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# Capacity design (EN)

Questo articolo è disponibile in:
By following it step-by-step, you will learn how to design and perform seismic code-check of a structural steel joint using Capacity design analysis type based on CBFEM in IDEA StatiCa Connection.
$$1 New projectLaunch IDEA StatiCa (download the newest version) and open the source project file. The joint design is finished, and it is prepared for the standard Stress/Strain analysis.2 Calculation and checkStart the stress/strain analysis by the Calculate button in the ribbon. The analysis model is automatically generated, the calculation is performed, and you can see the overall check results in the left top corner of the scene. You can see that based on the stress/strain analysis, the joint is well designed and passed all the checks.To keep these results, copy this project item. 3 Capacity checkIn the new project item (CON2), change the analysis type to CD – Capacity design. The dissipative item has to be selected. It can be added by the command from the top ribbon or by the mouse right-click in the tree in the scene. A member or a plate where a plastic hinge is expected to occur should be chosen as a dissipative item. The material overstrength factor and strain-hardening factor are applied to the chosen item. In this example, select the member IPE360 as the dissipative item and confirm the selection with the spacebar/enter key/right-click. In the properties Members, the parameters of IPE360 have to be adjusted: Set the Model type to N-Vz-My, because the connection can resist the bending moment in the vertical plane only and bending around minor beam axis must be restrained. Switch the Forces in parameter to Position, because then the exact acting force position can be defined. The position of the plastic hinge is similar to the position of the acting force: X = 365 mm. How to know the right position of the plastic hinge? The engineer needs to decide where it will occur. Usually, the plastic hinge is determined on the beam. In this example, it will occur just behind the face of the last stiffener. It’s handy to read the position from the application (wireframe view). In the next step, the load effects have to be defined. Loads for seismic analysis are code dependent (the material overstrength factor, strain-hardening factor) and influenced also by yield strength, geometrical characteristic of the cross-section, etc.Loads for this example were calculated by this procedure:$M_{\textrm{Ed}} = \gamma_{\textrm{sh}} \cdot f_{\textrm{y,ov}} \cdot W_{\textrm{p}l} = 1.2 \cdot 447.75 \cdot 10^6 \cdot 1.0218 \cdot 10^{-3} = 544.12 \, \textrm{kNm}$$$\gamma_{\textrm{sh}} = 1.2$$ $$f_\textrm{y} = 355 \, \textrm{MPa}$$$$f_{\textrm{y,ov}} = f_\textrm{y} \cdot \gamma_{\textrm{ov}} = 355 \cdot 1.25 = 443.75\, \textrm{MPa}$$$$\gamma_{\textrm{ov}} = 1.25$$ $$W_{\textrm{pl,IPE360}} = 1.0218 \cdot 10^6 \, \textrm{mm}^3$$ $V_{\textrm{Ed}} = \frac{2 \cdot M_{\textrm{Ed}}}{L_{h}} = 2 \cdot \frac{544.12}{7.32} = 148.67 \, \textrm{kN}$$$L_{h} = 7.32 \, \textrm{m} \, -$$ distance between plastic hinges on the beamAdd the calculated shear force and bending moment as a new Load effect (LE). The shear force and bending moment must be with appropriate signs so that the bending moment is decreasing on the beam in the direction away from the node.Copy this LE and change the orientation of the acting forces so that the second LE will act in the opposite direction. Now the capacity analysis can be started by Calculate command. You can see from the results that the joint didn’t pass the code check. Some changes in design are required.Increase the thickness of the endplate to 25 mm to prevent it from collapsing. To increase column load capacity, add a doubler to its web (add Stiffening plate manufacturing operation). The doubler is welded by butt welds to the web of the column, weld to the flanges need to be defined as well. Four welds have to be added to weld doublers on both sides of the column to both flanges. The stiffeners at the column web need to be cut and welded to the doublers by the Cut of the plate manufacturing operation. Repeat the cut of plate to connect all four stiffeners to the doublers. All design actions are done now, run Calculate in the Check tab. You can see that all components (like welds and bolts) passed the code-check. The plastic strain of the dissipative item plates does not influence overall results. The occurrence of the plastic hinge can be explored if the Plastic strain. The plastic hinge appeared at the expected location and this joint passed the checks required by the capacity design.For a better understanding of the results, see the Theoretical Background.4 ReportAt last, you can review the Report. IDEA StatiCa offers a fully customizable report to print out or save in an editable format. You have performed a capacity design check of a structural steel joint according to Eurocode (EN). File di esempio Apri nel Viewer Capacity design (EN)$$


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## No review this summer

Posted in Books, Statistics, University life with tags , , , , , , , , on September 19, 2019 by xi'an

A recent editorial in Nature was a declaration by a biologist from UCL on her refusal to accept refereeing requests during the summer (or was it the summer break), which was motivated by a need to reconnect with her son. Which is a good enough reason (!), but reflects sadly on the increasing pressure on one’s schedule to juggle teaching, research, administration, grant hunting, society service, along with a balanced enough family life. (Although I have been rather privileged in this regard!) Given that refereeing or journal editing is neither visible nor rewarded, it comes as the first task to be postponed or abandoned, even though most of us realise it is essential to keep science working as a whole and to make our own papers published. I have actually noticed an increasing difficulty in the past decade to get (good) referees to accept new reviews, often asking for deadlines that are hurting the authors, like six months. Making them practically unavailable. As I mentioned earlier on this blog, it could be that publishing referees’ reports as discussions would help, since they would become recognised as (unreviewed!) publications, but it is unclear this is the solution. If judging from the similar difficulty in getting discussions for discussed papers. (As an aside, there are two exciting papers coming up for discussion in Series B, ‘Unbiased Markov chain Monte Carlo methods with couplings’ by  Pierre E. Jacob, John O’Leary and Yves F. Atchadé and in Bayesian Analysis, Latent nested nonparametric priors by Frederico Camerlenghi, David Dunson, Antonio Lijoi, Igor Prünster, and Abel Rodríguez). Which is surprising when considering the willingness of a part of the community to engage into forii discussions, sometimes of a considerable length as illustrated on Andrew’s blog.

Another entry in Nature mentioned the case of two University of København tenured professors in geology who were fired for either using a private email address (?!) or being away on field work during an exam and at a conference without permission from the administration. Which does not even remotely sound like a faulty behaviour to me or else I would have been fired eons ago..!

## I’m getting the point

Posted in Statistics with tags , , , , , , on February 14, 2019 by xi'an

A long-winded X validated discussion on the [textbook] mean-variance conjugate posterior for the Normal model left me [mildly] depressed at the point and use of answering questions on this forum. Especially as it came at the same time as a catastrophic outcome for my mathematical statistics exam.  Possibly an incentive to quit X validated as one quits smoking, although this is not the first attempt

## exam question

Posted in Kids, Statistics, University life with tags , , , , , , , , , on June 24, 2016 by xi'an

A question for my third year statistics exam that I borrowed from Cross Validated: no student even attempted to solve this question…!

And another one borrowed from the highly popular post on the random variable [almost] always smaller than its mean!

## done! [#1]

Posted in Kids, pictures, University life with tags , , , , , , on January 16, 2016 by xi'an

After spending a few hours grading my 127 exams for most nights of this week, I am finally done with it! One of the exam questions was the simulation of XY when (X,Y) is a bivariate normal vector with correlation ρ, following the trick described in a X validated question asked a few months ago, namely that

XY≡R{cos(πU)+ρ}

but no one managed to establish this representation. And, as usual, some students got confused between parameters θ and observations x when writing a posterior density, since the density of the prior was defined in the exam with the dummy x, thereby recovering the prior as the posterior. Nothing terrible and nothing exceptional with this cohort of undergraduates. And now I still have to go through my second pile of exams for the graduate course I taught on Bayesian computational tools…

Posted in Kids, pictures, Statistics, University life with tags , , , , , , on January 19, 2015 by xi'an

Now my grading is over, I can reflect on the unexpected difficulties in the mathematical statistics exam. I knew that the first question in the multiple choice exercise, borrowed from Cross Validation, was going to  be quasi-impossible and indeed only one student out of 118 managed to find the right solution. More surprisingly, most students did not manage to solve the (absence of) MLE when observing that n unobserved exponential Exp(λ) were larger than a fixed bound δ. I was also amazed that they did poorly on a N(0,σ²) setup, failing to see that

$\mathbb{E}[\mathbb{I}(X_1\le -1)] = \Phi(-1/\sigma)$

and determine an unbiased estimator that can be improved by Rao-Blackwellisation. No student reached the conditioning part. And a rather frequent mistake more understandable due to the limited exposure they had to Bayesian statistics: many confused parameter λ with observation x in the prior, writing

$\pi(\lambda|x) \propto \lambda \exp\{-\lambda x\} \times x^{a-1} \exp\{-bx\}$

$\pi(\lambda|x) \propto \lambda \exp\{-\lambda x\} \times \lambda^{a-1} \exp\{-b\lambda\}$

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# American Institute of Mathematical Sciences

