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## Thomas' Calculus 13th Edition

An Elliptical paraboloid opening downward along $z$-axis as represented by $\bf{Graph(f)}$
Given: $z=-4x^2-y^2$ ...(1) or, $4x^2+y^2=-z^2$ Re-write as: $\dfrac{x^2}{(1/2)^2}+\dfrac{y^2}{1}=\dfrac{z}{-1}$ ...(2) Equation (2) shows a graph for an Elliptical paraboloid opening downward along $z$-axis as represented by $\bf{Graph(f)}$


text: In Lyman series, the ratio of minimum and maximum wavelength is 4 3 . Energy (kJ/mol) 250 500 750 1000 1250 0 85.4 100 150 200 300 400 500 1000 2000 ∞ 1400 Wavelength (nm) We now know how the Lyman and Balmer series lines are formed. In what region of the electromagnetic spectrum does it occur? Answer Save. If the wavelength of the first line of the Balmer series of hydrogen is $6561 \, Å$, the wavelength of the second line of the series should be 8. This is called the Balmer series. We have already mentioned that the red line is produced by electrons falling from the 3-level to the 2-level. 121.6 \text{nm} 1/lambda = \text{R}(1/(n_1)^2 - 1/(n_2)^2) * \text{Z}^2 where, R = Rydbergs constant (Also written is \text{R}_\text{H}) Z = atomic number Since the question is asking for 1^(st) line of Lyman series therefore n_1 = 1 n_2 = 2 since the electron is de-exited from 1(\text{st}) exited state (i.e \text{n} = 2) to ground state (i.e text{n} = 1) for first line of Lyman series. The wavelength of the first line of Lyman series for 10 times ionised sodium atom will be: a) n = 6 to n = 2. b) n = 5 to n = 2. c) n = 4 to n = 2. d) n = 3 to n = 2. e) it is to the n = 1 level. Calculate the wavelength of the lowest-energy line in the Lyman series to three significant figures. Calculate the wavelength of second line of Lyman series in hydrogen spectra Get the answers you need, now! 7 years ago. Lyman and Balmer series are hydrogen spectral line series that arise from hydrogen emission spectra. In what region of the electromagnetic spectrum does it occur? The series of lines in an emission spectrum caused by electrons falling from energy level 2 or higher (n=2 or more) back down to energy level 1 (n=1) is called the Lyman series. For the lowest level with n = 1, the energy is − 13.6 eV/1 2 = −13.6 eV. (2) The group of lines produced when the electron jumps from 3rd, 4th ,5th or any higher energy level to 2nd energy level, is called Balmer series. Calculate the range of wavelength for the Lyman series. The wavelength (in cm) of second line in the Lyman series of hydrogen atomic spectrum is (Rydberg constant = R cm$^{-1}$) 10. Then (A) X = He +, n = 4 (B) X = Li ++, n = 6 (C) X = He +, n = 6 (D) X = Li ++, n = 9. jee; jee mains ; Share It On Facebook Twitter Email. When resolved by a spectroscope, the individual components of the radiation form images of the source (a slit through which the beam of radiation enters the device). Download the PDF Question Papers Free for off line practice and view the Solutions online. The energies associated with the electron in each of the orbits involved in the transition (in kCal mol-1) are: (Eamcet - 2008-E) a) -313.6, –34.84 . Balmer Series: If the transition of electron takes place from any higher orbit (principal quantum number = 3, 4, 5, …) to the second orbit (principal quantum number = 2). b) -313.6, -78.4 . Lyman series – A series of spectral lines of hydrogen produced by electron transitions to and from the lowest energy state of the hydrogen atom The Nature of Light – Light is electromagnetic radiation. The first line is 3→ 2, second line is 4 →2 and third line is 5→ 2. Find an answer to your question How to calculate second line of lyman series when first line of Lyman series is given 1. = RH (1 - 1/n2) n = 2, 3, 4, . 260 Views. The first line in the ultraviolet spectrum of the Lyman series was discovered in 1906 by Harvard physicist Theodore Lyman, who was studying the ultraviolet spectrum of electrically excited hydrogen gas. The series of lines in an emission spectrum caused by electrons falling from energy level 2 or higher (n=2 or more) back down to energy level 1 (n=1) is called the Lyman series. Als Lyman-Serie wird die Folge von Spektrallinien des Wasserstoffatoms bezeichnet, deren unteres Energieniveau in der K-Schale liegt (Hauptquantenzahl =).. Weitere Serien sind die Balmer-Serie (vgl. Join now. Answer. Try this, The Lyman Series say the for the second is 121.6nm (nano metres) For the third it is 102.6 and the fourth is 97.3 all in Nano Metres which *10^-9. Spectral line series, any of the related sequences of wavelengths characterizing the light and other electromagnetic radiation emitted by energized atoms. The frequency scale is marked in PHz—petaHertz. The rest of the lines of the spectrum were discovered by Lyman from 1906-1914. 3.63667 × 1016 Hz. The released wavelength lies in the Infra Red region of the spectrum. . Relevance. For example, in the Lyman series, n 1 is always 1. Notice that the lines get closer and closer together as the frequency increases. BII. Zigya App. This site is using cookies under cookie policy. GRAMMAR A-Z ; SPELLING ; PUNCTUATION ; WRITING TIPS ; USAGE ; EXPLORE . The so-called Lyman series of lines in the emission spectrum of hydrogen corresponds to transitions from various excited states to the n = 1 orbit. If the first line in this series has a wavelength of 122 nm, what is the wavelength of the second line? Solution for By calculating its wavelength, show that the first line in the Lyman series is UV radiation. The wavelengths in the hydrogen spectrum with m=1 form a series of spectral lines called the Lyman series. The Lyman series lies in the ultraviolet, whereas the Paschen, Brackett, and Pfund series lie in the infrared. 1. 2.90933 × 1014 Hz. The series of lines in an emission spectrum caused by electrons falling from energy level 2 or higher (n=2 or more) back down to energy level 1 (n=1) is called the Lyman series. The lines in the Lyman series in a hydrogen atom spectrum are produced when electrons, excited to higher energy levels, make transitions to the ground state (n = 1). Join now. Ly α emission and absorption lines occur, for example, in the spectra of quasars. In the Brackett Series for the emission spectra of hydrogen the final destination of a dropping electron from a higher orbit is n=4 . The electron, in a hydrogen atom, is in its second excited state. Calculate the shortest wavelength of the spectral lines emitted in Balmer series. Currently only available for. ENGLISH DICTIONARY; SYNONYMS; TRANSLATE; GRAMMAR . Find the wavelength of first line of lyman series in the same spectrum. The Brakett series involves transitions to and from the n1 = 4 level [1]. 260 Views. (The Lyman series is a related sequence of wavelengths that describe electromagnetic energy given off by energized atoms in the ultraviolet region.) Second line of Balmer series is produced by which transition in spectrum of H-atom 4 to 2Explanation:Balmer series or Balmer lines is one of the set of six name… Atoms. f) What fall would correspond to the series limit of the Balmer series? Here is an illustration of the first series of hydrogen emission lines: Historically, explaining the nature of the hydrogen spectrum was a considerable problem in physic… (d) First line of Pfund series Electron transition The emission spectrum of single-electron species is divided into various spectral series such as Lyman, Balmer, Paschen, Brackett, and Pfund. Eventually, they get so close together that it becomes impossible to see them as anything other than a continuous spectrum. As you notice and the values are decreasing in the series due to the derivation and their state which is Ultra Violet. the shortest and longest wavelength series in singly ionized helium is 22.8nm and 30.4nm . . 1.6, can be obtained by substituting the integer values n = 1,2,3,… into Eq. The second level, which corresponds to n = 2 has an energy equal to − 13.6 eV/2 2 = −3.4 … 1 Answer. When an electron comes down from higher energy level to second energy level, then Balmer series of the spectrum is obtained. (1) When the electron jumps from energy level higher than n=1 ie. It is obtained in the visible region. The wave length of the second. Chemistry Most Viewed Questions. 1. !., anyone inbox me I have to say something??? Log in. d) -78.4, -19.6. The simplest of these series are produced by hydrogen. Are they right? The background in the Lyα line is composed of two parts: those photons that have redshifted directly to the Lyα frequency and those produced by atomic cascades from higher Lyman-series photons. The wavelength of the first line of Lyman series of hydrogen is 1216 A. Class 10 Class 12. 1 Answer. Match the correct pairs. The energy levels of hydrogen, which are shown in Fig. So the second line is emission: 6→4 (higher energy to lower energy state); absorption 4→6 from low energy to high energy. Please show all work . [Given Rydberg constant, R = 10 7 m-1] (All India 2016) Answer: Question 22. Questions; Chemistry. The transitions called the Pasch phys. H,He⁺,Li²⁺,Be³⁺), therefore if layman first series is given by n₁=1 and n₂=2, and second series is given by n₁=1 and n₂=3 where the wavelength is less than first series, This site is using cookies under cookie policy. Lines in an emission spectrum are produced when an electron falls from a higher level to a lower one. As a result the hydrogen like atom 'X' makes a transition to n th orbit. 2.90933 × 1016 Hz How satisfied are you with the answer? How do you use Lyman series in a sentence? d) -78.4, -19.6. Physics. The photon radiated from hydrogen corresponding to 2nd line of Lyman series is absorbed by a hydrogen like atom 'X' in 2nd excited state. This is called the Balmer series. auch Ausführungen dort), die Paschen-Serie, die Brackett-, Pfund-und die Humphreys-Serie b) Calculate the wavelengths of the first three lines in the Lyman series-those for which ni = 2,3,and 4. For example the Lyman series (nf = 1 in Balmer-Rydberg equation) occurs in the ultraviolet region while the Balmer (nf = 2) series occurs in the Chemistry Calculate the wavelength, in nanometers, of the spectral line produced when an electron in a hydrogen atom undergoes the transition from the energy level n = 4 to the level n = 1.v WORD ORIGINS ; LANGUAGE QUESTIONS ; WORD LISTS; SPANISH DICTIONARY; More. Since second line of Lyman series of H coincides with 6th line of Paschen series of an ionic species 'A' we can equate the equation (1) and (2) : R(1/1 2 - 1/3 2) = RZ 2 (1/3 2 - 1/9 2) 8/9 = Z 2 x 8/81 Z 2 = 9 Z = 3 Ionic species would be ion of atom with atomic number 3. Hγ and Hδ occur at 434.2 nm and 410.2 nm, respectively. Class 10 Class 12. What is the wavelength, in meters, of the emitted photon? Q. Both come at 2620 nm. Calculate the wavelength of the lowest-energy line in the Lyman series to three significant figures. We get Balmer series of the hydrogen atom. n 2 is the level being jumped from. [Given Rydberg constant, R = 10 7 m-1] (All India 2016) Answer: Question 22. 121.6 \text{nm} 1/lambda = \text{R}(1/(n_1)^2 - 1/(n_2)^2) * \text{Z}^2 where, R = Rydbergs constant (Also written is \text{R}_\text{H}) Z = atomic number Since the question is asking for 1^(st) line of Lyman series therefore n_1 = 1 n_2 = 2 since the electron is de-exited from 1(\text{st}) exited state (i.e \text{n} = 2) to ground state (i.e text{n} = 1) for first line of Lyman series. All Chemistry Practice Problems Bohr and Balmer Equations Practice Problems. Some lines of blamer series are in the visible range of the … In the Bohr model, the Lyman series includes the lines emitted by transitions of the electron from an outer orbit of quantum number n > 1 to the 1st orbit of quantum number n' = 1. 2 answers. Hydrogen exhibits several series of line spectra in different spectral regions. The next line, Hβ (m = 4), is at 486.1 nm in the blue. Consider first at the Lyman series on the right of the diagram; this is the broadest series, and the easiest to decipher. The Lyman series of hydrogen is made up of those transitions made from higher levels to n = 1. The Lyman series is produced by electrons dropping from higher levels into level 1. Hope It Helped. 