirkiness of the title; I am not aware of a name for this concept, therefore I just made a few up.
This blogpost is about a particular way to construct what I call a "pseudorandom permuter". It is distinct from a regular pseudorandom number generator (PRNG) in that it is constructed such that the sequence will never repeat until all $M = 2^k$ integers in its machine-integer state space (32- or 64-bit) has been used. In other words, it's a way to produce a permutation of $[0, M)$ that's (hopefully) statistically indistinguishable from a truly random permutation.
The generator is made up of two parts, a "discrete Weyl sequence", and an "avalanching perfect hash function".
A "discrete Weyl sequence" is defined by a parameterized function $W_{s, \gamma}: [0, M) \rightarrow [0, M)$ where the $i$-th member of the sequence is $W_{s, \gamma}(i) = (s + \gamma \cdot i) \mod M$. The two parameters can be chosen using a standard random number generator, with the caveat that $\gamma$ must be odd (this can be easily ensured by a | 1 instruction). This ensures it is coprime with the machine integer size $M$.
The advantage of the discrete Weyl sequence is that as it is parameterized, it is hard to predict by an adversarial test (whether by the problemsetter or by the "hacking" mechanic). The disadvantage however, is that it is extremely regular, and this regularity may be exploited by adversarial tests.
In comes the second part. An "avalanching perfect hash function" is a function that is "perfect" in the sense that it maps $[0, M) \rightarrow [0, M)$ without collisions, i.e. $a \neq b \Leftrightarrow f(a) \neq f(b)$, and exhibits the avalanche property, which is that flipping one bit in the input should flip roughly half the bits of the output without a regularly predictable pattern. Note that our discrete Weyl sequence fits the definition of a perfect hash function, however it doesn't exhibit the avalanche property.
Constructing a good APHF is more difficult, however since it doesn't have to be parameterized, we can simply use a fixed function. One option is to use the "multiply-xorshift" construction from the "splitmix64" generator:
function splitmix64_aphf(z: uint64): uint64 {
z = (z ^ (z >> 30)) * 0xbf58476d1ce4e5b9;
z = (z ^ (z >> 27)) * 0x94d049bb133111eb;
return z ^ (z >> 31);
}
where ^ denotes the binary xor operator, and >> the unsigned right-shift operator. I also use another option for 32-bit integers, taken from this offsite blogpost by Chris Wellons:
function int32_aphf(z: uint32): uint32 {
z = (z ^ (z >> 16)) * 0x7feb352d;
z = (z ^ (z >> 15)) * 0x846ca68b;
return z ^ (z >> 16);
}
We thus define our final hash function $f_{s, \gamma}(x) = A(W_{s, \gamma}(x))$, where $A$ is our chosen APHF. This combines the advantages of both parts; easy parametrization ("seedability") and good avalanche properties that make it hard to predict, as long as the adversary doesn't have access to the parameters or the generated values. (However, it is possible for an adversary with access to the generated values to reverse the function to find the parameters and thus predict the function; thus this construction is unsuitable for cryptographic purposes.)
Note that the actual splitmix64 generator is actually a construction of this type with a fixed $\gamma$ value. Hence my alternate name, "Splitmixes", as this construction allows easy generation of an unpredictable hash function with similar properties.
## Applications
This generator should not be used like a regular random-number generator, as its "no-collision until period" property will fail statistical tests that test the "birthday paradox" properties of a PRNG, i.e. generating a large number of values and checking if the number of collisions conforms to the statistically expected value.
However, it is this very property that makes it more useful than regular PRNGs for, e.g.,
• hash maps with integer keys (use $f_{s, \gamma}(x)$ for the hash-key of $x$)
• priority values in treaps (simply have a counter variable and use $f_{s, \gamma}(0), f_{s, \gamma}(1),$ etc, and you're guaranteed never to produce the same priority unless you generate more than 4-billion-plus values)
• +38
By Spheniscine, history, 2 years ago,
Spoiler
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### F – Random Max
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• +49
By Spheniscine, history, 2 years ago,
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### F – Heights and Pairs
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• +57
By Spheniscine, history, 2 years ago,
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### F – Contrast
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• +51
By Spheniscine, history, 3 years ago,
This is an extension of my article proposing the "ultimate" NTT. Here I will present a simplified and optimized variant of the Barrett reduction algorithm on that page. As it relaxes certain bounds of the standard parameters of the algorithm, I will also present a proof of the correctness of the variant, at least for this specific case.
First, the algorithm:
function mulMod(a: Long, b: Long): Long {
xh = multiplyHighUnsigned(a, b) // high word of product
xl = a * b // low word of product
g = multiplyHighUnsigned( (xh << 2) | (xl >>> 62) , BARR_R)
t = xl - g * MOD - MOD
t += (t >> 63) & MOD
return t
}
Notice that the middle routine has been greatly simplified; instead of computing the high and middle words of $xr$, we simply take the top 64 bits of $x$, in effect computing $\displaystyle \left\lfloor \left\lfloor \frac{x}{2^{62}} \right\rfloor \frac{r}{2^{64}} \right\rfloor$ in place of $\displaystyle \left\lfloor \frac {xr}{4^{63}} \right\rfloor$. Experimentally, this seems to work (and indeed, this variant is how Barrett reduction is normally implemented for multi-word cases), however a naive adjustment of the bounds of the Barrett reduction algorithm would relax the pre-normalization bound of $t$ to $[-m, 2m)$. This is bad news, as there is a possibility of overflowing into the sign bit, thus forcing the use of a modulus one bit shorter (e.g. $4611686018326724609$, $g = 3$) so that we could easily correct the result with two normalization rounds.
So the following is a proof of correctness for this relaxed algorithm, at least for the very specific use case in the described NTT:
On the Project Nayuki article on Barrett reduction (note that I will use $n$ instead of $m$ for the modulus in accordance to the notation on that page), one important lemma is the inequality
$\displaystyle\frac{x}{n} - 1 < \frac{xr}{4^k} ≤ \frac{x}{n} \implies \frac{x}{n} - 2 < \left\lfloor \frac{x}{n} - 1 \right\rfloor ≤ \left\lfloor \frac{xr}{4^k} \right\rfloor ≤ \frac{x}{n}$
We shall show that this inequality holds even if $\dfrac{xr}{4^k}$ is replaced by $\dfrac{(x-\delta) r}{4^k}$, with $0 ≤ \delta < 2^{62}$, the bits that have been omitted.