September  2015, 35(9): 4611-4638. doi: 10.3934/dcds.2015.35.4611

## Fields of extremals and sensitivity analysis for multi-input bilinear optimal control problems

 1 Dept. of Electrical and Systems Engineering, Washington University, St. Louis, Missouri, 63130-4899 2 Dept. of Mathematics and Statistics, Southern Illinois University Edwardsville, Edwardsville, Illinois, 62026-1653

Received  May 2014 Revised  October 2014 Published  April 2015

Optimal control problems with fixed terminal time are considered for multi-input bilinear systems with the control set given by a compact interval and the objective function affine in the controls. Systems of this type have been widely used in the modeling of cell-cycle specific cancer chemotherapy over a prescribed therapy horizon for both homogeneous and heterogeneous tumor populations. Necessary conditions for optimality lead to concatenations of bang and singular controls as prime candidates for optimality. In this paper, the method of characteristics will be formulated as a general procedure to embed such a controlled reference extremal into a field of broken extremals. Sufficient conditions for the strong local optimality of a controlled reference bang-bang trajectory will be formulated in terms of solutions to associated sensitivity equations. These results will be applied to a model for cell cycle specific cancer chemotherapy with cytotoxic and cytostatic agents.
Citation: Heinz Schättler, Urszula Ledzewicz. Fields of extremals and sensitivity analysis for multi-input bilinear optimal control problems. Discrete and Continuous Dynamical Systems, 2015, 35 (9) : 4611-4638. doi: 10.3934/dcds.2015.35.4611
##### References:
 [1] A. Agrachev, G. Stefani and P. L. Zezza, A Hamiltonian approach to strong minima in optimal control, in Differential Geometry and Control, (G. Ferreyra, et al., Eds.), American Mathematical Society, (1999), 11-22. [2] A. Agrachev, G. Stefani and P. L. Zezza, Strong optimality for a bang-bang trajectory, SIAM J. Control and Optimization, 41 (2002), 991-1014. doi: 10.1137/S036301290138866X. [3] M. S. Aronna, J. F. Bonnans, A. V. Dmitruk and P. A. Lotito, Quadratic conditions for bang-singular extremals, Numerical Linear Algebra, Control and Optimization, 2 (2012), 511-546. doi: 10.3934/naco.2012.2.511. [4] V. G. Boltyansky, Sufficient conditions for optimality and the justification of the dynamic programming method, SIAM J. Control, 4 (1966), 326-361. doi: 10.1137/0304027. [5] B. Bonnard and M. Chyba, Singular Trajectories and their Role in Control Theory, Springer Verlag, Series: Mathematics and Applications, Vol. 40, 2003. [6] U. Boscain and B. Piccoli, Optimal Syntheses for Control Systems on 2-D Manifolds, Mathématiques & Applications, 43, Springer Verlag, Paris, 2004. [7] A. Bressan and B. Piccoli, Introduction to the Mathematical Theory of Control, American Institute of Mathematical Sciences, Springfield, Mo, 2007. [8] A. Dmitruk, Jacobi type conditions for singular extremals, Control and Cybernetics, 37 (2008), 285-306. [9] U. Felgenhauer, On stability of bang-bang type controls, SIAM J. Control and Optimization, 41 (2003), 1843-1867. doi: 10.1137/S0363012901399271. [10] U. Felgenhauer, Lipschitz stability of broken extremals in bang-bang control problems, in: Large-Scale Scientific Computing (Sozopol 2007), (I. Lirkov et al., Eds.), Lecture Notes in Computer Science, vol. 4818, Springer (2008), 317-325. doi: 10.1007/978-3-540-78827-0_35. [11] U. Felgenhauer, L. Poggiolini and G. Stefani, Optimality and stability result for bang-bang optimal controls with simple and double switch behaviour, Control and Cybernetics, 38 (2009), 1305-1325. [12] H. Gardner-Moyer, Sufficient conditions for a strong minimum in singular control problems, SIAM J. Control, 11 (1973), 620-636. doi: 10.1137/0311048. [13] M. Kimmel and A. Swierniak, An optimal control problem related to leukemia chemotherapy, Scientific Bulletins of the Silesian Technical University, 65 (1983), 120-130. [14] U. Ledzewicz, K. Bratton and H. Schättler, A $3$-compartment model for chemotherapy of heterogeneous tumor populations, Acta Applicandae Mathematicae, 135 (2015), 191-207. doi: 10.1007/s10440-014-9952-6. [15] U. Ledzewicz, H. Maurer and H. Schättler, Sufficient conditions for strong local optimality in optimal control problem with $L_{2}$-type objectives and control constraints, Discrete and Continuous Dynamical Systems, Series B, 19 (2014), 2657-2679. doi: 10.3934/dcdsb.2014.19.2657. [16] U. Ledzewicz and H. Schättler, Analysis of a cell-cycle specific model for cancer chemotherapy, J. of Biological Systems, 10 (2002), 183-206. doi: 10.1142/S0218339002000597. [17] U. Ledzewicz, H. Schättler, M. Reisi Gahrooi and S. Mahmoudian Dehkordi, On the MTD paradigm and optimal control for multi-drug cancer chemotherapy, Mathematical Biosciences and Engineering (MBE), 10 (2013), 803-819. doi: 10.3934/mbe.2013.10.803. [18] H. Maurer and N. Osmolovskii, Quadratic sufficient optimality conditions for bang-bang control problems, Control and Cybernetics, 33 (2003), 555-584. [19] J. Noble and H. Schättler, Sufficient conditions for relative minima of broken extremals, J. of Mathematical Analysis and Applications, 269 (2002), 98-128. doi: 10.1016/S0022-247X(02)00008-2. [20] N. Osmolovskii and H. Maurer, Applications to Regular and Bang-Bang Control: Second-Order Necessary and Sufficient Optimality Conditions in Calculus of Variations and Optimal Control, SIAM, Philadelphia, USA, 2012. doi: 10.1137/1.9781611972368. [21] B. Piccoli and H. Sussmann, Regular synthesis and sufficient conditions for optimality, SIAM J. on Control and Optimization, 39 (2000), 359-410. doi: 10.1137/S0363012999322031. [22] L. Poggiolini and M. Spadini, Strong local optimality for a bang-bang trajectory in a Mayer problem, SIAM J. Control and Optimization, 49 (2011), 140-161. doi: 10.1137/090771405. [23] L. Poggiolini and G. Stefani, Sufficient optimality conditions for a bang-bang trajectory, in Proceedings of the 45th IEEE Conference on Decision and Control, (2006), 6624-6629. doi: 10.1109/CDC.2006.376760. [24] L. S. Pontryagin, V. G. Boltyanskii, R. V. Gamkrelidze and E. F. Mishchenko, The Mathematical Theory of Optimal Processes, MacMillan, New York, 1964. [25] A. Sarychev, Morse index and sufficient optimality conditions for bang-bang Pontryagin extremals, in: System Modeling and Optimization, Lecture Notes in Control and Information Sciences, 180 (1992), 440-448. doi: 10.1007/BFb0113311. [26] A. Sarychev, First and second order sufficient optimality conditions for bang-bang controls, SIAM J. on Control and Optimization, 35 (1997), 315-340. doi: 10.1137/S0363012993246191. [27] H. Schättler and U. Ledzewicz, Geometric Optimal Control: Theory, Methods and Examples, Springer Verlag, 2012. doi: 10.1007/978-1-4614-3834-2. [28] H. E. Skipper, On mathematical modeling of critical variables in cancer treatment (goals: better understanding of the past and better planning in the future), Bulletin of Mathematical Biology, 48 (1986), 253-278. doi: 10.1007/BF02459681. [29] H. J. Sussmann, The structure of time-optimal trajectories for single-input systems in the plane: the $C^{\infty}$ nonsingular case, SIAM J. Control Optimization, 25 (1987), 433-465. doi: 10.1137/0325025. [30] H. J. Sussmann, The structure of time-optimal trajectories for single-input systems in the plane: the general real analytic case, SIAM J. Control Optimization, 25 (1987), 868-904. doi: 10.1137/0325048. [31] H. J. Sussmann, Regular synthesis for time-optimal control of single-input real analytic systems in the plane, SIAM J. Control Optimization, 25 (1987), 1145-1162. doi: 10.1137/0325062. [32] A. Swierniak, Optimal treatment protocols in leukemia - modelling the proliferation cycle, IMACS Ann. Comput. Appl. Math., Baltzer, Basel, 5 (1989), 51-53. [33] A. Swierniak, Cell cycle as an object of control, J. of Biological Systems, 3 (1995), 41-54. doi: 10.1142/S0218339095000058. [34] A. Swierniak, U. Ledzewicz and H. Schättler, Optimal control for a class of compartmental models in cancer chemotherapy, Int. J. Applied Mathematics and Computer Science, 13 (2003), 357-368. [35] A. Swierniak, A. Polanski and M. Kimmel, Optimal control problems arising in cell-cycle-specific cancer chemotherapy, Cell Proliferation, 29 (1996), 117-139. doi: 10.1046/j.1365-2184.1996.00995.x. [36] T. E. Wheldon, Mathematical Models in Cancer Research, Boston-Philadelphia: Hilger Publishing, 1988.

