text:
find the length of the arc on a circle of radius r intercepted by a central angle θ

This time, you must solve for theta (the formula is s = rθ when dealing with radians): Plug in what you know to the radian formula.

The picture below illustrates the relationship between the radius, and the central angle in radians.

Radius r Central Angle θ 120o Need Help?Read It Find the length of the arc, s, on a circle of radius r intercepted by a central angle θ. If the radius of a circle is r then this is the hypotenuse of the right angled triangle so we can write the equation as: x 2 + y 2 = r 2. Lv 7. Express arc length in terms of pi. 7. Senior High School. Find the length of the arc on a circle of radius r intercepted by a central angle. Why is my calculator giving me a huge number for sin(3.140625)? We could also use the central angle formula as follows: In a complete circular pizza, we know that the central angles of all the slices will add up to 2π radians = 360°. To find the length of an arc with an angle measurement of 40 degrees if the circle has a radius of 10, use the following steps: Assign variable names to the values in the problem. An arc measure is an angle, the arc makes at the center of a circle, whereas the arc length is the span along the, arc. You can try the final calculation yourself by rearranging the formula as: L = θ * r. Then convert the central angle into radians: 90° = 1.57 rad, and solve the equation: L = 1.57 * … So if you need to find the length of an arc, you need to figure out what part of the whole …

Use integers -1,-2,-3,-4,-5,-6,-7,-8,-9 so that the sum along each side of the triangle formed is -20? pizza. The formula is $$S = r \theta$$ where s represents the arc length, $$S = r \theta$$ represents the central angle in radians and r is the length of the radius. That, curved piece of the circle and the interior space is called a sector, like a slice of. Do you want to solve for. nobodyinterestingnobodyinteresting.

This time, you must solve for theta (the formula is s = rθ when dealing with radians): Plug in what you know to the radian formula. Demonstration of the Formula S = r θ The interative demonstration below illustrates the relationship between the central angle of a circle, measured in radians, and the length of the intercepted arc. Where does the central angle formula come from? Also recall that a circle has 360 degrees. Since each slice has a central angle of 1 radian, we will need 2π / 1 = 2π slices, or 6.28 slices to fill up a complete circle. n Saturday. Then round your answer two decimal places. Bonus challenge - How far does the Earth travel in each season? If the circle is centred at the point (a,b), the equation of the circle is: (x - a) 2 + … A central angle is an angle with a vertex at the centre of a circle, whose arms extend to the circumference. Use the formula for S = r Θ and calculate the intercepted arc: 4Π. Radius, r = 6 inches; Central angle, theta = 175º ... Find the length of the arc, s, on a circle of radius r intercepted by a central angle theta. A.1 3           B. Real World Math Horror Stories from Real encounters. M= 1radian divided by 360 multiplied by 2pi(3) Simplify the problem by assuming the Earth's orbit is circular (.

Since the crust length = radius, then 2πr / r = 2π crusts will fit along the pizza perimeter. The length of an arc is directly proportional to the circumference of the circle and is dependent on both the central angle and the radius of the circle. with r representing the radius. Find the measure of the central angle of a circle in radians with an arc length of.

Also recall that a circle has 360 degrees. Use the formula for S = r Θ and calculate the solution. and a radius of 16. So if you need to find the length of an arc, you need to figure out what part of the whole circumference (or what fraction) you’re looking at. Answer to: Find the length of an arc intercepted by a central angle theta in a circle of radius r .

Even easier, this calculator can solve it for you. 1 4            C. 1 6           D.1 2? 2 Answers. Relevance. Then round your answer to two decimal places. Math. The Earth is approximately 149.6 million km away from the Sun. Question 292066: Find the length of the arc on a circle of radius r intercepted by a central angle. Since the problem defines L = r, and we know that 1 radian is defined as the central angle when L = r, we can see that the central angle is 1 radian. If the Earth travels about one quarter of its orbit each season, how many km does the Earth travel each season (e.g., from spring to summer)? This is the equation of a circle in standard form in Cartesian coordinates. …. The interative demonstration below illustrates the relationship between the central angle of a circle, measured in radians, and the length of the intercepted arc. There is a formula that relates the arc length of a circle of radius, r, to the central angle, $$\theta$$ in radians. find the second derivative at the point (1,3)? Express the arc length in terms of π.? radius, r=6 feet; Central angle, θ = 250° Your formula looks like this: Reduce the fraction. What patterns do you notice in 3, 8, 15, 24, 35, 48? Calculate the measure of the arc length S in the circle pictured below? Justify your answer. with r representing the radius. Find the measure of the central angle of a circle in radians with an arc length of . a. parking is not free on Sunday but the store is open o An arc can come from a central angle, which is one whose vertex is located at the center of the circle. Then round your… Find the length of the arc, s, on a circle of radius r intercepted by a central angle θ. (Round your answer to two decimal places.) Find the length of the arc, s, on a circle of radius r intercepted by a central angle θ. Get answers by asking now. Have you ever wondered how to find the central angle of a circle? Then round your answer to two decimal places. You can also use the arc length calculator to find the central angle or the circle's radius. The formula is S = r θ where s represents the arc length, S = r θ represents the central angle in radians and r is the length of the radius. Read It Submit AnswerSave Progress Practice Another Version -/1 points LarPCalc 10 4.1.052. When we cut up a circular pizza, the crust gets divided into arcs. Mary Jane Sterling aught algebra, business calculus, geometry, and finite mathematics at Bradley University in Peoria, Illinois for more than 30 years. The answers are  Use the central angle calculator to find arc length. The formula is $$S = r \theta$$ where s represents the arc length, $$S = r \theta$$ represents the central angle in radians and r is the length of the radius. Use the formula for S = r Θ and calculate the intercepted arc: 6Π.

The arc length for an angle measurement of 40 degrees. The central angle calculator is here to help; the only variables you need are the arc length and the radius.

The radius is 10, which is r. The figure shows what this arc looks like.

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Zhurnal Vychislitel'noi Matematiki i Matematicheskoi Fiziki
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 Zh. Vychisl. Mat. Mat. Fiz., 1997, Volume 37, Number 7, Pages 799–815 (Mi zvmmf2046)

Algorithms based on bilinear finite elements for iterative methods with split boundary conditions for a Stokes-type system in a strip under the periodicity condition

B. V. Pal'tsev, I. I. Chechel'

Moscow

Full text: PDF file (2194 kB)
References: PDF file   HTML file

English version:
Computational Mathematics and Mathematical Physics, 1997, 37:7, 775–791

Bibliographic databases:
UDC: 519.634
MSC: Primary 65N30; Secondary 76D07, 76M10
Revised: 02.08.1996

Citation: B. V. Pal'tsev, I. I. Chechel', “Algorithms based on bilinear finite elements for iterative methods with split boundary conditions for a Stokes-type system in a strip under the periodicity condition”, Zh. Vychisl. Mat. Mat. Fiz., 37:7 (1997), 799–815; Comput. Math. Math. Phys., 37:7 (1997), 775–791

Citation in format AMSBIB
\Bibitem{PalChe97} \by B.~V.~Pal'tsev, I.~I.~Chechel' \paper Algorithms based on bilinear finite elements for iterative methods with split boundary conditions for a Stokes-type system in a strip under the periodicity condition \jour Zh. Vychisl. Mat. Mat. Fiz. \yr 1997 \vol 37 \issue 7 \pages 799--815 \mathnet{http://mi.mathnet.ru/zvmmf2046} \mathscinet{http://www.ams.org/mathscinet-getitem?mr=1471694} \zmath{https://zbmath.org/?q=an:1058.65513} \transl \jour Comput. Math. Math. Phys. \yr 1997 \vol 37 \issue 7 \pages 775--791

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Citing articles on Google Scholar: Russian citations, English citations
Related articles on Google Scholar: Russian articles, English articles

This publication is cited in the following articles:
1. B. V. Pal'tsev, I. I. Chechel', “On some methods for enhancing the convergence speed for the higher harmonics of bilinear finite element implementations of iterative methods with boundary-condition splitting for a Stokes-type system”, Comput. Math. Math. Phys., 38:6 (1998), 916–929
2. B. V. Pal'tsev, I. I. Chechel', “Real properties of bilinear finite element implementations of methods with the splitting of boundary conditions for a Stokes-type system”, Comput. Math. Math. Phys., 38:2 (1998), 238–251
3. B. V. Pal'tsev, I. I. Chechel', “Bilinear finite element implementations of iterative methods with incomplete splitting of boundary conditions for a Stokes-type system on a rectangle”, Comput. Math. Math. Phys., 39:11 (1999), 1755–1780
4. N. A. Meller, B. V. Pal'tsev, E. G. Khlyupina, “On some finite element implementations of iterative methods with splitting of boundary conditions for Stokes and Stokes-type systems in a spherical layer: Axially symmetric case”, Comput. Math. Math. Phys., 39:1 (1999), 92–117
5. Belash V.O., Pal'tsev B.V., “On bicubic finite element implementations of iterative methods with splitting of boundary conditions for a singularly perturbed system of the Stokes type”, Russian J Numer Anal Math Modelling, 14:5 (1999), 397–401
6. A. S. Lozinskii, “On the acceleration of finite-element implementations of iterative processes with splitting of boundary conditions for a Stokes-type system”, Comput. Math. Math. Phys., 40:9 (2000), 1284–1307
7. V. O. Belash, B. V. Pal'tsev, “On the spectral and approximating properties of cubic finite-element approximations of the Laplace and first-derivative operators: The periodic case”, Comput. Math. Math. Phys., 40:5 (2000), 718–738
8. B. V. Pal'tsev, I. I. Chechel', “Exact estimates of the convergence rate of iterative methods with splitting of the boundary conditions for the Stokes-type system in a layer with a periodicity condition”, Comput. Math. Math. Phys., 40:12 (2000), 1751–1764
9. A. S. Lozinskii, “Finite-element realization of iterative processes with splitting of boundary conditions for a Stokes-type system in nonconcentric annuli”, Comput. Math. Math. Phys., 41:8 (2001), 1145–1157
10. Belash V.O., Pal'tsev B.V., Chechel I.I., “On convergence rate of some iterative methods for bilinear and bicubic finite element schemes for the dissipative Helmholtz equation with large values of a singular parameter”, Russian J Numer Anal Math Modelling, 17:6 (2002), 485–520
11. V. O. Belash, B. V. Pal'tsev, “Bicubic finite-element implementations of methods with splitting of boundary conditions for a Stokes-type system in a strip under the periodicity condition”, Comput. Math. Math. Phys., 42:2 (2002), 188–210
12. B. V. Pal'tsev, I. I. Chechel', “Increasing the rate of convergence of bilinear finite-element realizations of iterative methods by splitting boundary conditions for Stokes-type systems for large values of a singular parameter”, Comput. Math. Math. Phys., 44:11 (2004), 1949–1967
13. Pal'tsev B.V., Chechel I.I., “Finite-element linear second-order accurate (up to the poles) approximations of Laplace–Beltrami, gradient, and divergence operators on a sphere in R-3 in the axisymmetric case”, Doklady Mathematics, 69:2 (2004), 200–207
14. B. V. Pal'tsev, I. I. Chechel', “Second-order accurate (up to the axis of symmetry) finite-element implementations of iterative methods with splitting of boundary conditions for Stokes and stokes-type systems in a spherical layer”, Comput. Math. Math. Phys., 45:5 (2005), 816–857
15. Pal'tsev B.V., Stavtsev A.V., Chechel I.I., “Improved bicubic finite-element approximation of the Neumann problem for Poisson's equation”, Doklady Mathematics, 77:2 (2008), 258–264
16. M. K. Kerimov, “Boris Vasil'evich Pal'tsev (on the occasion of his seventieth birthday)”, Comput. Math. Math. Phys., 50:7 (2010), 1113–1119
17. M. B. Soloviev, “On numerical implementations of a new iterative method with boundary condition splitting for solving the nonstationary stokes problem in a strip with periodicity condition”, Comput. Math. Math. Phys., 50:10 (2010), 1682–1701
18. M. B. Soloviev, “Numerical implementations of an iterative method with boundary condition splitting as applied to the nonstationary stokes problem in the gap between coaxial cylinders”, Comput. Math. Math. Phys., 50:11 (2010), 1895–1913
19. Pal'tsev B.V., “On an Iterative Method with Boundary Condition Splitting as Applied to the Dirichlet Initial-Boundary Value Problem for the Stokes System”, Doklady Mathematics, 81:3 (2010), 452–457
20. Solov'ev M.B., “On Numerical Implementations of a New Iterative Method with Boundary Condition Splitting for the Nonstationary Stokes Problem”, Doklady Mathematics, 81:3 (2010), 471–475
21. B. V. Pal'tsev, M. B. Soloviev, I. I. Chechel', “On the development of iterative methods with boundary condition splitting for solving boundary and initial-boundary value problems for the linearized and nonlinear Navier–Stokes equations”, Comput. Math. Math. Phys., 51:1 (2011), 68–87
22. M. B. Solov'ev, “Numerical implementation of an iterative method with boundary condition splitting for solving the nonstationary stokes problem on the basis of an asymptotically stable two-stage difference scheme”, Comput. Math. Math. Phys., 54:12 (2014), 1817–1825
•  Number of views: This page: 134 Full text: 54 References: 27 First page: 1


text: Upon re-reading your original post I think you're confused about the use of SAGE_KEEP_BUILT_SPKGS=yes. Your confusion is understandable--it shouldn't be outputting that message. That's just a generic message for some make target failures, but it's not applicable to the Sage documentation build.


