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• CommentRowNumber1.
• CommentAuthorDavid_Corfield
• CommentTimeJul 4th 2018

I put the opening material into an Idea section, and replaced several ’?’s with ∞.

• CommentRowNumber2.
• CommentAuthorUrs
• CommentTimeFeb 20th 2020
• (edited Feb 20th 2020)

In diuscussion with laymen I noticed that this old entry didn’t explain very much at all. I have now expanded the Idea-section as follows:

$\infty Grpd$ is the (∞,1)-category of ∞-groupoids, i.e. of (∞,0)-categories. This is the archetypical (∞,1)-topos, the home of classical homotopy theory.

Equivalently this means all of the following:

1. $\infty Grpd$ is the simplicial localization of the category Top${}_k$ of (weakly Hausdorff) locally compact topological spaces at the weak homotopy equivalences. As such it is the ∞-category-enhancement of the classical homotopy category: $\tau_0(\infty Grpd) \simeq$ Ho(Top), itself presented by the classical model structure on topological spaces: $\infty Grpd \simeq L_{whe} Top_k$.

2. $\infty Grpd$ is the simplicial localization of the category sSet of simplicial sets at the simplicial weak homotopy equivalences. As such it is the ∞-category-enhancement of the classical homotopy category: $\tau_0(\infty Grpd) \simeq$ Ho(sSet), itself presented by the classical model structure on simplicial sets: $\infty Grpd \simeq L_{whe} sSet$.

Hence, as a Kan-complex enriched category (a fibrant object in the model structure on sSet-categories) $\infty Grpd$ is the full sSet enriched-subcategory in sSet on those that are Kan complexes.

3. $\infty Grpd$ is the full sub-(∞,1)-category of (∞,1)Cat on those (∞,1)-categories that are ∞-groupoids.

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## College Algebra 7th Edition

Published by Brooks Cole

# Chapter 1, Equations and Graphs - Section 1.2 - Graphs of Equations in Two Variables; Circles - 1.2 Exercises: 71

#### Answer

center: $(-3, 4)$ radius: $5$ units Refer to the image below for the graph.

#### Work Step by Step

RECALL: The circle $(x--h)^2+(y-k)^2=r^2$ has its center at $(h, k)$ and a radius of $r$ units. The given equation can be written as $(x+3)^2+(y-4)^2=5^2$, which is in the same form as the one in the recall part above. Thus, the circle has: its center at $(-3, 4)$; a radius of 5 units Plot the center. Then, plot the following points: 5 units above the center: $(-3, 9)$ 5 units below the center: $(-3, -1)$ 5 units to the left of the center: $(-8, 4)$ 5 units to the right of the center: $(2, 4)$ Connect the four points (not including the center) using a curve to form a circle.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.

format: plain


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# Tesla’s Math Spiral Diagram and Prime numbers

### Tesla’s Math Spiral Diagram and Prime numbers

According Tesla’s Math Spiral Diagram, all prime numbers (excluding 2 and 3) belong to 1/3 part of natural numbers set:
$$M1(n) = 1 + (n -1) * 12;$$
$$M5(n) = 5 + (n -1) * 12;$$
$$M7(n) = 7 + (n -1) * 12;$$
$$M11(n) = 11 + (n -1) * 12;$$

Statement. Any not prime number in these sets has the prime number(s) as its divisor(s)/factor(s).

Can this statement be proofed or not?
Mblacob

Posts: 6
Joined: Thu Jun 14, 2018 6:58 am
Reputation: 1

### Re: Tesla’s Math Spiral Diagram and Prime numbers

According Tesla’s Math Spiral Diagram these two sets of natural odd numbers:

M3(n)=3+(n−1)∗12;
M9(n)=9+(n−1)∗12;

don’t have any Prime numbers because ALL their numbers have 3 as the prime factor.
Guest

### Re: Tesla’s Math Spiral Diagram and Prime numbers

[quote="Mblacob"]According Tesla’s Math Spiral Diagram, all prime numbers (excluding 2 and 3) belong to 1/3 part of natural numbers set:
$$M1(n) = 1 + (n -1) * 12;$$
$$M5(n) = 5 + (n -1) * 12;$$
$$M7(n) = 7 + (n -1) * 12;$$
$$M11(n) = 11 + (n -1) * 12;$$
Guest

### Re: Tesla’s Math Spiral Diagram and Prime numbers

To which sequence of odd numbers belongs any given odd number N > 11?
If N has 3 as its primary factor then N belongs to sequences started from 3 and 9.
If mod(N+11, 12) = 0 N belongs to sequence started from 1.
If mod(N+7, 12) = 0 N belongs to sequence started from 5.
If mod(N+5, 12) = 0 N belongs to sequence started from 7.
If mod(N+1, 12) = 0 N belongs to sequence started from 11.

if we use N as some kind of "seed" then N +12k or N - 12k (where k = 1,2,3,...) will give us number which can be Prime.
Guest


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# Geometry/Angles

An angle is the union of two rays with a common endpoint, called the vertex. The angles formed by vertical and horizontal lines are called right angles; lines, segments, or rays that intersect in right angles are said to be perpendicular.

Angles, for our purposes, can be measured in either degrees (from 0 to 360) or radians (from 0 to ${\displaystyle 2\pi }$). Angles length can be determined by measuring along the arc they map out on a circle. In radians we consider the length of the arc of the circle mapped out by the angle. Since the circumference of a circle is ${\displaystyle 2\pi }$, a right angle is ${\displaystyle {\frac {\pi }{2}}}$ radians. In degrees, the circle is 360 degrees, and so a right angle would be 90 degrees.

## Naming Conventions

Angles are named in several ways.

• By naming the vertex of the angle (only if there is only one angle formed at that vertex; the name must be non-ambiguous) ${\displaystyle \angle B}$
• By naming a point on each side of the angle with the vertex in between. ${\displaystyle \angle ABC}$
• By placing a small number on the interior of the angle near the vertex.

Hai

## Classification of Angles by Degree Measure

Acute Angle

• an angle is said to be acute if it measures between 0 and 90 degrees, exclusive.

Right Angle

• an angle is said to be right if it measures 90 degrees.
• notice the small box placed in the corner of a right angle, unless the box is present it is not assumed the angle is 90 degrees.
• all right angles are congruent

Obtuse Angle

• an angle is said to be obtuse if it measures between 90 and 180 degrees, exclusive.

## Special Pairs of Angles

• adjacent angles are angles with a common vertex and a common side.
• adjacent angles have no interior points in common.
• complementary angles
• complementary angles are two angles whose sum is 90 degrees.
• complementary angles may or may not be adjacent.
• if two complementary angles are adjacent, then their exterior sides are perpendicular.
• supplementary angles
• two angles are said to be supplementary if their sum is 180 degrees.
• supplementary angles need not be adjacent.
• if supplementary angles are adjacent, then the sides they do not share form a line.
• linear pair
• if a pair of angles is both adjacent and supplementary, they are said to form a linear pair.
• vertical angles
• angles with a common vertex whose sides form opposite rays are called vertical angles.
• vertical angles are congruent.


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Encyclopedia > John Herschel

Article Content

John Herschel

John Frederick William Herschel (March 7, 1792 - May 11, 1871) English mathematician and astronomer.

Son of astronomer William Herschel.

John Herschel orginated the use of the Julian date system in astronomy.

All Wikipedia text is available under the terms of the GNU Free Documentation License

Search Encyclopedia
 Search over one million articles, find something about almost anything!

Featured Article
 Quadratic formula ... fraction; the common denominator is 4a2. We get $\left(x+\frac{b}{2a}\right)^2=\frac{-4ac+b^2}{4a^2}=\frac{b^2-4ac}{4a^2}.$ Taking square ...


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## Return to Question

3 added 680 characters in body; deleted 1 characters in body

What is the dual of a norm that is the sum of two-norms? Specifically, say we have the following norm for $\mathbf{x}\in \mathbb{R}^n$ and $\mathbf{A}_i \in \mathbb{R}^{m \times n}$

$\|\mathbf{x}\| = \displaystyle{ \sum_{i=0}^{k} \|\mathbf{A}_i \cdot \mathbf{x} \|_2}$.

How would you then find

$\|\mathbf{y}\|_* = \underset{\mathbf{x}}{\mathrm{max}} \left\{ \mathbf{y}^T \mathbf{x} \;\; \mathrm{s.t.} \;\; \|\mathbf{x}\| \leq 1\right\}$?

I've tried solving for the convex conjugate looking for hints, but was unable to come up with anything meaningful.

Also, if anyone has recommendations for packages that I could use (preferably matlab-based) to solve the above numerically for systems as small as $10^3$ and as large as $10^6$, I'd greatly appreciate it. CVX, of which I am admittedly a novice and a hack, will not maximize convex functions.

Edit: So using the advice in the below comments, I end up with an eigen equation for the critical point $\mathbf{x}_*$:

$\displaystyle{ \sum_{i=0}^{k} {\|\mathbf{A}_i \mathbf{x}_* \|_2 } } \cdot \mathbf{y} = \displaystyle{ \sum_{i=0}^{k} { {\mathbf{A}_i^T \mathbf{A_i} } \over{ \|\mathbf{A}_i \cdot \mathbf{x}_* \|_2 } } } \mathbf{x}_* \mathbf{x}_*^{T} \cdot \mathbf{y}$

The only other idea I have had is that we know that $\|y\|_*$ is the function such that $\underset{\mathbf{x}}{\sup} \{ \mathbf{y}^T \mathbf{x} - \|\mathbf{x}\|\}$ is zero whenever $\|y\|_* \leq 1$ and is $\infty$ otherwise. I have not been able to use this in any meaningful way however.

2 deleted 8 characters in body

What is the dual of a norm that is the sum of two-norms? Specifically, say we have the following norm for $\mathbf{x}\in \mathbb{R}^n$ and $\mathbf{A}_i \in \mathbb{R}^{m \times n}$

$\|\mathbf{x}\| = \displaystyle{ \sum_{i=0}^{k} \|\mathbf{A}_i \cdot \mathbf{x} \|_2}$.

How would you then find

$\|\mathbf{y}\|_* = \underset{\mathbf{x}}{\mathrm{max}} \left\{ |\mathbf{y}^T \cdot mathbf{y}^T \mathbf{x}| mathbf{x} \;\; \mathrm{s.t.} \;\; \|\mathbf{x}\| \leq 1\right\}$?

I've tried solving for the convex conjugate looking for hints, but was unable to come up with anything meaningful.

Also, if anyone has recommendations for packages that I could use (preferably matlab-based) to solve the above numerically for systems as small as $10^3$ and as large as $10^6$, I'd greatly appreciate it. CVX, of which I am admittedly a novice and a hack, will not maximize convex functions.

1

# Dual Norm For Sum of 2-Norms

What is the dual of a norm that is the sum of two-norms? Specifically, say we have the following norm for $\mathbf{x}\in \mathbb{R}^n$ and $\mathbf{A}_i \in \mathbb{R}^{m \times n}$

$\|\mathbf{x}\| = \displaystyle{ \sum_{i=0}^{k} \|\mathbf{A}_i \cdot \mathbf{x} \|_2}$.

How would you then find

$\|\mathbf{y}\|_* = \underset{\mathbf{x}}{\mathrm{max}} \left\{ |\mathbf{y}^T \cdot \mathbf{x}| \;\; \mathrm{s.t.} \;\; \|\mathbf{x}\| \leq 1\right\}$?

I've tried solving for the convex conjugate looking for hints, but was unable to come up with anything meaningful.