show all references

##### References:
 [1] A. Agrachev, G. Stefani and P. L. Zezza, A Hamiltonian approach to strong minima in optimal control, in Differential Geometry and Control, (G. Ferreyra, et al., Eds.), American Mathematical Society, (1999), 11-22. [2] A. Agrachev, G. Stefani and P. L. Zezza, Strong optimality for a bang-bang trajectory, SIAM J. Control and Optimization, 41 (2002), 991-1014. doi: 10.1137/S036301290138866X. [3] M. S. Aronna, J. F. Bonnans, A. V. Dmitruk and P. A. Lotito, Quadratic conditions for bang-singular extremals, Numerical Linear Algebra, Control and Optimization, 2 (2012), 511-546. doi: 10.3934/naco.2012.2.511. [4] V. G. Boltyansky, Sufficient conditions for optimality and the justification of the dynamic programming method, SIAM J. Control, 4 (1966), 326-361. doi: 10.1137/0304027. [5] B. Bonnard and M. Chyba, Singular Trajectories and their Role in Control Theory, Springer Verlag, Series: Mathematics and Applications, Vol. 40, 2003. [6] U. Boscain and B. Piccoli, Optimal Syntheses for Control Systems on 2-D Manifolds, Mathématiques & Applications, 43, Springer Verlag, Paris, 2004. [7] A. Bressan and B. Piccoli, Introduction to the Mathematical Theory of Control, American Institute of Mathematical Sciences, Springfield, Mo, 2007. [8] A. Dmitruk, Jacobi type conditions for singular extremals, Control and Cybernetics, 37 (2008), 285-306. [9] U. Felgenhauer, On stability of bang-bang type controls, SIAM J. Control and Optimization, 41 (2003), 1843-1867. doi: 10.1137/S0363012901399271. [10] U. Felgenhauer, Lipschitz stability of broken extremals in bang-bang control problems, in: Large-Scale Scientific Computing (Sozopol 2007), (I. Lirkov et al., Eds.), Lecture Notes in Computer Science, vol. 4818, Springer (2008), 317-325. doi: 10.1007/978-3-540-78827-0_35. [11] U. Felgenhauer, L. Poggiolini and G. Stefani, Optimality and stability result for bang-bang optimal controls with simple and double switch behaviour, Control and Cybernetics, 38 (2009), 1305-1325. [12] H. Gardner-Moyer, Sufficient conditions for a strong minimum in singular control problems, SIAM J. Control, 11 (1973), 620-636. doi: 10.1137/0311048. [13] M. Kimmel and A. Swierniak, An optimal control problem related to leukemia chemotherapy, Scientific Bulletins of the Silesian Technical University, 65 (1983), 120-130. [14] U. Ledzewicz, K. Bratton and H. Schättler, A $3$-compartment model for chemotherapy of heterogeneous tumor populations, Acta Applicandae Mathematicae, 135 (2015), 191-207. doi: 10.1007/s10440-014-9952-6. [15] U. Ledzewicz, H. Maurer and H. Schättler, Sufficient conditions for strong local optimality in optimal control problem with $L_{2}$-type objectives and control constraints, Discrete and Continuous Dynamical Systems, Series B, 19 (2014), 2657-2679. doi: 10.3934/dcdsb.2014.19.2657. [16] U. Ledzewicz and H. Schättler, Analysis of a cell-cycle specific model for cancer chemotherapy, J. of Biological Systems, 10 (2002), 183-206. doi: 10.1142/S0218339002000597. [17] U. Ledzewicz, H. Schättler, M. Reisi Gahrooi and S. Mahmoudian Dehkordi, On the MTD paradigm and optimal control for multi-drug cancer chemotherapy, Mathematical Biosciences and Engineering (MBE), 10 (2013), 803-819. doi: 10.3934/mbe.2013.10.803. [18] H. Maurer and N. Osmolovskii, Quadratic sufficient optimality conditions for bang-bang control problems, Control and Cybernetics, 33 (2003), 555-584. [19] J. Noble and H. Schättler, Sufficient conditions for relative minima of broken extremals, J. of Mathematical Analysis and Applications, 269 (2002), 98-128. doi: 10.1016/S0022-247X(02)00008-2. [20] N. Osmolovskii and H. Maurer, Applications to Regular and Bang-Bang Control: Second-Order Necessary and Sufficient Optimality Conditions in Calculus of Variations and Optimal Control, SIAM, Philadelphia, USA, 2012. doi: 10.1137/1.9781611972368. [21] B. Piccoli and H. Sussmann, Regular synthesis and sufficient conditions for optimality, SIAM J. on Control and Optimization, 39 (2000), 359-410. doi: 10.1137/S0363012999322031. [22] L. Poggiolini and M. Spadini, Strong local optimality for a bang-bang trajectory in a Mayer problem, SIAM J. Control and Optimization, 49 (2011), 140-161. doi: 10.1137/090771405. [23] L. Poggiolini and G. Stefani, Sufficient optimality conditions for a bang-bang trajectory, in Proceedings of the 45th IEEE Conference on Decision and Control, (2006), 6624-6629. doi: 10.1109/CDC.2006.376760. [24] L. S. Pontryagin, V. G. Boltyanskii, R. V. Gamkrelidze and E. F. Mishchenko, The Mathematical Theory of Optimal Processes, MacMillan, New York, 1964. [25] A. Sarychev, Morse index and sufficient optimality conditions for bang-bang Pontryagin extremals, in: System Modeling and Optimization, Lecture Notes in Control and Information Sciences, 180 (1992), 440-448. doi: 10.1007/BFb0113311. [26] A. Sarychev, First and second order sufficient optimality conditions for bang-bang controls, SIAM J. on Control and Optimization, 35 (1997), 315-340. doi: 10.1137/S0363012993246191. [27] H. Schättler and U. Ledzewicz, Geometric Optimal Control: Theory, Methods and Examples, Springer Verlag, 2012. doi: 10.1007/978-1-4614-3834-2. [28] H. E. Skipper, On mathematical modeling of critical variables in cancer treatment (goals: better understanding of the past and better planning in the future), Bulletin of Mathematical Biology, 48 (1986), 253-278. doi: 10.1007/BF02459681. [29] H. J. Sussmann, The structure of time-optimal trajectories for single-input systems in the plane: the $C^{\infty}$ nonsingular case, SIAM J. Control Optimization, 25 (1987), 433-465. doi: 10.1137/0325025. [30] H. J. Sussmann, The structure of time-optimal trajectories for single-input systems in the plane: the general real analytic case, SIAM J. Control Optimization, 25 (1987), 868-904. doi: 10.1137/0325048. [31] H. J. Sussmann, Regular synthesis for time-optimal control of single-input real analytic systems in the plane, SIAM J. Control Optimization, 25 (1987), 1145-1162. doi: 10.1137/0325062. [32] A. Swierniak, Optimal treatment protocols in leukemia - modelling the proliferation cycle, IMACS Ann. Comput. Appl. Math., Baltzer, Basel, 5 (1989), 51-53. [33] A. Swierniak, Cell cycle as an object of control, J. of Biological Systems, 3 (1995), 41-54. doi: 10.1142/S0218339095000058. [34] A. Swierniak, U. Ledzewicz and H. Schättler, Optimal control for a class of compartmental models in cancer chemotherapy, Int. J. Applied Mathematics and Computer Science, 13 (2003), 357-368. [35] A. Swierniak, A. Polanski and M. Kimmel, Optimal control problems arising in cell-cycle-specific cancer chemotherapy, Cell Proliferation, 29 (1996), 117-139. doi: 10.1046/j.1365-2184.1996.00995.x. [36] T. E. Wheldon, Mathematical Models in Cancer Research, Boston-Philadelphia: Hilger Publishing, 1988.
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calculate cyclotomic polynomial - Maple Help

Home : Support : Online Help : Mathematics : Group Theory : Number Theory : numtheory/cyclotomic

numtheory[cyclotomic] - calculate cyclotomic polynomial

 Calling Sequence cyclotomic(n, t)

Parameters

 n - non-negative integer t - variable

Description

 • The function cyclotomic(n, t) returns the nth cyclotomic  polynomial in t.
 By definition, cyclotomic(n, t) =

$\prod _{\mathrm{\zeta }}\left(t-\mathrm{\zeta }\right):\mathrm{\zeta }\mathrm{is}a\mathrm{primitive}n\mathrm{th}\mathrm{root}\mathrm{of}\mathrm{unity}$

 • With the exception of small values of n, for which a table lookup is used, $\mathrm{cyclotomic}\left(n,t\right)$ is computed using an algorithm based on the fact that $\mathrm{cyclotomic}\left(n,t\right)=\frac{{t}^{n}-1}{\prod _{d=\mathrm{divisors}\left(n\right)\setminus \left\{n\right\}}\phantom{\rule[-0.0ex]{5.0px}{0.0ex}}\mathrm{cyclotomic}\left(d,t\right)}$.
 • The degree of the nth cyclotomic polynomial is given by the totient function numtheory[phi].