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# CBSE Class 10 - Science

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CBSE X - Set -3
SCIENCE PAPER 2012

Time allowed: 180 minutes; Maximum Marks: 90

 General Instructions: 1) All questions are compulsory. 2) The question paper consists of 42 questions divided into 4 sections A, B, C and D. Section A comprises questions of 01 mark each, Section B comprises questions of 02 marks each, Section C comprises questions of 03 marks each and Section D comprises questions of 05 marks each. 3) All questions in Section A are to be answered in one word, one sentence or as per the exact requirement of the question. 4) There is no overall choice. However, internal choice has been provided wherever necessary.You have to attempt only one of the alternatives in all such questions. 5) In question on construction, drawing should be near and exactly as per the given measurements. 6) Use of calculators is not permitted.

SECTION A

### Question 1

1. Write the name and formula of the first member of the carbon compounds having functional group − CHO.

### Question 2

2. State one role of ciliary muscles in the human eye.

### Question 3

3. Write the name and formula of a molecule made up of three atoms of oxygen.

### Question 4

4. Name two decomposers operating in our ecosystem.

### Question 27

27. The following figure illustrates binary fission in Amoeba in an incorrect sequence.

The correct sequence is

Options:
1. III, II, IV, I
2. III, IV, II, I
3. II, III, IV, I
4. IV, III, II, I

### Question 28

28. From the following diagrams, select the correct ones showing stages of binary fission in amoeba:

Options:
1. I, II, III
2. IV, II, III
3. V, II, III
4. IV, I, III

### Question 29

29. Following diagrams were drawn by different student on having seen prepared slides of budding in yeast.

Correct diagrams are

Options:
1. I, II, III
2. II, III, IV
3. III, IV, V
4. I, IV, V

### Question 30

30. In which of the following figures is budding not shown?

Options:
1. I
2. II
3. III
4. IV

### Question 31

31. A student weighed some raisins and recorded the weight as ‘x’. She then soaked the raisins in distilled water. After about 2 hours she removed the raisins, wiped them dry and weighed again and recorded that as ‘y’. The percentage of water absorbed by raisins may be determined using the relationship:

Options:
1. $\frac{y-x}{y}×100$
2. $\frac{y-x}{x}×100$
3. $\frac{y-x}{x}×\frac{1}{100}$
4. (y−x)×100

### Question 32

32. In the experiment for determining the percentage of water absorbed by raisins, we do the final weighing of the raisins after keeping them dipped in water for about one hour. For the accuracy of the result, the extra water from the surface of the soaked raisins is removed by

Options:
1. Rubbing with cotton cloth
2. Hot air blower
3. Dry cotton wool
4. Filter paper

### Question 33

33. On adding 2 mL acetic acid to 2 mL of water in a test tube, it was observed that

Options:
1. a clear and transparent solution is formed
2. a white precipitate is formed almost immediately
3. two separate layers were formed
4. a colourless and odourless gas is evolved

### Question 34

34. On adding acetic acid to sodium hydrogen carbonate in a test tube, a student observes

Options:
1. No reaction
2. A colourless gas with pungent smell
3. Bubbles of a colourless and odourless gas
4. A strong smell of vinegar

### Question 35

35. Which one of the following are the correct observations about acetic acid?

Options:
1. It turns blue litmus red and smells like vinegar
2. It turns blue litmus red and smells like burning sulphur
3. It turns red litmus blue and smells like vinegar
4. It turns red litmus blue and has a fruity smell

### Question 36

36. The aqueous solutions of copper sulphate and zinc sulphate appear

Options:
1. Blue and green respectively
2. Green and colourless respectively
3. Blue and brown respectively
4. Blue and colourless respectively.

### Question 37

37. Solutions of copper sulphate, iron sulphate and zinc sulphate are prepared and marked I, II and III respectively. Few pieces of aluminum are added to each solutions. After some time a change will be observe in.

Options:
1. I and II
2. II and III
3. III and I
4. All the three

### Question 38

38. Four students showed the following traces of the path of a ray of light passing through a rectangular glass slab.

The trace most likely to be correct is that of student

Options:
1. I
2. II
3. III
4. IV

### Question 39

39. While performing the experiment on tracing the path of a ray of light through a rectangular glass slab, in which of the following experimental set-ups is a student likely to get best results? P1 and P2 are the positions of pins fixed by him.

Options:
1. I
2. II
3. III
4. IV

### Question 40

40. If you are to determine the focal length of a convex lens, you should have

Options:
1. a convex lens and a screen
2. a convex lens and a lens holder
3. a lens holder, a screen holder and a scale
4. a convex lens, a screen, holder for them and a scale

### Question 41

41. A student obtained a sharp inverted image of a distant tree on a screen placed in front of the concave mirror. He then removed the screen and tried to look into the mirror. He would now see

Options:
1. a very blurred image on the wall opposite to the mirror
2. an erect and magnified image of the tree in the mirror
3. no image as the screen has been removed
4. a highly diminished inverted image of the tree at the focus of the mirror

### Question 42

42. A student has to determine the focal length of a concave mirror by obtaining the image of a distant object on a screen. For getting best result he should focus

Options:
1. a distant tree or an electric pole
2. a well-illuminated distant building
3. well-lit grills of the nearest window
4. a burning candle placed at the distant edge of the laboratory table

SECTION B

### Question 5

5. Why do all the elements of the (a) same group have similar properties, (b) same period have different properties?

### Question 6

6. An element E has following electronic configuration:

KLM

2 8 6

(a) To which group of the periodic table does element E belong?

(b) To which period of the periodic table does element E belong?

(c) State the number of valence electrons present in element E.

(d) State the valency of the element E.

### Question 7

7. Why is vegetative propagation practiced for growing some types of plants? List two plants which are grown by this method.

### Question 8

8. State the role of placenta in the development of embryo.

### Question 9

9. To construct ray diagram we use two light rays which are so chosen that it is easy to know their directions after reflection from the mirror. List these two rays and state the path of these rays after reflection. Use these rays to locate the image of an object placed between centre of curvature and focus of a concave mirror.

### Question 10

10. What is the colour of the clear sky during day time? Give reason for it.

### Question 11

11. Draw a labelled ray diagram to illustrate the dispersion of a narrow beam of white light when it passes through a glass prism.

### Question 12

12. List the products of combustion of fossil fuels. What are their adverse effects on the environment?

### Question 13

13. List three problems which arise due to construction of big dams. Suggest a solution for these problems.

### Question 24

24. What are hydrocarbons? Write the name and general formula of (i) saturated hydrocarbons, (ii) unsaturated hydrocarbons, and draw the structure of one hydrocarbon of each type. How can an unsaturated hydrocarbon be made saturated?

### Question 25

25. What are detergents chemically? List two merits and two demerits of using detergents for cleansing. State the reason for the suitability of detergents for washing, event in the case of water having calcium and magnesium ions.

SECTION C

### Question 14

14. State the meaning of inherited traits and acquired traits. Which of the two is not passed on to the next generation? Explain with the help of an example.

### Question 15

15. How are fossils formed? Describe in brief, two methods of determining the age of fossils.

### Question 16

16. If we pure-bred tall (dominant) pea plant with pure-bred dwarf (recessive) pea plant we will get pea plants of F1 generation. If we now self-cross the pea plant of F1 generation, then we obtain pea plants of F2 generation.

(a) What do the plants of F2 generation look like?

(b) State the ratio of tall plants to dwarf plants in F2 generation.

(c) State the type of plants not found in F1 generation but appeared in F2 generation, mentioning the reason for the same.

### Question 17

17. A 5 cm tall object is placed perpendicular to the principle axis of a convex lens of focal length 12 cm. The distance of the object from the lens is 8 cm. Using the lens formula, find the position, size and nature of the image formed.

### Question 18

18. State the types of mirrors used for (i) headlights and (ii) rear view mirror, in cars and motorcycles. Give reason to justify your answer in each case.

### Question 19

19. An old man cannot see objects closer than 1 m from the eye clearly. Name the defect of vision he is suffering from. How can it be corrected? Draw ray diagram for the (i) defect of vision and also (ii) for its correction.

### Question 20

20. Name the oxidizing agent used for the conversion of ethanol to ethanoic acid. Distinguish between ethanol and ethanoic acid on the basis of (i) litmus test, (ii) reaction with sodium hydrogen carbonate.

### Question 21

21. List and explain in brief three methods of contraception.

### Question 22

22. Na, Mg and Al are the elements having one, two and three valence electrons respectively. Which of these elements (i) has the largest atomic radius, (ii) is least reactive? Justify your answer stating reason for each.

SECTION D

### Question 23

23. List the new Cartesian sign convention for reflection of light by spherical mirror. Draw a diagram and apply these conventions for calculating the focal length and nature of spherical mirror which forms a 1/3 times magnified virtual image of an object placed 18 cm in front of it.

OR

With the help of a ray diagram, state what is meant by refraction of light. State Snell’s law for refraction of light and also express it mathematically.

The refractive index of air with respect to glass is 2/3 and the refractive index of water with respect to air is 4/3. If the speed of light in glass is 2 × 108 m/s, find the speed of light in (a) air, (b) water.

### Question 26

26. Distinguish between unisexual and bisexual flowers giving one example of each. Draw a diagram showing process of germination of pollen grains on stigma and label the following parts:

(a) Female germ cell

(b) Male germ cell

(c) Ovary

OR

Draw a diagram of human female reproductive system and label the part.

(i) That produces eggs.

(ii) Where fusion of egg and sperm take place.

(iii) Where zygote is implanted.

What happens to human egg when it is not fertilised?

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text:
# Alternate Proofs that $\inf A = - \sup\left(-A\right)$ from Rudin

I've been struggling a bit with this proof from Rudin. I've written two proofs, both of which seem to differ rather considerably from other proofs of this I've seen. I'm hoping someone can look these over. I'm going to assume as a lemma a theorem proved in Rudin which establishes the existence of the infimum.

Theorem. Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A = - \sup\left(-A\right)$.

Proof 1. By the least-upper-bound property, any nonempty subset of $\mathbb{R}$ that is bounded above has a least upper bound.

Lemma. (Proved in Rudin's text, and I'll take it as given for the purpose of this proof). Suppose $S$ is an ordered set with the least-upper-bound property, $B \subset S$, $B$ is not empty, and $B$ is bounded below. Let $L$ be the set of all lower bounds of $B$. Then $\alpha = \sup L$ exists in $S$, and $\alpha = \inf B$.

So, by the aformentioned lemma, any nonempty subset of $\mathbb{R}$ that is bounded below has a greatest lower bound. Since $A$ is bounded below and is nonempty, $\inf A$ exists, and has the property that \begin{align*} \forall x \in A, \forall y \in \mathbb{R}, \left(x \geq \inf A \wedge x \geq y \implies y \leq \inf A \right). \end{align*} Since $A$ is nonempty, there exists at least one $x \in A, x \geq \inf A$, a property that is true for all such $x$. This implies that $-x \leq - \inf A$, for all $-x \in -A$ and we conclude that, $- \inf A$ bounds $-A$ above. Assume that there exists some other upper bound, $-y$, of $-A$, so $-y \geq -x$ for all $-x \in -A$. Note that \begin{align*} -y \geq -x \implies y \leq x \implies y \leq \inf A \implies -y \leq - \inf A, \end{align*} So, $-y \leq - \inf A$, meaning that $- \inf A$ is also the greatest lower bound of $-A$, so $- \inf A = \sup\left(-A\right)$, which implies that $\inf A = - \sup\left(-A\right)$.

Remark. I worry about beginning with the infimum, rather than the supremum. Upon working through this proof more, it seems that the existence of the supremum follows directly from the least-upper-bound property, whereas the infimum does not. I worry I could be assuming the conclusion of the proof in positing its existence, which is something I don't do in this second proof. I would be very interested in any critiques and comments on whether what I did here is mathematically valid.