Also, if anyone has recommendations for packages that I could use (preferably matlab-based) to solve the above numerically for systems as small as $10^3$ and as large as $10^6$, I'd greatly appreciate it. CVX, of which I am admittedly a novice and a hack, will not maximize convex functions.


text:
# Mod-$\phi$ convergence

Delbaen, Freddy; Kowalski, Emmanuel; Nikeghbali, Ashkan (2015). Mod-$\phi$ convergence. International Mathematics Research Notices, 2015(11):3445-3485.

## Abstract

Using Fourier analysis, we study local limit theorems in weak-convergence problems. Among many applications, we discuss random matrix theory, some probabilistic models in number theory, the winding number of complex Brownian motion and the classical situation of the central limit theorem, and a conjecture concerning the distribution of values of the Riemann zeta function on the critical line.

Using Fourier analysis, we study local limit theorems in weak-convergence problems. Among many applications, we discuss random matrix theory, some probabilistic models in number theory, the winding number of complex Brownian motion and the classical situation of the central limit theorem, and a conjecture concerning the distribution of values of the Riemann zeta function on the critical line.

## Citations

1 citation in Web of Science®
1 citation in Scopus®

## Altmetrics

Detailed statistics

Item Type: Journal Article, refereed, original work 07 Faculty of Science > Institute of Mathematics 510 Mathematics English June 2015 14 Jan 2016 11:47 05 Apr 2016 19:18 Oxford University Press 1073-7928 https://doi.org/10.1093/imrn/rnu035
Permanent URL: https://doi.org/10.5167/uzh-111518

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# Please find the ROE, RNOA, NOPM, NOAT, AT, ROA, Times Interest Earned, Cash Flow from Operations ...

Please find the ROE, RNOA, NOPM, NOAT, AT, ROA, Times Interest Earned, Cash Flow from Operations to Total Debt, Free Cash Flow to Total Debt, Current Ratio, Quick Ratio, Liabilities to Equity Ratio, Total Debt to Equity Ratio, Turnover Ratios for Cash, A/R, Inventory, and A/P, and Z-Score for all years. Also, please show work!

 Consolidated Balance Sheet - USD ($)$ in Millions Dec. 31, 2015 Dec. 31, 2014 Dec. 31, 2013 Dec. 31, 2012 Current Assets Cash and cash equivalents $8,524$ 7,441 $15,621$13,451 Short-term investments 4,903 8,278 1,865 2,690 Accounts receivable (net of allowance for doubtful accounts of $165 in 2015 and$153 in 2014) (excludes accounts receivable of $10 in 2015 and$80 in 2014 classified in Other assets - see Note 5) 6,484 6,626 7,184 7,672 Inventories (excludes inventories of $1,569 in 2015 and$1,664 in 2014 classified in Other assets - see Note 6) 4,700 5,571 6,226 6,535 Other current assets 5,153 4,689 0 0 Deferred income taxes and other current assets 0 0 4,789 4,509 Total current assets 29,764 32,605 35,685 34,857 Investments 13,039 13,515 9,770 7,305 Property, Plant and Equipment (at cost) Land 490 541 550 591 Buildings 12,154 13,101 13,627 13,196 Machinery, equipment and office furnishings 14,261 16,050 17,106 17,188 Construction in progress 1,525 1,448 1,811 2,440 Property, Plant and Equipment, Gross 28,430 31,140 33,094 33,415 Less: accumulated depreciation 15,923 18,004 18,121 17,385 Property, Plant and Equipment, Net, Total 12,507 13,136 14,973 16,030 Goodwill 17,723 12,992 12,301 12,134 Other Intangibles, Net 22,602 20,386 23,801 29,083 Other Assets 6,144 5,533 9,115 6,723 Total Assets 101,779 98,167 105,645 106,132 Current Liabilities Loans payable and current portion of long-term debt 2,585 2,704 4,521 4,315 Trade accounts payable 2,533 2,625 2,274 1,753 Accrued and other current liabilities 11,216 10,523 9,501 9,737 Income taxes payable 1,560 1,237 251 1,200 Dividends payable 1,309 1,308 1,321 1,343 Total current liabilities 19,203 18,397 17,868 18,348 Long-Term Debt 23,929 18,699 20,539 16,254 Deferred Income Taxes and Other Noncurrent Liabilities 13880 12,079 14,912 16,067 Merck & Co., Inc. Stockholders’ Equity Common stock, $0.50 par value Authorized - 6,500,000,000 shares Issued - 3,577,103,522 shares in 2015 and 2014 1,788 1,788 1,788 1,788 Other paid-in capital 40,222 40,423 40,508 40,646 Retained earnings 45,348 46,021 39,257 39,985 Accumulated other comprehensive loss (4,148) (4,323) -2,197 -4,682 Stockholders' equity before deduction for treasury stock 83,210 83,909 79,356 77,737 Less treasury stock, at cost: 795,975,449 shares in 2015 and 738,963,326 shares in 2014 38,534 35,262 29,591 24,717 Total Merck & Co., Inc. stockholders’ equity 44,676 48,647 49,765 53,020 Noncontrolling Interests 91 144 2,561 2,443 Total equity 44,767 48,791 52,326 55,463 Total Liabilities and Equity$ 101,779 $98,167$105,645 $106,132  Consolidated Statement of Income - USD ($) $in Millions 12 Months Ended Dec. 31, 2015 Dec. 31, 2014 Dec. 31, 2013 Dec. 31, 2012 Income Statement [Abstract] Sales$ 39,498 $42,237$ 44,033 $47,267 Costs, Expenses and Other Materials and production 14,934 16,768 16,954 16,446 Marketing and administrative 10,313 11,606 11,911 12,776 Research and development 6,704 7,180 7,503 8,168 Restructuring costs 619 1,013 1,709 664 Equity income from affiliates 0 0 0 -642 Other (income) expense, net 1,527 (11,613) 411 1,116 Total Costs, Expenses and Other 34,097 24,954 38,488 38,528 Income Before Taxes 5,401 17,283 5,545 8,739 Taxes on Income 942 5,349 1,028 2,440 Net Income 4,459 11,934 4,517 6,299 Less: Net Income Attributable to Noncontrolling Interests 17 14 113 131 Net Income Attributable to Merck & Co., Inc.$ 4,442 $11,920$ 4,404 $6,168 Basic Earnings per Common Share Attributable to Merck & Co., Inc. Common Shareholders (in dollars per share)$ 1.58 $4.12$ 1.49 $2.03 Earnings per Common Share Assuming Dilution Attributable to Merck & Co., Inc. Common Shareholders (in dollars per share)$ 1.56 $4.07$ 1.47 $2  Consolidated Statement of Cash Flows - USD ($) $in Millions 12 Months Ended Dec. 31, 2015 Dec. 31, 2014 Dec. 31, 2013 Dec. 31, 2012 Cash Flows from Operating Activities Net income$ 4,459 $11,934$ 4,517 $6,299 Adjustments to reconcile net income to net cash provided by operating activities: Depreciation and amortization 6,375 6,691 6,988 6,978 Intangible asset impairment charges 162 1,222 765 200 Foreign currency devaluation related to Venezuela 876 0 140 0 Net charge related to the settlement of Vioxx shareholder class action litigation 680 0 0 0 Gain on divestiture of Merck Consumer Care business 0 (11,209) 0 0 Gain on AstraZeneca option exercise 0 (741) 0 0 Loss on extinguishment of debt 0 628 0 0 Equity income from affiliates (205) (257) (404) -642 Dividends and distributions from equity method affiliates 50 185 237 291 Deferred income taxes (764) (2,600) (330) 669 Share-based compensation 299 278 276 335 Other 757 (95) 259 28 Net changes in assets and liabilities: Accounts receivable (480) (554) 436 349 Inventories 805 79 (365) -482 Trade accounts payable (37) 593 522 -302 Accrued and other current liabilities (8) 1,635 (397) -717 Income taxes payable (266) (21) (1,421) -34 Noncurrent liabilities (277) 190 (132) -1,747 Other (5) (98) 563 -1,203 Net Cash Provided by Operating Activities 12,421 7,860 11,654 10,022 Cash Flows from Investing Activities Capital expenditures (1,283) (1,317) (1,548) -1,954 Purchases of securities and other investments (16,681) (24,944) (17,991) -12,841 Proceeds from sales of securities and other investments 20,413 15,114 16,298 7,783 Divestiture of Merck Consumer Care business, net of cash divested 0 13,951 0 0 Dispositions of other businesses, net of cash divested 316 1,169 46 0 Proceeds from AstraZeneca option exercise 0 419 0 0 Acquisition of Cubist Pharmaceuticals, Inc., net of cash acquired (7,598) 0 0 0 Acquisition of Idenix Pharmaceuticals, Inc., net of cash acquired 0 (3,700) 0 0 Acquisitions of other businesses, net of cash acquired (146) (181) (246) 0 Acquisition of Bayer AG collaboration rights 0 (1,000) 0 0 Cash inflows from net investment hedges 139 195 350 39 Other 82 (80) (57) 168 Net Cash Used in Investing Activities (4,758) (374) (3,148) -6,805 Cash Flows from Financing Activities Net change in short-term borrowings (1,540) (460) (159) 624 Payments on debt (2,906) (6,617) (1,775) -22 Proceeds from issuance of debt 7,938 3,146 6,467 2,562 Purchases of treasury stock (4,186) (7,703) (6,516) -2,591 Dividends paid to stockholders (5,117) (5,170) (5,157) -5,116 Other dividends paid 0 (77) (120) -120 Proceeds from exercise of stock options 485 1,560 1,210 1,310 Other 56 208 60 86 Net Cash Used in Financing Activities (5,270) (15,113) (5,990) -3,267 Effect of Exchange Rate Changes on Cash and Cash Equivalents (1,310) (553) (346) -30 Net Increase (Decrease) in Cash and Cash Equivalents 1,083 (8,180) 2,170 -80 Cash and Cash Equivalents at Beginning of Year 7,441 15,621 13,451 13,531 Cash and Cash Equivalents at End of Year$ 8,524 $7,441$ 15,621 \$13,451


text:
## Arkiv för Matematik

• Ark. Mat.
• Volume 37, Number 1 (1999), 171-182.

### Norm convergence of normalized iterates and the growth of Kœnigs maps

#### Abstract

Let ϕ be an analytic function defined on the unit disk D, with ϕ(D)⊂D, ϕ(0)=0, and ϕ′(0)=λ≠0. Then by a classical result of G. Kœnigs, the sequence of normalized iterates Φnn converges uniformly on compact subsets of D to a function σ analytic in D which satisfies σ°φ=λσ. It is of interest in the study of composition operators to know if, whenever σ belongs to a Hardy space Hp, the sequence Φnn converges to σ in the norm of Hp. We show that this is indeed the case, generalizing a result of P. Bourdon obtained under the assumption that ϕ is univalent.

When ϕ is inner, P. Bourdon and J. Shapiro have shown that σ does not belong to the Nevanlinna class, in particular it does not belong to any Hp. It is natural to ask, how bad can the growth of σ be in this case? As a partial answer we show that σ always belongs to some Bergman space L ${}_{a}^{p}$ .

#### Note

The author is partially supported by NSF Grant DMS 97-06408 and wishes to thank Professor P. Bourdon for sharing results and conjectures.

#### Article information

Source
Ark. Mat., Volume 37, Number 1 (1999), 171-182.