Examples

 > $\mathrm{with}\left(\mathrm{numtheory}\right):$
 > $\mathrm{cyclotomic}\left(1,x\right)$
 ${x}{-}{1}$ (1)
 > $\mathrm{cyclotomic}\left(20,z\right)$
 ${{z}}^{{8}}{-}{{z}}^{{6}}{+}{{z}}^{{4}}{-}{{z}}^{{2}}{+}{1}$ (2)


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# Cutting a subset in many pieces with controlled perimeter

Let me premise that I am in no way an expert in the subject of this question.

Let's say we have a measure space $X$ with measure $m$ and a reasonable notion of perimeter for (nice enough) subsets of $X$. We denote $p(F)$ the perimeter of $F\subseteq X$.

I am interested in knowing classes of spaces for which the following property (P) holds:

(P) There is a constant $C<+\infty$ such that given any subset $F$ (nice enough) and any natural number $K$, we can split $F$ in disjoint pieces $F_1,\dots,F_K$ such that $m(F_j)\leq C \frac{m(F)}{K}$ and $p(F_j)\leq Cp(F)$ for any $j=1,\dots,K$.

Here nice enough may refer to having finite measure and perimeter and maybe some boundedness/compactness assumption.

Of course the question (if it is not stupid, because of some gross misunderstanding on my part) sounds absurdly vague, so let me point out that it seems to me that the property holds on euclidean spaces. If $F\subseteq\mathbb{R}^d$ one can cut $F$ using parallel hyperplanes in $K$ pieces having exactly $1/K$ of the total volume and everything should work with constant $C=1$.

I am especially interested in the graph case, i.e. $X=(V,E)$ is a (say countable and bounded-degree) graph with counting measure on vertices and perimeter of a subset $F$ measured by the number of edges connecting $F$ with its complement. Is $(P)$ a reasonable property? Are there interesting examples/counterexamples in the graph realm?

-

I'll take the chance to mention the following nice paper of Aronov and Hubard. In it they answer a question of Nandakumar and Rao of whether a convex body can be partitioned into $n$ equal pieces of equal volume and equal perimeter. Their proof shows that this is always possible (in any dimension) when $n$ is prime. The prime condition is admittedly a bit strange and is likely just an artifact of the proof.


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# When can we unprepare a circuit using Clifford + measurements?

I have seen sometimes that in some protocols, one can uncompute a circuit using only Clifford + measurements. See for example

Erase the two qubits containing $$p_{w,r}$$ and $$p_{w,s}$$, then the register containing component w of p. These erasures can be done with measurements and Clifford gates.

before eq 73 in Su et al. For context, these are qubits that were "copied" via C-Nots from another register, and have been used as a control in a C-Z gate.

When can one erase registers using measurements and Clifford gates in general?

I'm not sure what the exact general requirement is, but a sufficient requirement is:

1. The qubit's computational basis must be a function of other qubits' computational basis value. For example, the state should have the form:

$$\sum_k \alpha_k |k\rangle |\dots\rangle |f(k)\rangle$$

where $$f$$ is some classical function that outputs a single bit.

2. The fixup operation $$U$$ which performs $$U(\sum_k \alpha_k |k\rangle) = \sum_k (-1)^{f(k)} \alpha_k |k\rangle$$ must be possible to do using only stabilizer operations. Because you're going to measure the qubit storing $$f(k)$$ in the X basis, and if the result is $$|-\rangle$$ then you need to apply $$U$$ to finish the uncomputation.

• Thanks! Would $U$ not amount to an unimportant global phase now that we have measured $f(k)$ (the recovery operation does no longer depend on $k$)? Dec 22, 2021 at 22:27
• @Pablo No, because it has negated states matching $f$ relatives to states that don't. Dec 22, 2021 at 22:43
• For example, if $f$ was "an even number of bits are ON", then $U$ would negate the state $|101\rangle$ but not the state $|111\rangle$. Dec 22, 2021 at 22:45
• I think I get it: starting from $|\psi\rangle|k\rangle|f(k)\rangle$, - If $f(k) = 0$, then one applies X and gets $|\psi\rangle|k\rangle(|+\rangle+|-\rangle)$. Since controlling on $k$ the operation $U$ does nothing (because $f(k) = 0$), it stays the same in either case. - If $f(k)=1$ one would have $|\psi\rangle|k\rangle(|+\rangle-|-\rangle)$. I guess the idea is that if one measures $|-\rangle$ then applying $f(k)$ controlled on register $|k\rangle$ recovers the original state. I guess the main limitation is the need to control on $|k\rangle$. Is this right? Dec 22, 2021 at 22:51
• @Pablo Yes, that's the main limitation. It's usually easier to do $|k\rangle \rightarrow (-1)^f(k) |k\rangle$ than it is to do $|k\rangle \rightarrow |k\rangle|f(k)\rangle$, e.g. you'll be able to apply a CZ to the qubits of $k$ instead of a Toffoli controlled by the qubits of $k$. Dec 23, 2021 at 0:20


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# [Pw_forum] ph.x high I/O frequency

Paolo Giannozzi giannozz at democritos.it
Wed Oct 17 08:45:37 CEST 2012

On Oct 17, 2012, at 5:05 , Jiseok Kim wrote:

> So I tried to reduce the I/O by setting the option 'reduce_io=.true.'.

I don't think it has much effect on the phonon code. It only affects
the first step, the self-consistent calculation.

The current version of the phonon code follows an old paradigm:
all large arrays depending upon the k point (Kohn-Sham orbitals
\psi_k, perturbations \Delta V\psi_k, linear responses \Delta\psi_k)
are computed and stored to file. Only a single k-point is kept in
memory. This behavior can be changed only by modifying the code.

One can reduce the amount of load on the system by writing to a
local filesystems instead of a parallel file system. I am not sure it
works in the phonon case, though. In any case, the files end up
scattered among different processors so there are many problems
and limitation (no possibility to restart, for sure).

In conclusion: there are many possible ways to improve the behavior
of the phonon code with respect to I.O load, but none is simple, most
require some serious work.

P.
---
Paolo Giannozzi, Dept of Chemistry&Physics&Environment,
Univ. Udine, via delle Scienze 208, 33100 Udine, Italy
Phone +39-0432-558216, fax +39-0432-558222


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Problem 60
One corner of a leaf of width a is folded over so as just to reach the opposite side of the page. Find the width of the part folded over when the length of the crease is a minimum. See Figure 41.

Solution:

Problem 61
Solve Problem 60 above if the area folded over is to be a minimum.

Solution:

0 likes


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# domain and range

• August 26th 2008, 03:25 AM
sjenkins
domain and range
Find domain and range

f(x){-1,x<=1
3x+2,-1<x<1
7-2x, x>=1}

• August 26th 2008, 03:39 AM
Moo
Hello,
Quote:

Originally Posted by sjenkins
Find domain and range

f(x){-1,x<=-1
3x+2,-1<x<1
7-2x, x>=1}

For the domain, it's just the values x can take.

For the range... if x<=-1, then f(x)=-1. Thus the range contains -1.

f(x)=3x+2 if -1<x<1, that is to say -1<3x+2<5. This is part of the range too.

f(x)=7-2x if x>=1, that is to say 7-2x<=5. This actually gives the whole range.
• August 26th 2008, 03:49 AM
sjenkins
Functions have always been a particularly weak spot for me, I am not quite understanding. At first, I thought the domain was any number, but now I'm thinking -1,1. For the range, if 7-2x<=5, x would have to be 1 or less, right?
• August 26th 2008, 03:55 AM
Moo
Quote:

Originally Posted by sjenkins
Functions have always been a particularly weak spot for me, I am not quite understanding. At first, I thought the domain was any number, but now I'm thinking -1,1. For the range, if 7-2x<=5, x would have to be 1 or less, right?

$f(x)=\left\{ \begin{array}{lll} -1 \quad & \quad x \le -1 \\ 3x+2 \quad & -1

$x \le -1$ is like $]-\infty~,-1]$, $-1 is like $(-1~,1)$ and $x \ge 1$ is like $[1~,+\infty[$

So x can be any number in $\mathbb{R}$ (the set of real numbers).

Why would the range have been only -1 and 1 ? On the contrary, it's the values that would have been controversary because we're not sure the function is continuous at these points. In fact, it is.

You're told that if x>=1, then f(x)=7-2x. So find out what values 7-2x can take if x>=1. This gives 7-2x<=5
• August 26th 2008, 04:47 AM
sjenkins
Okay, maybe I should start from the beginning, for the given function, isn't the first number (or expression) the lowest number that can be used for the domain? And the second, the largest? That's why I picked -1 and 1 because in the first expression the number is -1 and in the last it is 1. I'm just really not sure. Isn't the range the result of what you put into the function?
• August 26th 2008, 04:54 AM
Moo
Quote:

Originally Posted by sjenkins
Okay, maybe I should start from the beginning, for the given function, isn't the first number (or expression) the lowest number that can be used for the domain? And the second, the largest? That's why I picked -1 and 1 because in the first expression the number is -1 and in the last it is 1. I'm just really not sure. Isn't the range the result of what you put into the function?

Yes. This means that you want the different values that f(x) can have.

For any value of x that is inferior to -1, the function is -1
For any value of x that is superior to -1 and inferior to 1, we have :
$-1. So the different values of f(x) while -1<x<1 are all the values within -1 and 5.