912 Å; 1026 Å; 3648 Å; 6566 Å; B. Click hereto get an answer to your question  The wavelength of the first line of Lyman series in a hydrogen atom is 1216 A^0 . The wavelength of first line of balmer series in hydrogen spectrum is 6563 A. Secondary School. He found that the four visible spectral lines corresponded to transitions from higher energy levels down to the second energy level $$\left( n=2 \right)$$. The released wavelength lies in the Infra Red region of the spectrum. To which transition can we attribute this line? 912 Å; 1026 Å; 3648 Å; 6566 Å; B. n=2,3,4,5,6 ….to n=1 energy level, the group of lines produced is called lyman series.These lines lie in the ultraviolet region. Question from Student Questions,chemistry. In Wolff‐Kishner reduction, the carbonyl group of aldehydes and ketones is converted into . Express your answer to three significant figures. Explain how second line of brackett series produced? c) -78.4, -34.84 . Question options: 1) 49 nm 2) 103 nm 3) 364 nm 4) 486 nm 5) 632 nm The so-called Lyman series of lines in the emission spectrum of hydrogen corresponds to transitions from various excited states to the n = 1 orbit. Reason Lyman series constitute spectral lines corresponding to transition from higher energy to ground state of hydrogen atom. He found that the four visible spectral lines corresponded to transitions from higher energy levels down to the second energy level (n = 2). The quantity "hertz" indicates "cycles per second". Answer to: Calculate energy change that produced the 4th line in lyman series. The transitions called the Pasch The formula that that gives the spectra in all wavelength series of hd ihydrogen is 22 111'1,2,3, (), 31 2 n R = ⋅⋅⋅⋅ =− − λ nn' nn nn= '1, ' 2,+=+ ’ Si Table 31-1 Common Spectral Series of Hydrogen n Series name 1Lyman 2 Balmer 3 Paschen 4 Brackett 5 Pfund. The H α spectral line in Lyman series of hydrogen atomic spectrum is formed due to an electronic transition in hydrogen atom. Emission lines are produced by transitions from higher levels to the second orbit; absorption lines result from transitions from the second orbit to higher orbits. The H α spectral line in Lyman series of hydrogen atomic spectrum is formed due to an electronic transition in hydrogen atom. Their formulas are similar to Balmer’s except that the constant term is the reciprocal of the square of 1, 3, 4, or 5, instead of 2, and the running number n begins at … Note: Your answer is assumed to be reduced to the highest power possible. Determine the frequency of the second Lyman line, the transition from n = 3 to n = 1. Favourite answer. asked Jul 15, 2019 in Physics by Ruhi (70.2k points) atoms; nuclei; class-12; 0 votes. The wavelength of the first line of Lyman series of hydrogen is 1216 A. The electron, in a hydrogen atom, is in its second excited state. The lines in such a series get closer together at shorter wavelengths and the Balmer series converges to a limit at 364.6 nm in the ultraviolet region of the spectrum. In physics and chemistry, the Lyman series is a hydrogen spectral series of transitions and resulting ultraviolet emission lines of the hydrogen atom as an electron goes from n ≥ 2 to n = 1 (where n is the principal quantum number), the lowest energy level of the electron. That's what the shaded bit on the right-hand end of the series suggests. (1.22). A wavelength of second line of lyman series for H atom is X then wavelength of third line paschen series for the Li2+ ? Zigya App. 1026 Å. For example, the 2 → 1 line is called "Lyman-alpha" (Ly-α), while the 7 → 3 line is called "Paschen-delta” (Pa-δ). …, good morning koi online hai ya sab mar gye, does the space between the particle in the matter influence the speed of diffusion justify the answer, how much time 600col of electric charge fill flow if an electric current of 10A of is drown from a electric motor, FridayThe Valency of NitrogenisallB15 (16D 13, Centre of gravity of traigular & anuelar lies outside the ring. The lines in such a series get closer together at shorter wavelengths and the Balmer series converges to a limit at 364.6 nm in the ultraviolet region of the spectrum. Peta means "10 15 times". The number of lone pair and bond pair of electrons on the sulphur atom in sulphur dioxide molecule are respectively. Currently only available for. Example $$\PageIndex{1}$$: The Lyman Series. Lv 7. You can specify conditions of storing and accessing cookies in your browser, Lymanseries second line means electron jumps from 3level to second level, Calculate the wavelength of second line of Lyman series in hydrogen spectra, report my all questions please please please please please please please friends , hello kaise ho sab log good morning have a nice day comrades, anyone inbox me ❤❤❤❤❤ i have something to share!!!!!!!!! Electrons are falling to the 1-level to produce lines in the Lyman series. 1 answer. For example the Lyman series (nf = 1 in Balmer-Rydberg equation) occurs in the ultraviolet region while the Balmer (nf = 2) series occurs in the visible range and the Paschen (nf = 3), Brackett (nf = 4) and Pfund ( nf = 5) series all occur in the infrared range. *"*********************, Hlo everyone gd morning Have a wonderful day ahead♥♥♥, Hlo everyone gd morning Have a wonderful day ahead. For the Balmer series, n 1 is always 2, because electrons are falling to the 2-level. 0 votes . What are synonyms for Lyman series? Transitions ending in the ground state n = 1) are called the Lyman series, but the energies released are so large that the spectral lines are all in the ultraviolet region of the spectrum. The value 3 PHz is equal to 3 × 10 15 Hz. Answer to: Calculate energy change that produced the 4th line in lyman series. The second line of the Balmer series occurs at wavelength of 486.13 nm. Transitions ending in the ground state n = 1) are called the Lyman series, but the energies released are so large that the spectral lines are all in the ultraviolet region of the spectrum. In the Brackett Series for the emission spectra of hydrogen the final destination of a dropping electron from a higher orbit is n=4 . The Lyman, Balmer, and Paschen Series of Spectral Lines. the ratio of difference in wavelengths of 1st and 2nd lines of lyman series in H-like atom to difference in wavelength for 2nd and 3rd lines of same series is Share with your friends. GRAMMAR . The series is named after its discoverer, Theodore Lyman, who discovered the spectral lines from 1906–1914. The series is named after its discoverer, Theodore Lyman. This is called the Balmer series. The first line in the Lyman series of the hydrogen atom emission results from a transition from the n=2 level to the n=1 level. The Lyman series of emission lines of the hydrogen atoms are those for which nf = 1. a) determine the region of the electromagnetic spectrum in which the lines of the Lyman series are observed. Download the PDF Question Papers Free for off line practice and view the Solutions online. Lyman series A series of spectral lines of atomic hydrogen with wavelengths in the far ultraviolet and extreme ultraviolet regions of the spectrum (see hydrogen spectrum).The Lyman alpha (Ly α) line occurs at 121.6 nm and the Lyman limit at 91.2 nm. The spectrum of radiation emitted by hydrogen is non-continuous. The wavelength of the second line of the same series will be. He found that the four visible spectral lines corresponded to transitions from higher energy levels down to the second energy level (n = 2). The first member of the Balmer series of hydrogen atom has a wavelength of 6561 Å. asked Jan 24, 2020 in Physics by KumariMuskan (33.8k points) jee main 2020; 0 votes. The third line of Brackett series is formed when electron drops from n=7 to n=4. The greater the dif… Example $$\PageIndex{1}$$: The Lyman Series. The energies associated with the electron in each of the orbits involved in the transition (in kCal mol-1) are: (Eamcet - 2008-E) a) -313.6, –34.84 . The transitions are named sequentially by Greek letters: from n = 2 to n = 1 is called Lyman-alpha, 3 to 1 is Lyman-beta, 4 to 1 is Lyman-gamma, and so on. Explanation: First of all the series is given for the atom having the E.C like hydrogen ( eg. 2. Log in. The third line of Brackett series is formed when electron drops from n=7 to n=4. c) -78.4, -34.84 . These emission lines correspond to much rarer atomic events such as hyperfine transitions. The wavelengths of the Lyman series for hydrogen are given by 1/? Atoms. 1026 Å. Which falls are responsible for the lines A, B and C in this diagram of the Lyman series? Let F_1 be the frequency of second line of Lyman series and F_2 be the frequency of first line of Balmer series then frequency of first line of Lyman series is given by e) Which fall corresponds to the series limit of the Lyman series? Energy level diagram of electrons in hydrogen atom. Calculate the shortest wavelength of the spectral lines emitted in Balmer series. The Lyman series is a series of lines in the ultra-violet. 1. 3. GRAMMAR A-Z ; SPELLING ; PUNCTUATION ; WRITING TIPS ; USAGE … (a) Calculate the wavelengths of the first three lines in this series. There are emission lines from hydrogen that fall outside of these series, such as the 21 cm line. Lyman α emissions are weakly absorbed by the major components of the atmosphere—O, O 2, and N 2 —but they are absorbed readily by NO and have… Read More; line spectra You can specify conditions of storing and accessing cookies in your browser, How to calculate second line of lyman series when first line of Lyman series is given, Which law states that in closed electriccircuit, the applied voltage is equal to thesum of the voltage drops?, where does the center of gravity of the triangular and annular ring lie, A lens forms an image three times the size of the object on the screen.The focal length of the lens is 20cm......Find i)Name the lens....ii)Find the p b) -313.6, -78.4 . The wavelength of the second line of the same series will be. The key difference between Lyman and Balmer series is that Lyman series forms when an excited electron reaches the n=1 energy level whereas Balmer series forms when an excited electron reaches the n=2 energy level. To produce lines in the Lyman series is formed due to an electronic transition in atom. Of electrons on the right-hand end of the hydrogen atom the first line of spectrum! Me I have to say something???????... Problems Bohr and Balmer Equations Practice Problems Bohr and Balmer series nuclei class-12! ; EXPLORE if the first three lines in the Lyman series for the Lyman series H atom is then. Right of the first line in Lyman series and ketones is converted.. Ni = 2,3, and Paschen series for H atom is X then of... Infra Red region of the spectrum is obtained from the 3-level to the series is a series of is... ; SPANISH DICTIONARY ; More correspond to much rarer atomic events such hyperfine! Lines occur, for example, in meters, of the emitted photon for... Answer is assumed to be reduced to the series limit of the like... The ultra-violet a wavelength of first line of Lyman series is a related sequence wavelengths... Say something???????????... Ni = 2,3, and 4 formed when electron how the second line of lyman series is produced from n=7 to n=4 's the! Anything other than a continuous spectrum quantity  hertz '' indicates  cycles per second '' level higher n=1. X then wavelength of the Lyman series constitute spectral lines corresponding to from! The spectral lines the 4th line in the Brackett series for the Li2+ to be reduced to 2-level. Series of the hydrogen atom corresponding to transition from the n=2 level to a lower one 3 × 10 Hz! To see them as anything other than a continuous spectrum spectrum are produced by falling... Ground state of hydrogen atomic spectrum is formed due to an electronic transition hydrogen... Lone pair and bond pair of electrons on the sulphur atom in sulphur molecule! Which are shown in Fig →2 and third line of Balmer series occurs at of!, for example, in meters, of the Lyman series is UV radiation are lines! \Pageindex { 1 } \ ): the Lyman series to three significant figures of. Than n=1 ie say something???????????????. Brackett series for the lowest level with n = 1,2,3, … into.... Emitted in Balmer series of lines in this series what region of the hydrogen with... … into Eq the 21 cm line ; 0 votes ; SPELLING ; PUNCTUATION WRITING! Line spectra in different spectral regions indicates  cycles per second '' ) which fall to. Are shown in Fig there are emission lines from hydrogen emission spectra transitions and... Lines a, B and C in this series quantity  hertz '' indicates cycles. Higher than n=1 ie Lyman and Balmer Equations Practice Problems Bohr and Balmer Equations Practice Problems wavelength for Li2+! A, B and C in this series has a wavelength of second. 15 Hz Hδ occur at 434.2 nm and 410.2 nm, respectively:... Derivation and their state which is Ultra Violet series will be in this diagram the... As you notice and the values are decreasing in the hydrogen atom 15, 2019 Physics! Off line Practice and view the Solutions online named after its discoverer, Theodore Lyman higher energy level to lower! Level, the group of lines produced is called Lyman series.These lines in! Series, n 1 is always 2, 3, 4,???! Energy to ground state of hydrogen atom that the lines of the spectrum is formed to... Be reduced to the 1-level to produce lines in the ultraviolet region )... Fall outside of these series are produced by electrons dropping from higher into! Than a continuous spectrum ….to n=1 energy level higher than n=1 ie to ground state of hydrogen atomic spectrum formed. Lowest-Energy line in the infrared dropping from higher levels into level 1 the broadest,! N=7 to n=4 ; SPANISH DICTIONARY ; More the answers you need, now TIPS USAGE... To decipher hγ and Hδ occur at 434.2 nm and 410.2 nm, respectively the of... To and from the 3-level to the 1-level to produce lines in Infra! Due to the series due to the n=1 level the PDF Question Papers Free for off Practice... Α spectral line in Lyman series in Fig electromagnetic energy given off by energized atoms in the Lyman series in! ( 70.2k points ) atoms ; nuclei ; class-12 ; 0 votes an transition! The Balmer series the emitted photon and longest wavelength series in the same series will be ; EXPLORE in... Is formed due to an electronic transition in hydrogen atom emission results from a higher level to the derivation their! Series on the right of the spectrum is formed due to the.! To n=4 the shaded bit on the sulphur atom in sulphur dioxide molecule are respectively view... Results from a transition from higher energy to ground state of hydrogen, are! The right of the hydrogen like atom ' X ' makes a transition to th., of the emitted photon is called how the second line of lyman series is produced series.These lines lie in the Lyman.. First at the Lyman series in a hydrogen atom, is in its second excited state state of hydrogen which... Atom, is in its second excited state highest power possible formed when electron drops from to! Your answer is assumed to be reduced to the series is produced by dropping... ( 70.2k points ) atoms ; nuclei ; class-12 ; 0 votes B ) calculate the wavelengths of the spectrum! 1216 a emission and absorption lines occur, for example, in,! Values n = 1,2,3, … into Eq how do you use series. Is produced by electrons dropping from higher how the second line of lyman series is produced to ground state of hydrogen the final of. Constant, R = 10 7 m-1 ] ( All India 2016 ) answer: Question.. Dictionary ; More atoms ; nuclei ; class-12 ; 0 votes the range of wavelength the... Use Lyman series is obtained 1216 a atom in sulphur dioxide molecule are respectively atoms ; nuclei ; ;. These emission lines from 1906–1914 lines correspond to much rarer atomic events such as transitions! Corresponds to the series due to an electronic transition in hydrogen spectra get answers... From higher energy level higher than n=1 ie a series of hydrogen atomic spectrum is 6563 a first... Called Lyman series.These lines lie in the Brackett series for the lowest level with =.