First we need to obtain a tighter bound on $r$. Given that $r$, $k$, and $n$ are fixed in our algorithm, we can verify that $\dfrac {4^k} n - r < 2^{-9}$. Let this value be $\varepsilon$. We thus obtain $\displaystyle \frac{4^k}{n} - \varepsilon < r < \frac{4^k}{n}$
Multiply by $x-\delta \geq 0$: $\displaystyle (x-\delta)\left(\frac{4^k}{n} - \varepsilon\right) ≤ (x-\delta)r ≤ (x-\delta)\frac{4^k}{n}$
Given that $\delta$ is nonnegative we can relax the right bound to $x\dfrac{4^k}{n}$.
Divide by $4^k$: $\displaystyle (x-\delta)\left(\frac{1}{n} - \frac \varepsilon {4^k}\right) ≤ \frac {(x-\delta)r}{4^k} ≤ \frac{x}{n}$
Recompose the leftmost expression: $\displaystyle \frac x n - \left(\frac {\varepsilon x} {4^k} + \frac \delta n\right) + \frac{\delta\varepsilon} {4^k} ≤ \frac {(x-\delta)r}{4^k} ≤ \frac{x}{n}$
$\displaystyle\frac{\delta\varepsilon} {4^k} \geq 0$, so relax the bound: $\displaystyle \frac x n - \left(\frac {\varepsilon x} {4^k} + \frac \delta n\right) ≤ \frac {(x-\delta)r}{4^k} ≤ \frac{x}{n}$
$x < n^2 < 4^k \implies \dfrac{x}{4^k} < 1$, so further relax the bound: $\displaystyle \frac x n - \left({\varepsilon} + \frac \delta n\right) < \frac {(x-\delta)r}{4^k} ≤ \frac{x}{n}$
$\delta < 2^{62}$ while $n$ is very slightly below $2^{63}$, so it can be verified that $\dfrac \delta n$ must be $< \dfrac34$.
$\dfrac34 + \varepsilon < 1$, therefore $\displaystyle \frac x n - 1 < \frac {(x-\delta)r}{4^k} ≤ \frac{x}{n}$
We can thus follow the rest of the proof in the Nayuki article to prove that our $t$ is in $[-n, n)$.
#### Samples (Kotlin, so YMMV on benchmarks for other languages)
74853038 for 993E - Nikita and Order Statistics, 328 ms faster
74852214 for 986D - Perfect Encoding, 452 ms faster
There are other applications of this proof besides NTT. For example, I found a safe prime $9223372036854771239$, which is a good property for a rolling-hash modulus to have, as any number in $[2, m-2]$ will have multiplicative order at least $\dfrac{m-1}{2}$. It seems that $\varepsilon$ and $\dfrac\delta n - \dfrac 1 2$ will both be rather small for moduli so close below a power of 2.
• +47
By Spheniscine, history, 3 years ago,
Prerequisites: you need to be familiar with both modular arithmetic and Fast Fourier Transform / number theoretic transform. The latter can be a rather advanced topic, but I personally found this article helpful. Note that there are some relatively easy optimizations/improvements that can be done, like precalculating the modular inverse of $n$, the "bit-reversed-indices" (can be done in $O(n)$ with DP), as well as the powers of $\omega_n$. Also useful is modifying it so that the input array length can be any power of two $\leq n$; some problems require multiplying polynomials of many different lengths, and you'd rather the runtime be $O(n \log n)$ over the sums of lengths, rather than (number of polynomials * maximum capacity).
Also helpful is knowing how to use NTT to solve problems modulo $998244353$, like 1096G - Lucky Tickets and 1251F - Red-White Fence. Note that for some problems it's easier to think of FFT not as multiplying polynomials, but of finding multiset $C$ as the pairwise sums of multisets $A$ and $B$, in the form of arrays $A[i] =$ number of instances of $i$ in multiset $A$. This is equivalent to multiplying polynomials of the form $\sum _{i=0} ^n A[i]x^i$.
Note that $\omega_n$ can be easily found via the formula $g ^ {(m-1) / n} \ \text{ mod } m$, provided that:
1. $m$ is prime
2. $g$ is any primitive root modulo $m$. It is easiest to find this before hand and then hardcode it in the submission. You can either implement the algorithm yourself or use Wolfram Alpha to find it via the query PrimitiveRoot[m]. (Spoiler alert, $g = 3$ works well for $998244353$)
3. $n$ divides $m-1$ evenly. As $n$ is typically rounded up to the nearest power of 2 for basic FFT implementations, this is easiest when $m$ is of the form ${a \cdot 2^{k} + 1}$ where $2^k \geq n$. This is why $998244353$ is a commonly-appearing modulus; it's $119 \cdot 2^{23} + 1$. Note that this modulus also appears in many problems that don't require FFT/NTT; this is a deliberate "crying-wolf" strategy employed by puzzle writers, so that you can't recognize immediately that a problem requires FFT/NTT via the given modulus.
Now onto the main topic, the "ultimate" NTT.
Rationale: There are a few problems like 993E - Nikita and Order Statistics that require FFT, however the results aren't output with any modulus, and indeed may exceed the range of a 32-bit integer type. There are several usual solutions for these types of problems:
1. Do NTT with two moduli and restore the result via Chinese Remainder Theorem. This has several prominent disadvantages:
1. Slow, as the NTT routine has to be done twice.
2. Complicated setup, as several suitable moduli must be found, and their primitive roots calculated
3. Restoring the result with CRT requires either brute force or multiplications modulo $pq$, which may overflow even 64-bit integer types.
2. Do FFT with complex numbers and floating point types. Disadvantages are:
1. Could be slow due to heavy floating-point arithmetic. Additionally, JVM-based languages (Java, Kotlin, Scala) suffer complications here, as representing complex numbers with object-based tuples adds a significant overhead.
2. Limited precision due to rounding error. Typically the problems are constructed such that it won't be a problem if care is taken in the implementation, but won't it be nice to just not to have to worry about it?
To solve these problems, I propose the "ultimate" NTT solution — just use one huge modulus. The one I use is $m = 9223372036737335297 = 549755813881 \cdot 2^{24} + 1, g = 3$. This is just over a hundred million less than $2^{63} - 1$, the maximum value of a signed 64-bit integer.
However, this obviously raises the issue of how to safely do modular arithmetic with such huge integers. Addition is complicated by possible overflow into the sign bit, thus the usual if(x >= m) x -= m won't work. Instead, first normalize $x$ into the range $[-m, m)$; this is easily done with subtracting $m$ before any addition operation. Then do x += (x >> 63) & m. This has the effect of adding $m$ to $x$ if and only if $x$ is negative.