So far, the range is $[-1~,5)$.

For any value of x that is superior to 1, we have :
$x \ge 1 \implies 2x \ge 2 \implies -2x \le -2 \implies \underbrace{7-2x}_{f(x) \text{ for } x \ge 1} \le 5$. So the different values of f(x) while $x \ge 1$ are all the values inferior to 5.

So we have $f(x) \le 5$ if $x \ge 1$, which $f(x) \in (-\infty~,5]$

The range will be the union of these two intervals :
$R=[-1~,5) \cup (-\infty~,5]=(-\infty~,5]$

Is it clear enough ? (Worried)
• August 26th 2008, 05:17 AM
sjenkins
Where did the 5 come from in -1<3x+2<5? I'm sorry, I really am trying to understand this.
• August 26th 2008, 05:27 AM
Moo
Quote:

Originally Posted by sjenkins
Where did the 5 come from in -1<3x+2<5?

I added 2 to each side (3+2), like I did for 7-2x :)

Quote:

I'm sorry, I really am trying to understand this.
You don't have to be sorry, it's a good thing ! (Wink)


text:
# What is the general solution of the differential equation dy/dx=6e^x?

Oct 7, 2017

$y = 6 {e}^{x} - 4$

#### Explanation:

We have:

$\frac{\mathrm{dy}}{\mathrm{dx}} = 6 {e}^{x}$

This is a FIrst Order Separable ODE so we can just separate the variables to get:

$\int \setminus \mathrm{dy} = \int \setminus 6 {e}^{x} \setminus \mathrm{dx}$

Both sides are directly integrable so if we integrate we get:

$y = 6 {e}^{x} + C$

Applying the initial Conditions, $y = 2$ when $x = 0$ we have:

$2 = 6 {e}^{0} + C \implies C = - 4$

$y = 6 {e}^{x} - 4$


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Question
Point charges of $25.0 \textrm{ }\mu\textrm{C}$ and $45.0 \textrm{ }\mu\textrm{C}$ are placed 0.500 m apart. (a) At what point along the line between them is the electric field zero? (b) What is the electric field halfway between them?
1. $0.214 \textrm{ m}$ from the $25.0 \textrm{ }\mu\textrm{C}$ charge.
2. $2.88 \times 10^6 \textrm{ N/C}$ away from the $45.0 \textrm{ }\mu\textrm{C}$ charge.
Solution Video

# OpenStax College Physics Solution, Chapter 18, Problem 55 (Problems & Exercises) (5:41)

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This is College Physics Answers with Shaun Dychko. We're going to find the position between these two charges such that the electric field is zero. So charge a on the left is 25 micro Coulombs and charge b is 45 micro Coulombs. And our position is going to be here somewhat to the left of center so that we're closer to this smaller charge and because the charge is smaller, we need to be closer to it in order to get a greater electric field from it versus this charge b, which has a greater magnitude and we're a bit farther away from it. So these two charges are separated by distance l which is 0.5 meters and for this position here, we'll say it's distance r from qa. And we'll express the distance from qb in terms of l and r. And so we'll write this distance as l for full separation minus r, the distance from qa. So the electric field due to charge a is Coulombs Constant times qa divided by r squared. And then, that's going to equal the electric field due to charge b which is going to be k times qb over l minus r squared. And we'll cancel the Coulombs Constant of both sides and then take the reciprocal of both sides which is the same as raising it to the exponent negative one and we get our unknown at the top. So we have r squared over qa equals l minus r squared over qb. Then, we'll multiply both sides by qa and then take the square root of both sides and we get r equals square root qa over qb times l minus r to the power of one. And then, we'll distribute the square root into the brackets there and we'll multiply each term inside by the factor here. So we have r equals l times square root qa over qb minus r times square root qa over qb. And then collect the r terms on the left side by adding r times square root qa over qb to both sides. And we get r plus r times square root qa over qb equals l times square root qa over qb. Then we factor out the common factor r, and we get this line here. And then divide both sides by this bracket. And then divide the right hand side by that, too. Then, we get the distance r is l times square root qa over qb divided by one plus square root qa over qb. So that's 0.5 meters separation between the charges times the square root of charge a which is 25 micro Coulombs divided by charge b which is 45 micro Coulombs. And normally I would change these units into base units and write times ten to the minus six Coulombs but in this particular case that isn't necessary because we're dividing here and when you divide units that are the same, they cancel. And so the only important thing is that they are the same units. And so you could've gone times ten to the minus six Coulombs on for each of this units and that will be fine, but it's just unnecessary writing. And so I didn't do that. And then we divide by one plus square root 25 micro Coulombs over 45 and that gives 0.214 meters which is from the 25 micro Coulomb charge between the charges. So, this position here which is 0.241 meters from charge a, this is where the electric field will be zero. So it's about approximately one quarter of the way between the two charges. It's almost half way, actually. Part b says, what would the electric field be exactly in the middle. Well, the net electric field will be that due to charge a plus electric field due to charge b and the sign here will indicate the direction. And so we're going to take positive to the right and so the field due to charge a is going to be positive and the field due to charge b will be negative since charge b will be creating a field to the left. And so we have kqa over the distance which is the separation divided by two squared minus kqb over l over two squared. Well the denominator here, it becomes l squared over four and then we can multiply top and bottom by four and then this four here is cancel and this one and we're left with four in the numerator. And that's a common factor between the two terms along with k and l squared in the denominator which we can factor out. So we have four k over l squared times the difference between the charges qa minus qb. So we're substitute in numbers now four times Coulombs Constant 8.988 times ten to the nine Newtons meter squared per Coulomb squared divided by 0.5 meters squared times the difference between the charges. And I did change the units into times ten to the minus six now because we're mixing Coulombs here with the Coulombs there. And so we have to be consistent with our units. So 25 times ten to the minus six Coulombs minus 45 times ten to the minus six Coulombs. This works out to negative 2.88 times ten to the six Newtons per Coulomb. This negative sign indicates the direction to the left. And so electric field is 2.88 times ten to the six Newtons per Coulomb away from the 45 micro Coulomb charge.


text: ... a critical component to supply chain success. In this topic we shall see an important method for evaluating many complicated integrals. f(z) = √z g(z) = 5z − 8. f ( z) = √ z g ( z) = 5 z − 8. then we can write the function as a composition. ex2+5x,cos(x3 +x),loge (4x2 +2x) e x 2 + 5 x, cos. . could really just call the reverse chain rule. Using less parcel shipping. the derivative of f. The derivative of f with respect to x, and that's going to give you the derivative of g with respect to x. Here is a general guide: u Inverse Trig Function (sin ,arccos , 1 xxetc) Logarithmic Functions (log3 ,ln( 1),xx etc) Algebraic Functions (xx x3,5,1/, etc) The Chain Rule The chain rule (function of a function) is very important in differential calculus and states that: (You can remember this by thinking of dy/dx as a fraction in this case (which it isn’t of course!)). Cauchy's Formula gives the result of a contour integration in the complex plane, using "singularities" of the integrand. Feel free to let us know if you are unsure how to do this in case ð, Absolute Value Algebra Arithmetic Mean Arithmetic Sequence Binomial Expansion Binomial Theorem Chain Rule Circle Geometry Common Difference Common Ratio Compound Interest Cyclic Quadrilateral Differentiation Discriminant Double-Angle Formula Equation Exponent Exponential Function Factorials Functions Geometric Mean Geometric Sequence Geometric Series Inequality Integration Integration by Parts Kinematics Logarithm Logarithmic Functions Mathematical Induction Polynomial Probability Product Rule Proof Quadratic Quotient Rule Rational Functions Sequence Sketching Graphs Surds Transformation Trigonometric Functions Trigonometric Properties VCE Mathematics Volume. Integration’s counterpart to the product rule. x, times f prime of x. to be the anti-derivative of that, so it's going to be taking something to the third power and then dividing it by three, so let's do that. (Use antiderivative rule 7 from the beginning of this section on the first integral and use trig identity F from the beginning of this section on the second integral.) As a rule of thumb, whenever you see a function times its derivative, you may try to use integration by substitution. with u-substitution. take the anti-derivative here with respect to sine of x, instead of with respect how does this relate to u-substitution? Which is essentially, or it's exactly what we did with the integral of g prime of f of x, g prime of f of x, times f prime of x, dx, well, this If you're seeing this message, it means we're having trouble loading external resources on our website. So let's say that we had, and I'm going to color code it so that it jumps out at you a little bit more, let's say that we had sine of x, and I'm going Integration by substitution is the counterpart to the chain rule for differentiation. This rule allows us to differentiate a … Well f prime of x in that circumstance is going to be cosine of x, and what is g? Times cosine of x, times cosine of x. Differentiate algebraic and trigonometric equations, rate of change, stationary points, nature, curve sketching, and equation of tangent in Higher Maths. Integrating functions of the form f(x) = 1 x or f(x) = x − 1 result in the absolute value of the natural log function, as shown in the following rule. Our mission is to provide a free, world-class education to anyone, anywhere. For definite integrals, the limits of integration can also change. which is equal to what? The Chain Rule and Integration by Substitution Suppose we have an integral of the form where Then, by reversing the chain rule for derivatives, we have € ∫f(g(x))g'(x)dx € F'=f. This skill is to be used to integrate composite functions such as. u-substitution, we just did it a little bit more methodically So I encourage you to pause this video and think about, does it Have Fun! So in the next few examples, \begin{aligned} \displaystyle \frac{d}{dx} \cos{3x^3} &= -\sin{3x^3} \times \frac{d}{dx} (3x^3) \\ &= -\sin{3x^3} \times 9x^2 \\ &= -9x^2 \sin{3x^3} \\ \end{aligned} \\ (b)    Integrate $$x^2 \sin{3x^3}$$. you'll get exactly this. a little bit faster. Integration by Parts. the sine of x squared, the typical convention It explains how to integrate using u-substitution. This is because, according to the chain rule, the derivative of a composite function is the product of the derivatives of the outer and inner functions. € ∫f(g(x))g'(x)dx=F(g(x))+C. ( ) … Just rearrange the integral like this: ∫ cos (x 2) 6x dx = 3 ∫ cos (x 2) 2x dx. here now that might have been introduced, because if I take the derivative, the constant disappears. The most important thing to understand is when to use it and then get lots of practice. In calculus, and more generally in mathematical analysis, integration by parts or partial integration is a process that finds the integral of a product of functions in terms of the integral of the product of their derivative and antiderivative. So when we talk about So what's this going to be if we just do the reverse chain rule? Substitute into the original problem, replacing all forms of , getting . The chain rule provides us a technique for finding the derivative of composite functions, with the number of functions that make up the composition determining how many differentiation steps are necessary. (a)    Differentiate $$\log_{e} \sin{x}$$. Strangely, the subtlest standard method is just the product rule run backwards. Type in any integral to get the solution, steps and graph Required fields are marked *. 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# The aptitude test scores of applicants to a university graduate program are normally...