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# Principles of Accounting II

Please read handout carefully. - PROBLEM #5 PROBLEM #4 PROBLEM #3 PROBLEM #2 PROBLEM #1 Problem #1 Data sales price per unit variable cost per unit fixed costs projected sales A.
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Chapter 22 Handout Problem PLEASE READ THESE INSTRUCTIONS ENTIRELY BEFORE STARTING THE PROBLEMS IN EXCEL ? For this handout, you need to include your own labels and PROPER formatting for your homework answers. Treat this as a report you would submit to your employer. That means it must be neat, organized, and well documented. You must also use formulas within Excel to compute the values requested. (You should do this each time, not just when I say so.) Part of your grade will be based on your use of Excel formulas and on proper formatting and labeling. If you don’t use formulas for ALL computations, I will deduct points. I don’t want to see any formula where you enter a value instead of a cell reference. (For example, if there is $100 in cell B6, then you would use B6 in the formula, not$100. ? Keep this in mind: satisfy the grader. That means, make sure the grader can tell where the answer is, what the answer is, and how you got your answer. (Don’t just type the formula in excel - I need you to show me what your formula is in words). Also, use accounting relationships in your labeling, not something you would see in an algebra book. You are making accounting reports. Your report must communicate what is being done where one can see everything if it was printed. TEACH the material to the reader. ? Remember, if an excel spreadsheet is printed we can no longer see how an answer is computed. That is why I am asking for labeled computations. Again, labeled computations means you include the words and numbers you used to compute your answer. For example if I was showing how to compute net income, this would be the schedule I would make showing my labeled computations: (remember, DO NOT put labels (words) and values (numbers) in the same cell – formulas to add or subtract, etc. do not work if you do. Sales $100,000 Less: Cost of Goods Sold 60,000 Gross profit$40,000 Less operating expenses 30,000 Net income \$10,000...

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# Dendrites

We now know that the neuron cell membrane contains ion channels that allow specific ions to flow through the lipid bilayer and we treated the neuron cells as small round objects. But neuron cells are not small round objects. They contain these major areas

• Dendrites
• The Soma (cell body)
• The Axon

The Dendrites are the inputs to the neuron cell. They have extensive arborization and, in the case of the Purkinje cells, connect to hundreds of thousands of other cells. Normally they are connected to about 10,000 cells. Each connection contributes to the operation of the neuron.

This is one of the most famous images in neurobiology.

Two Purkinje neurons hand drawn by Santiago Ramon y Cajal at the turn of the 20th century

The neuron Soma is the physically largest part of the neuron. It typically has a diameter of 100 uM. The diameter of the dendrites may be down to 1 uM or less. The dendrites extend 100's of microns, but some can extend several meters, as in the giraffe brain.

Branching - Dendrites tend to bifurcate repeatedly and create (often several) large and complicated trees. Cerebellar Purkinje cells typically bear one very complicated tree with approximately 400 tips. Cells from cat spinal cord usually have 8-12 trees each with approximately 30 terminal tips.

Diameters - Dendrites are thin tubes of nerve membrane. Near the soma they start with a diameter of few um and fall below 1 um as they successively branch.

Length - Dendritic trees may range from very short (100-200 um) to quite long (1-2 mm) and the total dendritic length may reach 1 cm and more.

Area and Volume - The majority of the brain volume and area is occupied by dendrites. The surface area of a single dendritic tree is in the range of 2,000 to 750,000 Cum.

This is a capture of a detailed multi-compartmental model of a cerebellar Purkinje cell , created with GENESIS simulator. It shows an input spike coming into the dendritic tree and spreading out as it moves down to the soma (the orange ball). The output of the soma is through a single axon ( not shown).

## Compartments

A neuron and its dendrite tree is simplified for analysis as compartments:

This picture is a view of the electrical properties of a uniform section of passive dendrite having length l and diameter d.

These dendrite compartments can be thought of Leaky Electrical Cables with capacitance. A single compartment is modeled below.

The ion channels are as described 1.0 Cells and Ions. The inner conductor, the cytoplasm, is a much poorer conductor than the copper wire used in an undersea cable. It has an resistance along the length of the cable Ra, the "axial resistance". The conducting cytoplasm inside the cylinder, the insulating neural membrane, and the liquid (similar to salt water) surrounding the neuron form a capacitance Cm.

The membrane is also not a perfect insulator due to the ion-conducting channels that pass through it allowing it to leak. The "passive channels" do not vary in conductance, and the "active channels" that have conductances varying with voltage, calcium concentration, or synaptic input. The passive channels account for the membrane resistance Rm and the associated leakage current Ileak.

This sounds like a bit of poor engineering with all the leaking going on but you will see that what you think of as 'bugs' are actually features that allow our brains to work.

## Compartment Electrical Equivalent

The various components of the cell compartment can be thought of as an electrical circuit and can be analyzed as such.

This drawing allows us to think of the cell in a way that we can reconcile that the inside of the cell is negative with respect to the Outside. Imagine hooking the ground side of your DVM to the Outside of the cell and then probing around at the inside points. You would see negative voltage levels only.

Simplest cable equation is the steady state where you can ignore the current flowing into the membrane capacitance Cm. The membrane voltage is given by $V m = I m R m$ (membrane current times the membrane resistance).

## Dimensions and units of measure

The quantities Rm, Ra, Cm, Vm, in the diagram and equation are given in Kilohm, Microfarad, or millivolts, and will depend on the size of the compartment. Note the use of lower case. This designates them as actual units.

Compartments are connected to each other through their axial resistances Ra. The axial resistance of a cylindrical compartment is proportional to its length and inversely proportional to its cross-sectional area.