The elephant in the room however is multiplication. The usual method requires computing a 126-bit product, then doing a modulo operation over a 63-bit integer; this could be slow and error-prone to implement. C++ users could rejoice, as Codeforces recently implemented support for 128-bit integers via this update. However, before you celebrate too early, there are still issues: this may not be available on other platforms, and I can't imagine that straight 128-bit modulo 64-bit is exactly the fastest operation in the world, so the following might still be helpful even to C++ users.
A rather simple-to-code general-case method is to "double-and-add" similar to modular exponentiation, however it is too slow for the purposes of this article; FFT implementation speed is heavily bound by the speed of multiplications.
My favored technique for this problem is called Barrett reduction. The explanation is (adapted from this website)
• Choose integer $k$ such that $2^k > m$. $k = 63$ works for our modulus.
• Precompute $\displaystyle r = \left\lfloor \frac {4^k} m \right\rfloor$. For our modulus this is $9223372036972216319$, which just overflows a signed 64-bit integer. You can either store it as unsigned, or the as the signed 64-bit representation $-9223372036737335297$ (yes, this just happens to be $-m$, which is quite a neat coincidence)
• Multiply the two integers. Note that we need all 126 bits of this product. The low bits can be easily obtained via straight multiplication, however the high bits need some bit-shifting arithmetic tricks to obtain. Adapt the following Java code, taken from here for your preferred language:
multiplyHighUnsigned code
• Let the product be $x$. Then calculate $\displaystyle t = x - \left\lfloor \frac{xr}{4^k} \right\rfloor m - m$. This requires some adaptation of grade-school arithmetic; pseudocode in spoiler below. $t$ is guaranteed to be in the range $[-m, m)$, so the t += (t >> 63) & m trick should work to normalize it. In the following pseudocode, BARR_R $= r$ and MOD $= m$. Also, ^ represents bitwise xor, not exponentiation.
mulMod pseudocode
Update: I have further optimized the Barrett reduction step, however it requires a new proof of correctness, so I'm keeping the old variant on this page. For info on the optimized version, read this article.
And that's it for this blogpost. This should be sufficient to solve most integer-based FFT problems in competitive programming. Note that if you need negative numbers in your polynomials, you can input them modulo $m$, then assume any integer in the output that exceeds $m/2$ is negative; this works as long as no number in the output has an absolute value greater than $m/2$, yet another advantage of using such a huge modulus.
Note that for arbitrary polynomials of length $n$ with input values not exceeding $a$, the maximum possible value in the product polynomial is around $a^2n$.
#### Samples
74641269 for 986D - Perfect Encoding — this is particularly interesting, because this huge modulus allows 6 digits per word in the custom big-integer implementation instead of the recommended 3, cutting down the FFT size (as well as the size of other arithmetic operations) by half. With this tight a time limit, I'm not sure this problem is solvable in Kotlin otherwise.
There are a small minority of problems that may ask to perform FFT with an inconvenient modulus like $10^9 + 7$, or an arbitrarily given one in the input (do any exist on Codeforces?). $a^2n$ in this case could overflow even this large modulus, but it can be handled with the "multiplication with arbitrary modulus" method listed in the CP-algorithms article. (I heard this is called MTT; also known as "FFT-mod") Even then, the large modulus this article covers might be helpful, as it reduces the number of decompositions you must make.
It seems this method is significantly faster when run on a 64-bit system:
130750094 (1808 ms, Rust ran on 32-bit before an update circa November, IIRC)
140255354 (576 ms, after 64-bit update)
It even beats out my implementations using floats (140229918, 717 ms) and CRT/Garner's algorithm with two 32-bit moduli (140230689, 842 ms)
Unfortunately Java and Kotlin, at this time of writing, still seems to run on a VM that's effectively 32-bit.
There is also another interesting prime modulus, as noted in this offsite blogpost: $18446744069414584321$ (hex: 0xffff_ffff_0000_0001, $g = 7$). Due to its bitpattern, you can use some adds, subtracts, and shifts to do the reduction under this modulus, instead of the multiplies needed by Barrett reduction.
However my implementation of it ended up slightly slower (140308499, 639 ms). It seems multiplies are just that good on 64-bit systems. It might still be worth considering for its slightly higher range, huge capacity (can theoretically support FFTs up to $2^{32}$ in length, though you'd likely MLE first). It might also be friendlier to 32-bit systems (needs testing). This modulus, as noted in the blog post, also has other interesting properties; chiefly that $2$ is a $192^{\text{nd}}$ root of unity; this allows some roots of unity to be expressed in powers of two, which allows using shifts instead of multiplication in some cases. However I've not figured out how to efficiently use this fact (it might be more helpful for mixed-radix variants of FFT, which I've not learned how to implement).
Either modulus can also be used to emulate convolutions in other arbitrary moduli, such as $10^9 + 7$ (https://judge.yosupo.jp/submission/70563) and $2^{64}$ (https://judge.yosupo.jp/submission/70564), by splitting each coefficient into $k$ "place values" for a particular base (I use $2^{16}$ in these examples), then assigning $2k - 1$ slots for each coefficient (the extra slots allow the convolution to "spill over"), essentially emulating multivariate convolution with a single FFT. The speed overhead of this method, however, is quite significant; my implementation using CRT/Garner's (https://judge.yosupo.jp/submission/70678) is much faster in this case.
• +183
By Spheniscine, history, 3 years ago,
I have noticed some of my older blogs and comments have some of their $\TeX$ mathematical expressions incorrectly rendered in their own line instead of in the line of text, which can be really distracting to read.
Examples: https://codeforces.com/blog/entry/72593 ($\mathbb Z /p \mathbb Z$, and the large numbers in the spoilers)
I'm not sure what the cause of this is, and there is no discernible pattern as to which expressions and comments are affected.
Upd: It also affects others' comments too, like: https://codeforces.com/blog/entry/73211?#comment-575152
• +11
By Spheniscine, history, 3 years ago,
Advent of Code is a website that releases programming puzzles every December from the 1st to the 25th. The puzzles have two parts, and the second part isn't revealed until you solve the first part.
This year, there have been many questions regarding part 2 of day 22, as it involves modular arithmetic in a way that hasn't been seen in previous puzzles there. I have been asked for a tutorial for it on Reddit, but I'm posting it here due to better support for mathematical notation.