Dates
First available in Project Euclid: 31 January 2017

https://projecteuclid.org/euclid.afm/1485898621

Digital Object Identifier
doi:10.1007/BF02384832

Mathematical Reviews number (MathSciNet)
MR1673430

Zentralblatt MATH identifier
1029.30017

Rights

#### Citation

Poggi-Corradini, Pietro. Norm convergence of normalized iterates and the growth of Kœnigs maps. Ark. Mat. 37 (1999), no. 1, 171--182. doi:10.1007/BF02384832. https://projecteuclid.org/euclid.afm/1485898621

#### References

• [B] Bourdon, P., Convergence of the Kœnigs sequence, in Studies on Composition Operators (Jafari, F., MacCluer, B. D., Cowen, C. C. and Porter, A. D., eds.), Contemp. Math. 213, pp. 1–10, Amer. Math. Soc., Providence, R. I., 1998.
• [BS] Bourdon, P. and Shapiro, J., Mean growth of Kœnigs eigenfunctions, J. Amer. Math. Soc. 10 (1997), 299–325.
• [CG] Carleson, L. and Gamelin, T., Complex Dynamics, Springer-Verlag, Berlin-Heidelberg-New York, 1993.
• [D] Duren, P., Theory of HpSpaces, Academic Press, New York, 1970.
• [G] Garnett, J., Bounded Analytic Functions, Academic Press, Orlando, Fla., 1981.
• [K] Korenblum, B., An extension of the Nevanlinna theory, Acta Math., 135 (1975), 187–219.
• [P1] Poggi-Corradini, P., The Hardy class of Kœnigs maps, Mich. Math. J. 44 (1997), 495–507.
• [P2] Poggi-Corradini, P., The Hardy class of geometric models and the essential spectral radius of composition operators, J. Funct. Anal., 143 (1997), 129–156.
• [R] Rudin, W., Function Theory in the Unit Ball ofCn, Springer-Verlag, Berlin-Heidelberg-New York, 1980.
• [S] Shapiro, J., Composition Operators and Classical Function Theory, Springer-Verlag, Berlin-Heidelberg-New York, 1993.


text:
# Donggyu Kim (김동규) gave a talk on 𝝘-graphic delta-matroids and their algorithmic applications at the Discrete Math Seminar

On October 26, 2021, Donggyu Kim (김동규) from KAIST & IBS Discrete Mathematics Group gave a talk at the Discrete Math Seminar, introducing 𝝘-graphic delta-matroids and showing their algorithmic applications. The title of his talk was “𝝘-graphic delta-matroids and their applications“.


text:
## Electronic Communications in Probability

### Fragmenting random permutations

#### Abstract

Problem 1.5.7 from Pitman's Saint-Flour lecture notes: Does there exist for each $n$ a fragmentation process $(\Pi_{n,k}, 1 \leq k \leq n)$ such that $\Pi_{n,k}$ is distributed like the partition generated by cycles of a uniform random permutation of $\{1,2,\ldots,n\}$ conditioned to have $k$ cycles? We show that the answer is yes. We also give a partial extension to general exchangeable Gibbs partitions.

#### Article information

Source
Electron. Commun. Probab. Volume 13 (2008), paper no. 44, 461-474.

Dates
Accepted: 14 August 2008
First available in Project Euclid: 6 June 2016

https://projecteuclid.org/euclid.ecp/1465233471

Digital Object Identifier
doi:10.1214/ECP.v13-1402

Mathematical Reviews number (MathSciNet)
MR2430713

Zentralblatt MATH identifier
1189.60022

Subjects
Primary: 60C05: Combinatorial probability
Secondary: 05A18: Partitions of sets

Rights

#### Citation

Goldschmidt, Christina; Martin, James; Spano, Dario. Fragmenting random permutations. Electron. Commun. Probab. 13 (2008), paper no. 44, 461--474. doi:10.1214/ECP.v13-1402. https://projecteuclid.org/euclid.ecp/1465233471

#### References

• Berestycki, Nathanaël; Pitman, Jim. Gibbs distributions for random partitions\n generated by a fragmentation process. J. Stat. Phys. 127 (2007), no. 2, 381–418.
• Elias, P.; Feinstein, A.; Shannon, C.. A note on the maximum flow through a network.\n Institute of Radio Engineers, Transactions on Information Theory IT-2 (1956),\n 117–119.
• Ford, L. R., Jr.; Fulkerson, D. R. Maximal flow through a network. Canad. J. Math. 8\n (1956), 399–404.
• Gnedin, A.; Pitman, J. Exchangeable Gibbs partitions and Stirling triangles. Zap.\n Nauchn. Sem. S.-Peterburg. Otdel. Mat. Inst. Steklov. (POMI) 325 (2005), Teor. Predst. Din.\n Sist. Komb. i Algoritm. Metody. 12, 83–102, 244–245; translation in J. Math. Sci. (N. Y.)\n 138 (2006), no. 3, 5674–5685
• Gnedin, Alexander; Pitman, Jim. Poisson representation of a Ewens fragmentation\n process. Combin. Probab. Comput. 16 (2007), no. 6, 819–827.
• Granovsky, Boris; Erlihson, Michael. On time dynamics of coagulation-fragmentation\n processes. arXiv:0711.0503v2[math.PR] (2007).
• Hall, P.. On representatives of subsets. J. London Math. Soc. 10 (1935),\n 26–30.
• Hoggar, S. G. Chromatic polynomials and logarithmic concavity. J. Combinatorial Theory\n Ser. B 16 (1974), 248–254.
• Kamae, T.; Krengel, U.; O'Brien, G. L. Stochastic inequalities on partially ordered\n spaces. Ann. Probability 5 (1977), no. 6, 899–912.
• Pitman, J. Combinatorial stochastic processes. Lectures from the 32nd Summer School on\n Probability Theory held in Saint-Flour, July 7–24, 2002. With a foreword by Jean Picard.\n Lecture Notes in Mathematics, 1875. Springer-Verlag, Berlin, 2006. x+256 pp. ISBN:\n 978-3-540-30990-1; 3-540-30990-X
• Pitman, Jim; Yor, Marc. The two-parameter Poisson-Dirichlet distribution derived from\n a stable subordinator. Ann. Probab. 25 (1997), no. 2, 855–900.
• Sagan, Bruce E. Inductive and injective proofs of log concavity results. Discrete\n Math. 68 (1988), no. 2-3, 281–292.


text:
### JEE Mains Previous Years Questions with Solutions

4.5
star star star star star
1

### AIEEE 2003

A strip of copper and another of germanium are cooled from room temperature to $80K.$ The resistance of
A
each of these decreases
B
copper strip increases and that of germanium decreases
C
copper strip decreases and that of germanium increases
D
each of these increases

## Explanation

The resistance of metal (like $Cu$) decreases with decrease in temperature whereas the resistance of a semi-conductor (like $Ge$) increases with decrease in temperature.
2

### AIEEE 2002

The part of a transistor which is most heavily doped to produce large number of majority carriers is
A
emmiter
B
base
C
collector
D
can be any of the above three

## Explanation

Emitter sends the majority charge carries towards the collector. Therefore emitter is most heavily doped.
3

### AIEEE 2002

The energy band gap is maximum in
A
metals
B
superconductors
C
insulator
D
semiconductor

## Explanation

The energy band gap is maximum in insulators.
4

### AIEEE 2002

By increasing the temperature, the specific resistance of a conductor and a semiconductor
A
increases for both
B
decreases for both
C
increases, decreases
D
decreases, increases

## Explanation

Specific resistance is resistivity which is given by

$\rho = {m \over {m{e^2}\,\tau }}$

where $n=no.$ of free electrons per unit volume and $\tau$ $=$ average relaxation time

For a conductor with rise in temperature $n$ increases and $\tau$ decreases. But decrease in $\tau$ is more dominant than increase in $n$ resulting an increase in the value of $\rho$

For a semiconductor with rise in temperature, $n$ increases and $\tau$ decreases. But the increase in $n$ is more dominant than decrease in $\tau$ resulting in decrease in the value of $\rho .$


text:
# Preserving inequality through inner product

Let $f, g, h \in H$ be functions mapping from $X$ to $\mathbb{R}$. Define an inner product $\langle \cdot, \cdot \rangle_H:H \times H \to F$. Under what conditions (if any) does $$0<f(x) < g(x), \quad 0<h(x), \quad\forall x \in X$$ imply $$\langle f, h \rangle_H < \langle g, h\rangle_H$$ ?

Assume for all $u,v\in H$ such that $u>0,v>0$ we have $$\langle u,v\rangle_H > 0$$
Then $g-f > 0$, $g-f \in H$ and $$\langle g,h \rangle_H - \langle f,h \rangle_H = \langle g-f,h \rangle_H > 0$$
and we can prove $$\langle g,h\rangle_H > \langle f,h\rangle_H$$
Assume that there exist an $u>0$ and a $v>0$ such that $\langle u,v\rangle_H \leq 0$, then take $h=u,f=u/2,g=v$ then $h,f,g\in H$, $0<f<g$ and $$\langle g-f ,h\rangle_H = \langle u/2 ,v\rangle_H = \frac{1}{2}\langle u ,v\rangle_H \leq 0$$ Therefore the condition is nessesary and sufficient
• Great answer. I think in your last line you didn't mean to put the $>$ symbols there. – Student Jun 1 '18 at 7:29