Proof 2. Since $A$ is nonempty, there exists at least one $x \in A \subset \mathbb{R}$. So, $-A \subset \mathbb{R} = \{-x: x \in A\}$. Since $A$ is bounded below, there exists some real number, $\beta$, such that $\forall x \in A, x \geq \beta$. This implies that $\forall - x \in -A, -x \leq - \beta$. Thus, $- \beta$ is an upper bound of $-A$, so by the least-upper-bound property, $\sup \left(-A\right)$ exists, and has the properties that \begin{align*} \forall y \in \mathbb{R}, \forall -x \in -A, \left(-x \leq \sup \left(-A\right) \wedge -x \leq y \implies y \geq \sup\left(-A\right) \right). \end{align*}

We now need to establish that $- \sup \left(-A\right)$ is the greatest lower bound of $A$. First, we show that it is a lower bound of $A$. Let $x \in A$. First, we have that for all $-x \in -A$, $-x \leq \sup\left(-A\right)$, which implies that $x \geq -\sup\left(-A\right)$, so $-\sup\left(-A\right)$ bounds $A$ below. Then, take some $-y \in \mathbb{R}$ that is also a lower bound of $A$, so $\forall x \in A, x \geq -y$. We then have that
\begin{align*} x \geq -y \implies -x \leq y \implies y \geq \sup \left(-A\right) \implies -y \leq - \sup\left(-A\right), \end{align*} which follows from the definition of the supremum of $-A$. So, any other lower bound, $-y$, of $A$ is no larger than $- \sup\left(-A\right)$, so $-\sup\left(-A\right)$ is both an upper bound of $A$ and the greatest such upper bound, so $\inf A = - \sup\left(-A\right)$.

Remark. This proof seems a bit smoother and draws on the same principles used earlier, but with the inequalities reversed. Hopefully the quantifiers are written correctly and I haven't overlooked anything.

So, my questions are:

(1) How does these proofs look? Have I made any mistakes or leaps of logic?

(2) Is one proof preferable to the other, e.g., am I wrong to draw on a lemma from the text to establish the greatest-lower-bound property, which is arguably what I'm trying to prove? Or, is this result from the text merely prove its existence, which is obviously needed to prove this result regardless?

1.

• Your first proof is not entirely correct. I follow you up to this point: $$y \leq \inf A \implies - y \leq - \inf A.$$ This should read: $$y \leq \inf A \implies -y \geq -\inf A.$$ (Note the change in direction of the inequality.) You should then conclude that $-\inf A$ is the least upper bound of $-A$, not the greatest lower bound, as you do. (I believe that this may have been a small mistake; you make the correct conclusion in the following sentences.)

• Your second proof looks correct to me.

After fixing the small mistakes, I have one point of advice: There is something to be said for brevity and clarity. For example, using English ("for all" versus $\forall$) over logical symbols is good style. It makes your proofs easier to parse for everyone.

As a concrete example, the way you state the properties of $\sup (-A)$ in your second proof is hard to digest. It would perhaps be easier to simply state that $\sup(-A)$ exists, and then quote its properties (with English!) as they arise. (Of course, when you are being extremely careful and making sure that your logic is correct, logical quantifiers are a wonderful thing to use! This is especially if a course requires it of you.)

Here is an example proof that is condensed into one paragraph, and uses hardly any symbols:

Since $\inf A$ is a lower-bound for $A$, it is clear that $- \inf A \geq -a$ for all $-a \in -A$. Thus $-\inf A$ is an upper bound of $-A$. If $x \in \mathbb{R}$ is such that $x < -\inf A$, then $-x > \inf A$. By definition of the greatest lower bound, this implies some $a \in A$ with $-x > a \geq \inf A$. But then $x < -a$, and $-a \in -A$. Thus $x$ is not an upper bound for $-A$, and so it follows that $\sup(-A) = -\inf A$.

1. I don't think that either proof is "preferable" to the other. In particular, I don't think that your "big picture" reasoning is ever circular, so both proofs are valid. They are merely different ways to go about proving the theorem.

I happen to like the idea of the first one, but that's just because I think Rudin's lemma is cool.

• +1 for the advice about using English. Some textbooks make the subject of analysis boring and difficult by using excessive symbolism. – Paramanand Singh Sep 3 '18 at 4:11


text:
# Given $$f\left( z \right) = \frac{{{z^2}}}{{{z^2} \;+ \;{a^2}}}$$. Then

1. z = ia is a simple pole and $$\frac{ia}{2}$$ is a residue at z = ia of f(z)
2. z = ia is a simple pole ia is a residue at z = ia of f(z)
3. z = ia is a simple pole and $$-\frac{ia}{2}$$ is a residue at z = ia of f(z)
4. none of the above

Option 1 : z = ia is a simple pole and $$\frac{ia}{2}$$ is a residue at z = ia of f(z)

## Residue Theorem MCQ Question 1 Detailed Solution

Concept:

Pole:

The value for which f(z) fails to exists i.e. the value at which the denominator of the function f(z) = 0.

When the order of a pole is 1, it is known as a simple pole.

Residue:

If f(z) has a simple pole at z = a, then

$$Res\;f(a)=\mathop {\lim }\limits_{z \to a} \left( {z - a} \right)f\left( z \right)$$

If f(z) has a pole of order n at z = a, then

$$Res\left( {at\;z = a} \right) = \frac{1}{{\left( {n - 1} \right)!}}{\left\{ {\frac{{{d^{n - 1}}}}{{d{z^{n - 1}}}}\left[ {{{\left( {z - a} \right)}^n}f\left( z \right)} \right]} \right\}_{z = a}}$$

Calculate:

Given:

$$f\left( z \right) = \frac{{{z^2}}}{{{z^2} \;+ \;{a^2}}}$$

For calculating pole:

z2 + a2 = 0

∴ (z + ia)(z - ai) = 0

∴ z = ai, -ai.

∴ z has simple pole at z = ai and -ai.

Residue:

If f(z) has a simple pole at z = a, then

$$Res\;f(a)=\mathop {\lim }\limits_{z \to a} \left( {z - a} \right)f\left( z \right)$$

For pole at z = ai

$$Res\;f(ai)=\mathop {\lim }\limits_{z \to ai} \left( {z - ai} \right)\left ( \frac{z^2}{z^2\;+\;a^2} \right )$$

$$Res\;f(ai)=\mathop {\lim }\limits_{z \to ai} \left( {z - ai} \right)\left ( \frac{z^2}{(z-ai)(z+ai)} \right )$$

$$Res\;f(ai)=\frac{(ai)^2}{2ai}\Rightarrow\frac{ai}{2}$$

For pole at z = -ai

$$Res\;f(-ai)=\mathop {\lim }\limits_{z \to -ai} \left( {z + ai} \right)\left ( \frac{z^2}{(z-ai)(z+ai)} \right )$$

$$Res\;f(-ai)=\frac{(-ai)^2}{-2ai}\Rightarrow\frac{ai}{2}$$

z has a simple pole at z = ai and $$\frac{ia}{2}$$ is a residue at z = ia of f(z)

# Residue at z = 2 of $$f\left( z \right) = \frac{{2z + 1}}{{{z^2} - z - 2}}$$ is

1. $$\frac{{5}}{{3}}$$
2. $$\frac{{1}}{{3}}$$
3. $$\frac{{3}}{{5}}$$
4. $$\frac{{2}}{{3}}$$

Option 1 : $$\frac{{5}}{{3}}$$

## Residue Theorem MCQ Question 2 Detailed Solution

Concept:

Pole – a point on which functional value is infinite.

If z = a is a simple pole of f(z) then the residue of f(z) at z = a is given by,
$$Re{s_{z\; =\; a}}f\left( z \right) = \mathop {\lim }\limits_{z \to a} \left( {z - a} \right)f\left( z \right)$$

Calculation:

Given:

$$f\left( z \right) = \frac{{2z + 1}}{{{z^2} - z - 2}}$$

We have to find residue at z = 2.

$$f\left( z \right) = \frac{{2z + 1}}{{{z^2} - z - 2}}\;=\;\frac{2z \;+\;1{}}{{(z\;-\;2)(z\;+\;1)}}$$

Value of f(z) is infinite at x = 2. hence z = 2 is a simple pole.

Residue of f(z) at z = 2,

$$Re{s_{z\;=\; 2}}f\left( z \right) = \mathop {\lim }\limits_{z \to 2} \left( {z - 2} \right) \times \frac{{2z + 1}}{{\left( {z - 2} \right) \times \left( {z + 1} \right)}}$$

$$= \frac{{2 \times 2 + 1}}{{2 + 1}} = \frac{5}{3}$$

# The value of integral $$\mathop \smallint \nolimits_C \frac{{\cos \pi {z^2}}}{{\left( {z - 1} \right)\left( {z - 2} \right)}}dz$$ where C : |z| = 3 (Positively oriented) is equal to

1. 2πi
2. 0
3. -4πi
4. 4πi

Option 4 : 4πi

## Residue Theorem MCQ Question 3 Detailed Solution

Concept:

Cauchy’s Theorem:

If f(z) is an analytic function and f’(z) is continuous at each point within and on a closed curve C, then

$$\mathop \oint \limits_C f\left( z \right)dz = 0$$

Cauchy’s Integral Formula:

If f(z) is an analytic function within a closed curve and if a is any point within C, then

$$f\left( a \right) = \frac{1}{{2\pi i}}\mathop \oint \limits_C \frac{{f\left( z \right)}}{{z - a}}dz$$

$${f^n}\left( a \right) = \frac{{n!}}{{2\pi i}}\mathop \oint \limits_C \frac{{f\left( z \right)}}{{{{\left( {z - a} \right)}^{n + 1}}}}dz$$

Residue Theorem:

If f(z) is analytic in a closed curve C except at a finite number of singular points within C, then

$$\mathop \smallint \limits_C f\left( z \right)dz = 2\pi i \times \left[ {{\rm{sum\;of\;residues\;at\;the\;singualr\;points\;within\;C}}} \right]$$

Formula to find residue:

1. If f(z) has a simple pole at z = a, then

$$Res\;f\left( a \right) = \mathop {\lim }\limits_{z \to a} \left[ {\left( {z - a} \right)f\left( z \right)} \right]$$

2. If f(z) has a pole of order n at z = a, then

$$Res\;f\left( a \right) = \frac{1}{{\left( {n - 1} \right)!}}{\left\{ {\frac{{{d^{n - 1}}}}{{d{z^{n - 1}}}}\left[ {{{\left( {z - a} \right)}^n}f\left( z \right)} \right]} \right\}_{z = a}}$$

Application:

$$\mathop \smallint \nolimits_C \frac{{\cos \pi {z^2}}}{{\left( {z - 1} \right)\left( {z - 2} \right)}}dz$$

Simple poles: z = 1, 2

Both the poles lie within the given region.

Residue at z = 1,

$$\mathop {\lim }\limits_{z \to 1} \frac{{\cos \pi {z^2}}}{{\left( {z - 1} \right)\left( {z - 2} \right)}} \cdot \left( {z - 1} \right) = \frac{{\cos \pi }}{{\left( { - 1} \right)}} = 1$$

Residue at z = 2

$$\mathop {\lim }\limits_{z \to 2} \frac{{\cos \pi {z^2}}}{{\left( {z - 1} \right)\left( {z - 2} \right)}}\left( {z - 2} \right) = \cos 4\pi = 1$$

Value of integral = 2πi [1 + 1] = 4πi

# An integral I over a counterclockwise circle C is given by$$I = \mathop \oint \limits_C^\; \frac{{{z^2} - 1}}{{{z^2} + 1}}{e^z}dz.$$ If C is defined as |z| = 3, then the value of I is

1. -πi sin (1)
2. -2πi sin (1)
3. -3πi sin (1)
4. -4πi sin (1)

Option 4 : -4πi sin (1)

## Residue Theorem MCQ Question 4 Detailed Solution

Concept:

For a given complex function with poles, the complex integral $$I = \mathop \oint \limits_C f\left( z \right)dz$$ is given by

Residue theorem as;

$$I = \mathop \oint \limits_C f\left( z \right)dz = 2\pi \;i \times \left\{ {Sum\;of\;residue\;of\;poles\;inside\;or\;on\;C} \right\}$$

Calculation:

Given C, |Z|= 3

|x + iy| = 3

$$\sqrt {{x^2} + {y^2}} = 3$$

x2 + y2 = 9

$$f\left( z \right) = \left( {\frac{{{z^2} - 1}}{{{z^2} + 1}}} \right){e^z}$$

Poles are defined by z2 + 1 = 0

(z + i)(z - i) = 0

z = i, -i (simple poles lying inside of closed path)

Residue of f(z) at z = i $$\Rightarrow \mathop {\lim }\limits_{z \to i} \left( {z - i} \right).\frac{{{z^2} - 1}}{{\left( {z + i} \right)\left( {z - i} \right)}}{e^z}$$

$$\Rightarrow \frac{{\left( {{i^2} - 1} \right){e^i}}}{{2i}} \to - \frac{{2{e^i}}}{{xi}} = i{e^i}$$