Part 2 summary
First off, if you are unfamiliar with modular arithmetic, I encourage you to read my other blog post, Modular Arithmetic for Beginners. The most important thing to understand from there is how the four basic operations (addition, subtraction, multiplication, division) can be redefined to work in modular ($\mathbb Z / p \mathbb Z$ field) arithmetic. I also recommend you try some of the puzzles linked (This one is appropriately Christmas-themed too). The terminology and notations I'll use will also be similar.
It is possible to solve this without explicitly invoking modular arithmetic (see this post for an example) but they basically amount to rediscovering certain properties of modular arithmetic anyway, as such that's the "language" I'll use for this tutorial.
The first thing to notice is that each of the given instructions amount to a transformation on the position of each card. Let $m$ represent the number of cards in the deck.
• "deal into new stack" moves cards from position $x$ to position $m - x - 1$. We can write this as $f(x) = m - x - 1$.
• "cut $n$" moves cards from position $x$ to position $x - n\ \text{ mod } m$ (note how indices "wrap around" to the other side), Thus $f(x) = x - n\ \text{ mod } m$. Note this also works for the version with negative $n$.
• "deal with increment $n$" moves cards from position $x$ to position $n \cdot x\ \text{ mod } m$. Thus, $f(x) = n \cdot x\ \text{ mod } m$
The next thing to notice is that each transformation can be rewritten in the form of $f(x) = ax + b\ \text{ mod } m$. This is called a linear congruential function, and forms the basis for linear congruential generators, a simple type of pseudorandom number generator.
• "deal into new stack": $f(x) = -x - 1\ \text{ mod } m$, so $a = -1, b = -1$
• "cut $n$": $f(x) = x - n\ \text{ mod } m$, so $a = 1, b = -n$
• "deal with increment $n$": $f(x) = n \cdot x\ \text{ mod } m$, so $a = n, b = 0$
The next step is to see what happens when you compose two arbitrary linear congruential functions $f(x) = ax + b\ \text{ mod } m$ and $g(x) = cx + d\ \text{ mod } m$. So what do you get if you evaluate $g(f(x))$? By substitution:
$g(f(x)) = c(ax + b) + d\ \text{ mod } m$
As I established in Modular Arithmetic for Beginners, algebra works with modular residues similarly to real numbers, so let's expand:
$g(f(x)) = acx + bc + d\ \text{ mod } m$
An alternate notation for $g(f(x))$ is $f\ ;g(x)$ ("first apply $f$, then apply $g$"). Further more, we can abstract all linear congruential functions (LCF) as its own data type, consisting of a tuple $(a, b)$. Thus we can implement a compose operation between two LCFs:
$(a, b)\ ; (c, d) = (ac \text{ mod } m, bc + d\ \text{ mod } m)$
We store all LCF coefficients modulo $m$, to avoid them growing too big and slowing down our program or taking all available memory. Also note the possible pitfall: when multiplying two residue values, it's possible to overflow even the 64-bit data type, as our modulus for part 2 exceeds $2^{32}$. If you use a language with an arbitrary-size number type (e.g. Java's BigInteger), that'd do. If you don't, you might consider either importing a third-party library, or you can implement "multiplication by doubling" via a process similar to "exponentiation by squaring" described on the Modular Arithmetic for Beginners page. It's not the most efficient way to get around this problem, but for our purposes it'd work fine.
With these methods, you can translate all the steps in your puzzle input into LCFs, then compose them successively ($f_1\ ; f_2\ ; f_3\ ;...$) into a single LCF. Let's call it $F$. Recall that we've stored it as a tuple $(a, b)$ representing the equation $F(x) = ax + b\ \text{ mod } m$.
$F$ now represents a single shuffle as directed by your input. You can test your implementation by re-solving part 1 with it. (note that you have to re-compose it with a different modulus for each part.)
However, we require over a hundred trillion shuffles (we'll call this number $k$), so we can't just compose $F$ into itself naively. There are two approaches to solving this problem:
Method 1:
Compose $F$ into itself $k$ times algebraically. Do a few rounds by hand on paper and you'll notice a pattern emerge:
$F^k(x) = a^k x + (a^{k-1} + a^{k-2} + ... + a^1 + 1) b\ \text{ mod } m$. The more-formal notation would be $\displaystyle F^k(x) = a^k x + \sum_{i=0}^{k-1} ba^i\ \text{ mod } m$.
The summation forms a geometric series. According to that Wikipedia page, we can thus transform the expression to: $\displaystyle F^k(x) = a^k x + \frac{ b(1 - a^k) } { 1 - a } \ \text{ mod } m$
Exponentiation by squaring is required to perform the exponentiations, and modular multiplicative inverse to perform the division.
Method 2:
Notice that composing LCFs are associative, i.e. $(f\ ; g)\ ; h = f\ ; (g\ ; h)$.
Proof: $h(g(f(x))) = h(f\ ; g(x)) = g\ ; h(f(x)) = f\ ; g\ ; h(x)$
Note that this doesn't uniquely apply to LCFs; composition of any "pure functions" (i.e. always returns the same output for a given input) is associative. This is computationally useful for any function that can be composed in a similar manner to what we do here with LCFs before invoking them with actual parameters.
The "exponentiation by squaring" method works on all associative operations, so it can be adapted to the compose operation as well. We can thus obtain $F^k$, $F$ composed into itself $k$ times.
For example, we could adapt the pseudocode for exponentiation by squaring (from Modular Arithmetic for Beginners) like this:
function pow_compose(f: lcf, k: int64) → lcf:
g := lcf(1, 0)
while k > 0:
if k is odd:
g := compose(g, f)
k := ⌊k/2⌋
f := compose(f, f)
return g
Notice how multiplication is replaced with the compose function we used to compose the steps in the input, and the initial multiplicative identity of 1 is replaced with the "identity LCF" $f(x) = x \text{ mod } m$, i.e. $(1, 0)$
Continued:
Now for the last step. Notice that we're not asked for where card $x$ ends up, but rather what card is in position $x$. To do that, we need to invert the function $F^k$. Let $F^k(x) = Ax + B\ \text{ mod } m$.
We invert it by substitution: $x = A \cdot F^{-k}(x) + B\ \text{ mod } m$, therefore $F^{-k}(x) = \dfrac {x - B} A\ \text{ mod } m$. Again this requires an application of modular multiplicative inverse.
Plug in $2020$ for $x$, and we should finally get our solution. Whew.