text:
[NTG-context] MKIV caption separator, revisited

Meer, Hans van der H.vanderMeer at uva.nl
Mon Oct 10 12:29:49 CEST 2016

```Thanks I did not know the correct designations for the various parts of the captionlabel: the prefixconnector and the numberstopper. It is clear now.
Hans van der Meer

On 10 Oct 2016, at 10:40, Wolfgang Schuster <schuster.wolfgang at gmail.com<mailto:schuster.wolfgang at gmail.com>> wrote:

Wolfgang Schuster<mailto:schuster.wolfgang at gmail.com>
10. Oktober 2016 um 10:39
prefixconnector is the separator between the section number (=prefix) and the float number (=figure number), the symbol at the end of the float number is called numberstopper. In your example you don’t have a section which can act as prefix number no did you change which prefixes should be shown (by default only chapters are used as prefix).

The suffix key you use (together with suffixconnector and suffixstopper) is used to show the number for subfigures, e.g. figure 3.a.
Here is a working example for the use of the prefixconnector key.

\starttext

\section{This is a section}

\setupcaptions[prefixsegments=section,prefixconnector=-]

\startplacefigure[title=\tex{setupcaptions[prefixconnector=-]}]
\externalfigure[dummy][width=.5\textwidth]
\stopplacefigure

\stoptext

Wolfgang
___________________________________________________________________________________
If your question is of interest to others as well, please add an entry to the Wiki!

maillist : ntg-context at ntg.nl<mailto:ntg-context at ntg.nl> / http://www.ntg.nl/mailman/listinfo/ntg-context


text:
+0

# Pre-calc help

+1
263
2
+295

Find the area of a triangle if a = 15ft b = 44ft C = 66°. Round the answer to 2 decimal places.

Deathstroke_rule  Apr 25, 2017

### Best Answer

#2
+7225
+2

Find the area of a triangle if a = 15ft b = 44ft C = 66°. Round the answer to 2 decimal places.

$$A=\frac{1}{2}absin\gamma=\frac{1}{2}\cdot 15ft\cdot44ft\cdot sin\ 66°$$

$$A=301.47\ ft^2$$

!

asinus  Apr 25, 2017
edited by asinus  Apr 25, 2017
Sort:

### 2+0 Answers

#1
+84362
+2

The area  =

(1/2) (15) (44) sin (66) ≈ 301.47 square feet

CPhill  Apr 25, 2017
#2
+7225
+2
Best Answer

Find the area of a triangle if a = 15ft b = 44ft C = 66°. Round the answer to 2 decimal places.

$$A=\frac{1}{2}absin\gamma=\frac{1}{2}\cdot 15ft\cdot44ft\cdot sin\ 66°$$

$$A=301.47\ ft^2$$

!

asinus  Apr 25, 2017
edited by asinus  Apr 25, 2017

### 24 Online Users

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text: dc.creator Ashizawa, Nolitaka en_US dc.creator Suganuma, Yoshitake en_US dc.creator Kowaka, Yasuyuki en_US dc.creator Baba, Masaaki en_US dc.date.accessioned 2010-07-12T13:57:54Z dc.date.available 2010-07-12T13:57:54Z dc.date.issued 2010 en_US dc.identifier 2010-MG-06 en_US dc.identifier.uri http://hdl.handle.net/1811/46129 dc.description Author Institution: Graduate School of Science, Kyoto University, Sakyo-ku, Kyoto; 606-8502, Japan en_US dc.description.abstract Perylene is one of the prototypical compact polycyclic aromatic hydrocarbons (PAHs). The fluorescence quantum yield in the $S_1$ state is considerably high. It is strange because radiationless transitions such as intersystem crossing (ISC) to the triplet state and internal conversion (IC) to the $S_0$ state are expected to be fast due to high density of coupling levels. We observed the ultrahigh-resolution spectrum of the $S_1 \;^1B_{2u}$ $\leftarrow$ $S_0 \;^1A_g$ transition of jet-cooled perylene, and determined the rotational constants in these two states. , {\em J. Mol. Spectrosc.}, {\bf 260}, 72 (2010)} The inertial defect is positive and small, so that the molecule is considered to be planar with $D_{2h}$ symmetry. The obtained rotational constants of the $S_1$ state is very similar to those of the $S_0$ state. It indicates that the structural change upon electronic transition is very small. We performed $ab$ $initio$ calculation, and the calculated geometrical molecular structure and vibrational energies were in good coincidence with the experimental results. The observed high fluorescence quantum yield shows that radiationless transitions are all slow in the $S_1$ state of perylene. The slow IC can be attributed to the small scale of its structural change. We also observed the ultrahigh-resolution spectrum in the magnetic field of 0.5 Tesla. Zeeman broadening was very small, indicating that the singlet-triplet interaction is very weak, and that ISC is minor in the $S_1$ state. In specific vibronic levels, the fluorescence lifetimes are remarkably short. We conclude that the main process is IC, and the accepting mode is in-plane ($a_g$) deformation vibration in the $S_1$ state. en_US dc.language.iso en en_US dc.publisher Ohio State University en_US dc.title STRUCTURE AND EXCITED-STATE DYNAMICS OF PERYLENE : ULTRAHIGH-RESOLUTION SPECTROSCOPY AND AB INITIO CALCULATION en_US dc.type Article en_US dc.type Image en_US dc.type Presentation en_US


text:
# [Microelectronic Circuits 6th Ed.] P2.27 (Regular Ed.) / P2.23 (International Ed.)

Discussion in 'Homework Help' started by Starhowl, Dec 14, 2013.

1. ### Starhowl Thread Starter New Member

Dec 3, 2013
25
0
Can someone please explain to me what I did wrong in trying to solve this task? Or maybe I did everything right and there's an error in both the international and regular edition?

File size:
1.2 MB
Views:
32
2. ### Starhowl Thread Starter New Member

Dec 3, 2013
25
0
As far as I understand my A is in truth $A_{2}$, which instead should be $A_1$ - I still don't get it. There seems to be no way to get to that form; when looking at the solution under
http://www.***********/doc/19231397/Microelectronic-Circuits-5thEd-Solution-Manual
2.27 the author of the solution suggests, that $\Delta|G|$ seems to linearly dependent on $\Delta A$, which in fact is not true, so the solution invalid?

3. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
513
I haven't worked through your algebra, but algebra involving moduli is more tricky than ordinary algebra.

One equation becomes two or three as you have to consider three situations separately

Say you have to do some algebra with a modulus such as |x| then you must consider

x>0 then |x| = x
x<0 then |x| = -x
x=0 then |x| = 0

and put these separately into your equation and work it through three times.

Here is an example that actually a full professor of maths got wrong on another website!

post#6 there is correct.

4. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,297
482
That does not seem right.

x>0, then |x|=x
x<0, then |-x|=x
x=0, then |0|=0

5. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
513
Why not?

x<0, then |-x|=x

This is clearly wrong. if x = -15 what is |x|?

Did you read the worked example?

Last edited: Dec 14, 2013
6. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
Some of the links don't open for me. In any event I make the minimum open loop gain to be Aol=20,098. Does that tally?

Note:
The value was based on the assumption that the ratio R2/R1 was selected to produce a gain of exactly -100 at the aforementioned amplifier open loop gain of 20,098. The gain falls to -99.5 when the amplifier open loop gain falls to 10,049. If the ratio of R2/R1 is preset to a nominal value of 100 then the answer would be different & easier to deduce.

Last edited: Dec 15, 2013
7. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,297
482
Absolute value of negative fifteen is fifteen. In equation form I would write it the following way:
|-15|=15

8. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
513
Well yes, if the x is -15 then the modulus is +15.

So how can the modulus equal x, which is -15.


text:
J. Meteor. Res.  2018, Vol. 32 Issue (5): 829-838 PDF
http://dx.doi.org/10.1007/s13351-018-8024-7
The Chinese Meteorological Society
0

#### Article Information

DONG, Wanlin, Hang YU, Lizhen ZHANG, et al., 2018.
Asymmetric Ridge–Furrow and Film Cover Improves Plant Morphological Traits and Light Utilization in Rain-Fed Maize. 2018.
J. Meteor. Res., 32(5): 829-838
http://dx.doi.org/10.1007/s13351-018-8024-7

### Article History

in final form June 12, 2018
Asymmetric Ridge–Furrow and Film Cover Improves Plant Morphological Traits and Light Utilization in Rain-Fed Maize
Wanlin DONG1,2,3, Hang YU6, Lizhen ZHANG2,5, Ruonan WANG2,5, Qi WANG2,5, Qingwu XUE7, Zhihua PAN2,5, Zhigang SUN1,4, Xuebiao PAN2,5
1. Key Laboratory of Ecosystem Network Observation and Modeling, Institute of Geographic Sciences and Natural Resources Research, Chinese Academy of Sciences, Beijing 100101, China;
2. Agricultural Meteorological Department, College of Resources and Environmental Sciences, China Agricultural University, Beijing 100193, China;
3. China Meteorological Administration Training Centre, Beijing 100081, China;
4. College of Resources and Environment, University of Chinese Academy of Sciences, Beijing 100190, China;
5. Wuchuan Scientific and Observing Experimental Station of Agro-Environment, Ministry of Agriculture and Rural Affairs, Hohhot 011700, China;
6. Yushu Meteorological Service of Jilin Province, Changchun 130061, China;
7. Texas A&M AgriLife Research and Extension, Amarillo, Texas 79106-1769, USA
ABSTRACT: Light is one of the most important natural resources for plant growth. Light interception (LI) and use efficiency (LUE) are often affected by the structure of canopy caused by growing pattern and agronomy managements. Agronomy practices, such as the ridge–furrow system and plastic film cover, might affect the leaf morphology and then light transmission within the canopy, thus change light extinction coefficient (k), and LI and LUE. The objective of this study is to quantify LI and LUE in rain-fed maize (Zea Mays L.), a major cropping system in Northeast China, under different combinations of ridge–furrow and film covering ratios. The tested ridge–furrow system (DRF: " double ridges and furrows”) was asymmetric and alternated with wide ridge (0.70 m in width and 0.15 m in height), narrow furrow (0.10 m), narrow ridge (0.40 m in width and 0.20 m in height), and narrow furrow (0.10 m). Field experiments were conducted in 2013 and 2014 in Jilin Province, Northeast China. Four treatments were tested: no ridges and plastic film cover (control, NRF), ridges without film cover (DRF0), ridges with 58% film cover (DRF58), and ridges with 100% film cover (DRF100). DRF0 significantly increased LI by 9% compared with NRF, while film cover showed a marginal improvement. Specific leaf area in DRF experiments with film cover was significantly lower than in NRF, and leaf angle was 16% higher than in NRF, resulting in a 4% reduction in k. LUE of maize was not increased by DRF0, but was significantly enhanced by covering film in other DRF experiments, especially by 22% in DRF100. The increase of LUE by film cover was due to a greater biomass production and a lower assimilation portioning to vegetative organs, which caused a higher harvest index. The results could help farmers to optimize maize managements, especially in the region with decreased solar radiation under climate change.
Key words: light interception     light use efficiency     film mulching     plant morphology     ridge and furrow cultivation
1 Introduction

Maize (Zea Mays L.) is a major crop in Northeast China under rain-fed condition. Maize yield in this area is 13% higher than the national average (National statistics yearbook, 1980–2008). However, the yield gap is still large (Liu et al., 2012). The major limiting factors on maize growth and grain yield in Northeast China are frequent droughts in summer and low temperatures in spring. Applying ridge–furrow and plastic film cover has been a dominant strategy to alleviate the negative impact of climate variability (Liu Y. et al., 2010; Liu et al., 2014) and increase crop yield (Dong et al., 2017) in dry land agriculture (Zhou et al., 2009; Qin et al., 2014).

The ridge–furrow system combined with film cover is developed and used in North China recently, where the total precipitation and heat resource are not sufficient for growing high yield crops such as maize (Zhou et al., 2009; Eldoma et al., 2016) and potato (Hu et al., 2014). This system increases crop yield by collecting water (Gosar et al., 2010; Wang X. K. et al., 2015) and increasing topsoil temperature (Zhou et al., 2009; Li et al., 2013). Maize in Northeast China often suffers drought (Zhang et al., 2016; Cai et al., 2017) and chilling injury (Li et al., 2012; Fan et al., 2013), especially under extremes of climate change (Piao et al., 2010; Chen et al., 2012). The asymmetric ridge–furrow system, alternating with a wide ridge and a narrow furrow (Eldoma et al., 2016), or a narrow ridge and a narrow furrow, and often covered with plastic film on wide ridges, is helpful practice to offset the negative effects of climate change and further shrink the crop yield gap in Northeast China (Dong et al., 2017).