Residue of f(z) at z = -i $$\Rightarrow \mathop {\lim }\limits_{z \to - i} \left( {z + i} \right).\frac{{{z^2} - 1}}{{\left( {z + i} \right)\left( {z - i} \right)}}{e^z}$$

$$\Rightarrow \frac{{\left( {i - 1} \right)}}{{ - 2i}}.{e^{ - i}} \to - i{e^{ - i}}$$

Now, using Cauchy’s Residue Theorem,

$$I = \mathop \oint \limits_c \frac{{{z^2} - 1}}{{{z^2} + 1}}{e^z}dz = 2\pi i\;\left( {Sum\;of\;residue} \right)$$

= 2πi (iei – ie-i)

= (2πi) (i) (ei – e-i)

= 2πi2 (2i sin (1))

⇒ -4πi sin (1)

# Integration of the complex function $$f\left( z \right) = \frac{{{z^2}}}{{{z^2} - 1}}$$ in the counterclockwise direction, around |z – 1| = 1, is

1. -πi
2. 0
3. πi
4. 2πi

Option 3 : πi

## Residue Theorem MCQ Question 5 Detailed Solution

Concept:

Cauchy’s Theorem:

If f(z) is an analytic function and f’(z) is continuous at each point within and on a closed curve C, then

$$\mathop \oint \limits_C f\left( z \right)dz = 0$$

Cauchy’s Integral Formula:

If f(z) is an analytic function within a closed curve and if a is any point within C, then

$$f\left( a \right) = \frac{1}{{2\pi i}}\mathop \oint \limits_C \frac{{f\left( z \right)}}{{z - a}}dz$$

$${f^n}\left( a \right) = \frac{{n!}}{{2\pi i}}\mathop \oint \limits_C \frac{{f\left( z \right)}}{{{{\left( {z - a} \right)}^{n + 1}}}}dz$$

Residue Theorem:

If f(z) is analytic in a closed curve C except at a finite number of singular points within C, then

$$\mathop \smallint \limits_C f\left( z \right)dz = 2\pi i \times \left[ {{\rm{sum\;of\;residues\;at\;the\;singualr\;points\;within\;C}}} \right]$$

Formula to find residue:

1. If f(z) has a simple pole at z = a, then

$$Res\;f\left( a \right) = \mathop {\lim }\limits_{z \to a} \left[ {\left( {z - a} \right)f\left( z \right)} \right]$$

2. If f(z) has a pole of order n at z = a, then

$$Res\;f\left( a \right) = \frac{1}{{\left( {n - 1} \right)!}}{\left\{ {\frac{{{d^{n - 1}}}}{{d{z^{n - 1}}}}\left[ {{{\left( {z - a} \right)}^n}f\left( z \right)} \right]} \right\}_{z = a}}$$

Application:

Given function is $$f\left( z \right) = \frac{{{z^2}}}{{{z^2} - 1}}$$

Poles: z = 1, -1

|z – 1| = 1

⇒ |x – 1 + iy| = 1

$$\Rightarrow \sqrt {{{\left( {x - 1} \right)}^2} + {y^2}} = 1$$

The given region is a circle with the centre at (1, 0) and the radius is 1.

Only pole z = 1, lies within the given region.

Residue at z = 1 is, $$\mathop {\lim }\limits_{z \to 1} \frac{{{z^2}}}{{\left( {z + 1} \right)}} = 0.5$$

The value of the integral = 2πi × 0.5 = πi

# The residue of $$f(z)=\dfrac{z^3}{(z-1)^4(z-2)(z-3)}$$ at z = 3 is

1. -8
2. $$\dfrac{101}{16}$$
3. 0
4. $$\dfrac{27}{16}$$

Option 4 : $$\dfrac{27}{16}$$

## Residue Theorem MCQ Question 6 Detailed Solution

Concept:

Residue Theorem:

If f(z) is analytic in a closed curve C except at a finite number of singular points within C, then

$$\mathop \smallint \limits_C f\left( z \right)dz = 2\pi i \times \left[ {{\rm{sum\;of\;residues\;at\;the\;singualr\;points\;within\;C}}} \right]$$

Formula to find residue:

1. If f(z) has a simple pole at z = a, then

$$Res\;f\left( a \right) = \mathop {\lim }\limits_{z \to a} \left[ {\left( {z - a} \right)f\left( z \right)} \right]$$

2. If f(z) has a pole of order n at z = a, then

$$Res\;f\left( a \right) = \frac{1}{{\left( {n - 1} \right)!}}{\left\{ {\frac{{{d^{n - 1}}}}{{d{z^{n - 1}}}}\left[ {{{\left( {z - a} \right)}^n}f\left( z \right)} \right]} \right\}_{z = a}}$$

Calculation:

z = 3 is a simple pole of f(z)

The residue at z = 3 is,

$$= \mathop {\lim }\limits_{z \to 3} \left( {z - 3} \right)\frac{{{z^3}}}{{{{(z - 1)}^4}\left( {z - 2} \right)\left( {z - 3} \right)}}$$

$$= \frac{{{3^3}}}{{{{\left( {3 - 1} \right)}^4}\left( {3 - 2} \right)}} = \frac{{27}}{{16}}$$

# The residues of a complex function $${\rm{X}}\left( {\rm{z}} \right) = \frac{{1 - 2{\rm{z}}}}{{{\rm{z}}\left( {{\rm{z}} - 1} \right)\left( {{\rm{z}} - 2} \right)}}{\rm{\;}}$$at its poles are

1. $$\frac{1}{2},{\rm{\;}} - \frac{1}{2}{\rm{\;and\;}}1$$
2. $$\frac{1}{2}, - \frac{1}{2}{\rm{\;and}} - 1$$
3. $$\frac{1}{2},{\rm{\;}}1{\rm{\;and}} - \frac{3}{2}$$
4. $$\frac{1}{2},{\rm{\;}} - 1{\rm{\;and}}\frac{3}{2}$$

Option 3 : $$\frac{1}{2},{\rm{\;}}1{\rm{\;and}} - \frac{3}{2}$$

## Residue Theorem MCQ Question 7 Detailed Solution

Concept:

Residue Theorem:

If f(z) is analytic in a closed curve C except at a finite number of singular points within C, then

$$\mathop \smallint \limits_C f\left( z \right)dz = 2\pi i \times \left[ {{\rm{sum\;of\;residues\;at\;the\;singualr\;points\;within\;C}}} \right]$$

Formula to find residue:

1. If f(z) has a simple pole at z = a, then

$$Res\;f\left( a \right) = \mathop {\lim }\limits_{z \to a} \left[ {\left( {z - a} \right)f\left( z \right)} \right]$$

2. If f(z) has a pole of order n at z = a, then

$$Res\;f\left( a \right) = \frac{1}{{\left( {n - 1} \right)!}}{\left\{ {\frac{{{d^{n - 1}}}}{{d{z^{n - 1}}}}\left[ {{{\left( {z - a} \right)}^n}f\left( z \right)} \right]} \right\}_{z = a}}$$

Calculation:

Given:

$${\rm{X}}\left( {\rm{z}} \right) = \frac{{1\;-\;2{\rm{z}}}}{{{\rm{z}}\left( {{\rm{z}}\;-\;1} \right)\left( {{\rm{z}}\;-\;2} \right)}}$$

Poles are simple and located at z = 0, z = 1, and z = 2

$$Res\;f\left( a \right) = \mathop {\lim }\limits_{z \to a} \left[ {\left( {z - a} \right)f\left( z \right)} \right]$$

At z = 0, the residue is:

$${{\rm{Res}\;f(0)}} = \mathop {\lim }\limits_{z \to 0} \left ( z-0 \right )\left \{ \frac{1-2z}{z(z-1)(z-2)} \right \}$$

$$Res\;f(0) = \frac{{1\;-\;2 \times 0}}{{\left( {0\;-\;1} \right)\left( {0\;-\;2} \right)}} = \frac{1}{2}$$

At z = 1, the residue is:

$${{\rm{Res}\;f(1)}} = \mathop {\lim }\limits_{z \to 1} \left ( z\;-\;1 \right )\left \{ \frac{1-2z}{z(z-1)(z-2)} \right \}$$

$$Res\;f(1) = \frac{{1\;-\;2 \times 1}}{{1\left( {1\;-\;2} \right)}} = 1$$

At z = 2, the residue is:

$${{\rm{Res}\;f(2)}} = \mathop {\lim }\limits_{z \to 2} \left ( z\;-\;2 \right )\left \{ \frac{1-2z}{z(z-1)(z-2)} \right \}$$

$$Res\;f(2)= \frac{{1\;-\;2 \times 2}}{{2\left( {2\;-\;1} \right)}} = - \frac{3}{2}$$

# If f(z) has a pole of order n at z = a, then residue of function f(z) at a is

1. $$\dfrac{1}{(n)!} \left\lbrace \dfrac{d^{n-1}}{dz^{n-1}}\left((z-a)^{n-1}f(z)\right)\right\rbrace_{z=a}$$
2. $$\dfrac{1}{(n-1)!} \left\lbrace \dfrac{d^{n-1}}{dz^{n-1}}\left((z-a)^{n-1}f(z)\right)\right\rbrace_{z=a}$$
3. $$\dfrac{1}{(n)!} \left\lbrace \dfrac{d^{n-1}}{dz^{n-1}}\left((z-a)^{n}f(z)\right)\right\rbrace_{z=a}$$
4. $$\dfrac{1}{(n-1)!} \left\lbrace \dfrac{d^{n-1}}{dz^{n-1}}\left((z-a)^{n}f(z)\right)\right\rbrace_{z=a}$$

Option 4 : $$\dfrac{1}{(n-1)!} \left\lbrace \dfrac{d^{n-1}}{dz^{n-1}}\left((z-a)^{n}f(z)\right)\right\rbrace_{z=a}$$

## Residue Theorem MCQ Question 8 Detailed Solution

Residue Theorem:

If f(z) is analytic in a closed curve C except at a finite number of singular points within C, then

$$\mathop \smallint \limits_C f\left( z \right)dz = 2\pi i \times \left[ {{\rm{sum\;of\;residues\;at\;the\;singualr\;points\;within\;C}}} \right]$$

Formula to find residue:

1. If f(z) has a simple pole at z = a, then

$$Res\;f\left( a \right) = \mathop {\lim }\limits_{z \to a} \left[ {\left( {z - a} \right)f\left( z \right)} \right]$$

2. If f(z) has a pole of order n at z = a, then

$$Res\;f\left( a \right) = \frac{1}{{\left( {n - 1} \right)!}}{\left\{ {\frac{{{d^{n - 1}}}}{{d{z^{n - 1}}}}\left[ {{{\left( {z - a} \right)}^n}f\left( z \right)} \right]} \right\}_{z = a}}$$

Important Points:

Cauchy’s Theorem:

If f(z) is an analytic function and f’(z) is continuous at each point within and on a closed curve C, then

$$\mathop \oint \limits_C f\left( z \right)dz = 0$$

Cauchy’s Integral Formula:

If f(z) is an analytic function within a closed curve and if a is any point within C, then

$$f\left( a \right) = \frac{1}{{2\pi i}}\mathop \oint \limits_C \frac{{f\left( z \right)}}{{z - a}}dz$$

$${f^n}\left( a \right) = \frac{{n!}}{{2\pi i}}\mathop \oint \limits_C \frac{{f\left( z \right)}}{{{{\left( {z - a} \right)}^{n + 1}}}}dz$$

# The contour C given below is on the complex plane z = x + jy, where $${\rm{j}} = \sqrt { - 1}$$The value of the integral $$\frac{1}{{{\rm{\pi j}}}}\mathop \oint \limits_{\rm{c}} \frac{{{\rm{dz}}}}{{{{\rm{z}}^2} - 1}}$$ is _______

## Residue Theorem MCQ Question 9 Detailed Solution

Given integral: $$\frac{1}{{\pi j}}\oint \frac{{dz}}{{{Z^2} - 1}}$$

From residue theorem:

$$\smallint f\left( z \right)dz = 2\pi i\;\left[ {sum\;of\;residues} \right]$$

Residue at z = 1

$$\begin{array}{*{20}{c}} {lt}\\ {z \to 1} \end{array}\;\left( {z - 1} \right)\frac{1}{{\left( {z - 1} \right)\left( {z + 1} \right)}} = \frac{1}{2}$$

Residue at z = -1 (Here the contour is traveled in clockwise sense, hence extra Negative sign will come)

$$\begin{array}{*{20}{c}} {lt}\\ {z \to -1} \end{array}\;\left( {z + 1} \right)\frac{-1}{{\left( {z + 1} \right)\left( {z - 1} \right)}} = \frac{1}{2}$$

Sum of residues $$= \frac{1}{2} + \frac{1}{2} = 1$$

Hence: $$\frac{1}{{\pi j}}\smallint \frac{{dz}}{{{z^2} - 1}} = 2\left( 1 \right) = 2$$