• +40
By Spheniscine, history, 3 years ago,
## Introduction
If you're new to the world of competitive programming, you may have noticed that some tasks, typically combinatorial and probability tasks, have this funny habit of asking you to calculate a huge number, then tell you that "because this number can be huge, please output it modulo $10^9 + 7$". Like, it's not enough that they ask you to calculate a number they know will overflow basic integer data types, but now you need to apply the modulo operation after that? Even worse are those that say you need to calculate a fraction $\frac pq$ and ask you to output $r$ where $r \cdot q \equiv p \pmod m$... not only do you have to calculate a fraction with huge numbers, how in the world are you going to find $r$?
Actually, the modulo is there to make the calculation easier, not harder. This may sound counterintuitive, but once you know how modular arithmetic works, you'll see why too. Soon you'll be solving these problems like second nature.
## Terminology and notation
For convenience, I will define the notation $n \text{ mod } m$ (for integers $n$ and $m$) to mean $n - \left\lfloor \dfrac nm \right\rfloor \cdot m$, where $\lfloor x \rfloor$ is the largest integer that doesn't exceed $x$. (This should always produce an integer between $0$ and $m-1$ inclusive.) This may or may not correspond to the expression n % m in your programming language (% is often called the "modulo operator" but in some instances, it's more correct to call it the "remainder operator"). If -8 % 7 == 6, you're fine, but if it is -1, you'll need to adjust it by adding $m$ to any negative results. If you use Java/Kotlin, the standard library function Math.floorMod(n, m) does what we need.
Also for convenience, I will also define the $\text{ mod }$ operator to have lower precedence than addition or subtraction, thus $ax + b\ \text{ mod } m \Rightarrow (ax + b) \text{ mod } m$. This probably does not correspond with the precedence of the % operator.
The value $m$ after the modulo operator is known as the modulus. The result of the expression $n \text{ mod } m$ is known as $n$'s residue modulo $m$.
You may also sometimes see the notation $expr_1 \equiv expr_2 \pmod m$. This is read as "$expr_1$ is congruent to $expr_2$ modulo $m$", and is shorthand for $expr_1 \text{ mod } m = expr_2 \text{ mod } m$.
## "Basic" arithmetic
First off, some important identities about the modulo operator:
$(a \text{ mod } m) + (b \text{ mod } m)\ \text{ mod } m = a + b\ \text{ mod } m$
$(a \text{ mod } m) - (b \text{ mod } m)\ \text{ mod } m = a - b\ \text{ mod } m$
$(a \text{ mod } m) \cdot (b \text{ mod } m)\ \text{ mod } m = a \cdot b\ \text{ mod } m$
These identities have the very important consequence in that you generally don't need to ever store the "true" values of the large numbers you're working with, only their residues $\text{ mod } m$. You can then add, subtract, and multiply with them as much as you need for your problem, taking the modulo as often as needed to avoid integer overflow. You may even decide to wrap them into their own object class with overloaded operators if your language supports them, though you may have to be careful of any object allocation overhead. If you use Kotlin like I do, consider using the inline class feature.
But what about division and fractions? That's slightly more complicated, and requires a concept called the "modular multiplicative inverse". The modular multiplicative inverse of a number $a$ is the number $a^{-1}$ such that $a \cdot a^{-1}\ \text{ mod } m = 1$. You may notice that this is similar to the concept of a reciprocal, but here we don't want a fraction; we want an integer, specifically an integer between $0$ and $m-1$ inclusive.
But how do you actually find such a number? Bruteforcing all numbers to a prime number close to a billion will usually cause you to exceed the time limit. There are two faster ways to calculate the inverse: the extended GCD algorithm, and Fermat's little theorem. Though the extended GCD algorithm is more versatile and sometimes slightly faster, the Fermat's little theorem method is more popular, simply because it's almost "free" once you implement exponentiation, which is also often a useful operation in itself, so that's what we'll cover here.
Fermat's little theorem says that as long as the modulus $m$ is a prime number ($10^9 + 7$ is prime, and so is $998\ 244\ 353$, another common modulus in these problems), then $a^m \text{ mod } m = a \text{ mod } m$. Working backwards, $a^{m-1} \text{ mod } m = 1 = a \cdot a^{m-2}\ \text{ mod } m$, therefore the number we need is $a^{m-2} \text{ mod } m$.
Note that this only works for $a \text{ mod } m \neq 0$, because there is no number $x$ such that $0 \cdot x\ \text{ mod } m = 1$. In other words, you still can't divide by $0$, sorry.
Multiplying $m-2$ times would still take too long; therefore a trick known as exponentiation by squaring is needed. It's based on the observation that for positive integer $n$, that if $n$ is odd, $x^n=x( x^{2})^{\frac{n - 1}{2}}$, while if $n$ is even, $x^n=(x^{2})^{\frac{n}{2}}$. It can be implemented recursively by the following pseudocode:
function pow_mod(x, n, m):
if n = 0 then return 1
t := pow_mod(x, ⌊n/2⌋, m)
if n is even:
return t · t mod m
else:
return t · t · x mod m
Or iteratively as follows:
function pow_mod(x, n, m):
y := 1
while n > 0:
if n is odd:
y := y · x mod m
n := ⌊n/2⌋
x := x · x mod m
return y
Now that you know how to compute the modular multiplicative inverse (to refresh, $a^{-1} = a^{m-2} \text{ mod } m$ when $m$ is prime), you can now define the division operator:
$a / b\ \text{ mod } m = a \cdot b^{-1}\ \text{ mod } m$
This also extends the $\text{ mod }$ operator to rational numbers (i.e. fractions), as long as the denominator is coprime to $m$. (Thus the reason for choosing a fairly large prime; that way puzzle writers can avoid denominators with $m$ as a factor). The four basic operations, as well as exponentiation, will still work on them as usual. Again, you generally never need to store the fractions as their "true" values, only their residues modulo $m$.
Congratulations! You have now mastered $\mathbb Z / p \mathbb Z$ field arithmetic! A "field" is just a fancy term from abstract algebra theory for a set with the four basic operators (addition, subtraction, multiplication, division) defined in a way that works just like you've learned in high-school for the rational and real numbers (however division by zero is still undefined), and $\mathbb Z / p \mathbb Z$ is just a fancy term meaning the set of integers from $0$ to $p - 1$ treated as residues modulo $p$.
This also means that algebra works much like the way you learned in high school. How to solve $3 = 4x + 5\ \text{ mod } 10^9 + 7$? Simply pretend that $x$ is a real number and get $x = -1/2\ \text{ mod } 10^9 + 7 = 500000003$. (Technically, all $x$ whose residue is $500000003$, including rationals, will satisfy the equation.)