In a previous study (Dong et al., 2017), we found that maize yield could be increased by 33% in an asymmetric ridge–furrow system with full film cover. The effects of ridge and film on maize growth parameters were significantly different between years. When the water supply was sufficient, the positive effect of film cover and ridge was reduced on rain-fed condition crops in Northeast China. The ridge-furrow with film cover improves soil moisture and temperature (Zhou et al., 2009; Zhao et al., 2012) and therefore increases crop yield. However, we have limited data to answer the questions such as whe-ther this asymmetric ridge–furrow and film cover system could also improve light interception and light use efficiency and thus contribute to yield increase. As we know, higher above-ground biomass in traditional film cover and/or ridge and furrow system commonly increases crops’ light use efficiency (Jia et al., 2018).

The ridge–furrow system affects potential leaf expansion (Salah and Tardieu, 1997), leaf area index, and leaf morphology (e.g., angle and thickness) (Liu F. D. et al., 2010). The plant morphological plasticity can be affected by the spatial arrangements, such as row space (Mattera et al., 2013) and so on. Leaf morphological changes, i.e., specific leaf area (SLA) and leaf angle (Hirose and Werger, 1995; Valladares et al., 2002), would pose an impact on light extinction coefficient and its distribution within the canopy. Light interception (LI) of crop is determined by leaf area index and light extinction coefficient (k) (Monsi and Saeki, 1953). Biomass production depends on the interception of incoming photosynthetic active radiation (PAR) and the conversion efficiency into dry matter, i.e., the light use efficiency (LUE) (Sinclair and Horie, 1989; Willey, 1990; Idinoba et al., 2002). However, light distribution and use efficiency of crop by the ridge and film cultivation were seldom reported, especially in relation to plant morphological plasticity.

The slope of the regression between crop biomass and LI is well known as LUE (Dreccer et al., 2000; Yang et al., 2004), which is strongly affected by the biomass allocation to the organs, harvest index (Tadesse et al., 2001), water availability (Bajgain et al., 2015), and row spacing (Mattera et al., 2013). Ridges and film cover improved crop growth rate (Dong et al., 2017) and increased soil temperature and soil water content (Zhou et al., 2009) by reducing evaporation loss, collecting rainfall into the place of growing crop, and enhancing the harmony of crop growth and natural resources (rainfall and heat). We hypothesize that leaf morphology might be changed by the asymmetric ridges and film cover, which further impacts on LI and LUE.

The objectives of this study are to (i) quantify LI in relation to the plant morphological traits under the asymmetric ridge–furrow and film cover system, and (ii) quantify LUE of maize under different combinations of ridges/furrows and film cover ratios in response to biomass partitioning.

2 Material and methods 2.1 Experimental site

Field experiments were conducted in 2013 and 2014 at Shuangyang (43°33′N, 125°38′E; 219.5 m in altitude), Jilin Province, Northeast China. The soil was loam, with a pH of 5.6, a bulk density of 1.43 g cm–3, and a total N content of 1.15 g kg–1 on average of 0–100 cm soil depth. The annual mean air temperature was 5.8 ± 0.13°C and 620 ± 27 mm for precipitation from 1980 to 2010. The average non-frost period was 149 days. The average annual pan evaporation was 1408 mm. Total rainfall from April to September (the crop growing season) was 737 mm in 2013, much higher than that in 2014 (365 mm). Annual mean air temperature during the maize growing period was 17.4°C in 2013 and 17.9°C in 2014. The climate data (Table 1) were obtained from Shuangyang Meteorological Experimental Station, 2 km away from the experimental field.

Table 1 Monthly mean maximum (Tmax) and minimum (Tmin) air temperatures, total rainfall, and sunshine hours in 2013 and 2014 at Shuangyang, Jilin Province, China
 Month Tmax (°C) Tmin (°C) Rainfall (mm) Sunshine hours (h) 2013 2014 2013 2014 2013 2014 2013 2014 April 8.6 17.9 –0.5 10.6 48.7 5.7 205 255 May 24.1 20.8 11.6 15.0 38.8 96.6 265 192 June 26.6 27.4 16.3 21.8 159.1 80.0 225 237 July 27.5 28.1 19.1 22.8 288.3 105.1 211 236 August 26.9 28.2 18.3 21.8 142.1 32.6 206 251 September 22.7 22.6 9.3 14.8 59.6 44.9 243 243 Average/Total 22.7 24.2 12.4 17.8 736.6 364.9 1356 1414
2.2 Experiment design

An asymmetric ridge–furrow system (DRF, “double ridges and furrows”) in the experiments was alternated with a wide ridge (0.70 m in width, 0.15 m in height), a narrow furrow (0.10 m), a narrow ridge (0.40 m in width, 0.20 m in height), and a narrow furrow (0.10 m). Maize was planted in furrows (Fig. 1). Four treatments were tested in this study, including (1) no ridge and plastic film cover as a control (NRF); (2) DRF without plastic film cover (DRF0); (3) DRF with wide ridge covered by plastic film, i.e., 58% of the ground was covered by film (DRF58); and (4) DRF with plastic film covering full ground (DRF100). The experiments were replicated four times in a complete randomized block design. The size of the experiment plot was 6 m in width and 10 m in length. Plastic film was white and transparent with a thickness of 0.008 mm.

 Figure 1 Layout of the asymmetric ridge–furrow system and film cover. Black curve indicates the place where plastic film is covered.

The sowing dates were 8 May 2013 and 23 April 2014. The harvest dates were 26 September of both years. Maize cultivars were Fuyou 968 in 2013 and Tunyu98 in 2014, which were cultivars used by local farmers with similar genetic traits. The compound fertilizer (N 25%, P2O5 10%, and K2O 12%) of 500 kg ha–1 was applied once in the furrow at sowing time in 2013, and 450 kg ha–1 in 2014. There was no irrigation given in both years. Other managements were practiced according to local farmers.

2.3 Measurements 2.3.1 Yield and dry matter

To measure maize grain yield (g m–2), the ear density (number of ears per m2 ground area), kernel number per ear, and 1000-kernel weight were determined at harvesting time in a sampling area of 4.4 m wide × 1 m long (row) in each plot. The final maize grain yield was calculated as the product of yield components, i.e., ear density, kernel number per ear, and kernel weight in each plot, and standardized to 14% water content.

To measure maize dry matter (g m–2) and partitioning of dry matter (e.g., ratio of vegetative organs over total dry matter), three plants in a sub-sampling area were randomly sampled in each plot at each sampling dates: 5 times on 31 May, 7 July, 1 August, 27 August, and 26 September (harvesting) 2013 and 6 times on 28 May, 21 June, 15 July, 5 August, 29 August, and 26 September (harvesting) 2014. The sub-sampling area was 1 m2 each and at least 0.5 m away from the plot edges and previous sub-sampling area to avoid subsequent sampling effects. Samples were subdivided into leaf, stem, and reproductive organ, and then oven-dried at 105°C for 1 h, then at 80°C for 48 h to obtain a constant weight.

Harvest index (HI), the ratio of the grain yield to above-ground dry matter, was calculated by grain yield and final dry matter.

2.3.2 Leaf area index

Leaf area of maize was measured by the product of leaf length, maximal width, and an empirical shape factor 0.75 (McKee, 1964). The sampling dates and number of plants were the same as dry matter samplings. Leaf area index (LAI) was calculated by multiplying total leaf area per plant and plant density.

2.3.3 Morphological traits

Leaf angle as the insertion angle between stem and a leaf was measured on 15 July (flowering stage) and 29 August (grain filling stage) 2014 by using a transparent plastic protractor with an accuracy of 0.1°, when the leaves were fully expanded. Leaf thickness presented as specific leaf area (SLA; m2 g–1), the ratio of the area to the mass for an individual leaf, was measured on 5 August 2014. For both leaf angle and SLA, all leaves in three randomly selected plants in the sub-sampling area per plot were measured. The plant samples were the same as those used in measuring dry matter.

2.3.4 Light intensity within canopy

To measure the light intensity within the canopy, the PAR sensors were placed across the rows for each plot with a LI-191 Line Quantum Sensor (Li-Cor Co. Ltd, Lincoln, Nebraska, USA) within the canopy and a LI-250A Light Meter (Li-Cor) at the top of the canopy. The measurement was carried out during 1000 to 1400 local time on 5 August 2014 (a sunny day, canopy fully closed the field). Eight canopy layers were measured, i.e., 0 cm (soil surface), 100, 160, 190, 220, 250, 280, and 360 cm (top of the canopy).

2.4 Data analyses 2.4.1 Light interception and use efficiency

Daily incoming solar radiation was calculated by using Angström–Prescott formula (Almorox and Hontoria, 2004).

 $H = {H_0} \;\times\; \left( {a + b \;\times\; \frac{S}{{{S\!_{\rm{L}}}}}} \right),$ (1)

where H and H0 are daily solar radiation (MJ m–2 day–1) and extraterrestrial solar irradiance (MJ m–2 day–1), respectively; S is actual daily sunshine hours (h), SL is day length (the maximum possible sunshine hours, h); a and b are empirically determined regression constants, 0.25 for a and 0.50 for b (Almorox and Hontoria, 2004). Incoming PAR was calculated by an empirical equation (Zhu et al., 2010) as follows,

 ${\rm{PAR}} = \mu \;\times\; H,\quad\;\;\;$ (2)
 $\mu = c + d \;\times\; {\rm{ln}}\frac{H}{{{H_0}}},$ (3)

where μ is the photosynthetic effective coefficient; c and d are empirical coefficients, 0.3549 for c and –0.0491 for d in the situation of Northeast China (Zhu et al., 2010).

Daily light interception of crop canopy (LI; MJ m–2 day–1) for all plots in 2013 and 2014 was computed by Eq. (4). Total LI was the sum of daily LI during the growing period.

 ${\rm{LI}} = {\rm{PAR}} \;\times\; \left[ {1 - \exp \left( { - k \;\times\; {\rm{LAI}}} \right)} \right].$ (4)

Light use efficiency (g MJ PAR–1) for yield (LUEY) was calculated by final yield divided by total LI during the whole growing season. LUE for biomass (LUEDM; g MJ–1) was computed as the slope of the regression between dry matter and light interception during each crop growing period (Monteith, 1977).

2.4.2 Light extinction coefficient

Light extinction coefficient (k) was derived by an exponential relationship between light transmission within a canopy layer i (Ii/I0) and the leaf area index of this layer(LAIi) measured for each plot in August 2014 when crop fully covered the field,

 ${I_i} = {I_0} \;\times\; \exp \left( { - k\; \times \;{\rm{LA}}{{\rm{I}}_i}} \right),$ (5)

where I0 is the PAR that reaches the top of crop canopy while Ii is the PAR that reaches the ith canopy layer.

2.4.3 Statistical analyses

The effects of LUE, LI, leaf angle, LAI, and SLA were analyzed by General Linear Model Univariate in SPSS 17.0. The least significant difference (LSD) test was used to compare the differences in mean values at α = 0.05 between those treatments.

3 Results 3.1 Leaf area index

Film cover significantly (P < 0.05) increased LAI, across two years and two film treatments (DRF 58 and DRF100), by 90% at 60 days after sowing (DAS) comparedwith NRF. At the time of canopy closure (85 DAS), LAI in DRF100 was 7% higher than that in NRF and 16% higher than that in DRF0 across two years (Fig. 2). Ridge–furrow without film cover (DRF0) increased LAI at 60 DAS in 2013 but not in 2014 compared with NRF.

 Figure 2 Leaf area index (LAI) under ridge and film cover treatments in (a) 2013 and (b) 2014. Error bar indicates the standard error for replicates.
3.2 Morphological traits and light extinction coefficient

Leaf angle was significantly (P < 0.05) increased by 16% in DRF experiments compared with NRF. Film cover treatments did not affect leaf angle compared with DRF 0 (Fig. 3).