The aptitude test scores of applicants to a university graduate program are normally distributed with mean 500 and standard deviation 60.
1. Applicants need a test score higher than 530 to be admitted into the graduate program. What proportion of applications qualify?
[Marks 2]
1. If the university wishes to set the cutoff score for graduate admission so that only the top 10% of applicants qualify for admission, what is the required cutoff score?
[Marks 2]
1. What percentage of applicants have test scores within two standard deviations of the mean?
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Question 1 The aptitude test scores of applicants to a university graduate program are normally distributed with mean 500 and standard deviation 60. Applicants need a test score higher than 530 to be admitted into the graduate program. What proportion of applications qualify? [Marks 2] If the university wishes to set the cutoff score for graduate admission so that only the top 10% of applicants qualify for admission, what is the required cutoff score? [Marks 2] What percentage of applicants have test scores within two standard deviations of the mean? [Marks 3] Question 2 Government officials in Canberra have recently expressed concern regarding overruns on military contracts. These unplanned expenditures have been costing Australians millions of dollars every year. The prime minister impanels a committee of experts to estimate the average amount each contract costs the government over and above the amount agreed upon. The committee has already determined that the standard deviation in overruns is $17.5 million, and that they appear normally distributed. If a sample of 25 contracts is selected, how likely is it the sample will overestimate the population mean by more than$10 million? [Marks 3] The prime minister will accept an error of \$5 million in the estimate of µ. How likely is he to receive an estimate from the committee within the specified range? [Marks 4] Question 3 You have just graduated with a post graduate degree in business and have obtained a position with a large manufacturing firm. The director of marketing has asked you to estimate the mean time required to complete a particular unit of the manufacturing process. A sample of 600 units yields a mean of 7.2 days. Since the population standard deviation is unknown, the sample standard deviation of s=1.9 days must be used. Calculate and interpret the 90 percent interval for the mean completion time for the manufacturing process. If this mean time is estimated to be in excess of 7 days,...

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## On Mori chamber and stable base locus decompositions

Antonio Laface
Abstract:
The effective cone of a Mori dream space [2] admits two decompositions into chambers called Mori chambers and stable base locus chambers (these are not necessarily convex cones). The former decomposition is a refinement of the latter, because Mori equivalent divisors have the same stable base locus. In this talk I will discuss both decompositions from a combinatorical viewpoint using the language of Cox rings [1,
$\S 3.3$
] and provide examples when the two are different. At the end I will state a criterion to establish whether the two decompositions coincide for a Mori dream space of Picard rank two.
This is joint work [3] with A. Massarenti and R. Rischter.
References
[1] Ivan Arzhantsev, Ulrich Derenthal, Jürgen Hausen, and Antonio Laface, Cox rings, Cambridge Studies in Advanced Mathematics, vol. 144, Cambridge University Press, Cambridge, 2015.
[2] Yi Hu and Sean Keel, Mori dream spaces and GIT, Michigan Math. J. 48 (2000), 331–348.
[3] Antonio Laface, Alex Massarenti, and Rick Rischter, On Mori chamber and stable base locus decompositionsarXiv:1805.10925.


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# Two Chain-Rules in one?

Printable View

• March 13th 2010, 04:56 PM
Zanderist
Two Chain-Rules in one?
To solve something like:

$f(x)= (8x^2 + 2)^6 * (3x^2-9)^(14)$

Would I use the chain rule twice?
• March 13th 2010, 05:04 PM
Riyzar
Multiply $14(3x^2-9)$ out and then use the product rule.

$[f(x)g(x)]' = f(x)g'(x)+g(x)f'(x)$

You'll have to use the chain rule once.

EDIT: Looking at your code did you mean $(3x^2-9)^{14}$?
• March 13th 2010, 05:07 PM
Zanderist
Quote:

Originally Posted by Riyzar
Multiply $14(3x^2-9)$ out and then use the product rule.

$[f(x)g(x)]' = f(x)g'(x)+g(x)f'(x)$

You'll have to use the chain rule once.

That's (3x^2-9) raised to the 14th
• March 13th 2010, 05:11 PM
Riyzar
Ah ok, then yes you will.

You should use the chain rule to find the derivatives of $(8x^2+2)^6$ and of $(3x^2-9)^{14}$

And then use the product rule.
• March 13th 2010, 05:18 PM
Zanderist
So that ends up being:

$48(8x^2+2)^5*(3x^2-9)^14 + (8x^2+2)^6 * 42(3x^2-9)^13$
• March 13th 2010, 05:27 PM
Riyzar
Close, look at this: $[(8x^2+2)^6]' = 6(8x^2+2)^5(16x)$
• March 13th 2010, 05:29 PM
Zanderist
oh god, I added in that 48 by mistake.
• March 13th 2010, 05:34 PM
Riyzar
Easy mistake (Happy) you did the same thing for the derivative of $(3x^2-9)^{14}$ So just make sure you check that and then you should be good!
• March 13th 2010, 06:16 PM
Zanderist
Quote:

Originally Posted by Riyzar
Easy mistake (Happy) you did the same thing for the derivative of $(3x^2-9)^{14}$ So just make sure you check that and then you should be good!

I enter in:

$(48x^2+12)^5 *(3x^2-9)^{14} + (8x^2+2)^6 * (42x^2-126)^{13}$

And it comes back as incorrect, what do I need to do next?

Wait that's not my question.

I now want to know what to do with the 16x

in

$f'(x)= 6(8x^2+2)^5*(16x)$

Math anxiety
• March 13th 2010, 06:45 PM
ione
Quote:

Originally Posted by Zanderist
I enter in:

$(48x^2+12)^5 *(3x^2-9)^{14} + (8x^2+2)^6 * (42x^2-126)^{13}$

And it comes back as incorrect, what do I need to do next?

Wait that's not my question.

I now want to know what to do with the 16x

in

$f'(x)= 6(8x^2+2)^5*(16x)$

Math anxiety

$f'(x)= 6(8x^2+2)^5*(16x)=96x(8x^2+2)^5$

$g'(x)= 14(3x^2-9)^{13}*(6x)=84x(3x^2-9)^{13}$

$
[f(x)g(x)]' = f(x)g'(x)+g(x)f'(x)$

$=(8x^2+2)^6(84x)(3x^2-9)^{13}+(3x^2-9)^{14}(96x)(8x^2+2)^5
$

$=12x(8x^2+6)^5(3x^2-9)^{13}(7(8x^2+2)+8(3x^2-9))$

$=12x(8x^2+6)^5(3x^2-9)^{13}(56x^2+14+24x^2-72)$

$=12x(8x^2+6)^5(3x^2-9)^{13}(80x^2-58)$

$=24x(8x^2+6)^5(3x^2-9)^{13}(40x^2-29)$
• March 13th 2010, 08:00 PM
Zanderist
Quote:

Originally Posted by ione
$f'(x)= 6(8x^2+2)^5*(16x)=96x(8x^2+2)^5$

$g'(x)= 14(3x^2-9)^{13}*(6x)=84x(3x^2-9)^{13}$

$
[f(x)g(x)]' = f(x)g'(x)+g(x)f'(x)$

$=(8x^2+2)^6(84x)(3x^2-9)^{13}+(3x^2-9)^{14}(96x)(8x^2+2)^5
$

$=12x(8x^2+6)^5(3x^2-9)^{13}(7(8x^2+2)+8(3x^2-9))$

$=12x(8x^2+6)^5(3x^2-9)^{13}(56x^2+14+24x^2-72)$

$=12x(8x^2+6)^5(3x^2-9)^{13}(80x^2-58)$

$=24x(8x^2+6)^5(3x^2-9)^{13}(40x^2-29)$

Alright, so what I've learned for this problem, is to treat it as two chain rules, then use the product rule.