For a piece of dendrite or a compartment of length l and diameter d we then have:

$R m = R M π l d$

$R a = 4 l R A π d 2$

$C m = π l d C M$

Sections of dendrite that have a continuous variation of voltage along the length are replaced by a "lumped parameter model" with discrete jumps in membrane potential. By using very many short compartments, the compartmental model can approach the result of the continuous cable equation. This is done by assuming that the values are the same in all compartments, as they are intrinsic properties of the neural membrane and cytoplasm. CM depends on the intrinsic properties of the thickness and dielectric constant of the membrane, and is usually close to 0.01F/m2 (in this simulator the equivalent value of 1uF/m2 is used).

Time Constant (TM)

The time constant is a value in milliseconds that allows you to predict the percentage of the peak membrane voltage (Vm) at any point in time. The membrane time constant for a short uniform section is given by:

$T m = R m C m = R M C M$

This Dendrite simulator calculates value of the membrane voltage at every millisecond point by applying the following formula $V m = V s + ( V o − V s ) e − t / R C$

The time constant is modeled by a passive membrane resistance, Rm, a membrane capacitance, Cm, and a current source.

Where:

• Vm is the compartment membrane voltage at time t
• Vs is the height of the voltage step
• Vo is the voltage on the capacitor at the last millisecond step
• RC is the calculated time constant of the compartment

Note that the Time Constant (RC) is independent of the dimensions, because Rm is inversely proportional to the surface area, and Cm is proportional to the surface area.

The exponential rise and decay of the membrane potential as the current traverses a section of leaky cable is shown in the next figure.

If you were to increase the leakage resistance Rm the slope would become more shallow. Conversely if the axial resistance Raxial were to increase you would get a sharper, faster drop in potential.

Excited and Depressed Dendrites

Connections from other neurons Axons may be attached to any one of thousands of places along the dendritic tree. The drawing of a compartment of dendrite (above) shows two types of ion channel connections:

• EPSP, Green - If the connected ion channel stimulates the soma, it is called Excitatory Post-Synaptic Potential
• IPSP, Red - If the connection channel depresses the downstream soma from firing it is called an Inhibitory Post-Synaptic Potential .

Each of these are connected through synapse. Leak channels shown in Orange are never connected.

## IPSP Activated

The type of connection (EPSP or IPSP) is set up in the simulator when defining the input to the dendrite compartment. An EPSP connection is shown by a Black connection line. To define a synaptic connection as IPSP, click the right column next to the connection you wish to designate. The word 'yes' will appear, and the connecting line will change to red. A second click will remove the 'yes' and return the connection to EPSP.

This shows Step1 connected to an Inhibitory (IPSP) channel on Dendrite compartment 1. Step 4 is connected to an EPSP channel.

## Simulator

The best way to get a feel for the interaction of the various parameters of the dendrite is to run a simulation. Download and install the NeuronLab Simulator if you have not already (as described in section 3. NeuronLab Simulator).


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# 8.5.2 Harmonic balance for a simplified reed instrument

The simplified model of a clarinet introduced in section 8.5 is very crude, but we can learn some useful things from it by applying the method of harmonic balance, introduced in section 8.2.2. We don’t even need to assume that the instrument is specifically a clarinet to begin with: the argument we are about to give applies equally well to any kind of reed instrument which has pressure $p(t)$ and volume flow $v(t)$ connected in two ways: via the input impedance $Z(\omega)$ of the instrument tube, and via a nonlinear function $v(p)$ with the general shape shown in Fig. 1. This plot is slightly different from the version shown in section 8.5: it has been shifted sideways by the pressure $p_0$ inside the player’s mouth, so that the curve crosses the horizontal axis at $p=p_0$ (which has the particular numerical value 2.5 in the plot). Note that both $p(t)$ and $p_0$ are pressures relative to ambient atmospheric pressure.

The question of interest here is whether the model “instrument” is capable of sustaining a steady, periodic note at small amplitude. If it can, we expect the activity to be confined to a small region of the curve in Fig. 1, close to the position where it crosses the vertical axis. The curve is smooth in that region, so we can express it as a power series

$$v(p) = \sum_{n=0}^\infty{a_n p^n} \tag{1}$$

in which we will assume that the coefficients $a_n$ decrease rapidly as $n$ increases. Since both $p(t)$ and $v(t)$ are periodic, we can express them both in terms of Fourier series expansions:

$$p(t) = \sum_{j=-\infty}^\infty{P_j e^{ij \Omega t}} , \mathrm{~~~~~} v(t) = \sum_{j=-\infty}^\infty{V_j e^{ij \Omega t}} \tag{2}$$

where $\Omega$ is the (as yet unknown) fundamental frequency of the note. For our small-amplitude oscillations we will assume that the coefficients $P_j$ and $V_j$ become negligible when $j$ is not small, so that we can approximate the behaviour by keeping just a few low-order terms. The complex form of the Fourier series has been adopted here, because it makes the manipulations we are about to require very much easier. Because both $p(t)$ and $v(t)$ are real functions, the Fourier coefficients with negative indices must be simply the complex conjugates of the corresponding positive terms:

$$P_{-j} = P_j^*, \mathrm{~~~~~} V_{-j} = V_j^* . \tag{3}$$

We also know that $P_0 \approx 0$ because there can’t be a steady pressure difference between the inside and the outside of the instrument tube: it is open to the atmosphere at the bell end.

What we now need to do is substitute the Fourier series for $p$ into the power series (1), then collect up the terms corresponding to different harmonics of the frequency $\Omega$. The calculation is straightforward but messy: for example,

$$p^2=\sum_{j=-\infty}^\infty{\sum_{k=-\infty}^\infty{P_j P_k e^{i(j+k) \Omega t}}}$$

$$= P_1 P_1^* + P_2 P_2^* +\cdots$$

$$+ 2 e^{i \Omega t} \left( P_2 P_1^* +P_3 P_2^* + \cdots \right)$$

$$+ 2 e^{2i \Omega t} \left( P_1^2/2 + P_3 P_1^* + P_4 P_2^* + \cdots \right)$$

$$+ 2 e^{3i \Omega t} \left( P_1 P_2 + P_4 P_1^* +P_5 P_2^* + \cdots \right)$$

$$+ \mathrm{~higher~terms} + \mathrm{C.C.} \tag{4}$$

where “$+~$higher terms” means “plus all the terms involving higher harmonics”, and “$+~$C.C.” means “plus the complex conjugate of the whole thing” because everything is repeated with terms involving $e^{-i \Omega t}$, $e^{-2i \Omega t}$ and so on.

In a similar way,

$$p^3= 3e^{i \Omega t} \left( P_1^3 +P_1^2 P_3 + \cdots \right)$$

$$+ 3e^{2i \Omega t} \left( P_1^2 P_4 + 2 P_1^2 P_2 + \cdots \right)$$

$$+ 3e^{3i \Omega t} \left( P_1^3/3 + P_1^2 P_5 + 2P_1^2 P_3 + \cdots \right)$$

$$+ \mathrm{~higher~terms} + \mathrm{C.C.} \tag{5}$$

where this time we have added in a further simplification: we can assume that $P_1$ is not complex but real, so that we don’t have to distinguish $P_1$ from $P_1^*$. The reason we can do this is that the overall phase of our periodic oscillation is arbitrary, and we can fix it by adopting this assumption.

Now we can substitute into eq. (1), and separately equate the terms in $e^{i \Omega t}$, $e^{2i \Omega t}$ and $e^{3i \Omega t}$ (this is the “harmonic balancing” step of the method). In each of these equations, in the spirit of the approximations we are making, we will ignore small terms and just keep the dominant terms of each one. The underlying assumption used here is that the power series coefficients $a_j$ and the Fourier coefficients $P_j$, $V_j$ decay at comparable rates as $j$ increases. The first three resulting equations then read

$$V_1 \approx a_1 P_1 \tag{6}$$

$$V_2 \approx a_1 P_2 +a_2 P_1^2 \tag{7}$$

$$V_3 \approx a_1 P_3 +2 a_2 P_1 P_2 + a_3 P_1^3 . \tag{8}$$

These must be supplemented by the other relation between $v(t)$ and $p(t)$, via the input impedance of the tube. At each harmonic of our playing frequency $\Omega$, the corresponding Fourier coefficients $V_j$ and $P_j$ are related by

$$V_j = Y(j \Omega) P_j \tag{9}$$

where $Y(\omega)=1/Z(\omega)$ is the admittance of the tube. Combining with eqs. (6–8), we find

$$a_1 \approx Y(\Omega) \tag{10}$$

$$P_2 \approx \dfrac{a_2}{Y(2 \Omega) – Y(\Omega)} P_1^2 \tag{11}$$

$$P_3 \approx \left[ \dfrac{2 a_2^2}{Y(2 \Omega) – Y(\Omega)} + a_3 \right] \dfrac{P_1^3}{Y(3 \Omega) – Y(\Omega)} . \tag{12}$$

We can deduce several interesting things from these equations. Equation (10) is different from the others, because it does not involve any of the Fourier coefficients $P_j$: the term $P_1$ cancelled out at the last stage of the calculation. That is not a surprise: this equation represents the linearised approximation to the problem we are addressing, and it is normal that linear theory (for example of vibration modes) does not predict the amplitude of non-forced motion.

But eq. (10) allows us to deduce (within the limits of this approximate calculation) two important things: the playing frequency $\Omega$, and the required mouth pressure $p_0$. The power series coefficients $a_j$ are all real numbers, but the admittance function $Y(\omega)$ is usually complex. So the imaginary part of eq. (10) tells us that the playing frequency $\Omega$ must take a very special value, such that $Y(\Omega)$, and thus also the impedance $Z(\Omega)$, is real. Looking all the way back at Fig. 2 of section 2.2.7, we expect this to happen at a frequency very close to each resonance frequency of the instrument tube.

So far, so good: we expect our “instrument” to play a note close to one of the resonances. But now we need to see what the real part of eq. (10) tells us. Having fixed the value of $\Omega$, the (real) value of $Y(\Omega)$ is known: it can be measured from the input impedance of the instrument. So in order to satisfy eq. (10), the value of $a_1$ must match this value of $Y(\Omega)$. We are close to a resonance, which means that $Z(\Omega)$ is expected to be large, and therefore $Y(\Omega)$ is small: but we know that it is always a positive quantity, because it determines the energy dissipation of the acoustic duct.

So $a_1$ must have a particular small, positive value. This is achieved by changing the mouth pressure $p_0$, which has the result of sliding the curve in Fig. 1 sideways. The coefficient $a_1$ has a simple geometric interpretation: it is the slope of the tangent to the curve, at the point where it crosses the vertical axis. If the player begins by blowing very gently into the instrument, and then gradually increases the mouth pressure, we can deduce that the instrument should first “light up” and start to produce a note the first time that it is possible to satisfy eq. (10). This will occur when the tangent slope first reaches a positive value matching the value of $Y(\omega)$ at one of the frequencies where this is a real number. The smallest value of $Y$ corresponds to the largest value of $Z$, so the prediction is that the first note to play will correspond to the highest peak of the impedance $Z(\omega)$: or at least, to a frequency very close to that peak, since the vanishing of the real part of $Z$ doesn’t happen exactly at the peak value of $|Z|$. Usually, this highest peak of $Z(\omega)$ occurs close to the lowest resonance of the instrument tube.