 Figure 3 Leaf angles of maize under ridge and film treatments at (a) flowering and (b) grain filling stages measured on 15 July and 29 August 2014, respectively. Leaf angle was averaged for all leaves. Error bar indicates the standard error for replicates. Same small letters (a or b) indicate no significant difference between treatments.

SLA within the 0–100 cm canopy layer (below ear position) in DRF100 was 0.01 m2 g–1, 46% lower than in NRF and 35% lower than in DRF0, but similar to that in DRF58 (Fig. 4). The effect of film cover on SLA reduced in the upper canopy layer (above 160 cm). Without film cover, ridge and furrow treatment (DRF0) also decreased SLA of maize.

 Figure 4 Specific leaf area (SLA) within maize canopy under ridge and film treatments at grain filling stage measured on 5 August 2014. Error bar indicates the standard error for replicates.

Light extinction coefficient (k) derived from exponential regression between I/I0 and LAI ranged from 0.42 to 0.52 in all treatments (Fig. 5). Compared with NRF, ridges (DRF0) increased k by 16%; however, film cover decreased k by 4%, as indicated by the two film treatments.

 Figure 5 Light extinction coefficient (k) derived from exponential regression between I/I0 and leaf area index (LAI) under ridge and film treatments in 2014. Error bar indicates the standard error for replicates.
3.3 Light interception and use efficiency

LI was significantly (P = 0.002) affected by DRF treatments (Table 2). LI in DRF0 was 9% higher than in NRF in 2013/14, but not further increased by film cover.

Table 2 Maize yield, final above-ground dry matters (DM), light interception (LI), light use efficiency for yield (LUEY), and harvest index (HI) as well as associated variance analysis results under ridge–furrow and film cover treatments in 2013 and 2014 in Jilin Province, China
 Year Pattern Yield DM LI LUEY HI t ha–1 t ha–1 MJ m–2 g MJ PAR–1 g g–1 2013 DRF100 13.9a 24.9a 677ab 2.06ab 0.56a DRF58 14.0a 24.8a 669b 2.09a 0.56a DRF0 12.5a 27.0a 702a 1.78b 0.47a NRF 11.7a 25.3a 655b 1.78b 0.46a s.e. 0.76 0.69 10.0 0.12 0.04 2014 DRF100 17.3a 29.7a 669a 2.59a 0.59a DRF58 14.0b 25.9b 683a 2.05b 0.54ab DRF0 12.8b 25.6b 692a 1.85b 0.50ab NRF 11.6b 25.4b 628b 1.86b 0.46b s.e. 0.96 0.96 14.6 0.15 0.04 Mean DRF100 15.6a 27.3a 673a 2.33a 0.58a DRF58 14.0ab 25.4a 676a 2.07ab 0.55ab DRF0 12.6bc 26.3a 697a 1.82b 0.49bc NRF 11.6c 25.3a 642b 1.82b 0.46c s.e. 0.64 0.82 8.8 0.10 0.03 P Ridge 0.092 0.128 0.002 0.928 0.399 Ridge × year 0.762 0.737 0.514 0.859 0.634 Film 0.023 0.177 0.092 0.010 0.123 Film × year 0.177 0.021 0.500 0.137 0.895 The P-value of ridge indicates the significance of an F-test comparing DRF0 and NRF. The P-value of film indicates the significance of an F-test comparing the three treatments with DRF. Same small letters (a, b, c, ab, etc.) indicate no significant difference between all four treatments. Data were given in mean values of replicates. The s.e. indicates a marginal standard error for all treatments with replicates within the same year.

LUEY was significantly (P = 0.01) increased by film cover. LUEY of DRF100 ranged from 2.06 to 2.59 g MJ PAR–1 in 2013 and 2014, 27% higher than that of NRF. LUEDM ranged from 3.69 to 4.95 g MJ–1 in all treatments (Fig. 6). Similar to LUEY, LUEDM of DRF0 was not different with that of NRF. LUEDM of DRF100 was 4.52 g MJ PAR–1 across two years and 17% higher than that of NRF. LUEDM of DRF100 was 8% higher than that of DRF58. The slight difference between LUEY and LUEDM was due to the changes in maize HI. The HI was slightly (not significantly) increased (5%) by ridge–furrow (DRF0), but significantly increased by 23% in film cover treatments (DRF58 and DRF100), compared with NRF (Table 2). The highest HI was observed in DRF100.

 Figure 6 The relationships between light interception and above-ground dry matter in (a) 2013 and (b) 2014. The slope of the linear regression is light use efficiency (LUE) for biomass. Error bar indicates the standard error for replicates.
3.4 Dry matter partitioning to leaves

Compared with NRF, the ridge–furrow with film co-ver (DRF58 and DRF100) reduced the ratio of vegetative organs over total above-ground dry matter by 6%, while DRF0 had no effect on this ratio (Figs. 7a, b). Film cover caused less biomass allocated into leaves. The ratio of leaves over total vegetative dry matter in DRF100 was 15% lower than that in NRF (Figs. 7c, d).

 Figure 7 (a, b) The ratio of vegetative organs to total above-ground dry matter and (c, d) the ratio of leaf dry matter to total vegetative organs, under ridge and film treatments in (a, c) 2013 to (b, d) 2014. Error bar indicates the standard error for replicates.
4 Discussion

Higher LAI and earlier time reaching its maximum under ridges and film treatments resulted in a higher LI (Mattera et al., 2013). This was caused by the improved growth and development of maize under ridges and film cover (Dong et al., 2017). The result was consistent with previous studies on maize (Flénet et al., 1996; Barbieri et al., 2000) and sunflower (Zaffaroni and Schneiter, 1989). However, k was higher in DRF0 than in NRF, which would cause a higher LI. But measured LI in DRF0 was lower than in NRF in our study. This might be explained by a greater LAI in DRF0 that has offset the negative effect caused by an increased k on LI. In addition, white film increased radiative reflection and would affect crop canopy microclimate environment (Oebker and Hopen, 1974; Ding et al., 2013), especially at the early stage (Fan et al., 2017). The lower k value in DRF100 and DRF58 implied a possibility to increase optimal plant density in this region.

By the structural plasticity of maize in ridge–furrow and film cover, characterized with thinner leaves (higher LAI) and lower k, maize intercepted more light in DRF than in NRF. Smaller SLA in ridges and film treatments was likely an adaptation for more efficient light interception (Niinemets et al., 2001) due to an increase of intra-specific competition. Konôpka et al. (2016) indicated that highest SLA values were found at lowest light availability under the canopy. Previous studies have shown that canopy architecture played an important role in increase of crop yield (Ku et al., 2016; Wang et al., 2017). The morphological traits such as leaf angle (Wang N. et al., 2015) and SLA were closely associated with LI. Our results found that leaf angle was increased by DRF. This suggested that the optimal plant density under ridge–furrow systems could be higher than flat cultivation. Plant morphological features (SLA and leaf angle) were often responsive to soil water conditions (French and Turner, 1991; White and Scott, 2006). Ridges with film cover increased soil moisture content (Wang et al., 2011; Gao et al., 2014) and affected plant architecture, thus improving photosynthetic capacity (Pierce et al., 1994), light utilization, and grain yield.

LUE was higher in DRF with film cover than in other treatments, which might also be due to the improved soil water availability by reducing soil evaporation, especially at early stage (Han et al., 2008; Liu F. D. et al., 2010). Film cover increased top soil temperature (Zhou et al., 2009) and accelerated the process of maize growth in the ridge with film cover (Dong et al., 2017), resulting in a higher LUE. The increase of LUEY (21% more relative to NRF) was greater than LUEDM (13%) in DRF58 and DRF100 when film cover was applied. This was mainly due to an increase of biomass allocation to reproductive organs, as indicated by an increasing HI in this study. The partitioning of biomass could be affected by water and nutrient availability (Vieira et al., 2004; Marcelis et al., 2006). The effect of film cover on biomass in the studied region was not significant due to the sufficient rainfall in experimental years. The film effect was stronger in a dry year (2014) than in a wet year (2013), especially for full film cover treatment. This result was coincidence with Zhang et al. (2018), who reported that the average yield increased 66% by using film cover in the area with rainfall less than 600 mm but only increased 20% in the area with precipitation greater than 600 mm. Using ridges in the rain-fed maize could harvest rainwater into furrows where plants grow, and crop yields increase by 15%, compared with no ridge control (Dong et al., 2017). Without plastic film, the ridge–furrow system in this study had no effect on LUE for both yield and biomass. Our results suggested that the effect of rainwater harvesting on LUE by ridge cultivation was probably much smaller compared with the effects of reducing soil evaporation and increasing soil temperature by film cover.

LUE for biomass in NRF (traditional practice without ridge and film) in our study was 3.84 g MJ PAR–1 across two years, which was consistent with previous studies ranging from 3.2 to 3.4 g MJ PAR–1 (Sinclair and Muchow, 1999) and 3.84 g MJ PAR–1 under optimal growing conditions (Lindquist et al., 2005). The high LUE in our study was due to the improved plant morphological traits and soil water availability (unpublished data), which promoted maize growth and development under ridge with film cover (Dong et al., 2017).

5 Conclusions

Light interception was significantly increased by asymmetric ridge and furrow system (DRF), but was not further increased by film cover. Light use efficiency was not increased by DRF0, but significantly enhanced by adding plastic film on DRF, especially with 100% film cover (DRF100). The increase of LI was due to the improvement of plant morphological traits (i.e., thin and erect leaves). The changes in leaf morphology also suggested a possibility to increase optimal plant density under ridge and film cover. The asymmetric ridge–furrow system combined with film cover might be a useful agronomy practice to alleviate the negative effect of climate change (e.g., drought and low temperature) globally. It is necessary to integrate our results at field level with crop models such as the Agricultural Production System sIMulator (APSIM); thus, better understanding on how to reduce climate risks and to alleviate the negative impacts of climate change through optimizing crop managements at the regional level could be achieved.

References


text:
# Chapter 13, Trigonometric Ratios and Functions - 13.4 Evaluate Inverse Trigonometric Functions - 13.4 Exercises - Skill Practice - Page 878: 16

$-0.64$ radians and $-36.87^{\circ}$

#### Work Step by Step

In order to get the answer in radians, we need to put the calculator in radians mode. Then, our result will be: $-0.64$ radians In order to get the answer in degrees, we need to put the calculator in degrees mode. Then, our result will be: $-36.87^{\circ}$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


text:
# [texhax] "Bug?" in a very stubborn table reference

Martin Heller mr_heller at yahoo.dk
Wed Jul 18 09:55:42 CEST 2007

Charla Mustard-Foote skrev:
> I have a \tref{} which seems to resolve correctly (at least it resolves
> to the expected table).  The table has the expected number.  However,
> the cross reference citation shows up as 3.1.2 instead of 3.1.  *Not*
> 3.1.1, which might be slightly logical.   It's like -- I don't know what
> it's like -- the princess of metaphors has none.
>
> Has anyone seen or fixed this?  It feels like one of those, "hardly ever
> happens and the fix has nothing obvious to do with the problem."
> situations.  I have gone through every table and figure in this 451 page
> book and everything else is fine.  It shows up correctly in the List of
> Tables as well.
>
> Any ideas?

Put \label *after* \caption. If this is not the problem you can
try pinpointing the problem by creating a minimal example code
that creates similar problems.
<http://www.tex.ac.uk/cgi-bin/texfaq2html?label=minxampl>


text:
Algebra 1: Common Core (15th Edition)

The answer is $(2\sqrt6-2\sqrt11)\div5$
We can simplify the expression as follows: $= -2\div(\sqrt6+\sqrt11)$ $= -2\div(\sqrt6+\sqrt11)*(\sqrt6-\sqrt11)\div(\sqrt6-\sqrt11)$ (It is always preferred to keep the denominator after simplifying the radicals, so multiply and divide by $(\sqrt6-\sqrt11)$) $= -2*(\sqrt6-\sqrt11)\div(\sqrt6)^{2}-(\sqrt11)^{2}$ (Since $(a+b)(a-b)=a^{2}-b^{2}$) $= -2\sqrt6+2\sqrt11\div(6-11)$ $= -2\sqrt6+2\sqrt11\div(-5)$ $= (2\sqrt6-2\sqrt11)\div5$ (Multiply and divide by -1)


text:
# Short Questions (Question 7 & 8)

Question 7:
Diagram below shows a circle with centre O and radius 12 cm.