You can also now take advantage of combinatoric identities, like $\displaystyle \binom{n}{k} = \frac{n!}{k! (n-k)!}$. The factorials can be too big to store in their true form, but you can store their modular residues instead, then use modular multiplicative inverse to do the "division".
There are only a few things you need to be careful of, like:
• divisions through modular multiplicative inverse would be slower than the other operations ($O (\log m)$ instead of $O(1)$), so you may want to cache/memoize the inverses you use frequently in your program.
• comparisons (once you represent a number by its modulo residue, comparisons are generally meaningless, as your $1$ might "really" be $m + 1$, $10^{100} m + 1$, $-5m + 1$, or even $\dfrac {1} {m+1}$)
• exponentiation (when evaluating $x^n \text{ mod } m$, you can't store $n$ as $n \text{ mod } m$. If $n$ turns out to be really huge, you need to calculate it modulo $\varphi(m)$ instead, where $\varphi$ stands for Euler's totient function. If $m$ is prime, $\varphi(m) = m - 1$. Note that this new modulus will then usually not be prime, thus "division" in it will not be reliable (you can still use the extended GCD algorithm, but only for numbers coprime to the new modulus), but you can still use the other three operators. In abstract algebra theory, $\mathbb Z / n \mathbb Z$ is a "ring" rather than a "field" when $n$ isn't prime due to this loss). Do be careful about the special case $0^0$, which should typically be defined as 1, while $0^{\varphi(m)}$ would still be $0$.
## Puzzles
Here are some simpler puzzles that require a modulo answer:
1281C - Cut and Paste
1279D - Santa's Bot
1178C - Tiles
1248C - Ivan the Fool and the Probability Theory
935D - Fafa and Ancient Alphabet
300C - Beautiful Numbers
• +46
By Spheniscine, history, 3 years ago,
I pretty much exclusively use Kotlin for competitive programming, mostly because it's the language I'm currently most comfortable with. Here are some scattered notes and tidbits about my experience which I think might be useful to others; if you have any tips/suggestions, feel free to let me know.
### Primer
• Kotlin has an official primer for competitive programming. However, the IO code suggested there is only so-so; it's definitely better than Scanner, but you definitely can save a lot of runtime in heavy input problems by using the classic Java combination of BufferedReader + StringTokenizer
My current IO template
### Useful features
• A lot less boilerplate than Java. Members are public by default. Type inference means a lot less "Pokémon speak". Variables and functions can be declared straight in the top-level of the file. (basically the equivalent of static functions). Fields have implicit getters and setters that can easily be overridden when necessary.
• PHP-like string templates, e.g. "$id$cost"
• Extension functions – syntactic sugar for static functions; gives more natural "afterthought" syntax, as well as allowing direct access to public members of the receiver
• data classes – basically custom tuples. Allows convenient destructuring declarations too.
• Has access to the data structures in the Java standard library (TreeMap, HashMap, PriorityQueue etc.), and also can use BigInteger and BigDecimal if needed
• Functional idioms for collection manipulation – map, fold, filter, etc.
• Sequences – lazy sequence generators, potentially infinite, has standard collection manipulation functions too. Using the sequence { ... } block function allows building a sequence using a scoped yield(value) function, reminiscent of Python
• inline classes – allows the creation of a new type that wraps over a base type, but that is represented by an underlying type at runtime. Especially useful for "modulo $10^9 + 7$" problems, as I keep code for a ModInt class that overloads the arithmetic operators appropriately, but is represented as a plain int in JVM runtime. Keep in mind that they are experimental as of Kotlin 1.3, but that's fine for CP in my opinion
• unsigned integer types in the standard library that use the inline class feature. Not used very often, but handy if needed
• inline functions – tells the compiler to inline the function to call sites. Useful for higher-order functions (JVM won't need to create a new object for the lambda) as well as small functions that are called very often; basically, anything you might use a macro for in C++, you probably want to use an inline fun or inline val for
• tailrec fun – tail recursion optimization
• run block function – great way to write code that needs to shortcut (e.g. return@run "NO") without having to write a new function and pass every relevant argument
• functions in functions – functions can be defined within e.g. the main function, so again, no having to pass lots of arguments or global variables. Keep in mind that these are represented as objects during runtime. It's too bad they can't be inline as of yet
### Potential pitfalls
• Generic wrappers for JVM primitive types can cause TLE for some problems. Use primitive arrays (IntArray etc.) whenever possible to avoid this, but see next point
• Inherits the hack-prone quicksort from Java for primitive arrays. Easiest solution is to use generic arrays or lists instead, but due to the performance benefit of primitive arrays, I've took the trouble to write code that shuffles them randomly before sorting.
• For Kotlin versions < 1.3.30, there is a bug that will throw an exception when using .asJavaRandom() on instances of kotlin.random.Random, including Kotlin's default instance. Either use Java's own Random class, or steal this wrapper:
class JavaRandom(val kt: Random) : java.util.Random(0) {
override fun next(bits: Int): Int = kt.nextBits(bits)
override fun setSeed(seed: Long) {}
}
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# NDSolve with Euler method
I want to solve this equation with NDSolve[] using the Euler method:
x'[t] == 0.5*x[t]-0.04*(x[t])^2
with initial condition and step size
x[0] == 1, h == 1
and final time = 10.
I also don't know how to make a table including x'[t] and x[t] and time.
-
NDSolve has a slew of options that allow you to control the method. You can find the standard reference here.
There, we learn how to access Euler's method using NDSolve:
Clear[x];
x = x /. First[
NDSolve[{x'[t] == 0.5*x[t] - 0.04*(x[t])^2, x[0] == 1}, x, {t, 0, 10},
StartingStepSize -> 1, Method -> {"FixedStep", Method -> "ExplicitEuler"}]
];
grid = Table[{t, x[t]}, {t, 0, 10, 1}]
ListLinePlot[grid]
It is also quite easy to program this from scratch, if you prefer.
-
No need for "FixedStep"; since Mathematica already "knows" that the Euler method is a fixed step method, one can do this instead: NDSolve[{x'[t] == 0.5*x[t] - 0.04*(x[t])^2, x[0] == 1}, x, {t, 0, 10}, StartingStepSize -> 1, Method -> "ExplicitEuler"] – Ｊ. Ｍ. Oct 11 '12 at 23:05
As Mark wrote "It is also quite easy to program this from scratch, if you prefer."
For a solution without NDSolve try..