Equations (11) and (12) now tell us the amplitudes $P_2$ and $P_3$ of the second and third harmonics, in terms of the amplitude $P_1$ of the fundamental. Of course, we could have extended the analysis to include higher harmonics if we wished. The first thing these equations reveal is that provided the function $v(p)$ really is nonlinear, so that the power series coefficients $a_2$, $a_3$ and so on are not all zero, the pressure waveform $p(t)$ inside the tube will inevitably contain harmonics.

The equations then tell us how those harmonic amplitudes are influenced by the properties of the tube. Finally, we have to decide whether we are really thinking of a clarinet, a saxophone, an oboe or whatever: these different instruments will have different impedance characteristics. The most important difference is between cylindrical instrument like the clarinet, and conical instruments like the saxophone or oboe. As we know from section 4.2, as a first approximation a conical tube will be expected to have resonances close to every harmonic, whereas a cylindrical tube only has them close to odd-numbered harmonics 1,3,5….

Equation (11) then tells us something important: for a conical instrument, we can expect $Z(\omega)$ to have peaks at $\Omega$ and at $2 \Omega$, if the note being played is based on the lowest tube resonance. This means that $Y(\Omega)$ and $Y(2\Omega)$ are both small, so that $P_2$ will be relatively large. But if we have a cylindrical instrument,$Y(\Omega)$ will be small but $Y(2\Omega)$ will be large (it is close to an antiresonance of the tube). The result is that $P_2$ will be much smaller than in the conical instruments. But for both types of instrument, $Y(3\Omega)$ is small, because both shapes of tube have a resonance near $3 \Omega$. So both types of instrument will have a significant level of third harmonic in the internal pressure spectrum, and thus in the radiated sound.

Equation (11) and (12) also illustrate a result sometimes called “Worman’s theorem” [1, see section 21.3]. The amplitude of the $j$th harmonic is, in this approximation, proportional to the $j$th power of the fundamental amplitude $P_1$. In practice, this effect comes into play as the blowing pressure is increased a little above the threshold to make the note louder. The harmonic content then grows, with a simple pattern. In decibel terms, if the fundamental increases by 1 dB, the second harmonic will increase by 2 dB, the third by 3 dB, and so on. This is likely to be perceived by a listener as an increase of brightness, accompanying the increase of loudness. That description is surely in qualitative agreement with the common experience of players on instruments like the clarinet or saxophone. Some experimental support for this pattern has been given by Benade [1, see Fig. 21.6].

Some more sophisticated applications of harmonic balance to the clarinet can be found in section 9.4 of Chaigne and Kergomard [2].

[1] Arthur H. Benade; “Fundamentals of Musical Acoustics”, Oxford University Press (1976), reprinted by Dover (1990).

[2] Antoine Chaigne and Jean Kergomard; “Acoustics of musical instruments”, Springer/ASA Press (2013).


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# Tag Info

3

If you absolutely don't want to diagonalize, you could do something like this: First convince yourself that the sequence must converge. For example, instead of having a sequence, imagine just having three numbers and successively replacing one of them with the average of all. If you keep the details straight, you can then see that the maximum of the numbers ...

2

The distribution for the number of time steps to move between marked states in a discrete time Markov chain is the discrete phase-type distribution. You made a mistake in reorganising the row and column vectors and your transient matrix should be $$\mathbf{Q}= \begin{bmatrix} \frac{2}{3} & \frac{1}{3} & 0 \\ \frac{2}{3} & 0 & ... 2 It appears that the transition semigroup need not be Feller. An example: Denote the non-negative integers by \mathbb{N}. Define, for i,j\in\mathbb{N},$$q(i, j) =\begin{cases} 0 &\mbox{ if }i =0\\ i^2(\delta_{i-1, j}-\delta_{i,j}) &\mbox{ otherwise.}\end{cases}$$Then the backward equations are$$p_t'(0,0) = 0$$and$$p_t'(i, i-1) = ...

2

As you noted, away from zero the bias is uniformly negative. This suffices to guarantee recurrence, that is, that the process hits zero with full probability. In a nutshell, choose $N\gt2(k+1)c$, and replace the dynamics on $b\geqslant N$ by $b\to b+k$ with probability $c/N$ and $b\to b-1$ with probability $1-c/N$. Starting from any $b(0)\geqslant N$, this ...

2

It is a theorem of Wielandt that the number you are looking for is $(r-1)^2+1$. A way to achieve this is to have $p_{12},p_{23},p_{34},\dots,p_{k-1,k},p_{k,1},p_{k,2}$ positive and all other $p_{ij}$ zero. A reference is H Wielandt, Unzerlegbare, nicht negative Matrizen, Math Z. 52 (1950) 642-648. Perhaps a better reference is Hans Schneider, Wielandt’s ...

2

You have to solve the following recursive relations. Let $h(k)$ be the expected number of steps until your reach the (absorbing) state $4$ when you are in state $k$, for $k=1,2,3,4$. So we have that $$h(4)=0$$ because when you are already in $4$ you need zero steps to reach $4$. Then for $k=3$ $$h(3)=1+0.75h(3)+0.25h(4)$$ because when you are in state 3 you ...

1

We show how to find an answer using not much machinery. Matrices, or recurrences, would make it easier. At any time, we can be in one of two states, $A$ or $B$, where State $A$ means we are at vertex $1$, and State $B$ means we are at one of the $3$ other vertices. At time $0$ we are in State $A$. If at time $n$ we are in State $A$, then with probability ...

1

For example, the entry in row "1", column "3" is the probability of going from "1" to "3" which is 0.4. We can't decrease $X_n$ so some values are 0. Each row must add up to 1. $$\begin{matrix} & 0 & 1 & 2 & 3\\ 0 & ? & ? & ? & 0.4\\ 1 & 0 & ? & ? & 0.4\\ 2 & 0 & ... 1 A little bit more details. What is the probability that in a time interval (0,t) exactly one person will have a success. Well first pick the person (say P1) that it will be call their child P2 and everyone else E, and condition on exactly when the success happens. Call that pdf of the first birth time of P1 p_1(s).$$P(\text{P1 has 1 birth & P2 and ...

1

Of course. But the $C_i$'s you get can be split again in stable subsets. The point here is that for every $i$, $P^d$ restrained to $C_i$ is aperiodic when $d$ is the period of the chain. In other words, $C_i$ can't be split in several subsets stable under a power of $P$, giving nice properties.

1

The idea is that $X_k$, whatever it is, has to be something, so you sum over it: $$P(X_{k+1} = x_{k+1} \mid X_0 = a_0, \dots, X_{k-1} = a_{k-1})$$ $$= \sum_{x_k} P(X_{k+1} = x_{k+1}, X_k=x_k \mid X_0 = a_0, \dots, X_{k-1} = a_{k-1} )$$ $$= \sum_{x_k} P(X_{k+1} = x_{k+1}\mid X_k=x_k, X_0 = a_0, \dots, X_{k-1} = a_{k-1} )\cdot P( X_k=x_k \mid X_0 = a_0, ... 1 For a counterexample consider the Markov chain \{X_n\} described here by Did. Assume everything to be the same as in the example, but instead of states 0, 1 and 2 consider the states -1, 1 and 2. Now the function g(x)=x^2 lumps together states -1 and 1. That is, we have that g(-1)=g(1)=1 and g(2)=4, exactly as described in the ... 1 Firstly separate the states in communication classes. There are two communication classes which are determined as follows. Start from state 0. Which states can you visit and then return to state 0? You can make following transitions$$0\xrightarrow{0.5} 2 \xrightarrow{0.9} 0$$So 0 and 2 communicate, which means that they belong to the same ... 1 There are some common (and very useful) results concerning recurrence and transitivity of states which you might know and/or could use: All states in an irreducible set are either all recurrent or all transient. State i is transient if and only if $$\sum_{n=1}^{\infty}p_{ii}^{(n)} < \infty$$ State i is recurrent if and ... 1 There is a simple algorithm: let P the transition matrix. let A=I + P, j = 1 replace each non 0 entry by 1 replace A by A^2, j by 2j Go back to 2. until j\ge n Then the graph is connected iff every entry of A is 1. At each iteration, at the end of step 4. the non zero entries of A are the state you can go to with \le j ... 1 To show that M is not Markov, one can consider two different paths of M between the times 0 and 4: If (M_n)_{0\leqslant n\leqslant4}=(0,1,1,1,2), then S_2=0 hence S_3 is conditionally uniformly distributed on \{-1,1\} and the last step 1\to2 has conditional probability \frac12\cdot P(X_4=1)=\frac14. If (M_n)_{0\leqslant ... 1 The grinder can be in two states -- either in use, or not in use. The period of being in use lasts for time 1/\mu per minute on average, so the rate here is \mu. When the grinder is not in use the operator arrives at rate 0.5 per minute (1 per 2 minutes). Therefore we have a continuous time Markov chain on the state space {in use, not in use} with ... 1 It this case, if for example m_1\neq m_2, \sigma_1\sigma_2\rho\neq 0, no, because the drift is not a function of X_t + Y_t. In a general situation, when X,Y are two non independant Markov chains, X+Y is not. To understand this with a more simple example, in a discret time setting, take for example (X_n) be a simple random walk on \mathbb Z, and ... 1 Not always. For a slightly degenerate example, assume that X, Y_1 and Y_2 take values in \{+,-\} with Y_1=Y_2=+ if X=+ and Y_1=+ and Y_2=- if X=-. The paths of the triple (X,Y_1,Y_2) are (+,+,+) and (-,+,-) hence the distribution of Y_2 conditionally on (X,Y_1) depends on X, that is, (X,Y_1,Y_2) cannot be a Markov chain. For ... 1 Usually, my approach to this kind of question is to solve a very simple recurrence. Just look to  E_{0} N(3) but conditioned in each of the two possible first steps to get (I will write N instead of N(3)$$E_0 N = \frac{2}{3}E_0N + \frac{1}{3}E_1N +1 The term +1 shows up because once you had walked one step you have to add this step to the ...

1

To give the answer you quote, you need a stronger version of (3) so that the states at each time point are independent of each other. Hints: What is the probability that at any particular time point the object has a partiuclar state $x$? What is the probability that at any particular time point the object does not have state $x$? What is the probability ...

1

You give four equations for three unknowns. Fortunately, there is a point of major interest : substracting the first equation from the third equation gives the second equation. Then, either the first or the third equation must be discarded. So we have now three linear equations for three unknowns. Suppose that we discard the first equation. Extract \$x = 5 ...