# Evaluate $$\underset{C}{\mathop \int }\,\frac{dz}{z\sin z}$$, where C is x2 + y2 = 1.

1. 1
2. 2
3. 0
4. -1

Option 3 : 0

## Residue Theorem MCQ Question 10 Detailed Solution

Concept:

$$\mathop \smallint \nolimits_C f\left( z \right)dz = 2\pi i\left( {{\rm{sum\;of\;residues}}} \right)$$

Where (residue at z = 0) = Coefficient of  in laurent’s expansion of f(z)

Calculation:

We have given $$\mathop \smallint \nolimits_C \frac{{dz}}{{z\;sin\;z}}$$; C is x2 + y2 = 1 (i.e. |z| = 1)

Now, $$\frac{1}{{z\sin z}} = \frac{1}{{z\left[ {z - \frac{{{z^2}}}{{3!}} + \frac{{{z^5}}}{{5!}} - \frac{{{z^7}}}{{7!}} + \ldots } \right]}} = \frac{1}{{\left( {{z^2}} \right)\left[ {1 - \frac{{{z^2}}}{{3!}} + \frac{{{z^4}}}{{5!}} + \frac{{{z^6}}}{{7!}} \ldots } \right]}}$$

$$= \frac{1}{{{z^2}}}\;{\left[ {1 - \left[ {\frac{{{z^2}}}{6} - \frac{{{z^4}}}{{120}} + \frac{{{z^6}}}{{7!}}} \right]} \right]^{ - 1}}$$

$$= \frac{1}{{{z^2}}}\left\{ {1 + \left( {\frac{{{z^2}}}{6} - \frac{{{z^4}}}{{120}}} \right) + {{\left( {\frac{{{z^2}}}{6} - \frac{{{z^4}}}{{120}}} \right)}^2} + \ldots } \right\}$$

{since (1 - x)-1 = 1 + x + x2 + x3 + … if x ≤ 1}

$$\frac{1}{{{\rm{z}}\sin z{\rm{\;}}}} = \frac{1}{{{z^2}}} + \frac{1}{6} + \frac{7}{{360}}{z^2} + \ldots$$

Above equation shows that if has z = 0 pole of order 2 & residue at the pole = 0.

Coefficient of $$\frac{1}{z} = 0$$

Sum of Residue

$$\mathop \smallint \nolimits_C \frac{1}{{z\sin z}} = 2\pi \;\left( {residue} \right) = 0$$

# The residues of a function $$f\left( z \right) = \frac{1}{{\left( {z - 4} \right){{\left( {z + 1} \right)}^3}}}$$ are:

1. $$- \frac{1}{{27}}\;and - \frac{1}{{125}}$$
2. $$\frac{1}{{125}}\;and - \frac{1}{{125}}$$
3. $$- \frac{1}{{27}}\;and\frac{1}{5}$$
4. $$\frac{1}{{125}}\;and - \frac{1}{5}$$

Option 2 : $$\frac{1}{{125}}\;and - \frac{1}{{125}}$$

## Residue Theorem MCQ Question 11 Detailed Solution

Concept:

For simple poles at z = a, b, c…

Residue of $${\{ f\left( z \right)\} _{z = a}} = \mathop {\lim }\limits_{z \to a} \left\{ {\left( {z - a} \right) \times f\left( z \right)} \right\}$$

For multiple poles at z = a, a, a … n times

{Residue of f(z)}z=a $$= \frac{1}{{\left( {n - 1} \right)!}}{\left\{ {\frac{{{d^{n - 1}}}}{{d_z^{n - 1}{{\left( {z - a} \right)}^n}}} - f\left( z \right)} \right\}_{z = a}}$$

Calculation:

Given, $$f\left( z \right) = \frac{1}{{\left( {z - 4} \right){{\left( {z + 1} \right)}^3}}}$$

For a simple pole at z = 4

Residue of {f(z)}z=4 $$= \mathop {\lim }\limits_{z \to 4} \left\{ {\left( {z - 4} \right).\frac{1}{{\left( {z - 4} \right){{\left( {z + 1} \right)}^3}}}} \right\}$$

$$= \mathop {\lim }\limits_{z \to 4} \frac {1}{\left( {z + 1} \right)^3} \Rightarrow \frac{1}{{{5^3}}} = \frac{1}{{125}}$$

For multiple pole (n = 3) at z = -1

Residue will be

$$\Rightarrow \frac{1}{{\left( {3 - 1} \right)!}}\left\{ {\frac{{{d^2}}}{{d{z^2}}}{{\left( {z + 1} \right)}^3}.\frac{1}{{\left( {z - 4} \right){{\left( {z + 1} \right)}^3}}}} \right\}$$

$$\Rightarrow {z_{ \to - 1}}\frac{{1\left( { - 1} \right)\left( { - 2} \right)}}{2}\frac{1}{{{{\left( {z - 4} \right)}^3}}}$$

$$\Rightarrow {z_{ \to - 1}}\frac{1}{{{{\left( {z - 4} \right)}^3}}} \Rightarrow \frac{1}{{{{\left( { - 5} \right)}^3}}} = - \frac{1}{{125}}$$

# The residue of the function $${\rm{f}}\left( {\rm{z}} \right) = \frac{1}{{{{\left( {{\rm{z}} + 2} \right)}^2}{{\left( {{\rm{z}} - 2} \right)}^2}}}{\rm{\;at\;z}} = 2{\rm{\;is}}$$

1. $$- \frac{1}{{32}}$$
2. $$- \frac{1}{{16}}$$
3. $$\frac{1}{{16}}$$
4. $$\frac{1}{{32}}$$

Option 1 : $$- \frac{1}{{32}}$$

## Residue Theorem MCQ Question 12 Detailed Solution

$${\rm{Res\;f}}{\left( {\rm{z}} \right)_{{\rm{z}} = {\rm{a}}}} = \frac{1}{{\left( {{\rm{n}} - 1} \right)!}}\frac{{{{\rm{d}}^{{\rm{n}} - 1}}}}{{{\rm{d}}{{\rm{z}}^{{\rm{n}} - 1}}}}{\left( {{\rm{z}} - {\rm{a}}} \right)^{\rm{n}}}{\left. {{\rm{f}}\left( {\rm{z}} \right)} \right]_{{\rm{z}} = {\rm{a}}}}$$

Here we have n = 2 and a = 2

Thus Res $${\rm{f}}{\left( {\rm{z}} \right)_{{\rm{z}} = 2}} = \frac{1}{{\left( {2 - 1} \right)!}}\frac{{\rm{d}}}{{{\rm{dz}}}}{\left[ {{{\left( {{\rm{z}} - 2} \right)}^2}\frac{1}{{{{\left( {{\rm{z}} - 2} \right)}^2}{{\left( {{\rm{z}} + 2} \right)}^2}}}} \right]_{{\rm{z}} = 2}}$$

$$\begin{array}{l} = \frac{{\rm{d}}}{{{\rm{dz}}}}{\left[ {\frac{1}{{{{\left( {{\rm{z}} + 2} \right)}^2}}}} \right]_{{\rm{z}} = 2}} = {\left[ { - \frac{2}{{{{\left( {{\rm{z}} + 2} \right)}^3}}}} \right]_{{\rm{z}} = 2}}\\ = - \frac{2}{{64}} = - \frac{1}{{32}} \end{array}$$

# Let (-1 - j), (3 - j), (3 + j) and (-1 + j) be the vertices of a rectangle C in the complex plane. Assuming that C is traversed in counter-clockwise direction, the value of the countour integral $$\displaystyle\oint_C \dfrac{dz}{z^2 (z-4)}$$ is

1. 0
2. jπ/16
3. jπ/2
4. -jπ/8

Option 4 : -jπ/8

## Residue Theorem MCQ Question 13 Detailed Solution

Concept:

Residue Theorem:

If f(z) is analytic in a closed curve C except at a finite number of singular points within C, then

cf(z) dz = 2πj × [sum of residues at the singular points within C]

Formula to find residue:

1. If f(z) has a simple pole at z = a, then

$$Res\;f\left( a \right) = \mathop {\lim }\limits_{z \to a} \left[ {\left( {z - a} \right)f\left( z \right)} \right]$$

2. If f(z) has a pole of order n at z = a, then

$$Res\;f\left( a \right) = \frac{1}{{\left( {n - 1} \right)!}}{\left\{ {\frac{{{d^{n - 1}}}}{{d{z^{n - 1}}}}\left[ {{{\left( {z - a} \right)}^n}f\left( z \right)} \right]} \right\}_{z = a}}$$

Application:

Given (-1 - j), (3 - j), (3 + j) and (-1 + j) are the vertices of a rectangle C in the complex plane

f(z) from the given data is,

$$f(z)=\dfrac{1}{z^2 (z-4)}$$

Poleas of f(z) is

z = 0 of order n = 2, lies in side the closed curve.

z = 4 of order n = 1, lies outside the closed curve.

∴ $$\displaystyle\oint_C \dfrac{dz}{z^2 (z-4)}=2π j\;Res\;f\left( 0 \right)$$

⇒ $$Res\;f\left( 0 \right) = \frac{1}{{\left( {2 - 1} \right)!}}{\left\{ {\frac{{{d^{2 - 1}}}}{{d{z^{2 - 1}}}}\left[ {{{\left( {z } \right)}^2}\dfrac{1}{z^2 (z-4)}} \right]} \right\}_{z = 0}}$$

$$Res\;f(0)=\dfrac{d}{dz}\left[\dfrac{1}{z-4}\right]_{z=0}=-\dfrac{1}{16}$$

$$\displaystyle\oint_C \dfrac{dz}{z^2 (z-4)}=2π j\left(\dfrac{-1}{16}\right)=\dfrac{-jπ }{8}$$

Cauchy’s Theorem:

If f(z) is an analytic function and f’(z) is continuous at each point within and on a closed curve C, then

$$\mathop \oint \limits_C f\left( z \right)dz = 0$$

Cauchy’s Integral Formula:

If f(z) is an analytic function within a closed curve and if a is any point within C, then

$$f\left( a \right) = \frac{1}{{2\pi i}}\mathop \oint \limits_C \frac{{f\left( z \right)}}{{z - a}}dz$$

$${f^n}\left( a \right) = \frac{{n!}}{{2\pi i}}\mathop \oint \limits_C \frac{{f\left( z \right)}}{{{{\left( {z - a} \right)}^{n + 1}}}}dz$$

# The value of $$\oint \frac{1}{{{z^2}}}dz,$$ where the contour is the unit circle traversed clockwise, is

1. -2πi
2. 0
3. 2πi
4. 4πi

Option 2 : 0

## Residue Theorem MCQ Question 14 Detailed Solution

$$\oint \frac{1}{{{z^2}}}dx$$

By Cauchy’s residue theorem,

$$\oint f\left( z \right)dz = 2\pi i\;\left[ {sum\;of\;residues} \right]$$

$$f\left( z \right) = \frac{1}{{{z^2}}}$$

Poles are, Z = 0

Residue at Z = a, is given by

$$\mathop {\lim }\limits_{z \to a} \frac{1}{{\left( {n - 1} \right)!}}\frac{{{d^{n - 1}}{{\left( {z - a} \right)}^n}}}{{d{z^{n - 1}}}}f\left( z \right)dz$$

Where n is the order pole.

Here, z = 0 is second order pole.