MyEuler[start_, end_, initialvalue_, nrOfsteps_] :=Module[
{a = start, b = end, j, m = nrOfsteps},
h = (b - a)/m; (* fixed step-size *)
T = Table[ a + (j - 1) h, {j, 1, m + 1}];
Y = Table[ initialvalue, {j, 1, m + 1}];
For[ j = 1, j <= m, j++,
Y[[j + 1]] = Y[[j]] + h f[T[[j]], Y[[j]]];
];
Transpose@{T, Y}]
Testing
f[t_, x_] = 0.5*x - 0.04*x^2;(* rhs of your ODE *)
pts = MyEuler[0.0, 10.0, 1.0, 10];
ListLinePlot[pts, Mesh -> All, MeshStyle -> Red, Frame -> True]
-
something like myEuler[t0_, t1_, initial_, steps_] := With[{h = (t1 - t0)/steps}, FoldList[#1 + h f[#2, #1] &, initial, Range[t0 + h, t1, h]]] could do – belisarius Oct 11 '12 at 20:53
@belisarius Sure! But the OP wanted some Table. I guess by table he meant the array to store the discrete states as commonly done in most textbooks. I felt that mundane For will also better suit a classic textbook Euler .. – PlatoManiac Oct 11 '12 at 21:09
You could also go one step further and show how to exten NDSolve with your method. – user21 Oct 11 '12 at 23:24
@ruebenko I think J.M already has done exactly that while we were sleeping in Europe..right? – PlatoManiac Oct 12 '12 at 7:04
yes, I think he must have posted this a few minutes after the comment... – user21 Oct 12 '12 at 10:19
show 1 more comment
Had the Euler method not been built-in, one could still use NDSolve[]'s method plug-in framework, which enables NDSolve[] to "know" how to use Euler's method.
Here's how to "teach" NDSolve[] the Euler method:
Euler[]["Step"[rhs_, t_, h_, y_, yp_]] := {h, h yp};
Euler[___]["DifferenceOrder"] := 1;
Euler[___]["StepMode"] := Fixed;
Plugging in the "new" method into NDSolve[] is a snap:
xa = x /. First @ NDSolve[{x'[t] == 0.5*x[t] - 0.04*(x[t])^2, x[0] == 1},
x, {t, 0, 10}, Method -> Euler, StartingStepSize -> 1];
Getting the corresponding table is easily done, thanks to the special methods for accessing the internals of an InterpolatingFunction[]:
pts = Transpose[Append[xa["Coordinates"], xa["ValuesOnGrid"]]]
{{0., 1.}, {1., 1.46}, {2., 2.10474}, {3., 2.97991}, {4., 4.11467}, {5., 5.49478},
{6., 7.03447}, {7., 8.57235}, {8., 9.91912}, {9., 10.9431}, {10., 11.6246}}
Showing the InterpolatingFunction[] and the points together in one plot is also easily done:
Plot[xa[t], {t, 0, 10}, Epilog -> {AbsolutePointSize[4], Red, Point[pts]}, Frame -> True]
If you wanted the derivatives as well in your table, it's easy to modify pts:
phs = Append[#, xa'[#[[1]]]] & /@ pts;
You can use these for the "phase plot" of your differential equation:
ParametricPlot[{xa[t], xa'[t]}, {t, 0, 10}, AspectRatio -> 1/GoldenRatio,
Epilog -> {AbsolutePointSize[4], Red, Point[Rest /@ phs]},
Frame -> True]
The phase plot looks ugly here, since the step size was not very small and the interval of integration was not sampled well. Due to this, the InterpolatingFunction[] is unable to make a smooth-looking derivative. A smaller value of StartingStepSize (say, $1/20$) would have resulted in something that looks smoother:
but of course at the expense of more evaluations of the right-hand side.
-
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# Help with sinusoidal function rule
• June 14th 2012, 05:09 PM
danny88
Help with sinusoidal function rule
Tthis is a sinusoidal function and the following information is given, the lenght of a day was 12 hours on may 1st
The length of a day kept going up until it reached its max at 15 hours and 30 mins. It then continued going down until its minimum at 8 hours and 30mins. It finally went back up to 12 hours on may 1st. What is the function rule?
Thanks
• June 15th 2012, 12:39 AM
richard1234
Re: Help with sinusoidal function rule
Let May 1 correspond with $x = 0$ and $x = 365$. The "amplitude" is 3.5 hours. Hence our equation could look something like
$f(x) = 12 + 3.5 \sin {\frac{x}{365}}$
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# Math Help - Rings and modules
1. ## Rings and modules
Hi everyone. I'm preparing for a test tomorrow and there are concepts and methods, which I seem to have forgotten or never understood properly.
For eksample there are these exercises, which I'm having difficulties with, if someone could help me throught them, it would be nice :
1) Explain why the $\mathbb{Z}$-modules $\mathbb{Z}/8\mathbb{Z}$ and $\mathbb{Z}/36\mathbb{Z}$ have finite lenght and write a chain of composition.
2) Find the associated primes and the support of the above $\mathbb{Z}$-modules
3) k is a field. Explain why $(x)$ is the only prime ideal in $R=k[X]/(X^4)$. Is $(x^2)$ a simple R-module?
4) Are some of the three $k[X, Y]$-modules
$k[X, Y]/(XY)$ , $k[X, Y]/(X^2)$ , $k[X, Y]/(Y^2)$ isomorphic? Are some of them isomorphic as rings?
Now, in 1) I would say that $8=2^3$ and therefore
$(0)\subset 4\mathbb{Z}/8\mathbb{Z}\subset 2\mathbb{Z}/8\mathbb{Z} \subset \mathbb{Z}/8\mathbb{Z}$ correct?
and $36=2^2 3^2$ so $\mathbb{Z}/36\mathbb{Z}\cong \mathbb{Z}/3\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$ so $(0)\subset \mathbb{Z}/2\mathbb{Z} \subset \mathbb{Z}/3\mathbb{Z} \subset \mathbb{Z}/36\mathbb{Z}$? So thus the lenght is three in both cases and thus finite?
In (4) I would say that the two last are isomorphic, because you can make an ismomorphism from $k[X, Y]/(X^2)$ to $k[X, Y]/(Y^2)$ by sending X to Y, Y to X and 1 to 1. Correct?
2. Originally Posted by stephi85
Hi everyone. I'm preparing for a test tomorrow and there are concepts and methods, which I seem to have forgotten or never understood properly.