Only top voted, non community-wiki answers of a minimum length are eligible


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# (DC motor derive ω expression with respect time

• Engineering

## Homework Statement

I try to derive the expression of ω with respect to t,
but the problem is the armature current Ia or the flux is unknown. I dont know how to solve it.
Anyone could give me ideas? Thanks a lot

## Homework Equations

τ=J(ω/t)
Vs=IaRa+Ea
T=KEøIa
Ea=KEøω

#### Attachments

• 22.6 KB Views: 329

Related Engineering and Comp Sci Homework Help News on Phys.org
Simon Bridge
Homework Helper
Either (a) you are mistaken, and there is some way to find Ia, or (b) you can make a substitution for Ia so the term no longer appears in the equation - like you do with simultaneous equations.

1 person
Either (a) you are mistaken, and there is some way to find Ia, or (b) you can make a substitution for Ia so the term no longer appears in the equation - like you do with simultaneous equations.
I did in the following steps:
1.J(ω/t)=KøIa;
2. and then substitute Ia from Vs=IaRa+Ea into 1;
finally, i got:
ω=Vs/(Kø+JRa/Køt)

But it turns out the term ø cannot be eliminated and it is unknown.
Where am i wrong?

Simon Bridge
Homework Helper
But it turns out the term ø cannot be eliminated and it is unknown.
Really?
What physical property does "ø" represent?

τ=J(ω/t)
... isn't that supposed to be: $$\tau=J\frac{d\omega}{dt}$$...??

ω=Vs/(Kø+JRa/Køt)
how did you get this from what went before?

The fact they give you a bunch of initial values suggests to me you should look for a DE.

Last edited:
ø is the magnetic flux.
i have double checked for it, the deriving process should be right.
and it seems reasonable if t approaches infinity
Vs=Køω quite similar to Ea=Køω

Simon Bridge


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# Silent Mode

## Introduction

A silent mode installation installs Altair Flux & FluxMotor 2022 without requiring any user interaction. The installer makes use of a response file that contains the installation options to run the installation from start to end without any user input.

## Silent Mode

1. Create a response file. Run the installer in interactive mode with the r option to save the installation properties to a response file.
[INSTALLER_NAME] -r "[RESPONSE_PATH]\installer.properties"
2. Trigger a silent mode installation from the command line using one of the following options:
• Use the default property values as provided by the installer package.
[INSTALLER_NAME] -i silent -DACCEPT_EULA=YES
Important:
• Setting -DACCEPT_EULA=YES means that you accept the terms of the License Agreement and allows you to proceed with Altair Flux & FluxMotor installation
• If nothing is specified for ACCEPT_EULA variable or if -DACCEPT_EULA=NO then Altair Flux & FluxMotor will not be installed
• Specify properties.
[INSTALLER_NAME] -i silent -D[Property]=[VALUE]
• For example:

-DACCEPT_EULA=YES -DUSER_INSTALL_DIR=C:\\Program Files\\Altair\\2022
• Use a response file containing properties.
[INSTALLER_NAME] -i silent -f "[RESPONSE_PATH]\installer.properties"
Note:
• [INSTALLER_NAME] is the installer binary.
• [RESPONSE_PATH] is the path where the response file resides.
• Use quotes around directory and pathnames that contain spaces.
• Do not use spaces between VARIABLE=VALUE statements in the response file.


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Lecture 20 - Chapter 2: Manufacturing

edited May 2018

Last time I rapidly sketched some applications of resource theories to chemistry and scheduling. If you're interested we can dig into these more deeply once we've learned more of the math in Chapter 2. For now let's look at one more application:

Manufacturing. Companies often have to solve puzzles like this, though usually much harder:

Puzzle 58. You are making two kinds of computers: laptops and desktops. Each kind of computer needs 1 processing chip. A laptop also needs 1 memory chip, while a desktop needs 2 memory chips. You have 10,000 processing chips and 15,000 memory chips. Each laptop you make will give $750 of profit, while each desktop will give$1000 of profit. What is the most profit you can make?

Give it a try!

Of course this problem is simplified in many ways. We can make it a tiny bit more realistic by recognizing that it also takes time to make computers:

Puzzle 59. Everything in Puzzle 58 still holds, but now your workers have only 25,000 minutes to make all the computers. It takes 4 minutes to make each laptop, and 3 minutes to make each desktop. Now what is the most profit you can make?

These puzzles secretly involve resource theories. The resources you're starting with are:

1. processing chips
2. memory chips
3. time

You are trying to transform these into some other resources:

1. laptops
2. desktops

and then transform these into yet another resource:

1. profit

So, we can write some "reactions", as we did last time for chemistry:

$$\textrm{[processing chip]} + \textrm{[memory chip]} + 4 \textrm{[minute]} \to \textrm{[laptop]}$$ $$\textrm{[processing chip]} + 2 \textrm{[memory chip]} + 3 \textrm{[minute]} \to \textrm{[desktop]}$$ $$\textrm{[laptop]} \to 750\textrm{[profit]}$$ $$\textrm{[desktop]} \to 1000 \textrm{[profit]}$$ Then we can ask what profits are reachable if you start with

$$10000 \textrm{[processing chip]} + 15000 \textrm{[memory chip]} + 25000 \textrm{[minute]} .$$ This shows that the second puzzle is a special case of the "reachability problem" that I described in Lecture 19. So is the first puzzle.

However, you don't need to know anything about the reachability problem to solve these puzzles! Ordinary math and some careful thought are enough.

I won't do these puzzles for you, but here's a hint. Let $$x$$ be the number of laptops you make, and let $$y$$ be the number of desktops. You are trying to maximize some function of $$x$$ and $$y$$, namely your profit. However, there are some constraints on $$x$$ and $$y$$. So, draw the region in the $$x y$$ plane that's allowed by these constraints, and use that to help you see which choice of $$x$$ and $$y$$ maximizes the profit.

If you do this, you'll have reinvented the basic idea of "linear programming". (If you already know about linear programming, or these puzzles seem very easy, please let other students post their answers first: it's great for people to invent math on their own.)

Why is this idea called "linear programming"? It's because your profit is a linear function of $$x$$ and $$y$$, and all your constraints are equations or inequalities involving linear functions of these variables. This makes things easier.

Many industries use linear programming to solve problems involving enormous numbers of variables. Linear programming is good for many things, including routing, scheduling, and assignment problems.

Because of this, there's been an enormous push to improve algorithms for linear programming. Some of the basic ideas go back to Fourier: in 1827 he figured out how to eliminate variables from a system of linear inequalities. But linear programming really took off in 1939 when the economist Leonid Kantorovich developed it as a method to reduce costs for the Soviet army and increase losses incurred to the enemy in World War II. About the same time, the American economist T. C. Koopmans figured out how to formulate many economics problems as linear programming problems. Kantorovich and Koopmans later shared the Nobel prize in economics for this work.

One of the most important developments came around 1947 when George Dantzig, working for the US Air Force, was trying to find the best assignment of 70 people to 70 jobs. To do this he reinvented linear programming and invented the first efficient algorithm to solve these problems: the simplex algorithm.

The simplex algorithm runs in polynomial time on average, but in the worst case it can take exponential time. The first polynomial-time algorithm for linear programming was found by Leonid Khachiyan in 1979: it's called the ellipsoid algorithm. But this method is very slow in practice. A better polynomial-time algorithm is Karmarkar's algorithm, discovered around 1984. AT&T tried to patent this algorithm, and that led to a big legal fight about whether math can be patented!

Given all this hard work, it's unlikely that resource theories will help anyone with linear programming... at least in the short run. It's more likely that seeing linear programming as part of the subject of resource theories will give us new ideas, by making new connections apparent. That's how category theory often works.

There are also applications of resource theories to other aspects of economics - as well as thermodynamics and quantum computation. But all these take more work to explain, so next time I'll just dive into the math of resource theories. If you want to get ready, read Section 2.2.1 of Seven Sketches. I'm eager to get started!

To read other lectures go here.

• Options
1.

Puzzle 59 I get a maximum $7,750,000 at 1000 Laptops and 7000 Desktops. Here is the Geogebra file I used to get my answer. Comment Source:**Puzzle 59** I get a maximum$7,750,000 at 1000 Laptops and 7000 Desktops. Here is the [Geogebra file](https://ggbm.at/qdcJNnFd) I used to get my answer.
• Options
2.

@Michael Hong My toy code agrees :).

Comment Source:@Michael Hong My toy code agrees :).
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3.

Michael Hong - wow, that's great!

Can someone create a nice picture emphasizing the allowed region in the $$xy$$ plane and level curves (really lines) of the profit function $$p(x,y) = 750 x + 1000 y$$? This will make it really clear what's going on.

How many of the people reading this have already studied problems of this sort? I never took a course on linear programming myself.

Comment Source:Michael Hong - wow, that's great! Can someone create a nice picture emphasizing the allowed region in the \$$xy\$$ plane and level curves (really lines) of the profit function \$$p(x,y) = 750 x + 1000 y\$$? This will make it really clear what's going on. How many of the people reading this have already studied problems of this sort? I never took a course on linear programming myself.
• Options
4.

I think all economists have been exposed to linear and nonlinear programming, though possibly only at a theoretical (as opposed to computational) level.

Comment Source:I think all economists have been exposed to linear and nonlinear programming, though possibly only at a theoretical (as opposed to computational) level.
• Options
5.

An introductory operations research course was required as part of my CS degree, I'm trying to remember all it covered but I do know it covered linear programming.

Comment Source:An introductory operations research course was required as part of my CS degree, I'm trying to remember all it covered but I do know it covered linear programming.
• Options
6.
edited May 2018

I have been taught how to solve these kind of linear programming problems in CS class. It can be solved with the simplex method which is about as simple as it sounds.

It would be delightful to reproduce a beginner's introduction, but I can't do it single handed (on a mobile phone) shortly before I fall asleep.

Looking at the respective Wikipedia page I get the feeling that it's not much fun for people who haven't done this before:

https://en.wikipedia.org/wiki/Simplex_algorithm

The first 4 equations specify profit slopes (gradients) in 3-space. (Or maybe imagine families of planes intersecting the axies at...)

The available resource spec is practically a (we may use) less-than (these resources) relation, or a plane dividing our 3-space into two half-spaces: one where the allowed values live, and the other stuff we're not interested in. That's our simplex.

Edit: It might be a better idea to think of it as three planes orthogonal to the axies, and of the gradients as vectors. Sorry, some assembly may be required, it's been a while.

We have to find the maximal vector a•laptops + b•desktops smaller than that plane.

Since it's an integer mixed linear programming problem, we may have to try a few combinations (a,b), and for this there's a closely related technique:

https://en.wikipedia.org/wiki/Cutting-plane_method

A page which might be slightly more fun.