Residue at z = 0,

$$\frac{{Lim}}{{z \to a}}\;\frac{1}{{\left( {2 - 1} \right)!}}\;\frac{d}{{dz}}{\left( z \right)^2}\frac{1}{{{Z^2}}}\; = 0$$

# If C is a circle |z| = 4 and $$f\left( z \right) = \frac{{{z^2}}}{{{{\left( {{z^2} - 3z + 2} \right)}^2}}}$$, then $$\oint f\left( z \right)dz$$ is

1. 1
2. 0
3. -1
4. -2

Option 2 : 0

## Residue Theorem MCQ Question 15 Detailed Solution

$$f\left( z \right) = \frac{{{z^2}}}{{{{\left( {{z^2} - 3z + 2} \right)}^2}}}$$

$$= \frac{{{z^2}}}{{{{\left( {z - 1} \right)}^2}{{\left( {z - 2} \right)}^2}}}$$

Poles of f(z) are z = 1, 2

Both Z = 1 and 2 are inside the curve |z| = 4

$$\mathop \smallint \limits_C f\left( z \right)dz = 2\pi i$$ [sum of residues at z = 1 and z = 2]

If f has a pole of order n at z = a, then Residue of f(z) at z = a is

$$= \frac{1}{{\left( {n - 1} \right)!}}\frac{{{d^{n - 1}}}}{{d{z^{n - 1}}}}{\left[ {{{\left( {z - a} \right)}^n}f\left( z \right)} \right]_{z = a}}$$

Residue at z = 1,

$$\mathop {{\rm{It}}}\limits_{z \to 1} \frac{d}{{dz}}\left[ {\frac{{{z^2}.\;{{\left( {z - 1} \right)}^2}}}{{{{\left( {z - 1} \right)}^2}{{\left( {z - 2} \right)}^2}}}} \right]$$

$$= \frac{{{{\left( {1 - 2} \right)}^2}\left( 2 \right) - {{\left( 1 \right)}^2}\left( {2\left( { - 1} \right)} \right)}}{{{{\left( {1 - 2} \right)}^4}}} = 4$$

Residue at z = 2,

$$= \mathop {{\rm{lt}}}\limits_{z \to 2} \frac{d}{{dz}}\left[ {\frac{{{z^2}}}{{{{\left( {z - 1} \right)}^2}}}} \right]$$

$$= \;\mathop {{\rm{lt}}}\limits_{z \to 2} \left[ {\frac{{{{\left( {z - 1} \right)}^2}\left( {2z} \right) - {z^2}\left( {2\left( {z - 1} \right)} \right)}}{{{{\left( {z - 1} \right)}^4}}}} \right]$$

$$= \frac{{{{\left( {2 - 1} \right)}^2} - 4\left( {2\left( 1 \right)} \right)}}{{{{\left( {2 - 1} \right)}^4}}} = - 4\;$$

$$\mathop \smallint \limits_C f\left( z \right)dz = 2\pi i\left( {4 - 4} \right) = 0$$

# What is the residue of the function $$\frac{{1 - {e^{2z}}}}{{{z^4}}}$$ at its pole?

1. $$\frac{4}{3}$$
2. $$\frac{{ - 4}}{3}$$
3. $$\frac{{ - 2}}{3}$$
4. $$\frac{2}{3}$$

Option 2 : $$\frac{{ - 4}}{3}$$

## Residue Theorem MCQ Question 16 Detailed Solution

$$f\left( z \right) = \frac{{1 - {e^{2z}}}}{{{z^4}}}$$

The function has multiple pole at z = 0

If z = a is a pole of order ‘m’, then residue of f(z) at z = a is,

$$\frac{1}{{\left( {m - 1} \right)!}}\begin{array}{*{20}{c}} {lim}\\ {z \to a} \end{array}\frac{{{d^{m - 1}}}}{{d{z^{m - 1}}}}\left[ {{{\left( {z - a} \right)}^m}f\left( z \right)} \right]$$

Residue at z = 0 is

$$= \frac{1}{{\left( {4 - 1} \right)!}}\begin{array}{*{20}{c}} {lim}\\ {z \to 0} \end{array}\frac{{{d^3}}}{{d{z^3}}}\left( {{z^4}\frac{{\left( {1 - {e^{2z}}} \right)}}{{{z^4}}}} \right)$$

$$= \frac{1}{6}\begin{array}{*{20}{c}} {lim}\\ {z \to 0} \end{array}\frac{{{d^3}}}{{d{z^3}}}\left( {1 - {e^{2z}}} \right)$$

$$= \frac{1}{6}\begin{array}{*{20}{c}} {lim}\\ {z \to 0} \end{array}\left( { - 8{e^{2z}}} \right)$$

$$= \frac{1}{6} \times \left( { - 8} \right) = - \frac{4}{3}$$

# Let z = x + iy be a complex variable. Consider that contour integration is performed along the unit circle in anticlockwise direction. Which one of the following statements is Not True?

1. The residue of $$\frac{{\rm{z}}}{{{{\rm{z}}^2} - 1}}{\rm{at\;z}} = 1{\rm{\;is\;}}1/2$$
2. $$\mathop \oint \limits_{\rm{c}}^{\rm{\;}} {{\rm{z}}^2}{\rm{dz}} = 0$$
3. $$\frac{1}{{2{\rm{\pi i}}}}\mathop \oint \limits_{\rm{c}}^{\rm{\;}} \frac{1}{{\rm{z}}}{\rm{dz}} = 1$$
4. $${\rm{\bar z}}$$ (Complex conjugate of z) is an analytical function

Option 4 : $${\rm{\bar z}}$$ (Complex conjugate of z) is an analytical function

## Residue Theorem MCQ Question 17 Detailed Solution

Residue of $$\frac{{\rm{z}}}{{{{\rm{z}}^2} - 1}}{\rm{\;at\;z}} = 1{\rm{\;is}} = \begin{array}{*{20}{c}} {{\rm{lim}}}\\ {{\rm{z}} \to 1} \end{array}\left( {{\rm{z}} - 1} \right).\frac{{\rm{Z}}}{{{{\rm{z}}^2} - 1}} = \begin{array}{*{20}{c}} {{\rm{lim}}}\\ {{\rm{z}} \to 1} \end{array}{\rm{\;}}\frac{{\rm{z}}}{{{\rm{z}} + 1}} = \frac{1}{2}$$

$$\begin{array}{l} \mathop \oint \limits_{\rm{c}}^{\rm{\;}} {{\rm{z}}^2}.{\rm{dz}} = 0\\ \frac{1}{{2{\rm{\pi i}}}}{\rm{\;}}\oint \frac{1}{{\rm{z}}}.{\rm{dz}} = \frac{1}{{2{\rm{\pi i}}}}.2{\rm{\pi i}}\left( 1 \right) = 1\\ {\rm{z\;}} = {\rm{\;x\;}} + {\rm{\;iy\;\;and\;\bar z}} = {\rm{x}} - {\rm{iy\;i}}.{\rm{e}}.{\rm{\;u}} = {\rm{x\;and\;v}} = - {\rm{y}}\\ {{\rm{u}}_{\rm{x}}} = 1{\rm{\;and\;}}{{\rm{v}}_{\rm{y}}} = - 1 \end{array}$$

$${{\rm{u}}_{\rm{X}}} \ne {{\rm{v}}_{\rm{y}}} \to {\rm{\bar z}}$$ is not Analytic Analytic

# Let $$f\left( z \right) = \frac{1}{{z + a}},\;a > 0$$. The value of the integral ∮ f(z) dz over a circle C with center (-a, 0) and radius R > 0 evaluated in the anti-clock wise direction is

1. 0
2. 2πi
3. -2πi
4. 4πi

Option 2 : 2πi

## Residue Theorem MCQ Question 18 Detailed Solution

Calculation:

$$f\left( z \right) = \frac{1}{{z + a}},\;a > 0$$

Centre = (-a, 0)

Singularly z + a = 0

⇒ z = -a

Singularly (z = -a) lies inside a circle C.

Here, the order of singularly n = 1

At z = -a, residue of f(z) $$= \frac{{ - 1}}{{\left( {1 - 1} \right)!}}\;{\left[ {{{\left( {z - a} \right)}^1}f\left( z \right)} \right]_{z = - a}}$$

$$= \left( {z - a} \right)\;\frac{1}{{\left( {z - a} \right)}} = 1$$

$$\mathop \smallint \limits_C f\left( z \right)dz = 2\pi \times Res\;of\;f\left( z \right)$$

= 2πi × 1 = 2πi

# The value of the integral $$\frac{1}{{2\pi i }}\mathop \smallint \limits_C^\; \frac{{{z^2} + 1}}{{{z^2} - 1}}dz$$ where z is a complex number and C is a unit circle with centre at 1 + 0j in the complex plane is __________ .

## Residue Theorem MCQ Question 19 Detailed Solution

$$\frac{1}{{2\pi i}}\mathop \smallint \limits_C \frac{{{z^2} + 1}}{{{z^2} - 1}}dz$$

Simple poles, z = ±1

According to Cauchy’s residue theorem,

$$\mathop \smallint \limits_C f\left( z \right)dz = 2\pi i\;\left[ {sum\;of\;residues} \right]$$

$$f\left( z \right) = \frac{{{z^2} + 1}}{{{z^2} - 1}}$$

At, z = +1, residue is

$$\begin{array}{*{20}{c}} {{\rm{lim}}}\\ {z \to 1} \end{array}\left( {z - 1} \right)\frac{{{z^2} + 1}}{{\left( {z + 1} \right)\left( {z - 1} \right)}} = 1$$

At, z = -1 residue is zero as z = -1 lies outside the curve C.

$$\frac{1}{{2\pi i}}\mathop \smallint \limits_C \frac{{{z^2} + 1}}{{{z^2} - 1}}dz = \frac{1}{{2\pi i}} \times 2\pi i\left[ 1 \right] = 1$$

# The value of the contour integral $$\frac{1}{{2\;\pi j}}\oint {\left( {z + \frac{1}{z}} \right)^2}dz$$ evaluated over the unit circle |z| = 1 is _______.

## Residue Theorem MCQ Question 20 Detailed Solution

$$\frac{1}{{2\pi j}}\oint {\left( {z + \frac{1}{z}} \right)^2}dz$$

$$\Rightarrow \frac{1}{{2\pi j}}\oint \left( {{z^2} + \frac{1}{{{z^2}}} + 2} \right)dz$$

$$\Rightarrow \frac{1}{{2\pi j}}\left[ {\oint {z^2}dz + \oint \frac{1}{{{z^2}}}dz + \oint 2dz} \right]$$

The first and third term are analytic. Hence integral = 0

For integral

$$\oint \frac{1}{{{z^2}}}dz$$

$$\begin{array}{*{20}{c}} {{\rm{lim}}}\\ {z \to {z_0}} \end{array}\frac{1}{{\left( {n - 1} \right)!}}\frac{{{d^{n - 1}}}}{{d{z^{n - 1}}}}\;{\left( {z - {z_0}} \right)^n}f\left( z \right)$$

$$\begin{array}{*{20}{c}} {{\rm{lim}}}\\ {z \to {z_0}} \end{array}\frac{1}{{\left( {2 - 1} \right)!}}\frac{d}{{dz}}{\left( {z - 0} \right)^2}\frac{1}{{{{\left( z \right)}^2}}}$$

$$\begin{array}{*{20}{c}} {{\rm{lim}}}\\ {z \to {z_0}} \end{array}\frac{1}{{1!}}\frac{d}{{dz}}\left( 1 \right)$$

= 0

Hence, $$\mathop \smallint \limits_{2\pi i}^{} {\left( {z + \frac{1}{z}} \right)^2}dz = 0$$

|z| = 1


text:
American Institute of Mathematical Sciences

Dynamics at infinity and Jacobi stability of trajectories for the Yang-Chen system

 1 Guangxi Colleges and Universities Key Laboratory of Complex System Optimization, and Big Data Processing, Yulin Normal University, Yulin 537000, China 2 College of Science, Guangxi University for Nationalities, Guangxi, 530006, China 3 School of Mathematics and Physics, China University of Geosciences (Wuhan), Wuhan, Hubei 430074, China 4 Zhejiang Institute, China University of Geosciences, Hangzhou, Zhejiang 311305, China

* Corresponding author: weizhouchao@163.com

Received  January 2020 Revised  May 2020 Published  August 2020

Fund Project: The first author is supported by National Natural Science Foundation of China (Grant No. 11961074), Natural Science Foundation of Guangxi Province (Grant Nos. 2018GXNSFDA281028, 2017GXNSFAA198234), the High Level Innovation Team Program from Guangxi Higher Education Institutions of China (Document No. [2018] 35), and the Science Technology Program of Yulin Normal University (Grant No. 2017YJKY28). The second author is supported by the Postgraduate Innovation Program of Guangxi University for Nationalities (Grant No. GXUN-CHXZS2018042). The third author is supported by National Natural Science Foundation of China (Grant No. 11772306), Zhejiang Provincial Natural Science Foundation of China under Grant (No.LY20A020001), and the Fundamental Research Funds for the Central Universities, China University of Geosciences (CUGGC05)

The present work is devoted to giving new insights into a chaotic system with two stable node-foci, which is named Yang-Chen system. Firstly, based on the global view of the influence of equilibrium point on the complexity of the system, the dynamic behavior of the system at infinity is analyzed. Secondly, the Jacobi stability of the trajectories for the system is discussed from the viewpoint of Kosambi-Cartan-Chern theory (KCC-theory). The dynamical behavior of the deviation vector near the whole trajectories (including all equilibrium points) is analyzed in detail. The obtained results show that in the sense of Jacobi stability, all equilibrium points of the system, including those of the two linear stable node-foci, are Jacobi unstable. These studies show that one might witness chaotic behavior of the system trajectories before they enter in a neighborhood of equilibrium point or periodic orbit. There exists a sort of stability artifact that cannot be found without using the powerful method of Jacobi stability analysis.