For eksample there are these exercises, which I'm having difficulties with, if someone could help me throught them, it would be nice :
1) Explain why the $\mathbb{Z}$-modules $\mathbb{Z}/8\mathbb{Z}$ and $\mathbb{Z}/36\mathbb{Z}$ have finite lenght and write a chain of composition.
Now, in 1) I would say that $8=2^3$ and therefore
$(0)\subset 4\mathbb{Z}/8\mathbb{Z}\subset 2\mathbb{Z}/8\mathbb{Z} \subset \mathbb{Z}/8\mathbb{Z}$ correct?
and $36=2^2 3^2$ so $\mathbb{Z}/36\mathbb{Z}\cong \mathbb{Z}/3\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$ so $(0)\subset \mathbb{Z}/2\mathbb{Z} \subset \mathbb{Z}/3\mathbb{Z} \subset \mathbb{Z}/36\mathbb{Z}$? So thus the lenght is three in both cases and thus finite?
well, they're both finite modules and thus obviously Artinian (and Noetherian). so they have finite length. your solution for $\mathbb{Z}/36 \mathbb{Z}$ is wrong.
2) Find the associated primes and the support of the above $\mathbb{Z}$-modules
recall that $\text{Supp}(R/I)=V(I)=\{P \in \text{Spec}(R): \ \ I \subseteq P \}.$ so $\text{Supp}(\mathbb{Z}/8\mathbb{Z})=\{2\mathbb{Z} \}$ and $\text{Supp}(\mathbb{Z}/36\mathbb{Z})=\{2\mathbb{Z}, \ 3\mathbb{Z} \}.$
let $\text{AP}(M)$ be the set of associated primes of R module M. if $p$ is a prime number, then $\text{AP}(\mathbb{Z}/p^k \mathbb{Z})=\{p\mathbb{Z} \}.$ so $\text{AP}(\mathbb{Z}/8\mathbb{Z})=\{2\mathbb{Z} \}.$
also $\text{AP}(\mathbb{Z}/36 \mathbb{Z})=\text{AP}(\mathbb{Z}/4\mathbb{Z}) \cup \text{AP}(\mathbb{Z}/9\mathbb{Z})=\{2\mathbb{Z}, \ 3\mathbb{Z} \}.$
3) k is a field. Explain why $(x)$ is the only prime ideal in $R=k[X]/(X^4)$. Is $(x^2)$ a simple R-module?
$k[X]$ is a PID. so a prime ideal of R is in the form $(f(X))/(X^4),$ where $f(X)$ is an irreducible element of $k[X]$ and $(X^4) \subseteq (f(X)).$ but $(X^4) \subseteq (f(X))$ if and only if $f(x) \mid X^4,$ which is possible
only if $f(X)=X$ because $f(X)$ is irreducible. the answer to the second part of your question is no because $(X^3)/(X^4)$ is a non-zero submodule of $(X^2)/(X^4).$
4) Are some of the three $k[X, Y]$-modules
$k[X, Y]/(XY)$ , $k[X, Y]/(X^2)$ , $k[X, Y]/(Y^2)$ isomorphic? Are some of them isomorphic as rings?
In (4) I would say that the two last are isomorphic, because you can make an ismomorphism from $k[X, Y]/(X^2)$ to $k[X, Y]/(Y^2)$ by sending X to Y, Y to X and 1 to 1. Correct?
the only possible isomorphism here is that "as rings" we have $k[X,Y]/(X^2) \cong k[X,Y]/(Y^2).$ the isomorphism sends $X$ to $Y$ and $Y$ to $X.$ to see that there are no other isomorphisms let
$R=k[X,Y], \ I=(X^2), \ J=(Y^2), \ K=(XY).$ suppose $f: R/I \longrightarrow R/K$ is a "ring" isomorphism. let $f(X+I)=p(X,Y) + K.$ then $0=(f(X+I))^2=(p(X,Y))^2+K.$
so $(p(X,Y))^2 \in K$ and thus $p(X,Y) \in K.$ (why?) therefore $f(X+I)=0$ and, since $f$ is an isomorphism, we get $X+I=0.$ that means $X \in I=(X^2),$ which is clearly false. so there's no
such isomorphism. let me show you how to prove that $R/I$ and $R/K$ are not isomorphic as "R-mouldes": suppose $g: R/I \longrightarrow R/K$ is an R-module isomorphism. let $g(q(X,Y)+I)=X+K.$
then $0=Yg(q(X,Y)+I)=g(Yq(X,Y)+I).$ so $Yq(X,Y) \in I=(X^2)$ and hence $q(X,Y) \in (X^2)=I.$ therefore $q(X,Y)+I=0,$ which gives us $X+K=g(q(X,Y)+I)=0.$ that means
$X \in K=(XY),$ which is obviously nonsense!
Originally Posted by NonCommAlg
your solution for $\mathbb{Z}/36 \mathbb{Z}$ is wrong.
Allright, but could you please explain what's wrong with it then? Is it because I only have to include one of the fields in the chain?
4. Originally Posted by stephi85
Allright, but could you please explain what's wrong with it then? Is it because I only have to include one of the fields in the chain?
here's a composition series for $\mathbb{Z}/36\mathbb{Z}: \ (0) \subset 12 \mathbb{Z}/36\mathbb{Z} \subset 4\mathbb{Z}/36\mathbb{Z} \subset 2\mathbb{Z}/36\mathbb{Z} \subset \mathbb{Z}/36\mathbb{Z}.$ do you see how i found this series?
5. Ok, thanks. I think I see the pattern now; You say $36=2^2 3^2$ and then take combinations of the powers $12=2^2 3^1$, $4=2^2 3^0$, $2=1^2 3^0$ and arrange the fields after decreasing number value.
But how come, then, that 6 and 3 aren't included?
6. Originally Posted by stephi85
Ok, thanks. I think I see the pattern now; You say $36=2^2 3^2$ and then take combinations of the powers $12=2^2 3^1$, $4=2^2 3^0$, $2=2^1 3^0$ and arrange the fields after decreasing number value.
But how come, then, that 6 and 3 aren't included?
the quotient of any two consecutive modules in our series has to be a simple moulde. that's the definition of a composition series. so you can't just put anything you want in there.
another composition series for $\mathbb{Z}/36\mathbb{Z}$ is: $(0) \subset 18 \mathbb{Z}/36\mathbb{Z} \subset 9\mathbb{Z}/36\mathbb{Z} \subset 3\mathbb{Z}/36\mathbb{Z} \subset \mathbb{Z}/36\mathbb{Z}.$ can you find another one?
so a module can have more than one composition series but the length of all composition series of a module, if they exist, are equal.
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