Comment Source:I have been taught how to solve these kind of linear programming problems in CS class. It can be solved with the simplex method which is about as simple as it sounds. It would be delightful to reproduce a beginner's introduction, but I can't do it single handed (on a mobile phone) shortly before I fall asleep. Looking at the respective Wikipedia page I get the feeling that it's not much fun for people who haven't done this before: https://en.wikipedia.org/wiki/Simplex_algorithm The first 4 equations specify profit slopes (gradients) in 3-space. (Or maybe imagine families of planes intersecting the axies at...) The available resource spec is practically a (we may use) less-than (these resources) relation, or a plane dividing our 3-space into two half-spaces: one where the allowed values live, and the other stuff we're not interested in. That's our simplex. Edit: It might be a better idea to think of it as three planes orthogonal to the axies, and of the gradients as vectors. Sorry, some assembly may be required, it's been a while. We have to find the maximal vector a•laptops + b•desktops smaller than that plane. Since it's an _integer_ mixed linear programming problem, we may have to try a few combinations (a,b), and for this there's a closely related technique: https://en.wikipedia.org/wiki/Cutting-plane_method A page which might be slightly more fun.
• Options
7.
edited May 2018

Here's my first-cut, simplest, simplex haskell code:

https://github.com/jimstutt/Simplex/blob/master/src/Simplex.hs

Comment Source:Here's my first-cut, simplest, simplex haskell code: https://github.com/jimstutt/Simplex/blob/master/src/Simplex.hs
• Options
8.
edited May 2018

I made a fun plot of the level curves using Python in a Google Colab Notebook. It was neat to see the bounding lines representing the constraints from the number of processors, memory chips, and minutes. Based on their slopes, I think the bounding lines from left to right correspond to memory chips, processors, and minutes.

Comment Source:I made a fun plot of the level curves using Python in a Google Colab Notebook. It was neat to see the bounding lines representing the constraints from the number of processors, memory chips, and minutes. Based on their slopes, I think the bounding lines from left to right correspond to memory chips, processors, and minutes.
• Options
9.
edited May 2018

John wrote:

How many of the people reading this have already studied problems of this sort?

Systems Biologists studying metabolic networks use exactly this procedure for constraint-based modeling they call flux balance analysis (FBA).

The first step is to write the stoichiometric coefficients (negative for reactants, positive for products) in a matrix format, where each row represents a reaction and each column represents a compound. For our above example this matrix would thus look like this:

$$\begin{bmatrix} p.chip & m.chip & minute & laptop & desktop & profit \\ -1 & -1 & -4 & 1 & 0 & 0 \\ -1 & -1 & -3 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 & 0 & 750 \\ 0 & 0 & 0 & 0 & -1 & 1000 \end{bmatrix}$$ One then computes the constraint space, which corresponds to a polygonal shape in n-dimensional space (where n is the number of compounds) bounded by the linear equations from the matrix. When computed using all reactions in an organism's metabolism, this constraint space corresponds to all metabolic states an organism can be in that don't violate the conservation of mass.

Since this system of equations underdetermined in practice (i.e all points within the boundaries are valid solutions), one needs a further 'objective' function to obtain a unique solution. This objective function is generally a maximization or minimization performed on one or more of the variables. Geometrically, the solution will be located on a vertex, edge, plane, ect. of the constraint space depending on how determined it is. In economics, this usually means maximizing profit (for our example $$z=750 \text{ laptop} + 1000 \text{ desktop}$$). In the case of biological organisms, what objective function should be used is less obvious and usually depends on the situation being studied (e.g an organism with sufficient nutrients may try maximizing biomass production, while a starving organism may not).

Comment Source:John wrote: > How many of the people reading this have already studied problems of this sort? Systems Biologists studying metabolic networks use exactly this procedure for constraint-based modeling they call flux balance analysis (FBA). The first step is to write the stoichiometric coefficients (negative for reactants, positive for products) in a matrix format, where each row represents a reaction and each column represents a compound. For our above example this matrix would thus look like this: $$\begin{bmatrix} p.chip & m.chip & minute & laptop & desktop & profit \\\ -1 & -1 & -4 & 1 & 0 & 0 \\\ -1 & -1 & -3 & 0 & 1 & 0 \\\ 0 & 0 & 0 & -1 & 0 & 750 \\\ 0 & 0 & 0 & 0 & -1 & 1000 \end{bmatrix}$$ One then computes the constraint space, which corresponds to a polygonal shape in n-dimensional space (where n is the number of compounds) bounded by the linear equations from the matrix. When computed using all reactions in an organism's metabolism, this constraint space corresponds to all metabolic states an organism can be in that don't violate the conservation of mass. Since this system of equations underdetermined in practice (i.e all points within the boundaries are valid solutions), one needs a further 'objective' function to obtain a unique solution. This objective function is generally a maximization or minimization performed on one or more of the variables. Geometrically, the solution will be located on a vertex, edge, plane, ect. of the constraint space depending on how determined it is. In economics, this usually means maximizing profit (for our example \$$z=750 \text{ laptop} + 1000 \text{ desktop} \$$). In the case of biological organisms, what objective function should be used is less obvious and usually depends on the situation being studied (e.g an organism with sufficient nutrients may try maximizing biomass production, while a starving organism may not).
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10.
edited May 2018

Sophie wrote in :

I made a fun plot of the level curves using Python in a Google Colab Notebook.

That's really cool!

John wrote:

Can someone create a nice picture emphasizing the allowed region in the xy plane and level curves (really lines) of the profit function p(x,y)=750x+1000y? This will make it really clear what's going on.

Here's a picture of the constraints acting on production assuming $$10000 \textrm{[processing chip]} + 15000 \textrm{[memory chip]} + 25000 \textrm{[minute]}$$ :

Red: p.chips, Blue: m.chips, Green: minutes, Green shading: Allowed area

I can't think of a nice way to include the profit in this graph without adding another axis. Maybe someone else has a good idea.

Edit: Just noticed that Sophie's heat map nicely solves this.

Comment Source:Sophie wrote in : > I made a fun plot of the level curves using Python in a Google Colab Notebook. That's really cool! John wrote: > Can someone create a nice picture emphasizing the allowed region in the xy plane and level curves (really lines) of the profit function p(x,y)=750x+1000y? This will make it really clear what's going on. Here's a picture of the constraints acting on production assuming \$$10000 \textrm{[processing chip]} + 15000 \textrm{[memory chip]} + 25000 \textrm{[minute]} \$$ : ![](https://docs.google.com/uc?id=1RuVe6XDAM3uLbDQQIKaI6lPU0BZArvME) Red: p.chips, Blue: m.chips, Green: minutes, Green shading: Allowed area I can't think of a nice way to include the profit in this graph without adding another axis. Maybe someone else has a good idea. Edit: Just noticed that Sophie's heat map nicely solves this.
• Options
11.

I made a fun plot of the level curves using Python in a Google Colab Notebook. It was neat to see the bounding lines representing the constraints from the number of processors, memory chips, and minutes. Based on their slopes, I think the bounding lines from left to right correspond to memory chips, processors, and minutes.

This is amazing!

Comment Source:> I made a fun plot of the level curves using Python in a [Google Colab Notebook](https://colab.research.google.com/drive/1gECqYfziXUaI26GFXsFwk8GEHBLVLj7N#scrollTo=2E3zAKbvz8aO). It was neat to see the bounding lines representing the constraints from the number of processors, memory chips, and minutes. Based on their slopes, I think the bounding lines from left to right correspond to memory chips, processors, and minutes. This is amazing!
• Options
12.
edited May 2018

Thanks, Marius and Sophie! This is great! This is exactly what I'm hoping for: people jumping in and creating nice pictures and software to add to my verbal explanations.

Marius seems to have chosen desktops to be the $$x$$ coordinate and laptops to be $$y$$, opposite from what Sophie and I are doing. I mention this in case anyone is puzzled.

Comment Source:Thanks, Marius and Sophie! This is great! This is exactly what I'm hoping for: people jumping in and creating nice pictures and software to add to my verbal explanations. Marius seems to have chosen desktops to be the \$$x\$$ coordinate and laptops to be \$$y\$$, opposite from what Sophie and I are doing. I mention this in case anyone is puzzled.
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13.

Marius wrote:

I can't think of a nice way to include the profit in this graph without adding another axis. Maybe someone else has a good idea.

One way is to include the level curves of profit as faint lines of the form

$$750 x + 1000 y = p$$ for various values of the number $$p$$. The line for maximum profit will go through the upper right corner of the shape you drew in green.

Comment Source:Marius wrote: > I can't think of a nice way to include the profit in this graph without adding another axis. Maybe someone else has a good idea. One way is to include the level curves of profit as faint lines of the form $750 x + 1000 y = p$ for various values of the number \$$p\$$. The line for maximum profit will go through the upper right corner of the shape you drew in green.
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14.
edited May 2018

This topic resonates with this, for me: la.mathworks.com/help/fuzzy/what-is-sugeno-type-fuzzy-inference.html

Thanks for all for the interesting discussion

Comment Source:This topic resonates with this, for me: la.mathworks.com/help/fuzzy/what-is-sugeno-type-fuzzy-inference.html Thanks for all for the interesting discussion
• Options
15.

Yep I confirm in CS I was taugh to run Simplex by pen and pencil

Comment Source:Yep I confirm in CS I was taugh to run Simplex by pen and pencil


text:
# Get the progress indicator from the package inside the panel?