Citation: Yongjian Liu, Qiujian Huang, Zhouchao Wei. Dynamics at infinity and Jacobi stability of trajectories for the Yang-Chen system. Discrete & Continuous Dynamical Systems - B, doi: 10.3934/dcdsb.2020235
References:

show all references

References:
Attractors of Yang-Chen system with (a) $a = 10$, $b = 8/3$, and $c = 16$; (b) $a = 35$, $b = 3$, and $c = 35$
Dynamics of the Yang-Chen system near the sphere at infinity in the local charts $U_1$ for (blue) $(a, b, c) = (0.5, 1.01, 1)$ with initial conditions $(z_1(0), z_2(0), z_(0)) = (0.03, 0.03, -0.03)$, (red) $(a, b, c) = (1, 1.01, 1)$ with initial conditions $(z_1(0), z_2(0), z_(0)) = (0.03, 0.03, -0.03)$, (black) $(a, b, c) = (0.1, 1.01, 1)$ with initial conditions $(z_1(0), z_2(0), z_(0)) = (0.03, 0.03, -0.01)$, respectively
Dynamics of the Yang-Chen system near the sphere at infinity in the local charts $V_1$ (blue) $(a, b, c) = (0.5, 1.01, 1)$ with initial conditions $(z_1(0), z_2(0), z_(0)) = (0.03, 0.03, 0.03)$, (red) $(a, b, c) = (1, 1.01, 1)$ with initial conditions $(z_1(0), z_2(0), z_(0)) = (0.03, 0.03, 0.03)$, (black) $(a, b, c) = (0.1, 1.01, 1)$ with initial conditions $(z_1(0), z_2(0), z_(0)) = (0.03, 0.03, 0.01)$, respectively
Phase portrait of the system (10), which corresponds to the phase portrait of the Yang-Chen system at infinity in the local charts $U_2$
Phase portrait of the system (12), which corresponds to the phase portrait of the Yang-Chen system at infinity in the local charts $U_3$
Phase portrait of system (1) at infinity
Time variation of the deviation vector and its curvature near $E_{1}$, for $a = 35$, $b = 3$
Time variation of instability exponent $\delta(E_{1})$ for $a = 35$, $b = 3$, and different values of $c$
Time variation of the deviation vector and its curvature near $E_{2, 3}$ with $a = 35, b = 3$. Initial conditions used to integrate deviation equations are $\xi_{1}(0) = \xi_{2}(0) = 0$, $\dot{\xi}_{1}(0) = \dot{\xi}_{2}(0) = 10^{-6}$
Time variation of curvature $\kappa_{0}$ of deviation vector near equilibrium points $E_{1}$ with $a = 35, b = 3$
Time variation of curvature $\kappa_{0}$ of deviation vector near equilibrium points $E_{2, 3}$ with $a = 35, b = 3$
A large version of Fig. 11 at time $0.25$ to $0.55$
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# Seminars (MFEW02)

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Event When Speaker Title Presentation Material
MFEW02 2nd December 2013
09:15 to 10:15
Trapping of Rossby waves in the equatorial betaplane model
In this talk we shall report on joint works with Christophe Cheverry, Thierry Paul and Laure Saint-Raymond in which we study an equatorial shallow water system under the betaplane approximation. We prove that in some asymptotic regimes, Rossby waves are trapped around the equator while Poincaré waves disperse. This involves the use of microlocal techniques related to semiclassical analysis, such as Mourre estimates, and some ODE methods.
MFEW02 2nd December 2013
10:15 to 11:00
J Gibbon Rescaled vorticity moments in the 3D Navier-Stokes equations
Co-authors: D. D. Donzis (Texas A and M), A. Gupta (University of Rome Tor Vergata), R. M. Kerr (University of Warwick), R. Pandit (Indian Institute of Science Bangalore), D. Vincenzi (CNRS, Universite de Nice)

The issue of intermittency in numerical solutions of the 3D Navier-Stokes equations is addressed using a new set of variables whose evolution has been calculated through three sets of numerical simulations. These variables are defined on a periodic box $[0,\,L]^{3}$ such that $D_{m}(t) = \left(\varpi_{0}^{-1}\Omega_{m}\right)^{\alpha_{m}}$ where $\alpha_{m}= 2m/(4m-3)$ \& the set of frequencies $\Omega_{m}$ for $1 \leq m \leq \infty$ are defined by $\Omega_{m}(t) = \left(L^{-3}\I |\mbox{\boldmath$\omega$}|^{2m}dV\right)^{1/2m}$\,; the fixed frequency $\varpi_{0} = \nu L^{-2}$. All three simulations unexpectedly show that the $D_{m}$ are ordered for $m = 1\,,...,\,9$ such that $D_{m+1} MFEW02 2nd December 2013 11:30 to 12:15 Recent observations of mesoscale structures in intense cyclones The impact of intense cyclones is often associated with coherent mesoscale structures embedded within them. Although the formation of cold fronts is to some extent described by semi-geostrophic theory, the dynamics of many of these features are not well understood. For example, whether they could be described by a form balance appropriate on the mesoscale, or in the presence of non-conservative processes. Some recent detailed observations of cyclones and fronts are presented and some of the unknown aspects of their dynamics highlighted. MFEW02 2nd December 2013 14:00 to 14:45 P Bartello Between quasigeostrophic and stratified turbulence While it is well-established that the frequency disparity between vortical and wave motion is key to understanding the quasigeostrophic limit, i.e. strong rotation and stratification, the starting point for this contribution is that it has recently been established that there is no such frequency disparity in stratified turbulence without rotation. It remains to ask what happens in between these two limits, long held as the prevailing dynamics between deformation-scale eddies and the microscale where isotropy is recovered. To do this, ideas from numerical weather prediction were borrowed in order to explore numerically the nonhydrostatic Boussinesq equations starting from initial conditions that are close to our current fuzzy notions of balance for a variety of Rossby and Froude numbers. It is found that evolution is immediately away from this balance in the small scales, and from steep to much more shallow spectra. It will be argued that this conclusion is robust to unce rtainties in the definition of balance. MFEW02 2nd December 2013 14:45 to 15:30 M Petcu Exponential Decay of the Power Spectrum and Finite Dimensionality for Solutions of the Three Dimensional Primitive Equations The purpose of this talk is to estimate the number of modes, volumes and nodes, sufficient to describe well the solution of the three dimensional primitive equations; the physical meaning of these estimates is also discussed. We also study the exponential decay of the spatial power spectrum for the three dimensional primitive equations. MFEW02 2nd December 2013 16:00 to 16:45 D Straub Energy cascades in the baroclinic ocean wind-driven double gyre problem Co-author: David Straub We consider the classic baroclinic quasigeostrophic wind-driven ocean double gyre problem over a range of deformation radii, wind stress amplitudes, and bottom friction coefficients with an aim of understanding transfer of energy across scales. In this beta-plane basin setting, we find significant differences from classic studies of gestrophic turbulence, which generally assume zonal periodicity. In a basin geometry, the beta term (related to a latitudinal dependence in the Corioils parameter) can play a key role. For example, it can be the dominant term allowing for energy transfer between the basin scale and the baroclinic mesoscale. We also find that barotropization of baroclinic mesoscale energy forces the barotropic mode at scales where the barotropic mode is most energetic. Related to this, the barotropic nonlinear inverse energy cascade does not extend between mesoscale injection and large scale dissipation wavenumbers, as is often assumed. Instead, it is part of a double cascade" of barotropic energy in which the nonlinear inverse cascade is nearly offset by a forward cascade associated with the beta term. This is particularly evident in weak bottom drag simulations, for which a time eddy-mean flow decomposition reveals the double cascade to beassociated with the eddy-only terms. MFEW02 3rd December 2013 09:15 to 10:15 Mathematical Study of Certain Geophysical Models: Global Regularity and Finite-time Blowup Results The basic problem faced in geophysical uid dynamics is that a mathematical description based only on fundamental physical principles, the so-called the \Primitive Equations", is often prohibitively expensive computationally, and hard to study analytically. In this talk I will discuss the main obstacles in proving the global regularity for the three-dimensional Navier-Stokes equations and their geophysical counterparts. However, taking advantage of certain geophysical balances and situations, such as geostrophic balance and the shallowness of the ocean and atmosphere, geophysicists derive more simplied and manageable models which are easier to study analytically. In particular, I will present the global well-posedness for the three-dimensional Benard convection problem in porous media, and the global regularity for a three-dimensional viscous planetary geostrophic models. Even though the primitive equations look as if they are more dicult to study analytically than the three-dimensional Navier-Stokes equations I will show, on the one hand, that the viscous primitive equations have a unique global (in time) regular solution for all initial data. On the other hand, I will show that in the non-viscous (inviscid) case there is a one-parameter family of initial data for which the corresponding smooth solutions develop nite-time singularities (blowup). MFEW02 3rd December 2013 10:15 to 11:00 RM Temam Change of phase for the humid atmosphere In this lecture we will recall the atmospheric equations of water vapor with saturation. In their simplest form, these equations form a nonlinear system of partial differential equations with discontinuities. We will address the issue of the modelling of the system in the presence of singularities, and some questions on the existence, uniqueness and regularity of these solutions. MFEW02 3rd December 2013 11:30 to 12:15 L Chumakova Leaky lid: new dissipative modes in the troposphere Much of our understanding of tropospheric dynamics is based on the concept of discrete internal modes. Internal gravity waves, such as those associated with convective systems, propagate at definite speeds, typically associated with the first to third baroclinic vertical modes. These waves are the dynamical backbone of the tropospheric dynamics, even though their nature and speed can be altered significantly by nonlinearity, moist convection, mean wind shear, etc. These discrete modes are a signature of systems of finite extent, and are derived in a case when the atmosphere is bounded above by a rigid lid. In reality, the atmosphere does not have a definite top, and, some argue, should be modeled as semi-infinite, leading to a continuous spectrum. Are the discrete rigid lid modes then just a fallacy of overly simplified theoretical models? We present a correction to the rigid lid by using a boundary condition at the top of the troposphere, that allows for a fraction of waves to escape to the stratosphere. The new discrete leaky” modes decay with characteristic time-scales, which are in the ballpark of many atmospheric phenomena. We present both the non-rotating and rotating cases. MFEW02 3rd December 2013 14:00 to 14:45 JG Esler Adaptive stochastic trajectory modelling of transport in geophysical flows Motivated by the goal of improving and augmenting stochastic Lagrangian models of particle dispersion in turbulent geophysical flows, techniques from the theory of stochastic processes are applied to a model transport problem. The aim is to find an efficient and accurate method to calculate the total tracer transport between a source and a receptor when the flow between the two locations is weak, rendering direct stochastic Lagrangian simulation prohibitively expensive. Two methods are found to be useful. The first is Milstein's measure transformation method', which involves adding an artificial velocity to the trajectory equation, and simultaneously correcting for the weighting given to each particle under the new flow. Various difficulties associated with making an appropriate choice for the artificial velocity field are detailed and addressed. The second method is a variant of Grassberger's go-with-the-winners' branching process, which acts to remove particles unlikely to contribute to the net transport, and reproduces those that will contribute. A simple solution to the problem of defining a `winner' for flows in a high Peclet number chaotic advection regime is proposed. It is demonstrated that, used independently or together, the two methods can act to reduce the variance of estimators of the total transport by several orders of magnitude compared with direct simulation. MFEW02 3rd December 2013 14:45 to 15:30 Modelling the interactions of near-inertial waves and vortical motion in the ocean Co-author: Eric Danioux (University of Edinburgh) Wind forcing of the ocean generates a spectrum of inertia-gravity waves that is sharply peaked near the local inertial (or Coriolis) frequency. The corresponding near-inertial waves (NIWs) make a dominant contribution to the vertical velocity and vertical shear in the ocean; they therefore play an important role for mixing, biological productivity, pollutant dispersion and, arguably, the thermohaline circulation. An asymptotic model proposed by Young and Ben Jelloul describes the slow evolution of NIWs that results from weak dispersion and from their interactions with the quasi-two-dimensional vortical motion. We derive this YBJ model by applying a form of Whitham averaging to the variational formulation of the primitive equations for a rotating stratified fluid. This provides a direct route to the YBJ equation and elucidates its variational structure and conservation laws. We then consider the effect of turbulent vortical motion (modelled as a homogeneous random field) of a scale similar to that of the waves. Specifically, we derive a transport equation for NIWs that describes their scattering by the vortical motion and show how this scattering leads to an isotropization of the NIW field. Direct numerical simulations of the YBJ equations are used to test the predictions of the transport equation. Possible models of the two-way coupling between NIWs and vortical motion are also discussed. MFEW02 3rd December 2013 16:00 to 16:45 Direct Statistical Simulation of Jet Formation in Local and Global Geometries Co-author: Brad Marston (Brown University) We present Direct Statistical Simulation (DSS) of jet formation. We consider the simplest barotropic model both on a spherical surface and a local beta-plane. DSS involves the direct solution of the low-order statistics via an expansion in cumulants. In both cases we compare the results of our DSS with statistics obtained from long DNS simulations. We discuss in what circumstances truncating the cumulant expansion at second order (thereby including eddy – mean-flow interaction but neglecting eddy-eddy interactions for the fluctuating fields) gives a good description of the dynamics of the flow. We demonstrate that this depends on the degree of lack of statistical equilibrium in the flow (as measured by the Zonostrophy parameter). We discuss briefly how to proceed to higher order to include eddy-eddy interactions and the possibility of forward and inverse cascades. MFEW02 4th December 2013 09:15 to 10:15 D Holm Slice models Co-author: Colin Cotter (Imperial College) A variational framework is defined for vertical slice models with three dimensional velocity depending only on horizontal$x$and vertical$z$. The models that result from this framework are Hamiltonian, and have a Kelvin-Noether circulation theorem that results in a conserved potential vorticity in the slice geometry. These results are demonstrated for the incompressible Euler--Boussinesq equations with a constant temperature gradient in the$y$-direction (the Eady--Boussinesq model), which is an idealised problem used to study the formation and subsequent evolution of weather fronts. We then introduce a new compressible extension of this model for testing compressible weather models running in a vertical slice configuration. (Joint work with CJ Cotter, Imperial College). MFEW02 4th December 2013 10:15 to 11:00 From simple particle models to PDE dynamics One often aims to describe systems out of equilibrium by the governing energy E and entropy S, as well as the corresponding evolution laws for E and S. How can we derive these ingredients of the macroscopic evolution from particle models? In recent years, a dynamic scale-bridging approach has been developed and applied to a number of problems; large deviation theory plays an important role. The talk will present some of these results, focussing on the derivation of the Wasserstein-entropy formulation of diffusion and the Vlasov-Fokker-Planck equation as a system driven by energy and entropy. Time permitting, an approach of deriving stochastic equations mimicking the fluctuations in underlying mesoscopic models will be sketched. This is joint work with Hong Duong, Rob Jack and Mark. A. Peletier. MFEW02 4th December 2013 11:30 to 12:15 B Marston Direct Statistical Simulation of a Two-Layer Primitive Equation Model Co-authors: Wanming Qi (Brown University), Steve Tobias (University of Leeds) Low-order statistics of the large-scale circulation of planetary atmospheres may be directly accessed by solving the equations of motion for the equal-time statistics. We implement such Direct Statistical Simulation of a two-layer primitive equation model by systematic expansion in the cumulants. The first cumulant is the zonally averaged vorticity, divergence, and temperature as a function of latitude and level, and the second cumulant contains information about nonlocal teleconnections. At second order (CE2) the expansion retains the eddy – mean-flow interaction but neglects eddy-eddy interactions and is realizable. Eddy-eddy interactions appear at third (CE3) order, but care must be taken to maintain realizability with a non-negative probability distribution function. The cumulant expansion is conservative, order-by order, in the total angular momentum, total energy, and mean-squared potential temperature. An intermediate approximation, CE2.5, is related to the Edd y-Damped Quasi-Normal Markovian (EDQNM) approximation and maintains realizability at the expense of the introduction of a phenomenological timescale for eddy damping. First and second cumulants accumulated by time-integration of the two-layer primitive equations are compared with those obtained at the fixed points found at CE2, CE2.5, and CE3 levels of approximation, and against statistics obtained from reanalysis of the mid-level atmosphere of the Earth. CE2 reproduces qualitative features of the zonal mean general circulation such as the mid-latitude jets. CE2.5 and CE3 improve quantitative agreement in both the zonal means, and in the teleconnections. MFEW02 4th December 2013 14:00 to 14:45 Navier-Stokes equations on a rotating sphere We showed that, as the rotation rate$1/\epsilon$increases, the solution of the 2d Navier-Stokes equations on a rotating sphere becomes zonal, in the sense that the non-zonal component of the energy becomes bounded by$\epsilon$. This is obtained by estimating near-resonant interactions in the nonlinear term. As a consequence, the global attractor reduces to a single stable steady state when the rotation is fast enough (but still finite). MFEW02 4th December 2013 14:45 to 15:30 The influence of fast waves and fluctuations on the evolution of three slow solutions of the Boussinesq equations Co-authors: Jared P. Whitehead (Brigham Young University), Terry Haut (Los Alamos National Laboratory) We present results from a study of the impact of the non-slow (typically fast) components of a rotating, stratified flow on its slow dynamics. We examine three known slow limits of the rotating and stratified Boussinesq equations: strongly stratified flow ($Fr \rightarrow 0, Ro \approx O(1)$), strongly rotating flow ($Ro \rightarrow 0, Fr \approx O(1)$) and Quasi-Geostrophy ($Ro \rightarrow 0, Fr \rightarrow 0, Fr/Ro = f/N\$ finite).