Given that A, B and C are points such that OA = AB and  ∠OAC = 90°, find
(b)  the area, in cm2, of the shaded region.

Solution:
(a) For triangle OAC,
cos  ∠AOC = 6/12

(b)
= Area of BOC – Area of triangle AOC
½ (12)2 (1.047) – ½ (6) (12) sin 1.047 (change calculator to Rad mode)
= 75.38 – 31.17
= 44.21 cm2

Question 8:
Diagram below shows a sector QOR of a circle with centre O.

It is given that PS = 8 cm and QP = PO= OS = SR = 5 cm.
Find
(a) the length, in cm, of the arc QR,
(b) the area, in cm2, of the shaded region.

Solution:
(a) Length of arc QR = θ = 10 (1.75) = 17.5 cm

(b)
= Area of sector QOR – Area of triangle POS
½ (10)2 (1.75) – ½ (5) (5) sin 1.75 (change calculator to Rad mode)
= 87.5 – 12.30
= 75.2 cm2

# 9.7 Second-Order Differentiation, Turning Points, Maximum and Minimum Points (Examples)

Example 1 (Maximum Value of Quadratic Function)
Given that y = 3x (4 – x), calculate
(a) the value of x when y is a maximum,
(b) the maximum value of y.

Solution:
(a)

(b)

Example 2 (Determine the Turning Points and Second Derivative Test)
Find the coordinates of the turning points on the curve y = 2x3 + 3x2 – 12x + 7 and determine the nature of these turning points.

Solution:

6x2 + 6x – 12 = 0
x2 + x – 2 = 0
(x – 1) (x + 2) = 0
x = 1 or x = –2

When x = 1
y = 2(1)3 + 3(1)2 – 12(11) + 7
y = 0
(1, 0) is a turning point.

When x = –2
y = 2(–2)3 + 3(–2)2 – 12(–2) + 7
y = 27
(–2, 27) is a turning point.

Hence, the turning point (1, 0) is a minimum point.

Hence, the turning point (–2, 27) is a maximum point.

# SPM Practice Question 2

Question 2:
The third term and the sixth term of a geometric progression are 24 and $7\frac{1}{9}$ respectively. Find
(a) the first term and the common ratio,
(b) the sum of the first five terms,
(c) the sum of the first n terms with n is very big approaching rn ≈ 0.

Solution:
(a)

(b)

(c)

Therefore, sum of the first n terms with n is very big approaching rn ≈ 0 is 162.

# Short Questions (Question 7 & 8)

Question 7:
The masses of mangoes in a stall have a normal distribution with a mean of 200 g and a standard deviation of 30 g.
(a) Find the mass, in g, of a mango whose z-score is 0.5.
(b) If a mango is chosen at random, find the probability that the mango has a mass of at least 194 g.

Solution:
µ = 200 g
σ = 30 g
Let X be the mass of a mango.

(a)
$\begin{array}{l}\frac{X-200}{30}=0.5\\ X=0.5\left(30\right)+200\\ X=215g\end{array}$

(b)
$\begin{array}{l}P\left(X\ge 194\right)\\ =P\left(Z\ge \frac{194-200}{30}\right)\\ =P\left(Z\ge -0.2\right)\\ =1-P\left(Z>0.2\right)\\ =1-0.4207\\ =0.5793\end{array}$

Question 8:
Diagram below shows a standard normal distribution graph.

The probability represented by the area of the shaded region is 0.3238.
(a) Find the value of k.
(b) X is a continuous random variable which is normally distributed with a mean of 80 and variance of 9.
Find the value of X when the z-score is k.

Solution:
(a)
P(Z > k) = 0.5 – 0.3238
= 0.1762
k = 0.93

(b)
µ = 80,
σ2 = 9, σ = 3
$\begin{array}{l}\frac{X-80}{3}=0.93\\ X=3\left(0.93\right)+80\\ X=82.79\end{array}$

# Short Questions (Question 5 & 6)

Question 5:
Diagram below shows the graph of a binomial distribution of X.

(a) the value of h,
(b) P (X ≥ 3)

Solution:
(a)
P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) = 1
$\begin{array}{l}\frac{1}{16}+\frac{1}{4}+h+\frac{1}{4}+\frac{1}{16}=1\\ h=1-\frac{5}{8}\\ h=\frac{3}{8}\end{array}$

(b)
P (X ≥ 3) = P (X = 3) + P (X = 4)
$P\left(X\ge 3\right)=\frac{1}{4}+\frac{1}{16}=\frac{5}{16}$

Question 6:
The random variable X represents a binomial distribution with 10 trails and the probability of success is ¼.
(a) the standard deviation of the distribution,
(b) the probability that at least one trial is success.

Solution:
(a)
n = 10, p = ¼
$\begin{array}{l}\text{Standard deviation}=\sqrt{npq}\\ \text{}=\sqrt{10×\frac{1}{4}×\frac{3}{4}}\\ \text{}=1.875\end{array}$

(b)
$\begin{array}{l}P\left(X=r\right)=C{}_{r}{\left(\frac{1}{4}\right)}^{r}{\left(\frac{3}{4}\right)}^{10-r}\\ P\left(X\ge 1\right)\\ =1-P\left(X<1\right)\\ =1-P\left(X=0\right)\\ =1-C{}_{0}{\left(\frac{1}{4}\right)}^{0}{\left(\frac{3}{4}\right)}^{10}\\ =0.9437\end{array}$

# SPM Practice 2 (Question 11 & 12)

Question 11 (3 marks):
Diagram 6 shows the graph of a straight line

Diagram 11

Based on Diagram 6, express y in terms of x.

Solution:

Question 12 (3 marks):
The variables x and y are related by the equation $y=x+\frac{r}{{x}^{2}}$ , where r is a constant. Diagram 8 shows a straight line graph obtained by plotting

Diagram 12

Express h in terms of p and r.

Solution:

# SPM Practice 2 (Linear Law) – Question 1

Question 1 (10 marks):
Use a graph to answer this question.
Table 1 shows the values of two variables, x and y, obtained from an experiment. A straight line will be obtained when a graph of is plotted.

(a) Based on Table 1, construct a table for the values of

(c) Using the graph in 1(b)
(i) find the value of y when x = 2.7,
(ii) express y in terms of x.

Solution:
(a)

(b)

(c)(i)

(c)(ii)

# SPM Practice 3 (Linear Law) – Question 6

Question 6
The table below shows the values of two variables, x and y, obtained from an experiment. The variables x and y are related by the equation $\frac{a}{y}=\frac{b}{x}+1$ , where k and p are constants.

(a) Based on the table above, construct a table for the values of $\frac{1}{x}$ and $\frac{1}{y}$ . Plot $\frac{1}{y}$ against $\frac{1}{x}$ , using a scale of  2 cm to 0.1 unit on the $\frac{1}{x}$ - axis and  2 cm to 0.2 unit on the $\frac{1}{y}$ - axis. Hence, draw the line of best fit.
(b) Use the graph from  (b)  to find the value of
(i)  a,
(ii)  b.

Solution

Step 1 : Construct a table consisting X and Y.

Step 2 : Plot a graph of Y against X, using the scale given and draw a line of best fit

Step 3 : Calculate the gradient, m, and the Y-intercept, c, from the graph

Step 4 : Rewrite the original equation given and reduce it to linear form

Step 5 : Compare with the values of m and c obtained, find the values of the unknown required

# SPM Practice 3 (Linear Law) – Question 5

Question 5
The following table shows the corresponding values of two variables, x and y, that are related by the equation $y=p{k}^{\sqrt{x}}$ , where p and k are constants.

(a) Plot $log 10 y$ against $x$  .  Hence, draw the line of best fit

(b) Use your graph in (a) to find the values of p and k.

Solution
Step 1 : Construct a table consisting X and Y.

Step 2 : Plot a graph of Y against X, using the scale given and draw a line of best fit

Step 3 : Calculate the gradient, m, and the Y-intercept, c, from the graph

Step 4 : Rewrite the original equation given and reduce it to linear form

Step 5
:
Compare with the values of m and c obtained, find the values of the unknown required

# SPM Practice 2 (Linear Law) – Question 4

Question 4
The table below shows the corresponding values of two variables, x and y, that are related by the equation $y=qx+\frac{p}{qx}$ , where p and q are constants.

One of the values of y is incorrectly recorded.
(a) Using scale of 2 cm to 5 units on the both axis, plot the graph of xy against ${x}^{2}$  .  Hence, draw the line of best fit

(i) State the values of y which is incorrectly recorded and determine its actual value.
(ii) Find the value of p and of q.

Solution
Step 1 : Construct a table consisting X and Y.

Step 2 : Plot a graph of Y against X, using the scale given and draw a line of best fit


text:
# NX¶

Tag: Integration

This integration type block allows to apply optimization and other data mining techniques to models created in Siemens NX. The block maps model parameters to input and output ports, so they can be changed in run-time by receiving data from other blocks. On each change of input parameters, the block starts a new Siemens NX process to calculate characteristics which are then sent to block output ports.

## Compatibility¶

To run, the NX block requires Siemens NX installed. The block was tested with NX 10.0 and NX 11.0 on Windows 7 and Windows Server 2012 as well as with NX 11.0 on Linux (CentOS 7); other versions of Siemens NX are also compatible, but were not actively tested.

## Files Configuration¶

Document and export files configuration uses the generic file configuration component. Refer to the File block documentation when you need specific details.

### Document File¶

To select a NX document, click in the Document File pane. Note that some of the possible file configurations are invalid:

• The document cannot be a file of the project or sandbox origin with initial data.
• The document cannot be a file of the temporary origin.

The block supports Siemens NX part files (.prt) and ZIP archives (.zip). If the document references other files, use Send to compressed(zipped) folder in Windows Explorer context menu or any other archive tool to provide a .zip file with all dependencies of the model.

### Export File¶

To select an export file, click in the Export File pane. Note that the export file cannot contain initial data.

The block supports export to the native Siemens NX format (.prt), STEP (.step), IGES (.igs), and Parasolid (.x_t) formats. The document file can also be saved as a ZIP archive.

## Document Tree¶

After a document is selected, you can click on the Document Tree pane to build the parameter tree — get the list of dimensions, attributes, expressions and mass properties from the document.

Note that the NX block actually launches Siemens NX when you build the tree, so this configuration step requires Siemens NX installed.

In the tree you can select certain nodes to map them to input and output ports. When you select a node, a new corresponding variable appears in the Variables list. The Ports column there shows which ports (input, output or both) are created for the corresponding tree node.

### Structure¶

The NX block allows to extract dimensions from parts and assemblies (including contained subassemblies and parts), mass properties of any component in different configurations, expressions and part attributes. The document tree structure is as follows:

• dimensions
• part or assembly name
• feature name
• dimension name
• ... more features
• ... more parts and assemblies
• expressions
• part or assembly name
• expression name
• ... more parts and assemblies
• attributes
• part or assembly name
• attribute name
• ... more parts and assemblies
• mass properties
• part or assembly name
• mass
• volume
• area
• center of mass X
• center of mass Y
• center of mass Z
• moment of inertia XX
• moment of inertia XY
• moment of inertia YY
• moment of inertia YZ
• moment of inertia ZX
• moment of inertia ZZ
• ... more parts and assemblies

### Node Properties¶

Hovering a tree node shows a tooltip with node’s properties: type, value and read-only state.

Type

The majority of dimensions and mass properties are of the RealScalar type, but some dimensions are IntScalar. Expressions and attributes can be of the following type: RealScalar, IntScalar, StringScalar, BoolScalar. When you select a node in the document tree, it is mapped to a port of this type.

Value

Dimensional values that NX reads from and writes to ports are in units of the document units properties. It can be Millimeters or Inches. Units of expressions or attributes you can find out by opening the model in Siemens NX.

If you select a read-only node in the document tree, it is always mapped to an output port. Other, writeable nodes are mapped to input ports by default.

• mass properties,

Note that if you map a read-only node (for example, “mass”) to an input port, a validation warning will appear in the Issues pane:

"mass" is read-only but mapped to input port.

The block can also raise this error at the run-time.

## Options¶

Error handling behavior

The action to perform if the block encounters an error.

Value: stop workflow, output defaults and signal, output signal only stop workflow

When set to stop workflow, the block simply reports an error and the workflow interrupts.

If set to output defaults and signal, the block suppresses the error. In this case output ports send values assigned to them (the defaults), and the done port outputs False (the failure signal). Note that the block automatically assigns values to output ports that are created when you select document tree nodes. These values are:

The output signal only behavior means to output only the False value to done; nothing is output to other ports.