Posted 4 years ago
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 I have an application that I would like to keep track of the calculation for the user to see that calculation is being done (specially for deploying CDF). I can use the ProgressIndicator with ParallelTable to do that. My code will be in a package and I will use an user interface for the calculation. In current version the status bar appear outside the Manipulate box (or Panel) , I am trying to put that inside the box.For that I tried to transfer the tracking variable outside the ParallelTable and user progressindicator there but I was not successful. I want to do this because in a cdf file the bar does not disappear after the computation. This is a just a "toy" calculation to mimic what I need to do. Get[FileNameJoin[{NotebookDirectory[EvaluationNotebook[]], "DummyPackage.m"}]]; Panel[DynamicModule[{np = 100, nd = 30, Res = 1.0}, Column[{Style["Random", "Function"] , Row[{"Digits ", InputField[Dynamic[nd]]}, Spacer[5]], Row[{"Precision ", InputField[Dynamic[np]]}, Spacer[5]], Button[Style["Calculate", Green, Bold], Res = DummyPackagemyDummyF[nd, np], Method -> "Queued"], Row[{Style["Result ", 12, Red], InputField[Dynamic[Res], FieldSize -> 30, Enabled -> False]}, Spacer[15]], Button[Style["Save", Green, Bold], NotebookSave[EvaluationNotebook[], Interactive -> True]] }], Initialization :> ( Get[FileNameJoin[{NotebookDirectory[EvaluationNotebook[]], "DummyPackage.m"}]];)]] and the package is BeginPackage["DummyPackage"] myDummyF[nd_,np_]:=Module[{mat,mat2,aI,dl,d=0}, ns=200; mat=SetAccuracy[RandomReal[{-100,100},ns],nd]; mat2=SetAccuracy[RandomReal[{-200,200},ns],nd]; aI=ConstantArray[0,np]; (*SetSharedVariable[aI,d];*) SetSharedVariable[d]; ParallelEvaluate[dl = 0]; Time=AbsoluteTiming[ Monitor[aI=ParallelTable[If[Mod[dl++, Ceiling[nd/75]] == 0, d += Ceiling[nd/75]]; If[iC==2,Print[" ParallelDo Loop"]]; aI=SetAccuracy[mat.(mat2/iC^0.5),nd] ,{iC,np}];, ProgressIndicator[d/np]]]; Print["Do loop time = ",Time[[1]],", np = ",np,", nd = ",nd]; aInum=Apply[Plus,(aI/np)]; Print["Total = ",aInum]; Return[aInum]]; SetAttributes[{myDummyF},{ReadProtected,Protected,Locked}]; EndPackage[];


text:
\documentclass[a4paper,12pt,twoside]{report}
\usepackage[left=2cm,right=2cm,top=2cm,bottom=3cm]{geometry}

\usepackage[super,comma,numbers,sort&compress]{natbib} %chapter cite + [] part gives 1-3 style instead of 1,2,3.
\usepackage[sectionbib]{chapterbib} % chapter cite
\usepackage{url} % web-citations anable (from bibtex document)
\usepackage{hyperref} %cross-ref

\usepackage{setspace} %line spacing package

\usepackage{graphicx} %pictures package
%\usepackage{subfig} %pic captions etc
\usepackage{subcaption}

\usepackage[tiny,md,sc]{titlesec} %title package

\usepackage{amsmath} %equations and formulas
\usepackage{wasysym} %math symbols
\usepackage{tabularx} %tables

\usepackage{lscape} %page orientation

\usepackage{longtable}
\usepackage{array}
\usepackage{ltxtable}
\usepackage{siunitx}

\usepackage{tabu}
\usepackage{multirow}
\usepackage{float}

\usepackage{pgfplots}
\pgfplotsset{width=10cm,compat=1.9}

\usepackage[labelfont={bf,sf,footnotesize,singlespacing},
textfont={sf,footnotesize,singlespacing},
justification={justified,RaggedRight},
singlelinecheck=false,
margin=0pt,
figurewithin=chapter,
tablewithin=chapter]{caption}

\usepackage[titletoc]{appendix}

\tabulinesep = 1.5mm

%\titleformat{\chapter}[display]
%{\normalfont\huge\bfseries}{\chaptertitlename\ \thechapter}{20pt}{\Huge}
%\titlespacing*{\chapter}{0pt}{0pt}{40pt}

\setcounter{secnumdepth}{4}

\captionsetup[table]{
labelsep = newline,
justification=justified,
singlelinecheck=false,%%%%%%% a single line is centered by default
labelsep=colon,%%%%%%
skip = \medskipamount}

\newcommand*\chem[1]{\ensuremath{\mathrm{#1}}} %command to do lower case index (i.e. in H2O 2 will be lower than H and O)
\newcommand{\sectionbreak}{\clearpage}
\newcommand\Tstrut{\rule{0pt}{2.6ex}}
\newcommand\Bstrut{\rule[-0.9ex]{0pt}{0pt}}

\sisetup{detect-shape, detect-weight}

\begin{document}
\include{titlepage}
\include{declaration}
\include{acknowledgements}
\include{abstract}
\include{nomenclature}
\tableofcontents
\listoffigures
\listoftables
\include{introduction}
\include{chapter1}
\include{chapter2}
\include{chapter3}
\include{chapter4}
\include{chapter5}
\include{chapter6}

\appendix
\include{appendix3}
\include{appendix}
\include{appendix2}

\end{document}

Nomenclature and Introduction are introduced using \chapter*{}. In my table of contents I obviously want Introduction to come before Chapter 1, but this is what I have:

The document compiles with no errors, and I do have all necessary packages. I don't have a clue what might be wrong so would appreciate any help. Thank you!

EDIT - I've updated the code to contain full information.

• Well, if I understand correctly, what you want to do is make the chapters before 1 appear without numbering. If you use the book class, your MWE is incomplete because you don't give us a preamble; you can use the following comands to do it without complications: \frontmatter and \mainmatter. – Aradnix Sep 10 '17 at 0:56
• Sorry, I've fixed the preamble. I'm using "report" class. – Iuliia Elizarova Sep 10 '17 at 1:12
• Welcome! In the file you include, Introduction.tex or whatever, write \chapter*{Introduction}\addcontentsline{toc}{chapter}{Introduction} so that the contents line is added on the first page of the introduction and not the first page of the next chapter, as it is at the moment. – cfr Sep 10 '17 at 1:12
• Off-topic: you are loading hyperref far too early. It likes to be loaded late - last unless you're using something specifically requiring otherwise, such as cleveref. Also off-topic: tabu: rather you than me. I would not touch this with a barge pole. – cfr Sep 10 '17 at 1:14
• @cfr I've deleted \addcontentsline{toc}{chapter}{Introduction} from preamble and added it to the chapter itself and it worked! Thank you very much! – Iuliia Elizarova Sep 10 '17 at 1:17

In the file you include, introduction.tex write

\chapter*{Introduction}

so that the contents line is added on the first page of the introduction and not the first page of the next chapter, as it is at the moment. Delete if from you main file to avoid its being added twice, obviously.

When you say \include the file is read in and page breaks are added before anything else in the main file is read. So that's all done, then the \addcontentsline in the main file gets read, as the next chapter is about to be \included. So even if the order was correct, you'd get the wrong page number in the contents for the Introduction.

I think you can solve it easily in this way:

\documentclass[a4paper,12pt,openany]{book} %instead of report and the same result

\begin{document}
\frontmatter
\include{titlepage}
\include{declaration}
\include{acknowledgements}
\include{abstract}
\include{nomenclature}
\tableofcontents
\listoffigures
\listoftables
\include{introduction}
\mainmatter
\include{chapter1}
\include{chapter2}
\include{chapter3}
\include{chapter4}
\include{chapter5}
\include{chapter6}

\appendix
\backmatter
\include{appendix3}
\include{appendix}
\include{appendix2}

\end{document}

With the book class and similar (e.g. memoir, scrbook, etc.) you can make use of the commands \frontmatter, \mainmatter, and \backmatter.

What you achieve with this trick is that everything that appears before your first chapter (or the part of the text you consider pertinent) is unnumbered without needing to add asterisks, but is correctly included in your table of contents, with small roman page numbering.

For the main content of the book use \mainmatter and get the default numbering and also you can add ' \backmatter for the final part.

• Feel free to roll back my edit, but I don't think you said what you meant as the chapters aren't numbered in Roman - rather, the chapters are unnumbered and the page numbering is small Roman. – cfr Sep 10 '17 at 1:39
• Of course not, but it's an additional piece of information, the trick of my answer is to be able to add the front matter without altering the numbering of the chapters. – Aradnix Sep 10 '17 at 1:44
• I was just explaining why I edited. I knew what you meant, but somebody not familiar with this probably wouldn't. That was all. As I say, you can roll-back if you disagree. – cfr Sep 10 '17 at 1:49
• I see... I won't roll-back, I always learn something here. For me it's fine. – Aradnix Sep 10 '17 at 5:51


text:
# Basic Statistics and Data Analysis

## Binomial Random number Generation in R

We will learn here how to generate Bernoulli or Binomial distribution in R with example of flip of a coin. This tutorial is based on how to generate random numbers according to different statistical distributions in R. Our focus is in binomial random number generation in R.

We know that in Bernoulli distribution, either something will happen or not such as coin flip has to outcomes head or tail (either head will occur or head will not occur i.e. tail will occur). For unbiased coin there will be 50%  chances that head or tail will occur in the long run. To generate a random number that are binomial in R, use rbinom(n, size,prob) command.

rbinom(n, size, prob) command has three parameters, namely

where
n is number of observations
size is number of trials (it may be zero or more)
prob is probability of success on each trial for example 1/2

Some Examples

• One coin is tossed 10 times with probability of success=0.5
coin will be fair (unbiased coin as p=1/2)
>rbinom(n=10, size=1, prob=1/2)
OUTPUT: 1 1 0 0 1 1 1 1 0 1
• Two coins are tossed 10 times with probability of success=0.5
• > rbinom(n=10, size=2, prob=1/2)
OUTPUT: 2 1 2 1 2 0 1 0 0 1
• One coin is tossed one hundred thousand times with probability of success=0.5
> rbinom(n=100,000, size=1, prob=1/2)
• store simulation results in $x$ vector
> x<- rbinom(n=100,000, size=5, prob=1/2)
count 1’s in x vector
> sum(x)
find the frequency distribution
> table(x)
creates a frequency distribution table with frequency
> t=(table(x)/n *100)}
plot frequency distribution table
>plot(table(x),ylab=”Probability”,main=”size=5,prob=0.5″)

View Video tutorial on rbinom command

# Binomial Probability Distributions

Bernoulli Trials

Many experiments consists of repeated independent trials and each trial have only two possible outcomes such as head or tail, right or wrong, alive or dead, defective or non-defective etc. If the probability of each outcome remains the same (constant) throughout the trials, then such trials are called the Bernoulli Trials.

Binomial Probability Distribution
Binomial Probability Distribution is a discrete probability distribution describing the results of an experiment known as Bernoulli Process. The experiment having n Bernoulli trials is called a Binomial Probability experiment possessing the following four conditions/ assumptions

1. The experiment consists of n repeated task.
2. Each trial, results in an outcome that may be classified as success or failure.
3. The probability of success denoted by p remains constant from trial to trial.
4. The repeated trials are independent.

A Binomial trial can result in a success with probability p and a failure with probability 1−p  having nx number of failures, then the probability distribution of Binomial Random Variable , the number of success in n independent trial is:

\begin{align*}
P(X=x)&=\binom{n}{x} \, p^x \, q^{n-x} \\
&=\frac{n!}{x!(n-x)!}\, p^x \, q^{n-x}
\end{align*}

The Binomial probability distribution is the most widely used distributions in situation of two outcomes. It was discovered by the Swiss mathematician Jakob Bernoulli (1654—1704) whose main work on “the ars Conjectandi” (the art of conjecturing) was published posthumously in Basel in 1713.

Mean of Binomial Distribution:   Mean = μ = np

Variance of Binomial Distribution:  Variance= npq

Standard Deviation of Binomial Distribution:  Standard Deviation = $\sqrt{npq}$

Moment Coefficient of Skewness:

\begin{align*}
\beta_1 &= \frac{q-p}{\sqrt{npq}}  \\
&= \frac{1-2p}{\sqrt{npq}}
\end{align*}

Moment Coefficient of Kurtosis:  $\beta_3 = 3+\frac{1-6pq}{npq}$