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# Proof of Union and Intersection with nothing

## Homework Statement

Prove:
$A \cup \varnothing = A$
$A \cap \varnothing = \varnothing$

## The Attempt at a Solution

Intuitively both are true. The first is true because union with nothing will eventually return the original set. The second is true because there is no element that can be in a set and at the same time belongs to nothing.

First proof:
$A \cup \varnothing = A$
Suppose $x \in (A \cup \varnothing)$, then $x \in A$ or $x \in \varnothing$, the only true implication under such condition is, $x \in A$. Hence $x \in A$ for all $x$.

$A \cap \varnothing = \varnothing$
Suppose $x \in (A \cap \varnothing)$, then $x \in A$ and $x \in \varnothing$. It can be seen that there is no $x$ that can satisfy the definition, hence the intersection will be equal to $\varnothing$.

The thing I'm worried about the last proof is by using the fact that $x \in A$ and $x \in \varnothing$ which will be always false, you can use that statement to prove anything you want. Say, like this:

Suppose $x \in (A \cap \varnothing)$, then $x \in A$ and $x \in \varnothing$, hence $x \in A$.

But it will be always vacuously true?

Last edited:

Related Calculus and Beyond Homework Help News on Phys.org
Dick
Homework Helper

## Homework Statement

Prove:
$A \cup \varnothing = A$
$A \cap \varnothing = \varnothing$

## The Attempt at a Solution

Intuitively both are true. The first is true because union with nothing will eventually return the original set. The second is true because there is no element that can be in a set and at the same time belongs to nothing.

First proof:
$A \cup \varnothing = A$
Suppose $x \in (A \cup \varnothing)$, then $x \in A$ or $x \in \varnothing$, the only true implication under such condition is, $x \in A$. Hence $x \in A$ for all $x$.

$A \cap \varnothing = \varnothing$
Suppose $x \in (A \cap \varnothing)$, then $x \in A$ and $x \in \varnothing$, hence $x \in \varnothing$.

The thing I'm worried about the last proof is by using the fact that $x \in A$ and $x \in \varnothing$ which will be always false, you can use that statement to prove anything you want. Say, like this:

Suppose $x \in (A \cap \varnothing)$, then $x \in A$ and $x \in \varnothing$, hence $x \in A$.

But it will be always vacuously true?
There is NO x such that $x \in \varnothing$, so you can't conclude $x \in A$.

There is NO x such that $x \in \varnothing$, so you can't conclude $x \in A$.
The proof will not be correct yes. But will the statement still stand logically if you follow the truth table of implication? The antecedent is certainly false, but then I can add new consequence which is true. This will cause the whole implication to be true also.

On top of that I just edited the first post to add slight revision to the proof.

Dick
Homework Helper
The proof will not be correct yes. But will the statement still stand logically if you follow the truth table of implication? The antecedent is certainly false, but then I can add new consequence which is true. This will cause the whole implication to be true also.

On top of that I just edited the first post to add slight revision to the proof.
Yes, the antecedent is always false. So the implication is a true statement. But that doesn't prove the consequent. If p->q and p is always false, p->q is always true. But that gives you no information about whether q is true or false.

Yes, the antecedent is always false. So the implication is a true statement. But that doesn't prove the consequent. If p->q and p is always false, p->q is always true. But that gives you no information about whether q is true or false.
Ok I get your point for the latter part, but do you have any problem with the union proof? All of the options for 'or' will be false except if x belongs to A. So it is a must if we want to follow the definition, that the union will always be A. Is this sort of reasoning correct?

Dick
Homework Helper
Ok I get your point for the latter part, but do you have any problem with the union proof? All of the options for 'or' will be false except if x belongs to A. So it is a must if we want to follow the definition, that the union will always be A. Is this sort of reasoning correct?
It looks fine to me. If a or b is true, and b is false, then a must be true.

vela
Staff Emeritus
Homework Helper
You've shown that $A \cup \emptyset \subset A$. You still need to show the opposite direction before you can conclude equality.

You've shown that $A \cup \emptyset \subset A$. You still need to show the opposite direction before you can conclude equality.
Let $x \in A$, hence $x \in A$ or $x \in \varnothing$, thus $x \in A \cup \varnothing$. So $A \subset (A \cup \varnothing)$

Is that sufficient?

vela
Staff Emeritus


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# If $i=\sqrt{-1}$, then what can be the value of $(i^{32})^{64}$
$\begin{array}{1 1} (a)\; i\\(b)\; 1 \\(c)\; -i \\(d)\; -1 \end{array}$


text:
# Homework Help: Wheel and turntable with different angular velocity

1. Oct 19, 2004

### physicsss

The axle of a wheel is mounted on supports that rest on a rotating turntable as shown in Fig. 10-50. The wheel has angular velocity w1 = 50.0 rad/s about its axle, and the turntable has angular velocity of w=35.0 rad/s about a vertical axis. (Note arrows showing these motions in the figure.)

here's the picture: http://www.geocities.com/sinceury/10-50.gif

What is the magnitude and direction of the angular acceleration of the wheel at the instant shown? Take the z axis vertically upward and the direction of the axle at the moment shown to be the x axis pointing to the right.

2. Oct 20, 2004

### Tide

The angular velocity of the wheel is

$$\omega_{wheel} \left(\cos \omega_{table} t \hat i - \sin \omega_{table} t \hat j \right) + \omega_{"table"} \hat k$$

from which you can find the angular acceleration by differentiating with respect to time. (I didn't check the signs carefully so check them - this will give you a start!)

3. Oct 20, 2004

### physicsss

I don't understand...

4. Oct 20, 2004

### Tide

If the table were not turning the angular velocity of the wheel would be a vector with only x and y components. What I did was to take that angular velocity and rotated it as a function of time about the z axis. That's the first part of the expression I wrote. In fact, I chose the rotation rate to be that of the table upon which the wheel sits. (The $\omega$'s are angular speeds - i.e. not vectors!.)

Imparting a rotation of the wheel about the z axis provides an additional component of the angular velocity of the wheel. That is the second term I wrote.

5. Oct 21, 2004

### physicsss

The angular acceleration acceleration is 50(cos35-sin35)...?

6. Oct 21, 2004

### Tide

No. The expression I wrote is the angular velocity. To find the angular acceleration you will need to differentiate with respect to time. Don't forget the angular acceleration is a vector meaning it has both direction and magnitude.


text:
A calibrated classifier is one where the predicted probability of a class closely matches the rate at which that class occurs, e.g. for data points which are assigned a predicted probability of class A of .8, approximately 80 percent of such points should belong to class A if the classifier is well calibrated. This is estimated empirically by grouping data points with similar predicted probabilities for each class, and plotting the rate of each class within each bin against the predicted probability bins.

generateCalibrationData(obj, breaks = "Sturges", groups = NULL, task.id = NULL)

## Arguments

obj (list of Prediction | list of ResampleResult | BenchmarkResult) Single prediction object, list of them, single resample result, list of them, or a benchmark result. In case of a list probably produced by different learners you want to compare, then name the list with the names you want to see in the plots, probably learner shortnames or ids. (character(1) | numeric) If character(1), the algorithm to use in generating probability bins. See hist for details. If numeric, the cut points for the bins. Default is “Sturges”. (integer(1)) The number of bins to construct. If specified, breaks is ignored. Default is NULL. (character(1)) Selected task in BenchmarkResult to do plots for, ignored otherwise. Default is first task.

## Value

CalibrationData. A list containing:

proportion

data.frame with columns:

• Learner Name of learner.

• bin Bins calculated according to the breaks or groups argument.

• Class Class labels (for binary classification only the positive class).

• Proportion Proportion of observations from class Class among all observations with posterior probabilities of class Class within the interval given in bin.

data

data.frame with columns:

• Learner Name of learner.

• truth True class label.

• Class Class labels (for binary classification only the positive class).

• Probability Predicted posterior probability of Class.

• bin Bin corresponding to Probability.

Other generate_plot_data: generateCritDifferencesData(), generateFeatureImportanceData(), generateFilterValuesData(), generateLearningCurveData(), generatePartialDependenceData(), generateThreshVsPerfData(), plotFilterValues()
Other calibration: plotCalibration()


text:
# solution of difference equation

I am trying to solve the following difference equation:

$$-\frac{\epsilon}{h^2}U_{n+1}+\left(\frac{2\epsilon}{h^2}+\frac{1}{h}\right)U_{n}-\left(\frac{\epsilon}{h^2}+\frac{1}{h}\right)U_{n-1}=0,\mbox{ }\mbox{ }\mbox{ }\mbox{ }U_0=1,\mbox{ }U_1=0.$$

I try $U_{n}=Aw^n$ then I get

$$w_{1,2}=\frac{\left(\frac{2\epsilon}{h^2}+\frac{1}{h}\right)\pm\sqrt{\left(\frac{2\epsilon}{h^2}+\frac{1}{h}\right)^2-4\frac{\epsilon}{h^2}\left(\frac{\epsilon}{h^2}+\frac{1}{h}\right)}}{2\frac{\epsilon}{h^2}}.$$

This seems a bit far from what I want to get. I am trying to verify that the solution is

$$U_n=\dfrac{1-(1+\rho)^{n-N}}{1-(1+\rho)^{-N}},$$

where $0\leq n\leq N$ and $\rho=h/\epsilon$.