Export/STEP

STEP version number for exported documents.

Value: 203 or 214 203

Specifies the STEP version to use when exporting files to the STEP (.step) format.

NX operation timeout

Maximum time that the block can wait for a response from NX process.

Value: non-negative integer 300

The block launches NX process at each step during the workflow run as well as when you select the document file or build the document tree. Zero value means no limit. If another value is set, the block raises an error when the timeout is exceeded. This error is handled according to the Error handling behavior setting.

NX wait timeout

The time in seconds to wait for the NX application to become available in case it is currently used by another NX block.

Value: non-negative integer 300

Several NX blocks in a workflow can start simultaneously — this is typically the case when there is a parallelized Composite block containing an NX block. In this case, only one of the NX blocks can connect to the NX application at a time (this is a NX limitation). Other blocks have to wait until NX becomes available, and will do so until the timeout is exceeded.

Exceeding the timeout raises an error which always stops the workflow. That is, the timeout error is never suppressed regardless of the Error handling behavior setting. The workflow is not terminated immediately: all blocks that are currently running continue their execution, and the workflow stops after they finish (soft stop). Note that setting the timeout to 0 will always trigger workflow stop if an NX block with zero timeout starts while there is another NX block running. The running block will be allowed to finish execution in this case, and the workflow will be stopped by the timeout error from the second block.

Path to NX libraries

Absolute path to Siemens NX libraries.

Value: string empty

The path to the directory which contains libufun.dll on Windows or libufun.so on Linux. If this option is set to empty value, the NX block searches NX libraries using the path from environment variable UGII_BASE_DIR on Windows and LD_LIBRARY_PATH on Linux.

If Siemens NX is installed to the default folder on Windows, the path is:

NX 10: C:\Program Files\Siemens\NX 10.0\UGII

NX 11: C:\Program Files\Siemens\NX 11.0\NXBIN

Trace NX API calls

If enabled, the block will output all NX API calls to the debug log.

Value: True or False False

## Known Issues¶

This section describes certain issues related to using the NX block.

### General¶

• Sometimes error messages from the NX block contain truncated text. This text is in fact an error message received by pSeven from the NX application. In some cases Siemens NX limits the length of message text to 132 characters and discards the rest, so unfortunately the truncated text cannot be recovered. In such cases it is recommended to look up the error in the NX documentation by searching for the available part of the message.

### Windows¶

• Using non-English names of document parameters is not recommended.

The support for non-English parameter names in Siemens NX for Windows is known to depend on the format selected in regional settings. Siemens NX assumes that the document language follows the selected format; if they do not match, the NX block may be unable to build the document tree because some parameter names are incorrectly interpreted by Siemens NX. This case can be identified by an error message like this:

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### Linux¶

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If this error appears, you will have to rename parameters using only English characters in order to load the document correctly with the NX block.


text:
# Stokes theorem

1. Aug 9, 2005

### daftjaxx1

I'm trying to solve this problem:

Compute $$\oint_c(y+z)dx + (z-x)dy + (x-y)dz$$ using Stoke's theorem, where c is the ellipse $$x(t) = asin^2t, \ y(t) = 2asintcost, z(t) = acos^2t, 0\leq t \leq \pi$$

The version of stoke's theorem I learned is:
$$\int_c \overrightarrow{F} \cdot d\overrightarrow{r} = \int_s curl \overrightarrow{F} \cdot d\overrightarrow{S} =\iint_s curl \overrightarrow{F}\cdot \overrightarrow{n} \cdot dS$$

where S is the elliptical surface bounded by the curve c, F is a vector field and n is the unit vector pointing out at that point.

In this case, $$F = <y+z, z-x, x-y>$$, and I calculated curl F to be $$<-2, 0, -2>$$.

So we have to find

$$\iint_s <-2, 0, -2> \cdot \overrightarrow{n} \cdot dS$$

How would I find $$\overrightarrow {n}$$ and dS, and also the bounds of integration for the double integral?

2. Aug 9, 2005

### HallsofIvy

Staff Emeritus
Fortunately, since the curl is a constant vector, you don't have to do any integration! Just find $$curl \overrightarrow{F}\cdot \overrightarrow{n}$$ which will be a constant, of course, and multiply by the area of the ellipse.

I don't know that this is the easiest way but here's what I did off the top of my head: When t= 0, a point on the ellipse is (0, 0, a). When t= $$\frac{\pi}{4}$$, a point on the ellipse is (a/2, a, a/2). When t= $$\frac{\pi}{2}$$, a point on the ellipse is (a, 0, 0). Since the interior of the ellipse is a plane, the vectors from one of those points to the other 2 lie in that plane and so perpendicular to any normal to the plane: Take the cross product of the two vectors to find a perpendicular (and then, of course, divide by its length to get a unit perpendicular).
Take the dot product of that unit perpendicular with the curl you have already calculated and "integrate" that over the ellipse. Since, as I said before, the integrand is a constant, that is just the constant times the area of the ellipse. It should be easy to see, by looking at the points for t= 0, t= $$\frac{\pi}{4}$$, t= $$\frac{\pi}{2}$$ , and t= $$\frac{3\pi}{4}$$ that the ellipse has semi-axes of length a and a/2. You can calculate the area of the ellipse from that.


text:
First Law Thermodynamics Problem

1. Sep 4, 2013

cytochrome

When asked to calculate the change in kinetic energy of a system, do you calculate the kinetic of the entire system as if it was moving? Or do you calculate the kinetic energies of the various subsystems within the systems and then add it up so that you have the total kinetic energy?

Last edited: Sep 4, 2013
2. Sep 5, 2013

clamtrox

Usually kinetic energy is defined through the motion of the center of mass. Motion of subsystems with respect to the center is then classified to angular momentum, temperature, etc., which are not included in K.E.


text:
# Gaussian processes on discrete data structures¶

This example illustrates the use of Gaussian processes for regression and classification tasks on data that are not in fixed-length feature vector form. This is achieved through the use of kernel functions that operates directly on discrete structures such as variable-length sequences, trees, and graphs.

Specifically, here the input variables are some gene sequences stored as variable-length strings consisting of letters ‘A’, ‘T’, ‘C’, and ‘G’, while the output variables are floating point numbers and True/False labels in the regression and classification tasks, respectively.

A kernel between the gene sequences is defined using R-convolution 1 by integrating a binary letter-wise kernel over all pairs of letters among a pair of strings.

This example will generate three figures.

In the first figure, we visualize the value of the kernel, i.e. the similarity of the sequences, using a colormap. Brighter color here indicates higher similarity.

In the second figure, we show some regression result on a dataset of 6 sequences. Here we use the 1st, 2nd, 4th, and 5th sequences as the training set to make predictions on the 3rd and 6th sequences.

In the third figure, we demonstrate a classification model by training on 6 sequences and make predictions on another 5 sequences. The ground truth here is simply whether there is at least one ‘A’ in the sequence. Here the model makes four correct classifications and fails on one.

1

Haussler, D. (1999). Convolution kernels on discrete structures (Vol. 646). Technical report, Department of Computer Science, University of California at Santa Cruz.

Out:

/home/circleci/project/sklearn/gaussian_process/kernels.py:420: ConvergenceWarning: The optimal value found for dimension 0 of parameter baseline_similarity is close to the specified lower bound 1e-05. Decreasing the bound and calling fit again may find a better value.
warnings.warn(
/home/circleci/project/examples/gaussian_process/plot_gpr_on_structured_data.py:174: UserWarning: You passed a edgecolor/edgecolors ((0, 1.0, 0.3)) for an unfilled marker ('x').  Matplotlib is ignoring the edgecolor in favor of the facecolor.  This behavior may change in the future.
plt.scatter(

print(__doc__)

import numpy as np
import matplotlib.pyplot as plt
from sklearn.gaussian_process.kernels import Kernel, Hyperparameter
from sklearn.gaussian_process.kernels import GenericKernelMixin
from sklearn.gaussian_process import GaussianProcessRegressor
from sklearn.gaussian_process import GaussianProcessClassifier
from sklearn.base import clone

class SequenceKernel(GenericKernelMixin, Kernel):
"""
A minimal (but valid) convolutional kernel for sequences of variable
lengths."""

def __init__(self, baseline_similarity=0.5, baseline_similarity_bounds=(1e-5, 1)):
self.baseline_similarity = baseline_similarity
self.baseline_similarity_bounds = baseline_similarity_bounds

@property
def hyperparameter_baseline_similarity(self):
return Hyperparameter(
"baseline_similarity", "numeric", self.baseline_similarity_bounds
)

def _f(self, s1, s2):
"""
kernel value between a pair of sequences
"""
return sum(
[1.0 if c1 == c2 else self.baseline_similarity for c1 in s1 for c2 in s2]
)

def _g(self, s1, s2):
"""
kernel derivative between a pair of sequences
"""
return sum([0.0 if c1 == c2 else 1.0 for c1 in s1 for c2 in s2])

if Y is None:
Y = X

return (
np.array([[self._f(x, y) for y in Y] for x in X]),
np.array([[[self._g(x, y)] for y in Y] for x in X]),
)
else:
return np.array([[self._f(x, y) for y in Y] for x in X])

def diag(self, X):
return np.array([self._f(x, x) for x in X])

def is_stationary(self):
return False

def clone_with_theta(self, theta):
cloned = clone(self)
cloned.theta = theta
return cloned

kernel = SequenceKernel()

"""
Sequence similarity matrix under the kernel
===========================================
"""

X = np.array(["AGCT", "AGC", "AACT", "TAA", "AAA", "GAACA"])

K = kernel(X)
D = kernel.diag(X)

plt.figure(figsize=(8, 5))
plt.imshow(np.diag(D ** -0.5).dot(K).dot(np.diag(D ** -0.5)))
plt.xticks(np.arange(len(X)), X)
plt.yticks(np.arange(len(X)), X)
plt.title("Sequence similarity under the kernel")

"""
Regression
==========
"""

X = np.array(["AGCT", "AGC", "AACT", "TAA", "AAA", "GAACA"])
Y = np.array([1.0, 1.0, 2.0, 2.0, 3.0, 3.0])

training_idx = [0, 1, 3, 4]
gp = GaussianProcessRegressor(kernel=kernel)
gp.fit(X[training_idx], Y[training_idx])

plt.figure(figsize=(8, 5))
plt.bar(np.arange(len(X)), gp.predict(X), color="b", label="prediction")
plt.bar(training_idx, Y[training_idx], width=0.2, color="r", alpha=1, label="training")
plt.xticks(np.arange(len(X)), X)
plt.title("Regression on sequences")
plt.legend()

"""
Classification
==============
"""

X_train = np.array(["AGCT", "CGA", "TAAC", "TCG", "CTTT", "TGCT"])
# whether there are 'A's in the sequence
Y_train = np.array([True, True, True, False, False, False])

gp = GaussianProcessClassifier(kernel)
gp.fit(X_train, Y_train)

X_test = ["AAA", "ATAG", "CTC", "CT", "C"]
Y_test = [True, True, False, False, False]

plt.figure(figsize=(8, 5))
plt.scatter(
np.arange(len(X_train)),
[1.0 if c else -1.0 for c in Y_train],
s=100,
marker="o",
edgecolor="none",
facecolor=(1, 0.75, 0),
label="training",
)
plt.scatter(
len(X_train) + np.arange(len(X_test)),
[1.0 if c else -1.0 for c in Y_test],
s=100,
marker="o",
edgecolor="none",
facecolor="r",
label="truth",
)
plt.scatter(
len(X_train) + np.arange(len(X_test)),
[1.0 if c else -1.0 for c in gp.predict(X_test)],
s=100,
marker="x",
edgecolor=(0, 1.0, 0.3),
linewidth=2,
label="prediction",
)
plt.xticks(np.arange(len(X_train) + len(X_test)), np.concatenate((X_train, X_test)))
plt.yticks([-1, 1], [False, True])
plt.title("Classification on sequences")
plt.legend()

plt.show()

Total running time of the script: ( 0 minutes 0.264 seconds)

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