In order to understand how the flow approaches and interacts with the slow dynamics we decompose the full solution into a component that is projected onto the null space of the fast operator and everything else. We use this decomposition to find evolution equations for the flow (and corresponding energy) on and off the slow manifold.

Numerical simulations indicate that for the geometry considered (triply periodic) and the type of forcing applied, the fast waves act as a conduit, moving energy onto the slow manifold. This decomposition clarifies how the energy is exchanged when either the stratification or the rotation is weak. In the quasi-geostrophic limit the energetics are less clear, however it is observed that the energy off the slow manifold equilibrates to a quasi-steady value.

At the end I will discuss how greater understanding of flow/fast dynamics could impact emerging numerical algorithms designed for future computer architectures.

MFEW02 4th December 2013
16:00 to 16:45
Resonant phenomena in the wave dynamics in the presence of boundaries in GFD
Co-author: Grigory Reznik (In-t of Oceanography, Moscow)

I will be discussing resonant interactions of coastal and shelf waves with free inertia-gravity waves and mean coastal currents in GFD.

MFEW02 5th December 2013
09:15 to 10:15
Use of the semi-geostrophic model in understanding large-scale atmosphere and ocean flows
The semi-geostrophic model is an accurate approximation to the Navier-Stokes equations when the Lagrangian Rossby number is small and the aspect ratio is less than f/N (where f is the Coriolis parameter and N the Brunt-Vaisala frequency). In practice this leads to a horizontal scale of around 1000km in the atmosphere and 100km in the ocean. The approximation is second order accurate in the limit epsilon tends to zero where the Rossby number is O(epsilon) and the Froude number O(sqrt(epsilon)). This approximation has a big advantage over the quasi-geostrophic approximation because it allow O(1) variations of the static stability, Coriolis parameter and orographic height. These features are essential in describing large-scale flows. The stability of the solutions of this system is consistent with the observed persistence of large-scale anomalies in the atmosphere and ocean. The failure of the approximation on smaller scales is associated with the lack of such structures on smal ler scales in the observed system. The talk will demonstrate how the semi-geostrophic model can be used to validate numerical methods, and how it can be extended to include a realistic model of the atmospheric boundary layer.
MFEW02 5th December 2013
10:15 to 11:00
M Feldman Lagrangian solutions for semigeostrophic system in physical space

In this introductory talk, we review results on existence of solutions to semigeostrophic system in physical and dual spaces. In particular, we show that Lagrangian solutions in physical space can be constructed for initial data satisfying a strict convexity condition. We also briefly discuss the recent joint work with A. Tudorascu, in which we relax the notion of Lagrangian solution to obtain existence for all convex initial data in physical space.

MFEW02 5th December 2013
11:30 to 12:15
Renormalized relaxed Lagrangian solutions for SG in physical space
A new, relaxed notion of Lagrangian solutions for SG in physical space will be introduced. The main motivation is the search for physical space solutions in the case of singular data in dual space. Existence and a weak stability result will be proved, along with an energy conservation result which comes as a consequence of the renormalization property. This presentation is based on joint work with M. Feldman (U. Wisconsin-Madison).
MFEW02 5th December 2013
14:00 to 14:45
Balance relations for rotating fluid flow
We discuss different nonlinear elliptic balance relations for rotating shallow water flow and describe computational tests which compare their utility as an initialization or diagnostic tool in nearly geostrophic situations.
MFEW02 5th December 2013
14:45 to 15:30
Generalized Large Scale Semigeostrophic Equations: geometric structure and global well-posedness
Co-authors: Marcel Oliver (Jacobs University), Mahmut Calik (Jacobs University)

We derive and study a family of approximate Hamiltonian balance models (called GLSG) for rotating shallow water in the semigeostrophic limit with spatially varying Coriolis parameter and non-trivial bottom topology. The models can be formulated in terms of an advected potential vorticity with a nonlinear vorticity inversion relation and include L_1 and LSG models proposed by R. Salmon as special cases.

We prove existence and uniqueness of global classical solutions to the GLSG equations for certain members of the family and study the PV invertibility as a function of the parameters.

MFEW02 5th December 2013
16:00 to 16:45
B Pelloni The semi-geostrophic system for large-scale atmospheric flows
Co-author: Mike Cullen (Met Office)

I could present results on the existence of solutions in 3D, free boundary setting as well as some preliminary results on the validity of the model as a reduction of the Euler system.

MFEW02 6th December 2013
09:00 to 09:45
G Vallis The Range of Planetary Circulations Described by the Dry Primitive Equations
Co-authors: Jonathan Mitchell (UCLA), Sam Potter (Princeton University)

The dry primitive equations can, with appropriate forcing and dissipation, provide a reasonable simulation of the large-scale features of the Earth's atmosphere and ocean. In this talk I will describe the behaviour of these PDEs when they are taken out of the terrestrial parameter regime. In n particular, I will describe their behaviour when the thermal Rossby number, Ekman number and a radiative relaxation timescale are varied considerably, moving into a parameter regime more appropriate for Mars or Titan.

I will pay particular attention to the formation of zonal jets, and in particular of equatorial superrotation, which is a feature of some other planets. It is well-known that zonal jets robustly arise in rotating atmospheres if there is a wavemaker at a particular latitude. Rossby waves are then generated that propagate away, and eastward momentum converges on the source region producing a zonal jet. The Earth's jet stream is, in part, formed this way. However, on slowly rotating atmospheres it seems unlikely that superrotation is produced by that mechanism. Rather, simulations indicate that, at small thermal Rossby number, a mechanism involving equatorial Kelvin waves is involved.

MFEW02 6th December 2013
09:45 to 10:30
I Roulstone Differential Geometry of the Semi-Geostrophic and Euler Equations
The role of contact and symplectic geometry of the semi-geostophic (SG) equations, in describing their Legendrian and Hamiltonian properties, will be reviewed. Using the geometry of 2-forms in 4 dimensions and the geometry of 3-forms in 6 dimensions, we show that the incompressible Euler equations in 2 and 3 dimensions admit geometric structures akin to the those present in the SG equations.
MFEW02 6th December 2013
11:00 to 11:45
O Bokhove On Time Integration and the Use of Clebsch Variables in Shallow Water Equations
Two topics will be covered in this lecture. The shallow water equations will be used as a test bed to introduce the ideas.

(i) For forced variational systems such as the potential flow shallow water wave equations, variational and symplectic time integrators will be extended using a new finite element approach. Here, a standard variational finite element discretization will be applied in space.

(ii) The shallow water equations formulated in terms of Clebsch variables will be discussed. The advantage of Clebsch variables is that they lead to canonical Hamilton's equations for shallow water dynamics, in the Eulerian framework. A disadvantage is that the the system, is less compactly expressed in comparison to the usual formulation in terms of the velocity and fluid depth. I will make a link between a symmetry in the Hamiltonian and the associated mass weighted potential vorticity conservation law, also within the Eulerian framework. This will be done in two dimensions (2D) and in a quasi-2D symmetric form.

MFEW02 6th December 2013
11:45 to 12:30
Monge-Ampère equations: geometry, invariants, and applications in 3D meteorological models
I describe a symplectic geometric approach to Monge-Ampère equations, discuss some classical and modern geometric structures related to this class of non-linear equations, their invariants and their role in 3D meteorological models. My talk is based on joint works with B. Banos and I . Roulstone.


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# Center text in Game engine with python

I'm trying to get a bitmap text centered in my game. The text is generated at start, so I wan't to scale & translate it based in its dimensions after the text has been set. This my current code.

text = str(scenesInfo[i][1])
tempBotonText['Text'] = text
textScaleX = 1/len(text)
tempBotonText.localScale[0] *= textScaleX
tempBotonText.localPosition[0] -=len(texto)  * 0.008

But as you can see, the number I am using in the translation is just a guess, and it doesn't work well with any length.

Is there a way I can know the dimensions of the text (it is a bitmap text) after its creation?

• It's hard to figure our what you're doing. What is tempBotonText? – dr. Sybren Dec 23 '14 at 22:52
• I've always worked around this by using monospaced fonts. But you might have some luck with blf.dimensions(fontID, "text here"). It returns a 2-float tuple containing width and height. blender.org/api/blender_python_api_2_72_1/blf.html – pqftgs Dec 24 '14 at 6:00