-
In your third term in your difference equation, did you mean $U_{n-1}$? –  Cameron Williams Jul 4 at 18:19
If you want to verify a solution you already have, just substitute it in the equation and check it is true, and also check the initial conditions. –  ABC Jul 4 at 18:39
Note that the homogeneous equation has solution $U_n=1$, equivalently you have a solution $w_1=1$. This makes life a lot easier. –  Mark Bennet Jul 4 at 18:39
@CameronWilliams, yes and thanks for that. –  Vaolter Jul 5 at 7:40
@MarkBennet, could you elaborate. –  Vaolter Jul 5 at 7:42
show 1 more comment


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# Flat torsionfree connection in Kaehler manifold

If $\nabla$ is a flat torsionfree connection and $J$ is a complex structure, we define \begin{align} \notag d^{\nabla}J(X,Y)=(\nabla_{X}J)Y-(\nabla_{Y}J)X. \end{align} Why flatness of $\nabla$ means $d^{\nabla}=0$, and $\nabla$ is torsionfree means $(d^{\nabla })^{2}(\mathrm{id})=0$?

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What is the source where these meanings are used? Looks like a typo. Usually, $d^{\nabla} Id$ can be interpreted as the torsion of $\nabla$, and $d^{\nabla} \circ d^{\nabla}$ is the curvature... –  Yuri Vyatkin Feb 12 '13 at 19:43
I found it in this paper: arxiv.org/abs/hep-th/9712042 on page 4. So $d^{\nabla}(X,Y)=\nabla_{X}Y-\nabla_{Y}X-[X,Y]$? –  Novak Djokovic Feb 14 '13 at 4:56
I'm afraid that in your last equation you have made a typo again. I posted an answer hoping that it may clarify the things to you. –  Yuri Vyatkin Feb 14 '13 at 21:46
The formula for $\mathrm{d}^{\nabla} J$ that you give in the question holds only for torsion free connections. I have added this point to my answer. –  Yuri Vyatkin Feb 16 '13 at 6:42

Short answer. In general (see the extended answer), the formula for $\mathrm{d}^{\nabla} J$ would be $$d^{\nabla}J(X,Y)=\nabla_{X} \left( J(Y) \right) - \nabla_{Y} \left( J(X) \right) - J([X,Y])$$ because the almost complex structure $J \colon TM \to TM$ is regarded as a $TM$-valued $1$-form (it has to satisfy $J^2 = - \mathrm{id}$ but this is irrelevant for the question we consider).

Torsion free means $T^{\nabla} = \mathrm{d}^{\nabla} \mathrm{id} = 0$, see equation (1.4) in the paper you mention.

Flat means $R^{\nabla} = \mathrm{d}^2_{\nabla} = \mathrm{d}^{\nabla} \circ \mathrm{d}^{\nabla} = 0$, see the cited paper, next line after equation (1.3)

Now, when the connection $\nabla$ is torsion free, we can use the fact that in this case the Lie bracket is expressed simply as $$[X,Y] = \nabla_X Y - \nabla_Y X$$ and using (see e.g. here) the notation $$(\nabla_X J) Y := \nabla_X (J(Y)) - J(\nabla_X Y)$$ the general expression for $\mathrm{d}^{\nabla} J$ is simplified to $$\mathrm{d}^{\nabla} J (X,Y) = (\nabla_X J) Y - (\nabla_Y J) X$$ as it is given, for instance, in this paper, Remark 1 on p.5.

The invariant formula \begin{align} \mathrm{d} \omega(X_0,...,X_k) = & \sum_{0 \le i \le k}(-1)^{i} X_i \left(\omega(X_0, \ldots, \hat{X_i}, \ldots, X_k)\right) \\ & +\sum_{0 \le i<j \le k}(-1)^{i+j} \omega([X_i, X_j], X_0, \ldots, \hat{X_i}, \ldots, \hat{X_j}, \ldots, X_k) \end{align} for the exterior derivative of a $k$-form $\omega \in \Omega^k(M)$ on a manifold $M$ can be used as a definition, and also it can be extended to $k$-forms with values in a vector bundle $E \to M$ almost without a change (formally): \begin{align} \mathrm{d}^{\nabla} \omega(X_0,...,X_k) = & \sum_{0 \le i \le k}(-1)^{i} \nabla_{X_i} \left(\omega(X_0, \ldots, \hat{X_i}, \ldots, X_k)\right) \\ & +\sum_{0 \le i<j \le k}(-1)^{i+j} \omega([X_i, X_j], X_0, \ldots, \hat{X_i}, \ldots, \hat{X_j}, \ldots, X_k) \end{align} where $\nabla \colon \Omega^0(E) \to \Omega^1(E)$ is a connection in vector bundle $E$.

Now let us make a few applications of this formula to the most basic cases.

For $\omega \in \Omega^0(E) \equiv \Gamma(E)$ we have $\mathrm{d}^{\nabla} \omega (X) = \nabla_X \omega$

When $\omega \in \Omega^1(E) \equiv \Omega^1(M) \otimes \Gamma(E)$ one gets $$\mathrm{d}^{\nabla} \omega (X,Y) = \nabla_X \omega(Y) - \nabla_Y \omega(X) - \omega([X,Y]) \tag{*}$$

The identity map $\mathrm{id} \colon TM \to TM$ can be naturally seen as a $TM$-valued $1$-form, and the latter identity applied to it reads as $$\mathrm{d}^{\nabla} \mathrm{id} (X,Y) = \nabla_X Y - \nabla_Y X - [X,Y]$$ where on the right hand side we see what is known as the torsion $T(X,Y) = T^{\nabla}(X,Y)$ of the connection $\nabla$

No surprise that torsion free ($T^{\nabla}(X,Y) = 0$) connections are in favour: we would like that the identity map has zero derivative!

Let us turn to the connection $\nabla$ itself. For $\omega \in \Omega^0(E)$ we regard $\mathrm{d}^{\nabla} \omega = \nabla \omega$ as a $E$-valued $1$-form, and so we can use the identity (*) again: \begin{align} \mathrm{d}^{\nabla} \mathrm{d}^{\nabla} \omega & = \nabla_X \mathrm{d}^{\nabla} \omega(Y) - \nabla_Y \mathrm{d}^{\nabla} \omega (X) - \mathrm{d}^{\nabla} \omega([X,Y]) \\ & = \nabla_X \nabla_Y \omega - \nabla_Y \nabla_X \omega - \nabla_{[X,Y]} \omega \end{align} and again we recognize in RHS the curvature operator $R_{X Y}$ acting on a section $\omega \in \Gamma(E)$ of a vector bundle $E$ with a connection $\nabla$

Informally speaking, $\mathrm{d}^{\nabla} \circ \mathrm{d}^{\nabla} = R^{\nabla}$

A connection is called flat if $R^{\nabla} = \mathrm{d}^{\nabla} \circ \mathrm{d}^{\nabla} = 0$

We like flat connections because the derivatives commute nicely.

This is precisely the story that can be read off from the equation (1.3) in the cited paper

All this is, indeed, very standard stuff. A more detailed treatment with everything made precise one can find e.g. in I.Madsen, J.Torenhave "From Calculus to Cohomology".

For a gentle introduction I would recommend the Wikipedia's article and the references therein (especially the book of R.W.R. Darling "Differential forms and connections").

Of course, all this is usually mentioned in the books and lectures on Kähler geometry in the sections on background. See, for instance, W.Ballmann's "Lectures on Kähler Manifolds", p.4.

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text:
# How do you find the slope of the line that passes through the points (3, 6) And (1, -4)?

###### Question:

How do you find the slope of the line that passes through the points (3, 6) And (1, -4)?

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text:
# GET_STEP_INFO

Utility/Data Access SubroutineThis subroutine returns information about the current simulation step in a MotionSolve simulation.

## Use

This function-style subroutine can be called from within any user-written subroutine to obtain information about the current simulation step for a MotionSolve model, or about the current solver step in a transient analysis. This utility can be used to check, for example, whether the current step is a predictor step, a corrector step or a converged step, for example. If this function is called during any other analysis, a value of -1 is returned.

## Format

Fortran Calling Syntax
call GET_STEP_INFO(FLAG)
C/C++ Calling Syntax
c_get_step_info(flag)
Python Calling Syntax
flag = py_get_step_info()
MATLAB Calling Syntax
flag = m_get_step_info()

## Output

FLAG
[integer]
The integer value returned by this function:
0
Indicates the current step is a converged step.
1
Indicates the current step is a predictor step.
2
Indicates the current step is a corrector step.
-1
Indicates that the current analysis is not a transient analysis.